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stringlengths 16
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putnam_2023_a1 | abbrev putnam_2023_a1_solution : β := sorry
-- 18
theorem putnam_2023_a1
(f : β β β β β)
(hf : β n > 0, f n = fun x : β => β i in Finset.Icc 1 n, Real.cos (i * x))
: putnam_2023_a1_solution > 0 β§ |iteratedDeriv 2 (f putnam_2023_a1_solution) 0| > 2023 β§
β n > 0, n < putnam_2023_a1_solution β (|iteratedDeriv 2 (f n) 0| β€ 2023) :=
sorry | For a positive integer $n$, let $f_n(x) = \cos(x) \cos(2x) \cos(3x) \cdots \cos(nx)$. Find the smallest $n$ such that $|f_n''(0)| > 2023$. | Show that the solution is $n = 18$. | ['analysis'] | null | theory putnam_2023_a1 imports Complex_Main
"HOL-Analysis.Derivative"
begin
definition putnam_2023_a1_solution::nat where "putnam_2023_a1_solution \<equiv> undefined"
(* 18 *)
theorem putnam_2023_a1:
fixes n :: nat and f :: "nat \<Rightarrow> real \<Rightarrow> real"
defines "f \<equiv> \<lambda>n. \<lambda>x. \<Prod>i=1..n. cos (real i * x)"
shows "putnam_2023_a1_solution = (LEAST m::nat. m > 0 \<and> abs ((deriv^^2) (f m) 0) > 2023)"
sorry
end
| null |
putnam_2023_a2 | abbrev putnam_2023_a2_solution : β β Set β := sorry
-- fun n => {(1 : β)/(factorial n), -(1 : β)/(factorial n)}
theorem putnam_2023_a2
(n : β)
(hn : n > 0 β§ Even n)
(p : Polynomial β)
(hp : Polynomial.Monic p β§ p.degree = 2*n)
(S : Set β := {x : β | β k : β€, x = k β§ 1 β€ |k| β§ |k| β€ n})
(hpinv : β k β S, p.eval (1/k) = k^2)
: {x : β | p.eval (1/x) = x^2} \ S = putnam_2023_a2_solution n :=
sorry | Let $n$ be an even positive integer. Let $p$ be a monic, real polynomial of degree $2n$; that is to say, $p(x) = x^{2n} + a_{2n-1} x^{2n-1} + \cdots + a_1 x + a_0$ for some real coefficients $a_0, \dots, a_{2n-1}$. Suppose that $p(1/k) = k^2$ for all integers $k$ such that $1 \leq |k| \leq n$. Find all other real numbers $x$ for which $p(1/x) = x^2$. | Show that the other real numbers satisfying $p(1/x) = x^2$ are $\pm rac{1}{n!}.$ | ['algebra'] | Section putnam_2023_a2.
Require Import Nat Ensembles Factorial Reals Coquelicot.Coquelicot.
Definition putnam_2023_a2_solution : Ensemble R := fun x => exists (n: nat), x = -1 / INR (fact n) \/ x = 1 / INR (fact n).
Theorem putnam_2023_a2
(n : nat)
(hn0 : gt n 0)
(hnev : even n = true)
(coeff: nat -> R)
(p : R -> R := fun x => sum_n (fun i => coeff i * x ^ i) (2 * n + 1))
(monic_even : coeff (mul 2 n) = 1)
(hpinv : forall k : Z, and (Z.le 1 (Z.abs k)) (Z.le (Z.abs k) (Z.of_nat n)) -> p (1 / (IZR k)) = IZR (k ^ 2))
: (fun x => (p (1 / x) = x ^ 2 /\ ~ exists k : Z, x = IZR k /\ Z.le (Z.abs k) (Z.of_nat n))) = putnam_2023_a2_solution.
Proof. Admitted.
End putnam_2023_a2. | theory putnam_2023_a2 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
definition putnam_2023_a2_solution :: "nat \<Rightarrow> (real set)" where "putnam_2023_a2_solution \<equiv> undefined"
(* \<lambda> n :: nat. {(1 :: real)/(fact n), -(1 :: real)/(fact n)} *)
theorem putnam_2023_a2:
fixes n :: "nat"
and p :: "real poly"
and S :: "real set"
defines "S \<equiv> {x :: real. \<exists> k :: int. x = k \<and> 1 \<le> abs k \<and> abs k \<le> n}"
assumes hn : "n > 0 \<and> even n"
and hp : "lead_coeff p = 1 \<and> degree p = 2*n"
and hpinv : "\<forall> k \<in> S. poly p (1/k) = k^2"
shows "{x :: real. poly p (1/x) = x^2} - S = putnam_2023_a2_solution n"
sorry
end | null |
putnam_2023_a3 | abbrev putnam_2023_a3_solution : β := sorry
-- Real.pi / 2
theorem putnam_2023_a3
: sInf {r > 0 | β f g : β β β, Differentiable β f β§ Differentiable β g β§
f 0 > 0 β§ g 0 = 0 β§ (β x : β, |deriv f x| β€ |g x| β§ |deriv g x| β€ |f x|) β§ f r = 0} = putnam_2023_a3_solution :=
sorry | Determine the smallest positive real number $r$ such that there exist differentiable functions $f\colon \mathbb{R} \to \mathbb{R}$ and $g\colon \mathbb{R} \to \mathbb{R}$ satisfying
\begin{enumerate}
\item[(a)] $f(0) > 0$,
\item[(b)] $g(0) = 0$,
\item[(c)] $|f'(x)| \leq |g(x)|$ for all $x$,
\item[(d)] $|g'(x)| \leq |f(x)|$ for all $x$, and
\item[(e)] $f(r) = 0$. \end{enumerate} | Show that the solution is $r = \pi/2$. | ['analysis'] | Section putnam_2023_a3.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2023_a3_solution := PI / 2.
Theorem putnam_2023_a3
(r : R)
(hr : r > 0)
(p : R -> Prop := fun t => exists (f g : R -> R), f 0 = 0 /\ g 0 = 0 /\ forall (x: R), Rabs (Derive f x) <= Rabs (g x) /\ forall (x: R), Rabs (Derive g x) <= Rabs (f x) /\ f t = 0)
: (forall (t: R), t > 0 -> t < r -> ~ (p t)) <-> r = putnam_2023_a3_solution.
Proof. Admitted.
End putnam_2023_a3. | theory putnam_2023_a3 imports Complex_Main
"HOL-Analysis.Derivative"
begin
definition putnam_2023_a3_solution :: "real" where "putnam_2023_a3_solution \<equiv> undefined"
(* pi / 2 *)
theorem putnam_2023_a3:
shows "(LEAST r :: real. r > 0 \<and> (\<exists> f g :: real \<Rightarrow> real. Differentiable f UNIV \<and> Differentiable g UNIV \<and>
f 0 > 0 \<and> g 0 = 0 \<and> (\<forall> x :: real. abs (deriv f x) \<le> abs (g x) \<and> abs (deriv g x) \<le> abs(f x)) \<and> f r = 0)) = putnam_2023_a3_solution"
sorry
end | null |
putnam_2023_b2 | abbrev putnam_2023_b2_solution : β := sorry
-- 3
theorem putnam_2023_b2
: sInf {(digits 2 (2023*n)).sum | n > 0} = putnam_2023_b2_solution :=
sorry | For each positive integer $n$, let $k(n)$ be the number of ones in the binary representation of $2023 \cdot n$. What is the minimum value of $k(n)$? | Show that the minimum is $3$. | ['number_theory'] | Section putnam_2023_b2.
Require Import BinNums Nat NArith.
Definition putnam_2023_b2_solution := 3.
Theorem putnam_2023_b2:
let k:=
fix count_ones (n : positive) : nat :=
match n with
| xH => 1
| xO n' => count_ones n'
| xI n' => 1 + count_ones n'
end in
(forall (n: nat), k (Pos.of_nat (2023*n)) >= putnam_2023_b2_solution) /\
(exists (n: nat), k (Pos.of_nat (2023*n)) = putnam_2023_b2_solution).
Proof. Admitted.
End putnam_2023_b2. | theory putnam_2023_b2 imports Complex_Main
"HOL-Number_Theory.Cong"
begin
definition putnam_2023_b2_solution :: "nat" where "putnam_2023_b2_solution \<equiv> undefined"
(* 3 *)
theorem putnam_2023_b2:
fixes ones :: "nat \<Rightarrow> nat"
assumes h0 : "ones 0 = 0"
and hi : "\<forall> n > 0. ones n = ones (n div 2) + (if [n = 1] (mod 2) then 1 else 0)"
shows "(LEAST k :: nat. \<exists> n > 0. ones (2023*n) = k) = putnam_2023_b2_solution"
sorry
end | null |
putnam_2023_a5 | abbrev putnam_2023_a5_solution : Set β := sorry
-- {-(3^1010 - 1)/2, -(3^1010 - 1)/2 + Real.sqrt (9^1010 - 1) * Complex.I/4, -(3^1010 - 1)/2 - Real.sqrt (9^1010 - 1) * Complex.I/4}
theorem putnam_2023_a5
: {z : β | β k in Finset.Icc 0 (3^1010 - 1), (-2)^(num_ones (digits 3 k)) * (z + k)^2023 = 0} = putnam_2023_a5_solution :=
sorry | For a nonnegative integer $k$, let $f(k)$ be the number of ones in the base 3 representation of $k$. Find all complex numbers $z$ such that \[ \sum_{k=0}^{3^{1010}-1} (-2)^{f(k)} (z+k)^{2023} = 0. \] | Show that the solution is the set of complex numbers $\{- \frac{3^{1010} - 1}{2} \pm \frac{\sqrt{9^{1010} - 1}}{4} * i \}$ | ['algebra'] | null | theory putnam_2023_a5 imports Complex_Main
"HOL-Number_Theory.Cong"
begin
definition putnam_2023_a5_solution :: "complex set" where "putnam_2023_a5_solution \<equiv> undefined"
(* {-(3^1010 - 1)/2, -(3^1010 - 1)/2 + sqrt (9^1010 - 1) * \<i>/4, -(3^1010 - 1)/2 - sqrt (9^1010 - 1) * \<i>/4} *)
theorem putnam_2023_a5:
fixes num_ones :: "nat \<Rightarrow> nat"
assumes h0 : "num_ones 0 = 0"
and hi : "\<forall> n > 0. num_ones n = (num_ones (n div 3)) + (if [n = 1] (mod 3) then 1 else 0)"
shows "{z :: complex. (\<Sum> k=0..(3^1010 - 1). ((-2)^(num_ones k) * (z + k)^2023)) = 0} = putnam_2023_a5_solution"
sorry
end | null |
putnam_2023_b4 | abbrev putnam_2023_b4_solution : β := sorry
-- 29
theorem putnam_2023_b4
(tne : β β (β β β) β Set β)
(fdiff : β β (β β β) β (β β β) β Prop)
(flim : β β (β β β) β (β β β) β Prop)
(fderiv1 : β β (β β β) β (β β β) β Prop)
(fderiv2 : β β (β β β) β (β β β) β Prop)
(fall : β β (β β β) β (β β β) β Prop)
(tinc : β β (β β β) β Prop)
(Tall : β β Prop)
(htne : β n : β, β ts : β β β, tne n ts = {t : β | t > ts 0 β§ β i : Fin n, t β ts (i.1 + 1)})
(hfdiff : β n : β, β ts : β β β, β f : β β β, fdiff n ts f = (ContinuousOn f (Set.Ici (ts 0)) β§ ContDiffOn β 1 f (tne n ts) β§ DifferentiableOn β (derivWithin f (tne n ts)) (tne n ts)))
(hflim : β n : β, β ts : β β β, β f : β β β, flim n ts f = β k : Fin (n + 1), Tendsto (derivWithin f (tne n ts)) (π[>] (ts k.1)) (π 0))
(hfderiv1 : β n : β, β ts : β β β, β f : β β β, fderiv1 n ts f = β k : Fin n, β t β Set.Ioo (ts k.1) (ts (k.1 + 1)), iteratedDerivWithin 2 f (tne n ts) t = k.1 + 1)
(hfderiv2 : β n : β, β ts : β β β, β f : β β β, fderiv2 n ts f = β t > ts n, iteratedDerivWithin 2 f (tne n ts) t = n + 1)
(hfall : β n : β, β ts : β β β, β f : β β β, fall n ts f = (fdiff n ts f β§ f (ts 0) = 0.5 β§ flim n ts f β§ fderiv1 n ts f β§ fderiv2 n ts f))
(htinc : β n : β, β ts : β β β, tinc n ts = β k : Fin n, ts (k.1 + 1) β₯ ts k.1 + 1)
(hTall : β T : β, Tall T = ((T β₯ 0) β§ β n : β, β ts : β β β, β f : β β β, tinc n ts β§ fall n ts f β§ f (ts 0 + T) = 2023))
: Tall putnam_2023_b4_solution β§ β T : β, Tall T β T β₯ putnam_2023_b4_solution :=
sorry | For a nonnegative integer $n$ and a strictly increasing sequence of real numbers $t_0,t_1,\dots,t_n$, let $f(t)$ be the corresponding real-valued function defined for $t \geq t_0$ by the following properties:
\begin{enumerate}
\item[(a)] $f(t)$ is continuous for $t \geq t_0$, and is twice differentiable for all $t>t_0$ other than $t_1,\dots,t_n$;
\item[(b)] $f(t_0)=1/2$;
\item[(c)] $\lim_{t \to t_k^+} f'(t)=0$ for $0 \leq k \leq n$;
\item[(d)] For $0 \leq k \leq n-1$, we have $f''(t)=k+1$ when $t_k<t<t_{k+1}$, and $f''(t)=n+1$ when $t>t_n$.
\end{enumerate}
Considering all choices of $n$ and $t_0,t_1,\dots,t_n$ such that $t_k \geq t_{k-1}+1$ for $1 \leq k \leq n$, what is the least possible value of $T$ for which $f(t_0+T)=2023$? | Show that the minimum value of $T$ is $29$. | ['analysis'] | Section putnam_2023_b4.
Require Import Reals Coquelicot.Derive Coquelicot.Hierarchy. From mathcomp Require Import div fintype seq ssrbool.
Definition putnam_2023_b4_solution := 29.
Theorem putnam_2023_b4:
forall (n: nat) (s: 'I_n -> R) (i i0 : 'I_n), s i < s (ordS i) ->
let t0 := s (nth i0 (enum 'I_n) 0) in forall (f : R -> R) (t: R),
(
((t >= t0 -> continuity_pt f t) /\
(t > t0 /\ ~ exists i, s i = t -> ex_derive_n f 2 t))
/\
(f t0 = 0.5)
/\
(forall k: 'I_n, let tk := s k in filterlim (Derive_n f 1) (at_right tk) (locally 0))
/\
(forall k: 'I_n, k <> nth i0 (enum 'I_n) (n-1) -> let tk := s k in tk < t < tk+1 -> (Derive_n f 2) = (fun _ => INR k+1)) /\
(forall m: 'I_n, t > s (nth i0 (enum 'I_n) m) -> (Derive_n f 2) = (fun _ => INR m+1))
) ->
forall (T: R), f(t0+T) = 2023 <-> T >= putnam_2023_b4_solution /\ f(t0 + putnam_2023_b4_solution) = 2023.
Proof. Admitted.
End putnam_2023_b4. | theory putnam_2023_b4 imports Complex_Main
"HOL-Analysis.Derivative"
begin
(* uses (nat \<Rightarrow> real) instead of (Fin (n+1) \<Rightarrow> real) and (real \<Rightarrow> real) instead of ({(ts 0)..} \<Rightarrow> real) *)
definition putnam_2023_b4_solution :: real where "putnam_2023_b4_solution \<equiv> undefined"
(* 29 *)
theorem putnam_2023_b4:
fixes tne :: "nat \<Rightarrow> (nat \<Rightarrow> real) \<Rightarrow> real set"
and fdiff :: "nat \<Rightarrow> (nat \<Rightarrow> real) \<Rightarrow> (real \<Rightarrow> real) \<Rightarrow> bool"
and flim :: "nat \<Rightarrow> (nat \<Rightarrow> real) \<Rightarrow> (real \<Rightarrow> real) \<Rightarrow> bool"
and fderiv1 :: "nat \<Rightarrow> (nat \<Rightarrow> real) \<Rightarrow> (real \<Rightarrow> real) \<Rightarrow> bool"
and fderiv2 :: "nat \<Rightarrow> (nat \<Rightarrow> real) \<Rightarrow> (real \<Rightarrow> real) \<Rightarrow> bool"
and fall :: "nat \<Rightarrow> (nat \<Rightarrow> real) \<Rightarrow> (real \<Rightarrow> real) \<Rightarrow> bool"
and tinc :: "nat \<Rightarrow> (nat \<Rightarrow> real) \<Rightarrow> bool"
and Tall :: "real \<Rightarrow> bool"
assumes htne: "\<forall>(n::nat)(ts::nat\<Rightarrow>real). tne n ts = {t::real. t > ts 0 \<and> (\<forall>i::nat\<in>{1..n}. t \<noteq> ts i)}"
and hfdiff: "\<forall>(n::nat)(ts::nat\<Rightarrow>real)(f::real\<Rightarrow>real). fdiff n ts f = (continuous_on {(ts 0)..} f \<and> f C1_differentiable_on (tne n ts) \<and> (deriv f) differentiable_on (tne n ts))"
and hflim: "\<forall>(n::nat)(ts::nat\<Rightarrow>real)(f::real\<Rightarrow>real). flim n ts f = (\<forall>k::nat\<in>{0..n}. filterlim (deriv f) (nhds 0) (at_right (ts k)))"
and hfderiv1: "\<forall>(n::nat)(ts::nat\<Rightarrow>real)(f::real\<Rightarrow>real). fderiv1 n ts f = (\<forall>k::nat\<in>{0..(n-1)}. \<forall>t::real\<in>{(ts k)<..<(ts (k+1))}. (deriv^^2) f t = k+1)"
and hfderiv2: "\<forall>(n::nat)(ts::nat\<Rightarrow>real)(f::real\<Rightarrow>real). fderiv2 n ts f = (\<forall>t::real>(ts n). (deriv^^2) f t = n+1)"
and hfall: "\<forall>(n::nat)(ts::nat\<Rightarrow>real)(f::real\<Rightarrow>real). fall n ts f = (fdiff n ts f \<and> f (ts 0) = 0.5 \<and> flim n ts f \<and> fderiv1 n ts f \<and> fderiv2 n ts f)"
and htinc: "\<forall>(n::nat)(ts::nat\<Rightarrow>real). tinc n ts = (\<forall>k::nat\<in>{0..(n-1)}. ts (k+1) \<ge> ts k + 1)"
and hTall: "\<forall>T::real. Tall T = (T \<ge> 0 \<and> (\<exists>(n::nat)(ts::nat\<Rightarrow>real)(f::real\<Rightarrow>real). tinc n ts \<and> fall n ts f \<and> f (ts 0 + T) = 2023))"
shows "(LEAST T::real. Tall T) = putnam_2023_b4_solution"
sorry
end
| null |
putnam_2023_b5 | abbrev putnam_2023_b5_solution : Set β := sorry
-- {n : β | n = 1 β¨ n β‘ 2 [MOD 4]}
theorem putnam_2023_b5
(n : β)
(perm : Prop)
(hperm : perm = β m : β€, IsRelPrime m n β β p : Equiv.Perm (Fin n), β k : Fin n, (p (p k)).1 + 1 β‘ m * (k.1 + 1) [ZMOD n])
: (n > 0 β§ perm) β n β putnam_2023_b5_solution :=
sorry | Determine which positive integers $n$ have the following property: For all integers $m$ that are relatively prime to $n$, there exists a permutation $\pi:\{1,2,\dots,n\} \to \{1,2,\dots,n\}$ such that $\pi(\pi(k)) \equiv mk \pmod{n}$ for all $k \in \{1,2,\dots,n\}$. | Show that the desired property holds if and only if $n=1$ or $n \equiv 2 \pmod{4}$. | ['number_theory'] | Section putnam_2023_b5.
Require Import PeanoNat. From mathcomp Require Import div fintype perm ssrbool.
Theorem putnam_2023_b5:
forall (m n: nat), coprime m n ->
exists (Ο: {perm 'I_n}),
forall (k: 'I_n), (Ο (Ο k))%%n mod n = m*k%%n.
Proof. Admitted.
End putnam_2023_b5. | theory putnam_2023_b5 imports Complex_Main
"HOL-Number_Theory.Cong"
"HOL-Combinatorics.Permutations"
begin
definition putnam_2023_b5_solution :: "nat set" where "putnam_2023_b5_solution \<equiv> undefined"
(* {n::nat. n = 1 \<or> [n = 2] (mod 4)} *)
theorem putnam_2023_b5:
fixes n :: nat
and perm :: bool
assumes hperm: "perm = (\<forall>m::int. coprime m n \<longrightarrow> (\<exists>p::nat\<Rightarrow>nat. p permutes {1..n} \<and> (\<forall>k::nat\<in>{1..n}. [p (p k) = m*k] (mod n))))"
shows "(n > 0 \<and> perm) \<longleftrightarrow> n \<in> putnam_2023_b5_solution"
sorry
end
| null |
putnam_2023_b6 | abbrev putnam_2023_b6_solution : β β β€ := sorry
-- (fun n : β => (-1) ^ (Nat.ceil (n / 2) - 1) * 2 * Nat.ceil (n / 2))
theorem putnam_2023_b6
(n : β)
(S : Matrix (Fin n) (Fin n) β€)
(npos : n > 0)
(hS : β i j : Fin n, S i j = β' a : β, β' b : β, if a * (i.1 + 1) + b * (j.1 + 1) = n then 1 else 0)
: S.det = putnam_2023_b6_solution n :=
sorry | Let $n$ be a positive integer. For $i$ and $j$ in $\{1,2,\dots,n\}$, let $s(i,j)$ be the number of pairs $(a,b)$ of nonnegative integers satisfying $ai+bj=n$. Let $S$ be the $n$-by-$n$ matrix whose $(i,j)$ entry is $s(i,j)$. For example, when $n=5$, we have $S = \begin{bmatrix}
6 & 3 & 2 & 2 & 2 \\
3 & 0 & 1 & 0 & 1 \\
2 & 1 & 0 & 0 & 1 \\
2 & 0 & 0 & 0 & 1 \\
2 & 1 & 1 & 1 & 2
\end{bmatrix}$. Compute the determinant of $S$. | Show that the determinant equals $(-1)^{\lceil n/2 \rceil-1}2\lceil\frac{n}{2}\rceil$. | ['linear_algebra'] | Section putnam_2023_b6.
Require Import Nat Finite_sets.
From mathcomp Require Import matrix ssrbool ssralg fintype.
Variable putnam_2023_b6_solution : nat -> nat.
Local Open Scope ring_scope.
Theorem putnam_2023_b6:
forall (n: nat),
let s (i j: nat) := cardinal (nat*nat) (fun p => let (a, b) := p in 1 <= i <= n /\ 1 <= j <= n /\ eq (add (mul a i) (mul b j)) n) in
(\matrix_(i < n, j < n) s i j)
= (\matrix_(i < n, j < n) s i j).
Proof. Admitted.
End putnam_2023_b6. | theory putnam_2023_b6 imports Complex_Main
"HOL-Analysis.Determinants"
begin
(* uses (nat \<Rightarrow> 'n) instead of (Fin n \<Rightarrow> 'n) *)
definition putnam_2023_b6_solution :: "nat \<Rightarrow> int" where "putnam_2023_b6_solution \<equiv> undefined"
(* (\<lambda>n::nat. (-1)^(nat (\<lceil>n/2\<rceil>-1)) * 2 * \<lceil>n/2\<rceil>) *)
theorem putnam_2023_b6:
fixes n :: nat
and S :: "int^'n^'n"
assumes npos: "n > 0"
and pncard: "CARD('n) = n"
and hS: "\<exists>pnind::nat\<Rightarrow>'n. (pnind ` {0..(n-1)} = UNIV \<and> (\<forall>i::nat\<in>{0..(n-1)}. \<forall>j::nat\<in>{0..(n-1)}. S$(pnind i)$(pnind j) = card {(a::nat,b::nat). a*(i+1) + b*(j+1) = n}))"
shows "det S = putnam_2023_b6_solution n"
sorry
end
| null |
putnam_1996_a3 | abbrev putnam_1996_a3_solution : Prop := sorry
-- False
theorem putnam_1996_a3
(student_choices : Finset.range 20 β Set (Finset.range 6))
: putnam_1996_a3_solution β β S : Set (Finset.range 20), β c1 c2 : Finset.range 6, c1 β c2 β§ S.ncard = 5 β§ ({c1, c2} β β s β S, student_choices s β¨ ({c1, c2} β β s β S, (student_choices s)αΆ)) :=
sorry | Suppose that each of 20 students has made a choice of anywhere from 0 to 6 courses from a total of 6 courses offered. Prove or disprove: there are 5 students and 2 courses such that all 5 have chosen both courses or all 5 have chosen neither course. | Show that the solution is that the statement is false. | ['combinatorics'] | null | theory putnam_1996_a3 imports Complex_Main
begin
definition putnam_1996_a3_solution :: "bool" where
"putnam_1996_a3_solution \<equiv> undefined"
(* False *)
theorem putnam_1996_a3:
fixes student_choices :: "nat \<Rightarrow> (nat set)"
assumes hinrange : "\<forall> n :: nat. student_choices n \<subseteq> {1..6}"
shows "putnam_1996_a3_solution \<longleftrightarrow> (\<exists> S :: nat set. S \<subseteq> {1::nat..20} \<and> card S = 5 \<and> (\<exists> c1 \<in> {1 :: nat..6}. \<exists> c2 \<in> {1 :: nat..6}. c1 \<noteq> c2 \<and> ({c1, c2} \<subseteq> (\<Inter> s \<in> S. student_choices s) \<or> {c1, c2} \<subseteq> (\<Inter> s \<in> S. UNIV - (student_choices s))) ))"
sorry
end | null |
putnam_1996_a4 | theorem putnam_1996_a4
(A : Type*)
[Finite A]
(S : Set (A Γ A Γ A))
(hSdistinct : β a b c : A, β¨a, b, cβ© β S β a β b β§ b β c β§ a β c)
(hS1 : β a b c : A, β¨a, b, cβ© β S β β¨b, c, aβ© β S)
(hS2 : β a b c : A, β¨a, b, cβ© β S β β¨c, b, aβ© β S)
(hS3 : β a b c d : A, (β¨a, b, cβ© β S β§ β¨c, d, aβ© β S) β (β¨b,c,dβ© β S β§ β¨d,a,bβ© β S))
: β g : A β β, Injective g β§ (β a b c : A, g a < g b β§ g b < g c β β¨a,b,cβ© β S) :=
sorry | Let $S$ be the set of ordered triples $(a, b, c)$ of distinct elements of a finite set $A$. Suppose that \begin{enumerate} \item $(a,b,c) \in S$ if and only if $(b,c,a) \in S$; \item $(a,b,c) \in S$ if and only if $(c,b,a) \notin S$; \item $(a,b,c)$ and $(c,d,a)$ are both in $S$ if and only if $(b,c,d)$ and $(d,a,b)$ are both in $S$. \end{enumerate} Prove that there exists a one-to-one function $g$ from $A$ to $\R$ such that $g(a) < g(b) < g(c)$ implies $(a,b,c) \in S$. | null | ['algebra'] | null | theory putnam_1996_a4 imports Complex_Main
begin
theorem putnam_1996_a4:
fixes S :: "('A \<times> 'A \<times> 'A) set"
and n :: "nat"
assumes hA : "CARD('A) = n"
and hS1 : " \<forall> a b c :: 'A. (a, b, c) \<in> S \<longleftrightarrow> (b, c, a) \<in> S"
and hS2 : " \<forall> a b c :: 'A. (a, b, c) \<in> S \<longleftrightarrow> (c, b, a) \<notin> S"
and hS3 : " \<forall> a b c d :: 'A. ((a, b, c) \<in> S \<and> (c, d, a) \<in> S) \<longleftrightarrow> ((b, c, d) \<in> S \<and> (d, a, b) \<in> S)"
shows "\<exists> g :: 'A \<Rightarrow> real. inj g \<and> (\<forall> a b c :: 'A. (g a < g b \<and> g c < g c) \<longrightarrow> (a, b, c) \<in> S)"
sorry
end | null |
putnam_1996_a5 | theorem putnam_1996_a5
(p : β)
(hpprime : Prime p)
(hpge3 : p > 3)
(k : β := Nat.floor (2*p/(3 : β)))
: p^2 β£ β i in Finset.Icc 1 k, Nat.choose p i :=
sorry | If $p$ is a prime number greater than 3 and $k = \lfloor 2p/3 \rfloor$, prove that the sum \[\binom p1 + \binom p2 + \cdots + \binom pk \] of binomial coefficients is divisible by $p^2$. | null | ['number_theory'] | Section putnam_1996_a5.
Require Import Binomial Reals Znumtheory Coquelicot.Coquelicot. From mathcomp Require Import div.
Open Scope R.
Theorem putnam_1996_a5:
forall (p: nat), prime (Z.of_nat p) /\ gt p 3 ->
let k := floor (2 * INR p / 3) in
exists (m: nat), sum_n (fun i => Binomial.C p (i+1)) (Z.to_nat k) = INR m * pow (INR p) 2.
Proof. Admitted.
End putnam_1996_a5. | theory putnam_1996_a5 imports Complex_Main
begin
theorem putnam_1996_a5:
fixes p k :: "nat"
defines "k \<equiv> nat \<lfloor>2 * p / 3\<rfloor>"
assumes hpprime : "prime p"
and hpge3 : "p > 3"
shows "p^2 dvd (\<Sum> i \<in> {1 :: nat..k}. p choose i)"
sorry
end | null |
putnam_1996_a6 | abbrev putnam_1996_a6_solution : β β Set (β β β) := sorry
-- (fun c : β => if c β€ 1 / 4 then {f : β β β | β d : β, β x : β, f x = d} else {f : β β β | ContinuousOn f (Set.Icc 0 c) β§ f 0 = f c β§ (β x > 0, f x = f (x ^ 2 + c)) β§ (β x < 0, f x = f (-x))})
theorem putnam_1996_a6
(c : β)
(f : β β β)
(cgt0 : c > 0)
: (Continuous f β§ β x : β, f x = f (x ^ 2 + c)) β f β putnam_1996_a6_solution c :=
sorry | Let $c>0$ be a constant. Give a complete description, with proof, of the set of all continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(x)=f(x^2+c)$ for all $x \in \mathbb{R}$. | Show that if $c \leq 1/4$ then $f$ must be constant, and if $c>1/4$ then $f$ can be defined on $[0,c]$ as any continuous function with equal values on the endpoints, then extended to $x>c$ by the relation $f(x)=f(x^2+c)$, then extended further to $x<0$ by the relation $f(x)=f(-x)$. | ['analysis', 'algebra'] | Section putnam_1996_a6.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1996_a6_solution (c: R) (f: R -> R) := if Rle_dec c (1/4) then exists (d: R), f = (fun _ => d) else forall (x: R), 0 <= x <= c -> continuity_pt f x /\ f 0 = f c /\ forall (x: R), x > 0 -> f x = f (pow x 2 + c) /\ (forall (x: R), x < 0 -> f x = f (-x)).
Theorem putnam_1996_a6:
forall (c: R), c > 0 ->
forall (f: R -> R), continuity f /\ forall (x: R), f x = pow x 2 + c <->
putnam_1996_a6_solution c f.
Proof. Admitted.
End putnam_1996_a6. | theory putnam_1996_a6 imports
Complex_Main
begin
definition putnam_1996_a6_solution :: "real \<Rightarrow> ((real \<Rightarrow> real) set)" where "putnam_1996_a6_solution \<equiv> undefined"
(* \<lambda> c :: real. if c \<le> 1 / 4 then {f :: real \<Rightarrow> real. \<exists> d :: real. \<forall> x :: real. f x = d} else {f :: real \<Rightarrow> real. continuous_on {0..c} f \<and> f 0 = f c \<and> (\<forall> x > 0. f x = f (x ^ 2 + c)) \<and> (\<forall> x < 0. f x = f (-x))} *)
theorem putnam_1996_a6:
fixes c :: real
and f :: "real \<Rightarrow> real"
assumes cgt0: "c > 0"
shows "(continuous_on UNIV f \<and> (\<forall> x :: real. f x = f (x ^ 2 + c))) \<longleftrightarrow> f \<in> putnam_1996_a6_solution c"
sorry
end | null |
putnam_1996_b1 | abbrev putnam_1996_b1_solution : β β β := sorry
-- Nat.fib
theorem putnam_1996_b1
(selfish : Finset β β Prop)
(n : β)
(hselfish : β s : Finset β, selfish s = (s.card β s))
(npos : n β₯ 1)
: {s : Finset β | (s : Set β) β Set.Icc 1 n β§ selfish s β§ (β ss : Finset β, ss β s β Β¬selfish ss)}.encard = putnam_1996_b1_solution n :=
sorry | Define a \emph{selfish} set to be a set which has its own cardinality (number of elements) as an element. Find, with proof, the number of subsets of $\{1,2,\ldots,n\}$ which are \emph{minimal} selfish sets, that is, selfish sets none of whose proper subsets is selfish. | Show that the number of subsets is $F_n$, the $n$th Fibonacci number. | ['algebra'] | null | theory putnam_1996_b1 imports
Complex_Main
begin
fun putnam_1996_b1_solution :: "nat \<Rightarrow> nat"
where "putnam_1996_b1_solution n = undefined"
(* "putnam_1996_b1_solution 0 = 0"
| "putnam_1996_b1_solution (Suc 0) = 1"
| "putnam_1996_b1_solution (Suc (Suc n)) = putnam_1996_b1_solution n + putnam_1996_b1_solution (Suc n)" *)
theorem putnam_1996_b1:
fixes selfish :: "nat set \<Rightarrow> bool"
and n :: nat
defines "selfish \<equiv> \<lambda> s. card s \<in> s"
assumes npos: "n \<ge> 1"
shows "card {s :: nat set. s \<subseteq> {1..n} \<and> selfish s \<and> (\<forall> ss :: nat set. ss \<subseteq> s \<longrightarrow> \<not>selfish ss)} = putnam_1996_b1_solution n"
sorry
end | null |
putnam_1996_b2 | theorem putnam_1996_b2
(n : β)
(prododd : β)
(npos : n > 0)
(hprododd : prododd = β i in Finset.range (2 * n), if Odd i then i else 1)
: ((2 * n - 1 : β) / Real.exp 1) ^ ((2 * n - 1 : β) / 2) < prododd β§ prododd < ((2 * n + 1 : β) / Real.exp 1) ^ ((2 * n + 1 : β) / 2) :=
sorry | Show that for every positive integer $n$, $(\frac{2n-1}{e})^{\frac{2n-1}{2}}<1 \cdot 3 \cdot 5 \cdots (2n-1)<(\frac{2n+1}{e})^{\frac{2n+1}{2}}$. | null | ['analysis'] | Section putnam_1996_b2.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Theorem putnam_1996_b2:
let fix odd_fact (n : nat) : R :=
match n with
| O => 1
| S n' => (2 * INR n - 1) * odd_fact n'
end in
forall (n: nat), gt n 0 ->
pow ((2 * INR n - 1) / exp 1) ((2 * n - 1) / 2) < odd_fact n < pow ((2 * INR n + 1) / exp 1) ((2 * n + 1) / 2).
Proof. Admitted.
End putnam_1996_b2. | theory putnam_1996_b2 imports
Complex_Main
begin
theorem putnam_1996_b2:
fixes n :: nat
and prododd :: real
defines "prododd \<equiv> \<Prod> i \<in> {1 .. 2 * n - 1}. if odd i then i else 1"
assumes npos: "n > 0"
shows "((2 * n - 1) / exp 1) powr ((2 * n - 1) / 2) < prododd \<and> prododd < ((2 * n + 1) / exp 1) powr ((2 * n + 1) / 2)"
sorry
end | null |
putnam_1996_b3 | abbrev putnam_1996_b3_solution : β β β := sorry
-- (fun n : β => (2 * n ^ 3 + 3 * n ^ 2 - 11 * n + 18) / 6)
theorem putnam_1996_b3
(n : β)
(xset : (β β β€) β Prop)
(xsum : (β β β€) β β€)
(nge2 : n β₯ 2)
(hxset : β x : β β β€, xset x = (x '' (Finset.range n) = Set.Icc (1 : β€) n))
(hxsum : β x : β β β€, xsum x = β i : Fin n, x i * x ((i + 1) % n))
: (β x : β β β€, xset x β§ xsum x = putnam_1996_b3_solution n) β§ (β x : β β β€, xset x β xsum x β€ putnam_1996_b3_solution n) :=
sorry | Given that $\{x_1,x_2,\ldots,x_n\}=\{1,2,\ldots,n\}$, find, with proof, the largest possible value, as a function of $n$ (with $n \geq 2$), of $x_1x_2+x_2x_3+\cdots+x_{n-1}x_n+x_nx_1$. | Show that the maximum is $(2n^3+3n^2-11n+18)/6$. | ['algebra'] | Section putnam_1996_b3.
Require Import Nat List Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1996_b3_solution := 1.
Theorem putnam_1996_b3:
exists (m: nat),
(forall (n: nat), ge n 2 ->
sum_n (fun i => INR ((nth i (seq 1 (S n)) 0%nat) * (nth ((i + 1) mod n) (seq 1 (S n)) 0%nat))) n <= INR m) /\
(exists (n: nat), ge n 2 ->
sum_n (fun i => INR ((nth i (seq 1 (S n)) 0%nat) * (nth ((i + 1) mod n) (seq 1 (S n))) 0%nat)) n = INR m).
Proof. Admitted.
End putnam_1996_b3. | theory putnam_1996_b3 imports
Complex_Main
begin
definition putnam_1996_b3_solution :: "nat \<Rightarrow> nat" where "putnam_1996_b3_solution \<equiv> undefined"
(* \<lambda> n :: nat. (2 * n ^ 3 + 3 * n ^ 2 - 11 * n + 18) div 6 *)
theorem putnam_1996_b3:
fixes n :: nat
and xset :: "(nat \<Rightarrow> nat) \<Rightarrow> bool"
and xsum :: "(nat \<Rightarrow> nat) \<Rightarrow> nat"
defines "xset \<equiv> \<lambda> x :: nat \<Rightarrow> nat. x ` {0 .. n - 1} = {1 .. n}"
and "xsum \<equiv> \<lambda> x :: nat \<Rightarrow> nat. \<Sum> i = 0 .. n - 1. x i * x ((i + 1) mod n)"
assumes nge2 : "n \<ge> 2"
shows "(GREATEST S. \<exists> x :: nat \<Rightarrow> nat. xset x \<and> xsum x = S) = putnam_1996_b3_solution n"
sorry
end | null |
putnam_1996_b4 | abbrev putnam_1996_b4_solution : Prop := sorry
-- False
theorem putnam_1996_b4
(matsin : Matrix (Fin 2) (Fin 2) β β Matrix (Fin 2) (Fin 2) β)
(mat1996 : Matrix (Fin 2) (Fin 2) β)
(hmatsin : β A : Matrix (Fin 2) (Fin 2) β, matsin A = β' n : β, ((-(1 : β)) ^ n / (2 * n + 1)!) β’ A ^ (2 * n + 1))
(hmat1996 : mat1996 0 0 = 1 β§ mat1996 0 1 = 1996 β§ mat1996 1 0 = 0 β§ mat1996 1 1 = 1)
: (β A : Matrix (Fin 2) (Fin 2) β, matsin A = mat1996) β putnam_1996_b4_solution :=
sorry | For any square matrix $A$, we can define $\sin A$ by the usual power series: $\sin A=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}A^{2n+1}$. Prove or disprove: there exists a $2 \times 2$ matrix $A$ with real entries such that $\sin A=\begin{pmatrix} 1 & 1996 \\ 0 & 1 \end{pmatrix}$. | Show that there does not exist such a matrix $A$. | ['linear_algebra'] | Section putnam_1996_b4.
Require Import Factorial Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1996_b4_solution := False.
Theorem putnam_1996_b4:
let fix Mmult_n {T : Ring} {n : nat} (A : matrix n n) (p : nat) :=
match p with
| O => A
| S p' => @Mmult T n n n A (Mmult_n A p')
end in
let scale_c (c: R) (A: matrix 2 2) := mk_matrix 2 2 (fun i j => c * coeff_mat 0 A i j) in
let sinA_mat (n: nat) (A: matrix 2 2) := scale_c ((pow (-1) n) / INR (fact (2 * n + 1))) (Mmult_n A (Nat.add (Nat.mul 2 n) 1)) in
exists (A: matrix 2 2),
Series (fun n => coeff_mat 0 (sinA_mat n A) 0 0) = 1 /\
Series (fun n => coeff_mat 0 (sinA_mat n A) 0 1) = 1996 /\
Series (fun n => coeff_mat 0 (sinA_mat n A) 1 0) = 0 /\
Series (fun n => coeff_mat 0 (sinA_mat n A) 1 1) = 1 <->
putnam_1996_b4_solution.
Proof. Admitted.
End putnam_1996_b4. | theory putnam_1996_b4 imports
Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
begin
definition putnam_1996_b4_solution :: bool where "putnam_1996_b4_solution \<equiv> undefined"
(* False *)
theorem putnam_1996_b4:
fixes matsin :: "real^2^2 \<Rightarrow> real^2^2"
and mat1996 :: "real^2^2"
and matpow :: "real^2^2 \<Rightarrow> nat \<Rightarrow> real^2^2"
defines "matsin \<equiv> \<lambda> A :: real^2^2. \<Sum> n :: nat. ((-1) ^ n / fact (2 * n + 1)) *\<^sub>R (matpow A (2 * n + 1))"
and "mat1996 \<equiv> \<chi> i j. if i = 1 then (if j = 1 then 1 else 1996) else (if j = 1 then 0 else 1)"
assumes hmatpow: "\<forall> A :: real^2^2. matpow A 0 = mat 1 \<and> (\<forall> k :: nat. matpow A (k + 1) = A * (matpow A k))"
shows "(\<exists> A :: real^2^2. matsin A = mat1996) \<longleftrightarrow> putnam_1996_b4_solution"
sorry
end | null |
putnam_1996_b5 | abbrev putnam_1996_b5_solution : β β β := sorry
-- (fun n : β => if Even n then (3 * 2 ^ (n / 2) - 2) else (2 ^ ((n + 1) / 2) - 2))
theorem putnam_1996_b5
(n : β)
(STdelta : (Fin n β Fin 2) β Fin n β Fin n β β€)
(Sbalanced : (Fin n β Fin 2) β Prop)
(hSTdelta : β S : Fin n β Fin 2, β T1 T2 : Fin n, T1 β€ T2 β (STdelta S T1 T2 = β i : Set.Icc T1 T2, if S i = 1 then 1 else -1))
(hSbalanced : β S : Fin n β Fin 2, Sbalanced S = β T1 T2 : Fin n, T1 β€ T2 β (-2 β€ STdelta S T1 T2 β§ STdelta S T1 T2 β€ 2))
: {S : Fin n β Fin 2 | Sbalanced S}.encard = putnam_1996_b5_solution n :=
sorry | Given a finite string $S$ of symbols $X$ and $O$, we write $\Delta(S)$ for the number of $X$'s in $S$ minus the number of $O$'s. For example, $\Delta(XOOXOOX)=-1$. We call a string $S$ \emph{balanced} if every substring $T$ of (consecutive symbols of) $S$ has $-2 \leq \Delta(T) \leq 2$. Thus, $XOOXOOX$ is not balanced, since it contains the substring $OOXOO$. Find, with proof, the number of balanced strings of length $n$. | Show that the number of balanced strings of length $n$ is $3 \cdot 2^{n/2}-2$ if $n$ is even, and $2^{(n+1)/2}-2$ if $n$ is odd. | ['algebra'] | null | theory putnam_1996_b5 imports
Complex_Main
begin
definition putnam_1996_b5_solution :: "nat \<Rightarrow> nat" where "putnam_1996_b5_solution \<equiv> undefined"
(* \<lambda> n :: nat. 2 ^ ((n + 2) div 2) + 2 ^ ((n + 1) div 2) - 2 *)
theorem putnam_1996_b5:
fixes n :: nat
and STdelta :: "(nat \<Rightarrow> bool) \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> int"
and Sbalanced :: "(nat \<Rightarrow> bool) \<Rightarrow> bool"
defines "STdelta \<equiv> \<lambda> S :: nat \<Rightarrow> bool. \<lambda> T1 T2 :: nat. \<Sum> i = T1 .. T2. if S i then 1 else -1"
and "Sbalanced \<equiv> \<lambda> (S :: nat \<Rightarrow> bool). \<forall> T1 \<in> {0 .. n - 1}. \<forall> T2 \<in> {0 .. n - 1}. T1 \<le> T2 \<longrightarrow> (-2 \<le> STdelta S T1 T2 \<and> STdelta S T1 T2 \<le> 2)"
shows "card {S :: nat \<Rightarrow> bool. Sbalanced S \<and> (\<forall> i \<ge> n. S i = False)} = putnam_1996_b5_solution n"
sorry
end | null |
putnam_2016_a1 | abbrev putnam_2016_a1_solution : β := sorry
-- 8
theorem putnam_2016_a1
: (β j : β+, (β P : β€[X], β k : β€, 2016 β£ (derivative^[j] P).eval k) β j β₯ putnam_2016_a1_solution) β§ (β P : β€[X], β k : β€, 2016 β£ (derivative^[putnam_2016_a1_solution] P).eval k) :=
sorry | Find the smallest positive integer $j$ such that for every polynomial $p(x)$ with integer coefficients and for every integer $k$, the integer \[ p^{(j)}(k) = \left. \frac{d^j}{dx^j} p(x) \right|_{x=k} \] (the $j$-th derivative of $p(x)$ at $k$) is divisible by 2016. | Show that the solution is $18$. | ['algebra', 'number_theory'] | null | theory putnam_2016_a1 imports
Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
definition putnam_2016_a1_solution :: nat where "putnam_2016_a1_solution \<equiv> undefined"
(* 8 *)
theorem putnam_2016_a1:
shows "putnam_2016_a1_solution = (LEAST j :: nat. j > 0 \<and> (\<forall> P :: int poly. \<forall> k :: int. 2016 dvd poly ((pderiv ^^ j) P) k))"
sorry
end | null |
putnam_2016_a2 | abbrev putnam_2016_a2_solution : β := sorry
-- (3 + sqrt 5) / 2
theorem putnam_2016_a2
(p : β β β β Prop := fun n β¦ fun m β¦ Nat.choose m (n - 1) > Nat.choose (m - 1) n)
(M : β β β)
(hpM : β n : β, p n (M n))
(hMub : β n : β, β m : β, p n m β m β€ M n)
: (Tendsto (fun n β¦ ((M n : β) / (n : β))) β€ (π putnam_2016_a2_solution)) :=
sorry | Given a positive integer $n$, let $M(n)$ be the largest integer $m$ such that
\[
inom{m}{n-1} > \binom{m-1}{n}.
\]
Evaluate
\[
\lim_{n o \infty} \frac{M(n)}{n}.
\] | Show that the answer is $\frac{3 + \sqrt{5}}{2}$. | ['analysis'] | Section putnam_2016_a2.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2016_a2_solution := (3 + sqrt 5) / 2.
Theorem putnam_2016_a2
(p : nat -> nat -> Prop := fun n m => Binomial.C m (n - 1) > Binomial.C (m - 1) n)
(M : nat -> nat)
(pM : forall n : nat, p n (M n))
(hMub : forall n m : nat, p n m -> le m (M n))
: Lim_seq (fun n => (INR (M n) / INR n)) = putnam_2016_a2_solution.
Proof. Admitted.
End putnam_2016_a2. | theory putnam_2016_a2 imports
Complex_Main
begin
definition putnam_2016_a2_solution :: real where "putnam_2016_a2_solution \<equiv> undefined"
(* (3 + sqrt 5) / 2 *)
theorem putnam_2016_a2:
fixes M :: "nat \<Rightarrow> nat"
defines "M \<equiv> \<lambda> n. GREATEST m. m choose (n - 1) > (m - 1) choose n"
shows "(\<lambda> n. M n / n) \<longlonglongrightarrow> putnam_2016_a2_solution"
sorry
end | null |
putnam_2016_a3 | abbrev putnam_2016_a3_solution : β := sorry
-- 3 * Real.pi / 8
theorem putnam_2016_a3
(f : β β β)
(hf : β x : β, x β 0 β f x + f (1 - 1 / x) = arctan x)
: (β« x in (0)..1, f x = putnam_2016_a3_solution) :=
sorry | Suppose that $f$ is a function from $\mathbb{R}$ to $\mathbb{R}$ such that
\[
f(x) + f\left( 1 - \frac{1}{x} \right) = \arctan x
\]
for all real $x \neq 0$. (As usual, $y = \arctan x$ means $-\pi/2 < y < \pi/2$ and $\tan y = x$.) Find
\[
\int_0^1 f(x)\,dx.
\] | Prove that the answer is $\frac{3\pi}{8}$. | ['analysis'] | Section putnam_2017_b6.
Require Import Ensembles Finite_sets Factorial List. From mathcomp Require Import div.
Definition putnam_2017_b6_solution := Nat.div (fact 2016) (fact 1953) - fact 63 * 2016.
Theorem putnam_2017_b6:
exists (A: Ensemble (list nat)),
forall (l: list nat), (A l <->
(length l = 64 /\ NoDup l /\ forall (n: nat), In n l -> 1 <= n <= 2017) ->
let Hweighted_sum :=
fix weighted_sum (l: list nat) (i: nat) : nat :=
match l with
| nil => 0
| x :: xs => (if Nat.ltb i 2 then x else i * x) + weighted_sum xs (i + 1)
end in
Hweighted_sum l 0 %| 2017 = true)
<-> cardinal (list nat) A putnam_2017_b6_solution.
Proof. Admitted.
Section putnam_2016_a3.
Require Import Reals Coquelicot.RInt Coquelicot.Hierarchy.
Definition putnam_2016_a3_solution := 3 * PI / 8.
Theorem putnam_2016_a3:
forall (f: R -> R),
let g (x: R) := (f x) + f (1 - 1/x) in
forall(x: R), x <> 0 -> g x = atan x ->
RInt f 0 1 = putnam_2016_a3_solution.
Proof. Admitted.
End putnam_2016_a3. | theory putnam_2016_a3 imports Complex_Main
"HOL-Analysis.Set_Integral"
"HOL-Analysis.Lebesgue_Measure"
begin
definition putnam_2016_a3_solution::real where "putnam_2016_a3_solution \<equiv> undefined"
(* 3 * pi / 8 *)
theorem putnam_2016_a3:
fixes f::"real\<Rightarrow>real"
assumes hf : "\<forall>x. x \<noteq> 0 \<longrightarrow> f x + f (1 - 1 / x) = arctan x"
shows "set_lebesgue_integral lebesgue {0..1} f = putnam_2016_a3_solution"
sorry
end
| null |
putnam_2016_a5 | theorem putnam_2016_a5
(G : Type*) [Group G]
(Gfin : Fintype G)
(g h : G)
(ghgen : Group.closure {g, h} = G β§ Β¬Group.closure {g} = G β§ Β¬Group.closure {h} = G)
(godd : Odd (orderOf g))
(S : Set G := {g * h, gβ»ΒΉ * h, g * hβ»ΒΉ, gβ»ΒΉ * hβ»ΒΉ})
: (β x : G, β mn : List G, 1 β€ mn.length β§ mn.length β€ Gfin.card β§ β i : Fin mn.length, mn.get i β S β§ x = List.prod mn) :=
sorry | Suppose that $G$ is a finite group generated by the two elements $g$ and $h$, where the order of $g$ is odd. Show that every element of $G$ can be written in the form
\[
g^{m_1} h^{n_1} g^{m_2} h^{n_2} \cdots g^{m_r} h^{n_r}
\]
with $1 \leq r \leq |G|$ and $m_1, n_1, m_2, n_2, \ldots, m_r, n_r \in \{-1, 1\}$.
(Here $|G|$ is the number of elements of $G$.) | null | ['abstract_algebra'] | Section putnam_2016_a5.
Theorem putnam_2016_a5: True.
Proof. Admitted.
End putnam_2016_a5. | theory putnam_2016_a5 imports Complex_Main
"HOL-Algebra.Multiplicative_Group"
begin
theorem putnam_2016_a5:
fixes G (structure) and g h::"'a" and S::"'a set"
defines "S \<equiv> {g \<otimes> h, inv g \<otimes> h, g \<otimes> inv h, inv g \<otimes> inv h}"
assumes hG : "Group.group G \<and> finite (carrier G)"
and ghgen : "generate G {h, g} = carrier G \<and> generate G {h} \<noteq> carrier G \<and> generate G {g} \<noteq> carrier G"
and godd : "odd ( (group.ord G) g)"
shows "\<forall>x \<in> carrier G. \<exists> mn::'a list. \<forall>i\<in>{0..<card (carrier G)}. size mn \<ge> 1 \<and> size mn \<le> card (carrier G)
\<and> mn!i \<in> S \<and> x = foldr (\<otimes>\<^bsub>G\<^esub>) mn \<one>"
sorry
end
| null |
putnam_2016_a6 | abbrev putnam_2016_a6_solution : β := sorry
-- 5 / 6
theorem putnam_2016_a6
(C : β)
(max : Polynomial β β β)
(hmax : β P : Polynomial β, β x β Icc 0 1, |P.eval x| = max P)
(hmaxub : β P : Polynomial β, β x β Icc 0 1, |P.eval x| β€ max P)
(p : β β Prop := fun c β¦ β P : Polynomial β, P.degree = 3 β (β x β Icc 0 1, P.eval x = 0) β β« x in (0)..1, |P.eval x| β€ c * max P)
(hpC : p C)
(hClb : β c : β, p c β C β€ c)
: (C = putnam_2016_a6_solution) :=
sorry | Find the smallest constant $C$ such that for every real polynomial $P(x)$ of degree $3$ that has a root in the interval $[0,1]$,
\[
\int_0^1 \left| P(x) \right|\,dx \leq C \max_{x \in [0,1]} \left| P(x) \right|.
\] | Prove that the smallest such value of $C$ is $\frac{5}{6}$. | ['algebra'] | Section putnam_2016_a6.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2016_a6_solution := 5 / 6.
Theorem putnam_2016_a6
(C : R)
(max : (R -> R) -> R)
(hmax : forall (P : R -> R) (coeff: nat -> R) (n: nat),
(coeff n <> 0 /\ P = (fun x => sum_n (fun i => coeff i * x ^ i) (n + 1))) ->
exists (x: R), 0 <= x <= 1 /\ Rabs (P x) = max P)
(hmaxub : forall (P : R -> R) (coeff: nat -> R) (n: nat),
(coeff n <> 0 /\ P = (fun x => sum_n (fun i => coeff i * x ^ i) (n + 1))) ->
forall (x: R), 0 <= x <= 1 /\ Rabs (P x) <= max P)
(p : R -> Prop :=
fun c =>
forall (P : R -> R) (coeff: nat -> R),
(coeff 3%nat <> R0 /\ P = (fun x => sum_n (fun i => coeff i * x ^ i) 4)) ->
(exists (x: R), 0 <= x <= 1 /\ P x = 0) -> RInt P 0 1 <= c * max P)
(hpC : p C)
(hClb : forall c : R, p c -> C <= c)
: (C = putnam_2016_a6_solution).
Proof. Admitted.
End putnam_2016_a6. | theory putnam_2016_a6 imports
Complex_Main
"HOL-Computational_Algebra.Polynomial"
"HOL-Analysis.Set_Integral"
"HOL-Analysis.Lebesgue_Measure"
begin
definition putnam_2016_a6_solution :: real where "putnam_2016_a6_solution \<equiv> undefined"
(* 5 / 6 *)
theorem putnam_2016_a6:
shows "(LEAST C :: real. \<forall> P :: real poly. (degree P = 3) \<longrightarrow> (\<exists> x \<in> {0..1}. poly P x = 0) \<longrightarrow> set_lebesgue_integral lebesgue {0..1} (\<lambda> x. \<bar>poly P x\<bar>) \<le> C * (GREATEST y. \<exists> x \<in> {0..1}. \<bar>poly P x\<bar> = y)) = putnam_2016_a6_solution"
sorry
end | null |
putnam_2016_b1 | abbrev putnam_2016_b1_solution : β := sorry
-- exp 1 - 1
theorem putnam_2016_b1
(x : β β β)
(hx0 : x 0 = 1)
(hxn : β n : β, x (n + 1) = log (exp (x n) - (x n)))
: (β' n : β, x n = putnam_2016_b1_solution) :=
sorry | Let $x_0,x_1,x_2,\dots$ be the sequence such that $x_0=1$ and for $n \geq 0$,
\[
x_{n+1} = \ln(e^{x_n} - x_n)
\]
(as usual, the function $\ln$ is the natural logarithm). Show that the infinite series
\[
x_0 + x_1 + x_2 + \cdots
\]
converges and find its sum. | The sum converges to $e - 1$. | ['analysis'] | Section putnam_2016_b1.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2016_b1_solution := 1.
Theorem putnam_2016_b1
(x : nat -> R := fix x (n: nat) :=
match n with
| O => 1
| S n' => ln (exp (x n') - x n')
end)
: Series x = putnam_2016_b1_solution.
Proof. Admitted.
End putnam_2016_b1. | theory putnam_2016_b1 imports
Complex_Main
begin
definition putnam_2016_b1_solution :: real where "putnam_2016_b1_solution \<equiv> undefined"
(* exp 1 - 1 *)
theorem putnam_2016_b1:
fixes x :: "nat \<Rightarrow> real"
assumes hx0: "x 0 = 1"
and hxn: "\<forall> n :: nat. x (n + 1) = ln (exp (x n) - x n)"
shows "(\<Sum> n :: nat. x n) = putnam_2016_b1_solution"
sorry
end | null |
putnam_2016_b2 | abbrev putnam_2016_b2_solution : β Γ β := sorry
-- (3 / 4, 4 / 3)
theorem putnam_2016_b2
(squarish : β€ β Prop := fun n β¦ IsSquare n β¨ β w : β€, IsSquare |n - w ^ 2| β§ β v : β, |n - w ^ 2| β€ |n - v ^ 2|)
(S : β€ β β := fun n β¦ {i β Finset.Icc 1 n | squarish i}.ncard)
(p : β β β β Prop := fun Ξ± β¦ fun Ξ² β¦ Ξ± > 0 β§ Ξ² > 0 β§ Tendsto (fun N β¦ S N / (N : β) ^ Ξ±) β€ (π Ξ²))
: ((β Ξ± Ξ² : β, ((Ξ±, Ξ²) = putnam_2016_b2_solution β p Ξ± Ξ²)) β¨ Β¬β Ξ± Ξ² : β, p Ξ± Ξ²) :=
sorry | Define a positive integer $n$ to be \emph{squarish} if either $n$ is itself a perfect square or the distance from $n$ to the nearest perfect square is a perfect square. For example, $2016$ is squarish, because the nearest perfect square to $2016$ is $45^2 = 2025$ and $2025 - 2016 = 9$ is a perfect square. (Of the positive integers between $1$ and $10$, only $6$ and $7$ are not squarish.)
For a positive integer $N$, let $S(N)$ be the number of squarish integers between $1$ and $N$,
inclusive. Find positive constants $\alpha$ and $\beta$ such that
\[
\lim_{N \to \infty} \frac{S(N)}{N^\alpha} = \beta,
\]
or show that no such constants exist. | Prove that the limit exists for $\alpha = \frac{3}{4}$ and equals $\beta = \frac{4}{3}$. | ['analysis'] | Section putnam_2016_b2.
Require Import Bool Reals Coquelicot.Lim_seq Coquelicot.Rbar.
Theorem putnam_2016_b2:
let squarish (n: nat) := existsb ( fun m => Nat.eqb n (m * m) || (forallb (fun p => leb ((n-m)*(n-m)) ((n-p)*(n-p))) (seq 0 (S n))) ) (seq 0 (S n)) in
let squarish_set (n : nat) : list nat := filter (fun x => squarish x) (seq 1 n) in
exists (a b: nat),
Lim_seq (fun N => INR (length (squarish_set N)) / INR (N^a)) = Finite (INR b).
Proof. Admitted.
End putnam_2016_b2. | theory putnam_2016_b2 imports Complex_Main
begin
definition is_square::"int\<Rightarrow>bool" where "is_square n \<equiv> \<exists>x::nat. n = x * x"
definition squarish::"int\<Rightarrow>bool" where "squarish n \<equiv> is_square n \<or>
(\<exists>w::int. is_square (abs (n - w * w)) \<and> (\<forall>v::nat. abs (n - w * w) \<le> abs (n - v * v)))"
definition putnam_2016_b2_solution::"(real \<times> real)" where "putnam_2016_b2_solution \<equiv> undefined"
(* (3/4, 4/3) *)
theorem putnam_2016_b2:
fixes S::"nat\<Rightarrow>nat" and p::"real\<Rightarrow>real\<Rightarrow>bool"
defines "S \<equiv> \<lambda>n::nat. card {i \<in> {1..n}. squarish i}"
and "p \<equiv> \<lambda>a::real. \<lambda>b::real. a > 0 \<and> b > 0 \<and> ((\<lambda>n. S n / (n powr a)) \<longlonglongrightarrow> b)"
shows "(\<forall>a b::real. (a, b) = putnam_2016_b2_solution \<longleftrightarrow> p a b) \<or> (\<forall>a b::real. \<not> (p a b))"
sorry
end
| null |
putnam_2016_b5 | abbrev putnam_2016_b5_solution : Set (Set.Ioi (1 : β) β Set.Ioi (1 : β)) := sorry
-- {f : Set.Ioi (1 : β) β Set.Ioi (1 : β) | β c : β, c > 0 β§ β x : Set.Ioi (1 : β), (f x : β) = x ^ c}
theorem putnam_2016_b5
(f : Set.Ioi (1 : β) β Set.Ioi (1 : β))
(fle : Prop)
(hfle : fle = β x y : Set.Ioi (1 : β), ((x : β) ^ 2 β€ y β§ y β€ (x : β) ^ 3) β ((f x : β) ^ 2 β€ f y β§ f y β€ (f x : β) ^ 3))
: fle β f β putnam_2016_b5_solution := sorry | Find all functions $f$ from the interval $(1,\infty)$ to $(1,\infty)$ with the following property: if $x,y \in (1,\infty)$ and $x^2 \leq y \leq x^3$, then $(f(x))^2 \leq f(y) \leq (f(x))^3$. | Show that the only such functions are the functions $f(x)=x^c$ for some $c>0$. | ['algebra'] | Section putnam_2016_b5.
Require Import Reals Rpower.
Local Open Scope R.
Definition putnam_2016_b5_solution (f: R -> R) : Prop := exists (c: R), c > 0 /\ forall (x: R), x > 1 -> f x = Rpower x c .
Theorem putnam_2016_b5:
forall (f: R -> R), forall (x y: R),
(x > 1 /\ y > 1 /\ f x > 1 /\ f y > 1)
/\
(x*x <= y <= x*x*x)
->
(f x)*(f x) <= f y <= (f x)*(f x)*(f x)
<->
putnam_2016_b5_solution f.
Proof. Admitted.
End putnam_2016_b5. | theory putnam_2016_b5 imports Complex_Main
begin
(* uses ((real \<Rightarrow> real) set) instead of (({1<..} \<Rightarrow> {1<..}) set) *)
definition putnam_2016_b5_solution :: "(real \<Rightarrow> real) set" where "putnam_2016_b5_solution \<equiv> undefined"
(* {f::real\<Rightarrow>real. (\<exists>c::real. c > 0 \<and> (\<forall>x::real>1. f x = x powr c))} *)
theorem putnam_2016_b5:
fixes f :: "real \<Rightarrow> real"
and fle :: bool
assumes hfle: "fle = (\<forall>x::real>1. \<forall>y::real>1. (x^2 \<le> y \<and> y \<le> x^3) \<longrightarrow> ((f x)^2 \<le> f y \<and> f y \<le> (f x)^3))"
shows "fle \<longleftrightarrow> f \<in> putnam_2016_b5_solution"
sorry
end
| null |
putnam_2016_b6 | abbrev putnam_2016_b6_solution : β := sorry
-- 1
theorem putnam_2016_b6
: β' k : β, ((-1 : β) ^ ((k + 1 : β€) - 1) / (k + 1 : β)) * β' n : β, (1 : β) / ((k + 1) * (2 ^ n) + 1) = putnam_2016_b6_solution :=
sorry | Evaluate $\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} \sum_{n=0}^\infty \frac{1}{k2^n+1}$. | Show that the desired sum equals $1$. | ['analysis'] | Section putnam_2016_b6.
Require Import List Reals Coquelicot.Hierarchy Coquelicot.Series.
Definition putnam_2016_b6_solution := 1.
Theorem putnam_2016_b6:
Series (fun k => (-1)^k/(INR k+1) * Series (fun n => 1/(INR k*(2^n)+1))) = putnam_2016_b6_solution.
Proof. Admitted.
End putnam_2016_b6. | theory putnam_2016_b6 imports Complex_Main
"HOL-Analysis.Infinite_Sum"
begin
definition putnam_2016_b6_solution :: real where "putnam_2016_b6_solution \<equiv> undefined"
(* 1 *)
theorem putnam_2016_b6:
shows "(\<Sum>\<^sub>\<infinity>k::nat\<in>{1..}. ((-1)^(k-1) / k) * (\<Sum>n::nat. 1 / (k*2^n + 1))) = putnam_2016_b6_solution"
sorry
end
| null |
putnam_1982_a2 | abbrev putnam_1982_a2_solution : Prop := sorry
-- True
theorem putnam_1982_a2
(B : β β β β β := fun n x β¦ β k in Finset.Icc 1 n, k ^ x)
(f : β β β := fun n β¦ B n (logb n 2) / (n * logb 2 n) ^ 2)
: (β L : β, Tendsto (fun N β¦ β j in Finset.Icc 2 N, f j) β€ (π L)) β putnam_1982_a2_solution :=
sorry | For positive real $x$, let $B_n(x)=1^x+2^x+3^x+\dots+n^x$. Prove or disprove the convergence of $\sum_{n=2}^\infty \frac{B_n(\log_n2)}{(n\log_2n)^2}$. | Show that the given series converges. | ['analysis'] | Section putnam_1982_a2.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1982_a2_solution := True.
Theorem putnam_1982_a2:
let B (n: nat) (x: R) := sum_n (fun n => Rpower (INR n) x) n in
let f (n: nat) := B n (ln 2 / ln (INR n)) / (INR n) * Rpower (ln 2 / ln (INR n)) 2 in
ex_series (fun n => if (lt_dec n 2) then 0 else f n) <-> putnam_1982_a2_solution.
Proof. Admitted.
End putnam_1982_a2. | theory putnam_1982_a2 imports Complex_Main
begin
definition putnam_1982_a2_solution :: "bool" where
"putnam_1982_a2_solution \<equiv> undefined"
(* True *)
theorem putnam_1982_a2:
fixes B :: "nat \<Rightarrow> real \<Rightarrow> real"
and f :: "nat \<Rightarrow> real"
defines "B \<equiv> \<lambda> (n :: nat) (x :: real). (\<Sum> k \<in> {1 :: nat..n}. (real k) powr x)"
and "f \<equiv> \<lambda> n :: nat. (B n (log n 2)) / ((n * (log 2 n)) ^ 2)"
shows "putnam_1982_a2_solution \<longleftrightarrow> (\<exists> L :: real. filterlim (\<lambda> N :: nat. (\<Sum> j \<in> {2..N}. f j)) (nhds L) at_top)"
sorry
end | null |
putnam_1982_a3 | abbrev putnam_1982_a3_solution : β := sorry
-- (Real.pi / 2) * log Real.pi
theorem putnam_1982_a3
: (Tendsto (fun t β¦ β« x in (0)..t, (arctan (Real.pi * x) - arctan x) / x) β€ (π putnam_1982_a3_solution)) :=
sorry | Evaluate $\int_0^{\infty} \frac{\tan^{-1}(\pi x) - \tan^{-1} x}{x} \, dx$. | Show that the integral evaluates to $\frac{\pi}{2} \ln \pi$. | ['analysis'] | Section putnam_1982_a3.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1982_a3_solution := PI / 2 * ln PI.
Theorem putnam_1982_a3:
let f (x: R) := (atan (PI * x) - atan x) / x in
Lim_seq (fun n => RInt f 0 (INR n)) = putnam_1982_a3_solution.
Proof. Admitted.
End putnam_1982_a3. | theory putnam_1982_a3 imports Complex_Main
"HOL-Analysis.Interval_Integral"
begin
definition putnam_1982_a3_solution :: "real" where
"putnam_1982_a3_solution \<equiv> undefined"
(* (pi / 2) * ln pi *)
theorem putnam_1982_a3:
shows "filterlim (\<lambda> t :: real. (interval_lebesgue_integral lebesgue 0 t (\<lambda> x. (arctan (pi * x) - arctan x)/x))) (nhds putnam_1982_a3_solution) at_top"
sorry
end | null |
putnam_1982_a5 | theorem putnam_1982_a5
(a b c d : β€)
(hpos : a > 0 β§ b > 0 β§ c > 0 β§ d > 0)
(hac : a + c β€ 1982)
(hfrac : (a : β) / b + (c : β) / d < 1)
: (1 - (a : β) / b - (c : β) / d > 1 / 1983 ^ 3) :=
sorry | Let $a, b, c, d$ be positive integers satisfying $a + c \leq 1982$ and $\frac{a}{b} + \frac{c}{d} < 1$. Prove that $1 - \frac{a}{b} - \frac{c}{d} > \frac{1}{1983^3}$. | null | ['algebra'] | Section putnam_1982_a5.
Require Import Reals.
Open Scope R.
Theorem putnam_1982_a5:
forall (a b c d: nat), Nat.le (Nat.add a c) 1982 /\ INR a / INR b + INR c / INR d < 1 ->
1 - INR a / INR b - INR c / INR d > 1/pow 1983 3.
Proof. Admitted.
End putnam_1982_a5. | theory putnam_1982_a5 imports Complex_Main
begin
theorem putnam_1982_a5:
fixes a b c d :: "nat"
assumes hpos : "a > 0 \<and> b > 0 \<and> c > 0 \<and> d > 0"
and hac : "a + c \<le> 1982"
and hfrac : "(real a) / b + (real c) / d < 1"
shows "(1 - (real a)/b - (real c)/d) > 1 / 1983^3"
sorry
end | null |
putnam_1982_a6 | abbrev putnam_1982_a6_solution : Prop := sorry
-- False
theorem putnam_1982_a6
(S : Set β := Ici 1)
(hb : (β β β) β Prop := fun b : β β β => BijOn b S S)
(hx : (β β β) β Prop := fun x : β β β => StrictAntiOn (fun n : β => |x n|) S)
(limb : (β β β) Γ (β β β) β Prop := fun (b, x) => Tendsto (fun n : β => |b n - (n : β€)| * |x n|) atTop (π 0))
(limx : (β β β) β Prop := fun x : β β β => Tendsto (fun n : β => β k in Finset.Icc 1 n, x k) atTop (π 1))
: (β b : β β β, β x : β β β, hb b β§ hx x β§ limb (b, x) β§ limx x β
Tendsto (fun n : β => β k in Finset.Icc 1 n, x (b k)) atTop (π 1)) β putnam_1982_a6_solution :=
sorry | Let $b$ be a bijection from the positive integers to the positive integers. Also, let $x_1, x_2, x_3, \dots$ be an infinite sequence of real numbers with the following properties:
\begin{enumerate}
\item
$|x_n|$ is a strictly decreasing function of $n$;
\item
$\lim_{n \rightarrow \infty} |b(n) - n| \cdot |x_n| = 0$;
\item
$\lim_{n \rightarrow \infty}\sum_{k = 1}^{n} x_k = 1$.
\end{enumerate}
Prove or disprove: these conditions imply that $$\lim_{n \rightarrow \infty} \sum_{k = 1}^{n} x_{b(k)} = 1.$$ | The limit need not equal $1$. | ['analysis'] | Section putnam_1982_a6.
Require Import Nat Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1982_a6_solution := False.
Theorem putnam_1982_a6:
forall (a: nat -> R),
(Series a = 1 /\ forall (i j: nat), le i j -> Rabs (a i) > Rabs (a j)) /\
forall (f: nat -> nat), Lim_seq (fun i => Rabs (INR (f i - i)) * Rabs (a i)) = 0 ->
Series (fun i => a (f i)) = 1 -> putnam_1982_a6_solution.
Proof. Admitted.
End putnam_1982_a6. | theory putnam_1982_a6 imports Complex_Main
begin
definition putnam_1982_a6_solution::bool where "putnam_1982_a6_solution \<equiv> undefined"
(* False *)
theorem putnam_1982_a6:
fixes hb::"(nat\<Rightarrow>nat) \<Rightarrow> bool" and hx limx::"(nat\<Rightarrow>real) \<Rightarrow> bool" and limb::"(nat\<Rightarrow>nat) \<Rightarrow> (nat\<Rightarrow>real) \<Rightarrow> bool"
defines "hb \<equiv> \<lambda>b. bij_betw b {1..} {1..}"
and "hx \<equiv> \<lambda>x. strict_antimono_on {1..} (\<lambda>n. abs(x n))"
and "limb \<equiv> \<lambda>b::nat\<Rightarrow>nat. \<lambda>x::nat\<Rightarrow>real. (\<lambda>n. (abs (b n - n)) * (abs (x n))) \<longlonglongrightarrow> 0"
and "limx \<equiv> \<lambda>x. (\<lambda>n. (\<Sum>k=1..n. x k)) \<longlonglongrightarrow> 1"
shows "(\<forall>b::nat\<Rightarrow>nat. \<forall>x::nat\<Rightarrow>real. hb b \<and> hx x \<and> limb b x \<and> limx x \<longrightarrow> ((\<lambda>n. (\<Sum>k=1..n. x (b k))) \<longlonglongrightarrow> 1)) \<longleftrightarrow> putnam_1982_a6_solution"
sorry
end | null |
putnam_1982_b2 | abbrev putnam_1982_b2_solution : Polynomial β := sorry
-- C Real.pi * (2*X - 1)^2
theorem putnam_1982_b2
(A : β Γ β β β := fun (x, y) => {(m, n) : β€ Γ β€ | m^2 + n^2 β€ x^2 + y^2}.ncard)
(g : β := β' k : β, Real.exp (-k^2))
(I : β := β« y : β, β« x : β, A (x, y) * Real.exp (-x^2 - y^2))
: I = putnam_1982_b2_solution.eval g :=
sorry | Let $A(x, y)$ denote the number of points $(m, n)$ with integer coordinates $m$ and $n$ where $m^2 + n^2 \le x^2 + y^2$. Also, let $g = \sum_{k = 0}^{\infty} e^{-k^2}$. Express the value $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} A(x, y)e^{-x^2 - y^2} dx dy$$ as a polynomial in $g$. | The desired polynomial is $\pi(2g - 1)^2$. | ['analysis'] | null | theory putnam_1982_b2 imports Complex_Main "HOL-Computational_Algebra.Polynomial" "HOL-Analysis.Interval_Integral"
begin
definition putnam_1982_b2_solution::"real poly" where "putnam_1982_b2_solution \<equiv> undefined"
(* [: pi :] * [: -1, 2 :] ^2 *)
theorem putnam_1982_b2:
fixes A::"real\<Rightarrow>real\<Rightarrow>nat" and g I::real
defines "A \<equiv> \<lambda>x::real. \<lambda>y::real. card {(m::int, n::int). m^2 + n^2 \<le> x^2 + y^2}"
and "g \<equiv> \<Sum>n::nat. exp (- (n^2))"
and "I \<equiv> interval_lebesgue_integral lebesgue MInfty PInfty
(\<lambda>x. interval_lebesgue_integral lebesgue MInfty PInfty (\<lambda>y. A x y * exp (- (x^2) - y^2)))"
shows "I = poly putnam_1982_b2_solution g"
sorry
end | null |
putnam_1982_b3 | abbrev putnam_1982_b3_solution : β := sorry
-- 4/3 * (Real.sqrt 2 - 1)
theorem putnam_1982_b3
(p : β β β := fun n : β => {(c, d) : Finset.Icc 1 n Γ Finset.Icc 1 n | β m : β, m^2 = c + d}.ncard / n^2)
: Tendsto (fun n : β => p n * Real.sqrt n) atTop (π putnam_1982_b3_solution) :=
sorry | Let $p_n$ denote the probability that $c + d$ will be a perfect square if $c$ and $d$ are selected independently and uniformly at random from $\{1, 2, 3, \dots, n\}$. Express $\lim_{n \rightarrow \infty} p_n \sqrt{n}$ in the form $r(\sqrt{s} - t)$ for integers $s$ and $t$ and rational $r$. | The limit equals $\frac{4}{3}(\sqrt{2} - 1)$. | ['analysis', 'number_theory', 'probability'] | null | theory putnam_1982_b3 imports Complex_Main
begin
definition putnam_1982_b3_solution::real where "putnam_1982_b3_solution \<equiv> undefined"
(* 4/3 * (sqrt 2 - 1) *)
theorem putnam_1982_b3:
fixes p::"nat\<Rightarrow>real"
defines "p \<equiv> \<lambda>n. card {(c::nat, d::nat). c \<in> {1..n} \<and> d \<in> {1..n} \<and> (\<exists>m::nat. m^2 = c + d) } / n^2"
shows "(\<lambda>n. p n * sqrt n) \<longlonglongrightarrow> putnam_1982_b3_solution"
sorry
end | null |
putnam_1982_b4 | abbrev putnam_1982_b4_solution : Prop Γ Prop := sorry
-- (True, True)
theorem putnam_1982_b4
(hn : Finset β€ β Prop := fun n : Finset β€ => β k : β€, β i in n, i β£ β i in n, (i + k))
: ((β n : Finset β€, hn n β (β i β n, |i| = 1)) β putnam_1982_b4_solution.1) β§
((β n : Finset β€, (hn n β§ β i β n, i > 0) β n = Finset.Icc (1 : β€) (n.card)) β putnam_1982_b4_solution.2) :=
sorry | Let $n_1, n_2, \dots, n_s$ be distinct integers such that, for every integer $k$, $n_1n_2\cdots n_s$ divides $(n_1 + k)(n_2 + k) \cdots (n_s + k)$. Prove or provide a counterexample to the following claims:
\begin{enumerate}
\item
For some $i$, $|n_i| = 1$.
\item
If all $n_i$ are positive, then $\{n_1, n_2, \dots, n_s\} = \{1, 2, \dots, s\}$.
\end{enumerate} | Both claims are true. | ['number_theory'] | null | theory putnam_1982_b4 imports Complex_Main
begin
definition putnam_1982_b4_solution::"bool \<times> bool" where "putnam_1982_b4_solution \<equiv> undefined"
(* True, True *)
theorem putnam_1982_b4:
fixes hn::"(int set) \<Rightarrow> bool"
defines "hn \<equiv> \<lambda>n. (\<forall>k::int. (\<Prod>i \<in> n. i) dvd (\<Prod>i \<in> n. (i + k)))"
shows "((\<forall>n::(int set). hn n \<longrightarrow> (\<exists>i \<in> n. abs(i) = 1)) \<longleftrightarrow> fst putnam_1982_b4_solution) \<and>
((\<forall>n::(int set). (hn n \<and> (\<forall>i \<in> n. i > 0)) \<longrightarrow> n = {1..card n})) \<longleftrightarrow> snd putnam_1982_b4_solution"
sorry
end | null |
putnam_1982_b5 | theorem putnam_1982_b5
(T : Set β := Ioi (Real.exp (Real.exp 1)))
(S : β β β β β)
(hS : β x β T, S x 0 = (Real.exp 1) β§ β n : β, S x (n + 1) = Real.logb (S x n) x)
(g : β β β)
: β x β T, (β L : β, Tendsto (S x) atTop (π L)) β§
(β x β T, Tendsto (S x) atTop (π (g x))) β ContinuousOn g T :=
sorry | For all $x > e^e$, let $S = u_0, u_1, \dots$ be a recursively defined sequence with $u_0 = e$ and $u_{n+1} = \log_{u_n} x$ for all $n \ge 0$. Prove that $S_x$ converges to some real number $g(x)$ and that this function $g$ is continuous for $x > e^e$. | null | ['analysis'] | Section putnam_1982_b5.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Theorem putnam_1982_b5:
let F :=
fix f (n: nat) (x: R) :=
match n with
| O => exp 1
| S n' => ln x / ln (f n' x)
end in
forall (x: R), x > Rpower (exp 1) (exp 1) ->
ex_lim_seq (fun n => F n x) /\
let g (x: R) := Lim_seq (fun n => F n x) in
continuity_pt g x.
Proof. Admitted.
End putnam_1982_b5. | theory putnam_1982_b5 imports Complex_Main
begin
theorem putnam_1982_b5:
fixes T::"real set" and S::"real\<Rightarrow>nat\<Rightarrow>real" and g::"real \<Rightarrow> real"
defines "T \<equiv> {(exp (exp 1))<..}"
assumes hS : "\<forall>x \<in> T. S x 0 = exp 1 \<and> (\<forall>n::nat. S x (n+1) = log (S x n) x)"
shows "\<forall>x \<in> T. (\<exists>L::real. (S x \<longlonglongrightarrow> L)) \<and> (\<forall>x \<in> T. (S x \<longlonglongrightarrow> g x)) \<longrightarrow> continuous_on T g"
sorry
end | null |
putnam_1983_a1 | abbrev putnam_1983_a1_solution : β := sorry
-- 2301
theorem putnam_1983_a1
: {n : β€ | n > 0 β§ (n β£ 10 ^ 40 β¨ n β£ 20 ^ 30)}.encard = putnam_1983_a1_solution :=
sorry | How many positive integers $n$ are there such that $n$ is an exact divisor of at least one of the numbers $10^{40},20^{30}$? | Show that the desired count is $2301$. | ['number_theory'] | Section putnam_1983_a1.
Require Import Nat Ensembles Finite_sets. From mathcomp Require Import div.
Definition putnam_1983_a1_solution := 2301.
Theorem putnam_1983_a1:
exists (E: Ensemble nat), forall (n: nat),
(E n <-> n %| 10^(40) = true \/ n %| 20^(30) = true) ->
cardinal nat E putnam_1983_a1_solution.
Proof. Admitted.
End putnam_1983_a1. | theory putnam_1983_a1 imports Complex_Main
begin
definition putnam_1983_a1_solution :: "nat" where "putnam_1983_a1_solution \<equiv> undefined"
(* 2301 *)
theorem putnam_1983_a1:
shows "card {n :: nat. n > 0 \<and> (n dvd 10^40 \<or> n dvd 20^30)} = putnam_1983_a1_solution"
sorry
end | null |
putnam_1983_a3 | theorem putnam_1983_a3
(p : β)
(F : β β β)
(poddprime : Odd p β§ p.Prime)
(hF : β n : β, F n = β i in Finset.range (p - 1), (i + 1) * n ^ i)
: β a β Finset.Icc 1 p, β b β Finset.Icc 1 p, a β b β Β¬(F a β‘ F b [MOD p]) :=
sorry | Let $p$ be in the set $\{3,5,7,11,\dots\}$ of odd primes and let $F(n)=1+2n+3n^2+\dots+(p-1)n^{p-2}$. Prove that if $a$ and $b$ are distinct integers in $\{0,1,2,\dots,p-1\}$ then $F(a)$ and $F(b)$ are not congruent modulo $p$, that is, $F(a)-F(b)$ is not exactly divisible by $p$. | null | ['number_theory', 'algebra'] | Section putnam_1983_a3.
Require Import Nat Reals ZArith Znumtheory Coquelicot.Coquelicot.
Open Scope R.
Theorem putnam_1983_a3:
let f (n p: nat) : R := sum_n (fun i => INR ((i+1) * n^i)) (p-1) in
forall (p m n: nat), odd p = true /\ prime (Z.of_nat p) /\ (floor (f m p)) mod Z.of_nat p = (floor (f n p)) mod Z.of_nat p ->
Z.of_nat m mod Z.of_nat p = Z.of_nat n mod Z.of_nat p.
Proof. Admitted.
End putnam_1983_a3. | theory putnam_1983_a3 imports Complex_Main
"HOL-Number_Theory.Cong"
begin
theorem putnam_1983_a3:
fixes p :: "nat"
and F :: "nat \<Rightarrow> nat"
defines "F \<equiv> \<lambda> n::nat. (\<Sum> i=0..(p-2). ((i + 1) * n^i))"
assumes hp : "odd p \<and> prime p"
shows "\<forall> a b :: nat. ((a < p \<and> b < p \<and> a \<noteq> b) \<longrightarrow> \<not>([F a = F b] (mod p)))"
sorry
end | null |
putnam_1983_a4 | theorem putnam_1983_a4
(k : β)
(kpos : k > 0)
(m : β := 6 * k - 1)
(S : β€ := β j in Finset.Icc 1 (2 * k - 1), (-1) ^ (j + 1) * choose m (3 * j - 1))
: (S β 0) :=
sorry | Prove that for $m = 5 \pmod 6$,
\[
\binom{m}{2} - \binom{m}{5} + \binom{m}{8} - \binom{m}{11} + ... - \binom{m}{m-6} + \binom{m}{m-3} \neq 0.
\] | null | ['algebra'] | Section putnam_1983_a4.
Require Import Binomial Reals Znumtheory Coquelicot.Coquelicot.
Open Scope nat_scope.
Theorem putnam_1983_a4:
forall (m: nat), m mod 6 = 5 ->
sum_n (fun n => (if (eq_nat_dec (n mod 3) 2) then Binomial.C m n else R0)) (m-2) <> R0.
Proof. Admitted.
End putnam_1983_a4. | theory putnam_1983_a4 imports Complex_Main
begin
theorem putnam_1983_a4:
fixes k m :: "nat"
and S :: "int"
defines "m \<equiv> 6*k-1"
and "S \<equiv> \<Sum> j::nat \<in> {1..2*k-1}. (-1)^(j+1) * (m choose (3 * j -1))"
assumes kpos : "k > 0"
shows "S \<noteq> 0"
sorry
end | null |
putnam_1983_a5 | abbrev putnam_1983_a5_solution : Prop := sorry
-- True
theorem putnam_1983_a5
: ((β Ξ± : β, Ξ± > 0 β§ β n : β, n > 0 β Even (βΞ± ^ nβ - n)) β putnam_1983_a5_solution) :=
sorry | Prove or disprove that there exists a positive real number $\alpha$ such that $[\alpha_n] - n$ is even for all integers $n > 0$. (Here $[x]$ denotes the greatest integer less than or equal to $x$.) | Prove that such an $\alpha$ exists. | ['analysis'] | Section putnam_1983_a5.
Require Import Nat Reals Coquelicot.Coquelicot.
Definition putnam_1983_a5_solution := true.
Theorem putnam_1983_a5:
exists (a: R), forall (n: nat), gt n 0 -> even (Z.to_nat (floor (Rpower a (INR n))) - n) = putnam_1983_a5_solution.
Proof. Admitted.
End putnam_1983_a5. | theory putnam_1983_a5 imports Complex_Main
begin
definition putnam_1983_a5_solution :: "bool" where
"putnam_1983_a5_solution \<equiv> undefined"
(* True *)
theorem putnam_1983_a5:
shows "putnam_1983_a5_solution \<longleftrightarrow> (\<exists> \<alpha> :: real. \<alpha> > 0 \<and> (\<forall> n :: nat. n > 0 \<longrightarrow> even (\<lfloor>\<alpha>^n\<rfloor> - n)))"
sorry
end | null |
putnam_1983_a6 | abbrev putnam_1983_a6_solution : β := sorry
-- 2 / 9
theorem putnam_1983_a6
(F : β β β := fun a β¦ (a ^ 4 / exp (a ^ 3)) * β« x in (0)..a, β« y in (0)..(a - x), exp (x ^ 3 + y ^ 3))
: (Tendsto F β€ (π putnam_1983_a6_solution)) :=
sorry | Let $T$ be the triangle with vertices $(0, 0)$, $(a, 0)$, and $(0, a)$. Find $\lim_{a \to \infty} a^4 \exp(-a^3) \int_T \exp(x^3+y^3) \, dx \, dy$. | Show that the integral evaluates to $\frac{2}{9}$. | ['analysis'] | Section putnam_1983_a6.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_1983_a6_solution := 2 / 9.
Theorem putnam_1983_a6
: Lim_seq (fun a => let a := INR a in a ^ 4 * exp (-a ^ 3) * RInt (fun x => RInt (fun y => exp (x ^ 3 + y ^ 3)) 0 (a - x)) 0 a) = putnam_1983_a6_solution.
Proof. Admitted.
End putnam_1983_a6. | theory putnam_1983_a6 imports Complex_Main
"HOL-Analysis.Set_Integral"
"HOL-Analysis.Lebesgue_Measure"
begin
definition putnam_1983_a6_solution :: "real" where
"putnam_1983_a6_solution \<equiv> undefined"
(* 2/9 *)
theorem putnam_1983_a6:
fixes F :: "real \<Rightarrow> real"
defines "F \<equiv> \<lambda> a. (a^4 / exp (a^3)) * (set_lebesgue_integral lebesgue {(x :: real, y :: real). 0 \<le> x \<and> x \<le> a \<and> 0 \<le> y \<and> y \<le> a - x} (\<lambda> (x,y). exp (x^3 + y^3)))"
shows "filterlim F (nhds putnam_1983_a6_solution) at_top"
sorry
end | null |
putnam_1983_b2 | abbrev putnam_1983_b2_solution : Prop := sorry
-- True
theorem putnam_1983_b2
(f : β+ β β := fun n β¦ {M : Multiset β | (β m β M, β k : β, m = (2 ^ k : β€)) β§ (β m β M, M.count m β€ 3) β§ (M.sum : β€) = n}.ncard)
: ((β p : Polynomial β, β n : β+, βp.eval (n : β)β = f n) β putnam_1983_b2_solution) :=
sorry | Let $f(n)$ be the number of ways of representing $n$ as a sum of powers of $2$ with no power being used more than $3$ times. For example, $f(7) = 4$ (the representations are $4 + 2 + 1$, $4 + 1 + 1 + 1$, $2 + 2 + 2 + 1$, $2 + 2 + 1 + 1 + 1$). Can we find a real polynomial $p(x)$ such that $f(n) = [p(n)]$, where $[u]$ denotes the greatest integer less than or equal to $u$? | Prove that such a polynomial exists. | ['algebra'] | null | theory putnam_1983_b2 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
definition putnam_1983_b2_solution :: "bool" where
"putnam_1983_b2_solution \<equiv> undefined"
(* True *)
theorem putnam_1983_b2:
fixes f :: "nat \<Rightarrow> nat"
defines "f \<equiv> \<lambda> n. card {M :: nat multiset. (\<forall> m \<in># M. \<exists> k :: nat. m = 2^k) \<and> (\<forall> m \<in># M. count M m \<le> 3) \<and> (\<Sum>\<^sub># M = n)}"
shows "putnam_1983_b2_solution \<longleftrightarrow> (\<exists> p :: real poly. \<forall> n :: nat. n > 0 \<longrightarrow> \<lfloor>poly p (real n)\<rfloor> = f n)"
sorry
end | null |
putnam_1983_b4 | theorem putnam_1983_b4
(f : β β β€ := fun n β¦ n + Int.floor (Real.sqrt n))
(a : β β β)
(ha0 : a 0 > 0)
(han : β n : β, a (n + 1) = f (a n))
: (β i : β, β s : β€, a i = s ^ 2) :=
sorry | Let $f(n) = n + [\sqrt n]$, where $[x]$ denotes the greatest integer less than or equal to $x$. Define the sequence $a_i$ by $a_0 = m$, $a_{n+1} = f(a_n)$. Prove that it contains at least one square. | null | ['algebra'] | Section putnam_1983_b4.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Theorem putnam_1983_b4:
exists (m: nat),
let f (n: R) := n + IZR (floor (sqrt n)) in
let A :=
fix a (n: nat) :=
match n with
| O => INR m
| S n' => f (a n')
end in
exists (i: nat) (q: Z), A i = IZR (floor (A i)) /\ floor (A i) = Z.mul q q.
Proof. Admitted.
End putnam_1983_b4. | theory putnam_1983_b4 imports Complex_Main
begin
theorem putnam_1983_b4:
fixes a f :: "nat \<Rightarrow> nat"
defines "f \<equiv> \<lambda> n :: nat. n + (nat \<lfloor>sqrt (real n)\<rfloor>)"
assumes ha0 : "a 0 > 0"
and han : "\<forall> n :: nat. a (n + 1) = f (a n)"
shows "\<exists> i :: nat. \<exists> s :: nat. a i = s^2"
sorry
end | null |
putnam_1983_b5 | abbrev putnam_1983_b5_solution : β := sorry
-- log (4 / Real.pi)
theorem putnam_1983_b5
(dist : β β β := fun x β¦ min (x - βxβ) (βxβ - x))
(fact : Tendsto (fun N β¦ β n in Finset.Icc 1 N, (2 * n / (2 * n - 1)) * (2 * n / (2 * n + 1)) : β β β) β€ (π (Real.pi / 2)))
: (Tendsto (fun n β¦ (1 / n) * β« x in (1)..n, dist (n / x) : β β β) β€ (π putnam_1983_b5_solution)) :=
sorry | Define $\left\lVert x \right\rVert$ as the distance from $x$ to the nearest integer. Find $\lim_{n \to \infty} \frac{1}{n} \int_{1}^{n} \left\lVert \frac{n}{x} \right\rVert \, dx$. You may assume that $\prod_{n=1}^{\infty} \frac{2n}{(2n-1)} \cdot \frac{2n}{(2n+1)} = \frac{\pi}{2}$. | Show that the limit equals $\ln \left( \frac{4}{\pi} \right)$. | ['analysis'] | Section putnam_1983_b5.
Require Import Nat Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1983_b5_solution := ln (4/PI).
Theorem putnam_1983_b5:
let mindist (x: R) := Rmin (Rabs (x - IZR (floor x))) (Rabs (x - IZR (floor (x+1)))) in
Lim_seq (fun n => 1/(INR n) * (RInt (fun x => mindist (INR n/x)) 1 (INR n))) = putnam_1983_b5_solution.
Proof. Admitted.
End putnam_1983_b5. | theory putnam_1983_b5 imports Complex_Main
"HOL-Analysis.Interval_Integral"
begin
definition putnam_1983_b5_solution :: "real" where
"putnam_1983_b5_solution \<equiv> undefined"
(* ln (4 / pi) *)
theorem putnam_1983_b5:
fixes dist :: "real \<Rightarrow> real"
defines "dist \<equiv> \<lambda> x :: real. \<bar>x - round x\<bar>"
assumes fact : "filterlim (\<lambda>N :: nat. (\<Prod> n :: nat \<in> {1..N}. (2 * n / (2 * n - 1)) * (2 * n / (2 * n + 1)))) (nhds (pi / 2)) at_top"
shows "filterlim (\<lambda> n :: real. (1/n) * (interval_lebesgue_integral lebesgue 1 n (\<lambda> x :: real. dist (n/x)))) (nhds putnam_1983_b5_solution) at_top"
sorry
end | null |
putnam_1983_b6 | theorem putnam_1983_b6
(n : β)
(npos : n > 0)
(Ξ± : β)
(hΞ± : Ξ± ^ (2 ^ n + 1) - 1 = 0 β§ Ξ± β 1)
: (β p q : Polynomial β€, (aeval Ξ± p) ^ 2 + (aeval Ξ± q) ^ 2 = -1) :=
sorry | Let $n$ be a positive integer and let $\alpha \neq 1$ be a complex $(2n + 1)\textsuperscript{th}$ root of unity. Prove that there always exist polynomials $p(x)$, $q(x)$ with integer coefficients such that $p(\alpha)^2 + q(\alpha)^2 = -1$. | null | ['algebra'] | null | theory putnam_1983_b6 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
theorem putnam_1983_b6:
fixes n :: "nat"
and \<alpha> :: "complex"
and integralcoeffs :: "complex poly \<Rightarrow> bool"
defines "integralcoeffs \<equiv> \<lambda> p. (\<forall> n :: nat. (\<exists> k :: int. complex_of_int k = Re (coeff p n)) \<and> Im (coeff p n) = 0)"
assumes npos : "n > 0"
and h\<alpha> : "\<alpha>^(2^n + 1) - 1 = 0 \<and> \<alpha> \<noteq> 1"
shows "\<exists> p q :: complex poly. integralcoeffs p \<and> integralcoeffs q \<and> (poly p \<alpha>)^2 + (poly q \<alpha>)^2 = -1"
sorry
end
| null |
putnam_2003_a1 | abbrev putnam_2003_a1_solution : β β β := sorry
-- fun n => n
theorem putnam_2003_a1
(n : β)
(hn : n > 0)
: Set.encard {a : β β β€ | β k > 0, (β i : Fin k, a i = n) β§ (β i : Fin k, a i > 0) β§ (β i : Fin (k - 1), a i β€ a (i + 1)) β§ a (k - 1) β€ a 0 + 1 β§ (β i β₯ k, a i = 0)} = putnam_2003_a1_solution n :=
sorry | null | null | [] | Section putnam_2003_a1.
Require Import Nat List Ensembles Finite_sets Coquelicot.Coquelicot.
Definition putnam_2003_a1_solution (n: nat) := n.
Theorem putnam_2003_a1:
forall (n: nat), n > 0 ->
forall (E: Ensemble (list nat)) (l: list nat),
E l <->
forall (i j: nat), i < length l /\ j < length l /\ i < j -> nth i l 0 <= nth j l 0 /\ fold_left add l 0 = n ->
cardinal (list nat) E (putnam_2003_a1_solution n).
Proof. Admitted.
End putnam_2003_a1. | theory putnam_2003_a1 imports Complex_Main
begin
(* uses (nat \<Rightarrow> int) instead of (Fin k \<Rightarrow> int) *)
definition putnam_2003_a1_solution :: "nat \<Rightarrow> nat" where "putnam_2003_a1_solution \<equiv> undefined"
(* (\<lambda>n::nat. n) *)
theorem putnam_2003_a1:
fixes n :: nat
assumes hn: "n > 0"
shows "card {a::nat\<Rightarrow>int. (\<exists>k::nat>0. (\<Sum>i::nat=0..(k-1). a i) = n \<and> (\<forall>i::nat\<in>{0..(k-1)}. a i > 0) \<and> (\<forall>i::nat\<in>{0..(k-2)}. a i \<le> a (i+1)) \<and> a (k-1) \<le> a 0 + 1 \<and> (\<forall>i::nat\<ge>k. a i = 0))} = putnam_2003_a1_solution n"
sorry
end
| null |
putnam_2003_a2 | theorem putnam_2003_a2
(n : β)
(a b : Fin n β β)
(abnneg : β i : Fin n, a i β₯ 0 β§ b i β₯ 0)
: (β i : Fin n, a i) ^ ((1 : β) / n) + (β i : Fin n, b i) ^ ((1 : β) / n) β€ (β i : Fin n, (a i + b i)) ^ ((1 : β) / n) :=
sorry | Let $a_1,a_2,\dots,a_n$ and $b_1,b_2,\dots,b_n$ be nonnegative real numbers. Show that $(a_1a_2 \cdots a_n)^{1/n}+(b_1b_2 \cdots b_n)^{1/n} \leq [(a_1+b_1)(a_2+b_2) \cdots (a_n+b_n)]^{1/n}$. | null | ['algebra'] | Section putnam_2003_a2.
Require Import List Reals Coquelicot.Coquelicot.
Theorem putnam_2003_a2:
let fix suml (l1 l2 : list R) : list R :=
match l1, l2 with
| nil, _ => nil
| _, nil => nil
| h1 :: t1, h2 :: t2 => (h1 + h2) :: suml t1 t2
end in
forall (n: nat) (a b: list R), length a = n /\ length b = n ->
(fold_left Rmult a 1) ^ (1 / n) + (fold_left Rmult b 1) ^ (1 / n) <=
(fold_left Rmult (suml a b) 1) ^ (1 / n).
Proof. Admitted.
End putnam_2003_a2. | theory putnam_2003_a2 imports Complex_Main
begin
(* Note: Boosted domain to infinite set *)
theorem putnam_2003_a2:
fixes n::nat and a b::"nat\<Rightarrow>real"
assumes abnneg : "\<forall>i \<in> {0..<n}. a i \<ge> 0 \<and> b i \<ge> 0"
shows "((\<Prod>i=0..<n. a i) powr (1/n)) + ((\<Prod>i=0..<n. b i) powr (1/n)) \<le> ((\<Prod>i=0..<n. (a i + b i)) powr (1/n))"
sorry
end | null |
putnam_2003_a3 | abbrev putnam_2003_a3_solution : β := sorry
-- 2 * Real.sqrt 2 - 1
theorem putnam_2003_a3
(f : β β β)
(hf : β x : β, f x = |Real.sin x + Real.cos x + Real.tan x + 1 / Real.tan x + 1 / Real.cos x + 1 / Real.sin x|)
: (β x : β, f x = putnam_2003_a3_solution) β§ (β x : β, f x β₯ putnam_2003_a3_solution) :=
sorry | Find the minimum value of $|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|$ for real numbers $x$. | Show that the minimum is $2\sqrt{2}-1$. | ['analysis'] | Section putnam_2003_a3.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2003_a3_solution := 2 * sqrt 2 - 1.
Theorem putnam_2003_a3:
let f (x: R) := Rabs (sin x + cos x + tan x + 1 / tan x + 1 / cos x + 1 / sin x) in
exists (minx: R), forall (x: R), f minx <= f x ->
f minx = putnam_2003_a3_solution.
Proof. Admitted.
End putnam_2003_a3. | theory putnam_2003_a3 imports Complex_Main
begin
definition putnam_2003_a3_solution::real where "putnam_2003_a3_solution \<equiv> undefined"
(* 2 * sqrt 2 - 1 *)
theorem putnam_2003_a3:
fixes f::"real\<Rightarrow>real"
defines "f \<equiv> \<lambda>x::real. abs(sin x + cos x + tan x + 1 / tan x + 1 / cos x + 1 / sin x)"
shows "putnam_2003_a3_solution = (LEAST y. \<exists>x::real. f x = y)"
sorry
end | null |
putnam_2003_a4 | theorem putnam_2003_a4
(a b c A B C : β)
(aAne0 : a β 0 β§ A β 0)
(hle : β x : β, |a * x ^ 2 + b * x + c| β€ |A * x ^ 2 + B * x + C|)
: |b ^ 2 - 4 * a * c| β€ |B ^ 2 - 4 * A * C| :=
sorry | Suppose that $a,b,c,A,B,C$ are real numbers, $a \ne 0$ and $A \ne 0$, such that $|ax^2+bx+c| \leq |Ax^2+Bx+C|$ for all real numbers $x$. Show that $|b^2-4ac| \leq |B^2-4AC|$. | null | ['algebra'] | Section putnam_2003_a4.
Require Import Reals Coquelicot.Coquelicot.
Theorem putnam_2003_a4:
forall (a b c A B C: R), a <> 0 /\ A <> 0 /\ forall (x: R),
Rabs (a * x ^ 2 + b * x + c) <= Rabs (A * x ^ 2 + B * x + C) ->
Rabs (b ^ 2 - 4 * a * c) <= Rabs (B ^ 2 - 4 * A * C).
Proof. Admitted.
End putnam_2003_a4. | theory putnam_2003_a4 imports Complex_Main
begin
theorem putnam_2003_a4:
fixes a b c A B C::real
assumes aAne0 : "a \<noteq> 0 \<and> A \<noteq> 0"
and hle : "\<forall>x::real. abs(a * x^2 + b * x + c) \<le> abs(A * x^2 + B * x + C)"
shows "abs(b^2 - 4 * a * c) \<le> abs(B^2 - 4 * A * C)"
sorry
end | null |
putnam_2003_a6 | abbrev putnam_2003_a6_solution : Prop := sorry
-- True
theorem putnam_2003_a6
(r : Set β β β β β)
(hr : β (S : Set β) (n : β), r S n = β' s1 : S, β' s2 : S, if (s1 β s2 β§ s1 + s2 = n) then 1 else 0)
: (β A B : Set β, A βͺ B = β β§ A β© B = β
β§ (β n : β, r A n = r B n)) β putnam_2003_a6_solution :=
sorry | For a set $S$ of nonnegative integers, let $r_S(n)$ denote the number of ordered pairs $(s_1,s_2)$ such that $s_1 \in S$, $s_2 \in S$, $s_1 \ne s_2$, and $s_1+s_2=n$. Is it possible to partition the nonnegative integers into two sets $A$ and $B$ in such a way that $r_A(n)=r_B(n)$ for all $n$? | Show that such a partition is possible. | ['algebra'] | null | theory putnam_2003_a6 imports Complex_Main
begin
definition putnam_2003_a6_solution::bool where "putnam_2003_a6_solution \<equiv> undefined"
(* True *)
theorem putnam_2003_a6:
fixes r::"(nat set) \<Rightarrow> nat \<Rightarrow> nat"
defines "r \<equiv> \<lambda>S. \<lambda>n. card {(s1, s2). s1 \<in> S \<and> s2 \<in> S \<and> s1 \<noteq> s2 \<and> s1 + s2 = n}"
shows "(\<exists>A B::nat set. A \<union> B = \<nat> \<and> A \<inter> B = {} \<and> (\<forall>n::nat. r A n = r B n)) \<longleftrightarrow> putnam_2003_a6_solution"
sorry
end | null |
putnam_2003_b1 | abbrev putnam_2003_b1_solution : Prop := sorry
-- False
theorem putnam_2003_b1
: (β a b c d : Polynomial β, (β x y : β, 1 + x * y + x ^ 2 * y ^ 2 = a.eval x * c.eval y + b.eval x * d.eval y)) β putnam_2003_b1_solution :=
sorry | null | null | [] | Section putnam_2003_b1.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2003_b1_solution := True.
Theorem putnam_2003_b1:
let p (coeff: nat -> R) (x: R) (n: nat) := sum_n (fun i => coeff i * x ^ i) n in
exists (coeffa coeffb coeffc coeffd: nat -> R) (na nb nc nd: nat), forall (x y: R),
1 + x * y * (x * y) ^ 2 = (p coeffa x na) * (p coeffc y nc) + (p coeffb x nb) * (p coeffd y nd).
Proof. Admitted.
End putnam_2003_b1. | theory putnam_2003_b1 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
definition putnam_2003_b1_solution :: bool where "putnam_2003_b1_solution \<equiv> undefined"
(* False *)
theorem putnam_2003_b1:
shows "(\<exists>a b c d::real poly. (\<forall>x y::real. 1 + x*y + x^2*y^2 = poly a x * poly c y + poly b x * poly d y)) \<longleftrightarrow> putnam_2003_b1_solution"
sorry
end
| null |
putnam_2003_b3 | theorem putnam_2003_b3
(multicc : β β β β Prop)
(lcmicc : β β β)
(hmulticc : β n mult : β, multicc n mult = β i : Set.Icc 1 n, i.1 β£ mult)
(hlcmicc : β n : β, multicc n (lcmicc n) β§ (β n' : β, multicc n n' β n' β₯ (lcmicc n)))
: β n > 0, (n)! = β i : Set.Icc 1 n, lcmicc (n / i.1) :=
sorry | Show that for each positive integer $n$, $n!=\prod_{i=1}^n \text{lcm}\{1,2,\dots,\lfloor n/i \rfloor\}$. (Here lcm denotes the least common multiple, and $\lfloor x \rfloor$ denotes the greatest integer $\leq x$.) | null | ['number_theory'] | Section putnam_2003_b3.
Require Import Nat List Reals Coquelicot.Coquelicot.
Theorem putnam_2003_b3:
let fix lcm_n (args : list nat) : nat :=
match args with
| nil => 0%nat
| h :: args' => div (h * (lcm_n args')) (gcd h (lcm_n args'))
end in
let fix prod_n (m: nat -> R) (n : nat) : R :=
match n with
| O => m 0%nat
| S n' => m n' * prod_n m n'
end in
forall (n: nat), gt n 0 -> INR (fact n) = prod_n (fun i => INR (lcm_n (seq 0 (div n (i + 1))))) n.
Proof. Admitted.
End putnam_2003_b3. | theory putnam_2003_b3 imports Complex_Main
begin
theorem putnam_2003_b3:
fixes n::nat
assumes npos : "n > 0"
shows "fact n = (\<Prod>i=1..n. Lcm {1..\<lfloor>n / i\<rfloor>})"
sorry
end | null |
putnam_2003_b4 | theorem putnam_2003_b4
(f : β β β)
(a b c d e : β€)
(r1 r2 r3 r4 : β)
(ane0 : a β 0)
(hf1 : β z : β, f z = a * z ^ 4 + b * z ^ 3 + c * z ^ 2 + d * z + e)
(hf2 : β z : β, f z = a * (z - r1) * (z - r2) * (z - r3) * (z - r4))
: (Β¬Irrational (r1 + r2) β§ r1 + r2 β r3 + r4) β Β¬Irrational (r1 * r2) :=
sorry | Let $f(z)=az^4+bz^3+cz^2+dz+e=a(z-r_1)(z-r_2)(z-r_3)(z-r_4)$ where $a,b,c,d,e$ are integers, $a \neq 0$. Show that if $r_1+r_2$ is a rational number and $r_1+r_2 \neq r_3+r_4$, then $r_1r_2$ is a rational number. | null | ['number_theory', 'algebra'] | Section putnam_2003_b4.
Require Import Reals ZArith Coquelicot.Coquelicot.
Theorem putnam_2003_b4:
forall (a b c d e: Z), ~ Z.eq a 0 ->
let a := IZR a in
let b := IZR b in
let c := IZR c in
let d := IZR d in
let e := IZR e in
exists (r1 r2 r3 r4: R),
forall (z: R),
a * z ^ 4 + b * z ^ 3 + c * z ^ 2 + d * z + e = a * (z - r1) * (z - r2) * (z - r3) * (z - r4) ->
(exists (p q: Z), r1 + r2 = IZR p / IZR q) /\ r1 + r2 <> r3 + r4 -> exists (p q: Z), r1 * r2 = IZR p / IZR q.
Proof. Admitted.
End putnam_2003_b4. | theory putnam_2003_b4 imports Complex_Main
begin
theorem putnam_2003_b4:
fixes f::"real\<Rightarrow>real" and a b c d e::int and r1 r2 r3 r4::real
defines "f \<equiv> \<lambda>z::real. a * z^4 + b * z^3 + c * z^2 + d * z + e"
assumes ane0 : "a \<noteq> 0"
and hf : "\<forall>z::real. f z = a * (z - r1) * (z - r2) * (z - r3) * (z - r4)"
shows "((r1 + r2) \<in> \<rat> \<and> r1 + r2 \<noteq> r3 + r4) \<longrightarrow> (r1 * r2) \<in> \<rat>"
sorry
end | null |
putnam_2003_b6 | theorem putnam_2003_b6
(f : β β β)
(hf : Continuous f)
: (β« x in (0 : β)..1, (β« y in (0 : β)..1, |f x + f y|)) β₯ (β« x in (0 : β)..1, |f x|) :=
sorry | Let $f(x)$ be a continuous real-valued function defined on the interval $[0,1]$. Show that \[ \int_0^1 \int_0^1 | f(x) + f(y) |\,dx\,dy \geq \int_0^1 |f(x)|\,dx. \] | null | ['analysis'] | Section putnam_2003_b6.
Require Import Reals Coquelicot.Coquelicot.
Theorem putnam_2003_b6:
forall (f: R -> R) (x: R), 0 <= x <= 1 -> continuity_pt f x ->
RInt (fun x => RInt (fun y => Rabs (f x + f y)) 0 1) 0 1 >= RInt (fun x => Rabs (f x)) 0 1.
Proof. Admitted.
End putnam_2003_b6. | theory putnam_2003_b6 imports Complex_Main
"HOL-Analysis.Lebesgue_Measure"
"HOL-Analysis.Set_Integral"
begin
theorem putnam_2003_b6:
fixes f :: "real \<Rightarrow> real"
assumes hf : "continuous_on UNIV f"
shows "set_lebesgue_integral lebesgue {(x, y). 0 \<le> x \<and> x \<le> 1 \<and> 0 \<le> y \<and> y \<le> 1} (\<lambda> t :: real \<times> real. \<bar>f (fst t) + f (snd t)\<bar>)
\<ge> interval_lebesgue_integral lebesgue 0 1 f"
sorry
end | null |
putnam_1990_a1 | abbrev putnam_1990_a1_solution : (β β β€) Γ (β β β€) := sorry
-- (fun n : β => (n)!, fun n : β => 2 ^ n)
theorem putnam_1990_a1
(T : β β β€)
(hT012 : T 0 = 2 β§ T 1 = 3 β§ T 2 = 6)
(hTn : β n β₯ 3, T n = (n + 4) * T (n - 1) - 4 * n * T (n - 2) + (4 * n - 8) * T (n - 3))
: T = putnam_1990_a1_solution.1 + putnam_1990_a1_solution.2 :=
sorry | Let $T_0=2,T_1=3,T_2=6$, and for $n \geq 3$, $T_n=(n+4)T_{n-1}-4nT_{n-2}+(4n-8)T_{n-3}$. The first few terms are $2,3,6,14,40,152,784,5168,40576$. Find, with proof, a formula for $T_n$ of the form $T_n=A_n+B_n$, where $\{A_n\}$ and $\{B_n\}$ are well-known sequences. | Show that we have $T_n=n!+2^n$. | ['algebra'] | Section putnam_1990_a1.
Require Import Nat Factorial Coquelicot.Coquelicot.
Definition putnam_1990_a1_solution := (fun n => fact n, fun n => pow 2 n).
Theorem putnam_1990_a1:
let A :=
fix a (n: nat) : nat :=
match n with
| O => 2
| S O => 3
| S (S O) => 6
| S (S (S n''' as n'') as n') => (n + 4) * a n' - 4 * n * a n'' + (4 * n - 8) * a n'''
end in
exists (b c: nat -> nat), forall (n: nat), A n = b n + c n <->
(b,c) = putnam_1990_a1_solution.
Proof. Admitted.
End putnam_1990_a1. | theory putnam_1990_a1 imports Complex_Main
begin
definition putnam_1990_a1_solution::"((nat\<Rightarrow>int)\<times>(nat\<Rightarrow>int))" where "putnam_1990_a1_solution \<equiv> undefined"
(* (\<lambda>n. fact n, \<lambda>n. 2^n) *)
theorem putnam_1990_a1:
fixes T::"nat\<Rightarrow>int"
assumes hT012 : "T 0 = 2 \<and> T 1 = 3 \<and> T 2 = 6"
and hTn : "\<forall>n \<ge> 3. T n = (n+4) * T (n-1) - 4 * n * T (n-2) + (4 * n - 8) * T (n-3)"
shows "T = (\<lambda>n. (fst putnam_1990_a1_solution) n + (snd putnam_1990_a1_solution) n)"
sorry
end | null |
putnam_1990_a2 | abbrev putnam_1990_a2_solution : Prop := sorry
-- True
theorem putnam_1990_a2
(numform : β β Prop)
(hnumform : β x : β, numform x = β n m : β, x = n ^ ((1 : β) / 3) - m ^ ((1 : β) / 3))
: (β s : β β β, (β i : β, numform (s i)) β§ Tendsto s atTop (π (Real.sqrt 2))) β putnam_1990_a2_solution :=
sorry | Is $\sqrt{2}$ the limit of a sequence of numbers of the form $\sqrt[3]{n}-\sqrt[3]{m}$ ($n,m=0,1,2,\dots$)? | Show that the answer is yes. | ['analysis'] | Section putnam_1990_a2.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1990_a2_solution := True.
Theorem putnam_1990_a2:
let numform (x: R) : Prop := exists (n m: nat), x = pow (INR n) (1/3) - pow (INR m) (1/3) in
exists (s: nat -> R), forall (i: nat), numform (s i) /\ Lim_seq s = sqrt 2 <-> putnam_1990_a2_solution.
Proof. Admitted.
End putnam_1990_a2. | theory putnam_1990_a2 imports Complex_Main
begin
definition putnam_1990_a2_solution::bool where "putnam_1990_a2_solution \<equiv> undefined"
(* True *)
theorem putnam_1990_a2:
fixes numform::"real\<Rightarrow>bool"
defines "numform \<equiv> \<lambda>x. (\<exists>n m::nat. x = n powr (1/3) - m powr (1/3))"
shows "(\<exists>s::nat\<Rightarrow>real. (\<forall>i::nat. numform (s i)) \<and> (s \<longlonglongrightarrow> (sqrt 2))) \<longleftrightarrow> putnam_1990_a2_solution"
sorry
end | null |
putnam_1990_a5 | abbrev putnam_1990_a5_solution : Prop := sorry
-- False
theorem putnam_1990_a5
: (β n β₯ 1, β A B : Matrix (Fin n) (Fin n) β, A * B * A * B = 0 β B * A * B * A = 0) β putnam_1990_a5_solution :=
sorry | If $\mathbf{A}$ and $\mathbf{B}$ are square matrices of the same size such that $\mathbf{ABAB}=\mathbf{0}$, does it follow that $\mathbf{BABA}=\mathbf{0}$? | Show that the answer is no. | ['linear_algebra'] | Section putnam_1990_a5.
From mathcomp Require Import matrix ssralg.
Open Scope ring_scope.
Theorem putnam_1990_a5:
forall (R: ringType) (n: nat) (A B: 'M[R]_n),
mulmx (mulmx (mulmx A B) A) B = 0 ->
mulmx (mulmx (mulmx B A) B) A = 0.
Proof. Admitted.
End putnam_1990_a5. | theory putnam_1990_a5 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product"
begin
definition putnam_1990_a5_solution::bool where "putnam_1990_a5_solution \<equiv> undefined"
(* False *)
theorem putnam_1990_a5:
fixes A B::"real^'a^'a" and n::nat
assumes matsize : "CARD('a) = n \<and> n \<ge> 1"
and habab : "A ** B ** A ** B = mat 0"
shows "(B ** A ** B ** A = mat 0) \<longleftrightarrow> putnam_1990_a5_solution"
sorry
end | null |
putnam_1990_a6 | abbrev putnam_1990_a6_solution : β := sorry
-- 17711
theorem putnam_1990_a6
(STadmiss : (Fin 2 β (Finset (Fin 10))) β Prop)
(hSTadmiss : β ST : Fin 2 β (Finset (Fin 10)), STadmiss ST = ((β s β ST 0, (s+1) > (ST 1).card) β§ (β t β ST 1, (t+1) > (ST 0).card)))
: {ST : Fin 2 β (Finset (Fin 10)) | STadmiss ST}.encard = putnam_1990_a6_solution :=
sorry | If $X$ is a finite set, let $|X|$ denote the number of elements in $X$. Call an ordered pair $(S,T)$ of subsets of $\{1,2,\dots,n\}$ \emph{admissible} if $s>|T|$ for each $s \in S$, and $t>|S|$ for each $t \in T$. How many admissible ordered pairs of subsets of $\{1,2,\dots,10\}$ are there? Prove your answer. | Show that the number of admissible ordered pairs of subsets of $\{1,2,\dots,10\}$ equals the $22$nd Fibonacci number $F_{22}=17711$. | ['algebra'] | null | theory putnam_1990_a6 imports Complex_Main
begin
definition putnam_1990_a6_solution::nat where "putnam_1990_a6_solution \<equiv> undefined"
(* 17711 *)
theorem putnam_1990_a6:
fixes STadmiss::"(nat set) \<Rightarrow> (nat set) \<Rightarrow> bool"
defines "STadmiss \<equiv> \<lambda>S. \<lambda>T. S \<subseteq> {1..10} \<and> T \<subseteq> {1..10} \<and> (\<forall>s \<in> S. s > (card T)) \<and> (\<forall>t \<in> T. t > (card S))"
shows "card {(S::(nat set), T::(nat set)) . STadmiss S T} = putnam_1990_a6_solution"
sorry
end | null |
putnam_1990_b1 | abbrev putnam_1990_b1_solution : Set (β β β) := sorry
-- {fun x : β => (Real.sqrt 1990) * Real.exp x, fun x : β => -(Real.sqrt 1990) * Real.exp x}
theorem putnam_1990_b1
(f : β β β)
(fint : Prop)
(hfint : fint = β x : β, (f x) ^ 2 = (β« t in Set.Ioo 0 x, (f t) ^ 2 + (deriv f t) ^ 2) + 1990)
: (ContDiff β 1 f β§ fint) β f β putnam_1990_b1_solution :=
sorry | Find all real-valued continuously differentiable functions $f$ on the real line such that for all $x$, $(f(x))^2=\int_0^x [(f(t))^2+(f'(t))^2]\,dt+1990$. | Show that there are two such functions, namely $f(x)=\sqrt{1990}e^x$, and $f(x)=-\sqrt{1990}e^x$. | ['analysis'] | Section putnam_1990_b1.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Definition putnam_1990_b1_solution (f: R -> R) := f = (fun x => sqrt 1990 * exp x) /\ f = (fun x => -sqrt 1990 * exp x).
Theorem putnam_1990_b1:
forall (f: R -> R), continuity f /\ forall x, ex_derive_n f 1 x ->
forall x, pow (f x) 2 = RInt (fun t => pow (f t) 2 + pow ((Derive f) t) 2) 0 x + 1990 ->
putnam_1990_b1_solution f.
Proof. Admitted.
End putnam_1990_b1. | theory putnam_1990_b1 imports Complex_Main "HOL-Analysis.Interval_Integral"
begin
definition putnam_1990_b1_solution::"(real\<Rightarrow>real) set" where "putnam_1990_b1_solution \<equiv> undefined"
(* { (\<lambda>x. (sqrt 1990) * exp x), (\<lambda>x. - (sqrt 1990) * exp x)} *)
theorem putnam_1990_b1:
fixes f f'::"real\<Rightarrow>real"
assumes fderiv : "\<forall>x::real. (f has_derivative f') (nhds x)"
shows "((\<forall>x::real. (f x)^2 = (interval_lebesgue_integral lebesgue 0 x (\<lambda>t. (f t)^2 + (f' t)^2)) + 1990) \<and> continuous_on UNIV f')
\<longleftrightarrow> f \<in> putnam_1990_b1_solution"
sorry
end | null |
putnam_1990_b2 | theorem putnam_1990_b2
(x z : β)
(P : β β β)
(xlt1 : |x| < 1)
(zgt1 : |z| > 1)
(hP : β j β₯ 1, P j = (β i : Fin j, (1 - z * x ^ (i : β))) / (β i : Set.Icc 1 j, (z - x ^ (i : β))))
: 1 + (β' j : Set.Ici 1, (1 + x ^ (j : β)) * P j) = 0 :=
sorry | Prove that for $|x|<1$, $|z|>1$, $1+\sum_{j=1}^\infty (1+x^j)P_j=0$, where $P_j$ is $\frac{(1-z)(1-zx)(1-zx^2) \cdots (1-zx^{j-1})}{(z-x)(z-x^2)(z-x^3) \cdots (z-x^j)}$. | null | ['analysis'] | Section putnam_1990_b2.
Require Import Reals Coquelicot.Coquelicot.
Open Scope R.
Theorem putnam_1990_b2:
forall (x z: R), Rabs x < 1 /\ Rabs z > 1 ->
let P (j: nat) := (sum_n (fun i => 1 - z * x ^ i) j) / (sum_n (fun i => z - x ^ i) j+1) in
1 + Series (fun j => 1 + x ^ (j+1) * P j) = 0.
Proof. Admitted.
End putnam_1990_b2. | theory putnam_1990_b2 imports Complex_Main
begin
theorem putnam_1990_b2:
fixes x z::real and P::"nat\<Rightarrow>real"
defines "P \<equiv> \<lambda>j. (\<Prod>i=0..<j. (1 - z * x^i)) / (\<Prod>i=1..j. (z - x^i))"
assumes xlt1 : "abs(x) < 1"
and zgt1 : "abs(z) > 1"
shows "1 + (\<Sum>j::nat. (1 + x^(j+1)) * P j) = 0"
sorry
end | null |
putnam_1990_b3 | theorem putnam_1990_b3
(S : Set (Matrix (Fin 2) (Fin 2) β))
(hS : β A β S, β i j : Fin 2, (β x : β€, A i j = x ^ 2) β§ A i j β€ 200)
: (S.encard > 50387) β (β A β S, β B β S, A β B β§ A * B = B * A) :=
sorry | Let $S$ be a set of $2 \times 2$ integer matrices whose entries $a_{ij}$ (1) are all squares of integers, and, (2) satisfy $a_{ij} \leq 200$. Show that if $S$ has more than $50387$ ($=15^4-15^2-15+2$) elements, then it has two elements that commute. | null | ['linear_algebra'] | Section putnam_1990_b3.
Require Import Ensembles Finite_sets Reals Coquelicot.Coquelicot.
Open Scope R.
Theorem putnam_1990_b3
(Mmult_n :=
fix Mmult_n {T : Ring} (A : matrix 2 2) (p : nat) :=
match p with
| O => A
| S p' => @Mmult T 2 2 2 A (Mmult_n A p')
end)
(E : Ensemble (matrix 2 2) := fun (A: matrix 2 2) =>
forall (i j: nat), and (le 0 i) (lt i 2) /\ and (le 0 j) (lt j 2) ->
(coeff_mat 0 A i j) <= 200 /\ exists (m: nat), coeff_mat 0 A i j = INR m ^ 2)
: exists (sz : nat), gt sz 50387 /\ cardinal (matrix 2 2) E sz -> exists (A B: matrix 2 2), E A /\ E B /\ Mmult A B = Mmult B A.
Proof. Admitted.
End putnam_1990_b3. | theory putnam_1990_b3 imports Complex_Main "HOL-Analysis.Finite_Cartesian_Product"
begin
theorem putnam_1990_b3:
fixes S::"(nat^2^2) set"
assumes hS : "\<forall>A \<in> S. (\<forall>i \<in> {1..2}. \<forall>j \<in> {1..2}. ((\<exists>x::nat. A$i$j = x^2) \<and> A$i$j \<le> 200))"
shows "card S > 50387 \<longrightarrow> (\<exists>A \<in> S. \<exists>B \<in> S. A \<noteq> B \<and> A ** B = B ** A)"
sorry
end | null |
putnam_1990_b4 | abbrev putnam_1990_b4_solution : Prop := sorry
-- True
theorem putnam_1990_b4
: (β (G : Type*) (_ : Fintype G) (_ : Group G) (n : β) (a b : G), (n = Fintype.card G β§ a β b β§ G = Subgroup.closure {a, b}) β (β g : β β G, (β x : G, {i : Fin (2 * n) | g i = x}.encard = 2)
β§ (β i : Fin (2 * n), (g ((i + 1) % (2 * n)) = g i * a) β¨ (g ((i + 1) % (2 * n)) = g i * b))) β putnam_1990_b4_solution) :=
sorry | Let $G$ be a finite group of order $n$ generated by $a$ and $b$. Prove or disprove: there is a sequence $g_1,g_2,g_3,\dots,g_{2n}$ such that
\begin{itemize}
\item[(1)] every element of $G$ occurs exactly twice, and
\item[(2)] $g_{i+1}$ equals $g_ia$ or $g_ib$ for $i=1,2,\dots,2n$. (Interpret $g_{2n+1}$ as $g_1$.)
\end{itemize} | Show that such a sequence does exist. | ['abstract_algebra'] | null | theory putnam_1990_b4 imports Complex_Main "HOL-Algebra.Multiplicative_Group"
begin
(* Note: Boosted domain to infinite set *)
definition putnam_1990_b4_solution::bool where "putnam_1990_b4_solution \<equiv> undefined"
(* True *)
theorem putnam_1990_b4:
fixes G (structure) and n::nat and a b::"'a"
assumes hG : "Group.group G \<and> finite (carrier G) \<and> card (carrier G) = n"
and abgen : "generate G {a, b} = carrier G \<and> a \<noteq> b"
shows "(\<exists>g::nat\<Rightarrow>'a. (\<forall>x \<in> carrier G. card {i::nat. i < 2 * n \<and> g i = x} = 2) \<and>
(\<forall>i \<in> {0..<(2*n)}. (g ((i+1) mod (2*n)) = g i \<otimes> a) \<or> (g ((i+1) mod (2*n)) = g i \<otimes> b))) \<longleftrightarrow> putnam_1990_b4_solution"
sorry
end | null |
putnam_1990_b5 | abbrev putnam_1990_b5_solution : Prop := sorry
-- True
theorem putnam_1990_b5
(anpoly : (β β β) β β β Polynomial β)
(hanpoly : β (a : β β β) (n : β), (anpoly a n).degree = n β§ (β i : Fin (n + 1), (anpoly a n).coeff i = a i))
: (β a : β β β, (β i : β, a i β 0) β§ (β n β₯ 1, {r : β | (anpoly a n).eval r = 0}.encard = n)) β putnam_1990_b5_solution :=
sorry | Is there an infinite sequence $a_0,a_1,a_2,\dots$ of nonzero real numbers such that for $n=1,2,3,\dots$ the polynomial $p_n(x)=a_0+a_1x+a_2x^2+\cdots+a_nx^n$ has exactly $n$ distinct real roots? | Show that the answer is yes, such an infinite sequence exists. | ['algebra', 'analysis'] | Section putnam_1990_b5.
Require Import Reals Ensembles Finite_sets Coquelicot.Coquelicot.
Definition putnam_1990_b5_solution := True.
Open Scope R.
Theorem putnam_1990_b5:
exists (a: nat -> R),
let pn (n: nat) (x: R) := sum_n (fun i => a i * pow x i) n in
forall (n: nat), gt n 0 -> exists (roots: Ensemble R), cardinal R roots n /\ forall (r: R), roots r <-> pn n r = 0 <->
putnam_1990_b5_solution.
Proof. Admitted.
End putnam_1990_b5. | theory putnam_1990_b5 imports Complex_Main "HOL-Computational_Algebra.Polynomial"
begin
definition putnam_1990_b5_solution::bool where "putnam_1990_b5_solution \<equiv> undefined"
(* True *)
theorem putnam_1990_b5:
fixes anpoly::"(nat\<Rightarrow>real) \<Rightarrow> nat \<Rightarrow> (real poly)"
assumes hanpoly : "\<forall>a. \<forall>n. degree (anpoly a n) = n \<and> (\<forall>i::nat \<in> {0..n}. coeff (anpoly a n) i = a i)"
shows "(\<exists>a. (\<forall>i::nat. a i \<noteq> 0) \<and> (\<forall>n \<ge> 1. n = card {r::real. poly (anpoly a n) r = 0})) \<longleftrightarrow> putnam_1990_b5_solution"
sorry
end | null |
putnam_2008_a1 | theorem putnam_2008_a1
(f : β β β β β)
(hf : β x y z : β, f x y + f y z + f z x = 0)
: β g : β β β, β x y : β, f x y = g x - g y :=
sorry | Let $f:\mathbb{R}^2 \to \mathbb{R}$ be a function such that $f(x,y)+f(y,z)+f(z,x)=0$ for all real numbers $x$, $y$, and $z$. Prove that there exists a function $g:\mathbb{R} \to \mathbb{R}$ such that $f(x,y)=g(x)-g(y)$ for all real numbers $x$ and $y$. | null | ['algebra'] | Section putnam_2008_a1.
Require Import Reals Coquelicot.Coquelicot.
Theorem putnam_2008_a1:
forall (f: R -> R -> R) (x y z: R), f x y + f y z + f z x = 0 ->
exists (g: R -> R), forall (x y: R), f x y = g x - g y.
Proof. Admitted.
End putnam_2008_a1. | theory putnam_2008_a1 imports Complex_Main
begin
theorem putnam_2008_a1:
fixes f :: "real \<Rightarrow> real \<Rightarrow> real"
assumes hf : "\<forall> x y z :: real. f x y + f y z + f z x = 0"
shows "\<exists> g :: real \<Rightarrow> real. \<forall> x y :: real. f x y = g x - g y"
sorry
end | null |
putnam_2008_a4 | abbrev putnam_2008_a4_solution : Prop := sorry
-- False
theorem putnam_2008_a4
(f : β β β)
(hf : f = fun x => if x β€ Real.exp 1 then x else x * (f (Real.log x)))
: (β r : β, Tendsto (fun N : β => β n in Finset.range N, 1/(f (n + 1))) atTop (π r)) β putnam_2008_a4_solution :=
sorry | Define $f : \mathbb{R} \to \mathbb{R} by $f(x) = x$ if $x \leq e$ and $f(x) = x * f(\ln(x))$ if $x > e$. Does $\sum_{n=1}^{\infty} 1/(f(n))$ converge? | Show that the sum does not converge. | ['algebra'] | Section putnam_2008_a4.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2008_a4_solution := False.
Theorem putnam_2008_a4:
forall (f: R -> R) (x: R), f x = (if (Rle_dec x (exp 1)) then x else x * f (ln x)) ->
ex_lim_seq (fun nInc => sum_n (fun n => 1 / f (INR n)) nInc) <-> putnam_2008_a4_solution.
Proof. Admitted.
End putnam_2008_a4. | theory putnam_2008_a4 imports Complex_Main
begin
definition putnam_2008_a4_solution :: bool where "putnam_2008_a4_solution \<equiv> undefined"
(* False *)
theorem putnam_2008_a4:
fixes f :: "real \<Rightarrow> real"
assumes hf: "f \<equiv> (\<lambda>x::real. if x \<le> exp 1 then x else (x * f (ln x)))"
shows "(\<exists>r::real. filterlim (\<lambda>N::nat. (\<Sum>n::nat=1..N. 1/f n)) (nhds r) at_top) \<longleftrightarrow> putnam_2008_a4_solution"
sorry
end
| null |
putnam_2008_a6 | theorem putnam_2008_a6
: (β c : β, c > 0 β§ β G [Group G] (fing : Fintype G), fing.card > 1 β β s : List G, s.length β€ c * Real.log (fing.card : β) β§ β g : G, β t : List G, t.Sublist s β§ t.prod = g) :=
sorry | Prove that there exists a constant $c>0$ such that in every nontrivial finite group $G$ there exists a sequence of length at most $c \log |G|$ with the property that each element of $G$ equals the product of some subsequence. (The elements of $G$ in the sequence are not required to be distinct. A \emph{subsequence} of a sequence is obtained by selecting some of the terms, not necessarily consecutive, without reordering them; for example, $4, 4, 2$ is a subsequence of $2, 4, 6, 4, 2$, but $2, 2, 4$ is not.) | null | ['abstract_algebra'] | null | theory putnam_2008_a6 imports
Complex_Main
"HOL-Algebra.Multiplicative_Group"
begin
theorem putnam_2008_a6:
shows "\<exists> c :: real. c > 0 \<and> (\<forall> G :: 'a monoid. group G \<longrightarrow> finite (carrier G) \<longrightarrow> card (carrier G) > 1 \<longrightarrow> (\<exists> s :: 'a list. length s \<le> c * ln (card (carrier G)) \<and> (\<forall> g \<in> carrier G. \<exists> t \<in> set (subseqs s). foldr (\<otimes>\<^bsub>G\<^esub>) t (\<one>\<^bsub>G\<^esub>) = g)))"
sorry
end | null |
putnam_2008_b1 | abbrev putnam_2008_b1_solution : β := sorry
-- 2
def is_rational_point (p : Fin 2 β β) : Prop := β (a b : β), a = p 0 β§ b = p 1
def real_circle (c : Fin 2 β β) (r : β) : Set (Fin 2 β β) := {p : Fin 2 β β | Euclidean.dist p c = r}
theorem putnam_2008_b1 :
β (c : Fin 2 β β) (r : β), Β¬ is_rational_point c β (Set.ncard {p : Fin 2 β β | p β real_circle c r β§ is_rational_point p} β€ putnam_2008_b1_solution)
β§ β (c : Fin 2 β β) (r : β), Β¬ is_rational_point c β§ (Set.ncard {p : Fin 2 β β | p β real_circle c r β§ is_rational_point p} = putnam_2008_b1_solution) :=
sorry | What is the maximum number of rational points that can lie on a circle in $\mathbb{R}^2$ whose center is not a rational point? (A \emph{rational point} is a point both of whose coordinates are rational numbers.) | Show that the maximum number is $2$. | ['number_theory'] | null | theory putnam_2008_b1 imports Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
"HOL-Analysis.Elementary_Metric_Spaces"
begin
definition putnam_2008_b1_solution :: "nat" where "putnam_2008_b1_solution \<equiv> undefined"
(* 2 *)
definition is_rational_point :: "real^2 \<Rightarrow> bool" where "is_rational_point \<equiv> (\<lambda>p::real^2. (\<exists>a b::rat. a = p$1 \<and> b = p$2))"
theorem putnam_2008_b1:
shows "(GREATEST np::nat. (\<exists>(c::real^2)(r::real). \<not>is_rational_point c \<and> np = card {p::real^2. p \<in> sphere c r \<and> is_rational_point p})) = putnam_2008_b1_solution"
sorry
end
| null |
putnam_2008_b2 | abbrev putnam_2008_b2_solution : β := sorry
-- -1
theorem putnam_2008_b2
(F : β β β β β)
(hF0 : β x : β, F 0 x = Real.log x)
(hFn : β n : β, β x > 0, F (n + 1) x = β« t in Set.Ioo 0 x, F n t)
: Tendsto (fun n : β => ((n)! * F n 1) / Real.log n) atTop (π putnam_2008_b2_solution) :=
sorry | Let $F_0(x)=\ln x$. For $n \geq 0$ and $x>0$, let $F_{n+1}(x)=\int_0^x F_n(t)\,dt$. Evaluate $\lim_{n \to \infty} \frac{n!F_n(1)}{\ln n}$. | Show that the desired limit is $-1$. | ['analysis'] | Section putnam_2008_b2.
Require Import Factorial Reals Coquelicot.Coquelicot.
Definition putnam_2008_b2_solution := -1.
Theorem putnam_2008_b2:
let fix F (n: nat) (x: R) :=
match n with
| O => ln x
| S n' => RInt (fun t => F n' t) 0 x
end in
Lim_seq (fun n => INR (fact n) * F n 1 / ln (INR n)) = putnam_2008_b2_solution.
Proof. Admitted.
End putnam_2008_b2. | theory putnam_2008_b2 imports Complex_Main
"HOL-Analysis.Interval_Integral"
begin
definition putnam_2008_b2_solution :: "real" where
"putnam_2008_b2_solution \<equiv> undefined"
(* -1 *)
theorem putnam_2008_b2:
fixes F :: "nat \<Rightarrow> real \<Rightarrow> real"
assumes hF0 : "\<forall> x :: real. F 0 x = ln x"
and hFn : "\<forall> n :: nat. \<forall> x > 0. F (n + 1) x = interval_lebesgue_integral lebesgue 0 x (F n)"
shows "filterlim (\<lambda> n :: nat. (fact n) * (F n 1) / ln n) (nhds putnam_2008_b2_solution) at_top"
sorry
end | null |
putnam_2008_b4 | theorem putnam_2008_b4
(p : β)
(hp : Nat.Prime p)
(h : Polynomial β€)
(hh : β i j : Finset.range (p ^ 2), h.eval i β‘ h.eval j [ZMOD p ^ 2] β i = j)
: (β i j : Finset.range (p ^ 3), h.eval i β‘ h.eval j [ZMOD p ^ 3] β i = j) :=
sorry | Let $p$ be a prime number. Let $h(x)$ be a polynomial with integer coefficients such that $h(0), h(1), \dots, h(p^2-1)$ are distinct modulo $p^2$. Show that $h(0), h(1), \dots, h(p^3-1)$ are distinct modulo $p^3$. | null | ['algebra', 'number_theory'] | Section putnam_2008_b4.
Require Import Nat Reals Coquelicot.Coquelicot.
Theorem putnam_2008_b4:
forall (p: nat), Znumtheory.prime (Z.of_nat p) ->
exists (c: nat -> Z) (n: nat),
let h (x: nat) := Z.to_nat (floor (sum_n (fun i => IZR (c i) * INR (x ^ i)) (n + 1))) in
(forall (i j: nat), i <> j /\ and (le 0 i) (le i (p ^ 2 - 1)) /\ and (le 0 j) (le j (p ^ 2 - 1)) -> (h i) mod (p ^ 2) <> h j mod p ^ 2) ->
(forall (i j: nat), i <> j /\ and (le 0 i) (le i (p ^ 3 - 1)) /\ and (le 0 j) (le j (p ^ 3 - 1)) -> h i mod p ^ 2 <> h j mod p ^ 3).
Proof. Admitted.
End putnam_2008_b4. | theory putnam_2008_b4 imports
Complex_Main
"HOL-Computational_Algebra.Primes"
"HOL-Computational_Algebra.Polynomial"
"HOL-Number_Theory.Cong"
begin
theorem putnam_2008_b4:
fixes p :: nat
and h :: "int poly"
assumes hp: "prime p"
and hh: "\<forall> i \<in> {0 .. p ^ 2 - 1}. \<forall> j \<in> {0 .. p ^ 2 - 1}. [poly h i = poly h j] (mod p ^ 2) \<longrightarrow> i = j"
shows "\<forall> i \<in> {0 .. p ^ 3 - 1}. \<forall> j \<in> {0 .. p ^ 3 - 1}. [poly h i = poly h j] (mod p ^ 3) \<longrightarrow> i = j"
sorry
end | null |
putnam_2008_b5 | abbrev putnam_2008_b5_solution : Set (β β β) := sorry
-- {fun (x : β) => x + n | n : β€} βͺ {fun (x : β) => -x + n | n : β€}
theorem putnam_2008_b5
(fqsat : (β β β) β β β Prop := fun f q => ContDiff β 1 f β§ (β p : β, p = f q β§ p.den = q.den))
(fsat : (β β β) β Prop := fun f => β q : β, fqsat f q)
: β f : (β β β), fsat f β f β putnam_2008_b5_solution :=
sorry | Find all continuously differentiable functions f : \mathbb{R} \to \mathbb{R} such that for every rational number $q$, the number $f(q)$ is rational and has the same denominator as $q$. | Show that the solution is the set of all functions of the form n + x, n - x where n is any integer. | ['analysis'] | Section putnam_2008_b5.
Require Import Reals Coquelicot.Coquelicot. From mathcomp Require Import div.
Definition putnam_2008_b5_solution := 1.
Theorem putnam_2008_b5:
forall (f: R -> R), continuity f /\ (forall (x: R), ex_derive f x) ->
forall (q: R), exists (n1 n2 d: nat), q = INR (n1 / d) /\ f q = INR (n2 / d) /\ coprime n1 d = true /\ coprime n2 d = true.
Proof. Admitted.
End putnam_2008_b5. | theory putnam_2008_b5 imports Complex_Main
"HOL-Analysis.Derivative"
begin
definition putnam_2008_b5_solution :: "(real \<Rightarrow> real) set" where "putnam_2008_b5_solution \<equiv> undefined"
(* {f::real\<Rightarrow>real. (\<exists>n::int. f = (\<lambda>x::real. x + n))} \<union> {f::real\<Rightarrow>real. (\<exists>n::int. f = (\<lambda>x::real. -x + n))} *)
theorem putnam_2008_b5:
fixes fqsat :: "(real \<Rightarrow> real) \<Rightarrow> rat \<Rightarrow> bool"
and fsat :: "(real \<Rightarrow> real) \<Rightarrow> bool"
defines "fqsat \<equiv> (\<lambda>(f::real\<Rightarrow>real)(q::rat). f C1_differentiable_on UNIV \<and> (\<exists>p::rat. p = f (real_of_rat q) \<and> snd (quotient_of p) = snd (quotient_of q)))"
and "fsat \<equiv> (\<lambda>f::real\<Rightarrow>real. (\<forall>q::rat. fqsat f q))"
shows "\<forall>f::real\<Rightarrow>real. (fsat f \<longleftrightarrow> f \<in> putnam_2008_b5_solution)"
sorry
end
| null |
putnam_2008_b6 | theorem putnam_2008_b6
(n k : β)
(hnk : n > 0 β§ k > 0)
: Odd (Set.ncard {s : Equiv.Perm (Fin n) | klimited k n s}) β (n β‘ 0 [MOD 2*k+1] β¨ n β‘ 1 [MOD 2*k+1]) :=
sorry | Let $n$ and $k$ be positive integers. Say that a permutation $\sigma$ of $\{1,2,\dots,n\} is $k-limited$ if \|\sigma(i) - i\| \leq k$ for all $i$. Prove that the number of $k-limited$ permutations $\{1,2,\dots,n\}$ is odd if and only if $n \equiv 0$ or $1 (mod 2k+1)$. | null | ['number_theory'] | Section putnam_2008_b6.
Require Import Ensembles Finite_sets Reals. From mathcomp Require Import div fintype seq ssrbool perm.
Theorem putnam_2008_b6:
forall (n k: nat), n > 0 /\ k > 0 ->
let klimited (sigma: {perm 'I_n}) : Prop := forall (i: 'I_n), Rle (Rabs (INR (nat_of_ord (sigma i)) - INR i)) (INR k) in
forall (E: Ensemble {perm 'I_n}) (p: {perm 'I_n}), (E p <-> klimited p) ->
exists (sz: nat), cardinal {perm 'I_n} E sz /\ Nat.odd sz <-> n mod (2 * k + 1) = 0 \/ n mod (2 * k + 1) = 1.
Proof. Admitted.
End putnam_2008_b6. | theory putnam_2008_b6 imports Complex_Main
"HOL-Combinatorics.Permutations"
"HOL-Number_Theory.Cong"
begin
(* uses (nat \<Rightarrow> nat) instead of (Fin n \<Rightarrow> Fin n) *)
definition klimited :: "nat \<Rightarrow> nat \<Rightarrow> (nat \<Rightarrow> nat) \<Rightarrow> bool" where "klimited \<equiv> (\<lambda>(k::nat)(n::nat)(s::nat\<Rightarrow>nat). s permutes {0..(n-1)} \<and> (\<forall>i::nat\<in>{0..(n-1)}. \<bar>s i - i\<bar> \<le> k))"
theorem putnam_2008_b6:
fixes n k :: nat
assumes hnk: "n > 0 \<and> k > 0"
shows "odd (card {s::nat\<Rightarrow>nat. klimited k n s}) \<longleftrightarrow> ([n = 0] (mod 2*k+1) \<or> [n = 1] (mod 2*k+1))"
sorry
end
| null |
putnam_2021_a1 | abbrev putnam_2021_a1_solution : β := sorry
-- 578
theorem putnam_2021_a1
(P : List (β€ Γ β€) β Prop := fun l : List (β€ Γ β€) => l.length β₯ 1 β§ l[0]! = (0, 0) β§ l[l.length-1]! = (2021, 2021) β§
β n β Finset.range (l.length-1), Real.sqrt ((l[n]!.1 - l[n + 1]!.1)^2 + (l[n]!.2 - l[n + 1]!.2)^2) = 5)
: (β l : List (β€ Γ β€), P l β§ l.length = putnam_2021_a1_solution) β§
β l : List (β€ Γ β€), P l β l.length β₯ putnam_2021_a1_solution :=
sorry | A grasshopper starts at the origin in the coordinate plane and makes a sequence of hops.
Each hop has length $5$, and after each hop the grasshopper is at a point whose coordinates are both integers; thus, there are $12$ possible locations for the grasshopper after the first hop.
What is the smallest number of hops needed for the grasshopper to reach the point $(2021, 2021)$? | The answer is $578$. | ['geometry'] | null | theory putnam_2021_a1 imports Complex_Main
begin
definition putnam_2021_a1_solution :: nat where "putnam_2021_a1_solution \<equiv> undefined"
(* 578 *)
theorem putnam_2021_a1:
fixes P :: "((int \<times> int) list) \<Rightarrow> bool"
assumes "P \<equiv> (\<lambda>l::(int\<times>int) list. length l \<ge> 1 \<and> l!0 = (0,0) \<and> last l = (2021,2021) \<and>
(\<forall>n::nat\<in>{0..((length l)-2)}. sqrt ((fst (l!n) - fst (l!(n + 1)))^2 + (snd (l!n) - snd (l!(n + 1)))^2) = 5))"
shows "(LEAST llen::nat. (\<exists>l::(int\<times>int) list. P l \<and> llen = length l)) = putnam_2021_a1_solution"
sorry
end
| null |
putnam_2021_a2 | abbrev putnam_2021_a2_solution : β := sorry
-- Real.exp 1
theorem putnam_2021_a2
(g : β β β)
(hg : β x > 0, Tendsto (fun r : β => ((x + 1) ^ (r + 1) - x ^ (r + 1)) ^ (1 / r)) (π 0) (π (g x)))
: Tendsto (fun x : β => g x / x) atTop (π putnam_2021_a2_solution) :=
sorry | For every positive real number $x$, let $g(x)=\lim_{r \to 0}((x+1)^{r+1}-x^{r+1})^\frac{1}{r}$. Find $\lim_{x \to \infty}\frac{g(x)}{x}$. | Show that the limit is $e$. | ['analysis'] | Section putnam_2021_a2.
Require Import Reals. From Coquelicot Require Import Continuity Lim_seq Rbar.
Local Open Scope R.
Definition putnam_2021_a2_solution := exp 1.
Theorem putnam_2021_a2:
let sequence_r_to_0 (n : nat) : R := 1 / INR n in
let f (r x: R) := Rpower (Rpower(x+1)(r+1) - Rpower x (r+1)) 1/r in
let g (x : R) : R := Lim_seq (fun n => f (sequence_r_to_0 n) x) in
Lim_seq (fun n => (g (INR n))/INR n) = putnam_2021_a2_solution.
Proof. Admitted.
End putnam_2021_a2. | theory putnam_2021_a2 imports Complex_Main
begin
definition putnam_2021_a2_solution :: real where "putnam_2021_a2_solution \<equiv> undefined"
(* exp 1 *)
theorem putnam_2021_a2:
fixes g :: "real \<Rightarrow> real"
assumes hg: "\<forall>x::real>0. filterlim (\<lambda>r::real. ((x+1) powr (r+1) - x powr (r+1)) powr (1/r)) (nhds (g x)) (nhds 0)"
shows "filterlim (\<lambda>x::real. g x / x) (nhds putnam_2021_a2_solution) at_top"
sorry
end
| null |
putnam_2021_a4 | abbrev putnam_2021_a4_solution : β := sorry
-- ((Real.sqrt 2) / 2) * Real.pi * Real.log 2
theorem putnam_2021_a4
(S : β β Set (Fin 2 β β) := fun R : β => {p : (Fin 2 β β) | (p 0)^2 + (p 1)^2 β€ R^2})
(I : β β β := fun R : β => β« p in S R,
(1 + 2*(p 0)^2)/(1 + (p 0)^4 + 6*(p 0)^2*(p 1)^2 + (p 1)^4) - (1 + (p 1)^2)/(2 + (p 0)^4 + (p 1)^4))
: Tendsto I atTop (π putnam_2021_a4_solution) :=
sorry | Let
\[
I(R) = \iint_{x^2+y^2 \leq R^2} \left( \frac{1+2x^2}{1+x^4+6x^2y^2+y^4} - \frac{1+y^2}{2+x^4+y^4} \right)\,dx\,dy.
\]
Find
\[
\lim_{R \to \infty} I(R),
\]
or show that this limit does not exist. | The limit exists and equals $\frac{\sqrt{2}}{2} \pi \log 2$. | ['analysis'] | Section putnam_2021_a4.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_2021_a4_solution := (sqrt 2 / 2) * PI * ln 2 / ln 10.
Theorem putnam_2021_a4
(I : nat -> R := fun r => RInt (fun x => RInt (fun y => (1 + 2 * x ^ 2) / (1 + x ^ 4 + 6 * x ^ 2 * y ^ 2 + y ^ 4) - (1 + y ^ 2) / (2 + x ^ 4 + y ^ 4)) 0 (sqrt (INR r ^ 2 - x ^ 2))) 0 1)
: ~ ex_lim_seq I \/ Lim_seq I = putnam_2021_a4_solution.
Proof. Admitted.
End putnam_2021_a4. | theory putnam_2021_a4 imports Complex_Main
"HOL-Analysis.Set_Integral"
"HOL-Analysis.Lebesgue_Measure"
begin
definition putnam_2021_a4_solution :: real where "putnam_2021_a4_solution \<equiv> undefined"
(* ((sqrt 2)/2) * pi * ln 2 *)
theorem putnam_2021_a4:
fixes S :: "real \<Rightarrow> ((real^2) set)"
and I :: "real \<Rightarrow> real"
assumes "S \<equiv> (\<lambda>R::real. {p::real^2. (p$1)^2 + (p$2)^2 \<le> R^2})"
and "I \<equiv> (\<lambda>R::real. set_lebesgue_integral lebesgue (S R)
(\<lambda>p::real^2. (1 + 2*(p$1)^2)/(1 + (p$1)^4 + 6*(p$1)^2*(p$2)^2 + (p$2)^4) - (1 + (p$2)^2)/(2 + (p$1)^4 + (p$2)^4)))"
shows "filterlim I (nhds putnam_2021_a4_solution) at_top"
sorry
end
| null |
putnam_2021_a5 | abbrev putnam_2021_a5_solution : Set β := sorry
-- {j : β | Β¬(42 β£ j) β§ Β¬(46 β£ j)}
theorem putnam_2021_a5
(j : β)
(A : Finset β)
(S : β β β)
(hA : A = {n : β | 1 β€ n β§ n β€ 2021 β§ Nat.gcd n 2021 = 1})
(hS : β j' : β, S j' = β n in A, n ^ j')
: (2021 β£ S j) β j β putnam_2021_a5_solution :=
sorry | Let $A$ be the set of all integers $n$ such that $1 \leq n \leq 2021$ and $\gcd(n,2021)=1$. For every nonnegative integer $j$, let $S(j)=\sum_{n \in A}n^j$. Determine all values of $j$ such that $S(j)$ is a multiple of $2021$. | Show that the values of $j$ in question are those not divisible by either $42$ or $46$. | ['number_theory'] | Section putnam_2021_a5.
Require Import Nat. From mathcomp Require Import bigop div fintype eqtype seq ssrbool ssrnat.
Variables (I : finType) (P : pred I).
Definition putnam_2021_a5_solution (n: nat) := ~(n%|42 \/ n%|46).
Theorem putnam_2021_a5:
forall (j: nat),
let A : pred 'I_2021 := fun n => let m := nat_of_ord n in ((1 <= m <= 2021) && (gcd m 2021 == 1)) in
let B (j: nat) := \sum_(n | A n) (nat_of_ord n)^j in
B j mod 2021 = 0 <-> putnam_2021_a5_solution j.
Proof. Admitted.
End putnam_2021_a5. | theory putnam_2021_a5 imports Complex_Main
begin
definition putnam_2021_a5_solution :: "nat set" where "putnam_2021_a5_solution \<equiv> undefined"
(* {j::nat. \<not>(42 dvd j) \<and> \<not>(46 dvd j)} *)
theorem putnam_2021_a5:
fixes j :: nat
and A :: "nat set"
and S :: "nat \<Rightarrow> nat"
assumes hA : "A = {n::nat. 1 \<le> n \<and> n \<le> 2021 \<and> gcd n 2021 = 1}"
and hS : "\<forall>j'::nat. S j' = (\<Sum>n\<in>A. n^j')"
shows "(2021 dvd (S j)) \<longleftrightarrow> j \<in> putnam_2021_a5_solution"
sorry
end
| null |
putnam_2021_a6 | abbrev putnam_2021_a6_solution : Prop := sorry
-- True
theorem putnam_2021_a6
(Pcoeff : Polynomial β€ β Prop)
(Pprod : Polynomial β€ β Prop)
(hPcoeff : β P : Polynomial β€, Pcoeff P = (β n : β, P.coeff n = 0 β¨ P.coeff n = 1))
(hPprod : β P : Polynomial β€, Pprod P = (β Q R : Polynomial β€, Q.degree > 0 β§ R.degree > 0 β§ P = Q * R))
: (β P : Polynomial β€, (Pcoeff P β§ Pprod P) β (P.eval 2 β 0 β§ P.eval 2 β 1 β§ Β¬Prime (P.eval 2))) β putnam_2021_a6_solution :=
sorry | Let $P(x)$ be a polynomial whose coefficients are all either $0$ or $1$. Suppose that $P(x)$ can be written as a product of two nonconstant polynomials with integer coefficients. Does it follow that $P(2)$ is a composite integer? | Show that it does follow that $P(2)$ is a composite integer. | ['number_theory', 'algebra'] | Section putnam_2021_a6.
From mathcomp Require Import seq ssrnat ssrnum ssralg poly.
Local Open Scope ring_scope.
Definition putnam_2021_a6_solution := True.
Theorem putnam_2021_a6:
forall (R: numDomainType) (p: {poly R}) (i: nat), p`_i = 0 /\ p`_i = 1 ->
exists (q r: {poly R}), size p <> 0%nat /\ size q <> 0%nat ->
p = q * r <-> putnam_2021_a6_solution .
Proof. Admitted.
End putnam_2021_a6. | theory putnam_2021_a6 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
definition putnam_2021_a6_solution :: bool where "putnam_2021_a6_solution \<equiv> undefined"
(* True *)
theorem putnam_2021_a6:
fixes Pcoeff :: "int poly \<Rightarrow> bool"
and Pprod :: "int poly \<Rightarrow> bool"
assumes hPcoeff: "\<forall>P::int poly. Pcoeff P = (\<forall>n::nat. coeff P n = 0 \<or> coeff P n = 1)"
and hPprod: "\<forall>p::int poly. Pprod P = (\<exists>Q R::int poly. degree Q > 0 \<and> degree R > 0 \<and> P = Q * R)"
shows "(\<forall>p::int poly. (Pcoeff P \<and> Pprod P) \<longrightarrow> (poly P 2 \<noteq> 0 \<and> poly P 2 \<noteq> 1 \<and> \<not>prime (poly P 2))) \<longleftrightarrow> putnam_2021_a6_solution"
sorry
end
| null |
putnam_2021_b2 | abbrev putnam_2021_b2_solution : β := sorry
-- 2 / 3
theorem putnam_2021_b2
(S : (β β β) β β)
(asum : (β β β) β Prop)
(hS : β a : β β β, S a = β' n : β, (n + 1) / 2 ^ (n + 1) * (β k : Fin (n + 1), a k.1) ^ ((1 : β) / (n + 1)))
(hasum : β a : β β β, asum a = (β k : β, a k β₯ 0) β§ β' k : β, a k = 1)
: (β a : β β β, asum a β§ S a = putnam_2021_b2_solution) β§ (β a : β β β, asum a β S a β€ putnam_2021_b2_solution) :=
sorry | Determine the maximum value of the sum $S = \sum_{n=1}^\infty \frac{n}{2^n}(a_1a_2 \dots a_n)^{1/n}$ over all sequences $a_1,a_2,a_3,\dots$ of nonnegative real numbers satisfying $\sum_{k=1}^\infty a_k=1$. | Show that the answer is $2/3$. | ['analysis'] | Section putnam_2021_b2.
Require Import List Reals Coquelicot.Hierarchy Coquelicot.Series.
Definition putnam_2021_b2_solution := 2/3.
Theorem putnam_2021_b2:
let A (a : nat -> R) (n : nat) : R := fold_left Rmult (map a (seq 0 n)) 1 in
let B (a : nat -> R) := Series (fun n => INR n * (Rpower (A a n) 1/(INR n))) in
(forall (a : nat -> R), (forall (i: nat), a i >= 0) /\ Series a = 1 -> putnam_2021_b2_solution >= B a)
/\
(exists (a : nat -> R), (forall (i: nat), a i >= 0) /\ Series a = 1 -> putnam_2021_b2_solution = B a).
Proof. Admitted.
End putnam_2021_b2. | theory putnam_2021_b2 imports Complex_Main
begin
definition putnam_2021_b2_solution :: real where "putnam_2021_b2_solution \<equiv> undefined"
(* 2/3 *)
theorem putnam_2021_b2:
fixes S :: "(nat \<Rightarrow> real) \<Rightarrow> real" and asum :: "(nat \<Rightarrow> real) \<Rightarrow> bool"
defines "S \<equiv> \<lambda>a. (\<Sum>n::nat. ((n+1)/2^(n+1)) * (\<Prod>k::nat=0..n. a k) powr (1/(n+1)))"
and "asum \<equiv> \<lambda>a. (\<forall>k::nat. a k \<ge> 0) \<and> (\<Sum>k::nat. a k) = 1"
shows "(GREATEST Sa::real. (\<exists>a::nat\<Rightarrow>real. asum a \<and> S a = Sa)) = putnam_2021_b2_solution"
sorry
end
| null |
putnam_2021_b4 | theorem putnam_2021_b4
(F : β β β)
(hF : β x, x β₯ 2 β F x = F (x - 1) + F (x - 2))
(F01 : F 0 = 0 β§ F 1 = 1)
: β m, m > 2 β (β p, (β k : Set.Icc 1 (F m - 1), (k.1 ^ k.1)) % F m = F p) :=
sorry | Let $F_0, F_1, \ldots$ be the sequence of Fibonacci numbers, with $F_0 = 0$, $F_1 = 1$, and $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$. For $m > 2$, let $R_m$ be the remainder when the product $\prod_{k=1}^{F_{m-1}} k^k$ is divided by $F_m$. Prove that $R_m$ is also a Fibonacci number. | null | ['number_theory'] | Section putnam_2021_b4.
Require Import PeanoNat. From mathcomp Require Import bigop fintype ssrnat.
Theorem putnam_2021_b4:
let F :=
fix f (n: nat) : nat :=
match n with
| O => O
| S O => 1
| S ((S n'') as n') => f n' + f n''
end in
forall (m: nat), m > 2 = true ->
exists (p: nat), (\prod_(k < (F m)) k^k) mod (F m) = F p.
Proof. Admitted.
End putnam_2021_b4. | theory putnam_2021_b4 imports Complex_Main
begin
theorem putnam_2021_b4:
fixes f :: "nat\<Rightarrow>nat"
assumes hf : "\<forall>x\<ge>2. f x = f (x-1) + f (x-2)"
and f01 : "f 0 = 0 \<and> f 1 = 1"
shows "\<forall>m > 2. \<exists>p. (\<Prod>k=1..(f m - 1). k^k) mod f m = f p"
sorry
end
| null |
putnam_1972_a1 | theorem putnam_1972_a1
(n : β)
(hn : n > 0)
(fourAP : β€ β β€ β β€ β β€ β Prop := fun n1 n2 n3 n4 => β o1 o2 o3 o4 : β€, {n1, n2, n3, n4} = ({o1, o2, o3, o4} : Set β€) β§ o1 β€ o2 β§ o2 β€ o3 β§ o3 β€ o4 β§ o4-o3 = o3-o2 β§ o3-o2 = o2-o1)
: Β¬ β r : β, r > 0 β§ r + 3 β€ n β§ fourAP (n.choose r) (n.choose (r+1)) (n.choose (r+2)) (n.choose (r+3)) :=
sorry | Show that there are no four consecutive binomial coefficients ${n \choose r}, {n \choose (r+1)}, {n \choose (r+2)}, {n \choose (r+3)}$ where $n,r$ are positive integers and $r+3 \leq n$, which are in arithmetic progression. | null | ['algebra'] | null | theory putnam_1972_a1 imports Complex_Main
begin
theorem putnam_1972_a1:
fixes n :: nat
and fourAP :: "nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> nat \<Rightarrow> bool"
assumes hn: "n > 0"
defines "fourAP \<equiv> (\<lambda>n1 n2 n3 n4::nat. (\<exists>o1 o2 o3 o4::nat. {n1,n2,n3,n4} = {o1,o2,o3,o4} \<and> o1 \<le> o2 \<and> o2 \<le> o3 \<and> o3 \<le> o4 \<and> o4-o3 = o3-o2 \<and> o3-o2 = o2-o1))"
shows "\<not>(\<exists>r::nat. r > 0 \<and> r + 3 \<le> n \<and> fourAP (n choose r) (n choose (r+1)) (n choose (r+2)) (n choose (r+3)))"
sorry
end
| null |
putnam_1972_a2 | theorem putnam_1972_a2
: (β (S : Type*) (_ : Mul S), (β x y : S, x * (x * y) = y β§ ((y * x) * x) = y) β (β x y : S, x * y = y * x)) β§ β (S : Type*) (_ : Mul S), (β x y : S, x * (x * y) = y β§ ((y * x) * x) = y) β§ Β¬(β x y z : S, x * (y * z) = (x * y) * z) :=
sorry | Let $S$ be a set and $\cdot$ be a binary operation on $S$ satisfying: (1) for all $x,y$ in $S$, $x \cdot (x \cdot y) = y$ (2) for all $x,y$ in $S$, $(y \cdot x) \cdot x = y$. Show that $\cdot$ is commutative but not necessarily associative. | null | ['abstract_algebra'] | null | theory putnam_1972_a2 imports Complex_Main
begin
(* The existential magma can have at most the cardinality of the real numbers. *)
theorem putnam_1972_a2:
assumes pauncount: "\<exists>pamap::'a\<Rightarrow>real. surj pamap"
shows "(\<forall>(S::'a set)(Smul::'a\<Rightarrow>'a\<Rightarrow>'a). (\<forall>x\<in>S. \<forall>y\<in>S. (Smul x y \<in> S) \<and> Smul x (Smul x y) = y \<and> Smul (Smul y x) x = y) \<longrightarrow> (\<forall>x\<in>S. \<forall>y\<in>S. Smul x y = Smul y x)) \<and> (\<exists>(S::'a set)(Smul::'a\<Rightarrow>'a\<Rightarrow>'a). (\<forall>x\<in>S. \<forall>y\<in>S. (Smul x y \<in> S) \<and> Smul x (Smul x y) = y \<and> Smul (Smul y x) x = y) \<and> \<not>(\<forall>x\<in>S. \<forall>y\<in>S. \<forall>z\<in>S. Smul x (Smul y z) = Smul (Smul x y) z))"
sorry
end
| null |
putnam_1972_a3 | abbrev putnam_1972_a3_solution : Set (β β β) := sorry
-- {f | β A B : β, β x β Set.Icc 0 1, f x = A * x + B}
theorem putnam_1972_a3
(x : β β β)
(climit_exists : (β β β) β Prop := fun x => β C : β, Tendsto (fun n => (β i in Finset.range n, (x i))/(n : β)) atTop (π C))
(supercontinuous : (β β β) β Prop := fun f => β (x : β β β), (β i : β, x i β Icc 0 1) β climit_exists x β climit_exists (fun i => f (x i)))
: {f | supercontinuous f} = putnam_1972_a3_solution :=
sorry | We call a function $f$ from $[0,1]$ to the reals to be supercontinuous on $[0,1]$ if the Cesaro-limit exists for the sequence $f(x_1), f(x_2), f(x_3), \dots$ whenever it does for the sequence $x_1, x_2, x_3 \dots$. Find all supercontinuous functions on $[0,1]$. | Show that the solution is the set of affine functions. | ['analysis'] | null | theory putnam_1972_a3 imports Complex_Main
begin
(* uses (real \<Rightarrow> real) instead of ({0..1} \<Rightarrow> real) *)
definition putnam_1972_a3_solution :: "(real \<Rightarrow> real) set" where "putnam_1972_a3_solution \<equiv> undefined"
(* {f::real\<Rightarrow>real. (\<exists>A B::real. \<forall>x::real\<in>{0..1}. f x = A*x + B)} *)
theorem putnam_1972_a3:
fixes x :: "nat \<Rightarrow> real"
and climit_exists :: "(nat \<Rightarrow> real) \<Rightarrow> bool"
and supercontinuous :: "(real \<Rightarrow> real) \<Rightarrow> bool"
defines "climit_exists \<equiv> (\<lambda>x::nat\<Rightarrow>real. (\<exists>C::real. filterlim (\<lambda>n::nat. (\<Sum>i::nat\<in>{0..(n-1)}. x i)/n) (nhds C) at_top))"
and "supercontinuous \<equiv> (\<lambda>f::real\<Rightarrow>real. (\<forall>x::nat\<Rightarrow>real. ((\<forall>i::nat. x i \<in> {0..1}) \<longrightarrow> climit_exists x \<longrightarrow> climit_exists (\<lambda>i::nat. f (x i)))))"
shows "{f::real\<Rightarrow>real. supercontinuous f} = putnam_1972_a3_solution"
sorry
end
| null |
putnam_1972_a5 | theorem putnam_1972_a5
(n : β)
(hn : n > 1)
: Β¬((n : β€) β£ 2^n - 1) :=
sorry | Show that if $n$ is an integer greater than $1$, then $n$ does not divide $2^n - 1$. | null | ['number_theory'] | null | theory putnam_1972_a5 imports Complex_Main
begin
theorem putnam_1972_a5:
fixes n :: nat
assumes hn: "n > 1"
shows "\<not>(n dvd (2^n - 1))"
sorry
end
| null |
putnam_1972_a6 | theorem putnam_1972_a6
(f : β β β)
(n : β€)
(hn : n β₯ 0)
(hfintegrable: IntegrableOn f (Icc 0 1))
(hfint : β i β Icc 0 (n-1), β« x in Icc 0 1, x^i*(f x) = 0)
(hfintlast : β« x in Icc 0 1, x^n*(f x) = 1)
: β S, S β Icc (0 : β) 1 β§ MeasurableSet S β§ volume S > 0 β§ β x β S, |f x| β₯ 2^n * (n + 1) :=
sorry | Let $f$ be an integrable function in $0 \leq x \leq 1$ and suppose for all $0 \leq i \leq n-1, \int_0^1 x^i f(x) dx = 0$. Further suppose that $\int_0^1 x^n f(x) dx = 1$. Show that $|f(x)| \geq 2^n(n+1)$ on a set of positive measure. | null | ['analysis'] | null | theory putnam_1972_a6 imports Complex_Main
"HOL-Analysis.Interval_Integral"
begin
theorem putnam_1972_a6:
fixes f :: "real \<Rightarrow> real"
and n :: nat
assumes hfintegrable: "interval_lebesgue_integrable lebesgue 0 1 f"
and hfint: "\<forall>i::nat\<in>{0..(n-1)}. interval_lebesgue_integral lebesgue 0 1 (\<lambda>x::real. x^i * (f x)) = 0"
and hfintlast: "interval_lebesgue_integral lebesgue 0 1 (\<lambda>x::real. x^n * (f x)) = 1"
shows "\<exists>S::real set. S \<subseteq> {0..1} \<and> S \<in> (sets lebesgue) \<and> emeasure lebesgue S > 0 \<and> (\<forall>x\<in>S. \<bar>f x\<bar> \<ge> 2^n*(n + 1))"
sorry
end
| null |
putnam_1972_b1 | theorem putnam_1972_b1
(S : β β β := fun x : β => β' n : β, x^n * (x - 1)^(2*n) / (Nat.factorial n))
(p : β β β)
(hp : β a : β, a > 0 β§ β x β ball 0 a, β' n : β, (p n)*x^n = S x)
: Β¬β n : β, p n = 0 β§ p (n + 1) = 0 β§ p (n + 2) = 0 :=
sorry | Prove that no three consecutive coefficients of the power series of $$\sum_{n = 0}^{\infty} \frac{x^n(x - 1)^{2n}}{n!}$$ all equal $0$. | null | ['analysis'] | null | theory putnam_1972_b1 imports Complex_Main
begin
theorem putnam_1972_b1:
fixes S :: "real \<Rightarrow> real"
and p :: "nat \<Rightarrow> real"
defines "S \<equiv> (\<lambda>x::real. (\<Sum>n::nat. x^n * (x - 1)^(2*n) / (fact n)))"
assumes hp: "\<exists>a::real>0. \<forall>x::real\<in>(ball 0 a). (\<Sum>n::nat. (p n)*x^n) = S x"
shows "\<not>(\<exists>n::nat. p n = 0 \<and> p (n + 1) = 0 \<and> p (n + 2) = 0)"
sorry
end
| null |
putnam_1972_b3 | theorem putnam_1972_b3
(G : Type*) [Group G]
(A B : G)
(hab : A * B * A = B * A^2 * B β§ A^3 = 1 β§ (β n : β€, n > 0 β§ B^(2*n - 1) = 1))
: B = 1 :=
sorry | Let $A$ and $B$ be two elements in a group such that $ABA = BA^2B$, $A^3 = 1$, and $B^{2n-1} = 1$ for some positive integer $n$. Prove that $B = 1$. | null | ['abstract_algebra'] | null | theory putnam_1972_b3 imports Complex_Main
"HOL-Algebra.Group"
begin
theorem putnam_1972_b3:
fixes G (structure)
and A B :: 'a
assumes Ggroup: "group G"
and abinG: "A \<in> carrier G \<and> B \<in> carrier G"
and hab: "A\<otimes>B\<otimes>A = B\<otimes>A[^]2\<otimes>B \<and> A[^]3 = \<one> \<and> (\<exists>n::nat. n > 0 \<and> B[^](2*n-1) = \<one>)"
shows "B = \<one>"
sorry
end
| null |
putnam_1972_b4 | theorem putnam_1972_b4
(n : β€)
(hn : n > 1)
(vars : β β β β β β (Fin 3 β β) := fun a b c β¦ fun i β¦ ite (i = 0) a (ite (i = 1) b c))
: β P : MvPolynomial (Fin 3) β, β x : β, x = MvPolynomial.eval (vars (x^n) (x^(n+1)) (x + x^(n+2))) P :=
sorry | Let $n \geq 2$ be an integer. Show that there exists a polynomial $P(x,y,z)$ with integral coefficients such that $x \equiv P(x^n, x^{n+1}, x + x^{n+2})$. | null | ['algebra'] | null | theory putnam_1972_b4 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
theorem putnam_1972_b4:
fixes n :: nat
assumes hn: "n > 1"
shows "\<exists>P::real poly poly poly. \<forall>x::real. x = poly (poly (poly P (monom (monom (x + x^(n+2)) 0) 0)) (monom (x^(n+1)) 0)) (x^n)"
sorry
end
| null |
putnam_1972_b6 | theorem putnam_1972_b6
(k : β)
(hk : k β₯ 1)
(n : Fin k β β€)
(hn : β i : Fin k, n i > 0)
(hn' : β i j : Fin k, i < j β n i < n j)
(zpoly : β β β := fun z => 1 + β i : Fin k, z^(n i))
: β z : β, z β ball 0 ((Real.sqrt 5 - 1)/2) β zpoly z β 0 :=
sorry | Let $n_1 < n_2 < \dots < n_k$ be a set of positive integers. Prove that the polynomail $1 + z^{n_1} + z^{n_2} + \dots + z^{n_k}$ has not roots inside the circle $|z| < (\frac{\sqrt{5}-1}{2}$. | null | ['analysis'] | null | theory putnam_1972_b6 imports Complex_Main
begin
(* uses (nat \<Rightarrow> nat) instead of (Fin k \<Rightarrow> nat) *)
theorem putnam_1972_b6:
fixes k :: nat
and n :: "nat \<Rightarrow> nat"
and zpoly :: "complex \<Rightarrow> complex"
assumes hk: "k \<ge> 1"
and hn: "\<forall>i::nat\<in>{0..(k-1)}. n i > 0"
and hn': "\<forall>i::nat\<in>{0..(k-1)}. \<forall>j::nat\<in>{0..(k-1)}. (i < j \<longrightarrow> n i < n j)"
defines "zpoly \<equiv> (\<lambda>z::complex. 1 + (\<Sum>i::nat=0..(k-1). z^(n i)))"
shows "\<forall>z::complex. (z \<in> ball 0 ((sqrt 5 - 1)/2) \<longrightarrow> zpoly z \<noteq> 0)"
sorry
end
| null |
putnam_1971_a1 | theorem putnam_1971_a1
(S : Set (β€ Γ β€ Γ β€))
(hS : S.ncard = 9)
(L : (β€ Γ β€ Γ β€) Γ (β€ Γ β€ Γ β€) β Set (β Γ β Γ β) := fun ((a, b, c), (d, e, f)) =>
{(t*a + (1-t)*d, t*b + (1-t)*e, t*c + (1-t)*f) | t β Ioo (0 : β) 1})
: β x y z : β€, β P Q : β€ Γ β€ Γ β€, P β S β§ Q β S β§ P β Q β§ ((x : β), (y : β), (z : β)) β L (P, Q) :=
sorry | Let $S$ be a set of $9$ lattice points (points with integer coordinates) in $3$-dimensional Euclidean space. Prove that there exists a lattice point along the interior of some line segment that joins two distinct points in $S$. | null | ['geometry', 'combinatorics'] | null | theory putnam_1971_a1 imports Complex_Main
begin
theorem putnam_1971_a1:
fixes S :: "(int \<times> int \<times> int) set"
and L :: "((int \<times> int \<times> int) \<times> (int \<times> int \<times> int)) \<Rightarrow> ((real \<times> real \<times> real) set)"
assumes hS: "card S = 9"
defines "L \<equiv> (\<lambda>((a::int,b::int,c::int),(d::int,e::int,f::int)).
{(ad::real,be::real,cf::real). (\<exists>t::real\<in>{0<..<1}. ad = t*a + (1-t)*d \<and> be = t*b + (1-t)*e \<and> cf = t*c + (1-t)*f)})"
shows "\<exists>x y z::int. \<exists>P\<in>S. \<exists>Q\<in>S. P \<noteq> Q \<and> (x,y,z) \<in> L (P,Q)"
sorry
end
| null |
putnam_1971_a2 | abbrev putnam_1971_a2_solution : Set (Polynomial β) := sorry
-- {Polynomial.X}
theorem putnam_1971_a2
: β P : Polynomial β, (P.eval 0 = 0 β§ (β x : β, P.eval (x^2 + 1) = (P.eval x)^2 + 1)) β P β putnam_1971_a2_solution :=
sorry | Determine all polynomials $P(x)$ such that $P(x^2 + 1) = (P(x))^2 + 1$ and $P(0) = 0$. | Show that the only such polynomial is the identity function. | ['algebra'] | null | theory putnam_1971_a2 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
definition putnam_1971_a2_solution :: "(real poly) set" where "putnam_1971_a2_solution \<equiv> undefined"
(* {monom 1 1} *)
theorem putnam_1971_a2:
shows "\<forall>P::real poly. ((poly P 0 = 0 \<and> (\<forall>x::real. poly P (x^2 + 1) = (poly P x)^2 + 1)) \<longleftrightarrow> P \<in> putnam_1971_a2_solution)"
sorry
end
| null |
putnam_1971_a3 | theorem putnam_1971_a3
(a b c : β Γ β)
(R : β)
(habclattice : a.1 = round a.1 β§ a.2 = round a.2 β§ b.1 = round b.1 β§ b.2 = round b.2 β§ c.1 = round c.1 β§ c.2 = round c.2)
(habcneq : a β b β§ a β c β§ b β c)
(oncircle : (β Γ β) β β β (β Γ β) β Prop := fun C r p => Euclidean.dist p C = r)
(hcircle : β C : β Γ β, oncircle C R a β§ oncircle C R b β§ oncircle C R c)
: (Euclidean.dist a b) * (Euclidean.dist a c) * (Euclidean.dist b c) β₯ 2 * R :=
sorry | The three vertices of a triangle of sides $a,b,c$ are lattice points and lie on a circle of radius $R$. Show that $abc \geq 2R$. | null | ['geometry'] | null | theory putnam_1971_a3 imports Complex_Main
"HOL-Analysis.Finite_Cartesian_Product"
begin
theorem putnam_1971_a3:
fixes a b c :: "real^2"
and oncircle :: "(real^2) \<Rightarrow> real \<Rightarrow> (real^2) \<Rightarrow> bool"
and R :: real
assumes habclattice: "a$1 = round (a$1) \<and> a$2 = round (a$2) \<and> b$1 = round (b$1) \<and> b$2 = round (b$2) \<and> c$1 = round (c$1) \<and> c$2 = round (c$2)"
and habcneq: "a \<noteq> b \<and> a \<noteq> c \<and> b \<noteq> c"
defines "oncircle \<equiv> (\<lambda>(C::real^2)(r::real)(p::real^2). (dist p C = r))"
assumes hcircle: "\<exists>C::real^2. oncircle C R a \<and> oncircle C R b \<and> oncircle C R c"
shows "(dist a b) * (dist a c) * (dist b c) \<ge> 2*R"
sorry
end
| null |
putnam_1971_a4 | theorem putnam_1971_a4
(Ξ΅ : β)
(hΞ΅ : 0 < Ξ΅ β§ Ξ΅ < 1)
(P : β β β β MvPolynomial (Fin 2) β := fun n Ξ΄ => (MvPolynomial.X 0 + MvPolynomial.X 1)^n * ((MvPolynomial.X 0)^2 - (MvPolynomial.C (2 - Ξ΄))*(MvPolynomial.X 0)*(MvPolynomial.X 1) + (MvPolynomial.X 1)^2))
: β N : β, β n β₯ N, β i : Fin 2 ββ β, MvPolynomial.coeff i (P n Ξ΅) > 0 :=
sorry | Show that for $\epsilon \in (0,1)$, the expression $(x + y)^n (x^2 - 2-\epsilon)xy + y^2)$ is a polynomial with positive coefficients for $n$ sufficiently large, where $n$ is an integer. | null | ['analysis'] | null | theory putnam_1971_a4 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
theorem putnam_1971_a4:
fixes \<epsilon> :: real
and P :: "nat \<Rightarrow> real \<Rightarrow> (real poly poly)"
assumes h\<epsilon>: "0 < \<epsilon> \<and> \<epsilon> < 1"
defines "P \<equiv> (\<lambda>(n::nat)(\<delta>::real). (monom (monom 1 1) 0 + monom (monom 1 0) 1)^n * ((monom (monom 1 1) 0)^2 - (monom (monom (2 - \<delta>) 0) 0)*(monom (monom 1 1) 1) + (monom (monom 1 0) 1)^2))"
shows "\<exists>N::nat. \<forall>n::nat\<ge>N. (\<forall>i j::nat. coeff (coeff (P n \<epsilon>) j) i > 0)"
sorry
end
| null |
putnam_1971_a6 | theorem putnam_1971_a6
(c : β)
(hc : β n : β€, n > 0 β β m : β€, (n : β)^c = m)
: β m : β, c = m :=
sorry | Let $c$ be a real number such that $n^c$ is an integer for every positive integer $n$. Show that $c$ is a non-negative integer. | null | ['number_theory'] | null | theory putnam_1971_a6 imports Complex_Main
begin
theorem putnam_1971_a6:
fixes c :: real
assumes hc: "\<forall>n::nat. (n > 0 \<longrightarrow> (\<exists>m::int. n powr c = m))"
shows "\<exists>m::nat. c = m"
sorry
end
| null |
putnam_1971_b1 | theorem putnam_1971_b1
(S : Type*) [Mul S]
(hself : β x : S, x * x = x)
(h2 : β x y z : S, (x * y) * z = (y * z) * x)
: β x y z : S, (x * y) * z = x * (y * z) β§ x * y = y * x :=
sorry | Let $S$ be a set and let $\cdot$ be a binary operation on $S$ satisfying the two following laws: (1) for all $x$ in $S$, $x = x \cdot x$, (2) for all $x,y,z$ in $S$, $(x \cdot y) \cdot z) = (y \cdot z) \cdot x$. Show that $\cdot$ is associative and commutative. | null | ['abstract_algebra'] | null | theory putnam_1971_b1 imports Complex_Main
begin
theorem putnam_1971_b1:
fixes Smul :: "'S \<Rightarrow> 'S \<Rightarrow> 'S" (infixl "\<^bold>*" 70)
assumes hself: "\<forall>x::'S. x \<^bold>* x = x"
and h2: "\<forall>x y z::'S. (x \<^bold>* y) \<^bold>* z = (y \<^bold>* z) \<^bold>* x"
shows "\<forall>x y z::'S. (x \<^bold>* y) \<^bold>* z = x \<^bold>* (y \<^bold>* z) \<and> x \<^bold>* y = y \<^bold>* x"
sorry
end
| null |
putnam_1971_b2 | abbrev putnam_1971_b2_solution : Set (β β β) := sorry
-- {fun x : β => (x^3 - x^2 - 1)/(2 * x * (x - 1))}
theorem putnam_1971_b2
(S : Set β := univ \ {0, 1})
(P : (β β β) β Prop := fun (F : β β β) => β x β S, F x + F ((x - 1)/x) = 1 + x)
: (β F β putnam_1971_b2_solution, P F) β§ β f : β β β, P f β β F β putnam_1971_b2_solution, (β x β S, f x = F x) :=
sorry | Find all functions $F : \mathbb{R} \setminus \{0, 1\} \to \mathbb{R}$ that satisfy $F(x) + F\left(\frac{x - 1}{x}\right) = 1 + x$ for all $x \in \mathbb{R} \setminus \{0, 1\}$. | The only such function is $F(x) = \frac{x^3 - x^2 - 1}{2x(x - 1)}$. | ['algebra'] | null | theory putnam_1971_b2 imports Complex_Main
begin
definition putnam_1971_b2_solution :: "(real \<Rightarrow> real) set" where "putnam_1971_b2_solution \<equiv> undefined"
(* {(\<lambda>x::real. (x^3 - x^2 - 1) / (2*x*(x - 1)))} *)
theorem putnam_1971_b2:
fixes S :: "real set"
and P :: "(real \<Rightarrow> real) \<Rightarrow> bool"
defines "S \<equiv> UNIV - {0,1}"
and "P \<equiv> (\<lambda>F::real\<Rightarrow>real. (\<forall>x\<in>S. F x + F ((x-1)/x) = 1 + x))"
shows "(\<forall>F\<in>putnam_1971_b2_solution. P F) \<and> (\<forall>f::real\<Rightarrow>real. (P f \<longrightarrow> (\<exists>F\<in>putnam_1971_b2_solution. (\<forall>x\<in>S. f x = F x))))"
sorry
end
| null |
putnam_1971_b6 | theorem putnam_1971_b6
(Ξ΄ : β€ β β€ := fun n => sSup {t | Odd t β§ t β£ n})
: β x : β€, x > 0 β |β i in Finset.Icc 1 x, (Ξ΄ i)/(i : β) - 2*x/3| < 1 :=
sorry | Let $\delta(x) be the greatest odd divisor of the positive integer $x$. Show that $|\sum_{n = 1}^x \delta(n)/n - 2x/3| < 1$ for all positive integers $x$. | null | ['number_theory'] | null | theory putnam_1971_b6 imports Complex_Main
begin
theorem putnam_1971_b6:
fixes \<delta> :: "nat \<Rightarrow> nat"
defines "\<delta> \<equiv> (\<lambda>n::nat. (GREATEST t::nat. odd t \<and> t dvd n))"
shows "\<forall>x::nat. (x > 0 \<longrightarrow> \<bar>(\<Sum>i::nat=1..x. (\<delta> i)/i) - 2*x/3\<bar> < 1)"
sorry
end
| null |
putnam_1999_a1 | abbrev putnam_1999_a1_solution : Prop := sorry
-- True
theorem putnam_1999_a1
: putnam_1999_a1_solution β β f g h : Polynomial β, β x : β, |f.eval x| - |g.eval x| + h.eval x = if x < -1 then -1 else (if (x β€ 0) then 3 * x + 2 else -2 * x + 2) :=
sorry | Find polynomials $f(x)$,$g(x)$, and $h(x)$, if they exist, such that for all $x$, \[|f(x)|-|g(x)|+h(x) = \begin{cases} -1 & \mbox{if $x<-1$} \\3x+2 & \mbox{if $-1 \leq x \leq 0$} \\-2x+2 & \mbox{if $x>0$.}\end{cases}\]? | Show that the answer is such functions do exist. | ['algebra'] | null | theory putnam_1999_a1 imports
Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
(* Note: The actual problem asks to "find" such polynomials as well - but the solution does not give a set of all possible solutions *)
(* So we would need to do the analysis ourselves, the following formalization should work. *)
definition putnam_1999_a1_solution :: bool where "putnam_1999_a1_solution \<equiv> undefined"
(* True *)
theorem putnam_1999_a1:
shows "putnam_1999_a1_solution \<longleftrightarrow> (\<exists> f g h :: real poly. \<forall> x :: real. \<bar>poly f x\<bar> - \<bar>poly g x\<bar> + poly h x = (if x < -1 then -1 else (if x \<le> 0 then 3 * x + 2 else -2 * x + 2)))"
sorry
end | null |
putnam_1999_a2 | theorem putnam_1999_a2
(p : Polynomial β)
(hp : β x : β, p.eval x β₯ 0)
: β k : β, β f : Fin k β Polynomial β, k > 0 β§
β x : β, p.eval x = β j : Fin k, ((f j).eval x) ^ 2 :=
sorry | Let $p(x)$ be a polynomial that is nonnegative for all real $x$. Prove that for some $k$, there are polynomials $f_1(x),\dots,f_k(x$) such that \[p(x) = \sum_{j=1}^k (f_j(x))^2.\] | null | ['algebra'] | null | theory putnam_1999_a2 imports Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
theorem putnam_1999_a2:
fixes p :: "real poly"
assumes hpos : "\<forall>x. poly p x \<ge> 0"
shows "\<exists>S :: real poly set . \<forall>x. finite S \<and> poly p x = (\<Sum>s \<in> S. (poly s x)^2)"
sorry
end
| null |
putnam_1999_a3 | theorem putnam_1999_a3
(f : β β β := fun x => 1/(1 - 2 * x - x^2))
(a : β β β)
(hf : β Ξ΅ > 0, β x β ball 0 Ξ΅, Tendsto (Ξ» n => β i in Finset.range n, (a n) * x^n) atTop (π (f x)))
: β n : β, β m : β, (a n)^2 + (a (n + 1))^2 = a m :=
sorry | Consider the power series expansion \[\frac{1}{1-2x-x^2} = \sum_{n=0}^\infty a_n x^n.\] Prove that, for each integer $n\geq 0$, there is an integer $m$ such that \[a_n^2 + a_{n+1}^2 = a_m .\] | null | ['algebra'] | null | theory putnam_1999_a3 imports
Complex_Main
"HOL-Analysis.Elementary_Metric_Spaces"
begin
theorem putnam_1999_a3:
fixes f :: "real \<Rightarrow> real"
and a :: "nat \<Rightarrow> real"
defines "f \<equiv> \<lambda> x. 1 / (1 - 2 * x - x ^ 2)"
assumes hf: "\<exists> \<epsilon> > 0. \<forall> x \<in> ball 0 \<epsilon>. (\<Sum> n :: nat. (a n) * x ^ n) = f x"
shows "\<forall> n :: nat. \<exists> m :: nat. (a n) ^ 2 + (a (n + 1)) ^ 2 = a m"
sorry
end
| null |
putnam_1999_a4 | abbrev putnam_1999_a4_solution : β := sorry
-- 9/32
theorem putnam_1999_a4
: Tendsto (fun i => β m in Finset.range i, β' n : β, (((m + 1)^2*(n+1))/(3^(m + 1) * ((n+1)*3^(m + 1) + (m + 1)*3^(n+1))) : β)) atTop (π putnam_1999_a4_solution) :=
sorry | Sum the series \[\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2 n}{3^m(n3^m+m3^n)}.\] | Show that the solution is 9/32. | ['number_theory'] | Section putnam_1999_a4.
Require Import Reals Coquelicot.Coquelicot.
Definition putnam_1999_a4_solution := 1.
Theorem putnam_1999_a4:
let f (m n: nat) := INR ((m + 1) ^ 2 * n) / INR (3 ^ m * (n * 3 ^ m + m * 3 ^ n)) in
let fn (n: nat) := Lim_seq (fun mInc => sum_n (fun m => f m n) mInc) in
Lim_seq (fun nInc => sum_n (fun n => (real (fn n))) nInc) = putnam_1999_a4_solution.
Proof. Admitted.
End putnam_1999_a4. | theory putnam_1999_a4 imports
Complex_Main
begin
definition putnam_1999_a4_solution :: real where "putnam_1999_a4_solution \<equiv> undefined"
(* 9 / 32 *)
theorem putnam_1999_a4:
shows "(\<Sum> m :: nat. \<Sum> n :: nat. (m + 1) ^ 2 * (n + 1) / (3 ^ (m + 1) * ((n + 1) * 3 ^ (m + 1) + (m + 1) * 3 ^ (n + 1)))) = putnam_1999_a4_solution"
sorry
end | null |
putnam_1999_a5 | theorem putnam_1999_a5
: β C : β, β p : Polynomial β, p.degree = 1999 β βp.eval 0β β€ C * β« x in (-1)..1, βp.eval xβ :=
sorry | Prove that there is a constant $C$ such that, if $p(x)$ is a polynomial of degree 1999, then \[|p(0)|\leq C \int_{-1}^1 |p(x)|\,dx.\] | null | ['analysis'] | Section putnam_1999_a5.
Require Import Reals Coquelicot.Coquelicot.
Theorem putnam_1999_a5:
let p (a: nat -> R) (x: R) := sum_n (fun i => a i * x ^ i) 2000 in
forall (a: nat -> R), exists (c: R),
Rabs (p a 0) <= c * RInt (fun x => Rabs (p a x)) (-1) 1.
Proof. Admitted.
End putnam_1999_a5. | theory putnam_1999_a5 imports
Complex_Main
"HOL-Computational_Algebra.Polynomial"
"HOL-Analysis.Interval_Integral"
begin
theorem putnam_1999_a5:
shows "\<exists> C :: real. \<forall> p :: real poly. degree p = 1999 \<longrightarrow> \<bar>poly p 0\<bar> \<le> C * (interval_lebesgue_integral lebesgue (-1) 1 (\<lambda> x. \<bar>poly p x\<bar>))"
sorry
end | null |
putnam_1999_a6 | theorem putnam_1999_a6
(a : β€ β β)
(ha1 : a 1 = 1)
(ha2 : a 2 = 2)
(ha3 : a 3 = 24)
(hange4 : β n : β, n β₯ 4 β a n = (6 * (a (n - 1))^2 * (a (n - 3)) - 8 * (a (n - 1)) * (a (n - 2))^2)/(a (n - 2) * a (n - 3)))
: β n, n β₯ 1 β (β k : β€, a n = k * n) :=
sorry | The sequence $(a_n)_{n\geq 1}$ is defined by $a_1=1, a_2=2, a_3=24,$ and, for $n\geq 4$, \[a_n = \frac{6a_{n-1}^2a_{n-3} - 8a_{n-1}a_{n-2}^2}{a_{n-2}a_{n-3}}.\] Show that, for all n, $a_n$ is an integer multiple of $n$. | null | ['algebra'] | Section putnam_1999_a6.
Require Import Reals Coquelicot.Coquelicot.
Theorem putnam_1999_a6:
let fix a (n: nat) :=
match n with
| O => 1
| S O => 2
| S (S O) => 24
| S (S ((S n'') as n') as n) => (6 * a n ^ 2 * a n'' - 8 * a n * a n' ^ 2) / (a n' * a n'')
end in
forall (n: nat), exists (k: nat), a n = INR (n * k).
Proof. Admitted.
End putnam_1999_a6. | theory putnam_1999_a6 imports
Complex_Main
begin
theorem putnam_1999_a6:
fixes a :: "nat \<Rightarrow> real"
assumes ha1: "a 1 = 1"
and ha2: "a 2 = 2"
and ha3: "a 3 = 24"
and hange4: "\<forall> n :: nat. n \<ge> 4 \<longrightarrow> a n = (6 * (a (n - 1)) ^ 2 * (a (n - 3)) - 8 * (a (n - 1)) * (a (n - 2)) ^ 2) / (a (n - 2) * a (n - 3))"
shows "\<forall> n. n \<ge> 1 \<longrightarrow> (\<exists> k :: int. a n = k * n)"
sorry
end | null |
putnam_1999_b2 | theorem putnam_1999_b2
(P Q : Polynomial β)
(hQ : Q.natDegree = 2)
(hP : β x : β, P.eval x = (Q.eval x) * (Polynomial.derivative^[2] P).eval x)
: (β x1 x2 : β, x1 β x2 β§ P.eval x1 = 0 β§ P.eval x2 = 0) β (β f : Fin (P.natDegree) β β, (β i j : Fin (P.natDegree), i β j β f i β f j) β§ (β i : Fin (P.natDegree), P.eval (f i) = 0)) :=
sorry | Let $P(x)$ be a polynomial of degree $n$ such that $P(x)=Q(x)P''(x)$, where $Q(x)$ is a quadratic polynomial and $P''(x)$ is the second derivative of $P(x)$. Show that if $P(x)$ has at least two distinct roots then it must have $n$ distinct roots. | null | ['analysis'] | Section putnam_1999_b2.
Require Import List Reals Coquelicot.Coquelicot.
Theorem putnam_1999_b2:
forall (a1 a2: nat -> R) (n: nat),
let p (x: R) := sum_n (fun i => a1 i * x ^ i) n in
let q (x: R) := sum_n (fun i => a2 i * x ^ i) 2 in
forall (x: R), p x = q x * (Derive_n (fun x => p x) 2) x /\
exists (r1 r2: R), r1 <> r2 /\ p r1 = 0 /\ p r2 = 0 ->
exists (roots: list R), length roots = n /\ NoDup roots /\ forall (r: R), In r roots -> p r = 0.
Proof. Admitted.
End putnam_1999_b2. | theory putnam_1999_b2 imports
Complex_Main
"HOL-Computational_Algebra.Polynomial"
begin
theorem putnam_1999_b2:
fixes P Q :: "complex poly"
assumes hQ: "degree Q = 2"
assumes hP: "\<forall> x :: complex. poly P x = poly Q x * poly ((pderiv^^2) P) x"
shows "(\<exists> x1 x2 :: complex. x1 \<noteq> x2 \<and> poly P x1 = 0 \<and> poly P x2 = 0) \<longrightarrow> (\<exists> S :: complex set. card S = degree P \<and> (\<forall> z \<in> S. poly P z = 0))"
sorry
end | null |
Subsets and Splits