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Proof of a summation of $k^4$ I am trying to prove $$\sum_{k=1}^n k^4$$ I am supposed to use the method where $$(n+1)^5 = \sum_{k=1}^n(k+1)^5 - \sum_{k=1}^nk^5$$ So I have done that and and after reindexing and a little algebra, I get $$(n+1)^5 = 1+ 5\sum_{k=1}^nk^4 + 10\sum_{k=1}^nk^3 + 10\sum_{k=1}^nk^2 + 5\sum_{k=1}^nk + \sum_{k=1}^n1$$ So then $$\sum_{k=1}^nk^4$$ is in my formula, and I solve for that and use the formulas for the sums that we already have which are $k^3,k^2,k, and 1$, but I cannot figure out the solution from where I am. Here is where I have simplified to. I think it is just my algebra that I can't figure out. $$(n+1)^5 - (n+1) -{10n^2(n+1)^2 \over 4} -{10n(n+1)(2n+1) \over 6} - {5n(n+1) \over 2} = 5\sum_{k=1}^nk^4$$ which I can get down to: $$n(n+1)(6n^3 + 24n^2 + 36n + 24 - (15n^2 + 35n + 25)) = 30\sum_{k=1}^nk^4$$ $$n(n+1)(6n^3 + 9n^2 + n - 1) = 30\sum_{k=1}^nk^4$$ But that doesn't seem right. Where am I messing up? Thank you! If my question is missing something please let me know and I will fix it. I have put in a lot of work with this question so please don't downvote it.

Here's an alternative approach using binomial coefficients. Firs we express $k^4$ as a linear combination of $\binom{k+a}{4}$ where $a=0,1,2,3$, i.e. $$k^4={k+3\choose 4}+11{k+2\choose 4}+11{k+1\choose 4}+{k\choose 4}$$. Summing this from $1$ to $n$ and using the hockey stick summation identity, i.e. $\sum_{r=0}^m {r\choose b}={m+1\choose {b+1}}$, we have $$\begin{align} \sum_{k=1}^nk^4&=\sum_{k=1}^n\left[{k+3\choose 4}+11{k+2\choose 4}+11{k+1\choose 4}+{k\choose 4}\right]\\ &={n+4\choose 5}+11{n+3\choose 5}+11{n+2\choose 5}+{n+1\choose 5}\\ \end{align}$$ which is convenient for numerical evaluation. This reduces to $$\begin{align} &\frac{(n+4)^\underline{4}+11(n+3)^\underline{4}+11(n+2)^\underline{4}+(n+1)^\underline{4}}{5!}\\ &=\frac 1{5!}n(n+1)\biggr[(n+4)(n+3)(n+2)+(n-1)(n-2)(n-3)+11(n+2)(n-1)\bigg((n+3)+(n-2)\biggr)\biggr]\\ &=\frac 1{30}n(n+1)(2n+1)(3n^3+3n-1) \end{align}$$ where $x^\underline{n}$ represents the falling factorial and is given by $\underbrace{x(x-1)(x-2)\cdots(x-n+1)}_{n\text{ terms}}$.

Calculus 1: Find the limit as x approaches 4 of $\frac{3-\sqrt{x+5}}{x-4}$ I understand how to find limits, but for some reason I cannot figure out the algebra of this problem. I tried multiplying by the conjugate and end up with 0/0. When I check on my calculator, or apply L'Hopital's rule I get -1/6. Is there an algebra trick that I am missing on this one? $\displaystyle\frac{3-\sqrt{x+5}}{x-4}$ I have solved similar problems with the square root by multiplying by the conjugate, but it doesn't seem to work for this one.

$$\begin{array}{rcl}\lim_{x\to 4} \frac{3-\sqrt{x+5}}{x-4} & = & \lim_{x\to 4} \frac{(3-\sqrt{x+5})(3+\sqrt{x+5})}{(x-4)(3+\sqrt{x+5})}=\lim_{x\to 4} \frac{9-(x+5)}{(x-4)(3+\sqrt{x+5})} \\ & = & \lim_{x\to 4} \frac{4-x}{(x-4)(3+\sqrt{x+5})} =\lim_{x\to 4} \frac{-1}{3+\sqrt{x+5}}=- \frac{1}{6}.\end{array}$$

solve indefinite integral I have this indefinite integral $\int 3 \sqrt{x}\,dx$ to solve. My attempt: $$\int 3 \sqrt{x}\,dx = 3 \cdot \frac {x^{\frac {1}{2} + \frac {2}{2}}}{\frac {1}{2} + \frac {2}{2}}$$ $$\int 3 \sqrt{x}\,dx = 3 \frac{x^{\frac {3}{2}}}{\frac {3}{2}} = \frac{2}{3} \cdot \frac{9}{3} x^{\frac {3}{2}}$$ $$\int 3 \sqrt{x}\,dx = \frac{18}{3} x^{\frac{3}{2}} = 6 x^{\frac{3}{2}}$$ But according to wolframalpha the answer should be $2 x^{\frac {3}{2}}$ Where did I make a error in my calculation? Thanks!

In your penultimate step, $\frac 23$.$\frac 93$ $=\frac {18}{9}$ $=2$, not 6.

Without actually calculating the value of cubes find the value of $(1)^3+(2)^3+2(4)^3+(-5)^3+(-6)^3$. Also write the identity used Without actually calculating the value of cubes find the value of $(1)^3+(2)^3+2(4)^3+(-5)^3+(-6)^3$. Also write the identity used

You can use the identity that if $a + b + c=0$, then $a^3 + b^3 + c^3 = 3abc$. So we can say that: * *$2^3 + 4^3 + (-6)^3 = 3\cdot 2\cdot 4\cdot -6 = -144$. *$1^3 + 4^3 + (-5)^3 = 3\cdot 1\cdot 4 \cdot -5 = -60$. Adding both, we get: $1^3 + 2^3 + 2 (4)^3 + (-5)^3 + (-6)^3 = -204$

Solve $x^3 - x + 1 = 0$ Solve $x^3 - x + 1 = 0$, this cannot be done through elementary methods. Although, this is way out of my capabilities, I would love to see a solution (closed form only). Thanks!

Let $x=u+v$ and note that $(u+v)^3=3uv(u+v)+(u^3+v^3)$ This has the required form if $3uv=1$ and $u^3+v^3=-1$ So $u^3v^3=\frac 1{27}$, and $u^3, v^3$ are roots of $y^2+y+\frac 1{27}=0$ $u,v= \sqrt[3] {\frac {-1\pm \sqrt{1-\frac 4{27}}}2 }$ $x=u+v=\sqrt[3] {\frac {-1+ \sqrt{\frac {23}{27}}}2 }+\sqrt[3] {\frac {-1- \sqrt{\frac {23}{27}}}2 }$

Find series expansion of 1/cosx Find the series expansion of 1/cosx from basic series expansions. I tried to find 1/cosx from the expansion of cosx but was unsure how to continue. When I found 1/cosx from the basic formula for finding series expansions I didn't get the same answer.

(going to the fifth term for an example purpose) Using the basic expansions of cos(x) gives us $$ \frac{1}{\cos(x)} = \frac{1}{1-\frac{x^2}{2}+\frac{x^4}{24} + \cdots} $$ of the form $ \frac{1}{1-X} $ which has a known and easy expansion : $$ 1+X+X^2+X^3+X^4+X^5+\cdots $$ where $ X = \frac{x^2}{2} + \frac{x^4}{24} $ (approching $0$ when $x$ approaches $0$). The smaller $x$ term is $x^2$, so we don't need to take more terms than $X^2$ in the above expansion (otherwise terms would exceed $x^5$ and be negligeable). Hence, $$ \frac{1}{\cos(x)} = 1+(\frac{x^2}{2}−\frac{x^4}{24})+(\frac{x^2}{2})^2+o_{x\to0}(x^5) $$ $$ = 1+\frac{x^2}{2}+\frac{5x^4}{24} $$

Repeated substitution gone wrong It was an exam question. $$ f(n)= \begin{cases} 0 & \mbox{if } n \leq 1 \\ 3 f(\lfloor n/5 \rfloor) + 1 & \mbox{if } n > 1 \\ \end{cases}$$ So by calculating some I have $f(5) = 1$, $f(10) = 4$, $f(50) = 13$. I had to solve this recurrence. So to get rid of the floor operator I said lets $n$ be $5^k$ so a multiple of $5$. So started to solve this. \begin{equation} \begin{split} f(5^k) & = 3\cdot f(5^{k-1}) + 1\\ & = 3\cdot[ 3f(5^{k-2}) + 1 ]+ 1\\ & = 3^2f(5^{k-2}) + 3 + 1\\ & = 3^2\cdot[ 3f(5^{k-3}) + 1 ]+ 3 + 1\\ & = 3^3f(5^{k-3}) + 9 + 3 + 1 \\ & = ...\\ & = 3^jf(5^{k-j}) + \sum_{j=0}^k 3^j\\ & = 3^jf(5^{k-j}) + \frac{3^{k+1} - 1}{2}\\ \text{let $j = k$}\\ & = 3^kf(5^0) + \frac{3^{k+1} - 1}{2}\\ & = 3^kf(1) + \frac{3^{k+1} - 1}{2}\\ & = 3^k\cdot 0 + \frac{3^{k+1} - 1}{2}\\ & = \frac{3^{k+1} - 1}{2}\\ \end{split} \end{equation} And now I'm stuck here trying to express my $3^{k+1}$ as a function of $n$. I'm pretty I probably made some error up there, if I did I don't see them.

I didn't check your whole argument, but assuming it's correct, you can use the fact that if $n$ is $5^k$, then $k$ is $\log_5 n$ to express the result in terms of $n$. (Side Question: are the $x$'s in the function definition supposed to be $n$'s?)

A question on a nonnegative quadratic form Denote $x,y,z$ as variables, and $a,b,c$ as coefficients. Suppose $a\leq b\leq 0\leq c$ and $a+b+c=0$. Could anyone help me prove whether the following quadratic form positive semi-definite? \begin{equation*} \begin{split} I(x,y,z)=&(a^2+4b^2+4c^2)a^2x^2+(4a^2+b^2+4c^2)b^2y^2\\ +&(4a^2+4b^2+c^2)c^2z^2\\ -&2ab(a^2+b^2+c^2+3ab)xy\\ +&2ac(a^2+b^2+c^2+3ac)xz\\ +&2bc(a^2+b^2+c^2+3bc)yz. \end{split} \end{equation*}

Well, I showed you how to try to attack these problems a couple of days ago, and you can use exactly the same strategy here. A problem on positive semi-definite quadratic forms/matrices Once again using MATLAB Toolbox YALMIP to compute a sum-of-squares certificate. Ordering of the variables is not required, but the equality appears to be necessary to exploit sdpvar a b c = -a-b I=(a^2+4*b^2+4*c^2)*a^2*x^2+(4*a^2+b^2+4*c^2)*b^2*y^2+(4*a^2+4*b^2+c^2)*c^2*z^2-2*a*b* (a^2+b^2+c^2+3*a*b)*x*y+2*a*c*(a^2+b^2+c^2+3*a*c)*x*z+2*b*c*(a^2+b^2+c^2+3*b*c)*y*z; % Floating-point decomposition [~,v,Q] = solvesos(sos(I)) sdisplay(v{1}) Q{1} % Integer decomposition Q = intvar(7,7); Match = coefficients(I-v{1}'*Q*v{1},[a b x y z])==0; optimize([Match, Q >=0]) value(Q) >> sdisplay(v{1}) ans = 'z*b^2' 'z*a*b' 'z*a^2' 'y*b^2' 'y*a*b' 'x*a*b' 'x*a^2' >> value(Q) ans = 5 6 1 1 1 -2 0 6 12 6 1 -1 -1 1 1 6 5 0 -2 1 1 1 1 0 5 4 -2 -2 1 -1 -2 4 8 -3 -2 -2 -1 1 -2 -3 8 4 0 1 1 -2 -2 4 5

Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result R when tested with various values of $a$ ( where $0<a<1$). $$ D_1 \, =\, \int_0^{2\pi}f_1\,\mathrm{d}\theta \, =\, \int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta \,=\, \frac{3a\pi}{(1-a^2)^{5/2}} \, =\,R $$ and $$D_2\, =\,\int_0^{2\pi}f_2\,\mathrm{d}\theta \, =\, \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta \, =\, \frac{3a\pi}{(1-a^2)^{5/2}} \, =\,R$$ The hypothesis: $D_1$ = $D_2$ has been proved in a separate question Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta = \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta$ . The remaining hypotheses $D_1$ = $R$ and $D_2$ = $R$ have not been proved. So the question is:- Prove $D_1$ = $R$ or $D_2$ = $R$. Only one proof is required because the other can then be obtained from $D_1$ = $D_2$. For information WolframAlpha computes expressions for the indefinite integrals $I_1,I_2$ as follows:- $$I_1 \, =\, \int\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta \,=\, $$ $$constant1 + \frac {a\,\sqrt{a^2-1}\sin\theta\,[-(2a^3+a)\cos^2\theta+3(a^2+1)cos\theta+a(2a^2-5)]} {2(a^2-1)^{5/2}(a\cos\theta-1)^3} $$ $$-\frac {6a\,(a\cos\theta-1)^3\,\tanh^-1 \left( \frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}} \right) } {(2(a^2-1)^{5/2}\,(a\cos\theta-1)^3} $$ and $$I_2 \, =\, \int\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta \, =\, $$ $$constant2 - \frac {2a^2\sin\theta-sin\theta} {2(a^2-1)^2(a\cos\theta-1)} -\frac {\sin\theta} {2(a^2-1)(a\cos\theta-1)^2} $$ $$ -\frac {3a\tanh^-1\left(\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}\right)} {(a^2-1)^{5/2}} $$ Note that the final terms of each expression ( i.e. the terms involving $\tanh^{-1} $ and $\tan$ ) are equivalent to each other. Also, note that $$\int\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta= \frac{-\sin\theta}{(1-a\cos\theta)^3} +\int \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta. $$ Written with StackEdit. UPDATE 20141028 I have accepted TenaliRaman's answer. I don't yet understand all the steps but his helpful exposition gives me confidence that with time I can understand it because the methods cited (binomials, series) are ones I have learned (at high school). The answer of M.Strochyk also appears to give a good proof. But the residue method is too advanced for me to understand at present. UPDATE 20220713 I have now accepted Quanto's answer (because it is simple enough for me to understand). I have also added an answer based on Quanto's but with the intermediate steps written out.

With $$ \int_{0}^{2\pi} \frac{1}{1-a \cos \theta} d \theta=\frac{2\pi}{\sqrt{1-a^2}} $$ evaluate the following integrals successively \begin{align}\int_{0}^{2\pi} \frac{1}{(1-a \cos \theta)^2} d\theta =&\frac{d}{da} \int_{0}^{2\pi} \frac{a}{1-a \cos \theta} d \theta=\frac{2\pi}{(1-a^2)^{3/2}}\\ \int_{0}^{2\pi} \frac{\cos \theta}{(1-a \cos \theta)^3} d\theta =&\frac{d}{da} \int_{0}^{2\pi} \frac{1}{2(1-a \cos \theta)^2 }d \theta=\frac{3a\pi}{(1-a^2)^{5/2}}\\ \int_0^{2\pi} \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta =&\int_0^{2\pi} \sin\theta \ d\left(-\frac{1}{(1-a\cos \theta)^3}\right)\\ \overset{ibp}=&\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}d\theta=\frac{3a\pi}{(1-a^2)^{5/2}}\\ \end{align}

Why can't prime numbers satisfy the Pythagoras Theorem? That is, why can't a set of 3 prime numbers be a Pythagorean triplet? Suppose $a$, $b$ and $c$ are three prime numbers. How to prove that $a^2 + b^2 \neq c^2$?

The sum of two odd numbers are even, so one of the numbers must be $2$. If $a$ or $b$ are $2$ we have $a^2+4=c^2$ or $4=(c+a)(c-a)$ Since $c-a$ and $c+a$ have the same parity, this is impossible. If $c=2$ we have $a^2+b^2=4$ but since $a$ and $b$ are positive, both must be $1$, but $1^2+1^2=2$.

Show complex solutions exist Let A be a complex number and B a real number. Show that the equation $\,\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B = 0\,$ has a solution iff $\,\lvert A^2\rvert \geq 4B$. If this is so, show that the solution set is a circle or a single point. Well i am trying to do the first part first. So assuming the equation has a solution that would mean $z = x+iy$ satisfies the equation. I was going to let $A = s+it$ for a complex number, but it is not working out for me. Wrong step?

We have $$ 0=\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B = z\overline{z}+\frac{1}{2}(Az+\overline{Az})+\frac{1}{4}\lvert A\rvert^2-\frac{1}{4}\lvert A\rvert^2+B=\left\lvert z+\frac{1}{2}A\right\rvert^2-\frac{1}{4}\lvert A\rvert^2+B $$ If $4B>\lvert A\rvert^2>0$, then $\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B=\left\lvert z+\frac{1}{2}A\right\rvert^2-\frac{1}{4}\lvert A\rvert^2+B>0$, and hence no solutions. If $4B>\lvert A\rvert^2\le 0$, set $C=\frac{1}{2}\sqrt{\lvert A\rvert^2-4B}$, and our equation is equivalent to $$ \left\lvert z+\frac{1}{2}A\right\rvert^2=C^2, $$ and hence equivalent to $$ \left\lvert z+\frac{1}{2}A\right\rvert=C, $$ the set of solutions of which is the circle centered at $-A/2$ with radius $C$.

How to find the sum of the series $\sum_{k=2}^\infty \frac{1}{k^2-1}$? I have this problem : $$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1}$$ My solution $$S_n=\sum_{k=2}^\infty \frac{1}{k^2-1} = -\frac{1}{2}\sum_{k=1}^\infty \frac{1}{k+1} -\frac{1}{k-1} = -\frac{1}{2}[(\frac{1}{3}-1)+(\frac{1}{4}-\frac{1}{2})+(\frac{1}{5}-\frac{1}{3})+(\frac{1}{6}-\frac{1}{4})+...]$$ I think that the sum should be $\frac{1}{2}$ since the limit of : $$-\frac{1}{2}[-1+\frac{1}{k}+...+\frac{1}{n}] = \frac{1}{2}$$ But that wrong. Any ideas?

Let $S_n=\displaystyle\sum_{k=1}^n\frac{1}{k^2-1}=\sum_{k=1}^n\frac{1}{2}\left[\frac{1}{k-1}-\frac{1}{k+1}\right]$ $\displaystyle\hspace{.2 in} =\frac{1}{2}\left[\bigg(\frac{1}{1}-\frac{1}{3}\bigg)+\bigg(\frac{1}{2}-\frac{1}{4}\bigg)+\bigg(\frac{1}{3}-\frac{1}{5}\bigg)+\cdots+\bigg(\frac{1}{n-2}-\frac{1}{n}\bigg)+\bigg(\frac{1}{n-1}-\frac{1}{n+1}\bigg)\right]$ $\hspace{.2 in}\displaystyle=\frac{1}{2}\left[1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}\right]$, so $\displaystyle S=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{1}{2}\left[1+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}\right]=\frac{3}{4}$.

Find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$ How to find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$ Let $y^2-xy-2x^2 =0...(1)$ and $y^2=x-2...(2)$ In equation (1) coefficient of $x^2 =-2; y^2=1, 2xy =\frac{-1}{2}$ We know that a second degree equation where $ab-h^2 =0 $ represent a parabola, $ab-h^2>1$ represent an ellipse $ab-h^2 <0$ represent a hyperbola Here $ab-h^2 <0$ therefore equation (1) represents hyperbola and equation (2) is parabola. ( where a,b, h are coefficients of $x^2,y^2,xy$ respectively. Now how to get the minimum distance between the two curves please suggest .. thanks.

$$y^2= 2x^2+xy \iff (y-2x)(x+y)=0 \iff y = -x \text{ or } y = 2x$$ so that is a couple of lines. Hence you want to find the shortest distance from $y=-x$ or $y=2x$ to the parabola $y^2=x-2$. Let this be denoted by the line segment with end points $(a, -a)$ or $(a, 2a)$ and $(b^2+2, b)$. Then we need the min of $(a-b^2-2)^2+(a+b)^2$ or $(a-b^2-2)^2+(2a-b)^2$. Case 1: We have by Cauchy Schwarz $$((a-b^2-2)^2+(a+b)^2)((-1)^2+1^2) \ge (-a+b^2+2+a+b)^2= ((b+\tfrac12)^2+\tfrac74)^2 \ge \frac{7^2}{4^2}$$ so we get a minimum distance of $\frac{7}{4\sqrt2}$. Case 2: We have again by Cauchy Schwarz $$((a-b^2-2)^2+(2a-b)^2)((-2)^2+1^2) \ge (-2a+2b^2+4+2a-b)^2= (2(b-\tfrac14)^2+\tfrac{31}8)^2 \ge \frac{31^2}{8^2}$$ so we get a minimum distance of $\frac{31}{8\sqrt5}$. As the second case happens to be higher, the shortest distance is $\dfrac7{4\sqrt2}$.

An equation for the third powers of the roots of a given quadradic polynomial The roots of the equation $3x^2-4x+1=0$ are $\alpha$ and $\beta$. Find the equation with integer coefficients that has roots $\alpha^3$ and $\beta^3$. GIVEN SR: $\alpha + \beta = \frac43$ PR: $\alpha\beta = \frac13$ REQUIRED SR: $\alpha^3 + \beta^3 = (\alpha + \beta) (\alpha^2 + 2\alpha \beta + \beta^2)$ $\alpha^3 + 3\alpha^2 \beta + \alpha \beta^2 + \alpha^2 \beta + 3\alpha \beta^2 + \beta^3$ $\alpha^3 + 3\alpha^2 \beta + \alpha \beta^2 + \alpha^2 \beta + 3\alpha \beta^2 + \beta^3 = (\alpha + \beta)^3$ $\alpha^3+ \beta^3 = (\alpha + \beta)^3 - 3(\alpha \beta)^2 - (\alpha \beta)^2$ $\alpha^3 + \beta^3 = (\frac43)^2 - 3(\frac13)^2 - (\frac13)^2$ $\alpha^3 + \beta^3 = \frac{16}{9} - \frac13 - \frac19$ $\alpha^3 + \beta^3 = \frac43$ PR: $\alpha^3 \beta^3$ $(\alpha \beta)^3 = (\frac13)^3 = \frac{1}{27}$ EQUATION $x^2 - \frac43 x + \frac{1}{27} = 0$ $27x^2 - 36x + 1 = 0$ I am not sure if it's suppose to be: $(\alpha + \beta) (\alpha^2 + 2\alpha \beta + \beta^2)$ Or $(\alpha + \beta) (\alpha^2 + 3\alpha \beta + \beta^2)$ Also sorry about the formatting, if you don't mind fixing it for me. Not sure how to do it. Thank You.

Using the quadratic formula, we have $$\alpha, \beta = \frac{-(-4) \pm \sqrt{(-4)^2-4(3)(1)}}{6} \\ = \frac{4 \pm \sqrt{16-12}}{6} \\ = \frac{4 \pm 2}{6} \\ = 1, \frac{1}{3} $$ Hence a polynomial you could use with roots at $\alpha^3 = 1^3 = 1$ and $\beta^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$ would be $$f(x) = (x-1)\left(x-\frac{1}{27}\right) \\ = x^2 -\frac{28x}{27}+\frac{1}{27}$$ Or in general, for any constant $ C \neq 0$, then $C\cdot f(x)$ will have zeroes at $\alpha^3, \beta^3$

find common ratio of $\sum_{k=1}^\infty \frac{1}{k(k+1)}$ I have this problem, I need to find the sum. $$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\cdots+\frac{1}{k(k+1)}$$ The problem is that the ratio is not conclusive, Any idea how to find the ratio? Thanks!

$$\sum_{k=1}^\infty \frac{1}{k(k+1)} = \frac{1}{2}+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+\cdots+(\frac{1}{k}-\frac{1}{k+1})$$ Do you see how to do it now?

How would one solve this system $$12x^2=6z\\2y=-z\\6x-y=7$$ It's been many years since I've dealt with system equations, and now find myself in need to solve them. I am not quite sure what to do; I am interested in finding $x$ and $y$, so assuming that $z \neq 0$ (can I do this?), I isolate it and get $$2x^2 = z = -2y \\ -x^2 = y \\ 6x - (-x^2) = 7 \\ 6x + x^2 = 7 $$ which gives $$x^2 + 6x - 7 = 0 \\ (x+7)(x-1)=0$$ so $$ x = -7 \\or \\x = 1$$ which gives $$y = -(-7)^2 = -49 \\or \\ y = -(1)^2= -1 $$ But the answer is supposed to be $(1,-1)$ and $(7,35)$ ?? Where's my mistake?

Given is $f(x,y)=4x^3+y^2$ with the constraint $y=6x-7$. The easiest way to solve this, is plugging in $y$ into the function, which gives $$f(x)=4x^3+(6x-7)^2=4x^3+36x^2-84x+49$$ and we are asked to find the extrema. Differentiating: $$f'(x)=12x^2+72x-84=0$$ Dividing by $12$ gives $f'(x)=x^2+6x-7=(x+7)(x-1)=0$. From this follow the solutions $x=-7$ and $x=1$. Another method is with Lagrange multipliers. The gradient of the constraint is $\left(\begin{array}{c}6\\-1\end{array}\right)$. The gradient from the function $f(x,y)$ is $2\left(\begin{array}{c}6x^2\\y\end{array}\right)$. These two gradients must be parallel to have an extremum at a given point $(x,y)$. From this follows that $\frac{6x^2}{y}=\frac{6}{-1}$ and $x^2=-y$. Plugging this expression for $y$ into the constraint $6x-y=7$ gives $x^2+6=-7$, which is the same quadratic equation as before. To summarize; there is a local maximum at $f(-7,-49)=1029$ and a local minimum at $f(1,-1)=5$. To answer the question, your approach is correct, so maybe the answer model is incorrect.

how to integrate $\mathrm{arcsin}\left(x^{15}\right)$? Integral by parts: $$ I = x\sin^{-1}\left(x^{15}\right) - \int\frac{15x^{15}}{\sqrt{1-x^{30}}}dx $$ then what? The answer by wolfram gives an answer contains hypergeometric ${}_2F_1$ function,because it has no elementary answer. The question I want to know is, how can we find the integral of $$ \frac{15x^{15}}{\sqrt{1-x^{30}}} $$ in terms of hypergeometric function?

$\int\dfrac{15x^{15}}{\sqrt{1-x^{30}}}dx$ $=\int_0^x15t^{15}(1-t^{30})^{-\frac{1}{2}}~dt+C$ $=\int_0^{x^{30}}15t^\frac{1}{2}(1-t)^{-\frac{1}{2}}~d(t^\frac{1}{30})+C$ $=\dfrac{1}{2}\int_0^{x^{30}}t^{-\frac{7}{15}}(1-t)^{-\frac{1}{2}}~dt+C$ $=\dfrac{1}{2}\int_0^1(x^{30}t)^{-\frac{7}{15}}(1-x^{30}t)^{-\frac{1}{2}}~d(x^{30}t)+C$ $=\dfrac{x^{16}}{2}\int_0^1t^{-\frac{7}{15}}(1-x^{30}t)^{-\frac{1}{2}}~dt+C$ $=\dfrac{15x^{16}}{16}~_2F_1\left(\dfrac{1}{2},\dfrac{8}{15};\dfrac{23}{15};x^{30}\right)+C$

derivative of $y=\frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$ $y=\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$ The answer is $\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5} \left(\dfrac{2}{x}+\dfrac{1}{2(x+1)}-\dfrac{1}{x+2}-\dfrac{5}{x-3}\right)$ I know that the quotient rule is used but I don't know how to do this problem. Would you multiply together all the terms and then differentiate?

It's a long derivative. First use the quotient rule: $$\frac{\frac{d}{dx}\left(x^2\sqrt{x+1}\right)(x+2)(x-3)^5- x^2\sqrt{x+1}\frac{d}{dx}\left((x+2)(x-3)^5\right)}{((x+2)(x-3)^5)^2}.$$ Then do the derivatives.

Question about recurrence relation problem. solve the following recurrence relation, subject to given initial conditions. $a_{n+1} = 6a_n -9,$ $a_0 = 0,$ $a_1 = 3.$ Here is what I have done. $a_{n+1} - 6a_n +9 = 0$ $a_n = r^n$ $r^{n+1} - 6r^n + 9 = 0$ $r + 6 = 0$ $r = -6$ $a_n$ = $X(-6)^n$ $a_1 = 3 = X(-6)^1$ $X = -1/2$ Therefore $a_n = -1/2(-6)^n$ Am I on the right track or are there any mistakes I made somewhere? Edited: added the capture of the problem

$$\begin{align} a_{n+1}&=6a_n-9\\ a_{n+1}-\frac95&=6a_n-\frac{54}5\\ &=6\left(a_n-\frac95\right)\\ u_{n+1}&=6u_n\\ u_n&=6u_{n-1}=6^2u_{n-2}=\cdots=6^{n-1}u_1\\ &=6^{n-1}\left(a_1-\frac95\right)\\ &=6^{n-1}\left(3-\frac95\right)\\ a_n-\frac95&=\frac65(6^{n-1})\\ &=\frac{6^n}5\\ a_n&=\frac15\left(9+6^n\right)\qquad\blacksquare \end{align}$$ Check: $$\begin{align} 6a_n-9&=\frac65(9+6^n)-9\\ &=\frac95+6(6^n)\\ &=\frac15(9+6^{n+1})\\ &=a_{n+1} \end{align} $$ Hence formula is correct.

Solve $\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$ for $y$ I have the equation: $\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$, where $k$ and $c$ are arbitrary constants. How do I go about simplifying this and solving for $y$ in terms of $x$, excluding the obvious solution $y=-x$

Treat $x$ as a constant as well. We could use the standard quadratic formula, but since you noticed that $y = -x$ is a solution, let's try factoring instead. We obtain: \begin{align*} 0 &= \tfrac{1}{2}ky^{2} - \tfrac{1}{2}kx^{2} + cy + cx \\ &= \tfrac{1}{2}k(y^{2} - x^{2}) + c(y + x) \\ &= \tfrac{1}{2}k(y - x)(y + x) + c(y + x) \\ &= (\tfrac{1}{2}k(y - x) + c)(y + x) \\ \end{align*} Thus, the other solution can be obtained by setting the first factor equal to zero, yielding: \begin{align*} \tfrac{1}{2}k(y - x) + c &= 0 \\ \tfrac{1}{2}k(y - x) &= -c \\ y - x &= \tfrac{-2c}{k} \\ y &= x - \tfrac{2c}{k} \end{align*}

taking the inverse of power series I am working with solution to near regular singular points. I started with: $$y_1(x)=x^\frac{1}{2}\left[1-\frac{3}{4}x+\frac{9}{64}x^2-\frac{3}{256}x^3+\cdots\right] $$ Then I squared it: $$y_1^2(x) = x\left(1-\frac{3}{2}x+\frac{27}{32}x^2-\frac{15}{64}x^3+\cdots\right)$$ Why is the inverse: $$\frac{1}{x}\left[1+\frac{3}{2}x+\frac{45}{32}x^2+\frac{69}{64}x^3+\cdots\right] \text{ ?}$$ I cannot seem to see how this works out. Any pointers

If $$y=\sqrt{x} \left(1-\frac{3 x}{4}+\frac{9 x^2}{64}-\frac{3 x^3}{256}+\cdots\right)$$ then effectively $$y^2=x \left(1-\frac{3 x}{2}+\frac{27 x^2}{32}-\frac{15 x^3}{64} +\cdots\right)$$ Now you want to compute $\frac{1}{y^2 }$. You can write $$\frac{x}{y^2 }=\frac{1}{1-\frac{3 x}{2}+\frac{27 x^2}{32}-\frac{15 x^3}{64} +\cdots}$$ and perform the long division. Another way is to consider the rhs as $\frac{1}{1-z}$ and expand it as $$\frac{1}{1-z}=1+z+z^2+z^3+\cdots$$ and reuse $$z=\frac{3 x}{2}-\frac{27 x^2}{32}+\frac{15 x^3}{64}+\cdots$$ to affectively arrive, after substitution and simplifications, to $$\frac{x}{y^2 }=1+\frac{3 x}{2}+\frac{45 x^2}{32}+\frac{69 x^3}{64}+\cdots$$ and finally $$\frac{1}{y^2 }=\frac{1}{x}\Big(1+\frac{3}{2}x+\frac{45}{32}x^2+\frac{69}{64}x^3+\cdots\Big)$$ From a semantic point of view, I shall not call that operation the inverse of a power series (which is something different).

How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$ Question: let $x,y,z>0$ and such $xyz=1$, show that $$x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$$ My idea: use AM-GM inequality $$x^3+x^3+1\ge 3x^2$$ $$y^3+y^3+1\ge 3y^2$$ $$z^3+z^3+1\ge 3z^2$$ so $$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$ But this is not my inequality,so How prove it? I know this condition is very important.but how use this condition? and this inequality is stronger

Here is a possible solution: (although it is not the most elegant one) I will employ Mixing Variables technique here. Since the inequality is symmetric, WLOG let $x=\min(x,y,z)$. Therefore $t^2:=yz \ge 1$. Let $$f(x,y,z)=x^3+y^3+z^3-2(x^2+y^2+z^2)$$ I wish to show $$f(x,y,z)\ge f(x,\sqrt{yz},\sqrt{yz}) = f(\frac1{t^2},t,t) \ge -3$$ Let us put $p^2=x, q^2=y, r^2=z$. The first inequality in the above chain is equivalent to $$q^6+r^6-2q^3r^3 \ge 2(q^4+r^4-2q^2r^2)$$ $$\iff (q^3-r^3)^2 \ge 2(q^2-r^2)^2$$ $$\iff (q^2+qr+r^2)^2 \ge 2(q+r)^2$$ This is true since $$(q^2+qr+r^2)^2 \ge q^4+r^4+2q^2r^2+2qr(q^2+r^2) \ge 4q^2r^2+2(q^2+r^2) \ge 2(q+r)^2$$ Therefore it enough to prove $f(\frac1{t^2},t,t)\ge -3$ for $t>0$ which is equivalent to $$(t-1)^2((t^7-2t^5+t^3)+(t^7+t-2t^4)+(t^4-t^3+t^2)+t+1)\ge 0$$ Each term in the brackets of the large factor is greater than zero by AM-GM. The last part is little tedious to do by hand. But you always know that $(t-1)$ has to be factor (possibly with multiplicity $2$) of that thing. That helps in simplification.

Prove that: $\lim_{x\to 0}\frac{x}{\sin^2(x) + 1} = 0$ Prove $$\displaystyle \lim_{x\to 0} \frac{x}{\sin^2(x) + 1} = 0$$ The proof: Let $$|x| \le 1 \implies -1 \le x \le 1$$ $$\displaystyle \frac{|x|}{|\sin^2(x) + 1|} < \epsilon\text{ for }\displaystyle |x| < \delta$$ $$-1 \le x \le 1 \\\implies \sin(-1) \le \sin(x) \le \sin(1) \implies -\sin(1) \le \sin(x) \le \sin(1) \\\implies \sin^2(1) \le \sin^2(x) \le \sin^2(1) \implies |\sin^2(x) + 1| = |\sin^2(1) + 1| \implies \displaystyle |\frac{1}{\sin^2(x) + 1}| = |\frac{1}{\sin^2(1) + 1}|$$ (1) $$|x| < \delta_1$$ (2) $$\displaystyle |\frac{1}{\sin^2(x) + 1}| = |\frac{1}{\sin^2(1) + 1}|$$ (3) $$\displaystyle \frac{|x|}{|\sin^2(x) + 1|} < \frac{\delta_1}{|\sin^2(1) + 1|}$$ (4) $$\displaystyle \frac{|\delta_1|}{|\sin^2(1) + 1|} = \epsilon \implies \delta_1 = (|\sin^2(1) + 1|)(\epsilon) $$ Finally, $\delta = \min(1, (|\sin^2(1) + 1|)(\epsilon)) \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \blacksquare$ Thoughts? EDIT: The original proof was indeed terrible, here's a new approach. Let $|x| < 1 \implies -1 < x < 1$ $\sin^2(-1) + 1 < \sin^2(x) + 1 <\sin^2(1) + 2$ $\implies \displaystyle \frac{1}{\sin^2(-1) + 1} > \frac{1}{\sin^2(x) + 1} > \frac{1}{\sin^2(1) + 1}$ $\implies \displaystyle \frac{1}{\sin^2(-1) + 1} > \frac{1}{\sin^2(x) + 1} \implies \frac{1}{|\sin^2(-1) + 1|} > \frac{1}{|\sin^2(x) + 1|} \implies \frac{1}{|\sin^2(x) + 1|} < \frac{1} {|\sin^2(-1) + 1|} $ $(1) |x| < \delta_1$ $(2) \displaystyle \frac{1}{|\sin^2(x) + 1|} < \frac{1} {|\sin^2(-1) + 1|}$ $(3) \displaystyle \frac{|x|}{|\sin^2(x) + 1|} < \frac{\delta_1} {|\sin^2(-1) + 1|}$ Finally, $\epsilon(\sin^2(-1) + 1) = \delta_1$ Therefore, $\delta = \min(1,\epsilon \cdot (\sin^2(-1) + 1)) \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \blacksquare$

This is way too complicated, don't you think? Why not just say that $$ \left| \frac{x}{1+\sin^2 x} \right| \le |x| \le \epsilon $$ as soon as $|x|<\delta = \epsilon$?

Inequality proof by induction, what to do next in the step I have to prove that for $n = 1, 2...$ it holds: $2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}}$ Base: For $n = 1$ holds, because $2\sqrt{2}-2 < 1$ Step: assume holds for $n_0$. $2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n + 1}}$. But I do not know what to do next? How this can be proved?

For induction step, it's enough to prove $\frac{1}{\sqrt{n+1}}>2(\sqrt{n+2}-\sqrt{n+1})$. $$2(\sqrt{n+2}-\sqrt{n+1})=\frac{2}{\sqrt{n+2}+\sqrt{n+1}}<\frac{2}{2\sqrt{n+1}}=\frac{1}{\sqrt{n+1}}$$

Finding the positive integer numbers to get $\frac{\pi ^2}{9}$ As we know, there are many formulas of $\pi$ , one of them $$\frac{\pi ^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}...... $$ and this $$\frac{\pi ^2}{8}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}......$$ Now,find the positive integer numbers $(a_{0}, a_{1}, a_{2}....)$ to get $$\frac{\pi^2 }{9}=\frac{1}{a_{0}^2}+\frac{1}{a_{1}^2}+\frac{1}{a_{2}^2}....$$

Hint: $\frac 8 9 = \frac 1 2 + \frac 1 3 + \frac 1 {18}$ Now try using your second formula.

Trigonometric equation, missing some solutions I'm missing part of the answer, and I'm not quite sure why. The given answer doesn't even seem to hold... Solve for x: $$\tan 2x = 3 \tan x $$ First some simplifications: $$\tan 2x = 3 \tan x $$ $$\tan 2x - 3 \tan x = 0$$ $$\frac{\sin 2x}{\cos 2x} - \frac{3 \sin x}{\cos x} = 0$$ $$\frac{2 \sin x \cos^2x - 3 \sin x \cos 2x}{\cos 2x \cos x} = 0$$ $$\frac{\sin x(2 \cos^2x - 3 \cos 2x)}{\cos 2x \cos x} = 0$$ $$\frac{\sin x(2 \cos^2x - 3 (\cos^2 x - \sin^2 x))}{\cos 2x \cos x} = 0$$ $$\frac{\sin x(2 \cos^2x - 3\cos^2 x + \sin^2 x)}{\cos 2x \cos x} = 0$$ $$\frac{\sin x(\sin^2 x - \cos^2 x)}{\cos 2x \cos x} = 0$$ $$\frac{\sin x(\sin^2 x - \cos^2 x)}{(\sin^2 x - \cos^2 x) \cos x} = 0$$ $$\frac{\sin x}{\cos x} = 0$$ Looks much simpler. Now solving for x, since $\frac{\sin x}{\cos x} = 0 $ when $\sin x = 0$ and $\sin x = 0$ for every half rotation, the answer must be $k\pi$. Alas, according to my answer sheet, I'm missing two values: $\frac{\pi}{6} + k\pi$ and $\frac{5\pi}{6} + k\pi$. But since $\frac{\sin(\frac{\pi}{6})}{\cos(\frac{\pi}{6})} = \frac{\sqrt3}{3}$, I'm not sure where these answers come from. Furthermore, this is the kind of mistake I'm making all over these exercises, I'd like to avoid that, but how can I be sure I have ALL the answers needed?

Setting $\tan x=t$ we have $$\frac{2t}{1-t^2}=3t\iff2t=3t(1-t^2)\iff t(2-3+3t^2)=0$$ If $t=0,\tan x=0, x=n\pi$ where $n$ is any integer $2-3+3t^2=0\iff 3t^2=1\implies\cos2x=\dfrac{1-t^2}{1+t^2}=\dfrac12=\cos\dfrac\pi3$ $\implies2x=2m\pi\pm\dfrac\pi3$ where $m$ is any integer

Linear independence of $\sin^2(x)$ and $\cos^2(x)$ The Wronskian for $\sin^2x, \cos^2x$ is \begin{align} & \left| \begin{array}{cc} \sin^2 x & \cos^2 x \\ 2\sin x\cos x & -2\cos x\sin x \end{array} \right| \\[8pt] = {} & -2\sin^2x \cos x \sin x - 2 \cos^2 x \sin x \cos x, \end{align} with $x = \frac{π}{6},$ this is $=$ $$ -\sqrt{\frac{3}{2}}dx $$ Does this mean $\sin^2x, \cos^2x$ are linearly independent on the interval from $(-∞, ∞)$?

It suffices to show that the Wronskian is not zero for a single value of $x$. We have: $$W(x) = \begin{vmatrix} \sin^2x & \cos^2x \\ 2\sin x \cos x & -2 \sin x \cos x\end{vmatrix} = -2\sin^3x \cos x - 2\sin x \cos^3 x$$ $$W(x) = -2\sin x \cos x = -\sin(2x)$$ Then, $W(\pi/4) = -1 \neq 0$, so the functions are linearly independent.

How to integrate the dilogarithms? $\def\Li{\operatorname{Li}}$ How can you integrate $\Li_2$? I tried from $0 \to 1$ $\displaystyle \int_{0}^{1} \Li_2(z) \,dz = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)}$ $$\frac{An + B}{n^2} + \frac{D}{n+1} = \frac{1}{n^2(n+1)}$$ $$(An + B)(n+1) + D(n^2) = 1$$ Let $n = -1, \implies D = 1$ Let $n = 0, \implies B = 1$ Let $n = 1, \implies A = -1$ $$\frac{-n + 1}{n^2} + \frac{1}{n+1} = \frac{1}{n^2(n+1)}$$ $$= \sum_{n=1}^{\infty} \frac{-n + 1}{n^2} + \frac{1}{n+1} = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)} = \sum_{n=1}^{\infty} \frac{1}{n^2} - \frac{1}{n} + \frac{1}{n+1} $$ The $1/n$ is the problem, it is the harmonic series, which diverges.

Maybe you should look at your decomposition as $$\frac1{n^2 (n+1)} = \frac1{n^2} - \frac1{n (n+1)}$$ The sum over the second term is easy, given that the indefinite sum is telescoping, i.e., $$\sum_{n=1}^N \frac1{n (n+1)} = 1-\frac1{N+1}$$ We take the limit as $N \to \infty$ and then we may view this as the infinite sum. (Otherwise, as you say, there are convergence issues.) Thus the sum in question is $$\frac{\pi^2}{6}-1$$

Clarification: Proof of the quotient rule for sequences My Problem I am currently looking for a proof for the quotient rule for sequences: $a_n$ and $b_n$ are two sequences with the limes a,b. So: When $ a_n \rightarrow a$ and $ b_n \rightarrow b$ Then: $\frac {a_n}{b_n} \rightarrow \frac{a}{b}$ Awesome stuff, but how do i prove it? Solving attempts I found a proof in my textbook, but i have a hard time understanding it. It goes as follows. To prove the quotient rule, we have to acknowledge: $b \ne 0$.Then $b_n \rightarrow b > 0 $ Thus, for every $\epsilon : \frac{|b|}{2}$, There is a $n_0$, so that $|b_n|>|b|-\epsilon= \frac{|b|}{2}$ for every $n>n_0$. For those n's (This is the thing i don't get) You can say: $$| \frac {1}{b_n}-\frac {1}{b}|=|\frac{b-b_n}{b_nb}| \le \frac {2}{|b|^2}|b-b_n|$$ Because of the factor rule(Which is the equivalent of the quotient rule, just for factors), the right side of the inequality goes towards 0. Because of the rule 22.3 you can conclude that $\frac{1}{b_n} \rightarrow \frac{1}{b}$, and if you apply the product rule again, also that $\frac{a_n}{b_n} \rightarrow \frac {a}{b}$. Rule 22.3: $\alpha_n $is a null sequence. If the inequality $|a_n- a|\le \alpha_n$ is valid from a certain point with a limited number of exeptions, then $\alpha_n \rightarrow n$ I understand the beginning of the proof, but not how we get to the right side of the inequality. And i mean i understand why this thing on the right is a null sequence. But how does this prove our point. If somebody could clarify or point me to another proof of this rule, that is maybe easier to understand, i would be very grateful.

If, for eny $\epsilon > 0 \in \mathbb{R}$, there exists an $n \in \mathbb{Z}$ such that $\forall i > n, |a_i - c| < \epsilon $, then $c$ is defined as the limit of the sequence of $a_i$. Assuming the limit of sequence $a_i$ is $a > 0$, and the limit of $b_i$ is $b > 0$, we want to prove the limit of $\dfrac{a_i}{b_i} = \dfrac{a}{b}$. If we apply the limit definitions to $a_i$ and $b_i$ at $\delta < \frac{b}{2}$, how different can $\dfrac{a_i}{b_i}$ be from $\dfrac{a}{b}$? We apply the limit definition and get a $n_a$ such that $\forall i \ge n_a, |a_i - a| < \delta$, and we get a $n_b$ such that $\forall i \ge n_b, |b_i - b| < \delta$. Let $n' = \max(n_a, n_b)$. We know that for all $i > n', $ both $|a_i - a| < \delta$ and $|b_i - b| < \delta$. We can also say this as $a - \delta \le a_i \le a + \delta$ and $b - \delta \le b_i \le b + \delta$. So $\dfrac{a - \delta}{b + \delta} \le \dfrac{a_i}{b_i} \le \dfrac{a+\delta}{b-\delta}$. How far do these minimum and maximum bounds stray from $\dfrac{a}{b}$? $$\begin{align} \frac{a}{b} - \frac{a-\delta}{b+\delta} & = \frac{a(b+\delta) - (a-\delta)b}{b(b+\delta)}\\ & = \frac{ab + a\delta - ab + \delta b}{b(b+\delta)} \\ & = \frac{(a+b)\delta}{b(b+\delta)}\\ & < \frac{a+b}{b^2}\delta \end{align}$$ $$\begin{align} \frac{a + \delta}{b - \delta} - \frac{a}{b} & = \frac{(a + \delta)b - a(b - \delta)}{b(b - \delta)}\\ & = \frac{ab + \delta b - ab + a\delta}{b(b - \delta)}\\ & = \frac{a + b}{b(b - \delta)}\delta\\ & < \frac{a + b}{b\frac{b}{2}}\delta && \text{remember $\delta < \frac{b}{2}$}\\ \frac{a + \delta}{b - \delta} - \frac{a}{b} & < \frac{a + b}{b^2}2\delta \end{align} $$ So now we can prove the limit of $\dfrac{a_i}{b_i}$. When we are given an $\epsilon > 0$, we can use the limits of $a_i$ and $b_i$ for the convenient $\delta = \min(\frac{b}{2}, \frac{b^2}{2(a+b)}\epsilon)$ to get $n_a$ and $n_b$. Then we determine $n' = \max(n_a,n_b)$. Then (omitting the trivial $\frac{b}{2} < \frac{b^2}{2(a+b)}\epsilon $ case) we know that for all $i > n'$, $$\begin{align} |\frac{a_i}{b_i} - \frac{a}{b}| & \le \frac{2(a+b)}{b^2}\delta \\ & \le \frac{2(a+b)}{b^2}\frac{b^2}{2(a+b)}\epsilon \\ & \le \epsilon \end{align}$$ And so we fulfill the definition of the limit.

Solve to find $y(x)$ of the $\frac{1}{\sum_{n=0}^{\infty }y^n}-\sum_{n=0}^{\infty }x^n=0$ Solve the equation to find the $y$ as a function to respect $x$ without $n$ $$\frac{1}{\sum_{n=0}^{\infty }y^n}-\sum_{n=0}^{\infty }x^n=0$$

Assuming $|x| < 1$ and $|y| < 1$, the two geometric series simplify to: $$\frac{1}{\frac{1}{1 - y}} = \frac{1}{1 - x}.$$ Consequently: $$(1 - x)(1 - y) = 1.$$ Solving for $y$ in terms of $x$, you get: $$1 - y = \frac{1}{1 - x}$$ which implies that $$y(x) = 1 - \frac{1}{1 - x} = \frac{x}{x - 1}.$$ Note that $|y(x)| < 1$, so that $|x/(x - 1)| < 1$. This implies that: $$|x| < |x - 1|,$$ which further implies that $$x^2 < (x - 1)^2 = x^2 - 2x + 1.$$ Consequently, $$x < \frac{1}{2}.$$

Sum of 6 cards being multiple of 6 I pick 6 cards from a set of 13 (ace-king). If ace = 1 and jack,queen,king = 10 what is the probability of the sum of the cards being a multiple of 6? Tried so far: I split the numbers into sets with values: 6n, 6n+1, 6n+2, 6n+3 like so: {6}{1,7}{2,8}{3,9}{4,10,j,q,k}{5} and then grouped the combinations that added to a multiple of 6: (5c4)(1c1)(2c1) + (2c2)(2c2)(2c2) + (5c4)(2c2) + (5c2)(1c1)(1c1)(2c1)(2c1) / (13c6) = 10/1716 I am almost certain I am missing combinations but am having trouble finding out which.

Your combinations are $4\cdot 4+1\cdot 0+1\cdot 2, 2\cdot 1+2\cdot2+2\cdot3, 4 \cdot 4+2\cdot 1,+$ something that doesn't make sense because there are three places you choose from $1$ and only $0,5$ qualify. The left number is the number of cards of that value $\pmod 6$ . You could also have $3\cdot 4+1 \cdot 5+1 \cdot 1+1 \cdot 0, 3 \cdot 4+ 2 \cdot 2 + 1 \cdot 0$ and others

Inequality of sides of triangle If a,b,c are the sides of a triangle ABC then prove $$ a/(b+c) + b/(a+c) + c/(a+b) < 2 $$I tried to solve this by triangle inequality but i was not able to get to the solution.

Hint: let $x=a+b-c$, $y=a+c-b$, $z=b+c-a$. Clearly $$a=\frac{x+y}2,\ b=\frac{x+z}{2},\ c=\frac{y+z}{2},\ a+b=\frac{2x+y+z}{2},\cdots$$ hence the original inequality reads $$\frac{x+y}{x+y+2z}+ \frac{x+z}{x+2y+z}+\frac{y+z}{2x+y+z}<2.$$ Note that $$\frac{x+y}{x+y+2z}+ \frac{x+z}{x+2y+z}+\frac{y+z}{2x+y+z}<\frac{x+y}{x+y+z}+ \frac{x+z}{x+y+z}+\frac{y+z}{x+y+z}\cdots$$

How does does exponent property work on $\left(xe^{\frac{1}{x}}-x\right)$ How does $\left(xe^{\frac{1}{x}}-x\right)$ become $\frac{\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x}}$

With more intermediate steps : $$\left(xe^{\frac{1}{x}}-x\right)=$$ $$=x\left(e^{\frac{1}{x}}-1\right)$$ Let $x=\frac{1}{t}$ Replace $x$ by $\frac{1}{t}$ $$=\frac{1}{t}\left(e^{\frac{1}{x}}-1\right)$$ $$=\frac{\left(e^{\frac{1}{x}}-1\right)}{t}$$ $t=\frac{1}{x}$ Replace $t$ by $\frac{1}{x}$ $$=\frac{\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x}}$$ Another nice method : $$\left(xe^{\frac{1}{x}}-x\right)=$$ $$=\frac{x\left(e^{\frac{1}{x}}-1\right)}{1}$$ Mulptiply the numerator and the denominator by the same term $\frac{1}{x}$ $$=\frac{\frac{1}{x}x\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x}}$$ Simplify $\frac{1}{x}x=1$ $$=\frac{\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x}}$$

Evaluation of $\int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$ Evaluation of Integral $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$ $\bf{My\; Try::}$ First we will convert $x^4+x^3+x^2+x+1$ into closed form, which is $\displaystyle \left(\frac{x^5-1}{x-1}\right)$ So Integral is $\displaystyle \int_{0}^{\infty}\frac{\sqrt{x-1}}{\sqrt{x^5-1}}dx$ now i did not understand how can i solve it Help me Thanks

The answer in terms of elliptic integrals turns out to be very simple – but it requires a very sneaky trick. As pointed out in Jack's answer, the integrals over $[0,1]$ and $[1,\infty]$ are the same, so $$I=\int_0^\infty\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx=2\int_0^1\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx$$ The sneaky trick is to substitute $u=x+\frac1x$, which gives $$I=2\int_2^\infty\frac1{\sqrt{(u^2-4)(u^2+u-1)}}\,du$$ This changes all the denominator quartic's roots from complex to real, which will make the sequel easier. Furthermore, the roots are very simple: $-2,-\varphi,1/\varphi,2$ where $\varphi$ is the golden ratio. We can now apply Byrd and Friedman 258.00 directly, which says that if $d<c<b<a<y$, $$\int_a^y\frac1{\sqrt{(t-a)(t-b)(t-c)(t-d)}}\,dt=gF(\psi,m)\\ \text{where }g=\frac2{\sqrt{(a-c)(b-d)}},\psi=\sin^{-1}\sqrt{\frac{(b-d)(y-a)}{(a-d)(y-b)}},m=\frac{(b-c)(a-d)}{(a-c)(b-d)}$$ ($m=k^2$ is the parameter, as used by Mathematica and mpmath.) We have $d=-2,c=-\varphi,b=1/\varphi,a=2$, but since $y=\infty$ we have to take the limit of the expression for $\psi$ as $y\to\infty$, which therefore becomes $$\psi=\sin^{-1}\sqrt{\frac{b-d}{a-d}}=\sin^{-1}\sqrt{\frac{1/\varphi+2}4}=\sin^{-1}\frac\varphi2=\frac{3\pi}{10}$$ The other key numbers follow: $$g=\frac2{\sqrt{(2+\varphi)(1/\varphi+2)}}=\frac2{\sqrt{4\varphi+3}}$$ $$m=\frac{4(1/\varphi+\varphi)}{(2+\varphi)(1/\varphi+2)}=\frac{4(2\varphi-1)}{4\varphi+3}=8\varphi-12$$ Finally we have the answer in terms of elliptic integrals: $$\boxed{I=\frac4{\sqrt{4\varphi+3}}F\left(\frac{3\pi}{10},8\varphi-12\right)}$$ #!/usr/bin/env python3 from mpmath import * mp.dps = 150 print(quad(lambda t: 1/sqrt(polyval((1,1,1,1,1),t)), [0, inf])) g = 4/sqrt(4*phi+3) m = 8*phi-12 print(g*ellipf(3*pi/10, m)) P.S. A certain Dr. Sonnhard Graubner gave an answer here, but it's incorrect.

If $f(1)=1\;,f(2)=3\;,f(3)=5\;,f(4)=7\;,f(5)=9$ and $f'(2)=2,$ Then sum of all digits of $f(6)$ $(1):$ If $P(x)$ is a polynomial of Degree $4$ such that $P(-1) = P(1) = 5$ and $P(-2)=P(0)=P(2)=2\;,$Then Max. value of $P(x).$ $(2):$ If $f(x)$ is a polynomial of degree $6$ with leading Coefficient $2009.$ Suppose further that $f(1)=1\;,f(2)=3\;,f(3)=5\;,f(4)=7\;,f(5)=9$ and $f'(2)=2,$ Then sum of all digits of $f(6)$ is $\bf{My\; Try\; For\; (1):}$ Given $x=-2\;,x=0\;,x=+2$ are the roots of $P(x)=0.$ So $(x+2)\;,(x-0)\;,(x-2)$ are factors of $P(x)=0$. So we can write $P(x)$ as $P(x) = A\cdot x\cdot (x-2)\cdot (x+2)(x-r)\;,$ So we have calculate value of $A$ and $r$ Now put $x=-1\;,$ we get $P(-1)=-3A\cdot (1+r)\Rightarrow -3A\cdot (1+r)=5............................(1)$ Now put $x=1\;,$ we get $P(1)=-3A\cdot (1-r)\Rightarrow -3A\cdot (1-r)=5..................................(2)$ So from $(1)$ and $(2)\;,$ We get $r=0$ and $\displaystyle A=-\frac{5}{3}.$ So Polynomial $\boxed{\boxed{\displaystyle P(x)=-\frac{5}{3}\cdot x^2\cdot (x^2-4)}}$ $\bf{My\; Try\; For \; (2):}$Let $f(x)=2x-1\;\forall\; x=1\;,2\;,3\;,4\;,5.$ So we can say that $(x-1)\;,(x-2)\;,(x-3)\;,(x-4)\;,(x-5)$ are the roots of $f(x)-2x+1=0$ So $f(x)-2x+1=2009\cdot \underbrace{(x-1)\cdot(x-2)\cdot (x-3)\cdot (x-4)\cdot (x-5)}\cdot\underbrace{(x-r)}$ Now How can i solve after that Help me and plz explain me, is my $(1)$ Try is right or not Thanks

For part 2, here's one way to simplify the differentiation. We have $$f(x)-2x+1=2009\cdot (x-1)\cdot(x-2)\cdot (x-3)\cdot (x-4)\cdot (x-5)\cdot(x-r)$$ Substituting $x + 3 \to x$, $$f(x + 3) - 2(x+3)+1 = 2009(x+2)(x+1)(x)(x-1)(x-2)(x+3-r)$$ $$f(x + 3) -2x - 5 = 2009x(x^2-1)(x^2-4)(x + 3 -r)$$ $$f(x + 3) - 2x - 5 = 2009(x^5 - 5x^3 + 4x)(x + 3 -r)$$ Applying the chain rule to LHS and product rule to RHS, $$f'(x+3) - 2 = 2009(5x^4 - 15x^2 + 4)(x + 3 - r) + 2009(x^5 - 5x^3 + 4x)$$ Substituting $x = -1$, $$f'(2) - 2 = 2009(5 - 15 + 4)(2-r) + 2009(1-5+4)$$ $$0 = 2009\cdot(-6)\cdot(2-r)$$ to give $r=2$. Hence, $$f(6) - 12 + 1 = 2009\cdot5!\cdot4$$ $$f(6) = 964331$$

Evaluating $\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx$ What starting point would you recommend me for the one below? $$\int_0^1 x \tan(\pi x) \log(\sin(\pi x))dx $$ EDIT Thanks to Felix Marin, we know the integral evaluates to $$\displaystyle{\large{\ln^{2}\left(\, 2\,\right) \over 2\pi}}$$

$\def\Li{{\rm{Li}_2\,}}$Denote the considered integral as $I$ and set $y=\pi x$, we have \begin{equation} I=\frac{1}{\pi^2}\int_0^\pi \frac{y\sin y}{\cos y}\,\ln(\sin y)\,dy \end{equation} Perform integration by parts by taking $u=y$, we have \begin{align} I&=-\left.\frac{y}{2\pi^2}\int\frac{\ln\left(1-\cos^2y\right)}{\cos y}\,d(\cos y)\right|_0^\pi+\frac{1}{2\pi^2}\int_0^\pi\int\frac{\ln\left(1-\cos^2y\right)}{\cos y}\,d(\cos y)\,dy\\ &=\left.\frac{y}{4\pi^2}\Li\left(\cos^2y\right)\,\right|_0^\pi-\frac{1}{4\pi^2}\int_0^\pi\Li\left(\cos^2y\right)\,dy\\ &=\frac{\Li\left(1\right)}{4\pi}-\frac{1}{4\pi^2}\int_0^{\pi}\sum_{k=1}^\infty\frac{\cos^{2k}y}{k^2}\,dy\quad\Rightarrow\quad\color{red}{\mbox{use series representation of dilogarithm}}\\ &=\frac{\pi}{24}-\frac{1}{4\pi^2}\sum_{k=1}^\infty\int_0^{\pi}\frac{\cos^{2k}y}{k^2}\,dy\quad\Rightarrow\quad\color{red}{\mbox{justified by Fubini-Tonelli theorem}}\\ &=\frac{\pi}{24}-\frac{1}{2\pi^2}\sum_{k=1}^\infty\frac{1}{k^2}\int_0^{\pi/2}\cos^{2k}y\,\,dy\quad\Rightarrow\quad\color{red}{\mbox{by symmetry argument}}\\ &=\frac{\pi}{24}-\frac{1}{4\pi^2}\sum_{k=1}^\infty\frac{\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{k^2\,\Gamma\left(k+1\right)}\quad\Rightarrow\quad\color{red}{\mbox{Wallis' integrals}}\\ &=\frac{\pi}{24}-\frac{1}{4\pi}\sum_{k=1}^\infty\frac{(2k)!}{4^k\,k^2\,(k!)^2}\tag1 \end{align} $\def\arctanh{{\rm{\,arctanh}\,}}$Here is the tedious part (and also the hardest part). I use Mathematica to help me out to find generating function of the following series. Let us start with \begin{equation} \sum_{k=0}^\infty\frac{\Gamma\left(k+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)x^k}{\Gamma\left(k+1\right)}=\pi\sum_{k=0}^\infty\frac{(2k)!\,\,x^k}{4^k\,(k!)^2}=\frac{\pi}{\sqrt{1-x}} \end{equation} Divide by $x$ and then integrate it, we have \begin{equation} \sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k\,(k!)^2}=\int\left[\frac{1}{x\,\sqrt{1-x}}-\frac{1}{x}\right]\,dx=-2\arctanh\left(\sqrt{1-x}\,\right)-\ln x+C_1 \end{equation} Taking the limit as $x\to0$, we obtain \begin{equation} C_1=\lim_{x\to0}\left(2\arctanh\left(\sqrt{1-x}\,\right)+\ln x\right)=\ln4 \end{equation} Hence \begin{equation} \sum_{k=1}^\infty\frac{(2k)!}{4^k\,k\,(k!)^2}x^k=-2\arctanh\left(\sqrt{1-x}\,\right)-\ln x+\ln4 \end{equation} Repeat the process once more, we obtain \begin{align} \sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k^2\,(k!)^2}=&\,-2\int\frac{\arctanh\left(\sqrt{1-x}\,\right)}{x}\,dx-\int\frac{\ln x}{x}\,dx+\ln4\int\frac{dx}{x}\\ =&\,\,2\arctanh\left(\sqrt{1-x}\,\right)\left[\arctanh\left(\sqrt{1-x}\,\right)-2\ln\left(\frac{\sqrt{1-x}+1}{2}\right)\right]\\&\,-2\,\Li\left(\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right)-\frac{\ln^2x}{2}+\ln4\ln x+C_2\\ \end{align} Taking the limit as $x\to0$, we obtain \begin{equation} C_2=-2\ln^22 \end{equation} Hence \begin{align} \sum_{k=1}^\infty\frac{(2k)!\,\,x^k}{4^k\,k^2\,(k!)^2} =&\,\,2\arctanh\left(\sqrt{1-x}\,\right)\left[\arctanh\left(\sqrt{1-x}\,\right)-2\ln\left(\frac{\sqrt{1-x}+1}{2}\right)\right]\\&\,-2\,\Li\left(\frac{\sqrt{1-x}-1}{\sqrt{1-x}+1}\right)-\frac{\ln^2x}{2}+\ln4\ln x-2\ln^22\tag2 \end{align} Thus, by putting $x=1$ to $(2)$ then $(1)$ becomes \begin{equation} I=\frac{\pi}{24}-\frac{1}{4\pi}\left(\frac{\pi^2}{6}-2\ln^2 2\right)=\frac{\ln^22}{2\pi} \end{equation} and we are done.

How do I prove that a matrix is a rotation-matrix? I have to prove that this matrix is a rotation-matrix $$\begin{pmatrix} \frac12 & 0 & \frac{\sqrt{3}}{2} \\ 0 & 1 & 0 \\ \frac{\sqrt{3}}{2} & 0 & \frac12 \end{pmatrix}$$ How do I do this? My idea is to multiplicate it with $\begin{pmatrix} x \\ y \\ z\end{pmatrix}$ and show that one component will remain unchanged . Is this enough? Do non-rotational transformations exist, which leave one component unchanged ?

I think there is a minus sign missing. As it is, the determinant is not $1$. After fixing, this specific case is easy. $$\begin{pmatrix} \frac12 & 0 & -\frac{\sqrt{3}}{2} \\ 0 & 1 & 0 \\ \frac{\sqrt{3}}{2} & 0 & \frac12 \end{pmatrix} = \begin{pmatrix} \cos \frac{\pi}{3} & 0 & -\sin \frac{\pi}{3} \\ 0 & 1 & 0 \\ \sin \frac{\pi}{3} & 0 & \cos \frac{\pi}{3} \end{pmatrix}$$ It is a rotation of $\pi/3$ around the $y$-axis.

Prove $ \lim_{x\to -\infty } \frac{x+8}{x+3} = 1 $ using only the definition of a limit I need to prove this limit: $$ \lim_{x\to -\infty } \frac{x+8}{x+3} = 1 $$ I started with: $|\frac{x+8}{x+3}-1|< ϵ $ $|\frac{x+8-x-3}{x+3}|=|\frac{5}{x+3}|=\frac{5}{|x+3|} <ϵ$ $\frac{5}{\epsilon} < |x+3|$ How do I proceed from here? how the hell can I extract the expression inside the absolute value? Am I in the wrong direction? Thanks a lot!

$$\begin{align*} \frac5\epsilon &< \left|x+3\right|\\ \frac5\epsilon &< -\left(x+3\right)\\ x &< -\frac5\epsilon-3 \end{align*}$$ Hence, given $\epsilon>0$, we can take $M= -\frac5\epsilon-3$ such that $x<M$ implies $$\left|\frac{x+8}{x+3}-1\right|<\epsilon$$

How does $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)}$ simplify? $\sum_{n = 1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + ...$ $= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... = 1$ My professor wrote this the other day. But I'm wondering...how does the series become $= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ...$?

It is because $\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}$

What's wrong with this argument? (Limits) In order to compute $\displaystyle \lim_{x \to \infty} \sqrt{9x^2 + x} - 3x$ we can multiply by the conjugate and eventually arrive at a limit value $1/6$. But what about the line of reasoning below, what is wrong with the argument and why? I can't think of a simple explanation, I had one involving the limit definition but I believe there should be a less complicated one. Here's the argument: Clearly for large $x$ we can say $\sqrt{9x^2 + x} \approx \sqrt{9x^2} = 3x$. Hence $$ \lim_{x \to \infty} \sqrt{9x^2 + x} - 3x = \lim_{x \to \infty} 3x - 3x = 0 \ . $$ So the limit ought to be zero, easy! What goes wrong and why?

The symbol $\approx $ is often misused. It can mean numerical approximation or it can mean asymptotically equal. For example it is not true that (n+1) 2 $\approx $ n 2 in both senses. This you can see from the difference ${(n + 1)^2} - {n^2} = 1 + 2n$ . You can observe that the difference is not small. Indeed it tends to infinity as n tends to infinity. In $\mathop {\lim }\limits_{x \to \infty } \sqrt {9{x^2} + x} - 3x$ you are taking the limit as x tends to infinity. You are looking for the difference of $\sqrt {9{x^2} + x} $ and $ 3x $ as x gets larger and larger. The limit exists if this difference gets closer and closer to a certain value. In example in variable $n$ above, the difference tends to infinity and so no limit. If you now look at the difference $\sqrt {9{x^2} + x} - 3x = \frac{{\left( {\sqrt {9{x^2} + x} - 3x} \right)}}{{\sqrt {9{x^2} + x} + 3x}}\left( {\sqrt {9{x^2} + x} + 3x} \right) = \frac{x}{{\sqrt {9{x^2} + x} + 3x}}$, you see that $\sqrt {9{x^2} + x} = 3x + \frac{x}{{\sqrt {9{x^2} + x} + 3x}}$ and $\frac{x}{{\sqrt {9{x^2} + x} + 3x}} \ge \frac{x}{{\sqrt {16{x^2}} + 3x}} = \frac{1}{7}$ no matter how large x gets and $1/7$ is definitely not a small value. So you cannot say $\sqrt {9{x^2} + x} \approx 3x$ . It is precisely this difference that gives you the limiting value. This question is to let you know it is erroneous to think of estimating large value as approximation. It is possible two functions need not have the same value as x gets larger and larger and it is possible that the two functions for large values of x are always a fixed quantity apart, or gets further and further apart or their difference may oscillate about a fixed quantity. For example x 2 and (x+1)2, x2 and x2+sin(x), x2 and x2+6. More precislely for this question, you want to show that no matter how small $\epsilon $ is you can find number $K$ so that for all $x > K$ , the difference $\left| {\sqrt {9{x^2} + x} - 3x - \frac{1}{6}} \right| = \left| {\frac{x}{{\sqrt {9{x^2} + x} + 3x}} - \frac{1}{6}} \right| < \varepsilon $ This is equivalent to $\mathop {\lim }\limits_{x \to \infty } \frac{x}{{\sqrt {9{x^2} + x} + 3x}} = \frac{1}{6}$ . This I think you can prove. Now you can say $\sqrt {9{x^2} + x} \approx 3x + \frac{1}{6}$ or more precisely $\sqrt {9{x^2} + x} $ is asymptotically equal to $3x + \frac{1}{6}$.

How can I evaluate $\lim_{n \to\infty}\left(1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)\right)/\left(1^2+2^2+3^2+\dots+n^2\right)^2$? How can I evaluate this limit? Give me a hint, please. $$\lim_{n \to\infty}\frac{1\cdot2\cdot3+2\cdot3\cdot4+\dots+n(n+1)(n+2)}{\left(1^2+2^2+3^2+\dots+n^2\right)^2}$$

Hint: $$1\cdot 2\cdot 3 + \dots +n(n+1)(n+2) \le n(n+2)^3 \le 8n^4$$ $$(1^2 + 2^2 + \dots + n^2)^2 = \left(\frac{n(n+1)(2n+1)}{6}\right)^2\ge \frac{1}{9}n^6$$ Hence $$\frac{1\cdot 2\cdot 3 + \dots +n(n+1)(n+2)}{(1^2 + 2^2 + \dots + n^2)^2} \le \frac{8n^4}{1/9\cdot n^6}=\frac{72}{n^2}\to 0$$

Double Integral $\int\limits_0^a\int\limits_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$ How to solve this integral? $$\int_0^a\!\!\!\int_0^a\frac{dx\,dy}{(x^2+y^2+a^2)^\frac32}$$ my attempt $$ \int_0^a\!\!\!\int_0^a\frac{dx \, dy}{(x^2+y^2+a^2)^\frac{3}{2}}= \int_0^a\!\!\!\int_0^a\frac{dx}{(x^2+\rho^2)^\frac{3}{2}}dy\\ \rho^2=y^2+a^2\\ x=\rho\tan\theta\\ dx=\rho\sec^2\theta \, d\theta\\ x^2+\rho^2=\rho^2\sec^2\theta\\ \int_0^a\!\!\!\int_0^{\arctan\frac{a}{\rho}}\frac{\rho\sec\theta}{\rho^3\sec^3\theta}d\theta \, dy= \int_0^a\!\!\!\frac{1}{\rho^2}\!\!\!\int_0^{\arctan\frac{a}{\rho}}\cos\theta \, d\theta \, dy=\\ \int_0^a\frac{1}{\rho^2}\sin\theta\bigg|_0^{\arctan\frac{a}{\rho}} d\theta \, dy= \int_0^a\frac{1}{\rho^2}\frac{x}{\sqrt{x^2+\rho^2}}\bigg|_0^ady=\\ \int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy$$ Update: $$\int_0^a\frac{a}{(y^2+a^2)\sqrt{y^2+2a^2}}dy=\frac{\pi}{6a}$$

Both the function that you are integrating as the region over which you are integrating it get unchanged if you exchange $x$ with $y$. Therefore, your integral is equal to$$2\int_0^a\int_0^x\frac1{(a^2+x^2+y^2)^{3/2}}\,\mathrm dy\,\mathrm dx.$$You can compute this integral using polar coordinates: $\theta$ can take values in $\left[0,\frac\pi4\right]$ and, for each $\theta$, $r$ can take values in $\left[0,\frac a{\cos\theta}\right]$. And\begin{align}\int_0^{\pi/4}\int_0^{a/\cos(\theta)}\frac r{(a^2+r^2)^{3/2}}\,\mathrm dr\,\mathrm d\theta&=\int_0^{\pi/4}\frac{1-\frac1{\sqrt{\sec ^2(\theta)+1}}}a\,\mathrm d\theta\\&=\frac1a\left(\frac\pi4-\int_0^{\pi/4}\frac1{\sqrt{\sec ^2(\theta)+1}}\,\mathrm d\theta\right)\\&=\frac\pi{12a}.\end{align}Note that the final equality is equivalent to$$\int_0^{\pi/4}\frac1{\sqrt{\sec^2(\theta)+1}}\,\mathrm d\theta=\frac\pi6.$$This can be justified as follows: you do $\theta=\arccos\left(\sqrt x\right)$ and $\mathrm d\theta=-\frac1{2\sqrt{x-x^2}}\,\mathrm dx$. Doing this, you will get\begin{align}\int_0^{\pi/4}\frac1{\sqrt{\sec^2(\theta)+1}}\,\mathrm d\theta&=\int_1^{1/2}-\frac1{2\sqrt{1-x^2}}\\&=\frac12\int_{1/2}^1\frac1{\sqrt{1-x^2}}\,\mathrm dx\\&=\frac12\left(\arcsin\left(1\right)-\arcsin\left(\frac12\right)\right)\\&=\frac12\left(\frac\pi2-\frac\pi6\right)\\&=\frac\pi6.\end{align}

Integral on sphere and ellipsoid Let $a,b,c \in \mathbb{R},$ $\mathbf{A}=\left[\begin{array}{*{20}{c}} \mathbf{a}&{0}&{0}\\ {0}&\mathbf{b}&{0}\\ {0}&{0}&\mathbf{c} \end{array}\right] , ~~\det A >1$ Let $~D = \{(x_1,x_2,x_3): x_1^2 + x_2^2 +x_3^2 \leq 1 \}~$ and $~E = \left\{(x_1,x_2,x_3): \frac{x_1^2}{a^2} + \frac{x_2^2}{b^2} + \frac{x_3^2}{c^2} \leq 1 \right\}~.$ Then for a compactly supported continuous function $f$ on $\mathbb{R}^3$, could anyone tell me which of the following are correct? 1.$\int_D f(Ax)dx = \int_E f(x)dx $ 2.$\int_D f(Ax)dx = \frac{1}{abc} \int_D f(x)dx $ 3.$\int_D f(Ax)dx = \frac{1}{abc} \int_E f(x)dx $ 4.$\int_{\mathbb{R}^3} f(Ax)dx = \frac{1}{abc} \int_{\mathbb{R}^3} f(x)dx $

A function $~f~$ is said to be compactly supported if it is zero outside a compact set. Let $~x=(x_1,x_2,x_3)\in\mathbb R^3~,$ be any arbitrary vector. $$\therefore~~Ax=\left[\begin{array}{*{20}{c}} \mathbf{a}&{0}&{0}\\ {0}&\mathbf{b}&{0}\\ {0}&{0}&\mathbf{c} \end{array}\right]\left[\begin{array}{*{20}{c}} {x_1}\\{x_2}\\{x_3} \end{array}\right]=(ax_1,bx_2,cx_3)$$ Now $$\int_D f(Ax)dx =\iiint_{x_1^2+x_2^2+x_3^2\le1} f(ax_1,bx_2,cx_3)~dx_1~dx_2~dx_3\tag1$$ Putting $~~ax_1=y_1,~~bx_2=y_2,~~cx_3=y_3 \implies x_1=\dfrac{y_1}{a},~~x_2=\dfrac{y_2}{b},~~x_3=\dfrac{y_3}{c}$ $$\implies dx_1=\dfrac{dy_1}{a},~~dx_2=\dfrac{dy_2}{b},~~dx_3=\dfrac{dy_3}{c}$$ So $$x_1^2+x_2^2+x_3^2\le1\implies \dfrac{y_1^2}{a^2}+\dfrac{y_2^2}{b^2}+\dfrac{y_3^2}{c^2}\le1$$ So from $(1)$, $$\int_D f(Ax)dx =\iiint_{\frac{y_1^2}{a^2}+\frac{y_2^2}{b^2}+\frac{y_3^2}{c^2}\le1} f(y_1,y_2,y_3)~\dfrac{dy_1}{a}~\dfrac{dy_2}{b}~\dfrac{dy_3}{c}$$ $$=\dfrac{1}{abc}\int_E f(y) dy=\dfrac{1}{abc}\int_E f(x) dx$$ $($ as $\det A>1$, therefore $abc>1$ $)$ Thus option $(3)$ is correct but option $(1)$ and $(2)$ are not correct. Similarly, we can show that $$\int_{\mathbb R^3} f(Ax)dx =\dfrac{1}{abc}\int_{\mathbb R^3} f(x) dx$$ Hence option $(3)$ and option $(4)$ are the only correct options.

how many positive integer solutions to the following equation? $a^2 + b^2 + 25 = ab + 5a + 5b$ I have tried looking for a factorisation that could solve this question but couldn't find anything useful - found $(a+b+5)^2$ - don't know if this is useful The equation does look similar to an equation of a circle - can you use this idea?

The intuition is that the right-hand side is, with a few possible exceptins, smaller than the left-hand side. Note that $(a-b)^2\ge 0$, so $ab\le \frac{a^2+b^2}{2}$. Thus the right-hand side is $\le \frac{a^2+b^2}{2}+5a+5b$. It follows that $$(a^2+b^2+25)-(ab+5a+5b)\ge \frac{a^2+b^2}{2}+25-5a-5b.$$ The right=hand side above is $$\frac{1}{2}\left(a^2+b^2-10a-10b+50\right).$$ This is $$\frac{1}{2}\left((a-5)^2+(b-5)^2\right),$$ which is $\gt 0$ unless $a=b=5$. Remark: We can make the proof more myaterious by writing the magic identity $$2\left[(a^2+b^2+25)-(ab+5a+5b)\right]=(a-5)^2+(b-5)^2+(a-b)^2,$$ and concluding that the left-hand side is $0$ precisely if $a=b=5$.

Irrational number inequality : $1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$ it is easy and simple I know but still do not know how to show it (obviously without simply calculating the sum but manipulation on numbers is key here. $$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}>\sqrt{3}$$

$2 \lt 3 \lt 4$ so $\sqrt{2} \lt \sqrt{3} \lt 2$ and $$1+\frac1{\sqrt{2}}+\frac1{\ \sqrt{3}} \gt 1+\frac12+\frac12 = 2 \gt \sqrt{3}.$$ Of course that does not generalise in any nice way. However $2 \lt \frac{100}{49}$ with $3 \lt \frac{49}{16}$ and $5 \lt \frac{81}{16}$ does yield $$1+\frac1{\sqrt{2}}+\frac1{\ \sqrt{3}} \gt 1+\frac7{10}+\frac47=\frac{159}{70} =\frac{636}{280}\gt \frac{630}{280}=\frac{9}{4} \gt \sqrt{5}.$$

How to simpify $\cos x - \sin x$ How does one simplify $$\cos x - \sin x$$ I tried multiplying by $\cos x + \sin x$, but that just gets me $$\cos x - \sin x = \frac{\cos 2x}{\cos x + \sin x}$$ which is worse. Yet wolframalpha gives me $\cos x - \sin x = \sqrt{2}\sin\left(\dfrac{\pi}{4}-x\right)$. How does one obtain this algebraically?

$$ s = \cos x - \sin x \\ s^2 = \cos^2 x - 2 \cos x \sin x + \sin^2 x = 1 - \sin 2x \\ = 1 - \cos (\frac{\pi}2 -2 x)\\ = 1 - \left(1 - 2 \sin^2(\frac{\pi}4 - x)\right)\\ =2 \sin^2(\frac{\pi}4 - x) $$ so $$ s = \pm \sqrt{2} \sin(\frac{\pi}4 - x) $$ and evaluating at $x=0$ shows that the positive sign must be taken

Proving of $\frac{\pi }{24}(\sqrt{6}-\sqrt{2})=\sum_{n=1}^{\infty }\frac{14}{576n^2-576n+95}-\frac{1}{144n^2-144n+35}$ This is a homework for my son, he needs the proving.I tried to solve it by residue theory but I couldn't. $$\frac{\pi }{24}(\sqrt{6}-\sqrt{2})=\sum_{n=1}^{\infty }\frac{14}{576n^2-576n+95}-\frac{1}{144n^2-144n+35}$$

You can apply the residue theorem after a bit of playing with the sums: \begin{align*}&\sum_{n=1}^\infty\frac{14}{576n^2-576n+95}-\sum_{n=1}^\infty\frac4{576n^2-576+140}=\\&\sum_{n=1}^\infty\left(\frac1{24n-19}-\frac1{24n-5}\right)-\sum_{n=1}^\infty\left(\frac1{24n-14}-\frac1{24n-10}\right)=\\&\sum_{n=1}^\infty\left(\frac1{24n-19}+\frac1{24(1{-}n)-19}\right)-\sum_{n=1}^\infty\left(\frac1{24n-14}+\frac1{24(1{-}n)-14}\right)=\\&\sum_{n=-\infty}^\infty\frac1{24n-19}-\sum_{n=-\infty}^\infty\frac1{24n-14}=\sum_{n=-\infty}^\infty\frac5{(24n-19)(24n-14)}\end{align*} Now consider $$\lim_{n\to\infty}\int_{\varphi_{n+1/2}}\frac{5\pi\cot\pi z}{(24z-19)(24z-14)}dz=0,$$ where $\varphi_{n+1/2}$ is the circle of radius $n{+}\small 1/2$. The sum of all residues of the integrated function is also $0$ and the residues at points of $\mathbb Z$ gives us the terms of the sum. But there are $2$ more, the residues at $\frac{19}{24}$ and $\frac{7}{12}$ are $\frac{5\pi\cot\frac{19}{24}\pi}{24\cdot(19-14)}$ and $\frac{5\pi\cot\frac7{12}\pi}{(14-19)\cdot 24}$ respectively, so your sum is equal to $$-\left(\frac{\pi\cot\frac{19}{24}\pi}{14}-\frac{\pi\cot\frac7{12}\pi}{24}\right)=\frac{\pi}{24}\left(\cot\tfrac{7}{12}\pi-\cot\tfrac{19}{24}\pi\right)=\ldots=\frac{\pi}{24}(\sqrt6-\sqrt2).$$

Find a closed form for the equations $1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$ This is the assignment I have: Find a closed form for the equations $1^3 = 1$ $2^3 = 3+5$ $3^3 = 7+9+11$ $4^3 = 13+15+17+19$ $5^3 = 21+23+25+27+29$ $...$ Hints. The equations are of the form $n^3 = a1 +a2 +···+an$, where $a_{i+1} = a_i +2$ and $a_0 =n(n−1)+1$. My reasoning: We have to find a formula that give us $n^3$ summing operands. (why is this useful?) We know that the first operand (or term) of the sum is $a_0 =n(n−1)+1$. In fact, if you put $n = 3$, then $a_0 = 3(3 − 1) + 1 = 3*2 + 1 = 7$, which is exactly the first number of sum. Then I notice that each $n$ sum has $n$ operands, and each operand differs from one another of 2. Thus I came out with this formula: $$ \sum\limits_{i=0}^{n-1} a_0 + 2 \cdot i $$ where $a_0 =n(n−1)+1$ For example, if $n = 3$, then we have $(n(n−1)+1 + 2 \cdot 0) + (n(n−1)+1 + 2 \cdot 1) + (n(n−1)+1 + 2 \cdot 2) \equiv$ $\equiv (7 + 0) + (7 + 2) + (7 + 4) \equiv$ $\equiv 7 + 9 + 11$ Which is what is written as third example. I don't know if this is correct form or even if this is a closed form, that's why I am asking...

$$n^3=\sum_{k=0}^{n-1}(n^2-(n-1)+2k)$$ Since $\sum_{k=0}^{n-1}(n^2-(n-1)+2k)=n^3-n(n-1)+2(\frac{n(n-1)}{2})=n^3$ So it is the summation of $n$ consequitive odd number.

Understanding a step in a double series proof I'm really confused, how do they get from the first line to the second line ? $$\begin{align*} S&=\frac12\left[\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2n}{3^m(n\cdot3^m+m\cdot3^n)}+\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{n^2m}{3^n(n\cdot3^m+m\cdot3^n)}\right]\\\\ &=\frac12\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{mn}{3^{m+n}} \end{align*}$$ Can anyone explain this step?

Making a common denominator and factoring, observe that: \begin{align*} \frac{m^2n}{3^m(n \cdot 3^m + m \cdot 3^n)} + \frac{n^2m}{3^n(n \cdot 3^m + m \cdot 3^n)} &= \frac{m^2n \cdot 3^n + n^2m \cdot 3^m}{3^m3^n(n \cdot 3^m + m \cdot 3^n)} \\ &= \frac{mn(m \cdot 3^n + n \cdot 3^m)}{3^{m+n}(n \cdot 3^m + m \cdot 3^n)} \\ &= \frac{mn}{3^{m+n}} \\ \end{align*}

How to evaluate the following integral $\int_0^{\pi/2}\sin{x}\cos{x}\ln{(\sin{x})}\ln{(\cos{x})}\,dx$? How to evaluate the following integral $$\int_0^{\pi/2}\sin{x}\cos{x}\ln{(\sin{x})}\ln{(\cos{x})}\,dx$$ It seems that it evaluates to$$\frac{1}{4}-\frac{\pi^2}{48}$$ Is this true? How would I prove it?

Find this $$I=\int_{0}^{\frac{\pi}{2}}\sin{x}\cos{x}\ln{(\cos{x})}\ln{(\sin{x})}dx$$ Solution Since $$\sin(2x) = 2\sin(x)\cos(x)$$ then $$I=\dfrac{1}{8}\int_{0}^{\frac{\pi}{2}}\ln{(\sin^2{x})} \ln{(\cos^2{x})}\sin{(2x)}dx$$ Let $\cos{(2x)}=y$, and since $$\cos(2x) = 2\cos^2x - 1 = 1 - 2\sin^2x$$ we get $$I=\dfrac{1}{16}\int_{-1}^{1}\ln{\left(\dfrac{1-y}{2}\right)} \ln{\left(\dfrac{1+y}{2}\right)}dy$$ Let $\dfrac{1-y}{2}=z$, then we have \begin{align*}I&=\dfrac{1}{8}\int_{0}^{1}\ln{z}\ln{(1-z)}dz=\dfrac{-1}{8}\sum_{n=1}^{\infty}\dfrac{1}{n} \int_{0}^{1}z^n\ln{z}dz\\ &=\dfrac{1}{8}\sum_{n=1}^{\infty} \dfrac{1}{n(n+1)^2}=\dfrac{1}{8}\sum_{n=1}^{\infty} \left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)-\dfrac{1}{8}\sum_{n=1}^{\infty} \dfrac{1}{(n+1)^2}\\ &=\dfrac{1}{4}-\dfrac{\pi^2}{48} \end{align*}

How can I differentiate this equation? $y = \sqrt[4]{\frac{(x^3+2\sqrt{x})^2(x-sinx)^5}{(e^{-2x}+3x)^3}}$ $y = \sqrt[4]{\frac{(x^3+2\sqrt{x})^2(x-sinx)^5}{(e^{-2x}+3x)^3}}$ I tried removing the root but that got me no where

the solution of the given problem should be this here $$\frac{\left(x^3+2 \sqrt{x}\right) (x-\sin (x))^4 \left(-3 \left(3-2 e^{-2 x}\right) \left(x^3+2 \sqrt{x}\right) (x-\sin (x))+5 \left(3 x+e^{-2 x}\right) \left(x^3+2 \sqrt{x}\right) (1-\cos (x))+2 \left(3 x+e^{-2 x}\right) \left(3 x^2+\frac{1}{\sqrt{x}}\right) (x-\sin (x))\right)}{4 \left(3 x+e^{-2 x}\right)^4 \left(\frac{\left(x^3+2 \sqrt{x}\right)^2 (x-\sin (x))^5}{\left(3 x+e^{-2 x}\right)^3}\right)^{3/4}}$$

Integral of $\frac{1}{\sqrt{x^2-x}}dx$ For a differential equation I have to solve the integral $\frac{dx}{\sqrt{x^2-x}}$. I eventually have to write the solution in the form $ x = ...$ It doesn't matter if I solve the integral myself or if I use a table to find the integral. However, the only helpful integral in an integral table I could find was: $$\frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}} \ln \left|2ax + b +2\sqrt{a\left({ax^2+bx+c}\right)}\right|$$ Which would in my case give: $$\frac{dx}{\sqrt{x^2-x}} = \ln \left|2x -1 + 2\sqrt{x^2-x}\right|$$ Which has me struggling with the absolute value signs as I need to extract x from the solution. All I know is that $x<0$ which does not seem to help me either (the square root will only be real if $x<-1$). Is there some other formula for solving this integral which does not involve absolute value signs or which makes extracting $x$ from the solution somewhat easier? Thanks!

$$\int \frac{dx}{\sqrt{x^2-x}} = \int \frac{dx}{\sqrt{\left(x-\frac{1}{2}\right)^2 - \frac{1}{4}}}$$ Setting $ x - \frac{1}{2} = t $ $$ \int \frac{dx}{\sqrt{\left(x-\frac{1}{2}\right)^2 - \frac{1}{4}}} = \int \frac{dt}{\sqrt{t^2 - \frac{1}{4}}} $$ It's possible to do this using a trig substitution, but if you want the inverse function, a better way is to use a hyperbolic substitution. Let $t = \frac{1}{2}\cosh u \Rightarrow dt = \frac{1}{2} \sinh u$ $$ \int \frac{dt}{\sqrt{t^2 - \frac{1}{4}}} = \int \frac{\frac{1}{2}\sinh u \, du}{\sqrt{\frac{1}{4}(\cosh^2 u - 1)}} = \int \frac{\frac{1}{2}\sinh u \, du}{\frac{1}{2}\sinh u} = \int du = u + C $$ Working backwards is easy since you already have $t$ as a function of $u$ $$t = \frac{1}{2}\cosh u \Rightarrow x = \frac{1}{2}\cosh u + \frac{1}{2}$$ Where $u$ is the integral in question

Show that $x^a+x-b=0$ must have only one positive real root and not exceed the $\sqrt[a]{b-1}$ If we take the equation $$x^3+x-3=0$$ and solve it to find the real roots, we will get only one positive real roots which is $(x=1.213411662)$. If we comparison this with $\sqrt[3]{3-1}=1.259921$, we will find that $x$ is less than $\sqrt[3]{3-1}$.This always happens with any value of $a$ so that $a$ any positive real number and $b$ is a positive real number. So we can ask the following: 1-prove that $x^a+x-b=0$ must have only one positive real root if $a$ positive real number and $b$ is a positive real number greater than $1$. 2-The value of this roots must be less than $\sqrt[a]{b-1}$ 3-What happens when $a$ is a complex value.I mean this thing stays right or not?

If $f(x)=x^a+x-b$ then $f'(x)=ax^{a-1}+1>0,$ since $a\in\mathbb{R}_+.$ Thus $f$ is strictly increasing and so it must take the value $0$ at most one time. Now, since $f(0)=-b<0$ and $f(b)>0$ it follows from the Intermediate Value Theorem that there exists a root in the interval $(0,b).$ So, $f$ has exactly one root. Finally, if $f(x)=x^1+x-3=2x-3$ has as root $x=3/2.$ However, $\sqrt[a]{b-1}=(b-1)^{1/a}=2>3/2.$

Finding Exact Value $7\csc(x)\cot(x)-9\cot(x)=0$ The values for $x$ on $[0,2\pi)$ solving $7\csc(x)\cot(x)-9\cot(x)=0$ are? I think that $\dfrac{\pi}2$ is one but I can't find the others. what are the others?

$7\csc x\cot x-9\cot x=0\implies7\cos x-9\cos x\sin x=0\implies (\cos x)(7-9\sin x)=0$, so $\cos x=0\implies x=\frac{\pi}{2}$ or $x=\frac{3\pi}{2}$ and $\sin x=\frac{7}{9}\implies x=\sin^{-1}\frac{7}{9}$ or $x=\pi-\sin^{-1}\frac{7}{9}$.

Evaluation of $ \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ Evaluation of $\displaystyle \lim_{x\to 0}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ where $\lfloor x \rfloor $ represent floor function of $x$. My Try:: Here $\displaystyle f(x) = \frac{x^2}{\sin x\cdot \tan x}$ is an even function. So we will calculate for $\displaystyle \lim_{x\to 0^{+}}\left\lfloor \frac{x^2}{\sin x\cdot \tan x}\right\rfloor$ Put $x=0+h$ where $h$ is a small positive quantity, and using series expansion So limit convert into $\displaystyle \lim_{h\to 0}\left\lfloor \frac{h^2}{\sin h\cdot \tan h}\right\rfloor = \lim_{h\to 0}\left\lfloor \dfrac{h^2}{\left(h-\dfrac{h^3}{3!}+\dfrac{h^5}{5!}- \cdots\right)\cdot \left(h+\dfrac{h^3}{3}+\dfrac{2}{15}h^5+ \cdot\right)}\right\rfloor$ Now how can i solve after that, Help me Thanks

Let me continue where you stopped in your post. Expand the denominator for a few terms and perform the long division. You should arrive to $$\frac{x^2}{\sin x\cdot \tan x} =1-\frac{x^2}{6}-\frac{7 x^4}{120}+O\left(x^5\right)$$ which is definitely smaller than $1$. I am sure that you can take from here.

Finding matrix for given recurrence For the recurrence relation: $f(0)=1$ $f(1)=1$ $f(2)=2$ $f(2n)=f(n)+f(n+1)+n$ $f(2n+1)=f(n)+f(n−1)+1$ How to find square matrices $M_0, M_1$ and vectors $u, v$ such that if the base-2 expansion of $n$ is given by $e_1 e_2 \cdots e_j$, then $$f(n) = u M_{e_1} \cdots M_{e_j} v.$$ ??

This is how I would approach this question, and I guess my answer would not be unique. For each $n = e_1e_2\cdots e_j$, there are $7$ numbers I should store (as a row vector) to calculate $f(2n)$ and $f(2n+1)$: $$uM_{e_1}M_{e_2}\cdots M_{e_j} = w(n) = \pmatrix {1&n&f(n-2)&f(n-1)&f(n)&f(n+1)&f(n+2)}.$$ To calculate $f(2n)$, the matrix $M_0$ is appended to the above product: $$w(2n) = \pmatrix{1\\2n\\f(2n-2)\\f(2n-1)\\f(2n)\\f(2n+1)\\f(2n+2)}^T =\pmatrix{1\\2n\\f(n-1)+f(n)+n-1\\f(n-1)+f(n-2)+1\\f(n)+f(n+1)+n\\f(n)+f(n-1)+1\\f(n+1)+f(n+2)+n+1}^T = w(n)M_0$$ To calculate $f(2n+1)$, the matrix $M_1$ is appended to the above product: $$ w(2n+1) = \pmatrix{1\\2n+1\\f(2n-1)\\f(2n)\\f(2n+1)\\f(2n+2)\\f(2n+3)}^T =\pmatrix{1\\2n+1\\f(n-1)+f(n-2)+1\\f(n)+f(n+1)+n\\f(n)+f(n-1)+1\\f(n+1)+f(n+2)+n+1\\f(n+1)+f(n)+1}^T = w(n)M_1$$ So $M_0$ can be written as $$M_0 = \pmatrix{1&0&-1&1&0&1&1\\ 0&2&1&0&1&0&1\\ 0&0&0&1&0&0&0\\ 0&0&1&1&0&1&0\\ 0&0&1&0&1&1&0\\ 0&0&0&0&1&0&1\\ 0&0&0&0&0&0&1}$$ And similarly for $M_1$. I imagine the base case to be $$w(2) = uM_1M_0 = \pmatrix{1&2&1&1&2&3&7}$$ For the rest, I have not tried out the calculation yet, but I suppose this involves solving the equation $$uM_1M_0 = w(2)$$ to solve for $uM_1$ and then $u$, and verify if that satisfies $w(0)$, $w(1)$ and $w(3)$. Lastly $$v = \pmatrix{0\\0\\0\\0\\1\\0\\0}$$ Edit: Just found out $uM_1M_0 = w(2)$ is inconsistent and there is no $uM_1$ which satisfy the equation. So hope this answer could give you some idea on other ways to solve this.

Solving $2\cos^2 x-2\sin^2 x-2\cos x=0$ $$f(x) = 2\cos^2 x-2\sin^2 x-2\cos x$$ Need values of x that which make $f(x) = 0$ Tried $a^2-b^2 = (a+b)(a-b)$ with no luck Really just need a hint that could bring me in the right direction Thanks EDIT: Solution thanks to everyones help! :D $$f(x) = 2\cos^2 x-2\sin^2 x-2\cos x$$ $$0 = 2\cos^2 x-2 + 2\cos^2 x-2\cos x$$ $$0 = 4\cos^2 x-2\cos x - 2$$ $$\cos x = 2\pm \sqrt {-2^2-4(4)(-2)\over8}$$ $$\cos x = {2 \pm 6\over 8}$$ $$\cos x = {-1\over2}, 1$$ $$x = 2n\pi \pm{2\pi\over3}$$ Thank you!

Hint: Use $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\sin^2x=1-\cos^2x$ Update: $$\cos x=-\frac{1}{2},1$$ $$\cos x=-\dfrac12=\cos\left(\frac{2\pi}3\right)\implies x=2k\pi\pm\dfrac{2\pi}3$$ $$\cos x=1\implies x=2k\pi$$

Simplifying $\frac{x^6-1}{x-1}$ I have this: $$\frac{x^6-1}{x-1}$$ I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$ Edit : I was wondering how to do this if I didn't know that it was the same as that.

Ill show an "tricky" method. $\displaystyle \frac{x^6 - 1}{x-1}$ $= \displaystyle \frac{x^6 -x + x - 1}{x-1} = \frac{x^6 - x}{x-1} + 1 = \frac{x^6 - x^5 + x^5 - x}{x-1} + 1 = x^5 + 1 + \frac{x^5 - x}{x-1} = \frac{x^5 - x^4 + x^4 - x}{x-1} + x^5 + 1 = \frac{x^4(x - 1) + x^4 - x}{(x-1)}$ Do you see the pattern? This is simply to show how you can manipulate expressions; its a trick.

Evaluate the integral by using Gauss divergence theorem. Evaluate $\int\int_SF.dS$ where $F=(xz,yz,x^2+y^2)$ by using the Gauss divergence theorem. Where $S$ is the closed surface obtained from the surfaces $x^2+y^2\leq 4,z=2,x^2+y^2\leq 16,z=0$ on the top and the bottom and $z=4-\sqrt{x^2+y^2}$ on the side. My calculation shows that answer is $\frac{40\pi}{3}$ but the answer should be $\frac{88\pi}{3}$.What is I am doing wrong? I set $$\iint_S \bf{F}\cdot dS=\iiint_{V}\text{div}FdV=2\iiint_VzdV$$ In cylindrical coordinates the limits are $$0\leq z\leq 4-r,2\leq r\leq 4,0\leq \theta\leq 2\pi$$ What is wrong with these limits?

Anyway, this is what I get: $$2 \int\int\int z dV = 2\int_0^2 \int_0^{4-z} \int_0^{2\pi} z r\ d\theta dr dz = 4\pi \int_0^2 \int_0^{4-z} zr\ dr dz$$ $$ = 2\pi \int_0^2 z(4-z)^2 dz = 2\pi \int_0^2 \bigg(z^3 - 8z^2 + 16z\bigg)dz$$ $$ = 2\pi \bigg( \frac{2^4}{4} - \frac{16(2^3)}{3} + \frac{16(2^2)}{2}\bigg) = \frac{88\pi}{3}$$

Evaluate $\lim_{x→0}\left(\frac{1+\tan x}{1+\sin x}\right)^{1/x^2} $ I have the following limit to evaluate: $$ \displaystyle \lim_{x→0}\left(\frac{1+\tan x}{1+\sin x}\right)^{1/x^2} $$ What's the trick here?

$$\lim_{x\to 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\frac{1}{x^2}}$$ $$=\exp\left(\lim_{x\to 0}{\frac{1}{x^2}}\left(\frac{1+\tan x}{1+\sin x}-1\right)\right)$$ $$=\exp\left(\lim_{x\to 0}{\frac{1}{x^2}}\left(\frac{\tan x-\sin x}{1+\sin x}\right)\right)$$ $$=\exp\left(\lim_{x\to 0}{\frac{1}{x^2}}\left(\frac{\frac{2tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}-\frac{2tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}}{1+\frac{2tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}}\right)\right)$$ $$=\exp\left(\lim_{x\to 0}{\frac{1}{x^2}}\left(\frac{2\tan\frac{x}{2}\left(1+\tan^2\frac{x}{2}-1+\tan^2\frac{x}{2}\right)}{\left(1-\tan^2\frac{x}{2}\right) \left(1+\tan^2\frac{x}{2}+2\tan\frac{x}{2}\right)}\right)\right)$$ $$=\exp\left(\lim_{x\to 0}{\frac{1}{x^2}}\left(\frac{4\tan^3\frac{x}{2}}{\left(1-\tan^2\frac{x}{2}\right) \left(1+\tan^2\frac{x}{2}+2\tan\frac{x}{2}\right)}\right)\right)$$ $$=\exp\left(\lim_{x\to 0}\left(\tan\frac{x}{2}\right)\lim_{x\to 0}\left(\frac{\left(\frac{\tan\frac{x}{2}}{\frac{x}{2}}\right)^2}{\left(1-\tan^2\frac{x}{2}\right) \left(1+\tan^2\frac{x}{2}+2\tan\frac{x}{2}\right)}\right)\right)$$ $$=e^{\left(\left(0\right)\left(\frac{1}{1\times1}\right)\right)}=e^{0}=1$$

Closed-form of sums from Fourier series of $\sqrt{1-k^2 \sin^2 x}$ Consider the even $\pi$-periodic function $f(x,k)=\sqrt{1-k^2 \sin^2 x}$ with Fourier cosine series $$f(x,k)=\frac{1}{2}a_0+\sum_{n=1}^\infty a_n \cos2nx,\quad a_n=\frac{2}{\pi}\int_0^{\pi} \sqrt{1-k^2 \sin^2 x}\cos 2nx \,dx.$$ This was considered in this earlier question, and in comments an observation was made: The Fourier coefficients all appear to be of the form $a_n= A_n(k) K(k)+B_n(k) E(k)$ where $K,E$ are the complete elliptic integrals of the first and second kind and $A_n(k),B_n(k)$ are rational functions of $k$. This is made plausible by the fact that $f(x,k)$ is the first $x$-derivative of the incomplete elliptic integral of the second kind $E(x,k)=\int_0^x f(x',k)\,dx'$. Moreover, this conjecture for $a_k$ can be confirmed by examining the Fourier series of $E(x,k)$; this appears in a 2010 paper by D. Cvijovic, with full text available on ResearchGate. Something we may conclude from this observation is that, since the Fourier coefficients are linear combinations of complete elliptic integrals, $f(x,k)$ itself must be of the form $A(x,k)K(k)+B(x,k)E(k)$ where $A(x,k), B(x,k)$ are even $\pi$-periodic functions whose Fourier coefficients are rational functions of $k$. Such a Fourier expansion of does appear in the paper noted above, but the functions themselves are not found in closed-form. Hence my question: Can $A(x,k)$, $B(x,k)$ be obtained in closed-form in terms of known special functions?

We have: $$ \frac{\pi}{2}\,a_n=\int_{0}^{\pi}\sqrt{1-k^2\sin^2\theta}\cos(2n\theta)\,d\theta =\frac{k^2}{4n}\int_{0}^{\pi}\frac{\sin(2\theta)\sin(2n\theta)}{\sqrt{1-k^2\sin^2\theta}}\,d\theta.\tag{1}$$ If we set: $$ b_m = \int_{0}^{\pi}\frac{\cos(2m\theta)}{\sqrt{1-k^2\sin^2\theta}}\,d\theta,\qquad c_m = \int_{0}^{\pi}\cos(2m\theta)\sqrt{1-k^2\sin^2\theta}\,d\theta $$ we have: $$ b_m = \int_{0}^{\pi}\frac{\cos(2\theta)\cos((2m-2)\theta)}{\sqrt{1-k^2\sin^2\theta}}\,d\theta-\int_{0}^{\pi}\frac{\sin(2\theta)\sin((2m-2)\theta)}{\sqrt{1-k^2\sin^2\theta}}\,d\theta$$ and expressing $\cos(2\theta)$ as $1-2\sin^2\theta$ we get: $$ b_m = \frac{2}{k^2}\int_{0}^{\pi}\frac{(k^2/2-k^2\sin^2\theta)\cos((2m-2)\theta)}{\sqrt{1-k^2\sin^2\theta}}\,d\theta-\frac{4(m-1)}{k^2}c_{m-1}$$ so: $$ b_m = \frac{2}{k^2} c_{m-1} + \frac{k^2-2}{k^2}b_{m-1}-\frac{4m-4}{k^2}c_{m-1} = \frac{k^2-2}{k^2}b_{m-1}-\frac{4m-2}{k^2}c_{m-1}. $$ Moreover, integration by parts gives: $$ c_m = \frac{k^2}{8m}\left(b_{m+1}-b_{m-1}\right), \tag{2}$$ hence: $$ b_m = \frac{k^2-2}{k^2}b_{m-1}-\frac{2m-1}{4m-4}(b_m-b_{m-2}), $$ $$ \frac{6m-5}{4m-4} b_m = \frac{k^2-2}{k^2}b_{m-1}+\frac{2m-1}{4m-4}b_{m-2}, $$ or: $$ b_m = \frac{4m-4}{6m-5}\cdot\frac{k^2-2}{k^2}b_{m-1}+\frac{2m-1}{6m-5}b_{m-2}.\tag{3} $$ Since $b_0 = 2 K(k) $ and $c_0 = 2 E(k)$ we have $$ b_1 = \frac{1}{k^2}\left((2k^2-4)K(k)+4E(k)\right).$$ In order to have explicit forms for $A(\theta,k)$ and $B(\theta,k)$, it is sufficient to find a closed-form expression for the recursion given by $(3)$, since, by $(1)$ and $(2)$: $$\frac{\pi}{2}a_n = c_n = \frac{k^2}{8n}\left(b_{m+1}-b_{m-1}\right).$$ By setting: $$ B(x)=\sum_{n\geq 0} b_n x^n $$ recursion $(2)$ can be converted into a first-order ODE for $B(x)$, giving that $B(x)$ behaves like the reciprocal of the square root of a cubic polynomial, and the radius of convergence of $B(x)$ is $\geq 2.485$. $A(\theta,k)$ and $B(\theta,k)$ can be recovered from $\Re\left(B(e^{2i\theta})\right)$.

Prove by induction that an expression is divisible by 11 Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$. I am rather confused by this question. This is my attempt so far: For $n = 2$ $2^5 + 5\cdot 9 = 77$ $77/11 = 7$ We assume that there is a value $n = k$ such that $2^{3k-1} + 5\cdot 3^k$ is divisible by $11$. We show that it is also divisible by $11$ when $n = k + 2$ $2^{3k+5} + 5\cdot 3^{k+2}$ $32\cdot 2^3k + 5\cdot 9 \cdot3^k$ $32\cdot 2^3k + 45\cdot 3^k$ $64\cdot 2^{3k-1} + 45\cdot 3^k$ (Making both polynomials the same as when $n = k$) $(2^{3k-1} + 5\cdot 3^k) + (63\cdot 2^{3k-1} + 40\cdot 3^k)$ The first group of terms $(2^{3k-1} + 5\cdot 3^k)$ is divisible by $11$ because we have made an assumption that the term is divisible by $11$ when $n=k$. However, the second group is not divisible by $11$. Where did I go wrong?

Keep going! $64\cdot 2^{3k-1} + 45\cdot 3^k = 9(2^{3k-1} + 5\cdot3^k) + 55\cdot2^{3k-1}$

Evaluate $\int_0^{\pi/2}x\cot{(x)}\ln^4\cot\frac{x}{2}\,\mathrm dx$ How to evaluate the following integral ?: $$ \int_{0}^{\pi/2}x\cot\left(\, x\,\right)\ln^{4}\left[\,\cot\left(\,{x \over 2}\,\right)\,\right]\,{\rm d}x $$ It seems that evaluate to $$ {\pi \over 16}\left[\, 5\pi^{4}\ln\left(\, 2\,\right) - 6\pi^{2}\zeta\left(\, 3\,\right) -{93 \over 4}\,\zeta\left(\, 5\,\right) \,\right] $$ Exactly ?.

Let $$J = \int_0^1 {\frac{{\arctan x{{\ln }^4}x}}{x}dx} \qquad K = \int_0^1 {\frac{x{\arctan x{{\ln }^4}x}}{{1 + {x^2}}}dx}$$ Then by M.N.C.E.'s comment, $$\tag{1}I = \int_0^{\pi /2} {x\cot x{{\ln }^4}\left( {\cot \frac{x}{2}} \right)dx} = 2J - 4K$$ Here is a symmetry of the integrand that we can exploit: $$\begin{aligned} K &= \int_0^1 {\frac{{x\arctan x{{\ln }^4}x}}{{1 + {x^2}}}dx} \\&= \int_0^1 {\frac{{\arctan x{{\ln }^4}x}}{x}dx} - \int_0^1 {\frac{{\arctan x{{\ln }^4}x}}{{x(1 + {x^2})}}dx} \\&= J - \int_1^\infty {\frac{{x\left( {\frac{\pi }{2} - \arctan x} \right){{\ln }^4}x}}{{1 + {x^2}}}dx} \\& = J - \int_1^\infty {\frac{1}{x}\left( {\frac{\pi }{2} - \arctan x} \right){{\ln }^4}xdx} + \int_1^\infty {\frac{1}{{x(1 + {x^2})}}\left( {\frac{\pi }{2} - \arctan x} \right){{\ln }^4}xdx} \\& = J - J + \frac{\pi }{2}\int_1^\infty {\frac{{{{\ln }^4}x}}{{x(1 + {x^2})}}dx} - \int_1^\infty {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx} \\&= \frac{\pi }{2}\int_0^1 {\frac{{x{{\ln }^4}x}}{{1 + {x^2}}}dx} - \int_0^\infty {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx} + \int_0^1 {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx} \\& = \frac{\pi }{2}\int_0^1 {\frac{{x{{\ln }^4}x}}{{1 + {x^2}}}dx} - \int_0^\infty {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx} + \int_0^1 {\frac{{{{\ln }^4}x\arctan x}}{x}dx} - \int_0^1 {\frac{{x{{\ln }^4}x\arctan x}}{{1 + {x^2}}}dx} \\& = \frac{\pi }{4}\int_0^1 {\frac{{x{{\ln }^4}x}}{{1 + {x^2}}}dx} - \frac{1}{2}\int_0^\infty {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx} + \frac{J}{2} \end{aligned}$$ The fact that exponent $4$ is even is paramount here. Plugging into $(1)$, the $J$ miraculously cancelled: $$I = 2\underbrace{\int_0^\infty {\frac{{\arctan x}}{{x(1 + {x^2})}}{{\ln }^4}xdx}}_{L} - \pi \underbrace{\int_0^1 {\frac{{x{{\ln }^4}x}}{{1 + {x^2}}}dx}}_{45\zeta(5)/64} $$ The crux of the problem is, indeed, evaulating the remaining integral. Note that $$ L = \int_0^{\frac{\pi }{2}} {x\cot x{{\ln }^4}(\tan x)dx} $$ and we have the following formula: For $-2<p<1$, $$\int_0^{\frac{\pi }{2}} {x{{\tan }^p}xdx} = \frac{\pi }{4}\csc \frac{{p\pi }}{2}\left[ {\psi ( - \frac{p}{2} + 1) - 2\psi ( - p + 1) - \gamma } \right]$$ Hence the value of $L$ follows from it by differentiating four times and set $p=-1$: $$L = -\frac{3 \pi ^3 \zeta (3)}{16}-\frac{3 \pi \zeta (5)}{8}+\frac{5}{32} \pi ^5 \ln 2$$ Finally, we obtain $$I = 2L - \pi \frac{45\zeta(5)}{64} = \color{blue}{-\frac{3 \pi ^3 \zeta (3)}{8}-\frac{93 \pi \zeta (5)}{64}+\frac{5}{16} \pi ^5 \ln 2}$$

Diagonalization and find matrix that corresponds to the given condition Diagonalize the matrix $$ A= \begin{pmatrix} 1 & 2\\ 0 & 3 \end{pmatrix} $$ and find $B^3=A$. I derived $A \sim \text{diag}(1,3)$ but I have problem finding any $B$. I tried to solve it by writing $B= \begin{pmatrix} 1 & x\\ 0 & 3\end{pmatrix}$, but is it okay to solve the problem in this way?

The eigen values are $1,3 $ clearly. So it is diagonalizabe(distinct eigen values). And so, there exits $P$ such that $$A=P\left(\begin{array}{cc}1& 0\\ 0& 3\end{array}\right)P^{-1}.$$ Now we need $B$ such that $B$ such that $B^3=A$ Supose there exists such a $B$ then, $$B^3=A=P\left(\begin{array}{cc}1& 0\\ 0& 3\end{array}\right)P^{-1}$$ Therefore, $$B=P\left(\begin{array}{cc}1& 0\\ 0& 3\end{array}\right)^{\frac 1 3}P^{-1}$$ $$=P\left(\begin{array}{cc}1^{\frac 1 3}& 0\\ 0& 3^{\frac 1 3}\end{array}\right)P^{-1}$$

Probability of selecting a red ball first An urn contains 3 red and 7 black balls. Players A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B, and so on. There is no replacement of the balls drawn.) How do I calculate this probability? I tried using the total probability rule without success. I used the $P(A)=\frac{3}{10}+P(A_2\mid B_1)$ and so on, where $B_i$=Player B doesn't get a red ball. The answer should be $0.0888$

Here's what I was thinking. $$\color{RED}R+BB\color{RED}R+BBBB\color{RED}R+BBBBBB\color{RED}R$$ $$P(A)=\frac{3}{10}+\frac{7}{10}\frac{6}{9}\frac{3}{8}+\frac{7}{10}\frac{6}{9}\frac{5}{8}\frac{4}{7}\frac{3}{6}+\frac{7}{10}\frac{6}{9}\frac{5}{8}\frac{4}{7}\frac{3}{6}\frac{2}{5}\frac{3}{4}=\frac{7}{12}$$

Question about sines of angles in an acute triangle Let $\triangle ABC$ be a triangle such that each angle is less than $ 90^\circ $. I want to prove that $\sin A + \sin B + \sin C > 2$. Here is what I have done: Since $A+B+C=180^{\circ}$ and $0 < A,B,C < 90^\circ$, at least two of $A,B,C$ are in the range 45 < x < 90, without loss of generality, let these angles be $A$ and $B$. $\sin A + \sin B + \sin C = \sin A + \sin B + \sin(180^\circ-A-B) = \sin A + \sin B + \sin(A+B)$ Since $45^\circ < A,B < 90^\circ$ it follows that $2^{0.5} < \sin A + \sin B < 2.$ Am I near the answer?

i am able to simplify $$ \sin A + \sin B + \sin (A + B) = \sin A + \sin B + \sin A \cos B + \sin B \cos A \\ = (1+\cos B)\sin A + (1 + \cos A)\sin B \\ = 4\cos^2 B/2\sin A/2 \cos A/2 + 4\cos^2 A/2 \sin B/2 \cos B/2 \\ = 4\cos B/2 \cos A/2(\sin A/2 \cos B/2 + \sin B/2 \cos A/2) \\ = 4\cos B/2 \cos A/2\sin (A/2 + B/2) = 4\cos A/2 \cos B/2 \cos C/2 $$ 4now, using the fact that $A/2 < 45^\circ, B/2 < 45^\circ$ and $C/2 < 45^\circ,$ i can only conclude $$\sin A + \sin B + \sin C > \sqrt 2.$$ i am going to try to improve the bound. introduce $0 <\alpha, \beta < 45^\circ$ so that $A/2 = 45^\circ - \alpha, B/2 = 45^\circ - \beta, C/2 = \alpha + \beta.$ in terms of these new variables, $$ 4\cos A/2 \cos B/2 \cos C/2 = 2(\cos \alpha + \sin \alpha)(\cos \beta + \sin \beta)\cos(\alpha + \beta) \\ = 2(\cos(\alpha + \beta) + \sin(\alpha + \beta))\cos(\alpha + \beta) \\ = 1 + \cos(2\alpha + 2\beta) + \sin(2\alpha + 2\beta) = 1 + \sqrt 2 \sin(2\alpha + 2\beta + 45^\circ)$$ since $0 < 2\alpha + 2 \beta < 90^\circ,$ we get the desired bound $$2 < 1 + \sqrt 2\sin(2\alpha + 2 \beta) < 1 + \sqrt 2 $$

Is this a legitimate proof? If not, how to prove? Question: Determine all natural numbers $n$ such that: $7 \mid \left(3^n - 2\right) \implies3^{n}\equiv 2\pmod{7}$ Multiply both sides by 7 $7 \cdot 3^{n}\equiv 7\cdot2\pmod{7}$ Divide both sides by seven, since $\gcd(7,7) = 7$, we have to divide modulus by $7$ $\implies3^{n}\equiv 2\pmod{7/7}$ $\implies3^{n}\equiv 2\pmod{1}$ Therefore $n$ is any natural number, since one divides everything. But I made a mistake somewhere since the original equation doesn't work for $n = 1$

Noting that $n=1$ does not work, let $n \ge 2$. Then as $3^2 \equiv 2 \pmod 7$, we have the equivalent statement $$2\cdot 3^{n-2} \equiv 2 \pmod 7 \iff3^{n-2}\equiv 1 \pmod 7$$ Now that has solutions $n = 6k+2$ as $3^6$ is the smallest positive power of $3$ that is $\equiv 1 \pmod 7$, so the solution is for natural number s.t. $n = 2 \pmod 6$.

How many positive integers of n digits chosen from the set {2,3,7,9} are divisible by 3? I'm preparing myself for math competitions. And I am trying to solve this problem from the Romanian Mathematical Regional Contest “Traian Lalescu’', $2003$: Problem $\mathbf{7}$: How many positive integers of $n$ digits chosen from the set $\{2,3,7,9\}$ are divisible by $3$? Solution. Let $x_n,y_n,z_n$ be the number of all positive integers of $n$ digits $2,3,7$ or $9$ which are congruent to $0,1$ and $2$ modulo $3$. We have to find $x_n$. Consider $\varepsilon=\cos\dfrac{2\pi}3+i\sin\dfrac{2\pi}3$. It is clear that $x_n+y_n+z_n=4^n$ and $$x_n+\varepsilon y_n+\varepsilon^2z_n=\sum_{j_1+j_2+j_3+j_4=n}\varepsilon^{2j_1+3j_2+7j_e+9j_4}=(\varepsilon^2+\varepsilon^3+\varepsilon^7+\varepsilon^9)^n\;.$$ It follows that $x_n-1+\varepsilon y_n+\varepsilon^2z_n=0$. Applying Proposition $4$ in Subsection $2.2.2$ we obtain $x_n-1=y_n=z_n=k$. Then $3k=x_n+y_n+z_n-1=4^n-1$, and we find $k=\dfrac13(4^n-1)$. Finally, $x_n=k+1=\dfrac13(4^n+2)$. Please help me with the solution, I don't understand it well enough, especially the displayed line. Are there any other solutions for this problem?

Here is an alternative approach. Let $x_n,y_n$, and $z_n$ be as in the argument given in the question; clearly $x_1=2$, and $y_1=z_1=1$. For $n\ge 1$ let $X_n,Y_n$, and $Z_n$ be the sets of $n$-digit numbers using only the digits $2,3,7$, and $9$ and congruent modulo $3$ to $0,1$, and $2$, respectively (so that $x_n=|X_n|$, $y_n=|Y_n|$, and $z_n=|Z_n|$). Finally, recall that an integer is congruent modulo $3$ to the sum of its digits. Now let $n>1$, and let $k\in X_n\cup Y_n\cup Z_n$. Let $\ell$ be the $(n-1)$-digit number obtained by removing the last digit of $k$, and let $d$ be the last digit of $k$. If $d=3$ or $d=9$, then $k\equiv\ell\pmod3$; if $d=7$, then $k\equiv\ell+1\pmod3$; and if $d=2$, then $k\equiv\ell+2\pmod3$. Thus, $k\in X_n$ iff either $d\in\{3,9\}$ and $\ell\in X_{n-1}$, or $d=2$ and $\ell\in Y_{n-1}$, or $d=7$ and $\ell\in Z_{n-1}$. These three cases are exhaustive and mutually exclusive, so $$x_n=|X_n|=2|X_{n-1}|+|Y_{n-1}|+|Z_{n-1}|=2x_{n-1}+y_{n-1}+z_{n-1}\;.$$ Similar reasoning shows that $$y_n=2y_{n-1}+x_{n-1}+z_{n-1}$$ and $$z_n=2z_{n-1}+x_{n-1}+y_{n-1}\;.$$ But clearly $x_{n-1}+y_{n-1}+z_{n-1}=4^{n-1}$, so these recurrences reduce to $$\left\{\begin{align*} x_n&=x_{n-1}+4^{n-1}\\ y_n&=y_{n-1}+4^{n-1}\\ z_n&=z_{n-1}+4^{n-1}\;. \end{align*}\right.$$ If we set $x_0=1$, the first of these recurrences holds for $n=1$ as well as for $n>1$, and we have $$x_n=1+\sum_{k=0}^{n-1}4^k\;.$$ If this isn’t immediately clear, note that $$\begin{align*} x_n&=4^{n-1}+x_{n-1}\\ &=4^{n-1}+4^{n-2}+x_{n-2}\\ &\;\vdots\\ &=4^{n-1}+4^{n-2}+\ldots+4^0+x_0\\ &=1+\sum_{k=0}^{n-1}4^k\;, \end{align*}$$ and the formula can be proved rigorously by induction on $n$. Finally, this is just a geometric series, so $$x_n=1+\frac{4^n-1}{4-1}=1+\frac13(4^n-1)=\frac13(4^n+2)\;.$$

Calculate the distance between intersection points of tangents to a parabola * *Question Tangent lines $T_1$ and $T_2$ are drawn at two points $P_1$ and $P_2$ on the parabola $y=x^2$ and they intersect at a point $P$. Another tangent line $T$ is drawn at a point between $P_1$ and $P_2$; it intersects $T_1$ at $Q_1$ and $T_2$ at $Q_2$. Show that $$\frac{|PQ_1|}{|PP_1|} + \frac{|PQ_2|}{|PP_2|} = 1$$ * *My attempt at the question I include a possible scenario of the graph for convenience'(I hope) sake: The outer two tangents are tangents $T_1$ and $T_2$, and the inner tangent is tangent $T$. * *Since points $P_1$ and $P_2$ are points on the parabola, I can give them coordinates as follows, $$\tag 1 P_1(P_{1x}, P^2_{1x})$$ and $$\tag 2 P_2(P_{2x}, P^2_{2x})$$ *Using $y\prime = 2x$, I calculate the equations for the tangents $T_1$ and $T_2$ respectively, they are, $$T_1 = y = 2P_{1x}(x - P_{1x}) + P^2_{1x}$$ and $$T_2 = y = 2P_{2x}(x - P_{2x}) + P^2_{2x}$$ * *By setting $T_1 = T_2$ and then solving for $x$ I show that the two tangents intersect at a point $x = \frac{P_{1x} + P_{2x}}{2}$, which in words is the two tangents to the parabola is halfway between points $P_1$ and $P_2$. Then substituting $x = \frac{P_{1x} + P_{2x}}{2}$ into any of the tangent line equations I get the $y$ coordinate of the tangent lines' intersection, which is $y = P_{1x}\cdot P_{2x}$ Now I have the coordinates for point $P$, that is $$\tag 3 P\Big(\frac{P_{1x} + P_{2x}}{2}, P_{1x}\cdot P_{2x}\Big)$$ *To get coordinates for points $Q_1$ and $Q_2$ I will substitute $Q_{1x}$ in to the equation of tangent $T_1$ and substitute $Q_{2x}$ in to the equation of tangent $T_2$. That yields the following for coordinates: $$\tag 4 Q_1(Q_{1x}, \,\,2P_{1x}Q_{1x} - P_{1x}^2)$$ $$\tag 5 Q_2(Q_{2x}, \,\,2P_{2x}Q_{2x} - P_{2x}^2)$$ *Since I have all the points necessary to calculate $\frac{|PQ_1|}{|PP_1|} + \frac{|PQ_2|}{|PP_2|}$, I feel inclined to apply the distance formula. Doing so yielded the following: $$\tag 6 |PQ_1| = \frac{\sqrt{(4P_{1x}^2 + 1)(P_{1x} + P_{2x} - 2Q_{1x})^2}}{2}$$ $$\tag 7 |PP_1| = \frac{\sqrt{(4P_{1x}^2 + 1)(P_{1x} - P_{2x})^2}}{2}$$ $$\tag 8 |PQ_2| = \frac{\sqrt{(4P_{2x}^2 + 1)(P_{1x} + P_{2x} - 2Q_{2x})^2}}{2}$$ $$\tag 9 |PP_2| = \frac{\sqrt{(4P_{1x}^2 + 1)(P_{1x} - P_{2x})^2}}{2}$$ * *Now I calculate $\frac{|PQ_1|}{|PP_1|} + \frac{|PQ_2|}{|PP_2|}$ using the above: $$\frac{|PQ_1|}{|PP_1|} + \frac{|PQ_2|}{|PP_2|} = \frac{\frac{\sqrt{(4P_{1x}^2 + 1)(P_{1x} + P_{2x} - 2Q_{1x})^2}}{2}}{\frac{\sqrt{(4P_{1x}^2 + 1)(P_{1x} - P_{2x})^2}}{2}} + \frac{\frac{\sqrt{(4P_{2x}^2 + 1)(P_{1x} + P_{2x} - 2Q_{2x})^2}}{2}}{\frac{\sqrt{(4P_{2x}^2 + 1)(P_{1x} - P_{2x})^2}}{2}}$$ $$\tag {10} =\frac{\sqrt{(P_{1x} + P_{2x} - 2Q_{1x})^2} + \sqrt{(P_{1x} + P_{2x} - 2Q_{2x})^2}}{\sqrt{(P_{1x} - P_{2x})^2}}$$ *I can't seem to find a way to show that $(10)$ is equal to $1$. I have, however tested a few instances and it held up, for what it's worth. But for now, I'm at a loss as to how to proceed. Any hints, suggestions, or alternative approaches?

Suppose that the third tangent is drawn at a point $A$ with coordinates $A(a, a^2)$. Then its tangent intersects $T_{i}$ at $$ Q_{i}\left(\frac{P_{ix} + a}{2}, P_{ix}a \right) $$ using your equation $(3)$. In other words, $$ 2(Q_{1x} - Q_{2x})=P_{1x}-P_{2x} $$ Therefore, in the argument of the square root in the numerator of the first term of (10), \begin{align*} P_{1x} + P_{2x} - 2Q_{1x} &= P_{1x} + P_{2x} - 2[(P_{1x} - P_{2x})/2 + Q_{2x}] \\ &= 2(P_{2x} - Q_{2x}) \end{align*} In general, $\sqrt{x^2} = |x|$ so we need to figure out the relative size of all of this. Assume without loss of generality that $P_{2x} > P_{1x}$ so the denominator of $(10)$ is $P_{2x} - P_{1x}$. Clearly $P_{2x} > Q_{2x}$ so the first term in the numerator is $2(P_{2x} - Q_{2x})$. Finally, \begin{align*} P &< Q_{2x} \\ (P_{1x} + P_{2x})/2 &< Q_{2x} \\ P_{1x} + P_{2x} - 2Q_{2x} &< 0 \end{align*} so the second term in the numerator is $2Q_{2x} - P_{1x} - P_{2x}.$ Putting this all together: \begin{align*} \frac{|P Q_{1}|}{|P P_{1}|} + \frac{|P Q_{2}|}{|P P_{2}|} &= \frac{2(P_{2x} - Q_{2x})+ (2Q_{2x} - P_{1x} - P_{2x})}{P_{2x} - P_{1x}} \\ &= \frac{(2P_{2x} - P_{2x}) - P_{1x} + (2Q_{2x} - 2Q_{2x})}{P_{2x} - P_{1x}} \\ &= \frac{P_{2x} - P_{1x}}{P_{2x} - P_{1x}} \\ &= 1 \end{align*}

How to solve a linear system in matrix form using Laplace transform? How to solve this linear system using Laplace transform? $$\mathbf X'(t)=\left[\begin{array}{r,r,r}-3&0&2\\1&-1&0\\-2&-1&0\end{array}\right]\mathbf X(t); ~~~~~~~~\mathbf X(0)=\left[\begin{array}{r}4\\-1\\2\end{array}\right]$$ I am struggling with this problem. I tried by write it as $$\begin{cases}x_1' &= -3x_1+2x_3,\quad x_1(0)=4\\ x_2'&= -1x_1+1x_2, \quad x_2(0)=-1\\ x_3'&= -2x_1 -1x_2, \quad x_3(0)=2 \end{cases}$$ Then Laplace transform both sides but I have hit a dead end. Is there any better way to solving this problem? I would be grateful for help. I did only simple problems so far. This is very complicated for me.

We are given: $$X'(t) = \begin{bmatrix} -3 & 0 & 2 \\ 1 & -1 & 0\\ -2 & -1 & 0\end{bmatrix} \begin{bmatrix} x(t) \\ y(t)\\ z(t)\end{bmatrix}, ~~ X(0) = \begin{bmatrix} 4 \\ -1\\ 2\end{bmatrix}$$ We can write this as: $$\tag 1 \begin{align} x' &= -3x + 2z \\ y' &= x-y \\ z' &= -2x - y \end{align}$$ Taking the Laplace transform of $(1)$ yields: $$\begin{align} s x(s) - x(0) &= -3 x(s) + 2 z(s) \\ s y(s) - y(0) &= x(s) - y(s) \\ s z(s) - z(0) &= -2 x(s) - y(s) \end{align}$$ This reduces to the system: $$\begin{bmatrix} s+3 & 0 & -2 \\ -1 & s+1 & 0\\ 2 & 1 & s\end{bmatrix} \begin{bmatrix} x(s) \\ y(s)\\ z(s)\end{bmatrix} = \begin{bmatrix} -4 \\ 1\\ -2 \end{bmatrix}$$ All that is needed is to solve for $x(s),y(s),z(s)$ and then find the inverse Laplace Transform. You could have also used many methods to solve this system, including eigenvalues/eigenvectors, matrix exponential, etc. Update If we find the inverse of the matrix on the left, we get: $$\begin{bmatrix} \frac{s^2+s}{s^3+4 s^2+7 s+6} & -\frac{2}{s^3+4 s^2+7 s+6} & \frac{2 s+2}{s^3+4 s^2+7 s+6} \\ \frac{s}{s^3+4 s^2+7 s+6} & \frac{s^2+3 s+4}{s^3+4 s^2+7 s+6} & \frac{2}{s^3+4 s^2+7 s+6} \\ \frac{-2 s-3}{s^3+4 s^2+7 s+6} & \frac{-s-3}{s^3+4 s^2+7 s+6} & \frac{s^2+4 s+3}{s^3+4 s^2+7 s+6} \\ \end{bmatrix}$$ Multiplying that by the column vector on the right yields: $$\begin{bmatrix} x(s) \\ y(s)\\ z(s)\end{bmatrix} = \begin{bmatrix} -\frac{2 (2 s+2)}{s^3+4 s^2+7 s+6}-\frac{4 \left(s^2+s\right)}{s^3+4 s^2+7 s+6}-\frac{2}{s^3+4 s^2+7 s+6} \\ -\frac{4 s}{s^3+4 s^2+7 s+6}+\frac{s^2+3 s+4}{s^3+4 s^2+7 s+6}-\frac{4}{s^3+4 s^2+7 s+6} \\ \frac{-s-3}{s^3+4 s^2+7 s+6}-\frac{4 (-2 s-3)}{s^3+4 s^2+7 s+6}-\frac{2 \left(s^2+4 s+3\right)}{s^3+4 s^2+7 s+6} \\ \end{bmatrix}$$ We now want to find the inverse Laplace Transform of each component, for example: $$x(t) = \mathscr{L^{-1}} \left(-\frac{2 (2 s+2)}{s^3+4 s^2+7 s+6}-\frac{4 \left(s^2+s\right)}{s^3+4 s^2+7 s+6}-\frac{2}{s^3+4 s^2+7 s+6} \right)$$ Thus, $$x(t) = -e^{-2 t} \left(-\sqrt{2} e^t \sin \left(\sqrt{2} t\right)+2 e^t \cos \left(\sqrt{2} t\right)+2\right) $$ Hopefully, this gives you enough to do the other two, else you will not learn.

Evaluating $\int_{0}^{\pi/2}\frac{x\sin x\cos x\;dx}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}$ How to evaluate the following integral $$\int_{0}^{\pi/2}\frac{x\sin x\cos x}{(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}dx$$ For integrating I took $\cos^{2}x$ outside and applied integration by parts. Given answer is $\dfrac{\pi}{4ab^{2}(a+b)}$. But I am not getting the answer.

let $x=\dfrac{t}{2}$, we have $$I=\dfrac{1}{2}\int_{0}^{\pi}\dfrac{\dfrac{t}{2}\sin{\dfrac{t}{2}}\cos{\dfrac{t}{2}}}{\left(a^2\sin^2{\dfrac{t}{2}}+b^2\cos^2{\dfrac{t}{2}}\right)^2}dt=\dfrac{1}{2}\int_{0}^{\pi}\dfrac{t\sin{t}}{[(a^2+b^2)+(a^2-b^2)\cos{t}]^2}dt$$ So \begin{align*}I&=-\dfrac{1}{2(a^2-b^2)}\int_{0}^{\pi}t\;d\left(\dfrac{1}{(a^2+b^2)+(a^2-b^2)\cos{t}}\right)\\ &=-\dfrac{1}{2(a^2-b^2)}\dfrac{1}{(a^2+b^2)+(a^2-b^2)\cos{t}}\Bigg|_{0}^{\pi}+\int_{0}^{\pi}\dfrac{1}{(a^2+b^2)+(a^2-b^2)\cos{t}}dt\\ &=-\dfrac{\pi}{4b^2(a^2-b^2)}+\int_{0}^{\pi}\dfrac{1}{(a^2+b^2)+(a^2-b^2)\cos{t}}dt \end{align*}

Series sum $\sum 1/(n^2+(n+1)^2)$ In an exercise, I caculate the Fourier expansion of $e^x$ over $[0,\pi]$ is $$e^x\sim \frac{e^\pi-1}{\pi}+\frac{2(e^\pi-1)}{\pi}\sum_{n=1}^\infty \frac{\cos 2nx}{4n^2+1}+\frac{4(1-e^\pi)}{\pi}\sum_{n=1}^\infty \frac{n\sin 2nx}{4n^2+1}.$$ From this, it is easy to deduce $$\sum_{n=1}^\infty \frac{1}{4n^2+1}=\frac{\pi}{4}\frac{e^\pi+1}{e^\pi-1}-\frac{1}{2}.$$ However, I could not find the following sum $$\sum_{n=1}^\infty \frac{1}{(2n-1)^2+1},$$ from which we can calculate the sum $\sum 1/(n^2+1)$.

We can approach such kind of series by considering logarithmic derivatives of Weierstrass products. For instance, from: $$\cosh z = \prod_{n=1}^{+\infty}\left(1+\frac{4z^2}{(2n-1)^2\pi^2}\right)\tag{1}$$ we get: $$\frac{\pi}{2}\tanh\frac{\pi z}{2} = \sum_{n=1}^{+\infty}\frac{2z}{z^2+(2n-1)^2}\tag{2},$$ so, evaluating in $z=1$: $$\sum_{n=1}^{+\infty}\frac{1}{(2n-1)^2+1}=\color{red}{\frac{\pi}{4}\tanh\frac{\pi}{2}}.\tag{3}$$ With the same approach, but starting from the Weierstrass product for $\frac{\sinh z}{z}$, we can compute $\sum_{n\geq 1}\frac{1}{1+n^2}$, too: $$\sum_{n\geq 1}\frac{1}{n^2+1}=\frac{-1+\pi\coth\pi}{2}.\tag{4}$$

Proving $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}>2-\frac{2}{n}$ by induction for $n\geq 1$ I have the following inequality to prove with induction: $$P(n): \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots\frac{1}{\sqrt{n}}>2-\frac{2}{n}, \forall n\in \mathbb{\:N}^*$$ I tried to prove $P(n+1)$: Let $S = \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}\Rightarrow$ $$P(n+1):S>2-\frac{2}{n+1}$$ I got into this point and I think it is all wrong, but I'll write it here also: $$S>\frac{2(n+1) + n\sqrt{n+1}}{n(n+1)}$$ and I don't know what to do next... Could anybody help me, please? I would also like to know if there's any other smarter way of solving this kind of exercises.

Initial comment: Begin by noting that, for all $n\geq 1$, we have that $$ n(\sqrt{n}-2)+2>0\Longleftrightarrow n\sqrt{n}-2n+2>0\Longleftrightarrow \color{red}{\sqrt{n}>2-\frac{2}{n}}.\tag{1} $$ Thus, it suffices for us to prove the proposition $P(n)$ for all $n\geq 1$ where $$ P(n): \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\geq \sqrt{n}.\tag{2} $$ If we can prove $(2)$, then we will have proven $$ \color{blue}{\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}}\color{red}{\geq\sqrt{n}}\color{blue}{> 2-\frac{2}{n}}, $$ as desired. I'm sure you can handle the proof of $(1)$ quite easily. Claim: For $n\geq 1$, let $P(n)$ denote the statement $$ P(n): \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}}\geq \sqrt{n}. $$ Base step: $P(1)$ holds since $1\geq\sqrt{1}$ is true. Before induction step: Consider the following inequality for any $x\geq 1$: $$ \sqrt{x}+\frac{1}{\sqrt{x+1}}>\sqrt{x+1}\tag{3}. $$ Briefly, observe that for $x\geq 1, \sqrt{x(x+1)}>x$; thus, $\sqrt{x(x+1)}+1>x+1$. Dividing by $\sqrt{x+1}$ proves $(3)$. The purpose of $(3)$ is to streamline the calculations below in the inductive step. Inductive step: Fix some $k\geq 1$ and suppose that $P(k)$ is true. Then \begin{align} \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} &\geq \sqrt{k}+\frac{1}{\sqrt{k+1}}\tag{by $P(k)$}\\[1em] &> \sqrt{k+1},\tag{by $(3)$} \end{align} which shows that $S(k+1)$ follows. This concludes the inductive step. Thus, for all $n\geq 1, P(n)$ is true. $\blacksquare$

In triangle ABC, Find $\tan(A)$. In triangle ABC, if $(b+c)^2=a^2+16\triangle$, then find $\tan(A)$ . Where $\triangle$ is the area and a, b , c are the sides of the triangle. $\implies b^2+c^2-a^2=16\triangle-2bc$ In triangle ABC, $\sin(A)=\frac{2\triangle}{bc}$, and $\cos(A)=\frac{b^2+c^2-a^2}{2bc}$, $\implies \tan(A)=\frac{4\triangle}{b^2+c^2-a^2}$ $\implies \tan(A)=\frac{4\triangle}{16\triangle-2bc}$. But the answer is in the form, $\frac{x}{y}$ where $x$ and $y$ are integers. Any help is appreciated. Thanks in advance.

HINT: $$16\triangle=(b+c+a)(b+c-a)$$ $$\iff16rs=2s(b+c-a)$$ $$8r=b+c-a$$ Using this and $a=2R\sin A$ etc., $$8\cdot4R\prod\sin\dfrac A2=2R\cdot4\cos\dfrac A2\sin\dfrac B2\sin\dfrac C2$$ $$\implies4\sin\dfrac A2=\cos\dfrac A2$$ as $0<B,C<\pi,\sin\dfrac B2\sin\dfrac C2\ne0$

Inverse Laplace Transformation I have a question about laplace transformation. $\frac{8s+4}{s^2+23}$ I tried to split them. $\frac{8s}{s^2+23}$ is the image of a cosine and $\frac{4}{s^2+23}$ is the image of a sine. Here is what I did : $\frac{8s}{s^2+(\sqrt{23})^2}$ is the image of $8\cos(\sqrt{23}t)$ and $\frac{4}{s^2+(\sqrt{23})^2}$ is the image of $\frac4{\sqrt{23}}\sin(\sqrt{23}t)$ But according to homework, I am wrong. But I am sure I have the correct answer, there is no mistake.

If you don't mind some Residue theory, we can check use that to check your solution. \begin{align} \mathcal{L}^{-1}\biggl\{\frac{8s+4}{s^2+23}\biggr\}&=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{8s+4}{s^2+23}e^{st}ds\\ &=\sum\text{Res} \end{align} The poles in the $s$-plane occur at $s=\pm i\sqrt{23}$ both of order one. Then we have \begin{align} \mathcal{L}^{-1}\biggl\{\frac{8s+4}{s^2+23}\biggr\}&= \lim_{s\to i\sqrt{23}}(s-i\sqrt{23})\frac{8s+4}{s^2+23}e^{st}+\lim_{s\to -i\sqrt{23}}(s+i\sqrt{23})\frac{8s+4}{s^2+23}e^{st}\\ &= \frac{8i\sqrt{23}+4}{2i\sqrt{23}}e^{it\sqrt{23}}+\frac{-8i\sqrt{23}+4}{-2i\sqrt{23}}e^{-it\sqrt{23}}\\ &=4e^{it\sqrt{23}}+4e^{-it\sqrt{23}}+\frac{2}{i\sqrt{23}}e^{it\sqrt{23}}-\frac{2}{i\sqrt{23}}e^{-it\sqrt{23}}\\ &=8\cos(t\sqrt{23})+\frac{4}{\sqrt{23}}\sin(t\sqrt{23}) \end{align} Thus, the answer agrees with your by tables.

Evaluating $\int_0^{2} \frac{dx}{\sqrt[3]{2x^2-x^3}}$ How to calculate this integral? $$\int_0^{2} \frac{dx}{\sqrt[3]{2x^2-x^3}}$$ I suppose that it should be parted like this: $$\int_0^{1} \frac{dx}{\sqrt[3]{2x^2-x^3}} + \int_1^{2} \frac{dx}{\sqrt[3]{2x^2-x^3}}$$ but I have no idea how to calculate these two. Thank you for help in advance.

$\sqrt[3]{2x^2-x^3} = x\sqrt[3]{\dfrac{2}{x}-1} \to u = \sqrt[3]{\dfrac{2}{x} - 1} \to u^3 = \dfrac{2}{x} - 1 \to x = \dfrac{2}{u^3+1}$. Can you take it from here?

Evaluate this infinite product involving $a_k$ Let $a_0 = 5/2$ and $a_k = a_{k-1}^{2} - 2$ for $k \ge 1$ Compute: $$\prod_{k=0}^{\infty} 1 - \frac{1}{a_k}$$ Off the bat, we can seperate $a_0$ $$= -3/2 \cdot \prod_{k=1}^{\infty} 1 - \frac{1}{a_k}$$ Lets see: $a_0 = 5/2$ $a_1 = 25/4 - 2 = 17/4$ $a_2 = 289/16 - 2 = 257/16$ $$P = (-3/2)\cdot(-13/4)\cdot(-241/16).....$$ Lets compute the first three for $P_3$ $P_1 = -3/2 = (-192/128)$ $P_2 = (-3/2)(-13/4) = (39/8) = (624/128)$ $P_3 = (39/8)(-241/16) = (-9399/128)$ But it is difficult to find a pattern for $P_k$ Help? Thanks =)

$\textbf{Hint:}$ show this by induction on $n\ge0$ $$a_n=2^{2^n}+2^{-2^n},\quad n=0,1,2\dots$$ and notice that $a_k+1=a_{k-1}^2-1=(a_{k-1}-1)(a_{k-1}+1),$ $$1-\frac{1}{a_k}=\frac{a_k-1}{a_k}=\frac{a_{k+1}+1}{a_k+1}\cdot\frac{1}{a_k}$$

3D coordinates of circle center given three point on the circle. Given the three coordinates $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, $(x_3, y_3, z_3)$ defining a circle in 3D space, how to find the coordinates of the center of the circle $(x_0, y_0, z_0)$?

In this formula the center $O$ of a circumscribed circle of $\triangle ABC$ is expressed as a convex combination of its vertices in terms of coordinates $A,B,C$ and corresponding side lengths $a,b,c$, suitable for both 2d and 3d: \begin{align} O&= A\cdot \frac{a^2\,(b^2+c^2-a^2)}{((b+c)^2-a^2)(a^2-(b-c)^2)} \\ &+B\cdot \frac{b^2\,(a^2+c^2-b^2)}{((a+c)^2-b^2)(b^2-(a-c)^2)} \\ &+C\cdot \frac{c^2\,(b^2+a^2-c^2)}{((b+a)^2-c^2)(c^2-(b-a)^2)} . \end{align}

Solve the equation $2x^2+5y^2+6xy-2x-4y+1=0$ in real numbers Solve the equation $2x^2+5y^2+6xy-2x-4y+1=0$ The problem does not say it but I think solutions should be from $\mathbb{R}$. I tried to express the left sum as a sum of squares but that does not work out. Any suggestions?

You can solve for $x$: $(2)x^2+(6y-2)x+(5y^2-4y+1)=0\implies$ $x_{1,2}=\frac{-(6y-2)\pm\sqrt{(6y-2)^2-4\cdot2\cdot(5y^2-4y+1)}}{2\cdot2}=\frac{-6y+2\pm\sqrt{-4y^2+8y-4}}{4}=\frac{-6y+2\pm\sqrt{-4(y-1)^2}}{4}=\frac{-6y+2\pm2i(y-1)}{4}$ Then the only real solution is with $y=1$, hence $x=-1$.

No. of different real values of $x$ which satisfy $17^x+9^{x^2} = 23^x+3^{x^2}.$ Number of different real values of $x$ which satisfy $17^x+9^{x^2} = 23^x+3^{x^2}.$ $\bf{My\; Try::}$Using Hit and trial $x=0$ and $x=1$ are solution of above exponential equation. Now we will calculate any other solution exists or not. If $x\geq 2\;,$ Then $17^x+9^{x^2}>9^{x^2} = (6+3)^{x^2}>6^{x^2}+3^{x^2} = (6^x)^x+3^{x^2}>23^x+3^{x^2}\;,$ bcz $(6^x>23)\; \forall x\geq 2.$ So no solution in $x\in \left[2,\infty\right)$ Now i did not understand how can i calculate in $x<0$ and $0<x<1$. Help me, Thanks

Using derivatives, is studying functions $ f, g: R \rightarrow R, f(x)= 9^{x^2}-3^{x^2}, g(x)=23^x-17^x$ and is found: * *$f$ has a minimum point in the interval $(0, 1)$ and limits to $+\infty$,$-\infty$ are equal with $+\infty$; *$g$ has a negative minimum point and limited to $-\infty$ is $0$ and to $+\infty$ is $+\infty$. For these reasons and noting that $f$ grows faster than $g$ infinite, it follows that graphs their only two points in common. Conclusion: The equation has exactly two real roots.

Prove $4^k - 1$ is divisible by $3$ for $k = 1, 2, 3, \dots$ For example: $$\begin{align} 4^{1} - 1 \mod 3 &= \\ 4 -1 \mod 3 &= \\ 3 \mod 3 &= \\3*1 \mod 3 &=0 \\ \\ 4^{2} - 1 \mod 3 &= \\ 16 -1 \mod 3 &= \\ 15 \mod 3 &= \\3*5 \mod 3 &= 0 \\ \\ 4^{3} - 1 \mod 3 &= \\ 64 -1 \mod 3 &= \\ 21 \mod 3 &= \\3*7 \mod 3 &= 0\end{align} $$ Define $x = \frac{4^k - 1}{3}$. So far I have: $$k_1 \to 1 \Longrightarrow x_1 \to 1 \\ k_2 \to 2 \Longrightarrow x_2 \to 5 \\ k_3 \to 3 \Longrightarrow x_3 \to 21 \\ k_4 \to 4 \Longrightarrow x_4 \to 85$$ But then it's evident that $$4^{k_n} = x_{n+1} - x_n$$ I don't know if this helps, these are ideas floating in my head.

Hint: $$4 \equiv 1 \mod 3 \Rightarrow 4^n \equiv 1^n \mod 3 \Rightarrow 4^n \equiv 1 \mod 3$$

Power series for the rational function $(1+x)^3/(1-x)^3$ Show that $$\dfrac{(1+x)^3}{(1-x)^3} =1 + \displaystyle\sum_{n=1}^{\infty} (4n^2+2)x^n$$ I tried with the partial frationaising the expression that gives me $\dfrac{-6}{(x-1)} - \dfrac{12}{(x-1)^2} - \dfrac{8}{(x-1)^3} -1$ how to proceed further on this having doubt with square and third power term in denominator.

The most simple way to prove your identity, IMHO, is to multiply both sides by $(1-x)^3$. This leads to: $$ 1+3x+3x^2+x^3\stackrel{?}{=}(1-3x+3x^2+x^3)\left(1+\sum_{n\geq 1}(4n^2+2)\,x^n\right).\tag{1}$$ If we set $a_n=(4n^2+2)$, for any $n\geq 4$ the coefficient of $x^n$ in the RHS is given by $a_n-3a_{n-1}+3a_{n-2}+a_{n-3}$ that is zero, since we are applying three times the backward difference operator to a polynomial in $n$ having degree two. So we just have to check that the first four coefficients, $[x^0],[x^1],[x^2],[x^3]$, match.

AlgebraII factoring polynomials equation: $2x^2 - 11x - 6$ Using the quadratic formula, I have found the zeros: $x_1 = 6, x_2 = -\frac{1}{2}$ Plug the zeros in: $2x^2 + \frac{1}{2}x - 6x - 6$ This is where I get lost. I factor $-6x - 6$ to: $-6(x + 1)$, but the answer says otherwise. I am also having trouble factoring the left side. Could someone please explain to me why the answer to the question was: $(x - 6)(2x + 1)$. How does $-\frac{1}{2}$ become $1$?

When you factor out the equation $2x^2-11x-6$, you get $(x-6)(2x+1)$ (David Peterson did the factoring process). This shows that the functions has two zeros in the graph. Thus, we have to sent $(x-6)(2x+1)$ equal to $0$: $$(x-6)(2x+1)=0.$$ Then we find the zeros: $$(x-6)(2x+1)=0$$ $$x-6=0$$ $$\boxed{x=6}$$ $$2x+1=0$$ $$2x=-1$$ $$\boxed{x=-\frac{1}{2}}.$$ Therefore, we can clearly see that by factoring and sovling for $x$, we get $$\boxed{x=6,-\frac{1}{2}}.$$

prove that $a^2 b^2 (a^2 + b^2 - 2) \ge (a + b)(ab - 1)$ Good morning help me to show the following inequality for all $a$, $b$ two positive real numbers $$a^2 b^2 (a^2 + b^2 - 2) \ge (a + b)(ab - 1)$$ thanks you

Let $a+b=2u$ and $ab=v^2$, where $v>0$. Hence, we need to prove that $2v^4u^2-(v^2-1)u-v^6-v^4\geq0$, for which it's enough to prove that $u\geq\frac{v^2-1+\sqrt{(v^2-1)^2+8v^4(v^4+v^6)}}{4v^4}$ or $(4v^5-v^2+1)^2\geq(v^2-1)^2+8v^4(v^4+v^6)$ because $u\geq v$, or $(v-1)^2(v+1)(v^2+v+1)\geq0$. Done!

Simplification a trigonometric equation $$16 \cos \frac{2 \pi}{15} \cos\frac{4 \pi}{15} \cos\frac{8 \pi}{15} \cos\frac{14 \pi}{15}$$ $$=4\times 2 \cos \frac{2 \pi}{15} \cos\frac{4 \pi}{15} \times2 \cos\frac{8 \pi}{15} \cos\frac{14 \pi}{15}$$ I am intending in this way and then tried to apply the formula, $2\cos A \cos B$ but i think I might not get the answer. What to do now? the result will be 1.

Using the identity, $\cos\theta\cos2\theta\cos2^2\theta\cdots\cos2^{n-1}\theta=\dfrac{\sin2^n\theta}{2^n\sin\theta}$ By putting $n=4$ and $\theta=\dfrac{\pi}{15}$ you will get, $- \cos\dfrac{\pi}{15}\cos \dfrac{2 \pi}{15} \cos\dfrac{4 \pi}{15} \cos\dfrac{8 \pi}{15}=-\dfrac{\sin2^4\dfrac{\pi}{15}}{2^4\sin\dfrac{\pi}{15}} $ Multiply both sides by $16$ and you get the required identity , $\implies - 16\cos\dfrac{\pi}{15}\cos \dfrac{2 \pi}{15} \cos\dfrac{4 \pi}{15} \cos\dfrac{8 \pi}{15}=-16\dfrac{\sin16\dfrac{\pi}{15}}{16\sin\dfrac{\pi}{15}} $ $\implies -\dfrac{\sin\dfrac{16\pi}{15}}{\sin\dfrac{\pi}{15}}= \dfrac{\sin\dfrac{\pi}{15}}{\sin\dfrac{\pi}{15}}=1$

Irrational number not ocurring in the period of rational numbers Write each rational number from $(0,1]$ as a fraction $a/b$ with $\gcd(a,b)=1$, and cover $a/b $ with the interval $$ \left[\frac ab-\frac 1{4b^2}, \frac ab + \frac 1{4b^2}\right]. $$ Prove that the number $\frac 1{\sqrt{2}}$ is not covered. What I did was the following:- We define the given period as P. Assume that $\frac 1{\sqrt{2}}$ is not present in P. Hence set $\frac ab=k=\frac 1{\sqrt{2}}+x$. Therefore we have to prove that $$ \begin{align}x\gt\frac 1{4b^2}&\implies x \gt \frac {k^2}{4a^2}\\ &\implies4a^2x\gt\left(\frac 1{\sqrt{2}}+x\right)^2\\ &\implies4a^2x\gt x^2+x\sqrt{2}+\frac 12\\ &\implies x^2-x(4a^2-\sqrt{2})+\frac 12\lt 0\end{align} $$ Hence $$ \begin{align} D\lt 0 & \implies (4a^2-\sqrt{2})^2-4\cdot\frac 12 \lt 0 \\ & \implies(4a^2-\sqrt{2})^2-\sqrt{2}^2\lt 0 \\ & \implies(4a^2-2\sqrt{2})*4a^2\lt 0\\ &\implies4a^2\lt2\sqrt{2}\\ &\implies a^2\lt \frac 1{\sqrt{2}}\end{align} $$ This is impossible since $a$ is a natural number???? So, what do I do?

I got a good solution from one of my teachers...$$\left|\frac ab-\frac 1{\sqrt2}\right|\left(\frac ab+\frac 1{\sqrt2}\right)=\left|\frac {a^2}{b^2}-\frac 12\right|=\frac {|2a^2-b^2|}{2b^2}\gt\frac 1{2b^2}$$ Also we know that $\frac ab+\frac 1{\sqrt2}\lt2=>\left|\frac ab-\frac 1{\sqrt2}\right|\gt\frac 1{4b^2}$. Hence proved.

Find the value of : $\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$ I need to calculate the limit of the function below: $$\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$$ I tried multiplying by the conjugate, substituting $x=\frac{1}{t^4}$, and both led to nothing.

Multiply both numerator and denominator by $\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}$ You will get $$\dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}$$ Divide both numerator and denominator by $\sqrt{x}$ $$\dfrac{\sqrt{1+\dfrac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{1}{x\sqrt{x}}}}+1}$$ On finding the limit to infinity, you get $$\dfrac{\sqrt{1+0}}{\sqrt{1+0}+1} = \dfrac{1}{2}$$

Functional inequalities Let x,y,z be the lengths of the sides of a triangle, and let$$f(x,y,z)=\left|\frac {x-y}{x+y}+\frac {y-z}{y+z}+\frac {z-x}{z+x}\right|.$$ Find the upper limit of $f(x,y,z)$. I simply used the fact that $|x-y|\le z$ and the other 3 to prove that $f(x,y,z)\le \frac 18=0.125$. But the answer given is in terms of irrational numbers. Actually it is $f(x,y,z)\le \frac {8\sqrt2-5\sqrt5}{3}=0.04446.$ How do irrational numbers come into the picture...and well could you give the solution as well?

Solution:let $x\ge y\ge z$, First Note $$I=\dfrac{x-y}{x+y}+\dfrac{y-z}{y+z}+\dfrac{z-x}{z+x}=\dfrac{(x-y)(x-z)(y-z)}{(x+y)(y+z)(x+z)}$$ then let $$x=c+b,y=c+a,z=a+b,c\ge b\ge a>0$$ so $$I=\dfrac{(c-b)(c-a)(b-a)}{(2a+b+c)(2b+a+c)(2c+a+b)}<\dfrac{(c-b)cb}{(b+c)(2b+c)(2c+b)}=\dfrac{1}{F}$$ then we only find $F$ minimum let $\dfrac{c}{b}=k\ge 1$,then $$F=\dfrac{(1+k)(2+k)(1+2k)}{k(k-1)},k>1$$ since $$F'_{k}=0\Longrightarrow \dfrac{2(k^4-2k^3-7k^2-2k+1)}{(k-1)^2k^2}=0$$ since $k>1$,this equation have only one roots,so seewolf $$k=\dfrac{1}{2}(1+\sqrt{10}+\sqrt{7+2\sqrt{10}})=\dfrac{1}{2}(1+\sqrt{2}+\sqrt{5}+\sqrt{10})$$ and $$F(k)\ge F\left(\dfrac{1}{2}(1+\sqrt{2}+\sqrt{5}+\sqrt{10})\right)=\left(\dfrac{8\sqrt{2}-5\sqrt{3}}{3}\right)^{-1}$$ so $$\left|\dfrac{x-y}{x+y}+\dfrac{y-z}{y+z}+\dfrac{z-x}{z+x}\right|\le \dfrac{8\sqrt{2}-5\sqrt{3}}{3}$$

Recurrence Relation Involving the gamma Function I'm having some doubts about my approach to the following problem. I am given that the function $k(z)$ is defined such that, $$k(z)=\Gamma\left(\frac{1}{2}+z\right)\Gamma\left(\frac{1}{2}-z\right)\cos{\pi z}$$ I'm required to find the recurrence relation linking $k(z+1)$ and $k(z)$ and to then evaluate $k(z)$ for one specific integer value and thus find $k(z)$ for any real, integer value. My attempt was as follows. Note that $\Gamma(s+1)=s\Gamma(s)$ and so \begin{align*} k(z+1)&=\Gamma\left(\frac{1}{2}+z+1\right)\Gamma\left(\frac{1}{2}-z+1\right)\cos{(\pi z + \pi)} \\ &=\left(\frac{1}{2}+z\right)\Gamma\left(\frac{1}{2}+z\right)\left(\frac{1}{2}-z\right)\Gamma\left(\frac{1}{2}-z\right)\cos{(\pi z + \pi)} \end{align*} Then since $\cos{(\pi z + \pi)}=-\cos{\pi z}$ we have that $$k(z+1)=\left(z^2-\frac{1}{4}\right)k(z)$$ If we then consider the case $z=0$, we have $$k(0)=\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)=\pi$$ From this we see that \begin{align*} &k(1)=-\frac{1}{4}\pi\\ &k(2)=-\frac{3}{16}\pi\\ &k(3)=-\frac{45}{64}\pi\\ &\vdots \end{align*} I can't spot any pattern here other than the $4^z$ in the denominator which is making me think i've done something wrong, maybe in my use of $\Gamma(s+1)=s\Gamma(s)$? Any advice would be really appreciated.

You made a sign error. \begin{align} \frac{k(n+1)}{k(n)} &=\frac{\Gamma\left(\frac{1}{2}+n+1\right)\Gamma\left(\frac{1}{2}-n\color{red}{-1}\right)(-\cos(\pi n))}{\Gamma\left(\frac{1}{2}+n\right)\Gamma\left(\frac{1}{2}-n\right)\cos(\pi n)}\\ &=\frac{\left(\frac{1}{2}+n\right)\Gamma\left(\frac{1}{2}+n\right)\frac{\Gamma\left(\frac{1}{2}-n\right)}{\left(\frac{1}{2}-n-1\right)}(-1)}{\Gamma\left(\frac{1}{2}+n\right)\Gamma\left(\frac{1}{2}-n\right)}\\ &=-\frac{\frac{1}{2}+n}{-\frac{1}{2}-n}\\ &=1 \end{align} Hence $k(n)=k(0)=\pi$.

Is $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$? Plotting both functions $\ln(1+\frac{1}{x-1})$ and $\frac{1}{x}$ in $[2,\infty)$ gives the impression that $\ln(1+\frac{1}{x-1}) \ge \frac{1}{x}$ for all $x \ge 2$. Is it possible to prove it?

hint : $$f(x)=ln(1+\frac{1}{x-1})-\frac{1}{x}\\x≥2\\ f'<0$$f(x) is decreasing function ,but f(x) is above the x axis $$\\f(2)>0,f(\infty)>0\\f(x)>0\\so\\ln(1+\frac{1}{x-1})-\frac{1}{x}≥0\\ln(1+\frac{1}{x-1})≥\frac{1}{x}$$

How find the maximum possible length of OC, where ABCD is a square, and AD is the chord of the circle? Given a circle $o(O(0,0), r=1)$. How to find the maximum possible length of $OC$, where $ABCD$ is a square, and $AD$ is the chord of the circle? I have no idea how to do this, can this be proved with simple geometry?

With the diagram as labeled, we see that $$\begin{align} |\overline{OC}|^2 &= \cos^2\theta + ( \sin\theta+2\cos\theta )^2 \\ &= \cos^2\theta + \sin^2\theta + 4 \cos\theta\sin\theta + 4 \cos^2\theta \\ &= 3+2 \sin 2\theta + 2 \cos 2\theta \\ &= 3+2\sqrt{2}\left( \sin 2\theta \cos45^\circ + \cos 2\theta \sin45^\circ \right) \\ &= 3+2\sqrt{2}\sin(2\theta+45^\circ) \end{align}$$ This value is clearly maximized when $\sin(2\theta+45^\circ) = 1$ (which happens for $\theta = 22.5^\circ$), so that $$|\overline{OC}|^2 = 3 + 2\sqrt{2} = ( 1 + \sqrt{2} )^2 \quad\to\quad |\overline{OC}| = 1 + \sqrt{2}$$

Prove that $ \sum_{n=1}^\infty \ln\big(n\sin \frac{1}{n}\big)$ converges. Prove that $\displaystyle \sum_{n=1}^\infty \;\ln\left(n\sin\frac{1}{n}\right)$ converges. My Work: $$\left|\ln \left(n \sin \frac{1}{n}\right)\right| \leq\left|\ln \left(n \sin \frac{1}{n^{2}}\right)\right| \leq\left|\ln \left(\sin \frac{1}{n^{2}}\right)\right|$$ I was going to use comparison test. But now stuck. Please give me a hint.

Let's amply use Taylor series to give the leading orders. Note that $\sin\left( \frac{1}{n} \right) \approx \frac{1}{n} - \frac{1}{6n^3}$. So $n \sin \left( \frac{1}{n} \right) \approx 1 - \frac{1}{6n^2}$. Note also that $\ln(1 - x) \approx x + x^2 + \dots$, so that $$\ln\left( n \sin \frac{1}{n} \right) \approx \frac{1}{6n^2} + \frac{1}{36n^4} + \dots$$ This means that you are wondering about $$ \sum_{n \geq 1} \sum_{j \geq 1} \frac{1}{(6n^2)^j} \ll \sum_{n \geq 1} \frac{1}{n^2},$$ which converges.

Prove: If $a^2+b^2=1$ and $c^2+d^2=1$, then $ac+bd\le1$ Prove: If $a^2+b^2=1$ and $c^2+d^2=1$, then $ac+bd\le1$ I seem to struggle with this simple proof. All I managed to find is that ac+bd=-4 (which might not even be correct).

1st Method $\begin{align}\left(a^2+b^2\right)\left(c^2+d^2\right)=1& \implies (ac+bd)^2+(ad-bc)^2=1\\&\implies (ac+bd)^2\le1\end{align}$ 2nd Method $\begin{align}\left(a^2+b^2\right)+\left(c^2+d^2\right)=2& \implies \left(a^2+c^2\right)+\left(b^2+d^2\right)=2\\&\implies 2(ac+bd)\le 2\qquad \text{(by A.M.- G.M. Inequality)}\\&\implies (ac+bd)\le 1\end{align}$

How do I integrate: $\int\sqrt{\frac{x-3}{2-x}} dx$? I need to solve: $$\int\sqrt{\frac{x-3}{2-x}}~{\rm d}x$$ What I did is: Substitute: $x=2\cos^2 \theta + 3\sin^2 \theta$. Now: $$\begin{align} x &= 2 - 2\sin^2 \theta + 3 \sin^2 \theta \\ x &= 2+ \sin^2 \theta \\ \sin \theta &= \sqrt{x-2} \\ \theta &=\sin^{-1}\sqrt{x-2} \end{align}$$ and, $ cos \theta = \sqrt{(3-x)} $ $ \theta=\cos^{-1}\sqrt{(3-x)}$ The integral becomes: $$\begin{align} &= \int{\sqrt[]{\frac{2\cos^2 \theta + 3\sin^2 \theta-3}{2-2\cos^2 \theta - 3\sin^2 \theta}} ~~(2 \cos \theta\sin\theta)}~{\rm d}{\theta}\\ % &= \int{\sqrt[]{\frac{2\cos^2 \theta + 3(\sin^2 \theta-1)}{2(1-\cos^2 \theta) - 3\sin^2 \theta}}~~(2 \cos \theta\sin\theta)}~{\rm d}{\theta} \\ % &= \int\sqrt[]{\frac{2\cos^2 \theta - 3\cos^2 \theta}{2\sin^2 \theta - 3\sin^2 \theta}}~~(2 \cos \theta\sin\theta) ~{\rm d}\theta \\ % &= \int\sqrt[]{\frac{-\cos^2 \theta }{- \sin^2 \theta}}~~(2 \cos \theta\sin\theta) ~{\rm d}\theta \\ % &= \int \frac{\cos \theta}{\sin\theta}~~(2 \cos \theta\sin\theta)~{\rm d}\theta \\ % &= \int 2\cos^2 \theta~{\rm d}\theta \\ % &= \int (1- \sin 2\theta)~{\rm d}\theta \\ % &= \theta - \frac {\cos 2\theta}{2} + c \\ % &= \sin^{-1}\sqrt{x-2} - \frac {\cos 2(\sin^{-1}\sqrt{x-2})}{2} + c \end{align}$$ But, The right answer is : $$\sqrt{\frac{3-x}{x-2}} - \sin^{-1}\sqrt{3-x} + c $$ Where am I doing it wrong? How do I get it to the correct answer?? UPDATE: I am so sorry I wrote: = $\int 2\cos^2 \theta .d\theta$ = $\int (1- \sin 2\theta) .d\theta$ It should be: = $\int 2\cos^2 \theta .d\theta$ = $\int (1+ \cos2\theta) .d\theta$ = $ \theta + \frac{\sin 2\theta}{2} +c$ What do I do next?? UPDATE 2: = $ \theta + \sin \theta \cos\theta +c$ = $ \theta + \sin \sin^{-1}\sqrt{(x-2)}. \cos\cos^{-1}\sqrt{(3-x)}+c$ = $ \sin^{-1}\sqrt{(x-2)}+ \sqrt{(x-2)}.\sqrt{(3-x)}+c$ Is this the right answer or I have done something wrong?

$$I=\int\sqrt{\frac{x-3}{2-x}}~{\rm d}x$$ Integrating Let $x=2\cos^2t+3\sin^2t$, $dx=\sin2tdt$ $$I=\int\sqrt{\frac{-\cos^2t}{-\sin^2t}}\sin2tdt=\int2\cos^2tdt=\int(1+\cos2t)dt=t+\frac12\sin2t+c\\I=\underbrace{\cos^{-1}\sqrt{3-x}}_{\pi/2-\sin^{-1}\sqrt{3-x}}+\sqrt{x-2}\sqrt{3-x}+c\\I=\underbrace{\sqrt{x-2}\sqrt{3-x}}_{\sqrt{5x-x^2-6}}-\sin^{-1}{\sqrt{3-x}}+c'$$ Differentiating back $$I'=\frac1{2\sqrt{(x-2)(3-x)}}\cdot(5-2x)-\underbrace{\frac1{\sqrt{1-(\sqrt{3-x})^2}}}_{\sqrt{x-2}}\cdot\frac1{2\sqrt{3-x}}(-1)\\I'=\frac{2(3-x)}{2\sqrt{(x-2)(3-x)}}=\sqrt{\frac{3-x}{x-2}}=\sqrt{\frac{x-3}{2-x}}$$

Limit involving square roots, more than two "rooted" terms The limit is $$\lim_{x\to\infty} \left(\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\right)$$ which has a value of $\dfrac{27}{4}$. Normally, I would know how to approach a limit of the form $$\lim_{x\to\infty}\left( \sqrt{a_1x^2+b_1x+c_1}\pm\sqrt{a_2x^2+b_2x+c_2}\right)$$ (provided it exists) by using the expression's conjugate, but this problem has me stumped. I've considered using the conjugate $$\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}-\sqrt{x^2+7x+5}$$ and a term like this one, $$\sqrt{x^2+5x-2}+\sqrt{4x^2-3x+7}-\sqrt{x^2+7x+5}$$ but that didn't seem to help simplify anything. Edit: I stumbled across something at the last second that lets me use the conjugate approach. The expression can be rewritten as $$\sqrt{x^2+5x-2}-\sqrt{4x^2-3x+7}+\sqrt{x^2+7x+5}\\ \sqrt{x^2+5x-2}-2\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}+\sqrt{x^2+7x+5}\\ \left(\sqrt{x^2+5x-2}-\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}\right)+\left(\sqrt{x^2+7x+5}-\sqrt{x^2-\frac{3}{4}x+\frac{7}{4}}\right)$$ which approaches $$\frac{23}{8}+\frac{31}{8}=\frac{27}{4}$$

Rewrite it as: $$\left(\sqrt{x^2+5x-2} - \left(x+\frac{5}{2}\right)\right) -\\ \left(\sqrt{4x^2-3x+7}-\left(2x-\frac{3}{4} \right)\right) +\\ \left(\sqrt{x^2+7x+5}-\left(x+\frac{7}{2}\right)\right)+\left(\frac{5}{2}+\frac 34+\frac{7}2\right)$$ Or something like that. Each of the first three terms has limit zero...

How many $4\times 3$ matrices of rank three are there over a finite field of three elements? Let $M$ be the space of all $4\times 3$ matrices with entries in the finite field of three elements. Then the number of matrices of rank three in $M$ is A. $(3^4 - 3)(3^4 - 3^2)(3^4-3^3)$ B. $(3^4 - 1)(3^4 - 2)(3^4 - 3)$ C. $(3^4-1)(3^4-3)(3^4-3^2)$ D. $3^4(3^4 - 1)(3^4 - 2)$ Is there any specific formula to solve this type of problem? Is there any specific formula to calculate the number of matrices? I have no idea how to start this problem. Any guidance please.

Sorry for the previous post.the answer I am getting is c $(3^4-1)(3^4-3)(3^4-3^2)$ For the $1$st column we have $4$ places and $3$ elements to fill it.So $3^4$ choices but the elements can't be all zero.S0 we have $3^4-1$ choices .For the second column we have $3^4$ choices but the second column cant be linearly dependent with the $1$st .So $(3^4-3)$ choices .Similarly the $3$rd column is not linearly dependent to both $1$st and $2$ nd cloumn

Factorise a matrix using the factor theorem Can someone check this please? $$ \begin{vmatrix} x&y&z\\ x^2&y^2&z^2\\ x^3&y^3&z^3\\ \end{vmatrix}$$ $$C_2=C_2-C_1\implies\quad \begin{vmatrix} x&y-x&z\\ x^2&y^2-x^2&z^2\\ x^3&y^3-x^3&z^3\\ \end{vmatrix}$$ $$(y-x) \begin{vmatrix} x&1&z\\ x^2&y+x&z^2\\ x^3&y^2+xy+x^2&z^3\\ \end{vmatrix}$$ $$(y-x)(z-x) \begin{vmatrix} x&1&1\\ x^2&y+x&z+x\\ x^3&y^2+xy+x^2&z^2+xz+x^2\\ \end{vmatrix}$$ $$R_2=R_2-xR_1\implies\quad (y-x)(z-x) \begin{vmatrix} x&1&1\\ 0&y&z\\ x^3&y^2+xy+x^2&z^2+xz+x^2\\ \end{vmatrix}$$ $$R_3=R_3-x^2R_1\implies\quad (y-x)(z-x) \begin{vmatrix} x&1&1\\ 0&y&z\\ 0&y^2+xy&z^2+xz\\ \end{vmatrix}$$ factor $x$$$\implies\quad x(y-x)(z-x) \begin{vmatrix} 1&1&1\\ 0&y&z\\ 0&y^2+xy&z^2+xz\\ \end{vmatrix}$$ $$\implies\quad x(y-x)(z-x)(yz^2-zy^2)$$ $$\implies\quad xyz(y-x)(z-x)(z-y)$$ Also I'd like practical tips on using the factor theorem for these types of questions. My understanding is that the determinant is $f(x,y,z)$ so if we hold $y$ and $z$ constant we could apply it somehow to $f(x)$ alone. I'm not that great spotting difference of squares etc and want a more fail safe alternative. Thanks in advance.

What you did is correct. But there is an easier way. Remember that for polynomial $p(x)$, if $p(a)=0$ then $(x-a)$ is a factor of $p(x)$. Denote the determinant by $\Delta$. It is obviously a polynomial in $x,\ y$ and $z$. Now, note that: * *$x=0\implies \Delta = 0$, so $x$ is a factor of $\Delta$. Same for $y = 0$ and $z=0$. *$x=y\implies \Delta = 0$, so $(x-y)$ is a factor of $\Delta$. Similarly for $y=z$ and $z=x$ Finally note that $\Delta$ is degree $6$ polynomial. So it cannot have more than $6$ linear factors, and we have listed all of them above. Clearly $$\Delta=Cxyz(x-y)(y-z)(z-x)$$ where $C$ is some constant. Taking some values (eg. $x=1,\ y=2,\ z=3$), we get $C=1$.

Given $a+b+c$, Can I calculate $a^2+b^2+c^2$? I want to calculate $a^2 + b^2 + c^2$ when I am given $a+b+c$. It is known that a,b,c are positive integers. Is there any way to find that.

If $a+b+c = k$ then $(a+b+c)^2 = k^2 \implies a^2+b^2+c^2 = k^2 - 2(ab + ac + bc)$, hence you would have to know the value of $ab + ac + bc$.

Laplace tranform of $t^{5/2}$ It is asked to transform $t^{5/2}$. I did $t^{5/2}=t^3\cdot t^{-1/2}$. Then followed the table result $$L\{{t^nf(t)}\}=(-1)^n\cdot\frac{d^n}{ds^n}F(s)$$ However i got $\frac{1}{2} \cdot\sqrt\pi \cdot s^{-7/2}$ instead of $\frac{15}{8} \cdot\sqrt\pi \cdot s^{-7/2}$. Can you help me with the derivations? Thanks

For every real number $r>-1$, we have, $$L(t^r)(s)=\int_0^\infty e^{-st}t^rdt\\\hspace{60mm}=\int_0^\infty e^{-x}(\frac{x}{s})^r\frac{dx}{s}\hspace{10mm}\text{(Putting $x=st$)}\\\hspace{20mm}=\frac{1}{s^{r+1}}\int_0^\infty e^{-x}x^rdt\\\hspace{5mm}=\frac{\Gamma(r+1)}{s^{r+1}}.$$ So for $r=\frac{5}{2}$, we have, $$L(t^{\frac{5}{2}})(s)=\frac{\Gamma(\frac{5}{2}+1)}{s^{\frac{5}{2}+1}}=\Gamma(\frac{7}{2})s^{-\frac{7}{2}}$$ Now we get $$\Gamma(\frac{7}{2})=\Gamma(\frac{5}{2}+1)=\frac{5}{2}\Gamma(\frac{5}{2})=\frac{5}{2}\Gamma(\frac{3}{2}+1)=\frac{5}{2}\frac{3}{2}\Gamma(\frac{1}{2}+1)\\=\frac{5}{2}\frac{3}{2}\frac{1}{2}\Gamma(\frac{1}{2})=\frac{15}{8}\sqrt{\pi}.$$ $$\therefore L( t^{\frac{5}{2}})(s)=\frac{15}{8}\sqrt{\pi}\cdot s^{-\frac{7}{2}}.$$ For $p>0$ the Euler gamma function $\Gamma(p)$ is defined as, $$\Gamma(p)=\int_0^\infty e^{-x}x^{p-1}dx$$ A property of gamma function: $$\Gamma(p+1)=p\cdot\Gamma(p)$$ Also $\Gamma(\frac{1}{2})=\sqrt{\pi}$.

How to define an affine transformation using 2 triangles? I have $2$ triangles ($6$ dots) on a $2D$ plane. The points of the triangles are: a, b, c and x, y, z I would like to find a matrix, using I can transform every point in the 2D space. If I transform a, then the result is x. For b the result is y, and for c the result is z And if there is a given d point, which is halfway from a to b, then after the transformation the result should be between x and y halfway. I've tried to solve it according to NovaDenizen's solution, But the result is wrong. The original triangle: $$ a = \left[ \begin{array}{ccc} -3\\ 0\\ \end{array} \right] $$ $$ b = \left[ \begin{array}{ccc} 0\\ 3\\ \end{array} \right] $$ $$ c = \left[ \begin{array}{ccc} 3\\ 0\\ \end{array} \right] $$ The x, y, z dots: $$ x = \left[ \begin{array}{ccc} 2\\ 3\\ \end{array} \right] $$ $$ y = \left[ \begin{array}{ccc} 3\\ 2\\ \end{array} \right] $$ $$ z = \left[ \begin{array}{ccc} 4\\ 3\\ \end{array} \right] $$ I've created a figure: I tried to transform the (0, 0) point, which is halfway between a and b, but the result was (3, 3.5) instead of (3, 3) The T matrix is: $$\left[ \begin{array}{ccc} 1/3 & 1/6 & 0\\ 0 & -1/2 & 0\\ 3 & 3,5 & 1\\ \end{array} \right]$$

There is a neat formula for your case $$ \vec{P}(p_1; p_2) = (-1) \frac{ \det \begin{pmatrix} 0 & \vec{x} & \vec{y} & \vec{z} \\ p_1 & a_1 & b_1 & c_1 \\ p_2 & a_2 & b_2 & c_2 \\ 1 & 1 & 1 & 1 \\ \end{pmatrix} }{ \det \begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ 1 & 1 & 1 \\ \end{pmatrix} }, $$ where $p_1$ and $p_2$ are coordinates of the point you are mapping. Here's how it works (plugging in initial points) $$ \vec{P}(p_1; p_2) = (-1) \frac{ \det \begin{pmatrix} 0 & \vec{x} & \vec{y} & \vec{z} \\ p_1 & -3 & 0 & 3 \\ p_2 & 0 & 3 & 0 \\ 1 & 1 & 1 & 1 \\ \end{pmatrix} }{ \det \begin{pmatrix} -3 & 0 & 3 \\ 0 & 3 & 0 \\ 1 & 1 & 1 \\ \end{pmatrix} } = \frac{\vec{z} - \vec{x}}{6} p_1 + \frac{2 \vec{y} - \vec{x} - \vec{z}}{6} p_2 + \frac{\vec{x} + \vec{z}}{2} = $$ now I plug in the final points $$ = \frac{1}{6} \left[ \begin{pmatrix} 4 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \end{pmatrix} \right] p_1 + \frac{1}{6} \left[ 2 \begin{pmatrix} 3 \\ 2 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \end{pmatrix} - \begin{pmatrix} 4 \\ 3 \end{pmatrix} \right] p_2 + \frac{1}{2} \left[ \begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 4 \\ 3 \end{pmatrix} \right] $$ Simplification yields $$ \vec{P}(p_1; p_2) = \begin{pmatrix} 1/3 \\ 0 \end{pmatrix} p_1 + \begin{pmatrix} 0 \\ -1/3 \end{pmatrix} p_2 + \begin{pmatrix} 3 \\ 3 \end{pmatrix} $$ Or you can write that in canonical form $$ \vec{P}(p_1; p_2) = \begin{pmatrix} 1/3 & 0 \\ 0 & -1/3 \end{pmatrix} \begin{pmatrix} p_1 \\ p_2 \end{pmatrix} + \begin{pmatrix} 3 \\ 3 \end{pmatrix} $$ For more details on how this all works you may check "Beginner's guide to mapping simplexes affinely", where authors of the equation elaborate on theory behind it. In the guide there is exactly the same 2D example solved as the one you are interested in. The same authors recently published "Workbook on mapping simplexes affinely" (ResearchGate as well). They discuss many practical examples there, e.g. show that the same equation may be used for color interpolation or Phong shading. You may want to check it if you want to see this equation in action.