Solution for previous proposal

Hello Brilliants, here I wish to share my solution for the proposed problem.


Before we start with the problem we shall consider the integral of the form for all n1n\geq 1 I(n)=01xn1ln(1x)dx=1n01ddx(xn1)ln(1x)dxI(n) = \displaystyle \int_0^1 x^{n-1} \ln(1-x) dx =\frac{1}{n}\displaystyle \int_0^1 \frac{d}{dx}(x^{n}-1)\ln(1-x) dx and by integration by parts we yield I(n)=1n[(xn1)ln(1x)0]011n011xn1xdxI(n)=\frac{1}{n}\left[\underbrace{(x^{n}-1)\ln(1-x)}_{0}\right]_0^1-\frac{1}{n}\int_0^1\frac{1-x^{n}}{1-x}dx =1n01j=1nxj1=1nj=1n01xj1dx=1nj=1n1j=Hnn(1)=-\frac{1}{n}\int_0^1\sum_{j=1}^{n} x^{j-1} =-\frac{1}{n}\sum_{j=1}^{n}\int_0^1 x^{j-1}dx=-\frac{1}{n} \sum_{j=1}^n\frac{1}{j} =-\frac{H_n}{n}\cdots (1) Note that D(k)=1n2m+m2n+kmn=1mn01xm+n+k1dx=xk1mn01xm+ndxD(k)=\displaystyle \frac{1}{n^2m+m^2n+kmn} = \frac{1}{mn}\int_0^1 x^{m+n+k-1} dx=\frac{x^{k-1}}{mn}\int_0^1{x^{m+n}}dx and therefore 1nmD(k)=01xk1n=0m=0xnnxmmdx=01xk1ln2(1x)dx\sum_{1\leq n\leq m\leq \infty} D(k)=\int_0^1 x^{k-1} \sum_{n=0}^{\infty}\sum_{m=0}^{\infty} \frac{x^n}{n}\cdot \frac{x^m}{m}dx =\int_0^1 x^{k-1}\ln^2(1-x) dx Now we consider the latter integral I(k)=01xkln2(1x)dx I(k)=\displaystyle \int_0^1 x^{k}\ln^2(1-x) dx for k0k\geq 0 which further can be written as 1k+101ddx(xk+11)ln2(1x)dx\displaystyle \frac{1}{k+1}\int_0^1 \frac{d}{dx}(x^{k+1}-1)\ln^2(1-x) dx and hence on Integration by parts we see that I(k)=1k+1[(xk+11)ln2(1x)0]012k+1011xk+11xln(1x)dxI(k)=\frac{1}{k+1}\left[\underbrace{(x^{k+1}-1)\ln^2(1-x)}_{0}\right]_0^1-\frac{2}{k+1}\int_0^1\frac{1-x^{k+1}}{1-x}\ln(1-x)dx =2k+101n=1k+1xn1ln(1x)=2k+1n=1k+101xn1ln(1x)dx=-\frac{2}{k+1}\int_0^1\sum_{n=1}^{k+1} x^{n-1}\ln(1-x)=-\frac{2}{k+1}\sum_{n=1}^{k+1}\int_0^1 x^{n-1}\ln(1-x)dx Plugging the result from (1)(1) to last integral we have 2k+1n=1k+1Hnn=2k+1(Hk+12+Hk+1(2)2)=Hk+12+Hk+1(2)k+1\frac{2}{k+1}\sum_{n=1}^{k+1}\frac{H_n}{n}=\frac{2}{k+1}\left(\frac{H_{k+1}^2+H_{k+1}^{(2)}}{2}\right)=\frac{H_{k+1}^2 +H_{k+1}^{(2)}}{k+1} Further note that nth partial sum of Hk+1(2)=ζ(2)ψ1(k+2)H_{k+1}^{(2)} = \zeta(2) -\psi^1(k+2) giving us required result of D(k)=Hk+12ψ1(k+2)k+1+π26(k+1)D(k)=\frac{H_{k+1}^2-\psi^1(k+2)}{k+1}+\frac{\pi^2}{6(k+1)}

Now we notice that 1(q+b)q+b+1+(q+b+1)q+b=(q+b)q+b+1(q+b+1)q+b(q+b)q+b+1(1)\frac{1}{(q+b)\sqrt{q+b+1}+(q+b+1)\sqrt{q+b}}=\frac{(q+b)\sqrt{q+b+1}-(q+b+1)\sqrt{q+b}}{(q+b)\sqrt{q+b+1}(-1)}=1q+b1q+b+1=\frac{1}{\sqrt{q+b}}-\frac{1}{\sqrt{q+b+1}} and hence n=0q=0nxn(1q+b1q+b+1Telescoping)=n=0xn(1b1n+b+1)\sum_{n=0}^{\infty}\sum_{q=0}^n x^n\left(\underbrace {\frac{1}{\sqrt{q+b}}-\frac{1}{\sqrt{q+b+1}}}_{\text{Telescoping}}\right)=\sum_{n=0}^{\infty}x^n\left(\frac{1}{\sqrt{b}}-\frac{1}{\sqrt{n+b+1}}\right) and by hence by algebra of sum and for for x<1|x|<1 n=0xnbn=0xnn+b+1=1b(1x)Φ(x,12,1)\sum_{n=0}^{\infty}\frac{x^n}{\sqrt{b}}-\sum_{n=0}^{\infty}\frac{x^n}{\sqrt{n+b+1}}=\frac{1}{\sqrt{b}(1-x)}-\Phi\left(x, \frac{1}{2},1\right) and we deduce that α=2,β=1,γ=6 \alpha = 2, \beta =1 ,\gamma =6. Thus Φ(βα,β,(α+2βγ)1)=Φ(1,1,110)=n=0(1)n(n+110)=D\Phi\left(\beta-\alpha,\beta , (\alpha+2\beta \gamma)^{-1}\right)=\Phi\left(-1,1,\frac{1}{10}\right)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{\left(n+\frac{1}{10}\right)}=D D=n=01(2n+101)(2n+101+1)=14n=01(n+201)(n+201+21)D= \sum_{n=0}^{\infty}\frac{1}{\left(2n+10^{-1}\right)(2n+10^{-1}+1)}=\frac{1}{4}\sum_{n=0}^{\infty}\frac{1}{(n+20^{-1})(n+20^{-1}+2^{-1})} =21(ψ(12+110)ψ(120))=21(ψ(110)2ψ(120)2ln2)=2^{-1}\left(\psi\left(\frac{1}{2}+\frac{1}{10}\right)-\psi\left(\frac{1}{20}\right)\right)=2^{-1}\left(\psi\left(\frac{1}{10}\right)-2\psi\left(\frac{1}{20}\right)-2\ln 2\right)and finally we have D=ψ(110)ψ(120)ln2\displaystyle D=\psi\left(\frac{1}{10}\right)-\psi\left(\frac{1}{20}\right)-\ln 2. Here to evaluate this expression DD , we use the Gauss's-Diagamma theorem ie ψ(rm)=γln(2m)π2cot(rπm)+2q=1m12cos(2πqrm)lnsin(πqn)\psi\left(\frac{r}{m}\right)=-\gamma -\ln(2m)-\frac{\pi}{2}\cot\left(\frac{r\pi}{m}\right)+2\sum_{q=1}^{\left\lfloor \frac{m-1}{2}\right\rfloor}\cos\left(\frac{2\pi qr}{m}\right)\ln\sin\left(\frac{\pi q}{n}\right) plugging r=1,m=10,20 r=1, m=10 ,20 we have I=ψ(110)=γln(20)π2cot(π10)+2q=14cos(πq5)lnsin(πq10)I= \psi\left(\frac{1}{10}\right)=-\gamma -\ln(20)-\frac{\pi}{2}\cot\left(\frac{\pi}{10}\right)+2\sum_{q=1}^{4}\cos\left(\frac{\pi q}{5}\right)\ln\sin\left(\frac{\pi q}{10}\right) and K=ψ(120)=γln(40)π2cot(π20)+2q=19cos(πq10)lnsin(πq20)K= \psi\left(\frac{1}{20}\right)=-\gamma -\ln(40)-\frac{\pi}{2}\cot\left(\frac{\pi}{20}\right)+2\sum_{q=1}^{9}\cos\left(\frac{\pi q}{10}\right)\ln\sin\left(\frac{\pi q}{20}\right) and also D=Ikln2D=I-k -\ln 2 and on further simplification of the DD we obtain that D=π41sin(120)cos(120)+A(logcot(π20))+B(logcot(3π20))D = \frac{\pi}{4}\frac{1}{\sin\left(\frac{1}{20}\right)\cos\left(\frac{1}{20}\right)}+A\left(\log\cot\left(\frac{\pi}{20}\right)\right)+B\left(\log\cot\left(\frac{3\pi}{20}\right)\right) with A=5+52=ϕ+2A=\sqrt{\frac{5+\sqrt 5}{2}}=\sqrt{\phi+2} and B=552=211025B=\sqrt{\frac{5-\sqrt{5}}{2}}=2^{-1}\sqrt{10-2\sqrt{5}} and π41sin(π20)cos(π20)=π21sin(18)=2π51\displaystyle \frac{\pi}{4}\frac{1}{\sin\left(\frac{\pi}{20}\right)\cos\left(\frac{\pi}{20}\right)}\\ =\frac{\pi}{2}\frac{1}{\sin(18^{\circ})}=\frac{2\pi}{\sqrt 5-1} since sin(18)=41(51)\sin(18^{\circ} )=4^{-1}\left(\sqrt 5-1\right).

Further we have cot(π20)=cot(9)\displaystyle\cot\left(\frac{\pi}{20}\right)=\cot(9^{\circ}). To evaluate cot(9\cot(9^{\circ} we shall be using the fact that cot(3x)=cos(3x)sin(3x)=4cos3x3cosx3sinx4sin3x=cosx(14sin2x)sinx(34sin2x)=cotx(1234sin2x)\cot (3x)=\frac{\cos (3x)}{\sin (3x)}= \frac{4 \cos^3 x-3\cos x}{3\sin x-4\sin^3 x}=\frac{\cos x(1-4\sin ^2 x)}{\sin x(3-4\sin^2 x)}=\cot x\left(1-\frac{2}{3-4\sin^2 x}\right) setting x=3x=3 we have the expression for cot9\cot9^{\circ} but then we will show that sin3=1481025153\sin 3^{\circ}=\frac{1}{4}\sqrt{8-\sqrt{10-2\sqrt 5}-\sqrt{15}-\sqrt{3}} {Note}: This identity is proposed by Narendra Bhandari and is proved by Sergio Esteban beautiful geometry work) and Ahmed Hegazi(trigonometry work). From half angle formula we have that sinϕ2=1cosϕ2\sin\frac{\phi}{2}=\sqrt{\frac{1-\cos \phi}{2}} now set ϕ=6\phi = 6^{\circ}. Giving us sin3=1cos62=1cos(3630)2\sin3^{\circ} =\sqrt{\frac{1-\cos 6^{\circ}}{2}}= \sqrt{\frac{1-\cos(36^{\circ} -30^{\circ})}{2}} Now using compound angle formula we can get
cos(3630)=cos36cos30+sin36sin30=(5+14)32+(10254)12=1025+3+158\displaystyle \cos(36^{\circ} -30^{\circ}) = \cos36^{\circ}\cos30^{\circ}+\sin36^{\circ}\sin30^{\circ} \\= \left(\frac{\sqrt 5 +1}{4} \right) \frac{\sqrt 3}{2} +\left(\frac{\sqrt{10-2\sqrt5}}{4}\right)\frac{1}{2} =\frac{\sqrt{10-2\sqrt 5}+\sqrt 3 +\sqrt {15}}{8}putting in original equation sin3=12(11025+3+158)=14(81025315)\sin3^{\circ} =\frac{1}{\sqrt 2} \left( \sqrt{1- \frac{\sqrt{10-2\sqrt 5}+\sqrt 3 +\sqrt {15}}{8} }\right) = \frac{1}{4}\left(\sqrt{8-\sqrt{10-2\sqrt 5}-\sqrt 3 -\sqrt {15} }\right) and hence cos3=1sin23=148+1025+10+3\cos 3^{\circ} = \sqrt{1-\sin^2 3^{\circ}} =\frac{1}{4}\sqrt{8+\sqrt{10-2\sqrt 5} +\sqrt{10}+\sqrt{3}} and hence we deduce easily that cot9=cot3(184+1025+15+3)\cot9^{\circ}= \cot3^{\circ}\left(1-\frac{8}{4+\sqrt{10-2\sqrt 5}+\sqrt{15}+\sqrt{3}}\right) also cot(3π20)=cot(27=cot(303) \cot\left(\frac{3\pi}{20}\right)=\cot(27^{\circ}=\cot(30-3) and using compound angle formula for cot(ab)cot(a-b) we have cot27=cot3cot301cot3+cot30=3cot31cot3+3\cot27^{\circ}= \frac{\cot 3^{\circ} \cot30^{\circ}-1}{\cot 3^{\circ}+\cot 30^{\circ}}=\frac{\sqrt 3 \cot3^{\circ} -1}{\cot3^{\circ}+\sqrt 3} and hence we have D=Φ(βα,β,(α+2β+γ)1D=\Phi(\beta-\alpha,\beta , (\alpha +2\beta +\gamma)^{-1} =2π51+ϕ+2log(θ8θ4+1025+15+3)+211025log(3θ1θ+3)=\frac{2\pi}{\sqrt5 -1}+\sqrt{\phi+2}\log\left(\theta -\frac{8\theta}{4+\sqrt{10-2\sqrt 5}+\sqrt{15}+\sqrt {3}}\right)+2^{-1}\sqrt{10-2\sqrt 5}\log\left(\frac{\sqrt 3 \theta -1}{\theta +\sqrt 3}\right) and thus θ=8+1025+15+381025153\displaystyle \theta=\frac{\sqrt{8+\sqrt{10-2\sqrt{5}}+\sqrt{15}+\sqrt{3}}}{\sqrt{8-\sqrt{10-2\sqrt{5}}-\sqrt{15}-\sqrt{3}}} we are done


Other different approaches are highly appreciated. I wish to be corrected if any such typos or errors are spotted in the above solution. Thank you!

Note by Naren Bhandari
5 months, 2 weeks ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

I don't know calculus but I like your solutions (both of them!)

A Former Brilliant Member - 4 months, 2 weeks ago

Log in to reply

I'm glad to know, you like it.Thank you :)

Naren Bhandari - 4 months, 2 weeks ago

Log in to reply

Man I didn't understood the step where you pull out the x to the k-1 variable out from the integral.

Aruna Yumlembam - 3 months, 3 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...