A calculus problem by Dwaipayan Shikari

Calculus Level 5

( 1 2 − π ) 2 + ( 1 2 + π ) 2 + ( 1 6 − π ) 2 + ( 1 6 + π ) 2 + ( 1 10 − π ) 2 + ( 1 10 + π ) 2 + ⋯ \small \left(\frac 1{2 - \pi} \right)^2 + \left(\frac 1{2 +\pi} \right)^2 + \left(\frac 1{6 - \pi} \right)^2 + \left(\frac 1{6 +\pi} \right)^2 + \left(\frac 1{10 - \pi} \right)^2 + \left(\frac 1{10 +\pi} \right)^2 + \cdots

The sum above can be represented as π a 2 b sec ⁡ 2 ( π a b ) \dfrac{π^a}{2^b}\sec^2\left(\dfrac{π^a}{b}\right) , where a a and b b are integers. Find b − 2 a b-2a .

The problem is original


The answer is 0.

Dwaipayan Shikari by Dwaipayan Shikari , 17, India

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1 solution

Dwaipayan Shikari
Jan 26, 2021

tan ⁡ ( x ) = 1 π 2 − x − 1 π 2 + x + 1 3 π 2 − x − 1 3 π 2 + x + 1 5 π 2 − x − ⋯ \tan(x)= \frac{1}{\frac{π}{2}-x}-\frac{1}{\frac{π}{2}+x}+\frac{1}{\frac{3π}{2}-x}-\frac{1}{\frac{3π}{2}+x}+\frac{1}{\frac{5π}{2}-x}-\cdots π 2 tan ⁡ ( π x 2 ) = 1 1 − x − 1 1 + x + 1 3 − x − 1 3 + x + 1 5 − x − 1 5 + x + ⋯ \frac{π}{2}\tan(\frac{πx}{2})= \frac{1}{1-x}-\frac{1}{1+x}+\frac{1}{3-x}-\frac{1}{3+x}+\frac{1}{5-x}-\frac{1}{5+x}+\cdots After Differentiating respect to x x , we get π 2 4 sec ⁡ 2 ( π x 2 ) = 1 ( 1 − x ) 2 + 1 ( 1 + x ) 2 + 1 ( 3 − x ) 2 + 1 ( 3 + x ) 2 + ⋯ \frac{π^2}{4}\sec^2(\frac{πx}{2})= \frac{1}{(1-x)^2}+\frac{1}{(1+x)^2}+\frac{1}{(3-x)^2}+\frac{1}{(3+x)^2}+\cdots Take x = π 2 x=\frac{π}{2} The series becomes π 2 2 4 sec ⁡ 2 ( π 2 4 ) = ( 1 2 − π ) 2 + ( 1 2 + π ) 2 + ( 1 6 − π ) 2 + ⋯ \frac{π^2}{2^4}\sec^2(\frac{π^2}{4})= \left(\frac 1{2-π}\right)^2+\left(\frac 1{2+π}\right)^2+\left(\frac 1{6-π}\right)^2+\cdots Answer is b − 2 a = 0 \boxed{b-2a}=0 Proof of the above Proposition \color{#E81990}\large \textrm{Proof of the above Proposition}

c o s x = ( 1 − 2 x π ) ( 1 + 2 x π ) ( 1 − 2 x 3 π ) ( 1 + 2 x 3 π ) ( 1 − 2 x 5 π ) ( 1 + 2 x 5 π ) . . . . \mathrm{cosx}=\left(\mathrm{1}-\frac{\mathrm{2}{x}}{\pi}\right)\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\pi}\right)\left(\mathrm{1}-\frac{\mathrm{2}{x}}{\mathrm{3}\pi}\right)\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{3}\pi}\right)\left(\mathrm{1}-\frac{\mathrm{2}{x}}{\mathrm{5}\pi}\right)\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{5}\pi}\right).... l o g ( c o s x ) = l o g ( 1 − 2 x π ) + l o g ( 1 + 2 x π ) + l o g ( 1 − 2 x 3 π ) + l o g ( 1 + 2 x 3 π ) + . . \mathrm{log}\left(\mathrm{cosx}\right)=\mathrm{log}\left(\mathrm{1}-\frac{\mathrm{2}{x}}{\pi}\right)+\mathrm{log}\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\pi}\right)+\mathrm{log}\left(\mathrm{1}-\frac{\mathrm{2}{x}}{\mathrm{3}\pi}\right)+\mathrm{log}\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{3}\pi}\right)+.. Differentiating both sides Respect to x x − s i n x c o s x = − 2 π 1 − 2 π x + 2 π 1 + 2 π x − 2 3 π 1 − 2 3 π x + 2 3 π 1 + 2 3 π x − . . . -\frac{\mathrm{sinx}}{\mathrm{cosx}}=\frac{-\frac{\mathrm{2}}{\pi}}{\mathrm{1}-\frac{\mathrm{2}}{\pi}{x}}+\frac{\frac{\mathrm{2}}{\pi}}{\mathrm{1}+\frac{\mathrm{2}}{\pi}{x}}-\frac{\frac{\mathrm{2}}{\mathrm{3}\pi}}{\mathrm{1}-\frac{\mathrm{2}}{\mathrm{3}\pi}{x}}+\frac{\frac{\mathrm{2}}{\mathrm{3}\pi}}{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{3}\pi}{x}}-... t a n x = 1 π 2 − x − 1 π 2 + x + 1 3 π 2 − x − 1 3 π 2 + x + . . \boldsymbol{\mathrm{tanx}}=\frac{\mathrm{1}}{\frac{\boldsymbol{\pi}}{\mathrm{2}}-\boldsymbol{{x}}}-\frac{\mathrm{1}}{\frac{\boldsymbol{\pi}}{\mathrm{2}}+\boldsymbol{{x}}}+\frac{\mathrm{1}}{\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{2}}-\boldsymbol{{x}}}-\frac{\mathrm{1}}{\frac{\mathrm{3}\boldsymbol{\pi}}{\mathrm{2}}+\boldsymbol{{x}}}+..

1 Helpful 1 Interesting 2 Brilliant 0 Confused

2 2 and 4 4 are not coprime, because gcd ⁡ ( 2 , 4 ) = 2 \gcd(2,4) = 2 and not 1 1 . I have amended the problem statement..

You can use \pi for π \pi , \times for × \times , \dfrac \pi2 π 2 \dfrac \pi2 for proper size fraction, the braces { } are not necessary if there is only one character, and \left( \right) to have auto fit brackets ( 1 2 ) \left(\dfrac 12 \right) .

Chew-Seong Cheong - 4 months, 2 weeks ago

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Thanks sir .That was a mistake

Dwaipayan Shikari - 4 months, 2 weeks ago

I have never seen the series in the top row. Could you explain?

Isaac YIU Math Studio - 4 months, 2 weeks ago

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Yes ,I have added an explanation

Dwaipayan Shikari - 4 months, 2 weeks ago

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