.......nice ............calculus....... 𝛗=Σ_(n=1) ^∞ ((sin(nx))/n) =(π/2)−(x/2) 𝛗=−Im(1−e^(ix) )=−Imln{(1−cos(x)−isin(x))} =−Im{ln((√((1−cos(x))^2 +sin^2 (x))) +itan^(−1) (((−sin(x))/(1−cos(x))))} =−Im{(√(2−2cos(x)_ )) −itan^(−1) (tan((π/2)−(x/2)))} =tan^(−1) (tan((π/2)−(x/2))) ........ 𝛗=(π/2)−(x/2) ...✓.....⟨∗⟩ ^(′′) ∫ ^(′′) both sides of ⟨∗⟩ −Σ_(n=1) ^∞ ((cos(nx))/n^2 )dx+C=((πx)/2)−(x^2 /4) x=0 ⇒ C=(π^2 /6) ...✓ Σ_(n=1) ^∞ ((cos(nx))/n^2 )=−((πx)/2)+(x^2 /4)+(π^2 /6) ....✓ x=π(1+(√(1/3)) ) .... Σ((cos(nπ(1+(√(1/3)) )))/n^2 )=(1/2)(−π^2 (1+(√(1/3)) )+((π^2 (1+(√(1/3)) )^2 )/2)+(π^2 /3)) =(π^2 /2)(−1−((√3)/3) +((1+2.((√3)/3) +(1/3))/2)+(1/3)) =(π^2 /2)(−1−(((√3) )/3)+(4/6)+(((√3) )/3)+(1/3))=0..✓✓ ......prepared by mr (Dwaipayan)..... m.n#


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Question Number 137873 by mnjuly1970 last updated on 07/Apr/21
$.......nice ............calculus....... 𝛗=Σ_(n=1) ^∞ ((sin(nx))/n) =(π/2)−(x/2) 𝛗=−Im(1−e^(ix) )=−Imln{(1−cos(x)−isin(x))} =−Im{ln((√((1−cos(x))^2 +sin^2 (x))) +itan^(−1) (((−sin(x))/(1−cos(x))))} =−Im{(√(2−2cos(x)_ )) −itan^(−1) (tan((π/2)−(x/2)))} =tan^(−1) (tan((π/2)−(x/2))) ........ 𝛗=(π/2)−(x/2) ...✓.....⟨∗⟩ ^(′′) ∫ ^(′′) both sides of ⟨∗⟩ −Σ_(n=1) ^∞ ((cos(nx))/n^2 )dx+C=((πx)/2)−(x^2 /4) x=0 ⇒ C=(π^2 /6) ...✓ Σ_(n=1) ^∞ ((cos(nx))/n^2 )=−((πx)/2)+(x^2 /4)+(π^2 /6) ....✓ x=π(1+(√(1/3)) ) .... Σ((cos(nπ(1+(√(1/3)) )))/n^2 )=(1/2)(−π^2 (1+(√(1/3)) )+((π^2 (1+(√(1/3)) )^2 )/2)+(π^2 /3)) =(π^2 /2)(−1−((√3)/3) +((1+2.((√3)/3) +(1/3))/2)+(1/3)) =(π^2 /2)(−1−(((√3) )/3)+(4/6)+(((√3) )/3)+(1/3))=0..✓✓ ......prepared by mr (Dwaipayan)..... m.n#$
$. . . . . . . n i c e . . . . . . . . . . . . c a l c u l u s . . . . . . .$ $ϕ = \underset{n = 1}{\sum^{\infty}} \frac{s i n (n x)}{n} = \frac{π}{2} - \frac{x}{2}$ $ϕ = - I m (1 - e^{i x}) = - I m l n {(1 - c o s (x) - i s i n (x))}$ $= - I m {l n (\sqrt{{(1 - c o s (x))}^{2} + {s i n}^{2} (x)} + {i t a n}^{- 1} (\frac{- s i n (x)}{1 - c o s (x)})}$ $= - I m {\sqrt{2 - 2 c o s {(x)}_{}} - {i t a n}^{- 1} (t a n (\frac{π}{2} - \frac{x}{2}))}$ $= {t a n}^{- 1} (t a n (\frac{π}{2} - \frac{x}{2}))$ $. . . . . . . . ϕ = \frac{π}{2} - \frac{x}{2} . . . ✓ . . . . . ⟨ * ⟩$ $^{^{″}} \int^{^{″}} b o t h s i d e s o f ⟨ * ⟩$ $- \underset{n = 1}{\sum^{\infty}} \frac{c o s (n x)}{n^{2}} d x + C = \frac{π x}{2} - \frac{x^{2}}{4}$ $x = 0 \Rightarrow C = \frac{π^{2}}{6} . . . ✓$ $\underset{n = 1}{\sum^{\infty}} \frac{c o s (n x)}{n^{2}} = - \frac{π x}{2} + \frac{x^{2}}{4} + \frac{π^{2}}{6} . . . . ✓$ $x = π (1 + \sqrt{\frac{1}{3}}) . . . .$ $Σ \frac{c o s (n π (1 + \sqrt{\frac{1}{3}}))}{n^{2}} = \frac{1}{2} (- π^{2} (1 + \sqrt{\frac{1}{3}}) + \frac{π^{2} {(1 + \sqrt{\frac{1}{3}})}^{2}}{2} + \frac{π^{2}}{3})$ $= \frac{π^{2}}{2} (- 1 - \frac{\sqrt{3}}{3} + \frac{1 + 2 . \frac{\sqrt{3}}{3} + \frac{1}{3}}{2} + \frac{1}{3})$ $= \frac{π^{2}}{2} (- 1 - \frac{\sqrt{3}}{3} + \frac{4}{6} + \frac{\sqrt{3}}{3} + \frac{1}{3}) = 0 . . ✓ ✓$ $. . . . . . p r e p a r e d b y m r (D w a i p a y a n) . . . . .$ $You can't use 'macro parameter character #' in math mode$
Commented by Dwaipayan Shikari last updated on 07/Apr/21

$H a v e y o u s e e n t h a t s i r ?$
Commented by mnjuly1970 last updated on 07/Apr/21

$y e s m r p a y a n :: 1 : Q :: 137637 . . . . .$ $2 :: i n b r i l l i a n t m a t h :$ $t h i s p r o b l e m w a s b e l o n g t o$ $y o u i n b r i l l i a n t . . . ⇓⇓⇓$
Commented by mnjuly1970 last updated on 07/Apr/21

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