nice-calculus-n-1-sin-nx-n-pi-2-x-2-Im-1-e-ix-Imln-1-cos-x-isin-x-Im-ln-1-cos-x-2-sin-2-x-itan-1

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#

$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:…….{nice}\:\:…………{calculus}……. \\ $$$$\:\:\:\:\boldsymbol{\phi}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({nx}\right)}{{n}}\:=\frac{\pi}{\mathrm{2}}−\frac{{x}}{\mathrm{2}} \\ $$$$\:\:\:\boldsymbol{\phi}=−{Im}\left(\mathrm{1}−{e}^{{ix}} \right)=−{Imln}\left\{\left(\mathrm{1}−{cos}\left({x}\right)−{isin}\left({x}\right)\right)\right\} \\ $$$$\:\:\:\:=−{Im}\left\{{ln}\left(\sqrt{\left(\mathrm{1}−{cos}\left({x}\right)\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \left({x}\right)}\:+{itan}^{−\mathrm{1}} \left(\frac{−{sin}\left({x}\right)}{\mathrm{1}−{cos}\left({x}\right)}\right)\right\}\right. \\ $$$$=−{Im}\left\{\sqrt{\mathrm{2}−\mathrm{2}{cos}\left({x}\right)_{\:} }\:−{itan}^{−\mathrm{1}} \left({tan}\left(\frac{\pi}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)\right)\right\} \\ $$$$\:\:={tan}^{−\mathrm{1}} \left({tan}\left(\frac{\pi}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\right)\right)\:\: \\ $$$$\:\:\:\:\:\:\:……..\:\:\:\boldsymbol{\phi}=\frac{\pi}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}\:\:…\checkmark…..\langle\ast\rangle \\ $$$$\:\:\:\:\:^{''} \:\:\int\:\:^{''} \:\:{both}\:{sides}\:{of}\:\langle\ast\rangle \\ $$$$\:\:\:\:\:\:−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }{dx}+{C}=\frac{\pi{x}}{\mathrm{2}}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:{x}=\mathrm{0}\:\Rightarrow\:{C}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\:\:\:…\checkmark \\ $$$$\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left({nx}\right)}{{n}^{\mathrm{2}} }=−\frac{\pi{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:….\checkmark \\ $$$$\:\:\:\:\:\:\:\:{x}=\pi\left(\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{3}}}\:\right)\:…. \\ $$$$\:\:\:\:\:\:\:\:\Sigma\frac{{cos}\left({n}\pi\left(\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{3}}}\:\right)\right)}{{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\left(−\pi^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{3}}}\:\right)+\frac{\pi^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\frac{\mathrm{1}}{\mathrm{3}}}\:\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\left(−\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:+\frac{\mathrm{1}+\mathrm{2}.\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\left(−\mathrm{1}−\frac{\sqrt{\mathrm{3}}\:}{\mathrm{3}}+\frac{\mathrm{4}}{\mathrm{6}}+\frac{\sqrt{\mathrm{3}}\:}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{0}..\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\:\:\:……{prepared}\:{by}\:{mr}\:\left({Dwaipayan}\right)….. \\ $$$$\:\:{m}.{n}# \\ $$$$\:\:\:\: \\ $$

Commented by Dwaipayan Shikari last updated on 07/Apr/21

$${Have}\:{you}\:{seen}\:{that}\:{sir}? \\ $$

Commented by mnjuly1970 last updated on 07/Apr/21

yes mr payan:: 1:Q:: 137637 ..... 2 :: in brilliant math: this problem was belong to you in brilliant...⇓⇓⇓

$$\:{yes}\:{mr}\:{payan}::\:\mathrm{1}:{Q}::\:\mathrm{137637}\:….. \\ $$$$\:\mathrm{2}\:::\:{in}\:{brilliant}\:{math}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{this}\:{problem}\:{was}\:{belong}\:\:{to} \\ $$$${you}\:{in}\:{brilliant}…\Downarrow\Downarrow\Downarrow \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\: \\ $$$$\:\:\:\: \\ $$

Commented by mnjuly1970 last updated on 07/Apr/21

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