Question Number 142629 by qaz last updated on 03/Jun/21
![Σ_(n=1) ^∞ (−1)^n ∙((2n−1)/((2n)!))∙((π/2))^(2n) =Σ_(n=1) ^∞ ((2n−1)/((2n)!))∙(−((π/2))^2 )^n =(2xD−1)∣_(x=π/2) Σ_(n=1) ^∞ (((−x^2 )^n )/((2n)!)) =(2xD−1)∣_(x=π/2) [Σ_(n=0) ^∞ (((−x^2 )^n )/((2n)!))−1] =(2xD−1)∣_(x=π/2) (cos x−1) =(−2xsin x−cos x+1)∣_(x=π/2) =1−π where is wrong?](https://www.tinkutara.com/question/Q142629.png)
$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{n}} \centerdot\frac{\mathrm{2n}−\mathrm{1}}{\left(\mathrm{2n}\right)!}\centerdot\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2n}} \\ $$$$=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2n}−\mathrm{1}}{\left(\mathrm{2n}\right)!}\centerdot\left(−\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} \right)^{\mathrm{n}} \\ $$$$=\left(\mathrm{2xD}−\mathrm{1}\right)\mid_{\mathrm{x}=\pi/\mathrm{2}} \underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!} \\ $$$$=\left(\mathrm{2xD}−\mathrm{1}\right)\mid_{\mathrm{x}=\pi/\mathrm{2}} \left[\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!}−\mathrm{1}\right] \\ $$$$=\left(\mathrm{2xD}−\mathrm{1}\right)\mid_{\mathrm{x}=\pi/\mathrm{2}} \left(\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right) \\ $$$$=\left(−\mathrm{2xsin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}+\mathrm{1}\right)\mid_{\mathrm{x}=\pi/\mathrm{2}} \\ $$$$=\mathrm{1}−\pi \\ $$$$\mathrm{where}\:\mathrm{is}\:\mathrm{wrong}? \\ $$
Answered by Dwaipayan Shikari last updated on 03/Jun/21
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}\right)}{\left(\mathrm{2}{n}\right)!}\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}{n}} −\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}{n}} \\ $$$$=−\left(\frac{\mathrm{2}}{\mathrm{2}!}\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{4}!}\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{4}} +\frac{\mathrm{6}}{\mathrm{6}!}\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{6}} +…\right)+\left(\frac{\mathrm{1}}{\mathrm{2}!}\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}!}\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{4}} +..\right) \\ $$$$=−\frac{\pi}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}!}\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{3}} +..\right)−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}!}\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} −…\right)+\mathrm{1} \\ $$$$=−\frac{\pi}{\mathrm{2}}{sin}\left(\frac{\pi}{\mathrm{2}}\right)−{cos}\left(\frac{\pi}{\mathrm{2}}\right)+\mathrm{1}=\mathrm{1}−\frac{\pi}{\mathrm{2}} \\ $$
Commented by qaz last updated on 03/Jun/21
![i have another solution. Σ_(n=1) ^∞ (((−1)^n (2n−1))/((2n)!))x^(2n) ∣_(x=(π/2)) =Σ_(n=0) ^∞ (((−1)^(n+1) (2n+1))/((2n+2)!))x^(2n+2) ∣_(x=(π/2)) =−x^2 Σ_(n=0) ^∞ (((−1)^n )/((2n+2)(2n)!))x^(2n) =−x^2 ∫_0 ^1 yΣ_(n=0) ^∞ (((−1)^n )/((2n)!))∙(xy)^(2n) dy =−x^2 ∫_0 ^1 ycos (xy)dy =−x^2 {[(1/x)ysin (xy)+(1/x^2 )cos (xy)]_0 ^1 }_(x=(π/2)) =−x^2 {(1/x)sin x+(1/x^2 )(cos x−1)}_(x=(π/2)) =1−(π/2) −−−−−−−−−−−−−−−−−−− but Σ_(n=1) ^∞ (2n−1)(((−1)^n )/((2n)!))x^(2n) =(xD−1)(cos x−1), i still dont understand,why not LFT=(2xD−1)(cos x−1)???](https://www.tinkutara.com/question/Q142645.png)
$$\mathrm{i}\:\mathrm{have}\:\mathrm{another}\:\mathrm{solution}. \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{2n}−\mathrm{1}\right)}{\left(\mathrm{2n}\right)!}\mathrm{x}^{\mathrm{2n}} \mid_{\mathrm{x}=\frac{\pi}{\mathrm{2}}} \\ $$$$=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} \left(\mathrm{2n}+\mathrm{1}\right)}{\left(\mathrm{2n}+\mathrm{2}\right)!}\mathrm{x}^{\mathrm{2n}+\mathrm{2}} \mid_{\mathrm{x}=\frac{\pi}{\mathrm{2}}} \\ $$$$=−\mathrm{x}^{\mathrm{2}} \underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{2}\right)\left(\mathrm{2n}\right)!}\mathrm{x}^{\mathrm{2n}} \\ $$$$=−\mathrm{x}^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{y}\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!}\centerdot\left(\mathrm{xy}\right)^{\mathrm{2n}} \mathrm{dy} \\ $$$$=−\mathrm{x}^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ycos}\:\left(\mathrm{xy}\right)\mathrm{dy} \\ $$$$=−\mathrm{x}^{\mathrm{2}} \left\{\left[\frac{\mathrm{1}}{\mathrm{x}}\mathrm{ysin}\:\left(\mathrm{xy}\right)+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{cos}\:\left(\mathrm{xy}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \right\}_{\mathrm{x}=\frac{\pi}{\mathrm{2}}} \\ $$$$=−\mathrm{x}^{\mathrm{2}} \left\{\frac{\mathrm{1}}{\mathrm{x}}\mathrm{sin}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\left(\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)\right\}_{\mathrm{x}=\frac{\pi}{\mathrm{2}}} \\ $$$$=\mathrm{1}−\frac{\pi}{\mathrm{2}} \\ $$$$−−−−−−−−−−−−−−−−−−− \\ $$$$\mathrm{but}\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{2n}−\mathrm{1}\right)\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!}\mathrm{x}^{\mathrm{2n}} =\left(\mathrm{xD}−\mathrm{1}\right)\left(\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right), \\ $$$$\mathrm{i}\:\mathrm{still}\:\mathrm{dont}\:\mathrm{understand},\mathrm{why}\:\mathrm{not}\:\mathrm{LFT}=\left(\mathrm{2xD}−\mathrm{1}\right)\left(\mathrm{cos}\:\mathrm{x}−\mathrm{1}\right)??? \\ $$