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Question Number 98953 by Ar Brandon last updated on 17/Jun/20

$$\mathcal{G}\mathrm{iven}\mathrm{f}\left(\mathrm{x}\right)={\sin }^2\mathrm{x}\mathrm{find}\mathrm{the}\mathrm{expansion}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)$$$$\mathrm{up}\mathrm{to}\mathrm{the}{\mathrm{n}}^{\mathrm{th}}\mathrm{term}.$$

Answered by mr W last updated on 17/Jun/20

$$f\left(x\right)={\sin }^{2}x=\frac{1}{2}(1-\cos 2x)$$$$=\frac{1}{2}-\frac{1}{2}(1-\frac{{\left(2x\right)}^2}{2!}+\frac{{\left(2x\right)}^4}{4!}-...)$$$$=\frac{1}{2}-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{2}^{2n-1}{x}^{2n}}{\left(2n\right)!}$$

Answered by mathmax by abdo last updated on 17/Jun/20

$$\mathrm{f}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{\mathrm{f}}^{\left(\mathrm{k}\right)}\left(0\right)}{\mathrm{k}!}{\mathrm{x}}^{\mathrm{k}}+....\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{x}\right)={\sin }^2\mathrm{x}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=2\mathrm{sinx}\mathrm{cosx}\Rightarrow$$$${\mathrm{f}}^{\left(\mathrm{k}\right)}\left(\mathrm{x}\right)=2{\left(\mathrm{sinxcosx}\right)}^{(\mathrm{k}-1)}=2\sum _{\mathrm{p}=0}^{\mathrm{k}-1}{\mathrm{C}}_{\mathrm{k}-1}^{\mathrm{p}}{\sin }^{\left(\mathrm{p}\right)}\left(\mathrm{x}\right){\cos }^{(\mathrm{k}-\mathrm{p})}\left(\mathrm{x}\right)$$$${\mathrm{f}}^{\left(\mathrm{k}\right)}\left(\mathrm{x}\right)=2\sum _{\mathrm{p}=0}^{\mathrm{k}-1}\sin (\mathrm{x}+\frac{\mathrm{p}\pi }{2})\cos (\mathrm{x}+\frac{(\mathrm{k}-\mathrm{p})\pi }{2})\Rightarrow$$$${\mathrm{f}}^{\left(\mathrm{k}\right)}\left(0\right)=2\sum _{\mathrm{p}=0}^{\mathrm{k}-1}\sin \left(\frac{\mathrm{p}\pi }{2}\right)\cos \left(\frac{(\mathrm{k}-\mathrm{p})\pi }{2}\right)\Rightarrow$$$${\sin }^2\mathrm{x}=2\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}!}\left(\sum _{\mathrm{p}=0}^{\mathrm{k}-1}\sin \left(\frac{\mathrm{p}\pi }{2}\right)\cos \left(\frac{(\mathrm{k}-\mathrm{p})\pi }{2}\right)\right){\mathrm{x}}^{\mathrm{k}}+....$$

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