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Question Number 161567 by HongKing last updated on 19/Dec/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)\mathrm{be}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\ln (1-\mathrm{x})}$$$$\mathrm{prove}\mathrm{there}\mathrm{exists}\mathrm{a}\mathrm{sequence}\left\{{\mathrm{a}}_{\mathbf{k}}\right\}\mathrm{such}\mathrm{that}$$$$\mathrm{D}\left[\mathrm{f}\left(\mathrm{x}\right)\right]=\left[\underset{0}{\sum ^{\mathrm{\infty }}}{\mathrm{a}}_{\mathbf{k}}{\mathrm{x}}^{\mathbf{k}}\right]{\left[\mathrm{f}\left(\mathrm{x}\right)\right]}^2$$

Answered by mindispower last updated on 19/Dec/21

$$whatdoyoumeanByD\left[f\left(x\right)\right]?$$

Commented by HongKing last updated on 19/Dec/21

$$\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{my}\mathrm{dear}\mathrm{Sir}$$

Answered by mindispower last updated on 19/Dec/21

$$Df\left(x\right)=\frac{ln(1-x)+\frac{x}{1-x}}{{ln}^2(1-x)}$$$$=\frac{x^2}{{ln}^2(1-x)}(\frac{ln(1-x)}{x^2}+\frac{1}{x(1-x)})$$$$\frac{ln(1-x)}{x^2}+\frac{1}{x(1-x)}=-\frac{1}{x^2}\sum _{k⩾1}\frac{x^k}{k}+\frac{1}{x}\sum _{k⩾1}{x}^{k-1}$$$$-\frac{1}{x}\sum _{k⩾1}\frac{x^{k-1}}{k}+\frac{\sum _{k⩾1}{x}^{k-1}}{x}$$$$=\frac{1}{x}(\sum _{k⩾2}{x}^{k-1}-\sum _{k⩾2}\frac{x^{k-1}}{k})=\sum _{k⩾2}\left(\frac{k-1}{k}{x}^{\mathit{k}-2}\right)$$$$=\sum _{k⩾0}\frac{k+1}{k+2}{x}^k$$$$Df\left(x\right)=\left(\sum _{k⩾0}\frac{k+1}{k+2}{x}^k\right)f^2\left(x\right)$$

Commented by HongKing last updated on 19/Dec/21

$$\mathrm{A}\mathrm{nice}\mathrm{solution}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}$$

Commented by HongKing last updated on 20/Dec/21

$$\mathrm{My}\mathrm{dear}\mathrm{Sir},\mathrm{can}\mathrm{you}\mathrm{imagine},\mathrm{what}$$$$\mathrm{could}\mathrm{we}\mathrm{create}\mathrm{by}\mathrm{writing}\mathrm{the}\mathrm{derivative}$$$$\mathrm{of}\mathrm{a}\mathrm{function}\mathrm{in}\mathrm{the}\mathrm{way}\mathrm{I}\mathrm{proposed}?$$

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