let-f-x-be-f-x-x-ln-1-x-prove-there-exists-a-sequence-a-k-such-that-D-f-x-0-a-k-x-k-f-x-2-

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

let f (x) be

f (x) = \frac{x}{\ln (1 - x)}

prove there exists a sequence {a_{k}} such that

D [f (x)] = [\underset{0}{\sum^{\infty}} a_{k} x^{k}] {[f (x)]}^{2}

Answered by mindispower last updated on 19/Dec/21

w h a t d o y o u m e a n B y D [f (x)] ?

Commented by HongKing last updated on 19/Dec/21

\frac{d}{dx} my dear Sir

Answered by mindispower last updated on 19/Dec/21

D f (x) = \frac{l n (1 - x) + \frac{x}{1 - x}}{{l n}^{2} (1 - x)}

= \frac{x^{2}}{{l n}^{2} (1 - x)} (\frac{l n (1 - x)}{x^{2}} + \frac{1}{x (1 - x)})

\frac{l n (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} (\frac{k - 1}{k} x^{k - 2})

= \sum_{k ⩾ 0} \frac{k + 1}{k + 2} x^{k}

D f (x) = (\sum_{k ⩾ 0} \frac{k + 1}{k + 2} x^{k}) f^{2} (x)

Commented by HongKing last updated on 19/Dec/21

A nice solution my dear Sir thank you

Commented by HongKing last updated on 20/Dec/21

My dear Sir, can you imagine, what

could we create by writing the derivative

of a function in the way I proposed ?