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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-

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Question Number 161567 by HongKing last updated on 19/Dec/21
let f(x) be f(x) = (x/(ln(1 - x))) prove there exists a sequence {a_k } such that D[f(x)] = [Σ_0 ^( ∞) a_k x^k ] [f(x)]^2
$$\mathrm{let}\:\:\mathrm{f}\left(\mathrm{x}\right)\:\:\mathrm{be} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{x}}{\mathrm{ln}\left(\mathrm{1}\:-\:\mathrm{x}\right)} \\ $$$$\mathrm{prove}\:\mathrm{there}\:\mathrm{exists}\:\mathrm{a}\:\mathrm{sequence}\:\left\{\mathrm{a}_{\boldsymbol{\mathrm{k}}} \right\}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{D}\left[\mathrm{f}\left(\mathrm{x}\right)\right]\:=\:\left[\underset{\mathrm{0}} {\overset{\:\infty} {\sum}}\:\mathrm{a}_{\boldsymbol{\mathrm{k}}} \:\mathrm{x}^{\boldsymbol{\mathrm{k}}} \right]\:\left[\mathrm{f}\left(\mathrm{x}\right)\right]^{\mathrm{2}} \\ $$
Answered by mindispower last updated on 19/Dec/21
what do you mean By D[f(x)]?
$${what}\:{do}\:{you}\:{mean}\:{By}\:{D}\left[{f}\left({x}\right)\right]? \\ $$
Commented by HongKing last updated on 19/Dec/21
(d/dx) my dear Sir
$$\frac{\mathrm{d}}{\mathrm{dx}}\:\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$
Answered by mindispower last updated on 19/Dec/21
Df(x)=((ln(1−x)+(x/(1−x)))/(ln^2 (1−x))) =(x^2 /(ln^2 (1−x)))(((ln(1−x))/x^2 )+(1/(x(1−x)))) ((ln(1−x))/x^2 )+(1/(x(1−x)))=−(1/x^2 )Σ_(k≥1) (x^k /k)+(1/x)Σ_(k≥1) x^(k−1) −(1/x)Σ_(k≥1) (x^(k−1) /k)+((Σ_(k≥1) x^(k−1) )/x) =(1/x)(Σ_(k≥2) x^(k−1) −Σ_(k≥2) (x^(k−1) /k))=Σ_(k≥2) (((k−1)/k)x^(k−2) ) =Σ_(k≥0) ((k+1)/(k+2))x^k Df(x)=(Σ_(k≥0) ((k+1)/(k+2))x^k )f^2 (x)
$${Df}\left({x}\right)=\frac{{ln}\left(\mathrm{1}−{x}\right)+\frac{{x}}{\mathrm{1}−{x}}}{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)} \\ $$$$=\frac{{x}^{\mathrm{2}} }{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}\left(\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}\left(\mathrm{1}−{x}\right)}\right) \\ $$$$\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}\left(\mathrm{1}−{x}\right)}=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{x}^{{k}} }{{k}}+\frac{\mathrm{1}}{{x}}\underset{{k}\geqslant\mathrm{1}} {\sum}{x}^{{k}−\mathrm{1}} \\ $$$$−\frac{\mathrm{1}}{{x}}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{x}^{{k}−\mathrm{1}} }{{k}}+\frac{\underset{{k}\geqslant\mathrm{1}} {\sum}{x}^{{k}−\mathrm{1}} }{{x}} \\ $$$$=\frac{\mathrm{1}}{{x}}\left(\underset{{k}\geqslant\mathrm{2}} {\sum}{x}^{{k}−\mathrm{1}} −\underset{{k}\geqslant\mathrm{2}} {\sum}\frac{{x}^{{k}−\mathrm{1}} }{{k}}\right)=\underset{{k}\geqslant\mathrm{2}} {\sum}\left(\frac{{k}−\mathrm{1}}{{k}}{x}^{\boldsymbol{{k}}−\mathrm{2}} \right) \\ $$$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{k}+\mathrm{1}}{{k}+\mathrm{2}}{x}^{{k}} \\ $$$${Df}\left({x}\right)=\left(\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{{k}+\mathrm{1}}{{k}+\mathrm{2}}{x}^{{k}} \right){f}^{\mathrm{2}} \left({x}\right) \\ $$
Commented by HongKing last updated on 19/Dec/21
A nice solution my dear Sir thank you
$$\mathrm{A}\:\mathrm{nice}\:\mathrm{solution}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you} \\ $$
Commented by HongKing last updated on 20/Dec/21
My dear Sir, can you imagine, what could we create by writing the derivative of a function in the way I proposed?
$$\mathrm{My}\:\mathrm{dear}\:\mathrm{Sir},\:\mathrm{can}\:\mathrm{you}\:\mathrm{imagine},\:\mathrm{what} \\ $$$$\mathrm{could}\:\mathrm{we}\:\mathrm{create}\:\mathrm{by}\:\mathrm{writing}\:\mathrm{the}\:\mathrm{derivative} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{function}\:\mathrm{in}\:\mathrm{the}\:\mathrm{way}\:\mathrm{I}\:\mathrm{proposed}? \\ $$

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