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Question Number 55269 by maxmathsup by imad last updated on 20/Feb/19
let f(x) =(2x+1)ln(1−x^2 ) 1) calculate f^((n)) (x) and f^((n)) (0) 2)developp f at integr serie. 3) calculate ∫_0 ^1 f(x)dx .
$${let}\:{f}\left({x}\right)\:=\left(\mathrm{2}{x}+\mathrm{1}\right){ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right){developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:. \\ $$
Commented by turbo msup by abdo last updated on 02/Mar/19
1)leibniz gormula give f^((n)) (x)=Σ_(k=0) ^n C_n ^k (2x+1)^((k)) (ln(1−x^2 )^((n−k)) =(2x+1)(ln(1−x^2 )^((n)) +2n(ln(1−x^2 )^((n−1)) we have ln(1−x^2 )^((1)) =((−2x)/(1−x^2 )) ⇒ (ln(1−x^2 )^((n)) =(−((2x)/(1−x^2 )))^((n−1)) =(((2x)/(x^2 −1)))^((n−1)) =((1/(x+1)) +(1/(x−1)))^((n−1)) =((1/(x+1)))^((n−1)) +((1/(x−1)))^((n−1)) =(((−1)^(n−1) (n−1)!)/((x+1)^n )) +(((−1)^(n−1) (n−1)!)/((x−1)^n ))
$$\left.\mathrm{1}\right){leibniz}\:{gormula}\:{give} \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(\mathrm{2}{x}+\mathrm{1}\right)^{\left({k}\right)} \left({ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\left({n}−{k}\right)} \right. \\ $$$$=\left(\mathrm{2}{x}+\mathrm{1}\right)\left({ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\left({n}\right)} \:+\mathrm{2}{n}\left({ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\left({n}−\mathrm{1}\right)} \right.\right. \\ $$$${we}\:{have}\:{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\left(\mathrm{1}\right)} \:=\frac{−\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\left({ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\left({n}\right)} \:=\left(−\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)^{\left({n}−\mathrm{1}\right)} \right. \\ $$$$=\left(\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\left({n}−\mathrm{1}\right)} \:=\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\:+\frac{\mathrm{1}}{{x}−\mathrm{1}}\right)^{\left({n}−\mathrm{1}\right)} \\ $$$$=\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{\left({n}−\mathrm{1}\right)} \:+\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}\right)^{\left({n}−\mathrm{1}\right)} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}+\mathrm{1}\right)^{{n}} }\:+\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}−\mathrm{1}\right)^{{n}} } \\ $$
Commented by turbo msup by abdo last updated on 02/Mar/19
also (ln(1−x^2 ))^((n−1)) =(((−1)^(n−2) (n−2)!)/((x+1)^(n−1) )) +(((−1)^(n−2) (n−2)!)/((x−1)^(n−1) )) ⇒ f^((n)) (x) =(2x+1)(−1)^(n−1) (n−1)!{(1/((x+1)^n )) +(1/((x−1)^n ))} +2n(−1)^(n−2) (n−2)!{(1/((x+1)^(n−1) )) +(1/((x−1)^(n−1) ))}
$${also}\:\left({ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\right)^{\left({n}−\mathrm{1}\right)} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{2}} \left({n}−\mathrm{2}\right)!}{\left({x}+\mathrm{1}\right)^{{n}−\mathrm{1}} }\:+\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{2}} \left({n}−\mathrm{2}\right)!}{\left({x}−\mathrm{1}\right)^{{n}−\mathrm{1}} }\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:=\left(\mathrm{2}{x}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\left\{\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{{n}} }\:+\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{{n}} }\right\} \\ $$$$+\mathrm{2}{n}\left(−\mathrm{1}\right)^{{n}−\mathrm{2}} \left({n}−\mathrm{2}\right)!\left\{\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{{n}−\mathrm{1}} }\:+\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{{n}−\mathrm{1}} }\right\} \\ $$
Commented by maxmathsup by imad last updated on 02/Mar/19
⇒f^((n)) (0) =(−1)^(n−1) (n−1)!{1+(−1)^n } +2n(−1)^n (n−2)!{1+(−1)^(n−1) } =−(−1)^n (n−1)!{1+(−1)^n } +2n(−1)^n (n−2)!{1−(−1)^n } ⇒ f^((n)) (0) =−(−1)^n (n−1)(n−2)!{1+(−1)^n } +2n(−1)^n (n−2)!{1−(−1)^n } =(−1)^n (n−2)!{(1−n)(1+(−1)^n ) +2n{1−(−1)^n }} =(−1)^n (n−2)!{1+(−1)^n −n−n(−1)^n +2n−2n(−1)^n } =(−1)^n (n−2)!{ 1+(1−3n)(−1)^n +n}
$$\Rightarrow{f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\left\{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \right\}\:+\mathrm{2}{n}\left(−\mathrm{1}\right)^{{n}} \left({n}−\mathrm{2}\right)!\left\{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \right\} \\ $$$$=−\left(−\mathrm{1}\right)^{{n}} \left({n}−\mathrm{1}\right)!\left\{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \right\}\:+\mathrm{2}{n}\left(−\mathrm{1}\right)^{{n}} \left({n}−\mathrm{2}\right)!\left\{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \right\}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=−\left(−\mathrm{1}\right)^{{n}} \left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)!\left\{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \right\}\:+\mathrm{2}{n}\left(−\mathrm{1}\right)^{{n}} \left({n}−\mathrm{2}\right)!\left\{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \right\} \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \left({n}−\mathrm{2}\right)!\left\{\left(\mathrm{1}−{n}\right)\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \right)\:+\mathrm{2}{n}\left\{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} \right\}\right\} \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \left({n}−\mathrm{2}\right)!\left\{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} −{n}−{n}\left(−\mathrm{1}\right)^{{n}} \:+\mathrm{2}{n}−\mathrm{2}{n}\left(−\mathrm{1}\right)^{{n}} \right\} \\ $$$$=\left(−\mathrm{1}\right)^{{n}} \left({n}−\mathrm{2}\right)!\left\{\:\mathrm{1}+\left(\mathrm{1}−\mathrm{3}{n}\right)\left(−\mathrm{1}\right)^{{n}} \:+{n}\right\} \\ $$
Commented by maxmathsup by imad last updated on 02/Mar/19
f^′ (x)=2ln(1−x^2 ) +(2x+1)((−2x)/(1−x^2 )) ⇒f^′ (0)=0 and f(0)=0 f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(n=2) ^∞ (((−1)^n )/(n(n−1))){n+1+(1−3n)(−1)^n } x^n =Σ_(n=2) ^∞ (((n+1)(−1)^n +1−3n)/(n(n−1)))x^n
$${f}^{'} \left({x}\right)=\mathrm{2}{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:+\left(\mathrm{2}{x}+\mathrm{1}\right)\frac{−\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\:\Rightarrow{f}^{'} \left(\mathrm{0}\right)=\mathrm{0}\:{and}\:{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:=\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left({n}−\mathrm{1}\right)}\left\{{n}+\mathrm{1}+\left(\mathrm{1}−\mathrm{3}{n}\right)\left(−\mathrm{1}\right)^{{n}} \right\}\:{x}^{{n}} \\ $$$$=\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\left({n}+\mathrm{1}\right)\left(−\mathrm{1}\right)^{{n}} +\mathrm{1}−\mathrm{3}{n}}{{n}\left({n}−\mathrm{1}\right)}{x}^{{n}} \\ $$

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