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Question Number 57486 by Abdo msup. last updated on 05/Apr/19
let f(x) =((ln(1+x))/(2−x^2 )) 1)calculate f^((n)) (x) 2) calculate f^((n)) (0) 3)developp f(x) at integr serie.
$${let}\:{f}\left({x}\right)\:=\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{2}−{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right){calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:\: \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right){developp}\:{f}\left({x}\right)\:{at}\:{integr}\:{serie}. \\ $$
Commented by maxmathsup by imad last updated on 07/Apr/19
1) we have f(x)=−((ln(1+x))/((x−(√2))(x+(√2)))) =−(1/(2(√2))){(1/(x−(√2))) −(1/(x+(√2))))ln(1+x) = (1/(2(√2))){ ((ln(1+x))/(x−(√2))) −((ln(1+x))/(x+(√2)))} ⇒f^((n)) (x) =(1/(2(√2))){ (((ln(1+x))/(x−(√2))))^n −(((ln(1+x))/(x+(√2))))^((n)) } let determine (((ln(1+x))/(x−(√2))))^((n)) leibniz formula give (((ln(1+x))/(x−(√2))))^((n)) =Σ_(k=0) ^n C_n ^k (ln(1+x))^((k)) ((1/(x−(√2))))^((n−k)) (ln(1+x))^((1) ) =(1/(1+x)) ⇒(ln(1+x))^()k)) =((1/(1+x)))^((k−1)) =(((−1)^(k−1) (k−1)!)/((x+1)^k )) ⇒ (((ln(1+x))/(x−(√2))))^((n)) =ln(1+x)(((−1)^n n!)/((x−(√2))^(n+1) )) +Σ_(k=1) ^n C_n ^k (((−1)^(k−1) (k−1)!)/((x+1)^k )) (((−1)^(n−k) (n−k)!)/((x−(√2))^(n−k+1) )) =(−1)^n n!((ln(1+x))/((x−(√2))^(n+1) )) +Σ_(k=1) ^n (−1)^(n−1) (k−1)!(n−k)! ((n!)/(k!(n−k)!(x+1)^k (x−(√2))^(n−k+1) )) =(−1)^n n! ((ln(1+x))/((x−(√2))^(n+1) )) +(−1)^(n−1) n! Σ_(k=1) ^n (1/(k(x+1)^k (x−(√2))^(n−k +1) )) also we have (((ln(1+x))/(x+(√2))))^((n)) =(−1)^n n!((ln(1+x))/((x+(√2))^(n+1) )) +(−1)^(n−1) n! Σ_(k=1) ^n (1/(k(x+1)^k (x+(√2))^(n−k+1) )) so the value of f^((n)) (x) isdetermined .
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)=−\frac{{ln}\left(\mathrm{1}+{x}\right)}{\left({x}−\sqrt{\mathrm{2}}\right)\left({x}+\sqrt{\mathrm{2}}\right)}\:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{2}}}\:−\frac{\mathrm{1}}{{x}+\sqrt{\mathrm{2}}}\right){ln}\left(\mathrm{1}+{x}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}−\sqrt{\mathrm{2}}}\:−\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}+\sqrt{\mathrm{2}}}\right\}\:\Rightarrow{f}^{\left({n}\right)} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\:\:\left(\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}−\sqrt{\mathrm{2}}}\right)^{{n}} \:−\left(\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}+\sqrt{\mathrm{2}}}\right)^{\left({n}\right)} \right\} \\ $$$${let}\:{determine}\:\left(\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}−\sqrt{\mathrm{2}}}\right)^{\left({n}\right)} \:\:{leibniz}\:{formula}\:{give} \\ $$$$\left(\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}−\sqrt{\mathrm{2}}}\right)^{\left({n}\right)} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left({k}\right)} \left(\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{2}}}\right)^{\left({n}−{k}\right)} \\ $$$$\left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left(\mathrm{1}\right)\:} \:=\frac{\mathrm{1}}{\mathrm{1}+{x}}\:\Rightarrow\left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left.\right)\left.{k}\right)} \:=\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{\left({k}−\mathrm{1}\right)} \:=\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}+\mathrm{1}\right)^{{k}} }\:\Rightarrow \\ $$$$\left(\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}−\sqrt{\mathrm{2}}}\right)^{\left({n}\right)} \:={ln}\left(\mathrm{1}+{x}\right)\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−\sqrt{\mathrm{2}}\right)^{{n}+\mathrm{1}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}+\mathrm{1}\right)^{{k}} }\:\frac{\left(−\mathrm{1}\right)^{{n}−{k}} \left({n}−{k}\right)!}{\left({x}−\sqrt{\mathrm{2}}\right)^{{n}−{k}+\mathrm{1}} } \\ $$$$=\left(−\mathrm{1}\right)^{{n}} {n}!\frac{{ln}\left(\mathrm{1}+{x}\right)}{\left({x}−\sqrt{\mathrm{2}}\right)^{{n}+\mathrm{1}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\left({n}−{k}\right)!\:\frac{{n}!}{{k}!\left({n}−{k}\right)!\left({x}+\mathrm{1}\right)^{{k}} \left({x}−\sqrt{\mathrm{2}}\right)^{{n}−{k}+\mathrm{1}} } \\ $$$$=\left(−\mathrm{1}\right)^{{n}} {n}!\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\left({x}−\sqrt{\mathrm{2}}\right)^{{n}+\mathrm{1}} }\:\:+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}!\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\:\:\frac{\mathrm{1}}{{k}\left({x}+\mathrm{1}\right)^{{k}} \left({x}−\sqrt{\mathrm{2}}\right)^{{n}−{k}\:+\mathrm{1}} }\:\:{also}\:{we}\:{have} \\ $$$$\left(\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}+\sqrt{\mathrm{2}}}\right)^{\left({n}\right)} \:=\left(−\mathrm{1}\right)^{{n}} {n}!\frac{{ln}\left(\mathrm{1}+{x}\right)}{\left({x}+\sqrt{\mathrm{2}}\right)^{{n}+\mathrm{1}} }\:+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}!\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\:\:\frac{\mathrm{1}}{{k}\left({x}+\mathrm{1}\right)^{{k}} \left({x}+\sqrt{\mathrm{2}}\right)^{{n}−{k}+\mathrm{1}} } \\ $$$${so}\:{the}\:{value}\:{of}\:{f}^{\left({n}\right)} \left({x}\right)\:{isdetermined}\:. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 07/Apr/19
2) we have f^((n)) (0) =(1/(2(√2))){ (−1)^(n−1) n! Σ_(k=1) ^n (1/(k(−(√2))^(n−k+1) )) −(−1)^(n−1) n! Σ_(k=1) ^n (1/(k((√2))^(n−k +1) ))} =(((−1)^(n−1) n!)/(2(√2))){ Σ_(k=1) ^n (1/k)( (1/((−(√2))^(n−k +1) )) −(1/(((√2))^(n−k +1) )))} =(((−1)^(n−1) n!)/(2(√2))) Σ_(k=1) ^n (1/k)((((√2))^(n−k +1) −(−(√2))^(n−k+1) )/((−2)^(n−k +1) )) =(((−1)^(n−1) n!)/(2(√2)(−1)^(n+1) )) Σ_(k=1) ^n ((((√2))^(n−k +1) −(−(√2))^(n−k +1) )/2^(n−k +1) )(−1)^k ⇒ f^((n)) (0) =((n!)/(2(√2))) Σ_(k=1) ^n (−1)^k ((((√2))^(n−k+1) −(−(√2))^(n−k +1) )/2^(n−k +1) ) .
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\:\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}!\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\:\frac{\mathrm{1}}{{k}\left(−\sqrt{\mathrm{2}}\right)^{{n}−{k}+\mathrm{1}} }\right. \\ $$$$\left.−\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}!\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\:\:\frac{\mathrm{1}}{{k}\left(\sqrt{\mathrm{2}}\right)^{{n}−{k}\:+\mathrm{1}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}!}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{{k}}\left(\:\:\frac{\mathrm{1}}{\left(−\sqrt{\mathrm{2}}\right)^{{n}−{k}\:+\mathrm{1}} }\:−\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}\right)^{{n}−{k}\:+\mathrm{1}} }\right)\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}!}{\mathrm{2}\sqrt{\mathrm{2}}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}\frac{\left(\sqrt{\mathrm{2}}\right)^{{n}−{k}\:+\mathrm{1}} \:−\left(−\sqrt{\mathrm{2}}\right)^{{n}−{k}+\mathrm{1}} }{\left(−\mathrm{2}\right)^{{n}−{k}\:+\mathrm{1}} } \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}!}{\mathrm{2}\sqrt{\mathrm{2}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(\sqrt{\mathrm{2}}\right)^{{n}−{k}\:+\mathrm{1}} \:−\left(−\sqrt{\mathrm{2}}\right)^{{n}−{k}\:+\mathrm{1}} }{\mathrm{2}^{{n}−{k}\:+\mathrm{1}} }\left(−\mathrm{1}\right)^{{k}} \:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{{n}!}{\mathrm{2}\sqrt{\mathrm{2}}}\:\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:\:\:\frac{\left(\sqrt{\mathrm{2}}\right)^{{n}−{k}+\mathrm{1}} \:−\left(−\sqrt{\mathrm{2}}\right)^{{n}−{k}\:+\mathrm{1}} }{\mathrm{2}^{{n}−{k}\:+\mathrm{1}} }\:. \\ $$
Commented by maxmathsup by imad last updated on 07/Apr/19
3) f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(n=1) ^∞ ((f^((n)) (0))/(n!)) x^n (f(0)=0) f(x) =(1/(2(√2))) Σ_(n=1) ^∞ { Σ_(k=1) ^n (−1)^k ((((√2))^(n−k+1) −(−(√2))^(n−k+1) )/2^(n−k+1) )} x^n .
$$\left.\mathrm{3}\right)\:{f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\:\:\:\:\left({f}\left(\mathrm{0}\right)=\mathrm{0}\right) \\ $$$${f}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \left\{\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \:\frac{\left(\sqrt{\mathrm{2}}\right)^{{n}−{k}+\mathrm{1}} \:−\left(−\sqrt{\mathrm{2}}\right)^{{n}−{k}+\mathrm{1}} }{\mathrm{2}^{{n}−{k}+\mathrm{1}} }\right\}\:{x}^{{n}} \:. \\ $$
Commented by maxmathsup by imad last updated on 07/Apr/19
forgive error of sine f(x) =(1/(2(√2))){ ((ln(1+x))/(x+(√2))) −((ln(1+x))/(x−(√2)))}
$${forgive}\:{error}\:{of}\:{sine}\:\:\:{f}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\:\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}+\sqrt{\mathrm{2}}}\:−\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}−\sqrt{\mathrm{2}}}\right\} \\ $$

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