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Question Number 34863 by a.i msup by abdo last updated on 12/May/18
let f(x)= ((artan(x+1))/(1+2x)) developp f at integr serie .
$${let}\:{f}\left({x}\right)=\:\frac{{artan}\left({x}+\mathrm{1}\right)}{\mathrm{1}+\mathrm{2}{x}} \\ $$$${developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$
Commented by math khazana by abdo last updated on 13/May/18
for ∣x∣<(1/2) we have (1/(1+2x)) = Σ_(n=0) ^∞ (−2x)^n =Σ_(n=0) ^∞ (−2)^n x^n let put w(x)=arctan(x+1) w^′ (x) = (1/(1+(x+1)^2 )) = (1/((x+1)^2 −i^2 )) = (1/((x+1−i)(x+1+i))) =(1/(2i)) ( (1/(x+1−i)) −(1/(x+1+i))) ⇒ ⇒w^((n+1)) (x) =(1/(2i)){ ((1/(x+1−i)))^((n)) −((1/(x+1+i)))^((n)) } =(1/(2i)) (((−1)^n n!)/((x+1−i)^(n+1) )) −(1/(2i)) (((−1)^n n!)/((x+1+i)^(n+1) )) ⇒ w^((n+1)) (0) = (((−1)^n n!)/(2i(1−i)^(n+1) )) −(((−1)^n n!)/(2i(1+i)^(n+1) )) ⇒w^((n)) (0)= (((−1)^(n−1) (n−1)!)/(2i)){ (((1+i)^n −(1−i)^n )/2^n )} = (((−1)^(n−1) (n−1)!)/(2i .2^n )) 2iIm(1+i)^n =(((−1)^(n−1) (n−1)!)/2^n ) ((√2))^n sin(n(π/4)) we have w(x)= Σ_(n=0) ^∞ ((w^((n)) (0))/(n!)) x^n = 1+Σ_(n=1) ^∞ (((−1)^(n−1) (n−1)! ((√2))^n )/(2^n n!)) sin(n(π/4))x^n f(x)=( Σ_(n=0) ^∞ a_n x^n ) (Σ_(n=0) ^∞ b_n x^n ) =Σ c_n x^n with c_n = Σ_(i+j=n) a_i bj so tbe developpement of f(x) is known .
$${for}\:\mid{x}\mid<\frac{\mathrm{1}}{\mathrm{2}}\:\:{we}\:{have}\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}{x}}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{2}{x}\right)^{{n}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{2}\right)^{{n}} \:{x}^{{n}} \:\:\:{let}\:{put}\:{w}\left({x}\right)={arctan}\left({x}+\mathrm{1}\right) \\ $$$${w}^{'} \left({x}\right)\:=\:\:\frac{\mathrm{1}}{\mathrm{1}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −{i}^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}−{i}\right)\left({x}+\mathrm{1}+{i}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\:\left(\:\frac{\mathrm{1}}{{x}+\mathrm{1}−{i}}\:−\frac{\mathrm{1}}{{x}+\mathrm{1}+{i}}\right)\:\Rightarrow \\ $$$$\Rightarrow{w}^{\left({n}+\mathrm{1}\right)} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\left(\frac{\mathrm{1}}{{x}+\mathrm{1}−{i}}\right)^{\left({n}\right)} \:−\left(\frac{\mathrm{1}}{{x}+\mathrm{1}+{i}}\right)^{\left({n}\right)} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}+\mathrm{1}−{i}\right)^{{n}+\mathrm{1}} }\:−\frac{\mathrm{1}}{\mathrm{2}{i}}\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{n}!}{\left({x}+\mathrm{1}+{i}\right)^{{n}+\mathrm{1}} }\:\Rightarrow \\ $$$${w}^{\left({n}+\mathrm{1}\right)} \left(\mathrm{0}\right)\:=\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\mathrm{2}{i}\left(\mathrm{1}−{i}\right)^{{n}+\mathrm{1}} }\:−\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\mathrm{2}{i}\left(\mathrm{1}+{i}\right)^{{n}+\mathrm{1}} } \\ $$$$\Rightarrow{w}^{\left({n}\right)} \left(\mathrm{0}\right)=\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\:\:\frac{\left(\mathrm{1}+{i}\right)^{{n}} \:−\left(\mathrm{1}−{i}\right)^{{n}} }{\mathrm{2}^{{n}} }\right\} \\ $$$$=\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}\:.\mathrm{2}^{{n}} }\:\:\mathrm{2}{iIm}\left(\mathrm{1}+{i}\right)^{{n}} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}^{{n}} }\:\left(\sqrt{\mathrm{2}}\right)^{{n}} \:{sin}\left({n}\frac{\pi}{\mathrm{4}}\right)\:\:{we}\:{have} \\ $$$${w}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{w}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\: \\ $$$$=\:\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\:\left(\sqrt{\mathrm{2}}\right)^{{n}} }{\mathrm{2}^{{n}} \:\:{n}!}\:{sin}\left({n}\frac{\pi}{\mathrm{4}}\right){x}^{{n}} \\ $$$${f}\left({x}\right)=\left(\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}} {x}^{{n}} \right)\:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{b}_{{n}} {x}^{{n}} \right)\:=\Sigma\:{c}_{{n}} {x}^{{n}} \:\:{with} \\ $$$${c}_{{n}} =\:\sum_{{i}+{j}={n}} \:\:{a}_{{i}} {bj}\:\:\:{so}\:{tbe}\:{developpement}\:{of}\:{f}\left({x}\right)\:{is} \\ $$$${known}\:. \\ $$$$ \\ $$

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