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Question Number 89510 by Ar Brandon last updated on 17/Apr/20
Find the expansion of Xe^(1/x)
$${Find}\:{the}\:{expansion}\:{of}\:{Xe}^{\frac{\mathrm{1}}{{x}}} \: \\ $$
Commented by mathmax by abdo last updated on 18/Apr/20
we have e^u =Σ_(n=0) ^∞ (u^n /(n!)) with radius r=+∞ ⇒ xe^(1/x) =x Σ_(n=0) ^∞ (1/(n! x^n )) =x(1 +Σ_(n=1) ^∞ (1/(n! x^n )))=x+Σ_(n=1) ^∞ (1/(n! x^(n−1) )) =x +(1/(1!)) +(1/(2!x)) +(1/(3!x^2 )) +....but this expansion is not integr serie if we want developpement st integr serie we write xe^(1/x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!))x^n with f(x)=xe^(1/x) by leibniz f^((n)) (x)=Σ_(k=0) ^n C_n ^k (x)^k (e^(1/x) )^((n−k)) = x (e^(1/x) )^((n)) +C_n ^1 (e^(1/x) )^((n−1)) (e^(1/x) )^((1)) =−(1/x^2 )e^(1/x) and (e^(1/x) )^((2)) =−(−((−2x)/x^4 )e^(1/x) +(1/x^2 )×(−(1/x^2 ))e^(1/x) ) =(−(2/x^3 ) +(1/x^4 ))e^(1/x) =(((−2x+1)/x^4 ))e^(1/x) any way f^((n)) =((p_n (x))/x^(2n) )e^(1/x) ....
$${we}\:{have}\:{e}^{{u}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{u}^{{n}} }{{n}!}\:\:{with}\:{radius}\:{r}=+\infty\:\Rightarrow \\ $$$${xe}^{\frac{\mathrm{1}}{{x}}} \:={x}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!\:{x}^{{n}} }\:={x}\left(\mathrm{1}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}!\:{x}^{{n}} }\right)={x}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}!\:{x}^{{n}−\mathrm{1}} } \\ $$$$={x}\:+\frac{\mathrm{1}}{\mathrm{1}!}\:+\frac{\mathrm{1}}{\mathrm{2}!{x}}\:+\frac{\mathrm{1}}{\mathrm{3}!{x}^{\mathrm{2}} }\:+….{but}\:{this}\:{expansion}\:{is}\:{not}\:{integr}\:{serie} \\ $$$${if}\:{we}\:{want}\:{developpement}\:{st}\:{integr}\:{serie}\:{we}\:{write} \\ $$$${xe}^{\frac{\mathrm{1}}{{x}}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}{x}^{{n}} \:\:{with}\:{f}\left({x}\right)={xe}^{\frac{\mathrm{1}}{{x}}} \:\:{by}\:{leibniz} \\ $$$${f}^{\left({n}\right)} \:\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} {C}_{{n}} ^{{k}} \:\left({x}\right)^{{k}} \left({e}^{\frac{\mathrm{1}}{{x}}} \right)^{\left({n}−{k}\right)} \:=\:{x}\:\left({e}^{\frac{\mathrm{1}}{{x}}} \right)^{\left({n}\right)} \:+{C}_{{n}} ^{\mathrm{1}} \left({e}^{\frac{\mathrm{1}}{{x}}} \right)^{\left({n}−\mathrm{1}\right)} \\ $$$$\left({e}^{\frac{\mathrm{1}}{{x}}} \right)^{\left(\mathrm{1}\right)} =−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{e}^{\frac{\mathrm{1}}{{x}}} \:\:{and}\:\left({e}^{\frac{\mathrm{1}}{{x}}} \right)^{\left(\mathrm{2}\right)} \:=−\left(−\frac{−\mathrm{2}{x}}{{x}^{\mathrm{4}} }{e}^{\frac{\mathrm{1}}{{x}}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }×\left(−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){e}^{\frac{\mathrm{1}}{{x}}} \right) \\ $$$$=\left(−\frac{\mathrm{2}}{{x}^{\mathrm{3}} }\:+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right){e}^{\frac{\mathrm{1}}{{x}}} \:=\left(\frac{−\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{4}} }\right){e}^{\frac{\mathrm{1}}{{x}}} \:\:\:{any}\:{way}\:{f}^{\left({n}\right)} =\frac{{p}_{{n}} \left({x}\right)}{{x}^{\mathrm{2}{n}} }{e}^{\frac{\mathrm{1}}{{x}}} \:\:\:…. \\ $$
Commented by Ar Brandon last updated on 18/Apr/20
Thank you
$${Thank}\:{you}\: \\ $$
Commented by mathmax by abdo last updated on 18/Apr/20
you are welcome .
$${you}\:{are}\:{welcome}\:. \\ $$

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