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let-f-x-2-x-developp-f-at-integr-serie-and-give-the-radius-of-convergence-

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Question Number 34696 by abdo imad last updated on 10/May/18
let f(x) =(√(2−x)) developp f at integr serie and give the radius of convergence.
$${let}\:{f}\left({x}\right)\:=\sqrt{\mathrm{2}−{x}} \\ $$$${developp}\:{f}\:{at}\:{integr}\:{serie}\:{and}\:{give}\:{the}\:{radius}\:{of} \\ $$$${convergence}. \\ $$
Commented by math khazana by abdo last updated on 11/May/18
f(x)=(√2) (√(1−(x/2))) =(√2)(1−(x/2))^(1/2) but we know that (1+u)^α = 1+αu ((α(α−1))/(2!))u^2 +... =1+Σ_(n=1) ^∞ ((α(α−1)...(α−n+1))/(n!)) u^n and (1−u)^n =1+ Σ_(n=1) ^∞ α(α−1)...(α−n+1)(((−1)^n )/(n!)) u^n (1−(x/2))^(1/2) = 1 +Σ_(n=1) ^∞ (1/2)((1/2)−1)...((1/2)−n+1)(((−1)^n )/(n!)) (x^n /2^n ) =1 + Σ_(n=1) ^∞ (((−1)^n )/(n! 2^(n+1) ))(−(1/2))...(−(3/2))... (−((2n−3)/2)) x^n =1+ Σ_(n=1) ^∞ (((−1))/(n! 2^(n+1) )) .((1.3.5.....(2n−3))/2^(n−1) ) x^n = 1 + Σ_(n=1) ^∞ ((−1)/(n! 2^(2n) )) (1.3.5.....(2n−3))x^n ⇒ f(x) =(√2) −(√2) Σ_(n=1) ^∞ ((1.3.5....(2n−3))/(n! 2^(2n) )) x^n .
$${f}\left({x}\right)=\sqrt{\mathrm{2}}\:\sqrt{\mathrm{1}−\frac{{x}}{\mathrm{2}}}\:\:=\sqrt{\mathrm{2}}\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:{but}\:{we}\:{know}\:{that} \\ $$$$\left(\mathrm{1}+{u}\right)^{\alpha} \:\:=\:\mathrm{1}+\alpha{u}\:\frac{\alpha\left(\alpha−\mathrm{1}\right)}{\mathrm{2}!}{u}^{\mathrm{2}} \:+… \\ $$$$=\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\alpha\left(\alpha−\mathrm{1}\right)…\left(\alpha−{n}+\mathrm{1}\right)}{{n}!}\:{u}^{{n}} \:\:\: \\ $$$${and}\:\left(\mathrm{1}−{u}\right)^{{n}} \:=\mathrm{1}+\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\alpha\left(\alpha−\mathrm{1}\right)…\left(\alpha−{n}+\mathrm{1}\right)\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:{u}^{{n}} \\ $$$$\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:=\:\mathrm{1}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\right)…\left(\frac{\mathrm{1}}{\mathrm{2}}−{n}+\mathrm{1}\right)\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\frac{{x}^{{n}} }{\mathrm{2}^{{n}} } \\ $$$$=\mathrm{1}\:+\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\:\mathrm{2}^{{n}+\mathrm{1}} }\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)…\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)…\:\left(−\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}}\right)\:{x}^{{n}} \\ $$$$=\mathrm{1}+\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)}{{n}!\:\mathrm{2}^{{n}+\mathrm{1}} }\:.\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}…..\left(\mathrm{2}{n}−\mathrm{3}\right)}{\mathrm{2}^{{n}−\mathrm{1}} }\:{x}^{{n}} \\ $$$$=\:\mathrm{1}\:\:+\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{−\mathrm{1}}{{n}!\:\mathrm{2}^{\mathrm{2}{n}} }\:\left(\mathrm{1}.\mathrm{3}.\mathrm{5}…..\left(\mathrm{2}{n}−\mathrm{3}\right)\right){x}^{{n}} \:\Rightarrow \\ $$$${f}\left({x}\right)\:=\sqrt{\mathrm{2}}\:\:−\sqrt{\mathrm{2}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2}{n}−\mathrm{3}\right)}{{n}!\:\mathrm{2}^{\mathrm{2}{n}} }\:{x}^{{n}} \:\:. \\ $$$$ \\ $$

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