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sinx-x-b-n-0-1-n-2n-1-x-2n-1-x-b-prove-that-




Question Number 186941 by mathlove last updated on 12/Feb/23
((sinx)/x^b )=((Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))x^(2n+1) )/x^b )  prove that
$$\frac{{sinx}}{{x}^{{b}} }=\frac{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{x}^{\mathrm{2}{n}+\mathrm{1}} }{{x}^{{b}} } \\ $$$${prove}\:{that} \\ $$
Commented by mr W last updated on 12/Feb/23
x^b =x^b  needs no proof.  sin x=Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/((2n+1)!))   just apply taylor series for function  f(x)=sin x at point x=0.  ⇒see taylor series!
$${x}^{{b}} ={x}^{{b}} \:{needs}\:{no}\:{proof}. \\ $$$$\mathrm{sin}\:{x}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\: \\ $$$${just}\:{apply}\:{taylor}\:{series}\:{for}\:{function} \\ $$$${f}\left({x}\right)=\mathrm{sin}\:{x}\:{at}\:{point}\:{x}=\mathrm{0}. \\ $$$$\Rightarrow{see}\:{taylor}\:{series}! \\ $$

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