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Question Number 94334 by mathmax by abdo last updated on 18/May/20
let f(x) =((sinx)/x)if x≠0 and f(0)=1 1) findf^((n)) (x) and f^((n)) (0) 2)developp f at integr serie st x_0 =0 and x_0 =(π/2)
$${let}\:{f}\left({x}\right)\:=\frac{{sinx}}{{x}}{if}\:{x}\neq\mathrm{0}\:\:{and}\:{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{findf}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right){developp}\:{f}\:{at}\:{integr}\:{serie}\:{st}\:{x}_{\mathrm{0}} =\mathrm{0}\:{and}\:{x}_{\mathrm{0}} =\frac{\pi}{\mathrm{2}} \\ $$
Answered by abdomathmax last updated on 18/May/20
1) we have f(x) =((sinx)/x) ⇒f^((n)) (x) =Σ_(k=0) ^n C_n ^k ((1/x))^((k)) (sinx)^((n−k)) =(1/x)sin(x+((nπ)/2)) +Σ_(k=1) ^n C_n ^k (((−1)^k k!)/x^(k+1) )sin(x+(((n−k)π)/2)) 2) we have sinx =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))x^(2n+1) ⇒ ((sinx)/x) =Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!)) x^(2n)
$$\left.\mathrm{1}\right)\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{sinx}}{\mathrm{x}}\:\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)^{\left(\mathrm{k}\right)} \left(\mathrm{sinx}\right)^{\left(\mathrm{n}−\mathrm{k}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{x}}\mathrm{sin}\left(\mathrm{x}+\frac{\mathrm{n}\pi}{\mathrm{2}}\right)\:+\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{C}_{\mathrm{n}} ^{\mathrm{k}} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{k}!}{\mathrm{x}^{\mathrm{k}+\mathrm{1}} }\mathrm{sin}\left(\mathrm{x}+\frac{\left(\mathrm{n}−\mathrm{k}\right)\pi}{\mathrm{2}}\right) \\ $$$$\left.\mathrm{2}\right)\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{sinx}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)!}\mathrm{x}^{\mathrm{2n}+\mathrm{1}} \:\Rightarrow \\ $$$$\frac{\mathrm{sinx}}{\mathrm{x}}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)!}\:\mathrm{x}^{\mathrm{2n}} \\ $$

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