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Question Number 40260 by maxmathsup by imad last updated on 17/Jul/18
let f(x)=sin(2x) 1) find f^((n)) (x) and f^((n)) (0) 2) developp f at integr serie .
$${let}\:{f}\left({x}\right)={sin}\left(\mathrm{2}{x}\right) \\ $$$$\left.\mathrm{1}\right)\:{find}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$
Commented by maxmathsup by imad last updated on 18/Jul/18
1) we have f(x)=2sin(x)cos(x) ⇒f^((n)) (x)=2Σ_(k=0) ^n C_n ^k (sinx)^((k)) (cosx)^((n−k)) =2Σ_(k=0) ^n C_n ^k sin(x+((kπ)/2))cos(x +(((n−k)π)/2)) ⇒ f^((n)) (0) = 2 Σ_(k=0) ^n C_n ^k sin(((kπ)/2))cos((((n−k)π)/2)) =2 Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) sin((((2p+1)π)/2))cos((((n−2p−1)π)/2)) =2Σ_(p=0) ^([((n−1)/2)]) (−1)^p C_n ^(2p+1) cos( (((n−1)π)/2) −pπ) = 2 Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) cos((((n−1)π)/2)) 2) f(x) = Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =Σ_(n=1) ^∞ ((f^((n)) (0))/(n!)) x^n f(x)= Σ_(n=1) ^∞ (2/(n!)) cos((((n−1)π)/2)) (Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) ) x^n
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)=\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right)\:\Rightarrow{f}^{\left({n}\right)} \left({x}\right)=\mathrm{2}\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:\left({sinx}\right)^{\left({k}\right)} \left({cosx}\right)^{\left({n}−{k}\right)} \\ $$$$=\mathrm{2}\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:{C}_{{n}} ^{{k}} \:\:{sin}\left({x}+\frac{{k}\pi}{\mathrm{2}}\right){cos}\left({x}\:+\frac{\left({n}−{k}\right)\pi}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}}\right){cos}\left(\frac{\left({n}−{k}\right)\pi}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:\:{sin}\left(\frac{\left(\mathrm{2}{p}+\mathrm{1}\right)\pi}{\mathrm{2}}\right){cos}\left(\frac{\left({n}−\mathrm{2}{p}−\mathrm{1}\right)\pi}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\left(−\mathrm{1}\right)^{{p}} \:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:{cos}\left(\:\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}}\:−{p}\pi\right) \\ $$$$\:=\:\mathrm{2}\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:{cos}\left(\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}}\right) \\ $$$$\left.\mathrm{2}\right)\:{f}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \\ $$$${f}\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{2}}{{n}!}\:{cos}\left(\frac{\left({n}−\mathrm{1}\right)\pi}{\mathrm{2}}\right)\:\left(\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \right)\:{x}^{{n}} \\ $$$$ \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 18/Jul/18
2) another method we have sint =Im( e^(it) )=Im( Σ_(n=0) ^∞ (((it)^n )/(n!))) = Im( Σ_(n=0) ^∞ (((−1)^n t^(2n) )/((2n)!)) +iΣ_(n=0) ^∞ (((−1)^n t^(2n+1) )/((2n+1)!))) =Σ_(n=0) ^∞ (((−1)^n t^(2n+1) )/((2n+1)!)) ⇒ sin(2x) =Σ_(n=0) ^∞ (((−1)^n 2^(2n+1) x^(2n+1) )/((2n+1)!)) and the radius of convergence is R=+∞
$$\left.\mathrm{2}\right)\:{another}\:{method}\:\:{we}\:{have}\:{sint}\:={Im}\left(\:{e}^{{it}} \right)={Im}\left(\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left({it}\right)^{{n}} }{{n}!}\right) \\ $$$$=\:{Im}\left(\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{t}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}\:+{i}\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{t}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{t}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\:\Rightarrow \\ $$$${sin}\left(\mathrm{2}{x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \:{x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\:{and}\:{the}\:{radius}\:{of}\:{convergence}\:{is}\:{R}=+\infty \\ $$$$ \\ $$

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