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Question Number 34698 by abdo imad last updated on 10/May/18
let f(x)=e^x sinx .developp f at integr serie.
$${let}\:{f}\left({x}\right)={e}^{{x}} \:{sinx}\:\:.{developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$
Commented by abdo mathsup 649 cc last updated on 10/May/18
f(x)=Im( e^x e^(ix) ) = Im( e^((1+i)x) ) but e^((1+i)x) = Σ_(n=0) ^∞ (((1+i)^n )/(n!)) x^n = Σ_(n=0) ^∞ (({(√2) e^(i(π/4)) }^n )/(n!)) x^n =Σ_(n=0) ^∞ ((((√2))^n e^(in(π/4)) )/(n!)) x^n ⇒ f(x)= Σ_(n=0) ^∞ ((((√2))^n )/(n!)) sin(n(π/4)) x^n .
$${f}\left({x}\right)={Im}\left(\:{e}^{{x}} {e}^{{ix}} \right)\:=\:{Im}\left(\:{e}^{\left(\mathrm{1}+{i}\right){x}} \right)\:{but} \\ $$$${e}^{\left(\mathrm{1}+{i}\right){x}} \:\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(\mathrm{1}+{i}\right)^{{n}} }{{n}!}\:{x}^{{n}} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left\{\sqrt{\mathrm{2}}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \right\}^{{n}} }{{n}!}\:{x}^{{n}} \:\: \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(\sqrt{\mathrm{2}}\right)^{{n}} \:{e}^{{in}\frac{\pi}{\mathrm{4}}} }{{n}!}\:{x}^{{n}} \:\Rightarrow \\ $$$${f}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(\sqrt{\mathrm{2}}\right)^{{n}} }{{n}!}\:{sin}\left({n}\frac{\pi}{\mathrm{4}}\right)\:{x}^{{n}} \:. \\ $$$$ \\ $$
Commented by abdo mathsup 649 cc last updated on 10/May/18
another method we have?f(x)= Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n but leibniz formula give f^((n)) (x)= Σ_(k=0) ^n C_n ^k (sinx)^((k)) (e^x )^((n−k)) = Σ_(k=0) ^n C_n ^k sin(x+k(π/2)) e^x ⇒ f^((n)) (0) = Σ_(k=0) ^n C_n ^k sin( k(π/2))⇒ f(x) = Σ_(n=0) ^∞ (1/(n!))( Σ_(k=0) ^n C_n ^k sin(k(π/2))) x^n .
$${another}\:{method} \\ $$$${we}\:{have}?{f}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\:\:\:{but}\:{leibniz} \\ $$$${formula}\:{give}\:{f}^{\left({n}\right)} \left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left({sinx}\right)^{\left({k}\right)} \:\left({e}^{{x}} \right)^{\left({n}−{k}\right)} \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:{sin}\left({x}+{k}\frac{\pi}{\mathrm{2}}\right)\:{e}^{{x}} \:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:{sin}\left(\:{k}\frac{\pi}{\mathrm{2}}\right)\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}!}\left(\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{sin}\left({k}\frac{\pi}{\mathrm{2}}\right)\right)\:{x}^{{n}} \:\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/May/18
p=e^x cosx q=e^x sinx p+iq=e^x cosx+ie^x sinx =e^x .e^(ix) =e^(x(1+i)) =x(1+i)+((x^2 .(1+i)^2 )/(2!))+((x^3 (1+i)^3 )/(3!))+((x^4 (1+i)^4 )/(4!))+... =(x+ix)+(((i2x^2 ))/(2!))+x^3 (1+3i−3−i)/3! +x^4 (((−4))/(4−!))+.. so e^x sinx= x+((2x^2 )/(2!))+x^3 (2/(3!))+....
$${p}={e}^{{x}} {cosx}\:\:\:{q}={e}^{{x}} {sinx} \\ $$$${p}+{iq}={e}^{{x}} {cosx}+{ie}^{{x}} {sinx} \\ $$$$={e}^{{x}} .{e}^{{ix}} \\ $$$$={e}^{{x}\left(\mathrm{1}+{i}\right)} \: \\ $$$$={x}\left(\mathrm{1}+{i}\right)+\frac{{x}^{\mathrm{2}} .\left(\mathrm{1}+{i}\right)^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} \left(\mathrm{1}+{i}\right)^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} \left(\mathrm{1}+{i}\right)^{\mathrm{4}} }{\mathrm{4}!}+… \\ $$$$=\left({x}+{ix}\right)+\frac{\left({i}\mathrm{2}{x}^{\mathrm{2}} \right)}{\mathrm{2}!}+{x}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{3}{i}−\mathrm{3}−{i}\right)/\mathrm{3}!\:+{x}^{\mathrm{4}} \frac{\left(−\mathrm{4}\right)}{\mathrm{4}−!}+.. \\ $$$${so}\:{e}^{{x}} {sinx}= \\ $$$${x}+\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}!}+{x}^{\mathrm{3}} \frac{\mathrm{2}}{\mathrm{3}!}+…. \\ $$

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