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Question Number 31748 by abdo imad last updated on 13/Mar/18

1)find the value of Σ_(n=0) ^∞ (x^(3n) /((3n)!)) 2) find Σ_(n=0) ^∞ (8^n /((3n)!)) .

$$\left.\mathrm{1}\right){find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{3}{n}} }{\left(\mathrm{3}{n}\right)!} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{8}^{{n}} }{\left(\mathrm{3}{n}\right)!}\:\:. \\ $$

Commented by rahul 19 last updated on 14/Mar/18

solution for 1) part plz?? 2) it can be done by putting x=2.

$$\left.{solution}\:{for}\:\mathrm{1}\right)\:{part}\:{plz}?? \\ $$$$\left.\mathrm{2}\right)\:{it}\:{can}\:{be}\:{done}\:{by}\:{putting}\:{x}=\mathrm{2}. \\ $$

Commented by abdo imad last updated on 16/Mar/18

let put f(x)= e^x +e^(jx) +e^(j^2 x) we have f(x)= Σ_(n=0) ^∞ (x^n /(n!)) +Σ_(n=0) ^∞ ((j^n x^n )/(n!)) +Σ_(n=0) ^∞ ((j^(2n) x^n )/(n!)) (j=e^(i((2π)/3)) ) =Σ_(n=0) ^∞ (1+j^n +j^(2n) ) (x^n /(n!)) let study A_n =1+j^n +j^(2n) n=3k ⇒ A_n =1+j^(3k) +j^(6k) =3 n=3k+1 ⇒ A_n = 1+j^(3k+1) +j^(6k+2) =1+j +j^2 =0 n=3k+2 ⇒ A_n =1+j^(3k+2) +j^(6k+4) =1+j^2 +j=0?so f(x) =3Σ_(k=0) ^∞ (x^(3k) /((3k)!)) ⇒ Σ_(k=0) ^∞ (x^(3k) /((3k)!)) =(1/3)( e^x +e^(jx) +e^(j^2 x) ) . 2) let take x=2 inside f(x) we get Σ_(k=0) ^∞ (x^(3k) /((3k)!)) =(1/3)( e^2 +e^(2j) +e^(2j^2 ) ) but e^(2j) =e^(2(−(1/2)+i((√3)/2))) = e^(−1) (cos((√3))+isin((√3))) e^(2j^2 ) = e^(2(−(1/2)−i((√3)/2))) = e^(−1) (cos((√3) )−isin((√3))) Σ_(k=0) ^∞ (x^(3k) /((3k)!)) =(1/3) ( e^2 +2e^(−1) cos((√3))) .

$${let}\:{put}\:{f}\left({x}\right)=\:{e}^{{x}} \:+{e}^{{jx}} \:+{e}^{{j}^{\mathrm{2}} {x}} \:{we}\:{have} \\ $$$${f}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{{n}!}\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{j}^{{n}} \:{x}^{{n}} }{{n}!}\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{j}^{\mathrm{2}{n}} \:{x}^{{n}} }{{n}!}\:\:\:\left({j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(\mathrm{1}+{j}^{{n}} \:+{j}^{\mathrm{2}{n}} \right)\:\frac{{x}^{{n}} }{{n}!}\:\:{let}\:{study}\:{A}_{{n}} =\mathrm{1}+{j}^{{n}} \:+{j}^{\mathrm{2}{n}} \\ $$$${n}=\mathrm{3}{k}\:\Rightarrow\:{A}_{{n}} =\mathrm{1}+{j}^{\mathrm{3}{k}} \:+{j}^{\mathrm{6}{k}} \:=\mathrm{3}\:\: \\ $$$${n}=\mathrm{3}{k}+\mathrm{1}\:\Rightarrow\:{A}_{{n}} =\:\mathrm{1}+{j}^{\mathrm{3}{k}+\mathrm{1}} \:+{j}^{\mathrm{6}{k}+\mathrm{2}} \:=\mathrm{1}+{j}\:+{j}^{\mathrm{2}} =\mathrm{0} \\ $$$${n}=\mathrm{3}{k}+\mathrm{2}\:\Rightarrow\:{A}_{{n}} =\mathrm{1}+{j}^{\mathrm{3}{k}+\mathrm{2}} \:+{j}^{\mathrm{6}{k}+\mathrm{4}} \:=\mathrm{1}+{j}^{\mathrm{2}} \:+{j}=\mathrm{0}?{so} \\ $$$${f}\left({x}\right)\:=\mathrm{3}\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{3}{k}} }{\left(\mathrm{3}{k}\right)!}\:\:\Rightarrow\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{3}{k}} }{\left(\mathrm{3}{k}\right)!}\:=\frac{\mathrm{1}}{\mathrm{3}}\left(\:{e}^{{x}} \:+{e}^{{jx}} \:+{e}^{{j}^{\mathrm{2}} {x}} \right)\:. \\ $$$$\left.\mathrm{2}\right)\:{let}\:{take}\:{x}=\mathrm{2}\:{inside}\:{f}\left({x}\right)\:{we}\:{get} \\ $$$$\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{3}{k}} }{\left(\mathrm{3}{k}\right)!}\:=\frac{\mathrm{1}}{\mathrm{3}}\left(\:{e}^{\mathrm{2}} \:\:+{e}^{\mathrm{2}{j}} \:+{e}^{\mathrm{2}{j}^{\mathrm{2}} } \right)\:\:{but} \\ $$$${e}^{\mathrm{2}{j}} ={e}^{\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} \:=\:{e}^{−\mathrm{1}} \left({cos}\left(\sqrt{\mathrm{3}}\right)+{isin}\left(\sqrt{\mathrm{3}}\right)\right) \\ $$$${e}^{\mathrm{2}{j}^{\mathrm{2}} } \:=\:{e}^{\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} =\:{e}^{−\mathrm{1}} \:\left({cos}\left(\sqrt{\mathrm{3}}\:\right)−{isin}\left(\sqrt{\mathrm{3}}\right)\right) \\ $$$$\sum_{{k}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{3}{k}} }{\left(\mathrm{3}{k}\right)!}\:\:=\frac{\mathrm{1}}{\mathrm{3}}\:\left(\:{e}^{\mathrm{2}} \:\:+\mathrm{2}{e}^{−\mathrm{1}} {cos}\left(\sqrt{\mathrm{3}}\right)\right)\:. \\ $$

Commented by abdo imad last updated on 16/Mar/18

we have e^x +e^(jx) +e^(j^2 x) =e^x +e^(x(−(1/2)+i((√3)/2))) +e^(x(−(1/2)−i((√3)/2))) = e^x + e^(−(x/2)) (cos(x((√3)/2)) +isin(x((√3)/2))) +e^(−(x/2)) (cos(x((√3)/2))−isin(x((√3)/2))) ⇒Σ_(k=0) ^∞ (x^(3k) /((3k)!)) =(1/3)( e^x +2 e^(−(x/2)) cos(x((√3)/2))) .

$${we}\:{have}\:{e}^{{x}} \:+{e}^{{jx}} \:+{e}^{{j}^{\mathrm{2}} {x}} \:={e}^{{x}} \:+{e}^{{x}\left(−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} \:\:+{e}^{{x}\left(−\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)} \\ $$$$=\:{e}^{{x}} \:\:+\:{e}^{−\frac{{x}}{\mathrm{2}}} \left({cos}\left({x}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\:+{isin}\left({x}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right)\:+{e}^{−\frac{{x}}{\mathrm{2}}} \left({cos}\left({x}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−{isin}\left({x}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right) \\ $$$$\Rightarrow\sum_{{k}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{3}{k}} }{\left(\mathrm{3}{k}\right)!}\:=\frac{\mathrm{1}}{\mathrm{3}}\left(\:{e}^{{x}} \:\:+\mathrm{2}\:{e}^{−\frac{{x}}{\mathrm{2}}} \:{cos}\left({x}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right)\:\:. \\ $$

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