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Question-163467

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Question Number 163467 by Zaynal last updated on 07/Jan/22
Answered by mr W last updated on 07/Jan/22
e^x =1+(x/(1!))+(x^2 /(2!))+(x^3 /(3!))...=Σ_(n=0) ^∞ (x^(2n) /((2n)!))+Σ_(n=0) ^∞ (x^(2n+1) /((2n+1)!)) ...(i) e^(−x) =1−(x/(1!))+(x^2 /(2!))−(x^3 /(3!))+...=Σ_(n=0) ^∞ (x^(2n) /((2n)!))−Σ_(n=0) ^∞ (x^(2n+1) /((2n+1)!)) ...(ii) (i)+(ii): e^x +e^(−x) =2Σ_(n=0) ^∞ (x^(2n) /((2n)!)) ⇒Σ_(n=0) ^∞ (x^(2n) /((2n)!))=((e^x +e^(−x) )/2)=cosh x
$${e}^{{x}} =\mathrm{1}+\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}…=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\:\:…\left({i}\right) \\ $$$${e}^{−{x}} =\mathrm{1}−\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+…=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$${e}^{{x}} +{e}^{−{x}} =\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$\Rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}=\mathrm{cosh}\:{x} \\ $$
Commented by Tawa11 last updated on 07/Jan/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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