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Question Number 86598 by Rio Michael last updated on 29/Mar/20
 write out the general summation formula for   the maclaurin series expansion for  (1/2) (cos x + cosh x)
$$\:\mathrm{write}\:\mathrm{out}\:\mathrm{the}\:\mathrm{general}\:\mathrm{summation}\:\mathrm{formula}\:\mathrm{for} \\ $$$$\:\mathrm{the}\:\mathrm{maclaurin}\:\mathrm{series}\:\mathrm{expansion}\:\mathrm{for}\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\mathrm{cos}\:{x}\:+\:\mathrm{cosh}\:{x}\right) \\ $$
Commented by mathmax by abdo last updated on 31/Mar/20
cosx =Σ_(n=0) ^∞  (((−1)^n )/((2n)!))x^(2n)    and ch(x)=((e^x +e^(−x) )/2)  =(1/2)(Σ_(n=0) ^∞  (x^n /(n!)) +Σ_(n=0) ^∞  (((−1)^n )/(n!))x^n )  =(1/2)Σ_(n=0) ^∞ (1/(n!)){1+(−1)^n }x^n  =Σ_(n=0) ^∞  ((x^(2n)  ⇒)/((2n)!))  (1/2)(cosx +ch(x)) =(1/2){ Σ_(n=0) ^∞   (((−1)^n )/((2n)!))x^(2n)  +Σ_(n=0) ^∞  (1/((2n)!))x^(2n) }  =(1/2)Σ_(n=0) ^∞ ((1+(−1)^n )/((2n)!))x^(2n)  =Σ_(n=0) ^∞   (x^(4n) /((4n)!))
$${cosx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}{x}^{\mathrm{2}{n}} \:\:\:{and}\:{ch}\left({x}\right)=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{{n}!}\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}{x}^{{n}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{n}!}\left\{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} \right\}{x}^{{n}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{2}{n}} \:\Rightarrow}{\left(\mathrm{2}{n}\right)!} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({cosx}\:+{ch}\left({x}\right)\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}{x}^{\mathrm{2}{n}} \:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)!}{x}^{\mathrm{2}{n}} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}{x}^{\mathrm{2}{n}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{4}{n}} }{\left(\mathrm{4}{n}\right)!} \\ $$

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