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Question Number 177957 by infinityaction last updated on 11/Oct/22
let a > 0 find the sum of the infinite series 1 + (((loga)^2 )/(2!)) + (((loga)^4 )/(4!)) + (((loga)^6 )/(6!))...
$$\mathrm{let}\:\:\mathrm{a}\:>\:\mathrm{0}\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{infinite}\: \\ $$$$\mathrm{series} \\ $$$$\:\mathrm{1}\:+\:\frac{\left(\mathrm{loga}\right)^{\mathrm{2}} \:}{\mathrm{2}!}\:+\:\frac{\left(\mathrm{loga}\right)^{\mathrm{4}} \:}{\mathrm{4}!}\:+\:\frac{\left(\mathrm{loga}\right)^{\mathrm{6}} \:}{\mathrm{6}!}… \\ $$
Answered by mr W last updated on 11/Oct/22
e^x =1+(x/(1!))+(x^2 /(2!))+(x^3 /(3!))+... e^(−x) =1−(x/(1!))+(x^2 /(2!))−(x^3 /(3!))+... e^x +e^(−x) =2(1+(x^2 /(2!))+(x^4 /(4!))+...) ((e^x +e^(−x) )/2)=1+(x^2 /(2!))+(x^4 /(4!))+... with x=log a ((a+(1/a))/2)=1+(((log a)^2 )/(2!))+(((log a)^4 )/(4!))+... ⇒1+(((log a)^2 )/(2!))+(((log a)^4 )/(4!))+...=((a^2 +1)/(2a))
$${e}^{{x}} =\mathrm{1}+\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$${e}^{−{x}} =\mathrm{1}−\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$${e}^{{x}} +{e}^{−{x}} =\mathrm{2}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+…\right) \\ $$$$\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}=\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+… \\ $$$${with}\:{x}=\mathrm{log}\:{a} \\ $$$$\frac{{a}+\frac{\mathrm{1}}{{a}}}{\mathrm{2}}=\mathrm{1}+\frac{\left(\mathrm{log}\:{a}\right)^{\mathrm{2}} }{\mathrm{2}!}+\frac{\left(\mathrm{log}\:{a}\right)^{\mathrm{4}} }{\mathrm{4}!}+… \\ $$$$\Rightarrow\mathrm{1}+\frac{\left(\mathrm{log}\:{a}\right)^{\mathrm{2}} }{\mathrm{2}!}+\frac{\left(\mathrm{log}\:{a}\right)^{\mathrm{4}} }{\mathrm{4}!}+…=\frac{{a}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{a}} \\ $$
Commented by infinityaction last updated on 11/Oct/22
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by Tawa11 last updated on 11/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by Ar Brandon last updated on 11/Oct/22
1+(x^2 /(2!))+(x^4 /(4!))+(x^6 /(6!))+∙∙∙=cosh(x) ⇒1+(((loga)^2 )/(2!))+(((loga)^4 )/(4!))+(((loga)^6 )/(6!))+∙∙∙=cosh(loga) =(1/2)(e^(loga) +e^(−loga) )=(1/2)(a+(1/a))=((a^2 +1)/(2a))
$$\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+\frac{{x}^{\mathrm{6}} }{\mathrm{6}!}+\centerdot\centerdot\centerdot=\mathrm{cosh}\left({x}\right) \\ $$$$\Rightarrow\mathrm{1}+\frac{\left(\mathrm{log}{a}\right)^{\mathrm{2}} }{\mathrm{2}!}+\frac{\left(\mathrm{log}{a}\right)^{\mathrm{4}} }{\mathrm{4}!}+\frac{\left(\mathrm{log}{a}\right)^{\mathrm{6}} }{\mathrm{6}!}+\centerdot\centerdot\centerdot=\mathrm{cosh}\left(\mathrm{log}{a}\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{\mathrm{log}{a}} +{e}^{−\mathrm{log}{a}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+\frac{\mathrm{1}}{{a}}\right)=\frac{{a}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{a}} \\ $$
Commented by Tawa11 last updated on 11/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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