Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

Question Number 92923 by Ar Brandon last updated on 09/May/20

Solve  1+(x/(2!))+(x^2 /(4!))+(x^3 /(6!))+∙∙∙ =0

$$\mathrm{Solve} \\ $$$$\mathrm{1}+\frac{\mathrm{x}}{\mathrm{2}!}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}!}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}!}+\centerdot\centerdot\centerdot\:=\mathrm{0} \\ $$

Answered by mr W last updated on 09/May/20

if x>0, LHS>0 ⇒no solution  i.e. x<0  let x=−t^2   1+(x/(2!))+(x^2 /(4!))+(x^3 /(6!))+∙∙∙   =1−(t^2 /(2!))+(t^4 /(4!))−(t^6 /(6!))+....=Σ_(n=0) ^∞ (((−1)^n t^(2n) )/((2n)!))  =cos t=0  ⇒t=2kπ±(π/2)  ⇒x=−(2kπ±(π/2))^2  with k∈Z

$${if}\:{x}>\mathrm{0},\:{LHS}>\mathrm{0}\:\Rightarrow{no}\:{solution} \\ $$$${i}.{e}.\:{x}<\mathrm{0} \\ $$$${let}\:{x}=−{t}^{\mathrm{2}} \\ $$$$\mathrm{1}+\frac{\mathrm{x}}{\mathrm{2}!}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}!}+\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}!}+\centerdot\centerdot\centerdot\: \\ $$$$=\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{t}^{\mathrm{4}} }{\mathrm{4}!}−\frac{{t}^{\mathrm{6}} }{\mathrm{6}!}+....=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {t}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$=\mathrm{cos}\:{t}=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{2}{k}\pi\pm\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow{x}=−\left(\mathrm{2}{k}\pi\pm\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} \:{with}\:{k}\in\mathbb{Z} \\ $$

Commented by Ar Brandon last updated on 09/May/20

��yes !

Terms of Service

Privacy Policy

Contact: info@tinkutara.com