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Question Number 71761 by TawaTawa last updated on 19/Oct/19

$$\mathrm{Find}\mathrm{at}\mathrm{least}\mathrm{the}\mathrm{first}\mathrm{four}\mathrm{non}\mathrm{zero}\mathrm{term}\mathrm{in}\mathrm{a}\mathrm{power}$$$$\mathrm{series}\mathrm{expansion}\mathrm{about}\mathrm{x}=0\mathrm{for}\mathrm{a}\mathrm{general}\mathrm{solution}$$$$\mathrm{to}{\mathrm{z}}^{″}-{\mathrm{x}}^2\mathrm{z}=0$$

Commented by mathmax by abdo last updated on 19/Oct/19

$$letz=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n\Rightarrow z^{{}^{\prime }}=\sum _{n=1}^{\mathrm{\infty }}{na}_n{x}^{n-1}and{z}^{{}^{″}}=\sum _{n=2}^{\mathrm{\infty }}n(n-1)a_n{x}^{n-2}$$$$\left(e\right)\Rightarrow \sum _{n=2}^{\mathrm{\infty }}{n}_{(ch.n-2=k)}(n-1)a_n{x}^{n-2}-x^2\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=0\Rightarrow$$$$=\sum _{k=0}^{\mathrm{\infty }}(k+2)(k+1)a_{k+2}{x}^k-\sum _{k=0(ch.k+2=n)}^{\mathrm{\infty }}{a}_k{x}^{k+2}=0\Rightarrow$$$$\sum _{k=0}^{\mathrm{\infty }}(k+1)(k+2)a_{k+2}{x}^k-\sum _{n=2}^{\mathrm{\infty }}{a}_{n-2}{x}^n=0\Rightarrow$$$$2a_2+6a_{3}x+\sum _{n=2}^{\mathrm{\infty }}\{(n+1)(n+2)a_{n+2}-a_{n-2}\}{x}^n=0\Rightarrow a_2=a_3=0and$$$$\mathrm{\forall }n⩾2(n+1)(n+2)a_{n+2}-a_{n-2}=0\Rightarrow \mathrm{\forall }n⩾2$$$$a_{n+2}=\frac{a_{n-2}}{(n+1)(n+2)}\Rightarrow a_2=\frac{a_0}{1\times 2}=0\Rightarrow a_0=0$$$$a_3=\frac{a_1}{2\times 6}=0\Rightarrow a_1=0$$$$wehave{a}_{2n+2}=\frac{a_{2n}}{(2n+1)(2n+2)}and$$$$a_{2n+3}=\frac{a_{2n-1}}{(2n+2)(2n+3)}$$$$$$

Commented by mathmax by abdo last updated on 19/Oct/19

$$wehave\frac{a_{2n+2}}{a_{2n}}=\frac{1}{2(n+1)(2n+1)}\Rightarrow$$$$\prod _{k=2}^n\frac{a_{2k+2}}{a_{2k}}=\frac{1}{2^{n-2}}\times \prod _{k=2}^n\frac{1}{(k+1)(2k+1)}\Rightarrow$$$$\frac{a_6}{a_4}.\frac{a_8}{a_6}......\frac{a_{2n+2}}{a_{2n}}=\frac{1}{2^{n-2}}\times \prod _{k=2}^n\frac{1}{k+1}\prod _{k=2}^n\frac{1}{2k+1}$$$$=\frac{1}{2^{n-2}}\times \frac{1}{3.4....(n+1)}\times \frac{1}{5.7.....(2n+1)}$$$$=\frac{2}{2^{n-2}}\times \frac{1}{(n+1)!}\times \frac{3.2.6.....\left(2n\right)}{2.3.5.6.7.....\left(2n\right)(2n+1)}$$$$=\frac{2}{2^{n-2}(n+1)!}\times \frac{3.2^{n}n!}{(2n+1)!}=2^{n+1-n+2}\times \frac{n!}{(n+1)!(2n+1)!}$$$$=\frac{8n!}{(n+1)!(2n+1)!}....$$

Commented by TawaTawa last updated on 19/Oct/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}$$

Commented by mathmax by abdo last updated on 20/Oct/19

$$youarewelcome.$$

Commented by TawaTawa last updated on 23/Oct/19

$$\mathrm{Sir},\mathrm{will}\mathrm{i}\mathrm{just}\mathrm{put}\mathrm{values}\mathrm{of}\mathrm{n}=1,2,3,4,....$$$$\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{least}\mathrm{first}\mathrm{four}\mathrm{term}?$$

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