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I have this Ode $zS''+S'+zS=0$ $z\in\mathbb{C}$ and $S$ a 2 times differentiable function. I was looking into the formal power series $S(X)=\sum_{n\geq0}a_nX^n$ that verify this ODE and i'm supposed to show that the convergence radius is infinite.

I did some calculation and got this $$z\sum_{n \geq 2} n(n-1)a_nX^{n-2} +\sum_{n \geq 1} na_nX^{n-1} +z\sum_{n\geq 0} a_nX^n =0 \\ z\sum_{n\geq 0} (n+2)(n+1)a_{n+2}X^n+\sum_{n\geq 0} (n+1)a_{n+1}X^n+z\sum_{n\geq 0} a_nX^n =0 $$

By unicity of the coefficients $$z((n+2)(n+1)a_{n+2}+a_n)+(n+1)a_{n+1}=0$$

And i'm stuck here. I've tried separating the real and imaginary part, i get $$x((n+2)(n+1)a_{n+2}+a_n)+(n+1)a_{n+1}=0\\ y((n+2)(n+1)a_{n+2}+a_n)=0$$

But i don't know what to do to get further. the imaginary oe gets me a recursive relation : $$a_{n+2}=-\dfrac{a_n}{(n+2)(n+1)}$$

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I think your solution needs correction and you should try this $$z\sum_{n \geq 2} n(n-1)a_nz^{n-2} +\sum_{n \geq 1} na_nz^{n-1} +z\sum_{n\geq 0} a_nz^n =0 $$ $$\sum_{n \geq 2} n(n-1)a_nz^{n-1} +\sum_{n \geq 1} na_nz^{n-1} +\sum_{n\geq 0} a_nz^{n+1} =0 $$ and after simplification you have $a_1 =0 $ and $a_{n+2}=-\dfrac{a_n}{(n+2)^2}$. Thus odd terms are zero and the convergence radius follows by the ratio test $$\lim_\infty\frac{|a_{n+2}z^{2n+2}|}{|a_nz^n|}\to0$$

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  • $\begingroup$ You're right since formal series is purely algebric , i think we can even form an algebra with them, so it comes down to solving your equation which you perfectly did. Thank you ! $\endgroup$ – C.Patrick Mar 2 '17 at 15:26
  • $\begingroup$ Except there's a typo it's $z^{n+2}$ not $ 2n+2$ $\endgroup$ – C.Patrick Mar 2 '17 at 16:32
  • $\begingroup$ OK. Your right. thanks. $\endgroup$ – Nosrati Mar 2 '17 at 16:36
  • $\begingroup$ Is there a way to compute the coefficients easily? Because after showing $g(z)=\dfrac{1}{\pi}\int_0^\pi cos(zcos(t))dt $ is analytic on $\mathbb{C}$ which i did by showing its equal to the sum of power series with radius of convergence $\infty$, i need to show g is equal to S, the sum of the solution i found earlier, with $a_0=1$. I don't think i should calculate the coefficients since it's really hard by hand but the linearity throws me off. How do i show that? $\endgroup$ – C.Patrick Mar 3 '17 at 7:26
  • $\begingroup$ We can integrate term by term since the convergence radius is $\infty$ $g(z)= \dfrac{1}{\pi}\displaystyle\int_0^\pi cos(zcost(t)) dt = \dfrac{1}{\pi}\displaystyle\int_0^\pi \displaystyle\underset{n\geq 0}{\sum} \dfrac{z^{2n}cos^{2n}(t)}{(2n)!}dt= \dfrac{1}{\pi}\displaystyle\underset{n\geq 0}{\sum} \dfrac{z^{2n}}{(2n)!}\displaystyle\int_0^\pi cos^{2n}(t) dt= \displaystyle\underset{n\geq 0}{\sum} \dfrac{z^{2n}}{4^n(n!)^2}$ By using Wallis' integral formula $\endgroup$ – C.Patrick Mar 3 '17 at 7:46
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The solution is $S(z)=c_{1}J_{0}(z)+c_{2}Y_{0}(z)$. Where $J_{0}$ and $Y_{0}$ are Bessel's functions of 1s and second kind. For the convergence proof see Bessel function radius of convergence

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