I am solving via power series the ivp $$y'-2xy=0,\quad y(1)=2.$$ The "solution" is $$y(x)=2\left(1+2(x-1)+3(x-1)^2+\frac{10}{3}(x-1)^3+\frac{19}{6}(x-1)^4+\frac{26}{10}(x-1)^5+\cdots\right)$$ with coefficients generated by the recurrence: $$(n+1)a_{n+1}-2a_{n-1}-2a_n=0,\quad n>0$$ or, $$a_{n+1}=\frac{2(a_n+a_{n-1})}{n+1}.$$ Due to the variable coefficient I'm not sure how to determine the radius of convergence from the recurrence. My naive attempt was to use $$\frac{a_{n+1}}{a_n}=\frac{2\left(1+\frac{a_{n-1}}{a_n}\right)}{n+1}$$ and assume that $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=R,\quad 0<R<\infty$$ to obtain a contradiction indicating that the radius is either 0 or infinite, but that is not very helpful. The generating function methods don't seem particularly helpful, but I'm quite likely wrong about that. I know that the solution (valid for all $x$) is $$y(x)=2\mathrm{e}^{x^2-1}.$$

Question: How would one obtain the radius of convergence of this series from the recurrence. As an aside are there any general methods that might indicate that the radius is greater than zero?


All $a_n$ are $>0$. From $a_{n+1}\sim{2\over n+1}a_{n-1}$ we see that we gain a factor $\sim{2\over n}$ in two steps. This observation leads to the claim that $$a_n\leq{4^n\over\sqrt{n!}}\qquad(n\geq1)\ .\tag{1}$$ Proof. The statement is true for $n=1$ and $n=2$, by inspection. Assume that it holds for $n-1$ and $n$. Then $$\eqalign{a_{n+1}&\leq{2\over n+1}\left({4^n\over\sqrt{n!}}+{4^{n-1}\over\sqrt{(n-1)!}}\right)\leq {2\over \sqrt{n(n+1)}}\left({4^n\over\sqrt{(n-1)!}}+{4^n\over\sqrt{(n-1)!}}\right)\cr &={4^{n+1}\over\sqrt{(n+1)!}}\ .\cr}$$ Using $(1)$ and, e.g., Stirlings fomula it is then easy to show that $$\rho=\limsup_{n\to\infty}{1\over|a_n|}=\infty\ .$$ As to your second question: The standard proof of Picard's theorem in a complex analysis setting shows that the solution is analytic in a neighborhood of $1\in{\mathbb C}$.

  • $\begingroup$ Thanks Christian, I've worked through the details. Was there a general idea that led to your claim or was it intuition/experience? $\endgroup$ – dumbquestions May 17 '16 at 14:06
  • $\begingroup$ aha! $\sqrt{n!}\leq n!!$ (thanks for the edit) $\endgroup$ – dumbquestions May 17 '16 at 16:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.