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So I have a question that says find the radius of convergence after I have found the power series solution of a given differential equation. I know to find the radius of convergence you take

$$ p=\lim_{n \rightarrow \infty} \left\lvert\frac{C_n}{C_{n+1}}\right\rvert $$

but, I don't understand where $C_n$ and $C_{n+1}$ come from. Could someone please explain how I would find $C_{n}$ and $C_{n+1}$?

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  • $\begingroup$ Last time I saw the formula for the radius of convergence, it was not $$\lim_{n\to\infty}\left\lvert\frac{C_n}{C_{n+1}}\right\rvert$$ $\endgroup$ – user228113 May 24 '16 at 22:54
  • $\begingroup$ In my textbook it says Theorem 3 Radius of Convergence and then it givens that $\endgroup$ – user6259845 May 24 '16 at 22:56
  • $\begingroup$ Ah, ok. It appears that, if the aforementioned limit exists (and it might not), then it is the same. $\endgroup$ – user228113 May 24 '16 at 22:57
  • $\begingroup$ The $C_n$-s are the coefficients of the Taylor polynomial at $0$, which can be computed recursively according to te specifical instance of the ODE. $\endgroup$ – user228113 May 24 '16 at 23:00
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An elementary (and not particularly smart) approach could be:

Let's say the problem is $$\begin{cases}y'(x)=f(x,y(x))\\ y(0)=y_0\end{cases}$$

Since $C_n=\dfrac{y^{(n)}(0)}{n!}$, you "only" need to compute $y^{(n)}(0)$.

You know that $y'(0)=f(0,y_0)$.

Deriving in $x$ the first equation you get $$y''(x)=\frac{\partial f}{\partial x}(x,y(x))+y'(x)\frac{\partial f}{\partial y}(x,y(x))$$

Whence $y''(0)=\dfrac{\partial f}{\partial x}(0,y_0)+y'(0)\dfrac{\partial f}{\partial y}(0,y_0)$. Notice that you have already calculated $y'(0)$ the step before.

Keep deriving $$y^{(3)}(x)=\\= \dfrac{\partial^2 f}{\partial x^2}(x,y(x))+2y'(x)\dfrac{\partial^2 f}{\partial x\partial y}(x,y(x))+y''(x)\dfrac{\partial f}{\partial y}(x,y(x))+(y'(x))^2\dfrac{\partial^2 f}{\partial y^2}(x,y(x)) $$

Again, you can evaluate everything in $x=0,\ y=y_0$ and get $y^{(3)}(0)$.

The formulas rapidly worsen the more you derive, but perhaps the specific instance of the problem simplifies the calculations.

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$p(x) = \sum c_n x^n$

The ratio test: a series converges (power-series of otherwise) the ratio of all successive elements greater than some N are less than 1.

That is, if there exists an $N$ such that $n>N \implies |\frac{c_{n+1} x^{n+1}}{c_{n} x^{n}}|<1$ then the series converges.

$|\frac{c_{n+1} x^{n+1}}{c_{n} x^{n}}| = |\frac{c_{n+1}}{c_{n}}||x|$

$\lim_\limits{n\to\infty}$$|\frac{c_{n}}{c_{n+1}}| = r$ and $\frac {|x|}r < 1$

G. Sassatelli tells me that I might have misundrstood your problem.

How do you find the power series....

You say Suppose: $y = \sum_\limits{n=0}^{\infty} c_n x^n$

$y' = \sum_\limits{n=0}^{\infty} n c_n x^{n-1}$

And if necessary

$y'' = \sum_\limits{n=0}^{\infty} (n)(n-1) c_{n} n x^{n-2}$

Now you plug these into your original equation. Set $c_0 = y(0)$ find and equation such that $c_{n+1} = f(c_n)$ And then try to find a general formula that for $c_n$

A simple example

$y' = y, y(0)= 1\\ y' = \sum_\limits{n=0}^{\infty} n c_n x^{n-1}\\ y' = \sum_\limits{n=1}^{\infty} (n+1)c_{n+1} x^{n}\\ \sum_\limits{n=1}^{\infty} (n+1)c_{n+1} x^{n} = \sum_\limits{n=0}^{\infty} c_{n} x^{n}\\ (n+1) c_{n+1} = c_n\\ c_n = \frac{c_0}{n!}\\ y = \sum_\limits{n=0}^{\infty} \frac{x^{n}}{n!}$

Now, in this case the radius of convergence is infinite, but that is not always true.

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  • $\begingroup$ I think the question asks for something else: how do you compute the $C_n$-s from a given ODE? $\endgroup$ – user228113 May 24 '16 at 23:04

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