Determination of the Lower Bound on the Radius of Convergence of the Power Series Solution

convergence-divergenceordinary differential equationspower series

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Given the following theorem is true:

If $x_0$ is an ordinary point of the differential equation (1):
$$P(x)\ddot y + Q(x)\dot y + R(x)y = 0 $$
that is, if $p = Q/P$ and $q = R/P$ are analytic at $x_0$, then the general solution of Eq. (1) is
$$y = \sum_{n=0}^{\infty} a_n(x – x_0)^n = a_0y_1(x) + a_1y_2(x)$$
where $a_0$ and $a_1$ are arbitrary, and $y_1$ and $y_2$ are two power series solutions that are analytic at x_0. The solutions $y_1$ and $y_2$ form a fundamental set of solutions. Further, the radius of convergence for each of the series solutions $y_1$ and $y_2$ is at least as large as the minimum of the radii of convergence of the series for $p$ and $q$.

End Theorem

Now, given the following equation:
$$e^{-x}\ddot y + ln(1 + x)\dot y – xy = 0$$

Can you determine a lower bound on the radius of convergence and, if so, what is the value? I believe this is where the confusion manifested from in my previous question. I am looking for the lower bound on the radius of convergence, not the actual value.

Proceeding with the theorem:
$$p(x) = Q(x) / P(x) = \frac{ln(1+x)}{e^{-x}} = e^x*ln(1 + x)$$
$$q(x) = R(x) / P(x) = \frac{-x}{e^{-x}} = -xe^x$$

Now, I would say both $p(x)$ and $q(x)$ are analytic at $x_0 = 0$ and $P,Q,R$ are polynomials so the radius of convergence should be able to be determined.

absolute ratio test:
$$\left|x-x_0\right| \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n} \right | < 1$$

Series of $e^x$ about $x_0 = 0$:
$$e^x = \sum_{n=0}^{\infty} \frac {x^n}{n!}$$
Series of $ln(1+x)$ about $x_0 = 0$:
$$ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac {x^n}{n}$$

The rest of my question is nearly identical to my previous question.

How do you appropriately shift indices and multiply the two series together to then be able to apply the ratio test?

Best Answer

You should know about a statement like this:

Theorem 1: The solution is analytical (or an analytical solution exists) on any disk where the coefficients of the normalized equation are analytical.

See my answer in On the radius of convergence of solutions of analytic ODE's for a possible proof of this theorem, after transforming the second order equation into a first order system.

To apply this to composite coefficients, you do not need to actually explore the power series expansion of the full coefficients, there are rules for the composition of analytical functions like

Proposition 2: The product of two analytical functions is analytical on any disk where both factors are analytical.

The only expression in your equation that has limits on its regions of convergence is $\ln(1+x)$, which around $x_0=0$ is analytical on a disk with radius $1$.

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