Power Series derivation and recursion formula

Question

I am trying to solve a certain differential equation while using a power serias as a solution to it, but because I am not familiar with the rules when you apply derivation to a power series and how to shift the indices, I am unable to get the same result that there is in the book. This is a physics problem, but at the moment it's purely mathematics.

$$[\frac{d}{dx^2}-2x\frac {d}{dx} + (2\epsilon -1)]h(x)=0$$

where $(2\epsilon -1$ can be considered a constant.

In my notes, the suggested solution is of the form: $h(x)=\sum_{m=0}^{\infty}a_{2m}x^{2m+p}$

EDIT: $a_0\neq 0$

where (even though not said) $p$ is a constant.

Then I have:

I assume that when you derivative a power series, let's say once, the initial value that the index is incremented by 1, logically that should be the case.

$$\sum_{m=2}^{\infty}a_{2m}(2m+p)(2m+p-1)x^{2m+p-2}-2\sum_{m=1}^{\infty}a_{2m}(2m+p)x^{2m+p} + (2\epsilon -1)\sum_{m=0}^{\infty}a_{2m}x^{2m+p}$$

This can be written as:

$$\sum_{m=0}^{\infty}[(2m+p+2)(2m+p+1)a_{2m+2}-(4m+2p+2)a_{2m+1} + (2\epsilon -1)a_{2m}]x^{2m+p}=0$$

this means that:

$$(2m+p+2)(2m+p+1)a_{2m+2}-(4m+2p+2)a_{2m+1} + (2\epsilon -1)a_{2m}=0$$

But in the book it is:

$$(2m+p+2)(2m+p+1)a_{2m+2}=(4m+2p-2\epsilon +1)$$

This is completely different then what I get.

And let's say this is the correct result, the text says the following:

A normalized solution exists only if the power series breaks off.

What does this mean?

And it continuous by saying: The smallest term is $x^{p-2}$ but how is this possible. If you look in the 2nd equation, for $m=0$ you have $x^{2m+p}=x^{p}$

For $m=0$:

$$p(p-1)a_0=0$$ and from here we find that $p=0$ or $p=1$. But this is a total mess of an equation. First we consider the 2nd equation, even though the script gives us that last equation and on top of that , for no explained reason this is equal to zero. Can anyone help me understand what is going on here?

Answer 1

$\sum_{m=0}^{\infty}a_{2m}(2m+p)(2m+p-1)x^{2m+p-2}-2\sum_{m=0}^{\infty}a_{2m}(2m+p)x^{2m+p} + (2\epsilon -1)\sum_{m=0}^{\infty}a_{2m}x^{2m+p}$

If $p>1$ the $x^{p}$ term will not be eliminated by differentiation. So, we should not be re-indexing. If it turns out that $p = 0$ or $p = 1,$ we might need to circle back. Additionally, since every appearance of $m$ is in th context of $2m$ we would only be re-indexing the summation for the second derivative to $m = 1$

But when $m = 0$ we have an x^{p-2} term. And, $a_0(p)(p-1) x^{p-2} = 0$

We have been told that $a_0\ne 0$ so $p = 1$ or $p = 0,$ and some reindexing is in order.

If $p=1$

$\sum_{m=1}^{\infty}a_{2m}(2m+p)(2m+p-1)x^{2m+p-2}-2\sum_{m=0}^{\infty}a_{2m}(2m+p)x^{2m+p} + (2\epsilon -1)\sum_{m=0}^{\infty}a_{2m}x^{2m+p}$

To reindex that first summation replace $m$ with $m+1$ throughout.

$\sum_{m=0}^{\infty}a_{2(m+1)}(2(m+1)+p+1)(2(m+1)+p-1)x^{2(m+1)+p-2}-2\sum_{m=0}^{\infty}a_{2m}(2m+p)x^{2m+p} + (2\epsilon -1)\sum_{m=0}^{\infty}a_{2m}x^{2m+p}$

Giving the

$a_{2m+2}(2m + p + 2)(2m + p + 1) - (2(2m+p) + (2\epsilon - 1))a_{2m} = 0$ that the book shows.

If $p=0$ it doesn't require reindexing, but it does do something when we evaluate the constant term.

$a_0(2\epsilon - 1) = 0$