# Series soln to d.e. - Index of summation after differentiation

1. May 11, 2013

### LoA

1. The problem statement, all variables and given/known data

I am confused about what happens to the index of summation when I differentiate a series term by term. Let me show you two examples from my diff eq book (boyce and diprima) which are the primary source of my confusion:

2. Relevant equations

From page 268:

The function $f$ is continuous and has derivatives of all orders for $|x − x_{0} | < ρ$.
Further, $f ′ ,f ′′ , \cdots$ can be computed by differentiating the series termwise; that is,

$f'(x) = a_{1} + 2a_{2}(x-x_{0}) + \cdots + na_{n}(x-x_{0})^{n-1} + \cdots$

$= \sum^{\infty}_{n = 1} na_{n}(x-x_{0})^{n-1}$

$f''(x) = 2a_{2} + 6a_{3}(x-x_{0}) + \cdots + n(n-1)a_{n}(x-x_{0})^{n-2} + \cdots$

$= \sum^{\infty}_{n = 2} n(n-1)a_{n}(x-x_{0})^{n-2}$

Note especially the shift upward in the lower index of summation corresponding to the elimination of the $a_{0}, a_{1}$ terms in the process of differentiating the series.

But later in the book on page 279, in an example demonstrating the method of Frobenius for solving a d.e. around a regular singular point, the book calculates $y, y', y''$ as follows:

$y = \sum^{\infty}_{n = 0} a_{n}(x)^{r+n}$

$y' = \sum^{\infty}_{n = 0} (r+n)a_{n}(x)^{r+n-1}$

$y'' = \sum^{\infty}_{n = 0} (r+n)(r+n-1)a_{n}(x)^{n-2}$

3. The attempt at a solution

If they simply skipped the step of decrementing the indices, then shouldn't all of the $n$'s inside the summation be incremented accordingly? Since getting the degree's of the x terms and the indices of summation set up correctly are important steps to setting up the recurrence relation for solving these problems, I'm very confused about this apparent discrepancy. I've played around with it both ways and as far as I can tell it seems to make a difference in the outcome of the problem. Is there something I'm missing about how differentiation interacts with the sigma notation for a series? I'm not very good with series atm but I'm trying to improve, so any help or pointers in this general area are all very welcome.

EDIT: NVM I think I figured it out 30 seconds after clicking submit. Is it because the first term is $a_{0}x^{r},$ which doesn't go away after differentiating? It becomes $ra_{0}x^{r-1}$. Man I'm bad at this.

Last edited: May 11, 2013
2. May 11, 2013

### Zondrina

Could possibly be a typo, but the extra terms really don't matter. I have this book too.