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

LoA
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Homework Statement



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:

Homework 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}

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:
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LoA said:

Homework Statement



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:

Homework 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}

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.

Could possibly be a typo, but the extra terms really don't matter. I have this book too.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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