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

Click For Summary
SUMMARY

The discussion centers on the differentiation of power series and the implications for the index of summation. The user references "Differential Equations" by Boyce and DiPrima, specifically pages 268 and 279, to illustrate how the index shifts when differentiating series term by term. The confusion arises from the apparent discrepancy in the treatment of indices during differentiation, particularly in the context of the method of Frobenius. Ultimately, the user resolves their confusion by recognizing that the first term, a0xr, remains after differentiation, leading to a correct understanding of the index behavior.

PREREQUISITES
  • Understanding of power series and their convergence
  • Familiarity with differentiation techniques for series
  • Knowledge of the method of Frobenius in solving differential equations
  • Basic concepts of recurrence relations in differential equations
NEXT STEPS
  • Study "Differential Equations" by Boyce and DiPrima for deeper insights into series solutions
  • Learn about the method of Frobenius in detail, focusing on its application to regular singular points
  • Explore the implications of termwise differentiation on power series
  • Practice solving differential equations using power series to solidify understanding of index manipulation
USEFUL FOR

Students and educators in mathematics, particularly those focusing on differential equations, series solutions, and the method of Frobenius. This discussion is beneficial for anyone seeking clarity on the differentiation of series and its effects on indices.

LoA
Messages
12
Reaction score
0

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:
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
8
Views
3K
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K