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

In summary, the index of summation shifts upward when differentiating a series term by term, as demonstrated in the example on page 268. However, in a later example on page 279, the indices of summation do not shift, possibly due to a typo, but this does not affect the outcome of the problem.
  • #1
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 [itex]f[/itex] is continuous and has derivatives of all orders for [itex]|x − x_{0} | < ρ [/itex].
Further, [itex] f ′ ,f ′′ , \cdots [/itex] can be computed by differentiating the series termwise; that is,

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

[itex]= \sum^{\infty}_{n = 1} na_{n}(x-x_{0})^{n-1} [/itex]

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

[itex]= \sum^{\infty}_{n = 2} n(n-1)a_{n}(x-x_{0})^{n-2} [/itex]

Note especially the shift upward in the lower index of summation corresponding to the elimination of the [itex] a_{0}, a_{1} [/itex] 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 [itex]y, y', y'' [/itex] as follows:

[itex]y = \sum^{\infty}_{n = 0} a_{n}(x)^{r+n} [/itex]

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

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

The Attempt at a Solution



If they simply skipped the step of decrementing the indices, then shouldn't all of the [itex] n[/itex]'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 [itex] a_{0}x^{r}, [/itex] which doesn't go away after differentiating? It becomes [itex] ra_{0}x^{r-1} [/itex]. Man I'm bad at this.
 
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  • #2
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 [itex]f[/itex] is continuous and has derivatives of all orders for [itex]|x − x_{0} | < ρ [/itex].
Further, [itex] f ′ ,f ′′ , \cdots [/itex] can be computed by differentiating the series termwise; that is,

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

[itex]= \sum^{\infty}_{n = 1} na_{n}(x-x_{0})^{n-1} [/itex]

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

[itex]= \sum^{\infty}_{n = 2} n(n-1)a_{n}(x-x_{0})^{n-2} [/itex]

Note especially the shift upward in the lower index of summation corresponding to the elimination of the [itex] a_{0}, a_{1} [/itex] 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 [itex]y, y', y'' [/itex] as follows:

[itex]y = \sum^{\infty}_{n = 0} a_{n}(x)^{r+n} [/itex]

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

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

The Attempt at a Solution



If they simply skipped the step of decrementing the indices, then shouldn't all of the [itex] n[/itex]'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 [itex] a_{0}x^{r}, [/itex] which doesn't go away after differentiating? It becomes [itex] ra_{0}x^{r-1} [/itex]. Man I'm bad at this.

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

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

1. What is a series solution to a differential equation?

A series solution to a differential equation is an infinite sum of terms that satisfies the differential equation when substituted into it. It is often used to approximate the solution to a differential equation when an exact solution cannot be found.

2. How is a series solution to a differential equation found?

A series solution is found by assuming a solution in the form of a power series, plugging it into the differential equation, and solving for the coefficients of the series. The series is then checked for convergence to ensure that it is a valid solution.

3. What is the index of summation in a series solution?

The index of summation refers to the variable that is being summed over in the series. In the context of a series solution to a differential equation, it represents the variable that is being differentiated in each term of the series.

4. Why is differentiation involved in a series solution to a differential equation?

Differentiation is involved in a series solution because it allows us to express the higher order derivatives of the solution in terms of the lower order derivatives. This makes it easier to solve for the coefficients of the series.

5. How accurate is a series solution to a differential equation?

The accuracy of a series solution depends on the number of terms included in the series. The more terms that are included, the more accurate the solution will be. However, since it is an approximation, there will always be some degree of error in the solution.

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