# Power Series Following an example problem

In summary, the speakers are discussing the power series method for solving differential equations with variable coefficients. They are trying to determine the appropriate procedure for expanding the coefficients in the power series solution, and the conversation touches on the differences between constant and linear coefficients. They also discuss the use of Taylor series expansions and how they relate to polynomials.

## Homework Statement

I am following along in an example problem and I am getting hung up on a step. We are seeking a power series solution of the DE:

$$(x - 1)y'' + y' +2(x - 1)y = 0 \qquad(1)$$

With the initial values $y(4) = 5 \text{ and }y'(4) = 0$. We seek the solution in the form

$$y(x) = \sum_{n=0}^{\infty}a_n(x - x_o)^n\qquad(2)$$Here it is convenient to let xo = 4. So we seek$$y(x) = \sum_{n=0}^{\infty}a_n(x - 4)^n\qquad(3)$$

Here is the book text:

To put (1) in standard form we divide by x - 1 which yields

$$y'' + \frac{1}{x - 1}y' + 2y = 0 \qquad(4)$$

Essentially we put (1) in the standard form of (4) so that we can test for analyticity and we find that p(x) and q(x) in $y'' + p(x)y' +q(x)y = 0$ are analytic at xo = 4.

Ok that's all great. Here is where I lose them:
To proceed we can either use the form (1) or (4). Since we are expanding each term in the differential equation about x = 4, we need to expand (x - 1) and 2(x - 1) if we use (1), or the 1/(x - 1) factor if we use (4). The former is easier since $x - 1 = 3 + (x - 4)\qquad(5)$ is merely a two-term Taylor series, whereas 1/(x - 1) is an infinite series ...
I am not sure what is happening here. Why are we expanding the coefficients of the DE (1) ? When we had a simple constant coefficient DE, the procedure was simple: i. assume y(x) takes the form of (2); ii. differentiate (2) as many times as necessary and plug back into DE; iii. tidy up.

Now we have a DE with variable coefficients the procedure is changing a bit, but I am not seeing how.

Let's talk generally for a moment. If I have a DE of the form $A_1(x)y''(x) + A_2(x)y'(x) + A_3(x)y = 0 \qquad(6)$, and I seek the solution in the form of $y(x) = \sum_{n=0}^{\infty}a_n(x - x_o)^n\qquad(2)$, how does the procedure that I outlined above for the constant coefficient case (italicized) change for the case of (6) with the Ai(x) coefficients? Do I need to expand my Ai(x) ' s about xo as well? If yes, why? (I have a feeling I know the answer, but would like verification. I feel like it only makes sense that my coefficients that depend on x should 'track' my assumed y(x). I know that is not very rigorous, but it's all I am coming up with.)

The power series method ends with adding factors of (x - 4)n, and putting them equal to zero, for each n separately.

With constant coefficients, that's easy … an(x-4)n becomes nan(x-4)n-1 and so on, only one term at a time.

But with linear coefficients, eg p + q(x-4), you get pan and qan-1 in the same equation (ie for the same power of (x-4)).

With quadratic coefficients, you get three terms, and so on (and a Taylor expansion is of course even worse).

For a Taylor expansion, you'd get an infinite number of terms for each power.

Let's talk generally for a moment. If I have a DE of the form $A_1(x)y''(x) + A_2(x)y'(x) + A_3(x)y = 0 \qquad(6)$, and I seek the solution in the form of $y(x) = \sum_{n=0}^{\infty}a_n(x - x_o)^n\qquad(2)$, how does the procedure that I outlined above for the constant coefficient case (italicized) change for the case of (6) with the Ai(x) coefficients? Do I need to expand my Ai(x) ' s about xo as well? If yes, why? (I have a feeling I know the answer, but would like verification. I feel like it only makes sense that my coefficients that depend on x should 'track' my assumed y(x). I know that is not very rigorous, but it's all I am coming up with.)

(ooh, i like your use of "black"! )

It's not really a question of expanding the As about x0, rather of turning the As into a polynomial in (x - x0) … though of course it comes to the same thin.

If your y is a polynomial in (x - x0), then your As must be also.

tiny-tim said:

The power series method ends with adding factors of (x - 4)n, and putting them equal to zero, for each n separately.

With constant coefficients, that's easy … an(x-4)n becomes nan(x-4)n-1 and so on, only one term at a time.

But with linear coefficients, eg p + q(x-4), you get pan and qan-1 in the same equation (ie for the same power of (x-4)).

With quadratic coefficients, you get three terms, and so on (and a Taylor expansion is of course even worse).

For a Taylor expansion, you'd get an infinite number of terms for each power.

(ooh, i like your use of "black"! )

It's not really a question of expanding the As about x0, rather of turning the As into a polynomial in (x - x0) … though of course it comes to the same thin.

If your y is a polynomial in (x - x0), then your As must be also.

Ok. I think I follow you. Thanks tiny-tim! I think I will figure this out when I go to solve a problem. (I am actually working on one now that I need a hand with; so look out for my next post if you have time )

Hi tiny-tim again

I am working in another Power Series solution in which $x_o=0$. I also have that the coefficient of y is a polynomial. I am thinking that because it is polynomial, then the taylor series expansion it about zero is that polynomial. Is this true in general? If we take $$TS\left[f(x)\right]_{x_o}$$ to mean the Taylor series of f(x) expanded about xo, then if p(x) is some polynomial we have

$$TS\left[p(x)\right]_{x_o=0}=p(x)$$

Since by definition:

$$TS\left[f(x)\right]_{x_o} = f(x_o) + \frac{f'(x_o)}{1!}(x-x_o)+\frac{f''(x_o)}{2!}(x-x_o)+ \dots$$

Which will give:

$$TS\left[p(x)\right]_{x_o} = p(x_o) + \frac{p'(x_o)}{1!}(x)+\frac{p''(x_o)}{2!}(x)+ \dots$$

Which I am thinking is the same as p(x) ... but I am not sure what rule I can use to prove it ...
Any thoughts?

Yes, that's fine … the Taylor series of a polynomial about zero is itself.

The proof is trivial, isn't it? … f(n)xk|x=0 = k! if n = k, and = 0 otherwise.

(btw, I vaguely remember that that's used in some important proof about differentiating Taylor series … it's considered obvious)

Okie dokie! Thanks again!

## 1. What is a power series?

A power series is an infinite series of the form ∑n=0∞ cn(x-a)n, where cn are coefficients and a is a constant. It is a type of mathematical representation that expresses a function as a sum of powers of x.

## 2. How do you find the radius of convergence for a power series?

The radius of convergence for a power series can be found by using the ratio test or the root test. These tests involve taking the limit as n approaches infinity of the absolute value of the ratio or root of the (n+1)th and nth terms of the series. If the limit is less than 1, the series converges. The radius of convergence is the distance from the center point (a) to the nearest point where the series converges.

## 3. Can a power series converge for all values of x?

No, a power series may only converge for certain values of x within its radius of convergence. Beyond this radius, the series will either diverge or converge to a different function.

## 4. How do you determine the interval of convergence for a power series?

The interval of convergence for a power series is the range of values for x where the series converges. This interval can be found by testing the endpoints of the radius of convergence (a ± r) and checking for convergence or divergence using a different test, such as the limit comparison test or the integral test.

## 5. Can a power series be used to represent any function?

Not necessarily. A power series can only represent functions that have a corresponding power series representation, such as polynomials or exponential functions. Some functions may not have a power series representation, or the power series representation may only converge for a limited range of values.

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