[CalcII/DiffEq] Closed form expression for f(x) which the series converges

In summary, the power series Σn=0..∞ n(-1)nxn+1 converges to the function f(x) = -x2/(1+x)2. The values of x for which f(x) equals the power series are between -1 and 1.
  • #1
24karatbear
34
0

Homework Statement


Find a closed form expression for the function f(x) which the power series Σn=0..∞ n(-1)nxn+1 converges to and determine the values of x for which f(x) equals the given power series.

Homework Equations


N/A

The Attempt at a Solution


I'm actually not sure how to start. First, what does it mean by "closed form expression"? Does that just mean an expression that isn't expressed as a series (with summation notation)?

The only thing I can think of is writing the terms out. That's probably not right though. If anyone can give me a hint or nudge me in the right direction, I'd be extremely grateful. Thank you.
 
Physics news on Phys.org
  • #2
A couple of examples of closed form expressions for power series are $$xe^x=\sum\limits_{k=0}^\infty\frac{x^{k+1}}{k!}$$ and $$\frac{x^2}{x+1}=\sum\limits_{k=0}^\infty(-1)^kx^{k+2}$$

The left-hand sides there are the "closed forms". The idea is that you try to manipulate the sum you are given so that it looks like a power series that you "know". You may need to differentiate or take an antiderivative and/or possibly multiply/divide by a power of ##x## to make that happen.
 
  • Like
Likes 24karatbear
  • #3
Oh, that makes a lot of sense. Thank you! I will attempt this now and post an update.
 
  • #4
Okay, I narrowed the possible forms to either be a variation of the arctan function or ln(1+x).

It seems very similar to arctan, except the n isn't in the denominator. Is it even possible to invert the n? It doesn't seem like it...
 
  • #5
Okay, ignore the previous posts. I heeded gopher's advice and noticed that it looks similar to f(x) = 1/1-x.

f(x) = 1/1-x = Σn=0..∞ xn
f'(x) = 1/(1-x)2 = Σn=0..∞ nxn-1 = Σn=0..∞ nxnx-1
-x2 * f'(x) = -x2/(1-x)2 = Σn=0..∞ -nxn+1 = Σn=0..∞ -nxnx

Then, I believe my next step is to substitute -1x into the function:

-x2/(1-(-x))2 = Σn=0..∞ -n(-x)n(-x) = Σn=0..∞ n(-1)nxn+1

Thus, if I did this correctly, my expression should be: -x2/(1+x)2

If this is correct, then I am assuming my next step would be to find the interval of convergence of the equivalent series to this expression. If anyone can verify this for me, I'd be extremely grateful.
 
Last edited:
  • #6
24karatbear said:
Okay, ignore the previous posts. I heeded gopher's advice and noticed that it looks similar to f(x) = 1/1-x.

f(x) = 1/1-x = Σn=0..∞ xn
f'(x) = 1/(1-x)2 = Σn=0..∞ nxn-1 = Σn=0..∞ nxnx-1
-x2 * f'(x) = -x2/(1-x)2 = Σn=0..∞ -nxn+1 = Σn=0..∞ -nxnx

Then, I believe my next step is to substitute -1x into the function:

-x2/(1-(-x))2 = Σn=0..∞ -n(-x)n(-x) = Σn=0..∞ n(-1)nxn+1

Thus, if I did this correctly, my expression should be: -x2/(1+x)2

If this is correct, then I am assuming my next step would be to find the interval of convergence of the equivalent series to this expression. If anyone can verify this for me, I'd be extremely grateful.

This looks mostly correct. The idea certainly is. I didn't check it completely, though, and the only part where I am concerned is when you got around to plugging in ##-x##. It seems like you may have lost a minus sign somewhere. Then again, I could be wrong.

As for checking the interval of convergence, there is a theorem which I'm guessing (hoping?) you learned regarding the relationship between the interval of convergence of a power series and the intervals of convergence of the power series that you obtain by taking the term-by-term derivatives and antiderivatives. Essentially you don't gain or lose anything substantial when you do those things; you just need to check the endpoints. In the case of derivatives, you may lose endpoints, but you never gain them. For antiderivatives, it's the other way around.
 
  • #7
gopher_p said:
This looks mostly correct. The idea certainly is. I didn't check it completely, though, and the only part where I am concerned is when you got around to plugging in ##-x##. It seems like you may have lost a minus sign somewhere. Then again, I could be wrong.

As for checking the interval of convergence, there is a theorem which I'm guessing (hoping?) you learned regarding the relationship between the interval of convergence of a power series and the intervals of convergence of the power series that you obtain by taking the term-by-term derivatives and antiderivatives. Essentially you don't gain or lose anything substantial when you do those things; you just need to check the endpoints. In the case of derivatives, you may lose endpoints, but you never gain them. For antiderivatives, it's the other way around.

Thanks so much for having a look, gopher! After I plugged in -x, I got a (-x)n and a regular -x. I just multiplied the regular -x by the -1 that was originally there to get x (which I then combined with xn to get xn+1 after pulling out the (-1)n).

Okay, so for the original power series, the interval of convergence was (-1,1). With the ratio test, the interval of the convergence appears to be the same after testing the endpoints x=1 and x=-1 with the test for divergence. So would it then be correct to conclude that the values of x for which f(x) equals the power series is between -1 and 1?
 

What is a closed form expression?

A closed form expression is a mathematical expression that can be written in terms of a finite number of well-known functions, such as polynomials, exponential functions, and trigonometric functions.

How is a closed form expression related to convergence?

A closed form expression is important in determining whether a series converges, as it allows us to evaluate the series at any point without having to sum up an infinite number of terms. If a series has a closed form expression, it is easier to determine whether it converges or not.

What is the significance of a series converging?

A series converging means that the sum of its terms approaches a finite value as the number of terms increases. This is important in many areas of mathematics and science, as it allows for more accurate calculations and predictions.

How do we find a closed form expression for a series?

Finding a closed form expression for a series can be a difficult task and often requires advanced mathematical techniques. In general, we can use known properties and formulas for well-known functions to manipulate the series and find a closed form expression. However, not all series have a closed form expression.

Can a series have multiple closed form expressions?

Yes, it is possible for a series to have multiple closed form expressions. This is because there may be different ways to manipulate the series or express it in terms of well-known functions. However, all of these expressions should converge to the same value when evaluated at a specific point.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
640
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
22
Views
3K
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
29
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
3K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
Back
Top