Power Series Representation-Quick question

  • Thread starter bcjochim07
  • Start date
  • #1
bcjochim07
374
0

Homework Statement


Is there a way to check if a power series representation of a function is correct?

Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,263
621
If f(x) is the function, you might want to check if f(0) is equal to the value of the power series at x=0. You may also want to check if f'(0) is equal to the derivative of the power series at x=0. You can continue for higher derivatives if you are in doubt. If you check all of the infinite number of derivatives you KNOW you are right.
 
  • #3
bcjochim07
374
0
Ok so if my function is x/(1-x)^2 and my power series is sum from 1 to infinity of nx^n, I can check it using derivatives and values of x, but what do I put in for n?
 
  • #4
bcjochim07
374
0
Ok f(0) = the value of the power series at x=0

The first derivative of the function is (-x^2 + 1)/(1-x)^4 so f'(0)=1
The first derivative of the power series is sum from n=1 to infinity of n^2*x^(n-1) and if I plug in x, I will get zero. What am I not understanding?
 
Last edited:
  • #5
Dick
Science Advisor
Homework Helper
26,263
621
Ok f(0) = the value of the power series at x=0

The first derivative of the function is (-x^2 + 1)/(1-x)^4 so f'(0)=1
The first derivative of the power series is sum from n=1 to infinity of n^2*x^(n-1) and if I plug in x, I will get zero. What am I not understanding?

The power series is x+2x^2+3x^3+... The derivative of that at x=0 is most certainly 1. What went wrong with your reasoning above?
 
  • #6
bcjochim07
374
0
Oh, ooops I wasn't thinking right
 
  • #7
bcjochim07
374
0
I forgot the first x term
 
  • #8
HallsofIvy
Science Advisor
Homework Helper
43,021
970
However, in general checking that derivatives at a specific point are the same does NOT prove that a power series is equal to a function. That is only true if the function in question is "analytic" at the point which means, by definition, that it is equal to its Taylor Series expansion about the point.

For example, it is easy to show that the function
[tex]f(x)= \begin{array}{c}e^{-\frac{1}{x^2}} x\ne 0 \\ 0 x= 0\end{array}[/tex]
is infinitely differentiable, has all derivatives at 0 equal to 0 and so has Taylor series expansion about 0 [itex]\Sum 0\cdot x^n[/itex] but is not equal to that anywhere except at x= 0.
 
Last edited by a moderator:
  • #9
bcjochim07
374
0
So then is there a way to prove that a power series is equal to a function?
 
  • #10
bcjochim07
374
0
Anyone have thoughts on this?
 

Suggested for: Power Series Representation-Quick question

Replies
8
Views
367
Replies
11
Views
572
Replies
9
Views
895
Replies
2
Views
588
  • Last Post
Replies
1
Views
529
Replies
8
Views
225
  • Last Post
Replies
31
Views
1K
Replies
8
Views
592
  • Last Post
Replies
8
Views
601
  • Last Post
Replies
7
Views
279
Top