Taylor polynomial 1/(1-x^2)

In summary, the conversation is about finding the Taylor polynomial for 1/(1-x^2) of degree 2n+1 at 0. The initial attempt was 1 + x^2 + x^4 + x^6 + ... + (x^4)/(1-x^2), but the textbook answer is 1 - x^2 + x^4 - ... + (-1)^n x^2n. After realizing the mistake of the original question being 1/(1+x^2) instead of 1/(1-x^2), it is mentioned that the Taylor series polynomial for 1/(1-x) can be used with a simple substitution. It is confirmed that the remainder term does not
  • #1
holezch
251
0

Homework Statement



The question asks me to write out a taylor polynomial for 1/(1-x^2)
of degree 2n+1 at 0.



The Attempt at a Solution



My answer was 1 + x^2 + x^4 + x^6 + ... + (x^4)/(1-x^2) which I just got from using hte geometric series formula. The textbook answer however is this:

1 - x^2 + x^4 - .... + (-1)^n x^2n

Am I wrong? I thought so, but I took out a graphing calculator, and the larger I make my polynomial degree, the closer it looks to 1/(1-x^2). So my answer must be correct?

thanks
 
Physics news on Phys.org
  • #2
omg, silly me. The question was 1/(1 + x^2) , not 1 / ( 1- x^2 )
 
  • #3
so should I just keep differentiating ? :S
 
  • #4
If you know what the taylor series polynomial is for 1/(1-x) it's just a simple substitution
 
  • #5
Office_Shredder said:
If you know what the taylor series polynomial is for 1/(1-x) it's just a simple substitution

right. that was very blind of me.. thanks! I also don't need to include the remainder term in my answer right? I just stop at 2n, since that's all they want the function to be equal up to
 

1. What is a Taylor polynomial for 1/(1-x^2) at x=0?

A Taylor polynomial is a representation of a function using a finite number of terms of a power series. The Taylor polynomial for 1/(1-x^2) at x=0 is 1+x^2+x^4+x^6+...

2. How is a Taylor polynomial useful for approximating 1/(1-x^2)?

A Taylor polynomial can be used to approximate a function at a specific point by using a finite number of terms from its power series. In the case of 1/(1-x^2), the Taylor polynomial can be used to approximate the function for values of x close to 0.

3. What is the degree of the Taylor polynomial for 1/(1-x^2) at x=0?

The degree of the Taylor polynomial for 1/(1-x^2) at x=0 is infinite, as the power series for this function has an infinite number of terms.

4. How can the Taylor polynomial for 1/(1-x^2) be used to find derivatives of the function?

The coefficients of the Taylor polynomial for 1/(1-x^2) can be used to find the derivatives of the function at x=0. The coefficient of x^n in the polynomial corresponds to the nth derivative of the function at x=0.

5. Can the Taylor polynomial for 1/(1-x^2) be used to approximate the function for values of x outside the interval [-1,1]?

No, the Taylor polynomial for 1/(1-x^2) is only valid for values of x within the interval [-1,1]. This is because the power series for 1/(1-x^2) only converges within this interval.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
829
  • Calculus and Beyond Homework Help
Replies
1
Views
790
  • Calculus and Beyond Homework Help
Replies
18
Views
2K
  • Calculus and Beyond Homework Help
Replies
27
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
953
  • Calculus and Beyond Homework Help
Replies
3
Views
888
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
640
  • Calculus and Beyond Homework Help
Replies
17
Views
485
Back
Top