What power series represents 1/(1+x^2) on (-1,1)

Click For Summary
SUMMARY

The power series representation of 1/(1+x^2) on the interval (-1, 1) can be derived by substituting -x^2 into the geometric series formula 1/(1-x) = ∑_{n=0}^∞ x^n. This substitution transforms the series into ∑_{n=0}^∞ (-x^2)^n, which converges for |x| < 1. The discussion highlights the importance of correctly applying the geometric series and clarifies common misconceptions about power series and chain rule issues.

PREREQUISITES
  • Understanding of geometric series and convergence criteria
  • Familiarity with power series and their representations
  • Basic knowledge of calculus, including derivatives and antiderivatives
  • Ability to manipulate algebraic expressions and substitutions
NEXT STEPS
  • Study the derivation of geometric series and its applications
  • Learn about the convergence of power series and the radius of convergence
  • Explore term-by-term integration of power series
  • Investigate the Taylor series expansion for common functions like arctan(x)
USEFUL FOR

Students studying calculus, particularly those focusing on power series and their applications in mathematical analysis.

CanaryK
Messages
3
Reaction score
0

Homework Statement


Word for word, the book says "Find a power series that represents 1/(1+x^2) on (-1, 1)


Homework Equations


It's in the chapter that talks about power series, so I think they want me to use the fact that 1/(1-x) is a power series with a=1 and r=x, but if I just substitute in -x^2 for x, that makes chain rule issues. The chapter also talks about term-by-term integration, which confuses me more than little bit.


The Attempt at a Solution


Uhm... I don't really have much yet. I think this is probably an extremely basic question and I'm just missing some underlying technique or trick to solve it.
I know that the antiderivative of 1/(1+x^2) is arctan(x) of but we don't know the series for arctan(x), so that doesn't really help. The derivative of 1/(1+x^2) is -2x/(1+x^2)^2, which doesn't seem to help either.

Thanks for any help :)
 
Last edited:
Physics news on Phys.org
CanaryK said:

Homework Statement


Word for word, the book says "Find a power series that represents 1/(1+x^2) on (-1, 1)


Homework Equations


It's in the chapter that talks about power series, so I think they want me to use the fact that 1/(1-x) is a power series with a=1 and r=x, but if I just substitute in x^2 for x, that makes power series issues.
What do you mean by "power series issues"? Of course, 1/(1-x), itself, is not a "power series", it is the sum of the geometric series \sum_{n= 0}^\infty x^n. And, to make "1/(1- x)" look like "1/(1+x^2)" replace x by -x^2, not just x^2.

The chapter also talks about term-by-term integration, which confuses me more than little bit.


The Attempt at a Solution


Uhm... I don't really have much yet. I think this is probably an extremely basic question and I'm just missing some underlying technique or trick to solve it.
I know that the antiderivative of 1/(1+x^2) is arctan(x) of but we don't know the series for arctan(x), so that doesn't really help. The derivative of 1/(1+x^2) is -2x/(1+x^2)^2, which doesn't seem to help either.

Thanks for any help :)
 
Last edited by a moderator:
Wow I'm sorry, for "power series issues" I actually meant to type "chain rule issues". Not sure what I was thinking there haha! Multiple bad typos in that, I'll go back and fix them in a second.

I'm not quite understanding your answer - so I CAN just go back and sub in -x^2 (that's what I meant, I swear haha)? And it doesn't matter that x^2 is a different order from x?
 
Yes. 1/(1+r)=sum (-r)^n for n=0 to infinity if |r|<1. Sub away.
 
Thanks so much, Dick :)
Unfortunately, the test still completely kicked my butt. Haha oh well.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
8
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K