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

In summary, the book says to find a power series that represents 1/(1+x^2) on (-1, 1), but I'm still missing something basic.
  • #1
CanaryK
3
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 :)
 
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  • #2
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 [itex]\sum_{n= 0}^\infty x^n[/itex]. And, to make "[itex]1/(1- x)[/itex]" look like "[itex]1/(1+x^2)[/itex]" replace x by [itex]-x^2[/itex], not just [itex]x^2[/itex].

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:
  • #3
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?
 
  • #4
Yes. 1/(1+r)=sum (-r)^n for n=0 to infinity if |r|<1. Sub away.
 
  • #5
Thanks so much, Dick :)
Unfortunately, the test still completely kicked my butt. Haha oh well.
 

1. What is a power series representation?

A power series representation is an infinite polynomial expression used to represent a function. It is typically written as a sum of terms with increasing powers of a variable.

2. How can 1/(1+x^2) be represented as a power series?

The power series representation for 1/(1+x^2) is 1 - x^2 + x^4 - x^6 + ... + (-1)^n * x^(2n) + ... on the interval (-1,1).

3. What is the interval of convergence for the power series of 1/(1+x^2)?

The interval of convergence for the power series of 1/(1+x^2) is (-1,1). This means that the series will only converge for values of x within this interval.

4. How is the convergence of a power series determined?

The convergence of a power series is determined by the radius of convergence, which is the distance from the center of the series to the nearest point where the series diverges. In this case, the radius of convergence is 1, making the interval of convergence (-1,1).

5. Why is the power series representation of 1/(1+x^2) useful?

The power series representation of 1/(1+x^2) is useful because it allows us to approximate the function for values of x outside of the interval (-1,1). Additionally, it can be used to evaluate the function at specific points by substituting the value of x into the series.

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