Stuck At the Last Step Simplifying A Series

In summary, the conversation discusses finding a power series for a function centered at c and determining the interval of convergence. The given function is g(x) = 2 / (1-x^2) with c=0. By simplifying the function, it is shown that the power series is equal to Sum[2x^(2n), n=0 to infinity] and the interval of convergence is (-1,1). The book had a typo stating (1,1) as the interval of convergence.
  • #1
ross1219
9
0

Homework Statement


Find a power series for the function, centered at c, and determine the interval of convergence.

Homework Equations


g(x) = 2 / (1-x^2) , c=0

The Attempt at a Solution


2 / (1+x)(1-x) = 1/(1+x) + 1/(1-x)

1/(1-(-x)) => Sum [(-1)^n (x^n)], n=0 to infinity. Converges when abs(x) < 1, (-1,1)
1/(1-x) => Sum x^n, n=0 to infinity. Converges when abs(x) < 1, (-1,1)

So, 2/(1-x^2) = Sum[(-1)^n + 1]x^n, n=0 to infinity.

OK, here's the part that is probably SO simple, but I'm just not seeing it. The book shows the series above, Sum[(-1)^n + 1]x^n, n=0 to infinity is then equal to:

Sum 2x^(2n), n=0 to infinity.

How do you do that last simplification? Sorry for such a simple question, but I'm stuck. :)

Also, the book has the interval of convergence for the series as abs(x^2)<1 or (1,1), which I'm pretty sure is just a typo. It should be (-1,1), right?
 
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  • #2
Try writing out the first few terms of [itex]\displaystyle\sum_{n=0}^{\infty}((-1)^n+1)x^n[/itex] and see what kind of pattern emerges. Then convert it back into summation notation!

Also, I believe that was a typo that your book had. (1,1) technically isn't even a number, never mind a range of numbers.
 
  • #3
Thank you so much scurty! I should have tried that, but thought I was missing something basic algebraically. I truly appreciate your help!
 
  • #4
No problem! I always write series out to see if I find patterns because I'm not really good at visualizing it otherwise.
 
  • #5
Yes, I learned at least two good lessons here. Write out some terms of a series if necessary, and this forum rocks! Thanks again. :)
 

FAQ: Stuck At the Last Step Simplifying A Series

1. What is the last step in simplifying a series?

The last step in simplifying a series is to combine like terms and constants. This means grouping together terms with the same variable and adding or subtracting them together.

2. How can I tell if I am stuck at the last step of simplifying a series?

If you have combined all like terms and constants, and the resulting series cannot be simplified any further, then you are likely stuck at the last step. You can also check by comparing your series to the original series and making sure they are equivalent.

3. What should I do if I am stuck at the last step of simplifying a series?

If you are stuck at the last step, you can try rewriting the series in a different form, such as factoring out a common factor or using the distributive property. You can also double check your work to make sure you have combined all like terms and constants correctly.

4. Can I use a calculator to simplify a series?

Yes, you can use a calculator to simplify a series. However, it is important to understand the steps and concepts behind simplifying a series so that you can check the accuracy of the calculator's results.

5. Are there any shortcuts for simplifying a series at the last step?

There are no shortcuts for simplifying a series at the last step. It is important to carefully follow the steps for combining like terms and constants to ensure the correct simplification.

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