Two series I'm having problems with

1. Sep 21, 2009

manenbu

1. The problem statement, all variables and given/known data

Express the following as a power series at x=0 and find the interval of convergence:

first one:
$$f(x) = \sqrt{1+x^3}$$
second:
$$f(x) = \frac{x^2-1}{x^2+1} + \cos^{2}{\frac{x}{2}}$$
2. Relevant equations

Maclaurin series for the first, no idea about the second.

3. The attempt at a solution

For the first one, I can express the first few terms but I can't find a general summation formula, and because of that I can't determine the convergence.
This is what I got:
$$f(x) = \sqrt{1+x^3} = \left(1+(x^3)\right)^{\frac{1}{2}} = 1 + \frac{x^3}{2} - \frac{x^6}{2^2 \cdot 2!} + \frac{3x^9}{2^3 \cdot 3!} - \frac{15x^{12}}{2^4 \cdot 4!} + ...$$
But now what? I can't find what $\sum a_n$ is equal to.. So I have no way of determining the convergence.

As for the second series, I don't even know where to start.

Last edited: Sep 21, 2009
2. Sep 21, 2009

xaos

such type of questions rely on a very small set of paradigm examples upon which we use substitution, so rewrite what you have into a form that is more familiar.

3. Sep 21, 2009

LCKurtz

In your attempt at a solution, write out a couple more terms of the binomial expansion. And in the numerator don't simplify the constant. Starting with the x^6 term the constant in the numerator is 1, then 1*3, then 1*3*5, then... You should see a pattern emerging. It doesn't matter if the pattern works for the first few terms because they don't affect convergence.

For the second example I would write the fraction as (-1 + x^2)/ (1 + x^2) and divide it out long division, and I would use the trig identity for cos^2(x/2).

4. Sep 21, 2009

manenbu

Ok, so for the first one I get:

$$1+\frac{x^3}{2}+\sum_{n=1} (-1)^n x^{3n} \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1)}{2^n \cdot n!}$$

How do I treat the 1*3*5*...*(2n-1)? Can I just factor it outside of the summation and ignore it while making one of the limit test (probably ratio, since I have the factorial)?

5. Sep 21, 2009

manenbu

btw, can I use binomial expansion when $(1+m)^{\frac{1}{2}}$?

6. Sep 21, 2009

LCKurtz

No, because it depends on n. Just include it in the ratio test. Most of the terms will cancel out.

7. Sep 21, 2009

HallsofIvy

Staff Emeritus
Since it depends on n, not you cannot "factor it outside the summation"!
You can write it as 1*3*5*...*(2n-1)= 1*2*3**5*6*...*(2n-1)(2n)/(2*4*6*...*2n)= 2n!/(2*4*6*...*(2n))= 2n!/((2^n)(n!))
Now cancel.