Two series I'm having problems with

[manenbu]

Homework Statement

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}}

Homework Equations

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

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.

 

[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.

 

[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).

 

[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)?

 

[manenbu]

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

 

[LCKurtz]

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)?

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

 

[HallsofIvy]

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.