1. ## Taylor/Maclaurin series

Hello, I was wondering if someone could explain this problem:

Expand f(x)=x^3/(1+x^3) in powers of x for abs(x)<1.

I think it's a maclaurin expansion, but I've evaluated up to the 9th derivative at f(0), and I come up with 2 terms. How long do I have to expand it? or am I just doing it wrong.

Thanks.

2. Originally Posted by hasbro
Hello, I was wondering if someone could explain this problem:

Expand f(x)=x^3/(1+x^3) in powers of x for abs(x)<1.

I think it's a maclaurin expansion, but I've evaluated up to the 9th derivative at f(0), and I come up with 2 terms. How long do I have to expand it? or am I just doing it wrong.

Thanks.
There's an easier way to do this. First

$
f(x) = \frac{x^3}{1+x^3} = \frac{1+x^3-1}{1+x^3} = 1 - \frac{1}{1+x^3}
$

Then from the geometric power series

$
\frac{1}{1 + x} = 1 - x + x^2 - x^3 + \cdots
$

replace the $x$ with $x^3$ giving

$\frac{1}{1 + x^3} = 1 - x^3 + x^6 - x^9 + \cdots$

so putting these together gives

$
1 - \left( 1 - x^3 + x^6 - x^9 + \cdots\right) = x^3 - x^6 + x^9 - \cdots
$

3. Thank you that's a lot easier.

What would you do if the function was (x^3)/(1+(x^4))?

4. Originally Posted by hasbro
Thank you that's a lot easier.

What would you do if the function was (x^3)/(1+(x^4))?
Use

$\frac{1}{1+x^4} = 1 - x^4 + x^8 - x^{12} + \cdots$ and multiply by $x^3$ (even easier).

5. Indeed, you could have done $\frac{x^3}{1+ x^3}$ in exactly that (easier) way:
$\frac{1}{1+ x^3}= \frac{1}{1- (-x^3)}$ and so is equal to the geometric series $1- x^3+ x^6- x^9+ \cdot\cdot\cdot= \sum_{n=0}^\infty (-1)^n x^{3n}$ and then multiply each term by $x^3$
$\frac{x^3}{1+ x^3}= x^3- x^6+ x^9+ \cdot\cdot\cdot$ $= \sum_{n=0}^\infty (-1)^n x^{n+3}= \sum_{j=3}^\infty (-1)^j x^j$ where, in the last sum, j= n+3.