# Finding a Taylor Series from a function and approximation of sums

1. Feb 12, 2013

### Illania

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

$\mu = \frac{mM}{m+M}$

a. Show that $\mu = m$
b. Express $\mu$ as m times a series in $\frac{m}{M}$

2. Relevant equations

$\mu = \frac{mM}{m+M}$

3. The attempt at a solution

I am having trouble seeing how to turn this into a series. How can I look at the given function differently and see a series in it?

2. Feb 12, 2013

### LCKurtz

I don't see how that can be. Is there something you aren't telling us?
Maybe try writing$$\frac{mM}{m+M}= m\left( \frac M {M+m}\right) = m\left( \frac 1 {1+\frac m M}\right)$$and use long division.

3. Feb 13, 2013

### I like Serena

I'd go for writing it as a geometric series.
$${1\over 1+x}=1-x+x^2-x^3+...$$

4. Feb 13, 2013

### HallsofIvy

If $\mu= \frac{mM}{m+Mz}= m\frac{M}{m+ M}$ then $\mu= m$ if and only if $\frac{M}{m+ M}= 1$ which leads to $M= m+ M$ and then $m= 0$.

5. Feb 13, 2013

### Illania

Apologies everyone, it is actually $\mu \approx m$.

Also, I do see how I could turn LCKurtz suggestion into a geometric series: $m * \frac{1}{1-(- \frac{m}{M})}$ is the sum of the geometric series: $\Sigma^{\infty}_{0} (-1)^nm(\frac{m}{M})^n$

6. Feb 14, 2013

### Redbelly98

Staff Emeritus
Is there more that you haven't told us? Making this approximation depends on how large m and M are relative to each other. But the problem statement says nothing about their relative magnitude or what they are supposed to represent.