Finding a Taylor Series from a function and approximation of sums

[Illania]

Homework Statement

[itex]\mu = \frac{mM}{m+M}[/itex]

a. Show that [itex]\mu = m[/itex]
b. Express [itex]\mu[/itex] as m times a series in [itex]\frac{m}{M}[/itex]

Homework Equations

[itex]\mu = \frac{mM}{m+M}[/itex]

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?

 

[LCKurtz]

Homework Statement

[itex]\mu = \frac{mM}{m+M}[/itex]

a. Show that [itex]\mu = m[/itex]

I don't see how that can be. Is there something you aren't telling us?

b. Express [itex]\mu[/itex] as m times a series in [itex]\frac{m}{M}[/itex]

Homework Equations

[itex]\mu = \frac{mM}{m+M}[/itex]

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?

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.

 

[I like Serena]

use long division.

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

 

[HallsofIvy]

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

 

[Illania]

Apologies everyone, it is actually [itex]\mu \approx m[/itex].

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

 

[Redbelly98]

Apologies everyone, it is actually [itex]\mu \approx m[/itex].

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.