Finding a Taylor Series from a function and approximation of sums

  • Thread starter Illania
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  • #1
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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?
 

Answers and Replies

  • #2
LCKurtz
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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.
 
  • #3
I like Serena
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use long division.

I'd go for writing it as a geometric series.
$${1\over 1+x}=1-x+x^2-x^3+...$$
 
  • #4
HallsofIvy
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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].
 
  • #5
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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]
 
  • #6
Redbelly98
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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.
 

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