Series Expansion for V(r) in Quantum Final | Step-by-Step Solution

quasar987
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Homework Statement


I am ashamed to ask this, but in my quantum final, there was a little mathematically-oriented subquestion that asked to show that the function

V(r)=-\frac{V_0}{1+e^{(r-R)/a}}

(r in [0,infty)) can be written for r>R as

V_0\sum_{n=1}^{\infty}(-1)^ne^{-n(r-R)/a}

The Attempt at a Solution


:blushing:
 
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You know the series expansion of (1+x)^-1 for |x|<1, right? So use it (and don't tell me that exp{(r-R)/a} >1 for r>R, because I know that).
 
Yeah ok!

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Or (really the same thing) the "geometric series"
\sum_{n=0}^\infty ar^n= \frac{a}{1- r}
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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