Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a function $$ g_N(x) = \sum_{n=-N}^N \frac{1}{x+n} $$ How can I prove that this function is odd, thus $ g_N(-x) = -g_N(x)$ ?

share|improve this question
    
In addition to the algebraic answers, there is the simple intuition that your function consists of N + 1 poles located at -N, -N + 1, ..., 0, 1, ... N. These are all just horizontal translations of $1/x$ (a well-known odd function), symmetrically placed around 0, added together. –  Kaz Apr 9 '13 at 23:13
add comment

4 Answers

up vote 1 down vote accepted

$$-g_N(x)=-\sum_{n=-N}^N \frac{1}{x+n} = \sum_{n=-N}^N -\frac{1}{x+n} =\sum_{n=-N}^N \frac{1}{-x+n} $$ $$g_N(-x)=\sum_{n=-N}^N \frac{1}{-x+n} $$

share|improve this answer
add comment

Hint: $$\frac{1}{-x + n} = -\frac{1}{x - n}$$

share|improve this answer
add comment

HINT: Note that $$\sum_{n=M}^K a_n = \sum_{n=-K}^{-M} a_{-n}.$$

share|improve this answer
1  
+1 If I had any votes left, I think both our hints combine as a full answer –  muzzlator Apr 9 '13 at 22:13
    
@muzzlator Yep :) –  Hagen von Eitzen Apr 9 '13 at 22:16
add comment

Fold the sum in half at the middle ($n = 0$): $$ \begin{align} g_N(x) = \sum_{n=-N}^{n=N} \frac{1}{x+n} &= \frac{1}{x} + \sum_{n=1}^{n=N} \left( \frac{1}{x-n} + \frac{1}{x+n} \right) \\ &= \frac{1}{x} + \sum_{n=1}^{n=N} \frac{2x}{x^2 - n^2} \end{align} $$ Now, you can verify directly that $g_N(-x) = -g_N(x)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.