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$$\sum_{\text{odd }k} \frac{-2e^{-ikt}}{ik\pi} = \sum_{\text{odd }k>0} \frac{-2e^{-ikt}+2e^{ikt}}{ik\pi}$$

How does limiting all $k$ values to greater than zero, introduce the new term in the numerator?

I've been staring at this for an hour and I can't for the life of me figure out how positive $k$'s allow for a new term.

Thank you in advance for the help.

Andrei
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    On the left you have all odd integers, odd and even. You get the right by pairing off $k$ with $-k$. – Angina Seng Oct 19 '17 at 17:43
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    @LordSharktheUnknown: You probably meant "positive and negative" – Martin R Oct 19 '17 at 17:43
  • They aren't new terms. They are the old negative terms. But now added to the positive terms. Basically you are just combining the positive and negative terms together. – fleablood Oct 19 '17 at 18:49
  • Restricting to positive eliminates half of the terms. You add those that were eliminated back to the ones that were left. The "new" terms are just the "old" terms that were eliminated. – fleablood Oct 19 '17 at 18:51

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$\sum\limits_{k \text{ odd}} a_k= $

$\sum\limits_{k \text{ odd }>0} a_k + \sum\limits_{k \text{ odd }<0} a_k=$

$\sum\limits_{k \text{ odd }>0} a_k + \sum\limits_{k \text{ odd }>0} a_{-k}=$

$\sum\limits_{k\text{ odd} > 0} (a_k + a_{-k})$

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In this case if $j = -k < 0$ and $k = |j| = - j> 0$.

Then $\frac{-2e^{-ijt}}{ij\pi}=\frac{-2e^{ikt}}{-ik\pi} =\frac{2e^{ikt}}{ik\pi}$

So $\sum\limits_{k = \pm M} \frac{-2e^{-ikt}}{ik\pi}=$

$\sum\limits_{k = M}\frac{-2e^{-ikt}}{ik\pi} + \sum\limits_{k = -M}\frac{-2e^{-ikt}}{ik\pi}$

$\sum\limits_{k = M}\frac{-2e^{-ikt}}{ik\pi} + \sum\limits_{k = M}\frac{2e^{ikt}}{ik\pi}$

$\sum\limits_{k = M}(\frac{-2e^{-ikt}}{ik\pi}+ \frac{2e^{ikt}}{ik\pi})$

fleablood
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