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I'm working my way through a textbook on Fourier Analysis and came across this stumbling block during a worked example. The example obtains the following expression for the Fourier coefficients:

$$C_k = \frac{1}{N_o} \frac{1-e^{-jkL\frac{2\pi}{N_0}}}{1-e^{-jk\frac{2\pi}{N_0}}}$$

The next step is, to quote, "pull out a common term from the numerator to write it as:"

$$(e^{-jkL\frac{2\pi}{N_0}\frac{1}{2}}) (e^{jkL\frac{2\pi}{N_0}\frac{1}{2}} - e^{-jkL\frac{2\pi}{N_0}\frac{1}{2}})$$

Similarly, pulling out a common term from the denominator yields: $$(e^{-jk\frac{2\pi}{N_0}\frac{1}{2}}) (e^{jk\frac{2\pi}{N_0}\frac{1}{2}} - e^{-jk\frac{2\pi}{N_0}\frac{1}{2}})$$

Frankly I'm drawing a total blank here, I'm really confused as to how these common terms are obtained. Any guidance at all would be much appreciated!

McKendrigo
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The specific exponent is irrelevant. If you have $1-e^a,$ then you can factor it as $e^{a/2}(e^{-a/2} - e^{a/2}).$ This follows from basic arithmetic laws. Note that, distributing the product, we find $$e^{a/2}(e^{-a/2}-e^{a/2}) = e^{a/2}e^{-a/2} - e^{a/2}e^{a/2}.$$ But $e^{x+y}=e^xe^y,$ so this becomes $e^0 - e^a = 1 - e^a,$ so $$1-e^a = e^{a/2}(e^{-a/2}-e^{a/2}).$$

When learning calculus, it is useful to be very solid on your fundamentals of algebra.