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I noticed a question while reviewing Taylor series expansions that has been bugging me. The question is: How many nonzero terms with odd exponents does the Taylor series $\operatorname{Log} (1 + e^z)$ about $z = 0$ have?

I'm having trouble figuring out how to do this. Can anyone help?

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2 Answers 2

up vote 6 down vote accepted

Let $\displaystyle f(z) = \log (1 + e^z)$.

Now

$$f'(z) = \frac{e^z}{1+e^z} = \frac{e^z - 1}{2(e^z + 1)} + \frac{1}{2} = g(z) + \frac{1}{2}$$

Now we have that

$$g(-z) = \frac{e^{-z} - 1}{2(e^{-z} + 1)} = \frac{1 - e^z}{2(1 + e^z)} = -g(z) $$

Thus $\displaystyle g(z)$ is an odd function and its power series will only contain odd terms.

Since $\displaystyle f(z) = C + \frac{z}{2} + \int g(z)$, the only odd power in the series for $\displaystyle f(z)$ is $\displaystyle z$, and its coefficient is $\displaystyle \frac{1}{2}$.

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$$\log(1 + e^z) = \log(e^{z/2}(e^{z \over 2} + e^{-{z \over 2}}))$$ $$=\log(e^{z \over 2}) + \log(e^{z \over 2} + e^{-{z \over 2}})$$ $$= {z \over 2} + \log(2\cosh({z \over 2}))$$ Since $\cosh({z \over 2})$ is even, so is any function of $\cosh({z \over 2})$. So all terms of the Taylor series of $\log(2\cosh({z \over 2}))$ have even powers. Thus ${z \over 2}$ is the only term of $\log(1 + e^z)$'s Taylor series with odd powers of $z$ showing up.

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