Quick Maclarin expansion question... Watch

XShmalX
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#1
Report Thread starter 8 years ago
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How do I go about showing why e^x/(1+e^x) will have no even powers of x... I've done some expansion but I'm not sure how to show it will always be the case....
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nuodai
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I'd start by finding \dfrac{d^n}{dx^n} \left( \dfrac{e^x}{1+e^x} \right) and showing that, when evaluated at x=0, it is equal to zero when n is even (and greater than 0). You can find the general expression for \dfrac{d^n}{dx^n} \left( \dfrac{e^x}{1+e^x} \right) by differentiating a few times to spot a pattern and proving it's true by induction.

EDIT: Actually, here's a better way.

The constant term in the Maclaurin series is \dfrac{1}{2}, so consider the function g(x) = \dfrac{e^x}{1+e^x} - \dfrac{1}{2}. The Maclaurin series of this function has only odd powers of x if and only if it is an odd function, so if you can show that g(-x) = -g(x) then you're done. [You can get your original function back by adding \frac{1}{2} to both sides.]
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XShmalX
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Report Thread starter 8 years ago
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(Original post by nuodai)
I'd start by finding \dfrac{d^n}{dx^n} \left( \dfrac{e^x}{1+e^x} \right) and showing that, when evaluated at x=0, it is equal to zero when n is even (and greater than 0). You can find the general expression for \dfrac{d^n}{dx^n} \left( \dfrac{e^x}{1+e^x} \right) by differentiating a few times to spot a pattern and proving it's true by induction.

EDIT: Actually, here's a better way.

The constant term in the Maclaurin series is \dfrac{1}{2}, so consider the function g(x) = \dfrac{e^x}{1+e^x} - \dfrac{1}{2}. The Maclaurin series of this function has only odd powers of x if and only if it is an odd function, so if you can show that g(-x) = -g(x) then you're done. [You can get your original function back by adding \frac{1}{2} to both sides.]
Ahh Thanks soooo much!!!
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