Maclaurin series of cosh(x) = (e^x + e^(-x)) / 2
- Thread starter cheffy
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skeeter
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- Dec 15, 2005
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you should already know the Maclaurin series for e<sup>x</sup> ...
\(\displaystyle \L e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...\)
so ...
\(\displaystyle \L e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} + ...\)
the series for cosh(x) is just the average of the above two series ... add 'em up, divide by 2.
\(\displaystyle \L e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...\)
so ...
\(\displaystyle \L e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} + ...\)
the series for cosh(x) is just the average of the above two series ... add 'em up, divide by 2.