Maclaurin series of cosh(x) = (e^x + e^(-x)) / 2

cheffy

Junior Member
Joined
Jan 10, 2007
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73
Find the Maclaurin series of
\(\displaystyle \
\cosh (x) = \frac{{e^x + e^{ - x} }}{2}
\\)

Please help! I don't even know what cosh(x) is! Thanks.
 
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.
 
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