Answer to Question #292332 in Calculus for Mistura

Question

What is the coefficient of x^4 in y = cosh(x)

Answer 1

The coefficient of $x^4$ is given by $\frac{(\cosh x)^{(4)}(0)}{4!}$, where $(4)$ means the fourth derivative of $(\cosh x)$.

We have the following relations :

$(\cosh x)'=(\frac{e^x+e^{-x}}{2})' = \frac{e^x-e^{-x}}{2} =\sinh x$

$(\sinh x)' = (\frac{e^x-e^{-x}}{2})' = \frac{e^x+e^{-x}}{2}=\cosh x$

Therefore we deduce that $\cosh x = (\cosh x)'' = (\cosh x)^{(4)}$. This gives us $\frac{(\cosh x)^{(4)}(0)}{4!}=\frac{1}{4!}$ as $\cosh (0) = 1$. We could also arrive to the same result just using the fact that the coefficient of $x^4$ in $\cosh x$ is (coefficient of $x^4$ in $e^x$ + coefficient of $x^4$ in $e^{-x}$)/2.