# What is the McLaurin series of f(x) = sinh(x)?

Dec 26, 2016

sinhx =sum_(k=0)^oo x^(2k+1)/((2k+1)!)

#### Explanation:

We can derive the McLaurin series for $\sinh \left(x\right)$ from the one othe exponential function: as for every $n$:

${\left[\frac{{d}^{n}}{{\mathrm{dx}}^{n}} {e}^{x}\right]}_{x = 0} = {e}^{0} = 1$

the Mc Laurin series for ${e}^{x}$ is:

e^x=sum_(n=0)^oo x^n/(n!)

Now as:

$\sinh x = \frac{{e}^{x} - {e}^{- x}}{2}$

We have:

sinhx = 1/2[sum_(n=0)^oo x^n/(n!)-sum_(n=0)^oo (-x)^n/(n!)]

and it is easy to see that for $n$ even the terms are the same and just cancel each other, so that just the odd order terms remain:

sinhx = 1/2[sum_(k=0)^oo x^(2k+1)/((2k+1)!)-sum_(k=0)^oo (-1)^(2k+1)x^(2k+1)/((2k+1)!)] = 1/2[sum_(k=0)^oo x^(2k+1)/((2k+1)!)+sum_(k=0)^oo x^(2k+1)/((2k+1)!)] = sum_(k=0)^oo x^(2k+1)/((2k+1)!)#

We can reach the same conclusion directly, noting that:

$\frac{d}{\mathrm{dx}} \sinh x = \cosh x$

${d}^{2} / \left({\mathrm{dx}}^{2}\right) \sinh x = \frac{d}{\mathrm{dx}} \cosh x = \sinh x$

so that all derivatives of odd order equal $\cosh x$ and all derivatives of even order equal $\sinh x$

But $\sinh \left(0\right) = 0$ and $\cosh \left(0\right) = 1$ yielding the same result.