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

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

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Answer 1 · 2016-12-26T23:43:14

$sinh x = \infty \sum k = 0 \frac{x^{2 k + 1}}{(2 k + 1)!}$ $sinhx =sum_(k=0)^oo x^(2k+1)/((2k+1)!)$

Explanation:

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

${[\frac{d^{n}}{{d x}^{n}} e^{x}]}_{x = 0}^{x} = e^{0} = 1$ $[(d^n)/(dx^n) e^x ]_(x=0) = e^0=1$

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

$e^{x} = \infty \sum n = 0 \frac{x^{n}}{n!}$ $e^x=sum_(n=0)^oo x^n/(n!)$

Now as:

$sinh x = \frac{e^{x} - e^{- x}}{2}$ $sinhx = (e^x-e^(-x))/2$

We have:

$sinh x = \frac{1}{2} [\infty \sum n = 0 \frac{x^{n}}{n!} - \infty \sum n = 0 \frac{{(- x)}^{n}}{n!}]$ $sinhx = 1/2[sum_(n=0)^oo x^n/(n!)-sum_(n=0)^oo (-x)^n/(n!)]$

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

$sinh x = \frac{1}{2} [\infty \sum k = 0 \frac{x^{2 k + 1}}{(2 k + 1)!} - \infty \sum k = 0 {(- 1)}^{2 k + 1} \frac{x^{2 k + 1}}{(2 k + 1)!}] = \frac{1}{2} [\infty \sum k = 0 \frac{x^{2 k + 1}}{(2 k + 1)!} + \infty \sum k = 0 \frac{x^{2 k + 1}}{(2 k + 1)!}] = \infty \sum k = 0 \frac{x^{2 k + 1}}{(2 k + 1)!}$ $sinhx = 1/2[sum_(k=0)^oo x^(2k+1)/((2k+1)!)-sum_(k=0)^oo (-1)^(2k+1)x^(2k+1)/((2k+1)!)] = 1/2[sum_(k=0)^oo x^(2k+1)/((2k+1)!)+sum_(k=0)^oo x^(2k+1)/((2k+1)!)] = sum_(k=0)^oo x^(2k+1)/((2k+1)!)$

We can reach the same conclusion directly, noting that:

$\frac{d}{d x} sinh x = cosh x$ $d/(dx) sinhx = coshx$

$\frac{d^{2}}{{d x}^{2}} sinh x = \frac{d}{d x} cosh x = sinh x$ $d^2/(dx^2) sinhx = d/(dx)coshx = sinhx$

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

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

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

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