How do you write the Taylor series for f(x)=coshx?

Feb 17, 2016

$f \left(x\right) \approx 1 + {x}^{2} / 2 + {x}^{4} / 24 + {x}^{6} / 720 + \ldots$ for values of $x$ close to $0$.

Explanation:

The Taylor series of a function is defined as:

sum_(n=0)^oof^n(x_0)/(n!)(x-x_0)^n

Where the $n$ in only ${f}^{n} \left({x}_{0}\right)$ denotes the $n$th derivative of $f \left(x\right)$ and not a power.

If we wanted to find, for example, the taylor series of $\cosh \left(x\right)$ around $x = 0$ then we set ${x}_{0} = 0$ and use the above definition. It is best to lay out two columns, one with the derivative and the other evaluating the value of ${f}^{n} \left({x}_{0}\right)$ at the point we wish to expand around.

• $f \left(x\right) = \cosh \left(x\right)$ $f \left(0\right) = 1$
• $f ' \left(x\right) = \sinh \left(x\right)$ $f ' \left(0\right) = 0$
• $f ' ' \left(x\right) = \cosh \left(x\right)$ $f ' ' \left(0\right) = 1$
• $f ' ' ' \left(x\right) = \sinh \left(x\right)$ $f ' ' ' \left(0\right) = 0$
• ${f}^{I V} \left(x\right) = \cosh \left(x\right)$ ${f}^{I V} \left(0\right) = 1$
• ${f}^{V} \left(x\right) = \sinh \left(x\right)$ ${f}^{V} \left(0\right) = 0$
• ${f}^{V I} \left(x\right) = \cosh \left(x\right)$ ${f}^{V I} \left(0\right) = 1$

We could continue these columns indefinitely but we should get a good approximation here.

We now have our values for ${f}^{n} \left(0\right)$ so all that is left is to put them into the sum above and we get:

1/(0!)(x-0)^0+0/(1!)(x-0)^1+1/(2!)(x-0)^2+0/(3!)(x-0)^3+1/(4!)(x-0)^4+0/(5!)(x-0)^5+1/(6!)(x-0)^6+...

Simplifying this series gives us:

$1 + {x}^{2} / 2 + {x}^{4} / 24 + {x}^{6} / 720 + \ldots$

And thus we have the first four non - zero terms for $\cosh \left(x\right)$, (but remember the series carries on infinitely). This will give us a fairly good approximation for values of $\cosh \left(x\right)$ near $0$. If you need more accuracy then you need to find more derivatives and continue building up the series.

Also, if you wish to expand the series around a value of ${x}_{0}$ that is not 0 then it will not be as clean as this.