Question

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

Answer 1

`f(x) ~~1+x^2/2+x^4/24+x^6/720+...` 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(x_0)` denotes the `n`th derivative of `f(x)` and not a power.

If we wanted to find, for example, the taylor series of `cosh(x)` 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(x_0)` at the point we wish to expand around.

• `f(x) = cosh(x)` `f(0) =1`
• `f'(x) = sinh(x)` `f'(0)=0`
• `f''(x) = cosh(x)` `f''(0)=1`
• `f'''(x) = sinh(x)` `f'''(0)=0`
• `f^(IV)(x)=cosh(x)` `f^(IV)(0)=1`
• `f^(V)(x) = sinh(x)` `f^(V)(0)=0`
• `f^(VI)(x)=cosh(x)` `f^(VI)(0) = 1`

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

We now have our values for `f^n(0)` 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+...`

And thus we have the first four non - zero terms for `cosh(x)`, (but remember the series carries on infinitely). This will give us a fairly good approximation for values of `cosh(x)` 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.