Question

How do you find the Taylor polynomial of degree 4 for cos(x), for x near 0?

Answer 1

`T_4(x) = 1 -x^2/2+x^4/24`

Explanation:

The development of `cosx` in Taylor series around `x=0` is:

[1]...`cosx = sum_0^oo 1/(n!)[(d^((n))cosx)/(dx^n)]_(x=0)x^n`

Now, note that:

`(d^((n))cosx)/(dx^n) = cos((npi)/2+x)`

So [1] becomes:

`cosx = sum_0^oo x^n/(n!)cos((npi)/2)`

We can see that only terms with `n` even are non null, as:

`cos((npi)/2)=0` if `n` is odd,

and furthermore:

`cos((npi)/2)=(-1)^(n/2)` if `n` is even, so that:

`cosx = sum_0^oo (-1)^kx^(2k)/(2k!)`

The Taylor polynmial of degree 4 is just the partial sum of this series for `k=2`:

`T_4(x) = 1 -x^2/2+x^4/24`