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

Dec 5, 2016

${T}_{4} \left(x\right) = 1 - {x}^{2} / 2 + {x}^{4} / 24$

#### Explanation:

The development of $\cos x$ 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:

$\frac{{d}^{\left(n\right)} \cos x}{{\mathrm{dx}}^{n}} = \cos \left(\frac{n \pi}{2} + x\right)$

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:

$\cos \left(\frac{n \pi}{2}\right) = 0$ if $n$ is odd,

and furthermore:

$\cos \left(\frac{n \pi}{2}\right) = {\left(- 1\right)}^{\frac{n}{2}}$ 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 $k = 2$:

${T}_{4} \left(x\right) = 1 - {x}^{2} / 2 + {x}^{4} / 24$