Taylor Polynomials -- $\cos(x)$

Question

Show that

$$\forall x : \cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots+\frac{(-1)^nx^{2n}}{(2n)!}$$

I know that this is true because it is one of the most common Taylor polynomials.

But how can prove it?

Answer 1

For any analytic function $f$ (such as $cos(x))$:

$$f(x) = \sum_{n=0}^{\infty} f^{(n)}(x_0)\frac{(x-x_0)^n}{n!}$$

Where $x$ is a point near some point $x_0$ and $f^{(n)}(x_0)$ is the $n$th derivative of $f$ at $x_0$.

Use this formula with $x_0=0$.

Answer 2

This is by my experience simplest to prove by definition, since for any smooth function $f$, the taylor polynomial is $$\sum_{i=0}^{\infty}x^i\frac{f^{(i)}(x)}{i!}.$$ Now, simply calculate the $i$-th derivative of the cosine function. For example, the first five derivatives (and their values at $x=0$) are

• $\cos(x)$, $1$
• $-\sin(x)$, $0$
• $-\cos(x)$, $-1$
• $\sin(x)$, $0$
• $\cos(x)$, $1$

Can you see the pattern?