Q:

Taylor Series - general 'what' and 'why' questions

I am a little confused with the Taylor Series at the moment, so please forgive me for my very basic questions. If we were to approximate a function, say $cos(x)$, I let $f(x)=cos(x)$
And I have been shown that $f(x)≈f(x_0)+f'(x_0)(x-x_0)$
Just so I understand, when we say we are approximating it, are we simply saying that we can  find the original function's $y$ value when given a specific $x$ value by using another, easier polynomial function that we substitute that same $x$ value into?
Anyway we found that $P_2(x)=-\frac{1}{2}x^2+1$ is an approximation of $cos(x)$. This didn't make much sense though as in the first step, we substituted zero in as $x$, and found $cos(0)=1$ But if this was another number, we would have to find the cosine of that number which defeats the whole purpose? Any help would be appreciated

A:

The answer can be found in the following graph representing $\color{darkblue}{\cos x}$ and $\color{darkmagenta}{-\tfrac12x^2+1}:$
$\phantom{XXX}$
Near $0$, our polynomial approximation is pretty good, but when you start going far, this approximation becomes less accurate as the graph shows.

A:

The idea is that you can control the error. Indeed, we have $\cos x \approx 1 - \frac{x^2}{2}$, if $x$ is close to zero. For instance, if I want to know $\cos 0.1$, may be reasonable that $$\cos 0.1 \approx 1 - \frac{0.1^2}{2} = 1 - 0.005 = 0.995$$
We can use Lagrange's formula for the remainder, to estimate how much we're missing. If $$\sum_{k = 0}^{n} \frac{f^{(k)}(x_0)}{k!} (x - x_0)^k $$
is the Taylor expansion of order $n$ of f, around $x_0$, then the error can be estimated by $$\frac{f^{(k+1)}(c)}{(k+1)!} (x - x_0)^{k+1}$$
for some $c$ between $x$ and $x_0$. Ok?

