Estimate error using Taylor Series I was told to use Taylor Series to find the error of this but I am not confident in doing this. 
Estimate the error in approximating $\cos x$ by $1-(x^2/2)$ on the interval $[-0.5, 0.5]$.
 A: The Lagrange error bound states that the gap between $f(x)$ and its Taylor expansion in $x_0$ of order $k$ is
$$
R_k(x) = \frac{f^{(k+1)}\big(\xi(x)\big)}{(k+1)!} (x-x_0)^{k+1} 
$$
where $\xi(x)$ is a point lying in the interval $[x_0,x]$ or $[x,x_0]$ (depending whether $x_0\leq x$ or $x_0\geq x$). In your case,
$$
1-\frac{x^2}2
$$
is the 3rd order approximation of $\cos x$ in $x_0=0$ (since $\cos'''(0)=\sin(0)=0$), so
$$
R_3(x)
~=~
\cos x - \left(1 - \frac{x^2}2\right)
~=~
\frac{\cos{\big(\xi(x)\big)}}{24}x^4
$$
If $x$ ranges in $[-0.5,0.5]$, then so does $\xi(x)$. Therefore
$$
\max_{[0.5,0.5]}{\big|R_3(x)\big|}
~\leq~
\frac{\max_{[0.5,0.5]}{\big|\cos(x)\big|}}{24}
\max_{[0.5,0.5]}{\big|x^4\big|}
~=~
\frac{1}{24}\cdot\frac{1}{2^4}
~=~
\frac{1}{384}
~\approx~
0.0026\ldots
$$
A: A related problem. Hint:
$$ 
    |\sum_{k=0}^\infty(-1)^k\,a_k\,-\,\sum_{k=0}^m\,(-1)^k\,a_k|\le |a_{m+1}|. $$
A: The taylor development of cos x around 0 is just
$$
1-x^2/2+x^4/24-x^6/720+\dots
$$
When you subtract your given term, you get get an error of magnitude 
$$
O(x^4)
$$
