Maclaurin series of $e^{-x^2}$ Error The task is to first estimate the second degree Maclaurin series of $e^{-x^2}$ and thus estimate the integral of the function from $0$ to $0.5$. This part is no problem.
The following task is to estimate the error for this estimation. I used the regular approach with calculating the third derivative and using it with the formulae for the reminder of Taylor polynomials which apparently is wrong. However doing the same thing with $4$-th derivative works but I have no idea why.
I know that in maclaurin formulae the term including third derivative becomes $0$ however I don't know how this is connected to the error calculation.
Also I'm aware of the method with $t=-x^2$ substitution of in $e^t$ expansion, but I don't understand why the normal method does not work.
Sorry if I'm too vague.
 A: Let us first deal with the function itself. The MacLaurin expansion has the shape $1-\frac{x^2}{1!}+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots$. Integrating from $0$ to $x$, we find that the Maclaurin expansion for the area up to $x$ has Maclaurin expansion $x-\frac{x^3}{3\cdot 1!}+\frac{x^5}{5\cdot 2!}-\frac{x^7}{7\cdot 3!}+\cdots$.
For the function, you were asked to truncate just after the $\frac{x^2}{1!}$ term. You can estimate the error in terms of the third derivative, evaluated at some unknown place $\xi$ between, in this case, $0$ and $0.5$. Now in fact this $\xi$ is close to $0$, but you do not officially know that, and the upper bound on the error that you get is quite a bit too pessimistic.
However, as you observed, the coefficient of $x^3$ is $0$, and therefore the Maclaurin polynomial up to $x^2$ is exactly the same as the Maclaurin polynomial up to $x^3$. So the error can be expressed in terms of the fourth derivative, and this gives a nicer upper bound on the error.
There is another way of looking at things that does not use the Lagrange formula for the remainder. Note that the Maclaurin polynomial, at least for $x\le 1$, is an alternating series. So the truncation error is less, in absolute value, than the first "neglected" term. This term has absolute value $\frac{x^4}{2!}$. So the absolute value of the error when we evaluate at $x=0.5$ is less than $\frac{(0.5)^2}{2!}$.
The error estimation for the integral uses exactly the same ideas. 
