# How to find an upper bound on the error of this Taylor polynomial in $[0, 1]$?

I have this Taylor polynomial of order $N = 3$, $x_{0} = 0$:

$P_{N}(x) = \sum_{n=0}^{N = 3} \frac{f^{(n)}(0)\cdot x^n}{n!} = x - x^2 + \frac{x^3}{2}$

The function is $f(x) = xe^{-x}$

How do I find an upper bound on the error in the interval $[0, 1]$ ?

I think I should use the Taylor theorem about errors but I'm not sure

• Is this an alternating series? Do you know something about the truncation error of alternating series? – Eric Towers Jun 2 '16 at 16:48
• I had to find the Taylor polynomial of order $3$ around $x_{0} = 0$ (I did and posted here) and then to give an upper bound on the error in the interval $[0, 1]$ – Stephanie Jun 2 '16 at 16:53
• You will have to use the Lagrange's form of error term. Do you know what that is? – model_checker Jun 2 '16 at 17:24

## 3 Answers

The error term, $R_N$, is given by $$R_N = \frac{f^{(N+1)}(c)}{(N+1)!} x^{N+1}$$ for some $c$ on the interval $[0,1]$.

Here's an identity that will come in handy: $$\frac{d^N}{dx^N} (x e^{-x}) = (-1)^N e^{-x} (x-N)$$ In our case $N=3$ and according to Lagrange's error term we need the $N+1$ derivative of $f$, i.e. $$\frac{d^4}{dx^4} (x e^{-x}) = e^{-x} (x-4)$$

Thus the error is given by: $$R_N = \frac{f^{(N+1)}(c)}{(N+1)!} x^{N+1} = \frac{x^4}{4!} \frac{d^4}{dx^4} (x e^{-x}) = \frac{x^4}{4!} e^{-c} (c-4)$$ for some $c$ and $x$ in $[0,1]$.

We want to find the largest possible value this error can have on the interval $[0,1]$. Looking at the equation of $R_N$ we see this will happen when the numerator is maximized and the denominator is minimized, i.e. $x=1$ and $c=0$. Thus an upper bound for $|R_N|$ is: $$|R_N| \leq \left|\frac{1}{3!} \right| = \frac{1}{6}$$

Hint:

$R_n(x:0)=$$f^{n+1}(c)*x^{n+1}\over{(n+1)!}, where c\in[0,x]. Here n=3, find the 4th derivative or f^{3+1}(c) and try to establish a bound on it using the fact that c\in[0,x]. If it is a decreasing function then f^4(c) is maximum at c=0. Hope that works out. Put P_3(x)-f(x)=:G(x). One has$$G'(x)=P_3'(x)-f'(x)={2\over3} x^3 - {5\over24}x^4 + {1\over20}x^5 - {7\over720} x^6+\ldots\ ,$$which is an alternating series with decreasing absolute values of its terms when x\in\>]0,1]. It follows that G'(x) is positive in \>]0,1], whence G(x):=P_3(x)-f(x) is strictly increasing in this interval. As G(0)=0 we can conclude that$$0\leq P_3(x)-f(x)\leq P_3(1)-f(1)={1\over2}-{1\over e}\doteq0.132121\qquad(0\leq x\leq1)\ .$\$