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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

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  • $\begingroup$ Is this an alternating series? Do you know something about the truncation error of alternating series? $\endgroup$ Commented Jun 2, 2016 at 16:48
  • $\begingroup$ 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 $[0, 1]$ $\endgroup$
    – Stephanie
    Commented Jun 2, 2016 at 16:53
  • $\begingroup$ You will have to use the Lagrange's form of error term. Do you know what that is? $\endgroup$
    – Student
    Commented Jun 2, 2016 at 17:24

3 Answers 3

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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: $$|R_N| \leq \left|\frac{1}{3!} \right| = \frac{1}{6} $$

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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 $$0\leq P_3(x)-f(x)\leq P_3(1)-f(1)={1\over2}-{1\over e}\doteq0.132121\qquad(0\leq x\leq1)\ .$$

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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.

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