Taylor Remainder Theorem

**veronicak5678** · October 12th 2011, 06:11 PM

Find the 5th degree Taylor poly and a bound for its error.

I have attached a doc with what I have so far. I don't understand how to bound the error when it depends on some c we don't know anything about.

**chisigma** · October 12th 2011, 09:33 PM

Originally Posted by veronicak5678

Find the 5th degree Taylor poly and a bound for its error.

I have attached a doc with what I have so far. I don't understand how to bound the error when it depends on some c we don't know anything about.

Take into account that in the remainder formula...

$R_{5} (x)= \frac{\xi\ (\xi^{6}-15\ \xi^{4} +45\ \xi^{2} -15)}{6!\ \sqrt{2 \pi}}\ e^{-\frac{\xi^{2}}{2}}\ x^{6}= f(\xi)\ x^{6}$ (1)

...is $0<\xi<1$ . So You can find with usual approach the 'extreme values' of $f(\xi)$ in $\xi \in (0,1)$ and automatically You will have the 'bounds'...

Kind regards

$\chi$ $\sigma$

**veronicak5678** · October 13th 2011, 12:07 PM

Why is c between 0 and 1? I forgot to mention that -10 <= x <= 10.

I'm confused about finding the max values when I have two variables in the remainder term. How do I find the max value when I need to know the values of c and x?

**chisigma** · October 13th 2011, 10:13 PM

Originally Posted by veronicak5678

Why is c between 0 and 1? I forgot to mention that -10 <= x <= 10...

My wrong replay is due to the fact that I remembered a formula for the Lagrange remainder where $\xi$ is 'normalized' to $x-x_{0}$

...

Never mind!... if $x \in [-10,10]$ and $x_{0}=0$ You have to find the extreme values of $f(\xi)$ with $\xi \in (-10,10)$ ...

Kind regards

$\chi$ $\sigma$

**veronicak5678** · October 14th 2011, 07:42 AM

I have found that for c between -10 and 10 f has a max at c= 1.27144933065785
but what about the x^6? Do I need to worry about that?

**chisigma** · October 14th 2011, 11:55 AM

Originally Posted by veronicak5678

I have found that for c between -10 and 10 f has a max at c= 1.27144933065785
but what about the x^6? Do I need to worry about that?

The computation of...

$R_{5}= \frac{\xi\ (\xi^{6} - 15\ \xi^{4} + 45\ \xi^{2} -15)}{6!\ \sqrt{2\ \pi}}\ e^{-\frac{\xi^{2}}{2}}\ x^{6}$ (1)

... for $\xi = 1.27$ and $x= \pm 10$ produce, if no errors of me, $R_{5} \sim 7.15\ 10^{3}$ ... of course that is the 'maximally unlucky case'

...

Kind regards

$\chi$ $\sigma$

**veronicak5678** · October 14th 2011, 12:54 PM

I don't understand where you got that number. Am I correct that to get the error, we first find the max error for the part of the function involving c for c between -10 and 10, ans then you find the max error for the function involving x for x between -10 and 10 and multiply? Doing that, we get 1.2714e+006. What am I doing wrong?

**chisigma** · October 15th 2011, 12:04 AM

Originally Posted by chisigma

The computation of...

$R_{5}= \frac{\xi\ (\xi^{6} - 15\ \xi^{4} + 45\ \xi^{2} -15)}{6!\ \sqrt{2\ \pi}}\ e^{-\frac{\xi^{2}}{2}}\ x^{6}$ (1)
... for $\xi = 1.27$ and $x= \pm 10$ produce, if no errors of me, $R_{5} \sim 7.15\ 10^{3}$ ... of course that is the 'maximally unlucky case' ...
Kind regards
$\chi$ $\sigma$

I apologize with Veronika for the fact that in hand computation I'm very poor [

...], so that I decided to use my computer to solve her problem...

The first step is to find the extreme points of the function...

$f(\xi)= \frac{\xi\ (\xi^{6} - 15\ \xi^{4} + 45\ \xi^{2} -15)}{6!\ \sqrt{2\ \pi}}\ e^{-\frac{\xi^{2}}{2}}$ (1)

According to my computer the derivative of (1) vanishes for $\xi_{1}= \pm .328599589228...$ and $\xi_{2}= \pm 4.00372687633...$ and because is $|f(\xi_{1})|= .00177930867978...$ and $|f(\xi_{2})|= .000711887409732...$ it must be [tex] $|f(\xi)| \le .00177930867978...$ . Now is $R_{5}= f(\xi)\ x^{6}$ so that for $x \in [-10,10]$ is...

$|R_{5}| \le 1.7793\ 10^{3}$ (2)

Kind regards

$\chi$ $\sigma$

**veronicak5678** · October 15th 2011, 11:19 AM

Does this seem right? It looks like such a large value.

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