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.

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- Oct 12th 2011, 06:11 PMveronicak5678Taylor Remainder Theorem
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. - Oct 12th 2011, 09:33 PMchisigmaRe: Taylor Remainder Theorem
Take into account that in the remainder formula...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Oct 13th 2011, 12:07 PMveronicak5678Re: Taylor Remainder Theorem
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? - Oct 13th 2011, 10:13 PMchisigmaRe: Taylor Remainder Theorem
My wrong replay is due to the fact that I remembered a formula for the Lagrange remainder where $\displaystyle \xi$ is 'normalized' to $\displaystyle x-x_{0}$ (Headbang) ...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Oct 14th 2011, 07:42 AMveronicak5678Re: Taylor Remainder Theorem
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? - Oct 14th 2011, 11:55 AMchisigmaRe: Taylor Remainder Theorem
The computation of...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Oct 14th 2011, 12:54 PMveronicak5678Re: Taylor Remainder Theorem
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?

- Oct 15th 2011, 12:04 AMchisigmaRe: Taylor Remainder Theorem
I apologize with Veronika for the fact that in hand computation I'm very poor [(Doh) ...], so that I decided to use my computer to solve her problem...

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Oct 15th 2011, 11:19 AMveronicak5678Re: Taylor Remainder Theorem
Does this seem right? It looks like such a large value.