taylor's inequality

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November 3rd 2010, 10:43 PM
pirateboy

taylor's inequality

I just want to make sure I'm doing this right.

So we're given the funciton
$f(x) = x \ln(x)$

we're first asked to find the 4th degree taylor polynomial centered at $a=1$

we're then asked to use taylor's inequality to bound the error in the approximation

$f(x)\approx T_4(x) \text{ for } x\in\left[0.2,1.8\right]$

we recall

$\left|R_n(x)\right|\leq \frac{M}{(n+1)!}\left|x-a\right|^{n+1}$

well, $f^{(5)}(x) = \frac{-6}{x^4}$ so i'd want to use $x=.2$ ?

if i understand correctly, I want M to be as big as possible. and since this is centered at a=1, this would give me

$\displaystyle \left|R_4(x)\right|\leq \frac{f^{(5)}(.2)}{5!}\left|.2-1\right|^{5} = \frac{\frac{-6}{(.2)^4}}{5!}(.8)^5 = -10.24$

this just seems like quite a big maximum error to me.

if someone could confirm that this is the error i should be getting, that'd be great.
November 4th 2010, 01:14 AM
CaptainBlack

Quote:

Originally Posted by pirateboy

I just want to make sure I'm doing this right.

So we're given the funciton
$f(x) = x \ln(x)$

we're first asked to find the 4th degree taylor polynomial centered at $a=1$

we're then asked to use taylor's inequality to bound the error in the approximation

$f(x)\approx T_4(x) \text{ for } x\in\left[0.2,1.8\right]$

we recall

$\left|R_n(x)\right|\leq \frac{M}{(n+1)!}\left|x-a\right|^{n+1}$

well, $f^{(5)}(x) = \frac{-6}{x^4}$ so i'd want to use $x=.2$ ?

if i understand correctly, I want M to be as big as possible. and since this is centered at a=1, this would give me

$\displaystyle \left|R_4(x)\right|\leq \frac{f^{(5)}(.2)}{5!}\left|.2-1\right|^{5} = \frac{\frac{-6}{(.2)^4}}{5!}(.8)^5 = -10.24$

this just seems like quite a big maximum error to me.

if someone could confirm that this is the error i should be getting, that'd be great.

When you take absolute values you do it throughout, so the right hand side should be positive. When that correction is made you do have an upper bound on the error, just not a very good one. The actual absolute error is less than 0.04.

CB
November 4th 2010, 01:32 AM
pirateboy

ah, right. I thought about that when I was posting. I should've taken a bit more time and corrected it. so the upper bound comes from

$\left|f^{(5)}(.2)\right|\leq M$

so I should've had

$\displaystyle \left|R_4(x)\right|\leq \frac{\left|f^{(5)}(.2)\right|}{5!}\left|.2-1\right|^{5} = \frac{\left|\frac{-6}{(.2)^4}\right|}{5!}(.8)^5 = 10.24$

and out of curiosity how did you get the absolute error? did you use a calculator for the value and then compare it?

it appeared to me that this might actually be able to be broken down into an alternating series so that i then might use the alternating series estimation theorem on it....but i wasn't sure.
November 4th 2010, 03:56 AM
CaptainBlack

Quote:

Originally Posted by pirateboy

ah, right. I thought about that when I was posting. I should've taken a bit more time and corrected it. so the upper bound comes from

$\left|f^{(5)}(.2)\right|\leq M$

so I should've had

$\displaystyle \left|R_4(x)\right|\leq \frac{\left|f^{(5)}(.2)\right|}{5!}\left|.2-1\right|^{5} = \frac{\left|\frac{-6}{(.2)^4}\right|}{5!}(.8)^5 = 10.24$

and out of curiosity how did you get the absolute error? did you use a calculator for the value and then compare it?

it appeared to me that this might actually be able to be broken down into an alternating series so that i then might use the alternating series estimation theorem on it....but i wasn't sure.

It is only an alternating series on one side of the point of expansion, all the terms have the same sign on the other side.

.. and yes I found the actual error by ploting the Taylor polynomial minus the function itself.

CB