Taylor Polynomial Error

Collegeboy110 · #1 $Old$ May 9th, 2009, 02:54 PM

I cant figure this one out either. Thank You.

Find an upper bound for the error in using a 3rd degree Taylor Polynomial to approximate (1.3)^(1/3).

TheEmptySet · #2 $Old$ May 9th, 2009, 02:58 PM

Quote:

Originally Posted by Collegeboy110 $View Post$

I cant figure this one out either. Thank You.

Find an upper bound for the error in using a 3rd degree Taylor Polynomial to approximate (1.3)^(1/3).

We need more info. Where is the series centered? that will change the error. 0,1,8...

Collegeboy110 · #3 $Old$ May 9th, 2009, 03:02 PM

The question is exactly written like this. It does not give me more info.

TheEmptySet · #4 $Old$ May 9th, 2009, 03:16 PM

Quote:

Originally Posted by Collegeboy110 $View Post$

The question is exactly written like this. It does not give me more info.

so the taylor expansion looks like this

$T(x)=1+\frac{1}{3}(x-1)-\frac{1}{9}(x-1)^2+\frac{5}{81}(x-1)^3-\frac{10}{243}(x-1)^4+...$

Since this is an alternating series (after the 1st term)

The error can be no larger than the next term in the series so we get

$\frac{10}{243}(.3)^4 =0.000\bar3$

Collegeboy110 · #5 $Old$ May 9th, 2009, 03:41 PM

Can u explain it to me further please. I cannot connect the answer to what they are asking me. The 3rd degree polynomial would it not be 5/81 (x-1)^3

TheEmptySet · #6 $Old$ May 9th, 2009, 03:44 PM

Quote:

Originally Posted by Collegeboy110 $View Post$

Can u explain it to me further please. I cannot connect the answer to what they are asking me. The 3rd degree polynomial would it not be 5/81 (x-1)^3

$T_3(x)=1+\frac{1}{3}(x-1)-\frac{1}{9}(x-1)^2+\frac{5}{81}(x-1)^3$

Is the 3rd degree taylor expansion centered at 1.

You need all of the terms of degree $\le 3$

Collegeboy110 · #7 $Old$ May 9th, 2009, 03:51 PM

Ok, so for the formula i get that 1.3-1 = .3.. how did u get 10/243?

Quote:

Originally Posted by TheEmptySet $View Post$

so the taylor expansion looks like this

$T(x)=1+\frac{1}{3}(x-1)-\frac{1}{9}(x-1)^2+\frac{5}{81}(x-1)^3-\frac{10}{243}(x-1)^4+...$

Since this is an alternating series (after the 1st term)

The error can be no larger than the next term in the series so we get

$\frac{10}{243}(.3)^4 =0.000\bar3$

TheEmptySet · #8 $Old$ May 9th, 2009, 04:00 PM

Quote:

Originally Posted by Collegeboy110 $View Post$

Ok, so for the formula i get that 1.3-1 = .3.. how did u get 10/243?

It is the next term in the Taylor series

$\frac{10}{243}(x-1)^4$

the_doc · #9 $Old$ May 9th, 2009, 07:52 PM

"TheEmptySet" just computed the Taylor expansion of $T(x) = x^{1/3}$ centered at $x=1$ . So to compute the coefficients you do the usual thing of sequentially evaluating each derivative of $T(x)$ at $x=1$ . So to work out the coefficient you are struggling with compute the fourth derivative i.e. $(1/3)\cdot (-2/3) \cdot (-5/3) \cdot(-8/3) x^{-11/3} = -80/81 \, x^{-11/3}$ , evaluate at $x=1$ then finally divide by $4!$ to give you the desired coefficient.

Collegeboy110 · #10 $Old$ May 10th, 2009, 12:32 PM

So How would this series change if it is centered at x= 0?

Quote:

Originally Posted by the_doc $View Post$

"TheEmptySet" just computed the Taylor expansion of $T(x) = x^{1/3}$ centered at $x=1$ . So to compute the coefficients you do the usual thing of sequentially evaluating each derivative of $T(x)$ at $x=1$ . So to work out the coefficient you are struggling with compute the fourth derivative i.e. $(1/3)\cdot (-2/3) \cdot (-5/3) \cdot(-8/3) x^{-11/3} = -80/81 \, x^{-11/3}$ , evaluate at $x=1$ then finally divide by $4!$ to give you the desired coefficient.

TheEmptySet · #11 $Old$ May 10th, 2009, 12:43 PM

Quote:

Originally Posted by Collegeboy110 $View Post$

So How would this series change if it is centered at x= 0?

It cannot be centered at zero becuase $f(x)=x^{1/3}$ is not differentable at x=0

$f'(x)=\frac{1}{3}\frac{1}{x^{2/3}}$ is undefined at zero.

Collegeboy110 · #12 $Old$ May 10th, 2009, 12:49 PM

The question says Find an upper bound for the error in using the 3rd degree polynomial to approximate the cubed root of 1.3.. which I assume is (1.3)^(1/3).. i asked my proffesor he said centered at 0.

TheEmptySet · #13 $Old$ May 10th, 2009, 12:57 PM

ummmm... I don't want to contradict your prof. but there is not a taylor expansion at 0 for $f(x) = x^{1/3}$

A taylor polynomial centered at a is given by the formula

$T(x)=\sum_{n=0}^{\infty}\frac{f^{n}(a)}{n!}(x-a)^n$

if we try to center it a zero we get

$T(x)=\sum_{n=0}^{\infty}\frac{f^{n}(0)}{n!}(x)^n$

This is a big problem becuse

$f'(x)=\frac{1}{3}\frac{1}{x^{2/3}}$

Is undefined at zero and so is every derivative after this one

Collegeboy110 · #14 $Old$ May 10th, 2009, 01:29 PM

I attached a similar problem (0.5)^(1/3) at x= 0..

the_doc · #15 $Old$ May 10th, 2009, 01:45 PM

Firstly, centred at zero simply means it is in a power series of $x$ rather than $x$ minus something.
The above post is correct but I suspect that your Prof. is actually wanting you to find a Taylor expansion centered at $x=0$ for the function $T(x) = (1+x)^{1/3}$ hence there is no contradiction. This is just a simple binomial expansion i.e.
$(1+x)^{1/3} = 1 + \frac{1}{3} x + \left( \frac{1}{3} \right) \cdot \left(-\frac{2}{3} \right)\frac{x^2}{2!} + \left( \frac{1}{3} \right) \cdot \left(-\frac{2}{3} \right) \cdot \left(-\frac{5}{3} \right) \frac{x^3}{3!}$ $+ \left( \frac{1}{3} \right) \cdot \left(-\frac{2}{3} \right) \cdot \left(-\frac{5}{3} \right) \cdot\left(-\frac{8}{3} \right) \frac{x^4}{4!} +\ldots$
which after tidying up gives you
$(1+x)^{1/3} = 1 + \frac{1}{3} x - \frac{1}{9} x^2 + \frac{5}{81} x^3 - \frac{10}{243} x^4 +\ldots$
which as you can see has all the same coefficients and in fact numerically amounts to the same power series ( $x=0.3$ instead of 1.3)!

So in conclusion you can centre at 0 but you have to choose a function which is differentiable at 0.