# Taylor Polynomials

• Aug 11th 2010, 02:19 PM
Ulysses
Taylor Polynomials
Find an approximated value for $\displaystyle \sqrt[ ]{9.03}$ using a Taylors polynomial of third degree and estimate the error.

I thought of solving it by using

$\displaystyle f(x)=\sqrt[]{x}$ centered at $\displaystyle x_0=9$

So

$\displaystyle P_n(x)=3+\dysplaystyle\frac{(x-9)}{6}-\dysplaystyle\frac{(x-9)^2}{216}+\dysplaystyle\frac{3(x-9)^3}{3888}$

Then I evaluated it at x=9.03, so I get:

$\displaystyle P_n(x)=3+\dysplaystyle\frac{(0.3)}{6}-\dysplaystyle\frac{(0.3)^2}{216}+\dysplaystyle\fra c{3(0.3)^3}{3888}\approx{3.049604167}$

I don't know if this is right, I've tried with the calculator and it gives 3.00500.... Now, how do I estimate the error? just by resting to the value the calculator gives the one I get?
• Aug 11th 2010, 02:32 PM
skeeter
correct the cubic term ...

$\displaystyle \displaystyle P_3(x) = 3 + \frac{x-9}{6} - \frac{(x-9)^2}{216} + \frac{(x-9)^3}{3888}$

$\displaystyle P_3(9.03) = 3.00499584028...$

since the series is alternating, the error will be less than the first omitted term ...

$\displaystyle \displaystyle Error < \left|-\frac{5(.03)^4}{279936}\right| \approx 1.45 \times 10^{-11}$
• Aug 11th 2010, 02:34 PM
Mush
Should it not be 0.03 in your brackets rather than 0.3?

Regardless, the error of an nth order Taylor Polynomial using a stepsize of h is always of the order of $\displaystyle h^{n+1}$

So, in your case, I would estimate the error to be of the order of $\displaystyle 0.03^{4}$
• Aug 11th 2010, 02:59 PM
Ulysses
Thanks. Skeeter, how did you get the expression?

$\displaystyle \displaystyle Error < \left|-\frac{5(.03)^4}{279936}\right| \approx 1.45 \times 10^{-11}$

I think you're using the expression for the residual, that looks quiet similar to the expressions of the terms for the Taylors polynomial.

The thing is that the reminder is:

$\displaystyle R_n(f) = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-a)^{n+1}$

where $\displaystyle \xi$ is such $\displaystyle a<\xi<x$

mmm I think I'm getting it, so $\displaystyle 9<\xi<9.03$?

And then thats why the error is smaller than the expression for the reminder. Right? thank you very much Skeeter.
• Aug 11th 2010, 03:03 PM
skeeter
no need for computing an error bound ... for a strictly alternating series, the error is no greater than the first omitted term. In this case ...

$\displaystyle \displaystyle Error < \left|\frac{f^4(9)(x-9)^4}{4!}\right|$