
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
$\displaystyle P_n(x)=3+\dysplaystyle\frac{(x9)}{6}\dysplaystyle\frac{(x9)^2}{216}+\dysplaystyle\frac{3(x9)^3}{3888}$
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?

correct the cubic term ...
$\displaystyle \displaystyle P_3(x) = 3 + \frac{x9}{6}  \frac{(x9)^2}{216} + \frac{(x9)^3}{3888}$
$\displaystyle P_3(9.03) = 3.00499584028...$
since the series is alternating, the error will be less than the first omitted term ...
$\displaystyle \displaystyle Error < \left\frac{5(.03)^4}{279936}\right \approx 1.45 \times 10^{11}$

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} $

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:
$\displaystyle R_n(f) = \frac{f^{(n+1)}(\xi)}{(n+1)!} (xa)^{n+1}$
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.

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 ...
$\displaystyle \displaystyle Error < \left\frac{f^4(9)(x9)^4}{4!}\right$