estimating error Taylor polynomials

I have been asked to do the following:

Suppose that f is a function with the property that |$\displaystyle f^n(x)$|$\displaystyle \leq$3 for all n and for all x on [-1,1].

a. Estimate the error if the Taylor polynomial $\displaystyle p_{n}$(x) (in powers of x) with n=4 is used to approximate f($\displaystyle \frac{1}{2}$).

b. Find the least positive integer n such that $\displaystyle p_{n}$(x) (in powers of x) will approximate f on [-1,1] to within .001.

I don't know where to start on this problem. I have no f(x) to work with so what do I do?

Re: estimating error Taylor polynomials

Use $\displaystyle f(x)=p_4(x)+\dfrac{f^{(5)}(\xi)}{5!}x^5$ where $\displaystyle \xi$ is between $\displaystyle 0$ and $\displaystyle x$ .

Re: estimating error Taylor polynomials

Quote:

Originally Posted by

**FernandoRevilla** Use $\displaystyle f(x)=p_4(x)+\dfrac{f^{(5)}(\xi)}{5!}x^5$ where $\displaystyle \xi$ is between $\displaystyle 0$ and $\displaystyle x$ .

Do I substitute $\displaystyle \dfrac{1}{2}$ in for x? If so, I end up with $\displaystyle f(x)=p_4(\dfrac{1}{2})+\dfrac{f^{(5)}(\xi)}{5!} *(\dfrac{1}{2})^5$ .

Or do I substitute $\displaystyle \dfrac{1}{2}$ in for $\displaystyle \xi$? Which gives me $\displaystyle f(x)=p_4(x)+\dfrac{f^{(5)}(\dfrac{1}{2})}{5!} *x^5$ .

In either case, is it possible to get a number as a value for the error with the information I've been given?

Re: estimating error Taylor polynomials

$\displaystyle f(1/2)-p_4(1/2)=\dfrac{f^{(5)}(\xi)}{5!}(1/2)^5\Rightarrow |f(1/2)-p_4(1/2)|\leq (3/2^5)(1/5!)$ .

Re: estimating error Taylor polynomials

Quote:

Originally Posted by

**FernandoRevilla** $\displaystyle f(1/2)-p_4(1/2)=\dfrac{f^{(5)}(\xi)}{5!}(1/2)^5\Rightarrow |f(1/2)-p_4(1/2)|\leq (3/2^5)(1/5!)$ .

The error for $\displaystyle \p_{4}$ is .0007813 and the error for $\displaystyle \p_{3}$ is .007813, so n must be equal to 4 to remain within .001.

Thank you.