Let $f(x) = e^{2x} – x$, $x_0 = 1$, $x_1 = 1.25$, and $x_2 = 1.6$. Construct interpolation polynomials of degree at most one and at most two to approximate $f(1.4)$, and find an error bound for the approximation.
So I know how to construct the interpolation polynomials, but I'm just not sure how to find the error bound. I know that the formula for the error bound is:
$${f^{n+1}(\xi(x)) \over (n+1)!} \times (x-x_0)(x-x_1)…(x-x_n)$$
For the interpolation polynomial of degree one, the formula would be:
$${f^{2}(\xi(x)) \over (2)!} \times (x-1)(x-1.25)$$
So if I take the second derivative of the function, I would get $f''(x) = 4e^{2x}$. Since $f''$ is strictly increasing on the interval $(1, 1.25)$, the maximum error of ${f^{2}(\xi(x)) \over (2)!}$ will be $4e^{2 \times 1.25}/2!$. Plugging in $x=1.4$ in the formula above gives us $1.461899$. I was just wondering if this is the correct way to calculate the error bound, since I've seen examples where they would take the derivative and find critical points and then I would get lost.
Can someone please clarify? Thanks!
Best Answer
I will do the part with all three points and you can do the other with two points.
We are given that $f(x) = e^{2x} - x$, $x_0 = 1$, $x_1 = 1.25$, and $x_2 = 1.6$.
We are asked to construct the interpolation polynomial of degree at most two to approximate $f(1.4)$, and find an error bound for the approximation.
You stated that you know how to find the interpolating polynomial, so we get:
$$P_2(x) = 26.8534 x^2-42.2465 x+21.7821$$
The formula for the error bound is given by:
$$E_n(x) = {f^{n+1}(\xi(x)) \over (n+1)!} \times (x-x_0)(x-x_1)...(x-x_n)$$
Since we do not know where $\xi(x)$ is, we will find each error over the range and multiply those together, so we have:
$$\max_{(x, 1, 1.6)} |f'''(x)| = \max_{(x, 1, 1.6)} 8 e^{2 x} = 196.26$$
Next, we need to find:
$$\max_{(x, 1, 1.6)} |(x-1)(x-1.25)(x-1.6)| = 0.00754624$$
Thus we have an error bound of:
$$E_2(x) = \dfrac{196.26}{6} \times 0.00754624 \le 0.246838$$
If we compute the actual error, we have:
$$\mbox{Actual Error}~ = |f(1.4) - P_2(1.4)| = |15.0446 - 15.2698| = 0.2252$$