# Lagrange Interpolating Polynomials - Error Bound

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.

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:

$$\mbox{Actual Error}~ = |f(1.4) - P_2(1.4)| = |15.0446 - 15.2698| = 0.2252$$

• how did u calculate $$\max_{(x, 1, 1.6)} |(x-1)(x-1.25)(x-1.6)| = 0.00754624$$? Nov 16, 2015 at 15:22
• Inside $x\in[1,1.25]$, the product is bounded by $\max|(x-1)(x-1.25)|\cdot |1.6-1|\le (0.125)^2\cdot 0.6=0.009375$ and in the second interval $[1.25,1.6]$ by $|1-6-1|\cdot max|(x-1.25)(x-1.6)|\le 0.6\cdot (0.175)^2=0.018375$ which is not exactly the bound given but it was quickly obtained without finding polynomial roots of the derivative. Apr 27, 2018 at 14:13
• @amzoti i am getting $\max_{(x, 1, 1.6)} |(x-1)(x-1.25)(x-1.6)| = 0.0135$ which occur at $x=1.45$. how did you arrive at $0.00754624$ Jun 21, 2020 at 12:44
• @Amzoti Why are you finding $\max_{(x, 1, 1.6)} |(x-1)(x-1.25)(x-1.6)|$? Don't we have $$E(x) = \frac {1}{3!} f^{(3)}(\xi(x)) (x - 1)(x - 1.125)(x - 1.6)$$ and therefore $$|E(1.4)| \le \left( \max_{y \in [1, 1.6]} \frac 16 f^{(3)}(y) \right)| (1.4 - 1)(1.4 - 1.125)(1.4 - 1.6) |?$$ Why are you also finding the max of the cubic polynomial? Apr 14, 2021 at 0:02

I know this is super late, but they use critical points to find the maximum of the absolute value of $(x−x_0)(x−x_1)\ldots(x−x_n)$ since we know that this product equals zero if $x= x_0, x_1,\ldots,\text{ or }x_n$.

This means that we don't consider the endpoints when finding the max in that interval, so the only possible choices are the critical points in that interval. And we know that there has to exist a critical point between each of the zeros so comparing the norms of each of the critical points always gives us the max for that part of the error.

• This question already had a well-accepted answer. You have contributed nothing new. Please refrain from doing this for old questions since they are pushed to the top as a result of activity. Feb 11, 2016 at 13:57