# Finding the Bound of the Error in Interpolation

The function $$f(x)$$ has the following values known at the points: $$f(x_1) = 2, \quad f(x_2) = 8 \quad \text{and} \quad f(x_3) = -2$$.

I've also been told that I can assume that $$f$$ and its derivatives satisfy the following bounds for all $$x$$: $$|f(x)| \le 10, \,\, |f'(x)| \le 20, \,\, |f''(x)| \le 30, \,\, |f'''(x)| \le 50 \text{ and } |f''''(x)| \le 100.$$

I want to be able to bound the error in approximating the polynomial $$f(x)$$ at $$x=2$$, for two different methods I have used to approximate the polynomial:

• Lagrange Polynomial Interpolation: $$P(x) = -2x^2 + 11x -7$$
• Piecewise Linear Interpolation: $$S(x) = 3x - 1$$ upon the interval $$[1,3]$$

I know that the formula for the error in polynomial interpolation is given as: $$f(x) - P(x) = \frac{f^{(n)}(\xi)}{n!} \prod_{k=1}^n (x-x_k).$$

But how can I use this formula to $$\underline{bound}$$ the error for these two types of interpolations? Any guidance would be appreciated.

• There is a link between Newton Divided difference and the error term. You may be able to extract the individual $x_i$'s there. Once you have those getting the maximum of that error formula should be trivial. Mar 11, 2021 at 20:25

In your case $$n=3$$, so $$|f(x) - P(x)| = \frac{\left|f^{(3)}(\xi)\right|}{3!} \prod_{k=1}^3 |x-x_k|\le\frac {50}6 \prod_{k=1}^3 |x-x_k|$$
Since you know that $$1 \le x \le 3$$, you can also find bounds on the $$|x-x_k|$$ elements, but since you haven't told us what the $$x_i$$ are, we cannot take it any farther.