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.

  • $\begingroup$ 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. $\endgroup$ Mar 11, 2021 at 20:25

1 Answer 1


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.


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