How many points do you need to determine the equation of a polynomial of degree n?

Nov 8, 2015

$n + 1$

Explanation:

One point $=$ one constraint

A polynomial of degree $n$ has $n + 1$ terms, from ${x}^{n}$ down to ${x}^{0}$, each with a separate coefficient.

You cannot force the value of more than one coefficient with one point.

Conversely, $n + 1$ distinct points are sufficient. If you give me the points $\left(0 , f \left(0\right)\right)$, $\left(1 , f \left(1\right)\right)$, ... , $\left(n , f \left(n\right)\right)$, I can write out a sequence:

$\textcolor{b l u e}{f \left(0\right)} , f \left(1\right) , f \left(2\right) , \ldots , f \left(n\right)$

Then write out a sequence of differences of that sequence:

$\textcolor{b l u e}{\left(f \left(1\right) - f \left(0\right)\right)} , \left(f \left(2\right) - f \left(1\right)\right) , \ldots , \left(f \left(n\right) - f \left(n - 1\right)\right)$

Then a sequence of differences of those differences:

$\textcolor{b l u e}{\left(f \left(2\right) - f \left(1\right)\right) - \left(f \left(1\right) - f \left(0\right)\right)}$, ...

Repeating until only one term is left.

Then the first terms of each of these sequences form the coefficients of a direct formula for $f \left(x\right)$:

f(x) = color(blue)(f(0))/(0!) + color(blue)((f(1)-f(0)))/(1!)x + color(blue)(((f(2)-f(1))-(f(1)-f(0))))/(2!)x(x-1) +...

This is the unique polynomial of degree $n$ passing through these $n + 1$ points.