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

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Answer 1 · 2015-11-08T00:58:09

#n+1#

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 #(0, f(0))#, #(1, f(1))#, ... , #(n, f(n))#, I can write out a sequence:

#color(blue)(f(0)), f(1), f(2),...,f(n)#

Then write out a sequence of differences of that sequence:

#color(blue)((f(1) - f(0))), (f(2)-f(1)), ..., (f(n) - f(n-1))#

Then a sequence of differences of those differences:

#color(blue)((f(2)-f(1))-(f(1)-f(0)))#, ...

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(x)#:

#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.