Q: Let
$$ f(x) = -2x^7 + 54321x^5 50x^3 + 2000x^2 - 10x + 0.000001, $$
and
$$ x_0 = 0.1, x_1 = 0.2, x_2 = 0.25, x_3 = 0.4, x_4 = 0.46, x_5 = 0.48, \\ x_6 = 0.50, x_7 = 0.52, x_8 = 0.6, x_9 = 0.7, x_{10} = 0.8, x_{11} = 1.25.$$
Find a polynomial $P(x)$ of degree at most 11 that interpolates $f(x)$ at $x_i (i=0,1,...,11)$.
A: Since $f(x) = −2x^7 + 54321x^5 + 50x^3 + 2000x^2 −10x + 0.000001 $ is a polynomial of degree $7$, so, $P(x) = f(x) = −2x^7 + 54321x^5 +50x^3 +2000x^2 − 10x +0.000001$.
I am slightly confused as to why a polynomial of degree $7$ such as $f(x)$ will suffice for the approximation and interpolation of data points in the range of $x_0$ to $x_{11}$.
The interpolation polynomial has degree at most $11$ , but every smaller degree is possible. Consider the case, when all $y$-values coincide. Then , the degree is $0$.
More concrete : If pairwise different values $x_0,\cdots,x_n$ are given, then for every tuple $y_0,\cdots,y_n$ there exists a unique polynomial $f(x)$ with degree at most $n$, which satisfies $f(x_j)=y_j$ $j=0,\cdots ,n$.
In the case $y_0=\cdots=y_n$, the interpolating polynomial is $f(x)=y_0$. Every degree from $0$ to $n$ can occur.
The typical case is $deg(f)=n$, but not the only possible case.