Given some conditions-finding the minimum degree of a polynomial equation.

Question

On, the first look, I think the solution to this problem is a long one with all these conditions. Please give me hints on starting off with this problem because I have been struggling with this(maybe due to wrong method) for a while.

Answer 1

For a function $f \colon \mathbb{R} \to \mathbb{R}$, define

$$\Delta f(x) := f(x+1) - f(x),$$

and the iterated differences

$$\begin{align} \Delta^0 f &= f\\ \Delta^{n+1} f &= \Delta(\Delta^n f) \end{align}$$

for $n \in \mathbb{N}$. You can easily verify (by induction or otherwise) $\Delta^n(f+g) = \Delta^n f + \Delta^n g$ and $\Delta^n (c\cdot f) = c\cdot \Delta^n f$ for all $n\in\mathbb{N}$ and $c\in\mathbb{R}$.

Then you can prove that for a polynomial $f(x) = a_0 + \dotsc + a_n x^n$, you have

$$\Delta^n f(x) \equiv n!\cdot a_n.$$

Applying the iterated differences to the conditions

$$p(-2) = -15,\, p(-1) = 1,\, p(0) = 7,\, p(1) = 9,\, p(2) = 13,\, p(3) = 25$$

yields

$$\begin{gather} \Delta p(-2) = 16,\, \Delta p(-1) = 6,\, \Delta p(0) = 2,\, \Delta p(1) = 4,\, \Delta p(2) = 12\\ \Delta^2 p (-2) = -10,\, \Delta^2 p(-1) = -4,\, \Delta^2 p(0) = 2,\, \Delta^2 p(1) = 8\\ \dotsb \end{gather}$$

and in a few steps the minimal possible degree $n$ for the constraints.