Let $f$ be a twice differentiable function over real line. Let $a\in \mathbb{R}$ and $h\in \mathbb{R}^{+}$. Let $P_f$ denote the interpolating polynomial of degree $2$ of $f$, i.e., $$f(a)=P_f(a), \qquad f(a-h)=P_f(a-h), \qquad f(a+h)=P_f(a+h).$$ Let $g$ be any polynomial of degree $d$ such that $$g''(a)=P_g''(a).$$ What is the maximum possible value of $d$ such that the above conditions hold for all polynomials of degree $d$?
Answer- $3$
First I show that $d$ can not be an even integer greater than $2$. I took $a=0, h=1$ and $g(x)=x^d$, where $d\geq 4$ and $d$ is even. Let $P_g(x)=c_0+c_1x+c_2x^2$. By the given conditions, we have $$g(0)=P_g(0)\implies c_0=0.$$ Further, $$g(1)=P_g(1)\implies c_1+c_2=1\ \ \text{and}\ \ g(-1)=P_g(-1)\implies -c_1+c_2=1.$$ This means $c_1=0, c_2=1$, i.e., $P_g(x)=x^2$ for this $g$. So, the fourth condition is $$g''(0)=0\neq 1=c_2=P_g''(0).$$ Thus, $d$ can not be an even integer greater than $2$.
Next, I show that $d$ can not be an odd integer greater than $3$. Let $d$ be an odd integer greater than $3$. I took $$g(x)=x^d+x+1$$ and obtain the contradiction in the similar manner as for the previous case.
Next, I know that if $d=2$, then $g(x)\equiv P_g(x)$ (identically same). It remains to see that whether $d=3$ is possible or not?
How to proceed for this?
Is my approach correct or am I missing some direct result? Plese help.
If you use Newton form for $P_g$ interpolating at $a-h, a, a+h$, then
$$P_g''(a) = \frac{g(a+h) + g(a -h) - 2g(a)}{h^2}$$
Any cubic polynomial can be written as $g(x) = g(a) + g'(a)(x-a) + \frac{1}{2}g''(a) (x-a)^2 + \frac{1}{6} g'''(a)(x-a)^3$, you can bring $g(a+h)$, $g(a-h)$ into the formula for $P_g''$, then all other terms except second order are cancelled, and you get $g''(a)$.
Thus the formula holds for $d=3$.