Find formula for $f^{\prime\prime}(x_0)$ using $f(x_0−2h),f(x_0-h),f(x_0+2h)$ [closed]

I want to find an approximation for $f^{\prime\prime}(x_0)$ using the values of: $f(x_0−2h),\,f(x_0-h),\,f(x_0+2h)$

How to use the method of undetermined coefficients to solve the formula.

I'll get you started. Under certain famous assumptions, Taylor series give$$\begin{align}f(x_0-2h)&=f(x_0)-2hf^\prime(x_0)+2h^2f^{\prime\prime}(x_0)+O(h^3),\\f(x_0-h)&=f(x_0)-hf^\prime(x_0)+\frac12h^2f^{\prime\prime}(x_0)+O(h^3),\\f(x_0+2h)&=f(x_0)+2hf^\prime(x_0)+2h^2f^{\prime\prime}(x_0)+O(h^3).\end{align}$$If you seek a linear combination equal to $h^2f^{\prime\prime}(x_0)+O(h^3)$, you should obtain$$\frac16(3f(x_0-2h)-4f(x_0-h)+f(x_0+2h))$$by solving simultaneous equations. But now for the clever part: the conjecture$$f^{\prime\prime}(x_0)=\lim_{h\to0}\frac{3f(x_0-2h)-4f(x_0-h)+f(x_0+2h)}{6h^2}$$does not require the aforementioned assumptions; instead, it will follow for any twice continuously differentiable $f$ by e.g. using L'Hôpital's rule twice. See here for a similar calculation.