My professor gave us this question to solve, but I don't know have much familiarity with the topic.
Question
The forward difference is first order accurate and is defined to be $$ D_{+} = \frac{f(x+h) - f(x)}{h} $$ and the centered difference is $$ D_{0} = \frac{f(x+h)-f(x-h)}{2h}$$ Consider the finite difference approximation where $$ f'(x) = \frac{af(x+h) + bf(x) + cf(x-h)}{h}$$ where a,b,c are constants. The forward difference has constants $(a,b,c) = (1,-1,0)$ and the centered difference has constants $(a,b,c) = (\frac{1}{2}, 0, \frac{1}{2})$. Are there any constants (a,b,c) that achieve 3rd order accuracy?
Work
So my approach was to take a look at the Taylor Series expansion of these terms. \begin{align} &af(x+h) = af(x)+ ahf'(x) + ah^2\frac{f''(x)}{2} + \dots \\ &bf(x) = bf(x) \\ &cf(x-h) = cf(x) - chf'(x) + ch^2\frac{f''(x)}{2} - \dots \end{align}
We want to approximate the first order derivative, which means we need a,b, and c to eliminate the f(x), f''(x), f'''(x) and solve for f'(x). This gives us the matrix
$$ \begin{pmatrix} & a & b & c \\ f & 1 & 1 & 1 \\ f' & 1 & 0 & -1 \\ f'' & 1 & 0 & 1 \\ f''' & 1 & 0 & -1 \end{pmatrix} \times \begin{pmatrix} a \\ b \\ c\end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} $$
This setup must be wrong as the last line of my matrix has $a-c = 0$, but the second line has $a-c = 1$. Can someone help me with this?
I should add that when I look this up online I find that this does exist with weights $\frac{-11}{6}, \frac{-3}{2}, \frac{1}{3}$ Otherwise, I would write that it doesn't exist.
Thanks.
The professor actually posted solutions and his solution says the answer is no for the reason I noticed myself. We can't have $a-c = 1$ and $a-c = 0$ at the same time. That website must've been referring to a different case that I am not aware of.