Explained: General — Finite Difference Stencil (example) [cfd]

dkfdxk|x0≈∑i=1ncif(xi)d to the k-th power f over d x to the k-th power end-fraction evaluated at x sub 0 end-evaluation is approximately equal to sum from i equals 1 to n of c sub i f of open paren x sub i close paren are the or coefficients of the stencil. 2. Derivation Step-by-Step To find the coefficients

f(xi)=f(x0)+(xi−x0)f′(x0)+(xi−x0)22!f′′(x0)+…+(xi−x0)n−1(n−1)!f(n−1)(x0)f of open paren x sub i close paren equals f of open paren x sub 0 close paren plus open paren x sub i minus x sub 0 close paren f prime of open paren x sub 0 close paren plus the fraction with numerator open paren x sub i minus x sub 0 close paren squared and denominator 2 exclamation mark end-fraction f double prime of open paren x sub 0 close paren plus … plus the fraction with numerator open paren x sub i minus x sub 0 close paren raised to the n minus 1 power and denominator open paren n minus 1 close paren exclamation mark end-fraction f raised to the open paren n minus 1 close paren power of open paren x sub 0 close paren Explained: General Finite Difference Stencil (Example) [CFD]

-th derivative (and cancels out all other lower and higher-order derivatives up to the desired accuracy), the coefficients must satisfy a system of linear equations: dkfdxk|x0≈∑i=1ncif(xi)d to the k-th power f over d

, we use the based on Taylor series expansions. A. Expand using Taylor Series For each point in your stencil, expand around the target point expand around the target point