Explained: General Finite Difference Stencil (example) [cfd] 〈DIRECT • 2027〉

-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:

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]

Substitute these expansions into the general summation formula. To ensure the approximation equals the -th derivative (and cancels out all other lower

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 Derivation Step-by-Step To find the coefficients , we

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

(The sum of weights for the function value itself must be zero) (Weight for the 1st derivative) (The weight for the -th derivative must be 1) (Higher order terms cancelled for accuracy) This is often represented as a problem: is the vector of unknown weights. 3. Example: Second-Order Forward Difference for Suppose we want to find using three points: