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given five points of a function one can approximate the derivate of the function at some point. The standard five point formula is

im not sure how they expect me to expand that polynomial they gave because it would be a total mess!

how would one go about deriving this formula?

my idea is using taylor polynomials

[tex] f'(x_{0}) = \frac{f(x_{0}) - f(x_{0} -h)}{h} + \frac{f''(x_{0}}{2} h - \frac{f^{(3)}(x_{0})}{6} h^2 + \frac{f^{(4)}(x_{0})}{24} h^3 + O(h^4) [/tex]

but im not sure how ot proceed

what to subsitute... i know its gonna take some algebraic gymnastics to get the answer...

**Derive an O(h^4) five point formula to approximate f'(x0) that uses**

[itex] f(x_{0}-h), f(x_{0}), f(x_{0} +h),f(x_{0} +2h),f(x_{0} +3h) [/itex].[itex] f(x_{0}-h), f(x_{0}), f(x_{0} +h),f(x_{0} +2h),f(x_{0} +3h) [/itex].

*(Hint:Consider the expression [itex] Af(x_{0} -h) + Bf(x_{0} +h) + Cf(x_{0} + 2h) + Df(x_{0} + 3h) [/itex]. Expand in fourth Taylor polynomials and choose A,B,C and D appropriately.)*im not sure how they expect me to expand that polynomial they gave because it would be a total mess!

how would one go about deriving this formula?

my idea is using taylor polynomials

[tex] f'(x_{0}) = \frac{f(x_{0}) - f(x_{0} -h)}{h} + \frac{f''(x_{0}}{2} h - \frac{f^{(3)}(x_{0})}{6} h^2 + \frac{f^{(4)}(x_{0})}{24} h^3 + O(h^4) [/tex]

but im not sure how ot proceed

what to subsitute... i know its gonna take some algebraic gymnastics to get the answer...

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