- #1
stunner5000pt
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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
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].
(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 I am not sure how ot proceed
what to subsitute... i know its going to 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].
(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 I am not sure how ot proceed
what to subsitute... i know its going to take some algebraic gymnastics to get the answer...
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