OpenStudy (anonymous):

Undetermined Coefficients The question will follow...

3 years ago
OpenStudy (anonymous):

\[\large \text{Problem:}\] Derive the two point formula \[f'(x)\approx \frac{f(x+h)-f(x-h)}{2h}\] using the method of UC. \[\large \text{Solution: }\] \[f'(0)=\frac{1}{2h}[f(h)-f(-h)]\] Assume we want a solution of the form \[f'(0)\approx Af(h)+Bf(-h)+Cf(0)\] Choose base functions {1,x,x^2} Then follows:\[f(x)=1: \text{ }0=A+B+C \\f(x)=x: \text{ }1=(h)A+(-h)B+(0)C \\f(x)=x^2: \text{ }0=(h)^2A+(-h)^2B+(0)C\] Solving simultaneously, A=h/2, B=-h/2, C=0, giving the required equation. \[\large \large \text{Here comes my questions:}\] 1. Why do we choose x=0 at the beginning? 2. Where does the 1 come frome in the second substitution?

3 years ago
OpenStudy (anonymous):

@.Sam. maybe you can shed some light on this?

3 years ago
OpenStudy (valpey):

1.) Because for most whatever f is, the compare/contrast of f(h) with f(-h) will be illustrative. 2.) because if f(x) = x then f'(x) = 1 (no matter what x is. In part 3, f'(x) = 2x, but f'(0) is 2*0 = 0

3 years ago
OpenStudy (valpey):

And shouldn't it be A=1/2h, B = -1/2h, C=0?

3 years ago