Finding an approximate formula for f'(a)

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theBEAST
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Homework Statement


Typeset homework question 5: Given the values f(a), f(a + h), f(a + 3h), find a second order formula with error approximately Ch^2 for f'(a).


The Attempt at a Solution


Handwritten derivation on lined paper: Taylor expansion f(a+h) and f(a+3h) about a, a linear combination for f'(a), and three coefficient equations c0+c1+c2=0, h(c1+c2/3)=1, c1-(9/2)c2=0.


Alright so what I did was expanded f(a+h) and f(a+3h) using Taylor Series Expansion. I then said that f'(a) would be some linear combination of f(a), f(a+h) and f(a+3h).

I summed up and factored out the terms f(a), f'(a) and f''(a) from the Taylor Series Expansions.

Finally for equations 1, I know that I don't want f(a) so c0+c1+c2 = 0. For equation 2, I want to keep f'(a) so that equation is set to equal to 1. For equation 3, I don't want f''(a) so that equation is set to 0.

I solved for the coefficients in terms of h and ended up with f'(a) ≈ [f(a+3h) - 33f(a) + 27f(a+h)] / 20hI would like to know if my method and/or answer is correct. My answer in my opinion looks really weird.
 
You have a good idea, you're just calculating it wrong. The expression for f'(a) is not right. Do it again more carefully.

Another, slightly easier approach would be to write the Taylor expansions for f(a+h) and f(a+3h) up to second order and then using these two, eliminate f''(a) and solve for f'(a).
 
clamtrox said:
You have a good idea, you're just calculating it wrong. The expression for f'(a) is not right. Do it again more carefully.

Another, slightly easier approach would be to write the Taylor expansions for f(a+h) and f(a+3h) up to second order and then using these two, eliminate f''(a) and solve for f'(a).

Thanks! I got the correct answer. Out of curiosity, which term would be the error term? the sum of the f''(a) terms or the sum of the f'''(a) terms?