How to develop a numerical differentiation formula using Taylor expansion/series? I have a problem understanding a numerical differentiation problem, using Taylor series. 
It's an exercise and we should develop a 2nd order approximation formula, which uses just $f(x), f(x-h)$ and $f(x+3h)$. As help, the professor gave us a tip in the exercice, that we should start with: 
$$f'(x) = \frac{af(x+3h) + bf(x) + cf(x-h) }h $$
Attached here is the problem AND the solution (which I think is wrong, or he made something wrong, at least I hope), and a legend for my questions. 

I found other results, and here are my 3 questions:


*

*For point 1 (marked 1 on the screenshot), I found $27/6$ as factor there, not $9/2$

*My 2. question (marked 2 in screenshot) concerns the last term in the result. Shouldn't we multiply that big term with $h^3$? 

*3rd question (marked 3 in screenshot): Let's say the professor is right, and there are no typos or calculating errors. If we are dividing by $h$, these $h$'s on the nominator will get eliminated by 1 power. So the term (middle term), that he says should be 1, is multiplied by $h$, and if we divide by $h$ there won't be any $h$ anymore there. For the term after if, which is multiplied by $h^2$, and we are dividing by $h$, it will be then just $h$, which makes our formula of 1st order. Or did I understand something wrong? 


Thank you for any help! Exam is tomorrow, and I am really freaking out about this.  
 A: 1) $\frac {27}{6} = \frac 92$
2) yes there should be an $h^3$ factor multiplying that term.  However, that term ultimately gets dropped (actually becomes the error estimate). And is not relevant for finding optimal values of $a, b, c$
3) yes the $h's$ cancel. That is kind of the point.
$\frac 1h [(a+b+c)f(x) + (3a+c)f'(x)h + (\frac 92 a + \frac 12 c) f''(x)h^2+\epsilon h^3] = f'(x)$ 
Which can only be true if 
$(a+b+c) = 0$ and $(3a+c) = 1$ 
as for $\frac 92 a + \frac 12 c = 0$ This makes your error term decrease at quadratic speed rather than linear speed.
A: More generally,
suppose you want
to find
$f'(x) \approx \dfrac{af(x+uh) + bf(x+vh) + cf(x+wh) }h
$.
$f(x+rh)
=f(x)+rhf'(x)+r^2h^2f''(x)/2+r^3h^3f'''(x)/6+O(h^4)
$
so
$\begin{array}\\
af(x+uh) + bf(x+vh) + cf(x+wh)
&=(a+b+c)f(x)+(au+bv+cw)hf'(x)\\
&\quad +(au^2+bv^2+cw^2)h^2f''(x)/2+(au^3+bv^3+cw^3)h^3f'''(x)/6\\
&\quad+O(h^4)\\
\end{array}
$
Therefore you want
$0=a+b+c,
1=au+bv+cw,
0=au^2+bv^2+cw^2
$
and the error is
$(au^3+bv^3+cw^3)h^2f'''(x)/6
+O(h^3)
$.
This leads to Vandermonde determinants.
Your case is
$u=3, v=0, w=-1$,
so
$0=a+b+c,
1=3a-c,
0=9a+c
$.
Adding the second and third,
$a=1/12$,
so
$c = -9a
=-3/4$.
From the first,
$b = -a-c
=-1/12+3/4
=2/3
$.
The error coefficient is
$au^3+bv^3+cw^3
=27/12-3/4
=15/12
=5/4
$
so the error is
$(5/4)h^2f'''(x)/6+O(h^3)
=(5/24)h^2f'''(x)+O(h^3)
$.
