I am trying to determine the second forward difference $\Delta^2y$ for $y = (x-1)^4$ at $h = 1$.
I came up with the solution $\Delta^2y = 2(6x^2+1)$ both by using $y = (x-1)^4$ directly and by using its expansion $y = x^4 - 4x^3 +6x^2 - 4x +1$.
The textbook gives the answer as $\Delta^2y = 12(x-1)^2$. (I do see that this solution is $\frac{d^2 x}{dy^2}$, which can't be coincidental...)
I did the algebra longhand for $x^4 - 4x^3 +6x^2 - 4x +1$ but used Wolfram Alpha (online) for $(x-1)^4$ and they both agreed on the answer, so I believe it must be my conceptual understanding rather than my algebra which is faulty. Can someone enlighten me?
For the interested, this is problem 9.3.2 of Numerical Methods for Scientists and Engineers, Second Edition by R.W. Hamming. (The problem is on page 153 of the Dover reprint; unfortunately, this section/page is not in Google Books, but a pirated PDF is not too hard to locate elsewhere on the Web.)
The difference formulas given in the textbook are $\Delta y(x) = y(x + h) - y(x)$ and $\Delta^n y(x) = \Delta[\Delta^{n-1}y(x)]$.
All of what you wrote means that $$f''(x) \approx \frac{\Delta_h^2[f](x)}{h^2} = \frac{f(x+2h) - 2 f(x+h) + f(x)}{h^{2}}$$ which is the standard formula for the second order forward difference (see here).
Applied to $f(x)=(x-1)^4$, after development, simplifications and factoring, $${f(x+2h) - 2 f(x+h) + f(x)}=12 h^2 x^2-24h^2\left(1-h\right) x+h^2(12 -24 h+14 h^2)$$ making $$f''(x)\approx 12 x^2-24\left(1-h\right) x+(12 -24 h+14 h^2)=12(x-1)^2+24hx-(24-14h)h$$ For sure, if $h\to 0$, you recover $12(x-1)^2$; but, if this is not the case, say $h=1$, the result is quite different.