My problem:
I am tasked with showing $$ \Delta^i c_0 = \sum_{k=0}^i {i \choose k}(-1)^{i - k} c_k $$ where the $c_k$'s are control points of a Bézier curve. $\Delta^i c_0$ denotes the $i$th forward difference of $c_0$, that is $\Delta^i c_0 = \Delta ^{i-1} c_1 - \Delta^{i-1}c_0$, with $\Delta^0c_0 = c_0$.
My attempt:
I have tried proceeding by induction, where the base case $i = 1$ yields $\Delta c_0 = c_1 - c_0$ which certainly holds. The problem is that when trying to prove the inductive step, I need information about $\Delta^i c_{1}$, where my inductive assumption only gives me information about $\Delta^i c_0$. I have therefore, instead tried proving a more general claim, namely that $$ \Delta^i c_j = \sum_{k=0}^i {i \choose k}(-1)^{i - k} c_{k+j}. $$
Here, again the base case follows easily, with $i = 1$ it yields $\Delta c_j = c_{j+1} - c_{j}$. Assuming for the sake of induction that the claim holds for all natural numbers up to and including $i$, I wish to show that it also holds for $i + 1$. I have that $$ \Delta^{i+1} c_j = \Delta^ic_{j+1} - \Delta^ic_j = \sum_{k=0}^{i} {i \choose k}(-1)^{i - k} (c_{k+j+1} - c_{k+j}) = \sum_{k=0}^i {i \choose k} (-1)^{i-k}\Delta c_{k + j}. $$ In this last term I can use the induction hypothesis and write $$ \sum_{k=0}^i {i \choose k} (-1)^{i-k}\Delta c_{k + j} = \sum_{k=0}^i \left({i \choose k} (-1)^{i-k}\sum_{h = 0}^{1} {1 \choose h}(-1)^{1 - h}c_{h+k+j}\right). $$
Where i fall short:
At this point, I am stuck - It's been a while since I have manipulated binomial coefficients, and I am not sure how to proceed. My goal is to end up with an expression like $$ \Delta^{i+1} c_j = \sum_{k=0}^{i+1} {i + 1 \choose k}(-1)^{i + 1 - k} c_{k+j}. $$
Any hints in the right direction or an indication whether what I have done so far is sensible would be much appreciated.
Thanks in advance!
I think you should expand your sums in a different way
$\sum_{k=0}^i \binom{i}{k}(-1)^{i-k}(c_{k+j+1}-c_{k+j}) = \sum_{k=1}^{i+1} \binom{i}{k-1}(-1)^{i+1-k}c_{k+j} + \sum_{k=0}^{i} \binom{i}{k}(-1)^{i+1-k}c_{k+j} \\= c_{i+j+1} + \sum_{k=1}^i (\binom{i}{k-1}+\binom{i}{k})(-1)^{i+1-k}c_{k+j} + (-1)^{i+1}c_j \\= c_{i+j+1} + \sum_{k=1}^i \binom{i+1}{k}(-1)^{i+1-k}c_{k+j} + (-1)^{i+1}c_j \\ = \sum_{k=0}^{i+1} \binom{i+1}{k}(-1)^{i+1-k}c_{k+j}$