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| \emph{Proof.} The proof is by induction. However, let us |
\emph{Proof.} The proof is by induction. However, let us |
| first recall that sum on the right side is a |
first recall that sum on the right side is a |
| piece-wise defined function of the upper limit $N-1$. |
piece-wise defined function of the upper limit $N-1$. |
| In other words, if the upper limit is below the lower |
In other words, if the upper limit is below the lower |
| limit $0$, the sum is identically set to zero. |
limit $0$, the sum is identically set to zero. |
| Otherwise, it is an ordinary sum. |
Otherwise, it is an ordinary sum. |
| We therefore need to manually check the first two cases. |
We therefore need to manually check the first two cases. |
| For the trivial case $N=0$, both sides equal to $a_0 b_0$. |
For the trivial case $N=0$, both sides equal to $a_0 b_0$. |
| Also, for $N=1$ (when the sum is a normal sum), it is easy to verify that |
Also, for $N=1$ (when the sum is a normal sum), it is easy to verify that |
| both sides simplify to $a_0 b_0 + a_1 b_1$. |
both sides simplify to $a_0 b_0 + a_1 b_1$. |
| Then, for the induction step, suppose that the |
Then, for the induction step, suppose that the |
|
claim holds for some $N\ge 2$. For $N+1$, we then have
|
claim holds for $N\ge 2$. For $N+1$, we then have
|
| \begin{eqnarray*} |
\begin{eqnarray*} |
| \sum_{i=0}^{N+1} a_i b_i &=& \sum_{i=0}^{N} a_i b_i + a_{N+1} b_{N+1} \\ |
\sum_{i=0}^{N+1} a_i b_i &=& \sum_{i=0}^{N} a_i b_i + a_{N+1} b_{N+1} \\ |
| &=& \sum_{i=0}^{N-1}A_i(b_i-b_{i+1})+A_N b_N + a_{N+1} b_{N+1} \\ |
&=& \sum_{i=0}^{N-1}A_i(b_i-b_{i+1})+A_N b_N + a_{N+1} b_{N+1} \\ |
| &=& \sum_{i=0}^{N}A_i(b_i-b_{i+1})-A_{N}(b_{N}-b_{N+1})+ A_N b_N + a_{N+1} b_{N+1}. |
&=& \sum_{i=0}^{N}A_i(b_i-b_{i+1})-A_{N}(b_{N}-b_{N+1})+ A_N b_N + a_{N+1} b_{N+1}. |
| \end{eqnarray*} |
\end{eqnarray*} |
| Since $-A_{N}(b_{N}-b_{N+1})+ A_N b_N + a_{N+1} b_{N+1} = A_{N+1} b_{N+1}$, |
Since $-A_{N}(b_{N}-b_{N+1})+ A_N b_N + a_{N+1} b_{N+1} = A_{N+1} b_{N+1}$, |
| the claim follows. $\Box$. |
the claim follows. $\Box$. |
