Prove that ( ∑ j = 0 N a j ) ( ∑ j = 0 N b j ) = ∑ j = 0 2 N c j

Question

Prove that       (          ∑              j        =        0                    N                    a              j              )        (          ∑              j        =        0                    N                    b              j              )    =      ∑          j      =      0              2      N            c          j

Jake Mcpherson · Accepted Answer

From the definition of the Cauchy product, we get

\sum_{k = 0}^{2 N} c_{k} = \sum_{k = 0}^{2 N} \sum_{i = 0}^{k} a_{i} b_{k - i} = \sum_{k = 0}^{2 N} \sum_{i + j = k} a_{i} b_{j} = \sum_{i + j \leq 2 N} a_{i} b_{j} .

However, by using the fact that

a_{i} = 0

and

b_{j} = 0

for

i, j > N

, we can freely introduce the extra conditions

i \leq N

and

j \leq N

to the last sum. So

= \sum_{\begin{matrix} i, j \leq N \\ i + j \leq 2 N \end{matrix}} a_{i} b_{j} = \sum_{i, j \leq N} a_{i} b_{j} = (\sum_{i = 0}^{N} a_{i}) (\sum_{j = 0}^{N} b_{j}) .

In the second step, the condition

i + j \leq 2 N

is removed because it is implied by

i, j \leq N

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