3
$\begingroup$

I wanted to ask if I shifted this sum correctly. I basically substitute $k$ with $k+N$. I often make mistakes when shifting the indexes.

$$ \sum_{k=N}^{2N}(\sum_{j=0}^{k}a_{k-j}b_{j})x^k = \sum_{k=0}^{N}\sum_{j=0}^{k+N}(a_{k-j+N}b_{j})x^{k+N} $$

Any help would be great! Thanks!

$\endgroup$
1
  • 1
    $\begingroup$ My strategy with sums expressed with $\Sigma$'s is to write them out with the first few and last terms with an ellipsis between, perhaps for small values of the limits, and see what's going on. I haven't checked your answer. $\endgroup$ Oct 22, 2016 at 13:25

2 Answers 2

2
$\begingroup$

This look fine.

You did the right think, and you thought about switching all the $k$ by $k+N$ so no mistake here.

$\endgroup$
1
$\begingroup$

This looks correct to me. One way that I often check whether I have shifted things correctly is by finding the first and last terms in a series/sum.

For your sum, the first power of $x$ is $x^{N}$ on both sides, so that looks good.

In front of $x^{n}$ you have the series $$ \sum_{j=0}^k a_{k-j}b_ j = a_Nb_0 + \dots a_0b_N $$ and on the other side (for $x^N$) $$ \sum_{j=0}^{k+N} a_{k-j+N}b_{j} = a_{N-j+N}b_0 + \dots a_0b_{0 + N} $$

In general your strategy by just replacing the $k$ by $k+N$ works fine when you adjust the index on the sum as well.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .