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My questions:

  1. For the inner summation, When $j=i$ does that mean $j$ will equal $2$ because $i$ equals $2$?

  2. I'm confused and stuck on the yellow, because I thought it would just be $n$. NOT ($(n-i+n)$. I understandstand Line 1 of the problem somewhat. It's just the index of the inner j=i is throwing me off.

Yes
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Kapooky Handy
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  • Could you type out specifically what you're talking about in the question? It's a little harder to parse flipping tabs. – Alekos Robotis Nov 11 '15 at 05:41
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    $$\sum_{i=a}^{b}\sum_{j=i}^{b} f(i) = \left(\sum_{j=a}^b f(j)\right)+\left(\sum_{j=a+1}^b f(j)\right)+\ldots+\left(\sum_{j=b-1}^b f(j)\right)+\left(\sum_{j=b}^b f(j)\right)$$

    Double sum is just a shorthand notation to express this. The rule here is to work from inner sums towards the outer sums, while keeping the outer sum variables constant. Although, there are a few cases when this doesn't apply.

     – Prasun Biswas Nov 11 '15 at 05:49

2 Answers2

5

The meaning of summation notation $ \Sigma $ follows as: $$ \sum^{n}_{k=i}(\text{formula of }k) = \text{Let's sum a formula of }k\text{ when }k=i, i+1, i+2 \ldots n.$$

so for your question 1, j=i does not mean j=2, even if it is placed right after i=2.

Actually, $$ \sum^{n}_{k=i}(\text{formula of }k)$$ can be written equivalently as(but don't use this notation if your lecture giver insists) $$ \sum_{k \in \{i, i+1, i+2, ... n\}}(\text{formula of }k)$$ so an equal sign does not have the meaning you think.

So, for your question 2, $$ \sum^{n}_{j=i}1 = \sum_{j \in \{i, i+1, i+2 ... n\}} 1 \\ = \text{sum of }(n - i + 1) \text{ } 1\text{'s} \\ = n - i + 1.$$

Hope my answer helps.

Jeong Jinmyeong
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2

For each $2 \leq i \leq n$, we have $$ \sum_{j=i}^{n}3i = \underset{(n-i+1\ \text{times})}{3i + \cdots + 3i} = (n-i+1)3i. $$

For each $i \geq 1$ the number of the elements of the set $\{ i, i+1, \dots, n \}$ is simply $n-i+1$, as you can try to prove; for instance, if $i=2$ and if $n = 4$, then the set $\{ i, i+1, \dots, n\} = \{ 2, 3, 4 \}$ has $3 = n-i+1$ elements.

Yes
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  • (n-i )will give me the total amount of cycles that inner summation with look through.. why is there a +1? – Kapooky Handy Nov 11 '15 at 05:59
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    You’re making the classic “fencepost error.” If $j$ ranges from $i$ to $n$ inclusive it will take on $n-j+1$ values, not $n-j$. Try $n=1$ and $j=0$, for example. – amd Nov 11 '15 at 06:20