Summation Question (Properties)

[dogma]

I have a rather simple question, but my rusty brain needs a good, swift kick-start.

I start with:

[tex]\sum_{i=1}^k i[/tex]

and substitute in [tex]i=k-j[/tex] to get:

[tex]\sum_{k-j=1}^k (k-j)[/tex]

How do I get from this to the following?

[tex]\sum_{k-j=1}^k (k-j) \rightarrow \sum_{j=0}^{k-1} (k-j)[/tex]

Thanks in advance for your help.

dogma

 

[arildno]

I have a rather simple question, but my rusty brain needs a good, swift kick-start.

I start with:

[tex]\sum_{i=1}^k i[/tex]

and substitute in [tex]i=k-j[/tex] to get:

[tex]\sum_{k-j=1}^k (k-j)[/tex]

How do I get from this to the following?

[tex]\sum_{k-j=1}^k (k-j) \rightarrow \sum_{j=0}^{k-1} (k-j)[/tex]

Thanks in advance for your help.

dogma

You're in some confusion in how to interpret the summation limits.
Let's write it explicitly, to see how it follows:
[tex]\sum_{i\geq{1}}^{i\leq{k}}i=\sum_{(k-j)\geq{1}}^{(k-j)\leq{k}}(k-j)[/tex]

Now, rearrange the inequalities in the last expression:
[tex]\sum_{(k-j)\geq{1}}^{(k-j)\leq{k}}(k-j)=\sum_{(k-1)\geq{j}}^{0\leq{j}}(k-j)[/tex]
Which in standard notation is nothing else than:
[tex]\sum_{j\geq{0}}^{j\leq(k-1)}(k-j)=\sum_{j=0}^{k-1}(k-j)[/tex]

 

[dogma]

Thank you!

I completely understand now. I just need a good, swift kick.

Thanks again and take care!

dogma