## anonymous one year ago Why is this limit 1?!?!?!

1. anonymous

2. anonymous

It looks like it would be 0 to me...

3. anonymous

How is that summation computed??

4. jim_thompson5910

how many copies of 1 are being added up here?

5. jim_thompson5910

a number line may help |dw:1443247594593:dw|

6. anonymous

n-(-n) +1 terms.

7. anonymous

So 2n+1 copies of 1.

8. jim_thompson5910

so really 2n+1 terms of 1 being added, yes

9. anonymous

10. anonymous

$\sum_{-\infty}^{\infty} u[k]$

11. anonymous

That's the same as it going from 0 to infinity.

12. jim_thompson5910

how is the u[k] function defined?

13. anonymous

It's 1 when greater than 0 and 0 when less than 0.

14. anonymous

So we repalce by 1 and... Ohh wait...

15. anonymous

..........

16. jim_thompson5910

ok so the unit step function

17. anonymous

Thanks.

18. jim_thompson5910

you're welcome. I'm glad to be of help

19. anonymous

Wait question.

20. anonymous

@jim_thompson5910 , dosen't the 2N+1 just tell us there are 2N+1 terms?

21. anonymous

Not that there are 2N+1 copies of 1?

22. jim_thompson5910

we have 2n+1 copies of 1+1+1+....+1

23. jim_thompson5910

as a shortcut 1+1+1+....+1 = (2n+1)*1 = 2n+1

24. anonymous

Ooooo I see. Thank you!

25. jim_thompson5910

you might have this formula in your book $\Large \sum_{k = m}^{n} c = (n-m+1)*c$ c is a constant there are n-m+1 terms of c being added. So as a shortcut, we really have (n-m+1)*c if you start at m = 0, then you have (n+1)*c if you start at m = 1, then you have n*c which may be what you're used to

26. anonymous

Interesting. I actually did not know that. Thank you!

27. jim_thompson5910

no problem