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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It looks like it would be 0 to me...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How is that summation computed??

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2how many copies of 1 are being added up here?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2a number line may help dw:1443247594593:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 2n+1 copies of 1.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2so really 2n+1 terms of 1 being added, yes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohh I see it okay how about this one then??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{\infty}^{\infty} u[k] \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's the same as it going from 0 to infinity.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2how is the u[k] function defined?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's 1 when greater than 0 and 0 when less than 0.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we repalce by 1 and... Ohh wait...

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2ok so the unit step function

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you're welcome. I'm glad to be of help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 , dosen't the 2N+1 just tell us there are 2N+1 terms?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not that there are 2N+1 copies of 1?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2we have 2n+1 copies of 1+1+1+....+1

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2as a shortcut 1+1+1+....+1 = (2n+1)*1 = 2n+1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ooooo I see. Thank you!

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you might have this formula in your book \[\Large \sum_{k = m}^{n} c = (nm+1)*c\] c is a constant there are nm+1 terms of c being added. So as a shortcut, we really have (nm+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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Interesting. I actually did not know that. Thank you!
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