anonymous
  • anonymous
Why is this limit 1?!?!?!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
It looks like it would be 0 to me...
anonymous
  • anonymous
How is that summation computed??

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

jim_thompson5910
  • jim_thompson5910
how many copies of 1 are being added up here?
jim_thompson5910
  • jim_thompson5910
a number line may help |dw:1443247594593:dw|
anonymous
  • anonymous
n-(-n) +1 terms.
anonymous
  • anonymous
So 2n+1 copies of 1.
jim_thompson5910
  • jim_thompson5910
so really 2n+1 terms of 1 being added, yes
anonymous
  • anonymous
Ohh I see it okay how about this one then??
anonymous
  • anonymous
\[\sum_{-\infty}^{\infty} u[k] \]
anonymous
  • anonymous
That's the same as it going from 0 to infinity.
jim_thompson5910
  • jim_thompson5910
how is the u[k] function defined?
anonymous
  • anonymous
It's 1 when greater than 0 and 0 when less than 0.
anonymous
  • anonymous
So we repalce by 1 and... Ohh wait...
anonymous
  • anonymous
..........
jim_thompson5910
  • jim_thompson5910
ok so the unit step function
anonymous
  • anonymous
Thanks.
jim_thompson5910
  • jim_thompson5910
you're welcome. I'm glad to be of help
anonymous
  • anonymous
Wait question.
anonymous
  • anonymous
@jim_thompson5910 , dosen't the 2N+1 just tell us there are 2N+1 terms?
anonymous
  • anonymous
Not that there are 2N+1 copies of 1?
jim_thompson5910
  • jim_thompson5910
we have 2n+1 copies of 1+1+1+....+1
jim_thompson5910
  • jim_thompson5910
as a shortcut 1+1+1+....+1 = (2n+1)*1 = 2n+1
anonymous
  • anonymous
Ooooo I see. Thank you!
jim_thompson5910
  • 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
anonymous
  • anonymous
Interesting. I actually did not know that. Thank you!
jim_thompson5910
  • jim_thompson5910
no problem

Looking for something else?

Not the answer you are looking for? Search for more explanations.