Why is this limit 1?!?!?!

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- anonymous

Why is this limit 1?!?!?!

- schrodinger

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- anonymous

##### 1 Attachment

- anonymous

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

- anonymous

How is that summation computed??

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## More answers

- jim_thompson5910

how many copies of 1 are being added up here?

- jim_thompson5910

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

- anonymous

n-(-n) +1 terms.

- anonymous

So 2n+1 copies of 1.

- jim_thompson5910

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

- anonymous

Ohh I see it okay how about this one then??

- anonymous

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

- anonymous

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

- jim_thompson5910

how is the u[k] function defined?

- anonymous

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

- anonymous

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

- anonymous

..........

- jim_thompson5910

ok so the unit step function

- anonymous

Thanks.

- jim_thompson5910

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

- anonymous

Wait question.

- anonymous

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

- anonymous

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

- jim_thompson5910

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

- jim_thompson5910

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

- anonymous

Ooooo I see. Thank you!

- 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

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

- jim_thompson5910

no problem

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