Question

Why is this limit 1?!?!?!

Answer 1

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

Answer 2

How is that summation computed??

Answer 3

how many copies of 1 are being added up here?

Answer 4

a number line may help

Answer 5

n-(-n) +1 terms.

Answer 6

So 2n+1 copies of 1.

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

how is the u[k] function defined?

Answer 12

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

Answer 13

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

Answer 14

..........

Answer 15

ok so the unit step function

Answer 16

Thanks.

Answer 17

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

Answer 18

Wait question.

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

as a shortcut

1+1+1+....+1 = (2n+1)*1 = 2n+1

Answer 23

Ooooo I see. Thank you!

Answer 24

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

Answer 25

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

Answer 26

no problem