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anonymous

  • one year ago

Why is this limit 1?!?!?!

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    It looks like it would be 0 to me...

  3. anonymous
    • one year ago
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    How is that summation computed??

  4. jim_thompson5910
    • one year ago
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    how many copies of 1 are being added up here?

  5. jim_thompson5910
    • one year ago
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    a number line may help |dw:1443247594593:dw|

  6. anonymous
    • one year ago
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    n-(-n) +1 terms.

  7. anonymous
    • one year ago
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    So 2n+1 copies of 1.

  8. jim_thompson5910
    • one year ago
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    so really 2n+1 terms of 1 being added, yes

  9. anonymous
    • one year ago
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    Ohh I see it okay how about this one then??

  10. anonymous
    • one year ago
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    \[\sum_{-\infty}^{\infty} u[k] \]

  11. anonymous
    • one year ago
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    That's the same as it going from 0 to infinity.

  12. jim_thompson5910
    • one year ago
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    how is the u[k] function defined?

  13. anonymous
    • one year ago
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    It's 1 when greater than 0 and 0 when less than 0.

  14. anonymous
    • one year ago
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    So we repalce by 1 and... Ohh wait...

  15. anonymous
    • one year ago
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    ..........

  16. jim_thompson5910
    • one year ago
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    ok so the unit step function

  17. anonymous
    • one year ago
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    Thanks.

  18. jim_thompson5910
    • one year ago
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    you're welcome. I'm glad to be of help

  19. anonymous
    • one year ago
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    Wait question.

  20. anonymous
    • one year ago
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    @jim_thompson5910 , dosen't the 2N+1 just tell us there are 2N+1 terms?

  21. anonymous
    • one year ago
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    Not that there are 2N+1 copies of 1?

  22. jim_thompson5910
    • one year ago
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    we have 2n+1 copies of 1+1+1+....+1

  23. jim_thompson5910
    • one year ago
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    as a shortcut 1+1+1+....+1 = (2n+1)*1 = 2n+1

  24. anonymous
    • one year ago
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    Ooooo I see. Thank you!

  25. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    Interesting. I actually did not know that. Thank you!

  27. jim_thompson5910
    • one year ago
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    no problem

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