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anonymous

  • 5 years ago

amistre, need you for about a minute

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  1. anonymous
    • 5 years ago
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    homogeneous vs. non-homogeneous recap

  2. amistre64
    • 5 years ago
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    ack!!... lol what do we need to recap :)

  3. anonymous
    • 5 years ago
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    a was not homogeneouss

  4. amistre64
    • 5 years ago
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    recursion equations.....

  5. anonymous
    • 5 years ago
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    On worksheet 12A, part a) is non-homogeneous. I've pasted in below the definition of the terminology for homogeneous. I think that what you are missing is that they are implying, but not directly saying, that you put you x terms on the left, and your non-x terms on the right. So the equation in a) should be analyzed as xn+1 +xn-1 = - n^2, or more directly matching the expression highlighted in yellow below, we would write: xn+1 - (-1)xn-1 = -n^2 Clearly the non-x term, -n^2 is not worth zero.

  6. amistre64
    • 5 years ago
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    ok... 1a or 2a? 2a was definanlty not homogenous...

  7. anonymous
    • 5 years ago
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  8. anonymous
    • 5 years ago
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    number 1 had 4 parts

  9. amistre64
    • 5 years ago
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    the only thing that is squared to get zero is zero.... right?

  10. amistre64
    • 5 years ago
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    and I agree that -n^2 would definantly not be 0...all the time ;)

  11. anonymous
    • 5 years ago
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    yes

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  12. amistre64
    • 5 years ago
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    X{n+1} + n^2 + X{n-1} = 0 that n^2 makes me think its not linear; or do we only look at the X{..} parts for that?

  13. anonymous
    • 5 years ago
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    only have c as linear

  14. anonymous
    • 5 years ago
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    so would b also be nonhom

  15. amistre64
    • 5 years ago
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    well, then X{...} parts all come to one side and we are left with 0 over there for b0 right?

  16. anonymous
    • 5 years ago
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    yes

  17. amistre64
    • 5 years ago
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    X{n+1} - (pi)X{n} " (2^2)X{n} = 0 right?

  18. amistre64
    • 5 years ago
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    is that spoosed to be 2 [X{n}]^2?

  19. anonymous
    • 5 years ago
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    that was original rationale but when I emailed him that I got confused by his response

  20. anonymous
    • 5 years ago
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    yes

  21. amistre64
    • 5 years ago
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    what was his respone?

  22. anonymous
    • 5 years ago
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    No, you can't get the right side to zero by transforming it, causing all >>of >>the constants to go to the left side.

  23. amistre64
    • 5 years ago
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    lol...isnt that what the book does? the material says to put it all to the other side and equate it to zero :)

  24. amistre64
    • 5 years ago
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    what aint attached to an X{...} stays put

  25. anonymous
    • 5 years ago
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    so b is homo

  26. amistre64
    • 5 years ago
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    id say yes, but your material is not easy to parse :)

  27. amistre64
    • 5 years ago
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    you need to find help with recurrsion equations; and see if anyone can aid you in that way :) calling them difference equations is like speaking a foriegn language..

  28. anonymous
    • 5 years ago
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    c is nonhomo d???

  29. anonymous
    • 5 years ago
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    I've bee reading about them all weekend and the name does throw people off...

  30. anonymous
    • 5 years ago
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    d homo?

  31. amistre64
    • 5 years ago
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    i would say that d is left with a 1/2 on one side..... im guesing nonhomo

  32. anonymous
    • 5 years ago
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    ok bc cand divide to move yn and have 1/2 not 0 on the right side

  33. amistre64
    • 5 years ago
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    with b you should subtract everything over there becasue they involve X{....} leaving you with 0

  34. anonymous
    • 5 years ago
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    yes so I have b only homo

  35. amistre64
    • 5 years ago
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    as far as I can tell; yes :)

  36. anonymous
    • 5 years ago
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    ok thanks!

  37. amistre64
    • 5 years ago
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    wish i could be more confident on those; but they seem to elude me ;)

  38. anonymous
    • 5 years ago
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    no problem,,,thanks sooooo much for all of your help!!!!!!

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