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anonymous

  • 5 years ago

Hi Amistre...any luck

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  1. amistre64
    • 5 years ago
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    some... how about you? shall we dance :)

  2. anonymous
    • 5 years ago
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    SURE!...I've been just reading over it all to try and make sense of it...

  3. amistre64
    • 5 years ago
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    difference equations go by a more common name of recursion formulas

  4. anonymous
    • 5 years ago
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    yes!

  5. amistre64
    • 5 years ago
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    go ahead and post your "problem" page and lets see what we can do with it now :)

  6. anonymous
    • 5 years ago
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  7. amistre64
    • 5 years ago
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    I you tubed a few recurrsions and have a better idea of what they are trying to say..... that material of yours is simply atrocious to parse information out of :)

  8. amistre64
    • 5 years ago
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    so we are on 2a right?

  9. anonymous
    • 5 years ago
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    I've been googling and reading for the past couple of days but nothing seems to fully explain it or provide examples with solutions ; ) to guide me

  10. amistre64
    • 5 years ago
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    We have 2 parts to every recurrsion formula; a start and a function 1) X{0} = 0 2) X{n+1}= 1/2 X{n} -1

  11. anonymous
    • 5 years ago
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    ok

  12. amistre64
    • 5 years ago
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    These are also refered to as NEXT = NOW + C equations..... and they can be written as: X{n} = k X{n-1} + C as well; they both mean the same thing right?

  13. anonymous
    • 5 years ago
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    yes

  14. amistre64
    • 5 years ago
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    So lets go ahead and list the first 6 terms; X{0} = 0 X{1} = -1 X{2} = -1' 1/2 X{3} = -1' 3/4 X{4} = -1' 7/8 X{5} = -1' 15/16 X{6} = -1' 31/32

  15. amistre64
    • 5 years ago
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    that does 2a, now for 2b

  16. anonymous
    • 5 years ago
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    so I should start at 0 instead of -1

  17. anonymous
    • 5 years ago
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    I had 0 originally

  18. anonymous
    • 5 years ago
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    but you get -1 when you plug in 0 for x sub n

  19. amistre64
    • 5 years ago
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    X{n+1} = (1/2) X{n} is the homo part;

  20. amistre64
    • 5 years ago
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    the first term is 0, its our starting term and is the first term :)

  21. anonymous
    • 5 years ago
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    ok just kept going back and forth with the table

  22. amistre64
    • 5 years ago
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    We need to come up with a solution that will make the homo part "fit" the data.

  23. amistre64
    • 5 years ago
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    -1 = (1/2)(0) + C -1' 1/2 = (1/2)(-1) + C is what I believe it is getting at

  24. anonymous
    • 5 years ago
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    true...it is like trial and error

  25. anonymous
    • 5 years ago
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    -2

  26. anonymous
    • 5 years ago
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    which looks like the limit?

  27. anonymous
    • 5 years ago
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    oops missed negative so -1

  28. amistre64
    • 5 years ago
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    yeah, thats what it was to begin with, which throws me off, why do this part if its just the same as the last ;)

  29. anonymous
    • 5 years ago
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    suppose to just drop constant to make it 0--go figure

  30. anonymous
    • 5 years ago
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    then have to combine general and particular for complete solution

  31. anonymous
    • 5 years ago
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    http://media.pearsoncmg.com/aw/aw_gordon_functioning_2/Gordon_12_01.pdf

  32. amistre64
    • 5 years ago
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    yeah :) Now to solve for any value of "n" we do this: X{n} = X{0} + (n-1)d X{0} = 0 so we are left with (n-1)d = X{n} right?

  33. anonymous
    • 5 years ago
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    correct!

  34. amistre64
    • 5 years ago
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    So from our table above we can input the values and find a solution for "d" (n-1)d = X{n} keep in mind that this can be written as: (n)d = X{n+1}

  35. anonymous
    • 5 years ago
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    so is trial and error pretty much the method

  36. anonymous
    • 5 years ago
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    using the patterns as a guide

  37. amistre64
    • 5 years ago
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    (n)d = X{3} ; 2+1 = 3 (2)d = -1.75 d = -1.75/2 = .875 :: 7/8

  38. amistre64
    • 5 years ago
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    im not sure about the trial and error method; if you are given an equationand asked to find the constant, then there are ways to determine its value. The trail and error might apply but it is reduced considerably

  39. amistre64
    • 5 years ago
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    -7/8 , i dropped the sign ;)

  40. anonymous
    • 5 years ago
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    i caught it

  41. amistre64
    • 5 years ago
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    X{n} = (-7/8)(n-1) does this fit?

  42. anonymous
    • 5 years ago
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    checking on calc

  43. amistre64
    • 5 years ago
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    for n=1

  44. amistre64
    • 5 years ago
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    all sequences start at n=1

  45. anonymous
    • 5 years ago
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    checked it with 3...WORKS!

  46. anonymous
    • 5 years ago
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    of course it works for 1 was checking another ; )

  47. amistre64
    • 5 years ago
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    check for X{6} :)

  48. anonymous
    • 5 years ago
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    not getting answer

  49. amistre64
    • 5 years ago
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    -4.375 is what you get right?

  50. anonymous
    • 5 years ago
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    yes!

  51. amistre64
    • 5 years ago
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    and it shouldnt be less than -2....

  52. amistre64
    • 5 years ago
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    -2 + (1/2)^n try that :)

  53. anonymous
    • 5 years ago
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    -1.984375

  54. anonymous
    • 5 years ago
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    works for previous value

  55. anonymous
    • 5 years ago
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    so raised to n-1?

  56. amistre64
    • 5 years ago
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    If we want to know X{n}; we can use -2+(.5)^n X{0} = 0 -2 +(1/2)^0 = -1..... -2 + (1/2)^(0-1) = -2 + 2 = 0

  57. amistre64
    • 5 years ago
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    yes ^(n-1) looks good :)

  58. amistre64
    • 5 years ago
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    X{5} = -1' 15/16 = -2 + (1/16) = -2 + (1/2)^(n-1)

  59. amistre64
    • 5 years ago
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    How we work this backwards tho..... to see how we got to it :)

  60. amistre64
    • 5 years ago
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    -2 +(1/2)^(n-1) = X{n} = 2(X{n+1} -1)

  61. anonymous
    • 5 years ago
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    slowly but surely...can't say I'm comfortable with doing alone which is my goal

  62. amistre64
    • 5 years ago
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    -2 +(1/2)^(n-1) = X{n} = 2(X{n+1} +1) <- signs again lol

  63. amistre64
    • 5 years ago
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    -2 +(1/2)^(n-1) = 2(X{n+1} +1)

  64. amistre64
    • 5 years ago
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    -2 +(1/2)^(n-1) = 2X{n+1} +2 (1/2)^(n-1) = 2[X{n+1}] +4

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

  66. anonymous
    • 5 years ago
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    yes

  67. amistre64
    • 5 years ago
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    X{0} + X{n} * (1/2) (1/2)^(n-1)......is what we need to be getiing at I think

  68. amistre64
    • 5 years ago
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    X{n+1} - X{n} (1/2) [(1/2)^(n-1)] = 0 Does that work out?

  69. anonymous
    • 5 years ago
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    got -1.9375 but lost track let me try again

  70. anonymous
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    works for n=0 working on others

  71. anonymous
    • 5 years ago
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    get -2 for n=1

  72. amistre64
    • 5 years ago
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    (1/2)^(n-1) split into (1/2)^n ------- = 2[(1/2)^n] (1/2)

  73. amistre64
    • 5 years ago
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    X{n+1} - X{n} 2 [(1/2)^n] = 0

  74. anonymous
    • 5 years ago
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    I'm taking notes and trying to make sense of it and thinking about the final project I have to create...smh, lol

  75. amistre64
    • 5 years ago
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    :)

  76. anonymous
    • 5 years ago
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    all for 1 credit, lol

  77. anonymous
    • 5 years ago
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    that was b correct

  78. amistre64
    • 5 years ago
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    lol.....its so hard to tell these days; but yeah, i think that is what b is going to end up looking like; it is works out.

  79. amistre64
    • 5 years ago
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    x{n+1} = (1/2)x{n} * (D)

  80. anonymous
    • 5 years ago
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    what i had originally, lol

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  81. anonymous
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    just noticed it is blurry

  82. amistre64
    • 5 years ago
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    trying to go that route just messes me up lol....do youhave to follow a certain step by step from the book?

  83. amistre64
    • 5 years ago
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    -1/0 does not equal 0.....

  84. anonymous
    • 5 years ago
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    at this point...no just want to make sense of it however I can...more than one way to skin a cat, right?

  85. anonymous
    • 5 years ago
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    that is undefined

  86. amistre64
    • 5 years ago
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    -1 + 1 = 0; whicn means -(1/2)X{0} = 1

  87. amistre64
    • 5 years ago
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    .5 X{n} = X{n+1} ..... but that is nowhere near the answers we get in the table; do we just punch thru thill we get a parttern for 1/2^(n-1)?

  88. amistre64
    • 5 years ago
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    wait, I think I see it; X{n+1} = (1/2) * X{n} only works if we do: X{n+1} = (1/2) * X{n} - 1.....when we combine the 2 we get X{n+1} = (1/2)D -2 ;X{n} becomes the variable.... and -1 -1 = -2

  89. amistre64
    • 5 years ago
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    (1/2)D needs to equal (1/2)^(n-1)

  90. anonymous
    • 5 years ago
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    d=2

  91. amistre64
    • 5 years ago
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    X{n} = (1/2)D - 2; D = X{n-1}

  92. amistre64
    • 5 years ago
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    the general solution gives us a "family" of curves to choose from right?

  93. anonymous
    • 5 years ago
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    yes

  94. anonymous
    • 5 years ago
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    one for each initial value

  95. amistre64
    • 5 years ago
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    2[(1/2)^n] is our general soluton then right? maybe?

  96. anonymous
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    looks good

  97. amistre64
    • 5 years ago
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    we know we need it; and then we pinpoint it with -2 as a constant right?

  98. anonymous
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    yeah

  99. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=plot+%282+*+%281%2F2%29%5Ex%29+-+2+from+x%3D0+to+20

  100. amistre64
    • 5 years ago
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    this is the "family" of curves.... the general solution

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  101. anonymous
    • 5 years ago
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    nice!

  102. amistre64
    • 5 years ago
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    if we look at it head on its like this.... but how to get to the family lol

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  103. amistre64
    • 5 years ago
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    So: X{x+1} = (1/2)X{n} = 2[(1/2)^(n-1)]

  104. amistre64
    • 5 years ago
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    + a constant...

  105. amistre64
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    (1/2)X{n} = 2[(1/2)^(n-1)] + C .....should we bother to figure out the inbetweens :)

  106. anonymous
    • 5 years ago
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    no is that b

  107. amistre64
    • 5 years ago
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    B should be 2 * (1/2)^(n-1) + C as a general solution; but how to get to it the way they suggest is beyond me :)

  108. anonymous
    • 5 years ago
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    I know this works so no prob

  109. amistre64
    • 5 years ago
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    the general solution will be: 2*(1/2)^n

  110. amistre64
    • 5 years ago
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    3c) Find a particular solution P{n} of the difference equation X{n+1}=(1/2) X{n} -1 ; X0=0 from Part a. Show your work....

  111. amistre64
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    http://www.math.kth.se/math/student/courses/5B1203/F/200304/linrek.pdf might be helpful

  112. amistre64
    • 5 years ago
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    In order to solve such a recursion, we need only solve the corresponding homogeneous recursion and then find one particular solution, say apart n , to the inhomogeneous recursion. Then any solution can be written as: an = a{n}hom + a{n}part ;

  113. anonymous
    • 5 years ago
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    are you going to be on later...have to fix lunch for daughter and work on final presentation for another course...

  114. amistre64
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    I got 5 hours left online today....

  115. anonymous
    • 5 years ago
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    ok I will study tonight and hope to see you tomorrow...thanks! I'll email you

  116. amistre64
    • 5 years ago
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    After more reading; the general solution is: C * (1/2)^n that starting to look like the material your studying off of....

  117. anonymous
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    I had that...just didn't know how, lol

  118. anonymous
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    now I have a better understanding

  119. amistre64
    • 5 years ago
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    the rest of it there is still confusing me; I keep reading over the material, but it is quite frankly abismal..

  120. anonymous
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    I know thanks ...

  121. anonymous
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    must get get 2-4 some how

  122. amistre64
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    X{n+1-1} - (1/2) C(1/2)^(n-1) = -1 X{n} - C (1/2)(1/2)^(n-1)= -1 C(1/2)^n - C(1/2)(1/2)^(n-1) = -1 C(1/2)^n [1 - (1/2) (1/2)^-1] = -1 C(1/2)^n = -1/(1-1) = -1/0 ..... thats just upsetting lol C(1/2)^n

  123. amistre64
    • 5 years ago
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    the answer for 2c is: -2, but how to get it .....ack!!

  124. amistre64
    • 5 years ago
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    the answer for 2d is: X{n+1} = general homo + particular X{n+1} = (1/2)^(n) + (-2) but the instructions are not clear on how to get the particular solution to the nonhomo equation.... cause I end up with -1/0 which is absurd ;)

  125. amistre64
    • 5 years ago
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    the particular solution is simply a multiple of the nonhomo term: A(-1) ; where A = 2

  126. anonymous
    • 5 years ago
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    professor said to use trial solutions and figure out the constant---hardest part

  127. amistre64
    • 5 years ago
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    X{n+1} - (1/2) X{n} = A*(-1) is what the book say to do wo far... it then say s to do this: X{n} = A*(-1) but we have no "n" nalue over there; so it suggests elsewhere to multiply both sides by "n" first... i think lol

  128. amistre64
    • 5 years ago
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    n * X(n+1) = An *(-1) (n-1) * X{n} = A(n-1) *(-1)

  129. amistre64
    • 5 years ago
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    X{n} = A *(-1) then...... now what :)

  130. amistre64
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    or rather: (n+1) * X({n+1} = A(n+1)(-1)

  131. amistre64
    • 5 years ago
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    X{n+1} = A * (-1)...... this feels redundant lol X{n+1} - (1/2)A(-1) = -1... A(-1) - -A/2 = -1 A(-1 +1/2) = -1 A(-1/2) = -1 A = 2....

  132. amistre64
    • 5 years ago
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    that did it ;)

  133. amistre64
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    X{n+1} = (1/2)^(n) + 2(-1) X{n} = (1/2)^(n-1) -2 is what we got to begin with without going thru this headache lol

  134. anonymous
    • 5 years ago
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    WOw, yep

  135. amistre64
    • 5 years ago
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    took 3 months to get to the answer, an answer that we got the first time lol......

  136. anonymous
    • 5 years ago
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    Scary

  137. amistre64
    • 5 years ago
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    So for the next one; we can see if this setup works the same.. X{n+1} = (1/2) X{n} + n ; X{0} = 0 1/2(0) + 0 ; X{0}=0 1/2(0)+1 ; X{1}=1 1/2(1)+2 ; X{2}=2' 1/2 1/2(2.5)+3 ; X{3}= 4' 1/4 1/2(4.25)+4 ; X{4}= 6' 1/8 1/2(6.125)+5 ; X{5}= 8' 1/16

  138. amistre64
    • 5 years ago
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    the homo part is: X{n+1} = (1/2)*X{n} X{n} = C * (1/2)^n ....right?

  139. amistre64
    • 5 years ago
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    teh solution we are getting at tho is: 2n + (1/2)^(n-1) right?

  140. anonymous
    • 5 years ago
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    is this 3...my table is wrong

  141. amistre64
    • 5 years ago
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    X{n+1} = (1/2)^(n) + An.... X{n} = A *n X{n+1} = A (n+1); use this in our equation: A(n+1) - (1/2)(1/2)^(n) = n

  142. amistre64
    • 5 years ago
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    yeah, working on 3 now..

  143. amistre64
    • 5 years ago
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    A(n+1) - (1/2) A(n) = n An + A -An/2 = n 2An +2a -An ----------- = n 2

  144. amistre64
    • 5 years ago
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    2An -An +2A = 2n An + 2A = 2n A(n + 2) = 2n A = 2n/(n+2)

  145. amistre64
    • 5 years ago
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    If we stop at: An + 2A = 2n ; when=1 we get: A +2A = 2 3A = 2 A = 2/3 ...... which is not what I like lol

  146. amistre64
    • 5 years ago
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    or maybe it is :) let see..... (1/2)^(n-1) + 2n/3 ..... nah, that aint it

  147. amistre64
    • 5 years ago
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    when n = 0 tho we need n = -2 if we go this route... X{0} = (1/2)^-1 + n = 0 2 + n = 0; n=-2; so A(0) = -2? X{3} = (1/2)^2 + n = 4' 1/4 1/4 + n = 4' 1/4; A(3)= 4 A=4/3.... ugh.....

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