Arithmetic Sequence fail

Discussion in 'Space Junk' started by Higgs Boson, Dec 6, 2009.

Arithmetic Sequence fail

Discussion in 'Space Junk' started by Higgs Boson, Dec 6, 2009.

  1. Higgs Boson

    Higgs Boson New Member

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    Ok either the book is wrong or I just found out that I fail at basic arithmetic sequences...

    [​IMG]

    The book says -1845. I say 'huh'?
     
  2. LordKerwyn

    LordKerwyn New Member

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    The books wrong, what was your answer?
     
  3. Higgs Boson

    Higgs Boson New Member

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    I left my notes in school, hold on I'll re-calculate.
     
  4. LordKerwyn

    LordKerwyn New Member

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    When your done here is my answer:
    80
    Assuming we agree I would say it makes a better case for the book being wrong.
     
  5. Higgs Boson

    Higgs Boson New Member

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    Well I assume you are correct since you got higher mathematic education than me however my answer does not match yours:
    S10=5(2+36) . S10=5(18-27) ==> 190x-45 = -8550

    I want my money back for basic education.
     
  6. LordKerwyn

    LordKerwyn New Member

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    Don't assume I am right just yet, as I take another look at everything something seems wrong.
     
  7. LordKerwyn

    LordKerwyn New Member

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    The book is right I fail at basic multiplication. *LK glares at himself -3*4 is -12 not -7...*
     
  8. Higgs Boson

    Higgs Boson New Member

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    Is it? Do you care to show me your working? Since everyone I asked seems to fail at everything when it comes to solving this question so I would love to put an end to this madness.
     
  9. LordKerwyn

    LordKerwyn New Member

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    How are you solving the question by hand or with a calculator? The way I would do it is put sum((-3 4i)(12-3i),i,1,10) into a calculator and let it plug and chug (which produces the book's answer). I actually multiplied it out and created an error which I caught because my answer just seemed to small given how negative the function is suppose to get. If your doing it by hand than good luck. The way that I think would create the least number of errors is write out the values for each function (for i from 1 to 10) seperately then multiply them together one by one and then add all of the results together.
     
  10. LordKerwyn

    LordKerwyn New Member

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    Which would look like this:

    i ui vi ui*vi
    1 1 9 9
    2 5 6 30
    3 9 3 27
    4 13 0 0
    5 17 -3 -51
    6 21 -6 -121
    7 25 -9 -225
    8 29 -12 -348
    9 33 -15 -495
    10 37 -18 -666

    all added together -1845
     
  11. Higgs Boson

    Higgs Boson New Member

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    Yep just finished it and got the correct answer. Blargh I must be doing something wrong. Shouldn't finding sum for u1 and v1 and then multiplying the outcomes be sufficient? As in Eu1 . Ev1
     
  12. LordKerwyn

    LordKerwyn New Member

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    No, that doesn't work because the result of a multiplication grows faster than the result of an addition as the inputs grow, which is why there is an order of operations. Just like you have to do the multiplication in 3*4+1*2 (which equals 14) or you get a bad result (if you do the addition first you get 30) you have to do the multiplcation in a sumation before the sumation itself (the sumation operation is on the same operation level as addition). Basicly, sum(x+2x,x,1,10) = sum(x,x,1,10) + sum(2x,x,1,10) but sum(x*2x) != sum(x,x,1,10)*sum(2x,x,1,10).
     
  13. Higgs Boson

    Higgs Boson New Member

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    Yes that makes perfect sence. At first I did think the same thing however my teacher persuaded me that Eui . Evi should work. Since he nor the rest of the group could do it I assumed there is something more to it but apparently I was just wrong (and the rest of us). I'll have to do quick revision on multiplication of arithmetic sequences and get back on my teacher.
    Thanks.
     
  14. Supahboih

    Supahboih New Member

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    **looks at thread**
    ...
    And I think MY math is hard ><
     
  15. sniper64

    sniper64 New Member

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    Sigh.........i'm in 8th grade.........so um, this isn't at *all* confusing.
     
  16. jasmine

    jasmine New Member

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    To reiterate,

    SUM{ u_i * v_i } is NOT the same as SUM{u_i } * SUM{v_i}

    but,

    SUM{ a_i + b_i } IS the same as SUM{a_i } + SUM{b_i}



    So what one can do is expand the summation.

    u_i * v_i = (-3+4i)*(12-3i) = (-36 + 48i + 9i - 12i^2) = (-12i^2 + 57i - 36)

    SUM{u_i*v_i} = SUM{12i^2 + 57i - 36} = SUM{-12i^2} + SUM{57i} - SUM{36}


    = -12.SUM{i^2} + 57.SUM{i} - 36.SUM{1}

    = -12.{n(n+1)(2n+1)/6} + 57.{n(n+1)/2} - 36.{n} ; where n=10

    = -2n(n+1)(2n+1) + 57n(n+1)/2 - 36n

    we could insert n=10 there, or we could tidy it up a little more first,

    = (n/2).( -4(n+1)(2n+1) + 57(n+1) - 72 )

    = (n/2).(-8n^2 - 12n - 4 + 57n + 57 - 72)

    = (n/2).(-8n^2 + 45n - 19)

    letting n=10 now;

    = (5)* (-8*100 + 45*10 - 19)
    = 5 * (-800 + 450 - 19)
    = 5 * (-369)

    = -1845 :)
     
    Last edited: Dec 8, 2009
  17. ijffdrie

    ijffdrie Lord of Spam

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    all i understand is that they are talking about imaginary numbers
     
  18. jasmine

    jasmine New Member

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    no they are not. i is just a variable here, like x or y. :p
     
  19. ijffdrie

    ijffdrie Lord of Spam

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    Oh oops.

    what does the capital sigma mean?
     
  20. Higgs Boson

    Higgs Boson New Member

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    SUM the following