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This just seems so simple so I think I'm doing it wrong (Algebra)

  1. Dec 4, 2008 #1
    This is a problem from a take home final I turned in two day ago... I'm wondering if I got it right.

    For [tex]f= \sum_{i=0}a_{i}x^{i}[/tex], set [tex]f^{(s)}= \sum_{i=0}a_{i}x^{si}[/tex].

    4a. Prove that for integers [tex]n_{1}...n_{t}[/tex], [tex](f^{(n_{1}...n_{t-1})})^{(n_{t})} = f^{(n_{1}...n_{t})}[/tex].

    [tex] (f^{(n)})^{(m)}= \left( \sum_{i=0} a_{i}x^{ni} \right)^{(m)} = \sum_{i=0} a_{i}x^{mni} [/tex]

    So, [tex](f^{(n_{1}...n_{t-1})})^{(n_{t})} = f^{(n_{1}...n_{t})}[/tex].

    4b. Prove that if f divides [tex]f^{(n_{i})}[/tex] for each [tex]n_{i}[/tex], then f divides [tex]f^{(n_{1}...n_{t})}[/tex]

    Since every term is only assuming higher powers, using the result in 4b, we can factor out [tex]x^{(n_{1}...n_{i-1}n_{i+1}...n_{t})}[/tex] we have: [tex]f^{(n_{1}...n_{t})} = x^{(n_{1}...n_{i-1}n_{i+1}...n_{t})}(f^{(n_{i})})[/tex]. f divides [tex]f^{(n_{i})}[/tex] so f divides [tex]f^{(n_{1}...n_{t})}[/tex].


    It just seemed too simple... am I missing something here?
  2. jcsd
  3. Dec 4, 2008 #2
    Looks right to me.
  4. Dec 4, 2008 #3

    If [tex]f = 2+ x+ 4x^{3} = 2x^{0} + x^{1}+ 4x^{3}[/tex] then [tex]f^{(2)} = 2x^{0*2} + x^{1*2}+ 4x^{3*2}= 2x^{0} + x^{2}+ 4x^{6} = 2 + x^{2}+ 4x^{6}[/tex] NOT [tex]2x^{2} + x^{3}+ 4x^{5}[/tex]

    Ugggg... I think I messed this up big time. :(
  5. Dec 4, 2008 #4
    You're right! If i starts from 1, everything works out though.
  6. Dec 4, 2008 #5
    I think I'm going to fail this take home now... I should have done some examples before turning it it.
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