- #1
futurebird
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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?
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?