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

1. Dec 4, 2008

### futurebird

This is a problem from a take home final I turned in two day ago... I'm wondering if I got it right.

For $$f= \sum_{i=0}a_{i}x^{i}$$, set $$f^{(s)}= \sum_{i=0}a_{i}x^{si}$$.

4a. Prove that for integers $$n_{1}...n_{t}$$, $$(f^{(n_{1}...n_{t-1})})^{(n_{t})} = f^{(n_{1}...n_{t})}$$.

$$(f^{(n)})^{(m)}= \left( \sum_{i=0} a_{i}x^{ni} \right)^{(m)} = \sum_{i=0} a_{i}x^{mni}$$

So, $$(f^{(n_{1}...n_{t-1})})^{(n_{t})} = f^{(n_{1}...n_{t})}$$.

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

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

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It just seemed too simple... am I missing something here?

2. Dec 4, 2008

### e(ho0n3

Looks right to me.

3. Dec 4, 2008

### futurebird

WAIT

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

Ugggg... I think I messed this up big time. :(

4. Dec 4, 2008

### e(ho0n3

You're right! If i starts from 1, everything works out though.

5. Dec 4, 2008

### futurebird

I think I'm going to fail this take home now... I should have done some examples before turning it it.