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

[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})}.---

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

 

[e(ho0n3]

Looks right to me.

 

[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. :(

 

[e(ho0n3]

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

 

[futurebird]

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