another problem asdfasdfasdfasdf

by pythag011, Sep 24, 2008, 10:33 PM

If P(n) = $ n^2+1$, find all polynomials Q such that P(Q(x)) = Q(P(x))

Hmm...x and $ x^2 + 1$ and $ x^4 + 2x^2 + 2$ are solutions...are there any other ones?

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that's not a very hard problem but how do you now when you are done taking solutions?!

by Poincare, Sep 24, 2008, 11:16 PM

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Err...whatdo you mean its not very har?

by pythag011, Sep 25, 2008, 1:27 AM

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LOL YAY POLYNOMIAL FUNC EQS

ok let's see all nonzero coefficents have to be of even power-ed stuff

so with polynomials I think the idea is, after normal func eq solving stuff, to be like "lawl infinitely many roots implies 0-ness"

so let's see?

P(Q(x)) = Q(P(x)), P = x^2+1

so like... let's look at roots. Roots are fun. Ok so there is a root when Q(x) = $ \pm i$ which is... not fun.

OH YEAH ALSO DEGREES OF STUFF:

say you have like P(x) = $ a_n x^n + a_{n-1}x^{n-1} + \ldots + a_0$

so now RHS is

$ a_n x^{2n} + 2a_n x^n + a_n + \ldots + a_0$

LHS is

$ P(x)^2 + 1$ bleargh that has constant term $ a_0^2 + 1$

ok so now you know $ Q(1) = Q(0)^2 + 1$

that could be helpful

oh snap also $ a_n = (a_n)^2$ so $ a_n=1$ due to coeff of $ x^{2n}$

blah now look at coeff of $ x^2$ ? but wait I think it is too late so I am getting silly stuff... you can probably continue from here

I also bet you want polynomials with integer coeffs so you maybe shoudl say that

by not_trig, Sep 25, 2008, 4:00 AM

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Hmm...I think the original problem had integer coefficients. 3 of the solutions are x, $ x^2 + 1$, and $ x^4 + 2x^2 + 2.$ x is the only odd solution....

EDIT: I have solved it; define the sequence $ P_i$ as $ P_1 = x$ and for $ n \ge 1$ $ P_{n+1}(x) = P_n(x^2 + 1)$ those are the soluitons. So there are infinitelty many solutions.

by pythag011, Sep 25, 2008, 1:35 PM

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i did not understand ONE THING THAT NOT_TRIG SAID but inverse functions. That was the stupidest advice you will ever be given :)

by Poincare, Sep 29, 2008, 10:58 PM

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indeed umm inverse functions suck in this case because polynomials blah

by not_trig, Sep 30, 2008, 1:53 AM

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Hmm...my proof was basicly using odd and even functions, proving the only odd solutions was f(x)=x, and induction.

by pythag011, Sep 30, 2008, 2:08 AM

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Could someone post a link to this blog now that pythag has made it have no title?

by Mewto55555, Oct 12, 2008, 8:16 PM

Click here if you did not make white MOP (Fake copy of pythag's blog)

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