Author 
Message 

Chasney913
Member
Joined: 28 Aug 2007 Posts: 117

Posted: 05 Nov 2008 05:14:31 pm Post subject: 


Let's see. My teacher wants to introduce us to formal proofs, and has given us some small things based on that.
Anyway, I'm not really here about the proof itself. She decided to give a question that's for a "challenge", but is still due, so I'm not sure how that works.
I (obviously) don't want anyone to do the question for me. A push in the right direction would be great, though. My friends and I have been looking at it and we can't get a good foothold.
Without further ado:
f(x) is a polynomial, and has the following properties:
i. f(1) is 1
ii. f(2x)/f(x+1) = 8  56/(x+7) where both sides are not equal to zero
Prove:
f(x) is cubic
f(1) is 5/21
I don't think we need to actually find f(x), because that would make both portions really easy.
The only I did was to change the RHS of ii into (8x)/(x+7), but that doesn't get me anywhere, other than possibly f(2x) = 0 when x = 0, meaning f(0) = 0. That means the constant term of the cubic is 0, but that doesn't seem to help.
Hopefully someone can help me out. 

Back to top 


DarkerLine ceci n'est pas une 
Super Elite (Last Title)
Joined: 04 Nov 2003 Posts: 8328

Posted: 05 Nov 2008 05:52:48 pm Post subject: 


It's a pretty cool problem!
One thing you know is that all polynomials can be factored (at least, over the complex numbers). So you can write f(x)=a(xx_{1})(xx_{2})(...)(xx_{n})  actually, it stops at n=3, being cubic, but you've yet to prove that. See what you can do from there.
You also will end up finding f(x) to answer the second part of the question.
Last edited by Guest on 05 Nov 2008 06:02:49 pm; edited 1 time in total 

Back to top 


Chasney913
Member
Joined: 28 Aug 2007 Posts: 117

Posted: 05 Nov 2008 06:12:37 pm Post subject: 


Right, so I considered this, and I'll see if you agree. I said that there are three factors if cubic, and that one factor should be (x) if f(0) is 0. This leaves a quadratic, which I know how to factor. I also know those factors (conjugates) must multiply to give 1. Using that, I can show that the "a" coefficient and the "c" coefficient are equal. That leaves me with f(x) = ax^3 + bx^2 + ax.
Now, I need to solve for a and c. I'm thinking a system of equations, but I'm not sure what data to use. Maybe f(0) = 0 and f(1) = 1? I'll try that now.
EDIT: except for the fact that f(0) = 0 is not really a help in this case. a and b could be anything, and it'll still work.
Last edited by Guest on 05 Nov 2008 06:17:40 pm; edited 1 time in total 

Back to top 


DarkerLine ceci n'est pas une 
Super Elite (Last Title)
Joined: 04 Nov 2003 Posts: 8328

Posted: 05 Nov 2008 06:24:07 pm Post subject: 


The coefficients of x^3 and x are not equal, so I'm not sure what you're doing. Furthermore, I don't think I follow your logic at all.
"I said that there are three factors if cubic"  you're trying to prove that f(x) is cubic. Assuming that it's cubic isn't going to do you any favors.
Given the factored form of f(x), with the x_{i} unknown, what does that tell you about f(2x)/f(x+1)? 

Back to top 


Chasney913
Member
Joined: 28 Aug 2007 Posts: 117

Posted: 05 Nov 2008 06:44:57 pm Post subject: 


The only thing I can see is that because the result has only x terms, and no higher power terms, some terms must cancel out.
I see what I was doing wrong in the a=c thing, so no need for me to explain that.
Maybe I should wait to post  I see we have a number of people reading. Maybe they have something useful as well.
Last edited by Guest on 05 Nov 2008 06:45:22 pm; edited 1 time in total 

Back to top 


DarkerLine ceci n'est pas une 
Super Elite (Last Title)
Joined: 04 Nov 2003 Posts: 8328

Posted: 05 Nov 2008 11:34:58 pm Post subject: 


Final hint: where do you think the 8 in "8x/(x+7)" came from?
Last edited by Guest on 05 Nov 2008 11:35:06 pm; edited 1 time in total 

Back to top 


Chasney913
Member
Joined: 28 Aug 2007 Posts: 117

Posted: 06 Nov 2008 07:02:20 am Post subject: 


Well, it would need to be some multiplication of the top term 2x; that's the only way to get something on the top. At first it seems like it would be (2x) cubed, but that produces 8x^3, which is wrong. Although, it's possible it's the "a" term, or a combination between the factors and the "a" term. The "a" term gets canceled out in division, though, so it wouldn't really make a difference what it was. 

Back to top 


thornahawk μολών λαβέ
Active Member
Joined: 27 Mar 2005 Posts: 569

Posted: 06 Nov 2008 08:12:03 am Post subject: 


Let's see...
1.) It is given that f[1] == 1.
2.) Using the identity f(2x)/f(x+1) = 8  56/(x+7) and replacing x with 0 and 1/2, we find that f[0] == 0 and f[3/2] == 15/8
3. ) Since 8  56/(x+7) is singular at x == 7, f(2x)/f(x+1) should also be singular at x == 7, and thus f[6] == 0.
I think you already know what to do from there. ;)
thornahawk 

Back to top 


