Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Recently while coming up with an example for a paper I'm writing I find myself wanting something to say about how 'awful' the first positive root of the equation $$ 4\cdot2^{2p}\cdot3^p-4\cdot3^{2p}-7\cdot2^{3p}+8\cdot2^{2p}+8\cdot3^p-4=0 $$ is. Numerically I know it's about 1.576, but I suspect that it's no only irrational but cannot be solved for using elementary functions, e.g. $\log_2(3+\sqrt{5})$.

I've not much / any background in number theory of things like this and am kind of stumped. Are there any simple arguments in this direction?

Note: This is the simplest of the awful equations I could come up with. I'm also aware that it has a root at $p=2$

share|improve this question
2  
Let $x=2^p$, $y=3^p$, then $4x^2y-4y^2-7x^3+8x^2+8y-4=0$ and $y=x^{\log_23}$. The first equation is a quadratic in $y$ with coefficients involving $x$. Solve it for $y$, and equate the two expressions for $x$. You may get something that can be handled by standard theorems on irrationality/transcendence (or maybe not - I haven't tried it). –  Gerry Myerson Jun 28 '11 at 7:22
    
@Gerry Myerson. Thanks for spotting that. I'll have a further bash at it. –  Stephen Sanchez Jul 1 '11 at 5:54

1 Answer 1

up vote 2 down vote accepted

I suspect there aren't any simple arguments for showing that one or all of the solutions to an equation of this kind is not expressible using elementary functions. A 1999 paper of Tim Chow reviews one reasonable way to formally define "elementary" in this context, and points out that proving anything to be non-elementary seems to be very difficult.

share|improve this answer
    
Thanks. I thought as much, but like I said I have very little experience in this. Thanks for the link to the paper. –  Stephen Sanchez Jun 28 '11 at 6:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.