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.

Say we have two transcendental numbers, u and v. And u presumably can be obtained as a result of applying a rational function $Q$ with integer coefficients to v. Is it possible to find such rational function?

In other words we need to find two polynomials $P_1$ and $P_2$ with integer coefficients such that

$u=\frac{P_1(v)}{P_2(v)}$

share|improve this question
    
You would want to include the restriction that $P_1$ and $P_2$ have no common factors. –  J. M. Apr 30 '11 at 11:06
    
Yes, of course. I meant the simplest form of that function. –  Anixx Apr 30 '11 at 11:08
    
If we find at leat one pair of polynomials we can cancel the common factors out of course so the task is ti find at least one such pair or the function Q directly. –  Anixx Apr 30 '11 at 11:09
    
It looks as if even the case of $P_2=1$ is hard... –  J. M. Apr 30 '11 at 11:12
    
How are you given $u$ and $v$? –  Qiaochu Yuan Apr 30 '11 at 11:19
show 1 more comment

2 Answers

Let's ask a much simpler question: if we have two transcendental numbers, and their difference is a rational number, can we find that rational number? Seems to me it would depend a bit on what it means to "have" a transcendental number. For all we know, $\pi-e$ is rational. The more decimals we know in the expansions of $\pi$ and $e$, the better the lower bound we can put on the numerator and denominator of the rational, but how can we ever find the rational?

share|improve this answer
add comment

Let's write it as $P_1(v)u = P_2(v)$, or rather $$ \sum \alpha_i v^i u - \sum \beta_i v^i = 0. $$ Now use an integer relation algorithm.

share|improve this answer
1  
I'm sure Yuval knows this, but just for the record: these algorithms will either tell you there is no relation with small numbers (leaving open the possibility that there is a relation with bigger ones) or they will give a relation that works up to the accuracy to which you know $u$ and $v$ (but won't give a proof that the relation is exact). –  Gerry Myerson May 1 '11 at 0:20
add comment

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.