Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is it possible to construct a polynomial of degree N, with all of them as integer coefficient have a root as the given value. The root value provided is not necessarily a rational number.

For example, if the root is 28.552622898861801 we can have a polynomial of degree 10 whose one root will be the given value.

10000 x^10 - 280000 x^9 - 150000 x^8 - 220000 x^7 - 40000 x^6 - 790000 x^5 - 160000 x^4 - 320000 x^3 - 270000 x^2 - 250000 x - 251271 = 0

Wolfram alpha link for this equation

share|improve this question
    
If your rational number is $p/q$ you can take the polynomial $(px-q)^N$. Or did I misunderstand the question. –  Michael Bächtold Oct 20 '11 at 7:30
    
I meant $(qx-p)^N$. –  Michael Bächtold Oct 20 '11 at 7:32
    
oops. updated the description. –  Arif Oct 20 '11 at 7:32
1  
See en.wikipedia.org/wiki/Algebraic_number. Not all real numbers are algebraic. –  Michael Bächtold Oct 20 '11 at 7:34
2  
Your welcome. For the future you might want to try asking at math.stackexchange.com instead of this site which is more for research level questions. –  Michael Bächtold Oct 20 '11 at 8:39

1 Answer 1

up vote 2 down vote accepted

The problem can be solved by running some Integer Relation algorithm (e.g., PSLQ) on the numbers $1, r, r^2, \dots, r^N$ where $r$ is a given root.

See http://en.wikipedia.org/wiki/Integer_relation_algorithm

For example, here is computation in PARI/GP which gives a better result than the polynomial shown in question:

? r = 28.552622898861801; algdep(r,10)

%1 = 3*x^10 + 38*x^9 - 3695*x^8 + 4582*x^7 + 3016*x^6 + 1435*x^5 + 4552*x^4 - 1219*x^3 - 9920*x^2 - 2402*x + 3087

? subst(%1,x,r)

%2 = -2.7334689816478450022 E-24

share|improve this answer

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.