Hi,
If ind this. I don't know if this is what you want.
(%i1) display2d:false;
(%o1) false
(%i2) expr:(x^(p/(2^k))-x^(1/r));
(%o2) x^(p/2^k)-x^(1/r)
(%i3) radcan([first(solve(expr=e,k)),first(solve(expr=-e,k))]);
(%o3) [k = -(log(log(x^(1/r)+e))-log(log(x))-log(p))/log(2),k =
-(log(log(x^(1/r)-e))-log(log(x))-log(p))/log(2)]
PS: there is a missing parentesis in your original expression.
Best.
Laurent.
-----Message d'origine-----
De?: maxima-bounces at math.utexas.edu [mailto:maxima-bounces at math.utexas.edu] De la part de VicTT
Envoy??: mercredi 30 janvier 2013 22:54
??: maxima at math.utexas.edu
Objet?: [Maxima] How to solve this inequality?
I'll keep this short in the interest of preserving your time:
I'm trying to find min(k) such that, given integers: x,r and rational epsilon,
abs(x^(p/(2^k)-x^(1/r))<epsilon.
Then, given integers: x,r,min(k) and rational epsilon, find p in the above inequality.
How would I do this in Maxima? (first step is, I assume, rewriting this as an equality
*somehow*)
Also, if anyone knows any 3rd party add-ons that could solve this, I'd appreciate the
contribution. I'm missing something, I just can't tell what.
(This is neither homework, nor a task I've devoted less than 1 hour towards achieving.)
Thank you in advance,
Pirvu Paul Daniel