Re: Factor x^12+15/11*exp(1/3*x^3) as a product of 6 reals.
- From: Robert Israel <israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Sun, 27 Jan 2008 15:10:14 -0600
Gerry <GerryMrt@xxxxxxxxx> writes:
Hi all,
how easy or difficult is this REAL factoring challenge?
I could be totally wrong but it doesn't look easy to me at all.
After tinkering a bit with this i came to the conclusion that
x^12+15/11*exp(1/3*x^3)
can be written as a product of maximum 6 reals.
For example for x=2 we get:
2^12+15/11*exp(1/3*2^3)=4115.6253401297498556...
which can be written as a product of R1*R2*R3*R4*R5*R6 reals.
What in the world are you talking about? Any real number can be
written as the product of as many reals as you want.
The smallest factor being R1=0.69084445965123865356...
And R1 is t * (R1/t) for any nonzero real t. So what?
And i am sure that someone can show me
how easy it is to determine the other factors.
Only if you tell us what you're really trying to do.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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