Re: Factor x^12+15/11*exp(1/3*x^3) as a product of 6 reals.

On Jan 27, 10:10 pm, Robert Israel
<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
Gerry <Gerry...@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              isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics        http://www.math.ubc.ca/~israel
University of British Columbia            Vancouver, BC, Canada- Hide quoted text -

- Show quoted text -

Hi Robert,

you are probably right however consider that we have a degree 12
polynomial
which can be written as a product of prod(i=1,12,x-g(a*exp(b*x)))

This way i get 12 complex conjugate numbers when i set x=2.
These 12 generate 6 reals which exactly multiply into the number
2^12+15/11*exp(1/3*2^3)=4115.6253401297498556...

Probably i'm overcomplicating things but still the way i'm
factoring gives me a true notation for the 6 factors
which are also function of exp(b*x).

Here are the 6 numbers

R1=0.69084445965123865356..
R2=2.0176002384816131468..
R3=4.3156086566514380406..
R4=6.9691202143121870272..
R5=9.2671286324820119210..
R6=10.593884411312386414..

Regards

Gerry
.

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