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

Gerry <GerryMrt@xxxxxxxxx> wrote:

On Jan 28, 4:32 am, Gerry Myerson <ge...@xxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
 Gerry <Gerry...@xxxxxxxxx> wrote:
On Jan 28, 12:03 am, Gerry Myerson <ge...@xxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:

 Gerry <Gerry...@xxxxxxxxx> wrote:
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.

Hi Robert,

you are probably right however consider that we have a degree 12
polynomial

No, we don't - x^12+15/11*exp(1/3*x^3) is not a polynomial
of degree 12 or of any other degree.

which can be written as a product of prod(i=1,12,x-g(a*exp(b*x)))

Can it? What are g, a, b? Maybe you should give us a simpler example.
Can you show us how to write x^2 - e^x as a product of two things
of the form x - g( a exp( b x))? If you can do that, maybe we'll get
a clue as to what you are on about.

For you i have the following example :

-x^2-exp(x)

which can be written as a product of maximum 2 factors in this case.

Factor1=(x-F1)=-I + (-1 - 1/2*I)*x - 1/8*I*x^2 - 1/48*I*x^3 -
1/384*I*x^4 - 1/3840*I*x^5 - 1/46080*I*x^6 - 1/645120*I*x^7 -
1/10321920*I*x^8 - 1/185794560*I*x^9 + O(x^10)

Factor2=(x-F2)=I + (-1 + 1/2*I)*x + 1/8*I*x^2 + 1/48*I*x^3 +
1/384*I*x^4 + 1/3840*I*x^5 + 1/46080*I*x^6 + 1/645120*I*x^7 +
1/10321920*I*x^8 + 1/185794560*I*x^9 + O(x^10)

(x-F1)*(x-F2)=1 + x + 3/2*x^2 + 1/6*x^3 + 1/24*x^4 + 1/120*x^5 +
1/720*x^6 + 1/5040*x^7 + 1/40320*x^8 + 1/362880*x^9 + O(x^10)

Which is the series expansion (first 10 terms) of -x^2-exp(x)

I have some problems with one of the symbols you're using,
and I'm not sure how my x^2 - e^x became your - x^2 - exp(x),
and in any event it looks like your last formula is x^2 + exp(x),
but I'll let all that pass, for now. So you have found a way to write
a power series as a product of two power series. That's *a* way;
do you have any reason to think that's the *only* way to write
x^2 + exp(x) as a product of two power series? or that it is in
any way special among the many, many ways to write x^2 + exp(x)
as a product of two power series? Do you think there is a power
series that can't be written as a product of two power series? Do
you have an example? a proof?

So tell me why can i not see this as a second degree polynomial?

You can see it any way you like. The rest of us will continue to call
something a second degree polynomial if and only if it looks like
a x^2 + b x + c, with a, b, and c constants, a non-zero. We will
do this, not because we're stubborn, but because we reckon we
can prove lots more nice theorems with our definitions than with
yours (if you provide one).

well yes i should have corrected the sign you are right,
Lets assume that it is a way to find a product of a
powerseries.
I'm sure that the notation is unique simply
because it's based on a double product of exp Pi I and
some a,b,c,d like in the formula to solve a quadratic.

I'm fluent in both mathematics and English
but this seems to be neither.

What can it mean for notation to be unique?

I was suggesting you enquire as to whether your solution
to the problem of writing x^2 - e^x as a product of
two power series is unique. For example, you can write
x^2 - e^x = e^x times (x^2 e^{-x} - 1)
= (1 + x + (1/2)x^2 + (1/6)x^3 + ...)
times (-1 + x^2 - x^3 + (1/2)x^4 + ...).
What makes your way of factoring x^2 - e^x any better
than this way, or any of the infinitely many other ways?

And where is there a d in the quadratic formula?
(Other than near the middle of the word, "quadratic")

Now by making the variable d zero the notation
becomes eactly the quadratic formula.

In which case you surly agree that we have a unique notation
of the solution of the quadratic even in case it's possible to further
define the solutions itself which might be element of N,Q.

Definitely not English. Probably not mathematics.
I give up.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.

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