Finding Minimal Poly of a possible algebraic number?

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Guest

Posted: Mon Dec 11, 2006 1:56 am

Hello all,

I came across a constant which I think is algebraic,

x = 19.05962891397391285670091722808301086216...

I believe this is the root of either a quadratic or a quintic poly
P(x). Of course, with an approximate number x one can always find a
P(x) with large enough coefficients such that x is a root, but if you
can find one with "small" coefficients, then that might be it.

Any help will be appreciated. I know Maple has a Minpoly feature...

P.S. I tried the "Integer Relations" applet at
http://www.cecm.sfu.ca/~aszanto/IntegerRelations/ but either there was
a glitch in the server or my comp and I couldn't access it. :-(

-Titus

Proginoskes

Posted: Mon Dec 11, 2006 2:49 am

Guest

titus_piezas@yahoo.com wrote:

Quote:

Or, if you start increasing the maximum degree to look for, the degree
of your polynomial should stop increasing.

Quote:

Any help will be appreciated. I know Maple has a Minpoly feature...

Yes; you can get help by using the ?minpoly; help feature. The Maple
code I used is as follows:

] with(polytools):
] r := 19.05962891397391285670091722808301086216;
] for i from 2 to 20 do; print(minpoly(r,i)); end do;

minpoly(r,i) looks for a polynomial of degree <= i, which r
approximately satisfies. If
i = 6, you get a polynomial of degree 5:

] -142 - 76 x + 51 x^2 - 137 x^3 - 12 x^4 + x^5

If i is between 10 and 20, you get:

] 42 + 4 x + 68 x^2 + 47 x^3 + 34 x^4 + 16 x^5 - 20 x^6 + x^7

which looks plausible, but it just might be the limited precision.

Quote:

P.S. I tried the "Integer Relations" applet at
http://www.cecm.sfu.ca/~aszanto/IntegerRelations/ but either there was
a glitch in the server or my comp and I couldn't access it. Sad

You need to download a special plug-in to use that page, and/or enable
Java.

--- Christopher Heckman

Robert Israel

Posted: Mon Dec 11, 2006 3:03 am

Guest

In article <1165816566.805167.159310@j44g2000cwa.googlegroups.com>,
<titus_piezas@yahoo.com> wrote:

Quote:

I don't think there is one. According to Maple (working with Digits = 40),
the minimal polynomial of degree <= 5 for this is

-4132242-15369394*_X-17324271*_X^2+2547562*_X^3-277501*_X^4+10167*_X^5

which is not exactly a case of "small" coefficients.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

Robert Israel

Posted: Mon Dec 11, 2006 3:16 am

Guest

In article <1165819757.273878.294180@l12g2000cwl.googlegroups.com>,
Proginoskes <CCHeckman@gmail.com> wrote:

Quote:

titus_piezas@yahoo.com wrote:
Hello all,

I came across a constant which I think is algebraic,

x = 19.05962891397391285670091722808301086216...

I believe this is the root of either a quadratic or a quintic poly
P(x). Of course, with an approximate number x one can always find a
P(x) with large enough coefficients such that x is a root, but if you
can find one with "small" coefficients, then that might be it.

Or, if you start increasing the maximum degree to look for, the degree
of your polynomial should stop increasing.

Any help will be appreciated. I know Maple has a Minpoly feature...

Yes; you can get help by using the ?minpoly; help feature. The Maple
code I used is as follows:

] with(polytools):
] r := 19.05962891397391285670091722808301086216;
] for i from 2 to 20 do; print(minpoly(r,i)); end do;

minpoly(r,i) looks for a polynomial of degree <= i, which r
approximately satisfies. If
i = 6, you get a polynomial of degree 5:

] -142 - 76 x + 51 x^2 - 137 x^3 - 12 x^4 + x^5

If i is between 10 and 20, you get:

] 42 + 4 x + 68 x^2 + 47 x^3 + 34 x^4 + 16 x^5 - 20 x^6 + x^7

which looks plausible, but it just might be the limited precision.

If you use a higher setting of Digits (say 40) the root of this polynomial
near Titus's number is 19.05962890999997758244232108134530386866
which is off by almost 4e-8.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

Proginoskes

Posted: Mon Dec 11, 2006 3:44 am

Guest

Robert Israel wrote:

Quote:

In article <1165819757.273878.294180@l12g2000cwl.googlegroups.com>,
Proginoskes <CCHeckman@gmail.com> wrote:

titus_piezas@yahoo.com wrote:
Hello all,

I came across a constant which I think is algebraic,

x = 19.05962891397391285670091722808301086216...

I believe this is the root of either a quadratic or a quintic poly
P(x). Of course, with an approximate number x one can always find a
P(x) with large enough coefficients such that x is a root, but if you
can find one with "small" coefficients, then that might be it.

Or, if you start increasing the maximum degree to look for, the degree
of your polynomial should stop increasing.

Any help will be appreciated. I know Maple has a Minpoly feature...

Yes; you can get help by using the ?minpoly; help feature. The Maple
code I used is as follows:

] with(polytools):
] r := 19.05962891397391285670091722808301086216;
] for i from 2 to 20 do; print(minpoly(r,i)); end do;

minpoly(r,i) looks for a polynomial of degree <= i, which r
approximately satisfies. If
i = 6, you get a polynomial of degree 5:

] -142 - 76 x + 51 x^2 - 137 x^3 - 12 x^4 + x^5

If i is between 10 and 20, you get:

] 42 + 4 x + 68 x^2 + 47 x^3 + 34 x^4 + 16 x^5 - 20 x^6 + x^7

which looks plausible, but it just might be the limited precision.

If you use a higher setting of Digits (say 40) the root of this polynomial
near Titus's number is 19.05962890999997758244232108134530386866
which is off by almost 4e-8.

And that error is within the default precision (which is 8 digits).

What Titus should do, of course, is to say how he happened upon this
number.

--- Christopher Heckman

The Last Danish Pastry

Posted: Mon Dec 11, 2006 5:23 am

Guest

"Robert Israel" <israel@math.ubc.ca> wrote in message
news:elivrk$a2s$1@nntp.itservices.ubc.ca...

Quote:

In article <1165816566.805167.159310@j44g2000cwa.googlegroups.com>,
titus_piezas@yahoo.com> wrote:
Hello all,

I came across a constant which I think is algebraic,

x = 19.05962891397391285670091722808301086216...

I believe this is the root of either a quadratic or a quintic poly
P(x). Of course, with an approximate number x one can always find a
P(x) with large enough coefficients such that x is a root, but if
you
can find one with "small" coefficients, then that might be it.

I don't think there is one. According to Maple (working with Digits
= 40),
the minimal polynomial of degree <= 5 for this is

-4132242-15369394*_X-17324271*_X^2+2547562*_X^3-277501*_X^4+10167*_X^5

which is not exactly a case of "small" coefficients.

Let the original poster's number be P, and let the relevant root of
the above polynomial, found by Maple, be M.

We have:

P=19.05962891397391285670091722808301086216
M=19.05962891397391285670091722808301086216000354...

I am wondering if you have asked Maple to find a root which is
unjustifiably close to P.

--
Clive Tooth
http://www.shutterstock.com/cat.mhtml?gallery_id=61771

Guest

Posted: Mon Dec 11, 2006 8:10 am

Proginoskes wrote:

Quote:

What Titus should do, of course, is to say how he happened upon this
number.

--- Christopher Heckman

It is part of a broad conjecture. The hypergeometric function
2F1[a,b,c,d] is given by Mathematica as Hypergeometric2F1[a,b,c,d].
Define the ff functions:

h1[a,x] = Hypergeometric2F1[a,1-a,1,1-x]
h2[a,x] = Hypergeometric2F1[a,1-a,1,x]

For 0<(a,x)<1, given the parameter "a", find the unique real x of the
equation,

h1[a,x]/h2[a,x] = Sqrt[n]

for a given rational n>0. For a=1/2, the problem reduces to finding the
"elliptic modulus k" and in Mathematica this is given as
x=ModularLambda[Sqrt[-n]]. The question is: What about other a?

Here is my conjecture: As n-> infinity, then,

x = E^(-2Pi*Sqrt[n/v_a]) C_a

where v_a and C_a are constants dependent on the parameter "a".
Tabulating the first few using the convention {a, v_a, C_a}:

{1/2, 4, 16}
{1/3, 3, 27}
{1/4, 2, 64}
{1/5, (5-Sqrt[5])/2, 163.9....} and {2/5, (5+Sqrt[5])/2, 19.059...}
{1/6, 1, 432}

etc...

For example, for h1[2/5,x]/h2[2/5,x] = Sqrt[n], then x is
asymptotically

x = E^(-2Pi*Sqrt[n/v_a]) (19.059628...)

where 19.059... is the number in the post and v_a = (5+Sqrt[5])/2.

For a=1/5, I know its C_a up to 30+ digits but have the data at home.
Note that the v_a for a=1/5 and 2/5 satisfy the equation x^2-5x+5=0.
For a=1/7, v_a is the smallest real root of x^3-7x^2+14x-7=0 so I know
something interesting is going on. (And I suspect v_a of a=2/7 and 3/7
are the other two roots of the cubic in the same manner a=1/5 and 2/5
share a poly.) In general, for a=1/k where k is an odd integer, v_a is
quite easy to find (the poly is related to the kth root of unity.)

The tricky part is finding a closed-form soln to C_a, that's why I
thought C_(2/5) was the root of either a quadratic or a quintic.

So that's the story of that number. How plausible is my conjecture?

-Titus

Guest

Posted: Tue Dec 12, 2006 2:00 am

After more experimentation, here are some observations:

Given the asympotic soln x = E^(-2Pi*Sqrt[n/v_a])C_a to the
hypergeometric equation defined above, then,

C_(1/3) = 3^3
C_(1/5) C_(2/5) = 5^5
C_(1/7) C_(2/7) C_(3/7) = 7^7

etc. for prime denominators.

v_(1/3) is the root of v-3 = 0
v_(1/5) and v_(2/5) are the two roots of v^2-5v+5 = 0
v_(1/7), v_(2/7), v_(3/7) are the three roots of v^3-7v^2+14v-7 = 0

In general, the v_a's are the roots of
(y+Sqrt[y^2-1])^p+(y-Sqrt[y^2-1])^p with y =Sqrt[v]/2 for prime p.

Anybody can find a closed-form expression for the C_a's, esp if they
are algebraic like the v_a's?

-Titus

Gerry

Posted: Wed Dec 13, 2006 2:06 am

Guest

titus_piezas@yahoo.com schrieb:

Quote:

Hi Titus,

i tried to found a quadratic solution which looks like it could
generate your constant. x^2+101188x-1928969.

Regards

Gerry

Han de Bruijn

Posted: Wed Dec 13, 2006 3:42 am

Guest

Gerry wrote:

Quote:

titus_piezas@yahoo.com schrieb:

Hello all,

I came across a constant which I think is algebraic,

x = 19.05962891397391285670091722808301086216...

I believe this is the root of either a quadratic or a quintic poly
P(x). Of course, with an approximate number x one can always find a
P(x) with large enough coefficients such that x is a root, but if you
can find one with "small" coefficients, then that might be it.

Any help will be appreciated. I know Maple has a Minpoly feature...

Hi Titus,

i tried to found a quadratic solution which looks like it could
generate your constant. x^2+101188x-1928969.

Sure. x = 19.0596289139739128567009172280830108621699999999999999999999

Han de Bruijn

Guest

Posted: Wed Dec 13, 2006 4:54 am

Han de Bruijn wrote:

Quote:

titus_piezas@yahoo.com schrieb:

Hello all,

I came across a constant which I think is algebraic,

x = 19.05962891397391285670091722808301086216...

After all the posts of the people here, it may be the case that x is
not algebraic at all but may involve some transcendental constant.
Curiously though, as I pointed out, x (which is C_(2/5)) and its
partner C_(1/5) =163.98493217.... satisfy,

C_(1/5) C_(2/5) = 5^5

though C_(1/5) + C(2/5) = 183.04156109... does not seem to be algebraic
either. If it were, one can then easily solve for C_(1/5) and C_(2/5).

-Titus

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