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pluton
science forum beginner
Joined: 16 Feb 2006
Posts: 8
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 Posted: Tue Jul 18, 2006 4:54 am Post subject:
inverting a cubic polynomial as a series
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Hi all,
I have a polynomial of the form : P(x)=a x^3+b x^2+c x. Is there a way
to invert it exactly knowing a, b and c ? (invert means find x(t) such
as P(x(t))=t) If not, what about an infinite series ? It looks like it
does not converge over a wide range of t ? Any suggestion ? Thank you,
Pluton |
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mariano.suarezalvarez@gma
science forum addict
Joined: 28 Apr 2006
Posts: 58
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| Posted: Tue Jul 18, 2006 5:25 am Post subject:
Re: inverting a cubic polynomial as a series
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pluton wrote:
| Quote: | Hi all,
I have a polynomial of the form : P(x)=a x^3+b x^2+c x. Is there a way
to invert it exactly knowing a, b and c ? (invert means find x(t) such
as P(x(t))=t) If not, what about an infinite series ? It looks like it
does not converge over a wide range of t ? Any suggestion ? Thank you,
|
(The polynomial you mention can be factored into a product of
of x and a quadratic polynomial. I guess you meant some other
polynomial  )
The equation
a x^3 + b x^2 + c x + d = 0
defines x implicitely as a function of a, b, c, d near points
where the implicit function theorem applies (basically, away
from multiple roots). Moreover, since the left hand side of
the equation is analytic, the function implicitely defined by
it is analytic, so you can in fact develop x(a,b,c,d) as a
multiple power series of a, b, c and d.
This is the tip of the iceberg known as the theory of
algebraic functions, btw.
You will find information about the implicit function theorem in any
book on calculus of functions of several variables. To see how it is
that one can conclude that the function x(a,b,c,d) is in fact analytic,
an introductory text on complex analysis should be of help.
-- m |
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alex.lupas@gmail.com
science forum beginner
Joined: 23 Feb 2006
Posts: 47
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| Posted: Tue Jul 18, 2006 7:04 am Post subject:
Re: inverting a cubic polynomial as a series
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pluton wrote:
| Quote: | Hi all, I have a polynomial of the form : P(x)=a x^3+b x^2+c x. Is there a way
to invert it exactly knowing a, b and c ? (invert means find x(t) such
as P(x(t))=t) If not, what about an infinite series ? It looks like it
does not converge over a wide range of t ? Any suggestion ? Thank you,
Pluton
=========== |
Perhaps try Lagrange series as someone inform you.
Also of interest are the works by Gauss , and
Hans von Mangoldt ,
"Ueber die Darstellung Wurzeln einer dreigliedrigen algebraischen
Gleichung durch unendlichen Reihen",Dissertation,Berline,1878.
See also Umbral Calculus(Gian-Carlo Rota,Steven
Roman,Odlyzko,Kahane,...) |
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Achava Nakhash, the Lovin
science forum beginner
Joined: 14 Sep 2005
Posts: 35
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| Posted: Tue Jul 18, 2006 5:25 pm Post subject:
Re: inverting a cubic polynomial as a series
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pluton wrote:
| Quote: | Hi all,
I have a polynomial of the form : P(x)=a x^3+b x^2+c x. Is there a way
to invert it exactly knowing a, b and c ? (invert means find x(t) such
as P(x(t))=t) If not, what about an infinite series ? It looks like it
does not converge over a wide range of t ? Any suggestion ? Thank you,
Pluton
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One potential problem with inverting it exactly is that it might not be
1-1. You need to check that condition first. I rather doubt that you
can find an explicit inverse in terms, say, of radicals, except in
special cases, and without an explicit polynomial in front of me, I
can't really comment in that. In general, there is a formula for the
inverse of a power series that gives you another power series.
Obviously there are some restirictions on the applicability of this
result, but they are minimal. A cubic, or any other degree, polynomial
is simply a short power series as far as this theorem is concerned. It
is probably a big mess and might or might not be useful to you, but it
is not hard (if you know anything about power series and how their
coefficients are derived) to figure out the first several.terms of the
power series of your inverse.
Hope this helps,
Achava |
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David W. Cantrell
science forum Guru
Joined: 02 May 2005
Posts: 352
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| Posted: Wed Jul 19, 2006 2:54 am Post subject:
Re: inverting a cubic polynomial as a series
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"pluton" <plutonesque@gmail.com> wrote:
| Quote: | Hi all,
I have a polynomial of the form : P(x)=a x^3+b x^2+c x. Is there a way
to invert it exactly knowing a, b and c ?
|
Yes. It's been known for centuries. The cubic formula can be found, for
example, in section 3.8.2 of Abramowitz and Stegun:
<http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP?Res=150&Page=17&Submit=Go>
It really puzzles me why no previous respondents mentioned the formula!
| Quote: | (invert means find x(t) such as P(x(t))=t)
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Of course, that inverse relation is not necessarily single-valued. But
that's not a "problem", IMO.
| Quote: | If not, what about an infinite series ?
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Considering that there is an exact inverse in terms of radicals, I suppose
you will not be interested in using series. Nonetheless, if you were still
interested, you could search for "reversion of series".
| Quote: | It looks like it does not converge over a wide range of t ?
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An important observation, indeed. This would make it crucial to choose the
correct point about which to expand, before doing the reversion. But as I
said, I doubt that you will want to go down this path...
David |
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pluton
science forum beginner
Joined: 16 Feb 2006
Posts: 8
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| Posted: Thu Jul 20, 2006 2:12 am Post subject:
Re: inverting a cubic polynomial as a series
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ok thank you. I do not know why I did not think about it either. I was
more focusing on a function than finding the roots but it is a solution
as well.
Pluton |
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Paul Abbott
science forum addict
Joined: 19 May 2005
Posts: 99
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| Posted: Thu Jul 20, 2006 4:12 am Post subject:
Re: inverting a cubic polynomial as a series
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In article <1153198491.059962.290590@b28g2000cwb.googlegroups.com>,
"pluton" <plutonesque@gmail.com> wrote:
| Quote: | I have a polynomial of the form : P(x)=a x^3+b x^2+c x. Is there a way
to invert it exactly knowing a, b and c ? (invert means find x(t) such
as P(x(t))=t)
|
The exact solution to an arbitrary cubic,
p[x_] = a x^3 + b x^2 + c x + d;
is, of course, well known.
| Quote: | If not, what about an infinite series?
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Such series can be obtained using series reversion. For example, with
p[x_] = a x^3 + b x^2 + c x + d;
then (using Mathematica),
x[t_] = InverseSeries[ p[x] + O[x]^4, t]
(where the expansion point is x = 0), yields
2 2 3
(-d + t) b (-d + t) (2 b - a c) (-d + t) 4
-------- - ----------- + ---------------------- + O[-d + t]
c 3 5
c c
Check that this is the inverse series:
p[x[t]] // Normal // Simplify
t
| Quote: | It looks like it does not converge over a wide range of t?
|
You only require the inverse series at t = 0:
Normal[x[t]] /. t -> 0
2 2 3
d b d (2 b - a c) d
-(-) - ---- - ---------------
c 3 5
c c
This agrees with the expansion of the exact solution.
For more on this topic, see
http://library.wolfram.com/examples/quintic/
and, in particular,
http://library.wolfram.com/examples/quintic/main.html#series
where series methods are illustrated.
Cheers,
Paul
_______________________________________________________________________
Paul Abbott Phone: 61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
The University of Western Australia (CRICOS Provider No 00126G)
AUSTRALIA http://physics.uwa.edu.au/~paul |
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