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Dec 3, 2006, 6:40:26â€¯AM12/3/06

to

Based on this reference

Cardan Polynomials and the Reduction of Radicals (by T. Osler)

(see also references therein)

(you can download the paper here:

http://www.jstor.org/view/0025570x/di021218/02p0059q/0?currentResult=0025570x%2bdi021218%2b02p0059q%2b0%2c03&searchUrl=http%3A%2F%2Fwww.jstor.org%2Fsearch%2FBasicResults%3Fhp%3D25%26so%3DNewestFirst%26si%3D1%26Query%3DOsler

)

the following expression can be reduced to 1

z = (2 + Sqrt[5])^(1/3) + (2 - Sqrt[5])^(1/3)

Mathematica gives

N[%]

1.9270509831248424 + 0.535233134659635*I

This is because by default it returns a complex number for the cube

root of a negative number

List @@ z

N[%]

{(2 - Sqrt[5])^(1/3), (2 + Sqrt[5])^(1/3)}

{0.30901699437494756 + 0.535233134659635*I, 1.618033988749895}

However defining

mycuberoot[x_] := Block[{w}, w = w /. Solve[w^3 == 1][[3]]; If[Re[x] <

0, w*x^(1/3), x^(1/3)]]

Then

{2 - Sqrt[5], 2 + Sqrt[5]}

mycuberoot /@ %

FullSimplify[%]

Together[Plus @@ %]

{2 - Sqrt[5], 2 + Sqrt[5]}

{(-1)^(2/3)*(2 - Sqrt[5])^(1/3), (2 + Sqrt[5])^(1/3)}

{(1/2)*(1 - Sqrt[5]), (1/2)*(1 + Sqrt[5])}

1

Is there a particular reason why by default Mathematicas returns a

complex number for the cube root of a negative number or it is a matter

of choise?

Following the same procedure I prove that

(10 + 6*Sqrt[3])^(1/3) + (10 - 6*Sqrt[3])^(1/3)

is equal to 2. Indeed

{10 + 6*Sqrt[3], 10 - 6*Sqrt[3]}

mycuberoot /@ %

FullSimplify[%]

Together[Plus @@ %]

{10 + 6*Sqrt[3], 10 - 6*Sqrt[3]}

{(10 + 6*Sqrt[3])^(1/3), (-1)^(2/3)*(10 - 6*Sqrt[3])^(1/3)}

{1 + Sqrt[3], 1 - Sqrt[3]}

2

This behavior of Mathematica does not affect simplifications by e.g.

RootReduce?

I must admit that I have gaps on my knowledge in these symbolic aspects

(I start to be interested in after I try to solve the the secular

Rayleigh equation)

so more experienced members of the forum may forgive any possible

mistakes of mine!

Anyway I don't understand this difference in treating nested radicals

between literature and Mathematica.

I really appreciate any kind of insight/guideness/comments.

Regards

Dimitris

Dec 4, 2006, 6:56:51â€¯AM12/4/06

to

dimitris wrote:

> Is there a particular reason why by default Mathematicas returns a

> complex number for the cube root of a negative number or it is a matter

> of choise?

Mathematica gives you the principal value of the cubic root function.

In the case of negative numbers, it turns out that it is not a real

number.

Best regards,

Jose Carlos Santos

Dec 4, 2006, 6:57:52â€¯AM12/4/06

to

On 3 Dec 2006, at 20:26, dimitris wrote:

> Based on this reference

>

> Cardan Polynomials and the Reduction of Radicals (by T. Osler)

>

> (see also references therein)

>

> (you can download the paper here:

> http://www.jstor.org/view/0025570x/di021218/02p0059q/0?

> currentResult=0025570x%2bdi021218%2b02p0059q%2b0%2c03&searchUrl=http

> %3A%2F%2Fwww.jstor.org%2Fsearch%2FBasicResults%3Fhp%3D25%26so%

> 3DNewestFirst%26si%3D1%26Query%3DOsler

> )

>

> the following expression can be reduced to 1

>

> z = (2 + Sqrt[5])^(1/3) + (2 - Sqrt[5])^(1/3)

>

> Mathematica gives

>

> N[%]

> 1.9270509831248424 + 0.535233134659635*I

>

> This is because by default it returns a complex number for the cube

> root of a negative number

>

> List @@ z

> N[%]

>

> {(2 - Sqrt[5])^(1/3), (2 + Sqrt[5])^(1/3)}

> {0.30901699437494756 + 0.535233134659635*I, 1.618033988749895}

>

> However defining

>

> mycuberoot[x_] := Block[{w}, w = w /. Solve[w^3 == 1][[3]]; If[Re[x] <

> 0, w*x^(1/3), x^(1/3)]]

>

> Then

>

> {2 - Sqrt[5], 2 + Sqrt[5]}

> mycuberoot /@ %

> FullSimplify[%]

> Together[Plus @@ %]

>

> {2 - Sqrt[5], 2 + Sqrt[5]}

> {(-1)^(2/3)*(2 - Sqrt[5])^(1/3), (2 + Sqrt[5])^(1/3)}

> {(1/2)*(1 - Sqrt[5]), (1/2)*(1 + Sqrt[5])}

> 1

>

> Is there a particular reason why by default Mathematicas returns a

> complex number for the cube root of a negative number or it is a

> matter

> of choise?

>

> Following the same procedure I prove that

>

> (10 + 6*Sqrt[3])^(1/3) + (10 - 6*Sqrt[3])^(1/3)

>

> is equal to 2. Indeed

>

> {10 + 6*Sqrt[3], 10 - 6*Sqrt[3]}

> mycuberoot /@ %

> FullSimplify[%]

> Together[Plus @@ %]

>

> {10 + 6*Sqrt[3], 10 - 6*Sqrt[3]}

> {(10 + 6*Sqrt[3])^(1/3), (-1)^(2/3)*(10 - 6*Sqrt[3])^(1/3)}

> {1 + Sqrt[3], 1 - Sqrt[3]}

> 2

>

> This behavior of Mathematica does not affect simplifications by e.g.

> RootReduce?

>

> I must admit that I have gaps on my knowledge in these symbolic

> aspects

>

> (I start to be interested in after I try to solve the the secular

> Rayleigh equation)

> so more experienced members of the forum may forgive any possible

> mistakes of mine!

>

> Anyway I don't understand this difference in treating nested radicals

> between literature and Mathematica.

>

> I really appreciate any kind of insight/guideness/comments.

>

> Regards

> Dimitris

>

The issue of which branch of a multivalued function should be chosen

as the so called "principal branch" is, of course, a matter of

convention. Since Mathematica defines Power[x,y] as Exp[y Log[x]],

the issue of what is (-1)^(1/3) is equivalent to choosing the value

of Log[-1]. Mathematica chooses the value

Log[-1]

I*Pi

Of course once that is decided, everything else follows:

FullSimplify[(-1)^(1/3) - Exp[(1/3)*I*Pi]]

0

It seems to me (though it is not something that lies within the scope

of my "professional" interest), that before the advent of computer

algebra there no need was felt for a uniform way of choosing

principal values for various multivalued functions that occur in

algebra and analysis. In other words, the relation

x^y = Exp[y,Log[x]]

was not treated as the definition of x^y, but as a relation that held

only up to the choice of branches of the multivalued functions

involved. So it seems to me that it was always thought that the

natural choice for Log[-1] is I Pi, but before computer algebra

systems appeared it was not necessarily felt that the "principal

value" of x^(1/3) is the one that makes x^y = Exp[y,Log[x]] hold.

Note that to keep this relation true and to have the principal value

of the cube root of -1 equal to -1, one would have do choose 3 I Pi

as the principal value of Log[-1], which does not seem very natural.

RootReduce is somewhat different. Given an algebraic number it first

tries to find its minimal polynomial. This does not depend on any

particular representation of the algebraic number. After that, it

expresses the answer in terms of roots of that minimal polynomial,

using root isolation and the associated ordering. This is something

that can be done in several ways, with Mathematica alone having two

different ordering methods. Other systems use other methods, so the

the order of roots may not be the same for different systems

(although the real roots will always correspond).

It is only when algebraic numbers are expressed in terms of radicals

that conventions about "principal branches" come into play. I expect

that all symbolic algebra systems use the same convention based on

the definition x^y = Exp[y,Log[x]], but I have not tried to check

this. My own view is that radical representations of algebraic

numbers should now be considered as belonging only to the history of

mathematics.

Best regards

Andrzej

Dec 4, 2006, 7:00:54â€¯AM12/4/06

to

Search the archives for "branch cut" and/or "principal value"

z==(2 + Sqrt[5])^(1/3) + (2 - Sqrt[5])^(1/3);

#^3&/@(#-(2 + Sqrt[5])^(1/3)&/@%)

(-(2 + Sqrt[5])^(1/3) + z)^3 == 2 - Sqrt[5]

Reduce[%,z,Reals]//ToRules

{z -> 1}

z==(10 + 6*Sqrt[3])^(1/3) + (10 - 6*Sqrt[3])^(1/3)

z == (10 - 6*Sqrt[3])^(1/3) + (10 + 6*Sqrt[3])^(1/3)

#^3&/@(#-(10 + 6*Sqrt[3])^(1/3)&/@%)

(-(10 + 6*Sqrt[3])^(1/3) + z)^3 == 10 - 6*Sqrt[3]

Reduce[%,z,Reals]//ToRules

{z -> 2}

Bob Hanlon

Dec 5, 2006, 6:17:17â€¯AM12/5/06

to

That Mathematica gives value I Pi for Log[-1] is consistent with the

most common convention is that the principal argument, Arg, of a nonzero

complex number z satisfies -Pi < Arg[z] <= Pi.

most common convention is that the principal argument, Arg, of a nonzero

complex number z satisfies -Pi < Arg[z] <= Pi.

Then the usual definition of the principal logarithm, Log, is

Log[z] = Log[Abs[z]] + I Arg[z],

and the multi-valued argument function, arg, would be given as:

arg[z] = set of all Arg[z] + n 2 Pi I (n an integer)

The multi-valued logarithm, log, would be given as

log[z] = Log[Abs[z]] + I arg[z].

In this case one can define

z^w = Exp[w log[z]]

and the principal value of this as Exp[w Log[z]].

Andrzej Kozlowski wrote:

> ...

--

Murray Eisenberg mur...@math.umass.edu

Mathematics & Statistics Dept.

Lederle Graduate Research Tower phone 413 549-1020 (H)

University of Massachusetts 413 545-2859 (W)

710 North Pleasant Street fax 413 545-1801

Amherst, MA 01003-9305

Dec 6, 2006, 6:18:33â€¯AM12/6/06

to

Of course this is completely true. But perhaps it has somewhat

obscured the point I was trying to make, which was meant to answer

Dimitris's question: "why is Mathematica's convention about principal

parts in variance with those of classical algebraist's?" It seems to

me that the answer is simply that they (the classical algebraists and

perhaps some of their modern successors) did not care about the

relation with "principal values" of multivalued functions in analysis

and did not even know about them (Vieta, for one, live a whole

century before Euler). Even in certain modern texts on algebra the

relationship between radicals and logarithms is not relevant, so it

is not surprising that some authors even today may find it convenient

even today to use different "principal values' when working in a

purely algebraic setting. The point I was making about computer

algebra programs like Mathematica is that with their universal scope

why have to be consistent across both algebra and analysis in a way

that a book on, say, Galois theory, need not be.

obscured the point I was trying to make, which was meant to answer

Dimitris's question: "why is Mathematica's convention about principal

parts in variance with those of classical algebraist's?" It seems to

me that the answer is simply that they (the classical algebraists and

perhaps some of their modern successors) did not care about the

relation with "principal values" of multivalued functions in analysis

and did not even know about them (Vieta, for one, live a whole

century before Euler). Even in certain modern texts on algebra the

relationship between radicals and logarithms is not relevant, so it

is not surprising that some authors even today may find it convenient

even today to use different "principal values' when working in a

purely algebraic setting. The point I was making about computer

algebra programs like Mathematica is that with their universal scope

why have to be consistent across both algebra and analysis in a way

that a book on, say, Galois theory, need not be.

Andrzej Kozlowski

Dec 6, 2006, 6:22:35â€¯AM12/6/06

to

No disagreement!

Andrzej Kozlowski wrote:

> *This message was transferred with a trial version of CommuniGate(tm) Pro*

>> --Murray Eisenberg mur...@math.umass.edu

Dec 7, 2006, 6:39:31â€¯AM12/7/06

to

The use of principal roots has received some replies but since I liked

the Osler article above and wanted to comment I thought I'd revisit.

(1) As pointed out by others (A. Kozlowski, M. Eisenberg) use of

principal values for fractional roots means, among other things, that

Power can be defined in terms of Log, and they can share a branch cut.

This is useful in and of itself (try figuring out jumps in definite

integration with a proliferation of functions having unrelated btranch

cuts).

Another reason to like the definition a^b==Exp[b*Log[a]] is that for r>0

it makes f[x_]=(-r)^x differentiable in x. With a choice of negative

roots for x equal to 1/n, n an odd integer, such a function would fail

even to be continuous.

Another nice feature is that it becomes simple to recover "surds" (that

is, the full set of values for a radical a^(1/n)) simply by taking the

principal value and multiplying by powers of the principal nth root of

unity. Were we to have (-1)^(1/3), say, be simply -1, then one would be

forced to use explicit complex exponentials instead of root-of-unity

radicals in order to attain the principal value. But having (-1)^(1/3)

be the principal value means we can easily attain other roots such as -1

by multiplying by appropriate powers of this root of unity.

(2) Osler's paper discusses some ways to reduce certain radicals to

simpler forms. This is a special case of radical denesting. In this case

one can use polynomial algebra techniques coupled with a selection

procedure to remove parasite roots.

One example uses something resembling

(2+Sqrt[5])^(1/3) + (2-Sqrt[5])^(1/3)

EXCEPT with the convention that the cube of the negative is a negative

rather than principal value. To find the desired value one might make

new variables for radicals and polynomials to define them (in effect

giving the surds, or full sets of values), eliminate all variables other

than the one representing the value of interest, and then find the root

of the resulting polynomial (in that remaining variable) that lies in

the region of interest. For this example we might let

x=(2+Sqrt[5])^(1/3)

with defining polynomial

x^3-(2+z) (where y is given by z^2-5). Letting t be the value of

interest and continuing in this way, we would get

polys = {t-(x+y), x^3-(2+z), y^3-(2-z), z^2-5};

Now form a Groebner basis eliminating all but t, and solve for t.

roots = t /. Solve[First[GroebnerBasis[polys,t,{x,y,z}]]==0, t];

Last, select the root that is real valued.

In[36]:= Select[roots, Element[#,Reals]&]

Out[36]= {1}

Of course we could use direct built in functionality, provided we first

translate the expression as per note (1) above so that we are indeed

getting the negative root for the second summand. This summand thus

becomes (-1)^(2/3)*(2-Sqrt[5])^(1/3)

and we do

In[37]:= RootReduce[(2+Sqrt[5])^(1/3) + (-1)^(2/3)*(2-Sqrt[5])^(1/3)]

Out[37]= 1

The last example in the paper is a bit more complicated but can be

handled in exactly the same ways. It was from the dedication of a 1997

paper, commerating an anniversary of the birth of Ramanujan.

const = (32*(146410001/48400)^3 - 6*(146410001/48400));

polys = {t^6-(const+b), b^2-(const^2-1)};

roots = t /. Solve[First[GroebnerBasis[polys,t,b]]==0, t];

Now grab any root real and larger than 1.

In[41]:= Select[roots, Element[#,Reals]&&#>1&]

Out[41]= {110}

(3) Osler defines and uses "Cardan polynomials" to do radical reduction.

I cannot help but notice* that these have interesting combinatorial,

algebraic, and analytic properties, a few of which I'll describe. First

his definition:

Ca[n_,x_,y_] := Expand[2*y^(n/2)*ChebyshevT[n,x/(2*Sqrt[y])]]

For example:

In[45]:= InputForm[Ca[9,x,y]]

Out[45]= x^9 - 9*x^7*y + 27*x^5*y^2 - 30*x^3*y^3 + 9*x*y^4

We'll work with a closely related family wherein we take absolute values

of coefficients and also add a term y^n.

Da[n_,x_,y_] := Expand[-I^n*Ca[n,I*x,y]+y^n]

In[47]:= InputForm[Da[9,x,y]]

Out[47]= x^9 + 9*x^7*y + 27*x^5*y^2 + 30*x^3*y^3 + 9*x*y^4 + y^9

Finally we'll want to define some very familiar polynomials.

Ru[n_,x_,y_] := Expand[(x+y)^n]

In[49]:= InputForm[Ru[9,x,y]]

Out[49]=

x^9 + 9*x^8*y + 36*x^7*y^2 + 84*x^6*y^3 + 126*x^5*y^4 + 126*x^4*y^5 +

84*x^3*y^6 + 36*x^2*y^7 + 9*x*y^8 + y^9

Note that Ru[n,x,y] has coefficients from the nth row of the Pascal

triangle.

(A) If one looks at rows of coefficients from Da[n,x,y] as n increases

one sees an assymmetric triangle. For example, the 11th row would be

1 11 44 77 55 11 1

We can derive this from the 11th row of the Pascal triangle using a

one-sided differencing. That is, for the kth element of the mth row,

we'll take absolute value of alternating sums from the prior row, up to

the kth element.

Pa[n_,k_,0] := Binomial[n,k]

Pa[n_,k_,m_] := Abs[Sum[(-1)^(j-1)*Pa[n,j,m-1],{j,m,k}]]

Now let's look at a table of these values, for n=11.

assymmetrictable[n_] := Table[Pa[n,k,j], {j,0,(n-1)/2}, {k,0,n-1}]

In[146]:= InputForm[assymmetrictable[11]]

Out[146]//InputForm=

{{1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11},

{0, 11, 44, 121, 209, 253, 209, 121, 44, 11, 0},

{0, 0, 44, 77, 132, 121, 88, 33, 11, 0, 0},

{0, 0, 0, 77, 55, 66, 22, 11, 0, 0, 0},

{0, 0, 0, 0, 55, 11, 11, 0, 0, 0, 0},

{0, 0, 0, 0, 0, 11, 0, 0, 0, 0, 0}}

We have recovered the nontrivial "middle" coefficients for Da[11,m] (the

two "end" coefficients are simply unity) from the first nonzero element

of row m: 11, 44, 77, 55, 11.

There are nice combinatorial formulas for these coefficients; I wanted

to illustrate an algorithmic approach to recovering them from the

binomial coefficients of the Pascal triangle.

(B) The maps Ru[n,x,y] and Da[n,x,y] each posess a group invariance

property. And, returning to the original intent of the thread, thses

involve multiplication by principal roots of unity. For Ru[n,x,y] we

have the map given by the matrix rumat = {{e,0},{0,e}} where

e=(-1)^(2/n) is the principal nth root of unity.

This gives rise to a group action {z,w} --> {e*z,e*w}.

For Da[n,x,y] we use damat = {{e,0},{0,e^2}}

giving the group action {{z,w} --> {e*z,e^2*w}. So let's define these

actions explicitly.

groupAction[n_,m_,expr_,x_,y_] := expr /.

{x->x*(-1)^(2/n),y->y*(-1)^(2*m/n)}

Now we'll check the claimed invariance properties for n=9.

In[156]:= InputForm[groupAction[9,1,Ru[9,x,y],x,y]]

Out[156]//InputForm=

x^9 + 9*x^8*y + 36*x^7*y^2 + 84*x^6*y^3 + 126*x^5*y^4 + 126*x^4*y^5 +

84*x^3*y^6 + 36*x^2*y^7 + 9*x*y^8 + y^9

This is indeed simp,ly Ru[9,x,y]

In[157]:= InputForm[groupAction[9,2,Da[9,x,y],x,y]]

Out[157] x^9 + 9*x^7*y + 27*x^5*y^2 + 30*x^3*y^3 + 9*x*y^4 + y^9

which again is just Da[9,x,y]

(C) It turns out that both Ru[n,x,y] and Da[n,x,y] map the lines x+y=1

to 1. That is, if we replace y by 1-x the polynomials will evaluate to

unity.

For example:

In[161]:= Ru[9,x,1-x]

Out[161]= 1

In[162]:= Da[9,x,1-x]

Out[162]= 1

The case of Ru[n,x,1-x] should not be a surprise. After all Ru[n,x,y] is

simply (x+y)^n and this is of course 1 on x+y=1. That Da[n,x,1-x]=1 is a

bit more subtle.

Also: all polynomial maps with this property that are invariant under

the action of groupAction[n,1,...] can be obtained from straightforward

operations that amount to "tensoring", and inversion thereof, of this

basic polynomial Ru[n,x,y]. Similarly, all polynomial maps that take

x+y==1 to 1 and are invariant under groupAction[n,2,...] are obtained

from such operations on Da[n,x,y].

Clearly (he said,) there are no other 2x2 finite matrix group

representations for which there are invariant polynomials taking x+y=1

to 1 (except for those obviously equivalent to damat, as these can each

be represented in two ways).

Anyone else catch these?*

Daniel Lichtblau

Wolfram Research

*Just trolling. This stuff is far from obvious but was familiar from a

previous lifetime. "Pa" is for Pascal, "Ru" for Rudin, "Da" for

D'Angelo. Obviously Rudin's polynomials predate Rudin; his contribution

was to note that they both map the unit ball in C^2 to higher

dimensional balls (analogous to those polynomials taking x+y-1 to 1),

and satisfy a group invariance property.

As noted above, the Cardan polynomials from Osler's article are

D'Angelo's but with one term dropped and signs alternating. For odd n

D'Angelo had defined them in an article from 1988, showing how they give

maps from the unit ball in C^2 with properties similar to Rudin's, but

working with a less trivial group. He also conjectured no other such

groups could be used in this way, either for C^2 or higher dimension

domains. Some possible matrix groups had been ruled out around that time

by F. Forstneric. Ruling out the rest was interesting work, and I think

it's all covered in D'Angelo's book "Several Complex Variables and the

Geometry of Real Hypersurfaces". The cases of n even-valued were used in

an article by D'Angelo from the 90's to discuss mappings from the ball

to hyperquadrics that have similar invariance properties. We do not know

if these polynomials were defined in any context prior to that. Various

combinatorial properties of the D'Angelo polynomials appear in a number

of his articles. The connection of these to Chebyshev polynomials is

observed in an American Mathematical Monthly article by Dilcher and

Stolarsky (October 2005). That all polynomial maps invariant under that

groupAction[n,2,...] and taking x+y=1 to 1 can be obtained from

Da[n,x,y] was proved in a hammock.

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