Factor.....

[gtsa...@auth.gr]

How can i factor A^4 + 3 + y^2 (A,y reals) for example with
Mathematica.....?

(
In order to have:
(A^2 + SQRT(y^2+3) + A·SQRT(2*SQRT(y^2+3))) · (A^2 + SQRT(y^2+3) -
A·SQRT(2*SQRT(y^2+3)))
)

(
OR:
(y^2 - i·SQRT(A^4+3)) · (y^2 + i·SQRT(A^4+3))
)

Thanks....

[dimitris]

Unfortunately I could not get what you want with the built-in function Factor.
I hope someone else succeed.

Anyway here is my solution

Clear[f, x, y, myfactor]

myfactor[f_, x_] := Times @@ Apply[Plus, Solve[f == 0, x], 2]

f[A_, y_] := A^4 + 3 + y^2

(myfactor[f[A, y], #1] & ) /@ Variables[f[A, y]]

{(A - (-3 - y^2)^(1/4))*(A - I*(-3 - y^2)^(1/4))*(A + I*(-3 -
y^2)^(1/4))*(A + (-3 - y^2)^(1/4)),
(-Sqrt[-3 - A^4] + y)*(Sqrt[-3 - A^4] + y)}

Regards
Dimitris

[dh]

Hi,

you have to tell Mathematica what form you want.

Assume e.g. that we want the form (A^2 + a1 A + a0)(A^2 + b1 A + b0).

Then we expand, equate the coefficients of A and solve for a0,a1,b0,b1:

r1=CoefficientList[(A^2+A a1+a0)(A^2+b1 A+ b0)//Expand,A]

r2=CoefficientList[A^4+3+y^2,A]

Solve[Thread[r1==r2],{a0,a1,b0,b1}]

this gives several possible expansions.

Daniel

[ab_...@prontomail.com]

Along the same lines, we can define an extended SolveAlways function
taking a third argument that specifies which parameters to solve for:

mySolveAlways[$Leq_, $Lvar_, $Lpar_ : {}] := Module[
{Leq = $Leq, Lvar = $Lvar, Lpar = $Lpar, ans},
{Leq, Lvar, Lpar} = If[ListQ@ #, #, {#}]& /@
{Leq, Lvar, Lpar};
ans = Solve[!Eliminate[!And @@ Leq, Lvar], Lpar];
Select[ans, FreeQ[#, Alternatives @@ Lvar]&] /;
Head@ ans =!= Solve
]

In[2]:= (A^2 + a1*A + b1)*(A^2 + a2*A + b2) /.
Last@ mySolveAlways[
A^4 + 3 + y^2 == (A^2 + a1*A + b1)*(A^2 + a2*A + b2),
A, {a1, b1, a2, b2}]

Out[2]= (A^2 - Sqrt[2]*A*(3 + y^2)^(1/4) + Sqrt[3 + y^2])*(A^2 +
Sqrt[2]*A*(3 + y^2)^(1/4) + Sqrt[3 + y^2])

Maxim Rytin
m...@inbox.ru

[Paul Abbott]

In article <eicnmq$g3i$1...@smc.vnet.net>, gtsa...@auth.gr wrote:

> How can i factor A^4 + 3 + y^2 (A,y reals) for example with
> Mathematica.....?
>

> In order to have:
> (A^2 + SQRT(y^2+3) + A·SQRT(2*SQRT(y^2+3))) · (A^2 + SQRT(y^2+3) -
> A·SQRT(2*SQRT(y^2+3)))

Seeking quadratic factors (using coercion into Series),

rs = Solve[(a^2 + a c + b)(a^2 + a e + d) == (a^4 + y^2 + 3) + O[a]^5,
{b, c, d, e}]

we simplify the result using the fact that y is real.

rs = Simplify[rs, Element[y, Reals]]

One obtains 6 solutions as three conjugate pairs. Here are the three
quadratic factorizations:

(a^2 + a c + b)(a^2 + a e + d) /. rs // Union

{
(a^2 - I Sqrt[2] a (3 + y^2)^(1/4) - Sqrt[3 + y^2])*
(a^2 + I Sqrt[2] a (3 + y^2)^(1/4) - Sqrt[3 + y^2]),

(a^2 - I Sqrt[3 + y^2]) (a^2 + I Sqrt[3 + y^2]),

(a^2 - Sqrt[2] a (3 + y^2)^(1/4) + Sqrt[3 + y^2])*
(a^2 + Sqrt[2] a (3 + y^2)^(1/4) + Sqrt[3 + y^2])
}

The last solution is the one you were looking for.

> OR:
> (y^2 - i·SQRT(A^4+3)) · (y^2 + i·SQRT(A^4+3))

You should have y not y^2 here. Seek quadratic factors of the form

Solve[(y + b) (y + c) == (a^4 + y^2 + 3) + O[y]^3, {b, c}]

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

[Andrzej Kozlowski]

In fact, there is no need to define any new function, for this
particular problem anyway, since you can do this as easily using the
already existing ones (and you get more answers as a bonus ;-))

(A^2 + a1*A + b1)*(A^2 + a2*A + b2) //.
{ToRules[Reduce[LogicalExpand[A^4 + 3 + y^2 == (A^2 + a1*A + b1)*(A^2
+ a2*A + b2) + O[A]^4],
{a1, a2, b1, b2}]]}

{(A^2 - Sqrt[-y^2 - 3])*(A^2 + Sqrt[-y^2 - 3]), (A^2 - Sqrt[-y^2 - 3])
*(A^2 + Sqrt[-y^2 - 3]),
(A^2 - Sqrt[2]*(y^2 + 3)^(1/4)*A + Sqrt[y^2 + 3])*(A^2 + Sqrt[2]*(y^2
+ 3)^(1/4)*A + Sqrt[y^2 + 3]),
(A^2 - I*Sqrt[2]*(y^2 + 3)^(1/4)*A - Sqrt[y^2 + 3])*(A^2 + I*Sqrt[2]*
(y^2 + 3)^(1/4)*A -
Sqrt[y^2 + 3]), (A^2 - I*Sqrt[2]*(y^2 + 3)^(1/4)*A - Sqrt[y^2 + 3])*
(A^2 + I*Sqrt[2]*(y^2 + 3)^(1/4)*A - Sqrt[y^2 + 3]),
(A^2 - Sqrt[2]*(y^2 + 3)^(1/4)*A + Sqrt[y^2 + 3])*(A^2 + Sqrt[2]*(y^2
+ 3)^(1/4)*A + Sqrt[y^2 + 3])}

ExpandAll[%]

{A^4 + y^2 + 3, A^4 + y^2 + 3, A^4 + y^2 + 3, A^4 + y^2 + 3, A^4 +
y^2 + 3, A^4 + y^2 + 3}

Andrzej Kozlowski
Tokyo, Japan

On 6 Nov 2006, at 16:52, ab_...@prontomail.com wrote:

> Along the same lines, we can define an extended SolveAlways function
> taking a third argument that specifies which parameters to solve for:
>
> mySolveAlways[$Leq_, $Lvar_, $Lpar_ : {}] := Module[
> {Leq = $Leq, Lvar = $Lvar, Lpar = $Lpar, ans},
> {Leq, Lvar, Lpar} = If[ListQ@ #, #, {#}]& /@
> {Leq, Lvar, Lpar};
> ans = Solve[!Eliminate[!And @@ Leq, Lvar], Lpar];
> Select[ans, FreeQ[#, Alternatives @@ Lvar]&] /;
> Head@ ans =!= Solve
> ]
>
> In[2]:= (A^2 + a1*A + b1)*(A^2 + a2*A + b2) /.
> Last@ mySolveAlways[
> A^4 + 3 + y^2 == (A^2 + a1*A + b1)*(A^2 + a2*A + b2),
> A, {a1, b1, a2, b2}]
>
> Out[2]= (A^2 - Sqrt[2]*A*(3 + y^2)^(1/4) + Sqrt[3 + y^2])*(A^2 +
> Sqrt[2]*A*(3 + y^2)^(1/4) + Sqrt[3 + y^2])
>
> Maxim Rytin
> m...@inbox.ru
>
> dh wrote:

[ab_...@prontomail.com]

In this case equating series terms gives the same result as using
mySolveAlways:

In[2]:= expr = A^4 + 3 + y^2 - (A^2 + a1*A + b1)*(A^2 + a2*A + b2);

In[3]:= Lsol1 = Solve[expr + O[A]^5 == 0, {a1, b1, a2, b2}];
Lsol2 = List@ ToRules@ Reduce[expr + O[A]^5 == 0, {a1, b1, a2, b2},
Backsubstitution -> True];
Lsol3 = mySolveAlways[expr == 0, A, {a1, b1, a2, b2}];
SameQ @@ Map[Sort, {Lsol1, Lsol2, Lsol3}, 2]

Out[6]= True

I suppose the advantage of the algebraic method is that you don't have
to work out the expansion order. Also sometimes it will succeed when a
series expansion isn't possible, as with mySolveAlways[x^p == x, x]. On
the other hand, the series method is useful for working out identities
like Sin[x] == a1*E^(b1*x) + a2*E^(b2*x).

Maxim Rytin
m...@inbox.ru