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separable form should be (x - something in terms of x1,x2,x3,x4)*(x- something in terms of x1,x2,x3,x4)*(x -something in terms of x1,x2,x3,x4 )*(x - something in terms of x1,x2,x3,x4)

factor(f, x);
Error, (in factor) 2nd argument, x, is not a valid algebraic extension

how to factor into separable form?

 

ferrai := -x1^3*x2*x3*x4-x1^2*x2^2*x3^2-x1^2*x2^2*x4^2-x1^2*x3^2*x4^2-x1*x2^3*x3*x4-x1*x2*x3^3*x4-x1*x2*x3*x4^3-x2^2*x3^2*x4^2+x1^2*x2*x3*y+x1^2*x2*x4*y+x1^2*x3*x4*y+x1*x2^2*x3*y+x1*x2^2*x4*y+x1*x2*x3^2*y+x1*x2*x4^2*y+x1*x3^2*x4*y+x1*x3*x4^2*y+x2^2*x3*x4*y+x2*x3^2*x4*y+x2*x3*x4^2*y-x1*x2*y^2-x1*x3*y^2-x1*x4*y^2-x2*x3*y^2-x2*x4*y^2-x3*x4*y^2+y^3;
coeff(ferrai, y^3);
coeff(ferrai, y^2);
coeff(ferrai, y);
res := simplify(ferrai - coeff(ferrai, y^3)*y^3 - coeff(ferrai, y^2)*y^2 - coeff(ferrai, y)*y);

c2 := -coeff(ferrai, y^2)/coeff(ferrai, y^3);
sys1 := c1*c3 - 4*c4 = coeff(ferrai, y)/coeff(ferrai, y^3);
sys2 := -c3^2-(c1^2)*c4+4*c2*c4 = res/coeff(ferrai, y^3);

diff(lhs(sys1),c1) = diff(rhs(sys1),c1);
diff(lhs(sys1),c3) = diff(rhs(sys1),c3);
diff(lhs(sys1),c4) = diff(rhs(sys1),c4);

diff(lhs(sys2),c1) = diff(rhs(sys2),c1);
diff(lhs(sys2),c3) = diff(rhs(sys2),c3);
diff(lhs(sys2),c4) = diff(rhs(sys2),c4);

c1sol := solve(diff(lhs(sys2),c4) = diff(rhs(sys2),c4), c1)[1];
indets(subs(c1=c1sol,sys1));
indets(subs(c1=c1sol,sys2));
indets([subs(c1 = c1sol, sys1), subs(c1 = c1sol, sys2)]);
result := solve({subs(c1 = c1sol, sys1), subs(c1 = c1sol, sys2)},{c3,c4}, explicit);

c1sol := solve(diff(lhs(sys2),c4) = diff(rhs(sys2),c4), c1)[1];
c2sol := -coeff(ferrai, y^2)/coeff(ferrai, y^3);
c3sol := rhs(result[1][1]);
c4sol := rhs(result[1][2]);
f := x^4-c1sol*x^3+c2sol*x^2-c3sol*x+c4sol;
aa := factor(f);
aa := factor(f,x);

how to calculate the polynomial map for a system of  polynomials

assume system of polynomial is in terms of a,b,c

how to find polynomial map

(r - something in terms of a,b,c)

(u - something in terms of a,b,c)

(v - something in terms of a,b,c)

 

Hi!

I am comptuing the eigenvalues and the characteristic polynomial of a 8 by 8 symmetric matrix, say M. Thus, we define the matrix M, and compute its charast. plynm. by

 

 

and its eigenvalues with the command

 

 

Well, Maple returns the charast. polynm. an dthe eigenvalues. But, if we compute p(E[k]), for k=1,...,8, thats is, the values of the polynomial p(x) in the eingenvalues, Maple not turns cero!!! I'm really confused ... anyone know what could be happening?

 

Maple attached file with this example. Thank very much for your help!!

 

Download exam_eigenvalues.mw

Hi all

Assume that we have folowings:

Assume that we devide [0,1] to N subintervals and in each subinterval we have:

Also we want to approximate arbitrary function x(t) with following manner:

 

How can we produce these basic polynomials?

Thanks in advance

 

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

 

abc
bca

if starting position is at first position,

i have to write a very long substitution, and need a temporary variable t and t2

subs(t2=c,subs(t=a,subs(a=b,subs(b=t2,subs(c=t,a*b*c+a*b)))));

if starting position is at second position, it will be

abc
acb

is there any simple method to do this circular shift of variables for a polynomial

 

 

 

Hello everyone!

Suppose, we have a differential polynomial

P := u(x) + (D@@2)(u)(x)

Given this, I am looking for a procedure which gives coefficients depending on the order given as input, for example, lets say, procedure name is fun_coeff which depends on two parameters, original polynomial P and the order n, then

fun_coeff(P, 3) should give [1, 0, 1, 0]

where each entry corresponds to the coeff of

 [u(x), (D@@1)(u)(x), (D@@2)(u)(x), (D@@3)(u)(x)]

in the polynomial, similarly

fun_coeff(P,4) should give [1, 0, 1, 0, 0]

corresponding to

 [u(x), (D@@1)(u)(x), (D@@2)(u)(x), (D@@3)(u)(x), (D@@4)(u)(x) ]

Thank you all for your time :)

hi.for linear coupling equation

dsys3 := {-72.49829200*(diff(f1(x), x, x))+0.8377580411e-2*(diff(f2(x), x))-8.873545400*10^9*(diff(f3(x), x))+2.114533515*10^18*f1(x), -878.8477313*(diff(f2(x), x, x))+1.590065471*10^20*f2(x)-7.353421206*10^(-26)*(diff(f3(x), x, x))+4.891459762*10^10*f3(x), 4.027667395*10^(-20)*(diff(f3(x), x, x, x, x))-0.6274394007e-2*(diff(f3(x), x, x))+8.873545401*10^9*(diff(f1(x), x))-7.352113720*10^(-26)*(diff(f2(x), x, x))+4.904509456*10^10*f2(x)+1.208381068*10^19*f3(x)-2.499990383*10^26*omega*(diff(f3(x), x, x))}

by assuming and to expand these functions(f1-f2-f3) in polynomial form(e.g. Chebyshev, power polynomials, Legendre and etc).:

f1(x):=(∑)H[i] *(e)^(lambda[i] *x); f2(x):=(∑)alpha[i]*H[i] *(e)^(lambda[i] *x);f3(x):=(∑)GAMMA[i]*H[i] *(e)^(lambda[i] *x)

 

how i detemine lambda[i] which are roots of the  characteristic equations?in other word how i can build characteristic relation for coupling equation?

2)how i can gain value for alpha[i] and GAMMA[i]

by using equation

Q1 := subs(x = 0, sum(H[i]*exp(lambda[i]*x), i = 1 .. 8)); Q2 := subs(x = L, sum(H[i]*exp(lambda[i]*x), i = 1 .. 8)); Q3 := subs(x = 0, sum(alpha[i]*H[i]*exp(lambda[i]*x), i = 1 .. 8)); Q4 := subs(x = L, sum(alpha[i]*H[i]*exp(lambda[i]*x), i = 1 .. 8)); Q5 := subs(x = 0, sum(GAMMA[i]*H[i]*exp(lambda[i]*x), i = 1 .. 8)); Q6 := subs(x = L, sum(GAMMA[i]*H[i]*exp(lambda[i]*x), i = 1 .. 8)); M := diff(sum(GAMMA[i]*H[i]*exp(lambda[i]*x), i = 1 .. 8), x); Q7 := subs(x = 0, M); Q8 := subs(x = L, M)

????

thanks...

 

chebyshev.mw

I want to obtain polynomial representation of my data:

with(Statistics):

X:=Vector([huge data package-X],datatype=float):

Y:=Vector([huge data package-Y],datatype=float):

NonlinearFit(c*t^2+b*t+a, X, Y, t)

but I can't see any result. What's the problem?

My worksheet: He_p=f(t).mw

I have a polynomial expression that I would like to cast into a specific form. The expression is

and I know that it can be simplified into a form involving squares of (A[Qi]-Pi). It is trivial to do this on paper; how can I convince Maple to do this.

The solution I came up with was to use mtaylor and expand about the forms I know to be there:

mtaylor((3),[A[Q1]=P1,A[0]=P2*rho/(rho+Q1),A[Q3]=P3],6);

which is what I want (close to, anyway). Now, I consider this to be a bit of a dirty trick that works here as the expression is simple and no higher-order terms are present so in fact the solution is exact. But, are there methods along simplify and friends that can do this? I have not been successfull with those...

This is a part of a much longer worksheet and part of a lecture, so I need Maple to be able to do this. The mtaylor trick works, but I would not want to miss an obvious approach that may work where mtaylor would get confused.

Thanks,

M.D.

test.mw

I am not seeing any reference in help to TRDPolynomial_ring in PolynomialRing function in RegularChains. Though if I debug it, I can get in it. Is it that it is part of kernel ?

how to round all coefficients for a polynomial in short code way?

In the running of an example I faced to computation of radical ideal of the following ideal:

<-c*m*u+d*c*n+m*b*v+m*c*t>

 

I used from Radical command in PolynomialIdeals package. But I dno't now why it's computation is very hard and Time-consuming?

What I have to do? I think there is a bug, since this ideal is simple, apparently.

Of course it's easy to write a procedure to compute the ith Horner polynomial of a polynomial but I'm carious to know if there is any built-in command for doing this, I searched in help of Maple and found several ways but they convert or rewrite the input polynomial in Horner form so the output is last Horner polynomial which is equal to the input polynomial with this difference that it is written in a kind of factorization form which shows the Horner structure and ofcourse with looking at it you can read other Horner polynomials too but why not having exactly the ith Horner for the output? So is there any other command with output being the ith Horner?

I have a multivariate polynomial in x and y. How can I firstly collect the terms with respect to the powers of x and y and then simplifying their coefficients.

restart

with(RealDomain):

with(LinearAlgebra):

A := Matrix(5, 5, {(1, 1) = 0, (1, 2) = 1, (1, 3) = 1, (1, 4) = 1, (1, 5) = 1, (2, 1) = 1, (2, 2) = 0, (2, 3) = c__1, (2, 4) = y, (2, 5) = c__2, (3, 1) = 1, (3, 2) = c__1, (3, 3) = 0, (3, 4) = c__3, (3, 5) = x, (4, 1) = 1, (4, 2) = y, (4, 3) = c__3, (4, 4) = 0, (4, 5) = c__4, (5, 1) = 1, (5, 2) = c__2, (5, 3) = x, (5, 4) = c__4, (5, 5) = 0})

Matrix(%id = 4510803138)

(1)

B := Determinant(A)

-2*c__1^2*c__4+2*c__1*c__2*c__3+2*c__1*c__2*c__4-2*c__1*c__2*x+2*c__1*c__3*c__4-2*c__1*c__3*y-2*c__1*c__4^2+2*c__1*c__4*x+2*c__1*c__4*y+2*c__1*x*y-2*c__2^2*c__3-2*c__2*c__3^2+2*c__2*c__3*c__4+2*c__2*c__3*x+2*c__2*c__3*y-2*c__2*c__4*y+2*c__2*x*y-2*c__3*c__4*x+2*c__3*x*y+2*c__4*x*y-2*x^2*y-2*x*y^2

(2)

``

 

Download A.mw

I can calculate the quotient and remainder (r1) of a polinom (p1) divided by another (p2).

Unfortunately, I can't figure out how to divide r1 with p2, so that I get negative powers as well.

For example:

p1 = x^2+3

p2 = x+2

(x^2+3):(x+2) = x, and the remainder is: (-2x+3)

-----------

(-2x+3):(x+2) = -2, and the remainder is: (-1)

-----------

(-1):(x+2) = -x^-1, and the remainder is: (2x^-1)

-----------

(2x^-1):(x+2) = 2x^-2

...

As a result I get:

p1/p2 = x + -2 - x^-1 + 2x^-2 and so on...

How can I achieve this?

 

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