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hey everyone
i got a little problem here i really dont get it why my fsolve give me this error

Error, (in sqrfree) argument must be a polynomial or a rational function in {cf[-2], cf[-1], cf[0], cf[1]}
these are my variables :


cf[-2], cf[-1], cf[0], cf[1]}


thats part of my code :

res1;
(34.30563197 cf[-1] - 10.13498047 cf[0] + 5.197134649 cf[1]

   - 0.4714805434) (0.3095515346 cf[-2] + 0.3822521253) - (cf[-1](2.466022067

  ) + cf[0](-0.6605923783) + cf[1](1.076415647))^2

   - cf[-1](2.466022067) - cf[0](-0.6605923783)

   - cf[1](1.076415647) - 333.1166471 cf[-1] + 186.8672744 cf[0]

   - 128.6145289 cf[1] + 0.8737683108, (-39.73727883 cf[-1]

   + 26.25682759 cf[0] - 7.289811219 cf[1] - 0.3506934780)

  (0.4161980514 cf[-1] + 0.4402863507) - (cf[-1](0.3227083347)

   + cf[0](2.517826949) + cf[1](-0.7889944208))^2

   - cf[-1](0.3227083347) - cf[0](2.517826949)

   - cf[1](-0.7889944208) - 15.03436409 cf[-1]

   - 129.5665191 cf[0] + 68.25827819 cf[1] + 0.5888637790,

  (17.36111111 cf[-1] - 29.80788311 cf[0] + 17.36111111 cf[1]

   - 0.2500000000) (0.5000000000 cf[0] + 0.5000000000)

                                                            2
   - (cf[-1](-1.833333334) + cf[0](0.) + cf[1](2.333333334))

   - cf[-1](-1.833333334) - cf[0](0.) - cf[1](2.333333334)

   + 38.17276297 cf[-1] + 34.00261850 cf[0] - 77.23526300 cf[1]

   + 0.3750000000, (-3.548332053 cf[-1] + 12.78057007 cf[0]

   - 19.34221012 cf[1] - 0.1707008416) (0.5290914191 cf[1]

   + 0.5597136492) - (cf[-1](1.060931345) + cf[0](-1.540306477)

   + cf[1](-0.2297630675))^2 - cf[-1](1.060931345)

   - cf[0](-1.540306477) - cf[1](-0.2297630675)

   - 20.59209188 cf[-1] + 28.27859544 cf[0] + 40.45630289 cf[1]

         -10                      
   + 1 10    cf[-2] + 0.2254717519


fsolve({seq(res1[v] = 0, v = 1 .. 2*N+2)})     ( N here is 1 )

why i cant get the awenser and if you can plz solve it for me

i would be happy that anyone can give me an explanation with details

tnX a lot

Arian
 

 

Using convert(f,elsymfun) when f is a symmetric polynomial will write it in terms of elementary symmetric polynomials. For example, x^2+y^2 would become (x+y)^2-2x*y. For this command not to return an error, f must be symmetric, equivalently of type symmfunc(all indeterminants in expression).

I'd like to use this in a broader sense, when there are extra variables hanging around. For example, suppose f is a polynomial in a,b,c,x,y,z, and symmetric in x,y,z, so that type(f,symmfunc(x,y,z)) returns true. Then it is still possible to write this in terms of the elementary symmetric polynomials on x,y,z (with coefficients taken being rational polynomials in a,b,c).

For example, a+x^2+y^2 can be written as a+(x+y)^2-2x*y.

Is there a command available for this? Or is there a roundabout way to make Maple forget temporarily that a,b,c are indeterminants, before putting them back in after?

Thanks

suppose I have

g := (x-4)^2+(y-6)^2-144:

e:=expand(g)

                     e := x^2+y^2-8*x-12*y-92

How do I get from the equation of e back to g?

                         

In using Isolate in RootFinding to compute roots of a real polynomial, the output contains, say, z= some number.  How to get rid of the "z =" so that I can declare that "some number" to be some variable?

Hi all I am attempting to use the randpoly feature to generate a random polynomial of 1st degree with non-zero coeffs between -2 and 2.

I have tried the following:

factor1 := randpoly(x, dense, degree = 1, coeffs = rand(-2 .. 2))

unfortunately as you will find quite often this produces the polynomial "0".  Which it is very clear this is not wanted.

 

Any ideas would be appreciated.

 

Thanks,

Mark

Hello!, Is there any package to do arithmetic operations on finite field? 
 

Example: Z(5) - over field 5: calculate with remainders:

(x^8 + 2x^4 + 1) / (x^5 + 4x^3 + x^2 + 3x +3).

 

thanks.

I have the following Polynomial F. Computing the genus shows that this curve has negative genus and thus is reducible. But using AFactor doesn't produce a factorization. Any ideas?

with(algcurves):
F := z^9+(-3/2+(3/2*I)*sqrt(3))*y^3*z^6+(-3/2-(3/2*I)*sqrt(3))*x^3*z^6+(-3/2-(3/2*I)*sqrt(3))*y^6*z^3+(-3/2+(3/2*I)*sqrt(3))*x^6*z^3+y^9+(-3/2-(3/2*I)*sqrt(3))*x^3*y^6+(-3/2+(3/2*I)*sqrt(3))*x^6*y^3+x^9-3*(x*y*z)^3:
z:=1:
genus(F, x, y);
evala(AFactors(F));

Hello!

I am working with the Maple 18.02 version. I just want want to perform a basic polynomial expansion using the command "expand" and it does not respond as it should according to what Maple Programming Help says it would. For example:

Maple Programming Help says:

I get:

.

Also, one sees that this isn't even true, as x(x+2) + 1 = x^2 +2x +1, which is not equal to x^2 + 3x +2.

Moreover, maple tells me it is equal..:

What is going on here? I woul like to get the full expanded form (without factors). Also, this is obviously not true, or maybe Maple means something else by x(x+2) +1...

Thank you!

I have a list of univariate polynomials

P:=[x^2-7*x+10, x^2, x^2+2*x+1]

How do I get the gcd of all of these?

It doesn't like it when I do the following:

igcd(P)

 

Thanks

Hi everyone, i am using Maple 18 and i have a problem in converting a equation to a nice polynomial form (a cubic equation with a form of A*x^3+B*x^2+C*x+D), can anyone please help me on the command? Thanks in advance.

My equation is "  d := s*x*(E*K*q-K*r+K*sigma[1]+r*x)*(1-x*(E*K*q-K*r+K*sigma[1]+r*x)/(K*sigma[2]*L))/(K*sigma[2])+sigma[1]*x-x*(E*K*q-K*r+K*sigma[1]+r*x)/K  " or for simplicity is

 

Can someone please teach me on the command? Really appreciate the help!

How to get homogenous expressions from symmetric polynomial.

Example. Let P = (a^2 + 2)(b^2 + 2)(c^2 + 2), we have deg(P) = 6. I want to get polynomials of degree n on A[n] with n = 0, 1, ..., 6. Specific

P = a^2b^2c^2 + 2(a^2b^2 + b^2c^2 + c^2a^2) + 4(a^2 + b^2 + c^2) + 8.

And

A[0] = 8

A[1] = 0

A[2] = 4(a^2 + b^2 + c^2)

A[3] = 0

A[4] = 2(a^2b^2 + b^2c^2 + c^2a^2)

A[5] = 0

A[6] = a^2b^2c^2

Thank you very much.

Dragilev:=proc(Polynomials::depends(list(ratpoly(integer,Variables))),Variables::list(symbol),DEvar::symbol,DEsuffix::string)

The above procedure parameter Polynomials accepts a list of polynomials containing indeterminates contained in parameter Variables, but also accepts simple arithmetic expressions such as 34.

Is there any parameter qualifying coding which will only accept polynomials containing one or more of the indeterminates passed in parameter Variables?

Hi!

I a have a question about factorizing real polynomials.

Suppose I have a real polynomial p(x) with integer coefficients. If the degree of p(x) is less than or equal to 4, we can factorize it into linear radical factors. On the other hand, if we require the factorization to be real, theoretically we can factorize it into linear and irreducible quadratic factors.

My question is, if the input p(x) is real polynomial with integer coefficients, is there any Maple function that can give me factorization output with real linear and irreducible quadratic factors, with radical coeffs?

For example, I tried q := 20*x^3+10*x^2+4*x+1, it has one real root and 2 complex roots. I want a factorization like q(x) = 20*(x-r1)*(a*x^2 + b*x + c), with r1, a, b, c, all real radicals.

I compared functions: factors(), solve(), sqrfree(), Splits(), and none of them give what I want.

factors(q) gives: 
[20, [[x^3+(1/2)*x^2+(1/5)*x+1/20, 1]]]

 

factors(q, real)  gives: 
[20., [[x+.3423840948583691316993036540027816871936619136844427977504078911, 1], [x^2+.1576159051416308683006963459972183128063380863155572022495921089*x+.1460348209828001458360112632660894203743660942160039146818509889, 1]]]

solve(q)   gives:
-(1/30)*(350+105*sqrt(15))^(1/3)+7/(6*(350+105*sqrt(15))^(1/3))-1/6, (1/60)*(350+105*sqrt(15))^(1/3)-7/(12*(350+105*sqrt(15))^(1/3))-1/6+(1/2*I)*sqrt(3)*(-(1/30)*(350+105*sqrt(15))^(1/3)-7/(6*(350+105*sqrt(15))^(1/3))), (1/60)*(350+105*sqrt(15))^(1/3)-7/(12*(350+105*sqrt(15))^(1/3))-1/6-(1/2*I)*sqrt(3)*(-(1/30)*(350+105*sqrt(15))^(1/3)-7/(6*(350+105*sqrt(15))^(1/3)))

simplify(convert(Splits(q,x),radical))    gives:
[20, [[(1/30)*(350+105*sqrt(5)*sqrt(3))^(1/3)+1/6-7/(6*(350+105*sqrt(5)*sqrt(3))^(1/3))+x, 1], [-(1/60)*(I*sqrt(3)*(350+105*sqrt(5)*sqrt(3))^(2/3)+(35*I)*sqrt(3)+(350+105*sqrt(5)*sqrt(3))^(2/3)-60*x*(350+105*sqrt(5)*sqrt(3))^(1/3)-10*(350+105*sqrt(5)*sqrt(3))^(1/3)-35)/(350+105*sqrt(5)*sqrt(3))^(1/3), 1], [(1/60*I)*sqrt(3)*(350+105*sqrt(5)*sqrt(3))^(1/3)+(7/12*I)*sqrt(3)/(350+105*sqrt(5)*sqrt(3))^(1/3)-(1/60)*(350+105*sqrt(5)*sqrt(3))^(1/3)+1/6+7/(12*(350+105*sqrt(5)*sqrt(3))^(1/3))+x, 1]]]

None of them give me what I want. Is there any build-in function that can help me do that?

Thanks!

William

 

after solve a set of equations, i got a set of solutions, [A,B,C], if i remove one of solution [C]

is it possible to find this removed solution from solutions set [A,B]

how to convert a^2*b+c to func2(func1(func1(abc[1],abc[1]),abc[2]),abc[3])

when i use custom function func2 to represent plus, func1 to represent multiply

input 3 parameters,

one is a^2*b + c one is [func1, func2] and second is [abc[1],abc[2],abc[3]] corresponding to a, b, c
a^2*b + c = func2(func1(func1(abc[1],abc[1]),abc[2]),abc[3]);

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