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2017-04-14 16:15:59 +0200 | answered a question | Compute elementary Symmetrical functions on roots of a polynomial To compute the elem. sym pol. on some (calculated) roots
with then these coefficients appear in the form of a sum of elementary functions Then I identify "by hand" the |
2017-04-14 11:38:00 +0200 | received badge | ● Associate Editor (source) |
2017-04-14 10:44:00 +0200 | answered a question | How to find the multiplicative order of an element in a quotient ring over finite field ? You have (X+1)^2k = X^2k + 1 + 2(...) so ( X+1 )^2k = 0 mod ( X^2k +1 , 2 ) In other words (X+1)^n is nilpotent in any ring F2[X]/(X^n+1) when n is even, so it remains to concentrate on odd values of n try something like this (N rather small...) for i in range(2,10): A=R.quo(x^(2*i+1)+1);j=2 while A((x+1)^j)<>A(x+1): j=j+1 |
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2017-04-13 20:56:00 +0200 | commented answer | Compute elementary Symmetrical functions on roots of a polynomial Yes but this is not only symbolic computation as I have some calculated values for
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2017-04-13 16:57:55 +0200 | asked a question | Compute elementary Symmetrical functions on roots of a polynomial How to compute the elementary symmetric functions |
2016-02-13 12:56:43 +0200 | answered a question | Constraints on polynomial coefficients When I use several ideals, I don't have commutativity. Is this normal? R.<x,a,b>=QQ[] P1=(X-a)*(X-b) I=R.ideal(a+b) J=R.ideal(a*b) A=P.mod(I).mod(J) => X^2 - b^2 B=P.mod(J).mod(I) => X^2 A==B; => False |
2016-02-13 12:41:44 +0200 | commented answer | Constraints on polynomial coefficients When I use several ideals, I don't have commutativity. Is this normal? R.<x,a,b>=QQ[] P1=(X-a)*(X-b) I=R.ideal(a+b) J=R.ideal(a*b) A=P.mod(I).mod(J) => X^2 - b^2 B=P.mod(J).mod(I) => X^2 A==B; => False |
2016-02-12 11:30:13 +0200 | answered a question | Constraints on polynomial coefficients I still have 2 concerns if you can help! Q.<x,a>=QQ[]; P = 5 * X^2 + 3 * a^2 * X + 5 - a^2 If: I=Q.ideal(a^2-5) => 5X^2 + 15X # OK However, this doesn't work modulo any integer: P.mod(n) = 0 whatever n. Merci! Secondly, when I use several ideals, I don't have commutativity. Is this normal? R.<x,a,b>=QQ[] P1=(X-a)*(X-b) I=R.ideal(a+b) J=R.ideal(a*b) A=P.mod(I).mod(J) => X^2 - b^2 B=P.mod(J).mod(I) => X^2 A==B; => False |
2016-02-12 11:28:39 +0200 | answered a question | Variables in a polynomial I still have one concern if you can help! Q.<x,a>=QQ[]; P = 5 * X^2 + 3 * a^2 * X + 5 - a^2 If: I=Q.ideal(a^2-5) => 5X^2 + 15X # OK However, this doesn't work modulo any integer: P.mod(n) = 0 whatever n. Merci! |
2016-02-11 19:07:04 +0200 | commented answer | Constraints on polynomial coefficients Yes, you are perfectly right. I answered this because I was using coefficients defined on a number field, and the "mod" didn't work because I firstly defined the number field and then the ring. I discovered that if instead you define the ring and then the number field as an extension, the method works, at least for what I tried so far. I didn't have time to explain more my comment in between, and I thank you for your help. |
2016-02-11 16:12:25 +0200 | answered a question | Variables in a polynomial Perhaps this example might be more understandable. Take a field " E " generated by " r " (here, "r" is a root of X^3 - 2 ) and 3 variables " b0 , b1 , b2 ". I produce the element " A = b0 + r * b1 + r^2 * b2 " and it's square: A^2 = b0^2 + (2r) * b0 * b1 + (r^2) * b1^2 + (2r^2) * b0 * b2 + 4 * b1 * b2 + (2*r) * b2^2 Notice that "r" is in a number field. I want to express A and A^2 in terms of the basis "( 1 , r , r^2 )" so that the result becomes: A^2 = r^2 ( b1^2 + 2 b0 * b2 ) + r ( 2 b0 * b1 + 2 b2^2 ) + ( b0^2 + b2^2 + b1 * b2 ) I edit as I seem to have found a solution. I need to define the number field as an extension of the ring of coefficients and the other way round! Good. B=PolynomialRing(E,3,'b'); v=B.gens() E.<r>=B.extension(x^3-2) A^2=sum( [v[i] * r^i for i in range(3) ] ) gives me the solution: (b1^2 + 2b0b2)r^2 + (2b0b1 + 2b2^2)r + b0^2 + 4b1*b2 result expressed in terms of powers of "r". |
2016-02-11 10:44:44 +0200 | commented answer | Variables in a polynomial Perhaps this example might be more understandable. Take a field " E " generated by " r " (here, "r" is a root of X^3 - 2 ) and 3 variables " b0 , b1 , b2 ". I produce the element " A = b0 + r * b1 + r^2 * b2 " and it's square: A^2 = b0^2 + (2r) * b0 * b1 + (r^2) * b1^2 + (2r^2) * b0 * b2 + 4 * b1 * b2 + (2*r) * b2^2 I want to express A and A^2 in terms of the basis "( 1 , r , r^2 )" so that the result becomes: A^2 = r^2 ( b1^2 + 2 b0 * b2 ) + r ( 2 b0 * b1 + 2 b2^2 ) + ( b0^2 + b2^2 + b1 * b2 ) I haven't found a method so far. E.<r>=NumberField(x^3-2) B=PolynomialRing(E,3,'b');v=B.gens() A=sum( [v[i] * r^i for i in range(3) ] ) |
2016-02-11 09:24:16 +0200 | commented answer | Constraints on polynomial coefficients Perhaps I wasn't clear at all. I have a polynomial where the coefficients are parameters. I take the example of P = ( X - a0) * (X -a1) = X^2 + (-a0 - a1)*X + a0 * a1 I want to impose an equation on the parameters who belong to a field ("QQ" here), say "a0 + a1 = 0". The result must be: P = X^2 + a0 * a1 A=PolynomialRing(QQ,2,'a') v=B.gens() R.<x>=PolynomialRing(A) P=R( ( X - v[0] ) * ( X - v[1] ) ) ; # P = ( X - a0) * (X -a1) |
2016-02-10 21:29:23 +0200 | answered a question | Constraints on polynomial coefficients Observe that the result is " X^2 - b^2 " while in case of " a+b=0", you should obtain " X^2 - a*b ". Am still trying ! |
2016-02-09 19:24:49 +0200 | asked a question | Constraints on polynomial coefficients Given a polynomial P=(X-a)(X-b)(X-c) over Q[a,b,c][X] , how can I force the coefficients to fullfill constraints such as elementary symetric function a+b+c=0 or abc=2, some polynomial constraint on coefficients. It is a bit like using coefficients on a number field with the constraint of a minimal polynomial but with different constraints here. A simple example: if P = ( X - a ) ( X - b ) and a + b = 0 then I want to find P = X^2 - ab |
2016-02-08 15:16:29 +0200 | answered a question | Variables in a polynomial modified below |
2016-02-08 13:48:10 +0200 | asked a question | Variables in a polynomial I edited my question and put the answer I found, so this solves my problem! My problem was: take a field " E " generated by " r " (here, "r" is a root of X^3 - 2 ) and 3 variables " b0 , b1 , b2 ". I want to produce the element " A = b0 + r * b1 + r^2 * b2 " and it's square: A^2 = b0^2 + (2r) * b0 * b1 + (r^2) * b1^2 + (2r^2) * b0 * b2 + 4 * b1 * b2 + (2*r) * b2^2 and express A and A^2 in terms of the basis "( 1 , r , r^2 )" so that the result becomes: A^2 = r^2 ( b1^2 + 2 b0 * b2 ) + r ( 2 b0 * b1 + 2 b2^2 ) + ( b0^2 + b2^2 + b1 * b2 ) The solution was simply to define the number field as an extension of the ring of coefficients and the other way round! Good! B=PolynomialRing(E,3,'b'); v=B.gens() E.<r>=B.extension(x^3-2) A^2=sum( [v[i] * r^i for i in range(3) ] ) gives me the solution: (b1^2 + 2b0b2)r^2 + (2b0b1 + 2b2^2)r + b0^2 + 4b1*b2 result expressed in terms of powers of "r". |
2015-12-17 01:57:53 +0200 | commented question | multiplication matrix in number field Sorry for late response. Very interresting too & thanks as I alos need this approach for different usage. I might come back! |
2015-11-02 21:28:32 +0200 | received badge | ● Scholar (source) |
2015-11-02 21:28:17 +0200 | answered a question | multiplication matrix in number field Great & thanks for your prompt answer! Indeed, Sage kept using the primitive element z of K instead of k.gens() within an order as defined on L, and I couldn't go around that. I will have a closer look to the number field element page and the .matrix(). to understand your answer. |
2015-11-02 19:52:13 +0200 | received badge | ● Student (source) |
2015-11-02 19:40:47 +0200 | edited question | multiplication matrix in number field Hello
Given two integral numbers, say since Thus, given a basis (1,a,a²,b,ab,a²b) for Q(a,b), one can determine the matrix "by hand": each column is the image of each element times
One can define the extension => for example: how make the vector |