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I've developed a framework in Maple for playing with cyclic division algebras. I have methods for multiplication and passing to and from:

The standard basis over the base field in terms of x^i*y^j and:
The standard matrix basis for the central simple algebra when taking coefficients in the splitting field.

I have it done in the case of degree 7, but could generalise it to any prime degree if this is of use to anyone.

I am building an algebra 1 course for my school, and was wanting to create simple randomly generating practice sets for graphing points / lines / etc.

However, the Sketch answer type doesn't allow for any variable input like the other questions do. Is there a work around? 

Presuming a randomly generated $x and $y, is there any way to check to see if a student has plotted the point ($x, $y)?

Lets say you a country where the official government debt (not adjusted for who owns the debt) is 30% of GDP. The official government debt consist of 50% government bonds and the central banks owns 40% of these government bonds. Since the government finance the centralbank we have to adjust the official government debt and remove the 40% of the government bonds that the centralbank own to get the real government debt expresed as % of GDP. In my initial calculation I simply did like this:

Real government debt expressed as % of GDP = 0.3*(1-0.4) =18%

However, that such number does not take into consideration that the official government debt is only made up of 50% government bonds. I tried to play around with different types of equations but I found none that makes any logical sense and provides me with a reasonable number. I would expect the real government debt to around 23% of GDP. This might be a simple question but I must admit I am stuck and I am not making any progress.  Any help would be appreciated.

 

I'm creating a randomly generated question bank that generates the following STYLE or problem:

12x-4y2
3x6y-5

I'm currently trying to use the answer type "formula without simplification," as I'd like to avoid the students putting the questions in as the answer, and this has been driving me crazy for hours now.

I have tried the "maple" function to simplify. E.g.:

$ANS = maple("( $C1*( $V1^$A1 )*( $V2^$B1 ) ) / ( $C2*( $V1^$A2 )*( $V2^$B2 ) )");

But it always throws an error.

I have simply done the math in the algorithm section, so you end up with an answer variable like this:

$ANS = -4.0*(((z)^2.0)/(5.0*((p)^9.0)))

However, it will still count all answers submitted as incorrect.

Any help would be GREATLY appreciated =/.

 

 

Variable name clarification
I have $C1 and $C2, which are the constants of the numerator and denominator, respectively ("12" and "3" in the example).
I have $V1 and $V2 which are the first and second variable, respectively, (in the example, "x" and "y").
I have $A1 and $A2, which are the exponents for the variable $V1 in the numerator and denominator, respectively.
I have $B1 and $B2, which are the exponents for the variable $V2 in the numerator and denominator, respectively.

All these generate from some interesting conditions to create the problems I want (no variables named i, e, or o, for example,) but all properly initialize.

The following program hangs on the last command and a hard restart is required. The computation of a 2 x 2 matrix times a 2-vector is not that hard. Any ideas as to what is happening?

Another question: if v is a vector that depends on x and y say why does
>solve(v=0,{x,y})
not work?

It should only take a few lines of code to change v=0 to the system {components of v = 0}

A prime producing polynomial.

 

Observations on the trinomial n2 + n + 41.

 

by Matt C. Anderson

 

September 3, 2016

 

The story so far

 

We assume that n is an integer.  We focus our attention on the polynomial n^2 + n + 41.

 

Furthur, we analyze the behavior of the factorization of integers of the form

 

h(n) = n2 + n + 41                                          (expression 1)

 

where n is a non-negative integer.  It was shown by Legendre, in 1798 that if 0 ≤ n < 40 then h(n) is a prime number.

 

Certain patterns become evident when considering points (a,n) where

 

h(n) ≡ 0 mod a.                                             (expression 2)

 

The collection of all such point produces what we are calling a "graph of discrete divisors" due to certain self-similar features.  From experimental data we find that the integer points in this bifurcation graph lie on a collection of parabolic curves indexed by pairs of relatively prime integers.  The expression for the middle parabolas is –

 

p(r,c) = (c*x – r*y)2 – r*(c*x – r*y) – x + 41*r2.           (expression 3)

 

The restrictions are that 0<r<c and gcd(r,c) = 1 and all four of r,c,x, and y are integers.

 

Each such pair (r,c) yields (again determined experimentally and by observation of calculations) an integer polynomial a*z2 + b*z + c, and the quartic h(a*z2 + b*z + c) then factors non-trivially over the integers into two quadratic expressions.  We call this our "parabola conjecture".  Certain symmetries in the bifurcation graph are due to elementary relationships between pairs of co-prime integers.  For instance if m<n are co-prime integers, then there is an observable relationship between the parabola it determines that that formed from (n-m, n).

 

We conjecture that all composite values of h(n) arise by substituting integer values of z into h(a*z2 + b*z + c), where this quartic factors algebraically over Z for a*z2 + b*z + c a quadratic polynomial determined by a pair of relatively prime integers.  We name this our "no stray points conjecture" because all the points in the bifurcation graph appear to lie on a parabola.

 

We further conjecture that the minimum x-values for parabolas corresponding to (r, c) with gcd(r, c) = 1 are equal for fixed n.  Further, these minimum x-values line up at 163*c^2/4 where c = 2, 3, 4, ...  The numerical evidence seems to support this.  This is called our "parabolas line up" conjecture.

 

The notation gcd(r, c) used above is defined here.  The greatest common devisor of two integers is the smallest whole number that divides both of those integers.

 

Theorem 1 - Consider h(n) with n a non negative integer. 

h(n) never has a factor less than 41.

 

We prove Theorem 1 with a modular construction.  We make a residue table with all the prime factors less than 41.  The fundamental theorem of arithmetic states that any integer greater than one is either a prime number, or can be written as a unique product of prime numbers (ignoring the order).  So if h(n) never has a prime factor less than 41, then by extension it never has an integer factor less than 41.

 

For example, to determine that h(n) is never divisible by 2, note the first column of the residue table.  If n is even, then h(n) is odd.  Similarly, if n is odd then h(n) is also odd.  In either case, h(n) does not have factorization by 2.

 

Also, for divisibility by 3, there are 3 cases to check.  They are n = 0, 1, and 2 mod 3. h(0) mod 3 is 2.  h(1) mod 3 is 1. and h(2) mod 3 is 2.  Due to these three cases, h(n) is never divisible by 3.  This is the second column of the residue table.

 

The number 0 is first found in the residue table for the cases h(0) mod 41 and h(40) mod 41.  This means that if n is congruent to 0 mod 41 then h(n) will be divisible by 41.  Similarly, if n is congruent to 40 mod 41 then h(n) is also divisible by 41.

After the residue table, we observe a bifurcation graph which has points when h(y) mod x is divisible by x.  The points (x,y) can be seen on the bifurcation graph.

 

< insert residue table here >

 

Thus we have shown that h(n) never has a factor less than 41.

 

Theorem 2

 

Since h(a) = a^2 + a + 41, we want to show that h(a) = h( -a -1).

 

Proof of Theorem 2

Because h(a) = a*(a+1) + 41,

Now h(-a -1) = (-a -1)*(-a -1 +1) + 41.

So h(-a -1) = (-a -1)*(-a) +41,

And h(-a -1) = h(a).

Which was what we wanted.

End of proof of theorem 2.

 

Corrolary 1

Further, if h(b) mod c ≡ = then h(c –b -1) mod c ≡ 0.

 

We can observe interesting patterns in the “graph of discrete divisors” on a following page.

 

I have a simple algebraic problem, but Maple can't eliminate the exp(3P) in each term. Please help.

I get the following error:

Error, (in solve) cannot solve for an unknown function with other operations in its arguments


restart

R13eqn := -2*H*Ybar3*Zbar-H*Z1-H1*Z+H1*Zbar+H4*Ybar3-H41 = 0

-2*H*Ybar3*Zbar-H*Z1-H1*Z+H1*Zbar+H4*Ybar3-H41 = 0

(1)

H := exp(3*P)*(Z+Zbar)

exp(3*P)*(Z+Zbar)

(2)

H1 := 3*P1*exp(3*P)*(Z+Zbar)+exp(3*P)*(Z1+Zbar1)

3*P1*exp(3*P)*(Z+Zbar)+exp(3*P)*(Z1+Zbar1)

(3)

H4 := H*(Z4+Zbar4)/(Z+Zbar)

exp(3*P)*(Z4+Zbar4)

(4)

H41 := ((H1*(Z4+Zbar4)+H*(Z41+Zbar41))(Z+Zbar)-H*(Z4+Zbar4)(Z1+Zbar1))/(Z+Zbar)^2

((3*P1(Z+Zbar)*(exp(3*P))(Z+Zbar)*(Z(Z+Zbar)+Zbar(Z+Zbar))+(exp(3*P))(Z+Zbar)*(Z1(Z+Zbar)+Zbar1(Z+Zbar)))*(Z4(Z+Zbar)+Zbar4(Z+Zbar))+(exp(3*P))(Z+Zbar)*(Z(Z+Zbar)+Zbar(Z+Zbar))*(Z41(Z+Zbar)+Zbar41(Z+Zbar))-exp(3*P)*(Z+Zbar)*(Z4(Z1+Zbar1)+Zbar4(Z1+Zbar1)))/(Z+Zbar)^2

(5)

simplify(R13eqn)

(((-3*Z(Z+Zbar)*P1(Z+Zbar)-3*P1(Z+Zbar)*Zbar(Z+Zbar)-Z1(Z+Zbar)-Zbar1(Z+Zbar))*Zbar4(Z+Zbar)+(-3*P1(Z+Zbar)*Z4(Z+Zbar)-Z41(Z+Zbar)-Zbar41(Z+Zbar))*Zbar(Z+Zbar)-3*P1(Z+Zbar)*Z(Z+Zbar)*Z4(Z+Zbar)+(-Z1(Z+Zbar)-Zbar1(Z+Zbar))*Z4(Z+Zbar)-Z(Z+Zbar)*(Z41(Z+Zbar)+Zbar41(Z+Zbar)))*(exp(3*P))(Z+Zbar)+exp(3*P)*(Z+Zbar)*(Zbar4(Z1+Zbar1)+Z4(Z1+Zbar1)+(3*P1-2*Ybar3)*Zbar^3+((3*P1-4*Ybar3)*Z+Zbar1)*Zbar^2+((-3*P1-2*Ybar3)*Z^2-2*Z1*Z+(Z4+Zbar4)*Ybar3)*Zbar-3*P1*Z^3+(-2*Z1-Zbar1)*Z^2+(Z4+Zbar4)*Ybar3*Z))/(Z+Zbar)^2 = 0

(6)

Zbar41 := -2*Zbar*Zbar1

-2*Zbar*Zbar1

(7)

Z41 := -2*Z1*Z

-2*Z1*Z``

(8)

Z4 := -Z^2

-Z^2

(9)

Zbar4 := -Zbar^2

-Zbar^2

(10)

simplify(R13eqn)

((3*P1(Z+Zbar)*Zbar(Z+Zbar)^3+(3*Z(Z+Zbar)*P1(Z+Zbar)+Z1(Z+Zbar)+3*Zbar1(Z+Zbar))*Zbar(Z+Zbar)^2+3*(Z(Z+Zbar)*P1(Z+Zbar)+(2/3)*Z1(Z+Zbar)+(2/3)*Zbar1(Z+Zbar))*Z(Z+Zbar)*Zbar(Z+Zbar)+3*(Z(Z+Zbar)*P1(Z+Zbar)+Z1(Z+Zbar)+(1/3)*Zbar1(Z+Zbar))*Z(Z+Zbar)^2)*(exp(3*P))(Z+Zbar)-exp(3*P)*(Z+Zbar)*(Zbar(Z1+Zbar1)^2+Z(Z1+Zbar1)^2+3*(Z+Zbar)*((P1+(1/3)*Ybar3)*Z^2+((2/3)*Zbar*Ybar3+(2/3)*Z1+(1/3)*Zbar1)*Z-Zbar*((P1-Ybar3)*Zbar+(1/3)*Zbar1))))/(Z+Zbar)^2 = 0

(11)

expand(((3*P1(Z+Zbar)*Zbar(Z+Zbar)^3+(3*Z(Z+Zbar)*P1(Z+Zbar)+Z1(Z+Zbar)+3*Zbar1(Z+Zbar))*Zbar(Z+Zbar)^2+3*(Z(Z+Zbar)*P1(Z+Zbar)+(2/3)*Z1(Z+Zbar)+(2/3)*Zbar1(Z+Zbar))*Z(Z+Zbar)*Zbar(Z+Zbar)+3*(Z(Z+Zbar)*P1(Z+Zbar)+Z1(Z+Zbar)+(1/3)*Zbar1(Z+Zbar))*Z(Z+Zbar)^2)*(exp(3*P))(Z+Zbar)-exp(3*P)*(Z+Zbar)*(Zbar(Z1+Zbar1)^2+Z(Z1+Zbar1)^2+3*(Z+Zbar)*((P1+(1/3)*Ybar3)*Z^2+((2/3)*Zbar*Ybar3+(2/3)*Z1+(1/3)*Zbar1)*Z-Zbar*((P1-Ybar3)*Zbar+(1/3)*Zbar1))))/(Z+Zbar)^2 = 0)

-3*(exp(P))^3*Ybar3*Zbar^4/(Z+Zbar)^2-2*(exp(P))^3*Z^3*Z1/(Z+Zbar)^2-3*(exp(P))^3*P1*Z^4/(Z+Zbar)^2-(exp(P))^3*Z^3*Zbar1/(Z+Zbar)^2+3*(exp(P))^3*P1*Zbar^4/(Z+Zbar)^2+(exp(P))^3*Zbar^3*Zbar1/(Z+Zbar)^2-(exp(P))^3*Z*Zbar(Z1+Zbar1)^2/(Z+Zbar)^2-(exp(P))^3*Z*Z(Z1+Zbar1)^2/(Z+Zbar)^2-(exp(P))^3*Z^4*Ybar3/(Z+Zbar)^2-(exp(P))^3*Zbar*Zbar(Z1+Zbar1)^2/(Z+Zbar)^2-(exp(P))^3*Zbar*Z(Z1+Zbar1)^2/(Z+Zbar)^2+3*P1(Z+Zbar)*(exp(3*P))(Z+Zbar)*Z(Z+Zbar)^3/(Z+Zbar)^2+3*P1(Z+Zbar)*(exp(3*P))(Z+Zbar)*Zbar(Z+Zbar)^3/(Z+Zbar)^2+3*(exp(3*P))(Z+Zbar)*Z(Z+Zbar)^2*Z1(Z+Zbar)/(Z+Zbar)^2+(exp(3*P))(Z+Zbar)*Z(Z+Zbar)^2*Zbar1(Z+Zbar)/(Z+Zbar)^2+(exp(3*P))(Z+Zbar)*Zbar(Z+Zbar)^2*Z1(Z+Zbar)/(Z+Zbar)^2+3*(exp(3*P))(Z+Zbar)*Zbar(Z+Zbar)^2*Zbar1(Z+Zbar)/(Z+Zbar)^2-4*(exp(P))^3*Ybar3*Z^3*Zbar/(Z+Zbar)^2-8*(exp(P))^3*Ybar3*Z^2*Zbar^2/(Z+Zbar)^2-8*(exp(P))^3*Ybar3*Z*Zbar^3/(Z+Zbar)^2-4*(exp(P))^3*Z^2*Z1*Zbar/(Z+Zbar)^2-2*(exp(P))^3*Z*Z1*Zbar^2/(Z+Zbar)^2-6*(exp(P))^3*P1*Z^3*Zbar/(Z+Zbar)^2+6*(exp(P))^3*P1*Z*Zbar^3/(Z+Zbar)^2-(exp(P))^3*Z^2*Zbar*Zbar1/(Z+Zbar)^2+(exp(P))^3*Z*Zbar^2*Zbar1/(Z+Zbar)^2+3*P1(Z+Zbar)*(exp(3*P))(Z+Zbar)*Z(Z+Zbar)^2*Zbar(Z+Zbar)/(Z+Zbar)^2+3*P1(Z+Zbar)*(exp(3*P))(Z+Zbar)*Z(Z+Zbar)*Zbar(Z+Zbar)^2/(Z+Zbar)^2+2*(exp(3*P))(Z+Zbar)*Z(Z+Zbar)*Zbar(Z+Zbar)*Z1(Z+Zbar)/(Z+Zbar)^2+2*(exp(3*P))(Z+Zbar)*Z(Z+Zbar)*Zbar(Z+Zbar)*Zbar1(Z+Zbar)/(Z+Zbar)^2 = 0

(12)

solve(-(exp(P))^3*Z^2*Zbar*Zbar1/(Z+Zbar)^2+(exp(P))^3*Z*Zbar^2*Zbar1/(Z+Zbar)^2-4*(exp(P))^3*Ybar3*Z^3*Zbar/(Z+Zbar)^2-8*(exp(P))^3*Ybar3*Z^2*Zbar^2/(Z+Zbar)^2-8*(exp(P))^3*Ybar3*Z*Zbar^3/(Z+Zbar)^2-4*(exp(P))^3*Z^2*Z1*Zbar/(Z+Zbar)^2-2*(exp(P))^3*Z*Z1*Zbar^2/(Z+Zbar)^2-6*(exp(P))^3*P1*Z^3*Zbar/(Z+Zbar)^2+6*(exp(P))^3*P1*Z*Zbar^3/(Z+Zbar)^2+3*P1(Z+Zbar)*(exp(3*P))(Z+Zbar)*Z(Z+Zbar)^2*Zbar(Z+Zbar)/(Z+Zbar)^2+3*P1(Z+Zbar)*(exp(3*P))(Z+Zbar)*Z(Z+Zbar)*Zbar(Z+Zbar)^2/(Z+Zbar)^2+2*(exp(3*P))(Z+Zbar)*Z(Z+Zbar)*Zbar(Z+Zbar)*Z1(Z+Zbar)/(Z+Zbar)^2+2*(exp(3*P))(Z+Zbar)*Z(Z+Zbar)*Zbar(Z+Zbar)*Zbar1(Z+Zbar)/(Z+Zbar)^2-(exp(P))^3*Z^3*Zbar1/(Z+Zbar)^2+(exp(P))^3*Zbar^3*Zbar1/(Z+Zbar)^2-(exp(P))^3*Z*Zbar(Z1+Zbar1)^2/(Z+Zbar)^2-(exp(P))^3*Z*Z(Z1+Zbar1)^2/(Z+Zbar)^2-(exp(P))^3*Z^4*Ybar3/(Z+Zbar)^2-(exp(P))^3*Zbar*Zbar(Z1+Zbar1)^2/(Z+Zbar)^2-(exp(P))^3*Zbar*Z(Z1+Zbar1)^2/(Z+Zbar)^2+(exp(3*P))(Z+Zbar)*Z(Z+Zbar)^2*Zbar1(Z+Zbar)/(Z+Zbar)^2+(exp(3*P))(Z+Zbar)*Zbar(Z+Zbar)^2*Z1(Z+Zbar)/(Z+Zbar)^2-3*(exp(P))^3*Ybar3*Zbar^4/(Z+Zbar)^2-2*(exp(P))^3*Z^3*Z1/(Z+Zbar)^2-3*(exp(P))^3*P1*Z^4/(Z+Zbar)^2+3*(exp(P))^3*P1*Zbar^4/(Z+Zbar)^2+3*P1(Z+Zbar)*(exp(3*P))(Z+Zbar)*Z(Z+Zbar)^3/(Z+Zbar)^2+3*P1(Z+Zbar)*(exp(3*P))(Z+Zbar)*Zbar(Z+Zbar)^3/(Z+Zbar)^2+3*(exp(3*P))(Z+Zbar)*Z(Z+Zbar)^2*Z1(Z+Zbar)/(Z+Zbar)^2+3*(exp(3*P))(Z+Zbar)*Zbar(Z+Zbar)^2*Zbar1(Z+Zbar)/(Z+Zbar)^2 = 0, P1)

Error, (in solve) cannot solve for an unknown function with other operations in its arguments

 

NULL

``


Download Help_Maple_divide_an_Exp_on_both_sides.mwHelp_Maple_divide_an_Exp_on_both_sides.mw

Greetings, all!

I'm extremely new to Maple T.A., and I am building tools for my math department since I'm the only one with a programming background. I have a generic question, and sincerely apologize if I've missed it being answered elsewhere. It's a case that will probably come up multiple times for us, so I was hoping to find an answer =).

I have a problem where students are adding exponents of like bases. As an example, you get questions that look like this.

When the coefficient is 1, I'd like to accept a blank answer in the first response area as a correct solution. Is there an easy (or less easy =) way of making this happen?

Thanks in advance!

Dear All

I intend to convert Lie algebra into abstract Lie algebra with the aid of Maple packages DifferentialGeometry and Lie algebras. I want to do this for classification of Lie algebra for group inavriant solution, can anybody help me out.....?

Regards

Dear All

Is there anybody who is working on contruction of optimal Lie algebra using Maple packages like DifferentialGeometry and LieAlgebra, I tried to find commands for constructing algebra in these packages but could not find such commands. I am sure these are only package that might help me. Following is Lie algebra whose optimal system is required:

 

with(PDEtools, SymmetryCommutator, InfinitesimalGenerator):

S[1], S[2], S[3], S[4], S[5], S[6], S[7], S[8], S[9], S[10], S[11] := [_xi[x] = 0, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = 1], [_xi[x] = 0, _xi[y] = t, _xi[t] = 0, _eta[u] = 0, _eta[v] = x], [_xi[x] = 0, _xi[y] = y, _xi[t] = 2*t, _eta[u] = -2*u, _eta[v] = -v], [_xi[x] = 0, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = t], [_xi[x] = 0, _xi[y] = 0, _xi[t] = 0, _eta[u] = 1, _eta[v] = y], [_xi[x] = 0, _xi[y] = 0, _xi[t] = 1, _eta[u] = 0, _eta[v] = 0], [_xi[x] = 1, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = 0], [_xi[x] = t, _xi[y] = 0, _xi[t] = 0, _eta[u] = 1, _eta[v] = 0], [_xi[x] = y, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = 2*x], [_xi[x] = x, _xi[y] = 0, _xi[t] = -t, _eta[u] = 2*u, _eta[v] = 2*v], [_xi[x] = 0, _xi[y] = 1, _xi[t] = 0, _eta[u] = 0, _eta[v] = 0]

[_xi[x] = 0, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = 1], [_xi[x] = 0, _xi[y] = t, _xi[t] = 0, _eta[u] = 0, _eta[v] = x], [_xi[x] = 0, _xi[y] = y, _xi[t] = 2*t, _eta[u] = -2*u, _eta[v] = -v], [_xi[x] = 0, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = t], [_xi[x] = 0, _xi[y] = 0, _xi[t] = 0, _eta[u] = 1, _eta[v] = y], [_xi[x] = 0, _xi[y] = 0, _xi[t] = 1, _eta[u] = 0, _eta[v] = 0], [_xi[x] = 1, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = 0], [_xi[x] = t, _xi[y] = 0, _xi[t] = 0, _eta[u] = 1, _eta[v] = 0], [_xi[x] = y, _xi[y] = 0, _xi[t] = 0, _eta[u] = 0, _eta[v] = 2*x], [_xi[x] = x, _xi[y] = 0, _xi[t] = -t, _eta[u] = 2*u, _eta[v] = 2*v], [_xi[x] = 0, _xi[y] = 1, _xi[t] = 0, _eta[u] = 0, _eta[v] = 0]

(1)

G[1] := InfinitesimalGenerator(S[1], [u(x, y, t), v(x, y, t)]); 1; G[2] := InfinitesimalGenerator(S[2], [u(x, y, t), v(x, y, t)]); 1; G[3] := InfinitesimalGenerator(S[3], [u(x, y, t), v(x, y, t)]); 1; G[4] := InfinitesimalGenerator(S[4], [u(x, y, t), v(x, y, t)]); 1; G[5] := InfinitesimalGenerator(S[5], [u(x, y, t), v(x, y, t)]); 1; G[6] := InfinitesimalGenerator(S[6], [u(x, y, t), v(x, y, t)]); 1; G[7] := InfinitesimalGenerator(S[7], [u(x, y, t), v(x, y, t)]); 1; G[8] := InfinitesimalGenerator(S[8], [u(x, y, t), v(x, y, t)]); 1; G[9] := InfinitesimalGenerator(S[9], [u(x, y, t), v(x, y, t)]); 1; G[10] := InfinitesimalGenerator(S[10], [u(x, y, t), v(x, y, t)]); 1; G[11] := InfinitesimalGenerator(S[11], [u(x, y, t), v(x, y, t)])

proc (f) options operator, arrow; diff(f, v) end proc

 

proc (f) options operator, arrow; t*(diff(f, y))+x*(diff(f, v)) end proc

 

proc (f) options operator, arrow; y*(diff(f, y))+2*t*(diff(f, t))-2*u*(diff(f, u))-v*(diff(f, v)) end proc

 

proc (f) options operator, arrow; t*(diff(f, v)) end proc

 

proc (f) options operator, arrow; diff(f, u)+y*(diff(f, v)) end proc

 

proc (f) options operator, arrow; diff(f, t) end proc

 

proc (f) options operator, arrow; diff(f, x) end proc

 

proc (f) options operator, arrow; t*(diff(f, x))+diff(f, u) end proc

 

proc (f) options operator, arrow; y*(diff(f, x))+2*x*(diff(f, v)) end proc

 

proc (f) options operator, arrow; x*(diff(f, x))-t*(diff(f, t))+2*u*(diff(f, u))+2*v*(diff(f, v)) end proc

 

proc (f) options operator, arrow; diff(f, y) end proc

(2)

``

 

Download Lie_Algebra_Classification.mwLie_Algebra_Classification.mw

   
 

 

 

   

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(2)
 

``

Regards


I am trying to setup a Dual Quaternion Multiplication Table. I found the table on Wikki. I  need some help here.

Have set

x1  =1   x2 = i   x3  =j   x4   =k   x5 =e   x6 = ei   x7 = ej   x8 =ek

 

restart

                                                                                                              #    x1   x2    x3   x4    x5   x6    x7   x8

with(DifferentialGeometry):

NULL

 

StructureEquations := [[x1, x1] = x1, [x1, x2] = x2, [x1, x3] = x3, [x1, x4] = x4, [x1, x5] = x1*x5, [x1, x6] = x6, [x1, x7] = x7, [x1, x8] = x8, [x2, x1] = x2, [x2, x2] = -1, [x2, x3] = x4, [x2, x4] = -x3, [x2, x5] = x6, [x2, x6] = -x5, [x2, x7] = x8, [x2, x8] = -x7, [x3, x1] = x3, [x3, x2] = -x4, [x3, x3] = -1, [x3, x4] = x2, [x3, x5] = x7, [x3, x6] = -x8, [x3, x7] = -x5, [x3, x8] = x6, [x4, x1] = x4, [x4, x2] = x3, [x4, x3] = -x2, [x4, x4] = -1, [x4, x5] = x8, [x4, x6] = x7, [x4, x7] = -x6, [x4, x8] = -x5, [x5, x1] = x5, [x5, x2] = x6, [x5, x3] = x7, [x5, x4] = x8, [x5, x5] = 0, [x6, x1] = x6, [x6, x2] = -x5, [x6, x3] = x8, [x6, x4] = -x7, [x7, x1] = x7, [x7, x2] = -x8, [x7, x3] = -x5, [x7, x4] = x6, [x8, x1] = x8, [x8, x2] = x7, [x8, x3] = -x6, [x8, x4] = -x5]

[[x1, x1] = x1, [x1, x2] = x2, [x1, x3] = x3, [x1, x4] = x4, [x1, x5] = x1*x5, [x1, x6] = x6, [x1, x7] = x7, [x1, x8] = x8, [x2, x1] = x2, [x2, x2] = -1, [x2, x3] = x4, [x2, x4] = -x3, [x2, x5] = x6, [x2, x6] = -x5, [x2, x7] = x8, [x2, x8] = -x7, [x3, x1] = x3, [x3, x2] = -x4, [x3, x3] = -1, [x3, x4] = x2, [x3, x5] = x7, [x3, x6] = -x8, [x3, x7] = -x5, [x3, x8] = x6, [x4, x1] = x4, [x4, x2] = x3, [x4, x3] = -x2, [x4, x4] = -1, [x4, x5] = x8, [x4, x6] = x7, [x4, x7] = -x6, [x4, x8] = -x5, [x5, x1] = x5, [x5, x2] = x6, [x5, x3] = x7, [x5, x4] = x8, [x5, x5] = 0, [x6, x1] = x6, [x6, x2] = -x5, [x6, x3] = x8, [x6, x4] = -x7, [x7, x1] = x7, [x7, x2] = -x8, [x7, x3] = -x5, [x7, x4] = x6, [x8, x1] = x8, [x8, x2] = x7, [x8, x3] = -x6, [x8, x4] = -x5]

(1)

``

(2)

DQ := LieAlgebraData(StructureEquations, [x1, x2, x3, x4, x5, x6, x7, x8])

_DG([["LieAlgebra", "L1", [8, table( [ ] )]], [[[1, 2, 2], 1], [[1, 3, 3], 1], [[1, 4, 4], 1], [[1, 5, 1], x5], [[1, 5, 5], x1], [[1, 6, 6], 1], [[1, 7, 7], 1], [[1, 8, 8], 1], [[1, 2, 2], -1], [[2, 3, 4], 1], [[2, 4, 3], -1], [[2, 5, 6], 1], [[2, 6, 5], -1], [[2, 7, 8], 1], [[2, 8, 7], -1], [[1, 3, 3], -1], [[2, 3, 4], 1], [[3, 4, 2], 1], [[3, 5, 7], 1], [[3, 6, 8], -1], [[3, 7, 5], -1], [[3, 8, 6], 1], [[1, 4, 4], -1], [[2, 4, 3], -1], [[3, 4, 2], 1], [[4, 5, 8], 1], [[4, 6, 7], 1], [[4, 7, 6], -1], [[4, 8, 5], -1], [[1, 5, 5], -1], [[2, 5, 6], -1], [[3, 5, 7], -1], [[4, 5, 8], -1], [[1, 6, 6], -1], [[2, 6, 5], 1], [[3, 6, 8], -1], [[4, 6, 7], 1], [[1, 7, 7], -1], [[2, 7, 8], 1], [[3, 7, 5], 1], [[4, 7, 6], -1], [[1, 8, 8], -1], [[2, 8, 7], -1], [[3, 8, 6], 1], [[4, 8, 5], 1]]])

(3)

DGsetup(DQ)

`Lie algebra: L1`

(4)

MultiplicationTable(DQ, "AlgebraTable")

Error, (in DifferentialGeometry:-LieAlgebras:-MultiplicationTable) invalid input: DifferentialGeometry:-ChangeFrame expects its 1st argument, frame_name, to be of type {name, string}, but received _DG([["LieAlgebra", "L1", [8, table( [ ] )]], [[[1, 2, 2], 1], [[1, 3, 3], 1], [[1, 4, 4], 1], [[1, 5, 1], x5], [[1, 5, 5], x1], [[1, 6, 6], 1], [[1, 7, 7], 1], [[1, 8, 8], 1], [[1, 2, 2], -1], [[2, 3, 4], 1], [[2, 4, 3], -1], [[2, 5, 6], 1], [[2, 6, 5], -1], [[2, 7, 8], 1], [[2, 8, 7], -1], [[1, 3, 3], -1], [[2, 3, 4], 1], [[3, 4, 2], 1], [[3, 5, 7], 1], [[3, 6, 8], -1], [[3, 7, 5], -1], [[3, 8, 6], 1], [[1, 4, 4], -1], [[2, 4, 3], -1], [[3, 4, 2]...

 

NULL

 

Download Dual_Quaternion_Defining_Algebra.mw

 

I'd like to have an algorithm for computing the ring of invariants of a reductive group (i.e. G acts on V, compute C[V]^G). I know that there are algorithms in existance for this, e.g. see http://www.math.lsa.umich.edu/~hderksen/Publications/invar1999.pdf. Does anyone know if this has been implemented in Maple, or where to find code in another language? Any ideas on what sort of data structure would be appropriate

how to calculate Ferrari resolvent of x^4-c1*x^3+c2*x^2-c3*x+c4

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

Hello,

i need help to translate a system which is given below to a for loop.

Other wise i am writing it with myself. 

instead of doing it like that 

with(LinearAlgebra);

sys := [galerkin_funcs[1], galerkin_funcs[2], galerkin_funcs[3], galerkin_funcs[4], galerkin_funcs[5], galerkin_funcs[6], galerkin_funcs[7], galerkin_funcs[8], galerkin_funcs[9], galerkin_funcs[10]];

var := [w[1], w[2], w[3], w[4], w[5], w[6]];

Kmat, Fmat := GenerateMatrix(sys, var);

i want to do it like that.

N:=10

for i to N+4 do

sys(1,i) := galerkin_funcs[i] 

end do

for i to N do

var(1,i) := w[i] 

end do

After that i will generate matrix with this comman Kmat, Fmat := GenerateMatrix(sys, var);

But this for loop i wrote is not doing the i want to do.

Thanks for your help.

 

 

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