Complex conjugation of a product of Grassmannians is defined to reverse the order of the multiplicands (while also complex conjugating each multiplicand by itself, of course). That does not seem to be implemented in the Physics package: The code

with(Physics):
Setup(
   mathematicalnotation = true,
   anticommutativepre   = theta
):
a := theta1*theta2;
b := conjugate(a);
c := evalc(b);

produces for a, b, and c the following results:

In b and c, the order of the multiplicands should be the reverse, or, equivalently, the signs of b and c should be the opposite. Is it due to an error, or due to not being implemented yet, or have I simply misunderstood something?

The following code of mine, which no doubt may be improved, performs this generalized conjugation (note that reversing a product of c-number-valued, i.e., commutative, quantities is, of course, inconsequential, and thus usual complex conjugation is maintained):

Conj := proc(x::algebraic)
    local expr;
    expr := expand(x);
    if   type(expr,`+`) then map(Conj,expr)
    elif type(expr,`*`) then `*`(ListTools:-Reverse(map(Conj,[op(expr)]))[]);
    else conjugate(expr)
    end if
end proc:

I want to define a band matrix using n vectors. Detial given maple sheet. please help me

Help.mw

I want to define a matrix of M by M which have some condition. detail given in maple file. please help me.

Help_Maple-1.mw

I explain the required matrix by taking some example hope you will be understand. Please make it quickly... I will be very thank full. Help_Maple-1.mw

I wan to define a square matrix for M and r detial given in Maple sheet. Please Help....Help_Maple.mwHelp_Maple.mw

Is it possible that this expression has an elementary one (specifically the dilog's):

Y0:=(1/16)*(s*t*(exp(2*t)*s+exp(4*t)+1)*ln((exp(2*t)*s-(-s^2+1)^(1/2)+1)^16*(1+(-s^2+1)^(1/2))^16/((exp(2*t)*s+(-s^2+1)^(1/2)+1)^16*(1-(-s^2+1)^(1/2))^16))+s^3*t*(exp(4*t)+1)*ln((exp(2*t)*s-(-s^2+1)^(1/2)+1)^8*(1+(-s^2+1)^(1/2))^8/((exp(2*t)*s+(-s^2+1)^(1/2)+1)^8*(1-(-s^2+1)^(1/2))^8))+exp(2*t)*t*ln((exp(2*t)*s-(-s^2+1)^(1/2)+1)^32*(1+(-s^2+1)^(1/2))^32/((exp(2*t)*s+(-s^2+1)^(1/2)+1)^32*(1-(-s^2+1)^(1/2))^32))+4*((exp(4*t)+1)*s+2*exp(2*t))*(s^2+2)*dilog((-exp(2*t)*s+(-s^2+1)^(1/2)-1)/(-1+(-s^2+1)^(1/2)))-4*((exp(4*t)+1)*s+2*exp(2*t))*(s^2+2)*dilog((exp(2*t)*s+(-s^2+1)^(1/2)+1)/(1+(-s^2+1)^(1/2)))+((32*s^2*t+64*t)*exp(2*t)+16*(((t+1/8)*s^2+2*t+2)*exp(4*t)-(5/4)*s*exp(-2*t)-(1/8)*exp(-4*t)*s^2+(5/4)*s*exp(6*t)+(1/8)*s^2*exp(8*t)+(t-1/8)*s^2-2+2*t)*s)*arctanh((exp(2*t)-1)*(-1+s)/((-s^2+1)^(1/2)*(exp(2*t)+1)))+8*(-s^2+1)^(1/2)*((1/8)*s*(exp(4*t)+1)*ln((exp(4*t)*s+2*exp(2*t)+s)^12/s^12)+(1/8)*exp(2*t)*ln((exp(4*t)*s+2*exp(2*t)+s)^24/s^24)+(s^2-6*t-3)*exp(2*t)+((-(1/8)*s^2-3*t)*exp(4*t)+s*exp(-2*t)+(1/8)*exp(-4*t)*s^2+s*exp(6*t)+(1/8)*s^2*exp(8*t)-(1/8)*s^2-3*t)*s))/((s*exp(-2*t)+exp(2*t)*s+2)*(exp(4*t)*s+2*exp(2*t)+s)*((-s^2+1)^(1/2)+2*arctanh((-1+s)/(-s^2+1)^(1/2))))

Also I'm wondering since Y0 should solve the ode

-(diff(diff(y(t), t), t))+(4-12/(1+s*cosh(2*t))+8*(-s^2+1)/(1+s*cosh(2*t))^2)*y(t) = C/(1+s*cosh(2*t))

with some constant C but I only get rubbish.

I ask this because I found that in another context this seems to be correct:

f1:=-(1/12)*Pi^2*((-s^2+1)^(1/2)-arccosh(1/s))/(-s^2+1)^(3/2)+(1/12)*arccosh(1/s)^3/(-s^2+1)^(3/2)-(1/4)*arccosh(1/s)^2/(-s^2+1)

f2:=(1/2)*((-s^2+1)^(1/2)*(polylog(2, s/(-1+(-s^2+1)^(1/2)))+polylog(2, -s/(1+(-s^2+1)^(1/2))))-polylog(3, s/(-1+(-s^2+1)^(1/2)))+polylog(3, -s/(1+(-s^2+1)^(1/2))))/(-s^2+1)^(3/2)

and f1=f2

but maple doesnt convert it.

Also maple has trouble to convert

2*arctanh(sqrt((1-s)/(1+s)))=arccosh(1/s)

everywhere: 0<s<1

Aslam-u-Alikum. How are you. I want to make band matrix by using two vector. Detail given in Maple fine also. Please help me urgent.Help.mw

Is it possible to find the limit
limit(int(x*cos(x)/(1+3*sin(n*x)^2), x = 0 .. Pi), n = infinity)assuming n::posint
with Maple?
Calculations suggest the one equals -1.

Hello

I calculated following two expressions, x1,and x2.

x1:=map(f,a+b);

x2:=map(f,y=a+b);

The results of these are

f(a) + f(b)

f(y) = f(a + b)

for each. And, I can understand the logic of this.

If I want to derive the result of x2 as f(y)=f(a)+f(b), how should I do about x2?

Isn't there other way than to write

map(f, lhs(x2))=map(f,rhs(x2))

?

Please teach me this.

Thank you in advance.

Taro

How can I get maple to integrate this expression numerically.

For a specific value 0<s<1 it should be enough to integrate from -40..40 instead of -infinity..infinity

Anyway. My maple version always hangs up :-(

(1/2)*(-4*dilog(-(exp(2*t)*s-(-s^2+1)^(1/2)+1)/(-1+(-s^2+1)^(1/2)))*exp(4*t)+arctanh((-1+s)/(-s^2+1)^(1/2))*s^2+arctanh((exp(2*t)*s-exp(2*t)-s+1)/((exp(2*t)+1)*(-s^2+1)^(1/2)))*s^2+8*(-s^2+1)^(1/2)*exp(4*t)+4*dilog((exp(2*t)*s+(-s^2+1)^(1/2)+1)/(1+(-s^2+1)^(1/2)))*exp(4*t)+4*exp(4*t)*arctanh((-1+s)/(-s^2+1)^(1/2))-8*arctanh((exp(2*t)*s-exp(2*t)-s+1)/((exp(2*t)+1)*(-s^2+1)^(1/2)))*exp(4*t)*s^2*t-4*ln(1+(-s^2+1)^(1/2))*exp(4*t)*s^2*t+4*ln(1-(-s^2+1)^(1/2))*exp(4*t)*s^2*t-4*ln(exp(2*t)*s-(-s^2+1)^(1/2)+1)*exp(4*t)*s^2*t+4*ln(exp(2*t)*s+(-s^2+1)^(1/2)+1)*exp(4*t)*s^2*t+12*(-s^2+1)^(1/2)*exp(4*t)*t-16*arctanh((exp(2*t)*s-exp(2*t)-s+1)/((exp(2*t)+1)*(-s^2+1)^(1/2)))*exp(4*t)*t-8*ln(1+(-s^2+1)^(1/2))*exp(4*t)*t+8*ln(1-(-s^2+1)^(1/2))*exp(4*t)*t-8*ln(exp(2*t)*s-(-s^2+1)^(1/2)+1)*exp(4*t)*t+8*ln(exp(2*t)*s+(-s^2+1)^(1/2)+1)*exp(4*t)*t-(-s^2+1)^(1/2)*exp(2*t)*s+8*arctanh((exp(2*t)*s-exp(2*t)-s+1)/((exp(2*t)+1)*(-s^2+1)^(1/2)))*exp(2*t)*s+4*exp(2*t)*arctanh((-1+s)/(-s^2+1)^(1/2))*s-(-s^2+1)^(1/2)*exp(6*t)*s-8*arctanh((exp(2*t)*s-exp(2*t)-s+1)/((exp(2*t)+1)*(-s^2+1)^(1/2)))*exp(6*t)*s+4*exp(6*t)*arctanh((-1+s)/(-s^2+1)^(1/2))*s+2*dilog((exp(2*t)*s+(-s^2+1)^(1/2)+1)/(1+(-s^2+1)^(1/2)))*exp(4*t)*s^2+2*(-s^2+1)^(1/2)*exp(4*t)*s^2-arctanh((exp(2*t)*s-exp(2*t)-s+1)/((exp(2*t)+1)*(-s^2+1)^(1/2)))*exp(8*t)*s^2+exp(8*t)*arctanh((-1+s)/(-s^2+1)^(1/2))*s^2+2*exp(4*t)*arctanh((-1+s)/(-s^2+1)^(1/2))*s^2-6*(-s^2+1)^(1/2)*ln(exp(4*t)*s+2*exp(2*t)+s)*exp(4*t)+6*(-s^2+1)^(1/2)*ln(s)*exp(4*t)-2*dilog(-(exp(2*t)*s-(-s^2+1)^(1/2)+1)/(-1+(-s^2+1)^(1/2)))*exp(4*t)*s^2)/((-s^2+1)^(1/2)*exp(8*t)*s^2-2*arctanh((-s^2+1)^(1/2)/(1+s))*exp(8*t)*s^2+4*(-s^2+1)^(1/2)*exp(6*t)*s-8*arctanh((-s^2+1)^(1/2)/(1+s))*exp(6*t)*s+2*(-s^2+1)^(1/2)*exp(4*t)*s^2-4*arctanh((-s^2+1)^(1/2)/(1+s))*exp(4*t)*s^2+4*(-s^2+1)^(1/2)*exp(4*t)-8*arctanh((-s^2+1)^(1/2)/(1+s))*exp(4*t)+4*(-s^2+1)^(1/2)*exp(2*t)*s-8*arctanh((-s^2+1)^(1/2)/(1+s))*exp(2*t)*s+(-s^2+1)^(1/2)*s^2-2*arctanh((-s^2+1)^(1/2)/(1+s))*s^2)

I have an equation as follows:

By inspection one can see that the last three terms can be simplified (factored) to

How can I coerce Maple to do this? None of the available tools seem to be getting close to this. A partial solution is like this: Writ a procedure as follows:

Fac:=proc(xpr,a,b);
  tmp:=xpr+(a^2+2*a*b+b^2);
  return tmp-(a+b)^2;
end proc;

and then call it:

Fac(lhs(eq),k0,2*Pi*n/L)=rhs(eq);

to get

which is what I want. But procedure Fac() is not general at all; e.g. it fails if the overall sign of the polynomial terms are different. There does not seem to be any way in Maple to determine the sign of a term in the sum of lhs(eq), I can only find ways to determine signs of a simple indeterminate. I'd like to make this procedure more general (which is trivial enough for a human) but I just cannot find any tools in Maple to support this.

Any ideas out there?

Mac Dude.

after a matrix operation, the result is not exactly the matrix i want

there is around 0.0001 difference difference in all element in matrix

how to deal with this random difference in order to be exact?

In order to get a little acquainted with the Grassmannian capabilities of the Physics package, I presently consider the following simple setup:

with(Physics):
Setup(
   mathematicalnotation = true,
   anticommutativepre   = theta
):
A := Matrix(2,(i,j) -> theta||i||j);
B := A . A;
C := Expand(B);

producing for A, B, and C the following results:

To me the [2,2]-entry of B seems erroneous: the first addend has the wrong ordering of the theta's, or, equivalently, the wrong sign. Not so surprisingly, this error is then present also in the [2,2]-entry of C. But in C, the [1,2]- and [2,1]-entries seem erroneous as well: the sign of theta11 in the [1,2]-entry is wrong, and so, too, is the sign of theta22 in the [2,1]-entry.

Have I fundamentally misunderstood something?

Hi everyone,

I have a very complicated function y with only one independent variable x, and want to fit or approximate it by a simpler function, say polynomial. Many books or maple reference seem to tell how to fit a set of data instead of a given function. But the argument x in the function is assumed to be continuous other than discrete, so I don't know whether it is possible to express datax in form of x's range such as 0..1, and express datay in form of the function. After that , maybe I can fit the two created data sets by a polynomial function.

Or, does anyone have a better or more direct way to do the fitting linking two fucntions?

I am appreciated for your help.

Best,

GOODLUCK

I have the following expression (obtained from an earlier calculation):

I want to collect all the terms under one summation. So I define a rule:

collectf:=proc(f)
A::algebraic*f(a::algebraic)+B::algebraic*f(b::algebraic)\
 +C::algebraic*f(c::algebraic)+D::algebraic*f(d::algebraic)=f(A*a+B*b+C*c+D*d);
end proc:

and then

applyrule(collectf(Sum),%);

I get

Error, (in +) unable to identify A::algebraic

I used similar constructs before so I think the rule is constructed correctly. I should, however, mention that I use the Physics:-Vectors package and in fact the expression I start up with here reads, in 1-d Maple inputform:

Physics[Vectors][`+`](Physics[Vectors][`+`](Physics[Vectors][`+`](-y*(Sum((diff(a[n](r), r))/(exp(I*Pi*n/L))^2, n))/r, (2*I)*(Sum(a[n](r)/(exp(I*Pi*n/L))^2, n))*k0), y*(Sum(a[n](r)/(exp(I*Pi*n/L))^2, n))*k0^2), -y*(Sum((diff(a[n](r), r, r))/(exp(I*Pi*n/L))^2, n)))

Is my problem related to the use of Physics:-Vectors? If so, how can I get around that?

TIA,

Mac Dude