vv

13832 Reputation

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These are answers submitted by vv

solve(y>0);

or

Y:=numer(y)*denom(y);
SolveTools:-SemiAlgebraic([Y>0]);

obtain the answer in a few seconds (Maple 2017).

But there are 56 regions and it depends on your final aim to use them.

Edit. You may try to change the order of the variables and hope for simpler regions.

The Maple implementation NumberTheory:-Totient is of course faster (for large numbers), but the simplest one is probably:

eulerphi:=(n::posint) -> add(`if`(igcd(k,n)=1,1,0),k=1..n);

 

Should be

solve(SysEqE,{diff(phi[1](t),t),diff(phi[3](t),t)});

op(0,a);

But you should try to see how this appeared, because it is a nonsense (probably from a programming error).

I don't think it has a special name; it is a multiplication by a diagonal matrix.

But are you sure it is useful? AFAIK the eigenvector algorithms always do a row (or column) normalization.

Use
evalc(Re(expr));

 

restart;

eq1 := diff(u1(x), x, x)+diff(u2(x), x)+int(2*x*s*(u1(s)-3*u2(s)), s = 0 .. 1) = 6*x^2+3*x*(1/10)+8;
eq2 := diff(u1(x), x)+diff(u2(x), x, x)+int((3*(s^2+2*x))*(u1(s)-2*u2(s)), s = 0 .. 1) = 21*x+4/5;
bcs := u1(0)+(D(u1))(0) = 1, u2(0)+(D(u2))(0) = 1, u1(1)+(D(u1))(1) = 10, u2(1)+(D(u2))(1) = 7;

EQ1 := diff(u1(x), x, x)+diff(u2(x), x)+a*x = 6*x^2+3*x*(1/10)+8;
EQ2 := diff(u1(x), x)+diff(u2(x), x, x)+b*x+c = 21*x+4/5;

diff(diff(u1(x), x), x)+diff(u2(x), x)+int(2*x*s*(u1(s)-3*u2(s)), s = 0 .. 1) = 6*x^2+(3/10)*x+8

 

diff(u1(x), x)+diff(diff(u2(x), x), x)+int(3*(s^2+2*x)*(u1(s)-2*u2(s)), s = 0 .. 1) = 21*x+4/5

 

u1(0)+(D(u1))(0) = 1, u2(0)+(D(u2))(0) = 1, u1(1)+(D(u1))(1) = 10, u2(1)+(D(u2))(1) = 7

 

diff(diff(u1(x), x), x)+diff(u2(x), x)+a*x = 6*x^2+(3/10)*x+8

 

diff(u1(x), x)+diff(diff(u2(x), x), x)+b*x+c = 21*x+4/5

(1)

sol:=dsolve({EQ1,EQ2},{u1(x),u2(x)}):

U1:=unapply( eval(u1(x),sol),x):
U2:=unapply( eval(u2(x),sol),x):

sys:=eval([(lhs-rhs)(eq1),(lhs-rhs)(eq2),bcs],[u1=U1,u2=U2]):

map(coeffs,sys,x):

consts:=solve(%):

SOL:=simplify(eval(sol,consts));

{u1(x) = (-55897*exp(1-x)+27844*exp(2-x)+35538*exp(1+x)+(229194*x^2+1583820*x-1422696)*exp(1)+(-45192*x^2-385320*x+335340)*exp(2)-300822*x^2-1458780*x-55263*exp(x)+1319226)/(-49980*exp(2)+232200*exp(1)-250080), u2(x) = (-55897*exp(1-x)+27844*exp(2-x)-35538*exp(1+x)+(464400*x^3-681738*x^2+1399212*x-1095936)*exp(1)+(-99960*x^3+160164*x^2-309456*x+259476)*exp(2)-500160*x^3+669894*x^2-1398996*x+55263*exp(x)+1038390)/(-49980*exp(2)+232200*exp(1)-250080)}

(2)

U1:=unapply( eval(u1(x),SOL),x):             # Check
U2:=unapply( eval(u2(x),SOL),x):             #
simplify(eval({eq1,eq2,bcs},[u1=U1,u2=U2])); #

{1 = 1, 7 = 7, 10 = 10, 21*x+4/5 = 21*x+4/5, 6*x^2+(3/10)*x+8 = 6*x^2+(3/10)*x+8}

(3)

 


Download sysode-int.mw

 

n is supposed to be a positive integer.

We compute the indefinite integral.

 

 

J:=n -> Int((2*cos(Pi*x*n/T)^2-1)*sin(Pi*x*n/T)*cos(Pi*x*n/T)/(sin(Pi*x/T)*cos(Pi*x/T)), x);

proc (n) options operator, arrow; Int((2*cos(Pi*x*n/T)^2-1)*sin(Pi*x*n/T)*cos(Pi*x*n/T)/(sin(Pi*x/T)*cos(Pi*x/T)), x) end proc

(1)

value(combine(J(n+1)-J(n)));

(1/2)*T*sin(2*Pi*x*(2*n+1)/T)/(Pi*(2*n+1))

(2)

K:=unapply(%,n);

proc (n) options operator, arrow; (1/2)*T*sin(2*Pi*x*(2*n+1)/T)/(Pi*(2*n+1)) end proc

(3)

J(n)=value(J(1))+Sum(K(k),k=1..n-1);  # Answer

Int((2*cos(Pi*x*n/T)^2-1)*sin(Pi*x*n/T)*cos(Pi*x*n/T)/(sin(Pi*x/T)*cos(Pi*x/T)), x) = -x+2*T*((1/2)*cos(Pi*x/T)*sin(Pi*x/T)+(1/2)*Pi*x/T)/Pi+Sum((1/2)*T*sin(2*Pi*x*(2*k+1)/T)/(Pi*(2*k+1)), k = 1 .. n-1)

(4)

value(%);  # Version of the answer

int((2*cos(Pi*x*n/T)^2-1)*sin(Pi*x*n/T)*cos(Pi*x*n/T)/(sin(Pi*x/T)*cos(Pi*x/T)), x) = -x+2*T*((1/2)*cos(Pi*x/T)*sin(Pi*x/T)+(1/2)*Pi*x/T)/Pi+((1/8)*I)*T*(exp((2*I)*Pi*x*(2*n+1)/T)*LerchPhi(exp((4*I)*Pi*x/T), 1, n+1/2)-exp((6*I)*Pi*x/T)*LerchPhi(exp((4*I)*Pi*x/T), 1, 3/2)+exp(-(6*I)*Pi*x/T)*LerchPhi(exp(-(4*I)*Pi*x/T), 1, 3/2)-LerchPhi(exp(-(4*I)*Pi*x/T), 1, n+1/2)*exp(-(2*I)*Pi*x*(2*n+1)/T))/Pi

(5)

simplify(diff( (lhs-rhs)(%), x)); # Check

0

(6)

This is a maths problem, not a Maple one. See https://en.wikipedia.org/wiki/Cycles_and_fixed_points

Of course it is easy to implement any of the formulae.

Here is a direct solution (based on simplex) asked by the OP who has an older version of Maple.

FacetsOfPolyhedralCone:=proc(L::list)
local i,j,J,A,X,u;
X:=indets(L,name);
A:=LinearAlgebra:-GenerateMatrix(L,X)[1]:
J:={seq(1..nops(L))}:
for j in J do
  add(A[j]-u[i]*A[i],i=J minus {j}); convert(%,list)=~0;
  if simplex[minimize](0, %, NONNEGATIVE) <>{} then J:=J minus {j} fi;
od:
J, L[[J[]]];
end:

L := [y-z, 3*y-2*z, 2*y-2*z, x-2*y+z, x-y, 2*x-y, x-z, x+y-z, x, y, z]:  # >= 0

FacetsOfPolyhedralCone(L);
      
{3, 4, 11}, [2*y-2*z, x-2*y+z, z]

solve({x[5] = x[2]/x[1], x[6] = x[3]/x[2], x[7] = x[1]/x[4],
       x[8] = (2*x[2]+x[4])/(2*x[1]+x[3]+x[4])}, {x[1], x[2], x[3], x[4]}); 

         {x[1] = 0, x[2] = 0, x[3] = 0, x[4] = 0}

It is a bug. The result of eliminate should be:

Mathematically it should be also added: x[5]<>0, x[7]<>0, x[4]<>0,  x[6] <> -(2*x[7]+1)/(x[5]*x[7])

@mmcdara 

Let's do the computations for 3 digits.

restart;
Digits:=3:
a,b,c:= sqrt~([2,3,6])[]:
fa,fb,fc := evalf(%);

                 fa, fb, fc := 1.41, 1.73, 2.45
fa*fb;
                              2.44
# Do you expect  2.45?  How?
141*173;
                             24393

 

 

num1dsubspaces := (p,n) -> (p^n - 1)/(p-1)

 

ineqs:=
y-z >=0,
3*y-2*z >=0,
2*y-2*z >=0,
x-2*y+z >=0,
x-y >=0,
2*x-y >=0,
x-z >=0,
x+y-z >=0,
x>=0,
y>=0,
z>=0:
with(PolyhedralSets):
A := PolyhedralSet([ineqs]);
B := PolyhedralSet([ineqs[3..4],ineqs[-3..-1]]);

evalb(convert(A,string)=convert(B,string));

        true

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