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Hi all 

I have the following segment of maple program which belongs to time delay systems dynamic. here C=X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P, is a matrix(vector) which comes from reordering the system terms and my goal is to minimizing J:=X.E.Transpose(X)+U.E.Transpose(U), subject to constraint C=0, but i don't know how to do so.

I will be so grateful if anyone can guide me

best wishes

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department


restart:
with(Optimization):
with(LinearAlgebra):
macro(LA= LinearAlgebra):
L:=1:  r:=2:  tau:= 1:
interface(rtablesize= 2*r+1):

Z:= Matrix(
     2*r+1, 2*r+1,
     [tau,
      seq(evalf((L/(2*(iz-1)*Pi))*sin(2*(iz-1)*Pi*tau/L)), iz= 2..r+1),
      seq(evalf((L/(2*(iz-1-r)*Pi))*(1-cos(2*(iz-1-r)*Pi*tau/L))), iz= r+2..2*r+1)
      ],
     scan= columns,
     datatype= float[8]
);
                        
Dtau00:= < 1 >:
Dtau01:= Vector[row](r):
Dtau02:= Vector[row](r):
Dtau10:= Vector(r):
Dtau20:= Vector(r):

Dtau1:= LA:-DiagonalMatrix([seq(evalf(cos(2*i*Pi*tau/L)), i= 1..r)]):
Dtau2:= LA:-DiagonalMatrix([seq(evalf(sin(2*i*Pi*tau/L)), i= 1..r)]):
Dtau3:= -Dtau2:
Dtau4:= copy(Dtau1):

Dtau:= < < Dtau00 | Dtau01 | Dtau02 >,
         < Dtau10 | Dtau1  | Dtau2  >,
         < Dtau20 | Dtau3  | Dtau4  > >;
 
P00:= < L/2 >:
P01:= Vector[row](r):
P02:= Vector[row](r, j-> evalf(-L/j/Pi), datatype= float[8]):
P10:= Vector(r):
P20:= Vector(r, i-> evalf(L/2/i/Pi)):
P1:= Matrix(r,r):
P2:= LA:-DiagonalMatrix(P20):
P3:= LA:-DiagonalMatrix(-P20):
P4:= Matrix(r,r):

P:= < < P00 | P01 | P02 >,
      < P10 | P1  | P2  >,
      < P20 | P3  | P4  > >;

interface(rtablesize=2*r+1):    # optionally
J:=Vector([L, L/2 $ 2*r]):      # Matrix([[...]]) would also work here

E:=DiagonalMatrix(J);

X:=  Vector[row](2*r+1,symbol=a);
U:=Vector[row](2*r+1,symbol=b);

X0:= Vector[row](2*r+1,[1]);
G:=Vector[row](2*r+1,[1]);
C:=simplify(X-X0-G.Z-X.Dtau.P+X.Dtau.Z-U.P);

Z := Matrix(5, 5, {(1, 1) = 1., (1, 2) = 0., (1, 3) = 0., (1, 4) = 0., (1, 5) = 0., (2, 1) = 0., (2, 2) = 0., (2, 3) = 0., (2, 4) = 0., (2, 5) = 0., (3, 1) = 0., (3, 2) = 0., (3, 3) = 0., (3, 4) = 0., (3, 5) = 0., (4, 1) = 0., (4, 2) = 0., (4, 3) = 0., (4, 4) = 0., (4, 5) = 0., (5, 1) = 0., (5, 2) = 0., (5, 3) = 0., (5, 4) = 0., (5, 5) = 0.})

Dtau := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1., (2, 3) = 0, (2, 4) = 0., (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1., (3, 4) = 0, (3, 5) = 0., (4, 1) = 0, (4, 2) = -0., (4, 3) = -0., (4, 4) = 1., (4, 5) = 0, (5, 1) = 0, (5, 2) = -0., (5, 3) = -0., (5, 4) = 0, (5, 5) = 1.})

P := Matrix(5, 5, {(1, 1) = 1/2, (1, 2) = 0, (1, 3) = 0, (1, 4) = -.318309886100000, (1, 5) = -.159154943000000, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = .1591549430, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (3, 5) = 0.7957747152e-1, (4, 1) = .1591549430, (4, 2) = -.159154943000000, (4, 3) = 0, (4, 4) = 0, (4, 5) = 0, (5, 1) = 0.7957747152e-1, (5, 2) = 0, (5, 3) = -0.795774715200000e-1, (5, 4) = 0, (5, 5) = 0})

E := Matrix(5, 5, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 0, (2, 2) = 1/2, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1/2, (3, 4) = 0, (3, 5) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 1/2, (4, 5) = 0, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 0, (5, 5) = 1/2})

X := Vector[row](5, {(1) = a[1], (2) = a[2], (3) = a[3], (4) = a[4], (5) = a[5]})

U := Vector[row](5, {(1) = b[1], (2) = b[2], (3) = b[3], (4) = b[4], (5) = b[5]})

X0 := Vector[row](5, {(1) = 1, (2) = 0, (3) = 0, (4) = 0, (5) = 0})

G := Vector[row](5, {(1) = 1, (2) = 0, (3) = 0, (4) = 0, (5) = 0})

C := Vector[row](5, {(1) = 1.500000000*a[1]-2.-.1591549430*a[4]-0.7957747152e-1*a[5]-.5000000000*b[1]-.1591549430*b[4]-0.7957747152e-1*b[5], (2) = a[2]+.1591549430*a[4]+.1591549430*b[4], (3) = a[3]+0.7957747152e-1*a[5]+0.7957747152e-1*b[5], (4) = a[4]+.3183098861*a[1]-.1591549430*a[2]+.3183098861*b[1]-.1591549430*b[2], (5) = a[5]+.1591549430*a[1]-0.7957747152e-1*a[3]+.1591549430*b[1]-0.7957747152e-1*b[3]})

(1)

J:=X.E.Transpose(X)+U.E.Transpose(U);

J := a[1]^2+(1/2)*(a[2]^2)+(1/2)*(a[3]^2)+(1/2)*(a[4]^2)+(1/2)*(a[5]^2)+b[1]^2+(1/2)*(b[2]^2)+(1/2)*(b[3]^2)+(1/2)*(b[4]^2)+(1/2)*(b[5]^2)

(2)

Minimize(J,{C=0});






Error, (in Optimization:-NLPSolve) invalid arguments

 

#XP:=-.015+X[1]+add(X[l+1]*f1(l)+X[r+l+1]*f2(l), l= 1..r):
#plot([XP,T1], t= 0..1);#,legend= "Solution Of x(t) with r=50"):

 

 

 

 

 

 

Download work1.mwswork1.mws

This is a mini-course I gave in Brazil last week, at the CBPF (Brazilian Center for Physics Research). The material will still receive polishment and improvements, towards evolving into a sort of manual, but it is also interesting to see it exactly as it was presented to people during the course. This material uses the update of Physics available at the Maplesoft Physics R&D webpage.

BrasilComputacaoAlge.zip

Edgardo S. Cheb-Terrab
Physics, Maplesoft

Hi all

In matlab software we have a command namely fmincon which minimizes any linear/nonlinear algebric equations subject to linear/nonlinear constraints.

Now my question is that: what is the same command in maple?or how can we minimize linear/nonlinear function subject to linear/nonlinear constraints in maple?

thanks a lot

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

Pleas i want to do continued fractional expansion to transfer function G(s) with numerator and denomenator as a polynomial in Lplace operator s.

with my gratitude

 

what is the best maple book? could every one introduce books which have studied about maple !?? which is the best due to everyones opinion ? tnx for answering.

Hi MaplePrime-ers!

I've been using the Maple(17) toolbox in Matlab(2012b) to quickly enumerate systems of equations by: (i) solving them symbolically, (ii) using unapply to make them functions, (iii) then supplying the points (driver equations) to get the system solution.  Speed is a must, because there may be 3 million+ systems to solve.  Symbolics is also very important because I am evaluating topology, so the structure of the equations may change, and therefore a functional approach will not work.

I have had success (seen in the first code snippet).  I would like similiar behaviour in the second code snippet, but sometimes I get 'solutions may be lost' as an error message,  or 'Error, (in unapply) variables must be unique and of type name'

The system of equations include:  Linear equations, 5th order polynomials, absolute functions, and pieceiwse functions.

Here is code with a topology that solves:

#Interconnection Equations
eq2[1] := FD_T + EM2_T = 0;
eq2[2] := ICE_T + GEN_T = 0;
eq2[3] := EM2_A + GEN_A + BAT_A = 0;
eq2[4] := -FD_W + EM2_W = 0;
eq2[5] := -ICE_W + GEN_W = 0;
eq2[6] := -EM2_V + GEN_V = 0;
eq2[7] := -EM2_V + BAT_V = 0;

#ICE
eq_c[1] := ICE_mdot_g=((671.5) + (-21.94)*ICE_T + (0.1942)*ICE_W + (0.5113)*ICE_T^2 + (-0.01271)*ICE_T*ICE_W + ( -0.0008761)*ICE_W^2 + (-0.006071)*ICE_T^3 + (9.867e-07)*ICE_T^2*ICE_W + (5.616e-05)*ICE_T*ICE_W^2 + (1.588e-06)*ICE_W^3 + (3.61e-05)*ICE_T^4 + (8.98e-07)*ICE_T^3*ICE_W + (-2.814e-07)*ICE_T^2*ICE_W^2 + (-8.121e-08)*ICE_T*ICE_W^3 + ( -8.494e-08 )*ICE_T^5 + (-2.444e-09)*ICE_T^4*ICE_W + (-9.311e-10)*ICE_T^3*ICE_W^2 + ( 5.835e-10)*ICE_T^2*ICE_W^3 ) *1/3600/1000 * ICE_T * ICE_W;

#BAT
eq_c[2] := BAT = 271;

#EM2
EM2_ReqPow_eq := (-148.3) + (4.267)*abs(EM2_W) + (12.77)*abs(EM2_T) + (-0.0364)*abs(EM2_W)^2 + ( 1.16)*abs(EM2_W)*abs(EM2_T) + (-0.258)*abs(EM2_T)^2 + ( 0.0001181)*abs(EM2_W)^3 + (-0.0005994)*abs(EM2_W)^2*abs(EM2_T) + ( 0.0001171)*abs(EM2_W)*abs(EM2_T)^2 + (0.001739 )*abs(EM2_T)^3 + (-1.245e-07 )*abs(EM2_W)^4 + ( 1.2e-06)*abs(EM2_W)^3*abs(EM2_T) + ( -1.584e-06)*abs(EM2_W)^2*abs(EM2_T)^2 + ( 4.383e-07)*abs(EM2_W)*abs(EM2_T)^3 + (-2.947e-06)*abs(EM2_T)^4;
eq_c[3] := EM2_P = piecewise( EM2_T = 0, 0, EM2_W = 0, 0, EM2_W*EM2_T < 0,-1 * EM2_ReqPow_eq, EM2_ReqPow_eq);
eq_c[4] := EM2_A = EM2_P/EM2_V;

#GEN
GEN_ReqPow_eq:= (-5.28e-12) + ( 3.849e-14)*abs(GEN_W) + (-71.9)*abs(GEN_T) + (-1.168e-16)*abs(GEN_W)^2 +(1.296)*abs(GEN_W)*abs(GEN_T) + (2.489)*abs(GEN_T)^2 + (1.451e-19)*abs(GEN_W)^3 + (0.0001326)*abs(GEN_W)^2*abs(GEN_T) + (-0.008141)*abs(GEN_W)*abs(GEN_T)^2 + (-0.004539)*abs(GEN_T)^3 +(-6.325e-23)*abs(GEN_W)^4 + (-2.091e-07)*abs(GEN_W)^3*abs(GEN_T) + ( 3.455e-06)*abs(GEN_W)^2*abs(GEN_T)^2 + ( 2.499e-05)*abs(GEN_W)*abs(GEN_T)^3 + (-5.321e-05)*abs(GEN_T)^4;
eq_c[5] := GEN_P = piecewise( GEN_T = 0, 0, GEN_W = 0, 0, GEN_W*GEN_T < 0,-1 * GEN_ReqPow_eq, GEN_ReqPow_eq);
eq_c[6] := GEN_A = GEN_P/GEN_V;

#ASSUMPTIONS
assume(BAT_V::nonnegative);
assume(FD_W::nonnegative);

#FINAL EQUATIONS

sys_eqs2 := convert(eq2,set) union {eq_c[1],eq_c[2],eq_c[3],eq_c[4],eq_c[5],eq_c[6]};

#Selecting which variables to solve for:

drivers2:= { ICE_T,ICE_W,FD_T,FD_W};
symvarnames2:=select(type,indets(convert(sys_eqs2,list)),name);
notdrivers2:=symvarnames2 minus drivers2;


#Symbolic solve

sol2:=solve(sys_eqs2,notdrivers2) assuming real:
symb_sol2:=unapply(sol2,convert(drivers2,list)):


#Enumerate (there will generally be about 40, not 6)

count := 0;
for i1 from 1 to 40 do
     for i2 from 1 to 40 do
          for i3 from 1 to 40 do
               for i4 from 1 to 40 do
                    count := count + 1;
                    solsol2(count) := symb_sol2(i1,i2,i3,i4);
               od;
          od;
     od;
od;
count;



This works great!  I would like simliar output in my second code snippet, but this time with more inputs to symb_sol.  However, if I try and change the interconnection equations a little, and add a piecewise function, and another driver... (differences in bold)

#Interconnection Equations
eq1[1] := FD_T+EM2_T = 0;
eq1[2] := ICE_T+GBb_T = 0;
eq1[3] := GEN_T+GBa_T = 0;
eq1[4] := EM2_A+GEN_A+BAT_A = 0;
eq1[5] := -FD_W+EM2_W = 0;
eq1[6] := -GEN_W+GBa_W = 0;
eq1[7] := -ICE_W+GBb_W = 0;
eq1[8] := -EM2_V+GEN_V = 0;
eq1[9] := -EM2_V+BAT_V = 0;

#ICE
eq_c[1] := ICE_mdot_g=((671.5) + (-21.94)*ICE_T + (0.1942)*ICE_W + (0.5113)*ICE_T^2 + (-0.01271)*ICE_T*ICE_W + ( -0.0008761)*ICE_W^2 + (-0.006071)*ICE_T^3 + (9.867e-07)*ICE_T^2*ICE_W + (5.616e-05)*ICE_T*ICE_W^2 + (1.588e-06)*ICE_W^3 + (3.61e-05)*ICE_T^4 + (8.98e-07)*ICE_T^3*ICE_W + (-2.814e-07)*ICE_T^2*ICE_W^2 + (-8.121e-08)*ICE_T*ICE_W^3 + ( -8.494e-08 )*ICE_T^5 + (-2.444e-09)*ICE_T^4*ICE_W + (-9.311e-10)*ICE_T^3*ICE_W^2 + ( 5.835e-10)*ICE_T^2*ICE_W^3 ) *1/3600/1000 * ICE_T * ICE_W;

#BAT
eq_c[2] := BAT = 271;

#EM2
EM2_ReqPow_eq := (-148.3) + (4.267)*abs(EM2_W) + (12.77)*abs(EM2_T) + (-0.0364)*abs(EM2_W)^2 + ( 1.16)*abs(EM2_W)*abs(EM2_T) + (-0.258)*abs(EM2_T)^2 + ( 0.0001181)*abs(EM2_W)^3 + (-0.0005994)*abs(EM2_W)^2*abs(EM2_T) + ( 0.0001171)*abs(EM2_W)*abs(EM2_T)^2 + (0.001739 )*abs(EM2_T)^3 + (-1.245e-07 )*abs(EM2_W)^4 + ( 1.2e-06)*abs(EM2_W)^3*abs(EM2_T) + ( -1.584e-06)*abs(EM2_W)^2*abs(EM2_T)^2 + ( 4.383e-07)*abs(EM2_W)*abs(EM2_T)^3 + (-2.947e-06)*abs(EM2_T)^4;
eq_c[3] := EM2_P = piecewise( EM2_T = 0, 0, EM2_W = 0, 0, EM2_W*EM2_T < 0,-1 * EM2_ReqPow_eq, EM2_ReqPow_eq);
eq_c[4] := EM2_A = EM2_P/EM2_V;

#GEN
GEN_ReqPow_eq:= (-5.28e-12) + ( 3.849e-14)*abs(GEN_W) + (-71.9)*abs(GEN_T) + (-1.168e-16)*abs(GEN_W)^2 +(1.296)*abs(GEN_W)*abs(GEN_T) + (2.489)*abs(GEN_T)^2 + (1.451e-19)*abs(GEN_W)^3 + (0.0001326)*abs(GEN_W)^2*abs(GEN_T) + (-0.008141)*abs(GEN_W)*abs(GEN_T)^2 + (-0.004539)*abs(GEN_T)^3 +(-6.325e-23)*abs(GEN_W)^4 + (-2.091e-07)*abs(GEN_W)^3*abs(GEN_T) + ( 3.455e-06)*abs(GEN_W)^2*abs(GEN_T)^2 + ( 2.499e-05)*abs(GEN_W)*abs(GEN_T)^3 + (-5.321e-05)*abs(GEN_T)^4;
eq_c[5] := GEN_P = piecewise( GEN_T = 0, 0, GEN_W = 0, 0, GEN_W*GEN_T < 0,-1 * GEN_ReqPow_eq, GEN_ReqPow_eq);
eq_c[6] := GEN_A = GEN_P/GEN_V;

#GB
FiveSpeedGearbox_R := proc(ig)
local i ,eq;
i[1]:=3.32;
i[2]:=2;
i[3]:=1.36;
i[4]:=1.01;
i[5]:=0.82;
eq:= piecewise(ig=1,i[1],ig=2, i[2],ig=3,i[3],ig=4,i[4],ig=5,i[5],1);
return eq(ig);
end proc;


eq_c[7] := GBb_T = -1/GB_R * GBa_T;
eq_c[8] := GBb_W = GB_R * GBa_W;
eq_c[9] := GB_R = FiveSpeedGearbox_R(ig);

 

#System Equations
sys_eqs := convert(eq1,set) union convert(eq_c,set);

 

 #Solve for variables
symvarnames:=select(type,indets(convert(sys_eqs,list)),name);
drivers:= {ig, ICE_T,ICE_W,FD_T,FD_W};
not_drivers := symvarnames minus drivers;

#Assumptinons

assume(BAT_V::nonnegative);
assume(FD_W::nonnegative);

sol:=(solve(sys_eqs,not_drivers) assuming real);

symb_sol:=unapply(sol,convert(drivers,list)): ---> Error, (in unapply) variables must be unique and of type name

Subsequent parts don't work...

count := 0;
for i1 from 1 to 40 do
     for i2 from 1 to 40 do
          for i3 from 1 to 40 do
               for i4 from 1 to 40 do
                    for i5 from 1 to 40 do
                         count := count + 1;
                         solsol2(count) := symb_sol2(i1,i2,i3,i4,5);
                    od;
               od; 
          od;
     od;
od;
count;

While running the last line sol:, 1 of 2 things will happen, depending on the solver. Maple17 will take a long time (30+ minutes) to solve, then report nothing, or sol will solve, but will report "some solutions have been lost".

Afterwards, evaluating symb_sol(0,0,0,0,0) will return a viable solution (real values for each of the variables).  Whereas evaluating symb_sol(0,X,0,0,0), where X <> 0, will return and empty list [].

Does anyone know how to (i) speed up the symbolic solve time?  (ii) Return ALL of the solutions?

 

Thanks in advance for reading this.  I've really no idea why this isn't working.  I've also attached two worksheets with the code: noGB.mw   withGB.mw

 Adam

hi all.
i have a system of ODE's including 9 set of coupled OED's . 

i have  converted second deravaties to dd2 , in other words : diff(a[i](t),t,t)=dd2[i](t) . i =1..9 :

and i have set these 9 equations in form of vibrational equations such :  (M.V22)[i]+(K(t).V(t))[i]+P(t)[i] = eq[i] , where M is coefficient Matrix of second  derivatives , V22 is Vector of second derivaties , for example V22[1] = diff(a[1](t),t,t) , and  P(t) is the numeric part of equations ( they are pure number and do not contain any symbolic function ) and K(t).V(t) is the remaining part of equations such that : (K(t).V(t))[i] = eq[i] - (M.V22)[i] - P(t)[i]  , and V(t) are vector of a[i](t)'s which V(t)[1] = a[1](t) ,

i have used step by step time integration method (of an ebook which i have attachted that part of ebook here), when i set time step of solving process to h=0.01 , i can solve this system up to time one second or more, but when i choose h=0.001 or smaller, the answer diverges after 350 steps . i do not know whether the problem is in my ODS system, or maple can not handle this ?the answer about the time t=0.3 are the same in both steps, but after that, the one with stpe time h=0.001 diverges. my friend has solved this in mathematica without any problem, could any body help me ?! it is urgent for me to solve this problem,thnx everybody.


ebook.pdf  step_=_0.001.mw  step_=_0.01.mw 

Let  us consider the general case of symbolic values C(xC,yC). I make use of the idea suggested by edgar in http://www.mapleprimes.com/questions/97743-How-To-Prove-Morleys-Trisector-Theorem : no assumptions.

restart; with(geometry); point(A, 0, 0);
point(B, 1, 0);
point(C, xC, yC);
point(MA, (xC+1)*(1/2), (1/2)*yC);
point(MC, 1/2, 0);
point(MB, (1/2)*xC, (1/2)*yC);
point(E, (0+1+xC)*(1/3), (0+0+yC)*(1/3));# the center of mass
line(l1, x = 1/4, [x, y]);
The coordinates of the center of the first described circle are found as the solutions of the system of the equations of midperpendiculars.

midpoint(ae, A, E); coordinates(ae);


S1 := solve({x = 1/4, ((xC+1)*(1/3))*(x-(xC+1)*(1/6))+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

BTW, Maple can't create the midperpendiculars in this case.

point(O1, op(map(rhs, S1)));
                               O1

Simple details are omitted in the above. The coordinates of the centers of the two next described circles are found similarly.
coordinates(midpoint(mce, MC, E));

S2 := solve({x = 3/4, ((-1/2+xC)*(1/3))*(x-5/12-(1/6)*xC)+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

point(O2, op(map(rhs, S2)));

                               O2
coordinates(midpoint(bma, B, MA)); coordinates(midpoint(be, B, E));
  

                

S3 := solve({(xC-1)*(x-(xC+3)*(1/4))+yC*(y-(1/4)*yC) = 0, ((-2+xC)*(1/3))*(x-(4+xC)*(1/6))+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

point(O3, op(map(rhs, S3)));

                               O3

Now we find the equation of the circumference which passes through O1, O2, and O3.

eq := a*x+b*y+x^2+y^2+c = 0:
sol := solve({eval(eq, S1), eval(eq, S2), eval(eq, S3)}, {a, b, c});

A long output can be seen in the attached .mw file.

eq1 := eval(eq, sol);

  Now we find (in suspense)  the coordinates of the next center and verify whether it belongs to the sircumference O1O2O3.

coordinates(midpoint(mac, C, MA)); coordinates(midpoint(ec, E, C)); S4 := solve({(xC-1)*(x-(3*xC+1)*(1/4))+yC*(y-3*yC*(1/4)) = 0, ((2*xC-1)*(1/3))*(x-(4*xC+1)*(1/6))+(2*yC*(1/3))*(y-4*yC*(1/6)) = 0}, {x, y});

 point(O4, op(map(rhs, S4)));

                               O4
simplify(eval(eq1, S4));

                             0 = 0

Hope the reader will have a real pleasure to find the two residuary centers and to verify these on his/her own.

geom2.mw

 

 

 

 

It is well known that the medians of a triangle divide it into 6 triangles.
It is less known that the centers of their circumscribed circles belong to one circumference as drawn below

This remarkable theorem  was proved in the 21st century! Unfortunately, I lost its source.
I can't prove this difficult  theorem by hand. However, I can prove it with Maple.
The aim of this post is to expose these proofs. Everybody knows that it is scarcely possible
to construct a general triangle with help of the geometry package of Maple.
Without loss of generality one may assume that the vertex A is placed at the origin,
the vertex B is placed at (1,0), and the vertex C(xC,yC). We firstly consider the theorem
in the case of concrete values of xC and yC.

restart; with(geometry):with(plots):
point(A, 0, 0);
point(B, 1, 0);
xC := 15*(1/10); yC := sqrt(3); point(C, xC, yC);
triangle(T, [A, B, C]);
median(mA, A, T, MA);
median(mB, B, T, MB);
median(mC, C, T, MC);
line(m1, [A, MA]);
line(m2, [B, MB]);
intersection(E, m1, m2);
triangle(AEMB, [A, E, MB]);
circumcircle(c1, AEMB, 'centername' = C1);
circumcircle(c2, triangle(CEMB, [C, E, MB]), 'centername' = C2);
circumcircle(c3, triangle(CEMA, [C, E, MA]), 'centername' = C3);
circumcircle(c4, triangle(BEMA, [B, E, MA]), 'centername' = C4);
circumcircle(c5, triangle(BEMC, [B, E, MC]), 'centername' = C5);
circumcircle(c6, triangle(AEMC, [A, E, MC]), 'centername' = C6);
circle(CC, [C1, C2, C3]);
IsOnCircle(C4, CC);
                              true

IsOnCircle(C5, CC);
                              true
IsOnCircle(C6, CC);
                              true
display([draw([T(color = black), mA(color = black), mB(color = black), mC(color = black), C1(color = blue), C2(color = blue), C3(color = blue), C4(color = blue), C5(color = blue), C6(color = blue), CC(color = red)], symbol = solidcircle, symbolsize = 15, thickness = 2, scaling = constrained), textplot({[-0.5e-1, 0.5e-1, "A"], [.95, 0.5e-1, "B"], [xC-0.5e-1, yC+0.5e-1, "C"]})], axes = frame, view = [-.1 .. max(1, xC)+.1, 0 .. yC+.1]);

This can be done as a procedure in such a way.

restart; SixPoints := proc (xC, yC) geometry:-point(A, 0, 0); geometry:-point(B, 1, 0); geometry:-point(C, xC, yC); geometry:-triangle(T, [A, B, C]); geometry:-median(mA, A, T, MA); geometry:-median(mB, B, T, MB); geometry:-median(mC, C, T, MC); geometry:-line(m1, [A, MA]); geometry:-line(m2, [B, MB]); geometry:-intersection(E, m1, m2); geometry:-triangle(AEMB, [A, E, MB]); geometry:-circumcircle(c1, AEMB, 'centername' = C1); geometry:-circumcircle(c2, geometry:-triangle(CEMB, [C, E, MB]), 'centername' = C2); geometry:-circumcircle(c3, geometry:-triangle(CEMA, [C, E, MA]), 'centername' = C3); geometry:-circumcircle(c4, geometry:-triangle(BEMA, [B, E, MA]), 'centername' = C4); geometry:-circumcircle(c5, geometry:-triangle(BEMC, [B, E, MC]), 'centername' = C5); geometry:-circumcircle(c6, geometry:-triangle(AEMC, [A, E, MC]), 'centername' = C6); geometry:-circle(CC, [C1, C2, C3]); return geometry:-IsOnCircle(C4, CC), geometry:-IsOnCircle(C5, CC), geometry:-IsOnCircle(C6, CC), geometry:-draw([CC(color = blue), C1(color = red), C2(color = red), C3(color = red), C4(color = red), C5(color = red), C6(color = red), T(color = black), mA(color = black), mB(color = black), mC(color = black), c1(color = green), c4(color = green), c2(color = green), c3(color = green), c5(color = green), c6(color = green)], symbol = solidcircle, symbolsize = 15, thickness = 2) end proc;
SixPoints(1.5, 1.2);

true, true, true, PLOT(...)
 SixPoints(1.5, 1.2)[4];

See geom1.mw

To be continued (The general case will be considered in  part 2http://www.mapleprimes.com/posts/200210-Six-Points-On-Circumference-2 .).

 

 





 

Dear collegues

I wrote the following code

 


restart:
Digits := 15;
a[k]:=0;
b[k]:=7.47;
a[mu]:=39.11;
b[mu]:=533.9;
mu[bf]:=9.93/10000;
k[bf]:=0.597;
ro[p]:=3880 ;
ro[bf]:= 998.2;
c[p]:= 773;
c[bf]:= 4182;
#mu[bf]:=1;
Gr[phi]:=0; Gr[T]:=0;
#dp:=0.1;
Ree:=1;
Pr:=1;
Nbt:=cc*NBTT+(1-cc^2)*6;

#######################
slip:=0.1;         ####
NBTT:=2;           ####
lambda:=0.1;       ####
phi_avg:=0.02;    ####
#######################


eq1:=diff( (1+a[mu]*phi(eta)+b[mu]*phi(eta)^2)*diff(u(eta),eta),eta)+dp/mu[bf]+Gr[T]*T(eta)-Gr[phi]*phi(eta);
eq2:=diff((1+a[k]*phi(eta)+b[k]*phi(eta)^2)*diff(T(eta),eta),eta)+lambda*T(eta)/k[bf];
eq3:=diff(phi(eta),eta)+1/Nbt*diff(T(eta),eta);
Q:=proc(pp2,fi0) local res,F0,F1,F2,a,INT0,INT10;
global Q1,Q2;
print(pp2,fi0);
if not type([pp2,fi0],list(numeric)) then return 'procname(_passed)' end if:
res := dsolve({subs(dp=pp2,eq1)=0,eq2=0,eq3=0,u(0)=slip*D(u)(0),u(1)=-slip*D(u)(1),D(T)(0)=0,D(T)(1)=1,phi(0)=fi0}, numeric,output=listprocedure,continuation=cc);
F0,F1,F2:=op(subs(res,[u(eta),phi(eta),T(eta)])):
INT0:=evalf(Int(F0(eta),eta=0..1));
INT10:=evalf(Int(F0(eta)*F1(eta),eta=0..1));
a[1]:=evalf(Int(F0(eta),eta=0..1))-Ree*Pr;;
a[2]:=INT10/INT0-phi_avg;
Q1(_passed):=a[1];
Q2(_passed):=a[2];
if type(procname,indexed) then a[op(procname)] else a[1],a[2] end if
end proc;
Q1:=proc(pp2,fi0) Q[1](_passed) end proc;
Q2:=proc(pp2,fi0) Q[2](_passed) end proc;
Optimization:-LSSolve([Q1,Q2],initialpoint=[0.3,0.0007]);




se:=%[2];
res2 := dsolve({subs(dp=se[1],eq1)=0,eq2=0,eq3=0,u(0)=slip*D(u)(0),u(1)=-slip*D(u)(1),D(T)(0)=0,D(T)(1)=1,phi(0)=se[2]}, numeric,output=listprocedure,continuation=cc);
G0,G1,G2:=op(subs(res2,[u(eta),phi(eta),T(eta)])):
TTb:=evalf(Int(G0(eta)*G2(eta)*(G1(eta)*ro[p]*c[p]+(1-G1(eta))*ro[bf]*c[bf] ),eta=0..1))/evalf(Int(G0(eta)*(G1(eta)*ro[p]*c[p]+(1-G1(eta))*ro[bf]*c[bf] ),eta=0..1));
with(plots):
odeplot(res2,[[eta,phi(eta)/phi_avg]],0..1);
odeplot(res2,[[eta,T(eta)/TTb]],0..1);
odeplot(res2,[[eta,u(eta)/(Ree*Pr)]],0..1);

res2(1);
Nuu:=(1/TTb);
1/((1+a[k]*G1(1)+b[k]*G1(1)^2)/(1+a[k]*phi_avg+b[k]*phi_avg^2));
(1/TTb)*(((1+a[k]*G1(1)+b[k]*G1(1)^2)/(1+a[k]*phi_avg+b[k]*phi_avg^2)));
>

I want to run the code for the value of NBTT in the range of 0.2 to 10. this code gave the results in the range of 4-10 easily. So, I used the continuation which improve the range of the results between 2-10. However, I coudnt gave the results when 0.2<NBTT<2. Would you please help me in this situation.

Also, It is to be said that the values of phi should be positive. in some ranges, I can see that phi(1) is negative. Can I place a condition in which the values phi restricted to be positive.

Thanks for your attentions in advance

Amir

Hi,

I wrote the following code which is properly run

 


restart:

# parametrs

MUR:=(1-phi)^2.5:
RhoUR:=(1-phi+phi*rho[p]/rho[f]):
RhoCPR:=(1-phi+phi*rhocp[p]/rhocp[f]):
BetaUR:=(phi*rho[p]*beta[p]+(1-phi)*rho[f]*beta[f])/(RhoUR*rho[f])/beta[f]:

dqu3:=diff(h(x),x$1)-RhoUR*BetaUR*T(x);
dqu2:=5*diff(T(x),x$2)+k[f]/k[nf]*Pr*RhoCPR*f(x)*diff(T(x),x$1);
dqu1:=5/(MUR)*diff(f(x),x$3)
+ 2*(diff(h(x),x$1)*x-h(x))
+RhoUR*(3*f(x)*diff(f(x),x$2)-diff(f(x),x$1)^2);
rho[f]:=998.2: cp[f]:=4182: k[f]:=0.597:   beta[f]:= 2.066/10000:
rho[p]:=3380: cp[p]:=773: k[p]:=36:   beta[p]:= 8.4/1000000:

k[nf]:=((k[p]+2*k[f])-2*phi*(k[f]-k[p]))/((k[p]+2*k[f])+phi*(k[f]-k[p])):
rhocp[nf]:=rho[p]*cp[p]*phi+rho[f]*cp[f]*(1-phi):
rhocp[p]:=rho[p]*cp[p]:
rhocp[f]:=rho[f]*cp[f]:

phi:=0.00:
binfinitive:=6: Pr:=7: lambda:=0:


with(plots):
pppe:=dsolve( {dqu1=0,dqu2=0,dqu3=0,T(0)=1,T(binfinitive)=0,f(0)=0,D(f)(0)=lambda,D(f)(binfinitive)=0,h(binfinitive)=0}, numeric );
-pppe(0);
print(odeplot(pppe,[x,diff(f(x),x)],0..binfinitive,color=black,numpoints=400));
print(odeplot(pppe,[[x,diff(f(x),x)]],0..binfinitive,color=black,numpoints=400));
print(odeplot(pppe,[[x,T(x)]],0..binfinitive,color=black,numpoints=400));


However, in some range of parameters, I must increase the value of binfinitive (for example binfinitive=50). however, my code is doesnt converge for higher values of 10 (at most). Can anyone change this algorithm in a way that it insensitive to the value of binfinitive?

Many thanks for your attention in advance

 

Amir

Hello there,

i got a question regarding derivatives in Matlab.

 

I got a function for a example:

f:=f1(y)*f2(x)

 

from this function i need the partial derivatives with respect to x and y, i can easily get them with

diff(f,x)

diff(f,y)

 

now i want to compute the derivative of f with respect to time t, assuming that y and x both depend on t - how can i tell maple that y and x depend on t?

 

thanks

ben

I have the following nonlinear Differential Equation and don't know how to solve.  Can anyone give me any hints on how solvle for E__fd(t).  I don't even know the specific classification (other than nonlinear) of this DE can someone at least give me hint on that. Thanks.

 

.5*(diff(E__fd(t), t)) = -(-.132+.1*e^(.6*E__fd(t)))*E__fd(t)+0.5e-1

 

Thanks,

Melvin

Hi,

I have a problem with dsolve in the following code

restart;
>
n:=20;
m:=1;
cc:=-200;
zzeta:=0.1;
sefr1:=0.3;
sefr:=0.2;
MM:=0;
lambda:=0.1;
Br:=1;
nn:=3;
>
>
#u(tau):=tau;
u(tau):=421.7129935*tau-2217.587728*tau^2+8897.376593*tau^3-27612.59182*tau^4+64248.00336*tau^5-1.083977605*10^5*tau^6-10.57029600-1.080951714*10^6*tau^13+7.999517316*10^5*tau^14-4.788741005*10^5*tau^15+2.309563748*10^5*tau^16+26511.11102*tau^18-5959.001794*tau^19+1.148523882*10^5*tau^7-95.23809524*tau^21+4.545454545*tau^22-9435.563781*tau^8-2.587683745*10^5*tau^9+6.473880128*10^5*tau^10+948.0272727*tau^20-88660.41892*tau^17-1.008692404*10^6*tau^11+1.175504242*10^6*tau^12;
>
>
B := 1+(2*(1-zzeta))*Br*(int(tau*(diff(u(tau), tau))^2, tau = (1/2)*zzeta*(1-zzeta) .. 1/2*(1-zzeta)));
eq4 := 4*B*u(tau)-(1+zzeta)*(diff(tau*(diff(theta(tau), tau)), tau))/tau-(1+zzeta)*Br*(diff(u(tau), tau))^2;


theta(tau):=sum(p^ii*theta[ii](tau),ii=0..nn);
HH:= p*((4*(1+(2*(1-zzeta))*Br*(int(tau*(diff(u(tau), tau))^2, tau = (1/2)*zzeta*(1-zzeta) .. 1/2-(1/2)*zzeta))))*u(tau)-(1+zzeta)*(diff(theta(tau), tau)+tau*(diff(theta(tau), tau, tau)))/tau-(1+zzeta)*Br*(diff(u(tau), tau))^2)+(1-p)*(diff(theta(tau),tau$2)):
eq5:=simplify(HH):
eq6:=collect(expand(eq5),p);

eq7:=
convert(series(collect(expand(eq5), p), p, nn+1), 'polynom');


for ii to nn do
ss[ii] := (coeff(eq7, p^ii)) ;
print (ii);
end do;

ss[0]:=diff(theta[0](tau), tau, tau);

icss[0]:=theta[0](zzeta/(2*(1-zzeta)))=0, D(theta[0])(1/(2*(1-zzeta)))=1;

dsolve({ss[0], icss[0]});
theta[0](tau):= rhs(%);


for ii to nn do
ss[ii]:=evalf[5](ss[ii]);
icss[ii]:=theta[ii](zzeta/(2*(1-zzeta)))=0, D(theta[ii])(1/(2*(1-zzeta)))=0;
dsolve({ss[ii], icss[ii]});
theta[ii](tau):=rhs(%);
end do;

I would be most grateful if you help me to find this problem.

Thanks for your attention in advance

 

How to find the equation of the tangent line to f(x)=6x/squareroot(x^2+12) at the point (2,3) ?

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