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MaplePrimes Questions

plot solution of differential system...

Asked by: lucaud 45
plot
November 05 2016
0  4
sol_L := dsolve({de_L, ic});

    {x(t) = (-y0 - x0) exp(-2 t) + (y0 + 2 x0) exp(-t),
      y(t) = -2 (-y0 - x0) exp(-2 t) - (y0 + 2 x0) exp(-t)}
How i can plot this?thanks

Error with Physics[diff]...

Asked by: aamirkhan 60
physics diff
November 05 2016
1  7

Hi everybody;

I have a problem with Physics[diff] command. When I run the following code, error messages appear where the Physics[diff] command exist. What is the source of error? How can I fix it?

Thanks in advance

Q1.mw

FindDistribution command in Maple...

Asked by: Christopher2222 6030
November 05 2016
0  10

In response to Markiyan Hirnyks reference to Mathematica's FindDistribution command http://www.mapleprimes.com/questions/219668-Determine-The-Distribution-Function.  It would be nice to have something similar coded in Maple. 

Optimization of volumes solid with constraint...

Asked by: TheFixer 15
optimization volume lagrange-multipliers
November 05 2016
1  12

Hello.

I am taking an intermediate mathematics course. Now we are heading towards the finals and I have started to review all the topics we have been visiting during this semester.

Now I came across an excercise I cannot solve, taking into consideration what our lectures looks like and topics on the list my best bet is using lagrange multiplie method to optimize a multivariable function with constraints.

The task gives a shape that is drawn within the circle given by the equation: x^2+y^2=2.

The shape is a hexagon with 2 vertecies on the y-axsis +- the radius 2, the other 4 vertecies are the following [+-x,+-y].

I´m told that this hexagon is spinned around the y-axis to form a solid sylinder with 2 cones. The problem is to choose both radius and hight of the cylinder in order to maximize the volume.

The first problem that I dont know how i can plot this in maple, I would like to plot both the 2d hexagon and the solid spinned around the y-axsis

Also I´m not to confident what the constraint should look like.

I know how to use the lagrange multiplier by hand and can apply that inside maple, however I would like to use this opportunity to get to know the power of maple functionality more in detail.

https://i.gyazo.com/9d9585ddb8eb719d2a5bd24a1ba1671b.png

The link provoided is an image of the hexagon, i didnt find out how to use image tags.

how to configure maple to show every steps it run ...

Asked by: LeeHoYeung 535
November 05 2016
0  4

how to configure maple to show every steps it run when solve a equation or system?

if so,

can i run the steps again and return the same result as a solve function do?

solve Derivative in maple...

Asked by: 1red 10
derivative
November 05 2016
1  2

solve Derivative in maple

see in photo

 

generate a table ...

Asked by: VR88 20
table
November 04 2016
2  6

Hello! 

I want to generate a table in maple, but with generic content. For example, in Mathematica you can do this using 

table=Table[0,{n=10},{m=2}];
 

and this generates a table with 2 columns and 10 rows with zeros in all its elements. I want to do this exactly but in maple. 

Thanks for your time! 

How I can plot the following function?...

Asked by: cesarbioq 15
plot
November 04 2016
3  1

O(t) = 9.2-[(8.93*35.3)/(11.17-8.93)]*(exp(-8.93*t)-exp(-11.17*t))-8.3*exp(-11.17*t)

t betwen 0 and 1

Maplesoft application search - not working...

Asked by: Christopher2222 6030
application search
November 04 2016
0  5

I'd like to bring up all the Maple 2016 applications from the application center.

How do I do that?

Problem with shooting method...

Asked by: olivertwist 15
ode differential-equations
November 04 2016
1  3

Dear all,

I would like to ask you to help me with the following problem that I got error.

Thanks in advance

 

restart; Shootlib := "C:/Shoot9"; libname := Shootlib, libname; with(Shoot);
with(plots);
N1 := 1; N3 := .5; N2 := 5; Bt := 6; AA := N1*Bt; gamma1 := .2; blt1 := 1;
FNS := {f(eta), fp(eta), fpp(eta), g(eta), gp(eta), h(eta), hp(eta), i(eta), ip(eta), fppp(eta)};
ODE := {diff(f(eta), eta) = fp(eta), diff(fp(eta), eta) = fpp(eta), diff(fpp(eta), eta) = fppp(eta), diff(g(eta), eta) = gp(eta), diff(gp(eta), eta) = N1*(2*g(eta)+eta*gp(eta)+2*g(eta)*fp(eta)-2*f(eta)*gp(eta)+2*N2*N3*(h(eta)*ip(eta)-i(eta)*hp(eta))), diff(h(eta), eta) = hp(eta), diff(hp(eta), eta) = AA*(h(eta)+eta*hp(eta)-2*f(eta)*hp(eta)+2*h(eta)*fp(eta)), diff(i(eta), eta) = ip(eta), diff(ip(eta), eta) = AA*(2*i(eta)+eta*ip(eta)-2*f(eta)*ip(eta)+2*N2*h(eta)*gp(eta)/N3), diff(fppp(eta), eta) = N1*(3*fpp(eta)+(eta-2*f(eta))*fppp(eta)-(2*N2*N2)*(diff(hp(eta), eta)))};
IC := {f(0) = 0, fp(0) = gamma1*fpp(0), g(0) = 1+gamma1*gp(0), gp(0) = beta, h(0) = 0, hp(0) = beta1, i(0) = 0, ip(0) = beta2, fppp(0) = alpha};
BC1 := {f(blt1) = .5, fp(blt1) = gamma1*fpp(blt1), g(blt1) = gamma1*gp(blt1), h(blt1) = 1, i(blt1) = 1};
infolevel[shoot] := 1;
S := shoot(ODE, IC, BC1, FNS, [alpha = .1, beta = .2, beta1 = .3, beta2 = .4], maxfun = 50000);
 

collect of part of an expression...

Asked by: taro 505
collect subsindets frontend
November 03 2016
0  7

Hello people in mapleprimes,
I want to collect
aa:=sigma*omega*alpha = mu(H, H)*s*omega*gamma*L+omega^(-sigma+1)*tau*mu(F, F)*s*L-omega^(-sigma+1)*tau*mu(F, F)*s*L*gamma;
with omega^(-sigma+1);

I know that the above expression could be collected with tau, which
produces the same expression as the one collected with omega^(-sigma).
And, in collecting with omega^(-sigma+1), I know that freeze-thaw could be used.

But, I want to know another way to do the same thing, using frontend or 
subsindets. Is this possible?

Thanks in advance.

taro
 

Bug in dsolve numeric or fdiff?...

Asked by: Dr. Venkat Subramanian 431
dsolve numeric differential-equations
November 03 2016
0  3


To summarize,

dsolve numeric parametric form gives wrong answers when fdiff is used to calculate the Hessian. Doing dsolve twice gives the correct answer. Hope I am not making syntax or programming errors.

Also, fdiff is not compatible with vector form.

> 

restart;
> 

Digits:=15;

Digits := 15

 (1)
> 

sys:=diff(y1(t),t)=-(u+u^2/2)*y1(t),diff(y2(t),t)=u*y1(t);

sys := diff(y1(t), t) = -(u+(1/2)*(u^2))*y1(t), diff(y2(t), t) = u*y1(t)

 (2)

 The system is solved with initial conditions 1,0 and value of u being equal to u1 for t<=0.5, and u2 for t>0.5. The objective is y2 at t =1. There are errors when Hessian is calculated using parametric dsolve and fdiff

> 

sol1 := dsolve({sys, y1(0) = alpha, y2(0) = beta}, type = numeric, 'parameters' = [alpha, beta, u],maxfun=0,range=0..0.5):
> 

obj1:=proc(u1,u2)
local z1,s1,s2,z2;
global sol1;
sol1('parameters'=[1,0,u1]);
z1 := sol1(.5);
s1:=subs(z1,y1(t));
s2:=subs(z1,y2(t));
sol1('parameters'=[s1,s2,u2]):
z2:=sol1(0.5);
-subs(z2,y2(t));
end proc;

obj1 := proc (u1, u2) local z1, s1, s2, z2; global sol1; sol1('parameters' = [1, 0, u1]); z1 := sol1(.5); s1 := subs(z1, y1(t)); s2 := subs(z1, y2(t)); sol1('parameters' = [s1, s2, u2]); z2 := sol1(.5); -subs(z2, y2(t)) end proc

 (3)
> 

u0:=[0.8,1.8];

u0 := [.8, 1.8]

 (4)
> 

obj1(op(u0));

-.552540796143903

 (5)
> 

Hess1:=Matrix(2,2):for i from 1 to 2 do for j from 1 to 2 do Hess1[i,j]:=fdiff(obj1,[i,j],u0,workprec=1.0);od;od;Hess1;

Matrix(2, 2, {(1, 1) = 115.470000000000, (1, 2) = -1.11000000000000, (2, 1) = -1.11000000000000, (2, 2) = 2.22000000000000})

 (6)

 Hessian is wrong. It should be noted that when dsolve is called twice inside the objective function there is no error in Hessian calculation with fdiff. The parametric dsolve is efficient, how to get correct answers with that? Another question is how to use fdiff with vector form instead u1,u2 etc to make it generic for any number of variables?

> 

obj2:=proc(u1,u2)
local z1,s1,s2,z2,sol2;
sol2:=dsolve({op(subs(u=u1,[sys])), y1(0) = 1, y2(0) = 0}, type = numeric,maxfun=0,range=0..0.5);
z1 := sol2(.5);
s1:=subs(z1,y1(t));
s2:=subs(z1,y2(t));
sol2:=dsolve({op(subs(u=u2,[sys])), y1(0) = s1, y2(0) = s2}, type = numeric,maxfun=0,range=0..0.5):
z2:=sol2(0.5);
-subs(z2,y2(t));
end proc;

obj2 := proc (u1, u2) local z1, s1, s2, z2, sol2; sol2 := dsolve({op(subs(u = u1, [sys])), y1(0) = 1, y2(0) = 0}, type = numeric, maxfun = 0, range = 0 .. .5); z1 := sol2(.5); s1 := subs(z1, y1(t)); s2 := subs(z1, y2(t)); sol2 := dsolve({op(subs(u = u2, [sys])), y1(0) = s1, y2(0) = s2}, type = numeric, maxfun = 0, range = 0 .. .5); z2 := sol2(.5); -subs(z2, y2(t)) end proc

 (7)
> 

obj2(op(u0));

-.552540796143903

 (8)
> 

Hess2:=Matrix(2,2):for i from 1 to 2 do for j from 1 to 2 do Hess2[i,j]:=fdiff(obj2,[i,j],u0,workprec=1.0);od;od;Hess2;

Matrix(2, 2, {(1, 1) = .234295697631809, (1, 2) = 0.101868715004724e-1, (2, 1) = 0.101868715004724e-1, (2, 2) = 0.853480932660994e-1})

 (9)

 Doing dsolve twice gives correct answer compared to one parametric dsolve even though objective returns the same number (even gradient returns the expected answers).


 

Download Bugindsolveorfdiff.mws

a problem with fsolve...

Asked by: torabi 90
fsolve
November 03 2016
0  3

hi

please help me for fsolve algebric equations...

thanks

AGM.mw
 

> 

restart:
> 

F(eta):=sum(a[i]*eta^i,i=0..5):
> 

theta(eta):=sum(b[i]*eta^i,i=0..5):K(eta):=sum(c[i]*eta^i,i=0..5):Omega(eta):=sum(d[i]*eta^i,i=0..5):
> 

``
> 

U1:=diff(theta(eta), eta, eta)-3*Omega(eta)*(F(eta)*(diff(theta(eta), eta))-theta(eta)*(diff(F(eta), eta)))/(2*K(eta))+((diff(K(eta), eta))/K(eta)-(diff(Omega(eta), eta))/Omega(eta))*(diff(theta(eta), eta)) = 0:U2:= diff(F(eta), eta, eta, eta)+Omega(eta)*(3*F(eta)*(diff(F(eta), eta, eta))-(diff(F(eta), eta))^2)/(2*K(eta))+((diff(K(eta), eta))/K(eta)-(diff(Omega(eta), eta))/Omega(eta))*(diff(F(eta), eta, eta))+Omega(eta)/K(eta) = 0:U3:= diff(K(eta), eta, eta)+Omega(eta)*(1.5*F(eta)*(diff(K(eta), eta))-K(eta)*(diff(F(eta), eta)))/K(eta)+((diff(K(eta), eta))/K(eta)-(diff(Omega(eta), eta))/Omega(eta))*(diff(K(eta), eta))+(diff(F(eta), eta, eta))^2-Omega(eta)^2 = 0:U4:= diff(Omega(eta), eta, eta)+Omega(eta)*(3*F(eta)*(diff(Omega(eta), eta))+Omega(eta)*(diff(F(eta), eta)))/(2*K(eta))+((diff(K(eta), eta))/K(eta)-(diff(Omega(eta), eta))/Omega(eta))*(diff(Omega(eta), eta))+Omega(eta)*(diff(F(eta), eta, eta))^2/K(eta)-Omega(eta)^3/K(eta) = 0:
> 

F(eta):=unapply(F(eta),eta):
> 

theta(eta):=unapply(theta(eta),eta):K(eta):=unapply(K(eta),eta):Omega(eta):=unapply(Omega(eta),eta):
> 

U1:=unapply(U1,eta):U2:=unapply(U2,eta):U3:=unapply(U3,eta):U4:=unapply(U4,eta):
> 

 

> 

s1:=F(eta)(0) = 0:
> 

s2:=K(eta)(0) = 0:s3:=Omega(eta)(0) = 0:s4:=theta(eta)(0) = 1:s5:=theta(eta)(1) = 0:s6:=(D(F(eta)))(0) = 0:s7:=(D(K(eta)))(1) = 0:s8:=(D(Omega(eta)))(1) = 0:s9:=((D@@2)(F(eta)))(1) = 0

20*a[5]+12*a[4]+6*a[3]+2*a[2] = 0

 (1)
> 

s10:=U1(0):s11:=U2(0):s12:=U3(0):s13:=U4(0):        s14:=U1(1):s15:=U2(1):s16:=U3(1):s17:=U4(1):    s18:=D(U1)(0):s19:=D(U2)(0):s20:=D(U3)(0):s21:=D(U4)(0):     s22:=D(U1)(1):s23:=D(U2)(1):s24:=D(U3)(1):
> 

 

> 

 

> 

Q:=fsolve([s1,s2,s3,s4,s5,s6,s7,s8,s9,s10,s11,s12,s13,s14,s15,s16,s17,s18,s19,s20,s21,s22,s23,s24],{a[0],a[1],a[2],a[3],a[4],a[5],b[0],b[1],b[2],b[3],b[4],b[5],c[0],c[1],c[2],c[3],c[4],c[5],d[0],d[1],d[2],d[3],d[4],d[5]}):
> 

F(eta):=eval(sum(a[i]*eta^i,i=0..5),Q):

Error, invalid input: eval received S, which is not valid for its 2nd argument, eqns

  
> 

theta(eta):=eval(sum(b[i]*eta^i,i=0..5),Q):K(eta):=eval(sum(c[i]*eta^i,i=0..5),Q):Omega(eta):=eval(sum(d[i]*eta^i,i=0..5),Q):

Error, invalid input: eval received S, which is not valid for its 2nd argument, eqns

  
> 

plot(g(x),x=0..1,axes=boxed,color=green,thickness=2,labels=[x,g]):

Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct

  
> 

plot(f(x),x=0..1,axes=boxed,color=blue,thickness=2,labels=[x,f]):

Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct

  
> 

 


 

Download AGM.mw

 

Speed of calculation of the inverse of a matrix...

Asked by: petit loup 40
linear-algebra complexity
November 03 2016
0  8

I consider  100 .100 real matrices A,B=Matrix(100,100,(i,j)->rand()) (with 12 significant digits).  In general, ConditionNumber(A) is <10^5; also I choose Digits:=17.Theoretically, the complexity of the calculations of Determinant(A), CharacteristicPolynomial(A,x), A.B and MatrixInverse(A) are similar (~n^3). Yet, the times of these calculations are respectively: 0"13, 0"67, 0"60 and, what surprises me, 75" (moreover, I don't display any result).

My question: concerning the calculation of the inverse, where does this factor 100 come from ? Would Matlab  be 100 times faster ? I do not see why this would be the case; in particular, the standard methods for the calculation of the inverse are  easily programmable.

Thanks in advance.

 

fsolve 3 exponential equations with 3 unknown term...

Asked by: Giulianot 25
fsolve
November 02 2016
0  1


 

T := proc (t) options operator, arrow; 80+(-1)*1.4375*53*(1-exp((-1)*0.13775e-1*2.875*k*h__1*h__2*t/(1.4375^2*4190*(k*(h__1+h__2)+0.1e-2*h__1*h__2))))/2.875 end proc

proc (t) options operator, arrow; 80+(-1)*1.4375*53*(1-exp((-1)*0.13775e-1*2.875*k*h__1*h__2*t/((1.4375^2*4190*(k*(h__1+h__2)+0.1e-2*h__1*h__2)))))/2.875 end proc

 (1)

T(204.39)

53.50000000+26.50000000*exp(-8.094482719*k*h__1*h__2/(8658.242188*k*(h__1+h__2)+8.658242188*h__1*h__2))

 (2)

T(429.63)

53.50000000+26.50000000*exp(-17.01469059*k*h__1*h__2/(8658.242188*k*(h__1+h__2)+8.658242188*h__1*h__2))

 (3)

T(872.5)

53.50000000+26.50000000*exp(-34.55372656*k*h__1*h__2/(8658.242188*k*(h__1+h__2)+8.658242188*h__1*h__2))

 (4)

fsolve({53.50000000+26.50000000*exp(-34.55372656*k*h__1*h__2/(8658.242188*k*(h__1+h__2)+8.658242188*h__1*h__2)) = 56.15, 53.50000000+26.50000000*exp(-17.01469059*k*h__1*h__2/(8658.242188*k*(h__1+h__2)+8.658242188*h__1*h__2)) = 64.1, 53.50000000+26.50000000*exp(-8.094482719*k*h__1*h__2/(8658.242188*k*(h__1+h__2)+8.658242188*h__1*h__2)) = 72.05}, {h__1 = 0 .. 1000, h__2 = 0 .. 1000, k = 0 .. 1000})

fsolve({53.50000000+26.50000000*exp(-34.55372656*k*h__1*h__2/(8658.242188*k*(h__1+h__2)+8.658242188*h__1*h__2)) = 56.15, 53.50000000+26.50000000*exp(-17.01469059*k*h__1*h__2/(8658.242188*k*(h__1+h__2)+8.658242188*h__1*h__2)) = 64.1, 53.50000000+26.50000000*exp(-8.094482719*k*h__1*h__2/(8658.242188*k*(h__1+h__2)+8.658242188*h__1*h__2)) = 72.05}, {h__1, h__2, k}, {h__1 = 0 .. 1000, h__2 = 0 .. 1000, k = 0 .. 1000})

 (5)

``

it would be great if you could explain me why maple cant numerically solve this set of 3 ecuatios for h1, h2, k being my unknows terms, it works with an exponential function. there is a way to solve it?

thanks..
 

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