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Hi everyone. Can enuone help with small parameter method?

restart;

eq := diff(x(t), `$`(t, 2))-epsilon(alpha*x(t)^4-beta)*(diff(x(t), t))-x(t)+x(t)^3;

subs(x(t) = 1+epsilon*x[1](t)+epsilon^2*x[2](t), eq);

pls help with this...

Hello. I have an inequality and I need to prove or negate if it is true or false. This inequality has 8 variables. I simplify it and try to see if it is ture or false. I tried "test relation" in maple and it seems I can't say it is always true or false. For some values of the variables it is true and for some others its false. Is there a method I can show if this inequlity is hold under some assumptions? I mean I want to keep some variables as constant and prove it up to a point. My inequlity is below. Thank you for the help in advance.


(P[A]*(p-w)/(1-P[A])-c)*H[A]+(w-P[A]*(p-w)/(1-P[A]))*P[A]*H[A]+w[u]*P[B]*(1-P[A])*H[B] < (P[B]*(p-w)/(1-P[B])-c)*H[B]+(w-P[B]*(p-w)/(1-P[B]))*P[B]*H[B]+w[u]*P[A]*(1-P[B])*H[B]

And this is how it looks on maple:

Hi, I am completely new to Maple, and I need to use it to optimize my equations in order to make my PLC codes more compressed. I am calculating forward kinematics with the Denavit-Hartenberg method and as such I get long expressions. After a lot of google'ing and frustration, I thought I'd ask here in the hope that one of you might be able to assist me.

I have the following equations;

X := L10*cos(q5) - L16*(sin(q10)*(sin(q5)*sin(q8) - cos(q8)*(cos(q5)*cos(q6)*cos(q7) - cos(q5)*sin(q6)*sin(q7))) - cos(q10)*(sin(q9)*(cos(q8)*sin(q5) + sin(q8)*(cos(q5)*cos(q6)*cos(q7) - cos(q5)*sin(q6)*sin(q7))) + cos(q9)*(cos(q5)*cos(q6)*sin(q7) + cos(q5)*cos(q7)*sin(q6)))) - d2*(cos(q10)*(sin(q5)*sin(q8) - cos(q8)*(cos(q5)*cos(q6)*cos(q7) - cos(q5)*sin(q6)*sin(q7))) + sin(q10)*(sin(q9)*(cos(q8)*sin(q5) + sin(q8)*(cos(q5)*cos(q6)*cos(q7) - cos(q5)*sin(q6)*sin(q7))) + cos(q9)*(cos(q5)*cos(q6)*sin(q7) + cos(q5)*cos(q7)*sin(q6)))) + L15*(sin(q9)*(cos(q8)*sin(q5) + sin(q8)*(cos(q5)*cos(q6)*cos(q7) - cos(q5)*sin(q6)*sin(q7))) + cos(q9)*(cos(q5)*cos(q6)*sin(q7) + cos(q5)*cos(q7)*sin(q6))) - L11*cos(q5)*sin(q6) + d1*cos(q5)*cos(q6) - L13*sin(q5)*sin(q8) + L14*cos(q9)*(cos(q8)*sin(q5) + sin(q8)*(cos(q5)*cos(q6)*cos(q7) - cos(q5)*sin(q6)*sin(q7))) + L13*cos(q8)*(cos(q5)*cos(q6)*cos(q7) - cos(q5)*sin(q6)*sin(q7)) - L14*sin(q9)*(cos(q5)*cos(q6)*sin(q7) + cos(q5)*cos(q7)*sin(q6)) + L12*cos(q5)*cos(q6)*cos(q7) - L12*cos(q5)*sin(q6)*sin(q7);

Y := L10*sin(q5) - L9 + L16*(sin(q10)*(cos(q5)*sin(q8) - cos(q8)*(sin(q5)*sin(q6)*sin(q7) - cos(q6)*cos(q7)*sin(q5))) - cos(q10)*(sin(q9)*(cos(q5)*cos(q8) + sin(q8)*(sin(q5)*sin(q6)*sin(q7) - cos(q6)*cos(q7)*sin(q5))) - cos(q9)*(cos(q6)*sin(q5)*sin(q7) + cos(q7)*sin(q5)*sin(q6)))) + d2*(cos(q10)*(cos(q5)*sin(q8) - cos(q8)*(sin(q5)*sin(q6)*sin(q7) - cos(q6)*cos(q7)*sin(q5))) + sin(q10)*(sin(q9)*(cos(q5)*cos(q8) + sin(q8)*(sin(q5)*sin(q6)*sin(q7) - cos(q6)*cos(q7)*sin(q5))) - cos(q9)*(cos(q6)*sin(q5)*sin(q7) + cos(q7)*sin(q5)*sin(q6)))) - L15*(sin(q9)*(cos(q5)*cos(q8) + sin(q8)*(sin(q5)*sin(q6)*sin(q7) - cos(q6)*cos(q7)*sin(q5))) - cos(q9)*(cos(q6)*sin(q5)*sin(q7) + cos(q7)*sin(q5)*sin(q6))) + L13*cos(q5)*sin(q8) - L11*sin(q5)*sin(q6) + d1*cos(q6)*sin(q5) - L14*cos(q9)*(cos(q5)*cos(q8) + sin(q8)*(sin(q5)*sin(q6)*sin(q7) - cos(q6)*cos(q7)*sin(q5))) - L13*cos(q8)*(sin(q5)*sin(q6)*sin(q7) - cos(q6)*cos(q7)*sin(q5)) - L14*sin(q9)*(cos(q6)*sin(q5)*sin(q7) + cos(q7)*sin(q5)*sin(q6)) + L12*cos(q6)*cos(q7)*sin(q5) - L12*sin(q5)*sin(q6)*sin(q7);

Z := L15*(cos(q9)*(cos(q6)*cos(q7) - sin(q6)*sin(q7)) - sin(q8)*sin(q9)*(cos(q6)*sin(q7) + cos(q7)*sin(q6))) - L11*cos(q6) - L8 - d1*sin(q6) + L16*(cos(q10)*(cos(q9)*(cos(q6)*cos(q7) - sin(q6)*sin(q7)) - sin(q8)*sin(q9)*(cos(q6)*sin(q7) + cos(q7)*sin(q6))) - cos(q8)*sin(q10)*(cos(q6)*sin(q7) + cos(q7)*sin(q6))) - d2*(sin(q10)*(cos(q9)*(cos(q6)*cos(q7) - sin(q6)*sin(q7)) - sin(q8)*sin(q9)*(cos(q6)*sin(q7) + cos(q7)*sin(q6))) + cos(q8)*cos(q10)*(cos(q6)*sin(q7) + cos(q7)*sin(q6))) - L13*cos(q8)*(cos(q6)*sin(q7) + cos(q7)*sin(q6)) - L14*sin(q9)*(cos(q6)*cos(q7) - sin(q6)*sin(q7)) - L12*cos(q6)*sin(q7) - L12*cos(q7)*sin(q6) - L14*cos(q9)*sin(q8)*(cos(q6)*sin(q7) + cos(q7)*sin(q6));

 

I need to optimize these equations, but still keep them separate. I would like to use mutual expressions for the calculations within, but still as I said keep the outputs of X, Y and Z separate.

This is MATLAB code.

 

Thanks in advance for any help.

I'm trying to solve a system of four pdes and I know that the Newton method won't converge.

Are there other numerical methods that I can use?

Any help would be greatly appreciated!

Thanks,

Eve

hai everyone. i am currently trying to solve an integration of the following ∫g(η)dη . integrate from 0 to 10.

from the following odes.

f ''' +1-(f ')2 +ff ''=0,

g''-gf'+fg'=0,

with boundary conditions f(0)=0, f'(0)=λ, f'(∞)=1, g(0)=1,g(∞)=0

First, i solve the odes using the shooting method. then i used the trapezoidal rule to solve for the integration of g(eta) using the following codes

> with(student);
> trapezoid(g(eta), eta = 0 .. 10, 10);
> evalf(%);

it seems that it can not read the data from the shooting method. can anyone suggest why it is happening?

thank you verymuch for your concern :)

hi 

 

i solved my equation as follow , i want to know a compelet describtion that which method of pde/numeric methods is using .

 

thanks 


 

 

 

 

Hi all!

 

I do a small calculation and get a system of 6
nonlinear equations.
And "n" is the degree of the equation is float.

Here are the calculations that lead to the system.

 

restart;
 with(DirectSearch):
 B:=1: 
 q:=1: 
 l:=1: 
 n:=4.7:
 V:=0.05:
 N:=1200:
 
 
 kappa:=Vector(N+1,[]):
 theta:=Vector(N+1,[]):
 u:=Vector(N,[]):
 M:=Vector(N,[]):
 Z:=Vector(N,[]):
 
 M_F:=q*(6*l*(z-l)-z^2/2):
 M_1:=piecewise((z<l), l-z, 0):
 M_2:=piecewise((z<2*l), 2*l-z, 0):
 M_3:=piecewise((z<3*l), 3*l-z, 0):
 M_4:=piecewise((z<4*l), 4*l-z, 0):
 M_5:=piecewise((z<5*l), 5*l-z, 0):
 M_6:=6*l-z:
 M_finish:=(X_1,X_2,X_3,X_4,X_5,X_6,z)->M_1*X_1+M_2*X_2+M_3*X_3+M_4*X_4+M_5*X_5+M_6*X_6+M_F:
 
 
 kappa_old:=0:
 theta_old:=0:
 u_old:=0:
 M_old:=0:
 
 
 step:=6*l/N:
 u[1]:=0:
 kappa[1]:=0:
 theta[1]:=0:
 
 
 
 
 for i from 2 to N do
 
 z:=i*step:
 kappa_new:=kappa_old+B/V*(M_finish(X_1,X_2,X_3,X_4,X_5,X_6,z))^n*step:
 
 theta_new:=theta_old+1/2*(kappa_old+kappa_new)*step:
 
 u_new:=u_old+1/2*(theta_old+theta_new)*step:
 
 Z[i]:=z:
 kappa[i]:=kappa_new:
 theta[i]:=theta_new:
 u[i]:=u_new:
 kappa_old:=kappa_new:
 theta_old:=theta_new:
 u_old:=u_new:
 
 end do:
 
 So,my system:


 u[N/6]=0;
 u[N/3]=0;
 u[N/2]=0;
 u[2*N/3]=0;
 u[5*N/6]=0;
 u[N]=0;

 

I want to ask advice on how to solve the system.
I wanted to use Newton's method, but I don't know the initial values X_1..X_6.

Tried to set the values X_1..X_6 and to minimize the functional
Fl:=(X_1,X_2,X_3,X_4,X_5,X_6)->(u[N/6])^2+(u[N/3])^2+(u[N/2])^2+(u[2*N/3])^2+(u[5*N/6])^2+(u[N])^2:

with the help with(DirectSearch):
GlobalOptima(Fl);
But I don't know what to do next

Please, advise me how to solve the system! I would be grateful for examples!

 

Hi,

Thank you very much for your idea in previous discussion: this is the link.

http://www.mapleprimes.com/questions/202744-Calculs-Using-Maple

I asked how can I plot the phase portrait of this system

Sys1 := {diff(r(t),t) = r(t)^2*sin(theta(t)), diff(theta(t),t) = -r(t)^2*(-2*cos(theta(t))^2+1)};

I get this answer:

 

1) For me it's not necessary to give an initial condition to plot the pharse portrait.

2) What is the line in this phase portrait.

3) Here, I try a second method using Matlab code, I get:

 

and this is the code:

[r, theta] = meshgrid(0:pi/4:2*pi, 0:pi/4:2*pi);
rdot =r.^2.*sin(theta) ;
thetadot = r.^2.*(1-2*sin(theta).^2);
quiver(r,theta,rdot,thetadot)
xlabel('r')
ylabel('theta')

the arrow in the phase portrait are not the same.

Can someone give me more clear information about this problem.

Many thinks.

 

I am a problem with solve differential equation, please help me: THANKS 

g := (y^2-1)^2; I4 := int(g^4, y = -1 .. 1); I5 := 2*(int(g^3*(diff(g, y, y)), y = -1 .. 1)); I6 := int(g^3*(diff(g, y, y, y, y)), y = -1 .. 1); with(Student[Calculus1]); I10 := ApproximateInt(6/(1-f(x)*g)^2, y = -1 .. 1, method = simpson);

dsys3 := {I4*f(x)^2*(diff(f(x), x, x, x, x))+I5*f(x)^2*(diff(f(x), x, x))+I6*f(x)^3 = I10, f(-1) = 0, f(1) = 0, ((D@@1)(f))(-1) = 0, ((D@@1)(f))(1) = 0};

dsol5 := dsolve(dsys3, numeric, output = array([0.]));

              Error, (in dsolve/numeric/bvp) system is singular at left endpoint, use midpoint method instead

****************FORMAT TWO ********************************************************

g := (y^2-1)^2; I4 := int(g^4, y = -1 .. 1); I5 := 2*(int(g^3*(diff(g, y, y)), y = -1 .. 1)); I6 := int(g^3*(diff(g, y, y, y, y)), y = -1 .. 1); with(Student[Calculus1]); I10 := ApproximateInt(6/(1-f(x)*g)^2, y = -1 .. 1, method = simpson);
dsys3 := {I4*f(x)^2*(diff(f(x), x, x, x, x))+I5*f(x)^2*(diff(f(x), x, x))+I6*f(x)^3 = I10, f(-1) = 0, f(1) = 0, ((D@@1)(f))(-1) = 0, ((D@@1)(f))(1) = 0};

dsol5 := dsolve(dsys3, method = bvp[midrich], output = array([0.]));
%;
                                   Error, (in dsolve) too many levels of recursion

I DONT KNOW ABOUT THIS ERROR

PLEASE HELP ME

THANKS A LOT

 

Hi every body:

how can i solve this equations(without numerical method):

eq1 := (D[1, 1](eta11))(t, a*t, a^2*t)+1.326096634*10^8*Pi^2*eta11(t, a*t, a^2*t)-3.315241586*10^7*Pi*eta21(t, a*t, a^2*t) = 0

eq2 := 2.054901810*10^13*eta21(t, a*t, a^2*t)+(D[1, 1](eta21))(t, a*t, a^2*t)-8.219607239*10^13*Pi*eta11(t, a*t, a^2*t)+4.137421500*10^8*Pi^2*eta21(t, a*t, a^2*t) = 0

eq3 := (D[1, 1](eta31))(t, a*t, a^2*t)+4.137421500*10^8*Pi^2*eta31(t, a*t, a^2*t) = 0

Dear all,

I would like to solve the Fredholm Integral equation, using numerical method.
This is my code.

there is a problem with subs, does not working here.
# Then, we obtain from (9) the coeficient A[n] and B[n].

Then I woulk like to recompute (2), and then compute (1).
# Puting x=m*h, in (1), how can we generate a linear Matrix from (1).

 

Fred.mw

Thanks

Hi.

please, I need some code about stepsize adaptative Runge Kutta method ( any order ) or other methods.

Thanks

Dear all,

Thank you for your Help.

h: stepsize;

x in [0,x0];

I give all the step of my code, but I think there is a mistake. I wait for your Help.

I would like to compute the error between  Method Huen with step size h and step size 2h using the definition of epsilon given below:

 ## The error written epsilon(x0,h)= sqrt(1/(N+1) * sum_i=0^N  (y_i^{2h}-y_(2i)^h)^2 ), where y_i^(2h) is the approximation of y(i*2*h).

 ## We want : loglog epsilon versus h.

  epsilon:=(x0,h)->sqrt( 1/(N+1)*add( (Heun1(f,x0,i)-Heun2(f,x0,i))^2,i=0..N ) );

  f:=(x,y)=1/(1+cos(y)); 

  ode:=diff(y(x),x)=f(x,y);

ic:=y(0)=1;  h:=x0/(2*N);

## Method Heun with step size 2h

> Heun1 := proc (f, x0,)

local x, y, i, h, k;

y := Array(0 .. N);

x := Array(0 .. N);

h := evalf((1/2)*x0/N);

x[0] := 0;

y[0] := 1;

for i from 0 to N do

x[i+1] := (2*i+2)*h;

k[1] := f(x[i], y[i]);

k[2] := f(x[i]+h, y[i]+h*k[1]);

y[i+1] := y[i]+h*((1/2)*k[1]+(1/2)*k[2]);

end do;

[seq([x[i], y[i]], i = 0 .. N)];

end proc;

### Now Heun with step size h  ( the same h)

> Heun2 := proc (f, x0,)

local x, y, i, h, k;

y := Array(0 .. N);

x := Array(0 .. N);

h := evalf((1/2)*x0/N);

x[0] := 0;

y[0] := 1;

for i from 0 to N do

x[i+1] := (i+1)*h;

k[1] := f(x[i], y[i]);

k[2] := f(x[i]+h, y[i]+h*k[1]);

y[i+1] := y[i]+h*((1/2)*k[1]+(1/2)*k[2]);

end do;

[seq([x[2*i], y[2*i]], i = 0 .. N)];

end proc;

 

 

Thanks you for your help.


                                

                        

 

I can get the function to iterate as a recursive function by just reevaluating the x := evalf(x-(f1*j-i*g1)/(h*k-i*j), 25); y := evalf(y-(h*g1-f1*j)/(h*k-i*j), 25) portion of the function below but im trying beneath it to assign it as newt2d so that i can iterate it as newtons method in two variables like (newt2d@@10) and I can't seem to figure out what im doing wrong. Thanks for any help you can provide!

f := proc (x, y) options operator, arrow; x+y-cos(x)+sin(y-1) end proc; f1 := f(x, y)

(x, y) -> x + y - cos(x) + sin(y - 1)
x + y - cos(x) + sin(y - 1)

> g := proc (x, y) options operator, arrow; x^4+y^4-2*x*y end proc; g1 := g(x, y);

(x, y) -> x + y - 2 x y
x + y - 2 x y

> dh := D[1](f); h := dh(x, y);

(x, y) -> 1 + sin(x)
1 + sin(x)

> di := D[2](f); i := di(x, y);

(x, y) -> 1 + cos(y - 1)
1 + cos(y - 1)

> dj := D[1](g); j := dj(x, y);

(x, y) -> 4 x - 2 y
4 x - 2 y

> dk := D[2](g); k := dk(x, y);

(x, y) -> 4 y - 2 x
4 y - 2 x

x := .3; y := .8

0.3
0.8

> x := evalf(x-(f1*j-i*g1)/(h*k-i*j), 25); y := evalf(y-(h*g1-f1*j)/(h*k-i*j), 25);

0.2924403963319692595180140
0.8321243516906678979858730

> newt2d(.3, .8);

0.2577789764, 0.8333916830

> (newt2d@@5)(.3, .8);

Error, (in @@) invalid arguments



Dear all,

here, I propose two methods for Adams Moulton Methos, but which one can I used.

The n-step Adams Moulton method to solve y'(x)=F(x,y(x)) is defined by the stencil

y(x+h)=y(x)+h *sum_{j=-1}^{n-1} beta_j F( x-j*h, y(x-j*h) ) + O(h^{n+2})

I want a procedure with single argument ''n'' that calculates and return the ''beta_i'' coefficients

I get two Methods. Which one correspond to my question please, and I don't understand the procedure proposed.

For me; the first give the iterative schemae used, but don't return the vector coefficients ( beta_i) and this methode method an interpolation of the function.

The second method, there is a function f, how this function is maded, and the same for the matrux A and the vector b...

the First Method:

> Adamsmoulton := proc (k::posint)

local P, t, f, y, n;

P := interp([t[n]+h, seq(t[n]-j*h, j = 0 .. k-2)], [f[n+1], seq(f[n-j], j = 0 .. k-2)], x);

y[n+1] = y[n]+simplify(int(P, x = t[n] .. t[n]+h))

end proc;

 

Second Method:

f:=proc(x,y) if x =0 and y=0 then 1 else x**y fi end;

n:=3; A:=matrix(n,n,(i,j)->f(1-j,1-i));

b:=vector(n, i->1/i);

linsolve(A,b);

 

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