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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!



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,


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 :)



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







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.


 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):
 for i from 2 to N do
 end do:
 So,my system:



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

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

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



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

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);

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





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).



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


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 ) );



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


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


> 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);



Respected sir,

     Maple package is very useful to all the Mathematics students to work with. however, i came to know how to solve

system coupled non-linear ODEs with boundary condtions in Maple 12.0. But my doubt is, the tool 'Boundary value problem solver' which methods and submethods are used to solve those BVPs. Sir please, clear my doubt. your response is highly helpfull for me. Please kindly do the need full.

Thanking you.

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