Maple 18 Questions and Posts

These are Posts and Questions associated with the product, Maple 18

Dear all

I have a first-order complex ODE, I would like to use the polar coordinates to find the associated system of the ordinary differential equation then I determine the equilibrium point of the system and deduce that there is at least one limit cycle.

many thanks for your help

syestem_polar_cordinates.mw

 

I found that i can use  simplify under assumption to gain the result :

simplify((-p^3)^(1/3), assume = negative);
                               -p

simplify((p^3)^(1/3), assume = positive);
                               p


But confusing ! I expected the simplify command just with the option=symbolic works at the same manner ?


 

My purpose :

simplify((-p^7)^(1/7), assume = negative);

-p

(1)

simplify((p^7)^(1/7), assume = positive);

p

(2)

Without negative sign the simple symbolic result appears :

simplify((p^3)^(1/3), symbolic);

p

(3)

In power 3 the Imaginary part is included too !

simplify(((-p)^3)^(1/3), symbolic);

(1/2)*p*(I*3^(1/2)+1)

(4)

NOT WORKING !

simplify((p^7)^(1/7), symbolic);

p

(5)

simplify((-p^7)^(1/7), symbolic);

p*(-1)^(1/7)

(6)

simplify((-p^7)^(1/7), symbolic, radical);

(-p^7)^(1/7)

(7)

``

``


 

Download odd_negative_powers.mwodd_negative_powers.mw

 

n := 0
u[0] := x^3+(1/2)*A*x^2
for k from 0 to n do
A[k] := sum((Diff(u[i], x))*(Diff(u[k-i], x)), i = 0 .. k);
A[k] := sum((diff(u[i], x))*(diff(u[k-i], x)), i = 0 .. k)
end do;
 

it gives A[0]:=0 which is incorrect. why? and how it will give correct answer?

 Mathematical model for the temperature profiles of steel pipes quenched by water cooling rings

Hi

 Can anyone solve the given equations along with the boundary conditions analytically with Maple and draw the graphs ???????????

Dear maple users,

How to plot a figure for different values of rk?

like rk=5,10,15,20:

jb.mw

Thanking you,

Waiting for replay.

 

how can i use curve fitting tools for get m and n in this equation  for excel data

equation

i want to get m and n?

New_Microsoft_Excel_Worksheet.xlsx

 

Dear all

If we travel in straight lines in R^3. We begin at the point A=(1, 2, 3)
in the direction of the vector (1, 2, 2) and we end at the final point (10, 11, 12).

We made a single 90-degree turn.

Can we sketch a figure and we show the position where we take the turn?

 

Thanks

Dear all

I hope to solve a linear system AX=bk where A is a nxn matrix and X is a nx1 vector and b is a vector from the canonical basis of R^n ( for example in R^3 : b1=[ 1 0 0], b2 =[ 0 1 0] and b3=[ 0 0 1]

 

 

GramMat.mw

Thanks

Dear all

I have a function g that I compute its limit at x0 

but when I compute series(g(x),x=x0) then I compute the limit of the series I get different limit

 

series_limit.mw

 

Many thanks for your help

Dear all

Can we use maple to compute integral using the residual theorem

Compute_integral.mw

 

We can consider the contour a circle that contains one pole (Ipi/2)

Many thanks

 

Dear maple users,
Greetings.
I am solving an ode problem with an analytical solution.
programming running properly, but my plot not exact with the already existing article plot. 
how to get the exact plot.

Thanking you.

Code:JVB.mw
 

restart

N := 3;

3

 

1

(1)

dsolve(diff(f(x), `$`(x, 3)));

f(x) = (1/2)*_C1*x^2+_C2*x+_C3

(2)

Rf := 2*(diff(f[m-1](x), x, x, x))-(2*mh*mh)*(diff(f[m-1](x), x))+sum(f[m-1-n](x)*(diff(f[n](x), x)), n = 0 .. m-1)-bet*(sum(sum(2*f[m-1-n](x)*(diff(f[n-t](x), x))*(diff(f[t](x), x, x))+f[m-1-n](x)*f[n-t](x)*(diff(f[t](x), x, x, x))+x*(diff(f[m-1-n](x), x))*(diff(f[n-t](x), x))*(diff(f[t](x), x, x)), t = 0 .. n), n = 0 .. m-1));

2*(diff(diff(diff(f[m-1](x), x), x), x))-2*(diff(f[m-1](x), x))+sum(f[m-1-n](x)*(diff(f[n](x), x)), n = 0 .. m-1)-.2*(sum(sum(2*f[m-1-n](x)*(diff(f[n-t](x), x))*(diff(diff(f[t](x), x), x))+f[m-1-n](x)*f[n-t](x)*(diff(diff(diff(f[t](x), x), x), x))+x*(diff(f[m-1-n](x), x))*(diff(f[n-t](x), x))*(diff(diff(f[t](x), x), x)), t = 0 .. n), n = 0 .. m-1))

(3)

dsolve(diff(f[m](x), x, x, x)-CHI[m]*(diff(f[m-1](x), x, x, x)) = h*H*Rf, f[m](x));

f[m](x) = Int(Int(Int(CHI[m]*(diff(diff(diff(f[m-1](x), x), x), x))+2*h*(diff(diff(diff(f[m-1](x), x), x), x))-2*h*(diff(f[m-1](x), x))+h*(sum(f[m-1-n](x)*(diff(f[n](x), x)), n = 0 .. m-1))-(1/5)*h*(sum(sum(2*f[m-1-n](x)*(diff(f[n-t](x), x))*(diff(diff(f[t](x), x), x))+f[m-1-n](x)*f[n-t](x)*(diff(diff(diff(f[t](x), x), x), x))+x*(diff(f[m-1-n](x), x))*(diff(f[n-t](x), x))*(diff(diff(f[t](x), x), x)), t = 0 .. n), n = 0 .. m-1)), x), x)+_C1*x, x)+_C2*x+_C3

(4)

f[0](x) := 1-exp(x);

1-exp(x)

(5)

for m to N do CHI[m] := `if`(m > 1, 1, 0); f[m](x) := int(int(int(2*CHI[m]*(diff(f[m-1](x), x, x, x))-(2*h*H*mh*mh)*(diff(f[m-1](x), x))+h*H*(sum(f[m-1-n](x)*(diff(f[n](x), x)), n = 0 .. m-1)), x)-h*H*(sum(sum(2*f[m-1-n](x)*(diff(f[n-t](x), x))*(diff(f[t](x), x, x))+f[m-1-n](x)*f[n-t](x)*(diff(f[t](x), x, x, x))+x*(diff(f[m-1-n](x), x))*(diff(f[n-t](x), x))*(diff(f[t](x), x, x)), t = 0 .. n), n = 0 .. m-1))*bet, x)+_C1*x, x)+_C2*x+_C3; s1 := evalf(subs(x = 0, f[m](x))) = 0; s2 := evalf(subs(x = 0, diff(f[m](x), x))) = 0; s3 := evalf(subs(x = 1, f[m](x))) = 0; s := {s1, s2, s3}; f[m](x) := simplify(subs(solve(s, {_C1, _C2, _C3}), f[m](x))) end do:

f(x) := sum(f[l](x), l = 0 .. N);

1-0.7644444444e-1*exp(5.*x)*h^2*x-0.1333333333e-1*x^2*exp(5.*x)*h^2-2.675700596*exp(2.*x)*h^2*x-0.5876096022e-1*exp(6.*x)*h^3*x-0.9282030175e-2*x^2*exp(6.*x)*h^3+.9962792493*exp(3.*x)*h^3*x+.1647896790*exp(5.*x)*h^3*x+0.2066962962e-1*x^2*exp(5.*x)*h^3+3.357118680*exp(2.*x)*h^3*x-.3264340965*exp(4.*x)*h^3*x+0.3999999998e-1*exp(2.*x)*ln(exp(x))*h^2+58.61348006*h^3+1.023148148*h^2*x^3+0.1364197531e-1*ln(exp(x))*h^3*x^3-0.8954734530e-1*exp(2.*x)*h^3*x^4-.1353159884*x^3*exp(4.*x)*h^3+.7542645986*exp(3.*x)*h^3*x^2-0.2830138323e-1*x^3*h^3*exp(3.*x)-0.6455420536e-1*exp(x)*h^3*ln(exp(x))*x+0.4775858416e-1*exp(x)*h^3*ln(exp(x))*x^2+0.8888888887e-3*exp(x)*h^3*ln(exp(x))^2+8.400000000*h*exp(x)-exp(x)-0.6666666666e-1*h*ln(exp(x))+.1416666666*exp(4.*x)*h^2*x-.4790123458*exp(3.*x)*h^2*x+.1333333333*exp(3.*x)*h*x+.3791666665*exp(4.*x)*h^2-1.340020575*exp(3.*x)*h^2+.3111111109*exp(3.*x)*h+5.570191338*h^2*exp(2.*x)-.4500000000*h*exp(2.*x)-0.9874869443e-1*exp(6.*x)*h^3+.4125877323*exp(3.*x)*h^3-4.984787877*h^3*exp(2.*x)-.8010958741*exp(4.*x)*h^3+.3215641638*exp(5.*x)*h^3-5.930474628*h^2*x+36.04284024*exp(x)*h^3*x+8.324321524*x^2*h^2-.5362260993*h^3*x^3-6.207072379*exp(x)*x^2*h^3+1.664189246*exp(x)*h^3*x^3-8.237962963*h+.1200000000*exp(x)*h^2*ln(exp(x))+0.2222222222e-1*exp(3*x)*h*x+24.00299428*h^3*x-2.098561083*x^2*h^3-53.48457977*h^3*exp(x)+0.9949705035e-2*ln(exp(x))*h^3*x^4-0.7308641971e-2*ln(exp(x))*exp(4.*x)*h^3+0.8984910834e-2*ln(exp(x))*exp(3.*x)*h^3-0.3741666666e-1*ln(exp(x))*h^3*exp(2.*x)-.1188740741*exp(5.*x)*h^2-12.53662834*x^2*h+25.90916526*h^2*exp(x)-30.39962862*h^2-0.7499999999e-1*h*exp(2*x)+0.5185185185e-1*exp(3*x)*h+5.372840718*exp(x)*x^2*h^2-25.09181716*exp(x)*h^2*x+0.8976305409e-1*h^3*x^5+0.2158026099e-1*exp(7.*x)*h^3+0.8606919260e-1*h^3*x^4+0.5079365079e-3*x^3*exp(7.*x)*h^3-.3215468487*x^2*exp(4.*x)*h^3+0.1762236380e-1*exp(7.*x)*h^3*x+0.5048727639e-2*exp(7.*x)*x^2*h^3-3.116709690*exp(2.*x)*x^2*h^3+.1066289908*exp(2.*x)*h^3*x^3-8.527777777*h*x-0.2814814814e-2*ln(exp(x))*exp(4.*x)*h^3*x-0.1053497943e-2*ln(exp(x))*exp(3.*x)*h^3*x+0.4848332783e-1*h^3*x^6+.7462278773*h^2*x^4+.5519508187*exp(x)*h^3*x^4+0.9367631194e-1*exp(x)*h^3*ln(exp(x))+3.581893812*exp(2.*x)*x^2*h^2

 

 

NULL


 

Download JVB.mw

 

Analytical solution approach:

 

 

 

 

Dear all

I hope to find the supremum of the sequence of the function using maple 18, but when I run the code there is no results

maximize.mw

Many thanks

 

 

Hi, I have the following problem:

I want to plot the cone given by 1/16*(3x^2+10xz+3z^2-16y^2) and x>=0, z>=0. I tried it with

implicitplot3d([1/16*(3*x^2+10*x*z-16*y^2+3*z^2), x >= 0, z >= 0], x = -5 .. 15, y = -15 .. 15, z = -5 .. 15, grid = [30, 30, 30], style = surface);

But the result is one surface for each inequaility and not the cone.

If i restrict the range of x and z to be 0..15 and dismiss the additonal inequalities, a big part of the cone surface is missing somehow:

Does anybody know how to fix this? Do I have to use another plotcommand?

Thanks for your help!

;

= 2*1/10);
                               1
                               -
                               5
= -1;
                               -1

= 10;
                               10
= -25;
                              -25

C= 1;
                               1

= (1/12*sqrt6/sqrtbeta*lambda*mu);
                       1     (1/2)  (1/2)
                      --- I 6      2     
                       24                
= alpha/((10*sqrt-lambda*mu)*beta);
                             1  (1/2)
                           - - 5     
                             4       
A0]= (1/2*alpha/((10*sqrt-lambda*mu)*(1/12*beta*sqrt6/sqrtbeta*lambda*mu)));
                    1    (1/2)  (1/2)  (1/2)
                   -- I 5      6      2     
                    4                       
A1]= -(1/10*alpha/((1/12*beta*mu*sqrt6/sqrtbeta*lambda*mu));
                        1    (1/2)  (1/2)
                       -- I 6      2     
                        2                
A[2] := -(12*((112)*sqrt(6)/sqrt(beta*lambda*mu)))*lambda^2*alpha/(10*sqrt(-lambda*mu));
                    1     (1/2)  (1/2)  (1/2)
                   --- I 6      2      5     
                    20                       
H := ln(sqrt(lambda/(-mu))*tanh(sqrt(-lambda*mu)*(xi+C)));
              /1/2)     /1  (1/2         \\
              ln|- 5      tanh|- 5      ( 1)||
                \5            \5                //
xi := k*x-t*w;
                1     (1/2)  (1/2)     1    (1/2)
               --- I 6      2      x + t    
        2               4         
u[0] := A[0]+A[1]*exp(-H)+A[2]*exp(-H)*exp(-H);
  1    (1/2)  (1/2)  (1/2)
 -- I 5      6      2     
  4                       

                     1    (1/2)  (1/2)  (1/2)                
                     - I 6      2      5                     
                     2                                       
    - -------------------------------------------------------
         (1/2) / 1     (1/2)  (1/2)     1    (1/2)    \\
      tanh|- 5      |--- I 6      2      x + - t 5      + 1||
          \5        \ 24                     4             //

                      1    (1/2)  (1/2)  (1/2)                
                      - I 6      2      5                     
                      4                                       
    - --------------------------------------------------------
                                                             2
          (1/2) / 1     (1/2)  (1/2)     1    (1/2)    \\ 
      tanh|- 5      |--- I 6      2      x + - t 5      + 1|| 
               \ 24              4             // 


plot3d(Im(u[0]), x = -10 .. 10, t = -10 .. 10);

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