Maple 18 Questions and Posts

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

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);

Dear all

I solve the first-order PDE with a boundary condition contains a parameter  s

When I run the code there is no solution displayed using pdsolve

Many thanks for your help

 

 

 

PDEBCS.mw

How do I solve an overdetermined system of algebraic equations in Maple? solve command returns trivial solution for variables which are not actually trivial when I solve them by hand.

Hi all

When I solve using maple the first-order differential equation: diff(y,x)=0  for x in the closed interval [0,1] we obtain a constant function as a solution

but one can define the piecewise constant function see please the attached code

diff_piecewise.mw

why when we differentiate the piecewise function gives undefined derivative at point zero and a half.

Whats is the relationship between this example and Existence and uniqueness theorem for fist order ode

Many thanks

 

 

 

The solution to the logistic map .The solution now oscillates but doesn't appear to show any discernible pattern. The value of Xn seems to "jump around". This  called chaotic.

Dear all

I have an first-order ode

y'(x) =0, for x in [0,1] but x different to 1/2

and we define a function y(x)= 2   if 0<=x<0.5  and y(x)=0 for 0.5<=x <=1

The attached code define these function ( maybe piecewise function is not welle defined) and how can we show that y is solution or not of the differential equation

ode.mw

many thanks

 

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