solve([a = -(-y+1)/(x-y+2), b = -(-x^2+2*x*y-y^2-3*x+3*y-2)/(x-2*y+3), c = -(x*y-y^2-2*x+4*y-4)/(x*y-y^2-x+2*y-1), d = -(-x*y+y^2+x-2*y+1)/(x-2*y+3)], [x,y]);

i do not know whether multiple solutions lead no solutions in this.

if so, is it possible to show all possible solutions of x and y in terms of a,b,c,d ?

Download Bryson_sesson1_p6.mw

I am exploring the onset of chaos in the driven damped pendulum. I have set up the non-linear initial value problem that describes the situation in terms of phi(t), the angle of the pendulum from the vertical. I can solve the differential equation numerically for certain initial conditions using dsolve, and then graph that solution using odeplot. I can then explore what happens when I change the initial conditions ever so slightly. Again I can solve this new initial value problem numerically with dsolve, and then graph it with odeplot, Suppose the solution to the first problem is phi1(t), and the solution to the second initial value problem phi2(t). I would like to plot log | phi2(t) - phi1(t) | versus t, that is, the logarithm of the absolute value of the difference of the two solutions versus t. I’m interested in how sensitive the system is to initial conditions.

So, how can I disentangle the two solutions, phi1(t) and phi2(t), from dsolve and then plot the graph I want?

I have an implcit function to plot. I can plot it with the parameter but that puts the parameter on the x axis instead of the y axis. What is the best way to present the plot with the parameter on the y axis?

T; R0 := 0.1003183099e-5; T;
/ / -15 -34
- \0.00007363071999 h \3.947841762 10 h + 3.158273412 10

-7 \\// (1/2)
- 3.947841763 10 h R0// |/ 2 -7 \
\\-Pi h + 1.256637062 10 R0/

\
2 / -15 -7 \|
R0 \-1.256637062 10 + 1.256637062 10 R0//
0.000001003183099
1 / 20 /
- --------------------------------- \5.862184047 10 h \4.00
(1/2)
/ 2 -13\
\-Pi h + 1.260637062 10 /

-16 2 -13 -34\\
10 Pi h - 1.260637062 10 Pi h + 3.158273412 10 //
plot(T, h = 0.1e-8 .. 0.10e-7, title = Tension*vs*height*of*dome*above*plane, labels = [m, N/m]);

I would like the plot to be inverted to show the height of the dome as a function of tension;

Maple's isprime is not a definitive primality test. The input has to pass a "strong pseudo-primality test" and "one Lucas test". This is well documented. I thought I remembered that there is also a way to get Maple to perform a true primality test, but I don't remember how and don't see anything about this in the Maple help system.

Is my memory faulty, or is there no definitive primality test in Maple?

Thanks in advance,

Doug

---------------------------------------------------------------------
Douglas B. Meade  <><
Math, USC, Columbia, SC 29208  E-mail: mailto:meade@math.sc.edu
Phone:  (803) 777-6183         URL:    http://www.math.sc.edu

Hei, I'm trying to create a random walk in the plane, with constant step length (=1) and the angle between two consecutive steps are decided by a probability density function. I just can't seem to find out how I should implement the density function into my worksheet.

The probability density function is: p(phi)=(1/4)*cos(phi/2), on the interval [-Pi,Pi].
And  I think i managed to do it by selecting a random angle, but don't know how to generate a random angle given this probability function. Any ideas? It'd be highly appreciated!

Hi everyone, I'm working a problem

Given a dog and a man. At t=0, the position of the man and his dog is (0,0) and (0,h), respectively. Then, the man start moving along Ox with constant speed vm. The dog keep running toward its master with constant speed vd. Describe the movement of the dog?

Therefore, let say the position of the dog is (f1(x),f2(x)), I wrote these:

restart;

with(plots);

MD := proc (h, vm, vd) local x, y, ode, ics, sol;

ode := {(diff(f2(t), t))*(vm*t-f1(t))+(diff(f1(t), t))*f2(t), diff(f2(t), t) = -sqrt(vd^2-(diff(f1(t), t))^2)};

ics := {f1(0) = h, f2(0) = 0};

sol := dsolve(ode union ics);

x := unapply(eval(f1(t), sol), t);

y := unapply(eval(f2(t), sol), t);

plot([x(t), y(t), t = 0 .. 10], scaling = constrained) end proc;

MD(10, 1, 5);

However, "sol" is returned NULL, which means the equation has no solution. I supposed I completed the motion part of the problem correctly. Please help me or point out an another method

Thank you in advance for your time.

I can search for any phrase or symbol in my Maple Worksheet but "sign symbol".

I have a very big result that I am not sure how many minus it has. When I search for - sign, no match found. I can see some minus in the worksheet but Maple can not find them.

Please help me.

I really appreciate.

if there is a funciton like f(i,j)=i+2*j
and I have already print out a 3D picture use the commend
"plot3d(f,i=1..10,j=1..10);"
the 3D picture just like the link : http://imgur.com/eLYhYEs
how can I convert the datas in 3D picture to text file?
the type may be like as follow I think.
-------------------------------------
(i,j)=f(i,j)
(1,1)=3
(1,2)=5
(1,3)=7...
-------------------------------------
I had try
"writedata("C:\\desktop\\a.txt",3Ddatas)"
but it seems can only output one variable parameter in a function, what else can I do?
thanks a lot.

Dear all;

I need you to understand this problem...

when i plot the function, using the graph i see that the function is above the x-axis but when I compute some values of this function I get a negative values....like
evalf(y(99.6));  is a negative value, but in the graph it is possible... I don't  undertand the problem...

restart:
with(plots):

# funciton

y:=x->-4.1123583570*10^281*exp(-(2/3)*x^(3/2))/(x^(1/4)*sqrt(Pi))+1.6554662320*10^(-289)*exp((2/3)*x^(3/2))/(x^(1/4)*sqrt(Pi))+(16/153)*x^(7/6)*sqrt(Pi)*exp((2/3)*x^(3/2))+Pi*((1/2)*exp(-(2/3)*x^(3/2))*(-1+exp((2/3)*x^(2/3)))/(x^(1/4)*Pi)-(16/153)*x^(7/6)*exp((2/3)*x^(3/2))/sqrt(Pi)):

#I plot this function in the interval (a,b)

a:=99;b:=100; # interval (a,b)
 forget(evalf): Digits:=20:
P1:=plottools:-transform((x,y)->[x+a,y])(plot(expand(y(x+a)),x=0..1,color=blue)):
forget(evalf): Digits:=4000:
P2:=plot(ysol, a..b, style=point, adaptive=false, numpoints=25, symbol=solidcircle, symbolsize=20, color=blue):
Digits:=20:
plots:-display(P1,P2);
evalf(y(99.6)); 

 Thank you in advantage for your remarks

how_do_i_get_floating_point_answer.mw

Download how_do_i_get_floating_point_answer.mw

Ramakrishnan V

rukmini_ramki@hotmail.com

Dear all;

Thank you for helping me, to plot two function on the same graph.

restart;
with(plots):

ode := diff(y(x), x, x) = x*y(x)+x;
a:=2; b:=3;  # we work in the interval (a,b)
ics := y(a) = 0, y(b) = 1;
sol:=dsolve({ode,ics}, numeric):  # Solution of my ode

# First figure: solution of my ode in the interval (2,3)

odeplot( sol,[x, y(x)], x=a..b);  # I plot the solution in the interval (a,b)

# Here I define a second function

z:=x->-4.9354831550*exp(-(2/3)*x^(3/2))/(x^(1/4)*sqrt(Pi))+0.752447265e-1*exp((2/3)*x^(3/2))/(x^(1/4)*sqrt(Pi))+(16/153)*x^(7/6)*sqrt(Pi)*exp((2/3)*x^(3/2))+Pi*((1/2)*exp(-(2/3)*x^(3/2))*(-1+exp((2/3)*x^(2/3)))/(x^(1/4)*Pi)-(16/153)*x^(7/6)*exp((2/3)*x^(3/2))/sqrt(Pi));

# Here a code to plot this function z

forget(evalf):
Digits:=20:
P1:=plottools:-transform((x,z)->[x+a,z])(plot(expand(z(x+a)),x=0..1)):
forget(evalf): Digits:=4000:
P2:=plot(z, a..b, style=point,adaptive=false, numpoints=25):
plots:-display(P1,P2);  # Second figure

My question: I want one graph that shows the solution of my ode and the funciton z. ( i.e plot the two figures in only one graph).

Thanks for helping me.

A strange verical line in output of dsolve.Should be a comma in Jacobi SN function and comma's in invert Jacobi SN function.

Possible_Bug.mw

Can anyone help me confirm this?

Hello,

How to change the length underline in latex ?

For example in this environment :

$\underline{\mathcal{G}}$

Thank you,

Gérard.

I tried doing LinearSolve with my own matrices, thought maybe there was a mistake there so I copied the example from Maple's website and it won't execute. Instead of printing an output matrix, it just prints LinearSolve(input matrix 1, input matrix 2). I feel this is a very simple syntax mistake, but I can't figure it out!