Maple 2018 Questions and Posts

These are Posts and Questions associated with the product, Maple 2018

How could I substitute the (t1*e) into t2 at the index of e?  I am trying many ways to do that, but it doesn't work. please help me.
 

restart:

alias(epsilon = e, omega = w, omega[0] = w0, t[1] = t1, t[2] = t2, alpha[1] = a1, alpha[2] = a2, alpha[3] = a3, beta[1] = b1, beta[2] = b2, F[0] = f0); e := proc (t1, t2) options operator, arrow; e end proc; w0 := proc (t1, t2) options operator, arrow; w0 end proc; a := proc (t1, t2) options operator, arrow; a end proc; f := proc (t1, t2) options operator, arrow; f end proc; mu := proc (t1, t2) options operator, arrow; mu end proc

 

 

(1/8)*alpha[2](t[1], t[2])*b^3*exp((3*I)*omega*t[1])+(1/8)*alpha[2](t[1], t[2])*b^3*exp(-(3*I)*omega*t[1])+(1/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp((3*I)*t[1])+(1/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp(-(3*I)*t[1])+3*alpha[2](t[1], t[2])*b^3*exp(-I*omega*t[1])*(1/8)+3*alpha[2](t[1], t[2])*b^3*exp(I*omega*t[1])*(1/8)-I*alpha[1](t[1], t[2])*A(t[2])*exp(-I*t[1])*(1/2)+3*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(-I*omega*t[1])*(1/4)-I*(diff(A(t[2]), t[2]))*exp(-I*t[1])+3*alpha[2](t[1], t[2])*A(t[2])*exp(-I*t[1])*b^2*(1/4)-I*alpha[1](t[1], t[2])*b*omega*exp(-I*omega*t[1])*(1/2)+I*alpha[1](t[1], t[2])*b*omega*exp(I*omega*t[1])*(1/2)+3*alpha[2](t[1], t[2])*A(t[2])^3*exp(-I*t[1])*(1/8)+I*(diff(A(t[2]), t[2]))*exp(I*t[1])+3*alpha[2](t[1], t[2])*A(t[2])*b^2*exp(-I*t[1]-(2*I)*omega*t[1])*(1/8)+3*alpha[2](t[1], t[2])*A(t[2])*b^2*exp(-I*t[1]+(2*I)*omega*t[1])*(1/8)+q[1](t[1], t[2])+3*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(-(2*I)*t[1]-I*omega*t[1])*(1/8)+3*alpha[2](t[1], t[2])*A(t[2])^3*exp(I*t[1])*(1/8)+3*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(-(2*I)*t[1]+I*omega*t[1])*(1/8)+3*alpha[2](t[1], t[2])*A(t[2])*b^2*exp(I*t[1]-(2*I)*omega*t[1])*(1/8)+3*alpha[2](t[1], t[2])*A(t[2])*b^2*exp(I*t[1]+(2*I)*omega*t[1])*(1/8)+diff(q[1](t[1], t[2]), t[1], t[1])+3*alpha[2](t[1], t[2])*A(t[2])^2*b*exp((2*I)*t[1]-I*omega*t[1])*(1/8)+3*alpha[2](t[1], t[2])*A(t[2])^2*b*exp((2*I)*t[1]+I*omega*t[1])*(1/8)+I*alpha[1](t[1], t[2])*A(t[2])*exp(I*t[1])*(1/2)+3*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(I*omega*t[1])*(1/4)+3*alpha[2](t[1], t[2])*A(t[2])*exp(I*t[1])*b^2*(1/4);

(3/4)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(-I*omega*t[1])+(3/4)*alpha[2](t[1], t[2])*A(t[2])*exp(-I*t[1])*b^2+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2*exp(-I*t[1]-(2*I)*omega*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2*exp(-I*t[1]+(2*I)*omega*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(-(2*I)*t[1]-I*omega*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(-(2*I)*t[1]+I*omega*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2*exp(I*t[1]-(2*I)*omega*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2*exp(I*t[1]+(2*I)*omega*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp((2*I)*t[1]-I*omega*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp((2*I)*t[1]+I*omega*t[1])+(3/4)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(I*omega*t[1])+(3/4)*alpha[2](t[1], t[2])*A(t[2])*exp(I*t[1])*b^2-((1/2)*I)*alpha[1](t[1], t[2])*A(t[2])*exp(-I*t[1])+((1/2)*I)*alpha[1](t[1], t[2])*A(t[2])*exp(I*t[1])+I*(diff(A(t[2]), t[2]))*exp(I*t[1])+(1/8)*alpha[2](t[1], t[2])*b^3*exp((3*I)*omega*t[1])+(1/8)*alpha[2](t[1], t[2])*b^3*exp(-(3*I)*omega*t[1])+(1/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp((3*I)*t[1])+(1/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp(-(3*I)*t[1])+(3/8)*alpha[2](t[1], t[2])*b^3*exp(-I*omega*t[1])+(3/8)*alpha[2](t[1], t[2])*b^3*exp(I*omega*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp(-I*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp(I*t[1])-I*(diff(A(t[2]), t[2]))*exp(-I*t[1])+((1/2)*I)*alpha[1](t[1], t[2])*b*omega*exp(I*omega*t[1])-((1/2)*I)*alpha[1](t[1], t[2])*b*omega*exp(-I*omega*t[1])+diff(diff(q[1](t[1], t[2]), t[1]), t[1])+q[1](t[1], t[2])

(1)

 

 

eq:=combine(subs(w=3+e*sigma,%),exp);

(3/4)*alpha[2](t[1], t[2])*A(t[2])*exp(-I*t[1])*b^2+(3/4)*alpha[2](t[1], t[2])*A(t[2])*exp(I*t[1])*b^2-((1/2)*I)*alpha[1](t[1], t[2])*A(t[2])*exp(-I*t[1])+((1/2)*I)*alpha[1](t[1], t[2])*A(t[2])*exp(I*t[1])+(3/4)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(-I*(epsilon*sigma+3)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2*exp(-I*t[1]-(2*I)*(epsilon*sigma+3)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2*exp(-I*t[1]+(2*I)*(epsilon*sigma+3)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(-(2*I)*t[1]-I*(epsilon*sigma+3)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(-(2*I)*t[1]+I*(epsilon*sigma+3)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2*exp(I*t[1]-(2*I)*(epsilon*sigma+3)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2*exp(I*t[1]+(2*I)*(epsilon*sigma+3)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp((2*I)*t[1]-I*(epsilon*sigma+3)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp((2*I)*t[1]+I*(epsilon*sigma+3)*t[1])+(3/4)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(I*(epsilon*sigma+3)*t[1])+I*(diff(A(t[2]), t[2]))*exp(I*t[1])+(1/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp((3*I)*t[1])+(1/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp(-(3*I)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp(-I*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp(I*t[1])-I*(diff(A(t[2]), t[2]))*exp(-I*t[1])+(1/8)*alpha[2](t[1], t[2])*b^3*exp((3*I)*(epsilon*sigma+3)*t[1])+(3/8)*alpha[2](t[1], t[2])*b^3*exp(I*(epsilon*sigma+3)*t[1])+(1/8)*alpha[2](t[1], t[2])*b^3*exp(-(3*I)*(epsilon*sigma+3)*t[1])+(3/8)*alpha[2](t[1], t[2])*b^3*exp(-I*(epsilon*sigma+3)*t[1])+((1/2)*I)*alpha[1](t[1], t[2])*b*(epsilon*sigma+3)*exp(I*(epsilon*sigma+3)*t[1])-((1/2)*I)*alpha[1](t[1], t[2])*b*(epsilon*sigma+3)*exp(-I*(epsilon*sigma+3)*t[1])+diff(diff(q[1](t[1], t[2]), t[1]), t[1])+q[1](t[1], t[2])

(2)

expand(%);

(3/4)*alpha[2](t[1], t[2])*A(t[2])*exp(I*t[1])*b^2+((1/2)*I)*alpha[1](t[1], t[2])*A(t[2])*exp(I*t[1])+I*(diff(A(t[2]), t[2]))*exp(I*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp(I*t[1])+(1/8)*alpha[2](t[1], t[2])*A(t[2])^3*(exp(I*t[1]))^3+(1/8)*alpha[2](t[1], t[2])*A(t[2])^3/(exp(I*t[1]))^3+(3/8)*alpha[2](t[1], t[2])*A(t[2])^3/exp(I*t[1])-I*(diff(A(t[2]), t[2]))/exp(I*t[1])+diff(diff(q[1](t[1], t[2]), t[1]), t[1])+q[1](t[1], t[2])+(1/8)*alpha[2](t[1], t[2])*b^3*(exp(I*epsilon*sigma*t[1]))^3*(exp(I*t[1]))^9+(3/8)*alpha[2](t[1], t[2])*b^3*exp(I*epsilon*sigma*t[1])*(exp(I*t[1]))^3+(1/8)*alpha[2](t[1], t[2])*b^3/((exp(I*epsilon*sigma*t[1]))^3*(exp(I*t[1]))^9)+(3/8)*alpha[2](t[1], t[2])*b^3/(exp(I*epsilon*sigma*t[1])*(exp(I*t[1]))^3)+(3/4)*alpha[2](t[1], t[2])*A(t[2])*b^2/exp(I*t[1])-((1/2)*I)*alpha[1](t[1], t[2])*A(t[2])/exp(I*t[1])+(3/4)*alpha[2](t[1], t[2])*A(t[2])^2*b/(exp(I*epsilon*sigma*t[1])*(exp(I*t[1]))^3)+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2/((exp(I*epsilon*sigma*t[1]))^2*(exp(I*t[1]))^7)+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2*(exp(I*epsilon*sigma*t[1]))^2*(exp(I*t[1]))^5+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b/(exp(I*epsilon*sigma*t[1])*(exp(I*t[1]))^5)+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(I*epsilon*sigma*t[1])*exp(I*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2/((exp(I*epsilon*sigma*t[1]))^2*(exp(I*t[1]))^5)+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2*(exp(I*epsilon*sigma*t[1]))^2*(exp(I*t[1]))^7+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b/(exp(I*epsilon*sigma*t[1])*exp(I*t[1]))+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(I*epsilon*sigma*t[1])*(exp(I*t[1]))^5+(3/4)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(I*epsilon*sigma*t[1])*(exp(I*t[1]))^3+((3/2)*I)*alpha[1](t[1], t[2])*b*exp(I*epsilon*sigma*t[1])*(exp(I*t[1]))^3-((3/2)*I)*alpha[1](t[1], t[2])*b/(exp(I*epsilon*sigma*t[1])*(exp(I*t[1]))^3)+((1/2)*I)*alpha[1](t[1], t[2])*b*exp(I*epsilon*sigma*t[1])*(exp(I*t[1]))^3*epsilon*sigma-((1/2)*I)*alpha[1](t[1], t[2])*b*epsilon*sigma/(exp(I*epsilon*sigma*t[1])*(exp(I*t[1]))^3)

(3)

eq:=combine(subs({e^(t1*e)=e^t2},%),exp);

(3/4)*alpha[2](t[1], t[2])*A(t[2])*exp(-I*t[1])*b^2+(3/4)*alpha[2](t[1], t[2])*A(t[2])*exp(I*t[1])*b^2-((1/2)*I)*alpha[1](t[1], t[2])*A(t[2])*exp(-I*t[1])+((1/2)*I)*alpha[1](t[1], t[2])*A(t[2])*exp(I*t[1])+(1/8)*alpha[2](t[1], t[2])*b^3*exp((3*I)*t[1]*epsilon*sigma+(9*I)*t[1])+(3/8)*alpha[2](t[1], t[2])*b^3*exp(I*epsilon*sigma*t[1]+(3*I)*t[1])+(1/8)*alpha[2](t[1], t[2])*b^3*exp(-(3*I)*t[1]*epsilon*sigma-(9*I)*t[1])+(3/8)*alpha[2](t[1], t[2])*b^3*exp(-I*epsilon*sigma*t[1]-(3*I)*t[1])+I*(diff(A(t[2]), t[2]))*exp(I*t[1])+(1/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp((3*I)*t[1])+(1/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp(-(3*I)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp(-I*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^3*exp(I*t[1])-I*(diff(A(t[2]), t[2]))*exp(-I*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2*exp(-(2*I)*epsilon*sigma*t[1]-(7*I)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2*exp((2*I)*epsilon*sigma*t[1]+(5*I)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(-I*epsilon*sigma*t[1]-(5*I)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(I*epsilon*sigma*t[1]+I*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2*exp(-(2*I)*epsilon*sigma*t[1]-(5*I)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])*b^2*exp((2*I)*epsilon*sigma*t[1]+(7*I)*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(-I*epsilon*sigma*t[1]-I*t[1])+(3/8)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(I*epsilon*sigma*t[1]+(5*I)*t[1])+diff(diff(q[1](t[1], t[2]), t[1]), t[1])+q[1](t[1], t[2])+(3/4)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(-I*epsilon*sigma*t[1]-(3*I)*t[1])+(3/4)*alpha[2](t[1], t[2])*A(t[2])^2*b*exp(I*epsilon*sigma*t[1]+(3*I)*t[1])+((3/2)*I)*alpha[1](t[1], t[2])*b*exp(I*epsilon*sigma*t[1]+(3*I)*t[1])-((3/2)*I)*alpha[1](t[1], t[2])*b*exp(-I*epsilon*sigma*t[1]-(3*I)*t[1])+((1/2)*I)*alpha[1](t[1], t[2])*b*epsilon*sigma*exp(I*epsilon*sigma*t[1]+(3*I)*t[1])-((1/2)*I)*alpha[1](t[1], t[2])*b*epsilon*sigma*exp(-I*epsilon*sigma*t[1]-(3*I)*t[1])

(4)

How could I substitute the (t1*e) into t2 at the index of e?  I am trying many ways to do that, but it doesn't work. please help me.   (t1*e)=t2


 

Download question.mw

Hi all

Hope the best for all. 

I have a single product of two sum, where the second sum admits one variable, namely j, from first one.

how can I it via maple?

thanks for any help.

HELLO

I AM A STUDENT AND WAS LOOKING FOR USING MAPLE2018.

I REQUESTED FOR A MAPLE EVALUATION. MY QUESTIONS ARE:

FOR HOW LONG CAN I USE THIS EVALUATION VERSION?

SECONDLY, WHAT SOFTWARE FACILITIES I WOULD NOT GET WITH THIS EVALUATION VERSION? WHAT ARE THE DIFFERENCES BETWEEN THE PRICED ENTERPRISE VERSION AND MAPLESOFT EVALUATION VERSION?

THIRDLY, WHAT ARE THE DIFFERENCES BETWEEN MAPLESOFT STUDENT VERSION AND THE ULTIMATE PRICED ENTERPRISED VERSION?

PLEASE REVERT BACK TO ME AT THE EARLIEST.

is there a way to change the color of just the labels on the axis ticks only? 

I wanted to do this for DEplot, but I started with normal plot since it is easier. I could change the colors of the axis, but this changes not only the tick labels but the whole axis itself:

restart;
plot(sin(x),x=-Pi..Pi,axis=[tickmarks=[color=red]], color=blue);

So the axis itself changed color as well. It is OK for the ticks as well as the ticks labels (the numbers of the ticks). But prefer the axis itself not to be red.

Once there is a way to do the above, I wanted to apply to DEtools:-DEplot. Why? Becuase it is hard to read the Axis tick labels on DEplot. I thought that changing the color of the tick labels will it easy. But DEtools:-DEplot does not even accept axis command:

ode:= diff(y(x),x)=2*x:
DEtools:-DEplot(ode,y(x),x = -2 .. 2,y = -2 .. 2, [[0.1,0]],
               labels=["",""],
               linecolour = red,
               color = blue,
               'arrows' = 'medium',
               axesfont=['Times', 'bold', 12]
               );

 

You can see it is hard to read the axis tick labels above, as arrows go over some of them,. If I make them say RED, I think it will make it more clear.

Now when adding the axis options, it gives error

DEtools:-DEplot(ode,y(x),x = -2 .. 2,y = -2 .. 2, [[0.1,0]],
               linecolour = red,
               color = blue,
               arrows = MEDIUM,
               axesfont=['Times', 'bold', 10],
               axis=[tickmarks=[color=red]]
               );

Error is generated, axis is not supported command.

Any idea how to make tick labels red for DEtools:-DEplot? It will be OK if I can get both the axis and the ticks RED, at least it will be better than it is now.

One workaround way I currently use is this:

ode:= diff(y(x),x)=2*x:
p1:=DEtools:-DEplot(ode,y(x),x = -2 .. 2,y = -2 .. 2, [[0.1,0]],
               labels=["",""],
               linecolour = red,
               color = blue,
               'arrows' = 'medium'                    
               );
p2:=plot(0,x=-2..2,y=-2..2,axis=[tickmarks=[color=red]]);
plots:-display([p2,p1]);

 

But this seems like a hack to me. There should be a way to directly set tickmarks color in DEplot?

Help me extracting the data from the Maple figure. I was used the 'implicitplod3D' command to draw the figure.  I want to save the parameters of sigma, A  and f0 in Exce
 

af_eq := (3*alpha[2](t[1], t[2])*A^3*(1/8)-A*sigma)^2+(1/4)*alpha[1](t[1], t[2])^2*A^2 = (1/4)*f0^2

(0.4871250000e-1*A^3-A*sigma)^2+0.2500000000e-2*A^2 = (1/4)*f0^2

(1)

with(plots, implicitplot,implicitplot3d):

 alpha[1]:=0.1;alpha[2]:=0.1299;

.1

 

.1299

(2)

##3D

implicitplot3d(af_eq, sigma = -1.5 .. 1.5,f0 =0..0.5, A = 0 .. 5,numpoints = 20000,style=surfacecontour,contours = 15);

 

 

 


 

Download extract_data_from_the_figure.mw

l format. then, I can use the data draw a picture in other software. Many thanks.

Hello guys

I have a simple but crucial question about variation, especially in cosmology and gravity. suppose we have the action such as

S=int(sqrt(-g)*F(R,RµναβRµναβ),d^4x) where g is determinant metric gµν and d^4x is space in 4D and F is an arbitrary function of R (Ricci scalar) and scalar which created by Riemann tensor. How we can define this action in the maple 18 and take variation with respect to the metric.

thanks

Is there a place to read about rules for naming constants of integrations for solution of ODE so that Maple odetest will not have problem verifying solution of an ODE?

It seems Maple does not like some constant names. If I use C or C1 or C[1] or A etc... odetest does not give back zero, even thought the solution is exactly the same as Maple's, except Maple uses _C1.

I read one is not supposed to use _ constants in own code, since these are reserved for Maple.

So what is one supposed to use? Here is an example

restart;
ode:=2*sqrt(x)*diff(y(x),x)=sqrt(1-y(x)^2):
my_sol:=y(x)=sin(sqrt(x)+(1/2)*C1);

Compare to Maple own solution

maple_sol:=dsolve(ode,y(x));

You can see, there is no difference, except the name of the constant.  Yet, Maple does not odetest my solution

odetest(my_sol,ode);

I tried naming my constant to C, C[1], A, Z, etc.. I got tired trying. Only when I use _C1 does Maple give zero for odetest.

What is so special about _C1?  And if one can't use _ symbols as Maple says, what is one to to do? 


Maple 2018.1, Physics version `2018, July 21, 4:16 hours, version in the MapleCloud: 78, version installed in this computer: 78`

 

I would like to do the following operation:

Apparently, Maple does not do this operation.

Does anyone know how to do this?

Thanks in advance.

Hello,

When 3d-Plotting a function say f(z)=sqrt(z)=sqrt(x+I*y) then Maple automatically only plots the principal branch.

In the example above is it possible to plot the second surface too?

Is it also possible to plot the full continuous Riemann surface as in

https://en.wikipedia.org/wiki/Riemann_surface#/media/File:Riemann_sqrt.svg

or

http://mathworld.wolfram.com/RiemannSurface.html

 

???

From answer posted in https://www.mapleprimes.com/questions/225137-Internal-Symbol-Z1--Causes-Error-

I've learned to use

subsindets(expr, 'suffixed(_)', f->n);

To replace in a Maple generated solution which contains system suffix, such as _Z, _Z1, etc... returned by Solve and Maple calls, with a symbol that I find easier to read. The above works in some cases but not others. For example, it works on this result

restart;
expr:=signum(0, _Z, 1);
subsindets(expr, 'suffixed(_)', f->n);

                      signum(0, n, 1)

But not on this one

restart;
expr:=RootOf(signum(0, _Z, 1));
subsindets(expr, 'suffixed(_)', f->n);

             RootOf(signum(0, _Z, 1))

I do not want to use pattern matching, since the result is something I do not know what it will be. I just want to replace all those Maple symbols that start with _Z in the solution by another symbol. I can't use subs() either, since I do not know what the symbol will be.

What is the correct way to do this?

 

Should this generate an error?

restart;
r:=evalc(Im(1/ln(x)));
solve(r,{x});

 

This did not help. Same error as above.

restart;
r:=evalc(Im(1/ln(x))):
solve(r,{x}) assuming x::real;

This made Maple not give an error

restart;
r:=evalc(Im(1/ln(x))):
solve(r,{x}) assuming x::real, x>0;

       {x = x}

Also this worked, but NULL returned

restart;
r:=evalc(Im(1/ln(x))):
solve(r,{x}) assuming x::real, x<0;

Also this worked with NULL returned

restart;
r:=evalc(Im(1/ln(x))):
solve(r,{x}) assuming   x<0;

Is this how Maple generally works? i.e It generates an exception error when it is not able to solve something? Or does it normally echo back the input back if it can't solve something? Or at least return NULL if it can't solve it? I am havin hard time figuring which method to use to check if Maple is able to solve something or not, because each time it seems to do something different.

Should one then put a try/catch around each Maple call, and treat the try section as if Maple was not able to solve the equation whatever it was?

 

Essentially I have this trigonometric equation and I want to solve (get the roots of) it within the range -Pi..Pi:

v := a+b*cos(t)-c*(d*(1-(a+b*cos(t))^2-d^2*sin(t)^2)^(1/2)+e*sin(t)) = 0;
v1 := t > -Pi;
v2 := t < Pi;

Where t is the variable and a..e are constants.  At the moment I am trying the following:                  
solve({v, v1,v2}, t, allsolutions, explicit);


My problem is that Maple tries to solve this - I get Evaluating in the bottom left corner of the window - but never seems to return with a solution even after using 52,000s and 5GB of memory: I am using a late model macbook.

Can anyone see a way of re-framing this equation so that Maple can return an answer?

The following product gives 0

restart;

product(1-1/k, k = 2 .. infinity);


                               0

However when I expand the product

1 - 1/2 - 1/3 - 1/4 - ... + 1/2*1/3 + 1/2*1/4 + ... + 1/3*1/4 + ... + triple products + quadruple products + and so forth...

Now the double, triple, quadruple, and so forth sums of products converge.

The 1/2 + 1/3 + 1/4 + ... nevertheless diverges, so why does maple give me 0?

question.mw

Hello there,

I am trying to solve a pretty simple Cauchy Problem:

PDE := diff(u(t, x, y), t)+.5*(diff(u(t, x, y), x, x))+diff(u(t, x, y), x, y)+.5*(diff(u(t, x, y), y, y))+diff(u(t, x, y), x)+diff(u(t, x, y), y)+(1+(x+y)^2)*u(t, x, y) = 0

bc := u(0,x,y)=1;

with pdsolve and I get a result.

But surprisingly, Maple's pdetest applied to this result doesn't yield zero.

Hence, Maple solves my problem, but it also says that this solution is not right?

How is that possible?

 

Thanks a lot for your help. (btw: this is my first post here :))

Best regards,

utcyp

 

Maple 2018 memory usage increases as I try to display or manipulate the expression presented in  test_maple2018.mw. The same worksheet works perfectly in Maple 18

I've tried both in Maple 2018.1 and Command Line Maple 2018 obtaining the same result

Maple 2018.1, X86 64 WINDOWS, Jun 8 2018, Build ID 1321769

From command line maple 2018 when trying to evaluate

memory used=3.6MB, alloc=40.3MB, time=0.14
memory used=4.7MB, alloc=72.3MB, time=0.19
memory used=33.3MB, alloc=107.3MB, time=0.45
memory used=109.2MB, alloc=143.1MB, time=1.37
memory used=217.7MB, alloc=185.6MB, time=2.60
memory used=292.6MB, alloc=217.6MB, time=3.87
memory used=328.0MB, alloc=254.6MB, time=5.16
memory used=437.3MB, alloc=299.9MB, time=7.44
memory used=549.8MB, alloc=335.9MB, time=13.60
memory used=630.9MB, alloc=375.4MB, time=17.07
memory used=686.9MB, alloc=401.8MB, time=19.61
memory used=785.6MB, alloc=431.9MB, time=22.31
memory used=944.8MB, alloc=427.9MB, time=26.74
memory used=1150.1MB, alloc=427.9MB, time=32.37
Interrupted

I was wondering if my Maple 2018.1 installation is corrupted. Since I have no acces to other Maple licenses, can anyone try to execute it? test_maple2018.mw

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