hi friends am trying code for 1d heat equation but my code is not giving me result which i needed...although its running but still am in doubt its not picking boundary conditions and initial condition in last loop of discritzed equation due to which my result are totally wrong.. am attaching worksheet please let me know where am wrong...will_see.mw

Could you please look at the following piace of code:

>assume(d::complex):
>z:=t=a/d:
>additionally(d≠0):
>z:=z*d;

The result of last command is

z:=d~t=d~*a/d~

that is not z:=d~t=a.

If I skip the command additionally(d≠0): then everything is good. However I need this command. Could you help to handle this problem?

amu^4+bmu^3+cmu^2+dmu+e=0

The unknown is mu

> int(abs(x), x = -1 .. 1);
                               1
> int(sign(x)*abs(x), x = -1 .. 1);
                               1
> int('sign(x)'*abs(x), x = -1 .. 1);

Hi is it possible to plot both x and y on the same graph versus t, I can't get anywhere with it but my maple skills aren't great.:

x := sqrt(1-10^14*sin(10^10*t)^2/(3*Pi*t^2))

y := 10^7*sin(10^10*t)/(t*sqrt(3*Pi))

I would uninstall maple but i get the following error:

java.lang.IllegalArgumentException: No product for ID=16000347-aaa2-0000-99a9-b0cb0eddf46e

                at ZeroGgv.<init>(DashoA10*..)

                at ZeroGf3.c(DashoA10*..)

               ...

   Let GL(2,Z) be the  group of all the matrices of dimension 2 over the integers with the determinant equal to +/-1.
   A matrix M in GL(2,Z) is called primitive if M is not equal to K^n for any K in GL(2,Z) and any positive integer n >= 2.
   Is the matrix M:= Matrix([[27,5],[11,2]]) primitive? How to determine it in Maple?

Edit. GL(2,Z) instead of UL(2,Z).

Hello

I am solving an equation that has two piecewise functions on it. It seems that Maple ignore the second piecewise. If I include it or remove it  I will get the same solution. I tried to fix the problem but still cannot figure out why Maple ignores it.

This is Maple code:

restart;

with(plots);

with(DEtools);

eq := diff(x(t), t) = -piecewise(abs(x(t))-.5 > 0, abs(x(t))-.5, 0)-piecewise(0 > x(t) and x(t) > 1, 0, x(t) > 1, 1);

Salut,

Il est clair que la multiplicité d'intersection des courbes C1:=x^2-y*(x+y^2) et C2:=x*(x+y^2) au point (0:0:1) est égale à 7 ; I(C1,C2,(0:0:1))=7.

Mais l'extension student[intercept] fournie que 4 solution :

> with(student):
>            
> C1:=x^2-y*(x+y^2):
> C2:=x*(x+y^2):
> intercept(C1=0,C2=0,[x,y]);

   [[x = 0, y = 0...

Hello.

I have following peace of code:

>restart;
>assume(a::complex);
>z1:=t=conjugate(a)*b;
>z2:=a^2;
>a:=solve(z1,a);
>z2;

I need to get (conjugate(t))^2/b as a result of last operation but get a~^2. What do I have to change in code? Of course, I can skip

>assume(a::complex);

and it will get me exactly what I need. But I have to use this command for other reasons.

Thank you.

######################################

Hello,

First of all I just want to say that your forum is great and it helped me to find a lot of new stuff for Maple. I am yet still a beginner but …we all need to learn. So let’s get to my question. I have an equation with 2 variables - x and y. This is the equation

x = y*(25+2*x)^(2/3)/(25*(.566*(25*x)^(2/3)))

So… I know I can solve the equation for x and y and I can give it specific values for y and it solve it every time. When,...

Dear all,

I’m a beginner in Maple and I have problem in understanding the RootOf

my equations are shown below:

M1alone := (1/2)*hs*fm*sigma*mu/(Ps*sqrt(sqrt(M1/h0)/Ps+ts))-As*h0*mu^2/M1 = (1/2)*hs*mu^2/Ps-(1/2)*hs*mu:

> M1sol := isolate(M1alone, M1);

M1 = Ps^2*(-ts+RootOf((-hs*mu*Ps+hs*Ps^2)*_Z^5+hs*fm*sigma*_Z^4*Ps+(-2*hs*Ps^2*ts+2*hs*mu*ts*Ps)*_Z^3-2*hs*fm*sigma*ts*Ps*_Z^2+(hs*Ps^2*ts^2-2*As*mu-hs*mu*ts^2*Ps...

I need to solve elementary equation, where "t" is real and all other variables are complex values:

Bonjour,

Où se trouve l'extention ?algcurves,intersectcurves dans maple 12?

Merci d'avence,

Gérard.

r1:=2*(1-s^2)^(1/2)*arctanh((1+s)*tanh(x)/((1-s)*(1+s))^(1/2))/((1-s)*(1+s))^(1/2)+ln((1/2)*arccosh(1/s)-x)

with x=1/2*arccosh(1/s)-beta*u

is to be calculated to the first order in u.

in fact im only interested in the first order not in the zero order.

so when applying the series I get for the first order

-beta*u/sqrt(1-s^2)

whereas by hand I get:

-beta*u*s/sqrt(1-s^2)

 I'd really appreciate some idea since...