MaplePrimes Questions

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...

I want to draw a graph of this non-linear equation

d^2*y/dt^2+y^3-sin*t = 0

when:

y(0) =0

y'(0)=1

someone can help me ?? 

 

Bonjour,

Comment se calcule (bien sûr avec Maple) la multiplicité d'intersection de deux courbes algébriques d'équations f(x,y)=0,g(x,y)=0 ?

 

Merci d'avence

Gérard.

 

Maple can do:

assume(H>0)

this function:

2*arctanh((1+s)*tanh(x)/sqrt((1-s)*(1+s)))+ln((1/2)*arccosh(1/s)-x)

shall be calculated @ x=1/2*arccosh(1/s)

when taking the limit maple tells me its 0

whereas when calculating it by hand i get ln(sqrt(1-s^2)/s)

whats the problem?

PS: 0<s<1

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