I have a problem with radiobutton

i have 3 types of fricction, and each radio button has one type of it.

I have a problem when i select one type of friction don´t plot it

that is the code  RBFEstatica is a radiobutton of type static fricction

if GetProperty(RBFEstatica,value) = true then

DocumentTools[SetProperty](Label0,caption,estatica);

p1 := 0;
 p2 := 0.1;
 deltav1 := piecewise((abs(x2q1(t)) > 0 and abs(x2q1(t)) < 0), p1, p2);

Hello all,

Im trying to apply the working of the algorithm shown below to a more complex one(attached). The simple algorithm produces separate spreadsheets for different values of j. In the attached algorithm Im trying to produce different spreadsheets for different values of 'angle1'. I have applied the same structure but with no success. 

Can someone please have a look at my code and point me to my mistake?

Any help is really appreciated

Hello everybody,

   I have to solve the following differential equation numerically

dsys1 := {0.27e-2*(diff(x(t), [`$`(t, 2)])) = -3*10^(-4)*sqrt((diff(x(t), t))^2+(diff(y(t), t))^2+(diff(z(t), t))^2)*(diff(x(t), t))-(2.25*10^(-4)*8)*(diff(y(t), t)), 0.27e-2*(diff(y(t), [`$`(t, 2)])) = -3*10^(-4)*sqrt((diff(x(t), t))^2+(diff(y(t), t))^2+(diff(z(t), t))^2)*(diff(y(t), t))+(2.25*10^(-4)*8)*(diff(x(t), t)), 0.27e-2*(diff(z(t), [`$`(t, 2)])) = -3*10^(-4...

I am not able to use the Maple 15, because after it starts up, I receive an error ""the kernel loader cannot find Maple engine..."

What does it means and how to solve it?

Thanks

Consider the simple loop below:

> for q  from 1 by 1 to 12 do;

> 
> if inter[q] <= B then xz[q] := inter[q] end if;
> 
> end do;

inter[1] to inter[12] is defined previously

B is also calculated beforehand.

My problem is that Maple insists on looking for inter[13] and xz[13]. My range is between 1 and 12 so Maple shouldnt be doing this.

Hi all. This is porbably a very elementary question. Maple outputs something of the form  " X +  0. I" after a computation, where X is a number. I'm confused as to what 0. I is. My first thought is that it's 0.1 * I, but then squaring X +  0. I  gives back X^2 + 0. I instead of the expected X^2 - 0. 01 + 0.2 X I. Could someone explain to me what Maple means by that? Thanks!

X +  0. I

Dear Maple Users

I was looking forward to having the Units package improved in Maple 15. Units are indeed converted to SI Units when direct calculations are made. It is a disappointment though that this conversion doesn't occur when solving equations as the example below show:

Unfortunately Maplesoft mix up their announcements for news in version 15 of Maple with...

Hi all,

I want to save a plot into Array or other and re-plot it by pointplot.

restart:with(plots):

f:=x->cos(x)+cos(x)^2;

plot(f(x),x=0..2*Pi);

With Array, I want to determine dx (step) of the plot (Array) and the dimension of Array (number of line and column).

If Pk:=plot(f(x),x=0..2*Pi); is Array, I can reproduce the plot as:

ef := {seq([Pk[i,1], Pk[i,2]], i = 1 .. N)}: # Pk[i,1] is x and Pk[i,2] is f(x), N number of line of Array

Answering to that question, I posted several procedures finding minimal polynomials for the elements of finite fields. The best one was the following,

alias(a=RootOf(T^100+T^97+T^96+T^93+T^91+T^89+T^87+T^86+T^82+T^81+T^71+T^70+T^67+T^61+
T^60+T^57+T^54+T^53+T^52+T^49+T^48+T^45+T^44+T^42+T^39+T^36+T^33+T^32+T^31+T^29+T^28+T^27+
T^26+T^24+T^23+T^22+T^18+T^17+T^16+T^14+T^13+T^12+T^10+T^8+T^7+T^6+T^3+T+1)):

F:=GF(2,100,op(a)):
z:=F:-input(2):

MinPolyGF:=proc(x,y:=_X)
local A, i;
A:=Matrix(100,...

I would like to do the following.

Ive got variables a(z) where z(tau). Now I would like to do something like:

>> t = z(tau);

>> a := a(z);

>> diff(a,tau);

  == (da/dz)*(dz/dtau)

Hi,

I have the function:

restart: with(plots): with(OrthogonalExpansions): with(DiscreteTransforms):

f:=t->piecewise(t<0,0,t<1,t,t<2,2-t,0);

The Fourier coefficients a[0], a[i] and b[i] can be calculated by:

T:=2: Trig:=FourierSeries(f(t),t=0..T,n,'Coefficients');

a[0]:= op(1, Coefficients);a[i]:= op(1, op(2, Coefficients))[1];b[i]:= op(1, op(2, Coefficients))[2];

I want to calculate those coefficients a[0...

Defined a demand function:

x:=(p,w)->w/(p[1]+p[2]+p[3])*Vector([p[2]/p[1],p[3]/p[2],p[1],p[3]])

The Slutsky Matrix is calculated as Dpx(p,w) + Dwx(p,w) x(p,w)^T with Dpx(p,w) = Jacobian for p1, p2, p3 and Dwx(p,w) is a column vector of the derivate of x1, x2, x3 in respect to w.

Step by step:

Dxp := Jacobian(x(p,w),[p[1],p[2],p[3]])

yields the correct Jacobian.

For Dwx I've tried several different methods, i.e.

Dwx := Jacobian(x(p,w),[w...

I am having a problem with variable assignment.

output 15 should be:

test :=[1]

[2]

[1 1]

At some point, the array test is being overwritten even though in my proceedure I use only local variables.

> I found this information and code quite useful. Now my set t1 changes slightly to include I and -I. I also set up conditionl statements to extract the right elements according to the criteria on the value of beta*Q as adviced previously. I then test this out using the values for (beta,Q) as (1,1), (-1,1),(0,1).

When beta*Q is positive (one in this case) I is not selected...

I'm looking for a quick way, preferrably without using any loops to 'map' a function using several vectors.

In the single vector case, this is quite easy, but with multiple vectors, I'm not sure if map is the correct command to use.

For example:

u:=<1,1,1>;
v:=<2,3,4>;

f:=(x,y)->ln(x-y);

What I'd like to do is create a new vector from the map of the function onto both vectors,...