Suppose we have some simple animations. Our goal - to build a more complex animation, combining the original animations in different ways.
We show how to do it on the example of the three animations. The technique is general and can be applied to any number of animations.

Here are the three simple animations:

restart;
with(plots):
A:=animate(plot, [sin(x), x=-Pi..a, color=red, thickness=3], a=-Pi..Pi):
B:=animate(plot, [x^2-1, x=-2..a, thickness=3, color=green], a=-2..2): 
C:=animate(plot, [[4*cos(t),4*sin(t), t=0..a], color=blue, thickness=3], a=0..2*Pi):
 

In Example 1 all three animation executed simultaneously:

display([A, B, C], view=[-4..4,-4..4]);

In Example 2, the same animation performed sequentially. Note that the previous animation disappears completely when the next one begins to execute:

display([A, B, C], insequence);

Below we show how to save the last frame of every previous animation into subsequent animations:

display([A, display(op([1,-1,1],A),B), display(op([1,-1,1],A),op([1,-1,1],B),C)], insequence);

Using this technique, we can anyhow combine the original animations. For example, in the following example at firstly animations  A  and  B  are executed simultaneously, afterwards C is executed:

display([display(A, B), display(op([1,-1,1],A),op([1,-1,1],B),C)], insequence);

The last example in 3D I have taken from here:

restart;
with(plots):
A:=animate(plot3d,[[2*cos(phi),2*sin(phi),z], z =0..a, phi=0..2*Pi, style=surface, color=red], a=0..5):
B:=animate(plot3d,[[(2+6/5*(z-5))*cos(phi), (2+6/5*(z-5))*sin(phi),z], z=5..a, phi=0..2*Pi, style=surface, color=blue], a=5..10):
C:=animate(plot3d,[[8*cos(phi),8*sin(phi),z], z =10..a, phi=0..2*Pi, style=surface, color=green], a=10..20):
display([A, display(op([1,-1,1],A),B), display(op([1,-1,1],A),op([1,-1,1],B),C)], insequence, scaling=constrained, axes=normal);

AA.mw

Hello guys,

I have some system of differential equations,

How can i find  eigenvalues of this system?

If i have solution

res := evalf(dsolve(sys union ics, convert(x, list), type = numeric, method = rkf45))

and

sol := evalf(dsolve(sys union ics, convert(x, list), method = laplace))

AnSolution.mw

Thanks!

I am maple newbie. Sometimes when I look at Maple code to try to understand the algorithm (which is hard, since I do not know Maple well), I see the code puts ` ` around some keywords. And sometimes it does not. For example, sometimes I see something like (these are random samples) from Maple code shown using showstat()

return `if`(assigned(r),r,{})
r := `union`(r,{solve(op(1,expr) = 1,vs)})
v2 := `intersect`(vs,indets(op(1,expr),'name'));

But sometimes, they do not put ` ` around functions name or keywords, like this:

if nops(v2) = 1 then
t1 := remove(a -> has(a,RootOf),t1);
for t in expr do

and so on.

Can some expert please give what is the rule thumb to use? Should user adopt this method also? When to put ` ` and when not to put ` `?   I understand that ` ` prevents one-time evaluation (or rather, holds off immediate evaluation) and ``  `` prevents two times evaluation and so on. But when to use ` ` is what is confusing me.

Hello,

I was wondering if it is possible to use units in Maple so I can always check if the result I have at the end of calculation is the meter.  For example:

The answer is of course 3.10^8 m^3*kg/s^3

I try to do something with the units but I am unable to crreate something that will simplify the m/s ffactor to 1.

Any idea?

Thank you in advance for your help.

how to convert a nested for loop to iterative version with stack

I'm trying to solve the really simple system u+v = <-2,4> and u-v = <-3,6> for u and v using Maple. My attempt attached.

Download solve_error.mw

Why do I get this error? How might I get the solution?

I have recently been working on a problem using fractional calculus and have come across something in Maple's fracdiff  command that makes no sense to me.

fracdiff(1, x, 1/2) = 0

It should be:     1/(sqrt(x)*sqrt(Pi))

Thanks.

fracdiff.mw

I solved this (aq1) equation and got a set of answers, but when I want to solve another equation (aq) that is like (aq1)  it took a long time and I stopped progressing.

aq1 := -6.801867*10^(-32)*omega^16+(2.20054799*10^(-46)*I)*omega^23+(6.14329398*10^(-52)*I)*omega^25+(1.*10^(-11)*I)*omega^3+(2.*10^(-14)*I)*omega^5+(9.*10^(-10)*I)*omega+(2.*10^(-25)*I)*omega^11+5.95367451*10^(-12)*omega^8-2.10490578*10^(-16)*omega^10+3.6487095*10^(-21)*omega^12-3.4507372*10^(-26)*omega^14-4.53641375*10^(-54)*omega^26-1.844174702*10^(-48)*omega^24-.2318547310*omega^2+0.2767383695e-3*omega^4+15.23320543-2.607001427*10^(-43)*omega^22-1.252442537*10^(-38)*omega^20-1.58024603*10^(-34)*omega^18-(4.8*10^(-28)*I)*omega^13+(2.68604*10^(-29)*I)*omega^15+(1.4639509*10^(-32)*I)*omega^17+(1.21776770*10^(-36)*I)*omega^19+(2.77270182*10^(-41)*I)*omega^21-7.070170160*10^(-8)*omega^6

sd1 := solve(aq1)

aq := (2.626145*10^(-111)*I)*beta^41+(2.723460372*10^(-55)*I)*beta^25-1.125718*10^(-103)*beta^38-4.42696*10^(-96)*beta^36+(4.038976*10^(-119)*I)*beta^43+(1.897840*10^(-135)*I)*beta^47-1.4537*10^(-128)*beta^44-2.393897*10^(-75)*beta^30-5.345113*10^(-69)*beta^28-(8.88232*10^(-14)*I)*beta^7+(3.78162*10^(-127)*I)*beta^45+(1.87236321*10^(-39)*I)*beta^19-4.22943*10^(-111)*beta^40-5.98764*10^(-82)*beta^32+(1.73215*10^(-27)*I)*beta^13+(3.14576*10^(-144)*I)*beta^49+1.0000002*10^(-150)*beta^50+1.000483*10^(-142)*beta^48+(0.5707492e-4*I)*beta-1.860356732*10^(-56)*beta^26-1.202764308*10^(-50)*beta^24+0.1078870970e-3*beta^4-.1337634356*beta^2+(4.558807*10^(-82)*I)*beta^33+11.99907662+(1.552482870*10^(-49)*I)*beta^23+1.50289073*10^(-33)*beta^16+(4.738072*10^(-89)*I)*beta^35-2.560992731*10^(-45)*beta^22-1.821396189*10^(-40)*beta^20+1.*10^(-159)*beta^52-1.025045*10^(-118)*beta^42-2.708307310*10^(-27)*beta^14-2.880*10^(-137)*beta^46+3.616579272*10^(-22)*beta^12-2.775313578*10^(-17)*beta^10+(2.247453*10^(-75)*I)*beta^31+(4.317773*10^(-69)*I)*beta^29-(8.4742656*10^(-63)*I)*beta^27+(1.086756*10^(-103)*I)*beta^39+(2.875650*10^(-96)*I)*beta^37+(2.927689932*10^(-44)*I)*beta^21+(5.19084*10^(-18)*I)*beta^9+(3.3203077*10^(-35)*I)*beta^17-1.867365177*10^(-8)*beta^6+1.091287414*10^(-12)*beta^8-3.549248092*10^(-36)*beta^18-6.89128*10^(-89)*beta^34-(1.32011*10^(-22)*I)*beta^11+(8.67973*10^(-32)*I)*beta^15+(5.131768*10^(-10)*I)*beta^5-(6.362604*10^(-7)*I)*beta^3-(5.75387*10^(-153)*I)*beta^51

How can I solve (aq) ?

The writings in the variables palettes are overlapped. Is there any solution suggestion?

The first half of this work sheet deals with SHM of pendulum. In the second half of the work sheet I attempt to solve for the general case of a swinging pendulum. Maple introduces a place holder (correct me if I have used the incorrect termonology) " __a" which I do not understand. What variable(s) should I replace it with and is there an automatic way of doing so?
 

Download SHM_and_not_so_SHM.mw

Hi All,

I have problem how to formulate optimization problem with additional parameters.
For example let objective function is as follow:

ObjF:=proc(x,Delta)
    return sin(x) + cos(x) + Delta;
end proc;

In Matlab we can formulate the following optimization problem:

Delta = 10;
f=@(x)ObjF(x,Delta);
options = optimset('Algorithm','interior-point');
x = fmincon(f,0.1,[1;2],[],[],[],[],[],[],options);

where:

function y = ObjF(x,Delta)
    y = sin(x) + cos(x) + Delta;
end

In above example I can pass additional parameters (Delta) to objective function an then I can start optimization
to find solution. How to define above problem in Maple ? How to pass additional  parameters  befor we starat the optymalization?

Best,

Rariusz

#my testing for wildcard to one
#after testing, it loop a very long time and not stop
ppp := [[0,0,0,x],[0,0,1,0],[0,1,0,0],[1,0,0,0]]:
check1 := [seq(0,ii=1..nops(ppp))];
ttt1 := [[0,0,0,1],[0,0,1,0],[0,1,0,0],[1,0,0,0]]:
mmmeaght1 := [seq(0,ii=1..nops(ppp[1]))]:
bbb1 := [seq(0,ii=1..nops(ppp[1]))]:
emap := [(xx) -> if [xx < 0 assuming x > 0] then 0 else 1 end if, (xx) -> evalf(1/(1+exp(xx)))]:
#trace(perceptronrule1);
MM(ppp, ttt1, mmmeaght1, bbb1, check1, emap);
 

when test wildcard variable for input, would like to assume x > 0 then

i try assuming x > 0 , got error

I am practicing with some diff equations. I am having problem solving for one of the constants. I am having a pproblem assigning a value to derivatives or 2nd derivatives. What is a good general technique is this type of situation?
 

Download SHM.mw

I was trying to write a procedure that would compute a simple linear equation using the Extended Euclidean Algorithm. I was thinking of a procedure like the following:

solveEeaMatrix := proc (a::list, b::list) 
 local c::list;  
 c := a -iquo(a[1],b[1])*b;  
 print(c);  
 while (c[1] <> gcd(a[1],b[1]) do 
 ...

I am basically stuck at this part as

1) I don't know how to setup a multi-dimensional array that could dynamically grow(as a possible solution).

2) I can't come up with a recursive function that could possibly take care of this.

In short, if I am given for example an equation like: 84*x+203*y = 14

I will transform it into 2 linear equation as follow:
row0 := [203, 0, 1] row1 := [84, 1, 0] Subsequently, I will perform the following:

c := a -iquo(a[1],b[1])*b;  

Where aand b are both lists and arguments of the procedure and cbeing another list and a local variable.

But I don't know how to do the following programmatically:

row3 := row1-iquo(row1[1], row2[1])*row2;
row4 := row2-iquo(row2[1], row3[1])*row3;
row5 := row3-iquo(row3[1], row4[1])*row4;
and so on ...                  

Any hint would be appreciated.

I want to calculate the intersection between three circles.
I know that in this case i can calculate intersection of only the first and second equation, but I need this for a interactive component.

The command "intersection"[GEOMETRY] work only with 2 circles.

I did this but it doesn't work.

Thanks.