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11 years, 190 days

MaplePrimes Activity

These are replies submitted by Kitonum

with(plots): with(plottools):
A:=plot(L, color=brown, thickness=10):
B:=plot([op(L1),op(map(t->[-t[1],t[2]],ListTools:-Reverse(L1)))], color="Green", thickness=10):
C:=polygon([op(L1),op(map(t->[-t[1],t[2]],ListTools:-Reverse(L1)))], color=green):
Tree:=display([A, B, C], scaling=constrained, axes=none):
P:=()->pointplot(r(),color=["Blue"$2, "Yellow"$2,"Crimson"$2,"Cyan"$2,"Pink"$2],symbol=solidcircle,symbolsize=20):

S:=cat("Happy New Year 2020!   "$3):
N:=length(S): a:=0.77*Pi: h:=2*Pi/N:
Text:=display(seq(textplot([cos(a-k*h), sin(a-k*h),S[k+1]], 
        rotation=-Pi/2+a-k*h, color=red, 'font'=["times","bold",30]), k=0..N-4)):
display(seq(display(AnimTree[i],Text,homothety(Tree, 0.12, [0,-0.6]))$3, i=1..50),insequence, axes=none, size=[500,500]);


 vv 6618
A good idea! But it looks a little gloomy. And the middle is desirable to fill something. If you don't mind, I can offer my option.

@Carl Love  I understand that this is the design and it is not new, but it is very sad that it has not yet been fixed.

@Carl Love  I still consider this as a bug. Take a look at how Mathematica does the same calculation:

N[1014.1 - 1007, 2]

In fact, in this example we have an iteration of two functions( `-`  and  evalf ), if we write this calculation as  evalf(`-`(P3,P1), 2)  or  evalf[2](`-`(P3,P1)) . In accordance with generally accepted rules, the internal function must first be calculated, and then the external function. We see in this example that these evaluational rules themselves in Maple are erroneous.

Interestingly, if we write this calculation explicitly as an iteration of two functions, the result will be correct:


Thus we see 2 conflicting results:

evalf[2](`-`(1014.1,1007)) <> (evalf[2]@`-`)(1014.1,1007);

@Glowing  I use Maple 2018.2

Should be  exp(-5*t)  instead of  e^(-5*t) .
e is just a symbol in Maple not the number 2.71828...

@Carl Love  for this. I just didn’t know about this command and thought that  LinearAlgebra:-Equal  was meant.  By the way, what you wrote can be written easier

L1:=<2-4*t, 5+6*t>:  L2:= <-6-12*t, 17+18*t>: 


@tomleslie  It is interesting how, using the  Equal  command, you will establish the identity of the lines  L1  and  L2  defined by their  parametric equations:

L1:=<2-4*t, 5+6*t>L2:=<-6-12t, 17+18t> 

@rlopez  Unfortunately, the commands in this package do not distinguish between coincident and parallel straight lines.
The example below shows this (here the lines are given by a point and a direction vector):

l1 := Line([0, 2, 6], <10, 15, 20>):
l2 := Line([0, 2, 6], <2, 3, 4>):
AreParallel(l1, l2);


Obviously  l2  is the same line as  l1 .

@Jjjones98  In the code of  replace the name of matrix  A  with  M , because  A  is the name of the unknown A :

M,b:=GenerateMatrix(Sys1, [A,B,C,E,L,F]);


@Jjjones98  For  C=Sol[3]  should be


The command  `<|>`(seq(`if`(k=3,b,A(..,k)),k=1..6))
replaces the third column in the matrix  A  with a column of free members  b .

@vv A very simple and clear solution, vote up. The key here, of course, is the application of the  expand  command to  w^(2*sigma)  and  w^(sigma-1) , which then allows factorization to be done:
expand( w^(2*sigma)) = (w^sigma)^2
expand( w^(sigma-1)) = (w^sigma)/w

@Jjjones98  Yes you are right. Maple returns the value of  A=Sol[1], but for some reason does not return  Sol[2], and so on. But you can still get the values of the rest of the unknowns if you use explicit formulas, for example, Cramer's rule. So the meaning of  B=Sol[2]  will be
and so on. Unfortunately, the expressions are still very bulky, and I don’t know how to simplify them and whether this is possible in principle.


@Carl Love  In the general case, the result may depend on the order in which the substitutions occur. Compare:

S:={a=0, b=2};
eval(eval(A, S), x=0);
eval(A, S union {x=0});


@Yaongyaong  To find the surface area of a paraboloid lying inside a cone, you can use the standard formula through the double integral:

int(sqrt(1+diff(z,x)^2+diff(z,y)^2),[y=-sqrt(9-x^2)..sqrt(9-x^2), x=-3..3]);

 Output:                           37*sqrt(37)*Pi*(1/6)-(1/6)*Pi

Another way:

Student:-Calculus1:-SurfaceOfRevolution(12-x^2,x=0..3, axis =vertical, output=value);

Output:                                 (1/6)*(37*sqrt(37)-1)*Pi 

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