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These are replies submitted by Kitonum

@mmcdara  I agree, but this is not a fundamental difference. We can add the  collect  command:

x^(n-1)*(collect(expand((n*x^n - 2*n*x^(n - 1) + x^n)/x^(n-1)), x));


@lcz If you are interested in arbitrary set partitions, you can also take a look at my  post .  It was written long ago, when Maple didn't have the  Iterator package yet.

@dharr  OK. Vote up.

@ecterrab In fact, the exact solution is easily expressed explicitly (see below). But a natural question arises: why does this solution not continue to the right from  x~2.55 , as shown by the numerical solution?

ics:=y(0)=0, D(y)(0)=1:
dsolve({ode, ics}, y(x), implicit);
isolate(%, y(x));
sol := simplify(%);
odetest(sol, [ode, ics]);


@Thomas Dean  Using the P procedure, you can plot the body of revolution for any value of the parameter  a  (this is the angle of rotation).  P (2*Pi+Pi/2)  will be the whole body:





@Axel Vogt  A good idea! This can be done even without using the plottools package:

plot([erf(y),y, y=-3..3]);



@Christian Wolinski 

It takes approximately 10 sec in Maple 2018.2.

@zenterix  In 2d input use prefix (Polish) notation:

A:=<1,2;3,4>: B:=<6,7;8,9>:


@acer  Thank you! I tried the  convert(..., factorial)  command, but for some reason did not think to apply the  expand  one. In Maple 2018, your method also works.

In Maple 2018, the decision takes less than a second. There are probably syntax errors in your code.

M3:=Matrix(6, 6, [[1., -1., 1., -1., 1., -1.], [1., 1., 1., 1., 1., 1.], [-2., 1.618033989, 30.94427190, -153.8246851, 371.1559479, -572.9674774], [-2., 0.6180339876, 22.94427191, -41.15905356, -37.04759741, 149.3606798], [-2., -0.6180339876, 13.05572810, 22.82468509, -20.15594802, -81.03252254], [-2., -1.618033989, 5.055728096, 28.15905368, 68.04759752, 104.6393203]]);
v:=Vector[column](6, [-1/3, -1, 0, 0, 0, 0]);
a := LinearAlgebra:-LinearSolve(M3, v);


@mmcdara  You wrote "Do you think that the OP will keep asking us to do his homework?" I dont know. Maybe it's not homework. I think that ideally everyone should do the job they like and try to do it well. It is likely that the teacher who issued this assignment will not approve if the OP passes off someone else's solution as his own.

@Carl Love 

Rose:= (n::posint)-> plots:-polarplot(`if`(n=2,[sin(n/2*t),sin(n/2*t-Pi)],`if`(n::even,sin(n/2*t),sin(n*t))), filled, color= `if`(n::odd, pink, aquamarine)):

Rose(1); Rose(2); Rose(4);  Rose(3);  # Examples


@Carl Love  You wrote "Your procedure needs to be changed for n=1 ". Why?

Rose:= (n::integer)-> plots:-polarplot(`if`(n::even,sin(n/2*t),sin(n*t)), filled, color= `if`(n::odd, pink, aquamarine)):

Rose(1); # Example



@The function  We see that the solution is returned as an piecewise expression: if  p=2  then we have an empty list  [ ]  (no solutions), if  p=1  then the solutions are expressed in terms of  x3  , which can be arbitrary, (infinitely many solutions), and so on.

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