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9 years, 283 days
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ok...

vv 12453
May 24 2020
0 0

@Carl Love 

Your heuristic test looks OK to me (but I am not a statistician :-)).

runs...

vv 12453
May 24 2020
0 0

@Carl Love 

I think you used a heuristic approach. In the cited paper the computed probabilities are Prob(there are j runs of length i), but only for i=1,2,3. For i>3 (the most interesting for our case)  the probabilities are considered too hard to compute.

(Actually in the paper are constructed 3 tests, for each i=1,2,3).

I don't know what P(run length = k)  means; in general a string has several runs. [BTW, the expected number of runs is (n+1)/2].

 

 

Thanks again...

vv 12453
May 23 2020
0 0

@Carl Love 

I see that you have not used the exact probabilities for runs of lengths 1,2,... (from the cited paper) but the approximations seem to be good enough for a simple statistical test.

title...

vv 12453
May 22 2020
0 0

The OP's expression is in the title (!) 

?...

vv 12453
May 22 2020
0 0

@mmcdara 

I don't see any contradiction. A good PRNG (pseudorandom number generator) exists of course (in my opinion) in Maple. The question was about the existence of a statistical test in Maple able to discriminate the two lists.  

@Carl Love  Thanks. I was hoping t...

vv 12453
May 21 2020
0 0

@Carl Love 

Thanks. I was hoping that one of the general Statistics' tests can be used (more or less directly) to answer the question.
After all, the problem does not seem to be very special. Probably using run lengths the two sequences can be discriminated indeed.

@janhardo  If you need help, you s...

vv 12453
May 20 2020
0 0

@janhardo 

If you need help, you should post worksheets like this:


 

> 

 

> 

f := x -> x^sin(x);

proc (x) options operator, arrow; x^sin(x) end proc

 (1)
> 

g := x -> x^cos(x);

proc (x) options operator, arrow; x^cos(x) end proc

 (2)
> 

plot( [f,g], 1..15, color=[red,blue]);

 
> 

# area=?, please help.


Download 02xxx.mw

syntax...

vv 12453
May 17 2020
0 0

@ecterrab 

I don't quite understand about one and only one syntax:

In 1D input:  3!=6;    ==>  6 = 6
In 2D input:  3!=6;    ==>  3 <> 6

It's true that in 2D,  3!  =6;    ==>  6 = 6
but isn't this about syntax?

Classic...

vv 12453
May 15 2020
0 0

@janhardo 

If you apply what tomleslie said, you have nothing to learn, because Worksheet Mode + 1D-input is almost the same as Classic Maple (actually Classic Maple also had & has a 2D input). The disadvantage of the modern Worksheet Mode is that it starts slower and uses much more memory -- due to Java; but it's more flexible.

the "obvious" fact...

vv 12453
May 15 2020
0 0

@Preben Alsholm 

> 

restart;
local gamma;
beta:=1/6; gamma:=1; delta:=13/6;  # special case
 

gamma

  

1/6

  

1

  

13/6

 (1)
> 

##The system in the pdf-file. Right hand sides:
RHS:=[x*(1-x)-x*y,y*(delta-beta*y/x)-alpha*y/(gamma+y)];
ROsol:=solve(RHS=~0,{x,y});
RO:=op(indets(ROsol[2],specfunc(RootOf)));
E:=subs(ROsol[2],[x,y]);
 

[x*(1-x)-x*y, y*(13/6-(1/6)*y/x)-alpha*y/(1+y)]

  

{x = 1, y = 0}, {x = 1-RootOf(14*_Z^2+(-6*alpha+1)*_Z+6*alpha-13), y = RootOf(14*_Z^2+(-6*alpha+1)*_Z+6*alpha-13)}

  

RootOf(14*_Z^2+(-6*alpha+1)*_Z+6*alpha-13)

  

[1-RootOf(14*_Z^2+(-6*alpha+1)*_Z+6*alpha-13), RootOf(14*_Z^2+(-6*alpha+1)*_Z+6*alpha-13)]

 (2)
> 

x2y2,x3y3:=allvalues(E);

[29/28-(3/14)*alpha-(1/28)*(36*alpha^2-348*alpha+729)^(1/2), (3/14)*alpha-1/28+(1/28)*(36*alpha^2-348*alpha+729)^(1/2)], [29/28-(3/14)*alpha+(1/28)*(36*alpha^2-348*alpha+729)^(1/2), (3/14)*alpha-1/28-(1/28)*(36*alpha^2-348*alpha+729)^(1/2)]

 (3)
> 

da:=diff(E,alpha);
dxa,dya:=op(subs(RO=y,y=1-x,da));
J:=VectorCalculus:-Jacobian(RHS,[x,y]);
TR:=LinearAlgebra:-Trace(J);
 

da := [-(6*RootOf(14*_Z^2+(-6*alpha+1)*_Z+6*alpha-13)-6)/(28*RootOf(14*_Z^2+(-6*alpha+1)*_Z+6*alpha-13)-6*alpha+1), (6*RootOf(14*_Z^2+(-6*alpha+1)*_Z+6*alpha-13)-6)/(28*RootOf(14*_Z^2+(-6*alpha+1)*_Z+6*alpha-13)-6*alpha+1)]

  

dxa, dya := 6*x/(29-28*x-6*alpha), -6*x/(29-28*x-6*alpha)

  

Matrix(%id = 18446744074364419718)

  

19/6-2*x-y-(1/3)*y/x-alpha/(1+y)+alpha*y/(1+y)^2

 (4)
> 

TR0:=simplify(eval(TR, [x,y]=~x2y2));

(1/7)*((432*alpha^3+288*alpha^2-5508*alpha+1692)*(36*alpha^2-348*alpha+729)^(1/2)+2592*alpha^4-10800*alpha^3-45432*alpha^2+147708*alpha+45684)/((-29+6*alpha+(36*alpha^2-348*alpha+729)^(1/2))*(27+6*alpha+(36*alpha^2-348*alpha+729)^(1/2))^2)

 (5)
> 

alpha0:=solve(TR0); evalf(%);

-1/2+2*3^(1/2)

  

2.964101616

 (6)
> 

simplify(eval(diff(TR0,alpha), alpha=alpha0));

3/4

 (7)
> 

subs(x=x2y2[1], alpha=alpha0,  gamma*( (beta+delta)*x - beta )^2 / (alpha*x)^2):  
simplify(%); # the "obvious" fact: 3/4 <> 1/3

1/3

 (8)
> 

 

solve...

vv 12453
May 11 2020
0 0

@Carl Love 

Thanks. I know and I have also used SolveTools:-SemiAlgebraic. Both could be better and faster.

Risch...

vv 12453
May 07 2020

Thank you for this info!
I see that the Risch algorithm works indeed better when the radicals are expressed as RootOfs, and even in older versions (e.g. Maple 2015). So, I wonder why the conversion is not automatic (at least as an option in int).

parametric...

vv 12453
May 05 2020
0 0

@nm 

It is a standard parametric plot. plot( [sin(t),sin(t),  t=0..420]); So, x = sin(t), y = sin(t), t in the interval [0,420].
 

geometrically...

vv 12453
May 02 2020

Nice, vote-up.
Note that a geometric solution is much simpler:

1. An ellipse is the orthogonal projection of a circle.
2. A triangle inscribed in a circle has maximal area iff it is equilateral.
So, an animation is given by:

a:=4: b:=sqrt(3): B:=plot([a*cos(t),b*sin(t), t=0..2*Pi], scaling=constrained):
plots:-animate(plot,[[seq([a*cos(t+2*k*Pi/3),b*sin(t+2*k*Pi/3)], k=0..3)]],t=0..2*Pi, background=B)

 

 

ln(1+x)...

vv 12453
May 01 2020
0 0

@Carl Love 

I know, but for small x (otherwise ln1 is useless) even series is much better than calling ln.

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