If it never goes more than 50%, my guess would be that it has dual core, 32-bit OS, and 32-bit Maple version. I don't think that anything can be done in this configuration to make use of both processors working simultaneously. The radical solution would be to install 64-bit Linux and use 64-bit Maple for it (properly built.)

In Windows, the best choice, probably, is using Classic interface - the performance is not much worse than in command line version.

Alec

If it never goes more than 50%, my guess would be that it has dual core, 32-bit OS, and 32-bit Maple version. I don't think that anything can be done in this configuration to make use of both processors working simultaneously. The radical solution would be to install 64-bit Linux and use 64-bit Maple for it (properly built.)

In Windows, the best choice, probably, is using Classic interface - the performance is not much worse than in command line version.

Alec

The problem with that is that all possible trajectories would cover all 3-dimensional space - they don't give a surface. If you want a surface, you could fix one of parameters, and change another one, as I did in examples above. For y0=2x0 the surface is flat, and in other cases it is not.

Alec

The problem with that is that all possible trajectories would cover all 3-dimensional space - they don't give a surface. If you want a surface, you could fix one of parameters, and change another one, as I did in examples above. For y0=2x0 the surface is flat, and in other cases it is not.

Alec

For small h, f'(t) may be close to (f(t+h)-f(t))/h, but it is not equal to it, and value h=1, in general, very rarely is small enough for that.

Alec

For small h, f'(t) may be close to (f(t+h)-f(t))/h, but it is not equal to it, and value h=1, in general, very rarely is small enough for that.

Alec

Just noticed your comment about threshold 5. That doesn't work like that. Positive correlation will be with any value - just different one (because there still will be dependency). For negative correlation you could assign x[i] to1-x[i-1] in cases where you assigned it to x[i-1]. Also, it would be negative correlation with any value. No threshold.

Alec

Just noticed your comment about threshold 5. That doesn't work like that. Positive correlation will be with any value - just different one (because there still will be dependency). For negative correlation you could assign x[i] to1-x[i-1] in cases where you assigned it to x[i-1]. Also, it would be negative correlation with any value. No threshold.

Alec

This way it also can be done. Just few comments.

1. instead of 3 ifs you could use if .. elif .. else .. fi

2. You could add option remember; after proc(n). Then F(n-1) won't be calculated another time.

3. The standard way of referring to the procedure name inside of a procedure is procname, i.e. procname(n-1) instead of F(n-1). Then, if you rename your procedure later, you won't have to change the code inside it.

Alec

This way it also can be done. Just few comments.

1. instead of 3 ifs you could use if .. elif .. else .. fi

2. You could add option remember; after proc(n). Then F(n-1) won't be calculated another time.

3. The standard way of referring to the procedure name inside of a procedure is procname, i.e. procname(n-1) instead of F(n-1). Then, if you rename your procedure later, you won't have to change the code inside it.

Alec

A usual way of avoiding recursive assignments is using a loop,

interface(rtablesize=20):randomize():
coin:=rand(0..1):
c10:=rand(1..10):
x1:=Vector[row](19,i->c10());

  x1 := [9, 8, 7, 6, 6, 7, 8, 10, 2, 1, 7, 1, 2, 10, 8, 6, 2, 1, 8]

x2:=Vector[row](20,[coin()]);

  x2 := [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

for i from 1 to 19 do 
x2[i+1]:=`if`(x1[i]<4,coin(),x2[i]) od:
x2;

     [1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0]

That, generally, is what I did in the SampleWithCorr procedure.

from 1 can be omitted - it is the default value in Maple.

Alec

A usual way of avoiding recursive assignments is using a loop,

interface(rtablesize=20):randomize():
coin:=rand(0..1):
c10:=rand(1..10):
x1:=Vector[row](19,i->c10());

  x1 := [9, 8, 7, 6, 6, 7, 8, 10, 2, 1, 7, 1, 2, 10, 8, 6, 2, 1, 8]

x2:=Vector[row](20,[coin()]);

  x2 := [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

for i from 1 to 19 do 
x2[i+1]:=`if`(x1[i]<4,coin(),x2[i]) od:
x2;

     [1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0]

That, generally, is what I did in the SampleWithCorr procedure.

from 1 can be omitted - it is the default value in Maple.

Alec

As ?return help page says,

Use of the RETURN procedure is strongly discouraged :)

Alec

As ?return help page says,

Use of the RETURN procedure is strongly discouraged :)

Alec

It is, obviously, an exercise. Perhaps, one of the series of similar ones. Look at the beginning of this series of exercises. There should be additional information about sets A, B, C, D, and the universal set U. Also, it is not clear what P means.

Alec