Robert Israel

6577 Reputation

21 Badges

18 years, 215 days
University of British Columbia
Associate Professor Emeritus
North York, Ontario, Canada

MaplePrimes Activity


These are replies submitted by Robert Israel

I get that too.  But, after clicking "Always trust content from this publisher" and "Run", I find the things I tried don't work very well at all.

First I tried Integration.  At first I didn't get an entry box at all.  Then I got one, but anything I tried to integrate gave me "Error: enter your function in the input".  Now I tried integrating x^2, and I get 1/3*x^3 + x^2 + x.

I also tried Matrix Inverse, but the inverse of every matrix I try comes out as the identity matrix.

I'm using Firefox 3.0.8 under Windows Vista.

 

The special feature of this equation, I think, is that b^2 = a^2 - 1.  Then the equation

ln(y)/ln(a) = ln(y-1)/ln(b)

has the obvious solution y = a^2 (at least if Re(a) > 0). 

Somewhat more generally, if a^p = 1 + b^p, then you have the solution y = a^p
(if ln(a^p) = p ln(a), which is true if |Im(p ln(a))| < Pi).

Now for any a and b with 0 < b < a, there will be some p > 0 satisfying
a^p = 1 + b^p, but probably not expressible in "closed form".

 

 

The special feature of this equation, I think, is that b^2 = a^2 - 1.  Then the equation

ln(y)/ln(a) = ln(y-1)/ln(b)

has the obvious solution y = a^2 (at least if Re(a) > 0). 

Somewhat more generally, if a^p = 1 + b^p, then you have the solution y = a^p
(if ln(a^p) = p ln(a), which is true if |Im(p ln(a))| < Pi).

Now for any a and b with 0 < b < a, there will be some p > 0 satisfying
a^p = 1 + b^p, but probably not expressible in "closed form".

 

 

You still didn't answer the question, so I'll try to answer it for you.  I think you want a contourplot of the joint density f(x,y) of the X and Y coordinates for the distribution from which your 1000 points are a sample.
There's no way to get that exactly, but we can perhaps do a kind of approximation of it by "smearing out" each
sample pont, i.e. if the sample points are (x_i, y_i), take f(x,y) = sum_i h(x - x_i, y - y_i) for some nice smooth "bump" function h.  I'll take h(s,t) = exp(-s^2 - t^2).  You might try other choices.
 


> f:= (s,t) -> add(exp(-(s-E[i,1])^2 - (t - E[i,2])^2), i=1..1000);
   contourplot(f, -8 .. 5, -7 .. 7, labels=[x,y]);

 

You still didn't answer the question, so I'll try to answer it for you.  I think you want a contourplot of the joint density f(x,y) of the X and Y coordinates for the distribution from which your 1000 points are a sample.
There's no way to get that exactly, but we can perhaps do a kind of approximation of it by "smearing out" each
sample pont, i.e. if the sample points are (x_i, y_i), take f(x,y) = sum_i h(x - x_i, y - y_i) for some nice smooth "bump" function h.  I'll take h(s,t) = exp(-s^2 - t^2).  You might try other choices.
 


> f:= (s,t) -> add(exp(-(s-E[i,1])^2 - (t - E[i,2])^2), i=1..1000);
   contourplot(f, -8 .. 5, -7 .. 7, labels=[x,y]);

 

Do you mean this?

> C:= sum(sum(4/((2*n+1)*Pi) * sin((2*n+1)*2*Pi/T * t - k*z), n = 0 .. 4), z = 0 .. 4);

We need values for the constants T and k (at least I presume they are constants, and t is the variable).

> C1:= eval(C, {T=1, k = 2});

Then just plot:

> plot(C1, t = 0 .. 1);

 

 

Do you mean this?

> C:= sum(sum(4/((2*n+1)*Pi) * sin((2*n+1)*2*Pi/T * t - k*z), n = 0 .. 4), z = 0 .. 4);

We need values for the constants T and k (at least I presume they are constants, and t is the variable).

> C1:= eval(C, {T=1, k = 2});

Then just plot:

> plot(C1, t = 0 .. 1);

 

 

I also get it returning unevaluated, in Maple 10.00, Maple 10.06, and, for that matter, Maple 6, Maple 9.50 and Maple 11.02, so I'd find it really surprising that minimize works here in any recent version of Maple.

 

I also get it returning unevaluated, in Maple 10.00, Maple 10.06, and, for that matter, Maple 6, Maple 9.50 and Maple 11.02, so I'd find it really surprising that minimize works here in any recent version of Maple.

 

For multiplication, you need a multiplication sign * (or a space will do when using 2D math input).  If you just write xy, it  is considered as another variable name.  So it should be 2*x*y - 1.  As for 2(x*y), in 2D math that may or may not be OK, depending on your Typesetting settings.

For multiplication, you need a multiplication sign * (or a space will do when using 2D math input).  If you just write xy, it  is considered as another variable name.  So it should be 2*x*y - 1.  As for 2(x*y), in 2D math that may or may not be OK, depending on your Typesetting settings.

Of course that's not what the professor had in mind (and it wouldn't work in a harder case where differentiation doesn't produce an easy differential equation).  He wanted to iterate psi.

Something like this, perhaps.

> psi:= f -> unapply(evalf(1/2*(int(x*y/(y+1)*f(y),y=0..1)+x^3)),x);
> f[0]:= x -> x;
   for i from 1 to 20 do
      f[i]:= psi(f[i-1])
   end do;

... and you see that by i=12 this stabilizes to

x -> 0.03038815504*x+0.5000000000*x^3

Of course that's not what the professor had in mind (and it wouldn't work in a harder case where differentiation doesn't produce an easy differential equation).  He wanted to iterate psi.

Something like this, perhaps.

> psi:= f -> unapply(evalf(1/2*(int(x*y/(y+1)*f(y),y=0..1)+x^3)),x);
> f[0]:= x -> x;
   for i from 1 to 20 do
      f[i]:= psi(f[i-1])
   end do;

... and you see that by i=12 this stabilizes to

x -> 0.03038815504*x+0.5000000000*x^3

Also interesting is that evalm(A &* B &* (2*B) - B &* Id) does not simplify the B &* Id to B.

Apparently this worked once upon a time, but even in Maple 6 (which is the oldest version I have on this computer) it doesn't work.

Also interesting is that evalm(A &* B &* (2*B) - B &* Id) does not simplify the B &* Id to B.

Apparently this worked once upon a time, but even in Maple 6 (which is the oldest version I have on this computer) it doesn't work.

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