Axel Vogt

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20 years, 259 days
Munich, Bavaria, Germany

MaplePrimes Activity


These are replies submitted by Axel Vogt

(x^n)^4/x^2-(x^n)^4*R/x^4+1/(4*n^2); #to yield
eval(%, x=sqrt(z));
combine(%);

                    (2 n - 1)      (2 n - 2)    1
                   z          - R z          + ----
                                                  2
                                               4 n

Preben, thank you - I like it!

Preben, thank you - I like it!

Was playing a bit more with it and if m*phi = psi is real one can do it, here a sketch.

Write L:=eval(t2a, phi = psi/m), where t2a is the 2nd list.

Then that is periodic in psi and symmetric w.r.t. psi = Pi

'L=eval(L, psi = psi + 2*Pi)'; is(%);

                                 true

'eval(L, psi = Pi-psi) = eval(L, psi = Pi+psi)'; is(%);

                                 true

Plotting indicates, that the minimum for the imaginary part is at psi=Pi.

for j from 1 to nops(L) do
  L[j]:
  plot( [Re(%), Im(%)], psi=0 .. 2*Pi, title=j);
  print(%);
end do:

It is obvious (or can be proved), that this is a local extremum.  And the
selection is correct, if you can prove, that it is a global minimum.

Evaluating there you can select the desired elements.

Moreover in that range you can write cos(psi) = c between -1 and +1 and now
writing a product z - L[i] you get a polynomial in z, parametric in c, where
your entries are the roots.

That polynomial turns out to be in z^2, z = sqrt(Z), so it is polynomial in
Z (upper case) of degree=2, parametric in c.
Was playing a bit more with it and if m*phi = psi is real one can do it, here a sketch.

Write L:=eval(t2a, phi = psi/m), where t2a is the 2nd list.

Then that is periodic in psi and symmetric w.r.t. psi = Pi

'L=eval(L, psi = psi + 2*Pi)'; is(%);

                                 true

'eval(L, psi = Pi-psi) = eval(L, psi = Pi+psi)'; is(%);

                                 true

Plotting indicates, that the minimum for the imaginary part is at psi=Pi.

for j from 1 to nops(L) do
  L[j]:
  plot( [Re(%), Im(%)], psi=0 .. 2*Pi, title=j);
  print(%);
end do:

It is obvious (or can be proved), that this is a local extremum.  And the
selection is correct, if you can prove, that it is a global minimum.

Evaluating there you can select the desired elements.

Moreover in that range you can write cos(psi) = c between -1 and +1 and now
writing a product z - L[i] you get a polynomial in z, parametric in c, where
your entries are the roots.

That polynomial turns out to be in z^2, z = sqrt(Z), so it is polynomial in
Z (upper case) of degree=2, parametric in c.

It is not on me to allow or not. But please note it has some side effects.

So just post once.

Thanks.

There is no need to update this thread so frequently, that it always stays at the
last posted threads, while it is 3 days old.

Though there are nice pictures and some sense advertising (even for book) is
ok for me I really would appreciate if that is enough.

Michael Carter just does that - as the system reports it. It should be enough now.

map(about, [p1,p2,p3]);

or

p1=convert(p1, name); is(%);

                                false

map(about, [p1,p2,p3]);

or

p1=convert(p1, name); is(%);

                                false

@kh2n : done so far and the rest should be done by yourself

@kh2n : done so far and the rest should be done by yourself

Look:

temp6c1:= subs(parvals,subs(params,temp6b)):
indets(%, symbol);

                           {R3, S0, Sigma}

So you have a function as I already said and to make things more easy
let us take just such a function:

f:= (r,s, t) -> r^2 + exp(t) + s;

Now you say "I want to plot t(r,s)".

What should that be?

Look:

temp6c1:= subs(parvals,subs(params,temp6b)):
indets(%, symbol);

                           {R3, S0, Sigma}

So you have a function as I already said and to make things more easy
let us take just such a function:

f:= (r,s, t) -> r^2 + exp(t) + s;

Now you say "I want to plot t(r,s)".

What should that be?

You have a function F(Sigma,  R30,S0) of 3 variables and some range for 2 of them.

Now what is the question?

You have a function F(Sigma,  R30,S0) of 3 variables and some range for 2 of them.

Now what is the question?

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