Axel Vogt

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20 years, 224 days
Munich, Bavaria, Germany

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These are replies submitted by Axel Vogt

@Carl Love 

Oh ... that is worth to be posted as an extra thread.

Yes, multiply numerator and denominator by the conjugate here is the same as 'rationalize'.

Then the "complex division" vanishes and one proceedes as usual (guessing the x is some real), if one knows how to multiply complex numbers:

1/z = (1*conjugate(z)) / (z*conjugate(z)) = complex/real

1/(5*x-I): %= rationalize(%);

                            1       5 x + I
                         ------- = ---------
                         5 x - I       2
                                   25 x  + 1

 

Maple should stop that cancellations (except spam).

Please consider to introduce moving threads to some trash folder, but do not delete.

If one uses MultiSeries:-asympt(f, q, 4) for f = integrand the constant 8*p prevents convergence in infinity

... the following is meant (I use 1-D Math input [finding it better to handle] and do not use labels)
 

# https://www.mapleprimes.com/questions/233766-Why-Square-Roots-Are-Not-Simplifying
restart; interface(version);
with(PDEtools):

`Standard Worksheet Interface, Maple 2021.2, Windows 7, November 23 2021 Build ID 1576349`

(1)

DepVars := [u(x, y, t), U(xi, eta)];
alias(u = u(x, y, t));

[u(x, y, t), U(xi, eta)]

 

u

(2)

xi1 := 1/2*(x^2+y^2);
xi2 := t;
u := (h(t)+(x^2+y^2)*(1/2))*arccos(x/sqrt(x^2+y^2))/t+U(xi1, xi2);

(1/2)*x^2+(1/2)*y^2

 

t

 

(h(t)+(1/2)*x^2+(1/2)*y^2)*arccos(x/(x^2+y^2)^(1/2))/t+U((1/2)*x^2+(1/2)*y^2, t)

(3)

dd := '(diff(u, x))*(diff(u, y))';

(diff(u, x))*(diff(u, y))

(4)

'dd'= ``;
eval(dd, [t = eta]):
simplify(%) assuming x::real, 0 < y:
``=algsubs(1/2*(x^2+y^2) = xi, %):
simplify(%) assuming 0<xi: # else nothing is known about xi

%, xi = 1/2*(x^2+y^2), eta=t;

dd = ``

 

`` = (1/2)*(arccos((1/2)*x*2^(1/2)/xi^(1/2))*x*xi+(D[1](U))(xi, eta)*eta*x*xi-(1/2)*y*(xi+h(eta)))*(2*arccos((1/2)*x*2^(1/2)/xi^(1/2))*xi*y+2*(D[1](U))(xi, eta)*eta*xi*y+x*(xi+h(eta)))/(eta^2*xi^2), xi = (1/2)*x^2+(1/2)*y^2, eta = t

(5)

 

 


 

Download MP_233766.mw

@mskalsi try it using "eval" instead of "subs"

For example: ... assuming 0<p  and read the help for that.

Using a search engine gives his publications https://www.researchgate.net/scientific-contributions/Rafal-Ablamowicz-8237292 (where I do not want to register) Downloading a paper should give you his concurrent contact information. Likewise you can try similar for his co-author Fauser. Good luck.

Edit: or contact him using https://www.linkedin.com/in/rafal-ablamowicz-89396529 if you are a member of Linkedin

Why does one need "Operators" ?

@janhardo 

You seem to expect that the usual rules are correct if working with floating point calculation (this is what tomleslie means).

But this is only true with rounding errors and what you see is just that, that's all.

Here is a simple example where (a+b) + c and a + (b+c) are different:

restart; interface(version); Digits:=10;

`Classic Worksheet Interface, Maple 2017.3, Windows, Sep 27 2017, Build ID 1265877`

Digits := 10

(1)

eps:=10^(-Digits); evalf(%);

eps := 1/10000000000

0.1000000000e-9

(2)

 

(1+eps) - eps;

1

(3)

1+eps; evalf(%);
% - eps;

1 - %;

10000000001/10000000000

1.000000000

.9999999999

0.1e-9

(4)

 

Download example_rounding_error.mws

Edit: and here an example for multiplication giving an error in the last decimal place, (a*b) * c and a * (b*c) are different

Digits:=15;
2.00000000000001;
%/2;
%*2;

@janhardo 

May be you search for some lecture notes about Complex Analysis. It is not important which one, the contents are similar (though the style varies). Do not only try to see what Maple provides, some Math would help you for that.

@janhardo 

That list is certainly not complete (square roots and friends) - but what do you *actually* want to know / achieve?

@Christian Wolinski Unfortunately I do not have a specific reference (except classical books like Kraft or Dolgachev which have in mind the general theory). Perhaps it is better to look for lectures & exercises. Your group is the one generated by the permutation 1 -> 2, 2 -> 3, 3 -> 1 (which is of order 3 and cyclic - but you take only that permutation). May be you ask in a math forum. If you would have q=0 then there are known relations (the Newton identities)

Just a thought: with some luck you may find answers within (classical) invariant theory (for permutations?)

Let R be Kitonum's rational function. Then series shows the singularity using

MultiSeries:-series(R, y=1, 6) or
MultiSeries:-multiseries(R, y=1, 'exact_order')

It can be seen that S:= - ( (3/4/(1-y)^2) - 21/4/(1-y) ) is the singularity in y=1.

limit(R-S, y=1) gives 87/16 and confirms it (or use series again)

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