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seq - numelems and declaring a local variable in ...

Asked by: Scot Gould 637
procedure seq warning
March 14 2021
1  7

inert % does not obey order of operations...

Asked by: Stretto 235
inert precedence
March 14 2021
0  1

4 %+ 3 %* 2;
                        %*(%+(4, 3), 2)

= (4+3)*2

It makes using % very difficult in complex expressions because one has to constantly use a ton of paranethesis just to get things to work out.

1 + 4%^3  %* 2 %- 6 %/ 4;

1 + ((((4 %^ 3) %* 2) %- 6) %/ 4)

= 1 + (4^3*2 - 6)/4

pdsolve exception int/gbinthm/structure INVALID s...

Asked by: nm 8552
pde pdsolve regression
March 14 2021
2  2

I noticed number of pde's now fail in Maple 2021 with the error 

          int/gbinthm/structure INVALID subscript selector

but they do not fail in Maple 2020.2.

Here are few   examples

restart;
pde :=a*x^n*diff(w(x,y),x) + n*x^m*y*diff(w(x,y),y) =s*x^p*y^q+d;
pdsolve(pde,w(x,y));

restart;
pde :=a*x^n*diff(w(x,y),x) + n*x^m*y*diff(w(x,y),y)=c*x^k*y^s+d; 
pdsolve(pde,w(x,y)) 

restart;
pde :=a*x^n*diff(w(x,y),x)+b*x^m*y*diff(w(x,y),y) =  (c*x^k*y^s + d)*w(x,y);
pdsolve(pde,w(x,y))

restart;
pde :=  a*diff(w(x,y),x)+ y*diff(w(x,y),y) = b*w(x,y)+ c*x^n*y^m;
pdsolve(pde,w(x,y))

#etc..

Error, (in int/gbinthm/structure) invalid subscript selector
 

While in Maple 2020.2 they all work. Screen shot

Maple 2021

 

Maple 2020.2

 

Any idea why this happens? Do others see the same error?

 

 

Wirtinger derivatives in Maple 2021

by: ecterrab 13431 Maple
differentiation complex-analysis physics
March 14 2021
6  0

Wirtinger Derivatives in Maple 2021

Generally speaking, there are two contexts for differentiating complex functions with respect to complex variables. In the first context, called the classical complex analysis, the derivatives of the complex components ( abs , argument , conjugate , Im , Re , signum ) with respect to complex variables do not exist (do not satisfy the Cauchy-Riemann conditions), with the exception of when they are holomorphic functions. All computer algebra systems implement the complex components in this context, and computationally represent all of abs(z), argument(z), conjugate(z), Im(z), Re(z), signum(z) as functions of z . Then, viewed as functions of z, none of them are analytic, so differentiability becomes an issue.

 

In the second context, first introduced by Poincare (also called Wirtinger calculus), in brief z and its conjugate conjugate(z) are taken as independent variables, and all the six derivatives of the complex components become computable, also with respect to conjugate(z). Technically speaking, Wirtinger calculus permits extending complex differentiation to non-holomorphic functions provided that they are ℝ-differentiable (i.e. differentiable functions of real and imaginary parts, taking f(z) = f(x, y) as a mapping "`ℝ`^(2)->`ℝ`^()").

 

In simpler terms, this subject is relevant because, in mathematical-physics formulations using paper and pencil, we frequently use Wirtinger calculus automatically. We take z and its conjugate conjugate(z) as independent variables, with that d*conjugate(z)*(1/(d*z)) = 0, d*z*(1/(d*conjugate(z))) = 0, and we compute with the operators "(∂)/(∂ z)", "(∂)/(∂ (z))" as partial differential operators that behave as ordinary derivatives. With that, all of abs(z), argument(z), conjugate(z), Im(z), Re(z), signum(z), become differentiable, since they are all expressible as functions of z and conjugate(z).

 

 

Wirtinger derivatives were implemented in Maple 18 , years ago, in the context of the Physics package. There is a setting, Physics:-Setup(wirtingerderivatives), that when set to true - an that is the default value when Physics is loaded - redefines the differentiation rules turning on Wirtinger calculus. The implementation, however, was incomplete, and the subject escaped through the cracks till recently mentioned in this Mapleprimes post.

 

Long intro. This post is to present the completion of Wirtinger calculus in Maple, distributed for everybody using Maple 2021 within the Maplesoft Physics Updates v.929 or newer. Load Physics and set the imaginary unit to be represented by I

 

with(Physics); interface(imaginaryunit = I)

 

The complex components are represented by the computer algebra functions

(FunctionAdvisor(complex_components))(z)

[Im(z), Re(z), abs(z), argument(z), conjugate(z), signum(z)]

 (1)

They can all be expressed in terms of z and conjugate(z)

map(proc (u) options operator, arrow; u = convert(u, conjugate) end proc, [Im(z), Re(z), abs(z), argument(z), conjugate(z), signum(z)])

[Im(z) = ((1/2)*I)*(-z+conjugate(z)), Re(z) = (1/2)*z+(1/2)*conjugate(z), abs(z) = (z*conjugate(z))^(1/2), argument(z) = -I*ln(z/(z*conjugate(z))^(1/2)), conjugate(z) = conjugate(z), signum(z) = z/(z*conjugate(z))^(1/2)]

 (2)

The main differentiation rules in the context of Wirtinger derivatives, that is, taking z and conjugate(z) as independent variables, are

map(%diff = diff, [Im(z), Re(z), abs(z), argument(z), conjugate(z), signum(z)], z)

[%diff(Im(z), z) = -(1/2)*I, %diff(Re(z), z) = 1/2, %diff(abs(z), z) = (1/2)*conjugate(z)/abs(z), %diff(argument(z), z) = -((1/2)*I)/z, %diff(conjugate(z), z) = 0, %diff(signum(z), z) = (1/2)/abs(z)]

 (3)

Since in this context conjugate(z) is taken as - say - a mathematically-atomic variable (the computational representation is still the function conjugate(z)) we can differentiate all the complex components also with respect to  conjugate(z)

map(%diff = diff, [Im(z), Re(z), abs(z), argument(z), conjugate(z), signum(z)], conjugate(z))

[%diff(Im(z), conjugate(z)) = (1/2)*I, %diff(Re(z), conjugate(z)) = 1/2, %diff(abs(z), conjugate(z)) = (1/2)*z/abs(z), %diff(argument(z), conjugate(z)) = ((1/2)*I)*z/abs(z)^2, %diff(conjugate(z), conjugate(z)) = 1, %diff(signum(z), conjugate(z)) = -(1/2)*z^2/abs(z)^3]

 (4)

For example, consider the following algebraic expression, starting with conjugate

eq__1 := conjugate(z)+z*conjugate(z)^2

conjugate(z)+z*conjugate(z)^2

 (5)

Differentiating this expression with respect to z and conjugate(z) taking them as independent variables, is new, and in this example trivial

(%diff = diff)(eq__1, z)

%diff(conjugate(z)+z*conjugate(z)^2, z) = conjugate(z)^2

 (6)

(%diff = diff)(eq__1, conjugate(z))

%diff(conjugate(z)+z*conjugate(z)^2, conjugate(z)) = 1+2*z*conjugate(z)

 (7)

Switch to something less trivial, replace conjugate by the real part ReNULL

eq__2 := eval(eq__1, conjugate = Re)

Re(z)+z*Re(z)^2

 (8)

To verify results further below, also express eq__2 in terms of conjugate

eq__22 := simplify(convert(eq__2, conjugate), size)

(1/4)*(z^2+z*conjugate(z)+2)*(z+conjugate(z))

 (9)

New: differentiate eq__2 with respect to z and  conjugate(z)

(%diff = diff)(eq__2, z)

%diff(Re(z)+z*Re(z)^2, z) = 1/2+Re(z)^2+z*Re(z)

 (10)

(%diff = diff)(eq__2, conjugate(z))

%diff(Re(z)+z*Re(z)^2, conjugate(z)) = 1/2+z*Re(z)

 (11)

Note these results (10) and (11) are expressed in terms of Re(z), not conjugate(z). Let's compare with the derivative of eq__22 where everything is expressed in terms of z and conjugate(z). Take for instance the derivative with respect to z

(%diff = diff)(eq__22, z)

%diff((1/4)*(z^2+z*conjugate(z)+2)*(z+conjugate(z)), z) = (1/4)*(2*z+conjugate(z))*(z+conjugate(z))+(1/4)*z^2+(1/4)*z*conjugate(z)+1/2

 (12)

To verify this result is mathematically equal to (10) expressed in terms of Re(z) take the difference of the right-hand sides

rhs((%diff(Re(z)+z*Re(z)^2, z) = 1/2+Re(z)^2+z*Re(z))-(%diff((1/4)*(z^2+z*conjugate(z)+2)*(z+conjugate(z)), z) = (1/4)*(2*z+conjugate(z))*(z+conjugate(z))+(1/4)*z^2+(1/4)*z*conjugate(z)+1/2)) = 0

Re(z)^2+z*Re(z)-(1/4)*(2*z+conjugate(z))*(z+conjugate(z))-(1/4)*z^2-(1/4)*z*conjugate(z) = 0

 (13)

One quick way to verify the value of expressions like this one is to replace z = a+I*b and simplify "assuming" a andNULLb are realNULL

`assuming`([eval(Re(z)^2+z*Re(z)-(1/4)*(2*z+conjugate(z))*(z+conjugate(z))-(1/4)*z^2-(1/4)*z*conjugate(z) = 0, z = a+I*b)], [a::real, b::real])

a^2+(a+I*b)*a-(1/2)*(3*a+I*b)*a-(1/4)*(a+I*b)^2-(1/4)*(a+I*b)*(a-I*b) = 0

 (14)

normal(a^2+(a+I*b)*a-(1/2)*(3*a+I*b)*a-(1/4)*(a+I*b)^2-(1/4)*(a+I*b)*(a-I*b) = 0)

0 = 0

 (15)

The equivalent differentiation, this time replacing in eq__1 conjugate by abs; construct also the equivalent expression in terms of z and  conjugate(z) for verifying results

eq__3 := eval(eq__1, conjugate = abs)

abs(z)+abs(z)^2*z

 (16)

eq__33 := simplify(convert(eq__3, conjugate), size)

(z*conjugate(z))^(1/2)+conjugate(z)*z^2

 (17)

Since these two expressions are mathematically equal, their derivatives should be too, and the derivatives of eq__33 can be verified by eye since z and  conjugate(z) are taken as independent variables

(%diff = diff)(eq__3, z)

%diff(abs(z)+abs(z)^2*z, z) = (1/2)*conjugate(z)/abs(z)+z*conjugate(z)+abs(z)^2

 (18)

(%diff = diff)(eq__33, z)

%diff((z*conjugate(z))^(1/2)+conjugate(z)*z^2, z) = (1/2)*conjugate(z)/(z*conjugate(z))^(1/2)+2*z*conjugate(z)

 (19)

Eq (18) is expressed in terms of abs(z) = abs(z) while (19) is in terms of conjugate(z) = conjugate(z). Comparing as done in (14)

rhs((%diff(abs(z)+abs(z)^2*z, z) = (1/2)*conjugate(z)/abs(z)+z*conjugate(z)+abs(z)^2)-(%diff((z*conjugate(z))^(1/2)+conjugate(z)*z^2, z) = (1/2)*conjugate(z)/(z*conjugate(z))^(1/2)+2*z*conjugate(z))) = 0

(1/2)*conjugate(z)/abs(z)-z*conjugate(z)+abs(z)^2-(1/2)*conjugate(z)/(z*conjugate(z))^(1/2) = 0

 (20)

`assuming`([eval((1/2)*conjugate(z)/abs(z)-z*conjugate(z)+abs(z)^2-(1/2)*conjugate(z)/(z*conjugate(z))^(1/2) = 0, z = a+I*b)], [a::real, b::real])

(1/2)*(a-I*b)/(a^2+b^2)^(1/2)-(a+I*b)*(a-I*b)+a^2+b^2-(1/2)*(a-I*b)/((a+I*b)*(a-I*b))^(1/2) = 0

 (21)

simplify((1/2)*(a-I*b)/(a^2+b^2)^(1/2)-(a+I*b)*(a-I*b)+a^2+b^2-(1/2)*(a-I*b)/((a+I*b)*(a-I*b))^(1/2) = 0)

0 = 0

 (22)

To mention but one not so famliar case, consider the derivative of the sign of a complex number, represented in Maple by signum(z). So our testing expression is

eq__4 := eval(eq__1, conjugate = signum)

signum(z)+z*signum(z)^2

 (23)

This expression can also be rewritten in terms of z and  conjugate(z) 

eq__44 := simplify(convert(eq__4, conjugate), size)

z/(z*conjugate(z))^(1/2)+z^2/conjugate(z)

 (24)

This time differentiate with respect to conjugate(z),

(%diff = diff)(eq__4, conjugate(z))

%diff(signum(z)+z*signum(z)^2, conjugate(z)) = -(1/2)*z^2/abs(z)^3-z^3*signum(z)/abs(z)^3

 (25)

Here again, the differentiation of eq__44, that is expressed entirely in terms of z and  conjugate(z), can be computed by eye

(%diff = diff)(eq__44, conjugate(z))

%diff(z/(z*conjugate(z))^(1/2)+z^2/conjugate(z), conjugate(z)) = -(1/2)*z^2/(z*conjugate(z))^(3/2)-z^2/conjugate(z)^2

 (26)

Eq (25) is expressed in terms of abs(z) = abs(z) while (26) is in terms of conjugate(z) = conjugate(z). Comparing as done in (14),

rhs((%diff(signum(z)+z*signum(z)^2, conjugate(z)) = -(1/2)*z^2/abs(z)^3-z^3*signum(z)/abs(z)^3)-(%diff(z/(z*conjugate(z))^(1/2)+z^2/conjugate(z), conjugate(z)) = -(1/2)*z^2/(z*conjugate(z))^(3/2)-z^2/conjugate(z)^2)) = 0

-(1/2)*z^2/abs(z)^3-z^3*signum(z)/abs(z)^3+(1/2)*z^2/(z*conjugate(z))^(3/2)+z^2/conjugate(z)^2 = 0

 (27)

`assuming`([eval(-(1/2)*z^2/abs(z)^3-z^3*signum(z)/abs(z)^3+(1/2)*z^2/(z*conjugate(z))^(3/2)+z^2/conjugate(z)^2 = 0, z = a+I*b)], [a::real, b::real])

-(1/2)*(a+I*b)^2/(a^2+b^2)^(3/2)-(a+I*b)^4/(a^2+b^2)^2+(1/2)*(a+I*b)^2/((a+I*b)*(a-I*b))^(3/2)+(a+I*b)^2/(a-I*b)^2 = 0

 (28)

simplify(-(1/2)*(a+I*b)^2/(a^2+b^2)^(3/2)-(a+I*b)^4/(a^2+b^2)^2+(1/2)*(a+I*b)^2/((a+I*b)*(a-I*b))^(3/2)+(a+I*b)^2/(a-I*b)^2 = 0)

0 = 0

 (29)

NULL


 

Download Wirtinger_Derivatives.mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

n-dimensional metric in Physics package...

Asked by: mykola 15
metric physics
March 14 2021
1  3

Hi!

How to define n-dimensional metric for arbitrary n in Physics package?

Thanks

Derive the Euler Lagrange equations using field th...

Asked by: daljit97 65
lagrangian physics euler-lagrange
March 14 2021
2  1

I would like to compute the euler-lagrange equations using the index notation in quantum field theory. So for example given the Lagrangian here:

 

Is there a way to derive the Euler-Lagrange equation in maple?

Maple 2021 / Problem while saving the workbook...

Asked by: Anthrazit 750
workbook save
March 14 2021
5  8

I've experienced problems a couple of times when trying to save a workbook in Maple 2021.

At first I thought it was due to a unstable VPN connection, but this also happens on a local drive.

How to force Maple to express the derivatives of L...

Asked by: Heinrich 95
derivative recurrence legendre
March 14 2021
0  8

Hi,

Consider the following example:

 

>> Maple express the second derivative of Legendre polynomial as

 

This form cannot, for instance, be integrated if we set n, for example, to 2.

i was wondering whether there is a trick to force Maple to express those derivatives in term of P_n using the relevent recurrence relations. This will help for further manipulations.

 

Thanks a lot!

 

Heinrich,

 

Programming a loop for Trapezoid Rule evaluation o...

Asked by: Jtroy2 5
March 13 2021
1  4

I have a 2688x1 matrix of ground acceleration data that I converted to a forcing function F (also a 2688x1 matrix) that I need to evaluate using duhamels integral. Since the forcing function can not be represented my a real expression numberical integration is required. Am needing to write a loop as seen on bottom of attached picture. Essentially to perform trapezoidal rule to return  structural motion (u).

 

Am looking for advice on how to write this loop in Maple.

I have attempted the for program in Maple per below with;

[F]:= Forcing Matrix 
[t]:= time step matrix

Assigned initial conditions sim to pic below

for i from 1..2688 do

added the functions sim to bottom of pic for Ai, Bi, and ui 

end do

 

 

 


 

why the new documentation PDF's for Maple 2021 are...

Asked by: nm 8552
pdf documentation latex
March 13 2021
4  4

This is just an obervation. May be due to me missing some clarification.

My understanding is that in Maple 2021, one can now export a complete Maple worksheet/document to Latex successfuly.

According to https://www.maplesoft.com/products/maple/new_features/  it says under "Latex export"
  
  Export is available for individual expressions or the entire document, though the latex command

So I would have expected that now all the online PDF documents for the new features in Maple 2021 to be generated from Latex for better formatting.

But when going over number of documents, they do not appear to have converted to Latex before generating the PDF.

Even the PDF document about the new latex command does not appear to have been generated from Latex. I can quickly recognize a PDF that was generated from Latex, and these PDF files are clearly not.

Looking at sample PDF from Maple website, it says the generator is "iText" and not latex compiler.

It looks like these documents were just a direct export of the Maple worksheet to PDF, instead of being exporting to Latex first, and then compiled to PDF.

Am I missing something? Why not take advantage of the Maple Latex export, and convert all the documents to Latex first and then generate the PDF from Latex? This would make them look much better in terms of formatting and math rendering.

Here are some examples

  https://www.maplesoft.com/products/maple/new_features/Maple2021/PDFs/AdvancedMath.pdf  


  https://www.maplesoft.com/products/maple/new_features/Maple2021/PDFs/ODEsandPDEs.pdf


  https://www.maplesoft.com/products/maple/new_features/Maple2021/PDFs/GraphTheory.pdf


  https://www.maplesoft.com/products/maple/new_features/Maple2021/PDFs/Physics.pdf

  https://www.maplesoft.com/products/maple/new_features/Maple2021/PDFs/LaTeX.pdf

Why did not Maple documentation take advantage of the new export command of Maple's latex command? This would also have been a good way and an opportunity to show case the effectiveness of the new latex export command.

May be it was not done due to time constraint before the release date of maple 2021? Or are there other technical reasons?

 

 

How to draw a streamline by using stream function ...

Asked by: Madhukesh J K 140
ode homework numeric
March 13 2021
0  5

How to draw a streamlines by using stream function provided in pdf. In the presence and absence of lambda parameter.

please help me to obtain graphs

1.pdf
Download 1.mw

 

Wht doesn't this work with Maple 13?...

Asked by: mapleatha 160
differentiation surd operator
March 13 2021
0  5

This does not work with Maple 13. Can I still do something like this just using the linear operator D?

g:=x->x^(1/3);

f:=x->convert(g(x),surd);

D(f)(x)???

I need the derivative D(f)!!

Thank you!!!

mapleatha

how to solve this problem...

Asked by: adel-00 135
integration numeric
March 13 2021
0  3

Hi Mapke experts,Here is my try to solve the integration

int(x-t)^(-alpha)*a(t-bln(t+b/b),t from 0 to x)

I found it complicated to solve it symbolic, so \i try this:

restart:
assume(t,real):
x[0]:=0:a:=-0.1:alpha:=0.5:h:=0.001:b:=Gamma(2-alpha)/((1-alpha)*Gamma(1-alpha)):
for n from 0 to 0 do
t[n]:=n*h:
vo:=a*(x[n]-b*(ln((x[n]+b)/b))):
uo:=a*(t[n]-b*(ln((t[n]+b)/b))):
u1:=(int((x[n]-t[n])^(-alpha)*uo,t[n]=0..x[n])):
S[n]:=vo+u1:
od:

data:=[seq([t[n],S[n]],n=0..100)]:

Sum of degree of neighborhs...

Asked by: vs140580 490
graphtheory homework
March 13 2021
1  14

A function such that given a Graph G and vertex say v as input to a function

It should output the sum of the degrees of all the neighbors of that vertex v in that graph G 

 

Kind help thank you 

Bug in Integration of square of Jacobi sine functi...

Asked by: nitramkaroh 10
March 13 2021
2  4

Dear all,

I believe there is a bug in integration of square of Jacobi sine function.

Here is what I did:

u := int(JacobiSN(x, k)^2, x);

The result I got is
                  x - EllipticE(JacobiSN(x, k), k)
             u := --------------------------------
                                  k^2               
However, according to  Handbook of elliptic integrals for engineers and scientists by Byrd and Friedman Eq (310.02), it is supposed to be  

                                            
                  x - EllipticE(JacobiAM(x, k), k)
             u := --------------------------------
                                  k^2     

 

 Note that I tried also to integrate first and third power of JacobiSN a the results are in agreement with the book.

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