Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

I get the degree sequence of a graph, I want to create more graphs, but maple seems to only get the only graph.

Is there a good way to get more, a foolish dream, can we get all the graphs that satisfy this degree sequence condition?

with(GraphTheory):
L:=[7$1..22,6]:
IsGraphicSequence(L):
G := SequenceGraph(L):
DrawGraph(%)

 

 

 

 

What I mean by shortest distance between vertex and edge is

 

In the above petersen graph consider the vertex say 1 it's distance to

(1,6) will be 0 as 1 is part of the same edge

((1,2) will be 0 again

(1,5) it will zero again

(5,8) will be 1 that is (1,5)-(5,8)

(8,9) it will be 2 that is (1,5)-(5,8)-(8,9)

(5,4) it will be 1 that is (1,5)-(5,4)

(9,10) it will be 2 (1,5)-(5,4)-(9,10)

Etc for all edges the graph from a given vertex v given a Graph G

The order in which the edges are chosen to find the distances should be maintained so that each index in the list correspond to distance of that vertex from the same edge in different vertex lists in a order 

Kind help I apologize to take your time 

In output I need a list say T=[0, 0,1,2,..]

   Data given

   List Given

  L:=[A,B,A,C,A,B,D,E,A,F,G,H,H,G,I,P,Q,W,A]

 

    I want the output to be a frequency table

    Frequency says number of times each Data occurs in the list

    I want output in the form of a table

    Data need not be always A,B,C,D etc it can be any text.

Data

Frequency

A

5

B

2

C

1

D

1

E

1

F

1

G

2

H

2

I

1

P

1

Q

1

W

1

 

I want to export this outputed table to word or Excel

Anyone have a problem with non-appearance of the animation toolbar in Maple 2021????

 

Whilst coming up with a response for the problem here

https://www.mapleprimes.com/questions/231862-Have-You-Ever-Heard-Of-Vector-Asterisks-

I found an issue with the non-appearance of the animation toolbar in Maple 2021. This is rather difficult to illustrate without the use of screenshots, for which I apologise.

Normally(?) I would just select a plot and the animation toolbar appears "as if by magic"

First screen shot is using Maple 2020, The blue highlighting rectangle around the plot was visible when I initiated the snip, but disappeared when the snipping tool activated. However this shows that the animation toolbar is available (and works)

I do exactly the same thing in  Maple 2021 and I can't make the animation toolbar appear - see below. Aagain the plot wa highlighted when I initiated the snip but the highlighting disappeared when the snipping tool activated. Now there is no sign of the animation toolbar

It is still possible to do very basic animation in Maple 2021 by clicking on the plot and using the context menu - but this is very basic

The code used in the above plots is given supplied below

  restart;
  kernelopts(version);
  plots:-display
         ( [ seq
             ( plot
               ( Vector([1, 3, 4, 6]),
                 Vector([8, 6, 2, 5]),
                 style = point,
                 symbol = j,
                 symbolsize = 40,
                 color = blue
               ),
               j in [ asterisk, box, circle, cross, diagonalcross,
                      diamond, point, solidbox, solidcircle, soliddiamond
                    ]
             )
           ],
           insequence = true
         );

Is it just me?

I have defined a function pp(x).  It plots correctly 

but pp(2020); should return 410, and it returns 98000. Maybe I am doing something stupid, but I just can't figure out what!

a := 0.340699639252428*10^13;
b := -0.118229737138742*10^11;
c := 0.175816922773262*10^8;
d := -14523.7138112711;
e := 7.19782673654200;
f := -0.00214008825847444;
g := 0.353463615941623*10^(-6);

h := -0.250170509163729*10^(-10);

pp := x -> a + b*x + c*x^2 + d*x^3 + e*x^4 + f*x^5 + g*x^6 + h*x^7;

plot([pp(x)], x = 1960 .. 2100, gridlines = true, size = [900, 600])

 

Good day, all. I hope we are staying safe.

Please I need help with the regard to the following codes.

The major issue is with rho:=x[n]/h. The rho needs to change values as x[n] changes in the computation. This I think I have gotten wrong. Your professional modifications or/and corrections to the code would be appreciated.

Thank you and kind regards.

restart;
Digits:=20:
#assume(alpha>0,alpha < 1):
f:=proc(n)
	-y[n]-x[n]+(x[n])^2+(2*(x[n])^(2-alpha))/GAMMA(3-alpha)+((x[n])^(1-alpha))/GAMMA(2-alpha):
end proc:

e2:=y[n+2] = y[n]+(1/2)*((alpha^2-alpha)*rho^2+(2*alpha^2-2)*rho+(4/3)*alpha^2-(2/3)*alpha)*(rho*h)^alpha*GAMMA(2-alpha)*f(n)/rho-alpha*GAMMA(2-alpha)*((-1+alpha)*rho^2+(2*alpha-2)*rho+(4/3)*alpha-8/3)*(h*(rho+1))^alpha*f(n+1)/(rho+1)+(1/2)*((alpha^2-alpha)*rho^2+2*(-1+alpha)^2*rho+(4/3)*alpha^2-(14/3)*alpha+4)*(h*(rho+2))^alpha*f(n+2)*GAMMA(2-alpha)/(rho+2):
e1:=y[n+1] = y[n]+(1/4)*((alpha^2-alpha)*rho^2+(alpha^2+3*alpha-4)*rho+(1/3)*alpha^2+(4/3)*alpha)*GAMMA(2-alpha)*f(n)/((rho*h)^(-alpha)*rho)-(1/2)*GAMMA(2-alpha)*((alpha^2-alpha)*rho^2+(alpha^2+alpha-2)*rho+(1/3)*alpha^2+(1/3)*alpha-2)*f(n+1)/((h*(rho+1))^(-alpha)*(rho+1))+(1/4)*((-1+alpha)*rho^2+(-1+alpha)*rho+(1/3)*alpha-2/3)*alpha*GAMMA(2-alpha)*f(n+2)/((h*(rho+2))^(-alpha)*(rho+2)):


alpha:=0.25:
inx:=0:
iny:=0:
#x[0]:=0:
h:=1/20:
N:=solve(h*p = 1, p):
n:=0:
c:=1:
rho:=x[n]/h: 
err := Vector(round(N)):
exy_lst := Vector(round(N)):

for j from 0 to 2 do
	t[j]:=inx+j*h:
end do:
vars:=y[n+1],y[n+2]:

step := [seq](eval(x, x=c*h), c=1..N):

printf("%4s%15s%15s%16s\n", 
	"h","Num.y","Ex.y","Error y");
st := time():
for k from 1 to N/2 do

	par1:=x[n]=t[0],x[n+1]=t[1],x[n+2]=t[2]:
	par2:=y[n]=iny:
	res:=eval(<vars>, fsolve(eval({e||(1..2)},[par1,par2]), {vars}));

	for i from 1 to 2 do
		exy:=eval(-c*h+(c*h)^2):
		printf("%5.3f%17.9f%15.9f%15.5g\n", 
		h*c,res[i],exy,abs(res[i]-exy)):
		
		err[c] := abs(evalf(res[i]-exy)):
		exy_lst[c] := exy:
		numerical_y1[c] := res[i]:
		c:=c+1:
		rho:=rho+1:
	end do:
	iny:=res[2]:
	inx:=t[2]:
	for j from 0 to 2 do
		t[j]:=inx + j*h:
	end do:
end do:
v:=time() - st;
printf("Maximum error is %.13g\n", max(err));

numerical_array_y1 := [seq(numerical_y1[i], i = 1 .. N)]:

time_t := [seq](step[i], i = 1 .. N):

with(plots):
numerical_plot_y1 := plot(time_t, numerical_array_y1, style = [point], symbol = [asterisk],
				color = [blue,blue],symbolsize = 15, title="y numerical",legend = ["Numerical"]);
exact_plot_y1 := plot(time_t, exy_lst, style = [line], symbol = [box], 
				color = [red,red], symbolsize = 10, title="y exact",legend = ["Exact"]);

display({numerical_plot_y1, exact_plot_y1}, title = "Exact and Numerical Solution of Example 1 ");

 

how can i solve this operate differential equation

convert(expand((D[1] - 1)*(D[1]^2 + 2)*y(x) = 0), diff);
dsolve(%)

 

Consider an easy example of a table.

myTable:=table(["a"=1,"b"=-1,"c"=1]);

We can give an index to it and get its corresponding entry if exists.

myTable["b"];

We can also get the set of all indices, or the set of all entries. But what about receiving the index or set of indices with a specific entry. for example asking what indices have the entry `1`?

Of course I can define a search procedure myself, but I thought there might be an efficient way which is already implemented as a function/method on tables.

Good afternoon. How do I disable the numbering of _Z constants? Because
of this, you can't use the procedure in a loop
Is it possible to disable numbering or do something else?

restart; with(linalg); t := 3

3

(1)

S := proc (a, b, z) local L, y, s1, s2, sist, chl, lambda1; L := diff(y(x), x, x) = lambda*y(x); assume(lambda < 0); dsolve(L, y(x)); y := unapply(rhs(%), x); s1 := y(a) = 0; s2 := y(b) = 0; sist := {s1, s2}; genmatrix(sist, {_C1, _C2}); det(%); chl := combine(%); _EnvAllSolutions := true; solve(chl); sort(%); lambda1 := subs(_Z1 = k, %); print(lambda1); lambda1 := unapply(%, k); lambda1(z) end proc:

S(0, Pi, 3);

Warning, solve may be ignoring assumptions on the input variables.

 

-k^2

 

-9

(2)

for i to t do S(0, Pi, i) end do:

Warning, solve may be ignoring assumptions on the input variables.

 

-_Z2^2

 

Warning, solve may be ignoring assumptions on the input variables.

 

-_Z3^2

 

Warning, solve may be ignoring assumptions on the input variables.

 

-_Z4^2

(3)


 

Download help.mw

This is new exception generated by Maple pdsolve in 2021. Different from the last post I gave on pdsolve. So I thought it will be better to keep them separate since the causes are different.

interface(version)
restart;
pde :=  diff(w(x,y,z),x)+(a1*x^n1*y+b1*x^m1)*diff(w(x,y,z),y)+(a2*x^n2*y+b2*x^m1)*diff(w(x,y,z),z)= 0;
pdsolve(pde,w(x,y,z));

#another example

restart;
local gamma:
pde := diff(w(x,y,z),x)+(a1*x^n1*y+ b1*x^m1)*diff(w(x,y,z),y)+(a2*x^n2*y+b2*x^m2)*diff(w(x,y,z),z)=c2*x^k2*y+c1*x^k1*z;
pdsolve(pde,w(x,y,z));

 

Error, (in GAMMA) numeric exception: division by zero

The same PDE works in 2020.2. The answer it gives is large so will not show it all below.

Screen shots

Maple 2021

 

Maple 2020.2


All on windows 10.

Do other see the same error? What causes it?

 

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

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

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 "`&Ropf;`^(2)->`&Ropf;`^()").

 

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 "(&PartialD;)/(&PartialD; z)", "(&PartialD;)/(&PartialD; (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

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