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janhardo

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Line integral input ?...

janhardo 715 Maple 2024
calculus integration line-integral
May 04 2024
0 11

Can this be better done in Maple ? , see worksheet.

Cauchy-Riemann test of analyticity of a ...

janhardo 715 Maple 2024
complex-analysis
May 01 2024
2 12

The CauchyRiemann procedure (for older version  of Maple )doesn't work quite right in Maple 2024 .
Also ran the procedure through the AI for so-called code improvement and now it shows what the code stands for 
The output according to the original procedure would look like on the screenshot, but running original procedure does not give this output ? 
I also want to extend the procedure with a plot of the complex function. 
That differentiability of complex functions is not obvious even if the cauchy-riemann equation is satisfied ?

 

> 

restart

"maple.ini in users"

 (1)
> 

NULL
> 

CauchyRiemann:=proc(expr::algebraic) # original procedure
  local x, y, u, v, u_x, u_y, v_x, v_y, flag1, flag2;

  u:=evalc(Re(eval(expr, z=x+I*y)));
  v:=evalc(Im(eval(expr, z=x+I*y)));

  u_x:=diff(u,x);
  u_y:=diff(u,y);
  v_x:=diff(v,x);
  v_y:=diff(v,y);

  print('f(z)'=expr);
  printf("\n");
  
  print('u(x,y)'=u);
  print('u[x](x,y)'=u_x);
  print('u[y](x,y)'=u_y);
  printf("\n");

  print('v(x,y)'=v);
  print('v[x](x,y)'=v_x);
  print('v[y](x,y)'=v_y);
  printf("\n");

  if u_x=v_y then
    print('u[x]=v[y]');
    print(u_x=v_y);
    flag1:=true;
  else
    print('u[x]<>v[y]');
    print(u_x<>v_y);
    flag1:=false;
  end if;

  if u_y=-v_x then
    print('u[y]=-v[x]');
    print(u_y=-v_x);
    flag2:=true;
  else
    print('u[y]<>-v[x]');
    print(u_y<>-v_x);
    flag2:=false;
  end if;
  
printf("\n");
if flag1=true and flag2=true then
   print(`Fullfill the Cauchy-Riemann Equations`);
   print(`The derivative is:`='u[x]+I*v[y]');
   print('diff(f(z),z)'=u_x+I*v_y);
else
   print(`Cauchy-Riemann ?`);
end if

end proc:
> 

f(z):=1/(z+2):
CauchyRiemann(f(z))

f(z) = 1/(z+2)

  

 

  

u(x, y) = (x+2)/(y^2+(x+2)^2)

  

u[x](x, y) = 1/(y^2+(x+2)^2)-(x+2)*(2*x+4)/(y^2+(x+2)^2)^2

  

u[y](x, y) = -2*(x+2)*y/(y^2+(x+2)^2)^2

  

 

  

v(x, y) = -y/(y^2+(x+2)^2)

  

v[x](x, y) = y*(2*x+4)/(y^2+(x+2)^2)^2

  

v[y](x, y) = -1/(y^2+(x+2)^2)+2*y^2/(y^2+(x+2)^2)^2

  

 

  

u[x] <> v[y]

  

1/(y^2+(x+2)^2)-(x+2)*(2*x+4)/(y^2+(x+2)^2)^2 <> -1/(y^2+(x+2)^2)+2*y^2/(y^2+(x+2)^2)^2

  

u[y] <> -v[x]

  

-2*(x+2)*y/(y^2+(x+2)^2)^2 <> -y*(2*x+4)/(y^2+(x+2)^2)^2

  

 

  

`Cauchy-Riemann ?`

 (2)
> 

 

Also ran the procedure through the AI for so-called code improvement and now it shows what the code stands for

> 

restart;

# Improved and corrected version of the CauchyRiemann procedure :ASKED AI 
CauchyRiemann := proc(expr::algebraic)
    local x, y, u, v, u_x, u_y, v_x, v_y, CR1, CR2;

    # Assign real and imaginary parts of the function
    u := evalc(Re(eval(expr, z = x + I*y)));
    v := evalc(Im(eval(expr, z = x + I*y)));

    # Calculate partial derivatives
    u_x := diff(u, x);
    u_y := diff(u, y);
    v_x := diff(v, x);
    v_y := diff(v, y);

    # Properly format and print function details
    printf("f(z) = %a\n", expr);
    printf("u(x, y) = %a, u_x = %a, u_y = %a\n", u, u_x, u_y);
    printf("v(x, y) = %a, v_x = %a, v_y = %a\n", v, v_x, v_y);

    # Evaluate and print Cauchy-Riemann equations
    CR1 := u_x = v_y;
    CR2 := u_y = -v_x;
    printf("\nCauchy-Riemann Equations:\n");
    printf("u_x = v_y: %a\n", CR1);
    printf("u_y = -v_x: %a\n", CR2);

    # Check both equations
    if CR1 and CR2 then
        printf("The function is analytic (holomorphic) at this point.\n");
        printf("The derivative f'(z) is %a + I*%a\n", u_x, v_y);
    else
        printf("The function does not satisfy the Cauchy-Riemann equations and is not analytic.\n");
    end if;
end proc;

# Test the procedure with a specific function
f := z -> 1/(z + 2);
CauchyRiemann(f(z));

"maple.ini in users"

  

proc (expr::algebraic) local x, y, u, v, u_x, u_y, v_x, v_y, CR1, CR2; u := evalc(Re(eval(expr, z = x+I*y))); v := evalc(Im(eval(expr, z = x+I*y))); u_x := diff(u, x); u_y := diff(u, y); v_x := diff(v, x); v_y := diff(v, y); printf("f(z) = %a ", expr); printf("u(x, y) = %a, u_x = %a, u_y = %a ", u, u_x, u_y); printf("v(x, y) = %a, v_x = %a, v_y = %a ", v, v_x, v_y); CR1 := u_x = v_y; CR2 := u_y = -v_x; printf(" Cauchy-Riemann Equations: "); printf("u_x = v_y: %a ", CR1); printf("u_y = -v_x: %a ", CR2); if CR1 and CR2 then printf("The function is analytic (holomorphic) at this point. "); printf("The derivative f'(z) is %a + I*%a ", u_x, v_y) else printf("The function does not satisfy the Cauchy-Riemann equations and is not analytic. ") end if end proc

  

proc (z) options operator, arrow; 1/(z+2) end proc

  

f(z) = 1/(z+2)
u(x, y) = (x+2)/(y^2+(x+2)^2), u_x = 1/(y^2+(x+2)^2)-(x+2)/(y^2+(x+2)^2)^2*(2*x+4), u_y = -2*(x+2)/(y^2+(x+2)^2)^2*y
v(x, y) = -y/(y^2+(x+2)^2), v_x = y/(y^2+(x+2)^2)^2*(2*x+4), v_y = -1/(y^2+(x+2)^2)+2*y^2/(y^2+(x+2)^2)^2

Cauchy-Riemann Equations:
u_x = v_y: 1/(y^2+(x+2)^2)-(x+2)/(y^2+(x+2)^2)^2*(2*x+4) = -1/(y^2+(x+2)^2)+2*y^2/(y^2+(x+2)^2)^2
u_y = -v_x: -2*(x+2)/(y^2+(x+2)^2)^2*y = -y/(y^2+(x+2)^2)^2*(2*x+4)
The function does not satisfy the Cauchy-Riemann equations and is not analytic.

  

NULL

Download CAUCHY_RIEMANN_-FORUM_VRAAG.mw

"The Basel problem "in Student Basics...

janhardo 715 Maple 2024
sum step-by-step ai
April 04 2024
1 21

I just discoverd today the step solutions for series in Student package 

Now i try to solve this with my own steps here ..

Note: SummationSteps(Sum(1/n^2, n = 1 .. infinity))was not capable to get a closed form?

"maple.ini in users"

 (1)

NULL

Euler's Basel Problem
In the Student Basics package, there is a command :

 

SummationSteps

generate steps for evaluating summations

NULL

> 

help("SummationSteps")

The SummationSteps command accepts an expression that is expected to contain summations and displays the steps required to evaluate each summation given.

2024
with(Student[Basics])

 

> 

" restart; with(Student[Basics])"

"maple.ini in users"

  

[CompleteSquareSteps, CurveSketchSteps, ExpandSteps, FactorSteps, FractionSteps, GCDSteps, LCMSteps, LinearSolveSteps, LongDivision, ModuloSteps, OutputStepsRecord, PartialFractionSteps, PowerSteps, PracticeSheet, SimplifySteps, SolveSteps, SummationSteps, TrigSteps]

 (2)

Try this out this SummationStepscommand for the Basel problem series  ( p-series example)

> 

SummationSteps(Sum(1/n, n = 1 .. infinity))

"[[,,[]],["&bullet;",,"Apply the P-test on" (1)/(n)", which shows the summation diverges if" p<=1 "for" (&sum;)1/((n)^p)],[,,p=1],["&bullet;",,"Since" 0<1 "and" 1<=1", we get that the summation diverges"],[,,([[(&sum;)(1)/(n)" diverges"]])],["&bullet;",,"We know the summation diverges, so now we should find what it diverges to"],[,,[]],["&bullet;",,"Evaluate sum" (&sum;)1/n],[,,infinity]]"

 (3)

Now the Basel Problem from Euler

> 

SummationSteps(Sum(1/n^2, n = 1 .. infinity))

"[[,,[]],["&bullet;",,"Apply the P-test on" (n)^(-2)", which shows the summation diverges if" p<=1 "for" (&sum;)1/((n)^p)],[,,p=2],["&bullet;",,"Since" 1<2", we get that the summation converges"],[,,([[(&sum;)(n)^(-2)" converges"]])]]"

 (4)
> 

f := sum(1/n^2, n = 1 .. infinity)

(1/6)*Pi^2

 (5)

How do we get this value from Euler ( The Basel Problem)

> 

# Step 1: Define the series f
f := sum(1/n^2, n = 1 .. infinity);

# Step 2: Write the series as a product of terms (1 - 1/p)
g := convert(product(1 - 1/p, p = primes), hypergeom);

# Step 3: Compare with the Taylor series of the sine function
h := series(sin(x), x = 0, 10);

# Step 4: Set up equations between corresponding terms
eq := seq(coeff(h, x, 2*k)/k!, k = 1 .. 5) =
      seq(coeff(g, x, k), k = 1 .. 5);

# Step 5: Solve the equations to find the value of the series
sol := solve({eq, seq(coeff(g, x, k) = 0, k = 6 .. 10)});

# Step 6: Replace x with pi/2 to find the value
sol_pi := subs(x = Pi/2, sol);

# Step 7: Compute the value of the series
value := sol_pi[1][2];

value;

(1/6)*Pi^2

  

1-1/primes

  

series(x-(1/6)*x^3+(1/120)*x^5-(1/5040)*x^7+(1/362880)*x^9+O(x^11),x,11)

  

(0, 0, 0, 0, 0) = (0, 0, 0, 0, 0)

  

Error, invalid input: solve expects its 1st argument, eqs, to be of type {`and`, `not`, `or`, rtable, algebraic, relation(algebraic), relation({rtable, algebraic}), {list, set}({`and`, `not`, `or`, algebraic, relation(algebraic)})}, but received {0 = 0, (0, 0, 0, 0, 0) = (0, 0, 0, 0, 0)}

  

sol

  

Error, attempting to assign to `value` which is protected.  Try declaring `local value`; see ?protect for details.

  

value

 (6)

NULL

Download Het_Basel_Probleem_van_Euler.mw

system ode plot not working...

janhardo 715 Maple
ode procedure programming
March 31 2024
1 12

 

> 

"restart: with(plots):with(DEtools):with(Student[ODEs]):infolevel[Student]:=3 ;"NULL

"maple.ini in users"

  

3

 (1)
> 

NULL#  plot  solutions to a system  of DEs

> 

NULL
> 

sys := {diff(x(t), t) = y(t), diff(y(t), t) = -x(t) - 1/2*y(t)}

{diff(x(t), t) = y(t), diff(y(t), t) = -x(t)-(1/2)*y(t)}

 (2)
> 

DEplot(sys, [x(t), y(t)], t = 0 .. 15, [[x(0) = 1, y(0) = 0]])

 
> 

;

This is not working, seems to be simple ...
Have made an attempt to create a procedure from these two commands, but no success
Apparently, I need to learn to use a debugger to determine what is wrong, then?

> 

dv_systeemplot:= proc ( DV1,DV2 ,RV1,RV2 )
                 local t, sys,systeemplot;  
                 sys := {DV1 , DV2 }:
                 systeemplot:= DEplot(sys, [x(t), y(t)], t = 0 .. 15, [[RV1 , RV2]]);
                 return systeemplot;
end proc;

proc (DV1, DV2, RV1, RV2) local t, sys, systeemplot; sys := {DV1, DV2}; systeemplot := DEplot(sys, [x(t), y(t)], t = 0 .. 15, [[RV1, RV2]]); return systeemplot end proc

 (3)

NULL

> 

DV1 := diff(x(t), t) = y(t)

diff(x(t), t) = y(t)

 (4)
> 

DV2 := diff(y(t), t) = -x(t)-(1/2)*y(t)

diff(y(t), t) = -x(t)-(1/2)*y(t)

 (5)
> 

RV1 := x(0) = 1

x(0) = 1

 (6)
> 

RV2 := y(0) = 0

y(0) = 0

 (7)
> 

dv_systeemplot( DV1,DV2 ,RV1,RV2 );
> 

 

Error, (in DEtools/DEplot/CheckDE) inputs must be ODEs

  
> 

NULL

Ask AI to get some code: never i got the right Maple code ! :)

 

> 

"restart: with(plots):with(DEtools):with(Student[ODEs]):infolevel[Student]:=3 ;"``
 

"maple.ini in users"

  

3

 (8)
> 

restart;

systeem := proc(eq1, eq2, beginwaarden)
  local x, y, oplossing;
  x := unapply(rhs(dsolve({eq1, eq2, beginwaarden}, {x(t), y(t)})), t);
  y := unapply(rhs(dsolve({eq1, eq2, beginwaarden}, {x(t), y(t)})), t);
  oplossing := DEplot([x(t), y(t)], t = 0 .. 10, beginwaarden);
  oplossing;
end proc;

# Definieer de differentiaalvergelijkingen
eq1 := diff(x(t), t) = y(t);
eq2 := diff(y(t), t) = -x(t);

# Geef de beginwaarden op
beginwaarden := [x(0) = 1, y(0) = 0];

# Roep de procedure aan met de differentiaalvergelijkingen en beginwaarden
systeem(eq1, eq2, beginwaarden);

"maple.ini in users"

  

proc (eq1, eq2, beginwaarden) local x, y, oplossing; x := unapply(rhs(dsolve({eq1, eq2, beginwaarden}, {x(t), y(t)})), t); y := unapply(rhs(dsolve({eq1, eq2, beginwaarden}, {x(t), y(t)})), t); oplossing := DEplot([x(t), y(t)], t = 0 .. 10, beginwaarden); oplossing end proc

  

diff(x(t), t) = y(t)

  

diff(y(t), t) = -x(t)

  

[x(0) = 1, y(0) = 0]

  

Error, (in dsolve) invalid input: PDEtools/sdsolve expects its 1st argument, SYS, to be of type Or(set({`<>`, `=`, algebraic}),list({`<>`, `=`, algebraic}),`casesplit/ans`(list,list)), but received {diff(x(t),t) = y(t), diff(y(t),t) = -x(t), [x(0) = 1, y(0) = 0]}

  

Probably i do need to work with a debugger here?


Download dv_systeem_via_mapleprimes_.mw

Drawing a line element field using a pro...

janhardo 715 Maple
ode procedure dsolve
March 29 2024
1 6

Well , honestly I can't make sense of how to do this.

Interactive can be done step by step , but now generalize via a procedure , but can't get a handle on it
How to get the right procedure ?

"maple.ini in users"

 (1)

NULL

> 

" restart: with(plots):with(DEtools):with(Student[ODEs]):infolevel[Student]:=3 ;"

"maple.ini in users"

  

3

 (2)
> 

DV := y(x)*(diff(y(x), x)) = exp(x)

y(x)*(diff(y(x), x)) = exp(x)

 (3)
> 

opl_1 := dsolve({DV, RV1}, y(x))

y(x) = (1-2*exp(1)+2*exp(x))^(1/2)

 (4)
> 

opl_2 := dsolve({DV, RV2}, y(x))

y(x) = -(1-2*exp(1)+2*exp(x))^(1/2)

 (5)
> 

opl := plot({rhs(opl_1), rhs(opl_2)}, x = 0 .. 5, y = -10 .. 10, thickness = 3)
> 

lijnelement := dfieldplot(DV, y(x), x = 0 .. 5, y = -10 .. 10, title = "lijnelementveld met intergaalkromme door (1,1)(1,-1)")
> 

display({opl, lijnelement})

 
> 

NULL
> 

" restart: with(plots):with(DEtools):with(Student[ODEs]):infolevel[Student]:=3 ;"

"maple.ini in users"

  

3

 (6)
> 

NULL
> 

DV, RV1, RV2

y(x)*(diff(y(x), x)) = exp(x), y(1) = 1, y(1) = -1

 (7)
> 

``
> 

> 

NULL
> 

NULL
> 

"restart: with(plots):with(DEtools):with(Student[ODEs]):infolevel[Student]:=3 ; "

"maple.ini in users"

  

3

 (8)
> 

DVplot:=proc(DVA,RV1A,RV2A)
          local DV;RV1;RV2;opl_1;opl_2;opl;lijnelement;
          DV:
          RV1:
          RV2:
          opl_1 := dsolve({DV, RV1}, y(x)):
          opl_2 := dsolve({DV, RV2}, y(x)):
          opl:= plot({rhs(opl_1), rhs(opl_2)}, x = 0 .. 5, y = -10 .. 10, thickness = 3):
          lijnelement:=dfieldplot(DV, y(x), x = 0 .. 5, y = -10 .. 10, title = "lijnelementveld met                intergaalkromme door (1,1)(1,-1)"):
          display({opl, lijnelement});
end proc:
> 

NULL
> 

DVA := y(x)*(diff(y(x), x)) = exp(x)

y(x)*(diff(y(x), x)) = exp(x)

 (9)
> 

RV1A := y(1) = 1

y(1) = 1

 (10)
> 

RV2A := y(1) = -1

y(1) = -1

 (11)
> 

DVplot(DVA, RV1A, RV2A)

Error, (in dsolve) not a system with respect to the unknowns [y(x)]

  
 

NULL

Download DV_plotten-_procedure.mw

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