Alfred_F

Mr. Alfred Flaßhaar

485 Reputation

11 Badges

1 years, 197 days
Brandenburg, Germany
As a retired individual with degrees from German universities in mathematics/analysis and structural engineering, I spent my professional life in responsible positions in research, teaching, and practical application, working on the mathematical modeling of states and processes in real-world systems. Now I have the time to explore interesting mathematical problems using Maple. It is my professional curiosity that drives me.

MaplePrimes Activity


These are replies submitted by Alfred_F

@Rouben Rostamian

...

is interesting:

It is identical to every special solution to the initial value of the form y(u)=u, where the additional conditions for u and y(u) must be met.

@Carl Love 

I would never have thought of that. I need to take a closer look at the 1D and 2D input methods and their differences. Where can I find more information about that?

I have got it. :-)

@Carl Love 

Your method is, of course, much easier to apply than my experiments with parameterized real and imaginary parts. It's interesting to observe, using examples, how the zeros of polynomials "arise" and paths in the complex plane of polynomials are mapped into this plane.

@Christian Wolinski 

As a Maple beginner, I'm not yet familiar with all the commands You use and the logical structure of nested code. Therefore, I'd appreciate your advice regarding the error message in the attached file. It doesn't seem to be the parentheses...

test.mw

@acer @nm

Learned something new again. Now, playing with polynomials in the complex plane is enjoyable. I am currently interested in how polynomials map given paths in the complex plane to this plane.

@nm 

...how can the polynomial be factored using radicals instead of decimals?

restart

with(PolynomialTools)

term := factor(x^4-12*x-12, complex)

(x+.9360298293)*(x+.8098950470+2.088088515*I)*(x+.8098950470-2.088088515*I)*(x-2.555819923)

(1)

convert(term, radical)

(x+.9360298293)*(x+.8098950470+2.088088515*I)*(x+.8098950470-2.088088515*I)*(x-2.555819923)

(2)

NULL

Download test.mw

@Kitonum 

...are easily understood with pen and paper. It's enough to apply the rules for powers and logarithms. The original equation only looks dangerous, but after simplification it is not. I managed to do that without Maple before. Maple is, of course, much faster. A plot for variants of C also looks quite nice.

Currently, the "surd" command is very interesting to me, as I want to explore the complex plane with Maple, so thank You for that.

@dharr 

Your very first sentence, with the remark about "default behavior," has already solved the problem. I will begin with the "residue counter" (residue theorem) and refresh my memory of the good-old book by Hurwitz and Courant.

@dharr 

...gives in Maple2024.2: [o,undefined] also - strange. 

Assumption:

ic:=y(infinity)=5*pi/4 is during the insertion test interpreted in Maple as the limit x-->infinity. This limit is not uniquely determinable, analogous to lim(sin(x), x-->infinity). Therefore, y(infinity) can be anything according to pi/4+k*pi. In any case, ic can be varied mod pi without changing the solution.

BTW:

I forgot to mention that while a general solution exists in the domain for x>0 according to the Picard-Lindelöf theorem, the proof presupposes the finiteness of the abscissa for the initial value. And y(infinity), however, lies, in a sense, on the boundary, which is not covered by the existence and uniqueness theorems. It is therefore not surprising that Maple leads classical solution methods to strange results.

@dharr 

"For your solution, odetest(sol, [ode, ic], y(x)) also gives [0,undefined]. "

With Maple 2024.2 I have got the attached results (with which I had played around ;-) )test.mw
 

restart

with(RealDomain)

assume(x > 0)

#Die Beschränkung x<>0 wird nach Auflösung der DGL nach y´ offensichtlich. Wegen des hier gewählten Anfangswertes muß x>0 gelten. The restriction x<>0 becomes obvious after solving the differential equation for y'. Because of the initial value chosen here, x>0 must hold.

ode := x^2*(diff(y(x), x))+cos(2*y(x)) = 1NULL

x^2*(diff(y(x), x))+cos(2*y(x)) = 1

(1)

#Folgende Lösung "mysol" wurde extern berechnet zum Anfangswert y(infinity)=5*Pi/4. Dieser exotische Anfangswert ist als Grenzwert für x-->infinity zu verstehen. Da der Cotangens periodisch ist, führt dieser Anfangswert nicht zu einer eindeutigen Lösung. Jede Ordinate Pi/4+k*2*Pi erfüllt die Forderung zu "mysol". Daher wählt Maple z. B. für (4) die Ordinate Pi/4 (blaue Kurve).         The following solution, "mysol," was calculated externally for the initial value y(infinity) = 5π/4. This unusual initial value is to be understood as the limit as x approaches infinity. Since the cotangent is periodic, this initial value does not lead to a unique solution. Every ordinate, π/4 + k*2π, satisfies the requirement for "mysol". Therefore, Maple chooses, for example, the ordinate π/4 (blue curve) for (4).

mysol := cot(y(x)) = 2/x+1

cot(y(x)) = 2/x+1

(2)

solve(cot(y(x)) = 2/x+1, y(x))

arccot((x+2)/x)

(3)

limit(arccot((x+2)/x), x = infinity)

(1/4)*Pi

(4)

#Der folgende Befehl liefert die zum Anfangswert y(infinity)=5*Pi/4 die korrekte Lösung (rote Kurve).             The following command provides the correct solution (red curve) for the initial value      y(infinity)=5*Pi/4.

sol := dsolve({ode, y(infinity) = (5/4)*Pi}, y(x))

y(x) = (3/2)*Pi-arctan((x+2)/x)

(5)

NULL

``

odetest(sol, ode, y(x))

0

(6)

plot([3*Pi*(1/2)-arctan((x+2)/x), arccot((x+2)/x)], x = 0 .. 100, color = ["Red", "Blue"])

 

NULL


 

Download test.mw

.

@dharr 

...the initial value is the key. Every initial value ordinate pi/4+k*2*pi yields the same solution. Without a "special command," the initial value does not lead to a unique solution. It appears that "sol := dsolve({ode, y(infinity) = 5/4*Pi}, y(x))" is required.

@Kitonum 

Regarding the background of my original question:

The integral (1) in the file "test" is supposedly by Borwein and is quite intriguing ;-) . Therefore, during my exploration of the Maple world, I played around with it and applied the mean value theorem of integral calculus. This means that in the interval 0<x<1, we need to find an abscissa whose ordinate yields the mean value. This leads to equation (1) from the attached file "test1". I had hoped that the two real results could at least be represented using some elements of specific function classes. Hence the attempt to "identify" (6) from file "test1".test1.mw
 

restart

Gesucht sind gemäß Mittelwertsatz der Integralrechnung reelle Lösungen der Gleichung:

According to the mean value theorem of integral calculus, real solutions of the equation are sought:

NULL

Q2 := sin(Pi*x)/(x^x*(1-x)^(1-x)) = Pi/exp(1)

sin(Pi*x)/(x^x*(1-x)^(1-x)) = Pi/exp(1)

(1)

solve(Q2, x)

RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z))/Pi

(2)

"(->)"

.2340257795502385151002175791580229871350403567739388325733228478980460706709848394726222465477567339

(3)

NULL

allvalues(RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z))/Pi)

RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z), 2.406378984)/Pi, RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z), -2.284494230+1.013004722*I)/Pi, RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z), -2.284494230-1.013004722*I)/Pi, RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z), .7352136698)/Pi, RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z), .7352136698-0.*I)/Pi, RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z), -5.438328304+1.319627688*I)/Pi, RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z), 5.426086883-1.013004722*I)/Pi, RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z), -5.438328304-1.319627688*I)/Pi, RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z), 5.426086883+1.013004722*I)/Pi

(4)

Nur die erste und die vierte Lösung sind reell.

Only the first and fourth solutions are real.

evalf[100](RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z), 2.406378984)/Pi)

.7659742204497614848997824208419770128649596432260611674266771521019539293290151605273777534522432661

(5)

evalf[100](RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z), .7352136698)/Pi)

.2340257795502385151002175791580229871350403567739388325733228478980460706709848394726222465477567339

(6)

Die Summe der Lösungen ist gleich 1. Sie sind die Abszissen der Mittelwertsordinaten des Integralsatzes. Die Abszissen liegen erwartungsgemäß symmetrisch zu 1/2, da der Integrand symmetrisch ist.

The sum of the solutions is equal to 1. They are the abscissas of the mean value ordinates of the integral theorem. As expected, the abscissas are symmetrical about 1/2, since the integrand is symmetrical.

evalf[100](RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z), 2.406378984)/Pi+RootOf(-exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*Pi+exp(-(-ln(_Z/Pi)*_Z+ln((Pi-_Z)/Pi)*_Z+Pi)/Pi)*_Z+sin(_Z), .7352136698)/Pi)

1.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

(7)

Einsetzprobe beider Lösungen in (1):

Substitution test of both solutions in (1):

"Q3(x):=(sin(Pi x))/(x^x (1-x)^(1-x));"

proc (x) options operator, arrow, function_assign; sin(Pi*x)/(x^x*(1-x)^(1-x)) end proc

(8)

evalf[100](Q3(.7659742204497614848997824208419770128649596432260611674266771521019539293290151605273777534522432661))

1.155727349790921717910093183312696299120851023164415820499706535327288631840916939440188434235673559

(9)

evalf[100](Q3(.2340257795502385151002175791580229871350403567739388325733228478980460706709848394726222465477567339))

1.155727349790921717910093183312696299120851023164415820499706535327288631840916939440188434235673559

(10)

identify(1.155727349790921717910093183312696299120851023164415820499706535327288631840916939440188434235673559)

Pi*exp(-1)

(11)

NULL


 

Download test1.mw

 

@manasp 

I need to pay better attention :-)test.mw

restart

"c(x):=(&sum;)(1/(10^(k))&lowast;sin(10^(k)&lowast;x));"

proc (x) options operator, arrow, function_assign; sum(sin(10^k*x)/10^k, k = 1 .. 100) end proc

(1)

plot(c(x), x = -(1/4)*Pi .. (1/4)*Pi)

 

NULL

Download test.mw

@janhardo 

...f(x)= |x| is only not differentiable at one point. However, as I understand the initial question, it concerns functions that are continuous at every point in their domain, but not differentiable there. And that includes the good old Weierstrass function.

@vv 

...the question was raised as to how Maple handles continuous but not everywhere differentiable functions. This is reminiscent of the famous Weierstrass function (Hewitt/Stromberg, "Real and Abstract Analysis", page 258 and an example at https://mathe-vital.de/Analysis1/3-1.html).

[Moderator note: Wikipedia reference to Weierstrass function.]

To illustrate:

Hardy_Darboux_Riemann_Cellerier.mw

Unfortunately, the series converge very slowly. Crossing a boundary seems impossible in Maple. Perhaps a larger summand index could be achieved with more powerful computers and clever programming?

1 2 3 4 5 6 7 Last Page 1 of 17