Alfred_F

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These are replies submitted by Alfred_F

@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?

@acer 

...

I noticed this and couldn't resist mentioning the source.

The problem isn't so much about the Kovacic algorithm. Rather, it's about how to generate a simplified equation through differentiation. I remember an old trick that might help here:

It's assumed that a continuous second derivative exists for the function in question. After calculating y'', the term structure of y' is found/generated within the term structure of y''. The term structure of y' in y'' is then replaced by the symbol y'. This results in a simpler equation with a solution.

In

https://www.12000.org/my_notes/my_paper_on_kovacic/paper.htm

is given a Maple-example of the algorithm in 5. 

Unfortunately, I can only partially understand the theory in the description of points 1 to 4. The algorithm was developed long after my time as a student, and I haven't had any experience with it. However, its structure is very interesting. Often necessary tricks for solving tricky equations are systematically avoided with the help of this algorithm.

... from:

"PDEtools:-Solve(my_sol_1,y(x)); map(X->odetest(X,[ode,IC]),[%]) # [[0, 1], [0, undefined]]"

Maple points out that there is a problem with satisfying IC for the two explicit solutions. The theory of ordinary differential equations that I know only deals with equations in explicit form and, since the 1960s, also with so-called semi-implicit equations and ensures the existence and uniqueness of the solution. For general implicit equations, individual case investigations are therefore always necessary. This is obvious in the current case.

@dharr 

...for me, as a Maple beginner. In particular, solving the nonlinear system will keep me busy in the future.

Translation correction:

The initially confusing/complex Diophantine equation dio intrigued me because, after transforming the lhs into a sum of three squares, which equals zero, each term must be zero. This makes the solution obvious, or rather, results in three simple equations. This trick is well-known from competitions and always worth a try.

@dharr 

...for pointing out the * for commutative multiplication instead of . for non-commutative multiplication. I hadn't noticed that before and just typed it out of habit. Now expand works.

...sorry, I made a mistake when transferring the formula from the paper to Maple. It should be -1440*x*y^2 (the zero was missing).

@vv 

... "A sufficient condition for F(x,y)  to be Lipschitz wrt  y near (x0,y0)  is that D[2](F) be without singularities near (x0,y0). " in the current case, the classic proof: abs(f/x,u)-f(x,v))<=L*abs(u-v) is locally valid replaced. With which constant L should the proof be carried out locally and globally? And that is what my original question refers to.

@dharr 

Please excuse my mistake. I promise to do better.

...for both advices (dharr 8722, janhardo 750). Now I have a lot to practice, understanding how to apply the theory I know "from paper" to the maple world.

@dharr 

...I found my mistake.

@dharr @janhardo 

Several tests have shown that a,...,f = -17, -3, -7, 371, 113, 131 is causing problems. The solution should be (21; -5). I would appreciate some advice.

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