Alfred_F

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These are replies submitted by Alfred_F

@vv 

In the future I have some plans regarding symbolic calculation. I would like to find out where the limits are. As before, I have chosen tasks for this on my journey of discovery.

@nm 

Apparently I expressed myself imprecisely, sorry. The background to my question is the task:
Determine all matrices X that are commutable with every regular matrix with respect to matrix multiplication, i.e. A*X = X*A for all A. This does not exclude the possibility of random hits for X, for which the factors X and A are also commutable.

@Carl Love 

This is a strong solution. But I need a little more time to understand it. The brevity of the solution is impressive. But what happens in the "background" after calling the commands is invisible. A long time ago I solved this task in individual steps using the good old derive. This resulted in two solutions for the auxiliary values ​​(s; x), both of which lead to y=11. But these values ​​are not what the task is looking for.

Thank you very much :-).

@vv 

You have proved a theorem under more general conditions, from which Völler's theorem follows as a special case - congratulations. Please forgive my late comment. I wanted to look into it thoroughly first.

@vv 

Your work shows the imprecise formulation of Völler's theorem and its proof in the original work from 1858. Only when working through his proof can it be seen that important prerequisite for the proof are not mentioned or are only mentioned in passing/superficially. Today the theorem would be formulated differently.

Interpretation in the sense of the original proof (loosely formulated):

In the Euclidean plane, the convex or concave curve section of a four-fold continuously differentiable function f(x) and the points A and B on it as well as the tangents at these points with their intersection point C are given. F is the area of ​​the triangle ABC and G is the area of ​​the curve segment over the chord of length s = AB. The ratio G/F = 2/3 is to be proven for the limiting case B ---> A, i.e. s ---> 0. The function f(x) is assumed to be convex/concave positive with non-vanishing 1st and 2nd derivatives at the points A and B.

Therefore, to explain:

In the original work according to the source, it is tacitly assumed that with a general function approach f(x), the first and second derivatives do not disappear at points A and B and that f(x) is continuously differentiable up to the fourth derivative, so that the Bernoulli/L'Hospital rule can be applied according to the proof procedure. This can only be seen in the course of the proof, including in the sketch and at the end of the calculation. Non-degenerate circles of curvature are assumed at curve points A and B and local extrema are to be excluded there. The sketch used in the original proof shows this without comment, because this is the only way to obtain the non-degenerate position of the tangent intersection point C.

Under these conditions, it is sufficient to evaluate the 4th derivative.

@Rouben Rostamian  

The "red line" of your proof is exactly the same as the one Mr. Völler did in 1858 with the help of paper and pen. I did the same thing some time ago with the help of another software (MC14) and had great difficulty using symbolic calculations. Apparently there are no such problems with Maple.

As a Maple newcomer, I learned something new again - thank you very much.

BTW:
It would be interesting to pose the Völler question in R^n for differentiable (n-1)-dimensional "surfaces". But I still need time to learn how to use Maple.

@Kitonum The proof using Maple is not difficult. Vector calculus is used completely correctly. However, I am of the opinion that a reference to an elementary proof procedure should be allowed. And in this case, elementary knowledge of geometry is sufficient.
Regardless of this, as a Maple newbie, I learned something new from the vector proof (geometry package).

There is a simpler solution that can be described using elementary geometric means. The result 6/7 is of course correct.

@Carl Love Thank you for all the explanations and "poly.mw". It helps.

@Scot Gould

Embarrassing - embarrassing - now I see it too. But thanks to your help I've learned something - be more careful when typing. Now the world is OK again.

@Scot Gould 

To avoid misunderstandings:
As a newbie/beginner in Maple, I am asking for help and advice in this forum, solely to get to know Maple. I accept references to more in-depth mathematical backgrounds with all due respect and an understanding smile/grin - in my case they are unnecessary. To practice Maple, I have chosen tasks from various subject areas and different levels of difficulty. I hope to receive information on my application errors in this forum so that I can learn from them.

To 1.)
I would like to use the Hesse matrix and its determinant in the solution of the task. Its definiteness is decisive for the type of local extremum. Maple offers it as "Hessian", so I use this. I have recalculated the determinant in another independent way and found that the value in your file is correct. The value in my file is incorrect. So where is my mistake in entering the command for "Hessian"? It is interesting to note that your matrix and my matrix only differ in the second derivatives of the function g(u,v) on the main diagonal. I cannot explain this and would like your advice.

2.)
For professional reasons, it was necessary for me to study specialist literature in the original language. This meant that, in addition to school and university, I learned several foreign languages. I will try and hope that we will continue to communicate objectively.

3.)
Thank you again for your advice on settings in Maple. I have already written about the different values ​​of the determinants. Perhaps it is an input error on my part?

Best wishes for the weekend, Alfred_F

@dharr 

Thank you for both answers. As there is no general solution method for Diophantine equations, I had hoped that Maple would have a solution command for a special class of equations of this type. But the numerical solution by force (try-success-failure) is also a solution method that can be continued indefinitely. But as a newbie to Maple, I cannot program yet.

The elegant theoretical solution (paper and pen) to the problem is based on "infinite descent" (Fermat, Euler). Based on this, it is possible to construct recursions for the solution specifically for this problem. So there is also a constructive solution, albeit in the software I used previously. The asymptotic behavior of the solutions is also interesting. You get close to the integer solutions quite quickly.

@Alfred_F missed the attachment again.AF_20240904.mw
 

restart

Finden Sie den kleinsten Wert des Ausdrucks mit Logarithmen zur Basis a

sqrt[ 106 + log^2_a cos(a*x) + log_a cos^10(a*x) ] +
sqrt[ 58  + log^2_a sin(a*x) - log_a sin^6(a*x) ] +
sqrt[ 5 + log^2_a tan(a*x) + log_a tan^2(a*x) ]
NULL
und alle Paare (a,x), an denen das Minimum angenommen wird.

(Lösung: a=2, x=π/8+k*pi, Minimum=9*sqrt(5))

Lösung durch Substitution der Logarithmenterme und "tan=sin/cos" nach Anwendung binomischer Formel/quadratische Ergänzung.NULL

 

f(a, x) d √(81 + (LOG(COS(a·x), a) + 5)^2 ) + √(49 + (LOG(SIN(a·x), a) - 3)^2 ) + √(4 + (LOG(TAN(a·x), a) + 1)^2 )

 

          NULL

NULL

NULL

Mit diesen Substitutionen wird :

 

 

"g(u,v):=sqrt(81+(u+5)^(2))+sqrt(49+(v-3)^(2))+sqrt(4+(v-u+1)^(2))"

proc (u, v) options operator, arrow, function_assign; sqrt(81+(u+5)^2)+sqrt(49+(v-3)^2)+sqrt(4+(v-u+1)^2) end proc

(1)

NULL

Berechnung der 1. Ableitungen von g(u,v) als Komponenten des Gradienten:

 

diff(g(u, v), u)

(1/2)*(2*u+10)/(81+(u+5)^2)^(1/2)+(1/2)*(-2*v+2*u-2)/(4+(v-u+1)^2)^(1/2)

(2)

diff(g(u, v), v)

(1/2)*(2*v-6)/(49+(v-3)^2)^(1/2)+(1/2)*(2*v-2*u+2)/(4+(v-u+1)^2)^(1/2)

(3)

NULL

Nullsetzen der 1. Ableitungen als notwendige Bedingung für lokales Extremum:

 

(2*u+10)/(2*sqrt(81+(u+5)^2))+(-2*v+2*u-2)/(2*sqrt(4+(v-u+1)^2)) = 0

(1/2)*(2*u+10)/(81+(u+5)^2)^(1/2)+(1/2)*(-2*v+2*u-2)/(4+(v-u+1)^2)^(1/2) = 0

(4)

``

 

(2*v-6)/(2*sqrt(49+(v-3)^2))+(2*v-2*u+2)/(2*sqrt(4+(v-u+1)^2)) = 0

(1/2)*(2*v-6)/(49+(v-3)^2)^(1/2)+(1/2)*(2*v-2*u+2)/(4+(v-u+1)^2)^(1/2) = 0

(5)

NULL

Lösen des Systems (4) und (5):

solve({(2*u+10)/(2*sqrt(81+(u+5)^2))+(-2*v+2*u-2)/(2*sqrt(4+(v-u+1)^2)) = 0, (2*v-6)/(2*sqrt(49+(v-3)^2))+(2*v-2*u+2)/(2*sqrt(4+(v-u+1)^2)) = 0}, {u, v})

{u = -1/2, v = -1/2}

(6)

NULL

Falls eine Lösung existiert, ist u = v = -1/2. Damit wird wegen v - u = 0 = log(tan(a*x),a). Für jede Logarithmenbasis  a > 1 ist dann:

 

tan(a*x) = 1

tan(a*x) = 1

(7)

NULL

Bis auf Periodizität ist daher:

 

a*x = (1/4)*Pi

a*x = (1/4)*Pi

(8)

 

Daraus folgt cos(π/4) = 1/2*sqrt(2)und sin(π/4) = 1/2*sqrt(2) .

Kontrolle der Determinante für die Hessesche Matrix im Punkt (-1/2, -1/2):

 

with(Student[VectorCalculus])

[`&x`, `*`, `+`, `-`, `.`, `<,>`, `<|>`, About, ArcLength, BasisFormat, Binormal, ConvertVector, CrossProduct, Curl, Curvature, D, Del, DirectionalDiff, Divergence, DotProduct, FlowLine, Flux, GetCoordinates, GetPVDescription, GetRootPoint, GetSpace, Gradient, Hessian, IsPositionVector, IsRootedVector, IsVectorField, Jacobian, Laplacian, LineInt, MapToBasis, Nabla, Norm, Normalize, PathInt, PlotPositionVector, PlotVector, PositionVector, PrincipalNormal, RadiusOfCurvature, RootedVector, ScalarPotential, SetCoordinates, SpaceCurve, SpaceCurveTutor, SurfaceInt, TNBFrame, TangentLine, TangentPlane, TangentVector, Torsion, Vector, VectorField, VectorFieldTutor, VectorPotential, VectorSpace, diff, evalVF, int, limit, series]

(9)

H, d := Hessian(g(u, v), [u, v] = [-1/2, -1/2], determinant)

Matrix(%id = 36893491086728902164), (64/2480625)*405^(1/2)*245^(1/2)+(16/50625)*405^(1/2)*4^(1/2)*5^(1/2)+(16/30625)*5^(1/2)*245^(1/2)*4^(1/2)

(10)

NULL

simplify(64*sqrt(405)*sqrt(245)*(1/2480625)+16*sqrt(405)*sqrt(4)*sqrt(5)*(1/50625)+16*sqrt(5)*sqrt(245)*sqrt(4)*(1/306250))

176/4375

(11)

NULL

Da der Wert (11) der Determinante positiv ist, liegt im Punkt(-1/2, -1/2) hinreichend bewiesen ein lokales Minimum. Der Wert für x ist zu bestimmen. Dazu wird in eine der Anfangssubstitutionen eingesetzt und der Logarithmus zur Basis a umgerechnet in den natürlichen Logarithmus.Dadurch wird die zu berechnende Zahl a explizit greifbar.

 

-1/2 = log[a](cos((1/4)*Pi))

-1/2 = ln((1/2)*2^(1/2))/ln(a)

(12)

solve(-1/2 = ln((1/2)*sqrt(2))/ln(a), a)

2

(13)

Aus a*x = π/4 und a = 2 folgt x = π/8. Wegen tan(a*x) = 1 gilt auch tan(a*x+k*π) = 1 für k = 0, 1, 2, ...

 

tan(Pi*k+a*x)

tan(Pi*k+a*x)

(14)

 

 

expand(tan(Pi*k+a*x))

(tan(a*x)+tan(k*Pi))/(1-tan(a*x)*tan(k*Pi))

(15)

NULL

Als Periode wird aus der Entwicklung des Tangens zunächst k*π für k = 0, 1, 2, ... angenommen und in der Entwicklung des Cosinus und des Sinus überprüft.

 

cos(Pi*k+a*x)

cos(Pi*k+a*x)

(16)

expand(cos(Pi*k+a*x))

cos(k*Pi)*cos(a*x)-sin(k*Pi)*sin(a*x)

(17)

sin(Pi*k+a*x)

sin(Pi*k+a*x)

(18)

expand(sin(Pi*k+a*x))

sin(k*Pi)*cos(a*x)+cos(k*Pi)*sin(a*x)

(19)

NULL

Da Sinus und Cosinus positiv sind, wird folglich die Lösungsperiode k*π auf die geraden natürlichen Zahlen k begrenzt. Das Minimum wird aus g(u,v) berechnet:

 

 

g(-1/2, -1/2)

9*5^(1/2)

(20)

Lösung:NULL

 

Mit a = 2 sind alle a*x + k*π für gerade natürliche Zahlen k Lösung der Aufgabe. Daher ist x = π/8 + k*π für alle natürlichen Zahlen k Lösung. Das Minimum beträgt 9*"sqrt(5)."

NULL


 

Download AF_20240904.mw

 

@Scot Gould 

I have tried to learn something from your suggestions. I have attached a new attempt. I would appreciate any critical comments.

@Alfred_F Forgot attachmentAF_20240901.mw
 

Finden Sie den kleinsten Wert des Ausdrucks mit Logarithmen zur Basis a

sqrt[ 106 + log^2_a cos(a*x) + log_a cos^10(a*x) ] +
sqrt[ 58  + log^2_a sin(a*x) - log_a sin^6(a*x) ] +
sqrt[ 5 + log^2_a tan(a*x) + log_a tan^2(a*x) ]

und alle Paare (a,x), an denen das Minimum angenommen wird.

(Lösung: a=2, x=π/8+k*pi, Minimum=9*sqrt(5))

Lösung durch Substitution der Logarithmenterme und "tan=sin/cos" nach Anwendung binomischer Formel/quadratische Ergänzung.``

 

f(a, x) d √(81 + (LOG(COS(a·x), a) + 5)^2 ) + √(49 + (LOG(SIN(a·x), a) - 3)^2 ) + √(4 + (LOG(TAN(a·x), a) + 1)^2 )

 

         

 

g := sqrt(81+(u+5)^2)+sqrt(49+(v-3)^2)+sqrt(4+(v-u+1)^2)

(81+(u+5)^2)^(1/2)+(49+(v-3)^2)^(1/2)+(4+(v-u+1)^2)^(1/2)

(1)

"Berechnung der 1. Ableitungen nach u und v.  Dies sind die Komponenten des Gradienten."

diff(g, u)

(1/2)*(2*u+10)/(81+(u+5)^2)^(1/2)+(1/2)*(-2*v+2*u-2)/(4+(v-u+1)^2)^(1/2)

(2)

diff(g, v)

(1/2)*(2*v-6)/(49+(v-3)^2)^(1/2)+(1/2)*(2*v-2*u+2)/(4+(v-u+1)^2)^(1/2)

(3)

"Im lokalen Extremum sind die 1. Ableitungen =0.  `Auflösung` des Systems nach u und v."

(2*u+10)/(2*sqrt(81+(u+5)^2))+(-2*v+2*u-2)/(2*sqrt(4+(v-u+1)^2)) = 0

(1/2)*(2*u+10)/(81+(u+5)^2)^(1/2)+(1/2)*(-2*v+2*u-2)/(4+(v-u+1)^2)^(1/2) = 0

(4)

(2*v-6)/(2*sqrt(49+(v-3)^2))+(2*v-2*u+2)/(2*sqrt(4+(v-u+1)^2)) = 0

(1/2)*(2*v-6)/(49+(v-3)^2)^(1/2)+(1/2)*(2*v-2*u+2)/(4+(v-u+1)^2)^(1/2) = 0

(5)

solve({(2*u+10)/(2*sqrt(81+(u+5)^2))+(-2*v+2*u-2)/(2*sqrt(4+(v-u+1)^2)) = 0, (2*v-6)/(2*sqrt(49+(v-3)^2))+(2*v-2*u+2)/(2*sqrt(4+(v-u+1)^2)) = 0}, [u, v])

[[u = -1/2, v = -1/2]]

(6)

eval(g, [u = -1/2, v = -1/2]) = (1/4)*405^(1/2)*4^(1/2)+(1/4)*245^(1/2)*4^(1/2)+5^(1/2)NULL

simplify((1/4)*sqrt(405)*sqrt(4)+(1/4)*sqrt(245)*sqrt(4)+sqrt(5))

9*5^(1/2)

(7)

NULL

Zur Übung die Ableitungen gemäß Hessian und Gradient:

 

with(Student[VectorCalculus]); Hessian(Student[VectorCalculus]:-`+`(Student[VectorCalculus]:-`+`(sqrt(Student[VectorCalculus]:-`+`(81, Student[VectorCalculus]:-`+`(u, 5)^2)), sqrt(Student[VectorCalculus]:-`+`(49, Student[VectorCalculus]:-`+`(v, -3)^2))), sqrt(Student[VectorCalculus]:-`+`(4, Student[VectorCalculus]:-`+`(Student[VectorCalculus]:-`+`(v, Student[VectorCalculus]:-`-`(u)), 1)^2))), [u, v], determinant); Gradient(Student[VectorCalculus]:-`+`(Student[VectorCalculus]:-`+`(sqrt(Student[VectorCalculus]:-`+`(81, Student[VectorCalculus]:-`+`(u, 5)^2)), sqrt(Student[VectorCalculus]:-`+`(49, Student[VectorCalculus]:-`+`(v, -3)^2))), sqrt(Student[VectorCalculus]:-`+`(4, Student[VectorCalculus]:-`+`(Student[VectorCalculus]:-`+`(v, Student[VectorCalculus]:-`-`(u)), 1)^2))), [u, v])

Vector(2, {(1) = (1/2)*(2*u+10)/sqrt(81+(u+5)^2)+(1/2)*(-2*v+2*u-2)/sqrt(4+(v-u+1)^2), (2) = (1/2)*(2*v-6)/sqrt(49+(v-3)^2)+(1/2)*(2*v-2*u+2)/sqrt(4+(v-u+1)^2)})

(8)

simplify((196*(u^2+10*u+106)^(3/2)+3969*(u^2-2*u*v+v^2-2*u+2*v+5)^(3/2)+324*(v^2-6*v+58)^(3/2))/((u^2-2*u*v+v^2-2*u+2*v+5)^(3/2)*(u^2+10*u+106)^(3/2)*(v^2-6*v+58)^(3/2)))

196*((81/2)*(5/2+(1/2)*u^2+(-1-v)*u+(1/2)*v^2+v)*(u^2+(-2*v-2)*u+v^2+2*v+5)^(1/2)+(u^2+10*u+106)^(3/2)+(81/49)*(v^2-6*v+58)^(3/2))/((u^2+(-2*v-2)*u+v^2+2*v+5)^(3/2)*(v^2-6*v+58)^(3/2)*(u^2+10*u+106)^(3/2))

(9)

rationalize((196*(u^2+10*u+106)^(3/2)+3969*(u^2-2*u*v+v^2-2*u+2*v+5)^(3/2)+324*(v^2-6*v+58)^(3/2))/((u^2-2*u*v+v^2-2*u+2*v+5)^(3/2)*(u^2+10*u+106)^(3/2)*(v^2-6*v+58)^(3/2)))

(196*(u^2+10*u+106)^(3/2)+3969*(u^2-2*u*v+v^2-2*u+2*v+5)^(3/2)+324*(v^2-6*v+58)^(3/2))/((u^2-2*u*v+v^2-2*u+2*v+5)^(3/2)*(u^2+10*u+106)^(3/2)*(v^2-6*v+58)^(3/2))

(10)

with(LinearAlgebra)

[`&x`, Add, Adjoint, BackwardSubstitute, BandMatrix, Basis, BezoutMatrix, BidiagonalForm, BilinearForm, CARE, CharacteristicMatrix, CharacteristicPolynomial, Column, ColumnDimension, ColumnOperation, ColumnSpace, CompanionMatrix, CompressedSparseForm, ConditionNumber, ConstantMatrix, ConstantVector, Copy, CreatePermutation, CrossProduct, DARE, DeleteColumn, DeleteRow, Determinant, Diagonal, DiagonalMatrix, Dimension, Dimensions, DotProduct, EigenConditionNumbers, Eigenvalues, Eigenvectors, Equal, ForwardSubstitute, FrobeniusForm, FromCompressedSparseForm, FromSplitForm, GaussianElimination, GenerateEquations, GenerateMatrix, Generic, GetResultDataType, GetResultShape, GivensRotationMatrix, GramSchmidt, HankelMatrix, HermiteForm, HermitianTranspose, HessenbergForm, HilbertMatrix, HouseholderMatrix, IdentityMatrix, IntersectionBasis, IsDefinite, IsOrthogonal, IsSimilar, IsUnitary, JordanBlockMatrix, JordanForm, KroneckerProduct, LA_Main, LUDecomposition, LeastSquares, LinearSolve, LyapunovSolve, Map, Map2, MatrixAdd, MatrixExponential, MatrixFunction, MatrixInverse, MatrixMatrixMultiply, MatrixNorm, MatrixPower, MatrixScalarMultiply, MatrixVectorMultiply, MinimalPolynomial, Minor, Modular, Multiply, NoUserValue, Norm, Normalize, NullSpace, OuterProductMatrix, Permanent, Pivot, PopovForm, ProjectionMatrix, QRDecomposition, RandomMatrix, RandomVector, Rank, RationalCanonicalForm, ReducedRowEchelonForm, Row, RowDimension, RowOperation, RowSpace, ScalarMatrix, ScalarMultiply, ScalarVector, SchurForm, SingularValues, SmithForm, SplitForm, StronglyConnectedBlocks, SubMatrix, SubVector, SumBasis, SylvesterMatrix, SylvesterSolve, ToeplitzMatrix, Trace, Transpose, TridiagonalForm, UnitVector, VandermondeMatrix, VectorAdd, VectorAngle, VectorMatrixMultiply, VectorNorm, VectorScalarMultiply, ZeroMatrix, ZeroVector, Zip]

(11)

IsDefinite([`?`])

Error, mscrolltable is not a command in the Typesetting package

 

NULL


 

Download AF_20240901.mw

 

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