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Alfred_F

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Re: :-)...

Alfred_F 380
September 17 2024
0 0

@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.

ok ;-)...

Alfred_F 380
September 17 2024
0 0

@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).

It's easier......

Alfred_F 380
September 16 2024
0 0

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 ex...

Alfred_F 380
September 11 2024
0 0

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

@Scot Gould Embarrassing - embarrassing...

Alfred_F 380
September 08 2024
0 0

@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.

No lack ... ;-)...

Alfred_F 380
September 07 2024
0 0

@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

Thank You!...

Alfred_F 380
September 05 2024
0 0

@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.

p. s....

Alfred_F 380
September 04 2024
0 0

@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

 

New attempt...

Alfred_F 380
September 04 2024
0 0

@Scot Gould 

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

p.s....

Alfred_F 380
September 01 2024
0 0

@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

 

Thank You!...

Alfred_F 380
September 01 2024
0 0

@Scot Gould 

I already know these instructions and couldn't do much with them. My goal is to transform tasks that were previously solved with the help of pen and paper and other programs into the much more powerful Maple. I'm currently trying to do this as a practice exercise with an old exam question that was set for the Abitur in a distant country ;-). Since I don't know a structural overview of Maple, I find it difficult to understand why "packages" are often required before entering a command.

I'm currently trying to check the positive definiteness of the matrix generated using "Hessian". The "IsDefinite" command produces an incomprehensible error message. I'm asking for advice.

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