Advanced Search

Alfred_F

330 Reputation

9 Badges

0 years, 329 days
Contact Alfred_F

MaplePrimes Activity


These are replies submitted by Alfred_F

p. s....

Alfred_F 330
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 330
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 330
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 330
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.

First 10 11 12 Page 12 of 12