Analysis

A24

Grad[f_, x_] := 
      Transpose[{Table[D[f, x[[k]]], {k, Length[x]}]}]

f = (x^2)  + 4  x y

g = (x^2) y - V

L = f + λ g

ableitung = Grad[L, {x, y, λ}]

stationär = 
      Solve[ableitung == 0, {x, y, λ}]

f /. stationär

Eliminate[ableitung == 0, {λ}]

A 25

F = y - x (1 - y)^2

Needs["Graphics`ImplicitPlot`"]

ImplicitPlot[F == 0, {x, -2, 10}, {y, -2, 10}]

sol = Solve[F == 0, y]

Plot[Evaluate[y /. sol], {x, -2, 10}]

sol = sol /. y -> y[x]

diff = D[sol, x] // Simplify

diff /. x -> 2

F /. x -> 2 // Factor

Solve[F == 0 /. x -> 2, y]

sol /. {y[x] -> 1/2, x -> 2}

D[F /. y -> y[x], x] == 0

sol = Solve[D[F /. y -> y[x], x] == 0, y'[x]]

sol2 = -D[F, x]/D[F, y]

A 26

(∫_(-2)^2)(∫_(-1)^1) 
              (x^2) (y^4) \[DifferentialD]y \[DifferentialD]x

inneres Integral:

(∫_(-1)^1) (x^2) (y^4) \[DifferentialD]y

(∫_(-1)^1)(∫_(-2)^2) 
              (x^2) (y^4) \[DifferentialD]x \[DifferentialD]y

inneres Integral:

(∫_(-2)^2) (x^2) (y^4) \[DifferentialD]x

Lineare Algebra

L23

B = (1/3)*{{1, 2, -2}, {2, 1, 2}, {-2, 2, 1}}

Eigensystem[B]

Der Eigenraum zum Eigenwert 1 ist also gegeben durch

ebene = {{x}, {y}, {z}} == 
        t {{-1}, {0}, {1}} + u {{1}, {1}, {0}}

Normalenvektor: Eigenvektor zum Eigenwert -1

n = {{1}, {-1}, {1}}

Normalform der Ebene

Eliminate[ebene, {t, u}]

Dies weist nach, dass der Vektor n senkrecht zur Ebene liegt.

L24

Erzeugung einer Matrix mit ganzzahligen Eigenwerten

A = {{-1, 2, -3, 0}, {4, -3, -3, 0}, {-5, 2, 5, 
          2}, {-3, 2, 1, 1}}

CharacteristicPolynomial[A, λ]

Eigensystem[A]

L25

Unprotect[C, D];
A = {{0, 0}, {0, 0}};
B = {{0, 1}, {0, 0}};
C = {{0, 0, 0}, {0, 0, 0}, {0, 0, 7}};
D = {{0, 1, 0}, {0, 0, 0}, {0, 0, 7}};

Eigensystem[A]

Eigensystem[B]

Eigensystem[C]

Eigensystem[D]