Aufgabe 19

Länge[
        a_] := (Sqrt[(∑_(k = 1)^(Length[a]))a[[k]]^2])

akreuzb = Cross[{-1, 1, 0}, {3, 2, 1}]

abstand = 
      Abs[akreuzb . ({-1, 2, -3} - {1, 2, 3})]/
        Länge[akreuzb]

N[abstand]

Aufgabe 20

(a - b) . (a - b) + (a + b) . (a + b) - 
        2 (a . a + b . b) // Expand

a = {a1, a2}; b = {b1, b2};

(a - b) . (a - b) + (a + b) . (a + b) - 
      2 (a . a + b . b)

(a - b) . (a - b) + (a + b) . (a + b) - 
        2 (a . a + b . b) // Expand

Clear[a, b, c, d]

Aufgabe 21

(((1 + I)*(2 - 3  I))/((4 - 6  I) (1 + 2  I)))

Aufgabe 22

(((-(1/2) + ((Sqrt[3])/(2 I))))^3)

(((-(1/2) + ((Sqrt[3])/(2 I))))^3) // Expand

Aufgabe 23

Plot[
      Evaluate[Table[(∑_(k = 0)^n) ((x^k)/(k!)), {n, 0, 
            5}]], {x, -1, 1}]

Plot[
      Evaluate[Table[(∑_(k = 0)^n) ((x^k)/(k!)), {n, 0, 
            5}]], {x, -5, 5}]

Plot[(∑_(k = 0)^100) ((x^k)/(k!)), {x, -5, 5}]

Table[(((1/(k!)))/((((1/2)))^k)), {k, 1, 20}] // N

(∑_(k = 0)^∞) ((x^k)/(k!))

Aufgabe 24

Clear[B]
B[1] = 0; B[2] = 1; 
    B[n_] := (1/2) (B[n - 1] + B[n - 2])

Table[B[n], {n, 1, 10}]

Clear[B]
B[1] = 0; 
    B[n_] := B[n - 1] + ((((-1))^(n - 2))/(2^(n - 2)))

Table[B[n], {n, 1, 10}]

Needs["DiscreteMath`RSolve`"]

sol = 
      RSolve[{b[n + 1] - b[n] == ((((-1))^(n - 1))/(2^(n - 1))), 
          b[1] == 0}, b[n], n]

-2 sol[[1, 1, 2, 2, 2]]

Limit[-2 sol[[1, 1, 2, 2, 2]], 
      n -> ∞]

Aufgabe 25

Clear[a]

RSolve[{a[n + 1] == a[n]^2 + (1/4), a[1] == (1/4)}, a[n], 
      n]

Solve[x == (x^2) + (1/4), x]