Aufgabe 57

D[Cot[x], x]

D[ArcTan[x], x]

D[ArcSinh[x], x]

Aufgabe 58

Limit[((Cos[x] - 1)/(x^2)), x -> 0]

Limit[((1 - Cos[3  x])/(1 - Cos[x])), x -> 0]

D[Exp[-(1/(x^2))], x]

Limit[D[Exp[-(1/(x^2))], x], x -> 0]

Hospital[f_, g_, x_ -> x0_] := 
      If[Limit[f, x -> x0] == 0  &&  
          Limit[g, x -> x0] == 0, 
        Limit[D[f, x], x -> x0]/
          Limit[D[g, x], x -> x0], 
Limit[f, 
            x -> x0]/Limit[g, x -> x0]]

Hospital[Exp[x] - 1, x, x -> 0]

Hospital[Cos[x] - 1, (x^2), x -> 0]

HospitalRekursiv[f_, g_, x_ -> x0_] := 
      Module[{}, 
zähler = 
          zähler + 1; 
If[
          Limit[f, x -> x0] == 0  &&  
            Limit[g, x -> x0] == 
              0, 
HospitalRekursiv[D[f, x], D[g, x], 
            x -> x0], 
          Limit[f, x -> x0]/
            Limit[g, x -> x0]]
]

zähler = 0

HospitalRekursiv[Cos[x] - 1, (x^2), x -> 0]

zähler

zähler = 0

HospitalRekursiv[1 - Cos[3  x], 1 - Cos[x], x -> 0]

zähler

zähler = 0

HospitalRekursiv[(x^100), (x^100), x -> 0]

zähler

Aufgabe 59

f[x_] := (x^3) - x

plot1 = Plot[f[x], {x, -2, 2}]

t[x_] := f[b] + f'[b] (x - b)

t[x]

Solve[t[a] == f[a], b]

Expand[t[a] - f[a]]

Factor[t[a] - f[a]]

Sekante[
        f_, {x_, x1_, 
          x2_}] := (((f /. x -> x2) - (f /. 
                    x -> x1))/(x2 - x1)) (x - x1) + (f /. 
            x -> x1)

Sekante[f[x], {x, -1/2, 1}]

plot2 = 
      Plot[Evaluate[Sekante[f[x], {x, -1, (1/2)}]], {x, -2, 
          2}]

Show[plot1, plot2]

Aufgabe 60

(a)

Plot[Exp[-(x^2)], {x, -3, 3}]

LinksSumme[f_, {x_, a_, b_}, 
        n_] := 
(∑_(k = 0)^(n - 1))(f /. 
              x -> a + ((k (b - a))/n)) ((b - a)/n)

RechtsSumme[f_, {x_, a_, b_}, 
        n_] := (∑_(k = 1)^n)(f /. 
              x -> a + ((k (b - a))/n)) ((b - a)/n)

int = LinksSumme[Exp[-(x^2)], {x, -3, 3}, 15]

N[int]

N[(int^2)]

π // N

untere und obere Schranke

int = 2*RechtsSumme[Exp[-(x^2)], {x, 0, 3}, 15]

N[int]

N[(int^2)]

int = 2*LinksSumme[Exp[-(x^2)], {x, 0, 3}, 15]

N[int]

N[(int^2)]

und mit feinerer Unterteilung

int = 2*RechtsSumme[Exp[-(x^2)], {x, 0, 3}, 1000]

N[int]

N[(int^2)]

int = 2*LinksSumme[Exp[-(x^2)], {x, 0, 3}, 1000]

N[int]

N[(int^2)]

(b) Approximation von π durch (∫_0^((1/2)))(Sqrt[1 - 
                (x^2)]) dx

Plot[{(Sqrt[3]) x, (Sqrt[1 - (x^2)])}, {x, 0, (1/2)}, 
      AspectRatio -> Automatic]

12 ((∫_0^((1/2)))(Sqrt[1 - 
                      (x^2)]) \[DifferentialD]x - ((Sqrt[3])/8)) // 
      Simplify

Näherung

{12 (RechtsSumme[(Sqrt[1 - (x^2)]), {x, 0, (1/2)}, 10] - ((Sqrt[3])/8)), 
        12 (LinksSumme[(Sqrt[1 - (x^2)]), {x, 0, (1/2)}, 10] - ((Sqrt[3])/8))} // 
      N

und mit feinerer Unterteilung

{12 (RechtsSumme[(Sqrt[1 - (x^2)]), {x, 0, (1/2)}, 100] - ((Sqrt[3])/8)), 
        12 (LinksSumme[(Sqrt[1 - (x^2)]), {x, 0, (1/2)}, 100] - ((Sqrt[3])/8))} // 
      N

Aufgabe 61

M = {{1, 1, 2, 0}, {2, 1, 3, 1}, {1, 0, 1, 1}}

null = NullSpace [ M]

M . null[[1]]

M . null[[2]]

M . Transpose[null]

{{x}, {y}, {z}, {u}}

M . {{x}, {y}, {z}, {u}}

Solve[M . {{x}, {y}, {z}, {u}} == 0, {x, y, z}]