Lösungen / Solutions

Remove["Global`*"]

1

a

D[12  x^5 - 4 x^3 + 6 x^2 + 11 x - 3 , x]

D[12  x^5 - 4 x^3 + 6 x^2 + 11 x - 3, {x, 2}]

b

D[3 x^5 Tan[10 x] - x^(4 x + 1), x]

c

D[5 x^4/(3 Cos[x] + Log[x]) - Sin[x]^2, x]

D[5 x^4/(3 Cos[x] + Log[x]) - Sin[x]^2, x] // Simplify

D[5 x^4/(3 Cos[x] + Log[x]) - Sin[x]^2, x] // Simplify // 
      Together

d

D[E^(Cos[2  x + 4]) - Log[Sin[x^3 + 1]], x]

2

f[x_] := 5  x^3 + 2  x^2 + 8  x - 9

Plot[f[x], {x, -1, 3}];

D[f[x], {x, 2}]

Solve[D[f[x], {x, 2}] == 0, x]

% // N

f[-(2/15)]

% // N

s[t_] := t - (2/15)

g[t_] := f[s[t]] + (6779/675)

g[t]

g[t] // Expand

f[s[t]] + (6779/675)  // Expand

Ungerade, punktsymmetrisch zu O

Plot[g[t], {t, -2, 2}];

f1[x_] := (x + 1) (x + 6) (x - 3) (x - 5)

Plot[f1[t], {t, -6, 6}];

f1'[x] // Expand

f1'''[x] // Expand

Solve[-6 + 24 x == 0, {x}]

% // N

3

a

Funktion f definieren

f[x_] := Cos[x (x - 1) (x - 3)] - 3 Log[x]

Ableitung = 2.5

D[f[x], x] == 2.5

ExpandAll[D[f[x], x]] == 2.5

Plot von f und c=2.5 im Intervall (-5 , 5)

Plot[{-(3/x) - 3 Sin[3 x - 4 (x^2) + (x^3)] + 
            8 x Sin[3 x - 4 (x^2) + (x^3)] - 
            3 (x^2) Sin[3 x - 4 (x^2) + (x^3)], 2.5}, {x, -5, 
          5}];

Plot von f und c=2.5 im Intervall (-0.5 , 5)

Plot[{-(3/x) - 3 Sin[3 x - 4 (x^2) + (x^3)] + 
            8 x Sin[3 x - 4 (x^2) + (x^3)] - 
            3 (x^2) Sin[3 x - 4 (x^2) + (x^3)], 2.5}, {x, 0.5, 
          5}];

Plot von f und c=2.5 im Intervall (-2.5 , 3). Man sieht, dass keine Lösung (Schnittpunkt) existiert!

Plot[{-(3/x) - 3 Sin[3 x - 4 (x^2) + (x^3)] + 
            8 x Sin[3 x - 4 (x^2) + (x^3)] - 
            3 (x^2) Sin[3 x - 4 (x^2) + (x^3)], 2.5}, {x, 2.5, 
          3}];

Nullstelle suchen: g(x) = f(x) - 2.5 = 0.   Das entspricht der Lösung von f(x) = 2.5
Start bei 1.5. Mit der Newton-Methode kommt eine falsche Lösung heraus und dazu eine Warnmeldung, die man beachten sollte....

FindRoot[-(3/x) - 3 Sin[3 x - 4 (x^2) + (x^3)] + 
        8 x Sin[3 x - 4 (x^2) + (x^3)] - 
        3 (x^2) Sin[3 x - 4 (x^2) + (x^3)] - 2.5, {x, 1.5}]

Start von Newton bei 1.9.

FindRoot[-(3/x) - 3 Sin[3 x - 4 (x^2) + (x^3)] + 
        8 x Sin[3 x - 4 (x^2) + (x^3)] - 
        3 (x^2) Sin[3 x - 4 (x^2) + (x^3)] - 2.5, {x, 1.9}]

Start von Newton bei 3.5. Da gibt es eine Lösung!

FindRoot[-(3/x) - 3 Sin[3 x - 4 (x^2) + (x^3)] + 
        8 x Sin[3 x - 4 (x^2) + (x^3)] - 
        3 (x^2) Sin[3 x - 4 (x^2) + (x^3)] - 2.5, {x, 3.5}]

Plot[{-(3/x) - 3 Sin[3 x - 4 (x^2) + (x^3)] + 
            8 x Sin[3 x - 4 (x^2) + (x^3)] - 
            3 (x^2) Sin[3 x - 4 (x^2) + (x^3)], 2.5}, {x, 3.4, 
          3.5}];

b

Plot[-(3/x) - 3 Sin[3 x - 4 (x^2) + (x^3)] + 
          8 x Sin[3 x - 4 (x^2) + (x^3)] - 
          3 (x^2) Sin[3 x - 4 (x^2) + (x^3)], {x, -5, 5}];

Plot[-(3/x) - 3 Sin[3 x - 4 (x^2) + (x^3)] + 
          8 x Sin[3 x - 4 (x^2) + (x^3)] - 
          3 (x^2) Sin[3 x - 4 (x^2) + (x^3)], {x, 0.5, 3}];

FindRoot[-(3/x) - 3 Sin[3 x - 4 (x^2) + (x^3)] + 
        8 x Sin[3 x - 4 (x^2) + (x^3)] - 
        3 (x^2) Sin[3 x - 4 (x^2) + (x^3)], {x, 2.5}]

FindRoot[-(3/x) - 3 Sin[3 x - 4 (x^2) + (x^3)] + 
        8 x Sin[3 x - 4 (x^2) + (x^3)] - 
        3 (x^2) Sin[3 x - 4 (x^2) + (x^3)], {x, 3}]

FindRoot[-(3/x) - 3 Sin[3 x - 4 (x^2) + (x^3)] + 
        8 x Sin[3 x - 4 (x^2) + (x^3)] - 
        3 (x^2) Sin[3 x - 4 (x^2) + (x^3)], {x, 1.5}]

4

a

Remove["Global`*"]

y[1] =   0.5;
f[x_] :=  ( Pi + 1) Cot[x] - 
          1 E^x;
g[x_] := Evaluate[D[f[s], s]] /. s -> x;

Plot[f[x], {x, 0.25, Pi/2}];

Plot[g[s], {s, 0.25, Pi/2}];

g[s]

y[n_] := 
      s - Evaluate[f[s]/g[s]] /. s -> y[n - 1]; ?y

y[2]

Table[y[k], {k, 1, 5}] // N

y[1] =   0.5;
Table[y[k], {k, 1, 5}] // N

y[1] =   1.0;
Table[y[k], {k, 1, 5}] // N

x[1] = 0.5;
x[n_] := f[x[n - 1]] + x[n - 1];
Table[x[n], {n, 1, 8}]//N

Remove["Global`*"]

y[1, a_] := 0.5;
f[x_, a_] :=   (Pi + 1) Cot[x] - 
          a E^x;
g[s_, a_] := D[f[s, a], s];
y[n_, a_] := 
        s - Evaluate[f[s, a]/g[s, a]] /. s -> y[n - 1, a] // 
          N;

y[5, 1]

Plot[y[5, x], {x, 0.1, 5}];

Plot[y[10, x], {x, 0.1, 10}];

Plot[y[5, x], {x, 0.1, 5}];

Plot[Evaluate[D[(Pi + 1) Cot[x] - 1 E^x, x]], {x, -10, 10}, 
        PlotRange -> {-20, 20}];

Plot3D[
        Evaluate[D[(Pi + 1) Cot[x] - a E^x, x]], {x, -10, 
          10}, {a, -12, 12}];

Table[
        Plot[{(Pi + 1) Cot[x] - a E^x, 
            x}, {x, -Pi/2 - 0.2, -0.4}, 
          AspectRatio -> Automatic, 
          PlotRange -> {-Pi/2, Pi/2}], {a, -10, 0, 
          0.5}];

Table[
        Plot[{(Pi + 1) Cot[x] - a E^x, x}, {x, 0.4, Pi/2 - 0.2}, 
          AspectRatio -> Automatic, PlotRange -> {0, Pi/2}], {a, 0, 
          10, 0.5}];

Table[
        Plot[{(Pi + 1) Cot[x] - a E^x, x}, {x, 0.4, Pi/2 - 0.2}, 
          PlotRange -> {0, Pi/2}], {a, 0, 4, 0.5}];

5

b

h0[x_]:=2x^3-3x^2+1;
h1[x_]:=x^3-2x^2+x;
h2[x_]:=x^3-x^2;
h3[x_]:=1-h0[x];
pH[x_,y1_,m1_,m2_,y2_]:=y1 h0[x]+ m1 h1[x]+ m2 h2[x]+y2 h3[x];

Plot[{h0[x],h1[x],h2[x],h3[x]},{x,0,1}];

Plot[{h0[x],h1[x],h2[x],h3[x],pH[x,2,8,-2,3]},{x,0,1}];

xL[1]=1;
xL[2]=3;
xL[3]=5;
xL[4]=7;
xL[5]=9;
delta[k_]:=(xL[k+1]-xL[k]);

t[x_,k_]:=(x-xL[k])/(xL[k+1]-xL[k])

Table[t[x,k],{k,1,4}]

t[x,1]

par[x_]:=a x^2+b x + c; sol=Solve[{par[1]==3,par[3]==5,par[5]==0},{a,b,c}]//Flatten

par[x_] := a x^2 + b x + c /. sol;  par[x]

m1 = par'[x] /. x -> 1

m2 = (1 - 2)/(5 - 1)

pH[x_,y1_,m1_,m2_,y2_,k_]:=y1 h0[x]+delta[k] m1 h1[x]+delta[k] m2 h2[x]+y2 h3[x];

pH[t[x,1],2,m1,m2,5,1]//Expand//Simplify

Plot[pH[t[x,1],2,m1,m2,5,1],{x,1,3}];

Programm CM26.nb

tab1 = {{1, 3}, {3, 5}, {5, 0}, {7, 4}, {9, 2}};

f1Abl[
        n_] := (tab1[[n + 
                    2]][[2]] - tab1[[n]][[2]])/(tab1[[n + 
                    2]][[1]] - tab1[[n]][[1]])

tab2 = Table[f1Abl[n], {n, 1, Length[tab1] - 2}]

f2Abl[
        n_] := (tab2[[n + 2]] - 
            tab2[[n]])/(tab1[[n + 
                    3]][[1]] - tab1[[n + 
                    1]][[1]])

tab2 = Table[f2Abl[n], {n, 1, Length[tab2] - 2}]