Polynome im Affenkasten

Polynome 3. Grades         Parabeln        Panther

Dritten Grades

g[x_]:=(b/(2 a^3)) x^3+(m-((3b)/(2a)) )x

g[x]

Manipulate[Plot[(-((3 b)/(2 a))+m) x+((b (x^3))/(2 (a^3))) ,{x,-4,4} ,PlotRange->{-6,6}],{{a,2},-8,8},{{b,3},-3,3},{{m,0},-2,2}]

h[x_]:=r x^3-s x; h[x]

Fall m=(1/2)

{(b/(2a^3)),m+((-3b)/(2a))}/.{a->2,b->3,m->(1/2)}

Manipulate[Plot[-s  x+r x^3 ,{x,-4,4} ,PlotRange->{-6,6}],{{r,3/16},-8,8},{{s,7/4},-3,3}]

Quit

Fall1 m=0

g[x]

{a,b,r,s}

g[2a]=.

?g

{h[a], h'[a],h[2a]}

h[a]==-b; h'[a]==0;h[2a]==b;

Solve[{h[a]==-b, h'[a]==0,h[2a]==b},{r,s}]

Fall 2 m<>0

eins=h[a]==m a-b

h'[x]

zwei=h'[a]==m

h[2a]

drei=h[2a]==2m a+b

Solve[{eins,zwei,drei},{a,b,m}]

Solve[{eins,zwei,drei},{a,b}]

Solve[{eins,zwei,drei},{a,m}]

Tests für das Buch

Quit

p[x_]:=t(x+3a)x^2

p[x]

b=((p[a])/2)

b=.

t=(b/(2(a^3)))

f[x_]:=p[x-a]-b   ; f[x]

Collect[Expand[f[x]],x]

Flächen

p[x]

{a,b}={2,3};Integrate[((b (x^2) (3 a+x))/(2 (a^3))),{x,-3a,a}]
{a,b}=.;

Integrate[((b (x^2) (3 a+x))/(2 (a^3))),{x,-3a,a}]

Integrate[((b (x^2) (3 a+x))/(2 (a^3))),x]

Integrate[((b (x^2) (3 a+x))/(2 (a^3))),{x,-3a,0}]

((27 a b)/(8 4 a b))

((27 a b)/(8  6 a b))

Manipulate[Plot[((b (a (x^3)+((x^4)/4)))/(2 (a^3))),{x,-10,4}],{{a,2},-4,4},{{b,3},-3,3}]

Parabeln

a=0.1; b=100;Plot[a x (x-b), {x,-b,2b}, PlotRange->All, PlotStyle->{Thickness[0.02],Color=Red}]
{a,b}=.

Pantherkäfig

Quit

Expand[(x+r)x(x-w)(x-w-r)]

Collect[%,x]

Manipulate[Plot[0.1(((r^2) w+r (w^2)) x+(-(r^2)-r w+(w^2)) (x^2)-2 w (x^3)+(x^4)),{x,-8,8},PlotRange->{-3,3}],{{r,1.5},-8,8},{{w,2},-2,2}]

g[x_]:=t((1/12)(x^4)-(w/6)(x^3)+d x); g[x]

Bedingung für Symmetrie

g[w]

gs[x_]:=t((1/12)(x^4)-(w/6)(x^3)+((w^3)/12)   x); gs[x]

Factor[%]

(1/12) t (w-x) x ((w^2)+w x-(x^2))s

Polynome 4. Grades aufbauen

p[x_]:=t x (x-w); p[x]

f[x_]:=Integrate[p[x],x]; f[x]

(t/6)x^2 (2 x-3 w)+ t d

d t+(1/6) t (x^2) (-3 w+2 x )

Dieses ist die allgemeine Form.

Verschieben Wendepunkt in den Ursprung

t d-(1/2) t w (x^2)+((t (x^3))/3)/.x->x+(w/2) //Expand

Wenn dieses punktsymmetrisch zum Ursprung sein soll, muss d passende gewählt werden

d t-((t (w^3))/12)-(1/4) t (w^2) x+((t (x^3))/3)/.d->((w^3)/12)

Also: Punktsymmetrisch zu ((w/2),0)

d t+(1/6) t (x^2) (-3 w+2 x )/.d->((w^3)/12) //Expand

Dazu gehört das Polynom 4. Grades

Integrate[((t (w^3))/12)-(1/2) t w (x^2)+((t (x^3))/3),x]

Manipulate[Plot[(1/12) t (w^3) x-(1/6) t w (x^3)+((t (x^4))/12)+m x,{x,-2,5},PlotRange->{-3,7}],{{t,(3/2)},0,3},{{w,2},0,3},{{m,0.5},0,3}]

(t/12)((x^4)-2w (x^3)+(w^3)x)

Veränderung dann als Scherung durch Addition von y= m x