Schwarz Lemma
Investigations on the real line

Math 302: Fall 2010

Manipulate[Plot[{1,-1,(a+1) x / (1 + a x^2)}, {x,-1,1}], {a,0,2}]

Simplify[D[(a+1) x / (1 + a x^2), x]]

deriv[x_,a_] := ((1+a-a (x^2)-(a^2) (x^2))/(((1+a (x^2)))^2))

Note numerator above simplies  and the derivative equals
(1+a) (1 - a (x^2))  / (((1+a (x^2)))^2)
which is clearly positive (and thus our function is increasing) if 0 ≤ a < 1, and the derivative at x = 0 is just 1 + a.

Manipulate[Plot[{0, deriv[x,a]},{x,-1,1}],{a,0,2}]

h[x_,k_] := Tanh[k x] / Tanh[k]
D[h[x,k],x]
hderiv[x_,k_] := k Coth[k] ((Sech[k x])^2);
Manipulate[Plot[{1,-1,h[x,k]},{x,-2,2}], {k,1,100}]
Manipulate[Plot[{0,hderiv[x,k]},{x,-1,1}], {k,1,100}]
Series[Tanh[k x] / Tanh[k], {x,0,5}]

c=2;
Manipulate[Plot[{-1,1, x  + h[x,k]Exp[-c/(1-x^2)] },{x,-1,1}],{k,1,100}]
Series[D[1- Exp[-5/(1-x)], x],{x,0,4}]
Simplify[D[x  + h[x,k]Exp[-c/(1-x^2)],x]]
newderiv[x_,k_] := 1+(((E^(2/(-1+(x^2)))) Coth[k] ((Sech[k x])^2) (k (((-1+(x^2)))^2)-2 x Sinh[2 k x]))/(((-1+(x^2)))^2));
newderiv[0,k]

Manipulate[Plot[{1,-1,(a+1) x / Sqrt[1 + (a^2 + 2a) x^2]},{x,-1,1}], {a,0,100}]
Simplify[D[(a+1) x / Sqrt[1 + (a^2 + 2a) x^2],x]]
Manipulate[Plot[((1+a)/(((1+2 a (x^2)+(a^2) (x^2)))^(3/2))), {x,-1,1}],{a,0,100}]

Manipulate[Plot[{1,-1,(a+1)^n x / (1 + a x^2)^n}, {x,-1,1}], {a,0,2},{n,1,10}]

Simplify[D[(a+1)^n x / (1 + a x^2)^n,x]]

Manipulate[Plot[{0,(((1+a))^n) (((1+a (x^2)))^(-1-n)) (1+a (1-2 n) (x^2))}, {x,-1,1}], {a,0,2},{n,.1,10}]

Simplify[D[(a+1)^n x / (1 + a x^4)^n,x]]

DETAILED ANLAYSIS: GETS ARBITRARILY LARGE BUT LOSES ANALYTICITY

Consider the following, where a good choice of n is n = 1/4

Manipulate[Plot[{1,-1,(a+1)^n x / (1 + a x^4)^n}, {x,-1,1}], {a,0,2},{n,.1,10}]

Manipulate[Plot[{1,-1,(a+1)^(1/4) x / (1 + a x^4)^(1/4)}, {x,-1,1}], {a,0,20000}]

Simplify[D[(a + 1)^(1/4) x / (1 + a x^4)^(1/4), x]]

((((1+a))^(1/4))/(((1+a (x^4)))^(5/4)))

Note the first derivative is always positive and thus our function is increasing, which is exactly what we wish. Further, the first derivative at the origin is just (1+a)^(1/4), which we can make arbitrarily large! Unfortunately it is only analytic for |x| < 1/a^(1/4). Oh well.

Manipulate[Plot[((((1+a))^(1/4))/(((1+a (x^4)))^(5/4))), {x,-1,1}],{a,1,2000}]

DAVID THOMPSON

f[x_,k_] := Erf[k x] / Erf[k]
Manipulate[Plot[{1,-1,f[x,k]}, {x,-1,1}],{k,1,1000}]

D[ Erf[k x] / Erf[k],x]

deriverf[x_,k_] := ((2 (E^(-(k^2) (x^2))) k)/((Sqrt[π]) Erf[k]))

Manipulate[Plot[{0,deriverf[x,k]}, {x,-1,1}],{k,1,1000}]

deriverf[0,1000.]

Erf[-6767]/Erf[6767.]

g[x_,a_] := Exp[-x^a];
h[a_,bound_] := NIntegrate[g[x,a], {x,0,bound}]
h[1]

Manipulate[Plot[{1,h[a,b]/h[a,1]}, {b,0,1}],{a,.1,100}]

Plot[1/h[a,1],{a,.0001,1}]

Plot3D[{1  If[x^2 + y^2 ≤ 1, 0, 0],Abs[Erf[x + I y]] If[x^2 + y^2 ≤ 1, 1, 0]}, {x,0,1}, {y,0,1}]

ourfn[x_,k_]:= Erf[k x]/Erf[k];
Plot[{ourfn[x,1], ourfn[x,5], ourfn[x,10],ourfn[x,50]},{x,-1,1}]

Plot3D[{Abs[Erf[x + I y]] If[x^2 + y^2 ≤ 1, 1, 0]}, {x,0,1}, {y,0,1}]

Series[Exp[-1./(x-1/5)^2], {x,0,10}]