Taylor expansion of lambda-terms

T:
S0:
    

S = <[x]<x>x>([z]z)^2

Overview Syntax rules Examples
delta applied to identity (try).
 T = ([x](x)x)[z]z
S0 = [z]z
Kirstead term (try).
 T = ([f](f)[x](f)[d]x)[r](r)(r)*
S0 = *
delta applied to 2 (try).
 T = ([x](x)x)[f][x](f)(f)x
S0 = [f][x]<f><f><f><f>x
delta applied to 3 (try).
 T = ([x](x)x)[f][x](f)(f)(f)x
S0 = [f][x]<f><f><f><f><f><f><f><f><f><f><f><f><f><f><f><f><f><f><f><f><f><f><f><f><f><f><f>x
Fixed point combinator, level 1 (try).
 T = [f]([x](f)(x)x)[x](f)(x)x
S0 = [f]<f>{}
Fixed point combinator, level 2 (try).
 T = [f]([x](f)(x)x)[x](f)(x)x
S0 = [f]<f><f>{}
Fixed point combinator, level 3 (try).
 T = [f]([x](f)(x)x)[x](f)(x)x
S0 = [f]<f><f><f>{}
Fixed point combinator, level 4 (try).
 T = [f]([x](f)(x)x)[x](f)(x)x
S0 = [f]<f><f><f><f>{}
Fixed point combinator, level 10 (try).
 T = [f]([x](f)(x)x)[x](f)(x)x
S0 = [f]<f><f><f><f><f><f><f><f><f><f>{}
delta delta: this doesn't give a very interesting result but I'm sure you want to try... (try).
 T = ([x](x)x)[x](x)x
S0 =