binomio inverso

(x+y)^{i(n)} = x^{i(n)}+(1/n)·x^{i(n+(-1))}y^{i(1)}+...+(1/n)·x^{i(1)}y^{i(n+(-1))}+y^{i(n)}

x^{i(0)} = 1

x^{i(1)} = x

(x+y)^{i(1)} = x+y

(x+y)^{i(2)} = x^{i(2)}+(1/2)·xy+y^{i(2)}

x^{i(2)} = (x^{2}/4)

(x+y)^{i(3)} = x^{i(3)}+(1/3)·x^{i(2)}y+(1/3)·y^{i(2)}x+y^{i(3)}

x^{i(3)} = (x^{3}/(9·4))

(x+y)^{i(4)} = x^{i(4)}+(1/4)·x^{i(3)}y+(1/6)·x^{i(2)}y^{i(2)}+(1/4)·y^{i(3)}x+y^{i(4)}

x^{i(4)} = (x^{4}/(16·9·4))

d_{x}[x^{i(n)}] = (1/n)·x^{i(n+(-1))}

d_{x...x}^{n}[x^{i(n)}] = (1/n!)

d_{x}[y(x)]^{2} = ( n/(n+(-1)) )·y(x)·d_{xx}^{2}[y(x)] <==> ( y(x) = x^{n} or y(x) = x^{i(n)} ) 

x^{i(n)} = (1/(n^{2})!)·x^{n}

x·x^{i(n)} = (n+1)^{2}·x^{i(n+1)}

d_{x}[x^{2}+x^{i(2)}] = (5/2)·x

d_{x}[x^{3}+x^{i(3)}] = (37/12)·x^{2}