Here is another interesting fact:
The Taylor series of i[x] is the following:
\( i_x = \sum_{k=0}^{\infty} i^{(k)}_x x^k / k! = i_x + x i'_x + x^2 1/2 i''_x + x^3 1/6 i'''_x + ... \)
and
\( i_x^{o2} = i_{i_x} = \sum_{k=0}^{\infty} i^{(k)}_x / k! + x \sum_{k=0}^{\infty} i'^k_x i^{(k)}_x / k! + x^2 \sum_{k=0}^{\infty} i''^k_x i^{(k)}_x / k! + ... |_{x=0} \)
So i[i[x]] has a part as 1 looking like this:
\( i_{i_x} = 1 + x \sum_{k=0}^{\infty} i'^k_x i^{(k)}_x / k! + x^2 \sum_{k=0}^{\infty} i''^k_x i^{(k)}_x / k! + ... |_{x=0} \)
Therefor di[x]/dx cannot be 0, then what can it be?
The Taylor series of i[x] is the following:
\( i_x = \sum_{k=0}^{\infty} i^{(k)}_x x^k / k! = i_x + x i'_x + x^2 1/2 i''_x + x^3 1/6 i'''_x + ... \)
and
\( i_x^{o2} = i_{i_x} = \sum_{k=0}^{\infty} i^{(k)}_x / k! + x \sum_{k=0}^{\infty} i'^k_x i^{(k)}_x / k! + x^2 \sum_{k=0}^{\infty} i''^k_x i^{(k)}_x / k! + ... |_{x=0} \)
So i[i[x]] has a part as 1 looking like this:
\( i_{i_x} = 1 + x \sum_{k=0}^{\infty} i'^k_x i^{(k)}_x / k! + x^2 \sum_{k=0}^{\infty} i''^k_x i^{(k)}_x / k! + ... |_{x=0} \)
Therefor di[x]/dx cannot be 0, then what can it be?
Xorter Unizo
