2 added 2 characters in body

And this is another construction.

Let $\sigma(x)=\exp(x)-1$ From this paper http://arxiv.org/abs/0812.4047 we know that

$$\exp(\sigma^{[p]}(t))=\sum_{n=0}^{\infty}B_n^p\frac{t^n}{n!}$$

where $B_n^p$ are the Bell's numbers of p-th order.

So to find $\sigma^{[1/2]}(t)$ we have to generalize Bell's numbers to fractional order. We can easily do that by induction as follows:

$$A_0^x=1$$ $$A_{n+1}^x=\sum_{k=0}^{x-1} A_n^x\star A_n^k$$

And then $$B_n^x=A_{n-1}^{x+1}$$

where $f(n)\star g(n)$ is the binomial convolution as described by David Knuth:

$$f(n)\star g(n)=\sum_{k=0}^n \binom nkf(n-k)g(k)$$

To obtain the value for any real x, we can note that the right part in $A_{n+1}^x=\sum_{k=0}^{x-1} A_n^x\star A_n^k$ is a polynomial of x and k of degree n-1 and integer coefficients and we can take indefinite sum of it symbolically following the rule

$$\sum_x ax^n=\frac{B_{a+1}(x)}{a+1}$$ax^n=\frac{a B_{n+1}(x)}{n+1}$$Where B_a(x) are the Bernoulli polynomials. 1 And this is another construction. Let \sigma(x)=\exp(x)-1 From this paper http://arxiv.org/abs/0812.4047 we know that$$\exp(\sigma^{[p]}(t))=\sum_{n=0}^{\infty}B_n^p\frac{t^n}{n!}$$where B_n^p are the Bell's numbers of p-th order. So to find \sigma^{[1/2]}(t) we have to generalize Bell's numbers to fractional order. We can easily do that by induction as follows:$$A_0^x=1A_{n+1}^x=\sum_{k=0}^{x-1} A_n^x\star A_n^k$$And then$$B_n^x=A_{n-1}^{x+1}$$where f(n)\star g(n) is the binomial convolution as described by David Knuth:$$f(n)\star g(n)=\sum_{k=0}^n \binom nkf(n-k)g(k)$$To obtain the value for any real x, we can note that the right part in A_{n+1}^x=\sum_{k=0}^{x-1} A_n^x\star A_n^k is a polynomial of x and k of degree n-1 and integer coefficients and we can take indefinite sum of it symbolically following the rule$$\sum_x ax^n=\frac{B_{a+1}(x)}{a+1}

Where $B_a(x)$ are the Bernoulli polynomials.