4 added 191 characters in body

You can find the half-iterate of a function from known integer iterates by using Newton series, for example:

$$f^{[1/2]}(x)=\sum_{m=0}^{\infty} \binom {1/2}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{[k]}(x)$$

This does not converge for $f(x)=a^x$ where $a>e^{1/e}$ but since your function is somewhat different you can try this method.

Update. Here is a plot for x<0:

For positive x it seems the formula mostly does not converge.

3 added 4 characters in body

You can find the half-iterate of a function from known integer iterates by using Newton series, for example:

$$f^{[1/2]}(x)=\sum_{m=0}^{\infty} \binom xm {1/2}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{[k]}(x)$$

This does not converge for $f(x)=a^x$ where $a>e^{1/e}$ but since your function is somewhat different you can try this method.

2 deleted 12 characters in body

You can find the half-iterate of a function from known integer iterates by using Newton series, for example:

$$f^{[1/2]}(x)=\lim_{n\to\infty}\sum_{m=0}^{n} f^{[1/2]}(x)=\sum_{m=0}^{\infty} \binom xm \sum_{k=0}^m\binom mk(-1)^{m-k}f^{[k]}(x)$$

This does not converge for $f(x)=a^x$ where $a>e^{1/e}$ but since your function is somewhat different you can try this method.

1