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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. |
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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. |
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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. |
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