Sorry, I thought you already considered this type of formula or made a quick check about derivatives and such ... anyway, my "function" here is continuous and differentiable (as far as I understand what that means), but not infinitely often differentiable. The 3rd derivative has a slight peak near the integer points that I didn't manage to wipe away with a higher accuracy of my "parameter" 0.345627(2-x), or maybe Excel is just too inaccurate there. I've tried similar formulae before, but never got this close (e.g. had this peak already in the 2nd derivative).

Generally, for a better understanding, I was originally searching a function between x and x+1, such that log(f(x))/log(b) below x and b^f(x) above x+1 runs smoothly together to a continuous, monotonously increasing line on a graph. (I could explain more details, if you care to hear the whole story.)

I had in mind that this might be a formula, even if it only gives approximated values, that is not too complicated to calculate with. (Okay, I still would need to find a way to get a general formula to substitute this 0.345627-thingy for other bases.) Or, if there is a way to get that peak out of the 3rd derivative and eventually make it infinitely differentiable, whether the results are congruent to already existing results or they aren't.

Generally, for a better understanding, I was originally searching a function between x and x+1, such that log(f(x))/log(b) below x and b^f(x) above x+1 runs smoothly together to a continuous, monotonously increasing line on a graph. (I could explain more details, if you care to hear the whole story.)

I had in mind that this might be a formula, even if it only gives approximated values, that is not too complicated to calculate with. (Okay, I still would need to find a way to get a general formula to substitute this 0.345627-thingy for other bases.) Or, if there is a way to get that peak out of the 3rd derivative and eventually make it infinitely differentiable, whether the results are congruent to already existing results or they aren't.