Emerging E

   
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x c(x)
2 0.693147...
? 1.000000...
3 1.098613...
4 1.386295...
5 1.609439...

There is a reason why taking the square root halves the rubbish - it's because $(1+\epsilon/2)^2$ is $1+\epsilon+\epsilon^2/4$, and when $\epsilon$ is small enough, $\epsilon^2$ effectively vanishes. So $(1+\epsilon/2)^2\approx 1+\epsilon$, provided $\epsilon$ is small.

Hence $\sqrt{1+\epsilon} \approx 1+\epsilon/2$

Take 10, and square root many times over. After a while you get 1 + rubbish. Taking the square root of that halves the rubbish, so we can deduce that $10^\epsilon = 1+c_{10}\epsilon$.

I've put the subscript on the $c$ because if you do this again but start with 5 instead of 10 you get a different constant, $c_5$.

So now we have a function, $c(x)$ where you start with $x$, successively square root, compute the constant, and that's the value of the function.

If we do this there are lots of questions to ask about the function $c$:

and so on. But we can observe that $c(2)<1$ and $1<c(3)$, and probably $c(2)<c(x)<c(3)$ when $x$ is between 2 and 3, so is there a value of $x$ that gives $c(x)=1$?

The answer is yes, the answer is $e$.


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