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I've re-read this, and it's really very abbreviated and could do
with some nice charts, etc. I'm working on it ...
| x | c(x) |
| 2 | 0.693147... |
| ? | 1.000000... |
| 3 | 1.098613... |
| 4 | 1.386295... |
| 5 | 1.609439... |
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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.
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Hence $\sqrt{1+\epsilon} \approx 1+\epsilon/2$
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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 $c$:
- Is it well defined?
- Is it continuous?
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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