Editing EmergingE
You are currently not logged in.
To change this, fill in the following fields:
Username
Password
Who can read this page?
The World
Members
Council
Admin
You have been granted an edit lock on this page
until Fri Apr 19 01:10:52 2024.
Press
to finish editing.
Who can edit this page?
World editing disabled
Members
Council
Admin
[[[>50 There is a reason why taking the square root halves the rubbish - it's because $(1+\epsilon)^2$ is $1+2\epsilon+\epsilon^2$, and when $\epsilon$ is small enough, $\epsilon^2$ effectively vanishes. So $(1+\epsilon)^2\approx 1+2\epsilon$, provided $\epsilon$ is small. ]]] 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 $e$.