How did they find out what epsilon is equal to?

Plato

MHF Helper
Aug 2006
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For case I and case II, can you please explain how professor came to what epsilon was equal to?
https://imgur.com/a/BU09F6d
This may not help you at all. But I refuse to jump into someone else's proof.
We are given that $(\forall n)[S_n\ge 0]~\&~ S_n\to S$
1) We are to prove that $S\ge 0$ So let's suppose that $S<0$
Let $\varepsilon = \frac{{ - S}}{2}$. By convergence $(\exists N\in\mathbb{Z}^+)[n\ge N \Rightarrow |S - {S_n}| < \varepsilon$
Thus we have $\begin{gathered} \left| {{S_n} - S} \right| < \varepsilon \hfill \\ - \varepsilon < {S_n} - S < \varepsilon \hfill \\
{S_n} < \varepsilon + S < \frac{{ - S}}{2} + S = \frac{S}{2} < 0 \hfill \\ \end{gathered}$ But $S_N\ge 0$ so there is a contradiction.
Hence $S\ge 0$ as you were asked to prove for 1).
Now can you construct a proof for $\sqrt{S_n}\to\sqrt{S}~?$
 
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Jul 2015
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I'll get back to you in a couple hours on the 2nd part of the proof. Currently about to drive home.
 
Jul 2015
605
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1 question from result..I understand it and makes sense, however, Why did you put absolute value of Sn-S is less than espilon and absolute value S-Sn is less than espilon as well? I don't see that in any of my convergent definitions. Why is that a legal thing we can do?

Moreover, to prove: \(\displaystyle \sqrt{S_n}\to\sqrt{S}\)

Proof: we are given \(\displaystyle (S_{n})\geq0(\forall n)\& S_{n}\to S\)
WTS: \(\displaystyle \sqrt{S_n}\to\sqrt{S}\)

\(\displaystyle (\exists N\in\mathbb{Z}^+)[n\ge N \Rightarrow |{S_n - S| < \varepsilon\)

\(\displaystyle \sqrt{S_n}-\sqrt{S}=\frac{\|S_n - S|{\sqrt{S_n}+\sqrt{S}} \Leftrightarrow \frac{\\epsilon}{\sqrt{S}+\sqrt{S}\)

So \(\displaystyle |\sqrt{S_n}-\sqrt{S}| <= \frac{\\epsilon}{2\sqrt{S}} (\forall n) >= N.\) Hence proved by def. of limit.


Not sure why it is showing \displaystyle for 2 of my lines of TeX...I am using \(\displaystyle ....\) Anyone know why?
 
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Plato

MHF Helper
Aug 2006
22,490
8,653
1 question from result..I understand it and makes sense, however, Why did you put absolute value of Sn-S is less than espilon and absolute value S-Sn is less than espilon as well? I don't see that in any of my convergent definitions. Why is that a legal thing we can do?

Moreover, to prove: \(\displaystyle \sqrt{S_n}\to\sqrt{S}\)

Proof: we are given \(\displaystyle (S_{n})\geq0(\forall n)\& S_{n}\to S\)
WTS: \(\displaystyle \sqrt{S_n}\to\sqrt{S}\)

$(\exists N\in\mathbb{Z}^+)[n\ge N \Rightarrow |{S_n - S| < \varepsilon$

$\sqrt{S_n}-\sqrt{S}=\frac{\|S_n - S|{\sqrt{S_n}+\sqrt{S}} \Leftrightarrow \frac{\\epsilon}{\sqrt{S}+\sqrt{S}|

So \(\displaystyle |\sqrt{S_n}-\sqrt{S}| <= \frac{\\epsilon}{2\sqrt{S}} (\forall n) >= N.\) Hence proved by def. of limit.
Surely you understand that $\forall x~\&~y$ it is true that $|x-y|=|y-x|~???$
Thus $|S-S_n|$ is the same as $|S_n-S|$.
 
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Dec 2013
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For case I and case II, can you please explain how professor came to what epsilon was equal to?
That's scratch-work done prior to the proof. Note that $\epsilon$ is not "equal to" $\epsilon\sqrt s$, but his prior work showed that such a value is useful for demonstrating the limit he sought to prove.