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 December 5th, 2017, 05:14 PM #1 Member   Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2 Deeper Understanding of Limit Proofs In an "N epsilon proof of sequence convergence", we tend to assume the limit. For example: If we were proving that the sequence $\displaystyle 1/n$ as n goes from 1 to infinity, converges to 0, we would start with $\displaystyle |1/n−0|<ϵ$. As you can see we initially assume that the limit is 0. However, what if we thought the limit was $\displaystyle 1/1000$ or anything else? In other words, what would make an "N epsilon" proof with an incorrect limit assumption false?
 December 5th, 2017, 07:09 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra The assumption that you can find an $N$ for every $\epsilon$ such that $|\tfrac1n - \tfrac1{1000}| < \epsilon$. In every $N-\epsilon$ or $\delta-\epsilon$ proof, the first step is to do some scratch work to determine a scheme for $N$ (or $\delta$) that works for any given $\epsilon$. There isn't one for the example above because once $\epsilon < \tfrac1{2000}$ there are no such values of $N$ to find. Thanks from Antoniomathgini
 December 6th, 2017, 03:34 PM #3 Member   Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2 You're right. But what about a sequence $\displaystyle an$ where the limit is 1? We could assume that the limit is 0, and $\displaystyle |an − L|<ϵ$ would still hold for any ϵ.
 December 6th, 2017, 06:39 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra No it wouldn't. If $a_n = 1$ for every $n$ but we assume that the limit is zero, $|a_n - L| = 1$ for which we can find an $N$ that works only when $\epsilon > 1$. Thanks from Antoniomathgini
 December 6th, 2017, 07:18 PM #5 Member   Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2 I think I understand. For the sequence you gave, if we assume the limit to be 0, we get that $\displaystyle ϵ>1$ must hold. However, must be able to make $\displaystyle ϵ$ as small as we wish, otherwise, the definition of a limit is contradicted. Thank you very much for your help Thanks from v8archie
 December 6th, 2017, 08:41 PM #6 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics There is a more fundamental definition of convergence than the one you are working with and as an added bonus, it doesn't require knowing the limit at all. The basic idea is this: If a sequence converges to a limit, then at some point the values cluster around the value of the limit. However, this must mean that the values are simultaneously clustering around one another. Or to put it another way, your sequence can't get arbitrarily close to a limit without the values in your sequence eventually becoming arbitrarily close to one another. The make it a bit more formal, a (real) sequence, $\{a_n\}$ is called "Cauchy" if for any $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for EVERY $m,n > N$, the condition $|a_m - a_n| < \epsilon$ holds. Now, we have the following theorem which you should prove as an exercise: A sequence in $\mathbb{R}$ (or any complete metric space for that matter) converges if and only if it is Cauchy. Note there is no need to know what the limit is and this is very powerful. There are many sequences we know converge but have no idea what the limit is. To show your example converges, we would fix $\epsilon > 0$ and we want to determine $N$ so that $|\frac{1}{m} - \frac{1}{n}| < \epsilon$ if $m,n > N$. Setting $N > \frac{2}{\epsilon}$ yields $\frac{2}{N} < \epsilon$. Then we apply the triangle inequality to see that if $n,m > N$, then $\left| \frac{1}{m} - \frac{1}{n} \right| \leq \left| \frac{1}{m} \right| + \left| \frac{1}{n} \right| < \frac{2}{N} < \epsilon$ proving the sequence is Cauchy. By the theorem above, it converges and we never needed to know the limit. To learn more you can start here: https://en.wikipedia.org/wiki/Cauchy_sequence Thanks from greg1313 and Antoniomathgini

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