Recent content by identity1

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    Does the Limit of aCn/na Approach 1/a! as n Approaches Infinity?

    No, you're limit is correct. It's also not that weird. Consider the expansion of {n\choose a}=\frac{n(n-1)\ldots (n-a+1)}{a!}. Then, consider that \frac{n(n-1)\ldots (n-a+1)}{n^a}=1\cdot (1-\frac{1}{n})\cdot\ldots\cdot (1-\frac{a-1}{n}). Each of those factors goes to 1 as n\rightarrow\infty...
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    Alternative proof of a theorem in Rudin

    Actually, talking about it made me realize the answer :P In my correction, where I realized I forgot to asset that k_n>k_{n-1}, I'm implicitly using a type of weak induction. Induction is necessary, and he does it more completely in his proof.
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    Alternative proof of a theorem in Rudin

    Well, he starts by constructing p_{n_1} similarly, and defines \delta=d(q, p_{n_1}). Then he proceeds by induction: have having constructed p_{n_1}, \ldots, p_{n_{i-1}}, he constructs p_{n_i} so that n_i>n_{i-1}, d(q, p_{n_i})<2^{-i}\delta. My main point of confusion I guess, was that I wasn't...
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    Alternative proof of a theorem in Rudin

    OH. Wow. I should've written p_{k_n}. My bad xP. Barring that notational slip-up, how's the proof look to you.
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    Alternative proof of a theorem in Rudin

    If it helps you to move over that point, Rudin does the same thing in his proof as well. One thing I did forget to write is that we choose n_k>n_{k-1} so that the subsequence keeps moving forward (we are guaranteed to be able to do this also by the definition of convergence).
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    Alternative proof of a theorem in Rudin

    Yea, but p_{n_k} is indexed by n and k. And in my construction, I write for each q_n there is a p_{n_k} for which something holds. So each p_{n_k} explicitly corresponds to each q_n..
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    Alternative proof of a theorem in Rudin

    But that's not now I defined p_{n_k}. Each index, n_k, corresponds to a specific n; you can't fix p_{n_k} but then alter the index of the q in the expression like that.
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    Alternative proof of a theorem in Rudin

    Could you elaborate? I only see that my construction shows that d(p_{n_1}, q)<2, d(p_{n_2}, q)<1, \ldots, d(p_{n_k}, q)<\frac{2}{k}.
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    Conservative Vector Field: Understanding Circulation and Potential Function

    * I mean the potential function \phi(x, y)=\arctan(\frac{y}{x})+C.
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    Conservative Vector Field: Understanding Circulation and Potential Function

    Do you remember the trig identity \arctan(x)+\arctan(\frac{1}{x})=\frac{\pi}{2}? Your expressions are in fact consistent :P The potential function \arctan(x)+C works fine.
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    Alternative proof of a theorem in Rudin

    In Rudin, theorem 3.7 is that: The set of subsequential limits of a sequence \{p_n\} in a metric space X is a closed subset of X. I tried proving this myself, and my proof came out to be similar to the one in Rudin, yet to me, simpler. I wanted to post it and ask if it seems correct...
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