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How to understand problems in analysis better.

  1. Jun 2, 2014 #1
    Hello

    One problem I often think I have is that I am able to solve a problem, but not really understanding the idea behind the problem, or what is happening. Is there a way to become better at this, or to train yourself up?, are there any books that would help me with this?

    For instance:
    1. There might be a very hard convergence problem. But to prove convergence, you only need an epsilon proof. So often if you just try different things, I'll get a solution that is correct, but I don't feel I've understood why it works.

    2. This is not an analysis problem, more a differential equation, but it is to explain what I mean. you could prove Euler's method for differential equation, and prove that the error-term is smaller than a value etc.. But to understand what really has happened, you need to think about tangents and moving small distances forward on these tangents.

    In Calculus I never thought about this problem, because it is mostly very visual what you do. How did you better understand what the problems were?, not only proving what the problem asked for.
     
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  3. Jun 2, 2014 #2

    jbunniii

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    This is a very general question since there are many different types of convergence. It's hard to give advice which will be useful in all situations. Can you give a particular example of the sort of problem you don't fully understand?
     
  4. Jun 3, 2014 #3
    Hello

    It was maybe too general yes, sorry. For instance, lets say that you want to prove that a vector space is complete if it's unit sphere is complete. Then you start with a cauchy, and must prove somehow that the cauchy converges to a point. And when you sit down and try, and use that the unit sphere is complete, you can make a sequence in the unit-sphere etc by dividing on it's length etc., and you get convergence to a point. So it is very technical, and you get convergence to a point you find, but the idea behind it can be a little tricky I think, I mean, why does only completeness of a sphere imply completeness of an entire vector-space, why is it logical that it have to be like this?
     
  5. Jun 3, 2014 #4

    jbunniii

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    [Edited for clarification]

    Here's an attempt to give you some insight into why this works.

    We can motivate this result by first considering the case of finite-dimensional normed vector space. In that case, the key difference between the unit sphere and the entire space is that the unit sphere is compact. If we let ##S## denote the unit sphere with the same metric as the vector space, then ##S## is a compact metric space.

    The nice thing about a compact metric space is that it is complete. Proof: let ##x_n## be any Cauchy sequence in ##S##. Since ##S## is compact, ##x_n## has a convergent subsequence. Any Cauchy sequence with a convergent subsequence is itself convergent, to the same limit.

    The intuition here is that the only way a Cauchy sequence can fail to converge is if there is a "hole" where its limit should be. Either (1) the sequence and all of its subsequences are converging to the same point, or (2) they are all "converging" to the same hole. Convergence of even one subsequence ensures that (2) is not the case.

    In general, the unit sphere is not compact, but we can add the hypothesis "if the unit sphere is complete" and continue:

    The reason that working with the unit sphere is sufficient is because we can project any sequence onto the unit sphere: if ##x_n## is any sequence of nonzero vectors, then ##x_n/\|x_n\|## is a sequence on the unit sphere. If ##x_n## is Cauchy, then ##x_n/\|x_n\|## is Cauchy and lies in ##S##. (Exception: if ##x_n \rightarrow 0## then this is not the case, but this special case is trivial to handle separately.) Therefore ##x_n/\|x_n\|## converges to some point ##x## with ##\|x\| = 1##. It follows that ##x_n## converges to ##x(\lim_{n \rightarrow \infty}\|x_n\|)##. The limit in the parentheses exists because if ##x_n## is Cauchy then ##\|x_n\|## is also Cauchy, due to the inequality ##| \|a\| - \|b\| | \leq \|a - b\|##. And since ##\|x_n\|## is a sequence of real numbers, Cauchy implies convergent. (The real numbers are complete.)
     
    Last edited: Jun 3, 2014
  6. Jun 4, 2014 #5
    Thank you very much for your insight.
     
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