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Cauchy functions problem for Calculus II

  1. Jul 26, 2011 #1
    So I've apparently been given an assignment on Cauchy functions (it says here on the title), but I have no idea what that means. Nevertheless, here's my attempt to solve this problem:

    Given (1):
    [itex]f(x)=\frac{k}{\pi}\times\frac{1}{k^{2}+(x-\beta)^{2}}[/itex]

    and (2):
    [itex]\hat{\beta}_{k}=arg\ max_{\beta} (\frac{k}{\pi})^{N}\ \prod^{N}_{i=1}\frac{1}{k^{2}+(x_{i}-\beta)^{2}}[/itex]

    Describe the effect of k on (2) corresponding to (1) as shown here:http://img854.imageshack.us/img854/2724/graphjz.th.jpg [Broken]. The function in (2) can be read as the values of β that would maximize the function in (1) or argument that would maximize (1).

    I'm assuming that the question implies N=1 and hence k=1 for (2), and what [itex]arg\ max[/itex] means here is crudely what [itex]\beta[/itex] value would achieve the maximum point of the function. I'm convinced that [itex]\hat{\beta}_{k}[/itex] is somehow related to [itex]{arg\ min}_{\beta} \sum^{N}_{i=1} |\beta-x_{i}|[/itex] hence [itex]x_{i}\approx\hat{\beta}_{k}[/itex], since [itex]|\frac{1}{a}|<|\frac{1}{b}|[/itex] iff [itex]|a|>|b|[/itex] for all [itex]a,b\in\mathbb{R}[/itex] and [itex]a,b\not =0[/itex]; and that [itex]|a|<|a|+|b|[/itex] for all [itex]a,b\in\mathbb{R}[/itex] and [itex]b\not =0\ \Rightarrow\ b\rightarrow 0[/itex].

    Here's where I'm stuck, I don't see how k could affect [itex]\hat{\beta}_{k}[/itex], unless [itex]\hat{\beta}_{k}[/itex] is affected by the maximum point of the function. So how is the maximum point related to [itex]\hat{\beta}_{k}[/itex]?

    EDIT: Added the complete question and changed (2). There's a typo in the equation. Sorry for not doing it earlier.
     
    Last edited by a moderator: May 5, 2017
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  3. Jul 26, 2011 #2

    micromass

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    What exactly is the question? Something like

    Doesn't make any sense to me.

    Is the question to actually find [itex]\hat{\beta_k}[/itex]?? In that case, I'd start by taking logarithms.
     
  4. Jul 26, 2011 #3
    The complete question reads like this: "Describe the effect of k on (2) corresponding to (1). The function in (2) can be read as the values of β that would maximize the function in (1) or argument that would maximize (1)."

    There's an accompanying image:
    http://img854.imageshack.us/img854/2724/graphjz.th.jpg [Broken]

    From what I gather from this, I think it wants to find how k (not K) can change [itex]\hat{\beta}_{k}[/itex].
     
    Last edited by a moderator: May 5, 2017
  5. Jul 26, 2011 #4
    Major Update!

    So I've spotted an error in the question, namely (2), which states k as a constant and a variable at the same time. I've reconciled this with my lecturer and it's confirmed that there's a misprint in the equation. (2) should actually look like this:
    [itex]\hat{\beta}_{k}=arg\ max_{\beta} (\frac{k}{\pi})^{N}\ \prod^{N}_{i=1}\frac{1}{k^{2}+(x_{i}-\beta)^{2}}[/itex]
    The difference is in the [itex]\prod^{N}_{i=1}[/itex].

    I've changed post #1. Problem is I'm still stuck.
     
    Last edited: Jul 26, 2011
  6. Jul 27, 2011 #5
    Correct me if I'm wrong, but the figure attached seems to hint at [itex]K=max\ \{ (\frac{k}{\pi})^{N}\ \prod_{i=1}^{N}\ \frac{1}{k^2+(x_{i}-\beta)^2}\}[/itex].

    Then [itex]x_{i}\rightarrow\beta\ \Rightarrow K=(\frac{1}{k\pi})^{N}\ \Rightarrow k^{N}\propto \frac{1}{K}[/itex] and [itex]k\not = 0[/itex]. (I'm sure this is flawed; surely not all [itex]x_{i}\rightarrow\beta[/itex]. Is it possible to be more accurate?)

    The image indicates that different [itex]K[/itex] do give about different [itex]\hat{\beta}_{k}[/itex], but how? It does seem like it has something to do with the diferent values of [itex]\sqrt{|x_{i}-\hat{\beta}_{k}|^{2}+K^{2}}[/itex]... Perhaps something relating to [itex]arg\ min_{\beta}\ \sum_{i=1}^{N}\ \sqrt{|x_{i}-\beta|^{2}+K^{2}}[/itex] ?

    I'm going about in circles.

    EDIT: My lecturer says that one of the solutions involved the usage of Newton's Method. How I have the faintest idea.

    EDIT2: The assumption is no longer k=1 in post #1, since we have changed (2). Can't edit my first post so I'll just post it here.
     
    Last edited: Jul 27, 2011
  7. Jul 27, 2011 #6
    Lecturer revealed the solution:

    [itex]\hat{\beta}_k=arg\ max_\beta\ \prod_{i=1}^{N}\ \frac{1}{k^{2}+(x_{i}-\beta)^{2}}[/itex]
    [itex]\Rightarrow \hat{\beta}_k=arg\ min_\beta\ \prod_{i=1}^{N}\ k^{2}+(x_{i}-\beta)^{2}[/itex]
    [itex]\Rightarrow \hat{\beta}_k=arg\ min_\beta\ \sum_{i=1}^{N}\ ln\ (k^{2}+(x_{i}-\beta)^{2})[/itex]
    let [itex]g(\beta)=\sum_{i=1}^{N}\ ln\ [k^{2}+(x_{i}-\beta)^{2}][/itex]
    since [itex]f(x)=g(x)+h(x) \Rightarrow f'(x)=g'(x)+h'(x)[/itex]
    then [itex]g'(\beta)=\sum_{i=1}^{N}\ \frac{-2(x_{i}-\beta)}{k^{2}+(x_{i}-\beta)^2}[/itex]
    to find the critical point, [itex]g'(\beta)=0[/itex]
    hence [itex]\sum_{i=1}^{N}\ \frac{-2}{k^{2}+(x_{i}-\beta)^2}\cdot (x_{i}-\beta)=0[/itex]
    [itex]\Rightarrow \sum_{i=1}^{N}\ [\frac{-2}{k^{2}+(x_{i}-\beta)^2}\cdot x_{i} - \frac{-2}{k^{2}+(x_{i}-\beta)^2}\cdot \beta] = 0[/itex]
    [itex]\Rightarrow \sum_{i=1}^{N}\ [\frac{2}{k^{2}+(x_{i}-\beta)^2}\cdot \beta - \frac{2}{k^{2}+(x_{i}-\beta)^2}\cdot x_{i}] = 0[/itex]
    Using Newton's Method to approximate [itex]\hat{\beta}_{k}[/itex],
    [itex]\hat{\beta}_{n+1}=\frac{\frac{2}{k^{2}+(x_{i}-\beta_{n})^2}\cdot x_{i}}{\frac{2}{k^{2}+(x_{i}-\beta_{n})^2}}[/itex].

    I don't follow all of the steps, especially how step 2 implies step 3, and how the Newton's method is used here, but most among these confusion is how the cheese does the solution answer how [itex]k[/itex] affects [itex]\hat{\beta}_{k}[/itex]? The solution only seems to find what [itex]\hat{\beta}_{k}[/itex] is...
     
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