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Math Help - Comparison Test

  1. #1
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    Comparison Test

    I was wondering if anyone could help me with the following problems.

    I am working with the function f(x) = Σ ({n!*x})/n! where n goes from 0 to infintity and {x} = distance from x to the nearest integer.

    a) I need to use the comparison test to show that f(x) converges. I know that I can compare it to Σ 1/n!

    b) I need to show that f(x) is continuous at x=a. I know that I can chose an N so that Σ ({n!*x})/n! < ε (n goes from N+1 to infinity), and then use
    Σ ({n!*x})/n! - Σ ({n!*a})/n! (n goes from N to infinity).
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  2. #2
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    Quote Originally Posted by CoraGB View Post
    I was wondering if anyone could help me with the following problems.

    I am working with the function f(x) = Σ ({n!*x})/n! where n goes from 0 to infintity and {x} = distance from x to the nearest integer.

    a) I need to use the comparison test to show that f(x) converges. I know that I can compare it to Σ 1/n!

    Exactly. The function \{x\}\,,\,\,x\in\mathbb{R} is also called the fractional part of x function, and it is true that \forall\,x\in\mathbb{R}\,,\,\,0\leq \{x\}\leq 1, so \frac{\{n!x\}}{n!}\leq \frac{1}{n!} and since the series of the right hand sequence clearly converges (and everything's non-negative here) we get that our series converges.


    b) I need to show that f(x) is continuous at x=a. I know that I can chose an N so that Σ ({n!*x})/n! < ε (n goes from N+1 to infinity), and then use
    Σ ({n!*x})/n! - Σ ({n!*a})/n! (n goes from N to infinity).

    Even better, perhaps: you can choose \forall\,\epsilon>0\,\,N_{\epsilon}\in\mathbb{N}\,  \,s.t.\,\,\forall\,n>N_{\epsilon}\,,\,\,\sum\limit  s_{n=N_{\epsilon}+1}^\infty\frac{1}{n!}<\epsilon, and then:

    \sum\limits_{n=N_{\epsilon}+1}^\infty\frac{\{n!x\}  }{n!}-\sum\limits_{n=N_{\epsilon}+1}^\infty\frac{\{n!a\}  }{n!}= \sum\limits_{n=N_{\epsilon}+1}^\infty\frac{1}{n!}\  left(\{n!x\}-\{n!a\}\right)\leq\sum\limits_{n=N_{\epsilon}+1}^\  infty\frac{1}{n!}<\epsilon

    Tonio
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  3. #3
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    Thank you so much, especially with part b.

    I wwas wondering if you could then help me with showing that f(x) is not differentiable at x=0.

    All I know so far is that with a sequence a_n goes to zero,
    [g(a_n) g(0)]/[a_n 0] = g(a_n)/a_n≥ 1
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  4. #4
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    Quote Originally Posted by CoraGB View Post
    Thank you so much, especially with part b.

    I wwas wondering if you could then help me with showing that f(x) is not differentiable at x=0.

    All I know so far is that with a sequence a_n goes to zero,
    [g(a_n) g(0)]/[a_n 0] = g(a_n)/a_n≥ 1

    If the function's differentiable at zero then we have:

    f'(0):=\lim_{x\to 0}\frac{f(x)-f(0)}{x}=\lim_{x\to 0}\sum\limits_{n=0}^\infty\frac{\{n!x\}}{xn!} , and this limit must exist no matter how we choose to make x\rightarrow 0, so:

    1) Choose to evaluate the limit through the sequence \left\{\frac{1}{k}\right\}_{k=1}^\infty, and check what happens with the sum when n\geq k (you can do this since convergence/divergence doesn't change when we throw away a finite number of summands in a series)

    2) Now choose to evaluate the limit through the sequence \left\{\frac{\pi}{k}\right\}_{k=1}^\infty...

    Tonio
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