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Math Help - Sequence

  1. #1
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    Sequence

    \forall n\in\mathbb{Z}^+, let f_n be the function on the interval from [0,1] by f_n(x)=\frac{x^n}{1+x^n}. Which of the following statements are true?

    1. The sequence { f_n} converges pointwise on [0,1] to a limit function f.
    \lim_{n\to\infty}\frac{x^n}{1+x^n}=\lim_{n\to\inft  y}\frac{nx^{n-1}}{nx^{n-1}}=0=f(x)

    \lim_{n\to\infty}\frac{1^n}{1+1^n}=\frac{1}{2}=f(x  )
    True

    2. The sequence { f_n} converges uniformly on [0,1] to a limit function f.
    False

    3. \lim_{n\to\infty}\int_0^1f_n(x)dx=\int_0^1(\lim_{n  \to\infty}f_n(x))dx
    Don't know
    Last edited by dwsmith; May 22nd 2010 at 07:16 PM.
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  2. #2
    Senior Member roninpro's Avatar
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    Quote Originally Posted by dwsmith View Post
    \forall n\in\mathbb{Z}^+, let f_n be the function on the interval from [0,1] by f_n(x)=\frac{x^n}{1+x^n}. Which of the following statements are true?

    1. The sequence { f_n} converges pointwise on [0,1] to a limit function f.
    \lim_{n\to\infty}\frac{x^n}{1+x^n}=\lim_{n\to\inft  y}\frac{nx^{n-1}}{nx^{n-1}}=1
    True
    This isn't quite correct. You need to consider two cases: either x\geq 1 or x<1. Try recalculating your limit.

    Also, did you have any ideas on the other two?
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  3. #3
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    Quote Originally Posted by roninpro View Post
    This isn't quite correct. You need to consider two cases: either x\geq 1 or x<1. Try recalculating your limit.

    Also, did you have any ideas on the other two?
    If x is less than 1, there would be \frac{-x}{-x} which is still 1. Same situation if x is positive.
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  4. #4
    Senior Member roninpro's Avatar
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    Sorry - I meant to write |x|<1 and |x|\geq 1.

    The important bit is this: if x is fractional, then x^n\to 0. Therefore, \frac{x^n}{1+x^n}\to 0.
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    Quote Originally Posted by roninpro View Post
    Sorry - I meant to write |x|<1 and |x|\geq 1.

    The important bit is this: if x is fractional, then x^n\to 0. Therefore, \frac{x^n}{1+x^n}\to 0.
    If it converges to 0 for all 0<x<1, 1 is still true though.
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  6. #6
    Senior Member roninpro's Avatar
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    Yes, that's right. But your limit was incorrect and will affect the way you do part two and three.

    For part two, you might try considering that there is an asymptote at x=-1 when n is odd.
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  7. #7
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    Quote Originally Posted by roninpro View Post
    Yes, that's right. But your limit was incorrect and will affect the way you do part two and three.

    For part two, you might try considering that there is an asymptote at x=-1 when n is odd.
    What would be an appropriate Mn for two to use?
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  8. #8
    Senior Member roninpro's Avatar
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    What do you mean by Mn?
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  9. #9
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    Quote Originally Posted by roninpro View Post
    What do you mean by Mn?
    |f_n(x)|\leq M_n \ \forall x
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  10. #10
    Senior Member roninpro's Avatar
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    Are you trying to use the Weierstrass M-Test? It isn't applicable here since we are not looking at a function defined by an infinite series.

    You need to use the definition of uniform convergence: Given any \varepsilon>0, there is N so that whenever n>N, |f_n(x)-f(x)|<\varepsilon for all x\in [0,1]. (Now I realise that you are restricted to [0,1] - sorry for the confusion earlier.) Can you find such an N? You should take the point x=1 into consideration.
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  11. #11
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    Quote Originally Posted by roninpro View Post
    Are you trying to use the Weierstrass M-Test? It isn't applicable here since we are not looking at a function defined by an infinite series.

    You need to use the definition of uniform convergence: Given any \varepsilon>0, there is N so that whenever n>N, |f_n(x)-f(x)|<\varepsilon for all x\in [0,1]. (Now I realise that you are restricted to [0,1] - sorry for the confusion earlier.) Can you find such an N? You should take the point x=1 into consideration.
    If x=1, then f_n(1)=\frac{1}{2}.

    |\frac{1}{2}-0|<\varepsilon
    False.
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  12. #12
    Senior Member roninpro's Avatar
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    This is the right idea, but still not exactly the case. The function converges pointwise to 	<br /> <br />
f(x) = \begin{cases}<br /> <br />
  0 & \text{for $x<1$} \\<br /> <br />
  \frac{1}{2} & \text{for $x=1$}<br />
\end{cases}. So, it turns out for all n, |f_n(1)-f(1)|=0<\varepsilon.

    Have a look at a plot of the sequence below and give it another shot.
    Attached Thumbnails Attached Thumbnails Sequence-plot.gif  
    Last edited by roninpro; May 22nd 2010 at 07:08 PM. Reason: Fixed function, due to dwsmith
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  13. #13
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    Quote Originally Posted by roninpro View Post
    This is the right idea, but still not exactly the case. The function converges pointwise to     <br /> <br />
f(x) = \begin{cases}<br /> <br />
  0 & \text{for $x<1$} \\<br /> <br />
  1 & \text{for $x=1$}<br />
\end{cases}. So, it turns out for all n, |f_n(1)-f(1)|=0<\varepsilon.

    Have a look at a plot of the sequence below and give it another shot.
    If x<1, then the limit is 0 and if x=1, then the limit is .5.

    x=1
    |\frac{1}{2}-\frac{1}{2}|=0<\varepsilon

    [0,1)
    |[0,1)-0|=[0,1)<\varepsilon
    which isn't always true.

    Hence, 2 is false?
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  14. #14
    Senior Member roninpro's Avatar
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    Yes, part two is false.

    Your argument needs to be refined a little bit: you need to find a point that "floats" away from our function for any n. Since all of the f_n are continuous, there exists a point y_n\in [0,1] so that f_n(y_n)=\frac{1}{4}. Since this is the case, we have |f_n(y_n)-f(y_n)|=|\frac{1}{4}-0|=\frac{1}{4}. Therefore, we can never find n large enough so that |f_n(x)-f(x)|<\varepsilon=\frac{1}{4} for all x\in [0,1]. (i.e. Uniform continuity fails if we try to choose \varepsilon=\frac{1}{4}.)

    In part three, try actually evaluating the integral \lim_{n\to \infty} \int_0^1 \frac{x^n}{1+x^n} \text{ d}x by putting a convenient bound on it.
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