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    functions

    Let $\displaystyle f^{(n)}(x) := 2n $ when $\displaystyle x \in [1/2n, 1/n] $ and $\displaystyle f^{(n)}(x) := 0 $ for all other values of $\displaystyle x $. Why does $\displaystyle f^{(n)} $ converge pointwise to $\displaystyle f(x) := 0 $?
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    Quote Originally Posted by manjohn12 View Post
    Let $\displaystyle f^{(n)}(x) := 2n $ when $\displaystyle x \in [1/2n, 1/n] $ and $\displaystyle f^{(n)}(x) := 0 $ for all other values of $\displaystyle x $. Why does $\displaystyle f^{(n)} $ converge pointwise to $\displaystyle f(x) := 0 $?

    Pointwise convergence depends on both n and x.

    So for any fixed x if it is out side of $\displaystyle [0,1]$

    $\displaystyle f^{n}=0$ for all n

    so we are okay.

    Now for any $\displaystyle x \in [0,1]$

    There exits an N such that for all n> N $\displaystyle f^{n}=0$

    The Key point is that it depends both on x and n. I hope this helps
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    Quote Originally Posted by TheEmptySet View Post
    Pointwise convergence depends on both n and x.

    So for any fixed x if it is out side of $\displaystyle [0,1]$

    $\displaystyle f^{n}=0$ for all n

    so we are okay.

    Now for any $\displaystyle x \in [0,1]$

    There exits an N such that for all n> N $\displaystyle f^{n}=0$

    The Key point is that it depends both on x and n. I hope this helps
    I think you mean $\displaystyle [1/2, 1] $. Also why is this true: There exits an N such that for all n> N $\displaystyle f^{n}=0$? Because $\displaystyle f^{(100)}(x) := 200 $ for all $\displaystyle x \in [1/2, 1 ] $.
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    Quote Originally Posted by manjohn12 View Post
    I think you mean $\displaystyle [1/2, 1] $. Also why is this true: There exits an N such that for all n> N $\displaystyle f^{n}=0$? Because $\displaystyle f^{(100)}(x) := 200 $ for all $\displaystyle x \in [1/2, 1 ] $.
    pay attention to your interval. $\displaystyle [1/2,1]$ is the BIGGEST it will be. for your example, $\displaystyle x = 100$, we have the interval $\displaystyle [1/200, 1/100]$. this is why TES considers the interval $\displaystyle (0,1]$, because $\displaystyle [1/(2n), 1/n] \to \{ 0 \}$ (note, 0 is never actually in the interval)

    so as $\displaystyle n$ gets large, the interval for which $\displaystyle f^{(n)} \ne 0$ gets smaller and smaller...
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    Quote Originally Posted by manjohn12 View Post
    I think you mean $\displaystyle [1/2, 1] $. Also why is this true: There exits an N such that for all n> N $\displaystyle f^{n}=0$?
    I meant $\displaystyle (0,1]$

    What we have is a sequnce of intervals.

    When n=1 $\displaystyle [\frac{1}{2},1]$
    n=2 $\displaystyle [\frac{1}{4},\frac{1}{2}]$
    n=3 $\displaystyle [\frac{1}{6},\frac{1}{3}]$
    n=4 $\displaystyle [\frac{1}{8},\frac{1}{4}]$
    ...

    So eventally every x in $\displaystyle (0,1]$

    Is not in $\displaystyle [\frac{1}{2n},\frac{1}{n}]$

    so $\displaystyle f^{n}=0$
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