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Thread: differentiable function /bounded sequence

  1. #1
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    differentiable function /bounded sequence

    Suppose that the function$\displaystyle f:R \rightarrow R $ is differentiable and that there is a bounded sequence $\displaystyle \{x_n\}$ with $\displaystyle {x_n} \neq {x_m}$ if $\displaystyle {n \neq m}$, such that $\displaystyle f(x_n) =0$ for every index n. Show that there is a point $\displaystyle x_0$ at which $\displaystyle f(x_0) = 0$ and $\displaystyle f'(x_0) = 0 $.

    ---> According to Bolzano–Weierstrass theorem, a bounded sequence has a convergent subsequence, but How can this be used to prove the above question.
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  2. #2
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    Quote Originally Posted by serious331 View Post
    Suppose that the function$\displaystyle f:R \rightarrow R $ is differentiable and that there is a bounded sequence $\displaystyle \{x_n\}$ with $\displaystyle {x_n} \neq {x_m}$ if $\displaystyle {n \neq m}$, such that $\displaystyle f(x_n) =0$ for every index n. Show that there is a point $\displaystyle x_0$ at which $\displaystyle f(x_0) = 0$ and $\displaystyle f'(x_0) = 0 $.

    ---> According to Bolzano–Weierstrass theorem, a bounded sequence has a convergent subsequence, but How can this be used to prove the above question.

    You are on the right track:

    Let {$\displaystyle y_n $} be the convergent subsequence and suppose it converges to y. Then I claim 2 things:

    1. f(y) = 0
    2. f'(y) = 0

    1. is easy and follows from continuity of f.

    For 2, we use first principles:

    $\displaystyle f'(y) = \lim_{x\to y} \frac {f(x) - f(y)}{x-y} = \lim_{x\to y} \frac {f(x)}{x-y} $

    But we know a) The limit exists (by assumption) and b) there's a sequence {$\displaystyle y_n $} that converges to y and that f($\displaystyle y_n $) = 0 for every point...
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  3. #3
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    Quote Originally Posted by southprkfan1 View Post
    You are on the right track:

    Let {$\displaystyle y_n $} be the convergent subsequence and suppose it converges to y. Then I claim 2 things:

    1. f(y) = 0
    2. f'(y) = 0

    1. is easy and follows from continuity of f.

    For 2, we use first principles:

    $\displaystyle f'(y) = \lim_{x\to y} \frac {f(x) - f(y)}{x-y} = \lim_{x\to y} \frac {f(x)}{x-y} $

    But we know a) The limit exists (by assumption) and b) there's a sequence {$\displaystyle y_n $} that converges to y and that f($\displaystyle y_n $) = 0 for every point...
    Thank you so much southparkfan..

    So all I had to do was to show that for a point at which the subsequence converges, $\displaystyle f(y)=0$ and $\displaystyle f'(y) = 0$ right?
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  4. #4
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    Quote Originally Posted by serious331 View Post
    Thank you so much southparkfan..

    So all I had to do was to show that for a point at which the subsequence converges, $\displaystyle f(y)=0$ and $\displaystyle f'(y) = 0$ right?
    Well that was the question, right?
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