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Math Help - differentiable function /bounded sequence

  1. #1
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    differentiable function /bounded sequence

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

     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 {  y_n } that converges to y and that f(  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 {  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:

     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 {  y_n } that converges to y and that f(  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, f(y)=0 and 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, f(y)=0 and f'(y) = 0 right?
    Well that was the question, right?
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