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  1. #1
    Senior Member slevvio's Avatar
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    Functional

    Hi everyone I was wondering if anybody could help me with this question; I have done the first 2 parts and a seperate 4th part but part 3 is causing me mathematical pain.

    Suppose that $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R} $

    satisfies $\displaystyle f(t/2) = f(t)/2, t \in \mathbb{R} $.

    1) Find f(0).

    [easy to show that f(0) = 0 ]

    2) Show that $\displaystyle f(2^{-n} t) = 2^{-n}f(t) \forall t \in \mathbb{R} \forall n \in \mathbb{N} $

    [can be done easily by induction]

    Now suppose that f is differentiable at 0.

    3) Show that f(t) = t f'(0) for all $\displaystyle t \in \mathbb{R} $ [Hint: For fixed t, consider the sequence $\displaystyle \{ f(2^{-n}t) / (2^{-n} t) \} $

    OK, we know that f is differentiable at 0, so

    $\displaystyle \lim_{t \rightarrow 0} \frac{f(t)}{t} = f'(0) $

    and for a fixed t, the above sequence simplifies to $\displaystyle \{ \frac{f(t)}{t} \} $ which is a constant sequence, but how does this help me? I guess I can say that

    $\displaystyle \lim_{n \rightarrow \infty} \frac{f(t)}{t} = \frac {f(t)}{t} $


    Any help would be appreciated
    Last edited by slevvio; Nov 7th 2009 at 04:03 PM. Reason: accidentally posted
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  2. #2
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    Quote Originally Posted by slevvio View Post
    Hi everyone I was wondering if anybody could help me with this question; I have done the first 2 parts and a seperate 4th part but part 3 is causing me mathematical pain.

    Suppose that $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R} $

    satisfies $\displaystyle f(t/2) = f(t)/2, t \in \mathbb{R} $.

    1) Find f(0).

    [easy to show that f(0) = 0 ]

    2) Show that $\displaystyle f(2^{-n} t) = 2^{-n}f(t) \forall t \in \mathbb{R} \forall n \in \mathbb{N} $

    [can be done easily by induction]

    Now suppose that f is differentiable at 0.

    3) Show that f(t) = t f'(0) for all $\displaystyle t \in \mathbb{R} $ [Hint: For fixed t, consider the sequence $\displaystyle \{ f(2^{-n}t) / (2^{-n} t) \} $

    OK, we know that f is differentiable at 0, so

    $\displaystyle \lim_{t \rightarrow 0} \frac{f(t)}{t} = f'(0) $

    and for a fixed t, the above sequence simplifies to $\displaystyle \{ \frac{f(t)}{t} \} $ which is a constant sequence, but how does this help me? I guess I can say that

    $\displaystyle \lim_{n \rightarrow \infty} \frac{f(t)}{t} = \frac {f(t)}{t} $


    Any help would be appreciated
    If f is differentiable at 0 then, if we fix t, as $\displaystyle 2^{-n}t$ goes to 0 as $\displaystyle n$ goes to infinity we have

    $\displaystyle f'(0)=\lim_{n\to\infty}\frac{f(2^{-n}t)-f(0)}{2^{-n}t}=\lim_{n\to\infty}\frac{2^{-n}f(t)}{2^{-n}t}=\frac{f(t)}{t}$
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  3. #3
    Senior Member slevvio's Avatar
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    Is it correct to say this?

    $\displaystyle f'(0) = \lim_{t \rightarrow 0} \frac { f(t) - f(0) }{t-0} = \lim_{2^{-n}t \rightarrow 0} \frac { f(2^{-n}t) - f(0) }{2^{-n}t-0} = \lim_{n \rightarrow \infty} \frac { f(2^{-n}t) - f(0) }{2^{-n}t-0} $

    I'm getting confused about fixing t. When do we do this, is it at the start? I don't really understand because we are looking what happens when t approaches zero so how can we fix it?
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  4. #4
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    Quote Originally Posted by slevvio View Post
    Is it correct to say this?

    $\displaystyle f'(0) = \lim_{t \rightarrow 0} \frac { f(t) - f(0) }{t-0} = \lim_{2^{-n}t \rightarrow 0} \frac { f(2^{-n}t) - f(0) }{2^{-n}t-0} = \lim_{n \rightarrow \infty} \frac { f(2^{-n}t) - f(0) }{2^{-n}t-0} $

    I'm getting confused about fixing t. When do we do this, is it at the start? I don't really understand because we are looking what happens when t approaches zero so how can we fix it?
    No, it is not formally correct. Your problem is that you are mixing t as variable in the definition of derivative and t as a fixed value. Let say, if f is differentiable at 0 then

    $\displaystyle
    f'(0) = \lim_{h \rightarrow 0} \frac { f(h) - f(0) }{h-0}
    $

    This implies that for any zero sequence $\displaystyle (h_n)$ we have

    $\displaystyle
    f'(0) = \lim_{n \rightarrow \infty} \frac { f(h_n) - f(0) }{h_n-0}
    $

    Hence, for a fixed t, you can take $\displaystyle h_n=2^{-n}t$ as a particular zero sequence and apply the hypothesis of differentiability at 0 of $\displaystyle f$ to conclude

    $\displaystyle
    f'(0) = \lim_{n \rightarrow \infty} \frac { f(2^{-n}t) - f(0) }{2^{-n}t-0}
    $

    But as you stated in your post you are "discretisnig" the definition of derivative in a confusing way. For the convergence of the limit of a particular sequence you can't deduce differentiability, you need the existence of the limit for all the possible zero sequences.
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