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Constant on [epsilon, 2epsilon] implies f'(0)=K

  1. Dec 7, 2015 #1

    RBG

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    1. The problem statement, all variables and given/known data
    Warning: I realize the title is misleading... the function itself isn't what's constant.
    Mod note: Edited to fix the LaTeX

    If ##f## is a continuous at 0 such that ##\lim_{x \to 0}\frac{f(x)-f(g(x))}{g(x)}=M##, where ##g(x)\to 0## as ##x\to 0## does this generally mean that ##f'(0)=M##? I have been working on the case that ##g(x)=x/2##. It seems like this should be the case since you are saying the secant line on any ##[\epsilon, 2\epsilon]## is constant, so as ##\epsilon## tends to zero, the slope will approach the slope of the tangent line. I've been trying to fudge the definition of continuity to get this and show ##|f'(0)-h(0)|<\epsilon## where ##h(t) = \lim_{x\to t} \frac{f(x)-f(g(t))}{x-g(t)}##

    2. Relevant equations
    Definition of continuity - \epsilon-delta and limit
    Definition of differentiability
    Mean Value Theorem
    Sequences of functions?
     
    Last edited: Dec 7, 2015
  2. jcsd
  3. Dec 7, 2015 #2

    SammyS

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    Hello RBG. Welcome to PF !

    Code (Text):
    Use $$ (Double dollar sign.) to set off LaTeX on an individual line.
    Use ## to set off LaTeX  for in-line style.
    As follows:

    If ##f## is a continuous at 0 such that $$\underset{x\to 0}{\lim}\frac{f(x)-g(x)}{g(x)}=M\,,$$ does this generally mean that ##f'(0)=M##?
     
  4. Dec 7, 2015 #3

    RBG

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    Thank you! Also, I edited my question to what I meant. I am really more just interested in the ##g(x)=\frac{x}{k}## where ##k\in\mathbb{N}## case.
     
  5. Dec 7, 2015 #4

    PeroK

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    $$\underset{x\to 0}{\lim}\frac{f(x)-x/k}{x/k}=M \ \Rightarrow \ \underset{x\to 0}{\lim}\frac{f(x)}{x} = \frac{M+1}{k}$$
     
  6. Dec 7, 2015 #5

    Ray Vickson

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    Your condition does not imply ##f'(0) = M##. For example, take ##f(x) = (M+1) g(x) \, \forall \, x##, which certainly satisfies your limit condition. If you take ##g(x) = \sqrt{|x|}## you will see that ##f'(0)## need not even exist.

    For the case of interest, namely, ##g(x) = kx## you can compute ##f'(0)## and see what you get.
     
  7. Dec 7, 2015 #6

    RBG

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    Woops! Man, for a first post I am messing up a lot. I meant what my edits say, that is ##f(g(x))## not just ##g(x)## in the numerator!
     
  8. Dec 8, 2015 #7

    PeroK

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    ##f'(0) = \underset{x\to 0}{\lim}\frac{f(x/k)-f(0)}{x/k} \ne \underset{x\to 0}{\lim}\frac{f(x) - f(x/k)}{x/k}##

    For example. Try ##f(x) = x##:

    ##\underset{x\to 0}{\lim}\frac{f(x) - f(x/k)}{x/k} = k - 1 \ne f'(0)## (unless ##k = 2##)
     
  9. Dec 8, 2015 #8

    RBG

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    So if we're in the ##k=2## case and the only thing I know about ##f## is that it's continuous at ##0##, how would I show that ##f'(0)=M##?
     
  10. Dec 9, 2015 #9

    PeroK

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    See Samy_A's post below. ##k = 2## is a special case.
     
    Last edited: Dec 9, 2015
  11. Dec 9, 2015 #10

    Samy_A

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    Now I'm confused too.

    Lets assume that ##f## is differentiable at ##x=0##, and ##M=\underset{x\to 0}{\lim}\frac{f(x) - f(x/2)}{x/2}##
    ##\frac{f(x)-f(0)}{x}=\frac{f(x)-f(x/2)+f(x/2)-f(0)}{x}=\frac{f(x)-f(x/2)}{x}+\frac{f(x/2)-f(0)}{x}=\frac{1}{2}\frac{f(x)-f(x/2)}{x/2}+\frac{1}{2}\frac{f(x/2)-f(0)}{x/2}##
    Taking the limit for ##x \to 0##, we get ##f'(0)=\frac{1}{2}M+\frac{1}{2}f'(0)##, so ##f'(0)=M##.
     
  12. Dec 9, 2015 #11

    Samy_A

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    But I assumed that ##f## is differentiable at ##x=0##. What he wants to prove is that if ##M=\underset{x\to 0}{\lim}\frac{f(x) - f(x/2)}{x/2}## and ##f## is continuous at ##x=0##, then ##f## is also differentiable in 0.

    For now, I find neither a proof, nor a counterexample.
     
    Last edited: Dec 9, 2015
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