Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Not homework, just curious

  1. Sep 20, 2005 #1
    *Let [itex] f\left( x \right) [/itex] be a twice-differentiable function for which
    [tex] \; \mathop {\lim }\limits_{x \to 0^ - } f\left( x \right) = \infty [/tex]

    Then, is it true that
    [tex] \mathop {\lim }\limits_{x \to 0^ - } f\,{''}\left( x \right) > 0\;? [/tex]

    ----------------------------------------------
    Or a little differently,

    *Let [itex] f\left( x \right) [/itex] be a twice-differentiable function for which
    [tex]\mathop {\lim }\limits_{x \to 0^ - } f\left( x \right) = \infty \;{\text{and }}\forall x < 0,f\,'\left( x \right) > 0 [/tex]

    Then, is it true that
    [tex] \mathop {\lim }\limits_{x \to 0^ - } f\,{''}\left( x \right) > 0\;? [/tex]

    Just curious|:redface:
     
    Last edited: Sep 20, 2005
  2. jcsd
  3. Sep 20, 2005 #2

    VietDao29

    User Avatar
    Homework Helper

    You can prove it by contradiction.
    So let f(x) be a function that is continuous, increasing in [-δ, 0), its first and second derivatives are also continuous in [-δ, 0), where δ > 0, so f(-δ) <> +∞, f'(-δ) <> +/-∞, f''(-δ) <> +/-∞.
    So assume [tex]\lim_{x \rightarrow 0 ^ -} f''(x) <= 0[/tex]
    Now [tex]\lim_{x \rightarrow 0 ^ -} f''(x) = 0[/tex] means that there exists ε > 0, and 0 <= σ < ∞, such that, [itex]\forall x \in (-\varepsilon,\ 0), \ f'(x) \leq \sigma[/itex], where 0 <= σ < ∞
    Since f(-ε) <> +/-∞, so f(0) <= f(-ε) + σε < + ∞.
    Now [tex]\lim_{x \rightarrow 0 ^ -} f''(x) < 0[/tex] means that there exists ε > 0 such that, [itex]\forall x_1, x_2 \in (-\varepsilon,\ 0) \mbox{ and } \ x_1 < x_2 \rightarrow \ f'(x_1) > f'(x_2)[/itex]
    Let [tex]x_2 = \lim_{\beta \rightarrow 0 ^ -} 0 + \beta[/tex]
    Let α be the slope of the line that connects two points (x1, f(x1)), and (x2, f(x2)), so
    f'(-ε) > f'(x1) > α > f'(x2) > 0 (f(x) is increasing), so α <> +/-∞.
    (You can use: [tex]\forall \alpha , \ \beta | \alpha < \beta, \exists \gamma | \ \alpha < \gamma < \beta,\frac{f(\alpha) - f(\beta)}{\alpha - \beta} = f'(\gamma)[/tex] to prove that).
    So f(x2) = f'(x1) + α(x2 - x1) <> + ∞, f(x2) does not tend to infinity.
    (Q.E.D).
    So, if:
    [tex]\lim_{x \rightarrow 0 ^ -} f(x) = + \infty[/tex], then:
    [tex]\lim_{x \rightarrow 0 ^ -} f''(x) > 0[/tex].
    Note that it should be + ∞, (not - ∞).
    Do the same, and you will get:
    [tex]\lim_{x \rightarrow 0 ^ -} f(x) = - \infty[/tex], then:
    [tex]\lim_{x \rightarrow 0 ^ -} f''(x) < 0[/tex].
    To me the 0- is can be change to γ- for a more general case, where [itex]\gamma \neq \pm \infty[/itex]. It's still true.
    Viet Dao,
     
    Last edited: Sep 21, 2005
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Not homework, just curious
  1. Curious identities (Replies: 20)

Loading...