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Thread: Prove this derivative is continuous

  1. #1
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    Prove this derivative is continuous

    Let $\displaystyle U$ be a neighborhood of 0 in $\displaystyle \mathbb {R} $.

    Suppose that $\displaystyle f:U \rightarrow \mathbb {R} $ is twice differentiable , f' and f'' are continuous, and $\displaystyle f(0)=0$.

    Define $\displaystyle g: U \rightarrow \mathbb {R} $ by $\displaystyle g(x)= \frac {f(x)}{x} $ if $\displaystyle x \neq 0 $ and $\displaystyle \lim _{ h \rightarrow 0 } \frac {f(h)}{h} $ if $\displaystyle x = 0 $

    Show that g' exists and is continuous on U.

    Proof so far.

    Claim: g' exists on U.

    By the division theorem, we know that $\displaystyle f(x)=xg(x)$, implies that for $\displaystyle x \neq 0 $, we would have $\displaystyle f'(x)=xg'(x)+g(x)$, so we have $\displaystyle g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x} $.

    For $\displaystyle x=0$, we have $\displaystyle g'(0)= \lim _{h \rightarrow 0 } \frac {g(h)-g(0)}{h} = \lim _{h \rightarrow 0 } \frac { \frac {f(h)}{h} - \frac {f(h)}{h}}{h} = 0 $

    Claim: g' is continuous on U.

    Write $\displaystyle f(x)=f(0)+f'(0)x+f'(t)x^2 \ \ \ \ \ 0<t<x$, so we have $\displaystyle f(x)=f'(0)x+f''(xh)x^2 \ \ \ \ \ 0<h<1$

    In the case that $\displaystyle x \neq 0$:

    Given $\displaystyle \epsilon > 0 $, pick $\displaystyle \delta > 0 $, then for each $\displaystyle x \in U $ and $\displaystyle |x-x_0|< \delta $, we have $\displaystyle |g'(x)-g'(x_0)|=| \frac {f'(x)- \frac {f(x)}{x}}{x} - \frac {f'(x_0)- \frac {f(x_0)}{x}}{x}|$$\displaystyle = \frac {|f'(x)-f'(x_0)|}{x} + \frac {| \frac {f(x)}{x}- \frac {f(x_0)}{x_0}|}{x}$

    I have a feeling I'm very wrong as I'm stuck at this point, any hints? Thank you.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Let $\displaystyle U$ be a neighborhood of 0 in $\displaystyle \mathbb {R} $.

    Suppose that $\displaystyle f:U \rightarrow \mathbb {R} $ is twice differentiable , f' and f'' are continuous, and $\displaystyle f(0)=0$.

    Define $\displaystyle g: U \rightarrow \mathbb {R} $ by $\displaystyle g(x)= \frac {f(x)}{x} $ if $\displaystyle x \neq 0 $ and $\displaystyle \lim _{ h \rightarrow 0 } \frac {f(h)}{h} $ if $\displaystyle x = 0 $

    Show that g' exists and is continuous on U.

    Proof so far.

    Claim: g' exists on U.

    By the division theorem, we know that $\displaystyle f(x)=xg(x)$, implies that for $\displaystyle x \neq 0 $, we would have $\displaystyle f'(x)=xg'(x)+g(x)$, so we have $\displaystyle g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x} $.

    For $\displaystyle x=0$, we have $\displaystyle g'(0)= \lim _{h \rightarrow 0 } \frac {g(h)-g(0)}{h} = \lim _{h \rightarrow 0 } \frac { \frac {f(h)}{h} - \frac {f(h)}{h}}{h} = 0 $
    - Note that in fact $\displaystyle g(0)=f'(0)$ (the limit defining $\displaystyle g(0)$ exists because $\displaystyle f$ is differentiable at 0).
    - Your last limit is incorrect. You can't just replace $\displaystyle g(0)$ by $\displaystyle \frac{f(h)}{h}$. You could write instead: for any $\displaystyle h\neq 0$ in $\displaystyle U$,
    $\displaystyle \frac{g(h)-g(0)}{h}=\frac{\frac{f(h)}{h}-f'(0)}{h}=\frac{f(h)-hf'(0)}{h^2}$,
    and use Taylor's theorem to show that the previous quantity converges to $\displaystyle \frac{f''(0)}{2}$.

    Claim: g' is continuous on U.
    First, you wrote that $\displaystyle g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x} $, so it should be clear (from the hypotheses) that $\displaystyle g'$ is continuous on $\displaystyle U\setminus\{0\}$. The only thing to check is that $\displaystyle \lim_{h\to 0}g'(h)=g'(0)$ ($\displaystyle =\frac{f''(0)}{2}$ because of the previous question) .

    For any $\displaystyle h\neq 0$ (and $\displaystyle h\in U$), $\displaystyle g'(h)=\frac{hf'(h)-f(h)}{h^2}$, so you have to apply Taylor's theorem to $\displaystyle f$ and to $\displaystyle f'$.
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  3. #3
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    My work so far:

    1. Use the Taylor's Theorem to show that $\displaystyle \frac{g(h)-g(0)}{h}=\frac{\frac{f(h)}{h}-f'(0)}{h}=\frac{f(h)-hf'(0)}{h^2}$ converges to $\displaystyle
    \frac{f''(0)}{2}
    $

    Proof.

    By the Taylor's Theorem, write $\displaystyle f(x)=f(0)+f'(0)x+f''(t)x^2 \ \ \ \ \ 0<t<x $, since $\displaystyle f(0)=0$, we have $\displaystyle f(x)=f'(0)x+f''(t)x^2$, implies that $\displaystyle f(h)=f'(0)h+f''(t)h^2 \ \ \ \ \ 0<t<h$.

    Then $\displaystyle g'(0)= \lim _{h \rightarrow 0} \frac {f(h)-hf''(0)}{h^2}$
    $\displaystyle = \lim _{h \rightarrow 0} \frac {f'(0)h+f''(t)h^2-f''(0)h}{h^2}= \lim _{h \rightarrow 0}f''(t)$

    Now, I know that $\displaystyle t \rightarrow 0 $ as h approaches to 0.

    So I have $\displaystyle g'(0) = f''(0)$.

    But I don't have $\displaystyle \frac {f''(0)}{2}$, what did I do wrong?

    2. Show that g' is continuous on U.

    First, since
    $\displaystyle
    g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x}
    $, both f and f' are continuous, and $\displaystyle x \neq 0 $, so g' must be continuous on $\displaystyle
    U\setminus\{0\}
    $ since the sum of two continuous functions is also continuous. (Is this reason enough?)

    Now, to show that g' is continuous on $\displaystyle U \setminus \{ 0 \} $.

    By the Taylor's Theorem, write $\displaystyle f'(h)=f'(0)+hf''(s), \ \ \ \ \ 0<s<h$

    Then $\displaystyle \lim _{h \rightarrow 0 } g'(h) = \lim _{h \rightarrow 0 } \frac {hf'(h)-f(h)}{h^2}$$\displaystyle = \lim _{h \rightarrow 0 } \frac {h[f'(0)+hf''(s)]-[f(0)+hf'(0)+h^2f''(t)]}{h^2}$
    $\displaystyle = \lim _{h \rightarrow 0 } f''(s)-f''(t)- \frac {f(0)}{h^2} $

    So as h approaches 0, both s and t approaches 0 as they are sandwiched in between 0 and h, so $\displaystyle f''(s)-f''(t) \rightarrow f''(0)-f''(0) =0 $

    So all I have left is: $\displaystyle \lim _{h \rightarrow 0 } g'(h) = \lim _{h \rightarrow 0 } \frac {f(0)}{h^2} $

    I must have done something wrong again, because this expression doesn't look converging to g'(0).

    Thank you very much for your help!!!
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  4. #4
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    The only problem is a $\displaystyle \frac{1}{2}$ factor in the Taylor's theorem (cf. wikipedia or whereever):

    Quote Originally Posted by tttcomrader View Post
    My work so far:

    1. Use the Taylor's Theorem to show that $\displaystyle \frac{g(h)-g(0)}{h}=\frac{\frac{f(h)}{h}-f'(0)}{h}=\frac{f(h)-hf'(0)}{h^2}$ converges to $\displaystyle
    \frac{f''(0)}{2}
    $

    Proof.

    By the Taylor's Theorem, write $\displaystyle f(x)=f(0)+f'(0)x+{\color{red}\frac{1}{2}}f''(t)x^2 \ \ \ \ \ 0<t<x $, since $\displaystyle f(0)=0$, we have $\displaystyle f(x)=f'(0)x+{\color{red}\frac{1}{2}}f''(t)x^2$, implies that $\displaystyle f(h)=f'(0)h+{\color{red}\frac{1}{2}}f''(t)h^2 \ \ \ \ \ 0<t<h$.

    Then $\displaystyle g'(0)= \lim _{h \rightarrow 0} \frac {f(h)-hf'(0)}{h^2}$
    $\displaystyle = \lim _{h \rightarrow 0} \frac {f'(0)h+{\color{red}\frac{1}{2}}f''(t)h^2-f'(0)h}{h^2}= \lim _{h \rightarrow 0}{\color{red}\frac{1}{2}}f''(t)$

    Now, I know that $\displaystyle t \rightarrow 0 $ as h approaches to 0.

    So I have $\displaystyle g'(0) = {\color{red}\frac{1}{2}}f''(0)$ because $\displaystyle \color{red}f''$ is assumed to be continuous.
    2. Show that g' is continuous on U.

    First, since
    $\displaystyle
    g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x}
    $, both f and f' are continuous, and $\displaystyle x \neq 0 $, so g' must be continuous on $\displaystyle
    U\setminus\{0\}
    $ since the sum of two continuous functions is also continuous. (Is this reason enough?)

    Now, to show that g' is continuous at 0.

    By the Taylor's Theorem, write $\displaystyle f'(h)=f'(0)+hf''(s), \ \ \ \ \ 0<s<h$

    Then $\displaystyle \lim _{h \rightarrow 0 } g'(h) = \lim _{h \rightarrow 0 } \frac {hf'(h)-f(h)}{h^2}$$\displaystyle = \lim _{h \rightarrow 0 } \frac {h[f'(0)+hf''(s)]-[f(0)+hf'(0)+{\color{red}\frac{1}{2}}h^2f''(t)]}{h^2}$
    $\displaystyle = \lim _{h \rightarrow 0 } f''(s)-{\color{red}\frac{1}{2}}f''(t)- \frac {f(0)}{h^2} $
    then, don't forget $\displaystyle f(0)=0$, and you'll be done.
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