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Math Help - Prove this derivative is continuous

  1. #1
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    Prove this derivative is continuous

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

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

    Define g: U \rightarrow \mathbb {R} by g(x)= \frac {f(x)}{x} if x \neq 0 and  \lim _{ h \rightarrow 0 } \frac {f(h)}{h} if  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 f(x)=xg(x), implies that for x \neq 0 , we would have f'(x)=xg'(x)+g(x), so we have g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x} .

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

    In the case that x \neq 0:

    Given  \epsilon > 0 , pick  \delta > 0 , then for each  x \in U and |x-x_0|< \delta , we have |g'(x)-g'(x_0)|=| \frac {f'(x)- \frac {f(x)}{x}}{x} - \frac {f'(x_0)- \frac {f(x_0)}{x}}{x}| = \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 U be a neighborhood of 0 in  \mathbb {R} .

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

    Define g: U \rightarrow \mathbb {R} by g(x)= \frac {f(x)}{x} if x \neq 0 and  \lim _{ h \rightarrow 0 } \frac {f(h)}{h} if  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 f(x)=xg(x), implies that for x \neq 0 , we would have f'(x)=xg'(x)+g(x), so we have g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x} .

    For x=0, we have 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 g(0)=f'(0) (the limit defining g(0) exists because f is differentiable at 0).
    - Your last limit is incorrect. You can't just replace g(0) by \frac{f(h)}{h}. You could write instead: for any h\neq 0 in U,
    \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 \frac{f''(0)}{2}.

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

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

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

    Proof.

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

    Then g'(0)= \lim _{h \rightarrow 0} \frac {f(h)-hf''(0)}{h^2}
    = \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 t \rightarrow 0 as h approaches to 0.

    So I have  g'(0) = f''(0).

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

    2. Show that g' is continuous on U.

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

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

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

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

    So all I have left is: \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 \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 \frac{g(h)-g(0)}{h}=\frac{\frac{f(h)}{h}-f'(0)}{h}=\frac{f(h)-hf'(0)}{h^2} converges to <br />
\frac{f''(0)}{2}<br />

    Proof.

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

    Then g'(0)= \lim _{h \rightarrow 0} \frac {f(h)-hf'(0)}{h^2}
    = \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 t \rightarrow 0 as h approaches to 0.

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

    First, since
    <br />
g'(x)= \frac {f'(x)- \frac {f(x)}{x} }{x}<br />
, both f and f' are continuous, and x \neq 0 , so g' must be continuous on <br />
U\setminus\{0\}<br />
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 f'(h)=f'(0)+hf''(s), \ \ \ \ \ 0<s<h

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