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Thread: Piecewise function with Taylor's Theorem

  1. #1
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    Piecewise function with Taylor's Theorem

    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 $\displaystyle g(0)$ exists and that g' exist and is continuous on U. Find $\displaystyle g(0),g'(0)$

    Proof so far.

    Find $\displaystyle g(0)$:

    By the Taylor's Theorem, I can write $\displaystyle f(x)=f(0)+f'(0)x+f''(t)x^2=f'(0)x+f''(t)x^2$

    Now, $\displaystyle g(0)= \lim _{h \rightarrow 0 } \frac {f(h)}{h} = \lim _{h \rightarrow 0 } \frac {f(h)-f(0)}{h} = f'(0) $

    Is that right so far?

    And how do I find g'(0)? Thank you.
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  2. #2
    Grand Panjandrum
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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 $\displaystyle g(0)$ exists and that g' exist and is continuous on U. Find $\displaystyle g(0),g'(0)$

    Proof so far.

    Find $\displaystyle g(0)$:

    By the Taylor's Theorem, I can write $\displaystyle f(x)=f(0)+f'(0)x+f''(t)x^2=f'(0)x+f''(t)x^2$
    No you can't, there has to be a remainder term.

    Now, $\displaystyle g(0)= \lim _{h \rightarrow 0 } \frac {f(h)}{h} = \lim _{h \rightarrow 0 } \frac {f(h)-f(0)}{h} = f'(0) $

    Is that right so far?
    Since $\displaystyle f(0)=0$:

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

    CB
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  3. #3
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    Do I find g'(0) the same way then?

    $\displaystyle g'(0) = \lim _{h \rightarrow 0 } \frac {g(h)-g(0)}{h} = \lim _{h \rightarrow 0 } \frac { \frac {f(h)}{h}-f'(0)}{h} = \frac {g(0)-g(0)}{h} = 0 $

    Is this right?
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