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Math Help - Infinitely differentiable function

  1. #1
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    Infinitely differentiable function

    Prove that the following function is infinitely differentiable:

    f: \mathbb{R}\rightarrow \mathbb{R}\:\:\: x \mapsto  \lbrace \begin{array}{l}<br />
\nearrow\raise5pt\hbox{$\exp ({\frac{-1}{x^2}}) \:if \:x \ne0$}\\<br />
\searrow\lower5pt\hbox{$0\:if\: x=0}$}<br />
\end{array}

    Thank you!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zadir View Post
    Prove that the following function is infinitely differentiable:

    f: \mathbb{R}\rightarrow \mathbb{R}\:\:\: x \mapsto  \lbrace \begin{array}{l}<br />
\nearrow\raise5pt\hbox{$\exp ({\frac{-1}{x^2}}) \:if \:x \ne0$}\\<br />
\searrow\lower5pt\hbox{$0\:if\: x=0}$}<br />
\end{array}

    Thank you!
    Ha! I just mentioned this to someone. It's evident that f is differentiable everywhere except zero. For zero we merely note that f'(x)=\frac{2}{x^3}e^{\frac{-1}{x^2}} and by induction f^{(n)}(x)=p\left(\frac{1}{x}\right)e^{\frac{-1}{x^2}} for some polynomial. See if you can conclude from there.
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  3. #3
    MHF Contributor chisigma's Avatar
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    It is important to define exactly the concepts of 'derivative' and 'differenziable funtion'. The derivative of a function y(x) in x=x_{0} is the limit [if a limit exists]...

    \displaystyle y^{'}(x_{0}) = \lim_{h \rightarrow 0} \frac{y(x_{0} + h)-y(x_{0})}{h} (1)

    A function y(x) is differentiable in x=x_{0} if we can write...

    \displaystyle \delta y= y(x_{0} + h)-y(x_{0}) = p\ h + h\ \varepsilon(h) (2)

    ... where p is independent from h and \displaystyle \lim_{h \rightarrow 0} \varepsilon (h)=0. Now we consider the function...

    y(x)=\left\{\begin{array}{ll} e^{-\frac{1}{x^{2}}} ,\,\,x \ne 0\\{}\\0 ,\,\, x=0\end{array}\right. (3)

    ... in x=x_{0}. On the basis of the (1) the derivative in x=x_{0} exists and is y^{'} (0)=0 but the function isn't differentiable in x=x_{0} because we are not able to write a relation like (2)...

    Kind regards

    \chi \sigma
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chisigma View Post
    It is important to define exactly the concepts of 'derivative' and 'differenziable funtion'. The derivative of a function y(x) in x=x_{0} is the limit [if a limit exists]...

    \displaystyle y^{'}(x_{0}) = \lim_{h \rightarrow 0} \frac{y(x_{0} + h)-y(x_{0})}{h} (1)

    A function y(x) is differentiable in x=x_{0} if we can write...

    \displaystyle \delta y= y(x_{0} + h)-y(x_{0}) = p\ h + h\ \varepsilon(h) (2)

    ... where p is independent from h and \displaystyle \lim_{h \rightarrow 0} \varepsilon (h)=0. Now we consider the function...

    y(x)=\left\{\begin{array}{ll} e^{-\frac{1}{x^{2}}} ,\,\,x \ne 0\\{}\\0 ,\,\, x=0\end{array}\right. (3)

    ... in x=x_{0}. On the basis of the (1) the derivative in x=x_{0} exists and is y^{'} (0)=0 but the function isn't differentiable in x=x_{0} because we are not able to write a relation like (2)...

    Kind regards

    \chi \sigma
    Spoiler:


    Another way of putting differentiability, opposed to the one you put, is that f' exists at x_0 if

    \displaystyle\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}

    exists. Now, for our f evidently it is differentiable for x\ne 0 and at x=0 we see that

    \displaystyle \lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\frac{e^{\frac{-1}{x^2}}-0}{x}=\lim_{x\to0}\frac{1}{x}e^{\frac{-1}{x^2}}=0

    And thus f'(0) exists and equals zero. Thus,

    f'(x)=\begin{cases}\frac{2}{x^3}e^{\frac{-1}{x^2}} & \mbox{if}\quad x\ne0\\ 0 & \mbox{if}\quad x=0\end{cases}

    So, it's clear that f'' exists for x\ne0 and we see that

    \displaystyle \lim_{x\to0}\frac{f'(x)-f'(0)}{x-0}=\lim_{x\to0}\frac{\frac{2}{x^3}e^{\frac{-1}{x^2}}-0}{x}=\lim_{x\to0}\frac{2}{x^4}e^{\frac{-1}{x^2}}=0

    and thus f''(0) exists and equals zero. So,

    f''(x)=\begin{cases}e^{\frac{1}{x^2}}\left(\frac{4  }{x^6}-\frac{6}{x^4}\right) & \mbox{if}\quad x\ne0\\ 0 & \mbox{if}\quad x=0\end{cases}

    From here we can proceed by induction. Namely, assume that f^{(n)}(x) exists for all x. Then, show, by induction that

    f^{(n)}(x)=\begin{cases}p\left(\frac{1}{x}\right)e  ^{\frac{-1}{x^2}} & \mbox{if}\quad x\ne 0\\ 0 & \mbox{if}\quad x=0\end{cases}

    And thus showing that f^{(n+1)}(0) exists (since clearly f is n+1 times differentiable for x\ne0) reduces to showing that

    \displaystyle \lim_{x\to0}\frac{f^{(n)}(x)-f^{(n)}(0)}{x-0}=\lim_{x\to0}\frac{p\left(\frac{1}{x}\right)e^{\  frac{-1}{x^2}}-0}{x}=\lim_{x\to0}\frac{1}{x}p\left(\frac{1}{x}\ri  ght)e^{\frac{-1}{x^2}}

    exists and, in fact equals zero. Thus it follows by induction that f has derivatives of all orders.


    Remark: This shows that being C^{\infty} does not imply real analyticity since this f is infinitely differentiable at 0 but for any I around 0 we have that \displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(0)x^n}{n!}=\sum_{  n=0}^{\infty}\frac{0x^n}{n!}=0. Thus, the Taylor series at 0 of f does not agree with f on any open interval around 0

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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    Spoiler:


    Another way of putting differentiability, opposed to the one you put, is that f' exists at x_0 if

    \displaystyle\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}

    exists. Now, for our f evidently it is differentiable for x\ne 0 and at x=0 we see that

    \displaystyle \lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\frac{e^{\frac{-1}{x^2}}-0}{x}=\lim_{x\to0}\frac{1}{x}e^{\frac{-1}{x^2}}=0

    And thus f'(0) exists and equals zero. Thus,

    f'(x)=\begin{cases}\frac{2}{x^3}e^{\frac{-1}{x^2}} & \mbox{if}\quad x\ne0\\ 0 & \mbox{if}\quad x=0\end{cases}

    So, it's clear that f'' exists for x\ne0 and we see that

    \displaystyle \lim_{x\to0}\frac{f'(x)-f'(0)}{x-0}=\lim_{x\to0}\frac{\frac{2}{x^3}e^{\frac{-1}{x^2}}-0}{x}=\lim_{x\to0}\frac{2}{x^4}e^{\frac{-1}{x^2}}=0

    and thus f''(0) exists and equals zero. So,

    f''(x)=\begin{cases}e^{\frac{1}{x^2}}\left(\frac{4  }{x^6}-\frac{6}{x^4}\right) & \mbox{if}\quad x\ne0\\ 0 & \mbox{if}\quad x=0\end{cases}

    From here we can proceed by induction. Namely, assume that f^{(n)}(x) exists for all x. Then, show, by induction that

    f^{(n)}(x)=\begin{cases}p\left(\frac{1}{x}\right)e  ^{\frac{-1}{x^2}} & \mbox{if}\quad x\ne 0\\ 0 & \mbox{if}\quad x=0\end{cases}

    And thus showing that f^{(n+1)}(0) exists (since clearly f is n+1 times differentiable for x\ne0) reduces to showing that

    \displaystyle \lim_{x\to0}\frac{f^{(n)}(x)-f^{(n)}(0)}{x-0}=\lim_{x\to0}\frac{p\left(\frac{1}{x}\right)e^{\  frac{-1}{x^2}}-0}{x}=\lim_{x\to0}\frac{1}{x}p\left(\frac{1}{x}\ri  ght)e^{\frac{-1}{x^2}}

    exists and, in fact equals zero. Thus it follows by induction that f has derivatives of all orders.


    Remark: This shows that being C^{\infty} does not imply real analyticity since this f is infinitely differentiable at 0 but for any I around 0 we have that \displaystyle \sum_{n=0}^{\infty}\frac{f^{(n)}(0)x^n}{n!}=\sum_{  n=0}^{\infty}\frac{0x^n}{n!}=0. Thus, the Taylor series at 0 of f does not agree with f on any open interval around 0

    I am very grateful to your full and very clear prove. I really appreciate it. Thank you.
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  6. #6
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    Are you sure?
    Quoted from wiki: "In calculus (a branch of mathematics), a differentiable function is a function whose derivative exists at each point in its domain."
    See Differentiable function - Wikipedia, the free encyclopedia
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by xxp9 View Post
    Are you sure?
    Quoted from wiki: "In calculus (a branch of mathematics), a differentiable function is a function whose derivative exists at each point in its domain."
    See Differentiable function - Wikipedia, the free encyclopedia
    Sure of what? The function is differentiable at every point of its domain.
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  8. #8
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    I agree with you, but chisigma said it is not at 0. See above post from chisigma.
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  9. #9
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by xxp9 View Post
    Are you sure?
    Quoted from wiki: "In calculus (a branch of mathematics), a differentiable function is a function whose derivative exists at each point in its domain."
    See Differentiable function - Wikipedia, the free encyclopedia
    Im my opinion the 'key' of the question is that the fact that the derivative of a function f(x) exists in x=x_{0} doesn't mean that f(x) is differentiable in x=x_{0}. Wiki as other people may have a different opinion but in that case they have to justify the fact that a function like...

    y(x)=\left\{\begin{array}{ll} e^{-\frac{1}{x^{2}}} ,\,\,x \ne 0\\{}\\0 ,\,\, x=0\end{array}\right. (1)

    ... has the derivatives of all order in x=x_{0} and therefore its Taylor expansion around x=0 doesn't exist. In fact such Taylor expansion exists only if f(x) is differentiable in x=0 and that is not the situation...

    Kind regards

    \chi \sigma
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  10. #10
    MHF Contributor chisigma's Avatar
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    A confirm of what i posted is in...

    http://www.math.jhu.edu/~fspinu/405/...20notation.pdf

    2.4 The following statement are true…

    a) f is differentiable at a
    b) there exist a real number such that f(x)= f(a) + k\ (x-a) + o(x-a)

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    \chi \sigma
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  11. #11
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    A function is called to be analytic in a domain if its Taylor series converges to itself in the domain.
    So you're mixing the two concepts: differentiable and analytic.
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  12. #12
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by xxp9 View Post
    A function is called to be analytic in a domain if its Taylor series converges to itself in the domain.
    So you're mixing the two concepts: differentiable and analytic.
    How? I haven't the slightest idea what either of you are talking about.

    To say that f is differentiable on \mathbb{R} is to say that \displaystyle \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0} exists for all x_0\in\mathbb{R}. Thus, one can show that this is true for this function. Moreover, it is infinitely differentiable if it has dervatives of all orders, i.e. if \displaystyle \lim_{x\to x_0}\frac{f^{(n)}(x)-f^{(n)}(x_0)}{x-x_0} exists for all n\in\mathbb{N} and all x_0\in\mathbb{R}. I showed that. Only after I proved it was differentiable did I make a REMARK about analyticity.
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  13. #13
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    Quote Originally Posted by chisigma View Post
    Im my opinion the 'key' of the question is that the fact that the derivative of a function f(x) exists in x=x_{0} doesn't mean that f(x) is differentiable in x=x_{0}. Wiki as other people may have a different opinion but in that case they have to justify the fact that a function like...

    y(x)=\left\{\begin{array}{ll} e^{-\frac{1}{x^{2}}} ,\,\,x \ne 0\\{}\\0 ,\,\, x=0\end{array}\right. (1)

    ... has the derivatives of all order in x=x_{0} and therefore its Taylor expansion around x=0 doesn't exist. In fact such Taylor expansion exists only if f(x) is differentiable in x=0 and that is not the situation...

    Kind regards

    \chi \sigma
    chisigma, I totally fail to understand what you are getting at here. For a start, the statements "the derivative of the function f(x) exists at x=x_{0}" and " f(x) is differentiable at x=x_{0}" mean exactly the same thing. Secondly, the function f defined by

    f(x) = \begin{cases}e^{-1/x^2}&(x\ne0) \\ 0&(x=0)\end{cases}

    is infinitely differentiable (equivalently, it has derivatives of all orders) at every point x_0 including x_0=0. At x=0, every derivative is equal to 0. That does not mean that the Taylor expansion around x=0 doesn't exist. The Taylor expansion around x=0 does exist, but it is identically zero.
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  14. #14
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Opalg View Post
    chisigma, I totally fail to understand what you are getting at here. For a start, the statements "the derivative of the function f(x) exists at x=x_{0}" and " f(x) is differentiable at x=x_{0}" mean exactly the same thing. Secondly, the function f defined by

    f(x) = \begin{cases}e^{-1/x^2}&(x\ne0) \\ 0&(x=0)\end{cases}

    is infinitely differentiable (equivalently, it has derivatives of all orders) at every point x_0 including x_0=0. At x=0, every derivative is equal to 0. That does not mean that the Taylor expansion around x=0 doesn't exist. The Taylor expansion around x=0 does exist, but it is identically zero.
    Honestly for me is a little hard to undestand how is possible that this function...

    y(x)=\left\{\begin{array}{ll} e^{-\frac{1}{x^{2}}} ,\,\,x \ne 0\\{}\\0 ,\,\, x=0\end{array}\right. (1)

    ... that has all its derivatives equal to zero in x=0, can be computed 'somewhere around' x=0 with the Taylor expansion...

    \displaystyle y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n} (2)

    If all the y^{(n)} (0) are equal to 0 , it is evident that the (2) gives 0 no matter which is x... but y(x) is not equal to the 'nul function'... It seems to me that the situation is 'a little unconfortable'...

    Kind regards

    \chi \sigma
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  15. #15
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chisigma View Post
    Honestly for me is a little hard to undestand how is possible that this function...

    y(x)=\left\{\begin{array}{ll} e^{-\frac{1}{x^{2}}} ,\,\,x \ne 0\\{}\\0 ,\,\, x=0\end{array}\right. (1)

    ... that has all its derivatives equal to zero in x=0, can be computed 'somewhere around' x=0 with the Taylor expansion...

    \displaystyle y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n} (2)

    If all the y^{(n)} (0) are equal to 0 , it is evident that the (2) gives 0 no matter which is x... but y(x) is not equal to the 'nul function'... It seems to me that the situation is 'a little unconfortable'...

    Kind regards

    \chi \sigma
    But this is the WHOLE point! This is an example of an infinitely differentiable function f, and a point x_0 such that the Taylor series of f at x_0 converges...but not to f!

    This is why real analyticity is NOT just a fancy word for C^{\infty}.
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