Infinitely differentiable function

**zadir** · November 10th 2010, 12:12 PM

Prove that the following function is infinitely differentiable:

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

Thank you!

**Drexel28** · November 10th 2010, 12:16 PM

Originally Posted by zadir

Prove that the following function is infinitely differentiable:

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

**chisigma** · November 10th 2010, 01:45 PM

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

$\chi$ $\sigma$

**Drexel28** · November 10th 2010, 02:11 PM

Originally Posted by chisigma

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:

**zadir** · November 10th 2010, 08:22 PM

Originally Posted by Drexel28

Spoiler:

I am very grateful to your full and very clear prove. I really appreciate it. Thank you.

**xxp9** · November 10th 2010, 09:40 PM

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

**Drexel28** · November 10th 2010, 09:53 PM

Originally Posted by xxp9

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.

**xxp9** · November 10th 2010, 10:19 PM

I agree with you, but chisigma said it is not at 0. See above post from chisigma.

**chisigma** · November 11th 2010, 12:42 AM

Originally Posted by xxp9

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

$\chi$ $\sigma$

**chisigma** · November 11th 2010, 01:06 AM

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 $f(x)= f(a) + k\ (x-a) + o(x-a)$

Kind regards

$\chi$ $\sigma$

**xxp9** · November 11th 2010, 01:09 AM

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.

**Drexel28** · November 11th 2010, 06:07 AM

Originally Posted by xxp9

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.

**Opalg** · November 11th 2010, 10:43 AM

Originally Posted by chisigma

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.

**chisigma** · November 11th 2010, 01:31 PM

Originally Posted by Opalg

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

$\chi$ $\sigma$

**Drexel28** · November 11th 2010, 02:03 PM

Originally Posted by chisigma

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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