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$$ f(x) = \begin{cases} e^\frac{1}{x} &\text{für } x<0 \\ 0 &\text{für } x \geq 0 \end{cases} $$

Problem: I know that to first I have to show that the function is differentiable at $f(x) = 0$ and $f(x) = e^{\frac{1}{x}}$.

Difference quotient for $f(x)=0$ is for me clear, but for $f(x) = e^{\frac{1}{x}}$, I was only able to get to this point: $$ \lim_{h\to 0} \frac{f(h+x)-f(x)}{h} = \lim_{h\to 0} \frac{e^{\frac{1}{h+x}} - e^{\frac{1}{x}}}{h} ={} ? $$

And also I don't understand how to show in general that a function is infinitely continuously differentiable. I will be very happy for help.

metamorphy
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dmaxru
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    A very brief indication of one method is described in this answer. I'm pretty sure your specific function is dealt with several times in MSE, but I don't have time to look now. – Dave L. Renfro Jan 03 '23 at 23:30
  • Put $\varphi(h) = \frac{f(h)-f(0)}{h}$. Consider $\lim_{h\downarrow 0}\varphi(h)$ and $\lim_{h\uparrow 0}\varphi(h)$. – Andrew Jan 04 '23 at 00:02
  • @AndrewZhang why is it necessary to consider the differentiability of a function at the point x0=0, but not at any point x of the real numbers? – dmaxru Jan 04 '23 at 01:14
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    @Goodryl Because $f$ is clearly smooth on $(0,\infty)$. Likewise, $f$ is smooth on $(-\infty,0)$. To see this, recall that the composition of $C^{k}$ function is also $C^k$. Hence $f\in C^{\infty}_{ (-\infty,0)}$. The only point that does not proceed directly from derivative rules is the origin. – Andrew Jan 04 '23 at 01:29
  • $\ldots$ but I don't have time to look now --- OK, I have time now. See 1.1 Definition of the function and 1.2 The function is smooth in the Wikipedia article for Non-analytic smooth function. Note that this other function and your function differ only by the variable change $x \mapsto -x$ (i.e. the graph of your function is the reflection about the $y$-axis of the graph of the Wikipedia's function). – Dave L. Renfro Jan 04 '23 at 06:01
  • @DaveL.Renfro Thanks for the link, it helped me figure out how to prove that a function is infinitely continuously differentiable. I understood everything, but until the moment of proof that $f^{(n+1)}$ = $\frac{P_{n+1}(x)}{x^{2n}}\cdot e^{-\frac{1}{x}}$ . Can you explain how the first equality after $f^{(n+1)}$ in proof came about? – dmaxru Jan 28 '23 at 01:41
  • In case you get lost in the next sentence, note that if $f,$ $g,$ $h$ are differentiable functions, then the derivative of $\left(\frac{f}{g}\right)h$ is equal to $\left(\frac{f}{g}\right)'h + \left(\frac{f}{g}\right)h' = \left(\frac{f'g - fg'}{g^2}\right)h + \left(\frac{f}{g}\right)h'$ (product rule to get left side, then quotient rule to get right side). Basic short-cut derivative formulas can be applied to find the derivative of $\left(\frac{p_n(x)}{x^{2n}}\right)e^{-\frac{1}{x}}$ for $x>0,$ which leads to (continued) – Dave L. Renfro Jan 28 '23 at 17:38
  • $\left(\frac{p'_n(x)x^{2n} ; - ; p_n(x)\cdot 2nx^{2n-1}}{x^{4n}}\right)e^{-\frac{1}{x}} + \left(\frac{p_n(x)}{x^{2n}}\right) \cdot \left(\frac{1}{x^2}\right)e^{-\frac{1}{x}}.$ By cancelling some of the $x$'s (specifically, reduce the 1st fraction a bit by dividing numerator and denominator by $x^{2n-1};$ thus, subtract $2n-1$ from each of the three exponents of $x$ in the 1st fraction) we get $\left(\frac{p'_n(x)x ; - ; p_n(x)\cdot 2n}{x^{2n+1}}\right)e^{-\frac{1}{x}} + \left(\frac{p_n(x)}{x^{2n+2}}\right)e^{-\frac{1}{x}}.$ (continued) – Dave L. Renfro Jan 28 '23 at 17:38
  • Some forward-thinking rewriting based on what the recursion formula for $f^{(n)}(x)$ for $x>0$ looks like when $n$ is replaced by $n+1$ motivates us to get the denominators to be $x^{2(n+1)} = x^{2n+2}.$ So in the last stuff we got, multiply the numerator and denominator of the 1st fraction by $x$ to get a denominator of $x^{2n+2}.$ (Of course, we can combine these last two steps by initially dividing by $x^{2n-2}$ instead of dividing by $x^{2n-1},$ (continued) – Dave L. Renfro Jan 28 '23 at 17:40
  • but doing this is probably not something you would consider doing unless you're really good at "looking ahead" to see what it takes to get where you want to go.) This gives $\left(\frac{p'_n(x)x^2 ; - ; p_n(x)\cdot 2nx}{x^{2n+2}}\right)e^{-\frac{1}{x}} + \left(\frac{p_n(x)}{x^{2n+2}}\right)e^{-\frac{1}{x}}.$ Now factor out $e^{-\frac{1}{x}}$ and combine the two fractions to get $\left(\frac{p'_n(x)x^2 ; - ; p_n(x)\cdot 2nx ; + ; p_n(x)}{x^{2n+2}}\right)e^{-\frac{1}{x}}; = ; \left(\frac{x^2 p'_n(x) ; - ; (2nx-1)p_n(x)}{x^{2n+2}}\right)e^{-\frac{1}{x}}.$ (continued) – Dave L. Renfro Jan 28 '23 at 17:40
  • Notice that this last expression is equal to $\left(\frac{p_{n+1}(x)}{x^{2(n+1)}}\right)e^{-\frac{1}{x}},$ which is equal to what the Wikipedia formula tells us we should have for $f^{(n+1)}(x)$ when $x>0$ (i.e. in the formula for $f^{(n)}(x),$ replace $n$ with $n+1).$ – Dave L. Renfro Jan 28 '23 at 17:40
  • It's important to note that this is all way too much work to get what you want. It is enough to prove by induction that for $x>0$ the $n$'th derivative of $e^{-\frac{1}{x}}$ is a rational function times $e^{-\frac{1}{x}}.$ It doesn't matter what the rational fraction is, or even what its numerator and denominator degrees are. For example, see my answer to Examples of applying L'Hôpitals rule ( correctly ) leading back to the same state?, which shows what to do in a situation similar to what you have here. – Dave L. Renfro Jan 28 '23 at 17:47
  • @DaveL.Renfro Thank you very much for the step by step and detailed explanation! But how did the numerator after the last equality become equal to $\ p_{n+1}(x)$ ? – dmaxru Jan 29 '23 at 13:01
  • The Wikipedia article defines $;p_{n+1}(x);$ to be $;x^2 p'n(x) - (2nx-1)p_n(x),;$ so by substitution we have $;\frac{x^2 p'_n(x) ; - ; (2nx-1)p_n(x)}{x^{2n+2}} ; = ; \frac{p{n+1}(x)}{x^{2n+2}}.$ – Dave L. Renfro Jan 29 '23 at 18:22

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well we can use the chain rule to differentiate $e^{1/x}$: we get $\frac{d}{dx}[e^{1/x}] = - \left(\frac{1}{x^2}\right) * e^{1/x}$, and from here it is obvious that we can differentiate the function as many times as we want (infinitely differentiable), as each new derivative will contain the term $e^{1/x}$ times some expression with (1/x) raised to some power.

Harish Chandra Rajpoot
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Martin Westin
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