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January 11th, 2017, 04:58 PM   #1
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C-inf smooth function

Hi all, I was reading through Roger Penroe's "the road to reality" and I found the following function

\begin{eqnarray}
h(x)&=&e^{-1/x} & x>0\\
h(x)&=&0 & \text{elsewhere}
\end{eqnarray}
where it says this functions is $C^{\infty}$ smooth, meaning all derivatives to all orders are smooth (the singular point here is $x=0$). It says to give it a try but doesnt show any proof, according to the book the proof might be a bit difficult. I have tried unsuccesfully to tackle it, any ideas to solve this?

Thanks!
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January 11th, 2017, 07:58 PM   #2
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h(x) is infinitely differentiable because no matter how many times you differentiate you get a sum of differentiable products which is differentiable.
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January 11th, 2017, 09:33 PM   #3
SDK
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Quote:
Originally Posted by zylo View Post
h(x) is infinitely differentiable because no matter how many times you differentiate you get a sum of differentiable products which is differentiable.
This is only obvous when $x > 0$. Indeed the difficulty here is to show that it is differentiable at $x = 0$.

1. Convince yourself that it is continuous at $x=0$.

2. The derivative exists at $x=0$ if the following limit exists:
$$ \lim_{h\rightarrow 0} \frac{e^{-\frac{1}{h}}}{h} = \lim_{u\rightarrow \infty} ue^{-u}.$$ Convince yourself that $h'(0) = 0$.

3. Now, you compute $h'(x)$ piecewise and you argue that $h \in C^1$. With this established, it is immediate that $h \in C^{\infty}$.
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January 12th, 2017, 07:28 AM   #4
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Ok, I get the first point, that is easy to demonstrate, but I dont get the second. Is that the condition of continuity of the derivative or what? (Actually the derivative should be divided by $x^{2}$ and not by $x$, why are you using $h$ as variable here?)

I dont also see the 3rd point, since it is a proof that the first derivative exists, but not to the derivative to all orders, moreover, is not only that they exist, but that they are all smooth functions (no peaks or discontinuities).

Last edited by greg1313; January 12th, 2017 at 11:09 AM.
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January 14th, 2017, 11:31 AM   #5
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SDK is right, but you have to show h$\displaystyle ^{(n)}$(0) exists for all n to show C$\displaystyle ^{\infty}$ for x $\displaystyle \geq$ 0.
You can do so because h$\displaystyle ^{(n)}$(0) will consist of a sum of terms of the form:
$\displaystyle \lim_{\delta\rightarrow 0}\frac{e^{-\frac{1}{\delta}}}{\delta^{n}}=\lim_{X\rightarrow \infty}\frac{X^{n}}{e^{X}}=0$
by L'hospitals rule.
And the limit of a sum is the sum of the limits.
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