Editing Non-analytic smooth function (section)

===The function is smooth===

The function ''f'' has [[Continuous function|continuous]] [[derivative]]s of all orders at every point ''x'' of the [[real line]].  The formula for these derivatives is

:<math>f^{(n)}(x) = \begin{cases}\displaystyle\frac{p_n(x)}{x^{2n}}\,f(x) & \text{if }x>0, \\ 0 &\text{if }x \le 0,\end{cases}</math>

where ''p<sub>n</sub>''(''x'') is a [[polynomial]] of [[Degree of a polynomial|degree]] ''n''&nbsp;&minus;&nbsp;1 given [[recursion|recursively]] by ''p''<sub>1</sub>(''x'')&nbsp;=&nbsp;1 and

:<math>p_{n+1}(x)=x^2p_n'(x)-(2nx-1)p_n(x)</math>

for any positive [[integer]] ''n''.  From this formula, it is not completely clear that the derivatives are continuous at 0; this follows from the [[one-sided limit]] 

:<math>\lim_{x\searrow 0} \frac{e^{-\frac{1}{x}}}{x^m} = 0</math>

for any [[nonnegative]] integer ''m''.

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By the [[Exponential function#Formal definition|power series representation of the exponential function]], we have for every [[natural number]] <math>m</math> (including zero)

:<math>\frac1{x^m}=x\Bigl(\frac1{x}\Bigr)^{m+1}\le (m+1)!\,x\sum_{n=0}^\infty\frac1{n!}\Bigl(\frac1x\Bigr)^n
=(m+1)!\,x e^{\frac{1}{x}},\qquad x>0,</math>

because all the positive terms for <math>n \neq m+1</math> are added. Therefore, dividing this inequality by <math>e^{\frac{1}{x}}</math> and taking the [[One-sided limit|limit from above]],

:<math>\lim_{x\searrow0}\frac{e^{-\frac{1}{x}}}{x^m}
\le (m+1)!\lim_{x\searrow0}x=0.</math>

We now prove the formula for the ''n''th derivative of ''f'' by [[mathematical induction]]. Using the [[chain rule]], the [[reciprocal rule]], and the fact that the derivative of the exponential function is again the exponential function, we see that the formula is correct for the first derivative of ''f'' for all ''x''&nbsp;>&nbsp;0 and that ''p''<sub>1</sub>(''x'') is a polynomial of degree 0. Of course, the derivative of ''f'' is zero for ''x''&nbsp;<&nbsp;0.
It remains to show that the right-hand side derivative of ''f'' at ''x''&nbsp;=&nbsp;0 is zero. Using the above limit, we see that

:<math>f'(0)=\lim_{x\searrow0}\frac{f(x)-f(0)}{x-0}=\lim_{x\searrow0}\frac{e^{-\frac{1}{x}}}{x}=0.</math>

The induction step from ''n'' to ''n''&nbsp;+&nbsp;1 is similar. For ''x''&nbsp;>&nbsp;0 we get for the derivative
:<math>\begin{align}f^{(n+1)}(x)
&=\biggl(\frac{p'_n(x)}{x^{2n}}-2n\frac{p_n(x)}{x^{2n+1}}+\frac{p_n(x)}{x^{2n+2}}\biggr)f(x)\\
&=\frac{x^2p'_n(x)-(2nx-1)p_n(x)}{x^{2n+2}}f(x)\\
&=\frac{p_{n+1}(x)}{x^{2(n+1)}}f(x),\end{align}</math>

where ''p''<sub>''n''+1</sub>(''x'') is a polynomial of degree ''n''&nbsp;= (''n''&nbsp;+&nbsp;1)&nbsp;&minus;&nbsp;1. Of course, the (''n''&nbsp;+&nbsp;1)st derivative of ''f'' is zero for ''x''&nbsp;<&nbsp;0. For the right-hand side derivative of ''f''<sup>&nbsp;(''n'')</sup> at ''x''&nbsp;=&nbsp;0 we obtain with the above limit

:<math>\lim_{x\searrow0} \frac{f^{(n)}(x) - f^{(n)}(0)}{x-0} = \lim_{x\searrow0} \frac{p_n(x)}{x^{2n+1}}\,e^{-1/x} = 0.</math>
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