Exercise [06.02]

Sameed Zahoor · 16 May 2008, 12:16

Before giving the proof I need to state a few definitions.
Let D be the set of all differentiable functions.
Let
$P=\{p(x)|p(x)=\sum{a_{k}x^{-k}},k\in{Z^{+}},a_k,x\in{R},x\ is\ not\ 0\}$
The following properties about P can easily proved.
1.If $p_1,p_2\in{P}$ then $p_{1}p_{2}\in{P}$ .
$(\sum{a_{i}x^{-i}}\sum{b_{j}x^{-j}}=\sum{a_{i}b_{j}x^{-(i+j)}})$
2.If $p_1,p_2\in{P}$ then $p_{1}+p_{2}\in{P}$ .
$(\sum{a_{k}x^{-k}}+\sum{b_{k}x^{-k}=\sum{(a_{k}+b_{k})x^{-k}}})$
3.If $p(x)\in{P}$ then $p\in{D}and\ p'(x)\in{P}$ .
$(\frac{d}{dx}(\sum{a_{k}x^{-k}})=\sum{-ka_{k}x^{-(k+1)}})$
Now let
$S=\{f(x)|f(x)\in{D},f'(x)=f(x).p_{1}(x)where\ p_{1}(x)\in{P}\}$
We now prove an amazing theorem,
If $f\in{S}$ then,f can be differentiated infinitely many times.
PROOF:(BY INDUCTION)
$f\in{S}\Rightarrow\ f'(x)=f(x).p_{1}(x)$
Assume $f^{k}(x)=f(x).p_{k}(x)$ where $p_{k}(x)\in{P}$
As $f(x),p_{k}(x)\in{D},f(x).p_{k}(x)\in{D}$ or $f^{n+1}(x)$ exists.
Now,
$f^{n+1}(x)=f(x).p'_{k}(x)+p_{k}(x)f'(x)=f(x)\{p'_{k}(x)+p_{k}(x).p_{k}(x)\}$
$Let \ p_{k+1}=\{p'_{k}(x)+p_{k}(x).p_{k}(x)\}$
Using properties 1,2 and 3
$p_{k+1}\in{P}$
Hence
$f^{k+1}(x)=f(x).p_{k+1}(x)$
By induction the theorem is thus valid for all $n\in{Z^{+}}$

Now to the actual problem.We have the function
$h(x)=0\ for\ x\leq{0}$ Hence, h is $C^{\infty}$ smooth for non-positive values of x.
Also,
$h(x)=e^{\frac{-1}{x}}\ for\ x\ greater\ than\ 0$
In this case, $h(x)\in{S}$ and is again $C^{\infty}$ smooth for positive x.
Thus, $h(x)\ is\ C^{\infty}$ smooth $\forall{x\in{R}}$

Exercise [06.02]

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