View source for Example:A smooth function that is not real analytic

{{Stub page|grade=A*|msg=Important to do! Needs to be nice and neat before this is removed.}}
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==Example==
Let {{M|f:\mathbb{R}\rightarrow\mathbb{R} }} be defined as follows:
* {{MM|1=f:x\mapsto\left\{\begin{array}{lr}e^{-\frac{1}{x} } & \text{if }x>0\\ 0 & \text{otherwise}\end{array}\right.}}
We will show this [[function]] is ''not'' [[real-analytic]] at {{M|0\in\mathbb{R} }} but is [[smooth]].
==Proof==
===For {{M|x<0}}, {{M|f}} is smooth===
This is trivial, as {{M|f\vert_{\mathbb{R}_{<0} }\ident 0}}
* {{XXX|The derivative of a constant is 0, the derivative of that is 0, and so forth - link to corresponding pages}}
===For {{M|x>0}}, {{M|f}} is smooth===
We will show a slightly different result.
* Let {{M|P_m(y)}} be a [[polynomial]] of {{link|order|polynomial}} {{M|m}} in the invariant {{M|y}}. {{M|p_m:\mathbb{R}\rightarrow\mathbb{R} }} is the map with {{M|p_m:y\mapsto p(y)}} meaning {{link|evaluation|polynomial}} at {{M|y\in\mathbb{R} }}.
We claim that:
* {{MM|\frac{\mathrm{d}^kf}{\mathrm{d}x^k}\eq p_{2k}(\tfrac{1}{x})\mathrm{e}^{-\frac{1}{x} } }}
====Proof====
To avoid any ''potential'' ambiguities {{caveat|I'm not sure whether or not there'd be problems with a 0 {{link|degree|polynomial}} polynomial, but I want to sidestep it}} we shall prove this by induction starting from {{M|k\eq 1}}.

The proof shall proceed as follows:
# Show that {{MM|\frac{df}{dx}\eq p_2(\tfrac{1}{x})e^{-\frac{1}{x} } }}
# Assume that {{MM|\frac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} } }} holds
# Show that {{MM|\left(\frac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} }\right)\implies\left(\frac{d^{k+1}f}{dx^{k+1} }\eq p_{2(k+1)}(\tfrac{1}{x})e^{-\frac{1}{x} }\right)}}
#* I.E. show that the RHS of this is true (as by the nature of [[logical implication]] if the LHS is true - which we assume it is - then for it to imply the RHS the RHS must also be true.
'''Proof:'''
: {{Green highlight|msg=Remember that here we are explicitly dealing with the {{M|x\in\mathbb{R}_{>0} }} case. We are basically considering {{M|f:\mathbb{R}_{>0}\rightarrow\mathbb{R} }} by {{M|f:x\mapsto e^{-\frac{1}{x} } }}}}
# Show that {{M|\frac{df}{dx}\eq p_2(\frac{1}{x})e^{-\frac{1}{x} } }} {{Caveat|I'm experimenting with differentiation notation here}}<ref group="Note">The details are as follows:
* {{MM|\frac{df}{dx} }} is itself a function that takes {{M|x'}} to {{MM|\frac{df}{dx}\Big\vert_{x\eq x'} }}
* {{MM|\frac{df}{dx}\Big\vert_x}} is another way of writing {{MM|\frac{df}{dx} }} - anywhere where the variable we are differentiating with respect to is where we are differentiating at invokes this, and is actually a function in the "at" part.
* {{MM|\frac{df}{dx}\Big\vert_{x\eq z} }} is an expression for the derivative of {{M|f}} (wrt {{M|x}}) - at {{M|z}} - note that it is an expression - not a constant:
** For example {{MM|\frac{d}{dx}x^3\Big\vert_{x\eq t}\eq 3t^2}} so {{MM|\frac{d}{dt}\frac{d}{dx}x^3\Big\vert_{x\eq t}\Big\vert_{t}\eq 6t}}
We may also use:
* {{MM|\frac{d}{dt}\Big[\text{whatever}\Big]}} to mean {{MM|\frac{d}{dt}\text{whatever}\Big\vert_t}}
* {{MM|\frac{d}{dt}\Big[\text{whatever}\Big]_{t} }} to mean {{MM|\frac{d}{dt}\text{whatever}\Big\vert_t}}, and
* {{MM|\frac{d}{dt}\Big[\text{whatever}\Big]_{t\eq p} }} to mean {{MM|\frac{d}{dt}\text{whatever}\Big\vert_{t\eq p} }}
Rather than:
* {{MM|\frac{d}{dt}\Big[\text{whatever}\Big]\Big\vert_{t\eq p} }} for example</ref>
#* {{MM|\frac{df}{dx}\Big\vert_x\eq\frac{d}{dz}e^{z}\Big\vert_{z\eq -\frac{1}{x} }\cdot\frac{d}{dx}-(x^{-1})\Big\vert_x}}{{MM|\eq e^{-\frac{1}{x} }\cdot(-1)\cdot\frac{d}{dx}x^{-1}\Big\vert_x}}{{MM|\eq x^{-2}e^{-\frac{1}{x} } }}
#* We see {{MM|\frac{df}{dx}\eq \left(\tfrac{1}{x}\right)^2e^{-\frac{1}{x} } }} or {{MM|\frac{df}{dx}\eq p_2(\tfrac{1}{x})e^{-\frac{1}{x} } }}
# Now we assume {{M|\dfrac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} } }}
# We attempt to show that {{M|\left(\dfrac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} }\right)\implies\left(\dfrac{d^{k+1}f}{dx^{k+1} }\eq p_{2(k+1)}(\tfrac{1}{x})e^{-\frac{1}{x} } \right)}}
#* {{MM|\frac{d^{k+1}f}{dx^{k+1} }\eq\frac{d}{dx}\left[\frac{d^kf}{dx^k}\right]_x\eq\frac{d}{dx}\left[p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} }\right]}}{{MM|\eq e^{-\frac{1}{x} }\left(\frac{d}{dy}\Big[p_{2k}(y)\Big]_{y\eq-\frac{1}{x} }\cdot\frac{d}{dx}\Big[-x^{-1}\Big]\right)+p_{2k}(\tfrac{1}{x})\cdot\frac{d}{dx}\Big[e^{-\frac{1}{x} }\Big] }}
#** Well if we differentiate a polynomial of order {{M|2k}} we get a polynomial of order {{M|2k-1}}, so: {{Green highlight|Switching from {{M|p}} to {{M|P}} for the polynomial to make the subscripts more obvious}}
#* {{MM|\frac{d^{k+1}f}{dx^{k+1} }\eq e^{-\frac{1}{x} }P_{2k-1}(\tfrac{1}{x})\cdot x^{-2} + P_{2k}(\tfrac{1}{x})\cdot x^{-2}e^{-\frac{1}{x} } }} {{MM|\eq e^{-\frac{1}{x} }\left(P_{2k-1}(\tfrac{1}{x})\cdot x^{-2}+P_{2k}(\tfrac{1}{x})\cdot x^{-2}\right)}}
#** As {{M|x^{-1}\eq(\tfrac{1}{x})^2}} we see:
#* {{MM|\frac{d^{k+1}f}{dx^{k+1} }\eq e^{-\frac{1}{x} } \left(P_{2k+1}(\tfrac{1}{x}) + P_{2(k+1)}(\tfrac{1}{x}) \right)}} (note that {{M|2k+2\eq 2(k+1)}} in the subscript of {{M|P}})
#* Thus: {{MM|\frac{d^{k+1}f}{dx^{k+1} }\eq P_{2(k+1)}(\tfrac{1}{x})e^{-\frac{1}{x} } }} - as expected.
#* We have shown the induction hypothesis.
{{Requires work|grade=A*|msg=Important work for manifolds, will probably crop up again!}}

==Notes==
<references group="Note"/>
==References==
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