Difference between revisions of "Example:A smooth function that is not real analytic"

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We will show this [[function]] is ''not'' [[real-analytic]] at {{M|0\in\mathbb{R} }} but is [[smooth]].
 
We will show this [[function]] is ''not'' [[real-analytic]] at {{M|0\in\mathbb{R} }} but is [[smooth]].
 
==Proof==
 
==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==
 
==References==
 
<references/>
 
<references/>

Latest revision as of 04:25, 27 November 2016

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Example

Let [ilmath]f:\mathbb{R}\rightarrow\mathbb{R} [/ilmath] be defined as follows:

  • [math]f:x\mapsto\left\{\begin{array}{lr}e^{-\frac{1}{x} } & \text{if }x>0\\ 0 & \text{otherwise}\end{array}\right.[/math]

We will show this function is not real-analytic at [ilmath]0\in\mathbb{R} [/ilmath] but is smooth.

Proof

For [ilmath]x<0[/ilmath], [ilmath]f[/ilmath] is smooth

This is trivial, as [ilmath]f\vert_{\mathbb{R}_{<0} }\ident 0[/ilmath]

  • TODO: The derivative of a constant is 0, the derivative of that is 0, and so forth - link to corresponding pages

For [ilmath]x>0[/ilmath], [ilmath]f[/ilmath] is smooth

We will show a slightly different result.

  • Let [ilmath]P_m(y)[/ilmath] be a polynomial of order [ilmath]m[/ilmath] in the invariant [ilmath]y[/ilmath]. [ilmath]p_m:\mathbb{R}\rightarrow\mathbb{R} [/ilmath] is the map with [ilmath]p_m:y\mapsto p(y)[/ilmath] meaning evaluation at [ilmath]y\in\mathbb{R} [/ilmath].

We claim that:

  • [math]\frac{\mathrm{d}^kf}{\mathrm{d}x^k}\eq p_{2k}(\tfrac{1}{x})\mathrm{e}^{-\frac{1}{x} } [/math]

Proof

To avoid any potential ambiguities Caveat:I'm not sure whether or not there'd be problems with a 0 degree polynomial, but I want to sidestep it we shall prove this by induction starting from [ilmath]k\eq 1[/ilmath].

The proof shall proceed as follows:

  1. Show that [math]\frac{df}{dx}\eq p_2(\tfrac{1}{x})e^{-\frac{1}{x} } [/math]
  2. Assume that [math]\frac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} } [/math] holds
  3. Show that [math]\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)[/math]
    • 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:

Remember that here we are explicitly dealing with the [ilmath]x\in\mathbb{R}_{>0} [/ilmath] case. We are basically considering [ilmath]f:\mathbb{R}_{>0}\rightarrow\mathbb{R} [/ilmath] by [ilmath]f:x\mapsto e^{-\frac{1}{x} } [/ilmath]
  1. Show that [ilmath]\frac{df}{dx}\eq p_2(\frac{1}{x})e^{-\frac{1}{x} } [/ilmath] Caveat:I'm experimenting with differentiation notation here[Note 1]
    • [math]\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[/math][math]\eq e^{-\frac{1}{x} }\cdot(-1)\cdot\frac{d}{dx}x^{-1}\Big\vert_x[/math][math]\eq x^{-2}e^{-\frac{1}{x} } [/math]
    • We see [math]\frac{df}{dx}\eq \left(\tfrac{1}{x}\right)^2e^{-\frac{1}{x} } [/math] or [math]\frac{df}{dx}\eq p_2(\tfrac{1}{x})e^{-\frac{1}{x} } [/math]
  2. Now we assume [ilmath]\dfrac{d^kf}{dx^k}\eq p_{2k}(\tfrac{1}{x})e^{-\frac{1}{x} } [/ilmath]
  3. We attempt to show that [ilmath]\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)[/ilmath]
    • [math]\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][/math][math]\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] [/math]
      • Well if we differentiate a polynomial of order [ilmath]2k[/ilmath] we get a polynomial of order [ilmath]2k-1[/ilmath], so: Switching from [ilmath]p[/ilmath] to [ilmath]P[/ilmath] for the polynomial to make the subscripts more obvious
    • [math]\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} } [/math] [math]\eq e^{-\frac{1}{x} }\left(P_{2k-1}(\tfrac{1}{x})\cdot x^{-2}+P_{2k}(\tfrac{1}{x})\cdot x^{-2}\right)[/math]
      • As [ilmath]x^{-1}\eq(\tfrac{1}{x})^2[/ilmath] we see:
    • [math]\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)[/math] (note that [ilmath]2k+2\eq 2(k+1)[/ilmath] in the subscript of [ilmath]P[/ilmath])
    • Thus: [math]\frac{d^{k+1}f}{dx^{k+1} }\eq P_{2(k+1)}(\tfrac{1}{x})e^{-\frac{1}{x} } [/math] - as expected.
    • We have shown the induction hypothesis.
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Important work for manifolds, will probably crop up again!

Notes

  1. ↑ The details are as follows:
    • [math]\frac{df}{dx} [/math] is itself a function that takes [ilmath]x'[/ilmath] to [math]\frac{df}{dx}\Big\vert_{x\eq x'} [/math]
    • [math]\frac{df}{dx}\Big\vert_x[/math] is another way of writing [math]\frac{df}{dx} [/math] - 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.
    • [math]\frac{df}{dx}\Big\vert_{x\eq z} [/math] is an expression for the derivative of [ilmath]f[/ilmath] (wrt [ilmath]x[/ilmath]) - at [ilmath]z[/ilmath] - note that it is an expression - not a constant:
      • For example [math]\frac{d}{dx}x^3\Big\vert_{x\eq t}\eq 3t^2[/math] so [math]\frac{d}{dt}\frac{d}{dx}x^3\Big\vert_{x\eq t}\Big\vert_{t}\eq 6t[/math]
    We may also use:
    • [math]\frac{d}{dt}\Big[\text{whatever}\Big][/math] to mean [math]\frac{d}{dt}\text{whatever}\Big\vert_t[/math]
    • [math]\frac{d}{dt}\Big[\text{whatever}\Big]_{t} [/math] to mean [math]\frac{d}{dt}\text{whatever}\Big\vert_t[/math], and
    • [math]\frac{d}{dt}\Big[\text{whatever}\Big]_{t\eq p} [/math] to mean [math]\frac{d}{dt}\text{whatever}\Big\vert_{t\eq p} [/math]
    Rather than:
    • [math]\frac{d}{dt}\Big[\text{whatever}\Big]\Big\vert_{t\eq p} [/math] for example

References

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