Difference between revisions of "Thomae function"

From specialfunctionswiki
Jump to: navigation, search
(Properties)
Line 23: Line 23:
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 
<strong>Theorem:</strong> The Thomae function has a (strict) [[local maximum]] at each [[rational number]].
 
<strong>Theorem:</strong> The Thomae function has a (strict) [[local maximum]] at each [[rational number]].
 +
<div class="mw-collapsible-content">
 +
<strong>Proof:</strong> █
 +
</div>
 +
</div>
 +
 +
<div class="toccolours mw-collapsible mw-collapsed" style="width:800px">
 +
<strong>Theorem:</strong> The [[Thomae function]] $f(x)$ is not [[Riemann integrable]] but it is [[Lebesgue integrable]] and
 +
$$\displaystyle\int_0^1 f(x) dx = 0.$$
 
<div class="mw-collapsible-content">
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> █  
 
<strong>Proof:</strong> █  
 
</div>
 
</div>
 
</div>
 
</div>

Revision as of 20:37, 11 April 2015

Thomae's function is given by the formula $$f(x) =\begin{cases} 1 & \text{if } x= 0 \\ \tfrac1{q} & \text{if } x = \tfrac{p}{q}\\ 0 & \text{if } x \in \mathbb{R}-\mathbb{Q}. \end{cases}$$


Properties

Theorem: The Thomae function is continuous at all irrational numbers and discontinuous at all rational numbers.

Proof:

Theorem: The Thomae function has a (strict) local maximum at each rational number.

Proof:

Theorem: The Thomae function $f(x)$ is not Riemann integrable but it is Lebesgue integrable and $$\displaystyle\int_0^1 f(x) dx = 0.$$

Proof: