Difference between revisions of "Bolzano function"

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(Properties)
 
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=Properties=
 
=Properties=
 
[[Bolzano function is continuous]]<br />
 
[[Bolzano function is continuous]]<br />
 
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[[Bolzano function is nowhere differentiable]]<br />
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<strong>Theorem:</strong> The [[Bolzano function]] is [[nowhere differentiable]].
 
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<strong>Proof:</strong> █
 
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=Videos=
 
=Videos=

Latest revision as of 17:10, 23 June 2016

A Bolzano function $B \colon [a,b] \rightarrow [c,d]$ is is defined as the limit of a sequence $\{B_k\}$ of continuous functions. Let $B_1 \colon [a,b] \rightarrow [c,d]$ be defined by the formula $B_1(x)=c + \dfrac{d-c}{b-a}(x-a).$ We break the interval $[a,b]$ into four subintervals $\left[ a, a+\dfrac{3}{8}(b-a) \right]$, $\left[ a+\dfrac{3}{8}(b-a),\dfrac{a+b}{2} \right]$, $\left[ \dfrac{a+b}{2}, a+\dfrac{7}{8}(b-a) \right]$, and $\left[ a +\dfrac{7}{8}(b-a),b \right]$. Define $B_2(x)$ to be piecewise linear between the prescribed values $B_2(a)=c$, $B_2 \left( a +\dfrac{3}{8}(b-a) \right)=c + \dfrac{5}{8}(d-c)$, $B_2 \left(\dfrac{a+b}{2} \right)=c+\dfrac{d-c}{2}$, $B_2 \left( a + \dfrac{7}{8}(b-a) \right)=d + \dfrac{d-c}{8}$, and $B_2(b)=d$. To make $B_3(x)$ we carry out the procedure we did to make $B_2(x)$ on each of the subintervals we created on the last step (i.e. break each 4 of them up and assign endpoints similarly).

The Bolzano function is defined by the pointwise limit $$B(x)=\displaystyle\lim_{k \rightarrow \infty} B_k(x).$$

Properties

Bolzano function is continuous
Bolzano function is nowhere differentiable

Videos

Bolzano's Continuous but Nowhere Differentiable Function by Wolfram

References

[1]
[2]