Difference between revisions of "Bolzano function"
(Created page with "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 $B_1 \colon [a,b] \rightarrow [c,d]$...") |
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| − | 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 | + | 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= | ||
| + | <div class="toccolours mw-collapsible mw-collapsed"> | ||
| + | <strong>Theorem:</strong> The [[Bolzano function]] is everywhere [[continuous]]. | ||
| + | <div class="mw-collapsible-content"> | ||
| + | <strong>Proof:</strong> █ | ||
| + | </div> | ||
| + | </div> | ||
| + | |||
| + | <div class="toccolours mw-collapsible mw-collapsed"> | ||
| + | <strong>Theorem:</strong> The [[Bolzano function]] is [[nowhere differentiable]]. | ||
| + | <div class="mw-collapsible-content"> | ||
| + | <strong>Proof:</strong> █ | ||
| + | </div> | ||
| + | </div> | ||
=Videos= | =Videos= | ||
Revision as of 22:08, 31 December 2015
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
Theorem: The Bolzano function is everywhere continuous.
Proof: █
Theorem: The Bolzano function is nowhere differentiable.
Proof: █
Videos
Bolzano's Continuous but Nowhere Differentiable Function by Wolfram
