- #1
Dragonfall
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Does there exist a continuous function [tex]f:\mathbb{R}\rightarrow\mathbb{R}[/tex] such that f is nowhere constant and [tex]\{x:f(x)=0\}[/tex] is uncountable?
Moo Of Doom said:Take the function f(x) = 0 on [0,1]. Now replace f on the interval [1/3,2/3] with a triangle wave. Now replace f on the intervals [1/9,2/9] and [7/9,8/9] with a similar triangle wave. Repeat this process for every interval on which f is zero, and we have a continuous function that vanishes on the Cantor set (which is uncountable), and is nowhere constant. Replacing the triangle wave with a suitable [itex]C^{\infty}[/itex] function yields an infinitely differentiable function that vanishes uncountably many times but is not constant. If you want it to be continuous but nowhere differentiable, replace each triangle wave with such a function instead.
Moo Of Doom said:"if we infinitely repeat..." simply refers to the limit function of the sequence of functions I described.
Define
[tex]
f_1[a,b](x) = \left\{
\begin{array}{cc}
0 & x \in \left[a,\frac{2a+b}{3}\right]\cup\left[\frac{a+2b}{3},b\right]\\
x-\frac{2a+b}{3} & x \in \left[\frac{2a+b}{3},\frac{a+b}{2}\right]\\
\frac{a+2b}{3}-x & x \in \left[\frac{a+b}{2},\frac{a+2b}{3}\right]
\end{array}
[/tex]
and
[tex]
f_{n+1}[a,b](x) = \left\{
\begin{array}{cc}
f_n\left[a,\frac{2a+b}{3}\right](x) & x \in \left[a,\frac{2a+b}{3}\right]\\
x-\frac{2a+b}{3} & x \in \left[\frac{2a+b}{3},\frac{a+b}{2}\right]\\
\frac{a+2b}{3}-x & x \in \left[\frac{a+b}{2},\frac{a+2b}{3}\right]\\
f_n\left[\frac{a+2b}{3},b\right](x) & x \in \left[\frac{a+2b}{3},b\right]
\end{array}
[/tex]
Then [tex]f(x) = \lim_{n\to\infty}f_n[0,1](x)[/tex] is the function I just described.
The existence of uncountable zeros in continuous functions is significant because it challenges the concept of continuity. According to the intermediate value theorem, a continuous function cannot have a jump or break in its graph. However, the presence of uncountable zeros suggests that there are points in the function where the value remains constant, creating a discontinuity.
Yes, continuous functions can have uncountable zeros. This is because the definition of a continuous function only requires that the function be continuous at every point in its domain, not that it has a derivative or is differentiable at every point. Therefore, a function can be continuous and have uncountable zeros at the same time.
One way to prove the existence of uncountable zeros in a continuous function is by using the Bolzano-Weierstrass theorem. This theorem states that if a continuous function has opposite signs at two points, then there must be a point between them where the function equals zero. By using this theorem repeatedly, we can show that there are infinitely many points where the function equals zero, and potentially an uncountable number of points.
Yes, there are real-world applications of uncountable zeros in continuous functions, particularly in physics and engineering. For example, in fluid dynamics, the Navier-Stokes equations describe the motion of fluids and have solutions with uncountable zeros. In electronic circuit design, the transfer function of a circuit can have uncountable zeros, which affect the stability and performance of the circuit.
The existence of uncountable zeros can cause disruptions in the graph of a continuous function. At these points, the function may have a constant value, creating a flat line in the graph. This can also result in a discontinuity in the graph, as the function's value suddenly changes from one constant value to another. Additionally, the presence of uncountable zeros can make it challenging to accurately graph the function, as there may be infinitely many points to plot.