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Chapter 5: Problem 5

Set $$ \Psi_{0}(x)= \begin{cases}x, & \text { if } 0 \leq x \leq \frac{1}{2} \\ 1-x, & \text { if } \frac{1}{2} \leq x \leq 1\end{cases} $$ and extend this function to the entire real line so as to have period 1 . We denote the extended function by \(\varphi_{0}\). Further, let $$ \varphi_{n}(x)=\frac{1}{4^{n}} \varphi_{0}\left(4^{n} x\right) $$ The function \(\varphi_{n}\) has period \(4^{-n}\) and a derivative equal to \(+1\) or \(-1\) everywhere except at the points \(x=\frac{k}{2^{2 n+1}}, k \in \mathbb{Z} .\) Let $$ f(x)=\sum_{n=1}^{\infty} \varphi_{n}(x) $$ Show that the function \(f\) is defined and continuous on \(\mathbb{R}\), but does not have a derivative at any point. (This example is due to the well-known Dutch mathematician B.L. van der Waerden (1903-1996). The first examples of continuous functions having no derivatives were constructed by Bolzano (1830) and Weierstrass (1860).)

Short Answer

Expert verified
Based on the given step-by-step solution, the function \(f(x)\), which is defined as the sum of all functions \(\varphi_{n}(x)\) for \(n=1,2,3,...\), is continuous on the entire real line but does not have a derivative at any point. This is because the function is well-defined and converges for all \(x\in \mathbb{R}\), and each of the constituent functions \(\varphi_n(x)\) is continuous and periodic. However, the limit used to compute the derivative of \(f(x)\) does not exist at any point due to the behavior of the constituent functions at specific points.

Step by step solution

01

Extend \(\Psi_{0}(x)\) to \(\varphi_{0}(x)\) with period 1

To extend the function \(\Psi_{0}(x)\) to have period 1, we can define \(\varphi_{0}(x)\) by repeating the values \(\Psi_{0}(x)\) has in the interval \([0, 1]\) for every interval of length 1 on the entire real line. The function can be defined as follows: $$ \varphi_{0}(x) = \Psi_{0}(x - \lfloor x \rfloor), $$ where \(\lfloor x \rfloor\) denotes the greatest integer less than or equal to \(x\). This effectively "shifts" the values of \(\Psi_{0}(x)\) to every integer value of \(x\), thereby creating a periodic function with period 1. #Step 2: Compute the definition of \(\varphi_{n}(x)\)#
02

Define \(\varphi_n(x)\)

For each positive integer \(n\), we define the function \(\varphi_{n}(x)\) by $$ \varphi_{n}(x)=\frac{1}{4^{n}} \varphi_{0}\left(4^{n} x\right) $$ Since \(\varphi_{0}(x)\) is periodic with period 1, each \(\varphi_{n}(x)\) will also be periodic, but with period \(4^{-n}\). #Step 3: Show that \(f(x)\) is defined on the entire real line#
03

Define \(f(x)\) on the entire real line

To show that \(f(x)\) is defined on the entire real line, we need to show that the sum \(\sum_{n=1}^{\infty} \varphi_{n}(x)\) converges for all values of \(x \in \mathbb{R}\). Notice that for each \(n\), \(0 \leq \varphi_{n}(x) \leq \frac{1}{4} \cdot \frac{1}{4^n} = \frac{1}{4^{n+1}}\). Since the series \(\sum_{n=1}^{\infty}\frac{1}{4^{n+1}}\) converges (as it is a geometric series with a common ratio less than 1), the sum \(\sum_{n=1}^{\infty} \varphi_{n}(x)\) also converges for every real \(x\), by the Comparison Test. Therefore, \(f(x)\) is defined on the entire real line. #Step 4: Show that \(f(x)\) is continuous#
04

Prove that \(f(x)\) is continuous

To prove that \(f(x)\) is continuous, we need to show that for any \(x \in \mathbb{R}\) and \(\epsilon > 0\), there exists a \(\delta > 0\) such that for all \(|x - c| < \delta\), the inequality \(|f(x) - f(c)| < \epsilon\) holds. Since each \(\varphi_{n}(x)\) is continuous and periodic, it is also uniformly continuous on its domain. Thus, for each \(n\) and given \(\epsilon > 0\), we can find \(\delta_n > 0\) such that for all \(|x - c| < \delta_n\), \(|\varphi_{n}(x) - \varphi_{n}(c)| < \frac{\epsilon}{2^n}\). Take \(\delta = \min\{\delta_1, \delta_2, ..., \delta_n, ...\}\). Then for all \(|x - c| < \delta\), the inequality \(|f(x) - f(c)| \leq \sum_{n=1}^{\infty} |\varphi_{n}(x) - \varphi_{n}(c)| < \sum_{n=1}^{\infty} \frac{\epsilon}{2^n} = \epsilon\) holds, showing that \(f(x)\) is continuous. #Step 5: Show that \(f(x)\) does not have a derivative at any point#
05

Prove that \(f(x)\) does not have a derivative at any point

To prove that \(f(x)\) does not have a derivative at any point, we need to show that the limit $$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ does not exist for any \(x \in \mathbb{R}\). Notice that for each \(n\), there exist points \(x_n\) and \(y_n\) such that \(\varphi_{n}(x_n + h) - \varphi_{n}(x_n) = \pm \frac{h}{4^n}\). Therefore, as we take the limit as \(h \to 0\), the term \(|\frac{1}{h}(\varphi_{n}(x_n + h) - \varphi_{n}(x_n))|\) does not converge to 0 for every \(n\). This implies that the limit $$\lim_{h \to 0} \frac{f(x_n+h) - f(x_n)}{h}$$ does not exist either. Furthermore, since \(x_n\) and \(y_n\) can be picked arbitrarily, \(f(x)\) does not have a derivative at any point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Periodic Functions
A periodic function is one that repeats its values at regular intervals. The concept of periodicity is crucial in mathematics, especially in analyzing patterns and cycles.
When we extend \(\Psi_{0}(x)\) to \(\varphi_{0}(x)\), we ensure that the function repeats every time we move one unit along the x-axis. This means that \(\varphi_{0}(x)\) will have the same value for \(x = 1, 2, 3, \) and so on.
  • **Applications**: Periodic functions are used in various practical fields, including engineering, physics, and signal processing, as they help model cyclical phenomena like sound waves.
  • **Notation**: The period of a function is symbolized as \(T\), and for a function \(f(x)\), it satisfies \(f(x+T) = f(x)\) for all \(x\).
The function \(\varphi_{n}(x)\) also builds upon this concept by scaling and repeating at different intervals, showing the versatility and significance of periodicity in mathematical modeling.
Derivative
The derivative of a function represents its rate of change. It tells us how the function's output changes for a small change in the input.
In the context of the given functions, each \(\varphi_{n}(x)\) is simple enough to have a derivative everywhere except at specific points. The function behaves in such a way that its derivative at most points is either \(+1\) or \(-1\), indicating direct and uniform slopes.
  • **Understanding Slope**: If you think about a hill, the derivative is like checking the steepness at various points. If the slope is rising, the derivative is positive, and if it鈥檚 falling, the derivative is negative.
  • **Non-Existence**: For \(f(x)\), however, the derivative is nonexistent at any point, which is an intriguing feature of this function. This happens because, as you zoom in, the function appears too oscillatory to determine a single tangent line.
This shows how complex behavior can arise from the combination of simpler functions with well-defined derivatives.
Geometric Series
A geometric series is a sum of terms, each multiplied by a fixed ratio from the previous one. Geometric series play a crucial role in understanding convergence, which is pivotal to proving the definition of the function \(f(x)\).
In mathematics, a series is convergent if the sum of its terms approaches a specific value as more terms are added. The given exercise employed a geometric series in showing that the sum of the functions \(\varphi_{n}(x)\) converges to define \(f(x)\).
  • **Convergence**: When analyzing \(\sum_{n=1}^{\infty} \ rac{1}{4^{n+1}}\), we see it converges because the ratio \(\frac{1}{4}\) is less than 1, ensuring the terms get progressively smaller.
  • **Formula**: The sum of an infinite geometric series \(a + ar + ar^2 + \cdots\) is \(\frac{a}{1-r}\) when \(|r| < 1\).
This principle underlies the function \(f(x)\), confirming it's properly defined over the real numbers due to the convergent nature of the series.
Uniform Continuity
Uniform continuity is a stronger form of continuity for functions. It ensures that small changes in the input lead to small changes in the output, uniformly across the function's entire domain.
For the function \(f(x)\), because each \('\varphi_{n}(x)\)\' exhibits uniform continuity, the overall series \(f(x)\) inherits this property, meaning it is continuous everywhere on the real line.
  • **Key Property**: With uniform continuity, the choice of \(\delta\) depends only on \(\epsilon\) and not on the point \(x\), contrasting with standard continuity where \(\delta\) might vary.
  • **Implications**: This guarantees that \(f(x)\) doesn't have abrupt changes or jumps anywhere on its domain, crucial for mathematical functions modeling real-world scenarios smoothly.
Understanding this concept helps appreciate why certain complex functions still maintain a continuity that seems counterintuitive at first glance.

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Most popular questions from this chapter

The Legendre transform \(^{21}\) of a function \(f: I \rightarrow \mathbb{R}\) defined on an interval \(I \subset \mathbb{R}\) is the function $$ f^{*}(t)=\sup _{x \in I}(t x-f(x)) $$ Show that a) The set \(I^{*}\) of values of \(t \in \mathbb{R}\) for which \(f^{*}(t) \in \mathbb{R}\) (that is, \(\left.f^{*}(t) \neq \infty\right)\) is either empty or consists of a single point, or is an interval of the line, and in this last case the function \(f^{*}(t)\) is convex on \(I^{*}\). b) If \(f\) is a convex function, then \(I^{*} \neq \varnothing\), and for \(f^{*} \in C\left(I^{*}\right)\) $$ \left(f^{*}\right)^{*}=\sup _{t \in I^{*}}\left(x t-f^{*}(t)\right)=f(x) $$ for any \(x \in I .\) Thus the Legendre transform of a convex function is involutive, (its square is the identity transform). c) The following inequality holds: $$ x t \leq f(x)+f^{*}(t) \quad \text { for } x \in I \text { and } t \in I^{*} $$ d) When \(f\) is a convex differentiable function, \(f^{*}(t)=t x_{t}-f\left(x_{t}\right)\), where \(x_{t}\) is determined from the equation \(t=f^{\prime}(x) .\) Use this relation to obtain a geometric interpretation of the Legendre transform \(f^{*}\) and its argument \(t\), showing that the Legendre transform is a function defined on the set of tangents to the graph of \(f\). e) The Legendre transform of the function \(f(x)=\frac{1}{\alpha} x^{\alpha}\) for \(\alpha>1\) and \(x \geq 0\) is the function \(f^{*}(t)=\frac{1}{\beta} t^{\beta}\), where \(t \geq 0\) and \(\frac{1}{\alpha}+\frac{1}{\beta}=1 .\) Taking account of \(\left.\mathrm{c}\right)\), use this fact to obtain Young's inequality, which we already know: $$ x t \leq \frac{1}{\alpha} x^{\alpha}+\frac{1}{\beta} t^{\beta} $$ f) The Legendre transform of the function \(f(x)=\mathrm{e}^{x}\) is the function \(f^{*}(t)=\) \(t \ln \frac{t}{e}, t>0\), and the inequality $$ x t \leq \mathrm{e}^{x}+t \ln \frac{t}{\mathrm{e}} $$ holds for \(x \in \mathbb{R}\) and \(t>0\)

Using induction on \(k, 1 \leq k\), we define the finite differences of order \(k\) of the function \(f\) at \(x_{0}\) : $$ \begin{aligned} \Delta^{1} f\left(x_{0} ; h_{1}\right):=& \Delta f\left(x_{0} ; h_{1}\right)=f\left(x_{0}+h_{1}\right)-f\left(x_{0}\right) \\ \Delta^{2} f\left(x_{0} ; h_{1}, h_{2}\right):=& \Delta \Delta f\left(x_{0} ; h_{1}, h_{2}\right)=\\\ =&\left(f\left(x_{0}+h_{1}+h_{2}\right)-f\left(x_{0}+h_{2}\right)\right)-\\\ &-\left(f\left(x_{0}+h_{1}\right)-f\left(x_{0}\right)\right)=\\\ =& f\left(x_{0}+h_{1}+h_{2}\right)-f\left(x_{0}+h_{1}\right)-f\left(x_{0}+h_{2}\right)+f\left(x_{0}\right), \\\ \vdots & \\ \Delta^{k} f\left(x_{0} ; h_{1}, \ldots, h_{k}\right):=& \Delta^{k-1} g_{k}\left(x_{0} ; h_{1}, \ldots, h_{k-1}\right) \end{aligned} $$ where \(g_{k}(x)=\Delta^{1} f\left(x ; h_{k}\right)=f\left(x+h_{k}\right)-f(x)\). a) Let \(f \in C^{(n-1)}[a, b]\) and suppose that \(f^{(n)}(x)\) exists at least in the open interval \(] a, b\left[\right.\). Show that if all the points \(x_{0}, x_{0}+h_{1}, x_{0}+h_{2}, x_{0}+h_{1}+h_{2}, \ldots, x_{0}+\) \(h_{1}+\cdots+h_{n}\) lie in \([a, b]\), then inside the smallest closed interval containing all of them there is a point \(\xi\) such that $$ \Delta^{n} f\left(x_{0} ; h_{1}, \ldots, h_{n}\right)=f^{(n)}(\xi) h_{1} \cdots h_{n} $$ b) (Continuation.) If \(f^{(n)}\left(x_{0}\right)\) exists, then the following estimate holds: $$ \begin{aligned} &\left|\Delta^{n} f\left(x_{0} ; h_{1}, \ldots, h_{n}\right)-f^{(n)}\left(x_{0}\right) h_{1} \cdots h_{n}\right| \leq \\ &\quad \leq \sup _{x \in j a, b \mid}\left|f^{(n)}(x)-f^{(n)}\left(x_{0}\right)\right| \cdot\left|h_{1}\right| \cdots\left|h_{n}\right| \end{aligned} $$ c) (Continuation.) Set \(\Delta^{n} f\left(x_{0} ; h, \ldots, h\right)=: \Delta^{n} f\left(x_{0} ; h^{n}\right)\). Show that if \(f^{(n)}\left(x_{0}\right)\) exists, then $$ f^{(n)}\left(x_{0}\right)=\lim _{h \rightarrow 0} \frac{\Delta^{n} f\left(x_{0} ; h^{n}\right)}{h^{n}} $$ d) Show by example that the preceding limit may exist even when \(f^{(n)}\left(x_{0}\right)\) does not exist. Hint: Consider, for example, \(\Delta^{2} f\left(0 ; h^{2}\right)\) for the function $$ f(x)= \begin{cases}x^{3} \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{cases} $$ and show that $$ \lim _{h \rightarrow 0} \frac{\Delta^{2} f\left(0 ; h^{2}\right)}{h^{2}}=0 $$

A body that can be regarded as a point mass is sliding down a smooth hill under the influence of gravity. The hill is the graph of a differentiable function \(y=f(x)\). a) Find the horizontal and vertical components of the acceleration vector that the body has at the point \(\left(x_{0}, y_{0}\right)\). b) For the case \(f(x)=x^{2}\) when the body slides from a great height, find the point of the parabola \(y=x^{2}\) at which the horizontal component of the acceleration is maximal.

a) Show that the equation \(\mathrm{e}^{w}=z\) in \(w\) has the solution \(w=\ln |z|+i\) Arg \(z\). It is natural to regard \(w\) as the natural logarithm of \(z\). Thus \(w=\operatorname{Ln} z\) is not a functional relation, since Arg \(z\) is multi-valued. b) Find \(\operatorname{Ln} 1\) and \(\operatorname{Ln} i\). c) \(\operatorname{Set} z^{\alpha}=\mathrm{e}^{\alpha \operatorname{Ln} z} .\) Find \(1^{\pi}\) and \(i^{i}\). d) Using the representation \(w=\sin z=\frac{1}{2 l}\left(\mathrm{e}^{i z}-\mathrm{e}^{-i z}\right)\), obtain an expression for \(z=\arcsin w\) e) Are there points in \(\mathbb{C}\) where \(|\sin z|=2 ?\)

Write the formulas for approximate computation of the following values: a) \(\sin \left(\frac{\pi}{6}+\alpha\right)\) for values of \(\alpha\) near 0 ; b) \(\sin \left(30^{\circ}+\alpha^{\circ}\right)\) for values of \(\alpha^{\circ}\) near 0; c) \(\cos \left(\frac{\pi}{4}+\alpha\right)\) for values of \(\alpha\) near 0 ; d) \(\cos \left(45^{\circ}+\alpha^{\circ}\right)\) for values of \(\alpha^{\circ}\) near 0 .
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