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Let \(g: E^{1} \rightarrow E^{1}\) (real) and \(f: E^{1} \rightarrow E\) be relatively continuous on \(J=[c, d]\) and \(I=[a, b],\) respectively, with \(a=g(c)\) and \(b=g(d) .\) Let $$ h=f \circ g $$ Prove that if \(g\) is one to one on \(J,\) then (i) \(g[J]=I,\) so \(f\) and \(h\) describe one and the same arc \(A=f[I]=h[J]\); (ii) \(V_{f}[I]=V_{h}[J] ;\) i.e., \(\ell_{f} A=\ell_{h} A .\) [Hint for (ii): Given \(P=\left\\{a=t_{0}, \ldots, t_{m}=b\right\\}\), show that the points \(s_{i}=g^{-1}\left(t_{i}\right)\) form a partition \(P^{\prime}\) of \(J=[c, d],\) with \(S\left(h, P^{\prime}\right)=S(f, P) .\) Hence deduce \(V_{f}[I] \leq\) \(V_{h}[J]\) Then prove that \(V_{h}[J] \leq V_{f}[I],\) taking an arbitrary \(P^{\prime}=\left\\{c=s_{0}, \ldots, s_{m}=d\right\\},\) and defining \(P=\left\\{t_{0}, \ldots, t_{m}\right\\},\) with \(t_{i}=g\left(s_{i}\right) .\) What if \(\left.g(c)=b, g(d)=a ?\right]\)

Short Answer

Expert verified
The arc lengths of \(f[I]\) and \(h[J]\) are equal; thus \(V_{f}[I] = V_{h}[J]\).

Step by step solution

01

Understanding the Function Mapping

First, recognize that if \(g\) is one-to-one on \(J\), then it uniquely maps each point in \(J\) to a point in \(I\). Therefore, the image \(g[J] = I\). Since \(f\) maps \(I\) to the arc \(A=f[I]\), the composition \(h=f \circ g\) means \(h[J] = f[g[J]] = f[I] = A\), thus proving \(g[J]=I\).
02

Establishing Equal Variation using Partitions

To prove \(V_{f}[I]=V_{h}[J]\), consider any partition \(P=\{a=t_{0}, ..., t_{m}=b\}\) of \(I\). The inverse function \(g^{-1}\) will map these \(t_i\) back to points in \(J\), given by \(s_i=g^{-1}(t_i)\), forming a partition \(P'=\{c=s_0,...,s_m=d\}\) of \(J\). The term \(S(h, P')\), which represents the sum of the lengths of \(h(s_i) - h(s_{i-1})\), will conveniently match \(S(f, P)\) because \(h(s_i) = f(t_i)\). Thus, \(V_{f}[I] \leq V_{h}[J]\).
03

Proving the Reverse Inequality

Now, consider any arbitrary partition \(P'=\{c=s_0, ..., s_m=d\}\) of \(J\). Using the one-to-one property of \(g\), define a corresponding partition \(P=\{t_0, ..., t_m\}\) of \(I\) where \(t_i=g(s_i)\). The sum \(S(f, P)\) will equal \(S(h, P')\), as \(f(t_i)\) corresponds to the variations \(h(s_i)=f(g(s_i))\). This shows \(V_{h}[J] \leq V_{f}[I]\).
04

Combining Results to Conclude Arc Length Equality

Since we showed both \(V_{f}[I] \leq V_{h}[J]\) and \(V_{h}[J] \leq V_{f}[I]\), we can conclude that \(V_{f}[I] = V_{h}[J]\), ensuring that the arc lengths \(\ell_{f} A\) and \(\ell_{h} A\) are indeed equal.

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Key Concepts

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

Arc Length
The arc length is a key concept in real analysis, measured as the distance along a curve. In our problem, we're dealing with a mapping of functions over intervals that create arcs in a geometric sense. The arc length can be thought of as the "curvy" distance between two points, contrasting with the straight distance (which would be a simple line segment). In calculus, we often calculate arc length by integrating, but in our given context, it relates to function composition: the composition of two functions, \(h = f \circ g\), creates an arc known as \(A\). The arc length \(\ell\) associated with this is the same regardless of whether \(f\) directly or the composite function \(h\) is used because of the one-to-one property of \(g\) which preserves that length by merely reparametrizing the path on the curve. Hence, \(\ell_{f} A = \ell_{h} A\).

To calculate the arc length, it is crucial to understand this equality and how the curve \(A\) remains unchanged through different mappings but within the same span of points.
Variation of a Function
Variation of a function quantifies how much a function changes over an interval. Imagine measuring the "twists and turns" over a given section. When we talk about the variation of function \(f\) over \([a, b]\) or \(g\) over \([c, d]\), we are essentially summing all these changes. In our context, we use specific partitions of intervals to calculate this total change. Each partition describes how the interval is divided into smaller subintervals. By examining how the function values vary across these subintervals, we compute what is called the total variation.

The exercise offered showcases how these variations are equal when compared between the original function \(f\) and the composite \(h = f \circ g\). The variations are elucidated through equivalent partitions of the intervals \(I\) and \(J\), establishing \(V_{f}[I] = V_{h}[J]\). This equality is central to illustrating how the mapping and re-mapping, due to the one-to-one property of \(g\), preserve the nature of change.
One-to-One Function
A one-to-one function is a type of mapping where each element of the domain is paired with a unique element of the codomain. In simple terms, imagine every input has a distinct output, ensuring no overlaps in the outputs. This feature is vital when discussing functions like \(g\) in our context because it guarantees that the mapping is consistent and reversible, allowing for the inverse function \(g^{-1}\) to exist.

In the exercise, understanding that \(g\) is one-to-one ensures that there is a unique path from \(J\) to \(I\), and consequently from \(I\) to the arc \(A\). This property allows partitions of \(J\) to precisely match partitions of \(I\) without any ambiguity. It secures that both \(f\) and \(h\) describe one and the same arc, leading to the conclusion that variations and arc lengths are conserved.
Real Analysis
Real analysis is a branch of mathematics focused on the behavior and properties of real numbers. It deals with sequences, series, continuity, differentiation, integration, and more. This field provides the toolbox of rigorous methods we need to tackle and prove properties related to function mappings, like in our exercise.

The problem is posed in the realm of real analysis, as it examines the behavior of real-valued functions through function composition, variation, and arc length. Real analysis offers the precision needed to interpret these mathematical concepts, ensuring that functions \(f\), \(g\), and \(h\) are well-behaved in the mappings explored. With real analysis, one learns how to manage continuity and differentiability properties necessary for comprehending changes and transformations from \(J\) to \(I\), and consequently to the arc \(A\). The rigorous examination within this field ascertains that transformations retain the essence of function's properties, all providing an analytical lens through which to view the exercise.

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Most popular questions from this chapter

Prove that if \(f: E^{1} \rightarrow E^{*}\) is of class \(\mathrm{CD}^{1}\) on \([a, b]\) and if \(-\infty\frac{f(b)-f(a)}{b-a}\left(x_{0}-a\right)+f(a) ; $$ i.e., the curve \(y=f(x)\) lies above the secant through \((a, f(a))\) and \((b, f(b)) .\)

Prove that if \(f\) is differentiable at \(p\) then \(\lim _{x \rightarrow p^{+}} \frac{f(x)-f(y)}{x-y}\) exists, is finite, and equals \(f^{\prime}(p)\) $$ \begin{array}{l} \text { i.e., }(\forall \varepsilon>0)(\exists \delta>0)(\forall x \in(p, p+\delta))(\forall y \in(p-\delta, p)) \\ \left|\frac{f(x)-f(y)}{x-y}-f^{\prime}(p)\right|<\varepsilon \end{array} $$ Show, by redefining \(f\) at \(p,\) that even if the limit exists, \(f\) may not be differentiable (note that the above limit does not involve \(f(p))\).

(i) Let \(f:(a, b) \rightarrow E\) be finite, continuous, with a right derivative on \((a, b) .\) Prove that \(q=\lim _{x \rightarrow a^{+}} f_{+}^{\prime}(x)\) exists (finite) iff $$ q=\lim _{x, y \rightarrow a^{+}} \frac{f(x)-f(y)}{x-y} $$ i.e., iff $$ (\forall \varepsilon>0)(\exists c>a)(\forall x, y \in(a, c) \mid x \neq y) \quad\left|\frac{f(x)-f(y)}{x-y}-q\right|<\varepsilon $$ [Hints: If so, let \(y \rightarrow x^{+}\) (keeping \(x\) fixed) to obtain $$ (\forall x \in(a, c)) \quad\left|f_{+}^{\prime}(x)-q\right| \leq \varepsilon . \quad \text { (Why?) } $$ Conversely, if \(\lim _{x \rightarrow a^{+}} f_{+}^{\prime}(x)=q,\) then $$ (\forall \varepsilon>0)(\exists c>a)(\forall t \in(a, c)) \quad\left|f_{+}^{\prime}(t)-q\right|<\varepsilon $$ Put $$ M=\sup _{a

Let \(x=f(t), y=g(t),\) where \(t\) varies over an open interval \(I \subseteq E^{1},\) define a curve in \(E^{2}\) parametrically. Prove that if \(f\) and \(g\) have derivatives on \(I\) and \(f^{\prime} \neq 0,\) then the function \(h=f^{-1}\) has a derivative on \(f[I]\), and the slope of the tangent to the curve at \(t_{0}\) equals \(g^{\prime}\left(t_{0}\right) / f^{\prime}\left(t_{0}\right)\). [Hint: The word "curve" implies that \(f\) and \(g\) are continuous on \(I\) (Chapter \(4, \S 10)\), so Theorems 1 and 3 apply, and \(h=f^{-1}\) is a function. Also, \(y=g(h(x)) .\) Use Theorem 3 of \(\S 1 .]\)

Find \(\lim _{x \rightarrow 0^{+}} \frac{e^{-1 / x}}{x}\). [Hint: Substitute \(z=\frac{1}{x} \rightarrow+\infty\). Then use the rule.]

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