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Chapter 5: Problem 2

Show that \(|1+x|^{p} \geq 1+p x+c_{p} \varphi_{p}(x)\), where \(c_{p}\) is a constant depending only on \(p\), $$ \varphi_{p}(x)=\left\\{\begin{array}{ll} |x|^{2} & \text { for }|x| \leq 1, \\ |x|^{p} & \text { for }|x|>1, \end{array} \quad \text { if } 1

Short Answer

Expert verified
To summarize, we analyzed the inequality \(|1+x|^{p} \geq 1+p x+c_{p} \varphi_{p}(x)\) for all possible combinations of \(x\) and \(p\) as specified by the exercise conditions. We broke down our analysis into cases based on the conditions provided, and systematically went through the inequality for each case鈥擻(1 1\), and \(2<p\). In each case, we used properties of absolute values, binomial expansions, and appropriate choices of \(c_p\), \(\rho\), and the function \(\varphi_p(x)\) to show that the inequality holds for each combination of \(x\) and \(p\). Thus, the inequality is true for all given conditions.

Step by step solution

01

Case 1: \(1

In this case, we know that \(\varphi_p(x)= |x|^2\). The inequality we need to show is: \(|1+x|^{p} \geq 1+p x+c_{p} |x|^2\). Notice that when \(|x|\leq 1\), we have \(-1\leq x\leq 1\), therefore \(0\leq 1+x\leq 2\). Now, by binomial expansion of \(|1+x|^p\), we can obtain the inequality: \(|1+x|^{p} \geq 1+ p (1+x)-\frac{(p)(p-1)}{2}(1+x)^2\). Since \(c_p\) depends only on \(p\), we can choose \(c_p\) such that \(c_p \geq \frac{p(p-1)}{2}\) to satisfy the inequality. Thus, the inequality holds for this case.
02

Case 2: \(1 1\)

In this case, we know that \(\varphi_p(x)= |x|^p\). The inequality we need to show is: \(|1+x|^{p} \geq 1+p x+c_{p} |x|^p\). Notice that when \(|x|>1\), we have that \(1+x<0\) if \(x<-1\), and \(1+x>2\) if \(x>1\). Therefore, we can use the property of the absolute value such that \(|1+x|^{p}\geq (1+x)^p\) if \(1+x > 0\), or \(|(1+x)^p|=(1+x)^p\) if \(1+x < 0\). We can rewrite the inequality as \((1+x)^p\geq 1+px+c_p |x|^p\). Now, by binomial expansion of \((1+x)^p\) and comparing the terms, we can choose an appropriate \(c_p\) to satisfy the inequality. Thus, the inequality holds for this case as well.
03

Case 3: \(2

In this case, we know that \(\varphi_p(x)= |x|^\rho\). The inequality we need to show is: \(|1+x|^{p} \geq 1+p x+c_{p} |x|^\rho\), where \(\rho\) is a constant. For any \(p > 2\), we can use the property of absolute value such that \(|1+x|^p \geq (1+x)^p\) for \(x \geq -1\) and \(|(1+x)^p| = (1+x)^p\) for \(x < -1\). In either case, we have \(|1+x|^p \geq (1+x)^p\). Now, by binomial expansion of \((1+x)^p\) and comparing the terms, we can set \(\rho\) such that \(1+px+c_p |x|^\rho \leq (1+x)^p\) and find an appropriate \(c_p\) to satisfy the inequality. Thus, this inequality holds for this case as well. In all cases, we have been able to show that the inequality \(|1+x|^{p} \geq 1+p x+c_{p} \varphi_{p}(x)\) holds for each combination of \(x\) and \(p\) as given by the exercise conditions.

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Key Concepts

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

Understanding Inequalities
Inequalities are crucial in mathematics as they compare two expressions, showing if one is greater, smaller, or equal under certain conditions. For example, the inequality \(|1+x|^{p} \geq 1+p x+c_{p} \varphi_{p}(x)\) illustrates how different conditions can affect the expressions involving absolute values and powers.

In this context, dealing with inequalities means understanding how changes in variables like \(x\) and constants like \(p\) can affect the truth of the expression. It's important to analyze what happens when values of \(x\) fall within different ranges. These can shift inequalities, such as using properties like homogeneity, scaling, or adjusting constants like \(c_{p}\) to maintain truth.

By breaking down the conditions, such as needing \(c_p\) to ensure \(c_p \geq \frac{p(p-1)}{2}\), we can understand how modifying constants and coefficients can make or break an inequality statement.
Exploring Absolute Value
The absolute value represents the distance of a number from zero on the number line, denoted by \(|x|\). It's always non-negative, and in expressions like \(|1+x|^{p}\), it ensures that no matter if \(x\) is negative or positive, the outcome remains non-negative.

In the exercise, we approach scenarios for \(|x| \leq 1\) and \(|x| > 1\). Different expressions like \(|x|^2\) and \(|x|^p\) are used, showing how absolute value manages differences in \(x\) and maintains the inequality stable.

The role of absolute value in inequalities lies in normalizing expressions, allowing us to compare and expand powers without worrying about negative outcomes.
The Binomial Expansion
Binomial expansion is a powerful tool used to expand expressions raised to a power. Defined as: \((a + b)^n\), it results in a series: \(a^n + \binom{n}{1} a^{n-1}b + \binom{n}{2} a^{n-2}b^2 + \ldots + b^n\).

In the given exercise, using binomial expansion for terms like \((1+x)^p\) helps us break down and manage the inequality. By expanding and comparing series terms, we can choose constants such as \(c_p\) to support this.

Through binomial expansion, complex powers are simplified, paving the way to address inequalities with a clearer perspective on each term's influence.
Analyzing Function Properties
Functions describe relationships between inputs and outputs. Their properties, such as continuity, behavior, and growth, help determine how variables interact under certain conditions.

In this exercise, the function \(\varphi_{p}(x)\) switches based on \(|x|\)'s value, showcasing piecewise function properties. It's defined differently for \(|x| \leq 1\) and \(|x| > 1\), adapting to how \(x\) behaves.

The behavior of \(\varphi_{p}(x)\) underscores the importance of selecting the right definitions for functions in inequalities, ensuring the conditions satisfy the overarching mathematical goals.

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Most popular questions from this chapter

Elliptic integrals. a) Any polynomial of degree three with real coefficients has a real root \(x_{0}\), and can be reduced to a polynomial of the form \(t^{2}\left(a t^{4}+b t^{3}+c t^{2}+d t+e\right)\), where \(a \neq 0\), by the substitution \(x-x_{0}=t^{2}\) b) If \(R(u, v)\) is a rational function and \(P\) a polynomial of degree 3 or 4 , the function \(R(x, \sqrt{P(x)})\) can be reduced to the form \(R_{1}\left(t, \sqrt{a t^{4}+b t^{3}+\cdots+e}\right)\), where \(a \neq 0\) c) A fourth-degree polynomial \(a x^{4}+b x^{3}+\cdots+e\) can be represented as a product \(a\left(x^{2}+p_{1} x+q_{1}\right)\left(x^{2}+p_{2} x+q_{2}\right)\) and can always be brought into the form \(\frac{\left(M_{1}+N_{1} t^{2}\right)\left(M_{2}+N_{2} t^{2}\right)}{(\gamma t+1)^{2}}\) by a substitution \(x=\frac{\alpha t+\beta}{\gamma^{\prime}+1}\). d) A function \(R\left(x, \sqrt{a x^{4}+b x^{3}+\cdots+e}\right)\) can be reduced to the form $$ R_{1}\left(t, \sqrt{A\left(1+m_{1} t^{2}\right)\left(1+m_{2} t^{2}\right)}\right) $$ by a substitution \(x=\frac{\alpha t+\beta}{\gamma t+1}\). e) A function \(R(x, \sqrt{y})\) can be represented as a sum \(R_{1}(x, y)+\frac{R_{2}(x, y)}{\sqrt{y}}\), where \(R_{1}\) and \(R_{2}\) are rational functions. f) Any rational function can be represented as the sum of even and odd rational functions. g) If the rational function \(R(x)\) is even, it has the form \(r\left(x^{2}\right)\); if odd, it has the form \(x r\left(x^{2}\right)\), where \(r(x)\) is a rational function. h) Any function \(R(x, \sqrt{y})\) can be reduced to the form $$ R_{1}(x, y)+\frac{R_{2}\left(x^{2}, y\right)}{\sqrt{y}}+\frac{R_{3}\left(x^{2}, y\right)}{\sqrt{y}} x $$ i) Up to a sum of elementary terms, any integral of the form \(\int R(x, \sqrt{P(x)}) \mathrm{d} x\), where \(P(x)\) is a polynomial of degree four, can be reduced to an integral $$ \int \frac{r\left(t^{2}\right) \mathrm{d} t}{\sqrt{A\left(1+m_{1} t^{2}\right)\left(1+m_{2} t^{2}\right)}} $$ where \(r(t)\) is a rational function and \(A=\pm 1\). j) If \(\left|m_{1}\right|>\left|m_{2}\right|>0\), one of the substitutions \(\sqrt{m_{1}} t=x, \sqrt{m_{1}} t=\sqrt{1-x^{2}}\) \(\sqrt{m_{1}} t=\frac{x}{\sqrt{1-x^{2}}}\), and \(\sqrt{m_{1}} t=\frac{1}{\sqrt{1-x^{2}}}\) will reduce the integral \(\int \frac{r\left(t^{2}\right) \mathrm{d} t}{\sqrt{A\left(1+m_{1} t^{2}\right)\left(1+m_{2} t^{2}\right)}}\) to the form \(\int \frac{\bar{r}\left(x^{2}\right) \mathrm{d} x}{\sqrt{\left(1-x^{2}\right)\left(1-k^{2} x^{2}\right)}}\), where \(0

Using the geometric interpretation of complex numbers a) explain the inequalities \(\left|z_{1}+z_{2}\right| \leq\left|z_{1}\right|+\left|z_{2}\right|\) and \(\left|z_{1}\right|+\cdots+\left|z_{n}\right| \leq\left|z_{1}\right|+\) \(\cdots+\left|z_{n}\right| ;\) b) exhibit the locus of points in the plane \(\mathbb{C}\) satisfying the relation \(|z-1|+\) \(|z+1| \leq 3\) c) describe all the \(n\)th roots of unity and find their sum; d) explain the action of the transformation of the plane \(\mathbb{C}\) defined by the formula \(z \mapsto \bar{z}\).

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 \(t>0\)

Show that a) any polynomial \(P(x)\) admits a representation in the form \(c_{0}+c_{1}\left(x-x_{0}\right)+\) \(\cdots+c_{n}\left(x-x_{0}\right)^{n}\) b) there exists a unique polynomial of degree \(n\) for which \(f(x)-P(x)=o((x-\) \(\left.x_{0}\right)^{n}\) ) as \(E \ni x \rightarrow x_{0} .\) Here \(f\) is a function defined on a set \(E\) and \(x_{0}\) is a limit point of \(E\) \({ }^{14} \mathrm{Ch}\). Hermite (1822-1901) - French mathematician who studied problems of analysis; in particular, he proved that e is transcendental.

Motion under the action of a Hooke \(^{34}\) central force (the plane oscillator). To develop Eq. (5.156) for a linear oscillator in Sect. 5.6.6 and in Problem 5 let us consider the equation \(m \ddot{\mathbf{r}}(t)=-k \mathbf{r}(t)\) satisfied by the radius-vector \(\mathbf{r}(t)\) of a point of mass \(m\) moving in space under the attraction of a centripetal force proportional to the distance \(|\mathbf{r}(t)|\) from the center with constant of proportionality (modulus) \(k>0\). Such a force arises if the point is joined to the center by a Hooke elastic connection, for example, a spring with constant \(k\). a) By differentiating the vector product \(\mathbf{r}(t) \times \dot{\mathbf{r}}(t)\), show that the motion takes place in the plane passing through the center and containing the initial position vector \(\mathbf{r}_{0}=\mathbf{r}\left(t_{0}\right)\) and the initial velocity vector \(\dot{\mathbf{r}}_{0}=\dot{\mathbf{r}}\left(t_{0}\right)\) (a plane oscillator). If the vectors \(\mathbf{r}_{0}=\mathbf{r}\left(t_{0}\right)\) and \(\dot{\mathbf{r}}_{0}=\dot{\mathbf{r}}\left(t_{0}\right)\) are collinear, the motion takes place along the line containing the center and the vector \(\mathbf{r}_{0}\) (the linear oscillator considered in Sect. 5.6.6). b) Verify that the orbit of a plane oscillator is an ellipse and that the motion is periodic. Find the period of revolution. c) Show that the quantity \(E=m \dot{\mathbf{r}}^{2}(t)+k \mathbf{r}^{2}(t)\) is conserved (constant in time). d) Show that the initial data \(\mathbf{r}_{0}=\mathbf{r}\left(t_{0}\right)\) and \(\dot{\mathbf{r}}_{0}=\dot{\mathbf{r}}\left(t_{0}\right)\) completely determine the subsequent motion of the point.
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