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Chapter 7: Problem 21

Prove that \(|\sin x-\sin y| \leq|x-y|\) and \(|\cos x-\cos y| \leq|x-y|\) for all \(x, y \in \mathbb{R}\).

Short Answer

Expert verified
We have proven the inequalities by applying the mean value theorem to the functions \(f(x) = \sin x\) and \(g(x) = \cos x\), which are continuous and differentiable for all \(x \in \mathbb{R}\). For each function, we found a value \(c\) between \(x\) and \(y\) such that \(|\cos c| \leq \left|\frac{\sin x - \sin y}{x - y}\right|\) and \(|\sin c| \leq \left|\frac{\cos x - \cos y}{x - y}\right|\), respectively. We then used the properties of absolute values and the bound \(|\cos c|, |\sin c| \leq 1\) to obtain the desired inequalities \(|\sin x - \sin y| \leq |x - y|\) and \(|\cos x - \cos y| \leq |x - y|\) for all \(x, y \in \mathbb{R}\).

Step by step solution

01

Proving the inequality for sin(x)

To prove the inequality \(|\sin x - \sin y| \leq |x - y|\), let's first consider the function \(f(x) = \sin x\). This function is continuous and differentiable for all \(x \in \mathbb{R}\). According to the mean value theorem, for any two points x and y, there exists some value \(c\) between x and y such that \(f'(c) = \frac{f(x) - f(y)}{x - y}\), which implies \(\cos c = \frac{\sin x - \sin y}{x - y}\). Now, taking absolute values on both sides, we get \(|\cos c| = \left|\frac{\sin x - \sin y}{x - y}\right|\). We know that \(|\cos c| \leq 1\). Hence, \(\left|\frac{\sin x - \sin y}{x - y}\right| \leq 1\). Then, multiplying both sides by \(|x - y|\), we obtain the desired inequality: \(|\sin x - \sin y| \leq |x - y|\).
02

Proving the inequality for cos(x)

Next, let's prove the inequality \(|\cos x - \cos y| \leq |x - y|\). Let's consider the function \(g(x) = \cos x\). This function is also continuous and differentiable for all \(x \in \mathbb{R}\). According to the mean value theorem, for any two points x and y, there exists some value \(c\) between x and y such that \(g'(c) = \frac{g(x) - g(y)}{x - y}\), which implies \(-\sin c = \frac{\cos x - \cos y}{x - y}\). Now, taking absolute values on both sides, we get \(|-\sin c| = \left|\frac{\cos x - \cos y}{x - y}\right|\), or \(|\sin c| = \left|\frac{\cos x - \cos y}{x - y}\right|\). We know that \(|\sin c| \leq 1\). Hence, \(\left|\frac{\cos x - \cos y}{x - y}\right| \leq 1\). Finally, multiplying both sides by \(|x - y|\), we obtain the desired inequality: \(|\cos x - \cos y| \leq |x - y|\). Thus, we have proven that both inequalities hold for all \(x, y \in \mathbb{R}\).

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Most popular questions from this chapter

Let \(p, q \in(1, \infty)\) be such that \((1 / p)+(1 / q)=1\). (i) If \(f:[0, \infty) \rightarrow \mathbb{R}\) is defined by \(f(x):=(1 / q)+(1 / p) x-x^{1 / p}\), then show that \(f(x) \geq f(1)\) for all \(x \in[0, \infty)\). (ii) Show that \(a b \leq\left(a^{p} / p\right)+\left(b^{q} / q\right)\) for all \(a, b \in[0, \infty)\). (Hint: If \(b \neq 0\), let \(x:=a^{p} / b^{q}\) in (i).) (iii) (H枚lder Inequality for Sums) Given any \(a_{1}, \ldots, a_{n}\) and \(b_{1}, \ldots, b_{n}\) in \(\mathbb{R}\), prove that $$ \sum_{i=1}^{n}\left|a_{i} b_{i}\right| \leq\left(\sum_{i=1}^{n}\left|a_{i}\right|^{p}\right)^{1 / p}\left(\sum_{i=1}^{n}\left|b_{i}\right|^{q}\right)^{1 / q} $$ Deduce the Cauchy-Schwarz inequality as a special case. (iv) (H枚lder Inequality for Integrals) Given any continuous functions \(f, g:[a, b] \rightarrow \mathbb{R}\), prove that $$ \int_{a}^{b}|f(x) g(x)| d x \leq\left(\int_{a}^{b}|f(x)|^{p} d x\right)^{1 / p}\left(\int_{a}^{b}|g(x)|^{q} d x\right)^{1 / q} $$ (v) (Minkowski Inequality for Sums) Given any \(a_{1}, \ldots, a_{n}\) and \(b_{1}, \ldots, b_{n}\) in \(\mathbb{R}\), prove that $$ \left(\sum_{i=1}^{n}\left|a_{i}+b_{i}\right|^{p}\right)^{1 / p} \leq\left(\sum_{i=1}^{n}\left|a_{i}\right|^{p}\right)^{1 / p}+\left(\sum_{i=1}^{n}\left|b_{i}\right|^{p}\right)^{1 / p} $$ (Hint: The \(p\) th power of the expression on the left can be written as \(\sum_{i=1}^{n}\left|a_{i}\right|\left(\left|a_{i}+b_{i}\right|\right)^{p-1}+\sum_{i=1}^{n}\left|b_{i}\right|\left(\left|a_{i}+b_{i}\right|\right)^{p-1} ;\) now use (iii). \()\) (vi) (Minkowski Inequality for Integrals) Given any continuous functions \(f, g:[a, b] \rightarrow \mathbb{R}\), prove that $$ \left(\int_{a}^{b}|f(x)+g(x)|^{p} d x\right)^{1 / p} \leq\left(\int_{a}^{b}|f(x)|^{p} d x\right)^{1 / p}+\left(\int_{a}^{b}|g(x)|^{p} d x\right)^{1 / p}. $$

If \(x_{1}, x_{2} \in \mathbb{R}\) are such that none of \(x_{1}, x_{2}\), and \(x_{1}+x_{2}\) equals \((2 k+1) \pi / 2\) for any \(k \in \mathbb{Z}\), then show that $$ \tan \left(x_{1}+x_{2}\right)=\frac{\tan x_{1}+\tan x_{2}}{1-\tan \left(x_{1}+x_{2}\right)}. $$

Determine whether the following functions are algebraic or transcendental: (i) \(f(x)=\pi x^{11}+\pi^{2} x^{5}+9\) for \(x \in \mathbb{R}\) (ii) \(f(x)=\frac{e x^{2}+\pi}{\pi x^{2}+e}\) for \(x \in \mathbb{R}\), (iii) \(f(x)=\ln _{10} x\) for \(x>0\), (iv) \(f(x)=x^{\pi}\) for \(x>0\).

Let \(r\) be a positive real number and \(\theta \in(-\pi, \pi]\) and \(\alpha \in \mathbb{R}\) be such that \(\theta+\alpha \in(-\pi, \pi] .\) If \(P\) and \(P_{\alpha}\) denote the points with polar coordinates \((r, \theta)\) and \((r, \theta+\alpha)\), respectively, then find the Cartesian coordinates of \(P_{\alpha}\) in terms of the Cartesian coordinates of \(P\). [Note: The transformation \(P \mapsto P_{\alpha}\) corresponds to a rotation of the plane by the angle \(\alpha\).]

Let \(a, b \in(0, \infty)\). (i) Consider the functions \(f, g:(0, \infty) \rightarrow \mathbb{R}\) defined by \(f(x):=\log _{a} x\) and \(g(x):=\log _{b} x .\) Show that \(f\) and \(g\) have the same rate as \(x \rightarrow \infty\). (ii) Consider the functions \(f, g: \mathbb{R} \rightarrow \mathbb{R}\) defined by \(f(x):=a^{x}\) and \(g(x):=b^{x} .\) Show that the growth rate of \(f\) is less than that of \(g\) as \(x \rightarrow \infty\) if and only if \(a
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