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Chapter 12: Problem 45

At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$h(x, y)=\cos (x+y)$$

Short Answer

Expert verified
The function h(x, y) = cos(x + y) is continuous at all points (a, b) in 鈩澛�.

Step by step solution

01

Recall the definition of continuity for functions of two variables

A function \(f(x, y)\) is continuous at a point \((a, b)\) if for any \(\epsilon > 0\), there exists a \(\delta > 0\) such that whenever \(\sqrt{(x - a)^{2} + (y - b)^{2}} < \delta\), we have \(|f(x, y) - f(a, b)| < \epsilon\).
02

Use the definition to analyze the function's continuity

To show that \(h(x, y) = \cos(x + y)\) is continuous at all points \((a, b) \in \mathbb{R}^2\), we need to find a \(\delta > 0\) that makes the inequality \(|h(x, y) - h(a, b)| < \epsilon\) true, for any arbitrary \(\epsilon > 0\). Consider the difference, $$|h(x, y) - h(a, b)| = |\cos(x + y) - \cos(a + b)|.$$
03

Apply the sum-to-product identity for cosine

Using the sum-to-product formula for cosine, we get $$|\cos(x + y) - \cos(a + b)| = 2\left|\sin\left(\frac{(x+y)-(a+b)}{2}\right)\sin\left(\frac{(x+y)+(a+b)}{2}\right)\right|.$$ Now we have $$|h(x, y) - h(a, b)| = 2\left|\sin\left(\frac{(x-a)+(y-b)}{2}\right)\sin\left(\frac{(x+a)+(y+b)}{2}\right)\right|.$$
04

Apply the triangle inequality and bound sine function

The sine function is bounded by \(-1 \leq \sin(x) \leq 1\), and we can use the triangle inequality in the form \(|\sin(x)| \leq |x|\): $$|h(x, y) - h(a, b)| \le 2\left|\frac{(x-a) + (y-b)}{2}\right| = |x - a| + |y - b|.$$ Taking into account that \(\sin(x)\) is Lipschitz continuous we can further state that \(|h(x, y) - h(a, b)| \leq 2\sqrt{(x - a)^{2} + (y - b)^{2}}\).
05

Find the appropriate value of \(\delta\)

Given an \(\epsilon > 0\), we can choose \(\delta = \frac{\epsilon}{2}\). Therefore, whenever \(\sqrt{(x - a)^{2} + (y - b)^{2}} < \delta = \frac{\epsilon}{2}\), we have \(|h(x, y) - h(a, b)| \le 2\sqrt{(x - a)^{2} + (y - b)^{2}} < \epsilon\).
06

Conclude the function's continuity

Since we've found a suitable \(\delta\) for any arbitrary \(\epsilon > 0\), it follows that \(h(x, y) = \cos(x + y)\) is continuous at all points \((a, b) \in \mathbb{R}^2\).

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