where \(a_n, a_{n-1}, \cdots, a_0 \in\mathbb{R}\) is a continuous function on \(\mathbb{R}\). Such functions are called polynomial functions.
If \(g(x)\) and \(h(x)\) are polynomial functions such that \(h(x) \neq 0\), the function \(f(x)\) defined by
\[f(x) = \frac{g(x)}{h(x)}\]
is continuous on \(\{x \mid x \in \mathbb{R}, h(x) \neq 0\}\). Such functions are called rational functions. If \(h(x) = 1\) then \(f(x) = g(x)\) so any polynomial functions are also rational functions (i.e., polynomial functions are a special case of rational functions).
We can ``algebraically'' define functions other than polynomial or rational functions. For example, \(f(x) = \sqrt{x}\) is not a rational function, but satisfies an algebraic equality
\[[f(x)]^2 - x = 0.\]
Definition (Algebraic function)
The continuous function \(f(x)\) is said to be an algebraic function if there exist polynomial functions \(g_0(x), g_1(x), \cdots, g_n(x)\) such that the following identity is satisfied:
Example. Rational functions are a special case of algebraic functions. □
Example. Let us prove that the function \(f(x) = \sqrt{x}\) defined on \(x \geq 0\) is continuous.
First, consider the case when \(a > 0\). For all \(\varepsilon > 0\), let us define \(\delta= \varepsilon\sqrt{a}\). Then, if \(x \geq 0\) and \(0 < |x - a| < \delta\),
Next, consider the case when \(a = 0\). For any \(\varepsilon > 0\), take \(\delta = \varepsilon^2 > 0\). If \(0 < x < \delta\), then \(|\sqrt{x} - 0| = \sqrt{x} < \sqrt{\delta} = \varepsilon\). Thus, \(\lim_{x\to +0}\sqrt{x} = 0\). Therefore, \(f(x) = \sqrt{x}\) is continuous at all \(x\geq 0\). □
Exponential functions
Definition (Exponential function)
Let \(a > 0\) be a real number. The function defined by
\[f(x) = a^x\]
is called an exponential function with base \(a\). In particular, when we simply say the exponential function, the base is \(e\) (Napier's constant).
Exponential functions are continuous everywhere on \(\mathbb{R}\). If \(a > 1\), then \(a^x\) is a strictly increasing function. If \(0 < a < 1\), then \(a^x\) is a strictly decreasing function. If \(a = 1\), then \(a^x = 1\) for all \(x\in\mathbb{R}\).
But what do we mean by \(a^x\), exactly? Review how we introduced \(e^x\) through \(\exp(x)\).
First, we provide the following theorem without proof.
Theorem
Let \(f(x)\) be a continuous and strictly monotone increasing function defined on an interval \(I\). Then \(f(x)\) has the inverse \(f^{-1}(x)\) which is also continuous and strictly monotone increasing.
Let \(f(x)\) be a continuous and strictly monotone decreasing function defined on an interval \(I\). Then \(f(x)\) has the inverse \(f^{-1}(x)\) which is also continuous and strictly monotone decreasing.
Remark. If \(f(x)\) is a strictly monotone function (either increasing or decreasing) on an interval \(I\), then the function is a bijection from \(I\) to \(f(I)\). For every bijection, there exists an inverse map that is also bijective. □
Using this theorem, we can see that for each exponential function \(f(x) = a^x\), there is its inverse function \(f^{-1}(x)\) which we define as the logarithmic function with base \(a\) denoted \(f^{-1}(x) = \log_a(x)\). When the base is \(e\) (Napier's constant), \(\log_{e}(x)\), this is the logarithm we have defined earlier and we often omit the base to write simply \(\log (x)\) or \(\ln(x)\).
Let \( t = e^x - 1\). Then \(x = \log(1 + t)\) As \(x \to 0\), \(t \to 0\) so \[\lim_{x\to 0}\frac{e^x - 1}{x} = \lim_{t\to 0}\frac{t}{\log(1+t)} = 1\] using the result of Part 1.
■
Trigonometric functions
We already know \(\sin\) and \(\cos\). The tangent function is defined as
\[\tan x = \frac{\sin x}{\cos x}, ~ x \in \mathbb{R}\setminus\left\{\left(n + \frac{1}{2}\right)\pi\mid n\in \mathbb{Z}\right\}.\]
Quiz. Why do we exclude the points \(x = \left(n + \frac{1}{2}\right)\pi, n \in \mathbb{Z}\), from the domain of the tangent function? □
\(\sin\) and \(\cos\) have the fundamental period of \(2\pi\) whereas \(\tan\) has the fundamental period of \(\pi\). That is, for \(n\in\mathbb{Z}\),
Since \(\sin\), \(\cos\), and \(\tan\) are periodic functions (and hence not monotone), they don't have inverse functions. Nevertheless, by restricting their domains, we may define the inverse functions.
\(\sin x\) is strictly monotone increasing on the closed interval \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). Therefore it has an inverse function on this domain, which we define as \(\arcsin x\). In other words, we consider \(\sin x\) as a function
Similarly, we restrict the domain of \(\cos x\) to \([0, \pi]\) to define its inverse, which we call \(\arccos x\):
\[\cos: [0, \pi] \to [-1, 1]\]
and
\[\arccos: [-1, 1] \to [0, \pi].\]
We restrict the domain of \(\tan x\) to the open interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) to define its inverse, which we call \(\arctan x\):
Note that these definitions of inverse trigonometric functions are not unique. We could restrict the domains of the trigonometric functions differently. For example, we could restrict the domain of \(\tan\) as
Example. Let us find the value of \(\arcsin\left(\sin\frac{5\pi}{6}\right)\). First note that \(\sin\frac{5\pi}{6} = \sin\frac{\pi}{6}\) (why?) and \(\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). Therefore \(\arcsin\left(\sin\frac{5\pi}{6}\right) = \frac{\pi}{6}\). □
Hyperbolic functions
The hyperbolic cosine, hyperbolic sine, and hyperbolictangent are defined, respectively, by
\[
\begin{eqnarray}
\cosh x &=& \frac{e^x + e^{-x}}{2},\\
\sinh x &=& \frac{e^x - e^{-x}}{2},\\
\tanh x &=& \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}.
\end{eqnarray}
\]
The domain of these functions is \(\mathbb{R}\).
Why are their names similar to trigonometric functions? Note that
\[\cosh^2\theta - \sinh^2\theta = 1.\]
Recall that the equation \(x^2 - y^2 = 1\) represents the unit hyperbola on \(\mathbb{R}^2\).
and \(x^2 + y^2 = 1\) represents the unit circle on \(\mathbb{R}^2\).
Therefore, on \(\mathbb{R}^2\), while \((\cos\theta, \sin\theta)\) corresponds to a point on the unit circle centered at the origin, \((\cosh\theta, \sinh\theta)\) corresponds to a point on the unit hyperbola.
Also, compare the following relations with the definitions of the hyperbolic functions:
\[\begin{eqnarray}
\cos x &=& \frac{e^{ix} + e^{-ix}}{2},\\
\sin x &=& \frac{e^{ix} - e^{-ix}}{2i},\\
\tan x &=& \frac{\sin x}{\cos x} = \frac{e^{ix} - e^{-ix}}{i(e^{ix} + e^{-ix})}.
\end{eqnarray}\]
We can use multiple integrals to compute areas and volumes of various shapes. Area of a planar region Definition (Area) Let \(D\) be a bounded closed region in \(\mathbb{R}^2\). \(D\) is said to have an area if the multiple integral of the constant function 1 over \(D\), \(\iint_Ddxdy\), exists. Its value is denoted by \(\mu(D)\): \[\mu(D) = \iint_Ddxdy.\] Example . Let us calculate the area of the disk \(D = \{(x,y)\mid x^2 + y^2 \leq a^2\}\). Using the polar coordinates, \(x = r\cos\theta, y = r\sin\theta\), \(dxdy = rdrd\theta\), and the ranges of \(r\) and \(\theta\) are \([0, a]\) and \([0, 2\pi]\), respectively. Thus, \[\begin{eqnarray*} \mu(D) &=& \iint_Ddxdy\\ &=&\int_0^a\left(\int_0^{2\pi}rd\theta\right)dr\\ &=&2\pi\int_0^a rdr\\ &=&2\pi\left[\frac{r^2}{2}\right]_0^a = \pi a^2. \end{eqnarray*}\] □ Volume of a solid figure Definition (Volume) Let \(V\) be a solid figure in the \((x,y,z)\) space \(\mathbb{R}^3\). \(V\) is...
Consider integrating a function \(f(x,y)\) over a region \(D\) which may not be bounded or closed. In the case of a univariate function, this corresponds to the improper integral where we took the limits of the endpoints of a closed interval. In the case of multiple integrals, we adopt the notion of a "sequence of regions." Consider a sequence of regions \(\{K_n\}\) where each \(K_n\) is a subset of \(\mathbb{R}^2\) that satisfies the following conditions: (a) \(K_1 \subset K_2\)\(\subset \cdots \subset\) \(K_n \subset K_{n+1} \subset \cdots\). (b) For all \(n\in \mathbb{N}\), \(K_n \subset D\). (c) For all \(n \in\mathbb{N}\), \(K_n\) is bounded and closed. (d) For any bounded closed set \(F\) that is included in \(D\) (i.e., \(F \subset D\)), if \(n\) is sufficiently large, then \(F \subset K_n\). In other words: for all bounded closed \(F \subset D\), there exists some \(N\in \mathbb{N}\) such that, for all \(n\in \mathbb{N}\), if \(n \geq N\) then \(F \subset K_...
In a previous post, we defined the multiple integral on a (closed) rectangular region and saw that continuous functions were integrable on such regions. But more functions are integrable. Some functions are not necessarily defined on a rectangular region. Yet other functions may not be continuous everywhere. To describe a wider class of integrable functions, we define the notion of the "null" sets. Null sets Definition (Null set) The subset \(A\) of \(\mathbb{R}^2\) is said to be a null set if the following condition is satisfied: For all \(\varepsilon > 0\), there exist a finite number of open rectangular regions \[L_i = (a_i, b_i)\times (c_i, d_i) ~(a_i < b_i, c_i < d_i, i = 1, 2, \cdots, r)\] such that \[A \subset \bigcup_{i=1}^{r}L_i\] and \[\sum_{i=1}^{r}(b_i - a_i)(d_i - c_i) < \varepsilon.\] This means that if a set is a null set if it can be "covered" by a finite number of open rectangular regions, and the total area of...
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