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# 数学代写|实分析代写Real Analysis代考|Continuity of some important functions

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## 数学代写|实分析代写Real Analysis代考|Continuity of some important functions

Polynomial function.
Let $f(x)=a_0 x^n+a_1: x^{n-1}+\cdots+a_{n-1} x+a_n$ for all $x \in \mathbb{R}$, where $a_o, a_1, \ldots, a_n$ are real numbers. Then $f$ is a polynomial function.
$f$ is the sum of $n+1$ functions $f_0, f_1, f_2, \ldots, f_n$ where $f_i=a_i x^{n-i}, i=$ $0,1,2, \ldots, n$. Each $f_i$ is continuns on $\mathbb{R}$. Therefore by Theoren 8.1.5; $f$ is continuous on $\mathbb{R}$.
Rational function.
Let $p(x)$ and $q(x)$ be polynomial functions on $\mathbb{R}$.
There are at most a finite number of real roots, say $\alpha_1, \alpha_2, \ldots, \alpha_m$ of $q(x)$. If $x \neq \alpha_1, \alpha_2, \ldots, \alpha_m$ then we can define a function $f$ by $f(x)=\frac{m(x)}{\eta(x)}, x \neq \alpha_1, \alpha_2, \ldots, \alpha_m$.
By Theorem 8.1.4, if $q(c) \neq 0$ then $f$ is continuous at $c$.
That is, if $c$ be not a root of $q(x)$ then $f$ is continuous at $c$.
So a rational function is continuous for all $x \in \mathbb{R}$ for which the function is defined.
Trigonometric functions.
(a) Let $f(x)=\sin x, x \in \mathbb{R}$. Let $c \in \mathbb{R}$.
\begin{aligned} |\sin x-\sin c| & =2\left|\cos \frac{x+c}{2} \sin \frac{x-c}{2}\right| \ & \leq 2\left|\sin \frac{\pi-c}{2}\right|, \text { since }|\cos x| \leq 1 \ & \leq 2\left|\frac{x-c}{2}\right|, \text { since }|\sin x| \leq|x| \ & =|x-c| . \end{aligned}
Let us choose $\epsilon>0$.
Then $|\sin x-\sin c|<\epsilon$ for all $x$ satisfying $|x-c|<\epsilon$.
So $f$ is continuous at $c$. Since $c$ is arbitrary, $f$ is continuous on $\mathbb{R}$.
(b) Let $f(x)=\cos x, x \in \mathbb{R}$. Let $c \in \mathbb{R}$.
\begin{aligned} |\cos x-\cos c| & =2\left|\sin \frac{x+c}{2} \sin \frac{x-c}{2}\right| \ & \leq 2\left|\sin \frac{x-c}{2}\right|, \operatorname{since}\left|\sin \frac{x+c}{2}\right| \leq 1 \ & \leq 2\left|\frac{x-c}{2}\right|, \operatorname{since}|\sin x| \leq|x| \ & =|x-c| \end{aligned}
Let us choose $\epsilon>0$.
Then $|\cos x-\cos c|<\epsilon$ for all $x$ satisfying $|x-c|<\epsilon$.
So $f$ is continuous at $c$. Since $c$ is arbitrary, $f$ is continuous on $\mathbb{R}$.
(c) Let $f(x)=\tan x$.
$f$ is not defined at the points $(2 n+1) \frac{\pi}{2}(n$ being an integer) where the denominator $\cos x=0$.
Let $c \in \mathbb{R}$ and $c \neq(2 n+1) \frac{\pi}{2}$. Then $\lim _{x \rightarrow c} \tan x=\tan c$.
So $f$ is continuous at $c$ when $c \neq(2 n+1) \frac{\pi}{2}$.
Thus $f$ is continuous on its domain.
(d) The functions $\cot x, \operatorname{cosec} x, \sec x$ are continuous on their respective domains.

## 数学代写|实分析代写Real Analysis代考|Some composite functions

(a) Let $D \subset \mathbb{R}$ and $f: D \rightarrow \mathbb{R}$ be such that $f(x) \geq 0$ for all $x \in D$ and $f$ is continuous on $D$. Then $\sqrt{f}$ is continuous on $D$.
To prove this, let $g(x)=\sqrt{x}$.
Then the composite function $g f: D \rightarrow \mathbb{R}$ is defined by $g f(x)=$ $\sqrt{f(x)}, x \in D$

Since $f$ is continuous on $D$ and $g$ is continuous on $f(D)$, the composite function $g f$, i.e., $\sqrt{f}$ is continuous on $D$.
Worked Examples.
(i) Prove that the function $h(x)=\sqrt{x^2+3}, x \in \mathbb{R}$ is continuous on $\mathbb{R}$. $h$ is the composite function $g f$ where $f(x)=x^2+3, x \in \mathbb{R}$ and $g(x)=\sqrt{x}, x \geq 0 . f(x)>0$ for $x \in \mathbb{R}$. $f$ is continuous on $\mathbb{R}$ and $g$ is continuous on $f(\mathbb{R})$.
So $g f$ is continuous on $\mathbb{R}$. That is, $h$ is continuous on $\mathbb{R}$.
(ii) Prove that the function $h(x)=\sqrt{\sin x}, x \in[0, \pi]$ is continuous on $[0, \pi]$
$h$ is the composite function $g f$ where $f(x)=\sin x, x \in[0, \pi]$ and $g(x)=\sqrt{x}, x \geq 0$.
$f(x) \geq 0$ for $x \in[0, \pi] . \quad f$ is continuous on $[0, \pi]=D$, say. $g$ is continuous on $f(D)$.
So $g f$ is continuous on $[0, \pi]$. That is, $h$ is continuous on $[0, \pi]$.
(iii) Prove that the function $h(x)=\sqrt{x+\sqrt{x}}, x \geq 0$ is continuous on $[0, \infty)$
$h$ is the composite function $g f$ where $f(x)=x+\sqrt{x}, x \geq 0$ and $g(x)=\sqrt{x}, x \geq 0$.
$f(x) \geq 0$ for $x \geq 0$. $f$ is continuous on $[0, \infty)=D$, say. $g$ is continuous on $f(D)$.
So $g f$ is continuous on $[0, \infty)$. That is, $h$ is continuous on $[0, \infty)$.

## MATLAB代写

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