Library of continuous functions

Library of continuous functions


	Library of continuous functions

In this lecture we consider elementary examples of continuous functions. In fact, we build a simple

library of elementary continuous functions

Polynomial, rational and irrational functions

Let us start with the constant function

f(x) = const

where const is any real number. The definition of continuous function is satisfied. Hence, the constant function is continuous.

The function

f(x) = x

is continuous by definition.

Due to the fact that the product of any two continuous functions is continuous we come to the conclusion that any polynomial

$P_n(x)= a_0x^n + a_1 x^{n-1} + \dots + a_{n-1}x + a_n$

is continuous. Formally speaking, we can prove this statement by mathematical induction with respect to the degree of the polynomial.

Basis of mathematical induction. The polynomial

P₀(x) = a₀

is a constant function. Hence, it is continuous.

Step of mathematical induction. Suppose that the polynomial of degree 'n'

P_n(x)

is a continuous function. We need to show that the polynomial of degree 'n+1'

P_{n + 1}(x)

is a continuous function.

Since

$P_{n+1}(x) = x \cdot P_n(x) + a_{n+1}$

and the product together with the sum of continuous functions are both continuous we have that

P_{n + 1}(x)

is a continuous function.

It is known that the ration of two continuous functions is continuous as long as it is defined. Hence, the ratio of two polynomials

$\frac{a_0x^n + a_1 x^{n-1} + \dots + a_{n-1}x + a_n}{b_0x^n + b_1 x^{n-1} + \dots + b_{n-1}x + b_n }$

is continuous at any point x₀ such that

$b_0x_0^n + b_1 x_0^{n-1} + \dots + b_{n-1}x_0 + b_n \not= 0$

An inverse of a continuous one-to-one function also is continuous. Hence,

$(a_0x^n + a_1 x^{n-1} + \dots + a_{n-1}x + a_n)^{\frac{1}{2n+1}}$

is continuous at any x and

$(a_0x^n + a_1 x^{n-1} + \dots + a_{n-1}x + a_n)^{\frac{1}{2n}}$

is continuous at any x such that

$a_0x^n + a_1 x^{n-1} + \dots + a_{n-1}x + a_n > 0.$

Trigonometric functions

Trigonometric functions cosine and sine are defined as x and y coordinates of a point on the unit circle.

Due to this definition

$\cos(\alpha ) \;\;\mbox{ and } \;\; \sin(\alpha )$

are continuous functions.

The following list contains the most basic elementary properties of sine and cosine. They follow immediately from their definition.

$\cos^2(\alpha ) \; + \; \sin^2(\alpha ) = 1$
$\cos(-\alpha ) = \cos(\alpha ),\;\;\sin(-\alpha ) = -\sin(\alpha )$
$\cos(\alpha + \pi ) = -\cos(\alpha ),\;\;\sin(\alpha + \pi ) = -\sin(\alpha )$
$\cos(\alpha + 2\pi n ) = \cos(\alpha ),\;\;\sin(\alpha + 2 \pi n ) = \sin(\alpha )\;\;\forall n \in \mathbb{N}$
$\cos(\alpha + \frac{\pi}{2} ) = -\sin(\alpha ),\;\;\sin(\alpha + \frac{\pi}{2} ) =\cos(\alpha)$

We recommend the reader to prove these properties independently.

Now we prove that

$\cos(\alpha - \beta ) = \cos (\alpha ) \cdot \cos(\beta ) \; + \; \sin (\alpha ) \cdot \sin(\beta )$

The distance between A and B in (x,y)-coordinates is calculated as

$\| A - B\| = \sqrt{(\cos(\alpha) -\cos(\beta ))^2 + (\sin(\alpha) - \sin(\beta ))^2}$

On the other hand, the distance between A and B in

$(\tilde x, \tilde y) \mbox{-coordinates}$

is defined by

$\| A - B\| = \sqrt{(\cos(\alpha-\beta) -1)^2 + (\sin(\alpha -\beta ))^2}$

Hence,

$\sqrt{(\cos(\alpha) -\cos(\beta ))^2 + (\sin(\alpha) - \sin(\beta ))^2} = \sqrt{(\cos(\alpha-\beta) -1)^2 + (\sin(\alpha -\beta ))^2}$

and consequently,

(cos(Î±) âˆ’ cos(Î²))² + (sin(Î±) âˆ’ sin(Î²))² = (cos(Î± âˆ’ Î²) âˆ’ 1)² + (sin(Î± âˆ’ Î²))²

After using the identity

(a + b)² = a² + 2ab + b²

and taking into account that

$\cos^2(\alpha ) \; + \; \sin^2(\alpha ) = 1$

we obtain

$-2(\cos(\alpha)\cdot \cos(\beta ) + \sin(\alpha)\cdot \sin(\beta )) +2 = -2\cos(\alpha-\beta) + 2$

The last proves our statement that

$\cos(\alpha - \beta ) = \cos (\alpha ) \cdot \cos(\beta ) \; + \; \sin (\alpha ) \cdot \sin(\beta )$

This identity implies the following list of important trigonometric identities.

$\cos(\alpha + \beta ) = \cos (\alpha ) \cdot \cos(\beta ) \; - \; \sin (\alpha ) \cdot \sin(\beta )$
$\sin(\alpha + \beta ) = \sin (\alpha ) \cdot \cos(\beta ) \; + \; \cos (\alpha ) \cdot \sin(\beta )$
$\sin(\alpha - \beta ) = \sin (\alpha ) \cdot \cos(\beta ) \; - \; \cos (\alpha ) \cdot \sin(\beta )$

If we add the third identity to the second then

$\sin(\alpha + \beta ) + \sin(\alpha - \beta ) = 2 \cdot \sin (\alpha ) \cdot \cos(\beta )$

Thus, we prove the following statement

$\sin (\alpha ) \cdot \cos(\beta ) =\frac{1}{2}(\sin(\alpha + \beta ) + \sin(\alpha - \beta ))$

We recommend the reader to prove the following statements independently.

$\cos (\alpha ) \cdot \cos(\beta ) = \frac{1}{2} (\cos(\alpha + \beta ) + \cos(\alpha - \beta ))$
$\sin (\alpha ) \cdot \sin(\beta ) = \frac{1}{2} (\cos(\alpha - \beta ) - \cos(\alpha + \beta ))$

In conclusion to this section we justify that

$\lim_{x\to 0 } \frac{\sin(x) }{x} = 1.$

As one can observe the area

$\frac{1}{2} \sin (\alpha )$

for triangle

(O,A,B)

is smaller than the area

$\frac{1}{2} \alpha$

for the sector

(O,A,B)

In turn, the area of the sector is smaller than the area of the triangle

(O,C,B),

which is calculated as

$\frac{1}{2} \tan (\alpha).$

Hence, we obtain the inequalities

$\frac{1}{2} \sin (\alpha ) \le \frac{1}{2} \alpha \le \frac{1}{2} \tan (\alpha).$

that are valid for

$0 \le\alpha <\frac{\pi}{2}.$

On the other hand, we can divide all terms in these inequalities by

The inequality

$1 \le \frac{\alpha }{\sin (\alpha ) } \le \frac{1}{\cos (\alpha) }.$

is valid for

$0 < \alpha < \frac{\pi}{2}.$

Thus,

$\frac{\alpha }{\sin (\alpha ) } \to \; 1 \;\;\mbox{ as } \alpha\;\to\;0$

Exponential function

The exponential function

$e^x = \lim_{n \to \infty } (1 + \frac{x}{n})^n$

is continuous. In order to prove this statement we need some additional tools. First, we expand

$(1 + \frac{x}{n})^n$

with the help of binomial formula

$(1 + \frac{x}{n})^n = 1+\sum_{m=1}^n \frac{n\cdot (n-1)\dots (n-(m-1))}{m!} (\frac{x}{n})^m$

Then we rewrite this formula as

$(1 + \frac{x}{n})^n = 1+\sum_{m=1}^n (1- \frac{1}{n})\dots (1 - \frac{(m-1)}{n}) (\frac{x^m}{m!})$

Let us assume that

x > 0

Then

$\sum_{m=1}^n (1- \frac{1}{n})\dots (1 - \frac{(m-1)}{n}) (\frac{x^m}{m!}) \le \sum_{m=1}^\infty \frac{x^m}{m!}$

which is valid $\forall \;\;n \in \mathbb{N}$ . Hence,

$(1 + \frac{x}{n})^n \le 1+\sum_{m=1}^\infty \frac{x^m}{m!}\;\;\;\forall \;\;n \in \mathbb{N}$

Consider any fixed integer k. Then

$1+ \sum_{m=1}^k (1- \frac{1}{n})\dots (1 - \frac{(m-1)}{n}) (\frac{x^m}{m!}) \le (1 + \frac{x}{n})^n \le 1+\sum_{m=1}^\infty \frac{x^m}{m!}\;\;\;\forall \;\;n > k$

Now taking n to infinity we obtain

$1+ \sum_{m=1}^k \frac{x^m}{m!} \le e^x \le 1+\sum_{m=1}^\infty \frac{x^m}{m!}\;\;\;\forall \;\; k \in \mathbb{N}$

This leads us to

$e^x = 1+\sum_{m=1}^\infty \frac{x^m}{m!}$

As seen from this formula the exponential function is the limit of continuous functions, polynomials having the form

$f_n(x) = 1+\sum_{m=1}^n \frac{x^m}{m!}$

In other words,

$\lim_{n\to \infty } f_n(x) = e^x$

In general the limit of the sequence of continuous functions is not continuous. For example,

$\lim_{n\to \infty } \frac{x^n}{1+x^n}$

is not a continuous function. However, if the sequence

{f_n(x)}_n

satisfies certain special uniform conditions than we can prove that the limit is continuous.

Definition 1. The sequence of continuous functions

{f_n(x)}_n

is uniformly convergent on a set S if

$\forall \;\varepsilon > 0 \;\;\exists\;\delta >0\;\;\mbox{ and } \;\;M \in \mathbb{N}\;\;\;\mbox{such that}$

$\forall\;\;x,\;y\in S\;\;|x-y|<\delta \;\;\mbox{ and } \;\;n,\;m >M \;\;\mbox{ we have }\;\;|f_n(x) - f_m(y)|<\varepsilon$

Theorem 1. If a sequence of continuous functions

{f_n(x)}_n

is uniformly convergent in some neighborhood of x₀ then the limit

$f(x) = \lim_{n \to \infty } f_n(x)$

is continuous at x₀.

Proof. We need to show that

$\forall \;\varepsilon > 0 \;\;\exists\;\delta >0\;\;\mbox{such that}$

$|x-y|<\delta \;\;\Longrightarrow \;\;|f(x) - f(y)|<\varepsilon$

The sequence

{f_n(x)}_n

is uniformly convergent in some neighborhood of

x₀

Let

Nbr(x₀)

denote this neighborhood of x₀. Then

$\forall \;\varepsilon > 0 \;\;\exists\;\delta >0\;\;\mbox{ and } \;\;M \in \mathbb{N}\;\;\;\mbox{such that}$

$\forall\;\;x,\;y\in Nbr(x_0) \;\;|x-y|<\delta \;\;\mbox{ and } \;\;n,\;m >M \;\;\mbox{ we have }\;\;|f_n(x) - f_m(y)|<\varepsilon$

Taking n and m to infinity in

$|f_n(x) - f_m(y)|<\varepsilon$

We obtain

$|f(x) - f(y)|<\varepsilon$

That proves the statement of this theorem.

We leave for the reader as an exercise to show that the sequence of polynomials

$f_n(x) = 1+\sum_{m=1}^n \frac{x^m}{m!}\;\;\;n=1,\;2,\;3 \dots$

is uniformly convergent on any interval

$[0,a]\;\;\forall \;\;a\in\mathbb{R}$

That implies the continuity of exponential function

$e^x\;\;\forall\;\;x \ge 0$

Since

$e^{-x} = \frac{1}{e^x}$

we have that

$e^x\;\;$

also is continuous when x is negative. Hence, we proved that the exponential function is continuous on the real line $\mathbb{R}$ .