## Tag Archives: functions

### Intermediate value theorem

Intermediate value theorem.

Let $f:[a,b] \rightarrow \Re$ be continuous. Suppose for $x_{1},x_{2} \in [a,b]$, $f(x_{1}) \neq f(x_{2})$. If c is a real number between $f(x_{1})$ and $f(x_{2})$, then there is an $x_{0}$ between $x_{1}$ and $x_{2}$ such that $f(x_{0})=c$.

Proof.

Define $g:[a,b] \rightarrow \Re$ by $g(x)=f(x)-c$. Then, g is a continuous function such that $g(x_{1})$ and $g(x_{2})$ are of opposite signs. The assertion of the theorem amounts to saying that there is a point $x_{0}$ between $x_{1}$ and $x_{2}$ such that $g(x_{0})=0$. Without loss of generality, we may take $g(x_{1})>0$ and $g(x_{2})<0$ (otherwise replace g by -g). If $g(\frac{x_{1}+x_{2}}{2})>0$, we write $\frac{x_{1}+x_{2}}{2}=a_{1}$ and $x_{2}=b_{1}$; otherwise, write $a_{1}=x_{1}$ and $b_{1}=\frac{x_{1}+x_{2}}{2}$ so that we have $g(a_{1})>0$, $g(b_{1})<0$. Now if $g(\frac{x_{1}+x_{2}}{2})=0$, then $x_{0}=\frac{a_{1}+b_{1}}{2}$.

If $g(\frac{a_{1}+b_{1}}{2})>0$, write $\frac{a_{1}+b_{1}}{2}=a_{2}$ and $b_{1}=b_{2}$, otherwise write $a_{2}=a_{1}$ and $b_{2}=\frac{a_{1}+b_{1}}{2}$, so that we have $g(a_{2})>0$ and $g(b_{2})<0$. We could continue this process and find sequences $(a_{n})_{n=1}^{\infty}$,

$(b_{n})_{n=1}^{\infty}$ with $g(a_{n})>0$ and $g(b_{n})<0$ and

$a_{1} \leq a_{2} \leq \ldots \leq a_{n} \leq a_{n+1} \leq \ldots \leq x_{2}$,

$b_{1} \geq b_{2} \geq \ldots \geq b_{n} \geq b_{n+1} \geq \ldots \geq x_{1}$.

Since $(a_{n})_{n=1}^{\infty}$ is a monotonically non-decreasing sequence bounded above, it must converge. Suppose it converges to $\alpha$. Similarly, $(b_{n})_{n=1}^{\infty}$ is monotonically non-increasing, bounded below and therefore converges to, say, $\beta$. We further note that $b_{n}-a_{n}=\frac{x_{2}-x_{1}}{2^{n}} \rightarrow 0$ as $n \rightarrow \infty$ implying $\alpha=\beta$. Let us call this $x_{0}$. By the continuity of g, we have $\lim_{n \rightarrow \infty}g(a_{n})=g(x_{0})=\lim_{n \rightarrow \infty}g(b_{n})$, and since $g(a_{n})>0$ for all n, we must have $g(x_{0}) \geq 0$ and at the same time since $g(b_{n})<0$ for all n, we must also have $g(x_{0}) \leq 0$. This implies $g(x_{0})=0$. QED.

Corollary.

If f is a continuous function in an interval I and $f(a)f(b)<0$ for some $a,b \in I$, then there is a point c between a and b for which $f(c)=0$. (Exercise).

The above result is often used to locate the roots of equations of the form $f(x)=0$.

For example, consider the equation: $f(x) \equiv x^{3}+x-1=0$.

Note that $f(0)=-1$ whereas $f(1)=1$. This shows that the above equation has a root between 0 and 1. Now try with 0.5. $f(0.5)=-0.375$. So there must be a root of the equation between 1 and 0.5. Try $0.75$. $f(0.75)>0$, which means that the root is between 0.5 and 0.75. So, we may try $0.625$. $f(0.625)<0$. So the root is between 0.75 and 0.625. Now, if we take the approximate root to be 0.6875, then we are away from the exact root at most a distance of 0.0625. If we continue this process further, we shall get better and better approximations to the root of the equation.

Exercise.

Find the cube root of 10 using the above method correct to 4 places of decimal.

More later,

Nalin Pithwa

### Some elementary properties of limits

Some elementary properties of limits.

1) Suppose for functions f and g, $\lim_{x \rightarrow a}f(x)$ and $\lim_{x \rightarrow a}g(x)$ exist. Then, we have

(a) $\lim_{x \rightarrow a}(f(x)+g(x))=\lim_{x \rightarrow a}f(x)+\lim_{x \rightarrow a}g(x)$

(b) $\lim_{x \rightarrow a}(f(x)g(x))=\lim_{x \rightarrow a}f(x). \lim_{x \rightarrow b}g(x)$

(c) $\lim_{x \rightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a}g(x)}$, provided $\lim_{x \rightarrow a}g(x) \neq 0$

(2) Suppose f is continuous at a, then $\lim_{x \rightarrow a}f(x)$ is simply $f(a)$.

(3) Suppose f is defined in a deleted neighbourhood of $x_{0}$, g is defined in a neighbourhood of $x_{0}$ and is continuous. If $f(x)=g(x)$ for x in the deleted neighbourhood of $x_{0}$, then $\lim_{x \rightarrow a}f(x)=g(x)$.

From the above, it is easy to see that

(a) $\lim_{x \rightarrow x_{0}}p(x)=p(x_{0})$ fpr a polynomial p.

(b) $\lim_{x \rightarrow x_{0}}\sin{x}=\sin{x_{0}}$

(c) $\lim_{x \rightarrow x_{0}}\frac{p(x)}{q(x)}=\frac{p(x_{0})}{q(x_{0})}$, for polynomials p, q if $q(x_{0}) \neq 0$.

More later,

Nalin Pithwa

### Some properties of continuous functions

Theorem:

If $g,f : [a,b] \rightarrow \Re$ are continuous functions and c is a constant, then

a) $f+g$ is a continuous functions.

b) $f-g$ is a continuous functions.

c) cf is a continuous function.

d) fg is a continuous function.

Proof:

We shall only prove statement d. Choose and fix any $\varepsilon >0$. Since is continuous at $x_{0}$, we have that for the positive number $\frac{\varepsilon}{(2|g(x_{0})|+1}$ there exists a $\delta_{1}>0$ such that

$|f(x)-f(x_{0}|< \frac{\varepsilon}{2(|g(x_{0}|+1)}$ whenever $|x-x_{0}|<\delta_{1}$

Since $||f(x)|-|f(x_{0})|| \leq |f(x)-f(x_{0})|$, we conclude that

$|f(x)|<|f(x_{0})|+\frac{\varepsilon}{2(|g(x_{0})|+1)}$, whenever $|x-s_{0}|<\delta_{1}$. Let $|f(x_{0})|+\varepsilon 2(|g(x_{0})|+1)=M$. Also, since g is continuous at $x_{0}$, for the positive number $\frac{\varepsilon}{2M}$, there is a $\delta_{2}>0$ such that $|g(x)-g(x_{0})|< \frac{\varepsilon}{2M}$ whenever $|x-x_{0}|<\delta_{2}$. Put $\delta=min(\delta_{1},\delta_{2})$. Then, whenever $|x-x_{0}|<\delta$, we have

$|f(x)g(x)-f(x_{0})g(x_{0})|$ equals

$|f(x)g(x)-f(x)g(x_{0})+f(x)g(x_{0})-f(x_{0})g(x_{0})|$

$\leq |f(x)||g(x)-g(x_{0})|+|g(x_{0})(f(x)-f(x_{0}))|$

which equals

$|f(x)||g(x)-g(x_{0})|+|g(x_{0})||f(x)-f(x_{0})|$

$< M.\frac{\varepsilon}{2M}+|g(x_{0})|.\frac{\varepsilon}{2(|g(x_{0}|+1)}$

which equals $\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$

Observe that we have not claimed that the quotient of two continuous functions is continuous. The problem is obvious: $\frac{f(x)}{g(x)}$ cannot have any meaning at x for which $g(x)=0$. So, the question would be, if $g(x) \neq 0$ for every $x \in [a,b]$, is the function $h:[a,b] \rightarrow \Re$, defined by $h(x)=\frac{f(x)}{g(x)}$, continuous? The answer is yes. For a proof, we need a preliminary result.

Lemma.

if $g:[a,b] \rightarrow \Re$ is continuous and $g(x_{0}) \neq 0$, then there is an $m > 0$ and $\delta >0$ such that if $x_{0}-\delta, then $|g(x)|>m$.

Proof.

Let $|g(x_{0})|=2m$. Now, $m>0$. By continuity of g, there is a $\delta>0$ such that

$|g(x)-g(x_{0})| for $x_{0}-\delta

But, $|g(x)-g(x_{0})| \geq ||g(x)|-|g(x_{0})||$ and hence, $-m<|g(x)|-|g(x_{0})|<m$, giving us

$m=|g(x_{0})|-m<|g(x)|$ for $x_{0}-\delta. Hence, the proof.

The lemma says that if a continuous function does not vanish at a point, then there is an interval containing it in which it does not vanish at any point.

Theorem.

If $f,g :[a,b] \rightarrow \Re$ are continuous and $g(x) \neq 0$ for all x, then $h:[a,b] \rightarrow \Re$ defined by $h(x)=\frac{f(x)}{g(x)}$ is continuous.

The proof of the above theorem using the lemma above is left as an exercise.

Examples.

a) $f:\Re \rightarrow \Re$ defined by $f(x)=a_{0}$ for all $x \in \Re$, where $a_{0}$ is continuous.

b) $f:\Re \rightarrow \Re$ defined by $f(x)=x$ is continuous.

c) $g:\Re \rightarrow \Re$ defined by $g(x)=x^{2}$ is a continuous function because $g(x)=f(x)f(x)$, where $f(x)=x$. Since f is continuous by (b), g must be continuous.

d) $h:\Re \rightarrow \Re$ by $h(x)=x^{n}$, n being a positive integer, is continuous by repeated application of the above reasoning.

e) $p: \Re \rightarrow \Re$ defined by $p(x)=a_{0}+a_{1}x+\ldots +a_{n}x^{n}$, where $a_{0}, a_{1}, \ldots , a_{n}$ is also continuous. This is because of the fact that if

$f_{1}, f_{2}, f_{3} \ldots, f_{n}:\Re \rightarrow \Re$ are defined by $f_{1}(x)=x$, $f_{2}=x^{2}$, …, $f_{n}=x^{n}$, then $a_{1}f_{1}, a_{2}f_{2}, \ldots a_{n}f_{n}$ are also continuous functions. Hence,

$a_{0}+a_{1}f_{1}+ \ldots +a_{n}f_{n}=p$ is also a continuous function as the sum of continuous functions is a continuous function. Thus, we have shown that a polynomial is a continuous function.

f) Let p and q be polynomials. Let $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} \in \Re$ be such that $q(\alpha_{1})=q(\alpha_{2})=\ldots=q(\alpha_{n})$  and $q(\alpha) \neq 0$ for $\alpha \neq \alpha_{1}, \alpha \neq \alpha_{2}, \ldots, \alpha neq \alpha_{n}$.

Now, let $D = \Re - \{ \alpha_{1}, \alpha_{2}, \ldots , \alpha_{n}\}$.

Then, $h:D \rightarrow \Re$ defined by $h(x)=\frac{p(x)}{q(x)}$ is a continuous function. What we have said is that a rational function which is defined everywhere except on the finite set of zeroes of the denominator is continuous.

g) $f:\Re \rightarrow \Re$ defined by $f(x)=\sin{x}$ is continuous everywhere. Indeed, $f(x)-f(x_{0})=\sin{x}-\sin{x_{0}}=2\sin{\frac{x-x_{0}}{2}}\cos{\frac{x+x_{0}}{2}}$. Therefore,

$|f(x)-f(x_{0})|=2|\sin{\frac{(x-x_{0})}{2}}| |\cos{\frac{(x+x_{0})}{2}}|\leq |x-x_{0}|$ (because $|\sin{x}| \leq |x|$, where x is measured in radians)

h) $f:\Re \rightarrow \Re$ defined by $f(x)=\cos{x}$ is continuous since

$|f(x)-f(x_{0})|=|\cos{x}-\cos{x_{0}}|=2|\sin{\frac{(x_{0}-x)}{2}}\sin{\frac{x+x_{0}}{2}}| \leq |x-x_{0}|$

i) $f:\Re - \{ (2n+1)\frac{\pi}{2}: n \in \mathbf{Z}\} \rightarrow \Re$ defined by $f(x)=\tan{x}$ is continuous. We had to omit numbers like $\ldots, \frac{-3\pi}{2}, \frac{-\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \ldots$ from the domain of f as $\tan{x}$ cannot be defined for these values of x.

j) $f:\Re_{+} \rightarrow \Re$ defined by $f(x)=x^{1/n}$ is a continuous function. Indeed,

$f(x)-f(a)=x^{1/n}-a^{1/m}$ which equals

$\frac{(x-a)}{x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}\frac{1}{a^{n}}+\ldots +a^{\frac{n-1}{n}} }$

Choose $|x-a|<|a/2|$ to start with, so that $|a/2|<|x|<(3/2)|a|$. Thus,

$|x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}a^{1/n}+\ldots+a^{\frac{n-1}{n}}|>|a|^{\frac{n-1}{n}} \times ((1/2)^{\frac{n-1}{n}}+(1/2)^{\frac{n-2}{n}}+\ldots+1)$

Given an $\varepsilon >0$, let

$\delta=min\{\frac{|a|}{2}, \varepsilon \times |a|^{\frac{n-1}{n}} \times \left( (1/2)^{\frac{n-1}{n}}+\ldots+1 \right)\}$.

Then, for $|x-a|<\delta$, we have

$|f(x)-f(a)|=\frac{|x-a|}{|x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}} \times a^{1/n}+\ldots+a^{\frac{n-1}{n}}|}< \varepsilon$.

It can be shown that f defined by $f(x)=x^{r}$ is also a continuous function for every real $r \in \Re$.

k) Consider the function $f:\Re \rightarrow \Re$ defined by $f(x)=a^{x}$. Is f a continuous function? This is left as an exercise. (Hint: It will suffice to prove continuity at $x=0$. This would follow from $\lim_{m \rightarrow \infty}a^{1/m}$).

k) Suppose $f:\Re \rightarrow \Re$ is defined by $f(x)=1/x$, if $x \neq 0$ and $f(0)=0$. We can see that f is not continuous at 0 as $f(x)$ changes abruptly when x goes over from negative to positive values.

More later,

Nalin Pithwa

### Injections, surjections and bijections

A function $f: A \rightarrow B$ is called an injection if $f(a)=f(a^{'})$ implies $a=a^{'}$. There is another name for this kind of function. It is also called a one-to-one function or an injective map. A map $f:A \rightarrow B$ is one-one if two distinct elements of A have distinct images under f.

A function $f: A \rightarrow B$ is called a surjection if for every $b \in B$ there is an element $a \in A$ such that $f(a)=b$. In other words, $f(A)=B$. That is to say, every element of B is an image of some element of A under f. A surjective map is also called an onto map.

A map $f:A \rightarrow B$ which is both one-to-one and onto is called a bijection or a bijective map.

Examples.

1) Suppose $f: \Re \rightarrow \Re$ is defined by $f(x) =\cos{x}$. It is clear that this is neither one-to-one nor onto. Indeed because $f(x)=f(x+2\pi)$, it cannot be one-to-one. Since $f(x)$ never takes a value below -1 or above 1, it cannot be onto.

2) $f: \Re \rightarrow \Re$ defined by

$f(x)= x$, if $x \geq 0$ and $f(x)=x-1$, if $x<0$ is one-to-one but not onto as $-1/2$ is never attained by the function.

3) $f:\Re \rightarrow \Re$ defined by $f(x)=\frac{x}{1+|x|}$ is one-to-one but not onto.

Warning.

Trigonometric functions like sine and cosine are neither one-to-one nor onto. So, how does one define $\sin^{-1}{x}$ or $\cos^{-1}{x}$? Actually, there is ambiguity in defining these. If we write $\sin^{-1}{x}=\theta$, it means that $\sin{\theta}=x$. It is easily seen that there is no $\theta$ if x is more than 1 or less than -1. Thus, the domain of $\sin^{-1}{x}$ or $\cos^{-1}{x}$ must be $[-1.1]$. Then, again

$\sin{\theta}=x$ has many solutions $\theta$ for the same x. For example, $\sin{\pi/6}=\sin{5\pi/6}=1/2$. So, which of $\pi/6$ or $5\pi/6$ should claim to be the value of $\sin^{-1}{(1/2)}$? In such a case, we agree to take only one value in a definite way. For $0 \leq x \leq 1$, we choose $0 \leq \theta \leq \pi/2$ such that $\sin{\theta}=x$. It is obvious that there is only one such $\theta$. Similarly, for $-1 \leq x <0$, we choose $-\pi/2 \leq \theta < 0$ such that $\sin{\theta}=x$. Thus, this way of choosing $\theta$ such that $\sin{\theta}=x$ for $-1 \leq x <1$ has no ambiguity. such a value of the inverse circular function is called its principal value, though we could choose another set of values with equal ease.

Similar problems arise in the context of a function $f:\Re \rightarrow \Re$ defined by $f(x)=x^{2}$. The function f is neither one-to-one nor onto. But, if we take $f: \Re \rightarrow \Re_{+}$ as $f(x)=x^{2}$, then f is onto. We would like to define $f^{-1}:\Re_{+} \rightarrow \Re$ as a function. In order that we are able to define $f^{-1}$ as a function we must agree, once and for all, the sign of $f^{-1}{(x)}$. Indeed, since $f(-1)=f(1)=1$, which one would we call $f^{-1}$ ? In fact, $f^{-1}{(x)}$ is what we would like to denote by $\sqrt{x}$. But, we must decide if we are taking the positive value or the negative value. Once, we decide that, $f^{-1}$ would become a function.

More later,

Nalin Pithwa

### Composition of functions

Suppose A, B and C are sets and $f:A \rightarrow B$ and $g:B \rightarrow C$ are functions. We define a function

$h: A \rightarrow C$ by

$h(a)=g(f(a))$ for every $a \in A$.

It is easily seen that h is a well-defined function, as $f(a) \in B$ for every $a \in A$ and $g(b) \in C$ for every $b \in B$. The function h is called the composition of the functions g and f and is denoted by $g \circ f$.

Examples.

a) Suppose that in a forest, carnivorous animals sustain themselves by feeding only on herbivorous animals and the nutrition level of a herbivorous animal depends on the vegetation around the animal. So the nutrition level of a carnivorous animal ultimately depends on the vegetation around the population of herbivorous animals it feeds on. Thus, if V is the density of vegetation around the herbivorous animals n is the level of nutrition of the herbivorous animal (for simplicity measured by its weight, though there are often more parameters depicting the level of nutrition of an animal), $n: V \rightarrow \Re$ is a function. Similarly, if c is the level of nutrition of a carnivorous animal, then c is a function depending on the level of nutrition of herbivorous animals it feeds on:

$c: \Re \rightarrow \Re$

Thus, $c \circ n: V \rightarrow \Re$

is the level of nutrition of the carnivorous animal ultimately depending on the density of vegetation.

b) Take for example the force experienced by a moving charged particle in a magnetic field which is varying in time. We know that the force on the charged particle depends on the strength of the magnetic field in which it moves. Again, as the magnetic field strength varies with time, the force experienced is ultimately a function of time and position.

c) Suppose there is a lamp  in the room. The intensity of illumination at point in the room depends on the illuminating power of the lamp. The illuminating power of the lamp again depends on the voltage of the electricity supply which makes the lamp glow. So, ultimately, the intensity of illumination depends on the voltage of the power supply.

Exercise.

Give five more examples of composition of functions.

More later,

Nalin Pithwa

### Basic Set Theory: Functions

Among the kinds of relations we have been talking about, there are some which are very useful in mathematics. A relation $f \subset A \times B$ is called a function from A to B if

i) domain of $f=A$

ii) $(a,b) \in f$ and $(a,b^{'}) \in f \Longrightarrow b=b^{'}$.

Thus a function from A to B is a relation whose domain is the whole of A and which is not one-to-many. The set B is called the codomain or range of the function f.

If $f \subset A \times B$ is a function from A to B, then we write $f: A \rightarrow B$ and say that f is a function from A to B. We also say that f maps A into B and often a function is called a map or mapping. If

$(a,b) \in f$, we write

$b=f(a)$.

We call b the value of the function f at a or image of a under f. Observe that there can be no ambiguity in writing f(a) because it is impossible that $(a,b) \in f$ and $(a,b^{'}) \in f$ and $b \neq b^{'}$. So by definition of a function $f: A \rightarrow B$, we have for every $a \in A$ a unique $f(a) \in B$. Thus, a function would be completely determined if we knew $f(a) \in B$ for every $a \in A$. That is why sometimes a function is defined as a “rule” which associates to every element $a \in A$ a unique element $f(a)$ of B. We are obliged to put the word rule within quotation marks for the simple reason that the rule is often elusive. In fact, the aim in many situations is to “discover” the rule.

Examples

a) Suppose we are observing the position of a particle moving in a straight line. We know that at every instant of time the particle has a unique position on the line. If we agree that every point of the line corresponds to a real number with some convenient point on it decided to represent the real number zero, then the position of the particle at the time t is represented by the real number $x(t)$. We observe that our with our ordinary notion of time and position, we do not expect a particle to occupy two different positions at the same time (nothing prevents, though,the particle from occupying the same position at two different times. Indeed, if the particle is at rest for a certain length of time, then it would have the same position for the entire length of time.) Suppose every time is represented by a real number, zero being a certain time deemed as the present time, then all future times are represented by positive real numbers and past times by negative real numbers. Here, a real number is used as an intuitive object, say, as points on a line, postponing a mathematical discussion of real numbers later. Let $\Re$ be the set of real numbers, then the position of the particle moving in a straight line is, in fact, a function

$x: \Re \rightarrow \Re$

which associates to every time t, the position $x(t)$.

b) Look at a railway time table. Below a particular train there are times recorded in a column. It records the times of arrival and departure of a train at certain stations. We may look upon this as a function whose domain consists of disjoint intervals of time and range consists of certain cities where the train stops. Thus, for every time in the mentioned interval we have a city where the train is at that time. The table does not say anything about the train’s position at a time after its departure and before its arrival at the next. This does not disqualify it to be a function if we take the domain to be the intervals of time depicting the times of arrival and departure at certain stations.

c) Consider the distance of a particle, falling freely under gravity, measured from the point from where the particle started its fall. if we record the time since it began to fall, we know that the distance $x(t)$ at a time is given by the formula

$x(t)=(1/2)gt^{2}$

(assuming, of course that the gravity does not vary and there is no air resistance).

Here we have in fact a “rule” which tells us where the particle would be at any time. This certainly represents a function. However, we are often not so lucky to have a neat formula like this for other functions.

d) The incoming news on a particular day that tells us the temperature of the four metros recorded at 5.30 am on that day. Here, we have a definite temperature for every such metro at 5.30am of that day. So it is in fact a function whose domain consists of the set of four metros and range the real numbers quantifying temperature.

e) Look at your school time table. What does it record on a particular day? it records the subjects to be taught at different times of the day. A particular student has a particular subject being taught during a particular period,as a student is not expected to be in two different classes at the same time. So, for each student, the time table is a function whose domain is the set of periods and the range the set of subjects.

f) When we want to mail a letter enclosed in an envelope, we usually go to the post master with it to tell us the denomination of the stamp to be affixed. The post master weighs the letter and tells us the postage according to the weight. Here, we have a definite postage for a definite weight, we don’t have different rates for the same weight (for same kind of mail like registered or ordinary). The rate chart, with the post master, for a particular kind of mail, truly, is a function whose domain is the set of positive real numbers representing the weights of the mail and the range again consists of positive real numbers representing the postage. We write the chart as a function f such that

$p-f(a)$

meaning p is the postage to mail a letter of weight w.

g) We know that the solubility of a salt varies with temperature. That is to say that a given salt has a definite solubility at a definite temperature. So the relation which associates to every temperature the definite solubility of a salt is in fact a function.

h) Consider the population of a community of biological species at different times. If $N(t)$ is the population of the community at time t, then surely

$N: \Re \rightarrow \Re$

represents a function.

i) If we measure the voltage of an AC power supply over a period of time, we observe that it is different at different times. At every instant of time, it has a voltage Sometimes, it is positive, sometimes it is negative and sometimes zero. In fact, for ordinary domestic supply in India, the voltage varies between 330 volts and -330 volts. But this change from 330 volts to -330 volts occurs within an interval of $1/100$th of a second. (The ordinary DC voltmeters of your lab won’t be able to record this quick change in voltage). In fact, if $V(t)$ is the volrage at time t, we have

$V(t)$=E_{0}\sin{(\omega t + \phi)}

where $\omega=2\pi/50$ and $E_{0}=330$ and $\phi$ is a constant known as the phase.

j) Let S be a sphere resting on a horizontal plane H. Let n be the north pole and s be the southpole, where the sphere is touching the plane. Now let us join any other point $p \in S$ to n and extend it to meet the plane at some point q. Now to every point $p \in S - \{ n \}$, we have a unique point q on H. This defines a function

$f: S - \{ n \} \rightarrow H$. This function is called the stereographic projection of the sphere on the plane.

k) The trivial map or function $id: A \rightarrow A$ which sends every element of the set A to itself is called the identity map.

Look around, see and think…are there any functions that come to your notice? Let me know…

More later,

Nalin Pithwa