Monthly Archives: September 2019

Set Theory, Relations and Functions: Preliminaries: IV:

Problem Set based on previous three parts:

I) Solve the inequalities in the following exercises expressing the solution sets as intervals or unions of intervals. Also, graph each solution set on the real line:

a) |x| <2 (b) |x| \leq 2 (c) |t-1| \leq 3 (d) |t+2|<1 (e) |3y-7|<4(f) |2y+5|<1 (g) |\frac{z}{5}-1| \leq 1 (h) | \frac{3}{2}z-1| \leq 2 (i) |3-\frac{1}{x}|<\frac{1}{2} (j) |\frac{2}{x}-4|<3 (k) |2x| \geq 4 (l) |x+3| \geq \frac{1}{2} (m) |1-x| >1 (n) |2-3x| > 5 (o) |\frac{x+1}{2}| \geq 1 (p) |\frac{3x}{5}-1|>\frac{2}{5}

II) Quadratic Inequalities:

Solve the inequalities in the following exercises. Express the solution sets as intervals or unions of intervals and graph them. Use the result \sqrt{a^{2}}=|a| as appropriate.

(a) x^{2}<2 (b) 4 \leq x^{2} (c) 4<x^{2}<9 (d) \frac{1}{9} < x^{2} < \frac{1}{4} (e) (x-1)^{2}<4 (f) (x+3)^{2}<2 (g) x^{2}-x<0 (h) x^{2}-x-2 \geq 0

III) Theory and Examples:

i) Do not fall into the trap |-a|=a. For what real numbers a is the equation true? For what real numbers is it false?

ii) Solve the equation: |x-1|=1-x.

iii) A proof of the triangle inequality: 

Give the reason justifying each of the marked steps in the following proof of the triangle inequality:

|a+b|^{2}=(a+b)^{2}…..why ?

=a^{2}+2ab++b^{2}

\leq a^{2}+2|a||b|+b^{2}….why ?

\leq |a|^{2}+2|a||b|+|b|^{2}….why?

=(|a|+|b|)^{2}….why ?

iv) Prove that |ab|=|a||b| for any numbers a and b.

v) If |x| \leq 3 and x>-\frac{1}{2}, what can you say about x?

vi) Graph the inequality: |x|+|y| \leq 1

Questions related to functions:

I) Find the domain and range of each function:

a) f(x)=1-\sqrt{x} (b) F(t)=\frac{1}{1+\sqrt{t}} (c) g(t)=\frac{1}{\sqrt{4-t^{2}}}

II) Finding formulas for functions:

a) Express the area and perimeter of an equilateral triangle as a function of the triangle’ s side with length s.

b) Express the side length of a square as a function of the cube’s diagonal length d. Then, express the surface area  and volume of the cube as a function of the diagonal length.

c) A point P in the first quadrant lies on the graph of the function f(x)=\sqrt{x}. Express the coordinates of P as functions of the slope of the line joining P to the origin.

III) Functions and graphs:

Graph the functions in the questions below. What symmetries, if any, do the graphs have?

a) y=-x^{3} (b) y=-\frac{1}{x^{2}} (c) y=-\frac{1}{x} (d) y=\frac{1}{|x|} (e) y = \sqrt{|x|} (f) y=\sqrt{-x} (g) y=\frac{x^{3}}{8} (h) y=-4\sqrt{x} (i) y=-x^{\frac{3}{2}} (j) y=(-x)^{\frac{3}{2}} (k) y=(-x)^{\frac{2}{3}} (l) y=-x^{\frac{2}{3}}

IV) Graph the following equations ad explain why they are not graphs of functions of x. (a) |y|=x (b) y^{2}=x^{2}

V) Graph the following equations and explain why they are not graphs of functions of x: (a) |x|+|y|=1 (b) |x+y|=1

VI) Even and odd functions:

In the following questions, say whether the function is even, odd or neither.

a) f(x)=3 (b) f(x=x^{-5} (c) f(x)=x^{2}+1 (d) f(x)=x^{2}+x (e) g(x)=x^{4}+3x^{2}-1 (f) g(x)=\frac{1}{x^{2}-1} (g) g(x)=\frac{x}{x^{2}-1} (h) h(t)=\frac{1}{t-1} (i) h(t)=|t^{3}| (j) h(t)=2t+1 (k) h(t)=2|t|+1

Sums, Differences, Products and Quotients:

In the two questions below, find the domains and ranges of f, g, f+g, and f-g.

i) f(x)=x, g(x)=\sqrt{x-1} (ii) f(x)=\sqrt{x+1}, g(x)=\sqrt{x-1}

In the two questions below, find the domains and ranges of f, g, \frac{f}{g} and \frac{g}{f}

i) f(x)=2, g(x)=x^{2}+1

ii) f(x)=1, g(x)=1+\sqrt{x}

Composites of functions:

  1. If f(x)=x+5, and g(x)=x^{2}-5, find the following: (a) f(g(0)) (b) g(f(0)) (c) f(g(x)) (d) g(f(x)) (e) f(f(-5)) (f) g(g(2)) (g) f(f(x)) (h) g(g(x))
  2. If f(x)=x-1 and g(x)=\frac{1}{x+1}, find the following: (a) f(g(\frac{1}{2})) (b) g(f(\frac{1}{2})) (c) f(g(x)) (d) g(f(x)) (e) f(f(2)) (f) g(g(2)) (g) f(f(x)) (h) g(g(x))
  3. If u(x)=4x-5, v(x)=x^{2}, and f(x)=\frac{1}{x}, find formulas or formulae for the following: (a) u(v(f(x))) (b) u(f(v(x))) (c) v(u(f(x))) (d) v(f(u(x))) (e) f(u(v(x))) (f) f(v(u(x)))
  4. If f(x)=\sqrt{x}, g(x)=\frac{x}{4}, and h(x)=4x-8, find formulas or formulae for the following: (a) h(g(f(x))) (b) h(f(g(x))) (c) g(h(f(x))) (d) g(f(h(x))) (e) f(g(h(x))) (f) f(h(g(x)))

Let f(x)=x-5, g(x)=\sqrt{x}, h(x)=x^{3}, and f(x)=2x. Express each of the functions in the questions below as a composite involving one or more of f, g, h and j:

a) y=\sqrt{x}-3 (b) y=2\sqrt{x} (c) y=x^{\frac{1}{4}} (d) y=4x (e) y=\sqrt{(x-3)^{3}} (f) y=(2x-6)^{3} (g) y=2x-3 (h) y=x^{\frac{3}{2}} (i) y=x^{9} (k) y=x-6 (l) y=2\sqrt{x-3} (m) \sqrt{x^{3}-3}

Questions:

a) g(x)=x-7, f(x)=\sqrt{x}, find (f \circ g)(x)

b) g(x)=x+2, f(x)=3x, find (f \circ g)(x)

c) f(x)=\sqrt{x-5}, (f \circ g)(x)=\sqrt{x^{2}-5}, find g(x).

d) f(x)=\frac{x}{x-1}, g(x)=\frac{x}{x-1}, find (f \circ g)(x)

e) f(x)=1+\frac{1}{x}, (f \circ g)(x)=x, find g(x).

f) g(x)=\frac{1}{x}, (f \circ g)(x)=x, find f(x).

Reference: Calculus and Analytic Geometry, G B Thomas. 

NB: I have used an old edition (printed version) to prepare the above. The latest edition may be found at Amazon India link:

https://www.amazon.in/Thomas-Calculus-George-B/dp/9353060419/ref=sr_1_1?crid=1XDE2XDSY5LCP&keywords=gb+thomas+calculus&qid=1570492794&s=books&sprefix=G+B+Th%2Caps%2C255&sr=1-1

Regards,

Nalin Pithwa

 

Ceiling and floor functions: IITJEE mains training

Problem 1:

For what values of x, is (a) \lfloor x \rfloor =0 (b) \lceil x \rceil =0?

Problem 2:

Which real numbers x satisfy the equation \lfloor x \rfloor = \lceil x \rceil?

Problem 3:

Does \lceil (-x) \rceil = - (\lfloor x \rfloor) for all real x? Give reasons for your answer.

Problem 4:

Graph: f(x)=\lfloor x \rfloor when x \geq 0; and f(x) = \lceil x \rceil, when x <0.

Why is f(x) called the integer part of x? Discuss the continuity and differentiability of f(x).

Cheers,

Nalin Pithwa

 

Various proofs of important algebraic identity: a^{3}+b^{3}+c^{3}-3abc

We know the following factorization: a^{3}+b^{3}+c^{3}-3abc= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

Proof 1:

Let a, b, c be roots of a polynomial P(X) Then by fundamental theorem of algebra

P(X)=(X-a)(X-b)(X-c) = X^{3}-(a+b+c)X^{2}+(ab+bc+ca)X-abc.

Now, once again basic algebra says that as a, b, c are roots/solutions of the above:

P(a)=a^{3}-(a+b+c)a^{2}+(ab+bc+ca)a-abc=0

P(b)=b^{3}-(a+b+c)b^{2}+(ab+bc+ca)b-abc=0

P(c)=c^{3}-(a+b+c)c^{2}+(ab+bc+ca)c-abc=0$

Adding all the above:

0= a^{3}+b^{3}+c^{3}-(a+b+c)(a^{2}+b^{2}+c^{2})+(ab+bc+ca)(a+b+c)-3abc

So, we get a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

Also, the above formula can be written as a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(\frac{1}{2})((a-b)^{2}+(b-c)^{2}+(c-a)^{2})

Proof 2:

Consider the following determinant D: \left| \begin{array}{ccc} a & b & c\\c & a & b\\ b & c & a \end{array} \right|

On adding all three columns to the first column: we know that the value of the determinant is unchanged: So we get the following:

D = \left| \begin{array}{ccc} a+b+c & b & c \\a+b+c & a & b\\a+b+c & c & a \end{array} \right|. Note that columns 2 and 3 of the three by three determinant do not change.

On expanding the original determinant D, we get

D = a(a^{2}-bc)-b(ac-b^{2})+c(c^{2}-ab)

D= a^{3}-abc-bac+b^{3}+c^{3}-cab

D= a^{3}+b^{3}+c^{3}-3abc

Whereas we get from the other transformed but equal D:

D =(a+b+c) \left| \begin{array}{ccc} 1 & b & c \\ 1 & a & b \\ 1 & c & a \end{array}\right|

D=(a+b+c)((a^{2}-bc)-b(a-b)+c(c-a))

So, that we again get a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

Proof 3:

Now, let us consider E=a^{2}+b^{2}+c^{2}-ab-bc-ca as a quadratic in a with b and c as parameters.

That is, E= a^{2} -(b+c)a + b^{2}+c^{2}-bc

Then, the discriminant is given by

\triangle = (b+c)^{2}-4 \times 1 \times (b^{2}+c^{2}-bc), which in turn equals:

\triangle = (b^{2}+c^{2}+2bc)-4(b^{2}+c^{2}-bc) = -3b^{2}-3c^{2}+6bc=-3(b-c)^{2}

a_{1}, a_{2} = \frac{(b+c) \pm i\sqrt{3}(b-c)}{2}

a_{1}= b(\frac{1+i\sqrt{3}}{2})=c(\frac{1-i\sqrt{3}}{2}) = -b\omega - c\omega^{2}

a_{2}=b(\frac{1-i\sqrt{3}}{2})+c(\frac{1-i\sqrt{3}}{2})=-b\omega^{2}-c\omega

Hence, the factorization of the above quadratic in a is given as:

a^{2}+b^{2}+c^{2}-ab-bc-ca = (a+b\omega+c\omega^{2})(a+b\omega^{2}+c\omega)

So, the other non-trivial factorization of the above famous algebraic identity is:

a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a+b\omega+ c\omega^{2})(a+b\omega^{2}+c\omega)

where 1, \omega, \omega^{2} are cube roots of unity.

Proof 4: 

Factorize the expression a^{3}+b^{3}+c^{3}-3abc by (a+b+c).

Solution 4: 

We can carry out the above polynomial division by considering the dividend to be a polynomial in a single variable, say a, (and assuming b and c are just parameters; so visualize them as arbitrary but fixed constants); further arrange the dividend in descending powers of a; so also arrange the divisor in descending powers of a (well, of course, it is just linear in a; and assume b and c are parameters also in dividend).

Proof 5:

Prove that the eliminant of

ax+cy+bz=0

cx+by+az=0

bx+ay+cz=0

is a^{3}+b^{3}+c^{3}-3abc=0

Proof 5:

By Cramer’s rule, the eliminant is given by determinant \left| \begin{array}{ccc}a & c & b \\ c & b & a\\b & a & c \end{array}\right|=0.

On expansion using the first row:

a(bc-a^{2})-c(c^{2}-ab)+b(ac-b^{2})=0

a^{3}+b^{3}+c^{3}-3abc=0 upon multiplying both the sides of the above equation by (-1). Of course, we have only been able to generate the basic algebraic expression but we have done so by encountering a system of three linear equations in x, y, z. (we could append any of the above factorization methods to this further!! :-)))

Cheers,

Nalin Pithwa

The personality of Leonhard Euler

The portrait of Euler that emerges from his publications and letters is that of a genial man of simple tastes and conventional religious faith. He was even wealthy, at least in the second half of his life, but ostentation was not part of his lifestyle. His memory was prodigious, and contemporary accounts have emphasized this. He would delight relatives, friends, and acquaintances with a literal recitation of any song from Virgil’s Aenesis, and he would remember minutes of Academy meetings years after they were held. He was not given to envy, and when someone made an advance on his work his happiness was genuine. For example, when he learnt of Lagrange’s improvements on his work on elliptic integrals, he wrote to him that his admiration knew no bounds and then proceeded to improve upon Lagrange!

But, what is most characteristic of his work is its clarity and openness. He never tries to hide the difficulties from the reader. This is in stark contrast to Newton, who was prone to hide his methods in obscure anagrams, and even from his successor, Gauss, who very often erased his steps to present a monolithic proof that was seldom illuminating. In Euler’s writings there are no comments on how profound his results are, and in his papers one can follow his ideas step by step with the greatest of ease. Nor was he chary of giving credit to others; his willingness to share his summation formula with Maclaurin, his proper citations to Fuguano when he started his work on algebraic integrals, his open admiration for Lagrange when the latter improved on his work in calculus of variations are all instances of his serene outlook. One can only contrast this with Gauss’s reaction to Bolyai’s discovery of non-Euclidean postulates. Euler was secure in his knowledge of what he had achieved but never insisted that he should be the only one on top of the mountain.

Perhaps, the most astounding aspect of his scientific opus is its universality. He worked on everything that had any bearing on mathematics. For instance, his early training under Johann Bernoulli did not include number theory; nevertheless, within a couple of years after reaching St. Petersburg he was deeply immersed in it, recreating the entire corpus of Fermat’s work in that area and then moving well beyond him. His founding of graph theory as a separate discipline, his excursions in what we call combinatorial topology, his intuition that suggested to him the idea of exploring multizeta values are all examples of a mind that did not have any artificial boundaries. He had no preferences about which branch of mathematics was dear to him. To him, they were all filled with splendour, or Herrlichkeit, to use his own favourite word.

Hilbert and Poincare were perhaps last of the universalists of modern era. Already von Neumann had remarked that it would be difficult even to have a general understanding of more than a third of the mathematicians of his time. With the explosive growth of mathematics in the twentieth century we may never see again the great universalists. But who is to say what is and is not possible for the human mind?

It is impossible to read Euler and not fall under his spell. He is to mathematics what Shakespeare is to literature and Mozart to music: universal and sui generis.

Reference:

Euler Through Time: A New Look at Old Themes by V S Varadarajan:

Hindustan Book Agency;

http://www.hindbook.com/index.php/euler-through-time-a-new-look-at-old-themes;

Amazon India link:

https://www.amazon.in/Euler-Through-Time-Look-Themes/dp/9380250592/ref=sr_1_1?keywords=Euler+Through+Time&qid=1568316624&s=books&sr=1-1

 

 

Rules for Inequalities

If a, b and c are real numbers, then

  1. a < b \Longrightarrow a + c< b + c
  2. a < b \Longrightarrow a - c < b - c
  3. a < b \hspace{0.1in} and \hspace{0.1in}c > 0 \Longrightarrow ac < bc
  4. a < b \hspace{0.1in} and \hspace{0.1in}c < 0 \Longrightarrow bc < ac special case: a < b \Longrightarrow -b < -a
  5. a > 0 \Longrightarrow \frac{1}{a} > 0
  6. If a and b are both positive or both negative, then a < b \Longrightarrow \frac{1}{b} < \frac{1}{a}.

Remarks:

Notice the rules for multiplying an inequality by a number: Multiplying by a positive number preserves the inequality; multiplying by a negative number reverses the inequality. Also, reciprocation reverses the inequality for numbers of the same sign.

Regards,

Nalin Pithwa.

Set Theory, Relations, Functions Preliminaries: Part III

FUNCTIONS:

Functions as a special kind of relation:

Let us first consider an example where set A is a set of men, and B is a set of positive real numbers. Let us say f is a relation from A to B given by : f = \{ (x,y) : x \in A, y \hspace{0.1in} is \hspace{0.1in} the \hspace{0.1in} weight \hspace{0.1in} of \hspace{0.1in} the \hspace{0.1in} person \hspace{0.1in} x \}

Hence, f “relates” every man in set A to his weight in set B. That is,

i) Every man has some weight associated with him in set B. (ii) That weight is unique. That is, a person cannot have more than one weight (at a given time, of course) !! 🙂 This, of course, does not mean that two different persons, say P and Q may not have the same weight 100 kg ( the same element of set B). The only thing it means is that any one person, say P will have one and only one weight (100kg) at the time instant of measurement and not more than one weights (which would be crazy) at a time instant it is measured !!

Definition I (a function defined as a relation):

A function f from a set A (called domain) to a set B (called codomain) is a relation that associates or “pairs up” every element of domain A with a unique element of codomain B. (Note that whereas a relation from a set A to a set B is just a subset of the cartesian product A \times B).

Some remarks: The above definition is also motivated by an example of a function as a relation. On the other hand, another definition of a function can be motivated as follows:

We know that the boiling point of water depends on the height of water above sea level. We also know that the simple interest on a deposit in a bank depends on the duration of deposit held in the bank. In these and several such examples, one quantity, say y, depends on another quantity “x”.

Symbol: f: A \longrightarrow B; if x \in A, y \in B, then we also denote: f: x \longmapsto y; we also write y=f(x), read as “y is f of x”.

Here, y is called image of x under f and x is called the preimage of y under f.

Definition: Range: The set of all images in B is called the range of f. That is, Range = \{ f(x): x \in A\}

Note: (i) Every function is a relation but every relation need not be a function. (Homework quiz: find illustrative examples for the same) (ii) If the domain and codomain are not specified, they are assumed to be the set of real numbers.

In calculus, we often want to refer to a generic function without having any particular formula in mind. Leonhard Euler invented a symbolic way to say “y is a function of x” by writing

y = f(x) (“y equals f of x”)

In this equation, the symbol f represents the function. The letter x, called the independent variable, represents an input value from the domain of f, and y, the dependent variable, represents the corresponding output value f(x) in the range of f. Here is the formal definition of function: (definition 2):

function from a set D to a set \Re is a rule that assigns a unique element f(x) in \Re to each element x in D.

In this definition, D=D(f) (read “D of f”) is the domain of the function f and \Re is the range (or codomain containing the range of f).

Think of a function f as a kind of machine that produces an output value f(x) in its range whenever we feed it an input value x from its domain. In our scope, we will usually define functions in one of two ways:

a) by giving a formula such as y=x^{2} that uses a dependent variable y to denote the value of the function, or

b) by giving a formula such as f(x)=x^{2} that defines a function symbol f to name the function.

NOTE: there can be well-defined functions which do not have any formula at all; for example, let f(x) = 0 when x \in Q and f(x)=1, when x \in Q^{'}.

Strictly speaking, we should call the function f and not f(x) as the latter denotes the value of the function at the point x. However, as is common usage, we will often refer to the function as f(x) in order to name the variable on which f depends.

It is sometimes convenient to use a single letter to denote both a function and the dependent variable. For instance, we might say that the area A of a circle of radius r is given by the function : A(r)=\pi r^{2}.

Evaluation:

As we said earlier, most of the functions in our scope will be real-valued function of a real variable, functions whose domains and ranges are sets of real numbers. We evaluate such functions by susbtituting particular values from the domain into the function’s defining rule to calculate the corresponding values in the range.

Example 1:

The volume V of a ball (solid sphere) r is given by the function: V(r)=\frac{4}{3}\pi  r^{3}.

The volume of a ball of radius 3 meters is : V(3)=\frac{4}{3}\pi (3)^{3}=36 \pi m^{3}.

Example 2:

Suppose that the function F is defined for all real numbers t by the formula: F(t)=2(t-1)+3.

Evaluate F at the output values 0, 2, x+2, and F(2).

Solution 2:

In each case, we substitute the given input value for t into the formula for F:

F(0)=2(0-1)+3=-2+3=1

F(2)=2(2-1)+3=2+3=5

F(x+2)=2(x+2-1)+3=2x+3

F(F(2))=F(5)=2(5-1)+3=11

The Domain Convention

When we define a function y=f(x) with a formula and the domain is not stated explicitly, the domain is assumed to be the largest set of x-values for which the formula gives real x-values. This is the function’s so-called natural domain. If we want the domain to be restricted in some way, we must say so.

The domain of the function y=x^{2} is understood to be the entire set of real numbers. The formula gives a real value y-value for every real number x. If we want to restrict the domain to values of x greater than or equal to 2, we must write ” y=x^{2}” for x \geq 2.

Changing the domain to which we apply a formula usually changes the range as well. The range of y=x^{2} is [0, \infty). The  range of y=x^{2} where x \geq 2 is the set of all numbers obtained by squaring numbers greater than or equal to 2. In symbols, the range is \{ x^{2}: x \geq 2\} or \{ y: y \geq 4\} or [4,\infty)

Example 3:

Function : y = \sqrt{1-x^{2}}; domain [-1,1]; Range (y) is [0,1]

Function: y=\frac{1}{x}; domain (-\infty,0) \bigcup (0,\infty); Range (y) is (-\infty,0)\bigcup (0,\infty)

Function: y=\sqrt{x}; domain (0,\infty) and range (y) is (0,\infty)

Function y = \sqrt{4-x}, domain (-\infty,,4], and range (y) is [0, \infty)

Graphs of functions:

The graph of a function f is the graph of the equation y=f(x). It consists of the points in the Cartesian plane whose co-ordinates (x,y) are input-output pairs for f.

Not every curve you draw is the graph of a function. A function f can have only one value f(x) for each x in its domain so no vertical line can intersect the graph of a function more than once. Thus, a circle cannot be the graph of a function since some vertical line intersect the circle twice. If a is in the domain of a function f, then the vertical line x=a will intersect the graph of f in the single point (a, f(a)).

Example 4: Graph the function y=x^{2} over the interval [-2.2]. (homework).Thinking further: so plotting the above graph requires a table of x and y values; but how do we connect the points ? Should we connect two points by a straight line, smooth line, zig-zag line ??? How do we know for sure what the graph looks like between the points we plot? The answer lies in calculus, as we will see in later chapter. There will be a marvelous mathematical tool called the derivative to find a curve’s shape between plotted points. Meanwhile, we will have to settle for plotting points and connecting them as best as we can. 

PS: (1) you can use GeoGebra, a beautiful freeware for plotting various graphs, and more stuff https://www.geogebra.org/ (2) If you wish, you can use a TI-graphing calculator. This is a nice investment for many other things like number theory also. See for example,

https://www.amazon.in/Texas-Instruments-Nspire-Graphing-Calculator/dp/B004NBZAYS/ref=sr_1_2?crid=3JSHJUOZMDMUS&keywords=ti+nspire+cx&qid=1569334614&s=electronics&sprefix=TI+%2Caps%2C267&sr=1-2

Meanwhile, you need to be extremely familiar with graphs of following functions; plot and check on your own:

y=x^{3}, y=x^{2/3}, y=\sqrt{x}, y=\sqrt[3]{x}, y=\frac{1}{x}, y=\frac{1}{x^{2}}, y=mx, where m \in Z, y=x^{3/2}

Sums, Differences, Products and Quotients

Like numbers, functions can be added, subtracted, multiplied and divided (except where the the denominator is zero) to produce new functions. If f and g are functions, then for every x that belongs to the domains of BOTH f and g, we define functions: f+g, f-g, fg by the formulas:

(f+g)(x)=f(x)+g(x),

(f-g)(x)=f(x)-g(x)

(fg)(x)=f(x)g(x)

At any point D(f) \bigcap D(g) at which g(x) \neq 0, we can also define the function f/g by the formula:

(\frac{f}{g})(x)=\frac{f(x)}{g(x)}, where g(x) \neq 0

Functions can also be multiplied by constants. If c is a real number, then the function cf is defined for all x in the domain of f by (cf)(x)=cf(x)

Example 5:

Function f, formula y=\sqrt{x}, domain [0,\infty)

Function g, formula g(x)=\sqrt{(1-x)}, domain (-\infty, 1]

Function 3g, formula 3g(x)=3\sqrt{(1-x)}, domain (-\infty, 1]

Function f+g, formula (f+g)(x)=\sqrt{x}+\sqrt{(1-x)}, domain [0,1]=D(f) \bigcap D(g)

Function f-g, formula (f-g)(x)=\sqrt{x}-\sqrt{(1-x)}, domain [0.1]

Function g-f, formula (g-f)(x)=\sqrt{(1-x)}-\sqrt{x}, domain [0,1]

Function f . g, formula (f . g)(x)=f(x)g(x) = \sqrt{x(1-x)}, domain [0,1]

Function \frac{f}{g}, formula \frac{f}{g}(x)=\frac{f(x)}{g(x)}=\sqrt{\frac{x}{1-x}}, domain is [0,1)

Function \frac{g}{f}(x) = \frac{g(x)}{f(x)}=\sqrt{\frac{1-x}{x}}, domain (0,1]

Composite Functions:

Composition is another method for combining functions.

Definition:

If f and g are functions, the composite function f \circ g (f “circle” g) is defined by (f \circ g)(x)=f(g(x)). The domain of f \circ g consists of the numbers x in the domain of g for which g(x) lies in the domain of f.

The definition says that two functions can be composed when the image of the first lies in the domain of the second. To (f \circ g)(x) we first find g(x) and second find f(g(x)).

Clearly, in general, (f \circ g)(x) \neq (g \circ f)(x). That is, composition of functions is not commutative.

Example 6:

If f(x)=\sqrt{x} and g(x)=x+1, find (a) (f \circ g)(x) (b) (g \circ f)(x) (c) (f \circ f)(x) (d) (g \circ g)(x)

Solution 6:

a) (f \circ g)(x) = f(g(x))=\sqrt{g(x)}=\sqrt{x+1}, domain is [-1, \infty)

b) (g \circ f)(x)=g(f(x))=f(x)+1=\sqrt{x}+1, domain is [0, \infty)

c) (f \circ f)(x)=f(f(x))=\sqrt{f(x)}=\sqrt{\sqrt{x}}=x^{\frac{1}{4}}, domain is [0, \infty)

d) (g \circ g)(x)=g(g(x))=g(x)+1=(x+1)+1=x+2, domain is \Re or (-\infty, \infty)

Even functions and odd functions:

A function f(x) is said to be even if f(x)=f(-x). That is, the function possesses symmetry about the y-axis. Example, y=f(x)=x^{2}.

A function f(x) is said to be odd if f(x)=-f(-x). That is, the function possesses symmetry about the origin. Example y=f(x)=x^{3}.

Any function can be expressed as a sum of an even function and an odd function.

A function could be neither even nor odd.

Note that a function like y^{2}=x possesses symmetry about the x-axis !!

Piecewise Defined Functions:

Sometimes a function uses different formulas or formulae over different parts of its domain. One such example is the absolute value function:

y=f(x) = |x|=x, when x \geq 0 and y=-x, when x<0.

Example 7:

The function f(x)=-x, when x<0, y=f(x)=x^{2}, when 0 \leq x \leq 1, and f(x)=1, when x>1.

Example 8:

The greatest integer function:

The function whose value at any number x is the greatest integer less than or equal to x is called the greatest integer function or the integer floor function. It is denoted by \lfloor x \rfloor.

Observe that \lfloor 2.4 \rfloor =2; \lfloor 1.4 \rfloor =1; \lfloor 0 \rfloor =0; \lfloor -1.2 \rfloor =-2; \lfloor 2 \rfloor =2; \lfloor 0.2 \rfloor =0\lfloor -0.3 \rfloor =-1; \lfloor -2 \rfloor =-2.

Example 9:

The least integer function:

The function whose value at any number x is the smallest integer greater than or equal to x is called the least integer function or the integer ceiling function. It is denoted by \lceil x \rceil. For positive values of x, this function might represent, for example, the cost of parking x hours in a parking lot which charges USD 1 for each hour or part of an hour.

Cheers,

Nalin Pithwa

 

 

 

Set Theory, Relations, Functions Preliminaries: II

Relations:

Concept of Order:

Let us say that we create a “table” of two columns in which the first column is the name of the father, and the second column is name of the child. So, it can have entries like (Yogesh, Meera), (Yogesh, Gopal), (Kishor, Nalin), (Kishor, Yogesh), (Kishor, Darshna) etc. It is quite obvious that “first” is the “father”, then “second” is the child. We see that there is a “natural concept of order” in human “relations”. There is one more, slightly crazy, example of “importance of order” in real-life. It is presented below (and some times also appears in basic computer science text as rise and shine algorithm) —-

Rise and Shine algorithm: 

When we get up from sleep in the morning, we brush our teeth, finish our morning ablutions; next, we remove our pyjamas and shirt and then (secondly) enter the shower; there is a natural order here; first we cannot enter the shower, and secondly we do not remove the pyjamas and shirt after entering the shower. 🙂

Ordered Pair: Definition and explanation:

A pair (a,b) of numbers, such that the order, in which the numbers appear is important, is called an ordered pair. In general, ordered pairs (a,b) and (b,a) are different. In ordered pair (a,b), ‘a’ is called first component and ‘b’ is called second component.

Two ordered pairs (a,b) and (c,d) are equal, if and only if a=c and b=d. Also, (a,b)=(b,a) if and only if a=b.

Example 1: Find x and y when (x+3,2)=(4,y-3).

Solution 1: Equating the first components and then equating the second components, we have:

x+3=4 and 2=y-3

x=1 and y=5

Cartesian products of two sets:

Let A and B be two non-empty sets then the cartesian product of A and B is denoted by A x B (read it as “A cross B”),and is defined as the set of all ordered pairs (a,b) such that a \in A, b \in B.

Thus, A \times B = \{ (a,b): a \in A, b \in B\}

e.g., if A = \{ 1,2\} and B = \{ a,b,c\}, tnen A \times B = \{ (1,a),(1,b),(1,c),(2,a),(2,b),(2,c)\}.

If A = \phi or B=\phi, we define A \times B = \phi.

Number of elements of a cartesian product:

By the following basic counting principle: If a task A can be done in m ways, and a task B can be done in n ways, then the tasks A (first) and task B (later) can be done in mn ways.

So, the cardinality of A x B is given by: n(A \times B)= n(A) \times n(B).

So, in general if a cartesian product of p finite sets, viz, A_{1}, A_{2}, A_{3}, \ldots, A_{p} is given by n(A_{1} \times A_{2} \times A_{3} \ldots A_{p}) = n(A_{1}) \times n(A_{2}) \times \ldots \times n(A_{p})

Definitions of relations, arrow diagrams (or pictorial representation), domain, co-domain, and range of a relation:

Consider the following statements:

i) Sunil is a friend of Anil.

ii) 8 is greater than 4.

iii) 5 is a square root of 25.

Here, we can say that Sunil is related to Anil by the relation ‘is a friend of’; 8 and 4 are related by the relation ‘is greater than’; similarly, in the third statement, the relation is ‘is a square root of’.

The word relation implies an association of two objects according to some property which they possess. Now, let us some mathematical aspects of relation;

Definition:

A and B are two non-empty sets then any subset of A \times B is called relation from A to B, and is denoted by capital letters P, Q and R. If R is a relation and (x,y) \in R then it is denoted by xRy.

y is called image of x under R and x is called pre-image of y under R.

Let A=\{ 1,2,3,4,5\} and B=\{ 1,4,5\}.

Let R be a relation such that (x,y) \in R implies x < y. We list the elements of R.

Solution: Here A = \{ 1,2,3,4,5\} and B=\{ 1,4,5\} so that R = \{ (1,4),(1,5),(2,4),(2,5),(3,4),(3,5),(4,5)\} Note this is the relation R from A to B, that is, it is a subset of A x B.

Check: Is a relation R^{'} from B to A defined by x<y, with x \in B and y \in A — is this relation R^{'} *same* as R from A to B? Ans: Let us list all the elements of R^{‘} explicitly: R^{'} = \{ (1,2),(1,3),(1,4),(1,5),(4,5)\}. Well, we can surely compare the two sets R and R^{'} — the elements “look” different certainly. Even if they “look” same in terms of numbers, the two sets R and R^{'} are fundamentally different because they have different domains and co-domains.

Definition : Domain of a relation R: The set of all the first components of the ordered pairs in a relation R is called the domain of relation R. That is, if R \subseteq A \times B, then domain (R) is \{ a: (a,b) \in R\}.

Definition: Range: The set of all second components of all ordered pairs in a relation R is called the range of the relation. That is, if R \subseteq A \times B, then range (R) = \{ b: (a,b) \in R\}.

Definition: Codomain: If R is a relation from A to B, then set B is called co-domain of the relation R. Note: Range is a subset of co-domain.

Type of Relations:

One-one relation: A relation R from a set A to B is said to be one-one if every element of A has at most one image in B and distinct elements in A have distinct images in B. For example, let A = \{ 1,2,3,4\}, and let B=\{ 2,3,4,5,6,7\} and let R_{1}= \{ (1,3),(2,4),(3,5)\} Then R_{1} is a one-one relation. Here, domain of R_{1}= \{ 1,2,3\} and range of R_{1} is \{ 3,4,5\}.

Many-one relation: A relation R from A to B is called a many-one relation if two or more than two elements in the domain A are associated with a single (unique) element in co-domain B. For example, let R_{2}=\{ (1,4),(3,7),(4,4)\}. Then, R_{2} is many-one relation from A to B. (please draw arrow diagram). Note also that domain of R_{1}=\{ 1,3,4\} and range of R_{1}=\{ 4,7\}.

Into Relation: A relation R from A to B is said to be into relation if there exists at least one element in B, which has no pre-image in A. Let A=\{ -2,-1,0,1,2,3\} and B=\{ 0,1,2,3,4\}. Consider the relation R_{1}=\{ (-2,4),(-1,1),(0,0),(1,1),(2,4) \}. So, clearly range is \{ 0,1,4\} and range \subseteq B. Thus, R_{3} is a relation from A INTO B.

Onto Relation: A relation R from A to B is said to be ONTO relation if every element of B is the image of some element of A. For example: let set A= \{ -3,-2,-1,1,3,4\} and set B= \{ 1,4,9\}. Let R_{4}=\{ (-3,9),(-2,4), (-1,1), (1,1),(3,9)\}. So, clearly range of R_{4}= \{ 1,4,9\}. Range of R_{4} is co-domain of B. Thus, R_{4} is a relation from A ONTO B.

Binary Relation on a set A:

Let A be a non-empty set then every subset of A \times A is a binary relation on set A.

Illustrative Examples:

E.g.1: Let A = \{ 1,2,3\} and let A \times A = \{ (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}. Now, if we have a set R = \{ (1,2),(2,2),(3,1),(3,2)\} then we observe that R \subseteq A \times A, and hence, R is a binary relation on A.

E.g.2: Let N be the set of natural numbers and R = \{ (a,b) : a, b \in N and 2a+b=10\}. Since R \subseteq N \times N, R is a binary relation on N. Clearly, R = \{ (1,8),(2,6),(3,4),(4,2)\}. Also, for the sake of completeness, we state here the following: Domain of R is \{ 1,2,3,4\} and Range of R is \{ 2,4,6,8\}, codomain of R is N.

Note: (i) Since the null set is considered to be a subset of any set X, so also here, \phi \subset A \times A, and hence, \phi is a relation on any set A, and is called the empty or void relation on A. (ii) Since A \times A \subset A \times A, we say that A \subset A is a relation on A called the universal relation on A. 

Note: Let the cardinality of a (finite) set A be n(A)=p and that of another set B be n(B)=q, then the cardinality of the cartesian product n(A \times B)=pq. So, the number of possible subsets of A \times B is 2^{pq} which includes the empty set.

Types of relations:

Let A be a non-empty set. Then, a relation R on A is said to be: (i) Reflexive: if (a,a) \in R for all a \in A, that is, aRa for all a \in A. (ii) Symmetric: If (a,b) \in R \Longrightarrow (b,a) \in R for all a,b \in R (iii) Transitive: If (a,b) \in R, and (b,c) \in R, then so also (a,c) \in R.

Equivalence Relation: 

A (binary) relation on a set A is said to be an equivalence relation if it is reflexive, symmetric and transitive. An equivalence appears in many many areas of math. An equivalence measures “equality up to a property”. For example, in number theory, a congruence modulo is an equivalence relation; in Euclidean geometry, congruence and similarity are equivalence relations.

Also, we mention (without proof) that an equivalence relation on a set partitions the set in to mutually disjoint exhaustive subsets. 

Illustrative examples continued:

E.g. Let R be an equivalence relation on \mathbb{Q} defined by R = \{ (a,b): a, b \in \mathbb{Q}, (a-b) \in \mathbb{Z}\}. Prove that R is an equivalence relation.

Proof: Given that R = \{ (a,b) : a, b \in \mathbb{Q}, (a-b) \in \mathbb{Z}\}. (i) Let a \in \mathbb{Q} then a-a=0 \in \mathbb{Z}, hence, (a,a) \in R, so relation R is reflexive. (ii) Now, note that (a,b) \in R \Longrightarrow (a-b) \in \mathbb{Z}, that is, (a-b) is an integer \Longrightarrow -(b-a) \in \mathbb{Z} \Longrightarrow (b-a) \in \mathbb{Z} \Longrightarrow (b,a) \in R. That is, we have proved (a,b) \in R \Longrightarrow (b,a) \in R and so relation R is symmetric also. (iii) Now, let (a,b) \in R, and (b,c) \in R, which in turn implies that (a-b) \in \mathbb{Z} and (b-c) \in \mathbb{Z} so it \Longrightarrow (a-b)+(b-c)=a-c \in \mathbb{Z} (as integers are closed under addition) which in turn \Longrightarrow (a,c) \in R. Thus, (a,b) \in R and (b,c) \in R implies (a,c) \in R also, Hence, given relation R is transitive also. Hence, R is also an equivalence relation on \mathbb{Q}.

Illustrative examples continued:

E.g.: If (x+1,y-2) = (3,4), find the values of x and y.

Solution: By definition of an ordered pair, corresponding components are equal. Hence, we get the following two equations: x+1=3 and y-2=4 so the solution is x=2,y=6.

E.g.: If A = (1,2), list the set A \times A.

Solution: A \times A = \{ (1,1),(1,2),(2,1),(2,2)\}

E.g.: If A = \{1,3,5 \} and B=\{ 2,3\}, find A \times B, and B \times A, check if cartesian product is a commutative operation, that is, check if A \times B = B \times A.

Solution: A \times B = \{ (1,2),(1,3),(3,2),(3,3),(5,2),(5,3)\} whereas B \times A = \{ (2,1),(2,3),(2,5),(3,1),(3,3),(3,5)\} so since A \times B \neq B \times A so cartesian product is not a commutative set operation.

E.g.: If two sets A and B are such that their cartesian product is A \times B = \{ (3,2),(3,4),(5,2),(5,4)\}, find the sets A and B.

Solution: Using the definition of cartesian product of two sets, we know that set A contains as elements all the first components and set B contains as elements all the second components. So, we get A = \{ 3,5\} and B = \{ 2,4\}.

E.g.: A and B are two sets given in such a way that A \times B contains 6 elements. If three elements of A \times B are (1,3),(2,5),(3,3), find its remaining elements.

Solution: We can first observe that 6 = 3 \times 2 = 2 \times 3 so that A can contain 2 or 3 elements; B can contain 3 or 2 elements. Using definition of cartesian product of two sets, we get that A= \{ 1,2,3\} and \{ 3,5\} and so we have found the sets A and B completely.

E.g.: Express the set \{ (x,y) : x^{2}+y^{2}=25, x, y \in \mathbb{W}\} as a set of ordered pairs.

Solution: We have x^{2}+y^{2}=25 and so

x=0, y=5 \Longrightarrow x^{2}+y^{2}=0+25=25

x=3, y=4 \Longrightarrow x^{2}+y^{2}=9+16=25

x=4, y=3 \Longrightarrow x^{2}+y^{2}=16+9=25

x=5, y=0 \Longrightarrow x^{2}+y^{2}=25+0=25

Hence, the given set is \{ (0,5),(3,4),(4,3),(5,0)\}

E.g.: Let A = \{ 1,2,3\} and B = \{ 2,4,6\}. Show that R = \{ (1,2),(1,4),(3,2),(3,4)\} is a relation from A to B. Find the domain, co-domain and range.

Solution: Here, A \times B = \{ (1,2),(1,4),(1,6),(2,2),(2,4),(2,6),(3,2),(3,4),(3,6)\}. Clearly, R \subseteq A \times B. So R is a relation from A to B. The domain of R is the set of first components of R (which belong to set A, by definition of cartesian product and ordered pair)  and the codomain is set B. So, Domain (R) = \{ 1,3\} and co-domain of R is set B itself; and Range of R is \{ 2,4\}.

E.g.: Let A = \{ 1,2,3,4,5\} and B = \{ 1,4,5\}. Let R be a relation from A to B such that (x,y) \in R if x<y. List all the elements of R. Find the domain, codomain and range of R. (as homework quiz, draw its arrow diagram);

Solution: Let A = \{ 1,2,3,4,5\} and B = \{ 1,4,5\}. So, we get R as (1,4),(1,5),(2,4),(2,5),(3,4),(3,5),(4,5). domain(R) = \{ 1,2,3,4\}, codomain(R) = B, and range(R) = \{ 4,5\}.

E.g. Let A = \{ 1,2,3,4,5,6\}. Define a binary relation on A such that R = \{ (x,y) : y=x+1\}. Find the domain, codomain and range of R.

Solution: By definition, R \subseteq A \times A. Here, we get R = \{ (1,2),(2,3),(3,4),(4,5),(5,6)\}. So we get domain (R) = \{ 1,2,3,4,5\}, codomain(R) =A, range(R) = \{ 2,3,4,5,6\}

Tutorial problems:

  1. If (x-1,y+4)=(1,2), find the values of x and y.
  2. If (x + \frac{1}{3}, \frac{y}{2}-1)=(\frac{1}{2} , \frac{3}{2} )
  3. If A=\{ a,b,c\} and B = \{ x,y\}. Find out the following: A \times A, B \times B, A \times B and B \times A.
  4. If P = \{ 1,2,3\} and Q = \{ 4\}, find the sets P \times P, Q \times Q, P \times Q, and Q \times P.
  5. Let A=\{ 1,2,3,4\} and \{ 4,5,6\} and C = \{ 5,6\}. Find A \times (B \bigcap C), A \times (B \bigcup C), (A \times B) \bigcap (A \times C), A \times (B \bigcup C), and (A \times B) \bigcup (A \times C).
  6. Express \{ (x,y) : x^{2}+y^{2}=100 , x, y \in \mathbf{W}\} as a set of ordered pairs.
  7. Write the domain and range of the following relations: (i) \{ (a,b): a \in \mathbf{N}, a < 6, b=4\} (ii) \{ (a,b): a,b \in \mathbf{N}, a+b=12\} (iii) \{ (2,4),(2,5),(2,6),(2,7)\}
  8. Let A=\{ 6,8\} and B=\{ 1,3,5\}. Let R = \{ (a,b): a \in A, b \in B, a+b \hspace{0.1in} is \hspace{0.1in} an \hspace{0.1in} even \hspace{0.1in} number\}. Show that R is an empty relation from A to B.
  9. Write the following relations in the Roster form and hence, find the domain and range: (i) R_{1}= \{ (a,a^{2}) : a \hspace{0.1in} is \hspace{0.1in} prime \hspace{0.1in} less \hspace{0.1in} than \hspace{0.1in} 15\} (ii) R_{2} = \{ (a, \frac{1}{a}) : 0 < a \leq 5, a \in N\}
  10. Write the following relations as sets of ordered pairs: (i) \{ (x,y) : y=3x, x \in \{1,2,3 \}, y \in \{ 3,6,9,12\}\} (ii) \{ (x,y) : y>x+1, x=1,2, y=2,4,6\} (iii) \{ (x,y) : x+y =3, x, y \in \{ 0,1,2,3\}\}

More later,

Nalin Pithwa