# Set Theory, Relations, Functions: preliminaries: part 10: more tutorial problems for practice

Problem 1:

Prove that a function f is 1-1 iff $f^{-1}(f(A))=A$ for all $A \subset X$. Given that $f: X \longrightarrow Y$.

Problem 2:

Prove that a function if is onto iff $f(f^{-1}(C))=C$ for all $C \subset Y$. Given that $f: X \longrightarrow Y$.

Problem 3:

(a) How many functions are there from a non-empty set S into $\phi$\?

(b) How many functions are there from $\phi$ into an arbitrary set $S$?

(c) Show that the notation $\{ X_{i} \}_{i \in I}$ implicitly involves the notion of a function.

Problem 4:

Let $f: X \longrightarrow Y$ be a function, let $A \subset X$, $B \subset X$, $C \subset Y$ and $D \subset Y$. Prove that

i) $f(A \bigcap B) \subset f(A) \bigcap f(B)$

ii) $f^{-1}(f(A)) \supset A$

iii) $f(f^{-1}(C)) \subset C$

Problem 5:

Let I be a non-empty set and for each $i \in I$, let $X_{i}$ be a set. Prove that

(a) for any set B, we have $B \bigcap \bigcup_{i \in I}X_{i}=\bigcup_{i \in I}(B \bigcap X_{i})$

(b) if each $X_{i}$ is a subset of a given set S, then $(\bigcup_{i \in I}X_{i})^{'}=\bigcap_{i \in I}(X_{i})^{'}$ where the prime indicates complement.

Problem 6:

Let A, B, C be subsets of a set S. Prove the following statements:

(i) $A- (B-C)=(A-B)\bigcup(A \bigcap B \bigcap C)$

(ii) $(A-B) \times C=(A \times C)-(B \times C)$

🙂 🙂 🙂

Nalin Pithwa

# Set Theory, Relations, Functions: Preliminaries: Part IX: (tutorial problems)

Reference: Introductory Real Analysis, Kolmogorov and Fomin, Dover Publications.

Problem 1:

Prove that if $A \bigcup B=A$ and $A \bigcap B=A$, then $A=B$.

Problem 2:

Show that in general $(A-B)\bigcup B \neq A$.

Problem 3:

Let $A = \{ 2,4, \ldots, 2n, \ldots\}$ and $B= \{ 3,6,\ldots, 3n, \ldots\}$. Find $A \bigcap B$ and $A - B$.

Problem 4:

Prove that (a) $(A-B)\bigcap (C)=(A \bigcap C)-(B \bigcap C)$

Prove that (b) $A \Delta B = (A \bigcup B)-(A \bigcap B)$

Problem 5:

Prove that $\bigcup_{a}A_{\alpha}-\bigcup_{a}B_{\alpha}=\bigcup_{\alpha}(A_{\alpha}-B_{\alpha})$

Problem 6:

Let $A_{n}$ be the set of all positive integers divisible by $n$. Find the sets (i) $\bigcup_{n=2}^{\infty}A_{n}$ (ii) $\bigcap_{n=2}^{\infty}A_{n}$.

Problem 7:

Find (i) $\bigcup_{n=1}^{\infty}[n+\frac{1}{n}, n - \frac{1}{n}]$ (ii) $\bigcap_{n=1}^{\infty}(a-\frac{1}{n},b+\frac{1}{n})$

Problem 8:

Let $A_{\alpha}$ be the set of points lying on the curve $y=\frac{1}{x^{\alpha}}$ where $(0. What is $\bigcap_{\alpha \geq 1}A_{\alpha}$?

Problem 9:

Let $y=f(x) = $ for all real x, where $$ is the fractional part of x. Prove that every closed interval of length 1 has the same image under f. What is the image? Is f one-to-one? What is the pre-image of the interval $\frac{1}{4} \leq y \leq \frac{3}{4}$? Partition the real line into classes of points with the same image.

Problem 10:

Given a set M, let $\mathscr{R}$ be the set of all ordered pairs on the form $(a,a)$ with $a \in M$, and let $aRb$ if and only if $(a,b) \in \mathscr{R}$. Interpret the relation R.

Problem 11:

Give an example of a binary relation which is:

• Reflexive and symmetric, but not transitive.
• Reflexive, but neither symmetric nor transitive.
• Symmetric, but neither reflexive nor transitive.
• Transitive, but neither reflexive nor symmetric.

We will continue later, 🙂 🙂 🙂

PS: The above problem set, in my opinion, will be very useful to candidates appearing for the Chennai Mathematical Institute Entrance Exam also.

Nalin Pithwa

# Set Theory, Relations, Functions: Preliminaries: part VIIIA

(We continue from part VII of the same blog article series with same reference text).

Theorem 4:

A set M can be partitioned into classes by a relation R (acting as a criterion for assigning two elements to the same class) if and only R is an equivalence relation on M.

Proof of Theorem 4:

Every partition of M determines a binary relation on M, where $aRb$ means that “a belongs to the same class as b.” It is then obvious that R must be reflexive, symmetric and transitive, that is, R is an equivalence relation on M.

Conversely, let R be an equivalence relation on M, and let $K_{a}$ be the set of all elements $x \in M$ such that $xRa$ (clearly, $a \in K_{a}$, since R is reflexive). Then, two classes $K_{a}$ and $K_{b}$ are either identical or disjoint. In fact, suppose that an element c belongs to both $K_{a}$ and $K_{b}$, so that $cRa$ and $cRb$. But by symmetry of R, being an equivalence relation, we can infer that $aRc$ also and, further by transitivity, we say that $aRb$. If now, $x \in K_{a}$ then we have $xRa$ and hence, $xRb$ (since we already have $aRb$ and using transitivity).

Similarly, we can prove that $x \in K_{b}$ implies that $x \in K_{a}$.

Therefore, $K_{a}=K_{b}$ if $K_{a}$ and $K_{b}$ have an element in common. Therefore, the distinct sets $K_{a}$ form a partition of M into classes.

QED.

Remark:

Because of theorem 4, one often talks about the decomposition of a set M into equivalence classes.

There is an intimate connection between mappings and partitions into classes, as illustrated by the following examples:

Example 1:

Let f be a mapping of a set A into a set B and partition A into sets, each consisting of all elements with the same image $b=f(a) \in B$. This gives a partition of A into classes. For example, suppose f projects the xy-plane onto the x-axis by mapping the point $(x,y)$ into the point $(x,0)$. Then, the preimages of the points of the x-axis are vertical lines, and the representation of the plane as the union of these lines is the decomposition into classes corresponding to f.

Example 2:

Given any partition of a set A into classes, let B be the set of these classes and associate each element $a \in A$ with the class (that is, element of B) to which it belongs. This gives a mapping of A into B. For example, suppose we partition three-dimensional space into classes by assigning to the same class all points which are equidistant from the origin of coordinates. Then, every class is a sphere of a certain radius. The set of all these classes can be identified with the set of points on the half-line $[0, \infty)$ each point corresponding to a possible value of the radius. In this sense, the decomposition of 3-dimensional space into concentric spheres corresponds to the mapping of space into the half-line $[0,\infty)$.

Example 3:

Suppose that we assign all real numbers with the same fractional part to the same class. Then, the mapping corresponding to this partition has the effect of “winding” the real line onto a circle of unit circumference. (Note: The largest integer $\leq x$ is called the integral part of x, denoted by [x], and the quantity $x -[x]$ is called the fractional part of x).

In the next blog article, let us consider a tutorial problem set based on last two blogs of this series.

🙂 🙂 🙂

Nalin Pithwa

# Set Theory, Relations, Functions: Preliminaries: Part VIII

SETS and FUNCTIONS:

Reference: Introductory Real Analysis by A. N. Kolmogorov and S V Fomin, Dover books.

Operations on sets:

Let A and B be any two sets. Then, by the sum or union of A and B, denoted by $A \bigcup B$, is meant the set consisting of all elements which belong to at least one of the sets A and B. More generally, by the sum or union of an arbitrary number (finite or infinite) of sets $A_{\alpha}$ (indexed by some parameter $\alpha$), we mean the set, denoted by $\bigcup_{\alpha}A_{\alpha}$ of all elements belonging to at least one of the sets $A_{\alpha}$.

By the intersection $A \bigcap B$ of two given sets A and B, we mean the set consisting of all elements which belong to both A and B. For example, the intersection of the set of all even numbers and the set of all integers divisible by 3 is the set of all integers divisible by 6. By the intersection of an arbitrary number (finite or infinite) of sets $A_{\alpha}$, we mean the set, denoted by $\bigcap_{\alpha}A_{\alpha}$ of all elements belonging to every one of the sets $A_{\alpha}$. Two sets A and B are said to be disjoint if $A \bigcap B=\phi$, that is, if they have no elements in common. More generally, let $\mathscr{F}$ be a family of sets such that $A \bigcap B=\phi$ for every pair of sets A, B in $\mathscr{F}$. Then, the sets in $\mathscr{F}$ are said to be pairwise disjoint.

It is an immediate consequence of the above definitions that the operations $\bigcup$ and $\bigcap$ are commutative and associative, that is,

$A \bigcup B=B \bigcup A$ and $(A \bigcup B) \bigcup C= A \bigcup (B \bigcup C)$; $A \bigcap B=B \bigcap A$; and, $(A \bigcap B)\bigcap C = A \bigcap (B \bigcap C)$.

Moreover, the operations $\bigcup$ and $\bigcap$ obey the following distributive laws:

$(A \bigcup B)\bigcap C = (A \bigcap C) \bigcup (B \bigcap C)$….call this I

$(A \bigcap B)\bigcup C = (A \bigcup C) \bigcap (B \bigcup C)$….call this II.

For example, suppose that $x \in (A \bigcup B)\bigcap C$, so that x belongs to the left-hand side of I. Then, x belongs to both C and $A \bigcup B$, that is, x belongs to both C and at least one of the sets A and B. But then x belongs to at least one of the sets $A \bigcap C$ and $B \bigcap C$, that is, $x \in (A \bigcap C)\bigcup (B \bigcap C)$, so that x belongs to the right hand side of I. Conversely, suppose that $x \in (A \bigcap C)\bigcup (B \bigcap C)$. Then, x belongs to at least one of the two sets $A \bigcap C$ and $B \bigcap C$. It follows that x belongs to both C and at least one of the two sets A and B, that is, $x \in C$ and $x \in A \bigcup B$, or equivalently $x \in (A \bigcup B)\bigcap C$. This proves I, and II is proved similarly.

By the difference of two sets, $A - B$ between two sets A and B (in that order), we mean the set of all elements of A which do not belong to B. Note that it is not assumed that $A \supset B$. It is sometimes convenient (e.g., in measure theory) to consider the symmetric difference of two sets A and B, denoted by $A \delta B$ and defined as the union of the two differences $A-B$ and $B-A$: $A \delta B=(A-B)\bigcup (B-A)$.

We will often be concerned later with various sets which are all subsets of some underlying basic set R, for example, various sets of points on the real line. In this case, given a set A, the difference $R-A$ is called the complement of A, denoted by $A^{'}$.

An important role is played in set theory and its applications by the following duality principle:

$R - \bigcup_{\alpha}A_{\alpha}=\bigcap_{\alpha}(R-A_{\alpha})$…call this III

$R - \bigcap_{\alpha}A_{\alpha}=\bigcup_{\alpha}(R-A_{\alpha})$…call this IV.

In words, the complement of a union equal the intersection of the complements; and, the complement of an intersection equals the union of the complements. According to the duality principle, any theorem involving a family of subsets of a fixed set R can be converted automatically into another, “dual” theorem by replacing all subsets by their complements, all unions by intersections and all intersections by unions. To prove III, suppose that $x \in R - \bigcup_{\alpha}A_{\alpha}$….call this V.

Then, x does not belong to the union $\bigcup_{\alpha}A_{\alpha}$. …call this VI.

That is, x does not belong to any of the sets $A_{\alpha}$. It follows that x belongs to each of the complements $R - \bigcup_{\alpha}A_{\alpha}$, and hence, $x \bigcap_{\alpha}(R-A_{\alpha})$….call this VII.

Conversely, suppose that VII holds, so that x belongs to every set $R - A_{\alpha}$. Then, x does not belong to any of the sets $A_{\alpha}$, that is, x does not belong to the union VI, or equivalently V holds true. This proves 3.

Homework: Prove IV similarly.

Remark:

The designation “symmetric difference” for the set $A \Delta B$ is not too apt, since $A \Delta B$ has much in common with the sum $A \bigcup B$. In fact, in $A \bigcup B$ the two statements “x belongs to A” and “x belongs to B” are joined by the conjunction “or” used in the “either …or …or both…” sense, while in $A \Delta B$ the same two statements are joined by “or” used in the ordinary “either…or…” sense (as in “to be or not to be”). In other words, x belongs to $A \bigcup B$ if and only if x belongs to either A or B or both, while x belongs to $A \Delta B$ if and only if x belongs to either A or B but not both. The set $A \Delta B$ can be regarded as a kind of “modulo-two sum” of the sets A and B, that is, a sum of the sets A and B in which elements are dropped if they are counted twice (once in A and once in B).

Functions and mappings. Images and preimages:

(Of course, we have dealt with this topic in quite detail so far in the earlier blog series; but as presented by Kolmgorov and Fomin here, the flavour is different; at any rate, it helps to revise. Revision refines the intellect.) 🙂 🙂 🙂

A rule associating a unique real number $y=f(x)$ with each element of a set real numbers X is said to define a (real) function f on X. The set X is called the domain (of definition) of f, and the set Y of all numbers $f(x)$ such that $x \in X$ is called the range of f.

More generally, let M and N be two arbitrary sets. Then a rule associating a unique element $b=f(a) \in N$ with each element $a \in M$ is again said to define a function f on M (or a function f with domain M). In this more general context, f is usually called a mapping of f M into N. By the same token, f is said to map M into N(and a into b).

If a is an element of M, the corresponding element $b=f(a)$ is called the image of a (under the mapping f). Every element of M with a given element $b \in N$ as its image is called a preimage of b. Note that in general b may have several pre-images. Moreover, N may contain elements with no pre-images at all. If b has a unique pre-image, we denote this pre-image by $f^{-1}(b)$.

If A is a subset of M, the set of all elements $f(a) \in N$ such that $a \in A$ is called the image of A, denoted by $f(A)$. The set of all elements of M whose images belong to a given set $B \subset N$ is called the preimage of B, denoted by $f^{-1}(B)$. If no element of B has a preimage, then $f^{-1}(B)=\phi$. A function f is said to map M into N if $f(M)=N$, as is always the case, and onto N if $f(M)=N$. Thus, every “onto” mapping is an “into” mapping, but converse is not true.

Suppose f maps M onto N. Then, f is said to be one-to-one if each element $b \in N$ has a unique preimage $f^{-1}(b)$. In this case, f is said to establish a one-to-one correspondence between M and N, and the mapping $f^{-1}$ associating $f^{-1}(b)$ is called the inverse of f.

Theorem I:

The preimage of the union of two sets is the union of the preimages of the sets $f^{-1}(A \bigcup B)=f^{-1}(A) \bigcup f^{-1}(B)$.

Proof of Theorem I:

If $x \in f^{-1}(A \bigcup B)$, then $f(x) \in A \bigcup B$ so that $f(x)$ belongs to at least one of the sets A and B. But, then x belongs to at least one of the sets $f^{-1}(A)$ and $f^{-1}(B)$, that is, $x \in f^{-1}(A) \bigcup f^{-1}(B)$.

Conversely, if $x \in f^{-1}(A) \bigcup f^{-1}(B)$, then x belongs to at least one of the sets $f^{-1}(A)$ and $f^{-1}(B)$. Therefore, $f(x)$ belongs to at least one of the sets A and B, that is, $f(x) \in A \bigcup B$. But, then $x \in f^{-1}(A \bigcup B)$.

QED.

Theorem 2:

The preimage of the intersection of two sets is the intersection of the preimages of the sets:

$f^{-1}(A \bigcap B)=f^{-1}(A) \bigcap f^{-1}(B)$.

Proof of Theorem 2:

If $x \in f^{-1}(A \bigcap B)$, then $f(x) \in A \bigcap B$, so that $f(x) \in A$ and (meaning, simultaneously) $f(x) \in B$. But, then $x \in f^{-1}(A)$ and $x \in f^{-1}(B)$, that is, $x \in f^{-1}(A) \bigcap f^{-1}(B)$.

Conversely, if $x \in f^{-1}(A) \bigcap f^{-1}(B)$, then $x \in f^{-1}(A)$ and $x \in f^{-1}(B)$. Therefore, $f(x) \in A$ and $f(x) \in B$, that is, $f(x) \in A \bigcap B$. But, then $x \in f^{-1}(A \bigcap B)$.

QED.

Theorem 3:

The image of the union of two sets equals the union of the images of the sets $f(A \bigcup B) = f(A) \bigcup f(B)$.

Proof of theorem 3:

If $y \in f(A \bigcup B)$, then $y=f(x)$ where x belongs to at least one of the sets A and B. Therefore, $y=f(x)$ belongs to at least one of the sets $f(A)$ and $f(B)$. That is, $x \in f(A) \bigcup f(B)$.

Conversely, if $y \in f(A) \bigcup f(B)$, then $y=f(x)$. where x belongs to at least one of the sets A and B, that is, $x \in A \bigcup B$ and hence, $y=f(x) \in f(A \bigcup B)$.

QED.

Remark I:

Surprisingly enough, the image of the intersection of two sets is not so “well-behaved”. The image of the intersection of two sets does not necessarily equal the intersection of the images of the sets. For example, suppose the mapping f projects the xy-plane onto the x-axis, carrying the point $(x,y)$ into the $(x,0)$. Then, the segments $0 \leq x \leq 1$ with $y=0$, and $0 \leq x \leq 1$ with $y=1$ do not intersect, although their images coincide.

Remark 2:

In the light of above remark: consider the following: If a function is not specified on elements, it is important in general to check that f is well-defined. That is, unambiguously defined. For example, if the set A is the union of two subsets $A_{1}$ and $A_{2}$, and we are considering a function $f: A \longrightarrow B$, then one can try to specify a function from set A to the set B $\{ 0,1\}$ by declaring that f is to map everything in $A_{1}$ to 0 and is to map everything in $A_{2}$ to 1. This unambiguously defines f unless $A_{1}$ and $A_{2}$ have elements in common in which case it is not clear whether these elements should map to 0 or to 1. Checking that this f is well-defined therefore amounts to checking that $A_{1}$ and $A_{2}$ have no intersection.

Remark 3:

Theorems 1-3 continue to hold for unions and intersections of an arbitrary number (finite or infinite) of sets $A_{\alpha}$:

$f^{-1}(\bigcup_{\alpha})A_{\alpha}=\bigcup_{\alpha}f^{-1}(A_{\alpha})$

$f^{-1}(\bigcap_{\alpha}A_{\alpha})=\bigcap_{\alpha}f^{-1}(A_{\alpha})$

$f(\bigcup_{\alpha})A_{\alpha}=\bigcup_{\alpha}f(A_{\alpha})$

Decomposition of a set into classes. Equivalence relations.

Decompositions of a given set into pairwise disjoint subsets play an important role in a great variety of problems. For example, the plane (regarded as a point set) can be decomposed into lines parallel to the x-axis, three dimensional space can be decomposed into concentric spheres, the inhabitants of a given city can be decomposed into different age groups, and so on. Any such representation of a given set M as the union of a family of pairwise disjoint subsets of M is called a decomposition or partition of M into classes.

A decomposition is usually made on the basis of a certain criterion, allowing us to assign the elements of M to one class or another. For example, the set of all triangles in the plane can be decomposed into classes of congruent triangles, or classes of triangles of equal area, the set of all functions of x can be decomposed into classes of functions all taking the same value at a given point x, and so on. Despite the great variety of such criteria, they are not completely arbitrary. For example, it is obviously impossible to partition all real numbers into classes by assigning the number b to the same class as number a if and only if $b>a$. In fact, if $b>a$, b must be assigned to same class as a but then a cannot be assigned to same class as b, since $a. Moreover, since a is not greater than itself, a cannot be assigned to the class containing itself!! As another example, it is impossible to partition the points of the plane into classes by assigning two points to the same class if and only if the distance between them is less than 1. In fact, if the distance between a and b is less than 1, and if the distance between b and c is less than 1, it does not follow that distance between a and c is less than 1. (Hint: Think of triangle inequality). Thus, by assigning a to the same class as b and b to the same class as c, we may well find that two points fall in the same class even though the distance between them is greater than 1!

These examples suggest conditions that which must be satisfied by any criterion it it is to be used as the basis for partitioning a given set into classes. Let M be a set, and let certain ordered pairs (a,b) of elements of M be called “labelled.” If $(a,b)$ is a labelled pair, we say that a is related to b by the binary relation R and we write it symbolically as aRb. For example, if a and b are real numbers, $aRb$ might mean $a, while if a and b are triangles, $aRb$ might mean that a and b have the same area. A relation between elements of M is called a relation on M, if there exists at least one labelled pair $(a,b)$ for every $a \in M$. A relation R on M is called an equivalence relation (on M) if it satisfies the following three conditions:

• Reflexivity: $aRa$ for every $a \in M$.
• Symmetry: If $aRb$, then $bRa$.
• Transititivity: If $aRb$ and $bRc$, then $aRc$.

🙂 🙂 🙂

Nalin Pithwa, more later

# Euler Series question and solution

Question:

Mengoli had posed the following series to be evaluated:

$1+ \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \ldots$.

Some great mathematicians, including Liebnitz, John Bernoulli and D’Alembert, failed to compute this infinite series. Euler established himself as the best mathematician of Europe (in fact, one of the greatest mathematicians in history) by evaluating this series initially by a not-so-rigorous method. Later on, he gave alternative and more rigorous ways of getting the same result.

Prove that the series converges and gets an upper limit. Then, try to evaluate the series.

Proof:

Due Nicolas Oresine:

Consider the following infinite series: $\phi(s)=1 + \frac{1}{2^{s}} + \frac{1}{3^{s}} + \frac{1}{4^{s}} + ldots$

We can re-write the preceding series as follows: $\phi(s) = 1+ (\frac{1}{2^{s}}+\frac{1}{3^{s}}) + (\frac{1}{4^{s}} + \frac{1}{5^{s}} + \frac{1}{6^{s}} + \frac{1}{7^{s}}) + \ldots$, which in turn is less than

$1 + (\frac{2}{2^{s}}) + (\frac{4}{4^{s}}) + \ldots$. Now, the RHS of this can be re-written as

$1+(\frac{2}{2^{s}}) + (\frac{4}{4^{s}}) + \ldots=1 + \frac{1}{2^{(s-1)}}+ (\frac{1}{2^{(s-1)}})^{2} + \ldots$, which is a geometric series and it is given by

$\frac{1}{1-\frac{1}{2^{(s-1)}}}$.

Now, we can say that $\phi(s)$ will converge if $\frac{1}{2^{(s-1)}}<1 \Longrightarrow s >1$.

In order to prove what is asked, we start with $\phi(s)=1 + \frac{1}{2^{s}}+ \frac{1}{3^{s}}+ \frac{1}{4^{s}}+\ldots$

And, then multiply both sides by $\frac{1}{2^{s}}$ and then subtract the resulting equation from the preceding equation to get

$(1-\frac{1}{2^{2}})\phi(s)=1+\frac{1}{3^{s}}+\frac{1}{5^{s}}+\ldots$

where all the terms containing the reciprocals of the sth power of even numbers vanished.

Repeating this procedure with $\frac{1}{3^{s}}$ gives

$(1-\frac{1}{2^{s}})(1-\frac{1}{3^{s}})\phi(s)=1+\frac{1}{5^{s}}+ \ldots$

where all terms containing the reciprocals of the sth power of multiples of 3 vanished.

By continuing this with all prime numbers, we get

$\prod_{p}(1-\frac{1}{p^{s}})\phi(s)=1$, where p represents all prime numbers. Thus, we get

$\phi(s)=1 + \frac{}{} + \frac{}{} + \frac{}{} + \ldots =\frac{1}{\prod_{p}(1-\frac{1}{p^{s}})}$

This is a remarkable result because the LHS is concerned with only positive integers, whereas the RHS is concerned with only primes. This result is known as the “Golden Key of Euler”.

Riemann created his famous $\zeta-$ function by extending the variable s to the entire complex plane, except $s=1$ with

$\zeta(s)=1+ \frac{1}{2^{s}} + \frac{1}{3^{s}} + \ldots$.

This function is now very famous as the Riemann zeta function.

How can we apply the Golden Key of Euler to Mengoli’s question that we started with?

Ans. In the Golden Key of Euler, substitute $s=2$.

Hence, we get the upper limit of the given series is 2.

Euler’s proof (1775):

The proof ran as follows:

It is a little roundabout way of arriving at the correct answer from a known result. Consider McLaurin’s series expansion of sin x:

$\sin{(x)}=x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} -\frac{x^{7}}{7!} + \frac{x^{9}}{9!} + \ldots$

By dividing both sides by x and then substituting $y=x^{2}$ on the right side, we get the following:

$\frac{\sin{(x)}}{x} = 1-\frac{y}{3!} + \frac{y^{2}}{5!} - \frac{y^{3}}{7!} + \ldots$

By taking a special value of $x=n\pi$ (and, hence $y=n^{2}\pi^{2}$), we get the following:

$\frac{\sin (n\pi)}{(n\pi)}=0=1-\frac{y}{3!} + \frac{y^{2}}{5!} - \frac{y^{3}}{5!}+ \ldots$

Note  that preceding equation is not a polynomial, but an infinite series. But, Euler still treated it as a polynomial (that is why it was not accepted as a rigorous result) and observed that this “infinite” polynomial has roots equal to $y_{n}=n^{2}x^{2}$. Then, Euler had used the fact that the sum of the reciprocals of the roots is determined by the coefficient of the linear term (here, the y-term) when the constant is made unity. (check this as homework quiz, for a quadratic to be convinced). So, Euler had arrived at the following result:

$1-\sum_{i=1}^{\infty}\frac{6}{y_{n}}=0 \Longrightarrow \sum_{i=1}^{\infty}\frac{1}{y_{n}}=\frac{1}{6}$. With $y_{n}=n^{2}(\pi)^{2}$, we get the following:

$\sum_{i=1}^{\infty}\frac{1}{n^{2}(\pi)^{2}}=\frac{1}{6}$ or, $\sum_{1}^{n^{2}}\frac{1}{n^{2}}=\frac{(\pi)^{2}}{6}$.

Another proof also attributed to Euler that uses the series expansion of sin (x) goes as follows below:

$\sin {(x)}$ has roots given by 0, $\pm \pi$, $\pm 2\pi$, $\pm 3\pi$, …So does this polynomial that Euler reportedly constructed:

$x(1-\frac{x^{2}}{(\pi)^{2}})(1-\frac{x^{2}}{(2\pi)^{2}})(1-\frac{x^{2}}{(3\pi)^{2}})\ldots=0$

So, Euler considered the preceding equation to be equivalent to:

$\sin{(x)}=x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \ldots=0$

Then, he had equated the coefficient of $x^{3}$ in both to get the result:

$\sum_{n=1}^{\infty}\frac{1}{n^{2}(\pi)^{2}} = \frac{1}{3!} = \frac{1}{6}$.

Thus, $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{(\pi)^{2}}{6}$.

Later on, Euler had provided a few more alternate and rigorous proofs of this result.

Reference: Popular Problems and Puzzles in Mathematics by Asok Kumar Mallik, IISc Press, Foundation Books.

Hope you all enjoyed it — learning to think like Euler !! By the way, it did take a long time for even analysis to become so rigorous as it is now….You might like this observation a lot. 🙂 🙂 🙂

Nalin Pithwa.

# More Tripos Problems for Practice : RMO and INMO Algebra

If you want to be a Wrangler, you need to tackle the following (so also the problems in previous post :-)):

1. If $x(2a-y) = y(2a-z) = z(2a-u) = u(2a-x) = b^{2}$, show that $x = y = z= u$ unless $b^{2}=2a^{2}$, and that if this condition is satisfied, the equations are not  independent.
2. Out of n straight lines whose lengths are 1, 2, 3, $\ldots , n$ inches respectively, the number of ways in which four may be chosen which will form a quadrilateral in which a circle may be inscribed is $\frac{1}{48} \{ 2n(n-2)(2n-5)-3+3(-1)^{n}\}$.
3. For the expansion of $\frac{1+2x}{1-x^{3}}$, or otherwise, prove that

$1-3n+\frac{(3n-1)(3n-2)}{1.2}-\frac{(3n-2)(3n-3)(3n-4)}{1.2.3}+\frac{(3n-3)(3n-4)(3n-5)(3n-6)}{1.2.3.4} - etc. = (-1)^{n}$, where n is an integer, and the series stops at the first term that vanishes.

4. Find the real roots of the equations:

$x^{2}+v^{2}+w^{2}=a^{2}$, $vw + u(y+z)=bc$,

$y^{2}+w^{2}+u^{2}=b^{2}$, $wu + v(z+x)=ca$,

$z^{2}+u^{2}+v^{2}=c^{2}$, $uv + w(x+y) = ab$.

5. If the equation $\frac{a}{x+a} + \frac{b}{x+b} = \frac{c}{x+c} + \frac{d}{x+d}$ have a pair of equal roots, then either one of the quantities a or b is equal to one of the quantities c or d, or else $\frac{1}{a} + \frac{1}{b} = \frac{1}{c} + \frac{1}{d}$. Prove also that the roots are then $-a, -a, 0$; $-b, -b, 0$; or, $0, 0, \frac{-2ab}{a+b}$.

More challenges are on the way! Are you getting ready for RMO 2016? !

Nalin Pithwa

# Some Tripos Problems Practice for RMO

1. Solve the equation: $\frac{(x-a)(x-b)}{x-a-b} = \frac{(x-c)(x-d)}{x-c-d}$
2. Find the value of a for which the fraction $\frac{x^{3}-ax^{2}+19x-a-4}{x^{3}-(a+1)x^{2}+23x-a-7}$ admits of reduction. Reduce it to lowest terms.
3. An AP, a GP and an HP have a and b for their first two terms; show that their $(n+2)^{th}$ terms will be in GP if $\frac{b^{2m+2}-a^{2m+1}}{ba(b^{2n}-a^{2n})}=\frac{n+1}{n}$

-Nalin Pithwa