# calculus

# Set theory, relations, functions: preliminaries part 1

# Every function can be written as a sum of an even and an odd function

Let be any well-defined function.

We want to express it as a sum of an even function and an odd function.

Let us define two other functions as follows:

and .

Claim I: F(x) is an even function.

Proof I; Since by definition , so so that F(x) is indeed an even function.

Claim 2: G(x) is an odd function.

Proof 2: Since by definition , so so that G(x) is indeed an odd function.

Claim 3:

Proof 3: indeed.

# Tutorial on Basic Set Theory and Functions: for PRMO, RMO and IITJEE Mains maths

I) Prove that every function can be represented as a sum of an even function and an odd function.

II)Let A, B, C be subsets of a set S. Prove the following statements and illustrate them with Venn Diagrams:

2a) The famous DeMorgan’s laws in their basic forms: and . Assume that both sets A and B are subsets of Set S. In words, the first is: union of complements is the complement of intersection; the second is: intersection of two complements is the complement of the union of the two sets.

**Sample Solution:**

Let us say that we need to prove: .

Proof: It must be shown that the two sets have the same elements; in other words, that each element of the set on LHS is an element of the set on RHS and vice-versa.

If , then and . This means that , and and . Since and , hence . Hence, .

Conversely, if , then and . Therefore, and . Thus, and , so that . QED.

2b) .

2c)

III) Prove that if I and S are sets and if for each , we have , then .

**Sample Solution: **

It must be shown that each element of the set on the LHS is an element of the set on RHS, and vice-versa.

If , then and . Therefore, , for at least one . Thus, , so that .

Conversely, if , then for some , we have . Thus, and . Since , we have . Therefore, . QED.

IV) If A, B and C are sets, show that :

4i)

4ii)

4iii)

4iv)

V) Let I be a nonempty set and for each let be a set. Prove that

5a) for any set B, we have :

5b) if each is a subset of a given set S, then

VI) Prove that if , , and are functions, then :

VII) Let be a function, let A and B be subsets of X, and let C and D be subsets of Y. Prove that:

7i) ; in words, image of union of two sets is the union of two images;

7ii) ; in words, image of intersection of two sets is a subset of the intersection of the two images;

7iii) ; in words, the inverse image of the union of two sets is the union of the images of the two sets.

7iv) ; in words, the inverse image of intersection of two sets is intersection of the two inverse images.

7v) ; in words, the inverse of the image of a set contains the set itself.

7vi) ; in words, the image of an inverse image of a set is a subset of that set.

For questions 8 and 9, we can assume that the function f is and a set A lies in domain X and a set C lies in co-domain Y.

8) Prove that a function f is 1-1 if and only if for all ; in words, a function sends different inputs to different outputs iff a set in its domain is the same as the inverse of the image of that set itself.

9) Prove that a function f is onto if and only if for all ; in words, the image of a domain is equal to whole co-domain (which is same as range) iff a set in its domain is the same as the image of the inverse image of that set.

Cheers,

Nalin Pithwa

# Prof. Tim Gowers’ on recognising countable sets

https://gowers.wordpress.com/2008/07/30/recognising-countable-sets/

Thanks Dr. Gowers’. These are invaluable insights into basics. Thanks for giving so much of your time.

# Advanced Calculus video lecture 3: IITJEE Advanced Maths: Mathematics Hothouse

# Advanced Calculus video lecture 2: IITJEE Advanced Maths: Mathematics Hothouse

# Advanced Calculus video lecture 1: IITJEE Advanced Maths: Mathematics Hothouse

# Huygens’ problem to Leibnitz: solution

In the Feb 23 2018 blog problem, we posed the following question:

Sum the following infinite series:

.

**Solution:**

The sum can be written as:

, where .

Thus, . This is the answer.

*If you think deeper, this needs some discussion about rearrangements of infinite series also. For the time, we consider it outside our scope.*

*Cheers,*

Nalin Pithwa.

# RMO Training: taking help from Nordic mathematical contest: 1988

**Problem:**

Let be a smallest value of the function . Prove that when .

**Proof:**

For ,

.

From this, we see that for and . Consequently, attains its maximum value in the interval . On this interval

So, . But,

As , the first term on the right hand side tends to the limit . In the second term, the factor

of the numerator tends to zero because

.

So,

auf wiedersehen,

Nalin Pithwa.

**Reference: Nordic Mathematical Contest, 1987-2009.**