Categories

## NBHM 2017 Problem 5.2 Solution – Jensen’s Inequality NBHM grants postdoctoral fellowships, Ph. D. scholarships, M.A./M.Sc. scholarships for research/education in mathematics.
The image is a front cover of a book named Real & Complex Analysis by Walter Rudin. Here you can find Jensen’s Inequality also this book is very useful for the preparation of NBHM Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let (n\in\mathbb{N}), (n\ge 2). Let (x_1,x_2,…,x_n\in(0,\pi)). Set (x=\frac{x_1+x_2+…+x_n}{n}). Which of the following are true?

A) (\prod_{k=1}^{n} sinx_k \ge sin^nx )

B) (\prod_{k=1}^{n} sinx_k \le sin^nx )

C) Neither (A) or (B) is necessarily true.

## Discussion:

Well, the title of this post gives it away. You have to use Jensen’s inequality. But before that, notice that the first inequality is easily seen to be a false one. For I can have the midpoint of two points in ((0,\pi)) as (\pi/2) which will mean the right-hand side is one and the left-hand side is less than 1.

There are many different forms of what is called Jensen’s inequality, in measure theory context the inequality is stated in terms of integrals. But we will use the sum form as stated below.

Let (a_1,…,a_n) be non-negative real numbers such that (a_1+a_2+…+a_n=1). Suppose that (f) is a convex function. Then

$$f(a_1x_1+a_2x_2+…+a_nx_n) \le a_1f(x_1)+a_2f(x_2)+…+a_nf(x_n)$$

where (x_i  )s are points in the domain of (f).

Note that if I apply this to two points (x_1,x_2) then I will just get the definition of a convex function. This can serve as a check if confused about the direction of the inequality.

Coming back to our problem, observe that we can not directly apply this inequality to (sin) function on ((0,\pi)). Because (sin) is not convex but concave in ((0,\pi)). There are many ways to verify the concavity of (sin) in this part, the easiest one being drawing the graph. Alternately, (sin”(x)=-sin(x)<0) on ((0,\pi)), meaning that the derivative of (sin) is a decreasing function, meaning (sin) is concave in this region.

Okay, once we have a concave function (g) ((g(x)=sinx)), we have a convex function also. Namely (-g). Once again, the easiest way to see this is to draw the graph. Alternatively, we have (g((1-a)x_1+ax_2) \ge (1-a)g(x_1)+ag(x_2) ) for (a\in [0,1]). So (-g((1-a)x_1+ax_2) \le (1-a)(-g)(x_1)+a(-g)(x_2) ) which proves the convexity of (-g).

So, to sum up, we can apply Jensen’s inequality to (-sin) on ((0,\pi)).

This gives

$$sin(\frac{x_1}{n}+\frac{x_2}{n}+…+\frac{x_n}{n}) \ge \frac{sinx_1}{n}+\frac{sinx_2}{n}+…+\frac{sinx_n}{n}$$

Here we have used each (a_i=1/n) ,(i=1,…,n).

Now notice that the right-hand side has each term non-negative. So we are allowed to apply AM-GM inequality. This gives

$$\frac{sin(x_1)+…+sin(x_n)}{n} \ge ( \prod_{k=1}^{n} sinx_k )^{1/n}$$

Combining the last two inequalities we get the inequality

(\prod_{k=1}^{n} sinx_k \le sin^nx ).

Categories

## TIFR 2015 Problem 7 Solution -Increasing Function and Continuity TIFR 2015 Problem 7 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let $f$ and (g) be two functions from ([0,1]) to ([0,1]) with (f) strictly increasing. Which of the following statements is always correct?

A. If (g) is continuous, then (fog) is continuous

B. If  (f) is continuous, then (fog) is continuous

C. If (f) and (fog) is continuous, then (g) is continuous

D. If (g) and (fog) are continuous, then (f) is continuous

## Discussion:

A: Let (g(x)=x) for all (xin [0,1]).

(f(x)=x) for (xin [0,frac{1}{2}]) and (f(x)=5+x) for (xin (frac{1}{2},1]).

Then (fog=f) and (f) is not continuous.

So A is False.

B: Reverse (f) and (g) in A to show that B is False.

C: If (f) and (fog) are continuous then (f) is 1-1 (increasing), continuous map ([0,1]to [0,1]).

(A subset [0,1] ) be closed. Then (A) is compact. (Closed subsets of compact spaces are compact).

Therefore (f(A)) is compact. (continuous image of compact set is compact).

We have that (f(A)) is a compact subset of ([0,1]). Therefore (f(A)) is closed in ([0,1]). (compact subspace of Hausdorff space is closed).

Therefore, (f) is a closed map. So (f^{-1}) is continuous.

Hence (f^{-1}ofog=g) is continuous.

So, C is True.

D: Let (g(x)=frac{x}{4}) for all (xin [0,1]).

(f(x)=x) for (xin [0,frac{1}{2}]) and (f(x)=5+x) for (xin (frac{1}{2},1]).

Then (fog(x)=f(frac{x}{4})=frac{x}{4}) for all (xin [0,1]).

So (fog) is continuous but (f) is not continuous.

So, D is False.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Continuity,Closed Set, Compact Set
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert
Categories

## TIFR 2015 Problem 4 Solution – Groups without Commuting Elements TIFR 2015 Problem 4 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Topics in Algebra by I.N.Herstein. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program

## PROBLEM:

Let (S) be the collection of isomorphism classes of groups (G) such that every element of G commutes with only the identity element and itself. Then what is (|S|)?

## Discussion:

Given any $g\in G$, it commutes with few obvious elements: $e,g,g^2,…$ , i.e, all integral powers of (g).

So by given condition, this whole set ${(e,g,g^2,…)}$ is same as the set ${(e,g)}$. So any element in (G) must have order ( \le 2).

Now let us look at (e). The identity commutes with every element. But by given condition, (e) commutes with (e) only. That implies there is no other element in (G).

So, (G=(e)).

So, (|S|=1).

## HELPDESK

• What is this topic:Modern Algebra
• What are some of the associated concept: Isomorphism
• Book Suggestions: Topics in Algebra by I.N.Herstein
Categories

## TIFR 2015 Problem 3 Solution – Trace of Product of Matrix TIFR 2015 Problem 3 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Linear Algebra Done Right by Sheldon Axler. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem Type:

Let (A) be a (10\times 10) matrix with complex entries such that all its eigenvalues are non-negative real numbers, and at least one eigenvalue is positive. Which of the following statements is always false?

A. There exists a matrix B such that (AB-BA=B)

B. There exists a matrix B such that (AB-BA=A)

C. There exists a matrix B such that (AB+BA=A)

D. There exists a matrix B such that (AB+BA=B)

Discussion:

We know that for two square matrix (A) and (B) of same size, (Tr(AB)=Tr(BA)) ((TrM)=Trace of (M) ).

In other words, (Tr(AB-BA)=0) for any two square matrices of the same size.

Since trace of a square matrix is also the sum of its eigenvalues, and (A) has all eigenvalues non-negative with at least one positive eigenvalue, we have (Tr(A) > 0). Taking trace of both sides of (AB-BA=A) we get a contradiction. So there does not exist any (B) such that (AB-BA=A).

Take (B=0) the 10×10 zero-matrix. Then (AB-BA=B) is satisfied. So is (AB+BA=B).

Take (B=\frac{1}{2}I), where (I) is the 10×10 identity matrix. Then (AB+BA=A) is satisfied.

## HELPDESK

• What is this topic: Linear Algebra
• What are some of the associated concept: Trace,Eigenvalue
• Book Suggestions: Linear Algebra done Right by Sheldon Axler
Categories

## TIFR 2015 Problem 2 Solution -Image of continuous function TIFR 2015 Problem 2 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let (f: \mathbb{R} \to \mathbb{R} ) be a continuous function. Which of the following can not be the image of ((0,1]) under (f)?

A. {0}

B. ((0,1))

C. ([0,1))

D. ([0,1])

## Discussion:

If f is the constant function constantly mapping to 0, which is continuous, then the image set is {0}.

Suppose that (f((0,1])=(0,1)) . Then (f((0,1))=(0,1)- {f(1)} ). Now since (f(1)\in (0,1) ) the set ( (0,1)- {f(1)} ) is not connected. But ((0,1)) is connected, and we know that continuous image of a connected set is connected. This gives a contradiction. So ((0,1)) can not be the image of ((0,1]) under f.

Define (f(x)=1-x).  Then (f((0,1])= [0,1)).

Define (f(x)=0) for (x\in [0,\frac{1}{2}] ) and (f(x)= 2(x-\frac{1}{2}) ) for (x\in [\frac{1}{2} ,1 ] ). (f) is continuous on  ((0,\frac{1}{2}] ) and ( [\frac{1}{2} ,1 ] ) and (f) agrees on the common points, by pasting lemma (f) is continuous on ( [0,1] ) . And image of ((0,1] ) is ([0,1]).

TIFR 2015 Problem 2 Solution is concluded.

## Chatuspathi

• What is this topic: Real Analysis
• What are some of the associated concept: Continuous Function, Metric Space
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert
Categories

## TIFR 2015 Problem 1 Solution -Invertible Matrix with Sum of Each Row 1 TIFR 2015 Problem 1 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programe leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Linear Algebra by Gilbert Strang. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let (A) be an invertible (10 \times 10) matrix with real entries such that the sum of each row is 1. Then

A. The sum of the entries of each row of the inverse of A is 1

B. The sum of the entries of each column of the inverse of A is 1

C. The trace of the inverse of A is non-zero.

D. None of the above.

## Discussion:

The sum of each row of (A) is 1, means that the sum of the columns of A is the vector ((1,1,…,1)^T ) .

Note that i-th column of (A) is given by (Ae_i ). Therefore, (\sum_{i=1}^{10} Ae_i = (1,1,…,1)^T ).

Since left multiplication by (A) is a linear transformation, the left-hand side of the last expression can be written as (A(\sum_{i=1}^{10}e_i)).

Now, (\sum_{i=1}^{10}e_i = (1,1,…,1)^T ).

Hence we get (A(1,1…,1)^T = (1,1,…,1)^T ).

Another way of saying the last expression is that the vector ( (1,1…,1)^T ) is fixed by A.

Since A is invertible, applying (A^{-1}) on both sides of the last expression we get ((1,…,1)^T = A^{-1}(1,1,…,1)^T ).

By the linearity argument as above, this gives ( (1,1…,1)^T = \sum_{i=1}^{10} A^{-1} (e_i) ). And the i-th term in right-hand side expression is the i-th column of (A^{-1}). Therefore, sum of columns of (A^{-1}) is the vector ((1,1,…,1)^T ). This is same as saying that sum of entries of each row of (A^{-1}) is 1.

## Helpdesk

• What is this topic: Linear Algebra
• What are some of the associated concept: Linear Transformation,Invertible Matrix
• Book Suggestions: Introduction to Linear Algebra by Gilbert Strang
Categories

## TIFR 2014 Problem 23 Solution – Homomorphisms from $S_n$ TIFR 2014 Problem 23 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Contemporary Abstract Algebra by Joseph A. Gallian. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program

## Problem:True/False?

There exists an onto group homomorphism

A. from (S_5) to (S_4)

B. from (S_4) to (S_2)

C. from (S_5) to (\mathbb{Z}_5)

D. from (S_4) to (\mathbb{Z}_4)

## Discussion:

(S_2) is the permutation group on 2 letters. It has order 2, so it is isomorphic to (\mathbb{Z}_2).

And we know a group homomorphism from (S_n) to  (\mathbb{Z}_2), namely the signature map.

(\sigma \to 0 ) if ( \sigma ) is even

(\sigma \to 1 ) if ( \sigma ) is odd.

This map is onto. And it is a homomorphism. This is fairly well known fact so let us not prove it here.

So we know for sure that B is true.

How do we know that C is not true?

For that, suppose (\phi:S_5 \to \mathbb{Z}_5 ) be an onto group homomorphism. Apply first isomorphism theorem for groups, then (S_5/ker{\phi} ) is isomorphic to (\mathbb{Z}_5). That implies their orders are same. And (|G/H|=|G|/|H| ) so we get (|ker{\phi}| = 4! =24 ).

Now remember that (ker{\phi} ) is a normal subgroup of (S_5). Also, notice that any 4-cycle in (S_5) must go to the identity of (\mathbb{Z}_5) since the order of image must be divisible by the order of domain which in the case of 4-cycles are 4. So the image of any 4-cycle must have order 1,2 or 4 in (\mathbb{Z}_5). But out of these, only order 1 is possible in (\mathbb{Z}_5). Hence the image of any 4-cycle is the identity. So, the kernel must contain all the 4-cycles in (S_5). Now we count the number of 5-cycles in (S_5).

There are ({{5}\choose{4}} ) ways to choose 4 elements out of 1,2,3,4,5. Next, given any 4 elements, we can always write them in increasing order. Let’s say we chose 2,4,5,3. Then we write this as 2,3,4,5. Now to count the number of possible 4-cycles (distinct) we fix 2 in the first position and permute the rest. Each of these permutations will give a different 4-cycle. There are (3!) such permutations. So in total, there are ({{5}\choose{4}}\times 3! =30 ) 4-cycles in (S_5). This is more than our cardinality of (ker{\phi}). This is a contradiction.

So option C is false.

The same kind of argument will apply to option D as well. Here one can consider the 3 cycles in (S_4) and get a contradiction as above.

For option A, consider the 5-cycles in (S_5). They must map to identity (because of order reasons as above). And there are (4!=24) 5-cycles (start the cycles with 1, permute rest). But first isomorphism theorem will give the cardinality of the kernel as (\frac{5!}{4!}=5), a contradiction. This disproves A.

## Helpdesk

• What is this topic:Abstract Algebra
• What are some of the associated concept: Permutation Group, Homeomorphism
• Book Suggestions: Contemporary Abstract Algebra by Joseph A. Gallian
Categories

## TIFR 2014 Problem 30 Solution – Number of Maps TIFR 2014 Problem 30 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Topics in Algebra by I.N.Herstein. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program

## PROBLEM:

How many maps (\phi: \mathbb{N} \cup  {0} \to \mathbb{N} \cup  {0}) are there satisfying (\phi(ab)=\phi(a)+\phi(b)) , for all (a,b\in \mathbb{N} \cup  {0}) ?

## Discussion:

Take (n\in \mathbb{N} \cup  {0} ).

By the given equation (\phi(n\times 0)=\phi(n)+\phi(0)).

This means (\phi(0)=\phi(n)+\phi(0)).

Oh! This means (\phi(n)=0). (n\in \mathbb{N} \cup  {0}) was taken arbitrarily. So…

(\phi(n)=0) for all (n\in \mathbb{N} \cup  {0} ).

There is only one such map.

## HELPDESK

• What is this topic:Algebra
• What are some of the associated concept: Number of Function
• Book Suggestions: Topics in Algebra by I.N.Herstein
Categories

## TIFR 2014 Problem 29 Solution – Maps from compact spaces TIFR 2014 Problem 29 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let (f: X\to Y ) be a continuous map between metric spaces. Then (f(X)) is a complete subset of (Y) if

A. X is compact

B. Y is compact

C. X is complete

D. Y is complete

Discussion:

Let (X) be compact. Then (f(X)) is compact. (continuous image of compact space is compact)

Now, compact subset of any Hausdorff space is closed. So in particular, compact subset of any metric space is closed.

Let ((y_n)) be a Cauchy sequence in (f(X)). Then since (f(X)) is compact, ((y_n)) has a convergent subsequence (converging to a point in that compact set i.e, in (f(X)) ).

Suppose (y_{n_k} \to y \in f(X) ).

Then by triangle inequality, we have (d(y_n,y) \le d(y_n,y_{n_k}) + d(y_{n_k},y) \to 0+0=0 ) as (n\to \infty)

Here we have used that (y_n) is cauchy to conclude (d(y_n,y_{n_k}) \to 0 ).

So this implies (y_n \to y). Since (y\in f(X)) we conclude that (f(X)) is complete.

This proves A.

Let (Y=[0,2]) and (X=(0,1)). Take the inclusion map (i(x)=x) for all (x\in X). This example shows that even if we take (Y) to be compact, or complete, (f(X)) need not be complete. So this disproves B and D.

Now take (X=\mathbb{R}) and take (Y=(0,1)). We know there is a homeomorphism between these two sets where the metric is usual topology. So, in this case, the image of a complete set is not complete. This disproves option C.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Cauchy Sequence,Homeomorphishm, Compact Space
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert
Categories

## TIFR 2014 Problem 28 Solution – Continuous Functions from Discrete Space TIFR 2014 Problem 28 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let (X) be a topological space such that every function $f: X \to \mathbb{R}$ is continuous. Then

A. (X) has the discrete topology.

B. (X) has the indiscrete topology.

C. (X) is compact.

D. (X) is not connected.

Discussion:

We know that if (Y) is a discrete space then any function (g: Y \to Z ) is continuous.

Option A asks whether the converse to this is true in the case that (Z= \mathbb{R} ).

To prove/disprove whether (X) has the discrete topology or not it is enough to prove whether every singleton set is open or not.

If we can show that for every (x\in X) there exists a function (f_x :X \to \mathbb{R}) such that (f_x^{-1} (-1,1) = {x} ) then we are done. Because we are given that (f_x) if exists must be continuous, and since ((-1,1)) is open in (\mathbb{R}) we will have the inverse image of it open in (X), so ({x} ) will be open in (X).

Now, this target is easy to handle. We define for each (x\in X)

(f_x (x) = 0 ) and (f_x (y) =2) for (y \neq x ).

This (f_x) satisfies our desired property. So (X) is discrete.

Taking (X= \mathbb{Z}) (for example) shows that (X) does not need to be indiscrete nor does it have to be compact.

Taking (X= {0} ) shows that (X) may be connected. Ofcourse if (X) has cardinality more than 1, it is not connected.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Continuity, Discrete Space
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert