ISI MMath Solutions and Discussion 2018 : PMB

ISI MMath PMB 2018 Subjective Questions along with hints and solutions and discussions

Note: You must try the problems on your own before looking at the solutions. I have made the solutions so that you don’t have to join any paid program to get the solutions. After you have failed several times in your attempt to solve a question even with the hints provided, then only look at the solution to that problem. If you cannot solve a single question on your own then this exam is not for you. And if you like my work then please do share this among your friends and do comment below.  

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To view previous year’s question papers: Click Here

Problem 1.

Find the values of a > 0 for which the improper integral

    \[ \int^\infty_0 \frac{\sin x}{x^a}dx \]

converges.

Topic: Real Analysis
Difficulty level: Medium

Indian Statistical Institute, ISI MMath 2018 Problem 1 Hints along with Solution & Discussion

Hint 1

Full Solution

The integral is defined as the limit of proper intergals

    \[ \int^\infty_0 \frac{\sin x}{x^a}dx=\lim_{\delta\to 0, C\to\infty}\int^C_{\delta}\frac{\sin x}{x^a}dx= \]

    \[ =\lim_{\delta\to 0}\int^1_\delta \frac{\sin x}{x^a}dx+\lim_{C\to \infty}\int^C_1\frac{\sin x}{x^a}dx \]

 

Using the relation \lim_{x\to 0}\frac{\sin x}{x}=1 we find \epsilon>0 such that for all x\in (0,\epsilon) we have \frac{1}{2}<\frac{\sin x}{x}<2. Then for all \delta<\delta'<\epsilon we have a two-sided estimate

    \[ \left|\int^1_\delta \frac{\sin x}{x^a}dx-\int^1_{\delta'} \frac{\sin x}{x^a}dx\right|=\int^{\delta'}_\delta \frac{\sin x}{x^a}dx\leq 2 \int^{\delta'}_\delta x^{1-a}dx \]

and

    \[ \left|\int^1_\delta \frac{\sin x}{x^a}dx-\int^1_{\delta'} \frac{\sin x}{x^a}dx\right|=\int^{\delta'}_\delta \frac{\sin x}{x^a}dx\geq \frac{1}{2} \int^{\delta'}_\delta x^{1-a}dx \]

So, existence of the limit \lim_{\delta\to 0}\int^1_\delta \frac{\sin x}{x^a}dx is equivalent to finiteness of the integral \int^1_0 x^{1-a}dx. So,

    \[ \exists \lim_{\delta\to 0}\int^1_\delta \frac{\sin x}{x^a}dx\Leftrightarrow 1-a>-1\Leftrightarrow a<2. \]

Let 1<C<C'. We have an estimate

    \[ \left|\int^C_1 \frac{\sin x}{x^a}dx-\int^{C'}_1 \frac{\sin x}{x^a}dx\right|=\left| \int^{C'}_C\frac{\sin x}{x^a}dx\right|= \]

integrate by parts with u=x^{-a}, dv=\sin x dx, so that v=-\cos x, du=-ax^{-a-1}dx

    \[ =\left| \frac{\cos C}{C^a}-\frac{\cos C'}{(C')^a}-a\int^{C'}_C\frac{\cos x}{x^{a+1}}dx\right|\leq \]

    \[ \leq \frac{1}{C^a}+\frac{1}{(C')^a}+a\int^{\infty}_C\frac{dx}{x^{a+1}}=\frac{2}{C^a}+\frac{1}{(C')^a}\to 0, C,C'\to \infty. \]

So, the limit \lim_{C\to \infty}\int^C_1 \frac{\sin x}{x^a}dx exists for any a>0.

 

The improper integral \int^\infty_0 \frac{\sin x}{x^a}dx converges if and only if a<2.

 

Problem 2.

Let f : [0, 1] \to \mathbb{R} be a real-valued continuous function which is differentiable on (0, 1) and satisfies f(0) = 0. Suppose there exists a

constant c\in (0,1) such that

    \[ |f'(x)|\leq c|f(x)| \mbox{ for all } x\in (0,1). \]

Show that f(x) = 0 for all x\in [0,1] .

Topic: Real Analysis
Difficulty Level: Medium

Indian Statistical Institute, ISI MMath 2018 Problem 2 Hints along with Solution & Discussion

Problem 3.

Let G be an abelian group of order n.

 

 

(a) If f : G \to\mathbb{C} is a function, then prove that for all h\in G,

    \[ \sum_{g\in G}f(g)=\sum_{g\in G}f(hg). \]

(b) Let \mathbb{C}^* be the multiplicative group of non-zero complex numbers and suppose f : G\to \mathbb{C}^* is a homomorphism. Prove that

    \[ \sum_{g\in G}f(g)=0 \mbox{ or } \sum_{g\in G}f(g)=n. \]

(c) If f : G \to \mathbb{C}^* is any homomorphism, then prove that

    \[ \sum_{g\in G}|f(g)|=n. \]

Topic: Group Theory
Difficulty Level: Medium

Indian Statistical Institute, ISI MMath 2018 Problem 3 Hints along with Solution & Discussion

Full Solution

(a) For any h\in G the mapping g\to hg is a bijection of G. When g ranges through G, the product hg also ranges through G:

    \[ \sum_{g\in G}f(hg)=\sum_{g\in G}f(g). \]

(b) For any h\in G we have

    \[ f(h)\sum_{g\in G}f(g)=\sum_{g\in G}f(hg)=\sum_{g\in G}f(g) \]

Taking the sum in h\in G we get

    \[ \sum_{h\in G}f(h)\sum_{g\in G}f(g)=n\sum_{g\in G}f(g). \]

So,

    \[ \left(\sum_{g\in G}f(g)\right)^2=n\sum_{g\in G}f(g), \]

    \[ \sum_{g\in G}f(g)\left(n-\sum_{g\in G}f(g)\right)=0. \]

Either \sum_{g\in G}f(g)=0 or \sum_{g\in G}f(g)=n.

(c) Identity

    \[ |f(hg)|=|f(h)f(g)|=|f(h)||f(g)| \]

implies that g\to |f(g)| is also a homomorphism. By previous part, either

\sum_{g\in G}|f(g)|=0 or \sum_{g\in G}|f(g)|=n. For all g\in G we have f(g)\neq 0, |f(g)|>0 and \sum_{g\in G}|f(g)|>0. It follows that \sum_{g\in G}|f(g)|=n.

Problem 4.

(a) Is the ideal I = (X + Y, X - Y ) in the polynomial ring \mathbb{C}[X, Y ] a prime ideal? Justify your answer.

(b) Is the ideal I = (X + Y, X - Y ) in the polynomial ring \mathbb{Z}[X, Y ] a prime ideal? Justify your answer.

Topic: Ring Theory
Difficulty Level: Medium

Indian Statistical Institute, ISI MMath 2018 Problem 4 Hints along with Solution & Discussion

Full Solution

(a) Relations

    \[ X=\frac{1}{2}(X+Y)+\frac{1}{2}(X-Y), Y=\frac{1}{2}(X+Y)-\frac{1}{2}(X-Y) \]

imply that I=(X,Y). General element of I is of the form XP(X,Y)+YQ(X,Y), i.e.

    \[ I=\{P\in \mathbb{C}[X,Y]: P(0,0)=0\}. \]

If PQ\in I, then P(0,0)Q(0,0)=0 which is possible only when P(0,0)=0 or Q(0,0)=0. I is a prime ideal.

(b) Assume that I is a prime ideal in \mathbb{Z}[X,Y]. Consider polynomial P(X,Y)=2X=(X+Y)+(X-Y)\in I. Then either 2\in I or X\in I. General element of I is of the form

    \[ (X+Y)Q(X,Y)+(X-Y)R(X,Y), \ Q(X,Y),R(X,Y)\in \mathbb{Z}[X,Y]. \]

In particular, an element of I vanished when X=Y=0. So, 2\not\in I. It follows that X\in I and

    \[ X=(X+Y)Q(X,Y)+(X-Y)R(X,Y)=X(Q(X,Y)+R(X,Y))+Y(Q(X,Y)-R(X,Y)). \]

Then, Q(X,Y)=R(X,Y) and

    \[ X=2Q(X,Y)X, 2Q(X,Y)=1, \]

which is impossible in \mathbb{Z}[X,Y]. I is not a prime ideal.

Problem 5.

Let n \geq 2 and A be an n\times n matrix with real entries. Let \mbox{Adj} A denote the adjoint of A, that is, the (i, j)-th entry of \mbox{Adj} A is the

(j, i)-th cofactor of A. Show that the rank of \mbox{Adj} A is 0, 1 or n.

Topic: Linear Algebra
Difficulty Level: Medium

Indian Statistical Institute, ISI MMath 2018 Problem 5 Hints along with Solution & Discussion

Hint 2

Assume that the rank of A is \leq n-2. Then each (n-1)\times (n-1) minor of A is zero, and all elements of \mbox{Adj} A are equal to zero. In this case the rank of \mbox{Adj} A is equal to zero.

Hint 3

Assume that the rank of A is equal to n-1. Then there exists a nonzero (n-1)\times (n-1) minor of A, so the rank of \mbox{Adj} A is \geq 1. Denote by a^{(ij)} the (i,j)-th cofactor of A. Then (\mbox{Adj} A)_{ij}=a^{(ji)}. Consider the product

A\cdot \mbox{Adj} A. Its (i,j)-th entry is

    \[ \sum^n_{k=1}a_{ik}( \mbox{Adj} A)_{kj}=\sum^n_{k=1}a_{ik}a^{(jk)}=0 \eqno(1) \]

Full Solution

If \det A\ne 0, then the inverse of A is given by \mbox{Adj} A. In this case the rank of \mbox{Adj}A is n.

 

Assume that the rank of A is \leq n-2. Then each (n-1)\times (n-1) minor of A is zero, and all elements of \mbox{Adj} A are equal to zero. In this case the rank of \mbox{Adj} A is equal to zero.

 

Assume that the rank of A is equal to n-1. Then there exists a nonzero (n-1)\times (n-1) minor of A, so the rank of \mbox{Adj} A is \geq 1. Denote by a^{(ij)} the (i,j)-th cofactor of A. Then (\mbox{Adj} A)_{ij}=a^{(ji)}. Consider the product

A\cdot \mbox{Adj} A. Its (i,j)-th entry is

    \[ \sum^n_{k=1}a_{ik}( \mbox{Adj} A)_{kj}=\sum^n_{k=1}a_{ik}a^{(jk)}=0 \eqno(1) \]

Indeed, if i=j then the sum in (1) is the determinant of A. If i\ne j, then the sum in (1) is the determinant of the matrix A after the j-th row is replaced with the i-th row. Such determinant is zero as the matrix has two identical rows. We see that A\cdot \mbox{Adj} A=0. That is, each column of \mbox{Adj} A is in the kernel of A. But the rank of A is n-1, hence the dimension of the kernel of A is 1 and the dimension of the column range of \mbox{Adj} A is \leq 1. This proves that the rank of \mbox{Adj} A is equal to 1.

Problem 6.

Suppose an urn contains a red ball and a blue ball. A ball is drawn at random and a ball of the same colour is added to the urn along with

the one that was drawn. This process is repeated indefinitely.

Let X denote the random variable that takes the value n if the first n - 1 draws yield red balls and the n-th draw yields a blue ball.

(a) If n\geq 1, find P(X > n).

(b) Show that the probability of a blue ball being chosen eventually is 1.

(c) Find E[X].

Topic: Probability
Difficulty Level: Medium

Indian Statistical Institute, ISI MMath 2018 Problem 6 Hints along with Solution & Discussion

Hint 1

(a) We use induction on n. Let n=1. The event \{X>1\} means that the first draw yielded a red ball,

    \[ P(X>1)=\frac{1}{2}. \]

Hint 2

(b) If the blue ball is never chosen, then \{X>n\} for all n. Denote by N the event when the blue ball is never chosen. Then

    \[ P(N)\leq P(X>n)=\frac{1}{n+1}\to 0,\ n\to\infty. \]

So, P(N)=0.

Full Solution

(a) We use induction on n. Let n=1. The event \{X>1\} means that the first draw yielded a red ball,

    \[ P(X>1)=\frac{1}{2}. \]

Assume that event \{X>n\} happens. Then first n draws yielded red balls, and before the (n+1)-st draw there are n+1 red balls and 1 blue ball in the urn. The probability to draw a red ball is \frac{n+1}{n+2}. So,

    \[ P(X>n+1)=P(X>n)P(X>n+1|X>n)=\frac{n+1}{n+2}P(X>n) \]

If P(X>n)=\frac{1}{n+1}, then P(X>n+1)=\frac{1}{n+2}. For n=1 the equality P(X>n)=\frac{1}{n+1} is true. So, for all n\geq 1

    \[ P(X>n)=\frac{1}{n+1} \]

(b) If the blue ball is never chosen, then \{X>n\} for all n. Denote by N the event when the blue ball is never chosen. Then

    \[ P(N)\leq P(X>n)=\frac{1}{n+1}\to 0,\ n\to\infty. \]

So, P(N)=0.

(c) For each n\geq 1 we have

    \[ P(X=n)=P(X>n-1)-P(X>n)=\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}. \]

Then

    \[ E[X]=\sum^\infty_{n=1}nP(X=n)=\sum^\infty_{n=1}\frac{1}{n+1}=\infty. \]

 

Problem 7.

A real number x_0 is said to be a limit point of a set S\subseteq \mathbb{R} if every

neighbourhood of x_0 contains a point of S other than x_0. Consider the

set

    \[ S=\{0\}\cup \left\{\frac{1}{m}+\frac{1}{n}:m,n\in\mathbb{N}\right\} \]

(a) Show that S contains infinitely many limit points of S.

(b) Show that S is a compact subset of \mathbb{R}.

(c) Find all limit points of S.

Topic: Real Analysis
Difficulty Level: Medium

Indian Statistical Institute, ISI MMath 2018 Problem 7 Hints along with Solution & Discussion

Hint 3

(c) Considerations of previous two parts shows that the limit points of S are \{0\}\cup \{\frac{1}{n}:n\geq 1\}.

Full Solution

(a) Observe that \frac{1}{n}=\frac{1}{2n}+\frac{1}{2n}\in S. Let \epsilon>0 and m>\frac{1}{\epsilon}. Then

    \[ \frac{1}{n}<\frac{1}{n}+\frac{1}{m}<\frac{1}{n}+\epsilon. \]

Every neighborhood of \frac{1}{n} contains a point from S distinct from \frac{1}{n}. Each point \frac{1}{n} is a limit point of the set S that belongs to the set S.

 

If m>\frac{2}{\epsilon}, then \frac{1}{m}+\frac{1}{m} is a point of S in the \epsilon-neighborhood of 0. 0 is a limit point of S. We’ve proved that points

    \[ \{0\}\cup \left\{\frac{1}{n}:n\geq 1\right\} \]

are limit points of S.
(b)

The set S is bounded: S\subset [0,2]. It remains to prove that S is closed. Suppose that x\in \bar{S}\setminus S. Every neighborhood of x contains a point y \in S. But x\not\in S and y\ne x. Then x is a limit point of S. Since we assume x\not \in S, it follows that x>0. For every k\geq 1 find a point \frac{1}{m_k}+\frac{1}{n_k}\in S, such that

    \[ \left|x-\left(\frac{1}{m_k}+\frac{1}{n_k}\right)\right|<\frac{1}{k}. \]

If both sequences (m_k)_{k\geq 1} and (n_k)_{k\geq 1} are bounded, then there exist values m,n\geq 1 such that m_k=m and n_k=n for infinitely many values of k. But then x=\frac{1}{m}+\frac{1}{n}\in S, which is not the case. Hence we can assume that (n_k)_{k\geq 1} is unbounded. Let (n_{k_j})_{j\geq 1} be a subsequence such that \lim_{j\to\infty}n_{k_j}= \infty. If the sequence (m_{k_j})_{j\geq 1} is bounded, there exists value m\geq 1 such that m_{k_j}=m for infinitely many values of j. But then

x=\frac{1}{m}\in S. If the sequence (m_{k_j})_{j\geq 1} is unbounded, we can pass to another subsequence (m_{k'_j})_{j\geq 1} that converges to \infty. Then we have x=0. In any case x\in S and S is compact.

(c) Considerations of previous two parts shows that the limit points of S are \{0\}\cup \{\frac{1}{n}:n\geq 1\}.

Problem 8.

(a) Let \{f_n\}^\infty_{n=1} be a sequence of continuous functions on [0, 1] such

that \sum^\infty_{n=1} f_n converges uniformly on (0, 1]. Show that \sum^\infty_{n=1}f_n(0)

converges.

(b) Find the set D of all points x\in [0,1] such that the series

    \[ \sum^\infty_{n=1}e^{-nx}\cos(nx) \]

converges. Does this series converge uniformly on D? Justify your answer.

Topic: Real Analysis
Difficulty Level: Medium

Indian Statistical Institute, ISI MMath 2018 Problem 8 Hints along with Solution & Discussion

Hint 1

(a) Denote s_m(x)=\sum^m_{n=1}f_n(x). The sequence (s_m)_{m\geq 1} converges uniformly on (0,1]: there exists function s on (0,1] such that

    \[ \lim_{m\to\infty}\sup_{0<x\leq 1}|s_m(x)-s(x)|=0. \]

Full Solution

(a) Denote s_m(x)=\sum^m_{n=1}f_n(x). The sequence (s_m)_{m\geq 1} converges uniformly on (0,1]: there exists function s on (0,1] such that

    \[ \lim_{m\to\infty}\sup_{0<x\leq 1}|s_m(x)-s(x)|=0. \]

We verify that \lim_{x\to 0}s(x) exists. Let \epsilon>0. Find m\geq 1 such that

    \[ \sup_{0<x\leq 1}|s_m(x)-s(x)|<\frac{\epsilon}{3}. \]

Using continuity of s_m find \delta>0 such that for all x,y\in (0,\delta)

    \[ |s_m(x)-s_m(y)|<\frac{\epsilon}{3}. \]

Then for all x,y\in (0,\delta) we have

    \[ |s(x)-s(y)|\leq |s(x)-s_m(x)|+|s_m(x)-s_m(y)|+|s_m(y)-s(y)|<\epsilon. \]

Since \epsilon>0 is arbitrary, this shows that \lim_{x\to 0}s(x) exists. We denote the limit by s(0).

 

Now, let \epsilon>0. Find m_0\geq 1 such that for all m\geq m_0 and all x\in(0,1] we have

    \[ |s_m(x)-s(x)|\leq \epsilon. \]

Passing to the limit as x\to 0 we get

    \[ |s_m(0)-s(0)|\leq \epsilon. \]

Since \epsilon>0 is arbitrary, this shows that

    \[ s(0)=\lim_{m\to\infty}s_m(0)=\lim_{m\to\infty}\sum^m_{n=1}f_n(0), \]

i.e. \sum^\infty_{n=1}f_n(0)

converges and has the sum s(0).

(b) The series converges for all x\in (0,1]. Indeed,

    \[ \sum^\infty_{n=1}|e^{-nx}\cos(nx)|\leq \sum^\infty_{n=1}e^{-nx}=\frac{e^{-x}}{1-e^{-x}}<\infty. \]

For x=0 the series diverges, as it is the series 1+1+1+\ldots. So, D=(0,1]. If the series converges uniformly, then it should also converge at x=0. So, the series \sum^\infty_{n=1}e^{-nx}\cos(nx) does not converge uniformly on D.

To view ISI MMATH 2020 solutions, hints, and discussions: Click Here

To view ISI MMATH 2019 solutions, hints, and discussions: Click Here

To view ISI MMATH 2017 solutions, hints, and discussions: Click Here

To view ISI MMATH 2016 solutions, hints, and discussions: Click Here

To view previous year’s question papers: Click Here

4 Comments

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  • For Problem 2, Can I solve this like this?

    Let x \in [0,1] then by using MVT there exists t_1 \in (0,x) such that f(x) - f(0) = f'(t_1)x i.e |f(x)| = |f'(t_1)x| \leq c|f(t_1)|x. Similary there exists t_2 \in (0,t_1) such that f(t_1) = f'(t_2)t_1 and applying modulus we get |f(t_1)| = |f'(t_2)|t_1 \leq c|ft_2)|t_1 So with the above inequalities we have |f(x)| \leq c^2 |f(t_2)|t_1x \leq c^2|f(t_2)|. Proceeding this way we have |f(x)| \leq c^n |f(x_n)| but since f is continuous on [0,1] so is |f| so it attains the maximum say M then the inequality becomes |f(x)| \leq  c^n|f(t_n)|\leq c^n M and we know that c^n M \to 0 . so for any \epsilon>0 there exists n_0 \in \mathbb{N} such that for all n \geq n_0 we have c^n M < \epsilon In particular c^{n_0}M < \epsilon hence |f(x)| \leq c^{n_0}M \leq \epsilon this implies f(x)=0 and since x was arbitrary we have f(x) = 0 \forall x \in [0,1]

  • For Problem 2, Can I solve this like this?

    Let x \in [0,1] then by using MVT there exists t_1 \in (0,x) such that f(x) - f(0) = f'(t_1)x i.e |f(x)| = |f'(t_1)x| \leq c|f(t_1)|x. Similary there exists t_2 \in (0,t_1) such that f(t_1) = f'(t_2)t_1 and applying modulus we get |f(t_1)| = |f'(t_2)|t_1 \leq c|ft_2)|t_1 So with the above inequalities we have |f(x)| \leq c^2 |f(t_2)|t_1x \leq c^2|f(t_2)|. Proceeding this way we have |f(x)| \leq c^n |f(x_n)| but since f is continuous on [0,1] so is |f| so it attains the maximum say M then the inequality becomes |f(x)| \leq  c^n|f(t_n)|\leq c^n M and we know that c^n M \to 0 . so for any \epsilon>0 there exists n_0 \in \mathbb{N} such that for all n \geq n_0 we have c^n M < \epsilon In particular c^{n_0}M < \epsilon hence |f(x)| \leq c^{n_0}M \leq \epsilon this implies f(x)=0 and since x was arbitrary we have f(x) = 0 \forall x \in [0,1]

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