ISI MMath Solutions and Discussion 2016 : PMB

Fractions > ISI MMath Solutions and Discussion 2016 : PMB

ISI MMath PMB 2016 Subjective Questions, 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.  

Sorry, did not get time to post the hints. 

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 2018 solutions, hints, and discussions: Click Here

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

To view previous year’s question papers: Click Here

Group A

Problem 1.

Let \{x_n\}_{n\in\mathbb{N}} be a sequence of real numbers defined as follows: x_1=1 and for all n\in \mathbb{N},x_{n+1}=(3+2x_n)/(3+x_n).

(a)Show that there exists \lambda\in(0,1) such that for all n\geq 2,

    \[|x_{n+1}-x_n|\leq \lambda|x_n-x_{n-1}|.\]

(b) Prove that \lim_{n\to\infty} x_n exists and find its value.

Topic: Real Analysis, Difficulty level: Easy

Full Solution

(a) By induction we check that x_n\geq 1. Indeed, for x_1 this is true by definition. Assuming that x_n\geq 1 we have \frac{3}{3+x_n}\leq \frac{3}{4}, hence

    \[x_{n+1}=\frac{3+2x_n}{3+x_n}=2-\frac{3}{3+x_n}\geq 2-\frac{3}{4}=\frac{5}{4}>1.\]

Next, we estimate

    \[|x_{n+1}-x_n|=\bigg|\frac{3+2x_n}{3+x_n}-\frac{3+2x_{n-1}}{3+x_{n-1}}\bigg|=\]

    \[=\frac{|9+6x_n+3x_{n-1}+2x_nx_{n-1}-9-6x_{n-1}-3x_n-2x_nx_{n-1}|}{(3+x_n)(3+x_{n-1})}=\]

    \[=\frac{3|x_n-x_{n-1}|}{(3+x_n)(3+x_{n-1})}\leq \frac{3}{16}|x_n-x_{n-1}|\]

The needed estimate holds with \lambda=\frac{3}{16}<1.

(b) We have x_2=\frac{5}{4}. By induction we check

    \[|x_n-x_{n+1}|\leq \frac{1}{4}\lambda^{n-1}\]

For n=1 it is true:

    \[|x_1-x_2|=\frac{1}{4}=\frac{1}{4}\lambda^0.\]

Assuming that |x_n-x_{n-1}|\leq \frac{1}{4}\lambda^{n-2}, we deduce from (a) that

    \[|x_{n+1}-x_n|\leq \lambda|x_n-x_{n-1}|\leq \frac{1}{4}\lambda^{n-1}\]

For n<m we estimate

    \[|x_m-x_n|\leq \sum^{m-1}_{k=n}|x_{k+1}-x_k|\leq \frac{1}{4}\sum^{m-1}_{k=n}\lambda^{k-1}\leq \frac{1}{4}\sum^\infty_{k=n}\lambda^{k-1}=\frac{\lambda^{n-1}}{4(1-\lambda)}\]

Since 0<\lambda<1 we deduce that

    \[\sup_{m>n}|x_m-x_n|\to 0, \ n\to\infty,\]

i.e. \{x_n\}_{n\geq 1} is a Cauchy sequence and the limit x=\lim_{n\to\infty}x_n exists. Passing to the limit in the equation

    \[x_{n+1}=(3+2x_n)/(3+x_n)\]

we find that

    \[x=\frac{3+2x}{3+x}\]

    \[3x+x^2=3+2x, \ x^2+x-3=0,\]

    \[x=\frac{-1\pm\sqrt{13}}{2}.\]

But x_n\geq 1 for all n\geq 1. So, x\geq 1 and

    \[x=\frac{\sqrt{13}-1}{2}.\]

Problem 2.

Examine, with justification, whether the following limit exists:

    \[\lim_{N\to\infty}\int^{e^N}_N xe^{-x^{2016}} dx\]

.
If the limit exists, then find its value.

Topic: Real Analysis, Difficulty Level: Easy

Full Solution

Consider the elementary estimate

    \[e^{-x}\leq \frac{1}{x}, \ x>0.\]

Applying it to the integral we get

    \[\int^{e^N}_N xe^{-x^{2016}} dx\leq \int^{e^N}_N \frac{1}{x^{2015}}dx=\frac{1}{2014}\left(\frac{1}{N^{2014}}-\frac{1}{e^{2014N}}\right)\leq\]

    \[\leq \frac{1}{N^{2014}}\to 0, N\to \infty.\]

Answer: The limit exists and is equal to zero.

Problem 3.

Does there exist a continuous function f:\mathbb{R}\to\mathbb{R} that takes every real value exactly twice ? Justify your answer.

Topic: Real Analysis, Difficulty Level: Medium

Full Solution

Assume that such a function exists and let y_0 be any of its values. By assumption, there exist a<b such that

    \[f(a)=f(b)=y_0.\]

On the interval (a,b) the function f never takes the value y_0. Consequently, either f(x)>y_0 for all x\in (a,b) or f(x)<y_0 for all x\in (a,b). Without loss of generality we can assume that f(x)>y_0 for all x\in (a,b) (in the opposite case we just consider the function g(x)=2y_0-f(x); g is also continuous and takes each value exactly twice). f(x) reaches maximum on the segment [a,b] at some point x_0\in [a,b]. Let y_1=\max_{x\in[a,b]}f(x). For each x\in (a,b) we have y_1\geq f(x)>y_0=f(a)=f(b), so y_1=f(x_0)>y_0 and we necessarily have x_0\in (a,b). By assumption, there exists x_1\ne x_0 such that f(x_1)=f(x_0)=y_1. Observe that x_1\ne a, x_1\ne b. Cconsider two possibilities:

  • x_1\not\in [a,b]. From the Intermediate Value Theorem, any value u\in (y_0,y_1) is reached at least three times: once between x_1 and an end-point of [a,b], once between x_0 and a, and once between x_0 and b. This is impossible.
  • x_1\in (a,b). Then f(x_0)=f(x_1)=y_1=\max_{x\in [a,b]}f(x). If f is constant between x_0 and x_1, then the value y_1 is taken infinitely many times, which is impossible. If f is not constant, there exists x_2 between x_0 and x_1 such that f(x_2)=y_2<y_1. Fix u\in (\max(y_2,y_0),y_1). The value u is taken at least four times: between x_0 and the end-point of [a,b], between x_1 and the end-point of [a,b], between x_0 and x_2 and between x_2 and x_1.

In any case we get a contradiction, so there is no continuous function f:\mathbb{R}\to\mathbb{R} that takes every real value exactly twice.

Problem 4.

Suppose f:[0,1]\to\mathbb{R} is a bounded function such that f is Riemann integrable on [a,1] for every a\in(0,1). Is f Riemann integrable on [0,1]? justify your answer.

Topic: Real Analysis, Difficulty Level: Medium

Full Solution

Let C=\sup_{0\leq x\leq 1}|f(x)|. We will prove that f is Riemann integrable by showing that for every \epsilon>0 there exists a partition \lambda: 0=t_0<t_1<\ldots<t_n=1 of [0,1] such that for the upper Darboux sum U(f;\lambda) and the lower Darboux sum L(f;\lambda) the inequality

    \[U(f;\lambda)-L(f;\lambda)<\epsilon\]

holds.

Let a\in (0,1) be such that a<\frac{\epsilon}{4C}. Since f is Riemann integrable on [a,1] there exists a partition a=s_0<s_1<\ldots<s_n=1 of [a,1] such that

    \[\sum^n_{j=1}\left(\sup_{x\in [s_{j-1},s_j]}f(x)-\inf_{x\in [s_{j-1},s_j]}f(x)\right)(s_j-s_{j-1})<\frac{\epsilon}{2}.\]

Set t_0=0, t_1=a, t_j=s_{j-1} for 2\leq j\leq n+1, so that 0=t_0<t_1<\ldots<t_{n+1}=1. Observe that

    \[\sup_{x\in[t_0,t_1]}f(x)-\inf_{x\in[t_0,t_1]}f(x)\leq 2C.\]

So,

    \[\sum^{n+1}_{j=1}\left(\sup_{x\in [t_{j-1},t_j]}f(x)-\inf_{x\in [t_{j-1},t_j]}f(x)\right)(t_j-t_{j-1})=\]

    \[=a\left(\sup_{x\in [t_{0},t_1]}f(x)-\inf_{x\in [t_{0},t_1]}f(x)\right)+\sum^n_{j=1}\left(\sup_{x\in [s_{j-1},s_j]}f(x)-\inf_{x\in [s_{j-1},s_j]}f(x)\right)(s_j-s_{j-1})\leq\]

    \[\leq 2aC+\frac{\epsilon}{2}<\epsilon.\]

f is Riemann integrable on [0,1].

Problem 5.

Let h:\mathbb{R}^2\to\mathbb{R}^2 be a surjective function such that

    \[||h(x)-h(y)||\geq 3||x-y||\]

for all x,y\in\mathbb{R}^2. Here ||.|| denotes the Euclidean norm on \mathbb{R}^2. Show that the image of every open set (in \mathbb{R}^2) under the map h is an open set (in \mathbb{R}^2).

Topic: Metric Space, Difficulty Level: Medium

Full Solution

Let G\subset \mathbb{R}^2 be open. Fix y_0\in h(G), y_0=h(x_0) for some x_0\in G. There exists r>0 such that

    \[\|x-x_0\|<r\Rightarrow x\in G.\]

Let \|y-y_0\|<3r. By assumption, there exists x\in \mathbb{R}^2 such that h(x)=y. We estimate

    \[\|y-y_0\|=\|h(x)-h(x_0)\|\geq 3\|x-x_0\|,\]

so,

    \[\|x-x_0\|\leq \frac{\|y-y_0\|}{3}<r\]

and x\in G, y=h(x)\in h(G). We have proved that

    \[\|y-y_0\|<3r\Rightarrow y\in h(G)\]

which means that G is open.

Problem 6.

Suppose that g:[0,1]\times [0,1]\to \mathbb{R} is a continuous function and

    \[D=\{(x,y)\in\mathbb{R}^2:0<x<y<1\}.\]

Define a new function G:[0,1]\times [0,1]\to \mathbb{R} by

    \[G(x,v)=\int^x_0 g(u,v)du, \hspace{.5cm} (x,v)\in [0,1]\times [0,1].\]

Now define another function \psi:D\to\mathbb{R} by

    \[\psi(x,y)=\int^y_x G(x,v)dv, (x,y)\in D\]

Does \frac{\partial\psi}{\partial y} exist at every point in D? Justify your answer.

Topic: Partial Differential, Difficulty Level: High

Full Solution

At first we check that G:[0,1]\to [0,1]\to \mathbb{R} is continuous. We need two consequences of continuity of g:[0,1]\times [0,1]\to \mathbb{R}:

  • (boundedness) there exists C>0 such that |g(x,y)|\leq C for all x,y\in [0,1];
  • (uniform continuity) for any \epsilon>0 there exists \delta>0 such that

        \[x,y,x',y'\in [0,1], |x-x'|\leq \delta, |y-y'|\leq \delta \Rightarrow |g(x,y)-g(x',y')|\leq \epsilon. \eqno(*)\]

 

Given \epsilon>0 choose \delta>0 such that (*) holds. Let x,y,x',y'\in [0,1] be such that |x-x'|\leq \frac{\epsilon}{C}, |y-y'|\leq \delta. For any u relation (*) implies

    \[|g(u,y)-g(u,y')|\leq \epsilon.\]

So, we have

    \[|G(x,y)-G(x',y')|=\bigg|\int^x_0 g(u,y)du-\int^{x'}_0 g(u,y')du\bigg|\leq\]

    \[\leq \bigg|\int^x_0 (g(u,y)-g(u,y'))du\bigg|+\bigg|\int^{x'}_x g(u,y')du\bigg|\leq x\epsilon+|x-x'|C\leq \epsilon +C\frac{\epsilon}{C}=2\epsilon.\]

Continuity of G is proved.

For fixed x\in (0,1) the function v\to G(x,v) is continuous on [x,1]. By the Fundamental Theorem of Calculus, the integral

    \[y\to \int^y_x G(x,v)dv, \ x<y<1\]

is differentiable at y, and

    \[\frac{d}{dy}\int^y_x G(x,v)dv=G(x,y).\]

So, we`ve proved that

    \[\exists \frac{\partial \psi}{\partial y}=G(x,y).\]

Group B

Problem 7.

Let A be a 2\times 2 matrix with complex entries. Suppose that \det(A)=0 and trace(A)\neq 0. Show the following:

(a) Kernel(A) \cap Range(A)=\{0\}.

(b) \mathbb{C}^2=span(Kernel(A) \cup Range(A)).

Topic: Linear Algebra, Difficulty Level: Easy

Full Solution

(a) The characteristic polynomial of A is

    \[p(\lambda)=\det (A-\lambda I)=\lambda^2- \mbox{trace}(A)\lambda+\det(A)=\lambda(\lambda -\mbox{trace}(A)).\]

So, A has two eigenvalues \lambda_1=0 and \lambda_2=\mbox{trace}(A)\ne 0. Let e_1 and e_2 be corresponding eigenvectors. \{e_1,e_2\} is a basis of \mathbb{C}^2. Assume y\in \mbox{Kernel}(A)\cap \mbox{Range}(A). Then y=Ax for some x\in \mathbb{C}^2, and Ay=A^2x=0. Write x=x_1e_1+x_2e_2. Then

    \[Ax=x_1Ae_1+x_2Ae_2=x_2\lambda_2 e_2, A^2x=x_2\lambda^2_2 e_2=0.\]

Since \lambda_2\ne 0, we deduce x_2=0, and y=Ax=x_2\lambda_2e_2=0.

 

 

(b) Ae_2=\lambda_2e_2\ne 0, so \dim \mbox{Kernel}(A)<2. On the other hand Ae_1=0 and \dim \mbox{Kernel}(A)>0. It follows that

    \[\dim \mbox{Kernel}(A)=1.\]

From relation Ax=x_2\lambda_2e_2 it follows that the range of A is the one-dimensional subspace generated by e_2,

    \[\dim \mbox{Range}(A)=1.\]

Finally,

    \[\dim \mbox{span}(\mbox{Kernel}(A) \cup \mbox{Range}(A))=\]

    \[=\dim \mbox{Kernel}(A)+\dim \mbox{Range}(A)-\dim \mbox{span}(\mbox{Kernel}(A) \cap \mbox{Range}(A))=2.\]

    \[\mbox{span}(\mbox{Kernel}(A) \cup \mbox{Range}(A))=\mathbb{C}^2.\]

Problem 8.

Suppose that B is a nonzero 2\times 2 matrix with complex entries. Prove that B^2=0 if and only if the B and the matrix \begin{pmatrix} 0&0\\ 1&0 \end{pmatrix} are similar.

Topic: Linear Algebra, Difficulty Level: Medium

Full Solution

Assume that B^2=0. Consider the Jordan normal form J of the matrix B: there exists an invertibale matrix P such that

    \[PBP^{-1}=J\]

and either J=\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2\end{pmatrix} or J=\begin{pmatrix} \lambda & 1 \\ 0 & \lambda\end{pmatrix}. In the first case

    \[J^2=\begin{pmatrix} \lambda^2_1 & 0 \\ 0 & \lambda^2_2\end{pmatrix}=PB^2P^{-1}=0\]

Then \lambda_1=\lambda_2=0, and J=B=0. This is not the case.

Consequently,

    \[J=\begin{pmatrix} \lambda & 1 \\ 0 & \lambda\end{pmatrix}.\]

In this case

    \[J^2=\begin{pmatrix} \lambda^2 & 2\lambda \\ 0 & \lambda^2\end{pmatrix}=PB^2P^{-1}=0\]

and \lambda=0. We have proved that B is similar to the matrix J=\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}. Observe that for Q=\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix} we have

    \[QJQ^{-1}=\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}=\]

    \[=\begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}\]

So,

    \[QPB(QP)^{-1}=\begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}.\]

Conversely, if B is similar to the matrix \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}, then B^2 is similar to the matrix

    \[\begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}\begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix},\]

i.e. B^2=0.

Problem 9.

Let S_{17} be group of all permutation of 17 distinct symbols. How many subgroups of order 17 does S_{17} have? Justify your answer.

Topic: Group Theory, Difficulty Level: Medium

Full Solution

A subgroup of prime order is cyclic. Let
H be a subgroup of S_{17} of order 17. Every element of H distinct from the identity is the generator of H. The order of a permutation is the least common mulitple of orders of permutations in its cyckle decomposition. If g\in H is a generator, then it is of order 17 and its cycle decomposition consists of a single cycle of length 17:

    \[g=(1,x_2,x_3,\ldots,x_{17}).\]

Replacing g with g^j for 2\leq j\leq 16 (if needed), we may assume that

    \[g=(1,2,x_3,\ldots,x_{17}).\]

There are 15! such elements and they generate distinct subgroups of order 17 in S_{17}.

Answer: there are 15! subgroups of order 17.

Problem 10.

Suppose that H and K are two subgroups of a group G. Assume that [G:H]=2 and K is not a subgroup of H. Show that HK=G

Topic: Group Theory, Difficulty Level: Easy

Full Solution

Take element k\in K\setminus H. The cosets H and Hk are disjoint, so H\cup Hk=G and G\subset HK\subset G.

Problem 11.

For any ring R, let R[X] denote the ring of all polynomials with indeterminate X and coefficients from R. Examine, with justification, whether the following pairs of rings are isomorphic:

(a) \mathbb{R}[X] and \mathbb{C}[X].

(b) \mathbb{Q}[X]/(X^2-X)] and \mathbb{Q}\times \mathbb{Q}

Topic: Ring Theory, Difficulty Level: Medium

Full Solution

(a) Assume that there exists an isomorphism \phi:\mathbb{C}[X]\to \mathbb{R}[X]. Observe that in \mathbb{C}[X] we have

    \[i^2+1=0.\]

Let \phi(i)=a_0+a_1X+\ldots+a_nX^n for some a_0,\ldots,a_n\in \mathbb{R}. Then

    \[\left(a_0+a_1X+\ldots+a_nX^n\right)^2+1=0.\]

However, the free term on the left-hand side is a^2_0+1\ne 0.

(b) Consider the mapping f:\mathbb{Q}[X]\to \mathbb{Q}\times \mathbb{Q} given by

    \[f(P(X))=(P(0),P(1)).\]

The mapping f is a surjective homeomoprhism. Indeed, for any p,q\in \mathbb{Q} we have

    \[f(p+(q-p)X)=(p,q).\]

We find the kernel of f.

    \[f(P(X))=0\Leftrightarrow P(0)=P(1)=0.\]

X=0 and X=1 are roots of P(X), i.e. P(X) is divisible by X(X-1)=X^2-X:

    \[P(X)=Q(X)(X^2-X).\]

The kernel of f coincides with the ideal generated by X^2-X. By the isomoprhism theorem,

    \[\mathbb{Q}[X]/(X^2-X)=\mathbb{Q}[X]/\mbox{Kernel}(f) \mbox{ and } \mathbb{Q}\times \mathbb{Q}=f(\mathbb{Q}[X])\]

are isomorphic.

Problem 12.

For any \alpha \in\mathbb{R}\setminus \mathbb{Q}, let Q(\alpha) be the smallest subfield of \mathbb{R} containing \mathbb{Q}\cup \{\alpha\}. Find a basis for the vector space \mathbb{Q}(\sqrt{3}+\sqrt{5}) over \mathbb{Q}(\sqrt{15})

Topic: Field Theory, Difficulty Level: Hard

Full Solution

Set \alpha=\sqrt{3}+\sqrt{5}. Then

    \[\alpha^2=8+2\sqrt{15}.\]

\alpha is a root of the polynomial

    \[P(X)=X^2-(8+2\sqrt{15})\in \mathbb{Q}(\sqrt{15})[X].\]

If we prove that P(X) is irreducible in \mathbb{Q}(\sqrt{15})[X], we will verify that

    \[[\mathbb{Q}(\sqrt{3}+\sqrt{5}):\mathbb{Q}(\sqrt{15})]=2\]

and

    \[\{1,\sqrt{3}+\sqrt{5}\}\]

is a basis for the vector space \mathbb{Q}(\sqrt{3}+\sqrt{5}) over \mathbb{Q}(\sqrt{15}).

To prove that P(X) is irreducible in \mathbb{Q}(\sqrt{15})[X] it is enough to show that the roots \pm \alpha do not belong to \mathbb{Q}(\sqrt{15}). Assume that

    \[\alpha=a+b\sqrt{15}, \ a,b\in\mathbb{Q}.\]

Then

    \[\alpha^2=8+2\sqrt{15}=a^2+15b^2+2ab\sqrt{15},\]

    \[ab=1, a^2+15b^2=8\]

    \[b=\frac{1}{a}, \ a^2+\frac{15}{a^2}=8\]

    \[a^4-8a^2+15=0\]

    \[(a^2-5)(a^2-3)=0\]

which is impossible as a\in\mathbb{Q}. So, P(X) is irreducible in \mathbb{Q}(\sqrt{15}).

Answer: \{1,\sqrt{3}+\sqrt{5}\} a basis for the vector space \mathbb{Q}(\sqrt{3}+\sqrt{5}) over \mathbb{Q}(\sqrt{15}).

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 2018 solutions, hints, and discussions: Click Here

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

To view previous year’s question papers: Click Here

2 Comments

Leave a Reply to Arkabrata Das Cancel reply