ISI MMath Solutions and Discussion 2016 : PMB

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ISI MMath PMB 2016 Subjective Questions, solutions and discussions

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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}).

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