TIFR 2013 Problem 40 Solution -Convergence of alternating series


TIFR 2013 Problem 40 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:True/False?

The series (1-\frac{1}{\sqrt2}+\frac{1}{\sqrt3}-\frac{1}{\sqrt4}+...) is divergent.

Hint:

Recall the alternating series test (or the Leibniz test)

Discussion:

Let (a_n=\frac{1}{\sqrt{n}}). The alternating series test says that if we have a series like (a_1-a_2+a_3-a_4+...) then a sufficient condition for the convergence of this series is: (a_n) is decreasing and (a_n\to 0 ) as (n\to \infty ).

Here, (a_n) satisfies the above condition.

Therefore, the series converges.

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TIFR 2013 Problem 39 Solution - Rank and Trace of Idempotent matrix

TIFR 2013 Problem 39 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:True/False?

If (A) is a complex nxn matrix with (A^2=A), then rank(A)=trace(A).

Hint:

What are the eigenvalues of (A)? What is trace in terms of eigenvalues?

Discussion:

If (v) is an eigenvector of (A) with eigenvalue (\lambda) then (Av=\lambda v), therefore (\lambda v=Av=A^2v=\lambda Av =\lambda^2 v). Therefore, since any eigenvector is non-zero, (\lambda =0 or 1 ).

Sum of eigenvalues is trace of the matrix. So, trace(A)= number of non-zero eigenvalues= total number of eigenvalues - number of 0 eigenvalues

Since (A) satisfies the polynomial (x^2-x), the minimal polynomial is either (x) or (x-1) or (x(x-1)). This means the minimal polynomial breaks into distinct linear factors, so (A) is diagonalizable. Therefore, the algebraic multiplicity of an eigenvalue is same as its geometric multiplicity.

In total there are n eigenvalues (for A is nxn) and the number of 0-eigenvalues is the algebraic multiplicity of 0, which is same as the geometric multiplicity of 0, i.e, the dimension of the kernel of A.

Therefore, trace(A)(=n-)nullity(A).

By the rank-nullity theorem, the right hand side of the above equation is rank(A).

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TIFR 2013 Problem 38 Solution -Eigenvalue of differentiation

TIFR 2013 Problem 38 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:True/False?

Let (V) be the vector space of polynomials with real coefficients in variable (t) of degree ( \le 9). Let (D:V\to V) be the linear operator defined by (D(f):=\frac{df}{dt}). Then (0) is an eigenvalue of (D).

Hint:

If 0 were an eigenvalue, what would be its eigenvector?

Discussion:

There are several ways to do this. One possible way is to find out the matrix representation of (D) with respect to standard basis ( {1,t,t^2,...,t^n})( or any other basis) and observe that it is a (strictly) upper triangular matrix with all diagonal entries 0 and therefore the determinant of (D) is 0. This implies that D is not injective, so there is some nonzero vector to which when (D) is applied gives the zero vector. Therefore, (D) has 0 eigenvalue.

Another way to do this is by observing that (D(1)=0(1)), therefore 0 is an eigenvalue of (D) with 1 as an eigenvector.

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Checking irreducibility over \(\mathbb{R}\) - TIFR 2013 Sum 37

Let's discuss a problem based on Checking irreducibility over \(\mathbb{R}\) from TIFR 2013, Problem 37. Try this Problem and then read the solution.

Question: Checking irreducibility over \(\mathbb{R}\)

True/False?

The polynomial \(x^3+3x-2\pi \) is irreducible over \(\mathbb{R}\)

Hint:

When is a odd degree polynomial irreducible over \(\mathbb{R}\)?

Discussion:

A polynomial \(p(x)\) is irreducible over a field if it can not be written as product of two non-trivial polynomials having coefficients from the same field.

In other words, a polynomial \(p(x) \in \mathbb{F}[x] \) of degree \(k\) is irreducible over \(\mathbb{F}\) if there is no non-constant polynomial of degree \(<k\) dividing \(p(x)\) in \(\mathbb{F}[x]\).

For a polynomial \(p(x)\) of odd degree \( \ge 3 \) over \(\mathbb{R}\) there is always a real root \(\alpha\). This is because the complex roots always occur in conjugate pairs. Therefore, \((x-\alpha ) \) divides \(p(x)\).

Therefore, \(p(x)\) is not irreducible.

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Checking injectivity | TIFR 2013 problem 36

Let's solve a problem based on checking injectivity from TIFR 2013 problem 36. Try it yourself first, then read the solution here.

Question:

True/False?

The function \(f:\mathbb{Z} \to \mathbb{R} \) defined by \(f(n)=n^3-3n\) is injective.

Hint:

Check for small values!

Discussion:

Surely, checking for small values will give you that f is not injective.

For example, let us look at \(f(0)=0\), \(f(1)=-2\), \(f(-1)=2\), \(f(2)=2\).

So \(f(-1) = f(2)\).

Therefore, \(f\) is not injective.

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TIFR 2013 problem 35 | Sequence Bounded / Unbounded

Let's discuss a problem and find out whether a sequence is bounded or unbounded from TIFR 2013 Problem 35. Try before reading the solution.

Question: TIFR 2013 problem 35

True/False?

Let \( \{a_n\} \) be any non-constant sequence in \(\mathbb{R}\) such that \( a_{n+1}=\frac{a_n + a_{n+2} }{2} \) for all \(n \ge 1 \). Then \(\{a_n\} \) is unbounded.

Hint:

The given expression is same as \( a_{n+1}-a_n = a_{n+2} -a_{n+1} \).

Discussion:

The distance between two successive terms in the given sequence is constant. It is given by \( |a_{n+1}- a_n| = |a_n - a_{n-1}| = ... = |a_1-a_0| \).

So for the sequence to be non-constant, \( a_1 \ne a_0 \). Because otherwise, the sequence will have distance between any two successive terms zero, which is just another way of saying that the sequence is constant.

There are two cases:

Case 1: \(a_1 > a_0\).

Then \(a_{n+1} > a_n\), that is the sequence is increasing, and not only that, it is an arithmetic progression with common difference \(a_1-a_0 (> 0)\). Therefore, the sequence is unbounded above.

Case 2: \(a_1<a_0\).

Then as in the previous case the sequence this time will become a decreasing sequence, not only that, it is an arithmetic progression with common difference \( < 0 \). Therefore, the sequence is unbounded below.

Remark:

We don't really need to take the two cases. The key point is that the given recurrence relation is that of an arithmetic progression whose common difference is non-zero. Hence the sequence has to be unbounded.

 

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TIFR 2013 Problem 34 Solution -Countable or not?


TIFR 2013 Problem 34 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 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:True/False?

Let \(S\) be the set of all sequences \(\{a_1,a_2,...,a_n,...\} \) where each entry \(a_i \) is either \(0\) or \(1\). Then \(S\) is countable.

Hint:

What if instead of \(0\) and \(1\) the values were allowed to be any digit from \(0\) to 9? What would that correspond to?

Discussion:

Given a sequence  \(\{a_1,a_2,...,a_n,...\} \) , we can associate it to the binary number \(0.a_1a_2...a_n... \). This association (or function if you like) is one-one, and onto the set of all real numbers having binary expansion in the form of \(0.something\), which is same as the set \((0,1)\), which is uncountable.

Remark:

One could use Cantor's diagonalization argument as well to argue in this problem. If possible let \(x_1,x_2,...\) is an enumeration of the given set (of sequences) then consider the sequence \(\{b_1,b_2,...\} \) in \(S\) defined by \(b_i \ne x_i  \). We get a contradiction because this is a sequence which is not in the enumeration but is a member of \(S\).

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TIFR 2013 Problem 32 Solution - Limit of a Sequence


TIFR 2013 Problem 32 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:True/False

( \lim_{n\to \infty } (n+1)^{1/3} -n^{1/3} = \infty )

Hint:

Simplify the given expression.

Discussion:

We feel that ( (n+1)^{1/3} ) goes to infinity at the same speed as ( n^{1/3} ). So in fact, the above limit should be zero.

We make this little bit more rigorous.

(   (n+1)^{1/3} -n^{1/3} = \frac{n+1-n}{(n+1)^{2/3}+(n+1)^{1/3}n^{1/3}+n^{2/3} } )

(  =\frac{1}{(n+1)^{2/3}+(n+1)^{1/3}n^{1/3}+n^{2/3} } \to 0) as (n\to \infty ).

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TIFR 2013 problem 31 | Inequality problem

Let's discuss a problem from TIFR 2013 Problem 31 based on inequality.

Question: TIFR 2013 problem 31

True/False?

The inequality \( \sqrt{n+1}- \sqrt{n} < \frac{1}{ \sqrt{n} } \) is false for all n such that \( 101 \le n \le 2000 \)

Hint:

Simplify the given inequality

Discussion:

\( \sqrt{n+1}- \sqrt{n} =  \frac {n+1- n}{ \sqrt{n+1}+ \sqrt{n} } \)

\(=  \frac {1}{ \sqrt{n+1}+ \sqrt{n}}  < \frac {1}{ \sqrt{n}} \) This holds for any natural number \(n\).

So the inequality is actually true for all natural numbers.

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TIFR 2013 Problem 30 Solution -Infinite set of independent vectors taken 3 at a time

TIFR 2013 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 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

ProblemType:True/False?

There exists an infinite subset \(S\subset \mathbb{R}^3 \) such that any three vectors in \(S\) are linearly independent.

Hint:

Can you find a sequence of vectors in \(\mathbb{R}^3\) which satisfy the following property? What does it mean to be linearly independent in terms of 3x3 matrices and their determinants?

Discussion:

We focus on finding a sequence as mentioned above. But first, we recall that:

If three vectors are linearly independent in \(\mathbb{R}^3\) then the columns of the  matrix formed by these three vectors as columns are linearly independent. Which means the column rank (or rank) of this matrix is 3 (i.e, full rank), hence the matrix must be invertible and so determinant of this matrix should be zero.

So we have now a slightly different goal: To find a sequence in  \(\mathbb{R}^3\) such that if any three of those are taken as columns of a 3x3 matrix then the determinant would be zero.

Now we begin our search. We start with simple integer valued 'nice-looking' sequences and very soon arrive at the following: \( (1,n,n^2) \) where \(n \in \mathbb{N}\). (Note: This is really a trial and error, at least that's how I arrived at it.)

We just need to verify: \( \begin{vmatrix} 1 & 1 & 1 \\ n & m & l \\ n^2 & m^2 & l^2 \end{vmatrix} \neq 0 \) where \(n,m,l\) are three different natural numbers. This is easy to check. (Hint: determinant will be product of differences taken two at a time).

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