TIFR 2014 Problem 4 Solution -Checking for Uniform continuity


TIFR 2014 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 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:[0,\infty)\to \mathbb{R}) be defined by

( f(x)=x^{2/3}logx)  for (x>0 ) (f(x)=0) if (x=0)

Then

A. f is discontinuous at x=0

B. f is continuous on ([0,\infty)) but not uniformly continuous on ([0,\infty))

C. f is uniformly continuous on ([0,\infty))

D. f is not uniformly continuous of ([0,\infty) but uniformly continuous of ((0,\infty))

Discussion:

First, we find whether (f) is continuous or not. (f) is "clearly" continuous on ((0,\infty)).

Question is, what happens near (x=0)?

Take limit.

(lim_{x\to 0}\frac{logx}{x^{-2/3}}) is in (\frac{-\infty}{\infty}) form.

We use the L'Hospital Rule.

(lim_{x\to 0}\frac{logx}{x^{-2/3}})=(lim_{x\to 0}\frac{1/x}{-\frac{2}{3}x^{-2/3-1}})

=(lim_{x\to 0}-3/2{x^{2/3}})

=(0).

Okay! So we now know that (f) is at-least continuous.

Now we present the HINT s:

For (x>0), (f'(x)=\frac{2}{3}x^{-1/3}logx + x^{2/3}x^{-1}) (By the formula for calculating derivatives of product).

After simplification,

(f'(x)=\frac{2logx+3}{3x^{1/3}}).

We want the boundedness of (f') and we don't care about what happens close to zero as much. (Because we are going to use the first hint).

(lim_{x\to \infty}\frac{2logx+3}{3x^{1/3}}) is in (\frac{\infty}{\infty}) form.

Once again, we use the L'Hospital rule.

(lim_{x\to \infty}\frac{2logx+3}{3x^{1/3}}=lim_{x\to \infty}\frac{2/x}{3/3 x^{-2/3}})

(= lim_{x\to \infty} 2x^{-1/3}=0)

What does this tell us? This tells us that (f') is bounded on (say) ([1,\infty))

So (f) is uniformly continuous on ([1,\infty)). Also, (f) is uniformly continuous on ([0,1]).

So given (\epsilon>0) we have two "delta's" for these two intervals. (which satisfies the condition for uniform continuous in these intervals respectively). Take the delta which is less. And this gives the uniform continuity for (f).

Helpdesk

TIFR 2014 Problem 3 Solution -Function bounds from derivative limits


TIFR 2014 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 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 differentiable function such that (\lim_{x\to \infty} f'(x)=1) then

A. f is bounded

B. f is increasing

C. f is unbounded

D. f' is bounded

Discussion:

The easiest function to look at is the function which has derivative 1 at all points. That is (f(x)=x). This function is not bounded. So option A is wrong.

Now, the function (f) is increasing. But let's say we disturb the same (f) around 0 a little bit, so that the function is not increasing any more. We can do this disturbing so that the function still remains differentiable, but it loses the increasing part of it. This tells us that B is wrong.

We can disprove D in a similar manner, for even though (f') can be bounded in the positive side of x-axis, what if (f'\to \infty) as (x\to -\infty)? And this case in absolutely possible, because we have no information about the nature of (f) for -ve values of x. We can construct such an example easily by the help of drawing graphs, or constructing functions explicitly. (Use (x^2) when x is negative and some linear function/translation of x when x is positive, and make sure the function transition is smooth in between).

Now, we have eliminated all but option C. We prove option C.

What does the expression (\lim_{x\to \infty} f'(x)=1) mean?

It means that for large values of (x) (f'(x)) is close enough to 1.

That means, for a sequence going to (\infty) we have (f') at the sequence-points going to 1.

Take the sequence as (n\in \mathbb{N}). Then

(\lim_{n\to \infty} lim_{h \to 0} \frac{f(n+h)-f(n)}{h} =1).

Let's forget about (h\to 0). Let's say, h=1.

Then (f(n+1)-f(n)=f'(y_n)(n+1-n)) by the mean-value theorem. Where (y_n\in[n,n+1]). As (n\to \infty), (y_n \to \lim_{x\to \infty} f'(x)=1).

That means, given (\epsilon >0); for large values of (n)

(|f(n+1)-f(n)-1|=|f'(y_n)-1|< \epsilon ). This means that (f) increases approximately by 1 for an increment of 1 in the value of (x). This means that (f) behaves like (x) for large values of (x). Which proves that (f) must be unbounded.

Helpdesk

TIFR 2014 Problem 2 Solution -Continuous Bounded Function


TIFR 2014 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 bounded function. Then

A. f has to be uniformly continuous

B. there exists an (x\in \mathbb{R}) such that (f(x)=x)

C. f cannot be increasing

D. (\lim_{x\to \infty}f(x)) exists.

Discussion:

Let us discard some answers first.

Define (f(x)=sin(x^2)). Then (f) is bounded, continuous.

Take (x_1=\sqrt{n\pi+\pi/2}), (x_2=\sqrt{n\pi}).

Then (|x_1-x_2|=\frac{\pi/2}{\sqrt{n\pi+\pi/2}+\sqrt{n\pi}}<\frac{1}{\sqrt{n\pi}}).

Since (\frac{1}{\sqrt{n\pi}}\to 0), given (\delta >0), we can have

(|x_1-x_2|<\delta) for large values of (n).

But (|f(x_1)-f(x_2)|=1). So given (\epsilon =1/2) we can never find a (\delta >0) for which (|x_1-x_2|<\delta) would imply that (|f(x_1)-f(x_2)|<\epsilon).

So this (f) is not uniformly continuous. So (A) is false.

Also from this example, since (f) does not have limit as (x\to \infty), we conclude that (D) is false.

Increasing does not mean strictly increasing. So, you are allowed to take a constant function as an example of increasing function. And it is bounded,continuous. This disproves (C).

Now we are left only with (C). Since we disproved all of the others, and one option is correct, (C) has to be true. We inspect the proof just to be sure.

Since (f) is bounded, there exists (M>0) such that (|f(x)|<M).

That means (-M<f(x)<M) for all (x\in \mathbb{R}).

Look at (g(x)=f(x)-x). Finding fixed point of (f) is same as finding a zero of (g).

We know that (f) is continuous, therefore (g) is continuous. We would like to find two points where (g) takes values with opposite signs. Now

(g(M)=f(M)-M<0) and (g(-M)=f(-M)-(-M)=f(-M)+M>0).

Therefore, (g) must cut the x-axis. Here we are using the intermediate value theorem for continuous functions.

This proves (B).

Helpdesk

TIFR 2014 Problem 1 Solution - Negation


TIFR 2014 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 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 A,B,C be three subsets of (\mathbb{R}) The negation of the following statement: For every (\epsilon >1), there exists (a\in A) and (b\in B) such that for all (c\in C), (|a-c|< \epsilon) and (|b-c|>\epsilon ) is:

(A) There exists ( \epsilon \leq 1), such that for all (a \in A) and (b \in B) there exists (c \in C) such that (|a − c| \geq \epsilon) or (|b − c| \leq \epsilon) 

(B) There exists ( \epsilon \leq 1), such that for all (a \in A) and (b \in B) there exists (c \in C) such that (|a − c| \geq \epsilon) and (|b − c| \leq \epsilon) 

(C) There exists ( \epsilon > 1), such that for all (a \in A) and (b \in B) there exists (c \in C) such that (|a − c| \geq \epsilon) and (|b − c| \leq \epsilon) 

(D) There exists ( \epsilon > 1), such that for all (a \in A) and (b \in B) there exists (c \in C) such that (|a − c| \geq \epsilon) and (|b − c| \leq \epsilon) 

Discussion:

Before looking at the options, let us understand what rules we have to negate a statement.
(Please note this topic have discussed in Appendix portion of mentioned book in Helpdesk section.)

When we negate a for-all statement we get a there-exists statement. For example, the negation of (\forall x P(x)) is true is (\exists x) for which (P(x)) is not true.

A negation of a there-exists statement is for-all. Example: the negation of (\exists x) such that ( P(x)) is true is (\forall x) (P(x)) is not true.

To negate a slightly complicated such as that given in question we follow the following simple procedure:

  1. replace the \(\forall\)s with \(\exists\) and replace \(\exists\) by \(\forall\).
  2. and in the end, negate the conclusion.

Following this procedure we obtain:

(\exists \epsilon>1),such that (\forall a\in A),and (b\in B) (\exists c\in C), (|a-c|\ge \epsilon) or (|b-c|\le \epsilon).

Note that

(\exists \epsilon>1),such that (\forall a\in A),and (b\in B) (\exists c\in C) is the part that comes from by 1. And the part (|a-c|\ge \epsilon) or (|b-c|\le \epsilon) comes from the part 2.

We look at the question paper now, and indeed, this answer is the statement of option (d).

Helpdesk

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.

Helpdesk

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).

HELPDESK

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.

HELPDESK

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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