ISI MMath Solutions and Discussion 2016 : PMB
ISI MMath PMB 2016 Subjective Questions, solutions and discussions
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Let be a sequence of real numbers defined as follows: and for all .
(a)Show that there exists such that for all ,
(b) Prove that exists and find its value.
Topic: Real Analysis, Difficulty level: Easy
(a) By induction we check that Indeed, for this is true by definition. Assuming that we have hence
Next, we estimate
The needed estimate holds with
(b) We have By induction we check
For it is true:
Assuming that we deduce from (a) that
For we estimate
Since we deduce that
i.e. is a Cauchy sequence and the limit exists. Passing to the limit in the equation
we find that
But for all So, and
Examine, with justification, whether the following limit exists:
If the limit exists, then find its value.
Topic: Real Analysis, Difficulty Level: Easy
Consider the elementary estimate
Applying it to the integral we get
Answer: The limit exists and is equal to zero.
Does there exist a continuous function that takes every real value exactly twice Justify your answer.
Topic: Real Analysis, Difficulty Level: Medium
Assume that such a function exists and let be any of its values. By assumption, there exist such that
On the interval the function never takes the value Consequently, either for all or for all Without loss of generality we can assume that for all (in the opposite case we just consider the function is also continuous and takes each value exactly twice). reaches maximum on the segment at some point Let For each we have so and we necessarily have By assumption, there exists such that Observe that Cconsider two possibilities:
- From the Intermediate Value Theorem, any value is reached at least three times: once between and an end-point of once between and and once between and This is impossible.
- Then If is constant between and then the value is taken infinitely many times, which is impossible. If is not constant, there exists between and such that Fix The value is taken at least four times: between and the end-point of between and the end-point of between and and between and
In any case we get a contradiction, so there is no continuous function that takes every real value exactly twice.
Suppose is a bounded function such that is Riemann integrable on for every . Is Riemann integrable on ? justify your answer.
Topic: Real Analysis, Difficulty Level: Medium
Let We will prove that is Riemann integrable by showing that for every there exists a partition of such that for the upper Darboux sum and the lower Darboux sum the inequality
Let be such that Since is Riemann integrable on there exists a partition of such that
Set for so that Observe that
is Riemann integrable on
Let be a surjective function such that
for all . Here denotes the Euclidean norm on . Show that the image of every open set (in ) under the map is an open set (in ).
Topic: Metric Space, Difficulty Level: Medium
Let be open. Fix for some There exists such that
Let By assumption, there exists such that We estimate
and We have proved that
which means that is open.
Suppose that is a continuous function and
Define a new function by
Now define another function by
Does exist at every point in Justify your answer.
Topic: Partial Differential, Difficulty Level: High
At first we check that is continuous. We need two consequences of continuity of
- (boundedness) there exists such that for all
- (uniform continuity) for any there exists such that
Given choose such that (*) holds. Let be such that For any relation (*) implies
So, we have
Continuity of is proved.
For fixed the function is continuous on By the Fundamental Theorem of Calculus, the integral
is differentiable at and
So, we`ve proved that
Let be a matrix with complex entries. Suppose that and trace. Show the following:
(a) Kernel(A) Range(A).
(b) span(Kernel(A) Range(A)).
Topic: Linear Algebra, Difficulty Level: Easy
(a) The characteristic polynomial of is
So, has two eigenvalues and Let and be corresponding eigenvectors. is a basis of Assume Then for some and Write Then
Since we deduce and
(b) so On the other hand and It follows that
From relation it follows that the range of is the one-dimensional subspace generated by
Suppose that is a nonzero matrix with complex entries. Prove that if and only if the and the matrix are similar.
Topic: Linear Algebra, Difficulty Level: Medium
Assume that Consider the Jordan normal form of the matrix there exists an invertibale matrix such that
and either or In the first case
Then and This is not the case.
In this case
and We have proved that is similar to the matrix Observe that for we have
Conversely, if is similar to the matrix then is similar to the matrix
Let be group of all permutation of distinct symbols. How many subgroups of order does have? Justify your answer.
Topic: Group Theory, Difficulty Level: Medium
A subgroup of prime order is cyclic. Let
be a subgroup of of order Every element of distinct from the identity is the generator of The order of a permutation is the least common mulitple of orders of permutations in its cyckle decomposition. If is a generator, then it is of order and its cycle decomposition consists of a single cycle of length
Replacing with for (if needed), we may assume that
There are such elements and they generate distinct subgroups of order in
Answer: there are subgroups of order
Suppose that and are two subgroups of a group . Assume that and is not a subgroup of . Show that
Topic: Group Theory, Difficulty Level: Easy
Take element The cosets and are disjoint, so and
For any ring , let denote the ring of all polynomials with indeterminate and coefficients from . Examine, with justification, whether the following pairs of rings are isomorphic:
(a) and .
Topic: Ring Theory, Difficulty Level: Medium
(a) Assume that there exists an isomorphism Observe that in we have
Let for some Then
However, the free term on the left-hand side is
(b) Consider the mapping given by
The mapping is a surjective homeomoprhism. Indeed, for any we have
We find the kernel of
and are roots of i.e. is divisible by
The kernel of coincides with the ideal generated by By the isomoprhism theorem,
For any , let be the smallest subfield of containing . Find a basis for the vector space over
Topic: Field Theory, Difficulty Level: Hard
is a root of the polynomial
If we prove that is irreducible in , we will verify that
is a basis for the vector space over
To prove that is irreducible in it is enough to show that the roots do not belong to Assume that
which is impossible as So, is irreducible in
Answer: a basis for the vector space over