Moving to wordpress
by adityaguharoy, Feb 22, 2021, 5:13 PM
This is to announce that we have moved this blog to here on wordpress, as I find it easier to post and maintain a blogroll there.
In some instances some contents may be posted here, but anyone interested in updates from the blog must visit this link.
In some instances some contents may be posted here, but anyone interested in updates from the blog must visit this link.
Wish everyone safety from Covid-19
by adityaguharoy, Mar 28, 2020, 11:23 AM
Wish everyone safety from the novel coronavirus attacks.
Take precautions and care to prevent the spread and incidence of the virus.
Visit https://www.who.int/emergencies/diseases/novel-coronavirus-2019/advice-for-public for additional details.
Take precautions and care to prevent the spread and incidence of the virus.
Visit https://www.who.int/emergencies/diseases/novel-coronavirus-2019/advice-for-public for additional details.
Convex and concave functions in Real numbers -- Basic 1
by adityaguharoy, Mar 1, 2018, 1:44 PM
Convex functions
Let
be a function, and let
be two real numbers with
. Then we say that
is a convex function on the interval
if and only if the following is true :
Given any
, and , any
then,
And we say that
is strictly convex on
if the above inequality is strict whenever
and
.
Concave functions
Let
be a function, and let
be two real numbers with
. Then we say that
is a concave function on the interval
if and only if the following is true :
Given any
, and , any
then,
And we say that
is strictly concave on
if the above inequality is strict whenever
and
.
Quick exercises
Let
be a function and
be two real numbers with
. Then prove that
is convex on the interval
if and only if the function
is concave on
.
(here
is defined by
)
Let
be a function and
be two real numbers with
. Further let
be twice differentiable on
. Prove that
is convex on
if and only if
(the double derivative of
) is non-negative on
.
Derive a version of
(as above) for concave functions.
Let us celebrate
Let




![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
Given any
![$t \in [0,1]$](http://latex.artofproblemsolving.com/6/7/3/6735b925696750e153b8d293780a7b620449b778.png)
![$x_1 , x_2 \in [a,b]$](http://latex.artofproblemsolving.com/b/3/a/b3a60e6a3e809b381d3ae396425baf202335a943.png)


![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)


Concave functions
Let




![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
Given any
![$t \in [0,1]$](http://latex.artofproblemsolving.com/6/7/3/6735b925696750e153b8d293780a7b620449b778.png)
![$x_1 , x_2 \in [a,b]$](http://latex.artofproblemsolving.com/b/3/a/b3a60e6a3e809b381d3ae396425baf202335a943.png)


![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)


Quick exercises





![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)

![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
(here








![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)

![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)


![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)


Let us celebrate
This is also the 100-th post in this blog.
Just noticed it now.
Congratulations to all contributors, and thanks to every readers and appreciators and everyone who commented, shouted, and visited the blog.
Just noticed it now.
Congratulations to all contributors, and thanks to every readers and appreciators and everyone who commented, shouted, and visited the blog.
This post has been edited 2 times. Last edited by adityaguharoy, Mar 4, 2018, 7:25 AM
Relaxing the to-be satisfied conditions for Squeezing Monotone sequences
by adityaguharoy, Feb 28, 2018, 4:09 PM
The Sandwich theorem (or Squeeze Principle ) for sequences of real numbers says that :
If
,
and
be three sequences of real numbers which obeys all the following :

Both the sequences
and
converges

Then,
must converge, and also obey
Forgetting the third condition (as given above) can lead us to errorenous results.
For example : If we define
as
and, we define
as
, and we define the sequence
as
, then
and both the sequences
and
converges, but however,
diverges.
But, it seems we can relax the conditions for some special type of sequences. Let us see how, we can relax it for monotone sequences.
Note, that the Monotone Convergence Theorem (of Bolzano Weirestrass) says that :
Every monotone bounded sequence of real numbers must converge.
Thus, if
,
, and ,
be three monotone sequences such that

Both the sequences
, and ,
converges
Then,
The sequence
must converge , and ,

If











Then,


For example : If we define














But, it seems we can relax the conditions for some special type of sequences. Let us see how, we can relax it for monotone sequences.
Note, that the Monotone Convergence Theorem (of Bolzano Weirestrass) says that :
Every monotone bounded sequence of real numbers must converge.
Thus, if









Then,
The sequence


This post has been edited 1 time. Last edited by adityaguharoy, Mar 1, 2018, 1:41 PM
1. Algebra and sigma algebra on a set
by adityaguharoy, Feb 28, 2018, 3:28 PM
Definitions of algebra on a set and sigma algebra on a set
Let
be an infinite set. And let
be the collection of all subsets
of
such that either
or
is finite. Prove that
is an algebra on
, but not a sigma algebra on
.
Proof
A related exercise :
Let
be an uncountable set
be the collection of all subsets
of
such that either
is countable or
is countable.
Is
a
algebra on
?
Answer
Hint to sketch
Definition (of Algebra on a set )
Let
be an arbitrary set. Then a collection
of subsets of
is called an algebra on
if and only if all the following are true :

for each
the set 
for every finite sequence
of elements each
the union 
Definition (of sigma algebra on a set )
Let
be an arbitrary set. Then a collection
of subsets of
is called an
algebra on
if and only if all the following are true :

for each
the set 
for every infinite sequence
of elements each
the union 
Note that : since
, so we conclude that every
algebra on a set
is also an algebra on
.
Let













Definition (of sigma algebra on a set )
Let














Note that : since




Let









Proof
Note that given a set
then
is infinite if and only if there is a one one function from
to
.
Thus, since given that
is an infinite set so, there must be a one one function from
to
. Let
be such a function.
Then, note that since each of the sets
(
) is finite so they all belong to
.
But, if
be the union
then
is an infinite set and also
is an infinite set. Thus,
is not an element of
.
So, (as per definitions),
is not a
algebra on
.
However, since finite union of finite sets is finite and by de Moivre’s law, we get that,
has to be an algebra on
.
This completes the proof.




Thus, since given that




Then, note that since each of the sets



But, if






So, (as per definitions),



However, since finite union of finite sets is finite and by de Moivre’s law, we get that,


This completes the proof.
A related exercise :
Let

a set
is called to be uncountable if and only if there is no one one function from
to 
. Let 







Is



Answer
Yes.
Hint to sketch
Countable union of countable sets is countable. Show this ! and then use this to conclude the result.
This post has been edited 2 times. Last edited by adityaguharoy, Feb 28, 2018, 4:16 PM
Broken Stick Problem with 3 pieces forming a triangle
by adityaguharoy, Jan 22, 2018, 11:50 AM
Let there be a stick of length
unit. We break it at two random points to get three pieces of the stick. Find the probability that these pieces form a triangle.
Proof Sketch Work

Proof Sketch Work
We can prove this by the following :
Observe that the distribution of the lengths of the three pieces is same as that of the distances between the sides of an equilateral triangle and a randomly chosen point inside it.
Thus, we evaluate area of the locus of such point
inside an equilateral triangle
such that the distances between
and
,
and
and
and
form a triangle.
Required probability
and then using some Euclidean geometry this area
.
Observe that the distribution of the lengths of the three pieces is same as that of the distances between the sides of an equilateral triangle and a randomly chosen point inside it.
Thus, we evaluate area of the locus of such point








Required probability


Ramsey problem: 3 people in 6 pairwise knowing or strangers
by adityaguharoy, Jan 19, 2018, 11:57 AM
Given any collection of
people. Prove that there is either a collection of
people among them such that they know each other pairwise, or there is a collection of
people among them who are pairwise strangers.
Proof
In the language of arrows



Proof
Let the
people be
.
Fix any one of them say
.
Consider the relation between
and the other
people.
By Pigeonhole Principle
must know at least
of them or be stranger to at least
of them.
Without loss of generality let
know
of them.
Let
know
.
Now, consider the relationship pairwise among
.
If, they are pairwise strangers then
is a set of
pairwise strangers.
And if they are not pairwise strangers then at least two of them must know each other.
Let,
and
know each other.
Then, observe that
is a set of
people who know each other pairwise.
Done !!


Fix any one of them say

Consider the relation between


By Pigeonhole Principle



Without loss of generality let


Let


Now, consider the relationship pairwise among

If, they are pairwise strangers then


And if they are not pairwise strangers then at least two of them must know each other.
Let,


Then, observe that


Done !!
In the language of arrows
We state this in the language of arrows as
.

Happy New Year 2018 wishes
by adityaguharoy, Jan 3, 2018, 3:24 AM
Happy New Year to everyone.
........so late...........
........so late...........
I actually forgot to wish you here on 31st December and 1st January and later thought I had already done so. All this because of me being busy with work now. But
seems interesting and fun.
Let us all enjoy the year.

Let us all enjoy the year.
Christmas is coming -- Advance wishes
by adityaguharoy, Dec 20, 2017, 4:15 AM
The oldest, shortest words — "yes" and "no" — are those which require the most thought.
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