Supremum and Infimum: IIT JAM 2018 Problem 11

Understand the problem

$ a_n=\begin{cases} 2+\frac{\{-1\}^{\frac{n-1}{2}}}{n}, & \text{if n is odd}\\ 1+ \frac{1}{2^n}, & \text{if n is even} \end{cases}$ Which of the following is true? (a) sup {$a_n|n \in \mathbb{N}$}=3 and inf {$a_n|n \in \mathbb{N}$}=1 (b) lim inf ($a_n$)=lim sup ($a_n$)=$\frac{3}{2}$ (c) sup {$a_n|n \in \mathbb{N}$}=2 and inf {$a_n|n \in \mathbb{N}$}=1 (d) lim inf ($a_n$)= 1 lim sup ($a_n$)=3

Start with hints

Hint 1: $a_n=\begin{cases} 2+\frac{\{-1\}^{\frac{n-1}{2}}}{n}, & \text{if n is odd}\\ 1+ \frac{1}{2^n}, & \text{if n is even} \end{cases}$ Now the limit points of this set are those points which the set does not attain.So, they might be the sup and inf which are not attained by this set. Basically sup($a_n$)= max{ limit points, $a_n$ | n $\in$ $\mathbb{N}$} Limit points are $2,1$ and $a_1= 2+1=3, a_3= 2- \frac{1}{3} ; a_5= 2+\frac{1}{5} $ $a_0= 1+1=2 , a_2= 1+ \frac{1}{4} , a_3= 1+\frac{1}{8} $ Now you can calculate the supremum?

Hint 2: From the observation of Hint 2 we have sup $a_n$= max $\{2,1,3,2\}=3 $ Similarly, inf $a_n$= min$\{$ limit points, $a_n | n \in \mathbb{N}\}$ Can you calculate that by yourself? Hint 3: inf $a_n$= min {2,1,2 -$\frac{1}{3}$}=1 So, option A is correct. Now there is another question regarding lim sup and lim inf. We can observe that we have mainly $3$ subsequences , corresponding to $ n$ is even; $n=2k$ $n$= $4k+1$ $n=4k+3$

Can you calculate the corresponding subsequences and their limits?

Hint 4: For $n=2k$ we have $a_{2k}=1+ \frac{1}{2^{ek}} \longrightarrow 1 $ ask For $a_{4k+1}= 2+ \frac{1}{4k+1} \longrightarrow 2$ ask $a_{4k+3}= 2-\frac{1}{4k+3} \longrightarrow 2$ ask So, lim sup $a_n$=max$\{1,2\}=2$ Lim inf $a_n$=min$\{1,2\}=1$ Therefore, Option C is also correct

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Problem:

Let (S) be the set of all tuples ((x,y)) with (x,y) non-negative real numbers satisfying (x+y=2n) ,for a fixed (n\in\mathbb{N}). Then the supremum value of (x^2y^2(x^2+y^2)) on the set (S) is:

A. (3n^6)

B. (2n^6)

C. (4n^6)

D. (n^6)

Discussion:

Write the expression in terms of (x) only by substituting (y=2n-x).

Let (f(x)=x^2(2n-x)^2(x^2+(2n-x)^2)). Here, (x\in[0,2n]).

Note that for (x=0) or (x=2n) the function (f(x)=0). Also, (f) is positive everywhere else on the interval.

So we want to find (sup{f(x)|x\in (0,2n)}). Note that it exists because the interval is compact and (f) is continuous.

One can straightaway take derivative and compute, or one can do the following:

Take log. Note that now we are only working on the open interval ((0,2n)).

(log(f(x))=2logx+2log(2n-x)+log(x^2+(2n-x)^2))

Now take derivative.

(\frac{f'(x)}{f(x)}=\frac{2}{x}+\frac{-2}{2n-x}+\frac{2x-2(2n-x)}{x^2+(2n-x)^2})

(=\frac{4n-4x}{x(2n-x)}+\frac{4n-4x}{x^2+(2n-x)^2})

(=4(n-x)[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]).

Now, (f'(x)=0) if and only if (4(n-x)[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]=0).

Note that ([\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]>0) for all (x\in (0,2n)).

Therefore, (f'(x)=0) if and only if (x=n). If now, (x) is slightly bigger than (n) then (\frac{f'(x)}{f(x)}<0) and since (f(x)>0) we have (f'(x)<0) in that case. And if (x) is slightly smaller than (n) then (f'(x)>0).

This proves that indeed the point (x=n) is a point of maxima.

Therefore, the supremum value is (f(n)=2n^6). So the correct answer is option B.

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