Supremum and Infimum: IIT JAM 2018 Problem 11

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Warm Yourself With An MCQ 

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Understand the problem

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  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
   
  [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.1"]IIT JAM 2018 Question 11[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.1" open="off"]Infinum and Supremum[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.1" open="off"]Real Analysis By S K Mapa[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

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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?  

 

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1"]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?[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1"]

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?

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.1"]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[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Look At The Knowledge Graph

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Fun Facts : The modulus function is not differentiable but lets look at the graph its continuous 

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Similar Problems

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TIFR 2014 Problem 13 Solution - Supremum of Function


TIFR 2014 Problem 13 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.

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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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