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January 20, 2020

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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Connected Program at Cheenta

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Similar Problems

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