Continuous Function: TIFR 2017 Problem 10

Understand the problem

There exists a non-negative continuous function $f:[0,1] \rightarrow \mathbb{R}$ such that $\int_{0}^{1} f^{n} d x \rightarrow 2$ as $n \rightarrow \infty$ (a) TRUE (b) FALSE

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"Hint 1" : Rather I want to say that it is a comment on the question that here $f^{n} $ does not mean $f \circ f \circ . . . . . . \circ f $ ($n $ times). Here $ f^{n}= f \bullet f . . . . . \bullet f $ ($ n $ times). Now do you want to think again with this disclosure? "Hint 2" : Make two cases Case 1: $ 0<f(x) \leq 1\ \forall x \in [0,1]$ [Observe that $ f $ is non negative function] Case 2: $f(x) > 1 $ for some $x \in [0,1] $ Now prove for each case that $\int_{0}^{1} f^{n} dx \rightarrow 2 $ as $ n \to \infty $. So, by the last statement you have guessed the validity of the statement . It is a false statement!! In the next two cases, we will basically prove two cases. "Hint 3" : Case 1: If $f(x) \leq 1\ \forall\ x \in [0,1]$ then $f^{n}(x) \leq 1\ \forall\ x \in [0,1] $ So, $\int_{0}^{1} f^{n} dx \leq\ \int_{0}^{1} 1 dx = 1 $ $\rightarrow \lim_{n \to \infty} \int_{0}^{1} f^{n} dx \leq 1 $ So, $\int_{0}^{1} f^{n} dx \rightarrow 2 $ as $n \to \infty $ "Hint 4" : Case 2: Suppose $f(x) > 1 $ for some $y \in [0,1] $ Now, as $f(x) $ is continuous function We have $f(x) > 1+\epsilon\ \forall\ x \in (y - \delta, y + \delta) $ for some $\epsilon , \delta > 0 $ $ \rightarrow f^{n}(x) > (1+ \epsilon)^{n} $ then we have $\int_{0}^{1} f^{n}\ dx > \int_{y - \delta} ^{y + \delta} (1+ \epsilon)^{n}\ dx $ [If $ y \in [0,1]] $ or $\int_{1 - \delta}^{1} (1+ \epsilon)^{n}\ dx $ [If $ y=1 $] or $\int_{0}^{0 + \delta} (1+ \epsilon)^{n}\ dx $ [If $y=0 $] In either case , $ \int_{0}^{1} f^{n}\ dx > (1 + \epsilon)^{n}\ \delta \to \infty $ So, the statement is false.

Rank:IIT JAM 2018 PROBLEM 9

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Let's Try A Warm Up MCQ

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Consider the vector space V over $latex \mathbb{R} $ of the polynomial functions of degree less than or equal to 3 defined on $latex \mathbb{R} $. Let $latex T : V \longrightarrow V $ defined by $latex (Tf)(x) = f(x)-xf'(x). Then the rank of T is (a) 1 (b) 2 (c) 3 (d) 4 [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.1"][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="4.1" 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 Problem 9[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.1" open="off"]Vector Space [/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"]Abstract Algebra 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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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.1" hover_enabled="0"]Rank(T) = dim(Range(T)) There is one easy way to calculate rank of every linear transformation. Step 1: Take by basis $latex \beta= \{e_1,....,e_n\} $ of the vector space $latex V $. Step 2: Write down the matrix $latex [T]_{\beta}^{\beta} $ Step 3: Calculate the rank of the matrix $latex [T]_{\beta}^{\beta} $ Now can you follow these steps to get the answer? [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1" hover_enabled="0"]Standard Basis of $latex V $ is $latex \{1,x,x^{2},x^{3}\} = \beta$ $latex (Tf) (x) =f(x) - xf^{'}(x)$ $latex (T1) (x) = 1 - 0 = 1$; $latex (Tx) (x) = x - x = 0$; $latex (T x^{2}) (x)= x^{2} - 2x^{2} = -x^{2}$ ; $latex (T x^{3}) (x) = -2x^{3} $ So, $latex [T]_{\beta}^{\beta} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -2 \\ \end{pmatrix} $ Hence the rank is $latex 3 $[/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"]

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The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][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="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Series Convergence: IIT JAM 2018 Problem 12

Understand the problem

Let $a, b, c \in \mathbb{R} .$ Which of the following values of $a, b, c$ do NOT result in the convergence of the series $$\sum_{n=3}^{\infty} \frac{a^{n}}{n^{b}\left(\log _{e} n\right)^{c}} ?$$ (A) $|a|<1, b \in \mathbb{R}, c \in \mathbb{R}$ (B) $a=1, b>1, c \in \mathbb{R}$ (C) $a=1, b \geq 0, c<1$ (D) $a=-1, b \geq 0, c>0$

Start with hints

[Hint 1] One disclaimer: In this question you will see that for some option the series is clearly convergent and for some option it might be convergent and might not be. So the question wordings are not very clear. Now having that disclaimer, what we have to find is the options where we have the series might be or might not be convergent. I want to end this hint here to give you a bit more room to search. Look for Leibnitz rule for alternating series.[ In mathematics Leibnitz's test states that if ${u_n}$ be a monotone decreasing sequence of positive real numbers and lim $u_n = 0$ , then the alternating series $u_1 - u_2 + u_3 - u_4 + ...........$ is convergent.] [Hint 2] $\sum_{n=3}^{\infty} \frac{a^{n}}{n^{b} (log_e^{n})^{c}} $ Let us talk about option D first $a=-1, b \geq 0 , c<0 $ $\sum_{n=3}^{\infty} \frac{(-1)^{n}}{n^{b}(ln^{n})^{c}} $ where $\frac{1}{n^{b}(ln n)^{c}} \longrightarrow 0 $ as $n \longrightarrow \infty $ Hence the series is convergence in this case. So option D is rejected. Now look for the other option and see Cauchy condensation test. (For a non increasing sequence $f(n) $ of non-negative real numbers, the series $\sum_{n=1}^{\infty} f(n) $ converges if and only if the "condensed series" $\sum_{n=0}^{\infty} 2^{n} f(2^{n})$ converges. Moreover if they converge,the sum of the condensed series is no more than twice as large of the sum as original) and D’ Alembert’s test( Let $\sum u_n $ be a series of positive real numbers and let $lim \frac{u_n +1}{u_n} = l $ Then $\sum u_n $ is convergent if $l<1 $ , $\sum u_n $ is divergent if $ l>1 $. [Hint 3]Moving on to option c $ a=1 , b \geq 0 , c<1 $ $ \sum_{n=3}^{\infty} \frac{1}{n^{b} (log_{e}^{n})^{c}} := S $(say) Observe if we have $c=b=\frac{1}{3} $ thus $S = \sum_{n=3}^{\infty} \frac{1}{n^{\frac{1}{3}} (log_{e}^{n})^{\frac{1}{3}}} > \sum_{n=3}^{\infty} \frac{1}{n^{\frac{2}{3}}} \longrightarrow \infty $ So, by comparison test(Let $\sum u_n $ and $\sum v_n $ be two series of positive real numbers and there is a natural number m such that $u_n \leq kv_n $ for all $n \geq m,k $ being a fixed positive number. Then (i) $\sum u_n $ is convergent if $\sum v_n $ is convergent. we have S is divergent. (ii) $ \sum u_n $ is divergent if $\sum u_n $ is divergent.) Now the question is: Can we get some point where the series is convergent? The first bet would be making $ b>1 $ say $b=2 $ and make $x c $ smaller Let $ c= \frac{1}{2} $ Thus $S= \sum_{n=3}^{\infty} \frac{1}{n^{2} (log_{e}^{n})^{\frac{1}{2}}} $ Here $\sum_{n=3}^{\infty} \frac{1}{n^{2} (log n)^{\frac{1}{2}}} < \sum \frac{1}{n^{2}} < \infty $ So, $ S $ is convergent and c is one correct answer. Look for the others. [Hint 4]Option b $ a=1,b>1,c \in \mathbb{R}$ $S=\sum_{n=3}^{\infty} \frac{1}{n^{b}(ln n)^{c}}$ If $ c=2 $ clearly by comparison test $S $ will be convergent Now the question is that, can we find one example such that the series will be divergent? Observe that, if $ c \geq 0 $ then as $b>1 $ we will get that the series is convergent. What will happen if $c<0 $ Here Cauchy Condensation test comes into play Consider $a=2>1 $ thus $\sum_{n=3}^{\infty} \frac{2^{n}}{(2^{n})^{b} (ln 2^{n})^{c}} =\sum_{n=3}^{\infty} \frac{{2^{n}}^{1-b}} {n^{c} (ln 2)^{c}} =\frac{1}{(ln 2)^{c}} \sum_{n=3}^{\infty} \frac{1}{(2^{b-1})^{n} n^{c}}$ Now we have to use D’ Alembert’s Ratio test : Consider $ a_n = \frac{1}{(2^{b-1})^{n} n^{c}} $ Thus $\frac{a_{n+1}}{a_n} = \frac{(2^{b-1})^{n} n^{c}}{(2^{b-1})^{n+1} (n+1)^{c}} \longrightarrow \frac{1}{2^{b-1}} < 1 $ Hence the series is convergent and so the series is convergent for any value of $c $ and here the series is convergent always and that is why option b is not correct. [Hint 5]option a) $ |a| < 1, b \in \mathbb{R} , c \in \mathbb{R} $ Here if we consider $ a<0 , b<0 , c<0 $ The series is convergent by Leibnitz test . So, the question is whether we can find out some values of $a,b,c $ such that the series will be divergent. Consider $ a_n = \frac{a^{n}}{n^{b} (\log_e n )^{c}} $ $ \frac{a_{n+1}}{a_n} = \frac{a}{(1+ \frac{1}{n})^{b} (\frac{(log{n+1}}{log{n})}^{c}}$ Now we know that $ \frac{n+1}{n} \longrightarrow 1 $ We have to think about $ \frac{\log(n+1)}{\log n}$ Let us consider $ \lim_{x \to \infty} \frac{\log(x+1)}{\log(x)}=\lim_{x \to \infty} \frac{\frac{1}{x+1}}{\frac{1}{x}}=\lim_{x \to \infty} \frac{x}{x+1}=1$[using L'Hopital's rule which states that for function f and g which are differentiable on an open interval I except possibly at a point c contained in I if $ {\lim}_{x \to c} f(x) = {\lim}_{x \to c} g(x) = 0 $ or $ -\infty , +\infty , g'(x) \neq 0 $ for all x in I with $ x \neq c$ and ${\lim}_{x \to c} \frac{f'(x)}{g'(x)}$ exist then ${\lim}_{x \to c} \frac{f(x)}{g(x)} = {\lim} \frac{f'(x)}{g'(x)}$ ] So, $\frac{\log(n+1)}{\log n} \to 1$ And hence $ \lim{n \to \infty} |\frac{a_{n+1}}{a_n}|=|a|<1$. So,the series is convergent $\forall |a|<1,b,c \in \mathbb{R} $Hence c. is the only correct answer.

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Series Convergence: IIT JAM 2018 Problem 12

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Warm Yourself With An Mcq

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Let a, b, c $latex \in \mathbb{R} $ Which of the following values of a ,b, c do NOT result in the convergence of the series $latex \sum_{n=3}^{\infty} \frac{a^{n}}{n^{b} (log_en)^{c}} $ (a) $latex |a|<1 , b \in \mathbb{R} , c \in \mathbb{R} $ (b) $latex a=1 , b>1 , c \in \mathbb{R} $ (c) $latex a=1 , b \leq 1 , c<1 $ (d) $latex a=-1 , b \geq , c>0 $

[/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="4.1" 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 Problem 12 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.1" open="off"]Convergence of a seris [/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="4.1" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#8300e9" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.1" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="4.1"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.1"] One disclaimer: In this question you will see that for some option the series is clearly convergent and for some option it might be convergent and might not be. So the question wordings are not very clear. Now having that disclaimer, what we have to find is the options where we have the series might be or might not be convergent. I want to end this hint here to give you a bit more room to search. Look for Leibnitz rule for alternating series.[ In mathematics Leibnitz's test states that if $latex {u_n}$ be a monotone decreasing sequence of positive real numbers and lim $latex u_n = 0$ , then the alternating series $latex u_1 - u_2 + u_3 - u_4 + ...........$ is convergent.][/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1"] $latex \sum_{n=3}^{\infty} \frac{a^{n}}{n^{b} (log_e^{n})^{c}} $ Let us talk about option D first $latex a=-1, b \geq 0 , c<0 $ $latex \sum_{n=3}^{\infty} \frac{(-1)^{n}}{n^{b}(ln^{n})^{c}} $ where $latex \frac{1}{n^{b}(ln n)^{c}} \longrightarrow 0 $ as $latex n \longrightarrow \infty $ Hence the series is convergence in this case. So option D is rejected. Now look for the other option and see Cauchy condensation test. (For a non increasing sequence $latex f(n) $ of non-negative real numbers, the series $latex \sum_{n=1}^{\infty} f(n) $ converges if and only if the "condensed series" $latex \sum_{n=0}^{\infty} 2^{n} f(2^{n})$ converges.Moreover if they converge,the sum of the condensed series is no more than twice as large of the sum as original) and D’ Alembert’s test( Let $latex \sum u_n $ be a series of positive real numbers and let $latex lim \frac{u_n +1}{u_n} = l $ Then $latex \sum u_n $ is convegent if $latex l<1 $ , $latex \sum u_n $ is divergent if $latex l>1 $.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1"]Moving on to option c $latex a=1 , b \geq 0 , c<1 $ $latex \sum_{n=3}^{\infty} \frac{1}{n^{b} (log_{e}^{n})^{c}} := S $(say) Observe if we have $latex c=b=\frac{1}{3} $ thus $latex S = \sum_{n=3}^{\infty} \frac{1}{n^{\frac{1}{3}} (log_{e}^{n})^{\frac{1}{3}}} > \sum_{n=3}^{\infty} \frac{1}{n^{\frac{2}{3}}} \longrightarrow \infty $ So, by comparison test(Let $latex \sum u_n $ and $latex \sum v_n $ be two series of positive real numbers and there is a natural number m such that $latex u_n \leq kv_n $ for all $latex n \geq m,k $ being a fixed positive number. Then (i) $latex \sum u_n $ is convergent if $latex \sum v_n $ is convergent. we have S is divergent. (ii) $latex \sum u_n $ is divergent if $latex \sum u_n $ is divergent.) Now the question is: Can we get some point where the series is convergent? The first bet would be making $latex b>1 $ say $latex b=2 $ and make $latex c $ smaller Let $latex c= \frac{1}{2} $ Thus $latex S= \sum_{n=3}^{\infty} \frac{1}{n^{2} (log_{e}^{n})^{\frac{1}{2}}} $ Here $latex \sum_{n=3}^{\infty} \frac{1}{n^{2} (log n)^{\frac{1}{2}}} < \sum \frac{1}{n^{2}} < \infty $ So, $latex S $ is convergent and c is one correct answer. Look for the others.[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.1"]Option b $latex a=1,b>1,c \in \mathbb{R}$ $latex S=\sum_{n=3}^{\infty} \frac{1}{n^{b}(ln n)^{c}}$ If $latex c=2 $ clearly by comparison test $latex S $ will be convergent Now the question is that, can we find one example such that the series will be divergent? Observe that, if $latex c \geq 0 $ then as $latex b>1 $ we will get that the series is convergent. What will happen if $latex c<0 $ Here Cauchy Condensation test comes into play Consider $latex a=2>1 $ thus $latex \sum_{n=3}^{\infty} \frac{2^{n}}{(2^{n})^{b} (ln 2^{n})^{c}} =\sum_{n=3}^{\infty} \frac{{2^{n}}^{1-b}} {n^{c} (ln 2)^{c}} =\frac{1}{(ln 2)^{c}} \sum_{n=3}^{\infty} \frac{1}{(2^{b-1})^{n} n^{c}}$ Now we have to use D’ Alembert’s Ratio test : Consider $latex a_n = \frac{1}{(2^{b-1})^{n} n^{c}} $ Thus $latex \frac{a_{n+1}}{a_n} = \frac{(2^{b-1})^{n} n^{c}}{(2^{b-1})^{n+1} (n+1)^{c}} \longrightarrow \frac{1}{2^{b-1}} < 1 $ Hence the series is convergent and so the series is convergent for any value of $latex c $ and here the series is convergent always and that is why option b is not correct.[/et_pb_tab][et_pb_tab title="Hint 5" _builder_version="4.1"]option a) $latex |a| < 1, b \in \mathbb{R} , c \in \mathbb{R} $ Here if we consider $latex a<0 , b<0 , c<0 $ The series is convergent by Leibnitz test . So, the question is whether we can find out some values of $latex a,b,c $ such that the series will be divergent. Consider $latex a_n = \frac{a^{n}}{n^{b} (\log_e n )^{c}} $ $latex \frac{a_{n+1}}{a_n} = \frac{a}{(1+ \frac{1}{n})^{b} (\frac{(log{n+1}}{log{n})}^{c}}$ Now we know that $latex \frac{n+1}{n} \longrightarrow 1 $ We have to think about $latex \frac{\log(n+1)}{\log n}$ Let us consider $latex \lim_{x \to \infty} \frac{\log(x+1)}{\log(x)}=\lim_{x \to \infty} \frac{\frac{1}{x+1}}{\frac{1}{x}}=\lim_{x \to \infty} \frac{x}{x+1}=1$[using L'Hopital's rule which states that for function f and g which are differentiable on an open interval I except possibly at a point c contained in I if $latex {\lim}_{x \to c} f(x) = {\lim}_{x \to c} g(x) = 0 $ or $latex -\infty , +\infty , g'(x) \neq 0 $ for all x in I with $latex x \neq c$ and $latex {\lim}_{x \to c} \frac{f'(x)}{g'(x)}$ exist then $latex {\lim}_{x \to c} \frac{f(x)}{g(x)} = {\lim} \frac{f'(x)}{g'(x)}$ ] So, $latex \frac{\log(n+1)}{\log n} \to 1$ And hence $latex \lim{n \to \infty} |\frac{a_{n+1}}{a_n}|=|a|<1$. So,the series is convergent $latex \forall |a|<1,b,c \in \mathbb{R} $Hence c. is the only correct answer.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="4.1" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#e06100" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" hover_enabled="0" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Take A Look Into This Knowledge Graph

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[/et_pb_code][et_pb_text _builder_version="4.1" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0038f0" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][et_pb_text _builder_version="4.1" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#e09900" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Sum Of Series: IIT JAM 2018 Problem 13

Understand the problem

Let $a_{n}=n+\frac{1}{n}, n \in \mathbb{N} .$ Then the sum of the series $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{a_{n+1}}{n !}$ is - (A) $e^{-1}-1$ (B) $e^{-1}$ (C) $1-e^{-1}$ (D) $1+e^{-1}$

Start with hints

Do you really need a hint? Try it first!

[Hint 1] Consider $a_n = n + \frac{1}{n} , n \in \mathbb{N} $ We have to use $e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!}+.....$ Specifically $e^{1} = 1+ \frac{1}{1!} + \frac{1}{2!}+....$ And $ e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + .......$ Do you want to play with it. [Hint 2]$ \sum_{n=1}^{\infty} (-1)^{n+1} \frac{a_{n+1}}{n!} = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n+1 + \frac{1}{n+1}}{n!} $ = $ \sum_{n=1}^{\infty}[ (-1)^{n+1} \frac{1}{(n-1)!} + (-1)^{n+1} \frac{1}{n!} + (-1)^{n+1} \frac{1}{(n+1)!}]$ Now we will be breaking it term by term for the ease of calculation. Can you do it from here? [Hint 3] $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{a_{n+1}}{n!} $ = $[1-\frac{1}{1!} + \frac{1}{2!} - ......] + [\frac{1}{1!} - \frac{1}{2!} + \frac{1}{3!} - .....] + [\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} -.......]$ = $e^{-1} + [1-e^{-1}] + [e^{-1} + 1 - 1] $ = $e^{-1} + 1 $ So option (D) is our required answer.

Let's have a look into the graphs

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Directional Derivative : IIT JAM 2018 Problem 42

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][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"]

Please give a warm-up quiz

[/et_pb_text][et_pb_code _builder_version="4.1"][/et_pb_code][et_pb_text _builder_version="4.1" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Let $latex \phi (x,y,z) = 3y^2+3yz $ for $latex (x,y,z) \in \mathbb{R}^{3}$ Then the absolute value of the directional derivative of $latex \phi $ in the direction of the line $latex \frac{x-1}{2} = \frac{y-2}{-1}= \frac{z}{-2} $ , at the point $latex (1,-2,1)$ is _____

[/et_pb_code][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.1" 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[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.1" open="off"]Directional Derivative [/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"]

Generalized Directional Derivatives By Pastor Karel

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Start with hints

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.1" hover_enabled="0"] The rate of change of $latex f(x,y,z) $ in the direction of the unit vector $latex \vec u = <a,b,c>$ is called the directional derivative and is denoted by $latex D_{\vec u}f(x,y,z) $. The defination of directional derivative is $latex D_{\vec u}f(x,y,z) = \lim h \to 0 \frac{f(x+ah,y+bh,z+ch)-f(x,y)}{h} $. Here if you consider $latex u$ to be $latex (1,0,0), (0,1,0)$ and $latex (0,0,1)$ then we will get $latex D_{\vec u}f(x,y,z)$ to be $latex f_x$, $latex f_y$ and $latex f_z$ respectively. If you observe closely, this is nothing but differentiating the function $latex f$ w.r.t $latex x,y$ and $latex z$ keeping all the other thing constant.

Now $latex \nabla (f)=(f_x,f_y,f_z)$ by the above definition can you try to solve the problem ???????

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

The points on the straight line $latex \frac{x-1}{2} = \frac{y-2}{-1}= \frac{z}{-2} = k $ $latex \implies x=1+2k , y=2-k , z=-2k $ So any point on the straight line would be of the form $latex (x,y,z)= (1,2,0) + (2,-1,-2)k $ Here the direction of the straight line would be $latex (2,-1,-2) $

Now we have to consider the unit vector in that direction and so the unit vector would be $latex u= \frac{(2,-1,-2)}{\sqrt{4+1+4}} = (\frac{-2}{3},\frac{-1}{3},\frac{-2}{3}) $Here the absolute value of the directional derivative would be $latex \nabla \phi . u $ . Would you like to calculate this dot product?

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

$latex \nabla \phi = (\phi_{x},\phi_{y},\phi_{z}) $ i.e the co-ordinates involving partial derivative of $latex \phi $ . So, $latex \nabla \phi = (0, 6y+3z , 3y)|_{(1,-2,1)} = (0,-9,-6)$

Then $latex \nabla \phi . u= \frac{9}{3} + \frac{6 \times 2}{3} = 3+4 = 7 $ (Ans) .

[/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"]

A graphical view point

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Take A Look At This Knowledge Graph

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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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Sequences & Subsequences : IIT 2018 Problem 10

What are we learning?

Sequences, Subsequences are the key features in the field of real analysis. We will see how to imply these concepts in our problem

Understand the problem

Let $s_n$ = 1+$\frac{1}{1!}$+$\frac{1}{2!}$+........+$\frac{1}{n!}$ for n $\in$ $\mathbb{N}$ Then which of the following is TRUE for the sequence $\{s_{n}\}^\infty_{n=1}$: (a) $\{s_{n}\}^\infty_{n=1}$ converges in $(\mathbb{Q})$ . (b) $\{s_{n}\}^\infty_{n=1}$ is a Cauchy sequence but does not converges to $(\mathbb{Q})$. (c) The subsequence $\{s_{k^n}\}^\infty_{n=1}$ is convergent in $(\mathbb{R})$ when k is a even natural number. (d) $\{s_{n}\}^\infty_{n=1}$ is not a Cauchy sequence. Difficulty Level Easy Suggested Book

Calculus: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability – Vol 2

Tom M. Apostol

Start with hints

I am going to give you 3 clues in the beginning you try to work out using them. Then I will elaborate this clues in the following hints (I) Every convergent sequence is a Cauchy sequence (II)Every subsequence of a convergent sequence is convergent (III)Consider then term 1+$\frac{1}{1!}$+$\frac{1}{2!}$+........+$\frac{1}{n!}$ Does this remind you any well known series?

I wil start with (III) consider $e^x$=1+$\frac{x}{1!}$+$\frac{x^2}{2!}$+........+$\frac{x^n}{n!}$ Isn't the seris that we have to , is the value at x=1. Hence the given series$\rightarrow$ e $\in$ $\mathbb{R}$ \ $\mathbb{Q}$

So option (a) is incorrect.

Every subsequence of a convergent sequence is convergent so $\{s_{k^n}\}^\infty_{n=1}$ is convergent not only for even k, but for any $k \in \Bbb N$. So option (c) is incorrect.

Every convergent sequence is a Cauchy sequence so option (d) is incorrect and $e \in$ $\mathbb{R}$ so the given subsequence is convergent in $\mathbb{R}$. So only option (b) is correct.

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Cyclic Groups & Subgroups : IIT 2018 Problem 1

Understand the problem

Which one of the following is TRUE? (A) $\Bbb Z_n$ is cyclic if and only if n is prime
(B) Every proper subgroup of $\Bbb Z_n$ is cyclic
(C) Every proper subgroup of $S_4$ is cyclic
(D) If every proper subgroup of a group is cyclic, then the group is cyclic.

Start with hints

Hint 1:

We will solve this question by the method of elimination. Observe that if n is prime then $\mathbb{Z}_n$ is obviously cyclic as any of the subgroup <a> has order either 1 or n by Lagrange's theorem. Now if the order is 1 then a=id. So choose a($\neq$e) $\in \mathbb{Z}_n$ then |<a>|=n and <a> $\subseteq$ $\mathbb{Z}_n$ $\Rightarrow$ <a>= $\mathbb{Z}_n$. The problem will occur with the converse see $\mathbb{Z}_6$ is cyclic but 6 is not prime. In general $\mathbb{Z}_n$ = <$\overline{1}$> is always cyclic no matter what n is!! so option (A) is false. Can you rule out option (C)

Hint 2:

Consider option (C) every proper subgroup of $S_4$ is cyclic. Consider { e , (12)(34) , (13)(24) , (14)(23) } = G Observe that this is a subgroup and |G|=4. Moreover o(g)=2 $\forall$ g($\neq$e) $\in$ G So G is not cyclic. Hence option (C) is not correct. Can you rule out option (D)?

Hint 3:

Consider $\mathbb{Z}_2$*$\mathbb{Z}_2$ which is also known as Klein's 4 group then it is not cyclic but all of it's proper subgroups are {0}*$\mathbb{Z}_2$ , $\mathbb{Z}_2$*{0} and {0}*{0} which are cyclic. Hence we can rule out option (D) as well.

Hint 4:

So option (B) is correct. Now let prove that H $\leq$ $\mathbb{Z}_n$ = {$\overline{0}$,$\overline{1}$,.....,$\overline{n-1}$}. By well ordering principle H has a minimal non zero element 'm'. Claim: H=<m> clearly <m> $\subset$ H. For any r $\in$ H by Euclid's algorithm we have r=km+d where 0 $\leq$ d < m which $\Rightarrow$ d=r-km $\in$ H If d $\neq$ 0 then d<m which is a contradiction So, d=0 $\Rightarrow$ r=km $\Rightarrow$ H=<m> and we are done

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Acute angles between surfaces: IIT JAM 2018 Qn 6

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Warm yourself up with an MCQ

[/et_pb_text][et_pb_code _builder_version="4.0.9"][/et_pb_code][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_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="4.0.9" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]In $latex \Bbb R^3$ the cosine of acute angle between the surfaces $latex x^2+y^2+z^2-9=0$ and $latex z-x^2-y^2+3=0$ at the point $latex (2,1,2)$ is

$latex \frac{8}{5\sqrt{21}}$
$latex \frac{10}{5\sqrt{21}}$
$latex \frac{8}{3\sqrt{21}}$
$latex \frac{10}{3\sqrt{21}}$

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" custom_padding="|0px||||"][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="4.0.9" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="4.0.9" hover_enabled="0"]IIT JAM 2018 Qn no 6[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.0.9" hover_enabled="0" open="off"]Multivable calculus[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.0.9" hover_enabled="0" open="on"]Easy [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0.9" hover_enabled="0" open="off"]

Calculus: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability – Vol 2

Tom M. Apostol

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.0.9" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0.9"]If we are asked to give the angle between two lines then it is very easy to calculate but our forehead will get skinned whenever we will be asked to find out the acute angle between two lines and even worse if we are asked to find the angle between two surfaces. Surprisingly it is not very hard if think stepwise. Observe when you are asked to find out the angle between two lines you calculate it in terms slope. So basically you are firing putting the gun on someone else's shoulder. Here the question is to find that shoulder when it comes in finding the angle between two curves. Observe from the conception of the intersection of two curves that the tangent line of those curves also intersects and we have their corresponding slopes. Bingo! why not calculating the acute angle of the tangent lines and call them the angle between two curves. Now can you think how to calculate the acute angle between to surfaces?[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0.9"]The acute angle between two surfaces would be the acute angle between their tangent plane. You can stop here and try to do the problem by your own otherwise continue... The main idea of finding tangent planes revolves around finding gradient of the corresponding surfaces. (For more info see question no 5). Can you calculate the gradient of the surfaces $latex x^2+y^2+z^2-9$ and $latex z-x^2-y^2+3$ at $latex (2,1,2)$?[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0.9"]The gradient of the surfaces $latex f=x^2+y^2+z^2-9$ and $latex g=z-x^2-y^2+3$ at $latex (2,1,2)$ are $latex n_1=f_xi +f_yj+f_zk$ and $latex n_2=g_xi +g_yj+g_zk$ at $latex (2,1,2)$ which is $latex n_1=2xi+2yj+2zk=4i+2j+4k$ and $latex n_2=-2xi-2yj+k=-4i-2j+k$. Now given these two gradients, can you find out the angle between them?[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0.9"]$latex n_1=2xi+2yj+2zk=4i+2j+4k$ and $latex n_2=-2xi-2yj+k=-4i-2j+k$.

This follows the cosine angles between two gradient is $latex cos \theta=|\frac{n_1.n_2}{|n_1||n_2|}|=|\frac{-16-4+4}{\sqrt{36 \times 21}}|=\frac{8}{3\sqrt{21}}$

Explanation of hints with graph

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