Some real analysis exercises on:
– Cauchy sequences
– Limits and delta epsilon
– Continuos and uniformly continuous functions
– Differentiation functions
– Zeno's paradox and Dirichlets Test
– Multivariate limits, pointwise and uniform convergence
Cauchy sequences
$$ { \text{ Prove } \left \{ cos \left( \frac {1}{n} \right) \right \} \text{ is a Cauchy Sequence } }$$Cauchy definition
$$ { \text{ Recall, } X_n \text{ is Cauchy if } \forall \epsilon \gt 0 , \exists N \in \mathbb{Z}^+ \text{ s.t }\forall n,m \gt N , |a_n-a_m| \lt \epsilon }$$Scratch work
$$ { \epsilon \gt 0 , \left| cos \left( \frac {1}{n} \right) - cos \left( \frac {1}{m} \right) \right| }$$ $$ { \text{ * } cosu - cosv = -2sin \left( \frac {u+v}{2} \right) sin \left( \frac {u-v}{2} \right) }$$ $$ { \left| -2 sin \left( \frac { \frac {1}{n}+\frac {1}{m}}{2} \right) sin \left( \frac { \frac {1}{n}+\frac {1}{m}}{2} \right) \right| }$$ $$ { 2 \left| sin \left( \frac { \frac {1}{n}+\frac {1}{m}}{2} \right) \right| . \left| sin \left( \frac { \frac {1}{n}+\frac {1}{m}}{2} \right) \right| }$$ $$ { 2 \left| sin \left( \frac { \frac {1}{n}+\frac {1}{m}}{2} \right) \right| . 1 = \left| \frac {1}{n}+\frac {1}{m} \right| }$$ $$ { \text{*} \left| a+b \right| \leq \left| a \right| +\left| b \right| }$$ $$ { \leq \left| \frac {1}{n} \right|+ \left|\frac {1}{m} \right| = \frac {1}{n}+\frac {1}{m} \lt \epsilon \Rightarrow \frac {1}{n} \lt \frac {\epsilon}{2} }$$ $$ { 2 \lt n . \epsilon \Rightarrow { \frac {2}{3} } \lt n \text{ or } n \gt \frac {2}{\epsilon}. N \gt \frac {2}{\epsilon} }$$Proof
$$ { \text{ Let } \epsilon \gt 0. \text{ Choose } N \ \gt { \frac {2}{3} } \text{, Then } \forall n,m \gt N , \left| cos \left( \frac {1}{n} \right) - cos \left( \frac {1}{m} \right) \right|}$$ $$ { \left| -2 sin \left( \frac { \frac {1}{n}+\frac {1}{m}}{2} \right) sin \left( \frac { \frac {1}{n}+\frac {1}{m}}{2} \right) \right| }$$ $$ { \leq \left| \frac {1}{n} + \frac {1}{m} \right| \leq \frac {1}{n}+\frac {1}{m} \text{, Since } n \gt N \gt \frac {2}{\epsilon} \text{, so }n \gt \frac {2}{\epsilon} }$$ $$ { \frac {\epsilon}{2} \gt \frac {1}{n} \text{, so } \frac {1}{n} \lt \frac {\epsilon}{2} \text{. Likewise } m \gt \frac {2}{\epsilon} \text{, so } \frac {1}{m} \lt \frac {\epsilon}{2} }$$ $$ {\text{ Then} \left| cos \left( \frac {1}{n} \right) - cos \left( \frac {1}{m} \right) \right| \leq \frac {1}{n}+\frac {1}{m} \lt \frac {\epsilon}{2} + \frac {\epsilon}{2} = \epsilon \text{ ∎ } }$$Limits and delta epsilon
$$ { \text{ Prove } \lim_{x \to 2} (x-2)^7 . cos \left( \frac {2}{x-2} \right) = 0 }$$Limit definition
$$ { \text{ Recall, } \lim_{x \to c} f(x) = L, \text{ means } }$$ $$ { \forall \epsilon \gt 0 , \exists \delta \gt 0 \text{ s.t }\forall x \in \mathbb{R} \text{ with } 0 \lt |x-c| \lt \delta }$$ $$ { \text{ we have } \left| f(x)-L \right| \lt \epsilon }$$Scratch work
$$ { \epsilon \gt 0 \text{ we need } \delta }$$ $$ { 0 \lt |x-2| \lt \delta \Rightarrow \left| (x-2)^7 . cos \left( \frac {2}{x-2} \right) \right| =0 }$$ $$ { |(x-2)|^7 . \left| cos \left( \frac {2}{x-2} \right) \right| \leq \left|(x-2)\right| ^7 . 1 \lt \epsilon }$$ $$ { \lt \delta ^7 \lt \epsilon }$$ $$ { \delta = \sqrt[7]\epsilon }$$Proof
$$ { \epsilon \gt 0. \text{ Choose } \delta = \sqrt[7]\epsilon \text{ .Then } \forall x \in \mathbb{R} \text{with} }$$ $$ { 0 \lt |x-2| \lt \delta \text{, we have } }$$ $$ { \left| (x-2)^7 . cos \left( \frac {2}{x-2} \right) - 0 \right| = | \left(x-2)\right|^7 . \left|cos \left( \frac {2}{x-2} \right)\right| }$$ $$ { \leq |(x-2)|^7 \lt \delta ^7 = \left( \sqrt[7]\epsilon \right)^7 }$$ $$ { = \epsilon \text{ ∎ } }$$Continuous functions
$$ { \text{ Consider f : } \mathbb{R} {\rightarrow} \mathbb{R} \text{ defined by } }$$ $$ { f(x) = \begin{cases} 1, & \text{if $x$ is rational} \\ 0, & \text{if $x$ is irrational} \end{cases} }$$ $$ { \text{ Prove f is not continuous anywhere or is a Dirichlet function }}$$Proof
$$ { \text{ Spse. f is continuous at a.} }$$ $$ { \text{ Set } \epsilon = 1 \text{. Then } \exists \delta \gt 0 \text{ s.t }\forall x \in \mathbb{R} \text{ with } |x-a| \lt \delta \text{, we have } \left| f(x) - f(a)\right| \lt 1. }$$ $$ { \text{If a is rational then f(a)} = 1. } $$ $$ { \text{ By the density of the irrationals, pick } x \in \left( a - f\ , a + f\ \right) \text{, so f(x) = 0,} }$$ $$ { \text{ hence } \left| a - f\ , a + f\ \right| \lt 1 }$$ $$ { \left| 0 - 1 \right| \lt 1 }$$ $$ { \left| - 1 \right| \lt 1 }$$ $$ { - 1 \lt 1 \text{ } \Rightarrow \Leftarrow }$$ $$ { \text{If a is irrational then f(a)} = 0. }$$ $$ { \text{ By the density of the rationals pick } x \in \left( a - f\ , a + f\ \right) \text{, so f(x) = 1.} }$$ $$ { \text{ Hence } \left| a - f\ , a + f\ \right| \lt 1 }$$ $$ { \left| 1 - o \right| \lt 1 }$$ $$ { \left| 1 \right| \lt 1 }$$ $$ { 1 \lt 1 \text{ } \Rightarrow \Leftarrow }$$Uniformly continuous functions
$$ { \text{ Prove } f(x) = x^2 \text{ is uniformly continuous on } (0,1) }$$Recall
$$ { \text{ f is uni cont. on } (0,1) \text{ if } \forall \epsilon \gt 0 , \exists \delta \gt 0 \text{ s.t }}$$ $$ { \forall x,y \in (0,1) \text{ with } |x-y| \lt \delta \Rightarrow |f(x)-f(y)| \lt \epsilon }$$Scratch work
$$ { \text{ we need } \delta \text{ s.t }|x^2-y^2| \lt \epsilon }$$ $$ { |x^2-y^2| = |x-y| |x+y| }$$ $$ { \lt |x-y| . 2 \lt 2 \delta \lt \epsilon }$$ $$ { \delta = \epsilon/2 }$$Proof
$$ { \text{ Let } \epsilon \gt 0. \text{ Choose } \delta = \epsilon/2 }$$ $$ { \text{ Then } \forall x,y \in (0,1) \text{ with } |x-y| \lt \delta }$$ $$ { \text{ We have } |x^2-y^2| = |x-y| |x+y| }$$ $$ { \lt 2 \delta = 2 \delta = 2 (\epsilon/2) }$$ $$ { = \epsilon \text{ ∎ } }$$Differentiable functions
$$ { \text{ Prove that if f is differentiable at x=c then it is continuous at x = c } } $$Proof
$$ { \text{ Spse. f is diff. at }x =c \text{ . This means } \lim_{x\to c} \frac {f(x)-f(c)}{x-c} }$$ $$ { \text{NTS. } \lim_{x\to c} f(x) = f(c) \text{ or, } \lim_{x\to c} \left( f(x) - f(c) \right) = 0 } $$ $$ { \text{Then. } \lim_{x\to c} \left( f(x) - f(c)\right) = \left( \lim_{x\to c} \frac {f(x)-f(c)}{x-c} \right) .(x-c) = f'(c).0 =0 } $$ $$ { \lim_{x\to c}f(x) = \lim_{x\to c}(f(x) - f(c) + f(c) ) = \lim_{x\to c}(f(x) - f(c)) + \lim_{x\to c}f(c) } $$ $$ { 0 + f(c) = f(c) } $$ $$ { \text{ So, } \lim_{x\to c}f(x) = f(c) }$$ $$ { \dot{.\hspace{.075in}.}\hspace{.1in} \text{ f is cont. at } x = c \text{ ∎ } } $$ $$ { \text{ Shows where a function is differentiable at a point then it continuous there. } }$$Dirichlet Test
$$ { \text{ Prove } \sum\limits_{n=1}^{\infty} \frac {sin(n)}{n} \text{ converges.}} $$ $$ { \text{ Recall, Dirichlet test } } $$ $$ { \text{ 1. } A_k \searrow 0 \text{ as } k\to\infty }$$ $$ { \text{ 2. } | \sum\limits_{k=1}^{m} | \leq m \text{ for all positive integers m. Have m is constant. } }$$ $$ { \text{ Dirichlet test tells us the infinite sum of the product of } a_k \text{and } b_k \text{ converges, }\Big| \sum\limits_{k=1}^{\infty} a_k b_k \Big| \text{ conv.}}$$Proof
$$ { \text{ set } a_k=1/k \text{ and } b_k=sin(k) \text{ , Note }a_k=1/k \searrow 0 \text{ as } k\to\infty }$$ $$ { \text{ Consider } \sum\limits_{k=1}^{m} sin(k) }$$ $$ { \text{ Recall trig identity, } 2sin(a)sinb(b)= cos(a-b)-cos(a+b) }$$ $$ { \text{ Then } 2sin(1) \sum\limits_{k=1}^{m} sin(k) = \sum\limits_{k=1}^{m}2sin(1)sin(k) }$$ $$ { = \sum\limits_{k=1}^{m} \Big( cos(1-k)-cos(1+k) \Big) \text{ , As cosine is even } cos(1-k)=cos(-1(1-k)) = cosk(k-1) }$$ $$ { = \sum\limits_{k=1}^{m} \Big( cos(k-1)-cos(k+1) \Big) \text{ , inputting k } }$$ $$ { = (cos0- \require{cancel}\cancel{cos2})+(cos1-\cancel{cos3})+(\cancel{cos2}-\cancel{cos4})+(\cancel{cos3}-\cancel{cos5})+(\cancel{cos4}-\cancel{cos6})+\cancel{...}}$$ $$ { +\Big(\cancel{cos(m-2)}-cos(m)\Big)+\Big(\cancel{cos(m-1)}-cos(m+1)\Big) }$$ $$ { = 1 +cos1-cos(m)-cos(m+1) }$$ $$ { \Big| \sum\limits_{k=1}^{m} \Big| =\Big| \frac {1 +cos(1)-cos(m)-cos(m+1)}{2sin(1)} \Big| }$$ $$ { \leq \Big| \frac {4}{2sin(1)} \Big| = \frac {2}{2sin(1)} = M }$$ $$ { \text{By Dirichlet's, } \sum\limits_{n=1}^{\infty} \frac {sin(n)}{n} \text{ conv. ∎} }$$Sequence uniform convergence
$$ { \text{ Prove the sequence }fn(x)= \frac {x}{1+nx^2} \text{ converges uniformly on } \mathbb{R} \text{ to 0 } } $$ $$ { \text{ Recall definition, } fn\to \text{f uniformly on } \mathbb{R} \Leftrightarrow \forall \epsilon \gt 0 \exists N, \in \mathbb{Z}^+ \text{ s.t }\forall n \gt N , | fn(x) - f(x) | \leq \epsilon } $$Scratch work
$$ { \epsilon > 0 ,| fn(x) - 0 | = | \frac {x}{1+nx^2} |, }$$ $$ { \text{ maximise, } fn(x) = \frac {x}{1+nx^2} \text{ using the first derivative test, } }$$ $$ { fn'(x) = \frac {1.(1+nx^2)- x.2nx}{(1+nx^2)^2} = \frac {1 - 1+nx^2 }{(1+nx^2)^2}=0 }$$ $$ { 1 - nx^2 = 0 \Rightarrow 1 = nx^2 \text{ ,so } x^2 = \frac {1}{n} \text{ so, } x = \pm \sqrt{\frac {1}{n}} }$$test point input
$$ { fn'\left(-2 \sqrt{\frac {1}{n}}\right) = \frac {1- n(-2 \sqrt{\frac {1}{n}})^2}{(1+n(-2 \sqrt{\frac {1}{n}})^2)^2} = \frac {1- n.4 \frac {1}{n}}{(1+n.4.\frac{1}{n})^2} < 0}$$ $$ { fn'\left(2 \sqrt{\frac {1}{n}}\right) < 0}$$ $$ { fn'(0) = \frac {1-0}{(1+0)^2} > 0}$$ $$ {\text{ max at x = } \frac {1}{n} \text{ , } fn'\left(\sqrt{\frac {1}{n}}\right) =\frac {\sqrt{\frac {1}{n}}}{1+n(\sqrt{\frac {1}{n}})^2}= \frac {\sqrt{\frac {1}{n}}}{1+n{\frac {1}{n}}} =\frac {\sqrt{\frac {1}{n}}}{1+1} = \frac {(\frac {1}{\sqrt{n}})}{2} = \frac {1}{2 \sqrt{n}} }$$Scratch work cont..
$${\text{ find N where, } | \frac {x}{1+nx^2} | \leq \frac {1}{2 \sqrt{n}} < \epsilon }$$ $${ \frac {1}{2} < \epsilon \sqrt{n} \Rightarrow \sqrt{n} > \frac {1}{2 \epsilon} \Rightarrow n > \left(\frac {1}{2 \epsilon } \right)^2 }$$ $${\text{ by Archimedean property } N > \left(\frac {1}{2 \epsilon } \right)^2 }$$Proof
$$ { \text{ let } \epsilon > 0 \text{ Choose N > } \left( \frac {1}{2 \epsilon}\right)^2. \text{ Then } \forall n \gt N, \forall x \in \mathbb{R} } $$ $$ { | fn(x) - 0 | = \frac {x}{1+nx^2} = \frac {1}{2 \sqrt{n}} \text{ , since max} \left(\frac {x}{1+nx^2} \right) = \frac {1}{2 \sqrt{n}} } $$ $$ { \text{ Since } n > N > \left( \frac {1}{2 \epsilon} \right)^2 \text{ , so } n = \left(\frac {1}{2 \epsilon} \right)^2 \text{ and , } \sqrt{n}= \frac {1}{2 \epsilon} }$$ $$ { \Rightarrow \epsilon > \frac {1}{2 \sqrt{n}} \text{ , } \frac {1}{2 \sqrt{n}} < \epsilon \text{ , thus } | fn(x) - 0 | \leq \frac {1}{2 \sqrt{n}} < \epsilon \text{ ∎} }$$