leehuan's All-Levels-Of-Maths SOS thread (2 Viewers)

parad0xica · Mar 28, 2016

Paradoxica said:
That doesn't look correct. The corollary of the triangle inequality does not state |a-b| < |a|-|b|

the closest I know of is ||x|-|y||<|x-y|

But that contradicts what you suppose.

What about

|x + y| < |x| + |y|

Let y = -y,

|x - y| < |x| + |-y| = |x| + |y|?

Paradoxica · Mar 28, 2016

leehuan said:
$$So we're saying that $\\ \begin{align*}\frac { 1 }{ 2 } \left| \frac { 4\sin ^{ 2 }{ \frac { x }{ 2 } } }{ { x }^{ 2 } } -1 \right| & \le \frac { 1 }{ 2 } \left| \frac { 4\sin ^{ 2 }{ \frac { x }{ 2 } } }{ { x }^{ 2 } } \right| -\frac { 1 }{ 2 }\\ & \le \frac { 1 }{ 2 } \left| \frac { 1 }{ { x }^{ 2 } } \right| -\frac { 1 }{ 2 }\end{align*}$

$$Hence, suppose $\\ \begin{align*}\frac { 1 }{ 2 } \left| \frac { 1 }{ { x }^{ 2 } } \right| -\frac { 1 }{ 2 }& <\varepsilon\\\Leftrightarrow\left|\frac{4}{{x}^{2}}\right|&<2\varepsilon+1\\\Leftrightarrow{\left| x\right|}^{2} &<\frac{4}{2\varepsilon+1}\end{align*}$
$Hence, pick $\delta=\frac{2}{\sqrt{2\varepsilon+1}}\\ $As all the above steps are reversible, we complete our proof?$

Reciprocating reverses inequality signs...

This path does not look promising anymore...

parad0xica · Mar 28, 2016

Paradoxica said:
Reciprocating reverses inequality signs...

This path does not look promising anymore...

I see no reversing

nvm lol

leehuan · Mar 28, 2016

Please check this proof. I feel I might've accidentally assumed too much

$\\ $Show that if $f, \, g, $ and $h$ are three functions defined on an open interval $I$, such that \\ a) $f(x)\le g(x) \le h(x) \, \forall x\in I \\ $b) $f(a)=g(a)=h(a)$ for some $a\in I\\$c) $f$ and $h$ are continuous at $a$ \\ \\ then $g$ is also continuous at $a.$

$\\$Answer:$ \\ $Taking limits to $a$ for the expression in a): $\\ \lim_{x\rightarrow a}{f(x)} \le\lim_{x\rightarrow a}{g(x)} \le \lim_{x\rightarrow a}{h(x)} \\ $But given that $f$ and $h$ are continuous, by statement b)$\\f(a)\le\lim_{x\rightarrow a}{g(x)}\le h(a) \Leftrightarrow g(a)\le\lim_{x\rightarrow a}{g(x)}\le g(a)$

$\\$This directly allows us to infer that the left and right sided limits are equal by the squeeze theorem, thus the function must therefore be continuous at $a$

leehuan · Mar 28, 2016

Redo with the triangle inequality amended to not use the corollary:

$$So we're saying that $\\ \begin{align*}\frac { 1 }{ 2 } \left| \frac { 4\sin ^{ 2 }{ \frac { x }{ 2 } } }{ { x }^{ 2 } } -1 \right| & \le \frac { 1 }{ 2 } \left| \frac { 4\sin ^{ 2 }{ \frac { x }{ 2 } } }{ { x }^{ 2 } } \right| +\frac { 1 }{ 2 }\\ & \le \frac { 1 }{ 2 } \left| \frac { 1 }{ { x }^{ 2 } } \right| +\frac { 1 }{ 2 }\end{align*}$

$$Hence, suppose $\\ \begin{align*}\frac { 1 }{ 2 } \left| \frac { 1 }{ { x }^{ 2 } } \right| +\frac { 1 }{ 2 }& <\varepsilon\\\Leftrightarrow\left|\frac{4}{{x}^{2}}\right|&<2\varepsilon-1\\\Leftrightarrow{\left| x\right|}^{2} &<\frac{4}{2\varepsilon-1}\end{align*}$
$Hence, pick $\delta=\frac{2}{\sqrt{2\varepsilon-1}}\\ $As all the above steps are reversible, we complete our proof?$

EDIT: WOAH OK WAIT..

leehuan · Mar 28, 2016

leehuan said:
$$Hence, suppose $\\ \begin{align*}\frac { 1 }{ 2 } \left| \frac { 1 }{ { x }^{ 2 } } \right| +\frac { 1 }{ 2 }& <\varepsilon\\\Leftrightarrow\left|\frac{4}{{x}^{2}}\right|&<2\varepsilon-1\\\Leftrightarrow{\left| x\right|}^{2} &<\frac{4}{2\varepsilon-1}\end{align*}$

Ok that's bad. Line 3 is wrong.

But then I get |x| > ... and not |x| < ...

parad0xica · Mar 28, 2016

Sorry leehuan, I was careless

Paradoxica · Mar 28, 2016

Okay, here's what I have thought of... Something similar appeared in the lecture I crashed that day... IDK

$$\noindent$\left| \frac{1-\cos{x}}{x^2}-\frac{1}{2}\right|<\epsilon \\ \left|\frac{4\sin^2{\frac{x}{2}}}{x^2}-1 \right|<2\epsilon \\ \left|\frac{\sin{\frac{x}{2}}}{x}-1 \right| \left|\frac{\sin{\frac{x}{2}}}{x}+1 \right| < 2\epsilon$

leehuan · Mar 28, 2016

Paradoxica said:
Okay, here's what I have thought of... Something similar appeared in the lecture I crashed that day... IDK

$$\noindent$\left| \frac{1-\cos{x}}{x^2}-\frac{1}{2}\right|<\epsilon \\ \left|\frac{4\sin^2{\frac{x}{2}}}{x^2}-1 \right|<2\epsilon \\ \left|\frac{\sin{\frac{x}{2}}}{x}-1 \right| \left|\frac{\sin{\frac{x}{2}}}{x}+1 \right| < 2\epsilon$

It did but that looks so tedious to deal with... nor do I actually know HOW to deal with it

Paradoxica · Mar 28, 2016

leehuan · Mar 28, 2016

$$Looks more promising but we need$ \\ \left|x\right|<\delta$

Edit: Nvm if sin^-1(-x)=-sin^-1(x) then the result might fall out by just changing the last few steps.

But just remind me why we don't have to consider [-pi/2,0]?

Paradoxica · Mar 28, 2016

leehuan said:
$$Looks more promising but we need$ \\ \left|x\right|<\delta$

Edit: Nvm if sin^-1(-x)=-sin^-1(x) then the result might fall out by just changing the last few steps.

But just remind me why we don't have to consider [-pi/2,0]?

Similar arguments can be made.

leehuan · Mar 28, 2016

WLOG it is. Anyway bump

leehuan said:
Please check this proof. I feel I might've accidentally assumed too much

$\\ $Show that if $f, \, g, $ and $h$ are three functions defined on an open interval $I$, such that \\ a) $f(x)\le g(x) \le h(x) \, \forall x\in I \\ $b) $f(a)=g(a)=h(a)$ for some $a\in I\\$c) $f$ and $h$ are continuous at $a$ \\ \\ then $g$ is also continuous at $a.$

$\\$Answer:$ \\ $Taking limits to $a$ for the expression in a): $\\ \lim_{x\rightarrow a}{f(x)} \le\lim_{x\rightarrow a}{g(x)} \le \lim_{x\rightarrow a}{h(x)} \\ $But given that $f$ and $h$ are continuous, by statement b)$\\f(a)\le\lim_{x\rightarrow a}{g(x)}\le h(a) \Leftrightarrow g(a)\le\lim_{x\rightarrow a}{g(x)}\le g(a)$

$\\$This directly allows us to infer that the left and right sided limits are equal by the squeeze theorem, thus the function must therefore be continuous at $a$

+ New question (under a continuity topic, but I've done limits as well of course)

$$How many real zeroes does $p(x)={x}^{3}-12{x}^{2}+45x-51$ have?$

Edit nvm this question, just realised stationary points can be combined with the intermediate value theorem. Just check the proof please.

parad0xica · Mar 28, 2016

leehuan said:
WLOG it is. Anyway bump

+ New question (under a continuity topic, but I've done limits as well of course)

$$How many real zeroes does $p(x)={x}^{3}-12{x}^{2}+45x-51$ have?$

Edit nvm this question, just realised calculus can be combined with the intermediate value theorem. Just check the proof please.

Proof looks good.

leehuan · Mar 28, 2016

Just out of curiosity, is there a name we define for the family of polynomials f(x)=x^n (n in Z+) rather than f(x) = ax^n+bx^(n-1)+etc

InteGrand · Mar 28, 2016

leehuan said:
Just out of curiosity, is there a name we define for the family of polynomials f(x)=x^n (n in Z+) rather than f(x) = ax^n+bx^(n-1)+etc

Those are examples of monomials: https://www.mathsisfun.com/definitions/monomial.html .

They are also special cases of power functions where the index is a positive integer: https://en.wikipedia.org/wiki/Power_function .

Paradoxica · Mar 28, 2016

Power functions

leehuan · Mar 28, 2016

$\\$Let $f:\left[a,b\right]\rightarrow\mathbb R$ be a continuous function such that $f\left(x\right)>0\, \forall \, x\in\left[a,b\right].$ Show that there is a positive constant $c$ such that $f\left(x\right)\ge c \, \forall \, x\in\left[a,b\right]$

This sounds like it's asking me to prove the max-min theorem, or at least the latter. How does this work?

Drsoccerball · Mar 28, 2016

leehuan said:
$\\$Let $f:\left[a,b\right]\rightarrow\mathbb R$ be a continuous function such that $f\left(x\right)>0\, \forall \, x\in\left[a,b\right].$ Show that there is a positive constant $c$ such that $f\left(x\right)\ge c \, \forall \, x\in\left[a,b\right]$

This sounds like it's asking me to prove the max-min theorem, or at least the latter. How does this work?

$What I would do (which is probably wrong) :$

$$ Consider cases if $ f(x) \rightarrow 0 $ as $ x \rightarrow a $ and $ f(x) \rightarrow \infty $ as $ x \rightarrow b$

$$ Then there is a minimum value between $ (0,\infty ) $ since $ f(x) >0 $ there is such a y value $ c $ which $ f(x) \geq c \ \in c (b,a)$

$$ Similarly for if $ a \rightarrow b \ , b \rightarrow a$

seanieg89 · Mar 28, 2016

leehuan said:
$\\$Let $f:\left[a,b\right]\rightarrow\mathbb R$ be a continuous function such that $f\left(x\right)>0\, \forall \, x\in\left[a,b\right].$ Show that there is a positive constant $c$ such that $f\left(x\right)\ge c \, \forall \, x\in\left[a,b\right]$

This sounds like it's asking me to prove the max-min theorem, or at least the latter. How does this work?

Call it the extreme value theorem, the term "max-min" and "min-max" occur a ton in other contexts.

How you should prove it will depend on exactly what you have learned, typically you assume this theorem without proof in most first year courses. (For good reason, discussion of things like compactness distract a bit from learning calculus which is the main point of first year calculus course).

Have you heard of any of the following?

-Compactness
-Bolzano-Weierstrass theorem
-Heine Borel theorem

leehuan's All-Levels-Of-Maths SOS thread (2 Viewers)

Active Member

-insert title here-

Active Member

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

-insert title here-

Well-Known Member

-insert title here-

Well-Known Member

-insert title here-

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)