leehuan's All-Levels-Of-Maths SOS thread (1 Viewer)

Drsoccerball · Mar 28, 2016

Paradoxica said:
When both paradoxicas are smarter then you

Why you cut?

Nailgun · Mar 28, 2016

Paradoxica said:
When both paradoxicas are smarter then you

damn shots fired ahaha

leehuan · Mar 28, 2016

Drsoccerball · Mar 28, 2016

parad0xica said:
$\forall \epsilon > 0, \exists \delta > 0 $ such that $ 0 < |x-0| < \delta \Rightarrow \left|\frac{1- \cos x}{x^2} - \frac{1}{2}\right| < \epsilon$

Choose $\delta = \frac{1}{\sqrt{2\epsilon}}$ , then we have $0 < |x| < \frac{1}{\sqrt{2\epsilon}}$ .

Now
$\left|\frac{1- \cos x}{x^2} - \frac{1}{2}\right| \\\stackrel{A}{=} \left|\frac{\sin ^2 \left(\frac{x}{2} \right)}{2x^2} - \frac{1}{2} \right| \\\stackrel{B}{=} \frac{1}{2} \left| \frac{\cos ^2 \left(\frac{x}{2} \right)}{x^2}\right| \\\stackrel{C}{=} \frac{1}{2} \frac{1}{|x|^2} \\\stackrel{D}{<} \epsilon$

as required.

A: double angle
B: s^2 + c^2 = 1
C: cosine attains max at 1
D: hypothesis

Slight correction, at C that's where the inequality is introduced.

parad0xica · Mar 28, 2016

leehuan said:
$$I was just thinking that...$ \\ $Proposition A seems to have assumed $\cos{2x}=1-\frac{1}{2}\sin^{2}{x} \\ $and I'm a bit lost on B in terms of how$\\ \frac{1}{2}\left|\frac{1}{{x}^{2}}-\frac{\cos^{2}{x/2}}{2{x}^{2}}-1\right|=\frac{1}{2}\left|\frac{\cos^{2}{x/2}}{2{x}^{2}}\right|$

oh crap... I made an error!

Paradoxica · Mar 28, 2016

Drsoccerball said:
double angle formula bruh

$$\noindent$\frac{1-\cos{x}}{x^2} \equiv \frac{2\sin^2{\frac{x}{2}}}{x^2} \\\\ x/2 = y\\\\ \frac{2\sin^2{y}}{4y^2} \equiv \frac{\sin^2{y}}{2y^2} \\\\ \Rightarrow \frac{\sin^2{x}}{2x^2} \\\\ \mathcal{Q.E.D.}$

Drsoccerball · Mar 28, 2016

$\text{You have} |f(x)-L|=\frac{1}{2}| \frac{2\sin^2{\frac{x}{2}}}{x^2}-1 |$

$$ Upon the omission of the $ -1 $ we introduce an inequality $ |f(x)-L| < |\frac{sin^2{\frac{x}{2}}}{x^2}|$

$$ Eliminate the sin $ \ |f(x)-L| < \frac{1}{x^2}$

$|f(x)-L| < \frac{1}{M^2}= \epsilon $ For some $ x>M$

Rinse and repeat

parad0xica · Mar 28, 2016

Let's start here

$\left|\frac{\sin ^2 \left(\frac{x}{2} \right)}{2x^2} - \frac{1}{2} \right| = \frac{1}{2}\left|\frac{\sin ^2 \left(\frac{x}{2} \right)}{x^2} - 1 \right|$

Apply triangle-inequality to get

$\leq \frac{1}{2}\left|\frac{\sin ^2 \left(\frac{x}{2} \right)}{x^2}\right| +\frac{1}{2}\left| 1 \right|$

$\leq \frac{1}{2} \frac{1}{|x|^2} + \frac{1}{2}$

$< \epsilon$

where delta is something...

Paradoxica, use your magnificent algebraic manipulation skills

InteGrand · Mar 28, 2016

Drsoccerball said:
$\text{You have} |f(x)-L|=\frac{1}{2}| \frac{2\sin^2{\frac{x}{2}}}{x^2}-1 |$

$$ Upon the omission of the $ -1 $ we introduce an inequality $ |f(x)-L| < |\frac{sin^2{\frac{x}{2}}}{x^2}|$

$$ Eliminate the sin $ \ |f(x)-L| < \frac{1}{x^2}$

$|f(x)-L| < \frac{1}{M^2}= \epsilon $ For some $ x>M$

Rinse and repeat

How do we justify dropping the -1?

leehuan · Mar 28, 2016

@Drsoccerball that -1 can only be dropped for limits to infinity and we also have like some polynomial or something left.

parad0xica · Mar 28, 2016

Drsoccerball said:
Slight correction, at C that's where the inequality is introduced.

What do you mean?

Paradoxica · Mar 28, 2016

parad0xica said:
Let's start here

$\left|\frac{\sin ^2 \left(\frac{x}{2} \right)}{2x^2} - \frac{1}{2} \right| = \frac{1}{2}\left|\frac{\sin ^2 \left(\frac{x}{2} \right)}{x^2} - 1 \right|$

Apply triangle-inequality to get

$\leq \frac{1}{2}\left|\frac{\sin ^2 \left(\frac{x}{2} \right)}{x^2}\right| +\frac{1}{2}\left| 1 \right|$

$\leq \frac{1}{2} \frac{1}{|x|^2} + \frac{1}{2}$

$< \epsilon$

where delta is something...

Paradoxica, use your magnificent algebraic manipulation skills

That's not right. The two should be on top if you were using double angle.

$\frac{2\sin^2{\frac{x}{2}}}{x^2}$

And if you did the algebraic conversion, then it should be

$\frac{\sin^2{x}}{2x^2}$

It is only one or the other. Not both.

InteGrand · Mar 28, 2016

To deal with the -1, use the triangle inequality as shown by parad0xica.

leehuan · Mar 28, 2016

parad0xica said:
Let's start here

$\left|\frac{\sin ^2 \left(\frac{x}{2} \right)}{2x^2} - \frac{1}{2} \right| = \frac{1}{2}\left|\frac{\sin ^2 \left(\frac{x}{2} \right)}{x^2} - 1 \right|$

Apply triangle-inequality to get

$\leq \frac{1}{2}\left|\frac{\sin ^2 \left(\frac{x}{2} \right)}{x^2}\right| +\frac{1}{2}\left| 1 \right|$

$\leq \frac{1}{2} \frac{1}{|x|^2} + \frac{1}{2}$

$< \epsilon$

where delta is something...

Paradoxica, use your magnificent algebraic manipulation skills

G-d damn I keep forgetting that the triangle inequality is now an ASSUMED result at uni.

But wait you still didn't address the problem with statement A
http://www.wolframalpha.com/input/?i=1-cos(x)=2sin^2(x/2)

Though my instinct tells me that it will be negligible

Drsoccerball · Mar 28, 2016

InteGrand said:
How do we justify dropping the -1?

Since LHS in the expression is >1 subtracting by one reduces the value of the expression.

LOL jks we dont know this.

parad0xica · Mar 28, 2016

Paradoxica said:
That's not right. The two should be on top if you were using double angle.

$\frac{2\sin^2{\frac{x}{2}}}{x^2}$

And if you did the algebraic conversion, then it should be

$\frac{\sin^2{x}}{2x^2}$

It is only one or the other. Not both.

Thanks for reminding the second time. I thought I remembered my double angles...

leehuan said:
G-d damn I keep forgetting that the triangle inequality is now an ASSUMED result at uni.

But wait you still didn't address the problem with statement A
http://www.wolframalpha.com/input/?i=1-cos(x)=2sin^2(x/2)

Though my instinct tells me that it will be negligible

Thanks. Yep, it wouldn't do much harm because we will then use triangle-inequality!

Paradoxica · Mar 28, 2016

leehuan said:
G-d damn I keep forgetting that the triangle inequality is now an ASSUMED result at uni.

But wait you still didn't address the problem with statement A
http://www.wolframalpha.com/input/?i=1-cos(x)=2sin^2(x/2)

Though my instinct tells me that it will be negligible

Not negligible. It affects the final result.

leehuan · Mar 28, 2016

Paradoxica said:
Not negligible. It affects the final result.

Pardon me, didn't mean negligible. Meant doesn't affect the method, just the answer.

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}}\\ $As all the above steps are reversible, we complete our proof?$

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.

leehuan · 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 if I replaced that - with a + keeping in mind |-1|=1

And reverse signs everywhere else where appropriate

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