What is the cauchy-schwarz inequality? (1 Viewer)

[~shinigami~]

I was talking to this guy in my math class today about 4U inequalities as it's in our upcoming assessment.

He mentioned the "cauchy-schwarz inequality" and when I asked him about it, he would not explain it as he explained in his own words "We're enemies until after the trials so I can't show you". I was of course thinking, wtf?

Would anyone care to answer a few of my questions, please?

1. What is the cauchy-schwarz inequality?

2. What is it used for and how is it used?

3. How do you prove it?

Thanks in advance.

 

[Jono_2007]

look at the wikipedia site.

 

[pLuvia]

Here's some information about it
http://en.wikipedia.org/wiki/Cauchy-Schwarz_Inequality

This inequality you will be required to show how it exists, the actual name of the inequality isn't required to be known in the syllabus, but you will encounter it. Such as in the triangle inequality

 

[~shinigami~]

Here's some information about it
http://en.wikipedia.org/wiki/Cauchy-Schwarz_Inequality

This inequality you will be required to show how it exists, the actual name of the inequality isn't required to be known in the syllabus, but you will encounter it. Such as in the triangle inequality

Thanks for the info pLuvia.

I'm going to take a look at that wiki page.

 

[vulgarfraction]

Actually, never use wikipedia for maths stuff. Well, not never. But just.. don't. You'll end up confused because the people who put the articles there are at least uni level.

Google it or go to wolfram.

Cauchy-Schwarz in simplest form:

(a^2+b^2)(c^2+d^2) >= (ac+bd)^2

 

[~shinigami~]

Actually, never use wikipedia for maths stuff. Well, not never. But just.. don't. You'll end up confused because the people who put the articles there are at least uni level.

Google it or go to wolfram.

Cauchy-Schwarz in simplest form:

(a^2+b^2)(c^2+d^2) >= (ac+bd)^2

Wow, thank you.

I was getting a bit confused at the wiki page.

 

[simplistic]

We're enemies until after the trials so I can't show you'
geeze shouldnt it be the opposite ?
coz u have to work together as a group so that ur assessment marks in the school are high ?
you dont want slackers to pull ur marks down ?

 

[STx]

"The arithmetic mean of n positive real numbers is always greater than or equal to their geometric mean"

 

[jyu]

Cauchy-Schwarz Inequality |a.b|<=|a||b|

If a=pi+qj and b=mi+nj, then C-S I can be expressed as
|pm+qn|<=rt(p^2+q^2)rt(m^2+n^2)
(pm+qn)^2<=(p^2+q^2)(m^2+n^2)

This can be extended to 3 or higher dimensions:
(pm+qn+...)^2<=(p^2+q^2+...)(m^2+n^2+...).

Proof: Since 0<=|cos$|<=1, therefore |a||b||cos$|<=|a||b|,
hence |a.b|<=|a||b|.

 

[blackfriday]

this applies to any vector containing n elements. but since you kids will only work with vectors with components X=(a, b) and Y=(c, d)...

if the vectors represented complex numbers X=a+ib and Y= c+id:

the cauchy schwartz inequality comes from this: for vectors X and Y: X.Y=|X||Y|cos(x)
so |X.Y|<=|X||Y|

but from your complex numbers work, you would interpret this as
|X|=sqrt(a^2 + b^2)
|Y| =sqrt(c^2 + d^2)
|X.Y|=sqrt[(ac)^2 + (bd)^2 + 2(abcd)]

if you're smarter than me, please correct any mistakes i may have made

 

[Slidey]

He mentioned the "cauchy-schwarz inequality" and when I asked him about it, he would not explain it as he explained in his own words "We're enemies until after the trials so I can't show you". I was of course thinking, wtf?

1) He was showing off and
2) He didn't actually know how to show you.

 

[Affinity]

Second that.

... Idiots who think they are all that.

 

[Wackedupwacko]

just pwn his ass in every exam. that should show how *smart* he really is and no you dont need any of that... (unless you feel like some background reading....)

 

[~shinigami~]

1) He was showing off and
2) He didn't actually know how to show you.

Second that.

... Idiots who think they are all that.

That makes sense now. You guys are both right, he was probably just bs'ing.

just pwn his ass in every exam. that should show how *smart* he really is and no you dont need any of that... (unless you feel like some background reading....)

I'm not sure that I can beat him because he's actually pretty good at 4U and I'm not that good. I'll definately try.