Waste less time on Facebook — follow Brilliant.
×

Homogenisation in inequalities

Can someone help me understand homogenisation and constraints application in inequalities?

Note by Abhi Kumbale
11 months, 3 weeks ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

What do you currently understand about it?

Calvin Lin Staff - 11 months, 2 weeks ago

Log in to reply

I know it is based on the property that f (x,y,z)=t^(-n)f (tx,ty,tz). So any inequality can be written as a function of several variables and this property can be applied if the function is homogeneous. I would like to know it's various applications.

Abhi Kumbale - 11 months, 2 weeks ago

Log in to reply

The basic idea is that a homogeneous polynomial (IE polynomial whose terms all have the same degree) can be easier to deal with than a polynomial whose terms have different degrees. As such, if we're given a condition, we try and use that to make the terms equal.

For example, a common condition is of the form \( x + y + z = 1 \) (or some other constant). If we had to deal with the expression \( xy+yz+zx - xyz \), we can homogenize by multplying the degree 2 terms by \( x+y+z = 1 \), giving us

\[ xy+yz+zx - xyz \\ = (x+y+z)(xy+yz+zx) - xyz \\ = (x+y)(y+z)(z+x) \\ = (1-x)(1-y)(1-z) \\ \leq \left( \frac{2}{3} \right)^3 \]

This is a (constructed) example where the homogenization leads to a nice factoring, which leads to a nice way to deal with the expression.

Calvin Lin Staff - 11 months, 2 weeks ago

Log in to reply

@Calvin Lin I have seen problems where no such constraints as above is given but one assumes such a constraint then solves the problem. How is this possible?

Abhi Kumbale - 11 months, 2 weeks ago

Log in to reply

@Abhi Kumbale In those cases, an assumption like \( x + y + z = t \) is made, and then we introduce another variable.

In a sense, it is similar to "Richard Feymanns favourite integration trick of differentiating through the integral" (assuming you are familiar with how that works).

Calvin Lin Staff - 11 months, 2 weeks ago

Log in to reply

@Calvin Lin I understand that you can introduce another variable but I have seen places where x+y+z=1 is introduced. Like this solution

https://brilliant.org/profile/surya-ou9y8q/sets/olympiad-proof-problems/354510/olympiad-proof-problem-day-1/

How is this right? I have also seen a similar approach in Engel's book.

Abhi Kumbale - 11 months, 2 weeks ago

Log in to reply

@Abhi Kumbale The key phrase is

The given expression is homogeneous equation of degree \(0\). So, we can make a constraint \(a+b+c+d=1\).

Do you understand why that means?

Suppose that \( a + b + c + d = t \), then consider the change of variables \( A = \frac{a}{t} ,\) etc. What is \( A + B + C + D \)?

Why does "homogenous equation of degree 0" allow us to do this substitution? What happens if we have a "homogenous equation of degree 1"? What happens for a general polynomial say \( a^2b + cd + c + 2 \)?

Calvin Lin Staff - 11 months, 2 weeks ago

Log in to reply

Sorry for being unable to answer due to the fact I have no in depth knowledge of inequalities.
But, I advise you to read Calvin Sir's note on 'How to ask for help?'; unable to provide link due to an error. It is in no way to offend you and I apologise if it does.

Yatin Khanna - 11 months, 2 weeks ago

Log in to reply

@Calvin Lin @Surya Prakash @Aditya Raut @Mark Henning @Sharky Kesa

Abhi Kumbale - 11 months, 3 weeks ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...