Difference between revisions of "User:Lcz"
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<math>\implies f(x)=xf(1)</math>. Setting <math>f(1)=c</math>, we get the desired.<math>\blacksquare</math> | <math>\implies f(x)=xf(1)</math>. Setting <math>f(1)=c</math>, we get the desired.<math>\blacksquare</math> | ||
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for all non-negative integers <math>x,y</math>. Determine <math>f(4,1981)</math>. | for all non-negative integers <math>x,y</math>. Determine <math>f(4,1981)</math>. | ||
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==F.Extra== | ==F.Extra== |
Revision as of 10:23, 11 June 2020
Contents
Introduction
So I failed the AOIME. I don't really want to do any more AIME prep, so I have decided to go do some Oly prep :). Here's some oly notes/favorite problems
Disclaimer
About half of these problems (?) are from the OTIS excerpts. I try to make mines easier to understand though you should probably just read the OTIS excerpts. These notes are to help me learn, though I don't mind if you read them.
Inequalities
Yay! I love inequalities. Clever algebraic manipulation+thereoms is all you need. It all comes from experience though...
I.Basics
AM-GM, Cauchy (Titu's Lemma as well), Muirhead, and Holder's. These are cool, remember that these should only be used when the inequality is homogenized already.. These are all pretty easy to prove as well.
Example 1:
(Evan Chen) Let with
. Prove that
Solution:
We need to try to homogenize this somehow. Plugging in the expression for on the LHS for won't work. If we try to do something on the left side, we'll still have a degree
. Wait a second, why are they all
's? Let's try to get rid of the
's first. Well, if we add
to both sides of the given condition, we get
,
,
By AM-GM.
Obviously the trivial solution satisfies this, so we haven't made any silly mistakes. We still haven't homogenized, but now the path is clear. Multiplying both sides by
, we get
,
,
Which is true from AM-GM. We shall now introduce Muirhead's...
Example 2:
Let (Again Evan Chen) and
. Prove that
.
Solution: First we homogenize:
Which is true because majorizes
Cauchy:
Problem 1: (2009 usamo/4): For let
,
, ...,
be positive real numbers such that
Prove that
.
Try to solve this on your own! Very cute problem.
Note that you'll probably only ever need Holder's for variables...
Example 3 (2004 usamo/5)
Let ,
, and
be positive real numbers. Prove that
.
Solution:
1. The is cubed, so we try to use Holder's. The simplest way to do this is just to use
on the LHS.
2. Now all we have to prove is that , or
.
Now note that if
, this is true, if
, this is true, and if
, this is true as well, and as we have exhausted all cases, we are done.
I.More advanced stuff, learn some calculus
You will need to know derivatives for this part. It's actually quite simple. Derivative=Tangent.
Basically, the derivative of is
Adding and other stuff works in the same way.
You also probably need to know
-Product Rule:
-Quotient Rule:
-Summing:
The second derivative of a function is just applying the derivative twice. A function is convex on an interval if it's second derivative is always positive in that interval. A function is convex if it's second derivative is negative in that interval.
Jensen's inequality says that if is a convex function in the interval
, for all
in
,
The opposite holds if
is concave.
Karamata's inequality says that if is convex in the interval
, the sequence
majorizes
,, and all
are in
,
TLT (Tangent Line Trick) is basically where you either
a. take the derivative, and plug in the equality cases or
b. plugging in both equality cases to form a line.
Problem 2: Show that
Problem 3: Using Jensen's and Holder's, solve 2001 IMO/2:
Let be positive real numbers. Prove
.
Problem 4: (2017 usamo/6)
Find the minimum possible value of
given that
are nonnegative real numbers such that
.
Problem 5: (Japanese MO 1997/6) Prove that
for any positive real numbers ,
,
.
I.Extra
Ravi Substitution for triangles. .
Weighted AM-GM (This is kind of dumb imo, just use AM-GM lol)
The weighted form of AM-GM is given by using weighted averages. For example, the weighted arithmetic mean of and
with
is
and the geometric is
. (AoPS Wiki)
Problem 6: (1983 imo/6, using Ravi!)
Let ,
and
be the lengths of the sides of a triangle. Prove that
.
Determine when equality occurs.
Inequality Problems:
https://artofproblemsolving.com/wiki/index.php/Category:Olympiad_Inequality_Problems
https://artofproblemsolving.com/wiki/index.php/Category:Olympiad_Algebra_Problems
Function Equations
Oops...I kind of suck at these :P
~Lcz 6/9/2020 at 12:49 CST
-Function=anything.
-Symmetry
-Plug in
-Check for linear/constant solutions first. They are usually the only ones.
-fff trick is pro
-Pointwise trap?!?
-Be aware of your domain/range. ((Probably not))
F.Intro
injective: a-1, b-3, c-2, nothing-4
surjective: a-1, b-1, b-2, c-3
bijective: a-1, b-3, c-2
A function where its domain and range are same is an involution if
for every
in its range/domain. This function is bijective.
F.Ok these things are sort of cool
Example 1 (2002 usamo/4):
Let be the set of real numbers. Determine all functions
such that
for all pairs of real numbers and
.
Solution:
I claim that the sole solution is for some constant
.
First note that setting and
as zero respectively yields
and
. Letting
in the second equation means that
is odd.
Now, we can substitute into the original equation our findings:
where
.
Finally,
. Setting
, we get the desired.
F.They're not that good actually
Problem 1 (2009 imo/5):
Determine all functions from the set of positive integers to the set of positive integers such that, for all positive integers
and
, there exists a non-degenerate triangle with sides of lengths
and
.
This question...is...cool...?
Example 2 (ISL/2015): Determine all functions
with the property that
holds for all .
BOGUS Solution:
The answer is or
. These clearly work.
Claim: there exists a such that
.
Proof: set .
Then (*) where
. Now
for some constant
, or
.
The former yields , the latter works and yields
.
Why this is wrong: You can't just assume because or
. Remember, a function is anything. Maybe this function is surjective.
Solution (after that step):
Now we plug in (because we know that the
will turn into
, and therefore will be proving that
is linear).
(from (*) on where
is actually
.
(follows directly from (*)).
Therefore, this equation is linear.
Now, letting ,
From (*), we get
If , then
So . Otherwise,
is constant and plugging back into (*), we get the only solution as
Ay, ok. Heres a semi-recent one.
Problem 2 (2016 usamo/4):
Find all functions such that for all real numbers
and
,
Problem 3(One of my favorites, found it in Inter. Alg I think; 1981 imo/6)
The function satisfies
(1)
(2)
(3)
for all non-negative integers . Determine
.
F.Extra
Nothin much
https://artofproblemsolving.com/wiki/index.php/Category:Functional_Equation_Problems