Try to find a pattern in the first $ 2$ digits:

$ 6^1: 06$

$ 6^2: 36$

$ 6^3: 16$

$ 6^4: 96$

$ 6^5: 76$

$ 6^6: 56$

$ 6^7: 36$

$ ...$

Thus, $ 6^{351}$ must have the last two digits of $ 16$.

$ 3^11 \equal{} 177147$ (note: you can use the same method as above)

$ 16 \plus{} 47 \equal{} 63$

Thus, the answer is $ \boxed{6}$.

Thats the answer I got when I did the problem using the same technique. But the answer key says 2 is the answer...


Let $ABC$ ba a triangle.Let $A_{1}, B_{1}, C_{1}$ be points on $BC;CA;AB$ respectively.Prove that:.
$\frac{AC_{1}}{C_{1}B}.\frac{BA_{1}}{A_{1}C}.\frac{CB_{1}}{B_{1}A}=\frac{sin ACC_{1}}{sin C_{1}CB}.\frac{sin BAA_{1}}{sin A_{1}AC}.\frac{sin CBB_{1}}{sin B_{1}BA}$

Use sin theorem with triangle $ACC_{1}, BCC_{1}$, we have:
$\frac{AC_{1}}{C_{1}C}=\frac{sin\hat{ACC_{1}}}{sin A}$ and $\frac{CC_{1}}{C_{1}B}=\frac{sin B}{sin\hat{C_{1}CB}}$, or $\frac{AC_{1}}{C_{1}B}=\frac{sin\hat{ACC_{1}}}{sin\hat{C_{1}CB}}.\frac{sin B}{sin A}$
Similar, we have $\frac{BA_{1}}{A_{1}C}=\frac{sin \hat{BAA_{1}}}{sin\hat{A_{1}AC}}.\frac{sin C}{sin B}$ and $\frac{CB_{1}}{B_{1}A}=\frac{sin \hat{CBB_{1}}}{sin \hat{B_{1}BA}}.\frac{sin A}{sin B}$
$\Rightarrow \frac{AC_{1}}{C_{1}B}.\frac{BA_{1}}{A_{1}C}.\frac{CB_{1}}{B_{1}A}=\frac{sinACC_{1}}{sin C_{1}CB}.\frac{sin BAA_{1}}{sin A_{1}AC}.sin \frac{CBB_{1}}{B_{1}BA}$


calculus integration trigonometry function logarithms geometry Asymptote

Help me to solve the integral, thanks

$ \int\limits_{0}^{\pi /2}{{{\left( \tan x \right)}^{1/4}}.dx}$

Hint: use integration by substitution with $ t \equal{} \sqrt [4]{\tan x}$.

(Or if you know the beta function identity, you can show that $ \textstyle \int_{0}^{\pi / 2} \tan^{\alpha} \theta \, d\theta \equal{} \frac{\pi}{2} \sec \left( \frac{\alpha \pi}{2} \right)$ for any $ \minus{} 1 < \alpha < 1$.)

This is my solution, I dont know it is correct?
$ */\,\,I\equal{}\int{\sqrt[4]{\tan x}.dx}\equal{}\frac{1}{2.\exp \left( \frac{3i.\pi }{8} \right)}\left[ i.\ln \left| \frac{\sqrt[4]{\tan x}\minus{}\exp \left( \frac{5i.\pi }{8} \right)}{\sqrt[4]{\tan x}\plus{}\exp \left( \frac{5i.\pi }{8} \right)} \right|\plus{}\ln \left| \frac{\sqrt[4]{\tan x}\minus{}\exp \left( \frac{i.\pi }{8} \right)}{\sqrt[4]{\tan x}\plus{}\exp \left( \frac{i.\pi }{8} \right)} \right| \right]$

$ \plus{}\frac{1}{2}\exp \left( \frac{3i.\pi }{8} \right)\left[ i.\ln \left| \frac{\sqrt[4]{\tan x}.\exp \left( \frac{5i.\pi }{8} \right)\minus{}1}{\sqrt[4]{\tan x}.\exp \left( \frac{5i.\pi }{8} \right)\plus{}1} \right|\plus{}\ln \left| \frac{\sqrt[4]{\tan x}.\exp \left( \frac{i.\pi }{8} \right)\minus{}1}{\sqrt[4]{\tan x}.\exp \left( \frac{i.\pi }{8} \right)\plus{}1} \right| \right]$

$ \int\limits_{0}^{\pi /2}{{{\left( \tan x \right)}^{1/4}}.dx}\equal{}0$

and please give me the link of indentity Beta function, thanks.

Hi chuong01. Here is a link to Beta functions: http://en.wikipedia.org/wiki/Beta_function

As to the integral, a sketch suggests the area is positive and so the integral is not zero. I've got it as zero at the lower limit of zero but haven't dealt with the asymptote at $ \frac{\pi}{2}$ yet.

So why is my solution wrong?

$ */\,\,I \equal{} \int{\sqrt [4]{\tan x}.dx}$

$ \tan x \equal{} {{t}^{4}},\,\,x \equal{} \arctan \left( {{t}^{4}} \right),\,\,dx \equal{} \frac {4{{t}^{3}}.dt}{{{t}^{8}} \plus{} 1}\Rightarrow \int{\sqrt [4]{\tan x}.dx} \equal{} 4\int{\frac {{{t}^{4}}.dt}{{{t}^{8}} \plus{} 1}}$

$ \frac {{{t}^{4}}}{{{t}^{8}} \plus{} 1} \equal{} \frac {{{t}^{4}}}{\left( {{t}^{4}} \plus{} i \right)\left( {{t}^{4}} \minus{} i \right)} \equal{} \frac {A}{{{t}^{4}} \plus{} i} \plus{} \frac {B}{{{t}^{4}} \minus{} i}\Leftrightarrow {{t}^{4}} \equal{} A\left( {{t}^{4}} \minus{} i \right) \plus{} B\left( {{t}^{4}} \plus{} i \right)$

$ \Rightarrow A \plus{} B \equal{} 1,\,\, \minus{} A \plus{} B \equal{} 0\Rightarrow A \equal{} B \equal{} \frac {1}{2}$

$ \Rightarrow I \equal{} 4\int{\frac {{{t}^{4}}.dt}{{{t}^{8}} \plus{} 1}} \equal{} 4\left( \frac {1}{2}\int{\frac {dt}{{{t}^{4}} \plus{} i}} \plus{} \frac {1}{2}\int{\frac {dt}{{{t}^{4}} \minus{} i}} \right)$

$ \equal{} \frac {2}{4.\exp \left( \frac {3i.\pi }{8} \right)}\left[ i.\ln \left| \frac {t \minus{} \exp \left( \frac {5i.\pi }{8} \right)}{t \plus{} \exp \left( \frac {5i.\pi }{8} \right)} \right| \plus{} \ln \left| \frac {t \minus{} \exp \left( \frac {i.\pi }{8} \right)}{t \plus{} \exp \left( \frac {i.\pi }{8} \right)} \right| \right]$

$ \plus{} \frac {2}{4}\exp \left( \frac {3i.\pi }{8} \right)\left[ i.\ln \left| \frac {t.\exp \left( \frac {5i.\pi }{8} \right) \minus{} 1}{t.\exp \left( \frac {5i.\pi }{8} \right) \plus{} 1} \right| \plus{} \ln \left| \frac {t.\exp \left( \frac {i.\pi }{8} \right) \minus{} 1}{t.\exp \left( \frac {i.\pi }{8} \right) \plus{} 1} \right| \right]$

$ \Rightarrow I \equal{} \int{\sqrt [4]{\tan x}.dx} \equal{} \frac {1}{2.\exp \left( \frac {3i.\pi }{8} \right)}\left[ i.\ln \left| \frac {\sqrt [4]{\tan x} \minus{} \exp \left( \frac {5i.\pi }{8} \right)}{\sqrt [4]{\tan x} \plus{} \exp \left( \frac {5i.\pi }{8} \right)} \right| \plus{} \ln \left| \frac {\sqrt [4]{\tan x} \minus{} \exp \left( \frac {i.\pi }{8} \right)}{\sqrt [4]{\tan x} \plus{} \exp \left( \frac {i.\pi }{8} \right)} \right| \right]$

$ \plus{} \frac {1}{2}\exp \left( \frac {3i.\pi }{8} \right)\left[ i.\ln \left| \frac {\sqrt [4]{\tan x}.\exp \left( \frac {5i.\pi }{8} \right) \minus{} 1}{\sqrt [4]{\tan x}.\exp \left( \frac {5i.\pi }{8} \right) \plus{} 1} \right| \plus{} \ln \left| \frac {\sqrt [4]{\tan x}.\exp \left( \frac {i.\pi }{8} \right) \minus{} 1}{\sqrt [4]{\tan x}.\exp \left( \frac {i.\pi }{8} \right) \plus{} 1} \right| \right]$

And wikipedia link has no proof for that identity...

Please help me to check the solution, thanks.


group theory
abstract algebra
invariant
complex numbers
superior algebra
superior algebra solved

Let $ C^{*}$ be the multiplied group of non-zero complex numbers and H be the subgroup which has finite index in $ C^{*}$. Prove that $ H\equal{}C^{*}$.

$ \mathbb{C}^*$ is a divisible abelian group. so every finite quotient is finite, divisible, and thus trivial.

Hix, I have known any notion about divisible group. and the problem just request knowledge about group, subgroup, group homomorphism. So please tell me how to solve this problem in easy way.

He already did  :wink:
An (abelian) group $ G$ is called divisible, if for all $ x \in G$ and all $ n \in \mathbb N$, there is an $ y \in G$ such that $ y^n\equal{}x$. Now try to prove that quotients of divisible groups are divisible again, and that nontrivial divisible groups are not finite.

this is a bit small-minded ... it is clear that you HAVE to find invariants to compare groups, the basic notions of "subgroup" and "homomorphism" don't suffice at all. as ZetaX has pointed out, the definition of "divisible" is elementary and does not require any other previous knowledge.

Please tell me that if there is another solution without using divisible group.

Well, one could simply not use the word "divisible" and write down the same proof :lol:


[quote="pianoforte"]Mountain Dew is 30% caffeine. Coke is 15% caffeine. Dim Sum wants to make a 100 gallon mixture of soda that is 25% caffeine. How much of each should he/she mix in?[/quote]

i think this problem has something to do with me

[hide="answer"]
i say $66 \frac{2}{3}$ gallons of mountain dew and $33 \frac{1}{3}$ gallons of coke
[/hide]

Treething, you are right:
[hide="Solution"]Let $M$ equal the amount of Mountain Dew.
Let $C$ equal the amount of Coke.

Then
$M+C=100$

$\frac{3}{10}M+\frac{3}{20}C = 25$

$2M+C=\frac{500}{3}$

$M=\frac{200}{3}$

$C=\frac{100}{3}$[/hide]

[quote="anirudh"][quote="pianoforte"]Mountain Dew is 30% caffeine. Coke is 15% caffeine. Dim Sum wants to make a 100 gallon mixture of soda that is 25% caffeine. How much of each should he/she mix in?[/quote]

[hide]
66% mountain dew
33%coke[/hide][/quote]

where's the other $1\%$?  :rotfl:

oops.

2/3 mt dew
1/3 coke

or $66.\bar{6}$ mt dew and $33.\bar{3}$ coke


Marguerite drove 100 miles in 2.4 hours. If Sam drove for 3 hours at the same average rate as Marguerite, how many miles did he drive?

Let $ m\equal{}\text{the number of miles Sam drove}.$ We have the ratio \[ \frac{100\text{ miles}}{2.4\text{ hours}}\equal{}\frac{m}{3\text{ hours}}.\] Simplifying, \[ m\equal{}\frac{300\text{ miles}}{2.4\text{ hours}}\equal{}\boxed{125\text{ miles}}.\]


calculus
geometry
LaTeX
calculus computations

I have a problem that is fairly straight forward algebraicly, but I can't for the life of me figure out how to do it with calculus. Please help me figure this out.

The base of an aquarium with given Volume V is made of slate and the sides are made of glass. If slate costs five times as much (per unit of area) as glass, find the dimensions of the aquarium that will minimize the cost of the materials.

So I have:
$V=hwl$
$A_{g}=2hw + 2hl$
$A_{s}=wl$
$C=5A_{g}+A_{s}$

(This LaTex thing is pretty cool!)

You have the 5 in the wrong place. (This is one of those famously widespread mistakes.)

That should be $C=A_g+5A_s=2hw+2hl+5wl.$

This is a classic problem for the use of Lagrange multipliers:

Maximize $C=2hw+2hl+5wl$ subject to the constraint
$hwl=V.$

Set $\nabla C=\lambda\nabla V.$

$2w+2l=\lambda wl$
$2h+5l=\lambda hl$
$2h+5w=\lambda hw$

Subtract the 2nd and 3rd equations: $5(l-w)=\lambda h(l-w).$

So either $5=\lambda h$ or $l=w.$

Putting $5=\lambda h$ into the second equation leads to $2h+5l=5l,$ or $h=0.$ This is not a possible configuration for the aquarium, so we have $l=w.$

(Of course, you probably could have and would have guessed this: the the optimal shape of the aquarium will have a square base.)

We now have these two equations:

$4w=\lambda w^2$
$2h+5w=\lambda hw.$

Since $w\ne 0,$ we have $4=\lambda w.$ That leads to

$2h+5w=4h$
$5w=2h,$ or $h=\frac 52w.$

That's our only critical point, so it's the minimum cost. Our optimal shape has a square base and the height is 2.5 times the side of the square base. You can plug all of that into the equation $hwl=V$ to get the actual dimensions in terms of $V.$

Thank you! It's like a light just came on about Lagrange multipliers. Up until this point I wasn't really getting them! Thanks again.


2 letters are selected from the word MISSISSIPPI without replacement.  What is the probability that the letters are the same?

This is my first post, so I'm not sure if I'm right but...

[hide]Total possibilities possible where MCrawford wrote:2 letters (are) selected from the word MISSISSIPPI without replacement[/hide][hide]  and [/hide][hide]MCrawford wrote: that the letters are the same[/hide][hide]
 would be 110 (11*10).

The possibility that you select two same letters are
II:4*3=12
SS:4*3=12
PP:2*1=2
12+12+2=26
∴ The possibility of 2 letters selected from the word MISSISSIPPI (without replacement) is 26/110 (or 13/55)

You could also do...  Total possibilities possible where [/hide][hide]MCrawford wrote:2 letters (are) selected from the word MISSISSIPPI without replacement[/hide][hide] and[/hide][hide] ]MCrawford wrote:that the letters are the same[/hide][hide] would be 110.(i=11
                                                                                Σ
                                                                                i=1)The possibility that you select two same letters are
II:i=3
     Σ     =6
    i=1SS:i=3
      Σ     =6
     i=1PP:i=1
      Σ     =1
     i=16+6+1=13
∴ The possibility of 2 letters selected from the word MISSISSIPPI (without replacement) is 13/55[/hide]
I hope I got it right...

P.S. I also hope you can understand the sigma notations...[/list]

Very good!  One little thing: On the second solution you took the sum from i = 1 to i = 11 when you only needed to go to i = 10.

Are you a Mathcounts competitor?

Well I don't know what[quote="MCrawford"]Mathcounts competitor[/quote]is :?:
(I'm new here :oops: )

Mathcounts is a national competition (in the U.S.) for 6th, 7th, and 8th graders.  It's a fun program that starts with a local contest, then a state contest.  The top four students from each state (or D.C., Puerto Rico, etc.) compete as teams at the national competition.  Some scholarships are given out.  It is much like the Spelling Bee (though more educational in my opinion).

This area of the AoPS site is dedicated to solving problems at the Mathcounts level (though I try to push a little above that level at times).

If you are of the age to compete at Mathcounts, you might check out Mathcounts.org and talk to your school about competing.

2*12/110 + 2/110 = 13/55..

is there a better way using permutations?


What is the value of the largest mean among the following sets of whole numbers?

A: all multiples of 3 between 1 and 50,
B: all multiples of 4 between 1 and 50,
C: all multiples of 6 between 1 and 50,
D: all multiples of 7 between 1 and 50,
E: all multiples of 9 between 1 and 50.

Is it 28? I'm not sure, and i need it better explained to me. Thanks  :lol:

I think all you need to do is find the largest and smallest multiples, subtract them and divide by two. So it's $ \boxed{22.5}$...?

I think...

[hide="solution that is fairly long"] First, let's find the mean in general. So we want to find the mean of all multiples of m between 1 and 50.

The sum of the numbers is $ m(1 \plus{} 2 \plus{} ... \plus{} n)$ where mn is the largest multiple of m under 50. The total number of multiples is n. Thus, the mean is $ \frac {m(n)(n \plus{} 1)}{2n} \equal{} \frac {m(n \plus{} 1)}{2}$.

Now we can just go through.

A is $ \frac {(3)(17)}{2} \equal{} 25.5$, B is $ \frac {(4)(13)}{2} \equal{} 26$, C is $ \frac {(6)(9)}{2} \equal{} 27$, D is $ \frac {(7)(8)}{2} \equal{} 28$, and E is $ \frac {(9)(6)}{2} \equal{} 27$.

Our answer is 28.[/hide]

[hide="alternatively and much faster"] The arithmetic mean of an arithmetic series is just the arithmetic mean of the first and last number. Using that, we get the same values for the mean as above. Once again, it is 28.[/hide]

akhil0422, you [b]add[/b] the smallest and largest value, not subtract. The rest of your solution is fine.

EDITED to fix hide tags.

lol that was the ONLY part of my solution. im really dumb :(

hey thnx for the explanation xpmath. i get it now :o


Two of the opposite sides of a square are increased by $ 20\%$ and the other two sides are decreased by $ 25\%$. What is the percent of change in the area?


integration
trigonometry
calculus
linear algebra
matrix
limit
complex numbers

Please help me to check this solution
\[ \text{* }I \equal{} \int\limits_{0}^{ \plus{} \infty }{{{e}^{ \minus{} {{x}^{2}}}}.dx} \equal{} \frac {\sqrt {\pi }}{2}\]

\[ \Rightarrow \int\limits_{0}^{ \plus{} \infty }{{{e}^{ \minus{} i{{x}^{2}}}}.dx} \equal{} \frac {1}{\sqrt {i}}\int\limits_{0}^{ \plus{} \infty }{{{e}^{ \minus{} {{\left( \sqrt {i}.x \right)}^{2}}}}.d\left( \sqrt {i}.x \right)} \equal{} \frac {\sqrt {\pi }}{2\sqrt {i}} \equal{} \frac {1}{2}.\sqrt {\frac {\pi }{i}} \equal{} \frac {\sqrt { \minus{} i.\pi }}{2}\]

\[ \int\limits_{0}^{ \plus{} \infty }{{{e}^{ \minus{} i{{x}^{2}}}}.dx} \equal{} \int\limits_{0}^{ \plus{} \infty }{\left[ \cos \left( {{x}^{2}} \right) \minus{} i.\sin \left( {{x}^{2}} \right) \right].dx} \equal{} \frac {\sqrt { \minus{} i.\pi }}{2}\]

\[ \minus{} i \equal{} \cos \left( \minus{} \frac {\pi }{2} \right) \plus{} i.\sin \left( \minus{} \frac {\pi }{2} \right) \equal{} {{e}^{ \minus{} i.\pi /2}}\Rightarrow \sqrt { \minus{} i} \equal{} \exp i\left( \frac { \minus{} \pi /2 \plus{} k.2\pi }{2} \right) \equal{} {{e}^{ \minus{} i.\pi /4}}\text{ and }{{e}^{3i.\pi /4}}\]

\[ {{e}^{ \minus{} i.\pi /4}} \equal{} \cos \left( \frac {\pi }{4} \right) \minus{} i.\sin \left( \frac {\pi }{4} \right) \equal{} \frac {\sqrt {2}}{2} \minus{} i.\frac {\sqrt {2}}{2}\text{ }\]

\[ {{e}^{3i.\pi /4}} \equal{} \cos \left( \frac {3\pi }{4} \right) \plus{} i.\sin \left( \frac {\pi }{4} \right) \equal{} \minus{} \frac {\sqrt {2}}{2} \plus{} i.\frac {\sqrt {2}}{2}\]

\[ \Rightarrow \int\limits_{0}^{ \plus{} \infty }{\left[ \cos \left( {{x}^{2}} \right) \minus{} i.\sin \left( {{x}^{2}} \right) \right].dx} \equal{} \frac {\sqrt { \minus{} i.\pi }}{2} \equal{} \frac {\left( \sqrt {2} \minus{} i.\sqrt {2} \right).\sqrt {\pi }}{4}\]

\[ \Rightarrow \int\limits_{0}^{ \plus{} \infty }{\cos \left( {{x}^{2}} \right).dx} \equal{} \int\limits_{0}^{ \plus{} \infty }{\sin \left( {{x}^{2}} \right).dx} \equal{} \frac {\sqrt {2\pi }}{4}\]
We only choose positive value because the integral is non-negative

You continue to use complex numbers to make easy/medium problems way more difficult than they have to be.  The first integral (the Gaussian integral) can be evaluated in a manner similar to how you did the second one.


abstract algebra
group theory
superior algebra
superior algebra unsolved

1) Show that if $ |G| \equal{} 80$, then either $ m_5(G)\equal{}1$ or $ m_2(G)\equal{}1$. Is $ G$ solvable? Give an example of three non-isomorphic non-abelian group of order 80 each with a normal 5-Sylow subgroup.

( $ m_p(G) \equal{}$ number of p-Sylow subgroup of $ G$)

2) Let $ G$ be a group of order $ 231 \equal{} 3 \cdot 7 \cdot 11$. Show that $ G$ has a cyclic subgroup of order 33.

What have you tried so far? Where do you get stuck?


Please take AMC B day servey so I can make any necessary adjustment to the AMC B day level.

Thanks.

Some of them I see right away, but some of them I have to actually get out pencil and scratch paper and dig in. Some work on paper, but not too hard, which is good. I like to practice skills which I know but haven't mastered yet.

I echo White Horse King.  I think that the problems are very nice.  Some of them I can see instantly, while others I have to do a little work on. Then there are those that are easy, but that I don't know how to do (log problems) and with simple examples it is fairly easy to learn how to do those types of problems.  And with my first last and only chance to take the AMC 12 in just a couple weeks I need all the practice I can get...Keep it up!


hey guys, I just have a small problem in maths...I'm having problems isolating a variable (At university, when things get too obvious you get confused) and I'd rather ask here than send this to my teacher and make him mad:

6 * (8 - SJ - SUS) = 8 + 2(SJ) - SUS

isolate SJ in terms of SUS...take them both as 2 variables, x and y

The answer I got was:

SJ = (8 - SUS)/4

Well, I'll assume that $ SJ$ is really one variable and not a product of variables.
Same for $ SUS$.

Let's call $ SJ \equal{} x$ and $ SUS \equal{} y$.
So, rewriting the equation, we have
$ 6 \cdot (8 \minus{} x \minus{} y) \equal{} 8 \plus{} 2x \minus{} y$
We want x in terms of y, so the first step is to get all of our x's on the same side.
Let's expand the left hand side.
$ 48 \minus{} 6x \minus{} 6y \equal{} 8 \plus{} 2x \minus{} y$
Now let's subtract 8 from both sides.
$ 40 \minus{} 6x \minus{} 6y \equal{} 2x \minus{} y$
Now add y to both sides.
$ 40 \minus{} 6x \minus{} 5y \equal{} 2x$
Almost done: add 6x to both sides.
$ 40 \minus{} 5y \equal{} 8x$
Nearly done. All we need to do now is divide by 8.
$ 5 \minus{} \frac{5}{8}y \equal{} x$

Substituting our original variable names in, we have
$ 5 \minus{} \frac{5}{8}SUS \equal{} SJ$


symmetry
combinatorics open
combinatorics

Don't know if these are truly open, but there are no formulas for them in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url]:

1) Define a "standard urinal-choice permutation" of length $ n$ to be a permutation $ w \equal{} w_1 \ldots w_n$ of $ [n] \equal{} \{1, \ldots, n\}$ such that $ w_1$ is arbitrary and $ w_i$ is chosen so as to maximize $ \min \{|w_1 \minus{} w_i|, |w_2 \minus{} w_i|, \ldots, |w_{i \minus{} 1} \minus{} w_i|\}$.  (Equivalently, we choose each successive entry to be as far as possible from the nearest of all preceding entries.)  How many standard urinal-choice permutations are there of length $ n$?

For example, the 8 such permutations of length 4 are
1423
1432
2413
2431
3124
3142
4123
4132.
The associated sequence is [url=http://www.research.att.com/~njas/sequences/A095236]A095236[/url].

2) Define a "modified urinal-choice permutation" of length $ n$ to be a permutation $ w \equal{} w_1 \ldots w_n$ of $ [n]$ such that $ w_1$ is arbitrary and $ w_i$ is chosen so as to maximize $ |w_1 \minus{} w_i| \plus{} |w_2 \minus{} w_i| \plus{} \ldots \plus{} |w_{i \minus{} 1} \minus{} w_i|$.  (Equivalently, we choose each successive entry to be as far as possible on average from all preceding entries.)  How many modified urinal-choice permutations are there of length $ n$?

For example, the 6 such permutations of length 4 are
1423
1432
2413
3142
4123
4132.
The associated sequence is [url=http://www.research.att.com/~njas/sequences/A095698]A095698[/url].  It is conjectured that if $ a_n$ is the number of such permutations then for $ k\geq1$, $ a_{2k} \equal{} a_{2k \minus{} 1} \plus{} 2^{k \minus{} 1}$ and $ a_{2k \plus{} 1} \equal{} 2a_{2k \minus{} 1} \plus{} a_{2k}$, so it might be nice to both prove this and solve the recurrence.

Related sequences (also in need of formulas) are [url=http://www.research.att.com/~njas/sequences/A095239]A095239[/url] (circular permutations) and [url=http://www.research.att.com/~njas/sequences/A095240]A095240[/url] (exhibiting foresight).

This is the solution for modified urinal-choice permutations, henceforth abbreviated MUCPs.

Let $ U_n$ be the number of MUCPs of length $ n$, and let $ a_{n, i}$ be the number of MUCPs of length $ n$ beginning with $ i$.  Note that $ |1 - i| + |n - i| = n - 1$ for any $ i \in \{2, \ldots, n - 1\}$, so if at any point we've already included both $ 1$ and $ n$ in our permutation, the sequence of choices we can make from that point onwards is the same as if we erased both $ 1$ adn $ n$ and acted as if we were working on permutations of length $ n - 2$.

Note that if $ w$ is a MUCP and $ w_1 = 1$ then $ w_2 = n$, while if $ w_1 = n$ then $ w_2 = 1$.  Thus $ a_{n, 1} = a_{n, n} = U_{n - 2}$.

If $ w$ is a MUCP and $ w_1 = i$ with $ 1 < i < \frac {n + 1}{2}$ then $ w_2 = n$ and $ w_3 = 1$ and so $ a_{n, i} = a_{n - 2, i - 1}$.  Similarly, if $ \frac {n + 1}{2} < i < n$ then $ w_2 = 1$ and $ w_3 = n$ and so $ a_{n, i} = a_{n - 2, i - 1}$.

Finally, if $ w$ is a MUCP, $ n \geq 3$ is odd and $ w_1 = \frac {n + 1}{2}$ then we have $ \{w_2, w_3\} = \{1, n\}$ in either order.  Thus $ a_{n, i} = 2a_{n - 2, i - 1}$.  Since $ a_{1, 1} = 1$ we have $ a_{n, i} = 2^{i - 1}$ and so also $ a_{n, i} = a_{n - 2, i - 1} + 2^{i - 2}$.

Putting all this together, we have for $ n \geq 2$ that
\begin{align*} U_{2n} & = \sum_{i = 1}^{2n} a_{2n, i} \\
& = 2U_{2n - 2} + \sum_{i = 2}^{2n - 1} a_{2n - 2, i - 1} \\
& = 3U_{2n - 2} \end{align*}
and for $ n \geq 1$ that
\begin{align*} U_{2n + 1} & = \sum_{i = 1}^{2n + 1} a_{2n + 1, i} \\
& = 2U_{2n - 1} + 2^{n - 1} + \sum_{i = 2}^{2n} a_{2n - 1, i - 1} \\
& = 3U_{2n - 1} + 2^{n - 1}. \end{align*}
Since $ U_{2} = 2$ this immediately gives $ U_{2n} = 2 \cdot 3^{n - 1}$ for $ n \geq 1$.  Solving the recursion for the odd terms is only modestly more difficult and gives $ U_{2n + 1} = 2\cdot 3^n - 2^n$.

Also of interest may be circular urinal-choice permutations: the $ i$th entry is chosen to be the midpoint of the longest interval that does not contain any previously chosen entry, where we consider $ 1$ and $ n$ to be adjacent.  This is sequence [url=http://www.research.att.com/~njas/sequences/A095239]A095239[/url], which begins 1, 2, 6, 8, 40, 96, 168, 384, 1728, 15360, etc.  Ignoring cyclic symmetry, this is 1, 1, 2, 2, 8, 16, 24, 48, 192, 1536, ..., which is not given by any sequence in the OEIS.


[hide]How many ways are there to put 4 balls in 3 boxes if two balls are indistinguishably green, two are indistinguishably red, and the boxes are distinguishable?

We can have:

All balls in one box - 2 ways

three balls in one, one in the other - 4 ways

two in one, two in other: 3 ways

2+4+3=9[/hide]

[color=red] [size=85] [Do not quote the post directly before yours] [/size] [/color]
But there are [b]$3$[/b] boxes, not just $2$.  So that wouldn't work.

[hide]We should use casework on this:
[u]Case #1: All Balls in One Box[/u]
There are $3$ Ways since the boxes are distinguishable.

[u]Case #2: 2 Balls in One Box, 2 Other Balls in 1 Other Box[/u]
$3$ ways to choose the 2 boxes $\times$ $3$ combinations of balls=$9$

[u]Case #3: 2 Balls in One Box and 1 Ball in Each of the Other Boxes[/u]
Box 1: G,G Box 2: R Box 3: R
Box 1: R,G Box 2: G Box 3: G
Box 1: R,G Box 2: R Box 3: G

Since there are a total of $3$ ways this can happen with Box 1 having 2 balls, and there are $3$ ways we can pick the box that holds $2$ balls, there is a total of $9$ possibilities for this case.

Now we add all these cases up to arrive to the answer:
$3+9+9=18+3=\boxed{21}$[/hide]

[hide]Your math seems right, but you missed the '$3$ balls in one and $1$ ball in another' case. In  this case we either have a red ball left out, or a green ball. So thats $2$ times the $6$ ways to choose the boxes gives use $12$ plus the original $21$ gives us $\boxed{33}$.[/hide]

[hide]No offense, but all of you are wrong: ALCUMUS says that the answer is 36

For two indistinguishable green balls, we can place the balls in a box together or in separate boxes. There are 3 options to arrange them together (in box 1, 2, or 3) and 3 options for placing them separately (nothing in box 1, 2, or 3). So there are 6 ways to arrange the indistinguishable green balls. By the same reasoning, there are 6 ways to arrange the indistinguishable red balls, for a total of 6 times 6 arrangements of the 4 balls.

------------
Lakers will dominate once again.[/hide]

[hide="Casework"]
[b]Case #1: All in a Box[/b]
There are 3 ways; one in the first box, one in the second, and one in the third.

[b]Case #2: 2 and 2[/b]
There exist 6 ways if it's two greens and two reds, because there are three boxes and the color order can be switched. There exist 3 ways if it's red green and red green (switching the colors would result in the same thing). 6+3=9.

[b]Case #3: 3 and 1[/b]
There are 6 ways if the lone ball is red, and 6 ways if it's green. 6+6=12.

[b]Case #4: 1, 1, and 3[/b]
This case gets a little complicated. If the two lone balls are red, then there are 3 ways (3C2), if they're green, they are also 3 ways. If one is green and one is red, then there are 6 ways, because the red and green can switch places. 3+3+6=12.

Adding these all together, there are $3+9+12+12=\boxed{36}$[/hide]

[hide=Stars and Bars] Imagine this as a candy problem. We first need to split the two green balls which can be done in $\dbinom{2+3-1}{3-1}$. The same goes for the red candy. Thus there are $\dbinom{4}{2}\dbinom{4}{2}=6\times6=\boxed{36}$ ways.[/hide]

How many ways are there to put 5 balls in 3 boxes if the balls are not distinguishable but the boxes are?ans 81

[quote=Aiyan]is this the non-searchable

Alcumus**

 forum?[/quote]

It's the "official" one if that's what you're asking, and it is non searchable

[quote=kevinmathz]I did it on Alcumus. The correct answer is 10.[/quote]

10 is not correct.

[hide=The correct solution, in my opinion, is:]
\begin{align*}
&\text{We figure out how many ways we can put the red balls in the boxes, then we square that(the white balls are independent of the red).}\\
\text{We have casework:}\\
&\text{Case 1: The 2 red balls are in different boxes.}\\
\binom{3}2=3\\
&\text{Case 2: The 2 red balls are in the same boxes.}\\
\binom{3}1=3\\
(3+3)^2=36\\
&\text{Thus, } \boxed{36} \text{ is our correct answer.}\\
\end{align*}[/hide]

I think we've already established that. And, technically, there is no "right" solution to a problem.

We are all getting different answers because we are solving different questions. There's like a ton of these questions which each have different wordings. :|  :oops_sign:  :surrender:

Suppose that $a$ varies inversely with $b^2$. If $a=9$ when $b=2$, find the value of $a$ when $b=3$.

[hide][hide][hide][size=50]I think it is 4?[/size][/hide][/hide][/hide]

We will consider this as a composite of two problems with two indistinguishable balls and 3 distinguishable boxes. For two indistinguishable green balls, we can place the balls in a box together or in separate boxes. There are $3$ options to arrange them together (in box 1, 2, or 3) and  $3$ options for placing them separately (nothing in box 1, 2, or 3). So there are 6 ways to arrange the indistinguishable green balls. By the same reasoning, there are 6 ways to arrange the indistinguishable red balls, for a total of $6 \times 6 = \boxed{36}$ arrangements of the 4 balls.

if you do the alcumus solution, then say so.

[hide=Solution=Casework]
We look at each case and count the number of distinguishable permutations:
6: GG RR __
3: GG R R
3: RR G G
6: GR R G
3: GR RG __
6: RRG G __
6: GGR R __
3: GGRR __ __
Adding up the cases gives $6+3+3+6+3+6+6+3=\boxed{36}$.
[/hide]

[quote=Gauss4Euler]oh yay another level 23 problem on easy setting[/quote]

Yep. That's the sad part.

Okay. Phew. Only 6 day bump, but
[hide=Proposal]
Can there be more variety?
I have done at least 10 balls in boxes problems.
Maybe add a name?
Bob needs to put 4 indistinguishable balls in 3 distinguishable boxes?
Or, Bob needs to throw 4 distinguishable pieces of trash in 3 identical trash cans?[/hide]

[color=red][responding to post #2][/color]

I made the exact same mistake on this problem; read 3 boxes as 2 boxes.  :rotfl:

[hide = Solution]

We can find the arrangements of the green balls:

$(2G,0,0)$
$(0,2G,0)$
$(0,0,2G)$
$(G,G,0)$
$(0,G,G)$
$(G,0,G)$

That is 6 possibilities. We can do the same thing for the two red balls, and we find that the number of ways to put the 4 balls is $6 \cdot 6 = \boxed{36}$

[quote=abc2142][quote=sub_math]wait why is my solution wrong :(

so basically, first, we can find the number of ways to arrange the 'word' $GGRR$ which is equal to $6$.

Then, we can use stars and bars to find the number of ways to divide the balls into boxes. One example is shown:

G|R|RG

using stars and bars, the answer becomes $15$

Then we multiply the two to get $15*6=90$

why is this wrong?[/quote]

\bump I thought about that too, but in stars and bars, it's indistinguishable to distinguishable, and this isn't (memmed). Can someone explain why this doesn't work though?[/quote]

Haven't done the math yet, but you guys sure that you aren't overcounting stuff with stars and bars? I mean you have to remember that G|R|RG is the same as R|G|RG or in any order....

[quote=abc2142]Oh, thank you! What I thought was that GIRIRG is the same as GIRIGR.[/quote]

wait- thats right....    G|R|GR is the same as G|R|RG what is saying is that G|R|RG is also the same as R|G|GR and you might have overcounted :P

Oops yes that was what I meant lol. My communication skills are bad

Stars and Bars solution (didn't see this one up there hehe)

Since we are given an indistinguishable - distinguishable case, we can use stars and bars. However, our restriction is the two different colors of the balls. Regardless, we use stars and bars on the two colors separately and then multiply them (they are both independent scenarios) $$\binom{2+3-1}{3-1} \binom{2+3-1}{3-1} \implies \boxed{36}$$


algebra
polynomial
modular arithmetic
number theory unsolved
number theory

Hello,
I have an examination soon, can someone please tell me the answer for the following questions ?

1) Use Gauss's Lemma to solve ((a+5)/37)

2) Let p be any prime congrent to 1 (mod 8) and let g be a generator for Z*p.

a) Explain why there is an element x of order 8 in Zp and why x^2=x^(-2)

b) Now show that (x-x^(-1))^2=-2

c)Use part b to find all elements y=squareroot(-2) (mod p)

Thank You

Can you write it by Text.I think it is not difficult.

[quote="Joanna.W"]
2) Let p be any prime congrent to 1 (mod 8) and let g be a generator for Z*p.

a) Explain why there is an element x of order 8 in Zp and why x^2=x^(-2)

b) Now show that (x-x^(-1))^2=-2

c)Use part b to find all elements y=squareroot(-2) (mod p)

Thank You[/quote]

a) Er...do you mean $ x^{4}\equiv x^{\minus{}4}$?  As for proving there is an element of order $ 8$, you can consider the polynomial $ x^{8}\minus{}1$ and its zeroes.  If there are no elements of order $ 8$, then any zero of this polynomial has order dividing $ 4$, and therefore that zero will divide $ x^{4}\minus{}1$.  By Lagrange, it then follows that there are at most $ 4$ zeroes of $ x^{8}\minus{}1$.  But the polynomial $ x^{p\minus{}1}\minus{}1$ has exactly $ p\minus{}1$ zeroes by Fermat, and therefore, $ x^{8}\minus{}1$ has exactly $ 8$ zeroes.  We have a contradiction.

b) We have $ x^{4}\equiv\minus{}1\mod{p}$.  So

$ \minus{}1\minus{}2x^{2}\plus{}1\equiv\minus{}2x^{2}$
$ x^{4}\minus{}2x^{2}\plus{}1\equiv\minus{}2x^{2}$
$ (x^{2}\minus{}1)^{2}\equiv\minus{}2x^{2}$
$ (x\minus{}x^{\minus{}1})^{2}\equiv\minus{}2$

c) We have there are exactly two such $ y$: plus and minus something.  So take an element $ x$ of order $ 8$.  Then $ (x\minus{}x^{\minus{}1})$ satisfies this property, as does $ (x^{\minus{}1}\minus{}x)$.  So these are all solutions.  This yields an interesting result: if we take all $ x$ of order $ 8$ (of which there are $ \varphi(8)\equal{}4$, prove this) modulo p, then the values $ x\minus{}x^{\minus{}1}$ are all $ \pm a$ for some $ a$.

[quote]So take an element X of order 8. Then (x-x^(-1)) satisfies this property, as does (x^(-1)-x)[/quote]

So do I pick any element of order 8 to show it satisfies (x-x^(-1)) and (x^(-1)-x) in my working out ? What element would be a simple one to prove it ?

[quote]of which there are w(8)=4, prove this[/quote]

I don't know how to prove this, can you or anyone please show me ?

[quote] 2) Let p be any prime congrent to 1 (mod 8) and let g be a generator for Z*p.

a) Explain why there is an element x of order 8 in Zp and why x^2= -x^(-2)[/quote]

I still don't get this, can anyone please simple show me how to do this question ?

Thank you for your input.

[hide="start on 2"]
There exists a primitive root $ g$ in $ \mathbb{Z}_{p}$ (an element of order $ p\minus{}1$). Then let $ x \equal{} g^{(p\minus{}1)/8}$, which makes sense given $ p\equiv 1\pmod{8}$. Show that $ x$ has order 8. So $ x^{8}\equal{} 1$ but $ x^{4}\neq 1$. Now use the fact that $ \mathbb{Z}_{p}$ has no zero divisors to get something nice for $ x^{4}$.
[/hide]

Okay, I get this one now, thank you.

Can scorpius119 or anyone help me with my first question, I'll quote it up again below.

[quote]So take an element X of order 8. Then (x-x^(-1)) satisfies this property, as does (x^(-1)-x)[/quote]

So do I pick any element of order 8 to show it satisfies (x-x^(-1)) and (x^(-1)-x) in my working out ? What element would be a simple one to prove it ?

Thank you for your input.


Find all positive intergers which can  be written as fraction of the form
$\frac{mn+1}{m+n}$ where m,n are positive intergers

[b]Solution.[/b] Let $\frac{mn+1}{m+n}=k, k\in\mathbb{N}$.  Then we have
\[mn-km-kn+1=0\implies (m-k)(n-k)=(k-1)(k+1) \]
As a result, for all $k\in\mathbb{N}$, we can select $(m,n)=(2k-1,2k+1)$. $\blacksquare$

This problem seems to be too easy.  Did I misunderstand anything? :maybe:


How much tissue should you bring, as the paper said? Sorry for making a thread out of this if it's unnecessary.

[quote="alexhhmun"]There are going to be 4 people in your apartment, so you only need one or two.  What are the odds of catching a cold in the middle of the summer in TX?[/quote]

100%, Texas is known for its summer blizzards

hmmm i can vouch against that statement, i have a cold right now :O

The cold just lowers your immune system, which makes you more vulnerable to colds.  I've actually had plenty of summer colds, so I guess the heat works similarly?

Actually it turned out I DID get a cold. But I didn't bring any tissue lol.

I got a cold too, but that was because someone turned the thermostat to 50 in the middle of the night...

I think almost everybody got a cold. Too many people swapping germs every day.

Our room had tissues, though  :D

I got sick too... so sick that I actually had to miss class one day. Thank goodness I brought like 8 packs!


function
real analysis
real analysis theorems

let $ f: \mathbb{R}\to\mathbb{R}$ be monotone. Prove that

1) $ f(\mathbb{R})$ is a $ G_{\delta}$ set

2) $ f(\mathbb{R})\neq \mathbb{R}\setminus\mathbb{Q}$

Prove that for every closed $ A\subset\mathbb{R}$ there exists monotone function $ f: \mathbb{R}\to\mathbb{R}$ such that $ f(\mathbb{R}) \equal{} A$.


limit
logarithms
search
ratio
function
geometric sequence
real analysis

$ L_1\equiv\lim_{n\to\infty}\frac {1^n \plus{} 2^n \plus{} 3^n \plus{} \ldots \plus{} n^n}{n^n} \equal{} \frac {e}{e \minus{} 1}$ $ \mathrm {(\ Isac\ Schoenberg\ )}\ .$

$ L_2\equiv$ $ \lim_{n\to\infty}\left(\frac {1}{\sqrt {n^2 \plus{} 1}} \plus{} \frac {1}{\sqrt {n^2 \plus{} 2}} \plus{} \ldots \plus{} \frac {1}{\sqrt {n^2 \plus{} (n \minus{} 1)}} \plus{} \frac {1}{\sqrt {n^2 \plus{} n}}\right)^n \equal{}$ $ \frac {1}{\sqrt [4] e}\ .$

let me try to prove seccond and after first problem  :) :
let consider sum
$ \sum _{i \equal{} 1}^{n}\frac {1}{\sqrt {n^{2} \plus{} i}}$~$ \sum _{i \equal{} 1}^{n} \frac {1}{n}(1 \minus{} \frac {k}{2n^{2}}) \equal{} 1 \minus{} \frac {n^{2} \plus{} n}{4n^{3}}$ it's true because $ k < < n^{2}$ when $ n \minus{} > \plus{} oo$
and in finish we have
$ (\sum _{i \equal{} 1}^{n}\frac {1}{\sqrt {n^{2} \plus{} i}})^{n}$~$ e^{n \ln (1 \minus{} \frac {n^{2} \plus{} n}{4n^{3}} )}$~$ e^{ \minus{} \frac {1}{4}}$ done

and first:

let me consider sum
$ S_{n \minus{} 1} \equal{} \sum _{i \equal{} 1}^{n \minus{} 1} i^{n}$
therefore
$ S_{n} \equal{} \sum_{i \equal{} 1}^{n}i^{n}$ let's try to prove that this sum $ S_{n}$ is aproximately equal to $ n^{n}\frac {e}{e \minus{} 1}$
because by Cezaro-Stolt's theorem we have (check)
$ \frac {S_{n} \minus{} S_{n \minus{} 1}}{\frac {e}{e \minus{} 1} (n^{n} \minus{} (n \minus{} 1)^{n})} \equal{} \frac {1 \minus{} \frac {1}{e}}{1 \minus{} (1 \minus{} \frac {1}{n})^{n}} \minus{} > 1$ when $ n \minus{} > \plus{} oo$
OK
after we have $ \frac {S_{n}}{n^{n}}$~$ \frac {n^{n}e/(e \minus{} 1)}{n^{n}} \equal{} \frac {e}{e \minus{} 1}$ Done
good problem Virgil Nicula thanks  :)

\[ L_1 \equal{} \lim_{n\to\infty}\frac {1^n \plus{} 2^n \plus{} \ldots \plus{} (n \minus{} 1)^n \plus{} n^n}{n^n} \equal{}\lim_{n\to\infty}\frac {(n \plus{} 1)^n}{(n \plus{} 1)^n \minus{} n^n}\]
\[ \equal{}\lim_{n\to\infty}\frac {1}{1 \minus{}\left( \frac{n}{n\plus{}1}\right)^n}\equal{} \frac {e}{e \minus{} 1}\]

where the first equality is by the [url=http://en.wikipedia.org/wiki/Stolz-Ces%C3%A0ro_theorem]Stolz-Cesàro theorem[/url].

the exponent also changes, your application of stolz theorem is wrong.

Be sure to check Moubinool's post [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=181425151&t=1088]here[/url].

[quote]$ L\equiv \lim_{n\to\infty}\left(\frac {1}{\sqrt {n^2 \plus{} 1}} \plus{} \frac {1}{\sqrt {n^2 \plus{} 2}} \plus{} \ldots \plus{} \frac {1}{\sqrt {n^2 \plus{} (n \minus{} 1)}} \plus{} \frac {1}{\sqrt {n^2 \plus{} n}}\right)^n \equal{} \frac {1}{\sqrt [4] e}\ .$[/quote]
[color=darkblue][b][u]Proof.[/u] Define $ a_n \equiv \sum_{k \equal{} 1}^n\frac {1}{\sqrt {n^2 \plus{} k}}$. Observe that $ \frac {n}{\sqrt {n^2 \plus{} n}}\le a_n\le \frac {n}{\sqrt {n^2 \plus{} 1}}$ $ \implies$
$ \lim_{n\to\infty}a_n\equal{}1$ and we are in the indetermination $ 1^{\infty}$. Therefore, $ L \equal{} e^l$,  where
$ l \equal{} \lim_{n\to\infty}n\cdot \left(\sum_{k \equal{} 1}^n\frac {1}{\sqrt {n^2 \plus{} k}} \minus{} 1\right) \equal{}$ $ \lim_{n\to\infty}n\cdot\sum_{k \equal{} 1}^n\left(\frac {1}{\sqrt {n^2 \plus{} k}} \minus{} \frac 1n\right) \equal{}$
 $ \lim_{n\to\infty}\sum_{k \equal{} 1}^n\frac {n \minus{} \sqrt {n^2 \plus{} k}}{\sqrt {n^2 \plus{} k}} \equal{}$ $ \minus{} \lim_{n\to\infty}b_n$, where $ b_n\equiv\sum_{k \equal{} 1}^n\frac {k}{n\sqrt {n^2 \plus{} k} \plus{} n^2 \plus{} k}\ .$
Define $ c_n\equiv\sum_{k \equal{} 1}^n\frac {k}{n\sqrt {n^2 \plus{} n} \plus{} n^2 \plus{} n}$ and $ d_n\equiv\sum_{k \equal{} 1}^n\frac {k}{n\sqrt {n^2 \plus{} 1} \plus{} n^2 \plus{} 1}$. Observe that
$ c_n\le b_n\le d_n$ and $ c_n \equal{} \frac {1}{n\sqrt {n^2 \plus{} n} \plus{} n^2 \plus{} n}\cdot\sum_{k \equal{} 1}^nk \equal{} \frac {n(n \plus{} 1)}{2\left[n\sqrt {n^2 \plus{} n} \plus{} n^2 \plus{} n\right]}\rightarrow\frac 14$
and $ d_n \equal{} \frac {1}{n\sqrt {n^2 \plus{} 1} \plus{} n^2 \plus{} 1}\cdot\sum_{k \equal{} 1}^nk \equal{} \frac {n(n \plus{} 1)}{2\left[n\sqrt {n^2 \plus{} 1} \plus{} n^2 \plus{} 1\right]}\rightarrow\frac 14$.
In conclusion, $ b_n\rightarrow\frac 14$ and $ l \equal{} \minus{} \frac 14$, i.e. $ L \equal{} \frac {1}{\sqrt [4]e}\ .$[/b][/color]

Let $ a>0$

$ u(n)\equal{}\sum_{k\equal{}1}^{n}k^{na}$

Find an equivalent of $ u(n)$

[b][color=darkred]Thank you, Misan ![/color]

[color=darkblue]I learned now a new instruction in LaTeX. See here :[/color]

$ \mathrm {ALPHABET}\equiv\overbrace{A \plus{} B \plus{} C \plus{} \ldots \plus{} X \plus{} Y \plus{} Z}^{\mathrm {n\ of\ them}}\ \ \vee\ \ \mathrm {alphabet}\equiv\underbrace{a \plus{} b \plus{} c \plus{} \ldots \plus{} x \plus{} y \plus{} z}_{\mathrm {n\ of\ them}}$

[color=darkblue]It is very nicely ![/color][/b]

[quote="Virgil Nicula"]$ L_1\equiv\lim_{n\to\infty}\frac {1^n \plus{} 2^n \plus{} 3^n \plus{} \ldots \plus{} n^n}{n^n} \equal{} \frac {e}{e \minus{} 1}\ .$[/quote]
here is a quick and dirty way... :D

$ \lim_{n\to\infty}\frac {1^n \plus{} 2^n \plus{} 3^n \plus{} \ldots \plus{} n^n}{n^n}$

reverse the terms and write it as

$ \equal{}\lim_{n\to\infty}\frac {n^n \plus{} (n\minus{}1)^n \plus{} (n\minus{}2)^n \plus{} \ldots \plus{} 1^n}{n^n}$

$ \equal{}\lim_{n\to\infty}1 \plus{} \left(1 \minus{} \dfrac{1}{n}\right)^n \plus{} \left(1 \minus{} \dfrac{2}{n}\right)^n \plus{} \left(1 \minus{} \dfrac{3}{n}\right)^n \plus{} \ldots \plus{} \left(1 \minus{} \dfrac{n\minus{}1}{n}\right)^n$

$ \equal{}\lim_{k\to\infty}1 \plus{} e^{\minus{}1} \plus{} e^{\minus{}2} \plus{} e^{\minus{}3} \plus{} \ldots \plus{} e^{\minus{}k\plus{}1}$

this is a geometric progression with $ 1^{st}$ term $ \equal{}\,1$ and common ratio $ \equal{}\,e^{\minus{}1}\;\;(<1)$. and so,

$ L_1 \equal{} \dfrac{1}{1\minus{}e^{\minus{}1}} \equal{} \dfrac {e}{e\minus{}1} \;\;\;\quad\quad (\because\;sum\;\;of\;\;G.P.\;\;with\;\;|r|<1\;\;\equal{}\;\;\dfrac{a}{1\minus{}r})$

[quote="Virgil Nicula"][quote="misan"]$ \lim_{n\to\infty}\left[1 \plus{} \left(1 \minus{} \dfrac{1}{n}\right)^n \plus{} \left(1 \minus{} \dfrac{2}{n}\right)^n \plus{} \left(1 \minus{} \dfrac{3}{n}\right)^n \plus{} \ldots \plus{} \left(1 \minus{} \dfrac{n \minus{} 1}{n}\right)^n\right]$

$ \equal{} \lim_{k\to\infty}\left(1 \plus{} e^{ \minus{} 1} \plus{} e^{ \minus{} 2} \plus{} e^{ \minus{} 3} \plus{} \ldots \plus{} e^{ \minus{} k \plus{} 1}\right)$[/quote]
[color=darkblue][b]Why ? Please, are you more clearly ![/b][/color][/quote]

$ \lim_{n\to\infty}\left[1 \plus{} \left(1 \minus{} \dfrac{1}{n}\right)^n \plus{} \left(1 \minus{} \dfrac{2}{n}\right)^n \plus{} \left(1 \minus{} \dfrac{3}{n}\right)^n \plus{} \ldots \plus{} \left(1 \minus{} \dfrac{n \minus{} 1}{n}\right)^n\right]$

$ \equal{}\lim_{n\to\infty}\sum_{j\equal{}0}^{n\minus{}1}\;\left(1\minus{}\dfrac{j}{n}\right)^n\;\equal{}\;\lim_{k\to\infty}\sum_{j\equal{}0}^{k\minus{}1}\;\left\{\lim_{n\to\infty}\;\left(1\minus{}\dfrac{j}{n}\right)^n\right\}\;\equal{}\;\lim_{k\to\infty}\sum_{j\equal{}0}^{k\minus{}1}\;\left\{e^{\minus{}j}\right\}$

$ \equal{} \lim_{k\to\infty}\left(1 \plus{} e^{ \minus{} 1} \plus{} e^{ \minus{} 2} \plus{} e^{ \minus{} 3} \plus{} \ldots \plus{} e^{ \minus{} k \plus{} 1}\right)$

the above follows from the [u]monotone convergence theorem[/u]

[quote="Moubinool"]Let $ a > 0$

$ u(n) \equal{} \sum_{k \equal{} 1}^{n}k^{na}$

Find an equivalent of $ u(n)$[/quote]
answer is $ ...$~$ \frac{n^{an}}{1\minus{}e^{\minus{}a}}$
 (by Cezaro-Stolts theorem )

[quote]$ L\equiv\lim_{n\to\infty}\frac {1^n \plus{} 2^n \plus{} 3^n \plus{} \ldots \plus{} n^n}{n^n} \equal{} \frac {e}{e \minus{} 1}\ .$[/quote]
[hide="Lemma."] $ x\in [0,1)\implies \ln (1 \minus{} x)\le \minus{} x$ and $ \sum_{k \equal{} 1}^na_k \equal{} \sum_{k \equal{} 1}^na_{n \plus{} 1 \minus{} k}\ .$
[b]Proof.[/b] Denote $ f: [0,1)\rightarrow \mathcal R\ ,\ f(x) \equal{} x \plus{} \ln (1 \minus{} x)\ .$
Observe that $ f(0) \equal{} 0$ and $ f'(x) \equal{} 1 \minus{} \frac {1}{1 \minus{} x} \equal{} \minus{} \frac {x}{1 \minus{} x}\le 0\ .$
Thus, the function $ f$ is decreasing ($ \searrow$), i.e. $ 0\le x < 1$ $ \implies$
$ f(x)\le f(0)$ $ \implies$ $ f(x)\le 0$ $ \implies$ $ \ln (1 \minus{} x)\le \minus{} x$ with equality iff $ x \equal{} 0\ .$[/hide]
$ \blacktriangleright$ Particularly, for any $ k\in \overline {1,n \minus{} 1}$ , $ \ln\left(1 \minus{} \frac kn\right)\le \minus{} \frac kn$, i.e. $ \left(1 \minus{} \frac kn\right)^n\le e^{ \minus{} k}\ .$
Therefore,  $ \sum_{k \equal{} 1}^n\left(\frac kn\right)^n \equal{}$ $ 1 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1}\left(\frac kn\right)^n \equal{}$ $ 1 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1}\left(\frac {n \minus{} k}{n}\right)^n \equal{}$
$ 1 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1}\left(1 \minus{} \frac kn\right)^n\le$ $ 1 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1}e^{ \minus{} k} \equal{}$ $ \frac {1 \minus{} \frac {1}{e^n}}{1 \minus{} \frac 1e} < \frac {e}{e \minus{} 1}\implies$ $ \boxed {L\le\frac {e}{e \minus{} 1}}\ \ (1)\ .$

$ \blacktriangleright$ For a fixed $ m\in \overline {1,n}$ we have $ \sum_{k \equal{} 1}^n\left(\frac kn\right)^n \equal{} \sum_{k \equal{} 1}^n\left(\frac {n \plus{} 1 \minus{} k}{n}\right)^n\ge \sum_{k \equal{} 1}^m\left(\frac {n \plus{} 1 \minus{} k}{n}\right)^n\ .$
For $ n\to\infty$ obtain $ L\ge \sum_{k \equal{} 1}^m\lim_{n\to\infty}\left(\frac {n \plus{} 1 \minus{} k}{n}\right)^n \equal{} \sum_{k \equal{} 1}^m\frac {1}{e^{k \minus{} 1}} \equal{} \frac {1 \minus{} \frac {1}{e^m}}{1 \minus{} \frac 1e}\ .$
For $ m\to\infty$ obtain $ \boxed {L\ge \frac {e}{e \minus{} 1}}\ \ (2)\ .$ From the relations $ (1)$ and $ (2)$ obtain $ \boxed {L \equal{} \frac {e}{e \minus{} 1}}\ .$

[quote="misan"][quote="Virgil Nicula"][quote="misan"]$ \lim_{n\to\infty}\left[1 \plus{} \left(1 \minus{} \dfrac{1}{n}\right)^n \plus{} \left(1 \minus{} \dfrac{2}{n}\right)^n \plus{} \left(1 \minus{} \dfrac{3}{n}\right)^n \plus{} \ldots \plus{} \left(1 \minus{} \dfrac{n \minus{} 1}{n}\right)^n\right]$

$ \equal{} \lim_{k\to\infty}\left(1 \plus{} e^{ \minus{} 1} \plus{} e^{ \minus{} 2} \plus{} e^{ \minus{} 3} \plus{} \ldots \plus{} e^{ \minus{} k \plus{} 1}\right)$[/quote]
[color=darkblue][b]Why ? Please, are you more clearly ![/b][/color][/quote]

$ \lim_{n\to\infty}\left[1 \plus{} \left(1 \minus{} \dfrac{1}{n}\right)^n \plus{} \left(1 \minus{} \dfrac{2}{n}\right)^n \plus{} \left(1 \minus{} \dfrac{3}{n}\right)^n \plus{} \ldots \plus{} \left(1 \minus{} \dfrac{n \minus{} 1}{n}\right)^n\right]$

$ \equal{} \lim_{n\to\infty}\sum_{j \equal{} 0}^{n \minus{} 1}\;\left(1 \minus{} \dfrac{j}{n}\right)^n\; \equal{} \;\lim_{k\to\infty}\sum_{j \equal{} 0}^{k \minus{} 1}\;\left\{\lim_{n\to\infty}\;\left(1 \minus{} \dfrac{j}{n}\right)^n\right\}\; \equal{} \;\lim_{k\to\infty}\sum_{j \equal{} 0}^{k \minus{} 1}\;\left\{e^{ \minus{} j}\right\}$

$ \equal{} \lim_{k\to\infty}\left(1 \plus{} e^{ \minus{} 1} \plus{} e^{ \minus{} 2} \plus{} e^{ \minus{} 3} \plus{} \ldots \plus{} e^{ \minus{} k \plus{} 1}\right)$

the above follows from the [u]monotone convergence theorem[/u][/quote]

Sorry, how do you apply monotone convergence theorem here?


geometry
geometric transformation
rotation

[quote="CatalystOfNostalgia"]Equilateral triangle ABC has D on BC, E on CA, F on AB, such that BD=CE=AF. Show that the triangle formed by AD, BE, and CF is equilateral.[/quote]
[hide]Note that AFE, CED, and BDF are congruent. Let $ \angle CDE\equal{}\theta$ Then $ \angle CED\equal{}120\minus{}\theta$ and $ \angle FEA\equal{}\theta$ (by congruent triangles).
So, we see that DEF=60. We can do this for all three of the angles, and the result is obvious.[/hide]

[quote="CatalystOfNostalgia"]Equilateral triangle ABC has D on BC, E on CA, F on AB, such that BD=CE=AF. Show that the triangle formed by AD, BE, and CF is equilateral.[/quote]
Wrong triangle diophanteint!  Yeah, I earlier thought it was the same triangle too...
[hide="Solution"]
By the way, we can easily show that triangle DEF is congruent by noting that triangles AEF, CDE, and BFD are congruent and thus DE=EF=DF (we don't need to play with the angles).

Denote the intersection between AD and BE to be X, BE and CF to be Y, and CF and AD to be Z.
Firstly, it is quite trivial to show that $ \angle BAD \equal{} \angle ACF \equal{} \angle CBE$ by noting that triangles BEC, ADB, and CFA are congruent.  Call all these angles equal to x.  Now, it is easy to see that $ \angle ADB \equal{} 120 \minus{} x \equal{} \angle CFA \equal{} \angle BEC$.  Thus, by SAS congruence, we have that triangles BXD, CYE, and AZF are congruent (noting that AF=CE=BD and that angles XBD, ECY, and FAZ are equal).  Thus, $ BX \equal{} YC \equal{} ZA$.  Also, noting that BC=AC=AB, $ \angle YBC \equal{} \angle ZCA \equal{} \angle BAX$, and $ \angle YCB \equal{} \angle XBA \equal{} \angle ZAC$, we have that triangles YBC, ZCA, and XAB are congruent.  Thus, $ YB \equal{} ZC \equal{} XA$.  So, we have that $ YB \minus{} BX \equal{} CZ \minus{} CY \equal{} XA \minus{} ZA$ and so $ XY \equal{} YZ \equal{} XZ$.  [/hide]

[quote="The QuattoMaster 6000"]
By the way, we can easily show that triangle DEF is congruent by noting that triangles AEF, CDE, and BFD are congruent and thus DE=EF=DF (we don't need to play with the angles).

Denote the intersection between AD and BE to be X, BE and CF to be Y, and CF and AD to be Z.
Firstly, it is quite trivial to show that $ \angle BAD \equal{} \angle ACF \equal{} \angle CBE$ by noting that triangles BEC, ADB, and CFA are congruent.  Call all these angles equal to x.  Now, it is easy to see that $ \angle ADB \equal{} 120 \minus{} x \equal{} \angle CFA \equal{} \angle BEC$.[/quote]

After this step, you can just note that by $ \Delta BXD$, $ \angle{BXD}\equal{}180^{\circ}\minus{}(x\plus{}120^{\circ}\minus{}x)\equal{}60^{\circ}$, so $ \angle{ZXY}\equal{}60^{\circ}$, and similarly for the others.

Hello, a 120 degree rotation about the center of the triangle maps AD to BE, or whatever. So the angle between those lines is 60 so we are done.

Let the intersections of the lines  be X, Y, and Z. Notice that ABD and BCE are congruent and therefore $ \angle{BAD} \equal{} \angle{CBE} \equal{} \theta$. now look at $ \triangle{BDX}$ it shares $ \angle{ADB}$ with $ \triangle{ABD}$ and it also has one angle (CBE) which is congruent to BAD as we proved. Therefore the last angle has to be 60. Same thing for the rest of the angles and we are done

EDIT: ok i think this is the solution catalyst intended two posts above mine.


function
algebra
domain
polynomial
algebra open

Find all functions $ f: \mathbb{R^\plus{}}\rightarrow \mathbb{R^\plus{}}$ satisfying \[ f(xf(x)\plus{}f(y))\equal{}(f(x))^2\plus{}y\] for all $ x,y>0.$

Here is a solution. Hopefully someone can propose a shorter one.

Let $ f(a)\equal{}b$ with $ a,b>0$.
$ P(a,y): f(ab\plus{}f(y))\equal{}b^2\plus{}y$. (F1)
$ P(x,a): f(xf(x)\plus{}b)\equal{}f(x)^2\plus{}a$.  (F2)
Replace $ y$ in (F1) with $ ab\plus{}f(y)$, we get
$ f(ab\plus{}b^2\plus{}y)\equal{}ab\plus{}b^2\plus{}f(y)$.
Let $ c\equal{}ab\plus{}b^2$, we get
$ f(c\plus{}y)\equal{}f(y)\plus{}c$  (F3)

1) $ f(u\plus{}v)\equal{}f(u\plus{}b)\plus{}f(v\plus{}ab)\plus{}d,\forall u,v>0$, where $ d$ is some constant.
From (F1), we know $ f$ is injective and $ (b^2,\infty)\subset Im(f)$.
Then from (F2), we see $ f(xf(x)\plus{}b)\equal{}t$ has solution whenever $ t>b^4\plus{}a$.
Also, $ f(ab\plus{}f(y))\equal{}t$ has solution whenever $ t>b^2$.
From (F3), we have $ f(y\plus{}nc)\equal{}f(y)\plus{}nc$, where $ n\in\mathbb N$.
For any fixed $ u,v>0$, we can find $ n,m\in\mathbb N$ such that
$ f(u\plus{}b\plus{}nc)\equal{}f(u\plus{}b)\plus{}nc>b^4\plus{}a$ and let $ r$ solve $ f(rf(r)\plus{}b)\equal{}f(u\plus{}b\plus{}nc)$
and
$ f(v\plus{}ab\plus{}mc)\equal{}f(v\plus{}ab)\plus{}mc>b^2$ and let $ q$ solve $ f(ab\plus{}f(q))\equal{}f(v\plus{}ab\plus{}mc)$.

Since $ f$ is injective, we have
$ rf(r)\plus{}b\equal{}u\plus{}b\plus{}nc$ and $ ab\plus{}f(q)\equal{}v\plus{}ab\plus{}mc$.
$ P(r,q): f(rf(r)\plus{}f(q))\equal{}f(r)^2\plus{}q\equal{}f(rf(r)\plus{}b)\minus{}a\plus{}f(ab\plus{}f(q))\minus{}b^2$ implies
$ f(u\plus{}nc\plus{}v\plus{}mc)\equal{}f(u\plus{}b\plus{}nc)\plus{}f(v\plus{}ab\plus{}mc)\plus{}d$, where $ d\equal{}\minus{}a\minus{}b^2$
Use (F3), we get
$ f(u\plus{}v)\equal{}f(u\plus{}b)\plus{}f(v\plus{}ab)\plus{}d$.

2) $ f(u\plus{}v)\equal{}f(u)\plus{}f(v)\plus{}e,\forall u,v>0$, where $ e$ is constant
Let $ n\in\mathbb N$ be such that $ nc>b$, we can apply $ (u\plus{}nc,v)$ and $ (u\plus{}nc\minus{}b,v\plus{}b)$ to 1),
$ f(u\plus{}nc\plus{}v)\equal{}f(u\plus{}nc\plus{}b)\plus{}f(v\plus{}ab)\plus{}d\equal{}f(u\plus{}nc)\plus{}f(v\plus{}ab\plus{}b)\Rightarrow f(u\plus{}b)\minus{}f(u)\equal{}f(v\plus{}ab\plus{}b)\minus{}f(v\plus{}ab)$.
RHS does not depend on $ u$. So we have $ f(u\plus{}b)\minus{}f(u)\equal{}e_1$ with constant $ e_1$.
Similarly $ f(v\plus{}ab)\equal{}f(v)\plus{}e_2$ with constant $ e_2$.
Then 1) implies 2)

3) $ f(u\plus{}v)\equal{}f(u)\plus{}f(v),\forall u,v>0$
Let $ g(x)\equal{}f(x)\plus{}e$. We see $ g$ satisfies Cauchy equation.
So for $ r\in\mathbb Q$, $ g(r)\equal{}rg(1)$ or $ f(r)\equal{}\minus{}e\plus{}rg(1)\equal{}\minus{}e(1\minus{}r)\plus{}rf(1)>0$.
Obviously $ e\le0$ and $ g(1)>0$.

Suppose $ e<0$.
(F1) can be rewritten as $ g(g(x))\equal{}x\plus{}e_3$, where $ e_3$ is constant.
$ g(g(r))\equal{}rg(g(1))\equal{}r\plus{}e_3,\forall  r\in\mathbb Q$
Hence, $ g(g(1))\equal{}1,e_3\equal{}0$ and $ g(g(x))\equal{}x$
For $ r,q\in\mathbb Q$,
$ P(r,q): f(rf(r)\plus{}f(q))\equal{}g(rg(r)\minus{}re\plus{}g(q)\minus{}e)\minus{}e\equal{}f(r)^2\plus{}q\equal{}(g(r)\minus{}e)^2\plus{}q$
or
$ r^2g(g(1))\plus{}rg(\minus{}e)\plus{}g(g(q))\plus{}g(\minus{}e)\minus{}e\equal{}(rg(1)\minus{}e)^2\plus{}q$
or
$ r^2\plus{}rg(\minus{}e)\equal{}r^2g(1)^2\minus{}2eg(1)r\plus{}e^2\plus{}e\minus{}g(\minus{}e)$.
Now, we treat both sides as polynomial of $ r$.
$ g(1)\equal{}1,g(\minus{}e)\equal{}\minus{}2e,e^2\plus{}e\equal{}g(\minus{}e)$ implies
$ e\equal{}\minus{}3,g(3)\equal{}6\ne 3g(1)$. Contradiction! So $ e\equal{}0$ and 3) holds.

4) $ f(x)\equal{}x$
3) implies $ f(r)\equal{}rf(1)$. Then $ f(f(r))\equal{}rf(f(1))$.
Hence, (F1) implies $ rf(f(1))\plus{}f(ab)\equal{}b^2\plus{}r\Rightarrow f(ab)\equal{}b^2$
Since $ a$ is arbitrary, we have
$ f(xf(x))\equal{}f(x)^2$ (F4)
Then the 3) and (F1) imply $ f(f(y))\equal{}y$ (F5)
Replace $ x$ with $ f(x)$ in (F4), we get
$ f(xf(x))\equal{}f(f(x))^2\equal{}x^2\equal{}f(x)^2$.
Hence, $ f(x)\equal{}x$.

[quote="xxp2000"]Here is a solution. Hopefully someone can propose a shorter one.[/quote]

Hello!. I have one question  about your demo and a proposal for another one (not very short too)

[quote="xxp2000"] 2) $ f(u \plus{} v) \equal{} f(u) \plus{} f(v) \plus{} e,\forall u,v > 0$, where $ e$ is constant
Let $ n\in\mathbb N$ be such that $ nc > b$, we can apply $ (u \plus{} nc,v)$ and $ (u \plus{} nc \minus{} b,v \plus{} b)$ to 1),
$ f(u \plus{} nc \plus{} v) \equal{} f(u \plus{} nc \plus{} b) \plus{} f(v \plus{} ab) \plus{} d \equal{} f(u \plus{} nc) \plus{} f(v \plus{} ab \plus{} b)\Rightarrow f(u \plus{} b) \minus{} f(u) \equal{} f(v \plus{} ab \plus{} b) \minus{} f(v \plus{} ab)$.
RHS does not depend on $ u$. So we have $ f(u \plus{} b) \minus{} f(u) \equal{} e_1$ with constant $ e_1$.
Similarly $ f(v \plus{} ab) \equal{} f(v) \plus{} e_2$ with constant $ e_2$.
Then 1) implies 2)[/quote]

we miss a "$ \plus{}d$" at the end of  $ f(u \plus{} nc \plus{} v) \equal{} f(u \plus{} nc \plus{} b) \plus{} f(v \plus{} ab) \plus{} d \equal{} f(u \plus{} nc) \plus{} f(v \plus{} ab \plus{} b)$ (but no consequence).
But I dont understand "Similarly $ f(v \plus{} ab) \equal{} f(v) \plus{} e_2$ with constant $ e_2$"

Here is my own proposal :

1) Original assertion $ P0(x,y)$ : $ f(xf(x)\plus{}f(y))\equal{}(f(x))^2\plus{}y$

Let $ a\equal{}f(1)$ and $ b\equal{}f(f(1))$

2) $ f(x)$ is an injective function :
If $ f(y_1)\equal{}f(y_2)$, subtracting $ P0(x,y_1)$ from $ P0(x,y_2)$ implies $ y_1\equal{}y_2$.
Q.E.D.

3) new assertion $ P1(x,y)$ : $ f(xf(x)\plus{}a^2\plus{}y)\equal{}(f(x))^2\plus{}a\plus{}f(y)$
$ P0(x,f(1)\plus{}f(y))$ $ \implies$ $ f(xf(x)\plus{}f(f(1)\plus{}f(y)))\equal{}(f(x))^2\plus{}f(1)\plus{}f(y)$. But (using $ P0(1,y)$) : $ f(f(1)\plus{}f(y))\equal{}(f(1))^2\plus{}y$.
Q.E.D.

4) new assertion $ S(x)$ : $ f(x\plus{}a^2\plus{}a)\equal{}f(x)\plus{}a^2\plus{}a$
This is an immediate result from $ P1(1,x)$
Q.E.D.

5) new assertion $ Q(x)$ : $ (f(x))^2\equal{}f(xf(x))\plus{}a^2\minus{}b$
$ P1(x,a)$ $ \implies$ $ f(xf(x)\plus{}a^2\plus{}a)\equal{}(f(x))^2\plus{}a\plus{}b$
$ S(xf(x))$ $ \implies$ $ f(xf(x)\plus{}a^2\plus{}a)\equal{}f(xf(x))\plus{}a^2\plus{}a$
These two equalities have the same left member. So : $ (f(x))^2\plus{}a\plus{}b\equal{}f(xf(x))\plus{}a^2\plus{}a$
Q.E.D.

6) new assertion $ P2(x,y)$ : $ f(xf(x)\plus{}f(y))\equal{}xf(x)\plus{}y\plus{}a^2\minus{}b$
This is an immediate result from using $ Q(x)$ in right part of $ P0(x,y)$
Q.E.D.

7) new assertion $ R(x)$ : $ (f(x))^2 \equal{} xf(x) \plus{} a^2 \minus{} a$
$ P2(x,a)$ : $ f(xf(x)\plus{}b)\equal{}xf(x)\plus{}a\plus{}a^2\minus{}b$
$ P2(1,xf(x))$ : $ f(a\plus{}f(xf(x)))\equal{}a\plus{}xf(x)\plus{}a^2\minus{}b$

These two equalities have the same right member. So, since $ f(x)$ is injective : $ xf(x)\plus{}b\equal{}a\plus{}f(xf(x))$ and so :
$ 0\equal{}\minus{}f(xf(x))\plus{}xf(x)\minus{}a\plus{}b$
Adding this assertion to $ Q(x)$, we get the required result.
Q.E.D.

8) new assertion : $ f(x)\equal{}x$
$ R(x\plus{}a^2\plus{}a)$ $ \implies$ $ (f(x\plus{}a^2\plus{}a))^2 \equal{} (x\plus{}a^2\plus{}a)f(x\plus{}a^2\plus{}a) \plus{} a^2\minus{}a$.
But ($ S(x)$) : $ f(x\plus{}a^2\plus{}a)\equal{}f(x)\plus{}a^2\plus{}a$ and so $ (f(x))^2\plus{}2(a^2\plus{}a)f(x)\plus{}(a^2\plus{}a)^2$ $ \equal{} xf(x)\plus{}(a^2\plus{}a)x\plus{}$ $ (a^2\plus{}a)f(x)\plus{}(a^2\plus{}a)^2 \plus{} a^2\minus{}a$, so :
$ (f(x))^2 \equal{} xf(x)\plus{}(a^2\plus{}a)x \minus{}(a^2\plus{}a)f(x) \plus{} a^2\minus{}a$
Subtracting $ R(x)$ from this, we get :
$ (a^2\plus{}a)f(x)\equal{}(a^2\plus{}a)x$
Hence the result (since $ a^2\plus{}a>0$)

9) The only solution to this equation is $ f(x)\equal{}x$
This is a required condition (from 8) and it's easy to verify that this mandatory form fit the equation.
Q.E.D

I mean "similarly" by applying $ (u,v\plus{}mc)$ and $ (u\plus{}ab,v\plus{}mc\minus{}ab)$ to 1), where $ m\in\mathbb N$ such that $ mc>ab$.

Ok, quite clear, thanks

Next question (I hope I dont bother you : I try to understand the full demo) :

[quote="xxp2000"]
3) $ f(u \plus{} v) \equal{} f(u) \plus{} f(v),\forall u,v > 0$
Let $ g(x) \equal{} f(x) \plus{} e$. We see $ g$ satisfies Cauchy equation.
So for $ r\in\mathbb Q$, $ g(r) \equal{} rg(1)$ or $ f(r) \equal{} \minus{} e \plus{} rg(1) \equal{} \minus{} e(1 \minus{} r) \plus{} rf(1) > 0$.
Obviously $ e\le0$ and $ g(1) > 0$. [/quote]
OK

[quote="xxp2000"] Suppose $ e < 0$.
(F1) can be rewritten as $ g(g(x)) \equal{} x \plus{} e_3$, where $ e_3$ is constant.
$ g(g(r)) \equal{} rg(g(1)) \equal{} r \plus{} e_3,\forall r\in\mathbb Q$
Hence, $ g(g(1)) \equal{} 1,e_3 \equal{} 0$ and $ g(g(x)) \equal{} x$
[/quote]

Here, I have again a question : I dont understand why (F1) can be rewritten as $ g(g(x)) \equal{} x \plus{} e_3$
(F1) is $ f(ab\plus{}f(y))\equal{}b^2\plus{}y$. So, writing $ h(x)\equal{}f(x)\plus{}ab$, we get $ h(h(x))\equal{}b^2\plus{}ab\plus{}x$ but, IMHO,  this $ h(x)$ is different (with a constant) from $ g(x)$ and we dont have $ h(x\plus{}y)\equal{}h(x)\plus{}h(y)$

[quote="xxp2000"]Using 2),
$ g(g(x)) \equal{} f(f(x) \plus{} e) \plus{} e \equal{} f(f(x) \plus{} ab) \minus{} f(ab \minus{} e) \minus{} e \plus{} e \equal{} x \plus{} b^2 \minus{} f(ab \minus{} e)$.[/quote]

That's OK (and $ f(ab\minus{}e)$ is meaningful since $ e<0$, so $ ab\minus{}e>0$)

No more questions, thanks.
And I do agree with your demo !

:)

I figure out there is still one flaw in my post above. we have to make sure $ f(x)\plus{}e>0$, which can be shown below.
$ g(rx)\equal{}rg(x)\Rightarrow f(rx)\equal{}\minus{}e\plus{}rg(x)>0\Rightarrow g(x)>\frac er\Rightarrow g(x)\ge0,\forall x>0$. If $ g(x)\equal{}0$, then $ g(2x)\equal{}0$ and $ f(x)\equal{}f(2x)$ conflicts with injective. So $ g(x)>0,\forall x>0$.

Thanks for your comments. pco.


inequalities
calculus
inequalities proposed

I have already posted it in the Calculus section, but have no reply, maybe someone here will help me...

Solve  $x^{356}\le 2x^{90}-7x+1$ using only Young's Inequality.

P.S. Young inequality states: for all $a,b \in \mathbb{R}$ and $p,q\in \mathbb{R}$ such that $p > 1$ and $\displaystyle \frac1p + \frac1q = 1$ is true that $\left|ab\right|\le \frac1p\left|a\right|^p + \frac1q \left|b\right|^q$

sorry.i don't think your problem is right
just $x \to \infty$

In fact, when $x=1$, the inequality is already wrong. :o

[quote="mathmanman"]Huh?
When x = 1, LHS = 1, RHS = 2^90 - 7 + 1. And, LHS < RHS..

 :?[/quote]
RHS = 2^90 - 7 + 1？？？？？？？？？

Oooh!  :blush: My big mistake!

So sorry  :roll:

Maybe the good ineq. is with $(2x)^{90}$ then?

I do not want you to prove that this inequality is true always, in fact it isn't so, I want you to help me finding the interval (possibly exact) in which this inequality is satisfied, i.e solving the inequality, maybe I'm in the wrong place, I have already posted it in the calculus section, but had no reply. So if you can help me you're wellcome.

[quote="TomciO"]Haha, now I see that you wrote "solve". It is funny how we got used to proving inequalities instead of solving them (it's much more interesting anyway :)).[/quote] Well once you've found the interval, you can prove that the relation holds in it :) The problem is still there: to find that interval.


calculus
integration
calculus computations

Find $ \int (tan^{\minus{}1}(ax))^2dx$ , where $ a\in \mathbb R^*$

 :D

Could we solve it in complex analysis??? :roll: [/b]

"Any bounds you like" is impossible. The integral can only be done exactly with a few specific choices of bounds- and the most likely candidate $ \int_0^{\infty}$ diverges. That's what not having an elementary antiderivative means- we have no simpler way to write down the integral for most limits than the integral itself.

[quote="wingless"]If you are asking for $ \int\arctan^2(ax) \, dx$, it can't be expressed as elementary functions.[/quote]

How do you prove this?

[quote="|-.-|"]Having the bounds, it may be doable.[/quote]

Consider, for example, $ I \equal{} \int_0^1(\arctan x)^2\,dx$.


Let $\mathcal S$ be a set of points in $3D$ and $\alpha, \beta$ be two planes such that $\text{pr}_\alpha \left( \mathcal S \right)$ and $\text{pr}_\beta \left( \mathcal S \right)$ are both squares.
Prove that if one of the squares has a side parallel to a side of the other square, then the two squares have the same area.

PS: I'm not sure if $\text{pr}$ is standard notation, so, in any case, I'm referring to [url=http://en.wikipedia.org/wiki/Projection_%28linear_algebra%29]this[/url].

what shape may the object S have ?
(i know it is a set of points , but if it where smooth,what would it be ?)

Well, it could be a polyhedron with two adjacent square faces (in this case the conclusion is trivial).


ARML
algebra
polynomial
inequalities
Rational Root Theorem

I'm trying out for my local ARML team next month, and have been working on a lot of problem sets.  One question, though, has me stumped:

Represent the number (989 * 1001 * 1007 + 320) as a product of primes.

I have an answer, but that's not as important as figuring out how to do a problem like this if it should come up.

If anyone has a solution and explanation on how to solve a problem like this, please let me know (for those of you who want the answer, its hidden)

[hide] 991 * 997 * 1009 [/hide]

JB

Hmm, lets see what we can do here.
[hide]
Lets let x = 989.
Then we want to factorise $x(x+12)(x+18)+320 = x^3 + 30x^2 + 216x + 320$.
Playing around with that (ok, so it will take a while and we have no guarantees that this will work), we can use the rational root theorem, get lucky and find that (x+2) is a factor :) So it factorises to (x+2)(x+8)(x+20).
[/hide]

Not brilliant, but maybe it will give someone an idea of how to improve it.

edit -
[hide]
Actually, if we let x = 999 (the average of the three primes), then we get $x^3 - 84x + 160=0$. For any solutions to that, x is even so $x=2a, a^3-21a+20=0$. a=1 is an obvious root and we get the factorisation. Still guesswork to come up with that polynomial though.. but it looks nice.
[/hide]

Triple M -

Thank you for your solutuion idea.  Using inequality root properties, I was able to solve the problem in about 8 minutes.

[hide]Let x = 1001
The number descirbed in the problem then is (x - 12)(x)(x + 6) +320
Which equals x3 - 6x2 - 72x +320.

Our goal is to factor this into the form (x + a)(x + b)(x + c).  By root properties, we can discern that:
1) a*b*c = 320
2) ab + ac + bc = -72
3) a + b + c = -6

By finding the factors of 320, we can find the possible values of a, b, and c (all values are plus or minus):

1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 160, 320
Now, we can start eliminating values.  For instance, if a were to equal 320, then b and c would both have to equal one or negative one.  Obviously, these values do not add up to negative 6.  Using this trick, we can eliminate all the values from 320 to 20, so a, b, and c all equal plus or minus:

1, 2, 4, 8, 10, or 16
Because the sum of our variables is -6, then obviously, at least one of our variables is negative. The only way to proceed now is to plug in values for a.

if a = -16, then b*c = -20.  (b,c) can equal (-1, 20), (-2, 10), (-4, 5),   (-5, 4), (-10, 2), (-20, 1)
However, none of these values for a, b, and c add up to -6.

If a = -10, then b*c = -32.  (b,c) can equal (-1 , 32), (-2, 16), (-4, 8),
(-8, 4), (-16, 2), (-32, 1)

When looking over these ordered pairs, we find that
-10 + -4 +8 = -6.  Finall,y we plug this in to equation number 2
(-10*-4) + (-10*8) + (-4*8) = 40 - 80 -32 = -72.  Therefore, a = -10, b = -4 and c = 8, so our expression factors into (x - 10)(x-4)(x+8), which equals 991*997*1009.  All of these numbers are prime.

This solution seems to take a long time.  However, when on paper in front of you, it is realtively quick.
[/hide]

JB

FWIW  knowing that   this was a problem asked in a competetion (means have an  "easy" type solution) -  this is what the  first thing came in my mide...
[hide]1001 is middle number so the product looks like: (Assuming x = 1001)

 $(x-12) x (x+6) + 320$

Now  facotrs of 320 which jump out are 32*10, and x=4 we see   $(-8)*4*10 = -320$ hence x-4 or 997 is a factor, rest is easy.  (We could also have guessed x=10 too, which IMO might be even more easier to guess)

Also note that it is really not that difficult to guess ...

[hide](Numbers to try  for x are  factors of 320,  specially those where x+6 and x-12 also remains factors of 320) - thats even factors only ..etc.. (we could have tried 10, or 4 or even  -8  .. 10 being, IMO easiest to guess)

Als I said it is not that difficult to guess ...Notice how the sign of the expression changes and  we know there is one root less than -6, and two between 0 and 12, and  suppose you tried x= 8  (and saw the expression -ive) you would know that there is one between 8 and 12 (only possible answer is 10).. and one between 0 and 8. ..

Point here is[b] not [/b]to expand   the ( (three factors) + 320) but apply rational root theorem by guessing factors of 320 as possible roots -  and it is easier if you keep them as three factors.[/hide]

[hide]In other words, knowing  the factors  (luckily even only here)   of 320 (2,4,8,10  etc) , numbers you should try should  be { you could have called any one of {1001, or 1007 or 989 ) as x and   + or -  (2,4,8,10..)   valus for x .. Hope this is helpful)

Also if you are ever asked this type of question in a "count down" type situations (where you have only few seconds to answer ...  the  old trick ...  largest number and add two (or the smallest number and add   2) .. (That is if you knew  what parity you want)  ... you would be correct/lucky  more often than not :)  (By parity I mean noticing    '320' is even, and rest of the numbers are  not ...   and 2 being the smallest factor ,,,you can see why?  :) ..
.[/hide]

And finally  one more fun thing:

[hide]to factor;
 $(x-12) x (x+6) + 320$  and noticing that roots of $x$ in  $(x-12) x (x+6) + 320 = 0$  has to be even.

we reduce it to  $(y-6)y(y+3)+320/8 = (y-6)y(y+3) +40 =0 $ is even more easy to work with.!!
   [/hide]

[/hide]

hi,

I go to ARML.  If you get in, im the dude with black hair and 4'10".

good luck

-Johnny


inequalities
symmetry
inequalities proposed

[quote="s1ck_psych0"][quote="Mildorf"]Show that for all positive reals $a, b, c$,

\[ \frac{3a^2-2ab-b^2}{3a^2+2ab+3b^2} + \frac{3b^2-2bc-c^2}{3b^2+2bc+3c^2} + \frac{3c^2-2ca-a^2}{3c^2+2ca+3a^2} \ge 0  \][/quote]

 We put $f(x)=\frac{1}{3x^2-2x+3}$, and since $f(x)$ is concave, by Jensen  [/quote]

$f(x)=\frac{1}{3x^2-2x+3}$
$f''(x)=\frac{2(27x^2-18x-5)}{(3x^2-2x+3)^3}$ . It is not always concave.

[quote="Minh Can"]S1ck:your denominator got messed up. it should be 3x^2 + 2x + 3 and yours is: 3x^2 - 2x + 3. check again make sure you got it right before proceeding.[/quote]

  His solution isn't work too, I doubt.
  $f(x)= \frac{1}{3x^2+2x+3}$ then $f''(x)=\frac{2(27x^2+18x-5)}{(3x^2+2x+3)^3}$

[quote="Minh Can"]S1ck:your denominator got messed up. it should be 3x^2 + 2x + 3 and yours is: 3x^2 - 2x + 3. check again make sure you got it right before proceeding.[/quote]

Sorry, I was sleepy when I solved it. I'll correct it ASAP and i'll post my solution.

You can rewrite it as
\[ \sum_{cyc} \frac{3x^2-2x-1}{3x^2+2x+3} \ge 0,  \]
but you pick up the constraint $xyz=1$. If you want to try Jensen, you'd need to substitute $e^{x_0}=x$, etc., but
\[ f(x) = \frac{3e^{2x}-2e^x-1}{3e^{2x}+2e^x+3}  \]
has two points of inflection, since
\[ f''(x) = \frac{4e^x\left(-9e^{4x}-30e^{3x}+36e^{2x}+38e^x-3\right)} {(3e^{2x}+2e^x+3)^2}  \]
It's not especially clear that this is useful. We are also hiding a lot of calculation. There are less computational approaches. ;)

Bump. Almost a month of silence! Surely someone out there knows how to do this one.  :wink:

Again amazing  :omighty:  :o
Did you make all these computations by hand ??
Again the majic  $(a-b)^{2}(b-c)^{2}(c-a)^{2}$  but how did you come up with the trick by the $\frac{4}{3}$
and $\frac{104}{3}$  ??

Thank you Arqady . :lol: Very nice method.......

Expansion, either directly or after forcing symmetry, seems to be the only solution that has been discovered by solvers. But the problem was written with the following solution in mind: \begin{eqnarray*}\sum_{cyc}\frac{4(a+b)b}{3a^{2}+2ab+3b^{2}}& \le & \sum_{cyc}\frac{2b}{\sqrt{2a^{2}+2b^{2}}}\\ & \le & \sqrt{4\left(a^{2}+b^{2}+c^{2}\right)\left(\sum_{cyc}\frac{a^{2}}{(a^{2}+b^{2})(a^{2}+c^{2})}\right)}\le 3, \end{eqnarray*} which is equivalent to the desired, since \[\sum_{cyc}\frac{4(a+b)b}{3a^{2}+2ab+3b^{2}}= 3-\sum_{cyc}\frac{3a^{2}-2ab-b^{2}}{3a^{2}+2ab+3b^{2}}.\] I wonder if there are still other solutions.

Yes, there's another solution I think. Take $ x\equal{} \frac{b}{a}, y\equal{}\frac{c}{b}, z\equal{}\frac{a}{c}$, we have $ xyz\equal{}1$ and the inequality is equivalent to

$ \frac{4x(x\plus{}1)}{3\plus{}2x\plus{}3x^2} \plus{}\frac{4y(y\plus{}1)}{3\plus{}2y\plus{}3y^2}\plus{} \frac{4z(z\plus{}1)}{3\plus{}2z\plus{}3z^2} \le 3$

with the following obvious inequality

$ 3\plus{}2t\plus{}3t^2 \ge \frac{8}{3}(1\plus{}t\plus{}t^2)$

We just need to prove

$ \frac{x(x\plus{}1)}{1\plus{}x\plus{}x^2}\plus{} \frac{y(y\plus{}1)}{1\plus{}y\plus{}y^2}\plus{} \frac{z(z\plus{}1)}{1\plus{}z\plus{}z^2} \le 2$

Which is equivalent to

$ \frac{1}{x^2\plus{}x\plus{}1}\plus{} \frac{1}{y^2\plus{}y\plus{}1}\plus{} \frac{1}{z^2\plus{}z\plus{}1} \ge 1$

Take $ x\equal{}\frac{uv}{w^2}, y\equal{}\frac{vw}{u^2}, z\equal{}\frac{wu}{v^2}$ It's

$ \frac{u^4}{u^4\plus{}u^2vw\plus{}v^2w^2}\plus{} \frac{v^4}{v^4\plus{}v^2wu\plus{}w^2u^2}\plus{} \frac{w^4}{w^4\plus{}w^2uv\plus{}u^2v^2} \ge 1$

By Cauchy-Schwarz we have

$ \frac{u^4}{u^4\plus{}u^2vw\plus{}v^2w^2}\plus{} \frac{v^4}{v^4\plus{}v^2wu\plus{}w^2u^2}\plus{} \frac{w^4}{w^4\plus{}w^2uv\plus{}u^2v^2}\\
\ge \frac{(u^2\plus{}v^2\plus{}w^2)^2}{(u^4\plus{}u^2vw\plus{}v^2w^2)\plus{}(v^4\plus{}v^2wu\plus{}w^2u^2)\plus{}(w^4\plus{}w^2uv\plus{}u^2v^2)}\\
\ge 1$

Hence result.

To prove

$ \frac{1}{x^2\plus{}x\plus{}1}\plus{} \frac{1}{y^2\plus{}y\plus{}1} \plus{}\frac{1}{z^2\plus{}z\plus{}1} \ge 1$

It's doable to expand it (the coefficients are not huge) and it's equivalent to

$ x^2\plus{}y^2\plus{}z^2 \ge xy\plus{}yz\plus{}zx$

[quote="shalex"]Yes, there's another solution I think. Take $ x \equal{} \frac {b}{a}, y \equal{} \frac {c}{b}, z \equal{} \frac {a}{c}$, we have $ xyz \equal{} 1$ and the inequality is equivalent to

$ \frac {4x(x \plus{} 1)}{3 \plus{} 2x \plus{} 3x^2} \plus{} \frac {4y(y \plus{} 1)}{3 \plus{} 2y \plus{} 3y^2} \plus{} \frac {4z(z \plus{} 1)}{3 \plus{} 2z \plus{} 3z^2} \le 3$

with the following obvious inequality

$ 3 \plus{} 2t \plus{} 3t^2 \ge \frac {8}{3}(1 \plus{} t \plus{} t^2)$

We just need to prove

$ \frac {x(x \plus{} 1)}{1 \plus{} x \plus{} x^2} \plus{} \frac {y(y \plus{} 1)}{1 \plus{} y \plus{} y^2} \plus{} \frac {z(z \plus{} 1)}{1 \plus{} z \plus{} z^2} \le 2$

Which is equivalent to

$ \frac {1}{x^2 \plus{} x \plus{} 1} \plus{} \frac {1}{y^2 \plus{} y \plus{} 1} \plus{} \frac {1}{z^2 \plus{} z \plus{} 1} \ge 1$

Take $ x \equal{} \frac {uv}{w^2}, y \equal{} \frac {vw}{u^2}, z \equal{} \frac {wu}{v^2}$ It's

$ \frac {u^4}{u^4 \plus{} u^2vw \plus{} v^2w^2} \plus{} \frac {v^4}{v^4 \plus{} v^2wu \plus{} w^2u^2} \plus{} \frac {w^4}{w^4 \plus{} w^2uv \plus{} u^2v^2} \ge 1$

By Cauchy-Schwarz we have

$ \frac {u^4}{u^4 \plus{} u^2vw \plus{} v^2w^2} \plus{} \frac {v^4}{v^4 \plus{} v^2wu \plus{} w^2u^2} \plus{} \frac {w^4}{w^4 \plus{} w^2uv \plus{} u^2v^2} \\
\ge \frac {(u^2 \plus{} v^2 \plus{} w^2)^2}{(u^4 \plus{} u^2vw \plus{} v^2w^2) \plus{} (v^4 \plus{} v^2wu \plus{} w^2u^2) \plus{} (w^4 \plus{} w^2uv \plus{} u^2v^2)} \\
\ge 1$

Hence result.[/quote]

Fantastic!I like it very much! :)

Till now three proofs exist. One by arqady, one by Mildorf, and one by me. But there exists the 4th proof (also very nice) which is given by Mou Xiaosheng, who is in Chinese Team 2008.

The inequality is equivalent to

$ \quad\frac{(3a\plus{}b)(a\minus{}b)}{3a^2\plus{}2ab\plus{}3b^2} \plus{} \frac{(3b\plus{}c)(b\minus{}c)}{3b^2\plus{}2bc\plus{}3c^2} \plus{} \frac{(3c\plus{}a)(c\minus{}a)}{3c^2\plus{}2ca\plus{}3a^2} \ge 0\\
\Longleftrightarrow \left( \frac{(3a\plus{}b)(a\minus{}b)}{3a^2\plus{}2ab\plus{}3b^2}\minus{} \frac{a\minus{}b}{a\plus{}b}\right)\plus{} \left( \frac{(3b\plus{}c)(b\minus{}c)}{3b^2\plus{}2bc\plus{}3c^2} \minus{} \frac{b\minus{}c}{b\plus{}c}\right)\\
\plus{} \left( \frac{(3c\plus{}a)(c\minus{}a)}{3c^2\plus{}2ca\plus{}3a^2}\minus{} \frac{c\minus{}a}{c\plus{}a} \right) \ge \frac{b\minus{}a}{b\plus{}a}\plus{} \frac{c\minus{}b}{c\plus{}b}\plus{} \frac{a\minus{}c}{a\plus{}c}\\
\Longleftrightarrow \frac{2b(a\minus{}b)^2}{(3a^2\plus{}2ab\plus{}3b^2)(a\plus{}b)}\plus{} \frac{2c(b\minus{}c)^2}{(3b^2\plus{}2bc\plus{}3c^2)(b\plus{}c)}
\\\plus{} \frac{2a(c\minus{}a)^2}{(3c^2\plus{}2ca\plus{}3a^2)(c\plus{}a)} \ge \left| \frac{(a\minus{}b)(b\minus{}c)(c\minus{}a)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \right|$

Assume that $ a$ is the smallest. We have

$ 3a^2\plus{}2ab\plus{}3b^2 \le 4b(a\plus{}b),\ 3c^2\plus{}2ca\plus{}3a^2 \le 4c(c\plus{}a)$

So

$ \quad\frac{2b(a\minus{}b)^2}{(3a^2\plus{}2ab\plus{}3b^2)(a\plus{}b)}\plus{} \frac{2a(c\minus{}a)^2}{(3c^2\plus{}2ca\plus{}3a^2)(c\plus{}a)}\\
\ge \frac{1}{2}\left( \frac{a\minus{}b}{a\plus{}b} \right)^2\plus{} \frac{1}{2}\left( \frac{c\minus{}a}{c\plus{}a} \right)^2\\
\ge \left|\frac{(a\minus{}b)(c\minus{}a)}{(a\plus{}b)(c\plus{}a)}\right|\\
\ge \left| \frac{(a\minus{}b)(b\minus{}c)(c\minus{}a)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \right|$

[quote="shalex"]Yes, there's another solution I think. Take $ x \equal{} \frac {b}{a}, y \equal{} \frac {c}{b}, z \equal{} \frac {a}{c}$, we have $ xyz \equal{} 1$ and the inequality is equivalent to

$ \frac {4x(x \plus{} 1)}{3 \plus{} 2x \plus{} 3x^2} \plus{} \frac {4y(y \plus{} 1)}{3 \plus{} 2y \plus{} 3y^2} \plus{} \frac {4z(z \plus{} 1)}{3 \plus{} 2z \plus{} 3z^2} \le 3$

with the following obvious inequality

$ 3 \plus{} 2t \plus{} 3t^2 \ge \frac {8}{3}(1 \plus{} t \plus{} t^2)$

We just need to prove

$ \frac {x(x \plus{} 1)}{1 \plus{} x \plus{} x^2} \plus{} \frac {y(y \plus{} 1)}{1 \plus{} y \plus{} y^2} \plus{} \frac {z(z \plus{} 1)}{1 \plus{} z \plus{} z^2} \le 2$

Which is equivalent to

$ \frac {1}{x^2 \plus{} x \plus{} 1} \plus{} \frac {1}{y^2 \plus{} y \plus{} 1} \plus{} \frac {1}{z^2 \plus{} z \plus{} 1} \ge 1$

Take $ x \equal{} \frac {uv}{w^2}, y \equal{} \frac {vw}{u^2}, z \equal{} \frac {wu}{v^2}$ It's

$ \frac {u^4}{u^4 \plus{} u^2vw \plus{} v^2w^2} \plus{} \frac {v^4}{v^4 \plus{} v^2wu \plus{} w^2u^2} \plus{} \frac {w^4}{w^4 \plus{} w^2uv \plus{} u^2v^2} \ge 1$

By Cauchy-Schwarz we have

$ \frac {u^4}{u^4 \plus{} u^2vw \plus{} v^2w^2} \plus{} \frac {v^4}{v^4 \plus{} v^2wu \plus{} w^2u^2} \plus{} \frac {w^4}{w^4 \plus{} w^2uv \plus{} u^2v^2} \\
\ge \frac {(u^2 \plus{} v^2 \plus{} w^2)^2}{(u^4 \plus{} u^2vw \plus{} v^2w^2) \plus{} (v^4 \plus{} v^2wu \plus{} w^2u^2) \plus{} (w^4 \plus{} w^2uv \plus{} u^2v^2)} \\
\ge 1$

Hence result.[/quote]
Actually, my solution is similar to yours :). And here is my similar one
\[ \sum \frac{(a\minus{}b)(13a\plus{}5b)}{a^2\plus{}b^2} \ge 0\]
:)

[quote="can_hang2007"]  And here is my similar one
\[ \sum \frac {(a \minus{} b)(13a \plus{} 5b)}{a^2 \plus{} b^2} \ge 0
\]
:)[/quote]
The following statement is more general, nice, but hard.

Find the minimum value of $ p$ such that the inequality below holds for any positive numbers $ a,b,c$:
$ \sum \frac {(a\minus{}b)(pa\plus{}b)}{a^2\plus{}b^2} \ge 0$.

On the other hand, I think the following is easier:
$ \sum \frac {(a\minus{}b)(2a\plus{}b)}{(a\plus{}b)^2} \ge 0$.

[quote="Vasc"]
On the other hand, I think the following is easier:
$ \sum \frac {(a \minus{} b)(2a \plus{} b)}{(a \plus{} b)^2} \ge 0$.[/quote]
It is equivalent to the known one
\[ \sum \frac{a\plus{}3}{(a\plus{}1)^2} \ge 3\]
for all $ a,b,c \ge 0,abc\equal{}1$. :)

[quote="can_hang2007"][quote="can_hang2007"][quote="Vasc"]
On the other hand, I think the following is easier:
$ \sum \frac {(a \minus{} b)(2a \plus{} b)}{(a \plus{} b)^2} \ge 0$.[/quote]
It is equivalent to the known one
\[ \sum \frac {a \plus{} 3}{(a \plus{} 1)^2} \ge 3
\]
for all $ a,b,c \ge 0,abc \equal{} 1$. :)[/quote]
My solution for this inequality :)[/quote]

The inequality

$ \frac{1}{(1\plus{}a)^2}\plus{} \frac{1}{(1\plus{}b)^2} \ge \frac{1}{1\plus{}ab}$

mentioned in your proof is really important. But the original problem is obvious from the inequality above. Actually

$ \quad \sum \frac{a\plus{}3}{(a\plus{}1)^2}\equal{} \sum \frac{1}{a\plus{}1}\plus{} \sum \left( \frac{1}{(b\plus{}1)^2}\plus{} \frac{1}{(c\plus{}1)^2} \right)\\
\ge \sum \frac{bc}{1\plus{}bc} \plus{} \sum \frac{1}{bc\plus{}1}\equal{}3$

[quote="shalex"][/quote][quote="can_hang2007"][quote="can_hang2007"][quote="Vasc"]
On the other hand, I think the following is easier:
$ \sum \frac {(a - b)(2a + b)}{(a + b)^2} \ge 0$.[/quote]
It is equivalent to the known one
\[ \sum \frac {a + 3}{(a + 1)^2} \ge 3
\]
for all $ a,b,c \ge 0,abc = 1$. :)[/quote]
My solution for this inequality :)[/quote][quote="shalex"]

The inequality

$ \frac {1}{(1 + a)^2} + \frac {1}{(1 + b)^2} \ge \frac {1}{1 + ab}$

mentioned in your proof is really important. But the original problem is obvious from the inequality above. Actually

$ \quad \sum \frac {a + 3}{(a + 1)^2} = \sum \frac {1}{a + 1} + \sum \left( \frac {1}{(b + 1)^2} + \frac {1}{(c + 1)^2} \right) \\
\ge \sum \frac {bc}{1 + bc} + \sum \frac {1}{bc + 1} = 3$[/quote]
Thanks, shalex. :)
It is very nice.

[quote="shalex"]
$ \Longleftrightarrow \left( \frac {(3a \plus{} b)(a \minus{} b)}{3a^2 \plus{} 2ab \plus{} 3b^2} \minus{} \frac {a \minus{} b}{a \plus{} b}\right)$
[/quote]

but how did Mou came up with the idea using $ \frac {a \minus{} b}{a \plus{} b}$ to find $ (a\minus{}b)^2$

[quote="Vasc"]
The following statement is more general, nice, but hard.

Find the minimum value of $ p$ such that the inequality below holds for any positive numbers $ a,b,c$:
$ \sum \frac {(a \minus{} b)(pa \plus{} b)}{a^2 \plus{} b^2} \ge 0$.[/quote]
The minimum value of $ p$ is $ p\equal{}\frac {95\plus{}39\sqrt 3}{71}\approx 2.33169$.
We can prove this using LCF-Theorem.
For $ p\equal{}\frac 7{3}$, we get a nice and hard inequality:
$ \sum\frac{(a\minus{}b)(7a\plus{}3b)}{a^2\plus{}b^2} \ge 0$,
which is equivalent to
$ \sum \frac {b(2a\plus{}5b)}{a^2\plus{}b^2} \le \frac{21}{2}$.

[quote="Vasc"][quote="Vasc"]
The following statement is more general, nice, but hard.

Find the minimum value of $ p$ such that the inequality below holds for any positive numbers $ a,b,c$:
$ \sum \frac {(a \minus{} b)(pa \plus{} b)}{a^2 \plus{} b^2} \ge 0$.[/quote]
The minimum value of $ p$ is $ p\equal{}\frac {95\plus{}39\sqrt 3}{71}\approx 2.33169$.
We can prove this using LCF-Theorem.
[/quote]
Mixing Variables Method also helps here.


A uniform circular ring of radius R is fixed in place. A particle is placed on the axis of the ring at a distance much greater than R and allowed to fall towards the ring under the influence of the ring’s gravity. The particle achieves a maximum speed v. The ring is replaced with one of the same linear mass density, but radius 2R and the experiment is repeated. What is the new maximum speed of the particle?

The magnitude of the electric field along the axis is given by
$ E\equal{}\frac{\lambda xr}{2\epsilon_0 \left( x^2\plus{}r^2\right) ^{\frac{3}{2}}}$

Hence, the work done on the moving charge $ W\equal{}\frac{\lambda r}{2\epsilon_0}\int_0^{\infty}\frac{xdx}{\left( x^2\plus{}r^2\right) ^{\frac{3}{2}}}\equal{}\frac{\lambda}{2\epsilon_0}$, which is independant of the radius of the ring. Hence, the new maximum speed of the particle is still $ v$

Electric field? hmmm..

I dont see where that is mentioned in the problem.

As I see its gravitational field... :D

Soo sorry... somewhere along the way I switched from 'mass density' to 'charge density'  :oops:

Anyway, the method is the same,
$ \alpha \equal{} \frac {2\pi G\lambda xr}{\left( x^2 \plus{} r^2\right)^{\frac {2}{3}}}$
and work done $ W \equal{} 2\pi mG\lambda$
(where $ \alpha$ is the magnitude of the Gravitational field)

so, the answer remains the same!

Nice method. One quick clarification: wouldn't there be an 'm' in the work (the answer doesn't change since it would cancel with the m in KE later).


vector
real analysis
real analysis unsolved

[quote="Ace123"]An unknown vector can be determined if both its scalar product and its vector product with a known vector are given. Assuming that A is a known vector, determine the unknown vector X if both p and B are given, where p=A dot X and B=A cross X[/quote]

$ \overrightarrow{A}\times\overrightarrow{X} \equal{} \overrightarrow{B} \implies(\overrightarrow{A}\times\overrightarrow{X})\times\overrightarrow{A} \equal{} \overrightarrow{B}\times\overrightarrow{A}$

$ \implies (\overrightarrow{A}\cdot\overrightarrow{A})\overrightarrow{X} \minus{} (\overrightarrow{A}\cdot\overrightarrow{X})\overrightarrow{A} \equal{} \overrightarrow{B}\times\overrightarrow{A}$

$ \implies |\overrightarrow{A}|^2 \overrightarrow{X} \minus{} p\overrightarrow{A} \equal{} \overrightarrow{B}\times\overrightarrow{A}$

$ \boxed{\boxed{\implies \overrightarrow{X} \equal{} \frac {1}{|\overrightarrow{A}|^2} \left(p\overrightarrow{A} \plus{} \overrightarrow{B}\times\overrightarrow{A}\right)}}$


Give $ c \geq b \geq a \geq 0 $
Find the greatest value of:
[tex] f(a,b,c)= (b-a)(m-a^2)+(c-b)(m-c^2)+(c-a)(m-b^2) [/tex]
where m is positive real.
 :)

just let $c-b=x, b-a=y$ then $x,y\geq 0$
if my computation is not wrong, $f(a,b,c)$ will reduce to
$(x+y)(2m-\frac 1{2} (x^2+y^2) -\frac 1{2} (x+y)^2-2a(x+y)-2a^2)$
or $(x+y)(2m-\frac 1{2} (x^2+y^2) -\frac 1{2}(x+y-2a)^2)$
but it less than $(x+y)(2m-\frac {3}{4}(x+y)^2 )$
let $x+y=t \geq 0$
then we must find the minimal value of
$f(t)=2mt-\frac {3}{4}t^3$ where $m>0$
i guess it can be done by Cauchy, right ?
 :D

Congratulations!  It was my way too :lol: And for the last one.[tex] Max{f(t)}=2\sqrt{ \frac{32m^3}{81}} [/tex]


I just didn't get one thing that happened here. have a look:

[quote="Tiks"]Prove that if $x,y,z\in{R}^{+}$ and $xyz=1$,then
\[3+x^{3}+y^{3}+z^{3}+x^{3}y^{3}+x^{3}z^{3}+y^{3}z^{3}\geq\sum{ x^{2}(y+z+1) }\]
[/quote]

This inequality is symmetric in its variables. And the following one which we get by symmetric substitution,

[quote="conceive"]Setting $x=\frac{a}{b},y=\frac{b}{c},z=\frac{c}{a}$ (that's [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=104638]harazi's substitution[/url]) with arbitrary reals $a,b,c>0$ it's equivalent to:

$(a^{6}b^{3}+a^{3}b^{6}+b^{6}c^{3}+b^{3}c^{6}+c^{6}a^{3}+c^{3}a^{6})+3a^{2}b^{2}c^{2}\geq$
$\geq2(a^{5}b^{2}c^{2}+a^{2}b^{5}c^{2}+a^{2}b^{2}c^{5})+(a^{5}bc^{3}+a^{3}b^{5}c+ab^{3}c^{5})$

[/quote]

is not symmetric, but only cyclic one. How is that possible? We have a symmetric inequality, we use symmetric substitution and we we an inequality, which is cyclic, but not symmetric. Huh?

[quote="conceive"]by symmetric substitution[/quote]
I think it is not true. Consider the inequality $x+y+z\geqslant 3$ for $xyz=1$. If the Harazi's substitution was symmetric, the inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geqslant 3$ would be symmetric too, but it is only cyclic. Thus this substitution is not symmetric.


This (stupid) method seems to work:

First, note that $\sum_{\text{cyc}} \frac{x^3}{(1+y)(1+z)} \ge \frac{(x^2+y^2+z^2)^2}{x(1+y)(1+z)+(1+x)y(1+z)+(1+x)(1+y)z}$, by Cauchy, so we need to prove: $4(x^2+y^2+z^2)^2 \ge 3(x(1+y)(1+z)+(1+x)y(1+z)+(1+x)(1+y)z)$. Well, let $x = a^3, y = b^3, z = c^3, abc = 1$, and homogenize it to find that we have to prove: $\sum_{\text{cyc}} 4a^{12} + 8a^6b^6 \ge \sum_{\text{cyc}} 3a^6b^3c^3 + 6a^5b^5c^2 + 3a^4b^4c^4$, which is perfectly fine by Muirhead.

This one was also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?p=124421 . A really easy one.

  darij

yes,indeed,too bad that this "stupid" works at this kind of inequalities. :(  ;)


When will the scores for round 4 be posted

Probably around April 10, because they start doing the prizes in mid-April iirc.

I have been waiting for a long time, everyday I come home I check my email for it, but still nothing. I really hope they have it out soon.


function
calculus
derivative
real analysis
real analysis unsolved

Give an example of function differentiable in exactly one point but  everywhere countinuous.

Multiply a nowhere differentiable continuous function by $x$. That product will be differentiable only at zero.

Make that $x^{2}$ to be sure of the differentiability at zero.

No, $x$ is good enough- the function we're multiplying by is continuous at zero, so the new derivative is the old function's value.


What would happen if gave two solutions to a problem on USAMO and you got one of them wrong, would they give you the highest scoring solution?

 :maybe:

Generally, people don't do this, but if a person were to find two solutions to the same problem, (s)he should probably just pick the best one and go with it.  Your time would be better spent working on two problems than getting two solutions for the same problem, and if you do that anyway, it would be in your best interests to submit whichever solution you thought was better.


ARML AMC AIME quadratics geometry inradius trigonometry

Is there a list anywhere of essential formulas that are helpful for AMC/AIME?

Pretty much everyone on this forum will tell you that you should not just memorize a list of formulas, but you should learn their derivations and such.

I know that. I'm not going to just memorize the formulas. But is there a list of formulas that I should know? (I'm not going to just memorize them...I will learn about how they are derived and such)

i would definitely say pythagorean theorem. what do you want a list of formula's for if you're not gonna memorize them?

Guys, before we criticize the question (which is perfectly understandable), let's answer the question FIRST.

No, there is not a list of formulas devoted for AMC/AIME. But, yes, there is a list that is devoted for high school math competition such as ARML, AMC, AIME, and possibly USAMO. Having said that, if you want to look for it, go ahead. I did not lie about its existence.

Now, here is where I agree with previous posters. Regardless of your intent to visualize and learn how the formulas were derived, it is much more important to see how formulas are used on actual problems and perhaps, how they are derived (like... how did the guy think of the formula?). These two will build insights into problems in AMC/AIME.

This is why I'm not going to tell you where the formula is at. Do the problems. Learn your mistakes. And before you know it, you'll have all formulas you need.

[hide="If you are really stubborn"]
Here is what I think is good.
Pythagorean Theorem
Ptolemy's Theorem (not always)
Quadratic Formula
Binomial Theorem
[/hide]

I would say that there are a certain set of techniques that you have to be familiar with to even have a chance to do the harder problems. However, just knowing the formulas won't get you far. You need some sense of problem solving ingenuity that comes from talent and doing lots of problems.

[quote="Silverfalcon"]I did not lie about its existence.
[/quote]

?

Seriously?

I think you may have.

Depends on what list of formulas we're thinking of.

Whatever you are talking about sounds as elusive as the book...

123456789, I like your name, check your PMs.

Don't forget Principle of Inclusion and Exclusion. Not very hard to understand how you get it, so it's worth it.

 Also, every area formula for triangle except heron's formula (unless you REALLY can't think of anything to do). Know how they are derived because it can create shortcuts to long problems.

 Power of a Point (not very hard to remember or derive), Binomial Theorem, etc.

[quote="zephyredx"]Also, every area formula for triangle except heron's formula (unless you REALLY can't think of anything to do). [/quote]
Hey!  I like Heron's formula!  It is elegant and symmetric, and provides equally elegant and symmetric expressions for the inradius and exradii.

Since it is so different from the other area formulas, you can find interesting relations by equating them.  For instance, we have
\[ \tfrac{1}{2}ab \sin C = \sqrt {s(s - a)(s - b)(s - c)},
\]
so
\[ \sin C = \sqrt {\frac {4 s(s - a)(s - b)(s - c)}{(ab)^2} .
}\]
What about the inequality $ |\sin C| \le 1$?  Replacing $ (s - a), (s - b), (s - c)$ with $ x,y,z$, we have
\[ \sin C = \sqrt {\frac {4(x + y + z)xyz}{(y + z)^2(z + x)^2}} ,
\]
and thus derive the inequality
\[ 4(x + y + z)xyz \le (y + z)^2(x + z)^2
\]
for all nonnegative reals $ x,y,z$.  Can we prove this algebraically?  Of course!  How?  (Hint: consider the equality case.)

I can't really solve these geometry problems in the harder AMC 10 problems, like #21-25 usually. Are there some theorems they use that I don't know about? Like ptolemy for trapezoids? etc.

For triangles, similar triangles, spheres, etc., because these questions bother me (hopefully they AMC 10 won't have too many geo problems)

Same for me, but I don't think you need formulas. It's better to just know concepts and approaches. For example, many problem in the last few years needed similar triangles. Also, a few problem could be solved by fiding two ways to express it, like let's say you have two or three right triangles, with one side common. You might be able to solve for one side, and find others.

[quote="funcia"][quote="Silverfalcon"]I did not lie about its existence.
[/quote]

?

Seriously?

I think you may have.

Depends on what list of formulas we're thinking of.

[/quote]

Yes I am serious. There is a list and I found it again yesterday (I know what to type though  :roll: ).

But [i]seriously[/i], I don't understand the purpose of this entire thread. The formulas mean nothing. While you guys are pondering over the formulas, it's much better to solve another problem.

@Bicameral: They do sometimes use Ptolemy's but it's often more of special insight like drawing extra line or combination of trigonometry and geometry simultaneously.

@Boy Soprano II: I like Heron's too!  :D

definitely learn stewart's theorem (with which you can kill many triangle + cevian problems)
menelaus and ceva's might not really help for AMC

definitely trig identities, some combinatorial identites (sums of combinations etc.), binomial theorem, remainder theorem for factoring polynomials.

definitely memorizing the answers would help, but a priori, they might be hard to procure :)

also know some basic modular arithmetic stuff... like $ S(n)\equiv n \pmod{9}$ where $ S(n)$ is the sum of the digits of $ n$... :( (I missed 10A #25 last year because I was stupid and decided to completely brute-force instead of doing mod stuff)

also power of a point is good!

While people are still replying to the thread I'd like to throw in balls and urns.