floor function
arithmetic sequence
irrational number
advanced fields
advanced fields unsolved

We have tightly growing sequence of natural numbers which is neither arithmetic nor geometrical. Give an example of assumption, estimate for what kind of sequences, examples, the phrase is true: it is possible to choose from it an infinite arithmetic sequence with initial term 1+kr and the difference r?

A sufficient condition: the sequence contains all but finitely many natural numbers.  :ninja:

Ok, but can we give an other example of condition if the sequence doesn't contain all natural numbers? and why in this sequence  $ a_{n \plus{} 1} \equal{} 3\lfloor a_n \rfloor \plus{} a_n \plus{} 3$, a1 is arbitrary, is it true, how to prove that?

That sequence has integer elements after the first one, which simplifies it to $ a_{n\plus{}1}\equal{}4a_n \plus{}3$. It does not contain an infinite arithmetic progression.
Indeed, it fails an obvious necessary condition: the sequence $ a_{n\plus{}1}\minus{}a_n$ must be bounded.

[quote="mlok"]Indeed, it fails an obvious necessary condition: the sequence $ a_{n \plus{} 1} \minus{} a_n$ must be bounded.[/quote] Can we chose such arithmetic sequence if $ a_{n\plus{}1}\minus{}a_n$ is bounded, how to prove that?

Please contemplate the question "is it true?" in some depth before asking "how to prove that?"

An arithmetic progression either has all odd elements, or all even, or they are alternating between even and odd. For example, the sequence that consists of all
 even one-digit integers
 odd two-digit integers
 even three-digit integers
 odd four-digit integers
 even five-digit integers
....
does not contain an infinite arithmetic progression.

Another counterexample: Consider $ a_n\equal{}\lfloor n\sqrt{2}\rfloor$. We have $ m\equal{}a_n$ for some $ n$ if and only if $ 0\le s<\frac1{\sqrt{2}}$, where $ s$ is the fractional part of $ \frac{m}{\sqrt{2}}$. For any arithmetic sequence of integers $ b_k$ and any irrational number $ \alpha$, the fractional parts of $ \alpha b_k$ are dense in $ [0,1)$. Applied to this case, the fractional part of $ \frac{b_k}{\sqrt{2}}\equal{}\frac{1\plus{}kr}{\sqrt{2}}$ must be between $ \frac1{\sqrt{2}}$ and $ 1$ for some $ k$, even infinitely many.

That one has $ a_{n\plus{}1}\minus{}a_n\le 2$. Without any new ideas, I can make the density of missing elements arbitrarily close to zero, and the gaps between them arbitrarily large.

if we have $ a_{n \plus{} 1} \equal{} 3\lfloor \sqrt [n] {n} \rfloor \plus{} a_n \plus{} 3$, then $ a_{n \plus{} 1} \minus{} a_n$  is bounded so exist infinite arithmetic subsequence? How prove that?

That sequence can just as well be written as $ a_{n\plus{}1}\equal{}a_n\plus{}6$.

ok, but how prove then we can chose from it an infinite arithmetic sequence?

It is already an infinite arithmetic progression, with difference $ 6$.

i'm sorry, i want to write $ a_{n \plus{} 1} \equal{} 3 \lfloor  \sqrt [n]{a_n}  \rfloor \plus{} a_n \plus{} 3$, , how prove then we can chose from it an infinite arithmetic sequence?

You need to prove that $ a_n<2^n$ for sufficiently large $ n$. As long as that is true, the sequence becomes $ a_{n\plus{}1}\equal{}a_n\plus{}6$. I'll leave the question of why $ a_n<2^n$ for large $ n$ to someone else.

[quote="mlok"]What if the gaps between missing elements tend to infinity?[/quote]
The answer is still no.

Here is a necessary condition for a set $ A\subset\mathbb N$ to contain an infinite arithmetic progression: there is a constant $ M$ such that for any $ k\in \mathbb N$ there is $ j\in\{0,1,\dots,k \minus{} 1\}$ such that the gaps between consecutive elements of $ A\cap (k\mathbb N \plus{} j)$ do not exceed $ kM$.

The set $ \mathbb N\setminus\bigcup_{k \equal{} 1}^{\infty}\bigcup_{j \equal{} 0}^{k \minus{} 1}\{ki \plus{} j\colon i \equal{} I_{k,j},\dots , I_{k,j} \plus{} k\}$, where $ I_{k,j}$ grow rapidly, fails the above condition but has increasingly large gaps between missing elements.

I'm pretty sure that the above necessary condition is not sufficient.


integration
inequalities
advanced fields
advanced fields unsolved

Let ${u\in C^{2}(\mathbb{R}\times\mathbb{R}})$ be a solution of $u_{t}=u_{xx}-u^{3}$ such that $u(t,x+1)=u(t,x)$. Prove the following two implications:
1) $u$ bounded $\Longrightarrow u=0$
2) $\frac{\partial u}{\partial x}$ bounded $\Longrightarrow u=0$

Multiplying by $u$ and integrating by parts, we get $\frac d{dt}\int_0^1 u^2=-2\int_0^1 u_x^2-2\int_0^1 u^4\le -2\left(\int_0^1 u^2\right)^2$. Thus, if $Y(t)=\int_0^1 u^2$, we have the inequality $Y'\le -2Y^2$. It follows that if $Y$ is defined for all negative times, then $Y$ is identically $0$. Indeed, if $Y(T)>0$, then $Y(t)>0$ for all $t<T$ and we can rewrite our inequality as $\left(\frac 1Y\right)'\ge 2$ on $[t,T]$, so $\frac 1{Y(T)}\ge \frac 1{Y(t)}+2(T-t)\ge 2(T-t)$ and $Y(T)\le \frac 1{2(T-t)}\to 0$ as $t\to-\infty$. So, none of the assumptions (1) or (2) is actually needed to draw the conclusion.


geometry
incenter
Gauss
circumcircle
ratio
geometry proposed

Let $ABCD$ be a bicentric quadrilateral, $I$ its the incenter and $M,N$ are the midpoints of $AC,BD$, also $EF$ is the third diagonal. If $P$ is the cut point of $AC,BD$ further the line $IP$ cut $EF$ at $Q$ . Prove that: $\left[ABCD\right]=2\cdot MN\cdot\sqrt{EQ\cdot FQ}$

Hello JC,

   From Casey Book I, Prop.7  area of EMN ( or FMN)  is 1/4 of areas of ABCD.
   Line MN is the Gauss line , goes thru midpoint of EF. A circle on diameter EF
   goes thru I  and IQ is  Geo-Mean of EQ and FQ and it is equal to the height of
   each of EMN or FMN to side MN.

    Maj.   Pestich

Let us put some substance to pestich's proof.

If $O$ is the circumcenter of $ABCD,$ then $I,O,P$ lie on a perpendicular to $EF,$ since $EF$ is the polar of $P$ WRT $(O)$ and $(I).$ By Newton's theorem, we know that $MN$ passes through $I$ and the midpoint $L$ of $EF.$  Circle $(L),$ whose diameter connects the conjugate points $E,F$ WRT $(O),$ is then orthogonal to $(O)$ and the circle $\odot(PMN)$ with diameter $\overline{PO}$ $\Longrightarrow$ $LE^2=LM \cdot LN$ $\Longrightarrow$ $LE$ is tangent to $\odot(EMN),$ i.e. $(L)$ is the E-Apollonius circle of $\triangle EMN.$ But $\triangle EAC \sim \triangle EBD$ yields $\triangle EMC \sim \triangle END$ $\Longrightarrow$ $\angle MEC=\angle NED$ $\Longrightarrow$ $EI$ bisects $\angle NEM,$ thus $I \in (L).$ If  $T$ denotes the projection of $E$ on $MN,$ then $LI=LE$ implies that $ET=IQ=\sqrt{EQ \cdot FQ}.$

According to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=406453]Ratio of areas[/url], the area of $\triangle EMN$ is 1/4 of the area of $ABCD,$ thus

$[ABCD]=4 \cdot [EMN]= 2 \cdot  MN \cdot ET=2 \cdot MN \cdot \sqrt{EQ \cdot FQ}.$


If $ g(x) \equal{} 2x^2 \minus{} 3$ and $ h(x) \equal{} 4x^3 \plus{}1$, what is the value of $ g(h(\minus{}1))$?

We just plug in the number:

$ g(h(\minus{}1))\equal{}g(4(\minus{}1)^3\plus{}1)\equal{}g(1\minus{}4)\equal{}g(\minus{}3)\equal{}2(\minus{}3)^2\minus{}3\equal{}2(9)\minus{}3\equal{}18\minus{}3\equal{}15$

Hey, your avatar isnt a monkey anymore. You can join the tourney now. :lol:

YEAH, ITS HELLO KITTY!

And i dont wanna join the tourney, thank you very much.


calculus
geometry
trigonometry
conics
MATHCOUNTS
function
real analysis
\/closed

[quote="horselvr101"]Which math classes do the Art of Problem Solving books correspond to? I think Volume 1 is primarily Algebra I and Geometry. Correct me if I'm wrong. But how much of Volume 2 is Algebra II and how much, if any, is Pre-Calculus?

I am asking this because I am trying to decide when to take what in school. Thanks for your help![/quote]

Volume I:
Algebra I
Algebra II
Geometry
Basic Trig

Volume II:
Advanced Algebra II such as conics and matrices
Difficult geometry but not the ones that are found in school curriculum
Detailed trig
Calculus although it's not emphasized as much as other stuffs

The AoPS books are not meant to be used for school curriculum. They do help a lot but there are plenty of materials for non-school math because the books focus on contest problems more.

[quote="Silverfalcon"]
The AoPS books are not meant to be used for school curriculum. They do help a lot but there are plenty of materials for non-school math because the books focus on contest problems more.[/quote]

Oh, I wasn't planning to use them for school curriculm. I know they are for contest problems and I plan to use them as such. I asked because
a) I was curious
b) I have to decide what curriculm to take and I'm having some difficulty picking a math class

Thanks for your help!

[quote="ShoeFactory"]AoPS 1 is definitely not algebra 1.

An algebra 1 student probably wouldnt be able to touch 97% of the problems in there.[/quote]

Not a typical algebra I student maybe, but a student with a good pre-algebra class and problem solving background from a program like MATHCOUNTS should be able to work many of them.

[quote="Silverfalcon"]
Volume I:
Algebra I
Algebra II
Geometry
Basic Trig

Volume II:
Advanced Algebra II such as conics and matrices
Difficult geometry but not the ones that are found in school curriculum
Detailed trig
Calculus although it's not emphasized as much as other stuffs[/quote]

...plus a bit of college number theory and combinatorics... :)

[quote="Silverfalcon"]
Calculus although it's not emphasized as much as other stuffs[/quote]Uhh, there's Calculus in AoPS V. 2?  Where?

[quote="joml88"]Limits are discussed in the AoPS books but I personally don't consider limits to be calculus.[/quote]
Limits are fundemental to calculus. Once you have that down, and figure out how to integrate a step function, then integrate general ones by bounding them with step functions, you can pretty much derrive all of calculus.

See the book by  T.A. Apostol for more.

[quote="joml88"]Limits are discussed in the AoPS books but I personally don't consider limits to be calculus.[/quote]

w/o getting into stuff like lebesgue integration, basically everything in single and multivariable calculus is defined in terms of the limit, so i'm not sure how limits aren't calculus.

I think joml88 means that limits themselves aren't part of calculus, but they are its foundation.

The difficulty level in AoPS 1 problems is pretty varied.  Some of the needle and thread ones are really hard!


Does there exist $ a,b \in Z $ such that : $ 3a^2 + b^2 $ is divisible by 2003 ?

$2003$ is a prime number, the only solutions in positive integers are pairs $(a,b)$ such that $2003|a$ and $2003|b$, in other cases
the sum $3a^2+b^2$ can be divisible only if $(\frac {-3} {2003})=1$, but that is not true cause
$(\frac {-3} {p})=1$ only for $p$ prime such that $p = 1 (mod 6)$ and
$(\frac {-3} {p})=-1$ only for $p$ prime such that $p=5 (mod 6)$. As we know, $2003=5 (mod 6)$.


$a - \sqrt{a^2 - 1} = e + f - 2 \sqrt{ef}$
Are you allowed to deduce that $2\sqrt{ef}=\sqrt{a^2-1}$? :?
and $a=e+f$?

There are of course other ways in which this could be true, but it is one possible identification. If it was not a possible identification it would lead to something untrue.

Riddler, yes, since if there are integer solutions, $e + f$ is the only integral part of the RHS and $a$ is the only integral part of the LHS, likewise $-2\sqrt{ef}$ is the only irrational part on the RHS and $-\sqrt{a^2-1}$ is the only irrational part on the LHS, so they must be equal too.  Now, technically $\sqrt{ef}$ doesn't have to be irrational, but $\sqrt{a^2-1}$ is irrational for $a \not = 1$, so if we get elegant solutions from this way, we can assume they are equal.

nice solutions :D  beta please hide your solutions ;)


ARML
AMC
USA(J)MO
USAMO
function
number theory
relatively prime

Hello, all!

Some teams are taking the test tomorrow, so please do not discuss it on this forum until then.  After that, though, rip away!

Paul

FIRST PLACE, INDIVIDUALS

Three-way tie.  In the proverbial alphabetical order ...

Haitao Mao: Thomas Jefferson C
Palmer Mebane: Chesapeake Archimedes
Jonathan Mei: Buchholtz 1

TOP TEAMS: ARCHIMEDES DIVISION

$ \begin{tabular}{lllll} Place & Team & Location & Coach & Score \\
& & & & \\
1 & Thomas Jefferson C & Alexandria, VA & Pat Gabriel/Jen Allard & 138 \\
2 & Lehigh Valley North of Philly & Radnor, etc & Don Davis & 126 \\
3 & Wayzata Gold & Wayzata & Tom Kilkelly & 123 \\
4 & Chesapeake Archimedes & Chesapeake & Catherine Asaro & 117 \\
5 & Team 1 & Waterloo, Ontario & Will Ma & 116 \\
6 & Zum Zum & Troy, Michigan & R. P. Khetan & 115 \\
7 & San Marino HS & San Marino, CA & Tom Armbruster & 112 \\
8 & Lehigh Valley \minus{} west of Philly & Valley Forge & Don Davis & 111 \\
9 & Lehigh Valley \minus{} North Jersey & Montgomery & Don Davis & 110 \\
10T & Lehigh Valley \minus{} Bethlehem & Bethlehem & Don Davis & 107 \\
10T & Buchholz 1 & Gainesville, FL & Ziwei Lu & 107 \\
12 & Lehigh Valley \minus{} \minus{} Upper Dublin & Ft Washington & Don Davis & 106 \\
13 & Northwood HS (Team A) & Irvine, CA & Maria Casper & 103 \\
14 & Massey Team A & Windsor & Christopher Ing & 102 \\
15 & West High I & Salt Lake City & Tihomir Asparouhov & 99 \\
16T & Iowa City West ARRR!!!!! & Iowa City & James Kirpes & 98 \\
16T & Palos Verdes Peninsula HS & Rancho Palos Verdes, CA & Vararat Chaiyont & 98 \\
18T & Garfield High School & Seattle & Elissa Farmer & 97 \\
18T & Waterford & Salt Lake City & Tihomir Asparouhov & 97 \\
20T & Lassiter High School & Marietta, GA & Don Slater & 96 \\
20T & Thomas Jefferson A & Alexandria, VA & Pat Gabriel/Jen Allard & 96 \\
20T & Arcadia High School A & Arcadia, CA & Kerry King & 96 \\
\end{tabular}$

BERNOULLI DIVISION

$ \begin{tabular}{lllll} Place & Team & Location & Coach & Score \\
& & & & \\
1 & South Burlington High School & South Burlington & Jean Ohlson & 58 \\
2 & Chesapeake Bernoulli & Chesapeake & Catherine Asaro & 47 \\
3 & Christ's Household of Faith School & Saint Paul & Mike Reiners & 31 \end{tabular}$

So, how about that theme round? I'm pretty sure we (Chesapeake) had all 10 correct answers down in 20 of the 60 minutes. I don't think I've ever played games during a contest time limit before. Did anyone else find it rather easy?

The theme round was definitely the easiest round.

Here's a difficulty chart for the questions on the individual, team, and theme rounds for the 8 teams that competed at Long Beach. Listed is either the number of individuals (out of 47) or the number of teams (out of 8) that got this particular question right.

$ \begin{tabular}{ccccccccccc}Round & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
Indiv. & 41 & 41 & 28 & 18 & 14 & 27 & 14 & 1 & 13 & 3 \\
Team & 6 & 7 & 4 & 6 & 5 & 5 & 6 & 1 & 5 & 3 \\
Theme & 8 & 8 & 6 & 7 & 3 & 7 & 7 & 3 & 6 & 6\end{tabular}$

One of our teams (Northwood A) got 40 points on the theme round. And what is it with question 8 on all three rounds?

Oh, and apologies to the team which competed as "Richard Fung's Team": I entered your scores under the more prosaic name "San Marino HS." So that's how it shows up above.

Catherine: please re-check your "place" counter in the left column. Upper Dublin should be 12th, Northwood A 13th, and so on.

[quote="Kent Merryfield"]Catherine: please re-check your "place" counter in the left column. Upper Dublin should be 12th, Northwood A 13th, and so on.[/quote]

Got it!  I'm a math coach; adding is beyond my purview.  :-D

I do not think the theme round was that easy for people without much experience in number theory.  Fortunately, our team had one person not in that category.  Oh, well.

(Also, I think that the name of Thomas Jefferson A is misspelled.)

Yeah, the theme round was pretty easy (I was also on Chesapeake).  At least we got some handwriting training in.  :P

I'm sort of mad about the individual round . . . I was being stupid on #6 and wasted so much time on it that I didn't get #5, either.  Had I gotten both, I would have had a 9 (made a bad assumption on #8).

[quote="Boy Soprano II"]Also, I think that the name of Thomas Jefferson A is misspelled.[/quote]

Ack.  Fixed that.  I could say I'm a math coach, so spelling isn't my purview, either, but at this rate, my purview won't be anything but playing Cubis on the computer!

[quote="Walk Around The River"]Hmm, no one informed me about this.
Where can I find questions?[/quote]
http://www.arml.com/arml_local/index.php

I was on Lehigh Valley North of Philly.  The LV ARML coach decided to split all of us into teams of six based on region, and we ended up with a pretty good team.  I didn't think the theme round was all that easy - we were working for pretty much the whole time  :oops:.  I spent a whole lot of time reading all the stuff about the more advanced encryption system before realizing that I could have just solved the problems without reading a word.  :lol:

The individual round was much easier than normal individual rounds.  I got 9 right, and I messed up #7 because I can't read.  Seriously, more than 1/2 the ARML problems I miss at our practices are because of reading mistakes.  Argh.

At our site, there were 13 people, so I got stuck by myself. But I still managed 9/10 on the theme round and figured out the 10th problem about 5 mins later. I only got 6 on the team round though, so the theme round was particularly easy.

Our team name is wrong.  It should be "ARRR!!!!!" instead of "ARRR!"  lol.   :lol:
The extra 4 exclamation points are definitely important.

Though I think we did pretty well considering none of our USAMO qualifiers showed up for ARML Local...   :D

On the team round, on the sum of squares of diagonals on an octagon problem, I somehow managed to think that the hypotenuse of a right triangle was equal to the sum of it's legs, but I only made that mistake on finding the long diagonal.

On the theme round, first, I forgot 1 was relatively prime to 26.
Then, I only counted 27 instead of 28 pairs that remain the same when performed twice, because I forgot (1,13), which was explicitly stated at the beginning.
After that, I found the decryption function instead of the encryption function of the quote.

Somehow, I managed to get a 9 on the individuals (I got #8 wrong; Matt was the only one in the whole room at Lehigh Valley, I think, to get #8 right).

The one person in the room in Long Beach who got #8 right got a 6 on the whole round. We also had someone who missed only #9 for a 9.

I'm from the South Burlington Team.  We were not so great.  I was the highest scorer in indiv, and I got 6/10.  But, since there were only 3 in r division, we pulled off a win.

Who is responsible for this series of tests? They are quite nice!

Results? Isn't ARML at the end of May???

or is this some practice

Does anyone happen to have individual results as far as # of people with 10's, # of people with 9's, etc.?

I would have had a 9 had I remembered that the square root of -23 is not 23i, but whatever.

As for 10's, the official site has this:

[quote]Congratulations also go out to Jonathan Mei (Buchholz 1), Palmer Mebane (Chesapeake Archimedes), and Haitao Mao (Thomas Jefferson C) for their perfect scores on the individual round.[/quote]
So that's three. I don't know how many 9's and how many 8's there were. (There were one 9 and two 8's among the teams competing at Long Beach). Paul Dreyer would know this information.

I'm reviving this old thread with a bit of data. Of the individuals who competed in ARML Local at Long Beach, 19 of them joined the Southern California ARML team and competed at Las Vegas. The following is a chart of the scores these 19 students had on the two contests. The rows are for scores at ARML Local; the columns are for scores at ARML National.

$ \begin{tabular}{cccccccccc}Local&0&1&2&3&4&5&6&7&8\\ 3&1&&&1&&&&&\\ 4&&&&2&1&&&&\\ 5&&1&1&1&1&&&&\\ 6&&&&&1&3&1&&\\ 7&&&&1&&1&&1&\\ 8&&&&&&&1&&\\ 9&&&&&&&&&1\\ 10&&&&&&&&& \end{tabular}$

That's a pretty good correlation. The person at point $ (9,8)$ is Aubrey Yang.


Does anyone know how to solve this problem:

The Fibonacci sequence is defined as $a_1=1,a_2=1$ and $a_n=a_{n-1}+a_{n-2}$ for each $n > 2$.

Compute:

\[\displaystyle\sum_{j=1}^{2004} \Bigg(i^{2004-a_j}\Bigg)\]

where $i=\sqrt{-1}$.

Hint: [hide]Consider the Fibbonaccis mod 4.[/hide]

[hide]
Consider [tex]\displaystyle i^{a_j}[/tex]:

[tex]i^{1\bmod{4}}=i[/tex]
[tex]i^{2\bmod{4}}=-1[/tex]
[tex]i^{3\bmod{4}}=-i[/tex]
[tex]i^{0\bmod{4}}=1[/tex]

Then consider the Fibonacci sequence in [tex]\bmod{4}[/tex]
Let [tex]k\equiv 0\bmod{6}[/tex]
Then [tex]2004-a_k\equiv 3\bmod{4}[/tex]
[tex]2004-a_{k+1}\equiv 3\bmod{4}[/tex]
[tex]2004-a_{k+2}\equiv 2\bmod{4}[/tex]
[tex]2004-a_{k+3}\equiv 1\bmod{4}[/tex]
[tex]2004-a_{k+4}\equiv 3\bmod{4}[/tex]
[tex]2004-a_{k+5}\equiv 0\bmod{4}[/tex]

[tex]\displaystyle \sum_{j=1}^{2004} \Bigg(i^{2004-a_j}\Bigg)=334(-i-i+1+i-i-1)=-668i[/tex]
[/hide]

Um this is my first post in Pre-Olympiad so my answer is probably wrong but here goes:

[hide]So it starts with i^2003, which is i^3.  Then it goes i^2, i, i, i^4, and so on until i^2, then i^1, then i^(2004-2004) or i^0 which is 1.

since i^1 = sqrt(-1)
since i^2 = -1
since i^3 = -sqrt(-1)
since i^4 = 1

we can sum all these pairs of i^1 through i^4 and they all cancel each other out

so we are left with very last term which is i^0 or 1 so answer is 1[/hide]


function
linear algebra
matrix
calculus
algebra
system of equations

We're learning about matrices right now, and I have some questions (don't answer all of them unless you realllly want to).  Please look at the very last question first.
1. Prove this theorem: For any matrices [i]A[/i] and [i]B[/i], [i]A[/i]-[i]B[/i] = [i]A[/i] + (-[i]B[/i].
2. Prove that for any m*n matrices A and B, A+B=B+A
3. Is a determinant a function of an n*n matrix?
4. Write an argument to convince a classmate that Cramer's rule cannot be used to solve a system of equations where the number of variables does not equal the number of equations.
5. What is true of the determinant of a transposed matrix?
6. Is it sometimes, always, or never true that, if A and B both have nimensions n*n, AB=BA.
7. If for two matrices A and B, both AB and BA exist, what can you determine about A, B, and their products?
8. If the inverse of matrix A is a certain matrix, then is finding A the same as finding the inverse, given matrix A?
9. Find a counterexample to prove that AB=I does not imply that B=A^-1
10. Prove that (A^-1)^-1=A.

And lastly, when doing row operations on matrices, are there general rules you follow in order to get the identity matrix using the formula [A|I]----->[I|A^-1].  When I do row operations, it seems like I'm always doing it randomly...

 Thanks so much for your time.:)

I don't think it belongs to the Getting Started forum..  :?

I'll just simply answer 10 for now.  That is like a function.  $f^{-1}[f(x)]=x$ or vece versa; taking the inverse of the same function yields the independent variable, in our case, x.

EDIT: variable.

Masoud Zargar

[quote="brady00n"]Believe me, this is beginning math.  It's only Algebra II.  I'm sure pretty much everyone here knows calculus.[/quote]Wrong. I don't know Calculus  :blush:

[quote="brady00n"]Believe me, this is beginning math.  It's only Algebra II.  I'm sure pretty much everyone here knows calculus.[/quote]

No, you are absolutely wrong.  There are tons of people on this site who don't know Calculus.  Likewise, I don't consider matrices a Getting Started topic.  I'm moving this to the linear algebra forum where it is better suited.

1. Lets say that matrix A has elements $a_{11}, a_{12},a_{21}, a_{22}$ where the notation $a_{nm}$ indicated the element at the "address" of row n, column m.  Similarly for B.
$a_{11} a_{12} - b_{11} b_{22} = a_{11} a_{12} + -b_{11} -{b_12}$
$a_{21} a_{22}$   $b_{12} b_{22}$ $a_{21} a_{22}$ $-b_{21} -b_{22}$
Thats pretty much the best I can do for that one.

2. Basically $a_{nm}+b_{nm}=b_{nm}+a_{nm}$ Also the best I can do for that one.

3. Yes. You can only take the determinant of a nxn matrix

4. x+y=39 has how many solutions?

5. It is the same as the determinant of the original matrix.

6. Sometimes.

7.  In order for both AB and BA to be defined, they both have to be of the same order nxn.  Thus they also have determinant and the product is a square matrix nxn. Det(A)*Det(B)=Det(AB) I dont know how far they wanted to go with that one.

8. ((A^-1)^-1)=A. I'm not sure if thats what the question was asking about though, these questions seem oddly worded.

9. I can't think of any off the top of my head

10.  Try it for a general case. Should work out.

Row Operations... well... it still seems like I'm doing it randomly too.

[quote]7. In order for both AB and BA to be defined, they both have to be of the same order nxn. [/quote]
It's not that restrictive. You could have $A$ is $n\times m$ and $B$ is $m\times n$ and have both $AB$ and $BA$ defined, albeit of different sizes.

One thing that would be true even in this setting: $\text{Tr}(AB)=\text{Tr}(BA)$ where $\text{Tr}(C)$ is the trace of square matrix $C,$ that is, the sum of the elements on the main diagonal.

9. $\begin{bmatrix}1&-1&0\\0&1&1\end{bmatrix} \begin{bmatrix}0&1\\-1&1\\1&0 \end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}.$

Ignore what boxedexe wrote for 10.  This has nothing to do with functions.

You want to prove that $(A^{-1})^{-1}=A$

Well, we have $A(A^{-1})=I_n$ by definition.  This means that $A$ is the inverse of $A^{-1}$ (since if $A$ is the inverse of $B$, then $B$ is the inverse of $A$; recall that any one-sided inverse is a two sided inverse)

So we have $(A^{-1})^{-1}=A$ as desired.


logarithms
limit
calculus
calculus computations

Write (n + 4)/(n - 4) as 1 + 8/(n - 4).  Expand using the binomial theorem.  The limit of the sum is the sum of the limits of the individual terms in the series, each of which after simplifying is of the form 4^i/i! from i = 0 to infinity. This is simply the power series for e^x with x = 4, giving the e^4.

or from $\lim_{n\rightarrow\infty}(1+8/(n-4))^{n/2}$
substitute $u = n-4$
$u$ still approaches $\infty$
$\lim_{n\rightarrow\infty}(1+8/u)^{(u+4)/2}$
$\lim_{n\rightarrow\infty}(1+8/u)^{u/2+2}$
$\lim_{n\rightarrow\infty}(1+8/u)^{u/2}(1+8/u)^{2}$
$\lim_{n\rightarrow\infty}\sqrt{(1+8/u)^{u}}$
$\sqrt{e^{8}}$
$e^{4}$

[quote="Kent Merryfield"]Or take the logarithm. If $L$ is the desired limit, then

$\ln L = \lim_{n\to\infty}\frac{n}{2}\ln\left(1+\frac8{n-4}\right).$

Work with that, either by using the power series expansion of $\ln (1+x)$ or L'Hospital's rule.[/quote]
When working with this specific kind of limit, you can also write $\frac{1}{x}\ln(1+x) = \frac{\ln(1+x)-\ln 1}{x}$, the definition of the derivative. You get the limit without using Taylor or l'Hospital.

[quote="phils"]
When working with this specific kind of limit, you can also write $\frac{1}{x}\ln(1+x) = \frac{\ln(1+x)-\ln 1}{x}$, the definition of the derivative. You get the limit without using Taylor or l'Hospital.[/quote]
This is exactly L'Hopital's rule.

[quote="Kalle"]
This is exactly L'Hopital's rule.[/quote]
Isn't this a somewhat weaker case? In general, L'Hospital's rule requires use of the Mean Value Theorem (or the "Cauchy MVT"). I am not using anything like that here.


vector
Functional Analysis
advanced fields
advanced fields unsolved

Hello everyone, I would like you can help me to figure out if this proposition is always true:

Let $ V$ be a Banach space and $ L: V \rightarrow V$ be a continuous linear operator.  Let $ \sigma(L)$ be the spectrum of $ L$. Then, the following two conditions are equivalent:

(1) $ \mu \in \sigma(L)$
(2) For all $ \varepsilon >0$ there exists $ v \in V$ such that $ \|v\|\equal{}1$ and $ \|Lv\minus{}\mu v\| \leq \varepsilon$.

Of course (1) $ \Rightarrow$ (2) is trivial.   (2) $ \Rightarrow$ (1) is not so obvious. I proved that this is true when $ V$ has finite dimension.  In the proof, I used the fact that $ S^1\equal{} \{\lambda \in \mathbb{C}: |\lambda|\equal{}1\}$ is compact in order to assure the existence of a subsequence that converges to a vector in $ V$, which was the vector that proves $ \mu$ is an eigenvalue of $ L$(if someone wants to see it I can write the proof).   However, I cannot use the compactness of $ S^1$ when $ V$ has infinite dimension, so my proof does not work in that case.    Therefore, I would like to know if this is still true when $ V$ has infinite dimension.

I really appreciate any help,

Thank you !

Adriana

[quote="-oo-"]we may assume $ \mu \equal{} 0$ by translating $ L$.

(2) says that $ L$ is not bounded from below, and (1) that $ L$ is not invertible. according to the closed graph theorem, this is equivalent to: $ L$ is not bounded from below or $ rg(L)$ is not dense.

thus $ (2) \Rightarrow (1)$ is trivial, but $ (1) \Rightarrow (2)$ is false, in general.[/quote]

why is $ (1) \Rightarrow (2)$ false in general?  If I assume $ (1)$, then there exists $ v \neq 0$ such that $ Lv \equal{} \mu v$.   We can assume $ \|v\| \equal{} 1$ (just normalizing it), so for every $ \varepsilon > 0$ I can take that $ v$, so
\[ \|Lv \minus{} \mu v\| \equal{} 0 \leq \varepsilon
\]
is there something wrong on that?

About your proof, I am a little bit confused since the version of the closed graph theorem that I know is that an operator $ T$ is continuous if its graph is closed.   Furthermore, what does bounded from below mean? does this mean $ T$ is an bounded operator? and what does $ rg(L)$ mean?

Thank you for your help

Adriana
Adriana

[quote="Adriana"]If I assume $ (1)$, then there exists $ v \neq 0$ such that $ Lv \equal{} \mu v$[/quote]
no. "not invertible => not injective" is false.
[quote="Adriana"]
About your proof, I am a little bit confused since the version of the closed graph theorem that I know is that an operator $ T$ is continuous if its graph is closed.
[/quote]
a well-known collorary is: a bijective, continuous linear operator between banach spaces is open and thus invertible.
[quote="Adriana"]
Furthermore, what does bounded from below mean?
[/quote]
$ L$ is bounded from below if $ \inf_{\parallel{}x\parallel{} \equal{} 1} \parallel{}Lx\parallel{} > 0$.
[quote="Adriana"]
and what does $ rg(L)$ mean?
[/quote]
range of $ L$.

Perhaps the confusion here for Adriana is in the definition of spectrum?  In finite dimensions, the spectrum is just the set of eigenvalues, but for operators on Banach spaces, the definition is more general:

$ \mu\in\sigma(L)$ iff $ \mu I \minus{} L$ is not invertible

which is a more general condition than the condition that $ \mu I \minus{} L$ has nontrivial kernel.

[quote="Adriana"]Thank -oo- and yenlee.    I got why $ (2) \Rightarrow (1)$ was trivial.    You are right yenlee, I was confused about the definition of spectrum.     About $ (1) \Rightarrow (2)$, What kind of Hypothesis could I add to assure this implication?

This was a question in my exam and in fact the question was TO PROVE OR DISPROVE and in the latter to add some hypothesis to become the statement true. As you saw I could prove this in the finite case, but in the infinite case I did not the answer it ( in fact, I made a mistake because I supposed $ (1) \Rightarrow (2)$ was true  :(  )

Thank you !

Adriana[/quote]

I got it.    Suppose $ \mu \in \partial \sigma(L)$ (which exists since $ \sigma(L)$ is closed)  and suppose there exists $ \varepsilon >0$ such that

\[ \|Lv\minus{}\mu v\|>\varepsilon,\] for all $ v \in V$ with $ \|v\|\equal{}1$.  Hence, by the Closed Graph Theorem, $ (L\minus{}\mu I)V$ is a proper closed subspace of $ V$.    Then, there is $ u \in V$ such that

\[ dist(u,(L\minus{}\mu I)V)\equal{}\delta >0.\]

Define $ W\equal{}\{w \in V: \|w\|\leq 2\varepsilon^{\minus{}1}\|u\|\}$.   It is not hard to prove that

(1) $ |\alpha| < \frac{\varepsilon}{2} \Rightarrow u \not \in (L\minus{}(\mu\plus{}\alpha)I)[V\backslash W].$
(2)  $ |\alpha|<\frac{\delta \varepsilon}{2\|u\|} \Rightarrow u \not \in (L\minus{}(\mu\plus{}\alpha)I)W.$
 Hence, for any $ \alpha \in \mathbb{C}$ such that

\[ |\alpha| < \min \{ \frac{\varepsilon}{2} ,\frac{\delta \varepsilon}{2\|u\|}\},\]

 $ u \not\in (L\minus{}(\mu \plus{} \alpha)I) V$, then $ L\minus{}(\mu \plus{}\alpha)I$ is not an automorphism.    Therefore, $ \alpha \plus{}\mu \in \sigma(L)$.   This implies that $ \mu$ is not a boundary point of $ \sigma(L)$ which is a contradiction.


I can't imagine a scenario where I would want parentheses, brackets, etc. to be the wrong size, so I'm wondering if there is a way to make LaTeX automatically interpret these symbols as if I had used \left and \right, since it is annoying to have to type them all the time. Thanks in advance.

[quote="stevem"]Yes and no :) You would need to redefine ( and ) but when you do that you cause other things to break. For example
[code]\def\mathactdef#1{%
  \mathcode`#1="8000
  \begingroup
  \lccode`~=`#1
  \lowercase{\endgroup\def~}}

\mathactdef({\left(}
\mathactdef){\right)}[/code]works in a simple document, but once you use the amsmath package it breaks. So maybe you should just define your own commands like
[code]\newcommand{\lbr}{\left(}
\newcommand{\rbr}{\right)}[/code] and then use \lbr, \rbr instead of (, )[/quote]
I had a feeling there would be no way to do it without causing problems. I think I'll use the second method, though. Thanks.


function
calculus
derivative
integration
calculus computations

i haeve a question about this.
i have seen in some places that Rolle's theorem is stated that $f(a)=f(b)=0$
and in others it simply states that $f(a)=f(b)$.
either way if $a<b$ and $f(x)$ is defined on $[a,b]$ and differential on $(a,b)$ then there is a point $c\in[a,b]$ such that $f'(c)=0$.

what i am wondering is which of the 2 other conditions must be met. is it the $f(a)=f(b)=0$ or the $f(a)=f(b)$?

we need $f(a)=f(b)$ ($f(a)=f(b)=0$ is just a particular case ) and of course $f$ must be continuos over $[a,b]$  :)

This is a question you could have answered yourself.  Here's the clue: what does adding a constant to a function do to the derivative of that function?

If you think about what that looks like, it's exactly the same thing. Suppose you prove it for the particular case $f(a) = f(b) = 0$. Then you have some function $g$ but $g(a) = g(b) = c$. Then let $f(x) = g(x) - c$. Then $f(a) = f(b) = 0$, but $f'(x) = g'(x)$, so it's obviously true for $g(x)$ also.

You can use this to prove the mean value theorem by subtracting a linear function in a similar way. Can you see how? :)

JBL: yeah i know that adding a constant does nothing.
I was just wondering which source was right...since using either condition gets the same restult, well sorta.

The reason for $f(a)=f(b)=0$ (an obviously unnecessary constraint) is, I think, to really emphasize Rolle's theorem as an essential lemma for proving the Mean Value theorem (think of the proof of that result).

Well, here it is the proof avec Rolle for mean value theoem:

[b] Mean Value theorem [/b]

  Let $f: [a, b]\to\mathbb{R}$ be a continuous function. Ptove that there is $c\in (a, b)$ such that

                                    \[ \int_{a}^{b}f(x)dx=(b-a)f(c). \]

  [i] Proof. [/i]

      Let us consider the function $H: [a, b]\to\mathbb{R}$ defined by
                                  \[ H(t)=(t-a)\int_{t}^{b}f(x)dx+(t-b)\int_{a}^{t}f(x)dx. \]

      It is easy to see that \[ H'(t)=\int_{a}^{b}f(x)dx-(b-a)f(t). \]

      Since $H(a)=H(b)=0$, by Rolle theorem the conclusion follows.

[quote="blahblahblah"]The mean value theorem I referred to was:

If $f$ is continuous and differentiable on $[a,b]$, then there exists $c\in[a,b]$ such that

$f(b)-f(a)=(b-a)f'(c)$[/quote]

Yes, you are right. But to distinguish from the one I posted, I think we should call this Lagrange's theorem. ;)

[quote="JBL"]Those two theorems are equivalent by the Fundamental Theorem of Calculus -- just replace $f$ in Cezar's theorem by $f'$ in blahblahblah's.[/quote]

In spite of that, Cezar's is possible to prove without any deep facts about either integration or differentiation.


function
real analysis
real analysis unsolved

Hello everyone!:)
Let $(f_n; n \in N)$ be a sequence of measurable functions and $f_n\to f$ by $\mu$ and let $g$ satisfies the Lipschitz condition. Proof that a sequence  $(g \circ f_n)\to g \circ f$ by $\mu$.
Thanks for help!

Let $K$ be a Lipschitz constant for $g$. Hence $|g\circ f_n-g\circ f|\leq K|f_n-f|$. It follows that
\[ \{x: |g(f_n(x))-g(f(x))|\geq \epsilon\}\subset \{x: |f_n(x)-f(x)|\geq \epsilon/K\}  \]
and the measure of the right set is goes to zero.


Hello Cindy,

[quote="cindy"]:sqrt: (5x+1) -  :sqrt: (3X-5)=2[/quote]

I noticed that this is your first post, therefore I know the moderators will have merci of you, but this is FAR FAR AWAY of a Superior Algebra problem. Nevertheless, let's give it a try. Some times an easy way to solve an equation is just to follow the most naive way you may think. For example (Assuming that $X=x$, Because otherwise does not have much sence):

\[\sqrt{5x+1} -  \sqrt{3x-5}=2\]

\[\Rightarrow(\sqrt{5x+1} -  \sqrt{3x-5})^2=4\]

\[\Rightarrow 5x+1 -  2\sqrt{(3x-5)(5x+1)} + 3x=4\]

\[\Rightarrow 4(x+1) = \sqrt{(3x-5)(5x+1)}\]

(square both sides and solve)

\[\Rightarrow x^2-10x+21=(x-3)(x-7)=0\]

Then, you test the numbers, both work, and the solution is $\{3,7\}$. Remember, it is $\Rightarrow$ and not $\Leftrightarrow$, that is, to be root of the first equation it has to be root of the last one, but not necesarly the other way around, this is, YOU HAVE TO PLUG IN AND BE SURE THE ROOTS REALLY WORK.

Best regards,

Hey Kent, Hello Forum,

[quote="Kent Merryfield"][quote]I know the moderators will have merci of you[/quote]
We moderators mostly just try to put posts in the best places for them. So I need to figure out where to send this. It's definitely high school level (not middle school). My best guess as to the right level is "Intermediate Topics", so that's where it's going.[/quote]

I didn't try to imply anything else, it just came as a "Funny coment", I wanted to say, mostly, that it was obviusly missplaced, but that wouldn't be a problem. Also, probably if someone with at least a hundred of posted messages makes this kind of mistake, then probably would recive a kind of "Com'on fella, you are already grown up, you know this is misplaced" (may be in a more politically correct tone, but the idea is the same), but given that it was the fist post of the lady, it would be explained to her carefully.

I know that the moderators have three basic principle: Keep the Order of the Posts, Keep the Peace in the Forum and Help Disoriented and Confused Souls in the Forum, and let me say, they do it quite good (even when they have to deal with people that sometimes is a pain in the neck, and yes, I try not to be, but I know sometimes I am, I authorize to any moderator or administrator to tell me if I am in behaving that way).

Best regards,

Well, you never know. To that person, this could have been "Superior" algebra. So you can't blame them, considering it's his/her first post.

I think mods mostly worry about a problem being too hard for a forum than being too easy.

Going back to the problem, I would have done it in a slightly different way:

\[\sqrt{5x+1} - \sqrt{3x-5}=2\]

\[\sqrt{5x+1}=2+\sqrt{3x-5}\]

\[5x+1=4+4\sqrt{3x-5}+3x-5\]

\[x+1=2\sqrt{3x-5}\]

\[x^2+2x+1=12x-20\]

\[x^2-10x+21=0\]

To cindy: when you want to solve equations like those, generally you want to isolate the radical on one side, square the equation (or cube, or whatever power it is).  If there are still radicals, then isolate that and square, and keep on repeating until all radicals are gone.


function
calculus
integration
real analysis
real analysis unsolved

The function f has fisrt derigative on $\mathbb{R},$ and we have $f'(x)<\frac{|x|}{|x|+1},$ for all $x\in\mathbb{R}.$ Prove that there exists $c\in\mathbb{R},$ such that $f(c)=c.$
Do you think there is a solution with no using integrals? Thank you.

It's not a strong enough contraction to invoke the Banach theorem right away ($\frac{|x|}{|x|+1}$ tends to 1 at $\pm\infty$)- we need something like an integral or sum estimate to make the argument work.

Let's just work from zero:

$|f(x)-f(0)|\le\int_{0}^{|x|}\frac{t}{1+t}\,dt=|x|-\ln(1+|x|).$

Now find $A$ such that $\ln(1+A)\ge|f(0)|.$

If $|x|\le A,$ then $|f(x)|\le |f(0)|+|x|-\ln(1+|x|).$

$|f(x)|\le \ln(1+A)-\ln(1+|x|)+|x|\le|x|\le A.$

Hence, $f$ maps $[-A,A]$ into itself. And since $|f'(x)|\le\frac{A}{1+A}$ on this set, we do have that $f$ is a strict contraction on that set and we can invoke the Banach fixed point theorem.

The key point about our estimating function, $g(x)=\frac{|x|}{1+|x|}:$ we have that $0\le g(x)<1,$ but also $\int_{-\infty}^{\infty}1-g(x)\,dx=\infty.$ We used all of those features.

It's not the integral of $g$ that has to be infinite; it's the integral of $1-g$.

[color=green][Typo, now fixed - K.M.][/color]


Prove for each $ n$, that $ (\varepsilon \plus{} 1)^{n} >\equal{} 1 \plus{} n\varepsilon$ if $ \varepsilon > \minus{} 1$

But do you want to have n in $ \mathbb{N}$ or maybe $ \mathbb{R}_{\plus{}}$
If you want only the first case try normal induction. The second case I think can be proved with Jensen's inequality... :wink:

Well then just apply induction:
Suppose:
$ (x \plus{} 1)^{n} >\equal{} 1 \plus{} nx$
Proof:
$ (1\plus{}x)^{n\plus{}1}\equal{}(1\plus{}x)(1\plus{}x)^{n}\geq (1\plus{}x)(1\plus{}nx)\equal{}1\plus{}nx^{2}\plus{}x\plus{}nx\equal{}1\plus{}(n\plus{}1)x\plus{}nx^{2}\geq 1\plus{}(n\plus{}1)x$
Remember that we used $ x>\minus{}1\iff 1\plus{}x>0$ in the first transformation :wink:

If you knew how to spell "Bernoulli," you could have done a search of the forum and not posted something that's appeared dozens of times before.

Bonus question: [i]which[/i] of the Bernoulli's is this inequality named after?


Find all continuous functions $ f: (0,\infty)\rightarrow \mathbb{R},\: t^2f(t)\equal{}f\left(\frac{1}{t}\right)$

Let $ g(x) \equal{} e^xf(e^x)$, then your condition equavalent to
(1) $ g(x) \equal{} g( \minus{} x)$.
And for any $ g(x): R\to R$, satisfyed (1) we get solution $ f(t) \equal{} \frac {g(ln t)}{t}$.


Two candles of the same height are lighted at the same time. The first is consumed in $ 4$ hours and the second in $ 3$ hours. Assuming that each candle burns at a constant rate, in how many hours after being lighted was the first candle twice the height of the second?

$ \textbf{(A)}\ \frac {3}{4} \qquad\textbf{(B)}\ 1\frac {1}{2} \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 2\frac {2}{5} \qquad\textbf{(E)}\ 2\frac {1}{2}$

after $ k$ hours, the height of the first candle is $ 1\minus{}\frac{k}4$ of the original height, and the second is $ 1\minus{}\frac{k}3$. So we'll find $ k$ such that $ 1\minus{}\frac{k}4\equal{}2(1\minus{}\frac{k}3)$. It's easy to get $ k\equal{}2\frac25$.


w and z are complex numbers w^2 + z^2 = 7, and w^3 +z^3 = 10, find the maximum value of w+z?

Look here: http://www.mathlinks.ro/Forum/viewtopic.php?t=22180  :)

hm, i'm not sure but could it have something to do with newton's sums? so if w & z were complex roots of a quadratic equation ax^2+bx+c=0, then it would become 10a+7b+(w+z)=0. But i'm not sure, it might be a dead end :/

I just used this problem for yesterday's AIME a Day.  Here is my solution again but it's basically like those on the page that ZetaX posted.

Let $w+z=S$.  So then:
\[w^2+z^2=(w+z)^2-2wz=S^2-2P=7.\]

Now let $wz=P$ just to make things a little easier.  We also can find out that:

\[w^3+z^3=(w+z)^3-3wz(w+z)=S^3-3SP=10.\]

So now we can solve the first for $P$:

\[P=\frac{S^2-7}{2}.\]

We plug this into the other equation:

\[S^3-3S\left(\frac{S^2-7}{2}\right)=10\Rightarrow S^3-21S+20=0.\]

This factors as $(x-1)(x+5)(x-4).$  Thus the largest value is $\fbox{4.}$


[quote="orl"]Given integers $a_1, \ldots, a_{10},$ prove that there exist a non-zero sequence $\{x_1, \ldots, x_{10}\}$ such that all $x_i$ belong to $\{-1,0,1\}$ and the number $\sum^{10}_{i=1} x_i \cdot a_i$ is divisible by 1001.[/quote]

It's suprise to learn that this's in ISL 1988, the first time i saw this is in The Mathematics and Youth Magazine (a faumous Problem Solving Journal in seconder. level in Vietnam). Here's sol.:

Consider all sum in form: $\sum^{10}_{i=1} x_i \cdot a_i$ where $x_i\in\{0;1\}$. There're $2^{10}=1024>1001$ these sums, so there are two of them which congruent to each other. Their differnce has the form as the problem required!


calculus
integration
real analysis
real analysis unsolved

I stuck here, because I don't know why is this so obvious:
Let \[F (h) = \int_a^{a+h} {f(x)dx} - \frac{h}{2} (f (a) + f (a + h))\]
In the book stands we conclude that $F'(h) = F''(h) = 0$ - which is trivial, but next line is following:
\[{F(h) = -\frac{1}2} \int_0^h {f (h - t)f''(a + t)dt}\]
$f(x)$ is twice differenciable  and continuous function...

You're obviously not paying enough attention to details. That should be $F(0)=F'(0)=F''(0)=0$, and your second expression for $F$ is complete nonsense.

We actually have $F''(h)=-\frac h2f''(a+h)$, differentiating the definition by the Fundamental Theorem.
Letting $G$ be your second expression, $G'(0)=-\frac12f(0)f''(a)\neq0.$

\[{F(h) = -\frac{1}2} \int_0^h {t (h - t)f''(a + t)dt}\]
I have a typo :-)
I found second derivate, but still don't know how to finish the proof!

Start from the expression of $F$ given in your post #3 and integrate twice by parts.


Let a triangle $ABC$.  If we know that   $48S^2=(a^2+b^2+c^2)^2$

where $S$  is the area of the triangle, prove that the triangle is equitaletar

To elaborate on how to find that equality case,

[hide]
$(a+b-c)(a-b+c)(-a+b+c) \le \left(\frac{a+b+c}{3}\right)^3$ by AM-GM

$3(a+b+c)\left(\frac{a+b+c}{3}\right)^3 = \frac{(a+b+c)^4}{9} \le (a^2+b^2+c^2)^2$ by AM-QM.

So we have equality in those cases iff

$a+b-c = a-b+c = -a+b+c$ and $a=b=c$. So the triangle must be equilateral. [/hide]

I remember solving this problem after a final exam, in a friend's house.... :lol:  :D

[quote="deej21"]What is AM-QM?[/quote]

Arithmetic Mean-Quadratic Mean, a special case of Power Mean.

$\frac{a+b+c}{3} \le \sqrt{\frac{a^2+b^2+c^2}{3}} \Rightarrow (a+b+c)^2 \le 3(a^2+b^2+c^2)$,

which is also a result of Cauchy.


algebra
polynomial
logarithms
function
calculus
derivative
real analysis

Let $p(x)$ be a real polynomial of degree $n$. Show that the equation $p(x) = \ln x, \quad x > 0$ has at most $n+1$ real roots.

I would assume it has something to do with the intermediate value theorem, but I can't figure out how to do it. Any help appreciated! Thanks.

Here is a small [hide="Hint"]  If the equation $p(x)-\ln x=0$ has $n+2$ real roots, then the equation $p'(x)-\frac1x=0$, which is equivalent to $xp'(x)-1=0$,  has at least $n+1$ real roots [/hide]

[quote="Kent Merryfield"][quote="Kalle"]I would assume it has something to do with the intermediate value theorem,[/quote]
No. The Intermediate Value Theorem would be used in proving that an equation has [b]at least[/b] a certain number of roots. Proving that an equation (involving differentiable functions) has [b]at most[/b] a certain number of roots calls for the Rolle's Theorem / Mean Value Theorem family of arguments. (And arguing by using intervals on which a function is increasing or decreasing falls into that family.)

[Further hint: this comment is perfectly compatible with fedja's hint.][/quote]
Alright, thanks to both of you. We haven't come to the things you've mentioned yet, so I guess I'll wait with trying to solve it until then.

Is it as simple as this?

Since we know that $xp'(x) - 1 = 0$ has at most $n$ roots (seeing as it is a polynomial of degree $n$), we know that $f(x) = p'(x) - \ln x$ changes sign at most $n+1$ times.

Sounds almost too simple... I bet I am missing something.

Then the follow-up question is to prove that $p(x) = \ln |x|$ has at most $n+2$ roots. I suppose this has something to do with the fact that the difference between them isn't continuous over 0... but I can't formalize it. Feel like giving me another hint? =p

We hadn't actually started with derivates when I posted this (I didn't think the question had to do with that), so that's why I'm a little late :)

Yes, it is that simple. As to $p(x)=\ln|x|$, you are completely right about this "having something to do with discontinuity at $0$". Just note that between any two roots you must have either a point where the derivative is $0$ or a point where there is no derivative.   :)


Just much capacity[/u] can a single '[b]transmitter[/b] anthena[b]'[/b] ([b] Ex[/b].[b] Cellphone[/b] Transmitter[b] Anthena[/b].[b])[/b] transmit voluminous of messages[b] coming from a single[/b] cellphone[b] user located in a given place?([/b] Ex.From Los[/b] Angeles[b],[/b] to San[b] Diego[/b].[b])[/b]What specific mathematical[u] calculations are needed connected to the over-all calculation performance related to telecommunication relay transmission?

                                                                        Respectfully,
                                                                        A.Eddington[/u]

Do you put the [/b] and [/u] tags in random places or do you have some kind of algorithm? Other than that I don't quite understand, could try to reformulate the question?

'm pretty sure it's random. And what is this doing in Computer Science and Informatics?


[quote="erfan_Ashorion"] p(0,x) =>f(x)=f(1)/x[/quote]
$0\notin\mathbb R^+$

[quote="Djurre"][quote="pco"]
$ P(\frac 1{f(\frac {f(2)}x)},\frac {f(2)}x)$
[/quote]

How do you come up with this? Is there some way to figure this fast out?[/quote]
Just tests and trials in order to get $f(something)=x$ :

Setting $x=\frac 1{f(y)}$ we get $f(2)=yf(y+\frac 1{f(y)})$

Setting then $y=\frac{f(2)}z$ we get $f(2)=\frac{f(2)}zf(\frac{f(2)}z+\frac 1{f(\frac{f(2)}z)})$ and so $z=f(\frac{f(2)}z+\frac 1{f(\frac{f(2)}z)})$

It remains to merge these two steps in a unique one.

a new way to prove that $f$ is non-increasing
if we have $a<b$,$f(a)<f(b)$,then we find $x_{1},x_{2}$ such that $1+x_{1}a=1+x_{2}f(b)$
$x_{1}+a=x_{2}+b$,let $(a,b)$ be $(x_{1},a)$ and $(x_{2},b)$. a contradiction!

[quote="erfan_Ashorion"]we khnow that :
1)f(X) is surjective!!by the up solution
2)we proof that f(X) is injective:
suppose that f(x)=f(y) we have:
p(0,x) =>f(x)=f(1)/x
f(x)=f(y)=>f(1)/x=f(1)/y =>x=y ;)
3)we khnow that f(1)=1
4)the proof is end f(x)=1/x!!![/quote]

but we cannot plug in zero for x since the equation holds for positive reals

The injectivity was proved. Now plugging $x=1,y=1$ we get $f(1+f(1))=f(2)\Rightarrow f(1)=1$.

[quote="Leonhard Euler"]Find all functions $ \,f: {\mathbb{R^\plus{}}}\rightarrow{\mathbb{R^\plus{}}}\,$ such that
$ f(1\plus{}xf(y))\equal{}yf(x\plus{}y)\quad (1)$ $ \text{for all}\,x,y\in\mathbb{R^\plus{}}.$[/quote]
This is my solution:

Replacing $x$ by $\frac{x}{f(y)}$ in $(1),$ we have
\[f(1+x)=y\cdot f\left(\frac{x}{f(y)}+y\right),\quad \forall x ,\,y \in \mathbb R^+. \quad (2)\]
Next, we replace $y$ by $\frac{f(1+x)}{y}$ in $(2)$ to get
\[y=f\left(\frac{x}{f\left(\frac{f(1+x)}{y}\right)}+\frac{f(1+x)}{y}\right),\quad \forall x,\,y \in \mathbb R^+. \quad (3)\]
From $(3),$ it follows that $f$ is surjective. Now, we will prove that $f$ is decreasing. Replacing $x$ by $x+z$ in $(1),$ we get
\[f\big(1+(x+z)\cdot f(y)\big)=y\cdot f(x+y+z),\quad \forall x,\,y,\,z \in \mathbb R^+. \quad (4)\]
Replacing $y$ by $y+z$ in $(1),$ we also have
\[f\big(1+x\cdot f(y+z)\big)=(y+z)\cdot f(x+y+z),\quad \forall x,\,y,\,z \in \mathbb R^+. \quad  (5)\]
Dividing $(4)$ for $(5),$ side by side, we obtain
\[\frac{f\big(1+(x+z)\cdot f(y)\big)}{f\big(1+x\cdot f(y+z)\big)}=\frac{y}{y+z},\quad \forall x,\,y,\,z \in \mathbb R^+. \quad (6)\]
Now, assume that there exists a pair $(y,\,z)$ such that $f(y+z)>f(y).$ In this case, by choosing $x=\frac{z\cdot f(y)}{f(y+z)-f(y)}$ vào $(6),$ we obtain $y=y+z,$ which is a contradiction. So we must have
\[f(y+z) \le f(y),\quad \forall y,\,z \in \mathbb R^+. \quad (7)\]
Now, we will prove that $f$ is injective. Assume that there are two numbers $a,\,b$ such that $f(a)=f(b).$ Replacing $y=a$ and $y=b$ in $(1)$ respectively in $(1),$ we get
\[a\cdot f(x+a)=b\cdot f(x+b),\quad \forall x\in \mathbb R^+. \quad (8)\]
From this, it follows that
\[1+a(y-1)\cdot f(x+a)=1+b(y-1)\cdot f(x+b),\quad \forall x, \, y \in \mathbb R^+,\, y>1. \quad  (9)\]
Plugging this into $f$ and using $(1),$ we get
\[(x+a)\cdot f(x+ay)=(x+b)\cdot f(x+by),\quad \forall x,\,y \in \mathbb R^+ ,\, y>1. \quad (10)\]
From $(10),$ it follows that, for any $x,\,y,\,z \in \mathbb R^+,\, y>1,$
\[1+(xz+az)\cdot f(x+ay)=1+(xz+bz)\cdot f(x+by). \quad (11)\]
Again, we plug this into $f$ and using $(1).$ It follows that
\[(x+ay)\cdot f(x+ay+az+xz)=(x+by)\cdot f(x+by+bz+xz)\quad  (12)\]
for any $x,\,y,\,z \in \mathbb R^+$ and $ y>1.$ On the other hand, according to $(10),$ we also have
\[\big[(x+xz)+a\big]\cdot f\big( (x+xz)+a(y+z)\big)=\big[(x+xz)+b\big] \cdot f\big((x+xz)+b(y+z)\big),\]
or
\[(x+xz+a)\cdot f(x+ay+az+xz)=(x+xz+b)\cdot f(x+ay+az+xz). \quad (13)\]
Dividing $(12)$ for $(13),$ side by side, we obtain
\[\frac{x+ay}{x+xz+a}=\frac{x+by}{x+xz+b},\quad \forall x,\,y,\,z \in \mathbb R^+,\, y>1. \quad (14)\]
It is easy to deduce that $a=b$ here, so $f $ is injective. Now, replacing $x=y=1$ in $(1)$ with notice that $f $ is injective, we  have $f(1)=1.$ Since $f$ is strictly decreasing ($f$ is decreasing and injective), we have
\[f(x)<1,\quad  \forall x>1.\quad (15)\]
Now, let us consider the case $y>x.$ Replacing $y$ by $y-x$ in $(1)$ and using the above remark, we get
\[f(y)=\frac{f\big(1+x\cdot f(y-x)\big)}{y-x}<\frac{1}{y-x},\quad \forall x,\,y \in \mathbb R^+,\, x<y. \quad  (16)\]
In $(16),$ we let $x\to 0^+$ and obtain
\[f(y) \le \frac{1}{y},\quad \forall y >0. \quad (17)\]
Next, replacing $x$ by $x-1$ and $y$ by $f(y)$ in $(3),$ we get
\[y=\frac{x-1}{f\left(\frac{f(x)}{f(y)}\right)}+\frac{f(x)}{f(y)},\quad \forall x ,\, y \in \mathbb R^+,\, x >1. \quad (18)\]
Since $f\left(\frac{f(x)}{f(y)}\right) \le \frac{f(y)}{f(x)},$ from $(18),$ we deduce that
\[y\ge \frac{(x-1)\cdot f(x)}{f(y)}+\frac{f(x)}{f(y)},\]
or
\[y\cdot f(y) \ge x \cdot f(x),\quad \forall x,\,y \in \mathbb R^+,\, x>1. \quad (19)\]
Changin the position of $x$ and $y$ in $(19),$ we also have
\[x\cdot f(x) \ge y\cdot f(y),\quad \forall x,\,y \in \mathbb R^+,\, y>1. \quad (20)\]
From the inequalities $(19)$ and $(20),$ we can easily deduce that
\[f(x)=\frac{k}{x},\quad \forall x>1. \quad (21)\]
Now, taking $x=1$ in $(1)$ and using $(21),$ we have
\[\frac{1}{1+f(y)}=\frac{y}{1+y},\]
or
\[f(y)=\frac{1}{y},\quad \forall y\in \mathbb R^+. \quad (22)\]
Clearly, the function $f(x)=\frac{1}{x}$ satisfies our equation.

[quote=pco][quote="Leonhard Euler"]Find all functions $ \,f: {\mathbb{R^ \plus{} }}\rightarrow{\mathbb{R^ \plus{} }}\,$ such that
$ f(1 \plus{} xf(y)) \equal{} yf(x \plus{} y)$ $ \text{for all}\,x,y\in\mathbb{R^ \plus{} }.$[/quote]
Let $ P(x,y)$ be the assertion $ f(1 \plus{} xf(y)) \equal{} yf(x \plus{} y)$

1) $ f(x)$ is a surjective function
=====================
$ P(\frac 1{f(\frac {f(2)}x)},\frac {f(2)}x)$ $ \implies$ $ f(2) \equal{} \frac {f(2)}xf(\frac 1{f(\frac {f(2)}x)} \plus{} \frac {f(2)}x)$

And so $ x \equal{} f(\text{something})$
Q.E.D.

2) $ f(x)$ is an injective function
=====================
Let $ a > b > 0$ such that $ f(a) \equal{} f(b)$
Let $ T \equal{} b \minus{} a > 0$
Comparing $ P(x,a)$ and $ P(x,b)$, we get $ af(x \plus{} a) \equal{} bf(x \plus{} b)$ and so $ f(x) \equal{} \frac baf(x \plus{} T)$ $ \forall x > a$
And so $ f(x) \equal{} \left(\frac ba\right)^nf(x \plus{} nT)$ $ \forall x > a,n\in\mathbb N$

Let then $ y$ such that $ f(y) > 1$ (such $ y$ exists since $ f(x)$ is a surjection, according to 1) above)
Let $ n$ great enough to have $ y \plus{} nT \minus{} 1 > 0$

$ P(\frac {y \plus{} nT \minus{} 1}{f(y) \minus{} 1},y)$ $ \implies$ $ f(1 \plus{} \frac {yf(y) \plus{} (nT \minus{} 1)f(y)}{f(y) \minus{} 1}) \equal{} yf(\frac {y \plus{} nT \minus{} 1}{f(y) \minus{} 1} \plus{} y)$ which may be written :

$ f(\frac {yf(y) \plus{} nT \minus{} 1}{f(y) \minus{} 1} \plus{} nT) \equal{} yf(\frac {yf(y) \plus{} nT \minus{} 1}{f(y) \minus{} 1})$

and since $ f(\frac {yf(y) \plus{} nT \minus{} 1}{f(y) \minus{} 1} \plus{} nT) \equal{} \left(\frac ab\right)^nf(\frac {yf(y) \plus{} nT \minus{} 1}{f(y) \minus{} 1})$, we get $ y \equal{} \left(\frac ab\right)^n$ $ \forall n$, which is impossible
Q.E.D.

3) $ f(1) \equal{} 1$
===========
$ P(1,1)$ $ \implies$ $ f(1 \plus{} f(1)) \equal{} f(2)$ and so, since $ f(x)$ is injective, $ f(1) \equal{} 1$
Q.E.D.

4) The only solution is $ f(x) \equal{} \frac 1x$
==========================
$ P(1,x)$ $ \implies$ $ f(1 \plus{} f(x)) \equal{} xf(1 \plus{} x)$ and so $ f(1 \plus{} x) \equal{} \frac 1xf(1 \plus{} f(x))$

$ P(\frac x{f(\frac 1x)},\frac 1x)$ $ \implies$ $ f(1 \plus{} x) \equal{} \frac 1xf(\frac x{f(\frac 1x)} \plus{} \frac 1x)$

And so (comparing these two lines) : $ f(1 \plus{} f(x)) \equal{} f(\frac x{f(\frac 1x)} \plus{} \frac 1x)$

And so (using injectivity) : $ 1 \plus{} f(x) \equal{} \frac x{f(\frac 1x)} \plus{} \frac 1x$ and so $ f(\frac 1x) \equal{} \frac x{f(x) \plus{} 1 \minus{} \frac 1x}$

This implies (changing $ x\to\frac 1x$) : $ f(x) \equal{} \frac {\frac 1x}{f(\frac 1x) \plus{} 1 \minus{} x}$

And so $ f(x) \equal{} \frac {\frac 1x}{\frac x{f(x) \plus{} 1 \minus{} \frac 1x} \plus{} 1 \minus{} x}$

Which gives $ x^2f(x)^2 \minus{} 2xf(x) \plus{} 1 \equal{} 0$

And so $ \boxed{f(x) \equal{} \frac 1x}$, which indeed is a solution[/quote]
T=b-a<0 , and this will change the condition for n such that it be great enough to have y+nT-1>0 . Can you correct please ?  :blush:

How can we get the contradiction from  [quote=pco]$ y = \left(\frac ab\right)^n$   [/quote]   could you please explain this a bit more?

The contradiction is not obtained from $y=\left(\frac ab\right)^n$

It is obtained from $y=\left(\frac ab\right)^n$ $\boxed{\forall n}$
Since $a\ne b$, such equality can not be true $\forall n$. Is it really necessary to explain more ?

[quote=pco][quote="Leonhard Euler"]Find all functions $ \,f: {\mathbb{R^ \plus{} }}\rightarrow{\mathbb{R^ \plus{} }}\,$ such that
$ f(1 \plus{} xf(y)) \equal{} yf(x \plus{} y)$ $ \text{for all}\,x,y\in\mathbb{R^ \plus{} }.$[/quote]
Let $ P(x,y)$ be the assertion $ f(1 \plus{} xf(y)) \equal{} yf(x \plus{} y)$

1) $ f(x)$ is a surjective function
=====================
$ P(\frac 1{f(\frac {f(2)}x)},\frac {f(2)}x)$ $ \implies$ $ f(2) \equal{} \frac {f(2)}xf(\frac 1{f(\frac {f(2)}x)} \plus{} \frac {f(2)}x)$

And so $ x \equal{} f(\text{something})$
Q.E.D.

2) $ f(x)$ is an injective function
=====================
Let $ a > b > 0$ such that $ f(a) \equal{} f(b)$
Let $ T \equal{} b \minus{} a > 0$
Comparing $ P(x,a)$ and $ P(x,b)$, we get $ af(x \plus{} a) \equal{} bf(x \plus{} b)$ and so $ f(x) \equal{} \frac baf(x \plus{} T)$ $ \forall x > a$
And so $ f(x) \equal{} \left(\frac ba\right)^nf(x \plus{} nT)$ $ \forall x > a,n\in\mathbb N$

Let then $ y$ such that $ f(y) > 1$ (such $ y$ exists since $ f(x)$ is a surjection, according to 1) above)
Let $ n$ great enough to have $ y \plus{} nT \minus{} 1 > 0$

$ P(\frac {y \plus{} nT \minus{} 1}{f(y) \minus{} 1},y)$ $ \implies$ $ f(1 \plus{} \frac {yf(y) \plus{} (nT \minus{} 1)f(y)}{f(y) \minus{} 1}) \equal{} yf(\frac {y \plus{} nT \minus{} 1}{f(y) \minus{} 1} \plus{} y)$ which may be written :

$ f(\frac {yf(y) \plus{} nT \minus{} 1}{f(y) \minus{} 1} \plus{} nT) \equal{} yf(\frac {yf(y) \plus{} nT \minus{} 1}{f(y) \minus{} 1})$

and since $ f(\frac {yf(y) \plus{} nT \minus{} 1}{f(y) \minus{} 1} \plus{} nT) \equal{} \left(\frac ab\right)^nf(\frac {yf(y) \plus{} nT \minus{} 1}{f(y) \minus{} 1})$, we get $ y \equal{} \left(\frac ab\right)^n$ $ \forall n$, which is impossible
Q.E.D.

3) $ f(1) \equal{} 1$
===========
$ P(1,1)$ $ \implies$ $ f(1 \plus{} f(1)) \equal{} f(2)$ and so, since $ f(x)$ is injective, $ f(1) \equal{} 1$
Q.E.D.

4) The only solution is $ f(x) \equal{} \frac 1x$
==========================
$ P(1,x)$ $ \implies$ $ f(1 \plus{} f(x)) \equal{} xf(1 \plus{} x)$ and so $ f(1 \plus{} x) \equal{} \frac 1xf(1 \plus{} f(x))$

$ P(\frac x{f(\frac 1x)},\frac 1x)$ $ \implies$ $ f(1 \plus{} x) \equal{} \frac 1xf(\frac x{f(\frac 1x)} \plus{} \frac 1x)$

And so (comparing these two lines) : $ f(1 \plus{} f(x)) \equal{} f(\frac x{f(\frac 1x)} \plus{} \frac 1x)$

And so (using injectivity) : $ 1 \plus{} f(x) \equal{} \frac x{f(\frac 1x)} \plus{} \frac 1x$ and so $ f(\frac 1x) \equal{} \frac x{f(x) \plus{} 1 \minus{} \frac 1x}$

This implies (changing $ x\to\frac 1x$) : $ f(x) \equal{} \frac {\frac 1x}{f(\frac 1x) \plus{} 1 \minus{} x}$

And so $ f(x) \equal{} \frac {\frac 1x}{\frac x{f(x) \plus{} 1 \minus{} \frac 1x} \plus{} 1 \minus{} x}$

Which gives $ x^2f(x)^2 \minus{} 2xf(x) \plus{} 1 \equal{} 0$

And so $ \boxed{f(x) \equal{} \frac 1x}$, which indeed is a solution[/quote]
pco you have mistake in second part. at first T < 0. if you take f(y) > 1 than y<1 definitely. so you can not take such y that y+nT-1>0.

A needlessly long and complicated solution, but a solution nonetheless. As always, let $P(x,y)$ be the assertion $f(1+xf(y))=yf(x+y)$

Part 1: If $f(y) \neq 1$ and $y \neq 1$, then $y < 1 \iff f(y) > 1$
	[hide]
Assume otherwise. Then $\frac{y-1}{f(y)-1}$ is a positive real and

$P(\frac{y-1}{f(y)-1},y) \implies f(1+\frac{y-1}{f(y)-1} \cdot y) = yf(\frac{y-1}{f(y)-1} + y) \implies y = 1$

Therefore, if $f(y) \neq 1$, then $y > 1 \iff f(y) < 1$ and $y < 1 \iff f(y) > 1$. One important conclusion is that, if we don't move around the functions, then the $LHS$ in $P(x,y)$ is $\leq 1$, and so is the $RHS$.[/hide]

Part 2: $xf(x) \leq 1$

[hide]Assume $xf(x) = 1 + \epsilon$, where $\epsilon > 0$. Then

$P(x-\frac{x}{1+0.5\epsilon},\frac{x}{1+0.5\epsilon}) \implies f(1+(x-\frac{x}{1+0.5\epsilon})f(\frac{x}{1+0.5\epsilon})) = \frac{x}{1+0.5\epsilon}f(x-\frac{x}{1+0.5\epsilon}+\frac{x}{1+0.5\epsilon}) \implies something\ at\ most\ 1 = \frac{1}{1+0.5\epsilon} \cdot (1+\epsilon) = something\ greater\ than\ 1$, contradiction

Therefore, $xf(x) \leq 1$.[/hide]

Part 3: $f^{-1}(1)$ exists

[hide]$P(\frac{1}{f(f(2))},f(2)) \implies f(1+\frac{1}{f(f(2))} \cdot f(2)) = f(2)(\frac{1}{f(f(2))}+f(2)) \implies f(\frac{1}{f(f(2))} + f(2)) = 1 $[/hide]

Part 4: $f(1) = 1$

[hide]By part 2, we know that $f(1) \leq 1$. Assume $f(1) < 1$. Then

$P(\frac{1}{f(1)},1) \implies f(2) = f(\frac{1}{f(1)}+1)$

$P(\frac{1}{(f(1))^n},1) \implies f(1+\frac{1}{(f(1))^{n-1}})=f(\frac{1}{(f(1))^n}+1)$

Then by induction $f(2) = f(\frac{1}{(f(1))^n}+1)$ for all positive integers $n$.

Let $f(2) = c$. By part 2, we know that $c = \frac{1}{2} - \epsilon$ for some non-negative $\epsilon$ less than $\frac{1}{2}$. We also know by part 2 that $f(\frac{1}{(f(1))^n}+1) \leq \frac{(f(1))^n}{(f(1))^n+1} = 1-\frac{1}{(f(1))^n+1}$.

But $0 < f(1) < 1$ implies that $\lim_{n \to +\infty}(1-\frac{1}{(f(1))^n+1}) = 0$. Then we can pick $n$ sufficiently large such that $f(\frac{1}{f(1)^n}+1) \leq 1-\frac{1}{(f(1))^n+1} < \frac{1}{2} - \epsilon$, which contradicts the induction result.

Therefore, $f(1) = 1$[/hide]

Part 5: $f^{-1}(1) = \{1\}$

[hide]Let $f(\lambda)=1$. By part 2, we know that $\lambda \leq 1$. Assume there is a $\lambda$ value less than one such that $f(\lambda)=1$.

$P(1-\lambda,\lambda) \implies f(1+(1-\lambda)f(\lambda))=\lambda f(1-\lambda+\lambda) \implies f(2-\lambda) = \lambda$

$P(\frac{1-\lambda}{\lambda},2-\lambda) \implies f(1+\frac{1-\lambda}{\lambda}f(2-\lambda)) = (2-\lambda)f(\frac{1-\lambda}{\lambda}+2-\lambda) \implies \frac{\lambda}{2-\lambda}=f(1+\frac{1}{\lambda}-\lambda)$

$P(\frac{1}{\lambda}-\lambda,\lambda) \implies f(1+(\frac{1}{\lambda}-\lambda)f(\lambda))=\lambda f(\frac{1}{\lambda}-\lambda+\lambda) \implies \frac{\lambda}{2-\lambda} = \lambda f(\frac{1}{\lambda}) \implies f(\frac{1}{\lambda}) = \frac{1}{2-\lambda}$

But by part 2 we know that $f(\frac{1}{\lambda}) \leq \lambda$. Therefore $\lambda (2-\lambda) \geq 1$. But by part 2 as well we know that $(2-\lambda)f(2-\lambda)=(2-\lambda) \lambda \leq 1$. Therefore $\lambda (2-\lambda)=1$, which implies $\lambda=1$, contradiction.

Therefore, $f^{-1} = \{1\}$, and this implies that $f(x)=1 \iff x=1$. Furthermore, this strengthens part 1, because we may now say $y > 1 \iff f(y) < 1$ and $y < 1 \iff f(y) > 1$.[/hide]

Part 6: $f$ is strictly decreasing from $x > 1$

[hide]Assume $x > 1$. Then

$P(\frac{x-1}{f(f(x))},f(x)) \implies f(1+\frac{x-1}{f(f(x))}f(f(x))) = f(x)f(\frac{x-1}{f(f(x))}+f(x)) \implies f(\frac{x-1}{f(f(x))}+f(x)) = 1$

By part 5, we get that $\frac{x-1}{f(f(x))} = 1-f(x)$

Let $\epsilon > 0$. Then,

$P(\frac{x-1+\epsilon}{f(f(x))},f(x)) \implies f(1+\frac{x-1+\epsilon}{f(f(x))}f(f(x)))=f(x)f(1-f(x)+\frac{\epsilon}{f(f(x))}+f(x)) \implies f(x+\epsilon)=f(x)f(1+\frac{\epsilon}{f(f(x))})$

By the corollary of part 5, we get that $f(1+\frac{\epsilon}{f(f(x))}) < 1$. Therefore, $f(x+\epsilon)<f(x) \ \forall \epsilon > 0$.

Now assume there exists $x,y > 1$ where $x < y$ and $f(x) \leq f(y)$. Then, take $\epsilon=y-x$. Therefore, $f(x+\epsilon)=f(y)<f(x) \leq f(y)$, contradiction. Therefore, no such $x,y$ exists and $f$ is strictly decreasing from $x > 1$.[/hide]

Part 7: $xf(2x) \geq f(2)$

[hide]By part 2, $1+xf(x) \leq 2$. Then, by part 6, $f(1+xf(x)) \geq f(2)$. But

$P(x,x) \implies f(1+xf(x))=xf(2x)$

Therefore, $xf(2x) \geq f(2)$[/hide]

Part 8: $f(2) = \frac{1}{2}$

[hide]By the proof in part 2, $f(\frac{1}{f(f(2))} + f(2)) = 1$. By part 5, $\frac{1}{f(f(2))} + f(2) = 1$. But using part 2 we get $f(2) \leq \frac{1}{2}$. Putting this in the previous equation we get that $f(f(2)) \leq 2$.

However, using part 7 with $x = \frac{f(2)}{2}$ we get that $\frac{f(2)}{2} f(2\frac{f(2)}{2}) \geq f(2) \implies f(f(2)) \geq 2$. Therefore, $f(f(2))=2$. Plugging this in the previous equation yields $f(2)=\frac{1}{2}$.

A corollary is that we may strengthen part 7, because we now have that $xf(2x) \geq \frac{1}{2}$[/hide]

Part 9: $f(x) = \frac{1}{x}$

[hide]Using the corollary from part 8 and multiplying it by $2$ we get that $2xf(2x) \geq 1$. But, using part 2 we have that $2xf(2x) \leq 1$. Therefore, $2xf(2x) = 1$. We may reassign $2x \to z$ and we get $f(z) = \frac{1}{z}$, which clearly works, because

$f(1+xf(y))=f(1+\frac{x}{y})=\frac{y}{x+y}=yf(x+y)$[/hide]

The domain and co-domain of the function is only the set of positive real numbers .But you found that f(x)=1/x and the domain and  co-domain of this function is the set of real numbers which contradicts the condition.Will you please explain it?

Proving that $f(x)$ is of a certain form (and seeing that it verifies the given equation) does not imply anything about a well-defined domain and the codomain. If anything, the function has to be defined in the entire domain in order for it to be a function. While $f(x)=1/x$ works for positive and negative reals, proving $f(x)=1/x$ does not mean the domain has to include negative values.

[quote=ayan_mathematics_king][quote=pco][quote="Leonhard Euler"]Find all functions $ \,f: {\mathbb{R^ \plus{} }}\rightarrow{\mathbb{R^ \plus{} }}\,$ such that
$ f(1 \plus{} xf(y)) \equal{} yf(x \plus{} y)$ $ \text{for all}\,x,y\in\mathbb{R^ \plus{} }.$[/quote]
Let $ P(x,y)$ be the assertion $ f(1 \plus{} xf(y)) \equal{} yf(x \plus{} y)$

1) $ f(x)$ is a surjective function
=====================
$ P(\frac 1{f(\frac {f(2)}x)},\frac {f(2)}x)$ $ \implies$ $ f(2) \equal{} \frac {f(2)}xf(\frac 1{f(\frac {f(2)}x)} \plus{} \frac {f(2)}x)$

And so $ x \equal{} f(\text{something})$
Q.E.D.

2) $ f(x)$ is an injective function
=====================
Let $ a > b > 0$ such that $ f(a) \equal{} f(b)$
Let $ T \equal{} b \minus{} a > 0$
Comparing $ P(x,a)$ and $ P(x,b)$, we get $ af(x \plus{} a) \equal{} bf(x \plus{} b)$ and so $ f(x) \equal{} \frac baf(x \plus{} T)$ $ \forall x > a$
And so $ f(x) \equal{} \left(\frac ba\right)^nf(x \plus{} nT)$ $ \forall x > a,n\in\mathbb N$

Let then $ y$ such that $ f(y) > 1$ (such $ y$ exists since $ f(x)$ is a surjection, according to 1) above)
Let $ n$ great enough to have $ y \plus{} nT \minus{} 1 > 0$

$ P(\frac {y \plus{} nT \minus{} 1}{f(y) \minus{} 1},y)$ $ \implies$ $ f(1 \plus{} \frac {yf(y) \plus{} (nT \minus{} 1)f(y)}{f(y) \minus{} 1}) \equal{} yf(\frac {y \plus{} nT \minus{} 1}{f(y) \minus{} 1} \plus{} y)$ which may be written :

$ f(\frac {yf(y) \plus{} nT \minus{} 1}{f(y) \minus{} 1} \plus{} nT) \equal{} yf(\frac {yf(y) \plus{} nT \minus{} 1}{f(y) \minus{} 1})$

and since $ f(\frac {yf(y) \plus{} nT \minus{} 1}{f(y) \minus{} 1} \plus{} nT) \equal{} \left(\frac ab\right)^nf(\frac {yf(y) \plus{} nT \minus{} 1}{f(y) \minus{} 1})$, we get $ y \equal{} \left(\frac ab\right)^n$ $ \forall n$, which is impossible
Q.E.D.

3) $ f(1) \equal{} 1$
===========
$ P(1,1)$ $ \implies$ $ f(1 \plus{} f(1)) \equal{} f(2)$ and so, since $ f(x)$ is injective, $ f(1) \equal{} 1$
Q.E.D.

4) The only solution is $ f(x) \equal{} \frac 1x$
==========================
$ P(1,x)$ $ \implies$ $ f(1 \plus{} f(x)) \equal{} xf(1 \plus{} x)$ and so $ f(1 \plus{} x) \equal{} \frac 1xf(1 \plus{} f(x))$

$ P(\frac x{f(\frac 1x)},\frac 1x)$ $ \implies$ $ f(1 \plus{} x) \equal{} \frac 1xf(\frac x{f(\frac 1x)} \plus{} \frac 1x)$

And so (comparing these two lines) : $ f(1 \plus{} f(x)) \equal{} f(\frac x{f(\frac 1x)} \plus{} \frac 1x)$

And so (using injectivity) : $ 1 \plus{} f(x) \equal{} \frac x{f(\frac 1x)} \plus{} \frac 1x$ and so $ f(\frac 1x) \equal{} \frac x{f(x) \plus{} 1 \minus{} \frac 1x}$

This implies (changing $ x\to\frac 1x$) : $ f(x) \equal{} \frac {\frac 1x}{f(\frac 1x) \plus{} 1 \minus{} x}$

And so $ f(x) \equal{} \frac {\frac 1x}{\frac x{f(x) \plus{} 1 \minus{} \frac 1x} \plus{} 1 \minus{} x}$

Which gives $ x^2f(x)^2 \minus{} 2xf(x) \plus{} 1 \equal{} 0$

And so $ \boxed{f(x) \equal{} \frac 1x}$, which indeed is a solution[/quote]

In the part where you proved that $f(x)$ is injective can we do the following,
$af(x+a)=bf(x+b)$
This implies,$ \frac {f(x+a)}{f(x+b)}$= $ \frac ba  $.
Putting $x=0$ we get,
$ \frac {f(a)}{f(b)}$ = $ \frac ba $.
As,$f(a)=f(b)$ ,$a=b$
Hence,$f(x)$ is injective.
Can we do this to prove $f$ injective[/quote]

Problem condition does not think the same as you

[quote=Leonhard Euler]Find all functions $f: {\mathbb{R^\plus{}}}\to{\mathbb{R^\plus{}}}$ such that
\[ f(1\plus{}xf(y))\equal{}yf(x\plus{}y)\]
for all $x,y\in\mathbb{R^\plus{}}$.[/quote]

Firstly, we show $f$ is surjective. Plug $x=\tfrac{1}{f(y)}$; then $\tfrac{f(2)}{y} \in f(\mathbb{R}^{+})$ yielding the claim. Now we show $(y-1)(f(y)-1) \le 0$ for all $y \ne 1$. Let $t \ne 1$ and $(t-1)(f(t)-1)>0$. Plug $x=\tfrac{t-1}{f(t)-1}$ and $y=t$; then we obtain $t=1$ which is absurd. Next we show that $xf(x) \le 1$ for all $x>0$. Indeed; pick any $\varepsilon>0$ and let $x=\varepsilon, y=z-\varepsilon$; then $$yf(z)=f(1+\varepsilon f(z-\varepsilon)) \le 1 \implies f(z) \le \frac{1}{z-\varepsilon}$$ for all $z>0$; proving the assertion.

Now we prove that $f$ is injective. Suppose there exists $0<a<b$ with $f(a)=f(b)$. Plug $y=a, y=b$; subsequently, we obtain $af(x+a)=bf(x+b)$. Let $c=(b-a)$ and $r=\left(\tfrac{a}{b}\right)$ then for all $x>a$ we have $f(x+c)=rf(x)$. Now plug $y=k$ with $f(k)=2$. Then $$f(1+2x)=kf(x+k)$$ for all $x>a$. Observe that $$r^2kf(x+k)=r^2f(1+2x)=f(1+2(x+c))=kf(x+c+k)=rkf(x+k)$$ hence $r^2=r \implies r=1$; thus $a=b$ and the assertion is proved.

Finally, we show that $f(x)=\tfrac{1}{x}$ for all $x>0$; note that $f(1+f(1))=f(2) \implies f(1)=1$. Put $x=1$ so $f(1+f(y))=yf(1+y)$ for all $y>0$. Now observe that $$xf(y)yf(x+y)=xf(y)f(1+xf(y))=f(1+f(xf(y)))$$ for all $x,y>0$. Now plug $x=\tfrac{1}{yf(y)}$; so $f\left(\tfrac{1}{yf(y)}+y\right)=f\left(1+f\left(\tfrac{1}{y}\right)\right)$ so by injectivity $$y-\left(1-\tfrac{1}{yf(y)}\right)=f\left(\tfrac{1}{y}\right) \le y \implies \tfrac{1}{yf(y)} \le 1 \implies f(y) \ge \tfrac{1}{y}$$ proving the claim.

Clearly, $x \mapsto \tfrac{1}{x}$ is a solution and thus it is the only solution. $\blacksquare$

Is my solution correct?.
I get f(x) =1/x as solution.
LetP(x, y) =(0, y) be the assertion.
So, f(1+0.f(y))=yf(0+y)
    Or, f(1)=yf(y)
     Or, f(y)=f(1)/y
    Now f(1) will hold some constant value. Assume that f(1)=k(say)
Therefore our soln is in the form of f(y) =k/y
Now k=?
Simply put the soln in the functional eqn.
We get, f(1+xf(y))=yf(x+y)
              Or, k/(1+xf(y))=y.k/(x+y)
               Or, k/(1+xk/y))=yk/(x+y)
               Or, ky/(y+kx) =ky/(x+y)
                Or, y+kx=x+y
                Or, kx-x=0
                 Or, k=1        (x≠0)
So k=1.Now we get the solution f(y) =1/ y.
       We can check it by verification.
      f(1+xf(y))=1/(1+xf(y))
                       =1/(1+x/y)
                        =y/(x+y)
                         =y. 1/(x+y)
                         = yf(x+y)

You can only plug in positive numbers...

[quote=Leonhard Euler]Find all functions $f: {\mathbb{R^\plus{}}}\to{\mathbb{R^\plus{}}}$ such that
\[ f(1\plus{}xf(y))\equal{}yf(x\plus{}y)\]
for all $x,y\in\mathbb{R^\plus{}}$.[/quote]

This is a proper pre-monster.

Let $P(x, y)$ be the assertion.

[Claim = Red] $f$ is bijective [/color]

Suppose $y = \dfrac{f(k)}{t}$. Then we would like $1 + xf(y)$ to be equalling $k$ so that we can establish a form of surjectivity. Hence $x = \dfrac{k-1}{f(y)} = \dfrac{t(k-1)}{f(k)}$.

Hence, $P\left ( \dfrac{t(k-1)}{f(k)}, \dfrac{f(k)}{t} \right ) \implies t = f\left (\mathrm{Nyess}\right )$. Hence $f$ is surjective. We did this steps because $y$ is only term beneficial for surjectivity but equalling $f(1+xf(y))$ ans $f(x+y)$ doesn't make sense hence these substitution are not blind faith or logic but rather are directed.

Now suppose $f(a)=f(b)$ for $a \neq b$. Observe that comparing $P(k, a)$ and $P(k, b)$ again and again yields that $f(k) = \left ( \dfrac{b}{a}\right )^r f(x + r(b-a))$ holds for all naturals $r$.

Now, let $r$ be such that $z + r(a-b) > 1$, where $z \in \mathbb{R}$ such that $f(z) > 1$ and $z$ isn't a perfect power of $\dfrac{a}{b}$.

$P\left ( \dfrac{z+r(a-b)-1}{f(z)-1}, z\right ) \implies z = \left ( \dfrac{a}{b}\right )^r$, I'll not do the calculation for this step but it's straightforward.

How I did this one step is, you need injectivity and again $y$ term in $\mathrm{RHS}$ is still crucial, hence after playing for 5 min, came up with this construction.

Hence $a = b$ and $f$ is both injective and surjective, hence $f$ is bijective.

Now $f(1) = 1$ follows from $P(1, 1)$. $P(1, k) \implies f(1+f(k)) = kf(1+k)$. Now $\dfrac{1}{k}P\left (\dfrac{1}{kf(k)}, k\right ) \overset{injectivity}{\implies} f\left ( \dfrac{1}{k}\right ) \leq k$ and similarly $1- \dfrac{1}{kf(k)} \leq 1$ so $f(k) \leq \dfrac{1}{k}$, so $f(k) = \dfrac{1}{k}$ for all $k \in \mathbb{R}^+$.

Now that I see, my solution is kinda similar to pco and am09 Solution but that's fine I guess given the nature of this problem.


geometry
3D geometry
combinatorics solved
combinatorics

How can a 3 x 3 x 3 cube be put up into 20 cubes (not necessarily the same size) ?

Divide it naturally in 27 cubes and replace 8 cubes that form an 2x2x2 cube by the 2x2x2 cube.
We had 27 cubes, lost 8 and added 1, so we now have 20 cubes.
Kind of a first grade problem, not a challenging combinatorics problem.

Hi Mihai,

problems must not be hard but they should be attracting. The problem is from Bulgaria. It is was created for 11-aged students.


Prove that $ a^4\plus{}b^4\plus{}c^4\plus{}\sqrt2(a^3b\plus{}b^3c\plus{}c^3a) \ge 0$ ,for $ a,b,c \in R$

And a hard general problem: Find the best constant k

$ a^4\plus{}b^4\plus{}c^4\plus{}k(a^3b\plus{}b^3c\plus{}c^3a) \ge 0$ ,for $ a,b,c \in R$

[quote="dduclam"]Prove that $ a^4 \plus{} b^4 \plus{} c^4 \plus{} \sqrt2(a^3b \plus{} b^3c \plus{} c^3a) \ge 0$ ,for $ a,b,c \in R$

And a hard general problem: Find the best constant k

$ a^4 \plus{} b^4 \plus{} c^4 \plus{} k(a^3b \plus{} b^3c \plus{} c^3a) \ge 0$ ,for $ a,b,c \in R$[/quote]

I can only prove that $ a^4 \plus{} b^4 \plus{} c^4 \plus{} (a^3b \plus{} b^3c \plus{} c^3a) \geq 0$ for $ a,b,c \in R$
What I did is that I consider one fraction of the inequality,
$ ma^4 \plus{} na^3b \plus{} (1 \minus{} m)b^4 \geq 0$ true for all of $ a,b \in R$
Notice that this fraction inequality is homogenous, so we can assume $ b \equal{} 1$
Now, we consider the inequality
$ f(a) \equal{} ma^4 \plus{} na^3 \plus{} (1 \minus{} m) \geq 0$
$ f'(a) \equal{} 4ma^3 \plus{} 3na^2 \equal{} a^2 (4ma \plus{} 3n)$. Consider $ m \geq 0$
so $ f(a) \geq f(\frac { \minus{} 3n}{4m}) \equal{} \frac { \minus{} 27n^4 \plus{} 256m^3(1 \minus{} m)}{256m^3}$

Now we need to solve $ \minus{} 27n^4 \plus{} 256m^3(1 \minus{} m) \equal{} 0$ Notice that, in order to do so $ m$ must be least than 1
so $ 256m^3(1 \minus{} m) \equal{} 27n^4$ with condition $ 0 \leq m \leq 1$

But the maximum for $ m^3(1\minus{}m)$we can get is $ m \equal{} \frac {3}{4}$ or $ n \equal{} 1$ .


let $ x^2 \plus{} y^2 \plus{} 2gx \plus{} 2fy \plus{} c \equal{} 0$ be a circle. Find the locus of the feet of the perpendicular from the origin to a chord subtending right angles at the origin.Find also the midpoints of the chord.
(Can anyone give a geometric interpretation)

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244487[/url]


calculus
integration
limit
logarithms
calculus computations

Evaluate the integral between [1 and infinite] and also [0 and 1] of...
(ln x) . e^-x

$ \int_{1}^{\infty}\ln(x) e^{\minus{}x}\,dx \equal{}\frac{1}{e}\plus{}\int_{1}^{\infty}\frac{dx}{xe^{x}}\equal{}\frac{1}{e}\plus{}\lim_{c\to\infty}\left[\sum_{k \equal{} 1}^{\infty}(\minus{}1)^{k}\frac{x^{k}}{k\cdot k!}\plus{}\ln(x)\right]_{1}^{c}.$

EDIT: My Mathematica says, it's around 0.219384.

sorry, i couldn't really do it at this moment, all i know is that..... :D

$ \int_{0}^{\infty}\;\;\ln x\cdot e^{\minus{}x}\;\;\textbf dx\;\;\equal{}\;\;\boxed{\minus{}\,\gamma}$ :|

it appears that

$ \int_{1}^{\infty}\;\;\ln x\cdot e^{-x}\;\;\textbf dx\;\;=\;\;\boxed{\frac{\textbf G}{e}}$

and so, ${ \int_{0}^{1}\;\;\ln x\cdot e^{-x}\;\;\textbf dx\;\;=\;\;\boxed{-\gamma-\frac{\textbf G}{e}}}$

where $ \textbf G$ is called [b]gompertz[/b]'s constant :|

click here:
http://mathworld.wolfram.com/ExponentialIntegral.html


function
trigonometry
real analysis
real analysis unsolved

Let $ f$ be a continuous function such that $ f(2x^2\minus{}1)\equal{}2xf(x)$. Prove that $ f(x)\equiv{0}$ on $ [\minus{}1,1]$.

First $ f$ is odd, because for $ x\ne 0$ we have $ \minus{} f(x) \equal{} \frac {f(2x^2 \minus{} 1)}{ \minus{} x} \equal{} \frac {f(2( \minus{} x)^2 \minus{} 1)}{ \minus{} x} \equal{} f( \minus{} x)$.

Then, substitution of $ x \equal{} 0$ shows that $ f(1) \equal{} 0$.

When $ a \equal{} 2x^2 \minus{} 1$, then $ x \equal{} \sqrt {\frac {a \plus{} 1}{2}}$, and $ a < \sqrt {\frac {a \plus{} 1}{2}} < 1$ if $ 0 < a < 1$.
I will show that $ f(a) \equal{} 0$ for all $ 0 < a < 1$ (and thus for all $ a$).
Let $ 0 < a < 1$.
We may assume that $ f(a)\ge 0$ (otherwise, change $ f$ into $ \minus{} f$).
then $ f(a) \le \frac {f(a)}{\sqrt {\frac {a \plus{} 1}{2}}} \equal{} f(\sqrt {\frac {a \plus{} 1}{2}})$
Thus for $ 0 < a_1 < 1$, there is a $ a_2$ with $ a_1 < a_2 < 1$ with $ f(a_1) \le f(a_2)$.
We can now make a sequence $ a \equal{} a_1, a_2, \cdots$ with limit 1 and $ a \equal{} a_1 < a_2 < \cdots < 1$ with $ f(a) (\ge0) \equal{} f(a_1) \le f(a_2) \le \cdots$ with limit $ f(1) \equal{} 0$.
So, $ f(a) \equal{} 0$.

[hide="Hint"]
Hm.. how if we substitute $ x\equal{}cos(A)$ for any $ (A)$ in the set of real numbers, because for any $ x$ in $ [\minus{}1,1]$ we can be sure that there exist $ A$ such that $ x\equal{}cos(A)$, then :
$ f(cos(2A))\equal{}2.cos(A).f(cos(A))$
$ sin(A).f(cos(2A))\equal{}sin(2A).f(cos(A))$
$ \frac {f(cos(A))}{sin(A)}\equal{}g(A)$
Then ? I hope someone can solve the problem with my hint because I am nearly sure that my solution is also available.
Thanks for the solution before and sorry for my bad grammar.
[/hide]


inequalities
trigonometry
inequalities unsolved

Let ABC be an acute triangle.Prove the inequality
$ \sum\frac{cosAcosB}{cosC}\ge 2\sum (cosA)^2$

[quote="MrMinh"]Let ABC be an acute triangle.Prove the inequality
$ \sum\frac {cosAcosB}{cosC}\ge 2\sum (cosA)^2$[/quote]
I think, it's easy enough.
$ \sum_{cyc}\frac {\cos A\cos B}{\cos C}\ge 2\sum_{cyc} \cos^2A\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(b^2\plus{}c^2\minus{}a^2)(a^2\plus{}c^2\minus{}b^2)}{c^2(a^2\plus{}b^2\minus{}c^2)}\geq\sum_{cyc}\frac{(b^2\plus{}c^2\minus{}a^2)^2}{b^2c^2}.$
Let $ a^2\plus{}b^2\minus{}c^2\equal{}z,$ $ a^2\plus{}c^2\minus{}b^2\equal{}y$ and $ b^2\plus{}c^2\minus{}a^2\equal{}x.$ Then
$ \sum_{cyc}\frac{(b^2\plus{}c^2\minus{}a^2)(a^2\plus{}c^2\minus{}b^2)}{c^2(a^2\plus{}b^2\minus{}c^2)}\geq\sum_{cyc}\frac{(b^2\plus{}c^2\minus{}a^2)^2}{b^2c^2}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{xy}{z(x\plus{}y)}\geq\sum_{cyc}\frac{2x^2}{(x\plus{}y)(x\plus{}z)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^3y^3\minus{}x^3y^2z\minus{}x^3z^2y\plus{}x^2y^2z^2)\geq0,$ which is true by Schur.


I just took the AMC 12 B from 2002 that my teacher gave me, and I got 113.5.  How would this compare to an AMC 10 score?  I'm not sure if you can do that, since the AMC 10 has more geometry, but if you can, I'd like to know what an AMC 10 score would be, since I'm taking that and not the AMC 12.

Generally, I multiply my AMC 12 scores by 1.2 for AMC 10. For example, I got 110 on 2000 AMC 12. On the AMC 10 2001, I got 133ish.


Let $ ABCD$ be a convex quadrilateral. Let $ M,N$ be the midpoints of $ AB,AD$ respectively. The foot of perpendicular from $ M$ to $ CD$ is $ K$, the foot of perpendicular from $ N$ to $ BC$ is $ L$. Show that if $ AC,BD,MK,NL$ are concurrent, then $ KLMN$ is a cyclic quadrilateral.

May be this problem is not exactly! Let choose $ ABCD$ so that $ \widehat{BCD}\equal{}90^0$ and $ OA\equal{}OC$ with $ O\equal{}AC\cap BD$, by Geometer Skatchpad I see that it isn't true!

Hint: You first have to prove that AB is perpendicular to CD; MN then parallels to the side of the triangle that has the intersection of MK and NL as its orthocenter. The rest is straightforward.

Let $E$ the midpoint of $AC$,and the four lines concur at $H$. Since $NH\perp (NE\parallel BC),MH\ perp (NE\parallel BC)$, hence $H$ is the orthocenter of $MNE$. Since $H,L,C,K$ are concyclic, we have $\angle NMH=\angle AEN=\angle ACD=\angle KLN$ so $M,N,K,L$ are concyclic.

BTW this is also equivalent to the fact that the diagonals are perpendicular

[quote="nawaites"]Why angle of NMH equivalen with AEN???[/quote]

\[\begin{gathered}
  In\mathop {}\limits^{} tr.\mathop {}\limits^{} ADC\mathop {}\limits^{} : \hfill \\
  \left. \begin{gathered}
  {\rm A}{\rm N} = {\rm N}D \hfill \\
  AE = EC \hfill \\
\end{gathered}  \right\}\mathop {}\limits^{}  \Rightarrow \mathop {}\limits^{} NE||DC\mathop {}\limits^{}  \Rightarrow A\hat EN = E\hat CK \hfill \\
\end{gathered} \]


An arithmetic sequence of 10 consecutive positive integers has the property that the $n^{th}$ term in the sequence is a multiple of $11-n$.  The smallest possible first term is $k$.  Compute the remainder when $k$ is divided by 1000.

[hide="Solution"]
We have:
$11-1|k\Longrightarrow k\equiv 0\bmod{10}$
$11-2|k+1\Longrightarrow k\equiv 8\bmod{9}$
$11-3|k+2\Longrightarrow k\equiv 6\bmod{8}$
$11-4|k+3\Longrightarrow k\equiv 4\bmod{7}$
$11-5|k+4\Longrightarrow k\equiv 2\bmod{6}$
$11-6|k+5\Longrightarrow k\equiv 0\bmod{5}$
$11-7|k+6\Longrightarrow k\equiv 2\bmod{4}$
$11-8|k+7\Longrightarrow k\equiv 2\bmod{3}$
$11-9|k+8\Longrightarrow k\equiv 0\bmod{2}$

Combining the first, third, sixth, seventh, and ninth equations, we get $k\equiv 30\bmod{40}$.  Combining the second, fifth, and eigth terms, we get that $k\equiv 8\bmod{18}$.  If we combine these two, we see that $k\equiv 350\bmod{360}$.  Then, because $k\equiv 4\bmod{7}$, we have that $k\equiv 2510\bmod{2520}$.  Since $k$ is positive, the least possible value of $k$ is then $2510$, which is equal to $510\bmod{1000}$.

[/hide]

[quote="miyomiyo"]I might have misread your solution, but I believe you misread the question.  What I mean is that the first term is a multiple of 10 (is congruent to 0 mod 10), the second is a multiple of 9, third is a multiple of 8, and so on until the last is a multiple of 1.[/quote]

That is what he did - he did 10 consecutive numbers k through k+9 , each in its appropriate mod. If you check, 2510 does work.

Also, 2520, which is 10 more than 2510, is the smallest positive integer that divides 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10.

Oh sorry.  I did misread the solution.  My solution was just to notice that 10 more than the multiple of 10 was the same as 9 more than the multiple of 9 and 8 more than the multiple of 8, and so on.


superior algebra
superior algebra unsolved

suppose for all $ A$,$ B \le G$, with $ N \le A$ and $ N \le B$, then $ A \le B$ iff $ A/N \le B/N$

how do i prove this?

This is completely trivial... what don't you get about it? If there is $ a \in A$ not in $ B$, then $ a \plus{} N$ is not in $ B/N$ ...

can someone just show me how this is true?  In the textbook i am using, the book just says it's easy to see this is true.  But I don't see it.

then you have to recall the definitions and work with'em


The positive integers are written consecutively in the pattern
below. What integer will be the eighth entry in Row A?
\begin{tabular}[t]{cccccccccccccccc}
Row A &&&&4 &&&&&&10&\\
Row B &&&3 &&5 &&&&9 &&11 &&&&\cdots\\
Row C &&2 &&&&6 &&8 &&&&12 &&14\\
Row D &1 &&&&&&7 &&&&&&13
\end{tabular}

ther is a pattern

ROW A NUMBERS = n+6

ther r 7 intervals and 6*7=42

add 4, and the ans is [b]46[/b]


let x be a number such that x+(1/x)=4. find the value of x^3+(1/x^3)


$y=x-x^2+x^3-x^4+$......to infinity

Find x in terms of y.

[i]Mod Edit: Split and moved from MathCounts forum.[/i]

Or you could note it's a geometric sequence with first term x and ratio -x.
$y=\frac{x}{1+x}$
$y+yx=x$
$y=(1-y)x$
$x=\frac{y}{1-y}$

:o This has been shifted to intermediate!!This is one of the most common sums of those infinity type ones....really low level.

Here's my method-

[hide]$y=x-x(x-x^2+x^3......)$
$y=x-xy$ :D [/hide]

[quote="shadysaysurspammed"]:o This has been shifted to intermediate!!This is one of the most common sums of those infinity type ones....really low level.[/quote]

Let this be a learning experience then.  This is intermediate level and problems of similar difficulty are also intermediate.

[quote="peeta"][quote="seamusoboyle"]We seem to be ignoring the question of convergance??? for some y there could be no value of x that satisfies it? Convince me![/quote]

seamusoboyle is right...
one can only say $y=x-xy$ iff $y$ has a value, if $y\to \infty$ and e.g. $x>0$ it would mean $\infty = x-x \infty=-\infty$[/quote]

Well yes y does not have the exact value...but we are finding the closest possible,for the sake of getting a range(can't think of a better word)....Because the same arguement can be used in many questions involving infinity,like the basic ones.Who knows what infinity even is,we cannot assume any value but according to our assumption,the value of y is approximately obtained.

I hope i didn't post loads of crap in this post :blush:

The series only converges for $x \leq 1$

So find the value of $x$ that gives me $y=3$

It seems to me like it only converges for $-1<x<1$...


[b]Suppose a goat is tethered to a corner of a building which is in the shape of a regular $ n$-gon. THE length of a side of the building and length of the tether are each $ r$. Find the area of the region over which the goat can graze as a function of $ r$ and $ n$. [/b]

help with this problem would be appreciated!

[hide]
Angles are in degrees.
Angle the goat has between the two nearest sides, $ 180-\frac{360}{r}$.
The possible area he has is a circle $ A(r)= 2 \pi r^{2}$.
A simple proportion $ \frac{A(r)}{360}= \frac{A'(r)}{180-\frac{360}{r}}$.
$ \boxed{A'(r) = \frac{A(r) \cdot \left( 180-\frac{360}{r}\right) }{360}= \pi \left( r^{2}-2r \right) }$.
[/hide]


integration
calculus
calculus computations

Solve $ \frac {dy}{dx} \plus{} \frac {y}{x} \equal{} x^2y^3$.

Just for a sanity check, is the answer

[hide]
$ y \equal{} \frac{\pm 1}{\sqrt{Cx^2 \minus{} 2x^3}}$
[/hide]
?

hello, see here:
http://mathworld.wolfram.com/BernoulliDifferentialEquation.html
Sonnhard.

$ \displaystyle\frac {dy}{dx} \plus{} \frac {1}{x}y \equal{} x^2y^3$  This is Bernoulli's Differential Equation.
$ \displaystyle dy \plus{} \frac {1}{x}ydx \equal{} x^2y^3dx$, let $ v \equal{} y^{ \minus{} 2}\Rightarrow dv \equal{} \minus{} 2y^{ \minus{} 3}dy$
Thus $ \displaystyle \minus{} \frac {v^{ \minus{} \frac {3}{2}}}{2}dv \plus{} \frac {1}{x}v^{ \minus{} \frac {1}{2}}dx \equal{} x^2v^{ \minus{} \frac {3}{2}}dx$
Simplifying $ \displaystyle dv \minus{} \frac {2}{x}vdx \equal{} \minus{} 2x^2dx\cdots (1)$.
The integrating factor equals $ \displaystyle e^{ \minus{} 2\int\frac {1}{x}dx} \equal{} x^{ \minus{} 2}$.
Multiply both sides of (1) by $ x^{ \minus{} 2}$ to get $ \displaystyle \int d(x^{ \minus{} 2}v) \equal{} \int \minus{} 2 dx$.
We have $ \displaystyle v \equal{} \minus{} 2x^3 \plus{} Cx^2 \equal{} \frac {1}{y^2}$ or $ \displaystyle y \equal{} \pm\frac {1}{\sqrt {Cx^2 \minus{} 2x^3}}$


For $a,b,c$ positive reals such that $a^2+b^2+c^2=1$ prove the inequality

$\frac{a}{1-a}+\frac{b}{1-b}+\frac{1}{1-c} \geq \frac{3\sqrt{3}+3}{2}$

Manlio, for how many times do we need to discuss the same problem? Indeed, the idea is to minorize each term with a constant multiplied by $a^2, b^2, c^2$. This inequality is much much older, created by Klamkin. And we have already discussed it.


trigonometry
calculus
derivative
calculus computations

One of einstein's most amazing predictions was that light travelling from distant stars would bend around the sun on the way to earth. His calculations involved solving x in the equation:

sin x + b(1 + (cos(x))^2 + cos(x) = 0

where b is a very small constant.

Explain why the equation could have a solution for x which is near 0.

Dunno... I'm not Einstein, okay? And if $b$ is indeed small (like $0.01$), then the equation cannot have a root near $0$, where the $\cos x$ term dominates the rest.

Edit: the equation should probably read $\sin x+b(1+(\cos x)^{2}+\cos x) = 0$ in which case the statement is true. For $x=0$ the left side is positive, but when $\sin x<-3b$, it's negative.

where you put your parenthesis is important, for instance, which expression is being squared in your equation?

There are 4 open parentheses, but only 3 close parentheses
It's causing confusion

sorry guys it should be:

sin x + b(1 + (cos(x))^2 + cos(x) )= 0

thanks

[hide]
if $b$ is very small, then

$b(1+\cos^{2}{x}+\cos{x})$ is also very small, leaving

$\sin x = 0$ whose solution is $x = 0$

as as $b\to0$, the solution $x\to0$ as well
[/hide]

chickendude, your explanation is not quite complete: it also matters that $\sin x$ has a nonzero derivative at $x=0$.


In the east coast, that is. But then again, does anything else matter?

On the west coast, that's in a few hours.

Should this be in G&FF?

what I'm sure we can have a serious discussion about america being awesome

also I can't find topics in etc amid all the tug of wars and crap

God bless the United States of America!  :)

And now, for a recital of [i]The Defense of Fort McHenry[/i]...

And yeah, GFF really does seem like a Factory; repetitive threads get cranked out at regular intervals.

Then make GFF a better place. This thread does not fit in the Round Table.
I already said two years ago that someone should stop that "downfall" of GFF, but most people didn't care.

Happy Birthday to you,
Happy Birthday to you,
Happy Birthday dear America,
Happy Birthday to you!

Now you don't get your wish unless you blow out these 232 candles.

my birthday was yesterday

and g&ff is beast

like it or not

and that means we can spam, right?

SPAM SPAM

Woot America Day! 11 hrs 12 min and 45 seconds till its over in the centeral US :(

[quote="ZetaX"]I already said two years ago that someone should stop that "downfall" of GFF, but most people didn't care.[/quote]

Too bad I wasn't here two years ago...*goes on mass deletion spree in G&FF* - this forum [i]does[/i] give me a headache sometimes.

Happy 4th of July to everyone!  :)

[quote="Anaxerzia"][quote="pacman2812"]ZetaX is true... :(... but this topic is about America's day

Happy America Day, guys!!![/quote]

It's not called America Day, it's called Independence Day or Fourth of July Day :)[/quote]
or just plain 4th of July.

Along with this, I was surprised that I didn't see the movie Independence Day on TV tonight - it was usually a staple for American television. (Great movie, too).

365 days till independence day!!! hurray!!!!!

 :icecream:  :gathering:  :wow:


$ x \bullet y \equal{} xy \minus{} 10x \minus{}10y\plus{}110$. $ x \bullet y \equal{} y$ for all $ y$. Find $ x$.

We have that $ xy\minus{}10x\minus{}10y\plus{}110\equal{}y$

Let $ y\equal{}0$

Then $ 0\minus{}10x\minus{}0\plus{}110\equal{}0$

$ x\equal{}11$ is our only possibility. Seeing if it works:

$ 11y\minus{}110\minus{}10y\plus{}110\equal{}y$

Yay it works.

Or without plugging $ y\equal{}0$:

$ x(y\minus{}10)\equal{}11y\minus{}110\equal{}11(y\minus{}10)$

As $ y\equal{}\neq 10$ for at least 1 y, we get $ x\equal{}11$


To whom it may be concerned.

Could you please open Japanese Community?

kunny

That's a good idea, kunny :)

The only problem is that we have very few active Japanese AoPSers... Maybe we should find some people who want to join the AoPS. As Valentin Vornicu mentioned in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=269045#p269045]this post[/url], once we have at least 3-4 Japanese AoPSers, we can have a Japanese forum.

Does anyone else want to join the Japanese community??

Don't worry frt & kunny.
 I don't have political convictions,
 but, "new nationalist party"is just only 5.
 :D
 I completely agree with you.
 good night!

 00section shin kazama

Ok, now we have 3 people, but it seems that there's no more...

Is it possible to have a Japanese forum if we have only 3 people?

I am not sure, but I remember that we needed way more than 3 Dutch-speaking people to open a Dutch-speaking community.

I have checked the Members List.

There are 25 people with Japanese flag.

Here are members.

13        Tare
29        kunny
208      frt
359      litoxe2718281828
544      vuhung
1318    beaucoa
1965    ChaoSpectre
2632    shin kazama
2728    gypsy
3118    asnsakidrops
3251    cecman
3290    haseo
3436    toshihiro shimizu
3610    Ai
6628    clea
7011    takuya
7439    hm
7476    Takeshi
8033    snow
8086    Norifumi Seki
8707    ronosh
9076    vuongtandat
10024  b16a
10154  tetsuyaAnDO
11250  TAKEKEN

[quote="frt"]Wow, great work kunny :lol:[/quote]Ok, for this work you deserve a Japanese community :D

Thank you very much, Valentin Vornicu and for cooperating with me, especially frt. :)

kunny


geometry perimeter conics hyperbola symmetry ellipse

Given a triangle ABC, how would you use only a compass and straight edge to find a point P such that triangles ABP, ACP and BCP have equal perimeters?

This is an extremely tough problem that is harder than Pre Olympiad.

It is the intersection of 3 half-hyperbolas (hyperbolae?)...
Do those hyperbolae always intersect in a unique point though?

I think it is a classic problem, but I can find no solution :?

This is a type of problem that you need to do some exploring and conjecture. Try a few simple cases that you can work out algebraically, like an equilateral triangle, right triangles, etc. I suppose that it is obvious by symmetry that the centroid of an equilateral triangle is the desired point, see what you can find with a right triangle. Note I have not worked out anything on the problem besides thinking about it, however, this is how i would start working. Also could you explain the hyperbola idea. It seems to me that it would be an intersection of ellipses because perimeter would be a constant sum, not difference, as a hyperbola is.

also, has anyone figured out yet why there exist so many different points in a triangle with special properties?


probability
calculus
integration
symmetry
probability and stats

Consider random walk on $ \mathbb{Z}^2$ started at the origin.  What is the probability that you hit $ (0,1)$ before returning to the origin?

The answer is $ 1/2$.  It's equivalent to evaluating $ 1/(2\cdot \alpha(0,1))$, where $ \alpha$ is the potential associated with simple random walk on $ \mathbb{Z}^2$, and this can be evaluated by computing an integral or by symmetry considerations.

I don't know an elementary solution and it's not clear that there is one -- for example, the probability of hitting $ (2,1)$ before hitting $ (0,0)$ again is $ 1/(2\cdot(8/\pi\minus{}1))$, so ...