I tried saving the fancyhdr package as the AoPSWiki instructions say, but when I selected the directory in the "local respiratory" section of the MiKTeK 2.5 update, it said "Directory does not appear to be a local package respitory." What went wrong? I copied fancyhdr.sty into a fancyhdr dir in MiKTeK/tex/latex as instructed. I made sure it was fancyhdr.sty and not fancyhdr.sty.txt.

[quote="Temperal"]I tried saving the fancyhdr package as the AoPSWiki instructions say, but when I selected the directory in the "local respiratory" section of the MiKTeK 2.5 update, it said "Directory does not appear to be a local package respitory." What went wrong?[/quote]The local repository is where MiKTeX stores its installation files, somewhere in C:\Documents and Settings. That directory contains files for installation by MiKTeX [i]not[/i] for you to use.
There are 2 things you can do:
1. Go to MikTeX 2.5, Browse Packages, wait a while while it collects information right-click on fancyhdr and choose install
or
2. Having put fancyhdr in the right place you know have to tell MiKTeX that it's there. Go to MiKTeX 2.5, Settings and press Refresh FNDB

Rather than searching for files every time you compile a document LaTeX uses a database which makes it much quicker. So if you change something you must refresh the database. Hence method 2 and is done automatically in method 1.

fancyhdr is easy as it only requires one file but for other more complicated cases it's best to use method 1.

Thanks, but unfortunately when I try the second method, a window alert pops up saying
[quote]MiKTeX Problem Report
Message: The remote package repository may be corrupted.
Data: http://ctan.math.utah.edu/tex-archive/systems/win32/miktex/tm/packages/
Source: mpm.cpp
Line: 2089
MiKTeX: 2.5
OS: Microsoft Windows XP Professional Service Pack 2 (Build 2600)
SystemAdmin: yes
PowerUser: no
SharedSetup: unknown
BinDir: C:\Program Files\MiKTeX 2.5\miktex\bin
Root0: C:\Documents and Settings\Peter\Application Data\MiKTeX\2.5
Root1: C:\Documents and Settings\Peter\Local Settings\Application Data\MiKTeX\2.5
Root2: C:\Program Files\MiKTeX 2.5
Install: C:\Program Files\MiKTeX 2.5
UserConfig: C:\Documents and Settings\Peter\Application Data\MiKTeX\2.5
UserData: C:\Documents and Settings\Peter\Local Settings\Application Data\MiKTeX\2.5
[/quote]

In fact, it does this no matter what package I try to install in this method.

According to the [url=http://sourceforge.net/forum/forum.php?forum_id=33790]MiKTeX forum[/url] the repositories are unavailable at the moment so you'll have to wait until they have been fixed, or else download the complete installation (assuming that's not affected as well).


Hello! Is anyone interested in doing some interesting QM problems?
Or anyone who knows a nice problem and wants to share?

Well ok, thats a good idea!  :D
So let me be the first one to post a problem. Hmm let me think, so here is an easy one, i just figured out:

Consider a Harmonic oscillator in the following state:

\[ \psi_n\equal{} N ( | n \rangle\plus{}| n\plus{}2\rangle\plus{}| n\plus{}4\rangle \plus{} \dots \plus{} | n\plus{}k\rangle)\]

($ k$ is a finite, even number)

(a) determine the normalization constant $ N$

(b) find the standard deviation of the operators $ x$ and $ p$.

(c) Check if Heisenberg's uncertainty principle is satisfied.

just ask if you have any questions  :D


calculus
derivative
integration
geometry
AoPS Books

:?

I am new to the AoPS website, and have been glancing at previous AIME's, and other U.S. contests involving proofs for the past month or so. I was wondering what type of background one needs in math to answer such challenging proofs, one look at them for me and I have no clue where to go! Could anyone possibly list what type of classes would be needed as pre-requisites, and if their are any tips that can be used for going through them?

Could anyone give me a bit of a hand? To give a bit of a background on my education, this year I took and successfully passed Pre-Calculus and some of Calculus 1, meaning all of limits and derivatives, but only basic integration (our school has an odd curriculum where we learn the rest of Calc I next year and all of Calc 2).

I would suggest you solve lots of problems and buy some books that train your problem solving skills, such as AoPS books since you're just starting (I'm assuming) (I haven't read them though, so I don't know how good they are).  After that, you may start reading Problem Solving Strategies, Geometry Revisited, Problem Solving Through Problems, Mathematical Olympiad Challenges, et cetera. Basically, solve lots of problems and understand new techniques.

Masoud Zargar

Your advice will not go unheeeded  :D.
Thanks much, and it isn't too early for a middle schooler to start learning them, right?


geometry
trapezoid
algorithm
pigeonhole principle
HCSSiM

i presume we are allowed to discuss answers...

i might have miscounted some of the easy ones...

the early geometry was quite easy...like the trapezoid one?

the last two problems were nice; it took me forever to realize you just put a knight on all the white squares
the last one was a cool area argument...i showed that the triangle, if it were greater than 63/(7/2) in area, then it would have to jet out on the cut out edge, but that would happen at both

for the first one...i was like wtf a pell equation...like while there is an algorithm for those problems, it is a pain to do...and quite difficult...like i had solved all the other problems in like a day...then i realized it factored...i was like wow that was easier than i thought

They were good problems, but I suck at geometry. So I got the trapezoid one, because it was super easy, but not the other two :rotfl: . And I did a pretty weak job on number 10, but I'm pretty sure I got the rest.

I found test C really easy.  Also 9 is quite well-known.  I proved it with Pigeonhole, but after the test I saw that another way to do it is to replace each square of the chessboard with a vertex and draw a segment between any two squares from one of which a knight can reach the other.  This graph is hamiltonian (ie a knight's tour exists).  So any choice of 33 or more squares must have two adjacent squares in the tour.  Hence blah.

This argument is nice, except how do we show the graph is hamiltonian other than by actually tracing out a full knight's tour?

Well, I applied...and between me telling my mom the address and my mom mailing it, "Monaco Drive" became "Monroe Drive".  So I got it back today, with "Returned to Sender: No such number" stamped at the top.  Hmm.  I got into HCSSiM, though, and had already decided to go there anyway.  So it doesn't matter.  And we got the application fee back.

Though I'd like to know if I'd have gotten in...


$ a,b,k\in R^{+}.$
$ g(a,b)=\sqrt{\frac{a}{a+kb}}+\sqrt{\frac{b}{b+ka}}$
Find $ g(a,b)_{min}$ and $ g(a,b)_{max}$

I think that we can setting $ x=\frac{b}{a}$ and we need find min, max of $ g(x)=\sqrt{\frac{1}{1+kx}}+\sqrt{\frac{x}{x+k}}$. Use real analysys now!  :maybe:

A little approach:
$ \text{Claim.}$ If $ k\geq 3$, $ g\geq \frac{2}{\sqrt{k+1}}$
Let $ a=x^{2},b=y^{2}$,assume $ x+y=1$ we find that $ f(t)=\frac{1}{\sqrt{t}}$is convex,
thus by Jensen we have
$ xf(x^{2}+y^{2}k)+yf(y^{2}+x^{2}k)\geq f(x^{3}+xy^{2}k+x^{2}yk+y^{3})\geq \frac{2}{\sqrt{k+1}}$,
which is equivlant to
$ 1+(k-3)xy\leq \frac{k+1}{4}$, it is trival by AM-GM

$ \text{Claim2.}$ If $ k\leq 1, g\geq 1$
In deed, $ g\geq\frac{a}{a+bk}+\frac{b}{b+ak}\geq\frac{a}{a+b}+\frac{b}{b+a}$,
we can set $ a=0$ then the result.


If we draw all diagonals of a convex n-gon such that no three diagonals pass through a point, into how many parts T_n is the    n-gon divided

Similar problem solved here http://mathworld.wolfram.com/RegularPolygonDivisionbyDiagonals.html

They don't have the stipulation that no three diagonals pass through the same point though.

A similar question was discussed (but not solved) [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=350905545&t=151093]here[/url].  I strongly suspect this question has appeared on the forums before.


Let $a_1, b_1, c_1$ be the sides of triangle $A_1B_1C_1$ and $a_2,b_2,c_2$ be the sides of triangle $A_2B_2C_2$. Denote $a = \sqrt{\frac{a_1^2+a_2^2}{2}}, b = \sqrt{\frac{b_1^2+b_2^2}{2}}, c = \sqrt{\frac{c_1^2+c_2^2}{2}}$.

a) Show that $a, b, c$ are the sides of some triangle $ABC$.

b) Suppose $S, S_1, S_2$ be respectively the areas of triangles $ABC, A_1B_1C_1, A_2B_2C_2$. Show that $S \geq \frac{S_1+S_2}{2}$.

part (a) follows from Minkowski inequality.

part (b) follows from Pedoe's inequality. Here is the sketch of the proof:
Heron's area formula gives

$16S^2=(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$

A straight calculation shows that

(*) $16S^2 = \frac{1}{4} (16S_1^2 + 16S_2^2 + 2(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2) - 4(a_1^2a_2^2+b_1^2b_2^2+c_1^2c_2^2))$

But by Pedoe's inequality, we have:

(**) $(a_1^2+b_1^2+c_1^2)(a_2^2+b_2^2+c_2^2) - 2(a_1^2a_2^2+b_1^2b_2^2+c_1^2c_2^2) \geq 16S_1S_2$

Combining (*) and (**) gives $S \geq \frac{S_1+S_2}{2}$.


quadratics
modular arithmetic
algebra
polynomial
number theory proposed
number theory

It's surely well known:
Let $a,b,c,d,e$ be some integers, not all having the same sign. Prove there are infinitely many integer solutions $v,w,x,y,z$ to the equation \[a v^2+b w^2+c x^2+d y^2+e z^2=0\]

Additional statement: it is wrong when using only four numbers/variables instead, but true again if they are pairwise coprime.

If we aim to apply the Minkowski-Hasse Theorem, we should prove that every such quadratic form with $5$ indeterminates represents $0$ in every $\mathbb Q_p$ (i.e. for every prime $p$).

Every quadratic form $f(x_1,\ldots,x_n)=\sum a_ix_i^2$ over $\mathbb Q_p$ can be written as $f_1+pf_2\ (*)$, where $f_1(x_1,\ldots,x_k)=\sum a_ix_i^2,f_2(x_{k+1},\ldots,x_n)=\sum b_jx_j^2$ are quadratic forms with coefficients which are $p$-adic integers, none of which is divisible by $p$ (we may eventually have to relabel $x_1,\ldots,x_n$). Now, it's obvious that $f=f_1+pf_2$ represents $0$ (i.e. $f(x_1,\ldots,x_n)=0$ has a solution $(x_1,\ldots,x_n)$ in $p$-adic integers, not all of which are divisible by $p$) if at least one of $f_1,f_2$ represents $0$: if, say $f_1$ does, then take $x_1,\ldots,x_k$ to be the solution to $f_1$, and $x_{k+1},\ldots,x_n$ to be $0$.

For $p\ne 2$ we can prove that $a_1x_1^2+a_2x_2^2+a_3x_3^2=0\ (*)$ has a non-trivial solution in $p$-adic integers whenever $a_1,a_2,a_3$ are not divisible by $p$: it's well-known that $a_1x_1^2+a_2x_2^2+a_3x_3^2\equiv 0\pmod p$ has a non-trivial solution (here, the $a_i$'s are integers which are congruent with the original $a_i$'s modulo $p$), and if, say, $p\not |\ x_1$, then we can regard $(*)$ as a polynomial in $x_1$ and apply Hensel's Lemma to conclude that $(*)$ has a solution in $p$-adic integers. This concludes it for $p\ne 2$, since whenever we write $f=f_1+pf_2$, at least one of $f_1,f_2$ will have at least three unknowns.

For $p=2$ it has to be done separately, and it's atcually a bit boring, so I'll leave it out :). Three unknowns don't suffice there, so we have to start all over again with the five unknowns and play around with them :).


Square $ EFGH$ is inside the square $ ABCD$ so that each side of $ EFGH$ can be extended to pass through a vertex of $ ABCD$. Square $ ABCD$ has side length $ \sqrt {50}$ and $ BE \equal{} 1$. What is the area of the inner square $ EFGH$?
[asy]unitsize(4cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));

pair D=(0,0), C=(1,0), B=(1,1), A=(0,1);
pair F=intersectionpoints(Circle(D,2/sqrt(5)),Circle(A,1))[0];
pair G=foot(A,D,F), H=foot(B,A,G), E=foot(C,B,H);

draw(A--B--C--D--cycle);
draw(D--F);
draw(C--E);
draw(B--H);
draw(A--G);

label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$E$",E,NNW);
label("$F$",F,ENE);
label("$G$",G,SSE);
label("$H$",H,WSW);[/asy]$ \textbf{(A)}\ 25\qquad \textbf{(B)}\ 32\qquad \textbf{(C)}\ 36\qquad \textbf{(D)}\ 40\qquad \textbf{(E)}\ 42$

[hide="Solution"][b]C[/b]?
From the Pythagorean theorym, $HB^2=50-1$ so $HB=7$. $BE=1$ so $HE=6$ and the area of the inner square is $36$.[/hide]

[hide="Answer"]Labeling each side of the inner square as $x$, then by Pythagorean Theorem we have $50=1+(x+1)^2\Rightarrow x+1=7\Rightarrow x=6$, so $x^2=36\Rightarrow \boxed{C}$.[/hide]

IT took me a while to figure out the picture, but I finally got it....

[hide]There creates four right triangle with a leg of 1 and a hypotenuse of $\sqrt{50}$. THus each triangle's area is 3.5, and the four triangle's areas would be 14. Since the area of square ABCD is 50, the area of EFGH would be 50-14=36. [/hide]

Would someone please post a picture for this? I've spent a good 20 minutes and still don't get it! Thanks.

By the Pythagorean Theorem (and symmetry), $ BH \equal{} AG \equal{} DF \equal{} CE \equal{} \sqrt {50 \minus{} 1} \equal{} 7$, so

$ [EFGH] \equal{} 50 \minus{} 4\cdot\frac12\cdot1\cdot7 \equal{} 50 \minus{} 14 \equal{} \boxed{\textbf{(C)}\ 36}$.


A student added up the first 100 positive integers. However, instead of getting 5050 like he was supposed to, he accidentally switched the two digits of one of the integers and got 5068 instead. What is the largest possible integer whose digits could have been switched to get this erroneous result?

You need the difference in the two numbers to be 18.  That is $ \left|10x\plus{}y\minus{}10y\minus{}x\right|\equal{}\left|9x\minus{}9y\right|\equal{}18\Rightarrow\left|x\minus{}y\right|\equal{}2$.  Choose $ x\equal{}9$ and $ y\equal{}7$, so that the two numbers are 79 and 97.  So the largest number can be 79.


Let $ f: [0,1] \times [0,1] \rightarrow \mathbb{R}$ be separately continuous in its variables and taking 0 on $ \mathbb{Q} \cap [0,1] \times \mathbb{Q} \cap [0,1]$. Then f is identically 0.

Edited after fedja's post.

[quote="akech"] taking 0 on $ \mathbb{Q} \cap [0,1]$. [/quote]
You almost certainly meant [i]taking 0 on $ (\mathbb{Q} \cap [0,1])^2$[/i] ;)

Hello akech,

here is some hint :
[hide]
Fix $ x \in [0,1]$  fix $ q \in [0,1]\cap \mathbb{Q}$ and take $ q_n$ a sequence of rational ,approximating $ x$.
Using the hypothesis you will get $ 0\equal{}f(q_n,q) \longrightarrow f(x,q)$ ...
[/hide]


Factor the following algebraic expression where $a,b,c$ are real numbers

$(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$

Time required to find the solution : 1 minute :lol:

Set $b+c-a=x,c+a-b=y,a+b-c=z$,we can rewrite our equation to $(x+y+z)^3-x^3-y^3-z^3=3(x+y)(y+z)(z+x)=24abc$.


Hi i have a question: Whenever I try to post a reply it says
Invalid Session. Please resubmit form
Ive resubmitted about 15 times and nothing happened. I also tried logging out, rebooting, etc....

Here's my solution (I think) to AIME 15's AMC 10 problem:

a-7b+8c=4, 8a+4b-c=7
Square both eqq's:
a(a-7b+8c) -7b(a-7b+8c) +8c(a-7b+8c)=16,     8a(8a +4b -c) +4b(8a +4b -c) -c(8a +4b -c)=49
Simplifying the first eqq:
a^2 -7ab +8ac -7ab +49b^2 -56bc +8ac -56bc +64c^2=16
a^2 + 49b^2 +64 c^2 -14ab +16ac -112bc=16
Simplifying 2nd eqq:
64a^2 +32ab -8ac +32ab +16b^2 -4bc -8ac -4bc +c^2 =49
64a^2 +16b^2 +c^2 +64ab -16ac -8bc=49
Adding both eqq's:
65a^2 +65b^2 +65c^2 +50ab -120bc = 65
65a^2 -65b^2 +65c^2 = 65-130b^2-50ab+120bc
65(a^2 -b^2 +c^2)= 65 -130b^2 -50ab +120bc
a^2-b^2+c^2= 1-2b^2 -(10ab+24bc)/13
Since our answer is an integer, (10ab+24bc)/13 is also an integer
Thus 10ab +24bc is a multiple of 13
Thus 5ab +12bc is a multiple of 13
We can write this as:
5ab +12bc = 13n
We can recognize this as the Pythagorean triplet 5-12-13
Thus ab=bc=n
a=c
Going back to the first 2 eqq's:
a-7b +8c=4 and 8a +4b-c=7
Multiply the first eqq by 4 and the second by 7:
4a-28b+32c=16 and 56a+28b-7c=49
Add the eqq's together:
60a +25c=65
substitute c in for a:
85c=65
c=(65/85)=(13/17)=a
Substituting this in the first eqq:
(13/17) -7b +(8*13/17)=4
-7b=(68-13-104)/17
-7b=-49/17
b=7/17
Thus a^2 -b^2 +c^2=
(13/17)^2 -(7/17)^2 +(13/17)^2
[size=150]=B: 1[/size]

I get that same problem but if I click and re-click submit, it eventually goes through.  If that happens, be sure to copy your post so that if it logs you out, you don't need to re-write an entire post.  It's happened to me before and it's not fun.


Simplify $ \sin^2 10\plus{}\sin^2 20\plus{}\dots \plus{} \sin^2 90$. Everything is in degrees.

What's the no-brainer step that I'm missing? :maybe:

[hide="This one"] $ \sin \theta \equal{} \cos (90 \minus{} \theta)$ [/hide]

what about same problem with tangent instead :wink:

Well, notice that $ \tan 90$ is infinite...

oh yes it is then how about without 90 degree   :D

I couldn't think of a better way to do this, but:

[hide="Hint"]Let $ x \equal{} \tan 10 ^{\circ}$ and use the tangent addition/subtraction formula on the various terms.[/hide]
You should get:

[hide="Answer"]
$ \frac {136}{6}$[/hide]

[hide="Answer"]As [b]t0rajir0u[/b] mentioned, $ \sin \; \theta \equal{} \cos(90 \minus{} \theta)$, so $ \sin^{2} \theta \equal{} \cos^{2}(90 \minus{} \theta)$, which makes our problem a whole lot easier:

$ \sin^{2} 10\plus{}\sin^{2} 20\plus{}\dots \plus{} \sin^{2} 90 \equal{}\\
\sin^{2} 10 \plus{} \sin^{2} 20 \plus{} \sin^{2} 30 \plus{} \sin^{2} 40 \plus{} \cos^{2} (90 \minus{} 50) \plus{} \cos^{2} (90 \minus{} 60) \plus{} \cos^{2} (90 \minus{} 70) \plus{} \cos^{2} (90 \minus{} 80) \plus{} \sin^{2} 90$

Using the identity $ \sin^{2} \theta \plus{} \cos^{2} \theta \equal{} 1$, we can simplify the addition to

$ 1 \plus{} 1 \plus{} 1 \plus{} 1 \plus{} \sin^{2} 90 \equal{} \boxed {5}$[/hide]

if  $ x \equal{} \dfrac{1}{\sin^2{10}\cos^2{10}} \plus{} \dfrac{1}{\sin^2{20}\cos^2{20}} \plus{} \dfrac{1}{\sin^2{30}\cos^2{30}} \plus{} \dfrac{1}{\sin^2{40}\cos^2{40}} \minus{} 8$ than multiplying both side with 1/4 we get $ x/4 \equal{} \dfrac{1}{sin^2{20}} \plus{} \dfrac{1}{sin^2{40}} \plus{} \dfrac{1}{sin^2{60}} \plus{} \dfrac{1}{sin^2{80}} \minus{} 2$ and dont know what to do now!!
by the way congratulations 1=2 :)

It's the same sum when they were tangents.

I got five also. Just use the cofunction identity to establish four instances of $ \sin^2{\alpha}\plus{}\cos^2{\alpha}\equal{}1$. Then you just add on $ \sin^2{90}\equal{}1$ to get $ 5$.

yes we get 4 cases of sin(x) ^2 + cos(x)^2 and adding on another 1 we get [b]5[/b]
with tangents- not so easy

[hide]My solution for the case when the sum was tangents:
Using the identity: tg^2(a)+tg^2(60-a)+tg^2(60+a)=9.tg^2(3a)+6
To prove this nice identity above, we just need to expand the sum of the LHS and simplifying it, we will easily get the RHS, as necessary.
Now plugging in a=10,20, respectively, we will have: tg^2(10)+tg^2(50)+tg^2(70)=9.tg^2(30)+6(1)
                                                                       tg^2(20)+tg^2(80)+tg^2(40)=9tg^2(60)+6(2)
Adding (1)to (2) plus tg^2(30) and tg^2(60) we will get the result as: 10tg^2(30)+10tg^2(60)+12=10/3+30+12=136/3, which is slightly different from The Zuton Force's result above :roll: [/hide]


Find the number of integers which lie between between $1$ and $10^6$ and which have the sum of the digits is equal to $12$.

[quote="amcavoy"][hide]The number must have $2$, $3$, $4$, $5$, or $6$ digits.  Therefore you need solutions to $\sum_{n=1}^{k}x_{n}=12$, where $k\in\left\{2,3,4,5,6\right\}$.  This means there are

\[ \sum_{n=1}^{5}\binom{12+n}{n}=\boxed{8567}  \]

numbers.[/hide][/quote]

I think there's a problem with that... you've correctly counted the number of ways 12 can be split into 2,3,4,5, or 6 numbers, but some of those numbers may be larger than individual digits.  For instance, in your count of two digit numbers that sum to 12 you have also counted 0 + 12, 1 + 11, 2 + 10, 10 + 2, 11 + 1, 12 + 0 in addition to the proper set 39, 48, 57, 66, 75, 84, 93.  I'm not sure how to fix that though.... :?

Oh yeah good call.  I don't really know how to fix that either.  Any ideas?

I'm not sure if there are any "clean" method. but you can definitely get the answer by considering the combination for 2 digits, 3 digits.......till 6 digits numbers, e.g. $7$ integers among 2-digit numbers, $84$ integers among 3-digit integers. i'm a bit lazy to work on the rest. but with a pencil and paper,a little patience , it can be done. then add up all the answers.it'll be a little tedious, though. hope someone have a better method  :P


geometry
geometric transformation
rotation
analytic geometry
graphing lines
slope

How could one go about proving that the rotation of $ (x,y)$ 90 degrees counter-clockwise about the origin is $ (\minus{}y,x)$?

[hide="Proof"]
Let $ (x,y)$ be on the complex plane so that $ z\equal{}x\plus{}yi$.

It is known that if Point $ C(c)$ is the rotation of $ B(b)$ with respect to $ A(a)$ by the angle $ \theta$, we have

$ c\equal{}a\plus{}(b\minus{}a)\text{cis}\theta$.

In this case, since $ A$ is the origin, $ a\equal{}0$.

Hence, we have $ c\equal{}b\text{cis}\theta$.

$ z'\equal{}z\text{cis}90^{\circ} \implies z'\equal{}z\cdot i$

$ z'\equal{}(x\plus{}yi)\cdot i \implies z'\equal{}\minus{}y\plus{}xi$

Thus, coordinates of $ z'$ is $ (\minus{}y,x)$, and we are done.[/hide]

Not so basic, but still  :D

[quote="FantasyLover"]It is known that if Point $ C(c)$ is the rotation of $ B(b)$ with respect to $ A(a)$ by the angle $ \theta$, we have

$ c \equal{} a \plus{} (b \minus{} a)\text{cis}\theta$.[/quote]
The proof of this fact requires what you're trying to prove.

Rotation by $ 90^{\circ}$ counterclockwise is the same thing as pretending the positive $ y$-axis is the positive $ x$-axis and the negative $ x$-axis is the positive $ y$-axis.  Do you see how that gets you the result?  If not, try to prove that rotation is a linear transformation (there is a nice way to do this geometrically) and show that the claim holds for $ (1, 0)$ and $ (0, 1)$.

Knowing the result makes it easier

First suppose $ x,y > 0$ then $ (x,y)$ is in the first quadrant on the cartesian plane.
we perform the transformations $ (x,y)\mapsto ( \minus{} y,x)\mapsto ( \minus{} x, \minus{} y)\mapsto (y, \minus{} x)\mapsto (x,y)$
Notice that we are moving through the quadrants counter-clockwise.

Now, If $ c$ is the dictance between $ (x,y)$ and $ ( \minus{} y,x)$ then $ c^2 \equal{} (x \plus{} y)^2 \plus{} (y \minus{} x)^2 \equal{} 2(x^2 \plus{} y^2)$

Also, if $ a,b$ are the distances to the the points $ (x,y)$ and $ ( \minus{} y,x)$ from the origin, then

$ a^2 \equal{} x^2 \plus{} y^2$ and $ b^2 \equal{} x^2 \plus{} y^2$

So we have $ a^2 \plus{} b^2 \equal{} c^2$ which is pythagoras. This ensures that we have rotated $ 90^o$ around the origin clockwise.

Keep in mind that this is HSB; most of you are giving responses that anyone who doesn't know how to show that (-y,x) is a 90 degree rotation of (x,y) about the origin simply won't understand.

For a simpler, but less general, explanation, we can let A = (-y,x), B = (x,y) and O = (0,0).  The distance formula gives us AO = BO and slopes tell us that AO and BO are perpendicular.  So, A is a 90 degree rotation of B about O.  Considering quadrants as ocha suggests gives us the orientation of the rotation.


The average human heart rate is 70 beats per minute. At this rate determine the number of heartbeats a human heart would make in a lifetime of 80 years. Assume 365 days per year.

Is there a faster way to multiply the numbers?

Well $8 \cdot 126 = 1000 + 8 = 1008$ and $100800\cdot 365$ can be done by taking $36500000$ and the other number and adding. Then multiply by $8$, add a zero to the end.

Still kind of hard to answer within the 45 seconds though


Find the integer coefficients $ a, b, c$ of the cubic $ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} 0$ where $ x \equal{} \frac {3}{\sqrt [3]{7} \minus{} 2}$ is a solution.

[quote="123456789"]Find the coefficients $ a, b, c$ of the cubic $ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} 0$ where $ x \equal{} \frac {3}{\sqrt [3]{7} \minus{} 2}$ is a solution.[/quote]Are you sure that there is nothing else missing in this question? This would be so easy...

Expand $ \left(x\minus{}\frac {3}{\sqrt [3]{7} \minus{} 2}\right)(x\minus{}1)x\equal{}0$ and there you got a cubic...

$ a, b, c$ are rational.

[hide="Hint"] Isolate the $ \sqrt[3]{7}$ and cube it. [/hide]

When do I isolate it? I don't know what to do as my first step...

[hide="The first step"] $ \sqrt[3]{7}x \minus{} 2x \equal{} 3$ [/hide]
[hide="The second step"] $ \sqrt[3]{7} x \equal{} 2x \plus{} 3$ [/hide]
[hide="The third step"] $ 7x^3 \equal{} 8x^3 \plus{} 36x^2 \plus{} 54x \plus{} 27$ [/hide]
[hide="The final step"] $ x^3 \plus{} 36x^2 \plus{} 54x \plus{} 27 \equal{} 0$, so $ (a,b,c) \equal{} \boxed{36, 54, 27}$. [/hide]

[quote="t0rajir0u"][hide="The first step"] $ \sqrt [3]{7}x \minus{} 2x \equal{} 3$ [/hide]
[hide="The second step"] $ \sqrt [3]{7} x \equal{} 2x \plus{} 3$ [/hide]
[hide="The third step"] $ 7x^3 \equal{} 8x^3 \plus{} 36x^2 \plus{} 54x \plus{} 27$ [/hide]
[hide="The final step"] $ x^3 \plus{} 36x^2 \plus{} 54x \plus{} 27 \equal{} 0$, so $ (a,b,c) \equal{} \boxed{36, 54, 27}$. [/hide][/quote]

Oh wow. I can't believe I didn't know to just set the expression equal to $ x$.


algebra
polynomial
inequalities
ceiling function
induction
function
linear algebra

For a polynomial $f \in \mathbb{C}[x]$, let $n(f)$ denote the number of non-zero coefficients of $f$. Prove or disprove that, for any polynomial $f$, the following inequality holds:
\[
n (f^2) \geq n(f).
\]

Really? Then, what's a counterexample? I have spent some time thinking on this problem but weren't able to neither prove it nor to construct a counterexample.

 In fact, (if I am not mistaken) I can answer the question in the affirmative for $\deg f \leq 10$, so if there are any counterexamples, they must be of a large degree. I also have some strong theoretical evidence supporting the proposed inequality. Is there really a counterexample?

Apropos, here  are some relevent porblems on the number of non-zero coefficients (though they are much easier to prove.) Enjoy!

Problem 1. Prove that if a polynomial $f \in \mathbb{C}[x]$ has a non-zero zero $\alpha \neq 0$ of multiplicity $s$, then $n(f) \geq s+1$.

Problem 2. Prove that if a real polynomial $f \in \mathbb{R}[x]$ has $k$ real roots, then $n(f) \geq \lceil \frac{k+1}{2} \rceil$.

--Vesselin

$\alpha$ is supposed to be a root, right? (that's what the word "multiplicity" suggests, but you just say "a non-zero $\alpha\neq 0$")

Yes, $\alpha$ is a root. I actually said "non-zero zero," which may sound a bit weird, but could've equally said "non-zero root." Zero, root --- it's all the same.

Oh, here's one more correction. In Problem 2, I suppose that our polynomial has $k$ *distinct* real roots (multiplicities are not counted), for otherwise the polynomial $x^2$ would clearly be a counterexample. So, the problem is this: if our polynomial has at least $k$ distinct real zeros, then it has at least $\lceil \frac{k+1}{2} \rceil$ non-zero coefficients.

(1) is indeed not so hard: we use induction on $s$. Assume $f(0)\ne 0$, because otherwise we can just use the argument for $\frac f{x^n}$, with $n$ being the maximal exponent for which $x^n|f$.

The case $s=1$ is trivial, and then we assume we have shown it for $s$ and try to show it for $s+1$. If $f$ has a non-zero zero (:D) of multiplicity $s+1$ then $f'$ has a non-zero root (the same) of multiplicity $s$, and the argument is true for $f'$, so $n(f')\ge s+1\Rightarrow n(f)\ge s+2$ because we add the free term which disappears when we derive.

I didn't think yet about 2), but it doesn't seem easy. Anyway, what is that symbol which appears in the expression for $\frac{k+1}{2} $. Is it the integer part or the smallest integer greater than $\frac{k+1}{2}$? I have another approach for 1):
  Suppose f has at most s non-zero coefficients. Then write $ f(x)=a_1\cdot x^{k_1}+...+a_s\cdot x^{k_s} $ with $ k_1<...<k_s $ and $ a_i $ complex (possibly 0). Use the fact that $ f(a)=f'(a)=..=f^{(a-1)}(a)=0 $ and a little bit of triviality to show that this reduces to the system $ a_1\cdot (k_1)^j\cdot a^{k_1}+...+a_s\cdot (k_s)^j\cdot a^{k_s} =0$ for $ j=0,1,...,s-1 $. Now treat this as a linear system in $ a_1\cdot a^{k_1},...,a_s\cdot a^{k_s} $. The determinant is non-zero and so it has just the trivial solution. Thus, all $a_i$ are 0, false. Here I have put $ a=alfa$

I think $\lceil x \rceil$ is the ceiling function, i.e. the smallest integer which is not smaller than $x$.

I think the problem is equivalent to the following one:
  For any m and any $ x_1,...,x_{2m} $ different reals the matrix with rows $ (x_1)^{p_1},..., (x_1)^{p_m} (first row ),...., (x_{2m})^{p_1},...,(x_{2m})^{p_m} (last row) $ has rank at least m. But I have no idea how to prove this.

Come on, 2) is almost as easy as 1). As you've already noticed, it's convenient to use the derivative polynomial because we know everything for $n(f')$ in terms of $n(f)$. But also, we know that (by Rolle's theorem) the roots of the derivative polynomial are related to the roots of $f$. What more does one need?  :lol:

Anyway, these are relatively easy problems (but certainly very nice as facts). But what about the original problem, does $n(f^2) \geq n(f)$ always? Does anyone even have an idea?

I think I saw an example somewhere (a long time ago) for which $n(f^2) < n(f)$ but I'm not sure.

Both Arne and Vess are right. Vess when saying that contraexamples must be of high degree and Arne when saying we has seen a contrexample. It is Renyi who built a contraexample of degree 23 (I do not know this polynomial, I just read that in an article). And even more, Erdos proved that there exists a constant $c<1$ such that there are infinitely many polynomials f with $N(f^2)<N(f)^c$.


Let $ f(x)$ be a polynomial whose all the coefficents are integers.Knowing that $ f(a)\equal{}f(b)\equal{}1$ and $ f(c)\equal{}f(d)\equal{}\minus{}1$ where $ a,b,c,d$ are integers satisfying $ a<b,c<d,a<c$.Prove that $ a,b,c,d$ are consecutive integers.

This is $ a,c,d,b$ are consecutive integers.
Use that $ f(x) \equal{} (x\minus{}a)(x\minus{}b)Q(x) \plus{}1$ for some polynomial $ Q$ with integer coefficients.
From $ f(c)\equal{}f(d) \equal{} \minus{}1$ we deduce that $ (c\minus{}a)(c\minus{}b)$ and $ (d\minus{}a)(d\minus{}b)$ divide $ \minus{}2$ which leads to a finite numbers of cases you have to check.

Pierre.

Let $ f(x)\minus{}1\equal{}(x\minus{}a)(x\minus{}b)g(x)$, where $ g(x)$ is a polynomial with integer coefficients.

Since $ f(c)\equal{}f(d)\equal{}\minus{}1$ so $ (c\minus{}a)(c\minus{}b)g(x)\equal{}(d\minus{}a)(d\minus{}b)g(x)\equal{}\minus{}2$. But $ a<c<d$. So $ d\minus{}a>c\minus{}a>0$. Since both $ c\minus{}a$ and $ d\minus{}a$ divides -2, we have $ c\minus{}a\equal{}1,d\minus{}a\equal{}2$. So $ a,c,d$ are consecutive in that order. Now we shall prove $ b\equal{}d\plus{}1$.

Note that $ f(d)\equal{}(d\minus{}a)(d\minus{}b)g(x)\equal{}\minus{}2$, and hence $ d\minus{}b$ divides -1. If $ d\minus{}b\equal{}1$, then $ b\equal{}c$, and $ 1\equal{}f(b)\equal{}f(c)\equal{}\minus{}1$, contradiction. Thus $ d\minus{}b\equal{}\minus{}1$, as desired.


probability
ratio
LaTeX
geometry
geometric transformation
reflection
blogs

I never perfectly understood, why the average person cannot not walk away with at least some capital gain in the stock market at least like 95% of the time. I am asking because, usually after you buy a stock it will at some point get greater then the current calue. So if a person sells it later, they have a capital gain, thius can be done for all stocks that at least at some point increase their value compared to the value for which it was bought.

Why does this not occur?

[quote]Remind me the rules for capital gains.

It has to do with how long the stock is held, right?

An enormous amount of stock is bought for the short term for a variety of reasons: day trading, liquidity trading, hedging, etc.. These preclude capital gains, right?

Also, I'm not sure I agree with the assertion that 95% of the time stock goes up. In order to get a real feel for whether or not a stock buy will result in gains is to look at the investment horizon and the volatility of the stock. The shorter the investment horizon, the close the probability of a gain is to 50%.

For longer horizons, you can use a Sharpe ratio based on the volatility along with the interest rate earnings of the stock to determine that probability. You could also use stock options to gauge the implied volatility (options are priced almost entirely by volatility calculations) to get an idea of what the probability of gains would be.

Wait, why does capital gains have to do with how much it is held, you can make a capital gain even in 1 minute right?

I just randomly came up with 95% which I dont believe is correct, I just think that if you randomly chose a stock of a corporation and buy it at a random time, there will be some point in time later, when the value of the stock will be greater then when bought, am I correct? That is why I think that capital gains can be realized most of the time.

What exactly is an investment horizon?

I believe that a person can make a capital gain from stock investments most of the time.  I know that I have.

However, let's examine the relevant factors:

1.  The income tax system assumes that there is no inflation.  On average, the income tax system is accurate.  However, since WWII this country has had consistent inflation.  If one assumes a 3% annual inflation, one gets a yearly 3% capital gain, even if the company one invests in doesn't change it's finances or production signifcantly.

2.  There is a difference between asked and offered prices (not normally significant, but like a car you sell it wholesale and buy it retail)

3.  There are commissions that can be deducted from the value of the stock (one doesn't get a gain until the appreciation has covered the transaction cost.

4.  As said by someone else, one has to hold the stock for a period, a year, before a gain is a capital gain, taxed as a lower rate.

Does this help, or have I missed your questions?

I think we've both been confused by the capital gain reference.

I think ultimately, what hello is saying is, can't we continue to gamble until we win (unless the stock goes bust without ever reaching the buying price)?

The answer is of course yes.  However, we must also factor in that losses are larger if we gamble this way.  It's not unlike doubling a wager after each loss.  Sure, you'll win small almost all the time, but win you lose, you lose big.

Investing is more complex.

In order to gain from investing in stocks, your return would hve to be greater than the inflation rate. Generally, the value of shares does rise, but is it worthwhile? The purpose of using such finanial instruments should be to facilitate the smoothing of consumption over time as required. You want stocks that will rise in value in times when your income may be lower than expected.
Why do people hold stocks in the company they work for!

Yes, I never really understood why people hold stock in the companies they work for unless their companies are doing extremely well?

But inflation is maybe 2-3% in the US, stocks usually do better then that, at least the ones that do well. That should not be too tough to beat?

[quote="hello"]I think ultimately, what hello is saying is, can't we continue to gamble until we win (unless the stock goes bust without ever reaching the buying price)?

The answer is of course yes. However, we must also factor in that losses are larger if we gamble this way.

Mathew, can you explain to me why the losses are larger this way. Like where does my logic become faulty?[/quote]

It's one thing to say that most trades are in the black at some point, and it's another to presume that they are in the black in the end.

I'm not talking about the fact that interest rates should be considered as cost of capital.  I'm talking about the fact that every trade is liquidated at some point.  If the goal is to liquidate trades as soon as they are positive (or never), then the gains are only small by definition and the losses are sometimes enormous.  On the other hand, if we play the investment game by time horizons, then we can model outcomes as skewed brownian motion (skewed because stock values tend to fall faster than they rise, but they rise more often).

I'm trying to find a way to simplify something that has no nice simplification.  Thinking that because most trades are positive at some point in their life matters in any way would be a fallacy.  If a baseball player hit/miss with equal probably, you would expect there to be some moment during the season when they had higher than a .500 average.  But that doesn't mean that you can expect them to leave any particular game batting over .500 for the season.

Yes, I understand the baseball analogy, I just don't understand why the stock market works the same way (I have a hard time following that whole paragraph with brownian motion), can you pinpoint the exact thing(s) that is wrong with the following statement?

Corporation A is currently trading at $\$$10. Person A buys the stock at $\$$20 from the previous person who had it. Person B buys it from Person A at $\$$30, Person C from Person B at $\$$40, etc. Where does this begin to err and why?

It begins to err where your text is hard to read  ;)

lol, I dont know why the text was all messed up, where there any changes made in AOPS?

I said, that suppose that Corporation A has stock which person A bought at 10. Person B buys it from person A for 20. Person C buys from person B at 30, etc, etc. Therefore, all make capital gains.

I hope my text does not become hard to read again, what is wrong with my example?

Our forums are LaTeX enabled, so dollar signs begin/end mathmode.  I edited for you once I realized exactly what you did.

Your example is extremely oversimplified.  First, there is some chance that any of the future price points never occurs.  Second, there is some chance that those price points don't occur during a reasonable investment horizon (they are after all fairly steep increases in stock value).  It's not as if I can hold stock forever.  I have a limited life span after all and it's not likely that my preferences for the use of my money never ever changes over the course of my lifetime.

You have to factor in all possibilities, which means applying a probability curve to all possible outcomes within a defined investment horizon to begin to get an idea as to the real chances for profit under different scenerios.

Investment isn't a matter of making capital magically appear.  It's a matter of moving capital resources to where they can be most productive.  Additionally, nobody can predict the future entirely and to the extent that we can gauge possible outcomes, stock prices often reflect a great deal of information about those outcomes, hence without a great deal of information, we cannot assume that they will perform significantly differently than the risk free interest rate in aggregate (plus a risk premium based on volatility of returns).

I think I get what you are saying to a certain extent. But I believe that an individual who puts in a large amount of money for say like 20 years (and doesnt take any of it out) in a ok stock, cant they do well? Also in regards to probabilities, in the oversimplified example I made, there are no probabilities that are actually required. I also dont get the concept of investment horizon which you mention all the time, what exactly does it mean?
Btw, I just noticed 2 days ago that Berkshire Hathaway stock is almost 90,000. Omg, that must be for really rich people, right?

I'm not exactly sure but I think that the investment horizont that MCrawford mentions is the time how long you want to hold the stock and his point is that you probably don't want to hold it longer than, say, 100 years because you'll most likely die before you see your money again - did I get it right?
And the second part of the argument was that there is always some chance that your company will go bankrupt and you lose all your money - so if you earn $1 one hundred times and then lose $100 once you're still not making any money...

OK, the first argument I understand.

For the second argument, not too many companies go bankrupt and usually there is some indication that it will happen so that you can take out your money much before it does. Am I correct?

Sounds reasonable.... however, you may still get unlucky and buy a stock that is right now at its peak value and from now on its value will slowly go down and down... and the company will eventually go bankrupt after 10 years or so, or the stock will just lose most of its value and stay low forever.

But in that case, after you see it go down for some time, you just will sell it. Anyways, aren't we going into relatively rare case scenerios?

So on average, do small investors win or lose and approximately how much if any such data is available?

But this will break your whole strategy. How long is "for some time"? Stock that is currently going down is not such a rare case, how do you know if it's better to wait until its value goes over the purchase value again or if it's a stock that will go down for quite a long time now and it's better to sell it right away?

I don't know if it actually works though, perhaps somebody more experienced could help here??

I was mainy leaning on my idea of stock being over-valued. I cant see the deep logic error in stock going up to approaching infinetly many times a companies earnings. It doesnt go in to my mind, a pure deduction of why that cant happend assuming that a person lives and invests without wanting to take their money back (that is my only postulate, or axiom).

Stocks go up more than down.

The question is...by how much?

Interest rates and risk are the two most important factors.

If stocks went up 20% per year on average (with little deviation) and interest rates were 5%, banks would never lend money -- they would just invest in the stock market.

Ah, but that's where the misunderstanding lies.  Once investors begin pumping money into those great stocks, the prices rise, and thus the future returns become smaller...

In time, investors, with a certain amount of information, which, on the whole is a fairly good set of information given the standards, set the price of stock such that they are not so much better an investment than everything else in the world.  This is why people don't commonly borrow every cent they can get their hands on in order to make diversified bets and cash in (and who would lend the money?).

Stocks do provide greater returns over time than the risk free interest rate, but this is precisely because there is risk involved.  It's not just bankruptcy that lets us evaluate that risk.  Volatility on returns in fact erodes value.  If you do the calculations, you'll find that consecutive years of 0 and 20% returns is not as good as consecutive years of 10% returns.  This is why bonds are typically bought by investors who can't take long horizon risks (or, if they buy long bonds, are diversifying their portfolio in order to decrease that risk factor).

The Sharpe ratio helps us compute the actual value of an investment (or set of investments) according to returns and volatility.

So it's all more complex than just buy low, sell higher.  It wouldn't make sense otherwise.  Something in this universe would fall apart.

OK,I understand what you mean, thanks for your help Mathew. What do you suggest are some good books or online resources to read about investing (texts that are to be understood by a beginner)? Thanks

I'm not sure about good books to read.  I learned almost everything I know about economics by picking up single ideas from anything that made me think and then debating ideas out with people I knew.  Knowing smart people to debate ideas with is worth a lot more as a real life classroom than any book.  Find smart people and talk to them.

For some ideas worth thinking about, Milton Friedman writes well on a lot of the fundamental ideas behind economics that I think are indespensible to a general understanding of finance/economics.

It's also often good to debate ideas out with libertarians (who come in many varieties from Jeffersonian to highly conservative-even-if-minarchist to down right anarchistic).  Even if libertarians have a general skew toward certain kinds of ideas, they've typically spent more time considering points and counterpoints in detail.

In that same vein, Adam Smith is a good author to read.  He's hard to read and very basic, but the soul of most of his economic ideas rests on very plain, if insightful observation.

I also think it's a good idea to school yourself on game theory.  The older I get, the harder it is not to see game theory in almost any market situation, even where you'd think classical economic forces would dominate.

Thanks for your suggesstions, Mathew. I am not sure where I can find someone smart to talk to but I can say that I read a lot of the econ blogs (majority of them written by Libertarians and I generally consider myself a Libertarian too especially because of my strong support for free markets) and think that their ideas are unique. It also seems that libertarian economists are the ones that think on a high level, for some reason they have a lot of the libertarian bloggers from George Mason University. Something else I wanted to say but cant remember what, ok. Thats it for now.

Also, this is just a basic finance question that I am emberrased to ask because it is simple and I should know the answer, but is the preferred stock the same value as the common stock or are they traded seperatly, I know the differences in divided payments, but I never knew the answer to this basic question, I usually see only one stock value for a company, so my guess is they are traded together, but I am not sure. Thanks to whoever can help out.

Well, I know with dividends the common stock usually pays more, but the whole thing is I dont know about the actual value of the stock? In general, I know that common stock has greater return, but is it because of the dividend, or also because they are traded seperatly on the market?

If a 2.00 stock pays an .08 dividend, its value is then reduced to 1.92.  Otherwise, there would be an arbitrage on buying up stock right before dividends and reselling right after.  This accounts for apparent excess returns of non-dividend stocks.

There are some small or subtle valuation issues that seperate them as well, such as transaction costs or reinvestment, preference by older investors for regular cash payments (dividends) as a fixed income, and other little things (volatility considerations).

Wait, so you are saying that after dividends are declared, the value of the stock drops by the value of the dividend on the market? (or approximately at least).

I understand what you are saying in regards to arbitrage and the other differences that you have mentioned, but, would this imply that there is a second market for preferred stock that is similar to common stock but varies mainly with declaration (or is it payment date?) of dividends? Or is my logic faulty, and the only actual difference is the dividend itself, which actually seems to make slightly more sense, but I am not sure?

Also, what is it that I have seen with some form of coverting from one stock to another of $\$$2.50, not sure that I know what I am talking about myself, it has something to do with $\$$2.50.


trigonometry
calculus
calculus computations

$ y'' \plus{} 5y' \plus{} 4y \equal{} x^2 \plus{} sin 2x$

Look up the method of undetermined coefficients.

http://en.wikipedia.org/wiki/Method_of_undetermined_coefficients


$ \frac{x^3 \plus{} 2x^2 \minus{} 2x \plus{} 3}{x^2 \minus{} 9}$

I use synthetic division for the numerator in order to factor it, and i get -3 as a solution, after I factor it I get:

$ \frac{(x\plus{}3)(x^2 \minus{} x \plus{} 1)}{(x\plus{}3)(x\minus{}3)}$

I simplify and get:

$ \frac{x^2 \minus{} x \plus{} 1}{x\minus{}3}$

Can I continue any longer, or am I doing this wrong?

Um, it's the same process as you did with $ \frac {x^3 \plus{} 2x^2 \minus{} 2x \plus{} 3}{x \plus{} 3}$, except it's now $ \frac {x^2 \minus{} x \plus{} 1}{x \minus{} 3}$, and you don't get a remainder of $ 0$...

so if finding x, imagining the first division would = 0... the solutions would be all real numbers except 3?

I'm sorry, but I'm not exactly sure what you mean. If you don't know how to do synthetic division, look [url=http://en.wikipedia.org/wiki/Synthetic_division#Polynomial_division_by_x_.E2.88.92_r]here[/url] and [url=http://mathworld.wolfram.com/SyntheticDivision.html]here[/url], since that's all you need to simplify the last expression you got to.

But I think I sort of know what you mean. After you simplify $ \frac {x^2 \minus{} x \plus{} 1}{x \minus{} 3}$, you have to add that it's not defined for $ x \equal{} \pm3$, since they make the denominator $ 0$ in the original expression.


Here are 5 speculations as to how the Harry Potter book would end

a. It's not possible that both Harry and Voldmort will live
b. If Snipe or Malpoi help Harry, Harry will live
c. If Snipe will help Harry then Voldmort will die
d. Voldmort will die [b]only if[/b] Snipe helps Harry
e. Harry will live [b]if and only if[/b] Voldmort dies

define elementary lines for a,b,c,d,e using ~, ->, <->, V ^

for every one of the following statements say if correct or not, prove your statement

1. a is tautologically equivilent to e
2. {a,b}|= c
3. {a,c,d}|=b

this is confusing me very much , can you help ?

the answer is : Lena , stop being lazy, and write the entire $2^{4}$ truth table


Is this a correct solution to the following problem: Find the probability that if a fair coin is tossed infinitely and the results are recorded in a sequence, "HH" will occur before "THTH". Solution: Let us count the complement. We take cases, i.e. p=the probability that "THTH" occurs before "HH" and the first toss is a head, and q=the same probabilty except that the first toss is a tail. If the first toss is a head, the next toss must be a tail, otherwise we would have "HH" before "THTH". In that case, the p=q/2, as the probability the first toss is a head is 1/2, and afterwards the probability that the other condition will be satisfied will be q (the first head will not affect the other condition). We multiply to get p=q/2. Now, q=1/16+p/2+p/4+.... This comes from the fact that either we can have a "THTH" right off the top, with probability 1/16, or we can have a run of "T"s and then an H. We cannot have a run of "T"s and then 2 H's because then the second condition will be violated. In any case, the probability that we have a run of "T"s and then an H and the second condition holds is p/2+p/4+... by the same reasoning that q=p/2, except that we can have any number of T's before reverting to an H. Solving these two equations, we have q=p+1/16 and 2p=q, so p=1/16 and q=18 => the total probability will be 3/16. Since we want the complement of this, the answer is 13/16.

Mystic -- there may be a small problem with what you've done.

[quote]This comes from the fact that either we can have a "THTH" right off the top, with probability 1/16, or we can have a run of "T"s and then an H. We cannot have a run of "T"s and then 2 H's because then the second condition will be violated. In any case, the probability that we have a run of "T"s and then an H and the second condition holds is p/2+p/4+... by the same reasoning that q=p/2, except that we can have any number of T's before reverting to an H.[/quote]

Now, I'm not entirely sure of this, but it seems to me that you are calling this situation:
TTTTTT...TTTH
the same as this one:
H
but they clearly aren't the same, since you will, on average, get THTH sooner in the first case than in the second.  I couldn't tell if your method dealt with that or not.

Here's a way I like -- we can do it with a tree.

Clearly, we can never get back to an equivalent situation paralleling a starting H, (for the reason I mentionned above), so really we need two trees, one with a starting H and one with a starting T.  First the tree with the starting H:
It gives us a probability of 1/2 that we get HH, and a probability 1/2 that we move over to the tree with a starting T.

Now, for the tree with a starting T, which looks like this:
T
| \
H  [b]T[/b]
|\
T   [b]H[/b]
|\
H  [b]T[/b]

(the bolded letters are ends of branches -- explanation below)

Now, working from top to bottom, starting with the second level: the T on the left just brings us to the top of the tree again.  The H continues the tree.
Third level:  the H makes us done, with probability 1/4 (HH has prob. 1/4.)  The T is not the same as the top of the tree, so it continues the tree.
Fourth level:  The H brings us to THTH, with prbability 1/8.  The T brings us to the top of the tree.

So, of all outcomes, we have THTH with probability 1/8 and HH with probability 1/4.  Every other branch just brings us to the top of the tree.  That means the total probabilities for the tree are HH: (1/4)/(1/4 + 1/8) = 2/3 and THTH: (1/8)/(1/8 + 1/4) = 1/3.

So, lets look at the probability of HH first:  we have 1/2 we'll be in the H-first tree, times 1/2 we get HH first.  This is 1/4.  Then we have 3/4 probability we'll be in the T-first tree, which gives us HH first with probability 2/3, so that's 1/2 for HH.  This gives us a total of 3/4.

Just to check, the probability of THTH first should be the 3/4 chance we're in T-first tree times the 1/3 we win from there, which gives us 1/4.

1/4 + 3/4 = 1, so we've probably done this correctly.

If you want to do your method, you need to make a seperate variable for the situation TH compared with H-first, and maybe for the situation THT compared with T-first or TT.

Here's the way I looked at the problem.  It's similar to the tree method, but perhaps a little shorter.

Let a, b, c, and d be the probabilities that HH occurs before THTH, starting from H, T, TH, and THT, respectively.  We get three equations for b, c, and d by looking at the next coin flip: b = (b + c)/2, c = (1 + d)/2, and d = b/2.  Solving, we get b = c = 2/3 and d = 1/3.

Also a = (1 + b)/2, so a = 5/6.  The probability we want is (a + b)/2, which is 3/4.

Yeah, the tree method is a little long to write up in text.  It does work very well in infinite-sequence problems, though.  As does a-variable-for-every-case combined with solving a bunch of linear equations.

Here's a problem in this vein that we would sometimes ask in interviews at DE Shaw.

You pick a three sequence of flips, like, say HTH.  You announce your sequence.  Then I choose mine.  Then we flip until one of our sequences occurs as 3 consecutive flips among the flipping.  (For example, if I pick TTT and you pick HTH, and we flip THTH as the first 4 flips, the game is over and you win.)

I offer you 3-2 odds to play this game (meaning you bet $2 against my $3).  Do you play?  What sequence do you choose?

Wow, that's a tough problem for an interview.  I think I saw the problem in the book [i]Concrete Mathematics[/i], so I won't give it away.

If I recall correctly, I think that book gave an approach to these kind of problems using generating functions.  That approach may be simpler than my many-variable method or Joel's tree method, at least for large patterns.

Deleted: I posted a duplicate message by accident.

Yeah, needless to say, there weren't many immediately correct resopnses.  When I asked it, I was primarily looking for someone who pretty quickly grasped that the game wasn't fair and understood why.  It was less important to me that they could quickly come up with the actual fair 'price' for the game.  If it isn't immediately apparent to an interviewee that this game is not fair (in that I should be giving odds to get people to play), then the interviewee was less likely to be a good candidate for some of the styles of trading we did.

Not a perfect filter, to be sure.

Lets see..

[hide]

If we pick HHH, they pick THH and we win 1/8 of the time, because the only way for HHH to appear first is if there are no Ts tossed at all.

If we pick HHT, they pick THH and we win 1/4 of the time, we only win in the case of HHHHHH....T.

If we pick HTH, they pick HHT. Call the probability of us winning x - after the first head turns up, then half the time they win (if another head is tossed, HHT must come before HTH), the other half, if the toss after that is tails, the probability is x again, otherwise wewin, overall, we win 1/3 of the time (by solving x = 1/2 + 1/4 x).

Similarly if we pick HTT, they pick HHT and its basically identical to the above, we win 1/3 of the time.

So the odds should be 2:1..

[/hide]


How many 3-digit numbers are divisible by 11?

We simply have $ \left \lfloor \frac{900}{11}\right \rfloor\equal{}\boxed{81}$.

However this may not be the case. How many numbers are from 0 to 11 (inclusive)?

You solution is $ \lfloor{\frac{12}{11}}\rfloor\equal{}1$ when the correct answer is 2.

THe way to go about doing this is to note it ranges from 110 (11x10) to 990(11x90) so its 90-10+1=81.


Find the maximum value of $ \alpha$ such that :

$ \sum_{cyc}\frac{a^2}{b\minus{}4}\ge \alpha\ ,\ (\forall)a,b,c\in (4;\infty)$

Answer: $ \alpha _{max}\equal{}48$ when $ a\plus{}b\plus{}c\equal{}24$ by C.S. and AM-GM.

[quote="alex2008"]Find the maximum value of $ \alpha$ such that :

$ \sum_{cyc}\frac {a^2}{b \minus{} 4}\ge \alpha\ ,\ (\forall)a,b,c\in (4;\infty)$[/quote]
are there any other conditions for $ a,b,c$, for example,
the maximum value if $ a\plus{}b\plus{}c\equal{}1$ isn't the same value if $ a\plus{}b\plus{}c\equal{}3$.

[quote="kunny"]Answer: $ \alpha _{max} \equal{} 48$ when $ a \plus{} b \plus{} c \equal{} 24$ by C.S. and AM-GM.[/quote]
...Why AM-GM, I think we can prove it only by Cauchy:
$ \left(\sum \frac{a^2}{b\minus{}4}\right)(a\plus{}b\plus{}c\minus{}12)\geq(a\plus{}b\plus{}c)^2\equal{}576$
then; $ \sum \frac{a^2}{b\minus{}4}\geq 48\equal{}\max\{\alpha\}$
if you have another solution more nice post it please  :)

When $ a\equal{}b\equal{}c\equal{}8$ , kunny .

Let be $ s \equal{} a \plus{} b \plus{} c\ ,\ s > 12$

From Titu Andreescu we have that :

$ LHS\ge \frac {s^2}{s \minus{} 12}\ge \alpha\Leftrightarrow s^2 \minus{} s\alpha \plus{} 12\alpha\ge 0\ ,\ (\forall)s > 12$ .

$ \Delta_s\le 0\Rightarrow \Delta_s \equal{} \alpha^2 \minus{} 48\alpha\le 0\Rightarrow \alpha \le 48$ . Equality holds if $ a \equal{} b \equal{} c \equal{} 8$

If $ \Delta_s > 0\Rightarrow \alpha > 48\Rightarrow s^{\prime} \equal{} \frac {\alpha \plus{} \sqrt {\alpha^2 \minus{} 48\alpha}}{2} > 12$ , but exists $ s\in (12,s^{\prime})$ such that $ s^2 \minus{} s\alpha \plus{} 12\alpha < 0$

$ \Rightarrow \alpha \equal{} 48$

[quote="kunny"][quote="kunny"]Answer: $ \alpha _{max} \equal{} 48$ when $ a \plus{} b \plus{} c \equal{} 24$ by C.S. and AM-GM.[/quote]

Since $ a > 4,\ b > 4,\ c > 4$, we have

$ \frac {a^2}{b \minus{} 4} \plus{} \frac {b^2}{c \minus{} 4} \plus{} \frac {c^2}{a \minus{} 4}\geq \frac {(a \plus{} b \plus{} c)^2}{(a \minus{} 4) \plus{} (b \minus{} 4) \plus{} (c \minus{} 4)}$

$ \equal{} a \plus{} b \plus{} c \minus{} 12 \plus{} \frac {12^2}{a \plus{} b \plus{} c \minus{} 12} \plus{} 24$

$ \geq 2\sqrt {(a \plus{} b \plus{} c \minus{} 12)\cdot \frac {12^2}{a \plus{} b \plus{} c \minus{} 12}} \plus{} 24 \equal{} 48$

when the equality holds when $ a \plus{} b \plus{} c \minus{} 12 \equal{} \frac {12^2}{a \plus{} b \plus{} c \minus{} 12} > 0\Longleftrightarrow a \plus{} b \plus{} c \equal{} 24\ (a > 4,\ b > 4,\ c > 4)$.

Thus the minimum value of $ \frac {a^2}{b \minus{} 4} \plus{} \frac {b^2}{c \minus{} 4} \plus{} \frac {c^2}{a \minus{} 4}$ is 24, yielding $ \alpha _{max} \equal{} \boxed{48}$.

[quote="peine"][quote="kunny"]Answer: $ \alpha _{max} \equal{} 48$ when $ a \plus{} b \plus{} c \equal{} 24$ by C.S. and AM-GM.[/quote]
...Why AM-GM, I think we can prove it only by Cauchy:
$ \left(\sum \frac {a^2}{b \minus{} 4}\right)(a \plus{} b \plus{} c \minus{} 12)\geq(a \plus{} b \plus{} c)^2 \equal{} 576$
then; $ \sum \frac {a^2}{b \minus{} 4}\geq 48 \equal{} \max\{\alpha\}$
if you have another solution more nice post it please  :)[/quote]

I can't understand your solution.[/quote]
why? I think it's clear, I used Cauchy's inequality,
$ \left(\sum \frac {a^2}{b \minus{} 4}\right)(a \minus{}4 \plus{} b\minus{}4 \plus{} c\minus{}4 )\geq(a \plus{} b \plus{} c)^2 \equal{} 576$ and we have,$ a\minus{}4\plus{}b\minus{}4\plus{}c\minus{}4\equal{}24\minus{}12\equal{}12$
then; $ \sum \frac {a^2}{b \minus{} 4}\geq \frac{576}{12} \equal{} 48 \equal{} \max\{\alpha\}$
I think now it's more clear  :) ,
generalisation: $ \max\{\alpha\}\equal{}4s\sqrt{3}$

Another solution.
[hide]
By rearrengment inequality we have that $ \sum{\dfrac{a^2}{b\minus{}4}}\ge\sum{\dfrac{a^2}{a\minus{}4}}$ with equality only when $ a\equal{}b\equal{}c$.

Let $ x \equal{} a \minus{} 4$, $ y \equal{} b \minus{} 4$ and $ z \equal{} c \minus{} 4$. The right side of the former inequality becomes $ \sum{\dfrac{(x \plus{} 4)^2}{x}}$.

By AMGM
$ \sum{\dfrac{(x \plus{} 4)^2}{x}}\ge\sum{\dfrac{16x}{x}} \equal{} 48$

With equality when $ x \equal{} y \equal{} z \equal{} 4$, that means when $ a \equal{} b \equal{} c \equal{} 8$.

So the largest such $ \alpha$ is $ 48$.
[/hide]
:)

[quote="ElChapin"]Another solution.
[hide]
Let $ x \equal{} a \minus{} 4$, $ y \equal{} b \minus{} 4$ and $ z \equal{} c \minus{} 4$. The left side of the inequality becomes $ \sum{\dfrac{(x \plus{} 4)^2}{x}}$.

By AMGM
$ \sum{\dfrac{(x \plus{} 4)^2}{x}}\ge\sum{\dfrac{16x}{x}} \equal{} 48$

With equality when $ x \equal{} y \equal{} z \equal{} 4$, that means when $ a \equal{} b \equal{} c \equal{} 8$.

So the largest such $ \alpha$ is $ 48$.
[/hide]
:)[/quote]
Are you sure? $ LHS \equal{}\frac{a^2}{b\minus{}4}\plus{}\frac{b^2}{c\minus{}4}\plus{}\frac{c^2}{a\minus{}4}$ :wink:
Regards

Sorry, I changed the denominators. Fortunately it is not a big problem since by rearrengment inequality we have that
\[ \sum{\dfrac{a^2}{b \minus{} 4}}\ge\sum{\dfrac{a^2}{a \minus{} 4}}
\]
with equality when $ a \equal{} b \equal{} c$, so $ 48$ still is the best costant.

It is edited now... Thank you.

:)

[quote="kunny"][quote="kunny"]Answer: $ \alpha _{max} \equal{} 48$ when $ a \plus{} b \plus{} c \equal{} 24$ by C.S. and AM-GM.[/quote]

Since $ a > 4,\ b > 4,\ c > 4$, we have

$ \frac {a^2}{b \minus{} 4} \plus{} \frac {b^2}{c \minus{} 4} \plus{} \frac {c^2}{a \minus{} 4}\geq \frac {(a \plus{} b \plus{} c)^2}{(a \minus{} 4) \plus{} (b \minus{} 4) \plus{} (c \minus{} 4)}$

$ \equal{} a \plus{} b \plus{} c \minus{} 12 \plus{} \frac {12^2}{a \plus{} b \plus{} c \minus{} 12} \plus{} 24$

$ \geq 2\sqrt {(a \plus{} b \plus{} c \minus{} 12)\cdot \frac {12^2}{a \plus{} b \plus{} c \minus{} 12}} \plus{} 24 \equal{} 48$

when the equality holds when $ a \plus{} b \plus{} c \minus{} 12 \equal{} \frac {12^2}{a \plus{} b \plus{} c \minus{} 12} > 0\Longleftrightarrow a \plus{} b \plus{} c \equal{} 24\ (a > 4,\ b > 4,\ c > 4)$.

Thus the minimum value of $ \frac {a^2}{b \minus{} 4} \plus{} \frac {b^2}{c \minus{} 4} \plus{} \frac {c^2}{a \minus{} 4}$ is 24, yielding $ \alpha _{max} \equal{} \boxed{48}$.
[/quote]
We have $ LHS \ge 48$ ,the equality when $ a \equal{} b \equal{} c \equal{} 8$,then $ \alpha_{max} \equal{} 48$ :wink:

this is an other soluton without supposons that, $ a\plus{}b\plus{}c\equal{}24$
with rerangment inequality, $ S\equal{}\sum\frac{a^2}{b\minus{}4}\geq\sum\frac{a^2}{a\minus{}4}$
by AM-GM $ \frac{a^2}{16}\plus{}4 \geq a \Leftrightarrow a\minus{}4\leq\frac{a^2}{16}$ then,
$ S\geq \sum\frac{a^2}{\frac{a^2}{16}}\equal{}16\times3\equal{}48$ (equality if $ a\equal{}b\equal{}c\equal{}8$)
what abou this,
let $ a,b,c\in (\lambda,\inf)$ and $ n\geq 1 \in IR$ find the minimum of the expression,
$ \sum \frac{a^n}{b\minus{}\lambda}$


calculus
inequalities unsolved
inequalities

[quote="bjh7790"]Prove that

$ (1 \plus{} na)^{n \plus{} 1} > (1 \plus{} (n \plus{} 1)a)^n$

for positive real number a and positive integer n.

Thanks.[/quote]
Let $ f(x)\equal{}\frac{\ln(1\plus{}ax)}{x}.$
Hence, $ f'(x)\equal{}\frac{\frac{ax}{1\plus{}ax}\minus{}\ln(1\plus{}ax)}{x^2}.$
Let $ g(x)\equal{}\ln(1\plus{}ax)\minus{}\frac{ax}{1\plus{}ax},$ where $ x\geq 0.$ Hence, $ g'(x)\equal{}\frac{a^2x}{(1\plus{}ax)^2}>0.$
Thus, $ g(x)> g(0)\equal{}0.$ Id est, $ f'(x)<0$ for all $ x>0,$ which gives your inequality.

is there any solution not using calculus?

From Bernoulli you have:
$ (1\plus{}na)^{\frac{n\plus{}1}{n}}\ge 1\plus{}\frac{na(n\plus{}1)}{n}\equal{}1\plus{}(n\plus{}1)a$ Equality only when $ a\equal{}0$ or $ \frac{n\plus{}1}{n}\equal{}1$ which will never happen.


Consider a  function which maps $\mathbb{Z}$ to ${\mathbb{Z}}$ satisfies the following conditions:
(1) $f(f(n)) = 3n$
(2) $f(n+1) > f(n)$
Compute the value of $f(1458)$

There are many functions satisfyed your conditions. For example $f(x)=\sqrt 3 x$.

I forgot to write that that the function maps $\mathbb{Z}$ to $\mathbb{Z}$

What is mean?
[quote="ashwinkrishnan"]Consider a continuous real valued function [/quote]

I fail to understand how a function can be continuous
 if it maps from Integers to Integers???

About function $f: N\to N$ posted before.
Solution is unique: $f(n)=n+3^{k}$ if $3^{k}\le n\le 2*3^{k}$ and $f(n)=3n-3^{k+1}$ if $2*3^{k}\le 3^{k+1}$.

The proof of that is quite a beautiful bit.
 I've found three so far.

the actual problem i believe comes from an olympiad in
 which $f(2001)$ is asked.

I am extremely sorry for the mistake.It is true that the function is not continous . I have edited the problem itself.


LaTeX
superior algebra
superior algebra unsolved

I dunno what Notherian is but it|s certainly in a wrong section...
Here is $\LaTeX$ version of the problem:

$R=\left\{\left[\begin{array}{cc}a&b\\0&c\end{array} \right]\, a\in \mathbb{Z},b,c\in\mathbb{Q}\right\}$.
R is a ring. Whether R is left Notherian or right Notherian.


integration
calculus
real analysis
real analysis unsolved

For (2) you need see cases $a<1,1\leq a\leq 4, a>4$ and you can find $I(a)$ in any cases :)

My answer in the first problem is 2*(100)!!/(101)!!.
This can be easy obtained by using Eulier's integral and substitution
  x=sin(t).

[quote="Slizzel"](1). Compute $J=\int_{0}^{1}{(1-x^{2})^{50}}dx$.  It's easy to find that $I_{n}=\int_{0}^{1}{(1-x^{2})^{n}}dx=\prod_{i=1}^{n}{\frac{2i}{2i+1}}$ but i don't know to compute $I_{50}$

(2). If $I_{a}=\int_{1}^{4}{\frac{dx}{|x-a|+1}},\ a\in\mathbb{R}$, compute $I'(-1)+I'(2)+I'(5)$. If $a \in [1,4]$ it's not so hard to solve it and we obtain div by 0 when calculating $I'(5)$  :mad:

Any ideeas are wellcomed  :idea:[/quote]
In first inegral: use substitute $x: =sin(y)$ and then is easy


I hope this is in the right forum:

How many nonnegative integer solutions $(x_{1},x_{2},x_{3})$ are there to the inequality $x_{1}+x_{2}+x_{3}\le50$?

Isn't it $\frac{53!}{50!3!}=23426$?
The problem is equivalent to $x_1+x_2+x_3+x_4=50$, with $x_1,x_2,x_3,x_4$ nonnegative.  Bijection to permutations of a string of 50 0's and 3 dividers.

[quote="kunny"]Yes, you are right,themonster. :blush:

How about this one?

$x_1+x_2+\cdots +x_n \leq m\ (x_1\geq 1,x_2\geq 2,\cdots, x_n\geq n )$[/quote]

You may have to define conditions suitable.

Under all assumed ideal conditions,
the solution is equivalent to the number of non negative integral solutions to the eqauation
$(x_1  - 1) + (x_2  - 2) + (x_3  - 3) + ....(x_n  - n) + x_{n + 1}  = m - \frac{{n(n + 1)}}{2} = k(say)$

Hence the answer is ${}^{k + n}C_n$

if you have x+y+z=50(or less) then you add another variable, lets call it a so you now have a+x+y+z=50. this is because you let a equal 50-(x+y+z). this is much easier to work with. you can now use the formula which states that given x1+x2+x3+x4+...+xm=n there are (m+n-1)C(m-1). in this case you end up with (50+4-1)C(3).

In other words, if we write 50 as a bunch of ones, we would have
1 1 1 1 1 1 1 1 1 ... 1 1 1, where there are 50 ones.  As maokid7 said, we can write 50 as a+x1+x2+x3. If _ divides the ones,        1 1 1 _ 1 _ 1 1 _ 1 1 would mean 3 + 1 + 2 + 2. Thus, there are 53 spots to put the 3 _'s, so the answer is $\frac{53!}{3!50!}=23426$.

[quote="boxedexe"][quote="TESoccer66"]In other words, if we write 50 as a bunch of ones, we would have
1 1 1 1 1 1 1 1 1 ... 1 1 1, where there are 50 ones.  As maokid7 said, we can write 50 as a+x1+x2+x3. If _ divides the ones,        1 1 1 _ 1 _ 1 1 _ 1 1 would mean 3 + 1 + 2 + 2. Thus, there are 53 spots to put the 3 _'s, so the answer is $\frac{3!}{3!50!}=23426$.[/quote]

I believe it should be $\frac{53!}{50!3!}=23426$ or $\frac{53\cdot 52\cdot 51}{3!}=23426$ .

Masoud Zargar[/quote]

oh... thats what I meant. :o I'll fix it.


Prove that if n is an integer and not a multiple of 3, then the angle pi/n can be trisected with rulers (straightedge) and compasses.

[hide="solution"]
assume I don't have to explain the following constructions:
equilateral triangle
angle copying/addition/subtraction
Then if you have an angle $ \frac{\pi}n$, you can add it to itself over and over to get an angle of $ \frac{k\pi}n$
also, from an equilateral triangle, you can get angles of $ \frac{\pi}3$ and $ \frac{2\pi}3$. THUS, you can subtract angles to get new angles of measure
$ \frac{\pi}3\minus{}\frac{k\pi}n\equal{}\frac{(n\minus{}3k)\pi}{3n}$ and $ \frac{2\pi}3\minus{}\frac{k\pi}n\equal{}\frac{(2n\minus{}3k)\pi}{3n}$
so therefore, depending on n's residue mod 3 (which is not 0), we can pick k such that the measure of one of those angles is $ \frac{\pi}{3n}$, the trisection of the original angle.
[/hide]


Any pictures related to me that don't fit anywhere else.

From left to right in both rows:

Johannes (dutch, jpsteenstra on ML) - Thomas (flemish, Psycho on ML) - Fokko (dutch, team leader, Tasar on ML)- Sander (dutch) - Arno (dutch),
Peter (flemish, Peter VDD on ML) - Sjoerd (dutch) - Iris (dutch) - Esther (dutch) - Victor (dutch).

IMC 2006 report in dutch can be found [url=http://www.problem-solving.be/viewtopic.php?t=1412]here[/url].

Fokko was team leader for the dutch IMC team? :) Didn't know that ...

Yes he was, he quits his IMO-deputy role, and will only do IMC from now on. So still a yearly chance on seeing him if you like. :)


function
combinatorics proposed
combinatorics

Is it possible to colour all positive real numbers by 10 colours so that every two numbers with decimal representations differing in one place only are of different colours? (We suppose that there is no place in a decimal representations such that all digits starting from that place are 9's.)

[i]Proposed by A. Golovanov[/i]

Nice problem!
I have a solution that uses my beloved Axiom of Choice  :D

Let us define the equivalence relation between positive real numbers: "$a \equiv b$ if $a-b$ has only finitely non-zero digits". It is reflexive (a-a = 0 that doesn't have non-zero digits), symmetric (a-b = -(b-a) and they have the same digits) and transitive ( $(a-b)+(b-c) = (a-c)$ and the sum of two numbers of the form $\frac{n}{10^{k}}$ is of this form).
So, this equivalence relations partitions the set of positive reals in a family of disjoint sets (classes of equivalence).
Let $f(a)$ be the equivalence class containing a. Let $g$ be a choice function of the equivalence classes (and here we use AC).
Now we finally define $c: \mathbb{R}^{+}\rightarrow \mathbb{Z}_{10}$ such that c(r) is the sum of the decimal digits (mod 10) of $a-g(f(a))$. In fact, $a \equiv g(f(a))$, so their difference has just a finite number of nonzero digits, and we can calculate this sum.

If two numbers a,b differ only in one place of their decimal rappresentation, $c(a)-c(b)$ is the sum of the decimal digits of their difference, that by hypotesis has one and only one nonzero digit $\Rightarrow c(a) \neq c(b)$.

Is it clear?
Now I ask you either to prove this problem without the Axiom of Choice, or to prove (a weak form of) the Axiom of Choice using this problem  :wink:


linear algebra
matrix
inequalities proposed
inequalities

Let $ n$ be a positive integer. Find the smallest real number $ k\equal{}k(n)$ such that, for all real numbers $ x_1,x_2,\dots,x_n$,
$ \left|\prod_{i<j} x_i\minus{}x_j\right|\leq k\left(x_1^2\plus{}x_2^2\plus{}\dots\plus{}x_n^2\right)^{\frac{n(n\minus{}1)}{4}}$

[hide="Hint, possibly"]
My first guess would be to look at the associated Vandermonde matrix, since the LHS matches up exactly with the determinant.
[/hide]


function
inequalities
inequalities unsolved

Let $a,b>0$ with $a^2+b^2=1$. Prove that $a+b+ \frac{1}{ab} \geq 2+ \sqrt{2}$.

Nice and easy and with a fine generalization:
    We have a+b+1/ab>=2sqrt(ab)+1/ab. But ab<=1/2 and the function f(x)=2sqrt(x)+1/x is decreasing for x<=1/2, so f(ab)>=f(1/2)

I have noted that the equal sign in the inequality is true when a=b=1/ \sqrt 2, so this afternoon I have tried to solve it using AM-GM but I failed. Do you think it is possible to find a proof without using functions?

I am very interested in this question.

Thank you very much for your answer.

the answer is yes.

we have 1/ab \geq 2 > \sqrt 2 \geq a+b (from AM-GM),  and because (a+b)/ab = 1/a+1/b \geq 2 \sqrt ( 1/ab ) \geq 2 \sqrt 2 and using the well-known lemma which says that if a\geq b\geq c then if ad\geq bc we have a+d \geq b+c the problem is solved.

proof of the lemma is simple, take b=a+x, c=a+x+y, d=a+x+y+z, where x,y,z \geq 0, and simply see that ad \geq bc => z \geq x => a+d \geq b+c.

Very nice proof !!!!

Thank you very much, Valentin.

[quote="Valentin Vornicu"]the answer is yes.

we have $\frac{1}{ab}\geq 2 > \sqrt{2}\geq a+b$ (from AM-GM),  and because $\frac{a+b}{ab}= \frac1a+\frac1b \geq 2 \sqrt{\frac{1}{ab}}\geq 2 \sqrt{2}$ and using the well-known lemma which says that if $a\geq b\geq c$ then if $ad\geq bc$ we have $a+d \geq b+c$ the problem is solved.

proof of the lemma is simple, take b=a+x, c=a+x+y, d=a+x+y+z, where x,y,z \geq 0, and simply see that ad \geq bc => z \geq x => a+d \geq b+c.[/quote]
About the lemma: clearly it suffices to prove case $d=\frac{bc}{a}$ but we see that the condition $a\geq b\geq c$ is unnecessary compared to $a\geq \max\{b,c\}$ which is all required... right?

I understand the mechanics of the proof but something fishy is happening with the equality cases...


What is the difference between writing "Noetherian" or "noetherian". Milne and Matsumura use the first one, while Hartshorne uses the second. Is it a matter of fashion, way of thinking (i.e. algebraic vs. geometric) or just personal taste? Which one would you choose?

It seems to me that "Noetherian" when it is naming something, while "noetherian" is used as an adjective.  I could be wrong.

Here is a previous discussion of the same subject:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=181478
(I found this by searching for the word "capitalization.")


LaTeX
SFFT
AMC
AIME
analytic geometry
MATHCOUNTS
AIME I

I run [b]AGMath.com[/b], a site geared towards advanced middle school instruction and competition preparation at the middle-school level ... hopefully many of you have visited the site already.  If you haven't, I hope you will find that there is an enormous amount of useful competition preparation material available there.

This year I am sharing some of the competition preparation materials I use with anyone interested in challenging problem solving in the first annual [b]AGMath Challenge[/b].

Information on how to participate can be found on the newly created page of my site [url=http://www.agmath.com/57427/223285.html]here[/url].

If you would like to "sign-up" in this thread, I suspect it will help spur interest and participation if you post your interest here ... anyone is welcome as long as they are currently in 8th grade or below.

Thanks, and good luck!

Friday the 13th? Seems to contradict the "good luck" :)

On the website, it's probably better to link to [url=http://www.artofproblemsolving.com/Forum/privmsg.php?mode=post&u=50143]here[/url] rather than to your user profile.

You do not need to sign up ... send in answers after each set is posted and I will keep track of results for anyone who submits answers.

With that said, I would like to have an idea of how many are interested, so please post a quick "I'm interested" in this thread if you plan on completing the problem sets.

Thanks.

If I have time, I'll definitely be doing this (and I think I can - Thomas the Engine said that means I really can!) and it sounds great. Are there prizes?

I'm interested too. =)

@izzy, probably not. For one, what would it be and how would he get it to you?

[quote="isabella2296"]Are there prizes?[/quote]

No prizes ... I suppose I could offer a copy of my book to the overall winner, but chances are the overall winner will know most of the math in my book... maybe I will offer a random drawing ... and LaTeX, you will of course be participating =)

I'm definitely interested.
I'm in eight grade and I'm going to take the AMC8.
My average so far of 4 practice tests are 22 out of 25 tho :(

[quote="AIME15"]@15!!! ... see this:

[quote="jbatterson"] ... anyone is welcome as long as they are currently in 8th grade or below.[/quote][/quote]

So? i dont get it. Do you know for a fact that 15 is older than gr.8?
Just wondering

[quote="pi31415926"]Do you know for a fact that 15 is older than gr.8?[/quote]

Yes.

[quote="CDKing129"]Oh darn, i just heard of this. Well idc i will do Set 5.

By the way you have set amazing questions for the first 4 sets! Especially set 4.[/quote]

Thanks!  They were fun to write.

[quote="The Hobbit"]When is set 5 due again?[/quote]

Not until January 8th ... I allowed plenty of time because of the holiday, and so that schools/math teams can give it if they want the week after break.

Probably, even though you are 4 rounds late!  :wink:

[quote="KingSmasher3"]May I join?[/quote]

Anyone who wants can do the final round ... up until January 8th (of course, you probably won't make it onto the cumulative rankings ... but results are also posted for individual rounds).

[quote="CDKing129"][quote="Thunder365"]Alternitavely, Stewarts Theorem kills #6[/quote]

I dont see how it "kills" it. Since if it killed it then it would solve it immediately, yet is too much computation, especially in a 20 minute time.

I prefer coordinates, as said by AIME, i think that "kills" it.[/quote]

The length of median formula (derived by Stewart's) has not much computation - probably is shorter than coordinates.  Unfortunately, I forgot the formula...

Round 5 is due on Friday.  Thanks to all who participated.  Feedback is still welcome, and I hope everyone enjoyed these.  More to come ... eventually.

[quote="CRICKET229"]jbatterson do you have any other mathcounts resources? (im from charlotte btw heard that you teach @ ligon)[/quote]

Look on his website.  :)

[quote="The Hobbit"]What is going to happen after Friday?[/quote]

I don't have any more challenge sets written (and participation seems to have dropped way off on this last set, probably due to the long break).  I will post final results and round 5 results ... I believe I also promised in this forum to randomly draw a name of a student who participated in all five challenges to receive a copy of my book.  I will do that as well.  I have mostly completed a mock state+ level sprint that I will likely post in March for anyone interested.  I'll announce it in this forum when I decide.

[quote="professordad"]Will your students be included in the drawing?    [hide=" :) "] Interesting. [/hide][/quote]

I just drew the name ... I didn't include my students or anyone who finished in the top 20 (my book would be far less useful to them).  Jennie Zheng from CA won (I don't know if she is an AoPS'er).

I have a few corrections to make to the results but they are posted along with a very quickly typed-up set of solutions for the very difficult challenge 5.

Thanks to all who participated.  I am in the process of writing up a mock Sprint (and probably a Target and Team round).  A few of the top national coaches and I should be collaborating on it, giving it to our kids, then posting it for anyone else to try ... State+ level, high-quality (though I am not happy with what I have now, I will polish it up)  Expect to see it in March on my site.

Results, etc. [url=http://www.agmath.com/57427/223285.html]HERE[/url]


we are all familiar with the problem about 3 monks and 3 cannibalists, with a 2-seat raft. Then a question is eventually proposed, what is the least number of seats that the raft can have so that the solution for the problem in the general case is capable : n monks and n cannibalists

Four seats are enough for any number of monks and cannibals;  just send two monks and two cannibals over, and send a monk and a cannibal back.  Repeat the process until everyone is across.

A one-seat canoe obviously can't do anything.  Two seats can handle at most three of each.  Suppose there are four or more of each type.  The first time monks are transported across the river, there are at most two monks across, and the same number of cannibals.  Either we transport the monks back, or we send one monk and one cannibal back.  The first just reverses the process.   The second leaves us with at least three monks and three cannibals on the near side;  thus we can only transport one monk and one cannibal across.  This reverses the last trip again, so we get nowhere.

A three-seat canoe can transport at most 5 people, by almost the exact same reasoning as for the two-seat case.  For $n = 5$, send three cannibals across, send one back, send two across, send one back, send three monks across, send a monk and a cannibal back, send three monks across, send a cannibal back, send three cannibals across, send one back, send two cannibals across.


Let $ f: \mathbb{N} \to \mathbb{N}$ be a function such that $ f(\mathbb{N})$ is infinite.Prove that there is a partition $ (A_n)_{n \in \mathbb{N}}$ of $ \mathbb{N}$ with the following properties : i) $ \forall n \in \mathbb{N}$, $ A_n$ is infinite and ii) $ \forall n \in \mathbb{N}$, $ f |_{A_n}$ (f restricted to $ A_n$) is strictly increasing.

I need a hint one this one, I want to solve it on my own.

Consider $ A_i$ the set of numbers who contain in their factorization $ i$ primes. For example, the set $ A_3$ contains the elements: $ 8\equal{}2 \cdot 2 \cdot 2$, $ 12\equal{}2\cdot 2 \cdot 3$ ,...

The sets $ A_1,A_2,A_3,...$ are infinite. Because $ 1$ and $ 0$ are not primes, WLOG suppose that they are both in $ A_1$.

Take the function $ f(x)\equal{}x$.

The problem  is not to find $ f$. The function is given.

You may see solve the problem thinking at pile of stones that you are building :
$ 1$ is in Pile n°$ 1$.
If $ f(2) \leq f(1)$ the it goes to pile n°$ 2$, otherwise, it goes to Pile n°$ 1$ then $ 3$ goes to Pile n°$ 2$
Then again Pile n°1 then Pile n°2 then Pile n°3,
and so on...
according to the fact that $ f(k)$ is greater than the previous values or not...

Pierre.

Yes. I didn't know that the function was given. I would like to see a complete solution.  :D

There is better way to write a solution than using piles :wink:

Since we have an infinite set of natural numbers $ f(x)$, for any $ x_1$, we may find an infinite increasing sequence $ x_1,x_2,x_3, \cdots$ such that $ f(x_1),f(x_2),f(x_3), \cdots$ is also increasing (for each $ n$, since the set of values $ f(x)$ such that $ f(x) > f(x_{n\minus{}1})$ is nonempty, just consider $ x_n$ be the least integer greater than $ x_{n\minus{}1}$ such that $ f(x_n) > (f(x_{n\minus{}1})$).

Now, let's consider such a sequence with $ x_1 \equal{} 1$, and define $ A_1 \equal{} \{x_{2n\plus{}1} /n \geq 0\}$. This set is clearly infinite and $ f$ is increasing on $ A_1$.
Note that, since we only used "half" of the $ x_i$'s, both sets $ \mathbb{N} \minus{}A_1$ and $ f( \mathbb{N}) \minus{} f(A_1)$ are infinite.
Let $ y_1$ be the least element in $ \mathbb{N} \minus{}A_1$, and consider a sequence $ y_1,y_2,\cdots$ as above. Then, let $ A_2 \equal{}  \{y_{2n\plus{}1} /n \geq 0\}$.
And so on...

Clearly, the sets $ A_1,A_2, \cdots$ are infinite and pairwise disjoints. Since the "first" element of $ A_n$ is the least natural number which is not in $ A_1 \cup \cdots \cup A_{n\minus{}1}$, every natural number belongs to one of the $ A_i$'s. Therefore, the $ A_i$'s form a partition of $ \mathbb{N}$.
And $ f$ is increasing on each $ A_i$.

Pierre.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Suppose that a real square matrix A has minimal polynomial pa of degree m
Prove that A is inversible if only if pa has nono constant term. In this case prove that A^-1 may be written as polynomial of degree m-1  in A

if the constant term is null  if and only if det(A)=P_A(0)=0
In this case -P(On)=P(A)-P(On)=A*Q(A) where Q is a polynom , it's follows that A-1 is
 Q(A)/(-P(0))

Suppose $A^k+c_{k-1}A^{k-1}+\cdots+c_1A+c_0I=0$, where $p(x)=x^k+c_{k-1}x^{k-1}+\cdots+c_1x+c_0$ is the minimum polynomial of $A$.
If $c_0\neq0$, we have $A^{-1}=\frac1{c_0}\left(A^{k-1}+c_{k-1}A^{k-2}+\cdots+c_2A+c_1I\right)$.
If $c_0=0$, then since $p$ divides the characteristic polynomial by Cayley-Hamilton, the characteristic polynomial is zero at zero. This means that zero is an eigenvalue, and $A$ is not invertible.


function
logarithms
floor function
continued fraction
real analysis
real analysis unsolved

Let $ f : \mathbb{N} \rightarrow \mathbb{R}$ satisfy the following conditiona:

a) $ f(mn) \equal{} f(m) \plus{} f(n)$
b)$ f(n \plus{} 1) \minus{} f(n) \rightarrow 0$ as $ n \rightarrow \infty$

Prove that $ f(n) \equal{} c\cdot log \, n$.

[quote="michalkotowski"]Let $ f : \mathbb{N} \rightarrow \mathbb{N}$ satisfy the following conditiona:

a) $ f(mn) \equal{} f(m) \plus{} f(n)$
b)$ f(n \plus{} 1) \minus{} f(n) \rightarrow 0$ as $ n \rightarrow \infty$

Prove that $ f(n) \equal{} c\cdot log \, n$.[/quote]

Be careful, at least one of $ f(2)$ and $ f(3)$ is not well-defined.

We can extend $ f$ to a function $ g : \mathbb{Q^\plus{}} \to \mathbb{R}$ given by $ g(m/n) \equal{} f(m) \minus{} f(n)$.

For each integer $ m > 0$, we have that $ f(n\plus{}m) \minus{} f(n) \to 0$ as $ n \to \infty$. This means $ g(1 \plus{} m/n) \to 0$ as $ n \to \infty$. Therefore $ g$ is continuous at $ 1$. Then $ g(p\plus{}q) \minus{} g(q) \equal{} g(1 \plus{} p/q) \to 0$ as $ q \to 0$, so $ g$ is continuous at all rational $ p$. Therefore $ g$ admits a unique extension to a continuous function $ h : \mathbb{R} \to \mathbb{R}$. We have that $ h(e^{x\plus{}y}) \equal{} h(e^x) \plus{} h(e^y)$, so $ h(e^x)$ is a continuous solution to the Cauchy equation; we have then that $ h(e^x) \equal{} Cx$ for some constant $ C$, and so $ h(y) \equal{} C\log y$.

That's a nice argument. One additional detail:
We need to show that this extension $ g$ is well-defined; if $ \frac{m}{n}\equal{}\frac{m'}{n'}$, $ f(m)\minus{}f(n)\equal{}f(m')\minus{}f(n')$. The fractions are equal when $ m\cdot n'\equal{}m'\cdot n$, so $ f(m)\plus{}f(n')\equal{}f(m\cdot n')\equal{}f(m'\cdot n)\equal{}f(m')\plus{}f(n)$. Subtract $ f(n)\plus{}f(n')$ from both sides, and we get $ f(m)\minus{}f(n)\equal{}f(m)\minus{}f(n')$ as desired.

Oops I don't think my proof of continuity (of $ g$) is correct! Can someone fill it in?

You should have written $ g(p\plus{}q)\minus{}g(p) \rightarrow 0$ instead of $ g(p\plus{}q)\minus{}g(q) \rightarrow 0$. Very cool trick, thanks!

No, that's not the issue- it's that the numerators of fractions near zero aren't bounded. There's no easy fix.

I'll come back around with another approach. Let $ M\equal{}\sup_n f(n\plus{}1)\minus{}f(n)$.
For $ n>m$, if $ m$ and $ n$ are both odd,
$ |f(n)\minus{}f(m)|\le |f(n)\minus{}f(n\minus{}1)|\plus{}|f(m\plus{}1)\minus{}f(m)|\plus{}\left|f\left(2\cdot \frac{n\minus{}1}{2}\right)\minus{}f\left(2\cdot \frac{m\plus{}1}{2}\right)\right|$
$ \le 2M\plus{}\left|f\left(\frac{n\minus{}1}{2}\right)\minus{}f\left(\frac{m\plus{}1}{2}\right)\right|$
If either $ m$ or $ n$ is even, we have a similar expression; one or both of the first two terms is missing, and we still have that halving.

Iterating this process, $ |f(n)\minus{}f(m)|\le 2M\log_2|n\minus{}m|\plus{}M$. For the remainder of this proof, let "$ \log$" refer to the base 2 logarithm.

Now, let $ f(2)\equal{}a$; we immediately get $ f(2^k)\equal{}a\log(2^k)$. Choose some other positive integer $ n$, and let $ f(n)\equal{}b\log n$. Take powers of $ n$; $ f(n^i)\equal{}b\log (n^i)$. We can compare these to nearby powers of $ 2$; if $ \lfloor \log n^i\rfloor \equal{}j$, we have
$ |f(n^i)\minus{}f(2^j)|\equal{}|b\log(n^i)\minus{}a\log(2^j)|$
$ \ge |b\log(2^j)\minus{}a\log(2^j)|\minus{}|b\log(2^j)\minus{}b\log(n^i)|\ge j|b\minus{}a|\minus{}|b|$
On the other hand, we have the estimate above; $ |f(n^i)\minus{}f(2^j)|\le 2M\log(n^i\minus{}2^j)\plus{}M$. Actually, since dividing a power of $ 2$ by $ 2$ repeatedly will never lead to an odd number before we bring th difference down to $ 1$, we can improve the estimate to $ |f(n^i)\minus{}f(2^j)|\le M\log(n^i\minus{}2^j)\plus{}M$. How big is this? At worst, $ \log(n^i\minus{}2^j)<j$, since $ n^i<2^{j\plus{}1}$. Can we have $ j|b\minus{}a|\minus{}|b|<M(j\plus{}1)$ every time even if $ b\neq a$? Easily.
But then, that's the worst case. Sometimes, the powers are much closer together. Since $ \log n$ is irrational, we can find $ i$ making the fractional part $ \log(n^i)\minus{}j$ arbitrarily small. That makes $ \log(n^i\minus{}2^j)$ smaller than $ j\minus{}R$ for some constant $ R$. Uh, not good enough.
OK, how about a tighter estimate? If we draw from the continued fraction expansion of $ \log n$, we can find infinitely many pairs $ p,q$ such that $ 0<\log n\minus{}\frac pq<\frac1{q^2}$. That gives $ \log(n^q)\minus{}p<\frac1q$, so $ n^q\minus{}2^p\equal{}2^p\cdot (2^{\log(n^q)\minus{}p}\minus{}1)<2^p\cdot (2^{\frac1q}\minus{}1)$. Take a logarithm, and that looks like $ p\minus{}c\log p$. Not close to good enough as $ p$ and $ q$ grow.

OK, that didn't work. This question is [i]hard[/i].

[quote="jmerry"]
OK, that didn't work. This question is [i]hard[/i].[/quote]
Not in this version. Let $ n$ be an integer. Consider the numbers $ n_k\equal{}\left[\frac n{2^k}\right]$ with $ k\le K\equal{}[\log_2 n]$. We have $ |n_k\minus{}2n_{k\plus{}1}|\le 1$, so $ f(n_k)\approx f(2)\plus{}f(n_{k\plus{}1})$ with an additive error that is small when $ n_k$ is large. Thus, $ f(n)\equal{}Kf(2)\plus{}o(K)\equal{}f(2)\log_2n\plus{}o(\log_2n)$ as $ n\to\infty$. Now just apply it to $ n^m$ instead of $ n$ to get
\[ f(n)\equal{}\frac 1m f(n^m)\equal{}\frac 1m(f(2)m\log_2 n\plus{}o(m\log n))\equal{}f(2)\log_2 n\plus{}o(1)\]
as $ m\to\infty$.

So, let's try to figure out if the conclusion remains true if the sequence $ f(n)\minus{}f(n\plus{}1)$ is assumed just bounded. In this case the only thing we get immediately from the above argument is $ f(n)\equal{}O(\log n)$. ;)

in the proof of Nukular, even the continuity argument at $ 1$ isn't correct ! I'm I right ?

Yes, that is exactly where the problem was. You can have $ 1\plus{}\frac pq\to 1$ with both $ p,q\to\infty$.


geometry
graph theory
combinatorics theorems
combinatorics

hello to all mathlinkers. please recommend me a good reference for graph theory. thanks.

[url=http://www.math.uni-hamburg.de/home/diestel/books/graph.theory/GraphTheoryIII.counted.pdf]Diestel[/url] (also available in book form, but the electronic edition is a nice touch).

[url=http://www.amazon.com/Modern-Graph-Theory-Bela-Bollobas/dp/0387984887]Bollobas[/url]

Unless you're interested in more specialized topics such as algebraic graph theory or random graphs, this should cover most reference needs.

You can also find some combinatorics and graph theory books [url=http://kolho3.tiera.ru/#50]here.[/url]

[url=http://www.ecp6.jussieu.fr/pageperso/bondy/books/gtwa/gtwa.html]A[/url] must-read...

Regards,

How useful is graph theory for Olympiad math? Is it one of the more useful areas of Combinatorics, or should I study other areas first?

[quote]Diestel (also available in book form, but the electronic edition is a nice touch).[/quote]

Hey, that's just the first page of the book. Can anyone provide the link to the remaining part? Thanks in advance.

This is the link for Diestel's 3rd edition...
[url]http://www.math.sc.edu/~cooper/math776/diestel.pdf[/url]

I know it is a naivve question but can anyone tell me how much of graph theory is required for the Olympiads?