Hi,

I dunno if this is the right place to ask.

I have an equation:

[quote]

xmax = 0.06 [u](ho^4/3)[/u]
......................0.015

[/quote]

I want to rearrange so i can find out ho e.g ho = etc etc

If you could provide a solution I would be very grateful.

hello, if you mean
$ x_{max}\equal{}\frac{0.06h^{\frac{4}{3}}}{0.015}$ then you will get
$ h_0\equal{}\left(\frac{x_{max}}{4}\right)^{\frac{3}{4}}$
Sonnhard.

Can someone move this to high school basics?

OP: You have $ x_{\max}\equal{}\frac{0.06h_0^{\frac{4}{3}}}{0.015}$. Since $ \frac{0.06}{0.015}\equal{}4$, this is $ x_{\max}\equal{}4h_0^{\frac{4}{3}}$. Isolate $ h_0$ by dividing by 4: $ \frac{x_{\max}}{4}\equal{}h_0^{\frac{4}{3}}$, then take the 3/4'th power of both sides: $ \left(\frac{x_{\max}}{4}\right)^{\frac{3}{4}}\equal{}h_0$.


Let $a,b,c>0; a+b+c=1$, and let $n\in\mathbb{N}$. Determine the larget value of $T_{n}$ satisfying:

$\sum_{cyc}\frac{1}{a^{n}+b^{n}}\geq T_{n}$

(inspired by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=148506]this one[/url])

We have

       T1=9/2,Tn=5*2^(n-1) (n>=2).
See:

http://www.irgoc.org/bbs/dispbbs.asp?boardid=19&id=2396&star=1#2396


The developer of the server was probably making some changes.


Find all positive integrer pairs $(a,n)$ such that $\frac{(a+1)^n-a^n}{n}$ is an integrer.

Already posted: (just few topics below)

http://www.mathlinks.ro/Forum/viewtopic.php?t=80728


geometry
circumcircle
perpendicular bisector

The point $ P$ is a fixed point on a circle and $ Q$ is a fixed point on a line. The point $ R$ is a variable point on the circle such that $ P,Q,R$ are not collinear. The circumcircle of triangle $ PQR$ meets the lines again at $ V$. Show that the line $ VR$ passes through a fixed point.

[hide="I think I misunderstood it"]
...but isn't V the same point as Q?

[/hide]

Circles and lines typically intersect in 2 points.  The phrase "meets again" is intended to indicate that you take the other point of intersection.  Also, there was really no reason to put that text between "hide" brackets.

[hide="here goes nothing"]
Draw the perpendicular bisector of PQ, the center of the circumcircle of PQR will always be on this line.  Then draw an arbitrary R and the circumcenter of PQR, call it O.  Because R and V are circle O, OR=OV
Not sure where to go from there..

Alternatively you could draw a random R and R1, find VR and VR1, and find the intersection  :rotfl:

[/hide]

[quote="CircleSquared"]Alternatively you could draw a random R and R1, find VR and VR1, and find the intersection.[/quote]

That would be a good way to figure out what the point is.  After you've done that, you have to prove that it works.

[hide="Answer without proof"]Suppose the line $ PQ$ intersects the given circle again at $ S$.  (If $ PQ$ is tangent to the circle, take $ S \equal{} P$.)  Draw a line through $ S$, parallel to the given line.  This line intersects the given circle at $ T$, which is the point you're looking for.[/hide]


THREE PARALLEL LINES L1,L2 , L3  ARE DRAWN THROUGH THE VERTICES  A , B, C  OF A SQUARE   ABCD.IF THE DISTANCE BETWEEN L1 AND L2 IS 7 AND L2 ,L3 IS 12,THEN AREA OF ABCD IS

as notation of leonxlin:

if $ ABX \equal{} \alpha$ then $ AB \cos{\alpha} \equal{} 12$ and $ BC \sin{\alpha} \equal{} AB \sin{\alpha} \equal{} 7$; then $ S \equal{} AB^2 \equal{} (AB \cos{\alpha})^2 \plus{} (AB \sin{\alpha})^2$.

bye

Draw the perpendicular of $ l_1,l_2,l_3$ through $ B$ and call it $ l_4$. Let $ l_1 \cap l_4 \equal{} X$, $ l_3 \cap l_4 \equal{} Y$, $ l_2 \cap AD \equal{} Z$.

Both being sides of the square, $ AB \equal{} BC$.
Also $ 90^{\circ} \equal{} \angle AXB \equal{} \angle BYC$.

$ 90^{\circ} \equal{} \angle ZBA \plus{} \angle ZBC \equal{} \angle CBY \plus{} \angle ZBC \Rightarrow \angle ABZ \equal{} \angle CBY$ $ \angle BAX \equal{} \angle ABZ$ (alternate interior angles), so $ \angle BAX \equal{} \angle CBY$.

So $ \triangle ABX \cong BCY$ by AAS. It was given that $ XB \equal{} 7, BY \equal{} 12$. So $ AX\equal{}12$. So the side of the square is $ \sqrt{7^2 \plus{} 12^2}$ and the area is $ 7^2 \plus{} 12^2 \equal{} \boxed{193}$.

Lol school geometry style proof.


Is a generalization of Cauchy Swcahrz in n variables instead of 2.

Generalized Hölder...

You know you shouldn't ask this kind of question in the inequalities unsolved problems section!


1)Find an integer n such that n^2-2  has exactly 10 positive divisors
2) Find an integer n such that n^2-3  has exactly 10 positive divisors

Well I suppose you know the solutions structure only can be in the following way:

n^2-2=A^9,   or  n^2-2=AB^4   where A and B are differents primes, so you only need to find A and B.

same with n^2-3 .

Good Luck and best regards.

For case n^2-2    take   n=235   (A=23) (B=7),   and
For case n^2-3    take   n=4936 (A=13) (B=37).

RX the Math

But how did you find them? By brute force?

Pierre.

No, using congruences in a correct way! think in this, congruences into 3, 5 and 7.


Hi. I would like to know where can i find all the special factorization cases such as :

$ x^{4}\plus{}x^{3}\plus{}x^{2}\plus{}x\plus{}1\equal{}\frac{x^{5}\minus{}1}{x\minus{}1}$

$ a^{8}\plus{}a^{4}\plus{}1 \equal{} (a^{2}\minus{}a\plus{}1)(a^{2}\plus{}a\plus{}1)(a^{4}\minus{}a^{2}\plus{}1)$

I think knowing those cases could be very useful. Someone knows where to find them?

...or does the books of AoPS provide those formulas?

What are these special cases of?  There are literally countless factorization methods so no book could contain all of them.

The first one on your list is just a truncated form for the sum of a geometric series.  [url=http://en.wikipedia.org/wiki/Factorization]Here[/url] is a list of some factorization formulas.

You are right, now i see the case of the first one......but i have never seen before the case of the second one. What would be a generalization for making a polynomial such as that into 3 factors?

A common technique for factoring problems such as your second one is to write as a difference of squares.

For example, $ a^8 \plus{} a^4 \plus{} 1 \equal{} (a^4 \plus{} 1)^2 \minus{} a^4 \equal{} (a^4 \plus{} a^2 \plus{} 1)(a^4 \minus{} a^2 \plus{} 1).$

Applying the same method again to the first term in the parentheses gives $ a^4 \plus{} a^2 \plus{} 1 \equal{} (a^2 \plus{} 1)^2 \minus{} a^2 \equal{} (a^2 \plus{} a \plus{} 1)(a^2 \minus{} a \plus{} 1)$.

Thus, $ a^8 \plus{} a^4 \plus{} 1 \equal{} (a^2 \minus{} a \plus{} 1)(a^2 \plus{} a \plus{} 1)(a^4 \minus{} a^2 \plus{} 1)$, as you said.

Try using the same reasoning on $ a^4 \plus{} 4b^4$, which is the well known [url=http://www.artofproblemsolving.com/Wiki/index.php/Sophie_Germain_Identity]Sophie Germain Identity[/url].

:o  :o  :o  Thank you xpmath. Now i see.

Very kind xpmath =D


Putnam
analytic geometry
combinatorial geometry
college contests

Given $ n\geqslant 3$ distinct points in the plane, no three of which are on the same straight line, prove that there exists a simple closed polygon with these points as vertices.

You also need to pick it so it doesn't lie on any line passing through two of them.

[quote="MellowMelon"]Pick an arbitrary point in the convex hull of the $ n$ points, not equal to any of them. Write each point in polar coordinates with some arbitrary axis being $ \theta \equal{} 0$. Then order the points by their $ \theta$ coordinate value, and connect them in that order.[/quote]

can I just write on putnam what you just wrote and get points for it?


calculate the sum of all the irreducible fractions of the form
a/1991 with a integer and 0 <a/1991 <1991

[hide="If I got it right..."]

We should compute $ \frac {1}{1991} \plus{} \frac {2}{1991} \plus{} ... \plus{} \frac {1991^2 \minus{} 1}{1991}$?
$ \sum_{i \equal{} 1}^{1991^2 \minus{} 1} \frac {i}{1991} \equal{} \frac {1 \plus{} 2 \plus{} 3 \plus{} ... \plus{} (1991^2 \minus{} 1)}{1991} \equal{} \frac {1991^2\cdot(1991^2 \minus{} 1)}{2\cdot 1991} \equal{} 3946241640$

EDIT: Now from that sum we subtract the other sum which contains from $ a$-s such that $ 11\mid a$ or $ 181\mid a$ and then add sum of all fractions with $ a$-s such that $ 1991\mid a$. I won't write now the sum.
[/hide]


[hide] This is the [url=http://mathworld.wolfram.com/MultinomialCoefficient.html]multinomial coefficient[/url] $ {n^2 \choose n, n, n, ... }$. [/hide]

We can use the easily proved result that the product of n consecutive numbers is always divisible by n!

$ n^2!$ is the product of $ n^2$ consecutive numbers i.e. n sets of n consecutive numbers each set divisible by n!

Hence $ n^2!$ is divisible by $ n!^n$ or $ \frac{n^2!}{n!^n}$ is an integer

[quote="RisingFlame"]Prove that  $ \dfrac{n^2!}{(n!)^{n}}$ has an integer value, where $ n$ is any whole number.[/quote]
It is enough to prove that for any prime number $ p$,the following inequality holds:
$ \sum_{k \equal{} 1}^{\infty}[\frac {n^2}{p^{k}}]\geq n(\sum_{k \equal{} 1}^{\infty}[\frac {n}{p^{k}}])$.
It follows from adding up the following inequalities:
$ [\frac {n^2}{d}]\geq [\frac {n}{d}] \plus{} [\frac {n}{d}] \plus{} \dots \plus{} [\frac {n}{d}]$,for all $ d\equal{}p,p^2,p^{3}\dots$.
The last sums RHS contains exaclty $ n$ summands.

[quote="Johan Gunardi"][quote="hsbhatt"]$ n^2!$ is the product of $ n^2$ consecutive numbers i.e. n sets of n consecutive numbers each set divisible by n![/quote]I don't understand this part. :huh:  Can you explain it more clearly?[/quote]

CI = consecutive integers

$ \underbrace{1 \cdot 2 \cdot \ldots n^2}_{n^2 \text { CI}} \equal{} \underbrace{\underbrace{1 \cdot 2 \cdot \ldots n}_{\text {n CI}} \cdot \underbrace{(n \minus{} 1) \cdot (n \minus{} 2) \cdot \ldots 2n}_{\text {n CI}} \cdot \ldots \cdot \underbrace{(n^2 \minus{} n \plus{} 1) \cdot (n^2 \minus{} n \plus{} 2) \cdot \ldots n^2}_{\text {n CI}}}_{\text{n sets}}$
Therefore, $ (n^2)!$ can be divided into the product of $ n$ sub-products of $ n$ consecutive integers each.

[b]RisingFlame's claim:[/b] Each of these $ n$ sub-products is divisible by $ n!$

[b]Proof:[/b] $ (k\plus{}1) \cdot (k\plus{}2) \cdot \ldots \cdot (k\plus{}n) \equal{} \frac{(k\plus{}n)!}{k!} \equal{} \frac{(k\plus{}n)!}{k! n!} n! \equal{} \binom{k\plus{}n}{n} n!$, which is a multiple of $ n!$ for all $ k$.

In this case, let $ k\equal{}(m\minus{}1)n$ for some $ m \in \{1,2, \ldots n \}$, thus, the entire product must be divisible by $ (n!)^n$, since there are $ n$ sub-products.


conics
ellipse
advanced fields
advanced fields unsolved

A variable tangent to an ellipse intersects the tangents from vertices A and A' in T and T'.Show that the product AT*TT' is constant and that the segment TT' is seen from the focal points under right angles.

This looks sort of unclear to me: are A and A' on the ellipse? It seems unlikely (in case the answer to the question above is "yes"), that the angle TFT' is always [tex]\frac\pi2[/tex].


limit
integration
trigonometry
logarithms
calculus
calculus computations

Can you say how I can evaluate   $ \lim_{n\rightarrow\infty}n\int^{\pi/2}_{1}(1 \minus{} \sqrt [n]{\sin x})\,dx$

Just expand the integrand in $ n$ around $ \infty$ to find $ 1 \minus{} \sin(x)^\frac {1}{n} \equal{} \minus{} \frac {\log \sin x}{n} \plus{} \mathcal{O}\left(n^{ \minus{} 2}\right)$ and so,
\[ \lim_{n\to \infty}n\int_1^{\frac {\pi}{2}}1 \minus{} \sin (x)^\frac {1}{n}dx \equal{} \minus{} \int_1^\frac {\pi}{2}\log \sin x dx\]
which I'm not sure can be expressed in terms of some common constants, but it doesn't involve $ n$ so that's your answer.

Edit: Mathematica seems to evaluate the integral by parts and uses the sum form of the dilogarithm to come up with the value $ \frac {(\pi \minus{} 2) \log 2 \minus{} \sum_{k \equal{} 1}^\infty \sin (2k)k^{ \minus{} 2}}{2}$.


Each the numbers $ 1, 2, ..., 46$ is colored with one of the red, yellow or green. Prove that exist two numbers $ a > b$ are same colors and $ a \minus{} b$ is a perfect square number.

If the problem statement is false, there is a coloring where the difference of two numbers of the same color is never a perfect square.
In such a coloring, numbers $ k, k\plus{}9, k\plus{}25$ ($ k \leq 21$) must be pairwise different, since their pairwise differences are $ 9$, $ 16$ and $ 25$.
Same thing happens with numbers $ k, k\plus{}16, k\plus{}25$. Hence, numbers $ k\plus{}9$ and $ k\plus{}16$ must have the same color for all $ k \leq 21$.

W.l.o.g. we may suppose that number $ 10$ is green, $ 19$ is red and $ 35$ is yellow. From that, we already know that numbers

$ 10, 17, 24, 31, 38, 45$ are green;
$ 12, 19, 26, 33, 40$ are red;
$ 14, 21, 28, 35, 42$ are yellow.

Number $ 46$ can't be green nor yellow since $ 46\minus{}10\equal{}6^2$, $ 46\minus{}42 \equal{} 2^2$. Hence, $ 46$ is red.
Number $ 30$ can't be green nor yellow nor red since $ 31\minus{}30 \equal{} 1^2, 30\minus{}21\equal{}3^2$, $ 46\minus{}30 \equal{} 4^2$.

A contradiction $ \rightarrow$ there is no such coloring, and the problem statement is true.


Let ABC be a triangle, AB=c, BC=a, CA=b. Prove that
$ \displaystyle  \frac{MA}{a} + \frac{MB}{b} + \frac{MC}{c} \ge$ $ \sqrt{3} $
Have you got any idea?  :)

You can use The flowing inequality $ a\plus{}b\plus{}c \ge \sqrt{3(ab\plus{}bc\plus{}ca)}$ But I'm not sure since I don't know if $ M$ is any or not  :(

[quote]But I'm not sure since I don't know if M is any or not[/quote]
M is a point in the plane (ABC).
And I wonder: In some ways, is this problem related to the problem of Torricelli point  :?:


If $ U \equal{} [u_{ij}] \in \mathbb{C}^{n \times n}$ is a unitary matrix, show that

$ \sqrt {n} \leq \sum_{i,j} |u_{ij}| \leq \sqrt {n^3}$.

Well, we have that $ \sum_{i,j}|u_{ij}|^2\equal{}n.$

We have two well-known inequalities:

(1) $ \sum_{k\equal{}1}^m|a_k|\le \sqrt{m}\left(\sum_{j\equal{}1}^m|a_k|^2\right)^{\frac12}$ (Cauchy-Schwarz)

(2) $ \sum_{k\equal{}1}^m|a_k|\le\left(\sum_{k\equal{}1}^m\sqrt{|a_k|}\right)^2$

To prove (2), normalize to the right hand side being equal to $ 1$ and then use that $ |a_k|\le\sqrt{|a_k|}$ for $ |a_k|\le 1.$

Since then number of terms is $ m\equal{}n^2,$ the right hand inequality follows from (1) and the left hand inequality from (2).

The left-side inequality is weaker than it has to be; we can get $ \sum_{i,j}|u_{ij}|\ge n$, since $ \sum_{i}|u_{ij}|\ge \left(\sum_i \sqrt{|u_{ij}|}\right)^2\ge \left(\sum_i |u_{ij}|\right)^2\ge \sum_i |u_{ij}|^2\equal{}1.$
Equality in that improved version is achieved when $ U$ is the identity. Equality in the other half is achieved when $ U$ is a multiple of the Fourier matrix $ u_{jk}\equal{}\frac1{\sqrt{n}}e^{2\pi i\cdot jk/n}$

[quote="jmerry"]$ \sum_{i}|u_{ij}|\ge \left(\sum_i \sqrt {|u_{ij}|}\right)^2$[/quote]  I don't think so.

Oops. Let's fix that:

$ \sum_i |u_{ij}|\equal{}\sqrt{\left(\sum_i |u_{ij}|\right)^2}\ge \sqrt{\sum_i |u_{ij}|^2}\equal{}1$
Sum over $ j$, and $ \sum_{i,j}|u_{ij}|\ge n$.

I was trying to apply Kent's #2 to the wrong quantities.


Two different numbers are chosen from the sequence of consecutive odd integers 51, 53, 55, …,95, 97, 99. Which of the following could not be their sum?

a. 161 b. 168 c. 186 d. 196 e. NG

The list of consecutive odd integers can be written as $ 2n \plus{} 1$ for $ 25 \le n \le 49$. Hence, the sum of two numbers from this list are in the form $ 2(n \plus{} k) \plus{} 2$, where $ n, k$ both satisfy the restrictions listed above. When we set \[ 2(n \plus{} k) \plus{} 2 \equal{} 161\] it simplifies to \[ 2(n \plus{} k) \equal{} 159\] However, 159 isn't divisible by 2, so $ n \plus{} k$ can't be an integer. But we know $ n, k$ are both integers, so clearly this doesn't work for 161. Therefore, the answer is $ \boxed{\textbf{A}}$.


vector
linear algebra
matrix
linear algebra unsolved

Let $ M_n(\mathbb{C})$ denote the vector space of complex $ nxn$ matrices .
Let $ A \equal{} \left[\begin{array}{ccccccc}0 & 1 & 0 & 0 & ... & 0 & 0 \\
0 & 0 & 1 & 0 & ... & 0 & 0 \\
0 & 0 & 0 & 1 & ... & 0 & 0 \\
... & ... & ... & ... & ... & ... & ... \\
0 & 0 & 0 & 0 & ... & 0 & 1 \\
0 & 0 & 0 & 0 & ... & 0 & 0\end{array}\right]$
and let the linear transformation $ T: M_{n}(\mathbb{C}) \to M_n{\mathbb{C}}$ be defined by $ T(X) \equal{} AX \minus{} XA$
Find the [b]Jordan Canonical [/b]for $ T$

Let $ U_k\equal{}\{M\equal{}[m_{ij}]\in\mathcal{M}_n; if \;i\minus{}j\geq{k}\; then\; m_{ij}\equal{}0\}$. $ U_n\equal{}\mathcal{M}_n$ and $ U_{\minus{}n}\equal{}\{0\}$. $ T(U_k)\subset{U_{k\minus{}1}}$ then $ T^{2n\minus{}1}\equal{}\{0\}$. Moreover if $ E\equal{}[e_{ij}]$ satisfies $ e_{ij}\equal{}0$ except $ e_{n,1}\equal{}1$ then $ T^{2n\minus{}1}E\not\equal{}0$. Then  the order of nilpotence of $ T$ is exactly $ 2n\minus{}1$. If $ J_k$ refers to the nilpotent Jordan matrix of dimension $ k$ then the Jordan decomposition of $ T$ is in the form $ diag(J_{2n\minus{}1},?????)$.
Conjecture: I think that the result is $ diag(J_{2n\minus{}1},J_{2n\minus{}3},\cdots,J_5,J_3,0)$.

Read in my preceding post $ T^{2n\minus{}2}E\not\equal{}{0}$. More precisely, $ T^{2n\minus{}2}E\equal{}F\equal{}[f_{ij}]$ where $ f_{ij}\equal{}0$ except $ f_{1,n}\equal{}(\minus{}1)^{n\plus{}1}(_{n\minus{}1}^{2n\minus{}2})$.


Any comments on the solution way and method of the following problem? Thanks and regards.
[url=http://imageshack.us][img]http://img90.imageshack.us/img90/1206/limit10nb4.gif[/img][/url]

one can immediately see that the highest power on top, 4/3, is greater than the highest on bottom, 1/2, and thus the limit must be infinity


does anyone have any MOP homework problems that are not from 2002?

I'm not sure that's allowed? Maybe...? :huh:


Is it open? Why? We need only use $ a^{n\plus{}1}\minus{}b^{n\plus{}1}\equal{}(a\minus{}b)\sum_{i\equal{}0}^na^ib^{n\minus{}i}$.


algebra
polynomial
calculus
integration
pigeonhole principle
Polynomials

Prove that there is no nonconstant polynomial $f(x)$ with integral coefficients such that $f(n)$ is prime for all $n \in \mathbb{N}$.

Consider the set $p_0=f(1)$, $p_k$=$f(kp_0+1)$. Using the fact that $a-b|f(a)-f(b)$ in $\mathbb{Z}[x]$, $kp_0|p_k-p_0$ so $p_0|p_k$.

If $f(n)$ is prime for all n, then $p_i$ is prime for all $i$, so $p_0=p_k$ for all k, but this implies that the polynomial $f(x)-p_0$ has an infinity of roots, contradicting the fact that it is a polynomial (which must have a finite degree).

Here is a solution that constructs a slightly different contradiction. Suppose FTSOC, $f(n)$ is prime for all $n\in\mathbb{Z}^{+}$. We know that $(a-b)|(f(a)-f(b))$ for $a,b\in\mathbb{Z}$. WLOG, let $b=1$ and let $f(b)=p$, $p$ a prime. Then we have $(a-1)|(f(a)-p)$. Now, take $a\in\{jp+1\}_{j=1}^{j=k}$. Then, we have $jp|(jq_{j}-p)$ for all $1\le j\le k$ for arbitrary primes $q_1, \ldots, q_k$. Since $q_1$ is prime and $p|(q_1-p)$, it follows that $q_1=\pm p$. Similarly, $q_j = \pm p$ for all $1\le j\le k$. Thus, by pigeonhole we get $f(x)=f(y)$ for every pair of integers $(x,y)$, which means $f$ is constant, contradiction.


This is a really cool (and pretty hard) problem.

[hide="Here are a number of words in Arabic, Hebrew, and Aramaic, respectively."]he wrote: kataba, kāθav, kθav
day: yawm, yōm, ???
they sacrificed: ???, ???, ???
horn: ???, ???, qarnā
three: θalāθ, šālōš, tlāθ
we noted: ???, rāšamnū, ršamn
he sacrificed: ðabaħa, zāvaħ, dvaħ
he noted: ???, ???, ???
bull: θawr, šōr, tawrā
eye: ʕayn, ???, ???
snow: θalĝ, šeleɣ, talgā
seventh: sābiʕ, švīʕī, švīʕāyā
peace: salām, šālōm, šlāmā
we sacrificed: ðabaħnā, ???, ???
tomb: qabr, ???, ???
thid: ???, šlīšī, ???
we wrote: ???, ???, ???[/hide]
1. Fill in the blanks
2. What do you suppose "he buried" is in each of these languages?

[hide="Hints on this type of problem"]Make tables of correspondences between letters. Watch for grammatical trends in verb conjugation, parts of speech, etc. There are 3-5 forms that are EXTREMELY hard to get, so be aware that you probably aren't going to get all of it.[/hide]

[quote]bull: θawr, šōr, tawrā [/quote]
Interesting: it's a completely different family of languages, but that sure looks the Latin/Romance word: taurus, toro, and so on.

Uh-oh, I missed one statement in the problem...
katabū, kāθvū, kθav - they wrote
My apologies. I think the problem is next to impossible without this.

[hide="Verbs"][u]Arabic verbs[/u]
kataba, ?, katabū - he, we, they wrote
ðabaħa, ðabaħnā, ? - he, we, they sacrificed
?, ?, ? - he, we, they noted

[u]Hebrew verbs[/u]
kāθav, ?, kāθvū - he, we, they wrote
zāvaħ, ?, ? - he, we, they sacrificed
?, rāšamnū, ? - he, we, they noted

[u]Aramaic verbs[/u]
kθav, ?, kθav - he, we, they wrote
dvaħ, ?, ? = he, we, they sacrificed
?, ršamn, ? = he, we, they noted

[u]Conclusions[/u]
Verbs are formed by changing vowels and suffixes with fixed consonants. In Arabic, we have -a-a-a, -a-a-nā, -a-a-ū. In Hebrew, -ā-a-, -ā-a-nū, -ā--ū. In Aramaic, --a-, --a-n, --a- (we see that ršamn actually has root ršam because the roots need to be three letters and -n seems to be a generic third person plural). This allows us to finish up everything except "he, we, they noted" in Arabic.[/hide]
[hide="Consonant correspondences"]k, y, m, l, r, ħ, ʕ constant
t - θ - θ
b - v - v
θ - š - t
θ - š - θ
ð - z - d
ĝ - ɣ - g
s - š - š

We also know that q exists in both Arabic and Aramaic. It also exists in Hebrew - the only ways I can think of to justify this are (a) know English words from Hebrew with q in them, (b) guess that because otherwise the problem would be impossible, or (c) know Hebrew or at least its alphabet. So q is also constant.
[/hide]
[hide="Random notes"]We don't know "he, we, they noted" in Arabic, but r is constant, š in both Hebrew and Aramaic comes from s, and m is constant, so we hypothesize the stem "rsm", from which we get the desired forms.

Almost all nouns/adjectives in Aramaic end in -ā, the one exception being "three". Numerals are often irregular, so this is okay. We see that these words also tend to have the triconsonantal root.

From "bull", we see the pattern -aw-, -ō-, -aw-ā, which also appears to apply to "day", so we have that the Aramaic for "day" is yawmā.

Now we start getting into some very fuzzy arguments. There are 4 hard ones left. We see that the pattern --ī-ī seems to be universal for ordinals in Hebrew, so ordinals probably have similar form. This indicates Arabic -ā-i- and Aramaic --ī-āyā. Then "third" in Arabic is θāliθ (this stem comes from the stem for "three"), and in Aramaic is tlīθāya.

We only have two examples of the stem -a--ā in Aramaic. One is -a--, -e-e-, -a--ā (the pattern from "snow"). The other is "horn", which we only know in Aramaic. We are forced to guess that Arabic is qarn and Hebrew is qeren. This pattern is also the only explanation for Arabic "tomb-qabr", so we should have Hebrew "qever" and Aramaic "qavrā".

For the last word (eye), we need to venture a complete guess. Since Aramaic is often just Arabic with -ā, the Aramaic ʕaynā is not too far-fetched. The Hebrew is ʕēn, which can only be guessed by comparing the pattern aw - ō - aw and realizing that ay - ē is the most reasonable sound change.[/hide]
[hide="Solution"]he wrote: kataba, kāθav, kθav
day: yawm, yōm, yawmā
they sacrificed: ðabaħū, zāvħū, dvaħ
horn: qarn, qeren, qarnā
three: θalāθ, šālōš, tlāθ
we noted: rasamnā, rāšamnū, ršamn
he sacrificed: ðabaħa, zāvaħ, dvaħ
he noted: rasama, rāšam, ršam
bull: θawr, šōr, tawrā
they wrote: katabū, kāθvū, kθav
eye: ʕayn, ʕēn, ʕaynā
snow: θalĝ, šeleɣ, talgā
seventh: sābiʕ, švīʕī, švīʕāyā
peace: salām, šālōm, šlāmā
we sacrificed: ðabaħnā, zāvaħnū, dvaħn
tomb: qabr, qever, qavrā
third: θāliθ, šlīšī, tlīθāya
we wrote: katabnā, kāθavnū, kθavn

2. "He buried" very likely has the same stem as "tomb", especially since otherwise this task would be impossible. Hence it should be "qabara" in Arabic, "qāvar" in Hebrew, and "qvar" in Aramaic.[/hide]

Guiding principles of this solution are [u]sound changes are very regular[/u] and [u]it is possible to solve the problem using only the information given[/u] (recall that we actually made use of the latter fact a few times, such as for the second task).

I guess this would be extremely hard for someone who has no knowledge of any of these languages. It was a bit easier for me just because I know the triconsonantal root stuff and basic patterns in Hebrew... So yes, this problem is extremely biased (especially towards Jews and Muslims). I still love it!


function
algebra
polynomial
calculus
integration
linear algebra
linear algebra unsolved

Could anyone please do it?

To approximate a function g over the interval [0, 1] by a polynomial p of degree
n (or less), we minimize the criterion
  f(a) = integral from 0 to 1 [g(x) - p(x)] ^ 2 dx,
where p(x) = anx^n + аn-1 x ^ n-1 + ... + a0. Find the equations satisfied by
the optimal coefficients a = (a0, a1, + .... + an).

Consider $ E$ the prehilbertian space $ C^0([0,1],\mathbb{R})$ with the scalar product $ <f,g>{\equal{}}\int_0^1{f(x)g(x)dx}$. The seeked polynomial $ q$ is the orthogonal projection of $ g$ over $ \mathbb{R}_n[x]$. If you want the true equations then use the Lagrange's polynomials $ L_n$ (orthogonal system for this scalar product but over [-1,1]). cf. wikipedia.
Sketch of the result:
$ \forall{i}\equal{}\{0,\cdots,n\}: <g\minus{}q,P_i>{\equal{}0}$, with $ P_i(x)\equal{}L_i(2x\minus{}1)$.

[quote="loup blanc"]If you want the true equations then use the Lagrange's polynomials $ L_n$ (orthogonal system for this scalar product but over [-1,1]).[/quote]
For a finite interval and a constant weight, the orthogonal system would be an appropriately scaled version of the Legendre polynomials - at least that's the name I know them by.


My friend ask me about Bihari inequality. I have no idea. Can you have me to prove Bihari inequality?

Tthis would be useful. :maybe:

Take a look [url]http://mathstat.helsinki.fi/EMIS/journals/JIPAM/v1n2/004_00.html[/url]


What's R? Also, do you mean that BC is a common tangent to both circles, and the tangent points are B and C, respectively?

hmm his statement doesnt specify whether A is the nearer or farther intersection point from line BC....

It doesn't matter. Those two circumcircles have the same radius.

Right, it shouldn't (ideally) matter, and a proof would (almost certainly) cover both cases at the same time.

But the reason I'm posting this is that I'm wondering what happened to the image I posted -- can you guys still see it, or has it disappeared for you?

I see it too.  Out of curiosity, what program did you use to create it?

Joel - I would like to know what you created that in as well. Maybe it's just me, but if you join the centre of the smaller circle to C, the line doesn't exactly look like it's perpendicular to BC...

Diagram with some additional lines drawn in: (made with GSP 3)

Did you manage to solve it, Rep? Here's what I did: It's easy to see (from that last figure :)) that p=c/2sinB and q=b/2sinC, so pq=(c/2sinC)*(b/2sinB)=R2, because a/sinA=b/sinB=c/sinC=2R (sine thm).

I was using Geometer's Sketchpad, version 4.00.  I made the diagram, selected it, copied and pasted into Paint where I saved it as a .jpg.

To be honest, I just estimated the perpendiculars instead of constructing them, so it probably is off a little bit at least.


limit
logarithms
inequalities
calculus
calculus computations

Given that $ \lim_{n \to \infty}\sum_{r\equal{}1}^{n}\frac{\log {(r\plus{}n)} \minus{} \log{n}}{n} \equal{} 2 (\log{2}\minus{}\frac{1}{2})$

and

$ \lim_{n \to \infty} \frac{1}{n^k}[(n\plus{}1)^k \plus{} (n\plus{}2)^k.....(n\plus{}n)^k]^\frac{1}{n}]$

Its answer is $ (\frac{4}{e})^k$.

Please help. Thank you so much for giving me your time.

Set $ G_{n} \equal{} \prod_{r\equal{}1}^{n}\left(1 \plus{} \frac{r}{n}\right)^{\frac{1}{n}} \equal{} \left(\prod_{r\equal{}1}^{n}\left(1 \plus{} \frac{r}{n}\right)\right)^{\frac{1}{n}}$. Then $ G_{n}$ is just the geometric means of n positive real numbers: $ a_{r} \equal{} \frac{n \plus{} r}{n}$.

Consider $ A_{n} \equal{} \frac{a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n}}{n}$. Then $ A_{n} \geq G_{n}$ - the prove is not that hard.

Notice that $ n\cdot A_{n} \equal{} \sum_{r\equal{}1}^{n} \frac{n\plus{}r}{n} \equal{} n \plus{} \frac{1}{n} \cdot \frac{n(n\plus{}1)}{2}$.

So $ A_{n} \equal{} \frac{3}{2} \plus{} \frac{1}{2n}$.

Also, $ \left(\prod_{r\equal{}1}^{n} \left(1 \plus{}\frac{r}{n}\right)\right)^{\frac{1}{n}} \geq \left(1 \plus{} \sum_{r\equal{}1}^{n}\left(1 \plus{} \frac{r}{n}\right)\right)^{\frac{1}{n}} \equal{} \left(\frac{3}{2}(n\plus{}1)\right)^{\frac{1}{n}}$ - The first inequality here is a result of the trivial inequality: $ x^2 \geq 0$ together with induction.

So we have $ \frac{3}{2} \plus{} \frac{1}{2n} \geq G_{n} \geq \left(\frac{3}{2}(n\plus{}1)\right)^{\frac{1}{n}}$

Note that all you needed was this $ \lim_{n\to\infty} \log G_{n}$.


Let F be on base AB of trapezoid ABCD so DF=CF. Let E be the intersection of AC BD, and $O_{1}, O_{2}$ be the circumcenters of triangle ADF, BCF. Prove $FE \perp O_{1}O_{2}$

No one with a solution i wil lpost one soon


An alphabetical logo is formed by arranging letters in alphabetical
order from left to right. How many 1-, 2-, 3-, 4-, or 5-letter logos can
be formed from the letters in the word HORSE if repetitions are not
allowed?

The answer is 2^5 - 1.

In short, we are looking for how many subsets of a set of 5 letters.  That would be 2^5.  But we can't count the empty set because we must have at least one letter.  So we get 2^5 - 1 = 31

An alternate, less assuming explanation might be as follows.  You are simply looking for how many combinations, since each combination from n unique letters has only one alphabetic ordering. There are:

5C1 (5 things taken 1 at a time) combinations with 1 letter,
5C2 combinations for 2 letters
5C3 combinations for 3 letters
5C4 combinations for 4 letters
5C5 combinations for 5 letters

This is almost the complete sixth row of Pascal's triangle.  If we also added 5C0 we would have the sixth row of Pascal's triangle, which adds up to 2^5. (The nth row of Pascal's Triangle add up to 2^(n-1).)  So without 5C0, we get 2^5 - 1 = 31.


The answer depends since some infinities are larger than others.  :D

You get an indeterminate form 0 * infinity.

[quote="Devendra Patel"]Suppose we take any positive Infinite number and move equal spaces to an equallly infinitesemally small and infinitesamally large number and multiply them what will we get[/quote]This isn't really a rigorously defined question, per se, but I'd say if you take the limit, it should just be one; it depends on your definitions, and your exact question.

The answer is 42.

An infinite number: 42x where x-->infinity
An infitessimally small number: 1/x where x-->infinity

42x * 1/x = 42

(limits, of course)

[quote="Devendra Patel"]What will be the answer if we multiply an infinitisimely small number by a infinitisimally large number[/quote]

The answer is that we will take an infinitisimely infinite number. :what?:

The answer is that the question is meaningless unless you define the terms "infinitely large number" and "infinitesimal number," as neither exist in conventional number systems.

[quote="daine"][quote="Devendra Patel"]What will be the answer if we multiply an infinitisimely small number by a infinitisimally large number[/quote]
[hide]The infinitely large number will be $\infty$ and the infinintely small one will be $-\infty$ so $-\infty^{2}$ [/hide][/quote]

Um, yeah, except "small" doesn't necessarily mean negative.
Ex. -213543266 CAN be "bigger" than 23
Just definition thingys.  :)

[quote="Devendra Patel"]Suppose we take any positive Infinite number and move equal spaces to an equallly infinitesemally small and infinitesamally large number and multiply them what will we get[/quote]
Your proposition doesn't make a lot of sense, but I'll try to interpret it.  You're basically saying that we take some nebulous very large number $x$ and multiply it by an equally "very small" number, which I assume means $1/x$.  Then the product is obviously 1 for any finite $x$, and if you take a limit it's obvious that it's still 1 for an "infinite" $x$ (though I'm using the term loosely, since there's no such thing as a number that is infinite).  Otherwise, I think you're asking about something that doesn't have a general answer, namely

$\lim_{x \rightarrow \infty} f(x) g(x)$

where $\lim_{x \rightarrow \infty} f(x) = \infty$ and $\lim_{x \rightarrow \infty} g(x) = 0$.  There is no general solution to this limit - it totally depends on what the functions are, and often you'll have to use L'hopital's rule to evaluate it.

[quote="randomdragoon"][quote="pieterminate"]Ha, but we see that 42x>x, limit thingys, so pwned!!!!! :)[/quote]
What do you mean?? 42x and x approach infinity at comparable rates!!![/quote]
[u]
Comparable[/u]?
They won't be the same though, as in 42x approaches infiinity 42 times as fast...... :roll:
They approach 186,000 the same speed, not infinity!   :D   :roll:

Yes, [u]comparable[/u]. Now, something that would be not comparable would be something like $x^2$ and $x$...

Sorry guys if I have confused you all a lot rather for not being exact with the question.
Correct me if I am wrong.
If we divide any number by a very very very  small number, we get an infinitely large number, isnt it?
If I am not mistaken its from relativity,   m = m0 / sq. rt (i-v2/c2)
Pls correct me wherever I go wrong.
I was wondering for any value of mo if we consider v as c then we always get m as infinite . So m becomes infinite for v=c
Does my question now make sense

[quote="Devendra Patel"]
If we divide any number by a very very very  small number, we get an infinitely large number, isnt it?[/quote]

If we divide any finite number by a number much, much smaller than it, we get a very large number.  Expressed in the limit concept,

$\lim_{x \to 0} \frac{n}{x} = \infty$ for $n > 0$.

In your equation,

$\lim_{v \to c^{-} } \frac{ m_0 }{ \sqrt{1 - \frac{v^2}{c^2}} } = \infty$


function
probability
geometric sequence
arithmetic sequence
arithmetic series

Can anyone help with these counting problems (from A Path to Combinatorics for Undergrads).

1. Determine the number of functions f: {1,2,..., 1999} -> {2000,2001,2002,2003} satisfying the condition that f(1) + f(2) + ... + f(1999) is odd.

2. Find the number of two-digit positive integers that are divisible by both their digits.

3. There are ten girls and four boys in Mr. Fat's combinatorics class. In how many ways can these students sit around a circular table such that no boys are next to each other?

4. Find the number of ordered triples of sets (A,B,C) such that A :cup: B :cup: C = {1,2,...,2003} and A :cap: B :cap: C = {} (empty set).

5. Compute the number of sets of three distinct elements that can be chosen from the set {2 1 , 2 2 , 2 3 ,..., 2 2000 } such that the three elements form an increasing geometric progression.

6. How many five-digit numbers are divisible by three and also contain 6 as one of their digits.

Thanks.

[hide="4"]

this is same thing as:

how many ways can we put 2003 numbers into 3 groups...

so it could be like this:

1,2,3,4,..... / 2000 / 2001,2002,2003

where the slashes mark the end of a group. since we have to make two slashes between 2004 possible spots, it is 2004C2 =
1002 * 2003 = 2007006[/hide]

Hi, I have a solution to #3:
[hide]In a circular pattern, we start with GB|GB|GB|GGGGGGG, since in the first segment if we add any girls, in a circular pattern it still will remain the same, therefore we have alterations in the second segemnt such as GB|GGB|GB|GGGGGG, GB|GGGB|GB|GGGGG, and so on. and same with the 3rd. therefore we yield 7+7+1= 15.[/hide]

2. [hide]Dunno if it's a great solution, but here goes:

Let's say our two digit number is $10a+b$. Then, $a|10a+b$ and $b|10a+b$.

If $a|10a+b$, we know that $a$ must divide $10a$, and this means that $a|b$ as well.

As for $b|10a+b$, there might be a better way, but I got it to the point that either $b|a$, $b|2a$, or $b|5a$. ($b|10a$ doesn't need to be considered because smallest $a=1$ and  then $b=10>9$ so wouldn't work)

If $b|a$, then we also note that $a|b$ as well. Therefore, $a=b$ is a must, and we have $11, 22, 33, 44, 55, 66, 77, 88, 99$ obviously.

If instead, $b|2a$, then also taking note that $a|b$, we see that either $a=b$ or $b=2a$. Since we already have all the cases for $a=b$, we see the cases for $b=2a$. We have $12, 24, 36, 48$

Finally, we see for $b|5a$, then $a=1$ is a definite. $b$ must then divide 5 and therefore must be 5 itself, so $15$ is one that works.

Our final answers are $11,12,15,22,24,33,36,44,48,55,66,77,88,99$
[/hide]

5. [hide]We see that in order for the three numbers to be in a geometric progression, their exponents must from an arithmetic sequence. The exponents are {1,2,3,...,2002,2003}. Since there must be 3 numbers, the common difference must be less than or equal to $\frac{2002}{2}=1001$.

Because it's an arithmetic sequence, we only need to count the number of different ways we can choose the first term for each different common difference.

We see that the number of ways to choose the first term where $d$ is the common difference is $2003-2d$.

We are looking for $\sum_{d=1}^{1001} 2003-2d$

This simplifies to $(2003\cdot1001)-(2+4+6+...+2002)$.

The latter is an arithmetic series which can be solved by formula for arithmetic series $\frac{n(t_1+t_n)}{2}$ where $n$ is number of terms, $t_1$ is first term, $t_n$ is last term.

We need to find $n$ and we do so by noting that each term is the (term/2) term of the series, so:

$\frac{1001(2+2002)}{2}=1003002$

We solve for the sum and it turns out to be $1002001$

This was a long post, and I was solving while typing, so I'll reread it and hope it's correct :)[/hide]

xxreddevilzxx, you are forgetting that A and B can both have the same element, but not the same with C.

Therefore,

[hide]You can choose to put each number 1 thru 2003 in either just one of the sets or in two of the three sets.  Therefore, you have, for each element, (3C1 + 3C2)^2003 = (3+3)^2003 = 6^2003

The 6 diff possibilities are: A, B, C, A & B, B & C, A & C.
[/hide]

This is how I reasoned [hide="6"] There are 5 digits. Use modular numbers to represent whether or not a 5 digit number can be divisible by 3.

You come up with, and I'll use the expression [2] to say 2 mod 3:

[2] [2] [2] [2] [1]
[2] [2] [2] [0] [0]
[2] [2] [1] [1] [0]
[2] [1] [1] [1] [1]
[2] [1] [0] [0] [0]
[1] [1] [1] [0] [0]
[0] [0] [0] [0] [0]

There are 7 different representations of 5 digit numbers divisible by 3.

Now you must choose one with a number 0 mod 3 because that is the digit 6.

So eliminate 2 of them.

You have:
[2] [2] [2] [0] [0]
[2] [2] [1] [1] [0]
[2] [1] [0] [0] [0]
[1] [1] [1] [0] [0]
[0] [0] [0] [0] [0]

From there, the counting is rediculous... Is there a way to do it this way, or another way?

 [/hide]. But I can't finish it. Help.

A different approach to number 6, is correct but perhaps not fully rigorous:

[hide]There are 9*10*10*10*10 5-digit integers.  Of these, 8*9*9*9*9 do not contain a 6, leaving us with (90000 - 8*9^4) that do contain a 6.  Of these, one third should be divisible by 3.  (That's a little fudgy, but I think it's actually true.)  So we get (30000 - 8*3^7) = 12504 such numbers.[/hide]

Another approach might be to count depending on the number of 6s in the number -- 5 sixes is easy, 4 sixes is fairly easy, and so on.

JBL's approach works.  You just need to note this little thing:

[hide]A number is divisible by 3 if the sum of its digits is divisible by 3.  Since 6 is divisible by 3, the sum of the digits of the other numbers is divisible by 3.  So we can think of it just as a 4 digit number then.  How many 4 digit numbers are there such that the sum of their digits is divisible by 3?  That's the same as saying how many 4 digit numbers are there such that the number is divisible by 3... [/hide]

I think the answer to #1 is [hide="this"]2^1999*1000[/hide].  Solution to come later if it is right, I don't forget, and no one else posts it.

Probability, your answer seems to be too small to me.  I think the answer is approximately (to within, say, 7 orders of magnitude) the square of what you put there.

One can also extend number 1 as follows:  what if we require the sum to be divisible by 4, not just 2?

[quote="Zyloch"]Indeed the answer to number 6 is 12504. I couldn't quite get it mathematically, but I made a quick program and it came out as that.

probability1.01, I was trying that method with the 4 digits, but it got a little shady with inclusion-exclusion and stuff since if we find the number of 4 digit numbers, then there are 5 places for a 6 to go, and you multiply by 5, but then for all 4 digit numbers with 1 6, we overcounted 1, so we have to add that back in, etc... and it got a bit confusing to me lol.[/quote]

Nah, just think. Every third number is divisible by 3... :lol:

1. JBL--you're probably right, I misread the problem.  I'll look at it again.

2. whitehorseking knows what I'm talking about :)

Ahh! I totally messed up on #3. Here should be the real solution:
[hide]Assuming that all the boys and girls are different, we can say that in general, we can assume that there will be 10 groups that we can arrange by, "|" seperating each group. In general, one of the ways is looking at a random order of the list: GB|GB|GB|GB|G|G|G|G|G|G|, there would be 8!/8=7! of different combinations, according to circular permutations. But the girls and boys in the "GB" groups are interchangeable, there would be 6 times more boys can be arranged in the GB}GB|GB|GB|GB, and 10*9*8*7 in the girls order in GB|GB|GB|GB group, therefore, we multiplying that to 7!, the solutions is 7!*10*9*8*7*6.

Heh. That one was confusing..[/hide]


More questions from Munkres. Would appreciate if someone can provide link to the answers from the book. :lol:

Consider $ \mathbb{R}^2$.  Show that the following set is not path connected:
\[ \{(\frac{1}{n},y): n\in\mathbb{Z}_{>0}, y\in[0,1]\}
\cup\{(x,0): x\in[0,1]\}\cup (0,1).\]

Which problem number is this?  I can't find it in the book.

If you provide me the number, I'll find my solution and post it.

are you sure about the text? usually this is an example of space connected but not locally connected.
BTW if I got the text right, this set is indeed path connected.

bye

dphexen: I'm actually rephrasing section 25 Q10(c) and I assumed that set $ B$ is not path connected. Basically, i'm stuck with the qn nukular posted.  [quote]What is a path from $ (0,1)$ to $ (0,0)$?[/quote]

There is no path from (0,0) to (0,1). Assume there were. Let $ p: [0,1] \to \mbox{your space}$, where $ p(0) \equal{} (0,0)$ and $ p(1) \equal{} (0,1)$,  be the path. Because $ p$ is continuous, $ p$ is eventually in any arbitrarily small circle around the point $ (0,1)$. Therefore we can define a sequence $ x_n \in (0,1)$ as follows. Let $ x_n$ be the smallest number such that $ |f(x_n) \minus{} (0,1)| < 1/n$. Clearly $ f(x_n) \neq (0,1)$ since otherwise the image of $ p$ is not connected.

So, intuitively, the points $ f(x_n)$ lie on a sequence of "teeth" of the comb that get closer and closer to the y-axis. But moving from some $ f(x_n)$ to some $ f(x_n)$ closer to $ (0,1)$ must involve touching the x-axis. Since this happens infinitely times, the image of $ p$ is comprised of infinitely many segments of length, say greater than $ 1/2$. But the continuous image of the compact set $ [0,1]$ cannot map onto an infinite succession of fixed-length segments, which is not compact, a contradiction.


logarithms
complex numbers
imaginary numbers

how is $ 3\sqrt{2}$ imaginary?

my friend said so.

Your friend is wrong.  Both square roots of $ 2$ are real.

he just said:
[quote]well, an imaginary number is a+bi. in this case a=3rt2 and b=0. so all real numbers are imaginary.[/quote]

Then he has the definition a bit wrong.  A [i]complex number[/i] is in that form, $ a\plus{}bi$.  All real numbers can be written as [i]complex[/i] numbers with $ b\equal{}0$, or the imaginary part being equal to zero.  :)

[quote="abacadaea"]he just said:
[quote]well, an imaginary number is $ a\plus{}bi$. in this case $ a\equal{}3\sqrt2$ and $ b\equal{}0$. so all real numbers are imaginary.[/quote][/quote]

[quote="AstroPhys"]Then he has the definition a bit wrong.  A [i]complex number[/i] is in that form, $ a \plus{} bi$.  All real numbers can be written as [i]complex[/i] numbers with $ b \equal{} 0$, or the imaginary part being equal to zero.  :)[/quote]
A bit wrong? It's totally wrong!
An imaginary number is a complex number $ a\plus{}bi$ whose real part $ a\equal{}0$.
Examples: $ 2i,\pi i,\minus{}i,1.49875937595i,\minus{}3i\sqrt5,(e\minus{}1)i,\frac{\pi^{2008}i}{3\plus{}\log\pi}$ are all imaginary numbers.
Not imaginary: $ \pi,\minus{}\ln2,1\plus{}i$
Like AstroPhys said, all real numbers are complex numbers with $ b\equal{}0$; therefore, all real numbers are not imaginary numbers, except 0 because 0 belongs to both as $ 0\plus{}0i$.


How did the US team fare at the CGMO this week?

Anyone know the results of the the China Girls Math Olympiad 2009 yet?

Electron

The US Team got 2 golds, 3 silvers and 2 bronzes.

The questions are already out there somewhere:))

I've seen Problems 1, 2, 5, and 6. Haven't seen the other 4 yet though. The missing 4 are purportedly the tougher ones.

[quote="mathgeniuse^ln(x)"]Does anyone know when the questions will be released to the public?  Or does anyone know if the questions are already out there?[/quote]

here are the problems:

[url=http://www.mathlinks.ro/viewtopic.php?t=296177]P1[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=296178]P2[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=296182]P3[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=296407]P4[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=296186]P5[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=296197]P6[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=296201]P7[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=296202]P8[/url]


When 1 + 7 + 7^2 +...+ 7^2004 is divided by 1000, a remainder of N is obtained. Determine
the value of N.

[hide="not sure at all"] We can sum this geometric series to get $ \frac{7^{2005}\minus{}1}{6}$. Since $ \phi{1000}\equal{}400$, $ 7^{400}\equiv1\pmod{1000}$, this reduces to finding the remainder when $ \frac{7^5\minus{}1}{6}$ is divided by 1000, which we can find to be 801 (?).[/hide]

I'm pretty sure your solution is correct; I did the exact same thing. :)


The Colorado forum has been split into 2 forums -- this public forum, and a [url=http://www.artofproblemsolving.com/Forum/index.php?f=246]private forum[/url], exclusively for Coloradoans practicing for (or interested in) ARML.

Coloradoans who want to join the private ARML forum should send me a PM and visit the Colorado ARML site at [url]http://www.coloradomath.org[/url]

How can we get access to the private forum because this year I was practicing and will be next year for the ARML

Funny, I've done both and he still ignores me.

Hi,
  I am new this forum,i am able to see all the forums in this site.You said Colorado forums has been split into two forums Private and Public but i have no idea which forums are public and which are private.Can you say how to identify those forums, because i want my information to be confidential so is that private forum will help me to keep confidential of my messages.'
=============================================
alesian

[url=http://www.drugtreatments.com/colorado]Colorado Drug Treatment[/url]

Does colorado's arml t-shirt have a rectangle on the front?


let $ a$ , $ b$ , $ c$ $ \in ]0\plus{}\infty[$ , such that $ a\plus{}b\plus{}c\equal{}3$ , prove that :

$ \frac{a^9}{(b\plus{}a)(a\plus{}c)}\plus{}\frac{b^9}{(c\plus{}b)(b\plus{}a)}\plus{}\frac{c^9}{(b\plus{}c)(c\plus{}a)} \geq \frac{3}{4}$

[quote="y-a-s-s-i-n-e"]let $ a$ , $ b$ , $ c$ $ \in ]0 \plus{} \infty[$ , such that $ a \plus{} b \plus{} c \equal{} 3$ , prove that :

$ \frac {a^9}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^9}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^9}{(b \plus{} c)(c \plus{} a)} \geq \frac {3}{4}$[/quote]
apply Holder'inequality, we have

$ \frac {a^9}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^9}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^9}{(b \plus{} c)(c \plus{} a)}\ge \frac {(a \plus{} b \plus{} c)^9}{3^7[(b \plus{} a)(a \plus{} c) \plus{} (c \plus{} b)(b \plus{} a) \plus{} (b \plus{} c)(c \plus{} a)]}$

$ \Leftrightarrow \frac {a^9}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^9}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^9}{(b \plus{} c)(c \plus{} a)}\ge \frac {9}{(b \plus{} a)(a \plus{} c) \plus{} (c \plus{} b)(b \plus{} a) \plus{} (b \plus{} c)(c \plus{} a)}$ ( because $ a \plus{} b \plus{} c \equal{} 3$)

we need prove that $ \frac {9}{(b \plus{} a)(a \plus{} c) \plus{} (c \plus{} b)(b \plus{} a) \plus{} (b \plus{} c)(c \plus{} a)}\ge \frac {3}{4}$

setting $ p \equal{} a \plus{} b \plus{} c, q \equal{} ab \plus{} bc \plus{} ca$ and $ r \equal{} abc$

$ \Leftrightarrow \frac {9}{p^2 \plus{} q}\ge \frac {3}{4}$

$ \Leftrightarrow p^2 \plus{} q\le 12 \Leftrightarrow q\le 3$ which is true with $ p \equal{} a \plus{} b \plus{} c \equal{} 3$

[quote="y-a-s-s-i-n-e"]thank's

why $ q \le 3$ can you prove it please ?[/quote]
$ \Leftrightarrow ab\plus{}bc\plus{}ca\le 3 \Leftrightarrow ab\plus{}bc\plus{}ca\le \frac{(a\plus{}b\plus{}c)^2}{3}$ (because $ a\plus{}b\plus{}c\equal{}3$)

$ \Leftrightarrow (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\ge 0$

[quote="y-a-s-s-i-n-e"]let $ a$ , $ b$ , $ c$ $ \in ]0 \plus{} \infty[$ , such that $ a \plus{} b \plus{} c \equal{} 3$ , prove that :

$ \frac {a^9}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^9}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^9}{(b \plus{} c)(c \plus{} a)} \geq \frac {3}{4}$[/quote]
What about the following inequality?
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:
\[ \frac {a^{\frac {7}{4}}}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^{\frac {7}{4}}}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^{\frac {7}{4}}}{(b \plus{} c)(c \plus{} a)} \geq \frac {3}{4}
\]
Good luck!
[hide="remark"] The following inequality with a same conditions is wrong already.
\[ \frac {a^{\frac {3}{2}}}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^{\frac {3}{2}}}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^{\frac {3}{2}}}{(b \plus{} c)(c \plus{} a)} \geq \frac {3}{4}
\]
 [/hide]

[quote="arqady"][quote="y-a-s-s-i-n-e"]let $ a$ , $ b$ , $ c$ $ \in ]0 \plus{} \infty[$ , such that $ a \plus{} b \plus{} c \equal{} 3$ , prove that :

$ \frac {a^9}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^9}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^9}{(b \plus{} c)(c \plus{} a)} \geq \frac {3}{4}$[/quote]
What about the following inequality?
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:
\[ \frac {a^{\frac {7}{4}}}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^{\frac {7}{4}}}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^{\frac {7}{4}}}{(b \plus{} c)(c \plus{} a)} \geq \frac {3}{4}
\]
[/quote]
$ \frac {a^{\frac {7}{4}}}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^{\frac {7}{4}}}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^{\frac {7}{4}}}{(b \plus{} c)(c \plus{} a)} \geq \frac {3^{\frac {7}{4}}}{3^{\frac { \minus{} 1}{4}}(p^2 \plus{} q)} \equal{} \frac {9}{p^2 \plus{} q}$
then.... :wink:
[quote="arqady"]
The following inequality with a same conditions is wrong already.
\[ \frac {a^{\frac {3}{2}}}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^{\frac {3}{2}}}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^{\frac {3}{2}}}{(b \plus{} c)(c \plus{} a)} \geq \frac {3}{4}
\][/quote]
I don't think so :)  Can you write exam?

the original one is

$ n$ is integer such that $ n \geq 2$ and $ a$ , $ b$ , $ c$ positive reals

$ \frac {a^n}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^n}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^n}{(b \plus{} c)(c \plus{} a)} \geq \frac {3}{4}\left(\frac {a^{n \minus{} 1} \plus{} b^{n \minus{} 1} \plus{} c^{n \minus{} 1}}{a \plus{} b \plus{} c}\right)$

i have the result by proving that

$ \frac {a^8 \plus{} b^8 \plus{} c^8}{a \plus{} b \plus{} c} \geq \frac {(a \plus{} b \plus{} c)^7}{2187}$

[quote="bunhiacovski"]
$ \frac {a^{\frac {7}{4}}}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^{\frac {7}{4}}}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^{\frac {7}{4}}}{(b \plus{} c)(c \plus{} a)} \geq \frac {3^{\frac {7}{4}}}{3^{\frac { \minus{} 1}{4}}(p^2 \plus{} q)} \equal{} \frac {9}{p^2 \plus{} q}$
then.... :wink:[/quote]
I think, your inequality is wrong. Try $ a\equal{}2.8$ and $ b\equal{}c\equal{}0.1.$ :wink:
[quote="bunhiacovski"][quote="arqady"]
The following inequality with a same conditions is wrong already.
\[ \frac {a^{\frac {3}{2}}}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^{\frac {3}{2}}}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^{\frac {3}{2}}}{(b \plus{} c)(c \plus{} a)} \geq \frac {3}{4}
\]
[/quote]
I don't think so :)  Can you write exam?[/quote]
The same counterexample. :wink:

[quote="arqady"][quote="bunhiacovski"]
$ \frac {a^{\frac {7}{4}}}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^{\frac {7}{4}}}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^{\frac {7}{4}}}{(b \plus{} c)(c \plus{} a)} \geq \frac {3^{\frac {7}{4}}}{3^{\frac { \minus{} 1}{4}}(p^2 \plus{} q)} \equal{} \frac {9}{p^2 \plus{} q}$
then.... :wink:[/quote]
I think, your inequality is wrong. Try $ a \equal{} 2.8$ and $ b \equal{} c \equal{} 0.1.$ :wink:
[quote="bunhiacovski"][quote="arqady"]
The following inequality with a same conditions is wrong already.
\[ \frac {a^{\frac {3}{2}}}{(b \plus{} a)(a \plus{} c)} \plus{} \frac {b^{\frac {3}{2}}}{(c \plus{} b)(b \plus{} a)} \plus{} \frac {c^{\frac {3}{2}}}{(b \plus{} c)(c \plus{} a)} \geq \frac {3}{4}
\]
[/quote]
I don't think so :)  Can you write exam?[/quote]
The same counterexample. :wink:[/quote]
I'm sorry by profit is stupid of mine :blush:


[quote="SUPERMAN2"]Let $ b \in N*$ and $ a$ be a divisor of $ b^2 \plus{} 1$.Prove there exists $ x,y \in N$ such that $ x^2 \plus{} (3b^2 \plus{} 4)y^2 \equal{} 4a$.[/quote]

I'm affraid this is wrong : choose an odd $ b\geq 3$ and $ a\equal{}2$ :

If $ y\equal{}0$, you get $ x^2\equal{}8$, impossible.
If $ y>0$, $ x^2\plus{}(3b^2\plus{}4)y^2\geq 3b^2\plus{}4\geq 31 > 4a\equal{}8$


advanced fields
advanced fields unsolved

I'm looking for nice proof of the following theorem :

The only connected subsets of $ (\mathbb{R},\tau_u)$ are the intervals.($ \tau_u$ is $ \tau_{usual}$ ).

Is there anyone know how to do nice one? :wink:

Waiting your reply.

Thank you in Advanced.

isn't it very easy?

take $ A \subseteq \mathbb{R}$ connected. it's cleary enough to show that $ a<c<b , a \in A, b \in A$ implies $ c \in A$ (then by taking $ inf A$ and $ sup A$, you get the endpoints of $ A$). but if not, $ A$ is covered by the disjoint, open, non-empty subsets $ A \cap \mathbb{R}_{<c}, A \cap \mathbb{R}_{>c}$, a contradiction.

the other direction is more dirty: it is enough to show that $ [0,1]$ is connected. if not, assume $ A$ is a non-empty open closed subset, and take $ a \in A$. then take the supremum $ s$ of all $ a \leq b \leq 1$ such that $ [a,b] \subseteq A$. because $ A$ is closed, we have $ s \in A$. choosing a neighborhood of $ s$ in $ A$ and the definition of $ s$, we see that $ s\equal{}1$. thus $ [a,1] \subseteq A$, and analoguously $ [0,a] \subseteq A$, i.e. $ A \equal{} [0,1]$.

Thank you in advanced.

yes , it's easy.

 :)


My solution (in spoiler) :

[hide]

:Sigma:1/((n + 1):sqrt:n) (sum for n :ge: 1)

= :Sigma:1/:sqrt:n - :Sigma::sqrt:n/(n + 1)) (sum for n :ge: 1)

= 1 + :Sigma:((:sqrt:n - :sqrt:(n - 1))/n) (sum for n :ge: 2)

< 1 + :Sigma:((:sqrt:n - :sqrt:(n - 1))/:sqrt:(n(n-1))) (sum for n :ge: 2)

= 1 + :Sigma:(1/:sqrt:(n - 1) - 1/:sqrt:n) (sum for n :ge: 2)

Now you can check that :Sigma:(1/:sqrt:(n - 1) - 1/:sqrt:n) < 1 (sum for n :ge: 2)

so the whole expression is < 2.[/hide]

nice!

It's kinda hard to read with all those parentheses and square root symbols in there, so I can't tell if I solved it the way you did...my solution involved adding and subtracting 1/V(n+1) inside the summation sign, and using two telescoping sums....interesting problem tho.


calculus
integration
logarithms
calculus computations

$ \int_{0}^{\infty }e^{ \minus{} x}\ln x dx$
I hope you can do it in more ways.

Do you mean $ e^{\minus{}x}$ rather than $ e^x?$ Because $ \int_0^{\infty}e^x\ln x\,dx\equal{}\plus{}\infty.$

[quote="Kent Merryfield"]Do you mean $ e^{ \minus{} x}$ rather than $ e^x?$ Because $ \int_0^{\infty}e^x\ln x\,dx \equal{} \plus{} \infty.$[/quote]
Yeah ,I'm sorry! It's $ e^{\minus{}x}$

[hide="Hint"]$ \Gamma(z) \equal{} \int_0^\infty e^{\minus{}t}t^{z\minus{}1}dt$
$ \Gamma'(z) \equal{} \int_0^\infty e^{\minus{}t}t^{z\minus{}1} (\ln t) dt$

So, $ \int_0^\infty e^{\minus{}x} \ln x dx \equal{} \Gamma'(1)$[/hide]

[quote="Kouichi Nakagawa"][hide="Hint"]$ \Gamma(z) \equal{} \int_0^\infty e^{ \minus{} t}t^{z \minus{} 1}dt$
$ \Gamma'(z) \equal{} \int_0^\infty e^{ \minus{} t}t^{z \minus{} 1} (\ln t) dt$

So, $ \int_0^\infty e^{ \minus{} x} \ln x dx \equal{} \Gamma'(1)$[/hide][/quote]
Could you get any other way, I work it out only in this way.
Thank you very much !


[quote="jayme"]Dear Mathlinkers,
you will find on my site : http:/perso.orange.fr/jl.ayme
a enterely synthetic proof of the Seimiya's line and an own lemme which prove the Turner's line (first synthetic ?)
My result : the Seimiya's line of the Turner's triangle of a triangle, is the Turner's line of that triangle.
Sincerely
Jean-Louis[/quote]
could you tell me what is 	The Seimiya-Ayme-Turner's line   ?
edited
i thought it was readable(according to the research at cambridge University
sorry didnt want to spam it is a sincere question)


A rectangle has an area of 36 m$ ^2$ and a width of $ \frac{2}{3}$ meters.
What is the number of meters in the length of the rectangle?


calculus
integration
algebra
polynomial
number theory proposed
number theory

how can we say that 11-7 divides $f(11)-f(7)$ please explain that

[quote="Davron"]Tiks thank you very much. Do you know that Armenia is hosting the Mendeleev's International Chemistry Olympiads
2006 . Now my friends are there i mean in Erevan. :)

Davron[/quote]
You'r wellcome Davron ;) .Surely I know that,but I don't know chemistry at all  :blush:
,that's why I don't interesting about that olimpiad.
I hope our countery can organize a good olimpiad. :)

[quote="Davron"][i]There is no polynomial[/i] $f(x)$ [i]with integer coefficients, so that[/i] $f(7)=11,f(11)=13$.

Davron[/quote]

can the eoefficients be negative numbers? I think not.. but still wish to confirm.

So it's OK  :)  - he obtained wanted contradiction this way as he prior assumed that such a polynomial exists

[quote="Tiks"][quote="xxxyyyy"][quote="ZetaX"]They can be negative.[/quote]

I tackle the problem for +ve integers as follows:

Let $f(x) = a_0 + a_1x + a_2x^2 + ...$, then using $f(7)=11$ and $f(11)=13$, we get...

$a_0 + 7a_1 = 11$
and, $a_0 + 11a_1 = 13$

solving these gives, $a_1 = 0.5$

but cannot proceed like this for negative numbers :-([/quote]
$0.5$ isn't integer :lol:  :rotfl: .[/quote]

Thats right... $0.5$ isn't integer, so for positive integer coefficients no such polynimial can exist.

For negative... ?

[quote="ZetaX"]For all integers it's proved above.

But how do you get these two equations for $a_1$¿[/quote]

Yeah the above proof is nice... I was just rtying to get it in different way.

Call $f(x) = a_0 + a_1x + a_2x^2 + ...$

so, we have, $f(7) = a_0 + 7a_1 + 49a_2 + ...$ = $11$
since all $a_i$ are integers and I am considering just the positive coefficients, we should have all $a_2, a_3$ etc be $0$.

Thus, we have the first equation: $a_0 + 7a_1 = 11$

Similarly, we get $a_0 + 11a_1 = 13$

:-)

[quote="abdurashidjon"]Davron that question as follows
$a,b$ are integers then $\frac{f(a)-(b)}{a-b}$ is integer, too.  From here it follows it is for all integers true  :lol:[/quote]
What do you think,what I proved in the  post #4? :)

[quote="abdurashidjon"]Ticks really I did thr samr thing I have just Davron"s problem and answered it :)[/quote]
It's ok,I just would like to say that my nick is[size=150]$Tiks$[/size],and not Ticks  :lol:  .


function
algebra
functional equation
algebra unsolved

Let $ f : \mathbb{R} \rightarrow \mathbb{R}$ be such that \[ f(x) \plus{} f\left(\dfrac{x\plus{}1003}{x\minus{}1}\right) \equal{} 4024\minus{}x\] for all real $x$. Find the value $ f(2009)$.

[quote="Lousin Garckz"]Let $ f : \mathbb{R} \rightarrow \mathbb{R}$ be such that $ f(x) \plus{} f(\dfrac{x \plus{} 1003}{x \minus{} 1}) \equal{} 4024 \minus{} x$. Find the value $ f(2009)$.[/quote]

Let $ P(x)$ be the assertion $ f(x)\plus{}f(\frac{x\plus{}1003}{x\minus{}1})\equal{}4024\minus{}x$

I suppose that the given functional equation is true $ \forall x\neq 1$

Then no such function exists :

Let $ g(x)\equal{}\frac{x\plus{}1003}{x\minus{}1}$. Notice that we have $ g(g(x))\equal{}x$

Then $ P(x)$ is  $ f(x)\plus{}f(g(x))\equal{}4024\minus{}x$, true $ \forall x\neq 1$
Then $ P(g(x))$ $ \implies$ $ f(g(x))\plus{}f(x)\equal{}4024\minus{}g(x)$ true $ \forall x\neq 1$ too

And so $ g(x)\equal{}x$ $ \forall x\neq 1$, which is obviously wrong

thanks pco!!.... but i'm really surprised because this was a problem in a contest and it didn't say determine if there exists a function, the problem was to assume that exist and find the value, oh too bad!! :rotfl:

p/d: pco, can you solve the problem that i posted about the number of points in $ \mathbb{Z}_{p}^{2}$ such that $ x^{2}\plus{}y^{2} \equal{} a, a \in \mathbb{Z}_{p}$, please? you are great!

[quote="Lousin Garckz"]thanks pco!!.... but i'm really surprised because this was a problem in a contest and it didn't say determine if there exists a function, the problem was to assume that exist and find the value, oh too bad!! :rotfl:

p/d: pco, can you solve the problem that i posted about the number of points in $ \mathbb{Z}_{p}^{2}$ such that $ x^{2} \plus{} y^{2} \equal{} a, a \in \mathbb{Z}_{p}$, please? you are great![/quote]

If you want to show the contradiction in a very short way :

$ x\equal{}2$ $ \implies$ $ f(2)\plus{}f(1005)\equal{}4022$
$ x\equal{}1005$ $ \implies$ $ f(1005)\plus{}f(2)\equal{}3019$


Let $ S$ be a subset of $ \{1, 2, 3, \cdots, 1989\}$ such that no two members of $ S$ differ by 4 or 7. what is the largest number of elements $ S$ can have?

The solution starts like this: We first show that, given any set of 11 consecutive integers from $ \{1, 2, 3, \cdots, 1989\}$, at most 5 of these 11 can be elements of $ S$...

My question is: Why just 11 consecutive integers? 11 seems to appear out of nowhere.

You want to somehow break this massive list into units with a repeating pattern of elements in and not in the subset. 11 is a potentially good unit to consider because 4 and 7 are relatively prime, and because 4+7=11.


The augmented matrix is

\[ \left[\begin{array}{ccccc|c}1 & 2 & 0 & 0 & 4 & 9 \\ 0 & 0 & 1 & 0 & 3 & 1 \\ 0 & 0 & 0 & 1 & 2 & 4 \\ 0 & 0 & 0 & 0 & 0 & \minus{}12\end{array}\right],\]
so there is no solution.


[quote="joejia"]A nylon rope of length 140 feet is cut into three pieces. The longest piece is five times as long as each of the other two pieces. How many feet are in the length of the longest piece?[/quote]
[hide]
$x+x+5x=140$

$7x=140$

$x=20$

$\text{The length of the shortest piece is 20, so the length of the longest piece must be 100.}$[/hide]

[hide]
Let x be the shortest piece. $  x + x + 5x = 140  $ Which simplfies to

$ 7x = 140  $ So

$  x = 20  $
Since x is the shortest piece, the longest piece is $  5 * 20 = \framebox{100}  $[/hide]


algebra
polynomial
inequalities unsolved
inequalities

Let a,b,c be positive real number.Prove that (a^2+b^2+c^2)^2>=3(a^3*b+b^3*c+c^3*a).

[quote="tang zy"]Let a,b,c be positive real number.Prove that $ (a^2\plus{}b^2\plus{}c^2)^2>\equal{}3(a^3*b\plus{}b^3*c\plus{}c^3*a)$.[/quote]
It is not easy at all. See here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=6026

[quote="SomKim"][quote="spanferkel"][quote="tang zy"]Let a,b,c be positive real number.Prove that $ (a^2 \plus{} b^2 \plus{} c^2)^2 \ge 3(a^3*b \plus{} b^3*c \plus{} c^3*a)$.[/quote]
It is not easy at all. See here:
[/quote]
the ineq is equivalent to
$ 2[(a^2 \plus{} b^2 \plus{} c^2)^2 \minus{} 3(a^3 b \plus{} b^3 c \plus{} c^3 a)](a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca) \equal{} ({(b \minus{} c)}^2 b \minus{} {(c \minus{} a)}^2 c)^2 \plus{} ({(c \minus{} a)}^2 c \minus{} {(a \minus{} b)}^2 a)^2 \plus{} ({(a \minus{} b)}^2 a \minus{} {(b \minus{} c)}^2 b)^2)$[/quote]Nice! Another "Stefan type identity". :)
Then the polynomials
\[ \sum[a(a\plus{}b)^2\minus{}b^2c]^2\] and \[ \sum[a^2c(a\plus{}b)^2\minus{}b^3(b\plus{}c)^2]^2\] should also be multiples of \[ (a^2 \plus{} b^2 \plus{} c^2)^2 \minus{} 3(a^3 b \plus{} b^3 c \plus{} c^3 a).\] But I won't check it by hand  :maybe:  :lol:


Other people don't have rating or they make one, but I have one..and it's not fake..

Is it a good thing or bad thing?
Does it do anything?

[quote="DonkeyKong"]...or they make one...[/quote]
Everybody has a rating, you just don't see others' ratings. Some users add a "Rating: x.x" line in their location so it looks like a rating.

[quote="DonkeyKong"]Does it do anything?[/quote]
No, it's just an indication of how other people think of your posts. However, don't think that your posts are bad if you have a bad rating, it simply means others don't value them highly. If you don't like it, just ignore it.

A rating is a numerical representation of a weighted average of all of your posts that others have rated. Only you, moderators, and administrators, can view your own rating.

To display a fake rating, just add enough text to your "Location" field to fill one line, then add a space and "Rating: arr, this is my rating".

[quote="i_like_pie"][quote="DonkeyKong"]...or they make one...[/quote]
Everybody has a rating, you just don't see others' ratings. Some users add a "Rating: x.x" line in their location so it looks like a rating.[/quote]

It's extremely annoying to mods (says my sister) because there are two "Rating" lines in a user's profile.

Ignore them? :P

Anyway, I remember Mr. Vornicu saying something about users having an option to turn their ratings on/off (to display). Was this just an idea or is this going to happen some time? Just curious. :)

It's going to happen soon. Some new features are being added in the future regarding the rating system (so basically till now we're just collecting data so we can see how the system is being used by the users on the site).


logarithms
trigonometry
calculus
calculus computations

Determine wether the following series is convergent or divergent! :)

$ \sum^{\infty}_{n=1}\ln\left[1+\sin\left(\frac{1}{n\sqrt{n}}\right)\right]$

The series converges.

$ \ln(1+x)\leq x$ for $ x>0$ and hence

$ \sum\ln(1+\sin \frac{1}{n^{\alpha}})\leq \sum_{n=1}^{\infty}\sin \frac{1}{n^{\alpha}}$.

The later series behaves, based on the limit test, like $ \sum\frac{1}{n^{\alpha}}$ which converges for $ \alpha >1$. Hence

$ \sum \ln(1+\sin \frac{1}{n^{\alpha}})$ converges provided that $ \alpha >1$.


function
induction
calculus
derivative
algebra
functional equation
algebra unsolved

Find all $f: \mathbb R \to\mathbb R$ such that for all real numbers $x$, $f(x) \geq 0$ and for all real numbers $x$ and $y$, \[ f(x+y)+f(x-y)-2f(x)-2y^2=0.  \]

the function is the same as
$3f(x+y)+f(x-y)=3f(x)+f(x+2y)$

given $f(0)=k$Consider the function $g(x)=f(x)-k\Rightarrow g(0)=0,g(x+y)+f(x-y)-2f(x)-2y^2=0$
Given$x=0$ we have:$g(y)+g(-y)=2y^2$
Given$x=y,y=x$we have$g(x+y)+g(y-x)=2g(y)+2x^2$
$\Rightarrow 2g(x+y)+g(x-y)+g(y-x)=2g(x)+2g(y)+2x^2+2y^2$
$\Rightarrow g(x+y)=g(x)+g(y)+2xy$
$\Rightarrow g(x+y)-(x+y)^2=g(x)-x^2+g(y)-y^2$
Consider the function $h(x)=g(x)-x^2\Rightarrow h(x+y)=h(x)+h(y) (1)$
otherwise,we have:$f(x)\geq 0\Rightarrow g(x)\geq 0 \forall x>k+1$
$\Rightarrow g(x+y)\geq 2xy\Rightarrow g(x)\geq \frac{1}{2}x^2$.By induction we  have:
$g(x)\geq \frac{2^n-1}{2^n}x^2 \forall n\Rightarrow g(x)\geq x^2\Rightarrow h(x)\geq 0 \forall x>k+1$
$\Rightarrow h(x)$is monotone (2)
$(1)(2)\Rightarrow h(x)=ax\Rightarrow f(x)=x^2+ax+k$

[quote="nttu"]Why $f(x) \geq 0 \to g(x) \geq 0 , \forall x >k+1$ ?[/quote]
Because $f(x)\geq 0\Rightarrow g(x)\geq -k\Rightarrow g(x+1)\geq 2x-2k\geq 0 \forall x\geq k$

Let $f(0)=t,f(1)=s+t$

We have            $f(x+1)=2f(x)-f(x-1)+2$
use induction      $f(n)=ns+t+n^2-n$,n is a natural number
We also have     $f(y)+f(-y)=2y^2+2t$
then                  $f(n)=ns+t+n^2-n$, n is ainterger number.

We have            $f((n+1)t)+f((n-1)t)=2f(nt)+2t^2$
use induction      $f(nx)=nf(x)+(n^2-n)x^2-(n-1)t$,n is a natural number ,then all interger number.
So easy to know $f(x)=xs+t+x^2-x$,x is a rational number
easy to prove that f is continuous in real number set.
$f(x)=x^2+Ax+B,A^2-4B\leq 0$

My solution is probably wrong, but here it goes:

Define a new function as $g(x)=f(x)-x^2$ or $f(x)=g(x)+x^2$. Substituting this into the equation yields that \[ g(x+y)+g(x-y)=2g(x) \]

Since this implies that $g(x)=Ax+B$ (see proof below) for some constants $A,B$;
$f(x)=x^2+Ax+B$, which has at most one root, implying that the discriminant is nonpositive i.e. $A^2-4B \leq 0 \Rightarrow A^2 \leq 4B$.

To prove that $g(x) = Ax+B$, we will reduce the equation to Cauchy's functional equation $f(x+y)=f(x)+f(y)$ which is known to have the solution $f(x)=Ax$.
Setting $x=\frac{a+b}{2}$, $y=\frac{a-b}{2}$, this equation becomes $g(a)+g(b)=2g(\frac{a+b}{2})$.
Setting $b=0$ yields that $\frac{g(a)+g(0)}{2}=g(\frac{a}{2})$.
Now, if $a = n + m$, $g(n+m) + g(0) = g(n) + g(m)$.
Subtracting $2g(0)$ and setting $h(x) = g(x)-g(0)$, yields $h(n+m)=h(n)+h(m)$, implying $h(x)=Ax$, $g(0)=B$, so that $g(x)=Ax+B$.

Hopefully the solution is right.

[quote="vietnamesegauss89"]given $f(0)=k$Consider the function $g(x)=f(x)-k\Rightarrow g(0)=0,g(x+y)+f(x-y)-2f(x)-2y^2=0$
Given$x=0$ we have:$g(y)+g(-y)=2y^2$
Given$x=y,y=x$we have$g(x+y)+g(y-x)=2g(y)+2x^2$
$\Rightarrow 2g(x+y)+g(x-y)+g(y-x)=2g(x)+2g(y)+2x^2+2y^2$
$\Rightarrow g(x+y)=g(x)+g(y)+2xy$
$\Rightarrow g(x+y)-(x+y)^2=g(x)-x^2+g(y)-y^2$
Consider the function $h(x)=g(x)-x^2\Rightarrow h(x+y)=h(x)+h(y) (1)$
[/quote]

It is sufficient to see that $h$ is bounded from below on an interval $[a,b]$, since $h$ is aditive, implies that $h$ is of the form $h(x)=ax+b$.

[quote="xirti"]

It is sufficient to see that $h$ is bounded from below on an interval $[a,b]$, since $h$ is aditive, implies that $h$ is of the form $h(x)=ax+b$.[/quote]

Can you write that clearer, please? I don't understand very much.

[quote="vietnamesegauss89"][quote="nttu"] $g(x)\geq-k\Rightarrow g(x+1)\geq 2x-2k\geq 0 \forall x\geq k$[/quote][/quote]
why? More clearly, please!

I have 2 ideas for a calculus proof (since everything else I tried failed), but both rely on the given function being differentiable. If someone could just prove that f is differentiable, then the proofs would be complete.
Proof 1: Let g(x) be the derivative of f(x). Then putting y = 0 and differentiating wrt x we get g(x+y) + g(x-y) = 2g(x). Then we proceed like ddziabenko to show that g(x) = Ax+B and hence f(x) = $ Ax^2 + Bx + C $ and substituting in the original equation we get A=1. Of course, $ B^2 - 4C > $ or = 0.
Proof 2: Differentiate using first principles and use the functional equation to obtain g(x) = g(x-h) + 2h, where g(x) is the derivative of x and h is tending to 0. Differentiate again to obtain h(x) = 2 where h(x) is the derivative of g(x). Integrating twice, we get the previous solution again.

Plugging in $y=0$ I get $f(x)+f(-x)=2x^2+2f(0)$. Breaking down $f(x)$ into an odd and even function,$f(x)=h(x)+g(x)$($h$ even,$g$ odd) I get $h(x)=x^2+f(0)$(obviously $f(0) \geq 0$). Thus, $g(x+y)+g(x-y)=2g(x)$, plug in $x=y \rightarrow g(2x)=2g(x)$ and I get $g(x)+g(y)=g(x+y)$. This is a Cauchy Equation, but to get $g(x)=ax$ I need to prove that $g$ is continuous, or equivalently bounded in an interval. Since $f(x) \geq 0$ we have $g(x) \geq -x^2-f(0)$ and by plugging $-x$ we have $g(-x) \geq -x^2-f(0)$ which is equivalent to $g(x) \leq x^2+f(0)$(since $g(-x)=-g(x)$) , thus $ -x^2-f(0) \leq g(x) \leq x^2+f(0)$ ,it is bounded and we have $g(x)=ax$. Thus the solution is $f(x)=x^2+ax+b$, with $a^2-4b \leq 0$.


1.Find the values of $ a$ so that two of the roots of equation $ (a\minus{}1){(x^2\plus{}x\plus{}1)}^2 \equal{} (a\plus{}1)(x^4\plus{}x^2\plus{}1)$ are real and distinct?

2.For what real values of $ a$,do the roots of equation $ x^2\minus{}2x\minus{}(a^2\minus{}1)\equal{}0$ lie between the roots of equation $ x^2\minus{}2(a\plus{}1)x\plus{}a(a\minus{}1)\equal{}0$

Thanks in advance

$ (a\minus{}1)(x^2\plus{}x\plus{}1)^2 \equal{} (a\plus{}1)(x^4\plus{}x^2\plus{}1)$
$ (a\minus{}1)(x^2\plus{}x\plus{}1)^2 \equal{} (a\plus{}1)(x^2\plus{}x\plus{}1)(x^2\minus{}x\plus{}1)$
$ 0 \equal{} (x^2\plus{}x\plus{}1) \bigg((a\plus{}1)(x^2\minus{}x\plus{}1) \minus{} (a\minus{}1)(x^2\plus{}x\plus{}1)\bigg)$
$ 0 \equal{} 2(x^2\plus{}x\plus{}1) \bigg(x^2 \minus{}ax \plus{}1\bigg)$

$ x^2\plus{}x\plus{}1\equal{}0$ has no real solutions; The discriminant of $ x^2\minus{}ax\plus{}1$ is $ a^2\minus{}4$.  For this to have two real, distinct solutions we need $ a^2\minus{}4 > 0$ so $ a \in (\minus{}\infty, \minus{}2) \cup (2,\infty)$

#2)

1st, lets find the solutions of each equation with the quadratic formula.

$ x^2\minus{}2x\minus{}(a^2\minus{}1) \equal{} 0 \iff x\equal{}\frac{2\pm\sqrt{4\plus{}4(a^2\minus{}1)}}{2} \equal{} 1\pm a$
$ x^2\minus{}2(a\plus{}1)x\plus{}a(a\minus{}1)\equal{}0 \iff x\equal{}\frac{2(a\plus{}1)\pm\sqrt{4(a\plus{}1)^2\minus{}4a(a\minus{}1)}}{2} \equal{} a\plus{}1 \pm \sqrt{3a\plus{}1}$.

So we want (I) $ a\plus{}1\minus{}\sqrt{3a\plus{}1} < 1\minus{}a \le 1\plus{}a < a\plus{}1\plus{}\sqrt{3a\plus{}1}$ if $ a \ge 0$ and if $ \minus{}1/3 \le a \le 0$ we need (II) $ a\plus{}1\minus{}\sqrt{3a\plus{}1} < 1\plus{}a \le 1\minus{}a < a\plus{}1\plus{}\sqrt{3a\plus{}1}$.

I) Since $ a$ is nonnegative in this case $ 1\minus{}a \le 1\plus{}a$ is always true.  Therefore, we can break the inequality into two parts:

i) $ a\plus{}1\minus{}\sqrt{3a\plus{}1} < 1\minus{}a$
$ 2a < \sqrt{3a\plus{}1}$.  Since the numbers in this inequality must be positive, squaring both sides will preserve the order:
$ 4a^2 < 3a\plus{}1 \iff 4a^2 \minus{} 3a \minus{}1 < 0$
$ 4a^2\minus{}3a\minus{}1 \equal{} (4a\plus{}1)(a\minus{}1) < 0$ so $ \minus{}1/4 < a < 1$.  But this inequality was found on the assumption that $ a$ was nonnegative, so we have only shown that $ 0 \le a < 1$ works.

ii) $ 1\plus{}a < a\plus{}1\plus{}\sqrt{3a\plus{}1}$
$ 0 < \sqrt{3a\plus{}1}$
$ 0 < 3a\plus{}1$
$ a > \minus{}1/3$

Thus we can conclude that for nonnegative $ a$ the condition $ a <1$ suffices.

II) $ a\plus{}1\minus{}\sqrt{3a\plus{}1} < 1\plus{}a \le 1\minus{}a < a\plus{}1\plus{}\sqrt{3a\plus{}1}$ if $ \minus{}1/3 \le a \le 0$.  $ 1\plus{}a \le 1\minus{}a$ is implied by the bounds on $ a$ here.  So again, we can break this inequality up:

i) $ a\plus{}1\minus{}\sqrt{3a\plus{}1} < 1\plus{}a$
$ \minus{}\sqrt{3a\plus{}1} < 0$.  Here squaring reverses order:
$ 3a\plus{}1 > 0$
$ a > \minus{}1/3$ - no new information

ii) $ 1\minus{}a < a\plus{}1\plus{}\sqrt{3a\plus{}1}$
$ \minus{}2a < \sqrt{3a\plus{}1}$
Now since $ a$ is nonpositive here, $ \minus{}2a \ge 0$.  So again squaring preserves order.
$ 4a^2 < 3a\plus{}1 \iff 4a^2\minus{}3a\minus{}1 < 0$
$ (4a\plus{}1)(a\minus{}1) < 0$
$ \minus{}1/4<a<1$
Just as we found when we considered only positives.  But now we are considering only nonpositives so $ \minus{}1/4 < a \le 0$.

Putting (I) and (II) together, we can conclude that $ \minus{}\frac{1}{4} < a < 1$.

2. $ x^2\minus{}2x\minus{}(a^2\minus{}1) \equal{} 0 \Rightarrow (x\minus{}1)^2 \equal{} a$ or $ x \equal{} 1 \pm a$

For the second equation to have real and distinct roots it should have a positive discriminant which gives 3a+1>0

$ f(1\plus{}a) \equal{} (1\plus{}a)^2 \minus{} 2(1\plus{}a)^2 \plus{} a(a\minus{}1) \equal{} \minus{}(3a\plus{}1)<0$

$ \therefore \ f(1\minus{}a) \equal{} (1\minus{}a)^2 \minus{} 2(1\plus{}a)(1\minus{}a) \plus{} a(a\minus{}1) >0 \Rightarrow 4a^2\minus{}3a\minus{}1 >0$

$ \Rightarrow \minus{}\frac{1}{4}<a<1$ and $ a>\minus{}\frac{1}{3}$

Thus $ a \in (\minus{}\frac{1}{4},1)$

thanks to all :)

I got solution of 1st one earlier(I was missing that simple factorisation)

Thanks again


The population of a midwestern city is increasing at an annual rate of $ 13\%$. The present population is 11 000 000 people. If this increase continues at the same rate, what will the population be in 3 years?


vector
real analysis
real analysis unsolved

For which $ s\in\mathbb C$ does the sequense $ (n^{\minus{}s})_{n}\equal{}1^{\infty}$ belong to $ l^{2}$?

For which $ s\in\mathbb C$ does the sequense $ (n^{\minus{}s})_{n\equal{}1}^{\infty}$ belong to $ l^{2}$?

Hint: $ \left|\left(\frac{1}{n^{s}}\right)^{2}\right|\equal{}\frac{1}{n^{2\text{Re}s}}$

The question is simply one about the convergence of a series.

You tell me: what is the definition of $ (a_{n})\in\ell^{2}$ ?

$ l^{2}$ denotes the vector space over $ \mathbb C$ of all comples sequences $ x\equal{}(x_{n})_{n\equal{}1^{\infty}}$ which are square summable, with componentwise addition and scalar multiplication and with inner product given by

$ (x,y)\equal{}\sum_{n\equal{}1^{\infty}}{x_{n}{y_{n}}_{o}verline},$ where $ x\equal{}(x_{n}), y\equal{} (y_{n})$.

[quote="peteryellow"]... which are square summable ...[/quote]
"Square summable" has a specific, technical meaning, namely that $ \sum_{n}|x_{n}|^{2}<\infty.$

More generally, the space $ \ell^{p}$ consists of those sequences for which $ \sum_{n}|x_{n}|^{p}<\infty.$


Find a ten digit number in decimal representation such that, read from left to right, the number formed for the first two digits is divisible by 2, the one formed by the first three digits is divisible by 3, and so forth, and additionally, it uses each of the ten digits exactly once.

[quote="amirhtlusa"]what's the answer? :?[/quote]The idea is for you to try. A hint: Use the laws of division by 2, 3, 4, 5, 6,..., and remember that each digit will be used exactly once!

I don't know why everyone is afraid to try it, it took me about 15 minutes to find the number (unique, by the way) using just pen and paper, and about 5 minutes to program a subrutine in Scheme that checked if it was right (and it was). It is not hard at all, and if you think a few minutes, there are not that many operations that you need to know.

[quote="alan"]i vaguely remember this question...iirc, it was a CCNY problem of the day

[hide]let the number be $ABCDEFGHIJ$
right at the beginning, we can use reasoning to find $J=0$ and $E=5$
$3|D+F+5$ and $D,F$ are even numbers, so $D+F=4+6$, not necessarily respectively.
$H=2$ because $8|FGH$.  also, $G$ is either 3 or 7.
$B=8$ because it is the only unplaced even number.
$D=6$ because $4|CD$, so the number now looks like $A8C654G2I0$
$7|A8C654G$ so $1000000A+10000C+G+806540=0 \mod{7}=A+4C+G=14$
$C=1$ and $I=9$
$3|A+8+1$ so $A=3$ and $G=7$
the number is $\boxed{3816547290}$[/hide][/quote]REALLY??? well, the question is a little obvious, then, I don't clame I was the first one to think about it, but... I have never seen it before I just wonder about it!

Any way, Let's propose the following. Instead of being the first two digits divisible by 2, the first three divisible by 3, etc let's make it a digit shorter. Say, the first digit divisible by 2, the number formed by first two digits divisible by 3, and so on, until the number itself is divisible by eleven. Someone wants to give it a try?

[quote="Shogia"]It is sad if I did not make a mistake but there are no numbers that fit this pattern.  I have no elegant way of showing this because I placed the 5 and the 0 and then started at the beginning listing possibilities for the first along with the possibilities for the next number for each of the beginning first numbers.  I was only able to get the first 6 digits before it failed.[/quote]

Read alan's solution. It works perfectly.

I was referring to his second challenge in the post above my first.

[quote="Shogia"]It is sad if I did not make a mistake but there are no numbers that fit this pattern.  I have no elegant way of showing this because I placed the 5 and the 0 and then started at the beginning listing possibilities for the first along with the possibilities for the next number for each of the beginning first numbers.  I was only able to get the first 6 digits before it failed.[/quote]Well, you are right and wrong. You are right, there is no number that fits into the conditions, and there is a simple argument around it (someone? :? ), but you can go until 9 digits before it fails. Actually, I think there is only one pattern of 9 non repited digits that fit into the pattern!

[quote="Devendra Patel"]You wrote that there is no number which fits the pattern,
But whats wrong with the number
3816547290 :)
Can you tell me how is that I can send my answers but keep it hidden so that others can enjoy the problem[/quote]Read carefully, because the pattern that we were discussing there (not the proposed on the very first post in the thread, but the one some posts before) does not have a solution, and yours is obviusly wrong for that one (even when is the unique solution for the first one as already discussed) because 2 does not divide 3.


search
number theory unsolved
number theory

Can anyone help me to prove that
(i)l = ((s-t)^2+1)/(s+1)(t+1) is an ineger. Then l is 1.
(ii)k = (x^2 + y^2 + 1)/(xy) is an integer. Then k is 3.

1)$x=s+1,y=t+1$ give $l=k-2$, were $k=\frac{x^{2}+y^{2}+1}{xy}$.
Last posted before many times.

can this help you? :wink:
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=866137662&t=50131


summer program
MathPath
MathZoom
AwesomeMath
AMC

Irrespective of the answer to this question, I would hope that you would do something more worthwhile with your summer than prepare for a high school competition; [i]there is life after the AMCs![/i]  :)  You could be doing research, pursuing a non-academic interest, interning - and there's certainly something to be said for the value of a fun summer camp.

It's sort of like going to SAT prep camp (yes, it exists, I have friends who did it...).

Before I go on, I agree with t0rajir0u.

But to answer your question, there are MathPath and some other ones (I cannot remember their names off my head; I never went to it) to improve your problem solving. There is no AMC and AIME/USAMO preparation summer camp. There are classes (I highly recommend Art of Problem Solving ones) but I don't know any one just dedicated to AMC, at least not the well known ones.

So to answer your question, no, there isn't any summer camp to prepare. You can go to summer camp to improve your problem solving, which would then help your mindset and teach you better ways to approach problems. But there isn't really any camp that you can go to boost your score (which is why I somewhat disagree to comparing with SAT prep camp).

Anyway, I want to expand on what t0rajir0u said. Unless you are absolutely sure that this is what you want to do, do not go to this camp.  That is, if all you are interested is to score well in AMC to look good on your application, don't even bother applying because it's not worth it. Things like this will get you ahead in applying for college but what would you do if this is all you spent in your summer and the first thing that you realize in college is that you don't like math? Or even more commonly, you like math but that's not what you want to do with life?

The simpler thing is this: you're young so enjoy the high school years. We are not trying to discourage math camp (it is a great thing) because in the process, you get to learn and meet with peers at your level or higher. But make sure that this is what you want to do because you're young once, and you can do so much more fun things.

While I strongly agree with the posts above, you still deserve an answer. The three that come to mind are Mathzoom, AwesomeMath and (surprise) MOP, in order of challenge. The first two have forums on this site where you can find more; the last one has to be qualified for by the USAMO.

Now I will quote a post made earlier this year in another topic about summer camps (replace AwesomeMath with arbitrary contest prep camp).
[quote="JAM"]... Not to say that AwesomeMath doesn't do it's job well, it is a very good program for what it does, but I don't think I has the right focus. AwesomeMath is heavily geared towards "contest math," the sort of math that's used for high school math competitions. While this sort of stuff might seem important now, it really isn't. Your math competition career will end when you enter college. Yes I know there's still a few things like the Putnam while you're an undergrad, but it's nothing like high school. In 4 or 5 years, do you think it's really going to matter what you got on the AMC or the AIME (or even the USAMO)? Being sucked into this mindset of "you must do whatever it takes to succeed on these competitions" is a real mistake. Contrary to what you might think, most of the math you're learning for these competitions is not "real" math. It's not the sort of thing you'd really be using for any sort of higher math. All these tricks and formulas and quick calculation methods is not what a real mathematician would use. (How often do you think you're average mathematician uses something like Stewart's theorem?)

That being said, I'm certainly not against competitions. I think they can be a lot of fun. They can get people interested in math, and can expose them to some things they wouldn't see otherwise. And they can develop your problem solving skills, which can be useful throughout life. But remember, they are a means to an end, not an end in themselves. If you're giving up other, better opportunities to train for these competitions, I would say that you are making a mistake, and one which you will likely regret in a few years.

While you may get some short term benefits from going to a place like AwesomeMath (like higher scores on some competition or making the USAMO, or even getting into MOP) it will really be kind of a shallow experience. I can almost guarantee you that in 10 years, you won't be any better off for having gone to AwesomeMath. If you go to some place like Mathcamp (or even somewhere like Ross or PROMYS) you will get a much richer experience that will benefit you later in college or in the rest of your life.[/quote]


[u]1[/u].Prove that for positive reals $ p,q,r$ positive reals,
$ \sum_{cyclic}\frac{p^2}{q^2} \ge \sum_{cyclic}\frac {r}{p \plus{} q}$

[hide="Very nice. The inequality, though, is actually strictly greater than."]Solution:

By AM-GM, $ 3 \plus{} \sum_{cyclic}\frac {p^{2}}{q^{2}} \geq 2\sum_{cyclic}\frac {p}{q}$

Then, $ \sum_{cyclic}\frac {p}{q} \geq 3$ and $ \sum_{cyclic}\frac {p}{q} > \sum_{cyclic}\frac {p}{q \plus{} r}$

Thus $ \sum_{cyclic}\frac {p^{2}}{q^{2}} > \sum_{cyclic}\frac {p}{q \plus{} r}$ :)[/hide]

[hide="May be wrong"]
[hide="Lemma"]
$ \frac{b\plus{}c}{a}\plus{}\frac{a\plus{}b}{c}\plus{}\frac{c\plus{}a}{b}\ge4\left(\frac{a}{b\plus{}c}\plus{}\frac{b}{a\plus{}c}\plus{}\frac{c}{a\plus{}b}\right)$
If u want then i can prove this also..
[/hide]
Using the above lemma and substituting $ a\equal{}\frac{p}{q},b\equal{}\frac{q}{r},c\equal{}\frac{r}{p}$ we are left to prove that
$ 4(a^{2}\plus{}b^{2}\plus{}c^{2})\ge\left(\frac{1}{c}\plus{}\frac{1}{b}\plus{}\frac{1}{a}\plus{}a\plus{}b\plus{}c\right)$
with condition $ abc\equal{}1$.Also using $ a^{2}\plus{}b^{2}\plus{}c^{2}\ge ab\plus{}bc\plus{}ac$
We finally have$ 3(ab\plus{}bc\plus{}ac)\ge(a\plus{}b\plus{}c)$
[/hide]

[quote="Complex_Ninja"]Question: Lets say someone writes,$ p,q,r$ are reals.
Then $ \sum_{cyclic}\frac {p^{2}}{q^{2}} \equal{} \frac {p^2}{q^2} \plus{} \frac {q^2}{p^2} \plus{} \frac {p^2}{r^2} \plus{} \frac {r^2}{p^2} \plus{} \frac {r^2}{q^2} \plus{} \frac {q^2}{r^2}$. Sorry if it is stupid. I want to understand the notation? . Also what is the difference, if the use symmetry instead of cyclic?Thanks[/quote]Actually, what you gave is the symmetric, not the cyclic. The cyclic sum would be merely

$ \sum_{cyclic}\frac {p^{2}}{q^{2}} \equal{} \frac {p^2}{q^2} \plus{} \frac {q^2}{r^2} \plus{} \frac {r^2}{p^2}$

See: http://www.artofproblemsolving.com/Wiki/index.php/Symmetric_sum
http://www.artofproblemsolving.com/Wiki/index.php/Cyclic_sum

[quote="geniusbliss"]thanks elixir it is indeed strict but what if they are non-negative instead of positive reals?
this is how i got it -
$ \frac {p^2}{q^2} \ge \frac {p^2}{(p + q)^2}$ similarly for the other two terms,
now by AM-GM,
${ \sum_{cyclic}(\frac {p^2}{(p + q)^2} + \frac {p^2}{(p + q)^2} + \frac {r^2}{p^2} + \frac {r^2}{p^2} + \frac {r}{p + q}) \ge 5\cdot\sum_{cyclic}\frac {r}{p + q}}$ and then use the earlier observation :P[/quote]If they are nonnegative, they have to be positive. None can be $ 0$, otherwise $ \sum_{cyclic}\frac {p^{2}}{q^{2}}$ will not be defined (each of $ p^2,q^2,r^2$ appears in the denominator once).

[hide][quote="Maths Mechanic"]Is my solution correct???[/quote]For the most part, it is. (Your first lemma is rather well-known; there's no need for you to prove it.)

A small error I can see is at the end, where you wrote $ 3(ab \plus{} bc \plus{} ac)\ge(a \plus{} b \plus{} c)$ where I am certain you meant $ 3(a^2 \plus{} b^2 \plus{} c^2)\ge(a \plus{} b \plus{} c)$. This final statement is easy enough to prove using Cauchy-Schwarz and the fact that $ (a \plus{} b \plus{} c)\ge 3$, so $ 3(a^2 \plus{} b^2 \plus{} c^2)\ge(a \plus{} b \plus{} c)^2\ge (a \plus{} b \plus{} c)$. Is this what you meant?[/hide]

ooh! sry elixir i guess i noticed just the LHS  :blush:


There are $ 1995$ students attending a certain university. If two of them know each other, then the number of their acquaintances (within the university) is the same, otherwise the numbers of acquaintances are different. Prove that there is a student who has at least $ 62$ acquaintances. Can $ 62$ be replaced by $ 63$ in the statement?

Then there can be a group of $ x\plus{}1$ students where each student has $ x$ acquaintances.

ATTC that there isn't necessarily a student with 62 acquaintances. Therefore there must be a value of $ x$ corresponding to each group, with $ x$ ranging from zero to 61. That means that one group has one student, another has two students, etc., up to 62 students. $ 1\plus{}2\plus{}\cdots\plus{}62\equal{}1953$. 1995 is greater than 1953, so we have a contradiction.

By similar reasoning, 62 cannot be replaced by 63 in the statement.


Looks like there are some errors with post counts: http://www.artofproblemsolving.com/Forum/memberlist.php


linear algebra
matrix
Galois Theory
superior algebra
superior algebra unsolved

Let $ A$ be an $ nxn$ symmetric matrix over a field $ F$, and let $ \lambda$ be any eigenvalue of $ A$ with corresponding eigenvector $ v$. Let $ < x,y >$ be the standard bilinear form on $ F^n$. Let $ \phi$ be a non-trivial element of the Galois group of $ F(\lambda)$ over $ F$.

(a) Prove that $ < \phi(v),v > \equal{} 0$.

(b) If $ v$ and $ w$ are eigenvectors with distinct eigenvalues in $ F$, and $ E$ is any extension of $ F$, then $ < \phi(v),w > \equal{} 0$ for any element $ \phi$ of the Galois group of $ E$ over $ F$.

for a)
$ \phi(\lambda)\equal{}\mu$ where $ \mu$ is an algebraic $ \lambda$-conjugate and $ \lambda\not\equal{}\mu$. Then $ \phi(v)$ is an eigenvector associated to the $ A$-eigenvalue $ \mu$. $ \lambda\not\equal{}0$ or $ \mu\not\equal{}0$; for example $ \lambda\not\equal{}0$. $ <\phi(v),v>{\equal{}}<\phi(v),1/\lambda{Av}>{\equal{}}<A\phi(v),1/\lambda{v}>{\equal{}}\mu/\lambda{<\phi(v),v>}$. Thus $ <\phi(v),v>{\equal{}}0$.


limit
real analysis
real analysis unsolved

Let $ \displaystyle\lim_{n\to\infty}(a_n \plus{} a_{n \plus{} !}) \equal{} 0$, where $ a_n\in R$.
Show that $ a_n$ has no more than two limit points or it has infinitely many limit points.

Hint:
[hide]
1. Consider the sequence $ b_n \equal{} (\minus{}1)^n$. Note that $ \lim_{n\to \infty} |b_{n\plus{}1} \minus{} b_n| \equal{} 0$. Denote the set of limit points of $ b_n$ by $ C$. Prove that if $ a,b \in C$ then $ (a,b) \subset C$. Therefore, either $ C$ contains at most one point or it contains an infinite number of points.
2. Show that if $ b_n$ has at most one limit point, then $ a_n$ has at most two limit points; if $ b_n$ has infinitely many limit points then so does $ a_n$.
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I am considering taking an AoPS course online this summer, but I have other obligations (i.e. I'm taking summer chemistry,) and I'm wondering if that would interfere with the classes.

As I assume it is relevant, the time of the summer school chem is July 10 through August 10 from 8:00 to 3:20 each day.

I'm guessing from your signature that you are on the east coast.  All our classes are in the evening, east coast time, so you should be fine.  There are transcripts for the classes, too, so if you miss something, you can review it later.


floor function
number theory proposed
number theory

Find all $x\in\mathbb{Q}$ such that $x=\frac{p}{q}>1\ (p,q\in\mathbb{N})$ and exists $\alpha, N$ with the property that forall $n\geqslant N$ there is
$\left|\{x^{n}\}-\alpha\right|\leqslant\frac{1}{2(p+q)}$
Where $\{a\}=a-\lfloor a\rfloor$.

We consider the sequencens $a_{n}=[ x^{n+1}]-[ x^{n}]$
Take notice that :
 $(x-1). x^{n}-a_{n})=|{ x^{n+1}-{ x^{n}}| \leq | x^{n+1}-\alpha|+| x^{n}-\alpha | \leq frac{1}{p+q}}$
Because $(x-1). x^{n}$ have denominator divisible $p^{n+1}$ then : $|(x-1). x^{n}-a_{n}|< \frac{1}{p+q}$
 Hence : $|q. a_{n+1}-p. a_{n}| \leq 1$
and : $a_{n+1}= \frac{p}{q}. a_{n}$
As a result we came to the following conclusion :
 $q=1$ and $x$ be possitive numbers
Sorry for my bad English

Yes, that's close to the right answer. Actually this is not a hard problem but many people didn't solve it in the exam and considered it as the hardest one.


Let $O$ be the circumcenter of triangle $ABC$ and let $A_1,B_1,C_1$ be the midpoints of sides $BC,CA,AB$. Prove that

$\frac{a}{OA_1}+\frac{b}{OB_1}+\frac{c}{OC_1} \geq \frac{a+b+c}{r}$

Indeed simple.But every simple problem is nice in its solution.Here is my one
We have that
$\frac{a}{OA_1}+\frac{b}{OB_1}+\frac{c}{OC_1}=\frac{a^2}{a\*OA_1}+\frac{b^2}{b\*OB_1}+\frac{c^2}{c\*OC_1}\geq
\frac{(a+b+c)^{2}}{a\*OA_1+b\*OB_1+c\*OC_1}.$
Then we notice that $a\*OA_1+b\*OB_1+c\*OC_1=2\*S=(a+b+c)\*r$.The problem is solved.


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

Prove that any homomorphism from a profinite group to $GL_{n}(\mathbb{C})$ has finite image.

Can one change $\mathbb{C}$ to another field for which there is infinite image possible?

[quote="berti"]Prove that any homomorphism from a profinite group to $GL_{n}(\mathbb{C})$ has finite image.
[/quote]

No wonder I couldn't solve it.. :) It's not true.

Put $n=1$ (in that case, $\mbox{Gl}(n,\mathbb C)$ is just $(\mathbb C^{*},\cdot)$, the multiplicative group of non-zero complex numbers), and let $G$ be an abelian profinite group which is not torsion (for example $\widehat{\mathbb Z}$, the completion of $\mathbb Z$ with respect to the topology induced by its subgroups). Take an element $x\in G$ of infinite order, and send it to a complex number which is not a root of unity. This gives a homomorphism of the subgroup $\langle x\rangle$ of $G$ generated by $x$ into $\mathbb C^{*}$. Because $\mathbb C^{*}$ is divisible, this homomorphism can be extended to a homomorphism $f: G\to\mathbb C^{*}$, and we're done: the image of $f$ is infinite, because $f(\langle x\rangle)$ is infinite.

Oops, the homomorphism is assumed to be continuous.

Oy have to mention these "details" :).

Let $G$ be our profinite group, and let $f: G\to\mbox{Gl}(n,\mathbb C)$ be a continuous homomorphism. The image of $f$ is a compact subgroup of the Lie group $\mbox{Gl}(n,\mathbb C)$, so it's a Lie subgroup $K$. Let $V$ be a neighborhood of the identity in $K$ which contains no non-trivial subgroups (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=groups+lie&t=113233]this[/url] for the proof of the existence of such a neighborhood), and let $U$ be a neighborhood of the identity in $G$ such that $f(U)\subset V$.

$G$ has a normal open subgroup $H$ contained in $U$. From the construction of $U$ it follows that $H\subset\ker f$, and being an open sbgroup, $H$ has finite index. This finishes the proof.


how do you solve this problem.......

find remainder when 7^100 is divided by 25 [easy, remainder is 0]
find remainder when 2^603 x 3^201 is divided by 25 [easy, remainder is 0]
hence show that 29 x 7^100 + 54 x 2^603 x 3^201 is divisible by 25

hi thanks for your quick reply... but i don't understand the explaination

find the remainder when 7^100 is divided by 25
remainder = 0 isn't it? if not what is the remainder?

what does 1 mod 25 mean?

thanks

The remainder is the amount that is left over when you divide some number by another number.  For example, the remainder when 12 is divided by 5 is 2.  "mod" is a shorthand in modular arithmetic.  Sry, but I have to go now.  Do a search on "modular arithmetic" or "mods" and there should be some little tutorials already on the forums.  When I get back I'll try explaining a little better.

Uhhh...
Maybe I misread it, but I don't think $ 7^{100} $ is divisible by 25, nor is $ 2^{603} \cdot 3^{201} $.

As for the combination, already solved.


Let $f(x)=x^{3}-[x]\le [x]^{3}-[x]$. It is easy to show $f(x)<1$ if x<1, and $f(x)\ge 6$ if $x\ge 2$. Therefore if x is solution, then $[x]=1$ it give $x^{3}=4$ or $x=2^{2/3}$.