Salutare! Eu sunt nou pe acest forum! Am o problema compusa la care astept solutii.
Iata:
      Intr-o clasa sunt 30 de elevi. Fiecare dintre acestia isi pun in gand un numar natural oarecare. Daca doi copii isi pun in gand acelasi numar spunem ca formeaza un grup (un copil poate face parte din mai multe grupuri). Stiind ca in clasa exista 226 de grupuri, demonstrati ca exista cel putin 4 copii care si-au ales acelasi numar.   :)

Salut si bine ai venit :)

Puteai sa zici in loc de grup, pereche si se intelegea mai repede ce vroiai sa zici :D (prima daca m-am pacalit, crezand ca toti copii cu un singur numar sunt in acelasi grup :) ).

Problema este foarte slaba (226 se poate inlocui cu 31).
Anyway, [hide="Solutie"]
Daca nu exista 4 copii cu acelasi numar, atunci pentru fiecare numar exista cel mult 3 copii care l-au luat. Cum cu 3 copii cu acelasi numar se pot forma maxim 3 grupuri, inseamna ca in total ar rezulta cel mult 30 de grupuri (daca unele numere sunt luate de 2 copii avem cel mult un grup, iar daca un numar este luat de numai un copil, atunci acela nu poate intra in nici un grup  contradictie.
[/hide]

Se poate si generaliza, pentru $n$ copii si $k$ grupuri, dar lucrurile se complica la formula finala.

Mersi Valentin! :) Sorry!  Ai dreptate ca nu mi-am ales exprimarea corecta. Practic, ideea mea cu alesul numerelor nu e una fericita, pentru ca de la 31 de grupuri in sus deja e clar ca exista poate chiar mai mult de 4 elevi. Asa ca voi folosi alta legatura intre doi copii din grup si anume prietenia:

    Se da o clasa cu 30 de elevi si daca doi dintre ei sunt prieteni, atunci formeaza un grup. Stiind ca exista 226 de grupuri in clasa sa se demonstreze ca:
   a) Exista 3 copii care se cunosc fiecare cu fiecare
   b) Exista k>3 si elevii a(1),a(2)...a(k) astfel incat a(i) se cunoaste cu a(i+1) pentru orice i intre 1 si k-1 si a(k) se cunoaste cu a(1).

   In acest caz este nevoie de 226 de grupuri cel putin, schimbandu-se strategia de rezolvare:) Daca se poate sa se gaseasca si k minimal (asta nu am reusit sa gasesc pentru ca se pare ca nu este 4 :( ).

Partea a) este doar un caz particular a [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5650]teoremei lui Turan[/url] (numarul maxim de "grupuri" este $\left\lfloor \frac{30^2}4 \right\rfloor = 225$ mai exact :) ca sa nu existe triunghiuri, care a aparut sub diverse forme in probleme pe la olimpiade in anii 80-90).

Partea b) spune ca intr-un graf cu 226 de muchii si 30 de varfuri exista un ciclu de lungime mai mare ca 3.

[hide="Solutie"]
Fie $A,B,C$ varfurile triunghiului de la punctul a). Fie $N(x)$ toate varfurile din graf adiacente varfului $x$, mai putin varfurile $A,B,C$ (pentru orice $x$ din graf). Evident $N(A) \cap N(B) = \emptyset$ si analoagele (altfel obtinem un ciclu de ordin 4).

Asadar, daca $x = | N(A) |$, $y = | N(B) |$ si $z = |N(C)|$ atunci $x+y+z= 27$, si in fiecare dintre multimile $N(A)$, $N(B)$, $N(C)$ respectiv nu pot exista triunghiuri (pentru ca din nou am obtine un $K_4$ si deci un ciclu de lungime 4) si de asemenea ele in total contin $226-3-27=196$ muchii, deci cel putin unul contine mai multe muchii decat varfuri, si cum nu contine triunghiuri, contine cel putin un ciclu de ordin 4 (este cunoscut faptul ca daca intr-un graf cu $n$ varfuri exista cel putin $n$ muchii atunci avem un ciclu).
[/hide]

E ok. Totusi nu am inteles ceva.  x+y+z=27 inseamna ca A, B si C sunt conectate cu toate cele 27 de varfuri ramase. Why?

      Solutia mea se bazeaza pe un rezultat celebru care face referire tot la teorema lui Turan, dar mult mai intarita si anume ca intr-un graf cu 2n varfuri si n^2+1 muchii exista n triunghiuri. Pentru n=15 (revenind la problema) avem 15 triunghiuri cu varfurile in 30 de puncte. De aici o simpla analiza arata ca exista un ciclu de cel putin 4 muchii!


Are you sure there's enough information? I can't seem to figure out $a_{1}$, and I'm pretty sure that you need that.

instead of $a_{1}$, how about the much more useful $a_{n}$ ...

and a way of going about doing this would be..
[hide]
find a way to denote sums of the terms[/hide]

i wont make you solve it.  the numbers...not so great.  the idea is the most important part anyway

I know, that's why I didn't solve it.   :lol:


Reminds me of Who Wants To Be a Millionaire... but there, you can stop after you're given the question.  And that is given the fact that all three lifelines are gone.

Depends on how hard the questions were.. I'd stop after the first time I had to guess (probably around 100,000).

(probability I answer wrong)(money I get if I answer right)<(money I have won)

If it's multiple choice, I would calculate the probability if I take a random guess. If it's not multiple choice, then I just walk away when I don't know the answer

I'll put a few more restrictions on.

1. Multiple choice, 4 choices
2. 2 lifelines, none of which have been used. The lifelines narrow it down to 2 choices.
3. After you see the question, you CANNOT leave, you must guess.

psh. it is a game. just have fun.

I'll leave once i exhausted my lifelines.

No, just a hypothical game show. The rules are pretty standard, most shows just do a bit of a twist.

[quote="mustafa"]If you were in a game show with the following rules

Each round you win more money for answering trivia right, in this order

500
1000
2000
5000
7500
10000
12500
15000
20000
30000
50000
100000
250000
500000
1000000

And if you get a question wrong, you lose EVERYTHING, where would you stop, if ever?

This question is asked because I just saw a guy stop on a game show for the first time EVER! (I don't watch much tv). He stopped at 80000 dollars.[/quote]

just curious - how did he get to 80000?


calculus
integration
inequalities
triangle inequality

[hide]By Triangle Inequality, we have the following equations ($x$= third side)

$12+9>x \rightarrow 21>x$

$x+9>12 \rightarrow x>3$

We see $X$ must satisfy $3<x<21$, $x=\{4, 5, 6, \ldots 20\}$. EDIT: Eek, once again, my inability to read the problem has occured. Real Answer: $\boxed{17}$.[/hide]

[quote="bfrock2000"]a non-degenerate triangle has a side of length 12 centimeters and a side of length 9 centmeters. how many integral values are possible for the length measured in centimeters, of the third side?[/quote]

[hide]
I don't know what a non-degenerate triangle is but...

the thrid side can be from 4-20, so 17?[/hide]

[quote="i_like_pie"][quote="ragnarok23"][hide]By Triangle Inequality, we have the following equations ($x$= third side)

$12+9>x \rightarrow 21>x$

$x+9>12 \rightarrow x>3$

We see $X$ must satisfy $3<x<21$, $x=\{4, 5, 6, \ldots 20\}$. Thus, the answer is $\boxed{204}$.[/hide][/quote]
How on Earth did you come up with that?[/quote]
Sorry. I thought it asked for sum of the integral values, not how many there are. Edited.

[quote="bfrock2000"]so what is a non-degenerated triangle?[/quote]
A triangle that is not a straight line. Basically, one that follows the Triangle Inequality.

not quite  :wink:
a triangle in which $x+y=z$ is degenerate.
a triangle in which $x+y\ngeq z$ is not degenerate- its nonexistent

[quote="mtms5467"][quote="bfrock2000"]a non-degenerate triangle has a side of length 12 centimeters and a side of length 9 centmeters. how many integral values are possible for the length measured in centimeters, of the third side?[/quote]
A shortcut for these problems where you have a triangle of given sides $a$ , $b$ and trying to find the range of values for a third side $x$, you can do this

$a-b<x<a+b$[/quote]

What if $b>a$ ?

[hide="My Solution"]

For this triangle, I would say we should use the Triangle Inequality.

So the sides are 9 and 12. Then the third side must satisfy the following:

$x<21$

$x>-3$

$x>3$

While $x>-3$ may be ignored because it is intuitively obvious, we regard the others then we know that the numbers between 3 and 21 exclusive are 18 numbers.[/hide]


$ n! \le \left( {\frac{{n \plus{} 1}}{2}} \right)^n$ for $ n > 1$. We know this is true: $ \left( {1 \plus{} \frac{1}{{n \plus{} 1}}} \right)^{n \plus{} 1}  \ge 2$

As in your other post, why one would want to do this by induction is baffling to me.  The sensible way to solve this is with AM-GM (which is easier than proving the result you're "supposed" to use to solve this question).

If you're really determined to go about it by induction, of course, there's really only one possible way to go about it.  One starts by answering the question, "How do I get $ (n \plus{} 1)!$ from $ n!$?"


Let $a,b$ be two positive numbers such that $ab=1$.
Prove that $a^3+b^3+1\geq2a+b^2$
 :)

Substitute $b=\frac1a$, multiply by $a^3$, and you got muirhead for $[6,3,0]>[4,4,1]$.

This is also clearly one from the pre-muirhead era. :P

Personally I just did an elementary approach!! :)
Good exercise whatsover!
 :lol:

But could anybody tell me what is [b]muirhead's[/b] iequality?
As I see it is used very often in [i]mathlinks[/i] forums......


algebra
polynomial
calculus
derivative
linear algebra
matrix
vector

Suppose F is a field with characteristic 0. Show that a polynomial f(x) in F(x) is such that f'(x)=0 if and only if f(x)=c (ie a constant).
Note: Here we define differentiation in the usual way

how to use the fact that F has characteristic 0 in this proof.

[quote="LeoYard"]Note: Here we define differentiation in the usual way.[/quote]
What is that supposed to mean? The "usual way" as far as a calculus course is concerned is to define the derivative as the limit of the difference quotient. Over an arbitrary field, that notion of limit might not even be defined.

This problem does make sense over an arbitrary field if you define the derivative operator to be the linear operator defined on polynomials that sends $ x^n$ to $ nx^{n\minus{}1}.$ With this definition, we see the problem with fields of finite characteristic: in characteristic $ p,$ the polynomial $ x^p$ has derivative zero.

However, if we define things this way, this is no longer a calculus problem: it is a linear algebra problem. I'm moving it.

Sorry, Kent. Did i post my question in the wrong forum? It reads "Computations and Tutorials".

I'm sorry if i did.

You wrote:

"if you define the derivative operator to be the linear operator defined on polynomials that sends x^n to n*x^n-1"

Yes this is what I mean by diferentiating a polynomial.

You wrote:

"With this definition, we see the problem with fields of finite characteristic: in characteristic p the polynomial x^p has derivative zero."

Yes, thus we can prove one of the directions in the "if and only if" easily (mainly if "f" is a contant polynomial then its derivative is 0). But how are we to prove that "f" must be a constant polynomial if its derivative is 0.

You posted this in the calculus section; I moved it because I judged that it was not calculus.

The derivative operator, restricted to the space of polynomials of degree no greater than $ n,$ has the following matrix:
\[ \begin{pmatrix}0&1&0&0&\cdots&0\\
0&0&2&0&\cdots&0\\ 0&0&0&3&\cdots&0\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
0&0&0&0&\cdots&n\\ 0&0&0&0&\cdots&0\end{pmatrix}\]
On the whole vector space of polynomials, we can consider the operator to have an infinite matrix, which is the matrix above extended down and to the right.

Now: what is the null space of that matrix?

[In all of this, I am taking for granted that we have already proved that the set $ \{1,x,x^2,x^3,\dots\}$ is independent and thus may be used as a basis for the space of polynomials.]


let $ p\geq 5$ is prime.$ \sum_{i\equal{}1}^{p\minus{}1}\frac{1}{i^p}\equal{}\frac{m}{n}$  .(gcd(m,n)=1)
prove $ p^3|m$

(sorry if it posted before)...

\[ S \equal{} \sum_{i \equal{} 1}^{p \minus{} 1}\frac {1}{i^p} \equal{} \sum_{i \equal{} 1}^{p \minus{} 1}\frac {1}{(p \minus{} i)^p}.\]
Therefore $ 2S \equal{} \sum_{i \equal{} 1}^{p \minus{} 1}A(i)$,
\[ A(i) \equal{} [\frac {1}{i^p} \minus{} \frac {1}{i^p(1 \minus{} \frac {p}{i})^p}] \equal{} p^2\frac {1}{i^{p \plus{} 1}} \plus{} O(p^3).\]
Because
\[ p|\sum_{i \equal{} 1}^{p \minus{} 1}\frac {1}{i^{p \plus{} 1}}\]
, when $ p > 3$ (for p=3 $ p \minus{} 1|p \plus{} 1$), we jeld result.


algebra
polynomial
function
real analysis
real analysis unsolved

Here is a super nice result: let $K$ be a convex compact in $R^n$ and $A(r)$ the set of points that are at distance at most $r$ from $K$. Prove that the volume of $A(r)$ is a polynomial function of $r$.

maybe an idea. Looking for large $r$, it is quite clear that $A(r)$ must be a polynomial of degree $n$ (if it is a polynomial). Moreover it seems not that hard to prove it for polytope (at least in $R^2$, it seems true). So suppose that it is true for polytope, that is that $A(r)$ is a polynomial of degree $n$. Then one can approach the compact by a sequence of polytopes $P_n$. Then $A_n$, the function associated to $P_n$ is a polynomial of degree $n$. Now, if one fixes $r_0>0$, then $A_n$ is a sequence of polynomial of degree at most $n$ that converges uniformly towards $r\to A(r)$ on $r \in [0;r_0]$. Hence $A$ is a polynomial of degree at most $n$.
(Of course there are very big holes in this reasonning, but it seems to be encouraging ?)


search
geometry
analytic geometry
conics
parabola
calculus
calculus computations

Hi everyone. New here and need a little help. I did a search but found nothing. The problem is:

Distance between two towers of the Golden Gate Bridge is apx. 1280m. The cable sags 143m. How long is the cable assuming it has an approximately parabolic shape of the form y=kx^2.

Here is what I did. Since we are studying arc length and a volume and area of solids, I figured this was arc length. I solved x^2 for y=143 = 11.958, using, then, arc length formula from 0 to 11.958 . I got an answer of 144.09, multiplied that by 2 for each half of the parabola. Then I added that to 1280m. Result 1568.2m.

Does it look right, or am I completely off base?

Thanks so much for the help!
-Chad

When you solved $ x^2 \equal{} 143$, what is the meaning of $ x$?  I think it would be a good idea to draw a picture to figure out what all the variables mean, so that you're actually solving the problem rather than manipulating equations.

The first thing we need to do is set down some kind of coordinate axis.  A good idea in this case is to put the origin (the point $ (0,0$)) at the vertex of the cable (i.e. the lowest point of the cable).  Now in this coordinate system, where are the tops of the two towers of the bridge?  We know that the cable sags 143m, we know that the vertex of the cable must be exactly in between the two towers, and we know that the length of the bridge is 1280m.  Draw a picture with the coordinate system on it to help see that the tops of towers of the bridge must be at $ ( \minus{} 640,143)$ and $ (640,143)$.

So we know that the cable is a parabola going through the three points $ ( \minus{} 640,143)$, $ (0,0)$, and $ (640,143)$.  Now what?  Well we know that any parabola that has its vertex and minimum at $ (0,0)$ is of the form $ y \equal{} kx^2$ (why?).  So now all we need to do is find a $ k$ such the parabola $ y \equal{} kx^2$ goes through the point $ (640,143)$ (notice it will automatically also go through $ ( \minus{} 640,143)$; why?).  We solve this:
\[ 143 \equal{} k(640)^2 \\
 \\
\Rightarrow k \equal{} \frac {143}{640^2}
\]
Now, what is the formula for the arclength of a parabola $ y \equal{} kx^2$ from $ 0$ to $ d$?  I assume from what you wrote that you have this formula already, it is:
\[ s \equal{} \frac {d}{2}\sqrt {1 \plus{} 4d^2k^2} \plus{} \frac {\text{arcsinh}(2dk)}{4k}
\]
Now just substitute in $ d \equal{} 640$, $ k \equal{} \frac {143}{640^2}$ to find $ s\approx 660.7045$.  This is the arclength of half the parabola, so our answer is $ \boxed{2s\approx 1321.41}$.

Questions to think about: The equation of a parabola is not, in general, $ y \equal{} kx^2$.  What symmetries in this problem ensured that it would be of this form?  If we had chosen the origin $ (0,0$) to be anywhere other than the lowest point of the cable, would the equation of the parabola describing the cable still be of the form $ y \equal{} kx^2$?  If the heights of the two towers were different and we chose the origin to be the lowest point of the cable, how would it change the problem?  How would we go about solving this problem if we chose a different origin?  Would the [i]answer[/i] be changed by the choice of origin?

Thanks Joe.  That helps a lot. Sometimes my critical thinking skills decide to turn themselves off.

pls help!:D

The cable of the horizontal suspension bridge are supported by  two towers 150 ft. apart and 50 ft high.  If the cable is 5 ft. above the floor of the bridge at the center, what is the height of the cable 15 ft.  from the end of the bridge?

Ans.  y=  33.8  ft.


calculus
integration
Putnam
logarithms
calculus computations

Anyone have any interesting tricks or techniques for solving integrals that people don't usually see otherwise?  For example, the easy way to solve one of the Putnam problems this year was to make a substitution which then allows you to get an equation in I, where I is the value of the integral.

Anyone? Maybe I wasn't specific enough.  Basically, there are a number of things you can do to solve integrals, but only some methods are taught normally.  On the Putnam, one of the problems was
$I=\int_0^1 \frac{\ln(1+x)}{1+x^2}dx$ ; find I
Making the substitution x=\frac{1-t}{1+t} changes the integral to
$I=\int_0^1 \frac{\ln(\frac{2}{1+t})}{1+t^2}dt=\int_0^1 \frac{\ln2}{t^2+1} dt - I$
$2I=\frac{\pi \ln2}{4}$
$I=\frac{\pi \ln2}{8}$

The idea behind this was getting an equation in I and solving for I.
There's probably lots of other unusual integrals that can be solved with some tricks.  Anyone know some?


Determine the maximal possible length of the sequence of consecutive integers which are expressible in the form $ x^3\plus{}2y^2$, with $ x, y$ being integers.

We prove that we can't write integers  $ 8k\plus{}4$ and $ 8k\plus{}6$ as $ x^{3}\plus{}2y^{2}$ .
So assume that we write integer $ m$ in the form of $ 8k\plus{}4$ as $ x^{3}\plus{}2y^{2}$ so we have :
$ x^{3}\plus{}2y^{2}\equal{}8m\plus{}4$ which gives $ x\equal{}2k$ . put this and we have : $ 4k^{3}\plus{}y^{2}\equal{}4m\plus{}2$ , but $ y^{2}$ is $ 0,1 (mod4)$ . contradiction
Similiary we cant prove that we can't write integers $ 8k\plus{}6$ as $ x^{3}\plus{}2y^{2}$ .
so the maximal length of sequence of consecutive integers which are expressible in the form $ x^{3}\plus{}2y^{2}$ is $ 5$ and for prove this u can write the numbers $ \minus{}1,0,1,2,3$ in this form .


Let $D$ be an interior point of the triangle $ABC$.
$CD$ and $AB$ intersect at $D_{c}$,
$BD$ and $AC$ intersect at $D_{b}$,
$AD$ and $BC$ intersect at $D_{a}$.
Prove that there exists a triangle $KLM$ with orthocenter $H$ and the feet of altitudes $H_{k}\in LM, H_{l}\in KM, H_{m}\in KL$, so that
$(AD_{c}D) = (KH_{m}H)$
$(BD_{c}D) = (LH_{m}H)$
$(BD_{a}D) = (LH_{k}H)$
$(CD_{a}D) = (MH_{k}H)$
$(CD_{b}D) = (MH_{l}H)$
$(AD_{b}D) = (KH_{l}H)$

where $(PQR)$ denotes the area of the triangle $PQR$

[quote=darij grinberg]The problem is a trivial consequence of [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=138259#138259]http://www.mathlinks.ro/Forum/viewtopic.php?t=6134 post #2[/url] Lemma 3.

I didn't expect a competition to propose a problem which is more or less a theorem from advanced geometry textbooks...

  darij[/quote]

Can I ask which textbook it is?


Prove that if $a,b,c\geq0$, then
\[ \frac1{\sqrt{a^2+bc}}+\frac1{\sqrt{b^2+ca}}+\frac1{\sqrt{c^2+ab}}\geq \frac6{a+b+c}. \]

[quote="nhmtan"][quote="zhaobin"]Use Cauchy-Schwarz's inequality.Then it will reduce to:
\[ 3(a+b+c) \ge 2(\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+ab})  \]
It is post at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=63645]www.mathlinks.ro/Forum/viewtopic.php?t=63645[/url].
And Vasc give a nice solution there. :)[/quote]Can you write again the solution?Thank you![/quote]

Now, it is not very hard. It would be rewarding if you can do it yourself.

[quote="zhaobin"]Use Cauchy-Schwarz's inequality.Then it will reduce to:
\[ 3(a+b+c) \ge 2\left(\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+ab}\right)  \]
It is post at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=63645]www.mathlinks.ro/Forum/viewtopic.php?t=63645[/url].
And Vasc give a nice solution there. :)[/quote]

How is this implied? This is the same as:
$9 \ge \left(\frac{6}{a+b+c}\right) \cdot \left(\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+ab}\right) \text{ (1)}$

Cauchy on the LHS of the proposed ineq gives:
$\left(\sqrt{\frac{1}{a^2+bc}}+\sqrt{\frac{1}{b^2+ac}}+\sqrt{\frac{1}{c^2+ab}}\right)\left(\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+ab}\right) \ge (1 + 1 + 1)^2 = 9$

Multiplying the original gives:
$\left(\sqrt{\frac{1}{a^2+bc}}+\sqrt{\frac{1}{b^2+ac}}+\sqrt{\frac{1}{c^2+ab}}\right)\left(\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+ab}\right) \ge \left(\frac{6}{a+b+c}\right) \cdot \left(\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+ab}\right)$

I don't think that $A \ge B$ and $A \ge C$ implies $B \ge C$ or (1)?

**I think in your eagerness to use a cool new inequality you made a slight mistake zhaobin, no?

what？From Cauchy-Schwarz we have:
\[ (\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+ab}) \left(\frac1{\sqrt{a^2+bc}}+\frac1{\sqrt{b^2+ca}}+\frac1{\sqrt{c^2+ab}}\right) \ge 9 \]
Then we should only prove the inequality I wrote :)


analytic geometry
graphing lines
slope
calculus
derivative
rotation
real analysis

Find the shortest distance between the curves $y=e^{2x}$ and $y=x$.

let $(a,a)$ the point on the curve $y=x$ and $(b,e^{2b})$ on the curve $y=e^{2x}$. The shortest distance is when $(e^{2b})'=1$, where $1$ is the slope of $y=x$. Then by resolving we get $b$ and then apply the distance formula and continue ...

But why does the tangent line at the curve $y=e^{2x}$ has to be parallel to the curve $y=x$ in order to get the shortest distance?

Is that always true for all curves?

If the shortest distance between a differentiable curve and a line in the plane is achieved, then it occurs either at a point of intersection or at a point on the curve whose derivative is equal to the slope of the line.  This is equivalent to the statement that a differentiable curve in the plane is closest to the x-axis either at a root or at a point where the slope is zero (just rotate to see the equivalence).


Prove that if $ G$ is a graph having only vertices of odd degree, then every edge of $ G$ is contained in an even number of Hamiltonian cycles.

I don't know why this is true, but I'd love to see an answer!


Find all the positive primes $p$ for which there exist integers $m,n$ satisfying :
$p=m^2+n^2$ and $m^3+n^3-4$ is divisible by $p$.

p=2 is a trivial solution
assume that p>=3
$m^3+n^3-4=(m+n)(p-mn)-4$
so $mn(m+n)+4=0(mod p)$
note $a=m+n , b=mn$
then p=a²-2b and ab+4=0(mod p)
$a^3-2ab=0(mod p)$
$a^3+8=0(mod p)$
so p|(a+2)(a²-2a+4)
note that $a=m+n<\sqrt(2p)$ so a+2<p if p>=6 ( we can see that 3 or 5 does not satisfy the condition)
so p|(a²-2a+4)=(2b-2a+4) (mod p)
p| (b-a)+2 so we have 0<a<p and 0<b<p
so 0<(b-a)<p =>  2<(b-a)+2<p+2 =>(b-a)+2=p
=>a²-2b=b-a+2  => a²+a-2=3b  => a²-4b=2-(a+b)
we know that a²>4b so a+b<2  .... absurde
so 2 is the Only Solution of the problem
is it right the demo!?

I wonder this problem is from Silk Road Mathematical olympiad 2004

But p=5 and set m=2,n=1]

OK]

so p=2 and 5

p=13 is also a solution !! note that m, n are integers not necessarily positive


inequalities
function
inequalities unsolved

for all positive  $ \ x, y, z$  if  $ x\plus{}y\plus{}z\equal{}9$ then prove

\[ \frac{1}{x^2\plus{}15}\plus{}\frac{1}{y^2\plus{}15}\plus{}\frac{1}{z^2\plus{}15} \geq \frac1 8\]

[quote="kai_desmond11"]for all positive  $ \ x, y, z$  if  $ x \plus{} y \plus{} z \equal{} 9$ then prove
\[ \frac {1}{x^2 \plus{} 15} \plus{} \frac {1}{y^2 \plus{} 15} \plus{} \frac {1}{z^2 \plus{} 15} \geq \frac1 8
\]
[/quote]
Try $ x\equal{}y\equal{}4.45$ and $ z\equal{}0.1.$ :wink:

[quote="kai_desmond11"]for all positive  $ \ x, y, z$  if  $ x \plus{} y \plus{} z \equal{} 9$ then prove
\[ \frac {1}{x^2 \plus{} 15} \plus{} \frac {1}{y^2 \plus{} 15} \plus{} \frac {1}{z^2 \plus{} 15} \geq \frac1 8
\]
[/quote]
This inequality does not hold. But the following is true and nice
$ \frac{1}{2a^2\plus{}3}\plus{}\frac{1}{2b^2\plus{}3}\plus{}\frac{1}{2c^2\plus{}3} \ge \frac{3}{5}$
with $ a,b,c \ge 0,a\plus{}b\plus{}c\equal{}3$. :)

[quote="can_hang2007"][quote="kai_desmond11"]for all positive  $ \ x, y, z$  if  $ x \plus{} y \plus{} z \equal{} 9$ then prove
\[ \frac {1}{x^2 \plus{} 15} \plus{} \frac {1}{y^2 \plus{} 15} \plus{} \frac {1}{z^2 \plus{} 15} \geq \frac1 8
\]
[/quote]
This inequality does not hold. But the following is true and nice
$ \frac {1}{2a^2 \plus{} 3} \plus{} \frac {1}{2b^2 \plus{} 3} \plus{} \frac {1}{2c^2 \plus{} 3} \ge \frac {3}{5}$
with $ a,b,c \ge 0,a \plus{} b \plus{} c \equal{} 3$. :)[/quote]
Its just Am-Gm...
Nice inequality Can!

[quote="can_hang2007"]But the following is true and nice
$ \frac {1}{2a^2 \plus{} 3} \plus{} \frac {1}{2b^2 \plus{} 3} \plus{} \frac {1}{2c^2 \plus{} 3} \ge \frac {3}{5}$
with $ a,b,c \ge 0,a \plus{} b \plus{} c \equal{} 3$. :)[/quote]
Let $ a\leq b\leq c$ and $ f(x) \equal{} \frac {1}{2x^2 \plus{} 3}.$ Then $ f''(x) \equal{} \frac {12(2a^2 \minus{} 1)}{(2a^2 \plus{} 3)^3}.$
Id est, $ f$ is concave function on $ [0,\frac {1}{\sqrt2}]$ and $ f$ is convex function on $ [\frac {1}{\sqrt2},3].$
Case 1. $ a\geq\frac {1}{\sqrt2}.$ Then $ \sum_{cyc}f(a)\geq3f\left(\frac {a \plus{} b \plus{} c}{3}\right) \equal{} 3f(1) \equal{} \frac {3}{5}.$
Case 2. $ a\leq\frac {1}{\sqrt2}\leq b\leq c.$
Hence, $ \sum_{cyc}f(a)\geq f(a) \plus{} 2f\left(\frac {b \plus{} c}{2}\right) \equal{} f(a) \plus{} 2f\left(\frac {3 \minus{} a}{2}\right) \equal{} \frac {1}{2a^2 \plus{} 3} \plus{} \frac {4}{a^2 \minus{} 6a \plus{} 15}.$
Hence, it remains to prove in this case that $ \frac {1}{2a^2 \plus{} 3} \plus{} \frac {4}{a^2 \minus{} 6a \plus{} 15}\geq\frac {3}{5}.$
But $ \frac {1}{2a^2 \plus{} 3} \plus{} \frac {4}{a^2 \minus{} 6a \plus{} 15}\geq\frac {3}{5}\Leftrightarrow a(4 \minus{} a)(a \minus{} 1)^2\geq0,$ which true.
Case 3. $ a\leq b\leq\frac {1}{\sqrt2}\leq c$ and $ a \plus{} b\leq\frac {1}{\sqrt2}.$
Hence, $ \sum_{cyc}f(a)\geq f(0) \plus{} f(a \plus{} b) \plus{} f(c) \equal{} \frac {1}{3} \plus{} f(3 \minus{} c) \plus{} f(c) \equal{}$
$ \equal{} \frac {1}{3} \plus{} \frac {1}{2c^2 \minus{} 12c \plus{} 21} \plus{} \frac {1}{2c^2 \plus{} 3}.$
Thus, it remains to prove here that
$ \frac {1}{2c^2 \minus{} 12c \plus{} 21} \plus{} \frac {1}{2c^2 \plus{} 3}\geq\frac {4}{15},$ which equivalent to $ (2c \minus{} 3)^2(3 \plus{} 3c \minus{} c^2)\geq0,$ which true.
Case 4. $ a\leq b\leq\frac {1}{\sqrt2}\leq c$ and $ a \plus{} b > \frac {1}{\sqrt2}.$
Hence, $ \sum_{cyc}f(a)\geq f\left(a \plus{} b \minus{} \frac {1}{\sqrt2}\right) \plus{} f\left(\frac {1}{\sqrt2}\right) \plus{} f(c) \equal{}$
$ \equal{} \frac {1}{2\left(3 \minus{} \frac {1}{\sqrt2} \minus{} c\right)^2 \plus{} 3} \plus{} \frac {1}{4} \plus{} \frac {1}{2c^2 \plus{} 3}.$
It est, it remains to prove that
$ \frac {1}{2(c^2 \minus{} (6 \minus{} \sqrt2)c \plus{} 11 \minus{} 3\sqrt2)} \plus{} \frac {1}{2c^2 \plus{} 3}\geq\frac {7}{20}$ is true for
$ c\in[\frac {1}{\sqrt2}\leq c < 3 \minus{} \frac {1}{\sqrt2}),$ which easy to check out.


[size=150]The inscribed circle of triangle ABC meets with BC, CA, AB at D, E, F each.

This inscribed circle meets with the segment AD at the middle point X. In other words, AX=XD.

Line XB and XC again meets with the inscribed circle at Y and Z each. Prove EY=FZ[/size]

Let I be the incenter of triangle ABC, P be the intersection of FD and BI
Since BF and BD tangent to (I), we deduce XB is the symedian of triangle XFD, which means $ \angle PXD \equal{} \angle BXF$. So $ \angle YFB \equal{} \angle XAB$, hence $ YF\parallel XD$.
Similarly, we have $ ZE\parallel XD$, so $ YF\parallel ZE$. The conclusion follows


calculus
derivative
LaTeX
trigonometry
function
calculus computations

Find the derivative of $ \frac {1\plus{}\csc x^2}{1\minus{} \cot x^2}$.

I have tried and have gotten something each time and each one is HORRIBLY ugly. I can't even put it Latex here, it would take half an hour.

Any shortcuts that I'm missing here?

Using implicit differentiation, find the second derivative, $ \frac {d^{2}y}{dx^2}$.
[hide="First of all, let me make sure that I got the first derivative right"]
Is it $ \frac {dy}{dx}\equal{}\frac {y\minus{}2x}{2y\minus{}x}$?
[/hide]

First question: a four-term numerator is merely ordinarily ugly.

Second question: It's nonsense until you say what the equation involving $ x$ and $ y$ is.

You would do well to simplify the expression using trigonometric identities before differentiating.  (It simplifies a lot.)

hello, using the following identities are very useful:
1) $ \csc(x)\equal{}\frac{1}{\sin(x)}$
2) $ \cot(x)\equal{}\frac{\cos(x)}{\sin(x)}$
Sonnhard.

$ \frac {1\plus{}\csc(x^2)}{1\minus{} \cot(x^2)} \equal{} \frac{1 \plus{} \sin(x^2)}{\sqrt{2}\sin\left(x^2 \minus{} \frac{\pi}{4}\right)}$.

hello Carcul, i think it means
$ \frac{1\plus{}\left(\csc(x)\right)^2}{1\minus{}\left(\cot(x)\right)^2}$.
Sonnhard.

Then sorry about that. That's the result of not using Latex properly. Anyway,

$ \frac {1 \plus{} \csc^2x}{1 \minus{} \cot^2x} \equal{} \frac {1 \plus{} \sin^2x}{\sin^2x \minus{} \cos^2x} \equal{} \frac {1 \plus{} \frac12 [1 \minus{} \cos(2x)]}{ \minus{} \cos(2x)} \equal{} \minus{} \frac32 \sec(2x) \plus{} \frac12$.

hello, i have $ \frac{1\plus{}\csc^2(x)}{1\minus{}\cot^2(x)}\equal{}\frac{\cos^2(x)\minus{}2}{2\cos^2(x)\minus{}1}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello Carcul, i think it means
$ \frac {1 \plus{} \left(\csc(x)\right)^2}{1 \minus{} \left(\cot(x)\right)^2}$.
Sonnhard.[/quote]

No, it is what I typed.
$ \frac {1\plus{}\csc (x^2)}{1\minus{}\cot (x^2)}$

Sorry about #2, I was falling asleep...
The function is
$ x^2\minus{}xy\plus{}y^2\equal{}3$

Oh.  Then yes, you should expect an "ordinarily ugly" answer.  That's not a very nice function.

hello, let $ f(x)=\frac{1+\csc(x^2)}{1-\cot(x^2)}$. Using the quotient and the
chaine rule we get
$ f^'(x)=\frac{2x\cot(x^2)\csc(x^2)}{-1+\cot(x^2)}+\frac{2x(1-\csc(x^2))\left(\csc(x^2)\right)^2}{\left(-1+\cot(x^2)\right)^2}$.
Simplifying this we have
$ f^'(x)=-2\sqrt{2}x\cos\left(\frac{x^2}{2}\right)\left(\csc(\frac{\pi}{4}-x^2)\right)^2\sin\left(\frac{\pi}{4}+\frac{x^2}{2}\right)$.
Sonnhard.


calculus
integration
algebra unsolved
algebra

[hide]Do partial fractions bring something to mind ? WHat about integration [/hide]

The answer is $1/n$.

Clearly, it suffices to show that

$\binom{n}{0}-\left(1+\frac {1}{2}\right)\binom{n}{1}+\ldots+(-1)^{n+2}\left(1+\frac {1}{2}+\ldots+\frac {1}{n+1}\right)$ and so on sums to

$\frac {1}{n+1}-\frac {1}{n}$

We may rewrite the above as

$\binom{n}{0}-\sum_j (-1)^j \left(\sum_i \frac 1i\right)\binom{n}{j}-\sum_k \frac {(-1)^k}{k+1}\binom{n}{k}$

We know that the first summand is $1$, the middle summand is $1/n$, and the last summand, via a calculus trick, is $n/(n+1)$

So that whole mess evaluates to

$1-\frac {1}{n}+\frac {n}{n+1}=\frac {1}{n+1}-\frac {1}{n}$

which allows us to finish easily by induction.

There must be an easier solution.

As far as I know, there is just the calculus based soln you have given which is equivalent to partial fraction.

I can evaluate that last sum without calculus, it's just that it takes a lot longer.  So there is a solution without calculus.  I was experimenting with a huge double summation in which I think things cancelled out very nicely, but I couldn't finish it off.

this may be stupid, but how did you reduce the summation to the sum you showed. Also how did you find 2nd sum.

Okay.  We are working by induction.

Define $\sum_j (-1)^j \left(\sum_i \frac 1i\right)\binom{n}{j}$, with appropriate bounds on $i,j$, to be $S_n$.  We claim that $S_n=1/n$ for each $n>0$.

If we add

$\displaystyle \binom{n}{0}-\left(1+\frac {1}{2}\right)\binom{n}{1}+\ldots+(-1)^{n+2}\left(1+\frac {1}{2}+\ldots+\frac {1}{n+1}\right) \ \ \ \ \ \ \ (*)$

to $S_n$, by using the identity

$\binom {n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}$

repeatedly, we get $S_{n+1}$.

So we only need to show that that $(*)$ sums to $1/(n+1)-1/n$, and then a trivial inductive step finishes things

In fact, we see that if we rearrange the terms in a certain way, we can extract the original sum from it, plus a few extra terms of the form

$\pm \frac {1}{i+1}\binom {n}{i}$

So we rearrange the terms of $(*)$ into

$\displaystyle \binom{n}{0}-\sum_j (-1)^j \left(\sum_i \frac 1i\right)\binom{n}{j}-\sum_k \frac {(-1)^k}{k+1}\binom{n}{k}$

The first term is just $1$, the middle term is $S_n=1/n$, the last term is evaluated via the trick I mentioned earlier, the details of which I am too lazy to state.

reducing the original sum to your sum was justa bit manipulation right?


The equation 2000x^6+100x^5+10x*3+x-2=0 has just two real roots.  Find them.

Is there a trick to finding the roots of this problem.  I didn't like how Kalva did it by just somehow managing to factor this out.  Does anyone know a better solution?


Regions B and C are disjoint sections of the circles shown.  The area of overlap is increased by 50%.  By what percent does (B - C) change?

[hide]Woldn't it be 0, or would it be impossible to determine, or am I just not thinking right?[/hide]

[hide="Youre right"]It would be zero...because as one circle edges in on the other it eats the same amount of "circle" from both sections...
If you don't see this, try choosing values for B, C, and the intersection to prove it to yourself.[/hide]

[hide]I think it depends on HOW the overlap was increased.
was it increased by enlarging the circles? If so, it would change (slightly)
or was it increased by just moving them closer together (in that case i'd prob. say it stayed the same)[/hide]

I think it's increased by moving them closer together.
[hide]Thanks for the help! :) [/hide]

[quote="falconwing64"]Wouldn't B-C get smaller?
I'm lost.[/quote]

[hide]No because it's still in proportion.[/hide]

Here is an animation demonstrating what is occuring.

Ignore the fact that when one circle expands...the difference does actually change...I did that just to show that in either case (when one is larger than the other or they are equal) you get the same answer.

But really try plugging in values.

Thanks for the animation to prove my point.


Find the smallest possible value of $C=15x+\frac{2}{x^{3}y(x-2000y)}$ for $x>2000y>0$

[hide="Hint 1"]Treat $x$ as a (positive) constant; choose $y$ in the given range to minimize your expression, which is equivalent to maximizing a certain quadratic polynomial.[/hide]

[hide="Hint 2"]The result should now be ready for a clever application of AM-GM, or another inequality of your choice.[/hide]

$C=6(x-2000y)+12000y+3x+3x+3x+\frac{2}{x^{3}y(x-2000y)}$
now , let use AM-GM to get the desired result :D

Call me stupid but I've never heard of AM-GM.  Can someone explain or continue to solve this problem using it?

AM-GM is the Arithmetic Mean - Geometric Mean inequality.  There are many posts on the forums about it -- try a search.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=7942]Link[/url]
This may be a useful link.
$C=6(x-2000y)+12000y+3x+3x+3x+\frac{2}{x^{3}y(x-2000y)}\geq \sqrt[6]{\frac{6(x-2000y) \times 12000y \times 3x \times 3x \times 3x \times 2}{x^{3}y(x-2000y)}}= \sqrt[6]{6 \times 12000 \times 3 \times 3 \times 3 \times 2}= \sqrt[6]{3888000}$


abstract algebra
vector
linear algebra
superior algebra
superior algebra unsolved

Give an example of a nonzero element in $ HOM_Z(\Pi_{n\in N}Z_n,Q)$.
 Thanks in advance

seems to be difficult. it can be shown that $ \mathbb{Q}$ is an injective module so that we can extend $ \langle (1,1,...) \rangle \to \mathbb{Q}, (1,1,...) \to 1$ to a homomorphism $ \prod_n \mathbb{Z}/n \to \mathbb{Q}$, which is nontrivial. but the proof that divisible $ \mathbb{Z}$-modules  such as $ \mathbb{Q}$ are injective requires zorn's lemma. is it really possible to avoid this here? note that every homomorphism has to vanish on $ \oplus_n \mathbb{Z}/n$.

edit: in the case of $ \mathbb{Q}$ the injectivity can be seen with (infinite-dimensional) linear algebra: because of $ Hom(A \otimes \mathbb{Q},\mathbb{Q}) \equal{} Hom(A,\mathbb{Q})$ and $ \mathbb{Q}$ is flat (it's a localization), it's enough to extend a $ \mathbb{Q}$-linear map $ U \to \mathbb{Q}$ to a $ \mathbb{Q}$-linear map $ V \to \mathbb{Q}$, where $ U$ is a subspace of a vector space $ V$ over $ \mathbb{Q}$. to do this, choose a basis of $ U$ (zorn's lemma!) and extend it to a basis of $ V$ (again!).

so in our case, we have to find a basis in $ \prod_n \mathbb{Z}/n \otimes \mathbb{Q}$ ... well after all, I don't think that this works explicitely.

[quote="-oo-"]seems to be difficult. it can be shown that $ \mathbb{Q}$ is an injective module so that we can extend $ \langle (1,1,...) \rangle \to \mathbb{Q}, (1,1,...) \to 1$ to a homomorphism $ \prod_n \mathbb{Z}/n \to \mathbb{Q}$, which is nontrivial.[/quote]

Thanks for your attention,but I think if you want to use Baer's Criterion you must show that $ Q$ as a $ \prod_n \mathbb{Z}/n$ module is(if true) an injective.

No, you need an injective $ \mathbb Z$-module. Bear is useless here and not used. See http://en.wikipedia.org/wiki/Injective_module .


let f(x)=x^3+17. Prove that for each natural number
n>=2, there is a natural number x for which f(x) is
divisible by 3^n but not by 3^(n+1).

I can't seem to paste my complete solution from an already written document, so here are the most importatn ideas.
1. If 3^n|f(x) then 3^n|f(x+3^n-1).
2. If 3^n|f(x) then for exactly one of the 3 numbers x, x+3^n-1 and x+2*3^n-1, f(that number) is divisible by 3^n+1.
3. Construct the solution by induction on n.


The first and the third are simply the imaginary and real parts of a geometric sum.

[quote="L_Euler"]Find the value of:
$ \prod\limits_{i = 1}^n {\sin {\rm{ix}}} = {\rm s}\nolimits {\rm{inx}}{\rm{.}}\sin 2x.\sin3x.........\sin (n - 1)x.{\rm s}\nolimits {\rm{innx}}$
[/quote]

I don't know is it true or false: $ \prod\limits_{i = 1}^n {x_i }  = x_1 .x_2 .x_3 .......x_{n - 1} .x_n$  :(  :(  :(  :(

hello, prove by induction that
$ \sum_{i\equal{}1}^n\sin(ix)\equal{}\frac{1}{2}\frac{\sin(x)\cos((n\plus{}1)x)}{\cos(x)\minus{}1}\minus{}\frac{1}{2}\sin((n\plus{}1)x)\minus{}\frac{1}{2}\frac{\sin(x)\cos(x)}{\cos (x)\minus{}1}\plus{}\frac{1}{2}\sin(x)$.
Sonnhard.

Let $ A\equal{}\sum_{i \equal{} 1}^n \cos{ix}$ and $ B\equal{}\sum_{i \equal{} 1}^n {\sin{ix}}$ .

$ A\plus{}\sqrt{\minus{}1}B\equal{}\sum_{i \equal{} 1}^n (\cos{ix}\plus{}\sqrt{\minus{}1}\sin{ix})\equal{}\sum_{i \equal{} 1}^n e^{ix\sqrt{\minus{}1}}\equal{}\frac{e^{x\sqrt{\minus{}1}}\minus{}e^{(n\plus{}1)x\sqrt{\minus{}1}}}{1\minus{}e^{x\sqrt{\minus{}1}}}$
$ \equal{}\frac{(e^{x\sqrt{\minus{}1}}\minus{}e^{(n\plus{}1)x\sqrt{\minus{}1}})(1\minus{}e^{\minus{}x\sqrt{\minus{}1}})}{(1\minus{}e^{x\sqrt{\minus{}1}})(1\minus{}e^{\minus{}x\sqrt{\minus{}1}})}\equal{}\frac{e^{x\sqrt{\minus{}1}}\minus{}1\minus{}e^{(n\plus{}1)x\sqrt{\minus{}1}}\plus{}e^{nx\sqrt{\minus{}1}}}{2\minus{}e^{x\sqrt{\minus{}1}}\minus{}e^{\minus{}x\sqrt{\minus{}1}}}$
$ \equal{}\frac{\cos{x}\minus{}1\minus{}\cos{(n\plus{}1)x}\plus{}\cos{nx}}{2\minus{}2\cos{x}}\plus{}\sqrt{\minus{}1}\frac{\sin{x}\minus{}\sin{(n\plus{}1)x}\plus{}\sin{nx}}{2\minus{}2\cos{x}}$

$ A\equal{}\frac{\cos{x}\minus{}1\minus{}\cos{(n\plus{}1)x}\plus{}\cos{nx}}{2\minus{}2\cos{x}}$
$ B\equal{}\frac{\sin{x}\minus{}\sin{(n\plus{}1)x}\plus{}\sin{nx}}{2\minus{}2\cos{x}}$


Hi guys,
I liked your website; it's great.
But, why there is no place for computational mathematics or things related to computer science? Some sub forum for the theory of computation , algorithms, numerical computing etc... It would be good.

Yes, there is one already. [url=http://www.mathlinks.ro/Forum/index.php?f=331]The Computer Science forum[/url] :)


ratio
quadratics
arithmetic sequence
number theory solved
number theory

I don't know how to solve the following one:
   Find all r such that one can find an arithmetic progression (a_n)n>=1 consisting of positive integers and with the property that for all n>=1 the number (a_1)^r+(a_2)^r+...+(a_n)^r is the square of an integer number.

[i]unknown source[/i]

I don't have a solution, but something is still better than nothing, right?

Let me denote by r the ratio of the progression and by k what you called r, so we have

a k + (a+r) k +..+(a+(n-1)r) k  must always be a perfect square (let's call this expression x_n).

First of all k must obviously be odd, because if it's even we work modulo 4 (if all the terms of the progression are even we divide everything by
2 maximum possible ) and we get some squares = 4t+2 try it and see).

Let's assume (a,r)=d. We divide everything by d and a/d=a1, r/d=r1 with (a1,r1)=1. Then
d k *a1 k + (a1+r1) k +..+(a1+(n-1)r1) k must always be a perfect square. Now let's take a prime p s.t the exponent of p in d is odd, which means that
p|a1 k + (a1+r1) k +..+(a1+(n-1)r1) k , which means that p|(a1,r1), which is false, so d is a perfect square, so
a1 k + (a1+r1) k +..+(a1+(n-1)r1) k must be a p.s. for all n, where (a1,r1)=1, so the problem can be reduced to (a,r)=1. Let's switch a1 and r1 back to a and r, where we assume (a,r)=1.

Now take a prime p which divides r => p doesn't divide a, so
(a+tr) k has the same residue mod p, which is different from 0, for all t, so x_n will eventually rech every residue mod p, so every residue mod p is a quadratic residue, which is false, since there are only (p-1)/2 quadratic residues mod p, so there are no primes which divide r, so r=1. This is, I think, really important.

So we have reduced it to k odd, r=1 and a=t 2 (because x_1=a k is a perfect square).

I wasn't able to continue from here.

Nobody bothered to look at this one, huh? I think it's done now:

I used something which is obviously true and can be quickly dismissed by using Dirichlet's theorem, but maybe it can be shown without any tough theorems which I don't know how to prove (like Dirichlet's :D ). What I used is this: there are infinitely many primes p s.t. (p-1,k)=1. Btw, Dirichlet's theorem: If (a,b)=1 there are infinitely many primes of the form ak+b.

Here's how we use that: Let's assume a=t 2 >1. We find a prime p>t^2 s.t. (p-1,k)=1. p is a term of the progresion because p>t 2 and the ratio is 1, so p k is the difference between 2 consecutive perfect squares x_(m+1) and x_m, so p k = (u+v)(u-v), so 2v is the difference of 2 powers of p, so p | v, which is (t 2 ) k + (t 2 +1) k +..+(p-1) k.

Since (p-1,k)=1, it means that 1 k , 2 k, .., (p-1) k are the same residues as 1, 2, .., p-1 in a different order. Since p|1+2+..+(p-1) it means that p|1 k +..+(p-1) k , and, since p|v, p divides the difference between these 2, so p|1 k +..+(t 2 -1) k.

This happens for infinitely many p s.t. (p-1,k)=1, so 1 k +..+(t 2 -1) k has infinitely many prime divisors, which is absurd, so t=1.

The prb now looks like this:
Find all natural k s.t. 1 k+2 k +..+n k is a perfect square for each n>=1. It's easy to see k=3 (I think this is what everybody guessed).

Can someone please check it and tell me if it's Ok?

So you want to found all the integers k such taht for all positive integer n, 1^k+2^k+...+n^k is a square.
So 1+2^k must be a square, i.e. 2^k=a^2-1=(a-1)(a+1).
So a-1=2^l et a+1=2^j
we must have 2^l=a-1=(a+1)-2=2^j-2=2(2^(j-1)-1)
2^(j-1)-1 must be of the form 2^p so 2^(j-1)-1=1 and j=2  and l=1 and a=3 so k=3.
So k=3 is the only solution.

Yup! That's exactly what I meant when I said it was easy to see k=3 is the only solution. I had already written so much I really didn't feel like writing this part as well... :D


modular arithmetic
number theory unsolved
number theory

Let $a,n$ be the positive integer and $n$ be the square-free and $n\mid a^n-1$ prove that $n^2\mid a^n-1$ :read:  :wacko:

Let $n=p_1\ldots p_k$ be the prime decomposition of $n$, and let $P_i=\frac n{p_i}$. Since $p_i|(a^{P_i})^{p_i}-1$, it means that $p_i|a^{P_i}-1$, i.e. $a^{P_i}=tp_i+1$, and the conclusion follows immediately (just expand $(tp_i+1)^{p_i}$ and observe that it's $1\pmod {p_i^2}$).


The product of three consecutive integers is 720. What is the largest of these integers?

My way.  Estimate.  You know that $ \sqrt[3]{729}$ is $ 9$.  This should be the middle number.

So, the largest integer is $ \boxed{10}$.

[quote="AIME15"]Because 8*9=72, 8*9*10=720.  Thus, the largest integer is 720.[/quote]

the largest integer would be 10
you could use ernie's way, or just do what AIME15 did

however, i'm not sure why you put 720 as largest integer......


Is it possible to block certain users from viewing your blog? If so, how? I don't see anything under "Blog Options."

Thanks!

All I know is to allow guests, members or friends.  I don't think there is an option to block people though...

[quote="Temperal"]Blocking Google is as simple as blocking guests. After that, only the GoogleBot AOPS account can access your blog, which it probably won't.[/quote]
Wait, really? But I've had my blog set on "Registered Users" for everything for a long time, and when I log out I can't see my newest entries, but they show up on Google anyway.

[quote="Temperal"]Then GoogleBot's found your blog, then, I guess.

Maybe you could go to California and force AoPS to edit robots.txt.  :P[/quote]

He's already at california!

@yunhua98: You are new, so you might not know this. You have posted comments on some topics from a year ago. I think the problems are already solved, so you can just leave them. I am a little new, so I don't know if that is considered spam.


What is the probability that when 8 knights are put on a chessboard, they attack every square occupied or not?

[hide] 0.

A knight can atmost attack 8 squares. 8 knights * 8 squares = 64 squares. So you need all the knights to attack all the squares, without any overlapping. But for a knight to attack a corner square, it needs to be like this: (x's are the squares and a big x is where the knight is)

x  x  x  x  x

x  x  x  x  x

x  [b]X[/b]  x  x  x ...
...

The knight then isn't attacking 8 squares, so it's 0.

[/hide]

duh!

I knew that! twin!!!!

GO TWIN!!!!

YAY!!!!!

ya i like I_would_say's explanatiion and hes my twin!

[quote="dwx314"]just a question, but these 8 knights are on the same team, so they can't kill each other, right?
Otherwise, it could be thrown all off[/quote]
what???

You guys have hearing problems. He said,

"just a question, but these 8 knights are on the same team, so they can't kill each other, right?
Otherwise, it could be thrown all off."

Can't you guys pay attention?  :roll:

 :rotfl:  :rotfl: lol


It is possible to earn $ 0,1,3,4$ or $ 1$0 points with each shot in the game of Blippy. How many positive scores less than $ 30$ cannot be achieved in three shots?

[hide="ANS"]7[/hide]

**Is there a faster way to do this problem besides checking all numbers $ 1 \leq x \leq 30$ ?**

[quote="Ihatepie"]I got the same answer[/quote]

HOW did you do the problem?

Thanks.

[quote="happyme"]2 Tips:

Instead of looking for ones you CAN'T do, just look for ones you CAN and subtract.

(Hint, you can find that using some functions...)

Look by primes (If doing by  hand)

These ways optimize the time you spend, instead of checking every number, you check in groups.[/quote]

Can you be more detailed, i.e., the full solution?

Thanks.


function
trigonometry
calculus
derivative
limit
real analysis
real analysis unsolved

corollary :If f is differentiable on [a,b], then f' cannot have any simple discontinuities on [a,b]

this is a corollary from the theorem : suppose f is a real differentiable function on [a,b] and suppose f'(a) <c<f'(b) then there is a point x in [a,b] s/t f '(x)=c.

Can any pls explain the corollary using the theorem? thxxx. =) or maybe give an example? :)

also, can any1 pls explain how f ' CAN have discontinuities of the second kind and STILL SATISFYING the theorem: "suppose f is a real differentiable....."  ?

thxx =)

The mean value theorem does not assume continuous differentiability, but rather just differentiability.  The difference is subtle, but nevertheless there is a difference because differentiability does not imply continuous differentiability.  The classical example is
\[ f(x) \equal{} \begin{cases}x^2\sin \frac {1}{x} & x\neq 0 \\
0 & x \equal{} 0 \end{cases}
\]
This function is differentiable everywhere, but has an essential discontinuity at $ x \equal{} 0$.

An example of a differentiable function with a discontinuous derivative: $ x^2\sin\frac1x$, extended to be zero at zero. The derivative is $ \cos\frac1x\plus{}2x\sin\frac1x$ away from zero and zero at $ x\equal{}0$. In any interval around zero, this takes on every value in $ [\minus{}1,1]$, and a bit more.

Now, what question are you asking? Do you want a proof of the theorem, or of the simpler fact that a derivative can't have a jump discontinuity if it exists at the point of discontinuity?

ohh thxx for the explaination. It really helps me understand.

But can you prove that corollary mathematically?
let's say, by assuming f ' has a discontinuity at point q in [a,b], and use the theorem to reach a contradiction... i understand the example but just can't see how we can prove it mathematically.

Also, the example you gave is discontinuity of second kind right so i got that part. :) thxx!

The assumption is not a discontinuity- it's a [b]jump[/b] discontinuity. We have $ \lim_{x\to q^\minus{}}f'(x)\equal{}c$ and $ \lim_{x\to q^\plus{}}f'(x)\equal{}d$. If $ c$ and $ d$ aren't the same, then some subinterval of $ (c,d)$ isn't covered in the local values of $ f'$.


calculus
integration
calculus computations

can anyone help me find...

given: integral of f(x) dx  from 0 to 4 is = 10

find integral of 2* f(x) dx from 0 to 2

please show work also,

Thank you

You can't do it, unless you want to find $\int^2_0 2f(2x)dx$ instead...

Because the original f(x) wasn't specified, then you shouldn't be getting a specific answer.

For example, if f(x) = (4/5)x here, then the answer is 16/5.

However, the nature of f(x) is not specified in the problem.

I'm pretty sure that that the original post is just a typo (since $f(x)$ assumes the same values over $[0,4]$ as $f(2x)$ over $[0,2])$.


integration
logarithms
calculus
limit
calculus computations

Prove that
   $ \sum_{n \equal{} 1}^{\infty} \frac{\text H_n}{n^3} \equal{} \frac{5}{4} \zeta (4)$

[hide="Idea"]Try to evaluate:
$ \int_{0}^{1}\frac{\ln x (\ln(1 \minus{} x))^{2}}{x}\ dx$
in two different ways.[/hide]

[hide="A little more precisely"]One way - just expand $ \frac{\ln x (\ln(1 \minus{} x))^{2}}{x}$ into power series and integrate it.
Other way (more interesting) - start with the integration by parts.[/hide]

[hide="Solution"]Let $ S \equal{} \sum_{n \equal{} 1}^{\infty}\frac{\mathrm{H}_{n}}{n^{3}}, \quad I \equal{} \int_{0}^{1}\frac{\ln x (\ln(1 \minus{} x))^{2}}{x}\ dx$
Since we have, for $ x \in (0, 1]$:
$ \ln (1 \plus{} x) \equal{} \sum_{n \equal{} 1}^{\infty}\frac{(\minus{}1)^{n\plus{}1}}{n}x^{n}$
then:
$ \ln (1 \minus{} x) \equal{} \minus{}\sum_{n \equal{} 1}^{\infty}\frac{x^{n}}{n}\\
(\ln(1\minus{}x))^{2} \equal{} \sum_{n\equal{}1}^{\infty}\sum_{k \equal{} 1}^{n}\frac{1}{k(n \minus{} k \plus{} 1)}x^{n} \equal{}\\ \equal{}\sum_{n \equal{} 1}^{\infty}\frac{1}{n \plus{} 1}\sum_{k \equal{} 1}^{n}\left(\frac{1}{k} \plus{} \frac{1}{n \minus{} k \plus{} 1}\right)x^{n} \equal{} 2\sum_{n \equal{} 1}^{\infty}\frac{\mathrm{H}_{n}}{n\plus{}1}x^{n}$
hence:
$ I \equal{} 2\int_{0}^{1}\sum_{n \equal{} 1}^{\infty}\frac{\mathrm{H}_{n}}{n \plus{} 1}x^{n \minus{} 1}\ln x\ dx \equal{} 2\sum_{n\equal{}1}^{\infty}\frac{\mathrm{H}_{n}}{n\plus{}1}\int_{0}^{1}x^{n}\ln x \ dx \equal{} \\ \equal{}\minus{}2\sum_{n\equal{}1}^{\infty}\frac{\mathrm{H}_{n}}{(n\plus{}1)^{3}} \equal{} \minus{}2(S \minus{} \zeta(4))$
because $ \int_{0}^{1}x^{n}\ln x\ dx \equal{} \left(\frac{1}{n \plus{} 1}x^{n \plus{} 1}\ln x\right)\bigg|_{0}^{1} \minus{}\frac{1}{n \plus{} 1}\int_{0}^{1}x^{n}\ dx \equal{}\minus{}\frac{1}{(n\plus{}1)^{2}}$

On the other hand:
$ I \equal{} \int_{0}^{1}\ln x \ln (1 \minus{} x)\frac{\ln (1 \minus{} x)}{x}\ dx\equal{}\\
\equal{}\left(\ln x \ln (1 \minus{} x) \int_{0}^{x}\frac{\ln (1 \minus{} t)}{t}\ dt\right) \bigg|_{0}^{1} \minus{} \int_{0}^{1}\left(\frac{\ln(1 \minus{} x)}{x} \minus{} \frac{\ln x}{1 \minus{} x}\right)\int_{0}^{x}\frac{\ln(1 \minus{} t)}{t}\ dt \ dx \equal{} \\
\equal{} 0 \minus{} \int_{0}^{1}\frac{\ln(1 \minus{} x)}{x}\int_{0}^{x}\frac{\ln(1 \minus{} t)}{t}\ dt \ dx \plus{} \int_{0}^{1}\frac{\ln x}{1 \minus{} x}\int_{0}^{x} \frac{\ln(1 \minus{} t)}{t}\ dt \ dx \equal{}\\
\equal{} \minus{}\frac{1}{2}\left(\int_{0}^{x}\frac{\ln(1 \minus{} t)}{t}\ dt\right)^{2}\bigg|_{0}^{1} \minus{} \int_{0}^{1}\left(\sum_{n \equal{} 0}^{\infty}x^{n}\ln x\right)\cdot \left(\int_{0}^{x}\sum_{n \equal{} 1}^{\infty}\frac{t^{n \minus{} 1}}{n}\ dt\right)\ dx \equal{} \\
\equal{} \minus{}\frac{1}{2}\left(\sum_{n\equal{}1}^{\infty}\frac{x^{n}}{n^{2}}\right)^{2}\bigg|_{0}^{1} \minus{}\int_{0}^{1}\left(\sum_{n \equal{} 0}^{\infty}x^{n}\ln x\right)\cdot \left(\sum_{n \equal{} 1}^{\infty}\frac{x^{n}}{n^{2}}\right)\ dx \equal{} \\
\equal{} \minus{}\frac{1}{2}\left(\sum_{n \equal{} 1}^{\infty}\frac{1}{n^{2}}\right)^{2} \minus{} \int_{0}^{1}\sum_{n \equal{} 1}^{\infty}\sum_{k \equal{} 1}^{n}\frac{\ln x}{k^{2}}x^{n} \equal{} \\
\equal{} \minus{}\frac{1}{2}\left(\sum_{n \equal{} 1}^{\infty}\frac{1}{n^{2}}\right)^{2} \minus{} \sum_{n \equal{} 1}^{\infty}\sum_{k\equal{}1}^{n}\frac{1}{k^{2}}\int_{0}^{1}x^{n}\ln x \ dx \equal{}\\
\equal{}\minus{}\frac{1}{2}\left(\sum_{n \equal{} 1}^{\infty}\frac{1}{n^{2}}\right)^{2} \plus{} \sum_{n \equal{} 1}^{\infty}\frac{1}{(n \plus{} 1)^{2}}\sum_{k \equal{} 1}^{n}\frac{1}{k^{2}} \equal{}\\
\equal{}\minus{}\frac{1}{2}\zeta(4) \minus{} \frac{1}{2}\sum_{n \equal{} 1}^{\infty}\frac{2}{(n \plus{} 1)^{2}}\sum_{k \equal{} 1}^{n}\frac{1}{k^{2}} \plus{} \sum_{n \equal{} 1}^{\infty}\frac{1}{(n \plus{} 1)^{2}}\sum_{k \equal{} 1}^{n}\frac{1}{k^{2}} \equal{}  \minus{}\frac{1}{2}\zeta(4)$
because
$ \left(\sum_{n \equal{} 1}^{\infty}\frac{1}{n^{2}}\right)^{2} \equal{} \left(\lim_{m \to \infty} \sum_{n \equal{} 1}^{m}\frac{1}{n^{2}}\right)^{2} \equal{} \lim_{m \to \infty} \left(\sum_{n \equal{} 1}^{m}\frac{1}{n^{2}}\right)^{2} \equal{}\\
\equal{} \lim_{m\to\infty}\left(\sum_{n \equal{} 1}^{m}\frac{1}{n^{4}} \plus{} \sum_{n \equal{} 1}^{m\minus{}1}\frac{2}{(n \plus{} 1)^{2}}\sum_{k \equal{} 1}^{n}\frac{1}{k^{2}}\right) \equal{}\\
\equal{} \lim_{m\to\infty}\left(\sum_{n \equal{} 1}^{m}\frac{1}{n^{4}}\right) \plus{} \lim_{m\to\infty}\left(\sum_{n \equal{} 1}^{m \minus{} 1}\frac{2}{(n \plus{} 1)^{2}}\sum_{k \equal{} 1}^{n}\frac{1}{k^{2}}\right)\equal{} \\
\equal{}\zeta(4) \plus{} \sum_{n \equal{} 1}^{\infty}\frac{2}{(n \plus{} 1)^{2}}\sum_{k \equal{} 1}^{n}\frac{1}{k^{2}}$
since all this limits exists (and they are all finite).

Finally:
$ \minus{}\frac{1}{2}\zeta(4) \equal{} I \equal{} \minus{}2(S \minus{} \zeta(4))\\
S \equal{} \frac{5}{4}\zeta(4)$[/hide]


Let $ a,b,c$ be real numbers. Prove that
\[ (b\plus{}c)^4\plus{}(c\plus{}a)^4\plus{}(a\plus{}b)^4 \ge \frac{4}{7}(a^4\plus{}b^4\plus{}c^4\plus{}(a\plus{}b\plus{}c)^4)\]
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}\minus{}3b\equal{}\minus{}3c$ or any cyclic permutations.
:)

Yes, it's equivlent to:
$ \sum\ (a\plus{}b\plus{}2c)^2(a\minus{}b)^2 \geq\ 0$

[quote="can_hang2007"]Let $ a,b,c$ be real numbers. Prove that
\[ (b\plus{}c)^4\plus{}(c\plus{}a)^4\plus{}(a\plus{}b)^4 \ge \frac{4}{7}(a^4\plus{}b^4\plus{}c^4\plus{}(a\plus{}b\plus{}c)^4)\](1)
Equality holds if and only if $ a\equal{}b\equal{}c$ or $ a\equal{}\minus{}3b\equal{}\minus{}3c$ or any cyclic permutations.
:)[/quote]

$(1) \Leftrightarrow 28\left( {x^4  + y^4  + z^4 } \right) \geqslant \left( {x + y + z} \right)^4  + \left( {y + z - x}\right)^4  + \left( {z + x - y} \right)^4  + \left( {x + y - z} \right)^4, (x = \frac{{a + b}}{2},y = \frac{{b + c}}{2},z = \frac{{a + c}}{2})$
.$\begin{gathered}
  \left( {x + y + z} \right)^4  + \left( {y + z - x} \right)^4  + \left( {z + x - y} \right)^4  + \left( {x + y - z} \right)^4  = 4\left( {x^4  + y^4  + z^4 } \right) + 24\left( {x^2 y^2  + z^2 y^2  + x^2 z^2 } \right) \hfill \\
  x^2 y^2  + z^2 y^2  + x^2 z^2  \leqslant x^4  + y^4  + z^4  \hfill \\
   \Rightarrow  \hfill \\ \end{gathered} $
$  28\left( {x^4  + y^4  + z^4 } \right) \geqslant \left( {x + y + z} \right)^4  + \left( {y + z - x}\right)^4  + \left( {z + x - y} \right)^4  + \left( {x + y - z} \right)^4 $


calculus
integration
combinatorics unsolved
combinatorics

Prove that some integral power of $ 2$ has decimal expansion which starts with the digits $ 1999$.           $ 2^n \equal{} 1999....$

It has appeared many times, most recently here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170823


Find all solutions to $2^n-7=x^2$

I can't even think of a way to approach it.  :?

Is this supposed to be integer solutions only? The general solution by moving the seven over and logging both sides is \[ n=log_2(x^2+7)  \]

To get integer solutions x must be odd so that \[ x^2+7  \] is even. Upon further investigation, one finds that for x=1, 3, 5, x^2+7 is a power of 2 After that, however, such powers become more scarce, the next one being at x=11 and not occuring up to at least 63 (I didn't test beyond it). The answer to your question, then, if it is for all real numbers is \[ n=log_2(x^2+7)  \] but for all integers I'm not sure whether there are an infinite number of solutions of finite, but the first few of these are x=1, 3, 5, and 11.

[hide="The solutions are..."]
$(3, 1), (4, 3), (5, 5), (7, 11), (15, 181)$
[/hide]

[hide="Long proof"]
Rewrite as $2^{n}-8 = x^{2}-1 = (x+1)(x-1) = 8(2^{n-3}-1)$.  The first obvious solution comes when $n-3 = 0, x-1 = 0$, which gives us $\boxed{n = 3, x = 1}$.  Next, we notice that for all larger solutions, $x^{2}-1$ must be divisible by 8, which occurs if either $x+1$ or $x-1$ is divisible by 4.  We rewrite $x = 4k \pm 1$ and divide by 8, giving us $2^{n-3}-1 = k(2k \pm 1)$.

A little bit of calculation gives the first few solutions to this new equation as $\boxed{n=4, k=1, x=3, n=5, k=1, x=5}$ and $\boxed{n=7, k=3, x=11}$.  For all other solutions, therefore, we can assume $n > 7$.

EDIT:  This next paragraph is [b]incorrect[/b].  The negative case actually gives us $2^{n-4}= 4m^{2}+3m+1$ rather than $4m^{2}+3m-1$ and so it is not so easy to solve as I thought.   :blush:  Apologies that it took so many months to catch this mistake.

Taking the equation $\bmod 2$, then, tells us that $k$ must be odd.  We write $k = 2m+1$, giving us $2^{n-4}= (4m \pm 1)(m+1) \pm 1$.  If we replace $\pm$ with $-$, we get the equation $2^{n-4}= (4m-1)(m+1)$, which cannot have any solutions (if one of the factors is a power of two, the other one cannot be a power of two.)

We now know that $x-1$ was the factor divisible by 4.  Expanding gives us $2^{n-4}= 4m^{2}+5m+2$.  Taking the equation $\bmod 4$ tells us that $m \equiv 2 \bmod 4$, so we write $m = 4q+2$ and plug in.

This gives us $2^{n-6}= 16q^{2}+21q+7$.  Taking the equation $\bmod 16$ tells us that $0 \equiv 5q+7 \bmod 16$, from which we find $q \equiv 5 \bmod 16$, so we rewrite $q = 16a+5$.

This gives us, after considerable simplification, $2^{n-6}= 2^{12}a^{2}+2^{9}(5) a+2^{4}(21) a+512$.  Plugging in $a = 0$ gives us our next solution $2^{n-6}= 512 = 2^{9}\implies \boxed{n = 15, x = 181}$.  For all other values of $a$, we have that the RHS $> 2^{14}$, so we can safely assume $n > 20$.

This means that we can divide our equation by $2^{4}$, giving us, again after considerable simpification, $2^{n-10}= 2^{8}a^{2}+181a+32$.  Taking this equation $\bmod 32$ gives us $0 \equiv 21a \bmod 32$, so we rewrite $a = 32b$.

This gives us (more simplification) $2^{n-15}= 2^{13}b^{2}+181b+1$.  Taking $\bmod 32$ again (remember, $n > 20$) gives us $0 \equiv 21b+1 \bmod 32$, giving us, after some calculations, $b \equiv 3 \bmod 32$, so we rewrite $b = 32c+3$.  (Almost there!   :) )

Simplifying (A LOT) gives $2^{n-20}= 2^{17}c^{2}+2^{12}3 c+2^{8}3^{2}+198$, which does not have a solution for $c = 0$, so we can again safely assume $n > 36$.

A final division by two:  $2^{n-21}= 2^{16}c^{2}+2^{12}3c+2^{7}3^{2}+99$.  We have already safely assumed that $n > 36$, meaning the LHS is even.  However, the RHS is odd.  [b]Contradiction.[/b]

Hence, there are no solutions $x > 181$.  Q.E.D.

(Whew.)
[/hide]


Show that the number of ordered partitions of $n$ is $2^{n-1}$.

Example:
$n=3 \rightarrow (1,1,1),(1,2),(2,1),(3) \rightarrow 2^{3-1}=4$ partitions

Edit: No cheap induction, please. :)

break the number $n$ in $n$ ones.  you have $n$ ones:  $11111\cdots11$
there are $n-1$ spaces. now you can put a divider among these ones to make different numbers, or not. so there are $2^{n-1}$ ways.

Just an hour ago I solved the same problem. :lol:
same method as you did amirhtlusa.

This was on the Int. Probability midterm.

I got it from OUSMI 2004.
URL: http://personalwebs.oakland.edu/~echeng/

Teaching
APM 405 Combinatorics Eunemeration
Assignment 2


vector
linear algebra
linear algebra unsolved

Hello, let E a finite dimensional real vector space and let $ \theta \in \bigwedge^1(E)$ and $ \alpha \in \bigwedge^k(E)$.

Prove that if $ \theta \wedge \alpha \equal{} 0$ then exists $ \beta \in \bigwedge^{k\minus{}1}(E)$ such that $ \alpha \equal{} \theta \wedge \beta$.

thanks

Assuming $ \theta\neq 0$, take an explicit basis $ B$ of $ E$ containing $ \theta$ and consider the basis of $ \wedge^k E$ induced by $ B$.


AMC
AIME
MATHCOUNTS
trigonometry
graph theory

soo it looks like stevenmeow already started a thread for his

well, my mock test is already written and you have more time to do it!

it incorporates many different aspects of many different tests:

[size=200]DUE: DECEMBER 14TH[/size]

RULES:

10 story problems 10 min (*)must be 8th grade or less, must not have made AIME(*)
5 relay problems 10 min (*)open to all(*)
10 short answer 12 min (*)open to all who have not made MOP(*)
1 essay 25 min (*) open to all(*)
15 mental math problems 4 min (*)open to all(*)

You can find the problems here:
http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=241495
also, by joining this mock test, it is assumed that your results will be posted and ranked in a excel document

if you do not wish that your score is published, then please note on the PM when you send your solutions.

No signups are necessary, but please reply if you are going to do it to get a jist of number of people doing it.

if there are a lot of people participating, then i may need another grader (currently me and pythag011)

I'll probably do relay, short answer, essay, and mental math [b]if i have time[/b].  I can just do short answer, right?  Or just relay and mental math?

you can mix & match whatever you like as long as you meet the criteria

I'll do mental math tomorrow morning.

Do we need solutions for mental math, or just answers?

Err...Is this MathCounts level?  I mean you use sin in there, and trig not in MC.

[quote="dannyhamtx"]hmm well there isn't a better place to post this anyways and the story thing is kinda MC level
(you should have seen my last tourney here. the problems went to IMO and unsolved problem level, that COULD be solved with MC techniques)

and yes, submit answers only for all of them except essay.[/quote]

Yeah, but I don't think Hall's Thereom is mc level. More like IMO level techniques. And I can't find a RIGOROUS proof that does not use Hall's Thereom.

[quote="dannyhamtx"]you can definitely prove hall's marriage lemma without much difficulty of using advanced theorems, you know...... (if you have bunch of lemmas, that is)
and then there's also the unsolved problem (which i asked WOOT people and rruszyck) and it just was too hard[/quote]

Oh the last one? I managed to solve it using computer programming. I can prove my program works...

Also, GRAPH THEORY isn't mc level. Hall's Thereom definitely isn't.

But I understand your point. There are a huge number of things that can be proven using mc techniques, though some of them take a ridiculously long time. There are many problems on Olympiads that require no standard techniques...Russian Olympiad has a lot of these problems.

Anyways, is this forum just for mc level tournaments? Because I was wondering whether I should put my Mock Olympiad here...


inequalities
algebra
polynomial
algebra proposed

This is just Landau's inequality on the Mahler measure of a polynomial, that was posted by Vess very long time ago. You have there some solutions.http://www.mathlinks.ro/Forum/viewtopic.php?highlight=measure+mahler&t=19242


calculus
integration
function
real analysis
real analysis unsolved

Let P>0 and real

Prove that the integral with respect to x of the function log(1+P(Tanx)^2) between the lower limit 0 and upper limit pi/2

is pi.log[1+sqrt(p)]


1. A cave contains 2 brown bats and 3 gray bats. The bats fly out of the cave one at a time. What is the probability that the fourth bat to fly out of the cave is gray? Express your answer as a common fraction.

2. What is the probability (from the same cave) that the second AND fourth bats are gray?

3. If all the bats are distinct individuals, how many ways can they fly out of the cave?

4. What if the grey bats are indistinguishable from each other and the brown bats are indistinguishable from each other?

5. What is the probability of drawing from a deck of cards an even numbered card followed by an odd numbered card (counting A as 1) followed by a face card, if all draws are without replacement?

6. What if the draws were with replacement?

7. What if the draws were with replacement, but we could draw the cards in any order?

8. What if the draws were without replacement, but we could draw the cards in any order, and the face card had to be a king?


hi !!

i have spent a lot time with this inequality but i haven't found the

demonstration .

\[ a,b,c \geq 0 \] and \[ ab + bc + ca = 3 \]

prove that : \[ \sqrt{a+3}+\sqrt{b+3}+\sqrt{c+3}\geq 6 \]

thanks advanced .

See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=69301
posts #4,6 ;)

sorry,i don't see any thing useful .please,can any one

help me to prove that inequalitie ?

[quote="toetoe"]
$a,b,c \geq 0$ and $ab + bc + ca = 3$
prove that :$\sqrt{a+3}+\sqrt{b+3}+\sqrt{c+3}\geq 6$
[/quote]
For you, toetoe. :)
$\sqrt{a+3}+\sqrt{b+3}+\sqrt{c+3}\geq3\cdot\sqrt[6]{(a+3)(b+3)(c+3)}.$
Remain to prove that $3\cdot\sqrt[6]{(a+3)(b+3)(c+3)}\geq6.$
But $3\cdot\sqrt[6]{(a+3)(b+3)(c+3)}\geq6\Leftrightarrow(a+3)(b+3)(c+3)\geq64\Leftrightarrow$
$\Leftrightarrow abc+3(ab+ac+bc)+9(a+b+c)+27\geq64\Leftrightarrow9(a+b+c)+abc\geq28\Leftrightarrow$
$\Leftrightarrow81(a+b+c)^2+18abc(a+b+c)+a^2b^2c^2\geq784\Leftrightarrow$
$\Leftrightarrow81(a^2+b^2+c^2)+18abc(a+b+c)+a^2b^2c^2\geq298\Leftrightarrow$
$\Leftrightarrow9(a^2+b^2+c^2)(a+b+c)^2+6abc(ab+ac+bc)(a+b+c)\geq\frac{298(ab+ac+bc)^3}{27}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(243a^4b^2+243a^4c^2+486a^4bc-298a^3b^3-$
$-246a^3b^2c-246a^3c^2b-182a^2b^2c^2)\geq0.$ The last inequality is obviously true.

I think if $a,$ $b,$ $c$ are non-negative real numbers and $ab+ac+bc=3$ then
$13(a+b+c)+6abc\geq45.$ ;)


MATHCOUNTS
geometry
geometric transformation
blogs

By the time a party is over, $ 28$ handshakes have occurred. If everyone shook everyone else's hand once, how many people attended the party?

[hide="Click for solution"]
If everyone shakes everyone else's hands, then each pair of people produces a handshake.  Thus, $ \binom{n}{2}=28$, where $ n$ is the number of people in the party.  Thus:

\begin{align*}
\binom{n}{2}&=28
\\ \frac{n(n-1)}{2}&=28
\\ n(n-1)&=56
\\ n&=\boxed{8}
\end{align*}

There are $ \boxed{8}$ people in the party.
[/hide]

This was on their TST?
Their [b][i]IMO[/i][/b] TST?

This could be on a MATHCOUNTS School Sprint round, arguably in the first 5-10 problems.

Unfortunately yes, it's a bit sad but you have to understand Puerto Rico's Math Olympics is Relatively new, like 8 or 9 maybe 10 years old, from what I believe, this problem is on the test so that the 7th and 8th graders don't feel discouraged from the rest of the test that they might not be able to get, (note 7th-12th graders take the same test). While the Students in the higher levels fight for the best solution.

yeah, but still.... itd be cool to go to imo!
i wish i lived in puerto rico.

[quote="grn_trtle"]Hmm... even so, they're kind of banking on the best students to make it through. If a beginner manages a perfect score on the TST and makes the IMO team, he/she won't be able to keep up with anyone else at the actual IMO.[/quote]

Haven't you seen our scores? That's mostly what happens (we can't keep up with other teams). Mathwise, we're way below average. Still, I enjoyed owning that test. :)

I wonder if I can travel to Puerto Rico, make IMO, and come back to the States and write that in my college applications...

It'd be pretty hard, first off normally they only have enough money to send 3 students to the IMO, maybe that's changing this year but I'm not sure, and plus the first round is always in spanish(From what I have seen), all the subsequent tests have translations to english, but this test was easier than other years, I would post the other years, but right now i'm too tired. Metafor do you want to post them?(I know you have them in your blog)

You will find last year's "TST" [url=http://www.mathlinks.ro/Forum/weblog_entry.php?t=216695]here[/url].


That's wrong.  If you show your work i can tell you what you did wrong. :|

[quote="colts18"][hide] 3.
the 1st one $\approx$ 1, the 2nd one $\approx$ 2. 1+2 $\approx$ 3 [/hide][/quote]

  colts18, if you're going to edit your posts please say something like EDIT: I've changed the answer, otherwise the rest of us think we've got the problem wrong too.

[quote="colts18"][hide] 3.
the 1st one $\approx$ 1, the 2nd one $\approx$ 2. 1+2 $\approx$ 3 [/hide][/quote]
you wrote the first one is
[hide]about 1 but it is one!
same with the second.
and 1+2 is 3
not approximatly 3[/hide]

lol math92, why do you have to be like that?  :P  [hide]720 * 6 *4 is the same thing as 6! * 4!... interesting.[/hide]


AMC
AIME
analytic geometry
geometry
3D geometry

can't give u the Q coz it's in Chinese and I can't translate

it gives you three coordinates which are P(7,12,10) Q(8,8,1) R(11,3,9)
and ask u the surface area of the "??"
I'm sure it's not a cube coz my answer is wrong

the answer is 294

It actually is a cube. We find that $PQ = QR = RP$, which means the lengths have to be face diagonals (try arranging it in a different way, it's not possible). So the surface area is $6\left(\frac{PQ}{\sqrt{2}}\right)^2 = 3(PQ)^2 = 3(1^2+4^2+9^2) = 3(98) = 294$.

thx...
I just didn't realize it's diagonal....


From a random image search on AoPS:

[img]http://www7.big.or.jp/~katsurao/fp/img/f-aops.jpg[/img]

Super indeed.   :D

Man...that realli puts the energy drink market into perspective. And I thought Red Bull had a cool logo...

[quote="Valentin Vornicu"]I think I actually drank one of those in Tokio, two summers ago :)[/quote]

Was it good?  :P

Best regards,


Is it possible to do this problem with using sin and cosin?
Meaning, can we do this problem, by only knowing that the ratio of the opposite to the adjacent sides in any triangle with an angle of interest measuring 10 degrees is always 0.1736:

Find the adjacent and opposite sides in a triangle with an angle of interest measuring ten degrees and a hypotenus measuring 0.1777.


Let me get one thing straight: I DO NOT HATE MEMMING just those who use it to ensure victory and change the game from one of math skill to one of memorization.
All of us use memming occasionally. I DO TOO, BUT only if the question is beyond my knowledge and i am not ready to learn it. I look everyday i am here as people are getting answers in under 1 sec. Although a 1.5 rule has been added, I still see many people memorizing answers. To stop this i am designing a club for those of use who oppose those who have significantly high rating due to "mem cheating"
and those who when they mem deny it always. So I plan that we will make those people ashamed by creating a vast club of people who visit aops regularly or at least visit it occasionally. We will ignore constant memmers at all turns. Using the  /ignore command we will make them stop just to be payed attention to. Now we will not stop from talking to them if we want to but at least not playing any games with them. Maybe this will be locked or deleted but i hope that many will join my cause to make an end of memming! Here are the requirements:

highest rating of no more then 1700 unless approved by me or the council i plan to make.
a recommendation of one of the first 20 i approve.
for the first 20 pm me and I will investigate and see if you are worthy.

being in this club does not mean you will stay. If ever i notice that you start to mem a lot you will be rejected and you must try to come back.

I do no mean to offend anyone by this but i see a need to stop those who will ruin the experience of a newcomer here.

I DO NOT deny memming.  Isn't this sorta a copy of CTC with a different theme?

Anyway, memming is now a PART OF THE GAME.  A way to stop memming is to have an infinite supply of problems, that NEVER repeat.  Obviously, this is humanely impossible.

The only other way to stop memming is to stop playing the game for a while.  Who wants to do that?

Memming isn't cheating.  There is no rule that explicitly states or even hints at "no memming".

Anyway, shouldn't this thread be in the FTWOS forum?

[quote="mathblitz"]Let me get one thing straight: I DO NOT HATE MEMMING just those who use it to ensure victory and change the game from one of math skill to one of memorization.
All of us use memming occasionally. I DO TOO, BUT only if the question is beyond my knowledge and i am not ready to learn it.[/quote]

You mem questions on purpose? I would say this is cheating. Almost everyone I know including myself, only mems when we recognize problems. Use me as an example. In a period of 6 months, I have played almost 4000 games. I play FTW around 4 hours a day. Wouldn't you expect I would recognize quite a few problems? But I never mem questions without knowing how to solve them. If I remember the answer but don't understand how to solve it, I try to solve it first. You shouldn't mem questions just because you don't know how to solve it. [b]That is true cheating, and liable for banning.[/b]

[quote]I look everyday i am here as people are getting answers in under 1 sec. Although a 1.5 rule has been added, I still see many people memorizing answers. To stop this i am designing a club for those of use who oppose those who have significantly high rating due to "mem cheating"[/quote]

You see, everyday, people getting questions under 1 second? I'm on for 4 hours a day, and even before the 1.5 rule, I rarely saw under 1 second answers. And didn't you just say you memorized answers on purpose? Also, levans changed the rating system around 2 months ago, so that's why so many people have "significantly high ratings".

[quote]and those who when they mem deny it always.[/quote]

Do we? Everytime I mem a question, I always say "sry mem"(to the point where it's my trademark. :P )

[quote] So I plan that we will make those people ashamed by creating a vast club of people who visit aops regularly or at least visit it occasionally. We will ignore constant memmers at all turns. Using the  /ignore command we will make them stop just to be payed attention to. Now we will not stop from talking to them if we want to but at least not playing any games with them. Maybe this will be locked or deleted but i hope that many will join my cause to make an end of memming![/quote]

Isn't this a little far? I've played the most games on FTW, and I only mem about 1 or 2 questions a game, especially since a lot of new questions were added to a game.

[quote]I do no mean to offend anyone by this but i see a need to stop those who will ruin the experience of a newcomer here.[/quote]

It doesn't ruin the experience of a newcomer. When I first started, in March, I was a newcomer. Within 3 months my rating was decent, about 1300. And if you're memorizing answers on purpose....

I mem a lot. I know I do, around 10% of problems. Is it truly my fault for playing 4000 games? Of course not. I [b]never[/b] mem on purpose. If I did, I would probably have been caught a long time ago, as would many, many others. You, however, have just admitted you memorize answers on purpose when they're too hard. Do you think anyone's going to join a club headed by the worst kind of memmer?

[quote="isabella2296"][quote="mathblitz"]Let me get one thing straight: I DO NOT HATE MEMMING just those who use it to ensure victory and change the game from one of math skill to one of memorization.
All of us use memming occasionally. I DO TOO, BUT only if the question is beyond my knowledge and i am not ready to learn it.[/quote]

You mem questions on purpose? I would say this is cheating. Almost everyone I know including myself, only mems when we recognize problems. Use me as an example. In a period of 6 months, I have played almost 4000 games. I play FTW around 4 hours a day. Wouldn't you expect I would recognize quite a few problems? But I never mem questions without knowing how to solve them. If I remember the answer but don't understand how to solve it, I try to solve it first. You shouldn't mem questions just because you don't know how to solve it. [b]That is true cheating, and liable for banning.[/b]

[quote]I look everyday i am here as people are getting answers in under 1 sec. Although a 1.5 rule has been added, I still see many people memorizing answers. To stop this i am designing a club for those of use who oppose those who have significantly high rating due to "mem cheating"[/quote]

You see, everyday, people getting questions under 1 second? I'm on for 4 hours a day, and even before the 1.5 rule, I rarely saw under 1 second answers. And didn't you just say you memorized answers on purpose? Also, levans changed the rating system around 2 months ago, so that's why so many people have "significantly high ratings".

[quote]and those who when they mem deny it always.[/quote]

Do we? Everytime I mem a question, I always say "sry mem"(to the point where it's my trademark. :P )

[quote] So I plan that we will make those people ashamed by creating a vast club of people who visit aops regularly or at least visit it occasionally. We will ignore constant memmers at all turns. Using the  /ignore command we will make them stop just to be payed attention to. Now we will not stop from talking to them if we want to but at least not playing any games with them. Maybe this will be locked or deleted but i hope that many will join my cause to make an end of memming![/quote]

Isn't this a little far? I've played the most games on FTW, and I only mem about 1 or 2 questions a game, especially since a lot of new questions were added to a game.

[quote]I do no mean to offend anyone by this but i see a need to stop those who will ruin the experience of a newcomer here.[/quote]

[b]It doesn't ruin the experience of a newcomer. When I first started, in March, I was a newcomer. Within 3 months my rating was decent, about 1300. And if you're memorizing answers on purpose....

I mem a lot. I know I do, around 10% of problems. Is it truly my fault for playing 4000 games? Of course not. I never[/b] mem on purpose. If I did, I would probably have been caught a long time ago, as would many, many others. You, however, have just admitted you memorize answers on purpose when they're too hard. Do you think anyone's going to join a club headed by the worst kind of memmer?[/quote]

Especially if it's an ANTI-MEMMERS CLUB?!?!?!?
I didn't see that part:
[quote="mathblitz"]I DO TOO, BUT only if the question is beyond my knowledge and i am not ready to learn it.[/quote]
Or else I woulda got you yesterday with that one.  Seriously, memming on purpose?  Hard problems shouldn't just be MEMMED, you should learn to solve them!

EDIT:  31^2 posts

i'm gonna make this short and simple, since i'm at the library and i only have 9 minutes left

memming is unfortunately part of the game now
it is impossible to not mem, because many of the problems have already been seen by pplz

the only possible way is to not play for a few months (which i have done)

memming on purpose is a different thing
that is basically cheating and you should not mem just because you don't know how to solve it
ask on the forum or ask someone in the lobby to explain how to solve the problem

with this said, i'm going to leave this thread open for discussion, as i would like to hear the thoughts of other people
plz note that if and when this thread starts getting to be more of a fight than a discussion, it will be locked

Ok, this is the dumbest thing I've ever heard.  You're avoiding memmers?  And you said it yourself.  You mem too.  So [i]if[/i] I was a mod, I would lock this up.  But I'm not.  Someone else wanna do the honors?

Hmm, I guess I will lock. This is a rather pointless discussion. Oh, and mathblitz? I highly recommend that you stop memming on purpose. Just a bit of advice... :wink:

Locked.


Call a number heavy if each digit is less than or equal to the on its immediate right. For exampe, 6421 is a heavy number. How many 4 digit heavy numbers exist?

[hide]
if the first digit (on the left) is 1, we have 4 heavy number, 1000, 1100,1110,1111. divide those number in this way: if the second digit is 0, we have just 1 possibility, if the second digit is 1, we have (1+2) possibilities. If the first digit is 2, we must add 6 possible heavy numbers, cause if the second digit is 2, we have (1+2+3) more heavy numbers, and so on until 9. So the number of 4 digits heavy numbers is $9 \cdot 1+9 \cdot (1+2)+8 \cdot (1+2+3)+...+1 \cdot (1+2+...+10)=714$

we  can see it following this scheme:

first digit----------heavy numbers
1------------------1+(1+2)
2------------------1+(1+2)+(1+2+3)
3------------------1+(1+2)+(1+2+3)+(1+2+3+4)
.
.
.
9------------------1+(1+2)+(1+2+3)+(1+2+3+4)+...+(1+2+3+...+10)

[/hide]

EDIT

I am probably wrong.
[hide]
Okay lets see how many heavy numbers have $4$ distinct digits. This is $\binom{10}{4}= 210$ because there is only one way of arranging $4$ distinct digits to make a heavy number.

Now lets see how many heavy numbers have $3$ different digits. There are $\binom{10}{3}= 120$ ways to choose 3 digits but for digits $x > y > z$, there are $3$ ways to arrange them such that one of them has $2$ digits. We can have xxyz, xyyz, or xyzz. So there are $120*3 = 360$ ways to create a heavy number from $3$ different digits.

Now lets try $2$ digits. $\binom{10}{2}= 45$. But there are 3 ways to arrange them to create a heavy number. For digits x > y, we can have xxyy xxxy or xyyy so the total number of ways to create a heavy number is $45*3 = 135$.

Now for $1$ digit, we have $9$ ways because we cant have 0000.

The final answer is $210+360+135+9 = \boxed{714}$
What did i do wrong?
[/hide]

hydro, i follow your reasoning but something seems wrong with the final sum. try writing out the intermediate terms. it can't follow that pattern. the last term would be $1(1+2+\ldots+10)$ right? I think that if you do the sum the way you intended to, you get shoefactory's answer, [hide]$714$[/hide], which is what I got.  :wink:

shoefactory, your method is very similar to mine.

[hide]$714$[/hide] is the answer

I made this problem up but it seems like it could be a good AIME problem (in the easy range)

yes I just got wrong with intermediate terms, the last one is the one you wrote, so my answer's right :wink:

Much easier method:
[hide]
We have 4 slots and 10 different digits.

So we want to arrange 4 C's (which represent "I choose this digit") and 9 S's (which represent "Skip this digit and go to the next"). We don't need one for the last digit, or 0, just a C if we choose it.

We will have something like this:

$SCCSSCSSSCSSS$
Which represents $2247$

$\binom{13}{9}= 715$
Now subtract $1$ to kill $0000$ so the final answer is $\boxed{714}$[/hide]


[quote="mewto55555"][quote="peregrinefalcon88"]go mewto5555[/quote]

I would like to debate this claim, do you have any sound evidence of this "fact"?[/quote]

"Go mewto" isn't a fact.

Oh, and this thread is not only spam, it's also spreading lies.  :P

ok...
a few things i would like to point out

1. this is spam
there is no use to it
and worse yet, you created a new topic for this spam, filling up FTW other stuff, as well as the topic

2. you spelled mew's name wrong (there are five 5's)

3. it's pretty obvious you are just posting for a higher post count, otherwise, you wouldn't do this
do now post just for the heck of it
only post if you think it is necessary

the point of this post is to tell you pplz not to just post because you feel like it
remember, to chat on FTW, you not only need to have a high post count, you have to have a good post rating
posting spam will not get you a high rating

You could get banned for posting this kind of spam.

I'm serious. :P


i) $ \{a_{i}|\ i\in \mathbb{N}^{*}\}\subset \mathbb{Z}$
ii) $ |\{a_{i}|a_{i}>0\}|=+\infty$ and $ |\{a_{i}|a_{i}<0\}|=+\infty$
iii) $ \forall (i,j,n)\in \mathbb{N}^{\ 3},\ 0<i<j\le n \Longrightarrow \ n\not| a_{i}-a_{j}$

prove that $ f: \mathbb{Z}\to \mathbb{Z}$ such that $ f(i)=a_{i}$ is bijectiv
and find $ f$.

are you talking about this IMO 2005 problem
http://www.mathlinks.ro/Forum/viewtopic.php?p=281572#281572 ?


linear algebra
matrix
vector
algebra
linear equation

Well, I'm back with another question, and i'll probably be back with more questions later since I'm trying to study for finals.

anyways...

How should I go about finding a basis for a Symmetric 2x2 Matrix.

Hints:

1. What is the dimension (as a vector space) of the set of all $2\times 2$ matrices?

2. Can you write the condition "$A$ is symmetric" as a linear equation or a set of linear equations? (A homogeneous system of linear equations, I should say.) How many such equations? Each such equation you need should subtract 1 from the dimension of your solutions set.

Knowing the dimsension ("number of degrees of freedom") of the set you seek should help you in looking for a basis.

Sorry I've been trying to rap my head around your hints and I think i've lost myself even more now.

A 2x2 matrix consist of 4 entries, so the space of all 2x2 matrices is 4-dimensional. E.g. you need 4 matrices in a base to span this entire space. Usually they'll be:

$A = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right]$, $B = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$, $C = \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right]$, $D = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right]$.

So any 2x2 matrix can be written as a linear combination like so:

$\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] = aA + bB + cC + dD$.

A 2x2 symmetric matrix essentially only consist of 3 entries, since the last will be determined uniquely. My suggestion for a base will be:

$P = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right]$, $Q = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right]$, $R = \left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right]$.

Now, any 2x2 symmetric matrix can be written as:

$\left[ \begin{array}{cc} p & q \\ q & r \end{array} \right] = pP + qQ + rR$.

the diagonal entries need not be the same.

Whoops ...  :blush:

My finals must have got to me at last.

(Fixed it)

The space of symmetric $3\times 3$ matrices will be 6-dimensional. Here's a basis for that:

$\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix}, \begin{bmatrix}0&0&0\\0&1&0\\0&0&0\end{bmatrix}, \begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix},$

$\begin{bmatrix}0&1&0\\1&0&0\\0&0&0\end{bmatrix}, \begin{bmatrix}0&0&1\\0&0&0\\1&0&0\end{bmatrix}, \begin{bmatrix}0&0&0\\0&0&1\\0&1&0\end{bmatrix}$

In general, the dimension of the space of $n\times n$ symmetric matrics will be $n+{n\choose 2}=\frac{n(n+1)}2,$ and the same idea as above can be used to find the basis.

Exercise: find a basis for the set of antisymmetric $3\times 3$ matrices, where "antisymmettric" means that $A^T=-A.$


vector
function
linear algebra
matrix
linear algebra unsolved

Let $ V$ be a vector space with inner product, $ L(V,V)\equal{}\{ T: V\rightarrow V ; T$ is linear$ \}$, and let $ *: L(V,V) \rightarrow L(V,V)$ be the function such that if $ T\in L(V,V)$ then $ *(T)\equal{}T^*$    ($ T^*$ is the adjoint operator of $ T$) .

i) Prove that $ *$ is a linear transformation (easy)
ii) Construct an inner product on $ L(V,V)$ such that $ *$ is self-adjoint

Since the question is posted in Matrix - Linear Algebra forum, and you don't mention the continuity of operatos under consideration, I supose you are dealing with finite-dimensional spaces. In this case operators can be considered as $ n \times n$ matricies (in a fixed orthonormal basis) and the requested inner product of $ A\equal{}(a_{k,j})$ and $ B\equal{}(b_{k,j})$ is
\[ \langle A, B \rangle \equal{} \sum_{k,j} a_{k,j} \overline{b_{k,j}}.\]

[quote="volodya"] inner product of $ A \equal{} (a_{k,j})$ and $ B \equal{} (b_{k,j})$ is
\[ \langle A, B \rangle \equal{} \sum_{k,j} a_{k,j} \overline{b_{k,j}}.\]
[/quote]

Write it as: $ \mathrm{tr}(AB^*)$ to see that the result does not depend on the choice of basis.


[quote="TALENTO"]In each box, the following figure, distinct positive integers are placed such that the sum of two numbers of neighboring cells is a prime number. What is the smallest possible sum of all numbers written?
Clarification: Two cells are neighbors if they have a side in common.[/quote]

Since $ 2$ cant be obtain as sum of two neighbour cases, all primes are odd and adjacent cases are odd/even. So we have either $ 6$ odd cases and $ 4$ even cases, either $ 6$ even cases and  $ 4$ odd cases.

The fewest sum possible would then be with $ 6$ odd cases ($ 1,3,5,7,9,11$) and $ 4$ even cases ($ 2,4,6,8$), which give the sum $ 56$

And this sum may be reached : apply the numbers (left to right, upper line to lower line) in the following manner :
Upper line : $ 6,7,5$
Middle line : $ 4,1,2,9$
Bottom line : $ 8,3,11$

Hence the answer : $ 56$


How would you think the results on American Idol would be different if viewers voted for their least favorite contestant rather than their most favorite?  Is this just a pointless quesion not worth thinking about or is there anything mathematical about it?

[quote="paradox"]Now if only you had made this into a show before [url=http://www.thewb.com/Shows/GenericShow/0,11116,169159,00.html]The WB[/url] did.... Though, I concede, "worst" is not necessarily also the least favorite.[/quote]
yup, that's true.
that's a pretty good question though. it'd benefit the people who are pretty much in the middle and don't get a ton of votes, but who knows?

I think it may also help the people who are "under the radar", but I agree, who knows.

yea...argh...Fantasia should've been voted off yesterday...*mutters*

Poor Jasmine...*sigh*

are you crazy?!?!?! fantasia is way better than jasmine

Based on the pervious day's singing fantasia should of won, becauae of that flat note by diana, but i feel that diana is a better singer over all. "I believe" is a waaaaaaaaay better song for diana than fantasia.

[quote]It was said early on that none of them are as goodly as recent competitors. Personally, I think Fantasia's voice is raspy.[/quote]

wow, i thought i was the only one o.O she sounds pretty bad on the radio single...[/quote]

[quote]bah Jasmine should've won. period.[/quote]

Is that vote consistent with your race?

I'd personally be interested in seeing how people voted when the top three were voted on earlier. I'm curious to see if a majority of whites voted for Diana, a majority of black people for Fantasia and a majority of asian/pacific/islandrs voted for Jasmine.


integration
limit
real analysis
real analysis unsolved


function
algebra
polynomial
limit
complex analysis
calculus
derivative

Hi everybody,

I submit you theses problems which seem interesting :

1) Let  $ f \in \mathcal{H}(\mathbb{C})$ such that $ f^{ \minus{} 1}(\mathbb{R}) \equal{} \mathbb{R}$ show that $ \exists a \in \mathbb{R},b\in \mathbb{R}$ with $ f(z) \equal{} az \plus{} b$.

2)Let $ f,g \in \mathcal{H}(\mathbb{C})$ such that $ f^{ \minus{} 1}(\mathbb{R}) \equal{} g^{ \minus{} 1}(\mathbb{R})$.

Do we have a $ (a,b,c,d) \in \mathbb{R}^4$ $ a \neq 0, c \neq 0$, and $ \tau_1(z) \equal{} az \plus{} b$ and $ \tau_2(z) \equal{} cz \plus{} d$ such that $ f(z) \equal{} \tau_1(g(\tau_2(z)))$ ?

3) What can we say about an entire map $ f$ such that $ f^{ \minus{} 1}(\mathbb{Z}) \equal{} \mathbb{Z}$

(in particular , $ f$ might be of the form $ f(z) \equal{} \pm z \plus{} n$ ...)

I have solved the problem 1) , and i did not succeed in solving 2) and 3). could someone help me  :D

For #3 you need to get some estimates for the growth of $ f$ (I surmise that is how you started #1 too). It [b]is[/b] possible to show that an entire function $ f$ satisfying $ f^{\minus{}1}(\mathbb Z)\subset\mathbb Z$ must be a polynomial mimicking one of the proofs of the Picard theorem but before showing you how to do it, I'd like to know what is the best growth estimate you can get yourself. :)

Hello fedja,

Concerning 1), i am interested in seeing how you can get some estimates,what i was unable to do until now.

 Nevertheless, i managed solving the problem in this way :

Let $ f$ an entire function such that $ f^{\minus{}1}(\mathbb{R})\equal{}\mathbb{R}$. Suppose we have $ f(a)\equal{}f(b)$ for some $ a,b \in \mathbb{R}$ , then the restriction to $ [a,b]$ of $ f$ admits a local extremum $ x_0$ in $ ]a,b[$. Let 's say a maximum for instance.
 We fix $ \alpha$ sufficiently small, $ f(B(0,\alpha))$ is an open set containing $ f(x_0)$ so containing also $ f(x_0)\plus{}\epsilon$ for some $ \epsilon >0$ . But by hypothesis $ f(x_0)\plus{}\epsilon$ must be reached on $ ]a,b[$ which contradicts the fact that $ f(x_0)$ is a local maximum. (indeed $ f$ is not constant so is an open map)
Consequently, $ f$ is injective on $ \mathbb{R}$,but in virtue of M.Picard theorem if $ f$ was not polynomial , there would exist $ c \in \mathbb{R}$ reached infinitely many times. Then $ f$ is a polynomial map, and if his degree was more than 2, $ P(z)\equal{}x,x \in \mathbb{R}$ would admits more than two solutions contradicting the injectivity.

Now, concerning 3),

let $ f$ an entire map such that $ f^{\minus{}1}(\mathbb{Z})\subset \mathbb{Z}$, equivalently $ f: \mathbb{C}\minus{}\mathbb{Z} \longrightarrow \mathbb{C}\minus{}\mathbb{Z}$.
Now $ f^{n}\equal{}fofofo...f$ must be a normal on $ \mathbb{C}\minus{}\mathbb{Z}$ family using strong montel theorem.
This leads to three cases

1)$ f^n$ converges to $ \infty$ on every compact of $ \mathbb{C}\minus{}\mathbb{Z}$, we use that if $ f$ is not affine, $ fof$ admits a fixed point $ z_0$ (which follows from little picard theorem studying $ \frac{fof(z)\minus{}z}{f(z)\minus{}z}$ ). In this case can notice that $ z_0 \in \mathbb{Z}$ because $ f^n(z_0)$ would tend to infinity otherwise.
Let $ p\equal{}z_0$, and $ q \in \mathbb{Z}\minus{}\{p,f(p)\}$, we considere $ g_n(z) \equal{}\frac{1}{f^n(z)\minus{}q}$ on $ B(p,0.5)$. $ g_n$ is well defined, because $ f^n(z)\equal{}q$ implies $ z \in B(p,0.5)\cap \mathbb{Z}$, and so $ z\equal{}p$. But using cauchy formula we get that $ \lim_{n \to \infty} g_n(p)\equal{}0$, which is contradictory.
Consequently, $ fof$ has no fixed point and then $ f$ is affine.

2) there exists $ n_k$ ang $ g$ holomorphic non constant on $ \mathbb{C}\minus{}\mathbb{Z}$ such that $ f^{n_k}$ converge to $ g$ uniformely on every compact subset of $ \mathbb{C}\minus{}\mathbb{Z}$. Note that the convergence and $ g$ can be extended in the whole complex plane by using  cauchy formulas around the integers. Now $ f^{n_{k\plus{}1}\minus{}n_k}$ is also a normal family, and the only possible limit $ h$ satisfies $ hog(z)\equal{}g(z)$ and so $ h(z)\equal{}z$. We deduce that $ f$ is an entire injective map, so affine.

3)There exists $ n_k$ such that $ f^{n_k}$ converge to some constant c ,on every compact of $ \mathbb{C}\minus{}\mathbb{Z}$, which can be extended to every compact of the complex plane by using cauchy formulas. $ c$ has to be a fixed point of $ f$, and must be trivially the only fixed point.
Now, we notice that $ g(z)\equal{}f(z\plus{}1)$ satisfies the same hypothesis.
By using 1) and 2) above, if $ g$ is not affine, $ g$ admits an only fixed point.
$ h(z)\equal{}f(z)\minus{}z$ admits $ 0$ and $ \minus{}1$ as exceptional picard values which is impossible except if $ f$ is polynomial. We can easily see that such a polynom would be an affine one.

I don't have a clue to get some estimates on $ f$, could you please explain how to do this?

(Searching for some estimate lead me to the following problem i could not solve :
if $ f$ and $ g$ are not polynomiales, how to prove that $ fog$ increases faster than $ f$ in some sense.)
Best regards,
popolux :)

You have quite an ingenious proof for #3.  :thumbup:

[quote="popolux"]
(Searching for some estimate lead me to the following problem i could not solve :
if $ f$ and $ g$ are not polynomials, how to prove that $ fog$ increases faster than $ f$ in some sense.)
[/quote]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89087]This[/url] may give you some ideas.

[quote="popolux"]
I don't have a clue to get some estimates on $ f$, could you please explain how to do this?
[/quote]
Are you familiar with Schottky's lemma (if a function is analytic in a disk and the derivative at the center is large, then the image contains a big disk)?

Hello Fedja! sorry to be so long to answer  :blush:

I didn't know your lemma, the lemma i know is the theorem of bloch which asserts things similar.
Could you please give me a reference to your lemma?