I assume what robinhe put are the questions:
[hide="numero uno"]
Simplify each square root:
$12\sqrt{2}-25\sqrt{2}-24\sqrt{5}+14\sqrt{5}$
Now factor:
$(12-25)\sqrt{2}(-24+14)\sqrt{5}$
Sumplify:
$\boxed{\boxed{\boxed{\boxed{\boxed{-13\sqrt{2}-10\sqrt{5}}}}}}$[/hide]
[hide="numero dos"]
$(-2)^2=4$
$-4^2=-16$
$\frac{4}{-16}=-\frac 14$
[/hide]
[hide="numero tres"]
Factor each term:
$2x-10\rightarrow2(x-5)$
$x^2-25\rightarrow(x-5)(x+5)$
$x^2-2x-15\rightarrow(x-5)(x+3)$
$x^2-4x-5\rightarrow(x-5)(x+1)$
Now put these back intot the original equation.
$\frac{2(x-5)}{(x-5)(x+5)}\div\frac{(x-5)(x+3)}{(x-5)(x+1)}$
Find the inverse of the second one to change it to multiplication:
$\frac{2(x-5)}{(x-5)(x+5)}\times\frac{(x-5)(x+1)}{(x-5)(x+3)}$
Now cancel out the common factors:
$\frac{2}{x+5}\times\frac{x+1}{x+3}$
Put that into one fraction:
$\frac{2(x+1)}{(x+5)(x+3)}$[/hide]
[hide="numero quatro"]
o geez this problem is ridiculous[/hide]

[quote="robinhe"]In the first one, are those square root signs? If so, here are the problems:

#1 $3\sqrt{32}-5\sqrt{50}-8\sqrt{45}+7\sqrt{20}$

#2 $\frac{(-2)^2}{-4^2}$

#3 $\frac{2x-10}{x^2-25} \div \frac{x^2-2x-15}{x^2-4x-5}$

#4 $\frac{y^0x^{-8}}{x^{-7z}}\left(\frac{\left(y^5\right)^{-2}}{\left(z^0\right)^9}-\frac{x^{11}}{y^4}\right)$

[hide="Number 1"]$12\sqrt{2}-25\sqrt{2}-24\sqrt{5}+14\sqrt{5}=-13\sqrt{2}-10\sqrt{5}$[/hide]

[hide="Number 2"]It becomes $\frac{4}{-16}=-\frac14$.[/hide]

[hide="Number 3"]$\frac{2(x-5)}{(x+5)(x-5)} \times \frac{(x-5)(x+1)}{(x-5)(x+3)}=\frac{2(x+1)}{(x+3)(x+5)}$[/hide]

[hide="Number 4"]It simplifies to $x^{7z-8}\left(y^{-10}-\frac{x^{11}}{y^4}\right)$ but I'm too lazy to multiply even more.  :sleeping: [/hide]

I probably made several typos.  :wacko:[/quote]

[hide] 1) $-13\sqrt{2}-10\sqrt{5}.$
2)$-1/4$[/hide]


I never learned this in school. Some other people tell me they learned this but I never heard of it. Is it important that I should learn it? I don't even know what it stands for. Thanks.

Can you tell us is it a short name for something (never heard of LAM)?

I assume it is (as the OP states "I don't even know what it stands for"-so, linguistically deducing, it should be an abbreviation). I'd like to know in which area of mathematics is the term used, because a thorough web search I've done in past couple of minutes didn't help me find out anything about it.

Perhaps you're mishearing them, and they're saying LEM? That could be Law of the Excluded Middle, which is basically this:
If you're given the statement "A is B" then either A IS B, or A IS NOT B.
This is pretty trivial, but if I remember right, there are some other interesting properties that extend to propositional logic.

Acronymfinder doesn't return any explicitly mathematical results.  Are you sure you heard correctly?  In what context was the term used?

No real context in when they say it, just heard it mentioned a couple of times. Didn't know it was that obscure.
Could it be "Liberal Arts Math"? I found that on some searches but still can't find any information about what it is.

And not to get off topic but...
[quote]If you're given the statement "A is B" then either A IS B, or A IS NOT B.
[/quote]
I don't understand that. If A is B, how could A ever not be B? Unless the "statement" was stated by a liar, which doesn't really make sense to even factor.

As it's name states, LEM excludes "the middle possibility": A is B or A isn't B is possible, but "A both is and isn't B", or "A is a little bit B" isn't possible.

For instance, you have two different reals: a and b. Using Principium tertii exclusi (LEM) we can say: it is either a>b or a<b. There's no third possibility.

That's not a consequence of the law of the excluded middle; depending on how you construct the reals, they may be defined to have that property.

What I hope is a more relevant example is the following:  the statement that "this sentence is false" is only a paradox if one assumes that the sentence can only be either true or false.  In other words, any proof by contradiction relies implicitly on the law of the excluded middle (which I believe is responsible for rrusczyk's disapproval of proof by contradiction).

[quote]That's not a consequence of the law of the excluded middle; depending on how you construct the reals, they may be defined to have that property.
[/quote]

I was very well aware of that while writing it here. In fact, I thought about the axiomatic construction of reals in which this property is defined. In that case, this property is often called "the exclusion of third"-not being a consequence of LEM, but retaining the original sense of it.

Is LEM getting confused with the Trichotomy Law?

So does Liberal Arts Math ring a bell for anyone? I'd still like to know what it is.

A search gives things about math for liberal arts students.


The points $ A$, $ B$ and $ C$ lie on the sides of a square of side 1 cm and no two points lie on the same side. Show that the length of at least one side of the triangle $ ABC$ must be less than or equal to $ (\sqrt{6} \minus{} \sqrt{2})$ cm.

[hide]

I forgot the exact numbers, but there were lots of square roots in there.
The only thing I used is Pythagoras.

Suppose that there is a triangle with all sides longer than $ \sqrt {6} \minus{} \sqrt {2}$.
On each side of the "square" there is exactly one point.
However, points in the blue zone and in the red zone are less than $ \sqrt {6} \minus{} \sqrt {2}$ apart, so on the left side (WLOG) the point lies below the red zone.
This means, however, that no point can lie in the green zone (because all green points are at a distance less than $ \sqrt {6} \minus{} \sqrt {2}$ from every point below the red zone).
So there must be a point to the right of the green zone.
However, all points of the right side lie at a distance less than $ \sqrt {6} \minus{} \sqrt {2}$ from the points to the right of the blue zone.[/hide]

OR:

make the largest possible equilateral triangle, and show that it's sides are no longer than sqrt6-sqrt2.

just pythagorus and some algebra

abcak: I considered that, but it is a convincing proof?


induction
number theory solved
number theory

In fact , we see that $1897=7\cdot 271$

And by taking mod 7 and mod 271 , it can be easily seen that they are both $\equiv 0$  :)


For any three positive reals a, b, c, we have

$\frac{a^3+abc}{b+c}+\frac{b^3+abc}{c+a}+\frac{c^3+abc}{a+b}\ge\frac{\sqrt{2}}{2}\left(a\sqrt{a^2+bc}+b\sqrt{b^2+ca}+c\sqrt{c^2+ab}\right)$

After some computations, the inequality is equivalent to
\[
\sum {\left( {a^2  - b^2 } \right)\left( {\frac{{a\sqrt {a^2  + bc} }}
{{\sqrt {2(a^2  + bc)}  + b + c}} - \frac{{b\sqrt {b^2  + ca} }}
{{\sqrt {2(b^2  + ca)}  + c + a}}} \right)}  \geqslant 0
\]
Now, when $a \ge b$,
\[
\begin{gathered}
  \frac{{a\sqrt {a^2  + bc} }}
{{\sqrt {2(a^2  + bc)}  + b + c}} \geqslant \frac{{b\sqrt {b^2  + ca} }}
{{\sqrt {2(b^2  + ca)}  + c + a}} \hfill \\
   \Leftrightarrow \frac{a}
{{\sqrt 2  + \frac{{b + c}}
{{\sqrt {a^2  + bc} }}}} \geqslant \frac{b}
{{\sqrt 2  + \frac{{c + a}}
{{\sqrt {b^2  + ac} }}}} \hfill \\
   \Leftrightarrow a\left( {\sqrt 2  + \frac{{c + a}}
{{\sqrt {b^2  + ac} }}} \right) \geqslant b\left( {\sqrt 2  + \frac{{b + c}}
{{\sqrt {a^2  + bc} }}} \right) \hfill \\
\end{gathered}
\]
$a \ge b$ is obvious, and
\[
a\left( {\frac{{c + a}}
{{\sqrt {b^2  + ac} }}} \right) \geqslant b\left( {\frac{{b + c}}
{{\sqrt {a^2  + bc} }}} \right)
\]
is indeed true, as
\[
\begin{gathered}
  a\left( {\frac{{c + a}}
{{\sqrt {b^2  + ac} }}} \right) \geqslant b\left( {\frac{{b + c}}
{{\sqrt {a^2  + bc} }}} \right) \hfill \\
   \Leftrightarrow a\left( {c + a} \right)\sqrt {a^2  + bc}  \geqslant b\left( {b + c} \right)\sqrt {b^2  + ac}  \hfill \\
   \Leftrightarrow \left( {a + c} \right)\sqrt a \sqrt {a^3  + abc}  \geqslant \left( {b + c} \right)\sqrt b \sqrt {b^3  + abc}  \hfill \\
\end{gathered}
\]
So we conclude that
\[ {\left( {a^2  - b^2 } \right)\left( {\frac{{a\sqrt {a^2  + bc} }}
{{\sqrt {2(a^2  + bc)}  + b + c}} - \frac{{b\sqrt {b^2  + ca} }}
{{\sqrt {2(b^2  + ca)}  + c + a}}} \right)}  \geqslant 0 \]
and the result follows.

I think it is

$\displaystyle
\sum {\left( {a^2 - b^2 } \right)\left( {\frac{{a\sqrt {a^2 + bc} }}
{{(b+c)(\sqrt {2(a^2 + bc)} + b + c)}} - \frac{{b\sqrt {b^2 + ca} }}
{{(c+a)(\sqrt {2(b^2 + ca)} + c + a)}}} \right)} \geqslant 0$

but it is ininfluential for your proof, siuhochung, so the proof is OK.

oh, thank you. :lol: you are very careful~


integration
Putnam
Functional Analysis
advanced fields
advanced fields unsolved

$E$ a Hilbert space, $S,T \in L(H)$, $S$ is a normal operator, $ST=TS$. Prove that $S^*T=TS^*$.

The Spectral Theorem can be avoided (there are proofs which do not use it, I think), but using it makes the assertion look obvious :).

Let $E$ be a resolution of the identity corresponding to the normal operator $S$. An operator commutes with $S$ iff it commutes with every projection $E(M)$, where $M$ is a Borel subset of the spectrum $\sigma(S)$ of $S$. Let $\mu_{x,y}$ be the Borel measure defined on $\sigma(S)$ by $\mu_{x,y}(M)=(E(M)x,y)$. $T$ commutes with $E(M)$ iff for all $x,y$ we have $(TE(M)x,y)=(E(M)Tx,y)$, which is equivalent, in terms of the measures $\mu$ defined above, to $\mu_{x,T^*y}=\mu_{Tx,y},\ \forall x,y\ (*)$. We have $(TS^*x,y)=(S^*x,T^*y)=\int_{\sigma(S)}\overline{\lambda}\ \mbox d\mu_{x,T^*y}(\lambda)$, and $(S^*Tx,y)=\int_{\sigma(S)}\overline{\lambda}\ \mbox d\mu_{Tx,y}(\lambda)$. This, together with $(*)$, implies that these two are equal, i.e. $(TS^*x,y)=(S^*Tx,y),\ \forall x,y$, and the conclusion follows.

Edit:

Here's a slight generalization, which is, I believe, called the Putnam-Fuglede Theorem:

If $A,B$ are normal operators on a (complex) Hilbert space $H$, and if $T$ is another operator such that $AT=TB$, then $A^*T=TB^*$.


quadratics
number theory unsolved
number theory

Prove that no square can be of the form $ x^3 \plus{} 7$

If $ x$ is even or $ x\equiv3\bmod{4}$ then $ x^3\plus{}7$ is not $ 0$ or $ 1$ $ mod4$
So $ x\equiv1\bmod{4}$
Adding $ 1$ to both sides of the equation yields:
$ x^3\plus{}8\equal{}y^2\plus{}1$

  $ (x\plus{}2)(x^2\minus{}2x\plus{}4)\equal{}y^2\plus{}1$
But then $ (x\plus{}2)\equiv3\bmod{4}$ so it has a prime divisor of the form $ p\equal{}4k\plus{}3$
$ y^2\equiv\minus{}1\bmod{p}$ so this implies $ \minus{}1$ is a quadratic residue $ mod p$. However, $ \minus{}1$ is a residue  $ mod p$ if $ p\equal{}2$ or $ p\equiv1\bmod{4}$
 So we reach a contradiction and the problem is solved

Daniel


inequalities
function
number theory
research

for an inequality between arithmetic functions to hold for every $ \epsilon > 0$ except for a set of density $ 0$?

I encountered this notion recently while perusing the paper [i]A conjecture in elementary Number Theory[/i] by L. Alaoglu and P. Erdös. Had you guys heard of it before?

Thanks in advance for your insighful responses.

Well, I was not asking for the definition of density, [b]Soarer[/b].

In any case, your post makes as much sense (to me) as that of Hugh Jass that when asked what the notion of [i]almost everywhere[/i] meant provided a link to the Merriam-Webster site where the definition of [i]almost[/i] was to be found... Does any of you guys feel the same way I do?  :noo:

Well, it's certainly not clear what question you [i]were[/i] intending to ask, then.  At least, I don't know what else about that phrase you might be failing to understand, and you've given close to 0 information about what problems you're having with it.

(Unrelatedly, I also think this is an extremely odd part of the forum in which to have placed this question.)


if ,(a+1)(b+2)(c+3)(d+4)=18
 then, find the maximum value of abcd ?

i think its a similar problem:
problem 2:if a+b+c=18 ,then find the max value of (a^2*b^3*c^4) ???
 can you please work out the solution for both,please give the problem no. before the solutions.
sorry for posting 2 problems in a single thread.

Well $ 18 \equal{} \minus{}1\cdot \minus{}2\cdot \minus{}3\cdot \minus{}3$. If we let $ d \plus{} 4 \equal{} \minus{}3\implies d \equal{} \minus{}7$, $ c \plus{} 3 \equal{} \minus{}3\implies \equal{} c \equal{} \minus{}6$, $ b \plus{} 2 \equal{} \minus{}2\implies b \equal{} \minus{}4$, and $ a \plus{} 1 \equal{} \minus{}1\implies a \equal{} \minus{}2$. Thus the maximum value of $ abcd$ is $ \boxed{336}$.


let (C) and (C')be two circles.

find the locus of point M inwich:

                                             the power of M to (C) _ the power of M to (C')=stable

It's just any line parallel to the radical axis...?

ofcourse it is ;) .as u know we can see it beter,if we change circles to points.and the locus will be the  mediator.

you are so cool ashegh,and i think you should be in the other forum becasue you are so cool. :D

oh,ure welcome dear jensen.lets solve some problems :D


algebra
polynomial
linear algebra
matrix
combinatorics open
combinatorics

Can the chromatic polynomial of a graph $ G$ be realized as the characteristic polynomial of a natural linear operator on the vertex set of $ G$?  (Equivalently, as the characteristic polynomial of a matrix such as the adjacency matrix or the Laplacian?)

In case it helps anyone to think, I ask because the [url=http://mathworld.wolfram.com/BerahaConstants.html]Beraha constants[/url] look like the largest eigenvalues of the adjacency matrices of [url=http://en.wikipedia.org/wiki/Cycle_graph]cycle graphs[/url] - either you add two self-loops to each vertex or you square the adjacency matrix.  If the chromatic polynomial could be realized as the characteristic polynomial of an operator naturally related to cycle graphs, it might partially explain the Beraha phenomenon.


[quote="Pirkuliyev Rovsen"]My example:⇓
Solve  the  equation
e^(x+2008) +e^(x+y-1)=2x+y+2009.
[[/quote]

Let $ u\equal{}x\plus{}2008$ and $ v\equal{}x\plus{}y\minus{}1$. This equation is $ e^u\plus{}e^v\equal{}u\plus{}v\plus{}2$

Let then $ f(x)\equal{}e^x\minus{}x\minus{}1$. The equation becomes $ f(u)\plus{}f(v)\equal{}0$
But it is easy to show that $ f(x)>0$ $ \forall x\neq 0$ and $ f(0)\equal{}0$

So the only solution is $ u\equal{}v\equal{}0$ and so $ (x,y)\equal{}(\minus{}2008,2009)$


Let $ a, b, c \ge 0$ and $ a\plus{}b\plus{}c\equal{}1$. Find min of $ f(a,b,c)\equal{}$ $ \frac{1}{a^2\plus{}b^2\plus{}c^2} \plus{} \frac{1}{ab}\plus{} \frac{1}{bc}\plus{} \frac{1}{ca}$

This is my solution:

$ 100=\displaystyle \left ( \sqrt{a^2+b^2+c^2}.\frac {1}{\sqrt{a^2+b^2+c^2}}+3\sqrt{ab}.\frac{1}{\sqrt{ab}} +3\sqrt{bc}.\frac{1}{\sqrt{bc}} +3\sqrt{ca}.\frac{1}{ca} \right)^2$ $ \le (a^2+b^2+c^2+9ab+9bc+9ca) \displaystyle \left( \frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+ \frac{1}{bc}+\frac{1}{ca}\right)$

$ \Longrightarrow 100 \le [(a+b+c)^2+7(ab+bc+ca)]. \displaystyle \left ( \frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+ \frac{1}{bc}+\frac{1}{ca} \right)$

But $ ab+bc+ca \le \frac{1}{3}(a+b+c)^2=\frac{1}{3}$

$ \Longrightarrow \frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+ \frac{1}{bc}+\frac{1}{ca} \ge 30$.


Given two bright spot and the intension of this are given by $\frac{a}{r^{2}}$ and $\frac{b}{r^{2}}$

so the question is wher will the intension it most between the two bright spots where the distantions is given by s.

So I try $\frac{a}{x^{2}}+\frac{b}{(s-x)^{2}}$ then I derivate this and I claim that it will be zero $0=-\frac{2b}{(x-s)^{2}}-\frac{2a}{x^{3}}$

But I don't com out the solotion must be $\frac{{(as)}^{\frac{1}{3}}}{{a}^{\frac{1}{3}}+{b}^{\frac{1}{3}}}$

What wil be the problem? Greets Thanks a lot.

[quote]Wouldn't it be
[img]http://www.mathlinks.ro/Forum/latexrender/pictures/189ecbb9967d356f595fd756a43dcbe1.gif[/img][/quote]

this is the solution of my book.

I think it must be possible to eliminate the x.

Greets.

[quote="Immanuel Bonfils"]JBL: wrong sign.[/quote]

No, not the wrong sign.  Your answer is correct, however.

Bert, the word "greets" (a shortening of "greetings") is used in English when you are greeting someone.  The way you use it is a little odd.

[quote]No, not the wrong sign. Your answer is correct, however.
[/quote]

oké the book is incorect? And my solotion?

[quote]Bert, the word "greets" (a shortening of "greetings") is used in English when you are greeting someone. The way you use it is a little odd.
[/quote]

I know that my english isn't wonderful but I will improve so thanks for comment.

Who can you  say 'greets' in a more modern way?

[quote="Bert"]Who can you  say 'greets' in a more modern way?[/quote]'Cheers' is cool.

oké but is the solution in my book fault.

Cheers  :wink:

[quote="Bert"]Who can you  say 'greets' in a more modern way?[/quote]

Just say hi or hello if you want to greet someone.

If you want to close post just thanks is enough in my opinion.

[quote]I too think "cheers" is good. It has the same tone and sounds more natural in this usage.

If the book gives the answer you wrote in your first post, the book is wrong.
[/quote]

Oké thanks.

Oké I have also find an other solution:

[img]http://www.wetenschapsforum.nl/mods/latexrender/pictures/933dce1676e0d8e984892b156d4fb8c2.gif[/img]

then [img]http://www.wetenschapsforum.nl/mods/latexrender/pictures/bcf4fca1a94f951c534ddc66e3453e27.gif[/img]

Thank you very wel Cheers!!  :)


1. Show that there exists positive integers $m,n$ such that  $3n^{2}+3n+1=m^{3}$

2. Find all of its solutions.

I'm blind. The equation is equivalent to $(n+1)^{3}-n^{3}=m^{3}$.

The equation is equivalent to $m^{3}+n^{3}=(n+1)^{3}$ .Because of Fermat's Theorem $\rightarrow$ the only solutions are n=0 and m=1

hi,

I think its ask to much...  :blush:  but I would like if someone give me one diofantine equation and its solution to do a kind of joke to a special friend of mine... I already use some problems of old imo's (pre-historic ones) but he always find very quickly the solution... which is very irritating...  :furious:

one more request, the solution has to be a finite group (not an infinite group)

thank you

i hope there still exist, in this world, real gentlemen to save me

before i will leave you in peace i want to apologize for my "english" (if we can call it that...) but i'm not english...


$a,$ $b,$ $c$ and $d$ are non-negative real numbers, such that $a+b+c+d=4.$
Prove that $\sum_{cyc}a^{\frac{3}{5}}\geq\frac{2}{3}\cdot\sum_{sym}ab.$ :)

And generalization:
Let $a_i\geq0,$ where $i=1,2,...,n$ and $n>1,$ such that $\sum_{i}a_i=n.$ Prove that
$\sum_{i}a_i^{\alpha}\geq\frac{2}{n-1}\cdot\sum_{i<j}a_ia_j$$,$ where $\alpha=\frac{2n-5}{2n-3}.$ :lol:


geometry
geometric transformation
rotation
rhombus

Points in the plane are each colored with one of three colors: red,
green, or blue. Prove that, for a given distance d, there always exist
two points of the same color at the distance d from each other.

[hide="The standard hint"] Consider equilateral triangles. [/hide]

Hm, I'm not sure if your hint works here :/ It's easy to color a grid formed of equilateral triangles at least. So some kind of trick must be used.

[u]Solution.[/u] Assume to the contrary that there exists a distance $ d>0$ that is not assumed by any of the colors.  Take a red point and construct an equilateral triangle of side length $ d$ with the red point as one of its vertices.  By the assumption, the other two vertices must be blue and green.  The construction of a congruent triangle on the side opposite to the initial red vertex constructs a red vertex for the additional conguent triangle.  A rotation of the rhombus about the initial red vertex must then force the creation of a red circle.  Contradiction! $ \Box$

[u]Remark.[/u]  It can be proved that there exists a color that assumes all distances $ d>0$.

Ah, excellent! Thanks for a nice solution! :)

(I wonder how I didn't see it o_0)


function
induction
algebra
functional equation
algebra proposed

Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that:

1-$f(x+y)+f(x-y) = 2(f(x)+f(y))$ for all $x$ and $y$ in $\mathbb{R}$

I think your analysis is valid except for this
[quote="abdurashidjon"]$f(2x)=4f(x)$ by induction it is easy to prove that $f(nx)=n^2f(x)$[/quote]
I think what you get by induction is $f(2^n x)=4^n f(x)$. And with a similar analysis, you get what I said:
$f(2^q)=4^qf(1)$ for every $q\in\mathbb{Q^+}$.
The only continuous solution is $f(x)=Cx^2$ for any $C\in\mathbb{R}$.

[u][i][b]Problem:[/b][/i][/u] [color=red]Find all functions $f : \mathbb{Q}\to\mathbb{Q}$ such that

$f(x+y)+f(x-y)=2f(x)+2f(y)$ for all $x,y\in\mathbb{Q}$[/color]

[u][i][b]Solution:[/b][/i][/u]

The only such functions are $f(x)=kx^2$ for rational$k$. Any such function works, since
$f(x+y)+f(x-y)= k(x+y)^2+k(x-y)^2 = kx^2+2kxy+ky^2+kx^2-2kxy+ky^2 = 2kx^2+2ky^2 = 2f(x)+2f(y)$

Now suppose $f$ is any function satisfying
$f(x+y)+f(x-y)=2f(x)+2f(y)$.

Then letting $x=y=0$ gives $2f(0)=4f(0)$, so $f(0)=0$

We will prove by induction that $f(nz)=n^2f(z)$ for any postitive integer n and any rational number z.

The claim holds for $n=0$ and $n=1$; let $n\geq2$ and suppose the claim holds for $n-1$ and $n-2$.

Then letting $x=(n-1)z,y=z$ in the given equation we obtain
$f(nz)+f((n-2)z)=f((n-1)z+z)+f((n-1)z-z)=2f((n-1)z)2f(z)$
so

$f(nz)= 2f((n-1)z)+2f(z)-f((n-2)z) = 2(n-1)^2f(z)+2f(z)-(n-2)^2f(z) =(2n^2-4n+2+2-n^2+4n-4)f(z) = n^2f(z)$
and the claim holds by induction.

Letting $x=0$ in ther given equation gives
$f(y)+f(-y)=2f(0)+2f(y)=2f(y)$ so$f(-y)=f(y)$ for all raional y; thus $f(nz)=n^2f(z)$ for all integers n.

Now let $k=f(1)$; then for any rational number $x=\frac{p}{q}$,
$q^2f(x)=f(qx)=f(p)=p^2f(1)=kp^2 so f(x)=\frac{kp^2}{q^2}=x^2$.

Thus the functions $\boxed{f(x)=kx^2}, k\in\mathbb{Q}$ are the only solutions.

davron

Thanks Davron: your solution is good for the problem in $\mathbb{Q}$ that you gave.

But the problem that i gave is in $\mathbb{R}$.
And sorry, i didn't say that the functions needed are not defined in zero.
Now, i correct it:
 find the function f , not only the continuous one, vérifying the given relation but defined as that:
$f: \mathbb{R}$ \{0} $\to \mathbb{R}$

There is another proof for rational.
Take $x=0$ you will see that $f(x)=f(-x)$.
Set $x=y$ then $f(2x)=4f(x)$ by induction it is easy to prove that $f(nx)=n^2f(x)$ $(*)$ and it works for natural, as we found our function is even so it works for integers, too. $x=\frac{m}{n} (**) \rightarrow x\times n=m\times 1$ then $n^2f(x)=f(n\times x)=f(m\times 1)=m^2f(1)\rightarrow n^2f(x)=m^2f(1)$ $(***)$ this all from $(*)$, and from $(***)$ follows that
$f(x)=\frac{m^2}{n^2} f(1)$ and from $(**)$  $f(x)=x^2f(1)$, let $k=f(1)$ then $f(x)=x^2\times k$

[quote="abdurashidjon"]$f(2x)=4f(x)$ by induction it is easy to prove that $f(nx)=n^2f(x)$[/quote]
I think the induction gives $f(2^nx)=4^nf(x)$.

Anyway, the problem has changed. With what afouzi said, we can't use $x=0$ or $y=0$ anymore.

I'll try to solve it without using $x=0$ or $x=y$ (they use $f(0)$).
$x=z+t$, $y=z-t$ $\to$ $f(2z)+f(2t)=2(f(z+t)+f(z-t))=4(f(z)+f(t))$ for every $t,z\in\mathbb{R^*}$.
$z$ and $t$ are independent, so we must have $f(2z)=4f(z)$ for every $z\in\mathbb{R^*}$.
And I think we had this functional equation before.

Now I will show it for any odd, I think if I show it for any odd it's enough. It follows as:
Set $x$ as $2x$ and $y$ as $x$ then
$f(3x)+f(x)=2(f(2x)+f(x))$ as we know $f(2x)=4f(x)$ , so
$f(3x)=9f(x)$ as  desired.
Abdurashid

Be careful, $3x$ is not any odd number. There are ways to prove it better. I recommend you to study Davron's solution.
And proving for odd numbers (or all integer numbers, or all rational numbers) is not enough. We should prove it for every non-zero real number, which is much more difficult.
I think at our level we can only prove it if the problem had said they only wanted the continuous functions, so that we can extend rationals to reals.


In the function f(x) = (x^c)(x^1/c) c is a nonzero constant.

If f(2) = 4√2, what is the value of c?

(2^c)(2^1/c) = 4√2

I'm completely lost on this one. Plus, fractional exponents really confuse me.
This one is a bit over my head, but I still want to try it.

[hide="Hints"]
$ x^a \cdot x^b \equal{} x^{a\plus{}b}$
$ 4 \sqrt 2\equal{}2^2 \cdot2^{\frac{1}{2}}$[/hide]

[quote="Brut3Forc3"]In general, $ x^{\dfrac{a}{b}} \equal{} \sqrt [b] {x}^a \equal{} \sqrt [b] {x^a}$.
In your example, $ x^{\dfrac{1}{c}} \equal{} \sqrt [c] {x}$.[/quote]

Fractional exponents are a little clearer, thank you. I'm still not sure how to set up the equation to solve the problem.

In this problem, you can guess that $ c=2$ or $ c=\frac{1}{2}$ by noting that $ 4=2^2$ and $ \sqrt{2}=2^{\frac{1}{2}}$. (You get $ c=2$ as a more obvious answer, but replacing $ c$ with $ \frac{1}{c}$ in the expression will yield the same result, so $ c=\frac{1}{2}$ as well.)

To be more rigorous, use the hints sdbro11 gave:
$ 2^c \cdot 2^{\frac{1}{c}} = 2^{c+ \frac{1}{c}} = 2^{\frac{c^2+1}{c}}\\
4\sqrt2 = 2^2 \cdot 2^{\frac{1}{2}} = 2^{\frac{5}{2}}$
For these two values to be equal to each other, their exponents must be equal, so $ \frac{c^2+1}{c}=\frac{5}{2}$. From there, you can cross-multiply and solve for $ c$.

[quote="Brut3Forc3"]In this problem, you can guess that $ c = 2$ or $ c = \frac {1}{2}$ by noting that $ 4 = 2^2$ and $ \sqrt {2} = 2^{\frac {1}{2}}$. (You get $ c = 2$ as a more obvious answer, but replacing $ c$ with $ \frac {1}{c}$ in the expression will yield the same result, so $ c = \frac {1}{2}$ as well.)

To be more rigorous, use the hints sdbro11 gave:
$ 2^c \cdot 2^{\frac {1}{c}} = 2^{c + \frac {1}{c}} = 2^{\frac {c^2 + 1}{c}} \\
4\sqrt2 = 2^2 \cdot 2^{\frac {1}{2}} = 2^{\frac {5}{2}}$
For these two values to be equal to each other, their exponents must be equal, so $ \frac {c^2 + 1}{c} = \frac {5}{2}$. From there, you can cross-multiply and solve for $ c$.[/quote]

Thank you.


Let $ a,b,c \geq 0$ satisfying $ a \plus{} b \plus{} c \equal{} 1$. Prove that:
$ \frac {a}{b^3 \plus{} c^3} \plus{} \frac {b}{c^3 \plus{} a^3} \plus{} \frac {c}{a^3 \plus{} b^3} \geq 6 \plus{} \frac {a^3}{b^3 \plus{} c^3} \plus{} \frac {b^3}{c^3 \plus{} a^3} \plus{} \frac {c^3}{a^3 \plus{} b^3}$  :D

i think this has already been solved in the forum nefore,  u better recheck,  i will try to provide the link.


vector
algebra
polynomial
linear algebra

1. Let $ P \equal{} {(x, y, z) \in R^3 | x \plus{} y \minus{} z \equal{} 0}$. Suppose $ A : R^3 \to R^3$ is a linear transformation satisfying $ A(v) \equal{} 0$ for all $ v \in P$ and also $ A(0, 0, 1) \equal{}$ [b]0[/b] (here [b]0[/b] denotes the vector $ (0, 0, 0)$). Then

(A) The dimension of the null space of $ A$ is $ 2$.
(B) $ A$ is the zero linear transformation.
(C) Image $ A \equal{} R^3$.
(D) The dimension of the image of $ A$ is $ 2$.

2. Suppose $ A : R^2 \to R^2$ is a linear transformation such that $ A^3 \equal{} I$, where $ I$ is the identity transformation. Then

(A) All eigenvalues of $ A$ have to be real.
(B) The product of the eigenvalues of $ A$ must be $ 1$.
(C) Necessarily $ A \equal{} I$.
(D) $ A$ need not be an invertible matrix.

Let $ v_1 \equal{} (1,0), v_2 \equal{} (1,\minus{}1)$ and $ v_3 \equal{} (0,1)$. How many linear transformations $ T : R^2 \to R^2$ are such that $ Tv_1\equal{}v_2,Tv_2\equal{}v_3$ and $ Tv_3\equal{}v_1$ ?

(a) $ 3!$
(b) $ 3$
(c) $ 1$
(d) $ 0$

For your first question, the null space of $ A$ contains the two-dimensional subspace $ P$ as well as another vector, $ w \equal{} < 0,0,1 >$, that is linearly independent of P. So what must be the dimension of the null space?

For your second question, what is the characteristic polynomial for A? Which coefficient of a polynomial tells you about the product of the roots of a polynomial?

For your last question, $ Tv_1 \equal{} v_2$. Applying $ T$ to both sides gives $ T^2v_1 \equal{} Tv_2 \equal{} v_3$. Applying $ T$ to both sides again gives $ T^3v_1 \equal{} Tv_3 \equal{} v_1$. This follows for all three vectors.
Therefore $ T^3v_k \equal{} v_k$ for [i]all three[/i] of the given vectors. So each of them is an eigenvector of $ T^3$. Anything about that fact strike you as strange?

Thanks for replying......

For question 3 since $ T^3$ can have only 2 eigenvectors so then answer is 0.

Plz tell if i am correct.

Also please explain questions 1 and 2.I am not able to decode what they try to say.


function
complex analysis
probability and stats

easy and funny:
suppose that $ X$ is a bounded random variable such that for every $ n>0$, there exists $ n$ random variables $ X_{1,n},X_{1,n}, \ldots, X_{n,n}$, independant and identically distributed such that $ X \equal{} X_{1,n} \plus{} X_{1,n} \plus{} \ldots \plus{} X_{n,n}$. Find the distribution of $ X$.

Since $ X$ is bounded, the characteristic function ([i]i.e.[/i] Fourier transform) of $ X,$ call it $ f(z),$ must be an entire function that grows at $ \infty$ no faster than $ e^{c|z|}$ for some $ c.$ For each $ n,$ the $ n$th root of $ f$ must also be the characteristic function of a bounded random variable, meaning the $ n$th root of $ f$ must also be entire. Hence $ f(z)$ can never equal $ 0,$ hence we can write $ f(z)\equal{}e^{g(z)}$ for some entire function $ g.$ Since $ g$ grows no faster than some $ c|z|,$ we can write $ g(z)\equal{}a\plus{}bz$ and hence $ f(z)\equal{}e^{a\plus{}bz}.$ But we must also have $ f(0)\equal{}1$ (true for all characteristic functions), so $ a\equal{}0.$ And we must have $ |f(x)|\le 1$ for real $ x,$ which forces $ b$ to be pure imaginary. So $ f(x)\equal{}e^{imx}$ for some real $ m.$

Conclusion: $ X$ is a point mass. There is a number $ m$ such that $ P(X\equal{}m)\equal{}1.$

cool  :)

Ps: there is an elementary solution without any complex analysis etc .. just the definition of what is an expectation and a variance.

Ah, yes, elementary.

Suppose $ |X|\le M$ almost surely. Then for identical independent $ X_{i,n}$ such that $ \sum_{i\equal{}1}^nX_{i,n}\equal{}X,$ we must have $ |X_{i,n}|\le\frac{M}{n}$ almost surely.

$ \text{Var}(X_{i,n})\le E(X_{i,n}^2)\le\frac{M^2}{n^2}.$

$ \text{Var}(X)\equal{}\sum_{i\equal{}1}^n\text{Var}(X_{i,n})\le \frac{M^2}{n}.$

Since this is true for all $ n,$ we have that $ \text{Var}(X)\equal{}0,$ which implies that $ X$ is a point mass.

Take away the requirement that the $ X$ be bounded and many other infinitely divisible distributions appear. These include normal distributions, Poisson distributions, and gamma distributions. I don't know a full characterization, not off the top of my head.

yes, there are a lot of infinitely divisible distribution (one for every Levy process). A full description is given by the levy khintchine formula (this description describes the form that the characteristic function should have for the random variable to be infinitely divisible)

Suppose that you want to describe a process $ X_t$ such that the increments $ X_{t}\minus{}X_{s}$ are independant and only depend on $ t\minus{}s$ (Levy processes). Then $ X_1$ is infinitely divisible since:
$ X_1 \equal{} (X_{\frac{1}{n}} \minus{} X_0) \plus{} (X_{\frac{2}{n}} \minus{} X_{\frac{1}{n}}) \plus{} \ldots \plus{} (X_{\frac{n}{n}} \minus{} X_{\frac{n\minus{}1}{n}})$
Take a brownian motion and you will get the normal distribution, take a Poisson process and you will get a Poisson random variable, etc ..  :)


induction
number theory proposed
number theory

Find general caculas for this equation
$ x_{1}\equal{} 1$
$ x_{n\plus{}1}\equal{} x_{n}\plus{}[\frac{x_{n}}{n}]\plus{}2$

you want an expression for $ x_{n}$ ?
what's $ a_{n}$?

I think it is a nice problem but now it is solved by any people.

I think it is not nice. Solution (by induction) is
$ x_{m(k)\plus{}l}\equal{}\frac{(k\minus{}1)k(k\plus{}1)}{3}\plus{}k\minus{}7\plus{}l(k\plus{}1),l\equal{}0,1,.., k\minus{}1$, were $ m(k)\equal{}\frac{k(k\minus{}1)}{2}\minus{}1,k\equal{}3,4,...$

Can you induction because I consider a subset sequence it is very well than Induction and it is --->Induction.

Or $ x_{n}\equal{}(n\plus{}2)k\minus{}8\plus{}\frac{k(k\minus{}1)(k\minus{}2)}{6}, n\ge 2, k\equal{}[\frac{3\plus{}\sqrt{8n\plus{}9}}{2}].$

You can consider
$ x_{n\plus{}1}\minus{}x_{n}$ it is very well.


geometry
geometric transformation
reflection
algebra unsolved
algebra

[quote="prowler"]This is a problem given by Mietek Dabkwoski for Mathematical Reflections journal:

Let $ \omega_{n}$ be an $ n$-th primitive root of the unity and let
\[ H_{n}=\underset{i=0}{\overset{n-1}{\prod}}(1+\omega_{n}^{i}-\omega_{n}^{2i}). \]
Show that $ H_{n}$ is an integer and give a formula for $ H_{n}$ which uses only integers.[/quote]

Let $ \phi_{1}$ and $ \phi_{2}$ the two real roots of equation $ x^{2}-x-1=0$. Then $ H_{n}=\prod_{k=0}^{k=n-1}(-1)(\phi_{1}-\omega_{n}^{k})(\phi_{2}-\omega_{n}^{k})$

But, since $ \omega_{n}$ is a primitive root of unity, then $ \{\omega_{n}^{k}\}_{k=0,n-1}$ is the set $ A_{n}$ of all n-th roots of unity. So $ H_{n}=\prod_{r\in A_{n}}(-1)(\phi_{1}-r_{k})(\phi_{2}-r_{k})$

So $ H_{n}=(-1)^{n}\prod_{r\in A_{n}}(\phi_{1}-r_{k})\prod_{r\in A_{n}}(\phi_{2}-r_{k})$

So ${ H_{n}=(-1)^{n}(\phi_{1}^{n}-1})(\phi_{2}^{n}-1)$

So $ H_{n}=(-1)^{n}((-1)^{n}+1-\phi_{1}^{n}-\phi_{2}^{n})$

So $ H_{n}=(-1)^{n+1}K_{n}+(-1)^{n}+1$ where $ K_{n}$ is the n-th element of the Fibonacci sequence begining with $ K_{1}=1$ and $ K_{2}=3$

My idea give same result, if calculate akkurate.
$ (1-\omega_{n}^{k}+\omega_{n}^{-k})(1-\omega_{n}^{-k}+\omega_{n}^{k})=3-\omega_{n}^{2k}-\omega_{n}^{-2k}=x^{-1}(x-\omega_{n}^{2k})(x-\omega_{n}^{-2k})$,
were $ x=\phi^{2}, \phi=\frac{1+\sqrt 5}{2}$.
I think your solution is better.


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

A matrix in M_2(R), a(A)=det(A+I)+det(A-I)
1) Suppose that detA=/=0 proove that a(A)>2
2)Proove there exist an infinity A such that a(A)=2
Lucian Dragomir, lista scurta ONM 2003
-----------------------------------------------------
If you have another solution i am interested.

1) For X,Y \in M_2(R), det(X+Y)+det(X-Y)=2(detX + detY)
a(A)=2(detA + 1)=2(detA)+2 > 2 since (detA)>0
2) a(A)=2 if and only if detA=0, consider
A=
(0 k)
(0 0) any k \in R

M.Omarjee

I think I  have another solution:
consider the characteristic polinomial P(x)=det(A-xI_2)=x^2-ax+b
so a(A)=P(-i)*P(i)+P(1)P(-1)=2b^2+2 and because b<>0 then it is obious that a(A)>2.
and for the 2nd part there exists an infinity of matrices because there exists an infinity of singular matrices!

cheers! :D :D


probability
combinatorics theorems
combinatorics

Maybe this is not the right place to post this, but here it goes:

My question is about definitions of sample spaces.

Suppose we roll one dice, and we catch the result. Of course, the sample space consists on all of the possible results:

$\Omega = \{1,2,3,4,5,6 \}$

If we roll two dice, we have

$\Omega = \{(x,y)|x,y \in \{1,2,3,4,5,6\}\}$

But if our experiment is to roll one dice $n$ times? Is $\Omega$ simply all the six possible numbers or $\Omega$ consists of all sequences of $n$-tuples which every term is so that it belongs to $\{1,2,3,4,5,6\}$.

Rolling a single die n times is equivalent to rolling n dice once, so I believe your second idea is correct.


1.  Please note that on the back page of the 2007 AIME contest pamphlet it says: "Your Exam Manager will receive a copy of the 2007 AIME Solution Pamphlet with the scores."  It is NOT an accidental oversight that the answers were not inlcuded with the contests.

2.  The scores will be mailed back to the School Exam Manager in about 2 weeks.  Your School Exam Manager will be the first to receive your scores.  Check with your School Exam Manager to find out your scores.  (Again the limiting factor is receiving the mailed answer sheets from the schools, we cannot score what we have not received.  We cannot tabulate meaningful statistics until we have received the answer forms.)

3.  No decision will be made on USAMO invitation levels and scores until after the Alternate AIME (March 28, 2007) has been scored.  Practically that means that about April 6, I will be able to announce the USAMO invitation scores levels.  This is precisely the same as in the USAMO Teachers' Manual, page 4, Section I.

4.  The 3-character answers for the 2007 AIME will be posted at the AMC website today (Wed, March 14, 2007).

Steve Dunbar
Director, American Mathematics Competitions

Thanks Steve for (as usual) keeping us very well-informed.  :thumbup:

Let me add a follow-up to point #3.  [b]Please[/b] do not ask, either us or the AMC, "When will the USAMO cut-off score and/or names be announced"?  When the scores and names are announced, we will let you know, but it is very annoying to be constantly pestered about it.  The AMC works extra-hard to double-check all the names and the scores in advance of the announcement, so that nobody gets mistakenly left out.  Please allow them the time to do this -- even if it delays the announcement of the USAMO qualifiers by a day or two -- so that they can complete their work accurately and without disturbance.

I recall that a couple of years ago, on the day qualifiers came out, people were flooding the office with calls asking when the list would be posted.  Rumors of what one secretary had said half an hour ago flew around AoPS and conflicted with rumors about what another secretary said 20 minutes ago.

Not only were the people at AMC not pleased with this (or so I imagine), but no one found out anything, and the qualifiers list didn't come out any earlier.

Yeah, people were making false web pages too...What an annoying year.  Probably because the expansion kept many people hoping.

So are we going with Tuesday as the latest release day (as said in the USAMO teacher manual), or should we go with around April 6th?? (I mean release on line)

Thanks Very Much.

I just really hope that I don't miss the cutoff this year by an AIME Question or 1/2 a point Index.

Have the AIME I scores been sent out to the teachers yet? My teacher hasn't received anything, and its been a week since this post suggests they will be received.


calculus
integration
linear algebra
matrix
algebra
polynomial
algorithm

Let L/K be a algebraic field extension and f(x) in L[x] then there exist g(x) in L[x] with g(x) unequel 0 and f(x)g(x) in K[x].
A proof of that use only elementary linear algebra or elementary galois theory if char(K)=0 (if char(K)=p a field theoretic prof use minor modifications).
What about this statement if we assume L/K to be a integral ring extension? What about an matrix-analoge of this statement: Let L/K be a algebraic field extension (integral ring extension) and A in M_n(L).Is there always B in M_n(L) with B unequel 0 and AB in M_n(K)?

Nice questions!! I will be brief since I don't have much time.

I rename all of your $ x$'s into $ X$, since I don't like polynomial indeterminates being denoted by lowercase letters.

Every time I speak of a "ring", I mean a commutative ring with unity.

[quote="Patrix"]What about this statement if we assume L/K to be a integral ring extension?[/quote]

First show that $ f\left(X\right)$ is integral over $ K\left[X\right]$ (you can use the standard facts about integral extensions here, like that the sum and the product of integral elements is integral again). Now it all bends down to the following fact:

[b](*)[/b] If $ A\subseteq B$ is a ring extension, and $ u\in B$ is integral over $ A$, then there exists a nonzero $ v\in B$ such that $ uv\in A$.

In fact, let $ n$ be the smallest positive integer such that there exists a monic polynomial $ P\in A\left[Y\right]$ of degree $ n$ satisfying $ P\left(u\right) \equal{} 0$. Let $ P\left(Y\right) \equal{} \sum_{i \equal{} 0}^{n \minus{} 1}a_iY^i \plus{} Y^n$. Then, set $ v \equal{} \sum_{i \equal{} 1}^{n \minus{} 1}a_iu^{i \minus{} 1} \plus{} u^{n \minus{} 1}$. Notice how $ v\neq 0$ follows from the minimality of $ n$.

[quote="Patrix"]What about an matrix-analoge of this statement: Let L/K be a algebraic field extension (integral ring extension) and A in M_n(L).Is there always B in M_n(L) with B unequel 0 and AB in M_n(K)?[/quote]

This again follows from [b](*)[/b]; you just have to set $ A \equal{} K$ and $ B \equal{} L\left[A\right]$ (oops, sorry, now I have two different $ A$'s here - the first one is the ring $ A$, the second one is the matrix $ A$). And $ u \equal{} A$ (the matrix $ A$), of course. The integrality of $ u$ over the ring $ A$ then follows from the Cayley-Hamilton theorem (okay, let's be more precise: the Cayley-Hamilton theorem shows that $ u \equal{} A$ (the matrix $ A$) is integral over the ring $ L$, but $ L$ is in turn integral over the ring $ A$, so we conclude that $ u$ is integral over the ring $ A$). As a consequence, you get not only $ AB\in\mathrm{M}_n\left(K\right)$, but even $ AB\in K$ (where $ K\subseteq\mathrm{M}_n\left(K\right)$ has to be understood as the embedding $ K\to\mathrm{M}_n\left(K\right),\ \ \ \ \ \ \lambda\mapsto\lambda\cdot I_n$).

  darij

See also my posts [1],[2] for further remarks on rationalizing denominators
and computing inverses via (minimal) polynomials, extended Euclidean
algorithm, Grobner bases, resultants, norms, etc.

[1] http://google.com/groups?selm=y8zwtid4b0u.fsf%40nestle.csail.mit.edu
[2] http://google.com/groups/search?q=group:sci.math+dubuque+rationalize

There is a view of logic that solve all such kind questions and could avoid most algebra computation.
$ L \equal{} K(\alpha)$, $ H(x) \equal{} irr(\alpha,K)$, H is monic.

If $ L/K \equal{} K(\alpha)$ is a field extension, then the conjugation $ f_i$ can be defined since conjugation of $ \alpha$ can be defined. Though this can't be done in ring but $ F \equal{} \prod_{i \equal{} 1}^{n \minus{} 1} f_i$,
 ($ f \equal{} f_n$),  can be defined since all coefficients of F are polynomial of coefficients of H and f .(by poly of H and f's coefficients it means Z[U]) This can be proved by showing $ \sigma_{i}(\alpha_1,\cdots,\alpha_{n \minus{} 1})$ ($ \alpha\equal{}\alpha_n$) is polynomial of H's coefficients. They are $ \sum_{j \equal{} 0}^{i} h_{j \plus{} n \minus{} i}\alpha^{j}$. (However this computation can be avoided. I will come to this later.) $ Ff \equal{} G$, coefficients of G are polynomial of only H and f's coefficients. This is also true in the comutative ring, since it can be shown only by "algebraic operation", which can be conducted all the same in the comutative ring (this is why the ring should be comutative).
To "prove" all coefficients of F are polynomial of coefficients of H and f, suppose one of them is $ q \equal{} \frac {u}{v}$, u,v are poly of H and f's co. Since the coefficients of H and f are arbitrary, choose such elements that v=0 and $ u\neq 0$, F can still be defined in the same way while the coefficients q is invailid, a contradiction.

In general, a proposition is stated in structure M but is easily proved in M'', an extension of M. And prove P in M by showing "If P is true in M'' then P can be proved in M', M' contain neccessary structure used in the proof of P", which may not need actually show the proof. In the above example, M'=comutative ring with unity, M''=field, and showing coefficients of G are poly of H and f's poly since it needs only "algebraic operation".

thank you darij :)
What about an "simultaneous analogue" of this problems for polynomials and matrices?
To be precise: Let f_1,...,f_n be in L[x]. When there exist a g in L[x] with g unequel 0 and f_1*g,...,f_n*g in K[x]?
It is easy to see that this is the case if and only if f_i/f_j is in K(x) for all i,j. But what about this problem for matrices
T_1,...,T_n?

Patrick


Circles with radii $ r_1; r_2; r_3$ are externally tangent to each other. Two other circles, with radii $ R; r$ are tangent to all previous three circles. Prove that
$ Rr\geq \frac{r_1r_2r_3}{r_1 \plus{} r_2 \plus{} r_3}$

Let $ A,B,C$ be a tangent points of first three circle and radius of its circumcircle is $ d$, then it is easy to show that $ d^{2}$ is same as RHS,
Think about inversion wrt circumcircle of $ ABC$. small circle which radius is $ r$ will be a big circle... now $ Rr\ge d^{2}$ is obvious..


geometry
cyclic quadrilateral
geometry unsolved

Let $ ABC$ be a triangle such that $ \angle ABC\equal{}90^{\circ}$. The incircle of triangle $ ABC$ touches sides $ BC,CA,AB$ at $ D,E,F$ respectively. Let $ CF$ intersect the incircle at $ P$. If $ \angle APB\equal{}90^{\circ}$, prove that $ CP\plus{}CD\equal{}PF$.

[quote="livetolove212"][b]The generalization:[/b]
Given triangle ABC and its incircle $ (I)$.Let $ D,E,F$ be the tangencies of $ (I)$ and $ BC,CA,AB$,respectively. $ AD\cap (I) \equal{} \{G\}$. If $ \angle BGC \equal{} 90^o$ then prove that $ GD \equal{} AG \plus{} AF$.

[b]Solution:[/b]
Denote $ K$ the intersection of $ EF$ and $ BC$. Since $ GFDE$ is harmonic then $ KG$ is the tangent of $ (I)$. Because $ (KDBC) \equal{} \minus{} 1$ and $ \angle BGC \equal{} 90^o$ we get $ GB$ is the bisector of $ \angle KGD$.
Denote $ J$ the second intersection of $ BG$ and $ (I)$ then $ \angle JDG \equal{} \angle KGJ \equal{} \angle JGD$. Hence $ JD \equal{} JG$.
On the other side, $ JFGD$ is harmonic. Therefore $ JF.GD \equal{} FG.JD$.
Applying Ptolemy's theorem for cyclic quadrilateral $ FJDG$ we obtain $ FD.JG \equal{} 2GF.DJ \Rightarrow FD \equal{} 2GF$. Since $ \Delta AFG\sim \Delta ADF$ then $ AF \equal{} 2AG. (1)$
But $ AD.AG \equal{} AF^2 \Rightarrow AF^2 \minus{} AG^2 \equal{} AG.GD$ $ \Leftrightarrow (AF \minus{} AG)(AF \plus{} AG) \equal{} AG.GD (2)$
From $ (1)$ and $ (2)$ we get $ GD \equal{} AF \plus{} AG$ (QED)[/quote]sorry, i'm new to this harmonic quadrilateral thing. please explain why GFDE is harmonic.

why $ GFDE$ is harmonic:

because $ \triangle AEG \sim \triangle ADE$, we have $ \frac{EG}{DE}\equal{}\frac{AG}{AE}$. Similarly, $ \frac{GF}{FD}\equal{}\frac{AG}{AF}$. Since $ AE\equal{}AF$, we then get $ \frac{EG}{DE}\equal{}\frac{GF}{FD}$, i.e. $ EG\cdot FD \equal{} GF\cdot DE$.


LaTeX AoPSwiki counting distinguishability

Kim has 10 identical lamps and 3 identical tables. How many ways are there for her to put all the lamps on the tables?

This is simply $ \frac {10!}{3!\times7!}$

I don't know how to do the combination binomial thingy in Latex

I disagree. They are identical, therefore, we can list them:

10-0-0
9-1-0
8-2-0
8-1-1
7-3-0
7-2-1
6-4-0
6-3-1
6-2-2
5-5-0
5-4-1
5-3-2
4-4-2
4-3-3

There are 14 of them.

To do the binomial, just type:

[code]$\binom {n}{r}$[/code]

And it comes up as:
$ \binom {n}{r}$

Uh-huh, PowerofPi is correct.  Simply list them out and you will get the correct answer.

Could you do Balls and Urns with indistingushable objects?

I think you mean indistinguishable to distinguishable. Also, you could notice a pattern for the specific problem, but sometimes that might take even longer than listing.

It's not hard by systematic counting. Just count by the greatest number of lamps in a table.

[quote=SuperNerd123]

I don't know how to do the combination binomial thingy in Latex[/quote]

[code]$\binom{x}{y}$[/code]

[quote=jhp76621]$\binom {10}{3}$[/quote]

That is incorrect, since both the lamps and the tables are identical. This is a problem asking for the different combinations of numbers of lamps on each table, where order does not matter.

Yeah, when you have indistinguishable boxes (tables) and balls (lamps), you have to use casework.

[hide=Here's what I did:]
$\text{Imagine the triples as sets. Either the smallest was }0,1,2, \text{or } 3. \text{ Now, we have casework:}$
\begin{align*}
\text{Case 1: Smallest is 0. }&\text{The largest number is from 10 to 5, or 6 possibilities.}\\
\text{Case 2: Smallest is 1. }&\text{The largest number is from 8 to 5, or 4 possibilities.}\\
\text{Case 3: Smallest is 2. }&\text{The largest number is from 6 to 4, or 3 possibilities.}\\
\text{Case 4: Smallest is 3. }&\text{The largest number is 4, or 1 possibility.}\\
&6+4+3+1=\boxed{14}
\end{align*}

[quote=saagarjha]I don't get how this is a level 24. This is pretty easy.[/quote]

Wait what, It's 23 for me :maybe:

[quote=Eyed]Nah its fourteen[/quote]

You must be thinking of a different problem....

[quote=Eyed]Nah its fourteen[/quote]

Yes, that is the answer, but I think they meant that the level of the problem was 23.

Yes, I meant that the answer is 14, not the level.

[hide=Slightly Less Bashy Solution]In other words, we want to find the number of nonnegative integer solutions $a\le b\le c$, such that
$$a+b+c=10$$
Note that we must have $a\le3$, otherwise, $a+b+c\ge12$, which is too big. Now, we do casework depending on the value of $a$.

[b]Case I.[/b] $a=0$.
$$b+c=10$$
The number of solutions to this is $6$, where $b=0,1,...,5$.

[b]Case II.[/b] $a=1$.
$$b+c=9$$
The number of solutions to this is $4$, where $b=1,...,4$.

[b]Case III.[/b] $a=2$.
$$b+c=8$$
The number of solutions to this is $3$, where $b=2,3,4$.

[b]Case IV.[/b] $a=3$.
$$b+c=7$$
The number of solutions to this is $1$, where $b=3$.

The answer is $6+4+3+1=\boxed{14}$.

[quote=jxl28]According to the [url=http://www.artofproblemsolving.com/Wiki/index.php/Balls_and_urns]AoPSWiki[/url] on Balls and Urns, you can only count out the ways for indistinguishable to indistinguishable.  In other words, there isn't a formula (that I know of at least).

EDIT:
Sorry, looks like I wasn't very clear.  I was trying to say that when its indistinguishable to indistinguishable, you can only count out (list out) the ways, like PoP did.  I was not saying that Balls and Urns was indistinguishable to indistinguishable.  Balls and Urns is indistinguishable to distinguishable, like PoP said.[/quote]

Wait, what if you just divide by 3 to take out the over counts?

@above I thought the same thing at first but not all possibilities are counted 3 times with Balls and Urns. So 5,5,0 is counted three times but 5,3,2 is counted six times.

how i did it (i think its a bit better than writing them all out)[hide]
lets find all of the possibilites but have their order in greatest to least to not overcount and stuff. if the 1st number was 10, theres 1 way. if it was 9 there would also be 1 way. but as you go on, theres alot more cases for each. so i did mine this way: if the 1st number was x, then the other 2 numbers needed to sum to 10-x. u can use balls and sticks for finding the number of ways to arrange the last 2 numbers, then divide it by 2 to place them in greatest to least order. Also for some u have to add 0.5 to it because 1 case could be the same number for both (like x, (10-x)/2, (10-x)/2), which balls and sticks counts as only 1 case. So doing that i got
10: 1
9: 1
8: 2
7: 2
6: 3
5: 3
4(you have to find the cases yourself bc the 2nd number could be 5 or 6 and be bigger than it which would make us overcount): 2
1 + 1 + 2 + 2 + 3 + 3 + 2 = 14(did i do them right?)
[/hide]

[quote=Gauss4Euler]I have a problem, my settings are at easy, but I'm getting level 18-25 problems each time. Can someone help me?[/quote]

[quote=Red_Fox_1293]Same here! My settings are on easy, and I got 1 level 7 problem and then all 14-25 levels. ?????[/quote]

[quote=Eyed]That probably means you're really good at Alcumus then :D[/quote]

its been 3 years since these posts but
no that is incorrect.
the only reason why you are getting high level problems is because all problems in this topic are high leveled. less than one fifth the problems in this topic are below level 20, so it is very likely to have high leveed problems even though you are on easy mode.

for these kind of indistinguishable to indistinguishable problems there is no other way that is not the same or similar to just brute force

(sorry for the bump) can someone explain why this isn't stars and bars?

shouldn't it be 66??? shouldn't it be stars and bars?? i mean like x1+x2+x3=10 with non-negative integer solutions, which should be 12 choose 2 = 66

Is there any more time-efficient, less annoying method?

Counting is the solution to basically everything in this topic, and I'm trying to find other solutions on the internet.


function
induction
strong induction
absolute value

Determine whether there is a subset $X$ of the integers with the following property: For any integer $n$ there is exactly one solution of $a+2b=n$ with $a,b, \in X$.

[hide] For positive integers, generating functions handles the problem easily. We let the set be
\[F(x) = x^{a_{0}}+x^{a_{1}}+... \]
Then
\[F(x^{2}) = x^{2a_{0}}+x^{2a_{1}}+... \]
So $F(x)F(x^{2}) = \sum_{i,j}x^{a_{i}+2a_{j}}= 1+x+x^{2}... = \frac{1}{1-x}$. Also,
\[F(x^{2})F(x^{4}) = \frac{1}{(1-x)(1+x)}\]
And $F(x) = (1+x)F(x^{4})$, so $F(x) = (1+x)(1+x^{4})(1+x^{4^{2}})...$
so $\{a_{i}\}$ is the set of numbers whose representation in base $4$ consist of only 0 and 1. But is this generalizable to negative integers?
[/hide]

[hide]
to generalize the base 4 to all integers, use base -4  :P

No, seriously, base -4 works because any integer has a unique representation as
\[\sum_{k=0}^\infty a_{k}(-4)^{k}\]
where each $a_{k}$ is 0,1,2, or 3. Uniqueness is simple (if two expressions give the same number and first differ in the $(-4)^{k}$ place, then taking modulo $4^{k}$ would be a problem).

For existence for each integer $m$, use strong induction on $|m|$. Let $r$ be (nonnegative) remainder on dividing by 4, then let $m^\prime =-\frac{m-r}{4}$ which has absolute value less than that of $m$ unless $m=0$ or $m=-1$, and these are easy to deal with.

Now that we have base -4, let $X$ be the set of integers that have only 0's and 1's in base -4. Then for any $a,b\in X$, no carrying occurs when multiplying $b$ by 2 and when adding $a+2b$. So when considering solutions to $a+2b=n$, we can look at each digit separately:

Each digit of 0 in $n$ means $a,b$ both have 0's in that place.

Each digit of 1 in $n$ means $a$ has 1, $b$ has 0.

Each digit of 2 in $n$ means $a$ has 0, $b$ has 1.

Each digit of 3 in $n$ means $a,b$ both have 1's.

This determines the unique solution to $a+2b=n$.
[/hide]

Hmmm also I think you can sort of trivialize it by constructing a set inductively.

[quote="Potato Theory"]Okay, I get it.
But I was trying to extend this to some other cases, and I rephrased it such that $a+2b=n$ has exactly [i]two[/i] cases instead of one, which would mean that:

$F(x)F(x^{2}) = \frac{2}{1-x}$.  The only problem is, when following your method, I get that:

$F(x) = (1+x)F(x^{4})$, which was the exact same recursion for the case where there was exactly [i]one[/i] solution, so it's contradictory!  If anyone could clarify my problem, I'd appreciate it.[/quote]

It's the same recursion, but a different function since now $F(0)=\sqrt{2}$. So it turns out that
\[F(x)=\sqrt{2}\prod_{k=0}^\infty (1+x^{4^{k}})\]
but of course, this is nonsense when trying to use it to construct a subset $X$ of nonnegative integers such that $a+2b=n$ has exactly 2 solutions with $a,b\in X$ for each $n$. i think this in fact shows no such set exists. (Then again, just consider $n=0$  :wink: )


linear algebra
matrix
linear algebra unsolved

I was having trouble figuring this out, so I would be extremely grateful for any help!

A matrix B is similar to a matrix A if there exists some (nonsingular) matrix P such that PAP^-1 = B.

Show that if A and B are similar, then they are both square matrices of the same size.

Suppose $A$ and $B$ are similar, then there exists a $P$ such that $PA=BP$. Suppose $A$ is $n\times m$ and $B$ is $r\times s$. Then by dimensional analysis (probably misuse of that term) we see that $P$ must be of size $s\times n$ for multiplication to be possible, but this gives resultant matrices of size $s\times m$ on the left and $r\times n$ on the right. So $s=r$ and $m=n$, they are square. But this immediately implies that $r=n$ and $s=m$.


I think you use (a+b+c)(1/a + 1/b+1/c) >=9 for the right member and then Jensen again for the roght member. now it`s enough to prove that

a log(bc)a + b log(ac)b + c log(ab)c>= (a+b+c)/2. but i don`t know if that it`s true. now i don`t have time to finish it, but i will solve this today :lol:


Counting with Casework Part 1

We solve two problems using casework, and introduce a great moment in geek history!

This is a great introductory video to casework. I found it highly useful, and just wanted to say thanks!

Nice intro.  Try writing faster.  It got a little boring to watch when he was writing the cube problem.

Great Video. Very intuitive ways of explaining methods. Great moment in "geek History". Thank you.

One of the cubes is 10 and 7 does anyone know what the second one is?
The question is:
1729 can be expressed as the sum of two different cubes.  What are they?

We'd have to do the same casework involved in the lesson.  I won't get into all that LaTeX, but I'll do it on paper and assume that, after watching the video, anyone can get it.  (Acually it is in the book, making a table, etc.)

First we find out that 12 is the largest cube below 1729.  $12^3$=1728, so it is very easy to see that $\boxed{12^3+1^3=1729}$

This video is highly recomended. It was also very helpful.

How to find the numbers that can be expressed as the sum of two different positive cubes up to 1000.

[hide="Using casework"] We first have to find the highest number we can get:

$9^3=729$
$10^3=1000$

We would have to have another number to go with $10^3$ and that is not possible.

So we have to count the cases in the set of numbers:
$\mathbb{P}={1^3, 2^3, 3^3, 4^3, 5^3, 6^3, 7^3, 8^3, 9^3}$

$1^3$: We notice how any positive integer plus $1^3$ would be $<1000$ so we have the set of numbers that go with $1^3$:
$\mathbb{P}={2^3, 3^3, 4^3, 5^3, 6^3, 7^3, 8^3, 9^3}$ or $\boxed{8}$ integers.

$2^3$:  We once again notice how that if you add any number cubed, it would still be $<1000$, but we already have the set: $\mathbb{P}={1^3, 2^3}$, so the set of integers that work are:
$\mathbb{P}={3^3, 4^3, 5^3, 6^3, 7^3, 8^3, 9^3}$ or $\boxed{7}$ integers.

$3^3$:  $3^3=27$ and our highest possible cubed number (729)+27 is still less than 1000, so we would have the set of integers:
$\mathbb{P}={4^3, 5^3, 6^3, 7^3, 8^3, 9^3}$ or $\boxed{6}$ integers.

$4^3$:  $4^3=64$ and $9^3=729$  $729+64<1000$  Our set of integers are:
$\mathbb{P}={5^3, 6^3, 7^3, 8^3, 9^3}$ or $\boxed{5}$ integers.

$5^3$:  $5^3=125$ and we know $9^3=729\implies 729+125<1000$  Our set of integers are:
$\mathbb{P}={6^3, 7^3, 8^3, 9^3}$ or $\boxed{4}$ integers.

$6^3$:  $6^3=216\implies 216+729\implies 945<1000$  Our set of integers are:
$\mathbb{P}={ 7^3, 8^3, 9^3}$ or $\boxed{3}$ integers.

$7^3$: $7^3=343\implies 343+729>1000$  So, we have to test out our second biggest cube, which is $8^3=512$  $343+512<1000$  Our set of integers are:
$\mathbb{P}={8^3}$ or $\boxed{1}$ integer.

$8^3$:  $8^3=512$  We already have the case $8^3+7^3$ accounted for, and $8^3+9^3>1000$ so $\boxed{0}$ integers work for 8 and 9.

We add up all our cases to get:
$8+7+6+5+3+1\implies 15+6+5+3+1\implies 21+5+3+1\implies 26+3+1\implies 29+1\implies \boxed{30}$ integers.  [/hide]


Alright, yo, Alabama's already here at the Las Vegas site, and we're bored (the strip ain't all that). So we should all have a meeting here. Alabama's staying Tenapu...Tenophoh,... Tenapoh hall, or something like that. So we should have a meeting. Common room seems cool, so meet there ok? 9PM today, dunno about tomorrow, we'll figure it out later.

It appears likely that all teams are staying in Tonopah Hall. It's a large, high-rise dorm. SFBA's three teams should arrive some time Friday morning. The three Southern California teams will arrive probably about 1:00 p.m. on Friday. I don't know what Washington's plans look like, but I suspect a Friday arrival for them as well.

Since I haven't been to UNLV before, I don't know what the physical layout of Tonopah Hall looks like. We'll all just have to improvise after we get there.

The team and power rounds should start at 7:00 p.m. on Friday.

The layout's pretty much the same as Penn state...there's common room on the bottom, and some computers, and a big TV. And then the rest is floors with rooms..

It's a big building

Washington is split. Some people are going today (already left) but most will be there by 1? or 1:30pm tomorrow. See you there!

SFBA leaves in 2 hours (8 PM PDT).  I think the plan is to check in to Tonopah at 8.  There are also a few people on the team that are either flying or driving themselves, so they'll be getting in at an unknown time on Friday.  I suspect we'll be kind of dead on Friday morning from the bus ride, though.

P.S.  Let us hope that the layout of Tonopah is not as confusing as that of Burge at Iowa.  It took me four years to really figure out how to get around that place.

P.P.S. Does anyone know if the Florida team ended up coming to Vegas?  Or are they at Penn State?


How can I participate in the AMC12 if my school does not offer it? Can it be done independently at another school that offer it?

The preferred solution is to convince your school to offer it. If that doesn't work, one option is to take it at a college site; here's the [url=http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-CU-list.shtml]list[/url] from 2007.

You can take it at another school that offers it, but this is an administrative hassle. If you have access to a college site, they might be more receptive, since they're dealing with the same hassle for everybody.


Source: thoughtprovokers book

Here's a question that doesn not make any sense logically:

A string is wrapped around the earths equator and the two ends on the string just touch.  Now suppose that another string is tied to the original string so that it is 100 feet longer.  If thes new string is placed around the equaotr and pulled tight so that it is suspended in air, how high will the string be above the ground:
a) Not even high enough to squeeze an atom underneath the string
b) Just high ebiugh to roll a bowling ball underneath the string
c) Just high enough to drive a Mack truck underneath the string
d) One hundred feet high.

Logically you would think that the answer would be a.  But:

C (circumference) = 2(pi)r  (r = the radius of the circle whose circumference is the equator.)

C + 100 = 2(pi)r +100

$\frac{C+100}{2\pi} = \frac{2\pir = 100}{2\pi}$

$\frac{C}{2\pi} = \frac{100}{2\pi} = r = \frac{100}{2\pi}$

Since $\frac{C}{2\pi} = r$

The part that was added to the radius is $\frac{100}{2\pi}$

Which is roughly 15 feet.

So the answer would be c.

But the radius could be anything, since we used a variable, 2 inches or 2000 miles.  Also, would 100 feet make a lot of difference in the circumference of the Earth?

Can anyone explain this, or is the math just misleading?

your two objections cancel each other out. *Since* the radius can be anything, it does not matter that 100 ft is nothing compared to the radius of the earth.

[quote]... Also, would 100 feet make a lot of difference in the circumference of the Earth?
[/quote]
Logically it makes sense - look it this way 100 feet is not much , practically negligible, as compared to cicumference of the earth (about 40,000 Km) so we only get 15 feet, not much, practically negligible, as copared to the[b] radius of the earth[/b] (about 6,400 Km)...."These two 'practically negligible' are similar!(In fact, speaking as a ration, exactly same)

I asked this as part of a local contest in 2001. Unfortunately, I don't have the item analysis for the results of that contest. In my version, there were 10 extra feet in the  string, and my multiple choice answers for what could fit under it were: a small molecule; an amoeba; a mouse; a human on hands and knees; an adult human walking upright.

I dug up the old contest and asked it again at an ARML practice this fall: of 32 students there, many of whom will be on the ARML team, only 12 answered it correctly (human on hands and knees).


There are n students seated around in a circle. Their teacher decide to give them all chocolates. He gives the first student a chocolate . Then he skips one student and gives a chocolate to the 3rd student . Then he skips 2 students and gives a chocolate to the 6th student and so on..........................
Find all values of n for which all students receive atleast one chocolate .

This is a IMO problem one of the few I solved myself  :D


Sa se arate ca pentru orice numar natural, $n=\geq 2$, exista $n$ puncte distincte in plan, nu toate coliniare, astfel incat distanta intre oricare $2$ este exprimabila printr-un numar intreg.

   Nu "se baga" nimeni? :D

solutia nu imi apartine, ci o stiu de la un curs
[hide="solutie"]
fie $y$ un numar cu suficient de multi divizori, par, si pentru $y=2a_{k}b_{k}$, alegem $x_{k}=a_{k}^{2}-b_{k}^{2}$ ($a_{k}>b_{k})$. $x_{k}^{2}+y^{2}$ este un patrat perfect, deci distanta dintre punctul $(x_{k},0)$ si $(0,y)$ este un numar intreg.
deci multimea de puncte formata din $(x_{k},0)$ si $(0,y)$ satisfac relatia din ipoteza (deoarece $x_{k}\in\mathbb{Z}$ si nu sunt toate coliniare), si daca $y$ are suficient de multi divizor, se pot gasi oricat de multe astfel de puncte.
[/hide]


calculus integration geometry trapezoid derivative function quadratics

If f(x)= x^2-3 ,   -1[u]<[/u]X[u]<[/u]5 find the Riemann sum with n=6 using right endpoints, left endpoints, then midpoints.

I missed the lecture on this so could you please do the whole problem.[/u]

Drawing the graph will help you :)
we take $\Delta x = \frac{5-(-1)}{6}$ since $n=6$ and so we have six partition $[-1,0]$, $[0,1]$, $[1,2]$, $[2,3]$,$[3,4]$,$[4,5]$.

For the leftsum we first start from the point $x_0= -1$ and then use the left endpoint:
$x_1=0$, $x_2=1$, $x_3=2$, $x_4=3$ and $x_5=4$ . The Rienmann sum is in the form $\sum_{i=0}^5 f(x_i) = -2-3-2+1+6+13=13$

For the rightsum we first start from the point $x_0=0$ and then use the right endpoint: $x_1=1$, $x_2=2$, $x_3=3$, $x_4=4$ and $x_5=5$ the rienmaan sum is  $\sum_{i=0}^5 f(x_i) = 37$

For the midpoint we choose the midpoint from each partition that is $x_0=-\frac{1}{2}$,$x_1=\frac{1}{2}$, $x_2=\frac{3}{2}$, $x_3=\frac{5}{2}$, $x_4=\frac{7}{2}$ and $x_5=\frac{9}{2}$ then the Rienmann sum is  $\sum_{i=0}^5 f(x_i) = 23.5$

and so we have leftsum < midlesum < rigthsum. if we do the definite integral then we have $24$. The midpoint approximation is better. For very large $n$ the approximation will getting better.

If you simply averaged the left-hand Riemann sum and the right-hand Riemann sum in this instance, you would get $\frac{13+37}2=25.$ This estimate goes by another name: the trapezoid rule.

For equal-width partitions (which is what we're talking about here) both the left-endpoint and right-endpoint Riemann sums are "first-order" methods: The error depends on the first derivative (they would be exact for constant functions) and the size of the error is proportional to $\frac1n$ where $n$ is the number of intervals in the partition. Beginner said "For very large $n$ the approximation will getting better." If we're using either right or left endpoint Riemann sums, doubling the number of steps would cut the error in half. Doing ten times as much work would gain us one decimal digit of accuracy.

Both the midpoint rule and the trapezoid rule (where the trapezoid rule is the simple average of the right and left endpoint rules) are "second order" methods. Their error depends on the second derivative (meaning they would be exact for straight lines), and the size of the error is proportional to $\frac1{n^2}.$ If the second derivative of the function is positive (which it is in this case), the midpoint rule will be below the actual integral and the trapezoidal rule will be above the actual integral. The error in the trapezoidal rule tends to be about twice as big as the error in the midpoint rule. The size of the error estimate means that by doing 10 times as much work, you would gain two decimal digits of accuracy.

Could we average the trapezoidal rule and the midpoint rule to get something better? Since the error in the trapezoidal rule tends to be twice as big as that for the midpoint rule, the right way to do this is a weighted average: if $T$ is the trapzoidal rule estimate and $M$ the midpoint rule estimate,

new estimate $=\frac{2M+T}3.$

In this particular case, you would get $\frac{2\cdot23.5+25}3=24.$

You might expect this to be a "third order" method in which the error depends on the third derivative (and that alone would account for us getting the exact answer for this quadratic polynomial), but detailed analysis shows that we actually did better than that: it's a "fourth order" method. The error depends on the fourth derivative, and the size of the error is proportional to $\frac1{n^4}$ This time, doing ten times as much work will gain us about 4 extra decimal digits of accuracy.

Of course that last method has a name: it's called Simpson's Rule.


[quote="b-flat"]When I was looking at n^4+4's profile, I went to see all of his posts, and I got the error

[quote]Information

No topics or posts met your search criteria

Click Here to return to the Index

 [/quote]

I know that he has several posts that I can see (several in the NC forum), so I was wondering why it won't let me view them all  :maybe:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=118602]example post[/url][/quote]
Perhaps its because his name has under two letters.

[quote="solafidefarms"]Maybe because of that ^ in his username?[/quote]
What about E^(pi*i)=-1, his posts are fine.

Search method will be changed in the (near) future (going after user_id rather than username which causes problems).


induction
combinatorics solved
combinatorics
combinatorial geometry

Let f(n) (n \in N) denote the smallest number of distinct points P_1, P_2, ..., P_f(n) in the plane such that for each k = 1,2,,...,n there is a straight line g_k which contains exactly k points of the set X_n = {P_1, P_2, ..., P_f(n)}. Determine f(n) !

We have f(2n) = n(n+1) and f(2n+1) = (n+1)^2

This is problem 2674 of Crux Mathematicorum. A solution appeared in the november 2002 issue (p. 472-473).

Not everyone has the Crux Mathematicorum.
It will nice, if you put the proof of Crux on the forum

Your formula is right but also might be written differently:

f(n) = ((n+1)/2)^2 for all n \in N and n odd

f(n) = n/2 * (n+2)/2 for all n \in N and n even

The formulae also might be combined as:

f(n) = li((n+1)/2) * li((n+3)/2) where li(x) denotes the greatest integer \leq x

But why don't you provide some arguments how one might come up with these formulas please ? Let's see whether the Crux proof is different from the one I have.

I didn't know that the problem was also used in Crux. Considering all the math journals like American Mathematical Monthly, Crux, Kvant, Kmal and others I guess there are quite a few interferences. Then olympiad problems are taken from problem sections of various math journals and are reused in another olympiad (maybe slighlty changed). And then someone else submits the olympiad problems again for the problem section of a math journal... So you see it is an infinite recurrence, also called a never ending story.

You are right Orl, we use it in the training for the French team, after I took it from Crux.

Hmm...here is how I did it (I think the solution published in Crux is similar) :

Let E be a 'good' point set for n.
For each k with 1 =< k =< n, there exists a line d_k with exactly k points from E. Thus, if k =/= k', we have d_k =/= d_k', from which we deduce that d_k and d_k' have at most 1 point of E in common.
It follows that, d_n uses n points form E
d_(n-1) uses at least n-2 other points from E
d_(n-2) uses at least n-4 more points from E
and so on...
Then E contains at least S(n) = n+(n-2)+(n-4)+..+e, where e = 0 or 1 according to the parity of n.
Computing this sum, it easily follows that
f(2p) >= p(p+1) and f(2p+1) >= (p+1)^2.

Conversely, we easily have f(1)=1.
Suppose, by induction, that for some n >= 1 you have a good set point E with S(n) elements.
For k=1,...,n, let d_k be a line with exactly k points from E (according to the induction hypothesis). Since there is a finite number of lines d_k, they determine a finite number of directions and a finite number of intersection points. Thus we may choose a line d which contains none of the points in E and which is not parallel to any of the d_k.

- If n = 2p is even : Les E' be the union of E and the set of the intersection points between d and d_k for k= p+1,...2p.
For i >= p+1, the line d_i contains exactly i+1 points from E'.
For i =< p, the line E' contains exactly i points from E'.
Moreover the line d contains exactly p points from E'.
Thus, in order to obtain a good set points E'' for n+1, it suffices to add an arbitrary point in E' which belongs to d but not to any other d_i.
Thus this set E'' contains exactly p(p+1) + p+1 = (p+1)^2 points.

- If n = 2p+1 is odd, the reasoning is the same as above, using the lines d_k for k = p+1 to 2p+1, but with no extra point at the end.

This ends the induction steps and the proof.

Pierre.


Determine all functions :$ f: R$ \ $ \{ 0,1{\text{\} }}$ $ \to R$
 satisfying :
$ f(x) \plus{} f\left( {\frac{1}
{{1 \minus{} x}}} \right) \equal{} \frac{{2(1 \minus{} 2x)}}
{{x(1 \minus{} x)}}$,$ \forall x \ne 0,x \ne 1$


I use cyclinder cups  :P
On other note, I think half is half filled for me
And the trapezoid clearly takes 1/3
cup is 2/3.
Unless I am wrong, I dont see an optical illusion...
Can you please be more clear?

[quote="RobRoobiks"]
 >> And the trapezoid clearly takes 1/3 cup is 2/3. <<
Unless I am wrong, I dont see an optical illusion...
Can you please be more clear?[/quote]

RobRoobiks, what is the highlighted sentence supposed to
be when typed out in full?  (There are missing words, and
you've got "trapezoid" and "cup" in the same sentence.
It's as if you had the complete thought in your head, but
typed an incomplete sentence.)

The idea of optical illusions here is that a person thinks the
dividing line of certain areas of the trapezoids is at certain places,
but someone's guesses are off.  Also, the dividing line in a cup
for certain volumes is off compared to someone's guesses.


algorithm
number theory
Euclidean algorithm
greatest common divisor
number theory theorems

[color=indigo]Question No1.[/color] What is [b]Mersene sequence[/b].?

[color=indigo]Question No2.[/color] It is "well-known" that \[ gcd(a^m-b^m,a^n-b^n)=a^{gcd(m,n)}-b^{gcd(m,n)}  \], but what if numbers $a$ and $b$ are not integer and $a^m-b^m$ and $a^n-b^n$ are i.e. Fibonacci numbers...
\[ F_n=\frac1{\sqrt5}((\dfrac{1+\sqrt5}2)^n-(\dfrac{1-\sqrt5}2)^n)  \]
one can easily prove that if $k|m$ then $F_k|F_m$, just put $(\dfrac{1+\sqrt5}2)=\alpha$ and  $(\dfrac{1-\sqrt5}2)=\beta$ divide them and there you are... but it also stands that if $k=gcd(m,n)$ then $F_k=gcd(F_m,F_n)$, so if one can prove that $(\alpha^m-\beta^m,\alpha^n-\beta^n)=\alpha^{(m,n)}-\beta{(m,n)}$ the problem I think will be solved but as I asked can it be proven when as in this case $a$ and $b$ are not integer?

There is Mersenne sequence? :huuh:  I only know Mersenne number and prime.

Mersenne number : $2^n-1$, where n is a natural number, is a Mersenne number.
Mersenne prime : Mersenne number that is prime

[url=http://mathworld.wolfram.com/]Mathworld[/url] doesn't have it either. Maybe the sequence is the sequence of one of these two? I don't know. :?

I already knew what Mersen number is but thank you anyway... I found someone talking about Mersene squence in some post from a few years ago, here is link so go there and se it....  http://www.mathlinks.ro/Forum/viewtopic.php?p=6800#p6800

Actually, that statement is true in a much more general context.

So, let $R$ be any commutative ring with 1(that means, we got a commutative addition with identity 0, a commutative multiplication with identity 1 which is distributive with respect to addition). Take any two elements $a,b\in R$ such that the ideal $(a,b)$ generated by them(that is, the set of all elements of $R$ of the form $xa+yb$ with $x,y\in R$) equals $R$. That means every element of $R$ is a linear combination of these two elements, in particular the $1=xa+yb$ for some $x,y\in R$. But if we can express $1=xa+yb$, then every element of $R$ is in $(a,b)$, since for $r\in R$, we have $r=r\cdot 1=r(xa+yb)=(rx)a+(ry)b$. Therefore we may just impose the condition that there are $x,y\in R$ such that $xa+yb=1$.
For your example, you may take the ring $R=\{u+v\dfrac{1+\sqrt{5}}2\ {\sf with}\ u,v\in \mathbb{Z}\}$ with addition and multiplication definde in the obvious way. Taking $a=\dfrac{1+\sqrt{5}}2$ and $b=\dfrac{1-\sqrt{5}}2$, you may choose $x=y=1$.

Then the following statement holds:
$(a^n-b^n,a^m-b^m)=(a^{\gcd(m,n)}-b^{\gcd(m,n)})$ for all $m,n\geq 1$.

Here the left-hand side is again the ideal generated by the two elements $a^n-b^n$ and $a^m-b^m$, the right-hand side is the ideal generated by $a^{\gcd(m,n)}-b^{\gcd(m,n)}$; that means, it is the set of multiples of that element. The identity means that the two sets agree.

The proof is the same as for $\mathbb{Z}$ with some easy additional steps; i guess it is a good exercise to do this ;).

In fact, the ideal $(a^n-b^n,a^m-b^m)$ is the analogon of the greatest common divisor in $\mathbb{Z}$ - in fact, for $\mathbb{Z}$, both things agree, as the Euclidean Algorithm ensures.

For your example, you are still working in the ring $R$, so the ideal generated by $a^n-b^n$ and $a^m-b^m$ in $R$ might be different from that they generate in $\mathbb{Z}$; however, i guess, you may finish the argument.

I guess a Mersenne sequence is a sequence $a_1,a_2,...$ of positive integers, such that $\gcd(a_m,a_n)=a_{\gcd(m,n)}$.

I just read this quickly and I will read it more cearfuly later but be sure that I will ask a lot of questions about it because it seems to me that this is more than I can take :D :D


linear algebra
matrix
inequalities
induction
linear algebra unsolved

For given $A_1,...,A_k$ matrices with positive or 0 eigenvalues, with real entries, what is the maximal value of $ det(x_1 A_1+...+x_n A_n)$ where $x_i$ are real numbers between -1 and 1?

Moubi, how many times should I repeat? A simple answer says nothing! Example: what are the numbers n such that $n^2$ divides $2^n+1$. The answer 3 worths 0 points. So don't be lazy and write solution, please.   ;)

If all matrices $A_k\in M_n(R)$ are trigonalisable

with $A_kA_g=A_gA_k$ then are simultaneously trigonalisable

$det(x_1A_1+...+x_nA_n)=(x_1w(1,1)+...+x_nw(n,1))....(x_1w(1,n)+...+x_nw(n,n))$

$w(k,p)\geq 0$ are eigenvalues of $A_p$

the  maximum is $det(A_1+...+A_n)$

What is your opinion professor harazi ?

Hmmm, I don't quite understand... I never said they commute.  Or I don't see obvious things?  :)

From the fact that all the eigenvalues of our matrices are real and non-negative we can conclude that our matrices are hermitian and positively defined, therefore there must be unitary matrix $U$ s.t. $UA_1U^{-1}=D$- diagonal and denote $UA_2U^{-1}=M$ -positively definite matrix. Hence $det(A_1+xA_2)=det(D+xM)=\sum_{p=0}^{n}x^p\sum_{i_1<i_2<\cdots<i_n}M(i_1,i_2,...,i_p;  i_1,i_2,...,i_p) dj_1dj_2...dj_{n-p}$ where the set $(j_1,...,j_{n-p})$ complements the set $(i_1,i_2,...,i_p)$ to $(1,...,n)$ and $M(i_1,i_2,...,i_p;  i_1,i_2,...,i_p)$ means the minor of the matrix $M$ which is obtained by the extracting rows and columns with indices $i_1,i_2,...,i_p$. From positively deniniteness of $M$ and $D$ it follows that $M(i_1,i_2,...,i_p;  i_1,i_2,...,i_p) \geq0$ and $d_j \geq0$. Thus
$det(A_1+xA_2) \leq \sum_{p=0}^{n}|x|^p\sum_{i_1<i_2<\cdots<i_n}M(i_1,i_2,...,i_p;  i_1,i_2,...,i_p) dj_1dj_2\cdots dj_{n-p} = \sum_{p=0}^{n}1\sum_{i_1<i_2<\cdots<i_n}M(i_1,i_2,...,i_p;  i_1,i_2,...,i_p) dj_1dj_2\cdots dj_{n-p}=det(D+M)=det(A_1+A_2)$.
So, we've proved that $det(x_1A_1+x_2A_2) \leq det(A_1+A_2)$ The result Moubinool alluded follows from afore mentioned inequality by induction
Am i right?

Eugene, please edit your post to make it readable.

Ecxuse me, i've edited only now beacuse of my very very bad connection

A single dumb question: how do you prove it by induction? It can happen that $x_1 A_1+x_2 A_2$ is not positive.

Ok,I'll try to explain.
Let's now prove that $det(A_1+...x_nA_n) \leq det(A_1+...+A_n)$. Let $A=A_1$ and $A'=x_2A_2+...+x_{k+1}A_{k+1}$. Since $A$ hermit there must be unitary matrix $U$ s.t. $UAU^{-1}=D$ diagonal; let's denote $M_j=UA_jU^{-1}$, $M=UA'U^{-1}$-non-negative defined, therefore $det(A+A')=det(D+M)\leq\sum_{p=0}^{n}\sum_{i_1<i_2<\cdots<i_n}M(i_1,i_2,...,i_p;  i_1,i_2,...,i_p) dj_1dj_2\cdots dj_{n-p}$. Since $M=x_2M_2+...+x_{k+1}M_{k+1}$ by the induction hypthesis $M(i_1,...i_p;i_1,...,i_p) \leq det(\sum_{j=2}^{k+1}M_j(i_1,...i_p;i_1,...,i_p))$.In order to finish the prood we need to note that $\sum_{p=0}^{n}dj_1dj_2\cdots dj_{n-p}det(\sum_{j=2}^{k+1}M_j(i_1,...i_p;i_1,...,i_p))=det(D+M_2+...+M_{k+1})=det(A_1+...+A_n)$

harazi what is your solution for this one

Here it is: put $ A=(1+x_1)A_1+...+(1+x_n)A_n$ and $ B=(1-x_1)A_1+...+(1-x_n)A_n$. They are both combinations with nonnegative coefficients of semi-positive matrices, thus they are semi-positive. It remains to show that for two semi-positive matrices $A,B$ we have $ det(A+B)\geq det(A-B)$. This is trivial since we can find invertible matrix $C$ such that $ A=C^{*}D_1C$ and $ B=C^{*}D_2 C$ for some diagonal matrices $ D_1,D_2$ with nonnegative elements. Is it good?

[quote="harazi"]It remains to show that for two semi-positive matrices $A,B$ we have $ det(A+B)\geq det(A-B)$. This is trivial since we can find invertible matrix $C$ such that $ A=C^{*}D_1C$ and $ B=C^{*}D_2 C$ for some diagonal matrices $ D_1,D_2$ with nonnegative elements. Is it good?[/quote]

Why the invertible matrix $C$ is the same for $A$ and $B$ ?
it seems it happen if $A,B$ commute

You still didn't say what means "ruse". ;)    For the other property, suppose $A,B$ are positive definite. We know that in this case there exists non-invertible $ T$ such that $T^{*}AT=I$. But then the matrix $ T^{*}BT$ is normal matrix, thus we can diagonalize it in an orthonormal base, that is there exists unitary matrix $U$ such that $ U^{*}T^{*}ATU=D$ for $D$ a diagonal matrix. It suffices to take $ TU$ as common matrix. Note that I use the fact that they are strictly positive definite. In the contrary case, it suffices to take sequences of strictly positive definite matrices converging to the two matrices and to use the continuity of the determinant. Note that there exists another proof of the inequality, which uses the more simple fact that all eigenvalues of $A^{-1}B$ are positive numbers if $A,B$ are strictly positive definite.


limit
function
calculus
calculus computations

let $ {x_{n}}$ be a sequence such that

$ x_{n\plus{}1}\equal{}\sqrt{\frac{x_{n}\plus{}\frac{1}{x_{n}}}{2}}$
for all $ n\geq 1$
given $ x_{1}\in R$
find limit $ \lim_{n\to\infty}(x_{1}x_{2}...x_{n})$ (as a function of $ x1$)

Clearly x(2) in positive, if it exists.
The for AM-GM x(n)>=1 for every n, and as (n+1)<x(n) if x(n)>1, then the limit will be 1 ^^

First, we should insist that $ x_{1}> 0;$ otherwise this doesn't make sense as a real sequence.

mitchan88 is essentially correct, as far as he went, which isn't all the way to the end of the problem. We do have that $ x(2) > 1$ and that after that, $ 1 < x_{n\plus{}1}< x_{n}.$ Therefore this is a bounded monotone sequence and converges. Given that, we have that its limit satisfies

$ x \equal{}\sqrt{\frac{x\plus{}\frac{1}{x}}{2}}.$

Square this and clear the fractions to get $ 2x^{3}\minus{}x^{2}\minus{}1 \equal{} 0.$ This cubic equation has only one real root, namely $ x \equal{} 1.$

So $ \lim_{n\to\infty}x_{n}\equal{} 1,$ as mitchan88 claimed.

However, go look at the original post. The question is really about the infinite product $ \prod_{n \equal{} 1}^{\infty}x_{n}.$

That $ x_{n}\to1$ is a necessary condition for the convergence of the infinite product but not a sufficient condition. We need for the convergence to be fast enough. (Hint: it is fast enough.) The problem is not finished yet.


Let $ x_1,x_2,..x_n$ are real numbers in the inteval $ (0;1)$.Prove that:
$ \frac {(x_1 \plus{} x_2 \plus{} ... \plus{} x_n)^2}{\sqrt {1 \minus{} x_1} \plus{} ... \plus{} \sqrt {1 \minus{} x_n}} \geq \frac {(\sqrt {x_1^2 \minus{} x_1^3} \plus{} ... \plus{} \sqrt {x_n^2 \minus{} x_n^3})^2}{\sqrt {1 \minus{} x_1}^3 \plus{} ... \plus{} \sqrt {1 \minus{} x_n}^3}$

Suppose that $ x_1 \geq x_2 \geq ... \geq x_n$
Then $ \sqrt {1 \minus{} x_1} \le \sqrt {1 \minus{} x_2} \le ... \le \sqrt {1 \minus{} x_n}$

So, by Chevychev inequality:
$ (\sqrt {x_1^2 \minus{} x_1^3} \plus{} ... \plus{} \sqrt {x_n^2 \minus{} x_n^3})^2 \le \frac{(x_1\plus{}x_2\plus{}...\plus{}x_n)^2(\sqrt {1 \minus{} x_1} \plus{} ... \plus{} \sqrt {1 \minus{} x_n})^2}{n^2} \hspace{2cm}(1)$

And we know that:
$ \frac{ \sqrt {1 \minus{} x_1}^3 \plus{} ... \plus{} \sqrt {1 \minus{} x_n}^3}{n} \geq \left( \frac{\sqrt {1 \minus{} x_1} \plus{} ... \plus{} \sqrt {1 \minus{} x_n}}{n}\right)^3$
$ \implies \frac{1}{ \sqrt {1 \minus{} x_1}^3 \plus{} ... \plus{} \sqrt {1 \minus{} x_n}^3} \le \frac{n^2}{(\sqrt {1 \minus{} x_1} \plus{} ... \plus{} \sqrt {1 \minus{} x_n})^3} \hspace{2cm} (2)$

Then from the product of $ (1)$ and $ (2)$:
$ \frac {(x_1 \plus{} x_2 \plus{} ... \plus{} x_n)^2}{\sqrt {1 \minus{} x_1} \plus{} ... \plus{} \sqrt {1 \minus{} x_n}} \geq \frac {(\sqrt {x_1^2 \minus{} x_1^3} \plus{} ... \plus{} \sqrt {x_n^2 \minus{} x_n^3})^2}{\sqrt {1 \minus{} x_1}^3 \plus{} ... \plus{} \sqrt {1 \minus{} x_n}^3}$


can someone please provide all the properties of limits

[tex]
\displaystyle
\lim_{x \to y} f(x) + g(x) = \lim_{x \to y} f(x) + \lim_{x \to y} g(x)

\displaystyle \lim_{x \to y} \frac{f_1(x)}{f_2(x)} = \frac{\lim_{x \to y} f(x)}{\lim_{x \to y} g(x)}

\displaystyle \lim_{x \to \infty} f(x) = \lim_{x \to 0} f\left(\frac{1}{x}\right)
[/tex]

I know there are some others, but my mind is blank right now.

[quote="nr1337"][tex]
\displaystyle
\lim_{x \to y} f(x) + g(x) = \lim_{x \to y} f(x) + \lim_{x \to y} g(x)
[/tex][/quote]

[tex]
\displaystyle
\lim_{x \to 0} (\frac{1}{x} + \frac{x-1}{x}) \neq \lim_{x \to 0} \frac{1}{x} + \lim_{x \to 0} \frac{x-1}{x}
[/tex]

the properties that he put assume that each limit exists as a finite number... in Joel's counterexample, [tex]\displaystyle \lim_{x\to0}\frac{1}{x}[/tex] does not exist, so the properties dont hold in that case.

For all of these, "as n approaches  :inf: ".

lim (c) = c.

lim (1/n) = 0.

lim (n + m) = lim (n) + lim (m)

lim (n*m) = lim (n)*lim (m)

lim (n/m) = (lim (n))/(lim (m))

So to find the limit of (2n+11)/(5n-1):

lim (2n+11)/(5n-1)
lim (2+(11/n))/(5-(1/n))            divide top and bottom by n.
(lim (2+(11/n)))/(lim (5-(1/n)))  use the division property
(lim2 + lim(11/n) ) / ( lim5 - lim(1/n)) addition property
(2+ (11)(0) ) / (5 - (1)(0) )          the first two properties.
2/5                        is the limit.

So any problem like that, the limit is the coefficients of the n's.
I'm sure there are more, but i need to get off now. Hope that example wasn't too messy.


How do you do this?:

$\sum_{i=0}^{14}(14$ choose $i)^{2}$

I'm stuck... tried finite differences but it's pointless calculating values for n > 6.

You mean
$\sum_{i=0}^{14}\left(C_{14}^{i}\right)^{2}=?$
If so...i'll post a more general solution:

$\sum_{k=0}^{n}\left(C_{n}^{k}\right)^{2}=?$
In order to prove this one we may consider the following identity $(1+x)^{2n}=(1+x)^{n}\cdot(1+x)^{n}$
The coefficient of $x^{n}$ from the left side is $C_{2n}^{n}$ ,and the coefficient from the right side of $x^{n}$ is: $C_{n}^{0}\cdot C_{n}^{n}+C_{n}^{1}\cdot C_{n}^{n-1}...C_{n}^{n}\cdot C_{n}^{0}$ ,But using the formula $C_{n}^{k}=C_{n}^{n-k}$ we have that the coefficient from the right side is $\sum_{k=0}^{n}\left(C_{n}^{k}\right)^{2}$ and it's equal to the coefficient from the left side.
Therefore
\[\sum_{k=0}^{n}\left(C_{n}^{k}\right)^{2}=C_{2n}^{n}\]

[b]Solution.[/b] Consider the following combinatorial argument.  Suppose that you have $n$ girls and $n$ boys.  You want to choose $n$ persons from them.  Clearly, the number of ways to do so is $\binom{2n}{n}$.  You can also select them in the sum of the number of ways to obtain each of the following cases: $(\text{\# of girls},\text{\# of boys})=(0,n),(1,n-1),(2,n-2),...(n,0)$, which is \[\sum_{k=0}^{n}\binom{n}{k}\binom{n}{n-k}=\sum_{k=0}^{n}\binom{n}{k}^{2}.\]

And the conclusion follows. $\Box$

I just realized that it's an instance of the Vandermonde identity. Doh!

Or you could just be like $\sum_{i = 0}^{n}\binom{n}{i}^{2}= \binom{2n}{n}.$ So the answer's $\binom{28}{14},$ or $40116600.$ Another way you could prove it is by block-walking... if you draw out Pascal's Triangle, then the number of ways to "walk" to $\binom{n}{k}$ from $\binom{0}{0}$ is just $\binom{n}{k}.$ In order to walk to $\binom{2n}{n},$ you must walk through the $n$th row, and the 0th, 1st, 2nd, 3rd, etc. entry at that. So, we can say that

$\[\binom{n}{0}^{2}+\binom{n}{1}^{2}+\cdots+\binom{n}{n}^{2}= \binom{2n}{n}\]$

[quote="Phelpedo"]Somehow I feel like that's what all the other posts have said  :wink:[/quote]

Heh, I was trying to prove it another way myself and then upload my proof... just doing it one part at a time  :D


abstract algebra
algebra proposed
algebra

If $d\in \mathbb{Q}$, is there always an $\omega \in \mathbb{C}$ such that $\omega ^n=1$ for some $n\in \mathbb{N}$ and $\mathbb{Q}(\sqrt{d})\subseteq \mathbb{Q}(\omega)$?

Yes, this is a special case of the Kronecker-Weber-Theorem that states that every finite abelian extension of $\mathbb{Q}$ is contained in a cyclotomic extension.

For this case the gaussian sums $\left(\sum_{n=1}^p \zeta^{n^2}\right)^2 = \pm p$ ($p$ being prime, $\zeta$ being a $p$-th root of unity) suffice to show that there is an extension containing $\sqrt{p}$, thus there is an extension containing $\sqrt n$ for any integer $n$, and the result for rationals gets trivial.

[quote="ZetaX"]$\left(\sum_{n=1}^p \zeta^{n^2}\right) = \pm p$ ($p$ being prime, $\zeta$ being a $p$-th root of unity)[/quote]

I think that you mistyped something (or am I missing something  :!: )
The lefthandside's norm is always less than $p$.

Sorry, I forgot the square after the brackets  :blush:


geometry
geometric transformation
reflection
incenter
angle bisector
geometry proposed

Let $ABC$ a triangle for which $C=60^{\circ}+A$. Denote the points $M\in (AC)$, $m(\widehat {ABM})=\frac 14 B$ and $N\in (AB)$, $m(\widehat {ACN})=30^{\circ}$. Prove that $m(\widehat {BMN})=30^{\circ}.$

Remark. For $A: =20^{\circ}$ we get the old problem
(see http://www.mathlinks.ro/Forum/viewtopic.php?t=31301 )

Reflect $N$ through $AC$ to $N'$. Connect $N'N$, $N'B$, $N'A$ and $N'C$.

Since $\angle{BCN}=30^o+\angle{A}=\angle{NCA}+\angle{A}=\angle{BNC}$. Hence $BN=BC$. Because $\angle{NCN'}=60^o$ and $NC=N'C$, so $\triangle{NCN'}$ is an equilateral triangle, thus $NN'=N'C$. Therefore $BNN'C$ is a kite, so $N'B$ is the angle bisector of $\angle{ABC}$, so $\angle{ABN'}=\frac{\angle{B}}{2}$, but since we have $\angle{ABM}=\frac{\angle{B}}{4}$, so $BM$ is the angle bisector of $\angle{ABN'}$.

Because we also have $MA$ bisects $\angle{BAN'}$, so $M$ is the incenter of $\triangle{BAN'}$. Therefore $N'M$ bisects $\angle{AN'B}$. Thus:
\[ \angle{NMB}+\frac{\angle{B}}{4}=\angle{ANM}=\angle{AN'M}=\frac{\angle{AN'B}}{2}=\frac{180^o-\angle{A}-\frac{\angle{B}}{2}}{2} \]
Hence: $2\angle{NMB}=180^o-2\angle{A}-\angle{B}=\angle{A}+\angle{C}-2\angle{A}=\angle{C}-\angle{A}=60^o$, therefore: $\angle{NMB}=30^o$.

[hide="Here is the author-proof."] Denote $G\in AC$ so that $m(\widehat {ABG})=60^{\circ}$. Prove easily that $BC=BG=BN=NG=MG$ a.s.o.[/hide]


algebra
polynomial
number theory unsolved
number theory

1. For each integer $k\geq4$, prove that if $f(x)$ is a polynomial with integer coefficients satisfying the condition $0\leq f(c)\leq k$ for every $c=0,1,...,k+1$, then $f(0)=f(1)=...=f(k+1)$.

2. Find all positive integers $n$ such that the equation $x^3+y^3+z^3=nx^2y^2z^2$ has positive integer solutions.

1) is from ISL 97 :

http://www.kalva.demon.co.uk/short/soln/sh9710.html

Beware there is a solution with th link.

Pierre.


If I try to change the color scheme of my blog, it takes a couple days for the results to show up!!!

Aha

Izzy told me the answer.
You just have to refresh after you change the settings.

Yes your browser caches the css stylesheet, thus you have to tell your browser to re-cache it (usually by refreshing the page F5).


There are 3 different numbers: A, B, and C. Their sum adds up to 100. A/B produces an answer of 5 with remainder 1. B/C gets the same result. What is C?

[hide]We can write three equations:

$ 5B \plus{} 1 \equal{} A$
$ 5C \plus{} 1 \equal{} B$
$ A \plus{} B \plus{} C \equal{} 100$

Putting $ A$ and $ B$ in terms of $ C$ gives us $ 25C \plus{} 6 \plus{} 5C \plus{} 1 \plus{} C \equal{} 100.$ Solving, $ 31C \plus{} 7 \equal{} 100,$ so $ C \equal{} \boxed{3}.$  [/hide]

[hide]Basically, you just make equations and get C=3[/hide]


For $ a,b,c$ positive reals such that $ a\plus{}b\plus{}c \equal{} 3$  prove that

$ 4 \geq abc \plus{} \sum a^2b$

P.S. I think it was created by Hungkhtn, but I am not sure.
       It cannot be solved by SOS, as at the equivalent form

       $ \sum (a\minus{}b)^2(13a \minus{} 5b \plus{} c) \geq 0$

      cannot be applied the known criteria.

     I tried to solve it using the inequality

     $ \sum ab \sum a^2b \leq 9$

     already proved in the forum, but it is not enough strong to solve it.

    So, I ask for you which is the key to solve it?

    Sorry, but I don't remember if it is an old inequality already proved in the forum. :(

[quote="manlio"]For $ a,b,c$ positive reals such that $ a \plus{} b \plus{} c \equal{} 3$  prove that

$ 4 \geq abc \plus{} \sum a^2b$

[/quote]
It's equivalent to $ \frac{a}{4a\plus{}4b\plus{}c}\plus{}\frac{b}{4b\plus{}4c\plus{}a}\plus{}\frac{c}{4c\plus{}4a\plus{}b}\le \frac{1}{3}$
and see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147258


In a school of 100 students, there are 90 students who take English, 75 who take Spanish, and 42 who take French. If each student must take at least one course, what is the least number of students that take all three courses?

10 who DON'T take English
25 who DON'T take Spanish
58 who DON'T take French

So if we want to minimize the number of people who take all three languages, we have 100-(10+25+58)=[b]7[/b]


It takes 18 men 6 hours to dig a trench 720 feet long. Working at the same rate,
how many feet of trench can be dug by 24 men in 8 hours?

[hide=" click to reveal hidden content"]

$ 18$ men, $ 6$ hours, $ 720$ feet.

$ 6$ men, $ 18$ hours, $ 720$ feet.

$ 12$ men, $ 9$ hours, $ 720$ feet.

$ 12$ men, $ 1$ hour, $ 80$ feet.

$ 12$ men, $ 8$ hours, $ 640$ feet.

[/hide]

The increased rate yields $ 720\times{\frac {24}{18}}\times{\frac {8}{6}}\implies 720(\frac {4}{3})^2 \implies 720(\frac {16}{9}) \equal{} \boxed{1280}$.

$ 18$ men, $ 6$ hours, $ 720$ feet.

$ 6$ men, $ 18$ hours, $ 720$ feet.

$ 12$ men, $ 9$ hours, $ 720$ feet.

$ 12$ men, $ 1$ hour, $ 80$ feet.

$ 12$ men, $ 8$ hours, $ 640$ feet.

$ 24$ men, $ 8$ hours, $ 1280$ feet.

Just finishing last step.  :wink:


In a competition 15 teams are participating. Every team plays exactly once against each other team. For a victory the winning team gets 3 points and the defeated team 1 point. A tie results into 2 points for each team. After the end of the competition all teams received pairwise distinct points, none of them less than 21. Prove that the winner of the competition had at least one tie.

[i]German IMO Selection Test 1983, problem 3[/i]

Let e_i be the number of points of the team T_i, whith 21  \leq e_1 < e_2 < ... < e_15.
Since the e_i's are integers, it follows that e_i \geq 20 + i for each i.
Thus e_1 + ...  + e_15 \geq 420. (1)

Since each match gives a total of 4 points and since there are 15*14/2 matchs, we deduce that :
e_1 + ....+ e_15 = 420.
Then, we are in equality case of (1), and it follows that e_i = 20 + i for each i. In particular, e_15 = 35.

Let v_i, d_i, t_i be the numbers of victories, defaites or ties of T_i, repsectively.
Then :
v_i + d_i + t_i = 14, and 3v_i + d_i + 2t_i = e_i = 20 + i.
Now, suppose that t_15 = 0. Then v_15 + d_15 = 14 and 3v_i + d_i = 35, which leads to v_i = 21/2. A contradiction.

Thus, t_i5 \geq 1, and we are done.

Pierre.


Euler
superior algebra
superior algebra unsolved

Let  $ p$ be an odd prime. Show that if $ [a]_p$ generates $ (\mathbb Z/p)^*$ then $ \left(\frac{a}{p}\right) \equal{}\minus{}1.$ I am thinking to use Eulers Criterium,

You're exactly right.  It's easier to prove the contrapositive:  if $ \left( \frac{a}{p} \right) \equal{} 1$ then $ a^{ \frac{p\minus{}1}{2} } \equal{} 1$ and $ a$ can't generate the multiplicative group.

Yes.  Euler's criterion is exactly what I was using.


p is a prime number and it not equals to 3.
a,b are integer numbers that p|a^2+b^2 and p^2|a^3+b^3.
prove that p^2|a+b or p^3|a^3+b^3

Romanian Junior Balkan Team Selection Test 2008

You got these problems in Resources section, along with many others from other competitions.

Resources: http://www.mathlinks.ro/resources.php?c=142&cid=78&year=2008

Topic about that problem: http://www.mathlinks.ro/viewtopic.php?p=1116458#p1116458

Isn't the condition: $ p\mid a\plus{}b$?

If it's the condition I wrote, that's the above problem.


superior algebra
superior algebra solved

Why in a ring with char 0, we have $k=1+1+1..+1$ is invertible, for any $k>0$?? I think it is not true, but look at the post http://www.mathlinks.ro/Forum/viewtopic.php?p=357641# . Here, they have a ring with char 0, and something like $3x=0$, and Valentins says: "Since R has char. 0, element 3 is invertible", so $x=0$. Why??

You are right. But can you please explain why in that problem he says that??

it doesn't have to be invertible to simplify with it.. just to see that it is not a zero divisor...

But why isn't $k$ o zero divisor in a ring with char 0?  :?

Good question. For Z, it is true. Is it true in general?

Does the ring in the IMC problem have $1$?

[quote="DiaDi"]$kl=1+1=1...+1$ (kl times)[/quote]
I don't understand. We have $kl=l+l+...+l$ (k times). And that is NOT necesary a sum of the form $1+1+1+..$

:oops: I made a mistake. I've also considered $l=1+1+...+1$. sorry about that.... :blush:

[quote="xirti"]Since it has $0$ char... Char is not the order of $1$? And however.. If it has.. How do you prove that if a ring which has 1 and char 0, then $k=1+1+..+1$ is not zero divisor.[/quote]

As I know it, the definition of the characteristic of a ring $A$, it is the minimal non-zero positive integer $n$ such that $na=0$ for all $a\in A$.

[quote="DiaDi"]when a ring has $char=0$ then it means $ord(1)=\infty$. hence $k=1+1+1...+1$ is not a divisor of zero beacuse $\forall l$ in that ring you will have $kl=1+1=1...+1$ (kl times). since $ord(1)=\infty$ $kl$ connot be zero, hence k isn't a zero divisor.[/quote]

I know a mistake has been found in the proof for, but to convince that the result is false, consider $\mathbb Z\times (\frac{\mathbb Z}{2\mathbb Z})$ with the canonical ring structure. $2$ is a zero divisor.

The characteristic of a ring $A$ is the minimal non-zero positive integer $n$ so that $n1=0$.  This is the definition I have found in various sources. So it refers strictly to $ord(1)$ .

The $ord(1)$ thing is just a special case, the other definition just generalizes to any ring, and they are identical on unitary rings (most sources don't consider non-unitary rings).

So, as a conclusion: In a ring with characteristic 0, we cannot say that $nx=0\Rightarrow x=0$. So, the solution in the link is wrong.