Let $f$ and $g$ be defined on the integers with values in the same set such that for all $n$ we have $|f(n+2)-f(n)|\leq 2$ and $f(x)+f(y)=g(x^2+y^2)$ for all $x$ and $y$. Then the set $M=\{f(n)\mid n=1,2,3,... \}$ has at most 6 elements.

Open question: what is the maximum number of elements of $M$? I would also be interested in your opinion about the problem: trivial,etc...

[i]Edited by Myth[/i]

Solution by my great supervisor :)
We see that $f(2m)+f(k)=f(2n)+f(k+2)$ for $k=m^2-n^2-1$. It follows that $|f(2m)-f(2n)|\leq 2$, so $|\{f(2\mathbb Z)\}|\leq 3$. Analogously for $\{f(2\mathbb Z+1)\}$. :)

You, that's it. I like it, it has such a short and nice solution...


Can anyone give a thorough explanation as to how to do it?  :)

I can't, but [url=http://www.cubefreak.net/F2L.html]this[/url] can.


function
inequalities
combinatorics unsolved
combinatorics

let $ A$ be a finite subset of $ \mathbb{Z}$ prove $ A$ has a subset $ B$ that $ |B| > \frac 54\sqrt [3]{|A|} \plus{} 1$ and if for $ x,y,z,w\in B$ we have $ x \plus{} y \equal{} z \plus{} w$ then $ \{x,y\} \equal{} \{z,w\}$ .

An almost full solution:
I think $ x \equal{} y$ or $ z \equal{} w$ are not allowed, otherwise for $ A \equal{} \{1,2,3,4,5\}$, we would need $ B$ to have at least $ 3.1$ elements, that is at least $ 4$ elements which is impossible. So assume that the imposed condition is: if $ x,y,z,w\in B$ and $ x\neq y$, $ z\neq w$ and $ x \plus{} y \equal{} z \plus{} w$ then $ \{x,y\} \equal{} \{z,w\}$.

Attempt to build $ B$ so that no $ x,y,z,w$ with $ \{x,y\}\neq\{z,w\}$ and $ x \plus{} y \equal{} z \plus{} w$ exist. Assume we cannot extend $ B$, that is for every $ a\in A$, there are different $ x,y,z\in B$ so that $ a \equal{} x \plus{} y \minus{} z$.  Let $ |A| \equal{} a$. For each $ t\in A\setminus B$ there is at least one triple $ (x,y,z)\in B^3$ so that $ t \equal{} x \plus{} y \minus{} z$. Call a triple $ (x,y,z)$ good if $ x < y$ ($ x,y$ are interchangeable) and if $ z\notin\{x,y\}$. Let $ M\subseteq B^3$ be the set of all good triples.
Now build a function $ f: (A\setminus B)\rightarrow M$ which assigns each $ t\in A\setminus B$ the triple $ (x,y,z)$ (or one of them, if several) so that $ t \equal{} x \plus{} y \minus{} z$ and $ x\neq y$. This triple must exist, since $ B$ cannot be expanded. Since $ x,y$ are interchangeable, we may assume $ x < y$. Since $ t\notin B$, $ z\notin\{x,y\}$. So we ensure that $ f(t)\in M$. Clearly $ f$ is injective, implying $ |M|\ge|A\setminus B|$, or $ |M|\ge a \minus{} b$.
It is easy to see that that $ |M| \equal{} \frac12b(b \minus{} 1)(b \minus{} 2)$.
So, $ |M| \equal{} \frac12\left[b(b \minus{} 1)(b \minus{} 2)\right]$. Denote $ k \equal{} b \minus{} 1$. Then $ |M| \equal{} \frac12\left[k^3 \minus{} k\right]$.
Hence $ k^3 \minus{} k\ge2(a \minus{} b) \equal{} 2a \minus{} 2k \minus{} 2$, or $ k^3 \plus{} k \plus{} 2\ge2a$.

If we had $ k^3\ge2a$, then $ k\ge\sqrt [3]{2a} > \frac54\sqrt [3]a$ and we would be done. Since we have $ k^3 \plus{} k \plus{} 2\ge2a$, we obtain the inequality $ k\ge\sqrt [3]{2a}$ only for sufficiently large $ a$, specifically for, $ \sqrt [3]a\ge t$, where $ t$ is root of $ 80t \plus{} 128 \minus{} 3t^3 \equal{} 0$. That's not a huge value, I guess somewhat about $ 216$. Still the cases $ a<\approx 216$ are left.
What am I missing?

I should comment that this topic and also this [url=http://www.mathlinks.ro/viewtopic.php?p=1042266#1042266]link[/url] are not problem of Iranian National Olympiads. They are problems of Sharif University contests which was held yesterday and today. I will post the other problems. Can the moderators change "Iran 2008" to "Sharif University contest 2008"?


For all nonnegative real numbers $ a,b$ and $ c$, no two of which are zero,
\[ \frac{1}{(a\plus{}b)^2}\plus{}\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}
\ge \frac{3\sqrt{3abc(a\plus{}b\plus{}c)}(a\plus{}b\plus{}c)^2}{4(ab\plus{}bc\plus{}ca)^3}\]

It's obviously trues because of Am-Gm, we have:
$ (\sum\ a^2 \plus{} \sum\ ab)^2. \sum\ ab \leq\ 108$ with $ a \plus{} b \plus{} c \equal{} 3$
Your ineq is nice and good result, ductrung  :)

[quote="ductrung"]For all nonnegative real numbers $ a,b$ and $ c$, no two of which are zero,
\[ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \ge \frac {3\sqrt {3abc(a \plus{} b \plus{} c)}(a \plus{} b \plus{} c)^2}{4(ab \plus{} bc \plus{} ca)^3}
\]
[/quote]
Replacing $ a,b,c$ by $ \frac{1}{a},\frac{1}{b},\frac{1}{c}$ respectively, we have to prove that
$ \sum\frac{a^2b^2}{(a\plus{}b)^2} \ge \frac{3\sqrt{3(ab\plus{}bc\plus{}ca)}(ab\plus{}bc\plus{}ca)^2}{4(a\plus{}b\plus{}c)^3}.$
Now, using Cauchy Schwarz Inequality, we have
$ \sum \frac{a^2b^2}{(a\plus{}b)^2} \ge \frac{(ab\plus{}bc\plus{}ca)^2}{(a\plus{}b)^2\plus{}(b\plus{}c)^2\plus{}(c\plus{}a)^2} \equal{}\frac{(ab\plus{}bc\plus{}ca)^2}{2(a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)}.$
It suffices to prove that
$ \frac{(ab\plus{}bc\plus{}ca)^2}{2(a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)} \ge \frac{3\sqrt{3(ab\plus{}bc\plus{}ca)}(ab\plus{}bc\plus{}ca)^2}{4(a\plus{}b\plus{}c)^3}$
or equivalently,
$ 2(a\plus{}b\plus{}c)^3 \ge 3\sqrt{3(ab\plus{}bc\plus{}ca)}(a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca),$
that is
$ 4(a\plus{}b\plus{}c)^6 \ge 27(ab\plus{}bc\plus{}ca)(a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)^2$
By AM-GM, we see that
$ 27(ab\plus{}bc\plus{}ca)(a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)^2 \le \frac{1}{2} \left( 2(ab\plus{}bc\plus{}ca)\plus{}(a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)\plus{}(a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)\right)^3 \equal{}4(a\plus{}b\plus{}c)^6.$
Therefore, our proof is completed. :)


Find all polynomials $P(x)$ that have only real roots, such that \[ P(x^2-1)=P(x)P(-x). \]

$Tipe$

I am gonna give my solution in brief- just in general lines because I am too lazy to write in details. I think that what I give is enough to understand the problem.

The hook is to see that only few numbers can be roots of P.
Actually, if $\phi_1$ and $\phi_2$ are the roots of $x^2-x-1=0$ then only the numbers:
0,-1, $\phi_1$ ,$\phi_2$ ,$-\phi_1$ ,$-\phi_2$ can be roots of P. Otherwise will get an infinite sequence of roots of P.

Also if $P=x^a(x+1)^bQ$ we can prove that a=b and what is left is to show that
if $P=x^a(x+1)^a(x-\phi_1)^k(x-\phi_2)^l(x+\phi_1)^m(x+\phi_2)^n$ then P satisfies the condition, which is TRUE!

I think that this is enough for a sketch of the proof:D

i'm not agree with u.the P solutions are:
P(x)=(x-µ_1)^m.(x-µ_2)^n        µ_1,µ_2 are root of x^2-x-1=0 and m+n is an even natural number

But $P(x)=x(x+1)$ is a solution.

$Tipe$

ooopss :blush:
actually $-\phi_1$ and $-\phi_2$ cannot be roots :)
sorry for the mistake i overlooked it.
so finally the answer should be
$P=(x^2 +x)^a(x-\phi_1)^b(x-\phi_2)^c$ and $b+c$ is even ;)

thanks marocleverness for the correction :)


Find all $ x,y\in N$ such that
$ \frac {x^3 \plus{} y^3 \minus{} x^2y^2}{(x \plus{} y)^2}$
is positive integer
thx for your nice solution  :)

[hide]If $ \gcd(x,y) \equal{} d$ then $ x \equal{} da,y \equal{} db$ and $ \gcd(a,b) \equal{} 1$ . Now try to prove that $ (a \plus{} b)|d$ [/hide]


How does someone become a moderator on AoPS?

Active participation on the boards for several months will usually give you the right to request a moderator spot. Sometimes there are none available and often, other requirements apply.

The best scenario would be to wait until moderators are needed (I don't believe they currently are). Admin will usually contact individuals they believe would do a good job.


Basically show:
$\frac na-\frac{n-1}{a+1}$ cancels to something nice.
$=\frac{n(a-1)-(n-1)a}{a(a+1)}=\frac{a-n}{a(a+1)}$
I'm not sure, I'll post soon with the answer.


rotation
geometry
circumcircle
trigonometry
geometry proposed

Given equilateral triangle $ABC,\gamma_{m},\gamma_{n}$ are two circles with the same center $O$, a ray through $O$ cuts $\gamma_{m},\gamma_{n}$ at $M,N$ prove that sum of square length of projection of segment $MN$ to the lines $BC,CA,AB$ is const when the ray rotate around $O$
 :D

What is $O$? Is circumcenter of triangle $ABC$.?

$O$ is an arbitrary point my good friend :D  :D

Denote $MN=x$, then $x$ is constant (This is nice space in your problem :P ) . Let $(MN,BC)=\alpha$, we need prove sum $x^{2}(\cos^{2}{\alpha}+\cos^{2}{(\alpha-\frac{\pi}{3})}+\cos^{2}{(\alpha+\frac{\pi}{3})})$ is constant. But this is easy problem :D .

[quote="N.T.TUAN"]we need prove sum $x^{2}(\cos^{2}{\alpha}+\cos^{2}{(\alpha-\frac{\pi}{3})}+\cos^{2}{(\alpha+\frac{\pi}{3})})$ is constant. But this is easy problem :D .[/quote]
Actually it isn't easy ,but you were right ,good solution :D


What is the positive difference between 5% of 600 and 4% of 500?

[hide]$ \text{5}\text{ percent of 600 is: 30} $

$ \text{4}\text{ percent of 500 is: 20} $

$ 30 - 20 = 10 $

$ \text{The positive difference is 10.} $[/hide]

[hide]
.05*600=30 and .04*500=20
So the answer is $\framebox{10}$[/hide]


vector
parameterization
geometry
3D geometry
absolute value

I have a dataset which I have generated using two different methods.  What's more important, though, is that I have two sets of observations (1volume and 2volume - each of these is essentially a 357-element vector).  I have another measurement (heritability - another 357-element vector) that I can use to test whether or not I have a good set of volumes, by simply finding the pearson correlation coefficient between my volume dataset and heritability dataset.

I used the following method to try to maximize the correlation between a mixed observation vector and the heritability vector:

My goal is to find the biggest correlation coefficient (either positive or negative) between a new volume vector and the heritability vector.  The new volume vector (nvolume) will be determined by using a combination of elements from vectors 1volume and 2volume (i.e. the first element, nvolume(1) could either be the first element from 1vector or the first element from 2volume, nvolume(2) could be either 1vector(2) or 2vector(2), etc.)

I started my iterative calculation by picking the 1volume or 2volume vector with the highest absolute correlation with the heritability vector, call this the winning vector and the other vector the losing vector.  Next in my script, I replace each element of the winning vector with the the corresponding element from the losing vector and find the correlation between all 357 new vectors and the heritability vector.  I then choose the vector with the best correlation, make this the new 'winning vector', and continue this process until the highest correlation is reached.

whew.

(1) I don't know if I'm getting the best combination of elements to maximize this correlation between the volume and heritability data. (2) I can't find another statistical method to solve this problem. (3) The option of using all possible combinations of elements to find the highest correlation coefficient is not viable, because this would require 2^357 calculations taking 2-3 seconds each.  Even on my best resource (100 processor supercomputer), this would take about 10^98 years to do...

Any help here or pushes in the right direction would be appreciated.

Thank you

I cannot tell how well your approach works but here is an alternative scheme that is guaranteed to give you almost the best answer in a reasonable time. Instead of asking what the largest coefficient is, we'll try to see if a given coefficient $ \lambda$ can be achieved (if I understand you right, you do not care about its sign, just want the largest absolute value, which seems to be rather crucial for the method below). Then you can do the usual bisection to figure out what the best $ \lambda$ is with any precision. Let your observations be $ a_j\pm b_j$ and let $ c_j$ be the heritability vector. I'll assume that $ c_j$ is already normalized so that $ \sum_j c_j\equal{}0$. The problem then reduces to the following:

Does there exist a choice of signs $ \delta_j\equal{}\pm 1$ such that
\[ \lambda^{\minus{}2}\left(\sum_j (a_j\plus{}\delta_j b_j)c_j \right)^2\plus{}\left[\frac 1n\left(\sum_j(a_j\plus{}\delta_jb_j)\right)^2\minus{}\sum_j(a_j\plus{}\delta_jb_j)^2\right]\left(\sum_j c_j^2\right)\]
is non-negative?

After opening some parentheses and regrouping the terms, we see that we really need to maximize the expression
\[ \lambda^{\minus{}2}\left(\sum_j \delta_j b_jc_j \right)^2\plus{}C^2\left(\sum_j \delta_j b_j\right)^2\plus{}\sum \delta_j R_j\]
with $ C^2\equal{}n^{\minus{}1}\sum_j c_j^2$ and some easy to compute $ R_j$ and to compare this maximum to some easy to compute constant.

The first observation is that this expression is convex in each $ \delta_j$, so it doesn't hurt to replace the discrete parameters $ \delta_j$ by the continuous parameters $ t_j\in[\minus{}1,1]$. Now the problem reduces to the following: given 3 linear forms
$ A(t)\equal{}\sum A_j t_j$, $ B(t)\equal{}\sum B_jt_j$ and $ R(t)\equal{}\sum_j R_jt_j$, find the maximum of $ A(t)^2\plus{}B(t)^2\plus{}R(t)$ over the unit cube.

Thus, your task reduces to finding the image of the unit cube $ [\minus{}1,1]^n$ under the mapping $ t\mapsto(A(t),B(t),R(t))$. The resulting body is a zonotope (Minkowski sum of $ n$ intervals) and a nice thing about it is that it may have only about $ n^2$ vertices.

I'm pretty sure that there are a lot of sources on the web about how to compute vertices of zonotopes fast.  You can look at some references [url=http://www.eurecom.fr/util/publidownload.en.htm?id=1238]here[/url], for instance. If you still have trouble with something, feel free to ask.

Hope this helps a bit :).


1) 3 butene-1-ol + hcl-> ?
2) 4 penten-2-ol + hcl->?

with reasons

[hide="Answer"]The answer depends on the exact reaction conditions. If it is aqueous HCl and some heat, dehydration is very favourable since it yields a conjugated unsaturated system in each case. If it is concentrated HCl, then addition to the double bond is the probable path.[/hide]

no conditions were mentioned. also the products are weird. for (a) it is a mixture of 1 chloro 3 butene and cyclopropylmethyl chloride. for (b) it is only 2 chloro 4 butene.


on this that was posted elsewhere too  :D
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=193741

I was looking for a different sonotion Using inclusion-conclusion

I think you mean inclusion-exclusion.  If you know what PIE (not that pie) is and how it works, then you should be able to do it.


The water in Ron's basement is 7 feet deep. The area of his basement floor is 1250 square feet. He has a pump that will remove 95 gallons of water per minute. How many hours will it take to pump all the water out of his basement? (Note: 7.5 gallons of water is equal to 1 cubic foot.) Express your answer as a decimal to the nearest tenth.

The volume of water is 1250 x 7 = 8750. To find the number of gallons we multiply by 7.5. This numbers 65625. We then divide by 95 and then by 60 because the question said hours and this was given in minutes. The answer is 11.5.


function
calculus
derivative
topology
algebra
polynomial
Functional Analysis

Hi every one!
I've read Baire theorem in a book and i know it's a very important theorem,specially in function analysis.
Would some one mind showing me some applications of this theorem,or some books about it?
Thanks very much.

Actually I have seen two different versions of Baire's theorem. I don't know which one have you learnt. Anyway, here are some problems:

Using the Baire's theorem, one can prove
1. the open mapping theorem.
2. the closed graph theorem.
3. the uniform boundedness principle.
4. that every complete metric space with no isolated points is uncountable.

5. Let $f: \mathbb{R}\to\mathbb{R}$ be a function having the derivatives of every order. For every $x\in \mathbb{R}$ there exists $n(x)\in\mathbb{N}$ such that $f^{(n(x))}(x)=0$. Show, by using Baire's theorem, that $f$ is a polynomial.

One that appeared here recently: an infinite-dimensional Banach space cannot have a countable spanning set.

One more recent application: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=118138]there is a null set[/url] $E\subset \mathbb R$ such that $E\cap h(E)\ne\varnothing$ for any homeomorphism $h\colon\mathbb R\to\mathbb R$

Some other nice facts:

  1. There exists no function $f$ from $R$ to $R$ such that $|f(x)-f(y)|\geq |x-y|^{a}$ if $0<a<1$.

  2. There exists a dense set $A$ of the set of continuous functions of period $2\pi$ such that for all $f\in A$, the Fourier series of $f$ diverges on a dense set.

  3. The continuous nowhere differentiable functions form a dense subset of the set of continuous real functions defined on $[0,1]$.

  4. (this is a really nice one) Let $E$ be a complex Banach space and let $T$ be a continuous endomorphism.  If for all $x\in E$ there exists a polynomial $f$ such that $f(T)(x)=0$, then there exists a polynomial $g$ such that $g(T)=0$.


When composing a message, sometimes I want to copy small portions of text from several previous posts and everything works fine but the dollar signs around $ \text{\LaTeX}$ codes disappear. So, I have to either type them back myself, or to open a second browser window, click "quote" there, and transfer the text I need. Both options are somewhat irritating. Is it possible to make the dollar signs to be copied as well?

The same thing happens with BBCode, like hide tags. I don't think it is possible though.


probability symmetry algebra function domain percent

I have a coin. I flip the coin. The coin come up heads. Thus I conclude the coin has 100% chance of coming up heads.

What is wrong with this logic?

uh...it's wrong.

cheers from audience: *it don't get better than that!*

retorting protests from others in the audience: *you fools we don't exist so shut up already!*

(sarcastic tone) *Wow, what a great solution you got there* *best explanation an ape could give* *woot* </jk>

an ape could probably come up with a slightly more intelligent solution. apes are quite smart really, no? like for example, it could eat some food. that would be smarter. therefore, i disagree.

Well, if you want to know that that particular coin NOW has chance being heads, it is 100%.  However, you tried to generalize that one coin to be every coin.  That cannot be done.

Interesting...can you go into details about how the probability can change over time? Do anyone want to graph this? (On the y-axis plot the range of probability the coin could have at any given point and on the x-axis have time: the number of time you flip the coin) However this will not be such an interesting graph...try to use a software/graphing calculator so that probability most likely to be observed be in dark colors and the probability least likely to happen in light colors. If you have time assign a color to each of the probability (For example if there is 10 ways to observe a probability then color it in a greenish tone) and try to make a colorful design. Have fun!

As for the problem can you prove that the probability of this particular coin coming up heads isn't 100%? In other words how can you tell this coin will ever come up tails? Can you figure out what the probability is if you tried this experiment infinite times solely from the set of givens?

well, it would be logical if the coin was a trick coin and had 2 heads.

All the coin toss tells you is that the probability of a head is somewhere in the range (0,1]. No matter how many times you toss it, it could still be anywhere in this range..

You may like to draw a graph as Tare said though.. there is a very nice result about the number of times during the tosses it has switched from "heads leading" to "tails leading".. and stuff like that. I think its more likely that the lead (more heads than tails or vice versa) [b]never[/b] changes rather than any other option. But thats just from memory..

Also after you graph it (and try to tweak it to make it pretty) answer the following questions:
1.) How many line of symmetry are there?
2.) What is the range and domain of the graph?
3.) As x increases what happens?

Well, in Las Vegas they use mercury to weight the coins so that, for example, there is a 100 percent chance to have it, say, heads.  But... I digress, yes, Dark_Shadow is correct, in my narrow way of thinking.  The dude in the example is thinking that 1=infinity.  Because he is assuming that one coin example will cover all examples.


induction
combinatorics proposed
combinatorics

Prove that a triangle-free planar graph can be colored with at most 4 colours.

(Note: It actually turns out you can do it with 3, but that's significantly harder to prove :))

let $ f_i$ denote the number of faces bounded by exactly $ i$ edges, and let $ V$, $ E$, $ F$ denote the usual. clearly, $ f_3\equal{}0$, so $ 4f_4\plus{}5f_5\plus{}\cdots \equal{} 2E$, since each edge is shared by exactly 2 faces. this implies that $ 4F\leq 2E$, that is, $ F\leq \frac{E}{2}$. then from Euler's relation we have

\[ E\plus{}2 \equal{} F\plus{}V \leq \frac{E}{2}\plus{}V \Rightarrow \frac{E}{2}\leq V\minus{}2 \Rightarrow E\leq 2V\minus{}4
\]

now we show that some vertex has degree 3 or less. indeed, if all the vertices had degree 4 or more then the number of edges would be at least $ \frac{4V}{2} \equal{} 2V$, contradicting the above inequality. now, to color the graph, pick a vertex $ Z$ of degree 3 or less and remove it from the graph. by induction, we may color the resulting graph with 4 colors, and then pick a color different from the (at most 3) colors of $ Z$'s neighbors to color $ Z$ with.


What is the sum of the reciprocals of the positive integer factors of 28?

The sum of the reciprocals of the factors of a perfect number is always $ \boxed{2}$.

if we didn't know $ 28$ was a perfect number, we could just add the fractions up.
$ 1 \plus{} \frac{1}{2} \plus{} \frac{1}{4} \plus{} \frac{1}{7} \plus{} \frac{1}{14} \plus{} \frac{1}{28}$. putting everything over a common denominator gives
$ \frac{28\plus{}14\plus{}7\plus{}4\plus{}2\plus{}1}{28} \equal{} \frac{56}{28} \equal{} 2$

[quote="davidkim2106"]What is a perfect number?[/quote]

It's a number that is equal to the sum of its proper divisors, i. e. all the positive divisors except itself.


logarithms
calculus
integration
calculus computations

Series A diverges by the integral test (or thinking of it as a divergent Bertrand series). For series B, n ln(ln n) < n ln(n), and so it also diverges. For C a similar argument (comparision with A) shows that it also diverges.
Series D diverges by limit comparision with the harmonic series.
Series E is a convergent Bertrand series.
For series F, $\ln n \cdot \ln(\ln n) < \ln n \cdot \ln n = \ln^{2}n \Rightarrow  \frac{1}{\ln n \cdot \ln(\ln n)}> \frac{1}{\ln^{2}n}$, and so the series diverges.
Series G diverges by using the Cauchy Condensation Test and then by limit comparion with series D.
Series H diverges by limit comparision with the divergent series A.
For series I, $\ln n \cdot \ln^{2}(\ln n) < \ln^{3}n$, and so the series diverges because $\frac{1}{\ln^{3}n}$ is also divergent.
Series J diverges by the Cauchy Condensation Test.
For series K, $\ln n \cdot \ln(\ln(\ln n)) < \ln^{2}n$ and so it also diverges.
Series L diverges by comparision with series K.
Series M diverges by comparision with $\frac{1}{\ln^{3}n}$.
Series N diverges by the Cauchy Condensation Test and then comparision with $\frac{1}{\ln^{2}n}$.
Series O diverges by the Cauchy Condensation Test and then comparision with series J.
For series P we use the Cauchy Condensation Test and then we have
$\ln^{2}(\ln(n \ln 2)) < \ln^{2}n$ and so the series also diverges.

So only series E seems to be convergent.

:what?: Was that a multiple-choice question? Must have been a really wide bubble form. I don't remember such tests from my days at WUSTL: this is either a recent innovation or a version of calculus that I was not involved in.


geometry
linear algebra
matrix
area of a triangle
calculus
calculus computations

Give a necessary and sufficient condition for the oriented area of a triangle $ (0,0), (x_1,y_1), (x_2,y_2)$ to be positive.

So $ y \minus{} y_1 \equal{} \frac{y_1}{x_1}x$. Then what would I do?

Also if I want to find the value of the 2-form $ dy \ dz \plus{} dz \ dx \plus{} dx \ dy$ on the oriented triangle with vertices $ (1,1,2), \ (3,5,\minus{}1), \ (4,2,1)$ would I just project this triangle on the respective planes, find those areas, and then add them up?

The condition is quite simple: $ \det\begin{pmatrix}x_1&y_1\\x_2&y_2\end{pmatrix}>0.$

The oriented area is half of that determinant.

So to evaluate the actual area, its just half the determinant? You dont have to project it on the planes? Why does the determinant have to be greater than 0?


linear algebra
matrix
group theory
abstract algebra
inequalities
modular arithmetic
superior algebra

Let $ a,b,c$ be elements of a group such that $ a^2 \equal{} b^3 \equal{} c^5 \equal{} (abc)^{ \minus{} 1}$. Prove that $ c^{610} \equal{} 1$.

Ok, this one is pretty tough.

First $ z: \equal{}a^2\equal{}b^3\equal{}c^5$ commutes with $ a,b,c$, and therefore lies in the centre of $ G: \equal{}\langle a,b,c: a^2\equal{}b^3\equal{}c^5\equal{}(abc)^{\minus{}1}\rangle$.
We need, that $ G/\langle z\rangle \equal{}\langle a,b,c: a^2\equal{}b^3\equal{}c^5\equal{}(abc)^{\minus{}1}\equal{}1\rangle\cong\langle a,b: a^2\equal{}b^3\equal{}(ab)^5\equal{}1\rangle$ has order $ \le 60$.

I have seen several technical proofs for this fact, which show $ H: \equal{}\langle a,b: a^2\equal{}b^3\equal{}(ab)^5\equal{}1\rangle\cong A_5$,
but my favorite and the most straight forward uses the [url=http://en.wikipedia.org/wiki/Coset_enumeration]coset enumeration[/url].
This gives only $ 12$ possible cosets of $ \langle ab\rangle$ in $ H$, namely
$ \langle ab\rangle ,\langle ab\rangle a,\langle ab\rangle b,\langle ab\rangle ba,\langle ab\rangle bab,\langle ab\rangle baba,\langle ab\rangle babab,\langle ab\rangle bababa\equal{}\langle ab\rangle bab^2,$
$ \langle ab\rangle babab^2,\langle ab\rangle babab^2a,\langle ab\rangle babab^2ab,\langle ab\rangle babab^2ab^2$.
So $ |\langle ab\rangle|\le 5$ and $ |H: \langle ab\rangle|\le 12$ gives $ |H|\le 60$.

But now $ G/\langle z^2\rangle \equal{}\langle a,b,c: a^2\equal{}b^3\equal{}c^5\equal{}(abc)^{\minus{}1}, a^4\equal{}1\rangle$ has order $ \le 120$.
On the other hand using the universal property we find a homo $ G/\langle z^2\rangle\to SL_2(5)$,
which sends $ a\mapsto A: \equal{}\begin{pmatrix}&1\\\minus{}1&\end{pmatrix}, b\mapsto B: \equal{}\begin{pmatrix}&\minus{}1\\1&1\end{pmatrix},c\mapsto C: \equal{}\begin{pmatrix}\minus{}1&1\\&\minus{}1\end{pmatrix}$.
But $ SL_2(5)\ge\langle A,B,C\rangle$ has order divisible by $ 4\cdot 3\cdot 5\equal{}60$ and therefore index $ \le 2$ in $ SL_2(5)$.
This gives $ \langle A,B,C\rangle\unlhd SL_2(5)$, and in fact $ \langle A,B,C\rangle\equal{} SL_2(5)$, as $ SL_2(5)$ is [url=http://en.wikipedia.org/wiki/Perfect_group]perfect[/url].

So $ SL_2(5)$ is epimorphic image of $ G/\langle z^2\rangle$ of order $ 120$,
and $ |G/\langle z^2\rangle|\le 120$ gives $ G/\langle z^2\rangle\cong SL_2(5)$ perfect.
Then looking at the second [url=http://en.wikipedia.org/wiki/Commutator_subgroup]commutator subgroup[/url] we get $ G/\langle z^2\rangle\equal{}(G/\langle z^2\rangle)''\equal{}G''\langle z^2\rangle/\langle z^2\rangle$
and therefore $ G\equal{}G''\langle z^2\rangle$ and $ G/G''\cong \langle z^2\rangle/G''\cap\langle z^2\rangle$ abelian.
This gives $ G'\equal{}G''$ perfect and $ G\equal{}G'\langle z^2\rangle$.
Also $ Z: \equal{}G'\cap \langle z^2\rangle$ is central in $ G'$ with $ G'/Z\cong G/\langle z^2\rangle\cong SL_2(5)$.
So $ G'$ is a [url=http://en.wikipedia.org/wiki/Group_extension]perfect central extension[/url] of $ SL_2(5)$.

But $ SL_2(5)$ has all Sylow subgroups cyclic or [url=http://en.wikipedia.org/wiki/Quaternion_group]quaternion[/url],
so $ SL_2(5)$ must have [url=http://en.wikipedia.org/wiki/Schur_multiplier]Schur multiplier[/url] $ 1$ and no nontrivial perfect central extension.

This gives $ Z\equal{}G'\cap \langle z^2\rangle\equal{}1$ and $ G'\cong SL_2(5)$ and $ G\equal{}G'\times \langle z^2\rangle$.
Then $ a\equal{}\tilde{a}z^{2n},b\equal{}\tilde{b}z^{2m},c\equal{}\tilde{c}z^{2k}$ for some $ \tilde{a},\tilde{b},\tilde{c}\in G'$ and $ n,m,k\in\mathbb{Z}$.
Then the relation $ z\equal{}a^2\equal{}b^3\equal{}c^5\equal{}(abc)^{\minus{}1}$ gives
$ z^2\equal{}\tilde{a}^4z^{8n}\equal{}\tilde{b}^6z^{12m}\equal{}\tilde{c}^{10}z^{20k}\equal{}(\tilde{a}\tilde{b}\tilde{c})^{\minus{}2}z^{\minus{}4n\minus{}4m\minus{}4k}$.
But then the unique factorisation in $ G\equal{}G'\times \langle z^2\rangle$ gives
$ 1\equal{}\tilde{a}^4\equal{}\tilde{b}^6\equal{}\tilde{c}^{10}\equal{}(\tilde{a}\tilde{b}\tilde{c})^{\minus{}2}$ and $ z^2\equal{}z^{8n}\equal{}z^{12m}\equal{}z^{20k}\equal{}z^{\minus{}4n\minus{}4m\minus{}4k}$.
Then $ 1\equal{}(z^2)^{10k\minus{}1}$ gives $ \langle z^2\rangle$ of finite odd order $ N$ coprime to $ k$.

This gives $ 1\equiv 4n\equiv 6m\equiv 10k\equiv \minus{}2n\minus{}2m\minus{}2k\pmod{N}$
and $ 2n\equiv 3m\equiv 5k\equiv \minus{}n\minus{}m\minus{}k\pmod{N}$.
But then finally
$ 0\equiv 6(5k\plus{}n\plus{}m\plus{}k)\equal{}36k\plus{}6n\plus{}6m\equiv 36k\plus{}15k\plus{}10k\equal{}61k\pmod{N}$
gives $ N|61$ and $ c^{610}\equal{}(z^2)^{61}\equal{}1$.

In fact it's now easy to show $ G\cong SL_2(5)\times (\mathbb{Z}/61\mathbb{Z})$ of order $ 7320$ with isomorphism
$ a\mapsto \begin{pmatrix}&1\\\minus{}1&\end{pmatrix}\times (\minus{}15\plus{}61\mathbb{Z}), b\mapsto \begin{pmatrix}&\minus{}1\\1&1\end{pmatrix}\times (\minus{}10\plus{}61\mathbb{Z}),$ $ c\mapsto \begin{pmatrix}\minus{}1&1\\&\minus{}1\end{pmatrix}\times (\minus{}6\plus{}61\mathbb{Z})$.

amazing. but is there a direct calculation?

Of course, using coset enumeration on the trivial subgroup of $ G$.
$ 7320$ elements isn't that much. :lol:

Why can't we just multiply everything by $abc$ which gives $a^{3}bc=1$? Taking this to the $60$ power and expressing everything in terms of $c$ gives $c^{610}=1$.

If you did what I think you did, that is, $(a^3bc)^{60} = a^{180}b^{60}c^{60}$, the major flaw with doing that is that you don't know whether or not the group is actually abelian or not. However, it could work if you can show that $a^3, b, c$ lie in the center of $G$.


trigonometry
integration
calculus
calculus computations

Find

$\int^{\sqrt{p^{2}+q^{2}}}_{0}{\frac{p\cos{x}-q\sin{x}}{\cos{\left(x+\tan^{-1}{\frac{q}{p}}\right)}}\,dx}$

where $p,q\in\Re$ and $p,q\not=0$

[hide="HINT"]
Write $p\cos{x}-q\sin{x}$ in the form $r\cos{(x+\alpha)}$ where $r,\alpha\in\Re$[/hide]

It looks to me as the integral might be undefined when $cos(x+\tan^{-1}(p/q))=0$,  that is for $x=\pi/2-\tan^{-1}(p/q) < \sqrt{p^{2}+q^{2}}$. This is the case when $p=1, q=2$..

Otherwise the integral is $p^{2}+q^{2}$

Yes, as with most cases the integral is undefined when the integrand is also undefined.  However, excluding this, your answer is incorrect  :o


Diophantine equation
number theory unsolved
number theory

Prove that : the equation :
 $ x^3 + y^3 + z^3 + t^3 =1999 $
  has unbounded integer roots satisfying : $ /x/ , /y/ , /z/ , /t/ \geq 2^{2004} $

look for solutions of the form:

$(10-b)^3+(10+b)^3+(-1/2-d/2)^3+(-1/2+d/2)^3=1999$

wich is equivalent to:

$d^2-80b^2=1$

and it is a pell equation, prove $d$ is always odd, and that you can get bigger enough $b$ and $d$


inequalities
logarithms
calculus
calculus computations

Find for which $a \in R^{+}$
$\log_{a}x<x$
for all real numbers $x>0$.

Consider $f(x)=x-\log_{a}x$. We need values of a for which
f(x)>0 for all x>0.

If a<1, for x<a, f(x)<0 Therefore no soln. for a.

If a>1,
$f(x) \longrightarrow \infty$ as $x \longrightarrow 0$ and also as $x \longrightarrow \infty$.
Therefore it can only have an absolute minimum.
Setting
$f\prime(x) =0$ ,we get $x_{1}= \frac{1}{log_{e}a}$

We need values of a for which the value of $f(x_{1})$ is positive.
Solving for this condn. we get $a>e^{1/e}$


In a circle, a chord 12 cm long is parallel to a tangent and bisects the radius drawn to the point of tangency. How long is the radius?


The sequence $ (u_n)$ is defined by $ u_1\equal{}1,u_2\equal{}3,u_n\equal{}(n\plus{}1)u_{n\minus{}1}\minus{}nu_{n\minus{}2}$. Which members of the sequence which are divisible by 11.

$ u_n \equal{} \sum_{k\equal{}1}^n k!$
by PMI

[quote="Kouichi Nakagawa"]$ u_n \equal{} \sum_{k \equal{} 1}^n k!$
by PMI[/quote]
How did you get the idea?

hello, for example is $ u_4\equal{}33$ divisible by 11.
Sonnhard.


A uniform circular cylinder of mass $ m$ and radius $ r$ is given an initial angular velocity $ \omega_{0}$ and no initial translational velocity. It is placed in contact with a plane inclined at an angle $ \alpha$ to the horizontal. If there is a coefficient of friction $ \mu$ for sliding between the cylinder and plane, find the distance the cylinder moves up before sliding stops. Also, calculate the maximum distance it travels up the plane. Assume $ \mu > \tan \alpha$.

Answer:
[hide]$ d_{1} \equal{} \frac{r^2\omega_{0}^2(\mu \cos \alpha \minus{} \sin \alpha)}{2g(3\mu\cos \alpha \minus{} \sin \alpha)^2}$
$ d_{max} \equal{} \frac{r^2\omega_{0}^2(\mu \cos \alpha \minus{} \sin \alpha)}{4g\sin \alpha(3\mu\cos \alpha \minus{} \sin \alpha)}$[/hide]


geometry
ratio
geometric sequence
real analysis
real analysis solved

A square is inscribed in a circle of radius R, a circle is inscribed in this square, then a square in this circle and so on [i]n [/i]times. Find the limit of the sum of areas of all the squares as n tends to infinity.

the side length of the first inscribed square is $\sqrt2 R$.
Then there is a geometric progression with the common ratio of $\frac{1}{\sqrt 2}$.
Hence the areas of all squares form a geometric progression with the common ratio of $\frac{1}{2}$, the first term of which is $2R^2$.
Thus the sum of the areas $=\frac{2R^2}{1-\frac{1}{2}} = 4R^2$.


geometry
3D geometry
pigeonhole principle

Prove that no matter how we number the vertices of a cube with integers from 1 to 8, there exists two opposite vertices in the cube (e.g. they are the endpoints of a large diagonal of the cube), united through a broken line formed with 3 edges of the cube, such that the sum of the 4 numbers written in the vertices of this broken lines is at least 21.

this was really stupid...i tried to do it although i don't know if it's right...because i saw the solution and it was really different:with colors ;)

Colours? The way I used is the following: We put 8 and 7 on our path whereever they are. Then we try to avoid the numbers 1 and 2. There are 3 cases.

Case 1: 8 and 7 are diagonally opposite. We can find three distinct paths from 8 to 7 which each cover 2 two different numbers, distinct from each other. Thus, by pigeonhole, one of them will not have a 1 or 2.

Case 2: 8 and 7 and diagonally opposite on the same plane. We have choose one of the other corner of that plane - we choose the bigger one. Also choose the bigger of the two numbers which are diagonally opposite 8 and 7. The sum will be at least 8+7+2+4=21.

Case 3: 8 and 7 are side by side. If we have come this far, this case should be easy - millions of ways of proving this (=


linear algebra
matrix
vector
combinatorial geometry
linear algebra unsolved

Here is a magnificient way to find the convex hull of $O(n)$:
  a) Let $A$ be a matrix such that $|||A|||_2\leq 1$. Prove that for any matrix $M$ there is a matrix $U\in O(n)$ such that $tr(UM)\geq tr(AM)$.
  b) Deduce that $A$ is in the convex hull of $O(n)$.

Could you please specify what the norm $|||\cdot|||_2$ is? :?

I think it is: $sup(\frac{|AX|}{|X|}, X \in E-{0})$ where $|X|$ is the usual euclidian norm.

I was thinking about similar stuff but in the more general setting of Hilbert spaces, and I remembered this problem.

[b](a)[/b]

$A$ can be written as $A=P\tilde U$, where $P$ is a symmetric positive semidefinite matrix of norm $\le 1$ (in fact, it satisfies $P^2=AA^t$) and $\tilde U$ is an orthogonal matrix. Then we'll have $\mbox{tr}(AM)=\mbox{tr}(P\tilde UM)$. We can assume that $P$ has diagonal form $\mbox{diag}(\lambda_i)$, with $\lambda_i\in[0,1],\ \forall i$.

If, in general, for an $n\times n$ matrix $X$ we denote its $i$'th line by $X_i$ and its $j$'th column by $X^j$ and $\langle\cdot,\cdot\rangle$ is the usual scalar product in $\mathbb R^n$, then we have $\mbox{tr}(AM)=\sum_i\lambda_i\langle\tilde U_i,M^i\rangle$. For all $i$ we can find $\varepsilon_i\in\{\pm 1\}$ s.t. $\lambda_i\langle\tilde U_i,M^i\rangle\le\langle\varepsilon_i\tilde U_i,M^i\rangle$. We change the lines $\tilde U_i$ of $\tilde U$ into $U_i=\varepsilon_i\tilde U_i$ to get the desired orthogonal matrix $U$ with lines $U_i$.

[b](b)[/b]

In the real Hilbert $\mathcal M_n(\mathbb R)$ with the scalar product $\langle X,Y\rangle=\mbox{tr}(XY^t)$, all linear functionals are of the form $X\mapsto\mbox{tr}(XM^t)$ for some fixed $M$. [b](a)[/b] says that for any linear functional $f$ on $\mathcal M_n(\mathbb R)$ there is an orthogonal matrix $U$ s.t. $f(U)\ge f(A)$. From this we should conclude that $A$ lies in the closed convex set $O(n)$. This follows from a theorem of Mazur, I believe, which states that in a normed vector space, if $x$ is a point outside a closed convex set $\mathcal C,\ x$ can be separated from $\mathcal C$ by a hyperplane, or, in other words, there is a linear functional $f$ on the respective space satisfying $f(x)>f(y),\ \forall y\in\mathcal C$.


let $a,$ $b$ and $c$ are non-negative numbers such that $2(a^{2}+b^{2}+c^{2})+3abc=9.$
Prove that $a+b+c\leq3.$


are $ a_{i},b_{i}$ the sides of a $ 2^{i}^{+}^{2}$ inscribed and circumscribed
polygon to a circle of diameter 1?

[hide="And the right inequality."]

Let $ b_{n}=\tan \xi_{n}$, and you know what to do :)[/hide]


I'm having some trouble solving;understanding this question. Help would be greatly appreciated. Thnx in advance.

Find the smallest integral value of k such that the given equation has imaginary roots.

x^2-3x+k= 0

The equation will have imaginary roots when the discriminant, $ b^2 \minus{} 4ac$, is less than 0. Thus, we want to solve $ 9 \minus{} 4 \cdot 1 \cdot k < 0$, or

$ 9 \minus{} 4k < 0 \implies 4k > 9 \implies k > \frac 94 \equal{} 2.25$

With this information, we can conclude that the smallest integral value is then $ \boxed 3$.


When will the list of courses for this year be up? Will we not get it until the camp actually starts?

Being not an alumnus, I am not sure, but from what I understand, we will get the courses for the week at the beginning of each week of camp.

[quote="Hamster1800"]As far as I know, the courses weren't released before camp last year. At the beginning of week 1, we got a detailed schedule for week 1, and an approximate week 1-4 schedule. Then we got a detailed schedule for week 2 at the beginning of week 2, etc. Week 5 is special. You'll see when you get to camp ;)[/quote]

That is correct.  You will receive the detailed schedule for week 1 and a tentative schedule for weeks 1-4 on arrivals day.  You will then receive the definite schedule for each of weeks 2 through 4 a few days before that week begins.  For week 5, we will propose many more courses than we can possibly teach, and students will vote.  You will also have an assigned academic advisor to help you choose among the long list of classes.

[quote="JSteinhardt"]I would totally be in favor of a class on Algebraic Graph Theory.[/quote]

Marisa Debowski is a graph theorist who taught classes on Introductory Graph Theory and Topological Graph Theory last year, so there is a good chance she might teach Algebraic Graph Theory this year.


Consider the pair of $(x,y,z)$ satisfying the following condition

$(A): \ x,y,z$ are positive integers such that $x^2+y^2+z^2=xyz$ and $x\leq y\leq z.$

(1) Find all the pairs $(x,y,z)$ satisfying the condition $(A)$ and $y\leq 3.$

(2) If the pair of $(a,b,c)$ satisfy the condition $(A)$, then prove that there exist $z$ such that the pair of $(b,c,z)$ satisfy the condition $(A).$

(3) Prove that there are infinitely many pairs of $(x,y,z)$ satisfying the condition $(A).$

1) One can verify without much difficulty that there are no triples $(x,y,z)$ for $y\leq 2$, and that for $y=3$, we must have $(x,y,z)=(3,3,3)$ (just check every possible value of $x$, and see that $x^2+y^2+z^2 > xyz$).

The only soln is $(3,3,3)$.

2) $(x,y,z)$ solution $\longrightarrow$ $x^2 - x(yz) + (y^2+z^2)=0$ $\longrightarrow$
$(yz-x)^2-(yz-x)yz+(y^2+z^2)=(yz-x)(yz-x-yz)+(y^2+z^2)=x(x-yz)+(y^2+z^2)=0$
$\longrightarrow$ some rearrangement of $(y,z,yz-x)$ is solution.
We thus just have to prove that $yz-x\geq z$, that is $(y-1)z\geq x$, which is true because $z \geq x$ and $y \geq 2$. In fact $y>2$, so $yz-x > z$, thus $(y,z,yz-x)$ is different from $(x,y,z)$.

3) (I suppose you meant 'there are infinetly many (x,y,z) solution')
There is at least one ($(3,3,3)$). Suppose there were finetly many. Take the one with biggest $z$. $(y,z,yz-x)$ give a solution 'bigger' than $(x,y,z)$.


trigonometry
calculus
geometry proposed
geometry

Let ABC be a triangle and D the midpoint of BC. Suppose that <BAD=<ACB
and <DAC=15. Determine <ACB.

$\triangle ABD\sim \triangle CBA$, we have $BD/AB=BA/BC$ thus $AB=\sqrt{BD\cdot BC}=\sqrt{2}BD$.
In triangle $ABD$, we have from Sine's law that
\[
\frac{BD}{\sin DAB} = \frac{AB}{\sin ADB}
\]
or (denote $\angle DAB=\alpha$)
\[
\frac{1}{\sin \alpha} = \frac{\sqrt2}{\sin (\alpha+15^\circ)}
\]
i.e,
\[
\sin (\alpha+15^\circ) = \sqrt2\sin \alpha
\]
It is easy to observe that $\alpha=30^\circ$ is a solution. We need only to prove that it is the unique solution in $(0,\pi/2)$.
It is sufficient to show that \[f(\alpha) = \frac{\sin (\alpha+15^\circ)}{\sin \alpha}\] is monotone.
\[f'(\alpha) = \frac{\cos(\alpha+15^\circ)\sin\alpha-\sin(\alpha+15^\circ)\cos\alpha}{\sin^2 \alpha} = -\frac{\sin 15^\circ}{\sin^2 \alpha} < 0\]
when $\alpha\in(0,\pi/2)$.

[quote="liyi"]It is easy to observe that $\alpha=30^\circ$ is a solution. We need only to prove that it is the unique solution in $(0,\pi/2)$.
It is sufficient to show that \[f(\alpha) = \frac{\sin (\alpha+15^\circ)}{\sin \alpha}\] is monotone.
\[f'(\alpha) = \frac{\cos(\alpha+15^\circ)\sin\alpha-\sin(\alpha+15^\circ)\cos\alpha}{\sin^2 \alpha} = -\frac{\sin 15^\circ}{\sin^2 \alpha} < 0\]
when $\alpha\in(0,\pi/2)$.[/quote]

The same without barbarism:

$\displaystyle \frac{\sin \left( \alpha +15^{\circ }\right) }{\sin \alpha }=\frac{\sin \alpha \cos 15^{\circ }+\cos \alpha \sin 15^{\circ }}{\sin \alpha }=\cos 15^{\circ }+\cot \alpha \sin 15^{\circ }$,

what is obviously injective in $\left[ 0;\;\pi \right]$ (not just in $\left[ 0;\;\frac{\pi }{2}\right]$, calculus suckz).

  Darij

I don't think there's any need for the trigonometric bash: we can easily see that $30^{\circ}$ works, and then we observe that $A$ is an intersection pt of two circles, one centered at $B$ and having radius $\sqrt 2 BD$, and the other one passing through $DC$ s.t. $DC$ subtebds an angle of $15^{\circ}$, so it's unique because $A'$, the other point in which the two circles intersect, is on the other side of $BC$, so $\angle BA'C=165^{\circ}$.


Let $A, B$ symmetrics matrices $n\times n$ above $\mathbb R.$   ($a_{ij}=a_{ji},b_{ij}=b_{ji}$).
Prove that $tr A^2B^2\geq tr(AB)^2.$ ($tr X=\sum_{i=1}^nx_{ii}$) :)

Arqady, isn't this Cauchy-Buniakovski-Schwarz? :?

[quote="arqady"]Maybe. I can to prove without that.[/quote]
Hi,arqady,can you post it now,Thanks

$tr(AB-BA)^2=tr(AB)^2-trBA^2B-trAB^2A+tr(BA)^2.$
If $C=AB-BA$ then $C_{ij}=-C_{ji}.$ Hence, $tr(AB-BA)^2\leq0.$
But $tr(AB)^2=tr(BA)^2$ and $trAB^2A=trBA^2B=trA^2B^2.$
Hence, $tr(AB)^2\leq trA^2B^2.$ :)

Thanks :) .Very nice and strange you consider $C=AB-BA$ :)


In the finite sequence 7, 8, 6, 9, 5, 10, ... , 0, the odd-numbered terms decrease by 1 and the even-numbered terms increase by 1. The last term in the sequence is 0. What is the sum of the terms?

I don't understand, the even and odds would just counter each other and would be 67676767... because the even would increase and turn into an odd which would decrease back to the even

The even numbered [b]terms[/b] increase by one, not the even numbers.  Same idea with the odd numbered terms.

You can also think of this sequence as two different sequences, one with the odd numbered terms, with common difference -1 and first term 7, and another sequence with the even numbered terms with common difference 1 and first term 8, and that the new sequence goes: 1st term from the first sequence, 1st terms from the second sequence, second term of the first sequence, effectively yielding the sequence 7,8,6,9...

@above welcome to AoPS!

[quote=ab2024][quote=Alphaiscoco]$14! = 0+1+2+3+4+5+6+7+8+9+10+11+12+13+14$[/quote]

$14!$ is not the sum of all the numbers till 14.. It's the product![/quote]

oh yeah i just realized lol


function
induction
Euler
geometry
geometric transformation
rotation
algebra

Let $a_1,a_2,...$ be the sequence of positive integers satisfying :With each n we have
  $\sum a_d$=$2^n$ for all d/n.
Show that for all n, $a_n$ is divided by n.
 :maybe: I know its solution based on Mobius Inversion formula, hence,I hope somebody give me a full solution.Thanks!

i think i once saw a really freaky combinatorics solution to this ...

this is it:
consider the number of binary strings of length n which cannot be divided into more than 1 identical block. (e.g. 10101 would be such a string but 101010 wouldn't be.) Then its obvious that this function satisfies the recurrence you specified. But now given one such binary string, take the last digit and put it to first place, and we get another string. We can do this exactly n times - so out of any one sequence we get another n-1. So we divide the whole set into such sequences, and n divides a(n) as required.

You don't need Mbius' formula for this one. A simple induction leads to the result.

The result is clear for $n=1$ since $a_1=1$.
Let $n > 2$ be given, and assume that for eah $k$ such that $1 \leq h < n$ the integer $a_k$ is divisible by $k$.
Let $p$ be a prime divisor of $n$, and $ n = p^ {\alpha}k$ where $gcd(k,p)=1$.
We will prove that $p^{\alpha}$ divides $a_n$ which will leads to the result since $p$ is arbitrary.

The divisors of $n$ may be divided into to disjoint groups according they divide $ \frac {n} {p}$ or not. In the latter, they are the numbers of the form $p^ {\alpha}d$ where $d$ divides $k$
Thus :

$2^n = \sum_{d|p^{ \alpha - 1}k} a_d + \sum_{d|k} a_{p^ {\alpha}d} = 2^{p^ { \alpha -1}k} + a_n + \sum_{d|k,d<k} a_{p^{\alpha}d}$
Then
$a_n=2^n - 2^{p^{ \alpha - 1}k} - \sum_{d|k,d<k} a_{p^{\alpha}d}.$
But, the last sum is either over the empty set if $k=1$, and in that case it is equal to 0, or is a sum of terms each of whose are divisible by ${p^ \alpha}$ from the induction hypothesis. It follows that :
$a_n = 2^n - 2^{p^{ \alpha - 1 }k}$ mod $[p^{\alpha}] = 2^{p^{ \alpha - 1 }k} (2^{p^{ \alpha - 1 }k(p-1)} - 1)$ mod $[p^{ \alpha}]$.
- If $p=2$ :
It is easy to prove that $2^{i-1} \geq i$ for each integer $i > 0$. Then $p^{ \alpha - 1 }k \geq  \alpha$, from which $2^{p^{ \alpha - 1 }k} = 0$ mod $[2^ { \alpha}]$, and then $a_n=0 $ mod$[2^ {\alpha}]$.

- If $p$ is an odd prime :
Using that $\varphi$ (Euler's function) satisfies $\varphi (p^{ \alpha}) = p^{ \alpha -1}(p-1)$ and that using Fermat-Euler theorem $2^{p^{ \alpha - 1 }(p-1)} = 1$ mod $[p^ { \alpha}]$, we deduce that :
$2^{p^{ \alpha - 1 }k(p-1)} = 1 $ mod $[p^{ \alpha} ]$, from which it easily follows that $a_n = 0 $mod $[p^ { \alpha}]$.

Pierre.

Well we might not actually need Mobius inversion, but it yields such a short solution (not half as beautiful as the one Vinoth posted, of course, but about as short as that one :)).

We have $a_n=\displaystyle \sum_{d|n}2^d\mu(\frac nd)$. If we have $n=p^kt,\ p$ doesn't divide $t$, then the sum is made up of sums of two terms of the type $2^{p^kt_1}\mu(t_2)+2^{p^{k-1}t_1}\mu(pt_2)$, where $p$ doesn't divide $t_i$ because $t_1t_2=t$. We thus have $\mu(t_2)=-\mu(pt_2)$, and since $2^{p^k}\equiv 2^{p^{k-1}}\ (\mod p^k)$, each such sum of two terms is $0\ (\mod p^k)$, and we're done.

I think I can produce some more interpretations. If they are presented in proper order, it is obvious that they are equivalent.

   First:
   Consider all possible necklaces of $n$ black and white beans, fixed (by nails?) on somebody's neck... It is obvious that their number is $2^n$. Then take off the nails and begin rotating necklace on the neck. Then some necklaces may turn into itself when moved by $d$ beans. If we take the minimum of such $d$ it is obvious that $d$ divides $n$. Moreover, if $a_n$ is the number of necklaces without non-trivial rotations preserving them, then the number of necklaces for which minimum rotation is by $d$ beans equals $a_d$. So $a_n$ thus defined is the sequence in question. Now it is obvious that $a_n$ necklaces are divided into groups of $n$ necklaces transformed into each other by rotation.

   (Of course, this is only Vinoth's proof restated.)

   Second:
   Let $F_2$ be a field of residues modulo $2$. Then an irreducible polynomial of degree $n$ over $F_2$ must divide $x^{2^n}-x$.
   Now we note that $a_n$ is the degree of the product of all such polynomials. Indeed, if an irreducible factor of $x^{2^n}-x$ has degree $d$, then it divides $x^{2^d}-x$ and therefore $d$ divides $n$. If the products of all the factors of degrees $d<n$ have degrees $a_d$ (induction), the rest should have degree $a_n$. The case $n=1$ is checked immediately. Divisibility by $n$ is even more obvious than in the first proof.

   (On the equivalence: rotating of a necklace is hidden in the proof of the lemma.)

   Third:
   Consider all the elements of $F_{2^n}$, a field of $2^n$ elements. Each of them is an algebraic element of some degree $d$ over $F_2$. $d$ must divide $n$, since this element generates $F_{2^d}$, so $F_{2^d}$ is a subfield of $F_{2^n}$, and $2^n$ is a degree of $2^d$. Now we see, again by induction, that $a_n$ is the number of elements of degree $n$ (that is, primitive elements) of $F_{2^n}$. Their number is divided by $n$ either because we can form groups of an element and its conjugates, or (which seem more algebraically impressive) because we have exactly $n$ automorphisms of $F_{2^n}$ over $F_2$ :)

   Surely, this is the same proof again.
   Perhaps a can think of some more translations...


limit
integration
calculus
real analysis
real analysis unsolved

1) try to divided interval of integration in two parts $ [0,n^{ \minus{} \alpha}]$ and $ [n^{ \minus{} \alpha},1]$ and choose $ 0 < \alpha < 1$  :). (answer will be $ f(0)$ )
2) similary as first: in this case you need to divide interval in such way $ I_{1} : \equal{} [0, \minus{} n^{ \minus{} \alpha} \plus{} 1/4]$ , $ I_{2}: \equal{} [ \minus{} n^{ \minus{} \alpha} \plus{} 1/4 , n^{ \minus{} \alpha} \plus{} 1/4]$ ,
$ I_{3}: \equal{} [n^{ \minus{} \alpha} \plus{} 1/4, \minus{} n^{ \minus{} \alpha} \plus{} 3/4]$ , $ I_{4} : \equal{} [ \minus{} n^{ \minus{} \alpha} \plus{} 3/4, n^{ \minus{} \alpha} \plus{} 3/4]$ and $ I_{5} : \equal{} [n^{ \minus{} \alpha} \plus{} 3/4, 1]$
on the interval $ I_{1}, I_{3}, I_{5}$ limit will be zero by choosing well $ 1 > \alpha > 0$ so answer will be equal to somethiong like this $ C_{1}f(1/4) \plus{} C_{2}f(3/4)$
3) I think you can use here some methods from the theory complex analisys :)

in 3) you can also find solution here proposition 1 :)
http://arxiv.org/PS_cache/arxiv/pdf/1001/1001.1366v1.pdf


linear algebra
matrix
linear algebra unsolved

Dear friends,

I have two matrices $ A$ and $ B$ such that $ A\geq B$, then is it true that $ Trace(A)\geq Trace(B)$? If yes how to prove it? Any help? Thanks

Bhupala

[quote="miwalin"]What do you mean by $ A\geq B$? The Lowner order, or $ A \minus{} B$ is nonnegative?[/quote]

It means that $ A\minus{}B$ is positive semi definite matrix.

Sorry for that.

bhupala.

[hide]We have $ A \geq B$ then $ A \minus{} B \geq 0$ then $ (A\minus{}B)$ is Hermite and positive semidefinite
then the exist $ U$ is matrix unita such that $ A\minus{}B\equal{}Udiag(\lambda_1,...,\lambda_n)U^{*},\lambda_k \in \mathbb{R},\forall k \in${1,2,..,n} ,Put $ y\equal{}U^{*}x\equal{}\left[\begin{array}{c}y_1\\y_2\\y_3\\...\\y_n\end{array}\right]$
We have $ \forall x \in M_{n1}{(\mathbb{C})}$ ,$ x^{*}(A\minus{}B)x \geq 0$ or $ \forall y\in M_{n1}(\mathbb{C})$
 $ y^{*}diag(\lambda_1,..,\lambda_n)y\equal{}\sum_{k\equal{}1}^{n}{\lambda_k|y_k|^2} \geq 0$
Let $ y\equal{}\left[\begin{array}{c}1\\0\\0\\...\\0\end{array}\right],\left[\begin{array}{c}0\\1\\0\\...\\0\end{array}\right],...,\left[\begin{array}{c}0\\0\\0\\...\\1\end{array}\right]$
  We will have $ \lambda_k \geq 0 ,\forall k \in$ {1,2,..,n}
 $ Tr(A \minus{} B)\equal{}\sum_{k\equal{}1}^{n}{\lambda_k} \geq 0$ implies $ Tr(A) \geq Tr(B)$[/hide]


Let ABC be a triangle and Ma, Mb, Mc the midpoints of the sides BC, CA, AB, respectively. Show that the triangle with side lengths AMa, BMb, CMc has area 3/4 of that of triangle ABC.

This seems to be easy enough. Translate $M_c$ by the vector $\overrightarrow{BM_b}$. You get a point $D$ s.t. $M_cD=BM_b,\ CD=AM_a$, so the triangle we're interested in is $CDM_c$. If $P=CA\cap M_cD$, then $P$ is the midpoint of $M_cD$. Its area is $2S_{CM_cP}=2(S_{AM_cC}-S_{AM_cP})=2(\frac 12\cdot S_{ABC}-\frac 18\cdot S_{ABC})$. If you finish the conputations you'll see we get the desired result.

Construct a point $X$ such that AMcCX is a parallelogram. Then it's easy to show that AMaX is the triangle with sides AMa,BMb,CMc. As Ma is the midpoint of BC and AB//XC, the area of $\triangle AM_aX$ is half of $ABCX$.

Oh, it seems that grobber types faster than me :D

This problem indeed is a special case of the following problem:

http://www.mathlinks.ro/Forum/viewtopic.php?t=6396


geometry
circumcircle
function
geometry proposed

$o_1$ is a center of circumcircle and $o_2$ is a center of incircle of acute-angled triangle ABC. Find the smallest constant $k \in Z^{+}$ which will obey:

$max(|AB|,|BC|,|CA|)<k |o_1o_2|$

each side has different lenght.

Let p-c = x, p-b=y, p-a= z.  x ≥ y ≥ z,  max =x+y.
We can use that
   R = (x+y)(y+z)(x+z)/4√xyz(x+y+z)
   r  = √xyz/(x+y+z)
OI2/(x+y)2 = (x+y)(y+z)(x+z)(∑ x2y+y2x - 6xyz)/16xyz(x+y+z)
z  (0,y] function a think is convex so
1 case if z goes to 0 then obviously OI2/(x+y)2 goes to infinity.
2 case if z =y then we have OI2/(x+y)2 =
(x-y)2/4x(x+2y) by direct calculations.

then we note x = y+t and we see that OI2/(x+y)2= t2/4(y+t)(3y+t) so when t goes to 0 then OI2/(x+y)2 goes to 0 so
I think we can't find such constant.

Maybe I misunderstood something.

[quote="Megus"]$o_1$ is a center of circumcircle and $o_2$ is a center of incircle of acute-angled triangle ABC. Find the smallest constant $k \in Z^{+}$ which will obey:
$max(|AB|,|BC|,|CA|)<k |o_1o_2|$[/quote]
If $ABC$ is a regular triangle then $|O_1O_2|=0$.
Reverse sign is also wrong, since it is possible $\max(|AB|,|BC|,|CA|)\to 0$ while $|O_1O_2|\to+\infty$.

Maybe you mean obtuse triangle ABC?

[quote="Megus"]sorry for this but there should be
1) AB \= BC \= AC  [each side has different lenght]   :blush:
now I think it's not possible for any of values to reach near-zero. And problem might be solved for obtuse triangle as well[/quote]
Anyway $O_1O_2$ can be very small, while triangle's sides are very large.

[quote="Megus"]Well I've still must have missed some conditions

Myth: can you give some additional conditions for this problem to make it more interesting ? or anybody ...  :maybe:[/quote]
I have already given one: obtuse triangle. Though problem is rather simple in this case.

Any chances to complicate it "a little" bit ?  :maybe:


Giea!

 A = 0 , a1a2...am a1a2...am ... a1a2...am
-----------1----------2------------------n

Otan to n teinei sto apeiro o aritmos A poia timi pairnei?

opou kathe ai einai enas apo tous arithmous  0,1,2,..,9 lkai
m enas peperasmenos arithmos

g.p A = 0 , 235342624 235342624 ... 235342624
i allilouxia 235342624 epanalambanetai n fores

sef καλως ηρθες...
ΑΝ ΚΑΤΑΛΑΒΑ ΚΑΛΑ ΤΗΝ ΑΣΚΗΣΗ......,σκεφτηκα οτι:
Ο αριθμος Αειναι αρχικα φραγμενος στο (0,1)  και για την ακριβεια,στο
(0,00000....100000......1......   , 0,9999999999999999.........)
Επομενως σκεφτηκα οτι το οριο του Α οταν το ν τεινει στο απειρο,πιθανοτατα θα ειναι το 0 η το 1.....αλλα γενικα ,θα ειναι ενας αριθμος στο [0,1]....
Ομως οταν το ν τεινει στο απειρο,η ξαλυτερα οταν το ν γινεται ν+1, ο Α αυξανεται,γιατι ουσιστικα μιν μ-αδα μηδενικων αντικαθισταται απο μια μ-αδα αριθμων,για παραδειγμα....

0,1234=0,12340000<0,12341234

Συνεπως οσο αυξανεται το ν,αυξανεται και ο Α,οποτε το ζητουμενο οριο θα ειναι το ανω φραγμα ,δηλαδη το 1.
Παρολα αυτα διατηρω τις επιφυλαξεις μου οσων αφορα την ορθοτητα της απαντησης μου.....γιατι:μου φαινεται λιγο παραλογο το 0,23232323232..... να τεινει στη μοναδα οσες φορες και αν βαλουμε δεξια το 23.Δεν ξερω γιατι,αλλα αν η απαντση μου ειναι σωστη,το προβλημα αυτο μου θυμιζει την ακολουθια 1/2+1/3+1/4+1/5+..... που παρολο που αυξανεται απιστευτα αργα,δεν εχει οριο....Η μονη διφορα με το δικο μας προβλημα ειναι οτι εδω το Α ειναι φραγμενο και μονοτονο και αρα εχει οριο...
Sef,πες μου τουλαχιστον αν πλησιασα στην λυση.....

ΡΑΛΛΗΣ :D

[quote="RAKAR"]

Συνεπως οσο αυξανεται το ν,αυξανεται και ο Α,οποτε το ζητουμενο οριο θα ειναι το ανω φραγμα ,δηλαδη το 1.

Παρολα αυτα διατηρω τις επιφυλαξεις μου οσων αφορα την ορθοτητα της απαντησης μου.....γιατι:μου φαινεται λιγο παραλογο το 0,23232323232..... να τεινει στη μοναδα οσες φορες και αν βαλουμε δεξια το 23.\\
ΡΑΝΛΛΗΣ :D[/quote]

Σχεδόν ανακάλυψες το λάθος σου...  :) Δεν είναι αναγκαίο να τείνει στο άνω φράγμα
Ουσιαστικά η άσκηση σου ζητάει τον ρητό που έχει δεκαδική παράσταση
αυτής της μορφής... :wink:

Epeidi den eimai exoikiomenos me ti Latex tha grapso ena meros
tis lisis xoris epalitheusi
Pernontas tin akolouthia
An = 0 , a1a2...am ... a1a2...am (1)
opou i allilouxia a1a2...am epanalambanete n fores
opote (10^m)An = a1a2..am + 0 , a1a2...am ... a1a2...am (2)
alla sti sxesi (2) o dexios prostheos emfanizei tin allilouxia a1a2...am n - 1 fores
etsi, exoume (10^m)An = a1a2...am + An-1 (3)
Tora diasthitika to limAn kai to limAn-1 einai idia
opote (10^m)limAn = a1a2...am + limAn-1 (4)
kai epeidi limAn = limAn-1 = j
H (4) grafete (10^m)j = a1a2...am + j (5)
kai j(10^m - 1) = a1a2...am

Kai telika limAn=j = a1a2...am/10^m - 1

$A=0,a_{1}a_{2}...a_{m}a_{1}a_{2}...a_{m}...............a_{1}a_{2}...a_{m}$ (με την ακολουθία ψηφίων $a_{1}a_{2}...a_{m}$ να επαναλαμβάνεται n φορές)

$= \frac{a_{1}a_{2}...a_{m}}{10^{m}-1}\left(1-\frac{1}{10^{mn}}\right)$

Από αυτό βλέπουμε ότι όταν το $n$ τείνει στο άπειρο τότε o αριθμός Α τείνει στον αριθμό  $\frac{a_{1}a_{2}...a_{m}}{10^{m}-1}$.

Όπου $a_{1}a_{2}...a_{m}$ θεωρώ την ακολουθία των ψηφίων $a_{1},a_{2},...,a_{m}$ που στη σειρά σχηματίζουν έναν αριθμό (και όχι γινόμενο...  :wink:  ).

Stelios...

[quote="kostas"]νομιζω οτι δεν ειναι δυσκολο να βρεθει αυτος ο αριθμος. ραμανατζαν:!: αλλα ειναι παντα ρητος? :maybe:[/quote]

Όταν επαναλαμβάνεται περιοδικά ο ίδιος αριθμός ναι   :)

οκ,καταλαβα .ωραιες λυσεις . :)

O RAKAR μου είπε ότι ο αριθμός τείνει στον
a(1)...a(m)/(10^m-1) και όχι στο 1

για περαιτέρω διευκρινήσεις περιμένετε και ρωτήστε τον ίδιο.


probability
symmetry
probability and stats

Suppose that there are 100 seats in a saloon for 100 students. All students except one know their seat. First student (which is the one who doesn't know his seat) comes to the saloon and sits randomly somewhere. Then others enter the saloon one by one. Every student that enters the saloon and finds his seat vacant, sits there and if he finds his seat occupied he sits somewhere else randomly. Find the probability that last two students sit on their seats.

One can assume that the seats are labelled from $ 1$ to $ 100$ and that for $ 2 \leq k \leq 100$ the $ k$-th student is to sit on the seat numbered $ k$.

related problem:
Consider the process $ X_1, X_2, X_3, \ldots$ on $ \{1,2, \ldots, 100\}$:
$ X_1$ is uniformly distributed on $ \{1,2, \ldots, 100\}$ and $ X_{k \plus{} 1}$ is uniformly distributed on $ \{1\} \bigcup \{X_k \plus{} 1, X_k \plus{} 2, \ldots, 100\}$. Consider the event:

$ A$ = the set $ \{1\}$ is reached before the set $ \{100, 99\}$

Let's compute the probability that the event $ A$ happens:
if $ P_k \equal{} P[A | X_1 \equal{} 100 \minus{} k \plus{} 1]$ then one knows that $ P_{1} \equal{} P_{2} \equal{} 0$ and $ P_{100} \equal{} 1$ and for $ 3 \leq 99$:
\[ P_k \equal{} \frac {1}{k} (P_1 \plus{} P_2 \plus{} \ldots \plus{} P_{k \minus{} 1} \plus{} P_{100})
\]
so that
\[ P_{k \plus{} 1} \equal{} \frac {1}{k \plus{} 1} (P_{100} \plus{} P_1 \plus{} P_2 \plus{} \ldots \plus{} P_{k}) \equal{} P_k
\]
Finally: $ P(A) \equal{} \frac {1}{100}(P_1 \plus{} P_2 \plus{} \ldots \plus{} P_{100}) \equal{} \frac {1}{3}$. This is also the answer to the original problem, which suggests that there is a more clever approach  :lol: More generally, it also says that if there are $ N$ people instead of $ 100$ then the probability that the last $ n$ persons sit on their seats is $ \frac {1}{n \plus{} 1}$, which is independent of $ n$ !

An equivalent process: choose a permutation randomly, then reassign everybody not in the first student's cycle to their assigned seats. The last two students are in their seats if and only if neither is in the first cycle.

What is the probability of this? We build that first cycle by choosing uniformly from $ \{1,2,\cdots,N\}$, excluding numbers already chosen, and terminating when $ 1$ comes up. We seek the probability that this happens before any of $ \{N\minus{}n\plus{}1,N\minus{}n\plus{}2,\cdots,N\}$ come up. By symmetry, it is clear that the first of these $ n\plus{}1$ elements could be any of them with equal probability, and the probability that $ 1$ is first is $ \frac1{n\plus{}1}$.

[quote]IMS is a competition for undergraduate students held by Iranian Mathematical Society. The website which unfortunately contains the problems only in Persian is http://ims.ir .
	[/quote]

[note in the [url=http://www.mathlinks.ro/resources.php?c=2&cid=109]IMS page of the Resources section[/url]]

the first event/sample space to occur is
Std. 1 sits correctly (1/100)
Std. 1 sits incorrectly (irrelevant seat) (97/100)
Std. 1 sits incorrectly (one of last 2 students' seats) (2/100)

In the case of the second event, the process of random seating may begin. No matter what, there will always be one seat taken that should go to a student who has yet to sit down. This is because every time a student takes a seat, his/her correct seat is taken, but he/she takes a new seat. This new seat becomes the one that is displaced.

Because there will always be an incorrectly occupied seat after this, the trick is to find out when someone who randomly sits down in a seat randomly picks the first student's seat. That is the only way it's going to work out. Furthermore, it is an instant game-over if one of the last two students' seats is taken.

It seems like a complicated model that could be modeled like a geometric distribution... I'll think about it.


I don't know if this goes in the intermediate thread or not, but anyway...

Let's say you have the following recursive equation

\[ f(x)\equal{}2f(x\minus{}1)\plus{}f(x\minus{}2)\]

Now, how do I solve for $ f(x)$ without requiring another $ f(something else)$ in my equation.

Any help will be appreciated.

Don't post the same question twice.  I just answered this.


arithmetic sequence
number theory proposed
number theory

Let p>2 be a prime . the sequence {$a_n$}is defined by:$a_0=0,a_1=1,a_2=2,...,a_{p-2}=p-2$and for all $n\geq p-1$a_n is the last integer greater than $a_{n-1}$that doesn't form and arithmetic sequence of length pwith any of the proceeding terms.prove that for all $n,a_n$,is the number obtained by writing n in base $p-1$ and reading the result in base p.

[quote="sam-n"]Let p>2 be a prime . the sequence {$ a_n$}is defined by:$ a_0 \equal{} 0,a_1 \equal{} 1,a_2 \equal{} 2,...,a_{p \minus{} 2} \equal{} p \minus{} 2$and for all $ n\geq p \minus{} 1$a_n is the last integer greater than $ a_{n \minus{} 1}$that doesn't form and arithmetic sequence of length pwith any of the proceeding terms.prove that for all $ n,a_n$,is the number obtained by writing n in base $ p \minus{} 1$ and reading the result in base p.[/quote]
[quote]first let (h_n) such that $ h_n \equal{}$the number obtained by writing $ n$ in base $ p \minus{} 1$ and reading the result in base $ p$[/quote]

for $ x_1,x_2,...,x_n\in\mathbb{N}$ arbaitrary, let the property $ P(x_1,x_2,...,x_n)$
$ x\ check\ the\ property\ P(x_1,x_2,...,x_n)\ \iff :$
$ x\ge max(x_1,x_2,...,x_n)$ and $ x$ doesn't form an arithmetic sequence of length p with any of the terms $ x_1,x_2,...,x_n$

we take the application $ A: P(\mathbb{N})\to P(\mathbb{N})$, such that $ A(\{x_1,x_2,....,x_n\}) \equal{} \{x\in\mathbb{N}: \ x\ check\ the\ property\ P(x_1,x_2,...,x_n) \}$

NOw we can see that There is one and only one sequence that satisfies the property $ (Q)$:
$ \forall i\in\{0,1,2,...,p \minus{} 2\}: \ b_i \equal{} i$ and for $ i\ge p \minus{} 1: \ b_{i} \equal{} min(A(b_0,b_1,...,b_{i}))$

then if we want pove that $ a_n \equal{} h_n$
it is suffusante to prove that then sequence $ (h_n)$ is a solution $ (Q)$.
now i'm thinking for the rest  :wink:
[quote]$ h_0 \equal{} 0$
if $ n \equal{} kp^m \plus{} (p \minus{} 1)^m \minus{} 1: \ h_{n \plus{} 1} \equal{} h_n \plus{} p^m \minus{} \frac {(p \minus{} 2)(p^m \minus{} 1)}{p \minus{} 1}$[/quote]


calculus integration logarithms function geometry trapezoid derivative

This integral is too hard for me.
Scientific calculator gives result 3.05911645.
I think it's right but I can't solve.
Does anyone help me?
Thanks!

This integral cannot be done in closed form using elementary functions. This is not an unusual case, and you should not react as if it were unusual.

Note: it's the same as $ \int_1^2\frac {e^t}{t}\,dt.$

In that form, it's fairly well-known that there's no elementary antiderivative. But also in that form, power series methods become attractive. For instance, this can be written as

$ \ln 2\plus{}\sum_{n\equal{}1}^{\infty}\frac{2^n\minus{}1}{n\cdot n!}\approx 3.05911653964595$

As part of a wide-ranging review session yesterday for my calculus course going into the final, I borrowed this problem. Here's the way I attacked it in lecture:

First, the question: which is larger, $ \int_e^{e^2}\frac{dx}{\ln x}$ or $ 3?$

We start with crude estimates. We're integrating a function that takes values between $ \frac12$ and $ 1,$ over the interval from $ e$ to $ e^2$ (length about $ 5$ or so). That gives us a range of possible values for this integral, with $ 3$ deep in the interior of that range. We come away from this knowing that crude estimates won't do it and we must be precise.

I then ask whether we have any chance to compute this integral in closed form. Eventually, I get someone to suggest the substitution $ u\equal{}\ln t.$ Why not? It's worth trying.

But as soon as I see $ \int\frac{e^t}{t}\,dt,$ I say that that looks a little too familiar, and I'm pretty sure we aren't going to get this one in closed form.

The next thing I try is the series in my previous post, adding in terms one at a time and keeping track of the running sum (using a calculator). After not very many terms, the sum is greater than $ 3,$ which settles the question.

I wasn't emphasizing numerical methods of integration (trapezoid, Simpson, etc.) but I could have taken the lecture in that direction had I wanted to.

This reminds me that I put $ \int_1^2 \ln x\,dx$ on the final Calc 2 exam (too easy, I know) and got a bunch of papers with $ \int_1^2 \ln x\,dx\equal{}\frac{1}{x}\bigg|_1^2\equal{}\minus{}\frac{1}{2}$.  :o

Well, if it had been $ \int_1^2\frac1{x^2}\,dx,$ there would have been two families of wrong answers - those with a logarithm of something, and those with some multiple of $ \frac1{x^3}$ in them.

Not getting straight whether you're integrating or differentiation - that's quite common.

[quote="mlok"]This reminds me that I put $ \int_1^2 \ln x\,dx$ on the final Calc 2 exam (too easy, I know)[/quote]
 :o

May be one of your students thought $ \ln x\equal{}\int_1^x\frac1u\,du.$

Then a double integral avoids integration by parts.

mlok would not have discussed double integrals anywhere in this course. That's in the next course. And no, the mistake is not that sophisticated. The most sophisticated students in his class just know off the top of their heads that the antiderivatve of $ \ln x$ is $ x\ln x\minus{}x.$ They're not the ones doing this.

Even worse: what if the question had been $ \int_0^1 \ln x\,dx.$

Then you get things like $ \left.\minus{}\frac1x\right|_0^1\equal{}\minus{}1\minus{}0\equal{}\minus{}1.$

Don't think about that too hard - it will give you a headache.

[quote="Kent Merryfield"]This integral cannot be done in closed form using elementary functions. This is not an unusual case, and you should not react as if it were unusual.[/quote]
Unfortunately, the impression one gets after reading a typical calculus textbook is that the integrals are classified into about six types:
(1) doable by substitution;
(2) - by parts;
(3) - by trigonometric substitution;
(4) - by partial fractions;
(5) - by combinations of the above;
(6) $ \int e^{ \minus{} x^2}\,dx$.

The version I've taught also includes $ \int\frac{\cos x}{x}\,dx$, in the context of showing that the improper integral to $ \infty$ converges. If you make the substitution and squint hard enough, that's really the same as this thread's example.

Elliptic integrals (the arclength of an ellipse) might make a nice illustration of approximate methods- except that they work too well.


Let $p$ be a prime and $X$ be a finite set containing atleast
 $p$ elements. A collection of mutually disjoint $p$-element
 subsets is called a $p$-family. the empty collection itself is
 regarded as a $p$-family. Let $A$(respectively, $B$) denote the
 number of $p$-families having an even(respectively, odd)
 number of $p$-element subsets of $X$. Prove that $A$ and
 $B$ differ by a multiple of $p$.

I hope this makes sense.
Call $P_{k}$ the number of k-element p-families of a set $X$ of cardinality $qp+r$ with $r<p$. Each k-element family can be created choosing $p$ elements from $X$, and then choosing another $p$ from what remains, and so on $k$ times, and dividing afterwards for the number of times a family has been counted in this way, which is exactly the number of possible orders in which its elements can have been chosen, $k!$.

$P_{k}\equiv \frac{\prod_{i=0}^{k-1}\binom{n-ip}{p}}{k!}\equiv\frac{\prod_{i=0}^{k-1}\frac{((q-i)p+r)((q-i)p+r-1)..(q-i)p..((q-i)p+r-p+1)}{p.(p-1)!}}{k!}\equiv\frac{\prod_{i=0}^{k-1}\frac{(q-i)((q-i)p+r)((q-i)p+r-1)..((q-i)p+1)((q-i)p-1)..((q-i)p+r-p+1)}{(p-1)!}}{k!}\equiv\frac{\prod_{i=0}^{k-1}(q-i)}{k!}\equiv\binom{q}{k}$
$(mod\; p)$

So

$\sum_{i=0}^{\infty}P_{2i}\equiv \sum_{i=0}^{\infty}\binom{q}{2i}\equiv \sum_{i=0}^{\infty}\binom{q}{2i+1}\equiv \sum_{i=0}^{\infty}P_{2i+1}$

$(mod \; p)$


Hello everyone,

The [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=229]log[/url] from this year's Math Jam about Mathcamp, including information about admissions, financial aid, and what the summer is like, is now posted.

Thanks to everyone who participated!  If you have any questions about the camp, feel free to get in touch via our web page or by private message here.

Thanks, Dan, for holding the Math Jam.

I'd like to add that I will be a guest instructor at Mathcamp this summer, so if you're going to Mathcamp and you're an AoPS person, please come by and say hi and some point!

[quote="JSteinhardt"]Awesome! Which week will you be there?[/quote]I don't know yet.


A cute one my teacher showed me:

f,g:[a,b]->R are 2 Riemann integrable functions. Show that fg is also a Riemann integrable function.

It really surprised me when I saw  it because the text sounds so simple and I had never asked myself if this is true.

[i]unknown source[/i]

many mathematical analysis books includes this conclusion.

Here's something really cool which I didn't notice (my teachr showed me this):

[color=white]fg=((f+g)2-f2-g2)/2, so it's actually enough to show that if f is integrable then so is f2[/color]

Doesn't Lebesgue's criterion work ? If A is the set of discontinuity points for f and B for g , then if A and B have measure 0, so does A U B.

[quote="grobber"]fg=((f+g)2-f2-g2)/2, so it's actually enough to show that if f is integrable then so is f2[/quote]
well, i remember that some books use this to prove fg is integrable.

Oh, I forgot: Lebesgue's criterion works just fine, so try not to use it :D
It's kind of easy now, but I still enjoyed the problem.


[img]http://img92.imageshack.us/img92/3333/untitledpsdzz1.jpg[/img]
in the figure,angle AEF=angle AFE and E is the midpoint of CA.prove that BD/CD=BF/CE.


What about them? Of course they converge. Since the convergence is extremely rapid, they can be numerically estimated to high accuracy using ordinary techniques. Closed-form computations of the sums seem unlikely. What else is there to say?

May be to decide if they are rational (it seems not likely, I think I saw somewhere a proof that the first one is not), or to prove that they are transcendental, what is, I think, a little tedius and complicated.


Let f:N->N (without 0) a bijection such that [n,f(n)]<3(n,f(n)) for all n. Find f*f :)

The condition is equivalent to that either we have $f(n)=2n$ or $n=2f(n)$ or $f(n)=n$ for all $n$.
Suppose $o$ is an odd number.
Then we have for some $x$ we have $f(x)=o$ so either $x=o$ or $x=2o$. In the last case we have $f(o)=2o$.
So either $o$ goes to itself or $o$ and $2o$ change. If $f(o)=o$ do this for $2o$ and we will have either $f(2o)=2o$ or again $2o$ and $4o$ change place.
In the other case do this for $4o$. And continue this process.
So doing this every number like $n$ is either a fixed point or is in a $2$ cylce.
So we always have $f(f(n))=n$.

Nice solution Nima Ahmadi Pour :) Mine is similar...


inequalities
quadratics
inequalities proposed

[b]Let be a,b > 0. Proof that :[/b]

$(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}})\sqrt{a+b}\geq2 \sqrt{2}$

[hide]It's famous that [i]quadratic average [/i] >= [i]harmonic average[/i]

The quadratic average for $\sqrt{a}$ and $\sqrt{b}$ is :

$\sqrt{\frac{\sqrt{a^{2}}+\sqrt{b^{2}}}{2}}=\sqrt{\frac{a+b}{2}}=\frac{\sqrt{a+b}}{\sqrt{2}}$

The harmonic average for $\sqrt{a}$ and $\sqrt{b}$ is :

$\frac{2}{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}}$

So,

$\frac{\sqrt{a+b}}{\sqrt{2}}\geq\frac{2}{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}}\Longleftrightarrow\sqrt{a+b}(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}})\geq2\sqrt{2}$[/hide]

I think that [b]expanding[/b] provides us with quite a shorter proof in this case... Well, we need to lean on the fact that $x+\frac{1}{x}\geq 2$ for every $x > 0$, but that's like one of the most elementary inequalities in everyone's repertoire around here, isn't it? :yoda:


Do we still need a cover sheet like last year for each round's submission?

No.  You do have to submit an entry form, though.


Wat do you think about my site?
http://www.linkx.ro

Good website! I would like to know where do you learn to do such websites.

Thank you for sharing your website.

Greetings.

Meh.. its okayish on a professional level but pretty good if you're just a student. Tip: flash SUCKS for web design. Period,

(and yes I realize I'm responding to spam)


The following is a rather difficult problem regarding turing machines, to anybody who is familiar with them:

Suppose you could observe the configurations of a Turing machine as it
executes. Suppose that the tape head never moves past some fixed cell
while you are watching. In other words, the machine is computing with
a bounded quantity of cells. How long do you have to watch in order to
determine whether or not the machine is in an infinite loop? Your answer
should work for any Turing machine and any bound.

Hm... I'm actually not very familiar with Turing machines :) but I know that it is some kind of automata with a read/write tape and head that can move in both directions. My question is - is there any limitation on the number of different states of one cell on the tape? If there was, then the whole machine would have only finite number of configurations so after that number of steps you'd know it will never stop (because it will be in the same configuration as it was some time before).

It could be trickier - aren't Turing machines allowed to have internal states?  If those can't be observed, then I'm not sure it's very possible... if they can, an N-state turing machine working with n cells, each of which has m states, has Nnm total states.  But that seemed to easy, so I'm sure my ignorance about Turing machines shows.

[quote="inteluser"]But that seemed to easy, so I'm sure my ignorance about Turing machines shows.[/quote]Exactly what I thought about my solution :).

However, if you observe a [i]configuration[/i] of a Turing machine, you also see it's state (the internal state you've mentioned) - this should not be an issue.


geometry rectangle ratio calculus integration

[quote="GameBot"]Twenty-four circles are drawn inside a rectangular region as shown, such that every circle is tangent to its neighbor circles and the rectangle is tangent to 16 of the circles. What is the ratio of the total area of the 24 circles to the area of the rectangle? Express your answer as a common fraction in terms of $ \pi$.
[asy]import graph;
draw((0,0)--(0,8)--(12,8)--(12,0)--(0,0),linewidth(0.7));
for(int i = 1; i < 12; ++i){
 for(int j = 1; j < 8; ++j){
   draw(Circle((i,j),1),linewidth(0.6));
  ++j;
 }
 ++i;
}[/asy][/quote]

$ \frac{24\pi\r^2}{96r^2}=\frac{1}{4}$.

...is that a typo?  Shouldn't it be $ \frac{24r^2\pi}{96r^2}\equal{}\boxed{\frac{\pi}{4}}$?

Note that this density of circles does not change if the rectangle has integral dimensions.


Find all positive integer solutions (p,q,n)
   to:

         p^n - q^n = 2^100, with n>1.

We have $(p-q)\displaystyle\sum_{i=0}^{n-1}p^{n-1-i}q^i=2^k$, so $p-q$ is a power of $2$. $p-q=2^t\ge 2$ because $p,q$ must have the same parity when $k\ge 1$ ($k=1$ is easy to solve). We have $2^k=(q+2^t)^n-q^n$, and this is divisible by $2^t$ and not by $2^{t+1}$, which means that $k=t$, so $p^n-q^n=p-q$, which is impossible if $n>1$, so for $k=100$ (as in our case), there are no solutions.

$n$ can be even. This problem is still unsolved.

A solution for $k=4$ (contradicting grobber's argument): $p=5,q=3,n=2$.

Right...

Well, then write $n=2^{\alpha}\beta$ with odd $\beta$. It's some more work, but I think it leads to the result. I'm tempted to say that we get $4\not |\ n$, but I'll look stupid if it's wrong :D.

A solution: Let $p=2^rx,q=2^ry$ with $x,y$ odd. (by grobber's argument, the same power of 2 must divide both). Thex $x^n-y^n=2^{k-rn}$ with $k=100$ in the original case. If $n$ has an odd factor $s$ greater than 1, we can write $x^n-y^n=(x^{\frac ns}-y^{\frac ns})\times$(sum of $s$ odd numbers), so $x^n-y^n$ has an odd factor and cannot be a power of $2$. If $4|n$, we have $x^n-y^n=(x^2-y^2)(x^2+y^2)$(stuff), so $x^2+y^2$ must be a power of $2$. Since $x^2 \equiv y^2 \equiv 1 (\mod 4)$, this forces $x=y=1$, which is impossible.

This leaves only $n=2$, in which case $x-y$ and $x+y$ must both be powers of 2. This is only possible with $x-y=2$, so $x=2^s+1,y=2^s-1$ and $x^2-y^2=2^{s+2}=2^{k-2r}$, giving $s=k-2r-2$.
In the original problem, the solutions are $p=2^{98-r}+2^r, q=2^{98-r}-2^r,n=2$ for $0 \le r \le 48$.


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Let $ A$ be a Noetherian ring having an embedded associated prime. If $ a \in A$ is a non-zero divisor satisfying $ \bigcap_{n \equal{} 1}^{\infty} a^n A \equal{} (0)$, then $ A/(a)$ also has an embedded associated prime.

I've found a solution somewhere (other suggestions are still welcome):

Let $ P \in \textrm{Ass}(A)$ be such that $ ht(P) \geq 1$, and let $ \mathfrak{p_1},\ldots,\mathfrak{p}_n$ be the prime divisors of $ (a)$. If $ P \nsubseteq \mathfrak{p}_i$ for all $ i$, then there exists an $ x \in P$ such that $ (a): x \equal{} (a)$. This is a contradiction, since $ x$ is a zero-divisor, but if $ xy \equal{} 0$ then $ y \in \bigcap_n a^n A$. Hence $ P \plus{} (a) \subseteq \mathfrak{p}_i$ for some $ i$, and then $ ht(\mathfrak{p}_i) \geq 2$.


Each term of an infinite sequence of natural numbers is obtained from the previous term by adding to it one of its nonzero digits. Prove that this sequence contains an even number.

Let the sequence be $ a_{1}, a_{2}, a_{3},\ldots$. Define a new sequence $ b_{k}\equal{}\lfloor a_{k}/ 10\rfloor$. We note that $ a_{k}\plus{}10 > a_{k\plus{}1}$, so $ b_{k}\plus{}1\geq b_{k\plus{}1}$. Thus $ \{b_{i}\}$ only increases by one at a time. Furthermore, the sequence $ \{b_{i}\}$ is unbounded since $ \{a_{i}\}$ is unbounded. Combined, we see that $ \{b_{i}\}$ assumes every integer greater than or equal to $ b_{1}$.

Suppose $ b_{1}$ has $ m$ digits with decimal representation $ d_{0}d_{1}d_{2}\ldots d_{m}$. Now define $ d^{\prime}_{k}$ to be the least odd number greater than or equal to $ d_{k}$. It's clear that $ 0\leq d^{\prime}_{k}< 10$ still holds. So $ d^{\prime}_{0}d^{\prime}_{1}d^{\prime}_{2}\ldots d^{\prime}_{m}$ is a decimal representation of a number. Call this number $ b^{\prime}$. Notice that $ b^{\prime}$ has all odd digits, and $ b^{\prime}> b_{1}$. This means that at least one term in the sequence $ \{b_{i}\}$ is equal to $ b^{\prime}$, and thus at least one term in the sequence has all odd digits.

Let $ k$ be such that $ b_{k}$ has only odd digits. We prove that $ a_{k}$ is even or $ a_{k\plus{}1}$ is even. Suppose $ a_{k}$ is odd. Then $ a_{k}$ ends in an odd digit; call it $ d$. We have $ a_{k}\equal{} d\plus{}10\cdot b_{k}$. Note that this number has all odd digits. So $ a_{k\plus{}1}$ will be $ a_{k}$, an odd number, plus a digit of $ a_{k}$, also odd. Thus $ a_{k\plus{}1}$ is even. This finishes the proof.

Moreover, we can prove that we can find as long as we want sequences of odd integers.


$ ABCD$ rectangular , $ AXY$ Equilateral Triangle
 X∈BC , Y∈CD  and $ AB\equal{}33 , BC\equal{}21\sqrt 3$

find$ AX$

We are to costruct a convex cyclic quadrilateral.

[url=http://www.mathlinks.ro/viewtopic.php?t=52887&search_id=370192833]See here.[/url]


search
induction
combinatorics unsolved
combinatorics

No three diagonals of a convex $ 2000$-gon meet at a point. The diagonals (but not the sides) are each colored with one of 999 colors. Show that there is a triangle whose sides are on three diagonals of the same color.

It has been posted several times before, without proof.  See
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1524439891&t=4995
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1524439891&t=196972

[quote="probability1.01"][hide="somewhat big hint"]Prove using induction that for any $ n \minus{} 2$ diagonals of a convex $ n$-gon where $ n \ge 5$, some three form a triangle.[/hide][/quote]

Your "triangle" can include the sides as well?

see here:

http://www.mathlinks.ro/viewtopic.php?search_id=1110127026&t=6376


demonstrati ca daca $a_1,a_2...a_n$ sunt nr reale din intervalul $[\frac{1}{2},1]$ atunci

$\sum\frac{1}{a_1^3+a_2^3+...+a_n^3-a_i^3+a_1a_2...a_n}$ $\le\frac{1}{2^n}$
 pt i= 1,..n

Cred ca e prea simpla...sau daca nu, atunci sigur e una clasica. :D

daca_i  asa simpla chiar as vrea sa stiu si eu :?  ca mie nu mi_a iesit prea simplu ,am aplicat vreo trei ineg diferite  :blush:

Cred ca e chiar usoara...nu am apucat sa ma uit prea mult pe ea, dar imi mentin parerea.Ceea ce as face eu la prima vedere e folosesc inegalitatea
   Daca $\ x$>$\frac{1}{2}$, atunci cu siguranta $\frac{1}{x}$<$\ 2$.
Bineinteles, nu m-am referit chiar la problema, dar poti folosi asta si cred ca iese imediat.

bun, dar e suma , si nu cred ca se poate ajunge in acest fel  :?  ma uit si eu sa vad dak merge

Merge si cu suma, numai ca luam $\ x_i$ sa fie egal cu numitorul sumei.

[quote="stancioiu sorin"]Cred ca e prea simpla...[/quote]

Dar tu stiai ca inegalitatea este mai mult sau mai putin gresita? (membrul drept mai exact)

mda...e exact invers....si  intr-adevar e banala


Using combinatorial argument prove that for all integers n,m,k  where n,m>= k

then prove ...................(file attached....becoz i dont know how to use latex here :(  :( )

What is combinatorial argument?

[hide="my solution"]$ (1\plus{}x)^{m\plus{}n}\equal{}(1\plus{}x)^m (1\plus{}x)^n$
$ \equal{}(_{m}C_{0}\plus{}_{m}C_{1}x\plus{}_{m}C_{2}x^2\plus{}...)(_{n}C_{0}\plus{}_{n}C_{1}x\plus{}_{n}C_{2}x^2\plus{}...)$

So the coefficient of $ x^k\equal{}\sum_{j\equal{}0}^{k}(_{n}C_{j})(_{m}C_{k\minus{}j})\equal{}_{n\plus{}m}C_{k}$[/hide]

by combinatorial argument  he means  to reaso out rather than vigorous  meaning.  atleast thats what i think :D

Combinatorial arguments are just as rigorous as algebraic arguments.

The RHS is the number of ways to choose k objects out of n+m objects. Let's systematically count the number of ways to do this in the following way:

Break the n+m objects into one group of n objects (which I will call group N) and one group of m objects (group M). Now, we have the following non-overlapping cases:

Case 0: Pick 0 objects from group N and k objects from group M
Case 1: Pick 1 object from group N and k-1 objects from group M
...
Case j: Pick j objects from group N and k-j objects from group M
...
Case k: Pick k objects from group N and 0 objects from group M

Note that each of these cases corresponds to one term in the summation in the LHS.

The "combinatorial rigorous argument" takes the following two standard forms:  to prove that $ ab \equal{} c$ where $ a \equal{} |A|, b \equal{} |B|, c \equal{} |C|$ and $ A, B, C$ are sets, we exhibit an explicit bijection $ A \times B \to C$, and to prove that $ a \plus{} b \equal{} c$ we exhibit an explicit bijection $ A \sqcup B \to C$.

The explicit bijection is exactly what TZF has given.  This is the [url=http://www.google.com/search?q=decategorification&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a]decategorification[/url] of the generating functions argument.


Sobre los lados $AB$ y $BC$ de un triángulo $ABC$ construya, externamente al triángulo $ABC$, los cuadrados $ABED$ y $CBFG$.
Sean $M$, $N$ y $P$ los puntos medios de $EF$, $AC$ y $DG$, respectivamente.
Pruebe que:
$a)$ $NBMP$ es un paralelogramo.
$b)$ $DP=MN$.

BUENO AHORA POR FALTA DE TIEMPO SOLO POSTEARE LA 1ERA PARTE.
A).-
RECORDEMOS EL TEREMA DE VON AUBEL'S.
EN EL TRIANGULO ABC, SI PROLONGAMOS NB ESTE SERA PERPENDICULAR A EF EN EL PUNTO N', ANALOGAMENTE BM PERPENDICULAR A AC EN M'. AHORA SOLO FALTARIA DEMOSTRAR QUE PN PERPENDICULAR A AC.
SEAM D' EL PIE DE LA PERPENDICULAR BAJADA DESDE D HACIA LA RECTA QUE PASA POR A Y C, ANALOGAMENTE G'. POR LA CONGRUENCIA EN LOS TRIANGULOS BM'A Y AD'D, BM'=AD', ANALOGAMENTE BM'=CG'.
ENTONCES N ES PUNTO MEDIO DE D' Y G', POR LO TANTO AL UNIR PN, SERA BASE MEDIA DEL TRAPECIO G'D'DG, ENTONCES PN//DD'//GG'. POR LOTANTO PN PERPENDICULAR, ENTONCES PN//MB, ANALOGAMENTE MP//BN, POR LO TANTO PMBN, ES PARALELOGRAMO.

bueno para la segunda parte :

sean los centros de los cuadrados $ABED$ y $BCFG$ , $O_{1}$ y $O_{2}$ respectivamente ahora vemos que la distancia entre estos dos puntos es $DP$ por base media en el triangulo $BDG$ , ahora tenemos por teorema de von aubels que $O_{1}O_{2}= BO_{3}$ donde $O_{3}$ es el centro del cuadrado $ACHI$ ahora por la demostracion anterior teniamos que $MB=PN=AC/2$
luego trazamos $PN$ hasta $O_{3}$ y vemos el paralelogramo $MNO_{3}B$ en el cual tendremos que  $MN=O_{3}B$
por lo tanto $MN=DP$

QED


the equation $ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} 0$ has three roots :$ x_1 < x_2 < x_3$
Prove that $ (x_2 \minus{} x_1)^3 \ge \frac {2}{3.\sqrt {3}}(2a^3 \minus{} 9ab \plus{} 27c)$


Alright, I have a math final and one of the problems is write the inverse of the equation.

    y = .5x - (3/2)

Can anyone help?

Thank you so very much, I really appreciate, if I were to ask for help for other problems i'm not too familiar with would that be considered spamming?

You should open another topic for the rest of the problems, and try to write how far you've got solving them. That way you make it easier for the person who is helping you and showing a dose of seriousness on your side.

[quote="SnipedYou"][hide="Solution"]
$ y \equal{} \frac {x \minus{} 3}{2}$
Switch the $ x$ and $ y$, then solve for $ y$
$ x \equal{} \frac {y \minus{} 3}{2}$
$ 2x \equal{} y \minus{} 3$
$ y \equal{} 2x \plus{} 3$
[/hide][/quote]

Wait, where did the .5 in the original equation go?

:lol:  try writing it as $ \frac{1}{2}$ and you'll see it is still there.


It seems that when the site renders $\text{\LaTeX}$, it has a black border around it.

If I take a new latex formula:
$\text{\LaTeX} {}$

as opposed to an already rendered formula:
$\text{\LaTeX}$

It doesn't really affect things much though.

It looks like it's a problem only with Opera.

Hmm... This is what I see:

I was using Opera 9 preview 1.

Heh.  Maybe that was the problem.


inequalities
real analysis
real analysis unsolved

Hi
Let x1,x2... xn be real positive numbers.(a) Show that :

(x1x2.....xn )^(1/n) <= (x1+x2+.....+xn)/n

Thanks

Hello, I think that is already posted somewhere in this forum, but I can find it :?

Hi. You are asking for the proof of the Arithmetic-Geometric mean inequality. It is very well-known. You can do a google.


counting
distinguishability
geometry
3D geometry
tetrahedron
probability
parallelogram

link: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10497[/url]

i took this test, and i would like to know if my answers are correct, however, there are not answers for all of the questions, i don't need solutions, i just need answers, i took the test as a benchmark for how well prepared for the amc tests, are there any answers that i do not know about? i know this is like a year old topic, but i would like some answers

does anybody know if they exist?, i would really like to know how i did

I don't think there are any solutions..

How about we make one?  :) I'll go start with 5 problems. Please check if this is right. For convenience, I'll post the problem as well.

[quote]1. How many ordered triples $(a,b,c)$ satisfy the equation $(a^2-1)^2+(b^2-4)^2+(c^2-9)^2 = 0$[/quote]

I'm not sure why but none of these problems have choices.  :? Anyway..

Since whole thing is 0, we want the each () to be zero as well. Setting $a^2-1 = 0$, we'll get $a = \pm 1$. Similarly, $b = \pm 2, c = \pm 3$. Since there are two choices for a,b,and c, the total number of triples is 2*2*2 = $\fbox{8}$.

[quote]2. How many distinguishable numbers greater than 3000000 can be formed from the digits 1,1,1,2,2,3,3?[/quote]

Fix the first digit to be 3. Then we have 6 choices for the rest of 6 digits. This is just $6!$. But since we want the "distinguishable numbers," we have to erase the repeats. The process is simply:

$\frac{6!}{3!2!1!} = \fbox{60}$

[quote]3. Factor the equation $x^4+x^2y^2+y^4$ into two quadratics with real coefficents.[/quote]

We note the well-known expansion $(a+b)^2 = a^2+2ab+b^2$. Apply this in here and we'll see that above equals:

$(x^2+y^2)^2-(xy)^2$

This is in form of $a^2-b^2 = (a+b)(a-b)$ so our final answer is $(x^2-xy+y^2)(x^2+xy+y^2)$.

[quote]4. The legs of a right triangle are 8 and 15. Find the altitude to the hypotenuse.[/quote]

By Pythagorean Theorem (or your memeory.. You should know this triple), we know hypotenuse is 17. Since:

$[ABC] = 8 \cdot 15 \cdot 0.5 = a \cdot 17 \cdot 0.5$

We can set middle one and the last one equal. Result is $\boxed{a = \frac{120}{17}}$

[quote]5. Pirates have captured the captain, the first mate, the cook, the stowaway, and an oarsman, and are going to make each of them walk the plank. How many possible orders are there for walking the plank?[/quote]

This seems too obvious.  :huh: This is simple permutation of $5!$ which is 120.

Please tell me if any of the following is wrong. Also, you can post the solutions to the other problems (Let's try to keep them in order though).

i will continue with some of my answers, and i agree with your solutions on the first 5 problems

[quote]6. The pirates are at it again!  As in the previous problem, the captain, the first mate, the cook, the stowaway, and an oarsman are all being forced to walk the plank.  But just as the first sailor is about to jump, the captain cries ``I shall be first; `tis my captain's honor''.  Then the first mate shouts ``and I shall follow him; 'tis my duty''.  If the pirates comply with the sailors' requests, how many possible orders are there?[/quote]
since they captain has to go first, and then the first mate goes second, then the other 3 people that can jump off so the ways are $1*1*3*2*1 = \boxed{6}$ ways

[quote]7. How many ways are there to paint the sides of a tetrahedron with exactly $4$ colors?  (Each side gets its own color, and two colorings are the same if one can be rotated to get the other.)[/quote]

think of the tetrahedron that can spin at one vertex, the side opposite that vertex can be 4 colors, then the other 3 can be 3 colors, in any order, and since there are 3 ways to spin the tetrahedron along that vertex, the probability is $\frac{4*3*2*1}{3} = \boxed{8}$ ways

[quote]8. Find the length of the longest median of a triangle with sides 4, 7, and 9.[/quote]

think of the formula for a median of a triangle $2(a^2 + b^2) -c^2 = m^2$ where a,b,c are sides, and c is the side the median m is drawn to (this formula comes from the fact that in a parallelogram, sum of sides squared = sum of diagonals squared, and if you take the triangle drawn with one median and rotate it around the midpoint, you get a parallelogram)
so maximizing this obviously a and b should be 7 and 9, so
$2(9^2 + 7^2) - 4^2 = m^2$
so the maximum for the median is $m = \boxed{2\sqrt{61}}$

[quote]9. A rectangular solid with a square base has volume $40$ cubic meters and surface area $72$ square meters.  If its length is $a$, its width is $b$, and its height is $c$, where $a$, $b$, and $c$ are rational numbers, find $a + b + c$.[/quote]

since it is has a square base, without loss of generality, we can say that $a=b$
so for volume, $a^2c=40$, and for surface area $2(a^2 + 2ac) = 72 \quad a^2 + 2ac = 36$
substituting gives $\frac{40}{c} + 2c\frac{2\sqrt{10c}}{c} = 36$
rearranging gives you: $10c^3 - 81c^2 + 180c -100 = 0$, since we are given that the sides are rational, we use the rational root theorem to find that $c = 2.5$, the plugging that back in, we get $a = b= 4$ so
$a + b+ c =  4 +4 + 2.5 = \boxed{10.5}$

[quote]10. Three rugs have a combined area of $700 m^2$.  By overlapping the rugs to cover a combined area of $440 m^2$, the area which is covered by exactly $2$ layers is $240 m^2$.  What area of floor is covered by all $3$ layers of rug?[/quote]

let the areas of the rug parts be $A_a$, $A_{ab}$, $A_{abc}$, etc
we know that $A_a + A_b + A_ c = 700$
we know that $A_{ab} + A_{bc} + A_{ac} = 440$
since the combined area is the sum of the individual areas minus the over counting of the areas shared by 2 and 3 rugs so $440 = (A_a + A_b + A_ c) - (A_{ab} + A_{bc} + A_{ac}) - A_{abc}$
so substituting $ 440 = 700 + 240 - A_{abc}$ so
the area shared by all 3 rugs is $A_{abc} = \boxed{500}$

For No. 7 :
"Two colorings are the same coloring if one can be rotated to get the other."
We can rotate it in more ways than just by the top vertex.

Consider it this way.  Let us call the four colors 1, 2, 3, and 4.  We can select the side that has color 1, and the side that has color 2 will always be touching it.  Putting these in a fixed position, there are only [b]2[/b] ways in which we can arrange the others :  with 3 on the left and 4 on the right, or with 4 on the left and 3 on the right.

For No. 8 :
Your formula is incorrect.

We will assume, for now that we wish to draw the median to leg 4.  (A proof will be provided tomorrow evening.)  Note that the triangle is obtuse at the angle opposite angle 9 (by extension of the Pythagorean Theorem, or Law of Cosines :  your choice).  Using Heron's, we can discover that the area of the triangle is $6\sqrt{5}$, so the altitude to 4 must be $3\sqrt{5}$.  Using the Pythagorean Theorem, we can discover that the distance to the near end of the segment (where it intersects with segment 7) is 2, and (concurrently) that the distance to the far end of the segment (where it intersects with 9) is 6.  Therefore, the distance to the intersection with the median is 4, and we can again use the Pythagorean Theorem to discover the length of the median, which is $\displaystyle \sqrt{61}$.

yes you are correct, i forgot to divide by 2 because the median would be half of the diagonal formed by the parallelogram


Find all real $x$ such that
$x^{8}+x-1=0$

[quote="Elemennop"]it's equilateral trinagle in the complex plane[/quote]
explain please


Find all a,b,c ,such that there is a function f satisfying the following . $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\plus{}ax^2 \plus{}bxy \plus{}cy^2$
a,b,c=?

__________________________________________
Azerbaijan Land of the Fire :huh:

[quote="Pirkuliyev Rovsen"]Find all a,b,c ,such that there is a function f satisfying the following . $ f(x \plus{} y) \equal{} f(x) \plus{} f(y) \plus{} ax^2 \plus{} bxy \plus{} cy^2$
a,b,c=?  [/quote]

$ y \equal{} 0$ $ \implies$ $ 0 \equal{} f(0) \plus{} ax^2$ $ \implies$ $ a \equal{} 0$
$ x \equal{} 0$ $ \implies$ $ 0 \equal{} f(0) \plus{} cy^2$ $ \implies$ $ c \equal{} 0$

And then $ f(x) \equal{} \frac {bx^2}2$ is such that $ f(x \plus{} y) \equal{} f(x) \plus{} f(y) \plus{} bxy$

Hence the solution : $ (a,b,c) \equal{} (0,b,0)$


geometry
pigeonhole principle
combinatorics proposed
combinatorics

Let $S$ be a square with the side length 20 and let $M$ be the set of points formed with the vertices of $S$ and another 1999 points lying inside $S$. Prove that there exists a triangle with vertices in $M$ and with area at most equal with $\frac 1{10}$.

[i]Yugoslavia[/i]

Let's count the number of disjoint triangles made in this way. For 1 point we have 4 triangles, and for every point we add 2 triangles. (regardless wether the point is on a line or inside a triangle)
So for $1999$ points, we have $4+2.1998=4000$ disjoint triangles, and a total area of $400$, so by pigeonhole there is at least one with area below $\frac1{10}$.

See also this thread:

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=points&t=52202

can anybody explain why counting like peter is valid.
and it's no need to use the 4 vertices right?
and what about 3 points lie in a line?

edit: I'm just get it from the link above.
thanks very much :-D

//so dumb question..

[b]Solution.[/b]
By induction, we can conclude if there are $n$ points [b]inside[/b] the triangle, the square can be triangulated into $2n+2$ triangles. So we can triangulate $S$ with $2*1999+2=4000$ triangles. The average area is $$\frac{\text{Area Of the Square}}{\text{Number of Triangles}}=\frac{400}{4000}=\frac{1}{10}$$