1. In how many ways can the letters of the word PERMUTE be arranged if (i) the first and last places are occupied by consonants, (ii) the vowels and consonants occupy alternate places.

2. Find the number of arrangements of the letter in the word PENCILS if (i) E is next to I, (ii) E precedes I, (iii) there are three letters between E and I.

3. How many arrangements can be made by the letters of word DEFINITION (i) if the letters I do not occupy the first or last place, (ii) if the letters I are together?

4. A car can hold 3 people in the front seat and 4 in the back seat. In how many ways can 7 people be seated in the car if 2 particular people must sit in the back seat and 1 particular person is the driver?

5. In how many ways can 5 men and 5 women by arranged in a circle so that the men are separated? In how many ways can this be done if two particular women must not be next to a particular men.

6. In how many ways can 8 oarsmen be seated in a eight-oared boat if 3 can row only on the stroke side and 3 can row only on the bow side?

2.
[hide]Not too good with combinations, but if ya don't try ya never get better I s'pose :P

i)  E is next to I

Well this can happen in two ways; either IE or EI. We can analyze the situation by calling IE one letter and EI one letter and observing the outcomes in both and adding them.

So if IE were one letter there would be $6!$ combinations for IE and $6!$ combinations for EI giving us a total of $\fbox{1440}.$

ii) well if E precedes I then now we only have 1 choice EI, giving us $\fbox{720}$ choices.

iii) If there are 3 letters between E and I, we can break this into two situations - one where we have E,x,x,x,I and another with I,x,x,x,E - remember we only have to count one of these situations because they both will end up the same.

So.. this Can happen lke
E,x,x,x,I,x,x
x,E,x,x,x,I,x
x,x,E,x,x,x,I

In $3$ ways. So multiply that by 2 and get $\fbox6$

Probably wrong, but whateva :P[/hide]

3. [hide]How many arrangements can be made by the letters of word DEFINITION (i) if the letters I do not occupy the first or last place, (ii) if the letters I are together?

(i) Ok, we can just count out the possible combinations letter by letter Note: (10 letters - 3 I's - 2 N's)

First and last letters only have 7 and 6 possibilities due to the fact that 3 Is can't be chosen, and also the numbers between the first and the last each have a descending 8 choices, because we've already used 2 up between the first and the last letters. Giving us

$7\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\cdot6$

We're not done because we have to account for the 3 I's and the 2 N's so we divide by $3!$ and $2!$

$\frac{7\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\cdot6}{3!2!} = \fbox{141,120}$

(ii) Hmm.. are the I's indistinguishable, because they could be together 6 ways.. well let's not deal with that because all I's are the same and who's really gonna know what order each of the I's are in :P

So.. let's call the new letter III as W :)

That means we now have 8 letters.. so we'd think it'd just be $8!$, right? Wrong! Forgot about the 2 N's
So we have $\frac{8!}{2!} = \fbox{20,160}$[/hide]

1.
[hide]
Not quite sure if this is right but..
i) Since there are 4 consonants, we know that there are 4 possibilities for the first slot and 3 for the last.  Since this uses 2 numbers, there are 5 left which will go in the middle 5 places.  We then have to divide by $2!$ since there are two Es:
$\frac{4(5*4*3*2*1)3}{2}$
$\frac{1440}{2}$
$720$

ii) Since the word will look like this: cvcvcvc, we can permute them separately.  So we have $4!3!$, but must again divide by 2 since there are two Es:
$\frac{4!3!}{2}$
$\frac{144}{2}$
$72$[/hide]

Btw, I get the same answer as you WindSlicer for 2.i and 2.ii, but in iii I'm not sure whether both E's have to be at least 3 letters away or just 1.

[hide="4"]
there are $4$ people left, and  can be arrange in $4!$ ways, the four people that seat in the back can be seated in $4!$ ways, the two people that seat in the front(other than the driver) can seat in $2!$ ways so the total is:
$4!4!2!=1152$ ;) [/hide]

For #6 where do the other 2 oarsmen sit? Do we just assume there's a place for them that doesn't involve calculating anything?

Tarquin, I'm not sure what you did there.

I think that the 6 oarsmen who must row on one specific side are fixed.

That gives: [hide] $4! * 4! * 2 = 1152$ [/hide]

Tarquin did 8 choose 3 b/c there are 8 oarsmen and 3 spots on one side. Once those are seated there are 5 oarsmen left and 3 seats on the other side giving us 5 choose 3.


2005年中国数学奥林匹克将于2005年1月20日至25日在河??郑州举行.

Good luck to all of you. And hope that the problems can be posted soon.

请??供2005年中国数学奥林匹克试题中文版,谢谢.

大家好啊..
我在AOPS 混了这么久?知?有中国人..感动..T.T

[quote="yinjiaqiu"]大家好啊..
我在AOPS 混了这么久?知?有中国人..感动..T.T[/quote]

er...?度的??了你。

[quote="qxy"]2005中国数学奥林匹克一等奖???

（共23人）

张宇 北京大学附中

王晨 甘肃兰州一中

罗晔 江西师大附中

陈昕韫 上海中学

康嘉引 广东深圳中学

殷?平 湖北?黄冈中学

?璇 湖北??师一附中

?晗生 上海?东师大二附中

景泰? 黑龙江?哈师大附中

田轶平 湖北??师一附中

钟?? 湖?长沙市一中

王崇? 浙江镇海中学

邓煜 广东深圳高级中学

孙若愚 湖?师范大学附中

谌昭 湖?长沙市雅礼中学

刘正伟 ?林东北师大附中

陈国庚 陕西?安康市张滩中学

张宇辰 天津耀?中学

王枫 安徽?安徽师大附中

郭城 黑龙江?哈师大附中

金龙 ?林东北师大附中

邵烜程 上海?旦大学附中

任庆春 天津耀?中学[/quote]

陕西?的那个学校我都没?说过，里?还有很熟悉的人??，希望?好几届都是一等奖，实力很强，看连今年中国队必有他了。

wpolly,?喜你得满分,?你好?在以?的考试中.

[quote="liyi"][quote="zhaobin"]wpolly,?喜你得满分,?你好?在以?的考试中.[/quote]
你英语说得太多了  :D[/quote]

是啊.... hehe~  :D

[quote="wpolly"][quote="Arwen"]希望?好几届都是一等奖，实力很强，看连今年中国队必有他了。[/quote]

Which one?[/quote]

你是王晨？我说的是?师大附中的那个?晗生，我?三?希望?培训题的时候看?他4次一等奖 （那时候他高二）:D 所以记忆比较深刻……说中国队有他?是预计一下，因为希望?是基础的比赛，他那么稳定的说明基础很好呀，所以……

[quote="wpolly"]他是去年???第二.....[/quote]

什么??？我怎么?明白……什么????呀？你认识他？

我认识他!
???是指集训队的

基本上是这样的， :)
你是高几的？
今年集训队在沈阳

[quote="zxingtan"]基本上是这样的， :)
你是高几的？
今年集训队在沈阳[/quote]

我是高一的，?赛考的?好（水平?够）……我周围的人最好的?绩还离?赛区一等差20分左?呢～在沈阳哦?怎么?系?加???呢？

这个我?太清楚,当年是?师帮我的.
?以?你们?师去?系

[quote="shobber"][quote="Arwen"][quote="zxingtan"]我认识他!
???是指集训队的[/quote]

那是?是说如果我?赛没有得?赛区一等，但是??想?加集训交点钱就?以了？如果这样的?岂?是……一定有什么?制?件??或者说我的?解有问题。[/quote]

?赛区一等?那得多少人哪!怎么也得上百???......[/quote]
得一等的大多是高三的,很少有高三的学生去?(应该几乎没有)况且?是高三的学生的一等?一定都去
但有些没得一等也去,?过多数人考虑的因素是1,路程;2,去了会?会影?课内
一般?说确实有近百人

能说说你们是怎么学数学的吗？


Hi all. Can someone please help me with this one question that is giving me difficulty? Thanks.

1. A typical "light dimmer" used to dim the stage lights in a theater consists of a variable inductor L (whose inductance is adjustable between zero and Lmax) connected in series with a lightbulb B as shown in the diagram. The electrical supply is 120 V (rms) at 60.0 Hz; the lightbulb is rated as 120 V, 1300 W. (a) What Lmax is required if the rate of energy dissipation in the lightbulb is to be varied by a factor of 5 from its upper limit of 1300 W? Assume that the resistance of the lightbulb is independent of its temperature. (b) What would be the maximum resistance of a variable resistor (adjustable between zero and Rmax) which can be used in place of an inductor?

Thanks.

All currents and voltages are rms values. The resistance of the light bulb in ohms is $R = \frac{V}{I} = \frac{V^2}{UI} = \frac{V^2}{P}$, where V = 120 V and P = 1300 W. If inductance $L_m$ is connected in series with the resistance R, then the resistive reactance is $\hat X_R = R$ and the inductive reactance is $\hat X_L = \text i \omega L_m$, where $\omega = 2 \pi f$ and f = 60 Hz. The impedance is

$\hat Z = \hat X_R + \hat X_L = R + \text i \omega L_m$

$Z = |\hat Z| = \sqrt{R^2 + \omega^2 L_m^2}$

The total rms current is now $I' = \frac V Z = \frac{V}{\sqrt{R^2 + \omega^2 L_m^2}}$, the rms voltage across the resistor (light bulb) is $V_R = IR = \frac{VR}{\sqrt{R^2 + \omega^2 L_m^2}}$, and the dissipated rms power in the resistor is

$P' = I'V_R = \frac{V}{\sqrt{R^2 + \omega^2 L_m^2}} \cdot \frac{VR}{\sqrt{R^2 + \omega^2 L_m^2}} = \frac{V^2R}{R^2 + \omega^2 L_m^2}$

No power is dissipated in the (ideal) inductor. This dissipated is supposed to be equal to one fifth of the dissipated power P without the inductor, so that

$P' = \frac{V^2R}{R^2 + \omega^2 L_m^2} = \frac P 5$

Calculating $L_m$ and substituting $R = \frac{V^2}{P}$ on the right,

$R^2 + \omega^2 L_m^2 = \frac{5V^2R}{P} = 5R^2$

$L_m^2 = \frac{4 R^2}{\omega^2},\ \ L_m = \frac{2R}{\omega}$

Substituting $R = \frac{V^2}{P}$ and $\omega = 2 \pi f$,

$L_m = \frac{2V^2}{2 \pi f P} = \frac{V^2}{\pi f P} \approx 0.059\ \text H = 59\ \text{mH}$

On the other hand, if another resistor$R_m$ is connected in series with the bulb, the total resistance $R + R_m$, the total current $I'' = \frac{V}{R + R_m}$, the voltage across the light bulb is $V_R = I''R = \frac{VR}{R + R_m}$ and power dissipated in the light bulb is $P'' = I''V_R = \frac{V^2R}{(R + R_m)^2}$. If this is equal to $\frac P 5$ as before, then

$P'' =\frac{V^2R}{(R + R_m)^2} = \frac P 5$

$(R + R_m)^2 = \frac{5V^2R}{P} = 5R^2,\ \ R + R_m = R \sqrt 5$

$R_m = R(\sqrt 5 - 1) = \frac{V^2(\sqrt 5 - 1)}{P} \approx 13.7\ \Omega$

Thank you so much yetti. Have a great day!

Kindly note that in the future, please explain what YOU couldn't understand about the question or couldnt do in it before expecting others to solve your problem fully. I hope you understand.

Moderator.

Please refer to my last post regarding this issue.


AMC AIME MATHCOUNTS USA(J)MO USAMO calculus geometry \/closed

So this was my last year at MathCounts (6th Place State of Washington), I'm going to highschool next year
I am looking towards the AIME and USAMO
I qualified for AIME this year but I failed at it, only like 3/15

Would These Classes be beneficial for me?
Intermediate Trig/Complex Numbers
Algebra 3
Calculus

I'm also thinking about taking the AMC 10 problem series and the AIME problem series B over summer. Would it be beneficial to also take the AMC 12 course and the AIME series A?

Also, a question about the books,
I don't plan to take the classes to relate these books (if applicable) so how much will these books help if I just buy them and study over summer by myself?

Intro to Geometry
Intro to Number Theory
Intermediate C&P
AOPS Vol. 2

Thanks in advance for the help.

I'd recommend you do most of your problem solving work at the more advanced AMC and AIME levels.  For classes, this means Intermediate classes and the AMC/AIME problem series.  (Calculus won't help for the AMCs, but if you want to very thoroughly learn calculus, it will be a good way to do so.)

If you will be doing a lot of self-study with AoPS books and taking some of the Intermediate classes, you probably don't need the AMC 10 Problem Series.  The AMC 12 and AIME Problem Series would probably be a better fit for you (again, assuming that you'll be shoring up the basics of number theory and geometry with the intro books as you suggest -- these books are specifically designed for students to study from on their own, so you should be fine doing so).

Algebra 3 and the Interm Trig class (which we'll likely rename to Precalculus) will help with harder AMC 12 problems, AIME problems, beginning steps on the olympiad, and will give you a very thorough algebraic foundation on which to build more advanced mathematical studies.

If you are able to spend a great deal of time on math over the summer, you may be in a position to benefit from WOOT, but you should probably wait until a little later in the summer to make that decision.


Problems: This forum alone has tens of thousands.  You can also go into the [url=http://www.artofproblemsolving.com/Wiki/index.php/Main_Page]AoPSWiki[/url] to find more.  You can click on the [url=http://www.artofproblemsolving.com/Forum/resources.php]Contests[/url] link for specific contests, and go into the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Articles.php]Resources pages[/url] for some free articles.

Classes: [url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]Click here[/url] for upcoming free classes.

Click around some more, there's tons of stuff here.

[quote="light23"]I could'nt find anything valuable in AoPS for free.  :(
Does anyone know what resources are available for free internationally?  :)  (i.e. problems, online classes, books, etc.)[/quote]
What exactly are you looking for? You can get everything you must pay for free if you play your cards right and ask the right questions.

[quote="Phelpedo"]What are you talking about PenguinIntegral?[/quote]
The nice people on the boards can answer almost any question and teach you very useful things. You just need to ask the right questions.

I think he's saying that if you study the mass resource of problems on your own, ask for help here when you need it, and read other people's guides/posts on concepts in this forum, you can get most of what you would have gotten from classes for free.


[quote="mathwizarddude"]Prove that for all positive real numbers $ a,b,c$ it holds true that

$ \frac1{a^3 \plus{} b^3 \plus{} abc} \plus{} \frac1{b^3 \plus{} c^3 \plus{} abc} \plus{} \frac1{c^3 \plus{} a^3 \plus{} abc} \le \frac1{abc}.$[/quote]

Even though I don't really like inequalities,I will post a solution:
[b]Solution:[/b]
Let's prove that
\[ a^3 \plus{} b^3\geq a^2b \plus{} b^2a
\]
Indeed,it is equivalent to $ a^2 \minus{} ab \plus{} b^2\geq ab$,which is obvious.
Therefore,if we use this inequality to each term,we will obtain the following inequality:
$ \sum_{a,b,c}\frac {1}{a^3 \plus{} b^3 \plus{} abc}\leq\sum_{a,b,c}\frac {1}{a^2b \plus{} b^2a \plus{} abc} \equal{} \frac {1}{abc}$.
$ \blacksquare$.


trigonometry
complex analysis
complex analysis unsolved

I am sure this is well-known but I been able to prove it.
Find all the point such that $ \sum_{n\equal{}0}^{\infty}z^n$ converges.

Uh - that's just a geometric series. It converges for $ |z|<1,$ and for nowhere on the boundary, since the terms do not go to zero at any point on the boundary.

[quote="Kent Merryfield"]Uh - that's just a geometric series. It converges for $ |z| < 1,$ and for nowhere on the boundary, since the terms do not go to zero at any point on the boundary.[/quote]
I did it in a painful way of writing $ \sum \sin (n)$ and $ \sum \cos (n)$ and showing the series aren't convergent. What you did actually gives an easier proof that the series are divergent, I didn't think of that.  :oops:


We'll take a break with the GBML series for now after this one (4weeks=1 month). Thus the deadline for the first four problem sets (Week 1 through Week 4) will be due on March 7th. The final score will be tallied up on the 8th and the top 3 scorers [b]may be [/b]presented with an award, no promises :wink:

Anyways, here it is!

1'.) Find the only whole number that equals 12 times the sum of its digits.
2.) Find the set of all x, xER, which are the solution for {x l (x+2)/(x-2)>5}.
3'.) Four positive numbers form a geometric sequence. The sum of these four numbers divided by the sum of the first two numbers is 37. If the first number is a, find the fourth number in terms of a.
4'.) Given points A(6, -2), B(t+1, -4), and C(t, 4) such that angle BAC is a right angle, find all possible values for t.
5'.) A fraction has the value 12/13. If the sum of the numerator and denominator is 2000, find the fraction's denominator.
6'.) Given N is a factor of 2002 and N has exactly 4 factors. How many possible values for N are there?
7'.) Ticket prices for the afternoon movie are $5.50 for adults, $4.50 for children and $4.00 for senior citizens. If 100 tickets were sold, the proceeds were $465, and more senior citizens than children attended, find the most numbers of children that could have been at the afternoon movie.

Zhu ni hao yun.

theone853 wrote:[hide]1) 108
3) 6^3 or 1*6^3???? What does express in terms of a mean?
4) 2, 9
5) 1040
6) 6[/hide]

Close enough for 3. The correct answer would've had "a" after the number you gave as the answer but again close enough. 5 points added and scores are updated.

[hide]1. it can't be 1-digit or 3 digit nless its 108 or smaller, and 108 works. (i was going to explain why 3 digit numbers don't work, but i guess they do).

2.{x l (x+2)/(x-2)>5}.  we have either x+2>5x-10 and x-2>0 or x+2<5x-10 and x-2<0. For the first set the answers are 2<x<3 and the second one is 3<x<2 which has no solutions so the set is 2<x<3.

3. a*r^3+a*r^2+ar+a=37(a+ar)  so ar^3+ar^2=36a+36ar we get r^3+r^2=36+36r, the only solution for r is 6. thus the answer is a*6^3.

4. we know that BC has length sqrt 65 if BAC is right. So we can make a graph, grahping A, y=4 and y=-4, then draw perpendicular lines from A to those lines and setting it up so AB^2+AC^2=65. The number that work for t are 9 and 2, one on either side of A (relating to the x-axis).

5. 12a/13a and 25a=2000, a=80, denominator is 1040

6. 2002=2*7*13*11, if N has 4 factors, then it has 1 as a factor, itself as a factor and 2 more in between, thus its a*b and a and b are from the set of 4 factors of 2002. we get 4*3/2=6 possibilities.

7. there must be less then 50 children obviously. next let's check 25 children. If there are 40 children, and 41 or more seniors, we don't get enough money. niether does 35 children get 465.  34 children is impossible to get 465.while with 33 children, no arrangment works. niether does any arrangment with 32 children.  but with 31 children, 36 seniors and 33 adults it results in 100 tickets and 465 dollars. so the answer is 31.[/hide]

Yup, both of you got no nothing nada wrong, good job! :) Scores are updated.

Reminder: March 7th is the deadline for Weeks 1-4

Current leaders' board:
[b]ml2[/b]: 30 points
topper: 21 points
theone853: 14 points

I wonder why no one is doing any of the past problems...hmmm...


geometry
3D geometry
calculus
integration

Solve WITHOUT graphing/squeeze methods.

4^x = x^3 - (7/3)x

Good luck.

Is there any algebraic solution?  B/c this is a transcendental equation...

Doh! I must have copied it down wrong, now where is that sheet of paper....

OK I can't find it. But I think you can use Cauchy's Integral Thereom...


limit
integration
calculus
real analysis
real analysis unsolved

Calcular
$ \mathop {\lim }\limits_{n \to \infty } \frac{1}
{n}\sum\limits_{k \equal{} 1}^n {\left( {e^{\left( {1 \plus{} \frac{k}
{n}} \right)^2 }  \minus{} \frac{{3e^{\left( {1 \plus{} \frac{{3k}}
{n}} \right)} }}
{{2\sqrt {1 \plus{} \frac{{3k}}
{n}} }}} \right)}$

The result is 0

$ \mathop {\lim }\limits_{n \to \infty } \frac {1} {n}\sum\limits_{k \equal{} 1}^n {\left( {e^{\left( {1 \plus{} \frac {k} {n}} \right)^2 } \minus{} \frac {{3e^{\left( {1 \plus{} \frac {{3k}} {n}} \right)} }} {{2\sqrt {1 \plus{} \frac {{3k}} {n}} }}} \right)} \equal{} \int_0^1 e^{(1 \plus{} x)^2}dx \minus{} {3\over 2}\int_0^1 {e^{1 \plus{} 3x}\over \sqrt {1 \plus{} 3x}}dx.$ $ t \equal{} \sqrt {1 \plus{} 3x}$ in the second integral yields
$ \int_0^1 e^{(1 \plus{} x)^2}dx \minus{} \int_1^2 e^{t^2}dt \equal{} 0$ by changing $ 1 \plus{} x \equal{} t$ in the first integral


First of all this is question is from USATMS October 13, 2009 round 1 not the current round. I was just looking over the questions and solutions and I didn't understand on how they got the solution for  the problem 5/1/21 A. The cubic equation x3+2x-1 = 0 has exactly one real root r. Note that 0.4 < r < 0.5.
(a) Find, with proof, an increasing sequence of positive integers a1 < a2 < a3 < ...... such  1/2= r^a1+r^a2+r^a3+....
I did not understand the solution. How did they get 1/2=r/1-r^3=r(r(1 + r3 + r6 + r9 + ... ) = r + r4 + r7 + r10 + .....
Thanks for your help.

The sum of an infinite geometric series can be found by the formula $ \frac {t_1}{1 \minus{} r}$, where $ r$ is the common ratio and $ t_1$ is the first term.

So if we have $ \frac {1}{2} \equal{} \frac {r}{1 \minus{} r^3}$, the right hand side is in the form of an infinite geometric series with first term $ r$ and common ratio $ r^3$, which produces the infinite series of terms $ r \plus{} r^3 \plus{} r^6 \plus{} r^9 \dots$. So we can rewrite the entire thing as $ \frac {1}{2} \equal{} r \plus{} r^3 \plus{} r^6 \plus{} r^9 \dots$

Whats important about the fact that $ .4<r<.5$ is that the ratio of the series cant be greater than 1 (think about it-can we find the sum of 1+2+4+8+....?). So by restricting $ |r|<1$ we can produce an infinite series which eventually converges.


function
modular arithmetic
algebra proposed
algebra

Let $ f(x)$ and $ g(x)$ be $ \mathbb Z\to\mathbb Z$ functions such that \[ f(m)+f(n)=g(m^2+n^2)\] and \[ -2\le f(n+2)-f(n)\le 2\] for all integers $m$ and $n$. What is the greatest possible cardinality of $ f(\mathbb Z$)?

[quote="MJ GEO"]Let $ f(x)$ and $ g(x)$ funxtions from $ \mathbb Z\to\mathbb Z$ such that $ f(m)\plus{}f(n)\equal{}g(m^2\plus{}n^2)$ and $ \minus{}2\le f(n\plus{}2)\minus{}f(n)\le 2$.
What is the greatest cardinal possible for $ f(\mathbb Z$)? [/quote]

I did not succed up to now to end this problem. Please, MJ GEO, could you give us your solution ?

What I did : at most $ 6$ elements. But I only succeed to find a three elements solutions (so it remains to show that this is the maximum or to find a solution with $ 4,5$ or $ 6$).

1) We have at most 6 elements :
$ f(ab\plus{}1)\plus{}f(a\minus{}b)\equal{}g(a^2b^2\plus{}a^2\plus{}b^2\plus{}1)$
$ f(ab\minus{}1)\plus{}f(a\plus{}b)\equal{}g(a^2b^2\plus{}a^2\plus{}b^2\plus{}1)$

So $ f(ab\plus{}1)\minus{}f(ab\minus{}1)\equal{}f(a\plus{}b)\minus{}f(a\minus{}b)$ and, since $ |LHS|\le 2$, we get that $ |f(a\plus{}b)\minus{}f(a\minus{}b)|\le 2$ and so $ |f(n\plus{}2p)\minus{}f(n)|\le 2$
So $ f(2p)$ can take at most $ 3$ values and $ f(2p\plus{}1)$ can also take at most $ 3$ values.

2) there exists solutions with $ 3$ elements :
$ f(4p)\equal{}0$
$ f(4p\plus{}2)\equal{}1$
$ f(2p\plus{}1)\equal{}3$

$ m^2\plus{}n^2\pmod{16}\in\{0,1,2,4,5,8,9,10,13\}$ and :

$ g(16p)\equal{}0$
$ g(16p\plus{}1)\equal{}3$
$ g(16p\plus{}2)\equal{}6$
$ g(16p\plus{}4)\equal{}1$
$ g(16p\plus{}5)\equal{}4$
$ g(16p\plus{}8)\equal{}2$
$ g(16p\plus{}9)\equal{}3$
$ g(16p\plus{}10)\equal{}6$
$ g(16p\plus{}13)\equal{}4$

[quote="MJ GEO"]i will post my solution 3 days later.[/quote]

 :rotfl:

[quote="FelixD"]What's so funny, pco?^^[/quote]

MJ GEO posted many many problems for which he says he has solutions ($ 22$ this week-end in algebra subforum)
MJ GEO is in a great hurry to have alternative solutions for these problems (he posts a up on these problems 15-30 mn after no answer and begs many times for answers : "[i]Please help me.I need the solution of this quastion today.its realy important for me[/i]")
MJ GEO received answers to a lot of these problems (unfortunetly not all of them up to now) in 1/2h-4h delay (I personaly gave solutions to 13-14 of them)

But when I ask a hint / solution for one of them, he answers : "3 days".
So we must give him a second solution in 15 mn but he'll give us some help in 3 days.

I found this very funny...

... and begin to think that the class during which the solutions to his homework will be given is next wednesday ... :)

[quote="MJ GEO, on Sun Dec 06, 2009 1:53 pm,"]i will post my solution 3 days later.[/quote]
Please, could you post your solution now.

Thanks.


Find an explicit homeomorphism between $ X$ and $ \mathbb{R}^2$, where $ X\equal{}\left\{(x,y)\in\mathbb{R}^2|1<x^2\plus{}y^2 <9 \;\mathrm{and}\;|x|>0 \;\mathrm{if}\;y>0 \right\}$.

How about this: First convert $ (x,y)$ to polar coordinates $ (r,\theta),$ with $ 1<r<3$ and $ \minus{}\frac{3\pi}2<\theta<\frac{\pi}2.$

Then send the ordered pair $ (r,\theta)$ to $ \left(\frac{r\minus{}2}{\sqrt{1\minus{}(r\minus{}2)^2}},\frac{\theta\plus{}\frac{\pi}2}{\sqrt{\pi^2\minus{}\left(\theta\plus{}\frac{\pi}2\right)^2}}\right).$

The answer is not unique, of course.

Note that it will not be possible to do this as a conformal map.

Thank you, professor Kent.

The solution the book suggests is $ h(x,y)\equal{}\left( f(\sqrt{x^2\plus{}y^2}\minus{}2) ,\frac{x}{\sqrt{x^2\plus{}y^2}\minus{}y}\right),$ where $ f(x)\equal{}\frac{x}{\sqrt{1\minus{}x^2}}$, which seemed a bit confusing to me.

After reading your solution I also understand this one.


probability
geometry
3D geometry
MATHCOUNTS
limit
search

What is the probability that if you multiply two randomly-selected whole numbers, the result is greater than 100 and less than 200? Express answer as common fraction.

Is it me, or is it impossible because the numbers can go infinitely high.

Yeah, is it 2-digit numbers, numbers that are between 10 and 20, what's the group of numbers?

Whole numbers implies that they are integers greater than or equal to 0.

We know, but the answer is still infinity, as there is no limit as to how high it can go.

$ \infty \times \infty \equal{} \mbox{Something I don't know}$

No it isn't. The answer is 0... There are simply infinite possible choices, but the limit ends up at 0.

The answer isn't zero, the answer is

x\(infinity)

where x is the amount that work, and infinity is the total number of pairs.

Wait so you're saying that $ \frac{x}{\infty}\equal{}0$...

I'm pretty sure that $ \frac{n}{\infty}$ isn't equal to zero and is in fact something else that's too complicated for this forum.

[quote="happyme"]The answer [b] isn't [/b] zero, the answer is

x\(infinity)

where x is the amount that work, and infinity is the total number of pairs.[/quote]

 :ninja:

This topic is a good example of middle schoolers being consumed by common intuition. Sit down still, forget about infinite, start from scratch and think for a moment...

Ok, now... still thinking that the event space has an infinite number of whole numbers? Repeat above process until you actually convince yourself it cannot be infinite.

Good enough? Ok here we go...

[hide="It is not infinite"]Two whole numbers $ a,b$ are given such that $ 100 < ab < 200$. Think about it: what happens when one of them is a large number. Get the idea? If not, try harder. Now, go find the probability if you do get the idea. :P :D[/hide]

But it would still be over $ \infty$ because there are an infinite amount of pairs that can be created using all positive integers.

Ok the problem is pointless if even one of the numbers is at least 200, so assume that the event space is all whole numbers less than 200. Now, calculate the probability.

Assume whole numbers here are positive integers greater than or equal to 0.

[quote="darkmatter47"]Actually, the total possibilities (ways to choose two numbers) IS infinite

therefore the answer is zero. Think of it this way.

n/infinity = 0. Think of it this way. A cube represents infinity,, and the "n" is a point. What is the probability that a randomly selected point is point n? Zero.[/quote]

I like pie too.

[quote="darkmatter47"]Actually, the total possibilities (ways to choose two numbers) IS infinite

therefore the answer is zero. Think of it this way.

n/infinity = 0. Think of it this way. A cube represents infinity,, and the "n" is a point. What is the probability that a randomly selected point is point n? Zero.[/quote]

Um, as I said, I'm pretty sure $ \frac{n}{\infty}\neq 0$.  There is the limit
[hide="not mathcounts level"]
$ \lim_{x \to \infty} \frac{n}{x}$
[/hide]
which is zero, but $ \frac{n}{\infty}$ isn't a real number and shouldn't be treated as such.

I'll go PM Jongao and ask about the problem, see what range he's talking about and where he got this problem.

[quote="lotsofmath"]Just to tell you, dividing anything by infinity is 0, so the answer would be 0.[/quote]
$ \frac{0}{\infty}$ isn't. :wink:

[quote="danb80"]To find the answer [b]you were looking for[/b] and not more discussion about infinity, go to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=971354&search_id=593621080#971354[/url]
where this problem has been previously discussed.
It sometimes helps to search for the problem on AoPS before posting it.[/quote]Thank you very much, danb80. That settles the fact that Jongao missed out the 'two-digit' condition that was on the original problem.

About infinity... it has been discussed in this forum tons of times...

Summary: infinity is not a number, it's a concept. Strictly speaking, you cannot say 'a number divided by infinity,' that is only as a reminder for your intuitionistic approach on how to approach approximating or 'taking limits' on certain expressions.

If it were a number, consider this: you say $ \frac{n}{\infty}\equal{}0$. But hey! $ \frac{n}{\infty}\equal{}\frac{n}{\infty\plus{}1}\equal{}0$. Wait a minute... we know $ \infty<\infty\plus{}1$ and so $ \frac{n}{\infty}>\frac{n}{\infty\plus{}1}\equal{}0$, contradiction??? No, I wanted to say that $ \infty\equal{}\infty\plus{}1$... but, oh no! we get $ 0\equal{}1$!!!!, another contradiction???

So, the lesson learned here is the fact that even though intuition helps a lot how you look at the problem but it doesn't mean that's what you are going to put down on your solution, formalism wins over intuition when it comes to writing down solutions.

Still believing that anything over infinity is zero? You better leave this topic and come back to this after a few weeks to save your sanity. :D

[quote="10000th User"][quote="danb80"]To find the answer [b]you were looking for[/b] and not more discussion about infinity, go to [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=971354&search_id=593621080#971354[/url]
where this problem has been previously discussed.
It sometimes helps to search for the problem on AoPS before posting it.[/quote]Thank you very much, danb80. That settles the fact that Jongao missed out the 'two-digit' condition that was on the original problem.

About infinity... it has been discussed in this forum tons of times...

Summary: infinity is not a number, it's a concept. Strictly speaking, you cannot say 'a number divided by infinity,' that is only as a reminder for your intuitionistic approach on how to approach approximating or 'taking limits' on certain expressions.

If it were a number, consider this: you say $ \frac {n}{\infty} \equal{} 0$. But hey! $ \frac {n}{\infty} \equal{} \frac {n}{\infty \plus{} 1} \equal{} 0$. Wait a minute... we know $ \infty < \infty \plus{} 1$ and so $ \frac {n}{\infty} > \frac {n}{\infty \plus{} 1} \equal{} 0$, contradiction??? No, I wanted to say that $ \infty \equal{} \infty \plus{} 1$... but, oh no! we get $ 0 \equal{} 1$!!!!, another contradiction???

So, the lesson learned here is the fact that even though intuition helps a lot how you look at the problem but it doesn't mean that's what you are going to put down on your solution, formalism wins over intuition when it comes to writing down solutions.

Still believing that anything over infinity is zero? You better leave this topic and come back to this after a few weeks to save your sanity. :D[/quote]

Hey, more pie lovers!


geometry
geometric transformation
homothety
linear algebra
matrix
linear algebra unsolved

If $A \in \mathcal{M}_n(\mathbb{R})$, denote by $C_A$ the subset of $M \in \mathcal{M}_n(\mathbb{R})$ such that $AM=MA$ and $M$ is similar to $A$ ([i]i.e.[/i] there exist a $P\in \mathrm{GL}_n(\mathbb{R})$ and $M=PAP^{-1}$).

Denote by $\mathcal{F}_n=\{ A \in \mathcal{M}_n(\mathbb{R}) ~|~ C_A \quad \mathrm{is} \quad \mathrm{finite}\}$.

Question: What are the elements of $\mathcal{F}_n$?

I posted it and noone answered. It is a problem given at written exam at ENS in 1976 and I guess it must be a theorem discovered in that period. Anyway, I find it totally impossible.

OK, but did you get anywhere with this problem?

Nope, as I said nothing worked with this terrible problem. And unfortunately, it seems that nobody has the correction of that ens exam.

The  result is (I think): A is an homothety or A has only simple eigenvalues (over C).
1)A is diagonalizable over C. If not A=D+N with DN=ND, D  a real diagonalizable matrix over C, N a real non zero nilpotent matrix.
Then for all non zero real k: D+kN is in C(A).
 2) Suppose that  A has at least 2 eigenvalues over C  AND     A has at least 1 mutiple eigenvalue:
 l is an eigenvalue of order>=2 and m is another eigenvalue.
a) l and m are real. We may suppose
A=[l]+[l]+[m]+V as a direct sum. If W is a 2,2 matrix s.t. spectrum(W)={l,m} then W+[l]+V is in C(A); but there exist an infinity of such matrices W.
b) l is not real. l=u+Iv where v<>0. We may suppose
A=U+U+V where U is the 2,2 matrix U=[u,-v,v,u] (line by line).
Let J be the 2,2 identity and W be the 4,4 matrix
W=[aJ,bJ,cJ,dJ] (with blocks line by line) where a,b,c,d are real, a+d=u+v and ad-bc=u^2+v^2.Then W+V is in C(A) and there exist an infinity of such matrices W (tensorial product).
c) l is real and m is not; m=u+Iv where v<>0.We may suppose that A=U+l.J+V where U,J are as in b). Let W the 2,2 matrix   W=[a,b,c,d]  where a,b,c,d are as in b).Then l.J+W+V is in C(A) and there exist an infinity of such matrices W.
3) -If A=tId then C(A)={A}.
    -If A has only simple eigenvalues then -over C-:
A=diag(l1,...ln); AB=BA implies B=diag(s1,...,sn) and C(A), over C ,contains at the most n! elements.Thus C(A) over R is finite.


hi, right now im a freshman in high school so I think I have room for improvement in math

I'm not all that bad, my highest SAT score was a 760, but math contests are a lot more challenging

I was wondering how I could get better at math contests such as AMC or AIME

any suggestions from those who have scored really well and have gone on to state/national competitions?

Welcome to the forums!

The best way to learn is to do problems and read books.

For books: Start with the Art of Problem Solving.  You can order them on http://www.artofproblemsolving.com .  If you completely master those, you should be able to make the AIME and get a few right there if you don't make stupid mistakes under pressure :D .  The Art and Craft of Problem Solving (by Paul Zeitz) is good for the AIME preparation.  For olympiads, you just need a whole lot of practice.  Get problem books (some nice ones are available at http://www.unl.edu/amc) .  If you can't solve a problem, study the solutions.  Ask for help on the forums ( :) ).

Start at the getting started (high school) forums and move up once you can do the majority of the problems on one board.

Order old AMC tests and look at [url=http://www.kalva.demon.co.uk]old AIME and olympiads from various countries.[/url]

And finally, practice practice practice practice practice practice !!!!!! :)

I'll definitely get that book, thanks a lot

You're welcome!  You'll find a lot of people asking the same question if you browse through the American Mathematical Competitions Forum: http://artofproblemsolving.com/Forum/index.php?f=133 .

Depending on how much you already know, you may not want AoPS volume 1. I suggest you get volume 2 first, and if you are lost or find that easier topics are evading you, buy volume 1. But if this is inconvenient, go ahead and get both; that suggestion is kind of strange, I know.

Well since the combined price is lower than getting each separately, I'd say go and get both. That way, you can fresh your memory with the stuff in Book 1 if you already know them.... it is not a bad thing to sometimes go and review what you already know, as you might not remember ;)

[quote="YroK"]I was wondering how I could get better at math contests such as AMC or AIME[/quote]Welcome to AoPS! :welcomeani:

I would recommend to you our [url=http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesP.php#amc10]AMC 10 Problem Series[/url] or [url=http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesP.php#amc12]AMC 12 Problem Series[/url] class.  Both of these classes will start in mid-Autumn 2005.  (It is also not too late to [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]enroll[/url] in this summer's AMC 10 class; the AMC 12 class is not being offered this summer.)  We also have a variety of classes on specific subjects.

Obviously we recommend our own [i]Art of Problem Solving[/i] books -- you can order them [url=http://www.artofproblemsolving.com/Books/AoPS_B_About.php]here[/url].  We also have a list of [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Books.php]other books[/url] that we recommend.

If you have more specific questions, please post them!  Our forum members will no doubt be happy to share their advice.

Well...I found that I got better at them after just doing them.  I took the AIME in 8th grade, and when i walked in and saw the first AIME problem of my life, I almost threw up.  I'd never seen something like that.  I ended up messing up a bit and managing a 4.  However, I then got all of the previous AIMEs and did many of them over the summer and some through the winter.  When I took it again this year, I made some careless mistakes as usual, but had improved to an 8!  Meanwhile, I decided to look at some USAMO/IMO problems and actually attempt some.  I had no idea how to approach them when I first tried in September, and felt just like I did with my first AIME problems.  However, I worked at them and struggled, and actually did many of them (not under the pressure of time of course heh).  You just have to find new ways of thinking, new ways of looking at problems and making connections, new approaches, and new rules.  Don't worry...you'll get the hang of it and you'll soon be getting 150s on the AMCs, 15s on the AIME and 42 on the USAMO/IMO...
Another tip...when you do the problems or try them, thouroughly read the solutions given there.  You can learn from most of them (at least I did).

[quote="K81o7"]Don't worry...you'll get the hang of it and you'll soon be getting 150s on the AMCs, 15s on the AIME and 42 on the USAMO/IMO...
[/quote]

We wish, right...

[quote="K81o7"]Another tip...when you do the problems or try them, thouroughly read the solutions given there. You can learn from most of them (at least I did).[/quote]

I totally second this.  Besides the AoPS books, you can also order some AIMEs and AMCs from the AMC here: http://www.unl.edu/amc  I think it only costs a dollar per copy.  Practice is key, key, key![/quote]

[quote="bubala"]you can also order some AIMEs and AMCs from the AMC here: http://www.unl.edu/amc  I think it only costs a dollar per copy. [/quote]

why spend money when you dont have to! We have a ton of Mock AMCs and AIMEs. They are located in the stickies of the AMC forum. And most importantly, they ARE FREE!

The real AMCs are much more realistic and helpful for practice. Plus, you're supporting the AMC - never a bad thing.


limit
logarithms
real analysis
real analysis unsolved

Are you allowed to use L'Hospital's Rules? http://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

Well.. I guess I could but we haven't learned it yet, so it will look odd...
It is part of a homework assignment from my first "Math Intro For Physicists" lesson.
All the other questions were solved with pretty basic algebra/trigonometry.

\begin{eqnarray*}\lim_{x \to 0} \frac{{e^{ - ax}  - e^{ - bx} }}{x} &=& \lim_{x \to 0} \frac{{\left( {e^{ - ax}  - 1} \right) - \left( {e^{ - bx}  - 1} \right)}}{x}\\&=& \lim_{x \to 0} \frac{{e^{ - ax}  - 1}}{x} - \lim_{x \to 0} \frac{{e^{ - bx}  - 1}}{x}\\&=&  - a\lim_{x \to 0} \frac{{e^{ - ax}  - 1}}{{ - ax}} + b\lim_{x \to 0} \frac{{e^{ - bx}  - 1}}{{ - bx}}\\&=& b - a\,\blacksquare\end{eqnarray*}

The second one is pretty similar, just apply $ a^x=e^{x\ln a},$ the same for $ b^x.$

thanks.

I thought it would be something like this, I even got to your second line, but couldn't see the multiplication by a or b...

thanks for #2 as well...


If $ a$ is an odd positive  integer greater than 2.....and $ n \in$ natural numbers
Prove that ${ a \mid \displaystyle\sum_{i = 1} }^{\ (a - 1)}i^{a^{n}}$

[hide]
We have $ k^{a^n} \plus{} (a \minus{} k)^{a^n}\equiv k^{a^n} \plus{} ( \minus{} k)^{a^n}\equiv 0\pmod a$ since $ a^n$ is odd. The result follows immediately, since
\[ 1^{a^n}\plus{}2^{a^n}\plus{}\dots\plus{}(a\minus{}1)^{a^n}\]can be paired into $ \frac{a\minus{}1}2$ terms of the above form.[/hide]


[i]Casenatico, May 6, 2005[/i]
Prove that among any 18 consecutive positive integers not exceeding n=2005 there is at least one divisible by the sum of its digits. Is this true for n=2006?

Among any 18 consecutive numbers there is one divisible by 18. Denote this number by x.
Now look at the sum of digits of x. It can be 9, 18, or 27. If it is 9 or 18, we are done.
So the only trouble may be if the sum of digits of x is 27. Among numbers from 1 to 2005 there is only one such number: x=1998. But it is easy to check that 1998 is a multiple of 27.
BTW, the next problematic number is 2898. So it is safe to let n=2897.


If anyone has the list of the top 57, could they please post it? Please post the name and the state, I wanna see which states had most people in the top 57.

I don't think they're posting them this year because of the... contamination of results... :|

I think the coaches got them though. So maybe frost13 or Jboyd would be kind enough to post them? Pretty Please?

They do have them. And also the results of how many people got each question right.

Yeah, like l Bob l was the only one on our team who got #22 right, I was the one who lasted the longest without getting a question wrong, I answered 3 of the last 4 questions right (27, 28, 29), everybody got #9 wrong, etc.

I'll ask my teacher for a copy of it and I'll post it here (she was the state coach).

texas had all four in the top 57 (actually all in the top 44) =D
i'm surprised that the results haven't been posted on mathcounts.org, it was up almost immediately last year...

Just too many results cramped on the site. Something like that...

i have the pdf file of the results...am i allowed to post it?

I would assume so. Why not? It's not copyrighted or anything.

[quote="mathcrazed"]I forced the MC people not to post it cause I failed so much.  Really... I dunno. Don't post it yet... unless you get all 228 competitors' permission lol.[/quote]

Dood, you didn't fail.  I did.  Just post the results.

[quote="jgjgjmath"]About a post like this - here are the written round seedings that I know

Texas ({1}, 32, 37, 44), New Jersey (19, 20, 34, 50), and California (22, 26, 33, 41) had all four members in the top 57.

Teams with three members in the top 57:
Maryland ({2},{4},{5})
Massachusetts (CD, 13, 43)
Indiana ({6}, 18, 52)
Washington ({3}, 39, 47)
Illinois (15, 48, 53)
North Carolina (21, 38, 40)
New York (29, 49, 55)

not sure about the Massachusetts seeding[/quote]

The Massachusetts countdown kid was 11th.

You can also now find the top 57 on the MC site at [url]http://mathcounts.org/NETCOMMUNITY/Document.Doc?id=221[/url]

[quote="1=2"][quote="mihail911"][quote="gauss1181"]Hey, but I thought you're a seventh grader right now, and you're going to eighth grade next year. (Or are you skipping grades?)[/quote]

1=2 went to seventh grade twice i believe. once homeschooled, and once in regular school or something like that.[/quote]

I have been homeschooled since the first grade. Owned.[/quote]
You do realize that now you can't compete again as an 8th grader, right?

[quote="diophantient"][/quote][quote="1=2"][quote="mihail911"][quote="gauss1181"]Hey, but I thought you're a seventh grader right now, and you're going to eighth grade next year. (Or are you skipping grades?)[/quote]

1=2 went to seventh grade twice i believe. once homeschooled, and once in regular school or something like that.[/quote]

I have been homeschooled since the first grade. Owned.[/quote][quote="diophantient"]
You do realize that now you can't compete again as an 8th grader, right?[/quote]

uh.... he already said he wasn't competing next year.

[quote="shtsxc12"][quote="frost13"]
You can also now find the top 57 on the MC site at [url]http://mathcounts.org/NETCOMMUNITY/Document.Doc?id=221[/url][/quote]

frost13, which menu navagations took you to this url, or you used search feature ?[/quote]

I just clicked on "2008 National Competition Winners" at the bottom left of the page.  Once you get to the results page where they only list top 12 click the link at the very bottom of the results to download the full list.

[quote="gauss1181"]He said he had many excuses about not feeling well and having a bloody nose :([/quote]
lol. this competition never happened... XD

[quote="tinytim"]Nothing ACTUALLY happened. MC nats was all a dream...[/quote]
... that happened to happen to several people at a time...

[quote="math154"]...in a fantasy place called REALITY...

edit: yes -0.5 rating points in 1 hour!!![/quote]
Well of course reality is a dream...  :P  then what are nightmares?  :P


How do you cross out bracketed term:

ie

\frac{(x-2)}{(x+2)(x-2)}

I want the (x-2) terms crossed out - like high-school elimination
a line through it starting bottom left and moving upwards
TIA

I want to be able to see a strike through the bracketed exr.
How do you do it in LaTeX???
I do know how to simplify!!!

$\not{x}$

$\not{xy}$

I don't know how to make the slash longer...

Use the cancel package (in a document though not on this forum) so you get for example
[img]http://sixthform.info/steve/wordpress/latexrender/pictures/e30b736d192e659406addc397b89b390.gif[/img]
The LaTeX code for this is [code]\displaystyle
\frac{32}{64}
=\frac{2\times4\times4}{4\times4\times4}
=\frac{\cancelto{1}{2}\times\cancel{4}\times\cancel{4}}{\cancelto{2}{4}\times\cancel{4}\times\cancel{4}}
=\frac{1}{2}[/code]

thxs Stevem,  I saw info on the cancel pakage over at tug.  I was looking for a way of doing the same without using a package.

Thanks anyway.  I think I need to reinstall tex stuff.

http://www.tex.ac.uk/cgi-bin/texfaq2html?label=cancellation

is the link I found

so do I
need
\usepackage{cancel}   ??????

Yes you do need \usepackage{cancel} though you need to check that the cancel package is installed in your system.


induction
combinatorics unsolved
combinatorics

Given a positive integer $ n\geq 2$, let $ B_{1}$, $ B_{2}$, ..., $ B_{n}$ denote $ n$ subsets of a set $ X$ such that each $ B_{i}$ contains exactly two elements. Find the minimum value of $ \left|X\right|$ such that for any such choice of subsets $ B_{1}$, $ B_{2}$, ..., $ B_{n}$, there exists a subset $ Y$ of $ X$ such that:
 (1) $ \left|Y\right| \equal{} n$;
 (2) $ \left|Y \cap B_{i}\right|\leq 1$ for every $ i\in\left\{1,2,...,n\right\}$.

With $ |X| \equal{} 2n \minus{} 2$, by choose $ B_1 \equal{} \{1,2\}, B_2 \equal{} \{3,4\},..,B_{n \minus{} 1} \equal{} \{2n \minus{} 3,2n \minus{} 2\},B_n \equal{} \{1,3\}$ then not exist $ Y$ satisfy $ |Y| \equal{} n$ and $ |Y \cap B_i| \leq 1$ for all $ i \equal{} 1...n$

So $ |X| \geq 2n \minus{} 1$. We will prove that $ \min|X| \equal{} 2n \minus{} 1$

Indeed, choose subset $ Y$ such that $ |Y|$ max and $ |Y \cap B_i| \leq 1$ for all $ i \equal{} 1...n$

If $ |Y| \leq n \minus{} 1$ so $ |X \minus{} Y| \geq n$. Because  with $ a \in (X \minus{} Y)$, exist $ 1 \leq t(a) \leq n$ and $ x_a \in Y$ satisfy $ (a,x_a) \in B_{t(a)}$ ( if not we can add $ a$ in $ Y$ and get $ |Y'| > |Y|$)

Then if $ a \neq b$ so $ B_{t(a)} \neq B_{t(b)}$. Other hand $ |X \minus{} Y| \geq n$ so we have at least $ n$ subset $ B_{t(a)}$ different. So, $ |X \minus{} Y| \equal{} n$

And we can see subset $ X \minus{} Y$ satisfy $ |(X \minus{} Y) \cap B_i|\equal{}1$ for all $ i \equal{} 1..n$ and $ |X \minus{} Y| \equal{} n$ (absurd)

So exist $ Y$ satisfy condition of problem

General, if $ |B_i|\equal{}m$ for all $ i\equal{}1..n$ ,we will get the result  $ \min |X|\equal{}mn\minus{}1$. It's easy to prove  :)

I think you mean $ \min \left|X\right| \equal{} m\left(n \minus{} 1\right) \plus{} 1$, right?

And in your solution, I don't see why $ a\neq b$ should imply $ B_{t(a)}\neq B_{t(b)}$. We could have $ x_a\equal{}x_b$...

  darij

[quote="darij grinberg"]I think you mean $ \min \left|X\right| \equal{} m\left(n \minus{} 1\right) \plus{} 1$, right?

And in your solution, I don't see why $ a\neq b$ should imply $ B_{t(a)}\neq B_{t(b)}$. We could have $ x_a \equal{} x_b$...

  darij[/quote]

Yes  :blush: .$ \min |X|\equal{}(n\minus{}1)m\plus{}1$

If $ a\neq b$ so $ B_{t(a)} \equal{} \{a,x_a\},B_{t(b)} \equal{} \{b,x_b\}$. Because $ a,b \in X \minus{} Y, x_a,x_b \in Y$ so $ B_{t(a)} \neq B_{t(b)}$. What's rong?? :lol:

[quote="tanlsth"]Because $ a,b \in X \minus{} Y, x_a,x_b \in Y$ so $ B_{t(a)} \neq B_{t(b)}$.[/quote]

How do you draw this conclusion?

  darij

tanlsth, can you please try to fix your solution? Because mine is 3 pages long, after it has undergone a lot of error fixing (apparently it is very easy to make mistakes in this problem), and I still am not as sure of its validity as I usually am when I write down a detailed (unreadable) proof. You can find my proof at http://www.cip.ifi.lmu.de/~grinberg/subsetcond.pdf ; here are its cornerstones:

[color=blue][b]Theorem 1.[/b] Let $ X$ be a set. Let $ n$ and $ m\geq1$ be two nonnegative integers such that $ \left\vert X\right\vert\geq m\left( n \minus{} 1\right) \plus{} 1$. Let $ B_{1}$, $ B_{2}$, ..., $ B_{n}$ be $ n$ subsets of $ X$ such that $ \left\vert B_{i}\right\vert \leq m$ for every $ i\in\left\{1,2,...,n\right\}$. Then, there exists a subset $ Y$ of $ X$ such that $ \left\vert Y\right\vert \equal{} n$ and $ \left\vert Y\cap B_{i}\right\vert \leq1$ for every $ i\in\left\{ 1,2,...,n\right\}$.[/color]

[i]Sketch of proof.[/i] Case 1 (when the union of the $ B_i$ covers all of $ X$) is very easy (by a counting argument, every set $ B_i$ contains an element which is not contained in any other $ B_j$, so you get your $ Y$), while Case 2 (when there is some $ x\in X$ which does not lie in any of the $ B_i$) is the actual argument, which mainly consists of replacing every $ B_i$ by

$ B_i\setminus \left(B_{n \plus{} 1}\setminus\left\{\text{some fixed element of }B_{n \plus{} 1}\right\}\right)\setminus\left\{x\right\}$

and applying induction over $ n$ (noticing that $ B_{n \plus{} 1}$ itself collapses to a one-element set, and one-element sets can be ignored).

Ah, and of course, it is obvious that $ m\left(n \minus{} 1\right) \plus{} 1$ is indeed the minimum (for $ m\left(n \minus{} 1\right)$, a counterexample can be easily found).

[b]PS.[/b] tanlsth, please tell me your real name in case you want it to be mentioned in the PDF.

  darij


function
complex numbers
complex analysis

I couldn't find a good explanation for the equation below. Is the explanation beyond high school mathematics?

[img]http://img46.imageshack.us/img46/8403/complexnumberdn4.jpg[/img]

Same problem but more elegant version of it:

    [img]http://img118.imageshack.us/img118/2110/d6dec9e9817c270d7f1f703tf3.gif[/img]

$ \sqrt{ab}\equal{}\sqrt{a}*\sqrt{b}$ is not necessarily true.

I see that. But what are the conditions your equation (not) holds (ovbiously not hold for complex numbers but are there other examples?).. Why shouldn't it not hold for complex numbers?

As Altheman said $ \sqrt{ab} \equal{} \sqrt{a} \cdot \sqrt(b)$ is not always true, either $ a$ or $ b$ must be positive.

The error is in the second line. Every square root of a NEGATIVE REAL or COMPLEX number has two possible values (as opposed to square roots of non-negative reals, which are taken to be non-negative), and when you wrote $ \sqrt { \minus{} 4}\cdot\sqrt { \minus{} 9} \equal{} 2i\cdot 3i$, you chose a wrong one. If you make $ 2i$ into $ \minus{} 2i$ or $ 3i$ into $ \minus{} 3i$, everything is ok.

The same with the second example: $ \sqrt { \minus{} 2}\cdot\sqrt { \minus{} 2} \equal{} (i\sqrt{2})\cdot(\minus{}i\sqrt{2})$

[quote="Farenhajt"]The error is in the second line. Every square root of a NEGATIVE REAL or COMPLEX number has two possible values (as opposed to square roots of non-negative reals, which are taken to be non-negative), and when you wrote $ \sqrt { \minus{} 4}\cdot\sqrt { \minus{} 9} \equal{} 2i\cdot 3i$, you chose a wrong one. If you make $ 2i$ into $ \minus{} 2i$ or $ 3i$ into $ \minus{} 3i$, everything is ok.

The same with the second example: $ \sqrt { \minus{} 2}\cdot\sqrt { \minus{} 2} \equal{} (i\sqrt {2})\cdot( \minus{} i\sqrt {2})$[/quote]

A get a bit confused. First of all I feel(!) that the first line of the equation should be true. (Also based on the comments of AstroPhys and Altheman second line should be wrong.) It is very logical that  $ \sqrt { \minus{} 4}$ has two roots $ 2i$ and $ \minus{} 2i$. How do i know which root to choose? I have to choose one negative and one positive?

$ \sqrt { ab}\equal{} \sqrt {a}\cdot\sqrt {b}$
Is there a proof that this is not hold for complex numbers?

The general question of which of multiple values of a multivalued function is a question of [b]complex analysis[/b], which is too complicated to get into here.

[quote="gonulate"]$ \sqrt { ab} \equal{} \sqrt {a}\cdot\sqrt {b}$
Is there a proof that this is not hold for complex numbers?[/quote]

What you need to grasp is this: if $ a,b$ are non-negative reals, then notation $ \sqrt{a},\sqrt{b}$ represents a unique value, by convention. But if they are negative or complex, those represent SETS of values, with no value having precedence over any other. Therefore, the most you can tell is this: set of values for $ \sqrt{ab}$ is the same as the set of values for $ \sqrt{a}\cdot\sqrt{b}$


calculus
integration
number theory solved
number theory

Solve the following diophantine:
x^2+x-9=y^4  (x,y are integrals)


Taste... is a matter of taste  :D

BTW, Orl, I see you have started to practice real British humor...  :D

  dg


inequalities
logarithms
number theory open
number theory

it is known that:

(1)   $ p_n < 36n\ln n$ for all $ n > 1$

(2)   $ p_n < 2n\ln n$ for sufficiently large $ n$.

Is it true that $ p_n\le 2 \plus{} 2n\ln  n$ for all $ n$ ?

There are numerous results similar to those you stated.  For example,
$ p_n \le n \log n \plus{} n(\log \log n \minus{} 0.9385)$ for $ n\ge 7022$. (cf. An Upper Bound on the nth Prime by John H. Jaroma)
Thus, $ p_n \le 2 \plus{} 2n\log n$ as you desired for $ n\ge 7022$.
Also, there are better results (with smaller $ n$) available.  If I remember correctly, Rosser and Schoenfeld showed something like
$ p_n \le n(\log n \plus{} \log \log n)$ for $ n\ge 30$.
So, yes this should be possible to show.


geometry
ARML
trigonometry
geometric transformation
reflection
congruent triangles
similar triangles

I am really having trouble on this problem:

Write a two column proof: The sum of the lengths of the segments drawn from any point in the base of an isosceles triangle perpendicular to the elgs is equal to the length of the altitude drawn to one leg.

I have no idea how to approach this problem.  I need the answer ASAP!!!!

thanks, math92

[hide="Hint"]
Divide the isosceles triangle into two smaller triangles, and express their areas in terms of the length of the leg and the perpendicular segments ;)

BTW, [url=http://www.nctm.org/high/challenge/oct05chall.htm]NCTM-ARML challenge problem #3[/url] uses the same idea.
(Don't look at the link before you figure it out... It has a solutoin that pretty much explains everything.)[/hide]

I'm not sure what you mean by a two column proof, but here's how you should approach it.  Usually when you're given a problem with altitudes and perpendicular lines within triangles, you can deal with it by area.  Try drawing it out and labeling sides and segments.  Adding a line or dividing a triangle often helps.   :)

[hide="solution if you give up"]
Label the triangle ABC.

$\overline{AB} = \overline{BC} = a$

Call the altitude $h$.  The arbitrary point you pick on the base can be called $D$.  The area of the triangle is then:

$A = \frac{ah}{2}$

The key is to see another way to represent the area.  Draw in $\overline{BD}$.  Now we have two triangles.  The two perpendiculars drawn from $D$ to the legs will be called $p_1$ and $p_2$.  So:

$A = \frac{ap_1}{2} + \frac{ap_2}{2} = \frac{a(p_1 + p_2)}{2}$

$\frac{ah}{2} = \frac{a(p_1 + p_2)}{2} \rightarrow h = p_1 + p_2$
[/hide]

can anyone give me a simpler solution....i doubt that's what my teacher is looking for becuase we are learning basic Euclidian geometry.  Its a nice solution, but i need a simpler one. :)

It is possible to prove it using "simple" concepts (congruent triangles, angles, etc.), but the proof becomes much longer. So I guess that is the quickest proof...

It's fine to use his solution. It's not like it contains any trig or anything.  :P

I used more advanced topics when I was in geometry and my teacher welcomed it, so unless your teacher is how a math teacher should be, don't worry about it. :)

[quote="math92"]can someone give the simpler proof?[/quote]

No offense but this is a "homework" problem. Normally, I would've locked the topic right away but since this was a proof, I let it go for a while.

If you see above, frt gave you hint and dakyru kindly posted the solution (bit complicated, you said). But if you see the time, you asked for the proof 4 minutes after dakyru posted his solution. Here's a nice idea: Why don't you try for more time and post the progress instead?

We can't help you on every single step, especially a homework problem. The main reason that the moderators block homeworks is that homeworks are for students to develop their problem solving skills outside the classroom. Although geometry proof is a tough topic for beginning people, you need to try harder.

Please repost only if you made a progresses. I never said you didn't try. But try again and post what you found. That way, we can help you.

experimentation gives some cool outputs.
[hide]
reflect one leg and its perpendicular segment about the base. Then you get one solid segment which is just the length of $p_1$ and $p_2$ added, which, if you shift it, is just the altitude...
[/hide]
heh.. this could be one of the proofs without words, although a more rigorous one would use the similar triangles argument..

How do you know that $p_1$ reflected will be collinear with $p_2$??

Isosceles triangle.  The two smaller triangles are similar, and there are vertical angles formed from the reflection.

this is quite simple... think of the similiar triangles that are formed by the right angles and the base angles of the triangle

let the base be AB, and the other vertex is C, let N be on AB

let AN=x, the perpendicular from N to AC=a

let BN=y, the perpendicular from N to BC=b

let the perpendicular from B to AC=c

notice the similiar triangles formed by the base angles A and B, and the right angles, and this gives us

$\frac{x+y}{c}=\frac{x}{a}=\frac{y}{b}=\frac{x+y}{a+b}$

$c=a+b$

(note that the last step was using the proportion property if $\frac{a}{b}=\frac{c}{d}$, then $\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}$)

[quote="math92"]i finally understood dakryu's solution.  And the link helped me alot, and i was able to come up with a simpler answer!  :)[/quote]

 :first:

Way to go!


$ a_n$ is an infinite sequence such that $ \frac{a_{n\plus{}1} \minus{} a_n}{2}$ tends to zero. Show that $ a_n$ tends to zero.

That’s not true, is it? For example, $ a_n \equal{} 1 \minus{} \frac {1}{n}$. :maybe:

No, something's really wrong about this statement (what if $ a_n$ is the harmonic series?)

...or $ a_n$ = 1 for all positive integers n?


dar mosalase $ABC$ agar noghat $M,N$ mozdavaj bashand sabet konid:
$\frac{AM.AN}{AB.AC}+\frac{BM.BN}{BA.BC}+\frac{CN.CM}{CA.CB}=1$

agar nesbat be noghteye $A$ enekaase makoos bezani, masale enghadr khoshgel hal mishe ke baavar kardanesh sakhte... :)


Let

$ x(4 \minus{} 3x) \plus{} y(4 \minus{} 3y) \equal{} 3xy$

Prove that

$ 0 \le x\plus{}y \le \frac{16}{9}$

Notice that for every arbitary reals $ x,y$ we have $ x^2\plus{}y^2\plus{}xy\geq 0$

Because $ x^2\plus{}y^2\plus{}xy\equal{}(x\plus{}\frac{1}{2}.y)^2\plus{}\frac{3}{4}y^2$  :wink:


A box contains 11 balls, numbered 1,2,3,....,11. If 6 balls are drawn simultaneously at random, what is the probability that the sum of the numbers on the balls drawn is odd?

A. $\frac{100}{231}$
B. $\frac{115}{231}$
C. $\frac{1}{2}$
D. $\frac{118}{231}$
E. $\frac{6}{11}$

The number of odd numbers drawn can only be one,three or five.There are five even numbers and six odd numbers.
So the posilibity is:
(1*6+10*20+5*6)/((11*10*9*8*7*6)/(1*2*3*4*5*6))=118/231

[hide]odd sum must be even+odd.  There are six odd numbers and 5 even numbers so the numerator is 6x5. There are 11 choose 2 different sums. So the probability of having an odd sum is $\frac{6*5}{55}=\frac{6}{11}=E$[/hide]

[hide]even numbers: $2, 4, 6, 8, 10$

odd numbers: $1, 3, 5, 7, 9, 11$

To get an odd sum we must have: $ \{ \begin{array}{lll} 1 \textrm{\,odd and\,} 5 \textrm{\,even} \\ 3 \textrm{\,odd and\,} 3 \textrm{\,even} \\ 5 \textrm{\,odd and\,} 1 \textrm{\,even} \end{array}$

Therefore our probability is $\frac{ \displaystyle{ {6 \choose 1} \cdot {5 \choose 5} + {6 \choose 3} \cdot {5\choose 3} + {6\choose 5} \cdot {5\choose 1}}} {\displaystyle {{11 \choose 6}}}=\frac{118}{231}$

Your answer is right [b]qmdtc[/b].[/hide]

qmdtc, please hide your answer for those of us who want to solve it on our own.

[hide="Answer"]Since two odds will sum to an even, we must have an odd number of odds. Since there are only five evens, there are six combinations with only one odd, $\binom{5}{3}\cdot \binom{6}{3}=200$ combinations with three odds, and $\binom{5}{1}\cdot \binom{6}{5}=30$ combinations with five odds, meaning a total of $236$ possible combinations. The number of total combinations is $\binom{11}{6}=462$, so our answer is $\frac{236}{462}=\frac{118}{231}\Rightarrow \boxed{D}$.[/hide]


Em có download mấy cái file có đuôi là ps,dvi nhưng không mở ra được.Xin mấy anh chị cho em biết dùng phần mềm nào để mở ra và cho em link download.Cảm ơn nhiều.


Find all positive primes $ p$ such that

$ 8p^4 \minus{} 3003$ is also a prime number.

*200th post yay!

[hide]For $ p\geqslant 7$ we can write $ p \equal{} 30k\pm 1,\pm 7,\pm 11,\pm 13$, hence $ p^2\equiv 1, \minus{} 11,1, \minus{} 11\pmod{30}$ and $ p^4\equiv 1\pmod{30}$. So we get $ 8p^4 \minus{} 3003\equiv 5\pmod{30}$, hence can't be prime.

So we must check $ 2,3,5$ - $ 2$ is too small, $ 3$ obviously doesn't satisfy, and for $ p \equal{} 5$ we get $ 8p^4 \minus{} 3003 \equal{} 1997$, which is prime. Hence the only such number is $ p \equal{} 5$.

NOTE: If "negative primes" are allowed, then $ p \equal{} 2$ is also a solution, since $ 8p^4 \minus{} 3003 \equal{} \minus{} 2971$ would also be "prime".[/hide]

[hide="another solution"]
Suppose that $ 8p^4\minus{}3003$ is prime, with $ p$ a prime.

$ 8p^4\minus{}3003\equiv 3(p^4\minus{}1)\mod 5$

If $ p\ne 5$ then $ (p,5)\equal{}1$ and, by Fermat's little theorem, $ 5|8p^4\minus{}3003$. Then $ 8p^4\minus{}3003\equal{}5$ or equivalently $ p^4\equal{}376$, a contradiction.

So $ p\equal{}5$, then $ 8p^4\minus{}3003\equal{}1997$.

Since $ 1997$ is a prime, we conclude that $ p\equal{}5$ is the only solution to the problem.[/hide]

:)


group theory
abstract algebra
combinatorics unsolved
combinatorics

Suppose G(V,E) is a simple undirected graph where all the nodes have degree higher then 2 (3 or more).
F is a subgroup of V where any circle in G must pass through at least one node of F.
S is the rest of the nodes = V-F.
denote x - number of edges in  the graph that the nodes of S span
z - number of edges in G
Prove that  2x<z

The graph spanned by $S$ (which will also be denoted by $S$) is a cycle-less graph, so it's a forest. We may as well assume it's connected, and thus a tree, because then we can reason in a similar fashion for each connected component of $S$.

Basicly, we have to prove this: given a tree $S$ on $n$ vertices, if $1_S$ is the number of vertices of $S$ with degree $1$, and $2_S$ is the number of vertices of $S$ with degree $2$, then $2\cdot 1_S+2_S\ge n\ (*)$. This is because we want to prove that the LHS in $(*)$ is larger than the number of edges of $S$, and this number is $n-1$, since $S$ is a tree on $n$ vertices. We imagine that we add two edges for each vertex with degree $1$, and $1$ edge for each vertex with degree $2$, and these added edges, as they will be called, have their other endpoints somewhere outside our tree.

This can be done by induction. Assume we've shown it for all trees with at most $n-1$ vertices, and we have to prove it for a tree $S$ with $n$ vertices. Remove a terminal vertex from $S$ (and the edge adjacent to it). The resulting tree $S'$ has $n-2$ vertices, and $2\cdot 1_{S'}+2_{S'}\ge n-1$. By adding a vertex and an edge to form $S$, we have modified the degree of one vertex, the one which will be adjacent to the newly added terminal vertex: if the old vertex had degree $1$, then it accounted for $2$ added edges, and now it has degree $2$, so it accounts for only $1$ added edge now, while the newly added vertex accounts for $2$ added edges, so, in total, we have removed $2$ added edges, and we have added $3$, meaning that the number of added edges has increased by $1$, as desired.  Pretty much the same happens in the other cases (when the vertex whose degree we change previously had degree $2$ or $>2$).

[quote="Valentin Vornicu"]By circle I assume you mean cycle in fact, and by x do you mean the sum of the degrees of the vertices in S, or the number of edges between points from S?[/quote]
x is the number of edges between the  points from S.


calculus
integration
algebra
function
domain
Ring Theory
topology

Let $ R$ be an integral domain and let $ Q\supseteq R$ be the field of quotients.  If $ \sigma: R\to R$ is an automorphism, show that there is a unique automorphism $ \bar{\sigma}: Q\to Q$ that satisfies $ \bar{\sigma}(r)\equal{}\sigma(r)$ for all $ r\in R$.

I was thinking of perhaps defining $ \bar{\sigma}(q)\equal{}\sigma(q)$ for $ q\in R$, but is there a way to extend this so that it is an automorphism onto $ Q$?  And how to show uniqueness?

$ \bar{\sigma}(r)\equal{}\sigma(r)$  Implies  $ \bar{\sigma}(1/r)\equal{}1/\sigma(r)$ as they are automorphisms.  So it follows that  $ \bar{\sigma}(a/b)\equal{} \sigma(a)/\sigma(b)$

Necessarily, $ \overline{\sigma}(a/b) \overline{\sigma}(b) \equal{} \overline{\sigma}((a/b)b) \equal{} \overline{\sigma}(a)$. Therefore $ \overline{\sigma}(a/b) \equal{} \overline{\sigma}(a) / \overline{\sigma}(b) \equal{} \sigma(a) / \sigma(b)$, necessarily.

In general $ Q$ obeys the universal property that if $ \alpha : R \to F$ is an injective ring homomorphism, and $ F$ is a field, then there is a unique ring homomorphism $ \beta : Q \to F$ such that $ \alpha \equal{} \beta \iota$, where $ \iota : R \to Q$ is the inclusion map. (This is a property of localizations even more generally).

If you have this lemma, then the general nonsense argument would be: $ \iota \sigma : R \to Q$ has a unique extension $ \beta : Q \to Q$ such that $ \beta \iota \equal{} \iota \sigma$. We wish to show $ \beta$ is an automorphism. However we also have a unique morphism $ \gamma : Q \to Q$ such that $ \gamma \iota \equal{} \iota \sigma^{ \minus{} 1}$. Then $ \gamma \beta \iota \equal{} \gamma \iota \sigma \equal{} \iota \sigma^{ \minus{} 1} \sigma \equal{} \iota$. Therefore $ \gamma \beta$ is the unique endomorphism of $ Q$ extending the identity map $ R \to R$, and so $ \gamma \beta \equal{} 1$. Similarly $ \beta \gamma \equal{} 1$, so $ \beta$ is an isomorphism.

I don't see that result in the textbook we're using.  Do you happen to know if it's in Dummit and Foote? (we're not using that book but I still have it).

And thanks for the proof.

If you are looking for the result; you will probably find it in the following form:

If $ A,B$ are (commutative, w/ identity) rings, and $ S \subset A$ is a multiplicatively closed set, then a morphism $ \varphi : A \to B$ has a unique extension to a morphism $ S^{ \minus{} 1} A \to B$ whenever $ \varphi(S) \subset B^\times$. (Here $ B^\times$ denotes the group of units of $ B$).

So what this says is that localization is the universal example of a ring where the elements of $ S$ are invertible.

The quotient field of a domain is the localization at $ S \equal{} A \setminus \{0\}$.

I've always seen this result mentioned very soon after localization is defined in any commutative algebra text.

Yes, it looks like D&F covers that result, or at least something very similar to it in chapters 7 and 15.  Thanks.


See this thread for the rules. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151259[/url]

Players for this game:

1. Ignite168
2. mustafa
3. Treething
4. perfect628
5.
6.

Since I'm #1 on the list I guess I'll start

G

So if Treething doesn't go within 24 hours, he gets a "G" and the next person goes?

oui the stupid The message is too small. Please make the message longer before submitting thing is SO ANNOYING!

shoulda gone the GRO[b]P[/b] route
oh well

he should be at least

what are the odds he doesnt show up

It has now been 24 hours since the last letter was put down, and so Herald of doom gets a G and it's lingomaniac's turn, I believe.

Well, if HoD forfeits his turn...

GROU[b]N[/b]

[quote="probability1.01"]Maybe you should give Herald of Doom a second chance... I heard he tried really hard to participate but couldn't.[/quote]

What do you mean? Has he been banned yet?


function
real analysis
real analysis unsolved

so lets get $ f \in K([a,b]), f([a,b]) \subset [c,d] , g \in BV([c,d])$

and try to have $ g(f(x)) \in BV ([a,b])$

what kind of set do we have to restrict $ K([a,b])$ to be?

for instance if its the set of monotone functions, then do we get the desired result....?

if we relieve that to the set of bounded variation functions... does it still work?

edit: even my easy example was wrong =(

For suppose that $ g$ is Lipshtz continuous on [a,b], then there exists a constant $ C$ such that $ \mid g(x) \minus{} g(y) \mid \leq C*\mid x \minus{} y \mid$ for all $ x,y \in [a,b]$ then for any partition of $ [a.b]$ we have that $ \sup_P  \sum_{i\equal{}1}^n \mid g(f(x_{i\plus{}1})) \minus{} g(f(x_i)) \mid \leq$ $ \sup_P  C* \sum_{i\equal{}1}^n \mid f(x_{i\plus{}1}) \minus{} f(x_i) \mid < \infty$


linear algebra
matrix
invariant
linear algebra unsolved

Let $ A=(a_{ij})\in M_{n}(\mathbb{R})$ such that $ a_{ij}=i+j\forall i,j=\overline{1,n}$. Find $ \text{rank}(A)$.

For $ n=1$ we have $ \text{rank}(A)=1$ and for $ n\geq2$ we have $ \text{rank}(A)=2$.

The first observation is trivial.  To see the second, just use elementary row operations.  For $ i\geq 2$, add minus column $ i-1$ to column $ i$.  This will produce a matrix like this:

$ \left[ \begin{array}{ccccc}2&1&1&1&\cdots\\\noalign{\medskip}3&1&1&1&\cdots\\\noalign{\medskip}4&1&1&1& \cdots\\\noalign{\medskip}5&\vdots&\vdots&\vdots&\ddots\end{array}\right]$

Now clearly the first two columns are linearly independent and all the other columns are dependent on column 2, so the matrix has rank 2.

(If you do not like column operations, you could just use row operations on the transpose of $ A$ and  note that rank is invariant under transposition).

... or you can just use row operations. This matrix is symmetric, after all.


I'm trying to remember the name/equation of a curve.  I think there's a nice polar form of it as well.

Basically, it can look like 2 seperate circles, then when you vary a constant the 2 circles stay the same except they get closer, then they meet and "merge", so to speak (the outside parts of the circle stay the same but the touching parts merge).  And eventually it looks like an ellipse (but isn't) then just one single circle.  Can anyone understand what I'm saying much less know what I'm talking about?

Could [url=http://astronomy.swin.edu.au/~pbourke/surfaces/egg/]this[/url] possibly be it?

Yes, grobber is right. It's Cassini's curve. It's like an ellipse, but instead of the sum of the distances from the foci fixed, it's the product.

There is a very interesting problem associated with Cassini's ovals.

Let F  and G are two "focal points" and Let P is a point on Cassini's oval.
P is defined as a locus  when (P F )*(P G) is constant.

Now consider right bisectors of  P F and  P G and call the point Q where they intersect.  What is the locus of point Q?

A chemistry picture prompted me to remember this curve.  It has to do with 2 hydrogen atoms approaching one another.  When they're apart, they're electron density is partially biased towards one-another.  Then when they meet it looks like the Cassini Oval.  I conjecture that it IS a Cassini Oval, but don't know where I'd get the data to see if I'm right.

I can't find a good picture on the web.  You can probably use your imagination, and the good pictures provided by the website on Cassini's oval, to see what I mean.  Here is one that is sort of good:
[url]http://www.molecularuniverse.com/MMMGIF/h2_diff.gif[/url]  The red/yellow part demonstrates the Cassini oval part.  The other colors could probably be described by a Cassini oval, too.

The Cassini Ovals are sections of a torus.  So while we call "conic sections" that very name we could call them "toric sections".


How can I get the arrow of a vector $ \vec{AB}$ to cover the whole top of the vector?
Is a somewhat short arrow acceptable?

\overrightarrow{AB} $ \overrightarrow{AB}$
Must have amsmath package loaded.


Each year $ 8$ subjects are taught by $ 4$ teachers in a school. Every teacher teaches $ 2$ subjects. At the end of this year they will meet to decide the course allotmet for the next year. Find the no. of ways in which the course distribution can be done so that each teacher teaches two subjects and each teacher teaches at least one subject different from the subjects which he taught this year.

By the Inclusion Exclusion Principle, the required number of ways is :
$ \frac{8!}{(2!)^4} \minus{} 4\frac{6!}{(2!)^3} \plus{} 6 \frac{4!}{(2!)^2} \minus{} 4\frac{2!}{2!} \plus{} 1$


We roll a fair 6-sided die 5 times.  What is the probability that we get an odd number in exactly 4 of the 5 rolls?

This is simply:

\[ \binom{5}{4} \left (\frac{1}{2} \right )^5\equal{}\boxed{\frac{5}{32}}
\]

[hide = Solution]We realize that this is equivalent to flipping a coin 5 times and looking for the probability that there are 4 heads. This is $\frac{\binom{5}{4}}{32} = \frac{5}{32}$[/hide]


[hide]
(1) 1:1. All objects fall at the same speed assuming there are no factors affecting them (such as wind). This seems like more of a physics problem than a math problem though... [/hide]

[quote="Interval"]A 4.0-kilogram rock and a 1.0-kilogram stone fall freely from rest from a height of 100 meters. After they fall for 2.0 seconds, the ratio of the rock’s speed to the stone’s speed is

(1) 1:1 (3) 2:1
(2) 1:2 (4) 4:1[/quote]
Either you have to mention their sizes and then it becomes a physics problem, or there are no factors affecting them and the answer is (1).

Thanks for helping me.  I tried solving this question but it appears to be physics.  I have not taken a physics course in over 20 years.

Thanks!


I think he should be allowed to do the reading of the names so long as the soldiers' families don't have a problem with it.

Ted Koeppel is the guy who reported from Iraq recently and says "quite literally" a lot, right?


modular arithmetic
trigonometry
logarithms
calculus
calculus computations

What can you say about

$\mathop{\sum}_{n=2}^{\infty}\frac{sin{(n)}}{log{(n)}}$

It converges conditionally.

We can prove that if $b_{n}$ is a sequence such that for some $N,$ for $n>N\ b_{n}$ is decreasing and tends to zero, then

$\sum_{n=1}^{\infty}b_{n}\sin(nx)$ converges for all $x$ and

$\sum_{n=0}^{\infty}b_{n}\cos(nx)$ converges for all $x\not\equiv0\pmod{2\pi}.$

The proof, which involves summation by parts, can be found in several places on this forum.

On the other hand

$\sum \frac{\sin nx}{\ln n}$ is not a Fouries series

while

$\sum \frac{\cos nx }{\ln n}$ is.


1)
Let $p(x)$ be a polynomial with integer coefficients, such that $p(1)=3$ and $p(3)=1$. Give a description of the possible values of $p(4)$.

2)
Let $p(x)$ be a polynomial with integer coefficients. Determine which one of the following assertions allows us to exclude that $p(x)$ has an integer root:
i) $p(0)$ and $p(2003)$ are both even;
ii) $p(0)$ and $p(2002)$ are both even;
iii) $p(0)$ and $p(2003)$ are both odd;
iv) $p(0)$ is odd and $p(2003)$ is even.

I solved the first one assuming that $p(x)$ has degree two, but of course that's a partial solution. I can't go on, 'cause I'd have to deal with long and boring calculations... I think I didn't see the correct path...
For the second one I have no idea...

[quote="Ponnamperuma"]1)
Let $p(x)$ be a polynomial with integer coefficients, such that $p(1)=3$ and $p(3)=1$. Give a description of the possible values of $p(4)$.

2)
Let $p(x)$ be a polynomial with integer coefficients. Determine which one of the following assertions allows us to exclude that $p(x)$ has an integer root:
i) $p(0)$ and $p(2003)$ are both even;
ii) $p(0)$ and $p(2002)$ are both even;
iii) $p(0)$ and $p(2003)$ are both odd;
iv) $p(0)$ is odd and $p(2003)$ is even.
[/quote]

[hide="Hint for both"]
Mods.  :wink: [/hide]

Ok, thanks for the tip...
I'll show you my solution for the first exercise, even if I still have some doubts...

I notice that $ p(1)=\sum_{i=0}^{n}a_{i}=3\equiv 0 \pmod3$.
Now, easily $ 4^{k}\equiv 1 \pmod3$, so I get $ p(4)=\sum_{i=0}^{n}4^{i}a_{i}\equiv \sum_{i=0}^{n}a_{i}\equiv 0 \pmod3$.

I know that's the correct answer, it's just that I didn't use the fact that $p(3)=1$... Or is it just a misleading clue? It's possible, since that the problem comes from a multiple choice test... but I'm not sure...

Ehi, could anyone say something about my solution?! Please!...   :)


geometry
analytic geometry
area of a triangle
Heron\u0027s formula

I've gotten in way over my head and was hoping someone could bail me out.  Can anyone help me solve this using Heron's Formula.

What it the area of a triangle bounded by the following:

X-2Y=12
X+3Y=9
4X-Y=-16

I'm drowning in frustation.  Thanks.

i can tell you how to do it a really long way. find the intersection point of two of those equations. you will do that 3 times and those will be the coordinates of you triangle. now use the distance formula to find the distance between those vertices and just plug it into heron's formula.

A little shorter:

Find the intersection points, and then use this to find the area:

[url]http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf[/url]

good point, tjhance..the shoelace formula does make things faster...didnt' consider that...i thought he was asking for a solution using heron's formula only, but that works even better :)

You are correct, I was lookinging only to use Heron's Theory.


function
advanced fields
advanced fields unsolved

Hi there
Let S(R) be the set of all bijective mappings from R onto itself.where R is the real field.
Prove that Card(S(R))=2^c

Thanks for your help

To each subset of $\mathbb R$ which does not have finite complement, we can associate a bijection by taking it to fix all the elements of the set, and partitioning the remaining elements in pairs $(a_i,b_i)_i$ (we use the Axiom of Choice) and putting $f(a_i)=b_i,f(b_i)=a_i$. Since there are $2^c$ subsets of $\mathbb R$ with infinite complement, it means that $Card(S_{\mathbb R})\ge 2^c$.

On the other hand, each real function is uniquely determined by its graph, which is a subset of $\mathbb R^2$. This means that there are at most $2^c$ real functions, so there are at most $2^c$ real bijections, in particular: $Card(S_{\mathbb R})\le 2^c$.

These two results give the desired conclusion.

A suggestion :P :
For the second part it would be maybe more intuitive to say that Card(S(R)) <= c^c = 2^c, since c^c is the cardinality of all real valued functions over real arguments, so obviously Card(S(R)) <= c^c, and c^c = 2^c


quadratics
inequalities
inequalities proposed

Find the maximal value of $ a^2+b^2+c^2 $ when the positive numbers $ a,b,c$ verify $ (ab)^3+(bc)^3+(ca)^3+2(abc)^3=1 $.

Oh, sorry, the problem asks for the minimum value.  :blush:  :blush:

I think I have a soln.
If a is the minimum, then mix b,c into their quadratic mean and prowe we increase
a^3b^3+..
Using a^3<=1/2 after easy factorisations and replacing quadratic mean by simple mean where necessary, we yield a stronger inequality which reduces to (b+c)^3>=2/3, which is easy.
So we can cuppose b=c then the inequality transforms into 2a^3=1/b^3-1.
Now one can prove a^2+2b^2 is convex with respect to b, so maximal value is either when b tends to 1, either all are equal.
Ugly solution..

Well, this ugly solution is very nice and shows that the inequality is not trivial. But I have such a nice solution! 2 lines and allmost no computation! But I like how quick you are, iura!

Thank you fro appreciating my solution, though it doesn't deserve it..
BTW could you please show your short solution to us?
I always get more satisfaction if I see a beautiful soln than if I find myself an ugly one..

It is immediate that there exists positive numbers x,y,z such that $ a^3=\frac{x}{y+z} $ and so on. Now use the fact that $\frac{2(x+y+z)}{x}\geq 3\cdot\sqrt[3]{\frac{2(y+z)^2}{x^2}} $ and you will find the result.

note that a,b,c are all less than 1 then by the condition imposed
(ab)^3= (cos A)2 and similar relation hold as per the constraint (and A,B,C are angles of an acute angled triangle)
we have to find the minima of
(cosAcosBcosC)^(2/3) [(secA)2+(secB)2+(secC)2]
APPLY am-gm and get the minima as 3
am i correct


If an ellipse has a focus at $(3,2)$, and the corresponding directrix is $x=6$. $e=\frac{1}{2}$. Find the equation of the ellipse.

[hide="Click here for Solution"]
Let semi major axis be a,  semi major axis be b.

Distance from centre to focus = ae
Distance from centre to directrix=a/e
So,  Distance from focus to directrix= a/e - ae = 6-3

Solving, a=2

Using this value to calculate b using:

b*b=(a*a)(1-e*e)

we get b=sqrt(3);

x of centre = 3 - ae = 2
y of centre = 2

Thus equation of ellipse is:

(x*x - 2)/(a*a) + (y*y - 2)/(b*b) = 1
or
(x*x - 2)/4 + (y*y - 2)/3 = 1

[/hide][/hide]


calculus
geometry
perimeter
calculus computations

Just got a new old problem with the start of the new semester. I haven't had a calc or any other math class in 2 years and need help with this:

A  circle is inscribed in a square. The circumference of the circle is increasing at a constant rate of 6 inches per second. As the circle expands to maintain tangency, find...

I. Find the rate at which the perimeter of the square is increasing.
II. At the instant when the area of the circle is 25pi square inches, find the rate if increase the area enclosed between the circle and the square

If possible, please include every step since my new prof is quite the stickler for showing work. A detailed answer will be greatly appreciated :)


Let $ ABCD$ be a cyclic quadrilateral.Diagonals $ AC$ and $ BD$ meet at $ E$.
Let $ P$ be point inside the quadrilateral.Let $ O_1,O_2,O_3$ and $ O_4$ be circumcenters of triangles $ ABP,BCP,CDP$ and $ DAP$ respectively.Prove that lines $ O_{1}O_{3},O_{2}O_{4}$ and $ OE$ are concurrent.

See here: http://www.mathlinks.ro/Forum/viewtopic.php?p=550567#550567

P.S. Could you please send me all problems in MOSP 2007?  :D
[b]EDIT.[/b] I got it  :P


geometry
circumcircle
Euler
geometry unsolved

Let triangle $ABC$ , $AB$ $\neq$ $BC$ $\neq$ $CA$.
Let $X$,$Y$,$Z$ are change points in $BC$,$CA$,$AB$ respectively such that triangle $XYZ$ similar triangle $ABC$.
Find the locus of point $T$ is the centroid of triangle$XYZ$

All the triangles $\triangle XYZ$ are spirally similar with the similarity center $O$, their common orthocenter and the circumcenter of the original triangle $\triangle ABC.$ The locus of their circumcenters $P$ is a normal to the Euler line $OH$ of the triangle $\triangle ABC$ (where $H$ is its orthocenter) erected at the 9-point circle center $N,$ the midpoint of the segment $OH.$ See the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=29888]Locus!!![/url] problem in the high school section. Since $\frac{GH}{GO}= \frac{TO}{TP}= 2,$ the locus of the centroids $T$ of $\triangle XYZ$ is a normal to the Euler line $OH$ erected at the centroid $G$ of the triangle $\triangle ABC.$


let triangle ABC have BC=a,AC=b,AB=c prove  :Sigma:  :sqrt:ac(a+c-b) >= 3 :sqrt: abc ;)

I don't think so :)
Let $a=x+y$, $b=y+z$ and $c=z+x$. Divide both sides to $\sqrt{abc}$, then you will obtain
\[\sqrt\frac{2x}{y+z}+\sqrt\frac{2y}{x+z}+\sqrt\frac{2z}{x+y}\geq 3,\]
though, as you can see in http://www.mathlinks.ro/viewtopic.php?p=157192 , the correct lower bound is $2\sqrt 2$.

I sorry because my problem is mistest.I 'm very glad


Find fifth degree of polynomial p(x) such that p(x) + 1 is divisible by $(x-1)^{3}$ and p(x) - 1 is divisible by $(x+1)^{3}$

Very Very easy.We apply Bezout's Law => P(x)=...........

is the question asking to find just one fifth degree polynomial that satisfies the conditions or just the general equation for $p(x)$? Because we have
\[p(x)+1 = (x-1)^{3}\cdot a(x)\]
\[p(x)-1 = (x+1)^{3}\cdot b(x)\]
for some second degree functions $a(x)$ and $b(x)$. Solving for $p(x)$,
\[p(x)=\left(\frac{1}{2}\right)(x-1)^{3}\cdot a(x)+\left(\frac{1}{2}\right)(x+1)^{3}\cdot b(x)\]
which clearly has more than one solution...

[quote="NeverOddOrEven"]is the question asking to find just one fifth degree polynomial that satisfies the conditions or just the general equation for $p(x)$? Because we have
\[p(x)+1 = (x-1)^{3}\cdot a(x) \]

\[p(x)-1 = (x+1)^{3}\cdot b(x) \]
for some second degree functions $a(x)$ and $b(x)$. Solving for $p(x)$,
\[p(x)=\left(\frac{1}{2}\right)(x-1)^{3}\cdot a(x)+\left(\frac{1}{2}\right)(x+1)^{3}\cdot b(x) \]
which clearly has more than one solution...[/quote]
But you need to make sure that the $a(x)$ and $b(x)$ actually fit both equations, not just their sum.  :wink:
[hide]
Letting $P(x)=1+(x+1)^{3}Q(x)$ for some quadratic $Q$, we need $P(x)+1$ to be divisible by $(x-1)^{3}$. In particular, $P(1)=-1$ so $Q(1)=-\frac{1}{4}$.

Now let $Q(x)=(x-1)R(x)-\frac{1}{4}$ for some linear function $R$. Since $(x+1)^{3}-8=(x-1)(x^{2}+4x+7)$, we have
\[\frac{2+(x+1)^{3}Q(x)}{x-1}=(x+1)^{3}R(x)-\frac{x^{2}+4x+7}{4}\]
Now we need [i]this[/i] function to be divisible by a second factor of $x-1$, so $R(1)=\frac{3}{8}$. This means $R(x)=\frac{3}{8}+C(x-1)$ for some constant $C$. Dividing out the second factor of $x-1$ gives
\[\frac{4(x+1)^{3}R(x)-x^{2}-4x-7}{4(x-1)}=C(x+1)^{3}+\frac{3x^{2}+10x+11}{8}\]
Finally, to get the third factor of $x-1$, we have $C=-\frac{3}{8}$. Our polynomial is then
\[P(x)=1+(x+1)^{3}\left(-\frac{3}{8}(x-1)^{2}+\frac{3}{8}(x-1)-\frac{1}{4}\right)\]
[/hide]

We Have: P(x)-1 is divisible by (x+1)^3, so P(-x)-1 is divisible by (-x+1)^3
 Or P(-x)-1 is divisible by (x-1)^3.
Similarly, We have P(x)+1 is divisible by (x-1)^3 and P(-x)+1 is divisible by (x+1)^3.
Hence, P(x)+P(-x) is divisible by (x+1)^3 and (x-1)^3
or Q(x)=P(x)+P(-x) is divisible by (x+1)^3 . (x-1)^3=(x^2-1)^3
On the other hand, Q(x) is the fifth degree polynomial
So, Q(x)=0 -->P(x)=-P(-x)                             (*)
Now, Put P(x)=(a.x^2+b.x+c).(x+1)^3+1.         (**)
Since (*) -->P(0)=0 -->c=-1.
Then, combinning (*) and (**) We finded a and b -->P(x) ( :D )


function
vector
topology
real analysis
real analysis unsolved

Consider the set X = {f : [0, 1] → R | f ∈ C[0, 1]} with the metric

d(f, g) = sup|f(x) − g(x)|
(x∈[0, 1])

Prove that the set A = {f ∈ X | f(0) > 1} is open in (X, d).

C[0, 1] is the set of all real-valued continuous functions on [0, 1]

Have you some initial thoughts on this (i.e. what have you tried so far)?

I would consider starting with the definition of an open set. (If I recall correctly I think a set is open if it's complement is closed(?))

Consider the function $ T: X\to\mathbb{R}$ defined by $ T(f) \equal{} f(0).$ Note that the set you are trying to prove is open can be expressed as $ T^{ \minus{} 1}(\{y\in\mathbb{R}: y > 1\}).$

We know that the inverse image of an open set through a continuous function is open. So is the function $ T$ described above continuous?

(It is, and it's more than that: it is also linear. It's a "bounded", which is to say uniformly continuous, linear functional defined on a normed vector space.)

It is enough to prove that $ A^c$ is closed. Let $ g$ be a limit point of $ A^c$. Then there exists a sequence of function $ f_n \in A^c$ such that $ f_n \to g$ as $ n\to\infty$. Fix any $ \epsilon >0$, there is $ N$ so that for all $ n>N$, $ |f_n(0) \minus{}g(0)| \le \sup \limits_{x\in[0,1]}|f_n\minus{}g| < \epsilon$. Since $ \epsilon$ is arbitrary, $ A^c$ is closed. Hence $ A$ is open.


Barney Schwinn notices the odometer on his bicycle reads $ 1441$, a palindrome, because it reads the same forward and backward.  After riding $ 4$ more hours that day and $ 6$ the next, he notices that the odometer shows another palindrome, $ 1661$.  What was his average speed in miles per hour?

$ \textbf{(A)}\ 15\qquad
\textbf{(B)}\ 16\qquad
\textbf{(C)}\ 18\qquad
\textbf{(D)}\ 20\qquad
\textbf{(E)}\ 22$

[hide]

Subtracting $ 1661$ from $ 1441$, we get the traveled $ 220$ miles in $ 4\plus{}6\equal{}10$ hours, dividing gets the average speed of $ \boxed{\textbf{(E)}\ 22}$[/hide]


Prove that $a^{3}+b^{3}+4$ is never of the form $n^{3}$. a,b,n all elements of the natural number.

davron latipov


function
logarithms
probability and stats

[quote="Mestat"]Hey all,
Its known that the Cauchy distribution has no mean and variance defined!
The Cauchy distribution is the Student distribution with df equal to 1, right!
But, If I need to calculate the maximum likelihood estimators (mu and sigma) of a bivariate Cauchy distribution.
Is it possible to obtain this results... Or this is a trick question...
Thanks in advance,
Marcio[/quote]It looks like a trick question, at least for the univariate Cauchy distribution (which should be simpler anyway). If you set up the (log) likelihood function for let's say a Cauchy distribution with center $ \mu$,
\[ L(x_1,x_2,\dots,x_n;\mu) \equal{} \sum_{i\equal{}1}^n \minus{}\log(1\plus{}(x_i \minus{} \mu)^2) \minus{} n \log \pi
\]
you'll notice that this function is in general not concave in $ \mu$. For suitable choices of the $ x_i$, you'll get $ n$ local minima. The global minimum cannot be found analytically.

For the bivariate case where in addition some dispersion quantity has to be found, the situation is only worse. This is a minimization problem in 5 variables for a non-convex function with potentially many local minima. Computationally difficult.


Let $ f(x)=ax^{3}+bx^{2}+cx+d$ be the cubic polynomial with integer coefficient.
If $ f(0),\ f(1),\ f(2)$ can't be divided by 3, prove that the cubic equation $ f(x)=0$ doesn't have integer solution.

that is false
take the polynomial $ f(x)=x(x-1)(x-2)$. we have $ f(0)=f(1)=f(2)=0$ so they are divisible by $ 3$ and it obviously has an integer solution...

Sorry I made a typo. :oops:
I have just edited.

kunny

[quote="N.T.TUAN"]General problem : Assume that $ f\in\mathbb{Z}[x]$ and for some $ n\in\mathbb N$ : $ n\not |f(i)(i=0,1,...,n-1)$. Prove that $ f$ has not integer root.[/quote]
if $ f\in\mathbb{Z}[x]$ has not integer root $ a\in \mathbb{Z}$,
$ \Longrightarrow \exists Q\in\mathbb{Z}[x]: \ f(x)=(x-a)Q(x)$
$ \Longrightarrow \exists i\in \{0,1,2,...,n-1\}: \ n|(i-a)\Longrightarrow n|(i-a)Q(i)=f(i)$

then  $ \forall i\in \{0,1,2,...,n-1\}: \ n\not |f(i) \Longrightarrow f\ has\ not\ integer\ root$


Here's an exam that I have made yesterday for my 10th grade students. The exam is made up of the following single question:

To a mass $ m$ of liquid water at 77ºC is added a mass $ \frac{2m}{3}$ of ice at 0ºC. After the equilibrium has been attained, a small massive sphere of lead with a mass of 177 g is submerged is the resulting mixture. Suppose $ m$ is such that the sphere does not change the temperature of the mixture. After thermal equilibrium, the sphere is taken from the mixture and is submerged in 100 g of liquid ethanol at 0ºC contained in an aluminium jar with a mass of 200 g. After a new thermal equilibrium is reached, the sphere is taken from the ethanol and imediatly heated in an oven to 330ºC. The resulting liquid lead is molded in a cube.

(a) Find the equilibrium temperature of the sphere when submerged in ethanol.

(b) In the text above it is referred that the lead sphere is "small". Find how small is really the sphere by computing its radius.

(c) Calculate the total energy that was needed to heat the lead to 330ºC.

(d) After cooling, the cube is put into contact, through two opposite faces, with two jars with liquid water, one at 80ºC and the other at 17ºC. Find the amount of energy that crosses the cube in one minute.

Data:

Heat capacities (in $ Jkg^{\minus{}1}K^{\minus{}1}$): water, 4186; lead, 128; aluminium, 900; ethanol, 2440.

Heats of fusion: ice, $ 3.34 \times 10^5\,Jkg^{\minus{}1}$; lead, $ 4.77\,kJmol^{\minus{}1}$.

$ \rho_{lead} \equal{} 11.34\,gcm^{\minus{}3}$

Lead melting point: < 330ºC

Thermal conductivity of lead: $ 35.3\,Wm^{\minus{}1}K^{\minus{}1}$

Molar mass of lead: $ 207.2\,gmol^{\minus{}1}$

I think that is a bit advanced for tenth graders.  Most people don't get an advanced treatment of thermal physics until college.

You think wrong. In Portugal, we learn/teach in 10th grade some basic knowledge of Thermodynamics: system and surroundings, walls, processes, zeroth, first, and second law (brief account), ideas gases, transfer of energy as heat, work, and radiation, isolated systems, heat capacities, heats of phase changes, thermal machines, and thermal conductivity and related Fourier's Law. This is all what is needed to solve this problem: just simple concepts.

Definitely haven't covered that here in the states.  Maybe some basic stuff in chemistry not but not in physics.

Darn, I know so little thermodynamics, but at least I hope I know enough to solve this problem.

Call $ T$ the equilibrium temperature of the water and ice - then
\[ m(4186 \frac {J}{kgK})(77K \minus{} T) \equal{} \frac {2m}{3}(3.34\times10^5 \frac {J}{kg}) \plus{} \frac {2m}{3}(4186 \frac {J}{kgK})T
\]
And $ m$ conveniently cancels, allowing us to solve for $ T$, which I'm too lazy to do.  Then I assume that the lead sphere must also be at this initial temperature $ T$ (is that what the problem means?).  Then you can use a similar process to find the temperature of the sphere after being submerged in the ethanol.  Then you can find the volume of the lead using its density to find the dimensions of the cube, which you need to answer the last part.

At this point, I looked up the definition of thermal conductivity on Wikipedia (lol):
\[ \frac {dQ}{dt} \equal{} kA\frac {\Delta T}{x},
\]
where $ x$ is the side length of the cube.  Then I supposed you can plug in stuff (i.e., $ \Delta T \equal{} 80K \minus{} 17K$, etc.).

Someone else can finish the rest, because I need some sleep.  Did I do anything patently stupid?

EDIT: Yay, I have 42 posts.

That looks ok, although your expression for Fourier's Law is scientifically incorrect.

Yeah, maybe I should have written it like $ \displaystyle \frac{dQ}{dt} \equal{} kA\frac{dT}{dx}$ or $ \displaystyle \frac{\Delta Q}{\Delta t} \equal{} kA\frac{\Delta T}{\Delta x}$

This is all what is needed to solve this problem: just simple concepts.
[url=http://www.math123xyz.com]Interactive Math Resources[/url]

Then disregard the differential form of Fourier's law and use the delta form.  I just wanted to present the more general and correct version.

[quote]use calculus at ur home! [/quote]

Do you have some sort of problem with calculus? :wink:   Calc is fun.

Whoa, reviving this thread was totally pointless.  Unless you wanted to present a complete solution.

hey sorry bro!
i didnt intend to...........
i love calculus but just that i thought that at elementary level calculus might not help!
sorry once again!
 :blush:


If $<a_1,a_2,....,a_n>$ and $<b_1,b_2,....,b_n>$ be two sequences of positive integers such that {$a_i + a_j : 1\leq i < j \leq n$} and
{$b_i + b_j : 1\leq i < j \leq n$} are identical. Prove that $n$ is a power of $2$.


probability
Support
combinatorics unsolved
combinatorics

[b]One-fourth of the female freshman at a private liberal art college are out-of-state students. Assume the students are assigned at a random to the dormitories, 3 to a room.

1/ What is the probability that no students in certain dorm room are out-of-state?

2/ What is the probability thth more than 1 students is an out-of-state student in this room?

I found:

1/                    3! / (0!3!) = 0.42

2/                   1-p(x=0)= 0.58

Could you tell me if my answers are right or wrong please?  :?

Thanks a lot!!  :blush:
[/b]

The answer would appear to depend on the size of the freshman class. But your answers are wrong in any case.

(In fact., $0.42$ is a very good approximation, but your displayed work does not support that answer.)

Suppose there are $4m$ girls in the freshman class, $m$ of whom are from out of state. Then the probability that exactly $k$ of them $(0\le k\le3)$ are from out of state is
\[ \frac{{m \choose k}{{3m} \choose{3-k}}}{{4m\choose3}} \]
The $k=0$ (none from out of state) case of this would be:

$\frac{{{3m}\choose 3}}{{{4m}\choose 3}}=\frac{3m(3m-1)(3m-2)} {4m(4m-1)(4m-2)} =(\frac34)^3\frac{(m-\frac13)(m- \frac23)}{(m-\frac14)(m-\frac24)}).$

If there are 40 girls in the freshman class $(m=10),$ then this is about $0.4109.$ For 200 girls in the class $(m=50),$ this is about $0.4198.$ For 1000 girls in the class, this would be about $0.4215.$

There's an obvious limiting case. There are two ways to get to that limiting case. One way is simply to say that $m$ is "so large that it doesn't matter." (In other words, taking the limit as $m\to\infty.)$ The other way is to reinterpret the problem as follows: For each student admitted, the probability of being from out of state is $1/4.$ Students are admitted independently and then assigned to rooms at random.

Either of those two assumptions will change the hypergeometric distribution that I have been working with to the binomial distribution. In that case, the probability of $k$ out of state girls in the room is ${3\choose k}(\frac14)^k(\frac34)^{3-k}.$

For $k=0,$ that is  $0.421875$
For $k=1,$ that is  $0.421875$
For $k=2,$ that is  $0.140625$
For $k=3,$ that is  $0.015625$

And you misread the second question. It didn't say "one or more," it said "more than one." For that add up the $k=2$ and $k=3$ probabilities, getting $0.15625.$

Oh - and this doesn't belong in the forum you placed it in. I'll find somewhere else to put it.

[b]h=Thanks for your help, i appreciate it!  :lol: [/b]


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Show that if $ R$ is an integral Boole Ring then is isomorphic to $ \mathbb{Z}_2$.

Integral means has no 0 divisors.
Boole means that $ x^2\equal{}x$ for all $ x$.

HINT $ x(x\minus{}1)\equal{}0\implies ...$

Assume that there is another element in the ring (say $ 0$, $ 1$ and $ a$) and reach a contradiction (use above hint).
Then there are only 2 elements, so $ R$ is a finite integral domain, so it is a field, so it is isomorphic to $ mathbb{Z}_2$