function
integration
trigonometry
real analysis
calculus
topology
real analysis unsolved

Let $f: (0,\infty) \rightarrow \mathbb{R}$ be differentiable, and assume that $f(x)+f'(x) \rightarrow 0$ when $x \rightarrow \infty$. Show that $f(x) \rightarrow 0$ as $x \rightarrow \infty$.

We discussed a similar problem recently, with $f+f'+f''$.

Let $f(x)=e^{-x}g(x)$, then $f(x)+f'(x)=e^{-x}g'(x)$. Thus, for any $\epsilon>0$ there is $M$ such that $|g'(x)|<e^{x}\epsilon$ for all $x>M$. By Mean Value theorem $|g(x)-g(x-1)|\le e^{x}\epsilon$ for $x>M+1$. Iterating this estimate, obtain $|g(x)-g(x-n)|\le \frac{e^{x}\epsilon}{1-e^{-1}}$ for $x>M+n$. Choose $n$ so that $x-n\in [M,M+1]$. Since $g$ is bounded on $[M,M+1]$, it follows that $\limsup_{x\to\infty}e^{-x}|g(x)|\le \frac{\epsilon}{1-e^{-1}}$.

There ought to be a better proof.  :|

The solution in the book used $g(x) = f(x)+f'(x) \Rightarrow f(x) = e^{-x}\int_{a}^{x}e^{t}g(t)dt+Ce^{-x}$ (by using an integrating factor). Then we can say $|g(x)| < \epsilon$ for all $x > a$, so

$|f(x)| \le \epsilon (1-e^{a-x})+|Ce^{-x}| < 2\epsilon$

for sufficiently large $x$.

How do you know that $\int_{a}^{x}e^{t}g(t)dt$ is defined? Here is a classical example: $f(x)=x^{2}\sin\frac{1}{x^{3}}$ is a differentiable function but $f'(x)=2x\sin\frac{1}{x^{3}}-\frac{3}{x^{2}}\sin\frac{1}{x^{3}}$ fails to be integrable in either Riemann or Lebesgue sense.

The problem goes away if one uses the [url=http://en.wikipedia.org/wiki/Henstock-Kurzweil_integral]Henstock-Kurzweil integral[/url], which can recover any differentiable function from its derivative. But I don't know if your book introduces this integral.

I see. That's a good point. Larson does not seem to address that problem, nor does it say anything about integrals other than the Riemann integral.

By generalizing the above argument based on repeated application of Mean Value Theorem, you can prove the following

[b]Lemma[/b]. Suppose that $f$ is differentiable on an open interval containing $[a,b]$. Let $g$ be a continuous function on $[a,b]$ such that $|f'(x)|\le g(x)$ for all $x\in [a,b]$. Then $|f(b)-f(a)|\le\int_{a}^{b}g(x)dx$.

This may be helpful in other situations when you want to estimate a differentiable function from its derivative.

This also brings us to a topic of current research: if $f$ is defined on an abstract metric space, a function $g$ is called an "upper gradient" of $f$ provided that $|f(\gamma(b))-f(\gamma(a))|\le\int_{a}^{b}g(\gamma(t))dt$ for any rectifiable curve $\gamma\colon[a,b]\to X$ parametrized by arclength. The point here is that although we can't differentiate in a metric space, we can define something like the magnitude of the gradient, and for some purposes (like the Sobolev embedding) the magnitude is enough. Recommended reading: "Lectures on Analysis on Metric Spaces" by Heinonen.


logarithms
trigonometry
algebra
polynomial
calculus
calculus computations

Find the order and degree ( if defined ) for the below two differential equations .

1) $ \frac{d^{2} y}{d x^{2} } \equal{} x \ln(\frac{dy}{dx} )$

2 ) $ \frac{d^{2} y}{d x^{2} } \equal{} \sin (\frac{dy}{dx})$

please help me  and if the answers are not defined please tell me a reason for those.

hello
[hide="1"]
$ \frac{d^{2}y}{d x^{2}}\equal{} x\ln(\frac{dy}{dx})$
order=$ 2$
this can't be written as polynomial in all differential coefficient , the degrees is not defined. [/hide]

[hide="2"]
$ \frac{d^{2}y}{d x^{2}}\equal{}\sin (\frac{dy}{dx})$
again order=$ 2$
degree can't be defined .Same reason.
[/hide]
[hide="NOTE"]
Refer to NCERT .I believe they have explained it in details. [/hide]
thank u

Thanks Kabi, even i thought abt the same thing just wasn't sure.

thank u very much  :)


calculus
integration
greatest common divisor
number theory

The ordered pair of positive integers $ (x,y)\equal{}(1000,1)$ is one of the many solutions to the equation
\[ 2x\plus{}8y\equal{}2008.
\]
Among the $ 81$ ordered pairs of positive digits $ (a,b)$, how many of the equations
\[ ax\plus{}by\equal{}2008
\]
have solutions $ (x,y)$ which are ordered pairs of positive integers?

The equation comes out to be:
[size=18][/size][color=violet][/color][b] y^2+ x^2= 2008

Then you slove it like a circle equaton. :D [/b]

Does someone have a more complete solution?  I'm still stuck on this problem.

I'm sorry; I don't understand.  Are you talking about the intersection of ax+by=2008 and 2x+8y=2008 or are you just talking about an integral solution to ax+by=2008 (given a and b are positive digits)?

Can someone please provide a full solution?  I haven't been able to solve the problem yet.  Thanks.

I believe each of a and b are numbers from 1 to 9 inclusive and we are asked to find how many of the 81 possible (a,b) have solutions.

[hide="hint that I am not too sure of"] The greatest common factor of a and b must divide 2008.

For example, (7,7) does not work.[/hide]

[hide="solution?"] We will count the number of (a,b) that do not satisfy that hint.

$ 2008\equal{}251\cdot2^8$. We clearly do not need to worry about 251.

Now, we can use 3, 5, 7, 9 as the greatest common factors.

If a=3, then b can be 3, 6 or 9.
If a=5, then b must be 5.
If a=7, then b must be 7.
If a=9, then b must be 3, 6, or 9.

I count 8 that do not work, perhaps 73?[/hide]

I do not believe my solution is right.

Let $ a \equal{} gcd(a,b) \cdot p, b \equal{} gcd(a,b) \cdot q$

It's easy to see that $ gcd(a,b)|2008$ must be satisfied.

Then the question becomes we can find positive integer $ x$ such that $ ax \le 2008 \minus{} b.......(1)$ and $ px \equal{} \frac {2008}{gcd(a,b)}(mod q).........(2)$. As $ p,q$ coprime , we can find $ 0 < x \le q$ satisfies (2), and obviously the smallest solution $ x_1 \le q \le 9 < 2008 \minus{} 9 \le 2008 \minus{} b$. So $ gcd(a,b)|2008$ is a sufficient and necessary condition.

Possible values of $ gcd(a,b) \equal{} 1,2,3,4$ for $ a \neq b$ , among them only $ 3 \nmid 2008$

So only $ (3,6),(3,9),(6,9),(6,3),(9,3),(9,6)$ and $ (3,3),(5,5),(6,6),(7,7),(9,9)$ give no such solution. The answer $ \equal{} 81 \minus{} 11 \equal{} 70$

What about (a,b)=(5,5), (7,7)?

If $ 5a\plus{}5b\equal{}2008$, then 5 divides the left side but not the right, so there are no solutions.
Same for $ 7a\plus{}7b\equal{}2008$.

The correct answer [hide="is"]$ 70$[/hide], but I can't follow stephencheng's solution.  Could someone please explain why he did what he did? (It looks like a solution with a lot of number theory, and I'm quite weak on my number theory.) Thanks!

Ugh I missed (6, 6), (6, 3), and (6, 9).

stephencheng's solution is complicated. I believe mine is easier to understand, but it's somewhat less rigorous. Basically, you know gcf(a,b) must be 3, 5, 6, 7, or 9. 5, 6, 7, and 9 are easy to get rid of; only (5, 5), (6, 6), (7, 7), (9, 9) do not work.

The 3 requires some more work, but not too bad.


Wow, that's amazing that you did the sprint in 15 minutes.  Did you get the rest of the 28 correct?

-interesting_move

[quote="Rep123max"]This isnt really a question but it seems most people come to this forum so i thought i would ask it here.

do any of you guys do UIL (i dont know if its only in Texas). Number Sense, Calculator, Math, and Science. Ive done it for 2 years and I feel it helps and hurts on these contests. It helps because in number sense you look for quick shortcuts which really is valuable in some of these tests (AMC MATHCOUNTS....[/quote]

UIL (University Interscholastic League) is based in Austin at U. Texas - to my knowledge you need to be a resident of Texas to officially participate.  I did it during my 2 years of living in Texas, and definitely agree that the number sense problems helped immensely with other contests (especially the old AHSME - now AMC 12).

Yea rep123max, UIL of course exists in Texas. As far as I know though, in most major cities its only offered for High school (but correct me if im wrong). The State Round of MC was extremely unfair this year. Problems were way too easy and (if you will recall our state rankings) most of the top 20 were stuck within 2 pts of each other. I was devestated when I saw the problems b/c my specialty is making stupid errors on easy problems and getting harder problems correct that nobody else would know how to do.


What are some distinguished proof contests?(other than USAMO and IMO)

[url]http://www.artofproblemsolving.com/Forum/resources.php[/url]

these are the ones linked under contests at the top of each aops page.

USAMTS, TST (Team Selection Test), and various international olympiads (for example, the USAMO would count, as well as CMO (Canadian Math Olympiad) and others).

Specifics, please? I have checked the link, hurdler, and I will see which ones are proof contests, but any others?

There is a separate category of "mail-in" proof contests. USAMTS falls into this category, as do some more local things, like the Wisconsin and Pomona College "Talent Search" contests.


Here s a very simple geometric interpretation of our Nesbitt Inequality that I happened to notice:
In a triangle $ ABC$ if all letters have their usual meaning: then,
prove that:
$ \sum_{i \equal{} 1}^{3} \frac {h_{i} \plus{} r}{h_{i} \minus{} r} \ge 6$

Here s yet another related problem:
For positive reals $ x,y,z$ prove that
$ \sum \frac {2}{x \plus{} y} \ge \frac{9}{x\plus{}y\plus{}z}$

[quote="Euclidean Geometer"]
For positive reals $ x,y,z$ prove that
$ \sum \frac {2}{x \plus{} y} \ge 9$[/quote]
Try $ x\equal{}y\equal{}z\equal{}1.$ :wink:

OOPS!!! Sorry
Thanks Arqady, I mistyped it. The correct one is:

$ \sum \frac {2}{x \plus{} y} \ge \frac {9}{x \plus{} y \plus{} z}$

Editted in the original post as well.  :)

[quote="Euclidean Geometer"]Here s a very simple geometric interpretation of our Nesbitt Inequality that I happened to notice:
In a triangle $ ABC$ if all letters have their usual meaning: then,
prove that:
$ \sum_{i \equal{} 1}^{3} \frac {h_{i} \plus{} r}{h_{i} \minus{} r} \ge 6$[/quote]$ \frac{h_a\plus{}r}{h_a\minus{}r}\plus{}\frac{h_b\plus{}r}{h_b\minus{}r}\plus{}\frac{h_c\plus{}r}{h_c\minus{}r}\geq 6$
$ \Longleftrightarrow \frac{2A\plus{}ar}{2A\minus{}ar}\plus{}\frac{2A\plus{}br}{2A\minus{}br}\plus{}\frac{2A\plus{}cr}{2A\minus{}cr}\geq 6$
$ \Longleftrightarrow \frac{r(a\plus{}b\plus{}c)\plus{}ar}{r(a\plus{}b\plus{}c)\minus{}ar}\plus{}\frac{r(a\plus{}b\plus{}c)\plus{}br}{r(a\plus{}b\plus{}c)\minus{}br}\plus{}\frac{r(a\plus{}b\plus{}c)\plus{}cr}{r(a\plus{}b\plus{}c)\minus{}cr}\geq 6$
$ \Longleftrightarrow \frac{(b\plus{}a)\plus{}(c\plus{}a)}{b\plus{}c}\plus{}\frac{(a\plus{}b)\plus{}(c\plus{}b)}{a\plus{}c}\plus{}\frac{(a\plus{}c)\plus{}(b\plus{}c)}{a\plus{}b}\geq 6$ true by AM-GM.


IP adresses are composed of 4 8-digit base 2 numbers. How many total IP addresses are there?

Extra credit to anyone who can find a pattern for (2n-1)^m (some number of 1s to the something power; example: 111^11)

Technically, Tare, you're wrong :-p Several IPs are reserved for networks seperate from the internet - all starting in 192.168 for example. And I don't believe that 0.0.0.0 is a valid IP. Anyway, interesting problem for the kiddies.

edit: fixed the typo

Ignoring all restrictions:[hide] There are 2 possible values (0 & 1) for each digit). There are 32 digits, so 2^32, =4,294,967,296. It's a big Internet out there.[/hide]

Yup.

doo-whoop, I'm trying to introduce some people into the magnificent world of base 2...

[quote="Ravi B"]I thought it was 192.168, not 198.168.[/quote]
Eh, yeah. Typo.


function
combinatorics unsolved
combinatorics

In a round robin chess tournament each player plays every other player exactly once. The winner of each game gets $ 1$ point and the loser gets $ 0$ points. If the game is tied, each player gets $ 0.5$ points. Given a positive integer $ m$, a tournament is said to have property $ P(m)$ if the following holds: among every set $ S$ of $ m$ players, there is one player who won all her games against the other $ m\minus{}1$ players in $ S$ and one player who lost all her games against the other $ m \minus{} 1$ players in $ S$. For a given integer $ m \ge 4$, determine the minimum value of $ n$ (as a function of $ m$) such that the following holds: in every $ n$-player round robin chess tournament with property $ P(m)$, the final scores of the $ n$ players are all distinct.

Answer: $2m-4$

Consider an arbitrary tournament with property $P(m)$ with $n\ge2m-3$ players. We call a set $S$ of $m$ players "good" if there is a player who won all her games against the other $m-1$ players in $S$ and a player who lost all her games against the other $m-1$ players in $S$. Otherwise, call it "bad".

(i) There is at least one tie.
Assume $P$ tied with $Q$. If there are $m-2$ players who lost to $P$, then these players with $P$ and $Q$ form a bad set. So there are at most $m-3$ players who lost to $Q$. Similarly, there are at most $m-3$ players who won are tied with $P$. Thus there are at most $m-3+m-3+2=2m-4$ players, a contradiction.

(ii) There are three players $P,Q,R$ such that $P$ won against $Q$, $Q$ won against $R$, and $R$ won against $P$ (call this a cycle).
Similar as case (i), we find that at most $m-4$ players lost against $P$ and at most $m-4$ players won against $P$. Hence there are at most $m-4+m-4+3=2m-5$ players, a contradiction.

(iii) There is no tie and no cycle.
There is a player $P$ with the largest number of wins. If $Q$ won against $P$, then $Q$ beat all players who lost against $Q$, so $Q$ has more wins, a contradiction. So $P$ won against all other players. Among players other than $P$, there is a player who won against all other players. Continuing this, we can label the players as $P_1,P_2,\ldots,P_n$ such that $P_i$ won against $P_j$ for all $i>j$. Clearly their final scores are ditinct.

So for $n\ge2m-3$, the result holds. Now we give a construction to show that $2m-4$ is not enough: $P_1,P_2,\ldots,P_{2m-4}$ are such that $P_i$ won against $P_j$ for all $i>j$, except that $P_{m-1}$ tied with $P_{m-2}$.


A polyhedron (a solid with polygonal faces, not a regular one) has an equal number of quadrilaterals and pentagons as its faces. Exactly three faces meet at each vertex. How many faces does this polyhedron have?

[hide]Let the polyhedron have $ X$ quadrilaterals and $ X$ pentagons.
There are $ 2X$ faces in the polyhedron.
Since each edge is shared by $ 2$ faces, there are $ \frac {1}{2}(4X + 5X) = \frac {9}{2}X$ edges.
Since each vertex is shared by $ 3$ faces, there are $ \frac {1}{3}(4X + 5X) = \3X$ edges.
By Euler's identity, $ V - E + F = 2 \Rightarrow 3X - \frac {9}{2}X + 2X =  2 \Rightarrow X = 4$. So, the polyhedron has $ 2X = 8$ faces.[/hide]


Check this: local Lipshitz condition (I know I misspelled Lipschitz but the punishment I got doesn't match the crime :P)

I first saw this about [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=143623]six months ago[/url].
That was a poster that didn't know the correct spelling rather than a typo.


Please help me in solving the following question.

When all diagonals are drawn in a regular octagon, the interior is divided into how many regions?

Thanks in advance!

Trying what athunder said, we get

Square: $ 4$ regions
Pentagon: $ 11$ regions
Hexagon: $ 20$ regions.

And, then I cam up with this formula: $ r\equal{}s^2\minus{}2s\minus{}4$, which works. Now plugging in $ 8$, we get $ r\equal{}8^2\minus{}2(8)\minus{}4\equal{}64\minus{}16\minus{}4\equal{}64\minus{}20\equal{}44$, which is hopefully correct..... No way to prove though.....  :(

This is not right. The answer for octagon is 80. This problem is from 1986 National Team 5. The tricky part is some of the lines intersect so it not possible to generalize easily.

Also it is different for a polygon with odd sides and a polygon with even sides.

[url=http://mathworld.wolfram.com/RegularPolygonDivisionbyDiagonals.html]This[/url] might be useful.

Funny that that's not on Victor's formula sheet...

Any easier way to do it?
Can anyone produce a direct calculation?

Draw a large diagram and count up?

You can use symmetry for this problem - it is discussed more [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=210200562&t=202511]here[/url].


The Detroit Tigers and the Chicago White Sox play each other in the first round of the playoffs, a series which ends when one team gets a total of 3 wins.  The probability that each team wins is fixed across all games -- that is, the probability the Tigers win game 2 is the same as the probability that they win every other game -- and there are no ties.  After 3 games, Detroit is found to be leading the series 2 wins to 1, which is exactly what one would expect given theirs respective probabilities of winning.  What is the probability that Detroit goes on to win the series?

Now, in the World Series, Detroit plays the Chicago Cubs.  If it is found that after 4 games, Detroit leads 3-1 and this is exactly what one would expect given their relative chances of winning each game, what is the probability that Detroit goes on to win the series?

Why doesn't the same analysis work if instead this were a seven-game series and Detroit lead 3-2 after 5 games?

Joel, maybe these questions will activate this strand.

Clarification of wording and intent?

first,  by the phrase "Detroit is found to be leading the series 2 wins to 1, which is exactly what one would expect given theirs respective probabilities of winning" I inferred that you meant that their probability of winning each game is 2/3?  Is that your intent?

second, I don't understand your cryptic comment in the third question.  To confirm whether I'm on the right track, would you confirm if [hide]8/9[/hide] is correct for #1 and 'by same method' I get [hide]21/25[/hide] for #3.

By the way, hint: [hide]use a probablity tree diagram[/hide]

My intent was to say that given the probability that Detroit would win each game, after three games Detroit could have been expected to have 2 wins.  It is your job to figure out what the probability that Detroit won each game would be.

Oh, and I left something out of the second one -- the World Series is a "best of 7" series, so the first team to win 4 games wins.

Your answer to number 1 is [hide]correct[/hide], and your answer to number 3 is [hide]not.[/hide]

The same analysis doesn't work for the third one because:[hide]

For the first two you can see that the other team has to win all of the games so you can do

1-(1/3) :^2: =8/9

Because the probibility the other team wins is 1/3 and the other team has to win 2 games consecutivly to win the series

and for the second one you do

1-(1/4) :^3: =63/64

Because the probibility the other team wins is 1/4 and the other team has to win 3 games consecutivly to win the series

The same analysis doesn't work for the third one because the other team doesn't have to all of the rest of the games.

[/hide]

My answer for the third one is[hide]477/625[/hide]

Good thing you didn't say the Chicago CUBS, or I would have had to say that the probability is 0.

JMI wrote:[hide]
The same analysis doesn't work for the third one because the other team doesn't have to all of the rest of the games.
[/hide]
My answer for the third one is[hide]477/625[/hide]

They actually do have to win all games. if they lose a single game, the Detroit has 4 wins, and thus wins the series. My answer is [hide]1-(2/5) :^2: =1-4/25=21/25[/hide] So I'd say the same reasoning does work.

I've seen correct answers for 1 and 2.  I'm interested in where the more complicated-looking answer to number 3 came from, but I think it's wrong.  The "obvious" answer for number 3, [hide]21/25[/hide] is wrong, though.

Oh I just realized I completely messed up number 3 but I can't find the right way to do it either.

JBL wrote:which is exactly what one would expect given theirs respective probabilities of winning

JBL wrote:The "obvious" answer for number 3, [hide]21/25[/hide] is wrong, though.

It seems to me you have given the classic definition of expected value.  The probability of Detroit winning a game follows directly as p=3/5.

Your previous responses indicate that's incorrect.  So you must be supplying an expected value for some conditional probability.

I looked at the problem another way.  Let's assume your phrase "what one would expect" applies only to the series that are still in play.  The series may have ended 4-0, 4-1, 0-4, or 1-4.  The only series that are still in play after game 5 are those that now stand 3-2 or 2-3.

Consider the binomial expansion of [tex](p+q)^5[/tex], where p is the probability that Detroit wins and [tex]q=(1-p)[/tex].  The terms representing a series that's 3-2 and 2-3 after 5 games are given by:  [tex]{{5}\choose {3}}p^3q^2[/tex] and [tex]{{5}\choose {2}}p^2q^3[/tex].  Solving these two terms for p and q, such that the ratio of the terms is 3:2 also results in p=3/5, q=2/5.

You have stumped me.

[quote="JBL"]Now, in the World Series, Detroit plays the Chicago Cubs.  If it is found that after 4 games, Detroit leads 3-1 and this is exactly what one would expect given their relative chances of winning each game, what is the probability that Detroit goes on to win the series?

Why doesn't the same analysis work if instead this were a seven-game series and Detroit lead 3-2 after 5 games?[/quote]

Is the 3-2 in the third question exactly what one would expect given their relative chances of winning each game?

Okay:

Detroit has some winning percentage against the Cubs, p.  After 5 games [b][i]in a best of 7 series[/i][/b] with this chance of winning, it is expected that Detroit would have won 3 games.  I will tell you that actually solving for Detroit's chances of winning the series is hard; however, to explain why it isn't 21/25 is less hard.  I've given a hint in the content of this message.  Just in case you're wondering, the fault lies here:

[quote="rcv"]The probability of Detroit winning a game follows directly as p=3/5. [/quote]

I am using the traditional meaning of expected value that all of you are working with, though.  You need to figure out why it doesn't work out to 3/5.

But I thought E(X+Y) = E(X) + E(Y) with a slightly warped sense of "+", so wouldn't we have E(5 games) = 5*E(1 game), and so 3 = 5*E(1 game), and E(1 game) = 3/5?

That would be true, except that you are ignoring a very important part of the problem.  Let's look at the definition of (finite) expected value: the weighted average of all possible outcomes.  Now, the formula you are given is correct if we simply had the Cubs and Tigers play 5 games.  However, they aren't just playing 5 games -- they are playing a best of seven series.  This means that our available outcomes are limited -- in particular, Detroit is incapable of winning 5 games.  Thus, the expected value, were Detroit to win with probability 3/5, would actually be [i]smaller[/i] than 3 games in 5.  Detroit must have a chance of winning greater than three fifths in order to be expected to win three fifths of its games.

In fact, it is relatively easy to calculate that with a winning chance of 3/5, Detroit can expect to win 2.90688 of its first 5 games in the series.


Let $D$ represent a repeating decimal.  If $P$ denotes the $r$ figures of $D$ which do not repeat themselves, and $Q$ denotes the $s$ figures which do repeat themselves, then the incorrect expression is:
\[ \text{(A)}\ D = .PQQQ\cdots \qquad \text{(B)}\ 10^rD = P.QQQ\cdots \]
\[ \text{(C)}\ 10^{r+s}D = PQ.QQQ\cdots \qquad \text{(D)}\ 10^r(10^s-1)D = Q(P-1) \]
\[ \text{(E)}\ 10^r \cdot 10^{2s}D = PQQ.QQQ\cdots \]

That was my gut.. but not in the mood to try and work it.

A, B, C and E are correct.  C-B gives the correct left side of D but the incorrect right side - it should be P(Q-1).

D...because it doesn't look like a decimal :|

D, definitely.  I checked all the others and they worked.  Checked D and it didn't work, so I'm going with everybody else on this one. :)

It has to be D because I have no idea how they got that since there is nothing that repeats. Plus, everyone else is saying D because it looks really wierd and different from the other choices, so I pick D.  :?


calculus
integration
real analysis
real analysis unsolved

If  $a<b$ and  $w \in C[a,b]$ has the properties :
[color=green] [b] 1) [/b]  [/color]$\int\limits_{a}^{b} w(x)\; dx > 0$ , and
[color=green] [b] 2)[/b] [/color]  for every $f \in C[a,b]$ there is $\theta \in [a,b]$ such that  $\int\limits_{a}^{b}f(x)w(x)\; dx = f(\theta) \int\limits_{a}^{b} w(x)\; dx ,$
  [b] prove that $w \ge 0$  on $[a,b]$ . [/b]

We admit that statement of problem is not true, so $\exists x\in [a,b]$ such that $w(x)<0.$
Now $w\in C[a,b]$ so $w^{-1}(-\infty;0)$ is open in $[a,b]$, and $U=\{y\in [a,b] | w(y)<0\}=\bigcup_{k} I_k=\bigcup_{k} (\alpha_k;\beta_k)$ where $\alpha_{k+1}\ge\beta_{k}$ . And $x\in U$
Than $\exists l$ such that $x\in I_l$. (We assume that $\alpha_l,\beta_l \in (a,b)$. If this statement not true the solution is almost the  same. I think it's clear)
(In other way $\alpha_k=\inf_{y\in[a,b]} w(y)<0$ and $\beta_k=\sup_{y\in [a,b]} w(y)<0$.)
Than $w(\alpha_l)=w(\beta_l)=0,$ and $w(y)<0$ when $y\in (\alpha_l;\beta_l)$.
We define $f(y)$ as $w(y)$ when $y\in [\alpha_l,\beta_l]$ and $f(y)=0$ in other case. Than $f(y)\in C[a,b]$. (Even if $\alpha_k=a$ or $\beta_k=b$)Thus $f(y)\le 0$. And since $f(x)<0$ we have
\[ 0<\int\limits_{a}^{b}f(t)f(t)\; dt =\int\limits_{a}^{b}f(t)w(t)\; dt = f(\theta) \int\limits_{a}^{b} w(t)\; dt \le 0  \] because $\int\limits_{a}^{b} w(x)\; dx \ge 0$  and $f(\theta)\le 0 \forall \theta.$
Contradiction... And $w(y)\ge 0.$ :)

Assume that there exists $x_0$ such that $w(x_0)<0$. Take $f(x)=|\min(0,w(x))|$ and $g(x)=f(x)w(x)=-\min(0,w(x))^2<0$. Is easy to prove that $f \in C[a,b]$.
$\int_a^b f(x)w(x)dx=\int_a^b g(x)dx<0$.
On the other hand we have that
$f(\theta)\int_a^bw(x)dx>0$, since $f(x)>0,\forall x\in [a,b]$. Contradiction. So $w$ can take only nonegative values.

xirti
$f(\theta) \int\limits_{a}^{b} w(x)\; dx \ge 0$ and $f\ge 0$ but you solution correct in any way  :)

I also didn't prove that if g continuous , $g(x)\ge 0$ and $g(x_0)>0$ we have $\int_0^1g(x)dx >0$.
Since g is continuous someone can find a neibourghood $[c,d]$ of $x_0$ such that $g(x)>\epsilon>0,\forall x \in [c,d]$.
$\int_0^1g(x)dx=\int_0^cg(x)dx+\int_c^dg(x)dx+\int_d^1g(x)dx\ge\int_c^dg(x)dx\ge\epsilon (d-c)>0$


Show that $ i^{n}$ takes on one of four values $ \pm1$ or \pm{i} if n is an integral

[hide]
$ n\equal{}4k\Rightarrow i^{4k}\equal{}(i^{4})^{k}\equal{}1$
$ n\equal{}4k\plus{}1\Rightarrow i^{4k\plus{}1}\equal{}i\cdot i^{4k}\equal{}i$
$ n\equal{}4k\plus{}2\Rightarrow i^{4k\plus{}2}\equal{}i^{2}\cdot i^{4k}\equal{}i^{2}\equal{}\minus{}1$
$ n\equal{}4k\plus{}3\Rightarrow i^{4k\plus{}3}\equal{}i^{3}\cdot i^{4k}\equal{}i^{3}\equal{}\minus{}i$
[/hide]


why does sinx equal square root of (1-(cosx)^2)? isn't it +-square root of (1-(cosx)^2)?

hello, yes, it is $ \sin(x)\equal{}\pm\sqrt{1\minus{}\cos(x)^2}$ because $ \sqrt{x^2}\equal{}|x|$.
Sonnhard.

Are you asking this question because that substitution was used in a trigonometric proof? If so, can you show us the problem?

Otherwise, in general, it is as Sonnhard said.

NOTE: $ \sin{x} \equal{} \color{red}{\text{one of}\color{black}\pm\sqrt {1 \minus{} \cos^2{x}}}$it depends from interval of argument

[quote="batteredbutnotdefeated"]This is covered in AoPS volume 2 chapter 2. You can refer to it there.[/quote]

That's best for review. I haven't done any trig, can't wait till the AoPS trig comes out. What do I use before then?

Any Precalc book should work, if you want basics.

I haven't used it, but I've heard that it was good for trig:

Plane Trigonometry by S. L. Loney

Actually I think all you need for AMC 12 trig is right triangle trig and some identities


Let a,b,c are positive numbers. Prove that
$ 27(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2\ge 64abc(a\plus{}b\plus{}c)^3$
It have a nice solution. :D

[quote="thegod277"]Let a,b,c are positive numbers. Prove that
$ 27(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2\ge 64abc(a \plus{} b \plus{} c)^3$
It have a nice solution. :D[/quote]
\[ \left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right) \ge \frac{8}{9}\left( {a \plus{} b \plus{} c} \right)\left( {ab \plus{} bc \plus{} ca} \right)
\]
 :|

oh yes :) . Why don't I think it early? :maybe: >So, can you prove that this inequality
$ (a^2\plus{}ab\plus{}b^2)(b^2\plus{}bc\plus{}c^2)(c^2\plus{}ca\plus{}a^2)\ge abc(a\plus{}b\plus{}c)^3$

[quote="thegod277"]oh yes :) . Why don't I think it early? :maybe: >So, can you prove that this inequality
$ (a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2)\ge abc(a \plus{} b \plus{} c)^3$[/quote]
It is Holder. :wink:

[quote="ishfaq420haque"]can you describe how you used holder here...i am not used to use holder.please help.........[/quote]
\[ \left( {a^2  \plus{} ab \plus{} b^2 } \right)\left( {bc \plus{} c^2  \plus{} b^2 } \right)\left( {a^2  \plus{} c^2  \plus{} ac} \right) \ge \left( {\sqrt[3]{{a^4 bc}} \plus{} \sqrt[3]{{abc^4 }} \plus{} \sqrt[3]{{ab^4 c}}} \right)^3  \equal{} abc\left( {a \plus{} b \plus{} c} \right)^3
\]


number theory
relatively prime
prime numbers
algebra unsolved
algebra

Find all polunomials $P$ with integer coefficients which satisfay the property that, for any relatively prime integers $a,b$  the sequence $\left\{ P(an+b)\right\}$ for $n\geq 1$ contains an infite number of terms , any two of which  are relatively prime.

Could anyone post a solution ?Harazi could you give a hint or a solution ?

It is nesesary and sufficiently, that $gsd(a_0,a_1,\dots,a_n)=1,$ were $P(x)=a_0+a_1x+\dots+a_nx^n.$

Can you write a proof for this fact. Maybe is easy but I cannnot figure it out.  thank you!

Can you write a proof for this fact. Maybe is easy but I cannnot figure it out.  I can only see that it is necesary.  thank you!

I was wrong. Nesessary and sufficiently conditions are
1)$m_a=gcd(a_0,a_1,\dots ,a_n)=1,$
2)$m_b=(a,P(b))=1,$
3)for all primes $p\not|a$ (p-1<n) exist x,suth that $p\not|P(x).$
For example $3|(x^3-x)$ for all x.
Nesesary conditions is obviosly. Let we have conditions 1-3. If $p|P(ak+b)$ for all k we have p|a and p|P(b) - give contradition.
Let $p_{i,j}|P(x_i), x_i=ak_i+b,j=1,2,\dots l_i,i=1,2,\dots,m.$ We have $x_{i,j}$ suth that $p_{i,j}\not |P(x_{i,j})$. Because
$M_m=\prod_{i,j}p_{i,j},(a,M_m)=1$ we have $k_{m+1}$ suth that $(M_m,P(ak_{m+1}+b)=1$.
It give, that 1-3 sufficiently and nesessary conditions.

I think they can be characterized otherwise. Take $n$ and suppose that $n$ and $f(n)$ are relatively prime. Then $f(n+kf(n))$ is a multiple of $f(n)$ for all $k$ and therefore contradiction with the hypothesis. Therefore $f(n)$ and $n$ are not relatively prime for all $n$. Take prime numbers and see what you'll get.  ;)

2)(a,P(b))=1, for yuor sequence b=n,a=P(n) (a,P(b))=a.


Prove that for a given triangle with perimeter $P$ an equalateral triangle has the maximum area.

I'm not super experienced with proofwriting, but here goes:

[hide]
If the perimeter is $P$, let each side $=\frac{P}{3}$. By Herons, the area A--> $A^2=\frac{P}{2}(\frac{P}{2}-\frac{P}{3})^3$

So we need to prove that the value
($\frac{P}{6})^3$ is maximized for all P, with nothing added to or taken from any of the sides.

Now consider all cases where P is not equilateral. Let the two longer sides (isoscoles works in this case too) be $\frac{P}{3}+a$ and $\frac{P}{3}+b$, and then the short side is $\frac{P}{3}-(a+b)$.

Applying Herons again, the expansion turns out to be $\frac{P^3}{6^3}+\frac{Pab}{6}-\frac{Pa^2}{6}-\frac{Pb^2}{6}-ba^2-ab^2$.

To show that this is always smaller than the equlateral case, we need only show that all the negative terms (remember, a and b are 0 or positive) have an absolute value greater than $\frac{Pab}{6}$.

To do this we see that $\frac{Pab}{6}-\frac{P(a^2+b^2)}{6}$ must always be less than zero, because either a or b must be greater than the other (or equal), so $a^2$ or $b^2$ must be greater than ab (or they are equal and $a^2+b^2=2ab$).

Therefore, the case where the sides all equal $\frac{P}{3}$ gives the maximum area for any perimeter P.[/hide]


induction
real analysis
real analysis solved

Consider the real sequence (a_n)n>=1 defined by a1=a and a_(n+1)=a_n/n for all n>=1.
Prove that there is a unique a>0 such that lim(n-->oo) ((a_n)^(1/2^n))=1.

cheers! :D :D

Are you sure? Please modify your post so that it would be clear if an is a_n or a*n. Or if you use a_nn for a_(n*n) please make it clear.

If the text is the one I understand, then for every a>0 that limit is 1.

Prove by induction a_n=a/n!
lim(n->00)((a_n)^(1/2^n))=1 <=>
lim (1/2^n) ln(a_n)=0 <=>lim (lna/(2^n)-(lnn!)/2^n)=-lim lnn!/2^n=(Stolz-Cesaro) =-lim ln(n+1)/2^n=0. So you can see that if a>0, then it doesn't count in the limit, so forget about uniqueness.


Let $n \geq 3$ be an integer. Consider positive real numbers $a_1,..., a_n$ such that their product is 1. Show that: $\sum_{i=1}^n \frac{a_i+3}{(a_i+1)^2} \geq 3$

It can be done with Ipt, but what does mean SUV and GRT ?

[quote="harazi"]I have no idea how this problem arrived in UK, since I sent it to our team leader this year for the national olympiad and TST. Quite interesting if both people had exactly the same inequality in mind. Anyway, it's not that difficult, but I would really like to see a smart solution, without IPt, SUV, GRT and so on...[/quote]

oh, it was in the balkan shortlists for this year.

Please tell me what Ipt , SUV , GRT are !!

For future, do not ask also, what are "iran1996", or "manlio239.17.8b" inequalities unless you want to seem an ignoramus.  Every educated mathlinker must know such basic things.

There is a plenty of three-letters abbreviations for some very useful theorems devised by forum members, using which one may easily solve almost any inequality.

I guess these are various facts about minimazing a sum $\sum f(x_i)$ for a fixed $\sum x_i$, when $f$ is not convex or concave, but has not too much inflection points. But I think so just because it is a standard way to solve such inequalities.

[quote="harazi"]...Anyway, it's not that difficult, but I would really like to see a smart solution, without IPt, SUV, GRT and so on...[/quote]

I found it difficult...

As nobody posted a solution, let me post mine.
We can break the initial inequality into two parts:
1. $\sum_{i=1}^{n} \frac{1}{a_i+1} \ge 1$ and
2. $\sum_{i=1}^{n} \frac{1}{a_i+1}^2 \ge 1$.
Proof of 1.
Notice that \[
\begin{array}{l}
 \frac{1}{{x + 1}} + \frac{1}{{y + 1}} \ge \frac{1}{{xy + 1}} \\
  \Leftrightarrow (x + y + 2)(xy + 1) \ge xy + x + y + 1 \\
  \Leftrightarrow xy(x + y) + 2xy + x + y + 2 \ge xy + x + y + 1 \\
  \Leftrightarrow xy(x + y) + xy + 1 \ge 0 \\
 \end{array}
\]
which is true.
So, $\sum_{i=1}^{n} \frac{1}{a_i+1} \ge \frac{1}{a_1+1} + \frac{1}{a_2 \cdots a_n +1} = 1$.
Proof of 2.
Notice that
\[
\begin{array}{l}
 \frac{1}{{\left( {x + 1} \right)^2 }} + \frac{1}{{\left( {y + 1} \right)^2 }} \ge \frac{1}{{\left( {xy + 1} \right)^2 }} \\
  \Leftrightarrow \left( {x^2  + y^2  + 2x + 2y + 2} \right)\left( {x^2 y^2  + 2xy + 1} \right) \ge  \\
 x^2 y^2  + 2x^2 y + 2xy^2  + x^2  + y^2  + 4xy + 2x + 2y + 1 \\
  \Leftrightarrow \left( {x^2  + y^2  + 2x + 2y + 1} \right)x^2 y^2  + 2xy(x^2  + y^2  + x + y) + 1 \ge 0 \\
 \end{array}
\]
So,$\sum_{i=1}^{n} \frac{1}{(a_i+1)^2} \ge \frac{1}{(a_1+1)^2} + \frac{1}{(a_2+1)^2}+\frac{1}{(a_3+1)^2} + \frac{1}{(a_4 \cdots a_n)^2 +1}$.
It remains to prove if $xyzt=1$, then $\sum \frac{1}{(x+1)^2} \ge 1$.
We can then use $\frac{1}{(x+1)^2}+\frac{1}{(y+1)^2} \ge \frac{1}{xy+1}$ to finish it.

Then, we have proved the inequality for $n \ge 4$ by adding (1)+2(2).

For n=3, denote $f(x)=\frac{x+3}{(x+1)^2}$. Troublesome checking yields that $f(x)+f(y) \ge 2 f(\sqrt{xy})$ for $xy \ge 1$. WLOG assume $xy \ge 1$, then we have to prove $2f(\sqrt{xy})+f(z) \ge 3$ or $2f(\frac{1}{a})+f(a^2) \ge 3$ for $a \le 1$
This is equivalent to prove $\frac{2(a^2+3a)}{(a+1)^2}+\frac{a^2+3}{(a^2+1)^2} \ge 3$.
By cauchy, $L.H.S. \ge \frac{a^2+3a}{a^2+1}+\frac{a^2+3}{(a^2+1)^2}$
Expanding of $\frac{a^2+3a}{a^2+1}+\frac{a^2+3}{(a^2+1)^2} \ge 3$ yields $3a^3+3a \ge 2a^4+4a^2$ which is true by $a \le 1$ and $a^3+a \ge 2a^2$.
QED.

[quote="siuhochung"]
For n=3, denote $f(x)=\frac{x+3}{(x+1)^2}$. Troublesome checking yields that $f(x)+f(y) \ge 2 f(\sqrt{xy})$ for $xy \ge 1$. WLOG assume $xy \ge 1$, then we have to prove $2f(\sqrt{xy})+f(z) \ge 3$ or $2f(\frac{1}{a})+f(a^2) \ge 3$ for $a \le 1$
This is equivalent to prove $\frac{2(a^2+3a)}{(a+1)^2}+\frac{a^2+3}{(a^2+1)^2} \ge 3$.
By cauchy, $L.H.S. \ge \frac{a^2+3a}{a^2+1}+\frac{a^2+3}{(a^2+1)^2}$
Expanding of $\frac{a^2+3a}{a^2+1}+\frac{a^2+3}{(a^2+1)^2} \ge 3$ yields $3a^3+3a \ge 2a^4+4a^2$ which is true by $a \le 1$ and $a^3+a \ge 2a^2$.
QED.[/quote]

For n=3 you can put $a_1=1/x,a_2=1/y,a_3=1/z$ and use Cauchy.
I remember the similar method was given on 5'th mathlinks contest.($abc\ge 8... $)

[quote="jhaussmann5"][quote="hardsoul"]For n=3 you can put $a_1=1/x,a_2=1/y,a_3=1/z$ and use Cauchy.
I remember the similar method was given on 5'th mathlinks contest.($abc\ge 8... $)[/quote]

Could you make that more explicit?  I can't get Cauchy to work.[/quote]
Well,unfortunately I have lost the solution paper,and couldn't remember my solution by Cauchy(it seems it was wrong)
Here's another solution:

1)We can reduce the inequality to the case n=3 by repeadately using the inequality $\displaystyle\frac{x+3}{(x+1)^2}+\frac{y+3}{(y+1)^2}\ge \frac{xy+3}{(xy+1)^2}$ for $x\ge1,y\leq 1$
2)Put $a_1=1/a,a_2=1/b,a_3=1/c$ to get $\displaystyle\sum\frac{a+3a^2}{(a+1)^2}\ge 3$ that is equivalent to
$\displaystyle\sum\frac{2a^2-a-1}{(a+1)^2}\ge 0\Leftrightarrow\displaystyle\sum\frac{2a^2}{(a+1)^2}\ge \displaystyle\sum\frac{1}{a+1}$  Now use the inequality $\sum_{cyc}\left(\displaystyle\frac{a^2}{(a+1)^2}+\displaystyle\frac{b^2}{(b+1)^2}\right)\ge \displaystyle\sum_{cyc}\frac{1}{c+1}$

[quote="indybar"][quote="zhaobin"]I think we should only prove $n=3$
and $n=3$ implies $n\geq3$[/quote]

Why is this? Can you show?[/quote]

  Well, I think zhaobin said is right. Because
$ \sum_{i=1}^4 \frac{a_i+3}{(a_i+1)^2} \geq\sum_{i=1}^3 \frac{a_i+3}{(a_i+1)^2} $
So if $n=3$ is true. Obviously$ n=4$ is true too.
Is there anyone use Jensen to solve it?  :?

[quote="Gibbenergy"]So if $n=3$ is true. Obviously$n=4$ is true too.[/quote]

Why "obviously"?

It should be:
if $abc \le 1$ then:
$\sum \frac{a+3}{(a+1)^2} \ge 3$

For n number, we suppose that $a=a_1 \le b=a_2 \le c=a_3 \le a_4 \le ... \le a_n$ then $abc \le 1$

Apply that inequality we have done.
(Note that $abc \le 1$, not $abc=1$)


abstract algebra
number theory
prime numbers
group theory
number theory unsolved

Find all positive integers $n>1$ for which there exists a unique integer $a$ with $0 < a\leq n!$ such that $a^n+1$ is divisible by $n!$.

The answer is all prime numbers $n$.

Here is my not very beautiful but I hope correct solution:

$n=2$ is a solution, for $n\ge 4$ even, $a^{n}$ is a square and $3\mid n!$ but $3\nmid a^{n}+1$, so $n$ must be odd if $n\ge 3$.

Let $n!=p_{1}^{k_{1}}\ldots p_{r}^{k_{r}}$ with $p_{i}$ distinct primes. By Chinese Remainder Theorem, $a^{n}+1\equiv 0\mod n!$ has unique solution in $a$ iff $a^{n}+1\equiv 0 \mod p_{i}^{k_{i}}$ has a unique solution for all $i=1\ldots r$.

We shall prove first, that $a^{n}+1\equiv 0 \mod 2^{k}$ has a unique solution.
We have $a^{n}+1=(a+1)(\sum_{i=0}^{n-1}(-1)^{i}a^{i})\equiv 0 \mod 2^{k}$, but $\sum_{i=0}^{n-1}(-1)^{i}a^{i}$ is odd, so $a+1\equiv 0\mod 2^{k}$.
Thus, we do not need to worry about the oddest prime $2$ anymore.

First, let $n\ge 3$ be a prime. Clearly, $v_{n}(n!)=1$.
By FLT, $a^{n}\equiv a \equiv-1\mod n$, so there is a unique solution modulo $n$.
Now let $p\geq 3$ be a prime dividing $n!$ and let $k=v_{p}(n!)$ and $p\neq n$.
Furthermore let $g$ be a primitive root $\mod p^{k}$.
Then $-1\equiv g^{\frac{\phi(p^{k})}{2}}\mod p^{k}$ and $a\equiv g^{x}\mod p^{k}$ for some $x\in \{1,2,\ldots,\phi(p^{k})\}$.
So $a^{n}\equiv-1\mod p^{k}\Leftrightarrow g^{nx}\equiv g^{\frac{\phi(p^{k})}{2}}\mod p^{k}\Leftrightarrow nx\equiv \frac{\phi(p^{k})}2 \mod \phi(p^{k})$.
Hence there is an unique solution $a\mod p^{k}$ iff there is a unique solution $x\mod \phi(p^{k})$. But $nx\equiv \phi(p^{k})/2 \mod \phi(p^{k})$ has a unique solution in $x$ since $gcd(n,phi(p^{k}))=gcd(n,(p-1)p^{k-1})=1$.
Thus there is a unique $a\mod n!$ for $n\ge 3$ prime.

Now let $n$ be composite and let $p\ge 3$ be a prime dividing $n$ and different from $n$ (which exists since we assumed that $n$ is odd).
Clearly, $n\ge 2p$ so $v_{p}(n!)\ge 2$.
Now if $a$ would be unique $\mod n!$ then it would be unique $\mod p^{k}$ where $k=v_{p}(n!)\ge 2$.
Let $g$ be a primitive root $\mod p^{k}$. Thus again, $-1\equiv g^{\frac{\phi(p^{k})}2}\mod p^{k}$ and $a\equiv g^{x}\mod p^{k}$ for some $x\in\{1,\ldots,\phi(p^{k})\}$.
Now $a^{n}\equiv-1 \mod p^{k}\Leftrightarrow nx\equiv \phi(p^{k})/2 \mod \phi(p^{k})$, so by the assumption that $a$ is unique $\mod p^{k}$, there is a unique $x\mod \phi(p^{k})$, i.e. that $nx\equiv \phi(p^{k})/2 \mod \phi(p^{k}) \Leftrightarrow nx\equiv \frac{(p-1)p^{k-1}}2 \mod (p-1)p^{k-1}$ has a unique solution in $x$.
But this is a contradiction since $p\mid n$ and $p\mid (p-1)p^{k-1}$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=85073 also.

@Yimin: I agree to your signature :)

Ghm, in all solutions to this problem that I have seen the part where one shows that prime $n$ works is handled somewhat meticulously. Correct me if I am wrong but it follows immediately from the fact that mapping $x\rightarrow x^{a}$ is a bijection for the set $A=\{x|(x, b)=1\}\subset \mathbb{Z}_{b}$ if $(a, \phi(b))=1$ (which is the case for the problem since $(p, \phi(p!))=1$).

Take prime power $p^{r}\parallel b$ such that $d=(a, \phi(p^{r})) \geq 2$. Write $a=kd$ and let $g$ be the primitive root $\mod{p^{r}}$. We have:
$(g^{\frac{\phi(p^{r})}{d}})^{d}\equiv 1 \mod{p^{r}}$.
Because $g$ is the primitive root $g^{\frac{\phi(p^{r})}{d}}$ is not $1$.
Also, $g^{\frac{\phi(p^{r})}{d}}\equiv x^{k}$ for some $x$ because $(k, \phi(p^{r}))=1$. Therefore: $(x^{k})^{d}= x^{a}\equiv 1$. Now, write $x \equiv \frac{s}{t}$ and we have $s^{a}\equiv t^{a}$. Since $x \not \equiv 1$ $k \not \equiv t$ and this mapping isn't bijective for some highest prime power dividing $b$. so it isn't a bijective mapping $\mod{b}$.
Actually, the case $b=2^{k}$ should be treated seperately.

More general claim:

Let $G$ be any finite group. Then $x \mapsto x^{k}$ is bijective iff $\gcd(k,|G|)=1$.

Proof:
If they are coprime, we find by Bezout's theorem integers $a,b$ with $ak-b|G|=1$. By Lagrange we have $x^{|G|}=e$ for all $x \in G$, thus $x^{ak}= x^{1+b|G|}= x$.
Thus the map $x \mapsto x^{a}$ is an inverse mapping, thus both mappings are bijective.
Converse:
When they are not coprime, let $p$ be a common prime divisor of $|G|$ and $k$. On various ways we can get that $x \mapsto x^{p}$ is already not injective, so also not $x \mapsto x^{k}$. The most appealing one (since it is near to be equivalent) is using Cauchy's theorem (see http://www.mathlinks.ro/Forum/viewtopic.php?t=27036 , post #6 for a proof) which gives that there is an $x$ with order $p$, so an $x \neq e$ with $x^{p}=e$, showing that it is not injective.

Since my proof for the residue case was more like the one of TomciO (for the original problem one could avoid using primitive roots, but well...), I will not post it.

[quote="venkata"]Find all positive integers $n>1$ for which there exists a unique integer $a$ with $0 < a\leq n!$ such that $a^{n}+1$ is divisible by $n!$.[/quote]
If n is odd then $n!|(n!-1)^{n}+1 \ \ (a=n!-1)$.

@Rust: this post was completely useless.
Read the previous posts and especially the links given there.
And note that this helps as good as nothing ;)

[quote="ZetaX"]The converse holds also (it is only bijective iff they are coprime), giving the whole statement. Thats the way I proved it in our TST, with proving this equivalence
I think this was the source of the problem also.[/quote]

It seems as though I'm too dumb to understand your solution  :maybe:

How do you prove the problem with the converse? I don't really understand how you deduce from $f$ being not bijective that $-1$ is hit more than once? How do you prove that e.g. the case that only $f(-1)=-1$ and the rest is not bijective (e.g. that there are different $x,y$ so that $f(x)=f(y)\neq-1$ or whatever) cannot occur?

Edit: Here $f$ is of course the map $x\mapsto x^{k}$ you were talking about

Sorry for not mentioning/writting that last step:

The general one for groups again:
When $f: G \to H$ is a homomorphism of groups (so $f(ab)=f(a)f(b)$) but not injective, then $f(x)=a$ has never exactly one solution $x$ for given $a \in H$.

Proof:
At first, note that $f(t^{-1})=f(t)^{-1}$ for all $t \in G$ (simple to see when not clear from beginning).
There are two different $x_{1},x_{2}\in G$ with $f(x_{1})=f(x_{2})$. Thus setting $z: =x_{1}x_{2}^{-1}\neq e_{G}$ we get $f(z)=f(x_{1}x_{2}^{-1})=f(x_{1})f(x_{2}^{-1})=e_{H}$, the neutral element (this is one direction of the well known property $f$ injective $\iff$ $\ker(f)=\{e_{H}\}$).
Now let that $a$ be given. When there is $x$ with $f(x)=a$ (when there is none, we are done) we find for $y: =xz \neq x$ that $f(y)=f(xz)=f(x)f(z)=a e = a$.
So there are at least two different solutions, done.

Now the mapping $x \mapsto x^{k}$ is a homomorphism $G \to G$ at least when $G$ is abelian, which is the case here.

All in all, this seems to be pure group theory  :lol:

Or note that $n$ must be odd (on the contrary $-a\neq a$ satisfies the congruence as well) and deduce that for a congruence $a^{n}\equiv-1\mod{n!}$ to have unique solution the congruence $(-a)^{n}\equiv 1\mod{n!}$ must have unique solution. But if the mapping is not bijective then there exist $b\neq c\colon \ b^{n}\equiv c^{n}\mod{n!}$. But then $\frac{b}{c}\neq 1$ and $(\frac{b}{c})^{n}\equiv 1$. Hence $-1$ is not unique and we are done.


Consider C(O,R) a circle and C'(O',R)another circle sunh that O' is on C.Consider A and B the two intersection points between    the two circles.Let C be on the arc AB and on the other part of AB again on the AB arc a point D.
a)Find the maximum area of ADC
b)Find the maximum area of ABCD
                                                                                Luca V. Iliesiu

For a given $D \in \mathcal C'$, the triangle $\triangle ACD$ has maximum area, when C is the tangency point of is the circle $\mathcal C$ with the tangent $t \parallel AD.$ Then $CO \perp t,\ CO \perp AD$ is the C-altitude line. Let F be the foot of this altitude. Denote the directed angle $\angle O'AD = \phi$ (can be positive or negative). Since the circles $\mathcal C' \cong \mathcal C$ are congruent and $O' \in \mathcal C,$ the triangle $\triangle AOO'$ is equilateral and $\angle OAO' = 60^\circ.$ Then $\angle OAF \equiv \angle OAD = \angle OAO' + \angle O'AD = 60^\circ + \phi,$ $OF = OA \sin \widehat{OAF} = R \sin (60^\circ + \phi)$ and $CF = OC + OF = R[1 + \sin(60^\circ + \phi)].$ In addition, $AD = 2R \cos \phi.$ Area of the triangle $\triangle ACD$ is then

$S = |\triangle ADC| = \frac{CF \cdot AD}{2} = R^2 [1 + \sin(60^\circ + \phi)] \cos \phi$

For the maximum area, $\frac{\text dS}{\text d\phi} = 0.$

$\cos(60^\circ + \phi) \cos \phi - [1 + \sin(60^\circ + \phi)] \sin \phi] = 0$

$\sin \phi = \cos(60^\circ + 2\phi) = \sin(90^\circ - (60^\circ + 2\phi)] = \sin (30^\circ - 2 \phi)$

$\phi = 30^\circ - 2 \phi,\ \ \ \phi = 10^\circ$

Starting from an arbitrary point $C \in \mathcal C,$ we can similarly find $\angle OAC = 10^\circ.$ Hence, the triangle $\triangle ADC$ is isosceles with $CD \parallel OO'$ and $\angle CAD = \angle CAO + \angle OAO' + \angle O'AD = 80^\circ.$

The 2nd problem is trivial, even if the circles $\mathcal C, \mathcal C'$ are not congruent or $O'$ is not on $\mathcal C.$ The area of the quadrilateral ABCD is $|ABCD| = \frac 1 2 AB \cdot CD \sin\widehat{(AB, CD)}.$ AB is fixed and both CD and $\sin\widehat{(AB, CD)}$ reach their maximum values when $C \equiv OO' \cap \mathcal C,\ D \equiv OO' \cap \mathcal C',$ where C is outside of $\mathcal C'$ and D is outside $\mathcal C.$

[color=darkblue]Sorry, Yetti, your proof isn't clearly a bit for me. I will try to prove more clearly what you proved above. Please, see you if I made a mistake somewhere ![/color]

[color=darkred]Define $x\ .s.s.\ y$ if and only if $x=y$ or $xy>0$. I will suppose that: the point $D$ belongs to the little arc $\frown\above AB\end$ of the circle $C'(O',R')$; the point $C$ belongs to the little arc $\frown\above AB\end$ of the circle $C(O,R)$ so that $CO\perp AD$. Denote $m(\widehat {O'AD})=x$ and the projection $F$ of the point $C$ on the line $AD$. We observe that: $30^{\circ}\le x\le 90^{\circ}$;  $AD=2R'\cos x$ and $CF=R[1+\sin (x-60^{\circ}]$. Therefore, $[ACD]=AD\cdot CF=$ $R\cdot R'f(x)$, where
$\boxed {f(x)=\cos x[1+\sin (x-60^{\circ})]\mathrm {\ ,where\ }x\in \left[30^{\circ},90^{\circ}\right]}$.
$f'(x)=-\sin x+\cos \left(2x-60^{\circ}\right)=\cos \left(2x-60^{\circ}\right)-\cos \left(90^{\circ}-x\right)\Longrightarrow$
$f'(x)\ .s.s.\ \left[\left(90^{\circ}-x\right)-\left(2x-60^{\circ}\right)\right]=3\left(50^{\circ}-x\right)$, i.e. $\boxed {f'(x)\ .s.s.\ \left(50^{\circ}-x\right)}\ .$
We came to the conclusion that for any $x\in \left[30^{\circ},90^{\circ}\right]$
we have $f(x)\le f\left(50^{\circ}\right)$ a.s.o.[/color]

nice solution...well mine was similar when I made the problem...Just using trigonometry

[quote="Virgil Nicula"][color=darkblue]Sorry, Yetti, your proof isn't clearly a bit for me. I will try to prove more clearly what you proved above. Please, see you if I made a mistake somewhere ![/color]...[/quote]
Maybe you did not notice that the circles $\mathcal C \cong \mathcal C'$ are congruent, there is no $R' \neq R.$ The minor arcs AB of both circles $\mathcal C, \mathcal C'$ are to be ignored - if either C or D or both are on the minor arcs, then the line CD meets the corresponding circle at C' or D' and areas of the triangles $\triangle AC'D, \triangle ACD', \triangle AC'D'$ are larger than the area of $\triangle ACD.$ If D on the major arc of $\mathcal C'$ is arbitrary, with an angle $\angle O'AD = \phi$, then the base AD is fixed and to maximize the altitude, it must be $CO \perp AD.$ Both AD and the C-altitude $CO \equiv CF$ can be calculated in terms of $\phi = \angle O'AD.$ It does not matter how $\phi$ is directed, if we direct it the wrong way, it will come out negative. Acording to my direction,

$AD = 2R \cos \phi,\ \ \ CF = R[1 + \sin(60^\circ + \phi)], \ \ \ \phi \in (-30^\circ, +90^\circ).$

According to your direction, $x = -\phi$ and

$AD = 2R \cos x,\ \ \ CF = R[1 + \sin(60^\circ - x)],\ \ \ x \in (-90^\circ, +30^\circ).$

However, if $R' \neq R,$ the side AD and the altitude CF would be

$AD = 2R' \cos x,\ \ \ CF = CO + OF = R + R' \sin(60^\circ - x)$

$\frac{CF \cdot AD}{2} = RR' \cos x \left[1 + \frac{R'}{R}\ \sin(60^\circ - x)\right] \neq RR' f(x),$ where $f(x) = \cos x\ [1 + \sin(60^\circ - x)]$

and it would not be easy to solve $f'(x) = 0$ for x, except numerically. Assuming the problem condition $\mathcal C \cong \mathcal C',$ then maximizing $f(x) = f(-\phi)$ with respect to the angles $\phi$ or $x = -\phi$ gives $\sin (30^\circ - 2 \phi) - \sin \phi = 0$ or $\sin(30^\circ + 2x) + \sin x = 0.$ Instead of concluding $30^\circ - 2 \phi = \phi$ or $30^\circ + 2x = -x$ leading to $\phi = +10^\circ,\ x = -10^\circ,$ we could rigorously write

$\sin (30^\circ - 2 \phi) - \sin \phi = 2 \cos \frac{30^\circ - \phi}{2} \sin \frac{30^\circ - 3 \phi}{2} = 0$

$\sin (30^\circ + 2x) + \sin x = 2 \sin \frac{30^\circ + 3x}{2} \cos \frac{30^\circ + x}{2} = 0$

$x = -\phi \in (-90^\circ, +30^\circ)\ \Longrightarrow$

$\frac{30^\circ + x}{2} = \frac{30^\circ - \phi}{2} \in (-30^\circ, +30^\circ)\ \ \&\ \ \frac{30^\circ + 3x}{2} = \frac{30^\circ - 3 \phi}{2} \in (-120^\circ, +60^\circ)$

Thus $\cos \frac{30^\circ + x}{2} = \cos \frac{30^\circ - \phi}{2} \neq 0$ and sine has the only root 0 in $(-120^\circ, +60^\circ),$ hence $\frac{30^\circ + x}{2} = \frac{30^\circ - \phi}{2} = 0,$ $x = -10^\circ, \phi = +10^\circ.$ The maximum triangle area is then

$|\triangle ADC|_{\text{max}} = R^2 \cos 10^\circ (1 + \sin 70^\circ) = R^2 \cos 10^\circ (1 + \cos 20^\circ) =$

$= 2R^2 \cos^3 10 = \frac 1 2 (2R \cos 10^\circ)^2 \sin 80^\circ$

Having calculated the maximum triangle area, we can stop right here.

O.K. Thanks. I understood wrong the initial problem,.


you need a two in the partial fraction decomp or something...

cause i get 1/(x+2)^2+1/(x)^2-1/(x+2)^2

and not enough stuff cancels...

Wow, I posted this at 3 AM and it shows.  I rewrote the problem.  Altheman, you were on the right track.  Sorry!


inequalities
rearrangement inequality
algebra solved
algebra

Prove that

x^(a+1)/y^a  + y^(a+1)/z^a  + z^(a+1)/x^a  \geq x+y+z

where x,y,z,a are all >0.

It is very simple by rearrangement inequality so I have found a solution without using rearrangement inequality.

I think the beauty of inequality resides in the possibility to find different solutions.

Indeed, another solution is the following:
   Let us suppose that x+y+z=1. Then using the convexity of f(t)=t^(a+1) we find that y*f(x/y)+z*f(y/z)+x*f(z/x)>=f(y*x/y+z*y/z+x*z/x)=f(1)=1, which is excatly the desired inequality.


So I understand how to do stoichiometric calculations with chemical equations containing strong acids/bases, but how do I do neutralization/tritration calculations when a weak acid or base or both take part in the reaction? I don't know how to take into account for the incomplete dissociation of the weak acids/bases into the calculation. Do I just multiply the result by the percent of the weak acid/base that dissociates?

Here is a starting point:
When a weak acid is titrated by a strong base, or vice versa, the reaction between the two is assumed to proceed to completion, due to the constant stress placed on the system (this is in accord with the Le Chatelier principle). Therefore, when doing calculations with titration, start with the stoichiometric calculations and then move on to calculating the equilibrium concentrations.

This is an equilibrium problem. Don't talk about "percent of dissociation" - that won't get you anywhere. You need to know the pKa for each of the species involved.

You will almost certainly never see a problem involving titration both a weak acid/weak base.  I have never worked one myself, but I think you would need to do a lot of algebra.  One of the reason these problems never come up is that in practice, they are never used.  You always have a choice of which titrant to use, so why use a weak acid/base when you could make things more simple by using a strong acid/base?

For how to actually perform these calculations, you may find the following site useful:

[url]http://www.chembuddy.com/?left=pH-calculation&right=toc[/url]


symmetry
geometry
geometric transformation
reflection
similar triangles
angle bisector

Let AM be a median of triangle ABC, where M is the midpoint of BC. Point B, C are reflected over AM to B', C', respectively. AC' intersects BC at D. B'C' intersects AC at E. AE=6, EC=12, BD=10, find AB.

$\begin{array}{ccc}
&A&\\C&--------&D\\&B&\end{array}$

A is B, reflected over CD

I've got no clue. Sorry. So far I have:

[hide]Triangle ACC' is isocelis so AC'=AC=AE+EC=18[/hide]

I don't think it helps much, though.

I'm having a terrible time drawing this monstrosity. =D  Anyone wanna help me out?

Solution: [hide]First refer to diagram 1:

1. Prove that BNM, B'NM, CQM, C'QM are all congruent.

2. Then the shape NB'CQ must be rectangular. Since AE = 1/2 EC, we must have AM = 1/2 B'C by similar triangles. However NM = 1/2 B'C, so A and N are the same point.

Now refer to diagram 2:

3. Then by symmetry, BD = B'E. AE = 6, EC = 12, B'E = 10, by similar triangles EM = 5. Let AM = x, B'C = 2x, AB = y. Then

y2 + x2 = 152

y2 + 4x2 = 182

Then x2 = 33, y2 = 192, y = 8:rt3:.[/hide]

Nice!! my solution:

[hide]By definition of reflection, , so AD=AE=6, angle DAM=angle DAE. So AM is the angle bisector of triangle DAC.

By angle bisector theorem,

AC/AD=18/6=MC/DM=3, MC=3MD.

Since M is the midpoint of BC. BD+MD=MC,  10+MD=3MD, MD=5, MC=15, BC=30, CD=20

Apply Stewart's Theorem to ABC, we get

AB^2(CD)+AC^2(BD)=AD^2(BC)+(BD)(CD)(BC)

AB^2(20)+(18^2)(10)=6^2(30)+(10)(20)(30)

 AB^2=192, AB=8 :sqrt: 3 [/hide]


calculus
derivative
function
integration
real analysis
real analysis unsolved

In Evans' book "PDE", page 29, thm7:
$ u$ harmonic in $ U$,
prove: $ |D^{\alpha}u(x_0)|<\equal{}\frac{C_k}{r^{n\plus{}k}}\parallel{}u\parallel{}_{L_1(B(x_0,r))}$ for each ball $ B(x_0,r) \subset U$.

In the proof: case $ k\equal{}0$ already proved, for case $ k\equal{}1$,
let $ x \in \partial B(x_0,r/2)$, then $ x \in B(x,r/2) \subset B(x_0,r)$, so use case $ k\equal{}0$, we have:
$ u(x)<\equal{}\frac{C_0}{r^{n}}\parallel{}u\parallel{}_{L_1(B(x_0,r))}$.

I think it should be $ u(x)<\equal{}\frac{C_0}{r^{n}}\parallel{}u\parallel{}_{L_1(B(x,r/2))}$ by using case $ k\equal{}0$ ( take
$ x_0\equal{}x$). Did I miss something?

Maybe $ \parallel{}u\parallel{}_{L_1(B(x,r/2))}<\equal{}\parallel{}u\parallel{}_{L_1(B(x_0,r))}$ because $ B(x,r/2) \subset B(x_0,r)$? but why?

[quote="fedja"][quote="kalazealot"]
Maybe $ \parallel{}u\parallel{}_{L_1(B(x,r/2))} < \equal{} \parallel{}u\parallel{}_{L_1(B(x_0,r))}$ because $ B(x,r/2) \subset B(x_0,r)$? but why?[/quote]
Because the integral of a non-negative function $ |u|$ over a larger set is not smaller ;)[/quote]

 I see. :blush:


Are we allowed to use mathematica while solving a problem?

From the USAMTS Rules page ([url]http://www.usamts.org/About/U_AbRules.php[/url]):
[quote="USAMTS Rules"]Except where otherwise noted in problems, participants may use calculators and computers to solve problems. In this case, the participant must explain the program that was used to solve the problem, and must prove that the method used in the program does solve the problem. Symbolic manipulation calculators and mathematical programs such as Mathematica, Maple, and Matlab are considered programs for this purpose. Therefore, if a key step of a participant's solution uses these tools, the participant must demonstrate that the tool completed the step correctly.[/quote]


(a) Find all five-digit numbers which are reversed on multiplying by $ 4$.

(b) Find all five-digit numbers which are reversed on multiplying by $ 9$.

(c) Find all five-digit numbers which are reversed on multiplying by $ 8$.

[hide="a."] We can set bounds as $ 20000\le x < 25000$ (since our end result is also $ 5$ digits and no multiple of $ 4$ ends in $ 1$). We can then set further bounds at $ 21000\lex\le21999$ and $ 23000\lex\le23999$ (since $ 2$, $ 22$, and $ 42$ are not divisible by $ 4$). We can next eliminate the $ 23abc$ case by noting that when this is multiplied by $ 4$ the ten thousands digit will be $ 9$ ($ a$, $ b$, and $ c$ are digits). But any number that ends in $ 9$ when multiplied by $ 4$ will end in $ 6$. Thus we look at the $ 21ab8$ cases. By systematically checking (we can limit checking by noting that $ \frac {85000}{4} = 21250$, $ \frac {86000}{4} = 21500$, and $ \frac {87000}{4} = 21750$), we find that the only is $ \boxed{21978}$, since $ 21978\times{4} = 87912$.[/hide]

[hide="c."] I don't think there are any. We have that $ \frac{100000}{8}=12500$, so any number which reverses on multiplying by $ 8$ would need to be in the bounds $ 10000\lex\le12500$. But any multiple of $ 8$ must be even, and that cannot be accomplished with a $ 1$ as the last digit. So there are no numbers. [/hide]

[hide="b."] By using a similar process to the other problems, we can set bounds on $ 9$. These bounds are $ 10000\le\11111$. Additionally, our number must end in $ 9$ (since when it is multiplied by $ 9$ the number will begin with a $ 9$) , and since it is a multiple of $ 9$, the number must sum up to $ 9$. Running through the cases, we find that the only number that works is $ \boxed{10989}$, since $ 10989\times{9}=98901$.[/hide]


pLease heLp me..

h0w t0 use the trig0n0metric tabLe with aLL the degrees, minutes, and sec0nds..

t0gether with tan, csc, cos, sin..

pLease heLp me..

urgent pLease..

huhuhu

...?
A minute is $\frac{1}{60}$ of a degree, and a second is $\frac{1}{60}$ of a minute. There are 360 degrees in a circle.
But I'm not quite clear on what you're asking about.

in the t0pic..

s0Lving angLes inv0Lving 0bLique triangLes..

instead 0f using a caLcuLat0r..

y0u can use the said tabLe..

but h0w can i use the tabLe?..

pLease repLy..

And, as you can see, it distracts the rest of us from your actual problem.  :D

[jli edit]: ok...we dont need 4 ppl telling him that its annoying.  we only need 3...jk.  so i deleted all the other ones.


analytic geometry
geometry
graphing lines
slope

If a Cartesian coordinate grid were placed on top of the map of Math City, Algebra Lane would go through the points $ (\minus{}1,3)$ and $ (1,\minus{}1)$. Geometry Boulevard runs perpendicular to Algebra Lane and passes through the points $ (2,2)$ and $ (\minus{}2,y)$. What is the value of $ y$?

the slope of the algebra lane is $ \frac{\minus{}1\minus{}3}{1\minus{}(\minus{}1)}$, which is $ \minus{}2$.

and geometry boulevard's slope is 1/2 (perpendicular to -2)

$ \frac{y\minus{}2}{\minus{}2\minus{}2}\equal{}1/2$

-4=2y-4.
y=0

answer : 0


twenty five points are arranged in a 5x5 grid. If two distinct points are chosen at random from th egrid, what is the probability that the segment connecting these two points will not pass through any other points

Are you sure it is a probability problem? Re-read the problem.
 (I did this problem already)

hello Mathlearner2012. The question says what is the probability....?
So It is is it not? Anyway can you help me?

[quote="PowerOfPi"][b]Please post problems in the Alcumus Problem Discussion forum.[/b][/quote]

This forum is for issues with the program, not problems/solutions

To tell you the truth, you are partly right. This problem is a "basic probability as counting(no combinations)" problem.


calculus
integration
trigonometry
calculus computations

[color=blue]
Evaluate the following integrals:

1.  $ \int \frac {1}{\left(a\plus{}b\cos x \right)^{2}} dx$       where  $ a \ne  b$

2.  $ \int \frac {\left(\cos x \plus{}2\sin x \plus{}3\right)}{\left(4\cos x \plus{}5\sin x \plus{}6\right)^{2}} dx$
[/color]

[quote="kunny"]$ \tan \frac {x}{2} \equal{} t$[/quote]
it will work but very-very lengthy solution. Try something else or some elegant approach

[quote="Dr Sonnhard Graubner"]hello, i think the tan half angle substitution works here.
Sonnhard.[/quote]
[color=blue]
I do agree with you but i want to say that don't put it at the outset of the problem. First try to rewrite it to reduce the length of solution considerably then you should go for it otherwise what you people are saying is theoretically easy practically not advisable also very complex and lengthy solution. Please write your answers at least if not the solution it self.
[/color]

A slight correction thanks kunny,
[color=blue]
Evaluate the following integrals:

1.  $ \int \frac {1}{\left(a \plus{} b\cos x \right)^{2}} dx$       where  $ |a| \ne |b|$

2.  $ \int \frac {\left(\cos x \plus{} 2\sin x \plus{} 3\right)}{\left(4\cos x \plus{} 5\sin x \plus{} 6\right)^{2}} dx$
[/color]


I have 5 red plates and 4 blue plates. If I randomly select two plates to serve dinner on, what is the probability that they're both the same color?

Case 1: Both are red.

There is a  5/9 times 4/8 = 5/18 of this occurring.

Case 2: Both are blue.

There is a 4/9 times 3/8 = 1/6.

Adding these two fractions together gives 4/9.

Alternatively, the probability that the two plates are not the same color is $ 2\left(\frac59\cdot\frac48\right)\equal{}\frac59$, so the probability that they are is $ 1\minus{}\frac59\equal{}\boxed{\frac49}$.


[quote="keta"]ever since I saw this problem on the AMC 12, I have never understood it, so would someone please expain to be the logic behind the problem?

A polynomial of degree 4 w/ leading coefficient 1 and integer coefficients has 2 real roots, both of which are integers. Which of the following can also be a zero of the polynomial?

a. (1+i :sqrt: 11)/2

B. (1+i)/2

C. 0.5 + i

D. 1 + i/2

E. (1+i :sqrt: 13)/2[/quote]this is what i think:
let the two integer roots be a and b.  the equation is now P(X)=Q(N)(x-a)(x-b) where Q(N) is  a quadratic with integer coefficients.  so whichever the answer is, (lets say it is m+a) then (m+a)*(m-a) is an integer(follows from quadratic).

DISCLAIMER: i may be completely wrong

[quote="polymorphic"]yeah...except it's the first one (a) since the product is 3.[/quote]
What?

The product of (a) and its complex conjugate is 3.


For every triangle $ ABC$, let $ D,E,F$ be a point located on segment $ BC,CA,AB$, respectively. Let $ P$ be the intersection of $ AD$ and $ EF$. Prove that:
\[ \frac{AB}{AF}\times DC\plus{}\frac{AC}{AE}\times DB\equal{}\frac{AD}{AP}\times BC\]

[quote="wangsacl"]For every triangle $ ABC$, let $ D,E,F$ be a point located on segment $ BC,CA,AB$, respectively. Let $ P$ be the intersection of $ AD$ and $ EF$. Prove that:
\[ \frac {AB}{AF}\times DC \plus{} \frac {AC}{AE}\times DB \equal{} \frac {AD}{AP}\times BC\]
[/quote]
It is nice but easy!.
In the case $ AB<AC$
Let $ BI$;$ CJ$ be parallel to $ EF$ where $ I;J$ belong to $ AD$. Thales Theorem shows us $ \frac{AB}{AF}\equal{}\frac{AI}{AP}$;$ \frac{AC}{AE}\equal{}\frac{AJ}{AP}$;$ \frac{ID}{DB}\equal{}\frac{JD}{DC}\equal{}\frac{IJ}{BC}$. (1)
Notice that $ AI\equal{}AD\minus{}ID\equal{}AD\minus{}\frac{DB.IJ}{BC}$ thus $ \frac{AI.DC}{AP}\equal{}\frac{AD.DC}{AP}\minus{}\frac{IJ.DB.DC}{BC.AP}$
A same thing is $ \frac{AJ.DB}{AP}\equal{}\frac{AD.DB}{AP}\plus{}\frac{IJ.DB.DC}{BC.AP}$.
Now , it is immediate that $ \frac{AI.DC}{AP}\plus{}\frac{AJ.DB}{AP}\equal{}\frac {AD}{AP}\times BC$ since the fact $ DC\plus{}DB\equal{}BC$.
Now, the proof can be completed after we look at  again (1)

Yup. The official solution only uses ratios in length and area, which is very-very beautiful (and algebraic  :D )

Here is my solution
$ \frac{AD}{AP}.\frac{AB}{AF}.\frac{DC}{BC}\plus{}\frac{AD}{AP}.\frac{AC}{AE}.\frac{DB}{BC}\equal{}\frac{[ABD]}{[AFP]}.\frac{[ADC]}{[ABC]}\plus{}\frac{[ADC]}{[APE]}.\frac{[ABD]}{[ABC]}\equal{}\frac{[ABD].[ADC]}{[ABC]}(\frac{1}{[AFP]}\plus{}\frac{1}{[APE]})\equal{}\frac{[ABD].[ADC]}{[AFP].[APE]}.\frac{[AFE]}{[ABC]}\equal{}\frac{AB.AD}{AF.AP}.\frac{AD.AC}{AP.AE}.\frac{AF.AE}{AB.AC}\equal{}\frac{AD^2}{AP^2}$

So, $ \frac{AD}{AP}.\frac{AB}{AF}.\frac{DC}{BC}\plus{}\frac{AD}{AP}.\frac{AC}{AE}.\frac{DB}{BC}\equal{}\frac{AD^2}{AP^2}$, which is equivalent with the problem

$ QED$
 :)

it's (almost) Cristea's Theorem: let ABC be a scalene triangle and D a point on the segment BC. M belongs to the segment AD. E and F belong to the segment AB and AC ,respectively. Then EF passes through M if and only if:
 DC(EB/EA) + BD(FC/FA) = BC(MD/MA)

BTW, that Cristea's Theorem is not well-known to me. But this problem is just an old result from [url]http://www.mathlinks.ro/viewtopic.php?p=1143691#1143691[/url]

A simple proof is using vector.
We have $ BD. \vec{AC}\plus{}DC.\vec{AB}\equal{}BC.\vec{AD}$
$ \Rightarrow \frac{AC}{AE}.BD.\vec{AE}\plus{}\frac{AB}{AF}.DC.\vec{AF}\equal{}\frac{AD}{AP}.BC.\vec{AP}$ or $ x.\vec{AE}\plus{}y.\vec{AF}\equal{}z.\vec{AP}$
On the other side, $ FP.\vec{AE}\plus{}EP.\vec{AF}\equal{}EF.\vec{AP}$
So $ \frac{x}{FP}\equal{}\frac{y}{EP}\equal{}\frac{z}{EF}\equal{}\frac{x\plus{}y}{FP\plus{}EP}\equal{}\frac{x\plus{}y}{EF}$
$ \Rightarrow x\plus{}y\equal{}z$ (QED)

[quote="wangsacl"]For every triangle $ ABC$, let $ D,E,F$ be a point located on segment $ BC,CA,AB$, respectively. Let $ P$ be the intersection of $ AD$ and $ EF$. Prove that:
\[ \frac {AB}{AF}\times DC + \frac {AC}{AE}\times DB = \frac {AD}{AP}\times BC\]
[/quote]I am giving a solution using Menelaus Theorem.
Construct $ F',E'$ on $ AB, AC$ resp. such that $ F'E' \parallel BC$
Now Menelaus on $ \triangle AF'E'$ and transversal $ FPE$ gives,
\begin{align*} & \frac{AF}{FF'}\cdot \frac{F'P}{PE'} \cdot \frac{EE'}{EA}=1
\\ \iff & \frac{PE'}{PF'}=\frac{AF}{FF'} \cdot \frac{EE'}{EA} \\
\iff  & \frac{DC}{DB}\cdot \frac{AB}{AC}=\frac{EE'}{FF'}\cdot \frac{AF}{AE} \cdot \frac{AF'}{AE'}=\frac{\frac{AE'-AE}{AE\times AE'}}{\frac{AF-AF'}{AF\times AF'}}\\
 \iff & AB\times DC \left[\frac{1}{AF'}-\frac{1}{AF}\right]=AC\times DB \left[\frac{1}{AE}-\frac{1}{AE'}\right]
 \\ \iff & AB\times DC \times \frac{1}{AF'}+AC\times DB \times \frac{1}{AE'} = AB\times DC \times \frac{1}{AF}+AC\times DB \frac{1}{AE}
\end{align*}
But \begin{align*} \frac{AD}{AP}\times BC=& \frac{AD}{AP}\times (BD+DC)=\frac{AB}{AF'}\times DC+\frac{AC}{AE'}\times BD\\ =&\frac {AB}{AF}\times DC + \frac {AC}{AE}\times DB \end{align*}


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let the set of all permutations of (1,2,3,..,n) be S(n), the symmetric group. Find all the normal subgroups of S(n). How many are there?

This is straightforward if you know (or are allowed to assume) that $ A_n$ is simple for $ n \geq 5$ ($ A_n$ is the normal subgroup of $ S_n$ consisting of even permutations).

The answer is that $ S_n$ has a unique non-trivial proper normal subgroup when $ n \geq 5$, and this is $ A_n$. The cases with $ n < 5$ can be manually analyzed fairly easily.

What does it mean for a subgroup to be simple?

So how many normal subgroups does S(4) have?

A subgroup is simple if it it simple as a group, i.e. it is nontrivial and it has no proper nontrivial normal subgroup. The only normal subgroups of $ S_4$ are: the whole group, the trivial subgroup, $ A_4$, and the Klein-four group.


geometry
3D geometry
LaTeX
arithmetic series
arithmetic sequence

[hide="hint."]
$ a_1^3 \plus{} a_2^3 \plus{} ...a_n^3 \equal{} (\frac {(a_n)(a_n \plus{} 1)}{2})^2$

A conjecture.

EDIT: Oopy, forgot a set of parentheses.
[/hide]

[quote="CircleSquared"]$ a_1^3 \plus{} a_2^3 \plus{} \ldots \plus{} a_n^3 \equal{} \frac {(a_n)(a_n \plus{} 1)}{2}^2$[/quote]
$ a_1^3 \plus{} a_2^3 \plus{} \ldots \plus{} a_n^3 \equal{} \left(\frac {a_n\left(a_n \plus{} 1\right)}{2}\right)^2\equal{}\frac{a_n^2\left(a_n\plus{}1\right)^2}{4}$

[quote="i_like_pie"][quote="CircleSquared"]$ a_1^3 \plus{} a_2^3 \plus{} \ldots \plus{} a_n^3 \equal{} \frac {(a_n)(a_n \plus{} 1)}{2}^2$[/quote]
$ a_1^3 \plus{} a_2^3 \plus{} \ldots \plus{} a_n^3 \equal{} \left(\frac {a_n\left(a_n \plus{} 1\right)}{2}\right)^2 \equal{} \frac {a_n^2\left(a_n \plus{} 1\right)^2}{4}$[/quote]

$ a_1^3 \plus{} a_2^3 \plus{} \ldots \plus{} a_n^3 \equal{} \left(\frac {n\left(a_1 \plus{} a_n\right)}{2}\right)^2 \equal{} \frac {n^2\left(a_1 \plus{} a_n\right)^2}{4}$

This is general, not just $ 1^3\plus{}2^2\plus{}...\plus{}n^3$.

But how do you find the sum of the distinct prime factors?

Regarding the two expressions above for $ a_1^3 \plus{} a_2^3 \plus{} ... \plus{} a_n^3$ written by I_like_pie and Anaxerzia:

1) Are they both correct?

2) If not, which is the correct one?

3) And how can we show the correct expression?

[quote="chessplayingballer"]Let N be the sum of all integer cubes from 1(1 cubed) to 1000000(100 cubed). Find the sum of the distinct prime factors of N.[/quote]
[hide="Solution"]The sum of the distinct prime factors of $ \sum_{k \equal{} 1}^{100}k^3 \equal{} \left(\frac {100\cdot101}{2}\right)^2$ is the same as of $ \frac {100\cdot101}{2} \equal{} 5050 \equal{} 2\cdot5^2\cdot101$.

$ 2 \plus{} 5 \plus{} 101 \equal{} \boxed{108}$[/hide]
[quote="Carcul"]Regarding the two expressions above for $ a_1^3 \plus{} a_2^3 \plus{} \ldots \plus{} a_n^3$ written by I_like_pie and Anaxerzia:

1) Are they both correct?

2) If not, which is the correct one?

3) And how can we show the correct expression?[/quote]
Mine is definitely correct, and can be proven with induction. I'm not sure about the other one.

[quote="i_like_pie"][quote="Carcul"]Regarding the two expressions above for $ a_1^3 \plus{} a_2^3 \plus{} \ldots \plus{} a_n^3$ written by I_like_pie and Anaxerzia:

1) Are they both correct?

2) If not, which is the correct one?

3) And how can we show the correct expression?[/quote]
Mine is definitely correct, and can be proven with induction. I'm not sure about the other one.[/quote]

Are you talking about arithmetic sequences (actually, series)?

If so, then your formula is incorrect:

$ 153 \equal{} 1^{3} \plus{} 3^{3} \plus{} 5^{3} \neq (5\cdot 6/2)^{2} \equal{} 225$

Same with Anaxerzia:

$ 153 \equal{} 1^{3} \plus{} 3^{3} \plus{} 5^{3} \neq (3\cdot (1 \plus{} 5)/2)^{2} \equal{} 81$

So both of you are incorrect. However, the average of the two: $ (225 \plus{} 81)/2 \equal{} 153$ is correct... Maybe this is just a special case...

The only general form of arithmetic series I've seen is for first-order sum, but we can still calculate some higher-order series (in fact, only ones that have a common difference 1) by subtracting two sums:

Original form: $ 1^3 \plus{} 2^3 \plus{} ... \plus{} n^3 \equal{} (\frac {n(n \plus{} 1)}{2}) ^2$

Consider two series: $ S_1 \equal{} 1^3 \plus{} 2^3 \plus{} ... \plus{} (a_1 \minus{} 1)^3$ and $ S_2 \equal{} 1^3 \plus{} 2^3 \plus{} ... \plus{} a_n^3$

Series $ S \equal{} a_1^3 \plus{} (a_1 \plus{} 1)^3 \plus{} ... \plus{} a_n^3 \equal{} S_2 \minus{} S_1$

$ S_2 \minus{} S_1 \equal{} \frac {(a_n^2(a_n \plus{} 1)^2 \minus{} a_1^2(a_1 \minus{} 1)^2)}{4}$

In this case, we only need the original formula:

$ 1^3 \plus{} 2^3 \plus{} ... \plus{} 100^3 \equal{} (100*101/2) ^2$, which is exactly the same as @i_like_pie says.

The series @vishalarul stated...I did it like this:

Let $ S$ be the desired series, $ S_1 \equal{} 1^3 \plus{} 2^3 \plus{} ... \plus{} 6^3, S_2 \equal{} 1^3 \plus{} 2^3 \plus{} 3^3$ then
$ S \equal{} S_1 \minus{} 2^3S_2 \equal{} (6*7/2) ^2 \minus{} 8*(4*3/2) ^2 \equal{} 153$

Perhaps we can derive a general formula from this?

P.S.: does anyone know how to type sigma notation, or where can I find a manual for LaTeX?

[quote="i_like_pie"][quote="vishalarul"]Are you talking about arithmetic sequences (actually, series)?

If so, then your formula is incorrect:

$ 153 \equal{} 1^{3} \plus{} 3^{3} \plus{} 5^{3} \neq (5\cdot 6/2)^{2} \equal{} 225$[/quote]
The formula works only for consecutive integers from $ 1$ to $ n$, which the problem is about.

$ \sum_{k \equal{} 1}^{100}k^3 \equal{} \left(\frac {100\cdot101}{2}\right)^2 \equal{} 5050^2$[/quote]

The way you wrote it implied a general case, with an arithmetic sequence $ a_1$, $ a_2$, ... $ a_n$.


It's a case of a more general problem (Euler trails):

On a given graph, can you start at a vertex and traverse every edge exactly once?

Euler was the first (as with many other things) to give a criterion for all such graphs...it's pretty simple (try finding it).

[quote]On a given graph, can you start at a vertex and traverse every edge exactly once?[/quote]

Hmm...I am a little confused here.  Do you mean like the graph of y=x^2?  B/c that really doesnt have edges.  Or are you talking about a regular polygon?  If so then couldent you just start at the vertex and go all the way around once?  Could you clarify?  Thanks.

a graph can also mean a set of vertices with edges connecting some of them.  In the original bridge problem, the islands were the vertices and the bridges were the edges.

Read these. They explain it very well.

From Cut-the-Knot: http://www.cut-the-knot.org/do_you_know/graphs.shtml

From MathWorld: http://mathworld.wolfram.com/KoenigsbergBridgeProblem.html


$a=b$

$a^2=ab$

$a^2-b^2=ab-b^2$
factor
$(a+b)(a-b)=b(a-b)$

$a+b=b$

since a and b are interchangeable as $a=b$

$b+b=b$

$2b=1b$

$2=1$

EDIT: i see something similar has been posted before, that it has to be 0.. and dividing is impossible. any thoughts?

There are some inconsistencies in math after all.

ToBe do u think u cud post the link to the other post that u mentioned in your "edit"..so we can take a look

Because a=b,so a-b=0,the wrong thing is that the equation cannot be divided by (a-b). :P

oh god.....i can't believe i over looked that one..

i forgot that a-b = 0, if a=b!!  :!:

This has been posted numerous times already. Please use the search function.  ;)

Locked.


Let $ a,b,c \geq\ 0$, a+b+c=5
Find the max,min of $ a^3 \plus{} b^3 \plus{} c^5$ :D

What are $ x,y,z$ and $ a,b,c$??? :?:
Do you mean $ a^3\plus{}b^3\plus{}c^5$?? :huh:

Of course it would be $ a^3\plus{}b^3\plus{}c^5$. (When $ a\plus{}b\plus{}c\equal{}5$ & $ a,b,c$ are positive).

hello, i have found no maximum  in the region where the objective function is defined and the constraints are satisfied, i think the maximum is on the boundary $ 3125$ for $ a\equal{}0,b\equal{}0$ and $ c\equal{}5$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i have found no maximum  in the region where the objective function is defined and the constraints are satisfied, i think the maximum is on the boundary $ 3125$ for $ a \equal{} 0,b \equal{} 0$ and $ c \equal{} 5$.
Sonnhard.[/quote]
Can you show me more detail? :)
Do you know what is the min? :lol:


Hello everyone,

Could someone please check to see where I may have erred in my steps for the following question? I arrived at an answer different from the one given in my textbook.

Thank you.

---

[b]1. A light is mounted at the top of a 15-ft tall pole. A man 6 ft tall walks away from the pole at 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole? [/b]

---

Let x be the distance from the lamp pole to the man. Let y be the distance from the man to the tip of the shadow.

Therefore:

$ \frac{15}{6} \equal{} \frac{x \plus{} y}{y}$

$ 15y \equal{} 6x \plus{} 6y$

$ y \equal{} \frac{2x}{3}$

Differentiate both sides:

$ \frac{Dy}{dt} \equal{} \frac{2}{3} \times \frac{dx}{dt}$

Since $ \frac{dx}{dt} \equal{} 5 ft/s$:

$ \frac{dy}{dt} \equal{} \frac{10}{3}$

However, the correct answer is [hide]$ \frac{25}{3}$[/hide].

According to your setup, the rate at which the shadow is moving is $ \frac {d}{dt}(x \plus{} y)$ when $ x \equal{} 40$, so up until $ y \equal{} \frac {2}{3}x$ you are fine.

Now $ \frac {d}{dt}(x \plus{} y) \equal{} \frac {d}{dt}(x \plus{} \frac {2}{3}x) \equal{} \frac {5}{3}\frac {dx}{dt} \equal{} \frac {5}{3}(5) \equal{} \frac {25}{3}$ ft/s.

Thank you for your response, HilbertThm90!


The resulting solid figure is a rectangular prism with dimensions 3, 1, and 3.

Its volume is $ (3)(1)(3)\equal{}\boxed{9}$ cubic units.

Use the fourth column as the center.Fold along the two vertical lines on either side of the figure creating a $ 1\times3$ rectangle.

Then fold the rest of the protruding $ 1\times3$ rectangles down so they are perpendicular to the first rectangle.

You have created a rectangular prism with a square base with dimensions $ 3\times3\times1$.

Volume: $ 3\times3\times1\equal{}\boxed{9}$ cubic units.


trigonometry function Asymptote algebra domain

how do you do the tangent function so that the graph never shows above a certain y value? Is there a way to make the asymptotes dotted lines?

Also, how do you do the cosecant function? I tried 1/sinx but that divdes by zero. Is there a way to make holes in the domain?

Thanks

To make holes in the domain see the very long discussion on
http://sourceforge.net/forum/message.php?msg_id=6173868

For dotted lines see linetype, dotted functions in the asymptote documentation
(on page 39)
[asy]size(100,0);
pen Dotted=dotted+2bp;
draw((0,0)--(1,1),Dotted);[/asy]

Perhaps ylimits can help for your first question.

O.G.


Which of these is the fastest rate of speed?
A. 40 miles per hour
B. 15 yards per second
C. 1,000 inches per second
D. 200,000 feet per hour
E. .75 miles per minute

i don't know some of the conversions, so i cant tell you the answer.

Oops didn't see the decimal point :P

The correct conversions in SI units are as follows:
A 17.88
B 13.71
C 25.40
D 16.93
E 20.11

You guys are right ;)


I ordered:

1 of the Art of Problem Solving, Volume 1: the Basics

1 of the Art of Problem Solving, Volume 1: the Basics Solutions Manual

i got the email saying it was supposed to arrive in 2-3 weeks... i received the email for confirmation of my order on January 23. Now it's April 16, and I still haven't received the books. What do I do???

Please forward the order email to orders@artofproblemsolving.com and we'll take care of it.


FTW
combinatorics unsolved
combinatorics

GAME 1)
[b]Rule[/b]:
Suppose there is a 1xn (n>2) rectangular region which consists of n unit squares.
We start by putting three coins on the last three squares on the right.
Each player then takes turns to move any one of the coins to its left hand side.
The coins cannot overlap.
The first one who cannot move a coin loses.

[b]Remark[/b]:
 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [C] [C] [C]      ->       [ ] [C] [ ] [ ] [ ] [ ] [ ] [C] [ ] [C]
This is a legal move.

[b]For example[/b]:
When n = 4: [ ] [C] [C] [C]
If the first player move the right most coin to the left, i.e.  [C] [C] [C] [ ], then the first player win.

When n = 5:  [ ] [ ] [C] [C] [C]
If the first player move the left most coin to the left, either  [ ] [C] [ ] [C] [C] or  [C] [ ] [ ] [C] [C].
Then no matter how the second player move any one of the two coins on the right. The first player then move the remaining coin and win the game.

[b]Question[/b]: Does any one of the players has a winning strategy for all possible values of n? If so, what is it?

Personally I agree that one of the players has a winning strategy, by the Zermelo's Theorem, this game cannot end in a draw, result follows.
But then I cannot think of a winning strategy for all n. Any comment ?

=====

GAME 2)
[b]Rule[/b]:
This is the misère tic-tac-toe, i.e. the same as ordinary tic-tac-toe but the first player who completes a row, column, or diagonal with that player's marks loses.

[b]Question[/b]: Want to show each of the two players has a non-losing strategy.

A non-losing strategy for the first player is simple.
Just put the mark at the centre first.
Then no matter how the second player put his or her mark, the first player just need to put the mark opposite to that of the second player.

However, I cannot think of a non-losing strategy for the second player, or equivalently, to show the first player has no winning strategy.

=====

GAME 3)
[b]Rule[/b]:
This is the misère SIM, i.e. the same as ordinary SIM but the first player who completes a triangle with that player's color wins.

[b]Question[/b]: Want to show one of the two players has a winning strategy and find the winning strategy.

I am not sure if my argument works.
As in the proof that the ordinary SIM cannot end in a draw, so the misère SIM cannot end in a draw too.
By the Zermelo's Theorem, one of the two players has a winning strategy.

Now, as in the ordinary SIM, the second player has a winning strategy, that is to prevent himself or herself to complete a triangle earlier than the first player.
For the misère SIM, the first player can then adopt the same idea to prevent the second player to complete a triangle earlier than himself or herself.

[quote="zeb"]For Game Two, the second player can first pick a corner as close as possible to player one's first move (assume it's the upper left), and then for his second move there will be an empty spot a knight's move away from his first spot (assume it's the middle of the right side). Then he tries to make one of the marked moves (marked with an M):

[code](O) (M) ( )
(*) ( ) (O)
(M) (M) ( )[/code]

If he can get one of those squares, he is guaranteed to have a nonlosing fourth move. If the first player has already taken the three marked spots, then the player moves in the *'d square, guaranteeing a win :)

Brute force FTW[/quote]
Oh, really  :P  what I did is to contruct a game tree, and certainly not complete yet :(
What do you think about GAME one? :maybe:  I wonder if there are different strategy for different value of n.

[quote="jerryjerrychan"]
Oh, really  :P  what I did is to contruct a game tree, and certainly not complete yet :(
What do you think about GAME one? :maybe:  I wonder if there are different strategy for different value of n.[/quote]
By now I've solved GAME one.
For even n, the first player move the right most coin to the left most square.
In the remaining part of the game, the first palyer just need to keep the two remaining coins are next to each other, and the number of empty squares between the left most coin and the middle coin(after moving) is even.
For odd n, nearly the same, only difference is to move the left most coin to the left most square at the begining instead of moving the right most coin.


Given that $ m$ and $ n$ are digits, what is the sum of the values for $ m$ and $ n$ which yield the greatest six-digit number $ 5m5,62n$ that is divisible by 44?

Set n as the largest it can be divisble by 4, or 8.  Then use the divisibility rule to find that m = 9.  9+8=17

What divisibility rule?  Explain it, please (because I don't know it, and whoever asked this probably doesn't either).

Divisibility rule for 11:

Say you have number $ abcdefghijkl$

Then for that number to be divisible by 11, $ a\minus{}b\plus{}c\minus{}d\plus{}e\minus{}f\plus{}g\minus{}h\plus{}i\minus{}j\plus{}k\minus{}l\equal{}\text{multiple of 11}$

You just have to alternately add and subtraact digits. This has to do with the fact that 10 is -1mod11

You should probably use the divisibility rule for 4 as well:  it seems most straightforward to do it in these steps:
0:  To be divisible by 44, of course you need to be divisible by 4 and 11 at the same time.

1:  Divisibility test for Four:  Four divides the number abc,def,...,...,xyz  iff 4 divides the number yz.  This is so because 4 is certain to divide abc,def,...,x00, which is a multiple of 100.  Based on this step, n is either 0, 4 or 8.

2:  Divisibility test for Eleven, as described above.  That says n-2+6-5+m-5=n+m-6 is divisible by eleven.

3.  Since n+m is at most 18, n+m-6 is at most 12.  So n+m-6= 0 or 11.  Then n+m= 6 or 17.  To make 6, you'd use n=0 m=6 or n=4 m=2, but n=8 doesn't work.   To make 17, you can notice that 0 and 4 for n won't work!  That leaves you with n=8 m=9.  So, that's going to turn out to be the solution, to get the largest.

This problem could be a lot trickier if the pair (8,9) weren't already so high :p  In a similar problem you could have m=8 n=9 as a solution and m=9 n=0 as a solution... in that case the second solution is larger!

PS:  You probably know the divisibility test for Three:  If the sum of all the digits in the number is divisible by three, the original number is also.
There is also a divisibility test for 9:  If the sum of digits is divisible by 9, so is the original number!
PPS:  Those are the most common ones... simliar rules are NOT true for all numbers.


For a,b,c,d positive real numbers prove that

(a^4+b^4+c^4+d^4) /4abcd  + 4(abcd)^(1/4)/(a+b+c+d)  \geq 2

Using the inequalities between means of order 4 and 1, and AM/GM, we have :
(a^4+b^4+c^4+d^4)/4  \geq ((a+b+c+d)/4)
 4
and ((a+b+c+d)/4) 3  \geq (abcd) 3/4 .
Thus :
(a^4+b^4+c^4+d^4) /4abcd + 4(abcd)^(1/4)/(a+b+c+d)  \geq ((a+b+c+d)/4)/(abcd) 1/4 + 4(abcd)^(1/4)/(a+b+c+d)  \geq 2 since x+1/x  \geq 2 for each positive x.

Note that equality occurs iff a=b=c=d.

Pierre.


topology
real analysis
real analysis unsolved

$X$ is a metric space and $E\subset X$.

The following statements are equivlant:

(i) $E$ is sequentially compact
(ii) $E$ is limit point compact
(iii) $E$ is compact(here the compact is defined by finite-covering)
(iv) $E$ is totally bounded and complete.

(ii)=>(i) and (iii)=>(ii) are not hard.

How to deduce (iv) from (i) and (iv) => (iii)?

and why we need *totally* bounded in (iv)? What will be wrong if it is only bounded, not totally?

[quote]and why we need *totally* bounded in (iv)? What will be wrong if it is only bounded, not totally?[/quote]
The problem is that "only" bounded isn't a topological property, it's a property of a particular metric.

Take this example. Let the set be $\mathbb{R}$ and define the metric by $d(x,y)=\frac{|x-y|}{1+|x-y|}$. Then we have a complete metric space that is bounded. (All distances are less than 1; the whole thing is contained in a ball of radius 1 centered anywhere you please.) It is quite clearly not a compact set.

In fact, the obvious mapping is a homeomorphism between this metric space and $\mathbb{R}$ with the usual metric. This is another way of saying that "compact" is a topological property but "bounded" isn't.

I have a question regarding all this:

What does sequentially compact really mean? I've found two definitions: on Planetmath it says that a set is sequentially compact if all sequences have a convergent subsequence. However, I knew that it meant that all sequences have a subsequence converging to a point in the set. Which one is correct? (I tend to think the second is correct).

I think in a metric space both compacity and sequential compacity are equivalent to the fact that the set is closed and bounded $(*)$, and from this we can prove (i)$\iff$(iii). It's easy to see how (i) implies the completeness, and (iii) implies the total boundedness, so (i)$\Rightarrow$(iv). However, I'm not sure about equivalence $(*)$.

[quote="grobber"]I have a question regarding all this:

What does sequentially compact really mean? I've found two definitions: on Planetmath it says that a set is sequentially compact if all sequences have a convergent subsequence. However, I knew that it meant that all sequences have a subsequence converging to a point in the set. Which one is correct? (I tend to think the second is correct).
[/quote]

Yes, it does mean the second one.

[quote="grobber"]
I think in a metric space both compacity and sequential compacity are equivalent to the fact that the set is closed and bounded $(*)$, and from this we can prove (i)$\iff$(iii). It's easy to see how (i) implies the completeness, and (iii) implies the total boundedness, so (i)$\Rightarrow$(iv). However, I'm not sure about equivalence $(*)$.[/quote]

In Euclidean spaces, a set is compact if and only if it is both closed and bounded. However, it is not true for metric spaces in general...

I can proof (i)=>(iii) by assuming Linderlofness :blush: But I think that (i)=>Linderlofness..., more precisely, (i)=>seperable=>Linderlofness... I currently have no idea about the proof...

[quote="liyi"][quote="grobber"]
I think in a metric space both compacity and sequential compacity are equivalent to the fact that the set is closed and bounded $(*)$, and from this we can prove (i)$\iff$(iii). It's easy to see how (i) implies the completeness, and (iii) implies the total boundedness, so (i)$\Rightarrow$(iv). However, I'm not sure about equivalence $(*)$.[/quote]

In Euclidean spaces, a set is compact if and only if it is both closed and bounded. However, it is not true for metric spaces in general...
[/quote]

I really need to be enlightened regarding these issues :).

Is it at least true that if a set is closed and bounded then it's compact (or sequentially compact)? (in a metric space, of course)

[quote="grobber"]Is it at least true that if a set is closed and bounded then it's compact (or sequentially compact)? (in a metric space, of course)[/quote]

No..As Kerry Merryfield says, "boundness" tells you little. Suppose you have a metric $d$ and we define a new metric $d' = \min (d, 1)$. Compactness (even completeness) is same for $(X,d)$ and $(X,d')$. Consider the metric space $(R,d)$ where $d(x,y)= \min (|x-y|,1)$. $R$ is closed and bounded but not compact.

I worked out the proof later. Seperable=>Linderlofness is not difficult, and I here write the outline of the proof of "seq compact => separable"

Suppose $E$ is uncountable. Pick any $x_1$. Define $d_1 = \sup\{d(x,x_1) | x\in E\}$, choose $x_2$ such that $d(x_1,x_2) > \frac{d_1}{2}$.

Chosen $x_1,\dots,x_k$, define $d_k = \sup\{\min\limits_{i\leq k} d(x,x_i) | x\in E\}$, choose $x_{k+1}$ such that $\min\limits_{i\leq k} d(x_{k+1},x_i) > \frac{d_k}{2}$.

We get a sequence $x_1,x_2,\dots,x_n$. we need to prove that it is dense in $X$. If not, there exists $y$ which has a neighbour $O(y,\delta)$ such that $O(y,\delta) \cap \{x_n\} = \emptyset$. thus we have $delta < d_n, \forall n$ and
$d(x_m, x_n) > \frac{d_{m-1}}{2} > \frac{\delta}{2}$
From Cauchy's Convergence Principle, we know that $\{x_n\}$ doesn't have a convergent subsequence, which contradicts with sequentially compactness.

I see. Thanks for clearing that up.

We still need (iv)$\Rightarrow$(iii), right?

We can prove it by using (i)$\Rightarrow$(iii). Take any infinite sequence of $E$. One of the balls of a finite covering of $E$ with bals of radius $1$ (one such covering exists because of the total boundedness) contains infinitely many terms of the sequence. pick one, and then cover this ball with finitely many balls of radius $\frac 12$. One of these balls contains infinitely many terms of the sequence, so we can pick one again, and cover the ball with smaller balls, and so on. You can see that we have defined an infinite Cauchy sequence, which must converge to a point in $E$ (because of the completeness), so $E$ is sequentially compact (this is (i)).

Separability (and thus Lindelofness) is even easier to prove here: cover the set with finitely many balls of radius: first $1$, then $\frac 12$, then $\frac 13$,etc. Take a point in each of these balls. We have found a dense and countable subset of $E$.

I think (i)$\Rightarrow$(iii) isn't hard to show using Lindelofness, so we're sort of done.

Is this correct? I'm a topology newbie :).


Find all possible sets of 4 consecutive integers such that the sum of the cubes of the smallest three is the cube of the 4th.

Let  :alpha: be the smallest integer.
Then  :alpha:  :^3:  + ( :alpha: +1) :^3:  + ( :alpha: +2) :^3: = ( :alpha: +3) :^3:
Simply solve it and we get one real root  :alpha: = 3 and two complex roots which should be rejected.  :(


In poker, a straight flush--five consecutive cards of the same suit--beats four of a kind.  In each suit, a straight flush can start with an ace, a deuce (a two), or any other card up to a 10 (the ace may rank hight or low), making ten pssibilities in all.  Since there are four suits, there are four times ten, or fory, different hands that are straight flushes.  There are only thirteen different four-of-a-kind hands.  So, if there are thirteen four-of-a-kind hands and forty straight flushes, why does a straight flush beat four of a kind?

Posted by MrMcGeek

[hide]

Four of a kind is more probable because the fifth card in your hand doesn't matter, while it does for a straight flush.[/hide]

you fail to account for fifth card in the four of a kind. you have 4 suits * 13 possible sets of four cards * 48 other cards to complete the hand. so a total of 2496 four-of-a kind.

Tom Davis about poker.

http://www.geometer.org/mathcircles/poker.pdf

[quote="MrMcGeek"]re-read that, you made one mistake, but your idea is right

Posted by MrMcGeek[/quote]

In particular, there is only one way to choose four cards from a set of four cards.

This problem is for those of you who know the poker game called Texas Hold'em (If you don't know the game, don't bother to learn it until you are 18 ).

Problem If you are holding a pair, what is the probability of flopping a set (three of a kind) or better?

[hide]Somebody is actually talking about having a Texas Hold'em poker tournament at Mathcamp, he must be watching the travel channel too much. [/hide]


symmetry
geometry
parallelogram
geometry solved

ABCDE is a REGULAR pentagon
SHOW THIS
(AB/AC)+(AC/AB)=3

I think $\frac{AB}{AC}+\frac{AC}{AB}=\sqrt{5}$ :?

Indeed, what is correct is the following assertion: If ABCDE is a regular pentagon, then $\frac{AB}{AC}+\frac{AC}{AB}=\sqrt{5}$.

Here is the simplest proof of this: Let the diagonals AC and BD of the pentagon ABCDE intersect at T. By the symmetry properties of the regular pentagon ABCDE, we have AD || BC, and thus Thales yields $\frac{AT}{AD}=\frac{CT}{CB}$.

But the symmetries of the regular pentagon ABCDE imply DE || CA and EA || DB; in other words, DE || TA and EA || TD. Hence, EATD is a parallelogram, so that AT = DE. Thus, CT = AC - AT = AC - DE. Thus, the equation $\frac{AT}{AD}=\frac{CT}{CB}$ becomes $\frac{DE}{AD}=\frac{AC-DE}{CB}$. Since ABCDE is a regular pentagon, DE = AB, AD = AC and CB = AB; thus, this transforms into $\frac{AB}{AC}=\frac{AC-AB}{AB}$. Equivalently, $\frac{AB}{AC}=\frac{AC}{AB}-1$. In other words, $\frac{AB}{AC}-\frac{AC}{AB}=-1$. Hence,

$\left(\frac{AB}{AC}+\frac{AC}{AB}\right)^2=\left(\frac{AB}{AC}-\frac{AC}{AB}\right)^2+4\cdot\frac{AB}{AC}\cdot\frac{AC}{AB}=\left(-1\right)^2+4=5$,

so that $\frac{AB}{AC}+\frac{AC}{AB}=\sqrt{5}$, and the proof is complete.

  Darij

I am agree with kunny.
please chek the problem :!:  :!:


let a,b,c is real number
find a,b,c so that (a*b*c) 2  \geq ( a+b+c) ^3
a 2+b2 +c 2 = (a+b)(b+c)(c+a)
                  :)  :(


induction
number theory
prime numbers
number theory unsolved

we have 2n - 1 natural numbers we have to proove that we can choose n of them , such that their sum is a multiple of n .

We can use induction.

1) Prove the result for all prime numbers n.
2) If n is not prime, say n = pq, then prove it using the induction hypothesis for p, q.

Then prove the proble for primes by Chevale theorem


Show that any basic open set about a point on the "top edge," that is, a point of form $ (a, 1)$, where $ a<1$, must intersect the "bottom edge."

[b]Background:[/b]

[b]Definition-[/b] The [b]lexicographic square[/b] is the set $ X\equal{}[0,1] \times [0,1]$ with the dictionary, or lexicographic, order. That is $ (a, b)<(c, d)$ if and only if either $ a<b$, or $ a\equal{}b$ and $ c<d$. This is a linear order on $ X$, and the example we seek is $ X$ with the order topology.

We follow usual customs for intervals, so that $ [(a,b),(c,d))\equal{}\{ (x,y) \in X : (a,b) \leq (x,y)<(c,d) \}$. A subbase for the order topology on $ X$ is the collection of all sets of form $ [(0,0),(a,b))$ or of form $ [(a,b),(1,1)).$

[url=http://books.google.com/books?id=O2Hwaj2SqigC&pg=PA792&lpg=PA792&dq=lexicographic+square+topology&source=web&ots=xjqD8trdVa&sig=WCh2Zmyk-MIIVuCZnb8u45UkgTU&hl=en&sa=X&oi=book_result&resnum=9&ct=result#PPA793,M1]here[/url] is a description on how open sets look like in the lexicographic square.
Interesting questions about this square:
Show that it is not metrizable, it is first countable, connected but not path connected..


Do there exits two distinct positive integers $ a;b$ such that $ b^n\plus{}n$ is a multiple of $ a^n\plus{}n$ for every positive integer $ n$?

Let $ p|b$, then $ p|a^p\plus{}p\to p|a$. Therefore $ rad(b)|a$.
Let $ p\not |b$ and $ S(b,p)\equal{}\{\minus{}b^0\mod p, \minus{}b^1\mod p,...,\minus{}b^T\mod p\}$, T - period mod p. Then for $ s\in S(b,p)$ we find $ n\equal{}s\plus{}xp$, suth that $ p|b^{s\plus{}xp}\plus{}s$ or $ x\equal{}y\minus{}s\mod T$, were $ s\equal{}\minus{}b^y\mod p$. From $ b^n\plus{}n|a^n\plus{}n$ we get $ n\equal{}\minus{}b\mod p\plus{}p(1\plus{}b\mod p), \ p|a^n\minus{}b$. Therefore $ p\not |a$ and for big prime $ p>b\plus{}1$ it mean $ p|a^{p\plus{}1}\minus{}b$ or $ p|a^2\minus{}b$ contradition.