calculus
integration
geometry
3D geometry
function
probability
expected value

In an exam a few years ago, there was a question on the mean of a continuous graph. It is not in our course but was justified in the marking by expecting an intuitive answer.

Basically you had to intuitively work out the mean value of the graph (1/9)*(4 - x^2) between the bounds x=-1 and x=2. Can anyone suggest a way of doing it intuitively?

Using the integral formula it is equal to the integral of x.f(x) between 1 and -2, it comes out easily to 0.25 . Can anyone prove this formula? Ive read a proof but it sounded like the guy didnt know what he was doing and was quite confusing. Can anyone give a clear proof?

Cheers.

[hide="Hint"]
Solid Geometry
[/hide]

[hide="Solution"]
$ \frac{1}{9}(4\minus{}r^{2}) \equal{}\frac{4}{9}\minus{}\frac{r^{2}}{9}\equal{}\frac{4}{9}\minus{}\frac{1}{9\pi}(\pi r^{2})$

We thus need to find the mean value of $ \pi r^{2}$ for $ r\in [\minus{}1,2]$.
We can think of this as a double-cone, first starting with radius 2, then decreasing to 0, and increasing back to radius 1.

The average value of $ \pi r^{2}$ is $ V/h$. Let $ h \equal{} 3j$. Then, $ V \equal{}\frac{\pi}{3}(1)^{2}(j)\plus{}\frac{\pi}{3}(2)^{2}(2j) \equal{} 3j\pi \equal{} h\pi$
The average value of $ \pi r^{2}$ is then $ \pi$, and the average value of the function is $ \frac{4}{9}\minus{}\frac{1}{9\pi}(\pi) \equal{}\boxed{\frac{1}{3}}$
[/hide]

[hide="Alternate consideration"]
Consider the squares of all the numbers from $ \minus{}k$ to $ 2k$, divide by $ 3k\plus{}1$ to find the average value. Note that this is approximately $ k^{2}$ times the average value of $ k^{2}$ from $ k \equal{}\minus{}1$ to $ k \equal{} 2$, so we divide by $ k^{2}$. Note that as $ k$ get's larger, we get more and more accurate (essentially an integral).

The sum of the first $ n$ numbers is $ \frac{n(n\plus{}1)(2n\plus{}1)}{6}$, so we want $ \frac{\frac{k(k\plus{}1)(2k\plus{}1)}{6}\plus{}\frac{2k(2k\plus{}1)(4k\plus{}1)}{6}}{k^{2}(3k\plus{}1)}\equal{} 1\plus{}\frac{1}{2k}$

As $ k\to\infty$, clearly $ 1\plus{}\frac{1}{2k}\to 1$, and so the average value is $ \frac{1}{9}(4\minus{}1) \equal{}\frac{1}{3}$
[/hide]

Your answer makes sense, but the answer to the exam was the mean of x=0.25, and using the integral formula (where the expected value of x = E(x))

E(x) = integral of xf(x)
      = integral of x(4-x^2)/9  between the bounds -1 and 2.
      = 0.25

But is there a proof for this formula?

According to [url]http://mathworld.wolfram.com/ExpectationValue.html[/url], it's a definition. There's no proof for definitions.

mm, but surely theres some explanation as to why it is that equation. i just need an explanation.

Sorry, I misunderstood your question, I thought you meant the mean value  :oops:

Say you had a discrete version. Suppose I give you a $ \frac{1}{2}$ probability of getting $ 1$, a $ \frac{1}{3}$ probability of getting $ 2$, and a $ \frac{1}{6}$ probability of getting $ 3$. Clearly, the expected value is $ 1\cdot\frac{1}2\plus{}2\cdot\frac{1}3\plus{}3\cdot\frac{1}6$. In other words, if we set a function, $ f$ to be our probability distribution, the expected value is $ 1\cdot f(1)\plus{}2\cdot f(2)\plus{}3\cdot f(3)$. It's the same way if we have a continuous thing. We want $ x\cdot f(x)$ at each point.

I suppose that he is what is the motivation for the definition. As always, we choose definitions so they have useful properties. Expected value is useful for determining it a game is profitable where different outcomes are weighted with different amounts of return. Example: You and I flip a coin. If it is heads, I get 2 dollars, if it is tails, I loose 1 dollar. The expected value for me is a 50 cents.

As Hamster1800 said, the discrete version is easier to understand. Once you have that definition, it is reasonable to extend it to a continuous function.


$x$ doesn't exist. $\cos (\arctan x)=-1\Rightarrow \arctan x=\pi$, but the range of the arctangent is only between $-\frac{\pi}{2}$ and $\frac{\pi}{2}\Rightarrow \arctan x<\pi$.

well , unless you are not considering one-to-one function then there is a solution when $x=0$ by using $\tan^2\theta+1=\frac{1}{\cos^2\theta}$ , $\cos\theta=-1$ , $\tan\theta=x$ .  :P


A Buddhist monk got an errand from his teacher: to meditate for exactly for 45 minutes. He has no watch; instead he is given two inscense sticks, and he is told that each of those sticks would completely burn in one hour. The sticks are not identical and they burn with variant but unknown rates( they are hand made). So he has these two incense and some matches: can he arrange for exactly45 minutes of meditation?

[hide]Can he break the sticks precisely? Then he could break one stick into quarters, so that he has a hour long stick and a fifteen minute stick, burn them both at the same time, and start meditating when the 15 minute stick finishes[/hide]

@ mustafa - The quarters will burn for different amounts of time, as stated in the problem.

oh :blush:

I assumed that the sticks would burn at the same rate the whole way through, but at different rates. In that case, this is a fairly tricky one. He could just meditate for an hour and get extra credit...

Lol I like that idea.

What if this particular monk has an atomic clock in his brain? That would simplify things...

Or maybe he can exactly calculate how fast each stick will burn at each point?

In reality, however, I'm not sure if this is even possible without any cheap tricks.

Umm... I think that this is in the wrong thread

[hide]

Light one stick on both ends and light the other stick on only one end, and begin meditating. The stick with both ends lit will burn out in 30 minutes. The other stick will have 30 minutes of stick left. Light the other end of this remaining stick and it will burn out in 15 minutes, for a total of 45 minutes.

[/hide]


geometry
trigonometry
calculus
trig identities
Law of Cosines
Law of Sines
geometry unsolved

It was a MOP homework assignment and no one got it right that year.

You have $\triangle{ABC}$. There is an interior point $P$ such that $AP=1, BP=2, CP=3$, and $\angle{A}=60^\circ$, what's the maximum area of $\triangle{ABC}$.

yes, I know that formula, but isn't that just for equilateral triangles? This triangle is not necessarily equilateral ;)

See http://www.mathlinks.ro/Forum/viewtopic.php?t=5308 and http://www.mathlinks.ro/Forum/viewtopic.php?t=81 for the source of the problem...

  darij

[quote="Altheman"]well...

(all in degrees)

Let $\angle BAP=x$, then $\angle CAP=60-x$. $AC=b$ and $AB=c$

The area is $\frac{1}{2}bc\sin 60$, so we want to maximize $bc$
using the law of cosines, we get

$b^2-2b\cos x-3=0$ and
$c^2-2x\cos(60-x)-8=0$

isolate b,c

$b=\cos x+\sqrt{\cos^2x+3}$
$c=\cos(60-x)+\sqrt{\cos^2(60-x)+8}$

then take the product...

$bc=\left[\cos x+\sqrt{\cos^2x+3}\right]\left[\cos(60-x)+\sqrt{\cos^2(60-x)+8}\right]$

some messy calculus gives us that the maximum occurs when

$2\sin(60-x)=3\sin(x)$

which makes us happy since that is such a nice result, the circumradii of $ABP$ and $ACP$ are equal by the law of sines!

you should be able to take it from here  ;)

p.s. me +calculator > MOP... :lol: just kidding

p.p.s. i would love to see the geometric interpretation of this problem[/quote]

$\sqrt{3}\cos x-\sin x=3\sin x$
$\tan x=\frac{\sqrt{3}}{4} \implies\cos x=\frac{4}{\sqrt{19}}, \sin x=\frac{\sqrt{3}}{\sqrt{19}}$

$\cos 60-x=\frac{7}{2\sqrt{19}}$

using the above formulas,

$b=\frac{4}{\sqrt{19}}+\sqrt{\frac{73}{19}}$
$c=\frac{7}{2\sqrt{19}}+\frac{\sqrt{657}}{2\sqrt{19}}$

then $A=\frac{\sqrt{3}}{4}bc$
[hide="..."]
$\frac{\sqrt{3}}{4}*\frac{4+\sqrt{73}}{\sqrt{19}}*\frac{7+3\sqrt{73}}{2\sqrt{19}}$
$=\frac{\sqrt{3}}{8*19}*(28+19\sqrt{73}+3*73)$
$=\frac{\sqrt{3}}{8*19}*(247+19\sqrt{73})$
$=\frac{\sqrt{3}}{8}*(13+\sqrt{73})$
[/hide]
$=\boxed{\frac{13\sqrt{3}+\sqrt{219}}{8}}$


I'll be at the Vestavia Tournament next month.  If I don't already know you, please introduce yourself to me.

Tipton had me prepare a test for the tournament.  The problems are more like ARML and AIME problems (well, not the hardest ones -- that would take too long for a one hour test) than most local tournament exams.  I prepared a set of solutions that is something like 8 pages long for those who take it and I hope that it's helpful to a lot of students.

Thanks, I'm looking forward to the test and the solution guide.
Hopefully we start some awareness for ARML, HMMT, and AoPS for other students in Alabama.

Cool, I plan on taking it. I should probably introduce myself too since I never have formally introduced myself to you (only watched in confusion as you taught our class occasionally). Can't wait to see how the test is.

*bangs head on desk* NO ONE told us about this test during break. no one. gah!!!!

on a side note, i love how the vandy math tournament is on the same day as our prom. lovely.

[quote="bleumoose"]*bangs head on desk* NO ONE told us about this test during break. no one. gah!!!!
[/quote]

I'll post the test and the solutions guide online soon so that anyone can work the test.

The afternoon test was open to anyone regardless of grade.

[quote="LuCy4EvA"]was the test of high school people only? we never knew about the test.[/quote]

I'll be posting the test online soon, so you can at least see the problems and solutions.

Are you the same Lucy who won a copy of DPatrick's new book?


[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \sqrt {\frac {a}{(a \plus{} 2b)^3}} \plus{} \sqrt {\frac {b}{(b \plus{} 2c)^3}} \plus{} \sqrt {\frac {c}{(c \plus{} 2a)^3}} \ge \frac {1}{\sqrt {3}}. \frac {a \plus{} b \plus{} c}{a^2 \plus{} b^2 \plus{} c^2}$  :)[/quote]
By Holder's Inequality, we have
$ \left[\sum \sqrt{\frac{a}{(a\plus{}2b)^3}}\right]^2\left( \sum a^2\right) \ge \left( \sum \frac{a}{a\plus{}2b}\right)^3.\quad (1)$
Also, by the Cauchy-Schwarz Inequality,
$ \sum \frac{a}{a\plus{}2b} \ge \frac{(a\plus{}b\plus{}c)^2}{a(a\plus{}2b)\plus{}b(b\plus{}2c)\plus{}c(c\plus{}2a)}\equal{}1. \quad (2)$
From (1) and (2), we deduce that
$ \sum \sqrt{\frac{a}{(a\plus{}2b)^3}} \ge \frac{1}{\sqrt{a^2\plus{}b^2\plus{}c^2}}.$
Therefore, it suffices to prove that
$ \frac{1}{\sqrt{a^2\plus{}b^2\plus{}c^2}} \ge \frac{1}{\sqrt{3}} \cdot \frac{a\plus{}b\plus{}c}{a^2\plus{}b^2\plus{}c^2},$
which is true according to the Cauchy-Schwarz Inequality.

[quote="can_hang2007"][quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \sqrt {\frac {a}{(a \plus{} 2b)^3}} \plus{} \sqrt {\frac {b}{(b \plus{} 2c)^3}} \plus{} \sqrt {\frac {c}{(c \plus{} 2a)^3}} \ge \frac {1}{\sqrt {3}}. \frac {a \plus{} b \plus{} c}{a^2 \plus{} b^2 \plus{} c^2}$  :)[/quote]
By Holder's Inequality, we have
$ \left[\sum \sqrt {\frac {a}{(a \plus{} 2b)^3}}\right]^2\left( \sum a^2\right) \ge \left( \sum \frac {a}{a \plus{} 2b}\right)^3.\quad (1)$
Also, by the Cauchy-Schwarz Inequality,
$ \sum \frac {a}{a \plus{} 2b} \ge \frac {(a \plus{} b \plus{} c)^2}{a(a \plus{} 2b) \plus{} b(b \plus{} 2c) \plus{} c(c \plus{} 2a)} \equal{} 1. \quad (2)$
From (1) and (2), we deduce that
$ \sum \sqrt {\frac {a}{(a \plus{} 2b)^3}} \ge \frac {1}{\sqrt {a^2 \plus{} b^2 \plus{} c^2}}.$
Therefore, it suffices to prove that
$ \frac {1}{\sqrt {a^2 \plus{} b^2 \plus{} c^2}} \ge \frac {1}{\sqrt {3}} \cdot \frac {a \plus{} b \plus{} c}{a^2 \plus{} b^2 \plus{} c^2},$
which is true according to the Cauchy-Schwarz Inequality.[/quote]

very nice...

 :)


[hide="please check this"]Answer: $ f(n) \equal{} \lfloor n/3\rfloor$

Assume $ f(m)\equal{}k,f(m\minus{}1)<k$, i.e., $ m$ is the minimal number for which $ f(m)\equal{}k$. By the minimality of $ m$, there are two possible places after $ k\minus{}1$ moves among $ m$ cups. But after that, the coin can be moved again, so there are three possible places. Thus there are at least four and five possible places among $ m\plus{}1$ and $ m\plus{}2$ cups respectively, after $ k\minus{}1$ moves. Therefore there are six possible places among $ m\plus{}3$ cups after $ k\minus{}1$ moves. So the $ k$-th move cannot locate the coin, i.e., $ f(m\plus{}3)>k$. Since $ f(4)\equal{}1,f(5)\equal{}1,f(6)\equal{}2,f(7)\equal{}2,f(8)\equal{}2$, then we know that $ f(9)\ge3,f(10)\ge3,f(11)\ge3,\ldots,f(n)\ge\lfloor\frac{n}3\rfloor$.

Now consider the following moves:
-1,2,3,4
-4,5,6,7
-7,8,9,10
and so on

We can check easily that the above way require $ \lfloor\frac{n}3\rfloor$ moves, and clearly this way can locate the coin.[/hide]

[quote="Johan Gunardi"][hide="please check this"]Answer: $ f(n) \equal{} \lfloor n/3\rfloor$

Assume $ f(m) \equal{} k,f(m \minus{} 1) < k$, i.e., $ m$ is the minimal number for which $ f(m) \equal{} k$. By the minimality of $ m$, there are two possible places after $ k \minus{} 1$ moves among $ m$ cups. But after that, the coin can be moved again, so there are three possible places. Thus there are at least four and five possible places among $ m \plus{} 1$ and $ m \plus{} 2$ cups respectively, after $ k \minus{} 1$ moves. Therefore there are six possible places among $ m \plus{} 3$ cups after $ k \minus{} 1$ moves. So the $ k$-th move cannot locate the coin, i.e., $ f(m \plus{} 3) > k$. Since $ f(4) \equal{} 1,f(5) \equal{} 1,f(6) \equal{} 2,f(7) \equal{} 2,f(8) \equal{} 2$, then we know that $ f(9)\ge3,f(10)\ge3,f(11)\ge3,\ldots,f(n)\ge\lfloor\frac {n}3\rfloor$.

Now consider the following moves:
-1,2,3,4
-4,5,6,7
-7,8,9,10
and so on

We can check easily that the above way require $ \lfloor\frac {n}3\rfloor$ moves, and clearly this way can locate the coin.[/hide][/quote]

@ Johan. Thank you for your solution. In think first part is OK, but could you explain more detaily about second part? The coin can move from n to 1 and then to 2 --> 3 ... and you can not catch it.

I misread the problem statement. I thought they were arranged in a row. Sorry about that. But I'm sure the idea is similar. I'll post my solution as soon as I find it. :lol:

[quote="Namdung"]There are n cups arranged on the circle. Under one of cups is hiden a coin. For every move, it is allowed to choose 4 cups and verify if the coin lies under these cups. After that, the cups are returned into its former places and the coin moves to one of two neigbor cups. What is the minimal number of moves we need in order to eventually find where the coin is?

Really nice problem and I dont know how to solve it.[/quote]

I don´t know the solution yet, but friends from my school solved it and said to me that the quantities of moves are different for $ n$ odd and $ n$ even.I didn´t solved it yet, but i didn´t tried it that much.   :|

[quote="sandt1g3r"]is it $ \frac {1}{2}\left(n \minus{} 1 \right)$ if n is odd and $ \frac {1}{2}\left(n \minus{} 2\right)$ if n is even ?  :maybe:[/quote]

nope.test for $ 7$ coins.

Let us rotate the cups by 1 position clockwise after each move. Now we can assume the cups do not move and the coin moves by either 0 or 2 positions clockwise. Let us number the cups $1,2,\ldots,n$, clockwise. After each move we count the number of "good cups", that is the cups in which we know that the coin isn't there. Clearly the number of good cups increases by at most 4 after each move.

If there is a "bad cup" with odd number, then the next odd numbered cup (in clockwise order) will become bad after the coin moves. Therefore, the number of good cups can increase by 4 only if we know that all cups of a certain parity are all good. This situation can occur at most twice. Hence after $k$ moves we find at most $3(k-2)+4+4=3k+2$ bad cups. If we have found the coin after $k$ moves, then $3k+2\ge n-1$, we get $k\ge\left\lceil\frac{n-3}3\right\rceil$.

I haven't done the construction, but I think it's easy.

@Johan; I'm not sure that your bound is the best possible. For example, your appraoch gives $f(8)=3$, but in fact we have $f(8)=2$

Let the cups be $1,2,...,8$. In the first move we check $A=\{2,4,6,8\}$. If the coin is there, we are done. If not, the coin must be under one of $B=\{1,3,5,7\}$, but once we replace the cups the coin will be moved from $B$ to $A$, so if we select $A$ for a second time, we find the coin.

Okay, I have edited my solution above. I hope my third try doesn't fail again. :oops:

I'm sorry to say, that the bound is still not good  :(  For $n=9$ your bound give $k=2$, but we need $k=3$

[hide="why?"]
Consider the paths that the coin can take in two moves, that is $i\to i+1$ or $i+1\to i$ for some $i$. Now if $n=9,k=2$. Then the first move reveals $4$ cups, so there are five possible starting places for the coin. Each starting place can go to $2$ possible other landing place, giving $10$ possible paths. In the second move four more cups are revealed. Each cup can be a landing place to at least two paths, so at most $2\cdot 4=8$ of the $10$ paths are useless. This leaves at least $2$ paths which the coin can take undetected. After the two moves, you won't know which path it was under originally. Therefore we need at least $k=3$
[/hide]

In fact, i think your original bound is the best for $n\ge 9$. An idea for a solution would be: for sets $B_1,B_2,...,B_m$, where $B_i\subset M=\{1,2,...,n\}$ and $|B_i|=4$. Show by induction that if $n\ge 9$ and $m=\left\lceil \frac{n-3}{2}\right\rceil - 1$ then there exists a good sequence $(b_1,b_2,...,b_m)$ such that $b_i\in M\backslash B_i$ for $i=1,2,...,m$.

I think my bound is the best for some $n$, like $n\equiv2,4\pmod6$. Anyway, my bound doesn't work at all for odd $n$. Seems that v235711's friend is quite reliable after all. :D

[quote="nima-amini"]can a person post solution?[/quote]
//edit: Sorry I thought that the number of coins is twelve

I think it is possible with 4 searches (moves)

Color the cups with black & white

like this:

...  B   W   B  W  B  W  B  W ...
... 11 12 1  2  3  4  5  6 ...

assume that coin is under a white cup

so choose these:   2,4,6,8

at the next round it must be in a black cup (because white neighbours are black!)

if coin isn't under these cups so in the next round it must be in :  9,11,1

else the coin must be in a black cup at the first round

like previous we choose :

1,3,5,7
& then
8,10,12

Am I right?

It seems there is a very intuitive strategy, although I'm not sure if it's necessarily the best:

Start anywhere on the circle, pick up cups 1, 2, 3, 4 going clockwise. Next, pick up 4 5 6 7. The idea here is that you have eliminated the possibility of a coin being in either the 5th, 6th or 7th position before the first lift. So a coin cannot escape to the 4th position (ie move from 5 to 4) after the 2nd lift. Note you cannot cheat and lift up 4 6 7 8, because a coin may have been hidden in 6 before the first lift and moved to 5 before the 2nd lift. Just continue this method arond the circle.

So it's easy to see that the best strategy for the coin (if the coin can think) to escape capture is to start at position 0, and only move clockwise  (same direction as you). But eventually you will catch up to the coin. Your speed is 3 cups per move while the coin is 1 cup per move. So at worst you will catch up to the coin after 1.5 laps around the circle. So the number of lifts is approximately 1.5n/3 or n/2. I can't be bothered to give a precise integer bound so there.

not sure whether this is true

For $n=4k$,
$f(n)=2( \frac{n}{4} -1)$
For $n=4k+1$ and $4k+2$,
$f(n)=2( \left\lfloor\frac{n}{4}\right\rfloor -1)+1$
For $n=4k+3$,
$f(n)=2( \left\lfloor\frac{n}{4}\right\rfloor)$

$\lfloor\frac{n-3}{2}\rfloor$ would be the answer, I should post a solution in a few hours

The original statement: https://problems.ru/view_problem_details_new.php?id=1154
There are 100 cups in the original problem statement, so you can just solve it for even numbers.


Hi. I found out about the USABO this year, and I want to prepare for it. I read on the website that you should read Campbell's Biology, which I got, but do you really just memorize that book? Do you know where I might get some practice problems? Also, how are the tests set up (time limit, number of problems, etc.)? Thanks!

The mod has old USABO tests, I believe (correct me if I'm wrong, adid).  And its also not a bad idea to look at the old Australian IBO qualifying rounds and old IBOs

In addition to knowing cambpell's really well (like to the level of detail where you can tell me what type of epithelial tissues exist where and what type of WBC would be elevated in a parasite infection), you should also memorize the IBO taxonomy chart, esp. the plants and insects.  Campbell's, along with some other basic info, will get you to finals (someone can correct me on this; it did for me, but I don't know if its changed.

To do well in finals, you need all this plus some other info, which they teach you in the lab section of finals.  Memorize the handouts; I didn't, and I was slaughtered.

Thank you so much for the information! But where can I get the old USABO's or the old IBO's? I've looked for them, but I only found several example problems, not entire tests...

You may have to buy old tests; you can buy the Australian ones somewhere if you search on google.

Old USABO open and semifinal exams are floating around the internet.  I'll post my copy for download as soon as I find it.

i believe you need 3 textbooks:
campbell biology
lodish molecular cell biology
and some other book for ecology

that ought to cover you. for the mcb and ecology, you just have to read half the books, and that should get you all the way to finalist assuming you memorize campbell. all in all around 2000 pages (all of campbell + half of each, i didn't have an ecology textbook and regretted it), very doable in a 3-week cramming where you read 3 hours/day.

how hard is it to be a semifinalist (whats the score that generally needed)

do you think a person who has taken honors bio and self taught himself in AP bio during the summer and is dedicated and has some raw talent can become semifinalist?

Or are most of the stuff not generally covered in AP bio?

Raven is a good biology textbook. I find that Soloman is better. Very nice pictures and diagrams.

How does the AP exam compare to the USABO?

by Raven, mathguy probably meant Raven's Biology of Plants. As for a general textbook, Campbell is far better than Solomon.

The USABO Open Exam is doable with just AP knowledge, and even if you only self-study it, you can become a semifinalist. However, to do well on the semifinals and have a shot at finals, you should read more advanced molecular biology and plant textbooks.

As for the cutoff score, it has been 29 or 30 for the past couple years, but depends somewhat on the difficulty of the exam.

you can just sign up and take it right?

teachers dont have to "recommend" you and sign you up to take the test (they did that for the usnco this year i think)?

because that would suck since im not taking ap bio until senior year

and whats the first test out of?  thanks.

Technically, the teachers have to nominate you, though most let anyone who wants to take it (I've heard of some who only let certain students take it).

As for the Open, it's out of 50 questions. (also, no penalty for guessing)

- Kimball's Biology Pages is a helpful resource.
- Campbell and Reece's Biology (7th ed.)
- Biology OCW
- Perhaps you'll find something useful [url=http://www.4shared.com/network/search.jsp?sortType=1&sortOrder=1&sortmode=2&searchName=biology&searchmode=2&searchName=biology&searchDescription=&searchExtention=&sizeCriteria=atleast&sizevalue=10&start=0]here[/url]

Good luck.  :)

Is 7th edition necessary? or is any edition of campbell okay?

I believe any edition is fine - there's no "need" for a specific textbook.

7th edition has colorful diagrams, but that's about it. Also, I highly suggest reading lodish. Many questions on the 2008 semis were only answerable if you read that book. Granted, I read it (with campbell's) and just missed the cut-off...so I can't talk (although I epically failed on a few big part C questions which might explain this).

I advice to use Campbell Biology 8th ed.,solomon,raven,or life the science of biologyfor practise to USABO.

Getting to Semis is no sweat... What's harder is getting to finals. I suggest that you
A) Read Campbell. Read it and understand it.
B) Raven is pretty useful, but I got into Finals w/o it. The USABO people love to put questions from there on the test.
C) Never read Lodish, but I just barely squeaked into Finals, tho, so perhaps I shouldn't talk.
D) If you get to Finals: MEMORIZE THE HANDOUTS. That is basically the whole finals test.
E) As strange as it sounds, GRE tests and handbooks are really useful for studying.

AP bio is nothing compared to USABO.


Let  V and W be two vector spaces of countable dimensions .
Is dimension of L(V.W) also countable? If yes, how do we  prove it?

I'm just throwing this out since ive only dealt with finite dimensions but i think you'd do the same thing as in the proof for linear spaces of finite dimensions, except maybe use induction. if you know that the cartesian product of countable sets is countable then you should be fine

If $ V \equal{} \ell^1$ (over reals) and $ W \equal{} \mathbb R$, then $ L(V,W)$ is the dual of $ \ell^1$, which is a nonseparable space, namely $ \ell^{\infty}$.

[quote]if you know that the cartesian product of countable sets is countable then you should be fine[/quote]
Problem is, the cartesian product of countably many countable sets is not countable.

There's some ambiguity over what "dimension" means, and mlok used "countable Schauder basis in a normed space."

Here's another try, using "dimension" to mean "cardinality of a Hamel basis":

Let $ V$ be the vector space of finitely supported sequences over field $ K.$ Its Hamel basis is the "natural basis", where $ e_k$ has a $ 1$ in the $ k$th component and zeros elsewhere. Let $ W\equal{}K.$ Then $ L(V,W)$ is the vector space of [i]all[/i] sequences. And that does not have a countable Hamel basis.


I need to determine whether the following sequence converges uniformly on the given intervals:

$ \{\frac{n^2x}{1\plus{}n^3x^2}\}$ on the intervals:

[1,2]
[a,inf) a>0

For the first one, i noticed each function is decreasing on the interval, so the $ \sup|\frac{n^2x}{1\plus{}n^3x^2}|$ occurs when x=1. When x=1, the sequence goes to 0, proving uniform convergence.

I'm not sure what to do with the second one, since the functions may not all be decreasing on [a,b]

Any help?

By weighted AM-GM, $ u^pv^{1\minus{}p}\le pu\plus{}(1\minus{}p)v.$

Choose a value of $ p$ such that $ \frac23<p<1.$ Let $ u\equal{}\frac{n^3x^2}{p}$ and $ v\equal{}\frac{1}{1\minus{}p}.$ Then

$ 1\plus{}n^3x^2\ge \frac1{(1\minus{}p)^{1\minus{}p}}\cdot \frac{n^{3p}x^{2p}}{p^p}.$ Hence,

$ \frac{n^2x}{1\plus{}n^3x^2}\le (1\minus{}p)^{1\minus{}p}p^p\,\frac{n^2x}{n^{3p}x^{2p}}\equal{} (1\minus{}p)^{1\minus{}p}p^pn^{2\minus{}3p}x^{1\minus{}2p}.$

On the interval $ [a,\infty),$ this is less than $ Cn^{2\minus{}3p}a^{1\minus{}2p}$ which decreases as $ n\to\infty,$ so this tends uniformly to zero.

This argument does not work on $ [0,\infty).$


Inradius of a triangle is $ r$. Tangent lines parallel to the sides produce three smaller triangles, whose inradii are $ r_1,r_2,r_3$. Prove that $ r\equal{}r_1\plus{}r_2\plus{}r_3$.

The small triangle by vertex $ A$ is similar to the larger triangle, so the ratio of the inradii is equal to the ratio of the heights from vertex $ A$, if $ h_a$ denotes the height from $ A$ to $ BC$, and $ r$ is the in radius, then $ \frac{r_1}{r}\equal{}\frac{h_a\minus{}2r}{h_a}\equal{}1\minus{}\frac{2\frac{A}{s}}{\frac{2A}{a}}\equal{}1\minus{}\frac{2a}{a\plus{}b\plus{}c}$. Summing cyclical results finishes the problem.

How do the tangent lines parallel to the sides produce three smaller triangles?

Construct a triangle and its incircle. For each side construct a line parallel to that side on the opposite side of and tangent to the incircle.

[quote="Johan Gunardi"]Inradius of a triangle is $ r$. Tangent lines parallel to the sides produce three smaller triangles, whose inradii are $ r_1,r_2,r_3$. Prove that $ r = r_1 + r_2 + r_3$.[/quote]

Consider the following notations:
$ \|\begin{array}{cccc} A_1,A_2 \in (BC)\\
\\
B_1,B_2\in (CA)\\
\\
C_1,C_2 \in (AB) \end{array}$ such as $ C_1B_2,B_1A_2,A_1C_2$ are the tangent to the incircle $ C(I,r)$ and parallel with the triangles sides.

[color=red][b]Remark[/b][/color] They can be parallel and also [color=brown]antiparallel[/color]
and $ r_1,r_2,r_3$ ar the inradii of the triangles $ AC_1B_2,BA_1C_2,CB_1A_2$

[b]Proof:[/b]
The triangles $ ABC, AC_1B_2,BA_1C_2,CA_2B_1$ are similar triangles.The triangles $ AC_1B_2,BA_1C_2,CA_2B_1$ have the semiperimeters $ p_1,p_2,p_3$ and denote $ D\in BC$ , $ E \in AC$, $ F\in AB$ the tangent points of $ C(I,r)$ with the sides of the triangle $ ABC$
$ \frac{p_1}{p_2}=\frac{r_1}{r}$ $ \equiv$ $ \frac{p-a}{p}=\frac{r_1}{r}$,Analogously $ \frac{p-b}{p}=\frac{r_2}{r}$, $ \frac{p-c}{p}=\frac{r_3}{r}$. By summing we obtain $ r_1+r_2+r_3=r$

[quote="Altheman"]$ \frac {r_1}{r} \equal{} \frac {h_a \minus{} 2r}{h_a} \equal{} 1 \minus{} \frac {2\frac {A}{s}}{\frac {2A}{a}} \equal{} 1 \minus{} \frac {2a}{a \plus{} b \plus{} c}$. Summing cyclical results finishes the problem.[/quote]

How does $ 2r/h_a$ become ($ 2\frac{A}{s}$ divided by $ 2\frac{A}{a}$ )... and how does that become your expression with $ 2a$ as the numerator?

And then... what is cyclic summing and what is being added in this problem?

$ r\equal{}\frac{A}{s}$ and $ h_a\equal{}\frac{2A}{a}$ where $ A$ is the area of the triangle.

By summing cyclically, I mean do the same thing for $ b$ and $ c$ then add.

how about if the lines aren't parallel to the sides of the triangle?
does the equality still hold?

no it doesnt....just take extreme cases like when the lines are the sides of the triangle


geometry
geometric transformation
homothety
power of a point
radical axis
geometry unsolved

Consider a circle $ (O)$ and a chord $ AB$. Let circles $ (O_1),(O_2)$ be internally tangent to $ (O)$ and $ AB$ and let $ M,N$ their intersection. Prove that $ MN$ passes through the midpoint $ T$ of the arc $ AB$ which does not contain $ M$ and $ N$.

Let $ O_1$ and $ O_2$ be tangent to $ AB$ at $ C$ and $ D$, respectively, and tangent to $ O$ at $ E$ and $ F$, respectively.  Then the homothety about $ E$ which maps $ O_1$ to $ O$ maps $ C$ to $ T$: hence, $ E, C,$ and $ T$ are collinear.  Similarly, $ F, D,$ and $ T$ are collinear.  Now we have $ m\widehat{TA}\equal{}m\widehat{TB}$, and so $ m\angle{TCD}\equal{}m\angle{TFE}$.  Hence, $ \triangle{TCD} \sim \triangle{TFE}$, and so $ TC \cdot TE \equal{} TD \cdot TF$.  Hence, $ T$ has the same power with respect to the two circles, and hence lies on their radical axis, $ MN$, as desired.

Inversion with center $ T$ and power $ TA^2$ takes the circle $ (O)$ to the line $ AB$ and the line $ AB$ into the circle $ (O)$ $ \Longrightarrow$ it takes the circles $ (O_1), (O_2)$ tangent to both $ (O)$ and $ AB$ to themselves $ \Longrightarrow$ it takes their intersection $ M$ to the other intersection $ N.$ Point $ M$ and its inversion image $ N$ are collinear with the inversion center $ T.$

let $ C$and $ E$,be the points of tangency of$ (O),(O_1)$and $ (O),(O_2)$,respectively,again $ D,F$,are the points of tangency of $ O_1,O_2$,with $ AB$,respectively.
it is clear that $ ECDF$ is cyclic,and $ CD,EF$,intersect each other at $ T$,the midpoint of arc $ AB$,
then $ T$,lies on the radical axis of the two circles $ (O_1),(O_2)$,and has a constant power to
these two circles equal to $ TA^2$and doesnt depend on the position of $ M,N$ :wink:


trigonometry
calculus
geometry
circumcircle
trig identities
Law of Cosines

Find the exact value of cosine 108 degrees WITHOUT using a calculator.

One method is to use the Taylor Series but that requires calculus and the angle measure to be in radians.

[hide="Hint"]Take an isosceles triangle $ \Delta ABC$ with $ \alpha \equal{} \beta \equal{} 72^\circ$ and draw the bisector of $ \angle BAC$.[/hide]

$ \cos 108^\circ \equal{} \minus{} \cos 72^\circ \equal{} \minus{} \sin 18^\circ \equal{} \minus{} \frac {\sqrt {5} \minus{} 1}{4}$.

Let $ ABCDE$ be a regular pentagon with side length 1, we have $ \triangle ACD$ with $ \angle{ACD} \equal{} \angle{ADC} \equal{} 72^\circ$. If $ H$ is the foot of the perpendicular from $ A$, we have $ AH$ is perpendicular to $ CD$.
For $ \triangle{CAH}$, we have $ \angle{CAH} \equal{} 18^\circ$, $ \sin 18^\circ \equal{} \frac {CH}{AC} \equal{} \frac {\sqrt {5} \minus{} 1}{4}$, yielding $ \cos 108^\circ \equal{} \minus{} \frac {\sqrt {5} \minus{} 1}{4}$.

Note: length of diagonal of regular pentagon=$ \frac{1\plus{}\sqrt{5}}{2}*$ length of side length of the pentagon.

[quote="kunny"]$ \cos 108^\circ \equal{} \minus{} \cos 72^\circ \equal{} \minus{} \sin 18^\circ \equal{} \minus{} \frac {\sqrt {5} \minus{} 1}{4}$.

Let $ ABCDE$ be a regular pentagon with side length 1, we have $ \triangle ACD$ with $ \angle{ACD} \equal{} \angle{ADC} \equal{} 72^\circ$. If $ H$ is the foot of the perpendicular from $ A$, we have $ AH$ is perpendicular to $ CD$.
For $ \triangle{CAH}$, we have $ \angle{CAH} \equal{} 18^\circ$, $ \sin 18^\circ \equal{} \frac {CH}{AC} \equal{} \frac {\sqrt {5} \minus{} 1}{4}$, yielding $ \cos 108^\circ \equal{} \minus{} \frac {\sqrt {5} \minus{} 1}{4}$.

Note: length of diagonal of regular pentagon=$ \frac {1 \plus{} \sqrt {5}}{2}*$ length of side length of the pentagon.[/quote]

But how do you prove the last theorem?

[hide]
Place a regular pentagon with side length $ 1$ in its circumcircle and connet diagonals, and use Ptolemy's Theorem. Let the length of diagonal be $ x$, then
$ x^2 \equal{} x \plus{} 1$
$ x \equal{} \varphi$
Then in a triangle bounded by a diagonal and two sides, use Law of Cosines
$ \cos108^o \equal{} \frac {1 \plus{} 1 \plus{} \varphi^2}{2} \equal{} \frac {1 \minus{} \sqrt {5}}{4}$

[/hide]


function
calculus
derivative
trigonometry
calculus computations

I want to get the rate of change of y, which is an angle of a triangle in radians
cosy=100/x

Thansk,

Do you want it in terms of $ x$, or in terms of some unspecified other variable?

Either way, the technique is implicit differentiation (Interpretation may vary).

[hide="just cuz im bored"]

$ \minus{} \sin(y)y' \equal{} \frac { \minus{} 100}{x^{2}}$

$ y' \equal{} \frac { 100}{x^{2}\sin(y)}$

[/hide]

Yes I know but how can I differentiate y=(cos^-1)(100/x)?

Recall that $ (\cos^{ \minus{} 1}(x))' \equal{} \frac { \minus{} 1}{\sqrt {1 \minus{} x^{2}}}$. With $ \cos^{ \minus{} 1}\left(\frac {100}{x}\right)$, same idea but with the chain rule.


Hi, I'm having problems understanding and doing this problem:

For a fixed a ∈ R, determine the dimension of the subspace of P_n(R) defined by {f ∈ P_n(R)|f(a) = 0}.

Thanks!

Are you using P_n(R) for projective $ n \times n$ matrices in $ \mathbb{R}$?

I'm guessing he probably means polynomials. The only question is whether he means degree $ \le n$ or degree $ < n.$

Hi I'm sorry, I kind of just copied the problem from the book. I am referring to polynomial. and it is less than or equal to degree n.


For anyone who doesn't understand why (you never know, just in case  :) ):

$ \left( \frac{(x\plus{}1)^2(x^2\minus{}x\plus{}1)^2}{(x^3\plus{}1)^2}\right)^2*\left( \frac{(x\minus{}1)^2(x^2\plus{}x\plus{}1)^2}{(x^3\minus{}1)^2}\right)^2\equal{}\\
\left( \frac{((x\plus{}1)(x^2\minus{}x\plus{}1))^2}{(x^3\plus{}1)^2}\right)^2*\left(( \frac{((x\minus{}1)(x^2\plus{}x\plus{}1))^2}{(x^3\minus{}1)^2}\right)^2\equal{}\\
\left( \frac{(x^3\plus{}1)^2}{(x^3\plus{}1)^2}\right)^2*\left( \frac{(x^3\minus{}1)^2}{(x^3\minus{}1)^2}\right)^2\equal{}\boxed1$


geometry
3D geometry
quadratics
algebra
factorization
difference of cubes
special factorizations

The root $ r$ of an equation satisfies $ r^2 \plus{} 4r \plus{} 16 \equal{} 0$. Find $ r^6 \plus{} r^3 \plus{} 1$.

[quote="sunehra"]The root $ r$ of an equation satisfies $ r^2 \plus{} 4r \plus{} 16 \equal{} 0$. Find $ r^6 \plus{} r^3 \plus{} 1$,[/quote]
$ r \equal{} \frac { \minus{} 4 \pm \sqrt {4^2 \minus{} 4(16)}}{2}$
You get complex roots. perhaps substituting them in their places will give you value of $ r^6 \plus{} r^3 \plus{} 1$.

Notice the r^2 + 4r + 16 is part of the difference of cubes formula for r^3 - 64. Multiplying both sides of the quadratic equation by r-4, we get r^3 - 64 = 0, or r^3 = 64. Also notice that r^6 + r^3 + 1 is the same as r^3^2 + r^3 + 1. The answer is 64^2 + 64 + 1 = 4161.


geometry
Asymptote
area of a triangle
Heron\u0027s formula

Find the area of the triangle formed by the intercepts and the intersection of the asymptotes of the graph of \[ f(x) \equal{} \frac{3}{x\minus{}3}\minus{}3\]

[quote="RobRoobiks"]the horizontal asymptote is y=-3
so (4,0), (0,-4), (3,-3) would be the points
I would use Matrices to calcute the area since thats the only one I learnt
1/2 det |1 4 0 |
           |1 0 -4|
           |1 3 -3|
which gives you 4
Please better mny method for finding area of the triangle[/quote]

The area of the triangle is equal to the area of the larger triangle
having vertices (0, -4), (4, 0), and (4, -4) minus twice the area of
the triangle with vertices of (0, -4), (3, -3), and (3, -4) minus
the area of the unit square with vertices of (3, -4), (3, -3), (4, -3),
and (4, -4).

This equals
(0.5)(4)(4) - 2[(0.5)(3)(1)] - 1 =
8 - 2(1.5) - 1 =
8 - 3 - 1 =
4

4 square units for the area of the triangle with vertices at (0, -4), (4, 0),
and (3, -3)

where did the points (4,0), (0,-4), (3,-3) come from?

the y intercept
y=3/(x-3) -3
y=3/-3 -3
y=-1-3
y=-4
(0,-4)

x-intercept
y=3/(x-3) -3
0=3/(x-3) -3
4=x
(4,0)

asymtops: y=-3, x=3
so the points are
(3,-3)

It was there until Arrange his tan dlted his previous post  :lol:

you could also calculate the sides of the triangle with santa pit's teorem(pythagorean toerem)  and use heron's formula for the area of triangle


algorithm
induction
algebra
polynomial
superior algebra
superior algebra unsolved

Show that [(x^p^d] - x divides [(x^p^n] - x  if and only if d divides n.

Obviously wrong. What did you really mean?

The statement you wrote is clearly false.  jmerry was assuming that there is some true statement you were trying to ask.

The professor had a typo in the homework list, so I just edited the question.

It's still wrong, because you have the parentheses wrong.$ (x^p)^d\equal{}x^{pd}$.

Start with the field $ F_p$ of $ p$ elements. What are the splitting fields of $ x^{p^d}\minus{}x$ and $ x^{p^n}\minus{}x$ over $ F_p$?

A more elementary approach: show that $ gcd(p^a\minus{}1,p^b\minus{}1)\equal{}p^{gcd(a,b)}\minus{}1$, and also show that $ gcd(x^a\minus{}1,x^b\minus{}1)\equal{}x^{gcd(a,b)}\minus{}1$.  It's a simple induction.

As often happens, "more elementary" is also a little harder. Choose a level appropriate to what you know.

I should note that working in characteristic $ p$ as I recommended only tells you about divisibility mod $ p$. If $ x^{p^d}\minus{}x$ divides $ x^{p^n}\minus{}x$ mod $ p$, we still need to show divisibility over the integers, by constructing the quotient.

[quote="LydianRain"]A more elementary approach: show that $ gcd(p^a \minus{} 1,p^b \minus{} 1) \equal{} p^{gcd(a,b)} \minus{} 1$, and also show that $ gcd(x^a \minus{} 1,x^b \minus{} 1) \equal{} x^{gcd(a,b)} \minus{} 1$.  It's a simple induction.[/quote]

This may be viewed as a q-analog of the integer Bezout theorem. See my sci.math post linked below for a little-known slick proof from that viewpoint, which specializes to the integer Bezout theorem for q = 1.
http://google.com/groups?selm=y8zd5vf55k9.fsf%40nestle.csail.mit.edu

Induction?  It's a corollary of the integer Bezout theorem, as follows:  if $ d \equal{} \gcd(x^a \minus{} 1, x^b \minus{} 1)$ then $ x^a, x^b \equiv 1 \bmod d \Rightarrow x^{ax \plus{} by} \equiv 1 \bmod d \forall x, y \in \mathbb{Z} \Rightarrow x^{(a, b)} \equiv 1 \bmod d$.  On the other hand, clearly $ x^{(a, b)} \minus{} 1 | d$.  (This proof holds both in the polynomial sense and when $ x$ is replaced by an integer.)  The $ q$-analogue interpretation is quite nice, though.


complex analysis
complex analysis unsolved

What is the correct interpretation of [z^(n/m)]^(m/n) where z is complex with n and m relative prime (this is not necessary, though)?  What is the correct order of operation in this case?  Should we multiply the exponents first, or "do it inside-out"?  If we use the latter approach, z^(n/m) is multivalued with m possible values, and the exponent m/n gives another n; so the original expression has m*n different values!  If we use the former approach, it would yields only one value -- z^1 = z.

Also, does (-a)^(b/b) = a or -a for all positive real a and all positive even integer b?

You do it as written, and it's multivalued, with $ n$ values ($ z$ times an $ n$th rooth of unity).


ABCD is a quadrilateral. ABCD-A1B1C1D1 is a square prism.

AB=AD=2, DC=2*sqrt(3), AA1=sqrt(3).
AD is perpendicular to DC.
AC is perpendicular to BD and intersect at E.

1. Prove BD is perpendicular to A1C.

2. Determine the angle between the two planes A1BD and BC1D

3. Determine the angle formed by lines AD and BC1 which are in different planes.

This is where the problem is from:

http://news.bbc.co.uk/2/hi/uk_news/education/6589301.stm

$ AD \equal{} AB$ and $ AC \perp BD$ $ \Longrightarrow$ $ ABCD$ is a kite with $ AC \equal{} 4$ and equilateral $ \triangle BCD$. From this:
1. Plane $ (ACC_1) \perp BD$ $ \Longrightarrow$ $ AC_1 \perp BD.$
2. $ \angle((AB_1D), (BC_1D)) \equal{} \angle A_1EC_1$. $ \triangle A_1EC_1 \cong \triangle ABC$ $ \Longrightarrow  \angle A_1EC_1 \equal{} \angle ABC \equal{} 90^\circ.$
3. $ M$ is midpoint of $ CD$, $ BM \parallel AD$, $ BM \perp CD$, $ BM \equal{} 3$, $ MC_1 \equal{} \sqrt{6}$. $ \triangle BMC$, $ \triangle BMC_1$ are right at $ M$ $ \Longrightarrow$ $ \tan \angle(AD, BC_1) \equal{} \tan \angle MBC_1 \equal{} \frac{MC_1}{BM} \equal{} \sqrt{\frac{2}{3}}$.


[quote="Drunken_Math"][quote="joejia"]In triangle ABC, CD is the altitude to AB, and AE is the altitude to BC. If the lengths of AB, CD, and AE are known, the length of DB is:
(A) not determined by the information given
(B) determined only if A is an acute angle
(C) determined only if B is an acute angle
(D) determined only if ABC is an acute angle
(E) none of these is correct[/quote]
[hide]
If you mean (D) to be "determined only if ABC is an acute triangle" than that's the answer.

Because you can find the length of EB using the Pythagorean Theorum.

Then, if you find the length of CE, you can use the Pythagorean Theorum to find DB.

You can find the length of CE only if angle ACB is acute by using trigonomic functions (tangent)[/hide][/quote]
I agree we can find $BD$ using pythagoras theorem, as we know $BC$ and $CD$


parameterization
limit
trigonometry
calculus
calculus computations

Let $ a>b$ with $ a,b \in (4,9) \cap \mathbb{N}$.

Evaluate:

$ \lim_{n \to \infty} {\frac{\sin\left[(3\plus{}\sqrt{a})^n \cdot \pi \right]}{ \tan \left[(3\plus{} \sqrt{b})^n \cdot \pi \right]}}\equal{}?$.

$ \alpha_n \equal{} (3\plus{}\sqrt a )^n \mod 2\equal{}(\sqrt a\minus{}1)^n \mod 2$
$ \beta_n \equal{} (3\plus{}\sqrt b )^n \mod 2\equal{}(\sqrt b \minus{}1)^n \mod 2$

pretty sure that these guys will both get arbitrarily close to zero infinitely often, but not at the same time, therefore no limit?

The limit does exist; notice that $ \left(3 \plus{} \sqrt {a}\right)^n \plus{} (3 \minus{} \sqrt {a})^n\in \mathbb{N}$ for all $ n\in \mathbb{N}$ if $ a\in \{5,6,7,8\}$ and therefore,
\[ \lim_{n \to \infty} {\frac {\sin\left[(3 \plus{} \sqrt {a})^n \pi \right]}{\tan \left[(3 \plus{} \sqrt {b})^n \pi \right]}} \equal{} \lim_{n\to \infty}\frac {\sin \left [ (3 \minus{} \sqrt {a})^n \pi\right]}{\tan \left[ (3 \minus{} \sqrt {b})^n\pi \right]} \equal{} \lim_{n\to \infty}\frac {\left(3 \minus{} \sqrt {a}\right)^n}{\left(3 \minus{} \sqrt {b}\right)^n} \\
\equal{} \begin{cases} 0 & a > b \\
1 & a \equal{} b \\
\infty & a < b\end{cases}
\]


modular arithmetic
number theory unsolved
number theory

Let $p$ be a prime and $a_1, a_2, \dots, a_k$ ($k\geq3$) be integers such that $p$ does not divide $a_i$ for any $i$ and $a_i \not\equiv a_j \pmod{p}$ for $i\neq j$.

Let $S_n = \{n \vert 1 \leq n \leq p-1, (na_1)_p < (na_2)_p < \dots < (na_k)_p\}$, where $(b)_p$ denotes the smallest non-negative integer congruent to $b$ (mod $p$).

Show that $\vert s_n \vert < \frac{2p}{k+1}$.

Sorry for this post but the problem is not from Chinese TST 2005 I think

I got it from [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=45448]here[/url].


I did not put the number $ 2$ in my answer because I had $ 3\le x\le11$.  Will I be counted off?

Yes since 2 was one of the possible answers. What does $ x$ represent in "$ 3 < x < 11$"?

BTW: Don't make 4 duplicate (quadruplicate?) topics. That could be considered spamming.


Let a,b,c be real positive numbers. Prove that

 a/(2a+b) + b/(2b+c) + c/(2c+a) \leq 1

a not very smart solution:

brute force works just fine!

We have \sum a/(2a + b) \leq 1
<=> \sum a(2b + c)(2c + a) \leq (2a + b)(2b + c)(2c + a)
<=> \sum (4abc + 2a2b + 2ac2 + a2c) \leq 9abc + 4a2b + 4ac2 + 2a2c + 4b2c + 2ab2 + 2bc2
<=> 12abc + 4a2b + 4ac2 + 4b2c + a2c + ab2 + bc2 \leq ...
<=> a2b + b2c + c2a \geq 3abc
which is quite true by AM-GM. :D

Remember that Romanian inequality:

If xi are positive numbers with product 1 (i from 1 to n), then
\sum 1/(n-1+xi)<=1.

This problem reduces to the case n=3 for the problem mentioned above. You just simplify the 1'st fraction with a, the 2'nd one with b and the 3'rd one with c, and then put x=b/a etc.

It was just an innocent observation. It would be crazy to solve the general case in order to solve this easy problem, because the proof for the general case is really hard to find.

There's another even easier solution. Rewrite the inequality like so \sum b/(2a+b) \geq 1, or \sum b 2 /(2ab+b 2 )\geq 1, which follows immediately by Cauchy-Schwartz.


find a method for solving the equation $x^{n}-y^{n}=a$ with $x,y,n,a$ positive intergers. solusions are the pairs $(x,y)$


limit
trigonometry
factorial
real analysis
real analysis unsolved

What do you mean by $!$, factorial of $n$? If so, how can this limit possibly be defined. It will alter between large negative and large positive values.

$e$ is special. What is the fractional part of $n!e$?
This isn't the first time this has been posted, but it's a nice problem.

[color=green][b]Hint:[/b][/color] [b]  For each $n\in {\mathbb N}^{*}$ there exists at least a pont $\theta_{n+1} \in (0,1)$  such that $e=\sum\limits_{k=1}^{n}\frac{1}{k!} + \frac{e^{\theta_{n+1}}}{(n+1)! } =: S_n + \frac{e^{\theta_{n+1}}}{(n+1)! }=S_{n}+\frac{1}{(n+1)!}+ \frac{e^{\theta_{n+2}}}{(n+2)!} .\;$
This implies $en! = n!S_n + \frac{e^{\theta_{n+1}}}{n+1 } =: N_n + \frac{e^{\theta_{n+1}}}{n+1}= N_n+ \frac{1}{n+1}+ \frac{e^{\theta_{n+2}}}{(n+1)(n+2)}$  where    $N_n \in {\mathbb N}^{*} .$
 Therefore        $2\pi en! = 2\pi N_n + \frac{2\pi e^{\theta_{n+1}}}{n+1}=2\pi N_n+ \frac{2\pi}{n+1} +r_n$  with  $\lim\limits_{n\to \infty}nr_n=0 \; ,$  and  if  $X_n: = \frac{2\pi}{n+1}+ 2\pi r_n$
 \[ \begin{array}{c} n\cdot \sin\left(2 \pi e n!\right) = n\cdot \sin \left( \frac{2\pi e^{\theta_{n+1}}}{n+1}\right) = \\ =n\cdot \sin \left( \frac{2\pi}{n+1}+ 2\pi r_n\right) = nX_n\frac{\sin{ \left(X_n\right)}}{X_n}= \to 2\pi. \end{array}  \]
[/b]

[quote="flip2004"][color=green] [b] If $e$ is Napier's constant,  what about  [/b] [/color]  $\; \; \; \lim\limits_{n\to \infty} n^2\left( \sin{\left(2\pi e n !\right) -\frac{2\pi}{n}\right) \;}$   ?[/quote]

Professor flip2004,

I found $\pi$, is it correct ?

[quote="Moubinool"] ......I found $\pi$, is it correct ?.....[/quote]
Please post the complete solution.Thanks for interest.

It must be strange, but i've got $0$ as an answer (which strongly contradicts Moubinool's answer):
$\sin(\frac{2\pi}{n+1}+\frac{2\pi}{(n+1)(n+2)} )<sin(2\pi en!)=sin(2\pi n! \sum_{j=n+1}^{\infty}\frac{1}{j!})<sin(\frac{2\pi}{n+1}+\frac{2\pi}{(n+1)^2} )$
since $\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}<\sum_{j=n+1}^{\infty}\frac{n!}{j!}<\frac{1}{n+1}+\frac{1}{(n+1)^2}$, but both limits $\lim_{n\to\infty}n^2(sin(\frac{2\pi}{n+1}+\frac{2\pi}{(n+1)^2})-\frac{2\pi}{n})$ and $\lim_{n\to\infty}n^2(sin(\frac{2\pi}{n+1}+\frac{2\pi}{(n+1)(n+2)})-\frac{2\pi}{n})$ tend to zero if of course i haven't done stupid mistakes.

All of u have different answers , so whom should i believe if i have no solution? :lol:

Rustam, according to Moubinool and me-we've got the same answer : $0$

Finally i understood the solution of this beatifull problem .when i asked my teacher for hint he figured out interesting limits:
       If f(x) is continuous and there exist T such that f(x+T)=f(x) (is T called the period in eng I don't know exact) and f(0)=0 then
(i) if f is ones differencial then $\lim_{n\to\infty}nf(Ten!)=Tf'(0)$;
(ii) if f is twice differencial then $\lim_{n\to\infty}n(nf(Ten!)-Tf'(0))=Tf''(0)/2$;
(iii) if f is thrice differenciable then $\lim_{n\to\infty}n(n(nf(Ten!)-Tf'(0))-\frac{Tf''(0)}{2})=\frac{Tf'''(0)}{3!}$;
and so on.. :)  isn't it nice?


Are you sure you didn't mean $x_{1}+x_{2}+...+x_{n}$?  Because this problem is trivial.  By Jensen's we have minimum when $x_{i}$ are as close as possible (so $x_{i}\in \{ 66, 67 \}$; in fact, we have $x_{1}= ... = x_{7}= 66, x_{8}= ... = x_{30}= 67$ and $P \approx 5.45 \times 10^{54}$) and maximum when $x_{1}, x_{2}, ... x_{29}= 1, x_{30}= 1974, P = 1974$.


Solve the equation $ 2 \sin ^{2} x \equal{}5 \cos x \minus{} 1$ for $ x$ between $ 0^{\circ}$ and $ 360^{\circ}$.

[quote="AndrewTom"]Solve the equation $ 2 \sin ^{2} x = 5 \cos x - 1$ for $ x$ between $ 0^{\circ}$ and $ 360^{\circ}$.[/quote]
[hide]$ 2(1-\cos^{2}x)=5\cos x-1$
$ 2-2\cos^2x=5\cosh-1$
$ 2\cos^2x+5\cosh-3=0$
$ (2\cos x-1)(\cos x+3)=0$
$ \cos x=\frac{1}{2}$, or $ \cos x=-3$, but disregard the 2nd one because the minimum of $ \cos$ is $ -1$.
The first equation yields $ \frac{\pi}{3}$ or $ \frac{5\pi}{3}$, or $ 60^{\circ}$ and $ 300^{\circ}$[/hide]

[quote="ckck"][quote="AndrewTom"]Solve the equation $ 2 \sin ^{2} x \equal{} 5 \cos x \minus{} 1$ for $ x$ between $ 0^{\circ}$ and $ 360^{\circ}$.[/quote]
[hide]$ 2(1 \minus{} \cos^{2}x) \equal{} 5\cos x \minus{} 1$
$ 2 \minus{} 2\cos^2x \equal{} 5\cos x \minus{} 1$
$ 2\cos^2x \plus{} 5\cos x \minus{} 3 \equal{} 0$
$ (2\cos x \minus{} 1)(\cos x \plus{} 3) \equal{} 0$
$ \cos x \equal{} \frac {1}{2}$, or $ \cos x \equal{} \minus{} 3$, but disregard the 2nd one because the minimum of $ \cos$ is $ \minus{} 1$.
The first equation yields $ \frac {\pi}{3}$ or $ \frac {5\pi}{3}$, or $ 60^{\circ}$ and $ 300^{\circ}$[/hide][/quote]

I corrected a couple typos in ckck's post. The $ \cos x$ was missing in a couple lines; this happens in latex when you write \cosx instead of \cos x.

hello, the solutions are
$ x\equal{}\frac{\pi}{3}$
$ x\equal{}\frac{5\pi}{3}$
Sonnhard.


logarithms
real analysis
real analysis unsolved

We know the close approximation
\[ \frac{\pi^9}{e^8}\approx 9.999838813\approx 10.\]
Are there positive integers $ l,\; m$ such that $ \frac{\pi^l}{e^m}$ is closer to a positive integer than for the case given?

The values of $ l \ln \pi \bmod e$ are equidistributed in $ [0, e)$, so one would expect infinitely many such approximations.


If $ \minus{}1<a,b,c<1$, then
$ \sum\sqrt[4]{1\minus{}a^2}\plus{}\sum\sqrt[4]{1\minus{}a}\plus{}\sum\sqrt[4]{1\plus{}a}\leq9$

By AM-GM inequality we have

$ \sum_{cyc}\left(\sqrt [4]{1 \minus{} a^{2}} \plus{} \sqrt [4]{1 \minus{} a} \plus{} \sqrt [4]{1 \plus{} a}\right)\le \sum_{cyc}\left(\frac {(1 \minus{} a^2 \plus{} 1 \plus{} 1 \plus{} 1) \plus{} (1 \minus{} a \plus{} 1 \plus{} 1 \plus{} 1) \plus{} (a \plus{} 1 \plus{} 1 \plus{} 1 \plus{} 1)}{4}\right) \equal{} 9 \minus{} \frac{(a^2 \plus{} b^2 \plus{} c^2)}{4}\le 9$


Let $a_i$ be the i-th positive integer that does not have a 2 in its base ten representation. Prove that

$\sum\limits_{i = 1}^\infty {\frac{1} {{a_i }}}$

converges.

[hide]
There are $8 \cdot 9^{n-1}$ positive integers with $n$ digits such that none of the digits is $2$ ($8$ choices for first, $9$ for the rest). Each is at least $10^{n-1}$. So $\sum_{10^{n-1} \le a_i < 10^n} \frac{1}{a_i} \le \frac{8 \cdot 9^{n-1}}{10^{n-1}}$. Then

$\sum \frac{1}{a_i} \le \frac{8 \cdot 1}{1}+\frac{8 \cdot 9}{10}+\frac{8 \cdot 9^2}{10^2}+\cdots$.

But this is a geometric series with $|r| < 1$, so it clearly converges (to $80$).[/hide]


Find all integers $ n$ such that all factors of $ n$ except for 1 can be written in the form $ a^r \plus{} 1$ where $ a, r$ are positive integers and $ r > 1$

I'll try some numbers.

2=a^r+1, so a^r=1, a=1, r=2
3=a^r+1, so a^r=2, no solution
4=a^r+1, so a^r=3, no solution
5=a^r+1, so a^r=4, a=2, r=2
....

That gives me an idea. We can construct a [b]list of every possible value[/b] of a^r+1, like this:
a=1, r=2, a^r+1=2
a=2, r=2, a^r+1=5
a=3, r=2, a^r+1=10
a=4, r=2, a^r+1=17
...
a=1, r=3, a^r+1=2
a=2, r=3, a^r+1=9
a=3, r=3, a^r+1=28
a=4, r=3, a^r+1=65
...
a=1, r=4, a^r+1=2
a=2, r=4, a^r+1=17
...
...

After the list is completed, you can multiply these numbers and get all the possible values of n.

hope that helps.


calculus
integration
geometry
rotation
geometric transformation
function
rectangle

Q: Find the volume generated by area bounded by two curves rotated through 360 degrees about the y-axis:
a) y = x²
y = 2- x²

b) y = x2
y2 = 8x

For these two parts, which formulae should I use?

Volume generated

= π ∫ x² dy
or = π ∫ [f(y)]² – [g(y)]² dy ?

For part a, it will be simplier if I use the first formula where I simplify the y before using the first formula:

y = 2 - x² - x²
y = 2 - 2x²
x = √ (2 - y)/2

After using the first formula, my answer is π.

However, if I use the second formula, I got diferent answer:

y = x²
x = √y

y = 2 - x²
x = √(2 - y)

V = π ∫ [f(y)]² – [g(y)]² dy
= π ∫ [√y]² - [√(2 - y)]² dy
= .....
= π (4 - 4)
= 0

For part b, it seems that I can't use the first formula, as after simplify,
y = √(8x) - x²
it's hard to change to make x the subject in terms of y.

Using the second formula for part b, I got the answer 24/7π.

So how do we know which formula to be used for which question? Because the question is involving volume generated by area bounded by two curves, shouldn't we be using the second formula only? Why using both formulas I get different answers? Please help me check if my answer is correct.

Besides I just knew I can use the shell method but I am more familiar with the disk method? Which formula should I be using and why the answer differs if I use both the methods I mentioned?

Many thanks in advance!! smile

[url=http://mathdemos.gcsu.edu/mathdemos/shellmethod/]This[/url] site gives a good tutorial of how to revolve around the $ y$-axis.  Using the cylindrical shells method means you have to do the integral $ 2\pi\int_{a}^{b}x(f(x)\minus{}g(x))\,dx$, which I'll let you solve.  Note that $ 2\minus{}x^{2}\equal{}x^{2}\Rightarrow x\equal{}\pm 1$, but it's symmetric, so you can just go from 0 to 1.  Also note that $ 2\minus{}x^{2}\geq x^{2}$ on $ [0,1]$.

I see. Thanks! But can I use the Disk Method instead of the shell method? Because I thought Disk method should be able to solve the volume generated by area bounded by two curves?

I've just tried the Shell method on this question but I got stuck...

Q: Find the volume generated when region bounded by y=2x² + 5 and y=3x² + 1 rotated through 2 right angles about the x-axis.

For rotation about the x-axis, we have to change x the subject in terms of y right? Then I got 2π∫ y. √[(y-5)/2] - √[(y-1)/3] dy and got stuck there...

Some excellent general advice on solids of rotation:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=883719&search_id=673412012#883719
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1553980&search_id=673412012#1553980

You seem to be trying to write down formulas without first drawing pictures.  In particular, the last integral you've written down is a mishmash.  Draw and think, don't rely on the formulas.

Now let's get at those calculations. It's my understanding that any method can be used.
[hide="(A)"]for this problem I chose so sum the areas of the resulting disk as I work up the z axis. Obviously we have A=pi r^2. but r=x=f(y). As it is, we have two functions: x=y^(1/2) for y in (0,1) and x=(2-y)^(1/2) for y in (1,2). So we'll do two integrals. They are, \[ V\equal{}\pi\int^1_0 y\text{ d}y \plus{}\pi\int^2_1 (2\minus{}y)\text{ d}y\equal{}\pi\][/hide]
[hide="(B)"]If I got the equations for B correct then we are revolving the region bounded below by $ y\equal{}x^2$ and above by $ y\equal{}\sqrt{8x}$ about the y axis. I will use the same method, but this time I take the difference of disk areas to find the shaded regions revolved area as i proceed up the y axis. \[ V\equal{}\pi\int^4_0 y\text{ d}y \minus{}\frac{\pi}{64}\int^4_0 y^4\text{ d}y\equal{}\frac{24\pi}{5}\][/hide]

As for the shell method I have the following to say: you do integrate with respect to x. The idea is that you do 2pi r*A where r is given by distance from y axis (r=x) and the A is given by height times width of each rectangle (f-g dx).  JRav has given the collected formula for this method so long as you are revolving about the y axis. It seems a bit overkill for this problem however, the other method was quite easy.


Let $ a, b$ be arbitrary real numbers satisfying $ a^2 \plus{} b^2 \equal{} 10.$ The largest possible value of $ |a \plus{} 2b|$ is

$ (A) 7$
$ (B) 5$
$ (C) 3\sqrt {10}$
$ (D)\sqrt {50}$

[hide="my attempt"]
$ a^2 \plus{} b^2 \equal{} 10$
$ (a^2 \plus{} b^2)(1^2 \plus{} 1^2) \equal{} 20$
$ |a \plus{} b| \le \sqrt {20}$

Also, $ |b| \le \sqrt {10}$

$ \equal{} > |a \plus{} b| \plus{} |b| \le \sqrt {10} \plus{} \sqrt {20}$
$ \equal{} > |a \plus{} 2b| \le \sqrt {10} \plus{} \sqrt {20}$
Since, $ |x| \plus{} |y| \ge |x \plus{} y|$ [/hide]

EDIT :
[hide="got it"]
$ (a^2 \plus{} b^2)(1^2 \plus{} 2^2) \equal{} 50$
$ |a \plus{} 2b| \le \sqrt {50}$[/hide]

hello, i think the largest value is $ 5\sqrt{2}$ under the condition $ a^2\plus{}b^2\equal{}10$, we have
$ 0\le (5a^2\minus{}10)^2$
$ 0\le 25a^4\minus{}100a^2\plus{}100$
$ 16a^2(10\minus{}a^2)\le 100\plus{}9a^4\plus{}60a^2$
$ \minus{}3a^2\plus{}4a\sqrt{10\minus{}a^2}\le10$
$ a^2\plus{}4(10\minus{}a^2)\plus{}4a\sqrt{10\minus{}a^2}\le 50$
$ a^2\plus{}4b^2\plus{}4ab\le 50$
$ (a\plus{}2b)^2\le 50$
$ |a\plus{}2b|\le5\sqrt{2}$
The case with $ b\equal{}\minus{}\sqrt{10\minus{}a^2}$ is analog.
Sonnhard.

The Cauchy-Shwarz solution is the most natural for these types of problems. Here's one with only the trivial inequality:

\begin{align*} S^2 & = a^2 + 4ab + 4b^2 \\
& = 5(10 - a^2 - b^2) + a^2 + 4ab + b^2 \\
& = 50 - (b - 2a)^2 \\
& \leq 50 \end{align*}

$ \therefore \; S \leq 5 \sqrt {2}$, equality when $ b = 2a$.

I thought $ a \equal{} \sqrt{10}\cos \theta, b \equal{} \sqrt{10} \sin \theta \Rightarrow a\plus{}2b \equal{} \sqrt{10}\sqrt{5} \cos (\theta\plus{} \alpha)$ was the most natural

That is the same thing as using Cauchy-Schwarz.

In practice, what high schoolers usually use as Cauchy-Schwarz is really one particular instance of Cauchy-Schwarz, in which:
1. The two vectors are in Euclidean space
2. The inner product used is the standard dot-product
3. The inequality is written in terms of standard basis components

That's how you get from $ \langle x,y\rangle^2 \leq \langle x,x\rangle \cdot \langle y,y\rangle$ to $ \left(\sum_{i\equal{}1}^n x_i y_i\right)^2\leq \left(\sum_{i\equal{}1}^n x_i^2\right) \left(\sum_{i\equal{}1}^n y_i^2\right)$

If we instead just use (1) and (2), without converting into components, we get this:
$ x \cdot y \leq \left\|x\right\| \left\|y\right\|$

But we know we can write $ x \cdot y$ as $ \left\|x\right\| \left\|y\right\| \cos \theta$, where $ \theta$ is the angle of separation between $ x$ and $ y$. This is why your result ends up looking like $ \sqrt{50} \cos \phi \leq \sqrt{50}$.


Prove that there is a space $V$ of singular $nxn$ matrices (in any field) such that any other space which contains $V$ also contains an invertible matrix.

What am I misunderstanding here? :? Just take any space made up of singular matrices of maximal dimension.


For how many positive integers $ n$ is $ 2n \plus{}7 < 23$?

$ 2n<16$
$ n<8$

$ n$ would be the positive integers 1 to 7, inclusive, so this inequality would hold for $ \boxed{7}$ positive integers.


algebra
polynomial
geometry
induction
analytic geometry
vector
Ring Theory

Let $K$ be an infinite field (may or may not be algebraically closed), and $V\subset K^n$ a finite set of points. Show that its ideal $\mathcal{I}_K(V)$ in $K[X_1,...,X_n]$ is generated by $n$ polynomials.

[i]Hint[/i]: The source says, "use interpolation". I don't know how, though.

I find this problem incredibly annoying. It appears in the problems in the first chapter of Kunz's book and indeed it gives the hint of using interpolation. Fibonacci doesn't know how to use interpolation, I don't know what interpolation is in this case. The problem follows as an easy corollary of a hard theorem later in the book, but again it says that it is also an easy exercise.

First I will present the problem in a more elementary context so that people who haven't started the study of algebraic geometry can understand it:

[b]Let $K$ be an infinite field. Let $V\in K^n$, $V=\{v_1,\ldots,v_m\}$. Let $I(V)=\{f\in K[X_1,\ldots,X_n]|\ f(v_i)=0\ \forall\ i=\overline{1,m}\}$.
Prove that $I$ is generated by at most $n$ elements.[/b]

It can be proved that $I(V)=(X-v_1)\cap\ldots\cap(X-v_m)$, where $(X-v)$ denotes the ideal (which turns out to be maximal in $K[X_1,\ldots,X_n]$) $(x_1-w_1,x_2-w_2,\ldots,x_n-w_n)$ if $v=(w_1,\ldots,w_n)\in K^n$.
Interpolation may refer to finding a nice way of expressing $(X-v_1)\cap\ldots\cap(X-v_m)$.

We solve the problem by induction on $n$.
We first treat the case $n=1$. In this case we have $V$ a finite subset of $K$ and then $\mathcal I(V)=\left(\prod_{v\in V}(X-v)\right)$. Therefore $\mathcal I(V)$ is a principal ideal (generated by $1=n$ elements).

Assume we have proved the problem for $n$ and we want to prove it for $n+1$.
Let $V=\{v_1,\ldots,v_m\}$ be a finite subset of $K^{n+1}$. For all $i$ let $x_i\in K^n$ and $y_i\in K$ such that $v_i=(x_i,y_i)$.
We can choose a system of coordinates such that $x_i\neq x_j$ for all $i\neq j$. This is basically because $V$ is a finite set and because if $f: K^{n+1}\to K^{n+1}$ is a $K$-vector spaces automorphism, then $\mathcal I(f(V))=\{P\circ f^{-1}|P\in K[X_1,\ldots,X_{n+1}],\ P(v)=0\ \forall\ v\in V\}$.
There exists a polynomial $P\in K[X_1,\ldots,X_n]$ such that $P(x_i)=y_i$ for all $i=\overline{1,m}$. This is interpolation.

I will prove that $\mathcal I(V)=(X_{n+1}-P, P_1,\ldots, P_n)$ with $P_i\in K[X_1,\ldots,X_n]\subset K[X_1,\ldots,X_{n+1}]$ such that $(P_1,\ldots,P_n)=\mathcal I(x_1,\ldots,x_m)$ (ideals taken this time in $K[X_1,\ldots,X_n]$). Such polynomials exist by the induction hypothesis.

Let $J=(X_{n+1}-P, P_1,\ldots, P_n)$. For all $v_i\in V$ we have $(X_{n+1}-P)(v)=y_i-P(x_i)=0$ and $P_j(v_i)=P_j(x_i)=0$ for all $j=\overline{1,n}$. This proves that $J\subset\mathcal I(V)$.
Conversely, let $Q\in \mathcal I(V)$. It is easy to see that $Q\in \mathcal I(V)$ iff $\bar Q(X_1,\ldots,X_{n+1})=Q(X_1,\ldots,X_n,P(X_1,\ldots,X_n))\in\mathcal I(V)$. Then $\bar Q(v_i)=0$ for all $i$ which in view of the fact that the variable $X_{n+1}$ does not appear in $\bar Q$ is equivalent to $\bar Q(x_i)=0$ for all $i$. By the induction hypothesis $\bar Q\in (P_1,\ldots,P_n)$ in $K[X_1,\ldots,X_n]$. It is now easy to see that $Q\in J$. With this, we also have the reverse inclusion, therefore $J=\mathcal I(V)$.
$J$ is generated by $n+1$ polynomials and we are done.


So I want to prove that $ m^{2} \plus{} 1 \equiv 0\pmod{p}$ has no solutions if $ p\equiv 3\pmod{4}$.

I've been stuck on this for a while now... The only thing I managed to prove is that:
[hide]If $ p\equal{}4(3k)\plus{}3\equal{}12k\plus{}3$, then there are no solutions because if $ m^{2} \plus{} 1 \equal{} lp\equal{}l(12k\plus{}3)\equal{}3l(4k\plus{}1)$, then $ m^{2} \equiv \minus{}1\pmod{3}$, and that's impossible.[/hide]

I think this should be well known, but I don't know what it's called.

Yes, this is well known.

[hide="Hint"]Use the fact that $ m^{p\minus{}1}\equiv 1\pmod{p}$.[/hide]

The relevant term here is [url=http://en.wikipedia.org/wiki/Quadratic_residue]quadratic residue[/url], and the stronger statement is that $ m^2 \plus{} 1 \equiv 0 \bmod p$ has no solutions if and only if $ p \equiv 3 \bmod 4$ (for prime $ p$).

[quote="matt276eagles"]Yes, this is well known.

[hide="Hint"]Use the fact that $ m^{p \minus{} 1}\equiv 1\pmod{p}$.[/hide][/quote]

Thanks.

[hide="Solution"]Let's go by contradiction.
Since $ m^{p\minus{}1}\equiv 1\pmod{p}$, we have $ p|(m^{2}(1\plus{}m^{p\minus{}3}))$. Since $ p$ is a prime, either $ p|m^{2}$ or $ p|(1\plus{}m^{p\minus{}3})$.

If $ p|m^{2}$, then $ 1 \equiv m^{2} \plus{} 1 \equiv 0\pmod{p}$, which is impossible.

If $ p|(1\plus{}m^{p\minus{}3})$, then $ m^{p\minus{}3}\equiv \minus{}1\pmod{p} \implies m^{2p\minus{}6}\equiv \pmod{p}$. Let $ k$ be the order of $ m$ $ \pmod{p}$. Then $ k|(2p\minus{}6)$ and $ k|(2p\minus{}2)$ since $ m^{2p\minus{}2} \equal{} (m^{p\minus{}1})^{2} \equiv 1\pmod{p}$. Therefore, $ k|(2p\minus{}6 \minus{} (2p\minus{}2))$, implying $ k|4$.

If $ k\equal{}1$ or $ k\equal{}2$, then $ m^{2}\equiv 1\pmod{p}$. Since $ m^{2}\equiv \minus{}1\pmod{p}$, $ 1\equiv \minus{}1\pmod{p}$, which is impossible.

If $ k\equal{}4$, we have $ 4|(2p\minus{}6)$ and $ 4|(p\minus{}1)$, so $ 4|(p\minus{}5)$, which is impossible because $ p\equiv 3\pmod{4}$.

So we have our contradictions.[/hide]

Nice order argument, but you can use the same type of argument from the beginning - try it out.  :wink:

[hide="What he means is..."] Write $ m^{ p\minus{}1 } \equal{} \left( m^2 \right)^{ \frac{p\minus{}1}{2} } \equal{} (\minus{}1)^{ \frac{p\minus{}1}{2} }$. [/hide]

I think what t0rajir0u's trying to say is:

Except $ m \equiv 0\pmod{p}$,

$ \\ m^{ \frac{p\minus{}1}{2} } \equiv (\minus{}1)^{ \frac{p\minus{}1}{2} }\pmod{p}$

Also, a challenge: Prove that if p is not 3 mod 4, then p can be expressed as a sum of two perfect squares.

[quote="anonymous03"]Also, a challenge: Prove that if p is not 3 mod 4, then p can be expressed as a sum of two perfect squares.[/quote]
What about 6?

In this thread (and usually), $ p$ is prime. Should have been stated clearly though.

[quote="analytic"]u all have been giving quite complicated solutions, here's a simple one
[hide]p=3[mod4]   therefore, in equation m^2+1=0 [mod p],   m^2+1 = 3[mod4] as it would also leave same remainder as p . next step, m^2+1 will give remainder as 1+1 or 0+1[mod 4] , which would never give 3.

so,it has no solutions , hence proved as situation is not possible.[/hide][/quote]

Not quite. Just because $ x\equiv 0\mod p$ and $ p\equiv 3\mod 4$ does not mean that $ x\equiv 3\mod 4$. For instance, consider $ x\equal{}3p$. Then $ x\equiv 3p\equiv 3\cdot 3\equiv 1\mod 4$.

[quote="anonymous03"]Except $ m \equiv 0\pmod{p}$,

$ \\
m^{\frac {p \minus{} 1}{2} } \equiv ( \minus{} 1)^{\frac {p \minus{} 1}{2} }\pmod{p}$[/quote]
It is given that $ m^2 \equiv \minus{}1 \bmod p$, so neither that condition nor that implication are sensible as written.


Prove that
\[ 4(a^2\plus{}b^2\plus{}c^2)\minus{}3(ab\plus{}bc\plus{}ca)\plus{}8abc\sum \frac{1}{b\plus{}c} \ge \sum \sqrt{(a^2\plus{}4bc)(b^2\plus{}4ca)}.\]

a) If $ a,b,c$ are the sidelengths of a triangle.

b) If $ a,b,c$ are nonnegative real numbers.

The case a) with this hypothesis, we can find an incredible solution for this inequality. Can you find it? :)

[quote="can_hang2007"]Prove that
\[ 4(a^2 \plus{} b^2 \plus{} c^2) \minus{} 3(ab \plus{} bc \plus{} ca) \plus{} 8abc\sum \frac {1}{b \plus{} c} \ge \sum \sqrt {(a^2 \plus{} 4bc)(b^2 \plus{} 4ca)}.
\]
a) If $ a,b,c$ are the sidelengths of a triangle.

b) If $ a,b,c$ are nonnegative real numbers.

The case a) with this hypothesis, we can find an incredible solution for this inequality. Can you find it? :)[/quote]
This is the simple proof for a):
\[ 2b^2\minus{}3bc\plus{}2c^2 \plus{}\frac{8abc}{b\plus{}c} \ge \sqrt{(a^2\plus{}4bc)(b^2\plus{}4ca)}\]
;)


Calculate the sum: $\sum_{k=0}^{n}k^{2}\dbinom{n}{k}5^{k}$


Please help me solving this problem. Explain it very clear and write all the theorems you use (imn't very good at geometry).

For the following figure demonstrate that a+B=0(theta)

Any book you recommend me to become good at geometry? ...maybe Aops introduction to geometry book?

[img]http://i204.photobucket.com/albums/bb126/agerador/geometry2.jpg[/img]

Hi Chalao.

It seems from your diagram that since the angles $ \alpha$ and $ \beta$ stand on a diameter, they add to $ 90$ (as the angle in a semi-circled is a right-angle.

As for books, two I like are Geometry Revisited availble at http://www.amazon.co.uk/Geometry-Revisited-Mathematical-Association-Textbooks/dp/0883856190 and Crossing the Bridge available at http://www.mathcomp.leeds.ac.uk/publications/list/

I hope this is of some use to you.

Someone can tell me the whole solution please? :lol:

Hi Chalao.

I'm wondering if there's a typo in your question.


it's a very confusing story, but i think i have it.

[hide]does the door open out into the hallway, not towards him?[/hide]

[quote="Ihatepie"]"Lets return to the scene," said Shadow. Shadow and Roger Ragtop drove to the deserted storage house. They walked down one of the many halls on the 3rd floor and stopped in front of the ninth door on the left hand side. The area had been cordoned off with police tape. Shadow removed the tape and pulled the door open. They entered the room and Roger retold his story. "Sam swore he was going to kill me when he got out of jail. Last week he was released. He saw me on the street so I turned and ran to this old storage house. With so many rooms I didn't think he would find me. Then, to my horror, I saw the doorknob slowly turn. As the door started to swing towards me, I panicked and started to shoot. I din't mean to kill him but rather scare him off." Shadow looked up at Ragtop and replied, "That story will never hold up." Why not?[/quote]
[hide]that door thing i saw before[/hide]


real analysis
calculus
calculus computations

The title says it all; I need to know how to calculate the partial sums of the series 1/n.  Thanks guys.

See the [url=http://en.wikipedia.org/wiki/Harmonic_number]harmonic numbers[/url].

Thanks JRav,
This was for a problem in Rudin, Chapter 3, Numerical Sequences and Series.  I was expecting there to be a fairly simple algebraic way of calculating the partial sums (the harmonic numbers, as it were), but since there clearly isn't I assume I'm doing something wrong.  I'll give it another shot.

If it's a problem in Rudin, you probably want an estimate rather than a sum.

Maybe I'm just not looking hard enough, but which problem?


Prove that \[\left( 2^{2n}+2^{n+m}+2^{2m}\right) !\] is divisible by \[\left( 2^{n}! \right)^{2^{n}+2^{m-1}}\cdot \left( 2^{m}! \right)^{2^{m}+2^{n-1}},\] for all $m,n \in \mathbb N^\ast$.

[i]Dana Heuberger[/i]

[quote=Catalin]Sorry to revive this, but since it had no solution... :-)
Simply notice that the formula of Permutations with Repetitions is hidden here: $$2^{2n}+2^{n+m}+2^{2m}=2^n\cdot(2^n+2^{m-1})+2^m\cdot(2^m+2^{n-1})$$
so we can conclude that $$\frac{(2^{2n}+2^{n+m}+2^{2m})!}{\underbrace{2^n!\cdot 2^n! \cdot ... \cdot 2^n!}_{2^n+2^{m-1}} \cdot \underbrace{2^m! \cdot 2^m!\cdot ... \cdot 2^m!}_{2^m+2^{n-1}}}=\frac{(2^{2n}+2^{n+m}+2^{2m})!}{\left( 2^{n}! \right)^{2^{n}+2^{m-1}}\cdot \left( 2^{m}! \right)^{2^{m}+2^{n-1}}}$$ is a natural number, thus the conclusion.[/quote]

11 years had almost passed since it was posted... :P

Wait, the source says $RMO$. But surely that Regional Mathematics Olympiad does not have such a hard problem??!!

$\text {RMO} $ here stands for Romanian Mathematical Olympiad :-D


calculus
integration
inequalities
real analysis
real analysis unsolved

Hallo to everyone; I have two exercise that I'm not able to do; if someone can help me....
thanks very much.

1) Let be $ \sum_{n=1}^\infty a_n$ a series where ${{ a_n \in \mathbb{R}}, a_n>0 \qquad \forall n \in \mathbb{N}}$; we assume that exists a real postive terms serie $ (b_n)$ and a positive constant $ k$:
${ b_n \; \frac{a_n}{a_{n+1}}-b_{n+1}\geq k \qquad\forall n \in \mathbb{N}}$
Proof that:
$ \sum_{n=1}^\infty a_n < +\infty$

2) Let be ${ f \in C([0,\infty],\mathbb{R}})$ we assume that exists and is real the generalized integral:
$ \int_1^\infty \frac{f(t)}{t} dt$

Proof that for all ${ a,b, \in \mathbb{R}}, a,b>0 \qquad$ exists the integral

$ \int_0^\infty \frac{f(a \; t)-f(b \; t)}{t}dt$

and it is equal to $ f(0) \; Log\frac{b}{a}$

thanks to everybody, bye

For 1) just rewrite the given inequalities as $ ka_{n\plus{}1}\le b_na_n\minus{}b_{n\plus{}1}a_{n\plus{}1}$ and add them up.

For 2) consider the integral $ \int_m^M$ with small $ m>0$ and large $ M>0$, split it into the difference of 2 integrals, make the linear change of variable in each of them so that you would end up integrating $ \frac{f(t)}t$ in both, cancel the common part, and look attentively at what is left.

Thanks, anyway if you could tell me the solution of the second problem more in details, because I 'm not very well in english and I dont undarstand everything... thanks a lot.


modular arithmetic
combinatorics unsolved
combinatorics

Let $ a$ be an irrational number. Prove that the set $ A\equal{}\{ ma\plus{}n | m,n \in \mathbb{Z} \}$ has the property that for any interval $ I \subset \mathbb{R}$ there is $ b \in A$, so that $ b \in I$.

Given our irrational $ a$ We need to show that $ B \equal{} \{ ma \pmod 1 | m\in\mathbb{Z}\}$ is dense on the set $ [0,1]$

To do this it is sufficient to show that for any given $ \epsilon > 0$ , $ \exists m \in\mathbb{Z}$ such that $ ma < \epsilon \pmod 1$

Assume the hypothesis is untrue, and there is an $ \epsilon '$ such that $ \nexists m \in\mathbb{Z}$ such that $ ma < \epsilon ' \pmod 1$

Then for every $ m$ , $ \nexists m' \neq m$ such that $ m'a \in (ma \minus{} \epsilon' , ma \plus{} \epsilon') \pmod 1$ ,else we would have either  $ (m' \minus{} m)a$ or $ (m \minus{} m')a < \epsilon' \pmod 1$

It follows that the sets $ (a, a \plus{} \epsilon'), (2a, 2a \plus{} \epsilon'), (3a, 3a \plus{} \epsilon') ... \pmod 1$ are all disjoint

But all these sets are in $ [0,1]$, and their total length is $ \infty$ , contradiction.


search
combinatorics unsolved
combinatorics

I am trying to prove that
\[ \sum_{k \equal{} 0}^{m \plus{} 1}( \minus{} 1)^{k} {n \choose k}{n \plus{} m \minus{} k \choose m \plus{} 1 \minus{} k} \equal{} 0.
\]
It seems like the formula is correct. Need your help^

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=505547670&t=16482

Pierre.


trigonometry
geometry
geometric transformation
reflection
function
calculus
analytic geometry

I teach an SAT class and I am getting feedback that more advanced mathematical concepts like trigonometry (sine and cosine problems) and a great deal of Algebra II problems are showing up on the test.  The College Board's book of practice tests does not reflect this.  I would be interested in feedback from anyone who has taken the SAT in October, November or December 2005. Thank you.

No trigonometry is needed on the SAT.  I took both the October and November tests: No question required trig.

The practice tests in the [i]Official SAT Study Guide[/i] are pretty close in topics to the actual test.

Hmm... I took the December SAT and I don't remember any trig...

[quote="chilly_snow"]I took the December test and there were some Trig questions.
all I remember is that the last question was a Trig Equation.[/quote]

You must mean SAT II Math, not the SAT Reasoning Test Math. In the SAT Reasoning Test Math, you need to know everything up to Algebra II (functions, exponents, etc). We don't need to know trig, logs, or any Pre calculus math...

I took the May SAT II Math IIC and there was a question that required calculus. It was a simple question and I don't remember it, but I do remember that it was really easy to do with calculus but really hard without it. I think it had something to do with the slope of a quadratic in parametric form, but I don't remember.

[quote="tokenadult"]Expressing a quadratic equation in parametric form is a precalculus topic, I think.[/quote]

Really? I don' even know what a parametric is ( i have heard of it though), and I already took precalculus.


[quote="neworder"]Positive reals a,b,c satisfy $ab+bc+ca=3$. Prove: $a^{3}+b^{3}+c^{3}+6abc \geq 9$.[/quote]

Finally someone who posts this [i]after[/i] the deadline...  :)

See the posts at the end of http://www.mathlinks.ro/Forum/viewtopic.php?t=63308 .

  Darij

I remember that I gave a solution to this nice inequality [i] before[/i] the deadline :blush: , using Schur inequality.
Actually, I didn't know that the deadline was 5 December. Anyway, here is an alternative solution a little bit different
than the one I gave then. :

 [color=red]

   Problem.

       Let $a, b, c>0$ such that $ab+bc+ca=3$. Then the following inequality holds true

                                             \[ a^{3}+b^{3}+c^{3}+6abc\geq 9.  \]
 [/color]

 [i] Solution. [/i]

     Let us use this well-known identity: \[ a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca).  \]
     We deduce that $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-3).$
     So, the inequality given rewrites as $(a^{3}+b^{3}+c^{3}-3abc)+9abc\geq 9$.
     We have \[ (a+b+c)(a^{2}+b^{2}+c^{2}-3)+9abc\geq 9.  \] But If we apply this inequality( Schur),
     $2(ab+bc+ca)-(a^{2}+b^{2}+c^{2})\leq\frac{9abc}{a+b+c}$ we only need to prove that
     $a+b+c\geq 3$ which follows immediately from $(a+b+c)^{2}\geq 3(ab+bc+ca)$. :)  ;)

I too posted a solution before the deadline, at sci.math (oops!), which was identical to the one posted by 'death'.


If r = - .24 and y = 14 when x = 10, give a reasonable estimate for y when x = 20

42.  :D

But really, I don't quite understand the problem. Is there an equation involved, does r stand for something?

I'm not sure either. This is a problem that was given to me by my professor. How did you come up with 42?

I think $r$ stands for rate, so the relationship between $x$ and $y$ is linear, using the point slope formula, we see that $y=-.24x+16.4$, so when $x=20$, $y=11.6$


Consider a sequence of polynomials definded by:

This makes no sense.
If $ P_0 \equal{} 1,P_1,\ldots,P_n$ satisfy the above equations for $ k \equal{} 0,1,\ldots,n \minus{} 1$,
then so does $ P_0 \equal{} 1,P_1,\ldots,P_n,P_{n \plus{} 1}$  for $ k \equal{} 0,1,\ldots,n$
for any arbitrary $ P_{n \plus{} 1}.$
There is no way to calculate it.

Furtheremore
$ \deg P_n < \deg P_{n \minus{} 1} < \cdots < \deg P_1 < \deg P_0 \equal{} 0$
or
$ 0\equiv P_n\equiv P_{n \minus{} 1}\equiv \cdots\equiv P_1\equiv P_0\equiv 1$.
So there is no such sequence at all.

sorry,I corrected the statement now  :maybe:

[quote="olorin"]This makes no sense.
...
Furtheremore
$ \deg P_n < \deg P_{n \minus{} 1} < \cdots < \deg P_1 < \deg P_0 \equal{} 0$
or
$ 0\equiv P_n\equiv P_{n \minus{} 1}\equiv \cdots\equiv P_1\equiv P_0\equiv 1$.
So there is no such sequence at all.[/quote]
This doesn't change.


search
modular arithmetic
AoPSwiki
number theory
Pascal\u0027s Triangle
relatively prime

I don't clearly understand this concept yet. All I have gotten as an idea from reading previous problems that seemingly were faster when using mods or easier to see as an idea, are that they have something to do with division/ remainders? Does anyone have a clear explanation of what mods are?  :huh:

I feel kind of dumb asking this, but I'll live.

[url]http://mathworld.wolfram.com/ModularArithmetic.html[/url]

[url]http://en.wikipedia.org/wiki/Modular_arithmetic[/url]

or use the search function.

if $a\cong b (mod c)$ then the remainder when $a$ is divided by $c$ is $b$.

[code]a\equiv b\pmod{c}[/code]
$a\equiv b\pmod{c}$

...or...

[code]a\equiv b\bmod{c}[/code]
$a\equiv b\bmod{c}$

The thing with mod congruences (careful, not equations...) is that there are infinitely many solutions.

e.g.

$5 \cong 1 \mod 2$
$5+2 \equiv 1 \bmod 2$
$5+2+2 \equiv 1 \bmod 2$
$5+2+2+2 \equiv 1 \bmod 2$

The resulting thing *before* the mod is the remainder when you divide the original number by the number after the mod.

This is helpful when looking at problems that ask for the units (or tens and units) digits of a number.

For example, find the units digit of $2^{3^{2007}}$
If we look at this sort of problem in MOD 10, we look at the remainder of that number when divided by ten, or, in other words, the UNITS DIGIT.

This is just one application (by the way, if anyone wants to solve that, they get a cookie.)

[quote="DiscreetFourierTransform"]For example, find the units digit of $2^{3^{2007}}$[/quote]
[hide]$3^{2007}\equiv(-1)^{2007}\equiv-1\equiv3\pmod{4}$

$2^{3}=\boxed{8}$[/hide]

mods=moderators
jk

just search modular math/arithmetic on google. you get good results. mods are good for number theory and related things

There's a pretty thorough introduction in the AoPSWiki - just search for modular arithmetic there.

if you have AoPS volume 1 one of the chapters have a pretty good explanation of modular arithmetic

[quote="bpms"]Mods were part of the cirriculum in fifth grade, all we did though was modular multiplying, dividing, adding, subtracting, and solving (mod 29 mostly).
If 5th grade math we also covered pascal's triangle, and pascal's identity.[/quote]
dude what 5th grade were you in

in 5th grade we were still adding fractions

omg in fifth grade we did NOTHING
i learned about mods and pascal's triangle very recently.

Anyway, mods are like remainders, if $a\equiv b\mod n$ then the remainder of $a$ divided by $n$ is the same as the remainder of $b$ divided by $n$.

$0\equiv n\equiv 2n\equiv 3n\mod n$
$1\equiv n+1\equiv 2n+1\equiv 3n+1\mod n$
$2\equiv n+2\equiv 2n+2\equiv 2n+2\mod n$
etc...

Another way to look at it is that if $a\equiv b \mod n$, then $a-b$ is divisible by $n$

You can perform many standard operations with mods:

$a\equiv b\mod n$
$a+c\equiv b+c\mod n$ (addition)
$2a+2c\equiv 2b+2c\mod n$ (mutliplication)
$(2a+2c)^{2}\equiv (2b+2c)^{2}\mod n$ (squaring)

However, division does not always work in mod... you can divide by a number $d$ if it is $d$ is relatively prime to the modulus. For example,
$7x\equiv 7y\mod 9$
$x\equiv y\mod 9$
This was okay because 7 is relatively prime to 9. However,
$15x\equiv 15y\mod 9$
$3x\equiv 3y\mod 9$
$x\equiv y\mod 3$
Dividing by 5 was fine. However, when I divided by 3, since 3 divides the modulus 9, I had to divide the modulus by 3 as well.
So if I divide by a divisor $d$, I must also divide the modulus by any common factors between $d$ and the modulus.

[quote="bpms"]Mods were part of the cirriculum in fifth grade, all we did though was modular multiplying, dividing, adding, subtracting, and solving (mod 29 mostly).
If 5th grade math we also covered pascal's triangle, and pascal's identity.[/quote]

chess64 told me it was college leveled....

Dividing in [i]prime[/i] mods is okay though. (as long as it's not by 0 or something congruent to it, of course)

[quote="Klebian"]I'm... pretty sure he was kidding... dividing in mods is a no-no[/quote]
Dividing in mods is possible, it's just slightly more complicated than ordinary mods.
Anyways, I wasn't kidding.

The way we did mods we didn't do dividing, but we were told that a certain x had been tripled and then remodded and we needed to unmod it.

It was an awesome class, though I was the only 5th grader in that class every one else was a 6th grader.


4 men (Al, Bob, Carl, Dan, let's say) are seated around a table. A hat is put on each of the men, so that the men can see the other three men's hats but he cannot see his own hat. They are told that the hats on their heads are chosen from a group of two white hats, two black hats, and one red hat.

None of the men are given the red hat (but they don't know this)

In turn, each man either says "I know the color of my hat" or "I do not know the color of my hat". Then the next man does the same, and this continues around the table until no more information can be obtained.

The turn order, predictably, is Al->Bob->Carl->Dan->Al and so on.

Given exactly one man was not able to name the color of his hat, determine which man this was.

[hide="HIDDEN"]The first guy, Al, because he doesn't know whether he has the red hat, or the white or black hat that is left over from the other 3 hats. Once he says that he doesn't know Which hat he has, then the other guys know what they have.[/hide]

You might want to start using hide tags.
[hide]I think it is Al, because he doesn't have any information besides 3 other hats out 5, and when he says he doesn't know, the others have information of 4 out of 5 hats.[/hide]

No, but Al can state again after Dan states.

[hide]I think it's Carl. When A and B have black and C and D have white, C seems to be the only one who can not find out what he has.
I could have made a mistake since this problem requires to consider things like this: when D supposes that C has supposed that B has supposed that A supposed that he has red. :rotfl: [/hide]


see http://en.wikipedia.org/wiki/Kronecker_theorem  :D


hint:[hide]
add on to the long side to make a regular octagon with side length 1[/hide]

You can form different hexagons that fit the requirements, and you can even form a concave hexagon, so there could be infinitely many answers.


Let $x,y,z$ be real numbers such that $x+y+z=0$, prove that for $t\geq0$,
\[{t^{x-y}+t^{y-z}+t^{z-x}\geq t^{x}+t^{y}+t^{z}}.\]

$\iff ab+bc+ca\ge a+b+c \iff (1,1,0)\ge (\frac{4}{3},\frac{1}{3},\frac{1}{3})$
by Muirhead Inequality, this is true.

[quote="nayel"]setting a=t^x, b=t^y and c=t^z, our question becomes
for abc=1, show that a/b+b/c+c/a>=a+b+c.
now someone can proceed. can't they?? :P[/quote]

Homogenise the inequality to get

$\sum{a^{2}c}\geq \sum{a^\frac{5}{3}b^\frac{2}{3}c^\frac{2}{3}}$

which is true by a lot of inequalities :wink: .. say AM-GM because $\frac{2}{3}a^{2}c+\frac{1}{3}b^{2}a\geq a^\frac{5}{3}b^\frac{2}{3}c^\frac{2}{3}$


[b]Put a hot metal ball on parallel horizontal rails. The ball starts to move.  Why does this happen?[/b]

Is there any theoratical suggestions on HOW to make the ball move and WHY?
I have tried thousands of methods and still unsuccessful.

Let me guess (beware, what follows might be outrageous):

Due to heat conduction, the metal ball transfers heat to the rails.  The rails will expand (thermal expansion), and thus the ball, which is originally on a segment of the rail, is "pushed" along with the expansion of the rail.

Correct me if I'm wrong  :oops:

Now, it would be interesting if someone could make an attempt at quantifying the mvt of the ball.

but in a sense the expansion should be equal on both sides of rail for both rails.
So the ball still shouldn't move.
i.e. it's in a sense in an "equilibrium" as heat dissipates along both sides and same force on both sides of rails so in a way overall there is no unbalance so the ball [b]should't[/b] move and so far it [b]doesn't[/b] move :|

i think this must be true - on supplying heat the rails expand .due to the expansion of the rails the distance between the rails decreases .the higher the temperature of the ball the higher is the rate at which the rails absorb it.if the rate of absorption is high it expands quickly.this sudden expansion generates a force that results in the motion of the ball

but would we be able to predict the direction/speed/acceleration of the ball from how hot it is?

Also we come back to the same problem.  If the ball is hot and placed on rails, both rails will expand equally on both sides of the ball.  So if there is a force, there is an equal and opposite force on the other end.  So in a way it shouldn't move?

[hide="just a logical but funny proposition"](from what i observe the only obvious possibility is going left right up down.  If it can't go up and doesn't move left or right, it should move down? :P  :!: )[/hide]

[quote="aadil"]if the rails were frictionless then the heat energy is stored as the kinetic energy of the ball.[/quote]

Really? What about loss of heat due to radiation? :wink:

[quote="handrake"]
(from what i observe the only obvious possibility is going left right up down. If it can't go up and doesn't move left or right, it should move down?

[/quote]

Yes, that's the only thing possible, I feel, ie: the ball moving down. I don't see any reason why it should move sideways.

.

[quote="Anvay"]heat energy lost=kinetic energy gained
assuming ideal rails and ideal ball[/quote]

I don't understand how the ball will [b]gain[/b] kinetic energy.

So as to gain speed, it needs a push, an impulse, sideways. And there is nothing providing that.

And how do you find the heat energy of a solid ball? I have so far studied only about gases. :wink:

[quote="Anvay"]U CAN GIVE IT A MOLECULAR LEVEL APPROACH
THE PART IN CONTACT WITH THE COOL RAILS WILL GET COOLER SO THE MOLECULES IN THAT PART WILL DECREASE THEIR KINETIC ENERGY (THUS THEIR HEAT ENERGY) BY LAW OF CONESERVATION OF ENERGY................MY PREVIOUS VERSION[/quote]

Fine, but why would the ball gain speed? If this holds true, I guess everything in our kitchens would be running around. :rotfl:

[quote="integration"]
Due to heat conduction, the metal ball transfers heat to the rails. The rails will expand (thermal expansion), and thus the ball, which is originally on a segment of the rail, is "pushed" along with the expansion of the rail.

[/quote]

I remember seeing a model in a museum in Bangalore. It consisted of two cones welded together along their circumferences and placed on two horizontal, non-parallel rails. They moved towards the wider part (hope you can imagine how).

So the same applies here, get it?

IT MAY BE AN IDEAL CASE.....DO U HAVE BALLS IN UR KITCHEN???????I DNT THINK SO
OR ARE THEY HEATED ONES/??????COULD BE CANT SAY ABOUT UR KITCHEN..........,
IT IS CLEARLY ASSUMED THAT THERE ARE NO DISSIPATIVE FORCES(HEAT IS NOT LOST IN THE ENVRNMENT) THEN I SUPPOSE THE LOST HEAT FRM THE HEATED BALL WENT TO UR KITCHEN RITE???/WHER DOZ DA HEAT GO THEN>????????TELL ME

I was wondering when someone was going to post this IYPT problem on this forum.

[quote="aadil"]rajiv i was saying that the sudden expansion of the rails might create a force that makes the ball move[/quote]

How is that? I could provide an explanation to how the ball may move while its radius decreases and the distance between the rails increases, how can you support your claim?

Do you mean to say that the rails expand so fast that the ball is pushed? But will the rails not expand symmetrically with equal magnitude in both directions?

[quote="Anvay"]
but the rails are parallel who the hell told u that they are non parallel.

[/quote]

Remember, the rails are being heated, will that not cause some expansion and subsequent increase in the width of the rails, perhaps due to bending?

I'm not trying to be the contrary, but from what one can see here, you don't seem to adopt a logical path of reasoning. How can the energy possesed by the molecules of a body by virtue of heat ever be converted into kinetic energy of the body just by leaving it standing for some time? Can you give me an example so that I'm convinced?

Here is some maybe comments I have;

1. The ball may not be at uniform temperature throughout: thus, there might be convection currents.
2. If the end of the rails are mechanically held, and the rails expand sideways as well as lengthwise, the ball will rise (don't see how it will lead to rolling though).

I hope it might help.

From purely symmetry arguments, though, I can't see how there can be a net force anywhere. (Other than up or down.)

Hint: Try preventing horizontal expansion.


probability
real analysis
real analysis theorems

Suppose {$X_{n}, n\geq 1$} is an independent sequence of random variables,
how can I prove that if $S_{n}$ converges in distribution then $S_{n}$ converges in probability? where $S_{n}=X_{1}+...+X_{n}$.
This is a classical result but I don't know the proof.

I have found the proof in Loeve's book Probability Theory One.


geometry
angle bisector
similar triangles

[quote="digger"]Let a circle is tangent to AB of angle BAD at point B and intersects AD in points C and D. Bisector of the angle BAD intersects BC and BD in points E and F.
Prove that triangle FBE is isoscelent.[/quote]

hmm should be in high school basics.... :lol:

This should be in a high school forum (not sure which one...).
But I made a picture to make it clearer. I think this is what it's supposed to be.

Here is my solution.  Say that Angle ABC=$x$ and Angle BAC=$2y$.  Angle D is also equal to $x$, and Angle BAE=Angle EAC=$y$.  This implies Angle BEF equals $x+y$ and Angle BCD=$x+2y$.  From Triangle BCD, we have that Angle CBD=$180-x-2y-x=180-2x-2y$.  Now look at Triangle BEF.  Angle BFE=$180-(180-2x-2y)-(x+y)=x+y=$Angle BEF.  Hence, Angle BFE=Angle BEF.  Therefore, Triangle BEF is isoceles. (Sorry, this proof probably doesn't make any sense).

well, actualy we have BEF = FAD + FDA = x + y = BFE  :D

Here is my solution:
$ \angle ABC \equal{} \angle ADB$, so $ \triangle ABC$ and $ \triangle ABD$ are similar. Since AF is an angle bisector of both triangles, it breaks them up into similar pieces, so $ \angle AEC \equal{} \angle AFB$, and by vertical angles, $ \angle AEC \equal{} \angle BEF$, so $ \angle AFB \equal{} \angle BEF$

$ \angle BAC\equal{}2\theta$
$ \angle ABC\equal{}\angle BDA$
$ \angle BEF\equal{}\theta \plus{} \angle ABC$
$ \angle BFE\equal{}\theta \plus{} \angle BDA$
$ \angle BEF\equal{}\angle BFE$
$ \triangle BFE is isoceles$

[quote="PowerOfPi"]This thread is more than 2 years old, and two solutions are already given. Think about your post, was it necessary?[/quote]

Several users always repeat that, but the actual idea is [b]don't post in an old thread if you don't have something new to add.[/b]

New solutions are great to add to problems, as they show another way of solving a problem. More solutions=better. Yes, this post was completely useless.

EDIT: No, I meant [b]this[/b] post (my post) is useless, as it's not productive at all  :wink: .

What, how is my post useless  :( ? It does show a different way to solve it, and an interesting idea, in my opinion. It is not totally an angle chase, but uses that fact that the angle bisector of two similar triangles (or the same type of cevian, altitude, etc.) breaks up similar triangles into similar pieces.


function
floor function
number theory unsolved
number theory

Determine $ \sum_{n \leq x} \mu (n) \left\lfloor \frac{x}{n} \right\rfloor$.

There is a nice, more general result which applies to this.  From the generalized inversion formula:  if $ F(x)\equal{}\sum_{n\le x} f(n)$, then
\[ \sum_{n\le x}\sum_{d\mid n} f(d) \equal{} \sum_{n\le x} f(n) [x/n]\equal{}\sum_{n\le x} F(x/n).\]
So, if you believe the above result, then
\[ \sum_{n \leq x} \mu (n) \left\lfloor \frac{x}{n} \right\rfloor \equal{} \sum_{n \leq x} \sum_{d\mid n} \mu(d) \equal{} \sum_{n\le x} \left\lfloor \frac{1}{n} \right\rfloor\equal{}1\]


Given any 2n integers, show that there are n of them whose sum is divisible by n.

Can be sharpened to $ 2n-1$ numbers, see http://www.mathlinks.ro/Forum/viewtopic.php?t=57546 (my post) .


function logarithms calculus integration limit algebra domain

Can any one give me an example of functions on $(0,\infty)$ such that
1) $f\in L^p$ iff $p=1$
2) $f\in L^p$ iff $1\leq p\leq 2$

For #1, try $f(x)=\frac1{x(1+\ln^2x)}$ on $(0,\infty).$

If $p<1,$ the integral $\int_1^{\infty}[f(x)]^p\,dx$ diverges.

If $p>1,$ the integral $\int_0^1[f(x)]^p\,dx$ diverges.

The same basic idea should work for #2:

$f(x)=\begin{cases}\frac1{x(1+\ln^2x)}&x\ge 1\\ \frac1{\sqrt{x}(1+|\ln x|)}&0<x<1\end{cases}$

We see that
\[ \lim_{x\to\infty}\frac{x^{-p}(1+\ln^2x)^{-p}}{x^{-p}}=0  \]
So
\[ \int_1^\infty \frac{1}{x^p(1+\ln^2x)^p}\,dx<\infty \Leftrightarrow \int_1^\infty x^{-p}\,dx<\infty\Leftrightarrow p>1  \]
in particular it is divergen for $p=1$?

[quote="Xevarion"]$\|\frac{1}{x(1+ln(x)^2)}\|_{L^1} = \pi$, if I'm not mistaken.[/quote]

When I compute it directly for $p=1$, I also have the same result. So anybody can point me out why my argument was wrong??

[quote="Xevarion"]I don't see why you have both ways implication. If $\frac{1}{x(1+\ln(x)^2)}$ decays faster than $\frac{1}{x}$, then the second integrable from 1 to $\infty$ implies the same for the first, but not the other way around.[/quote]

OK thank you. I realized know that in order $\int_1^\infty f$ convergent iff $\int_1^\infty g$ convergent we need to show that
\[ \lim_{x\to\infty}\frac{f(x)}{g(x)}=\ell>0  \]

The whole idea of my examples was to use powers of $x$ to force the divergence of the integral at the $p$'s for which we want divergence. But if you just use powers of $x,$ then you get divergence at the borderline values of $p.$ I used the logarithms to tweak that so the borderline-$p$ integrals would converge. But the logarithm grows too slowly to outweigh a power of $x$ at anything but a borderline case.

Does that make the rationale behind the examples clear?

Here's another one: find $f$ so that $f\in L^p$ iff $1<p<2.$

This time I want divergence at the borderline cases, so I don't need the logarithmic tweak. But I still have to treat $x$ near zero and $x$ large differently.

$f(x)=\begin{cases}\frac1{\sqrt{x}}&0<x<1\\ \frac1x&x\ge 1\end{cases}$

Thanky you Kent, it is much more clear now.

BTW how do you prove that
\[ \int_1^\infty x^{-p}(1+\ln^2x)^{-p}\, dx  \]
is divergent for $p<1$. What I can see that for $p\geq 1$ the integral is convergent.

Suppose $p<1.$ Choose $q$ such that $p<q<1.$

Then for sufficiently large $x,\ x^{-p}(1+\ln^2x)^{-p}>x^{-q}.$

Now: can you find a function that is in $L^1$ but not in $L\log L?$

Let's make this a domain of finite measure - say, $(0,1).$ Find a function $f$ such that

$\int_0^1|f|<\infty$ but $\int_0^1|f|\ln(1+|f|)=\infty.$