[quote="Rep123max"]Your are probably sick of my putting so much political stuff on the board, but I have a question for all Bush supporters.

On one of the newest Bush ads, it says that Kerry voted against the Laci Peterson Law which protects pregnant women from violence. Now the Laci Peterson Law actually only recognizes the fetus as an unborn child. This does not protect pregnant women from violence, instead it makes it when a pregnant women dies, there are two murders.

Also, it says that while campaigning, Kerry missed 2/3 of the votes in the Senate. Well if we are measuring how much work people have done, shouldn't we mention how Bush has vacationed 43% of the time as president of the United States?[/quote]
Rep, do your homework before you post, please. The Laci Peterson Law does indeed make it so that there are two murders when a pregnant woman is killed - but it also makes me a murderer if I punch a pregnant woman in the stomach, causing her to lose the baby. Changing that crime from "assault" to "murder" certainly will help protect pregnant women. Also, somehow, I bet that you either made that last statistic up or got it from Michael Moore (which is no better).

[quote="confuted"]Also, somehow, I bet that you either made that last statistic up or got it from Michael Moore (which is no better).[/quote]

It was calculated by the Manchester Guardian at 42%, although that does include weekends at Camp David. His month-long vacation in August 2001, however, was the longest presidential vacation in 32 years.

[quote="h_s_potter2002"]That's like waiting to start on a major project, 2 days before it's due.  Would any of you do that?[/quote]

Bad example, I'd say.

[quote="h_s_potter2002"]I'm not exactly a Kerry fan...

But why is the president going on vacation that much, when this is one of the times we really need leadership and guidance?  We have finally realized that our nation can be badly attacked by terrorists, and we're trying very hard to make things right in Iraq and Afghanistan, and the President just goes to vacation 1/3 of the time?  That's like waiting to start on a major project, 2 days before it's due.  Would any of you do that?[/quote]
I certainly wouldn't start a major project two days before it's due. I would start it the night before it was due.

Weekends at Camp David are hardly "vacations" - the same can likely be said about many of his "vacations." It's not like he left the country without any form of leadership in the executive branch for a month; he likely made important decisions on a daily basis during that time - he just didn't do it from his office. I doubt that a US president EVER really "gets" a vacation while he's in office - he just gets some time away from the desk and makes his decisions elsewhere.

Rep, your posts are pathetic. Just stop.

[quote="Rep123max"]Once this election is done I'll stop. If Bush and his campaign feel it is okay to do the commercials they are doing, what I put was not only okay, but superior. It was exactly about his record. Unlike Bush's commericals, it is something called "factual." Well, I guess I might be <b>misunderestimating</b> Bush. Oh wait, no.[/quote]

And you say BUSH has communication issues?

[quote="Rep123max"]Well, I guess I might be misunderestimating Bush. Oh wait, no.[/quote]

 :lol:  :lol:  :lol:

Although the 42% does include Camp David, they still calculated that he spent about 16-17% at his home, which I believe is still more than any previous president. However, he might well have been working when at his ranch, so we don't really know whether the 42% is a relavent statistic or not. However, what really bothers me is the [i]month-long[/i] vacation he took in August 2001. He'd only been in office for five and a half months! I know that he may have done some work over that vacation, but I doubt that he would be nearly as productive at home as in the White House.

Also, confuted, I don't think calling Rep's posts pathetic is going to stop him from posting what he believes nor further your argument. I would like to hear the rationalization for Bush's month-long vacation, so if you could provide this I would appreciate it.

Come on Rep- Do you really think a president has any vacation time?  When he is at the Crawford Ranch every day he is briefed and has work to do- calls to make.  I think that should be obvious to anyone who really thinks about the job that there is very little if any down time when president!  I suppose what Clinton did in the Oval Office was down time (the blue dress does prove he had some).

As for staying in the classroom- even democrats have defended that.  They said that it was better than rushing out and scaring a group of little kids.  Also, he wasn't reading the book- he was probably thinking about what he had to do next.  They would also never have shot the plane down- the military hadn't gotten to the plane before it crashed (PA) and everything else happened fast.  Remember before this our country would most likely never have shot down a plane with passengers.

Now I'm sure that judging by what you say everything in the world is Bush's fault- does that mean the economy is his fault?  Or could it perhaps be because idiots crashed into our buildings and caused people to stop flying and if people don't fly then they don't need hotels and if they don't need hotels then they don't need restaurants etc....Yep, if Bush would just go out to eat more the economy would be thriving!

[quote="soilent green"]As for staying in the classroom- even democrats have defended that.  They said that it was better than rushing out and scaring a group of little kids.  Also, he wasn't reading the book- he was probably thinking about what he had to do next.  They would also never have shot the plane down- the military hadn't gotten to the plane before it crashed (PA) and everything else happened fast.  Remember before this our country would most likely never have shot down a plane with passengers.[/quote]

First of all, I'd like to know what democrats have defended this. Second, I don't think it's unreasonable to say that it's OK to scare a few kids when [i]national security is at stake![/i] Whether or not he was thinking of what he needed to do, what he should have done was left the room and gotten briefed by someone who knew the situation so that at least he could be in a position to decide what to do. Sitting silently in a room full of schoolchildren in a time of crisis accomplished nothing.

One Democrat that defended Bush was George Stephanopolis when interviewed on Good Morning America.  He was one of Clintons Cronies- if you remember from the first term (press secretary).  It was also stated that never before had a sitting president been informed about a huge tragedy while being filmed.  Presidents should be given this time so they don't make huge errors like using nuclear weapons.  Even Roosevelt when Japan bombed Pearl Harbor took time making a decision before commiting this country to war in the Pacific.

People want to see leaders calm under stress.

[quote="JSRosen3"]First of all, I'd like to know what democrats have defended this. Second, I don't think it's unreasonable to say that it's OK to scare a few kids when [i]national security is at stake![/i] Whether or not he was thinking of what he needed to do, what he should have done was left the room and gotten briefed by someone who knew the situation so that at least he could be in a position to decide what to do. Sitting silently in a room full of schoolchildren in a time of crisis accomplished nothing.[/quote]
Nobody knew the situation better than 'a plane just flew into the WTC' and then 'another!'. The briefing you mentioned easily could have been (and I'm sure was) whispered into his ear. We all know that very very little information was available in the first few minutes - and how could it have been?

Uh... Rep, when information, as confuted said, isn't available, doesn't that mean  you [b]can't[/b] go out and get the information?  What confuted said doesn't support your point at all, however much you may want it to.

[quote="Rep123max"]It means you don't wait in a classroom for it to come for you. You leave the classroom, get all your top aids, discuss what they know/don't know. Get on Air Force 1, and take it from there.[/quote]

Exactly. No matter how little information there was, the president needed to [i]put himself in a position[/i] to make a decision. That means leaving the classroom and taking charge of the situation.

Or maybe George Bush considered his image in front of the children to be more important than the security of United States.

20/1 hindsight, I say.  Even if he had been briefed, what possibly decisions could he have possibly made to react to this?  The argument put forth is that he should have been briefed so that he could make an informed decision.  A decision on what?  The terrorists' intentions weren't known until it would have been far too late to make a decision.  Sure, the President and his team could work on their reaction time, but it's just Bush's bum luck that 9/11 happened while he was trying to do something charitable for the American people.  How much experience with crisis has Bush had, anyway?  To expect for him to get it right on the first try without faltering really seems like asking a bit much.

[quote="Rep123max"]And can somebody tell me what Bush did pre 9/11 to fight terrorism?[/quote]

Rep, no one really thought there was a serious threat to U.S. untill 9/11.  The only reason there is this much hype for fighting terrorism is because of 9/11.

im glad rep isnt our president. to be a president, you have to be calm in the line of fire, not hotheaded.


Solve $ 5^{x}-3^{y}=2$ in $ \mathbb{N}^{+}$. I didn't find any topic regarding this diophantine. I tried some combinations of modules, but didn't find any contraditction. Also, I'd like to know if there is any elegant solution.

EDIT: see ZetaX two posts below.  :oops:

[hide="solution"]Look at it $ \mod(5)$

$ \rightarrow 3^{y}= 2 \mod(5)$
By casework and FLT then $ y=1 \mod(4)$
All we need is that $ y$ is odd.

Now $ \mod(3)$
$ \rightarrow 5^{x}= 2 \mod(3)$
Same argument we get $ x$ is odd.

But then $ 5^{x}-3^{y}=(5-3)(5^{x-1}+3.5^{x-2}+\cdots+3^{y-1})$

For $ x \text{ or }y >1$ the second bracket is $ >1$ thus $ x=y=1$ only solution.[/hide]

I find another solution :)
we can easily prove both $ x,y$ are odd . now if $ y=1$ then $ x=1$ .
now assume that $ y\geq2$ we have $ 5^{x}\equiv 2\mod[9]$ and we give $ x=6k+5$
know use $ mod7$ , we give $ 5^{6k+5}\equiv 3\mod[7]$ and it results $ 3^{y}\equiv 1\mod[7]$ and we give $ y=6m$ which is contradiction because $ y$ is odd .

[quote="seamusoboyle"]But then $ 5^{x}-3^{y}=(5-3)(5^{x-1}+3.5^{x-2}+\cdots+3^{y-1})$[/quote]
No, that's wrong (the second bracket makes in fact no sense if $ x \neq y$).


Let $ x_{1},x_{2},....$ be a sequence of positive real numbers .If $ a_{k}$ and $ g_{k}$ are the arithmetic and geometric means , respectively, of $ x_{1},....,x_{k}$, prove that

1)  $ \frac{a_{n}^{n}}{g_{k}^{n}}$ $ \ge$ $ \frac{a_{n\minus{}1}^{n\minus{}1}}{g_{k}^{n\minus{}1}}$

2) $ n(a_{n}\minus{}g_{n})$ $ \ge$ $ (n\minus{}1) (a_{n\minus{}1}\minus{}g_{n\minus{}1})$

2)
$ \dfrac{(n\minus{}1)g_{n\minus{}1}\plus{}x_n}{n}\ge\sqrt[n]{g_{n\minus{}1}^{n\minus{}1}x_n}\equal{}g_n$ (by AM-GM)
and $ x_n\equal{}na_n\minus{}(n\minus{}1)a_{n\minus{}1}$
$ \Rightarrow n(a_n\minus{}g_n)\ge(n\minus{}1)(a_{n\minus{}1}\minus{}g_{n\minus{}1})$

hey can anyone answer the first inequality??

Is the question wrong or very difficult??
[u]can no one  solve it??[/u]

I assume you want to prove $ \frac{a_n^n}{g_n^n}\ge\frac{a_{n\minus{}1}^{n\minus{}1}}{g_{n\minus{}1}^{n\minus{}1}}$.

$ a_n^n\equal{}(\frac{(n\minus{}1)a_{n\minus{}1}\plus{}x_n}n)^n\ge a_{n\minus{}1}^{n\minus{}1}x_n\equal{}a_{n\minus{}1}^{n\minus{}1}\frac{g_n^n}{g_{n\minus{}1}^{n\minus{}1}}$.


Matt scored $ 70\%$ on the first 60 questions of the 100-question math final. If each question has the same value, what percent of the remaining questions must he answer correctly to receive a total score of $ 80\%$?

Since he scored $ 70\%$ on the first $ 60$ questions, that means he answered $ .7\times{60} \equal{} 42$ of those questions right. In order to get $ 80\%$, he must have answered $ 80$ questions correctly. So, of the last $ 40$ questions, he must answer $ 80 \minus{} 42 \equal{} 38$ of those questions correctly. His percentage on those questions must be $ \frac {38}{40} \equal{} \boxed{95\%}$.


Prove that for $ \triangle ABC;A,B,C<\dfrac{\pi}{2} m\geq 2; m \in N$ we have
$ cos^{m}A \plus{} cos^{m}B \plus{} cos^{m}C \geq \dfrac{3}{2^{m}}$

[quote="desmrhai"]Prove that for $ \triangle ABC;A,B,C < \dfrac{\pi}{2} m\geq 2; m \in N$ we have
$ cos^{m}A \plus{} cos^{m}B \plus{} cos^{m}C \geq \dfrac{3}{2^{m}}$[/quote]

$ LSH\equal{}\sum (cos^2A)^{\dfrac m{2}}  \ge 3\left(\dfrac{\sum cos^2A}3\right)^{\dfrac m{2}}\ge 3\left(\dfrac{1}4\right)^{\dfrac m{2}}\equal{}\dfrac{3}{2^{m}}$  :)

[quote="desmrhai"][quote="dduclam"][quote="desmrhai"]Prove that for $ \triangle ABC;A,B,C < \dfrac{\pi}{2} m\geq 2; m \in N$ we have
$ cos^{m}A \plus{} cos^{m}B \plus{} cos^{m}C \geq \dfrac{3}{2^{m}}$[/quote]

$ LSH \equal{} \sum (cos^2A)^{\dfrac m{2}} \ge 3\left(\dfrac{\sum cos^2A}3\right)^{\dfrac m{2}}\ge 3\left(\dfrac{1}4\right)^{\dfrac m{2}} \equal{} \dfrac{3}{2^{m}}$  :)[/quote]
can you explain $ \sum (cos^2A)^{\dfrac m{2}} \ge 3\left(\dfrac{\sum cos^2A}3\right)^{\dfrac m{2}}$[/quote]

Dear friend,
It's the lemma: $ \frac{a^n\plus{}b^n\plus{}c^n}3\ge \left(\frac{a\plus{}b\plus{}c}3\right)^n$
Easy to prove this lemma by using AM-GM inequality. :)

Infact, your inequality's also true for all $ 2\le m\in R^*$. :)


I was just wondering if I have a chance of getting into Countdown or the top 4 in Virginia.  Right now, I'm averaging about a 37 or a 38 on practice tests I take.  Is this a decent score?  I don't know too much about the whole thing because even though I qualified for states the last two years, I had a scheduling conflict and couldn't make it.  Anyway, thanks for any help.

It all comes down to which state you live in. You live in Virginia so I assume that state is EXTREMELY competitive. I live in OK so I have pretty much no REAL competition. Since you live in the more populated states, I think the cutoff for state in Virginia would be above 40.   :lol:

[quote="Ihatepie"]Are you the type of person who does better at home than at the actual competition?[/quote]
My first year I was, but after that I pretty much got used to it and scored about the same score that I got on practice.

Are you in 7th or 8th? I'd guess that if you can get a couple more points, you're basically guaranteed for nats.

I'm in 8th :(, but thanks for giving me some info.  At least now I know I have a chance.

I actually think you have a good chance of getting in with your current score. But keep on studying.

I could only do one year of Mathcounts (learned about it in 8th grade) and didn't study enough. I got a decent 76th (top third  :lol:) at nationals, but I could've easily cracked 50 with more practice.

[color=blue] [size=59]and if one kid from my school hadn't been hiding mathcounts in 6th and 7th I could've easily made the top 30 and maybe even made countdown[/color][/size]

Yea a 37 or 38 will get you to countdown. Anything mid 30's should get you there. A 37 will not guarantee you top 4 but you should be around the middle. It also depends how hard this years test is.

Actually I was lying; you're not close to the cutoff.

jk  :D

I'm pretty sure Virginia is VERY competitive, so that score would probably get you into countdown, maybe. Study really hard!
If you were in my state, atleast last year, you would have won by a lot. My state is SOOO much better this year though. :(

I think you would be secure with a 37 in state written.
My state OK is horrible..last year number 1 got like a 28.

omg omg, i'm going to Va states too, i usually get around 20-24 in sprint, and usually a perfect on target. This is my first year going to states, and i'm in 8th grade :( . so, vamathlete, i guess i'll see u there. Keep practicing.

[quote="theprodigy"][quote="Ihatepie"]I doubt there is any state not counting Texas and California that 38 won't get you to countdown.

Is countdown official in your state? And how is it held?[/quote]

new york?
and yeah texas...i have a feeling 42/43+ required 4 countdown[/quote]

I believe the cutoff last year for going to nats last year in  texas was 41 actually. Getting to CD would probably require at least a 34 there, but not 42/43+. If that were the case, they'd be out of tiebreakers before the top 4 was decided  :D

I think I heard from someone that the cutoff for countdown in Texas was around a 35.

As for your question, try for a 46. I'm not sure if that'll get you in though.  :wink:


Just wondering, who were the two non-WOOT alumni that were USAMO winners this year? I think Sergei was one, don't know about the other. According to the WOOT Information page 10 of the 12 winners were alumni.

Everyone who makes mop is invited to WOOT next year, so Sergei might actually count as an alumn (don't know why he wouldn't enroll if it was free...) Brian and Sherry, on the other hand, probably didn't enroll.


Define the sequence of positive integers $\{x_{n}\}$ by $x_{1}=10^{999}+1$ and for all $n \geq 2$ the number $x_{n}$ is obtained from $11x_{n-1}$ by deleting the first digit.Is the sequence bounded?Why?

well, the sequence of the number of digits of $x_{i}$ is nonincreasing...


Right triangle $ABC$ is made of paper.  Hypotenuse $\overline{AB} = 36$.  Vertex $C$ is folded onto the midpoint of $\overline{AB}$, forming crease $\overline{XY}$ (where points $X$ and $Y$ are on the legs of the triangle).  Given that $XY = 20$, compute the area of triangle $XYC$.

[hide]
Let midpt of AB = M. CM= 18
Draw in X and Y. Quadrilateral CXMY is made up of two congruent right triangles, XCY and XMY.
XY=20 and is perpendicular bisector of CM. So area of triangle CXY = 9 x 20/2 = 90.[/hide]

Nice.  I don't know why it took me so long to figure out the distance from C to the midpoint of the hypotenuse.


analytic geometry
geometry
rectangle
combinatorics proposed
combinatorics

Consider an 8x8 board divided in 64 unit squares. We call [i]diagonal[/i] in this board a set of 8 squares with the property that on each of the rows and the columns of the board there is exactly one square of the [i]diagonal[/i]. Some of the squares of this board are coloured such that in every [i]diagonal[/i] there are exactly two coloured squares. Prove that there exist two rows or two columns whose squares are all coloured.

Ok, it id easier than I first thought :

1) Consider any diagonal and then its two coloured cells in different rows and columns. Assume that they are the ones with coordinates $(i,j)$ and $(r,s)$. Then consider the diagonal obtained from the one above by replacing the two coloured cells by the cells $(i,s)$ and $(r,j)$. Since this new diagonal has two coloured cells and that the initial one has no other coloured cells, it follows that these two new cells are coloured too.

2) There is no coloured cells which does not belong to row $j$ or $s$ and to columns $i$ or $r$, otherwise with the two first coloured cells above, we may find a diagonal with at least three coloured cells.

3) Wlog, assume that the four coloured cells in 1) are the ones in the upper left $2\times 2$ square, that are $(1,8),(2,8),(1,),(2,7).$
Now, assume that there exists a non-coloured cell in the upper row, say $(x,8)$ with $x>2.$.
By considering this cell and $(2,7)$ and $(1,y)$ with $y <7$, we may complete them to form a diagonal and, using 2), we deduce that $(1,y)$ is coloured. Thus all the first column is coloured.
We prove in the same way that the second column is coloured by considering $(2,y)$ instead of $(1,y).$
Therefore there are two coloured columns.

If all the upper row is coloured, then as in 1), we deduce that all the second row from above is coloured, and we are done.

Pierre.

[hide="Solution"]Label the columns $a,b,c\ldots$ and the rows $1,2,3\ldots$. Begin by considering the main diagonal $a1-h8$. It must have two colored squares. WLOG rearrange the rows and columns so that they are on $a1$ and $b2$. Then diagonals containing $a1$ and $b2$ can contain any square within the rectangle defined by $c3$ and $h8$, and since the diagonal must have exactly two colored squares ($a1$ and $b2$), none of the squares in that rectangle can be colored.

Now consider the diagonal containing $a3, b2, c1$ and any other 5 squares we know are not colored. We know $b2$ is colored, so either $a3$ or $c1$ is colored.

Assume $a3$ is colored. Then the diagonals containing $b1,a3$ and $b2,a3$ show that no square on columns $c-h$ is colored. And then by necessity the entire $a$ and $b$ columns must be colored to satisfy the conditions. If $c1$ is colored, the entire $1$ and $2$ rows are colored by a similar argument. This completes the proof.[/hide]


Ok, the time is rolling around to start picking a new Science Fair Topic that is required by the school. I'd really love to do a math topic but can't really think of any good ones. Any of you guys have ideas?
(Note: I come from a ghetto town and will be completely on my own in this endeavor so try not to post any really advanced/hardcore topics. Also, I will not plagiarize any ideas posted unless given explicit permission. I'm just looking for ideas to jump start my own.)

I was thinking about doing a research in math too...

but what exactly can we do???

What level of math do you know? Do you know calculus yet?

for me, I'm taking AB this year (junior)

I've taken BC already. Well, I've already come up with one I think. I'm going to look at the different definitions/approximations of the golden ratio and do a precision analysis, determine which one approaches the value of phi the "fastest". I'm going to try to come up with a few on my own and compare those against the standard ones out there.... (good luck to me, heh). It's still tentative but I may settle for that one.

I did a project applying the Collatz algorithm to different subsets of numbers and observing the results. It was a really EASY project, but everyone else thinks it was complicated. :lol: I'm going to the citywide fair this weekend.


Find all polynomials $ P(x)\in \mathbb R[x]$ such that
$$ P(x^2)=P(x)P(x+2),$$
for all $x\in \mathbb R$.

let $ a$ be a complex root of $ P$. so $ P(a)\equal{}0$
therefore $ a^2,a^4,a^8$ etc. have to be roots of $ P$ also. This leeds to contradiction because we get infinitely many roots and polinomial should be of infinite degree except if $ |a|\equal{}1;0$ not hard to exclude the $ 0$ part

but if we put $ a\minus{}2$ into the expression we get $ P((a\minus{}2)^2)\equal{}P(a\minus{}2)P(a)$ meaning $ P((a\minus{}2)^2)\equal{}0$ which means again infinitely many roots with the same reason as before. except if $ |a\minus{}2|\equal{}1$

the only $ a$ such that $ |a\minus{}2|\equal{}|a|\equal{}1$ is $ a\equal{}1$ and it is the only theoretical root of the polynomial

now we put $ a_n(x\minus{}1)^n$ into starting expression and conclude that all polinomials of form $ (x\minus{}1)^n$ for any natural $ n$ are solutions plus trivial solutions $ P(x)\equal{}0$ and $ P(x)\equal{}1$

[quote="radio"]Find all $ P(x)\in R[x]$ such that
$ P(x^2) \equal{} P(x)P(x \plus{} 2)$[/quote]

$ P(x)\equal{}0$ and $ P(x)\equal{}1$ are solution.
If $ P(x)$ is different from constant polynomial, then :
If $ z$ is a complex root, then so is $ z^2$ and so, since we have finite number of roots, all non zero roots have $ |z|\equal{}1$.
If $ 0$ is a root, $ P((\minus{}2)^2)\equal{}P(\minus{}2)P(0)$ and $ 4$ is a root, but $ |4|\neq 1$ and so $ 0$ is not a root.
If $ z$ is a complex root, then so is $ (z\minus{}2)^2$ and so all roots have $ |z|\equal{}1$ and $ |z\minus{}2|\equal{}1$

So the only complex root is $ 1$.

So $ P(x)\equal{}a(x\minus{}1)^n$ and it is easy to check that $ a\equal{}1$

So the solutions are :
$ P(x)\equal{}0$
$ P(x)\equal{}1$
$ P(x)\equal{}(x\minus{}1)^n$


$ \sqrt{x^2\plus{}3x} \geq 0 \\
\Rightarrow x^2\plus{}3x\geq 0\\
\Rightarrow x\geq 0 \ or \ x \leq \minus{}3........(1)$

now squaring we get
$ 4x^2\minus{}4x\plus{}1 \leq \ x^2\plus{}3x\\
\Rightarrow 3x^2\minus{}7x\plus{}1\leq 0\\
\Rightarrow x^2\minus{}7x/3\plus{}1/3 \leq 0\\
\Rightarrow \frac{7\minus{}\sqrt{37}}{6} \leq  x \leq \frac{7\plus{}\sqrt{37}}{6}.....(2)$
combining 1 and 2
$ \frac{7\minus{}\sqrt{37}}{6} \leq  x \leq \frac{7\plus{}\sqrt{37}}{6}$

[quote="implex"]
combining 1 and 2
$ \frac {7 \minus{} \sqrt {37}}{6} \leq x \leq \frac {7 \plus{} \sqrt {37}}{6}$[/quote]
Is this the final answer? It's not correct then.  :o

[hide="my solution"]
1) $ 0\le 2x\minus{}1 \Rightarrow x\ge \frac{1}{2} \dots (1)\\
2x\minus{}1\le \sqrt{x^2\plus{}3x}\\
4x^2\minus{}4x\plus{}1\le x^2\plus{}3x\\
3x^2\minus{}7x\plus{}1\le 0\\
\frac{7\minus{}\sqrt{37}}{6}\le x \le \frac{7\plus{}\sqrt{37}}{6}$
from (1):
$ \frac{1}{2}\le x \le \frac{7\plus{}\sqrt{37}}{6}$
2) $ 0 > 2x\minus{}1 \Rightarrow x<\frac{1}{2}\dots (2)\\
\minus{}\sqrt{x^2\plus{}3x}\le 2x\minus{}1\\
x^2\plus{}3x \le 4x^2\minus{}4x\plus{}1\\
0 \le 3x^2\minus{}7x\plus{}1\\
\frac{7\minus{}\sqrt{37}}{6} \le x \le \frac{7\plus{}\sqrt{37}}{6}$
from (2):
$ \Phi \text{ No Solution}$
the final solution:
$ \frac{1}{2}\le x \le \frac{7\plus{}\sqrt{37}}{6}$

[/hide]

Put $ x\equal{}0$. It works!

Put $ x\equal{}\minus{}100$. It works too!

So... what's going on here? Look closely at the domains and ranges.

ok :
$ \left[0,\frac {7 \plus{} \sqrt {37}}{6}\right]\cup ( \minus{} \infty,\minus{}3]$

[quote="mathson"]ok :
$ \left[0,\frac {7 \plus{} \sqrt {37}}{6}\right]\cup ( \minus{} \infty,3]$[/quote]A small typo, it should be $ \minus{}3$ ;)

[quote="10000th User"][quote="mathson"]ok :
$ \left[0,\frac {7 \plus{} \sqrt {37}}{6}\right]\cup ( \minus{} \infty,3]$[/quote]A small typo, it should be $ \minus{} 3$ ;)[/quote]

$ \text{thanks }10^4 \text{th User}$


inequalities
Cauchy Inequality
algebra unsolved
algebra

First, I can only guess the correct formulation of the problem. Here is my guess:

[color=blue][b]Problem.[/b] Let $a_1$, $a_2$, ..., $a_n$ denote positive numbers. Prove that $\sum_{i=1}^n\frac{a_i^2}{a_i+a_{i+1}}\geq\frac12\sum_{i=1}^n a_i$.[/color]

[i]Solution.[/i] First,

$2\sum_{i=1}^n\frac{a_i^2}{a_i+a_{i+1}}-\sum_{i=1}^n\frac{a_i^2+a_{i+1}^2}{a_i+a_{i+1}}=\sum_{i=1}^n\frac{2a_i^2-\left(a_i^2+a_{i+1}^2\right)}{a_i+a_{i+1}}=\sum_{i=1}^n\frac{a_i^2-a_{i+1}^2}{a_i+a_{i+1}}$
$=\sum_{i=1}^n\left(a_i-a_{i+1}\right)=0$.

Thus, $2\sum_{i=1}^n\frac{a_i^2}{a_i+a_{i+1}}=\sum_{i=1}^n\frac{a_i^2+a_{i+1}^2}{a_i+a_{i+1}}$. Now, for any two positive reals x and y, we have $\frac{x^2+y^2}{x+y}\geq\frac{x+y}{2}$, since $2\left(x^2+y^2\right)\geq\left(x+y\right)^2$ (because $2\left(x^2+y^2\right)-\left(x+y\right)^2=\left(x-y\right)^2\geq 0$). Thus,

$2\sum_{i=1}^n\frac{a_i^2}{a_i+a_{i+1}}=\sum_{i=1}^n\frac{a_i^2+a_{i+1}^2}{a_i+a_{i+1}}\geq\sum_{i=1}^n\frac{a_i+a_{i+1}}{2}=\sum_{i=1}^n a_i$,

so that $\sum_{i=1}^n\frac{a_i^2}{a_i+a_{i+1}}\geq\frac12\sum_{i=1}^n a_i$, and the problem is solved.

  darij

Using Cauchy inequality,
$\left ( \sum_{i=1}^{n} (a_{i+1}+a_{i+2}) \right ) \cdot \left (\sum_{i = 1}^n\frac{a_i^2}{a_{i + 1} + a_{i + 2}} \right ) \geq (\sum_{i = 1}^n a_i)^2$.
While $\sum_{i=1}^{n} (a_{i+1}+a_{i+2}) = 2\sum_{i=1}^{n}a_i$, the result follows.

Another proof.
Using AM-GM inequality.
Since $\frac{a_i^2}{a_{i + 1} + a_{i + 2}} + \frac{a_{i + 1} + a_{i + 2} }{4} \geq a_i,$ $i = 1, 2, ..., n.$
Adding such $n$ inequlities, and notice that  $\sum_{i=1}^{n}\frac{a_{i + 1} + a_{i + 2}}{4} = \frac{1}{2}\sum_{i=1}^{n}a_i.$
The result follows.

From these two proofs, we see that the denominator can be selected freely.


Try the more general sum $ \sum_{n\equal{}1}^{\infty}\frac{x^{2n}}{(2n)(2n\minus{}1)}$. (Your series evaluates this at $ x\equal{}\frac{1}2$)

What happens when you differentiate this series a couple of times?

Thank you for the help.

I differentiated the series twice, cancelled the common terms, and obtained a geometric series.
Then I evaluted the series and integrated to recover the original sum :)


Determine all rational numbers $ r$ such that all solutions of the equation:

$ rx^2\plus{}(r\plus{}1)x\plus{}(r\minus{}1)\equal{}0$ are integers.

[quote="moldovan"]Determine all rational numbers $ r$ such that all solutions of the equation:

$ rx^2 \plus{} (r \plus{} 1)x \plus{} (r \minus{} 1) \equal{} 0$ are integers.[/quote]

Clearly $ r\equal{}0$ is a solution, so for $ r \neq 0$, let $ r\equal{}\frac{q}{p}$, where $ \gcd(p,q)\equal{}1$, $ p$ is integer and $ q$ is positive integer, then the equation

$ \Leftrightarrow qx^2\plus{}(q\plus{}p)x\plus{}(q\minus{}p)\equal{}0$

Now let the solutions of the equation be integers $ a, b$.

Then $ q(x\minus{}a)(x\minus{}b) \equiv qx^2\plus{}(q\plus{}p)x\plus{}(q\minus{}p)$ gives $ q|q\plus{}p \Rightarrow q\equal{}1$

Now we are going to find $ p$ so that $ x^2\plus{}(p\plus{}1)x\plus{}(1\minus{}p)$ has only integer roots.

Rewrite the equation like this: $ x^2\plus{}x\plus{}1\equal{}p(1\minus{}x)$.
Clearly for $ x$ is integer, $ (1\minus{}x) \neq 0$
And $ (1\minus{}x)|\gcd(x^2\plus{}x\plus{}1,1\minus{}x)\equal{}\gcd(3,1\minus{}x)$
$ \Rightarrow 1\minus{}x\equal{}1,\minus{}1,3,\minus{}3$

$ 1\minus{}x\equal{}1 \Rightarrow p\equal{}1 \Rightarrow r\equal{}1$ as solution after checking.
$ 1\minus{}x\equal{}\minus{}1 \Rightarrow p\equal{}\minus{}7 \Rightarrow r\equal{}\minus{}\frac{1}{7}$ as solution after checking.
$ 1\minus{}x\equal{}\minus{}3 \Rightarrow p\equal{}\minus{}7$  again.
$ 1\minus{}x\equal{}3 \Rightarrow p\equal{}1$ again.

So $ r\equal{}0$ or $ r\equal{}1$ or $ r\equal{}\minus{}\frac{1}{7}$ are the only solutions.


It's a very old problem,I think it should be in mathlink.

[quote="Adalbert"]Find the maximum value of $A$ for all $x,y,z>0$:

$\frac{x}{\sqrt{y^2+z^2}} + \frac{y}{\sqrt{z^2+x^2}} + \frac{z}{\sqrt{x^2+y^2}} \geq A$[/quote]
Ibelieve that it should be:
Find the maximum value of $A$ for all $x,y,z>0$:

$\frac{x}{\sqrt{y^2+z^2}} + \frac{y}{\sqrt{z^2+x^2}} + \frac{z}{\sqrt{x^2+y^2}} > A$

if so then using $a\geq \frac{2a^2}{a^2+1}$
$\frac{x}{\sqrt{y^2+z^2}} + \frac{y}{\sqrt{z^2+x^2}} + \frac{z}{\sqrt{x^2+y^2}}\geq \frac{2x^2}{x^2+y^2+z^2}+\frac{2y^2}{x^2+y^2+z^2}+\frac{2z^2}{x^2+y^2+z^2}=2$ but we can not have equality ;)

i saw it   $\geq$  but it could be   $>$  i guess...nice solution, thanks

We can think of this inequality. $x^2+y^2+z^2 \ge 2x\sqrt{y^2+z^2}, \frac{x}{\sqrt{y^2+z^2}} \ge \frac{2x^2}{x^2+y^2+z^2}$, then we have $\sum{\frac{x}{\sqrt{y^2+z^2}} \ge 2}$ but here the inequality isn't equal to 2.


function algebra polynomial trigonometry domain

A function "slides" if f(x) = f(x+1) for all x in [0,1]. Give one non-polynomial function which slides and prove that a non-constant polynomial cannot slide.

[hide="Hint for the first part"] This is pretty obvious; sliding is a weaker condition than periodicity. [/hide]
[hide="Hint for the second part"] How many roots can the function $ g(x) \equal{} f(x) \minus{} f(x\plus{}1)$ can have? [/hide]

Can you elaborate on that? I would be pleased to see your solution.

The problems you've just posted look like exercises out of a textbook, so I'd prefer it if you showed us your progress so far instead.  What's confusing about the hints?

$ g(x) \equal{} f(x) \minus{} f(x\plus{}1)$ can't have infinite amount of roots unless it is a constant polynomial. One function that works could be $ f(x) \equal{} \minus{}2$ if $ x<0$, $ f(x) \equal{} 0$ if $ 0\leq x \leq2$ and $ f(x) \equal{} 2$ if $ 2\leq x$

Thanks, aznlord1337.
t0rajir0u, your assumptions are wrong: the questions are not from a textbook (someone was asked them at a university interview, which is why I thought they might be interesting); I have solved them (the first example I thought of is Sin (2pi x); I am not confused by your hints. Thanks anyway.

[hide="first part"]Since the sine function repeats ever $ 2\pi$, we could multiply $ x$ by $ 2\pi$ and it repeats every 1. Thus $ \sin{2\pi x}$ slides. So does $ \cos{2\pi x}$. You can also add any constant to a sliding function and it will still be a sliding function.[/hide]

[hide="second part"]Let $ e(x)\equal{}f(x)\minus{}f(x\plus{}1)$. Thus $ g(x)\equal{}0$ for all $ x\in [0,1]$. Thus $ g(x)$ has infinitely many roots, and $ g(x)$ is not a polynomial, and hence $ f(x)$ isn't either.[/hide]

[quote="1=2"]Let $ e(x) \equal{} f(x) \minus{} f(x \plus{} 1)$. Thus $ g(x) \equal{} 0$ for all $ x\in [0,1]$. Thus $ g(x)$ has infinitely many roots, and $ g(x)$ is not a polynomial[/quote]
This is not quite correct; $ g(x)$ could be the zero polynomial.  You then need to show that it is not possible for $ f(x \plus{} 1)$ to equal $ f(x)$ identically (if $ f$ is a nonconstant polynomial).

oh....

[hide="well then"]Note that the constant terms of $ f(x)$ and $ f(x\plus{}1)$ are the same. When we plug in $ x\plus{}1$ into $ f(x)$, we must not have it affect the constant term. But $ (x\plus{}1)^n$ contains a constant term for all $ n$. Thus we may not have a $ x^n$ term in $ f(x)$ for all $ n$. Thus $ f(x)$ isn't a polynomial.[/hide]

Is that right?

Alternatively, for 2 you could

[hide]
Polynomial functions are continuous, so f(x) is bounded on [0,1) which implies that it is bounded, which implies that it is the 0 function[/hide]

[quote="1=2"]Note that the constant terms of $ f(x)$ and $ f(x \plus{} 1)$ are the same. When we plug in $ x \plus{} 1$ into $ f(x)$, we must not have it affect the constant term. [/quote]
There's no contradiction here if $ f(0) \equal{} f(1)$.

One easy way to finish from here is to show that if $ f(x)$ is a polynomial of degree exactly $ n$ then $ f(x \plus{} 1) \minus{} f(x)$ is a polynomial of degree exactly $ n\minus{}1$, which is just done by comparing coefficients.  You could also do what pythag011 did.

And for a third alternative, $ f(x) \minus{} f(c)$ for any constant $ c$ has infinitely many roots.

Polynomial Case: If $ f(x)\equal{}f(x\plus{}1)$, then $ f(x)\equal{}f(x\plus{}n)$ for all $ n\in\mathbb{N}$. If we look at this algebraically, it is clear that we can extend the domain to $ \mathbb{C}$. If $ f$ constant, it obviously works. If $ f$ is non-constant, then it has a root, $ r$. Let $ deg(f)\equal{}m$. Observe that $ r$, $ r\plus{}1$, ..., $ r\plus{}m$ are $ m\plus{}1$ distinct roots of a non-constant polynomial. Thus $ deg(f)\ge m\plus{}1$ which is a contradiction.

The message here is that a lot of things go wrong if you require a nonconstant polynomial to also be periodic.  Here's another:  since polynomials are smooth, not only is it necessary that $ f(0) \equal{} f(1)$, but we must also have $ f'(0) \equal{} f'(1), f''(0) \equal{} f''(1)$, and so on until you get to a linear polynomial - which must then be constant.


probability
videos
Pascal\u0027s Triangle

Probability: Count *Equally Likely* Events!

We do two probability problems that teach us what it is important to count equally likely outcomes when counting successful and possible outcomes in a probability problem.

This video made it easy for me to see how to deal with probability

:lol: this is so great! i was horrible at this site before! :P

Probability of 0 yellows is $ \frac{1}{64}$
Probability of 1 yellow is $ \frac{3}{32}$
Probability of 2 yellows: $ \frac {\binom{6}{2}}{2^6} \equal{} \frac{12}{64} \equal{} \frac{3}{16}$
Probability of 3 yellows:$ \frac {\binom{6}{3}}{2^6} \equal{} \frac{20}{64} \equal{} \frac{5}{16}$
Probability of 4 yellows is equal to the probability of 2 blues so $ \frac {3}{16}$
Probability of 5 yellows is 1 blue, so $ \frac{3}{32}$.
Probability of 6 yellows is just $ \frac{1}{64}$
But summing gives$ \frac{29}{32}$, what did I do wrong?

[quote="brightzhu"]Probability of 0 yellows is $ \frac {1}{64}$
Probability of 1 yellow is $ \frac {3}{32}$
Probability of 2 yellows: $ \frac {\binom{6}{2}}{2^6} \equal{} \frac {12}{64} \equal{} \frac {3}{16}$
Probability of 3 yellows:$ \frac {\binom{6}{3}}{2^6} \equal{} \frac {20}{64} \equal{} \frac {5}{16}$
Probability of 4 yellows is equal to the probability of 2 blues so $ \frac {3}{16}$
Probability of 5 yellows is 1 blue, so $ \frac {3}{32}$.
Probability of 6 yellows is just $ \frac {1}{64}$
But summing gives$ \frac {29}{32}$, what did I do wrong?[/quote]

You mean pulling 6 balls from an infinite supply of yellow, blue balls?

EDIT: $ \binom{6}{2}\equal{}12$?  :)

oh i see, those numbers form the sixth row of pascal's triangle!    :D

This video made it very easy for me to understand probability. It was explained very well and his humor made it fun to watch.

These are the simplest of probability problems. These are usually found on the AMC 8.

[hide="My solution"]0 yellow faces:$\binom{6}{0}$
1 yellow face: $\binom{6}{1}$
2 yellow faces: $\binom{6}{2}$
3 yellow faces: $\binom{6}{3}$
4 yellow faces: $\binom{6}{4}$
5 yellow faces: $\binom{6}{5}$
6 yellow faces: $\binom{6}{6}$

We know the fact that:
$\binom{n}{0}+\binom{n}{1}+.....\binom{n}{n-1}+\binom{n}{n}=2^n$
We have $\binom{6}{0}+\binom{6}{1}+\binom{6}{2}+\binom{6}{3}+\binom{6}{4}+\binom{6}{5}+\binom{6}{6}\implies 2^6=64$

$\mathbb{P}=\frac{Success's}{\text{Total possibilities}}\implies \frac{Success's}{2^6}\implies \frac{Success's}{64}$

0 yellow faces: $\frac{\binom{6}{0}}{64}\implies \frac{1}{64}$
1 yellow face:  $\frac{\binom{6}{1}}{64}\implies \frac{6}{64}\implies \frac{3}{32}$
2 yellow faces: $\frac{\binom{6}{2}}{64}\implies \frac{\frac{6*5}{2}}{64}\implies \frac{15}{64}$
3 yellow faces: $\frac{\binom{6}{3}}{64}\implies \frac{\frac{6*5*4}{3*2}}{64}\implies \frac{20}{64}\implies \frac{10}{32}\implies \frac{5}{16}$
4 yellow faces: $\frac{\binom{6}{4}}{64}\implies \frac{\binom{6}{2}}{64}\implies \frac{\frac{6*5}{2}}{64}\implies \frac{15}{64}$
5 yellow faces: ${\frac{\binom{6}{5}}{64}\implies \frac{6}{1}}{64}\implies \frac{6}{64}\implies \frac{3}{32}$
6 yellow faces: $\frac{\binom{6}{6}}{64}\implies \frac{\binom{6}{0}}{64}\implies \frac{1}{64}$

Also, as Fersolve pointed out, that those numbers:
$\in{1,6, 15, 20, 15, 6, 1}$ are the sixth row of Pascal's triangle, and we know that the nth row of pascal's triangle is $2^n$ so we can find that the denominator is easily $2^6$[/hide]

Edit: YAY!  100th math post!  Click on the link in my signature to see what I am at now =)
Edit 2: [quote="brightzhu"]Probability of 0 yellows is $ \frac {1}{64}$
Probability of 1 yellow is $ \frac {3}{32}$
Probability of 2 yellows: $ \frac {\binom{6}{2}}{2^6} = \frac {12}{64} = \frac {3}{16}$
Probability of 3 yellows:$ \frac {\binom{6}{3}}{2^6} = \frac {20}{64} = \frac {5}{16}$
Probability of 4 yellows is equal to the probability of 2 blues so $ \frac {3}{16}$
Probability of 5 yellows is 1 blue, so $ \frac {3}{32}$.
Probability of 6 yellows is just $ \frac {1}{64}$
But summing gives$ \frac {29}{32}$, what did I do wrong?[/quote]
Probability of 2 yellow is $\frac{\binom{6}{2}}{2^6}$
$\binom{6}{2}=\frac{6*5}{2}\implies \frac{\cancel{6}*5}{\cancel{2}}\implies 3*5=\boxed{15}$, so the probability of 2 yellows= $\frac{15}{64}$ not $\frac{12}{64}$
You made that mistake in $\binom{6}{4}$ also.

Also:
$\frac{1}{64}+\frac{6}{64}+\frac{12}{64}+\frac{20}{64}+\frac{12}{64}+\frac{1}{64}$ using your method is:
$\frac{1+6+12+20+12+6+1}{64}\implies \frac{7+12+20+12+6+1}{64}\implies \frac{19+20+6+12+1}{64}\implies \frac{39+6+12+1}{64}\implies \frac{45+12+1}{64}\implies \frac{57+1}{64}\implies \frac{58}{64}\implies  \frac{\frac{58}{2}}{\frac{64}{2}}\implies \frac{29}{32}$ which is correct with your mistake, if your reasoning for $\binom{6}{2}=12$


trigonometry
calculus
calculus computations

The region bounded by $ y\equal{}\cos x$, $ y\equal{}x$, and $ x\equal{}0$ is revolved around the line $ y\equal{}\minus{}2$. Find the volume of the solid of revolution.
~~~

I have sketched the curves. From the origin to the point of intersection, $ y\equal{}\cos x$ is always above $ y\equal{}x$, so I'm thinking that the washer method can be used for this problem.

The problem is, I'm not sure of an exact answer to $ x\equal{}\cos x$. If I knew that, then I could solve the definite integral.

The solution to $ x\equal{}\cos{x}\approx 0.739085$.  It's not particularly easy to show that by hand, though.  Use a calculator.

$ x^2\plus{}2x\minus{}2\equal{}0$  gives an O.K. approximation in case you can't (don't want to) use a calculator.


Simple problem but it's driving me crazy.

[code]

\begin{array}{|l|l|}
\hline
\hspace{10pt} \sin(\theta) = \frac{O}{H} & \hspace{20pt} \cos(\theta) = \frac{A}{H} & {} \\[10pt] \hline
\hspace{10pt} \tan(\theta) = \frac{O}{A} & \hspace{20pt} \csc(\theta) = \frac{H}{O} & {} \\[10pt] \hline
\hspace{10pt} \sec(\theta) = \frac{H}{A} & \hspace{20pt} \cot(\theta) = \frac{A}{O} & {}
\end{array}

[/code]

I keep getting 2 error messages for no reason. What's going on?

There are 2 errors here:
1. array must be in maths mode so you need dollar signs as in $ \$$\begin{array} .... \end{array}$ \$$
2. You have 3 columns (the last one blank) yet you have only 2 columns setup in \begin{array}{|l|l|}

I'm not sure of the purpose of \hspace and there's no need for \\[10pt] as the way to stretch the distance between rows is to use \renewcommand{\arraystretch}{[i]factor[/i]} where factor=1 is normal, factor=1.5 is 1.5 times normal etc.

This looks OK to me:
[code]$\begin{array}{|c|c|c|}
\hline
\sin(\theta) = \frac{O}{H} & \cos(\theta) = \frac{A}{H} & \\ \hline
\tan(\theta) = \frac{O}{A} & \csc(\theta) = \frac{H}{O} & \\ \hline
\sec(\theta) = \frac{H}{A} & \cot(\theta) = \frac{A}{O} & \\ \hline
\end{array}$[/code]

Now the spacing of the array is being a pain in the rear end.

[code] \begin{mcen}$\begin{array}{|c|c|c|c|}
\hline
{} & {} & {} & {} \\
\sin(0) = 0 & \sin(\frac{\pi}{2}) = 1 & \hspace{-7.5pt} \sin(\pi) = 0 & \sin(\frac{3\pi}{2}) = -1 \\
\hline
{} & {} & {} & {} \\
\cos(0) = 1 & \cos(\frac{\pi}{2}) = 0 & \cos(\pi) = -1 & \cos(\frac{3\pi}{2}) = 0 \\
\hline
{} & {} & {} & {} \\
\tan(0) = 0 & \tan(\frac{\pi}{2}) = * & \hspace{-7.5pt} \tan(\pi) = 0 & \tan(\frac{3\pi}{2}) = * \\
\hline
{} & {} & {} & {} \\
\csc(0) = * & \csc(\frac{\pi}{2}) = 1 & \csc(\pi) = * & \csc(\frac{3\pi}{2}) = -1 \\
\hline
{} & {} & {} & {} \\
\sec(0) = 1 & \sec(\frac{\pi}{2}) = * & \hspace{-7.5pt} \sec(\pi) = -1 & \sec(\frac{3\pi}{2}) = * \\
\hline
{} & {} & {} & {} \\
\cot(0) = * & \cot(\frac{\pi}{2}) = 0 & \cot(\pi) = * & \cot(\frac{3\pi}{2}) = 0 \\
\hline
\end{array}$ \\[10pt] \end{mcen}
[/code]

where mcen stands for the "mini-center" command, which centers something without any other auto-spacing annoyance to worry about.

How can I manipulate the "white space" above and below the text in each box?

I don't really understand what you are trying to do. What is wrong with
[code]\renewcommand{\arraystretch}{2}
$\begin{array}{|l|l|l|l|}
\hline
\sin(0) = 0 & \sin(\frac{\pi}{2}) = 1 & \sin(\pi) = 0 & \sin(\frac{3\pi}{2}) = -1 \\
\hline
\cos(0) = 1 & \cos(\frac{\pi}{2}) = 0 & \cos(\pi) = -1 & \cos(\frac{3\pi}{2}) = 0 \\
\hline
\tan(0) = 0 & \tan(\frac{\pi}{2}) = * & \tan(\pi) = 0 & \tan(\frac{3\pi}{2}) = * \\
\hline
\csc(0) = * & \csc(\frac{\pi}{2}) = 1 & \csc(\pi) = * & \csc(\frac{3\pi}{2}) = -1 \\
\hline
\sec(0) = 1 & \sec(\frac{\pi}{2}) = * & \sec(\pi) = -1 & \sec(\frac{3\pi}{2}) = * \\
\hline
\cot(0) = * & \cot(\frac{\pi}{2}) = 0 & \cot(\pi) = * & \cot(\frac{3\pi}{2}) = 0 \\
\hline
\end{array}$[/code]


THE USAMO list is really up. I am the first person to post this topic!!!
I make it!!!
So far the index is XXX :D

http://www.unl.edu/amc

[quote="Go Beyond"][quote="pieterminate"]where? :rotfl:[/quote]

You are fooled so many times???
This time is real, and you don't believe???[/quote]

Well.....i didn't make it anyways, so.....

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/06usamoqualstate.html

7th grade: 2
8th grade: 15
9th grade: 47 <--All of you freshmen MOP aspirers, note this number carefully!
10th grade: 98
11th grade: 89
12th grade: 116

I might be off by 1 for some but these should be correct.

Good job Randy Jia! Amazing. One of 2 7th graders.


algebra
polynomial
calculus
integration
induction
number theory solved
number theory

Let [tex] P(x) [/tex] is a polynomial with integral coefficient .  [tex] P(x) \geq 0 [/tex] with every x . [tex] a_0 = 0 , a_1 = f(a_0) , a_2 = f(a_1) ,...., a_n= f(a_{n-1}) , .... [/tex] . Prove that [tex] (a_m , a_n ) = a_{(m,n)}[/tex] with m , n are natural numbers , m is different from n .

Let $P_k = PoPo \cdots oP$ $k$ times.
Then $a_{i+k} = P_k(a_i)$ for all $i,k$.

Let $P(x) = \sum^{i=0}^q \alpha_ix^i$.

Let $m \geq n$, and $d = gcd(a_m,a_n)$. Let $m=n+k$.
We have $a_m = P_k(a_n) = a_n(\sum^{i=1}^q \alpha{}_ia{}_n^{i-1} ) + P_k(0)$
It folows that $d$ divides $P_k(0) = P_k(a_0)=a_k=a_{m-n}$.
Thus, an easy induction leads to $d$ divides $a_{gcd(m,n)}$.

Conversely, for all $i$ and all $r \geq 1$, $a_i$ divides $a_{ri}$. This follows from another easy induction because if $a_i$ divides $a_{ri}$ then it divides $P_i(a_{ri} )= a_{(r+1)i}$ (since, from above, the coefficient of $x^0$ in $P_i$ is $a_i$).
Thus $a_{gcd(m,n)}$ divides both $a_m$ and $a_n$ so that it divides $d$.
Since everything is non-negative here, we deduce that $a_{gcd(m,n)}= d$, and we are done.

Pierre.


limit
logarithms
trigonometry
calculus
calculus computations

Because

$ \lim_{x \to 0_{+}}\sqrt{x}\ln x \sin \frac{1}{x^{2}}= 0$

and

$ \lim_{x \to 0_{+}}\frac{\sin \sqrt{x}}{\sqrt{x}}= 1$


function
real analysis
real analysis unsolved

Prove the Intermediate Value Theorem using uniform continuity :)

Because I'm going through problems in an Intro to Analysis book, and I'm stuck :(

I meant: why would anyone want to use uniform continuity here? In its general form the IVT is a purely topological statement: the image of a [url=http://en.wikipedia.org/wiki/Connected_space]connected set[/url] under a continuous map is connected. Whether the map is uniformly continuous is irrelevant.

If you are interested in the proof for continuous functions on real line, it's [url=http://en.wikipedia.org/wiki/Intermediate_value_theorem]here[/url].


That's ch.14 in vol.1. Angle chasing is basically trying to find the measure of a given angle. You can use many different tools for angle chasing,i.e, angle bisector, parallel/ perpendicular lines, sum of angles in a triangle,etc. To learn angle chasing, do lots of problems and then it will come easier. :)


Can we use paper, tape, and scissors to cut out and create 3D geometry figures to make it easier to visualize? For example when I had a picture of the 1985 AIME problem #15 I was able to solve it in nearly 2 seconds.

i believe you are allowed to use any eucliedean tools... ;)

[quote="joml88"]From the front cover of the 2005 AIME:

[quote]No aids other than scratch paper, ruler, compass, and protractors are permitted.  In particular, [b][u]calculators and computers are not permitted[/u][/b][/quote][/quote]Hmmm...then it doesn't seem they ban ripping the paper...

[quote="Elemennop"]I'm not entirely sure, but cutting out figures might be viewed as distracting to the other test-takers (this happened to me when I tried to make a paper cube on the AMC10 my freshman year).[/quote]

o_0
wouldn't that have been time consuming?
I don't think its against the rules or anything, i mean, the compass's point is quite sharp and can be used as a cutting tool : |

Hahahaha well tape and scissors are technically not permitted there ;-). But I guess if your folding/compass-cutting skills were good enough, you could get by without them. Still, it doesn't seem like it would help you that much (especially if you had to take the trouble to fold it and stuff). The figures shouldn't ever be too hard to visualize.

Hehe, this is where my advanced origami skills come into play.   :P


A pitcher contains $ 4\frac{1}{2}$ cups of water. A $ \frac{1}{3}$-cup scoop is used to
transfer water from the pitcher to a bowl. How many scoops are
needed until the bowl has more water than the pitcher?

Basically, we just need to do:

$ 4\frac {1}{2} \div \frac {1}{3}$

First we change $ 4\frac {1}{2}$ into an improper fraction, to get $ \frac {9}{2}$.

Second, we take the reciprocal of the other fraction, $ \frac {3}{1}$ and multiply it by $ \frac {9}{2}$, to get:

$ \frac {9}{2} \times \frac {3}{1} \implies \frac {9 \cdot 3}{2 \cdot 1} \implies 13.5$

However, we want the bowl to have more water than the pitcher, so our answer is $ \boxed{14}$.

EDIT: Oops, I see.

@izzy, the problem asked for something different

After 6 scoops, there $ 2\frac {1}{2}$ cups of water and 2 cups in the bowl.
When we scoop again, there is $ 2\frac {1}{6}$ cup left in the pitcher and $ 2\frac {1}{3}$ cups in the bowl. At this point there is more water in the bowl than in the pitcher, so the answer is 7.

We can also used izzy's solution and said that since she rounded up the number of scoops, the answer is the ceiling of the number of scoops divided by 2, or 7.


Let a_1, a_2,...,a_1999 be non-negative real numbers satisfying the following two conditions:

(a) \sum^{1999}_{k=1} a_k = 2,

(b) a_1*a_2 + a_2*a_3 + ... + a_1998*a_1999 + a_1991*a_1 = 1

Let S = \sum^{1999}_{k=1} (a_k)^2. Find the maximum and minimum possible values of S.

Also consider the ILL problem 3 at: http://www.kalva.demon.co.uk/short/sh82.html

The following fact is very interesting:
    We have already discussed the fact that (x_1+...+x_n))^2>=4(x_1*x_2+...+x_n*x_1) for all reals x_1,...,x_n. But the sequence in the problem verifies the above relation with equality. So, we should think about the cases of equality, which I believe it's not an easy job.

I think the condition (b) is : a_1*a_2+...+a_1999*a_1=1
[img]http://www.geocities.com/duyduca1/MAth1.gif[/img]

Yes, of course. I've been thinking about this problem, but I really don't know how to deduce the case of equality.


The only thing I know is that you can put  signs around something.

How do you do basics? Like drawing a triangle. making 20 rt 4 look real. and like showing squared. and stuff like that

and labeling things

Please have a look at the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ[/url]


calculus
function
geometry
analytic geometry
integration
derivative
topology

I came upon this question while browsing the Cornell site. It is quite funny.
So do you like mathematical analysis?
If you don't and you are somewhat forced to study it, do you think that "is there life after calculus?"

Yes Analysis IS fun. And it's often intuitive.

in my opinion the correct statement would sound "some analysis is fun, but not all of it" :D

I agree with Valentin. But this is a general statement. I don't know a serious field from math that is all fun.

I don't have a completely formed, life lasting opinion about analysis. Until now I didn't like it. But my future research projects include Analytical Number Theory, so analysis will have to be one of my tools.

Which part of Calculus do you hate most ??

it is said that the most difficult part in analysis (not a part of mathematical analysis) is Real Variable Functions Analysis which was developed greatly by Lesbegue.

[quote="Valentin Vornicu"]the part about double intregrals, analitic functions, etc. :D[/quote]

Don't tell me you prefer Analytic Geometry, which is so ugly - to my opinion - that is shouldn't be allowed to be part of our beloved Maths   :D

I also hate analytic geom, and I'm a big fan of synthetic geometry, but sometimes you really can't do without coordinates and calculations and ugly stuff like that  :D

As for calculus, I like it, but I don't think my mind is built for it ( :D ). In my opinion, to do well in calculus you have to be good at manipulating lots of abstract structures, but I'm more of a visual thinker. Would you believe I make drawings for most of the calculus problems I solve ? This isn't entirely accurate, since my interests include integrals now and it's harder to make drawings which suit this sort of problems, but back in 11'th grade, when my main interest was derivatives, I drew lots of tangents to curves and things like that. Despite writing a 'standard' solution, using  \varepsilon-delta definitions and stuff, my first step would be to draw something, this kind of thinking might be harmful when I start studying more advanced things which can't be solved by looking at a drawing   :D

i agree with df/dx
i also hate that problems which u don't have to think... u have to make ONLY calculus... a lot of calculus

For df/dx or d 2 f/d 2 x:
What's intuitive about this:
find a primitive for  \sqrt (2+cos(x))?

[quote="df/dx"]
Don't tell me you prefer Analytic Geometry, which is so ugly - to my opinion - that is shouldn't be allowed to be part of our beloved Maths   :D[/quote]
First of all I think that you ment Differential Geometry, cuz Analytic Geometry is in fact algebra. I don't like both of them.

as amfulger pointed out, analysis can be a real pain in the ass :) it most certainly has its beauty, but I for one tend to see its uglyness :)

It might look intuitive at the beginning, but when you get to more advanced stuff I think it loses all its intuitiveness (is that a word ? :D ).

So everybody seems to hate Analysis  :(  ...

Not me  :D !

it's not that we hate it. we just don't like most of it. :)
and now that you've mentioned it, who are you? :D

Yes. Analytic function seems to be appeared in Complex Variable number theory.

I don't like tensor analysis....

I think you all have a misinterpretation of "analysis" and what it really is. There are many beautiful and unintuitive things in analysis. The invention of calculus is one of the great inventions of mankind!! It forever changed the world and opened up whole new worlds of mathematics.

Problems in analysis can range from everything from foundational - the study of infinite sets and the construction of the real numbers (a truly beautiful subject). To the theory of measure and integration (which lays the foundation for proper understanding of probability). To the absolute gorgeousness of complex analysis (perhaps the most beautiful of all subjects). To fourier analysis (which can be used to solve a surprising range of problems from physics to number theory). To differential geometry and differential topology (deceptively subtle and powerful subjects).

I don't think it is good to cordon off any areas of mathematics as "good" or "bad", because a great mathematician should be able to recognize connections between seemingly separate areas and integrate them (no pun intended).

[quote="gauss202"]There are many beautiful and unintuitive things in analysis.[/quote]

I agree, how intuitive is it that the p-adic rationals (and integers) are homeomorphic to the Cantor sets?

[quote="gauss202"]I think you all have a misinterpretation of "analysis" [/quote]
Don't think I do.
For me not liking anlysis just means I will not chose it as a research field. You can't do research in all mathematics. You chose a NT problem. You realize you need some analysis. But it is just a tool for you. You apply a statement that fits the problem conditions. You don't even have to know a proof for what you use. You even might not have known what you have used until you grabbed an analysis book and found ouy what you where looking for (sound like a U2 song).
I saw a proof for Dirichlet's theorem (the one about the distribution of primes) that uses only real analysis. I was able of understanding it and I am not good in analysis (I don't solve very hard problemsin analysis).

Analysis was created for engineering purposes. They wanted to measure areas, so they invented the integrals.

[quote="amfulger"]
Analysis was created for engineering purposes. They wanted to measure areas, so they invented the integrals.[/quote]

No this is not true. I mean, Newton was inspired by physical problems in his discovery of calculus but that is not all it is for! Look at Cantor and Riemann, would you consider them "egineers"?

Analysis is really the study of the infinite, to put it broadly. Anything that involves infinite sets could be said to fit under the realm of analysis. That covers a tremendously broad range of things. It is not enough to just know the results of theorems in analysis. The more deeply you understand them, the greater your insight into mathematics as a whole. Math is not separated into boxes of different fields. It is tied together.

Look at Alaine Connes, the great french mathematician, for instance. He has proposed some very interesting lines of attack on the Riemann Hypothesis that are rooted in deep ideas from analysis. He could not have come up with these ideas without very good understanding and subtle insights into why things were true in analysis. And this is becoming increasingly true in Number Theory as a whole.

I'm not saying that all of analysis interest me, I'm just saying that there is more to the field than just mundane integrations and obscure inequalities.

[quote="gauss202"][quote="amfulger"]
Analysis was created for engineering purposes. They wanted to measure areas, so they invented the integrals.[/quote]
No this is not true. I mean, Newton was inspired by physical problems in his discovery of calculus but that is not all it is for! Look at Cantor and Riemann, would you consider them "egineers"?
I'm not saying that all of analysis interest me, I'm just saying that there is more to the field than just mundane integrations and obscure inequalities.[/quote]

I only wanted to say analysis started as a computational tool.
I know that there is more to this field than just computations. I verry much like the study of cardinal numbers.
I just say that I can't be an analyst because most of the analysis does not suite my representative way of thinking. I know mathematics is a whole. I've heard that until now it is impossible to avoid calculus in any proof for the Fundamental Theorem of Algebra.

analysis can nearly be seen everywhere in mathematics.
many things referring to functions are met with analysis.

proofs of the fundamental theorem of algebra can be roughly divided into three categories. first there are the topological proofs, which are vased on topological considerations such as the winding number of a curve in R^2 around 0. then, there are analytical proofs, which are related to Liouville's theorem: a nonconstant entire function on C is unbounded. finally, there are the algebraic proofs and these proofs use only the fact that every polynomial of odd degree with real coefficients has a real root and that every complex number has a square root. and recently, a proof based on linear algebra is published. the paper is 'The Fundamental Theorem of Algebra and Linear Algebra', MAA, Monthly 110.

But even all of those proofs rely on analysis on a subtle level. The Fundamental Theorem of Algebra is an analysis result. It relies essentially on the fact that the real numbers have no "gaps" in them, i.e. that they have the least upper bound property. This is, for example, what you need to prove that every odd degree polynomial has a real root. All proofs of the FTA rely on some result of that kind. A good discussion of the Fundamental Theorem of Algebra can be found in a book called [u]Numbers[/u] by Ebbinghaus, et. al. It is a very fun read.

Analysis is useful for differential equation wich is use in economy.
analysis is useful for probability modelisation of stock exchange (bank...)
Also in applied mathematics for planes, rockets, ...
One day I have seen a book of Debreu nobel prize of economy, in its book there was a lot topology, metric space, point fixed theorem.
John Nashe used analysis
Yeah could have great  life after calculus.

I think that analysis being developed as a tool fpr physics in the beginning has no importance, and I dare say that it would have been discovered and used in all those fields anyway. This is because people are just too complex to rely only on finite quantities. We were bound to tackle infinity-related problems sooner or later, and it just so happenes that the man who first used analysis was a physicist and discovered it for purposes related to physics.

It's not entirely true that calculus was "invented" by any one man. It really arose from a lot of contributions over a long period of time. Newton and Leibniz just unified a lot of these concepts and were the first to be able to solve a wide range of problems with them. Nor is it true that they used calculus only for physics problems. They were both concerned with Calculus from a pure mathematical standpoint.

There is a false perception of calculus from high school students that it's somehow the "end" of mathematics. It may be the end of high school mathematics, but it's just the beginning of what you might call modern mathematics. If you imagine all of the math topics that you study in high school - you really have only studied topics that were developed before the 18th century. In University you cover mainly topics developed during the 18th and 19th century, and maybe a few topics from the 20th century. Then, only when you get to graduate school, do you start to study really modern topics from the last century.

So you can think of calculus as your first taste of mathematics from the 18th century and beyond. Hardly an "end" to mathematics.

You know, I don't think "invented" is a good word. There's a difference between invented and discovered things. My opinion is that inventions are those things which probablt wouldn't have come to existence if it weren't for the inventors who first thought of them.

On the other hand, I think calculus is something we have discovered: someone would have found it sooner or later because, as I said in my previous post, it's only natural for beings as complex as we are to be concerned with the concept of infinity.

Even the fact that both Newton and Leibniz found nearly the same results in roughly the same period of time shows us that calculus was bound to "happen".

I don't think there is such a thing as "the end of mathematics", or "the end of physics" or anything like that. We' way too far from a comprehensive truth which explains what we see around us, and until we reach something like that, all sciences will/must continue to "grow" and develop. We may never reach such a unifying truth, but we can try getting closer and closer (asymptotically :D).

[quote="Anonymous"]In my opinion it doesn't exist life after calculus, and math is so great, it makes you 'think' :D ....and may sometimes be fun, or sometimes can be hard, but always will be interesting :)[/quote]

exactly my opinoin (kind of)... :D

df/dx, you forgot to answer a question asked by a very important person: our admin Valentin. He asked: WHO ARE YOU? Pls respond.

[quote="darij grinberg"]I just wanted to say I have nothing against calculus, but I have something against application of calculus to elementary theorems (especially when this application is unnecessary)...

  dg[/quote]

But why is that? In some cases, calculus may provide a more elegant proof of a theorem. For example, the Prime Number Theorem has a proof using only "elementary" mathematics, but it has a more elegant proof using Calculus.

Plus, many facts that are taken for granted in elementary mathematics are really built on a hidden calculus foundation. For example, the definition of e and natural logarithms, and the area of a circle.

I don't think that such a distinction should be made between calculus and the rest of mathematics; no more than a distinction should be made between geometry and algebra. The only reason it is done for competition purposes is to make ensure that results aren't needed that some high school students wouldn't have been exposed to yet. But there is nothing impure, or inelegant about calculus.

[quote="gauss202"]But why is that? In some cases, calculus may provide a more elegant proof of a theorem. For example, the Prime Number Theorem has a proof using only "elementary" mathematics, but it has a more elegant proof using Calculus.[/quote]

Well, I don't know what you mean by "prime number theorem" (if it is about density, then its formulation already uses calculus, since it depends on the notion of a limit). Yes, calculus often simplifies a proof or shows the "real" reason behind a fact. But calculus is different from elementary mathematics, it uses more axioms and often approaches things with very different arguments. So why don't also search for an elementary, non-calculus proof? Any alternative proof of a theorem already proven can be useful for the development of mathematics. Especially, there are lots of cases when calculus provides a straightforward but ugly proof of something (think about the use of differentiation to kill inequalities!).

[quote="gauss202"]Plus, many facts that are taken for granted in elementary mathematics are really built on a hidden calculus foundation. For example, the definition of e and natural logarithms, and the area of a circle.[/quote]

As for pi and e, I don't call them elementary mathematics. In my opinion, for instance, elementary geometry is the part of geometry where [b]R[/b] can be replaced by the set of all algebraic numbers. I know well that there are parts of calculus "hidden" in the high school curriculum, but the curriculum is not my criterion of distinction between elementary and advanced mathematics.

  Darij

Maybe but idk and irdc calc

Xdddddddddd


please u have to know where to pose each exercices.be carful
$a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ca)$ .then use Vieta's formula.


calculus
integration
modular arithmetic
symmetry

Suppose $ xz\minus{}2yt\equal{}3$ and $ xt\plus{}yz\equal{}1$, find the integral solutions to $ (x,y,t,z)$.

first, note that $ xz \equal{} 3\plus{}2yt$ so $ xz \equiv 1 \pmod{2}$, which tells us that both $ x$ and $ y$ are odd.  From the second equation, we can also see that $ gcd(x,y) \equal{} gcd(x,z) \equal{} gcd(y,t) \equal{} gcd(z,t) \equal{} 1$.

stuck.

Big Hint:
[hide]$ (xz \minus{} 2yt)^2 \plus{} 2(xt \plus{} yz)^2 \equal{} (x^2 \plus{} 2y^2)(z^2 \plus{} 2t^2)$.[/hide]

ok so $ 11 \equal{} (x^2\plus{}2y^2)(z^2\plus{}2t^2)$, and we can only factor $ 11$ as $ 1$ and $ 11$, so let, WLOG (b/c of symmetry)
$ x^2\plus{}2y^2 \equal{} 1$ which implies $ x\equal{}1$ and $ y\equal{}0$

which in our original equations gives $ xt\plus{}yz \equal{} 1 \implies t\equal{}1$ and $ xz\minus{}2yt\equal{}3 \implies z\equal{}3$

So our solutions are $ (1,0,1,3)$ and $ (3,1,0,1)$


The vertices of triangle ABC are lattice points and there is no smaller triangle similar to triangle BAC with its vertices at lattice points. Prove that the circumcenter of ABC is not a lattice point.


[quote="mentalgenius"]Isn't the answer 20?[/quote]

Yep, an ancient case of "answer the question that's asked".
Note that you could also replace the word "consecutive" with "distinct", and you'd still get the same answer by using an anti-greedy algorithm starting with $1$ and packing in the smallest positive odd integers that you can.

[hide="Solution"]
Q: What is the maximum number of consecutive odd positive integers that can be added together before the sum exceeds 401?
A: The formula for the sum of the first n perfect squares is $n^2$. $20^2$ is 400, so the sum of the first 20 odd numbers is 400. For the first 21 odd numbers, the sum is 441, which is greater than 401. Therefore, the maximum number of consecutive odd positive numbers is $20$.
[/hide]


Prove that there are infinitely many twin prime pairs iff there are infinitely many positive integers $ n$ that cannot be written in the form $ n \equal{} 6|ab|\plus{}a\plus{}b$ for any choice of integers $ a$ and $ b$.


abstract algebra
algebra
polynomial
analytic geometry
Ring Theory
superior algebra
superior algebra unsolved

Let A = k[x,y,z] / (xy - z^2)  (k is a field).

Let X, Y, Z denote the images of x, y, z in A. Show that p = (X, Z) is a prime ideal of A.

Now, I want to show that A / p is an ID. It looks like A / p is isomorphic to k[y].

...not 100% sure how to show that.

Can we consider a homomorphism from A to k[y] by sending X to 0 and Z to 0?? it's kernel is p = (X, Z) and it's onto...

your approach is correct. but to avoid a nasty calculation with elements to show that the kernel equals p, use the universal property of polynomial algebras and quotient algebras:

for every $ k$-algebra $ B$, we have natural bijections

$ Hom(A/p , B) \equal{} \{f \in Hom(A,B) : f(X) \equal{} f(Z) \equal{} 0\}$
$ \equal{} \{f \in Hom(k[x,y,z],B) : f(xy \minus{} z^2) \equal{} f(x) \equal{} f(z) \equal{} 0\}$
$ \equal{} \{f \in Hom(k[x,y,z],B) : f(x) \equal{} f(z) \equal{} 0\}$
$ \equal{} \{(a,b,c) \in B^3 : a \equal{} c \equal{} 0\} \equal{} B \equal{} Hom(k[y],B)$,

so the yoneda-lemma implies $ A/p \cong k[y]$. if you don't want to use the yoneda-lemma, you have to repeat its proof here (which is done very often, I wonder why ..): verify with the universal properties, that $ A \to k[y], X \to 0, Z \to 0, Y \mapsto y$ yields a well-defined $ k$-algebra homomorphism. the kernel contains $ p$, thus we get $ A/p \to k[y]$. now define $ k[y] \to A/p$ by $ y \mapsto Y$. now it's clear that $ k[y] \to A/p \to k[y]$ and $ A/p \to k[y] \to A/p$ are identity homomorphism for this can be checked on the variables $ y$ resp. $ X,Y,Z$.

as you can see, the proof with the yoneda-lemma uses exactly the same arguments, but it's more easy. besides it's very geometric: you intersect the variety $ xy \minus{} z^2 \equal{} 0$ with the variety $ x \equal{} z \equal{} 0$ and then you just get the affine line with coordinate $ y$. with this approach you can determine quotients of finitely generated $ k$-algebras without really thinking about it.  :)

hm, that's slick.....didn't actually know about yoneda lemma. Thanks!

One could also apply the fact that quotients "commute." Let $R$ be ring and $I_0,I_1,\dots\subseteq R$ be (a perhaps infinite set of) ideals. Take $\mathbb N$ to be the natural number object, i.e. the category generated by $$\textbf{0}\overset{f_0}{\longrightarrow} \textbf{1}\overset{f_1}{\longrightarrow} \textbf{2}\overset{f_2}{\longrightarrow}\cdots$$ Then the diagram $\mathcal Q:\mathbb N\to \text{RING}$ with $$\mathcal Q(\textbf{0})=R$$ $$\mathcal Q(\textbf{n+1})=\mathcal Q(\textbf{n})/\mathcal Q(f_{n-1})\circ \cdots \circ \mathcal Q(f_{0})``(I_n)$$ $$\text{And } \mathcal{Q}(f_n):\mathcal Q(\textbf{n})\to \mathcal Q(\textbf{n+1})\text{ the induced quotient morphism}$$ Has colimit $R/(I_1+I_2+\cdots)$. In the original problem, take $R=k[x,y,z]$, $I_0=(xy-z^2)$, and $I_1=I_2=\cdots=(x,z)$. Functor $\mathcal Q$ has image $$k[x,y,z]\overset{\eta_0}{\longrightarrow} k[x,y,z]/(xy-z^2)\overset{\eta_1}{\longrightarrow} k[x,y,z]/(xy-z^2)/\eta_0``((x,z))\overset{\text{id}}{\longrightarrow}k[x,y,z]/(xy-z^2)/\eta_0``((x,z))\overset{\text{id}}{\longrightarrow}\cdots$$ Which is equivalent to the diagram $$k[x,y,z]\overset{\eta_0}{\longrightarrow} k[x,y,z]/(xy-z^2)\overset{\eta_1}{\longrightarrow} k[x,y,z]/(xy-z^2)/\eta_0``((x,z))$$ And thus has colimit $A/p=k[x,y,z]/(xy-z^2)/\eta_0``((x,z))$ for purely categorical reasons. Appealing, however, to the above result, $\text{colim}(\mathcal Q)$ may also be written as $k[x,y,z]/(xy-z^2,x,z)= k[x,y,z]/(x,z)\cong k[y]$, and it follows that $$A/p\cong \text{colim}(\mathcal Q)\cong k[y]$$ At which point we may conclude that $p\subseteq A$ is prime, as noted in the OP $\blacksquare$


An ellipse is inscribed in triangle $ABC$. From a focus $F$, lines are drawn to vertices $A$, $B$, $C$ and to tangent point $D$ on side $AB$. Prove that $\angle AFD + \angle BFC = 180^{\circ}$

a hint woud be nice, or a proof.

[hide]By Poncelet we have:
$\angle AFD = \angle AFG = \alpha$
$\angle CFD = \angle CFE = \beta$
$\angle BFE = \angle BFG = \gamma$

Then...

$2\alpha + 2\beta + 2\gamma = 360^{\circ}$
$\alpha + \beta + \gamma = 180^{\circ}$

But...

$\angle BFC = \angle BFE + \angle CFE = \gamma + \beta$
and
$\angle AFD = \alpha$

Hence...

$\angle AFD + \angle BFC = \alpha + \beta + \gamma = 180^{\circ}$[/hide]


Let $ a,b,c\geq0$ s.t. $ abc\equal{}1$.Prove that :
$ (\dfrac{a\plus{}b\plus{}c}{3})^5\geq\dfrac{a^2\plus{}b^2\plus{}c^2}{3}$

@@ .So old
$ 3abc(a^2\plus{}b^2\plus{}c^2)\equal{}\frac{3abc(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)}{a\plus{}b\plus{}c} \le \frac{(ab\plus{}bc\plus{}ca)^2(a^2\plus{}b^2\plus{}c^2)}{a\plus{}b\plus{}c} \le (a\plus{}b\plus{}c)^5/27$
Done ;)


If the sum of all solutions to the equation $ (x^{log 3})^{2}\minus{}3^{log x}\minus{}2\equal{}0$ is $ a^{log_{b}c}$, find $ a\plus{}b\plus{}c$, where $ b$ and $ c$ are prime numbers and $ a$, $ b$, and $ c$ are all positive integers.

Since $ \log{(x^{\log 3})}\equal{}\log 3\log x\equal{}\log{(3^{\log x})}$, we see that $ x^{\log 3}\equal{}3^{\log x}.$ So you're equation becomes $ (3^{\log x})^2\minus{}3^{\log x}\minus{}2\equal{}0$. Solve this to obtain $ 3^{\log x}\equal{}2$ or $ \minus{}1$. $ \minus{}1$ is extraneous since $ 3^{\log x}$ is positive, so it must be $ 3^{\log x}\equal{}2$. Solve for $ x$ to get $ x\equal{}10^{\log_32}$. This is the same as $ 2^{\log_3{10}}$, but the problem says $ b$ and $ c$ are prime, and either way, the answer to you're problem is $ 15$.


geometry
3D geometry
algebra
polynomial
arithmetic sequence
More Sequences

[quote="Peter"]Show that if an infinite arithmetic progression of positive integers contains a square and a cube, it must contain a sixth power.[/quote]
http://www.kalva.demon.co.uk/short/soln/sh9715.html

Let $ r,s$ be coprime. If an arithmetic progression contains a $ r$-th power and a $ s$-th power, then it contains a $ rs$-th power.

Proof:
This is equivalent to: Let $ a,d$ be integers. If there are integers $ x,y$ such that $ a \equiv x^r \equiv y^s \mod d$, then there is an integer $ z$ such that $ a \equiv z^{rs} \mod d$.

First observation: we can assume that $ d\equal{}p^q$ is a prime power, $ p$ prime.
Indeed, by the chinese remainder theorem, an integer polynomial equation can be solved $ \mod n$ iff it can be solved $ \mod$ all prime powers dividing $ n$. Here, the equations are $ x^r\minus{}a , y^s\minus{}a, z^{rs}\minus{}a$.

Second observation: let $ d\equal{}p^q$ as said, then we can assume $ \gcd(a,p^q)\equal{}1$.
Indeed, if $ p|a$, we write $ x\equal{} p^u x', y\equal{} p^v y'$ with $ x', y'$ both not divisible by $ p$. Then $ x^r \equiv y^s \mod p^q$ gives $ a \equiv p^{ur} x'^r \equiv p^{vs} y'^s \mod p^q$.
Thus either $ a \equiv 0 \mod p^q$, in this case we can take $ z\equal{}0$, or we get that $ vs\equal{}v_p(x^r) \equal{} v_p(a) \equal{} v_p(y^s) \equal{} ur <q$. This gives that $ r|v$, thus $ vs\equal{}rsw$ for some $ w$, giving that we simply want to solve $ z'^{rs} \equiv \frac{a}{p^{rsw}} \mod p^{q\minus{}rsw}$ with $ \frac{a}{p^{rsw}}$ coprime to $ p$.

Third observation: the problem's statement is true if $ a$ is coprime to $ d$.
Indeed, if $ x^r \equiv a \equiv y^s \mod p^s$ and $ p \nmid x,y$, we write $ 1\equal{}ur\plus{}vs$ for some integers $ r,s$. Then we take $ z\equal{}x^v y^u$ because then $ z^{rs} \equal{} x^{vrs} y^{urs} \equiv a^{vs\plus{}ur} \equal{} a \mod d$.

Putting all together, we solved the problem.


Let  $f$ be a  function which satifies \[  f(x)+f\left( \frac 1{\sqrt[3]{1-x^3}}\right)=x^3 \]  for every $x \neq 1$  real numbers.
Find  $f(-1)$.

There's a hint:

[hide]
Put $x=-1,x=\frac{1}{\sqrt[3]{2}},x=\sqrt[3]{2}$.
You will have 3 equations and three variables.
[/hide]


calculus
derivative
function
analytic geometry
graphing lines
slope
limit

1. If you see a graph of a function, and you are asked to draw the derivative for this function, then how can you draw it? I know you have to find some points, and find their slopes, and then put it on the graph and connect the points, but how can I estimate the slopes. I can only figure out the slope of 0 and positive slope or negative slope. Other than that, I can't tell the slope. Hope you can tell me.
2. lim h-->0  [(1+h)^4-1]/h   How to find the answer? my teacher said since it looked like the derivative form lim f(a+h)-f(a)/h, but then I didn't know how to do it.
Thanks for help.

2)

You mean, $ \lim_{h\to 0}\frac{(1+h)^{4}-1}{h}$?

If so, let $ f(x) = x^{4}$. By definition we have that
\begin{eqnarray*}f^{\prime}(a) & = &\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\\ & = &\lim_{h\to 0}\frac{(a+h)^{4}-a}{h}.\end{eqnarray*}

Now, let $ a=1$  and we are done, aren't we?

0 slopes and positive/negative slopes are a good start. Let's call the original function whose graph is given $ f(x)$. First plot the points where the derivative is 0 (that is, where $ f^{\prime}(x)\equal{}0$). Draw a few tangent lines on the graph of $ f(x)$ (eyeball them) and guess the values of the slopes at these points (you know what slopes of $ \pm 1,\pm 2$ look like, so compare and guess,) then plot these points. Also you can tell if the derivative, $ f^{\prime}(x)$, is increasing or decreasing by checking the concavity of $ f(x)$ (by looking at the graph):

$ f(x)$ is concave up => $ f^{\prime}(x)$ is increasing ,  $ f(x)$ is concave down => $ f^{\prime}(x)$ is decreasing

On that note, the inflection points of $ f(x)$ (a point where the concavity of $ f(x)$ changes; such points occur where $ f^{\prime\prime}(x)\equal{}0$) are the critical points (critical point = a maximum, minimum, or saddle point) of $ f^{\prime}(x)$, in particular:

If $ x_{0}$ is an inflection point of $ f(x)$, and

[b]i.[/b] $ f(x)$ changes from concave up to concave down at $ x_{0}$, then $ f^{\prime}(x)$ has a max at $ x_{0}$

[b]ii.[/b] $ f(x)$ changes from concave down to concave up at $ x_{0}$, then $ f^{\prime}(x)$ has a min at $ x_{0}$

[b]iii.[/b] $ f(x)$ does not change concavity at $ x_{0}$ (even though $ f^{\prime\prime}(x_{0})\equal{}0$), then $ f^{\prime}(x)$ has a saddle point at $ x_{0}$

Note: An example of case [b]iii[/b] is $ f(x)\equal{}x^{4}$ with $ x_{0}\equal{}0$.

This will only give you the locations of the extrema/saddle points of $ f^{\prime}(x)$, for the values of $ f^{\prime}(x)$ at these points you will need to draw the tangent lines and guess the slopes (as above).


probability
analytic geometry
geometry
trapezoid
inequalities
graphing lines
slope

(a)  Two mathematicians take a lunch break seperately sometime between 12:00 and 1:00 and stay for $\frac{m}{n}$ minutes.  If the probability of them meeting each other is $.4$, compute $m+n$.

(b)  Three mathematicians take a lunch break seperately sometime between 12:00 and 1:00 and stay for $\frac{m}{n}$ minutes.  If the probability that two of them meet but the third doesn't meet with either of them is $.4$, compute $m+n$.

[quote="miyomiyo"][quote="cincodemayo5590"][hide="a"]
Say the first person enters at time $x$ and the second at time $y$, where $0 \leq x,y \leq 60$ is the amount of time elapsed since 12:00. Then, in order to have the mathematicians meeting each other, we need that
\[|x-y| \leq \frac{m}{n}=t \]
Graphing this on a coordinate axes, we see that it divides the square of side $60$ (i.e. the time frame of the problem) into two congruent isosceles triangles of side length $60-t$ and a trapezoid. We want that the area of the trapezoid to be $\frac{2}{5}\cdot 3600 = 1440$. This means that:
\[3600-2 \cdot (\frac{1}{2}\cdot (60-t)^{2}) =1440 \implies \boxed{ t= 60-12 \sqrt{5}}\]
But that doesn't fit with the answer format... Did I interpret the problem wrong?  Must each mathematician stay for the whole $t$ minutes [u]within[/u] the time frame? [/hide]

[hide="b"] Similar method. Take the inequality for the first two, then subtract the intersection of the inequalities with each of the first two and the third. Can we do Shoelace in 3-D ?   :D  [/hide][/quote]

I got something else for (a):

[hide="Possibly wrong solution"]
P(t) where t is time spent and P is probability of meeting is LINEAR, is it not?  Whic means that you can pick 2 times such as 10 and 20 to get the slope and then you can solve for .4.  This was my original thought, but now I think I'm wrong.  Is P(t) linear?
[/hide][/quote]

I think it's quadratic, since it's based on the area of the probability.


calculus
integration
trigonometry
calculus computations

Calculate

\[\int^{\frac{\pi}{2}}_{0}2x\sin ^2 x dx\]

Use $2\sin^2x=1-\cos (2x)$ and integration by parts


geometry
ratio
search
MATHCOUNTS
analytic geometry

I need some help on this first on, but I want to check if I got the second one. :yup:

23. A right triangle has side lengths 3,4 and 5 units. Another
triangle has sides of length 4, x and y units and has the same
area as the right triangle. If x<y, what is the smallest possible
value of y? Express your answer in simplest radical form. :what?:  :wacko:

On this one, I wanted to make sure... :heli:
 [geogebra]5ccfe7e98135c196ddb4797323876e83edb10480[/geogebra]
If the triangle ABC is a right triangle, the length of AD is 2, DB is 2, BE is 3, EC is 4, and find the area of triangle AFC.

I first solved for point F by finding the intersection of line AE and and line DC. The two equations would be:

y=-4/3 x+4 and y=-2/7 x+2

We subtract them to get 22/21 x=2
So x=21/11.  :arrow:
Now we can find the area of triangle ADF and triangle BCD and subtract that by the area of triangle ABC and we will have the area of triangle AFC, right?

We have 14-7-21/11=56/11. Is that how you solve it? :?:

For number 23, did you get [hide="this answer"]$ \sqrt{13}$[/hide]?

Hint for second question:
[hide="hint"]There are three ways to solve this problem: equation bash (the method you're using), area ratios by connecting line BF, and mass points bash.[/hide]

That's supposed to be the answer, but how do you solve for it?
 And for that 2nd question, what do you mean?!?

I recently answer #23 on the 2008 sprint questions thread a few threads below this. Try to just search.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1487198#1487198]Here it is[/url].

I believe the second was #10 on the 2008 nats team round? That's right.

Sorry for the short reply to your topic, CubeX.

I simply posted an answer for number 23 as I assumed you had already solved it. If not, remember to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1308079057&t=274775]search[/url] the forum before posting new questions.

For the second question, your solution works, and my answer agrees with yours. It is possible to find a solution (two, actually), that do not require analytic geometry; see if you can find one! :)

are you sure you worded #23 right? because it says $ x<y$ not $ x\le y$ :maybe:

Well, the original question said $ \le$, so I assumed you meant that. If the question asks for $ x < y$, then the answer would technically be $ \sqrt {13} \plus{} \epsilon$ where $ \epsilon$ is a small number greater than zero. Clearly such an answer is not something you'd expect for a MATHCOUNTS problem. :wink:

EDIT: Yeah, I guess I'm not really familiar with this stuff; pythag011 is probably right. :blush:

[quote="ThinkFlow"]Well, the original question said $ \le$, so I assumed you meant that. If the question asks for $ x < y$, then the answer would technically be $ \sqrt {13} \plus{} \epsilon$ where $ \epsilon$ is a small number greater than zero. Clearly such an answer is not something you'd expect for a MATHCOUNTS problem. :wink:[/quote]

Actually, the minimum would not exist.


1) Outline reasonable synthesis of o-chlorobromobenzene and m-chlorobromobenzene.

2) Identify compounds A and B in the scheme bellow:

$ m\minus{}chlorobromobenzene\,\, \plus{}\,\, Mg/THF \longrightarrow A$

$ A \plus{} CO_2/then\, H^\plus{} \longrightarrow B$

3) Identify compounds C, D, and E in the following scheme:

$ o\minus{}chlorobromobenzene\,\, \plus{} \,\, Mg/THF \longrightarrow C$

$ C \plus{} CH_2\equal{}CHCH_2Br \longrightarrow D$

$ D \plus{} Mg/Et_2O/then\, H_2O \longrightarrow E$

Properties of compound E: molecular formula $ C_9H_{10}$; gives a positive test with Br2/CCl4; when treated with hot, alcaline potassium permanganate, followed by acidification, phthalic acid is obtained.

Provide a mechanistic explanation for the formation of compound E.

i think A is metachloro magnesium bromide and B is metachloro benzoic acid.

this is because when compared to C-Cl bond C-Br bond is more weak.
so it forms the grignard with that. then carbondioxide replaces the C-(MgBr) bond to form the acid.

That is correct. How about parts 1) and 3)?

treat Benzene with Nitration mixture(2 moles) to get dinitro benzene(one NO2 is attached to the meta position). Reduce using Ammonium sulphide.this reduces one of the nitro groups. Diazotise the amine group and use CuCl /HCl to getthe chlorine at one of the 2 positions.
Then separately reduce the second group,diazotise and use CuBr/HBr to get the bromo substituent. i think this produces meta chloro bromo benzene with good yield.

Is this correct? :wink: ?????????????????

[quote="valeriummaximum"]First Nitrate benzene.Do chlorination with chlorine(1 mole quatitatively) in the presence of FeCl3 to get  meta chloro nitro benzene. treat this with acetic anhydride in the presence of pyridine to get the acetanilide. now brominate to get the Br group in the para position of the amino group. Now hydrolyse,to get back the amino group. Now diazotise with nitrous acid prepared in situ to get the diazonium chloride.Then treat with hypophosphorous acid/ethanol to get o-chloro bromobenzene..
IS THIS CORRECT  :wink: ????????[/quote]
i have a doubt first in this step how did you get -NH2 first of all you still have only -NO2 and then tell me a valid reason for you to do all this for preparing ortho chloro bromo benzene why cant you just do friedel crafts halogenation of bromine first and then protect the para postion using -SO3H and the do 1EQ of friedelcraft chlorination and there is it ???

[quote="srinitrs"]
i have a doubt first in this step how did you get -NH2 first of all you still have only -NO2 and then tell me a valid reason for you to do all this for preparing ortho chloro bromo benzene why cant you just do friedel crafts halogenation of bromine first and then protect the para postion using -SO3H and the do 1EQ of friedelcraft chlorination and there is it ???[/quote]

Well wouldnt the yield be lesser even if it did attach to the meta posn ?( as ortho is still a viable product )

i am sorry.....
first chlorinateand then reduce and then perform the above said reactions.................

The compound E is 3-phenyl prop-1-ene.
This is because when o-chlore bromo benzene is treated with Mg/THF,o-chloro Phenyl magnesium bromide.So the compound D is [size=150]o-chloro Phenyl magnesium bromide[/size]

[size=100]In the second step,Allyl carbocation is generated.this attaches itself to the phenyl group giving out MgBr2. [size=150]The compound E is[/size] (2-chloro)3-phenyl prop-1-ene[/size]

The third step is for eliminating the chlorine atom. Aryl Magnesium chloride is generated.this on hydrolysis gives you E which is nothing but [size=150]3-phenyl prop-1-ene.[/size]

Is this correct :wink: ?????????????????

[quote="Valeriummaximum"]treat Benzene with Nitration mixture(2 moles) to get dinitro benzene(one NO2 is attached to the meta position).(...) this produces meta chloro bromo benzene with good yield.[/quote]

Correct!

[quote="Valeriummaximum"]treat this with acetic anhydride in the presence of pyridine to get the acetanilide.[/quote]

How can that be possible? Unless, of course, you reduce first the nitro to amine. The remaining of the synthesis seems ok.

[quote="Srinitrs"]to do all this for preparing ortho chloro bromo benzene why cant you just do friedel crafts halogenation of bromine first and then protect the para postion using -SO3H and the do 1EQ of friedelcraft chlorination and there is it ???[/quote]

That is indeed a much better synthesis.

[quote="Valeriummaximum"]The compound E is 3-phenyl prop-1-ene.
This is because when o-chlore bromo benzene is treated with Mg/THF,o-chloro Phenyl magnesium bromide.So the compound D is o-chloro Phenyl magnesium bromide

In the second step,Allyl carbocation is generated.this attaches itself to the phenyl group giving out MgBr2. [size=150]The compound E is (2-chloro)3-phenyl prop-1-ene[/size]

The third step is for eliminating the chlorine atom. Aryl Magnesium chloride is generated.this on hydrolysis gives you E which is nothing but 3-phenyl prop-1-ene.

Is this correct [/quote]

That is all wrong.

Can you tell me the correct answer for the 3rd question????

Carcul -I think i am right in the 3rd answer..........

Refer to my file for further details.......

IF not, Then tell me where my mistake is :( ???

[quote="Valeriummaximum"]Can you tell me the correct answer for the 3rd question[/quote]

No. Start by asking yourself who is compound C, because that is the key to the answer.

[quote="Valeriummaximum"]Carcul -I think i am right in the 3rd answer[/quote]

I am sure you are wrong. Also, what does the Kumada coupling have to do with part 3? Do you see any transition metal salt in my scheme above?

I think compound c is a grignard...
C is o-chloro phenyl magnesium bromide..
D is 2 chloro 3-phenyl prop-1-ene
E is 3-phenyl prop-1-ene

[quote="Valeriummaximum"]I think compound c is a grignard...
C is o-chloro phenyl magnesium bromide..[/quote]

No, that's not correct. That's way your answers for D and E are also wrong.

ANSWERS
[hide]
[b][i]
C-BENZYNE
D-BROMO SUBSTITUTED INDENE
E-INDENE[/i][/b][/hide]

Hey chemrock great man i never thought of a diels alder reaction....
good ....
That was very nice.........
Nice question Carcul

[quote="Valeriummaximum"]is compound c a benzyne[/quote]

Yes.

[quote="Chemrock"]D-BROMO SUBSTITUTED INDENE
E-INDENE [/quote]

Unfortunately, your answer is wrong. Did you even noticed that the molecular formula of indene doesn't match the one I gave for compound E? Also, your answer lacks a mechanistic explanation.

[quote="Valeriummaximum"]Hey chemrock great man i never thought of a diels alder reaction....[/quote]

How can there be a Diels-Alder reaction if there is no diene?


inequalities
inequalities proposed
algebra
highschoolmath

$ a,b,c>0$ prove:

$ \sqrt[3]{\frac{a}{b\plus{}c}}\plus{}\sqrt[3]{\frac{b}{a\plus{}c}}\plus{}\sqrt[3]{\frac{c}{a\plus{}b}}>2$

[quote="zaya_yc"]$ a,b,c > 0$ prove:

$ \sqrt [3]{\frac {a}{b \plus{} c}} \plus{} \sqrt [3]{\frac {b}{a \plus{} c}} \plus{} \sqrt [3]{\frac {c}{a \plus{} b}} > 2$[/quote]

see here
http://www.mathlinks.ro/viewtopic.php?t=301485

[b]Inequalities Marathon: Problem 428[/b] : For nonnegative numbers a, b and c prove
$ \sqrt\frac{a}{b+c}+\sqrt\frac{b}{c+a}+\sqrt\frac{c}{a+b} \ge 2 $
[b]Remark on problem 428[/b]With same conditions$\sum_{cyc}\sqrt[n]{\frac{a}{b+c}} \ge 2 $ $(n \ge 2 )$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=299899&start=1140
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=397993&p=3476910#p3476910

[url=http://www.artofproblemsolving.com/community/c6h299899p2129359]Remark on problem 428[/url]

[hide=Thanks.][quote=arqady][quote=sqing]Let $a,b,c$ be  non negative real number  .Prove that$$ \sqrt[3]{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\frac{c}{a+b}>2.$$

[/quote]
It's wrong. Try $a=0$, $b=\sqrt[3]4$ and $c=1$.[/quote]
[/hide]

[hide=Thanks.][quote=thuanz123][quote=sqing]Let $a,b,c$ be  non negative real number  .Prove that$$ \sqrt[3]{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\frac{c}{a+b}>2.$$

[hide=h]Let $a,b,c$ be  non negative real number and $c(a+b)>0$ .[url=https://artofproblemsolving.com/community/c4h2041239p14439858]Find minimum of  [/url] $ \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\frac{c}{a+b}$[/hide][/quote]

May be

Let $a,b,c$ be non-negative real numbers. Prove or Disprove:
$$\sqrt[3]{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\frac{c}{a+b}\ge\frac{4\sqrt[4]{3}}{3}$$

[hide=Note]Equality occurs when $a=\frac{c\sqrt[4]{3}}{3},b=0$[/hide][/quote][/hide]
$$\frac{4\sqrt[4]{3}}{3}=1.7547...$$


function
logarithms
trigonometry
geometric series
real analysis
real analysis unsolved

Consider the function $ f(x)=\sum_{n\geq 1}{\frac{1}{n}sin(\frac{x}{4^n})}$. Prove that there is $c$ not depending on $x$ such that $ |f(x)|< c ln ln x$ for all $x>e$.

It seems fairly easy to me.  Split the sum into two parts: $n \le \log_4 x$ and $n > \log_4 x$.  The first part we can bound by a finite harmonic series, giving a bound of $O(\log\log x)$.  The second part we can bound by an infinite geometric series, using $|\sin x| \le |x|$, giving a bound of $O(1)$.

It is, but this was the first question. The second one was to prove that there is a sequence of real numbers $x_n$ which goes to infinity such that $inf_{n>0}{\frac{ |f(x_n)|}{lnln x_n}>0}$.

Okay, that's a little more interesting.  How about $x_k = 4^k \cdot 120^\circ$?  For the first $k$ terms of the series, the sine term will be $\sin 120^\circ = \sqrt{3}/2$, the same positive constant.  So again we can approximate the first part of the sum by a finite harmonic series.  The remaining part of the series we can again bound by $O(1)$.


How can one locate any known stars in any constellation[b]?[/b] How can one identify[b] any stars in the heavens? Is there any way of counting[/b] them? If so,how? Is there any truth that in counting stars,one must count[b] each[/b] stars[b] with an INTERVAL[/b] of[b] second[/b] each[b]?[/b]

There are several ways to locate and identify or to 'categorise' stars.
The oldest and simplest method is obviously actual names (like Vega or Rigel) and using memory for location. Johann Bayer made the first systematic catalogue now known as [url=http://en.wikipedia.org/wiki/Bayer_designation]Bayer's designation[/url], in which visible stars of a constellation are named by Greek letters in order of brightness (like $ \alpha$ Lyrae or $ \beta$ Orionis).  Another commonly used catalogue is the [url=http://en.wikipedia.org/wiki/Flamsteed_designation]Flamsteed's catalogue[/url] in which stars are numbered starting from the lowest RA (like 3 Lyrae or 19 Orionis).The most faint stars are catalogued in advanced surveys like Henry Draper Catalogue (HD/HDE) made with large telescopes (like HD 172167 or HD 34085)

As to locating stars, Astronomers use the most common-sense idea: [url=http://en.wikipedia.org/wiki/Celestial_coordinate_system]Coordinate system[/url], the most widely used among them being the [url=http://en.wikipedia.org/wiki/Equatorial_coordinate_system]"Celestial Coordinates"[/url]: or [b] RA-Dec[/b] system.

I don't rightly know what you mean by "counting with interval of a second" but if we have detailed stellar catalogues, there is hardly any need to count stars seperately. For the record, depending on your eye-sight and location on earth, about 6000 stars in the sky are visible to the naked eye and about 1000 of them have a Bayer designation.

Nice post.  :)

A current project that is attempting to image the night sky is SLOAN Digital Sky Survey.  Some of the mappings are used in planetarium software, uncluding GoogleSky, the GoogleEarth addition.

Try this... :wink:
http://www.rssd.esa.int/SA-general/Projects/Hipparcos/apps/SelectArea.html


geometry
3D geometry
quadratics
algebra
quadratic formula
complex numbers
AMC

Source: Columbus State University Math Tournament, 2001 Ciphering Round
Difficulty: Very Easy (Medium AMC10/12)

i) Find the sum of the solutions, both real and imaginary (potentially complex), of x:^3: + 8 = 0.

ii) On the interval 0 :le: x :le: 5:pi:, how many solutions does the equation tan(x) = -7 have?

iii) Find the exact value of (2x + 3y)/(5x + 7y) if y/x = 3.

iv) In how many distinct ways can 16 be written as the sum of four positive odd integers?

Again, show work for full credit.

i, ii, and iv are indeed all very easy.  The easiest way to do iii is probably to multiply the fraction by (1/x)/(1/x), so that all you have is constants or (y/x)'s.

Since no one wants to try my challenges, I think I will.

Quote:i) Find the sum of the solutions, both real and imaginary (potentially complex), of x:^3: + 8 = 0.

[hide]
x:^3:+8 is the sum of two cubes, and therefore factors to (x+2)(x^2+2x+4).  So one solution is x=-2.  To find the other two solutions, we will use the quadratic formula:
x = [-2 +- sqrt(4-4*1*4)]/2 = [-2 +- sqrt(-12)]/2 = -2 +- sqrt(-3) = -2 +- i:rt3:.  So the other solutions are -2 + i:rt3; and -2 - i:rt3:.  The sum of these three is -6.[/hide]

Quote:ii) On the interval 0 :le: x :le: 5:pi:, how many solutions does the equation tan(x) = -7 have?

[hide]tan(x) will be equal to -7 in the intervals (-pi/2, 0), (pi/2, pi), (3pi/2, 2pi), etc.  The intervals from ..., pi) to ..., 5pi) are within (0, 5pi).  Thus there are five x-values.[/hide]

Quote:iii) Find the exact value of (2x + 3y)/(5x + 7y) if y/x = 3.

[hide]y/x=3
y=3x

(2x + 3y)/(5x + 7y) = (2x + 3(3x))/(5x + 7(3x)) = 11x/26x = 11/26, which is a fairly easy method, too.[/hide]

Quote:iv) In how many distinct ways can 16 be written as the sum of four positive odd integers?

[hide]List (cannot use any number bigger than first number)

13 1 1 1

11 1 1 3

9 1 1 5 or 9 1 3 3

7 1 1 7 or 7 1 3 5 or 7 3 3 3

5 5 3 3

8 ways.

[/hide]

alternate solution to ii) (which is very similar to matfan's) [hide]the period of tangent is :pi:, as there are 5 periods in the interval (0,5:pi:), any real value will be assumed exactly 5 times on that interval.[/hide]

MysticTerminator wrote:mathfanatic wrote:Since no one wants to try my challenges, I think I will.

Quote:i) Find the sum of the solutions, both real and imaginary (potentially complex), of x:^3: + 8 = 0.

[hide]
x:^3:+8 is the sum of two cubes, and therefore factors to (x+2)(x^2+2x+4).  So one solution is x=-2.  To find the other two solutions, we will use the quadratic formula:
x = [-2 +- sqrt(4-4*1*4)]/2 = [-2 +- sqrt(-12)]/2 = -2 +- sqrt(-3) = -2 +- i:rt3:.  So the other solutions are -2 + i:rt3; and -2 - i:rt3:.  The sum of these three is -6.[/hide]
your factorization is incorrect; the second factor should be [hide]x:^2:-2x+4[/hide]. also, while using the quadratic formula, you made an algebraic error, forgetting to [hide]divide the real part by 2.[/hide] Once you take these into account, you get the answer of [hide]0, which you get also get simply by noting that the three roots will be the vertices of an equilateral triangle (symmmetric with respect to the origin) in the complex plane.[/hide] You could also get that answer by [hide]using Viete's formula, onse you ensure there are no double/triple roots.[/hide]

Oh, duh.  I feel stupid now.

Okay, so corrected, we have

[hide]x:^3:+8 is the sum of two cubes, and therefore factors to (x+2)(x^2-2x+4).  So one solution is x=-2.  To find the other two solutions, we will use the quadratic formula:

x = [2 +- sqrt(4-4*1*4)]/2 = [2 +- sqrt(-12)]/2 = 1 +- sqrt(-3) = 1 +- i:rt3:.  So the other solutions are 1 + i:rt3; and 1 - i:rt3:.  The sum of these three is 0.[/hide]

[quote="JBL"]Actually, Mystic, "The sum of the roots" implies that you take multiplicities counted multiple times.  That is, the sum of the roots of
(x - 1)^753 = 0  is 753, not 1.[/quote]

oh sorry didn't know that. hm. that seems a bit...um...nonintuitive, but it sure helps for using viete!


there are n points on the plane and we know that all of them are not on a line.
prove that there is a line that only two points are on it.

It's the classic "Sylvester's probem"  :D

Domi