Let $ f(x)\equal{}x^2\minus{}ax\plus{}b$, determine the values of $ a,\ b$ such that the maximum, munimum value of $ f(x)$ for $ 0\leq x\leq 2$ is 3, 2 respectively.

Can't calculas be used to solve this problem?

What would you use calculus for? The most you could do with it is find the vertex of the parabola.

[hide="Solution:"]
One of five things can happen:

1a. The minimum over the interval is achieved at the vertex, and the maximum is achieved at x=0
1b. The minimum over the interval is achieved at the vertex, and the maximum is achieved at x=2
1c. The minimum is achieved at the vertex, and the maximum is simultaneously achieved at x=0 and x=2
2a. The minimum is achieved at x=0 and the maximum at x=2
2b. The minimum is achieved at x=2 and the maximum at x=0

Now, you use the fact that all parabolas of the form $ x^2\plus{}ax\plus{}b$ are congruent to the parabola $ x^2$.

We first look at case 1, in which the vertex occurs between x=0 and x=2. Note that the maximum is 1 more than the minimum. We see from $ y\equal{}x^2$ that the parabola is 1 more than the minimum when you move out a distance of 1 in either direction. We find that if the vertex is at x=v, then the maximum will occur at x=v +/-1. Solving for v with possible maxima at x=0 and x=2, we find that v=1. Thus, one possibility is the parabola $ (x\minus{}1)^2\plus{}2 \equal{} x^2\minus{}2x\plus{}3$

Next, we look at case 2a. We have $ y\equal{}x^2\plus{}ax\plus{}b$, and that $ (0,2)$ and $ (2,3)$ satisfy. From the first, we find $ b\equal{}2$, and from the second, we get $ 4\plus{}2a\plus{}b\equal{}3 \implies a\equal{}\minus{}3/2$

Similarly in case 2b, we find that $ b\equal{}3$ from the point $ (0,3)$, and then plug in $ (2,2)$ to get $ a\equal{}\minus{}5/2$.

Thus, the solution set is $ \boxed{(a,b) \in \left \{ (\minus{}2,3), ;\ \left( \minus{} \frac{3}{2}, 2 \right), \; \left( \minus{} \frac{5}{2}, 3\right) \right \}}$[/hide]


The sum of the first $ n$ positive even integers is 90. What is the sum of the first $ n$ positive odd integers?

hello, we have
$ \sum_{i\equal{}0}^{n\minus{}1}2i\equal{}90$ from here we get $ n^2\minus{}n\equal{}90$, solving this equaton with respect to $ n$ we get
$ n\equal{}10$ and so we have to calculate $ \sum_{i\equal{}0}^{9}2i\plus{}1$ this gives $ 100$
Sonnhard.

actually it is n^2 + n so the answer is 9^2 which is 81.


limit
calculus
real analysis
real analysis solved

let {u_n} :  u_1 = 1 ; u_2 = -1 ; and u_(n+2) = (u_(n+1))^2 + 1/2*u_n
does this sequence have a limit ( <> 0 and <> oo ) ?

Thank Mobinool.
I used to see some one who solved this problem by using inequalies for the standard u_n but they don't calculus any u_i(i \leq  10) ,so that they failed.
To post this problem , I want to note to the other member that : don't forget any easiest things because in many problem , the solution is found from some little thing.
Sorry for my bad English

[quote="lephuocdaint1987"]... don't forget any easiest things because in many problem , the solution is found from some little thing...
[/quote]

I'll remember your advise, thx lephuocdaint1987.

Moubinool not Mobinool


I apologize - there is no way I can replicate the question which shows a series of patterns of hearts, clubs, spades, etc, and asks you to complete the pattern. It sounds easy, but it isn't. If anyone has read this book - the question was from a Raven's IQ test and is found on page 78 - I am stumped. They give the answer but not an explanation of how to derive the answer. Does anyone have any idea how to solve?

Do you have a scanner?  If so, you can scan the image, save it, and then upload it.


Given the solution is an integer,
find the square root of: (50*51*52*53+1)

[hide="solution"]
We try smaller numbers:
$ \sqrt{1\cdot 2\cdot 3\cdot 4\plus{}1}\equal{}5$
$ \sqrt{2\cdot 3\cdot 4\cdot 5\plus{}1}\equal{}11$
$ \sqrt{3\cdot 4\cdot 5\cdot 6\plus{}1}\equal{}19$
$ \sqrt{4\cdot 5\cdot 6\cdot 7\plus{}1}\equal{}29$

From here we can guess that \[ \sqrt{n(n\plus{}1)(n\plus{}2)(n\plus{}3)\plus{}1}\equal{}n^2\plus{}3n\plus{}1\], so we would get \[ \sqrt{50\cdot 51\cdot 52\cdot 53\plus{}1}\equal{}\boxed{2651}\]

It turns out that the general form is true, you can try to prove it yourself
[/hide]

[quote="Shakthi S"]Didn't look at any answers.[/quote]

over 80% of people who say "didnt look at any answers" yet show no work usually looked at the answers  :wink:

i know this is spamming , but aren't you spamming too?
 :D  :D  :D


Already stickied... had my moderator powers been granted to me, I'd close this.

had you pmed me i'd have delted this on my own.  i musthave missed this b/c it was sticked


Show that the set of real numbers x for which

1/(x - 1) + 2/(x - 2) + 3/(x - 3) + ... + 167/(x - 167) \geq 7

is the union of a finite number of disjoint intervals.
Determine the sum of the lengths of these intervals. (The nice part :))

[i]Adapted from IMO 1988[/i]

I found 2004, is it correct Arne ?
I 've seen you were on "math et dlires"

Yes, I modified the numbers a bit so that I would get 2004 :)

I was on "Maths et Dlires", indeed.

So it was you...i was wondering too...
Bienvenue sur 'maths et dlires', Arne...

Pierre.

let A={x\in R 1/(x - 1) + 2/(x - 2) + 3/(x - 3) + ... + 167/(x - 167)  \geq  7}

then long(A)=\sum_{j= 1 to 167}(x_j - j) where

\sum_{j= 1 to n}x_j  is the coefficient of x^166 in the polynomial

Q(x) = (1- 1/7\sum_{k=1 to n}(k/(x-k)).product_{j= 1 to 167}(x-j)

Here a new question what can we say about

T= \sum_{j=1 to n}(x_j-j)^2 =  \sum (x_j)^2 + \sum j^2 -2 \sum j.x_j

 \sum  (x_j)^2 =( \sum x_j)^2- \sum _{j<k}x_j.x_k can obtained with

coefficients of x^166 and x^165,   \sum j^2 is well known

What about  \sum j.x_j, values ? Estimation ?


If $ \frac{1}{2} \le x \le 1$, then  show that $ \cos^{-1} {x} + \cos^{-1}( {\frac{x}{2}+\frac{\sqrt{3-3x^{2}}}{2}})=\frac{\pi}{3}$

[hide]
My solution
I substituted $ x=cos \theta$ to get the given expression as  $ \cos^{-1}({\cos \theta})+ \cos^{-1} {\cos{(\theta-\frac{\pi}{3}})}$ and by this I am getting
 $ 2\cos^{-1} ({x})-\frac{\pi}{3}$ as the RHS
What is the problem
 [/hide]

There are many angles whose cosine is $ \cos\left(\theta\minus{}\frac{\pi}{3}\right)$, not just $ \theta\minus{}\frac{\pi}{3}$. Which one does the function $ \cos^{\minus{}1}$ pick out?

I believe you can just take the cosine of both sides.  Then you can draw a unit circle to help you with sin(cos^-1(that big mess))

[quote="Bictor717"]There are many angles whose cosine is $ \cos\left(\theta \minus{} \frac {\pi}{3}\right)$, not just $ \theta \minus{} \frac {\pi}{3}$. Which one does the function $ \cos^{ \minus{} 1}$ pick out?[/quote]

I think we have to take care of the fact that $ \frac{1}{2}\le x\le 1$ while evaluating the inverse of cosines but dont know how.

[hide="Weeeee~"]
$ x\equal{}\cos\theta$
Then
$ \theta \plus{} \cos^{\minus{}1}\left(\frac{\cos\theta \plus{} \sqrt{3(1\minus{}\cos^2\theta)}}{2}\right) \equal{} \frac{\pi}{3}$
$ \Leftrightarrow \theta \plus{} \cos^{\minus{}1}\left(\frac{\cos\theta \plus{} \sqrt{3}\sin\theta}{2}\right) \equal{} \frac{\pi}{3}$
$ \Leftrightarrow \frac{\cos\theta \plus{} \sqrt{3}\sin\theta}{2} \equal{} \cos\left(\theta\minus{}\frac{\pi}{3}\right)$
$ \Leftrightarrow \cos\theta \plus{} \sqrt{3}\sin\theta \equal{} 2\left(\cos\theta\cos\frac{\pi}{3}\plus{}\sin\theta\sin\frac{\pi}{3}\right)$
$ \Leftrightarrow \cos\theta \plus{} \sqrt{3}\sin\theta \equal{} 2\left(\frac{1}{2}\cos\theta\plus{}\frac{\sqrt{3}}{2}\sin\theta\right)$
$ \Leftrightarrow \cos\theta \equal{} \cos\theta$
So we are done (:
[/hide]

[quote="Complex_Ninja"][hide="Weeeee~"]
$ x \equal{} \cos\theta$
Then
$ \theta \plus{} \cos^{ \minus{} 1}\left(\frac {\cos\theta \plus{} \sqrt {3(1 \minus{} \cos^2\theta)}}{2}\right) \equal{} \frac {\pi}{3}$
$ \Leftrightarrow \theta \plus{} \cos^{ \minus{} 1}\left(\frac {\cos\theta \plus{} \sqrt {3}\sin\theta}{2}\right) \equal{} \frac {\pi}{3}$
$ \Leftrightarrow \frac {\cos\theta \plus{} \sqrt {3}\sin\theta}{2} \equal{} \cos\left(\theta \minus{} \frac {\pi}{3}\right)$
$ \Leftrightarrow \cos\theta \plus{} \sqrt {3}\sin\theta \equal{} 2\left(\cos\theta\cos\frac {\pi}{3} \plus{} \sin\theta\sin\frac {\pi}{3}\right)$
$ \Leftrightarrow \cos\theta \plus{} \sqrt {3}\sin\theta \equal{} 2\left(\frac {1}{2}\cos\theta \plus{} \frac {\sqrt {3}}{2}\sin\theta\right)$
$ \Leftrightarrow \cos\theta \equal{} \cos\theta$
So we are done (:
[/hide][/quote]

That's not fair you evaluated LHS and RHS seperately try to get RHS out of LHS only. :)

[hide="Much Easier"]
$ x =\cos\theta$
Then
LS = $ \theta+\cos^{-1}(\frac{\cos\theta+\sqrt{3(1-\cos^{2}\theta)}}{2})$
LS = $ \theta+\cos^{-1}(\frac{\cos\theta+\sqrt{3}\sin\theta}{2})$
LS = $ \theta+\cos^{-1}(\frac{1}{2}\cos\theta+\frac{\sqrt{3}}{2}\sin\theta)$
LS = ${ \theta+\cos^{-1}(\cos\frac{\pi}{3}\cos\theta+\sin\frac{\pi}{3}}\sin\theta)$
LS = ${ \theta+\cos^{-1}(\cos(\frac{\pi}{3}-\theta}))$
LS = $ \theta+ \frac{\pi}{3} -\theta = \frac{\pi}{3}$

[/hide]

[quote="Maths Mechanic"][quote="Bictor717"]There are many angles whose cosine is $ \cos\left(\theta \minus{} \frac {\pi}{3}\right)$, not just $ \theta \minus{} \frac {\pi}{3}$. Which one does the function $ \cos^{ \minus{} 1}$ pick out?[/quote]

I think we have to take care of the fact that $ \frac {1}{2}\le x\le 1$ while evaluating the inverse of cosines but dont know how.[/quote]

Good! Since $ \frac12\le x\le 1$, we have $ 0\le\theta\le\frac{\pi}{3}$, for $ \theta\equal{}\cos^{\minus{}1}x$. Does that help you fix your solution?
[hide]Therefore, $ \theta\minus{}\frac{\pi}{3}<0$ is not in the range of $ \cos^{\minus{}1}$. (When $ \theta\minus{}\frac{\pi}{3}\equal{}0$, the answer you obtained is equal to $ \frac{\pi}{3}$.) Cosine is an even function so $ \cos(\theta\minus{}\frac{\pi}{3})\equal{}\cos(\frac{\pi}{3}\minus{}\theta)$. Moreover, $ \frac{\pi}{3}\minus{}\theta$ is in the range of $ \cos^{\minus{}1}$. Therefore, $ \cos^{\minus{}1}(\cos(\theta\minus{}\frac{\pi}{3})) \equal{} \frac{\pi}{3}\minus{}\theta$.[/hide]


Well, there you all go.  :wink:
Told you they would come out soon.
Although I am quite surprised with the 8-9% figure on the 12B.

[quote="davidyko"]Well, there you all go.  :wink:
Told you they would come out soon.
Although I am quite surprised with the 8-9% figure on the 10B.[/quote]I think you mean the 12B.

Meep, yeah, sorry. *Edits post*
But really, the 12B was that easy? I would've took it if I'd known... :P

My sympathy to the-anxious-for-what-now-appears-to-be-in-vain now a ranger and all of the others who were spamming this forum numerous for polls and whatnot of qualifying score predictions. I'm not really surprised by this at all. We knew that the 10B was harder than normal, which is shown in the qualifying score of 115.5, and I heard many people say that 12B was easier than normal. I AM surprised, however, about the 8-9% above 100 in the 12B, although it really doesn't affect me since I took the 10

[quote="kstan013"]We knew that the 10B was harder than normal, which is shown in the qualifying score of 115.5, and I heard many people say that 12B was easier than normal. [/quote]

Actually most people on AoPS are saying that 12B was harder than usual.

[quote]Actually most people on AoPS are saying that 12B was harder than usual.[/quote]

Perhaps, but people on AoPS have higher expectations, so when they say it was harder, they probably meant they didnt get the score they wanted, but they still passed.  So, I bet if you loooked at the average score for the people that passed on the A and the average score for the people that passed on B, the avaerage score for the people that passed on the A might actaully have been higher, but the percentage of people that passed the A was lower.  So in essence, the B was easier to pass, but harder to get a perfect score.

And another big factor was the people that took it.  The A included more "average" students that took it than the B, so the A's 5% score was reduced as more students who had no shot at passing took the A, lowering the score.  So, for example out of 100 people that took the A, only 5 passed.  But on the B, lets say only 50 took it, but 4 passed.  The amount of people that passed the B was lower, but the percentage was higher because it didnt have people that had no shot at passing taking the B.

[quote="beta"]Actually most people on AoPS are saying that 12B was harder than usual.[/quote]
In truth, there's a disconnection here between two populations of test-takers. Those who try to do every problem, whose goals are those 130+ scores - they ran into some very difficult problems on the back page. The back page of the 12A wasn't that hard. So for those people, the 12B was harder than the 12A. Perfect papers and other extremely high scores are likely to be pretty rare on the 12B.

But there's another population: those whose goal is to get barely 100. They're going to manage their time so they never even see the back page - how difficult they think it is is largely determined by problems 7-18 or something like that. I already had some suspicion that the 12B was easier for these people. The AMC announcement confirms this suspicion.

[quote="kimyejin2000"]its over wow

i feel like im going to cry
i could have pass if i didnt do a stupid mistake
all i had to do was not solve one stupid problem
which i decided to do and got wrong

and miss the cutoff by a point

this is not the only thing
in other contest i recently took i missed by a point also

wats  going on with me????????????????????????????????????????????????[/quote]

lol careless mistakes is part of math competition. Actually the majority of the people who took AMC probably made some kind of careless mistake.

I really am saddened by the announcement of the 10B cutoff.

I got a 114, and the cutoff was 115.5  :(  :(

OH well, time for preparation for next year starting in the summer, and I really want to make USAMO next year.

[quote="madskillz1"]hahahaha-a dumass in my school got a 99(100 to pass) on amc 12b nad 115.5(117 to pass) on amc 10a (hmm who could that be)[/quote]
I don't think you should call people "dumbasses" for missing the cutoff by 1-2 points.  :mad: Try to imagine how you would feel if you were in his situation.

[quote="life=tennis+math"]Judging by his last phrase, he is referring to himself as a dumb---. Thus he knows how he would feel in this situation.
I also got a 99 :blush:  on my AMC 12B. I think the last 5 questions were harder, but it was a dumb mistake that screwed me up. Fortunately, I still make AIME through the 10A.[/quote]

And how did you do on that???

Did you misread #17 like me, and about 8 other people that took the 12b at my school or nearby that i know..people either misread the part about integer a or nonzero a..since we are going so fast

Darn it!!! I got 114. At least most likely, cause I'm sure of getting 19 correctly and  I don't remember my answers to the rest of them.

[quote="mathnerd314"][quote="kimyejin2000"]its over wow

i feel like im going to cry
i could have pass if i didnt do a stupid mistake
all i had to do was not solve one stupid problem
which i decided to do and got wrong

and miss the cutoff by a point

this is not the only thing
in other contest i recently took i missed by a point also

wats  going on with me????????????????????????????????????????????????[/quote]

:wallbash_red: The exact same thing happened to me. I transposed the last two digits on the first problem, and got 867 instead of 876.

I bet I also made stupid mistakes elsewhere. :censored:

If I'd only have skipped number 16 instead of guessing.[/quote]
Hey, I don't feel so bad now seeing that there are others who messed up the 876 thing...Man, could've gotten it right if it weren't multiple choice...

I'm not surprised at the 12B cutoff.  Many people said that it was harder, but I didn't feel that the first 20 problems were difficult at all.  Then again, I had more prep than for the A.  But for those just trying to qualify, it was much easier on this exam.  That, and more smart people took it compared to the A.

I guess that problem 17 messed a lot of people up.  I misread it at first, then just plugged in values.

[quote="superwizard188"]xD. I got exactly 115.5. I think. I hope!!

I actually made a miscalculation. I THOUGHT I left 3 blank, so I thought, there's no point in leaving only 3 blank, so I bubbled them all in. At the end of the test, I realized that I actually left 4 blank! But time was up so I didn't bubble in that last one or get to erase my other guesses. I got 19 right and I left one blank for a total of 115.5. I HOPE. (If my eyes weren't playing a creul, cruel, cruel, cruel, cruel trick on me that day). Sigh, I guess that's part of the test as well.[/quote]

that's what happened to me, except I had 3 blank and bubbled them all in. So 114.

Next time I take a test like this i'm going to make my own answer sheet so I can easily see which problems / how many problems I didn't do. That bubble-thing is so hard to read! xD


search
complex analysis
complex analysis unsolved

Hello,

here is a question i could not solve :

Let $ f \in \mathcal{H}(\mathbb{C})$ such that $ f(\mathbb{Q}) \subset \mathbb{Q}$ can we conclude that $ f$ is in $ \mathbb{Q}[X]$?

No. This has been discussed several times but I'm too lazy to search for the links now, so here is the key idea of the construction:

Enumerate all rationals as $ q_k$, $ k\equal{}1,2,3,\dots$ and set
\[ f(z)\equal{}\sum_j a_j\prod_{k\equal{}1}^{j\minus{}1}(z\minus{}q_k)\]
where $ a_j$ decay fast enough to make the series converge and are all rationals.

By the way, nice to see you again on AoPS :).

Thank you again for your help, this is a great solution.
Nice to see you too . :)


Do you know any good science fair project ideas?
I have some in mind, but i don't think they are good enough.

Prove the Riemann Hypothesis.

No really, I suppose it depends on what kind of science fair.  Also, it's not necessarily about [i]what topic you do[/i], but rather [i]what you do on that topic[/i].  Find something that interests you, and then explore it.

[quote="aanjohri"]Good idea. Thanks! but we have to have a hypothesis and procedure and data table and stuff[/quote]

Oh, like an experiment-based science fair?  Hm...so then you have to run an experiment.  I say that the easiest experiment to run is a physics experiment (of course a math experiment may be easier :)) but it depends on what is possible.

do u know any hypothesises (plural of hypothesis?)

If you want to make it far in the science fair, you'll need to internalize the research you do. From my experience with science fairs, finding a good question is part of the fun. I wouldn't want to deny you of that.

That being said, I can say a few things on topics:

- Math projects have to be tastefully chosen. Noone wants to see a project that beats around an unsolved problem with no progress. Good math projects may involve less "classical" things, such as algorithms/efficiency, sequences, random number generation, structures such as cellular automata, math in a CS context, etc.

- Plants are boring, and they will die on you. Don't do a plant project unless you really know you have an interesting idea.

- Animals are a lot of paperwork, and they will die on you.

- People are even more paperwork. People are also a pain in the $ \text{ @\$\$ }$ with scheduling. Also be careful of the fact that some contests don't like human-related experiments; I know that Siemens Westinghouse doesn't like anything directly or indirectly involving human behavior.

- Try to find out if engineering projects are allowed at the fair; that's a good option for people good at building things.

The Zuton Force, I can really related to your warnings. I did plants and people for my science fair once, the experiments were disasters. Nothing particularly interesting happen with the plants and it was more like they experimented my patience. People... well, not so much fun either.

Back to the original topic, you can try to contact a professor in the topic of your interest at a local university. They have a wealth of good suggestions about what's hot and what's not in the scientific community. The time that I did it, I ended up with knocking out pieces of gene along with a really devoted biology professor. I had exposure to really advanced project and it was really cool. It was in fact, too advanced for me, but I did not regret the experience at all.

On a similar topic, if we do a math project, do we have to have the hypothesis, procedure, independent variable, dependant variable, constants, etc that we need for, say, a biology project? I think this was asked but not clearly answered.

Look, the whole point of that is so that you have a way to go about the problem, and so that the reader (judge) can follow your logic.

That being said, when you're doing a math project, it probably won't be the generic:
 [i]The effect of[/i] [b]<dropdown menu of chemicals>[/b] [i]on[/i] [b]<dropdown menu of organisms>[/b]
project. Or anything like it. You need to make up for this, then, be finding a more relevant but equally logical approach to the problem. Much of the procedures will be the same, but it will be different in that you probably will not be turning one dial and watching the readouts change.

1. Do some research; explore fields of interests.

2. Find a gap in that field, or an area for potential contribution. How is it relevant to that field? What might its implications be?

3. In many cases, it would be silly to have a hypothesis. In science, the purpose of the hypothesis is to generate a specific conditional statement which you can test; not to throw up a wild guess. This is not necessarily the case in math; the statement you want to test might require no guesswork. In this case, you need to make sure you clearly define exactly what you are testing and why.

4. Procedure? Anything will have a procedure.

5. As for independent and dependent variables, and constants, their purpose is to isolate relationships between different variables without the interference of other variables. This is necessary so that the final conclusions follow logically and naturally from the data and data analysis. If this is not applicable to a particular math project, the equivalent would be to draw logical conclusions from your results.

6. Conclusions, applications, limitations, future extensions, and references should be part of every project. Acknowledgments where appropriate.

It's important to keep in mind the purpose behind the parts of the scientific method. Unless you're at a middle school science fair (or a low level science fair), chances are the judges won't spank you if you don't use those exact words. They're not "keywords" in any way. But you need to make sure that the appropriate content is there, and that the conclusions follow naturally and logically from the results (which are also logically produced).

[quote="Kevin K."]Wait, when doing a [b]real[/b] science fair project, do you have to do an experiment totally new (that no one, not even scientists) have ever done before?

Like in 5th grade, I did one on the effects of heat on magnetism.
If you google that, you'll get the answer, but i decided to do some experiments to show it.[/quote]

you can do extensions or generalizations of things that are known.

thanks for all the tips.
 :)
waaaah! i still don't have a testable question/hypiothesis.
do you know any good websites where you can find a list of project ideas?

[quote="The Zuton Force"]

... Good math projects may involve less "classical" things, such as algorithms/efficiency, sequences, random number generation, structures such as cellular automata, math in a CS context, etc... [/quote]
Hmm, what are some good algorithms for this?


[quote="Litlle 1000t"]$a,b,c$ are positive numbers such that $a+b+c=1$. Prove that
$(1+a)\sqrt{\frac{1-a}{a}}+(1+b)\sqrt{\frac{1-b}{b}}+(1+c)\sqrt{\frac{1-c}{c}}\ge{\frac{{3}\sqrt{3}}{4}}{\frac{(1+a)(1+b)(1+c)}{\sqrt{(1-a)(1-b)(1-c)}}}$[/quote]
[hide="hint"]
It follows from my inequality
\[\sum\frac{a(b+c)^{2}}{(1+bc)^{2}}\ge \frac{3\sqrt{3}}{4}\]
where $a,b,c$ be nonnegative real numbers such that $ab+bc+ca=1$. :)[/hide]


[hide="2"]$ 425\equal{}25\cdot 17$. It's pretty obviously $ 0\bmod 25$ and $ 1\bmod 17$. That is, $ 35^{150}\equal{}25k$

$ 25k\equiv 8k\equiv 1\equiv 120\implies k\equiv 15\bmod 17$.

So $ k\equal{}17j\plus{}15$ and $ 35^{150}\equal{}25(17j\plus{}15)\equal{}425j\plus{}\boxed{375}$[/hide]

[quote="bangvn"]1. Find the remainder of division $ 10^{10} \plus{} 10^{10^{2}} \plus{} ... \plus{} 10^{10^{10}}$ to 7

2. Find the remainder of division $ 35^{150}$ to 425[/quote]
[hide="1"]
$ 10^3 \equiv \minus{}1 \pmod{7}$
Since $ 10^k \equiv 1 \pmod{3}$
$ \Longrightarrow 10^{10^k} \equiv \minus{}1 \cdot 10 \equiv 4 \pmod{7}$
So summing it to $ 10$ terms...the remainder is $ 40 \equiv \boxed{5} \pmod{7}$
[/hide]


My reason to take it: To redeem myself since I was retarded on the AMC12A this year

My reason not to take it: I have to pay money

If I do take it, my strategy would be attempting the last five questions first and get at least 3 of them.

Why do the hardest questions first?

I'd pay the money. But of course I pay the money now for my son's test registrations. Then I try to get other parents to join in, so it isn't quite ALL my money.

Actually I have decided not to take it because it really doesn't matter for me, I can redeem myself by doing well on AIME this year and qualify for USAMO.

I want to save the money for AIME seminar :) .

I did the beginning and the end ok, but messed up the middle, getting an ok 123 (which is 10 points shy of what I now realize I need to count on a 200+ for our State qualification).   What do I do? -- :downto:

 :^o:  Take test at school means maybe All of $73.  But it also means giving myself a chance at it--said and done.  Also, there are 4 others at my school who I think should have qualified but didn't.  If they did on B, even our Team score would improve (:approx: 320 --> 350 or more).

 :^o:  Take test at different school means driving 3 hours, possibly paying some money, but Certainly needing to make that same 3 hours (at least in a good direction) for the AIME, USAMO.  My score wouldn't represent my school.  I would at least get to give a team in the State of Utah a chance at a really good (395+) team score that could mean nice things--and I like the other school; I already know a handful of their math team very well.

 :^o:  Don't bother taking the B means doubtful chance of qualifying.  I would at least get to keep the money (yes, for the seminar), spend the time on the Problem Series that I actually should be finishing, and really make a solid prep for the AIME.  BUT this also means that I don't get to go down with a State record score or at least something awesome, 140 range (I assume that's about all I'm looking at for a record based on the last few qualifiers and last year's published stats).

That just got longer and longer as I went.  :wacko:   Anyway, how can I change this:  :mad:  to this:  :) ?  All I want is to think I've done the most cool with the least blowing resources (which is sometimes called a perfect life  :huh: ).

Confused new mather

[quote="beta"]I want to save the money for AIME seminar :) .[/quote]

I like that answer, and I don't even work for the company except as a volunteer moderator.  ;)


function
real analysis
real analysis solved

Letthe continuous functions $f_n(x),n=1,2,\dots$,be defined on the interval $[a,b]$ such that every point of $[a,b]$ is a root of the $f_n(x)=f_m(x)$ for some $m\neq n$.prove that there exists a subinterval of $[a,b]$ on which two of the functions are equal.

I think I am missing something: consider the numerable family of closed sets $ A(m,n)$ for m,n differents where x is in A(m,n) if and only if $ f_m(x)=f_n(x)$. The union of the sets of the family is the interval $ [a,b]$ and thus by Baire's theorem one of the sets contains an interval. On this interval the corresponding $f_n, f_m$ are equal. I am sure it is wrong, I cannot solve your problems usually.

Wow! Sam-n, what is happening to you?  ;)

[quote] !i'm really idiot[/quote]
Don't say that!

It happens with all of us from time to time that we cannot solve something.


inequalities
function
inequalities unsolved

Prove that for positive numbers $ x_{1},x_{2},...,x_{n}$,  $ \ x_{1}\plus{}x_{2}\plus{}...\plus{}x_{n}\equal{}1$ holds :
$ \sum_{i\equal{}1}^{n} \sqrt{x_{i}} \cdot \sum_{i\equal{}1}^{n} \cfrac{1}{1\plus{} \sqrt{1\plus{}2x_{i}}} \leq \cfrac{n^{2}}{\sqrt{n\plus{}2}\plus{}\sqrt{n}}$

Not so hard ,but ugly .At last my solution is ugly (as always  :P )

By the inequality $ ab\leq\frac {(a \plus{} b)^2}{4}$ we have that $ LHS\leq\frac {(\sum_{i \equal{} 1}^n \sqrt {x_i} \plus{} \frac {1}{1 \plus{} \sqrt {1 \plus{} 2x_i}})^2}{4}$

Now you should just use Jensen to the concave function $ f(x) \equal{} \sqrt {x} \plus{} \frac {1}{1 \plus{} \sqrt {1 \plus{} 2x}}$
and done  :)


cf Wikipedia (in french) "méthode de Cardan" for the first polynomial and "méthode de Ferrari" for the second.


I've been thinking about non violent and violent independence movements. Which one do you think was better in the long run?

It all depends upon who you're facing. If you are facing reasonable people who could eventually be interested in your having independence (even if they aren't currently interested in that), and within a reasonable time frame, nonviolent independence is of course the better course by far. However, if you are facing some tenacious enemy who it is quite clear will not let go of you no matter what, then you'll naturally have to shoot your way out of the problem, preferably with a fast offensive designed to throw the enemy totally off his guard and force him to agree to what you want. If he recovers, and still fights on, then you'll have to switch to guerrilla tactics designed to cause the populace of whatever nation you are fighting against to completely loose their morale, which will basically mean that, sooner or later, they'll give you what you want.

Violent independence movements tend to replace one totalitarian government with another.

[quote="mdk"][quote="fredbel6"][quote="mdk"]I've been thinking about non violent and violent independence movements. Which one do you think [b]was[/b] better in the long run?[/quote]
What do you mean, "was"? :maybe:[/quote]

I mean which countries are now better off.[/quote]
There are still many peaceful independence movements as we speak : in Basque Country, in Scotland, in Flanders, in Catalonia,.....

There are plenty of cases of a colonial power simply deciding to let its colonies go; Canada and Brazil are two examples of this. If the colonizer doesn't want to let go, it usually takes violence somewhere; India's independence had a lot to do with World War II weakening Britain, even though there wasn't much violence between the British and the Indians.

[quote="roh3x2n"]Is there any single example of winning independence without violence ?[/quote]
In the republic of Macedonia, a referendum was held in 1991, and afterwards, they seceded from Yugoslavia, as far as I know without any violence.
Czechoslovakia was also split without any problems (but note : both parties wanted to split it up).

[quote="fredbel6"]Czechoslovakia was also split without any problems (but note : both parties wanted to split it up).[/quote]
I'm not sure what you mean by "both parties."  The Czechs and the Slovaks?  There certainly was opposition to the division: Vaclav Havel was opposed, for example, and resigned the presidency as a result.  It's also hard to characterize that as an "independence movement" (though I suspect the Slovaks would be much more likely to do so than the Czechs, although I have no direct evidence for this).

Most of the disolution of the Soviet Union was non-violent.  In the breakup of Yugoslavia, Macedonia (for you Greeks out there, the Former Yugoslav Republic of Macedonia) and Slovenia both escaped with relatively little in the way of military action (Slovenia was "invaded," if that's the right word, by the Yugoslav (Serbian) army for only 10 days -- I don't know details about Macedonia, but there was certainly no long military standoff).

[quote="youdonotknowme"]aren't canada still part of U.K....in a way, even though the government is separate.
isn't the king or queen of uk the king or queen of canada?[/quote]
Yes, the monarch of the United Kingdom is also the Head of State of Australia, New Zealand, Canada, Jamaica, the Bahamas,....
However, all of these countries became independent "in a friendly way".
What I find amazing is that Ireland still had the British king as their king as well, until they became a republic in 1949. :|


advanced fields
advanced fields theorems

Theorem: Let $ x$ be any elements of the multiplicative group $ G$. Then $ x^k \equal{} e$ if and only if the order of $ x$ is a divisor of $ k$.

Proof $ (\rightarrow)$

Let $ x^k \equal{} e$. Let $ N$ denote the finite order of group $ x$. To show that k divides evenly, we apply the Division Algorithm.

$ k\equal{}Nq \plus{} r$ where $ 0 \leq r < N$

We must show that the remainder $ r\equal{}0$
$ x^k \equal{} x^{Nq\plus{}r} \equal{} x^{Nq}x^r \equal{} ex^r \equal{} x^r$

[b]Note[/b]: I didn't get the following line. I didn't get how come $ r\equal{}0$.
By Hypothesis, $ x^k\equal{}e$ so $ x^r \equal{} e$. Since $ N$ is the least positive integer satisfying $ x^N\equal{}e$ and $ 0 \leq r < N$, $ r$ cannot be positive. That is, $ r\equal{}0$.

The number $ N$ is the smallest positive integer for which $ x^N \equal{} e$ for all $ x\in G$. Now you have that $ x^k \equal{} x^r \equal{} e$ . But as $ 0\leq r < N$, this can happen only if $ r \equal{} 0$.


Statement: given 3 matrices A, B and C their size is nXn and they are different from zero. Also A^4=0 => then ABC <> B.
Is it right?

Notational point: $ <>$ makes no sense on sets that don't have an order. $ \neq$ is a much more relevant symbol.

It's false. A $ 2\times 2$ counterexample: $ A\equal{}\begin{bmatrix}0&1\\0&0\end{bmatrix}$, $ B\equal{}\begin{bmatrix}0&0\\1&0\end{bmatrix}$, $ AB\equal{}C\equal{}ABC\equal{}\begin{bmatrix}1&0\\0&0\end{bmatrix}$.


hey guys,
right now i'm reading AoPS 2.
anyway, are there any other good books for me to work on after i'm done? thanks!

Typical answer: Zeitz's ACoPS, and Engel's Problem Solving Strategies.
I just ordered those yesterday.


let $n$be integer prove that this equation $x^{2}-y^{2}=n$ have solution in $\mathbb{Z}$if and only if $n\not\equiv 2 \pmod{4}$

[quote="Kurt Gödel"][quote="Andreas"][hide]Quadratic residues $\bmod 4$ are $0, 1$. Permutations and then the result.[/hide][/quote]

In fact, I think you should also prove that there are solutions when $n\equiv 0,1,3\mod{4}$. [hide="But..."] I didn't say it was difficult to prove, though...  :D
If $n\equiv 0\mod{4}$, then $\left(x,y\right) = \left(\frac{n}{4}+1,\frac{n}{4}-1\right)$ is a solution.
If $n\equiv 1\mod{4}$, then $\left(x,y\right) = \left(\frac{n-1}{2}+1,\frac{n-1}{2}\right)$ is a solution.
If $n\equiv-1\mod{4}$, then $\left(x,y\right) = \left(\frac{n+1}{2},\frac{n+1}{2}-1\right)$ is a solution.[/hide][/quote]
Yeah, this solution is complete because it addresses the "iff" part.


I'm starting a crack the code game here, since I'm bored.  I'll start it off with the code in my siggy.
Then another person cracks it, and posts a new one.
Yay...

[quote="mathnerd314"][hide="Having played several online riddles"]I recognize it as Base64

[quote]<1337s1g>
Anyone know how to crack THIS code?
</1337s1g>[/quote][/hide]

[code]
486578636F646520697320616C736F2066756E2E20446F657320616E796F6E65206B6E6F7720686F7720746F20726561642069743F00
[/code][/quote]

Easy.  Hex:

[hide]
Hexcode is also fun. Does anyone know how to read it?
[/hide]

How about this one:
lkldt vwzsr swwny wmivl fctld rdwfy wmivl fvjfv cdntt vwzsr uj.

RAUGH?

I remember this from 2006 Chapter Target Round

Yeah, our team did that one like a week ago.

It's actually LAUGH. I knew sailingdog would get it, though.

What? I didn't...
I would have though...
And I think it was "EQUALS"

Yrldoensuae!alr!
[code]Y r l d
o e n s
u a e !
a l r ![/code]
read down...

hhmm....

How about this one:

Ø˜ç´ ¨πø˜ å µˆ∂˜ˆ©˙† ∂®´å®¥≤ ∑˙ˆ¬´ ˆ πø˜∂´®´∂ ∑´å˚ å˜∂ ∑´å®¥≤
Ø√´® µå˜¥ å œ¨åˆ˜† å˜∂ ç¨®ˆ¨ß √ø¬¨µ´ øƒ ƒø®©ø††´˜ ¬ø®´


Integer from $1$ to $1979$ writing as $1234567891011121314...19781979$. What is the remainder when this number divided by $1980$ ?

$1980 = 9 \cdot 11 \cdot 20$

The remainder $\bmod 20$ is clearly $19$.

The remainder $\bmod 9$ is easy; it's just $\frac{1979(1980)}{2}\equiv 0 \bmod 9$.

The remainder $\bmod 11$ is slightly tricky, though.  Anyone have a really nice way of doing it, because it looks somewhat ugly to me...

Anyway, once that step is complete it's just CRT.

[quote="t0rajir0u"]
 because it looks somewhat ugly to me...
.[/quote]

You mean finding the alternating sum?

Yes; it's doable, but it involves a bit of casework on the number of digits, and I'm lazy right now...


calculus
integration
calculus computations

Derive:
$ \int_{\minus{}x}^{x^2} \frac{\cos(t)\cos(x)\plus{}\sin(t)\sin(x)}{4\plus{}t}dt$

[hide="Hint."]Writing that numerator as $ \cos(x\minus{}t)$ cleans up some of the algebra you'll have to do.[/hide]


number theory
greatest common divisor
inequalities
number theory unsolved

Hello,

is it true that we have always $ gcd\left(\left(\begin{array}{c} n\\ k\end{array}\right),n\right)\neq 1$?

Thanks.

Yes, its true (unless you allow $ k\equal{}n$)
let $ p$ be an arbitrary prime factor of $ n$. than the exponent of this prime in prime factorisation of $ \binom{n}{k}$ is given by $ \sum_{i\equal{}1}^{\infty}([\frac{n}{p^i}]\minus{}[\frac{k}{p^i}]\minus{}[\frac{n\minus{}k}{p^i}])$ which is (due to inequality $ [x\plus{}y]\geq [x]\plus{}[y]$) grater or equal to zero. actually always grater if there exist such $ p$ that would devide $ n$ more times than $ k$. therefore no such primes $ p$ can exist (or we are done). but that implies that $ n\equal{}k$ and $ \binom{n}{k}\equal{}1$


[quote="gemath"]ok but follow your solution when do we have equal ?[/quote]
For $x=y=\frac{1}{2}-\epsilon,$ $z=2\epsilon$ and $\epsilon\rightarrow0^{+}.$
Thus, no equal. :wink:

ah ok, actually it only has the condition $x,y,z\ge 0$ thus we will have equal when $(x,y,z)=(\frac{1}{2},\frac{1}{2},0)$ and others permutation.


number theory
greatest common divisor
number theory unsolved

Suppose $ n,p,k$ are positive integers with $ p$ prime such that

$ n|k^{n(p\minus{}1)}\minus{}1$.

Is it possible that $ \gcd (n,k^{p\minus{}1}\minus{}1)\equal{}1$?

[quote="jbmorgan"]Suppose $ n,p,k$ are positive integers with $ p$ prime such that

$ n|k^{n(p \minus{} 1)} \minus{} 1$.

Is it possible that $ \gcd (n,k^{p \minus{} 1} \minus{} 1) \equal{} 1$?[/quote]
[hide="Solution"]
We let $ k^{p \minus{} 1} \equal{} x$.  Furthermore, let the smallest prime factor of $ n$ be $ q$.  We have that $ x^n \minus{} 1\equiv 0\bmod q$.  If $ l$ is the order of $ x\bmod q$, then we have that $ l|(q \minus{} 1), n$.  Yet, $ n$ has all of its prime factors above $ q$ and $ q \minus{} 1$ has all of its prime factors below $ q$, so $ \gcd (q \minus{} 1, n) \equal{} 1$, meaning that $ l \equal{} 1$.  Then, $ x\equiv 1\bmod q$, so $ q|\gcd (n, k^{p \minus{} 1} \minus{} 1)$.  It follows that $ \gcd (n, k^{p \minus{} 1} \minus{} 1)\ge q > 1$ when $ n > 1$.  Therefore, as long as $ n > 1$, $ \gcd (n, k^{p \minus{} 1} \minus{} 1) > 1$. [/hide]


Given $x,y,z$, three different integers.
Prove that $((x-y)^5+(y-z)^5+(z-x)^5)$ is divisible by $5(x-y)(y-z)(z-x)$.


Can anyone help me solve this one :

$ (x\minus{}6y)y'\plus{}2x\plus{}y \equal{} 0$

Many thanks ! :)

It is homogeneous.

Rearrange and set $ y\equal{}vx$

[hide]
Since $ 2yy^{\prime}\equal{}\frac{d}{dx}[y^{2}]$, $ y\plus{}xy^{\prime}\equal{}\frac{d}{dx}[xy]$, we get
\[ \frac{d}{dx}[x^{2}\plus{}xy\minus{}3y^{2}]\equal{}0\]
So the solution is $ x^{2}\plus{}xy\minus{}3y^{2}\equal{}C$ for constant $ C$. (Of course, this was just lucky; it doesn't work very well for an equation like $ (5x\minus{}6y)y^{\prime}\plus{}2x\plus{}y\equal{}0$, although multiplying by $ e^{f(x)}$ for some function $ f(x)$ might help.)
[/hide]


[hide="hint"]
In order to prove the orginal inequality, we just show the stronger
\[ \frac{(4\minus{}a\minus{}b)(4\minus{}b\minus{}c)(4\minus{}c\minus{}a)}{\sqrt{(a\plus{}b)(b\plus{}c)(c\plus{}a)}} \ge 2\sqrt{2}
\]
:)
[/hide]


Since no one has posted the solution for your inequality {since it is so easy and deserves to be in the pre olympiad} , if u really need it here it is:
Now you can relate this to the Fermat Point of a Triangle where u can now get evrything in terms of the sides of a trianghle.
Moreover, know one thing which is improtant
$ xy\plus{}yz\plus{}zx \equal{} \frac{4}{\sqrt3} \triangle$ and $ r\equal{} \frac{abc}{4\triangle}$.

We have
\[ \sqrt {y^2  \plus{} yz \plus{} z^2 }  \plus{} \sqrt {z^2  \plus{} zx \plus{} x^2 }  \plus{} \sqrt {x^2  \plus{} xy \plus{} y^2 }  \equal{} \sum {\sqrt {\left( {y \plus{} \frac{z}{2}} \right)^2  \plus{} \frac{3}{4}z^2 } }  \ge \sqrt {\left( {\frac{3}{2}\left( {x \plus{} y \plus{} z} \right)} \right)^2  \plus{} \frac{3}{4}\left( {x \plus{} y \plus{} z} \right)^2 }  \ge 3\sqrt {xy \plus{} yz \plus{} xz}
\]


Hello,

I remember that the first time I solved this problem(almost ten years ago), I was very naive, and did a very complicated proof of it, even when afterwards I found a very easy solution. Nevertheless, I think that I learned more from doing the complicated solution, than from the easyone. Maybe some of you can find it ilustrative too.

Find the number of subsets of the set $\{1,2,\dots,2n\}$ on which the equation $x+y=2n+1$ has no solutions.

I hope you'll enjoy it.

Best regards,

Hello SnowStorm,

[quote="SnowStorm"][hide]
I'm pretty sure the answer is
$4n^2 - 3n$
I'm not quite sure how to prove it yet though, but I'm thinking.[/hide][/quote]

I would recomend you to do some experiments with small cases, and probably it would help you.
Ex: $n=1$ We want all the subset of $\{1,2\}$, such that there are no two elements $x$ and $y$ in such sets such that $x+y=3$. It is easy to see that those sets are $\emptyset, \{1\}, \{2\}$, that is, 3 subsets, but $4\cdot1^2-3\cdot1=1\neq3$.

Do $n=2$ and $n=3$ that are not that many, and for sure you will notice the pattern.

Best,

I thought it was for only subsets of two, guess i need to learn to read :-D.

Edit: nevermind thats wrong

Hello,

[quote="SnowStorm"]
Pattern seems to be
$2n^2 + 1$
[/quote]

Well, this looks like an actuion almost. Any other idea?

Best,

Hello,

[hide][quote="SnowStorm"]Its $3^n$[/quote][/hide]

Can you prove it (Probably yes  :D )

Best,


Find all functions $h: \mathbb{Z}\to \mathbb{Z}$ such that for all $x,y\in \mathbb{Z}$:  \[h(x+y)+h(xy)=h(x)h(y)+1.\]

Beautiful problem , solution :
assume that $ h(0) \equal{} c$ put $ x \equal{} y \equal{} 0$ so we have : $ 2c \equal{} c^{2}\plus{}1$ which gives $ c \equal{} 1$ , assume that $ h(1) \equal{} a$ and $ h(\minus{}1) \equal{} b$ , put $ x \equal{} 1 , y \equal{}\minus{}1$ we get :
$ ab \equal{} b$ which follows $ h(1) \equal{} a \equal{} 1$ or $ h(\minus{}1) \equal{} b \equal{} 0$ so we have 2 case :
1)$ h(1) \equal{} 1$ then put $ y \equal{} 1$ then we have $ h(x\plus{}1) \equal{} 1$ so we have $ h(x) \equal{} 1$ for all $ x$ .
2)$ h(\minus{}1) \equal{} 0$ then put $ y \equal{}\minus{}1$ then we have $ h(x\minus{}1)\plus{}h(\minus{}x) \equal{} 1$ so we have $ h(1)\plus{}h(\minus{}2) \equal{} 1\longrightarrow a\plus{}h(\minus{}2) \equal{} 1$
now put $ y \equal{} 1$ we have : $ h(x\plus{}1)\plus{}h(x) \equal{} ah(x)\plus{}1$ now put $ x \equal{}\minus{}2$ we have $ h(\minus{}1)\plus{}h(\minus{}2) \equal{} ah(\minus{}2)\plus{}1$ and we have $ h(\minus{}2) \equal{} ah(\minus{}2)\plus{}1$ now put $ h(\minus{}2) \equal{} 1\minus{}a$ so we have $ 1\minus{}a \equal{} a(1\minus{}a)\plus{}1\longrightarrow a^{2}\minus{}2a \equal{} 0$ so $ a \equal{} 0$ or $ a \equal{} 2$ so we have 2 case :
$ 1)a \equal{} 0$ then put $ y \equal{} 1\longrightarrow h(x\plus{}1)\plus{}h(x) \equal{} 1$ but $ h(0) \equal{} 1$ and $ h(1) \equal{} 0$ so we $ h(x) \equal{} 1$ if $ x \equal{} 2k$ and $ h(x) \equal{} 0$ if $ x \equal{} 2k\plus{}1$ . u can check this solution !
$ 2)a \equal{} 2$ , now put $ y \equal{} 1$ so we have : $ h(x\plus{}1) \equal{} h(x)\plus{}1$ and u can prove that $ h(x) \equal{} x\plus{}1$ very easily .
so there are 3 solutions !

Let $h(x)=f(x)$ and let $P(x,y)$ denote the given assertion. $P(0,0)\implies f(0)=1.$
Let $f(1)=c+1,$ where $c$ is any constant. $P(x,1)\implies f(x+1)=cf(x)+1.$
Making some basic substitutions one gets that $c=-1,0,1.$
a. $c=-1$: then $f(x)=1$ for even and $f(x)=0$ for odd, which works.
b. $c=0$: Then $f$ is identically 1, which works.
c. $c=1$: Then $P(1,-1)\implies f(-1)=0.$
Comparing $P(x,1)$ and $P(x,-1)$ we get $f(x)=x+1,$ which works.


Hello,

I just knew about Huygens Inequality but can someone please define it clearly and also a proof with one or two applications are preferred.

Thanks for your time


geometry
perimeter
trigonometry
rotation
angle bisector

ABCD is a square with side of length 4.

X is a point of CB,Y is a point of CD such that m∠XAY=45

find

and The perimeter of the Triangle CXY

We have:
${ 1 = \tan{45^{\circ}} \\
   \&= \tan{(\angle DAY + \angle BAX)} \\
   \&= \frac{\tan{\angle DAY} + \tan{\angle BAX}}{1-\tan{\angle DAY}\tan{\angle BAX}} \\
   \&= \frac{4(DY + BX)}{16-(DY)(BX)}}.$
So $ (DY+4)(BX+4)=32.$ Hence $ (8-YC)(8-XC)=32$ and $ 2(8YC+8XC-(YC)(XC))=64$. Now the perimeter $ p=YC+XC+\sqrt{YC^2+XC^2}$. This can be rearranged to $ p^2=2(p(YC)+p(XC)-(YC)(XC))$. So $ p=8$.

Because $m(\angle XAY)=90^\circ-m\left (\frac{\angle XCY}{2}\right)$ and $A$ lays on the angle bisector of $\angle XCY$, $A$ is the $A-$ excircle of $\triangle XAY$, hence if $Z$ is the projection of $A$ onto $XY$, then $AB=AZ=AY$, and $\triangle AZX\equiv\triangle ABX, \triangle AZY\equiv\triangle ADY$; consequently $XZ=XB, YZ=DY$ and the required perimeter is twice the side length of the square.

Best regards,
sunken rock


Does anyone know who got what from the Hong Kong team?

On Yip Chung (Bronze)
Jack Hui (Silver)
Ming Fung Poon (Silver)
Ho Chung Siu (Silver)
Yun Pui Tsoi (Gold)
Chiu Wai Wong


$ a,b,c$are positive numbers,find the best $ k$ such that
$ \frac{a^3b\plus{}b^3c\plus{}c^3a}{a^2b\plus{}b^2c\plus{}c^2a}\geq\frac{a\plus{}b\plus{}c}{k}$

Many thanks....

[quote="Jianxing113725"]$ a,b,c$are positive numbers,find the best $ k$ such that
$ \frac {a^3b \plus{} b^3c \plus{} c^3a}{a^2b \plus{} b^2c \plus{} c^2a}\geq\frac {a \plus{} b \plus{} c}{k}$

Many thanks....[/quote]
It's wrong for all $ k>0.$ :wink:

[quote="karis"]I think with k=3, ineq is right
Because, suppose $ a\geq b\geq c$
Case 1: $ b^2\geq ca\Rightarrow a^2b\geq b^2c\geq c^2a\Rightarrow\sum a^3b\geq\frac {1}{3}\sum a\sum a^2b$
Case 2: $ b^2\leq ca\Rightarrow a^2b\geq c^2a\geq b^2c$
we have $ a \plus{} c\geq 2b\Rightarrow\sum a^3b\geq\frac {1}{3}\sum a\sum a^2b$ :lol:[/quote]
Why suppose $ a\geq b\geq c$.I think only suppose $ a\equal{}max\{a,b,c\}$

[quote="math10"][quote="arqady"][quote="Jianxing113725"]$ a,b,c$are positive numbers,find the best $ k$ such that
$ \frac {a^3b \plus{} b^3c \plus{} c^3a}{a^2b \plus{} b^2c \plus{} c^2a}\geq\frac {a \plus{} b \plus{} c}{k}$

Many thanks....[/quote]
It's wrong for all $ k > 0.$ :wink:[/quote]
Why,can you post example?
I think it true with k=9(not best)[/quote]
Let $ c\rightarrow0^\plus{}$ and $ k>0.$ Hence, we obtain: $ (k\minus{}1)a^3b\geq a^2b^2$ or $ (k\minus{}1)a\geq b,$ which is wrong.

[quote="arqady"][/quote][quote="math10"][quote="arqady"][quote="Jianxing113725"]$ a,b,c$are positive numbers,find the best $ k$ such that
$ \frac {a^3b + b^3c + c^3a}{a^2b + b^2c + c^2a}\geq\frac {a + b + c}{k}$

Many thanks....[/quote]
It's wrong for all $ k > 0.$ :wink:[/quote]
Why,can you post example?
I think it true with k=9(not best)[/quote][quote="arqady"]
Let $ c\rightarrow0^ +$ and $ k > 0.$ Hence, we obtain: $ (k - 1)a^3b\geq a^2b^2$ or $ (k - 1)a\geq b,$ which is wrong.[/quote]
Yes,you had right :D .


Eh can I be on a relay team

also what is with K r i s h a n u R o y S a n k a r

why did he post the TST?

because he is weird

we should tell everyone at some point that those weren't actually the problems

since y'know they weren't

also dude yeah GUYS GUYS should zhai join our arml team moose

if zhai were willing to, i see no reason why he could not...and I am serious about this.  he should join our team, as then we would have a chance of a threepeat.  If Zhai wants to arrange some way to get to Penn State, then he could probably be on our ARML team next year - this year's team has already been registered.  The only conflict might be if whatever ARML team he would normally be on takes students from his area, as it states in the ARML Rules on ARML.com that no two teams may draw competitors from overlapping geographic areas.

Relay Teams (my prediction)

Jeremy, Yasha, Arnav - this should almost be a no-brainer
2. Vivek, Misha, John (maybe not this particular order)
3. Joe, Ray, Daniel
4. Bangchen, Drew, Steven
5. ?, Bryce, Damien

(? denotes whoever else makes it onto A-Team)

Teams absolutely may not overlap in geographic area. This is not a frivolous rule. It is a very important rule that should be established for other contests as well, and it keeps teams like the WOOTlings (from HMMT) from completely dominating. Besides, isn't it more fun to win when you're legitimate? Even if his area does not have an ARML team, we cannot randomly claim states far away.

I agree with those relay teams, except maybe switch Ray and Joe. That probably doesn't matter at all though. But YES that particular order for #2. That should not be changed, especially if it involves me leaving the #3 seat.

What... shouldn't Vitek be on there somewhere ?

yeah same with berman why is berman not on there

[quote="trigonometry1.01"]Hey guys,

I'm a n00b could someone tell me how to prove $\sum\frac a{b+c}\geq \frac{3}{2}$?

Thanks.[/quote]

multiply out to get

symsum a^3 + 2a^2b + abc >= symsum 3a^2b + abc

I will be pretty impressed if that is actually right since I did it in my head

but in any case that one is true

Errr what not really I think it's

cycsum a^3 + symsum a^2b + 3abc >= 3/2 symsum a^2b + 3abc

which is like
you know
pretty true

also it is a pretty triv inequality... but you are only allowed to multiply it out

[quote="Phelpedo"]Possibly john=john berman and daniel = daniel vitek. Actually I'm not even sure that's the right first name, but it sounds like it might be (maybe I should add this so as not to end with a preposition).[/quote]

What preposition? I don't see any prepositions besides "with a preposition", and that is obviously not what you're talking about.

[quote="trigonometry1.01"]Thanks Arnov! You really helped me understand the problem better![/quote]

hmm are you from north carolina because yasha b k is from connecticut

YES ZHAI I AM DANIEL VITEK.  JOHN = JOHN BERMAN = jb05.

I put "Daniel" instead of "Vitek" because I had received a complaint earlier that "Vitek" was easy to confuse with "Vivek", especially since me and vivek attend the same school.

i put "john" instead of "berman" to see how he would react.   :P

why does it matter where yasha is from?  NC is where he lives, fools! (as far as i know)

$\sum_{cyc}\frac{a}{b+c}= \sum_{cyc}\frac{a^{2}}{ab+ac}$

by titu's lemma, $\sum_{cyc}\frac{a^{2}}{ab+ac}\ge \frac{3}{2}\Leftrightarrow \frac{\left(a+b+c\right)^{2}}{2\left(ab+bc+ca\right)}\ge \frac{3}{2}\Leftrightarrow a^{2}+b^{2}+c^{2}+2\left(ab+bc+ca\right) \ge 3\left(ab+bc+ca\right) \Leftrightarrow a^{2}+b^{2}+c^{2}\ge ab+bc+ca$

using AM-GM to get $\frac{a^{2}+b^{2}}{2}\ge ab$ and adding cyclic permutations gives you this conclusion.  you can also probably get this from muirhead's inequality.

...but his name is berman...

BERMAN

BERMAN

BERMAN

Oh while I have this post to mess around with... can we stop actually being serious to trigonometry1.01 since he is obviuosly someone like zhai or... someone. I mean if you don't believe me, just look at his other posts (not in NC).

But the difference between you and us, trigonometry1.01, is that we know how to spell it.  Eh, Arnav?  :P

Sorry about that, jb05: edited.

might want to edit your quote, too.

and trigonometry1.01 stands for STFU AND SPELL ARNAV'S NAME RIGHT!!!

especially since you misspelled it right after he told you what his name was.

but yeah i would really like to be a #3 again this year...b/c i'm not really fast enough to be a #1.  maybe i could be a #2.

uhh ok

so um yasha is the #1 on my relay team?

er ok why is this a no-brainer

yasha might be better in a 2 or 3 spot

SINCE HE IS, like, YASHA B K

Dude yeah... what MysticTerminator said... in addition, he is from Connecticut.

@MysticTerminator, probability1.01 - sorry I did not make this clear, in my original relay prediction post on another topic I had explicitly stated that my relay predictions were not necessarily in optimal order.

@probability1.01 - Yasha is from Connecticut?  Really?  Gee, if he moves to Texas and becomes President, he could be like George Bush!

and also, maybe we should put somebody else on arnav's relay team because that group doesn't really have a good #1 spot.

dude yeah who is usually the #1 spot for jeremy and me? I am kinda confused by this


conics
parabola
symmetry
ellipse
hyperbola
geometry proposed
geometry

Show that the plane cannot be represented as the union of the inner regions of a finite number of parabolas.

Another idea: In a projective viewpoint, a parabola is a conic which is tangent to the line at $\infty$. If we apply a projective transformation taking the line at $\infty$ to an ordinary line and make sure all points of tangency go to finite points, the inner regions of the parabolas become the interiors of ellipses.

On http://www.mathlinks.ro/viewtopic.php?t=204665 it was told that some of these solutions are wrong, so let's correct them:

[quote="jmerry"]Another idea: In a projective viewpoint, a parabola is a conic which is tangent to the line at $ \infty$. If we apply a projective transformation taking the line at $ \infty$ to an ordinary line and make sure all points of tangency go to finite points, the inner regions of the parabolas become the interiors of ellipses.[/quote]

I don't see why the parabolas are mapped to ellipses and not to hyperbolas (or even parabolas) here.

[quote="maky"]and another idea :)
the intersection of a line with a parabola (with interior) can be empty, a point (when the line is tangent), a segment, or a semi-line (only when the line is the axis of symmetry).
because the axis of symmetry are only a finite number of lines, take a line in the plane that is not one of them. the intersections of this lines with the parabola are only points or segments and their number is finite, thus it can`t be covered.[/quote]

This is correct (I think) modulo a minor mistake: The intersection of a line with the interior of a parabola is a semi-line if the line is parallel to the axis of symmetry (and not only if the line is the axis of symmetry itself). Still the proof works.

  darij

Why is [b]maky[/b]'s correct,Darij? He clearly didn't check all the cases. I don't understand why you named the mistake [i]minor[/i].
If it is minor can you help in understanding the last case? i mean why does the proof work?

Thanks. :(

Darij (and maky too, probably) just wants you take a line that is not parallel to any of the symmetry axes.
That line intersects every parabola in only a segment.

Yeah.

Sorry again for my stupid question. :D


probability
limit
ratio
number theory unsolved
number theory

Please relocate if this is in the wrong place.

Was on mathworld.com here [url]http://mathworld.wolfram.com/RationalNumber.html[/url] and saw what was to me an interesting bit of info: "The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)."

Does anyone know why this is the case? Or is there a link to the original source?

The assertion is referenced, but the [url=http://www.inwap.com/pdp10/hbaker/hakmem/number.html#item54]reference[/url] is merely a restatement of the assertion. I haven't found a proof or explanation yet, so now I'm just as curious as you are. :P

As soon as some probability distribution is given, this should be just a bit of assymptotics  :wink:

I interpret this the following way: Let \[ Q_n \equal{} \{ (a,b) : |a| \leq n , 0 \leq b \leq n , \gcd(a,b)\equal{}1\}\] be the rational numbers with bounded numerator and denominator (by $ n$). Note that by construction, every rational corresponds to exactly one such pair.
Let \[ R_n \equal{} \{ (a,b) \in Q_n : 2|b \}\] be those with even denominator.
Claim: \[ \lim_{n \to \infty} \frac{|R_n|}{|Q_n|} \equal{} \frac 13.\]

Since the parity of the denominator does not change when we add or subtract an integer from a rational, I think it would be computationally simpler to restrict ourselves to $ 0 \le a < b$.  Then we're just looking at [url=http://en.wikipedia.org/wiki/Farey_sequence]Farey sequences[/url].  The proportion $ p_n$ of elements of the $ n^{th}$-order Farey sequence with even denominator is given by

$ p_1 \equal{} 0$
$ p_2 \equal{} \frac {1}{3}$
$ p_3 \equal{} \frac {1}{5}$
$ p_4 \equal{} \frac {3}{7}$
$ p_5 \equal{} \frac {3}{11}$
$ p_6 \equal{} \frac {5}{13}$
$ p_7 \equal{} \frac {5}{19}$

so it does appear that the proportion hovers around $ \frac {1}{3}$, at least for small values of $ n$.  Of course, it's not clear that the restriction $ 0 \le a < b$ keeps ZetaX's limit invariant.

The probability for a natural number to be divisible by x=1/x
For the denominator to be even,
numerator should be odd multiple of 2^k while denominator should be any multiple of 2^(k+1); where k=0,1,2,....
Probability for numerator to be odd multiple = (Probability for choosing an odd number)*(Probability for the number to be divisible by 2^k)=1/2*(1/2)^k
=(1/2)^(k+1)
Probability for denominator to be multiple of 2^(k+1)=(1/2)^(k+1)
So, probability for denominator to be even = Σ (1/2)^(k+1)*(1/2)^(k+1)= Σ (1/2)^(2k+2); 0≤k<infinity
Sum of infinite GP whose 1st term is 1/4 and common ratio is 1/4 = (1/4)/1-(1/4) = 1/3

It's not that easy. E.g. you didn't take care of cancelation. You counted pairs of integers with some property, not fractions.
The next problem is your view of probability: there is no way to choose an integer at random (as long as this is supposed to have the "natural" meaning).


Find all functions of $ R\to R$ with the property
$ f(xf(x) \plus{} f(y)) \equal{} (f(x))^2 \plus{} y$

- For $ x\equal{}0$, we have $ f(f(y)) \equal{} y\plus{}f^2(0)$, for all $ y$. (1)
It easily follows that $ f$ is bijective from $ \mathbb{R}$ onto $ \mathbb{R}$.

- Let $ a$ be such that $ f(a)\equal{}0$, which exists since $ f$ is surjective.
Then, for $ x\equal{}a$, we deduce that $ f(f(y))\equal{}y$ for all $ y$. (2)
With (1), it follows that $ f(0)\equal{}0$, which means that $ a\equal{}0$.

- Using (2) and for $ x\equal{}f(b), y\equal{}f(c)$ in the initial equation, we have $ f(bf(b)\plus{}c)\equal{}[f(f(b))]^2\plus{}f(c) \equal{} b^2\plus{}f(c)$ for all $ ,b,c$.
Thus, $ f(xf(x)\plus{}f(y))\equal{}x^2\plus{}y$ for all $ x,y$.
In particular, since $ f(0)\equal{}0$ and (2), we have $ f(xf(x)) \equal{} x^2 \equal{} f(f(x^2))$.
Since $ f$ is injective, it leads to $ xf(x)\equal{}f(x^2)$ for all $ x$. (3)

Then using (3) for $ f(x)$, we deduce that, for all $ x$, we have $ f(x)x\equal{}f(x)f(f(x)) \equal{} f(f^2(x))$. (4)
Therefore, since $ f$ is injective, we have $ f^2(x) \equal{} x^2$, which means that, for all $ x$, we have $ f(x)\equal{}x$ or $ f(x)\equal{}\minus{}x$.

Now, assume that $ xy \neq 0$ and $ f(x)\equal{}x,f(y)\equal{}\minus{}y$.
Thus, back to the initial equation, we have $ f(x^2\minus{}y)\equal{}f(xf(x)\plus{}f(y)) \equal{} x^2\plus{}y$.
If $ f(x^2\minus{}y)\equal{}x^2\minus{}y$ it gives $ y\equal{}0$, a contradiction.
If $ f(x^2\minus{}y) \equal{} y\minus{}x^2$ it gives $ x\equal{}0$, a contradiction.
Thus, $ f(x)\equal{}x$ for all $ x$, or $ f(x)\equal{}\minus{}x$ for all $ x$.

It is easy to verify that these two functions are solutions.

Pierre.

Let’s    f(0)=a   .   f(0f(0)+f(0))=(f(0))^2 +0 = a^2  then   f(a) =a^2
For every  x  in R   f(0f(0)+f(x)) =(f(0))^2 +x  so  f(f(x)) = a^2+x
For every  t  in R   let’s  s=t-a^2   ,f(f(s)) = t   .If u= f(s)  ,f(u) = t  ,so f is  surjective .
If   f(y1)=f(y2) ,f(f(y1)) = f(f(y2))   then   y1+a^2=y2+a^2   and y1=y2  .So f is injective .
Function f is surjective so there is a z such that  f(z) = 0
f(zf(z)+f(a))=(f(z))^2 +a . That  yields f(f(a)) = a     then  a^2+a =a   resulting a=0 .
Could someone continue that ?


Let $ l_{a}, l_{b}, l_{c}$ are angle bisectors of triangle ABC. Find the area of the triangle.

[quote="modeler"]I cannot get your question. The size of the triangle is not fixed by the angle bisector.[/quote]
Actually,if you know lengths of all angel bisectors,you can always find lengths of all sides of a triangle.Hence the area too.

Erken, I think triangle is defined by its angle bisectors but I cannot find a formulae for sides expressed with angle bisectors. I think they cannot not be expressed with elementary functions or if it is possilbe it is not easy at all.

Here is the formula you need:
$ l_a\equal{}\sqrt{\frac{4p(p\minus{}a)bc}{(b\plus{}c)^2}}$,where $ p\equal{}\frac{a\plus{}b\plus{}c}{2}$.

I asked for formula for $ a, b, c$ in terms of $ l_{a}, l_{b}, l_{c}$. Thank you anyway.

No i just want to show modeler that if you know the length of three angle bisectors,then we can completely define triangle.

Yes, I think so, but it may not be expressed in known functions or if it is possible it is not easy at all. You should be exremely good in math to prove the problem is not solvable or if it is solvable - to solve it.

Sorry  :blush:
I misunderstood $ l_{a}, l_{b}, l_{c}$ as three lines on a plane.
(To make your question clearer, I think you should write "the lengths of angle bisectors" instead of just "angle bisectors")

No problem, but it is too long process I cannot edit my initial post... it is forbidden.


Is there a formula or special method of counting for finding the number of nondegenerate triangles that can be formed from three points from an nxn array of lattice points?

For example, how many triangles can be formed from 3 points from a lattice of 8x8 points.

[hide="Idea/Beginning of Solution"]Choose any two points. There are $ \tbinom{n^2}{2}$ ways to do this. Now, all other points that are on the line connecting those two points are out of the question, but all others would form a triangle. The only problem is when the line connecting the two points isn't vertical/horizontal/diagonal.[/hide]

[hide="start"]There are $ \binom{64}{3}$ ways to choose 3 points. All we need to do now is to count how many lines we have...uh oh.[/hide]

[hide="generalization"]for an $ a\times{b}$ array of lattice points, there are $ \binom{ab}{3}\minus{}a\binom{b}{3}\minus{}b\binom{a}{3}\minus{}\frac{(a^2\minus{}2a\plus{}\frac{1\plus{}(\minus{}1)^{a\plus{}1}}{2})(b^2\minus{}2b\plus{}\frac{1\plus{}(\minus{}1)^{b\plus{}1}}{2})}{8}$ triangles. try and see why-i'll explain later if necessary.[/hide]

subtitute $ n\equal{}a\equal{}b$ to get

[hide]$ \binom{n^2}{3}\minus{}2n\binom{n}{3}\minus{}\frac{(n^2\minus{}2n\plus{}\frac{1\plus{}(\minus{}1)^{n\plus{}1}}{2})^2}{8}$.[/hide]

try and see why yourselves.

[quote="perfect628"]
subtitute $ n \equal{} a \equal{} b$ to get

[hide]$ \binom{n^2}{3} \minus{} 2n\binom{n}{3} \minus{} \frac {(n^2 \minus{} 2n \plus{} \frac {1 \plus{} ( \minus{} 1)^{n \plus{} 1}}{2})^2}{8}$.[/hide]

try and see why yourselves.[/quote]

I agree with your formula.
For those who don't understand it is the combinations of all three points in the nxn lattice minus the collinear points on the vertical and horizontal lines minus the collinear points on the diagonals.

I believe we can truncate your formula using the Hockeystick Identity.

$ \binom{n^2}{3} \minus{} 2n\binom{n}{3} \minus{} 4\binom{n\plus{}1}{4} \plus{} 2\binom{n}{3}$

More compact

$ \binom{n^2}{3} \minus{} (2n\minus{}2)\binom{n}{3} \minus{} 4\binom{n\plus{}1}{4}$


Να αποδείξετε ότι δεν υπάρχει συνάρτηση $f: \mathbb{R^+}\rightarrow\mathbb{R^+}$ η οποία  για κάθε $x,y\in\mathbb{R^+}$ να ικανοποιεί τη σχέση: $(f(x))^2\geq f(x+y)(f(x)+y).$

Bάζω μια λύση .Γράφουμε τη σχέση ως
$f(x)-f(x+y)\geq\frac{yf(x)}{y+f(x)}$ .Πρώτα θα αποδείξουμε ότι $f(x)-f(x+1)\geq\frac{1}{2}$ .Προφανώς η f δεν είναι αύξουσα .  Κρατάμε σταθερό το x και επιλέγουμε ένα $n$ ώστε
$nf(x+1)\geq 1$ .Όταν k=0,1,.....,n-1  παίρνουμε
$f(x+\frac{k}{n})-f(x+\frac{k+1}{n})\geq\frac{f(x+\frac{k}{n})\frac{1}{n}}{f(x+\frac{k}{n})+\frac{1}{n}}\geq\frac{1}{2n}$

Προσθέτοντας τις παραπάνω παίρνουμε $f(x)-f(x+1)\geq\frac{1}{2}$ .Τώρα έστω $m$  ώστε  $m\geq 2f(x)$ . Άρα
$f(x)-f(x+m)=\sum_{i=0}^{m-1}(f(x+i)-f(x+i+1))\geq\frac{m}{2}\geq f(x)$
Άρα  $f(x+m)\leq 0$ άτοπο .


limit
logarithms
real analysis
real analysis unsolved


calculus integration logarithms Euler function Gauss algebra

$\int_{[0,\infty]\times[0,\infty]} \ln(x)\ln(y)\exp\left(\frac{x^2+y^2-2xya}{-2(1-a^2)}\right)\,dx\,dy$

thank you very much!

I just edited your post for TeX. The big thing: for posting on this board, you have to put a dollar sign ($\$) before and after the formula in order for the formula to post in TeX. Other changes: \infty instead of inf, \ln insted of ln, \left( and \right) instead of ( and ) to properly size the parentheses.
I also deleted your other post, the one with the .pdf file. Better to concentrate answers in one place.
Please check to make sure it's what you intended.
I haven't looked at the integral yet, except to note that if $|a|<1$ it clearly converges, if $|a|=1$ it makes no sense, and if $|a|>1$ it clearly diverges. Hence we should assume that $|a|<1.$

Thank you very much, Kent, yes, the absolute value of a <1. I tried Methmatica, but it doesn't work.

I also tried Matlap, but failed, please help me you guys.

Here the value for $a=0$ with Fubini theorem we get

$\left(\int_{0}^{+\infty}\ln x . e^{-x^2/2}dx \right )^2$

Now use $\int_{0}^{+\infty}e^{-bx^2}\ln x dx = -1/4 \left (\gamma +\ln{4b}\right)\sqrt{\frac{\pi}{b}}$  with the condition $b>0$

$\gamma$ is Euler constant

yes, you are right.

but how about in the case -1<a<1?

thank you.

Hello, Guys, does it mean that there is no solutions of my question? :(

Thank you!

can anyone help me to get a numerical solution of this double integral for -1<a<1?

Don't be so impatient, rliu!  :)  Be sure, people will think of your problem in their free time but don't expect an immediate answer...

For $a$ not too close to $1$ in absolute value, the following trick may be useful. Make change of variable $x=\sqrt{1-a^2}\,s$, $y=\sqrt{1-a^2}\,t$. The integral becomes
$(1-a^2)\iint_{s,t>0}(\ln s+\ln\sqrt{1-a^2})(\ln t+\ln\sqrt{1-a^2}) e^{ast}e^{-\frac{s^2+t^2}2}\,ds\,dt$
Then decompose $e^{ast}=\sum_{n\geq 0}\frac{a^n}{n!}s^n t^n$. This will give you the series
$(1-a^2)\sum_{n\geq 0}\frac{a^n}{n!}\left(\int_0^\infty(\ln s+\ln\sqrt{1-a^2})s^n e^{-\frac{s^2}2}\,ds\right)^2$
$=(1-a^2)\sum_{n\geq 0}\frac{a^n}{n!}(P_n+\ln\sqrt{1-a^2}Q_n)^2$
where $P_n=\int_0^\infty s^n\ln s\,e^{-\frac{s^2}2}\,ds$, $Q_n=\int_0^\infty s^n \,e^{-\frac{s^2}2}\,ds$.  Essentially you need to know $P_0$, $P_1$, $Q_0$ and $Q_1$ becvause the other values can be found from recurrences $P_{n+2}=(n+1)P_n+Q_n$, $Q_{n+2}=(n+1)Q_n$. It remains to note that $Q_0=\sqrt{\frac\pi 2}$, $Q_1=1$, $P_0$ was told you by Moubinool. As to $P_1$, we have $P_1=\int_0^\infty s \ln s\,e^{-\frac{s^2}2}\,ds=(s^2=2t)=\frac12\int_0^\infty(\ln 2+\ln t)e^{-t}\,dt=\frac12(\ln 2-\gamma)$.
Hope this helps :).

Hello, fedja, thank you very much for your kind help, it is really a very interesting transformation, and it also took me some time to implement it,  but there are still have some problems,  that when $\abs{a}> 0.35$, this approximation is not very precise (I fixed n = 20), , especially when $\abs{a}> 0.8$ (I have some value for certain numbers of $a$ ) , the bias becomes bigger and biger. I hope to get more help from you, and many thanks again for your time! :?

by the way, basically I expect  the solution that I am trying to find,  is some function of $a$.

[quote="rliu"]when $\abs{a}> 0.35$, this approximation is not very precise (I fixed n = 20), , especially when $\abs{a}> 0.8$[/quote]
This is to be expected. The error is approximately $a^n$, and $0.8^{20}$ is not such a small number. Fortunately, this can be easily fixed for positive $a$: just expand your summation to something like $n=\frac{20}{1-a}$. The series consists of positive terms, so you'll always get a slightly smaller value than needed but the relative error of the answer due to round-offs is not worse than the relative error in each term. For $a<0$, the series is alternating and the terms cancel each other, so the method is not as good for $a\to -1$. Also, what is an acceptable accuracy for you? Is, say, 1% error acceptable?

[quote]
 (I have some value for certain numbers of $a$ ) ,
[/quote]
If not a secret, where did they come from (this is also a polite way to ask how reliable they are)

[quote]
by the way, basically I expect  the solution that I am trying to find,  is some function of $a$.[/quote]
It is almost definitely not an elementary function. BTW, do you really need the values for $a$ close to $1$? I'm asking because if your aim is to investigate the behavior very near $a=1$, it is a much simpler problem than finding a closed form for the integral.
:)

:) Hello, fedja, thank you very much again, as to the value for different $a$, I did some simulations by GAUSS; yes, you are right, it is not absolutely reliable, but I did 1000 times  Monto Carlo, the value converges with error within 0.0001.

 I don't need some certain values for vertain $a$, I expect to have a solution which is some function of $a$ for $\abs{a}<1$ . based my simulation values, I also did some polynomial and polynomial combine exponent function regression, but it seems not satisfying either.  :(

And I am also interested for the similar integral

$\int_{[0,\infty]\times[0,\infty]} (\ln(x))^2(\ln(y))^2\exp\left(\frac{x^2+y^2-2xya}{-2(1-a^2)}\right)%Error. "dxdy" is a bad command.
$, do you think it is possible that the both integrals can be described by some certain function of $a$ for $-1<a<1$ with approximately less than 0.00001 error?

Thank you very much!

There is one thing that may be discouraging here. Let us talk about the original integral with $a\to 1$ (for $|a|<0.9$ my power series method is both fast and reliable, so I would stick to it). Let $I(a)$ be the value of the integral, $R(a)=I(a)/\sqrt{1-a}$.

Rigorous theorem: $R(a)\to R=2\sqrt\pi\int_0^\infty(\ln x)^2e^{-x^2/2}\,dx= 7.273692\ldots$ as $a\to 1$.

Computation with error $0.000001$ or less:
$a=0.9\qquad I=0.852435\ldots\qquad R=2.69\ldots$
$a=0.99\qquad I=0.468359\ldots\qquad R=4.68\ldots$
$a=0.999\qquad I=0.187906\ldots\qquad R=5.94\ldots$
$a=0.9999\qquad I=0.066390\ldots\qquad R=6.63\ldots$
$a=0.99999\qquad I=0.022100\ldots\qquad R=6.98\ldots$
$a=0.999999\qquad I=0.007151\ldots\qquad R=7.15\ldots$
$a=0.9999999\qquad I=0.002248\ldots\qquad R=7.22\ldots$

All numbers are from the power series method with $n=12/(1-a)$, the last one took about 20 minutes to compute, so I decided to stop here. See if it agrees with what you get with GAUSS and with Monte-Carlo. If all those values are attainable by some method and if that method gives you the right asymptotics near $1$, just go for it.

My point is that you'll have to approximate these values and the power series method gets too slow before the asymptotical formula becomes more or less precise. (more accurate asymptotics near $1$ is $R\sqrt{1-a}-\frac12 (1-a)(\ln(1-a))^2$ but it also produces a more or less reliable result only beyond $0.999999$ where the error is about $0.00003$ or $0.5$%). As of now I have no idea as to what the next correction term may be.

Looking at these values and the asymptotical formula, it is easy to see that no polynomial or exponential regression can be good. You may try to do some approximation in terms of powers of $(1-a)$ for the remainder of the asymptotic formula but I don't know how reliable that will be.

Another point is that, when near $1$, you should know the value of $a$ with very high precision to get a decent precision for $I(a)$: to reach your goal of $0.00001$, one needs to know $a$ with precision $0.0000000001$. Are you sure you have it with this precision when coming to the task of computing $I(a)$?

The other integral you mentioned is even worse. So let me ask you: what are you going to do with your function that requires to know it very near $1$? Is it a part of some other expression? You see what am I driving at: I don't know what to do near $1$ but, maybe, it is not really necessary to know. :) Also, suppose you know that your function is given by some fancy formula. What will you do with it next? Do you just need to be able to compute it fast many times for different values of $a$ or your task is to find something like the maximum of a certain expression?

[quote="rliu"]do you think it is possible that the both integrals can be described by some certain function of a for -1<a<1 with approximately less than 0.00001 error? [/quote]
The most irritating thing is that the answer is "yes" (it is possible to find the asymptotics near the endpoints and far from the endpoints the function is analytic, so you can approximate it by a polynomial or something). Here is something encouraging:

For $a\in[0.999,1]$, you may use the formula
$I(a)\approx R\sqrt{1-a}-\frac 12(1-a)(\ln(1-a))^2+1.2241(1-a)[\ln(1-a)-8]$
The kind of the terms is predicted by theory, but the last 2 coefficients are experimental. The correct coefficients can be derived theoretically but it would take too much time and is of marginal interest. Check this formula against the other methods you have. I believe it is correct up to the requested $0.00001$.
I do not recommend playing with the coefficients in this formula experimentally because the main problem is to get everything right very near 1.

For $a\in[0.99,0.999]$ the formula
$I(a)\approx  R\sqrt{1-a}-\frac 12(1-a)(\ln(1-a))^2+1.28(1-a)[\ln(1-a)-7.3459]+7.3(a-0.99)(a-0.999)$
seems to work (again, you'd better check it against other methods). $0.999$ is within the range of the power series method with $n=15/(1-a)$, so here you may play with the coefficients to get the optimal precision on the interval. I obtained my coefficients more or less by intelligent trial and error.

If we figure out something that works from $0.9$ to $0.99$, the rest of the positive $a$ will be easy. :lol: Actually a sum of the first 150 terms of a power series is already an acceptable formula if you want it as a subroutine in some program but you may try to approximate the function by some nice splines. After all, it is an analytic function there... I'll edit this post when I get something new. Please tell me whether the above two expressions work as expected.

Hi,  :) fedja, yes, I am trying to write a subroutine, and my purpose is to estimate $a$ if given data, in last post, your fomulars  for different intervals definetly can improve the fitness after modifying some coefficients, but I am not able to put the interval functions into the  subroutine programme. so i am trying to figure out ONE fancy formula which is fit for all the entire interval -1<a<1; and I am still working on your first post and hope to find something new to share with.

many thanks again!

[quote="rliu"]Hi,  :) fedja, yes, but I am not able to put the interval functions into the  subroutine programme. so i am trying to figure out ONE fancy formula which is fit for all the entire interval -1<a<1[/quote]
My God! What kind of programming language is this that wouldn't allow you conditional expressions? :? As to [b]one[/b] fancy formula, I was very happy when I was able to cover the interval $[0.999,1]$ with one expression and my most optimistic hope was to get away with 6 to 10 pieces. Also you may always try explicit partitions of unity, though if you require precision $0.00001$, they are very clumsy and unnatural (like $\sqrt[10000001]x$ or $\frac{x}{\sqrt{x^2+0.000000001}}$, which, for all practical purposes are $-1$ for $x<0$ and $1$ for $x>0$). Will you call this an acceptable component of a formula? And doesn't your programming language have $\text{sign}$ function?  :? Because if the answer to either of these two questions is "yes", you can glue as many pieces as you want together and the "one formula" demand becomes meaningless. On the other hand, if you cannot use $\frac{x}{\sqrt{x^2+0.000000001}}$ in your subroutine, I am completely at loss about what "a formula" is.  :?

Hello, fedja, :)  it seems there is some misunderstanding,  the fact is that I tried to put the conditional function into the subroutine, which generates running errors, and the reference tells me that because parameter $a$ was generated from another subordinate combined with some other parameters, and then  transfer to another procedure (I use GAUSS); actually, the intuition is also clear because,  the subordinate incuding $a$ is not facing certain real numbers. that is why I need one analyical expression for the interval $-1<a<1$ Of couse, the conditional function can be recongnised if it is directly dealing with the real numbers. and of course i can use $\frac{x}{\sqrt{x^2+0.000000001}} $, and as to the answers for the both are "yes".

[quote="rliu"]Hello, fedja, :)  actually, the intuition is also clear because,  the subordinate incuding $a$ is not facing certain real numbers. [/quote]
Hi, rliu! I still do not quite understand this sentence. If it is not facing certain real numbers, what is it facing then? Do you mean that you are trying to do [b]symbolic[/b] computations and you should take $a$ and pass the value of the integral to another procedure not as numbers but as "formulas" (so your input is something like a symbolic expression $x^2-8$ and your output should be something like a symbolic expression $(x^2-8)^3-10$)?. Or do you mean that $a$ may be complex? At any rate, will the formula $a(1-\text{sign}\, a)+a^2(1+\text{sign}\, a)$ generate running errors? It is clearly made of two pieces but formally written as "one analytical expression". :?

I have one more good news: near $-1$ one can get a series decomposition of the form
$\sum_{n\geq 0}\frac{(-1)^n}{n!}\frac{(1+a)^{n+1}}{(1-a)^n}\bigl[ a_n(\ln(1+a))^2+ b_n\ln(1+a)+c_n  \bigr]$ where $a_n, b_n, c_n$ are certain constants expressible via some not very complicated one-dimensional definite integrals. This decomposition is as good as the power series I originally proposed except it works for $a<0$. So we may use it for $a<-1/2$ with about 40 terms to get a very good approximation (provided that you find some way to use different formulae on different intervals; I hate to say it, but to write truly [b] one [/b] expression may be just impossible...). I still do not know whether it is possible to write a nice series decomposition near 1.

Hello, fedja  :)  thank you for your news for the deccompositions for the case $a<0$, which seems quite  complicate, I prefer to leave the case $a<0$ at the moment.
let's go back to $0<a<1$, yes, it is exactly like what you said "aymbolic" computation, sorry for my not precisely explaination in last post, do you mean it is also not possible to write one expression even for $0<a<1$?

btw, I made a expression for the solution of $(1-a^2)\sum_{n\geq 0}\frac{a^n}{n!}\left(\int_0^\infty(\ln s+\ln\sqrt{1-a^2})s^n e^{-\frac{s^2}2}\,ds\right)^2 $ by MATHEMATICA;please see the attachment, I fixed $n=20$, with you method of$n = 12/(1-a)$, does it mean that for $a = 0.99$, we need $n = 1200$?

Hello, rliu,
[quote="rliu"] yes, it is exactly like what you said "symbolic" computation [/quote]
I see. Then I have only one stupid question left: what might the purpose of such a  project be? I mean, you start with something lovely like $\cos x$, run a couple of subroutines of this type and turn it into a 400 (or, rather, 10000)-term monster that occupies a lot of computer memory and cannot even be printed out. Doesn't make much sense to me but, probably, I am missing something as usual :?

[quote]do you mean it is also not possible to write one expression even for $0<a<1$?[/quote]
With precision $0.00001$? Well, I do not have a rigorous prood of impossibility, of course, but I do not see any possible approach at the moment except incorporating $\text{sign}$ functions or its approximants, which is just a fancy way to circumvent the "one formula" demand.

[quote] with you method of$n = 12/(1-a)$, does it mean that for $a = 0.99$, we need $n = 1200$?[/quote] Alas, yes. As I said, the error is about $a^n$ ($a^n/\sqrt n$, to be more precise), and $0.99^{1200}$ is merely $e^{-12}\approx 10^{-6}$.

my purpose is to estimate a vector of parameters, $a$ is one of the parameter, and the question of the double integral comes from the the density function of correlated Gaussian distribution (here $x$ and $y$ are two correlated Gaussians with correlation $a$),  in my case, I need a solution of the mean value of log(x) *log(y) --- the first double integral, and  solution of the mean value of (log(x))^2 *(log(y))^2 --- the second double integral, which are required in my subroutine.

[quote="rliu"]my purpose is to estimate a vector of parameters, $a$ is one of the parameter, ...[/quote] Yes, that is understood and makes perfect sense. What is unclear is why or what for you chose to do symbolic computation rather than numerical one, thus making the problem about 1000 times harder. It would make sense if one could come up with a neat short formula one could put in a textbook, but what arises is a monstrous expression that only a computer can handle anyway. It is, of course, an off-topic question and I do not doubt that you have some good reason for doing so, I just cannot imagine what that reason can be. Everything would be so simple and straightforward if the task were just to compute the integral with a given precision for a given real $a$ (as I initially thought it was). Right now it seems that you need not just [b]a[/b] formula, but a [b]short and neat[/b] formula. And what are you going to do if the simplest expression that has the required precision is 100 terms long and  quite ugly? Abandon the project altogether? Sorry for asking, it may be none of my business, but it seems to me that the idea of manipulating complicated formulae instead of simple real numbers may be not very good here and at some stage may ruin the entire project whatever it is. So there should be some very compelling reason to resort to symbolic computations, which is what I am curious about. I'll see what can be done anyway, but I do not really hope to get something  satisfactory. At least, not very soon. Some functions are just hard to approximate by simple expressions, you know...   :(


I scored 99 (I made a few arithmetic mistakes) last year on the AMC 12B and can generally solve the first 2/3 of the test. The last third gives me trouble and I usually can't figure out how to start attacking those.

I'm confident that I can make AIME next year, though my AMC score still won't be that high. If my goal is to do well on the AIME (say 7-8, I can only get around 2 right now, based on practice), should I take the AMC 12 series or the AIME class? I'm willing to take on a challenge, but I don't want it to be so hard that I can't even understand the explanations...

I'm also unsure of how the class actually works. Do the problem series classes just throw problems at you and ask you to solve them, or do they explain concepts beforehand? Thanks for any help.

I suggest you take the AMC 12 Problem Series because you will be able to answer more AIME problems if you are able to answer the last few AMC 12 problems. (There is a slight overlap.)

I am currently taking it now and what they do is introduce problems pertaining to a given topic that is being covered in class that week and teach you concepts while working through the problem. This way, you understand where the idea of the concept came from and an example of its use. If you want to get the most out of the class, make sure you do the message board problems. They are typically more challenging than the problems given in the class and make good practice. (There are not like the exercises you do after reading a lesson.) Also, you get to view your classmates' solutions and you can learn techniques from them, if they had a different solution than the one you had.

A supplement to the class would be Art of Problem Solving Vol. 1 & 2.


We have $ \left(\frac ab\plus{}\frac ba\right)^3\ge \frac{a^3\plus{}b^3}2\left(\frac 1a\plus{}\frac 1b\right)^3$ since if we expand we get $ \frac{(a\minus{}b)^4(a^2\plus{}ab\plus{}b^2)}{2a^3b^3}\ge0$. So we have $ \sqrt[3]{\frac{a^3\plus{}b^3}2}\left(\frac1a\plus{}\frac 1b\right)\plus{}2\le(a\plus{}b)\left(\frac1a\plus{}\frac1b\right)$ and thus $ \sqrt[3]{\frac{a^3\plus{}b^3}2}\le a\plus{}b\minus{}\frac2{\frac 1a\plus{}\frac 1b}$. Adding this cyclically we get \[ \sum_{cyc}\sqrt[3]{\frac{a^3\plus{}b^3}2}\le2\sum_{cyc}a\minus{}2\sum_{cyc}\frac1{\frac1a\plus{}\frac1b}\le 2\sum_{cyc}a\minus{}4\] because of the AM-HM. So that's it I think. (Of course equality iff $ a\equal{}b\equal{}c\equal{}d\equal{}1$.)

Ok i post my solution(it is also author solution) but I hope you will find others :wink:
Lemma: $ x,y>0$ so $ \frac{(x^3\plus{}y^3}{2})^{\frac{1}{3}}\le\frac{x^2\plus{}y^2}{x\plus{}y}$
Proof: its equivalent to $ 2(x^2\plus{}y^2)^3\minus{}(x\plus{}y)^3(x^3\plus{}y^3)\ge0$
or $ (x\minus{}y)^4(x^2\plus{}xy\plus{}y^2)\ge0$
so we must prove
$ 2(a\plus{}b\plus{}c\plus{}d)\minus{}(\frac{a^2\plus{}b^2}{a\plus{}b}\plus{}\frac{b^2\plus{}c^2}{b\plus{}c}\plus{}\frac{c^2\plus{}d^2}{c\plus{}d}\plus{}\frac{d^2\plus{}a^2}{d\plus{}a})\ge4$
but $ a\plus{}b\minus{}\frac{a^2\plus{}b^2}{a\plus{}b}\equal{}\frac{2ab}{a\plus{}b}\equal{}\frac{2}{\frac{1}{a}\plus{}\frac{1}{b}}$ so
$ \frac{2}{\frac{1}{a}\plus{}\frac{1}{b}}\plus{}\frac{2}{\frac{1}{b}\plus{}\frac{1}{c}}\plus{}\frac{2}{\frac{1}{c}\plus{}\frac{1}{d}}\plus{}\frac{2}{\frac{1}{d}\plus{}\frac{1}{a}}\ge4$ if $ x\equal{}\frac{1}{a}...$ and we use $ x\plus{}y\plus{}z\plus{}t\equal{}4$ we must prove $ 4\plus{}\frac{z\plus{}t}{x\plus{}y}\plus{}\frac{x\plus{}z}{y\plus{}t}\plus{}\frac{x\plus{}y}{z\plus{}t}\plus{}\frac{y\plus{}t}{x\plus{}z}\ge8$ which is obviously true :wink:


we're currently  working on the difference of green charcoal and wood charcoal.

we want to study the different concentration of air pollutants giving out during BBQ

but since the conc. are usually low we need a method to collect the pollutants from the burning charcoal,as well as experiments to test the amount of PAH and difference in the taste of food.

plus the experiment has to be carried out in a microscale....

can anyone plz help?:??????
 :?:

If you are just looking for a quick indication, such as total weight of emissions, I would suggest taking something like a filter paper (cotton ball will work), hook it up to a fan that draws the emissions through the media.  Do it for a fixed time and weigh the difference between the two charcoals.

If you can see a residue, you can measure it with an accurate enough scale or balance.

This is a poor test, because all of the vapors, such as CO, H2O, CO2, NOx are ignored.  Also the heavy tars might settle out on the surrounding grill and not really be part of an air pollution concern, but this will allow you to do the experiement and get your hands dirty.


Find the number of all integers $n>1$, for which the number $a^{25}-a$ is divisible by $n$ for every integer $a$

[quote="Rust"]n is divisor $2*3*5*7*13$.[/quote]
To prove!

[quote="fermat3"]if we use primitive root we find if p/a^24-a => p-1/24    :)[/quote]
 :D , $5|5^{24}-5$ but $5\not |24$

except if you want to say $p|a^{24}-a$ and $p^{2}\not|a^{24}-a$ => $p-1|24$ like $rust$

[quote="fermat3"]to know what mean primitive  root use surch please :mad:[/quote]
I do not need to know primitive  root for saying that this implication (if we use primitive  root  we find if p/a^24-a => p-1/24 ) is false.  :)

why is false let $a$ be the primitive root then $a^{24}\equiv1\mod(p)$.because $a$ is the primitive root of $p$ then $ord_{a}(p)=p-1$ then $p-1|24$ :lol:

[quote="fermat3"]if we use primitive root we find if p/a^24-a => p-1/24    :)[/quote]
you mut say
if we use primitive root we find if p/a^24-1 => p-1/24
 :D


The value of 11 + 22 + 33 + .......... + 1100 is _______.

I 'm quite confused about this kind of questions....

There are 100 terms total:  you can see this by dividing each term by 11:  you have 1,2,3,...,100.

The average value is 1100+11/2=555.5

Multiply by 100 terms, you get 55550.

$ 11(1\plus{}2\plus{}\cdots \plus{}100)$ follows a genious Gauss Method.

What's Gauss Method??
Can anyone explain??

also its a partial arithmetic sequence, so the sum is the median x number of terms, the method aime15 used.


topology
advanced fields
advanced fields unsolved

$X$ and $Y$ are locally compact Hausdorff spaces. $f: X\to Y$ is continuous and surjective. Prove that $f$ can be extended to be a continuous map from $X\cup\{\infty\}$ to $Y\cup\{\infty\}$, if and only if $f^{-1}(K)$ is compact in $X$ for every compact set $K\subseteq Y$.

My question is, how to prove the 'only if' part?
I know that $f^{-1}(K)$ is closed for compact set $K$, so
(1) $f^{-1}(K)$ is compact in $X$
or
(2) $f^{-1}(K)=D\cup\{\infty\}$, where $D$ is a closed set in $X$.
I get stuck here. I don't know how to deduce a contradiction from (2).

And please give an example that two spaces are not homoeomorphic but their one-point compactifications are homoeomorphic.

If I don't misunderstand something, the "only if" part is trivial if the entire space $Y$ is not compact (just notice that we must have $f(\infty)=\infty$ and use the definition of continuity at $\infty$) and false otherwise.

As to the example, it is very easy: take a half-open interval and a union of two half-open intervals. In both cases, the one point compactification is homeomorphic to a closed interval (only in one case $\infty$ is an "end point" and in the other case it is a "middle point").


Let $A=\left[\begin{array}{c}r_{1}\\r_{2}\\\vdots\\r_{n}\end{array}\right]_{n\times n}$ such that each $r_{i}$ is an $1\times n$ array whose elements are either one $1$ and one $-2$, or one $1$ and two $-1$s; all the other elements are $0$s. If $|A|=0$, prove that there exists $r\in\{0,1\}^{n}$, $r\neq 0$ which is a linear combination of $r_{i}$s.

Sorry, I forgot one more condition: in each column, there is exactly one $1$.


group theory
abstract algebra
linear algebra
linear algebra unsolved

1)for any $n\geq 2$ construct a sub-group of $SL_n(Z)$ with cardinal $2^{n-1}n!$

2) find the maximal cardinal of a finite sub-group of $SL_3(Z)$

1) Consider the subgroup of signed permutations of $\mbox{GL}_{n}(\mathbb{Z})$, which is $O_{n}(\mathbb{Z})$, the orthogonal group over the integer, whose order is $2^{n}n!$.

Then $O_{n}(\mathbb{Z})\cap \mbox{SL}_{n}(\mathbb{Z})$ is a subgroup of $\mbox{SL}_{n}(\mathbb{Z})$ whose order is $2^{n-1}n!$.

2) Using 1), it's at least 24, but in fact it's exactly 24.


search
LaTeX
inequalities unsolved
inequalities

This has already been posted.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=561406#561406
but has not been solved. :!:

It has been solved:

http://www.mathlinks.ro/Forum/viewtopic.php?t=55503

(took me half an hour to find)

  darij

[quote="darij grinberg"]
took me half an hour to find
[/quote]
With search or without search? Thank you.

[quote="arqady"]With search or without search?[/quote]

With search, in fact one cannot efficiently search for LaTeX strings right now so I had to try lots of keywords.

  darij

[quote="math227"]for $a,b,c\in\mathbb{R}$
prove:

$8abc \geq (a+b+c-\frac{ab}{c})(a+b+c-\frac{ac}{b})(a+b+c-\frac{bc}{a})$[/quote]
If $abc>0$ then it is equivalent to
$(a-b)^2(a-c)^2(b-c)^2+4\cdot\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0.$ ;)


[quote="Tiks"]Prove or disprove: For any prime number $p$ grater than $k+1$,where $k$ is an even positive integer, and for some $m$ and $n$ such that $\gcd(m;n)=1$ and
\[\frac{m}{n}=\sum_{i=1}^{p-1}\frac{1}{i^{k}}\]
we have that $p|m$.[/quote]

Hm, this is the problem that you want to post?

The denominator of RHS isn't multiple of p, so, you only want to know if $p|\sum_{i=1}^{p-1}\frac{(p-1)!^{k}}{i^{k}}$

But $\sum_{i=1}^{p-1}\frac{(p-1)!^{k}}{i^{k}}\equiv (-1)^{k}\sum_{i=1}^{p-1}(i^{-1})^{k}\equiv (-1)^{k}\sum_{i=1}^{p-1}i^{k}(mod. p)$.

If g is a primitive root of p, we have:

$\sum_{i=1}^{p-1}i^{k}\equiv \sum_{i=1}^{p-1}g^{ik}(mod.p)$.

And we can procede in the classic way..

http://www.mathlinks.ro/Forum/viewtopic.php?t=40171 (and we only need $(p-1) \nmid k$ there).

Thanks for replaying. Now let's try another question: Let $S$ be the set of all $k$s for which $\parallel m\parallel _{p}\geq 2$ for any prime $p\geq{k+1}$. Is the set $S$ finite?(the notations are the same as in the first post)

Not that set $S$ isn't empty because $2\in{S}$!


Let $ n \geq 2$ a natural number. For every permutation $ \sigma \in S_n$ note that $ m(\sigma)$ is the number of inversions in $ \sigma$. Evaluate:
\[ \sum_{\sigma \in S_n} m(\sigma).\]

This has what to do with linear algebra?  It's just a linearity of expectation problem: any pair $ (i < j)$ of positions gives an inversion exactly half the time, so the average permutation has $ \frac {1}{2} \binom{n}{2}$ and all permutations together have $ \frac {n!}{2} \binom{n}{2}$.  (This also works perfectly well for $ n \equal{} 0, 1$.)