Παιδιά σας παρακαλώ λίγη βοήθεια για τις παρακάτω ασκήσεις.
[url=http://img216.imageshack.us/my.php?image=mathimageec0.png][img]http://img216.imageshack.us/img216/7239/mathimageec0.th.png[/img][/url]

1. (αυμβολίζω με l το λ)
Έστω A,B,C οι αναπαραστάσεις των μιγαδικών στο μιγαδικό επίπεδο αντίστοιχα τότε:
$ \vec{AB} \equal{} k\vec{BC}(k\not \equal{} 0) \Leftrightarrow ( \minus{} 1, \minus{} 1) \equal{} k(4 \minus{} 2l,5 \minus{} l \minus{} 7)$ άρα
$ k(4 \minus{} 2l) \equal{} \minus{} 1$
και
$ \minus{} k(l \plus{} 2) \equal{} \minus{} 1$
άρα
$ 2l \minus{} 4 \equal{} l \plus{} 2 \Leftrightarrow l \equal{} 6$

2.έστω $ x \equal{} 2 \plus{} 3i$ τότε $ z \equal{} x^{4n} \plus{} ( \minus{} i\bar{x})^{4n} \equal{} x^{4n} \minus{} (\bar{x})^{4n} \equal{} x^{4n} \minus{} \bar{x^{4n}} \in \mathbb{R}$
ως άθροισμα συζηγών

4. $ |w| \equal{} 1\Leftrightarrow w\bar{w} \equal{} 1 \Leftrightarrow \frac {(z \plus{} i)(\bar{z} \minus{} i)}{(z \minus{} i)(\bar{z} \plus{} i)} \equal{} 1 \Leftrightarrow$

$ z\bar{z} \plus{} i(\bar{z} \minus{} z) \plus{} 1 \equal{} z\bar{z} \minus{} i(\bar{z} \minus{} z) \plus{} 1 \Leftrightarrow z \plus{} \bar{z} \equal{} \minus{} (z \plus{} \bar{z})\Leftrightarrow z \plus{} \bar{z} \equal{} 0 \Leftrightarrow z \in \mathbb{R}$

5. έστω $ w \equal{} z \plus{} \frac {1}{z}$ τότε $ w \plus{} \bar{w} \equal{} z \plus{} \frac {1}{z} \plus{} \bar{z} \minus{} \frac {1}{\bar{z}} \equal{}$

$ z \plus{} \bar{z} \plus{} \frac {z \plus{} \bar{z}}{z\bar{z}} \equal{} z \plus{} \bar{z} \plus{} \frac {z \plus{} \bar{z}}{|z|} \equal{} z \plus{} \bar{z} \plus{} z \plus{} \bar{z} \equal{}$
(επειδή $ z \in \mathbb{C} \Leftrightarrow z \equal{} \minus{} \bar{z}$) $ z \minus{} z \plus{} z \minus{} z \equal{} 0 \Leftrightarrow w \in \mathbb{R}$

6. $ |1 \minus{} z| \equal{} 1 \Leftrightarrow (1 \minus{} z)(1 \minus{} \bar{z}) \equal{} 1 \Leftrightarrow 1 \minus{} z \minus{} \bar{z} \plus{} z\bar{z} \equal{} 1 \Leftrightarrow$
$ 1 \minus{} z \minus{} \bar{z} \plus{} 1 \equal{} 1 \Leftrightarrow z \plus{} \bar{z} \equal{} 1 \Leftrightarrow (z \equal{} x \plus{} iy) x \plus{} iy \plus{} x \minus{} iy \equal{} 1 \Leftrightarrow x \equal{} \frac {1}{2}$

$ \Leftrightarrow y^2 \equal{} \frac{3}{4}$ άρα $ z\equal{}\frac{1\plus{}\sqrt{3}}{2}$ ή $ z\equal{}\frac{1\minus{}\sqrt{3}}{2}$

10. $ w \equal{} z \plus{} \frac {1}{z} \equal{} z \plus{} \frac {\bar{z}}{|z|} \equal{} z \plus{} \frac {\bar{z}}{2} \equal{}$

$ (z \equal{} x \plus{} iy) x \plus{} iy \plus{} \frac {x \minus{} iy}{2} \equal{} \frac {3}{2}x \plus{} i\frac {1}{2}y$

άρα $ w_x^2 \plus{} 9w_y^2 \equal{} 9 \Leftrightarrow \frac {w_x^2}{3^2} \plus{} \frac {w_y^2}{1^2} \equal{} 1$

άρα το $ w$ κείτεται πάνω σε έλλειψη

Επέλεξα τυχαία κάποιες για λύση. Θα συμπληρώσω τις υπόλοιπες αύριο αν στο μεταξύ δεν έχει ασχοληθεί κανένας. Γενικά πάντως αυτού του είδους οι ασκήσεις νομίζω ότι μετά από κάποια εντρίφηση μαζί τους φαντάζουν εύκολες. Το όλο θέμα είναι να βρεις το κόλπο.  :lol:

Andrew κάποιες ασκήσεις βρίσκονται οι ίδιες ή με μια μικρή αλλαγή στα νούμερα μέσα στο σχολικό βιβλίο.

Για παράδειγμα η άσκηση 10 βρίσκεται μέσα στο σχολικό ( σελ102 Ασκηση9)

Για την άσκηση 7 που ρωτάς υπάρχει η εφαρμογή 2 στη σελίδα 93 ( σχολικό )

Η άσκηση 5 που ρωτάς είναι στη σελίδα 101 άσκηση 3 ( σχολικό )

Να διορθώσω λίγο τον Μανώλη.

Για την άσκηση (1) που ρωτάς  αν πούμε Α(3,4) , Β(4,5), C( 2λ-4,λ+2) τις εικόνες των μιγαδικών z1, z2, z3 τότε

 $ \vec{AB}$ = (1,1),  και  $ \vec{BC}$ = ( 2λ-4,λ+2) .

Για να είναι οι εικόνες συνευθειακά σημεία πρέπει det$ (\vec{AB},\vec{BC})$=0 =>  λ = 6

Για την 6 που ρωτάς πρέπει να ισχύει ταυτόχρονα

|z|=1  kai  |z-1|=1  απ΄όπου αν θέσουμε  z=x+yi προκύπτει το σύστημα

$ x^2 \plus{} y^2 \equal{} 1$,  kai  $ (x \minus{} 1)^2 \plus{} y^2 \equal{} 1$ όπου δίνει λύσεις

(χ,y) = ( 1/2 , $ \sqrt {3}$/2 ) kai (χ,y) = ( 1/2 , $ \minus{} \sqrt {3}$/2 )

άρα την αρχική σχέση |z|=|z-1|=1 επαληθεύουν οι μιγαδικοί

  z1 = 1/2 +$ \sqrt {3}$/2  kai  z2 = 1/2 $ \minus{} \sqrt {3}$/2

Για την (4) που θέλεις πρέπει να ισχύει

$ \left| \frac {z \plus{} i}{z \minus{} i}\right| \equal{} 1$ <=> |z-i|=|z+i|

Οι εικόνες των μιγαδικών  z  που επαληθεύουν την εξίσωση αυτή όμως ανήκουν  στη μεσοκάθετη του ευθύγραμμου τμήματος με άκρα τα σημεία Α(0,1) ΚΑΙ Β(0,-1). Η μεσοκάθετη αυτή είναι ο άξονας χχ΄ συνεπώς ο z ανήκει στο R.

H άσκηση αυτή ( με διαφορετικά νούμερα ) βρίσκεται στο σχολικό βιβλίο στη σελίδα 101 , Ασκηση 5 Α΄ομάδας ερώτημα (α).

Γενικά να σου πώ ότι όλες οι ασκήσεις αυτές βρίσκονται οι ίδιες ή με μικρές παραλλαγές σε εξωσχολικά βοηθήματα. :)

Να βοηθήσω και στην άσκηση 9.  Κατ αρχήν να σου πώ ότι υπάρχει λάθος στην εκφώνηση.  Θα πρέπει το λ να είναι θετικο  ( λ>0). Αν το λ ανήκει στο R* τότε παρατήρησε ότι για λ = -1 προκύπτει $ |z_1\plus{}z_2|^2 \le 0$

Η αρχική γίνεται :

$ (z_1\plus{}z_2)(\bar{z_1}\plus{}\bar{z_2}) \le |z_1|^2\plus{}\lambda |z_1|^2 \plus{}|z_2|^2\plus{}\frac{1}{\lambda}|z_2|^2$ <=>

$ |z_1|^2\plus{}|z_2|^2\plus{}z_1\bar{z_2}\plus{}z_2\bar{z_1} \le |z_1|^2\plus{}\lambda |z_1|^2 \plus{}|z_2|^2\plus{}\frac{1}{\lambda}|z_2|^2$

διώχνουμε τα $ |z_1|^2,|z_2|^2$, και πολλ/ζουμε επί  λ>0 και μένει να δ.ο.

$ \lambda^2|z_1|^2 \plus{} |z_2|^2\minus{}\lambda z_1\bar{z_2}\minus{}\lambda z_2\bar{z_1} \ge0$  ποu ισχύει γιατί είναι ισοδύναμη με την

$ |\lambda z_1\minus{}z_2|^2 \ge 0$ :)

Πράγματι, έχω λάθος στην 6 (μπέρδεψα τα σύνολα C και I! Αν είναι δυνατόν!  :o ).  Για την 1, έχω πάρει τα διανύσματα με ανάποδη φορά αλλά είναι 2λ - ... = λ ... που το έκανα 3λ = ... (Είχα καιρό να κάνω κάτι τέτοιο να γελάσουμε λίγο.) Τώρα τα έχω διορθώσει.


topology
inequalities
geometric sequence
advanced fields
advanced fields unsolved

Let $X$ be a metric space. We say that $N = \{x_{i}| i \in I\}$ is an $\epsilon$-net on $X$ if
*for any distinct $x_{1}, x_{2}\in N$, $d(x_{1}, x_{2}) > \epsilon$
*the collection of balls $B_{i}= B(x_{i}, \epsilon)$ centered at $x_{i}$ with radius $\epsilon$ cover $X$.

Let $\#(Y, \epsilon)$ be the cardinality of an $\epsilon$-net on $Y$. Suppose there exist constants $C, r > 0$ (depending only on $X$, and $C$ depending on $r$) such that
$\#(B_{R}, R\epsilon) \le \frac{C}{\epsilon^{r}}$
for any ball $B_{R}$ of radius $R$ in $X$ and all $0 < \epsilon \le \frac{1}{2}$ (we consider $B_{R}$ as a subspace of $X$ and find a net in it).

Then let $\alpha = \inf\{ r | \text{the above holds}\}$. This is called the Assouad dimension of $X$.

Show that $\alpha \ge \rm{dim}_\mathcal{H}(X)$, the Hausdorff dimension. Is there a non-horrific example of a space where the inequality is strict?

edit: In fact, $\alpha$ is at least as big as the Minkowski dimension, and there are easy examples of sets where the Minkowski dimension is bigger than the Hausdorff dimension. But I think spaces exist where the Hausdorff dimension (and also the Minkowski dimension) are finite, but the inequality above does not hold for any $r$.

I also read that there exist sets where the Assouad dimension increases under projection! :o

Note: it is sometimes called the Bouligand dimension?

This is quite a fancy notion. Let's play just with sequences of points. Consider first a geometric progression $E=\{2^{-k}\}_{k\ge 0}$. It is not hard to check that $\alpha(E)=0$. On the other hand, if $E\subset \mathbb R$ contains arbitrarily long arithmetic progressions (the steps of the progressions do not matter), then $\alpha(E)=1$ (if $x_{0}+kd$, $k=1,\dots,N$ is our progression, just take $R=Nd$, $\varepsilon=1/N$ and conclude that we must have $N\le CN^{r}$ for all $N>0$).

This immediately gives an example of the situation when $\alpha$ increases under projection. Just consider any sequence $x_{k}$ satisfying $0<x_{k}<2^{-k}$ and containing arbitrarily long arithmetic progressions and the set $E\subset\mathbb R^{2}$ consisting of all points $(x_{k},2^{-k})$. Then $\alpha(E)=0$ but the projection of $E$ to the $x$-axis has $\alpha=1$.

According to Wikipedia, Bouligand dimension is the same as Minkowski dimension.

[quote="fedja"]According to Wikipedia, Bouligand dimension is the same as Minkowski dimension.[/quote]
I saw that, but in a review of a paper I read (I read the review, but not the paper, since I couldn't find the paper) the reviewer said that Bouligand dimension was Assouad dimension. In such an obscure matter I'm more inclined to trust something I read on an AMS page than on Wikipedia, but I still put the question mark just in case. ;)

Seems like you are right and Wikipedia is wrong. I was too lazy to dig up the original Bouligand paper of 1928 but quite a few authors of modern papers available online agree with you. :) Olson in [url=http://nyjm.albany.edu:8000/PacJ/p/2002/202-2-12.pdf]this paper[/url] discusses the properties of the  "Assouad-Bouligand" dimension in detail including the formula for the dimension of the Cartesian product and an example of a projection increasing the dimension.

Assouad, Assouad-Bouligand, whatever. There is also "Assouad-Nagata dimension" which is actually different. I'll try to motivate the definition.

A metric space $X$ is doubling if there is a constant $N$ such that any ball $B_{X}(x,R)$ can be covered by at most $N$ balls of radius $R/2$. This can be interpreted as being "finite dimensional" in a metric sense. For example, if $X$ has a bi-Lipschitz embedding into some $\mathbb R^{n}$, then $X$ is doubling. [The converse is false, but counterexamples are not obvious.] This is nice enough, but the doubling constant $N$ has no obvious relation to the embedding dimension $n$, and is very hard to compute even for $\mathbb R^{n}$ itself.

However, if we efficiently cover $B^{n}(x,R)$ with balls of radius $\epsilon R$, we should expect to have about $\epsilon^{-n}$ smaller balls. That is, the asymptotic behavior of the number of balls is geometrically meaningful, and this is where the Assouad dimension comes from. Although this dimension may increase under Lipschitz maps, it is bi-Lipschitz invariant.

In conclusion: a metric space is doubling iff its Assouad dimension is finite; the value of Assouad dimension is more interesting than the value of a doubling constant.

Ok, I think I see why the Assouad dimension is so big. Since the net is a collection of points in the space, you're a lot more restricted than in the other dimensions (Hausdorff dimension, Minkowski dimension) where you are covering with sets from a more general collection. So a net you use for the Assouad dimension will give you a covering that you could also use for Hausdorff or Minkowski dimension, so $\alpha \ge \dim_{M}$, but we don't necessarily have equality because it's possible that a covering you have for the Minkowski dimension won't translate easily to a net (as the centers of the boxes might be outside of the set if you are bilipschitz-embedding it in $\mathbb{R}^{n}$ and looking at it as a subset of $\mathbb{R}^{n}$, which is basically the way I think about this stuff...).

So, sets like the ones fedja was talking about are probably a good way to get examples of strict inequality. But I read that you can find a space with finite Hausdorff dimension and infinite/undefined Assouad dimension! I can't imagine that!  :huh:

You are slightly misleaded. The problem is not "designing a net" for the entire space as opposed to just "covering" but designing a net/covering at every scale and [b]for every ball[/b], not only for the whole space. This "locality" of the definition is what makes it really different from the Minkowski dimension. As to the example of a set with zero Hausdorff and infinite Assouad dimension, read Olson's article I mentioned in my previous post. I think it'll clear up a lot of things for you. ;)

The solution to $\sqrt{x}= 5$ is $x=25$.


integration
function
algebra
rational function
calculus
calculus computations

transform to solve y(s)
\[2\int_{0}^{t}(t-u)^{2}y(u)du+y'(s)=(t-1)^{2}\quad t\geq0 \]

I get (Y is transf of y) that $Y(s)(\frac{4}{s^{3}}+s)=2e^{-s}\frac{1}{s^{3}}$
this means that we must find function with transform
\[\frac{2e^{-2}}{4+s^{4}}\]

But I cant do that. Anyone who can, or have I made a mistake?? Please help.

Note that the function whose Laplace transform is $\frac{2e^{-s}}{s^{3}}$ is not $(t-1)^{2}$ for $t>0$ but rather $(t-1)^{2}H(t-1)$ where $H$ is the Heaviside unit step function.

If $f(t)=(t-1)^{2}=t^{2}-2t+1$ for $t>0,$ then

$F(s)=\frac2{s^{3}}-\frac2{s^{2}}+\frac1s.$

Now, as for finding the inverse transform of a function with $s^{4}+4$ in the denominator, use the factorization

$s^{4}+4=(s^{2}-2s+2)(s^{2}+2s+2).$

You can find in closed form the inverse Laplace transform of any rational function of $s$ under one condition: you must be able to factor the denominator.


AMC AIME search USA(J)MO USAMO geometry Support

I don't mean for this to sound like a complaint because I blame any of my mistakes on myself and did okay on the AIME, but I was just wondering what kind of a test environment everyone else has?  (To clarify, I'm not trying to use my situation as an excuse for my score.  I got what I got and don't really care about "if this hadn't happened...")

Personally, for the first half of the AIME II, I was in a room with a teacher who was incessantly and extremely animatedly talking on the phone for at least 20-30 minutes combined.  I didn't want to cause any trouble and seriously contemplated changing rooms, but due to the strange organization of my school (math wing right next to gym and connected to the main building by a bridge), I didn't want to waste time in search of an empty classroom, of which there aren't many.

At about halfway through, I had to change rooms, and the second half was a lot better, though some technology specialist came in for a few minutes to talk with the teacher.

So is my situation unique?  Or does everyone else have to suffer like this?  I remember last year I had to switch to a room with a horrible clock (There's nothing like a constant *tick*...*tock*... to drive you insane).  I just want to know whether or not I'm in a position to perhaps request a better room, "proctor," maybe even request a single room rather than a transition in the middle.  Thanks for any feedback.

My school generally has 10+ people taking it, so it's fine. We're in a basement room and there is a chair-moving party at the beginning of every period above us, but other than that I have no complaints.

We took the AIME II in the conference room of our library. The only disturbances were the bells that rang between every period, but they were useful in basically warning us of each passing hour.

i was in an office with one other asian kid (i didnt take it at my school, i took it at a ghetto school)
since it was an office, random teachers were talking, but i am not the easily distracted type

Our school had about 10 people taking it. It was pretty quiet but the room was cold or maybe it was just my nerves.

Conference room. There were a few distractions (one instance of drilling, morning announcements, some muted talking), but overall it was fine.

Classroom after school had ended. Almost complete silence.
Taking the AMC was a bit harder. We were inside a the school theater and had no desks, which was bad for me, because i use a lot of paper and like to spread it out.

Including me, there were two kids in my school taking it. We each had like three conference room desks to work with.

I took the AIME I in the principal's office, oddly enough, he wasn't there that day, so the math dept head decided to use it.  She was the proctor for almost the whole thing.  I had plenty of space, a large desk (only one other person was taking it).  As for distractions, the bells, the staff talking outside (the office is right off of the main office), and the proctor's knitting was about it.  I was pretty happy in general (way better than taking AMC, where I got less space and people were leaving intermittently).  Last year was better for me though, I took it in a conference room (of the superintendent's office, oddly enough), and I had a huge table to myself, though a test with one proctor and one student is somewhat odd.

I took it in the school library.  No distractions at all (except for bells and muted talking), and a big table all to myself.

Only 2 people from my school qualified (and we make up a 6th of the state) so we almost got stuck taking it on a 4 ft. diameter table, and about a square ft. desk respectively. Luckily, our teacher managed to get us the conference room, so we had more space, and it was quiet the whole time. I love that our school doesn't have bells.

[quote="barbieboy"]I took it in the school library, no distractions other than the usual police chasing car-jacking people and pointing guns at random people, and the announcements telling everyone to stay calm and not panic since the carjacking people were on the loose in our campus[/quote]

uhm, whao? I guess that sucks.

There were about 10 people that took the test at my school. It was in the basement, so it was pretty quiet in general, if I remember correctly (I'm used to distractions, though, so I can't be sure).

[b]Edit:[/b] I forgot about the morning announcements. Those were really annoying.

I took it in our library. It was fine for the most part except random bells, morning anouncements, student laughing loudly in the hall ways, people talking loudly in the library, and a false fire alarm which probably DQs our school.

[quote="MCTP"][quote="MCTP"]At local and state math contests I've been to, the conditions were far better than those I described in my earlier post, and the reason is easy enough to see: there were more people there. [/quote][/quote]

I completely agree.  The more people there are, the more that teachers and a school will probably care about them.  For some reason, however, many teachers don't even consider the AMC tests as math competitions, which is the main problem.  During the AIME, I asked my noisy "proctor" is she could speak a little softly (instead of "HAHAHAHAHAHAHA..." on the phone), and the next block she told her class, "I had this really arrogant student working on some stuff last block who told me to be quiet.  Well, this is my room, and I'll do what I want."  The teachers simply don't care about math, ironic because that's the subject that they teach.  I mean, if a teacher won't shut up that's one thing, but telling her class afterwards about an "arrogant student" who's just trying to take the AIME in peace?

I'd be pretty annoyed if that happened here. The teachers/administrators at my school have generally always been very accommodating to these kinds of things. I'm grateful to have teachers that care.

I posted earlier, but forgot to tell of someone I knew who went to a school with a dress code and even though they were sequestered in a room by themselves, they were "forced" to wear an oxford shirt and tie (and dress pants and shoes) the entire 2 days of the test! They protested to the administration of the school and lost. Now, how uncomfortable would any of you be if you had to test with these clothes??? I know that I'd be miserable if I couldn't wear comfy clothes...

Our school had about 120 students taking the AIME I...

(Korean Minjok Leadership Academy)

divided into 2 conference rooms, and testing environment was OK.

Was able to concentrate As much as I wanted to....

What matters for me is the USAMO environment.

Just like last year, as an international participant, I have to abide by the continental US time.

Which...makes my schedule[u][b]  1:30~6:00AM, 4/29~4/30.[/b][/u]

Very nice....very nice to test and improve my concentration skills :D

Well, I did it last year, and I'm doing it this year too, so....I'm hoping for the best of luck.

It's actually not too bad if you get some sleep during the day and get excuses.

As for the environment you were in, I think there MUST BE SOMETHING DONE, because AIME is not just

a trivial test. You must be guaranteed the best environment that suits you, esp. if you are one of the few

qualifiers.

Try contacting the school through official ways - After all, a AIME Qualifier is an honor for the school too!

[quote="physisnu"]Try contacting the school through official ways - After all, a AIME Qualifier is an honor for the school too![/quote]
Unfortunately, some schools don't see it this way and regard AIME qualifiers (USAMO qualifiers more so, I would imagine) as a nuisance. At least that's what I gather from this topic and from my own experience; if a school thought that it was an honor to have a student taking the AIME, they would treat them better.

[quote="MCTP"]
Unfortunately, some schools don't see it this way and regard AIME qualifiers (USAMO qualifiers more so, I would imagine) as a nuisance. [/quote]

I don't agree with it, but I get how some schools could think so, there's a lot of paperwork to wade through for this stuff (especially USAMO, the manual is 28 pgs worth of instructions and forms...)  For teachers who don't understand how much of an honor it is to be invited to take USAMO, it could appear as a nuisance...

[quote]the manual is 28 pgs worth of instructions and forms [/quote]

Don't scare people, it's not THAT bad, we're not the IRS!

The USAMO part is
  -- 3 pages (I admit two-column) about half just general info, about half instructions,
  - - then 3 pages of forms for the teacher,
       --- one before the USAMO, and
       --- one each day of the USAMO
  -- then two pages of info for the student and parents
       --- one form for the student and parent to return
  -- then 6 pages of blank templates for the answers.

We do make an effort to keep the directions complete, but concise, and we try not to ask for information we don't use.
I hope it's not a nuisance, it's an effort to make sure we can do our job for you quickly and efficiently.

Steve Dunbar
MAA Director, American Mathematics Competitions

I think two much larger reasons for the grouping of students are affluence and training. Affluent schools tend to have more resources to cater to gifted students, and so they get support from the administration, and the schools which have historically done well have developed programs to train their students.

[quote="physisnu"]

Just like last year, as an international participant, I have to abide by the continental US time.

Which...makes my schedule[u][b]  1:30~6:00AM, 4/29~4/30.[/b][/u]

[/quote]

wow. :O that really sucks. good luck!

At TJ, we usually have well over 100 qualifiers, so most years they place us in the cafeteria where it's silent and away from all the classrooms. Another nice thing is that we don't have any bell system at TJ, so that never bothers us :)

When I was taking the AIME I this year, distractions were minimal (it was taken in the lobby of a guidance office), namely people entering the guidance office, bells, and morning announcements. Room was ample, and paper was unlimited (very useful: I had about 25 pages of work). I definitely needed the good conditions that provided an equivalent of 177 minutes of distraction-free time, because in the last 10-20 minutes I was fervently rechecking my answer to problem number 14 (ironically, my final answer to that problem was not right but that affected nothing).

I believe the school I attend gave as close to an ideal testing environment as it had available, something for which I am glad.

[quote="herefishyfishy1"]My school had A LOT of AIME qualifiers (about 160), so we had to use the cafeteria. There weren't any distractions. At all. It was nice.[/quote]
what???????!!!!!!!!!!!!!!!
where do u go?

[quote="physisnu"]

Just like last year, as an international participant, I have to abide by the continental US time.

Which...makes my schedule[u][b]  1:30~6:00AM, 4/29~4/30.[/b][/u]

[/quote]

1:30 to 6:00 AM? You're not going to be going to school for the week after that... that's brutal.

At least ours (which is from 12:30 to 5:00, still not very comfortably fitting into the school day) is during the daytime hours.


[quote="drunner2007"][quote="Inspired By Nature"]Prove the Steiner-Lehmus theorem in as many ways as possible. :D[/quote]
Can you restate it? What is it?[/quote]
Sure. The Steiner-Lehmus theorem states that any triangle that has two equal angle bisectors (measured from the vertex to the opposite side) is isosceles. :)

[quote="neelnal"]Proof #1:

You see that the triangles formed by the bisectors are similar because their share the base of the original triangle. The bisectors are also equal, so by the SS similarity theorem, the triangles are similar. thus, the base angles are equal by the definition of similarity and the triangle is symettric.[/quote]

it's SAS simililarity, you did not prove the included angles are congruent

Oops. Sorry forgot that is was SAS.

Well, I'll work on that later. Another method you can use is indirect proof. Start backwards.

let a be the base, and b,c be the alleged legs, using the angle bisector length theorem, (after clearing the denominators)

$ac(a+b)^{2}(a+c-b)(a+c+b)=ab(a+c)^{2}(a+b-c)(a+b+c)$

let $p=a+b+c$

$f(p)=c(p-c)(p-2b)-b(p-b)(p-2c)=0$

$f(p)$ is a quadratic, $f(a+2b)=0$ and $f(a+2c)=0$ since those imply $a=b$, making the expression 0, there are only two roots, so it factors
$f(p)=k*(c-b)(b-c)=0$

so $b=c$


1. Let $ a,b,c$ be sidelengths of a triangle. Prove that
\[ \frac{a^2\minus{}bc}{\sqrt{4a^2\plus{}bc}}\plus{}\frac{b^2\minus{}ca}{\sqrt{4b^2\plus{}ca}}\plus{}\frac{c^2\minus{}ab}{\sqrt{4c^2\plus{}ab}}\le0\]
When equality occurs?

Clearly when a=b=c the inequality holds
and by SOS we can prove the inequality,i think

[quote="dduclam"]1. Let $ a,b,c$ be sidelengths of a triangle. Prove that
\[ \frac {a^2 \minus{} bc}{\sqrt {4a^2 \plus{} bc}} \plus{} \frac {b^2 \minus{} ca}{\sqrt {4b^2 \plus{} ca}} \plus{} \frac {c^2 \minus{} ab}{\sqrt {4c^2 \plus{} ab}}\le0\]
When equality occurs?[/quote]
Yes, it's easy SOS.
$ 2\sum_{cyc}\frac {a^2 \minus{} bc}{\sqrt {4a^2 \plus{} bc}}\equal{}\sum_{cyc}\frac{(a\minus{}b)(a\plus{}c)\minus{}(c\minus{}a)(a\plus{}b)}{\sqrt {4a^2 \plus{} bc}}\equal{}$
$ \equal{}\sum_{cyc}(a\minus{}b)\left(\frac{a\plus{}c}{\sqrt {4a^2 \plus{} bc}}\minus{}\frac{b\plus{}c}{\sqrt {4b^2 \plus{} ac}}\right)\equal{}$
$ \equal{}\sum_{cyc}\frac{c(a\minus{}b)^2(c^2\minus{}2(a\plus{}b)c\plus{}a^2\minus{}7ab\plus{}b^2)}{\sqrt{(4a^2\plus{}bc)(4b^2\plus{}ac)}\left((a\plus{}c)\sqrt{4b^2\plus{}ac}\plus{}(b\plus{}c)\sqrt{4a^2\plus{}bc}\right)}.$
Id est, it remains to prove that
$ c^2\minus{}2(a\plus{}b)c\plus{}a^2\minus{}7ab\plus{}b^2\leq0,$ which is obvious.

Nice, Arqady.  Some same form problems are

2. Let $ a,b,c$ be sidelengths of a triangle. Prove that
\[ \frac{a^{2}\minus{}bc}{\sqrt{15a^{2}\plus{}(b\plus{}c)^2}}\plus{}\frac{b^{2}\minus{}ca}{\sqrt{15b^{2}\plus{}(c\plus{}a)^2}}\plus{}\frac{c^{2}\minus{}ab}{\sqrt{15c^{2}\plus{}(a\plus{}b)^2}}\le0\]

3. Let $ a,b,c$ be sidelengths of a triangle. Prove that
\[ \frac{a^{2}\minus{}bc}{\sqrt{7a^{2}\plus{}b^2\plus{}c^2}}\plus{}\frac{b^{2}\minus{}ca}{\sqrt{7b^{2}\plus{}c^2\plus{}a^2}}\plus{}\frac{c^{2}\minus{}ab}{\sqrt{7c^{2}\plus{}a^2\plus{}b^2}}\le0\]

[quote="dduclam"]1. Let $ a,b,c$ be sidelengths of a triangle. Prove that
\[ \frac {a^2 \minus{} bc}{\sqrt {4a^2 \plus{} bc}} \plus{} \frac {b^2 \minus{} ca}{\sqrt {4b^2 \plus{} ca}} \plus{} \frac {c^2 \minus{} ab}{\sqrt {4c^2 \plus{} ab}}\le0\]
When equality occurs?[/quote]

Notice that we have the following Inequality with the same condition
\[ \frac {a^2 \minus{} bc}{4a^2 \plus{} 5bc} \plus{} \frac {b^2 \minus{} ca}{4b^2 \plus{} 5ca} \plus{} \frac {c^2 \minus{} ab}{4c^2 \plus{} 5ab}\ge0\]


aca un problema que parece conocido   :D

halle todas las ternas $(x,y,z)$ los cuales pertenecen a los reales positivos tales que :

         $x+y+z = 6$
    y
         $xyz= 20-(xy+yz+zy)$

[quote="mhuarancca"]  $xyz= 20-(xy+yz+zy)$-->??? no es así:
$xyz= 20-(xy+yz+zx)$[/quote].
considerando eso.
ahora agamos el siguiente cambio de variable:
$a=x+1, b=y+1, c=z+1. \Rightarrow abc=(1+x)(1+y)(1+z)=1+x+y+z+xy+yz+xz+xyz=1+6+20=27 \to abc=27$
aplicando $MA\geq MG$, a los terminos $a,b,c$.
$(a+b+c)/3\geq (abc)^{1/3}$
$(x+y+z+3)/3\geq (27)^{1/3}$->$x+y+z+3\geq 9$
$x+y+z\geq 6$, ->$x+y+z=6$, entonces se cumple la igualdad pero en medias se cumple la igualdad cuando se da la igualdad entre sus terminos por lo tanto $x=y=z$, remplasando $3x=6$, y, $x^{3}=20-3x^{2}$
de aqui $x=y=z=2$.

[quote="mhuarancca"]aca un problema que parece conocido   :D

halle todas las ternas $(x,y,z)$ los cuales pertenecen a los reales positivos tales que :

         $x+y+z = 6$
    y
         $xyz= 20-(xy+yz+xz)$[/quote]

Desde que $x^{2}+y^{2}+z^{2}\geq xy+yz+xz$ se sigue que $(x+y+z)^{2}\geq 3(xy+yz+xz)$ con lo que $36\geq 3(xy+yz+xz)$ y de ahi que $12\geq (xy+yz+xz)$. Del dato $xyz= 20-(xy+yz+xz)$ con lo que $xyz\geq 8$. De la desigualdad $MA-MG$ tenemos que $x+y+z\geq 3\sqrt[3]{xyz}$ con lo que $xyz\leq 8$ luego tenemos que $xyz=8$ con lo que $x=y=z=2$.


linear algebra
matrix
linear algebra unsolved

Let A be a matrix in Mn(C) s.t. Aii = n-1 and Aij Є {-1, 1} for i ≠ j.
If det(A) = 0 then A is symmetric.

If the determinant is 0 the coresponding system of ecuations attached to the matrix must have a non-trivial solution $(x_1,x_2,...x_n)$. Assume WLOG that $|x_1| \ge |x_i|\forall i$. Then, the equation $(n-1)x_1+x_2+x_3+...+x_n=0\Rightarrow \sum|x_i|\le(n-1)|x_1|=|\sum x_i|\le\sum|x_i|$. We can get easily that $x_1=x_2=...=x_n$. Hence, the matrix is symetric.


Determine all functions $ f: R\rightarrow R$ such that $ f(x \minus{} f(y) ) \equal{} f( f(y) ) \plus{} x f(y) \plus{} f(x) \minus{} 1$ for all $ x, y \in R$.

[color=green]Problem  : Find all functions $ f: \mathbb{R}\longrightarrow \mathbb{R}$ such that:
\[ f(x \minus{} f(y)) \equal{} f(f(y)) \plus{} xf(y) \plus{} f(x) \minus{} 1
\]
for all $ x,y$ in the real numbers.

Solution: Let $ P(x,y)$ the property $ f(x \minus{} f(y)) \equal{} f(f(y)) \plus{} xf(y) \plus{} f(x) \minus{} 1.$

Obviously $ \exists \;y_0$ such that $ f(y_0)\neq 0$.

Then, in $ P(x,y_0)$, let $ x\rightarrow \plus{} \infty$ : $ \text{RHS}$ is not bounded, and so $ \text{LHS}$ can't be bounded.

$ \star\;P(f(x),x)$ implies $ f(0) \equal{} 2f(f(x)) \plus{} f(x)^2 \minus{} 1$ and so
\[ f(f(x)) \equal{} \frac {f(0) \plus{} 1}{2} \minus{} \frac {f(x)^2}{2}
\]
Or
\[ f(x) \equal{} \frac {f(0) \plus{} 1}{2} \minus{} \frac {x^2}{2}, \forall x\in \{f(x), x\in\mathbb{R}\}
\]
$ \star\;P(2f(x),x)$ implies $ f(f(x)) \equal{} f(f(x)) \plus{} 2f(x)^2 \plus{} f(2f(x)) \minus{} 1$ and so

   Then
\[ f(2f(x)) \equal{} 1 \minus{} \frac {(2f(x))^2}{2}
\]
Or
\[ f(x) \equal{} 1 \minus{} \frac{x^2}{2}\;\text{for all x such that}\;\frac{x}{2}\in \{f(x), x\in\mathbb{R}\}
\]
Since $ f(x)$ is not bounded , and since the set of discontinuity points of $ f(x)$ is included in an interval $ (a,b)$, it is possible to find two values $ u$ and $ v$ such that $ f(v) \equal{} 2f(u)$

Because $ f(f(v)) \equal{} \frac {f(0) \plus{} 1}{2} \minus{} \frac {f(v)^2}{2}$

And  $ f(f(v)) \equal{} 1 \minus{} \frac {f(v)^2}{2}$

So $ \frac {f(0) \plus{} 1}{2} \equal{} 1$ or $ f(0) \equal{} 1$

Now we prove that : $ f(x) \equal{} 1 \minus{} \frac {x^2}{2}, \forall x\in\mathbb{R}$

Let $ g(x) \equal{} f(x) \minus{} (1 \minus{} \frac {x^2}{2})$. $ P(x,y)$ becomes : $ g(x \minus{} f(y)) \equal{} g(f(y)) \plus{} g(x)$

Let then $ x \equal{} f(y)$, we have $ g(f(y)) \equal{} \frac {g(0)}{2} \equal{} 0.$ And so
\[ g(x \minus{} y) \equal{} g(x) \forall x\in\mathbb{R}, \forall y\in \{f(x), x\in\mathbb{R}\}
\]
Since we know that it exists an intervall $ [ \minus{} \infty,x_1)$ or $ (x_2, \plus{} \infty)$ included in $ \{f(x), x\in\mathbb{R}\}$, we can conclude that $ g(x) \equal{} \text{contant} \equal{} 0$ or $ f(x) \equal{} 1 \minus{} \frac {x^2}{2}, \forall x\in\mathbb{R}$
And hence the result.

It took me 30 minutes to type  :wallbash_red:  :wacko:. [/color]

Since f(x) is not bounded , and since the set of discontinuity points of f(x) is included in an interval (a,b), it is possible to find two values u and v such that f(v) = 2f(u)

Can you clarify please?


every 20 numbers, the last two digits repeats. 1234/20 is something r 14. 'that should mean the last two digits is 69.

of course, i probably made a silly mistake.

[quote="math92"]What are the last two digits of 3 to the 1234 power?[/quote]

\[3^{1234} \equiv 3^{34} \pmod {100}\]
\[3^{34} \equiv (3^2)^{17} \pmod {100}\]
\[9^{17} \equiv 9^9 \cdot 9^8 \pmod {100}\]
\[9^9 \cdot 9^8 \equiv 89 \cdot 21 \pmod {100}\]
\[89 \cdot 21 \equiv \fbox {69} \pmod {100}\]

Why is this topic called "adress"?


I need help with these problems:

1) Consider an object that has a mass, $ m$, and a weight, $ W$, at the surface of the moon. If we assume
the moon has a nearly uniform density, which of the following would be closest to the object’s
mass and weight at a distance halfway between Moon’s center and its surface?

A) $ \frac12 m$ & $ \frac12W$     B) $ \frac14 m$ & $ \frac14 W$      C) $ 1 m$ & $ 1 W$      D) $ 1 m$ & $ \frac12 W$      E) $ 1 m$ & $ \frac14 W$

2)One object at the surface of the Moon experiences the same gravitational force as a second
object at the surface of the Earth. Which of the following would be a reasonable conclusion?

[A] both objects would fall at the same acceleration
[B] the object on the Moon has the greater mass
[C] the object on the Earth has the greater mass
[D] both objects have identical masses
[E] the object on Earth has a greater mass but the Earth has a greater rate of rotation.

3)Two artificial satellites, $ 1$ and $ 2$, are put into circular orbit at the same altitude above Earth’s
surface. The mass of satellite $ 2$ is twice the mass of satellite $ 1$. If the period of satellite $ 1$ is $ T$,
what is the period of satellite $ 2$?

A. $ T/4$ B. $ T/2$ C. $ T$ D. $ 2T$ E. $ 4T$

4)A hypothetical planet orbits a star with mass one-half the mass of our sun. The planet’s
orbital radius is the same as the Earth’s. Approximately how many Earth years does it take for
the planet to complete one orbit?

When solving problems like these, what formulas or concepts are important to remember?
And I would also appreciate it if someone points me toward good websites for more practice.

Thanks.

Here's some help on the first one:

What is mass?  Does mass ever change?  If so, when?  (Assuming stationary-it technically changes at relativistic speeds but let's ignore that).  How about weight?  What is it?  What does Newton's Law of Gravitation state?

If you can answer these questions, there's no reason you can't answer your question.

[quote="kcn2rivers"]Well I know that the mass of the object doesn't change in this case. Also, the weight of the object is the force of gravity acting between the object and the moon.
So, if the object is at the surface, the weight is $ W_1 \equal{} \frac {Gm_{obj}m_{moon}}{r^2}$.
If the object is halfway between the center and the surface, the weight is $ W_2 \equal{} \frac {Gm_{obj}m_{moon}}{(0.5r)^2} \equal{} 4W_1$. So, the weight would be $ 4$ times larger in if it's in the middle.
But that's not one of the answers. What am I doing wrong here?[/quote]
If the object is inside the moon,then the outer shell of it doesn't exert force on the object,so $ m_{moon}$ should be $ \frac {1}{8}$ as it used to be.Thus the force is $ \frac {1}{2}$ smaller. The answer's $ \boxed{D}$,isn't it?

Yes that's right, thank you.

Can someone tell me if this is right for #3?

$ R^3 \equal{} kT_1^2 \equal{} kT_2^2$, where $ R$ is the mean distance of the two satellites. $ k$ is the same value because the satellites are both orbiting around Earth. So, $ T_1\equal{}T_2$, $ \boxed{C}$ ?

If I am right, the mass of a satellite doesn't affect its period if the radius is kept constant?

[quote="kcn2rivers"]
Can someone tell me if this is right for #3?

$ R^3 \equal{} kT_1^2 \equal{} kT_2^2$, where $ R$ is the mean distance of the two satellites. $ k$ is the same value because the satellites are both orbiting around Earth. So, $ T_1 \equal{} T_2$, $ \boxed{C}$ ?

If I am right, the mass of a satellite doesn't affect its period if the radius is kept constant?[/quote]
I think it's right.


vector
analytic geometry
superior algebra
superior algebra solved

I've totally lost in this question, Any help is highly appreciated.

Given

          (U`,U``)+(V`,V``)=(U`+V`+1, U``+V``-1)

           k(U`,U``)=(kU`+k-1, kU``-k+1)

Let vecter U=(1,0),vecter V=(-1,0)

a) Show that {U,V} form a basis in V

b)Find the coordinate vectors of (2,0) abd (0,0) and (-1,1) relative to this basis.

[quote="Myth"]We have $kU=(2k-1,-k+1)$ and $lV=(-1,-l+1)$, $kU+lV=(2k-1,-k-l+1)$. Threfore if $kU+lV=(a,b)$ then $k=\frac{a+1}{2}$ and $l=-b-\frac{a+1}{2}+1$. We see that each $(a,b)$ can be uniquely expressed.[/quote]

How do you deduce such a scalar multiplication. It doesn't seem clear, at least not for me.

Thanks. Best,

[quote="Myth"]I don't know what you mean. I don't see any scalar multiplication in my post.[/quote]

What does it mean then $kV$, because if $kV=V+V+V+\cdots+V$ $k$ times, then you don't have a representation for every vector. Note that otherwise, you are multiplying a vector for an apriori real, not necessarely ingeter, scalar!!

Best regards,

[quote="Myth"]But who are speaking about integer multipliers? See problem: there is a rule for multiplication by any real number.[/quote]

OK, I missread, but anyway, the problem itself is written wrong, because it does not define a real vector space (a vector space has to have a scalar multiplication, that is, a rule of multiplication for elements on the field) but the one they propose does not hold with the distributive property, because, in general, $kX+kY\neq k(X+Y)$.

Best,

That's because this is an exercise with a silly disguise. Isomorphism from $V$ to $F^2,$ where $F$ is the field the elements are taken from: $(a,b)\mapsto (a+1,b-1)$


infinite grid is partitioned into dominoes. Coloring the set of dominoes in 2 colors is good
if any point belongs to at most 2
one-coloured dominoes. Is there a good coloring for any partition?


find real solutions of the system
$\sin x+2\sin(x+y+z)=0$
$\sin y+3\sin(x+y+z)=0$
$\sin z+4\sin(x+y+z)=0$

This should help:

[hide]sin(x+y+z) = sin(x)cos(y+z) + cos(x)sin(y+z)
               = sin(x)cos(y)cos(z) - sin(x)sin(y)sin(z) + cos(x)
                            sin(y)cos(z) + cos(x)cos(y)sin(z)

Therefore, we have the following equations:
sin(x) + 2cos(x)cos(y)sin(z) + 2cos(x)sin(y)cos(z)
         + 2sin(x)cos(y)sin(z) - 2sin(x)sin(y)sin(z) = 0
sin(y) + 3cos(x)cos(y)sin(z) + 3cos(x)sin(y)cos(z)
         + 3sin(x)cos(y)sin(z) - 3sin(x)sin(y)sin(z) = 0
sin(z) + 4cos(x)cos(y)sin(z) + 4cos(x)sin(y)cos(z)
         + 4sin(x)cos(y)sin(z) - 4sin(x)sin(y)sin(z) = 0

Through a simple rearrangement and combination,
sin(x)/2 = sin(y)/3 = sin(z)/4
           = -cos(x)cos(y)sin(z) - cos(x)sin(y)cos(z)
               - sin(x)cos(y)sin(z) + sin(x)sin(y)sin(z)[/hide]

Edit: I'm not a newbie anymore!!!  :lol:


algebra
polynomial
quadratics
number theory unsolved
number theory

Given some prime $p$, let $a, b \in \mathbb{Z}_p$. If the polynomial $x^3 + ax + b$ is irreducible over $\mathbb{Z}_p$, prove that $-4a^3 - 27b^2$ is a square in $\mathbb{Z}_p$.

Here's the advanced but easy way to do it:
Any irreducible polynomial of degree $n$ over $F_p$ splits completely in the field with $p^n$ elements. If the discriminant of a polynomial is not a square, the splitting field of that polynomial contains the square root of the discriminant, and therefore the splitting field contains a quadratic extension of the base field. An extension of degree $n$ can only contain an extension of degree $m$ if $m|n$. Since $2$ does not divide $3$, the splitting field of an irreducible cubic over $F_p$ cannot contain the square root of a nonsquare element of $F_p$, and the discriminant must already be a square.

In fact, a cubic polynomial with a nonsquare discriminant must factor as the product of a linear polynomial and an irreducible quadratic.

Is there a reasonable elementary solution?

[quote="perfect_radio"]what's the discriminant of a polynomial? i know what it is for $n=2$ ( $\Delta=b^2-4ac$ ) but how do we define it for bigger degrees?[/quote]
You need to know the general solution to a cubic equation.
$x^3+ax^2+bx+c=0$
Substitute $y=x+\frac{a}{3}$, the equation becpomes
$y^3+py+q=0$ where $p=\frac{a^2}{3}+b$ and $q=\frac{2a^3}{27}-\frac{ab}{3}+c$
The solution is
$y=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$.
(This is found by letting $y=u+v$ such that $3uv=-p$)
The discriminant is $\frac{q^2}{4}+\frac{p^3}{27}$.
Another definition of discriminant is given [url=http://en.wikipedia.org/wiki/Cubic_equation#The_nature_of_the_roots]here[/url].

Normally, you define the discriminant of a polynomial as $\left(\prod_{1 \leq i < j \leq n} (a_i-a_j)\right) ^2$ where the $a_k$ are the $n$ roots of the polynomial. E.g., the discriminant is $0$ iff two roots are identical.


At a party there were:

14 adults,
17 children,
12 males,
19 females.

Then X arrived and the number of man-woman couples possible became equal to the number of boy-girl couples possible. (e.g.: If there are 6 men and 8 women, there are 48 possible man-woman couples)

Is X a man, woman, boy, or girl?

X is a [hide]girl OR X is a woman[/hide]because of the following . . .
[hide]Let 12 - x = # of male children
Then x = # of male adults
Let 19 - y =  # of female children
Then y = # of female adults
=========================================

Case I -- Suppose X is a boy.

Add 1 to 12 - x.
(13 - x)(19 - y) = xy
This can become:
247 = 13y + 19x
Which can become:
13 - x = (13/19)y
No solutions in positive integers x and y.

------------------------------------------------

Case II -- Suppose X is a girl.

Add 1 to 19 - y.
(12 - x)(20 - y) = xy
This can become:
60 = 3y + 5x
Here, x = 3, y = 15; x = 6, y = 10; x = 9, y = 5
This supports X is a girl.

----------------------------------------------------

Case III -- Suppose X is a man.

Add 1 to x.
(12 - x)(19 - y) = (x + 1)y
This can become:
228 = 13y + 19x
Which can become:
12 - x = (13/12)y
No solutions in positive integers x and y

----------------------------------------------

Case IV -- Suppose X is a woman.

Add 1 to y.
(12 - x)(19 - y) = x(y + 1)
This can become:
57 = 3y + 5x
Here, x = 6, y = 9
This supports X is a woman.

\_/---\_/---\_/---\_/---\_/---\_/---\_/---\_/---\_/

Therefore, X can be either a girl or a woman.
[/hide]

Arrange your tan, you've neglected one of the constraints of the problem (that there be 14 adults and 17 children); as a result, 3 of your 4 proposed possible answers don't actually satisfy the requirements of the question.

Amending that, from my case II
(where one supposes the addition of a girl),
it would have to be that x = 9 and y = 5.

Then there would be 9 men, 5 women, 3 boys,
and 14 girls before the one girl was added.


abstract algebra
algebra
polynomial
induction
superior algebra
superior algebra unsolved

Let M = {f(x) in Q[x] | f(Z) is a subset of Z} (here Q is the rationals and Z is the integers). Show that M is a free Z-module of countably infinte rank with basis {(x i)|i=0,1,2,...} Here (x i) is the binomial coefficient.

It is readily that $M$ is $\mathbb{Z}$-module, so we actually just need to verify that the polynomials  form a base in this $\mathbb{Z}$-module. In order to do this we can use the following fact:
Let $p$ be a polynomial of degree $\leq k$ such that $p(\mathbb{Z}) \subset \mathbb{Z}$, then $p(x)= \sum_{i=0}^k c_0 \left(\begin{array}{c}x\\k-i \end{array} \right)$ (*) with $c_i \in \mathbb{Z}$.
Proof. It is easy to readily that $\left(\begin{array}{c}x\\k-i \end{array} \right), i=0,1,2...,$ form a basis in $\mathbb{R}[X]$ and thus there exist such $c_i$ as in (*), we only need to prove that $c_i$ are actually integers.
We can do this by induction on the degree of polynomial $k$. The base is trivial. Suppose this result is prove for all the polynomials of degree $\leg k$, and let $p(x)= \sum_{i=0}^{k+1} c_0 \left(\begin{array}{c}x\\k-i \end{array} \right)$, consider $\Delta p(x) =p(x+1)-p(x)=\sum_{i=0}^k c_0 \left(\begin{array}{c}x\\k-i \end{array} \right)$ - it takes integer values at integers piints, so by induction hypothesis $c_i,i=0,1,...,k \in \mathbb{Z}$ and $c_{k+1}=p(n)- \sum_{i=0}^k c_0 \left(\begin{array}{c}n\\k+1-i \end{array} \right)$ is also integer, so we are done.


In my Blog Options, here is the exact CSS in the "Custom CSS" box:

[code]#main {
background: none;
}

#wraper {
background: none;
}[/code]

(yes I am quite aware that it should be "#wrapper")

I want it to be

[code]#main {
background: transparent;
}

#wrapper {
background: transparent;
}[/code]

However, when I apply these changes and click "Submit", when I click "Click Here to set additional options for your blog," I go back and see that the changes were [b]not[/b] made and the original CSS is still there. Is this a bug or am I just doing something wrong?

Bug confirmed. Pretty sure it's a permission problem on the folders.


What is the smallest positive value of $\theta$ where $\sin \theta$ is the same in degrees and radians?

Draw it out. You'll see that when the radian graph comes back down, it must intersect with the degrees graph.

$\sin k-\sin (180-k)$

We have:
$\sin \theta = \sin \left( 180-\frac{\pi}{180}\theta \right) \implies \theta = \frac{180 \pi}{180+\pi}$

In fact its already [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=714637#p714637]here[/url] :wink:
I guess it never says there but thats AIME 2002 II #10

[quote="pkothari13"]This was an AIME problem from a few years back!
I think the question asked to find the two smallest values of theta.[/quote]
Oh wow, I didn't even know if this question was possible so I posted it here.
How do we know for sure that the two posted at the other link are the smallest?


quadratics
number theory
prime factorization

[b]WHAT IS THE LEAST POSSIBLE NUMBER OF POSITIVE INTEGERS SUCH THAT THE SUM OF THEIR SQUARES EQUALS1995 :?: [/b]

The sum of the squares of the FIRST $ n$ integers is $ \frac{n(n\plus{}1)(2n\plus{}1)}{6}$

So what we'll do is find an upper bound first.

$ n(n\plus{}1)(2n\plus{}1) \equal{} 11970$
$ n \equal{} 17.660604$

So our upper bound is 18 integers.

Can you modify this to reach a better result?

Sums of squares of consecutive positive integers have little to do with sums of squares of a collection of integers (and even if they did, the calculation you performed does not imply an upper bound).  The actual upper bound for [i]any[/i] positive integer is $ 4$ by the Four Square Theorem.

Hmm. Could you explain that theorem a little more?
EDIT: I can do it in three:

$ 37^2 \plus{} 25^2 \plus{} 1^2 \equal{} 1995$
I don't think it's possible to go lower than that.

Correct.  It's not possible to write $ 1995$ as a sum of two squares.

[hide="Generalizations"] A positive integer $ n$ is expressible as a sum of two squares iff any prime congruent to $ 3 \bmod 4$ in the prime factorization of $ n$ has even exponent.

A positive integer $ n$ is expressible as a sum of three squares iff it is not of the form $ 4^n (8k \plus{} 7)$ for non-negative integers $ n, k$.

[i]Every[/i] positive integer is expressible as a sum of four squares. [/hide]

[quote="t0rajir0u"]
A positive integer $ n$ is expressible as a sum of three squares iff it is not of the form $ 4^n (8k \plus{} 7)$ for non-negative integers $ n, k$.
[/quote]

Where'd you get THAT from?

That seems a little arcane to me. I followed the other two lines pretty well.

As I recall, the proof of that result is actually more difficult than the proof of the Four Square Theorem.  It's easy to verify that numbers congruent to $ 7 \bmod 8$ can't be written as the sum of three squares (since squares are congruent to $ 0, 1, 4 \bmod 8$).  It's also easy to verify that if a sum of three squares is congruent to $ 0 \bmod 4$ then each square is divisible by $ 4$.  In other words, the hard part is showing that numbers not of that form [i]are[/i] representable as a sum of three squares.

For the 3-squares-theorem, you either need a lot about quadratic forms, or the theorem of Hasse-Minkowski, or some work and Dirichlet's theorem on primes in arithmetic progression...
It's the hardest case, all other number of squares are much more elementary.

[quote="t0rajir0u"]A positive integer $ n$ is expressible as a sum of three squares iff it is not of the form $ 4^n (8k \plus{} 7)$ for non-negative integers $ n, k$. [/quote]

Can you provide an example of this? I'm not sure what $ k$ stands for.

$ k$ is a non-negative integer.  The set of positive integers that can't be written as a sum of three squares can be ordered first according to the value of $ k$ and then according to the value of $ n$.  In other words, the list of all such integers can be written

$ n \equal{} 0:  7, 15, 23, 31, 39, 47, 55, ...$
$ n \equal{} 1:  4 \cdot 7, 4 \cdot 15, 4 \cdot 23, 4 \cdot 31, 4 \cdot 39, 4 \cdot 47, 4 \cdot 55, ...$
$ n \equal{} 2:  16 \cdot 7, 16 \cdot 15, 16 \cdot 23, 16 \cdot 31, 16 \cdot 39, 16 \cdot 47, 16 \cdot 55, ...$
$ n \equal{} 3:  64 \cdot 7, 64 \cdot 15, 64 \cdot 23, 64 \cdot 31, 64 \cdot 39, 64 \cdot 47, 64 \cdot 55, ...$
$ n \equal{} ...$

and so forth.


A line segment of length 15 + 5sqroot of 3 is folded to create a 30-60-90 right triangle. Two of these are put together to form a equilateral triangle. Four of these equilateral triangles are put together to form a tetrahedron. Two of these are put together to form a hexahedron. What is the volume of this final hexahedron?[/img]

Dude, this was already posted.
Try figuring out some on your own, Chris.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=116080

Plagiarism=very bad. Reposting a problem when the original is in the top 10 topics=very bad.


5. Moe uses a mower to cut his rectangular 90-foot by 150-foot lawn. The swath
he cuts is 28 inches wide, but he overlaps each cut by 4 inches to make sure that
no grass is missed. He walks at the rate of 5000 feet per hour while pushing
the mower. Which of the following is closest to the number of hours it will take
Moe to mow his lawn?

$ (A) 0.75 \qquad (B) 0.8 \qquad (C) 1.35 \qquad (D) 1.5 \qquad (E) 3$

file attached

if you question, emal to me
thstjdus0629@naver.com

I had no clue what all that swath stuff meant, so I just did stuff. It would regularly take him 2.7 hours, and 1.35 is half of that, so I chose it.


No offense, but I don't get it at all...

$ \frac{1}{10}$

It's like the classic problem "I have two coins that have a total value of 15 cents.  One of the coins is not a dime.  What are the two coins?"  The answer is a nickel and a dime because the problem states that [b]one[/b] coin (the nickel) is not a dime.  It doesn't say anything about the other coin.

Hmm... surd raised to a surd implies that the surds are the same,... but $ (\sqrt 2)^{\sqrt 2} \neq 2$... Perhaps the surd you are looking for is $ 1.55961046946$ :)

Yongyi781,

no, it does not impy that the surds are the same!  The phrase
"a surd raised to a surd" literally means an irrational number
raised to an irrational number.  There is nothing in the English
that stated/states that they are the same surd.  You worked
the English, which was/is not ambiguous or otherwise, incorrectly.

One of the surds is the the square root of two.  The other
surd is the square root of two raised to the square root of
two.

No, the other surd is 2.
\[ (\sqrt 2)^2 \equal{} 2
\]
But 2 is not irrational, contradiction.

But, you can go the other way.
\[ \left(\sqrt 2^{\sqrt 2}\right)^{\sqrt 2} \equal{} 2
\]
Which gives the desired result.

[quote="QuantumTiger"]The only thing I find funny is that you are now banned from StripCreator.[/quote]
That doesn't make any sense, unless you know that being banned from
StripCreator is actually a badge of honor, because the users and moderators
of it are tyrants who want comics to made a certain way (cookie-cutter style),
and the moderators don't tolerate any dissent.  (If you want to keep going
off topic, away from the merits of the comic strip, then do so, but you'll not
be making any valid arguments that way.)

I'm sorry for going off topic, but I just happened to find that funny, just as you happened to find your comic funny. Sorry, but I know nothing about StripCreator. Please forgive me.  :roll:


group theory
abstract algebra
superior algebra
superior algebra unsolved

Any one have an idea?
:)
Here's the Question:
"Let $p$ and $q$ be distinct primes and suppose that $G$ is a finite group with exactly $p+1$ $p$-Sylow subgroups and exactly $q+1$ $q$-Sylow subgroups.  Prove that there exists a $p$-Sylow subgroup $P$ and a $q$-Sylow subgroup $Q$ of $G$ such that the direct product $P\times Q$ is also a subgroup of $G$."

Note: we are assuming that $p$ and $q$ are not $2$ and $3$.

I was given a hint to consider the normalizer of one of the Sylow subgroups (that is $N_G(P)$ or $N_G(Q)$) to help show there is an injective group homomorphism from $P\times Q$ into $G$.

Or to consider the action of $P$ on the conjugates of $Q$.  That either approach should help solve the problem.  But I'm not having much luck yet.

Any help would be appreciated.

Prove that there exists a $p$-Sylow subgroup $P$ and a $q$-Sylow subgroup $Q$ of $G$ such that the direct product $P\times Q$ is also a subgroup of $G$.

What exactly do you mean here, up to isomorphism?

I don't know if I understand your question.
In my question, we know there are exactly $p+1$, $p$-Sylows and $q+1$, $q$- Sylows of the group $G$.
And we have to show that the direct product of some $p$-Sylow and some $q$-Sylow is a subgroup of $G$.
So I need help because I don't know HOW to show that $P\times Q$, the direct product of a $p$-Sylow and a $q$-Sylow of $G$, is a subgroup of $G$.

I don't know if 'up to isomorphism' has anything to do with this problem.

Please let me know if any clarification is needed.

I would love some input on this problem.

Thanks!


yeah i know this problem was one of the problems in the 2001 imo short list - proposed by Bogdan Enescu, Romania.

a hint for the solution: first apply Cauchy and then try to re-arrange the terms to obtain a telescopical sum (i.e. a sum of which some consecutive terms have the sum 0).


geometry unsolved
geometry
BritishMathematicalOlympiad

Using Menelaos we obtain that DL=LC, where L is the intersection of AE and DC. That means that m(LAC)=m(LCA). Using that m(DAL)=m(LAD),we find DAC=90. :lol:

That's ok.
$\frac{CL}{CD}\frac{DA}{AB}\frac{BE}{EC}=1\Rightarrow CL=LD$.

[quote="PaulS"]In triangle $ABC$, $D$ is the midpoint of $AB$ and $E$ is the point of trisection of $BC$ nearer to $C$. Given that $\angle ADC=\angle BAE$ find $\angle BAC$.[/quote]

[b] A simpler solution,but the same nice ![/b]

 If $M$ is the midpoint of $BE$ then $BM=ME = EC$ and
   [color=red]a) [/color]from triangle $BAE$ : $DM // AE$
 [color=red]  b) [/color]from triangle $CDM$: $LD= LC$, since $EL // MD$ and $E$ is the midpoint of $MC$.
  But , if $L$ is midpoint of $CD$  , then  $LA=LD=LC$ , so triangle $BAC$ must be right at vertex $A$.

 [u]Babis[/u]


Euler
function
number theory
prime numbers

Calculate the Euler function of m for m= p (a prime number), where m=p1p2 where pi and p2 are prime numbers, m=95. Find alln that satisfies that The Euler function for n is even.

This is a very confused post.

[hide="Sorting out"] There are many common lemmas about $\varphi(n)$, two of which are that

$gcd(a, b) = 1 \implies \varphi(ab) = \varphi(a) \varphi(b)$
$\varphi(p) = p-1$

Hence

$\varphi(p_{1}p_{2}) = (p_{1}-1)(p_{2}-1)$

For $m = 95$ we have

$\varphi(5 \cdot 19) = 4 \cdot 18 = 72$

As for the last question, $\varphi(n)$ is even for all $n$ except $n = 2$.  This is because $gcd(a, n) = 1 \implies gcd(n-a, n) = 1$.  Since $a \neq \frac{n}{2}$ except for $n = 2$ it follows $a \neq n-a$. [/hide]


Find the remainder on dividing $ x^{100} \minus{} 2x^{51} \plus{} 1$ by $ x^2 \minus{} 1$

$ x^2\equiv 1\pmod{x^2\minus{}1}$, so we have $ 1\minus{}2x\plus{}1\equal{}2\minus{}2x$.

I don't think that's the right solution
This is what the book says but I don't understand.

[hide="Solution"]
$ x^{100} \minus{} 2x^{51} \plus{}1\equal{} (x^2 \minus{} 1)q(x) \plus{} ax \plus{} b$. Putting $ x \equal{} 1$ into this relation gives  $ b \equal{} 0$. Putting $ x \equal{} \minus{} 1$ gives $ a \equal{} \minus{} 4$. Thus the remainder is $ \minus{} 4x$
[/hide]

[quote="TheHeinrich"]I don't think that's the right solution
This is what the book says but I don't understand.

[hide="Solution"]
$ x^{100} \minus{} 2x^{51} \plus{} 1 \equal{} (x^2 \minus{} 1)q(x) \plus{} ax \plus{} b$. Putting $ x \equal{} 1$ into this relation gives  $ b \equal{} 0$. Putting $ x \equal{} \minus{} 1$ gives $ a \equal{} \minus{} 4$. Thus the remainder is $ \minus{} 4x$
[/hide][/quote]

Check the substitution. Putting $ x\equal{}1$ gives $ a\plus{}b\equal{}0$ and putting $ x\equal{}\minus{}1$ gives $ \minus{}a\plus{}b\equal{}\minus{}4$. Solve that simple system, and you get what I said.

$ f(x)\equal{}x^{100}\minus{}2x^{51}\plus{}1\equal{}(x^2\minus{}1)Q(x)\plus{}l(x)$, $ l(x)\equal{}ax\plus{}b$
$ \Longleftrightarrow f(x)\minus{}l(x)\equal{}(x\plus{}1)(x\minus{}1)Q(x)$, which means that $ y\equal{}f(x)$ has intersection points with $ y\equal{}l(x)$ at $ x\equal{}\pm 1$, or

$ y\equal{}l(x)$ passes through two points $ (\minus{}1,\ 4),\ (1,\ 0)$, yielding $ l(x)\equal{}\minus{}2x\plus{}2$.


[quote="GameBot"]How many integers belong to the arithmetic sequence 13, 20, 27, 34, ..., 2008?[/quote]

$ 2008\equal{}13\plus{}(n\minus{}1)7\implies n\equal{} \frac {2008\minus{}13}{7}\plus{}1\equal{}286$


inequalities
geometry
incenter
inequalities proposed

\[ \sum\frac{a}{b\plus{}c\minus{}a}\equal{}\frac{2R\minus{}r}{r}\]

\[ \sum\frac{b\plus{}c\minus{}a}{a}\equal{}\sum(\frac{a\plus{}b\plus{}c}{a}\minus{}2)\equal{}\frac{(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)}{abc}\minus{}6\equal{}\frac{p^2\plus{}r^2\plus{}4Rr}{2Rr}\minus{}6\equal{}\frac{p^2\plus{}r^2\minus{}8Rr}{2Rr}\]

It is known that $ p^2\le 4R^2\plus{}4Rr\plus{}3r^2$ so

\[ \sum\frac{b\plus{}c\minus{}a}{a}\equal{}\frac{p^2\plus{}r^2\minus{}8Rr}{2Rr}\le\frac{4R^2\minus{}4Rr\plus{}4r^2}{2Rr}\equal{}\frac{2R\minus{}r}{r}\minus{}\frac{R\minus{}2r}{R}\equal{}\sum\frac{a}{b\plus{}c\minus{}a}\minus{}\frac{R\minus{}2r}{R}\]

We are done, I haven't solve the second, but the first equivalent to $ p^2\le 4R^2\plus{}4Rr\plus{}3r^2$ so the second stronger than $ p^2\le 4R^2\plus{}4Rr\plus{}3r^2$, I think it isn't true?

Hi! I am new to geometric inequalities. Can you please show me how to prove "well-known" inequality:

\[ \sum \frac {a}{b \plus{} c \minus{} a}\ge\sum \frac {b \plus{} c \minus{} a}{a}
\]

Thank you!

P.S: in encyclopedia's solution, he uses:

\[ \sum \frac {a}{b \plus{} c \minus{} a}\equal{}\frac{2R\minus{}r}{r}
\]
How do you get this kind of identities?

[quote="shaaam"]Hi! I am new to geometric inequalities. Can you please show me how to prove "well-known" inequality:
\[ \sum \frac {a}{b \plus{} c \minus{} a}\ge\sum \frac {b \plus{} c \minus{} a}{a}
\]
Thank you!

P.S: in encyclopedia's solution, he uses:
\[ \sum \frac {a}{b \plus{} c \minus{} a} \equal{} \frac {2R \minus{} r}{r}
\]
How do you get this kind of identities?[/quote]
For the last question:
We have $ 4RT \equal{} abc$ and $ T \equal{} rs$, where $ T$ is the areal of the triangle with side $ a,b,c$. ($ T^2 \equal{} s(s\minus{}a)(s\minus{}b)(s\minus{}c)$ where $ s \equal{} \frac{a\plus{}b\plus{}c}{2}$)
Now $ \frac{2R\minus{}r}{r} \equal{} \frac{abc \minus{} 2(s\minus{}a)(s\minus{}b)(s\minus{}c)}{2(s\minus{}a)(s\minus{}b)(s\minus{}c)}$. And thus it comes down to proving that:
$ \sum \frac {a}{2(s\minus{}a)} \equal{} \frac{abc \minus{} 2(s\minus{}a)(s\minus{}b)(s\minus{}c)}{2(s\minus{}a)(s\minus{}b)(s\minus{}c)}$, which should be possible.

For the inequality:
Let $ a\equal{}x\plus{}y$, $ b\equal{}y\plus{}z$ and $ c\equal{}z\plus{}x$. Then $ x,y,z > 0$.
Then we just have to prove:
$ \sum_{cyc} \frac{z\plus{}x}{2y} \ge \sum_{cyc} \frac{2y}{z\plus{}x}$. Then just notice that $ z\plus{}x \ge 2\sqrt{zx}$. Thus it suffices to prove that:
$ \sum_{cyc} \frac{z\plus{}x}{2y} \ge \sum_{cyc} \frac{y}{\sqrt{xz}}$. Which follows by muirhead since $ (1,0,\minus{}1) \succ (1,\minus{}\frac{1}{2},\minus{}\frac{1}{2})$.

Thanks Mathias!  :)
we can also finish off your proof differently. we have to prove:

\[ \sum_{cyc} \frac{z\plus{}x}{2y} \ge \sum_{cyc} \frac{2y}{z\plus{}x}
\]

From Cauchy-Schwarz:
\[ \sum_{cyc} \frac{2y}{z\plus{}x}\leq \frac{1}{2}\cdot \sum_{cyc}\left(\frac{y}{x}\plus{}\frac{y}{z}\right)\equal{}\sum_{cyc} \frac{z\plus{}x}{2y}
\]

 :)

How you prove that $ p^2\leq 4R^2\plus{}4Rr\plus{}3r^2$

[quote="Flakky"]How you prove that $ p^2\leq 4R^2 \plus{} 4Rr \plus{} 3r^2$[/quote]
If $ I$ is incenter and $ H$ is orthocenter, then $ IH^2\equal{}4R^2\plus{}4Rr\plus{}3r^2\minus{}p^2\ge 0\ .$

[quote="Virgil Nicula"]The geometrical inequality $ \sum \frac {a}{b + c - a}\ge\sum \frac {b + c - a}{a}$ is [u]well-known[/u].

[b]Prove that[/b] $ \{\begin{array}{cc} 1\blacktriangleright & \sum\frac {a}{b + c - a}\ge\sum\frac {b + c - a}{a} + \frac {R - 2r}{R}\ . \\
 \\
2\blacktriangleright & \sum\frac {a}{b + c - a}\ge\sum\frac {b + c - a}{a} + \frac {R - 2r}{R}\cdot\frac {8R - r}{4(2R - r)}\ .\end{array}$[/quote]
We know the distancies $ \|\begin{array}{c} \mathrm {IH}^2 = 4R^2 + 4Rr + 3r^2 - s^2\ge 0 \\
 \\
\mathrm H\Gamma^2 = 4R^2[1 - \frac {2s^2(2R - r)}{R(4R + r)^2}]\ge 0\end{array}\|$ and we can show easily

that $ \|\begin{array}{c} \sum\frac {a}{b + c - a} = \frac {2R - r}{r} \\
 \\
\sum\frac {b + c - a}{a} = \frac {s^2 + r^2 - 8Rr}{2Rr}\end{array}\|$ . Thus, $ \boxed {\ s^2\stackrel {(*)}{\ \le\ } \frac {R(4R + r)^2}{2(2R - r)}\le 4R^2 + 4Rr + 3r^2\ }$

and $ \sum\frac {a}{b + c - a} - \sum\frac {b + c - a}{a} =$ $ \frac {2R - r}{r} - \frac {s^2 + r^2 - 8Rr}{2Rr} = \frac {4R^2 + 6Rr - r^2 - s^2}{2Rr}\stackrel {(*)}{\ \ge\ }$

$ \frac {4R^2 + 6Rr - r^2 - \frac {R(4R + r)^2}{2(2R - r)}}{2Rr} =$ $ \frac {2(2R - r)(4R^2 + 6Rr - r^2) - R(4R + r)^2}{4Rr(2R - r)} =$ $ \frac {(R - 2r)(8R - r)}{4R(2R - r)}\ge 0$

$ \implies$ $ \boxed {\ \sum\frac {a}{b + c - a}\ge\sum\frac {b + c - a}{a} + \frac {(R - 2r)(8R - r)}{4R(2R - r)}\ge\sum\frac {b + c - a}{a} + \frac {R - 2r}{R}\ge\sum\frac {b + c - a}{a}\ }\ .$


Here are the correct answers compiled by some kids at my school for the Barbra Nunn that Algebra II and Geometry students took today. If you disagree with an answer, post the problem number and your solution.

[list]
1. B
2. A
3. D
4. C
5. C
6. C
7. E
8. C
9. A
10. E
11. A
12. D
13. D
14. D
15. B
16. A
17. C
18. B
19. B
20. A
21. C
22. B
23. D
24. B
25. C
26. C
27. C
28. D
29. C
30. D
[/list]

Didnt take it.... better place to post this though is on the mualpha message board - just a suggestion.

[quote="WarpedKlown1335"]Farazy, don't you go to Seminole?[/quote]

Wassup Yannik, Nice to talk to you again, this is David.


Show that the sum of the inradii of a triangulated cyclic polygon is independant of the triangulation.

http://agutie.homestead.com/files/Geoproblem_B.htm#Carnot

See 6. Sangaku Problem

I never solved this prb, but I think using the fact that cosA+cosB+cosC=1+r/R works, combined with the fact that the polygon is cyclic. I'm not at all sure, though.


geometry
geometric transformation
rotation

[quote="ccy"]The Riverside Bed and Breakfast Inn serves three married couples breakfast at their round table which seats exactly six people. The hostess wants to seat guests so that no husband and wife sit next to each other and the guests alternate male and female. How many different arrangements are possible? A rotation of an arrangement is not considered a different arrangement.[/quote]

[hide]29? there are 6*6-7the[/hide]

Remember, it's a round table, so you want circular permutations.

[hide]There are two different combinations for the women to sit at. After they are seated, there are two combinations for the men to sit at. Answer:4[/hide]

[hide]Suppose a husband and his wife sits down, it doesn't matter where, but this stops the rotations of the table.

Next to the husband, women #2 and #3 can either sit on his left or his right, so there are a total of $\boxed{2}$ ways.[/hide]

[hide]Supposing the couples are (#1, #2) (#3, #4) (#5, #6)
#1 sits down first. Now on either side of him can only be #4 and #6.
After all the possibilitys, there are only two ways.
Possibilitys (clockwise) :
[b]#1, #4, #5, #2, #3, #6,
#1, #6, #3, #2, #5, #4,[/b] :lol:
  [/hide]

[hide]
Since we want changing between male and female and couples are together, we can have each couple be its own individual unit.
THen there are 3 units and for circular permutations the formula is $(n-1)!$
so the answer is $2!=2$[/hide]


http://www.artofproblemsolving.com/Forum/memberlist.php
Look at #1

[quote="moogra"]He has 100% of the posts.[/quote]

Yeah, it used to be three people. Now it's someone else.

yeah.. i saw three people with that many number of posts... :D

One was a test by me (Chromotron)  :wink:
This is just a bug, not harmful, just a bit disturbing.

Wow congratulations. [/sarcasm]

This is the $ n$th time this bug has been mentioned. The admins are fixing it.


If $ y \equal{} \log_{a}{x}$, $ a > 1$, which of the following statements is incorrect?

$\textbf{(A)}\ \text{If }x=1,y=0 \qquad\\
\textbf{(B)}\ \text{If }x=a,y=1 \qquad\\
\textbf{(C)}\ \text{If }x=-1,y\text{ is imaginary (complex)} \qquad\\
\textbf{(D)}\ \text{If }0<x<z,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero} \qquad\\
\textbf{(E)}\  \text{Only some of the above statements are correct}$

[hide]A is correct since $ a^0\equal{}1$. B is correct because $ a^1\equal{}a$. C is correct since if $ x\equal{}\minus{}1$, $ y$ must be imaginary. D is also correct...so E[/hide][/hide]

First of all, what is z? Assuming z is infinity, then A is right because numbers to the zeroth power are 1, and B is right because numbers to the first power are just themselves, and C is right because positive numbers to complex powers can be negative, but D is wrong because what if let's say, a is 2 and x is 4, then y is 2 and isn't negative, but it does decrease without limit as x approaches zero. Still, in this way, D is wrong, so E is right, so answer = D.

[quote=ItsAmeYushi]That is a typo, the original test has
$\textbf{(D)}\ \text{If }0<x<1,y\text{ is always less than 0 and decreases without limit as }x\text{ approaches zero} \qquad$

This is certainly true, if you are not convinced, you can look at a graph.[/quote]

OK this makes a lot more sense, if on D, 0<x<1, then D is right and E is wrong. Thanks!


search
number theory solved
number theory

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

Thanks.

[quote="pbornsztein"]The result still holds for a collection of $2n-1$ integers. This appeared many times here.
Search for Erdös-Gizburg-Ziv theorem.

Pierre.[/quote]

This is also  Ucraina Olympiad


The figure below shows a right triangle $ \triangle ABC$.
[asy]unitsize(15);
  pair A = (0, 4);
  pair B = (0, 0);
  pair C = (4, 0);
  draw(A -- B -- C -- cycle);
  pair D = (2, 0);
  real p = 7 - 3sqrt(3);
  real q = 4sqrt(3) - 6;
  pair E = p + (4 - p)*I;
  pair F = q*I;
  draw(D -- E -- F -- cycle);
  label("$A$", A, N);
  label("$B$", B, S);
  label("$C$", C, S);
  label("$D$", D, S);
  label("$E$", E, NE);
  label("$F$", F, W);[/asy]
The legs $ \overline{AB}$ and $ \overline{BC}$ each have length $ 4$.  An equilateral triangle $ \triangle DEF$ is inscribed in $ \triangle ABC$ as shown.  Point $ D$ is the midpoint of $ \overline{BC}$.  What is the area of $ \triangle DEF$?

[hide="Bash"]Let $ A \equal{} ( \minus{} 2,2), B \equal{} ( \minus{} 2,0), C \equal{} (2,0), D \equal{} (0,0)$, and let $ F \equal{} ( \minus{} 2,a)$. We rotate $ F$ about the origin $ D$ -60 degrees counterclockwise to find $ E \equal{} ( \minus{} 1 \plus{} \frac {\sqrt {3}}{2}a,\sqrt {3} \plus{} \frac {1}{2}a)$. $ E$ lies on the line $ AC$, which is $ x \plus{} y \equal{} 2$. Thus, we have $ \minus{} 1 \plus{} \frac {\sqrt {3}}{2}{a} \plus{} \sqrt {3} \plus{} \frac {1}{2}a \equal{} 2 \iff a \equal{} \frac {2(3 \minus{} \sqrt {3})}{\sqrt {3} \plus{} 1} \equal{} 4\sqrt {3} \minus{} 6$. Then $ DF^2 \equal{} ( \minus{} 2)^2 \plus{} (4\sqrt {3} \minus{} 6)^2 \equal{} 88 \minus{} 48\sqrt {3}$. The area of this triangle is then this times $ \frac {\sqrt {3}}{4}$, or $ 22\sqrt {3} \minus{} 36$.[/hide]

Your work was fine until the final answer.

Ah, found the error and edited it. Thanks.

[hide="Alternate Solution"]
Let $BF=f$, $DE=EF=FD=s$, and $EC=x$. Then, we get three equations, one from the Pythagorean Theorem and two from LoC: \[ \begin {eqnarray*} f^2 + 4 &=& s^2, \\ (4-f)^2 + (4\sqrt{2}-x)^2 - 2(4-f)(4\sqrt{2}-x)\frac{1}{\sqrt{2}} &=& s^2, \\ 4 + x^2 - \frac {4x}{\sqrt{2}} &=& s^2. \]It is a bit of a pain to solve this, but not impossible. After much bashing, we see that $ s = \pm 2 \cdot \sqrt {22 \pm 12\sqrt{3}} $. Obviously, the only non-extraneous solution is $ s = 2 \cdot \sqrt {22 - 12\sqrt{3}} $, so we get \[ [\triangle DEF] = \frac {\sqrt{3}s^2}{4} = \sqrt {3} \cdot \left( 22 - 12\sqrt{3} \right) = \boxed {22\sqrt{3} - 36}. \]
[/hide]

We could also use polar coordinates centered around D, noting that since $DE = DF$, we have in polar coordinates $E = (r, \theta)$ and $F = (r,\theta -60)$. Then use $r\cos(\theta)$ and $r\sin(\theta)$ along with the fact that for $E$, $y=-x+2$.


[quote="Christiano89"]Υπάρχει κάποιο βιβλίο το οποίο να λέει πως λύνουμε συναρτησιακές εξισώσεις??[/quote]

Δεν γνωριζω αν υπαρχει τετοιο βιβλιο, η θεωρια στις συναρτησιακες ειναι ελαχιστη, το παν ειναι να σκευτεσε τις σωστες αντικαταστασεις... υπαρχουν καποια ελαχιστα θεωρηματα με τα οποια λυνονται καποιες συναρτησιακες (συναρτησιακη Cauchy, Jensen κλπ) καθως και καποιες μεθοδολογιες (κυριως επαγωγες) για τα οποια υπαρχει ενα κεφαλαιο στο βιβλιο για ολυμπιαδες του Arthur Engel (Springer εκδοσεις)

οκ τότε... Θα βρω το βιβλίο και θα το ψάξω. Ευχαριστώ

Δώσε μου Gmail να σου στείλω αν μπορέσω βιβλία με συναρτησιακές σχέσεις ηλεκτρονικά


This contest seemed harder than the last one.

[hide="my results"]
bclgold     1  1  0  0  1  0  0  0  0  0
gourmet     *  *  *  *  *  *  t  *  t  t
sightsee    *  *  *  *  *  *  *  *  t  t
[/hide]

[hide="sightsee"]
Binary search on the answer, use Bellman-Ford to detect negative cycles.
[/hide]

[hide="gourmet"]
Greedy, use interval trees to make it NlogN.
[/hide]

[hide="bclgold"]
Use Rabin-Karp (polynomial) hashing, binary search on the agreement between prefix and suffix and then greedy.
[/hide]

I did:
bclgold     1  0  0  0  1  0  0  0  0  0
gourmet     *  *  *  *  *  x  t  t  t  t

bclgold I did the failed O(n) algo. Gourmet I did greedy, except I did it inefficiently [sort, then unnecessarily do #ofcows swaps, then start at the most expensive and work downwards, for guaranteed O(n^2) instead of more like O(nlogn).]

[code]
bclgold     1  1  1  1  1  1  1  1  4  4
gourmet     *  *  *  *  *  *  t  t  t  t
sightsee    *  x  x  x  x  x  x  x  t  t
[/code]

My sightsee isn't working correctly for some reason, and it's a bit slow in any case. 15 or so runs of O(VE) Bellman-Ford seems like it would time out, I think.

[code]
           1  2  3  4  5  6  7  8  9 10
bclgold     1  1  1  1  1  1  1  1  0  0
gourmet     *  *  *  *  *  x  x  t  t  t
[/code]

O(N^2) on bcl. I got 6/8 of the cases in 9 and 10, but it seems I get no credit for that :( Can someone explain the method of suffix arrays used in the analysis?

I had O(N log N) in gourmet, but I did not use long longs (64 bit) hence the two WA and coding errors (a comparison that is always true), made my set work in O(N) time instead of O(log N), hence my code was O(N^2). As of now, I have gotten my program to work through case 8, but I can't get 9 or 10.

To move up midcontest, you generally need all but one or two testcases at least. If you don't move up midcontest, 800 points generally will move you from bronze to silver and 700 points will generally move you from silver to gold (the cutoff was below 700 a lot last year).

[quote="nebula42"]It might be because of double imprecision. Try adding epsilons (like 1e-9 or something) in your comparisons in Bellman-Ford, and see if that fixes it.[/quote]

Actually, it turns out the problem was that in my Bellman-Ford, I used index i instead of j when looping through the edges. I still can't get the last test case to work in time though. Did anybody else get it?

Double imprecision shouldn't be a problem on sightsee because they seem to be accepting answers within $ .01$ of the correct answer...

I think the test data was a bit big. Richard Peng claims that if you stop updating when none of the values change, then it will run in time. Also, if that's too slow then try only updating edges coming out of vertices that were updated in the previous run. The largest data is random so this should be a significant improvement.

My guess is IOI scoring. (Maybe no one got the last couple cases on sightsee and they'll be weighted zero.)

[quote="Fraxia"]I'd discuss problems, but I'm only in Bronze Division  :P

Do you need to solve all the problems in one contest in order to be bumped up to Silver? I couldn't the details on moving between divisions.[/quote]
Since you did bronze, did you think my analyses were clear enough?  :)  (If you haven't seen them yet, you can find them [url=http://ace.delos.com/DEC07]here[/url].)

[quote="JSteinhardt"]Double imprecision shouldn't be a problem on sightsee because they seem to be accepting answers within .01 of the correct answer...

I think the test data was a bit big. Richard Peng claims that if you stop updating when none of the values change, then it will run in time. Also, if that's too slow then try only updating edges coming out of vertices that were updated in the previous run. The largest data is random so this should be a significant improvement.

My guess is IOI scoring. (Maybe no one got the last couple cases on sightsee and they'll be weighted zero.)[/quote]
I meant double imprecision in comparisons, as in doing
[code]if (a + b > c + epsilon)[/code]
instead of just
[code]if (a + b > c)[/code]
Also, it's definitely possible to get a fast enough solution on 'sightsee', since I currently have a working solution that takes at most 0.15 seconds on the contest data. I do agree that the limits were a bit high though, especially since the problem was basically hard enough already.  :|

well.. in the gold1 sightseeing problem, how can u make the node weight into edge weight? are u making node an edge? if so, then how can u handle no path goes through a node multiple time? will it be automatically handled?

You can prove that the optimal cycle will simple, so you don't have to worry about repeated vertices.


Show that the distance between opposite edges in an icosahedron is phi times the sidelength.

I thought that proofs weren't allowed in Getting Started?  :?

Edit: Oh, I see you moved it.  :)


hi...
let u v subspace in R^n and dimv=dimu=n-1 ,show it u+v=R^n

This is false if the two subspaces are equal, so you'd better include $ U\ne V$ as a condition.

If they're not equal, then either there's an $ x\in U\setminus V$ or there's an $ x\in V\setminus U.$ What can you do with that?

u and v different vector i need complet proof pleas

[quote="npare"]i need complet proof pleas[/quote]
This is not the recommended way to make the regular participants here want to answer you. We'd like to see some signs of effort on your part - including sometimes being satisfied with a hint.


calculus
integration
function
real analysis
derivative
algebra
functional equation

Find the continuous function $f(x)$ which satisfies for all real numbers $x$ the following equation.

\[\int^x_0 f(t)dt+\int^x_0 t f(x-t)dt=e^{-x}-1\]

This is a good one for the use of Laplace transforms. If we let $F(s)=\int_0^{\infty}e^{-sx}f(x)dx$ be the Laplace transform of $f$, it isn't too hard to show that the Laplace transform of the first term (the indefinite integral of $f$) is $\frac1sF(s).$ The second term is the convolution of $x$ with $f(x).$ The Laplace transform of the convolution is the product of the Laplace transforms, and the Laplace transform of $x$ is $\frac1{s^2}.$ Thus if we take the Laplace transform of the whole equation, we get

$\frac1sF(s)+\frac1{s^2}F(s)=\frac1{s+1}-\frac1s$

$F(s)=-\frac{s}{(s+1)^2}=-\frac1{s+1}+\frac1{(s+1)^2}$

From this, $f(x)=-e^{-x}+xe^{-x}=e^{-x}(x-1).$

Hello! Kent Merryfield.

That's the very [b]Laplace Transform [/b].You are right.
Regrettably,in Japan which you have been tired of hearing,(laghing!),senior high school student don't study Laplace Transform.I will post my solution later.
Please wait for a while.

kunny

We could use integration by parts.. :)

If we take $F(x)=\int^x_0 f(t)dt$ and $\mathcal F(x)=\int^x_0 F(t)dt$ we get, after changing the second intergal to $\int^x_0 (x-t)f(t)$, $F(x)+\mathcal F(x)=e^{-x}-1$, and left part is simply $\frac{[e^x\mathcal F(x)]'}{e^x}$

A second solution.

Note as a preliminary that convolution is commutative, and that with a change of variables we could write that
$\int_0^xtf(x-t)dt=\int_0^x(x-t)f(t)dt=x\int_0^xf(t)dt-\int_0^xtf(t)dt$
Differentiate this by the Fundamental Theorem of Calculus, using the product rule where appropriate. Its derivative is
$\int_0^xf(t)dt+xf(x)-xf(x)=\int_0^xf(t)dt.$

Now differentiate the original integral equation:

$f(x)+\int_0^xf(t)dt=-e^{-x}$

Putting $x=0$ in this gives us $f(0)=-1.$

Differentiate a second time:

$f'(x)+f(x)=e^{-x}.$

This is a first order linear differential equation; the technique of multiplying factors works easily for it. The solution becomes:

$e^xf(x)=x+C$. The initial condition $f(0)=-1$ implies $C=-1.$

$f(x)=e^{-x}(x-1).$

I posted the Laplace transform solution first just because the problem is so beautifully suited to transform methods.

[quote="kunny"]Regrettably,in Japan which you have been tired of hearing,(laghing!),senior highschool student don't study Laplace Transform.[/quote]
In the U.S., this would most likely be in a course intended for 2nd year (or sometimes 3rd year) college students in Differential Equations. It's not part of the AP Calculus syllabus that the top high school students follow; high school students typically would not have seen it. My second solution includes only topics that would be in the syllabus of AP Calculus BC.

Thank you for your solutions! grobber and Kent Merryfield.

My solution is exactly same as [b]Kent Merryfield's [/b]solution,which is the typical solution in Japanese Entrance Exam for university.To tell the truth,since 1997
differential equation,functional equation have been eliminated in curicurum of senior high school's mathematics,but next year these are making revival.
So is linear transformation instead of dissapearing Complex plane (Gaussian plane).

kunny


Does anyone know how to solve the following problem, or the origin, known results or anything related?  Thank you!

Let's call a triple of numbers good if one of them is the sum of the other two.  Call a number n admissible if the numbers 1,2,3,..., 3n can be partitioned into n good triples.  An obvious necessary condition that n be admissible is n to be 0 or 1 modulo 4.  Is the necessary condition also sufficient?

Some previous threads on this problem:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=218089

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1054571#1054571

I do not know if your conjecture is true. I think $ n\equal{}24$ is the smallest value I'm not sure about (that is, can you partition $ \{1,\ldots,72\}$ into 24 good triples?).


I am somewhat confused, but was is the difference between a hedge fund and an investment bank?
Also is there anywhere online that will provide me with a list of hedge funds since I am having no luck finding one.
Thanks.

[url]http://en.wikipedia.org/wiki/Hedge_fund[/url] has a list.

Also: please start a new topic for a new question, rather than posting a somewhat off-topic question in a long-dormant thread.  Thanks.

Thanks for the list, and sorry for the off topic question.


Let $ a,b,c,d\geq 0$ and $ \frac{1}{p}\plus{}\frac{1}{q}\equal{}1$, $ (p,q>1)$, and $ \alpha\plus{}\beta \equal{}1$, $ \alpha , \beta \geq 0$ then
\[ (\alpha a\plus{}\beta c)^{\frac{1}{p}}(\alpha b\plus{}\beta d)^{\frac{1}{q}}\geq \alpha a^{\frac{1}{p}}b^{\frac{1}{q}}\plus{}\beta c^{\frac{1}{p}}d^{\frac{1}{q}}\]


Let us play THE GAME. If you want to know the rules, ask at your own risk. If you know how to play THE GAME, I dare you to post a ddr emoticon telling me you know how the play THE GAME. You are allowed to annoy others. Begin![/i]

Don't lock, or else there will be no place to announce loss...

I just lost...

[Moderator: I propose PMing mathlete every loss]

Mods decide what gets locked, not the posters. Anyway, I won.
[img]http://imgs.xkcd.com/comics/anti_mind_virus.png[/img]
At least this topic can be locked in peace now.

[Moderator: Indeed!]
[hide=";)"][Moderator: Well you may start one in Fun Factory][/hide]


there are 4 points on a plane :A,B,C,D
prove that the length of all of AB,AC,AD,BC,...,CD (6 ones) are not  integer
unfortunately my english is not good but i hope you understand what i mean...

[quote="Rafikafi"]I think you have to add no 3 are on one line cause
$ (0|0) (3|0) (3|4) (6|0)$ would be a counterexample if i understood you right[/quote]
yes you are right
i told this to my friend and he edited his problem
[b] the number of the points is infinite and they are not all on a line [/b]


Find the point on hte graph of y=Square root of ( x+1) closet to the point (3,0)

This is a calculus problem again. You just want to minimize the distance between a point on the function and the given point. I don't think this kind of problem belongs in this forum.

Not necessarily calculus... but the same idea:

$d^2 = (x-3)^2+(x+1) = x^2-5x+10 = \left(x-\frac{5}{2}\right)^2+\frac{15}{4}$ which has a minimum at $(x,y) = \left(\frac{5}{2}, \frac{\sqrt{14}}{2}\right)$.


here is what i and Vihang(whose initiative this was) decide that why not we share our screen shots...the most beautiful.. i know there are lots of geeks out here....out of the 20 odd selected in INO! 4 visit AOPs/ML ...so why not
let me start with mine
[url=http://ubuntuforums.org/attachment.php?attachmentid=35672&d=1182110755]this[/url] is how mine looks like :)

he would have got the idea from long back  :D  And how come it says that i cannot view the page like i have to be registered?  :huh:   :)

that is why I WROTE that IT WAS HIS INITIATIVE AND HE WANTED ME TO START :mad:
taht's because  u aren't registered [url=http://ubuntuforums.org/]here[/url]
let me find another way out

[url=http://img120.imageshack.us/img120/4660/screenshotls7.png]Mine[/url]

Yeah that's XP ...
My FC6 looks like any other standard FC6 ... no point in putting that here.

so u have a mac leopard tranformation...gr8..
try compiz fusion if u can with FC6

nice idea vihang
[url=http://img257.imageshack.us/my.php?image=ubuntufinaleww5.png]here[/url] is mine
[url=http://img412.imageshack.us/my.php?image=screenshotyz8.png]here[/url] is  another


Hello people!

Welcome to [b]jain_harshi's one day tournament[/b]. This tournament's first round is made up of response games, 4 people each. Each game will have 10 questions. The winners move on to the semifinals, which is made up of countdowns. The final ends in a countdown. Please join!

I don't believe it is; we're already having a CTC Tourney scheduled for mid-August - we wouldn't be having two tourneys - it defies the point of CTC.

[quote="myyellowducky82"]is this CTC-certified?[/quote]

No, this tourney is illegal for all CTC members; it is not CTC-certified.

That pretty much makes it impossible, seeing as almost everyone is in CTC.

Sorry, but since this tournament is illegal, it has to canceled. I actually [b]forgot[/b] the CTC-certification rule.

It's not really illegal, it's just that no one in CTC is supposed to join it, which means few people will join it.

I am part of the CTC. It kind of like breaking a law.

It is breaking a law. Shut down this tournament or you are exiled from CTC.

I did. Can't you see? For sure, I WILL CLOSE IT AGAIN:

[size=200]

[b][i][u]This tournament is now closed, canceled, destroyed, etc.[/u][/i][/b][/size]

SURE I'LL JOIN. DON'T REALLY WANT TO BUT WHO CARES :rotfl:

But the [size=200]Tournament is closed[/size]. No one can now join


Is there a $3 \times 3$ magic square consisting of distinct Fibonacci numbers (both $f_{1}$ and $f_{2}$ may be used; thus two $1$s are allowed)?

no, it doesn't exist.

we have:

A  B  C
D  E  F
G  H  I.

$ f_n$ is the greatest number that we use.

CASE 1)$ f_n$ is in A (or simmetrically in C,G,I)
we must have A+B+C=C+F+I so A+B=C+I but C+I $ \le$ A < A+B, absurd.

CASE 2)$ f_n$ is in B(or simmetrically in D,F,H)
we must have B+E+H=D+E+F  so B+H=D+F but D+F $ \le$ B < B+H, absurd.

LAST CASE)$ f_n$ is in E.
we must have G+H+I=A+B+C=D+E+F>E. but $ \frac{(G\plus{}H\plus{}I)\plus{}(A\plus{}B\plus{}C)}{2} \le \frac{\sum_{i\equal{}n\minus{}6}^{n\minus{}1}{f_i}}{2}\equal{}\frac{(f_{n\minus{}1}\plus{}f_{n\minus{}2})\plus{}(\sum_{n\minus{}6}^{n\minus{}3}{f_i})}{2}< f_n$, absurde

bye
paolo

an easier approach in which we prove that even the sum of 3 different rows cannot be the same;

Let f(n+4) be the largest occuring Fibonacci number. Then there must be a row in which the largest Fibonacci number is no larger than
f(n+2). But the sum of that row is no larger than f(n) + f(n+1) + f(n+2) = f(n) + f(n+3) < f(n+2) + f(n+3) = f(n+4), impossible.

Note that the previous proof alsof easily generalizes to m*m magic squares; If we let f(m+n-1) be the largest occuring Fibonacci number, there must be a row in which the largest Fibonacci number is no larger than f(n). The sum in that row is smaller than or equal to: f(n-m+1) + f(n-m+2) + .. + f(n) < (m-1)f(n) < 1.5^(m-1)f(n) < f(n+m-1), contradiction.

Another approach; Let X = F(x) = the x-th fibonacci number. Let

A B C
D E F
G H I

be our 3*3 magic square. Then A + I = 2E = C + G. This implies that max(A, C, G, I) is at least f(e+2). But this is already larger than 2E, contradiction.


Of course it is.

Don't worry, another mod will respond to your problems (most of the time, anyway)  :D

BTW, here is another, if you want.

What are all possible solutions to $\prod_{k=0}^{\infty}x^{k}$?

[size=75][color=red]b-flat: Post in a new thread  :D [/color][/size]

None, the exponent grows without bounds and $0^\infty$ is an indeterminate form and therefore no real solutions.

Sorry, I meant to say, something more like what are all the things that $x\times x\times x\cdots$ tends to (for non-negative $x$)?

Sorry I didn't know that it was so complicated  :blush:

I thought that it was just if $\mid x\mid <1$, then it is 0, if $x=1$, then it is 1, and if $x>1$, then it will keep getting bigger and bigger, approaching infinity.

Sorry again.  :oops:

[quote="nebula42"][quote="Kent Merryfield"]Infinite product; the question of whether or under what circumstances it converges is an issue. I'd say it probably belongs in the "calculus computations and tutorials" forum, along with most other things involving the convergence of infinite sums or infinite products.

This infinite product tends to zero if $\mid x\mid <1.$ It converges to $1$ if $x=1.$ Otherwise, it diverges.

(There's a convention with infinite products that would say tht even though the manner in which this tends to zero for $0<\mid x\mid <1$ is certainly a convergent sequence, we nonetheless say that the product diverges to zero.)[/quote]
Just wondering, what happens if $x=-1$? Would you just say that it's undefined? (since it's either $1$ or $-1$, but it's impossible to tell which)[/quote]

Wouldn't this be called indeterminate? Since we can't determine what the product is?


Show that for any positive integer $n$, we can find a positive integer $f(n)$ such that $19^{f(n)}-97$ is divisible by $2^{n}$.

Let $k$ be natural number which satisfies $2^{n}\parallel 19^{k}-1$. Then we can define $f(n+1)=f(n)$ if $2^{n+1}| 19^{f(n)}-97$ or otherwise $f(n+1)=f(n)+k$. Such a $k$ exists because of the lemma if $2^{\alpha}\parallel a^{2}-1$ then for any $\beta \geq 0$: $2^{\alpha+\beta}|a^{n}-1$ iff $2^{\beta+1}|n$


[quote="bpms"][quote="randomdragoon"]In 12 years Martha age will be two times of Bill's age now. When Bill's age was 1/2 of Martha's age then, Bill was 16 years younger than Martha is now. How old is Martha?[/quote]
[hide]
$m+12=2b$
$\frac{m+12}{2}=m-16$
$m+12=2m-32$
$m=44$
martha was 44.
We don't need to find billy's age, but it is 28.[/hide][/quote]
Sorry, your solution isn't consistent with the last condition "When Bill's age was 1/2 of Martha's age then, Bill was 16 years younger than Martha is now." Please note the convoluted use of present and past tense in the same sentence.

(I'm not sure if I worded it properly -- what I mean is if Bill were 6 and Martha were 9, Bill was 1/2 of Martha's age then when Bill was 3 and Martha was 6 -- 3 years ago).