You are in a dark room. It is pitch black and you can't see anything. There is a table in the room, and on it are 100 quarters. 25 of them are heads up and 75 are tails up. You have to divide all the quarters into 2 groups so that each has the same number of heads in it.

Good Luck! This problem seems fairly easy once you know the answer, but is hard to figure out.

This belongs in puzzles and brainteasers.

Can you flip any of the coins over?

Also, if you can, can you telll if a coin is tails or head up depending  on how it feels?

you would have to be able to flip coins over, otherwise there would always be an odd number of heads

Yes, you can flip them over, but you have no way of telling whether it is heads or tails up.

[hide]Divide the coins into a pile of 25 and one of 75 (the one with 25 is on the left). Suppose that there are x heads in the left pile. Then there are 25-x heads in the right pile. Now flip over all the coins in the left pile. There are now x tails and 25-x heads, so there are an equal number of heads in both piles.[/hide]

The only reason I came up with this is because I read the solution in a book...

That's right! :lol: But next time try doing without a book!

[quote="lotrgreengrapes7926"]I think this was posted in Getting Started or Intermediate before.[/quote]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103044]Here[/url], for instance.

[quote="JBL"]lotrgreengrapes7926 wrote:
I think this was posted in Getting Started or Intermediate before.

Here, for instance.[/quote]

I suppose this is somewhat a special case of that particular problem.

Also, the method used for solving both problems could be extended to any number of heads, providing that the solver knows the number.

This problem was a problem of the day at MathPath.  The solution that Mathnerd3.14 presented was the same solution at MathPath.

You are right, once you know the solution it is obvious, but it is hard to figure it out.


Αν α θετικός ακέραιος, β και γ ακέραιοι και ισχύουν:

5<β+γ<30

και β^2-α^2-1=2(γ^2-αβ-2αγ)

να βρεθεί η μέγιστη τιμή του β και η ελάχιστη τιμή του γ.

(Ενώ δε χρειάζεται κάτι σπουδαίο, την κατασκεύασα ανάποδα και αυτό την κάνει αρκετά έως πολύ δύσκολη)

Η συνθήκη γράφεται .
$2(c-a)^{2}=(b+a)^{2}-1$ .

Όμως από την άλλη συνθήκη έχουμε $2(c-a)^{2}\geq 2(5-b-a)^{2}$
Άρα $(b+a)^{2}-1\geq 2(5-b-a)^{2}$ που είναι ισοδύναμη με
$(a+b-3)(a+b-17)\leq 0$

Synexizetai sto epomeno post

Σιλουανέ θα μπορούσες σε παρακαλώ να εξηγήσεις πώς προέκυψε ότι η δοσμένη ανίσωση ισοδυναμεί με την $2(c-a)^{2}\geq 2(5-b-a)^{2}$  :?: και πού χρησιμοποίησες ότι β,γ ακέραιοι...
Επίσης νομίζω πως παρότι πιθανόν η λύση σου να είναι και σωστή, δεν εξετάζεις κάποιες περιπτώσεις (οι οποίες βέβαια οδηγούν σε άτοπο αλλά αυτό είναι άλλο ζήτημα :P ).Π.χ. θα μπορούσε η περίπτωση $b+a=3$ να μας δίνει μεν ένα μικρότερο β αλλά και ταυτόχρονα να δίνει ένα "ελάχιστο" γ (βέβαια αυτή η περίπτωση δίνει $b+c=5$, άτοπο)
Τέλος πάντων θεωρώ πως θα έπρεπε να δουλέψουμε με βάση τη σχέση $2(c-a)^{2}=(b+a)^{2}-1$ στους ακεραίους και να βρούμε τα "μικρότερα" ζευγάρια των ακεραίων για τα οποία επιτυγχάνεται αυτό :wink:

Κώστα περιμένω και τη δική σου λύση για να δω τι είχες στο μυαλό σου :lol:

Βασικά αυτό <<ελάχιστο γ και μέγιστο β>>  με προβληματίζει....
Ίσως έπρεπε να γίνει καλύτερη διατύπωση kai parapano ekana lathos .Na to sosto
Όσο για την ανισότητα χρησιμοποιώ ότι $c\geq 5-b$   :wink:
Λοιπόν και για να γίνει πιο ξεκάθαρο .
Από την παραπάνω βρήκαμε ότι $b+a\in [3,17]$
Εξετάζοντας όλες τις ακέραιες τιμές οι μόνες που δίνουν ακέραια λύση για το c-a είναι οι $b+a=17$ kai $b+a=3$ ap;o tis opoies prokyptoun antistoixa oi
$c-a=12$ kai $c-a=2$ . H deuteri d;inei atopo afou dinei $b+c=5$ eno i alli dinei
$b+c=29$ . Και τώρα το μέγιστο b,c που δίνουν λύσεις είναι
$b=28$ , $c=1$

Σιλουανέ σωστά έχεις βρει ότι αναγκαστικά β+γ=29. Όμως γιατί το β να είναι 28 και το γ 1? Δεν μπορεί π.χ. να είναι β=30 και γ=-1? Δεν είπα θετικοί ακέραιοι... Είσαι πολύ κοντά πάντως.

Για να γίνω πιο συγκεκριμένος, η ισοδύναμη εξίσωση που έδωσες στην αρχή είναι προφανώς Pell και αν ψάξεις τις πρώτες λίγες λύσεις είναι α+β=3, α-γ=2
                α+β=17, α-γ=12
                α+β=99, α-γ=70

Λόγω της συνθήκης είναι δεκτή η τρίτη λύση άρα β+γ=29 (είναι πολύ εύκολο να αποδείξεις από τους τύπους λύσεων της PELL ότι η διαφορά των λύσεων αυξάνεται).

Άρα α+β=99 και α-γ=70.
Επειδή α θετικός ακέραιος για α=1 έχουμε β=98 και γ=-69 ως μέγιστη και ελάχιστη τιμή.

Ρε Κώστα τι θα γίνει με αυτά τα edit που κάνεις στα posts σου :mad: με ανάγκασες να διαγράψω ολόκληρο post αυτή τη φορά :P
Τέλος πάντων, Σιλουανέ το $c\geq 5-b$ είναι λάθος γιατί δεν ισχύει η ισότητα...γι' αυτό και το γεγονός ο φραγμός του $a+b$ αποδείχθηκε λάθος :wink:

Ναι έχω κάνει λάθος γιατί όταν ύψωσα στο τετράγωνο.....(δεν ήταν θετικά  :oops: )

[quote="Nick Rapanos"][quote="bilot4"]Ρε Κώστα τι θα γίνει με αυτά τα edit που κάνεις στα posts σου :mad: με ανάγκασες να διαγράψω ολόκληρο post αυτή τη φορά :P
Τέλος πάντων, Σιλουανέ το $c\geq 5-b$ είναι λάθος γιατί δεν ισχύει η ισότητα...αλλιώς η λύση σου θα ήταν απολύτως σωστή :wink:[/quote]

Pote ekane edit o kostas?[/quote]\

Στο προηγούμενο ποστ του πρόσθεσε τη λύση...και σε ένα άλλο στο topic της Α' Λυκείου αν δεν απατώμαι...

Καλά μην αρπάζεσαι. :wink:  Θα' βαζες τη λύση? Σόρρυ ρε. :blush:

Δεν "αρπάζομαι" (άκου λέξη:)) εξάλλου δεν είχα σκοπό να βάλω τη λύση( αφού δεν τη βρήκα :P ) απλά ένα τρόπο να προσεγγίσουμε τη λύση...τέλος πάντων

Mh mou thymizete tis palies epoxes despotaki-rakar....tha piaso ola ta post sas kai tha kano rate os spam!


There are n points on a circle. These can be joined in many ways to form an inscribed n-gon. From any point on the circle, perpendiculars are drawn to the sides to one of these n-gons. Prove that the product of lenghts of the n perpendiculars is the same for all n-gons

Ok, how can they be joined in many ways? Say I have five points around the circle. There's only one pentagon that they can make; all the others have intersecting sides.

Also, what exactly do you mean by "the perpendiculars"? It might be more clear if you say, for example, let $ A_{1}$, $ A_{2}$, $ A_{3}$, ... $ A_{n}$ be n points distributed around a circle, labeled in counterclockwise order, and then define whatever you want using those point labels.

sorry, but I 'm not really sure I can make it clearer than that.

Then you aren't trying very hard.  Where are the perpendiculars drawn from?

The perpendiculars are drawn from an arbitrary point on the circle.

[quote]
There are n points on a circle. These can be joined in many ways to form an inscribed n-gon. [b]From any point on the circle, perpendiculars are drawn to the sides to one of these n-gons.[/b] Prove that the product of lenghts of the n perpendiculars is the same for all n-gons
[/quote]

[quote="Ponnamperuma"]I'll try to clarify your doubts... please, mdk, say if I'm right!

I think that the biggest problem is the use of the letter $ n$. Perhaps the problem is:
There are $ n\geq 3$ points on a circle. These can be joined in many ways to form an inscribed convex $ k$-gon, where $ 3\leq k\leq n$. Let $ P$ be a point on that same circle. Draw the perpendiculars from $ P$ to the sides of the chosen $ k$-gon. Prove that the product of the lengths of the $ k$ perpendiculars (as the segment that connects $ P$ with a side of the $ k$-gon) is the same for all the possible choices of $ k$.[/quote]

you correct. Nice way to restate the problem.

Thanks, what's the source of the problem? It seems nice!... is it?  :wink:


why would you memorize (or look up) such a pointless identity when $\sin{3x}=\sin{(2x+x)}$...?

Other than that, yea it's fine

uh it's easy to memorize with a mnemonic

like

Shaq = 34

[quote="mysmartmouth"][quote="Elemennop"]
Shaq = 34[/quote]

Uh, you wanna explain that?  I guess by 34 you mean his age (his number is 32) so wouldn't that change every year?  And why would I remember how old Shaq is anyway....just wondering...[/quote]

Back in my day (when shaq was on the lakers) Shaq wore 34.  When he was on the Magic and now on the Heat he wore/wears 32.  On a good day, it's his free throw percentage as well.

Yeah, the last time I really kept up with basketball was back around when Shaq wore 34 and T-Mac was still on the Magic...yeah...

yea I remember his lakers number

anyways, shaq corresponds to sine, and the 34 corresponds to the coefficients (3sinx - 4sin^3x)

[quote="mysmartmouth"]Problem:
Compute $\sin (3\cos^{-1}(\frac{3}{5}))$.

I did this problem, but there was no answer in the book.  My method (is this correct?):

[hide="Answer"]
Starting with the inside, we rewrite to get $\cos x = \frac{3}{5}$.  This means that $\sin x = \frac{4}{5}$.  However, we are looking for $\sin 3x$.  So we use the identity:

$\sin 3x = 3 \sin x-4\sin^{3}x$.

Plugging in values:

$\sin 3x = 3(\frac{4}{5})-4(\frac{4}{5})^{3}$
$\sin 3x = \frac{12}{5}-4(\frac{64}{125})$
$\sin 3x = \frac{300}{25}-(\frac{256}{125})$
$\sin 3x = \boxed{\frac{44}{125}}$
[/hide][/quote]

Shouldn't $\sin x = \pm \frac45$?

[quote="mysmartmouth"]I didn't memorize it; I derived it from what you said.  I agree that would be pointless to learn.[/quote]

eh, well why did u spend extra time simplifying everything in terms of $\sin$?

and adorableaddict, it probably specified 1st quadrant

[quote="AdorableAddict"][quote="mysmartmouth"]Problem:
Compute $\sin (3\cos^{-1}(\frac{3}{5}))$.

I did this problem, but there was no answer in the book.  My method (is this correct?):

[hide="Answer"]
Starting with the inside, we rewrite to get $\cos x = \frac{3}{5}$.  This means that $\sin x = \frac{4}{5}$.  However, we are looking for $\sin 3x$.  So we use the identity:

$\sin 3x = 3 \sin x-4\sin^{3}x$.

Plugging in values:

$\sin 3x = 3(\frac{4}{5})-4(\frac{4}{5})^{3}$

Well
$\sin 3x = \frac{12}{5}-4(\frac{64}{125})$
$\sin 3x = \frac{300}{25}-(\frac{256}{125})$
$\sin 3x = \boxed{\frac{44}{125}}$
[/hide][/quote]

Shouldn't $\sin x = \pm \frac45$?[/quote]

Yes, it should, be that, but the book probably said something like positive.

[quote="lotrgreengrapes7926"][quote="AdorableAddict"]Shouldn't $\sin x = \pm \frac45$?[/quote]
There is only one value for $\cos^{-1}\frac{3}{5}$ and its $\sin$ is $\frac{4}{5}$. This is because the range of the principal inverse cosine ($\cos^{-1}$) is $[0,\pi)$. The $\sin$ of anything between $0$ and $\pi$ is nonnegative. :D

Consequently, $\sin(\cos^{-1}\theta)\ge0$ and $\cos(\sin^{-1}\theta)\ge0$ for any $\theta$.[/quote]

Yea.  Just consider a triangle with sides 3, 4, and 5, and this follows from the definitions of sin and cos.


function
logarithms
calculus
calculus computations

Let $ V(t)$ is a positive function, i.e., $ V(t)\ge 0,\forall t\ge 0$  such that

$ V(t)\le \gamma^n V(0), \forall t\ge nT_0$, $ n\equal{}\{0,1,2,\cdots\}$

where  $ 0< \gamma<1$. Then, prove that $ V(t)\rightarrow 0$ [b]exponentially fast [/b]as $ t\rightarrow 0$. Determine the rate of convergence.

I think this problem is simple. I can see the convergence result. But, I am not able to show that the convergence is exponential.

Thanks for all your help,
Ajoye

I think I was missing a big condition that $ \dot{V}(t) < 0, \forall t > 0$ and $ \dot{V}(0)\equal{}0$. Hence, $ V(t)$ is a monotonically decreasing function everywhere except origin.

I'm also pretty sure that "as $ t\to 0$ must be "as $ t\to\plus{}\infty$". ;)

As to the question itself, just write down the definition of exponential convergence and compare what is given to what you need to prove. If you want a more detailed hint, look
[hide="Here"]
$ \gamma^{[t/T_0]}\le \gamma^{\minus{}1}e^{(\log \gamma/T_0)t}$
[/hide]

[quote="fedja"]I'm also pretty sure that "as $ t\to 0$ must be "as $ t\to \plus{} \infty$". ;)
[/hide][/quote]

Yes it is. Thats embarrassing. Thanks Feda.

Ajoye


[quote="breez"]Does anyone know of a good online website that explains electromagnetic waves and radiation?[/quote]

Yes, you will find a complete understanding of Electromagnetic Energy and it's radiation process on the website of

[url]http://theuniversalcodeofliving.com[/url]

Also check out [b]The Second Theory of Everything[/b]

Cheers David


I'm wondering if there's a nice way to get Asy to recognize homemade TeX style files or non-standard packages. Here's an example.

I would like to color certain parts of captions in diagrams to make them stand out, but when I try to do {\color{green} text}, I get an error from Asy. I know that Asy has its own way of coloring text, but this extends to commands from other large TeX packages. Is there an easy way to include new TeX packages while still running Asy externally from WinEdt?  :maybe:

Use texpreamble command at the beginning of your .asy code. Like this:
[code]texpreamble ("\usepackage{x}");
texpreamble ("\usepackage{y}");[/code]


Determine whether the matrices are inverses:
[ 5 3 ]  , [  2 –3]
 3 2          -3  5

Mark if you didnt know the answer you could have just told me.  You didnt have to post it in the forum

Jenny maybe someone in the forum can help you?
Dont get mad at me

Determine whether the matrices are inverses:
[ 5 3 ]
  3 2
[ 2 –3]
  3 5

[quote="Mark9182"][ 5 3 ]
  3 2
[ 2 –3]
  3 5[/quote]
Any typo there? I think you forgot a minus before $3$ in the last line.
[quote="Hoborobo"]$\begin{pmatrix}5 & 3 \\ 3 & 2 \end{pmatrix}\begin{pmatrix}2 &-3 \\-3 & 5\end{pmatrix}= \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$

So yes, they are inverses.

Btw. It took me 10 minutes to figure out how to type matrices in latex.[/quote]
You also had to write that $\begin{pmatrix}2 &-3 \\-3 & 5\end{pmatrix}\begin{pmatrix}5 & 3 \\ 3 & 2 \end{pmatrix}= \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$ or $AB=I\implies (AB)^{T}=I\iff B^{T}A^{T}=I\iff BA=I$ for symmetric matrices $A,B$, because generally $AB=I\implies BA=I$ does not hold.


The game goes like this. One person starts with a book title, lets say [u]The Lightning Thief[/u] by Rick Riordan. The second person must post a book with a title that has at least one word in common from the first one, for example, [u]The Thief Lord[/u] by Cornelia Funke. Or, you can say a title who is by the same author, but with no words in common, for example [u]The Titan's Curse[/u] by Rick Riordan. You must state the author's name, and the word "the" does not count!!! If you post a book title that does not exist or is not by the author that you have stated, you lose a point. You are only allowed to use technique two, that is the same author but no words in common, three times, before you run out and must use technique one. Each book title can only be used once, or else it's going to be a big mess... Ten people can play at once. Everyone plays in order, for example, person 1, then person 2, then person 3... and so on. Everybody has 5 points to begin with, and when it is your turn and you do not post for TWO DAYS, then you lose a point. People are out if they lose all their points. Signups are here. The order in which play goes is in order of who signs up first. I will play this round, I'll be in slot 5. If you have any questions, p.m. me or post it on the forum.

Signup List:
1.
2.
3.
4.
5. Mr. Happy
6.
7.
8.
9.
10.

Sign-up list:

1. bluecarneal
2. monkeygirl13
3.
4.
5. Mr. Happy
6.
7.
8.
9.
10.

Ace of Diamonds, are you playing or not?

[quote="AceOfDiamonds"]I just lost the game.[/quote]
uhh...your point is?

The title makes me lose the game :(

1. bluecarneal
2. monkeygirl13
3.
4.
5. Mr. Happy
6.
7. isabella2296
8.
9.
10. MEEWHEEISSOSUPERAWESOMESHEHASTOLIKEPLAYTHEGAMEEVENTHOUGHSHELOSTTHEGAME! (AKA meewhee009)

Ok. I am experiencing some very odd responses.

1. bluecarneal
2. monkeygirl13
3.
4.
5. Mr. Happy
6. gogators
7. isabella2296
8.
9.
10. meewhee009

1. bluecarneal
2. monkeygirl13
3. ckck
4.
5. Mr. Happy
6. gogators
7. isabella2296
8.
9.
10. meewhee009

1. bluecarneal
2. monkeygirl13
3. ckck
4.
5. Mr. Happy
6. gogators
7. isabella2296
8. FlyAgaric
9.
10. meewhee009

me play, me is dilberto!!!! me number 9!!!!

Darn I wanted to play!  :(

Can you squeeze me inside?
Like number 11?

Ok, sure, one extra player won't hurt that much.

1. bluecarneal
2. monkeygirl13
3. ckck
4. jjfun1
5. Mr. Happy
6. gogators
7. isabella2296
8. FlyAgaric
9. dilberto
10. meewhee009
11. Scamper

Let's start. Bluecarneal, you'll be first.

Ok. Last one, no more entries!

1. bluecarneal
2. monkeygirl13
3. ckck
4. jjfun1
5. Mr. Happy
6. gogators
7. isabella2296
8. FlyAgaric
9. dilberto
10. meewhee009
11. Scamper
12. Cgyao15

Dear Mod,
Do we have to post in order?

Dear bluecarneal,
it's BRYAN Davis

god, that's REALLY hard. CANDLESTONE?  :(

How about that if there is no post within 24 hours that person is skipped and play resumes.

It's been 48 hours since bluecarneal posted so I'll resume gameplay :D
[u]Oracles of Fire[/u] by Bryan Davis

Sorry about this abrupt change, but for compound words, like candlestone, candle and stone are considered two different words.

Isn't it jjfun1's turn?
Maybe if a person isn't responding for like 2 turns, then we could PM him/her reminding him/her
about the game?

Sorry, I thought ckck had not posted yet. My bad.

[u]Dragons in Our Midst[/u] by Bryan Davis

Well it's Mr. Happy's turn. Please check this game Mr. Happy!

OK, since Mr. Happy isn't posting, it's gogators turn.

Peoples!

If you're not going to play then tell that you're dropping the game.
So that way you don't have people waiting a day or two for you!

I think you'll have to ask Mr. Happy about that.

The game has already gone overboard, so he might say no...

ME WANNA PLAY!

MR. HAPPY, THIS IS UR GAME AND U DON'T PLAY IT :cursing:

Move this to the Fun Factory and resume gameplay.

[u]The Book of Dragons[/u], by Michael Hague.

[quote="jjfun1"]
[u]Dragons in Our Midst[/u] by Bryan Davis[/quote]

That is the name of the series....

Cool story, bro!

If I broke a rule, just tell me which one. If I didn't, just keep playing.

Whoops, I misread the rules. >.<

Ignore Mr. Happy and let play continue. Don't prod people, and harshly enforce the 2 day posting period.


number theory
least common multiple
LaTeX
prime factorization
superior algebra
superior algebra unsolved

[color=green]Q: Decide whether it is true that if n is a positive integer and d is a divisor of φ(n), then there is an integer a with ord[size=75]n[/size]a=d.  Give reasons for your answer.[/color]

Don't know how to do this, please any one help??  Thanks very much.  :blush:

I remember encountering this problem at PROMYS.  I think it's not true, but I don't remember the counterexample.

This can be done for all divisors $d$ of $\phi(n)$ if and only if there is a primitive root mod $n$, and the group $(\mathbb{Z}/n\mathbb{Z})^{*}$ is cyclic.
Whenever $n$ is divisible by two different odd primes, the group is not cyclic. $-1$ has four square roots mod $15$, and that can't happen in a cycilc group.

A relatively quick proof shows that o(a mod n) = lcm[o(a mod p1^k1), o(a mod p2^k2), . . ., o(a mod pm^km)] where n = p1^k1*p2^k2*...*pm^km. Therefore, you can do it for any order o where o divides the lcm - that is, if each prime power in the prime factorization of o can be found in the phi of at least one of the pi.
Someone rewrite this for me in latex a bit later, please?
=Bmsiv-


inequalities
calculus
derivative
inequalities proposed

Prove for x,y,z in R,

$ xy^2z^3 \leq \frac {1}{3^{\frac {3}{2}} 4}(x^2 \plus{} y^2 \plus{} z^2)^3$

without using partial derivatives or Lagrange multipliers.

omg LOL.  Thanks.  Perhaps i should have posted this in the pre-olympiad thread.


If polynomials P(x)=Q(x) for infinitely many real x, then is it necessarily true that P(x)=Q(x) for all real x?

[quote="Myth"]Of course, we have $P(x)-Q(x)=0$ has infinitely many roots. If we suppose $P(x)-Q(x)\not\equiv 0$ then $P(x)-Q(x)=0$ should have at most $\deg (P-Q)$ roots -- contradiction.[/quote]

Uh, if we suppose $P(x)-Q(x)\not=0$, then $P(x)-Q(x)$ should have at most $\max\{\deg P,\deg Q\}$ roots  :D .  But obviously the rest is correct.

[quote="Myth"]I didn't understand the aim of your remark.  :?[/quote]

Sorry, I was confused.  For some reason I read $\deg(P-Q)$ as $\deg P-\deg Q$  :blush:

oh yeah, of course! Sorry, I guess I wasn't thinking that day...(it seems so obvious now, thanks Myth) :D


quadratics
algebra
quadratic formula
geometry
power of a point

i know a theorem that states is two chords intersect in a circle then the products of the measures of the segments of the chords are equal. i just need help writing an equation(and solvong it)

[hide="hint"] Let $AE = x$ and $EB = 17-x$. Then you'll get your equation.[/hide]

i got 60=x(17-x) or 60=17x-x(squared) how do i solve

Please do not double post. Also, you can write squared by using the ^ sign.

To solve, we have
$60=17x - x^2$
$\Rightarrow x^2-17x + 60=0$

From here, you should be apply any of the various methods for solving a quadratic.

Trying to do power of a point, but don't know how to solve quadratics?  You can use the [url=http://mathworld.wolfram.com/QuadraticFormula.html]Quadratic Formula[/url]

Also, I wonder what your birthdate is :)

i solved the qudratic formaula and i got 5 and 12 thanks for the help i didnt mean to double post(twice) and my birth date is 07/06/90 and im a sophomore


calculus
derivative
algebra
polynomial
trigonometry
calculus computations

Well there is only one real solution
at $-1.12373\ldots$ and we can be sure of that by
studying the derivative $2x+7x^{6}$ of the $RHS$ function.

But I assume you want an analytic solution.

[b]Semigroups[/b]: Do you have an solution for this, or do you at least know if it should be nice, that is non-numerical.

[quote="mszew"]
$\!\(\((1+x+x^{2})\)\ \((1-x+x^{2}-x^{4}+x^{5})\)\)$[/quote]
$x^{2}+x+1 = 0$ is solvable by quadradic formula ... baby stuff.

By I dont think the first one is nicely solvable.
Because that polynomial is irreducible over $\mathbb{Q}[x]$.

assume f'(x)=x^2+x^7+1
f(x)=x^3/3 +x^8/8 +x
Try to apply rolle's theorem between some 2 intervals x1 and x2 (the problem reduces to finding x1 and x2 such that f(x1)=f(x2)).

In fact I was originally thought to use Trignometry and complex number to solve this equation, but I didn'g try out which Trignometric form was usful, I personally think we could use Trigonometry to get a nice identity of this high power equation.

btw, How to use Excel to approximate the value of pi(=3.1415926535...)  by using the Euler's method to approximate y(1) of the equation: dy/dx=4/(1+x^2),y(0)=0 ?

 :)


How many different four-digit, positive integers are there where each digit is a prime number?

There are only 4 one-digit primes(2,3,5,7), so our answer is $ 4 \times 4 \times 4 \times 4 \equal{} \boxed{256}$.

There are 4 one digit prime numbers, 2,3,5, and 7.

Thus we have 4 possibilities for the thousands digit, 4 for the hundredas, 4 for the tens, and 4 for the ones digit. Thus it is 4x4x4x4=256.


topology
advanced fields
advanced fields unsolved

Prove that if $ J$ is uncountable, then $ \mathbb{R}^{J}$ (with the product topology) is not normal.

Now you say Munkres since it can be found in that text, but I believe that the proof should be attributed to A.H. Stone. Also, should this maybe be in the Advanced Fields section. It is a really hard proof.


number theory
relatively prime
number theory unsolved

find the integers x,y that:
\[ x^2\plus{}(x\plus{}1)^2\equal{}y^4\]

[quote="stvs_f"]find the integers x,y that:
\[ x^2 \plus{} (x \plus{} 1)^2 \equal{} y^4
\]
[/quote]

The equation $ x^2\plus{}(x\plus{}1)^2\equal{}y^2$ certainly no root  because $ (x,x\plus{}1)\equal{}1$ and because Don't exist Group 3 interger numbers Pytago $ (a,b,c)$ such that $ (a,b)\equal{}1$:D

[quote="ducpbc"]

The equation $ x^2 \plus{} (x \plus{} 1)^2 \equal{} y^2$ certainly no root  because $ (x,x \plus{} 1) \equal{} 1$ and because Don't exist Group 3 interger numbers Pytago $ (a,b,c)$ such that $ (a,b) \equal{} 1$:D[/quote]

check the pair (0,1).

[quote="randomgraph"]A less trivial solution to the original equation is $ 119^2 + 120^2 = 13^4$.

In fact, this is the only non-trivial solution. Here's a very rough sketch of a proposed proof.

If $ x^2 + (x + 1)^2 = y^4$, then $ (x, x + 1, y^2)$ is a primitive Pythagorean triple. Thus there are relatively prime integers $ u,v$ such that $ x = u^2 - v^2$ and $ x + 1 = 2uv$ or vice versa; in other words $ u^2 - v^2 - 2uv = \pm 1$ which can be written as $ (u - v)^2 - 2v^2 = \pm 1$. Setting $ u - v = z$ this is Pell's equation $ z^2 - 2v^2 = \pm 1$.

From the theory of Pell's equation we can characterize the possible $ (u,v)$ pairs as consecutive elements of the sequence $ 1,2,5,12,29,70, \ldots$ defined as $ a_1 = 1,a_2 = 2, a_{n + 1} = 2a_n + a_{n - 1}$. For example the solution $ z = 7, v = 5$ corresponds to $ u = 12, v = 5$ which indeed yields the solution I indicated above. Now, we have $ a_k^2 + a_{k + 1}^2 = a_{2k + 1}$ as one easily checks, so $ y^2 = u^2 + v^2$ is itself a member of that sequence. But [url=http://en.wikipedia.org/wiki/Pell_number]it is known[/url] that the only square in this sequence is $ 169$ (I'll dig up a link or find a way to prove this - I'm not sure how difficult this is, perhaps not at all). (EDIT: link added, see the section "Primes and Squares").

Therefore the only solution to the original equation satisfies $ z = 7, v = 5, u = 12, y^2 = 169, x = 12^2 - 5^2 = 119, x + 1 = 2 \times 5 \times 12 = 120$.[/quote]

sorry,i'm wrong :D  :blush:
Using Pell's equation is true,I think.


How many diagonals does a regular seven-sided polygon
contain?

Formula: n(n-3)/2. Plug in n=7, so 7*4/2=[b]14[/b], our answer.

[quote="tinspirecx"]Wait so are you saying that the formula for finding diagonals is :

$  \boxed{n(n-3)}  $    ?

If so then thanks.[/quote]

[hide="Formula"]
There's a $/2$ part.

The formula is $\frac{n(n-3)}{2}$

A way to remember is if you have 7 vertices, from one vertice, how many vertices can you pick to create a line? This is simply excluding the vertice your on and the two adjacent vertices, leaving $-3$. But we overcounted so we divide it by 2.

[/hide]


geometry
ratio
trigonometry
ARML
power of a point

[quote="Silverfalcon"]Points $ A,B,C,$ and $ D$ lie on the given circle. If $ AB \equal{} 8, AP \equal{} 2,$ and $ PC \equal{} 4,$ determine the ratio of the area of quadrilateral $ PAEC$ to the area of $ \triangle BAE$.[/quote]

[hide="..."]Power of a point gives $ CD \equal{} 1$...$ \triangle BED ~ \triangle AEC$...law of cosines on $ \triangle PBD$ and $ \triangle PAC$ gives $ \frac{BD}{AC}\equal{}\frac{5}{2}$...

By power of a point again $ AE \cdot DE \equal{} BE\cdot CE$...but $ \frac{AE}{BE}\equal{}\frac{CE}{DE}\equal{}\frac{2}{5}$...then $ \frac{AE}{DE}\cdot\frac{CE}{BE}\equal{}\frac{4}{25}$...

But $ \triangle AEB ~ \triangle CED$ and $ \frac{AB}{CD}\equal{}\frac{AE}{CE}\equal{}\frac{BE}{DE}\equal{}8$...then $ \frac{AE}{DE}\equal{}\frac{64CE}{BE}$, so $ \left(\frac{64CE}{BE}\right)\left(\frac{CE}{BE}\right)\equal{}\frac{4}{25}$, then $ \frac{CE}{BE}\equal{}\frac{[AEC]}{[ABE]}\equal{}\frac{1}{20}$

We ALSO KNOW THAT $ \frac{[APC]}{[AEC]\plus{}[DEC]}\equal{}4$ but $ [AEC]\equal{}\frac{[ABE]}{20}$ and $ [DEC]\equal{}\frac{[ABE]}{8}$ so $ \frac{[APC]}{[ABE]}\equal{}\frac{7}{10}$ AND FINALLY $ \frac{[APC]}{[ABE]}\plus{}\frac{[AEC]}{[ABE]}\equal{}\frac{[PAEC]}{[ABE]}\equal{}\frac{7}{10}\plus{}\frac{1}{20}\equal{}\boxed{\frac{3}{4}}$[/hide]

[quote="the future"]wait, ch1n353ch3s54a1l, I get 5/16 using a much simpler method...

I dont know if thats right though, its just similar triangles..and you split up the quadrilateral in another way.

but this is kinda odd because it was number 8 on individual and it only took me 15 seconds :maybe: (this coming frmo a person that never could even draw a picture for the number 8 on the 2007 arml-after 10 minutes)[/quote]

oh yeah, 5/16, i think that's right

yeah that's smart, just draw PE and do the ratio thingy, i had $ \frac {[ABE]}{[CED]}$ wrong, it's 64, not 8...(I think...)

[hide="Solution"]By Power of a Point, $ CD\equal{}1$.  We also have $ \triangle ABE ~ \triangle CDE$.  Let $ x\equal{}EC$, therefore $ AE\equal{}8x$.

Draw $ \overline{PE}$.  Note that $ \triangle PAE$ and $ \triangle BAE$ share an altitude that runs through $ E$.  Therefore $ \frac{[PAE]}{[BAE]}\equal{}\frac{PA}{BA}\equal{}\frac{1}{4}$.  Let $ K\equal{}[PAE]$, therefore $ [BAE]\equal{}4K$.

Since $ m\angle BAD\equal{}m\angle BCD$, we also have $ m\angle PAE\equal{}m\angle PCE$.  Let this angle be $ \theta$.  Applying $ K\equal{}\frac{1}{2}ab\sin\theta$ to $ \triangle PAE$, we get $ K \equal{} 8\sin\theta$.  Likewise, $ [PCE]\equal{}2\sin\theta\equal{}\frac{K}{4}$.  Therefore $ \frac{[PAEC]}{[BAE]}\equal{}\frac{K\plus{}\frac{K}{4}}{4K}\equal{}\frac{1.25}{4}\equal{}\boxed{\frac{5}{16}}$.[/hide]

Very well done. The answer was indeed 5/16.

All solutions were good but I especially liked lingomaniac88's solution because I didn't (or the solution) get to consider $ \triangle PAE$ and $ \triangle BAE$ as a pair that share an altitude. Nice approach there!

If anyone has any other method to this problem, feel free to share.  :roll:


probability
conditional probability
probability and stats

1. A student has managed to enter in a given university. It was informed that the probability of getting a scholarship is 40%. In the case of getting it, the probability of finish his course is 0.8, while in the case of not getting the scholarship that probability is only 0.4.

(a) What is the probability of the student finish the course?

(b) If, after some years from now you meet the student already with his course finished, what is the probability that he have gained a scholarship?

2. A box contains 2 red balls and 3 blue balls. Two balls are randomly taken out, without reposition. What is the probability of:

(a) The second ball being red?
(b) Both balls being red?
(c) The first being red and the second blue?

3. A box contains 9 white balls and 11 black balls. Two extractions of one ball are made. What is the probability of getting two balls with different colours (i) with, and (ii) without reposition?

4. In a bag there are 6 green balls and 4 blue balls. Three balls are sucessively taken out, seing the colour of the ball and puting it back in the bag.

(a) What is the probability of taking first a blue ball, secondly a blue ball, and then a green ball?
(b) What is the probability of taking at least one blue ball?

what are you doing ? copy/paste your last homework sheet ?

No, because I am not a student. I am just trying to learn Probability and Statistics by my own. My apologies if you feel scandalized because this is so easy to you.

nobody is scandalized here  :lol: I wish you all my best in your trying to understand these beautiful subjects. You may get more answers if you post these exercises in different threads and explain where you are stuck etc ... So what did you get so far ?

1. [b]a)[/b] The probability that he gets a scholarship and finishes is $ (0.4)(0.8)=0.32$. The probability that he does not get a scholarship but still finishes is $ (0.6)(0.6)=0.36$
Since they are mutually exclusive:
$ P(scholarship$ $ \&finishes\bigcap$ $ no$ $ scholarship$ $ \&finishes)$
$ =P(scholarship$ $ \&finishes)+P(no$ $ scholarship$ $ \&finishes)$
$ =0.32+0.36$
$ =0.68$

[b]b)[/b] We know that $ 68%
$ of the population has already finished, and we know that $ 32%
$ of the population has gained a scholarship and finished. Thus, the probability that the student had gotten a scholarship, knowing that he finished, is:
$ \frac{P(scholarship\bigcap finished)}{P(finished)}$
$ =\frac{0.32}{0.68}$
$ =\frac{8}{17}$

2. [b]a)[/b] Probability that the first ball is red is $ 0.4$. After taking the first ball out, the probability of picking a red ball changes.
Probability that the first ball is blue and the second ball is red is $ \left (\frac{3}{5}\right )\left (\frac{2}{4}\right )=\frac{3}{10}$.
Probability that the first is red and second is red is $ \left (\frac{2}{5}\right )\left (\frac{1}{4}\right )=\frac{1}{10}$.
Probability of the second ball being red is equal to $ \frac{1}{10}+\frac{3}{10}=\frac{2}{5}$
[b]b)[/b] This has already been established as being $ \frac{1}{10}$.
[b]c)[/b] The probability that the first is red and the second is blue is $ \left (\frac{2}{5}\right )\left (\frac{3}{4}\right )=\frac{3}{10}$.

Thank you both for your answers.

EastyMoryan, shouldn't the probability of finish but without a scholarship be 0.6x0.4?

Define the events B = "receive a scholarship" and T = "finish the course", and let $ p(A/C)$ denote the conditional probability of A knowing that C occurred, and $ \bar{A}$ denote the complementary event of A. The given data translates to

$ p(B) \equal{} 0.4$, $ p(T/B) \equal{} 0.8$, and $ p(T/\bar{B}) \equal{} 0.4$.

For (a), since the events $ T \cap B$ and $ T \cap \bar{B}$ are mutually exclusive, we have

$ p(T) \equal{} p(T \cap B) \plus{} p(T \cap \bar{B}) \equal{} p(B) \times p(T/B) \plus{} p( \bar{B}) \times p(T/\bar{B})$

$ <\equal{}> p(T) \equal{} 0.4 \times 0.8 \plus{} 0.6 \times 0.4 \equal{} 0.56$.

For (b), we want to compute $ p(B/T)$. But

$ p(T \cap B) \equal{} p(T) \times p(B/T) <\equal{}> p(B/T) \equal{} \frac{p(T \cap B)}{p(T)} \equal{} \frac{0.32}{0.56} \equal{} 0.57$.

Is this correct?


In triangle ABC, medians AM and CN are perpendicular to each other and
intersect each other at G. A median that is cut by another median is
divided in a 1 to 2 ratio, with the length of the median from the
midpoint to the intersection of the medians being the shorter
length. If GM = 2 units and CN = 8 units, what is the area of
triangle ABC?

Is there a formula?

Can you please explain how you got the area to be 32?

Thanks!

First, find AG, GM, CG, and GN. Once you find these lengths, can you find the area of CGM? How about AGC? And since a median divides a side of the triangle in half, wont those two combined areas equal half the total area?

Triangle CGM is right in G and the sides are:GM=2 and CG=16/3 .So CM is found and is half of BC.In the same manner triangle AGC is right and AG=4 ,CG=16/3 and we find AC.Triangle AGN is right and sides are AG=4,GN=8/3 and we find AN and so AB.The ares is find with (s(s-a)(s-b)(s-c))^.5 where s is the semiperimeter.


Source: Massachusetts Mathematics League December 2005

Al, Bob, and Carl each have different amounts of money. They share it as follows: Al gives Bob and Carl each 1/3 of his money. Bob then gives Al and Carl each 1/3 of his new total. Carl then gives Al and Bob each 1/3 of his new total. The result is that Al ends up with $45 more than Bob and Bob ends with $45 more than Carl. If after the first sharing Carl had $78, how much money did Al end up with?

I'm confuzzled (I've lost track of how many times I've said that on AoPS).  Can someone post a solution?


email
Mafia
calculus
integration
vector
rotation
linear algebra

Source: school

The game is kill the bird. You must save the bird from the previous poster and then attempt to kill it.

In the real version you have a drawing to go along with it, and you draw means of death as you go along.

*lights bird on fire*

*Shoves the bird into a tank to put out the flames and drown it*

*Sucks all the water out of tank with vacuum and attempts to squeeze bird to death*

blows air into the bird to the point of popping

*Squeezes air out of bird and attempts to squeeze to death again*

Careesses the air out of the bird and starts tickling it to insanity.

*Sends it to mental ward to cure it of it's insanity, then hires hitman to kill it*

bribes hitman with 2 gold pieces and then attempts to hang the bird

*Chops rope that is about to hang bird, then convinces CIA that it is a dangerous criminal and deserves the death penalty*

hypnotizes CIA and attempts to poison it with a non-antiobiotic lollipop

*Provides proper medication and keeps it in a luxury cage made of highly radioactive Plutonium*

Place heavy lead shield in cage which crushes the bird.

*uses oxy acetylene flame to destory the shield and the bird*

(and to Leao's signature: Pawn to E5!)

puts the bird into a tank of water and puts acid in the water.

So the bird has died. :stretcher:

FINALLY

It lived for 72 days.

I try to revive it back by casting 'Resurrection' but I misspell it to 'Animate Dead' so it lives again, but gonna die 4 minutes later  :D

it's stupid because its brain does not function because it is dead.


The base of a triangle is 15 inches.  Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into three equal areas.  The length of the parallel closer to the base is:
\[ \text{(A)}\ 5\sqrt{6}\ \text{inches} \qquad \text{(B)}\ 10\ \text{inches} \qquad \text{(C)}\ 4\sqrt{3}\ \text{inches} \qquad \text{(D)}\ 7.5\ \text{inches} \qquad \text{(E) none of these} \]

[hide]Let $b$ be the length of the parallel closest to the base, $h$ be the height of the similar triangle formed by $b$, and $x$ be the height of the entire triangle.

$\frac hb = \frac x{15}$

$\frac {bh}2  = 5x$

$b^2 = 150$

The correct answer is $5 \sqrt 6$[/hide]

i used tarquin's method and got[hide]a[/hide]


function
geometry
geometric transformation
rotation
superior algebra
superior algebra unsolved

We are given
$ \alpha$ =
$ \left(\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\
5 & 2 & 4 & 1 & 3 \end{array}\right)$ and $ \beta$ = $ \left(\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\
5 & 4 & 3 & 2 & 1 \end{array} \right)$

And the composition $ \alpha\beta$ = $ \left(\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\
3 & 1 & 4 & 2 & 5 \end{array} \right)$ = $ (1342)$ in cycle notation.

There is another way of doing this by writing out $ \alpha$ and $ \beta$ in cycle notation and then compose to get the same cycle notation. But can someone guide me through this process as it confuses me how's that's possible.

$ \alpha \equal{} (1534)$ and $ \beta \equal{} (15)(24)$ so $ \alpha\beta \equal{} (1534)(15)(24)$ = $ (1342)$
How is $ (1534)(15)(24)$ = $ (1342)$ ?

I usually just take each element and read through right to left.

For example, start with $ 1$.  $ (24)$ doesn't do anything, $ (15)$ sends $ 1$ to $ 5$, and $ (1534)$ sends $ 5$ to $ 3$.  Hence we have $ (13\ldots)$.  Then we see where $ 3$ goes... etc.

As far as I'm aware, multiplication of cycles works left to right.

Composition of functions is generally written right to left however, as in:
"gf" is read "g follows f".

However if one applies this to cycle multiplication directly, then:

(1,2)*(1,3) would be "transpose 1 and 3, then transpose 1 and 2", which resuts in (1,3,2) not (1,2,3) as is the correct answer usually.

I believe this may vary between authors, but at least in Sage/GAP, (1,2)*(1,3) = (1,2,3) which implies left-to-right, not right-to-left of usual composition.

Now as to the state question:

a=(1,5,3,4)
b=(1,5)(2,4)

Note that a decomposes into: (1,5)(1,3)(1,4)

When we multiply a*b we then get:
(1,5,3,4)*(1,5)(2,4)
= (1,5)(1,3,4)(1,5)(2,4)
= (5,3,4)(2,4)
= (4,5,3)(4,2)
= (4,5,3,2)
= (2,4,5,3)

Computing b*a yields the answer you posted originally.

Rules I use to compute cycles:
1) (a1, a2, ..., aN) = (a1, a2)(a1, a3)...(a1, aN)
2) transpositions are involutions; that is, they are their own inverses
3) a permutation conjugate is a kind of "translation", that is a^-1 * b * a is a kind of rewriting of b, so in the above, I recognize the part "(1,5)(1,3,4)(1,5)" as a translation of the (1,3,4) cycle.

So basically, my method is to rotate things around so that I can discover translations.

Edit: here's the above in sage to verify: http://www.sagenb.org/home/pub/116

Do notice that the original post is using right-to-left multiplication, however, which is quite common.

Hmm I usually see it that:
(a1,....,an)=(a1,a2)...(a1,an)

and not the reverse:
(a1,....,an)=(a1,an)...(a1,a2)

likely precisely because it is far more intuitive and straightforward to multiply cycle notation left-to-right.

I'm not sure that the RTL ordering is as common as you think, and as far as manual computation of cycle notation goes, LTR is certainly more natural for most.

Right-to-left is [i]much[/i] more natural when you want to talk about the underlying group action, otherwise you're forced to write $ \sigma(x)$ as $ x\sigma$ to make it consistent.  It's possible that it's less common than left-to-right, but it's still all over the place.  And again, the original post is definitely using right-to-left.

Just throwing my $ \$$0.02 in here, but I'm fairly sure that when I learned this, we did everything left to right.

I guess, like most things, if you're very clear about defining your terms and operations before you start, it doesn't really matter.

Strange, from the many people I've met (students and all professors related) in Waterloo it is assumed to be a composition of functions and hence read from right to left. $ \sigma\beta \equal{} \sigma(\beta(x))$

Maybe we can tally up  the results by general region and see if we can find a pattern.

The whole problem comes from the fact that the usual function notation $ f(x)$ is not consistent to the Latin alfabet left-to-right writing. An alternative(actually 3 alternatives) is to write images of functions as $ x\cdot f$, or $ xf$, or $ x^f$.


Im having problems with a question, here it is:

Find X

$ \log_{10} (\log_2 (\log_x (.16)))\equal{}0$

Just use properties of logs.  $ 10^{0}\equal{}\log_{2}(\log_{x}(.16))\equal{}1\rightarrow 2\equal{}\log_{x}(.16)\rightarrow x^{2}\equal{}.16\rightarrow x\equal{} .4$.


A prison has 2004 cells, numbered 1 through 2004. A jailer, carrying
out the terms of a partial amnesty, unlocked every cell. Next he locked
every second cell. Then he turned the key in every third cell, locking
the opened cells, and unlocking the locked ones. He continued this way,
on nth trip, turning the key in every nth cell, and he finished his mission
after 2004 trips. How many prisoners were released?

This is a very common type of problem. Notice that for any cell $ n$, it will be touched if and only if the current numbered turn divides $ n$. So initially they are closed. If $ n$ has an even number of divisors, then the jailer will switch door $ n$ an even number of times, and it will remain closed. On the other hand, if $ n$ has an odd number of divisors, then the jailer will switch door $ n$ an odd number of times, and it will be open.

The only numbers with an odd number of divisors are squares, and there are $ \lfloor \sqrt {2004} \rfloor \equal{} \boxed {44}$ squares, so there are $ \boxed {44}$ open doors after the jailer is finished, and thus $ \boxed{44}$ released prisoners.


so which comprehensive is the easiest? (the one i should go to )

Alright! Congrats on getting the NC forum you guys. :D

[quote="NeverOddOrEven"]which comprehensive[/quote]

There's only one comprehensive...

Only take it if you're done with Alg II and Geometry  :wink:

Either way, you need to go to a regional contest, preferably with your school, and (hopefully) qualify for the State Finals. There are 3 sites with AlgI/II and Geometry (All 3 at each site) and 1 site, NCSSM, for Comprehensive Finals.

[quote="not_trig"]I think you go to the one closest to you :wink:

They should an be about the same[/quote]

some are much harder to qualify through than others

e.g. elon seems to have become pretty tough

[quote] elon seems to have become pretty tough [/quote]

since east has basically taken over that competition?  :mad:

[quote="NeverOddOrEven"]no i mean between the qualifying sites for comprehensive like Elon, App state, UNC Charlotte, etc...[/quote]

If you go to a small site, there won't be as much competition (generally), but they won't take as many people.  The larger competitions have more competition, but take more people.  You might not want to go to the same competition that East Chapel Hill goes to, because they generally do well at contests. :D

Also, each site has different tests to take.

I qualified at App state.  Here, the contest consists of two parts - multiple choice (25 questions, 4 points each) and a written response where you have to explain what you did (3 questions, 8 points each).  They take 6 people, but 2 of those people are the top local people.  This year, except for the 2 locals and me, everyone was from Chapel Hill.

If you live within an hour or two of Charlotte, I would definitely say take the UNC Charlotte Exam.  The exam is written by Harold Reiter, but I don't remember how long the contest is, ect.  This site takes about 8 people, and I think it is a good site to go to.

yea ill probably go to either app state or unc charlotte since our school gives us a choice to go to any of the 3 sites i think.

hmm. charlotte definitely used to be one of the harder ones, I think (I failed to qualify from that site my first year), but that was back in the middle of the anders era

the charlotte test seems to be more difficult now than the previous years. the 2006 test only had high scores in about the 80 range compared to the previous perfect scores on the other tests.

This would probably be more appropriate as a separate topic.  Split.

[quote="MysticTerminator"]the middle of the anders era[/quote]

 :rotfl:

That is true though...

I'm stuck with Enloe at Wake Tech. Bad.......
I hope I won't get pwned that badly...

Hopefully Wake Tech improved from my time there - took alg II there once and they called time early when the pizza came rather than keeping with the announced finishing time.

Sorry to revive an old topic, but I didn't want to start a new one. :D

Could you please tell me the time limit and scoring methods for the Comprehensive contest? Thanks.

I think multiple choice are 4 for a right answer, 1 for a blank. Integer answers are 6 for a right answer. The time limit is 2 hours or thereabouts (1 hour, 45 minutes?).

I just found out Hoggard is doing its subject tests at wake tech this year, since ECU is during spring break. We'll be slaughtered. Though I guess the subject tests really don't have that much competition.

Thanks!

... but according to the website, the test was from like 9:30 to 11:00 last year... wouldn't that be 1.5 hours? Or did the schedule lie? :mad:

 :D

I need to make the A team this year... what score do I need for say, top 15 or something?

Thanks!

It may have been 1.5 hours... Why did you ask that if you knew the answer, anyway?

Score has no effect on A vs. B team. Go to the practices and do well on the AMC/AIME. Mr. Benton knows you, I would imagine. Besides, everybody knows the state math contest is a bad indicator of talent (I would know).

It will be hard to make the A team this year (for anyone), since there are only... what, 3?... spots left. Don't be sad if you don't make it.

I wasn't sure, and I didn't know how it was graded.

3 spots? Plus Noah, possibly, won't go, so I'll have a chance...
Not many contenders, but more than 4...

The only way the comprehensive will help is if you get in the top 3 or so.  That would definitely help for A team, but I think they care about AIME, AMC 10/12 and USAMO scores more than the comprehensive contest.

It obviously isn't the top 15 because I was 15th last year & not on the A team.

I believe they count the state math contest far more than you might expect.  In general, they consider USAMO qualifiers and AIME high-scorers definite team members.  From time to time these people do not score well on or even qualify for the state math contest.  However, there are typically fewer than 5 of those students, and so making top 10 on the state contest ought to guarantee a spot.

Like Jane Bae... USAMO qualifier, but nowhere at states... lol

She's good though, and she has a good chance of making the A team this year.


How many zeros does $ f(x)\equal{}\cos(\log(x))$ have on the interval $ 0<x<1$?

$ \textbf{(A)} \ 0 \qquad \textbf{(B)} \ 1 \qquad \textbf{(C)} \ 2 \qquad \textbf{(D)} \ 10 \qquad \textbf{(E)} \ \text{infinitely many}$

Thanks in advance!

[hide="probly wrong"]so $ \cos{\log{x}} = 0$, for $ \cos{\theta} = 0$,

$ \frac{\pi}{2} + \pi k$ for any integer.

since $ 0< x <1$

$ \log{x} = (-\infty, 0)$

so any negative integer value of $ k$ in $ \frac{\pi}{2} + \pi k$ is a possibility so i think the answer is $ \boxed{\infty}$[/hide]

[quote="Xantos C. Guin"][hide]$ f(x) \equal{} \cos(\log(x)) \equal{} 0 \Rightarrow \log(x) \equal{} \frac {2k \plus{} 1}{2}\pi$ for some $ k \in \mathbb{Z}$. Since $ \log(x)$ has a range of $ ( \minus{} \infty,0)$ for $ x \in (0,1)$, $ \log(x) \equal{} \frac {2k \plus{} 1}{2}\pi$, for some $ k \in \mathbb{Z}$, for infinitely many $ x$. [/hide][/quote]

Where did the $ \frac {2k \plus{} 1}{2}\pi$ come from?

$ \cos(x)\equal{}0$ for $ x \equal{} \ldots \minus{}\frac{5}{2}\pi, \minus{}\frac{3}{2}\pi, \minus{}\frac{1}{2}\pi, \frac{1}{2}\pi, \frac{3}{2}\pi, \frac{5}{2}\pi, \ldots$.

$ x\equal{}\frac{2k\plus{}1}{2}\pi$ for $ k \in \mathbb{Z}$ is a general expression for these zeros.


inequalities
function
inequalities unsolved

[b][i][size=150][color=#B93B8F]If $ a,b,c$ be positive real numbers , prove that
$ \frac{a^2}{b^2\plus{}c^2}$ $ \plus{}$ $ \frac{b^2}{c^2\plus{}a^2}$ $ \plus{}$ $ \frac{c^2}{a^2\plus{}b^2}$ $ \geq$ $ \frac{a}{b\plus{}c}$ $ \plus{}$ $ \frac{b}{c\plus{}a}$ $ \plus{}$ $ \frac{c}{a\plus{}b}$[/color][/size][/i] [/b]

In this equality ,we consider the function:
$ f(t)\equal{}\frac{a^t}{b^t\plus{}c^t}\plus{}\frac{b^t}{a^t\plus{}c^t}\plus{}\frac{c^t}{a^t\plus{}b^t}$
$ f'(t)\ge 0$ so $ f(2)\ge f(1)$.QED


inequalities
integration
function
advanced fields
advanced fields unsolved

Suppose $ u \in W^{2,p}(U) \cap W_0^{1,p}(U)$. For $ 2 \leq p$, is it correct that
\[ \int |\nabla u|^p dx \leq C (\int |u|^p)^{1/2} (\int |\nabla^2 u|^p)^{1/2}?\]

It's true (it's an exercise in Evans).  I think the idea is to establish it for $ C^{\infty}_c$ functions by integrating by parts and applying Cauchy Schwarz, and then taking sequences $ v_k\to u$ in $ W^{2,p}$ and $ w_k\to u$ in $ W^{1,p}_0$.


An iscoceles triangle with some amount of 30 degree angles has its base angles quintisected (5 parts).  We will call these new lines quintisections.  One of the quintisection intersections ( :) ) makes a 90 degree angle.  All the quintisections except the perpendicular ones are erased  At the point where each quintisection hits a leg of the triangle, a perpendicular so that you have an inner retangle formed.  If this rectangle's area is 1, find the area of the inverted kite created by the quintisections.

I'll make it easier - [i]m[/i] times the square root of [i]m[/i] plus [i]m[/i] is [i]n[/i] times the area of the kite.  If the previous expression is in lowest terms, compute the ordered pair ([i]m[/i], [i]n[/i])

sorry, what's an inverted kite?

[hide="lengthy construction"]
The $30^o$ degree angle is at the top angle of the isoceles triangle, so that each of the base angles is $75^o$. Quintisect the $75^o$ angle and you get $15^o$ each... where the second quintisection lines (counting from the bottom) cross is the $90^o$ deg angle, because you have an isoceles triangle with two $45^o$ angles. The rectangle referred to is a square because you have 3 right angles. Thus the side length of that square is 1.
well this is as far as i got...  :|

[/hide]

A kite is a quadrilateral where two pairs of adjacent sides are congruent, so an inverted kite is one where one of the diagonals is outside of it (sort of like the netlogo "turtle" if you're familiar with that programming language).

Would it be a sort of v shape?

Also the triangle doesn't have to be 30,75,75. It could be 30,30,120.

[quote="minsoens"]

[hide="lengthy construction"]
The $30^o$ degree angle is at the top angle of the isoceles triangle, so that each of the base angles is $75^o$. Quintisect the $75^o$ angle and you get $15^o$ each... where the second quintisection lines (counting from the bottom) cross is the $90^o$ deg angle, because you have an isoceles triangle with two $45^o$ angles. The rectangle referred to is a square because you have 3 right angles. Thus the side length of that square is 1.
well this is as far as i got...  :|

[/hide][/quote]

[hide="Much simpler way to do that"]
Isosceles triangles are symmetrical, therefore, of course it's a square  ;)
[/hide]


Εστω S το συνολο των πραγματικων εκτος του -1.Οριζουμε επι του S μια πραξη *,οπου: a*b=a+b+ab  ,a,b ανηκουν S.Να λυσεται την εξισωση         2*x*3=7. ελπιζω να ειναι καλη. (το * ειναι απλα αστερακι,ουτε δυναμη ουτε πολλαπλασιασμος).  :)

Παρατηρούμε ότι η πράξη * είναι προσεταιριστική .
Άρα 2*x*3=(2*x)*3 . Από εδώ και πέρα είναι εύκολο πλέον  ;)


Show that if $f$ is bounded and $G_{f}= \{ (x,f(x))|x\in R\}$ is closed in $R^{2}$, $f$ is continuous

Suppose that $f$ is not continuous in $y \in \mathbb R$ and consider a sequence $x_{n}\to y$ such that $f \left( x_{n}\right)$ has the limit $\ell \neq f(y)$ (we could use $\liminf_{x \to y}$ and/or $\limsup_{x \to y}$ to find such a sequence). But $G_{f}$ is closed and $\ell$ is finite, so $\left( y, \ell \right) \in G_{f}$, i.e. $f(y) = \ell$, contradiction. Thus, $f$ is continuous everywere.


Is there a nice geometric proof of tan(A+B) = (tanA + tanB)/(1 - tanA tanB)? Thanks..

The following diagram may or may not be useful: (RTS, RTQ, RQP are right angles)

I didn't use point P in your diagram.
Let RT=1, ST=a, TQ=b. So tanA=a and tanB=b
Area of triangle RSQ=a+b/2
RS=  :sqrt:  (1+a :^2: ), RS*h=a+b assuming h is the altitude lies inside the triangle. h=(a+b)/:sqrt:  (1+a^2).
RQ= :sqrt:  (1+b :^2: ).  :sqrt: (RQ :^2: -h :^2:) =l where l is the other leg of rt triangle with angle (A+B)=(1-ab)/:sqrt:  (1+a^2).
h/l=tan(A+B)=(a+b)/(1-ab)
Very similar proof when altitude h lies outside of the triangle, all you do is switch negative signs.

Yeah, I tried to use h but couldn't see it. Thanks!


calculus
integration
algorithm
national olympiad

Problem 1. Let $n$ be a positive integer. Find the largest nonnegative real number $f(n)$ (depending on $n$) with the following property: whenever $a_1,a_2,...,a_n$ are real numbers such that $a_1+a_2+...+a_n$ is an integer, there exists some $i$ such that  $|a_i-\frac{1}{2}|\ge f(n)$.

Problem 2. Prove that every positive integer can be written as a finite sum of distinct integral powers of the golden ratio.

Problem 3. Let $p\ge5$ be a prime and let $r$ be the number of ways of placing $p$ checkers on a $p\times p$ checkerboard so that not all checkers are in the same row (but they may all be in the same column). Show that $r$ is divisible by $p^5$. Here, we assume that all the checkers are identical.

Problem 4. Let $A,B$ be two distinct points on a given circle $O$ and let $P$ be the midpoint of the line segment AB. Let $O_1$ be the circle tangent to the line $AB$ at $P$ and tangent to the circle $O$. Let $l$ be the tangent line, different from the line $AB$, to $O_1$ passing through $A$. Let $C$ be the intersection point, different from $A$, of $l$ and $O$. Let $Q$ be the midpoint of the line segment $BC$ and $O_2$ be the circle tangent to the line $BC$ at $Q$ and tangent to the line segment $AC$. Prove that the circle $O_2$ is tangent to the circle $O$.

Problem 5. In a circus, there are $n$ clowns who dress and paint themselves up using a selection of 12 distinct colours. Each clown is required to use at least five different colours. One day, the ringmaster of the circus orders that no two clowns have exactly the same set of colours and no more than 20 clowns may use any one particular colour. Find the largest number $n$ of clowns so as to make the ringmaster's order possible.

haven't worked out the details for #2 yet, but the solution is obviously based on the relation $\phi^n = \phi f_n+f_{n-1}$, true for any $n \in \mathbb Z$ (after we extend the fibonacci sequence in the negative-index direction). it then amounts to writing an arbitrary positive integer in the form $\phi \sum f_{k(i)} + \sum f_{k(i)-1}$; for that we need $\sum f_{k(i)}$ to be 0. to arrange things such that $\sum f_i = 0$ we'll have to use the fact that $f_{-k} = (-1)^{k+1} f_k$. tying this with the fact that any positive integer can be written as the sum of different-index fibonacci numbers should give the solution. when i have time, i'll write a concrete proof. or maybe someone else will just complete it  :)

Problem 2:

Let t be the golden ratio. We know that t^n = t^(n-1) + t^(n-2) so we can deduce that t^n + t^n = t^(n+1) + t^(n-2). We want to writte an integer k as a sum of powers of t so we write k = k * t^0.
Then every step we erase t^n + t^n and write t^(n+1) + t^(n-2) for an integer n that is repeated in the sum. Every step we reduce the average of the exponents by 1/k. Supose that this algorythm never ends, after 3k^2 steps the average of the exponents will be -3k so the lower exponent may be lower than -3k. Now we divide the integers from -3k to 0 in k groups of three ((-1, -2, -3), (-4, -5, -6)...(-3k+2, -3k+1, -3k)). Before we reach a group we might have been in all the upper groups before, and once we reach a group we can't leave it because t^n leads to t^n-2, t^n-2 leads to t^n-1 and t^n-1 leads to t^n.
Now we have at least one power of t in each group but we have k powers and k groups so there are not two equal powers of t


Nim is played using 2 players. There are 3 postivie integers written on the board. Each player, on his turn, must erase one of the numbers are replace it with a smaller nonnegative integer. The player who has no legal moves left, loses.

I know the winning strategy, but I can't prove why it works. Anyone know?

The proof can be found in many places, including some olympiad books (Yaglom and Yaglom, [i]Challenging Mathematical Problems with Elementary Solutions[/i]).  For the game with $n$ piles two possible strategies for a proof come to mind.  (There is no major difference between $n=3$ and $n > 3$, but $n < 3$ have very simple winning strategies.)

One is to try to prove directly that the formula works, by induction on the sum of the piles.
If you just unwind carefully what is to be proved, it is some statement about binary representation and not that hard to check.   Some version of this is the proof in most books.

Another idea, which is more "correct" from the point of view of Nim (aka Sprague-Grundy) theory, is to reduce to a position with $n-1$ piles by proving that the game with two piles is equivalent to a game with one pile equal to their Nim-sum. So eventually one can reduce to a game with one pile equal to the Nim-sum.  I haven't seen any published version of this proof from first principles, and it should be a good exercise to work it out.

A good bit of information about Nim can be found on the site http://www.cut-the-knot.org/index.shtml Just do a search.

you could do a search im sure its somewhere on the internet


Is anyone else here trying out for their state's all-state orchestra (which I assume exists)/all state band?  More specifically, how about Missouri's all state orchestra?

I'm auditioning for All-State Band and Chorus, though don't expect to make either. :D

We have regionals before all-state.

So I'd have to make that first.

But I'm going to try.

For Ohio, Violin.

My friend (nomar)'s brother (murtonfan91) made all-state choir, I think, for Indiana.

I didn't make all-state band, but I'm hoping for NYSBDA (kinda like a second tier for New Yorkers who didn't make all-state).

Awesomely enough, three of the five trombonists in my school's concert band made all-state, and the other two (myself included) are trying for NYSBDA.

I'm only allowed to do one of them  :mad: , so I'm choosing Band, even though I like Chorus better.

Does anyone else have to pay $\$175$ if you get into All-State?


Is it possible for moderators to be able to grant modding powers to specific users over their own threads (for example, in mafia games), if they deem it necessary? This sounds like something very difficult to do from a limited perspective, though...

Yes I totally agree with this proposition. I like having The Mole thread locked during quizzes, but had to ask Klebian specifically not to lock it, so I won't have to wait to post results. Also, it would be nice to be able to edit the first post.

lol I can never understand why people spend so much time in the Fun and Games forum. It seems some members (obviously not you two) come only for that portion of the site  :P

Anyway, I'd like to bump this
A very good idea if possible to implement

bump

It would help a lot with lots of games... main example: Mafia.

Perfect told me to bump this.

And for good reason - I'm modding a few games where editing the first post would be very convinient.

And I do bascially only come the site for the G&FF!  :lol:

The issue with is that (from my knowledge of the technology on which AoPS is based) it would take a [b]lot[/b] of effort to implement.

[quote]I'll agree with worthawholebean - given my knowledge of being an admin of a forum (created some random ones before in the past) it would be pretty hard to do this - even with superior programming skills. I would recommend turning off the 48 hour edit rule as an alternative solution.
[/quote]

Yeah, that sounds like a good idea.

Why was the edit rule there for? I've never come upon a forum that stopped allowing you to edit your posts after a set limit, so I'm curious for it's intended purpose.


Let $a,b \in %Error. "bbmath" is a bad command.
{R}$, $a,b>1$. A function $f: (0,\infty) \to \mathbb{R}$ so that $g:%Error. "bbmath" is a bad command.
{R} \to %Error. "bbmath" is a bad command.
{R}$, $g(x)=f(a^x)-x$ is strictly increasing and $h:%Error. "bbmath" is a bad command.
{R} \to %Error. "bbmath" is a bad command.
{R}$, $h(x)=f(b^x)-x$ is strictly decreasing, exists if and only if $a>b$.

Go to :
http://www.mathlinks.ro/Forum/topic-29008.html
 :cool:


geometry
angle bisector
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14501[/img]

 :(

It's knon that those points are the Isodynamic Points, if you want to costruct them, call A' and A'' the feet of the internal and external angle bisector of $ \angle BAC$ and equally for B', B'', C', C''. The 3 circle (AA'A''), (BB'B''), (CC'C'') concurr at 2 point that are the Isodynamic Points

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14515[/img]

 :thumbup:


Could someone please explain to to me how to solve quadratic inequalities?  I have read several books, but cannot fully understand them.  Do not forget to include example problems.

Hi,

The introduction to algebra book from art of problem solving has a very thorough lesson on quadratic inequalities.

So here's an example from the book:

Find all values of r that satisfy $ r^2 \minus{} 8r < \minus{} 12.$

So our goal is to somehow find bounds for the values of $ r,$ but there is no way to solve the above inequality with the twelve on the right side. For instance, factoring out the $ r$ gives (from the LHS) $ r(r \minus{} 8) < \minus{} 12.$ But there is no direct way to tell when the LHS is going to be less than -12. Since -12 is negative, we know that r and r-8 must differ in signs, but that's not enough as even though r and r-8 are of different signs (which makes their product negative), the product could still be a negative greater than -12.

So the best way to do this would be to solve this as a quadratic bringing everything to one side so that we have a zero on the other side. Doing so gives us $ r^2 \minus{} 8r \plus{} 12 < 0.$ Factoring like a quadratic gives $ (r \minus{} 2)(r \minus{} 6) < 0.$ Since the product is less than zero, we know it's negative, thus the two terms r-2 and r-6 must differ in signs. Since r-6 < r-2, we know that r-6 must be negative and r-2 must be positive, which gives us r-6 < 0 => r < 6, and r-2 > 0 => r > 2. So r is greater than 2 but less than 6, but it cannot equal either, since that would give zero. So we have $ r \in (2,6).$

So the main strategy should be to make one side 0 (as 0 is where the negatives and the positives are divided up) of the inequality, and then compare the other side with the quadratics factored to the 0.


trigonometry
real analysis
real analysis unsolved

Find $x(t),y(t)$ derivables such that

$2x'(t)=(\cos(t)+\sin(t))x+(\cos(t)-\sin(t))y$
$2y'(t)=(\cos(t)-\sin(t))x+(\cos(t)+\sin(t))y$

$x(t)=\frac{1}{2}(Ae^{\sin t}+Be^{-\cos t})$

$y(t)=\frac{1}{2}(Ae^{\sin t}-Be^{-\cos t})$

where $A,B$ are arbitrary Constants.


This problem is from classic AHSME and it is very popular. If you have seen this before, leave this to beginners.

Evalute below:

${{{{\frac {1}{\log_{2} {100!}}} + \frac {1}{\log_{3} {100!}}} + \frac {1}{\log_{4} {100!}}} + .. + \frac {1}{\log_{100} {100!}}} $

Hint:

All logarithrm of $\log_{a} {b}$ can be rewritten as $\frac {\log{b}}{\log {a}}$

is this how to do it...

[hide]make them all (log 2) / (log 100!)... etc.  Then, the top numbers are multiplied together by a law of logarithms, so that it is (log 100!) / (log 100!) = 1?[/hide]

That looks right Jeremy.

Silverfalcon, I think you have your hint backwards.  It should say that $\log_ab=\frac{\log b}{\log a}$.

Thanks, Joe. Fixed now.

Can anyone tell me another approach to this question if:

$\log_{a} b + \log_{a} c = \log_{a} (bc)$


For all natural $ n$ prove that
\[ |\sin1|\plus{}|\sin2|\plus{}|\sin3|\plus{}...\plus{}|\sin2n|>\frac{4n}{5}\]

\[\implies |\sin1|\plus{}|\sin2|\plus{}|\sin3|\plus{}...\plus{}|\sin6m|>\frac{12m}{5}.\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=564300:
\[\left| {\sin 1} \right| + \left| {\sin 2} \right| +\left| {\sin 3} \right| + ... + \left| {\sin 3n} \right| > \frac{{8n}}{5}\]\[\implies |\sin1|\plus{}|\sin2|\plus{}|\sin3|\plus{}...\plus{}|\sin6m|>\frac{16m}{5}.\]
Where $m$ be a positive integer.


$ AL$, $ BM$, and $ CN$ are the medians of $ \triangle ABC$. $ K$ is the intersection of medians. If $ C,K,L,M$ are concyclic and $ AB \equal{} \sqrt 3$, then the median $ CN$ = ?

$\textbf{(A)}\  1 \qquad\textbf{(B)}\  \sqrt 3 \qquad\textbf{(C)}\  \frac {3\sqrt3}{2} \qquad\textbf{(D)}\  3 \qquad\textbf{(E)}\ \text{None}$

Here is my method, somewhat ugly:
By Power of a point from A, $ \frac{AC^2}{2}\equal{}\frac{2}{3}AL^2$. Similarly $ \frac{BC^2}{2}\equal{}\frac{2}{3}BM^2$.
By Stewart's theorem with cevian AL, we get $ 3\plus{}AC^2\equal{}2AL^2\plus{}\frac{BC^2}{2}$. Substituting for AL, we get
$ 3\equal{}\frac{AC^2}{2}\plus{}\frac{BC^2}{4}$. But by symmetry, we could use Stewart's to get this same equation but with AC and BC switched around. Thus, we must have $ AC\equal{}BC$, in which we find $ AC\equal{}BC\equal{}\sqrt{3}$. By Pythag, it's easy to see $ CN\equal{}\frac{3}{2}$


Sometimes, when I'm viewing a thread, there is more than one thing on the bottom that lets you jump directly to a forum.  Is this intentional?

Edit: if it helps, it seems to vary directly with the number of posts in a thread.  For instance, a thread with 5 posts will have 5 forum-jumping things.

It's interesting,

I have a few weeks to be inhere, but any way I believe that wasn't happening before, I noticed it just yesterday.

Best regards,

I noticed this when I was browsing a few minutes ago. It's exactly one per post.

The first time I noticed it, I wasn't logged in, and all 12 could only jump to "Keine". I don't know how to reproduce that yet.

After more testing:

When multiple jump boxes appear, only the first works properly; the others trigger a jump to wherever the first is pointing.

As for that "Keine" issue: if I'm not logged in, none of the forum jump boxes anywhere work; they all display "Keine" where the list of forums should be. This does not depend on the language selected.

Heh, minor mistake in my coding :D  :blush: It will be repaired :P

Done. A small } mistake :) Thanks for the tip that the jumpboxes number = the number of posts, that helped in identifying the problem quickly. You can now choose to hide the jumpbox in the options -> Board Layout menu :)

More options will soon come, so stay tuned! :P :D

Dunno about Keine issue :? ... but I am probably going to restrict the jumpbox to registred users only to avoid that.

Also, when you click "Announcements" from the drop down menu and then click "go", you get a "forum does not exist" message. Although, "News Flashes", which is apparently a subforum of "Announcements", does work.

Yes, that is how it is supposed to work. There is no bug there.


vector
linear algebra
linear algebra unsolved

Problem A:  Let v_1, ...,v_n be a basis for a vector space V. Show that this remains a basis even if
v_1 is replaced by v_1 - y,  where y is in the span of v_2,..., v_n.

You have two things to check: the span is the same, and the vectors are still linearly independent. In both cases, it's only linear combinations involving $ v_1$ that are different from what they were.


When you click the view newest posts button on an announcement from the blogs page, you get an error:

 Forum index
Information

Not_a_valid_blog

Click Here to return to the blogs page

Click Here to return to the Index

Powered by phpBB © 2001, 2002 phpBB Group

It works fine if you just click on the regular link, however.

I tried it again, and the same thing still happens.

Forgot to mention earlier, but I am using Windows XP, Firefox (1.5, the second newest).

It happens when you go directly to a specific page instead of just clicking the name of the announcement.

Bug is indeed confirmed. I will see what I can do about it :)


1. Find number of solutions in non-negative reals to the following equations:
\begin{eqnarray*}x_1 + x_n ^2 = 4x_n \\ x_2 + x_1 ^2 = 4x_1 \\ ... \\ x_n + x_{n-1}^2 = 4x_{n-1} \end{eqnarray*}

2.  The triangle $ABC$ has $AC = BC$. $P$ is a point inside the triangle such that $\angle PAB = \angle PBC$. $M$ is the midpoint of $AB$. Show that $\angle APM + \angle BPC = 180^{\cdot}$.

3.  The sequence $p_1, p_2, p_3, ...$ is defined as follows. $p_1$ and $p_2$ are primes. $p_n$ is the greatest prime divisor of $p_{n-1} + p_{n-2} + 2000$. Show that the sequence is bounded.

4.  $PA_1A_2...A_n$ is a pyramid. The base $A_1A_2...A_n$ is a regular n-gon. The apex $P$ is placed so that the lines $PA_i$ all make an angle $60^{\cdot}$ with the plane of the base. For which $n$ is it possible to find $B_i$ on $PA_i$ for $i = 2, 3, ... , n$ such that $A_1B_2 + B_2B_3 + B_3B_4 + ... + B_{n-1}B_n + B_nA_1 < 2A_1P$?

5.  For each $n \geq 2$ find the smallest $k$ such that given any subset $S$ of $k$ squares on an $n x n$ chessboard we can find a subset $T$ of $S$ such that every row and column of the board has an even number of squares in $T$.

6.  Show that the only polynomial of odd degree satisfying $p(x^2-1) = p(x)^2 - 1$ for all $x$ is $p(x) = x$.

Is Problem 4 easy?.I solved ,but the solution too long(1 pages).Do you give me its solution?Thanks

1. http://www.mathlinks.ro/Forum/viewtopic.php?p=372282#p372282
2. http://www.mathlinks.ro/Forum/viewtopic.php?p=155745#p155745
3. http://www.mathlinks.ro/Forum/viewtopic.php?p=372280#p372280
4. http://www.mathlinks.ro/Forum/viewtopic.php?p=372278#p372278
5. http://www.mathlinks.ro/Forum/viewtopic.php?p=155740#p155740
6. http://www.mathlinks.ro/Forum/viewtopic.php?p=372275#p372275


advanced fields
advanced fields unsolved
Discrete Mathematics
computer science
college contests
Miklos Schweitzer

Assume that $ R$, a recursive, binary relation on $  \mathbb{N}$ (the set of natural numbers), orders $ \mathbb{N}$ into type $ \omega$. Show that if $ f(n)$ is the $ n$th element of this order, then $ f$ is not necessarily recursive.

[i]L. Posa[/i]

We will explicitly construct a Turing machine $T$, that computes such relation $R$, so that $f$ is not recursive, i.e. there does not exist any Turing machine that computes $f$.
We will call a Turing machine $T'$ good if $T'(j) $($T'$ with input $j\in \mathbb{N}$) halts with output $f(j)\in \mathbb{N}$ for every $j\in \mathbb{N}$ and $f(j),j=1,2,\ldots$ is a rearrangement of $\{1,2,\ldots\}$.
Here is the construction:
$T$ starts with successively enumerating all Turing machines $T_1, T_2,\ldots$ and simulates $T_i$ running on the input $i$. Denote this process by $T_i(i)$. It can be done, as usual, by running a fixed number of cycles consecutively on each process $T_1(1); T_1(1),T_2(2);T_1(1),T_2(2),T_3(3);\ldots$
Some of the processes (e.g. the good Turing machines) $T_i(i)$ will halt with an output $i'\in \mathbb{N}$, the others will run forever.
Let $T_i(i)$ be the first process which halts with an output $i'$. If $i'\neq 1$, we place $i$ in the first place on a row and tag it with the number $i'$. If not, we just tag it and proceed to the next machine that halts. In that way, after a while, there will be a process $T_j(j)$ that halts with an output $j'\neq 1$. We place $j$ in the first place on the row and tag it with $j'$. The other tagged numbers are placed on the same row watching that the tag $i'$ not equals to the number of the place. If the first $i$ places are already filled, but $T_i(i)$ is still running we place $i$ in the first empty place and keep $T_i(i)$ running.
In this manner we consecutively put the natural numbers on the row.
Suppose at some moment a machine $T_i(i)$, still running, with the corresponfing number $i$, put at the $n$-th place in the row, halts with an output $i'$. We consider if $i'$ coincides with $n$. If it does, the next machine $T_j(j)$ that halts with an output $j'\,, j'\neq n$ , will be put in the $n$-th place and all placed numbers after that will move right by one position. If some of their tagged numbers match their new position we do the same thing, insert consecutively in their places some other machines that will halt and move them right by one position.
To sum up. We have constructed a machine $T$ that places the natural numbers in a row, i.e. orders $\mathbb{N}$ into type $\omega$. If denote by $f(i)$ the number of the position where $i$ is finally placed, we have $f(i) \neq T_i(i)$ for every good Turing machine $T_i$. It just means there does not exist any good Turing machine that computes $f$. On the other hand the relation $R$ of the order is computable by $T$, since once any two numbers are placed in the row, their precedence relationship never changes.

I searched for this thread after I had posted some thoughts on a movie about Alan Turing [url=https://dgrozev.wordpress.com/2020/02/28/the-imitation-game/]in my blog[/url]. The above solution needed some fixes, so I rewrote it, and also provided the "official" solution I found in the "Contests in Higher Mathematics" book. I also tried to add some useful notions from that area - recursively enumerable sets, recursive sets/function, Turing machine, and some basic properties they have. I suppose, it would be helpful.

[url=https://dgrozev.wordpress.com/2020/03/25/non-recursive-function-miklos-schweitzer-1976-problem-1-part-i/]Non recursive function, ..., Part 1[/url]
[url=https://dgrozev.wordpress.com/2020/03/26/non-recursive-function-miklos-schweitzer-1976-problem-1-part-ii/]Non recursive function,... , part 2[/url]


group theory
abstract algebra
superior algebra
superior algebra unsolved

An element $x$ of finite grope $G$ is said to be [i]selfdouble[/i] if there exist non necessarily distinct elements $u\ne e,v\ne e\in G$ such that $x=uv=vu.$ Prove that if $x\in G$ is not selfdouble then $x$ is an element of order $2$ and $G$ contains $2(2k-1)$ elements for some $k\in\mathbb{N}.$
(V. Mazorchuk)

If the order of $x$ were to be $k\ge 3$, then we could take $u=x^{2}\ne e,\ v=x^{k-1}\ne e$, so the order of $x$ must indeed be $2$.

In a $2$-group (i.e. a group whose order is a power of $2$) of order $\ge 4$ every subgroup of order $2$ is contained in a group of order $4$, which is either cyclic or the Klein group. Either way, it's easy to see that an element $x$ of order $2$ in a group of order $4$ is selfdouble. Since our $x$ is not selfdouble, this means that the Sylow subgroups of $G$ have order $2$, and this is what we wanted.


Some time ago we had a "printer-friendly" function that formatted every topic in a way it was compatible to printers in most cases (not in every case, e. g. long formulas still exceeded page margins, but at least not every line exceeded margins as it happens when printing a "normal" phpBB forum page). It is gone now, I am wondering whether Valentin is going to return it.

What I also would like to suggest is a more extreme version of this feature: a function that automatically converts any thread into a TeX file. I imagine this being a pretty simple routine, converting [ b ] tags into \textbf{}, [ quote ] or [ hide ] tags into "quote" or "paragraph" environments, [ code ] tags into "quote" environments with \texttt{}, \displaystyle'ing all formulas, separating different posts by horizontal lines etc.. No, I don't want smilies to be added as graphics ;) . The resulting TeX code should be directly downloadable and compilable using standard packages unless there are LaTeX mistakes in the posts themselves. Such a feature would serve not only the needs of some power users, but would also make this whole site look more serious (something quite useful when thinking about sponsorship of AoPS) and help people learn LaTeX. What do you think about this, Valentin?

  darij

[url=http://www.iwriteiam.nl/html2tex.html]html2tex[/url] might be useful.


abstract algebra
group theory
superior algebra
superior algebra unsolved

I'm new at group theory, so I apologize in advance for a question that's probably below the level of this forum...

Anyway, one of the exercises in my book (Lang, Undergraduate Algebra) was to prove that if H is a normal subgroup in G, G/H is abelian if and only if $ xyx^{ \minus{} 1}y^{ \minus{} 1}\in H$ for all $ x,y\in G$. My original solution is below, but now I'm worried that it doesn't work because in a later section the author uses the canonical homomorphism G -> G/H, and the proof is a lot longer.

So if we assume G/H is abelian, take two cosets aH, bH in G/H (note: the coset representatives a, b are arbitrary, so we can have aH=bH and a and b different). Then, abH=abHH=aHbH=bHaH=baH since H is normal. Thus, for all $ h\in H$, $ abh\in baH$, and multiplying on the left by $ a^{ \minus{} 1}b^{ \minus{} 1}$ gives $ aba^{ \minus{} 1}b^{ \minus{} 1}\in H$. Since $ a,b$ were arbitrary we have the 'only if' direction, and it should reverse easily.

Is this correct?

That is essentially correct. I would just write it in a slightly different order. Take any $ a, b\in G$. We want to show $ aba^{\minus{}1}b^{\minus{}1}\in H$.

So now we consider the cosets $ aH, bH\in G/H$. We get from abelian that $ aHbH\equal{}bHaH\Rightarrow abH\equal{}baH$ and that happens iff $ ab(ba)^{\minus{}1}\equal{}aba^{\minus{}1}b^{\minus{}1}\in H$.

Actually $ abH\equal{}baH\ \Leftrightarrow\ a^{\minus{}1}b^{\minus{}1}ab\in H.$ If you want it to be in the form $ aba^{\minus{}1}b^{\minus{}1}$ as given, it is better to use right cosets instead.

[list]$ G/H$ Abelian[/list]
$ \Leftrightarrow\ \forall\,x,y\in G,\ (Hx)(Hy) \equal{} (Hy)(Hx)$

$ \Leftrightarrow\ \forall\,x,y\in G,\ Hxy \equal{} Hyx$

$ \Leftrightarrow\ \forall\,x,y\in G,\ xyx^{ \minus{} 1}y^{ \minus{} 1}\in H$

If $ S \equal{} \{xyx^{ \minus{} 1}y^{ \minus{} 1}: x,y\in G\},$ the subgroup $ G' \equal{} \langle S\rangle$ is the commutator or derived subgroup of $ G.$ It is normal in $ G$ and is the smallest normal subgroup of $ G$ such that that $ G/G'$ is Abelian.


calculus integration derivative function trigonometry vector logarithms

In high school I somehow did well enough on the calc AP test to skip calculus at my university even though I knew virtually no calculus, and as a math major that's hurt me in college.  Now I have a semester off and I'm going back and teaching myself calculus (including multivariable) very thoroughly.  I'm going to make sure that I'm able to derive every formula in the textbook, but how much memorization would you say is necessary for a future professional mathematician?  Right now I'm planning on memorizing all of the thousands of identities and integrals in my textbook (and many that aren't in my textbook) so that I'll be able to identify them very quickly when I come across them in future work.  Do you think this is worth my time, or should I just memorize some basic identities and move on to an analysis textbook?

I don't even think it's a question of if you should. You can't memorize it all even if you wanted to. And even if you could, you wouldn't have any of the troubles you alluded to. I'll let someone else answer in more detail.

You should not spend your time memorizing formulas or integration identities. Only a few of the very very basic identities are necessary to memorize, for example: the definition of the derivative, the derivatives of the most basic elementary functions (power functions, the 6 trig functions, e^x, and ln x), the definition of the Definite Integral, the Mean Value Theorem, and the two versions of the Fundamental Theorem of Calculus. And even then, understanding why they work or the motivation behind them is equally as important as the formula or statement itself.

The most important thing is your understanding of the concepts -- which even proofs don't always illuminate. Your understanding (or misunderstanding) of concepts is best brought out by doing problems. So instead of memorizing things from the book, I would suggest finding a good book and working as many problems as you can from it. If you do enough problems, you will remember the most important formulas (and ideas) in the process.

Recommendations (in order from easy to hard): [u]Vector Calculus[/u] by Marsden and Tromba (for multivariable calculus), [u]Calculus[/u] by Spivak (for single variable calculus), and [u]Calculus, vols. I and II[/u] by Tom Apostol (vol. I for single variable, vol. II for multivariable calculus and linear algebra).

Memorization? I have mixed feelings. It's obviously a wrong approach to devote oneself to memorizing some great big formula sheet which contains All Truth, and it's easy to mock those who think that's a good approach.

But look around you at the people who do extraordinarily well on contests. Of course these people focus on concepts, and understanding, and imagination - but at the same time, I'll guarantee that they have a tremendous amount of mathematics, and even mathematical minutiae memorized. The more you already know, the more connections you can make. Having some things down to routine means that you don't have to devote your brain's processing power to routine manipulation and you're free to see and understand the whole of a problem.

Do you really need to know that $ \int \ln x\,dx \equal{} x\ln x\minus{}x\plus{}C?$ No of course not - that can be quickly re-derived. But if you do know that, you're a step faster. Do you really need to know that $ \int_0^{\frac{\pi}2}\cos^2x\,dx\equal{}\int_0^{\frac{\pi}2}\sin^2x\,dx\equal{}\frac12\cdot\frac{\pi}2\equal{}\frac{\pi}4\equal{}$ "half the box?" No, there are ways to find that - but if you know it, you're faster and you're a great deal more accurate. And that particular result easily extends to things like $ \int_0^{100}\cos^2(4\pi t)\,dt\equal{}50.$

A calculus student who can't accurately state the product rule for differentiation is in big trouble. A calculus student who sees "$ \ln 0$" (as part of the evaluation phase for an integral) and doesn't immediately pick up on the that that that's $ \minus{}\infty$ (and the original integral was improper, and possibly divergent) is someone who's going to struggle with finishing an exam in the time allotted.

Even thought I would never advocate "memorize a bunch of things" as a strategy, it's still true that the best students are usually the ones who do have the most things memorized, and the worst students are the ones who have the least accurate memories.

I recently gave as a first test in a calculus class a first question that involved just memorization. Here are the 10 parts of this question:

a)$ \int\sin x\,dx\equal{}?$  b) $ \frac{d}{dx}\,\sec x\equal{}?$ c) $ \int\sec x\,dx\equal{}?$ d) $ \frac{d}{dx}\,\arcsin x\equal{}?$  e) $ \sin^2\theta\equal{}?$ (formula should contain trig functions of $ 2\theta.$) f) $ \int\frac{1}{y}\,dy\equal{}?$ g) $ \int\frac{1}{y^2}\,dy\equal{}?$ h) $ \int\frac{dx}{1\plus{}x^2}\equal{}?$ i) $ \int\frac{dx}{4\plus{}x^2}\equal{}?$ j) $ \int e^x\,dx\equal{}?$

All pretty mindless, right? And yet the correlation coefficient between performance on that question and performance on the test as a whole was $ \plus{}.80.$

I think that memorization is a big part of intuition.  If you don't have a large enough list of results that you can recall almost immediately, it's difficult to draw parallels between problems and it's difficult to discard lines of reasons which are 'obviously wrong' based on a quick analysis.

This isn't saying that there's much to be gained by memorizing tables of integrals, but there certainly are some things that should be memorized.

Thank you everyone for your replies; it's great to hear what people have to say about this.  Dr. Merrryfield, I'd love to have the books by Apostol; they look great, but they cost a fortune!  My school uses Calculus and Analytic Geometry by Sherman K. Stein and Anthony Barcellos for single variable and Calculus 3 by Jerrold Marsden and Alan Weinstein for multivariable, so I have those, and I have Calculus by Salas, Hille and Etgen, 8th edition, which covers both.  I bought that one a few years ago at a bookstore for $ \$12$ because it was outdated.  That one is my favorite and the one I've been going through, but the problems in all of these are too trivial.  I'm about halfway through the book right now, learning integration techniques at the moment.  Do you think it would be worth a few hundred dollars for me to buy the ones by Apostol?  Can anyone direct me to free resources with some better problems?

Sorry, I thought Dr. Merrryfield was the one who suggested those textbooks.  I guess I was talking to you, gauss202.  Well, I went ahead and ordered Apostol, vol. 1 and 2.  I've started reading it online (the book's in the mail), and I like it very much.  I also found Rudin for about $ \$25$, so I went ahead and bought that one, too.  After I finish going through Apostol I'm going to get started on Rudin, and I'm planning to work through Harvard's Math 55 problem sets, though I don't actually know how difficult that will be, so we'll see.  I may be going back to school this spring or I may wait until the summer to start taking classes again, so I don't know exactly how much time I'm going to have.  Again, thanks everyone for all your input.

Over the years I have taught trigonometry dozens and dozens of times and I can never remember any of the usual theorems and formulas for the subject except, sin of x squared plus cos of x squared equals one.  With that,  the Pythagorean theorem,  and the formula of a unit circle with center the origin,  I can derive everything else,  while teaching the students to do the same..

I have friends with better memory and who can teach trigonometry,  from memory,  and know on the first day of class exactly what will be done in each class through the course.

I think my way is better,  for it teaches the students to learn to actually do  mathematics rather than memorize.  That same thing can be done with about any mathematics course,  all the way through undergraduate and graduate school.  Others disagree.  It does require much more effort on both teacher and student.

The more time one spends working with material, the more one memorizes without even trying to memorize. Certain ideas, certain formulas, even certain numbers appear as old friends.

An anecdote from my calculus class. Early in our section on series, I asked about the sum $ \sum_{n \equal{} 1}^{\infty}\frac {1}{n2^n}.$ We wer able to show that it converges. I then asked what it converges to. Now, since we hadn't done power series, we didn't yet have tools to compute that exactly, but the series converges reasonably rapidly and it's not hard to get a good approximation with a calculator.

The sum came up as about $ .693.$ I said that that is a well-known number, a number that can be simply named. I asked for its identification - and not a single person in the class recognized it. Not enough time playing with numbers to see it as an old friend.

$ .866\quad .7071\quad .618 \quad 2.718 \quad \frac {22}7\quad \frac {355}{113}\quad 1024$

I think you mean $ \sum_{n\equal{}1}^{\infty}\frac{1}{n\cdot 2^n}$. I have some power series memorized...

I did not recognize $ .693$  :maybe:  eventually had to guess (correctly).

Exponential growth and exponential decay problems lead to talk of doubling times or half lives, which leads to $ \ln 2.$


Find $ 2\times 2$ matrices $ A$ and $ B$ having real entries such that $ \lim_{m\rightarrow\infty}A^{m}$, $ \lim_{m\rightarrow\infty}B^{m}$, and $ \lim_{m\rightarrow\infty}(AB)^{m}$ all exist, but
$ \lim_{m\rightarrow\infty}(AB)^{m}\neq\left(\lim_{m\rightarrow\infty}A^{m}\right)\left(\lim_{m\rightarrow\infty}B^{m}\right)$
I'm thinking that there is probably some really easy example but I just can't seem to find it.

There are probably many different examples based on many different ideas. Here's one:

$ A \equal{} \frac1{\sqrt {2}}\begin{bmatrix}1 & 0 \\
1 & 0\end{bmatrix}$ and $ B \equal{} \frac1{\sqrt {2}}\begin{bmatrix}1 & 1 \\
0 & 0\end{bmatrix}.$

We have both $ \lim A^n \equal{} 0$ and $ \lim B^n \equal{} 0.$ However,

$ AB \equal{} \frac12\begin{bmatrix}1 & 1 \\
1 & 1\end{bmatrix}.$ And $ (AB)^2 \equal{} AB$ so that $ \lim (AB)^n \equal{} AB\ne 0.$

----

What I was thinking: if the eigenvalues of $ AB$ are the products of the eigenvalues of $ A$ and the eigenvalues of $ B,$ then we can't make this work. So $ A$ and $ B$ should not be simultaneously diagonalizable, so they shouldn't commute.

Okay, thanks.  I actually came up with an example yesterday but before I did, I was trying to figure out a way to use simultaneously diagonalizable matrices.


[b]THEOREMS[/b]
$ (f \pm g)(x) \equal{} f(x) \pm g(x)$
$ (f.g)(x) \equal{} f(x).g(x)$
$ \left(\frac {f}{g}\right)(x) \equal{} \frac {f(x)}{g(x)}$, if $ g(x) \neq 0$
$ f^n (x) \equal{} [f(x)]^n$
$ (\sqrt [n]{f})(x) \equal{} \sqrt [n]{f(x)}$

[b]PROBLEMS TO SOLVE[/b]
1. If two functions $ f(x)$ and $ g(x)$ satisfies $ (f \plus{} g)(x) \equal{} x^2 \minus{} 4x \minus{} 5$ and $ \left(\frac {f}{g}\right)(x) \equal{} x \minus{} 6$, find a linear function $ h(x)$ so that $ (g.h \minus{} f)(x) \equal{} 8x \plus{} 8$. Otherwise, find $ h( \minus{} 2)$.

2. We have some functions $ f(x) \equal{} \sqrt {\frac {3 \minus{} x}{3 \plus{} x}}$, $ g(x) \equal{} \frac {3}{x} \minus{} 1$, and $ h(x) \equal{} \sqrt {9 \minus{} x^2}$, then find the value of $ p$ that satisfies $ \left(h \plus{} \frac {f}{\sqrt {g}}\right)(p) \equal{} 4$.

3. It is known that $ f(2 \minus{} x) \equal{} 8x^2 \minus{} 32x \plus{} 30$ and $ (f \plus{} g)\left(\frac {1}{2}x\right) \equal{} 2x^2 \minus{} x \minus{} 1$. Find the domain of $ h(x)$ if $ h(x) \equal{} \left( \sqrt {\frac {f}{g} \plus{} f}\right)(x)$.

[hide]Atleast one of $ f(x)$ or $ g(x)$ has an $ X^2$ term. $ \left(\frac {f}{g}\right)(x) \equal{} x \minus{} 6$, So $ F(x)$ must have the $ x^2$ term.
$ (x\minus{}6)(x\plus{}A)\plus{}(x\plus{}A)\equal{}x^2 \minus{} 4x \minus{} 5\\
(X\minus{}5)(x\plus{}A)\equal{}(X\minus{}5)(X\plus{}1)$
$ F(X)\equal{}(X\minus{}6)(X\plus{}1)$ and $ G(X)\equal{}X\plus{}1$

$ ((X\plus{}1)h)\minus{}(X\minus{}6)(X\plus{}1)) \equal{} 8(x\plus{}1)$
$ ((h\minus{}(x\minus{}6))(X\plus{}1)\equal{}8(X\plus{}1)$
$ ((h\minus{}x\plus{}6)\equal{}8$
$ h\equal{}x\plus{}2$
$ h(\minus{}2)\equal{}0$
[/hide]


calculus
logarithms
function
number theory

How should I go about solving this?
Evaluate in terms of $ x$ when the sum exists.
$ \displaystyle\sum^\infty_{n\equal{}1}\frac{x^n}{n}$
I can do it the calculus way, but I want to avoid that.

[hide="Maybe this helps.."]Let $ S \equal{} \sum^\infty_{n \equal{} 1}\frac {x^{n}}{n}$, then

$ S \equal{} x \plus{} \frac {x^2}{2} \plus{} \frac {x^3}{3} \plus{} \frac {x^4}{4} \plus{} ...$

$ xS \equal{} x^2 \plus{} \frac {x^3}{2} \plus{} \frac {x^4}{3} \plus{} \frac {x^5}{4} \plus{} ...$

$ (x \minus{} 1)S \equal{} (0 \minus{} 1)x \plus{} (1 \minus{} \frac {1}{2})x^2 \plus{} (\frac {1}{2} \minus{} \frac {1}{3})x^3 \plus{} ...$

The coeffiecient of $ x^k$ is $ \frac {1}{k \minus{} 1} \minus{} \frac {1}{k} \equal{} \frac {1}{k(k \minus{} 1)}$, except when $ k \equal{} 1$.

$ (x \minus{} 1)S \equal{} \minus{} x \plus{} \frac {x^2}{2} \plus{} \frac {x^3}{6} \plus{} \frac {x^4}{12} \plus{} ...$

$ S \equal{} \frac {\sum_{k \equal{} 2}^{\infty} \frac {x^k}{k(k \minus{} 1)} \minus{} x}{x \minus{} 1} \equal{} \frac {x(\sum_{k \equal{} 1}^{\infty} \frac {x^k}{k(k \plus{} 1)} \minus{} 1)}{x \minus{} 1}$

Kinda stuck here..[/hide]

[quote="JBL"]There is no "non-calculus" way to do this: we're talking about a Taylor series for a logarithm here, and there's no way to understand what that means without calculus.[/quote]
Hm, thanks. This came up with a generating functions question in my number theory book and was hoping I could keep Calculus out of the scene, but I guess not. Thanks!


geometry
3D geometry
tetrahedron
octahedron
icosahedron

I don't know the word for the unit, but I'm talking about the one I know -- you take three of them and you make a tetrahedron with an equilateral-triangle base and three right-triangle faces, and you put them together to make a cube, or stellated octahedron or icosahedron usually.

Now, moving on:
If you make a cube, you can smash it into another cube by adding three more units after opening up one corner.  This is equivalent to stellating the figure that you get by joining two regular tetrahedra on one face.  You can keep adding tetrahedra (smashing on new cubes) in a variety of ways.  If you add ten of them in the proper way, you get an interesting thing -- you can wrap around so that they close up, and it turns out you've built the same shape that you get when you first make an icosahedron and then push in two opposite corners.  This raises two questions for me:
1) Is what I'm seeing exact or approximate?  That is, can one really situate 10 congruent regular tetrahedra in space so that each shares a side with two others and they all share a common vertex, or does it just work because paper is flimsy?
2) Is it possible to similarly wrap tetrahedra around in a way that leads to a toral shape with an actual hole (rather than a common vertex) at the center?

hi joel!
firstly, the unit is called sonobe
i'm not quite sure what you mean by mushing, i tried to follow it and what i got was an octahedron like thing with two faces that are pentagons (so i guess it's more like a decahedron?). i'll play around a bit more to see what should really come out, but i have an idea of what you're talking about
i have actually had that question in the past, but never voiced it... what you're seeing makes sense to me!
i'm thinking that in order to get a torus you could increase the number of cubes that you put together, maybe something along the lines of what is done in [url]http://www.merrimack.edu/~thull/combgeom/tori/torusnotes.html[/url] (as i see it, phizz unit and sonobe unit are in a way duals, you can make the dual of a shape and enclose it by the shape made of the other unit (if that made any sense at all))
i could also ask around and see what others may think if you want :)

Hi Maria,
Here's a picture that illustrates what I mean by smushing cubes together.

ok, so i finally got to play around with paper (yay, midterms over, lets not start thinking about finals now!)

so i managed to convince myself that what u get by smashing cubes is actually the same as what you get by pushing through the icosahedron. i was trying to model it by actual tetrahedra but i dont' have either the origami or the geometry ability to do it.

for the torus thing, i don't think it can be done with this same approach- need more complex figures to start with than a tetrahedron- though if you have arrived at a better conclusion feel free to share

So, I got half-way to an icosahedron -- 15 units, I guess -- and convinced myself that it really was the same, too, but I still haven't tried to figure out if that's exact or fudging.  Also, I think I convinced myself I could do a big torus with about 96 units, but I am not really very serious about trying with finals and all that coming up, I've only done about 30 or so, and then my roommate has crushed those.  So, basically I haven't made much progress on it  :)

i think a torus could be made with 90 pieces (or was it 120, i'm not sure) i made one a long time ago and then it got crushed and i never bothered to put it back together. also, i made it with the phizz unit and not sonobe- that's the one that makes the frame.

good luck on finals :)

so i just realized something- this "smashing" idea can be applied to any shape made of sonobe units- just connect them in that way. in order for it to not open up (to become the icosahedron in the case of the cubes) if the corners opened are not directly next to each other (adjacent at a side or at a vertex)


calculus
integration
complex analysis
complex analysis unsolved

$ \int_{C}\frac {\sin(iz)}{1 \plus{} z^{2}}dz$.   where C is a straight line from $ 2i$ to $ \minus{} 1 \minus{} 2i$.

i am quite confused.  please help me evaluating the above.
thanx

I'm pretty sure that it is a part of the problem, not its full text. If so, you have made a wrong reduction: this particular part is unsolvable in the sense that the answer is not expressible in terms of reasonable constants. I surmise that the full problem is to find the integral over the broken line going from $ 1\plus{}2i$ to $ 2i$ to $ \minus{}1\minus{}2i$.


It is proposed to partition the set of positive integers into two disjoint subsets $A$ and $B$ subject to the following conditions:
                (i)1 is in $A$;
                (ii)no two distinct members of $A$ have the sum of the form $2^{k}+2 (k=0,1,2...)$;
                (iii)no two distinct members of $B$ have a sum of that form.
Show that this partitioning can be carried out in a unique manner and determine the subsets to which $1987,1988,1989$ belong

$1988$ can not be in the same set as $62$ (since $1988+62=2^{11}+2$).  And $62$ can't be in the same set as $4$.  So $1988$ is in the same set as $4$ which is $A$.  Similarly for the other numbers.

It is easy to check that the sets exist and are unique.  The uniqueness comes from the fact that when you decide which set a number $n$ belongs to, you always narrow down you choice by playing the above game.  The existence comes from the fact that there is only one number to avoid when playing the game above because if $n+m=2^{k}+2$ where $m<n$ then $n\geq 2^{k-1}+2$ so we don't need to worry about $n+m'=2^{k-1}+2$ or less.  And $n\leq 2^{k}+1$ so we don't need to worry about $n+m'=2^{k+1}+2$ or more.

The problem makes more sense if you consider $A-1$ and $B-1$ and then the condition is to avoid sums equal to $2^{k}$.  Everything else is routine from that point, and the two sets (minus 1) turn out to be the numbers whose binary representation has an even or odd number of blocks of 0's and 1's, to the left of the final $1$.