I'm posting this for those who are curious about the last two problems. Discuss at your own peril  :P.

[b][i]8.  Amy, Brian, and CoScott are playing a game, with Dana as the judge. Dana puts eight stickers (four red and four blue) in a bowl, then draws two at random for each player and sticks them to their foreheads. Each player can see both of the other players' stickers, but not his or her own or the two left in the bowl. Dana then asks Amy whether she knows the colors of her stickers. If Amy does not, Dana asks Brian whether he knows his colors, then CoScott, then Amy again, then Brian again, and so on. The game ends if one of the players can logically deduce his or her colors. What is the probability that the game ends on one of Amy's turns? Brian's? CoScott's? What is the probability that it never ends at all?[/i][/b]

CoScott... :lol:

Is there a better way?
[hide="u. g. l. y."]
We will use R and B to denote the two colors of stickers.  To describe any particular assignment of stickers, we will use the notation $ XX\minus{}XX\minus{}XX$.  For example, the assignment $ RB\minus{}BB\minus{}RR$ means that Amy has one of each color, Brian has two blues, and CoScott has two reds.  Note that the order of colors on each person does not make a difference, so $ BR$ for one person is the same as $ RB$ for that person.

We will now list out the possible assignments, and compute the probabilities of each assignment occurring.

CASE 1 -- No person has two different colors:  The only possible arrangements are

\begin{verbatim}
RR-BB-BB    BB-RR-BB    BB-BB-RR
BB-RR-RR    RR-BB-RR    RR-RR-BB.
\end{verbatim}
Because of symmetry, each of these arrangements has an equal probability of occurring.  So we only need to find the probability of a particular one occurring, say RR-BB-BB.  First, the probability of choosing two R's for Amy is $ \frac{4}{8} \cdot \frac{3}{7} \equal{} \frac{3}{14}$.  Next, given Amy's choice, the probability of Brian getting two B's is $ \frac{4}{6} \cdot \frac{3}{5} \equal{} \frac{2}{5}$.  Finally, given Amy's and Brian's choice, the probability of CoScott getting two B's is $ \frac{2}{4} \cdot \frac{1}{3} \equal{} \frac{1}{6}$.  Thus, the total probability of getting this arrangement is \[ \frac{3}{14} \cdot \frac{2}{5} \cdot \frac{1}{6} \equal{} \frac{1}{70}.\]
So by symmetry, the probability for each arrangement in CASE 1 is $ \frac{1}{70}$.

CASE 2 -- Exactly one person has two different colors:  The only possible arrangements are

\begin{verbatim}
RB-RR-BB    RR-RB-BB    RR-BB-RB
RB-BB-RR    BB-RB-RR    BB-RR-BR
\end{verbatim}

Because of symmetry, each of these arrangements has an equal probability of occurring.  So we only need to find the probability of a particular one occurring, so for now we'll just use RB-RR-BB.  The probability of Amy getting RB \textit{in that order} is $ \frac{4}{8} \cdot \frac{4}{7}$, but she could have also gotten BR, so we need to multiply by two.  This gives the probability for Amy getting one of each color as $ \frac{4}{8} \cdot \frac{4}{7} \cdot 2 \equal{} \frac{4}{7}$.  Next, given Amy's choice, the probability of Brian getting $ RR$ is just $ \frac{3}{6} \cdot \frac{2}{5} \equal{} \frac{1}{5}$.  Finally, the probability of CoScott now getting $ BB$ is $ \frac{3}{4} \cdot \frac{2}{3} \equal{} \frac{1}{2}$.  Thus, the total probability of getting this arrangement is \[ \frac{4}{7} \cdot \frac{1}{5} \cdot \frac{1}{2} \equal{} \frac{2}{35}.\]  By symmetry, all arrangements in CASE 2 have probability $ \frac{2}{35}$.

CASE 3 -- Exactly two people have two different colors:  The only possible arrangements are

\begin{verbatim}
RB-RB-RR    RB-RR-RB    RR-RB-RB
RB-RB-BB    RB-BB-RB    BB-RB-RB
\end{verbatim}

We'll use the same strategy as before; this time we'll find the specific probability of RB-RB-RR.  The probability of Amy getting RB is $ \frac{4}{8} \cdot \frac{4}{7} \cdot 2 \equal{} \frac{4}{7}$.  Given Amy gets RB, the probability of Brian also receiving RB is $ \frac{3}{6} \cdot \frac{3}{5} \cdot 2 \equal{} \frac{3}{5}$.  Finally, the probability of CoScott getting RR from the remaining stickers is $ \frac{2}{4} \cdot \frac{1}{3} \equal{} \frac{1}{6}$.  Thus, the total probability of this arrangement, and by symmetry, all other arrangements in CASE 3, is \[ \frac{4}{7} \cdot \frac{3}{5} \cdot \frac{1}{6} \equal{} \frac{2}{35}.\]

CASE 4 (Nearly done!) -- All three people each have two different colors:  The only possible arrangement is
\begin{verbatim}
RB-RB-RB
\end{verbatim}
The probability of Amy receiving RB is $ \frac{4}{8} \cdot \frac{4}{7} \cdot 2 \equal{} \frac{4}{7}$, Brian also receiving RB is $ \frac{3}{6} \cdot \frac{3}{5} \cdot 2 \equal{} \frac{3}{5}$, and CoScott receiving RB as well is $ \frac{2}{4} \cdot \frac{1}{3} \cdot 2 \equal{} \frac{2}{3}$.  Thus, the probability of RB-RB-RB occurring is $ \frac{4}{7} \cdot \frac{3}{5} \cdot \frac{2}{3} \equal{} \frac{8}{35}$.

END OF CASES

Now we can actually start looking at the game.
Note that each arrangement has a unique possible outcome, because there is no luck involved in the game after the distribution of stickers.  Forst, let's look the arrangements from CASE 1:

RR-BB-BB \\
\textit{Amy}:  Since all of the blue stickers are evidently used up, I know I have both red stickers.  I win! \\
By symmetry, Amy also wins in BB-RR-RR.

BB-RR-BB \\
\textit{Amy}:  I could be anything... \\
\textit{Brian}: Haha, the four B's are all used up, so I have RR.  I win! \\
By symmetry, Brian also wins in RR-BB-RR.

BB-BB-RR \\
\textit{Amy}:  I have no clue what I am... \\
\textit{Brian}:  Hmm, if I was RR, then Amy would have known that she was BB.  But other than that, I don't know anything... \\
\textit{CoScott}:  Finally, I win.  Because all of the B's are gone, I must have two R's. \\
Similarly, CoScott also wins in RR-RR-BB.

That was pretty simple.  However, for the arrangements in CASE 2, we will need to think a little harder:

RB-RR-BB \\
\textit{Amy}:  I don't know anything... \\
\textit{Brian}:  I can't be BB, because three of the B's are already used up by Amy and CoScott.  However, I could equally well be RB or RR, because in both cases, Amy would still not have known her color... \\
\textit{CoScott}:  I can't be RR, because three R's are already used.  If I was either RB or BB, then Brian and Amy would have still not known what their colors were, so I can't tell whether I am RB or BB... \\
\textit{Amy}:  If I had been BB, then Brian would have known he had RR, because all four B's would be used on me and CoScott.  Similarly, if I had been RR, then CoScott would have known he was BB.  So I must be RB, therefore I win! \\
By symmetry, Amy wins in RB-BB-RR as well.

RR-RB-BB \\
\textit{Amy}:  I can't be BB, because three B's are already used... \\
\textit{Brian}:  I'm not BB, or else Amy would have known she was RR.  But I could equally well be RB as RR... \\
\textit{CoScott}:  I can't be RR, because three R's are already used up.  But if I was RB or BB, then Amy and Brian would have made the same responses, so I am not sure which I am... \\
\textit{Amy}:  I still have no idea what I am, except for the fact that I am not BB.  If I was RB or RR, then Brian and CoScott would have still not known their colors.  So, I'm stuck... \\
\textit{Brian}:  Aha!!! I already knew I couldn't be BB.  Now I know that I can't be RR, or else CoScott would have immediately deetermined his colors!  Hence, I am RB.  I win! \\
And, of course, Brian wins in BB-RB-RR as well, by symmetry.

RR-BB-RB \\
\textit{Amy}:  I can't be BB, because three B's are used by the boys, but apart from that I'm stuck... \\
\textit{Brian}:  I'm not RR, because the others have received three R's.  I could be RB or RR, but both would cause Amy to not know what she is, so I'm stuck... \\
\textit{CoScott}:  I'm not BB, or else Amy would have won, and I'm not RR, or Brian would have won, so I am RB and I win! \\
CoScott similarly wins BB-RR-RB.

Now for the arrangements in CASE 3  (Note.  As all of the players are perfect logicians, they could have worked out all of our previous results for this problem.  So we will see them making references to our previous results.  Do not be alarmed, there is no circular logic here.  It's just to save space so we don't have to write out all the logic for a previous result to incorporate it into something new):

RB-RB-RR \\
\textit{Amy}:  I'm not RR, because there are only four R's and three are already used.  Otherwise, I'm stuck... \\
\textit{Brian}:  I'm not RR, because three R's are used by others.  Unfortunately, there is nothing more I can deduce... \\
\textit{CoScott}:  No!  Both my opponents have RB's!  I have no idea.  No matter what I coud possibly be, Amy and Brian wouldn't have been able to deduce their colors, so I'm unable to do anything... \\
\textit{Amy}:  Still, I have no idea.  If I was BB, then the arrangement would be BB-RB-RR, which we all know is won by Brian on his second turn.  And if I was RB, everyone would still not be able to deduce their positions at this point.  So, I'm not sure whether I am RB or BB... \\
\textit{Brian}:  Hmm...  If I was BB, then the arrangement would be RB-BB-RR, which we all know is won by Amy on the second turn.  But alas!  Amy was tight-lipped this second round.  So I must not be BB, so I am RB!  I win! \\
By symmetry, Brian wins RB-RB-BB as well.

RB-RR-RB \\
\textit{Amy}:  I'm not RR, because there are only four R's total.  I can't say anything else... \\
\textit{Brian}:  Ugh, both opponents are RB's.  I can't determine anything on this first turn... \\
\textit{CoScott}:  I'm not RR, because three R's are used up.  If I was RB, then everyone's responses would be as they are now, and if I was BB, then we would have RB-RR-BB, in which we all know everyone remains tight-lipped for the first round.  So I cannot find any contradictions... \\
\textit{Amy}:  If I was BB, then we'd have BB-RR-RB in which everyone knows CoScott wins on his first turn.  However, he hasn't won, so I know I must be RB!  I win!

RR-RB-RB \\
\textit{Amy}:  I cannot say anything about what I have... \\
\textit{Brian}:  I'm not RR, but otherwise I cannot tell anything... \\
\textit{CoScott}:  I'm not RR.  If I was BB, then the arrangement would be RR-RB-BB, and everyone knows that in that case, nobosy wins on the first round.  If I was RB, then also no one would be able to win on the first turn... \\
\textit{Amy}:  I still cannot deduce anything, because both of my opponents have RB... \\
\textit{Brian}:  Yes!  If I was BB, then we would have RR-BB-RB, which we all know is won by CoScott in the first round.  But CoScott has not won, so I must not be BB.  Thus, I must be RB, and I win! \\
Similarly, Brian wins BB-RB-RB.

We're almost there...  Finally, we'll look at CASE 4:

RB-RB-RB \\
\textit{Amy}:  Eww, two RB opponents.  No conclusions... \\
\textit{Brian}:  Whoa, two RB opponents.  I can't say anything... \\
\textit{CoScott}:  Hm, in the cases RB-RB-RR and RB-RB-BB, no one wins in the first round, which is what's happening here.  If I was RB, then no one would talk either.  So, I cannot even eliminate one possiblity of my colors... \\
\textit{Amy}:  No ideas...  If I was RR or BB, then we'd have a CASE 3 arrangement, and everyone knows that no one talks in the first round of a CASE 3 arrangement, and that's what's happening right now...  In addition, if I was RB, then this would also happen... \\
\textit{Brian}:  Yay!  If I was RR or BB, then we'd have RB-RR-RB or RB-BB-RB, and it's common knowledge that Amy wins on her second turn in both of those cases.  Since she has remained silent,  I must conclude that I have RB.  I win!

Let's now organize our results:
\begin{verbatim}
RR-BB-BB Amy      BB-RR-BB Brian    BB-BB-RR CoScott
BB-RR-RR Amy      RR-BB-RR Brian    RR-RR-BB CoScott

RB-RR-BB Amy      RR-RB-BB Brian    RR-BB-RB CoScott
RB-BB-RR Amy      BB-RB-RR Brian    BB-RR-RB CoScott

RB-RB-RR Brian    RB-RR-RB Amy      RR-RB-RB Brian
RB-RB-BB Brian    RB-BB-RB Amy      BB-RB-RB Brian

                  RB-RB-RB Brian
\end{verbatim}

So Amy wins two arrangements from CASE 1, two arrangements from CASE 2, and two from CASE 3.  Thus, her total probability for winning is
\[ 2 \cdot \frac{1}{70} \plus{} 2 \cdot \frac{2}{35} \plus{} 2 \cdot \frac{2}{35} \equal{} \frac{9}{35}.\]
Brian wins two from CASE 1, two from CASE 2, four from CASE 3, and the one from CASE 4.  Thus, his total probability of winning is
\[ 2 \cdot \frac{1}{70} \plus{} 2 \cdot \frac{2}{35} \plus{} 4 \cdot \frac{2}{35} \plus{} 1 \cdot \frac{8}{35} \equal{} \frac{3}{5}.\]
CoScott wins two from CASE 1 and two from CASE 2.  So the probability that he wins is
\[ 2 \cdot \frac{1}{70} \plus{} 2 \cdot \frac{2}{35} \equal{} \frac{1}{7}.\]

With some patience, we have found the three probabilities!   Note that it is impossible for the game to never end, because each assignment of stickers has a unique winner.[/hide]

[quote="Darmani"]Yes, I got the same answer!

I managed to abstract out a fair amount of the talk about the game itself via a textual form of recursion. I love the term I came up with for the strategy to win this game -- "contradicted hypothetical canceling."

Here's the first portion of my solution (the rest talks about the combinatorics):

Any given distinct permutation of the combinations of stickers on the three players’ heads is a “position.”

In order to investigate the probability that any given player or none will win, it is imperative to examine the characteristics of the various positions that determine who, if anyone, will win.

Since it does not matter what order two stickers are placed on a player’s head, there ends up being three distinct possible combinations of stickers on any player’s head: solid red (two red stickers), solid blue (two blue stickers), and mixed (one red, one blue). For convenience, these adjectives will be applied to the players rather than to the formations on their head (e.g.:  a mixed player is one with a mixed combination of stickers).

Deducing what colors lie on one’s head or abstaining from guessing will be referred to as an “action.” Players must choose an action based on only two dynamic types of information: the combinations on other’s heads, and other player’s actions. Recursively applying the above, it means that other players actions will be based on the position from their perspective and yet other players actions, which are in turn based on the position from their perspective and yet other players’ actions, which are in turn based on the same things, et cetera ad infinitum until the base case where no actions have been made to observe. As a result, all actions based off observations of actions ultimately reduce to how the actions correspond with possible perspectives of various players.

Note that positions from one’s perspective, and therefore observations of actions, do not give any positive information about the stickers on one’s heads are; they only give negative information on what the stickers on one’s head could not be. For example, a player who sees that the other two players both have two red stickers knows he has two blue stickers only because, using knowledge of the total numbers of stickers, he knows that he cannot have two red stickers or a red sticker and a blue sticker. This strategy of using one’s perspective to directly determine what combinations cannot be on one’s head will be known as “canceling.” A position in which a player can win via canceling (i.e.: a position with three solid players) is a Layer 0 position.

However, if one of the players in a Layer 0 position is changed to mixed (forming a “Layer 1” position), that player will not immediately know if the position is Layer 0 or Layer 1. He must thus rely on observations of other’s actions, which will rely on their perspectives. Since he does not know their perspectives, he must mentally construct a “Layer 1 hypothetical position,” which will be Layer 0 and in which he is either solid blue or solid red, and then compare the actions of others to what they would do in these hypothetical perspectives. When neither player is able to win via canceling, he can rely thus conclude that both hypothetical perspectives in which he is solid are false, and thus deduce that he is mixed. Let this strategy of winning by watching others fail to win via canceling be known as contradicted hypothetical canceling; in this situation it is “Layer 1 contradicted hypothetical canceling.” Note that Layer 1 contradicted hypothetical canceling depends on watching one action from each other player. The first mixed player who can do so is the Layer 1 observer.

In a similar manner, in the “Layer 2” position, which is one containing two mixed players and a solid player, one mixed player can deduce that he is mixed using Layer 2 contradicted hypothetical canceling – by watching the other fail to win via Layer 1 contradicted hypothetical canceling. That is, he can construct a Layer 1 hypothetical position, which will be Layer 1, and then calculate what actions would be taken if another player constructs his own Layer 1 hypothetical position, thus resulting in a Layer 2 hypothetical position which will be Layer 0. Note that the target of observation of the winning mixed player will be the Layer 1 observer; the other mixed player will therefore be the Layer 2 observer.

In the same manner, in the Layer 3 position (i.e.: all three players are mixed), one player (the Layer 3 observer) can win via Layer 3 contradicted hypothetical canceling – watching others fail to win using Layer 2 contradicted hypothetical canceling. Since all positions can be classified as Layer 0, 1, 2 or 3, all positions are thus winnable, and [b]the probability that the game goes on forever is 0[/b].

An example of how contradicted hypothetical works is shown by the following possible inner monologue on Brian’s second turn to guess in a Layer 3 situation:

[i]Say I was solid either color. Amy would know that, if she was solid the other color, CoScott would have known that, since neither of us were able to immediately figure out our stickers, he would have to be mixed. But that didn’t happen, so Amy would have known she could not have been solid. But she didn’t guess, so therefore I am mixed.[/i]

One very important result of the fact that negative information is the only type available is that solid players can never beat mixed players, as, in a Layer n position, a Layer m hypothetical position constructed by a mixed player will be a Layer n-m position, while one constructed by a solid player will only be Layer n-m+1, and thus the mixed players can always be one step closer to forming a hypothetical position that is Layer 0 than a solid player; players must be able to form hypothetical positions that are Layer 0 to win, as only at Layer 0 can canceling be done, and contradicted hypothetical canceling requires hypothetical positions where canceling can be done so that contradictions to those hypothetical positions can allow real deductions to be made.[/quote]
Yeah, I started to realize about how recursion could clean some of it up by the time I was midway through layer 3 or something in the write up so I thought what the heck, let's keep going...

Yea this one was a messy problem. I did a similar concept and put everything using Latex in tables to demonstrate patterns. I think this solution was my longest one, roughly 3-4 pages in Latex. Is there a shorter way of doing this other than using the "cases" brute force method?


If $ a,b,c > 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ Prove that
1)
\[ A \equal{} \frac92 \minus{} \frac {1}{1 \minus{} ab} \minus{} \frac {1}{1 \minus{} bc} \minus{} \frac {1}{1 \minus{} ac}\geq 0
\]
\hfill (Vasile Cirtoaje)
2)
\[ B \equal{} \sum\limits_{cyc}(a \minus{} b)(a \minus{} c)\left[\frac {a^2}{2(1 \minus{} ab)(1 \minus{} ac)}\right]\geq 0
\]
3) and finally my inequality
\[ A \minus{} \frac {B}{1 \plus{} \frac12\sum\limits_{cyc}(a \minus{} b)(a \minus{} c)}\geq 0
\]
\hfill (Bold Sanchir)

[hide]
Dear Can I will send u proof.[/hide]

First one should be:
For all real numbers $ a,b,c$ with $ a^2\plus{}b^2\plus{}c^2\equal{}1$ prove that:
$ \frac{1}{1\minus{}ab}\plus{}\frac{1}{1\minus{}bc}\plus{}\frac{1}{1\minus{}ca}\leq \frac{9}{2}$. :)

[quote="Inequalities Master"]First one should be:
For all real numbers $ a,b,c$ with $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ prove that:
$ \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} ca}\leq \frac {9}{2}$. :)[/quote]
I edited!  :lol:

[quote="Inequalities Master"]First one should be:
For all real numbers $ a,b,c$ with $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ prove that:
$ \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} ca}\leq \frac {9}{2}$. :)[/quote]
$ \Leftrightarrow 3\minus{}5q\plus{}7pr\minus{}9r^2\ge 0$ (with $ p\equal{}a\plus{}b\plus{}c,q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$)
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1\Rightarrow p^2\equal{}2q\plus{}1$
by Schur and AM-GM we have:
$ \frac{p(p^2\minus{}2)}{9}\le r\le \frac{(p^2\minus{}1)^2}{12p}$
The first case : $ 7p\minus{}18r\ge 0$
$ \Rightarrow f'(r)\equal{}7p\minus{}18r\ge 0$
$ \Rightarrow f(r)\ge f(\frac{p(p^2\minus{}2)}{9})\equal{}\frac{1}{18}(3\minus{}p^2)(2p^4\minus{}16p^2\plus{}33)\ge 0$
The second case: $ 7p\minus{}18r\le 0$
$ \Rightarrow f'(r)\equal{}7p\minus{}18r\le 0$
$ \Rightarrow f(r)\le f(\frac{(p^2\minus{}1)^2}{12p})\equal{}\frac{(3\minus{}p^2)(3p^6\minus{}31p^4\plus{}101p^2\minus{}1)}{48p^2}\ge 0$
we have done  :lol:

[quote="bunhiacovski"][quote="Inequalities Master"]First one should be:
For all real numbers $ a,b,c$ with $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ prove that:
$ \frac {1}{1 \minus{} ab} \plus{} \frac {1}{1 \minus{} bc} \plus{} \frac {1}{1 \minus{} ca}\leq \frac {9}{2}$. :)[/quote]
$ \Leftrightarrow 3 \minus{} 5q \plus{} 7pr \minus{} 9r^2\ge 0$ (with $ p \equal{} a \plus{} b \plus{} c,q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$)
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1\Rightarrow p^2 \equal{} 2q \plus{} 1$
by Schur and AM-GM we have:
$ \frac {p(p^2 \minus{} 2)}{9}\le r\le \frac {(p^2 \minus{} 1)^2}{12p}$
The first case : $ 7p \minus{} 18r\ge 0$
$ \Rightarrow f'(r) \equal{} 7p \minus{} 18r\ge 0$
$ \Rightarrow f(r)\ge f(\frac {p(p^2 \minus{} 2)}{9}) \equal{} \frac {1}{18}(3 \minus{} p^2)(2p^4 \minus{} 16p^2 \plus{} 33)\ge 0$
The second case: $ 7p \minus{} 18r\le 0$
$ \Rightarrow f'(r) \equal{} 7p \minus{} 18r\le 0$
$ \Rightarrow f(r)\le f(\frac {(p^2 \minus{} 1)^2}{12p}) \equal{} \frac {(3 \minus{} p^2)(3p^6 \minus{} 31p^4 \plus{} 101p^2 \minus{} 1)}{48p^2}\ge 0$
we have done  :lol:[/quote]
and how about 2) and 3) ?


calculus
integration
LaTeX
calculus computations

Can anybody help get though this one:
\[ \int_0^1 \frac{16x^3(1+x^{8k})}{(1+x^4)^{2k+2}}\,dx\]
where $ k$ is a positive integer.

Thank you in advance.

[color=green]Moderator edit: I replaced the image by $ \text{\LaTeX}$ for you this time. Try to learn it, it's easy! :)[/color]

are you sure that's your integral, because it is to easy :D  ( the denominator is a constant and so this is the integral of a polynomial...)

Grrr ... you're right ... that should be an x^4 in the denominator

The answer is:

$ I \equal{} \frac{4}{1 \plus{} 2k}$

but it's too much LaTeX to do. But a hint, use substitution:

$ (1 \plus{} x^4)^{2k \plus{} 2} \equal{} u$

this should help.

[quote="milin"]The answer is:

$ I \equal{} \frac {4}{1 \plus{} 2k}$

but it's too much LaTeX to do. But a hint, use substitution:

$ (1 \plus{} x^4)^{2k \plus{} 2} \equal{} u$

this should help.[/quote]Isn't the substitution $ u\equal{}1\plus{}x^4$ easier than $ (1 \plus{} x^4)^{2k \plus{} 2} \equal{} u$?  :maybe:


calculus
integration
ratio
real analysis
real analysis unsolved

$ \frac{\int_0^1 (1\minus{}x^n)^m\,dx}{\int_0^1 (1\minus{}x^n)^{m\minus{}1}\,dx}\equal{}\frac{mn}{mn\plus{}1}$. For the specific numbers asked, we get $ \frac{5051}{5050}$.

I got there by integrating by parts a hundred times*. Does anyone have a clever interpretation that avoids that for the ratio?
*Okay, there are $ \dots$ involved.

On second thought, here's a version that doesn't calculate the whole thing:
$ I_m\equal{}\int_0^1 (1\minus{}x^n)^m\,dx\equal{}\left[(1\minus{}x^n)^m\cdot x\right]_0^1\plus{}\int_0^1 mnx^{n\minus{}1}(1\minus{}x^n)^{m\minus{}1}\cdot x\,dx$
$ \equal{}mn\int_0^1 x^n\cdot(1\minus{}x^n)^{m\minus{}1}\,dx$
$ I_{m\minus{}1}\minus{}\frac1{mn}I_m\equal{}\int_0^1 (1\minus{}x^n)^{m\minus{}1}\,dx\minus{}\int_0^1 x^n\cdot(1\minus{}x^n)^{m\minus{}1}\,dx$
$ \equal{}\int_0^1 (1\minus{}x^n)^m\,dx\equal{}I_m$
Solving for the ratio, we get $ I_{m\minus{}1}\equal{}\frac{mn\plus{}1}{mn}I_m$.

i thnk by parts applied to denominator wud end the question........


[quote]thank you very much!!!
can you resolve the equation?????[/quote]

I suppose that the equation is: [b][color=orange]thank you very much!!! [/color][/b]

$ \text{much}!!!$
That would be the [url=http://en.wikipedia.org/wiki/Factorial#Superfactorials]superduperfactorial[/url] of $ \text{much}$?  :rotfl:


number theory
prime numbers
prime factorization

Can somebody please thouroughly explain problem 2.19 on page 38 in the Number Theory Book. The problem is: What are the 5 smallest prime numbers greater than 1000. Thanks

Probably not the best way...but...
Clearly they are odd. list about teh first 20 odd numbers.
Cross out all multiples of 3
Cross out all multiples of 5
etc.
You only have to go with primes up to 37 since sqrt(1000)<37

Then, whatever is left is prime.

You could also guess and check...but i dont think that works too well either

Um, the questtion is

"greater than 1000" not "100".

[quote="math154"]We all know that you have to memorize the $ 25$ primes less than $ 100$, and then memorize to $ 200$ for MATHCOUNTS. Anyway, they are $ 101,103,107,109,113$.[/quote]

You do?

No. :P

Sorry for misreading the question :blush: ...anyway, it's $ 1009,1013,1019,1021,1031$, and there's not really a good way. Use your divisibility rules (although this won't work for a few of them) and the primality test mentioned by abacadaea. Also what may help is that $ 1001\equal{}7\cdot11\cdot13$, $ 17|1003$, $ 19|1007$, which makes it easy to test the primes under $ 20$; only $ 23,29,31$ remain (since $ 37^2\equal{}1369$ and we figure that it is much larger than the fifth prime above $ 1000$).

The primes less than $ 100$ are good enough for this problem; trial division works fine here as long as you know the primes up to $ 34$.  First, it's pretty easy to ignore numbers that end in $ 0, 2, 4, 5, 6, 8$ and numbers whose digits sum to a multiple of $ 3$; that gets rid of divisibility by $ 2, 3, 5$, so we only have to check for divisibility for $ 7, 11, 13, 17, 19, 23, 29, 31$.  In order for the following method to work quickly it is good to learn the $ 2$-digit multiples of each of these primes.  So, a sample:

$ 1001 \equal{} 7 \cdot 11 \cdot 13$; you know this if you memorized the prime factorization of $ 2002$, or you can write $ 1001 \equal{} 700 \plus{} 280 \plus{} 21 \equal{} 1100 \minus{} 99$.

$ 1003 \equal{} 17 \cdot 59$; you know this if you memorized the prime factorization of $ 2006$, or you can write $ 1003 \plus{} 17 \equal{} 1020 \equal{} 20 \cdot 51 \equal{} 60 \cdot 17$.

$ 1007$ is not divisible by $ 2, 3, 5, 7, 11, 13, 17$ by the above arguments, but $ 1007 \equal{} 950 \plus{} 57 \equal{} 19 \cdot 53$.

$ 1009$ is not divisible by $ 2, 3, 5, 7, 11, 13, 17, 19$ by the above arguments, and $ 1009 \equal{} 920 \plus{} 89$ so $ 23 \not | 1009$.  In addition, $ 1009 \minus{} 29 \equal{} 980$, so $ 29 \not | 1009$.  Finally, $ 1009 \equal{} 930 \plus{} 79$, so $ 31 \not | 1009$.

The rest are similar.


a, b, c are positive, real numbers satisfying [b]ab+bc+ac = 3[/b]
Prove that
[b]a^3 + b^3 + c^3 + 6abc >= 9[/b]

This one is from Polish Mathematical Olympiad which is still in progress - locked till 5th of December.

(Sorry [b]Together[/b] I had to delete your post)


trigonometry
calculus
integration
algebra
polynomial
algebra unsolved

can anyone tell me how to prove this:

$\displaystyle\sum_{k=1}^{6} \frac{\sin\left(\frac{k\pi}{7}\right)}{\sin\left(\left(\frac{3k+7}{21}\right)\pi \right)}=0$

Do you think Pre-Olympiad is appropriate place for such problem? ;)

i dont know, where do u think it should go?

[quote="Myth"]I think moderators should move it to Advanced Section->Algebra->Unsolved Problems[/quote]Done :) Still... this looks pretty ugly to me.

[hide]
\[{ \frac{\sin\theta}{\sin (\theta+\frac{\pi}{3})}=\frac{\sin\theta}{\frac{1}{2}\sin\theta+\frac{\sqrt{3}}{2}\cos\theta}}=\frac{2}{1+\sqrt{3}\cot\theta}\]
Pairing the $ \theta$ term with the $ \pi-\theta$ term,
\[ \frac{2}{1+\sqrt{3}\cot\theta}+\frac{2}{1-\sqrt{3}\cot\theta}=\frac{4}{1-3\cot^{2}\theta}\]
Expressing each angle in first quadrant, our sum is equal to
\[ 4\sum_{k=1}^{3}\frac{1}{1-3\cot^{2}\frac{\pi k}{7}}\]
Now for $ \alpha$ an integral multiple of $ \frac{\pi}{7}$ and nonzero, we have $ \sin 7\alpha =0$ so that $ (\cot\alpha+i)^{7}$ is real. So we have
\[ 0=-1+21\cot^{2}\alpha-35\cot^{4}\alpha+7\cot^{6}\alpha\]
Then the polynomial with roots $ \cot^{2}\frac{\pi}{7},\ \cot^{2}\frac{2\pi}{7},\ \cot^{2}\frac{3\pi}{7}$ is
\[ 7t^{3}-35t^{2}+21t-1\]
Now substitute $ u=\frac{1}{1-3t}$ so $ ut=\frac{u-1}{3}$. Multiplying by $ u^{3}$ gives
\[ 0=7(u-1)^{3}-35\cdot 3u(u-1)^{2}+21\cdot 9u^{2}(u-1)-27u^{3}\]
with $ u^{3}$ coefficient $ 7-35\cdot 3+21\cdot 9-27\neq 0$ (in fact, not a multiple of 3) and $ u^{2}$ coefficient
\[ -7\cdot 3+35\cdot 3\cdot 2-21\cdot 9=0\]
So the sum of the $ u$'s is zero, as desired.
[/hide]


i answered -3\sqrt[3]{3} but got it wrong.  Upon review, FTW said the right answer was $ -3%Error. "SQRT" is a bad command.
[3]{3}$.

is this a math question or a LaTeX question?

if the AoPS staff added this question to FTW, should they add the necessary LaTeX instructions to the FTW directions?
It's very nice that they added \sqrt{x} and \pi, but it seem we need more.  On no other FTW question was the $ ...$ required for a correct answer.


I'm looking for  high - school level problems with solutions in circles, triangles and other polygons.... Any good resources for this??? i'm going through the resource links in the branch...but then thought i might post it here as well to get a quicker response...also if some one has such solved questions as a e-book or document kindly mail it to me fizsle@yahoo.com

You could also check the AoPS Wiki, which is still sparse but will eventually include the resources section.  Look at the categories of geometry problems to find what's already up there and tagged.


[hide="Possible hint"] We have the inequality $\cos \frac{1}{n}> 1-\frac{1}{2n^{2}}$. [/hide]

I will use $\cos x \geq \sqrt{1-x^{2}}$.


function
probability
logarithms
complex numbers

Hi all! I want to know what is Riemann talking about in his hypothesis, and what the Zeta Function have to do with prime numbers... thanks!!!!!!.

The Zeta Function,

$\zeta(s) = \sum_{k=1}^{\infty} \frac{1}{k^s}$,

can be rewritten as

$\zeta(s) = \prod_{p \in \mathbb{P}}^{\infty} \left( \frac{p^s}{p^s - 1} \right)$,

And moreover it can be extended to complex values of $s$.  The Riemann Hypothesis is a hypothesis about the nature of the complex values of $s$ for which $\zeta(s) = 0$ - specifically, that for any non-trivial zero of the Zeta Function, it has a real part $\frac{1}{2}$.  (The "trivial zeroes" occur at $s = -2, -4, -6, ...$ when $\zeta(s)$ is extended sufficiently.)

t0rajir0u:

Thanks for the "definition". Now, can you explain it with your words??? Because that theorical definition I have already seen it; however, I still don't understand it... I want to know it like in a normal conversation, without technisism, if possible.

Well... from Wikipedia:

[quote]The zeros of $\zeta(s)$ are important because certain path integrals involving the function $ln( \frac{1}{\zeta(s)})$ can be used to approximate the prime counting function $\pi(x)$. These path integrals are computed with the residue theorem and hence knowledge of the integrand's singularities is required.[/quote]

There is a theorem called the Prime Number Theorem that says, roughly, that the probability of finding a prime close to an integer $N$ is about $\frac{1}{\ln N}$.  That's about as accurate as we can get to deciphering the distribution of primes.  But if we could prove the Riemann Hypothesis, we could make this bound a lot stronger, and have much more accurate estimates for the distribution of primes.

The Riemann hypothesis deals more with error terms in the prime number theorem, and several other similar functions (the Mertens function, the Liouville function, etc).  If RH holds, then not only do certain quantities behave asymptotically the way we want them to, but we can also make the error terms very small.


18 small, 8 medium, 2 large, so 28 total

[quote="Mewto55555"]A quick thing to notice here is that there is no triangle that crosses the border of the two triangles. Thus, you can simply find the number of triangles in one "big" triangle, and multiply it by 2.

This is simply 1+4+9=14, so 2x14=28.[/quote]

That is not right, and I almost made that same mistake. There are actually 1 + 3 + 9 triangles in each of the 2 big triangles. 2 of the medium triangles overlap


find the center and radius of the circumcircle of triangle ABC with coordinates $ ( \minus{} x,0), (x,0), (x_1,y_1)$

[hide="Method"]
The circumcenter is the intersection point of the perpendicular bisectors, the equations of which are easy to find.

The circumradius is the distance from the circumcenter to any of the vertices.

Incidentally, if you don't need the circumcenter, another way to find the radius would be to find the length of each side and the area using shoelace, then do $ [ABC]\equal{}\frac{abc}{4R}$.
[/hide]


linear algebra
matrix
algebra
polynomial
linear algebra theorems

Given a matrix say:  $ \begin{pmatrix}
1 & 1 \\
2 & 4 \\
\end{pmatrix}$ what is the procedure for converting it to jordan form?

If a matrix is diagonalizable, then the diagonal form is the Jordan form. This one is diagonalizable. Start by finding its eigenvalues. (Note that in this case, the eigenvalues are not rational.)

ok bad example, what about:$ \begin{pmatrix}
1 & 2 \\
\minus{}2 & 4 \\
\end{pmatrix}$ with only eigenvalue $ \minus{}3$

You still don't have the right example. The eigenvalues of that are $ \frac{5\pm i\sqrt{7}}2.$

Let's try one more time. Let $ A\equal{}\begin{bmatrix}1&2\\\minus{}2&5\end{bmatrix}.$ Characteristic polynomial $ \lambda^2\minus{}6\lambda\plus{}9,$ eigenvalues $ 3$ and $ 3.$

First, consider the homogeneous equation $ (A\minus{}3I)x\equal{}0.$ We get only a one-dimensional space, spanned by $ x\equal{}\begin{bmatrix}1\\1\end{bmatrix}.$ Since we didn't get a second independent eigenvector, $ A$ is not diagonalizable.

Now use this $ x$ as the right hand side of a new non-homogeneous equation: $ (A\minus{}3I)y\equal{}x.$ As an augmented matrix:

$ \begin{bmatrix}\minus{}2&2&1\\\minus{}2&2&1\end{bmatrix}.$

We need only one particular solution of this; $ \begin{bmatrix}0\\\frac12\end{bmatrix}$ will work.

Let $ P\equal{}\begin{bmatrix}1&0\\1&\frac12\end{bmatrix}.$ We compute that $ P^{\minus{}1}\equal{}\begin{bmatrix}1&0\\\minus{}2&2\end{bmatrix}.$

By direct calculation $ P^{\minus{}1}AP\equal{}\begin{bmatrix}3&1\\0&3\end{bmatrix}.$

That is the Jordan canonical form for $ A.$

Thanks Dr Merryfield, I appreciate the help.

So here is my understanding of things:
1. Compute Characteristic polynomial, if the number of distinct eigenvalues of our matrix is equal to the dimension of the VS it is acting on, then we have a diagonal Jordan form.
2. If not then we may still have a diagonal Jordan form, if we have dim(V) distinct eigenvectors. So we consider the Augmented matrix equations $ (A\minus{}\lambda I)x\equal{}0$ and preform ERO's to find a basis for the solution space. These are our independent eigenvectors.
3. If we did not get dim(v) eigenvectors from all our homogenous equations then $ A$ is not diagonalizable.
4. now I think we are finding a basis for our generalized eigenspaces? so we are looking for a basis of solution space of $ (A\minus{}\lambda_1 I)^{dim(v)}y\equal{}x$ where $ x$ is our eigenvector with eigenvalue $ \lambda_1$
5. now we create matrix $ P$ with column that are our basis for all the general eigenspaces. We also compute $ P^{\minus{}1}$
6. now compute $ P^{\minus{}1}AP$ which is our jordan form for $ A$

Questions:
is the summery correct?
How do we know our $ P$ will be invertible?
Why did we chose a basis of generalized eigenvectors to create $ P$?
Is there a simpler way to find just any matrix that is similar to $ A$ and upper triangular?

Yes, that summary looks ok.

[quote]How do we know our $ P$ will be invertible?[/quote]

Because its columns make up a basis for V and so they are linearly independent.

[quote]Why did we chose a basis of generalized eigenvectors to create $ P$?[/quote]

Because the (simple) eigenvectors were not enough to create a basis, and the generalized eigenvectors scheme is probably the simplest method to arrive at the Jordan Canonical Form.

[quote]Is there a simpler way to find just any matrix that is similar to $ A$ and upper triangular?[/quote]

Schur's lemma describes how to construct such a matrix.

I meant to ask how do we know the matrix $ P$ will be the one that gets us to the Jordan canonical form, and not to some other similar matrix that is just upper triangular.


limit
trigonometry
calculus
calculus computations

hi all
what is limit of[( (e^x) -1)/sinx]  (as x approaches to 0)  without L'HOPITAL[/img]

You seemed to make double posting, :lol: [color=green]The other one is gone now. - Mod[/color]

anyway, you can use formulae ${ \boxed{\boxed{\lim_{x\rightarrow 0} \frac {e^x - 1}{x} = 1,\ \lim_{x\rightarrow 0} \frac {\sin x}{x} = 1}}}$


What is the sum of the two perfect square integers that are closest to 2005?

Knowing square numbers up to ~50 is helpful ;)

We find the two closest square numbers are $ 44^2\equal{}1936$ and $ 45^2\equal{}2025$. Summing, we get $ \boxed{3931}$

By the way, to add perfect squares quickly, keep in mind that $ a^2 \plus{} b^2 \equal{} (a \plus{} b)^2 \minus{} 2ab$, which may be faster than just computing $ 45^2$ and $ 44^2$.

are you sure izzy?
I think $ 1936\plus{}2025$ is easier than $ (89)^2\minus{}2\times44\times45\equal{}7921\minus{}3960$ ;)

$ 89^2 \equal{} (90 \minus{} 1)^2$, so that could help. $ 44 \times 45 \equal{} (45 \minus{} 1)(45)$, which can help as well.

I think both methods could work in this problem...


Could anyone help me with these problems?

1. A CD player has 5 trays, numberd 1 through 5. Each tray holds one CD. In how many ways can the CD player be loaded from a collection of 100 different discs if exactly four discs are loaded into the player?

Here's my thinking: We can choose 4 CDs in $100\choose 4$ ways, and each subset of 4 CDs can be permuted in $4!$ ways. Now for every one of those arrangements, we have 5 choices for the empty tray (1-5). So would the final answer be ${100\choose 4}\cdot{4!}\cdot{5}$?

2. How many different ways can 9 different algebra books, 4 different geometry books, and 7 different calculus books be but on a shelf if the left-most book must be an algebra book?

Here's my thinking: We have 9 options for the left-most algebra book. For the remaining 19 books, there are $19!$ ways to permute them. So the answer is $9\cdot{19!}$. Is this correct?

3. A group of 15 boys and 15 girls are to be lined up. How many different lines are possible if no two boys stand next to each other in line?

Here's my thinking: The only way we can have no two boys next to each other is if we have the boys and girls alternate (what's the proof behind this-Pigeonhole?). Imagine the 15 girls standing in a line with blank's b/w every two consecutive girls. There are $15!$ ways to line up the girls, and $15!$ ways to line up the boys in the blanks. But either a boy or a girl could lead, so we have 2 options. So a total of $2\cdot{(15!)}^2$ possible lines?

I'm simply confused about whether I should multiply by the factor of 2. While it would make sense as to which gender starts the line, but must I also consider that 15 different people of each gender could lead, or have I already counted that?

Any help is vastly appreciated. Thanks.

They look right to me. With the third problem, you've already arranged the girls and boys, so you already know which people will be the first of their gender in line...right? So I think you've already counted the possibilities for the leaders.

No... For the third problem, let G represent a position for a girl, and _ as a position for a boy: so we can either have:

1. G_G_G_G_G_G_G_G_G_G_G_G_G_G_G_ (This is where a  girl leads, but this already takes into account the 15! ways the girls can stand in their positions).

2. _G_G_G_G_G_G_G_G_G_G_G_G_G_G_G (This is where a boy leads, note that the ${(15!)}^2$ only counts the 1st arrangement where the girls and boys are in their respective positions. The second arrangement dictates that we must add another ${(15!)}^2$ where a boy leads (imagine a girl can lead OR a boy can lead). Please correct me if I'm wrong, and thanks for the checkover LynnelleYe.

Yes, I believe there are two cases, 1. when a girl leads. 2. when a boy leads

It looks correct to me. ( so you should multiply by 2 )

Imagine there is a line of 15 boys and a line of 15 girls, each with $15!$ ways to arrange them.  Then there is $(15!)^2$ ways to arrange the the two lines [u]seperatly[/u]. Now the single line can be constructed by taking the first person from [u]either[/u] the boys' line or the girls' line, and alternating every other person.  There are two ways to do this and $(15!)^2$ ways to arrange the people, so there are a total of $2(15!)^2 = 3420024505448398848000000$ ways to do this.


inequalities
geometry
inequalities unsolved

For the 5-th time, please check in the open questions area, I proved there a stronger statement of Vasc.

Dear harazi,could you post it?I don't find it.Thanks very much.

I think I went on the wrong way again!

By AM-GM, we have: $a^2+b^2+c^2+d^2 \geq 2ab+2dc \geq 4 \sqrt{abcd}$
Hence $abcd+4\sqrt{abcd} \leq 5$, which gives: $\sqrt{abcd} \leq 1$.
So we have: $a^2+b^2+c^2+d^2 \geq 4$.

But by Cauchy, we know $(a+b+c+d)^2 \leq 4\cdot (a^2+b^2+c^2+d^2)$. And I don't know how to do from here...

then, we have
$ (a + b + c + d)^2 \leq 4 \cdot 4 $
$ a + b + c + d \leq 4 $

[quote="asjsd"]
$ (a + b + c + d)^2 \leq 4 \cdot 4 $
[/quote]
Why? I think the correct logic should be $(a+b+c+d)^2 \leq 4 (a^2+b^2+c^2+d^2) \leq 4 \cdot 4$. But we have $(a^2+b^2+c^2+d^2) \geq 4$.

Can anyone tell me how to prove this inequality?

Guys, I just can't understand you. I said pretty clear that you can find the solution in open questions area. I can't believe you cannot find it there, since there are at most 35 posts there.  :huh:  :huh:


function
real analysis
real analysis solved

Preparing my algebra exam, I found in a book the following super-problem:
   Let A the ring of all continuous functions from [0,1] to the reals. Find all morphisms from A to the reals.
   A truly beautiful problem.

More general we  consider A, a ring  of functions f:S-->R such that :
- the constant functions are in A
- if  f\inA and f(x)<>0 for all x\inS, then 1/f is in A
- there is g\inA such that for all real r, the equation g(x)=r has a finite set of solutions,
Manifestly M(f)=0, and M(f) = f(x0) for all f \in A are morphismes M:A-->R.
Let M be a non trivial morphisme (M<>0) . For all a\inR we note K_a the constant function : K_a(x) = a  for all x\inS
Put T(a) = M(K_a) . Manifestly T is a non-trivial morphisme T:R --> R , and it is well known that T(a) = a , so M(K_a)=a for a\inR
Let f be such that M(f)<>0 ===> M(f)=M(f)*M(K_1) ===> M(K_1)=1
Let f such that f(x)<>0 for all x\inS . So M(f )* M(1/f) = M(f*1/f) = M(K_1)=1 ===> M(f)<>0
So M(f)=0 ===> there is x\inS such that f(x)=0 . M(f)=M(K_M(f)) ===> M(f - K_M(f)) = 0 ===> there is x\inS s.t. f(x)=M(f)
Note Z(f) = {x / (x\inS)&(f(x)=M(f)}.
M((f - K_M(f))2  +  ((g - K_M(g))2) = (M(f - K_M(f)))2 + M(g - K_M(g)))2 = 0 ===> there is x\inS s.t. (f(x) - M(f))2 +  (g(x) - M(g))2 =0 ===> Z(f) \cap Z(g)<>\emptyset

Finally, because  there is g\inA with Z(g)  finite,  we have that V =  \cap (f\inA){Z(f)} is not empty.
Let x0 be s.t x0\inV , we obtain M(f) = f(x0)  for all f\in A .... (q.e.d.)

See you ..


function
induction
algebra
polynomial
limit
real analysis
real analysis unsolved

Define a function $ f$ by
\[ f(x) = \{ \begin{array}{cc} 0 & \text{if} \, x \leq 0 \\ e^{-1/x^2} & \text{if} \, x > 0 \\ \end{array}
\]
Prove rigorously (i.e. $ \epsilon - \delta$) that $ f$ is infinitely differentiable on $ \mathbb{R}$ and that $ f^{(n)}(0) = 0$ for all $ n \geq 1$.

Remark: This shows that infinitely differentiable functions are not necessarily analytic on $ \mathbb{R}$ (unlike in $ \mathbb{C}$).

[hide="hint"]Show by induction that for $ x > 0$, $ f^{(n)}(x)$ is of the form $ e^{ \minus{} \frac {1}{x^2}}\frac {p_n(x)}{q_n(x)}$ where $ p_n(x)$ is a polynomial of degree $ 2n \minus{} 2$ and $ q_n(x)$ is a polynomial of degree $ 3n$.  Conclude therefore that $ \lim_{x\to 0}f^{(n)}(x)\equal{}0$ for all $ n$.[/hide]

[quote="Phelpedo"]Remark: This shows that infinitely differentiable functions are not necessarily analytic on $ \mathbb{R}$ (unlike in $ \mathbb{C}$).[/quote]
Do you mean the following?
If the function is analytic, then because it is bounded, it should be a constant. So the function is not analytic on $ \mathbb{R}$.

No, that's wrong. An analytic function can be bounded on $ \mathbb{R}$; consider $ \frac1{1\plus{}x^2}.$

To see that this isn't analytic, go straight to the definition. What function does the power series based at zero converge to?


In attempting to fly from city A to city B, a distance of 300 miles, a pilot inadvertently took a course that was 12 degrees.

What new average speed should the pilot keep so that the total time of the trip is 100 minutes?

NOTE: The old average speed was 225 miles per hour.

When does the pilot notice that he took a wrong course? (After 300 miles?)


calculus
integration
derivative
symmetry
calculus computations

To save a little headache, I would introduce $ F(x,y)\equal{}\int_0^x \int_0^y e^{(uv)^2}dudv$. The partial derivatives of $ F$ aren't that hard to find, since you are in fact differentiating a single integral. The compute $ p',p''$ using the multivariable chain rule: $ p(t)\equal{}F(t,t)$.

[b]Here's a solution.[/b]

By symmetry, $ p(x)$ is twice the integral over the triangle bounded by $ u \equal{} v$, $ v \equal{} 0$ and $ u \equal{} x$, so $ p(x)\equal{}\int_{0}^{x}\int_{0}^{u}e^{u^2v^2}dvdu$. It follows that $ {p}'(x)\equal{}2\int_{0}^{x}e^{x^2v^2}dv$. Set $ \phi \equal{}xv$ so that $ dv\equal{}d\phi /x$ and $ {p}'(x)\equal{}(2/x)\int_{0}^{x^2}e^{\phi ^{2}}d\phi$. And, it is easy to show that $ {p}''(x)\equal{}\frac{\minus{}2}{x^2}\int_{0}^{x^2}e^{\phi^2} d\phi\plus{}2x\frac{2}{x}e^{x^4}$. Therefore, $ x{p}''(x)\plus{}{p}'(x)\equal{}4xe^{x^4}$ and hence $ 2010{p}''(2010)\plus{}{p}'(2010)\equal{}8040e^{2010^4}$. $ \square$

PS. This problem appeared in Viginia Tech Reg'l Math Competition.


1)Let S be a finite set of positive integers such that none of them
has a prime factor greater than three. Show that the sum of the
reciprocals of the elements in S is smaller than three.

2)Let x1; x2; ....... ; x5 be nonnegative real numbers such that x1 +
x2 + x3 + x4 + x5 = 5. Determine the maximum value of
x1x2 + x2x3 + x3x4 + x4x5.

3) Find all triples (x; y; z) in integers such that $x^2+y^2+z^2 = 2^{2004}.$

4) Suppose that n is a positive integer. Determine all the possible
values of the first digit after the decimal point in the decimal
expression of the number $\sqrt{n^3 + 2n^2 + n}.$

I didn't understand problem 1): if $S= \{3,3,3,3,3...\}$with for example 100 elements all=3. The reciprocal of all elements is $\frac{1}{3}$, and the sum of the reciprocal is $\frac{100}{3}>3$

A set can't contain the same element twice.

At the risk of giving it away, the point of problem 1 is to consider all positive integers of the form $2^m 3^n$, where $m$, $n \ge 0$.

[quote="hydro"]Thanks nsato!
I try number 1...

[hide]
Suppose that $S$ contains all elements that it can contains: $S=\{1,2,3,4,6... \}$ until a certain element $n$. Then we can create a new set $S_1$ that contains all powers of 2 that also $S$ contains, and for each element of $S$ that is not a power of 2, contains a new power of 2 that $S$ didn't contains. Clearly, the sum of the reciprocal of the elements of $S$ is $<$ than the one of $S_1$, because all elements of $S$ are $\geq$ than the elements of $S_1$. Now, take $\sum_{i=0}^n\frac{1}{q^i}$. This sum is maximum when $n \rightarrow +\infty$, but we know that
$\lim_{n \rightarrow + \infty}\sum_{i=0}^n\frac{1}{q^i}=\frac{q}{q-1}$, that for $q=2$ is equal to $2<3$, and we have the proof. [/hide][/quote]
I too tried using 2<3 as a bounding trick, but I got stuck. Your solution confuses me when it says "for each element of $S$ that is not a power of 2, contains a new power of 2 that $S$ didn't contains" I assume this new power of two would be different from the ones you already have, so it would have to be bigger, so say replacing 6 with 64. But then that would make the reciprocals smaller.  :?:  Or I am just not understanding, I do that sometimes.

[quote="PenguinIntegral"]1)Let S be a finite set of positive integers such that none of them
has a prime factor greater than three. Show that the sum of the
reciprocals of the elements in S is smaller than three.[/quote][hide]Let $U$ be the set of all integers that factor into the form \[ 2^m3^n,\qquad m,n\in\mathbb{Z}\qquad m,n\ge 0. \] Thus, \[ U=\{1, 2, 3, 4, 6, 8, 9, 12, 16, 18,\ldots\}. \] By definition, \[ S\subset U. \] Hence, the sum of the reciprocals of the elements in $S$ must be smaller than the sum of the reciprocals of the elements in $U$. But the sum of the reciprocals of the elements in $U$ is \[ \left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots\right)\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots\right)=\left(2\right)\left(\frac{3}{2}\right)=3. \] Therefore, the sum of the reciprocals of the elements in $S$ is smaller than three.[/hide]

[hide="Hint for #3"]Look at it mod 16, and keep reducing. :) [/hide]


is there exist $ a,b,c\in\mathbb{Z} , a,b>1$ such that for infinitely many natural $ n$:
$ a^n|b^n\plus{}c$ ?

oh...I miss the condition $ c\not\equal{}0$...
no one?

Mhmm can we say let $ a\equal{}2$, and $ b\equal{}2$,$ 2^n\equal{}c$
so we have $ 2^n | 2^{n\plus{}1}$ ?

No.  $ c$ stays constant as $ n$ varies.

[quote="ali666"]is there exist $ a,b,c\in\mathbb{Z} , a,b > 1$ such that for infinitely many natural $ n$:
$ a^n|b^n \plus{} c$ ?[/quote]
Assume that $ a \equal{} p$ is a prime. (If $ a \equal{} p_1 p_2 ... p_r$ satisfies the equation, so does $ a \equal{} p_1$).
Also see that $ (p,b) \equal{} 1$ Otherwise $ p^n \mid c$ which is obviously only possible for finitly many (natural) $ n$. (In the same manner see $ (p,c) \equal{} 1$)
Let $ b \equiv r \bmod p \iff b \equal{} pk \plus{} r$

So $ p^n \mid (pk \plus{} r)^n \plus{} c$. Obviuosly for this to happen we have that $ r \equiv \minus{} c \bmod p$ Since $ b^n$ is periodic $ \bmod p$, we kan say that $ n \equal{} l*m \plus{} q$ Set $ f \equal{} b^q$ and $ g \equal{} b^m$. So we have to find $ f$ and $ g$ such that $ p^n \mid f \cdot g^n \plus{} c$.

We easily see that $ g \equiv 1 \bmod p$ and $ f \equiv \minus{} c \bmod p$. Thus we have $ p^n \mid (pk \minus{} c)(pj \plus{} 1)^n \plus{} c$ (Sorry for using $ k$ again..)

We now "just" have too prove that this can't happen. Sorry for not finishing :P

i think this is hard.

I have read somewhere that if there exist infinitely many n such that $ \frac {b^n\minus{}1}{a^n\minus{}1}$ is integer then $ b\equal{}a^p$ for some $ p \in \mathbb{N}$ but i can't remember where  :D . It is the same kind of problem and i am very curious to read a solution

See my post called A masterpiece (it is one of those posts that repeat regularly...) for the solutions of this problem. There are also some other threads on variations of this problem, but I cannot remember their names. Sorry, I realize that the problem you asked is much more difficult than the one I thought. Unfortunately, if you want a solution for the problem you posted, you really have to be an expert of number theory: the result is very recent and very difficult. See the webpage of Yann Bugeaud for the article in question.

Really thank you for your help, there is indeed a good paper from Yann Bugeaud dealing with this kind of problem , i'll try to understang few things but it seems very (very) hard and complicated... :o


For $a,b,c$ real numbers prove that

$ 3(a^2-a+1)(b^2-b+1)(c^2-c+1) \geq (abc)^2 + abc +1 $

P.S.This inequality was already posted in this forum (section: Solved Problems)

as topic "More inequaly" but in hxtung prove there is a passage I think not correct

(can you please check it)  where he says that

$ \frac{1}{9}(a^2+a+1)^3 \geq a^6 +a^3+1 $  for $a\geq 1 $

so I have reposted it as I consider it very nice. Can anyone tell me the author,

please?

if he wrote what you cited here, then it is sure incorrect since the reverse inequality holds...

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13467&highlight=]An inequality in Kiran Kedlaya's notes[/url].


I want to create a document which is a combination of 5 other documents. I've tried to use \input{filename} but then only the first \input command works and the rest are ignored. If I use \include{filename} or \includeonly{filename1, filename2 ...} I get an error message and no output PDF file. Can anyone help me plz??

In the files that you include there shouldn't be any \begin{document} or \end{document} otherwise LaTeX will stop compiling at the first \end{document} it comes to.

Ok so do I include all the preamble stuff in the 1 document and get rid of them in the other 5?

That's exactly it. There can only be one preamble for each final document - you only want the text imported from the other files.

Ok thanks. Appears to be working now. Final question, how do you center pictures included in the document from the \includegraphics command?

Put \begin{center} ... \end{center} round the \includegraphics command:[code]\begin{center}
\includegraphics ....
\end{center}[/code]


inequalities
algebra
polynomial
geometry
geometric transformation
reflection
function

Let $a,$ $b$ and $c$ are positive real numbers, such that $abc=1.$
Prove that $\sum_{cyc}\frac{1}{a^2-2a+3}\leq\frac{3}{2}.$

By simplify and clearing denominators, one can write the equivalent inequality in the following form:

[(a-1)(b-1)(c-1)]^2 + f(x,y,z) >= 0.

With f(x,y,z) = A(x^2+y^2+z^2) - B(x+y+z) + C for xyz = 1 and
x,y,z are positive reals and A, B, and C are positive constants.

The next step is to do a 3-stage check.

stage 1: prove f(x,y,z) >= f(x,(yz)^(1/2), (yz)^(1/2)) for yz >= 1
stage 2: prove f(x,y,y) >= 0 for y >= 1 and xy^2 = 1 and this will generally reduces to showing:

Q(y) = (y-1)^2.R(y) where R(y) is an integer coefficient polynomial of degree 4.
stage 3: show that R(y) >= 0 and this can be a bit challenging but doable.
The worse can happen is we have to use cubic polynomial's roots formulae to find a critical points with three of them at most.

____________________________________________________________

[quote="Minh Can"]By simplify and clearing denominators, one can write the equivalent inequality in the following form:

[(a-1)(b-1)(c-1)]^2 + f(x,y,z) >= 0.

With f(x,y,z) = A(x^2+y^2+z^2) - B(x+y+z) + C for xyz = 1 and
x,y,z are positive reals and A, B, and C are positive constants.
[/quote]
No! You are wrong. :D:)
$f(a,b,c) = A(a^2+b^2+c^2) - B(a+b+c) + C+ D(ab+ac+bc).$

[quote="Jan"]The condition $abc=1$ is completely unnecessary I think:

$\sum_{cyc}{\frac{1}{a^2-2a+3}} = \sum_{cyc}{\frac{1}{(a-1)^2+2}} \leq \frac{3}{2}$

Using the fact that $x^2 \geq 0$. Done![/quote]
Thank you Jan! :)
I meant to $\sum_{cyc}{\frac{1}{a^2-2a+3}} \leq \frac{3}{2}\Leftrightarrow7(a-1)^2(b-1)^2(c-1)^2+4\sum_{cyc}(a+b-2)^2\geq0.$ :D

The first part is right. It is only an attack plan and not a completely worked out detailed solution yet. But the direction is correct though. One needs to take out the factor [(a-1)(b-1)(c-1)]^2. Once this is taken out the rest shouldn't be that hard. But the hardest it can get is to use the tactic you used the first place for the other problem.

More generally, anything of the form $\forall a,b,c: \sum_{cyc}f(a) \ge N$ can be solved in Jan's way.

I absolutely have doubt about your claim that " [i]any thing [/i]of the form

cyclic sum f(a) >  N  can be proved by certain tactic ". It might reflect

a personal opinion but not a mathematical one.

It not an opinion, but a theorem.

Even for $n$ variables: the statement $\sum_{cyc}^n f(a) \le N$ can only be true iff we have that $f(x)\le \frac{N}{n}$. These are simply equivalent statements.

The proof is simple: assume there was an $y$ for which $f(y)>\frac{N}{n}$, then putting all variables equal to $y$ would make $\sum_{cyc}^n f(a) > N$.

Of course, all this provided that we are not given restrictions on a,b,c. I can even say that this method [i]only[/i] works iff no conditions are needed. The fact that this method works shows that the conditions are unnecessary. (I cannot give this last statement in a rigorous theorem since it is hard to define 'unnecessary' rigorously in this context.)

[quote="mateivld"]$\frac{1}{a^2-2a+3}\leq\frac{1}{2}$ because $(a-1)^2\geq 0$ :lol:[/quote]
You are right. My miss.
Try it: $a,$ $b$ and $c$ are positive real numbers, such that $abc=1.$
Prove that $\sum_{cyc}\frac{1}{11a^2-5a+3}\geq\frac{1}{3}.$

Hello,
[quote]
$a,b$ and $c$ are positive real numbers, such that $abc=1.$
Prove that $\sum_{cyc} \frac{1}{11a^2-5a+3} \geq \frac{1}{3}$[/quote]
I think it can be proven if we let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ and use Jensen's inequality for the function $f(t)=\frac{1}{t}$. :?

[quote="kar"]Hello,
... and use Jensen's inequality for the function $f(t)=\frac{1}{t}$. :?[/quote]
Try. I meant to something another.


νδο
$ 6|abc(a \plus{} b \plus{} c)(a \plus{} b)(a \plus{} c)(b \plus{} c)$

ps:επειδη εχω καιρο να μπω στο forum δν ξερω αν υπαρχει ηδη αυτη η ασκηση,οποτε αν  υπαρχει ας σβησουν το τοπικ οι moderatorts

Arketa efkolo alla oraio. Den prepei na exei ksanampei sto forum (toulaxiston ego den to exo ksanadei)

Loipon, esto $ A \equal{} abc(a \plus{} b \plus{} c)(b \plus{} c)(c \plus{} a)(b \plus{} a)$, an enas apo tous $ a,b,c$ einai pollaplasio tou 3 tote o 3 diairei ton $ abc$ ara kai ton $ A$

An oloi afinoun ipolipo 2 me ton 3 tote $ a \plus{} b \plus{} c \equiv 6 \equiv 0 (mod3)$
An oloi afinoun ipolipo 1 me ton 3 tote $ a \plus{} b \plus{} c \equiv 3 \equiv 0 (mod3)$

An 2 apo tous $ a,b,c$ afinoun diaforetika mi midenika ipolipa me ton 3, esto logo simmetrias oti aftoi einai oi $ a,b$, tote pali $ a \plus{} b \equiv 3 \equiv 0 (mod3)$

Ara se kathe periptosi o 3 diairei ton $ A$

An tora enas apo tous 3 einai artios, tote o 2 diairei ton $ abc$ ara kai ton $ A$, an kai oi 3 einai perittoi tote o $ a \plus{} b$ einai artios epomenos pali o 2 diairei ton $ A$

Telika se kathe periptosi o 2 kai o 3 diairoun ton A, kai epidi $ gcd(2,3)\equal{}1$, kai o 6 diairei ton $ A$

[quote="NickNafplio"]Arketa efkolo alla oraio. Den prepei na exei ksanampei sto forum (toulaxiston ego den to exo ksanadei)

Loipon, esto $ A \equal{} abc(a \plus{} b \plus{} c)(b \plus{} c)(c \plus{} a)(b \plus{} a)$, an enas apo tous $ a,b,c$ einai pollaplasio tou 3 tote o 3 diairei ton $ abc$ ara kai ton $ A$

An oloi afinoun ipolipo 2 me ton 3 tote $ a \plus{} b \plus{} c \equiv 6 \equiv 0 (mod3)$
An oloi afinoun ipolipo 1 me ton 3 tote $ a \plus{} b \plus{} c \equiv 3 \equiv 0 (mod3)$

An 2 apo tous $ a,b,c$ afinoun diaforetika mi midenika ipolipa me ton 3, esto logo simmetrias oti aftoi einai oi $ a,b$, tote pali $ a \plus{} b \equiv 3 \equiv 0 (mod3)$

Ara se kathe periptosi o 3 diairei ton $ A$

An tora enas apo tous 3 einai artios, tote o 2 diairei ton $ abc$ ara kai ton $ A$, an kai oi 3 einai perittoi tote o $ a \plus{} b$ einai artios epomenos pali o 2 diairei ton $ A$

Telika se kathe periptosi o 2 kai o 3 diairoun ton A, kai epidi $ gcd(2,3) \equal{} 1$, kai o 6 diairei ton $ A$[/quote]

Ωραιος Νιck αυτο εναι!!!
Στην αποδειξη σουχρισημοποιηθηκε το γεγονος οτι αναμεσα σε ν+1 ακεραιους υπαρχουνυν τουλαχιστον 2 απο αυτους οι οποιοι ειναι ισουπολοιποι μοντουλο ν (απο την αρχη του περιστερωνα)


Let $ n$ be a natural numbers such that $ n \ge 1$. Prove that
\[ \frac{1}{n^2} \plus{}\frac{1}{n^2\plus{}1}\plus{}\cdots \plus{}\frac{1}{n^2\plus{}2n} \le \frac{2}{n}\]
:)

[quote="can_hang2007"]Let $ n$ be a natural numbers such that $ n \ge 1$. Prove that
\[ \frac {1}{n^2} \plus{} \frac {1}{n^2 \plus{} 1} \plus{} \cdots \plus{} \frac {1}{n^2 \plus{} 2n} \le \frac {2}{n}
\]
:)[/quote]
We have
\[ \left( {\frac{1}{{n^2 }} \plus{} ... \plus{} \frac{1}{{n^2  \plus{} n \minus{} 1}}} \right) \plus{} \left( {\frac{1}{{n^2  \plus{} n}} \plus{} ... \plus{} \frac{1}{{n^2  \plus{} n \plus{} n}}} \right) \le \frac{n}{{n^2 }} \plus{} \frac{{n \plus{} 1}}{{n^2  \plus{} n}} \equal{} \frac{2}{n}
\]
:)

[quote="Honey_S"][quote="can_hang2007"]Let $ n$ be a natural numbers such that $ n \ge 1$. Prove that
\[ \frac {1}{n^2} \plus{} \frac {1}{n^2 \plus{} 1} \plus{} \cdots \plus{} \frac {1}{n^2 \plus{} 2n} \le \frac {2}{n}
\]
:)[/quote]
We have
\[ \left( {\frac {1}{{n^2 }} \plus{} ... \plus{} \frac {1}{{n^2 \plus{} n \minus{} 1}}} \right) \plus{} \left( {\frac {1}{{n^2 \plus{} n}} \plus{} ... \plus{} \frac {1}{{n^2 \plus{} n \plus{} n}}} \right) \le \frac {n}{{n^2 }} \plus{} \frac {{n \plus{} 1}}{{n^2 \plus{} n}} \equal{} \frac {2}{n}
\]
:)[/quote]
That's it, Honey. :)
From this inequality, we can solve the following problem
If $ a_1 \ge a_2 \ge \cdots$ is a sequence positive real numbers such that
\[ \frac{a_1}{1}\plus{}\frac{a_4}{2}\plus{}\cdots \plus{}\frac{a_{k^2}}{k} \le 1\]
for all $ k \ge 1$. Then, for all $ n \ge 1$, we have
\[ \frac{a_1}{1}\plus{}\frac{a_2}{2}\plus{}\cdots \plus{}\frac{a_n}{n} \le 2.\]
(MOSP 2000)


This is from AoPS 1 Ch. 7.

139) Find the sum of the prime factors of $ 2^{16} \minus{} 1$

Alright, so when I was doing this, I factored it into $ (2^2 \plus{} 1)(2^4 \minus{} 2^2 \plus{} 1)(2^2 \minus{} 1)(2^4 \plus{} 2^2 \plus{} 1)$

Then, I found that 5 x 13 x 3 x 21 = $ 2^{16} \minus{} 1$ Therefore, I thought the prime factors would be 3, 5, 13, and 7, after changing 21 to 7 x 3. Why does this not work?

Why would you do it the other way, which seemingly works, factoring into $ (2^4 \plus{} 1)(2^2 \plus{} 1)(2^2 \minus{} 1)(2^8 \plus{} 1)$? I do not see why you should not factor $ 2^8 \plus{} 1$ using sum of cubes..

I tried to find out why this works, but I guess this is just a form of finding a prime factorization? The former looks like one as well.. and my instinct was to do that when I first read the problem.

$ 2^8\plus{}1$ is prime, unless I'm stupid.
8 isn't divisible by 3, so you can't use sum of cubes.

[quote="Bicameral"]
This is from AoPS 1 Ch. 7.

139) Find the sum of the prime factors of $ 2^{16} \minus{} 1$

Alright, so when I was doing this, I factored it into $ (2^2 \plus{} 1)(2^4 \minus{} 2^2 \plus{} 1)(2^2 \minus{} 1)(2^4 \plus{} 2^2 \plus{} 1)$

Then, I found that 5 x 13 x 3 x 21 = $ 2^{16} \minus{} 1$ Therefore, I thought the prime factors would be 3, 5, 13, and 7, after changing 21 to 7 x 3. Why does this not work?

[/quote]

That factorization doesn't work
$ 5\cdot13\cdot3\cdot21 \equal{} 2095 \not\equal{} 65534 \equal{} 2^{16}\minus{}1$

[quote]
Why would you do it the other way, which seemingly works, factoring into $ (2^4 \plus{} 1)(2^2 \plus{} 1)(2^2 \minus{} 1)(2^8 \plus{} 1)$? I do not see why you should not factor $ 2^8 \plus{} 1$ using sum of cubes..

I tried to find out why this works, but I guess this is just a form of finding a prime factorization? The former looks like one as well.. and my instinct was to do that when I first read the problem.[/quote]

This factorization results from repeated use of difference of squares.

[quote="alanchou"][quote="xpmath"]$ 2^8 \plus{} 1$ is prime, unless I'm stupid.
8 isn't divisible by 3, so you can't use sum of cubes.[/quote]

257 is prime :)[/quote]
Yay, I'm not stupid... kind of.


Let $ a,b,c\in R_ \plus{}$ and $ a \plus{} b \plus{} c \equal{} 1$, $ n\in N_ \plus{}$,
Show that $ \frac {a^{n \plus{} 1} \plus{} b}{b \plus{} c} \plus{} \frac {b^{n \plus{} 1} \plus{} c}{c \plus{} a} \plus{} \frac {c^{n \plus{} 1} \plus{} a}{a \plus{} b}\geq \frac {1 \plus{} 3^n}{2\times 3^{n \minus{} 1}}$.

It not correct. I have received a counterexample.


Do not write poor, badly written solutions to problems. Here are some guidelines to follow:

1. Use $ \text{\LaTeX}$. The last thing an asker needs to see is:

"We have x^2 + y^2 = 1 and x^4 + y^4 = 17/18. Square the first equation to get (x^2 + y2)^2 = 1^2 and simplify to get x^4 + 2x^2y^2 + y^4 = 1. We know what x^4+y^4 is - that's 17/18! So subtract 17/18 from both sides to get 2x^2y^2 = 1/18. Divide by 2 to get x^2y^2=1/36. Finally, take the square root of both sides to get xy=1/6. Thus that is our answer."

2. Use proper grammar and spell the best you can. If you ask a question, do you want to see this?

"u take the first equation and square it 2 get $ x^4 + 2x^2y^2 + y^4 = 1^2$. tehn u subtract 17/18 from both sides cuz we see what $ x^4 + y^4$ equals in the 2nd eq and we get $ 2x^2y^2 = \frac{1}{18}$ and tthen we divide by 2 on boht sides of teh eq. to get $ x^2y^2 = \frac{1}{36}$ and then take the sqrt of the both sides and we get a final ans of 1\6."

If you don't speak English well, that's okay, but do your best.

3. Don't rush your solutions, like in the following example.

"Square both sides of the first equation and subtract 17/18. Then divide by 2 and then take the square root of both sides. Thus the answer is 1/6."

If someone asked this question and didn't know how to solve it, most likely this would be a difficult solution to puzzle through.

----

Here is an ideal solution:

We are given the equations $ x^2 + y^2 = 1$ and $ x^4 + y^4 = \frac{17}{18}$ and we wish to find $ xy$. We could find $ x$ and $ y$ individually, but it doesn't look like they are integers, so let's try something else.

Take the first equation and square it, like this:

\[ (x^2 + y^2)^2 = 1^2 \]

\[ x^4 + 2x^2y^2 + y^4 = 1 \]

We know that $ x^4 + y^4$ is - it's $ \frac{17}{18}$!

\[ \frac{17}{18} + 2x^2y^2 = 1 \]

\[ 2x^2y^2 = \frac{1}{18} \]

\[ x^2y^2 = \frac{1}{36} \]

Now take the square root of both sides to get our final answer.

\[ xy = \boxed{\frac{1}{6}} \]

Thank you for taking the time to read this.

Izzy is absolutely right.

Just because there is a "loot" of problems and someone else is in the forum doesn't mean you have to take them all.

Do we have to LaTeX-ify every single fraction we use (for example: 1/2, 1/3, etc)?

Not at all, just make sure it's legible. For instance, you could say "and the answer is thus 1/6." but not something ugly like "a^2b/45 - (3/4)^4/5" and so on.  :)

But it's good to make a habit of using $ \text{\LaTeX}$.

the work shouldn't be cramped into one big paragraph, as i've seen some people do.....

[quote="math154"]Actually, some trivial steps can be skipped. The average reader just has to be able to understand it.[/quote]I'm in the minority here, but I like seeing the trivial steps. Helps me understand when I don't see something right away. Also, it's very helpful for the newer problem solvers.

See [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]the LaTeX tutorial[/url], so conveniently located on the navigation bar.  :)

I still dont understand how to use LaTeX

This is getting off topic.
Try posting it in the Q,A,S Forum.


[b][Putnam 1972].[/b] [i]Let $ *$ be a binary operation on a set $ S$ with the following properties:
\[ x*(x*y)\equal{}y\equal{}(y*x)*x\ \forall x,y\in S.\]
Prove that $ *$ is commutative.[/i]

- Take $ x(xy) \equal{} y$ and multiply it from the left by $ x$:

$ x[x(xy)] \equal{} xy$

- Apply $ u(uv) \equal{} (vu)u$:

$ [(xy)x]x \equal{} xy\qquad(*)$

- Switch $ x$ and $ y$ in $ (yx)x \equal{} y$:

$ (xy)y \equal{} x$

- Multiply that from the left by $ xy$:

$ (xy)[(xy)y] \equal{} (xy)x$

- Apply $ t(ty) \equal{} y$:

$ y \equal{} (xy)x\qquad(**)$

- Plug $ (**)$ into $ (*)$:

$ yx \equal{} xy$

QED


AMC
AMC 12
AMC 12 A
AMC 12 B
rotation
geometry
3D geometry

[quote="1=2"]What did you get?[/quote]

I'm pretty sure I got a 144... didn't have enough time on #25, so I just guessed 175. I did it afterwards, and I'm pretty sure that the answer is actually 225...

Nevermind, I messed up on 24, so a 138

[quote="Darmani"]That's even worse considering this is probably the easiest AMC 12 I've ever seen.
[/quote]

Really? I actually thought that this amc was harder than the AMC 12A...

I thought it was harder than the 12A... I got 135 I think (blank 22 [darn I finally know shoelace NOW], 25, missed 24 [I threw out angles because they didn't have integer degree measures... lol])? bleargh I should break my streak of failing to get >140 on AMC for ... 6 AMCs straight ><

unfortunately there was [i]too much combo[/i] on this one *cough*

144 again. Still aiming for the ever-elusive 150.

I got 24 wrong and guessed 25 correctly.

[quote="Darmani"]I epic failed, getting a 108. That's even worse considering this is probably the easiest AMC 12 I've ever seen.

(Among other things, I spent way too long on #11, and ended up messing up.)[/quote]

Bleh, I had horrible time management. On the AMC12A, I got 22 problems done in 40 minutes, and got two of those wrong. On the AMC12B, I got 16 done in 45 minutes, and got two of those wrong. After that I paniced. EEeep * joins the 108 club *

I think the 12B was quite difficult, compared to the normal. I bombed it, but fortunately my 10A score wasn't too bad.

I thought this AMC 12 was much easier than usual.  Maybe this is because I had seen many of the problems before.

After a 129 on AMC 12A, I thought I had gotten 130.5 on this test, but after discovering that I was misled on various problems, I ended up with a 118.5.

I got a 129, the same as on the 12A.  This one was definitely easier, though, I just wasn't thinking well.  I forgot the range of arcsin on 24, somehow convinced myself that there were no rotated squares on 25 . . . ugh.  I got 2 hours of sleep the night before because I needed to finish my philosophy paper - not the best idea.

136.5 on 12a, but 97.5 on 12b, [b]39[/b]pt drop. Win or fail?

I'm god awful at counting problems. Harder, imo. So far my amc12 scores have gone from 132 to 97.5 to 136.5 yo 97.5...  :rotfl:

Hm, 129 again; did 1-24 skipping #17(reading mistake) and got #11 and #24 wrong by off-by-1 errors.

I was seriously distracted during the test because I got my 12 A score before that and it was 10.5 points lower than I expected (in hindsight, it was STUPID of me to even ask for that score)...And this frightened me to spend the last 15 min just checking bubbles instead of checking answers...

Overall, I feel this test was comparable to 12 A (perhaps slightly easier). The last question seems very doable though I skipped it.

made 4 stupid mistakes and winded up with a 112.5  :rotfl:  (not making aime by amc10 standards)
EDIT; maybe if i had grabbed a 136.5 i would have gotten something out of this.

I got 3 problems wrong. That was not so good. On the other hand, here's an approach for #22 that doesn't involve weird analytic geometry facts. It's not too much computation - stuff cancels (at least compared to some of the other problems...) nicely.

[hide]We can compute the sine of the angle between the positive y-axis and y=kx, let it be theta, then the angle in the parallelogram is 45-x. Then the area of the parallelogram is just like, absintheta, where a and b are the two side lengths. We can get sin(45-x) pretty easily (just get sinx and cosx, use the angle addition thing). Like I said, stuff cancels. You get ab(k-1)=1,000,000 as before.[/hide]

[quote="CatalystOfNostalgia"]On the other hand, here's an approach for #22 that doesn't involve weird analytic geometry facts.[/quote]The parallelogram-area formula is not a "weird analytic geometry fact." In fact, shift your eyes left to the Proofs Without Words gallery. It's in there.

got #24 wrong, lol

144 sounds about right

[quote="DataBox"]Epic failure on 12B with a 133.5[/quote]

hello did you read the post that was, what, 3 posts above yours?

133.5 is hardly an "epic" fail. You can express discontent, or anger, or whatever, but you did not "epic fail." I suppose lots of people here would empathize with you hoping to get higher... but please think of those who scored lower!

This isn't directed just at you. I suppose this is hyperbole, venting, but trying to jockey for position all together... just like so many people on AoPS (I definitely used to as a middle schooler/freshman) do. it'll just keep happening... younger kids want to justify themselves I suppose?

Hmm...132 I think. I could've done a lot better though, so I'm disappointed. Actually I think I got 126. Blah. Depends on what I decided to bubble in for 21, which I forgot.

So basically, I was done with 1-17 in 25 minutes, and I look at the clock, and I am like "wow I am in good shape". Then I get to 18, and I read it incorrectly, and solve the wrong problem. My answer is not on the multiple choice. Uh oh...so I read the problem again and try to solve it, except I somehow still was solving the wrong problem, and my answer, again, was not on the choices. So I kinda wasted 30 minutes figuring out what was wrong, so I had about 7 minutes left to do the rest :O I think if I didn't do that I could've gotten 22 and one of the last few, but oh well. No use complaining.

I thought this test was pretty easy, but opinions seem to vary a lot.

[quote="not_trig"]133.5 is hardly an "epic" fail. You can express discontent, or anger, or whatever, but you did not "epic fail." I suppose lots of people here would empathize with you hoping to get higher... but please think of those who scored lower![/quote]Yeah. Epic failure would consist of, say, spending 15 minutes on the first problem. Or making stupid mistakes on 10 problems. Things like that.

or 118.5, which is what i got, after a 150 on the 10a...blargh...

i need to work on geometry...couldn't even figure out #16

EDIT: MISSED MY GOAL BY FREAKIN 1.5 POINTS.....so mad. but this was because I was stupid and can't do math...i got all the ones i did right, except i couldn't figure out a bunch.

Wow, I got a 118.5, much higher than I expected. This was by FAR the easiest amc12 I've ever seen. Just to compare, I got like 80something on the 12A.

But then It's because I LOVE probability and HATE geometry. I can't do the simplest geometry questions, but solved all the number theory/probability ones.

[quote="hookedonconics"]I did simple arithmetic errors on one.
94.5..
I doubt the cutoff will be that low.
When college apps come along, is it even worth it to report it?[/quote]

I think it is. The fact that you even took it sets you apart. Sure, not as much as it would if you had gotten a perfect score, but still...

This was my first year taking AMCs, or any math contest for that matter. (I heard of it from a friend at our middle school a few months ago). I just ordered and read AoPS vol1 and a bit of vol2 reeeaaaallllyy quickly, and did some past tests during winter break. I actually got a 124 on 12 for one year, but at the 2009 12B, I got only a 102 (Terrible time management....). I messed up on the millet seed and clock problems, dunno why...my mind just stopped functioning normally... Oh well, I've got 4 years to improve  :D

Other than reading the AoPS guide and reviewing previous tests, what would be the best method to study for the AMC 12?

[quote="DataBox"]Epic failure on 12B with a 133.5[/quote]
..... I got that on the AMC 12A and that was my highest score on the AMC 12A ever. As an 8th grader. When I see that I don't think you appreciate a very high score like that.


function
real analysis
real analysis unsolved

Solve the equation $ u^2_x\plus{}u^2_y\equal{}a^2, u(x,2x)\equal{}1$

That's a cutie. The idea is that if you follow the gradient curves, you should go away from the line at the maximal possible speed (otherwise, you have too large value too close to the line and the mean value theorem will tell you that you have too large gradient somewhere). So the function must be linear. Once you know that, the rest is trivial.

hello, i have found
$ u(x,y)\equal{}\minus{}\sqrt{a^2\minus{}C_2^2}x\plus{}C_1\plus{}C_2y\plus{}C_2$
Sonnhard.

Which means that you haven't read the original post attentively (there is also an initial condition there ;) )

hello, sorry, i have read the post very attentively but the proposer of the problem can solve the rest by himself.
Sonnhard.

Condition $ u(x,2x)\equal{}1$ not satisfied by your set of solutions, or am I missing something?

Yes, alekk, this time you are missing something :) (just reread the whole thread once more and look up other posts by Sonnhard ;) )


function
calculus
derivative
linear algebra
matrix
algebra
polynomial

(1)
f:R->R is a continuous function, derivable on R\{x_0}, having finite side derivatives at x_0. Show that there exists a derivable function g:R->R, a linear function h:R->R and a in {0,1,-1} s.t. f(x)=g(x)+a*|h(x)| for all real x.

Aurelian Gheondea

(2)
We consider a matrix A in M_n(C), with rank r, where n>=2 and 1<=r<=n-1.

a)Show that there exist B in M_(n,r)(C) and C in M_(r,n)(C), with rank B=rank C=r s.t. A=BC.
b)Show that the matrix A verifies a polynomial equation of degree r+1 with complex coefficients.

Mircea Becheanu and Ion Savu

(3)
Let f:R->[0,00) be a function with the property |f(x)-f(y)|<=|x-y| for all real x and y. Show that:

a)if lim{n->00}f(x+n)=00 for every x in R, then lim{x->00}f(x)=00.
b)if lim{n->00}f(x+n)=a, a in [0,00), for every x in R, then lim{x->00}f(x)=a.

Mihai Piticari and Sorin Radulescu

(4)
The continuous function f:[0,1]->R has the property
lim{n->00}n[f(x+1/n)-f(x)]=0 for every x in [0,1). Show that:

a)for every t>0 and a in (0,1) we have
sup{x in [0,a) | |f(x)-f(0)|<=tx}=a.
b)f is constant.

Cristinel Mortici

[quote="grobber"]
(2)
We consider a matrix A in M_n(C), with rank r, where n>=2 and 1<=r<=n-1.

a)Show that there exist B in M_(n,r)(C) and C in M_(r,n)(C), with rank B=rank C=r s.t. A=BC.
b)Show that the matrix A verifies a polynomial equation of degree r+1 with complex coefficients.

Mircea Becheanu and Ion Savu

[/quote]

a) If rankA =r then A is equivalent to J_r matrix defined by block
J_r=
(I_r 0)
(0  0)   with I_r is dentity matrix M_r(C)
geometrically J_r is a projection matrix (J_r)^2 =J_r
 there exist P,Q invertible such that
A=PJ_r.Q=(PJ_r)(J_rQ)
if you divide P also by blocks like J_r the product P.J_r has the last n-r columns equal to 0, so we can identify P.J_r has a matrix of M_(n,r)(C)
Same idea for J_r.Q has the last n-r row 0 it can be view as a matrix
of M_(n,r)(C)
B=PJ_r, C=J_rQ since P,Q are invertible they don't affect the rank
rank(B)=rank(C)=rank(J_r)=trace(J_r)=r

[quote="Moubinool"][quote="grobber"]
(2)
We consider a matrix A in M_n(C), with rank r, where n>=2 and 1<=r<=n-1.

a)Show that there exist B in M_(n,r)(C) and C in M_(r,n)(C), with rank B=rank C=r s.t. A=BC.
b)Show that the matrix A verifies a polynomial equation of degree r+1 with complex coefficients.

Mircea Becheanu and Ion Savu

[/quote]

[/quote]

For b) I think it is related to the minimal polynomial if rank(A)=r then
 \pi _A(x) minimal polynomial has degree r+1, and  \pi _A(A)=0

[quote="Moubinool"]
For b) I think it is related to the minimal polynomial if rank(A)=r then
 \pi _A(x) minimal polynomial has degree r+1, and  \pi _A(A)=0[/quote]

It is correct not moubi, here a counter exemple

Let A be
(-1 0 0 0 0)
(0 -1 0 0 0)
(0 0 -1 0 0)
(0 0 0  0 0)
(0 0 0 0 0)
rankA=r=3, minimal polynomial is x+x^2 its degree is 2 < r+1=4

[quote="grobber"]
(2)
We consider a matrix A in M_n(C), with rank r, where n>=2 and 1<=r<=n-1.

a)Show that there exist B in M_(n,r)(C) and C in M_(r,n)(C), with rank B=rank C=r s.t. A=BC.
b)Show that the matrix A verifies a polynomial equation of degree r+1 with complex coefficients.

Mircea Becheanu and Ion Savu

[/quote]

For b) The caracteristic polynomial of BC\in Mn(C) and CB \in Mr(C)
are related by P_(BC)(x)=x^(n-r)P_(CB)(x)
T(A)=0 where T(x)=x^(-n+r+1).P_(A)(x) where P_A is caracteristic polynomial of A, deg T = r+1

[quote="grobber"]
(3)
Let f:R->[0,00) be a function with the property |f(x)-f(y)|<=|x-y| for all real x and y. Show that:

a)if lim{n->00}f(x+n)=00 for every x in R, then lim{x->00}f(x)=00.
b)if lim{n->00}f(x+n)=a, a in [0,00), for every x in R, then lim{x->00}f(x)=a.

Mihai Piticari and Sorin Radulescu

[/quote]

b)The condition  |f(x)-f(y)| \leq |x-y| for all real x and y is called 1-Lispchitzien, so f is unformly continuous on R.
We can suppose a = 0 substitute f by f-a

We have

Lemma: Let a(n) an increasing sequence of positives reals,  \lim a(n)=+oo,  a(n+1)-a(n) is bounded, f:R ->R uniformly continuous.
Then ( for any x \in R+, \lim{n->+oo} f(x+a(n))=0)=>\lim{y->+oo} f(y)=0

Proof: M=sup(a(n+1)-a(n))
f uniformly continuous for any e>0, there exist h>0 for any x,y s.t |x-y|<h=> |f(x)-f(y)|<e

Let  p=sup{ n \in N; n.h \leq M}, for k \leq p

(there exist m(k) \in N) (for any n) n>m(k)=>|f(k.h+a(n))|<e

Let m = sup(m(1),m(2),...,m(p))

If y>a(m)+M there exist n>m and k \leq p s.t. y-a(n) \in [0,M] and |y-a(n)-kh|<h ,
n>m \geq m(k)=> |f(a(n)+kh)|<e

 |y-a(n)-kh|<h=>|f(y)-f(a(n)+kh)|<e
and for any y>a(m) + M => |f(y)| \leq |f(y)- f(a(n)+kh)|+|f(a(n)+kh)|<2e


[quote="Moubinool"]$u_n$ sequence real number such that $u_n^{2005}+nu_n-1=0$

Give an asymptotic expansion of order 2006, find $c_k$
$u_n=\sum_{k=1}^{2006}\frac{c_k}{n^k}+o(\frac{1}{n^{2006}})$[/quote]

This reduces to $x^{2005} + x = f(n) \to 0$ where $f(n) = n^{-2005/2004} =: ({x_n}/{u_n})$. The expansion is obtained by inverting the series $f = x + x^{2005}$ near $0$, i.e. $x_n = x = f - f^{2005} + \dots$.


calculus
integration
trigonometry
calculus computations

So I have this textbook example:

Given the indefinite integral: $ \int\dfrac{3}{(x + 1)^2 + 9}\: dx$

We'd let $ u = x + 1$
so $ du = dx$

Substituting: $ \int\dfrac{3}{u^2 + 9}\: du$

The next step changed that to $ \dfrac{1}{3}\int\dfrac{1}{u^2 + 9}\: du$, but I don't understand why it's not: ${ \int\left(3\cdot\dfrac{1}{u^2 + 9}\right)\: du} = 3\int\dfrac{1}{u^2 + 9}\: du$

Thanks.

You are most certainly correct, and your book is wrong.
Constant scaling factors are always brought outside of the integral with no weird reciprocal business.

Don't worry, you are right because of the linearity of the integral. This has to be a mistake in your textbook.

EtotheX

ah, okay thank you both. I was thinking maybe it was the book's mistake too :)

well, not really a book since I'm homeschooled online:

[img]http://i25.tinypic.com/qoc1g0.png[/img]

then it'd be $ 3\int\dfrac{1}{u^2 \plus{} 3^2}\: du \equal{} 3\cdot\dfrac{1}{3}\tan^{ \minus{} 1}\dfrac{u}{3} \plus{} C$ and the 3 and 1/3 would cancel out.. yay.


Hi,

Okay, it takes me quite a while to do the final 10 problems out of the 25. In fact, around #12/13 I usually start to feel a slow-down in my speed.

First off, does anyone have the AMC 12 from 2002?
It appears that 2003 had an extremely easy first 10 problems. 2001 is fairly tough. And the practice set http://www.unl.edu/amc/a-activities/a7-problems/practice1012p/02practice.html has some of the toughest problems for the first 10 I've seen.

Anyways, how long do these first problems take you on average? I got a bit discouraged today after doing the 2001 AMC 12A because I did horribly. For some reason, I still think that 2003 contains the easiest problems I've seen, but it may just be because I've looked at them so many times now (except for #25, which has the wierdest question yet; I hope this year's #25 is geometry).

First, in response to the practice being tough:
they are the taiwanese ones, so you should expect them to be tougher than the ones given in the US. They are awesome problems tho, if you kno the tricks, but still noticeably more difficult than the one given in the states.

Second, its natural to get a little slower when doing questions after 12/13, particularly since the first 12 or so questions on 12 are 10 material, and after 15 usually is the good, harcore 12 questions. So expect them to be tougher.

DOn't get discouraged tho. Just do lotsa math and u'll get it, eventually, i hope,.. hahah.

I didn't think the Taiwanese problems were harder than the American ones when I did them in preparation for the 2002 AMC12s. The reason is that the 2001 AMC12 was probably the hardest AMC12/AHSME for quite a few years, and so by comparison the AMC12P was rather easy.


topology
Functional Analysis
advanced fields
advanced fields unsolved

In [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60364]this topic[/url] it is asked to prove that a limit (in the norm topology) of a sequence of operators with finite rank on a Hilbert space is a compact operator. Prove that the converse is also true: every compact bounded linear operator on a Hilbert space is the limit of a sequence of bounded linear operators with finite rank.

P.S.

Apparently, the corresponding problem for Banach spaces is unsolved.

http://www.mathlinks.ro/Forum/viewtopic.php?t=82566 , just to "close" this unanswered post. :)


Prove that in any triangle $ABC$ the following inequality holds:

                             \[ \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\leq\frac{1}{4r^{2}}.  \] :)

Well, let $\displaystyle p=\frac{a+b+c}{2}$ and $x=p-a$, etc.

Your inequality turns into \[ \sum \frac{1}{(x+y)^2} \leq \sum \frac{1}{4xy} ,  \] which is simple ($(x+y)^2 \geq 4xy$).

Too easy (?)   :?

Some details:

[hide]See that $x+y+z=p$.

We have $S=pr$, so $r^2=\frac{S^2}{p^2}$. We know that $S^2 = p(p-a)(p-b)(p-c) = (x+y+z)xyz$. Thus, $r^2=\frac{xyz(x+y+z)}{(x+y+z)^2}=\frac{xyz}{x+y+z}$.

$\frac{1}{4r^2}=\frac{x+y+z}{4xyz} = \sum \frac{1}{4yz}$[/hide]


[hide]The last digit of powers of 2 cycle in 4's, so \[ 2^{2009}\equiv2^{1}\equiv2\pmod{10}\] which is the answer choice $ \fbox{\textbf{(B)}}.$[/hide]

2^1=2
2^2=4
2^3=8
2^4=16
2^5=32

The unit digits of powers of twos repeats every four powers, and since 2009 is one more than a multiple of 4 it is 2^1=2 so the answers is B.


algorithm
function
induction
graph theory
combinatorics unsolved
combinatorics

This problem was in the final (Combinatorics) contest of Iran Computer Olympiad for the choosing the members of the team (not for the team) :

How to color the edges of a "Complete graph with $2^n$ vertices" ( $K_{2^n}$ ) with minimum color, so in any hamiltonian cycle, there are at least $n$ color ?

I think, I have nice example for $n$ colors :)

i think with n color we cant coloring.
vertex is 1..n.

in these 4 hamilton tours we dont see leess than n+1 color for edges.and with thinking , i think we dont see leess than 3n/2+1 color for edges

hamilton tour 1

vertex 1 joind 2 2 joind 3 .... 2^n joind 1.

hamilton tour 2
vertex 2 joind 1 1 joind 4 4 joind 3 3 joind 5 5 joind 6 6 joind 8 ... 2^n joind 2.

hamilton tour 3
vertex 2 joind 1 1 joind 3 3 joind 4 4 joind 6 6 joind 7 7 joind 8 8 joind 10 ..... 2^n joind 2.

hamilton tour 4

vertex 2 joind 3 3 joind 1 1 joind 4 4 joind 6 6 joind 7 7 joind 5 5 joind 8 .... 2^n joind 2.

i have one algorithm for 2n colors.
algorithm =/ example.    :D

I don't see why your tours mean that we need at least $3n/2$ colors.
And I think my coloring works.

I tried to do precisely the same thing, but failed, so I'm also curious about the solution, if it's not too much trouble, Myth.

[quote="seshadri"]Myth,could you post your coloring-or is it very complicated to describe?
i expect that by labelling all the vertices as binary strings,$S_n$ of length $n$,one can describe a function on ${S_n\choose 2} \mapsto \{1,2,..,n\}$ which could work...[/quote]
I also was thinking about such function, but after all I found following coloring.
Divide all vertices into sets $A_0$, $A_1$, ..., $A_{n-1}$, where $|A_i|=2^i$ for $i=0,1,2,...,n-2$ and $|A_{n-1}|=2^{n-1}+1$.
Now color edge $(v,w)$ into color $i$ iff $v\in A_i$, $w\in A_j$ and $i\leq j$.
Consider arbitrary cycle, which passes all vertices. It is clear that for each color $i$ there exists edge $(v,w)$ in our cycle s.t. $v\in A_i$ and $w\in A_j$, $i\leq j$ (otherwise we immediately obtain contradiction, since $|A_0|+|A_1|+...+|A_{i-1}|<|A_i|$).

Is it correct? If yes, then it is nice, isn't it? :D

what you have done looks fine by me-inductively color $0$ must appear since every edge incident with $A_0$ is colored $0$.suppose $C$ is a hamiltonian circuit and colors $1,2,..,i-1$ appear in the circuit.consider the set of edges of $C$ incident with $A_i$.if there is an edge between a vertex of $A_i$ and $A_j$ with $j\geq i$,then clearly color $i$ too appears in $C$.Else it must be the case that every vertex of $A_i$ has as its neighbors, in $C$, vertices in  $\cup_{j=0}^{i-1} A_j$ but then this contradicts the observation you made and the induction is complete.
indeed,a nice proof! :)

Actually, you don't need induction. :)

Any comments from Grobber? ;)

Nothing to say here :). Except that it's beautiful, of course :).

Wait wait ;
coloring means (in this problem) that no two edge with a common edge have same color
so we know in a complete graph with n vertices, in edge coloring, n color is necessary (in every graph, DELTA(g) or DELTA(g)+1 color is necessary for edge coloring of graph and DELTA(g) means maximum degree in g)

The solution is in this graph with 2^n vertices, 2^n colors is necessary and sufficient ;
we can number vertices from 0 to 2^n-1 and color each edge uv with color (u xor v)
the proof of with this coloring each hamiltonian cycle have at least n color is not easy (but you can solve it)

Sorry for this mistake (if it is mine)


geometry
perpendicular bisector
angle bisector
geometry solved

$BC$ is diameter of circle. Point $A$ is lying on circle so that $\angle AOC > 60^{o}$. Perpendicular bisector of segment $AO$ cuts circle in points $E$ and $F$. Angle bisector of $\angle AOB$(smaller) cuts circle in point $D$. Line, which crosses O and is pararel to segment $AD$, meets chord $AC$ in $J$. Prove that $J$ is the centre of incircle triangle $CEF$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17318]IMO 2002 problem 2[/url]. An IMO problem could be nicer...

  Darij


At the beginning of a simple game, there is a board with two positive numbers (> 0). In each turn you can keep a positive difference (>0) of any two numbers already present on the board (no duplicates), and then the opponent plays the same way, followed by your turn….. So on. The loser is the player unable to make a move. What will be your strategy for any given +ve numbers?

Here's the deal. Let the numbers already on the board be $a$ and $b$. Then by the division algorithm, the only numbers you can make are going to be those of the form $k\cdot(a,b)$. The number of moves that can be made is $g=\displaystyle\frac{max(a,b)}{(a,b)}-2$. It doesn't matter what you do after the game starts, but you want $g$ to be odd if you go first, and even otherwise. Choose to go first or second accordingly.

The name of this game is the alquiot game. Because alquiot has something to do with the euclidian division algoritm.


As the number of sides of a polygon increases from $3$ to $ n$, the sum of the exterior formed by extending each side in succession:

$\textbf{(A)}\ \text{Increases} \qquad
\textbf{(B)}\ \text{Decreases} \qquad
\textbf{(C)}\ \text{Remains constant} \qquad
\textbf{(D)}\ \text{Cannot be predicted} \qquad\\
\textbf{(E)}\ \text{Becomes }(n-3)\text{ straight angles}$

the sum of all the exterior angles of any polygon is always 360 so it stays the same.


calculus
integration
LaTeX
logarithms
trigonometry
calculus computations

Can you solve?

∫_0^1▒〖arctan√x〗 dx
∫〖x^2 √(4-x^2 ) dx〗
∫〖x^2  ln((1-x)/(1+x)) 〗 dx

hello, can you use $ \text{\LaTeX}$ please?
Sonnhard.

$ \begin{gathered}
  1.{\text{ }}\int\limits_0^1 {\arctan \sqrt x } dx \equal{} \int\limits_0^1 {\left( x \right)^\prime  \arctan \sqrt x } dx \equal{} \left. {x\arctan \sqrt x } \right|_0^1  \minus{} \int\limits_0^1 x \left( {\arctan \sqrt x } \right)^\prime  dx \equal{}  \hfill \\
   \equal{} \frac{\pi }
{4} \minus{} \int\limits_0^1 {x\frac{{\left( {\sqrt x } \right)^\prime  }}
{{1 \plus{} \left( {\sqrt x } \right)^2 }}} dx \equal{} \frac{\pi }
{4} \minus{} \frac{1}
{2}\int\limits_0^1 {\frac{x}
{{\sqrt x \left( {1 \plus{} x} \right)}}} dx \equal{} \frac{\pi }
{4} \minus{} \frac{1}
{2}\int\limits_0^1 {\frac{{\sqrt x }}
{{1 \plus{} x}}} dx \equal{} \left\{ \begin{gathered}
  \sqrt x  \equal{} t, \hfill \\
  x \equal{} t^2 , \hfill \\
  dx \equal{} 2t \hfill \\
\end{gathered}  \right\} \equal{}  \hfill \\
   \equal{} \frac{\pi }
{4} \minus{} \int\limits_0^1 {\frac{{t^2 }}
{{1 \plus{} t^2 }}} dt \equal{} \frac{\pi }
{4} \minus{} \int\limits_0^1 {\left( {1 \minus{} \frac{1}
{{1 \plus{} t^2 }}} \right)} dt \equal{} \left. {\frac{\pi }
{4} \minus{} \left[ {t \minus{} \arctan t} \right]} \right|_0^1  \equal{} \left. {\frac{\pi }
{4} \minus{} \left[ {\sqrt x  \minus{} \arctan \sqrt x } \right]} \right|_0^1  \equal{}  \hfill \\
   \equal{} \frac{\pi }
{4} \minus{} \left[ {\left( {1 \minus{} \frac{\pi }
{4}} \right) \minus{} \left( {0 \minus{} 0} \right)} \right] \equal{} \frac{\pi }
{4} \minus{} \left( {1 \minus{} \frac{\pi }
{4}} \right) \equal{} \frac{\pi }
{4} \minus{} 1 \plus{} \frac{\pi }
{4} \equal{} \frac{\pi }
{2} \minus{} 1. \hfill \\
\end{gathered}$

Are remaining integrals $ \int {x^2 \sqrt {4 \minus{} x^2 } } dx$ and $ \int {x^2 \ln \left( {\frac{{1 \minus{} x}}{{1 \plus{} x}}} \right)dx}$ ???

Thank you :lol: , but this integral $ \int {x^2 \ln \left( {\frac {{1 \minus{} x}}{{1 \plus{} x}}} \right)dx}$ need for [b]light1889[/b].

$ \begin{gathered}
  \int {x^2 \sqrt {4 \minus{} x^2 } } dx \equal{} \left\{ \begin{gathered}
  x \equal{} 2\sin \left( t \right), \hfill \\
  dx \equal{} 2\cos \left( t \right) \hfill \\
\end{gathered}  \right\} \equal{} 8\int {\sin ^2 \left( t \right)\sqrt {4 \minus{} 4\sin ^2 \left( t \right)} \cos \left( t \right)dt \equal{} 16} \int {\sin ^2 \left( t \right)} \cos ^2 \left( t \right)dt \equal{}  \hfill \\
   \equal{} 16\int {\left( {\frac{{1 \minus{} \cos \left( {2t} \right)}}
{2}} \right)} \left( {\frac{{1 \plus{} \cos \left( {2t} \right)}}
{2}} \right)dt \equal{} 4\int {\left( {1 \minus{} \cos ^2 \left( {2t} \right)} \right)} dt \equal{} 4t \minus{} 4\int {\frac{{1 \plus{} 2\cos \left( {4t} \right)}}
{2}dt \equal{} }  \hfill \\
   \equal{} 4t \minus{} 2t \minus{} 2\int {\cos \left( {4t} \right)} dt \equal{} 2t \minus{} 2\int {\cos \left( {4t} \right)dt}  \equal{} 2t \minus{} \frac{1}
{2}\int {\cos \left( {4t} \right)d\left( {4t} \right)}  \equal{}  \hfill \\
   \equal{} 2t \minus{} \frac{{\sin \left( {4t} \right)}}
{2} \plus{} C \equal{} 2\arcsin \left( {\frac{x}
{2}} \right) \minus{} \frac{1}
{2}\sin \left( {4\arcsin \left( {\frac{x}
{2}} \right)} \right) \plus{} C. \hfill \\
  \left[ \begin{gathered}
  \frac{1}
{2}\sin \left( {4\arcsin \left( {\frac{x}
{2}} \right)} \right) \equal{} \sin \left( {2\arcsin \left( {\frac{x}
{2}} \right)} \right)\cos \left( {2\arcsin \left( {\frac{x}
{2}} \right)} \right) \equal{}  \hfill \\
   \equal{} 2\sin \left( {\arcsin \left( {\frac{x}
{2}} \right)} \right)\cos \left( {\arcsin \left( {\frac{x}
{2}} \right)} \right)\left[ {\cos ^2 \left( {\arcsin \left( {\frac{x}
{2}} \right)} \right) \minus{} \sin ^2 \left( {\arcsin \left( {\frac{x}
{2}} \right)} \right)} \right] \equal{}  \hfill \\
   \equal{} x\sqrt {1 \minus{} \left( {\frac{x}
{2}} \right)^2 } \left( {1 \minus{} \left( {\frac{x}
{2}} \right)^2  \minus{} \left( {\frac{x}
{2}} \right)^2 } \right) \equal{} \frac{x}
{2}\sqrt {4 \minus{} x^2 } \left( {1 \minus{} \frac{{x^2 }}
{2}} \right) \equal{}  \minus{} \frac{x}
{4}\left( {x^2  \minus{} 2} \right)\sqrt {4 \minus{} x^2 }  \hfill \\
\end{gathered}  \right] \hfill \\
  \boxed{\int {x^2 \sqrt {4 \minus{} x^2 } } dx \equal{} 2\arcsin \left( {\frac{x}
{2}} \right) \plus{} \frac{x}
{4}\left( {x^2  \minus{} 2} \right)\sqrt {4 \minus{} x^2 }  \plus{} C} \hfill \\
\end{gathered}$

Let $ x \equal{} 2\sin \theta$, $ \int (4 \minus{} x^2)^{\frac {3}{2}}\ dx \equal{} \int (4\cos ^ 2 \theta)\ \cdot 2\cos \theta d\theta$

$ \equal{} 16\int \cos ^ 4 \theta d\theta \equal{} 16\int (\cos ^ 2 \theta)^2 d\theta$

$ \equal{} 16\int \left(\frac {1 \plus{} \cos 2\theta}{2}\right)^2 \equal{} 4\int (1 \plus{} 2\cos 2\theta \plus{} \cos ^ 2 2\theta)d\theta$

$ \equal{} 4\int \left(1 \plus{} 2\cos 2\theta \plus{} \frac {1 \plus{} \cos 4\theta }{2}\right)\ d\theta$

$ \equal{} 4\left(\frac {3}{2}\theta \plus{} \sin 2\theta \plus{} \frac {1}{8}\sin 4 \theta\right)$

$ \equal{} 6\theta \plus{} 4\sin 2\theta \plus{} \frac {1}{2}\sin 4\theta \plus{} C$,

$ \therefore \int x^2\sqrt {4 \minus{} x^2}\ dx \equal{} \int x\cdot x\sqrt {4 \minus{} x^2}\ dx$

$ \equal{} \minus{} \frac {1}{3}x\sqrt {4 \minus{} x^2} \plus{} \frac {1}{3}\int (4 \minus{} x^2)^{\frac {3}{2}}\ dx$

$ \equal{} \boxed{2\sin ^{ \minus{} 1} \frac {x}{2} \plus{} \frac {4}{3}\sin \left(2\sin ^ { \minus{} 1} \frac {x}{2}\right) \plus{} \frac {1}{6}\sin \left(4\sin ^ { \minus{} 1} \frac {x}{2}\right) \minus{} \frac {1}{3}x\sqrt{4 \minus{} x^2} \plus{} C}$

Last edited.

[quote="kunny"]Let $ x \equal{} 2\sin \theta$, $ \int (4 \minus{} x^2)^{\frac {3}{2}}\ dx \equal{} \int (4\cos ^ 2 \theta)\ \cdot 2\cos \theta d\theta$

$ \equal{} 16\int \cos ^ 4 \theta d\theta \equal{} 16\int (\cos ^ 2 \theta)^2 d\theta$

$ \equal{} 16\int \left(\frac {1 \plus{} \cos 2\theta}{2}\right)^2 \equal{} 4\int (1 \plus{} 2\cos 2\theta \plus{} \cos ^ 2 2\theta)d\theta$

$ \equal{} 4\int \left(1 \plus{} 2\cos 2\theta \plus{} \frac {1 \plus{} \cos 4\theta }{2}\right)\ d\theta$

$ \equal{} 4\left(\frac {3}{2}\theta \plus{} \sin 2\theta \plus{} \frac {1}{8}\sin 4 \theta\right)$

$ \equal{} 6\theta \plus{} 4\sin 2\theta \plus{} \frac {1}{2}\sin 4\theta \plus{} C$,

$ \therefore \int x^2\sqrt {4 \minus{} x^2}\ dx \equal{} \int x\cdot x\sqrt {4 \minus{} x^2}\ dx$

$ \equal{} \minus{} \frac {1}{3}x\sqrt {4 \minus{} x^2} \plus{} \frac {1}{3}\int (4 \minus{} x^2)^{\frac {3}{2}}\ dx$

$ \equal{} \boxed{2\sin ^{ \minus{} 1} \frac {x}{2} \plus{} \frac {4}{3}\sin \left(2\sin ^ { \minus{} 1} \frac {x}{2}\right) \plus{} \frac {1}{6}\sin \left(4\sin ^ { \minus{} 1} \frac {x}{2}\right) \minus{} \frac {1}{3}(4 \minus{} x^2)^{\frac {3}{2}} \plus{} C}$
[/quote]

Why have you posted this integration

[quote]$ \int (4 \minus{} x^2)^{\frac {3}{2}}\ dx$
[/quote]

[quote="kunny"]Sorry, I can't understand what you say. :maybe:

kunny[/quote]

Sorry!
I mean that your posted solutions not for the light1889's need.