geometry
parallelogram
trigonometry
calculus
derivative
geometry unsolved

Let $M$ be a point lying between the side of an acute angle: $xOy.$ Let $d$ be a line passing through $M;\, A=d\cap Ox,\,B=d\cap Oy$.
Find the position of $d$ such that the area of triangle $OAB$ attains its least value.

Draw a line $e$ through $M$ parallel to $Oy,$ intersecting $Ox$ at $E.$ Also draw a line $f$ through $M$ parallel to $Ox,$ intersecting $Oy$ at $F.$ Now $OEMF$ is a parallelogram so $OE=FM=a$ and $OF=EM=b.$ $\angle MAE \cong \angle BMF$ (corresponding angles formed by parallel lines $Ox$, $MF$ and transversal $AB$) and $\angle AME \cong \angle MBF$ (parallel lines $Oy$, $ME$ cut by transversal $AB$) so $\triangle AME \sim \triangle MBF$ and \[\frac{AE}{EM}=\frac{MF}{FB}.\] Letting $AE=x$, $FB=y$, this may be written as \[\frac{x}{b}=\frac{a}{y}\text{ or }xy=ab.\]
Now the amount we are trying to minimize is \begin{eqnarray*}\text{Area }OAB&=&\frac{1}{2}\sin\angle O\cdot OA\cdot OB\\ &=&\frac{1}{2}\sin\angle O(a+x)(b+y)\\ &=&\frac{1}{2}\sin\angle O(ab+ay+bx+xy)\\ &=&\frac{1}{2}\sin\angle O(2ab+ay+bx)\\ \end{eqnarray*}
Since $a$, $b$ and $\angle O$ are fixed and only $x$ and $y$ depend on line $d$, this is equivalent to minimizing $ay+bx.$ To eliminate $a$ and $b$ I let $\frac{x}{a}=\frac{b}{y}=r$ whence $x=ar$ and $y=\frac{b}{r}.$ Then \begin{eqnarray*}ay+bx&=&a(\frac{b}{r})+b(ar)\\ &=&ab(\frac{1}{r}+r)\end{eqnarray*}
so the problem is reduced to minimizing $r+\frac{1}{r},$ $r>0.$
This may be accomplished by calculus (set the first derivative $1-\frac{1}{r^{2}}$ equal to 0, yielding $r=1$) or alternatively by the following algebraic transformation:\begin{eqnarray*}r+\frac{1}{r}&=&2+r-2+\frac{1}{r}\\ &=&2+(\sqrt{r})^{2}-2\sqrt{r}\frac{1}{\sqrt{r}}+(\frac{1}{\sqrt{r}})^{2}\\ &=&2+(\sqrt{r}-\frac{1}{\sqrt{r}})^{2}.\end{eqnarray*}
This value clearly cannot be less than 2. It will equal 2 iff
\begin{eqnarray*}\sqrt{r}-\frac{1}{\sqrt{r}}&=&0\\ \sqrt{r}&=&\frac{1}{\sqrt{r}}\\ \sqrt{r}\sqrt{r}&=&\sqrt{r}\frac{1}{\sqrt{r}}\\ r&=&1\end{eqnarray*}
Returning to the original problem, we have $x=ar=a,$ $y=\frac{b}{r}=b$. The requisite line $d$ is found by locating $A$ on ray $OE$ with $EA=a$, and drawing line $MA$ for $d.$ That $BF=b$ follows automatically since $\triangle AME \cong \triangle MBF$ by AAS.
If it is desired to construct line $d$ by ruler and compass only, I see no better way than to first construct $ME \parallel Oy$ and then lay off $EA=EO.$ As before, finish by joining $MA.$ Constructing $MF$ is unnecessary for finding this line.

[color=indigo][b] Solution.[/b] Draw a line $e$ through $M$ parallel to $Oy,$ intersecting $Ox$ at $E.$ Also draw a line $f$ through $M$ parallel to $Ox$ intersecting $Oy$ at $F$
We have:
$\frac{S(OEF)}{S(OAB)}=\frac{OE}{OA}\cdot\frac{OF}{OB}\le\frac{1}{4}\left(\frac{OE}{OA}+\frac{OF}{OB}\right)=\frac{1}{4}\left(\frac{BM}{AB}+\frac{AM}{AB}\right)^{2}=\frac{1}{4}\\ \Longrightarrow S(OAB)\ge 4S(OEF)$
Equality occurs iff $\frac{OE}{OA}=\frac{OF}{OB}\iff d\parallel EF$
Hence: $S(OAB)_{\text{min}}=4S(OEF)$. With equality iff $\frac{OE}{OA}=\frac{OF}{OB}\iff d\parallel EF$ [/color]


Suppose that the number of claims that may arise for a 'portfolio' of 300 insurance policies in a year is modeled as a poisson random variable, with mean m. Suppose that 30 claims occured from this portfolio in year 2005. Find the p-value of this result in a test of
H0:m=36 verses H1:m<36

I dont think what im doing is right...

P[X ≥ 30|µ=36]

P[(X-36)/(σ/√36) ≥ (30-36)/(σ/√36)]

P[Z ≥ -6/(σ/6)]

How do you find σ?

[quote="laney5"]Suppose that the number of claims that may arise for a 'portfolio' of 300 insurance policies in a year is modeled as a poisson random variable, with mean m. Suppose that 30 claims occured from this portfolio in year 2005. Find the p-value of this result in a test of
H0:m=36 verses H1:m<36

I dont think what im doing is right...

P[X ≥ 30|µ=36]
[/quote]
Compute this directly from the cdf. The result is 1-p.


does anyone know or have a good guess on what the qualifying index for USAJMO and USAMO will be this year? I just want a range of scores. (for example, like 200 to 220) or something like that.

In another forum on the AMC, it explains the whole thing on USAMO and USAJMO. It also states that the qualifying index is probably going to be around 210, so 200 to 220 would seem very reasonable.


Does anyone consider Problem-Solving Through Problems a good book? I've started reading it and it seems very poorly written.

I think it's a good book. I know the beginning chapter is ridiculously easy, but it does get better. But I'm talking about content and you're talking about how it's written- in what way do you think it's bad?


function
calculus
derivative
real analysis
real analysis unsolved

7. Let $a,b \in R$, $a<b$, and let $f$ be a differentiable real-valued function on an open subset of $R$ that contains $[a,b]$. Show that if $\gamma$ is any real number between $f'(a)$ and $f'(b)$ then there exists a number $c\in(a,b)$ such that $\gamma = f'(c)$. (Hint: Combine the mean value theorem with the intermediate value theorem for the function $(f(x_{1})-f(x_{2}))/(x_{1}-x_{2})$ on the set $\{(x_{1},x_{2})\in E^{2}\}: a\le x_{1}< x_{2}\le b\}$.

Can anyone help me with this proof?

We may assume $f'(a)>\gamma>f'(b)$. Let $g(x)=f(x)-\gamma x$. Being continuous, $g$ attains its maximum on $[a,b]$ at some point $c\in [a,b]$. Since $g'(a)>0$ and $g'(b)<0$, neither $a$ nor $b$ can be a point of maximum. Hence $c\in (a,b)$ and $g'(c)=0$.

Sorry if this is not hard enough.  :)

I meant that the problem is harder if you use the hint. Can anyone solve it with the hint?

Oh yeah, how can you prove that if $f'(a)>0, f'(b)<0$, then $\exists c\in (a,b)$ such that $f'(c) = 0$? Assume $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. This part is part of the proof.

I used two results as known: (a) a continuous function attains its maximum on a closed bounded interval; (b) if this happens at an interior point, then the derivative vanished at that point.


function
irrational number
real analysis
real analysis solved

Prove the assertion from the title of the topic.

Actually, it is not difficult at all: one may define the binary digits of the two Liouville numbers in question [i]explicitly[/i] as a function of the binary digits of the given number.

It suffices to show that every number in the interval $(0,1)$ is expressible a the sum of two Liouville numbers. Let $\alpha \in (0,1)$ be an arbitrary real number with base $2$ expansion
\[
\alpha = \sum_{i \geq 1} \alpha_i 2^{-i},  \quad \alpha_i \in \{0,1\}.
\]
We claim that $\alpha$ is the sum of two Liouville numbers in the interval $(0,1)$ that are well-approximated by subsequences of their truncated binary expansions. Start with the following trivial

[b]Observation. [/b] If $x \in (0,1), x = 0.x_1x_2\ldots _{(2)}$ is an irrational number with the property that for every integer $k$ for which there is an $l$ such that the binary sequence $x_{l+1},x_{l+2}, \ldots$ starts with a block of $kl$ zeros, then $x$ is a Liouville number.

With this observation in mind, it is sufficient to show that there exist
\[
\beta = \sum_{i \geq 1} \beta_i2^{-i}, \gamma = \sum_{i \geq 1} \gamma_i2^{-i}, \quad \beta_i,\gamma_i \in \{0,1\}
\]
such that $\alpha_i = \beta_i + \gamma_i$ for all $i$, and there exist two integer-valued functions $f,g : \mathbb{N} \to \mathbb{N}$ with the property that, for all $n$,
\[ (*)
\beta_{f(n)+1} = \beta_{f(n)+ 2} = \cdots = \beta_{nf(n)} = \gamma_{g(n) + 1} = \cdots = \gamma_{ng(n)} = 0.
\]
Our choices for $f,g$ will be $f(n) = (2n)!, g(n) = (2n+1)!$ (so that, for each $n$, the sets
\[
\{f(n)+1,\ldots,nf(n)\}, \quad \{g(m)+1,\ldots, mg(m) \} \qquad (m,n \geq 1)
\]
are pairwise disjoint).

Now, define, in addition to (*), $\beta_i = \alpha_i$ for $i \in  \bigcup_{n \geq 1} \{g(n) + 1, \ldots, ng(n)\}$ and $\gamma_i = \alpha_i$ for $i \in \bigcup_{n \geq 1} \{f(n) + 1, \ldots, nf(n)\}$ (the definitions are correct because of the disjointedness of the sets). In this way, we have defined the digits $\beta_i, \gamma_i$ on the set $S := \bigcup_{n \geq 1} (\{f(n) + 1,\ldots,nf(n)\} \cup \{g(n)+1,\ldots,ng(n)\})$ such that $\alpha_i = \beta_i + \gamma_i$ for $i \in S$. For $i \in \mathbb{N} \setminus S$, we may define $\beta_i, \gamma_i \in \{0,1\}$ arbitrarily in such a way that $\beta_i + \gamma_i = \alpha_i$. Then, $\beta_i + \gamma_i = \alpha_i$ for all $i$, and hence $\alpha = \beta + \gamma$ is the sum of two Liouville numbers, as claimed. QED

There is just one very minor problem with the above solution: we also have to ensure that infinitely many $\beta_i$ and infinitely many $\gamma_i$ are equal to $1$ (otherwise, $\beta$ or $\gamma$ will be rational). The above proof shows that every real number is the sum of two numbers that are either Liouville or rational. Actually, this is not a problem at all.  :) First, we may assume that infinitely many $\alpha_i$ are $1$s, because every non-zero real number has a binary representation with infinitely many $1$s. Second, if it happens that all but finitely many $i$ for which $a_i=1$ are contained in $S$ (otherwise we are done, because we have full freedom to define $\beta_i, \gamma_i$ on $\mathbb{N} \setminus S$), then we may suitably redefine $f$ and $g$ to get rid of this hurdle.

The assertion can also be proven using the fact that the Liouville numbers form a dense $G_{\delta}$-set.

--Vesselin

While that looks very cool, to my shame, I did not read it...  :blush:

I'll try to show the following (I guess this would be a proof based on the $G_{\delta}$-ness of the set of Liouville numbers, that vess mentioned :)):

If $A$ is the intersection of a countable collection of dense open real sets, then we can find some $x\in A$ s.t. $-x\in A$.

By Baire's Theorem, that is equivalent to $A$ being a dense $G_{\delta}$ set, which is the fact that vess mentioned.

We can assume that $A=A_1\cap A_2\cap A_3\cap\ldots$, where $A_1\supseteq A_2\supseteq A_3\supseteq\ldots$ are dense open sets. Given an interval $I$ (which doesn't contain $0$, for simplicity) in $A_1$, $-I$ must contain some element $x$ of $A_1$, because $A_1$ is dense. Now take an open interval $-I'$ around $x$, which lies in both $A_1$ and $-I$ (we can do that, since $A_1$ is open). We thus have two open intervals, $I',-I'$, both in $A_1$. From $I'$, extract a closed interval $I_1$. We thus have a compact interval $I_1$ s.t. both $I_1,-I_1$ are in $A_1$.

Now repeat the process, extracting an open interval $I$ from $I_1\cap A_2$, taking a point $x\in A_2\cap -I_1\cap -I$, then finding an open interval $I'$ s.t. both $I',-I'$ are in $A_2\cap I_1$. Now extract a compact interval $I_2$ from $I'$. We thus have $I_2,-I_2\subseteq A_2$. We do this for every $A_n$, and we obtain two descending sequences of closed intervals $I_1,I_2,\ldots$ and $-I_1,-I_2,\ldots$ with $I_j,-I_j\subseteq A_j,\ \forall j$. The intervals in a descending sequence of compact intervals have non-void intersection, and if we take $x\in I_1\cap I_2\cap\ldots$, then $x\in A$ and $-x\in A$, which is what we wanted to prove.

Now, I think it's clear how this applies to the problem we have here: there's nothing special about $-x$, meaning that we can find, by the same method, $x\in A$ s.t. $a-x\in A$ for any real $a$ we want, so that $x+a-x=a$.

Is it correct?

P.S. I think only a slight modification of this would also prove that every non-zero real is the product of two Liouville numbers, which is a problem proposed by vess in the "Analysis" section. Instead of working with $I,-I$, we work with $I$ and $\frac 1I=\{\frac 1x|x\in I\}$, where $I$ is some interval not containing $0$.


So, Gelbaum and Olmsted define the real numbers as a [b]complete ordered field[/b].  In their first counterexample, they assumed the fact that the product of two positive elements is positive.  I've tried for weeks to find a proof, but get hung up.  It's easy if one of them is rational, but if both are irrational, there's always a little irrational leftover that throws a wrench in the works.  I guess it's more fundamental than continuity and limits, so I quit bothering with those.

I wanted a more algebraic proof... now I'll take anything.  Thanks.

It really matters exactly what definitions you're working from.

Yes, indeed: it does matter. For instance, the way real numbers were defined in my Mathematics course was setting up a set of 15 axioms (first 14 were axioms of the ordered field, and 15th was the axiom of continuity).

Why am I bringing that up? Because axiom 14 was:
[i]
Compatibility of the relation $ \geq$ with multiplication[/i]: if $ x\geq 0$ and $ y\geq 0$, then $ xy\geq 0$.

To basically restate what hsiljak said: the axioms of an ordered field include compatability of the order and field relations (or otherwise an ordered field would just be a set with two completely unrelated structures on it), and "compatability" includes that the product of positive elements is positive.

Ah well, all the constructions I've seen were easily shown to be equivalent.  Here, you're given

1) The normal field axioms,
2) The trichotomy: there is a subset $ \mathcal{P}$ of so-called positive numbers such that for any $ x$, one of the following is true: $ \minus{}x\in \mathcal{P}$, $ x\equal{}0$, or $ x\in \mathcal{P}$,
3) every bounded set has an LUB.

(1) doesn't define an order relation and (3) needs that there already be an order relation in order to make sense.  From trichotomy, you can define an order relation by $ x > y$ if and only if $ x \minus{} y \in \mathcal{P}$, but without additional axioms I don't see how we can get a total order (and thus the real numbers) out of this.  ((3) only needs a partial order to make sense,* but I don't think anyone has ever used "ordered field" to mean "partially ordered field.")  For example, don't the real numbers ordered backwards (which is not an ordered field) satisfy all three conditions?

* Edit: I've changed my mind: (3) needs a notion of distance (for boundedness), and that probably requires that we have the total order.

I'm sorry!  The book defined $ \mathcal{P}$ to be closed under multiplication in the middle someplace.  Your example of ordering them backwards helped.  Thanks.


function
geometry
geometric transformation
rotation
projective geometry
calculus
calculus computations

how would one calculate the volume of revolution of a function when the centroid lies on the axis of rotation(since pappus's theorem would yeild zero since the centroid does not move, however it is obvious that a solid with volume forms)?

You have a definition problem here. If you have a two-dimensional area, parts of which lie on each side of the proposed axis of rotation, then it is highly likely that in one $360^{\circ}$ rotation, some regions of space will be swept through twice. What do you want to do about that? Double-count that volume? Or determine the intersection so you can subtract it off?

It seems safest to me to only define "solid of revolution" when the figure being rotated lies entirely on one side of the axis of rotation. The proofs of the theorem of Pappus that I am familiar with have that as an assumption, albeit possibly unstated.

Which in turn means that you would have to calculate the new centroids, of the areas which the rotation axis divides the given figure into.


abstract algebra
superior algebra
superior algebra unsolved

let P be a projective a-module. define Q the complement of P iff P+Q=F where f is free and + is direct sum. Prove that exist Q free A-module.

hint: try to prove it for A=Z6 and P=Z2 (I failed!!! :blush: )

I found that exist Q by definition of projective module, I know that P+kerf= B free for any epimorphism f:B-->P but I'm unable to prove that kerf is necessarily free.

Thanks a lot...

There's a language barrier here. I'm not sure exactly what you mean by "Prove that exist Q free A-module." If you mean that any $ Q$ such that $ P\oplus Q$ is free is itself free, this is false. Example: $ \mathbb{Z}/6\mathbb{Z}$ is isomorphic to $ \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}$.

If you mean that there is a pair $ Q,F$ such that both $ Q$ and $ F$ are free and $ P\oplus Q$ is free, there is a chance. In the $ \mathbb{Z}/6\mathbb{Z}$ and $ \mathbb{Z}/2\mathbb{Z}$ example, we can take $ Q$ free on infinitely many generators $ x_1,x_2,\cdots$, then finding $ F$ free on infinitely many generators $ (1,4x_1), (0,3x_1\plus{}4x_2),(0,3x_2\plus{}4x_3),\cdots$.

This might help you see how to construct an example in general.

Please be careful with quantifiers. They may not look like much, but their precise placement is essential to a question's meaning.

I meant exist one particular Q (obviously it cannot be true for all..),
thank you for tha answer..i'm not sure I can manage infinite genarators..but thank you very much!
could you advise me about a text to help me?
bye
Ellis


Xiaotou is a car salesman. He makes 100% profit on his cars, (Chinese people should know why) and sold 69 cars this week. The average car this week sold for $\$$16200. No car sold for less than $\$$5000 and no car sold for more than $\$$60000. The price of each car was distinct, and all the prices of the cars ended in two zeroes. What is the greatest possible median of the prices?

Out of curiosity, what's the big secret Chinese people know?
Just so this won't be spam:

[hide="Hint"]If you are stuck, use a CONSTRUCTIVE approach to this problem[/hide]

Xiaotou means thief :D meaning he steals the cars.

[hide="Hint"]Triangular numbers[/hide]

[hide="a stupid and slow method"]Since we want the median to be as big as possible, we want the prices to be 5000, 5100, 5200, ... unil the 34th number, 8300. Then we want it to be as big as possible. Using the sum formula, The last number is 27400. So the prices from the last to first are 27400, 27300, 27200, ... until the 34th number, 24100. Apply the sum formula to the 1st set of the 34 numbers and 2nd set of the 34 numbers, to get
1st set: 226100
2nd set: 875500
The sum of everything is 16200*69=1117800.
Thus the answer is 1117800-226100-875500=16200. The answer is $\boxed{\$16200}$, the average price.
I think I am right if I did not make any calculation mistakes.[/hide]


analytic geometry
calculus
ratio
trigonometry

This is a 2 dimential tug of war. There are four teams: A, S, T, C. A takes the first quadrant, S takes the second, T takes the third, C takes the fourth. Each team must get a point into their quadrant by the end of the week. The POINT startes at (0,0) and Each team MUST move the POINT to the nearest lattice point(I.E. The lattice points that are next to the POINT) and It can be in any direction. If the POINT is in the nth quadrant after a week where n is a positive integer less than 5, the team that ownes that quadrant wins. If The POINT is on a point that one of the coordinates is 0, it's a draw.  You MUST move one space parralell to one of the axis. I will make the teams now and each team must have the same number of players.
I will be Captain of team C. Three other captains may sign up now. after all the captains have signed up, the players will now sign up. We will pick and choose which players we want. I can only dub thee a captain, so you should hurry!

OKAY Choose one word accept "calculus". I chose that.

All students take calculus.

he is assigning quadrants.

ASTC is a common way to remember which sin,cos,tan things are positive.

All in Q1, Sin in Q2, Tan in Q3, Cos in Q4

[quote="Ignite168"]Umm I don't get what you're saying, are you saying make up a one word acceptance or say calculus?

And not all students take calculus, in fact, a large proportion don't[/quote]
He meant "except".

[quote="MysticTerminator"]WHAT THE HECK ARE YOU TALKING ABOUT YOU MUST BE INSANE

OHHHHHHH[/quote]

NO ARNAV, I AM NOT INSANE, I'M TALKING ABOUT....UM

ERR, UM

[quote="nutz_for2.718281828"]Basicly, You choose a word that is not in bold from the sentence "All students take [b]calculus[/b]" and I will assign you your quadrant.[/quote]

Awl of course.

How about "All students take cigarettes"? That is nearer the truth.

[u]From the factfile:[/u]
I don't take cigarettes.

Or:
"Another soccer team captain" (I made that up now)

I take soccer.

[quote="ShoeFactory"]ASTC is a common way to remember which sin,cos,tan things are positive.

All in Q1, Sin in Q2, Tan in Q3, Cos in Q4[/quote]

What do you mean by that?

Hey! It's only for making sure that if someone sees "4 wins" he doesn't think that "I'll move four paces!" A,S,T and C are good enough.

[quote="bubka"]Don't you know that? He means that all angles between $0$ and $\frac{\pi}2$ have ALL their trigonometric ratios positive, all angles between $\frac{\pi}2$ and $\pi$ have only their SINES (and COSECANTS) positive, all angles between $\pi$ and $\frac{3\pi}2$ have only their TANGENTS (and COTANGENTS) positive, and finally, angles between $\frac{3\pi}2$ and $2\pi$ have only their COSINES (and SECANTS) positive. For angles greater than $2\pi$, keep subtracting $2\pi$ till it is less than $2\pi$ and then apply these rules. This is one of the first things you learn in high school trigonometry.

Having said that, I think bringing these things in here is only showing off, nothing else. Quadrants 1, 2, 3, 4 would do fine.[/quote]

Apparently, I wasn't taught about the structure of their graphs yet (I went into that in my spare time).  I just didn't understand what you were saying.

EDIT: I think I just thought of an interesting way to expand on this, but it'd probably be a bit complicated.  Try to "pull" a point on a plane to either (10,10), (10,-10), (-10,-10), or (-10,10), depending on your team.  But you can only pull the point up to one unit away from where it was before, and it must always have rational $x$ and $y$ values that have 3 or less decimal places.  So you could start by moving it from $(0,0)$ to $(0.707,0.707)$.  Just an idea.

That was the idea of the game I proposed long back. These will be corners. But let mbabbitt2003 have his own rules. It is his thread. However in my game you can pull it either along the x-axis or along the y-axis for one unit. That saves a lot of complications.

That's what I'm talking about, bubka! Yu can move it exactly 1 unit parallel to the x or y axis. It has to be parraleel to at least one of the axis.

Do us all a favour by getting the game started.

We need one more coach!

[quote="nutz_for2.718281828"]

Okay, The captains are as thus.

|Ignite 168 |First quadrant
|bubka |Second quadrant
| |Third quadrant
|nutz_for2.718281828 |Fourth Quadrant

We need one more captain![/quote]

Also, signups can begin for being on certain teams.

This is taking too long to start, I QUIT.

I have three points to make.
1. GRRR!
2. Why isn't anybody signing up???!!!
3. We need people to sign up!

Your points are invalid.

Here's a game: If you're the last person to post when the topic reaches page 2, you win!

COME ON!
Ok. This is going to be free for all. You can choose which team you're on. I'll be a quadrant 4 guy.

omfg Im captain
edit:wtf this is a 200 year old thread?
editedit:what about ccy's 2d tow


Asymptote
real analysis
real analysis unsolved

Help me to find the asymptotes of $ x^2y\equal{}(2y\plus{}x)(y\minus{}2x)$. Pls give the method to find asymptotes for any curve.

Hello
Kindly see this-
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=247408
Thank u.


superior algebra
superior algebra unsolved

Suppose d is a negative integer. Show that there does not exist a cyclic Galois extension $ \mathbb{Q}\subset E$ with $ [E: \mathbb{Q}]\equal{}4$ and $ \sqrt{d}\in E$.

More generally: $ L : \equal{} K(\sqrt {d})$ can be embedded in a $ C_4$-extension iff $ d$ is a sum of two squares in $ K$.

Proof: $ G_K \to \mathbf{Z}/2$ can be factored through $ \mathbf{Z}/4$ iff $ H^1(K,\mathbf{Z}/4) \to H^1(K,\mathbf{Z}/2) \to H^2(K,\mathbf{Z}/2)$. But the boundary map is the Bockstein homomorphism, given by the cup product with oneself. $ \phi \cup \phi \equal{} \minus{} 1 \cup \phi$, so $ \phi$ is a sum of two squares.

(Here, you can just look what complex conjugation does.)


function
geometry
quadratics
symmetry
geometric transformation
reflection
integration

Find the volume of the wedge cut from the square column abs.(x + y) + abs.(x - y) <= 1 by the planes z = 0
and 3x + 3y + z = 3.

*abs. stands for absolute value.

$ |x\,+\,y|\,+\,|x\,-\,y|$ transitions across $ z\,=\,1$ when

\begin{eqnarray*}
|x\,+\,y|\,+\,|x\,-\,y|\,&=&\,1 \\
(|x\,+\,y|\,+\,|x\,-\,y|)^2\,&=&\,1^2 \hspace{1cm} \mbox{(both sides positive)}\\
(x\,+\,y)^2\,+\,2|x\,+\,y\parallel{}x\,-\,y|\,+\,(x\,-\,y)^2\,&=&\,1 \\
x^2\,+\,2xy\,+\,y^2\,+\,2|x^2\,-\,y^2|\,+\,x^2\,-\,2xy\,+\,y^2\,&=&\,1 \\
2x^2\,+\,2y^2\,+\,2|x^2\,-\,y^2|\,&=&\,1
\end{eqnarray*}

This amounts to

$ 4x^2\,=\,1 \hspace{1cm} \mbox{when } y\,\le\,x$
$ 4y^2\,=\,1 \hspace{1cm} \mbox{when } x\,\le\,y$

and we obtain the solutions

$ x\,=\,\pm0.5 \hspace{1cm} \mbox{when } y\,\le\,x$
$ y\,=\,\pm0.5 \hspace{1cm} \mbox{when } x\,\le\,y$

defining not a square but a horizontal strip of width 1 that turns down at [i]x[/i] = [i]y[/i]. The test point (0, 0) shows that the function is less than 1 inside the strip.

The plane $ 3x\,+\,3y\,+\,z\,=\,3$ can be rewritten $ z\,=\,3\,-\,3x\,-\,3y$, which transitions at 0 when $ x\,+\,y\,=\,1$, at the corner of the strip defined above.

The area of all sections cut by $ z\,=\,0$ and the plane above is infinite, and [b]therefore the problem has no solution.[/b]

If, however, the plane were $ 3x\,+\,3y\,+\,z\,=\,\frac{3}{2}$ and cut the axes at the half-markings at $ x\,+\,y\,=\,\frac{1}{2}$, it would give us a triangle below $ z\,=\,0$ of base area $ \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\,=\,\frac{1}{8}$ and height $ \frac{3}{2}$, so that we would obtain the answer

$ \frac{1}{2}\cdot\frac{1}{8}\cdot\frac{3}{2}\,=\,\frac{3}{32}.$

Correction: the base area would be $ \frac{1}{3}\cdot \frac{1}{8} \cdot \frac{3}{2}\,\equal{}\,\frac{1}{16}.$

Scott: squaring things was a bad idea. There's nothing quadratic about this problem. But you did arrive at the correct intermediate conclusion - you just misinterpreted the symmetry of the problem.

Consider one narrow special case, a 1/8 of the plane wedge: $ 0\le y\le x.$ On that wedge, we can write $ |x \plus{} y| \plus{} |x \minus{} y| \equal{} (x \plus{} y) \plus{} (x \minus{} y) \equal{} 2x.$ For that to be $ \le 1,$ we must have $ x\le\frac12.$

From there employ symmetry: the base is symmetric under the reflections $ x\mapsto \minus{} x, y\mapsto \minus{} y,$ and $ (x,y)\mapsto (y,x).$ Taking advantage of all of those, we see that the base is indeed a square and can be written as $ \minus{} \frac12\le x\le \frac12, \minus{} \frac12\le y\le \frac12.$

And the base area is 1.

We can also see that $ z \equal{} 3 \minus{} 3x \minus{} 3y\ge0$ on that square so the region described is well-defined. (It is also bounded and has a finite volume.) The volume can be computed as
\[ \int_{ \minus{} \frac12}^{\frac12}\int_{ \minus{} \frac12}^{\frac12} 3 \minus{} 3x \minus{} 3y\,dx\,dy
\]
But that in turn can be greatly simplified. $ \minus{} 3x$ integrates to zero over the square by its odd-in-$ x$ symmetry, and $ \minus{} 3y$ integrates to zero by its odd-in-$ y$ symmetry. So the only term to contribute anything nonzero to the integral is $ 3.$ Put another way, the [i]average[/i] height of the plane over the base is $ 3,$ which makes the volume $ 3.$

Thank you, Kent; I just realized where I made my first error.

I tried to deduce

$ 4x^2\,\equal{}\,1 \hspace{1cm} \mbox{when } y\,\le\,x$
$ 4y^2\,\equal{}\,1 \hspace{1cm} \mbox{when } x\,\le\,y$

from

$ 2x^2\,\plus{}\,2y^2\,\plus{}\,2|x^2\,\minus{}\,y^2|\,\equal{}\,1$

when I could only have deduced

$ 4x^2\,\equal{}\,1 \hspace{1cm} \mbox{when } y^2\,\le\,x^2$
$ 4y^2\,\equal{}\,1 \hspace{1cm} \mbox{when } x^2\,\le\,y^2.$

The faulty deduction led to an incorrect answer. It was a careless mistake, and I apologize.


Wow... i couldnt believe that i got number 126 (p. 101) of vol 2 so easily, so I wanted to make sure what i did was valid.
THe function was f(x+t) = f(x) + f(t) + 2 :sqrt: f(x)f(t)  .

I set x = t, and then got f(2x) = 2f(x) + 2 :sqrt: f(x)^2
which then has two values.... f(2x) = 2f(x)-2(f(x) and
f(2x) = 2f(x) + 2(f(x) = 4f(x). Fromt he first equation f(2x)= 0.
Then from the second f(x) = 0/4 = 0.
THat is the answer (surprisingly for me), but is that a valid way of doing this... just seemed too easy...?

it seems right to me, except that i can't really say that with full confidence since I only have book 1...what chapter was it in?

When you have the equation

f(2x) = 2f(x) + 2sqrt(f(x)^2)

all you can deduce is that for any given x, either

f(2x) = 2f(x) - 2f(x) (when f(x) <= 0)

or

f(2x) = 2f(x) + 2f(x) (when f(x) >= 0)

The only way both are true is for f(x) = 0.

Unfortunately, I think your solution assumes both are true for each x.  Keep at it.  You may be able to turn this into a solution yet (or find another attack).


calculus
integration
logarithms
calculus computations

Show that

$ (1) \qquad \int_{0}^{\infty} \log (1 \plus{} e^{\minus{}x} ) \, dx \equal{} \frac{1}{2} \zeta(2)$ and

$ (2) \qquad \int_{0}^{\infty} \log^2 (1 \plus{} e^{\minus{}x} ) \, dx \equal{} \frac{1}{4} \zeta(3)$,

where $ \log^2 x \equal{} (\log x)^2$.

$ \int_{0}^{\infty }{\int_{0}^{e^{\minus{}x}}{\frac{dt\,dx}{1\plus{}t}}}\equal{}\int_{0}^{1}{\int_{0}^{\minus{}\ln t}{\frac{dx\,dt}{1\plus{}t}}}\equal{}\minus{}\int_{0}^{1}{\frac{\ln t}{1\plus{}t}\,dt},$ last integral is well known.

Sorry, but I havent learned this integral yet. Please explain this to me, thanks.

$ \int\limits_{0}^{1}{\frac{\ln y}{1\plus{}y}\,dy}$

I just get this...

$ \ln y \equal{} \ln \left( y \plus{} 1 \minus{} 1 \right) \equal{} \sum\limits_{k \equal{} 1}^{n}{\left( \minus{} 1 \right)\frac {{{\left( y \plus{} 1 \right)}^{k}}}{k}}\Rightarrow \minus{} \frac {\ln y}{y \plus{} 1} \equal{} \sum\limits_{k \equal{} 1}^{n}{\frac {{{\left( y \plus{} 1 \right)}^{k \minus{} 1}}}{k}}$

$ \Rightarrow \minus{} \int\limits_{0}^{1}{\frac {\ln y}{1 \plus{} y}\,dy} \equal{} \int\limits_{0}^{1}{\sum\limits_{k \equal{} 1}^{n}{\frac {{{\left( y \plus{} 1 \right)}^{k \minus{} 1}}}{k}}.dy} \equal{} \sum\limits_{k \equal{} 1}^{n}{\left[ \frac {{{\left( y \plus{} 1 \right)}^{k}}}{{{k}^{2}}} \right]_{0}^{1}} \equal{} \sum\limits_{k \equal{} 1}^{n}{\frac {{{2}^{k}} \minus{} 1}{{{k}^{2}}}}$

that's wrong, i'm talkin' about power series, not summation, those are finite, these ones are infinite.

check http://en.wikipedia.org/wiki/Power_series

If you hint to me integrating by part first, maybe it will be more easier.


Prove the following inequality:
$\sum_{n=1}^{k}{\sum_{m=1}^{k}{\frac{a_{m}\cdot{a_{n}}}{m+n}}}\ge{0}$,where $a_{i}$ $(i=1,2...,k)$ are real numbers.

Write \[
f(x) = \sum\limits_{n = 1}^k {\sum\limits_{m = 1}^k {\frac{{a_m a_n }}{{m + n}}x^{m + n} } }
\]
Then,
\[
f'(x) = x\left( {a_1  + a_2 x + ... + a_k x^{k - 1} } \right)^2
\]
So f(x) is increasing for x>0 (equality holds when all a_i equals 0)
f(0) = 0
so f(1) >= 0


inequalities
function
logarithms
induction
inequalities proposed

Let $ a,b,c$ be positive real numbers. Prove that

$ \ \ \ \ \ (1)  \ \ (a\plus{}\frac 1{b})(b\plus{}\frac 1{c})(c\plus{}\frac 1{a}) \le (a\plus{}\frac 1{a})(b\plus{}\frac 1{b})(c\plus{}\frac 1{c})$;

 $ \ \ \ \ \ (2 \ \  \frac 1{a\plus{}\frac 1{b}}\plus{}\frac 1{b\plus{}\frac 1{c}}\plus{}\frac 1{c\plus{}\frac 1{a}} \ge \frac 1{a\plus{}\frac 1{a}}\plus{}\frac 1{b\plus{}\frac 1{b}}\plus{}\frac 1{c\plus{}\frac 1{c}}$.

(1)<=>
$ abc\plus{} \frac{1}{a}\plus{} \frac{1}{b}\plus{} \frac{1}{c}\plus{}a\plus{}b\plus{}c\plus{} \frac{1}{abc} \le abc\plus{} \frac{a}{bc}\plus{} \frac{b}{ac}\plus{} \frac{c}{ab}\plus{}\frac{ab}{c}\plus{} \frac{bc}{a}\plus{} \frac{ca}{b}\plus{} \frac{1}{abc}$
true because
$ \frac{a}{bc}\plus{} \frac{b}{ca}\plus{} \frac{c}{ab} \ge \frac{1}{a}\plus{} \frac{1}{b}\plus{} \frac{1}{c}$
$ \frac{ab}{c}\plus{} \frac{bc}{a}\plus{} \frac{ca}{b} \ge a\plus{}b\plus{}c$

(1) Multiplying both sides by $ abc$, the inequality becomes \[ (ab\plus{}1)(bc\plus{}1)(ca\plus{}1)\le(a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1)\] or \[ a^2b^2c^2\plus{}a^2bc\plus{}ab^2c\plus{}abc^2\plus{}ab\plus{}bc\plus{}ca\plus{}1\le a^2b^2c^2\plus{}a^2b^2\plus{}b^2c^2\plus{}c^2a^2\plus{}a^2\plus{}b^2\plus{}c^2\plus{}1.\] So it suffices to prove that $ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\ge a^2bc\plus{}ab^2c\plus{}abc^2$ and $ a^2\plus{}b^2\plus{}c^2\ge ab\plus{}bc\plus{}ca$ which are equivalent to $ (ab\minus{}bc)^2\plus{}(bc\minus{}ca)^2\plus{}(ca\minus{}ab)^2\ge0$ and $ (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\ge0$ respectively. Q.E.D.

The easy inequality (1) holds also for any positive $ a,b,c,d$:

$ \ \ \ \ \ \ \ (a \plus{} \frac 1{b})(b \plus{} \frac 1{c})(c \plus{} \frac 1{d})(d \plus{} \frac 1{a}) \le (a \plus{} \frac 1{a})(b \plus{} \frac 1{b})(c \plus{} \frac 1{c})(d \plus{} \frac 1{d})$.

Edit. (I knew something had to be wrong)

Inequalities Master, your last inequality is wrong.

[quote="nayel"]We have the following inequality
\[ \left(a + \frac 1a\right)\left(b + \frac 1b\right)\ge \left(a + \frac 1b\right)\left(b + \frac 1a\right),\]
which is equivalent to $ a/b + b/a\ge 2$, true by AM-GM.

Using the above inequality we have
\[ \left(a + \frac 1a\right)\left(b + \frac 1b\right)\ge \left(a + \frac 1b\right)\left(b + \frac 1a\right) \\
\left(b + \frac 1b\right)\left(c + \frac 1c\right)\ge \left(b + \frac 1c\right)\left(c + \frac 1b\right) \\
\left(c + \frac 1c\right)\left(d + \frac 1d\right)\ge \left(c + \frac 1d\right)\left(d + \frac 1c\right) \\
\left(d + \frac 1d\right)\left(a + \frac 1a\right)\ge \left(d + \frac 1a\right)\left(a + \frac 1d\right).\]
Multiplying, we get
\begin{align*}\prod_{cyc}\left(a + \frac 1a\right)^2 & \ge\frac {(ab + 1)^2(bc + 1)^2(cd + 1)^2(da + 1)^2}{a^2b^2c^2d^2} \\
& = \prod_{cyc}\left(\frac {ab + 1}{b}\right)^2 = \prod_{cyc}\left(a + \frac 1b\right)^2,\end{align*}
which was what we wanted.

Remark: The following inequality holds for any numbers $ a_1,a_2,\dots,a_n > 0$ and $ a_{n + 1} = a_1$:
\[ \prod_{i = 1}^n\left(a_i + \frac {1}{a_i}\right)\ge\prod_{i = 1}^n\left(a_i + \frac {1}{a_{i + 1}}\right)\]
[/quote]

Indeed, this is also my original proof for $ n$ numbers. :lol:

What about inequality (2) ? This is also true for $ n$ numbers. Can someone to prove it for $ n=3$?

Here is my proof for $ n\equal{}3$. For $ n$ variables, I used induction.

Assume that $ c\equal{}\max\{a,b,c\}$, we have
$ (b\plus{}\frac{1}{a})(c\plus{}\frac{1}{c}) \minus{}(b\plus{}\frac{1}{c})(c\plus{}\frac{1}{a})\equal{}\frac{(a\minus{}c)(b\minus{}c)}{ac} \ge 0,$
hence
$ \frac{1}{b\plus{}\frac{1}{c}} \plus{}\frac{1}{c\plus{}\frac{1}{a}} \ge \frac{1}{b\plus{}\frac{1}{a}}\plus{}\frac{1}{c\plus{}\frac{1}{c}}.$
So, it suffices to prove that
$ \frac{1}{a\plus{}\frac{1}{b}}\plus{}\frac{1}{b\plus{}\frac{1}{a}} \ge \frac{1}{a\plus{}\frac{1}{a}}\plus{}\frac{1}{b\plus{}\frac{1}{b}},$
or
$ a\plus{}b \ge \frac{a(ab\plus{}1)}{a^2\plus{}1}\plus{}\frac{b(ab\plus{}1)}{b^2\plus{}1},$
that is
$ \frac{(a\minus{}b)(a^2\minus{}b^2)}{(a^2\plus{}1)(b^2\plus{}1)}\ge 0,$
which is true.
This completes our proof.

[quote="Vasc"]Let $ a,b,c$ be positive real numbers. Prove that

$ \ \ \ \ \ (1) \ \ (a \plus{} \frac 1{b})(b \plus{} \frac 1{c})(c \plus{} \frac 1{a}) \le (a \plus{} \frac 1{a})(b \plus{} \frac 1{b})(c \plus{} \frac 1{c})$;

 $ \ \ \ \ \ (2 \ \ \frac 1{a \plus{} \frac 1{b}} \plus{} \frac 1{b \plus{} \frac 1{c}} \plus{} \frac 1{c \plus{} \frac 1{a}} \ge \frac 1{a \plus{} \frac 1{a}} \plus{} \frac 1{b \plus{} \frac 1{b}} \plus{} \frac 1{c \plus{} \frac 1{c}}$.[/quote]

We have
\[ (1) (a \plus{} \frac {k}{b})(b \plus{} \frac {k}{c})(c \plus{} \frac {k}{a}) \le (a \plus{} \frac {k}{a})(b \plus{} \frac {k}{b})(c \plus{} \frac {k}{c}) ;\]

\[ (2) \frac {1}{a \plus{} \frac {k}{b}} \plus{}  \frac {1}{b \plus{} \frac {k}{c}} \plus{} \frac {1}{c \plus{} \frac {k}{a}} \ge\frac {1}{a \plus{} \frac {k}{a}} \plus{} \frac {1}{b \plus{} \frac {k}{b}} \plus{} \frac {1}{c \plus{} \frac {k}{c}} .\]

[quote="fjwxcsl"] We have
\[ (1) (a \plus{} \frac {k}{b})(b \plus{} \frac {k}{c})(c \plus{} \frac {k}{a}) \le (a \plus{} \frac {k}{a})(b \plus{} \frac {k}{b})(c \plus{} \frac {k}{c}) ;\]

\[ (2) \frac {1}{a \plus{} \frac {k}{b}} \plus{} \frac {1}{b \plus{} \frac {k}{c}} \plus{} \frac {1}{c \plus{} \frac {k}{a}} \ge\frac {1}{a \plus{} \frac {k}{a}} \plus{} \frac {1}{b \plus{} \frac {k}{b}} \plus{} \frac {1}{c \plus{} \frac {k}{c}} .\]
[/quote]
Indeed Fjwxcls, if $ k>0$, then the inequality holds for $ n$ numbers and for any convex function. :lol:

[quote="Johan Gunardi"][quote="Vasc"]Very nice your solution, Can_hang2007.  :lol:
Actually, the following statement holds:

Let $ f$ be a convex function on $ (0,\infty)$. If $ a_1,a_2,...,a_n$ are positive real numbers, then

$ f(a_1 \plus{} \frac 1{a_2}) \plus{} f(a_2 \plus{} \frac 1{a_3}) \plus{} ... \plus{} f(a_n \plus{} \frac 1{a_1})\ge f(a_1 \plus{} \frac 1{a_1}) \plus{} f(a_2 \plus{} \frac 1{a_2}) \plus{} f(a_n \plus{} \frac 1{a_n})$.

For $ f(x) \equal{} \minus{} \ln x$ and $ f(x) \equal{} \frac 1{x}$, we get the original proposed inequalities, respectively.[/quote]how to prove it?[/quote]
By induction, considering $ a_{n\plus{}1}\equal{}\max\{a_1,a_2,...,a_n, a_{n\plus{}1}\}$ and using the following known lemma:

[b]Lemma.[/b] Let $ f$ be a convex function on $ I$. If $ x,y,z,w\in I$ such that  $ x\plus{}y\equal{}z\plus{}w$  and $ x\equal{}\max \{x,y,z,w\}$, then $ f(x)\plus{}f(y) \ge f(z)\plus{}f(w)$.

That lemma makes everything's easy now. Is it a new result you've just found or is it an old, well-known result?

Does the general form holds: if $ (a_1,a_2,\ldots,a_n)\succ(b_1,b_2,\ldots,b_n)$ then $ f(a_1)\plus{}f(a_2)\plus{}\ldots\plus{}f(a_n)\ge f(b_1)\plus{}f(b_2)\plus{}\ldots\plus{}f(b_n)$?

can someone  help me .I found it ina book ,it says it is from Vasc

a,b,c are positive. Prove that ∑a*(a^2+3bc)^1/2>=2(ab+bc+ca)

[quote="Vasc"]Very nice your solution, Can_hang2007.  :lol:
Actually, the following statement holds:

Let $ f$ be a convex function on $ (0,\infty)$. If $ a_1,a_2,...,a_n$ are positive real numbers, then

$ f(a_1 \plus{} \frac 1{a_2}) \plus{} f(a_2 \plus{} \frac 1{a_3}) \plus{} ... \plus{} f(a_n \plus{} \frac 1{a_1})\ge f(a_1 \plus{} \frac 1{a_1}) \plus{} f(a_2 \plus{} \frac 1{a_2}) \plus{} f(a_n \plus{} \frac 1{a_n})$.

For $ f(x) \equal{} \minus{} \ln x$ and $ f(x) \equal{} \frac 1{x}$, we get the original proposed inequalities, respectively.[/quote]
Dear Vasc, this is a very nice result. I also used induction to prove it, what about your proof? Thanks

[quote="Johan Gunardi"]That lemma makes everything's easy now. Is it a new result you've just found or is it an old, well-known result?

Does the general form holds: if $ (a_1,a_2,\ldots,a_n)\succ(b_1,b_2,\ldots,b_n)$ then $ f(a_1) \plus{} f(a_2) \plus{} \ldots \plus{} f(a_n)\ge f(b_1) \plus{} f(b_2) \plus{} \ldots \plus{} f(b_n)$?[/quote]
The inequality above with convex function and $ k\equal{}1$ (or $ k>0$) is a new result for me. I don't post on this forum old known inequalities.
I don't think we can prove this inequality without induction, using for example the Karamata inequality for n variables:  if $ (a_1,a_2,\ldots,a_n)\succ(b_1,b_2,\ldots,b_n)$ then $ f(a_1) \plus{} f(a_2) \plus{} \ldots \plus{} f(a_n)\ge f(b_1) \plus{} f(b_2) \plus{} \ldots \plus{} f(b_n)$.


You should all be familiar with the popular telescoping sum:
$\sum_{i=1}^{n} \frac{1}{i(i+1)}=\frac{n}{n+1}$

You can also find with partial fraction decompositions that:
$\sum_{i=1}^{n} \frac{1}{i(i+1)(i+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$
$\sum_{i=1}^{n} \frac{1}{i(i+1)(i+2)(i+3)}=\frac{n(n^2+6n+11)}{18(n+1)(n+2)(n+3)}$
$\sum_{i=1}^{n} \frac{1}{i(i+1)(i+2)(i+3)(i+4)}=\frac{n(n+5)(n^2+5n+10)}{96(n+1)(n+2)(n+3)(n+4)}$
and so forth.

What I want to know is, is there a general solution for:
$\sum_{i=1}^{n} \frac{1}{i(i+1)(i+2)...(i+m)}$ for any nonnegative integer m?

Here's a hint: Compute
\[ \frac{1}{4} - \frac{n(n+3)}{4(n+1)(n+2)} \]
and
\[ \frac{1}{18} - \frac{n(n^2+6n+11)}{18(n+1)(n+2)(n+3)}. \]

If you're using telescoping sums, you should be getting these expressions directly.

oh... i see it now. thanks.  :)

so the general answer would be something like this:

[hide]$\sum_{i=1}^{n} \frac{1}{i(i+1)(i+2)...(i+m)}=\frac{1}{m}(\frac{1}{m!}-\frac{n!}{(n+m)!})$[/hide]


function
analytic geometry
induction
USAMTS
algebra
domain
AMC

Bleh, failed. Somehow I got a 1 for #4 with the solution:

Let m=n=f(m)+f(n)

So f(2(m+n))=2(f(m)+f(n))

If m=n=0, then f(0)=4f(0) => f(0)=0.

Let n=0. f(2m)=2f(m).

This is a special case of Jenson, which is 2f(x+y)=f(2x)+f(2y) has solution set of linear functions.

Since y=0 in this case, we need only examine linear functions.

Then I solved for the function (and included f(x)=-x...

How did I get a 1 for that? I can't find anything wrong with it in the graing criteria.

Got a 5 for 1,2. Yay?

and #5 I got 2 for a completely wrong solution.

I got 4/2/2/3/5=16

#1: I'm such an idiot... For some reason I thought it only worked for powers of 2, not even numbers  :huh:
#2: I somehow got 13 instead of 9. I was expecting 0, so I'm pretty happy.
#3: Wrote a program and got a slightly correct answer, but again I was expecting 0, so whatever.
#4: I was so sure I'd get a 4 on this... Lost points for not rigorously proving $ f(0)\equal{}0$. Interestingly, I don't think they took points off for including $ f(x)\equal{}\minus{}x$.
#5: Coordinate bashed :D

I got 5/2/0/2/3=12. :(

This gives me a total year score of 55 (honorable mention) :( (with 11 on Round 1, 20 on Round 2, 12 on Round 3, 12 on Round 4) I feel so stupid cuz I omitted #3 on Round 4 for no reason >.<

And I went careless on #2...got 12 instead of 9 due to a stupid mistake in my graph theory process!!!! :mad:

The graders also said my #4 was unrigorous...

Darn it. I should not have completely given up on #3 for this round (which I left blank)

pppppppppppppphhhhhhhhhhhhhhhhhhhhaiiiiiiilll :mad:  :mad:

I made silver too!
I got a much higher score than I expected. This round I got 5/5/5/3/5=23

[quote="BOGTRO"]
Let m=n=f(m)+f(n)

So f(2(m+n))=2(f(m)+f(n))
[/quote]
Um, sorry what did you just do here? I'm assuming you mean something like "let $ m\equal{}n\equal{}f(x)\plus{}f(y)$ for some $ x,y\in Z_{\ge 0}$, but plugging that in gives you
$ f(2f(f(x)\plus{}f(y))) \equal{} 2(f(x)\plus{}f(y))$
which is very different from what you had.

5/5/5/4/5 = 24

I'm going to try to figure out #4 and what went wrong, but this is absolutely great!!!!

wow my grader was pretty lenient on my number 4. I basically had no clue how to solve it, and I jotted down something last second (something like letting m = 0 and plugging some 1 or 2 other random values, and absolutely nothing else), and just said it's clear that f(x)=x and f(x)=k both work. Obviously the second statement is wrong (like I said i I did this last second), but I somehow got a 2...was expecting a 0..... Apparently you shouldn't ever leave a problem blank on USAMTS. Also, I somehow I got 5's on the other 4 even though my solution to 3 was pretty disorganized.

ugghh, lost 1 point in number 1 cuz I said "by simple induction" $ a_n\equal{}r^{n\minus{}1}a_1$.  I thought it was pretty self-evident and felt like showing the induction would kill the elegant concision in my proof.

[quote="JAM"][quote="BOGTRO"]
Let m=n=f(m)+f(n)

So f(2(m+n))=2(f(m)+f(n))
[/quote]
Um, sorry what did you just do here? I'm assuming you mean something like "let $ m \equal{} n \equal{} f(x) \plus{} f(y)$ for some $ x,y\in Z_{\ge 0}$, but plugging that in gives you
$ f(2f(f(x) \plus{} f(y))) \equal{} 2(f(x) \plus{} f(y))$
which is very different from what you had.[/quote]
I have seen equations like the second one, but accompanied by a lot more explanation.

Let $ m\equal{}n$, then $ f(f(m)\plus{}f(n)) \equal{} m\plus{}n$ becomes both $ f(2f(m)) \equal{} 2m$ and $ f(2f(n))\equal{}2n$. Summing them gives $ f(2f(m))\plus{}f(2f(n)) \equal{} 2m \plus{}2n$. Applying $ f$ to both sides of the equation gives $ f(f(2f(m))\plus{}f(2f(n))) \equal{} f(2m\plus{}2n)$, and the properties of $ f$ simplify the left side to give $ 2f(m) \plus{} 2f(n) \equal{} f(2m\plus{}2n)$.

Unfortunately for BOGTRO, leaving out explanations was the main reason people lost points on Problem 4. Two points off.

[quote="BOGTRO"]This is a special case of Jenson, which is 2f(x+y)=f(2x)+f(2y) has solution set of linear functions.
[/quote]
I haven't seen that particular theorem by Jenson (nor by Jensen). Could you give a source? I gave credit for using the theorem when people stated the heart of it as BOGTRO did, but I am curious to know the entire theorem.

Unfortunately, the location of the 2's in BOGTRO's f(2(m+n))=2(f(m)+f(n)) do not match up with the location of the 2's in Jenson. I don't think he can validly use Jenson here. Another point off.

[quote="BOGTRO"]Then I solved for the function (and included f(x)=-x...
[/quote]
If BOGTRO did not rule out the $ f(x) \equal{} \minus{}x$ case at the end, that would be another point off.

[quote]How did I get a 1 for that? I can't find anything wrong with it in the grading criteria.
[/quote]
The rubric lists the points taken off for common errors. Unique errors are judged on a case by case basis.

BOGTRO's proof as stated here skips too many steps. Intelligent people, such as JAM, cannot follow it.

Erin Schram
USAMTS Grader

[quote="MellowMelon"][quote="BOGTRO"]f(2m)=2f(m).

This is a special case of Jenson, which is 2f(x+y)=f(2x)+f(2y) has solution set of linear functions.[/quote]
Only if your function is on the reals and continuous. Consider for example $ f(2^k) \equal{} 2^k$ and $ 0$ everywhere else, which satisfies $ f(2m) \equal{} 2f(m)$ but is not linear. A real-valued function taking reals that isn't linear is $ f(x) \equal{} x$ on rational $ x$ with denominator a power of $ 2$ in reduced form, and $ f(x) \equal{} 0$ everywhere else.

The temptation to be tricky on functional equations can be strong, but you really want to stick to basic algebra and variable substitutions, unless you have all sorts of things known like your function is continuous/differentiable/whatever.[/quote]
Yes, that is the case for the people who quoted the Cauchy Functional Equation. They applied a theorem meant for real, continuous functions to a discrete function. I don't know whether Jenson requires that or not.

Erin Schram
USAMTS grader

[quote="JAM"][quote="BOGTRO"]
Let m=n=f(m)+f(n)

So f(2(m+n))=2(f(m)+f(n))
[/quote]
Um, sorry what did you just do here? I'm assuming you mean something like "let $ m \equal{} n \equal{} f(x) \plus{} f(y)$ for some $ x,y\in Z_{\ge 0}$, but plugging that in gives you
$ f(2f(f(x) \plus{} f(y))) \equal{} 2(f(x) \plus{} f(y))$
which is very different from what you had.[/quote]

and f(f(m)+f(n)) is m+n, by the problem statement.

EDIT: [url]http://eqworld.ipmnet.ru/en/solutions/fe/fe3103.pdf[/url] y=0.

Your first point is trivial to show. Not nessacary (at all).

Your third point is valid. This is why I think I should get a 4.

What that page you cites says if that $ 2f(x\plus{}y) \equal{} f(2x) \plus{} f(2y)$ holds for some unspecified domain [b]for all $ x,y$ in the domain[/b] (likely the reals, but this detracts from the source's reliability; you HAVE to have the domain), then the function is linear. Actually this isn't even true on reals; you need $ f$ continuous, otherwise any solution of the Cauchy equation works and those are not all linear.

Furthermore, $ f(2m) \equal{} 2f(m)$ isn't a "special case" with $ y \equal{} 0$. That's saying that

$ f(2m) \equal{} 2f(m) \Rightarrow 2f(m\plus{}n) \equal{} f(2m) \plus{} f(2n)$

This implication is neither obvious nor true - it goes the other way. Nevermind the whole issue of the domain.

I already gave a counterexample to $ f(2m) \equal{} 2f(m)$ implying $ f$ linear, so I wouldn't try too hard to save this.

[quote="BOGTRO"][quote="JAM"][quote="BOGTRO"]
Let m=n=f(m)+f(n)

So f(2(m+n))=2(f(m)+f(n))
[/quote]
Um, sorry what did you just do here? I'm assuming you mean something like "let $ m \equal{} n \equal{} f(x) \plus{} f(y)$ for some $ x,y\in Z_{\ge 0}$, but plugging that in gives you
$ f(2f(f(x) \plus{} f(y))) \equal{} 2(f(x) \plus{} f(y))$
which is very different from what you had.[/quote]

and f(f(m)+f(n)) is m+n, by the problem statement.
[/quote]

Ok, I'll buy that, but that doesn't mean you can get away without showing your reasoning. Just because you understand your argument doesn't mean other people (i.e. the graders) will. When in doubt, give more detail not less (you will never lose points for explaining too much).

[quote="BOGTRO"]
EDIT: [url]http://eqworld.ipmnet.ru/en/solutions/fe/fe3103.pdf[/url] y=0.
[/quote]

Make sure you completely understand the statement of a theorem before you try blindly applying it. First off, I'm going to guess that thing also assumes that $ x$ and $ y$ range over the reals (not just the nonnegative integers), and it might also assume some additional constraint like $ f$ continuous, neither of which you have.

Also be careful about your quantifiers. A statement like this clearly means "if [b]for all[/b] $ x,y$ ($ \in \mathbb R$, probably) $ 2f(x \plus{} y) \equal{} f(2x) \plus{} f(2y)$ then..."

Here it does [b]not[/b] hold for all $ x$ and $ y$. You've only shown it for $ y \equal{} 0$, so you can't even use this.

A good general tip for this sort of thing: before you reach for some obscure theorem, you should just try to prove it on your own.

To finish your solution,
[hide]
Using $ f(2n) \equal{} 2f(n)$ and $ f(2(m \plus{} n)) \equal{} 2(f(m) \plus{} f(n))$ it's pretty easy to get
$ 2f(m \plus{} n) \equal{} 2(f(m) \plus{} f(n)) \Rightarrow f(m \plus{} n) \equal{} f(m) \plus{} f(n)$

You've already shown $ f(0) \equal{} 0$. Let $ c \equal{} f(1)$. A simple induction now shows $ f(n) \equal{} cn$ (if $ f(k) \equal{} ck$, $ f(k \plus{} 1) \equal{} f(k) \plus{} f(1) \equal{} (k \plus{} 1)c$). From here it's pretty easy to show $ c \equal{} 1$.

(Note: I did [b]not[/b] intend this to be a complete solution, your solution should [b]not[/b] look like this. See if you can fill in the details.)[/hide]

EDIT: Ok, MellowMelon kind of beat there...

[quote="Yongyi781"]Yes, I made silver! My total score is 81 and my cumulative score for the first three rounds was 56.[/quote]
Yeah me too!! collective silver WOOT!!

Yeah, third person I've seen so far on this thread w/ a perfect!

I almost lost points because I went straight from

$ f(0)\equal{}4f(0)$ to $ f(0)\equal{}0$

and $ f(0)\equal{}0$ to $ f(f(k))\equal{}k$ when the only transition I made was saying,

Plugging this [$ f(0)\equal{}0$] into the given equation, we get $ f(f(k))\equal{}k$.

85 for a Silver!!!!! yeah

Sweet I got a 5/0/0/0/0!
That's because I spent literally less than 10 minutes on this round.  I just did #1 and didn't look at the other four problems.  Why?  Because after getting a perfect on Round 1, I realized that USAMTS requires far more time than I have; I barely struggle to finish homework on most days, let alone prepare for contests or work on USAMTS.  But I had spent too much time already on Round 1 so I decided that I might as well try to crank out a prize, reading on the website that a 40 could get me Honorable Mention.  So then I did only problems 1 and 2 for Round 2 (giving me 10 points) because I solved those problems immediately.  Round 3 I just didn't do because I didn't have time and I couldn't find immediate solutions.  And I've already said that I did problem 1 for Round 4 because that's all I would need for Honorable Mention.

So hopefully I can get a good prize.  :D

Here's a different angle to take: if you are able to score perfect but spend too much time, then your goal for usamts could be to improve your thinking efficiency, which is good for both contests and beyond. I remember last year some person mentioning solving all five problems of a round in about 30 min (probably not including writing up); in contrast, I spent maybe two weeks to solve them (not all the time, for sure).

EDIT: I seems to remember that person is Matt Superdock, a Honorable Mention last year on USAMO, which kinda explains that kind of speed...(correct me if I'm wrong.)

[quote="gauss1181"]how's it possible that someone can solve an USAMTS/USAMO level problem in only a matter of several minutes??????? :maybe: That's genius :o[/quote]

It's not difficult to solve every round 4 problem except possibly #3 in a matter of minutes, and many of the other problems from this year are not at all hard either.

Being able to solve problems quickly is to some extent a function of natural talent, but it's more dependent on practice, especially in this case. There are problems on USAMTS occasionally which I have seen very similar ones to.

5/5/2/5/4

Ug...they took a point off #5 because i rounded...protest also declined...
Score overall: 16/14/21/21=72. I actually achieved my goal.

[quote="funtwo"]#3 was just bash, it was easy to solve in a few minutes. usually , usamts is hard for me, but this round was extremely trivial, with all problems requiring basically a single concept[/quote]

I found #3 more difficult than most of the problems this year (including all the ones I didn't even bother to read because I was too intimidated by them but later found out were easy), but oh well.

i thought the most intimidating problem was problem 5 on round 2. geometry. argh

My four-page proof wasn't clear enough so nobody could understand it.

[quote="tenniskidperson3"][quote="mathemonster"][quote="123456789"]I thought the hardest was Round 3 Problem 5. Just my opinion though.[/quote]
i got a 3/5 by bsing with beatty sequences haha[/quote]

Is that what they're called?  That's what I first thought of trying, because I saw the $ \lfloor na\rfloor$ and this reminded me of something I had read in t0rajir0u's blog once.  I never got it working, though, and got a 2/5.[/quote]

Actually, there is a solution using that; it immediately trivializes the case where the number is irrational.

[quote="benzi455"]I thought round 3 number 5 was easy.  I just visualized a number line in my head and thought, "What would it imply if there were only finitely many integral powers of $ b$?" and it didn't take too long to figure it out.  Basically, if there are only finitely many powers of $ b$, then from that point the powers of $ b$ have to always be within a certain distance from the nearest whole number, and that turns out to be impossible because repeatedly multiplying by $ b$ will increase that distance.

I thought round 2 number 5 was the hardest.  I trig-bashed, and I got twice the correct answer because I said the area of a triangle was $ ab \,\sin{C}$ >_<[/quote]

Agree completely with this, except that I didn't manage to get the correct answer through trig bashing.

i thought round 1 was by far the easiest round this year. or maybe i just had more time then.

anyway...

r1: 5/5/0/5/5=20
r2: 5/5/0/1/2=13
r3: 4/0/5/5/0=14
r4: 5/5/0/5/0=15

so total = 20/13/14/15=62. bronze yay!!!

Hmm, well I didn't do round 4, so r1: 24, r2: 20, r3:18, r4:0 = 62


Let P be a convex polygon in the plane with a prime number p of sides, all angles equal, and all sides of rational length. Show
that P is regular (i.e., all sides also have equal length)

There was a very interesting thread about this problem before, and [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=152452]a corresponding blog post[/url] containing the solution and then some extra musings.


Suppose that we have a row of 1000 integers $ a_{1},a_{2},a_{3},...,a_{1000}$. The next row will be $ f(a_{1}),f(a_{2}),...,f(a_{1000})$ where $ f(a_{i})$ is the number of occurances of $ a_{i}$ in the previous row. If we continue like that prove that there will be a 'constante' row.

I'm not sure what exactly $ f(a_{i})$ is.  What is "the occurance of $ a_{i}$ in the previous row"?

In other words, $ f(a_{i})$ is the number of occurrences of $ a_{i}$ in the previous row.  I also assume you mean not that the row itself will be eventually be composed of a single number (which is certainly false, see e.g. the previous post on this thread) but rather than it will eventually be fixed by this operation.

[hide="Hint"]Prove that after the first application of $ f$, we will have $ f(a_{i})\geq a_{i}$. Then since no element of the row can be greater than 1000, it has to reach a maximum in a finite number of steps.[/hide]


linear algebra
matrix
probability
vector
algebra unsolved
algebra

Hello

A matrix is filled with 0s and 1s, in each of its 4 rows and 4 columns.

What is the probability that its determinant will be odd?

Please help!!!

Move to the field with two elements. Now it's the question of how many elements $ GL_4(F_2)$ has- the answer is $ (2^4\minus{}1)(2^4\minus{}2^1)(2^4\minus{}2^2)(2^4\minus{}2^3)\equal{}15\cdot 7\cdot 3\cdot 2^6\equal{}\frac{8!}{2}\equal{}20160$

To turn that into a probability, divide by $ 2^{16}$. It's a little more than $ 30$ percent.

To derive the answer, count ordered bases for the space- we have $ 2^4\minus{}1$ choices for the first vector (which can be anything nonzero) and so on.

This would be at home in either the Linear Algebra or Superior Algebra sections of the college forums.


quadratics
algebra
quadratic formula
Diophantine Equations

Find all pairs of integers $(x, y)$ satisfying the equality \[y(x^{2}+36)+x(y^{2}-36)+y^{2}(y-12)=0.\]

[quote="Peter"]Find all pairs of integers $ (x, y)$ satisfying the equality
\[ y(x^{2} \plus{} 36) \plus{} x(y^{2} \minus{} 36) \plus{} y^{2}(y \minus{} 12) \equal{} 0.
\]
[/quote]
$ yx^2\plus{}x(y^2\minus{}36)\plus{}y^2(y\minus{}12)\plus{}36y$
$ \Delta\equal{}(y^2\minus{}36)^2\minus{}4y(y^3\minus{}12y^2\plus{}36y)$
           $ \equal{}\minus{}3y^4\plus{}48y^3\minus{}216y^2\plus{}36^2$
           $ \equal{}\minus{}3(y^4\minus{}16y^3\plus{}72y^2\minus{}432)$
Easy to find out $ y$ to $ \Delta\geq 0$
Try to find out $ y$ and solve with x.
Note: Are there any method? It is quite hard to carculas.

:?:  I don't think you proved anything here?

[quote="pavel kozlov"]Consider our equation as quadratic respectively $ x$:
$ yx^{2} \plus{} (y^{2} \minus{} 36)x \plus{} y(y \minus{} 6)^{2} \equal{} 0$.
It's discriminant is the next:
$ D \equal{} (y^{2} \minus{} 36)^{2} \minus{} 4y^{2}(y \minus{} 6)^{2} \equal{} (y \minus{} 6)^{2}((y \plus{} 6)^{2} \minus{} 4y^{2}) \equal{} \minus{} 3(y \minus{} 6)^{2}(y^{2} \minus{} 4y \minus{} 12) \equal{} \minus{} 3(y \minus{} 6)^{3}(y \plus{} 2)$.
But $ D\geq 0$, hence $ \minus{} 2\leq y\leq 6$.
1) $ y \equal{} \minus{} 2$: $ \minus{} 2x^{2} \minus{} 32x \minus{} 128 \equal{} 0$ :arrow: $ x^{2} \plus{} 16x \plus{} 64 \equal{} 0$ :arrow: $ x \equal{} \minus{} 8$;
2) $ y \equal{} \minus{} 1$: $ D \equal{} 21*7^{2}$ - is not perfect square;
3) $ y \equal{} 0$: $ \minus{} 36x \equal{} 0$ :arrow: $ x \equal{} 0$;
4) $ y \equal{} 1$: $ D \equal{} 3^{2}*5^{2}*5$ - is not perfect square;
5) $ y \equal{} 2$: $ D \equal{} 3*4^{4}$ - is not perfect square;
6) $ y \equal{} 3$: $ D \equal{} 3^{4}*5$ - is not perfect square;
7) $ y \equal{} 4$: $ 4x^{2} \minus{} 20x \plus{} 16 \equal{} 0$ :arrow: $ x^{2} \minus{} 5x \plus{} 4 \equal{} 0$ :arrow: $ x \equal{} 1,4$;
8) $ y \equal{} 5$: $ D \equal{} 21$ - is not perfect square;
9) $ y \equal{} 6$: $ 6x^{2} \equal{} 0$ :arrow: $ x \equal{} 0$.
[b]Answer: [/b] $ ( \minus{} 8, \minus{} 2); (0,0); (0,6); (1,4); (4,4)$.[/quote]

the equation mod x yields y = 0, 6 mod x. y = 0 implies y  = kx which leads to the cubic $(k^3+k^2+k)x^3+(-12k^3)x^2+(36(k-1))x = 0$. It can be seen from the quadratic formula that x = $\frac{6k^2\pm6\sqrt{k}}{k^3+k^2+k} \rightarrow k = 1, k =4$ (note that at k = 9 and beyond the denominator is too big to make x an integer) which implies x=y=4, x = 1, y = 4, and x=y=0 are solutions. For the case y = kx+6 we obtain the cubic $x^2(x(k^3+k^3+k)+6k^2+12k+6) = 0 \rightarrow x = \frac{-6(k+1)^2}{k^3+k^2+k} \rightarrow k = 1, -2 \rightarrow x = -8, 1, 0 \rightarrow y = -2, 4, 6$ checking yields that x=-8, y = -2 and x = 0, y = 6 are also solutions.


Is the problem $ \frac{4}{3}x\equal{}\frac{8}{9}\sqrt{3}$?

If so, multiplying boht sides by 3/4:

24/36 rt3=2rt3/3

[quote="cdymdcool"](4/3)x=(8/9)(square root of 3)
what's x?[/quote]

The other possibility that I can think of is

$ \left(\frac43\right)^x\equal{}\frac{8\sqrt3}9\implies x\equal{}\log_{\frac43}{\frac{8\sqrt3}9}\equal{}\log_{\frac{2^2}3}{\frac{2^3}{3^{\frac32}}}\equal{}\boxed{\frac32}$.

Judging from the title " log?", your comment is right. :blush:

Needless to say your answer is correct.


Does there exist an increasing function f from the positive integers with values in the same set such that f(f(f(n)))=2n+3?

I think such a function does exist. I'll define it synthetically (not explicitly) and I'll let you people figure out why it works (I'm not very sure of that myself  :D; I just took f(1)=2 and f(2)=3 and started building it from there)

Let's take f(0)=1 (I also defined it in 0). Then f is f:N->N*. Let g(n)=f(n)-n. It'll be easier to see what g(n) is. We take g(n) to be 1=2^0 for the first 3=2^1+1 numbers (including 0). Then take g(n) to be 2=2^1 for the next 5=2^2+1 numbers. Then g is 3 (just for 1 number); then g is 4=2^2 for the next 9=2^3+1 numbers; then g is 5,6,7 (for 1 number each; then g is 8=2^3 for 17=2^4+1 numbers, and so on. I'll write the first 'few' values:

1,1,1,
2,2,2,2,2,
3,
4,4,4,4,4,4,4,4,4,
5,
6,
7,
8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,
9,
10,
11,
12,
13,
14,
15,
it's 16=2^4 fo the next 33=2^5+1 numbers, then 17, 18, 19, .., 31,
then 32=2^5 for the next 2^6+1 numbers etc. I think you can see the pattern. Like I said, I'm not sure it works, but it can't be too hard to prove that it does (I'm too lazy to do that right now; you might say this just an empirical solution, which means it's no solution at all, but just an observation :D ).

Ok, my solution. Let us define f(1)=2, f(2)=3, f(3)=5. Then we put f(3*2^(n-1)+k)=4*2^(n-1)+k for k=-3, ..., 2^n-3 and f(5*2^(n-1)+k)=6*2^(n-1)+2k+3 for k=-3,...,2^(n-1)-3. Try to prove that this function is a solution. Prove also that there exists only one function which verifies the conditions of the problem.


$ A\equal{}\{\text{factors of 36}\}$, $ B\equal{}\{\text{perfect squares}\}$, and $ C\equal{}\{\text{odd numbers}>2\}$. How many numbers are in $ A\cap B\cap C$?

We want the intersection of these sets. List those in Set A, and find those which satisfy B, and then which are in C.

A: 1  2  3  4  6  9  12  18  36

A and B: 1  9  36

A and B and C: 9

This a one number, so the interesction consists of $ \boxed{1}$ elements.


Princeton
college
articles
Support
geometry
geometric transformation
rotation

The first beam circled LHC ([url=http://en.wikipedia.org/wiki/LHC]What is LHC?[/url]) today :roll: .
It is supposed to revolutionize particle physics.
Also there are concerns over safety issues ([url=http://en.wikipedia.org/wiki/Safety_of_the_Large_Hadron_Collider]Why?[/url]).
What do you think about it?

Some physicist (at Princeton I think) claimed in a NY Times article that the LHC would create dragons that would eat us.  Obvioulsy, that is quite unlikely, but if the LHC fails to produce any meaningful results, it would be considered a colossul waste of funds.

All the kids at my school were freaking out about the world ending. Someone said microscopic black holes couldn't suck anything and, (while resisting the obvious joke) I asked them what Hawking Radiation was. The response was kind of funny.

Mass hysteria is amusing until it gets frightening.

My opinion is that there is no risk. If the experts tell us that the world will not end due to a tiny black hole, then I trust them, because the alternative is to trust people who are not experts. I'm sure once the concern was raised, they carefully checked every possibility so thoroughly that we cannot even comprehend it.

Well black holes are formed when there is a tremendous amount of energy or matter. The size depends on how much, and the smaller it is the less time it takes to evaporate. The black holes made by the LHC are calculated to dissipate in extremely tiny fractions of a second, similar to the ones created by cosmic radiation bombarding the atmosphere.

On a related note -- a girl committed suicide fearing the end of the world: [url=http://www.wunderkabinett.co.uk/damndata/index.php?/archives/1537-Large-Hadron-Collider-claims-its-first-victim-plus-have-ritual-suicides-opened-the-Stargate.html]see here![/url]

[quote="paladin8"]The most straightforward answer: http://hasthelhcdestroyedtheearth.com/[/quote]
Come on. They didn't yet collide anything yesterday. Only a test run to circle the whole 27 km.
The high energy collisions are likely to start sometimes around 21st October. So there was no more possibility of the World Ending yesterday than today.

However I personally think the world won't end by LHC with a $ (100\minus{}10^{\minus{}5})\%$ chance. However can anyone rule out the entire possibility? After all it's something happening for the first time, and it's not like a space mission where people are expecting something fixed as a well-defined success and if everything works according to calculations there is no threat. There's supposed to be new things discovered, something that goes outside the estimates, right? And I know a bit about Hawking radiation, but remember it's not a tested fact.

And I don't support closing the project because it's worth taking the risk of $ 10^{\minus{}5}\%$, however someone might disagree, it's the sake of the entire world.

One more thing, forget about Doomsday, some explosion on the machine could cause huge loss of people and property.

However, I'm glad that many people (including some renowned newspaper) believes that everything happened yesterday and there's no more risk ahead. One of my teachers think that they were actually doing one single experiment (insteade of a decade long project) and it's over now.

[quote="JRav"]It's natural that people are scared since we are dealing with very large amounts of energy, but personally, I doubt that anything will happen.  But the physcists at my school are really excited about this, especially those who have projects there.[/quote]
I read that it's maximum power is about 760MJ concentrated in the beam, which actually isn't that much if you think about it...a few tanks of gasoline being burned, except instantaneously.

This is ridiculous. People actually think that the scientists who built the LHC are sufficiently incapable of risk analysis to go ahead with the project despite a non-infinitesimal chance that it would cause the world to end. "Even if there is a tiny tiny *tiny* chance a black hole gets out of control, it isn't worth it!" Then these same people get in cars and go on the highway. I don't understand it.

Also, the value of an experiment shouldn't be completely dependent on whether it yields (figuratively) explosive new evidence. If the LHC doesn't find anything, the fact that it didn't would still be useful information. A lot more useful than what we'd get from not building the LHC at all, at least.

really? even on a forum of mathematically intelligent people, all we can find to say about the LHC is whether or not it will produce black holes that will destroy the world?

guys I am somewhat disappointed

WHICH IS WHY I WOULD LIKE TO INTRODUCE YOU TO THE LHC RAP

BEHOLD

[youtube]j50ZssEojtM[/youtube]

That is brilliant.  I wish those were real CERN physicists doing the rapping and the dancing, but it's still hilarious.

I will admit that I have done no actual thinking about this. However, I believe the LHC is dangerous. Let us assume the survival of the human race is important. Then we should not risk destroying our only known habitation for the sake of 'finding another particle!!!111!'. Certainly it is believed that the risk is very low. But it is also believed that Hawking Radiation exists. We have not verified this, as far as I know. Thus, if our theory of Hawking radiation is wrong, and we make a black hole or many, we are fairly sunk as a race. If we weren't doing this on our only habitation, I wouldn't have a problem with it. (Further, suppose this turns out fine, so we do another experiment; eventually our theories will prove wrong because everyone is wrong sometimes, and we destroy the Earth and ourselves because we did dangerous things without a backup plan. Woot!)

(This is significantly biased by a dream I had about 6 years ago about minature black holes doing significant damage to the earth. I do not blame you for thinking this opinion resembles that of a cracked kettle.)

there was only one scientific error in it -- I was pretty impressed

also I think (not sure) that some of the background dancers may have been physicists.

Remember guys, they thought the atomic bomb would light the air on fire and destroy the world.

[quote="daboa"][quote="stuck"]Most obvious answer:

High-energy cosmic rays go through our atmosphere every second whose energies even the LHC will not reach, nor any collider in the near future. ($ 10^{20}$ vs. $ 10^{13}$) I think we would know if one of these stray rays caused a black hole by now.

EDIT (Clarification): A [i]stable, apocalyptic[/i] black hole.[/quote]
You are neglecting the fact that those (stable) black holes would have very high velocity and hence would vanish from earth very quickly whereas (stable) black holes created by LHC would have low velocity and probably be charged.[/quote]
So tell me how much quantum mechanics and physics do you know. I bet you must have a Ph.D and you're going to be the next Stephen Hawking.  :roll:

Damn that rap was pretty good.

Well, daboa has a point. I naively assumed that there would be no difference between the types of black holes generated. So it was only fair for me to look at the wikipedia article for Hawking radiation to see if there is significant difference. I must profess that I have never studied this in particular before, but at least intuitively my argument made sense.

From looking over the Wikipedia article, the equations they give are indeed for non-charged non-spinning black holes. However, I do not see a term for velocity in the time of dissipation for black holes. The only variable in the simplified (no charge, no rotation) case is initial mass.
Here is the relevant equation for those interested, from Wikipedia, keeping in mind that approximations have been made:
$ t_{\operatorname{ev}}\equal{}{5120\,\pi\,G^2M_0^{\,3}\over\hbar\,c^4}$

Still, looking over many sources, it seems to be that we only should be worried about black hole propagation in the case of a particular theory, in particular, large extra dimension theories. So if these are right, we will be getting black holes. Now, if so, are they harmful? According to [url=http://arxiv.org/abs/0806.3414]This Preprint[/url], no.

[quote="serialk11r"]So tell me how much quantum mechanics and physics do you know. I bet you must have a Ph.D and you're going to be the next Stephen Hawking.  :roll:[/quote]
I really shouldn't bother answering this but okay, here we go: I don't see what my non-existing physics education has to do with this. I wrote what a physics professor (he is involved in the ATLAS project) earlier had told me about the [b]difference[/b]* of the black holes (asuming one would be created). I do not know how much quantum mechanics he has been reading but I bet it's a lot. Do you want me to ask him?

* I only pointed out the difference between them. I never said we'll die or anything else like that, ok?


Well, it's the beginning of the year and I'm already stumped :-X.

The problem tells me to add, subtract, multiply and divide two given functions, but I am not sure how to continue since they are square roots.

Here are the functions:

f(x) = sq.root(1+x) and g(x) = sq.root(1-x)

Find: f+g, f-g, fg, and f/g

Any help is much appreciated. big_smile

[quote="Chaoticx"]Well, it's the beginning of the year and I'm already stumped :-X.

The problem tells me to add, subtract, multiply and divide two given functions, but I am not sure how to continue since they are square roots.

Here are the functions:

$f(x) = \sqrt{1+x} \text{ and } g(x) = \sqrt{1-x}$

Find: f+g, f-g, fg, and f/g

Any help is much appreciated. big_smile[/quote]

[hide]
$f+g=\sqrt{1+x}+\sqrt{1-x}$
$f-g=\sqrt{1+x}-\sqrt{1-x}$
Not sure if ^^^^^^^^^^^^^^^ can be simplified more.

$fg=\sqrt{1+x}*\sqrt{1-x}=\sqrt{(1+x)(1-x)}=\sqrt{1-x^2}$
$\frac{f}{g}=\frac{\sqrt{1+x}}{\sqrt{1-x}}=\frac{\sqrt{1+x}\sqrt{1-x}}{\sqrt{1-x}\sqrt{1-x}}=\frac{\sqrt{1-x^2}}{1-x}$

[/hide]

Thank you very much! Helps a lot. Would you happen to know how to find the domain of each of those functions? Thanks again!

well to find the domain of a function, we have to consider 2 things,
1) if the bottom of a fraction equals 0, then it is undefined
2) if the number under the square root symbol is negative, there is no real answer
    of the number under the square root is positive or 0 then we are fine
for an example
$f(x) = \frac{\sqrt{3-x}}{x-2}$
looking at the 2 rules that i stated above,
1) $x-2 \ne 0$
$x \ne 2$
2) $3-x \ge 0$
$3 \ge x$
so the domain would be $x \le 3$ and $x \ne 2$
i hope this makes sense, but the 2 above concepts you must get down


[quote="xyidan"]Let A be the mx(1,0,-1;0,2,0;-1,0,1). Find the distinct eigenvalues H of A and their multiplicities. Determine the dimensions of the eigenspaces N(|A - H|). Find orthonormal bases of these spaces. Combine these base into one orthonormal basis B of R3 and verify that the matrix of A relative to B is a diagonal matrix with entries the eigenvalues of A, each repeated as many times as its multiplicity as a root of the characteristic polynomial.
[/list][/code][/quote]

Could anyone help me about this question,please? Thanks!

It's been a while since I did this, but...

The characteristic equation is:

$\displaystyle \left| \begin{array}{ccc} 1-\lambda & 0 & -1 \\ 0 & 2-\lambda & 0 \\ -1 & 0 & 1-\lambda \end{array} \right| = 0$.

Developing the determinant along the second row gives:

$\displaystyle (2-\lambda)\left| \begin{array}{cc} 1-\lambda & -1 \\ -1 & 1-\lambda \end{array} \right| = (2-\lambda)((1-\lambda)^2-1)=(2-\lambda)(\lambda^2-2\lambda)=-\lambda(\lambda-2)^2=0$.

So the eigenvalues are $\lambda_1=0, \lambda_2=2$ with algebraic multiplicities $1,2$ respectively.

To find the eigenspaces, their dimensions and their bases we solve ($A$ is the matrix, $I$ is the identity matrix):

$(A-\lambda_1 I)v=0$, which gives $(s,0,s)$ for solution, so the dimension of this eigenspace is $1$, and an orthonormal base is $\lbrace{\frac{1}{\sqrt{2}}(1,0,1)\rbrace}$.

Similarly, $(A-\lambda_2 I)v=0$ gives $v=(s,0,-s)$, so this eigenspace also has dimension $1$ and the base is for instance ${\lbrace\frac{1}{\sqrt{2}}(1,0,-1)\rbrace}$.

Now, we can't make a base for $\mathbb{R}^3$ with these, because we lack a dimension, so either I've got it wrong or I don't understand what you're saying. I also don't know what you mean by "the matrix of A relative to B" either.

Thanks, you guys! Sorry about mis-writing A-$\lambda$I.

hi, Kent, I still didn't get how to get the orthonormal. I think I am confused between orthogonal and orthonormal. Could you explain that for me,please? Thanks agian!

[quote="Kent Merryfield"]TheBB: you've come to the conclusion that $A$ is not diagonalizable. This conclusion is wrong. For one thing, it has to be wrong: $A$ is a symmetric matrix, and a symmetric matrix is always diagonalizable. Orthogonally diagonalizable, to be specific.

The culprit is the eigenspace associated to $\lambda=2.$ This eigenspace is the null space of $A-2I.$ In other words, the null space of

$\begin{bmatrix}-1&0&-1\\0&0&0\\-1&0&-1\end{bmatrix}.$

This is rank 1 matrix; the dimsnsion of its null space is 2. Here's one possible basis:

$\left\{\begin{\bmatrix}1\\0\\-1\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix}\right\}$

That's even an orthogonal basis. It's not yet orthonormal, but you know how to fix that. And you should now see which vector you were missing.[/quote]

A set $\{v_1,v_2,\dots,v_k\}$ is orthogonal if the inner product $v_i\cdot v_j=0$ whenever $i\ne j$ and none of the $v_i$ is the zero vector. The set is orthonormal if in addition to being orthogonal, for each $j,\,||v_j||=\sqrt{v_j\cdot v_j}=1.$

Another way to say that the set is orthonormal is this:

$v_i\cdot v_j=\delta_{ij}=\begin{cases}1&\text{if }i=j\\0&\text{if }i\ne j\end{cases}.$

Given an orthogonal set, you can create an orthonormal set by dividing each vector by its own length. And given any independent set, by the Gram-Schmidt process, you can create an orthogonal set (orthonormal, if you prefer) with the same span.

One slightly confusing piece of terminology: a square matrix $A$ whose columns form an orthonormal set (a condition you can write as $A^TA=I$) is traditionally called an orthogonal matrix. Even though we probably should call that an orthonormal matrix, the terminology is too deeply ingrained to change.

from the top, the solution $lamda$/=2. When I was solving it, I got two solution for $lamda$/ which are 2 and 0. Do I need to worry about 0?

Now, I understand the difference between orthogonal and orthonormal, but I still don't know how to find all the possible basis in this question. Could you work it out in detail for me because a numerical example can make me understand,please?
Also, You said I should see now see which vector I was missing. I dont understand what you mean. Did I miss a vector?
Thanks for your helping!

xyidan: I just erased the quotes from your last two posts, becuase the quotes served no purpose. Please do not quote other posts without a reason for your quote, and, especially, please refrain from qouting long posts in their entirety.

[quote="xyidan"]Also, You said I should see now see which vector I was missing. I dont understand what you mean. Did I miss a vector? [/quote]
If you'll go back and read that post of mine carefully, you'll see that the "you" I was referring to there was TheBB. TheBB was missing one of the two vectors spanning the $\lambda=2$ eigenspace.

[quote="xyidan"]Do I need to worry about 0?[/quote]
Yes, of course. This matrix has two eigenvalues, 2 and 0. The eigenspace assoicated with $\lambda=2$ is two dimensional and contributes two independent (which can be made orthonormal) eigenvectors. The eigenspace associated with $\lambda=0$ is one dimensional and contributes one eigenvalue.

Take those three vectors, (orthonormal version). Write them as column vectors, and stack the three colums together into a $3\times 3$ matrix. Call that matrix $U.$

First (a check): multiply out $U^TU.$ That should be the identity matrix. If it's not, then something is wrong.

Then: multiply out $U^TAU.$

[quote="xyidan"]Could you work it out in detail for me because a numerical example can make me understand,please? [/quote]
No, you work it out. Follow the plan I gave in the previous paragraph.

Try not to push the patience of the people on this board too far. We respond to questions not out of any obligation, but because we want to. So if you want our response, you have to make us want to respond - and if you act too needy or too helpless, we'll eventually stop paying attention.

I am sorry about it, Kent. Because it was my first time to touch "eigenvalues", this concept was so confused for me.  Sorry about it!

[quote="Kent Merryfield"]TheBB: you've come to the conclusion that $A$ is not diagonalizable. This conclusion is wrong. (...)[/quote]

Right... sorry about that. It really HAS been a while.  :blush:


Does anyone have a copy of the 2004 USAMO they could send me?

It's online on the AMC website, i.e. [url]http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamo-test.html[/url].

Look before you leap.   :)

He looked, he leaped (just a dumb joke on my part) :) .

[quote="mathfanatic"]It's online on the AMC website, i.e. http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamo-test.html .

Look before you leap.   :)[/quote]

It works for me

All the questions of the 2004 olympiads (including USAMO) are present on the AoPS - MathLinks site.

USAMO 2004 questions (and solutions) can be found here
http://www.mathlinks.ro/Forum/viewtopic.php?t=5423

[quote="theone853"]mathfanatic, you added an extra "." at the end of the url[/quote]

Oops.  Fixed.

[quote="Valentin Vornicu"]I am surprised that many times users ask for different questions (USAMO, IMO Shortlists) that are present for download on the site, and many people worked for that to happen, however usually moderators point people to other (external sites) for the questions ... :)[/quote]

Sorry about that, Valentin.  But it also makes some sense to send people to the official USAMO site... :).  I'll remember that in the future, though.


USAMTS
probability
Vieta
function
geometry
analytic geometry
complex numbers

Which of the USAMTS round 1 questions was your favorite?

I personally liked Questions 2 and 4 the most, with 4 barely ahead.  Question 5 was cool due to its initial complexity, but actual simplicity upon further inspection.  3 was okay ... I love complex numbers but this problem became a bit too easy once you got the factorization.  1 was my least favorite because it was a little too simplistic.

What did you think?

JB

The only one I didn't like was 1, all the rest were A+!

I love number 3 - because instead of factorization, I use Vieta's to find that the imagery part of the roots must be an inverse function of the real part, with different constant $k$ for each two pairs of conjugates.  So if you plot all the possible roots for the equation on the complex plane you'll get 4 inver function.  And that's why they state specificly $r$ does not equal to $0$, because that's the point where all four inverse function is undefined.  This whole process is just so cool :lol:  :lol:  :lol:

I liked 5. 4 was pretty cool, but I made a dumb mistake and missed something at the very last step. >_< I can probably still get 4 points on it, though, since my reasoning up to that was good... but I just didn't have enough time to work out the end fully.

i liked #2 the best... i took the time to find the probability of each outcome.. then when i looked at the solution that was given in the mathjam... man... that was waaay more elegant than my solution

I liked #3.

#1 was too easy
#2 was annoying (I never much liked counting problems, but I love the recursive function approach.  Very clever)
#3 was extremely difficult, but I had a lot of fun solving it.
#4 was horribly tedious.  Thinking about it hurt my head, so I wrote a program to do it for me. :P
#5 was very creative in how tricky it seemed.  I spent the most time on this problem, by far.  About two hours before I submitted my solutions, I realized that it could be solved with similar triangles and the pythagorean theorem.  (That's it!) I can't believe how disguised that problem was!

I like them all, except #4 because I made a slight mistake and might get a point or two off :( .  My favorite, however, is #5.  Geometry is my weakest subject so this one took me a while.  I was stuck on the problem even though I was just one easy step away.  When I finally figured it out...ow, man, that's one of the greatest feelings in the world!

I thought #1 and #2 were pretty cool.  #1 was pretty easy...but only if you're experienced with mods and you realized to look at squares modulo the difference.  The recursive route wasn't my first choice on #2, but once I tried it I knew immediately how to proceed.  #3 took a bunch of time.  Pretty good problem though.

My favorite problem was number 4. I love logic-type stuff and thought it was neat because it made (well, it did for me) you scratch your head a little bit a first. For the others you could say at first glance: okay I'll need to factor this, i'll need to do some probability, i'll obviously need to draw this diagram, etc. When I read number 4 for the first time, I just though "huh?" After maybe 5-10 more minutes of reading the problem and thinking, I made some progress, and it didn't turn out to be that difficult, but I still loved the orgininal intrigue.

The weird thing is I didn't have much trouble w/ numbers 2 and 3, as a few other people seemed to (although perhaps i didn't have the most creative solutions), yet I couldn't figure out #1 at all, and everyone seems to think it was incredibly easy! Ahh! Although...for number 3, I wrote my solution as 8i instead of 8 for the area, a stupid mistake w/ forgetting that each imaginary coordinate is just 1, not 1i. Will this be that important in grading?

I really don't know which one i liked the most...didn't like 1 much though...
Zanttrang, the reason it works out is because their solution broke it up into six parts, and then found the expected value for each...i never got that solution either...

[quote="joml88"][quote="zanttrang"]I'll give the nod to number 2 - I never found the intended solution path, but I did manage to get the correct answer, and the actual solution is ingenius (at least to me; I still haven't quite grasped why if you add up the probability of each option, you get the expected value).[/quote]

Essentially you're taking the weighted sum of the outcomes.  Consider rolling one die.  What is the expected value of the number facing up?   You could add up all possibilities and divide by the number of possibilities: $\dfrac{1+2+3+4+5+6}{6}=3.5.$  But if you view this as $\frac 16(1)+\frac 16(2)+\cdots+\frac 16(6)$ you have the weighted sum.

If you think of expected value as the average value, I think that might make it easier.[/quote]
I understand that method fine - it's the other way of doing it that has tripped me up.  Basically what the easiest solution to #2 has done is to find the probability of a loop being tied on any particular turn, and to add up the probabilities for all of the turns, because the probability is the average part of a loop you'll tie on that turn, so adding them gives the total number of loops (possibly not a whole number) that you tie.  It makes sense now that I've thought about it for a while, but it took me a while to grasp.

That method is using the linearity of expected value.  Let $X_1$, $X_2$, $\ldots$, $X_n$ be random variables.  Then we have the following formula:
\[ E(X_1 + X_2 + \cdots + X_n) = E(X_1) + E(X_2) + \cdots + E(X_n) \, .  \]

In the problem at hand, $X_i$ is 1 if the $i$th turn is a loop, and is $0$ otherwise.  Each term on the right is a probability.

[quote="zanttrang"]I understand that method fine - it's the other way of doing it that has tripped me up.  Basically what the easiest solution to #2 has done is to find the probability of a loop being tied on any particular turn, and to add up the probabilities for all of the turns, because the probability is the average part of a loop you'll tie on that turn, so adding them gives the total number of loops (possibly not a whole number) that you tie.  It makes sense now that I've thought about it for a while, but it took me a while to grasp.[/quote]

Yea basically on any given turn you can make 0 or 1 loops. So when you add up the probabilities that you will make a loop on any given turn your essentially adding up each probability multiplied by 1 (1 loop.) You don't have to worry about adding the probabilities of making 0 loops since multiplying by 0 will give you 0.

I liked number 4!! The puzzle was really intriguing.  :)

What's Maxima? *googles it* IT'S FREE?? Wow! :D

[quote="Danbert"]#3.  I was feeling lazy, so I used Maxima to solve it for me.  When I wrote up the solution to that problem, I found that the answers it gave me were wrong!  It did the same thing repeatedly.  Odd.  Anyway, it gave me a warm fuzzy feeling to know that my computer can be wrong too. :P[/quote]

Just don't think about that next time you get on a plane :)

I liked number 3 the best, even though I never solved it.  Until the Math Jam, I kept thinking that the last term ( $(9r^8+10r^4+1)$ ) was actually $(9r^8+4r^4+1)$ for some weird reason :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red:  :wallbash_red: .  However, after I figured out my error, I saw through the problem and it became really easy.  :lol:

I thought that this competition was very well written overall.  Most were fairly simple! :D

[quote="zanttrang"]I'll give the nod to number 2 - I never found the intended solution path, but I did manage to get the correct answer, and the actual solution is ingenius (at least to me; I still haven't quite grasped why if you add up the probability of each option, you get the expected value).[/quote]
For each of the individual parts, the probability of forming a loop IS the expected number of loops for that part. The expected number of loops for all parts is therefore just the sum of the individual probabilities.

I liked all of the problems except for #1. Problem 1 was straightforward and very tedious ---> blah

My favorite was number 5 because [b]it was a geometry problem I could actually solve![/b] Number 1 was a bunch of casework, and so was number 4, so those were my least favorite. I never got #3, and it looked like just some big ugly equation that they made up just for the problem. I'll have to admit, I'm not a pure pure mathematician; I can sense a limit where a problem is totally impractical. Area of the quadrilateral formed by the roots in the complex plane? Give me a break! #2 was cool, though, since it comes out to a nice pattern (and is a concrete problem).

I found these significantly harder than last year's round 1, since they required much more than just staring at the problem for about 30 seconds, and exclaiming, "Oh, I've got it!".

I liked #5 for pretty much the same reason. #3 actually came out quite well, though. Area of the quadrilateral was $8$. ;)

The roots were a little weird, but not bad. IIRC, it was $\frac{1}{r}\pm 3ri$ and $\frac{1}{r}\pm ri$, but I'm not entirely sure on that.

[quote="SamE"]I never got #3, and it looked like just some big ugly equation that they made up just for the problem.[/quote]

No, it was actually a nice equation once you factored it... ;)


algebra
polynomial
calculus
integration
algebra unsolved

Find all monic integer polynomials $p(x)$ of degree two for which there exists an integer polynomial $q(x)$ such that $p(x)q(x)$ is a polynomial having all coefficients $\pm1$.

Note that the coefficient off $x^{2}$ and the constant term are both plus or minus $1$.

I think one could construct a $q(x)$ for each $p(x)=x^{2}+ax+1$, $p(x)=-x^{2}+ax+1$, $p(x)=x^{2}+ax-1$, $p(x)=-x^{2}+ax-1$, for integral $a$

see http://www.mathlinks.ro/Forum/viewtopic.php?t=82906 .

  darij

thanks darij for ur timely help. it was very helpful.

Holy crap those solutions are amazing!  :)

P.S. Is there any good polynomial training material?


algebra
polynomial
real analysis
real analysis unsolved

Let $a,b,c$ $ \in R$ so that $a+b+c=6$ and $ab+bc+ca=9$.

a) prove that $(a-1)^{2}+(b-1)^{2}+(c-1)^{2} = 9$.

b) prove that the monic polynomial, which has a,b,c as roots has the form:

$p'(2)p(x)= x^{3} - x^{2}+9x+m$; $m \in R$.

c) solve the equation $p'(x) = 0$.

d) prove that $0 \leq abc \leq 4$.

e) prove that if $a \leq b \leq c$ then we have $a \in [0,1]$, $b \in [1,3]$ and $c \in [3,4]$.

thank you very much!

and for e)
WLOG we can assume $a \le b \le c$ so
$ab+bc+ac=9$--->$c(a+b)+ab=9$----->$c(6-c)+ab=9$----->$c^2(6-c)+abc=9c$----->$abc=c(c-3)^2$
but we should have $c\ge 0$ thus $abc\ge0$
and similary  we can show that
$abc=a(a-3)^2$
but
$a+b+c=6$---->$a\le2$---->$a(a-3)^2\le4$---->$abc\le4$ and
$0\le abc\le 4$
is my proo right? :? f

Payman

[quote="Mitzah"][quote="payman_pm"]but we should have $c\ge 0$
[/quote]

why?  :blush:[/quote]

 Maybe because $c \geq b \geq a$ and $a +b +c = 6$  :blush:

becuse if $a\le b\le c$ and $c \le 0$ then $a+b+c\le0$!!


Following ineq solves problem:

a/b + a/c + b/a + b/c + c/a + c/b + 4(ab+bc+ca)/(a^2+b^2+c^2) >= 10

This ineq can be proven by following:

a/b + b/a - 2 = (a-b)^2/ab >= 2(a-b)^2/(a^2+b^2) > = 2(a-b)^2/(a^2+b^2+c^2)

Sum up the similar ineqs, we get

a/b + a/c + b/a + b/c + c/a + c/b - 6 >= 2[(a-b)^2+(b-c)^2+(c-a)^2]/(a^2+b^2+c^2) = 4 + 4(ab+bc+ca)/(a^2+b^2+c^2)

Namdung

You really are tough, Namdung. I also have a solution to this problem, but your solution is much more beautiful and natural than mine.


I have two nice problem about amout of primes among n and 2n:
1/Prove that:for all positive integer k there always exists n such that:Among n and 2n has k primes..
2/For N=p_1.p_2...p_n(p_i is i-th prime) then among N and 2N has >= (2^n-2) primes
   I think they're famous but i do not know their solution.....

1) Let $f(n) = \pi (2n) - \pi (n).$
Then :
$f(n+1)-f(n) \in {-1,0,1 }$, the exact value depending on the fact that $n+1$ and/or $2n+1$is prime.
According to the prime number theorem, we know that $f(n)$ goes to $+ \infty$ as $n$. Since $f1) = 1$, it follows that $f$ takes each positive integer value at least once. Hence result.

2) According to Sierpinski (see [1] ), it can be proved that $\prod_{k < p \leq 2k} p >2k$ for $n > 647$ (where $p$ stands for a prime).
Using that for $k = p_1 \cdots p_n \geq 2^n$ leads to the desired result for all $n$ except maybe for some of the first positive integers such that $p_1 \cdots p_n \leq 647$. These values can be checked by hands.

Reference :
[1] W. Sierpinski, Elementary theory of numbers, North-Holland, p.137-144.

Pierre.

Can you show me where we can see your prime number theorem($f(n)\to +%Error. "infinit" is a bad command.
$) and Sierpinski??
   Thanx....

PNT states that $\pi (x)$ is equivalent to $\frac x { Ln(x)}.$
Thus $\frac{ \pi (2n) } {\pi (n) }$ has limit $2$. It follows that for sufficientely large $n$, we have $\pi (2n) > \frac 3 2 \pi (n)$ so that $f(n) > \frac 1 2 \pi (n)$. Since $\pi(n)$ tends to $+ \infty$, so does $f(n).$

Try to find Sierpinski's book because I will not copy 7 pages.... :huh:

Pierre.

Tschebychefs theorem will do well:
there exist $0<a<1<b$ (e.g. $a=ln(2),b=ln(4)$) with
\[ a n < \pi(n) ln(n) < bn  \]
for all $n$.

[quote="{x}"]Sierpinski's book can be downloaded from here
http://matwbn.icm.edu.pl/kstresc.php?tom=42&wyd=10
It's in English ;).[/quote]
 Thank you for this website..It's really interesting for me and the results (there) is very nice and important..I've solved my problems
   Thanx...


calculus
integration
function
limit
real analysis
real analysis unsolved

Let $ f: [0,1]\rightarrow [0,1]$ be a strictly increasing differentiable function with $ f(1)\equal{}1$ and $ f'(1)\neq 0$.

Calculate the limit
$ \lim\limits_{n\rightarrow \infty}n^{2}\int\limits_{0}^{1}x^{n}\left(f(x^{n})\right)^{n}dx.$

Is the limit equal to 1? If not, i don't want to post my ugly attempt  :wink:

No, the limit is not equal to 1. Consider the function $ f(x)\equal{}x^{2}$.

just the behaviour in a neighboorhood of $ 1^\minus{}$ matters, and if $ \alpha\equal{}f'(1)$ and $ \epsilon>0$, one can find $ \eta>0$ such that:
$ x^{\alpha\plus{}\epsilon} \leq f(x) \leq x^{\alpha\minus{}\epsilon}$
on $ [1\minus{}\eta; 1]$. Because if $ f\equal{}x^{\beta}$ then $ n^2 \int_{[1\minus{}\eta; 1]} x^n f(x^n)^n dx \to \frac{1}{\beta}$, one can conclude that the desired limit is $ \frac{1}{f'(1)}$

by substitution $ x^{n} \equal{} z$ and dividing interval of integration like this $ [0,1 \minus{} \frac {1}{n^{\alpha}}]$ and $ [1 \minus{} \frac {1}{n^{\alpha}},1]$ it's easy to show that limit is equal to integral of $ (zf(z))^{n}$ on second interval (for some good $ \alpha$) . by Taylor approximation of $ zf(z)$  at point $ 1$  one can see that limit is equal to $ 1/f'(1)$


Compute the sum of all real numbers x such that
$ 2x^6 \minus{}3x^5 \plus{} 3x^4\plus{}x^3\minus{}3x^2\plus{}3x\minus{}1 \equal{} 0$

Let $ f(x) \equal{} 2x^6 \minus{} 3x^5 \plus{} 3x^4 \plus{} x^3 \minus{} 3x^2 \plus{} 3x \minus{} 1$.
$ f(1) \equal{} 2$
$ f( \minus{} 1) \equal{} 0$
$ f\left(\frac {1}{2}\right) \equal{} 0$
$ f\left( \minus{} \frac {1}{2}\right) \equal{} \minus{} \frac {49}{16}$
So, $ (x \plus{} 1)(2x \minus{} 1)|f(x)$.

Theredore, $ f(x) \equal{} (x \plus{} 1)(2x \minus{} 1)(x^4 \minus{} 2 x^3 \plus{} 3 x^2 \minus{} 2 x \plus{} 1)$.

And, $ x^4 \minus{} 2 x^3 \plus{} 3 x^2 \minus{} 2 x \plus{} 1 \equal{} x^2(x^2 \minus{} x \plus{} 1) \minus{} x(x^2 \minus{} x \plus{} 1) \plus{} (x^2 \minus{} x \plus{} 1) \equal{} (x^2 \minus{} x \plus{} 1)^2$.

Hence, $ f(x) \equal{} (x \plus{} 1)(2x \minus{} 1)(x^2 \minus{} x \plus{} 1)^2$.

Thus, $ x \equal{} \minus{} 1, \frac {1}{2}, \frac{1 \pm \sqrt{3} i}{2}$.