topology
vector
real analysis
calculus
calculus computations

Let $ F$ and $ K$ be disjoint, nonempty subsets of a metric space $ M$ with $ F$ closed and $ K$ compact.  Show that $ d(F,K)\equal{}\inf\{d(x,y): x\in F, y\in K\}>0$.  Show that this may fail if we assume only that $ F$ and $ K$ are disjoint closed sets.

Suppose to the contrary; then there exist sequences $ x_n \in F, y_n \in K$ such that $ d(x_n, y_n) < \frac {1}{n}$.  The sequence $ (y_n)$ has a convergent subsequence which converges to $ y \in K$ such that $ y$ is a limit point of $ F$, hence in $ F$; contradiction.

As for the second, let $ F, K \in \mathbb{Q}$ with $ F \equal{} \{ q : q^2 < 2 \}$ and $ K \equal{} \{ q : q^2 > 2 \}$.  $ \mathbb{Q}$ is totally disconnected, so $ F, K$ are both closed in $ \mathbb{Q}$.

this is a really silly question, but what's the default topology on $ \mathbb{Q}$?  the relative topology from $ \mathbb{R}$?

another example is something silly like $ A_1 \equal{} \{(x,y): y \geq e^x\}$ and $ A_2 \equal{} \{(x,y): y \leq \minus{} e^x\}$ (look at what happens along $ (t,0)$ as $ t\to \minus{} \infty$)

edit: oh never mind I didn't notice that we're working with metric spaces, it's pretty obvious what the topology the metric induces on $ \mathbb{Q}$ is :)

[quote="blahblahblah"]this is a really silly question, but what's the default topology...[/quote]
Perhaps I'm just not familiar with your terminology but what do you mean by default topology?  Metric?

By convention, when we talk about a topological space without specifying the topology there is some agreed-upon topology that we default to.  In this case, $ \mathbb{Q}$ is given the obvious (that is, not $ p$-adic!) metric and the corresponding topology.

(The confusion here is that topologies are not always metrizable.)

You can define a lot of topologies on a given set $ X$.  For example, you could take all subsets of $ X$ to be open, or just take the empty set and $ X$ to be open.  A topology is just more or less specifying the collection of open sets, subject to some compatibility conditions (closed under arbitrary unions, and contains the empty set and the whole space)

On the other hand, if you have a metric space $ (X,d)$, $ d$ induces a topology on $ X$ in the obvious way (a set $ U$ is open iff for every $ x\in U$, there exists $ \varepsilon > 0$ such that $ B(x,\varepsilon)\subset U$.  note that the empty set is vacuously open)

If you have metric spaces $ (X,d)$ and $ (X,d')$, it's entirely possible that $ d$ and $ d'$ will induce different topologies on $ X$.  One condition that guarantees that $ d$ and $ d'$ will induce the same topology is that they are equivalent.  In finite dimensional normed vector spaces, all norms (basically, metrics which are compatible with the vector space operations) are equivalent, but this is far from true in the infinite-dimensional situation.

In this case, what I was (stupidly) asking what the 'usual' topology on $ \mathbb{Q}$ was.  You can look at it as the collection of sets $ U\cap\mathbb{Q}$, where $ U$ is open in $ \mathbb{R}$ (the relative topology), or just as the topology induced on $ \mathbb{Q}$ by $ |\cdot|$ (the usual absolute value).  It's fairly easy to see that these are the same.

[quote="t0rajir0u"]By convention, when we talk about a metric space without specifying the topology there is some agreed-upon topology that we default to.  In this case, $ \mathbb{Q}$ is given the obvious (that is, not $ p$-adic!) metric and the corresponding topology.

(The confusion here is that topologies are not always metrizable.)[/quote]

you were very clear, I just clearly have been working in settings without a metric far too often lately :)

$ x \to d(x,F)$  is continuous, and hence assumes a (positive) minimum on $ K$.

Perhaps I phrased my question wrong.  What I meant to ask is, "What do you mean by 'topology'?  I have not taken a topology course so I'm more familiar with metric spaces than topological spaces.

blahblahblah explained it as specifying what the open sets are.

One of the things that I feel Rudin does a very bad job of is separating the properties of a metric space that are unique to metric spaces and the properties of a metric space that extend to all topological spaces.  A [url=http://en.wikipedia.org/wiki/Topology#Mathematical_definition]topology[/url] is a collection of subsets of some set $ S$ called the "open sets" satisfying certain axioms.  In a metric space, the topology is generated by the open balls about every point in $ S$, but a topology can be defined without a metric.

That is, Rudin defines an open subset of a metric space as a set with the property that every point is interior, but an alternate equivalent definition is as the [i]finite or infinite union of finite intersections of open balls[/i], and this is the definition that generalizes.

Words like "open," "closed," "compact," and "continuous" are all topological and not metric words.  ("Uniformly continuous," on the other hand, cannot be defined in an arbitrary topological space.)

Here is the counter example I used in my homework:

{(n,1/n)} and {(n,-1/n)}

[quote="t0rajir0u"]One of the things that I feel Rudin does a very bad job of is separating the properties of a metric space that are unique to metric spaces and the properties of a metric space that extend to all topological spaces.[/quote]
I'm actually not using Rudin but your description was still helpful.  So a topology is somehow determined by the "open" sets within, whatever "open" means in the sense.

Not "somehow"; all a topology is is a list of the open sets, and that list [b]defines[/b] what it means to be open.

So if I take some arbitary set, make a random list of sets, I can decide that these sets are "open" and make that the topology.

No, because there are certain properties your list must satisfy.
(1) The empty set and the full space are open.
(2) The union of any collection of open sets is open.
(3) The intersection of any two open sets is open.


calculus
derivative
algebra
system of equations
algebra unsolved

Solve the system of equations:
$ \sqrt{x^2\plus{}91}\equal{}\sqrt{y\minus{}2}\plus{}y^2$
$ \sqrt{y^2\plus{}91}\equal{}\sqrt{x\minus{}2}\plus{}x^2$

[hide="not completed solution"]
$ x,y \geq 2$
if $ x <y$ then  $ \sqrt{x\minus{}2}\plus{}x^2<\sqrt{y\minus{}2}\plus{}y^2$ or $ \sqrt{x^2\plus{}91} > \sqrt{y^2\plus{}91}$ then $ x > y$
if $ x >y$ similarly...
$ x\equal{}y$
solve the equation $ \sqrt{x^2\plus{}91} \equal{} \sqrt{x\minus{}2}\plus{} x^2$.[/hide]

[hide]
If y > x, then
$ \sqrt{y^2\plus{}91} > \sqrt{x^2\plus{}91} \equal{} \sqrt{y\minus{}2} \plus{}y^2 > \sqrt{x\minus{}2} \plus{} x^2$.

So x = y, and $ \sqrt{x^2\plus{}91} \equal{} \sqrt{x\minus{}2} \plus{} x^2$. x = 3 is a solution. Note that, for the the range the equation is defined on, the right side will increase faster than the left side. (Use derivatives to see this). Therefore, there can only be one solution, so x = 3 is the only solution.
[/hide]

It is a problem in M&Y
http://www.mathlinks.ro/viewtopic.php?t=168836
Please lock this topic!


topology
Ring Theory
advanced fields
advanced fields unsolved

The title says it all :):

Prove that a compact Hausdorff topological ring is the limit of a projective system of finite rings.

(all rings are assumed to be unital, and ring homomorphisms preserve multiplicative units; the rings are not necessarily commutative)

What about the $p$-adic integers? They're Hausdorff because they have a metric, and compact by homeomorphism to a Cantor set. They're not a limit of finite rings, since the $p$-adic integers have characteristic zero and have no finite subrings.

$\mathbb Z_{p}$ is the inverse limit of $\mathbb Z/p^{n}\mathbb Z$ (with the natural homomorphisms). Profinite rings have to have lots of finite [i]quotient rings[/i], not subrings :?.

Just an idea: show first that compact rings which are fields must be finite.


algebra
system of equations
algebra unsolved

Find $ x,y,z \in R$ from system of equations: $ \begin {cases} x^3\minus{}3x\minus{}3\plus{}\ln(x^2\minus{}x\plus{}1)\equal{}y \\ y^3\minus{}3y\minus{}3\plus{}\ln(y^2\minus{}y\plus{}1)\equal{}z \\ z^3\minus{}3z\minus{}3\plus{}\ln(z^2\minus{}z\plus{}1)\equal{}x \end {cases}$

More precisely the $ 9$ solutions in $ x$ are:
$ \{u_1\approx{2.18769},u_2\approx{\minus{}1.8324},u_3\approx{\minus{}0.6360},u_4\approx{\minus{}1.7764},u_5\equal{}f(u_4),u_6\equal{}
f(u_5),u_7\approx{\minus{}1.6226},u_8\equal{}f(u_7),u_9\equal{}f(u_8)\}$
where $ f(u_1)\equal{}u_1,f(u_2)\equal{}u_2,f(u_3)\equal{}u_3$ are the $ 3$ fixed point of $ f$.


logarithms
calculus
integration
trigonometry
calculus computations

Should we take that square root as the one coming from the principal branch cut of the complex logarithm?

hello, Maple 13 can't find a value of that integral.
Sonnhard.

The integral is only well defined on the interval $ \left[0, \frac{\pi}{2}\right]$, so perhaps there is a typo, and the integral in question is of the form

$ I \equal{} \int_0^{\pi/2} \frac{(\sin x)^a}{(\sin x)^a \plus{} (\cos x)^a}\,dx$,

in which case we have $ I \equal{} \frac{\pi}{4}$ for every a.

More harder problem which bgbgbgbg posted [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=305582[/url]

Mathematica 7.0 computed x/2 + 1/2 Log[
   Cos[x] - Sin[x]] + (Sqrt[2]
     Cos[x] ((1 -
         I) EllipticPi[-I, -ArcSin[Sqrt[Tan[x/2]]], -1] + (1 +
         I) EllipticPi[I, -ArcSin[Sqrt[Tan[x/2]]], -1] -
      EllipticPi[1 - Sqrt[2], -ArcSin[Sqrt[Tan[x/2]]], -1] -
      EllipticPi[1 + Sqrt[2], -ArcSin[Sqrt[Tan[x/2]]], -1]) Sqrt[
    Tan[x/2]])/(Sqrt[Cos[x]/(
    1 + Cos[x])] (Cos[x] -
      Sin[x]) ((Sqrt[
         Cos[x]] ((1 -
              I) EllipticPi[-I, -ArcSin[Sqrt[Tan[x/2]]], -1] + (1 +
              I) EllipticPi[I, -ArcSin[Sqrt[Tan[x/2]]], -1] -
           EllipticPi[1 - Sqrt[2], -ArcSin[Sqrt[Tan[x/2]]], -1] -
           EllipticPi[1 + Sqrt[2], -ArcSin[Sqrt[Tan[x/2]]], -1]) Sec[
          x/2]^2)/(2 Sqrt[2] Sqrt[Cos[x]/(1 + Cos[x])] Sqrt[Sin[x]]
          Sqrt[Tan[x/2]]) - (Cos[x]^(
         3/2) ((1 -
              I) EllipticPi[-I, -ArcSin[Sqrt[Tan[x/2]]], -1] + (1 +
              I) EllipticPi[I, -ArcSin[Sqrt[Tan[x/2]]], -1] -
           EllipticPi[1 - Sqrt[2], -ArcSin[Sqrt[Tan[x/2]]], -1] -
           EllipticPi[1 + Sqrt[2], -ArcSin[Sqrt[Tan[x/2]]], -1]) Sqrt[
         Tan[x/2]])/(Sqrt[2] Sqrt[Cos[x]/(1 + Cos[x])] Sin[x]^(
         3/
          2)) - (((1 -
              I) EllipticPi[-I, -ArcSin[Sqrt[Tan[x/2]]], -1] + (1 +
              I) EllipticPi[I, -ArcSin[Sqrt[Tan[x/2]]], -1] -
           EllipticPi[1 - Sqrt[2], -ArcSin[Sqrt[Tan[x/2]]], -1] -
           EllipticPi[1 + Sqrt[2], -ArcSin[Sqrt[Tan[x/2]]], -1]) Sqrt[
         Sin[x]] Sqrt[Tan[x/2]])/(Sqrt[2] Sqrt[Cos[x]] Sqrt[Cos[x]/(
         1 + Cos[x])]) - (Sqrt[
         Cos[x]] ((1 -
              I) EllipticPi[-I, -ArcSin[Sqrt[Tan[x/2]]], -1] + (1 +
              I) EllipticPi[I, -ArcSin[Sqrt[Tan[x/2]]], -1] -
           EllipticPi[1 - Sqrt[2], -ArcSin[Sqrt[Tan[x/2]]], -1] -
           EllipticPi[1 + Sqrt[2], -ArcSin[Sqrt[Tan[x/2]]], -1]) ((
           Cos[x] Sin[x])/(1 + Cos[x])^2 - Sin[x]/(1 + Cos[x])) Sqrt[
         Tan[x/2]])/(Sqrt[2] (Cos[x]/(1 + Cos[x]))^(3/2) Sqrt[
         Sin[x]]) + (Sqrt[2] Sqrt[Cos[x]] Sqrt[
         Tan[x/2]] (-(((1/4 + I/4) Sec[x/2]^2)/(
            Sqrt[1 - Tan[x/2]] (1 - I Tan[x/2]) Sqrt[Tan[x/2]] Sqrt[
             1 + Tan[x/2]])) - ((1/4 - I/4) Sec[x/2]^2)/(
           Sqrt[1 - Tan[x/2]] (1 + I Tan[x/2]) Sqrt[Tan[x/2]] Sqrt[
            1 + Tan[x/2]]) + Sec[x/2]^2/(

           4 Sqrt[1 - Tan[x/2]] Sqrt[Tan[x/2]] Sqrt[
            1 + Tan[x/2]] (1 - (1 - Sqrt[2]) Tan[x/2])) + Sec[x/2]^2/(
           4 Sqrt[1 - Tan[x/2]] Sqrt[Tan[x/2]] Sqrt[
            1 + Tan[x/2]] (1 - (1 + Sqrt[2]) Tan[x/2]))))/(Sqrt[
         Cos[x]/(1 + Cos[x])] Sqrt[Sin[x]])))

for the indefinite integral and the definite integral is $ \frac{\pi}{4}$.

Sorry I don't know why.


quadratics
abstract algebra
superior algebra
superior algebra unsolved

How to determine whether a quadratic ring has or hasn't a unique factorization???

By calculating it's class number (e.g. using the class number formula for abelian extensions). Or tackle them in somewhat general manner.
For special cases you may give counterexamples of uniqueness or may be able to show that the ring is euclidean.


geometry
3D geometry
combinatorics unsolved
combinatorics

Here are the problems I have tried to solve:

1. If there are 12 flavors of ice cream available and each cone has three scoops (order matters: chocolate on bottom is not the same as chocolate on top), how many different types of ice cream cones can be made if:

(a) There can be repeats of flavors
(b) Each scoop needs to be of a distinct flavor

2.
(a) In how many ways can three pints of ice cream out of 12 different flavors be chosen?
(b) In how many ways can three pints of ice cream can be chosen if the flavors don't need to be different?

My attempted solutions:

1.

(a) 12*12*12 = 1728 cones
(b) 12*11*10 = 1320 cones

2. This is the tricky one...does order matter here? If it does, then it's the same as above...if it doesn't, then I got:

(a) 12 C 3 = 12!/[(3!)(9!)]  = 6*11*10 = 660 ways
(b) 12 P 3 = 12!/(9!) = 12*11*10 = 1320 ways

Everything's good except for 2(b) -- 12 P 3 counts something completely different then 12 "multichoose" 3, and it's the latter that you want.

2. a) Assuming the flavors must be different, there are $ \binom{12}{3}\equal{}220$ ways. (You made an arithmetic mistake).

b) If all three flavors are selected, there are $ \binom{12}{3}$ choices. If two flavors are selected, there are $ \binom{2}{1}\binom{12}{2}$ choices (Choose $ 2$ flavors of the $ 12$, then choose which flavor to have two pints of). If a single flavor is selected, there are $ \binom{12}{1}$ choices.
Thus, the total ways to choose the pints of ice cream is $ \binom{12}{3}\plus{}\binom{2}{1}\binom{12}{2}\plus{}\binom{12}{1}\equal{}364$ choices.


function
real analysis
calculus
derivative
real analysis unsolved

Let $ f$ be a real valued function on $ (0,1)$ and $ M>0$

$ 1.$ Suppose that $ f$ is differentiable on $ (0,1)$ and $ |f'| \leq M$.
 Prove that $ f$ is bounded.

$ 2.$ Suppose that $ f$ is differentiable a.e. w/t Lebesgue measure on $ (0,1)$ and $ |f'|\leq M$ everywhere that $ f'$ id defined.
Does it follow that $ f$ is bounded? Please explain.

1) Mean value theorem.
2) Cantor staircase stretched to infinity near one or both ends.

1. It's a simple application of mean value theorem. You can even prove that $ f$ is in fact of bounded variation.

2. $ f$ needs not be bounded. We have counter-examples like $ \textstyle f \equal{} \sum \chi_{(0, \tfrac{1}{n}]}$. whose derivative is defined a.e. and vanishes whenever it exists. You can even find more exotic counter-examples which strictly increases to $ \infty$ on $ (0, 1)$ but nevertheless its derivative vanishes a.e. (some variant of Cantor-Lebesgue function would work).


Is it possible to put $ 3^{2005}$ apples in $ 4^{1997}$ boxes such that  each of the boxes have an odd number of apples?  :D

[quote="Elio (n)"]Is it possible to put $ 3^{2005}$ apples in $ 4^{1997}$ boxes such that  each of the boxes have an [b]add [/b]number of apples?  :D[/quote]

add?
do u mean odd?

no...
for every 2 boxes, the problem requires there to be an even number of apples. so we can group the boxes such that there are 2 boxes in every group. this is possible with none left over since there are 4^1997 boxes (which is even).
4^1997=2^3994 (since 2^2=4)
there are 2^3994/2=2^3993  groups of 2 boxes, each of which are ven. so we are adding 2^3993 even numbers together, which will end up being an even number.
therefore there must be an even number of apples for an odd number of apples to be in each of 4^1997 boxesbut 3^2005 is odd

alternatively, you can prove 3^2005 is less than 4^1997 and there must be at least 1 apple per box but there are not enough apples to make this happen

But can't there be a apple in a box in a box.

So $ 3^{2007}$ apples in one box and that box is in all the other boxes.
Every box has $ 3^{2007}$ apples which is odd.

YAY  :rotfl:


All of them.

AoPS's number 1 Halo Fan. SIGNED.

Since I have a Wii and not an Xbox 360, I only have Halo 1 for PC. Which I play nonstop. Which I never get bored of even after over a year.

Actually if anyone here has Halo 3 and Xbox Live I can organize a small tournament just for AoPSers.

[quote="ZhangPeijin"]How about you all get Halo Trial Version and we can play a nice CTF game on Blood Gulch?[/quote]

But it's on PC, which takes next to no skill. I'd rather play like straight up Slayer on Narrows on H3 Live.

[quote="ZhangPeijin"]But I don't have a xbox360.[/quote]

Do you have a friend who has one (plus Live)?

[quote="ZhangPeijin"]He has a 360 but no live.[/quote]

How can he have no Live? You get a free month with every account you create on your Xbox. Unless his parents are azn and he doesn't know how to hook up an ethernet cable...

[quote="ZhangPeijin"]How about you all get Halo Trial Version and we can play a nice CTF game on Blood Gulch?[/quote]

dude Blood Gulch gets SOOOOOO OLD after like... a month of halo trial

[quote="not_trig"]dude Blood Gulch gets SOOOOOO OLD after like... a month of halo trial[/quote]

Try playing BTB on Coagulation (Halo 2 remix of Blood Gulch) all night  :wink: .

His parents are too lazy to cancel the trial when it ends, and they think they may forget, so they will get charged more money...
Anyways... Yeah.
Get Halo Trial.

[quote="ZhangPeijin"]His parents are too lazy to cancel the trial when it ends, and they think they may forget, so they will get charged more money...[/quote]

Just tell him to make a new account...you'll start off with no achievements or anything but you get a free month just to play in the tournament. And once that live account ends they don't start charging, they ask first if you want to renew and you can always say no.

[quote="ZhangPeijin"]But I wanna play too.
 :([/quote]

Tell him to create a new account, hook up to Live. Then, when you go over to his house sign on as his guest...

n0vad3m0n (those are zeros). When he sends me a friend invite tell him to put "Peijin" in the text so I won't delete it thinking it's spam.

My sister can play too so it'll be a Doubles Match (unless you want to do a Free For All?).

Doesn't matter...
Though I've never played on xbox yet so I'll probably suck.


From a set of 8 women and 5 men, how may ways are there to form a committee if should have 4 people with 2 of each gender , and Mr. and Mrs. X  cannot be in the committee?

I presume you mean that Mr and Mrs X can't both be on the committee.

If Mr X is on the committee, then we are left with $ 8$ women and $ 4$ men: $ \binom{8}{2} \times \binom{4}{1}$.

If Mrs X is on the committee, then we are left with $ 7$ women and $ 5$ men: $ \binom{7}{1} \times \binom{5}{2}$.

If niether of them is on the committee, then we have : $ \binom{7}{2} \times \binom{4}{2}$.

I make this $ 28 \times 4 \plus{} 7 \times 10 \plus{} 21 \times 6 \equal{} 308$.


Let triangle ABC. $ I$ is incenter. $ E$ is center of Euler circle. Prove
$ IE\equal{}\frac{R^2}{2}\minus{}R$

[quote="duongptnkvn"]I think it's very easy.It's a problem in secondary school in Vietnam.
 We can use inverse to prove it.[/quote]
can you post the Proof???

[quote="mufc"]Let triangle ABC. $ I$ is incenter. $ E$ is center of Euler circle. Prove $ IE \equal{} \frac {R^2}{2} \minus{} R$[/quote]
This relation is no nicely because it has different dimensions left and right ...

$ G\in (OH)\cap (IN)$, where $ N$ is the [b]Nagel's point[/b].

Thus, $ \left\|\begin{array}{c} GO \equal{} 2\cdot GE \\
\ GN \equal{} 2\cdot GI\end{array}\right\|$ $ \implies$ $ \left\|\begin{array}{c} IE\parallel ON \\
\ 2\cdot IE \equal{} NO\end{array}\right\|$

The well-known relation $ \boxed {ON \equal{} R \minus{} 2r}$ $ \implies$  $ IE \equal{} \frac R2 \minus{} r\ .$


Let $ p(x) \equal{} x^6 \plus{} ax^5 \plus{} bx^4 \plus{} cx^3 \plus{} dx^2 \plus{} ex \plus{} f$ be a polynomial such that $ p(1) \equal{} 1, p(2) \equal{} 2, p(3) \equal{} 3, p(4) \equal{} 4, p(5) \equal{} 5,$ and $ p(6) \equal{} 6.$ What is $ p(7)$?

A. 0
B. 7
C. 14
D. 49
E. 727

[hide="hrm"]
Consider $ G(x) \equal{} p(x) \minus{} x$.  $ G(x)$ has roots at $ 1,2,3,4,5,6$, and it is a 6th degree polynomial.

So, $ G(x) \equal{} p(x) \minus{} x \equal{} (x \minus{} 1)(x \minus{} 2)(x \minus{} 3)(x \minus{} 4)(x \minus{} 5)(x \minus{} 6)$

$ G(7) \equal{} p(7) \minus{} 7 \equal{} 720$.

So, $ \boxed{p(7) \equal{} 727}$[/hide]


topology
analytic geometry
function
advanced fields
advanced fields unsolved

let $x$ be a Hausdorff space and let $f: X \to \mathbb{R}$ be such that $\{(x,f(x)); x \in X\}$ is a compact subset of $X\times \mathbb{R}$. Show $f$ is continuous.

Let $x_{n}\to a$ be an arbitrary sequence in $X$, we show that $f(x_{n})\to f(a)$ and since the space is hausdorff we will be done. $(x_{n},f(x_{n}))$ is a sequence in a compact set, so it has a convergent subsequence that converges to say, $(a,b)$. But $b$ is nothing but $f(a)$ since that compact set is closed.

[quote="MeKnowsNothing"]I thought the theorem is true also in non-metrizable spaces. There one can't use sequences so I was thinking a proof using open sets.[/quote]Note that covering compactness implies sequential compactness, in [u]general[/u].

Let's replace $\mathbb{R}$ by an arbitrary topological space $Y$, and define $F=\{(x,f(x))\}$. For any closed $C\subset Y$, $X\times C$ is closed in $X\times Y$, and thus $(X\times C)\cap F$ is relatively closed in $F$. A closed subset of a compact set is compact, so $(X\times C)\cap F$ is compact. Projection $\pi_{X}$ onto the first coordinate is a continuous map, so $\pi_{X}((X\times C)\cap F)=f^{-1}(C)$ is compact. Since $X$ is Hausdorff, this makes $f^{-1}(C)$ closed. $C$ was arbitrary, and this is the definition of continuity.

The first solution of the problem given by The Capitain is only valid if $X$ is a first countable space, that is, every point of $x\in X$ must have a countable open local basis.

For a counterexample on why this argument is not valid, take the space $\begin{bold}\Omega\end{bold}$ of ordinals less or equal to the first uncountable ordinal $\Omega_{0}$. Give $\begin{bold}\Omega\end{bold}$ the order topology (which is Hausdorff) and consider the function $f: \begin{bold}\Omega\end{bold}\to\mathscr{R}$ given by $f(\alpha)=0$ if $\alpha\in\begin{bold}\Omega\end{bold}-\{\Omega_{0}\}$ and $f(\Omega_{0})=1$. Clearly this function is not continuous because $f^{-1}(\{1\})=\{\Omega_{0}\}$ which is not open. There are two types of convergent sequences in $\begin{bold}\Omega\end{bold}$. One type is the constant sequence $(\Omega_{0})_{n\in\mathbb{N}}$ and the other is a sequence totally contained in $\begin{bold}\Omega\end{bold}-\{\Omega_{0}\}$(or at least it's eventualy in that set), which must converge to a point of $\begin{bold}\Omega\end{bold}-\{\Omega_{0}\}$ (this is an easy result on ordinals). Then it is clear that $f$ sends convergent sequences to convergent sequences, although $f$ is not continuous.

The solution provided by jmerry is more general. However, the sequence proof may be refrased in terms of nets and I think it would be correct.

[color=red]EDIT[/color]: Sorry,  :P the argument I just posted doesn't say the first proof is wrong. I just showed that the fact that convergent sequences are sent to convergence sequences  does not imply that the function is continuous, but in that proof the other implication is used, that is always true....

So what I really need to show is that compacity cannot be deduced from the fact that all sequences have an acumulation point. Consider the space $\begin{bold}\Omega\end{bold}-\{\Omega_{0}\}$. By the same argument as before, every sequence must be contained in some $[0,\alpha]$ for some $\alpha\in\begin{bold}\Omega\end{bold}-\Omega_{0}$. But $[0,\alpha]$ is compact (it is a closed subset of $\begin{bold}\Omega\end{bold}$ which is compact) so the sequence considered has an acumulation point in $[0,\alpha]$. However, the space $\begin{bold}\Omega\end{bold}-\{\Omega_{0}\}$ is not compact (consider the covering $\{[0,\alpha]: \alpha\in\begin{bold}\Omega\end{bold}-\{\Omega_{0}\}\}$).

 :rotfl: OK now it's OK I think...

Hi.

Let $X$ be non-compact and $Y$ hausdorff.  Is there always a map $f: X\to Y$ such that the graph $\{(x,f(x)) | x\in X\}\subset X\times Y$ is closed in $X\times Y$ but $f$ is not continous?

It looks quite similar to the following problem:
Show that a topological space $X$ is compact if and only if for every other space $Y$ the projection $\pi_{2}: X\times Y\to Y$ is closed.

Hanno

Compactness of $X$ doesn't mean anything. It's $Y$ that needs to be noncompact. In fact, if $X$ is locally compact and $Y$ is compact, you can show that any function with a closed graph is continuous by my argument earlier in the thread.
You do need some condition on $X$; if $X$ is discrete, everything is continuous automatically.

Hi.

Omg :o , that was nonsense. Another try: Suppose $Y$ is non-compact. Is there a space $X$ and a non-continous map $f: X\to Y$ such that the graph of $f$ is closed in $X\times Y$?

Hanno


Actually, it's just a mathcounts program that was cancelled.

Now that I have your attention, one of the top math programs in the country has been cancelled by an evil principal.  If you want to help out, write this principal a letter (this is at Mr. Rusczyk's request)

Refer to his blog for more information.

Address your letter to
Principal Chris Hargrave
Muirlands Middle School
1056 Nautilus Street
La Jolla, CA 92037

Hey!  I needed to sound like someone famous in my letter, so I said that I was the President of the Organization of Kids in Mathematics (ORKIM).  Although this started as a joke, it has the potential to become real.  I already appointed reddevilz as the Vice President of ORKIM.   If anyone else is interested in a job at ORKIM, please PM me.

SEE MY BLOG FOR MORE DETAILS!!!!!!

I repeat: SEE MY BLOG, "THINK GLOBALLY, ACT LOCALLY" FOR MORE DETAILS.

wait, what are you going to do?  SEE MY BLOG FOR MORE DETAILS.

Of course it was.

Oh, and HIRE ME!! I already posted in your blog why I'm perfect from the job...

[quote="eryaman"]To be honest, I'm not really sure it was a good idea to do that.[/quote]

Of course it wasn't.  He needs to smarten up a bit.

[quote="blahblahblah"]He needs to smarten up a bit.[/quote]

That would be the understatement of the year.

Not to pick on anyone, but seriously there's better, more professional ways to get things done. For those of you at home, that's why we have adults - and more specifically, parents.

I feel bad for my friends who go to that school. Our mathcounts program was cancelled last year and it stunk.

I'm a little bit scared to look at bubala's blog from this thread.  So is this thing serious or not? :?

Bubala has expressed his displeasure at the comments of some members of the community on this thread and requested that I delete it.  Rather than doing so, I am posting the text of his message to me.  I would also recommend that those individuals he mentions re-consider how they talk about others as I agree with him that their comments were impolite and unnecessary.

[quote="bubala"]Hi Joel,

I know that you moderate the Round Table. I am somewhat offended by the comments made about me on this thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=36653

I ask that you delete it. I do not believe the Round Table is a forum for sladering others. And it is slander, considering the following:

Eryaman: "To be honest, I'm not really sure it was a good idea to do that."
Blahblahblah: "Of course it wasn't. He needs to smarten up a bit."

My response? Here is Mr. Rusczyk's blog. http://www.artofproblemsolving.com/Forum/weblog.php?w=1

rrusczyk: "So, I'm writing the principal a letter. Perhaps you might, too...Here's my letter. Feel free to copy whatever you like."

I only did the what MR. RUSCZYK asked us to do. I don't think that's the wrong thing, and I don't deserve to hear "you need to smarten up a bit" because I did Mr. RUSCZYK's bidding.

In fact, MR. RUSCZYK said : "Thanks for pitching in," to me in his comments section.

I was even more upset by the following comment by MithsApprentice. when blahblahblah said "he needs to smarten up a bit," Miths wrote: "That would be the understatement of the year." If someone said this about him, I don't think he would be flattered. On a side note, did you see his reason?

"Not to pick on anyone, but seriously there's better, more professional ways to get things done. For those of you at home, that's why we have adults - and more specifically, parents."

I hate this kind of attitude. I think that the only thing stopping people from making a difference is themselves. If one has the attitude "there are lots of problems in the world, but there's nothing I can do," then how is one to make a positive difference and change things????

I really would appreciate if you would delete this thread. Obviously, "What is popular is not always right, and what is right is not always popular, " but that doesn't mean that people should say offensive things about me on the round table that have no factual basis. For example, if you said, "bubala, some of your posts are immature," I wouldn't be offended, because that's true. But if you say "Bubula, you need to smarten up a bit, don't do what MR. RUSCZYK says next time, there's other people who can get things done in this world," especially on a PUBLIC forum, I do take offense.

I know it's your decision whether or not to remove the thread, but I'd sincerely appreciate it if you would. Thanks so much.[/quote]


What is the value of $ y$ in the equation $ \frac{30}{50}\equal{}\sqrt{\frac{y}{50}}$?

Square both sides of the equation to get $ \frac{900}{2500}\equal{}\frac{y}{50}$. Then divide both sides by 50 to get $ \frac{900\cdot50}{2500}\equal{}y\equal{}\boxed{18}$

Don't you mean multiply both sides by 50?

No, izzy.  You can square both sides in an equation.  And I think that the actual answer was not stated.  But it is 18.

[quote="hunter34"]Then [b]divide both sides by 5[/b]0 to get $ \frac {900\cdot50}{2500} \equal{} y \equal{} \boxed{18}$[/quote]

I mean in that step. I think hunter34 meant to say multiply. I just wanted to clarify.

Actually I'm not so sure about it.  But I know that the answer is 18.

Yeah, I agree the answer is 18. But it appears to me that he/she [b]did[/b] multiply by 50 on both sides, not divide, so I just wanted to clarify for whoever asked it.

1) Square both sides of the equation. $ \frac{900}{2500}\equal{}\frac{y}{50}$
2) Simplify $ \frac{18}{50}\equal{}\frac{y}{50}$
3) Logic -apparently, y=18

DUDE- JUST STOP TYPING THE EXACT SAME ANSWER WITH THE EXACT SAME SOLUTION!

IF U WANNA DO IT OVER AGAIN, GET A PIECE OF PAPER AND FRIGGIN DO IT!

@dragon96

[b]Kindly do not keep posting the same solution. Thank you.[/b]

@ernie

Now, now, Ernie, be nice.  :P


Find all positive integers  $(a,m,n)$, ${a,m}\ge{2}$  such that ${a^{m}+1}\mid(a+1)^{n}$


There is a machine with 8 toys in it that each cost between 25 cents and 2 dollars, with each toy being 25 cents more expensive than the next most expensive one. Each time Sam presses the big red button on the machine, the machine randomly selects one of the remaining toys and gives Sam the option to buy it. Sam has enough money, he will buy the toy, the red button will light up again, and he can repeat the process. If Sam has 8 quarters and a ten dollar bill and the machine only accepts quarters, what is the probability that Sam has to get change for the 10 dollar bill before he can buy his favorite toy- the one that costs $ \$1.75$? Express your answer as a common fraction.

We use complementary probability.

There are two cases: he can get the 25 cent toy and then the 1.75 toy, or 1.75 toy first.

Then we have $ \frac {1}{8} \times \frac {1}{7} + \frac {1}{8} = \boxed\frac {1}{7}$.  So our required probability is $ \boxed{\frac{6}{7}}$.

The only cases in which he doesn't have to get change is if he gets his favorite toy on the first turn, or gets the cheapest toy and then his favorite toy, which have a total probability of $ \frac18\plus{}\frac18\cdot\frac17\equal{}\frac17$.

Thus, the probability that he has to get change is $ 1\minus{}\frac17\equal{}\boxed{\frac67}$.

[quote="AIME15"]We use complementary probability.

There are two cases: he can get the 25 cent toy and then the 1.75 toy, or 1.75 toy first.

Then we have $ \frac {1}{8} \times \frac {1}{7} \plus{} \frac {1}{8} \equal{} \boxed{\frac {1}{7}}$.  So our required probability is $ \boxed{\frac {6}{7}}$.[/quote]

Fixed  :roll:

I don't understand the problem. Shouldn't the answer be 7/8 because 175/200 = 7/8?

The thing that might make all of your answers wrong is that the $1.75 toy must be bought after the 8 quarters are used up.
There are only 5 ways (the number is the cost of the toy(s) in quarters)
1) 8 quarters
2)6 then 2
3)5 then 3
4) 5 then 1 then 2
5) 4 then 3 then 1

Then the prob is 1/8 + 2(2(1/8 *1/7) + 2(6(1/8 *1/7 *1/6)
which is [b][u]13/56[/u][/b]

Not true... if I buy something for 6 quarters, I am left with 2 quarters, which is not enough to buy the 1.75 toy.

We solve by complementary probability.

We have two possible cases for him to buy the toy before breaking his 10 dollar bill: Getting the 25 cent toy then the 1.75 one, or just getting the 1.75 one first.

We have a $\dfrac{1}{8}\cdot \dfrac{1}{7}$ chance of the first case, and a $\dfrac{1}{8}$ chance of the second one.  Thus, our probability is

\[1-(\dfrac{1}{56}+\dfrac{1}{8})\implies 1-(\dfrac{1}{7})\implies \boxed{\dfrac{6}{7}}\]

I just did it this way:

He won't need change if he either gets the 1.75 toy first or the .25 then 1.75

P(getting 1.75 toy first) = 1/8
P(getting 0.25 then 1.75 toys) = (1/8)(1/7) = 1/56

P(He won't need change) = 1 - (1/8 + 1/56) = $\boxed{\dfrac{6}{7}}$

I thought I wasn't in english class?
[hide=Solution]
There are only two cases he can get the $7$ quarter toy with $8$ quarters; he either gets it on the first try with probability $\frac18$ or gets the $1$ quarter toy first and then the $7$ quarter toy, with probability $\frac18 \cdot \frac17 = \frac{1}{56}$. The answer is what we don't want, e.g. $1 - \left(\frac18 + \frac{1}{56}\right) = \boxed{\frac67}$.
[/hide]


function
induction
algebra unsolved
algebra

[quote]Given a function $ f : \mathbb N \to\mathbb N$ such that every $ n\in\mathbb N$ satisfies
$ (f(2n) + f(2n + 1) + 1)(f(2n + 1) - f(2n) - 1) = 3(1 + 2f(n))$
and
$ f(2n) \geq f(n)$.
Denote $ M = \{m \in f(\mathbb N): m \leq 2008\}$
Find the number of elements of $ M$.[/quote]

obviously $ f(2n + 1) - f(2n) - 1 > 0$ therefore $ f(2n + 1) > f(2n) \Rightarrow{f(2n + 1) + f(2n) + 1 > 2f(2n) + 1\geq2f(n) + 1\Rightarrow{f(2n + 1) + f(2n) + 1 > 2f(n) + 1}}$
therefore
$ 3(1 + 2f(n)) = (f(2n + 1) + f(2n) + 1)(f(2n + 1) - f(2n) - 1) > (1 + 2f(n))(f(2n + 1) - f(2n) - 1)$

$ \Rightarrow {f(2n + 1) - f(2n) - 1 < 3}$ also we have $ f(2n + 1) - f(2n) - 1|3(1 + 2f(n))$
therefore $ f(2n + 1) - f(2n) - 1$ is odd and therefore for all $ n$ we have:
$ f(2n + 1) - f(2n) - 1 = 1\Rightarrow{f(2n + 1) = f(2n) + 2}$
and by put this realation in original equation we get that for all $ n$:
$ f(2n) = 3f(n)$
It seem that for all value of $ f(1)$ exist such function and maybe some thing in the problem is missing... :maybe:

[quote]$ f(2n \plus{} 1) \minus{} f(2n) \minus{} 1|3(1 \plus{} 2f(n))$[/quote]
it means that $ (f(2n \plus{} 1) \minus{} f(2n) \minus{} 1)$ divides $ 3(1 \plus{} 2f(n))$

[quote="ali666"]obviously $ f(2n \plus{} 1) \minus{} f(2n) \minus{} 1 > 0$ therefore $ f(2n \plus{} 1) > f(2n) \Rightarrow{f(2n \plus{} 1) \plus{} f(2n) \plus{} 1 > 2f(2n) \plus{} 1\geq2f(n) \plus{} 1\Rightarrow{f(2n \plus{} 1) \plus{} f(2n) \plus{} 1 > 2f(n) \plus{} 1}}$
therefore
$ 3(1 \plus{} 2f(n)) \equal{} (f(2n \plus{} 1) \plus{} f(2n) \plus{} 1)(f(2n \plus{} 1) \minus{} f(2n) \minus{} 1) > (1 \plus{} 2f(n))(f(2n \plus{} 1) \minus{} f(2n) \minus{} 1)$

$ \Rightarrow {f(2n \plus{} 1) \minus{} f(2n) \minus{} 1 < 3}$ also we have $ f(2n \plus{} 1) \minus{} f(2n) \minus{} 1|3(1 \plus{} 2f(n))$
therefore $ f(2n \plus{} 1) \minus{} f(2n) \minus{} 1$ is odd and therefore for all $ n$ we have:
$ f(2n \plus{} 1) \minus{} f(2n) \minus{} 1 \equal{} 1\Rightarrow{f(2n \plus{} 1) \equal{} f(2n) \plus{} 2}$
and by put this realation in original equation we get that for all $ n$:
$ f(2n) \equal{} 3f(n)$
It seem that for all value of $ f(1)$ exist such function and maybe some thing in the problem is missing... :maybe:[/quote]
Then,what is solution?

I think some think is missing in the problem...because exist infinite many $ f$...actually let $ a$ be a arbitary natural number and define $ f(1)=a$ and for $ n>1$:
$ f(n) = \left\{\begin{array}{cc}3f(\frac {n}{2}), & \mbox{ if } \text{n is even} \\
f(n - 1) + 2, & \mbox{ if }\text{n is odd}{}\end{array}\right.$
obviusly $ f$ satisfy these three condition:
$ f: N\to{N} \\
(f(2n) + f(2n + 1) + 1)(f(2n + 1) - f(2n) - 1) = 3(1 + 2f(n)) \\
f(2n)\geq{f(n)}$
and $ |M|$ is dependent on $ a$...for example if $ a > 2008$ then $ |M| = 0$...
please correct me if I'm wrong...

[quote="ali666"]I think some think is missing in the problem...because exist infinite many $ f$...actually let $ a$ be a arbitary natural number and define $ f(1) = a$ and for $ n > 1$:
$ f(n) = \left\{\begin{array}{cc}3f(\frac {n}{2}), & \mbox{ if } \text{n is even} \\
f(n - 1) + 2, & \mbox{ if }\text{n is odd}{}\end{array}\right.$
obviusly $ f$ satisfy these three condition:
$ f: N\to{N} \\
(f(2n) + f(2n + 1) + 1)(f(2n + 1) - f(2n) - 1) = 3(1 + 2f(n)) \\
f(2n)\geq{f(n)}$
and $ |M|$ is dependent on $ a$...for example if $ a > 2008$ then $ |M| = 0$...
please correct me if I'm wrong...[/quote]
We have:
$ f(2n)=3f(n)$
With $ n=0$ we have $ f(0)=0$
We have $ f(2n+1)=f(2n)+2$ so $ f(1)=2<2008$

[quote="chien than"][quote="ali666"]I think some think is missing in the problem...because exist infinite many $ f$...actually let $ a$ be a arbitary natural number and define $ f(1) = a$ and for $ n > 1$:
$ f(n) = \left\{\begin{array}{cc}3f(\frac {n}{2}), & \mbox{ if } \text{n is even} \\
f(n - 1) + 2, & \mbox{ if }\text{n is odd}{}\end{array}\right.$
obviusly $ f$ satisfy these three condition:
$ f: N\to{N} \\
(f(2n) + f(2n + 1) + 1)(f(2n + 1) - f(2n) - 1) = 3(1 + 2f(n)) \\
f(2n)\geq{f(n)}$
and $ |M|$ is dependent on $ a$...for example if $ a > 2008$ then $ |M| = 0$...
please correct me if I'm wrong...[/quote]
We have:
$ f(2n) = 3f(n)$
With $ n = 0$ we have $ f(0) = 0$
We have $ f(2n + 1) = f(2n) + 2$ so $ f(1) = 2 < 2008$[/quote]
I think there is diffrenet definition for $ N$...as I know $ 0$ is not natural number and $ N=\{1,2,....\}$
but it seems the problem means $ N=\{0,1,2,...\}$
thanks chien than... :)
Now,I will try complete the solution...

suppose that $ n$ display in base $ 2$ by:$ a_{k}a_{k \minus{} 1}...a_{1}a_{0}$ then $ f(n)$ is:
$ \sum_{i \equal{} 0}^{k}2a_i\times3^i$ actually $ f(n)$ is $ (2a_k)(2a_{k \minus{} 1})...(2a_{0})$ in base $ 3$...
(one can prove this fact by induction on $ n$)
therefore $ f(N)$ are the number that when display in base $ 3$ don't have digit $ 1$.
therefore we shoud find the number that are $ \leq2008$ and when display in base $ 2$ don't have $ 1$...rest is easy!

[quote="ali666"]obviously $ f(2n \plus{} 1) \minus{} f(2n) \minus{} 1 > 0$ therefore $ f(2n \plus{} 1) > f(2n) \Rightarrow{f(2n \plus{} 1) \plus{} f(2n) \plus{} 1 > 2f(2n) \plus{} 1\geq2f(n) \plus{} 1\Rightarrow{f(2n \plus{} 1) \plus{} f(2n) \plus{} 1 > 2f(n) \plus{} 1}}$
therefore
$ 3(1 \plus{} 2f(n)) \equal{} (f(2n \plus{} 1) \plus{} f(2n) \plus{} 1)(f(2n \plus{} 1) \minus{} f(2n) \minus{} 1) > (1 \plus{} 2f(n))(f(2n \plus{} 1) \minus{} f(2n) \minus{} 1)$

$ \Rightarrow {f(2n \plus{} 1) \minus{} f(2n) \minus{} 1 < 3}$ also we have $ f(2n \plus{} 1) \minus{} f(2n) \minus{} 1|3(1 \plus{} 2f(n))$
therefore $ f(2n \plus{} 1) \minus{} f(2n) \minus{} 1$ is odd and therefore for all $ n$ we have:
$ f(2n \plus{} 1) \minus{} f(2n) \minus{} 1 \equal{} 1\Rightarrow{\boxed{f(2n \plus{} 1) \equal{} f(2n) \plus{} 2}}$
and by put this realation in original equation we get that for all $ n$:
$ \boxed{f(2n) \equal{} 3f(n)}$
It seem that for all value of $ f(1)$ exist such function and maybe some thing in the problem is missing... :maybe:[/quote]
for here we have $ \boxed{f(2n) \equal{} 3f(n)}\boxed{f(2n \plus{} 1) \equal{} f(2n) \plus{} 2}$
for $ n\in\mathbb{N}$ if $ n\equal{}\overline{a_ma_{m\minus{}1}...a_1a_0}^{(base\ 2)}$ and $ X$ is $ (n\minus{}a_0)/2$ is base 3
if $ a_{0}\equal{}0$ then $ f(\overline{a_ma_{m\minus{}1}...a_1a_0}^{(base\ 2)})\equal{}3f(\overline{a_ma_{m\minus{}1}...a_1}^{(base\ 2)})\equal{} \overline{X0}^{(base\ 3)} \equal{} \overline{X(2a_0)}^{(base\ 3)}$
if $ a_{0}\equal{}1$ then $ f(\overline{a_ma_{m\minus{}1}...a_1a_0}^{(base\ 2)})\equal{}3f(\overline{a_ma_{m\minus{}1}...a_1}^{(base\ 2)})\plus{}2\equal{}\overline{X2}^{(base\ 3)}\equal{} \overline{X(2a_0)}^{(base\ 3)}$
so like say ali666 $ f(\overline{a_ma_{m\minus{}1}...a_1a_0}^{(base\ 2)})\equal{}\overline{(2a_m)(2a_{m\minus{}1})...(2a_1)(2a_0)}^{(base\ 3)}$

now we have to find $ |f(\mathbb{N})\cap \{0,1,....,2008\}|\equal{}|S\cap \{0,1,....,1004\}|$ where $ S\equal{}$numbers wihtout $ 2$ in representation in base $ 3$


limit
algebra
polynomial
ratio
system of equations
calculus
calculus computations

[quote="Carcul"]We must have
$5a-b+4c = 0$
$2a+b-3c = 0$
$-a+5b-c = 2$

By solving this system of equations we get:

$a =-\frac{2}{109}, b = \frac{46}{109}, c = \frac{14}{109}$.[/quote]
could someone explain Carcul' reply?

Thanks

For the limit to be 1, the highest power of the polynomial in the numberator and the highest power of the polynomial in the denominator must be equal, and the ratio of the leading coefficient on top to the leading coefficient on bottom must be 1.


geometry
3D geometry
real analysis
real analysis unsolved

What is the minimum sum of lengths of line segments which joins the vertexes of unit cubes?

What do you mean by "joins the edges"? :?. Should we understand the question as "Find the length of the shortest broken line that contains at least one point on each edge of the unit cube" ?

Sorry. I meant vertexes. I was looking the shortest broken line that contains the eight vertexes of the unit cube.


Find all functions $f:(0,+\infty)\to \mathbb{R}$ such that \[f(x)-f(y)=(y-x)f(xy)\]

Not hard at all, actually. Put y=1, then

$f(x)-f(1)=(1-x)f(x)$

$f(x)+(x-1)f(x)=f(1)$

$xf(x)=f(1)$

$f(x)=\frac{f(1)}{x}$

Since $f(1)$ is constant, we put $k=f(1)$ and then $f(x)=\frac{k}{x}$, where k is any real number. Plugging this back into the original equation, we need to have

$\frac{k}{x}-\frac{k}{y}=(y-x)\frac{k}{xy}$

which is clearly true upon multiplication by $xy$.$\Box$

[quote="pigfly"]I agree with [b]Saul[/b] that in the case[b] Pascual[/b] required, the problem is easy, but I think in the exam questions of Slovak, the problems must be as follow:
Find all $f :S \to \mathbb R $  (where $S=\{x/ x\in \mathbb R,x>1\}$)satisfying:
\[f(x)-f(y)=(y-x)f(xy)\]
In this case,this problem is not easy! ;)[/quote]
I also think that ths problem is rather hard.
I've got : $(x+1)f(x+1)=x^{2}f(x^{2}), f(x+1)=x^{2}f(x^{2}(x+1))$. This is all for the moment, can it help?

ah, I made a mistake, the real problem is the one stated by pigfly wich is hard

[quote="pigfly"]I solved this problem two years ago.The answer is $f(x)=\frac{C}{x}$ ( where [b]C=const  [/b]).
[b]Hint[/b]
You can use a arbitrary real number $z$ in the relationship to solve it:
 \[f(x)-f(y);f(y)-f(z);f(z)-f(x)\]
If you don't know my ideas,please tell me!(My E is so poor :( )[/quote]
Can you write the solution or write the hint more clearer?

Best regards.

[quote="pigfly"]I agree with [b]Saul[/b] that in the case[b] Pascual[/b] required, the problem is easy, but I think in the exam questions of Slovak, the problems must be as follow:
Find all $f :S \to \mathbb R $  (where $S=\{x/ x\in \mathbb R,x>1\}$)satisfying:
\[f(x)-f(y)=(y-x)f(xy)\]
In this case,this problem is not easy! ;)[/quote]
yes, i use f(2),f(4) to solve this pro
and what happend if  $S=\{x/ x\in \mathbb R,x>k\}$, k is real

[quote="Armo"]
Can you write the solution or write the hint more clearer?

[/quote]
OK,here is the solution:
\[(\frac {1}{yz}-\frac {1}{xy})f(xz)=(\frac {1}{xz}-\frac {1}{xy})f(yz)+(\frac {1}{yz}-\frac {1}{xz})f(xy),\forall x,y,z >1\]
Put $a=xy,b=xz,c=yz$.Then:
$f(c)=\frac {1}{c} \cdot (\frac{f(b)-f(a)}{\frac{1}{b}-\frac {1}{a}})
 +\frac{f(a)}{b}-\frac{f(b)}{a}$
Fixed $a,b$ so $f(c)=\frac{A}{c}+B$.
It's easy to check that $B=0$ and deduce that :$f(x)=\frac{A}{x}$.
Hey [b]nguyenquockhanh[/b], $k$ is not important in above solution. Can you post your solution which you use $f(2),f(4)$?Thanks!

okie: here is my solution: using the hypothesis we have
[tex]f(x)-f(y)=(y-x)f(xy)[/tex]
[tex]f(xy)-f(y)=y(1-x)f(xy^2)[/tex]
[tex]f(x)-f(y^2)=(y^2-x)f(xy^2)[/tex]
thus [tex]f(x)\frac{xy(x-y-1+\frac{1}{y})}{y(1-x)(y-x)}=f(y^2)+f(y)\frac{(x-y^2)(1+y-x)}{y(1-x)(y-x)}f(y)[/tex]
Now,setting [tex]x=\frac{y^2+y-1}{y}[/tex](>y>k) we obtain [tex]yf(y^2)=f(y)[/tex]
hence [tex]xf(x)=yf(y)[/tex] for all x><[tex]\frac{y^2+y-1}{y}[/tex]
and f(x)=[tex]\frac{c}{x}[/tex]for all x>k
NOTE: iff k=1 then y=2,[tex]\frac{y^2+y-1}{y}[/tex]=4

pigfly,i think you have some mistake: fixed a,b-> c<ab

[quote="nguyenquockhanh"]
pigfly,i think you have some mistake: fixed $a,b\Rightarrow c<ab$ [/quote]
My ideas
For all fixed point $a,b$ ,there exists $A,B$ sucth that:
\[f(c)=\frac{A}{c} +B,\forall 1<a<b<c<ab\]
so $B=0 \Rightarrow \frac{f(a)}{b}=\frac{f(b)}{a}$.Hence,$a\cdot f(a)=b\cdot f(b)\Rightarrow f(x)=\frac{k}{x}$.
Where are my mistakes?Thanks!
@Khanh:Are you pupil in Le Hong Phong school (Nam Dinh)?

[quote="Armo"][quote="nguyenquockhanh"]can you tell me how to check B=0????
@pigfly: i'm a pupil in LE Hong Phong-NAMDINH city, and who are you!@[/quote]
Just put $f(c)=\frac{A}{c}+B$ in original equation.[/quote]

oh, no- arno, you must note: c<ab so can't put $f(c)=\frac{A}{c}+B$ in original equation
okie
goodluck

[b]nguyenquockhanh[/b] told me he think the solution is right and above solution is not clearly,[b]nguyenquockhanh[/b]'s ideas:Let $a,b>k^2 \Rightarrow ab>k^4$ and put $c,d\to k^2$ in the original equation then $B=0$.

Nguyenquockhanh, I understand what you mean. I think pigfly's idea is correct and here is a way to make it clear.We have got that for fixed $a_{1},b_{1}$ there exist $A,B$ such that for every $c\in(1,a_{1}b_{1}) f(c)=\frac{A}{c}+B$. Now take $a_{2}, b_{2}$ such that $a_{2}b_{2}>a_{1}b_{1}$, we have that for every $c\in(1,a_{2}b_{2})$ there exist $C$ and $D$ such that $f(c)=\frac{C}{c}+D$.Now let's fix what we have:
$\forall c\in(1,a_{1}b_{1}), f(c)=\frac{A}{c}+B;
\forall c\in(1,a_{2}b_{2}) , f(c)=\frac{C}{c}+D;$ thus $C=A$ and $B=D$,so we can always make interval $(1,ab)$ larger and we always get the same function $f(x)=\frac{A}{x}+B$. Now put it in original equation to get $B=0$.QED
Is it clear now?
         :)

okie, sorry for stupid :D
and now, we should finish topic in here
goodluck

Just a conjecture: Do you think the condition in the problem implies the continity of the function f? Can we prove that f is bounded in any interval (a,b)?

I found another solution( but the asker gave the continuty to the function f :lolz)

It's here:
http://www.ddtoanhoc.net/forum/index.php?showtopic=2057&st=0&


I saw this inq 2 weeks ago but i can't remember the author
       1)Let a,b,c >0 anh abc=1.Prove that
          sum  \sqrt (8a 2 +1)<=3(a+b+c)

      I had the sol anh found a general result
       2)for 0<=k<=8.Prove that
          sum \sqrt (ka 2 +9-k)<=3(a+b+c)

       3)find the maximum of r such that
          for all 0<=k<=r<=9,we have inq 2
       (note: when r~9,for example r=8.99,the inq 2 is wrong)
      I haven't solved problem 3 yet.And i think it not simple

The first is my problem. The second is very interesting and I'm really interested in a solution. The third is (my opinion) incredibly hard.

This inq is really difficult.My sol is not nice.
  I don't have enough time,so i will prove problem 2 in some steps
  we can assume that a  \geq  b  \geq  c
  Step   1)a \geq 1 \geq b \geq c
    (a-1)(b-1)(c-1) \geq 0
     a+b+c \geq ab+bc+ca
     A= sum(9-k+ka 2 )bc
      =k(a+b+c)+(9-k)(ab+bc+ca) \leq 9(a+b+c)
     We have sum1/ab=a+b+c
     By C-C inq A*(sum 1/ab) \geq LHS 2
     qed

I also solved this case, with Cauchy:
  (sum sqrt(ka^2+9-k))^2<=(sum a)(k*sum a+(9-k)sum ab)<=3sum a.

I think you are interested in step 2.It's an important point in my solution
Step 2) a \geq b \geq 1 \geq c
   g(x)=3x- \sqrt (9-k)+kx 2
   I prove that g(a)+g(b) \geq 2g( \sqrt ab) for all a,b \geq 1
   g(x)=f(lnx),so f(x)=g(e x) for all x \geq 0
   We prove f(a)+f(b) \geq 2f((a+b)/2) by Jensen
   f''(x)=e x (3-(ke x/ \sqrt (9-k+ke 2x ))-(k(9-k)e x/ \sqrt (9-k+ke 2x ) 3 )))
   To prove f''(x)>0,we prove h(t)=(kt/ \sqrt (9-k+kt 2 ))
+k(9-k)t/ \sqrt (9-k+kt 2 ) 3  <3 for all t \geq 1   (by replace e x =t \geq 1)
   the equation h'(t) =0 can easily to solve
   So h(t)<3 for all t \geq 1
   So f''(x)>0 for all x \geq 1
 Notice that step 1 and step 2 are right when 0 \leq k \leq 9.So inq 2 is true iff this inq is true
     Let a 2 c=1 ,a \geq 1 \geq c>0.Prove that
     2 \sqrt (9-k+ka 2 )+ \sqrt (9-k+kc 2 ) \leq 6a+3c

In step 3 I will prove that  this inq is true for 0 \leq k \leq 8

[u]Step 3)[/u]fixed a \geq 1,I will prove that
   f(k)=2 \sqrt (9-k+ka 2) + \sqrt (9-k+k/a 4 ) \leq 6a +3/a 2  for 0 \leq k \leq 8
   f'(k)=(a 2 -1)/ \sqrt (9+k(a 2 -1)) + ((1/a 4) -1)/ \sqrt (9+k(1/a 4-1))
   the equation f'(k)=0 has one root
    K=9((2a 2 +1)(2a 4 +a 2 +1))/(a 2 +1)(4a 4+a 2 +1))
   since a \geq 1,we have K>8 so f'(k) =0 has no root when 0  \leq k \leq 8
   f'(0)>0
   So f(k) increase ,then f(k) \leq f(8)

    [u]Step 4  [/u] a \geq 1 .Prove that
    2 \sqrt (1+8a 2 )+ \sqrt (1+8/a 4) \leq 6a +3/a 2
       Square two times,we have
         16a^12-168a^10+288a^9+9a^8-360a^7+280a^6-138a^4+72a^3+1 \geq 0
    <=>  (a-1) 2 (16a^10+32a^9-120a^8+16a^7+161a^6-54a^5+11a^4+76a^3+3a^2+2a+1) \geq 0
      We have
           54a^6 \geq 54a^5
           107a^6+16a^10 \geq 2 \sqrt (...)*a^8  (AM-GM)
           32a^9+16a^7 \geq 2 \sqrt (...)a^8
  So this inq is right when k=8
  by step 3 and step 2,this inq is right for 0 \leq k \leq 8
Please check this solution.Thanks
Note :in step 3 the equation f'(k)=0 has only one root   8<K \leq 9
So put R=min K when a \geq 1
This inq is wrong for all k>R

Hmmmm,... I think I was right when I said it is an extremely difficult problem. I won't check your solution! Guess why. Anyway, congratulations, I still couldn't find a solution.


In this sum of two 2-digit numbers, each letter represents a distinct digit, but if a letter is used twice, it represents the same digit each time. What is the least possible value for A?

AB+BA=CDC

or

[hide]Q[/hide]AB

+BA

------

CDC

A can't be 1 since then B is 9 and that means 91+19=110 and doesn't work, so a is 2 and B is 9. 92+29=121


Inside a triangle $ABC$ there's a point $P$ such that $AP=3$,$BP=2$,$CP=1$, if $ABC$ has the maximun area possible, then $P$ is the:

$a)centroid$     $b)incenter$     $c)circumcenter$     $d)orthocenter$     $e)ex-center$

of triangle $ABC$.

If $ABC$ is the triangle with greatetst area, the line $l$ parallel to $AC$ wich contains $B$ intersects the circle $\Gamma$ with centre $B$ and radius $2$ in an unique point ($B$). Then $l$ is tangent to $\Gamma$ at $B$, and $BP$ is orthogonal to $AC$.
In the same way, $AP$ and $CP$ are orthogonals to $BC$ and $AB$ respectively.
In particular, $P$ is the orthocenter.

Sorry, a little correction:

If $ABC$ is the triangle with greatetst area, the line $l$ parallel to $AC$ wich contains $B$ intersects the circle $\Gamma$ with centre $P$ and radius $2$ in an unique point ($B$). Then $l$ is tangent to $\Gamma$ at $B$, and $BP$ is orthogonal to $AC$.
In the same way, $AP$ and $CP$ are orthogonals to $BC$ and $AB$ respectively.
In particular, $P$ is the orthocenter.


A large country has several villages and no proper roads. They decide to share the costs of building roads. Each village decides that it will finance the road to its nearest village. For each village, there is a unique 'nearest village'.  (It may so happen that two villages may each be the nearest to each other - in that case they may share the cost - but that doesn't matter here).

Based on above, can we solve the following:

$1.$ Is it possible that some of the roads form a closed polygon?
$2.$ Is it possible that two roads cross each other?
$3.$ What is the maximum number of villages that one village can get a direct road to?

Any idea?

im getting no, no, and 5.  I haven't finished the proofs yet though but just think circles.

the first one is true...consider a scalene triangle

the second one i believe is not true...

say we have segments AB, and CD, with AB>CD and them intersecting at X

draw circles centered at A,B with radii AB, the circles intersect at Y,Z

then C,D must like outside that area otherwise, the road would not be with CD, but YZ=AB, but to not be inside the circles, and intersect the line, CD>AB...contradiction

the last one

i believe this is associated with a regular hexagon, but i cannot prove it rigirously just yet...consider in that exact arrangement, there are not unique roads, but i cannot also show that we get a max when the center point has equal distances from the other villiages...

I still think 1 is no, heres my reasoning:

Start with points A and B with distance x.  In order for B to be A's closest city there can be no cities that are shorter than x distance away so draw a circle.

Now In order for B to have a closest city that isn't A it must have a city with distance y such that y<x  so make a circle with radius x around B and this time our next point must be in it. Call it point C.

Now if we want to attach it back to point A it must be less then y distance,  however since point  C cannot be within x distance of A and y<x then they cannot be attached.

ok and Number 3 I too started with a regular hexagon around one point:

Ok since if they were in a regular hexagon around the one point then they would also be closest to two other cities on the hexagon.  So lets remove a point and space them out slightly.

Now in order for the middle point to have a unique nearest city there must be one inside the hexagon (well now a pentagon).  So we'll put it in where the other one is gone.  In doing this we see that the two points on the outside are closer to this one now then the center, so lets remove another from the outside.

Now we have four on the outside and one on the inside and it works, however I have no idea as to how to prove thats the best arrangment


[quote="hardsoul"][quote="N.T.TUAN"]Let $a, b, c$ be positive real numbers. Prove that
\[1+\frac{3}{ab+bc+ca}\geq\frac{6}{a+b+c}. \]
[/quote]

 $1) 3(ab+bc+ca) \le (a+b+c)^{2}$
 $2) 1+\frac{9}{x^{2}}\ge \frac{6}{x}$[/quote]
nice sol....but with this method of yours,you cant solve my inequality,it has a different solution,try to solve that to...

1)$1+\frac{3}{ab+bc+ca}\geq\frac{6}{a+b+c}\longleftrightarrow (a+b+c)+\frac{3(a+b+c)}{ab+bc+ca}\geq 6$
now by AM-GM we have :
$(a+b+c)+\frac{3(a+b+c)}{ab+bc+ca}\geq 2\sqrt \frac{3(a+b+c)^{2}}{ab+bc+ca}$
we know that : $(a+b+c)^{2}\geq 3(ab+bc+ca)$ $\longrightarrow$ $(a+b+c)+\frac{3(a+b+c)}{ab+bc+ca}\geq 6$
2)$abc=1 , 1+\frac{3}{a+b+c}\geq\frac{6}{ab+bc+ca}\longleftrightarrow ab+bc+ca+\frac{3(ab+bc+ca)}{a+b+c}\geq6$
by AM-GM we have:
$ab+bc+ca+\frac{3(ab+bc+ca)}{a+b+c}\geq 2\sqrt \frac{3(ab+bc+ca)^{2}}{a+b+c}$
$(ab+bc+ca)^{2}\geq 3abc(a+b+c) \longrightarrow$ $(ab+bc+ca)^{2}\geq 3(a+b+c)$ $\longrightarrow$ $ab+bc+ca+\frac{3(ab+bc+ca)}{a+b+c}\geq 6$

fjwxcsl's attempt is almost comical for such a simple problem

[quote="BaBaK Ghalebi"]there is a similar problem:
$abc=1$,prove that $1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$[/quote]
let $a=\frac{1}{x}$ and $b=\frac{1}{y}$ and $c=\frac{1}{y}$.So we have $xyz=1$.So $1+\frac{3}{xy+yz+xz}\geq \frac{6}{x+y+z}$.Then like hardsoul.

[quote="N.T.TUAN"]Let $a, b, c$ be positive real numbers. Prove that \[1+\frac{3}{ab+bc+ca}\geq\frac{6}{a+b+c}.\][/quote]

we can use uvw-method to solve this problem easily
 $ a+b+c=3u,\ ab+bc+ca=3v ^2 $
then  $ u\geq v>  0 $
So the ineq above is equa to  $ 1+\frac{1}{v ^2}\geq \frac{2}{u}  $
because $ u\geq v>  0 $
$ 1+\frac{1}{v ^2}\geq \frac{1}{u ^2} $
So we only need to prove
 $ 1+\frac{1}{u ^2}\geq \frac{2}{u}  $
$ \ (1-\frac{1}{u })^2 \geq\ 0 $
that's sure

Very nice problem!
My solution:
$ a+b+c\ge\sqrt{3(ab+bc+ca)}\Leftrightarrow\frac{6}{a+b+c}\le\frac{6}{\sqrt{3(ab+bc+ca)}} $
It is enough to show that:
$ 1+\frac{3}{ab+bc+ca}\ge\frac{6}{\sqrt{3(ab+bc+ca)}} $
witch with substitution $ t=\sqrt{3(ab+bc+ca)} $ returns in
$ 1+\frac{9}{t^2}\ge\frac{6}{t}\Leftrightarrow (1-\frac{3}{t})^2\ge 0 $
Because the last inequality is obvious, the proof is finished!
__________
Sandu Marin

[u][b]My one liner[/b][/u]:

Applying [b]AM-GM[/b] inequality we get:
$1+\frac{3}{\sum ab}\ge 2\sqrt{\frac{3}{\sum ab}}\ge 2\sqrt{\frac{3}{{\frac{(\sum a)^2}{3}}}}=2\sqrt {\frac{9}{(\sum a)^2}}=\frac{6}{a+b+c}$

Maths is the doctor of science.

Sayantan

[quote=N.T.TUAN]Let $a, b, c$ be positive real numbers. Prove that \[1+\frac{3}{ab+bc+ca}\geq\frac{6}{a+b+c}.\][/quote]
Let $a, b, c,d$ be positive real numbers. Prove that$$1+\frac{6}{ab+ac+ad+bc+bd+cd}\geq\frac{8}{a+b+c+d},$$$$1+\frac{4}{ab+bc+cd+da}\geq\frac{8}{a+b+c+d}.$$

[quote=N.T.TUAN]Let $a, b, c$ be positive real numbers. Prove that \[1+\frac{3}{ab+bc+ca}\geq\frac{6}{a+b+c}.\][/quote]

Notice that by [b]AM-GM[/b] we have that:
$1+\frac{3}{ab+bc+ca}\ge 2\sqrt{\frac{3}{\sum_{cyc}ab}}$
$\Longrightarrow \frac{3}{\sum_{cyc}ab}\ge \frac{9}{(\sum_{cyc}a)^2}$ $\implies (\sum_{cyc}a)^2\ge 3\sum_{cyc}ab$
$\Longrightarrow (\sum_{cyc}a)^2 - 2\sum_{cyc}ab\ge \sum_{cyc}ab$
$\Longrightarrow \sum_{cyc}a^2\ge \sum_{cyc}ab$ which implies our result.
And this finishes our proof!
So we are done! :nhl:


function
algebra
functional equation
algebra proposed

Find all $ f: R\to R$ such that $ f[(x\plus{}1)f(y)]\equal{}y[f(x)\plus{}1]$

[hide="Attempt"]Making x = y = 0 we get $ f(f(0)) \equal{} 0$. Now make $ y \equal{} f(0)$: then we have $ f(0) \equal{} f(0)[f(x) \plus{} 1]$. Now, if $ f(0) \neq 0$, then this last expression implies $ f(x) \equal{} 0$ for all x, which does not verify the given functional equation. Therefore we must have $ f(0) \equal{} 0$. Now making x = 0 in the given functional equation yields $ f(f(y)) \equal{} y$, or $ f(f(x)) \equal{} x$. Making $ y \equal{} f(x)$ in the given equation then yields $ f[x(x\plus{}1)] \equal{} f(x) [f(x) \plus{} 1]$. By now its obvious that f(x) = x is a/the solution, but how to complete the solution? Any hint?[/hide]

[quote="Carcul"][hide="Attempt"]Making x = y = 0 we get $ f(f(0)) \equal{} 0$. Now make $ y \equal{} f(0)$: then we have $ f(0) \equal{} f(0)[f(x) \plus{} 1]$. Now, if $ f(0) \neq 0$, then this last expression implies $ f(x) \equal{} 0$ for all x, which does not verify the given functional equation. Therefore we must have $ f(0) \equal{} 0$. Now making x = 0 in the given functional equation yields $ f(f(y)) \equal{} y$, or $ f(f(x)) \equal{} x$. Making $ y \equal{} f(x)$ in the given equation then yields $ f[x(x \plus{} 1)] \equal{} f(x) [f(x) \plus{} 1]$. By now its obvious that f(x) = x is a/the solution, but how to complete the solution? Any hint?[/hide][/quote]
Yes,carcul,i found it,but like you,i cant comlete ,so i must post it here and hope any one can solve it
[hide]but,i think if we add condition,we only need $ f$ continuos[/hide]

$ f\left((x \plus{} 1)f(y)\right) \equal{} y\left(f(x) \plus{} 1\right) \hspace{0.5in} \forall \;x,y \in \mathbb{R}. \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} (*)$

$ \text{Substituting } y \equal{} 0$ $ \text{in } (*)$, $ \text{ gives } f\left((x \plus{} 1)f(0)\right) \equal{} 0 .$
$ \text{ Case 1: } f(0) \neq 0 \Rightarrow f \equiv 0 \Rightarrow \Leftarrow$

$ \text{ Case 2: } f(0) \equal{} 0 \Rightarrow f\left(f(y)\right) \equal{} y \Rightarrow f \text{ is a bijection. }$

$ \text{Substituting } y \equal{} f(y) \text{ in } (*), \text{ gives } \hspace{0.2in}f((x \plus{} 1)y) \equal{} f(y)\left(f(x) \plus{} 1\right) \minus{} \minus{} \minus{} \minus{} (**)$

$ \text{Substituting } y \equal{} 1 \text{ in } (**), \text{ gives } \hspace{0.2in}f(x \plus{} 1) \equal{} f(1)\left(f(x) \plus{} 1\right) \minus{} \minus{} \minus{} \minus{} (***)$

$ \text{Substituting }y \equal{} (x \plus{} 1)\text{ in } (**),\text{ gives } f((x \plus{} 1)(x \plus{} 1)) \equal{} f(x \plus{} 1)(f(x) \plus{} 1)$

$ \text{Using } (***) \text{ it simplifies to } f((x \plus{} 1)^2) \equal{} f(1)(f(x) \plus{}1)^2.$

$ \Rightarrow f(x^2 \plus{} 2x \plus{} 1) \equal{} f(1)(f(x)^2 \plus{} 2f(x) \plus{} 1)$

$ \Rightarrow f(1)(f(x^2 \plus{} 2x) \plus{} 1) \equal{} f(1)(f(x)^2 \plus{} 2f(x) \plus{} 1) \hspace{0.1in}(\text{ using } (***))$

$ \Rightarrow f(x^2 \plus{} 2x) \equal{} f(x)^2 \plus{} 2f(x) \hspace{0.1in}(\because f(0) \equal{} 0 \text{ and } f(1) \neq 0 \text{ since } f \text{ is a bijection})$

$ \Rightarrow f(x)(f(x \plus{} 1) \plus{} 1) \equal{} f(x)^2 \plus{} 2f(x) \hspace{0.1in}(\text{ using } x \equal{} x \plus{} 1 \text{ and } y \equal{} x \text{ in } (**))$

$ \text{For }x \neq 0, f(x) \neq 0. \text{ Hence, we get } f(x \plus{} 1) \equal{} f(x) \plus{} 1 \hspace{0.1in}\forall x \neq 0$.

$ \text{Comparing the above with }(***),\text{ we get } f(1)(f(x) \plus{} 1) \equal{} f(x) \plus{} 1 \hspace{0.1in}\forall x \neq 0$

$ \text{ Since } f \text{ is a bijection}, \exists x \text{ such that } f(x) \plus{} 1 \neq 0$.
$ \text{Hence}, f(1) \equal{} 1 \text{ and from } (***), \text{ we have }$

$ f(x \plus{} 1) \equal{} f(x) \plus{} 1 \hspace{0.1in}\forall x \in \mathbb{R}$

$ \text{ Using the above equation in }(**) \text{ and replacing }x \plus{} 1 \text{ by } x \text{ in }(**), \text{ we get}$

$ f(xy) \equal{} f(x)f(y) \hspace{0.1in}\forall x, y \in \mathbb{R}.$

$ \text{ We have }f(x^2) \equal{} f(x)^2. \text{ Hence, if }x > 0, f(x) > 0$.
$ \text{ For } 0 < y < x, \exists \alpha > 0 \text{ such that } (\alpha \plus{} 1) y \equal{} x$.
$ \text{ Now, we have }$

$ f(x) \equal{} f((\alpha \plus{} 1) y) \equal{} f(\alpha \plus{} 1)f(y) \equal{} (f(\alpha) \plus{} 1)f(y) \equal{} f(\alpha)f(y) \plus{} f(y)$
$ \hspace{0.2in} \equal{} f(\alpha y) \plus{} f(y) > f(y) (\because f(\alpha y) > 0 )$

$ \text{ Hence, } f \text{ is increasing for } x > 0$
$ f(x) < x \Rightarrow f(f(x)) < f(x) \Rightarrow x < f(x) \Rightarrow \Leftarrow$
$ f(x) > x \Rightarrow f(f(x)) > f(x) \Rightarrow x > f(x) \Rightarrow \Leftarrow$
$ \Rightarrow f(x) \equal{} x \hspace{0.1in} \forall x > 0$.

$ \text{ Since } f \text{ is bijective }, f(x) < 0 \hspace{0.1in} \forall x < 0$.
$ \text{ We can repeat the above argument for } x < y < 0 \text{ and } \alpha > 0$
$ \text{ and show that } f \text{ is increasing } \forall x < 0 \text{ and } f(x) \equal{} x \hspace{0.1in} \forall x < 0$.

$ \text{ Therefore} , f(x) \equal{} x \hspace{0.1in} \forall x$.

Really thank you,I missed the step
[quote="nicetry007"]\text{ Hence, } f \text{ is increasing for } x > 0
f(x) < x \Rightarrow f(f(x)) < f(x) \Rightarrow x < f(x) \Rightarrow \Leftarrow
f(x) > x \Rightarrow f(f(x)) > f(x) \Rightarrow x > f(x) \Rightarrow \Leftarrow
\Rightarrow f(x) = x \hspace{0.1in} \forall x > 0.
[/quote]
Thank for your smart solution


inequalities
search
inequalities unsolved

let $ (a_{n})_{n\geq 1}$ an increasising sequence with strictly positive real numbers

prove that:

$ \frac{1}{a_{1}}\plus{}\frac{2}{a_{1}\plus{}a_{2}}\plus{}\frac{3}{a_{1}\plus{}a_{2}\plus{}a_{3}}\plus{}...\plus{}\frac{n}{a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}} <4(\frac{1}{a_{1}}\plus{}\frac{1}{a_{2}}\plus{}...\plus{}\frac{1}{a_{n}})$

[quote="chien than"]This is a problem in Russian 1986
See at
http://www.mathlinks.ro/viewtopic.php?search_id=1191386986&t=173813 with stronger problem[/quote]
Why $ i^2\sum_{k \equal{} i}^{n}\frac {4k}{(k(k \plus{} 1))^2} < 2$ is true? :wink: :P  :)


Find $ x,y,z\in R$ from this equation:

$ 7x^2 \plus{} 3y^2 \plus{} 7z^2 \plus{} 3xy \minus{} 3yz \plus{} 11zx \minus{} x \minus{} 27y \plus{} 26z \plus{} 67 \equal{} 0$

$ 7x^2 \plus{} 3y^2 \plus{} 7z^2 \plus{} 3xy \minus{} 3yz \plus{} 11zx \minus{} x \minus{} 27y \plus{} 26z \plus{} 67 \equal{} 0$
$ 14x^2 \plus{} 6y^2 \plus{} 14z^2 \plus{} 6xy \minus{} 6yz \plus{} 22zx \minus{} 2x \minus{} 54y \plus{} 52z \plus{} 134\equal{} 0$
$ 11(x\plus{}z\plus{}1)^2\plus{}3(x\plus{}y\minus{}4)^2\plus{}3(y\minus{}z\minus{}5)^2\equal{}0$

$ x\plus{}z\plus{}1\equal{}0$
$ x\plus{}y\minus{}4\equal{}0$
$ y\minus{}z\minus{}5\equal{}0$
with solutions $ \boxed{(x,y,z)\equal{}(\minus{}1\minus{}z,5\plus{}z,z)}$


What is the best way to have columns of problems?  If I want to make simple fraction addition problems (say - 5 rows by 4 columns), should I use the tabular environment?  If so, could someone give me an example of how to do it?

Thanks!

     -- Pawel

You might want to use the matrix environment or the array environment (I don't really know how to use this one) if you don't want to get the grid lines.


#1

The digits of the three-digit integers $a,b$ and $c$ are the nine non-zero digits 1,2,3....9 each of them appearing exactly only once. Given that the ratio $a:b:c$ is $1:3:5$, determine $a,b$ and $c$.

Please show your work on this.

The answer was:

[hide="Key"]
129,387,645[/hide]

Okay, I'll admit I kind of brute-forced this with a really ugly solution, but still, I got it.
[hide]If the numbers are in ratio $1:3:5$, that means the smallest number must have a hundreds digit $1$.  If it was over $1$, then the largest number would not be $3$ digits.  For example, if the smallest number was $200$, then the largest would be $1000$, which is clearly not $3$ digits.  So next I considered the units digit of the smallest number.  First of all, we can eliminate all the even units digits, because multiplying them by $5$, for the largest number, would give units digit $0$ for the largest number, which isn't allowed. So we consider the odd numbers.
1: Because the hundreds digit of the smallest number is 1, as I previously determined, the units digit can't also be one.

3:Couldn't find anything prohibiting this

5:If the units digit was 5, then getting the 2nd largest number by multiplying 3 would end in a 5, since $5\times3$ ends in 5.  Since we can't repeat digits, 5 can't be the units digit.

7:If the units digit was 7, then getting the 2nd largest number by multiplying 3 would end in a $1$, which as explained above must be the hundreds digit of the smallest number.

9:couldn't find anything here

So testing all the numbers ending in 3 (in which the number tested is the smallest number) 123,143,154,163.....193 none worked. So, I began testing the nines, and 129 worked.  So the answer is $129, 387, 645$.[/hide]

[hide="Killing it with minimums and maximums..."]
a: _ _ _
b: _ _ _
c: _ _ _
The units digit of a must be odd, or else the units digit of c will be zero. This forces the units digits of all three numbers to be odd, and more specifically in the case of c, 5.
If the hundreds digit of a is not 1, then c will not be a three digit number. So right now, we have:
a: 1 _ A
b: _ _ B
c: _ _ 5 where A and B denote an odd digits.
If A=7, then B=1, which is not allowed since we have already used 1.
Thus, the possibilities are A=3, B=9, or A=9, B=7.

Let's try the first case:
a: 1 _ 3
b: _ _ 9
c: _ _ 5
The hundreds digit of b must be 4 since 100<a<200, and the digits 3 and 5 have already been used. If 400<b<500, then 133<a<167, but 3, 4, and 5 have already been used, so we have:
a: 1 6 3
b: 4 _ 9
c: _ _ 5
a=163 results in b=489 and c=815, which is not allowed since the digit 1 is repeated.

So our numbers must be like this instead:
a: 1 _ 9
b: _ _ 7
c: _ _ 5
The max value of c is 865, so the max value of b now is 519. The min value of a is 129, so the min value of b is 387. Thus, the hundreds digit of b must either be 3 or 4, since 5 is already used. We can knock out the case where the hundreds digit of b is 3 very quickly, since b<400 implies a<134, and the only possbility that satisfies is a=129, from which we see:
a: 1 2 9
b: 3 8 7
c: 6 4 5
How happy. However, I'm not sure that this is the only case, so now let's try b>400.
a: 1 _ 9
b: 4 _ 7
c: _ _ 5
The min value of b is 427, so the min value of c is 712, and we see that c>800 since the digit 7 has alraedy been used.
a: 1 _ 9
b: 4 _ 7
c: 8 _ 5
The min value of c is now 825, so a>165, and the only case is a=169. Then b=507, which is clearly not allowed.

Thus, the only possibility is a=129, b=387, c=645.
[/hide]

[hide]We'll try to find a way to set a limit on the bounds of each number. Without loss of generality, we can say $a<b<c$. It is easy to see that $123<a<198$ because each digit of $a$ is different and nonzero, and 198 is the largest such number that when multiplied by 5 to get $c$, will not be over 3 digits.

Now, let's say $a=def$ for some distince digits $d, e, f$. Likewise, $b=ghi$ and $c=jkl$ for some distinct digits $g, h, i, j, k, l$. Because $a$ must be multiplied by 5 to get $c$, $f$ must not be 5 (since $5\cdot5=5$) and $f$ must also be odd (since any even $5f$ would end in 0). $f \neq 7$ because otherwise $3a$ would repeat $d$ in $i$. Finally, $f \neq 1$ because we know that $d=1$ from the limits we set on $a$.

Thus, $f \in {3, 9}$. Now, knowing this, we can figure out that $l=5$ because $5a$ will always end in 5 for any odd $f$. If $f=9$, then $e$ must be even--for if $e$ is odd, then $k$ will repeat $f$. If $f=3$, $e$ must be odd or else $k$ will repeat $d$.

We'll consider the case where $f=3$. In this case, $e$ is odd, therefore $e=7$. This is because $i=9$. Thus, we have the $a=173, b=519, c=865$ which obviously does not work.

Therefore, $f=9$ and $e$ is even. From this, we know that $i=7$. We can also pseudo get $5a$ and find out $k=4$ from carrying. This means that $e \in {2,6,8}$. From carrying, it is easy to see that $h=2+\text{ units digit of } 3e$.

We notice if $e=6$, then the $h=0$ which is impossible, so therefore $e \in {2,8}$. It is clear that neither $g$ or $j$ can be 2, therefore $e=2$. This gives us the three numbers $\boxed{129, 387, 645}$ which work out perfectly. There's probably a better solution, but this is what I was thinking as I typed :)[/hide]


$a,b$ are reals numbers ,such that : $a^3 =3ab^2 +2$ and $b^3=3a^2b+11$
find: $a^2+b^2$

[quote="Rust"]$a^2=\frac{-85+\sqrt{8809}}{16},b^2=\frac{59+\sqrt{4449}}{16}.$[/quote]
So it should be...
But Rust, you know what I think about just posting the answer...


Randomly select a whole number $x$ such that $0 < x < 100$. What is the probability that $x$ equals the sum of two perfect squares? Express your answer as a fraction.

[hide="Click here for answer(s?)"]So the answer the book gave was 46/99, However, my team found only 43. We think the book double-counted the three sums that are duplicated (0+25 & 16+9;1+49 & 25+25;16+49 & 1+64). Can anyone think of a way the book is right? Or are they actually wrong?[/hide]

I'm inclined to agree that they didn't realize that some would duplicate. I counted them out and got 43.
[hide="If you did it algebraically"]You should have $10$ possibilities for perfect squares (0 -> 9), so you have $10+9+8+7+6+5+4+3+2+1=55$ possibilities. However, $8$ are too large and $1$ too small, so you're down to $46$, which is I think what they did, but then you eliminate $3$ for duplicates and you're left with $43$.[/hide]

I agree with your reasoning. I have seen a couple of problems in a handbook which they are occasionally incorrect. It is not impossible.

I got 43 too. I think the book is wrong.

I don't see any mention of this in mathcount's errata archive. Maybe someone should protest.


Compute: $ \dfrac{1}{1\plus{}\frac{1}{2}}$.  Express your answer as a common fraction.

hello, we get
$ \frac{1}{1\plus{}\frac{1}{2}}\equal{}\frac{1}{\frac{3}{2}}\equal{}\frac{2}{3}$.
Sonnhard.


According to the table below, how many dollars are in the median value of the 59 salaries paid to this company’s employees?
\begin{tabular}{|c|c|c|}
\hline
\textbf{Position Title}&\textbf{\# of Employees with this Title}&\textbf{Salary for Position}\\\hline
President&1&$ \$130,000$\\\hline
Vice-President&5&$ \$90,000$\\\hline
Director&10&$ \$75,000$\\\hline
Associate Director&6&$ \$50,000$\\\hline
Administrative Specialist&37&$ \$23,000$\\\hline
\end{tabular}

[quote="GameBot"]According to the table below, how many dollars are in the median value of the 59 salaries paid to this company’s employees?
\begin{tabular}{|c|c|c|} \hline \textbf{Position Title} & \textbf{\# of Employees with this Title} & \textbf{Salary for Position} \\
\hline President & 1 & \$130,000 \\
\hline Vice - President & 5 & \$90,000 \\
\hline Director & 10 & \$75,000 \\
\hline Associate Director & 6 & \$50,000 \\
\hline Administrative Specialist & 37 & \$23,000 \\
\hline \end{tabular}[/quote]
Since more than half of the employees are Administrative Specialists $ ($this is the reason Bugi noted that $ \frac {59}{2} < 37)$, that salary for that title is the median value.  This is simply $ \boxed{23000}$, as stated by Bugi.


blogs
inequalities proposed
inequalities

For any triangle with sides a,b,c prove that
$ a^2b(a\minus{}b)\plus{}b^2c(b\minus{}c)\plus{}c^2a(c\minus{}a)\ge0$

[quote="sophie germain"]For any triangle with sides a,b,c prove that
$ a^2b(a \minus{} b) \plus{} b^2c(b \minus{} c) \plus{} c^2a(c \minus{} a)\ge0$[/quote]
let a=x+y,b=y+z,c=z+x;
after expanding,we need to prove that:
$ \sum_{cyc}x^3y \geq (x\plus{}y\plus{}z)xyz$
obvious true

It's IMO 1983 problem 3; you can see my solution in my blog her,
http://mathsmaster.wordpress.com/2009/01/12/imo-1983-probleme-3/

take a look at this solution by me,
http://www.mathlinks.ro/weblog_entry.php?t=260554


AMC
AIME
rotation
analytic geometry
function
USAMTS
geometry

#1 I misread

#2-5 were basically the same as everyone else.

[quote="besttate"]Here are mine (in Word/.rar, let the stoning begin):

My number 1 was not that elegant, may only get a 4.
Number 2 uses identities rather than the AM-GM posted in the math jam.
My number 4 was a little "weird", I managed to get the special case zinger. Still a potential 5 though.
Number 5 I was able to generalize.

Hoping for a 25 to get me into the AIME, but I'll have to wait and see![/quote]
EDIT: Never mind.
I'm proud that I fit each solution onto 1 page. :D
Elegance doesn't count. I also used identities for 2 and got the generalization for 5, if you mean explicit $f(x)$.

Here are my solutions. However, I am wrong on number 1 because I decided to rotate the number of indistinguisible ways. Please feel free to make comments.

#1- I love my answer to number 1, because I used Burnside's Lemma, my favorite lemma  of all time.

#2-5 I used basically the same solution as everyone else, but I managed to keep my solutions very short yet still clear, every question fit onto 1 page.  I especially like my number 5, it consists of 6 lines.

Here are mine. As always they are as concise as I could make them.

#1 was hard to explain and I'm not sure if I'm going to get full credit on it.

My #2 was good--it's mostly the same idea as what everyone did but fit into half a page. It could possibly be commended.

I felt bad about using coordinates on #3 because I usually try to avoid it, but I felt a little better when I saw that everyone else used them too. Still, there's always a way to do it without coordinates and someday I will find the coordinate-free solution to this.

#4, my arguments were more or less the same as the ones in the math jam but I'm scared because my solution still seems a lot shorter. I think it's well-justified... hopefully the graders agree.

When I was working on #5 I found the explicit function before finding f(1) primarily through guess and check. I think that simply stating "The following function fits the description, and its f(1) equals so-and-so" would suffice as a complete proof, but I did the readers a courtesy of at least partially deriving the function instead of pulling it out of thin air.

My solutions use mostly different methods than the ones above.  I also favored explaining everything out as opposed to being concise.  It's weird... USAMTS graders say they want concise solutions in order to get a commended, but then you could also get points deducted if you fail to explain something... it's a thin tightrope to walk.

[quote="ehehheehee"]Could you explain burnside's lemma in words that I can understand?[/quote]Its a Group Theory formula for counting the number of orbits in a group.  An orbit happens when you start with one element, and repeatedly apply the same action to it.  Eventually you will get back to  the element you started with, and all the elements you went through along with way combine to form 1 orbit.

In terms of the problem, the group $G$ consists of the actions to perform on the cube (the various rotations), and the set $X$ consists of the different colorings of the cube (some of which are teh same up to a rotation).

An orbit would be a unique coloring.  You can rotate it repeatedly to get other elements of $X$, but its still the same coloring.  That's why I want to count orbits.

The number of orbits is equal to $\frac{1}{|G|}\sum_{g\in G}|X_{g}|$

$|X_{g}|$ is the number of elements of $X$ which are fixed by a given $g\in G$.  In terms of this problem, this would take a given rotation ($g$) and count the number of colorings of the cube which are unchanged after $g$ has been applied to them.

The wikipedia article on the lemma http://en.wikipedia.org/wiki/Burnside_lemma contains an OK explanation, along with an example of counting unique colorings of the face of a cube (very similar to what I did).

Thanks, I'll make sure to read that (and understand it) when I am more awake, but is all that college math?

[quote="ehehheehee"]Could you explain burnside's lemma in words that I can understand?[/quote]

I know this just got answered (I was beat to it) but here's my explanation.

Burside's lemma took me a good bit of weeks of thinking after reading about it to truly understand. You have to know about some group theory but I'll try to explain it. It might be a little bit confusing or flawed at times because I myself only have a very amateur understanding of group theory.

Basically, you start out with a group, G. A group has elements, and an operation. In this case, the group's elements are "ways to rotate a cube." So there are 24 elements. (One of the ways to rotate the cube is "do nothing." This is the identity element.) The group's operation is "followed by," meaning you can do one operation FOLLOWED BY another (think of this as being somewhat analogous to having one number MULTIPLIED BY another). For example, a rotation by 90 degrees FOLLOWED BY a rotation by 90 degrees = a rotation by 180 degrees, just like 2 MULTIPLIED BY 2 = 4. Get it? This isn't Burnside's lemma yet, just general group theory.

Now you have a set of objects X. In this case X is the set of all colorings where two touching edges aren't the same color (we haven't started considering rotations yet--for now, it's the set of all colorings, even when two colorings can be rotated into each other). Now, the elements of G ("ways to rotate a cube") can act upon the elements of X ("colorations of a cube"). Now, if an element of G acts upon an element of X, and gets that same element back as the result, then we say that the element of X is [b]fixed[/b] by that element of G. So, if you rotate a certain coloration but in the end, the cube still looks the exact same, then that coloration is "fixed" by that rotation. For example, the rotation "do nothing" fixes all the elements of X (no matter what a coloration is, if you do nothing to it, the result will be the original coloration. Makes sense?) But a rotation by 180 degrees can only fix a certain coloration if that coloration is symmetrical about the axis of rotation.

Now, even more group theory. Suppose that an element of G acts upon one element of X, and the result is another element of X. That means that those two elements of X are in the same [b]orbit[/b]. So, in this problem, suppose that a certain rotation acts upon a certain coloration in X, and you get another coloration in X. That means that those two colorations are in the same orbit, which in this case is the same thing as saying that they are indistinguishable because they can be rotated into each other. So, each orbit corresponds to a unique coloration, after rotations are considered. So the total number of unique colorations equals the total number of orbits.

Fortunately, Burnside's lemma tells us how to find the total number of orbits when a group G acts upon X. Here's how it works: Go through the elements of G. For each element of G, find out the total number of elements in X that are fixed by it. Then, find the average of those numbers. For example, suppose G had 3 elements: a, b, and c. And suppose that a fixed 10 elements of X, b fixed 0 elements of X, and c fixed 2 elements of X. Then the total number of orbits would be (10+0+2)/3 = 4.

So here's what renatzu did. He went through all the elements of  G. Notice that he clumped similar rotations together--rotations around vertices, rotations around faces by 90 degrees, rotations around faces by 180 degrees, etc., because each of these rotations would have the same number of fixed points due to symmetry. Then for each kind of rotation, he found out how many elements were fixed by it. He did this by starting out with "ok, if a coloration is fixed by this rotation, than these edges have to be the same color" then went on to find out how many colorations existed with that restriction.

Then he found the average of all of that and got the answer.

Yeah, it's college math. Meaning it's by no means required in the USAMTS. In fact, it's usually just the longer way to use it. But it's still good to know.

[quote="K81o7"]Read
Forgot to simplify on #3[/quote]

Interesting solution for #5. Nice.

[quote="ehehheehee"]Agreed.  I would have never seen that.  Where did you get the idea?[/quote]

For 5, it didn't take too long to figure out it had to be of the form $-\frac{1}{ax}$.  I fiddled around a little bit with the identities to get $f(x)<-\frac{1}{2x}$, then looked back at what I did exactly and I realized how to bound f closer and closer, eventually formalizing.


[url=http://www.princeton.edu/main/news/archive/S16/16/11S55/index.xml?section=prospective]Massey’s mentorship creates network of mathematicians[/url]


Find all real numbers satisfying the equation
(x − 4)(x − 1)x(x + 2)(x + 3)(x + 6) + 100 = 0.

there are supposed to be five real numbers satisfying the equation.

You're correct. Compute this to the end and u'll can see that :)


$ A \equal{} \left [\begin {array} {ccc} a & b & c \\
\minus{} a & \minus{} b & \minus{} c \\
a & b & c \end {array} \right] \equal{} CL$
where $ C \equal{} \left [\begin {array} {c} 1 \\
\minus{} 1 \\
1\end {array} \right],L \equal{} [a\ b\ c]$
not that $ LC \equal{} (a \plus{} b \minus{} c)\in\mathbb{R}$
then $ {A^{k} \equal{} (CL)(CL)(CL)...(CL)(CL) \equal{} C(LC)(LC)...(LC)L \equal{} (LC)^{k \minus{} 1}(CL) \equal{} (a \plus{} b \minus{} c)^{k \minus{} 1}A}$
the rest is easy :wink:

"... Ai yeu [b] BAC HO CHI MINH [/b]  hon thieu nhi VIET NAM ?..."
Thanks !
Here :  Rank(A)= 1 , eassy :$ A^{2}\equal{}Tr(A).A$ (Allway is a nice problem)


http://www.youtube.com/watch?v=Ooa8nHKPZ5k&NR=1

 :lol:

If you need the lyrics:

[i]What you know about math?

Hey, don’t you know I represent Math League
when I add shorty subtract
Freshmen backpack where I’m holding all my work at

What you know about math?
I know all about math

Answers 44
It’s real easy cuz of sig figs
You got 45 you rounded high
Your answers too big

What you know about math?
I know all about math

TI-84
Solar Edition you know I’m shining dawg
Extra memory, I look back to do my natural log

You know we multiply
while memorizing pi
take limits to the sky
be sure to simplify

graphing utility
that’s trinonometry
One hundred I’m math b
dont you cheat off me
distance is rate times time
the sine graph aint a line
exponential decline
but your score cant beat mine

We’re memorizing grades
for our mathly states
against the mathly greats
not getting many dates
I got to find a mate
but girls just playa hate
and always make me wait (can’t even integrate)

Dont you know I represent Math League
when I add shorty subtract
freshmen backpack where I’m holding all my work at

What you know about math?
I know all about $math$ (heh)[/i]

god, i cringe everytime i see this...

TI-84 solar edition?

Wow you guys are taking this seriously... :maybe:

uuhh.. the guy who was singing that I bet doesn't know about ARML.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147931

on the same page stupid

LOCK THIS CRAP

hey ignite

thanks, I didn't see that it was already there

but anyway, mine is different, because mine got lyrics

well, anyway, why did you insult me!? are u bullying me?! just because I can't see well, you aren't supposed to say "stupid" etc

that's not cool   :(

AoPS says you can't insult other members

you should apologise

hahahha

and don't bully me   :mad:

[quote="miyomiyo"][quote="Fabiomr1"]TI-84 Solar Edition[/quote]
[quote="Fabiomr1"]One hundred I’m math b[/quote]I'm pretty sure these aren't correct lyrics.[/quote]
I think the second one should say "on" instead of "I'm".

and what does the line "when I add shorty subtract" mean?


I'm not sure if this is the right place to post this but here goes anyway:

Suppose that we are trying to prove an identity for all numbers; for example the identity $ a(a\plus{}b)\equal{}a^2\plus{}ab$. Suppose that we have proved it by combinatorial/other means for all positive/non-negative integers $ a,b$. Is there an efficient way of concluding that the identity holds for all real/complex numbers $ a,b$?

Yes.  If a polynomial has infinitely many roots, then it is the zero polynomial.  Note that this doesn't work for blatantly non-polynomial "identities" such as $ \sin 2 \pi n \equal{} 0$.

Edit:  I'll be a little more specific since it's unclear that this extends directly to two dimensions.

[b]Theorem:[/b]  Suppose a polynomial $ f : \mathbb{C}^n \to \mathbb{C}$ takes the value zero on $ \mathbb{Z}^n$.  Then it is the zero polynomial.

[i]Proof.[/i]  We'll prove the contrapositive:  every nonzero polynomial takes a nonzero value somewhere on $ \mathbb{Z}^n$.  To see this, pick integer values for $ n\minus{}1$ of the variables so that the polynomial in the resulting variable (call it $ x$) is nonzero - this can be done by singling out the term of greatest degree in $ x$ (WLOG, such a term exists).  The resulting polynomial in one variable is nonzero and therefore clearly attains a nonzero value somewhere; just take $ x$ to be arbitrarily large.

Thank you very much for the help. It was really useful. :)


Erika, who is 14 years old, flips a fair coin whose sides are labeled 10 and 20, and then she adds the number on the top of the flipped coin to the number she rolls on a standard die. What is the probability that the sum equals her age in years? Express your answer as a common fraction.

The only way this case results is if we get a 10 on the coin and a 4 on the die. This is one of 6*2=12 cases, so our prob. is $ \boxed{1/12}$


OK, I need some serious help on this...i really dont get this

triangles had sides a,b,c and angles A,B,C opposite the corresponding lettered side.

a,b,c are side of a triangle in which a^4 + b^4 + c^4 = 2(a^2 + b^2) * c^2
what could angle C be?
132
133
134
135
136

if a triangle is given in the standard convention such that a^2 + b^2 = 1989 c^2, then the value of $ \frac{cot(c)}{cot(a)\plus{}cot(b)}$ is a 3 digit number. what is the sum of the 3 digits?
21
22
23
24
25

Thank you SO MUCH!!

[hide="number 1"]
Rearrangement of the cosine law yeilds
$ \cos(C) \equal{} \dfrac{a^2 \plus{} b^2 \minus{} c^2}{2ab}$
$ \Rightarrow \cos^2(C) \equal{} \dfrac{a^4 \plus{} b^4 \plus{} c^4 \minus{} 2c^2(a^2 \plus{} b^2)}{4a^2b^2} \plus{} \frac {1}{2}$

And it follows that $ \cos^2(C) \minus{} \frac {1}{2} \equal{} 0$
$ C \equal{} 45^o$ or $ 135^o$[/hide]

[hide="number 2"]
Taking the cosine law and; $ \bigtriangleup \equal{} \frac{1}{2}ab \sin(C)$
We find that $ \cot(C) \equal{} \dfrac{a^2 \plus{}b^2 \minus{} c^2}{4\bigtriangleup }$

$ \Rightarrow \dfrac{1}{\cot(A) \plus{} \cot(B)} \equal{} \frac{2\bigtriangleup}{c^2}$

$ \Rightarrow \dfrac{\cot(C)}{\cot(A) \plus{} \cot(B)} \equal{} \dfrac{a^2 \plus{}b^2 \minus{} c^2}{2c^2}$

$ \dfrac{1989c^2 \minus{} c^2}{2c^2} \equal{} 994$

Answer: $ 9\plus{}9\plus{}4 \equal{} 22$[/hide]


Hi, Rep, long time no see.

From a College Board Web page,

[url=http://www.collegeboard.com/prod_downloads/highered/ra/sat/SubjTestPercentileRanks.pdf]Subject Test Percentile Ranks[/url]

it appears that percentiles are usually based on the performance of test-takers in the previous high school graduation year. Your rank compares your standard score to a whole year's worth of previous test-takers.

Yes, everyone who receives an 800 (or any other specific scaled score) gets the same percentile.

The percentiles for SAT II's are of somewhat limited usefulness, since the pool of students who take each of these tests is so strongly self-selected. That is, each SAT II is largely taken by students who believe that the particular subject is a particular strength for them. Also, the colleges that require the SAT II or for which the SAT II is an advantage in the admissions process are more selective than those that do not, so the population of students who take any SAT II at all is already a better-qualified group on average than those who take the SAT I.

I have seen a very large number of test scores of incoming freshmen at my institution. For the same student who took both, it is clearly more common for the scaled (200-800) score on the SAT II Math Level 2 to be lower than the SAT I Math score than the other way around - and yet if the scaled scores were the same, the percentile would be lower for the SAT II than for the SAT I.


[hide]If I'm not wrong L'Hôpital gives $\frac{17}{60}$.[/hide]

the limit is equivalent to ,$lim_{x\rightarrow0}\frac{\sqrt[3]{1+3x}}{x}(\sqrt[5]{1+5x}-1)+\frac{\sqrt[3]{1+3x}-1+1-\sqrt[4]{1+4x}}{x}$.
now we will get ,$lim_{x\rightarrow0}\uparrow=1$ :huh:

Actually, L'H gives $\text{lim}\;=1$ also

[quote="nobody1"]What is L'Hôpital  :?:[/quote]
[url]http://mathworld.wolfram.com/LHospitalsRule.html[/url] gives a pretty good run-down.

I was wondering a while ago how to prove l'hopital, and couldnt seem to be able to think of any way to relate the function to its derivitive. But the proof is a three-liner! Give it a go if you dont already know it.


[quote="diophantient"]You have a circle with radius 9.51. An angle of 2.5 radians subtends an arc. What's its length?[/quote]Formula for arc length is $ s\equal{}r\theta$.

[quote="10000th User"][quote="diophantient"]You have a circle with radius 9.51. An angle of 2.5 radians subtends an arc. What's its length?[/quote]Formula for arc length is $ s \equal{} r\theta$.[/quote]
Yes, I'm aware of that, I just keep getting an answer different from my textbook after doing this, and think that I'm missing something obvious.

[quote="123456789"]What answer are u getting, and what's the answer in the textbook?[/quote]

Uh, I would guess that he got $ 9.51\cdot 2.5\equal{}23.775$.

When in doubt, assume the book is wrong. Unless you misread the problem or something.

One thing for sure is that a full circle has $ 2\pi$ radians and clearly $ 2.5 < 2\pi$ so there shouldn't be any problems... and don't forget that books aren't perfect, they are human products and it is OK that we make mistake, even in books.


Veo que nadie aun ha creado un topic para colgar los examenes de la IBERO,
pues este seria el lugar,

[b]Pascual ..
Leonardo .. [/b]

ustedes son las personas mas indicadas para editar las preguntas de los examenes en este forum.. espero sus mensajes con las preguntas para su discusion....

gracias

Carlos Bravo
Lima - Peru

Hola,

Yo hay creado un topic con los problemas de la Ibero 2005:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=53808[/url]

Pero está en Inglés pues no se escribir ó hablar español bien.

Ok!!

 Gracias por la informacion.... :lol:

 Carlos Bravo
Lima - Peru

Colgar los resultados de la IBERO,  :idea:

conseguire la informacion de la selecc de Peru para el lunes...

[b]Pascual, leonardo, cyshine,...[/b]
 ustedes estuvieron en la ceremonia.... pueden enviar informacion al respecto??

se los agradeceria muchisimo...!!!

Carlos Bravo
Lima - Peru  :lol:

Bueno, acabo de llegar a mi casa en Caracas. Aca coloco los resultados que recuerdo, seguramente alguna otra persona podria ser mas exacta:

Hubo en total 3 pruebas perfectas, las cuales son Pascual (Colombia), Guevara (Mexico) y Gabriel Bujokas (Brasil).
Aqui estan las medallas que recuerdo (no coloco los nombres porque no los recuerdo todos  :| )

Brasil: 4 oros
Mexico: 2 oros, 1 plata y 1 bronce
Argentina: 1 oro y 3 platas
Uruguay: 1 plata y 3 bronces
Colombia: 1 oro y una mencion
Ecuador: 1 bronce
Panama: 1 plata y 1 bronce (creo)
Cuba: 3 platas y un bronce
Chile: 1 plata y un bronce
Venezuela: 1 plata (yo :lol: ) y un bronce

Los cortes fueron 33 Oro, 22 Plata y 10 Bronce.
Ahora son los que recuerdo, disculpen que la informacion sea tan inexacta, me faltaron muchos paises por nombrar, pero no recuerdo con exactitud sus resultados. Espero que pronto alguien posea la informacion mas detallamente

Bueno,
he editado las preguntas de la Ibero (shine las colgo dias atras)
en un formato mas amigable
y para los chicos que tengan problemas con el idioma
esta en español

hago un pedido a cyshine que haga lo mismo, pero en portuguese..
para toda la gente de brazil y portugal que quisiera tener los examenes y tengan problemas con el idioma

gracias de antemano...

PD: Algun error en la traduccion haganmelo saber para su correccion!! Gracias

Carlos Bravo
Lima - Peru

Gracias Carlos, ya en la pagina oficial estan resultados y problemas tambien.

[url=http://olimpia.uanarino.edu.co/xxoim/resultados.htm]OLIMPIADA IBEROAMERICANA[/url]


In $\triangle ABC$, we are given that $\cos A\cos B+\sin A\sin B\sin C=1$.

What kind of triangle is $\triangle ABC$?

Doesnt really solve it but I guess $C=\frac{\pi}{2}$ and get $A=B=\frac{\pi}{4}$ which is an isosceles-right-angle triangle . :P

Your guess is correct ;)

Can anyone come up with a complete solution?

I think i got it  :)

using $\sin C\leq 1$ we obtain

$\cos A\cos B+\sin A\sin B\sin C\leq \cos A\cos B+\sin A\sin B=\cos (A-B)$

Hence $\cos (A-B)\geq 1$ $\implies A-B=0$ $\implies A=B$ and the equality occurs when $\sin C=1$ which is $C=\frac{\pi}{2}$ . Hence we are done .


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Just like signatures, whose user name do you like the most (the most creative)? we know some are just names and others a strange mix of numbers and letters.

h_s_potter2002

[hide] :rotfl:  :rotfl:  :rotfl:  :rotfl:
Sorry Shravas I couldnt help it[/hide]

Are you talking about some personal stuff I don't know about?

I dunno, I don't know about my name... a lot of people ask about it

in my opinion, if you're on a math forum, anything with numbers in it or related to math in any way is uninteresting. Sorry.

I see, well, I have both  :D
You're actually saying that more than half of the people of this site have bad user names.  ;)

i think mine is stupid... i like funny ones..

my mom made mine :D haha. so I guess mine is special.

Hello, deosnt anyone here think mine is cool? very original right? lol

I'm a holy knight :starwars: Or a Diablo II character. Or a unit from Age of Empires 2. Or a class from Final Fantasy Tactics (I think). Yeah, completely original :D

i have to say though, kids that make their username the same as their birthdate or related to their name have sort of no imagination.

Mine is just some random thing somebody suggested.  But I use it for everthing, including AIM and Xbox Live.

I thought so because your mouth isn't smart, or sort of..  :P

I just kinda liked the sound of it.  It had a ring to it.  mysmartmouth.... LOL

[quote="h78reg"]my username doesn't mean anything. i just pushed random buttons on my keyboard and somehow memorized them. :D[/quote]

 :rotfl:

I can tell, some of your letters are bunched up together on a keyboard.  :P

[quote="Drunken_Math"][quote="236factorial"]Hang on a second, when did MathPath start?[/quote]

June 26th... I believe.[/quote]

Oh, so your username came two days after it started. I see

[quote="Drunken_Math"]My screen name was the first thing that came to my mind when I first heard of this website at Mathpath. There's a rough reinactment of how my screenname came to be: (this was on the first day of Mathpath)

nr1337: Hi Treething! Wow! You don't really look like what people think of you on AoPS.

Treething: <inappropriate things for this forum>

Me: What's AoPS?

A+MATH: You don't know what Art of Problem Solving is? How'd you make it into this camp?

Me: What? I knew about the books...

A+MATH: Well, there's a website for it too... and there's a forum on it. You should join.

Treething: I used to be a moderator... <tells the story of how he lost his position as moderator>

Me: I should make a screen name... and it should have to do with math... unlike Treething...

IntrepidMath (my roommate): *pops out of the room* Man, I'm getting drunk off of sucking on ice cubes!

Me: That's it! My screen name will be Drunken_Math!

Counseler: What a weird name... but it sounds awesome.[/quote]

thats an awesome story... i also like you signature

mines just my first initial and my last name

I have the most uncreative name out of all you guys!

I've had plenty of user names!!!

CyberPower

Elven Rider

Elven Chaos

Shadow

Edvereste Wispress

Edvereste of Turandia

Edvereste's Shadow

Edvereste of the Nothing (Anyone seen the never ending story?)

Filosofa (portuguese for a female philosopher)

The list goes on and on... I like funny names, I had this one: I am me, you are not you, they are content. Nobody ever understod that! I have another one that is kind of an inside joke between me and a friend of mine. Because he loves "doink doink" from law and order, ever since we saw the preview for it. You can barely understood what he was saying one time and I repeated what he said: Ok, you're drawing frogs that say doink doink? So thus my screen name: I am a frog in a suit that croaks "doink doink"

The screen name I have is a mix between piano and one of my favorite bands: White Stripes.

236factorial, I listen to a lot of classical music too. NO idea who that guy is on iversonfan's picture... No knowledge of hip-hop, rap, etc.
I prefer classical, rock, classic rock, alt. rock, and some other stuff.

i have a username for a site i never use "theultimatemathnerdthatwillnotspeaktoyouifyousayyouhatemathchesssaladorfood

lol :rotfl:
i like math and i was born in '92 so i have math92 :)

[quote="jli"]mines just my first initial and my last name[/quote]

I can't read your username... I always thought your name had two i's until you changed your avatar.