Which seems most viable of the proposed theories for the end of the universe?
The Big Crunch, The Big Rip, or The Big Chill?

Personally, I think the Big Crunch makes the most sense.

The other two aren't unrealistic, bit I too think that a reversal of the Big Bang would be most likely.

You should add a poll to make it easier to see the general belief.

Interesting. Why do you think the big crunch makes the most sense? I thought there wasn't enough matter in the universe to counteract the "dark energy" carried by empty space, so wouldn't a big crunch be a little improbable? Hmm. I'd like to hear your thoughts.

Personally, I think the big rip seems most likely, even though it wouldn't technically be the end of the universe, would it? Time--or at least some interpretations of it--would continue. Anyway, with the way the universe is going, I think everything will separate eventually.

I'm no expert, and my opinion might not be so valuable, but I somehow came upon the conclusion that it was more probable than the others.

Maybe I was wrong?

Well I used to be a Big Rip proponent, but it seems impractical, all of the matter in the universe being torn apart, just impractical.  Now, why I like the Big Crunch, is because once dark matter becomes important (as in the Big Rip), that "importance" would provide a counteracting force, making the universe switch from expanding to collasping on itself.

As for the other, the Big Chill, it does exist, a universe that expands without bound making the universe as a whole "cool down".
Another theory, other then those mentioned, is one that states the universe well eventually stop expanding and then remain at the size forever.

About making this thread into a poll, when initially posted there was not an option for a poll  :maybe: .

I'm not sure I understand what you mean by the "importance" of dark matter. Are you saying the amount of dark matter increases with dark energy or somehow has a greater effect as the universe expands more? I'm confused. :huh:

[quote="AstroPhys"]About making this thread into a poll, when initially posted there was not an option for a poll  :maybe: .[/quote]
Strange... In most sub-forums, when you post a new topic, you get a window like this:

[hide="Too big, hidden"][img]http://img510.imageshack.us/img510/9371/forumtopicnormalwd8.png[/img][/hide]

Here, however, you get:

[hide="Too big, hidden"][img]http://img510.imageshack.us/img510/3681/forumtopicnopollux7.png[/img][/hide]

The only difference is the Poll option.

As for that theory about the universe stopping, that seems pretty unlikely because some reactions will inevitably expand and/or collapse it.

Thanks Valentin :lol: .

KrazyNerd, I meant that as the universe expands the amount of dark matter and energy theoretically increases providing a counteracting force to the expansion.
Remember, at times I may seem to babble and make absolutely no sense  :wink:  so quesiton me if I seem incoherent.

So, the amount of dark matter is increasing? Hmm. I guess you learn something new every day.

Actually, though, dark energy doesn't "counteract" the expansion; rather, I think it's the [i]cause[/i] of the expansion. As I understand it, dark energy is a property of empty space that opposes gravity.

I'm not sure which option to pick for the poll now... :| I guess I'll just have to do some research.

Add another option!
'None of the above'
The universe will never end, even if the Everything is destroyed, it will only be another big bang.javascript:emoticon(':(')

I don't know what these terms are but I think that Andromeda  (a neighboring galaxy) will combine together and destroy us all.

For the universe. It will expand forever till it certain point which it rips, causing a massive flow of gravity sucking everything in and starting over with the big bang.

How would a rip in the space-time continuum cause a massive flow of gravity? Unless you're talking about the idea that gravity could be "leaking" into our universe from somewhere else, which is why we observe it to be so weak in comparison to other forces.

Also, about the oscillating universe theory: even if gravity could somehow become strong enough to make all matter collapse into a singularity, how would it also "suck in" all of space-time? I thought most current theories said that massive objects only [i]bend[/i] space-time (or something to that effect), so how would space-time also collapse and start the cycle over again? We can't have big bang, big crunch, big bang, etc. with everything completely restarting, can we? Unless I'm missing something, space-time would keep on doing whatever while matter [i]alone[/i] oscillated. :huh:

Sorry if any of that's confusing.

For those who don't know any of the terms.

[url=http://en.wikipedia.org/wiki/Big_Crunch]Big Crunch[/url]

[url=http://en.wikipedia.org/wiki/Big_Rip]Big Rip[/url]

[url=http://en.wikipedia.org/wiki/Big_Freeze]Big Freeze(Big chill)[/url]

Though this option is extremely unlikely, and some results of quantum mechanics might not allow it, among other things, so I'm not sure if it's even possible, maybe another option should be included in a poll.

I'm not sure what the technical name for it is, but it would be that the expansion "slows down," approaching a certain point of expansion but never quite reaching it.

That is, a "static universe" theory.

[quote="QuantumTiger"]Do you mean expansion continues forever? That would be the Big Chill theory.

Personally, from what I know, the Rip seems most improbable. It depends on the amount of non-matter in the universe to figure out which one of the other two is correct, though.[/quote]

Close, but not quite.  It does continue forever.  However, it slows down over time, approaching a well-defined constant size with no expansion over time.  (Notice the usage of the word approaching.)  That is, if we made a "size vs. time" graph (not to say this would be easy and easy to even find a method to do so), it would approach a particular value as t (representing time) goes to infinity.

EDIT:

However, this isn't how I think the universe will end.  It's a possibility, but my actual opinion is The Big Chill.  I posted this in the topic "Infinite Universe," and here it is.

[quote="bdejean"]It's hard to actually define space being infinite with the expansion of the universe.  The observable universe can never be infinite, as eventually, when you get to a point of a certain distance from where we are observing from, the space between us and the light coming from that point is expanding faster than light, and so the light can never reach us.  Therefore, we can only see for a certain distance around us, and if we can't see something, then we can't see it, we can't test it, we can't do experiments on it, we can't even show that it's there.  I could be wrong here, but I think this should mean that by some principles of science, it shouldn't truly exist to us.  By this, from any reference point, the universe is finite.

Now, you might be thinking, "why not send a spaceship to observe something outside of our observable range, and have it tell us something about that something?"  I'm not sure if this is possible, but if it is, then the spaceship would be our of our observable range, and it couldn't tell us anything because any transmissions would never reach us.

I think of this as evidence for the "big chill" theory, as eventually, our "observable range" will have reached everything possible, but if you move a little to the left, then your "observable range" will shift a little to the left, and so you can see things a little more to the left than you would have been previously, and so you can't go back to the right, and so the "observable range" will be zero, as someone a little to the right won't be able to see you.  This sounds a bit like the "big chill."

I might not have recorded the ideas perfectly, but I hope you understand what I'm trying to say.

Anyway, if it turns out I'm wrong and space is infinite, go for however long needs be in some random direction, and eventually you'll land at a perfect copy of the solar system, though it would be incredibly far away, similarly to the "infinite monkeys on typewriters" theory about one eventually typing all the works of Shakespeare.[/quote]

EDIT 2:

Perhaps it's actually a combination of the three.  Eventually, the Big Chill occurs.  It eventually isolates every planet and star from every other.  At some point, the expansion accelerates to such a rate that any body's gravitational pull can't keep it from ripping apart.  Then, we have just molecules.  The expansion will rip apart the electrons from interacting with each other in these molecules, and so we have atoms.  The electrons get stripped from their atoms.  Next, even the strong nuclear force (technically a manifestation of the color force) can't hold the protons and neutrons of the nucleus together, and we're just left with isolated hadrons, leptons, bosons, and mesons in the universe.  This sounds a bit like the big rip.

Now, the hadrons cannot be ripped apart further because as the expansion pulls them apart, the potential energy from the color force produces three quark-antiquark pairs.  The three antiquarks bind with the quarks being pulled apart and produce three mesons, and the quarks bind to produce a new hadron for this to happen all over again.

For the mesons, a single quark-antiquark pair is produced.  The quark goes off with the antiquark from the meson, and the antiquark binds with the quark from the meson.

You'll notice that new mesons are getting created, and these have mass.  They are "pulling" this energy from whatever the heck is causing the universe to expand, but in any case, "new matter" is being created, and this has a gravitational pull.  This will eventually halt the expansion of the universe, starting to pull it together, resulting in a big crunch.

Everything following this assumes string theory to be the T.O.E.

Now, if we assume string theory to be the T.O.E., even though a bit of refinement needs doing first, the universe will appear to "bounce back" from having a size of the Planck length and reverse its deflation.  I suggest you read Brian Greene's The Elegant Universe for a full explanation of why this happens, but it will.  Long story why.

Anyway, this will start the universe expanding again, starting a universe for this to happen to again.

Now, you might be thinking, "why doesn't all this matter from the quark-antiquark pairs stop the expansion and cause a big crunch a whole lot earlier?  Not doing so would be a clear violation of conservation of energy/mass."  The answer is that energy doesn't have to be conserved at the moment the universe is at the size of the Planck length.  Here's why.

A certain mathematical theorem relates nearly every symmetry with a corresponding conservation law.  (I don't remember its name, but it's a good theorem to know.)  It says that conservation of energy only [b]has[/b] to hold if time-translational symmetry holds.  However, at this moment, time and space have deflated to nearly nothing.  Time has collapsed to where a string has just enough room to vibrate through time, but not enough to actually move through time at all.  Therefore, there's essentially no time for time-translational symmetry to hold, and so conservation of energy at this moment does not have to hold.

I've identified two flaws with this theory, being the fact that wherever the quark-antiquark pairs to create all these mesons to cause the eventual collapse of the universe come from should also have gravitational pull, and the other being that all the particles are isolated in their own little "mini-universes," and it's energy [b]density[/b] that matters for if the universe collapses, but the particles are so far apart (in their own universes, for goodness' sake!) that the density should be 0.

Nevertheless, I like this theory.  (I wish there was a mischievous smile smiley, then I could insert it here.)

[quote="ahaanomegas"]I have a question. What is the difference between the Big Crunch and the Big Rip. According to the Wikipedia articles on them, I see no difference.[/quote]

I'm assuming you know about the universe being "pulled back together" in the Big Crunch.

The Big Rip is the theory that eventually, the expansion of the universe will reach a rate such that the protons/neutrons are ripped apart from each other in the nucleus, tearing everything apart, I think.  Anyways, it's the theory that matter literally gets ripped apart by the expansion of the universe.

Personaly, I prefer the Big Chill, because expansion will cease when the amount of matter in the universe runs out.
Hence,  everything that exists will sort of fade away slowly.

[quote="ZetaX"]Actually it's more complicated. In laymans terms, the universe would expand faster than the speed of light, therefore making any interaction between anything impossible, essentially rendering the universe an accumulation of solitary particles.[/quote]

Basically what I meant.  I guess I didn't convey it correctly.  :P

The Big Crunch seems improbable because if this were to happen, there would have to be an equilibrium point by the Intermediate Value Theorem.

The intermediate value theorem cannot even be applied as we are not in one-dimensional settings ;)
But even if space would be that homogenous this is no argument: It's only an equlibrium of expansion, not of anthing else (especially not acceleration), thus it just passes. Let me ask you a question: If you throw a ball upwards, does it not (or very unlikly) come down again because it would need to be in an equlibrium first¿
Actually, I'm not even sure if such an event is called an equlibrium at all, as it gives the question of "what is equal" in it.

I think in this case we're referring to the one-dimensional radius of the Universe.

[quote="ZetaX"]...
But even if space would be that homogenous
...[/quote]

I honestly think big chill. The current value of omega is too small to reverse the expansion of the universe. Even if we included dark and anit matter, the estimated value of omega would still not be enough to reverse the expansion. So, for now, the universe is expanding and will most likely reach the Cosmic Whimper, or Big Chill.


[quote="aidan"]In how many ways can the vertices of a square pyramid be painted in 5 colours if adjacent vertices must have different colours?[/quote]

[hide]Let colors be $ 1,2,3,4,5$. Say the vertex which isn't the vertex of the square be color $ 1$. Then in clockwise (or counterclockwise, I don't think it matters) order around the vertices of the square, the colors can be:

1) 2,3,2,3
2) 2,3,2,4
3) 2,3,4,5

For 1) number of colorings is $ \binom{5}{3}\cdot 3\equal{}30$ since we just choose three colors and choose one to go on the special vertex (the one which is not a vertex of a square)
For 2) number of colorings is $ \binom{5}{4}\cdot 4\cdot 3\cdot 2\equal{}120$ - choose 4 colors, four colors possible to go on top, three colors possible to duplicate, and two possible distinct colorings
For 3) number of colorings is $ 5\cdot (4\minus{}1)!\equal{}30$ - five possible colors to go on top, and then arranging the last four colors around a circle

So hopefully $ 30\plus{}120\plus{}30\equal{}\boxed{180}$ is correct

Could [url]http://en.wikipedia.org/wiki/Cauchy-Frobenius_lemma[/url] apply here?[/hide]

I got that answer as well, but the official solution says the answer is 420 and assumes the pyramid can't be rotated  :D

I think the Burnside-Cauchy-Frobenius lemma is applicable, but I'm not quite well versed with how to apply it to problems like these yet.


My school only offers Calculus AB. Is it possible to study from an AP Calculus review book or study the entire text and be prepared for the Calculus BC exam?

Or, should I opt for distance learning? Any suggestions?

 :wink:

if you want credit from your school, i suggest distance ed. if you just want to take the tests..im guessing self-study will be fine. dont take my word for it, listen to some teh wiser peeps!

I learned calculus on my own but then still had to take the classes offered at my school. I think you should be able to pick up a good textbook on calculus and learn the stuff and ace the AP exam. Start with [i]Calculus the Easy Way[/i], which is very informal, and then when you finish that, get a normal textbook. My first proper calculus textbook was by Loomis, and I thought that one was quite good, and the one by Stewart is also good. Spivak is Paul Zeitz's recommendation, but it's a bit tougher.


If $A^{2}$ varies directly as $B$ and inversely as $C^{3}$. If $A=10$ when $B=200$ and $C=2$, then what is $B$ when $A=5$ and $C=4$?

[hide]$A^{2}=\frac{Bk}{C^{3}}$, where k is a constant. Plug in given values and solve for k: $100=\frac{200k}{8}$, so $k=4$. Now put in the other values: $25=\frac{4B}{64}$ so $\boxed{B=400}$[/hide]


How self-conscious are you?

I think that I am personally overly self-conscious. I always worry about my hair and whether it looks too ruffled or not (my hair is dense and doesn't stay down, so it is hard to comb). I also sometimes worry too much about what I say in front of people I don't know very well (especially girls), and about my appearance. I also sometimes worry whether people think I'm a loser, and whether people think I'm an introvert because I am not too talkative.

Ug. I really oughta loosen up a bit. Does anyone else have this problem?

I have this problem.  Whenever someone is laughing, I worry that they might be laughing at me.   I'm generally a "control freak" when it comes to my hair - it has to be kept down.  And sometimes I come up with useful things to say for various occasions ("how was your weekend?") because I otherwise may run out of things to say.

In summary, I am more self-conscious then I ought to be.

I'm not as self conscious as I used to be.  I think that this is because I am much more comfortable with my appearance now than before(i.e. I have changed a lot over the last 5 years).

I used to be really really shy.. like REALLY shy.. I'd be absolutely freaked whenever an oral presentation came along in class.. but over the years, I've definitely felt more comfortable with people.. I don't think I'm really that self-concious anymore.

Oh me... I dont even have to think about it. Im the least self-concious person ever. I dont care if u dont like me go away leave alone and eventually Ill be your friend. I am very friendly and I dont generally care about anyones opinion well except for someones :wink:. I care about myself but I dont care about what others think of me. Im just like whats best for me and what I like the best is what Im going to wear or whatever and I take little thought to who dislikes my "look".

I think teaching has helped make me much less self-conscious.  What really bugs me out is people speaking in foreign languages and laughing.  Particularly when they're watching me play a bad game of pool.  Ah, well ...

A lot of people in college end up dealing with stress and pressure (of whatever sort, whether study-induced or self-induced as yours is) by drinking.  When you go to college, be careful with that.

IM "SmarterChild" (without the quotation marks), type in "menu" then 3 then 7 then 2 then more then 7. There.

I'm self-conscious to the extent that I want respect from people. *shrug* As to personal appearance and so on, I don't really care - so as long as it's neat.

[quote]
Oh me... I dont even have to think about it. Im the least self-concious person ever. I dont care if u dont like me go away leave alone and eventually Ill be your friend. I am very friendly and I dont generally care about anyones opinion well except for someones . I care about myself but I dont care about what others think of me. Im just like whats best for me and what I like the best is what Im going to wear or whatever and I take little thought to who dislikes my "look".[/quote]
ha, i think i am exactly like that as well!

[quote]
What really bugs me out is people speaking in foreign languages and laughing. Particularly when they're watching me play a bad game of pool. Ah, well ... [/quote]  I gotta try that sometimes ;)

I'm not as bad as I used to be but I am really shy still.  I used to not be able to talk to anyone at all.  Now I have several very good friends, but I still don't talk too much and I'm really scared of doing things like oral presentations.

Oh yeah and to answer the original question: I totally didn't care about how I looked until say middle school. Then I started caring like whack, then I got mental and I got depressed and stuff so now I am moderated. Sure I scrub my mouth till it bleeds everytime I eat outside but then again I walk around with wood dust all over myself while having both my shirt and my pants either backwards or inside-out. meh.

i'm not self conscience at all. sometimes, i really think that i need to be more so. i don't really care about what clothes i wear if it's within the boundaries of normality (i.e. no barney apparal). I never comb my hair because it's so short, but i'll gel it once in a while. sometimes, i'm shy around new people, but to people whom i know, i'll say almost anything openly.

im really self-consious...i freak out when im talking to people and i say one little thing wrong, and i have probably the worst hair on earth, so that doesnt help me feel better about myself.  i think im so self-consious because im a perfectionist.

ill have to agree w/ fiery n mathfanatic.  they r exactly like me, self-conciouswise


Hello people, I have a question

A pair of fuzzy dice is hanging by a string from your rearview mirror. While you are accelerating from a stoplight to 20 m/s (in 5.0 seconds), what angle theta does the string make with the vertical?

I am asking this question because I have no idea how to do it.

[quote="mdk"]A pair of fuzzy dice is hanging by a string from your rearview mirror. While you are accelerating from a stoplight to 20 m/s (in 5.0 seconds), what angle theta does the string make with the vertical?[/quote]

Supposing that the acceleration is uniform, we have that $a = \frac{\Delta v}{\Delta t}= \frac{20 \, \frac{\textrm m}{\textrm s}}{5 \, \textrm s}= 4 \, \frac{\textrm m}{\textrm s^{2}}$.

Let $m$ be the mass of the pair of fuzzy dice. Two forces act on them: $G = mg$ downwards and $F = ma$ towards the back of the car. Now make a drawing. You'll see that $\tan \theta = \frac{ma}{mg}= \frac{a}{g}\approx 0.408$.


MATHCOUNTS
search
number theory
least common multiple
relatively prime

Eleven girls are standing around a circle. A ball is thrown clockwise around the circle. The first girl, Ami, starts with the ball, skips the next three girls and throws to the fifth girl, who then skips the next three girls and throws the ball to the ninth girl. If the throwing pattern continues, including Amis initial throw, how many total throws are necessary for the ball to return to Ami?

answer  11 throws

Since I'm too lazy, a search [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1916955636&t=210444]reveals[/url]...

0k soooo

Lets number the people 1-11 and call Ami 1

This ishow the ball goes

1 5 9 2 6 10 3 7 11 4 8  and after these 11 it goes back 1 or Ami

In general, if there are n people and you skip m-1 people in the pass, and if m and n are relatively prime, it takes n passes to get back to the start.

Therefore, 10 passes will be made to get it to the last person, and another to get it to the first person, which makes 11 throws.

Proof for the first part. Because m and n are relatively prime, and the throw always goes to the person with the number that is the next multiple of m (or the pth person, where p is the residue of the multiple of m mod n), and it will only go back to the first person when the multiple of m is a multiple of n, it will take n throws, because the LCM of m and n is mn, and thus we have the nth multiple of n, so it takes n passes.


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ k \geqslant 1$, $ n \geqslant 2$, and $ a_1, \ldots, a_n$ pairwise distinct integers.
Denote by $ P_k(X) \equal{} (X \minus{} a_1)^k\cdots(X \minus{} a_n)^k \plus{} 1$.
Show that $ P_2$, and $ P_4$ are irreducible over $ \mathbb{Q}[X]$.
Can we find all the $ k\geqslant 1$, s.t. $ P_k$ is irreducible over $ \mathbb{Q}[X]$?

$ P_k$ is only irreducible for $ n\equal{}k\equal{}1$ ...

-oo-
I made a typo, I forgot $ \plus{}1$.
It's now correct.


If $m_a,m_b,m_c$ are medians of a triangle $ABC$ prove that

$\sqrt 3(a^2+b^2+c^2)\geq a(m_b+m_c)+b(m_a+m_c)+c(m_a+m_b)$

by the cauchy-schwarz inequality, we have that
$\bigl(\sum_{cyc}(a^2+a^2)\bigl) \bigl(\sum_{cyc}(m_b^2+m_c^2)\bigl)\geq (a(m_b+m_c)+b(m_a+m_c)+c(m_a+m_b))^2 \\\Rightarrow 4\bigl(\sum_{cyc}a^2\bigl) \bigl(\sum_{cyc}m_a^2\bigl)\geq (a(m_b+m_c)+b(m_a+m_c)+c(m_a+m_b))^2$
but from the median theorem, we have that
$\sum_{cyc}m_a^2=\frac{1}{4}\bigl(\sum_{cyc}(2b^2+2c^2-a^2)\bigl) \\\Rightarrow 4\sum_{cyc}m_a^2=3(a^2+b^2+c^2) \\\Rightarrow 3(a^2+b^2+c^2)(a^2+b^2+c^2)\geq (a(m_b+m_c)+b(m_a+m_c)+c(m_a+m_b)^2 \\\Rightarrow \sqrt 3(a^2+b^2+c^2)\geq a(m_b+m_c)+b(m_a+m_c)+c(m_a+m_b)$


calculus
integration
combinatorics unsolved
combinatorics

How many possibilities are there such that a log with an integral length of n metres is cut into pieces of 1 metre ? At each step all the parts with a length greater than 1 must be cut into 2 pieces (of integral length).

reference: http://mathworld.wolfram.com/CatalanNumber.html

So we have the same recurrence as for Catalan numbers.
Indeed, let $P_{n}$ be desired number of possibilities. We can make arbitrary first cut obtaining pairs of pieces $(1,n-1)$, $(2,n-1)$, ..., $(n-1,1)$. Further we cut these pieces separately, thus
\[P_n=P_1P_{n-1}+P_2P_{n-2}+...+P_{n-1}P_1,\]
where $P_1=1$.

That's all.


For which $n$ is it possible, given any $n$ lines in general position in the plane, to label the points with the numbers $1,2,\ldots,n-1$ s.t. each line has points with all $n-1$ labels on it?

("lines in general position" means that no three are concurrent, and no two are parallel)

If such a labeling does exist then $1$ appear exactly $\frac k 2$ times (once for each line, and each intersection point is counted twice) so that $k$ has to be even.

Conversly, if $k = 2$ then it is trivial, and if $k=2p$ for $p \geq 2$ then consider the regular convex $k-1$-gon and label its vertices from $P_1$ to $P_{k-1}$. Since $k-1$ is odd, for each $1 \leq i <j \leq k-1$ the perpendicular bissector of the line segment $P_iP_j$ passes through exactly one vertex, say $P_t$. Clearly, for a given $i$, as $j$ varies we will reach each $t \neq i$ exactly once. Thus, label the intersection point of line number $i$ with line number $j$ by $t$.
In this manner, the lines numbered $i \leq k-1$ has point with each label except label $i$.
Now, label $i$ the intersection point of line $k$ with line $i$, and we are done.

Pierre.


logarithms function algebra domain calculus derivative

hello,

i) Solve in $ \mathbb{R}$ the equation : $ \displaystyle \sqrt[3]{x+6}+\sqrt{x-1}=x^2-1$

ii) Solve in $ \mathbb{R}$ the equation : $ {\Large}\displaystyle 2003x=(2004) 2003^\log_x{(2004)}$

[hide="1"]$ x \equal{} 2$ is a solution. We will show that it is the only solution.
Moving everything to the LHS,
$ 1 \minus{} x^2 \plus{} \sqrt [3]{x \plus{} 6} \plus{} \sqrt {x \minus{} 1} \equal{} 0$
and let
$ f(x) \equal{} 1 \minus{} x^2 \plus{} \sqrt [3]{x \plus{} 6} \plus{} \sqrt {x \minus{} 1}$.
Since $ 1 \minus{} x^2$, $ \sqrt [3]{x \plus{} 6}$, and $ \sqrt {x \minus{} 1}$ are all concave functions, their sum $ f(x)$ is concave throughout its domain. $ f'(x)$ is decreasing.
Assume for the sake of contradiction that the original equation has two or more solutions. Then, in $ [1, \infty)$, the domain of $ f$, there are two or more zeroes. Call the first two zeroes $ a$ and $ b$, where $ a < b$. Since $ f(1) \equal{} \sqrt [3]{7} > f(a) \equal{} 0$ and $ 1 < a$, $ f(x)$ must be decreasing at some value $ c \in (1,a)$. Thus $ f'(c) < 0$. Since $ f'(x)$ is decreasing, for all $ x > c$, $ f'(x) < 0$, and $ f(x)$ is decreasing. Since $ b > a > c$, $ f(b) < f(a) \equal{} 0$, contradiction.
[/hide]

[hide="Equation 1 (no derivatives necessary)"]First of all, $ x\geqslant 1$ because of the second term.

Write the equation as

$ \sqrt [3]{x \plus{} 6} \minus{} 2 \plus{} \sqrt {x \minus{} 1} \minus{} 1 \equal{} x^2 \minus{} 4$

$ \frac {x \minus{} 2}{\sqrt [3]{(x \plus{} 6)^2} \plus{} 2\sqrt [3]{x \plus{} 6} \plus{} 4} \plus{} \frac {x \minus{} 2}{\sqrt {x \minus{} 1} \plus{} 1} \equal{} (x \minus{} 2)(x \plus{} 2)$

Obviously, $ x \equal{} 2$ is a solution. Assume $ x\neq 2$. Then

$ \frac {1}{\sqrt [3]{(x \plus{} 6)^2} \plus{} 2\sqrt [3]{x \plus{} 6} \plus{} 4} \plus{} \frac {1}{\sqrt {x \minus{} 1} \plus{} 1} \equal{} x \plus{} 2$

The LHS is $ \leqslant {1\over 4} \plus{} {1\over 1}$, and the RHS is $ \geqslant 3$, so there can be no solution.

Therefore $ x \equal{} 2$ is the only solution.[/hide]

hey 2003 no es solucion, 2004 y 2003^-1si lo son


Find all real numbers $m$ so that the equation:
\[\left(\frac{1}{5}\right)^{|x^{2}-4x+3|}=m^{4}-m^{2}+1\]
has four roots.

$m^{4}-m^{2}+1\ge\frac{3}{4}$, and we want to find all $a$ such that

$|x^{2}-4x+3|=a$ has 4 [distinct] solutions

immediately, a>0, furthermore, if we look at this graphically, we have something like this:

\/\/, where there is a maximum in the middle, this is the same as the negative of the maximum of the nonabsolute value equation, the min is easily is -1, also there are two double roots for a=0, so we have $0<a<1$, and then

$0< \log_{\frac{1}{5}}(m^{4}-m^{2}+1)<1$
since that is a strictly decreasing function,
$1>m^{4}-m^{2}+1>\frac{1}{5}$, but we have the stronger equality for real $m$
$1>m^{4}-m^{2}+1\ge \frac{3}{4}$

now we have to solve for $m$, which is trivial


Prove that the ortocentre of a triangle whose vertices are $ (a \cos \alpha, a \sin \alpha)$, $ (a \cos \beta, a \sin \beta)$, $ (a \cos \gamma, a \sin \gamma)$ is
$ (a \cos \alpha \plus{} a \cos \beta \plus{} a \cos \gamma, a \sin \alpha \plus{} a \sin \beta \plus{} a \sin \gamma)$

This should a) be posted in the High school or pre-olympiad forum or
                b) be in the Theorems and Formulae section.


Can anybody tell me how to measure sines, cosines, and tangents?  I know the first two have to do with distance from axises, but I don't know much about them

Use a calculator? And I don't know what you mean by distance from axises, it has to deal with ratios of legs in an right triangle with two specific angles.

Imagine a unit circle and angle $\theta$...

http://en.wikipedia.org/wiki/Image:Circle-trig6.svg


[b]f(x)[/b][img]http://i21.tinypic.com/28irf2s.jpg[/img]
[b]$ f(x)^2$[/b][img]http://i21.tinypic.com/2zogj0w.jpg[/img]

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

Why is the line going through the horizontal asymptote and when your sketching $ f(x)]^2$ how would you identify that it goes the an asymptote

There is a point $ a$ on $ f$ such that $ f(a)\equal{}2$. For every $ n$ on the left side of $ a$, $ |f(n)|<2$, so $ f(n)^2<4$. However, for $ n$ between $ a$ and $ 1$,  $ f(n)>2$, so $ f(n)^2>$. Thus on the left side of $ a$, the square of the original function is less than 4 and under the asymptote, but between $ a$ and $ 1$, the square of the function is greater than 4 and is above the asymptote.

Remember that functions can pass through horizontal asymptotes, just not vertical ones.

A horizontal asymptote $ y \equal{} a$ exists when a function approaches $ a$ as $ x$ approaches either $ \infty$ or $ \minus{}\infty$. There is no stipulation that the function cannot pass through the asymptote; this is simply a property of vertical asymptotes when dealing with continuous functions. It's about the limits, not whether the function passes through the asymptote.

AHem!! what will be the equation of f(x)^2 like ?
That f(x) in such an eqn assumes a value 4 as x approaches -infinity and again it assumes a value of 4 at some point x<1....


If I am enrolling in the Mandelbrot for the first time this year, and I am unsure whether to do national or regional.  Any suggestions, or benchmarks as to give an idea of what to do, say based on AIME score, or AMC score?

If you got above a 6 or 7 on AIME..I'd do national...

i got a 6 on the AIME (should have been 8...), but the top 3 people on my team (besides me) are only at a 1-3 on the AIME, level.

should I go with regional?


[hide="Solution"]Let's use $a, b, c$ for less typing, and let $r = |a| = |b| = |c| \in \bf R$.  Since $a, b, c \in \bf R$ we have ${\bf R}\ni a+bc = \overline{a+bc}= \bar{a}+\bar{b}\bar{c}$.  Then ${\bf R}\ni (\bar a+\bar b \bar c)(b+ac) = \bar{a}b+r^{2}c+r^{2}\bar c+a\bar b r^{2}$.  Now, $r^{2}c+r^{2}\bar c \in \bf R$ so we have $\bar a b+a \bar b r^{2}\in \bf R$.  Considering the imaginary part of this expression we see that either $r^{2}= 1$ or $a\bar b \in \bf R$.

If $r^{2}= 1$ then $|a| = |bc| = 1$, so $a+bc \in \bf R$ implies either $bc = a^{-1}$ (in which case we're set) or that $bc =-a$.  Applying the same logic gives also that $ab =-c$.  Multiplying these two equations, we have $ab^{2}c = ac$, so $b^{2}= 1$.  Then by symmetry, $a^{2}= b^{2}= c^{2}= 1$ and so two of them are the same, contradicting distinctness.

If $a\bar b \in \bf R$ then $\bar b = k \cdot a^{-1}$ for some $k \in \bf R$.  Taking absolute values, $\bar b = r^{2}a^{-1}$ and so $b = r^{2}\bar{a^{-1}}= a$.  Then $a = b = c$, violating the assumption of distinctness, and we're done.[/hide]

Since $z_{1}, z_{2}, z_{3}$ lie on the same circle, $z_{1}+z_{2}z_{3}\in\mathbb{R}$ implies that they lie on the unit circle:
$z_{1}+z_{2}z_{3}\in\mathbb{R}\Rightarrow$ Im($z_{1}$)$=-$Im($z_{2}z_{3}$), a fortiori $|z_{1}|=|z_{2}z_{3}|=|z_{2}\parallel z_{3}|$ (owing to the fact that all the $z$ lie on the same circle) so that $|z_{1}|=1$. Similar result holds for each $i=2, 3$.

Furthermore, the above also implies that $z_{i}$ and $z_{j}z_{k}$ are conjugate pairs, where we have $\overline{z_{1}}=z_{2}z_{3}, \overline{z_{2}}=z_{1}z_{3}, \overline{z_{3}}=z_{1}z_{2}$.

The product $\overline{z_{1}}\overline{z_{2}}\overline{z_{3}}=\overline{z_{1}}(z_{1}z_{3})(z_{1}z_{2})=z_{1}z_{2}z_{3}$, using the conjugate-pair identities above, implies $z_{1}z_{2}z_{3}\in\mathbb{R}$, and hence equal $1$ or $-1$.

The same product using the conjugated identities (e.g. $z_{1}=\overline{z_{2}z_{3}}$) and expressed in the form $z_{1}z_{2}z_{3}=\overline{z_{1}}^{2}\overline{z_{2}}^{2}\overline{z_{3}}^{2}$ reveals that $z_{1}z_{2}z_{3}\geq0$. Therefore, the product is $1$.

[quote="zulu-yankee"]Im($z_{1}$)$=-$Im($z_{2}z_{3}$), a fortiori $|z_{1}|=|z_{2}z_{3}|=|z_{2}\parallel z_{3}|$ (owing to the fact that all the $z$ lie on the same circle) so that $|z_{1}|=1$.[/quote]
This is wrong.  For example, $z_{1}= 3+4i$, $z_{2}= 4+3i$ and $z_{3}= \frac{-12-4\sqrt{609}+(-16+3\sqrt{609})i}{25}$ have the property that $|z_{1}| = |z_{2}| = |z_{3}| = 5$ and that $z_{1}+z_{2}z_{3}\in \bf R$.

Even if that were right, this
[quote]Furthermore, the above also implies that $z_{i}$ and $z_{j}z_{k}$ are conjugate pairs, where we have $\overline{z_{1}}=z_{2}z_{3}, \overline{z_{2}}=z_{1}z_{3}, \overline{z_{3}}=z_{1}z_{2}$.[/quote]
would be wrong because it neglects the possibility that $z_{1}=-z_{2}z_{3}$.  It takes a short argument (see my previous post) to eliminate this possibility.

Oh yeah, totally neglected that possibility...

And, in fact, the frst part is so, so wrong...

My apologies...


function
Support
abstract algebra
real analysis
real analysis theorems

Hi,

I am really a dummy when it comes to analysis, so this might seem a basic question :

We are studying $\mathcal{D}(K)$ with K compactum, that is the set of all infinitely differentiable functions with support within K

now first question : my professor wants to consider tensor products of distributions ,
and he insists on calling $\mathbb{R}^{n+m}$ the tensor product of $\mathbb{R}^n$ and $\mathbb{R}^m$
Shouldn't that be direct sum?  He said no direct sum means two subspaces of the same space , I tried to point out that there is also an outer direct sum, closely related, but he insisted on calling it tensor product?

second, more important question, when he says
$\mathcal{D}(K_1) \otimes \mathcal{D}(K_2)$ is dense in $\mathcal{D}(K_1 \times K_2)$
what does he mean?
that every function can be approximated by products of functions?
or by finite sums of them?

All comments are welcome! :oops:

The tensor product of $\mathbb{R}^n$ and $\mathbb{R}^m$ over $\mathbb{R}$ is isomorphic to $\mathbb{R}^{nm}$. Your professor is wrong.
It would be acceptable to consider $\mathbb{R}^{n+m}$ as the (direct product/ Cartesian product/direct sum) of $\mathbb{R}^n$ and $\mathbb{R}^m$, but not a tensor product.

I'll try to make sense of that last statement:
For any set $S$, let $\mathbb{R}^S$ be the space of functions from $S$ to $\mathbb{R}$. There is a canonical injection from $\mathbb{R}^{S_1}\otimes\mathbb{R}^{S_2}$ to $\mathbb{R}^{S_1\times S_2}$, generated by $f\otimes g\to f*g$, where $f*g(x,y)=f(x)g(y)$. This is an isomorphism when $S_1$ and $S_2$ are finite.
This method also works nicely for tensor products of function spaces; the most important is the identification of $V^*\otimes W^*$ with a subspace of $(V\times W)^*$. In this case, we can map $\mathcal{D}(K_1)\otimes\mathcal{D}(K_2)$ into $\mathbb{R}^{K_1\times K_2}$; it is easily verified that the image of $f\otimes g$ is $C^\infty$ for each $f,g$, so this is a map into $\mathcal{D}(K_1\times K_2)$.

Since $\cal{D}$ has many different meaningful norms/topologies, we should specify which norm/topology this density result is based on.. The tensor product is defined as the space of finite linear combination of the $f\otimes g$, so it means that every function in the space can be approximated arbitrarily closely by a finite sum/linear combination of functions of the form $F(x,y)=f_1(x)f_2(y)$. We obviously can't do it with just one product.

Thanks, as I expected

But one question, do you mean it can be approximated by a sequence
$u_n=\sum_{i=1}^{n} f_n \otimes g_n$

or by a sequence $u_n=\sum_{i=1}^{n} f_{i,n}\otimes g_{i,n}$

I'd use $u_n=\sum_{i=1}^{N(n)} f_{i,n}\otimes g_{i,n}$. There's no reason to increase the number of terms so uniformly.
It doesn't really matter- all of the forms are equivalent with suitable trickery.


geometry
circumcircle
parallelogram
geometry unsolved

There is a right-angled trianlge ABC, with AB=AC, angle BAC=90degree, D,E are two points on side BC, the circumcircle of triangle ADE touches sides AB,AC on P,Q respectively, and BP+CQ=PQ, find the degree of angle DAE.

Is there a solution provided with forming a parallelogram?


probability
calculus
calculus computations

Two players play tennis . 1 match finishes when 1 player won 6 sets . It's known that : the winning probability of player 1 is $p$ and the winning probabilyty of player 2 is $1-p$ .
  Define the mean time of 1 match ?

I think it's  min(6/p, 6/1-p) because  E(R)=Np.

Can you explain your solution more clearly ?

my initial thought was it was a binomial distribution, which would offer Np, but i wasnt sure so i didnt post, the binomial distribution is:

$p(x) = \frac{N!}{x!(N - x)!}p^x(1-p)^{N-x}$

n is the number of trials, p is the probability of a success in one of the trials, and if i remember correctly p(x) the probability you will get x successes.

the mean would be $\sum_x x*p(x)$

Np is by definition the mean of a binomial distribution and i dont think you'd have to show work for it

but i would look into binomial distributions

I am not 100% sure that my solution is correct but here is my thought: Np is the mean for binomial distribution, so let's say p=0.2, then I would expect 6 wins for this player whenever 30 games are completed. 6=30*0.2
I am using this formula Np in back way, N is variable.
Another way could be take Exp. for both sides:

E(E(R))=E(N*p)
since p is const and E(E(R))=E(R) then
E(R)=E(N)*p   in our case we are interesting in E(R)=6
so E(N)=6/p      for this player.
Again, this is just my thought.


let n a positive integers if 4^n+2^n+1 is a prime than n is a power of 3 solve this problem man

http://www.mathlinks.ro/Forum/viewtopic.php?t=1489
http://www.mathlinks.ro/Forum/viewtopic.php?t=291
http://www.mathlinks.ro/Forum/viewtopic.php?t=47701
http://www.mathlinks.ro/Forum/viewtopic.php?t=27620

  darij


[color=darkblue]Given two intersecting circles C1,C2.(Let C1 be in the left)C1 intersects C2 in points M,N(let M be in the top).W is moving on MN.We draw a line such that passes H and intersects C1 in points A,C(Let A be in the left) and intersects C2 in points B,D(Let C be in the left).
Then we draw a tangent line from B that intersects C1 in points E,G(let E be in the top).After it we draw another tangent line from C that intersects C2 in points F,H(let F be in the top)[/color].
:spider:
[color=indigo]Is FEBC a inscribed quadrilateral?Prove your notation[/color].
[color=green](May be this forum is better than open questions forum fot this problem ;) )[/color]

I have solved it when CB=DA.Later i will give you my solution. :lol:

BH intersects GC in point S and DF intersects CE in point Q
[color=green]Lets have my solution when CB=DA:[/color]
\[ DC \times CB = FC \times CH (1) \]
\[ BC \times BA = BE \times BG (2) \]
\[ (1),(2) \rightarrow \frac{FC}{BE}=\frac{BG}{CH} \rightarrow
\frac{FC}{BG}=\frac{BE}{CH} (3)\]
\[ \triangle FQC \sim \triangle BSG \rightarrow \frac{FC}{BG}=\frac{FQ}{BS}=\frac{QC}{GS} \]
\[ \triangle CSH \sim \triangle EQB \rightarrow \frac{BE}{CH}=\frac{EQ}{CS}=\frac{BQ}{SH} \]
\[ (3) \rightarrow \frac{FQ}{BS}=\frac{QC}{GS}=
\frac{EQ}{CS}=\frac{BQ}{SH} (4) \]
\[ (4) \rightarrow \frac{QC}{BQ}=\frac{GS}{SH} (\star) \]
\[ \angle BHC=\angle EBF and   \angle BEC=\angle GCH (6) \]
\[ (6) \rightarrow \angle BSC=\angle BQC (\star \star) \]
\[ (\star \star),(\star) \rightarrow \triangle GSH \sim \triangle EQF (7) \]
\[ (7) \rightarrow \angle SHG=\angle QBC and   \angle SGH=\angle QCB (8) \]
\[ (8),(6) \rightarrow \angle EBC=\angle CHG \]
$ \rightarrow $[color=blue]CBGH and CFEB are inscribed quadrilaterals.[/color] :D

Can anybody solve this problem?
What sort of mathemathicans
 are you? ;)

We are mathematicians who like to solve problems which are formulated in a clear and nice way. That might be the reason why you didn't get any reactions yet.

The real mathemathican is who solve any questions and enjoy it! :roll:

Nice proof Jensen(yensene khodemoon)  :first:

so ur problem is true only for CB=DA .
U know why :P

i know it is not true in general.but how do you know it is true only when AC=BD? may be this is true with some conditions! :?:


$E$ and $F$ are interior points of convex quadrilateral $ABCD$ such that $AE = BE$, $CE = DE$, $\angle AEB = \angle CED$, $AF = DF$, $BF = CF$, $\angle AFD = \angle BFC$.  Prove that $\angle AFD + \angle AEB = \pi$.

There are lot of properties I think I found about this figure:

For example, the properties of $E$, implies the ones of $F$. Also it implies $AC=BD$, and also the lines trought midpoints of $AB$ and $CD$, and $AD$ and $BC$ are perpendicular!

This properties are not hard to prove. If someone is interested I can post them and show how they easily imply the result.

Ok can anyone show that property...

Davron

standard using complex nos. :)
take one pt. of intersection of perpendicular bisectors as origin,right given conditions using mod a\modb=a\b8eto the power ic,c-angle bet a and b.

I think there's a more basical solution, but I can't find it :D .

[b]Problem:[/b]
$E$ and $F$ are interior points of convex quadrilateral $ABCD$ such that $AE=BE$, $CE=DE$, $\angle AEB = \angle CED$, $AF=DF$, $BF=CF$, $\angle AFD = \angle BFC$. Prove that $\angle AFD+\angle AEB = \pi$.

[b]Solution:[/b]
Notice that, because $E$ lies in the interior of convex quadrilateral $ABCD$, \[\angle AEC = \angle AED+\angle CED;\]  \[\angle AEC = \angle AED+\angle AEB;\]  \[\angle AEC = \angle BED,\] and that $AE = BE$ and $CE = DE$, so $\triangle AEC \cong \triangle BED$.

Similarly, we show that $\triangle AFC \cong \triangle DFB$.

Let $S$ be the intersection of $AC$ and $BD$. Then $\angle EAS = \angle EBS$. As $ABCD$ is convex, and thus that $S$ (as well as $E$) lies in the interior of $ABCD$, $A$, $B$, $E$ and $S$ are concyclic. Thus $\angle AEB = \angle ASB$.

Similarly, we show that $\angle AFD = \angle ASD$.

Notice that $B$, $S$ and $D$ are collinear, so $\angle ASD+\angle ASB = \pi$. Thus $\angle AFD+\angle AEB = \pi$. $\Box$

Let $\angle BEA=x$ and $\angle AFC=y$. WLOG the vertices are labelled in anticlockwise order.

Consider the following transformation:

Anticlockwise rotation centered at $E$ with angle $x$, followed by anticlockwise rotation centered at $F$ with angle $y$.

This transformation is a rotation with angle $x+y$, mapping point $A$ to point $C$, and mapping point $C$ to point $A$.

Thus we see that $x+y$ is of the form $(180 + 360k)^\circ$, and a simple bound yields $x+y=180^\circ$.

My solution:

Let $ X $ be the intersection of $ AC $ and $ BD $ .

Easy to see $ \triangle ACE \cong \triangle BDE $ .
Since $ E $ is the center of spiral similar which map $ \triangle ACE \mapsto \triangle BDE $ ,
so we get $ X \in \odot (ABE) \Longrightarrow \angle CXB=2\angle CXE=2\angle EAB $ . ... $ (1) $
Similarly, we can prove $ X \in \odot (DAF) $ and $ \angle DXC=2\angle DXF=2\angle DAF $ . ... $ (2) $

Since $ \angle DXC+\angle CXB=\angle DXB=180^{\circ} $ ,
so from $ (1), (2) $ we get $ \angle EAB+\angle DAF =90^{\circ} \Longrightarrow \angle AFD+\angle AEB=180^{\circ} $ .

Q.E.D


Let $(C)$ and $(L)$ be a circle and a line. $P_{1},\dots,P_{2n+1}$ are odd number of points on $(L)$. $A_{1}$ is an arbitrary point on $(C)$. $A_{k+1}$ is the intersection point of $A_{k}P_{k}$ and $(C)$ ($1\leq k\leq 2n+1$). Prove that $A_{1}A_{2n+2}$ passes through a constant point while $A_{1}$ varies on $(C)$.

This is same?

http://www.mathlinks.ro/Forum/viewtopic.php?t=17033

Also this is the same:
http://www.mathlinks.ro/Forum/viewtopic.php?t=133465

(and I'm still waiting for Tiks' synthetical solution  :wink: )


[quote="pardesi"]$ \ce{CH2 \dbond CHCl} \plus{} \ce{HCl}$
$ \ce{CH2 \dbond CH \sbond CH2Cl} \plus{} \ce{HCl}$[/quote]
in the first case H+ will get attached to the CHCl part and Cl- will get attached to the CH2 part the reason is that when HCl approaches formation of carbocation is less preferred at CHCl part as Cl is electronwithdrawing in nature and destabilises the carbocation!!! :idea:
in your second question, when we add HCl it attacks the double bond rather than the allylic carbon cause the allylic carbocation is greatly destabilised by the chlorine attached to it so it must break the double bond but one doubt here is CH2Cl +I or -I group??? :maybe:  :?:

[quote="pardesi"]For the first the answer is a geminal dihalide and not vicinal
why should HCl attack the allylic Carbon atom??
yes it attacks the double bond and here the product is vicinal not geminal[/quote]
whats wrong with my reasoning in the first case??? :ninja:  :o

there is nothing wrong but that does not equally remove out the scope having a geminal dihalide

The reactivity of vinyl chloride can be understood if we make an analogy with the reactivity and orientation of halobenzenes: in these last compounds, the halogens make the ring less reactive (they deactivate it) but have an ortho-para orientating effect. With vinyl chloride the halogen is also linked to an unsaturated system (this time only one double bond) but the effect is the same: vinyl chloride reacts less readily than ethylene and the orientation is "ortho". In the vinyl chloride reaction the electrophile is H+, so a proton will make a bond with carbon 2 of vinyl chloride (which gives the most stable carbocation), and next a chloride ion will use a pair of electrons to make a bond with that cationic center: the final product is then 1,1-dichloroethane.

$ CH_2 \equal{} CH \minus{} Cl \plus{} H^ \plus{} \longrightarrow CH_3 \minus{} CH^ \plus{} \minus{} Cl$

$ CH_3 \minus{} CH^ \plus{} \minus{} Cl \plus{} Cl^ \minus{} \longrightarrow CH_3CHCl_2$.

The reaction of allyl chloride can be understood in a similar fashion (assuming similar behaviour): what is the effect of the substituent $ \minus{} CH_2Cl$ on a benzene ring? Well, the effect is the same of an halogen: deactivating (because of the electronwithdrawing inductive effect) and ortho-para director (because -CH2X can still stabilize a positive charge, this time by hyperconjugation). So, in allyl chloride the reaction with HCl is predicted to be slower compared with ethylene, but the electrophile is predicted to go to the "ortho" position,

$ CH_2 \equal{} CH \minus{} CH_2Cl \plus{} H^ \plus{} \longrightarrow CH_3 \minus{} CH^ \plus{} \minus{} CH_2Cl$

$ CH_3 \minus{} CH^ \plus{} \minus{} CH_2Cl \plus{} Cl^ \minus{} \longrightarrow CH_3CHCl \minus{} CH_2Cl$.

However, one may observe from experience that this is not the case: while in vinyl chloride the behaviour of halogen is the same on both unsaturated systems, in the case of allyl chloride the electronwithdrawing inductive effect of the $ \minus{}CH_2Cl$ group seems to be prevalent, and so case the chloromethyl group behaves as a deactivating and "meta" director:

$ CH_2 \equal{} CH \minus{} CH_2Cl \plus{} H^ \plus{} \longrightarrow CH_2^\plus{} \minus{}CH_2\minus{}CH_2Cl$

$ CH_2^\plus{} \minus{}CH_2\minus{}CH_2Cl \plus{} Cl^ \minus{} \longrightarrow ClCH_2CH_2CH_2Cl$.

thanks Carcul that was educative
first one is right for the secoond one it will be a 1,3 dichloro compouned and not 12 dichloro
sorry i ahd written that as vicinal  :(

In that case the electronwithdrawing inductive effect of $ \minus{}CH_2Cl$ will dominate (see the edited post above).

thanks carcul that post demanded rates  :lol:

why would the inductive effect be more prevalent? The secondary carbocation would be more stable due to hyper conjugation(5 alpha hydrogens) and shouldnt vicinal diol be formed?

[quote="Zimrock"]why would the inductive effect be more prevalent?[/quote]

Because that's what experimental evidence seems to show. And taking the inductive effect of the chloromethyl group then the secondary carbocation wouldn't be the most stable. Also, there were be only 4 alpha hydrogens for hyperconjugation.


geometry
geometric transformation
reflection

Geia se olous. An kai den ime se kamia periptosi sto diko sas epipedo sta mathimatika mou aresoun poli ta problhmata geometrias olympiadwn, ta opoia prospatho na liso me tis liges mou gnoseis. An mporite boithiste me sto parakato, giati den mporo na tp liso me tipota.

The circle Γ and the line e have no common points. Let AB be the
diameter of Γ perpendicular to e, with B closer to e than A. An arbitrary
point C (other than A,B) is chosen on Γ. The line AC intersects e at D. The line
DE is tangent to Γ at E, with B and E on the same side of AC. Let BE
intersect e at F, and let AF intersect Γ again at G (other than A). Prove that the
reflection of G in AB lies on the line CF.

Εστω $ S$ σημειο τομης $ (AB), (e)$ και $ T$ σημειο τομης της $ (CF)$ με τον (Γ) ,.
 Αρκει $ (GT)$ καθετο στο $ AB$ ή $ (GT)\parallel{}(e)$. Oμως $ \angle{TGA} \equal{} \angle{DCF}$
αρα αρκει να ειναι ομοια τα τριγωνα $ (DCF),(DFA)$ ωστε $ \angle{DCF} \equal{} \angle{DFA}$ και να προκυψει η ζητουμενη παραλληλια.
Oμως $ (DE)^2 \equal{} (DC)(AD)$ και $ DEF$ ισοσκελες λογω του εγγραψιμου $ DEAF$
 Συνεπως $ (DF)^2 \equal{} (DC)(DA)$, απο οπου κατευθειαν το ζητουμενο.

Telika, brika mia lisi pou xrisimopoiei elaxistes geometrikes gnosis (mexri kai eggrapsima), alla einai poli pio ektenis. Den tin vazo giati i diki sou ine pio suntomi kai kompsi.

Είναι η άσκηση  :
   ΙΜΟ , short list 2004 και μια λύση της έχουμε στο βιβλίο ''Στεργίου Χ. , Γεωμετρία για διαγωνισμούς - 3, άσκηση 1.72, σελίδα 67''.
 Πρόκειται όντως για όμορφο θέμα !

   Καλό καλοκαίρι !

Μπάμπης


Let $ a,b,c>0$ s.t. $ abc\equal{}1$.Prove that:
$ a) \frac{a\minus{}1}{b}\plus{}\frac{b\minus{}1}{c}\plus{}\frac{c\minus{}1}{a}\geq0$.
$ b) \frac{a\minus{}1}{b\plus{}c}\plus{}\frac{b\minus{}1}{c\plus{}a}\plus{}\frac{c\minus{}1}{a\plus{}b}\geq0$

[quote="Inequalities Master"]Let $ a,b,c > 0$ s.t. $ abc \equal{} 1$.Prove that:
$ a) \frac {a \minus{} 1}{b} \plus{} \frac {b \minus{} 1}{c} \plus{} \frac {c \minus{} 1}{a}\geq0$.[/quote]

equivalence $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} (\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})$

equivalence $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} (ab \plus{} bc \plus{} ca)$ (so $ abc \equal{} 1$)

we have $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\ge \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}} \equal{} a \plus{} b \plus{} c$

$ \rightarrow \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} (ab \plus{} bc \plus{} ca) \ge a \plus{} b \plus{} c \minus{} (ab \plus{} bc \plus{} ca)$

we need prove that $ a \plus{} b \plus{} c \minus{} (ab \plus{} bc \plus{} ca)\ge 0$

setting $ F(a,b,c) \equal{} a \plus{} b \plus{} c \minus{} (ab \plus{} bc \plus{} ca)$

$ F(a,b,c) \minus{} F(a,\sqrt {bc},\sqrt {bc}) \equal{} (\sqrt {b} \plus{} \sqrt {c})^2(1 \minus{} a)$

assume $ a \equal{} min(a,b,c) \rightarrow a\le 1$

thus, $ (\sqrt {b} \plus{} \sqrt {c})^2(1 \minus{} a)\ge 0 \rightarrow F(a,b,c)\ge F(a,\sqrt {bc},\sqrt {bc})$

by Mixing Variable we need prove $ \frac {1}{t^2} \plus{} 2t \minus{} \frac {2}{t} \minus{} t^2$ but it is true.

done~ :wink:

By AM-GM and from $ abc\equal{}1$

$ \frac {a}{b}\plus{}\frac {b}{c}\plus{}\frac {b}{c} \geq 3\sqrt[3] {\frac {ab}{c^2}}$

$ \frac {a}{b}\plus{}\frac {b}{c}\plus{}\frac {b}{c} \geq 3\sqrt[3] {(ab)^3} \equal{}3ab$

So we have  3 $ (\frac {a}{b}\plus{}\frac {b}{c}\plus{}\frac {c}{a}) \geq 3(ab\plus{}bc\plus{}ca)$

or $ \frac {a}{b}\plus{}\frac {b}{c}\plus{}\frac {c}{a} \geq \frac {1}{a}\plus{}\frac {1}{b} \plus{}\frac {1}{c}$

[quote="Inequalities Master"]Let $ a,b,c > 0$ s.t. $ abc \equal{} 1$.Prove that:
$ b) \frac {a \minus{} 1}{b \plus{} c} \plus{} \frac {b \minus{} 1}{c \plus{} a} \plus{} \frac {c \minus{} 1}{a \plus{} b}\geq0$[/quote]
assume $ a\ge b\ge c$

we have $ a\plus{}b\ge a\plus{}c\ge b\plus{}c \rightarrow \frac {1}{b \plus{} c} \ge \frac {1}{c \plus{} a} \ge \frac {1}{a \plus{} b}$

and $ a\minus{}1\ge b\minus{}1\ge c\minus{}1$

by chebyshev we have

$ \frac {a \minus{} 1}{b \plus{} c} \plus{} \frac {b \minus{} 1}{c \plus{} a} \plus{} \frac {c \minus{} 1}{a \plus{} b} \ge [(a\minus{}1)\plus{}(b\minus{}1)\plus{}(c\minus{}1)](\frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{a \plus{} b})$

but $ (a\minus{}1)\plus{}(b\minus{}1)\plus{}(c\minus{}1)\equal{}a\plus{}b\plus{}c\minus{}3$ and $ a\plus{}b\plus{}c\ge 3\sqrt[3]{abc}\equal{}3$

$ \rightarrow a\plus{}b\plus{}c\minus{}3\ge 0 \rightarrow (a\minus{}1)\plus{}(b\minus{}1)\plus{}(c\minus{}1) \ge 0$

thus $ [(a\minus{}1)\plus{}(b\minus{}1)\plus{}(c\minus{}1)](\frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{a \plus{} b})\ge 0$

done~ :wink:

[quote="rachid"][quote="bunhiacovski"]
we need prove that $ a \plus{} b \plus{} c \minus{} (ab \plus{} bc \plus{} ca)\ge 0$
[/quote]
Not always true.[/quote]
$ a \plus{} b \plus{} c \minus{} (ab \plus{} bc \plus{} ca)\ge 0$ true with $ abc\equal{}1$

[quote="bunhiacovski"][quote="rachid"][quote="bunhiacovski"]
we need prove that $ a \plus{} b \plus{} c \minus{} (ab \plus{} bc \plus{} ca)\ge 0$
[/quote]
Not always true.[/quote]
$ a \plus{} b \plus{} c \minus{} (ab \plus{} bc \plus{} ca)\ge 0$ true with $ abc \equal{} 1$[/quote]
$ a\equal{}100,b\equal{}100,c\equal{}\frac{1}{10000}$

[quote="MoonRider"][quote="bunhiacovski"]
assume $ a\ge b\ge c$

we have $ a \plus{} b\ge a \plus{} c\ge b \plus{} c \rightarrow \frac {1}{b \plus{} c} \ge \frac {1}{c \plus{} a} \ge \frac {1}{a \plus{} b}$

and $ a \minus{} 1\ge b \minus{} 1\ge c \minus{} 1$

by chebyshev we have

$ \frac {a \minus{} 1}{b \plus{} c} \plus{} \frac {b \minus{} 1}{c \plus{} a} \plus{} \frac {c \minus{} 1}{a \plus{} b} \ge [(a \minus{} 1) \plus{} (b \minus{} 1) \plus{} (c \minus{} 1)](\frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {1}{a \plus{} b})$

done~ :wink:[/quote]

But Chebychev cannot be used with negative real (some of  a-1,b-1,c-1 are negative reals )[/quote]
sorry, i think Chebychev can be used with negative real ,you should see your theorem


Δινεται ο πινακας

$ A\equal{}\left[\begin{array}{c c c }
2 & 2 & 1 \\
1 & 3 & 1 \\
1 & 2 & 2
\end{array}
\right]$

Να βρειτε τον $ A^{100}\minus{}A^{50}$

$ A=\left[\begin{array}{c c c }
2 & 2 & 1 \\
1 & 3 & 1 \\
1 & 2 & 2
\end{array}
\right]
B=\left[\begin{array}{c c c }
-1 & 2 & 0 \\
-1 & -1 & 1 \\
1 & 0 & -2
\end{array}
\right]
D=\left[\begin{array}{c c c }
5 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}
\right]
A=BDB^{-1} \to A^{100}-A^{50}=B(D^{100}-D^{50})B^{-1}$ plug and chug?

i hope the part i cant read doesnt say something crazy making my answer irrelevant.


vector
algebra
function
domain
complex analysis
complex numbers
complex analysis theorems

Hi, I was wondering what's the difference between a complex number and a vector in 2-D. Both are defined as a pair of numbers (a,b) and they both have the same rule for addition, and have the same formula for finding their length, etc.

I know that they are suppose to be different but I don't know why?

$\mathrm{Mathematics\ are\ a\ monstrous\ tautology\ .}$

$(x,y)\in \mathbb R^{2}$ $\Longleftrightarrow$ $P(x,y)\in \mathbb P$ (plane) $\Longleftrightarrow$ $z=a+b\cdot i\in\mathbb C$ $\Longleftrightarrow$ $\overrightarrow{OP}\in\mathbb V$ $\Longleftrightarrow$ $\ldots$ isomorphical structures !

We say in more languages about same thing (matter).

I disagree with that. I think that there is a very big difference (conceptually speaking).

On the one hand, vectors are part of [url=http://en.wikipedia.org/wiki/Vector_space]Vector Spaces[/url], which are the central object of study in linear algebra. We can have not only vectors in $\mathbb R^{2}$, but also in any $\mathbb R^{n}$.

[url=http://en.wikipedia.org/wiki/Complex_number]Complex numbers[/url] on the other hand form an algebraically closed field. That, among others, means that the very important operation $\cdot$ is defined on $\mathbb C$ (it's also defined for vectors, but it means something else, scalar product). We can't "construct" anything similar to them in "3D", but we can do it in 4D: [url=http://en.wikipedia.org/wiki/Quaternion]Quaternions[/url] (they are almost a field, but not quite).

$\mathrm{Mathematics\ are\ a\ monstrous\ tautology\ .}$ These words belong to a great Romanian mathematician , Gr. Moisil, who was my theacher at University. Generally, and I wasn't agree with this sentence. However, the above chain of equivalencies is, in my opinion, a tautology. It generates many methods for solving problems in geometry. [b]The tautology is a fertil spring of creation new domain of research,[/b] at least in mathematics.

[quote="perfect_radio"]On the one hand, vectors are part of [url=http://en.wikipedia.org/wiki/Vector_space]Vector Spaces[/url], which are the central object of study in linear algebra. We can have not only vectors in $\mathbb R^{2}$, but also in any $\mathbb R^{n}$.

[url=http://en.wikipedia.org/wiki/Complex_number]Complex numbers[/url] on the other hand form an algebraically closed field. That, among others, means that the very important operation $\cdot$ is defined on $\mathbb C$ (it's also defined for vectors, but it means something else, scalar product). We can't "construct" anything similar to them in "3D", but we can do it in 4D: [url=http://en.wikipedia.org/wiki/Quaternion]Quaternions[/url] (they are almost a field, but not quite).[/quote]
This makes a whole lot of sense.

[quote="Virgil Nicula"]$\mathrm{Mathematics\ are\ a\ monstrous\ tautology\ .}$[/quote]
What does this quote exactly mean?

[quote="Virgil Nicula"]$(x,y)\in \mathbb R^{2}$ $\Longleftrightarrow$ $P(x,y)\in \mathbb P$ (plane) $\Longleftrightarrow$ $z=a+b\cdot i\in\mathbb C$ $\Longleftrightarrow$ $\overrightarrow{OP}\in\mathbb V$ $\Longleftrightarrow$ $\ldots$ isomorphical structures ![/quote]
What's isomorphical structures?

The one bit of structure that $\mathbb{C}$ has which is different from the structure of the vector space $\mathbb{R}^{2}$ is the multiplication of complex numbers.

Under that multiplication, $\mathbb{C}$ is a field. That multiplication has geometric meaning not seen in just looking at vectors. (To multiply complex numbers, multiply the absolute values and add the arguments.)

All of the "magic" that makes complex analysis work is bound up in that complex multiplication.

There are many more smooth functions from $\mathbb{R}^{2}$ (or some open subset) to $\mathbb{R}^{2}$ than there are holomorphic functions from $\mathbb{C}$ (or some open subset) to $\mathbb{C}.$ The distinction: the former act locally like some linear mapping, but the latter, much more specifically, act locally like multiplication by a complex number.

[Yes, I'm basically just repeating what perfect_radio said.]

I seems from this discussion that the sets themselves isn't what distinguishes complex numbers from vectors but it is also their properties. Because initially I thought that 2-D vectors and complex number should be the same thing since they are defined as an ordered pair of real numbers (a,b). Any comments on this?

Yes, the operation of multiplication on compex numbers, distinguishes it from vectors in R2.


algebra polynomial function LaTeX calculus algorithm

How do I graph the two complex polynomial functions below WITHOUT using a graphing calculator?

(1) f(x) = x^4 + 3.45x^3 - 11.6639x^2 - 5.864241x - 0.69257738

(2) f(x) = -2x^4 - pi(x^2) + [sqrt{3}](x) + 2

[quote="13375P34K43V312"]Please use $\LaTeX$. It's really annoying to read all through the text. Also you've been here long enough, you should definitely be using it by now.

There's no easy way to graph these other than plugging in values of $x$ and manually calculating values of $y$. There's really no point of these problems, they don't seem to have any interesting results.

And the title "Complex Polynomial Functions" is kinda misleading, Complex generally implies complex numbers.[/quote]

Yeah, and these things are just normal cubics, so stop posting these types of questions please.

Latex is cool.  Try to use it, it is only a couple of dollar signs and braces.

[quote="mathgeniuse^ln(x)"][quote="13375P34K43V312"]Please use $\text{\LaTeX}$. It's really annoying to read all through the text. Also you've been here long enough, you should definitely be using it by now.

There's no easy way to graph these other than plugging in values of $x$ and manually calculating values of $y$. There's really no point of these problems, they don't seem to have any interesting results.

And the title "Complex Polynomial Functions" is kinda misleading, Complex generally implies complex numbers.[/quote]

Yeah, and these things are just normal cubics, so stop posting these types of questions please.

Latex is cool.  Try to use it, it is only a couple of dollar signs and braces.[/quote]

Well you should let people post whatever they want, especially if they are problems they think are interesting or challenging... you have no right to stop that :( perhaps moving forums might benefit the problems?

[quote="Interval"]How do I graph the two complex polynomial functions below WITHOUT using a graphing calculator?

\begin{eqnarray}f(x) &=& x^{4}+3.45x^{3}-11.6639x^{2}-5.864241x-0.69257738 \\ f(x) &=&-2x^{4}-\pi(x^{2})+\sqrt{3}(x)+2 \end{eqnarray}[/quote]

There, latex happy now.

Okay that we go the LATEX stuff out of the way, how do I graph such crazy functions?

Plug in values for x and plot one point at a time. You can use some calculus to find points of inflection or extrema. Make a sketch combining all of these elements.

I am not up to calculus yet.  How do I graph such polynomial functions without calculus and without a calculator?

It's been said several times, plot points, and connect them with a curve.

Are you coming up with these questions?

Also, for a graph by hand you can replace the coefficients of large powers with integers, and such.

Also, ignore all but the x term and constant term when graphing values with small magnitude.

Other than that I don't know why you would want to do this since it is not such a crazy function after all (if you want a crazy function, google "Busy Beaver Algorithm")


coloured balls are distributed in 3 bags as shown below .

BAG 1  :  White = 1 , Black = 2  , Red = 3
BAG 2  :  White = 2 , Black = 4  , Red = 1
BAG 3  :  White = 4 , Black = 5  , Red = 3

A bag is selected at random and then 2 balls are randomly drawn from the bag from the selected bag.They happen to be a red and a white.what is the probability that they came from BAG 2 .

please give the solution .

[hide]i get the answer as $ \frac{12}{61}$ [/hide]

I got something else:

by Baye's the probability is $ \frac{\frac{1}{3}\cdot\frac{1\cdot 2}{\binom{7}{2}}}{\frac {1}{3}\cdot \frac{1\cdot 2}{\binom{7}{2}}\plus{}\frac{1}{3}\cdot \frac{1\cdot 3}{\binom{6}{2}}\plus{}\frac{1}{3}\cdot \frac{4\cdot 3}{\binom{12}{2}}}\equal{}\frac{110}{1496}\equal{}\frac{5}{68}$

I agree with the LHS of pluristiq's computation, but I think he did the arithmetic wrong to get to the 2nd expression.

Unrelatedly, the person after whom this theorem is named is [url=http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Bayes.html]Thomas Bayes[/url], making it Bayes' theorem, not Baye's.

I thought the name looked weird when I typed it, heh


Given triangle $ ABC$ with Fermat point $ F$ prove that
\[ FA+FB+FC\le\frac{\sqrt{3}}{3}(a+b+c) \]

\[ FAFB+FBFC+FCFA\le\frac{1}{3}(ab+bc+ca) \]

\[ FA^{2}+FB^{2}+FC^{2}\ge\frac{1}{3}(a^{2}+b^{2}+c^{2}) \]
(well know)
Conjecture \[ (FA+FB+FC)^{2}\le ab+bc+ca\]

[quote="gemath"]Given triangle $ ABC$ with Fermat point $ F$ prove that
\[ FA+FB+FC\le\frac{\sqrt{3}}{3}(a+b+c) \]
[/quote]
Let $ FA=x,$ $ FB=y$ and $ FC=z$ and let $ \max\{\alpha,\beta,\gamma\}\leq120^{\circ}.$
If $ \alpha>120^{\circ}$ then $ F\equiv A$ and  the rest is easy.
Hence, it remains to prove that $ \sum_{cyc}\sqrt{x^{2}+xy+y^{2}}\geq\sqrt3(x+y+z).$
But $ \sum_{cyc}\sqrt{x^{2}+xy+y^{2}}\geq\sqrt3(x+y+z)\Leftrightarrow$
$ \Leftrightarrow2\cdot\sum_{cyc}\sqrt{(x^{2}+xy+y^{2})(x^{2}+xz+z^{2})}\geq\sum_{cyc}(x^{2}+5xy).$
We obtain $ (x^{2}+xy+y^{2})(x^{2}+xz+z^{2})-(x^{2}+\frac{x(y+z)}{2}+yz)^{2}=\frac{3x^{2}(y-z)^{2}}{4}\geq0.$
Thus, it remains to prove that $ 2\cdot\sum_{cyc}(x^{2}+\frac{x(y+z)}{2}+yz)\geq\sum_{cyc}(x^{2}+5xy),$
Which equivalent to $ \sum_{cyc}(x^{2}-xy)\geq0.$  :)
[quote="gemath"]
Conjecture
\[ (FA+FB+FC)^{2}\le ab+bc+ca \]
[/quote]
$ (FA+FB+FC)^{2}\le ab+bc+ca\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt{(x^{2}+xy+y^{2})(x^{2}+xz+z^{2})}\geq(x+y+z)^{2}.$
But $ \sum_{cyc}\sqrt{(x^{2}+xy+y^{2})(x^{2}+xz+z^{2})}\geq\sum_{cyc}(x^{2}+\frac{x(y+z)}{2}+yz)=(x+y+z)^{2}.$ :)


algebra
polynomial
greatest common divisor
algebra unsolved

Suppose polynomial of an odd degree $ n$ has rational coefficients and has exactly $ n$ real roots,that form geometric progression.
Prove that at least one of the roots is rational number.
Note:I think,that this problem is quite hard,as it was solved by only few students...

Let $ n\equal{}2k\plus{}1$.
Let $ a$ and $ q$ be real numbers such that the roots are $ a,aq,aq^2,\ldots,aq^{2k}$.

Consider a $ k$-element subset $ S$ of $ X\equal{}\{0,1,\ldots,2k\}$.
Let $ T$ denote the $ (k\plus{}1)$-element subset $ X\minus{}S$.
Let $ U$ denote the $ (k\plus{}1)$-element subset of $ X$ that contains $ x$ iff $ T$ contains $ 2k\minus{}x$.
By mapping $ S$ into $ U$, we define a bijection between the $ k$-element subsets and the $ (k\plus{}1)$-element subsets of $ X$.

Consider the sum of elements in $ S.T,U$:
If the sum of all elements in $ S$ equals $ s$, then the sum $ t$ of elements in $ T$ equals $ t\equal{}k(2k\plus{}1)\minus{}s$.
And the sum of elements in $ U$ is $ 2k(k\plus{}1)\minus{}t\equal{}k\plus{}s$.

Now let us return to the polynomial and its roots.
The coefficient $ c$ of $ x^{n\minus{}k}$ is the sum of the products of all $ k$-tuples of roots.
The coefficient $ d$ of $ x^{n\minus{}k\minus{}1}$ is the sum of the products of all $ (k\plus{}1)$-tuples of roots.
The above bijection shows that $ c\equal{}a^k*\sum q^i$ and $ d\equal{}a^{k\plus{}1}*\sum q^{i\plus{}k}$,
where in both sums the summation goes over the same indices.

If $ c\ne0$ then $ d/c\equal{}aq^k$ is rational, and the root $ aq^k$ is rational.
If $ c\equal{}0$, we have to do something else.

What I did above for $ k$-tuples and $ (k \plus{} 1)$-tuples can also be done generally for $ (k \plus{} 1 \minus{} \ell)$-tuples and $ (k \plus{} \ell)$-tuples.
Here $ \ell$ runs from 1 to $ k$.
If the corresponding coefficients of $ x^{k \plus{} \ell}$ and $ x^{k \plus{} 1 \minus{} \ell}$ are non-zero, we get that $ a^{2\ell \minus{} 1}q^{k(2\ell \minus{} 1)} \equal{} (aq^k)^{2\ell \minus{} 1}$ is rational.

Furthermore the constant term is $ (aq^k)^{2k\plus{}1}$ and is rational.

Now let $ L$ denote the set of all $ \ell$ for which these coefficients are non-zero.
To account for the constant term,  add $ k\plus{}1$ to $ L$.
Consider the greatest common divisor $ t$ of all numbers $ 2\ell\minus{}1$ with $ \ell\in L$.
If $ t\equal{}1$, then there exists a linear combination of the $ 2\ell\minus{}1$ with integer coefficients and value~1. This linear combination yields that $ (aq^k)^1$ is rational. Done.

If $ t>1$,  then consider the corresponding polynomial.
It can be shown that the polynomial is of the form $ Q(x^m)$ where $ Q$ has rational coefficients and where $ m>1$.
If $ m\ge3$ this means that the polynomial also has non-real roots. Contradiction.
Also the case $ m\equal{}2$ leads to a contradiction.


modular arithmetic
pigeonhole principle
number theory proposed
number theory

Demonstrate that there are no three consecutive odd integers such that each is the sum of two squares greater than zero.

Maybe i misunderstood the problem but...

[hide="click to reveal"]The sum of two squares, say $h^{2}+k^{2}\equiv 0,1,2\pmod 4$. On the other hand, for the pigeonhole principle, at least one among three consecutive odd integers must be $\equiv 3\pmod 4$. Thus we have finished.  :wink: [/hide]


analytic geometry
geometry unsolved
geometry

Let $ A$ and $ B$ be fixed distinct points on the $ X$ axis, none of which coincides with the origin $ O(0, 0),$ and let $ C$ be a point on the $ Y$ axis of an orthogonal Cartesian coordinate system. Let $ g$ be a line through the origin $ O(0, 0)$ and perpendicular to the line $ AC.$ Find the locus of the point of intersection of the lines $ g$ and $ BC$ if $ C$ varies along the $ Y$ axis. Give an equation and a description of the locus.

it's very simple
 [geogebra]595702fad33f0b0f39df574de6fee4fdff8ec65f[/geogebra]
Take Any two points (a,o) and (b,0) on x axis..(fixed)
take any pint (0,k) on y axis (variable)
y=mx
x+my=$
$=a
or $=a/k
x+ ay/k =a ..1
y= ax/k ...2
x +yb/k -b=0 ...3
eliminating k from 2 and 3 we get the locus of the point of intersection


analytic geometry
conics
parabola
graphing lines
slope
trigonometry
number theory

All three vertices of an equilateral triangle are on the parabola $ y \equal{} x^2$, and one of its sides has a slope of 2. The x-coordinates of the three vertices have a sum of $ m/n$, where $ m$ and $ n$ are relatively prime positive integers. What is the value of $ m \plus{} n$?

$ \textbf{(A)}\ 14\qquad
\textbf{(B)}\ 15\qquad
\textbf{(C)}\ 16\qquad
\textbf{(D)}\ 17\qquad
\textbf{(E)}\ 18$

This problem took me too long.  :(

Okay, the basis of this problem is these two facts:

1) If two lines with slopes $m_1$ and $m_2$ cross each other at an angle $\theta$, then $\tan(\theta) = \frac{m_1 - m_2}{1+m_1m_2}$

2) The slope between two points $(x_1,x_1^2)$ and $(x_2, x_2^2)$ is $x_1+x_2$

[quote="pkothari13"]1) If two lines with slopes $m_1$ and $m_2$ cross each other at an angle $\theta$, then $\tan(\theta) = \frac{m_1 - m_2}{1+m_1m_2}$[/quote]Really?  Cool, I never knew that.  Can someone post a proof?

I had to beat this question to death, but I eventually got it. Please, can someone provide a much prettier solution? :)

Let's call the coordinates of vertices $A, B, C$ as $(x_A,y_A),(x_B,y_B),(x_C,y_C)$.

Now, we can say $m_{AC}=\frac{y_C-y_A}{x_C-x_A}=2$. Since these points are both on $y=x^2$, then we can substitute and get \[ \frac{x_C^2-x_A^2}{x_C-x_A}=2 \] \[{ \frac{(x_C-x_A)(x_C+x_A)}{x_C-x_A}=2} \] \[ x_C+x_A=2 \] Note that we can use this same result to say that, in general, the slope between two points will be equal to $x_1+x_2$.

Now, we know that when two lines with slopes $m_1$ and $m_2$ cross each other at an angle $\theta$, then $\tan{\theta}=\frac{m_1-m_2}{1+m_1m_2}$ This equation will give us the slopes of the other two lines depending on whether we sub in $m_1=2$ or $m_2=2$. The two results are $m=\frac{\sqrt{3}+2}{1-2\sqrt3}$ and $m=\frac{2-\sqrt3}{1+2\sqrt3}$ Thanks to the earlier result, we can now say $(x_A+x_B)+(x_B+x_C)+(x_A+x_C)=2\frac{m}{n}$. Subbing in our previous results, \[ 2+\frac{\sqrt{3}+2}{1-2\sqrt3}+\frac{2-\sqrt3}{1+2\sqrt3}=2\frac{m}{n} \] \[ \frac{6}{11}=2\frac{m}{n} \] \[ \frac{m}{n}=\frac{3}{11} \] \[ m+n=11+3=14 \] which is $A$!

So ugly :( Anyone have anything better?

[quote="1234567890"]Well, slope is just arctan, isn't it? So it follows.[/quote]

Close...slope is tan.

\[ \text{slope}=\frac{\text{rise}}{\text{run}}=\frac{\Delta y}{\Delta x}=\tan \theta  \]

$\theta$ is the angle the line makes with a horizontal.

Edit: Whoops...yeah $\Delta$...not $\delta$ :P

Hmm you mean $\Delta$ instead of $\delta$, right?

[quote="eryaman"]Really?  Cool, I never knew that.  Can someone post a proof?[/quote]

It's angle subtraction. Take it from the horizontal, and you'll see. $\theta =m_1-m_2$, where $m_1$ is the upper line. All you need to do is plug those in to find the other slopes, then use the other point which pkothari listed.

[hide="Answer"]Bottom side of the triangle has slope: $\tan 60=\sqrt{3}=\frac{2-m_2}{1+2m_2}\Rightarrow m_2=\frac{5\sqrt{3}-8}{11}$. Third side of the triangle has slope: $\tan 60=\sqrt{3}=\frac{m_1-2}{1+2m_1}\Rightarrow m_1=\frac{-(5\sqrt{3}+8)}{11}$. Add together the three slopes and divide by two: $\frac{22+(5\sqrt{3}-8)-(5\sqrt{3}+8)}{22}=\frac{3}{11}$. This is also the sum of the $x$-coordinates, so our answer is $3+11=14\Rightarrow \boxed{A}$.[/hide]

[quote="chess64"][quote="JesusFreak197"]This is also the sum of the $x$-coordinates...[/quote]

Why? :?[/quote]

Because slope is $\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}$, but since we're talking about the parabola $y=x^2$ then slope is $\frac{x_2^2-x_1^2}{x_2-x_1}=\frac{(x_2-x_1)(x_2+x_1)}{x_2-x_1}=x_1+x_2$

Oh, I remember--I did that when I attempted the problem, but I forgot about it now :blush:


Is there anyone who knows how to download Mildorf's <Olympiad Inequalities> and < Topics in Inequalities>,Hojoo Lee(freeeeeeeeee~)
Thx                                       :?:

Well, you could do a little search on this site....
Anyway, here's Mildorf's page ( http://web.mit.edu/~tmildorf/www/ ), and see also this topic http://www.mathlinks.ro/viewtopic.php?t=88087 :wink:


quadratics
function
algebra unsolved
algebra

[quote="shobber"]Given a function $f(x)=x^{2}+ax+b$ with $a,b \in R$, if there exists a real number $m$ such that $\left| f(m) \right| \leq \frac{1}{4}$ and $\left| f(m+1) \right| \leq \frac{1}{4}$, then find the maximum and minimum of the value of $\Delta=a^{2}-4b$.[/quote]
$x^{2}+ax+b=\frac{(2x+a)^{2}-\Delta }{4}$. Let y=2x+a-1, then $|(y+1)^{2}-\Delta |\le 1, |(y-1)^{2}-\Delta |\le 1$. It give $0\le \Delta \le 2$.


1   if f(x) and g(x) are differentiable function at x=a then the function defined as h(x)=maximum of {f(x),g(x)}

(A)  IS ALWAYS differentiable at x=a

(B )  IS NEVER  differentiable  x=a

(C  ) is differentiable at x==a    provided f(a) not equal to g(a)

Q2  let h(x)  =1/q if x=p/q

                     0  if   x is irrational

then h(x)    is not discontinuous for all x in (0,infinity)

how
pl give a example regarding first question

For the first part consider $ f(x) \equal{} x \equal{} g(x)$ and $ f(x) \equal{} 2x, g(x) \equal{} x$ at $ x\equal{}0$. For your second question see the appendix [url=http://math.feld.cvut.cz/mt/txtb/4/txe3ba4s.htm]here[/url].


Problem 1

MNG2,3,4,6 get 7point, MNG1 gets 6point, MNG5 gets 5 point.

problem 4

MNG1  7 point, MNG2  7 point, MNG3  7 point, MNG5  3 point, MNG6  4 point

problem 5

MNG1  7 point, MNG3  7 point

problem 3

MNG3  1 point

PS. Oyuundariin hoyrdugaar bodlogyn bodolt buren baina. Margaash ogloo hoyor, zurgaadugaar bodlogyg hamgaalna.

[quote="uglysolutions"]baysa, have you seen another teams' scores? If so could you post them please?

Thank you very much and congrats to your team! They seem to have done very well  :)

P.S. I'm especially interestead in Latin American teams' scores[/quote]

[color=green]
Me too. Ask for the Argentinian results  :D[/color]

Problem2

MNG3 and MNG4 get 7 points, MNG1 gets 5 point, MNG2 gets 4 point, MNG5 gets 1 point.

PS. Temuge kvadrat tegshitegliinhee ahmad gishuunii omnoh koeffitsientiig martsan baina, zalruulaad bodohod shine shiid ogt ogohgui baina. Oyuundari alfa betagiin bairyg soliod bodson baina. Odoo zurgaadugaar bodlogo uldsen, tuun deer ih hatuu zasalt hiij baigaa tul manaihan barag onoo avahgui baih (avah gej oroldoh bolovch). Ingeed niileed zuun zurgaan onootoi bolloo. b.b.

Congratulations.

MNG1 and MNG3 get silver, MNG2 gets bronze.

PS. Byambajav, Oyuundari, Temuge nartaa bayarlalaa. Enkhzaya, Zorigoo hoyor ireh jil altan medali avna shuu. Oroi uulzaya. B.B.


All the best to all AOPS'ers, writing IIT-JEE 08. There's a huge list, so difficult to list all names here.

But

[b]ALL THE BEST!!!!!!!![/b] Come out with flying colours and lets all get the group of our choice. :)

Be calm be focussed. Don't strain yourself on the last day. Relax.... and dont get tensed looking at the paper.

yes yes....all the best to all the twelthies here :)

[quote="Chronoz"]Be calm be focussed. Don't strain yourself on the last day. Relax.... and dont get tensed looking at the paper.[/quote]

Thank you grandpa....
aapka naam jaroor roshan karunga....

[quote="Chapplikka"]Guys, Mythili and Shruthi :rotfl: ,
Wish you all the best for JEE. Wish all of us come out with flying colours..[/quote]

You forgot Soumya.... :huh: Never mind, she doesn't mind...

And I will simply come out flying from the exam hall without any colour.....

Anyway, best of luck to all the aspirants....(and non-aspirants too)

[quote="madness"]:huh: How come chappli forgot her.... erm ....... his ....... hes std. punch dialogue? :D

Anyway all the best to everyone here and everyone not here too. Hope that we have some top 10 rankers in this forum (ofc soumya is air 1, but apart from that) :D[/quote]

Ya thanks for acknowledging the inevitable... :rotfl:  :rotfl:  :P

[hide]Honestly, anything better than AIR 500 is ok 4 me...
[/hide]

[quote="SOUMYASHANT NAYAK"][hide]Honestly, anything better than AIR 500 is ok 4 me...
[/hide][/quote]

Hey, actually getting in a top 500 rank without much preparation is awesome you know :)

I would give anything to get that :)

hey ppl ALL THE BEST to everyone present here hope you all manage to clear the entrance in flying colours coming out with the satisfaction that you have succeeded in what you all have aimed at!!!!! amigo!!! esapanio meia cheers dio!!! :thumbup:  :coolspeak:

[quote="Albert Einstien"]amigo!!! esapanio meia cheers dio!!! [/quote]

Hmm... I can make that controversial if I want to :D People who understand that should be terrified of me now and some others might wanna help too :D :rotfl:

Yes, hope to see a lot of Top Rankers this time from AOPS. We are the cool nerds. :)

Yeah the Best of wishes from the "Rags".
 :D

[quote="hash_include"]All the best to everyone at AoPS (and those not) for the big day on 13th! :)
Hope we all get awesome ranks, or even better, i hope all of us can get converted to OBC's  :D[/quote]

Do you want to be prosecuted?? You are not supposed to call them OBC's. Hereafter by the order of Supreme court, they shall hereby be referred to only as SEBC - Socially and Economically backward Classes. Seriously.

super
best of luck
AOPS rox!!!!!!! :D  :D  :D

all the best to all here!!
i m sure everyone's gonna explode the exam paper like anything (thats a term used by my chemistry sir, paper phodke aana!!)

we will see single digit, double digit and 3 digit ranks!!!

[quote="phoenix76"]best of luck to jee for all
i know a lot of top 10 rankers will be from aops
they dont need any luck :rotfl:  :D[/quote]

There is actually a strict limit on the no. of top 10 rankers possible..... :rotfl:

All the Best  To all those who are attempting the JEE.
I hope every one in AOPS would come out with flying colors and rattle the rank list.

[quote="SOUMYASHANT NAYAK"]
There is actually a strict limit on the no. of top 10 rankers possible..... :rotfl:[/quote]
i don't know about jee but at other places(eg-GOLF) there is no limit of Top 10 :roll:

u see i mean hundreds can b T9 or even more..... :rotfl:  :rotfl:

[quote="Ram"]All The BeSt to all!![b]esp to tamilians,chennaities and smartans!!!![/b] :)[/quote]

This site is promoting regionalism :D  :P

BEST OF LUCK. I guess i will be the last one to wish :D

nope!
u r wrong my friend!!
 there's no one in this world who is more vella (slang for no gooder) than me!!
so i will be the last one to wish everyone luck!!!

i m --still about 20 mins to go so :roll:
BEST of LUCK :P  :rotfl:

[quote="Ram"]All The BeSt to all!!esp to tamilians,chennaities and smartans!!!! :)[/quote]

What do you mean by smartans?

[quote="madness"]The paper started at 9:00 am :P :rotfl:[/quote] :oops_sign:   :!:

[quote="pardesi"][quote="Euclidean Geometer"][quote="madness"]Iyers are called smartans :)[/quote]

was neone flattered by dat ?
i am an iyer also. who else here is ?
And in this code, what are iyengars gonna be called :rotfl:[/quote]
[hide="Chumma JUJUBI :P"]
Smartangars......but hey we do deserve that ....don't we :P  :P [/hide][/quote]

Sure sure da...  :rotfl:

[quote="anandghegde"][quote="Ram"]All The BeSt to all!![b]esp to tamilians,chennaities and smartans!!!![/b] :)[/quote]

This site is promoting regionalism :D [/quote]

None read that  :o  :D

[quote="Euclidean Geometer"][quote="Chronoz"][quote="pardesi"]
Smartangars......but hey we do deserve that ....don't we  [/hide][/quote]

 :rotfl:  :rotfl:

I am an Indian![/quote]

well all of us are da...
since u r kannadiga, u probably are unaware...
tamil of us are of two types: iyers and iyengars.
tthe former apparently followers of shiva and the latter of Vishnu or hari.[/quote]

Thank you very much. I've been in chennai all my life and know about the culture.. maybe much better than you especially considering my school.

[quote="Chronoz"][/quote][quote="Euclidean Geometer"][quote="Chronoz"][quote="pardesi"]
Smartangars......but hey we do deserve that ....don't we  [/hide][/quote]

 :rotfl:  :rotfl:

I am an Indian![/quote]

well all of us are da...
since u r kannadiga, u probably are unaware...
tamil of us are of two types: iyers and iyengars.
tthe former apparently followers of shiva and the latter of Vishnu or hari.[/quote][quote="Chronoz"]

Thank you very much. I've been in chennai all my life and know about the culture.. maybe much better than you especially considering my school.[/quote]

oh thats nice to hear...  :D
i ve lived in madras for sm 7 yrs.
bangalore since then.
and not dat i am totally ignorant.
dats not really possible considering my dad s from madurai and mum from kumbakonam and i ve a strong kumbakonam influence.

@Euclidean Geometer

Hey are you in Bangalore now? :)

yup anand, raghav is also from base and he's bsr's pet, and since u seem to be taking more in co geo too, you two will probably get along well. :D

Maybe they should be called the Co-Geo Brothers.  :rotfl:  :rotfl:

Varun, do all of you guys goto the same IIT Class?

yes we al attend d same institution da base by name
der's none other's  good enough 4 d requirement
but regards thier brotherhood (no offence intended)  frm above posts i find one's completely against regionalism :D  while d other is d embodiment of regionalism or rather religionalism :rotfl:


What is the difference between the Intel Science Talent Search (STS) and the Intel International Science & Engineering Fair (ISEF). The winners of both contests seem to be the same exact people, and they did the same exact research for each one.

They are different contests, but they both test roughly the same thing (student research).  They are quite different in structure, however, and you can read all about them by going to [url]http://www.sciserv.org[/url].

Can someone give me a brief summary of how you get started in these science fairs, specifically the Intel ones. The Intel website, compared to the Siemens Westinghouse website, is messy and has terrible organization, and I am having a hard time finding out information like due dates and the competitions process.

[quote="mathnovice"]Can you do a research project and submit it to several different competitions? For example, ISEF, STS, Siemens Westinghouse Competition, a state science fair, etc.[/quote]

Yes.

For your other question, the application date to Intel will be on the website somewhere, presumably in the [url=http://www.sciserv.org/sts/students/]Students[/url] section.  For the process of getting started, we've done a whole long conversation on this before -- try doing a search for it.


$ Question 1$
solve for integer values of $ r$, $ s$, $ t$, if
$ r^2 \plus{} s^2 \plus{} t^2 \equal{} 377$
$ r*s \plus{} s*t \plus{} t*r \equal{} 356$
$ r*s*t \equal{} 1248$

$ Question 2$
You are given 3 integers, $ x$,$ y$,$ z$ with $ x > y > z > 0$
You are also given that the 6 numbers, $ x \plus{} y$, $ x \minus{} y$, $ y \plus{} z$, $ y \minus{} z$, $ x \plus{} z$ and $ x \minus{} z$ are perfect squares.
Can you find $ x$, $ y$ and $ z$?

[hide="Question1"]$ (r \plus{} s \plus{} t)^2 \equal{} r^2 \plus{} s^2 \plus{} t^2 \plus{} 2rs \plus{} 2st \plus{} 2tr$

$ (r \plus{} s \plus{} t)^2 \equal{} 377 \plus{} 2(356) \implies r \plus{} s \plus{} t \equal{} 33$

thus $ r \plus{} s \plus{} t \equal{} 33$
$ rs \plus{} st \plus{} tr \equal{} 356$
$ rst \equal{} 1248$

then we can say that $ r, s, t$ are three roots of equation $ x^3 \minus{} 33x^2 \plus{} 356x \minus{} 1248 \equal{} 0$

solving for x, we have $ x \equal{} 8,12,13$

answer : $ (r,s,t) \equal{} (8,12,13)$ or its arrangements

[/hide]


if $ a < \minus{}1$ then $ 1 < \minus{}a$ then $ 1 < \minus{}a^9$ then
$ \frac{1}{a^3} < \minus{}a^3$ then $ a^3 < \minus{}a^{\minus{}3} < a^{\minus{}3}$

That's just total nonsense: in particular, the last (three-part) inequality you've claimed is obviously false.  (The middle term is positive, while the two outer terms are negative.)

If $ a < \minus{}1$ then $ a^3 \in (\minus{} \infty, \minus{}1)$ and $ a^{\minus{}3} \in (\minus{}1, 0)$.

[quote="Miccoh"]If $ a < - 1$ prove that $ a^3 < a^ - ^3$.[/quote]
I think: $ a^3 < a^ - ^3 \Leftrightarrow a^3 < \frac{1}{a^3} \Leftrightarrow a^6 <1$
true for all a<1!

[quote="NguyenDungTN"]I think: $ a^3 < a^ - ^3 \Leftrightarrow a^3 < \frac {1}{a^3} \Leftrightarrow a^6 < 1$
true for all a<1![/quote]
You're wrong.  $ - 2 < - 1$.  $ ( - 2)^6 = 64 > 1$.

[quote="1=2"]$ c < - 1$
$ b > - 1$
$ c > b$[/quote]
$ c < - 1 < b \Longrightarrow c < b$.  Also, if $ a = - 2 < - 1$ then $ b = ( - 2)^3 = - 8 < - 1$ and $ c = ( - 2)^{ - 3} = - \frac {1}{8} > - 1$.  You did three things wrong and they happened to cancel out.

[quote="JBL"]You did three things wrong and they happened to cancel out.[/quote]

I did TWO things wrong: c is greater than -1, and b is less than -1. That should fix my proof.

Sketching graphs of $ y\equal{}a^3$ and $ a\equal{}a^\minus{}3\ (a<\minus{}1)$ gives the proof.

Seeing is believing!

[quote="1=2"]I did TWO things wrong: c is greater than -1, and b is less than -1. That should fix my proof.[/quote]
No, you got two inequalities wrong ($ c < \minus{}1, b > \minus{}1$) and then you combined them incorrectly.

The errors in your solution (and those in the other two completely wrong solutions on this thread) could have been avoided by the substitution of [i]explanation[/i] for [i]assertion[/i] (and this is a principle that extends far beyond mathematics).

i dont quite understand. If we know something is true because they're asking us to prove it, why dont we just prove that the opposite is false. Say a>-1 like a=0. Then the inequality is false. ??

Just because the inequality is false when the problem condition is false does not mean that the inequality is true when the problem condition is true.  By your logic, we could prove that if $ a < \minus{}1$ then $ a^2 < 0$.

Hello, since $ a<\minus{}1$, $ a\ne 0$ obviously, so we can multiply both sides by $ a^6>0$...so

$ a^9<a^3\iff a^3(1\minus{}a)(a^2\plus{}a\plus{}1)(1\plus{}a)(a^2\minus{}a\plus{}1)>0$...then the result follows... since $ a^3<0$, $ 1\minus{}a>0$, $ a^2\plus{}a\plus{}1>0$, $ 1\plus{}a<0$, and $ a^2\minus{}a\plus{}1>0$.

Am I missing something?

We know that $ a^6 \equal{} (a^2)^3 > (( \minus{} 1)^2)^3 \implies a^6> 1$ and that $ a<0 \implies a^3 < 0$.

To divide by a negative, flip the sign: $ \frac {a^6}{a^3} < \frac {1}{a^3} \iff a^3 < a^{ \minus{} 3}$

If $ a<\minus{}1$ then $ a$ is negative.

\[ a^6>1\] for example, $ \minus{}2<\minus{}1 \Rightarrow  2^6>1$.

Then \[ a^6\equal{}a^3\cdot a^3>1 \, /: a^3\] since $ a$ is negative:

\[ a^3<\frac{1}{a^3}\equal{}a^{\minus{}3}\]


Suppose x and  y are positive real numbers such that x+2y=1.  prove that :

 1/x + 2/y>= 25/1+48xy^2

This isn't very pretty but it works:  Substitute $x=1-2y$ into the inequality and clear fractions to get the equivalent inequality
\[2(144y^{4}-168y^{3}+73y^{2}-14y+1) \geq 0.\]   But this factors as
\[2(1-3y)^{2}(1-4y)^{2}\geq 0,\]
which is clearly true.

I am trying to understand where your interesting inequality came from. So I did:
\[\frac1{x}+\frac1{y}+\frac1{y}\geq \frac{9}{x+2y}=9 \]
Now, \[x+2y\geq 3\sqrt[3]{xy^{2}}\implies 1\geq 27xy^{2}\implies 9\cdot 48xy^{2}\leq 16\iff \]
\[\iff 9(1+48xy^{2})\leq 25\iff 9\leq \frac{25}{1+48xy^{2}}\]
Thus, your inequality is sandwiched between the Cauchy
\[\frac1{x}+\frac1{y}+\frac1{y}\geq \frac{25}{1+48xy^{2}}\geq 9 \]

Ok here it is

\[\frac1{x}+\frac1{y}+\frac1{y}=\frac1{xy}\]

We have, inequality $\iff$ \[1+48xy^{2}\geq 25xy\iff 1\geq 24x^{2}y+xy\]

We treat case $x\geq y$:
$\iff 1\geq 3x\iff 9xy\geq 24x^{2}y+xy$
but, $1\geq 9xy\implies 1\geq 9xy\geq 24x^{2}y+xy=(25-48)xy\iff$ the inequality

The other case is left open :!:

[quote="me@home"]Ok here it is
\[\frac1{x}+\frac1{y}+\frac1{y}=\frac1{xy}\]
[/quote]

That isn't true...

$\frac{1}{x}+\frac{2}{y}=\frac{y+2x}{xy}$


$ x,y,z > 0$ then prove :

$ (x\plus{}y\plus{}z)^2(xy\plus{}yz\plus{}zx)^2 \le 3 (x^2\plus{}y^2\plus{}xy)(y^2\plus{}z^2\plus{}yz)(z^2\plus{}x^2\plus{}zx)$

[quote="FOURRIER"]$ x,y,z > 0$ then prove :

$ (x \plus{} y \plus{} z)^2(xy \plus{} yz \plus{} zx)^2 \le 3 (x^2 \plus{} y^2 \plus{} xy)(y^2 \plus{} z^2 \plus{} yz)(z^2 \plus{} x^2 \plus{} zx)$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194516


Prove:
$ 3m^2n^2p^2\plus{}m^4p^2\plus{}n^4m^2\plus{}p^4n^2\geq2(m^3n^2p\plus{}mn^3p^2\plus{}m^2np^3)$

I was proving inequality when I came to this and I don't know how to prove it.

let $ m^2p \equal{} a,p^2n \equal{} b,n^2m \equal{} c$ then your inequality is equivalent to:

$ 3\sqrt[3]{\left(abc\right)^2} \plus{} a^2 \plus{} b^2 \plus{} c^2\geq 2\left(ab \plus{} bc \plus{} ca\right)$

which is true for $ a,b,c\in\mathbb{R}$,see here:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=18430[/url]


Let $ (x_n)_{x\in Z_{+}}$ be a sequence such as ${ \lim_{n\to\infty}(x_{n+2}-x_n})=l \in R$.Prove that $ \lim_{n\to\infty}\frac{x_{n+1}-x_n}{n}=0$

I'm not too sure about this...
[hide]
As $ n\to \infty$ $ x_n \equal{} mn \plus{} c$ where $ m \equal{} \frac {l}{2}$ is the gradient of a straight line graph and $ c$ is a constant

Therefore as $ n\to \infty, x_{n \plus{} 1} \minus{} x_n \equal{} \frac {l}{2}$

$ \Rightarrow \lim_{n\to \infty} \frac {x_{n \plus{} 1} \minus{} x_n }{n} \equal{} \frac {l}{2n} \equal{} \frac {l}{\infty} \equal{} 0$[/hide]

[hide]Amounts to dividing $ (x_n)$ into two arithmetic subsequences. The worst that could happen is one of them is increasing and the other is decreasing, in which case $ (x_n\plus{}1 \minus{} x_n)$ forms another arithmetic sequence.[/hide]


Hi I just have a general question regarding the technical abilities of physicists in mathematics.  I'm hoping to hear some of the views of members on this board.  Thanks.

We all know that physicists use some very powerful and sophisticated mathematics, but do they really know how those mathematical arguments or results or theorems are actually proven or derived.

In my study of mathematics, my teachers and professors would never allow a student to use any mathematical result without the student knowing how it is proved.  The intention for this restriction is that the student must be able to understand why that result is fact, and the student must be able to defend his/her solution.

I'm wondering if there are physicists out there who use mathematical results without knowing their proofs.

Secondly, are physicists knowledgeable in the mathematics that they use to the level that they can teach those mathematics to a class of math majors? (I ask this question, because we all know that knowing how to do something yourself is completely different than teaching that concept to others).
For example can a physicist who uses tensor calculus or differential geometry from relativity theory be able to teach that stuff to math majors?
Another example, say a physicist who uses concepts from group theory to exploit symmetries, would he/she be knowledgeable to teach a course in abstract algebra to math majors?

Thirdly, is it just a myth that physicists usually lack the mathematical rigor of mathematicians?

If the former question is true (i.e. not a myth), then how about string theorists?  I have heard that the mathematical prowess of string theorists rival even the best mathematicians.  Is that true?

It depends on the physicist.  Those who also have degrees in math are almost certain to be able to understand the math behind it (and many of them do have BA's in math).  Those who just have degrees in physics probably couldn't tell you all of the real advanced theory.

That said, I would not trust a physicist to teach me advanced math.

The few encounters I've had with physicists who are fantastic mathematicians were string theorists. I would definitely trust their ability to teach me advanced things, say, about Calabi-Yau manifolds that lots of just purely math people probably don't study.

Yay for Calabi-Yau. :D

I think it also depends on what you want to study.  Chances are, a string theorist could probably explain some of the differential geometry to you, but I wouldn't trust, say, an astrophysicist, to teach me algebraic number theory.

There are also mathematical-physicists, who are pretty well-versed in both I assume.

[color=green](Broken quote tags fixed by moderator -- preview your posts, folks!)[/color]
[quote="Mitochondria"]We all know that physicists use some very powerful and sophisticated mathematics, but do they really know how those mathematical arguments or results or theorems are actually proven or derived.
...
I'm wondering if there are physicists out there who use mathematical results without knowing their proofs.
[/quote]
It's impossible to answer such a general question.  Some physicists understand some of the mathematical results that they use.  Some physicists do not understand the some of the mathematical results that they use.  Of course, both statements are still true if you replace the word "physicist" by "mathematician," so I'm not saying anything.

[quote="Mitochondria"]
Secondly, are physicists knowledgeable in the mathematics that they use to the level that they can teach those mathematics to a class of math majors?
[/quote]
Again, the question is still too general.  I assume that you're mainly talking about theoretical physicists, so in order to be definitive, let's just say you're talking about string theorists.  Most string theorists could probably teach at least some upper level undergraduate pure math classes, but (a) most wouldn't want to, and (b) they would bring a completely different point of view to the subject.  But you have to realize that you don't necessarily have to really *know* a subject off the top of your head in order to teach it.  Or in other words, professors don't actually know everything although it sometimes appears that way.  (Well, *some* profs do seem to know everything...)

[quote="Mitochondria"]
For example can a physicist who uses tensor calculus or differential geometry from relativity theory be able to teach that stuff to math majors?
Another example, say a physicist who uses concepts from group theory to exploit symmetries, would he/she be knowledgeable to teach a course in abstract algebra to math majors?
[/quote]
Again, yes, but the point of view could be very different.

[quote="Mitochondria"]
Thirdly, is it just a myth that physicists usually lack the mathematical rigor of mathematicians?
[/quote]

No, it's not a myth, because the nature of actual doing physics research requires you to give up some degree of mathematical rigor, because physics is not a purely mathematical theory.  There's a name for physicists who are unwilling to give up any rigor--they're called mathematical physicists (who are, somewhat confusingly, often classified as mathematicians).

[quote="Mitochondria"]
If the former question is true (i.e. not a myth), then how about string theorists?  I have heard that the mathematical prowess of string theorists rival even the best mathematicians.  Is that true?[/quote]

The question doesn't quite make sense, but I'll interpret it to mean something like, "Are string theorists as good at math as mathematicians?"  For the most part, the answer is no.  A typical string theorist does not have the training or inclination to conduct successful mathematical research, just as a typical mathematician does not have the training or inclination to conduct successful string theory research.  However, there are always some who work at the border of the two fields and can pass back and forth.  People say this bit about "the mathematical prowess of string theorists" probably because of Witten, who is the world's most famous string theorist *and* a Fields Medalist.  But there's really only one of him.

A side comment may be in order:  When someone in high school talks about being "good at math" or "mathematical ability," they're often talking about something like "mathematical aptitude" or "innate mathematical ability" (inasmuch as such things exist), but at the research level, the whole idea of mathematical aptitude is irrelevant and nonsensical.  So when I say "good at math" I'm talking about how much mathematical knowledge you have, and how able you are to leverage that knowledge into new mathematical knowledge (i.e. research results).

@HilbertThm90:  Mathematicians most certainly do study Calabi-Yau manifolds.