AMC USA(J)MO USAMO inequalities vector function

I request the users of this forum to post in this thread all the good and tough inequalities that they have come across(till now) which require a suitable application of holder,jensen,minkowski, schur , muirhead etc. (and post their solutions too if they have one)

Again if someone has a sound knowledge of the lagrange multipliers he or she may please give a detailed elementary explanation(as i don't have much knowledge about the lagrange multipliers)

And lastly, someone may please explain the idea of majorization and discuss the majorization inequality and show its application in some typical as well as hard problems.

PS I know it is very annoying to u all as i ask u quite a lot of favour,my friends. but i need ur help for sure and shall be happy to receive it.

"May the stars watch over u all!"
& "May the force be with u"    Thanks in advance. :D

why do you need them NOW?

err...i just know AM-GM, bernoulli, and cauchy.
bernoulli
$ (1\plus{}x)^n \ge 1\plus{}xn$

[quote="abacadaea"]why do you need them NOW?

err...i just know AM-GM, bernoulli, and cauchy.
bernoulli
$ (1 \plus{} x)^n \ge 1 \plus{} xn$[/quote]

thanks for the bernoulli, but do u have some good and tough problems for bernoulli(with solutions of course)?

and i need them urgently as i am preparing for the olympiads in my country and only two weeks left!!

thank u mellowmelon for ur help.

 but can any of u tell me about  majorization ,normalization& Homogenization and Lagrange multipliers

majorization is most probably related to vectors i think.

sorry, no idea about the rest :oops:

[quote="MellowMelon"]Majorization is discussed just before the twelve inequalities are introduced and also in the third solution to the first example problem. Homogenization is discussed in the second example problem. Lagrange multipliers are discussed in the twenty-eighth problem. It's all there.[/quote]

again thank's for the help.....yes it is all there but my concepts and fundamentals are still not clear about the above said topics. I can benefit from a detailed explanation from the mathlinkers....

Mr. Mildorf has mentioned everything alright, but it is appropriate for someone who has prior knowledge of the topics and wants to see some applications.
He has merely mentioned the rules and justified their usage in the respective problems.

But a stupid fella like me needs more elaborate explanations.

I list some of my unanswered queries:

1)What is the majorization inequality and its proof?

2)What the hell is the lagrange multiplier theorem?

3)some detailed explanation about making the degrees of both sides of an inequality same by homogenizing and the "scaling" thing.

4)isn't there any substitute for the so called brute force "dumbassing" method that is laborious multiplication and collection of terms done to preserve sharpness of an inequality?(i hate dumbassing)

5)how does one know what is the sharpness of an inequality that is how sharp an inequality is?

6)proof of muirhead?

7)proof of Maclaurin

waiting for someone's help......very eagerly.... :roll:

PS:i have also tried the article by Hojoo Lee but it didn't help either.

There is a [b]lot[/b] of useful discussion [url=http://www.artofproblemsolving.com/Forum/index.php?f=55]here[/url].  (Karamata is the majorization inequality.)  To answer one of your questions for now, the method of [url=http://en.wikipedia.org/wiki/Lagrange_multiplier]Lagrange multipliers[/url] is a general technique for maximizing a multivariate function given a constraint, but you should never need to use it on an Olympiad.

Ah, specific questions. Much better. Bear in mind I am no inequalities expert (go to the inequalities forums to see the ridiculous amount of theory that's there now), but I do know most of the fundamentals.

[quote]1)What is the majorization inequality and its proof?[/quote]
I know of two inequalities that might be called this: Muirhead and Karamata. I would suggest getting familiar with the applications of AM-GM and Jensen respectively before you start trying to invoke these. As a matter of fact Muirhead and Karamata are basically a repeated application of AM-GM and Jensen.

[quote]2)What the hell is the lagrange multiplier theorem?[/quote]
Basically what allows you to use Lagrange multipliers. My personal recommendation, if you are beginning inequalities, is that you do not learn Lagrange multipliers, especially since you say later you hate dumbassing. Lagrange multipliers are the king of brute force methods, and solutions using them are [i]ridiculously[/i] ugly.

[quote]3)some detailed explanation about making the degrees of both sides of an inequality same by homogenizing and the "scaling" thing.[/quote]
Let's suppose you're given a constraint like $ abc \equal{} 1$ ($ a,b,c > 0$) and you need to prove
$ a^2 \plus{} b^2 \plus{} c^2 \geq \frac1a \plus{} \frac1b \plus{} \frac1c$
The reason most inequalities are homogeneous is the following: if we have say $ f(a,b,c) \geq g(a,b,c)$ and $ f$'s degree is less than $ g$, then by making all of the variables huge, the value of $ g$ becomes much larger than the value of $ f$. Conversely, if the degree of $ f$ is larger, making all the variables small positive values near $ 0$ makes $ g$ become larger than $ f$ again. Neither of these can happen.

However, for constrained inequalities ($ a\plus{}b\plus{}c\equal{}1$ is a constraint in the inequality above), you can't do this thing where you make all of the variables really big or really small. Now the inequalities can have varying degrees. The idea of making them homogeneous is to make the degree of each term the same. In our example above $ a^2 \plus{} b^2 \plus{} c^2 \geq \frac1a \plus{} \frac1b \plus{} \frac1c$, the degree of each term on the right is $ \minus{}1$, and on the left it is $ 2$. If we want to homogenize it, we should increase the degree of the right by $ 3$, and so we multiply it by $ abc$, which is equal to $ 1$ and doesn't affect the inequality.

One reason you want homogeneous inequalities is because AM-GM and the like tend to work better on them. If you AM-GM a bunch of terms whose degrees are the same, you'll get something of the same degree back. If you have that same degree on the other side, you can pin down the inequality exactly by getting the other side to have those exact terms. In the example above, it's not clear how to AM-GM the $ a^2, b^2, c^2$ terms to get something that we know is bigger than something like $ \frac1a$. But homogenize, and we find that $ \frac1a \plus{} \frac1b \plus{} \frac1c \equal{} ab \plus{} bc \plus{} ca$. We know how to AM-GM $ a^2, b^2, c^2$ to be greater than or equal to terms like $ ab$. One solution is
$ \left( \frac{a^2 \plus{} b^2}{2} \right) \plus{} \left( \frac{b^2 \plus{} c^2}{2} \right) \plus{} \left( \frac{c^2 \plus{} a^2}{2} \right) \geq ab \plus{} bc \plus{} ca$.

Another reason to get a homogeneous inequality is that once you have one, you can often throw away the constraint completely and/or put one of your own in. Return to the inequality above after we converted it to $ a^2 \plus{} b^2 \plus{} c^2 \geq ab \plus{} bc \plus{} ca$. If this is true for all $ abc \equal{} 1$, then by scaling the variables by a factor of $ k$, we get $ abc \equal{} k^3$ and $ k^2(a^2 \plus{} b^2 \plus{} c^2) \geq k^2(ab\plus{}bc\plus{}ca)$. This is equivalent to the original inequality, and $ abc \equal{} k^3$ is basically allowing $ abc$ to be anything. Now you know that you can throw away the constraint.

(There are exceptions of course, like unconstrained inequalities that aren't homogeneous.)

[quote]4)isn't there any substitute for the so called brute force "dumbassing" method that is laborious multiplication and collection of terms done to preserve sharpness of an inequality?(i hate dumbassing)[/quote]
Dumbassing is a brute force method as you said, which means that in almost any contest problem it will not be required. In other words cleverness is a substitute.

[quote]5)how does one know what is the sharpness of an inequality that is how sharp an inequality is?[/quote]
It's more of an intuitive thing than a definition. If you find that by using a simple application of an inequality like AM-GM actually flips the inequality around, you probably have a sharp one. If you try testing values for the variables and find that the sides are rarely very different, you definitely have a sharp one. Sharpness is basically a measure of how clever you need to be*, and you should reach for algebra, substitutions, or some clever convex functions (for i.e. Jensen's) to break the sharp ones down.

* Actually, really weak inequalities can be surprisingly difficult sometimes, because you know you probably have to drop a whole bunch of terms and such, and you won't know which ones.

[quote]6)proof of muirhead?[/quote]
Muirhead is basically a bunch of applications of AM-GM, as said above. Formally proving it requires some messy induction and use of the majorization definitions. It would be sufficient and much more helpful to figure out on one's own what polynomial inequalities AM-GM can kill alone (knowing the majorization definition is a head start to the answer). Once you have a general idea of the criterion, you basically know Muirhead's inequality.

\[ {\rm{for}}\_a_i ,b_i  \in R,x \in \left[ {0,1} \right]\_prove\_the\_following
\]

 :roll:

i know the definition of majorization. So can any one tell me what exactly is karamata's inequality and its proof and if there is a thread about it in this site can anyone pls give that particular link?

Already did.

[quote="t0rajir0u"]There is a [b]lot[/b] of useful discussion [url=http://www.artofproblemsolving.com/Forum/index.php?f=55]here[/url]. [/quote]
One of the most prominent stickies on that page (if you had followed the link) is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14975]this[/url].  You'll also see darij's post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=231490]here[/url].


If you were asked to pick a card, look at it, put it back in the deck and then choose "another" card, would you assume to choose a different card?

if so...

then the solution would be

1- P(not choosing the 16 cards)

1- (36/52)(35/51)

Did anyone else approach it this way?

[quote="GameBot"]A card is chosen at random from a standard deck of 52 cards, and then it is replaced and another card is chosen. What is the probability that at least one of the cards is a diamond or an ace?[/quote]

The probability that none of the cards are diamonds or aces is $ \left(\frac9{13}\right)^2 \equal{} \frac{81}{169}$ (since there's replacement). Thus, our final answer is $ 1 \minus{} \frac{81}{169} \equal{} \boxed{\frac{88}{169}}$.

Nearly clowned another C&P problem by doing it without replacement. Thankfully got it on second try.

[hide=Solution]There are $13+4-1=16$ cards that are diamond or ace, so there are $36$ cards that are neither diamond nor ace. Our desired probability is hence $1-\left(\frac{36}{52}\right)^2=\boxed{\frac{88}{169}}$.[/hide]


problem goes like this:

A thin rod with mass [i]M[/i] and length [i]L[/i] lies along the y-axis with its center at the origin.  Find the gravitational force it exerts on a point-mass [b]m[/b]  located on the x-axis a distance [i]x[/i] away.

PS:  Could we use the intergral symbol in here?

Just use straight forward method (Integrate).  Hint: It is not hard.


SMO 2007, Open #15

In an acute-angled triangle $ABC$, points $D$, $E$ and $F$ are the feet of the perpendiculars from $A$, $B$ and $C$ onto $BC$, $AC$, and $AB$, respectively. $H$ is the intersection of $AD$ and $BE$. Suppose $\sin A = \frac{3}{5}$ and $BC = 39$, find the length of $AH$.

we have : $2R=\frac{a}{\sin{A}}$ $\longrightarrow$ $2R=65$
now we have $AH=2R\cos{A}$ (because $\angle A<90$) .
$cos^{2}{A}=1-sin^{2}{A}\longrightarrow$ $\cos{A}=\frac{4}{5}$ ($\cos{A}>0$)
$\longrightarrow AH=52$ .


what difficulty is olymon supposed to be at? Which contest is it comparable too? I'm having fun playing with the problems, it seems that there are at least a few problems a month I can make significant progress on, what do you guys think?

Olymon's difficulty extends from AIME level, to CMO level, with some IMO level questions.  Generally, problem sets by Prof. Barbeau will be at CMO level or higher.  Problem sets from other people will be at AIME or CMO level.

Oops, i had subbmited what I thought were the answeres, my bad.

I think you should ask Ed Barbeau, who runs Olymon.

[quote="Inspired By Nature"]Hi everybody,
I just had a problem. The thing is that I sent in my solution to a problem for Olymon 7:3(Problems for April). Now I checked the Olymon site, and they have issue 7:4(problems for June) available. But the problem is that I didn't get my solution marked and sent back to me. When should I expect a response? And will it be an E-mail or a letter? Thanks in advance.[/quote]

[quote="Inspired By Nature"][quote="Farnak"]:?  Um.....I downloaded the problems and it says "Please send your solution ... no later than March 28, 2006".
It really means July right?[/quote]

No I don't think so. I think it means June. Anyway, guys please help me with my problem (explained in detail in my post above). :lol:[/quote]
The Problems for June are now due August 15.

that's sweet
that means that i'll actually have time to do them
hey alex, i bet i'll beat you on this olymon
(thanks for your solutions btw  :rotfl: )

LOL
I gave you incorrect solutions. :rotfl: That's okay, Greg

And you actually might beat me. I am not that good at doing Olymon problems for some reason.
But I have a lot of time :)

[quote="rem"]
And you actually might beat me. I am not that good at doing Olymon problems for some reason.
[/quote]
you're not good at anything though  :rotfl:
jk...


function
combinatorial geometry
topology
real analysis
real analysis unsolved

$E$ a banach space. Does there exists a convex function that is not continuous ?

yes  :)  after that, try to prove that if this convex function is semi continuous inferior, then it is locally bounded and locally lipschitzian.

Assume first that we've already shown the function is locally bounded. Then we can prove that it's locally Lipschitz as follows:

Let $B=B(x_0,r)$ be the ball of radius $r$ around $x_0$, where $r$ is chosen such that $m\le f\le M$ in $B(x_0,2r)$. Now suppose that for each $n\ge 1$ we can find $x_n,y_n\in B$ such that $f(y_n)-f(x_n)>n\|y_n-x_n|$. Now take $z_n=y_n+\frac r{\|y_n-x_n\|}(y_n-x_n)$. We have $f(z_n)-f(x_n)>n\|z_n-x_n\|\ge nr$, so $f(z_n)>nr+m$. Since $z_n\in B(x_0,2r)$, this contradicts the boundedness of $f$ in this ball, and we have a contradiction.

Now let’s prove that $f$ is locally bounded. The lower semicontinuity already tells us that it’s locally bounded from below, so it suffices to prove that it’s locally bounded from above. First  we’ll prove that it’s locally bounded in at least one point.

Assume the contrary, and consider an arbitrary point $x_0$. By assumption, in the ball $B_0=B(x_0,r_0)$ there is at least one point $x_1$ with $f(x_1)>1$. Then $f>1$ in a ball $B_1=B(x_1,r_1)\subset B_0$, and in this ball we can find $x_2$ with $f(x_2)>2$. Again, by lower semicontinuity, $f>2$ in a ball $B_2=B(x_2,r_2)\subset B_1$, and by assumption, we can find $x_3\in B_2$ with $f(x_3)>3$, and so on. If we choose $r_n\to 0$, then, because of the completeness of our space, the intersection of the $B_n$ will be non-empty. However, if $x\in\bigcap B_n$, then $f(x)>n,\ \forall n$, which is absurd.

The above shows that there is an $x_0$ such that $f$ is bounded from above in $B=B(x_0,r)$. Now let $x,a$ be arbitrary points. A small enough neighborhood of $x$ is contained in the convex hull of $B\cup\{a\}$, so in this neighborhood $f$ is bounded from above by $\max\left(\sup_Bf,f(a)\right)$, which proves the local boundedness from above at the arbitrary point $x$.

I can't understand why one can find such a $x_2 \in B_1$ such that $f(x_2)>2$. One only knows that $f$ is not bounded in $B_0$, but maybe it is locally bounded in $B_1$. (you implicitely used the fact it is convex ?)

In that part of the proof I showed that $f$ is locally bounded in at least one point. I was asuming the contrary to this, i.e. that $f$ was not locally bounded from above in [i]any[/i] point. This means that in every open set we can find arbitrarily large values of $f$. In other words, that part of the proof shows that: a lower semicontinuous function must be locally bounded from above in at least one point.

I used the convexity in the next part of the proof, which shows that a convex function which is bounded from above in a certain open set is locally bounded from above everywhere.

ok, I got it  :)  my approach was almost the same. I showed it was locally bounded around one point, with Baire and the semicontinuity condition, and then I conclude like you, with convexity.


inequalities
induction
inequalities proposed

Let $ a_1, a_2,, a_n$ be positive real numbers smaller than $ n^2+1$ that add up to $n^3$. Prove that $\sum_{k=1,n}{\sqrt{\frac{n^2+1}{a_k}-1}\leq 1}$.

For $n\geq 4$ it is banal Jensen.
For $n=2$ we can apply easy direct checking.
For $n=3$ hmm... For $n=3$ we should wait another approach :)

Thanks, Myth, I did not know that for n at least 4 it is banal Jensen. I hate to compute derivatives. But at least for n=3 it does not work.  :)

Actually, my approach work for $n=3$, since it has troubles only if one number $7\leq a_i<7.5$, but in this case it is easy to make simple estimate.

By Cauchy we have

$\displaystyle \sum_{k=1,n}{\sqrt{\frac{n^2+1}{a_k}-1}\leq \sqrt{n( \sum_{k=1,n}{\frac{1}{a_k}})} }$

Is it correct this first step?

No, because it is possible $\sqrt{n( \sum_{k=1,n}{\frac{1}{a_k}})}>1$. Sorry.

I have tried also by AM-GM, by induction and by mixing variables, but it doesn't want to collapse. :(

Really very hard and very original as all Harazi inequalities :)

Can you, please, post your solution, Harazi?

Thank you very much.

Here is my idea:
$t_n=1+\frac{1}{n^2}$
Proof that ineq use intruction:
$\sum{\sqrt {\frac{k}{b_i}-1}} \le n\sqrt{k-1}$
For all $1\le k \le t_n$ and $\sum{b_i}=n$

n^2+1 isn't best const?Maybe it is $n^2\frac{(n+1)^3}{(n^3+1)}$?

Dear hungkhtn,

sorry but I don't understand where does it come

 $k=(u^2+t^2)(u+t)$

For $t$ not equal $u$ I obtain another result.

I obtained $k(t^2+ut+u^2)=(t+u)(t^2+u^2)$

and how to prove that this equation has not real solutions with

$1 \leq k \leq 1+\frac{1}{n^2}$ and $u \geq1 $

Thank you very much. ;)

I really sorry you because I made a mistake in this step.
I tried to prove that ,but I think the inequal:
$\sqrt{\frac{k}{t}-1}+n\sqrt{\frac{nt}{n+1-t}-1}\le (n+1)\sqrt{k-1}$
still true.
Waiting for me a time,I will try to prove it!

Harazi, can you please post your solution? :)

:)  :)  :)  :)  :)
Oh,now I'm realize my mistake.I'm wrong when caculate f'(x),very very stupiad error.
I've edit my solution,I belive it's true.

$x_{n+1}=n+1-\frac{n}{c},c\le 1$
$f(c)=n\sqrt{kc-1}+\sqrt{\frac{kc}{(n+1)c-n}-1}$
$2f'(c)=\frac{nk}{\sqrt{kc-1}}-\frac{nk}{((n+1)c-n)^2\sqrt{\frac{kc}{(n+1)c-n}}-1}$
...check it for me...
$f'(c)=0 \leftrightarrow ((n+1)c-n)^2\sqrt{\frac{kc}{(n+1)c-n}-1}=\sqrt{kc-1}$
Take $u=(n+1)c-n->c=\frac{u+n}{n+1}$
And $f'(c)=0 \leftrightarrow \frac{k(u+n)}{n+1}-1=\frac{ku^3(u+n)}{n+1}-1$
$\leftrightarrow u=1$ or
$\leftrightarrow \frac{k(u+n)}{n+1}(u^2+u+1)=u^3+u^2+u+1$
$\leftrightarrow \frac{k(u+n)-n-1}{n+1}=\frac{1}{u^2+u+1}$
Because $u\le 1 then k(u+n)-n-1=n(k-1)+uk-1\le (n+1)(k-1)$
Then $\frac{k(u+n)-n-1}{n+1}\le k-1 \le \frac{1}{n^2} \le \frac{1}{u^2+u+1}$
So $f'(c)=0 \leftrightarrow c=1$

 :)  :)  :)  :)  :)  :)

Ok, here is my solution. It is sufficient for n=3 since for n at least 4 is was destroyed by Myth. For n=3 a simple argument by absurd followed by a substituion shows that it is enough to prove that if $ a+b+c=1 $ then $\sum {\frac{1}{1+a^2}}\leq\frac{27}{10}$. I think we have already discussed this some long time ago and anyway if not a breaking of the form $\frac{1}{1+x^2}\leq a+bx$ will lead to solution. Note that I used this very same inequality to find a very simple solution to the famous Japan inequality 1997.

Thank you harazi.Can you show $a$ and $b$,Is $\sum{x}=1$?
 :(  :(  :(

Yes, it is that condition. I do not remember exactly the exact value of a,b but I am sure that you can find it in 5 minutes. Just impose the condition to have equality for $3x=1$ and then factorize everything and you will have a,b.

No, my friend, I thought I was clear: we work just with 3 variables, since for a greate number of variables Myth has proved it directly. So you can take n=3.

Oh,I see.Now I've really known it.
Thank you,harazi!

[quote="harazi"]$\sum {\frac{1}{1+a^2}}\leq\frac{27}{10}$. I think we have already discussed this some long time ago and anyway if not a breaking of the form $\frac{1}{1+x^2}\leq a+bx$ will lead to solution.[/quote]
In other words, we need tangent in point $x=\frac{1}{3}$. It is $\frac{9}{10}-\frac{200}{243}(x-1/3)$, hence result.

Harazi,can you explain better your  reasoning by "absurd"?

I am so stupid not to understand it.

Sorry very much :blush:  :blush:

You can take $x_k=\sqrt{\frac{n^2+1}{a_k}-1}$
And now $\frac{1}{1+x_k^2}=\frac{n^3}{n^2+1}$
We need to prove $\sum{x_k}\le 1$
It $\leftrightarrow$ If $\sum{x_k}=1$ then
$\sum{\frac{1}{x_k^2+1}} \le \frac{n^3}{n^2+1}$
I think there is it.

Ok, if $\sum{\sqrt{\frac{10}{a}-1}>1}$, then we can find $k>1$ and $x,y,z$ which add up to 1 such that $\sqrt{\frac{10}{a}-1}=kx,...$. Find am inpose condition for they add up to 1 and it is done.

Thank you very much.   Nice reasoning :)


calculus
AMC
AIME
Putnam
Gauss
geometry
AoPS Books

I'm so intimidated!!!

Well before, I considered myself to be reasonably good in maths as I was in the top 3 in my year (currently year 10).

I've always enjoyed challenging myself and thus I tried solving the AMC-type of questions.  And for doing that, I'm considered as a big nerd in my school.

Now I come here... and I get COMPLETELY intimidated by all aged 10 and 11 students who produce such simple, beautiful, flawless proofs and answers to the questions which completely baffled me.

Some youngsters solve AoPS books at age 10!

man... I feel so mediocre.

Please tell me that all of you smart ppl solve math questions everyday.
I simply don't get how you can be so witty at solving problems in a snap.

Argh.. help me out here!! Are there anyone else who felt completely downgraded after looking at some posts here?

PS.  Some Q's for aged 11~14 stumbled me.  Does that imply my math simply SUX?  I do calculus BC currently and I was proud of it; I was wrong

Hi there!

Glad to have you a board sailor :)
You should definitely not feel intimidated. It's not always possible to solve all problems, and usually the ones which post solutions have thought longer than you think about the problem, some knew them ahead, others are indeed geniuses, but not all of course.

The age guides were put in place so that older students (which used to enter and post quick answers to easy questions which were not addressed to them) stick with the Advanced forums. Certainly not all people which you see posting in Getting Started are 10 years old, and so on.

There were lots of users who felt like you, especially when AoPS merged with MathLinks, but all of them can tell you now that they got used with it and they have generally improved their problem solving skills.

If you say you enjoy challange, just think about this as a new challenge. Try to make some friends amongst the community here.

There are also a lot of stuff that you can do to better yourself in math for AMC / AIME and other such competitions. For example, you can take AoPS Classes, you can stick around the forum, browse through old threads, check the resources pages ...

PS Don't care too much about people calling you "nerd" at school, you'll forget it quickly in college :D

[quote="nodrog2002"]I'm so intimidated!!!

Well before, I considered myself to be reasonably good in maths as I was in the top 3 in my year (currently year 10).

I've always enjoyed challenging myself and thus I tried solving the AMC-type of questions.  And for doing that, I'm considered as a big nerd in my school.

Now I come here... and I get COMPLETELY intimidated by all aged 10 and 11 students who produce such simple, beautiful, flawless proofs and answers to the questions which completely baffled me.[/quote]

Some of that work is better than I would have posted, if I were inclined to spoil people's fun by giving answers. Some of the problems in the Pre-Olympiad section are beyond me, yet you youngsters still manage to answer them. I am 42 years old and an actively researching mathematician.

Being exceptionally good in mathematics is a talent (or skill, since it can be developed by hard work) that is hard to judge. If someone is better at mathematics than all his or her classmates and even the teachers, is that person a mathematical genius and should devote his or her life to the pursuit of mathematics? No. If someone is not even the best in his or her Mathematics for Gifted Students class, should he or she give up on hopes of being a mathematician? No. The main qualification for becoming a mathematician is wanting to become a mathematician (otherwise, all those proofs would bore the person to tears). Are there 10-year-old mathematical geniuses posting in the AoPS forum? Probably yes. Should you be intimidated by that? No.

I have worked with world-class genius mathematicians. It is rather fun playing with the level of ideas they create. I am not one of those geniuses, but I don't have to be in the top 1000 to do serious research. The number of mathematical questions that could give useful results if researched exceeds the number of questions that mathematicians will find time to research. Don't let someone being better at mathematics divert you from your ambition to master mathematics.

Reading the forum is not a good way to judge your current level of mathematical skill. Math competitions are a better way to measure yourself. Reading this forum leads to improving your own skill, instead.

And that skill will grow rapidly. I remember my high school experience. As a high school sophmore, I scored an honorable mention in the Michigan Mathematics Prize Competition. As a junior, I was in the top 100 in the MMPC. As a senior, I tied for second place in the MMPC. As a college freshman, after being on my college's Putnam team, I visited my favorite high school teacher when he was looking over that year's MMPC questions. I was capable of solving several in my head.

Many problems that seem impossible to you today will be within your abilities next year. Of course, those people whose work makes you feel mediocre will have improved too, but don't worry about that. Study math for the fun of it.

As for the problem of people calling you a nerd, I am afraid that that is outside my experience. I had an idyllic high school existence. I was teased, but no more than anyone else. No-one held my brainy inclinations or my simple-minded social skills against me. I think it was because I liked people, which tended to make them like me.

Erin Schram

In truth even if you are not a genius at math you can still do than others better later on as long as you develop your work happits and work hard, also don't give up. Just being smart isn't going to do every thing you need. So don't get to discouraged.

This was written by a 12 year old.   ;)

I know exactly how you feel. You are in your own school, a "big fish", to use a nice arrogant cliche, and then along you come here. And you go from being the guy who understands everything faster than everyone else, who understands and knows more than everyone else, to being the guy who, essentially, wastes the smart people's air. I don't know if all of that is true of you, but it certainly is of me. But as several others mentioned, this is a wonderful opportunity. Most people here are highly supportive. For  example, in my AMC 12 class, I'm generally the one who asks the questions everyone else knows the answer to. But no one mocks me, or anything like that. Rather, they sit down and try to hep me, by answering my questions, suggesting general strategies, sources, etc. So we really do learn a lot in a very positive way here.

Hey look at me. I'm 16 years old and in 10th grade and I still post in the beginner forum an look through intermidiate sometimes.
And I ever liked maths. But I just recognized last year what maths really is. Since then I started problem solving. Meanwhile I met Friedrich Feuerstein (who was a member of the German IMO team 2003) for the first time. I fellt so small and stupid and I was so jealous and facinated how someone could be so genuisly.....
Or last summer when I found mathlinks for the first time....
I didn't expect that here are so many people which are as good as Friedrich or better....
The time was quite hard when I realzied where my frontires are and that there are many people whose frontires are far beyond mine.
But hey I'm fine. We are all just humans and no one is perfect and we are all uniques. Hence everybody has diffrent mathskills. And I don't belive that here is someone who had been to the IMO without practicing a lot before...

Age should never be a limit. Intimidation is only in your mind. If you have that mindset, nothing, and truly nothing, can intimidate you. Just work hard and know what you want as a goal! Then, you can work to achieve.

I never really became serious about math until I was 12. In 7th grade.

So, I'm still behind most people here, but there's no reason to back off and give up or feel behind. Everyone here is really supportive, and in no case, will you feel dumb or stupid. Now I'm 14, and I'm only taking algebra II, because my school is 2 years behind in its math curriculum. But, I read ahead and work on extra problems, so It's all good.  :D :lol:  :D

you'll get better at this math stuff. but i do agree that it's not fair that 10 year olds are better at math than i'll ever be. :(

i really dont think there is a lot of ten year olds on this board.  If there was, then it would be amazing because the youngest person to score an 800 on the SAT was ten.  Gauss recognized that the sum of the numbers from 1 to 100 was 5050 when he was ten.  And he was known as the "infant prodigy."  Amc contests have those kinds of problems all the time, now i doubt that ten year olds could that.  We would hear about it in the papers.

Not really because times have changed now and there are a lot of really talented people these days.

[quote]Amc contests have those kinds of problems all the time, now i doubt that ten year olds could that. We would hear about it in the papers.[/quote]

The difference is that Gauss found the formula $1+2+ \ldots + n = \frac{n(n+1)}{2}$ by himself. A 10 year old today could easily use the formula, without understanding of it. far from discovering it himself.

there are more talented people these days.  How do you figure?

The concepts of archimedes, pascal and other ancients may seem of obvious to us now, but back then they were revolutionary.  Concepts of einstein are relatively new, are they going to become the same way?

AMC 12 contests stresses the ability of doing math rather than just plug into a formula.

You have an AMC 12 class? Wow.....

I'm taking geometry and I'm in 8th....At my school people call me smart and stuff but when I came to this site I was like "WHOA!"

I've also noticed that generally the people who call other people nerds are generally less intelligent than you. I mean, the other people in my geometry class don't think I'm a nerd, but yet others do.... :?


So, I saw the recursive formula for the number of rooted trees with n nodes, but can't quite figure out why exactly it is true.

The formula is (6) at http://mathworld.wolfram.com/RootedTree.html

Any explanations would be awesome

If you look at that page, it references Finch 2003.  If you go down to the bottom, in the bibliography, you'll see that Finch 2003 is an article titled, "Two Asymptotic Series," and it has a link next to it.  If you click on that link, you'll get taken to a page which includes a .pdf version of the article in question.  It doesn't look like easy reading -- have fun!

Definitely not easy reading. Although I can't say I fully understand all of that, I think the citation was incorrect because Finch does not give a recursive formula for T_n anywhere close to the form of mathematica.

What I was wondering is if someone could explain the recursion in terms of why it would work logically. That is, what is each sum counting and how does that add up to T_n.


inequalities
function
algebra
polynomial
combinatorial geometry
inequalities solved

Let $n$ be a positive integer. Let $x_{1}$, $x_{2}$, ..., $x_{n}$ be $n$ nonnegative real numbers such that $\sum_{i=1}^{n}x_{i}=1$. Prove that

$\sum_{i=1}^{n}\left(x_{i}\cdot\left(1-x_{i}\right)^{2}\right)\leq\left(1-\frac{1}{n}\right)^{2}$.

for any a,b in [0,1], we have a(1-a) 2 +b(1-b) 2 -2(a+b)/2[1-(a+b)/2] 2 =(a-b) 2 (3(a+b)/4-1)<0 (since a+b \leq 1<4/3).
If none of the xi's are >2/3 then by Jensen we have the result(f(x)=x(1-x) 2 is concave over [0,2/3]).
If there is an xi (at most 1 of them)>2/3 (WLOG let x1>2/3), By the lemma above we have f(x1)+f(x2)<2f((x1+x2)/2), and (x1+x2)/2<2/3. Hence by Jensen again we obtain the result.

this wuold follow form jensen's inequality, if only the function

\[
f(x)=x{(1-x)}^2
\]

were concave on [0,1]

unfortunately, $f''(x)=6x-4$ is $\leq 0$ for $x\leq \frac{2}{3}$

bad luck! ;)

maybe by induction...

That nearly finishes such a proof. Obviously, if all of the [tex]x_{i}[/tex] are between 0 and 2/3, then Jensen asserts that the maximum is when [tex]x_{1} = \cdots = x_{n} = \frac{1}{n}[/tex], and it is easy to see that this yields equality.  The remaining part is still greatly assisted by Jensen's inequality.  Since [tex]f''(x) = 6x - 4[/tex], [tex]f[/tex] is concave over [0, 2/3] and convex over [2/3, 1]. Therefore, the LHS is maximized if all [tex]x_{i} \in [0, \frac{2}{3}][/tex] are equal, and all [tex]x_{i} \in [\frac{2}{3}, 1][/tex] are as spread out as possible.  Thus, we need only consider the cases [tex]x_{1} = x_{2} = \cdots = x_{n-1} = x < \frac{1}{3(n-1)}[/tex] and [tex]x_{n} = 1 - (n-1)x > \frac{2}{3}[/tex].  These cases can be reduced to a cubic polynomial of one variable, and are easily shown to yield a LHS less than [tex]\left(1 - \frac{1}{n}\right)^{2}[/tex].

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

Alternatively, define a function [tex]g : [0, 1] \rightarrow \mathbb{R}[/tex] by

[tex]g(x) = f(x)[/tex] [tex]\forall x \in [0, .5][/tex] and [tex]g(x) = f(.5) + (x - .5)f'(.5)[/tex] [tex]\forall x \in [.5, 1][/tex]. [tex]g[/tex] is everywhere concave, and [tex]g(x) \ge f(x)[/tex] [tex] \forall x \in [0, 1][/tex] since [tex]g(.5) = f(.5) = \frac{1}{8}, f'(.5) = \frac{-1}{4}[/tex], and [tex]g(1) = f(1) = 0[/tex].  Notice, however, that [tex]g\left(\frac{1}{n}\right) = f\left(\frac{1}{n}\right)[/tex] [tex]\forall n \in \mathbb{Z^{+}}[/tex].

The problem is now straightforward; by Jensen:

[tex]\displaystyle{ \sum_{i=1}^{n} x_{i}(1 - x_{i})^{2} = \sum_{i=1}^{n} f(x_{i}) \le \sum_{i=1}^{n} g(x_{i}) \le n \cdot g\left(\frac{1}{n}\right) = \left(1 - \frac{1}{n}\right)^{2}}[/tex]

First, we can prove that for all $x \le 2(1-1/n)$, we have
\[ f(x) \le f(1/n) + f'(1/n) (x - 1/n).\]
That inequality is a cubic with a double root at 1/n, so it boils down to a linear inequality.

Once we have that inequality, we can apply it to all $x_i$, and finally sum.

We can prove that $x(1-x)^2 \leq ax+b$ where $\displaystyle a = \frac{(n-1)(n-3)}{n^2}$ and $\displaystyle b=\frac{2(n-1)}{n^3}$.

In fact, this inequality is equivalent to $(nx-1)^2(nx-2n+2) \leq 0$ which is obviously true when $0\leq x\leq 1$ and $n\geq 2$. Therefore,

\[ \sum_{i=1}^n x_i(1-x_i)^2 \leq a \sum_{i=1}^n x_i + nb = a+nb = \left(1-\frac{1}{n}\right)^2.\]

If you are interested how I found the constants $a$ and $b$, you can read:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3048

This is probably very close to what Ravi provided although I am not very clear how to actually carry out his method.

Cheers! :D  :D

Fuzzylogic, yes, our proof is essentially the same, although you explained it better.  My displayed inequality is equivalent to yours, and can also be proved by rearranging it into the product of a square and a linear factor.  Alternatively, because f is a cubic, we can use the exact Taylor expansion
\[ f(x) = f(1/n) + f'(1/n)(x-1/n) + \frac{1}{2}f''(1/n)(x-1/n)^2 + \frac{1}{6}f'''(1/n)(x-1/n)^3.\]
The sum of the last two terms is the product of $(x-1/n)^2$ and a linear factor, and is nonpositive in our domain.  So we get the inequality I mentioned.

[quote="Ravi B"] Alternatively, because f is a cubic, we can use the exact Taylor expansion
\[ f(x) = f(1/n) + f'(1/n)(x-1/n) + \frac{1}{2}f''(1/n)(x-1/n)^2 + \frac{1}{6}f'''(1/n)(x-1/n)^3.\]
The sum of the last two terms is the product of $(x-1/n)^2$ and a linear factor, and is nonpositive in our domain.  So we get the inequality I mentioned.[/quote]

Very nice! Taylor expansion is a more natural way to derive $f(x) \leq f(1/n) + f'(1/n)(x-1/n)$. Thanks.

Consider the mean value of the x's wich is 1/n.
Write every x as  : 1/n  +  y
Check that the sum of the y's is 0.
Replace in the sum.You'll get it easily.I suppose.

${\bf Problem}$
$x_{i}$ ($i=1,2,...,n$) are non-negative real number such that $\displaystyle\sum_{i=1}^n x_{i} = 1$. Prove that
\[\displaystyle\sum_{i=1}^n x_{i}(1-x_{i})^2 \le (1-\frac{1}{n})^2.\]

Without loss of generality suppose $x_{1} \ge ... \ge x_{n}$. Since if $a+b \le 1$, then $a-a^2 \ge b-b^2 \Leftrightarrow (a-b)(1-a-b) \ge 0 \Leftrightarrow a\ge b$, hence we have $x_{1}(1-x_{1}) \ge ... \ge x_{n}(1-x_{n})$ and $1-x_{1} \le ... \le 1-x_{n}$.
By the Chebysev's inequality,
\[\displaystyle\sum_{i=1}^n x_{i}(1-x_{i})^2 \le \frac{1}{n}(\displaystyle\sum_{i=1}^n x_{i}-\displaystyle\sum_{i=1}^n x_{i}^2)(\displaystyle\sum_{i=1}^n 1 - \displaystyle\sum_{i=1}^n x_{i})= \frac{n-1}{n}(1-\displaystyle\sum_{i=1}^n x_{i}^2).\]
By the Cauchy-Schwarz inequality,
\[-\displaystyle\sum_{i=1}^n x_{i}^2 \le -\frac{1}{n}(\displaystyle\sum_{i=1}^n x_{i})^2 = -\frac{1}{n},\]
so
\[\frac{n-1}{n}(1-\displaystyle\sum_{i=1}^n x_{i}^2) \le (1-\frac{1}{n})(1-\frac{1}{n})=(1-\frac{1}{n})^2.\]

One only needs the existence of a concave upper bound for $f(x) = x(1-x)^{2}$ which is equal to $f(x)$ at the relevant point (here $x = 1/n$).   The convex hull of the graph of $f(x)$ on $[0,1]$ would also do, or a linear extrapolation of $f$ over the "bad" interval $[2/3,1]$.

[quote="dondigo"]I don't understand the first line
$(1-\frac{1}{n})^2=\frac{(n-1)^2}{n^2}$ not $\frac{(n-1)^2}{n^3}$[/quote]
It is inside the summation.

AAA.... I see now. It's obvious... of course you're right  :blush:


If someone wouldn't mind explaining this one to me I'd be most grateful.

"A mass m is attached to a number of springs which have equilibrium lengths equal to
zero. The other ends of the springs are fixed at various points in space. The spring
constants are all the same. The mass sits at its equilibrium position and is then given a kick in an
arbitrary direction. Describe the resulting motion."

I can kind of get some intuition here but I have no idea how to show it rigorously...

Thanks.

The following explanation is assuming that all of the springs act with a restoring force directly proportional to the displacement (i.e. they follow Hooke's Law).

The mass $m$ is a point particle, and thus we can assume that the results of the kick do not depend on where you kicked $m$. You can ignore gravity, so let us pretend that there is just one spring.

If there is one spring connected to mass $m$, and you kick $m$ in a random direction, then what will be [hide="the resulting motion?"]
Simple harmonic oscillation on a straight line.
[/hide]
Easy enough. Now add another spring connected in a completely different angle. What is different about kicking $m$? The restoring force is stronger. However, the restoring force is still directly proportional to the displacement, and $m$ is still a point particle, so the net spring force after the kick follows Hooke's law.

The basic idea is that no matter how many springs you put, the net spring force follows Hooke's law.

Another way of thinking about it:
Each spring exerts a force $F_i(r) = k_i(r-r_i)$ where $r$ is the vector location of the mass and $r_i$ is the vector location of the fixed end of the spring.

Therefore, the net force on the mass due to all of the springs is of the form
$F(r) = k(r-R)$
$R$ is the location of the equilibrium point. In the coordinate system where $R=0$, we have
$F(r) = kr$.

So the mass acts as though it were attached to a single spring.

What is this Morrin book?

Thanks Schulmannerism and theska, that makes perfect sense... but I'm still curious as to how to describe it mathematically  if you have, say, n springs with constant k at angles $\theta_1 ... \theta_n$ to the direction of displacement.

And the Morin book is available online for free from [url=http://my.harvard.edu/icb/icb.do?course=fas-phys16&pageid=tk.page.phys16.dir.7f829e3adaebcfa703b7db42e0201022]here[/url].

Well it doesn't actually matter what the angles are--what matters is the center of mass of the fixed points of the springs. The force comes out to be

$F(r) = nk(r-R)$ where $R$ is the center of mass of the fixed ends of the springs.

If the mass is given a kick from its equilibrium position, or the line of the initial trajectory goes through the equilibrium point, then the mass will undergo simple harmonic motion along a straight line.

If the line of the initial trajectory does not go through the equilibrium point, you will get an elliptical orbit. Consider the plane containing the line of the initial trajectory and the equilibrium point. In this plane, the mass undergoes simple harmonic motion in it's x-coordinate and in it's y-coordinate, and the cycles have the same period, but there is a phase difference, hence an ellipse.

P.S. Wow that's an awesome book!!

Yep that makes sense given either a phase angle or a sine term. Center of mass certainly does make things clearer...

And yeah, it is pretty awesome. I'm using it for physics next year, I'm so excited :D


Third degree on top and geometric on the bottom.  Just multiply thru by $\frac{1}{4}$ and subtract a few times and the answer will pop up.

[hide] Ooh notice that the expresion is equal to \[\sum_{i=1}^{\infty}(\frac{i}{2^{i}}^{2})\] . Now to check that the series converges: $\frac{i^{2}}{2^{2i}}< \frac{1}{2^{i}}$ for all $i>0$, so it is less than a well known convergent series. So let's call this sum $S$.
Then $S = \frac{1}{4}+\frac{4}{16}+\frac{9}{64}... \\ \\ \frac{S}{4}= \frac{1}{16}+\frac{4}{16}...$

So let's subtract, $\frac{3S}{4}= \frac{1}{4}+\frac{3}{16}+\frac{5}{64}...$. Now for convenience let this new sum be $R = \frac{3S}{4}$.

$R = \frac{1}{4}+\frac{3}{16}+\frac{5}{64}... \\ \\ \frac{R}{4}= \frac{1}{16}+\frac{3}{64}...$

Subtractt again! $\frac{3R}{4}= \frac{1}{4}+\frac{2}{16}+\frac{2}{64}... = \frac{5}{12}\text{(I think)}$.

So $S = \frac{4}{3}R = \frac{5}{9}$.
[/hide]

Yea pretty much any polynomial over geometric series sum can be done through this method, just keep chugaluggin


Anyone else see the Survivor finale last night??? I thought it was pretty goode. Amber played a goode game, and the whole thing between her and rob was pretty coolio...him proposing and all. Who did you want to win?

They haven't canceled that damn show yet?  WTF?

[quote="Magnara"]They haven't canceled that - show yet?  WTF?[/quote]

tis what i thought too...except in different terminology...

heh.  Yea I watched the show only because my friend probally would have killed me if I didn't.  I thought Rupert should have won this is twice now that I think he should have won.  but thats my opinion.  Yes Amber played a very good game a lot of the players did.

yeah...Rupert was DEFINATELY my fave in both Pearl Islands and All-Stars...you can vote to give him a million on the vote at: http://www.cbs.com/survivor

[quote="aalindsey100"]Survivor?????????? I didnt know that was still going on??? I hate reality shows!!!! They just take up perfectly good time where something that is actually goode could be showing.[/quote]
true, so true.

"Reality," my foot.  <I>I've</I> never been on a desert island with a bunch of other people competing for a massive sum of money.  Have you?

So-called "reality" shows are proof positive that people can be just as overly dramatic in real life as on soap operas.


limit
real analysis
real analysis unsolved

How to prove that if $\mu(X)<\infty$, we have for all $f\in L^\infty (\mu)$ $\parallel f\parallel _\infty =\lim_{p\to\infty}\parallel f\parallel _{p}$?

Let $\|f\|=C$. We look at the sets $S_{x}=\{t: |f(t)|\ge x\}$. If $x>C$, $\mu(S_{x})=0$; use this to show that $\|f\|_{p}\le C$ for all $p$. If $x<C$, $\mu(S_{x})>0$; use this to create lower bounds for $\|f\|_{p}$. These estimates tend to $x$ as $p\to\infty$; since $x$ can be chosen arbitrarily close to $C$, the $\liminf$ of the $\|f\|_{p}$ is at least $C$.


Let's give $f(x) : (-2,2)\rightarrow R$ is a derivable function
Prove that we can find $x\in (-2,2)$ satisfying  $f'(x)<(f(x))^2+1$

[quote="Rust"]It is true for all interval (a,b), suth that $b-a>\pi$.[/quote]
Hmmm I'm not sure about this  :?  :?
Can you post your solution, please ?

If $y=f(x),y'\ge y^2+1$ in (a,b), then $d %Error. "arctg" is a bad command.
y\ge d.x$ It give contradition if $b-a\ge \pi.$

Here was a related problem:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33769]www.artofproblemsolving.com/Forum/viewtopic.php?t=33769[/url].


trigonometry
logarithms
integration
calculus
calculus computations

Find the radius of convergence of the following power series:

$ 1.\sum_{n\equal{}1}^{\infty}(n\tan\frac{1}{n})^{n^2}(x\minus{}1)^n$.

$ 2.\sum_{n\equal{}1}^{\infty}(n\arcsin\frac{1}{n})^{n^2}x^n$.

$ 3.\sum_{n\equal{}1}^{\infty}\sin\frac{1}{n}\sin\frac{2}{n}...\sin1\cdot (x\minus{}1)^n$.

$ \left(n\tan\frac1n\right)^{n^2} \equal{} \left(1 \plus{} \frac1{3n^2} \plus{} O(n^{ \minus{} 4})\right)^{n^2}$

tends to $ e^{1/3}$ as $ n\to\infty.$ Any power series whose coefficients tend to a constant has a radius of convergence of $ 1.$ Hence the first series converges for $ x\in(0,2)$ (with boundary points not yet explored.)

The second series is quite similar, and its radius of convergence is also $ 1.$

The third series is a little more complicated. If that coefficient were $ \frac1n\cdot\frac2n\cdot\frac3n\cdots1,$ then that would be $ \frac {n!}{n^n}$ and the radius of convergence would be $ e.$

We need to explore whether the sines make any difference, but $ e$ will serve as a first guess about the radius of convergence.

The sines do matter. The logarithm of that product is approximately $ n\int_0^1 \ln(\sin x)\,dx$, so the radius of convergence is $ \exp\left(\minus{}\int_0^1\ln(\sin x)\,dx\right)$.

I see no reason to expect that this integral can be calculated exactly.


Can you please help me solve this problem, I am confused??

In isosceles trapezoid ABCD, legs AD and BC are each 5 units. Diagonal AC is 12 units and is perpendicular to leg AD. What is the length of segment AB expressed as a mixed number?

Please don't post twice...
You already posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=322749]here[/url] and received a response.


algebra
polynomial
calculus
search
group theory
abstract algebra
Galois Theory

Compute Galois groups for the following extensions of $ \mathbb{Q}$ (i.e. extensions of $ \mathbb{Q}$ to the splitting fields of the polynomials):
a) $ f \equal{} x^{3}\minus{}x\minus{}1$
b) $ f \equal{} x^{5}\minus{}5$

a)
f is irreducible since has deg 3 and no rational roots and simple calculus checks show that has one real root and two complex conjugates ones. Let them $ a,z, \bar z$. The degree of the splitting field over Q must be 6 because it must divide it and must also be a multiple of two because the Galois group contains the automorphism $ z \mapsto \bar z$ which has order 2 and of course is not the only one in the group (this exclude galois group of order 2). Hence our Galois group is $ S_3$.

(there is a more general claim which says that every rational and irreducible poly of prime degree $ p$ with exactly two complex roots has $ S_p$ as Galois group)

b)
The generators of the splitting field are $ a \equal{} \sqrt[5]5$ and $ \omega$ a primitive 5th complex root of 1.
The order of Galois group is 20 (the degree of splitting field over Q, because this ext is galois) and each automorphism is determined once you know where $ a$ and $ \omega$ are mapped.
$ a$ can be mapped to any other root of $ f$, and $ \omega$ can be mapped to any other power of himself (which are the other roots of its minimum poly over Q which is $ x^4\plus{}x^3\plus{}x\plus{}1$)

bye

I don't understand this part:

$ a$ can be mapped to any other root of $ f$, and $ \omega$ can be mapped to any other power of himself.

Suppose $ a$ is mapped to $ \omega^{k}$ for some k - our automorphism has to be a bijection, so some $ \omega^{n}$ must be mapped to $ a$. WLOG assume that $ \omega$ is mapped to $ a$ (all powers of $ \omega$ are primitive roots) - then $ \omega^{2}$ must be $ a^{2}$, a contradiction. Whats going on? It isn't true that $ a$ can be mapped to any other root (5 possibilities) and $ \omega$ to any other power of itself (4 possibilities), so we would get 20 automorphisms: for example, if $ a$ is mapped to $ \omega$ and $ \omega$ stays in place (one of those 20 possibilities), this isn't even injective.

let g any rational polynomial, $ a$ his root in our extension, and $ \sigma$ in the galois group. It is $ g(a) \equal{} 0$. We apply $ \sigma$ to both sides to get $ 0 \equal{} \sigma(f(a)) \equal{} f( \sigma(a))$ and that proves that the image of an element in some field extension through some automorphism in the galois group is another root of the same poly.
Conversely it can be proved that for any other root $ b$ of the minimum poly of $ a$ there exists an automorphism in the galois group such that $ a \mapsto b$. That is why we seek possible images among the roots of the same minimum poly (which are called conjugates)

[quote]
for example, if $ a$ is mapped to $ \omega$ and $ \omega$ stays in place (one of those 20 possibilities), this isn't even injective.[/quote]

now you should understand that there are no automorphisms $ a \mapsto \omega$ :)

bye

If there are no automorphisms that map $ a$ to $ \omega^{k]}$, how is the order of this Galois group supposed to be 20? We know that if an extension is Galois, then its degree = order of its Galois group, and here the degree is 20, so the group is supposed to have 20 automorphisms, but where are they? I'm a bit confused.

it is $ \omega \mapsto \omega^i$ for i=1,2,3,4 which are 4

we must also decide the image of $ a$ which has 5 possibilities ($ a \omega^j$ for j=0,1,2,3,4) for each image of $ \omega$ decided before.

So a list of automorphisms is $ \sigma_{i,j}: (a,\omega) \mapsto (a \omega^i, \omega^j)$ for i=0,1,2,3,4 and j=1,2,3,4. That makes 20.

b) is a particular case of this one:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=801271825&t=137762

One more question about Galois groups:

I have extension of Q by $ x^4 \plus{} 2$. Let $ \epsilon$ is the 8th root of unity, then I put in $ 2^{\frac {1}{4}}\epsilon$, $ 2^{\frac {1}{4}}\epsilon^{3}$, $ 2^{\frac {1}{4}}\epsilon^{5}$, $ 2^{\frac {1}{4}}\epsilon^{7}$ to get my field. I see that the degree of this extension is 4, so the Galois group has 4 elements and I can easily find them.

Now, this group has 2 normal subgroups, so I'm supposed to have 2 normal extension between Q and my splitting field. But I don't know how to find them. These can't be just extension by some roots of my polynomial, because then it would either be the whole field or woudn't be normal, and my field admits only normal extensions.

The group you need is the semi-direct product of $ C(4)$ and $ C(2)\times C(2)$. The only normal subgroups are the $ C(4)$, the $ C(2)$ it contains, a copy of $ C(2)\times C(2)$, the semi-direct product of $ C(2)$ and $ C(2)\times C(2)$.

Thus, the intermediate Galois extensions are given by $ \mathbb{Q}(\zeta_8)$, $ \mathbb{Q}(\zeta_8, \sqrt{2})$, $ \mathbb{Q}(\zeta_4, \sqrt{2})$ and $ \mathbb{Q}(\sqrt{2})$.


2. In how many ways can the digits 1 through 9 be arranged into three 3-digit positive integers using each digit once such that the three integers add up to 1000?

zero?
the sum will be 0 mod 9
but 1000 is 1 mod 9
blah here is proof the sum is 1 mod 9
[hide="blah"]
yeah so 1+2+3+4+5+6+7+8+9 is 0 mod 9
and every time you move a digit, say the number 5, to another place value, the mod 9 doesnt change no matter what digit you move (because you multiply by 9 or 99 and then you get a 0 mod 9 and you add that and the mod 9 stays constant)
blah the sum will be 0 mod 9
[/hide]
blah that proof was silly

then you find $ \boxed{0}$ cases


How many days will it take 2 get
to 1,000,000 if the equation is n = 2^(x-1) and I have no idea how to do that using a calculator. I tried the button that looks like a exponent but it gives me the wrong answer; plus I dont know how to work it backwards
I could find the million if I knew how many days but the other way around confuses me....

What does n represent?

What does x represent?

What do we call this type of equation?

n represents the number
x represents the number of days
it is an exponential equation

$n = 2^{x-1}$
$\log n = (x-1)\log2$
$\frac{\log n}{\log2}+1 = x$
replacing n with 1000000 we find x to be equal to roughly $20.93$ meaning it takes 20.93 days for n to grow to 1 million.


find value of x:- sin inverse x + sin inverse 2x = pie/3
                only using expansion of tan inverse x+ tan inverse y or tan inverse x-tan inverse y

Did you forget to type what to prove?????  :wink:

Edit: Oops, you posted it now...... :blush:

:maybe:  :maybe:  :maybe:
I don't understand what u mean.....


Năm 1966, khi đất nước ta đang còn chiến tranh ác liệt, Hội Toán học Việt Nam đã được Chính phủ chính thức ra quyết định thành lập. Từ đó đến nay, Hội đã trải qua một chặng đường 40 năm - một quãng thời gian không nhỏ trong sự phát triển đầy biến động của đất nước.

Có lẽ cũng cần nhìn lại một cách khái quát về những gì chúng ta đã có, đang có với hy vọng hình dung được những gì cần có trong chặng đường trước mặt. Phải nói rằng, cộng đồng toán học Việt Nam là một cộng đồng giàu trí tuệ, đã có ít nhiều truyền thống, nhưng cũng đang đứng trước không ít khó khăn và thách thức. Trong phạm vi bài viết này, các tác giả phác hoạ vài nét về thực trạng tình hình toán học nước ta, từ đó gợi lên đôi điều đáng suy nghĩ.

Chưa bao giờ toán học có vai trò quan trọng như ngày nay

Sự phát triển của tất cả các ngành khoa học, kể cả khoa học cơ bản cũng như khoa học ứng dụng, và tất cả các ngành công nghiệp then chốt như dầu khí, viễn thông, hàng không... đều không thể thiếu toán học, và ngày càng gắn bó mật thiết hơn với toán học. Ngay sự ra đời và phát triển mạnh mẽ của công nghệ thông tin (CNTT), một lĩnh vực đang đưa lại những hiệu quả cực kỳ to lớn trong mọi lĩnh vực của đời sống xã hội, cũng khởi nguồn và chủ yếu dựa trên toán học. Cũng chính CNTT đã làm cho nhiều ứng dụng kỳ diệu của toán học thật sự bùng nổ, mà ví dụ tiêu biểu có thể thấy đó là việc thay thế có hệ thống phần lớn các thử nghiệm tốn kém và mất thời gian trong thiết kế bằng phương pháp mô hình hoá toán học vừa tiết kiệm, nhanh chóng, vừa cho phép rút ra những thông tin ẩn khuất nhờ những phép toán tinh vi từ những khối dữ kiện khổng lồ. CNTT cũng đã đưa toán học thâm nhập ngày càng sâu vào kinh tế và quản lý hiện đại. Việc kết hợp chặt chẽ quản lý+công nghệ+tổ chức+toán học nhằm nâng cao khả năng quản lý chiến lược trong môi trường biến động nhanh là nét chuyển hướng mới trong đào tạo các nhà quản lý cao cấp ở các nước. Ngày nay, toán học không đơn thuần chỉ là công cụ như nó đã từng như vậy trong nhiều năm trước đây, mà đã thực sự trở thành một bộ phận không tách rời của những phát minh mới nhất, giá trị nhất. Chẳng phải ngẫu nhiên mà phần lớn các giải Nobel về kinh tế từ những năm 70 của thế kỷ XX trở lại đây đều được trao cho các nhà nghiên cứu toán kinh tế hoặc các nhà toán học trong lĩnh vực này.

Kinh nghiệm của nhiều nước cho thấy, sự phát triển kinh tế của một đất nước không phải bao giờ cũng do tài nguyên thiên nhiên quyết định, mà nhiều khi lại phụ thuộc chủ yếu vào trình độ dân trí. Nhật Bản, Hàn Quốc, Singapore... đã trở thành những nước công nghiệp phát triển sau một thời gian ngắn, mặc dù các nước này khá nghèo về tài nguyên. Toán học có vị trí đặc biệt trong việc nâng cao và phát triển dân trí, góp phần tạo nên nguồn tài nguyên chất xám - nguồn tài nguyên quý nhất cho đất nước. Có lẽ không có ngành khoa học nào, tự nhiên cũng như xã hội, có thể so sánh được với toán học về mật độ xuất hiện của nó trong chương trình đào tạo ở mọi ngành và mọi bậc học. Đó là bởi vì toán học không chỉ cung cấp cho con người những kỹ năng tính toán cần thiết, mà còn (và đây là điều chủ yếu) rèn luyện cho con người khả năng tư duy logic, một phương pháp luận khoa học. Xã hội ngày nay đòi hỏi một lực lượng lao động có trình độ suy luận, biết so sánh, phân tích, ước lượng, tính toán, hiểu và vận dụng được những mối quan hệ định lượng hoặc lôgic, xây dựng và kiểm định các giả thuyết và mô hình để rút ra những kết luận có tính lôgic. Nói cách khác, trong thế giới hiện đại ngày nay, năng suất lao động phụ thuộc không nhỏ vào vốn hiểu biết định lượng, tức là vốn văn hoá tính toán của đội ngũ những người lao động, cái văn hoá do giáo dục toán học cung cấp. Thế nhưng vai trò quan trọng này của toán học lại thường ít được nhận biết, vì tuy nó thể hiện ở mọi nơi, mọi lúc, ở mỗi con người, nhưng lại không phải là một cái gì đó cụ thể, có thể nắm bắt, cân đong đo đếm!

Việt Nam ngày nay được đánh giá là có một nền giáo dục và nghiên cứu toán học tương đối tốt...

Xét trên bình diện quốc tế thì đây là hiện tượng “quý hiếm”, vì nước ta vốn là một nước nghèo, chậm phát triển, trải qua chiến tranh kéo dài và không có truyền thống toán học lâu đời.

Những ai đã từng chứng kiến thực trạng toán học Việt Nam những năm 60, khi đất nước mới ra khỏi cuộc chiến tranh chống Pháp, chắc sẽ thấy rõ sự phát triển và trưởng thành nhanh chóng của lĩnh vực này. Từ một tập thể với 10-15 cán bộ giảng dạy trong mỗi khoa toán của vài trường đại học thuở ấy như Sư phạm, Bách khoa, Tổng hợp, mà hầu hết chỉ mới tốt nghiệp đại học, chúng ta đã tiến dần lên có được hàng trăm nhà toán học, trong đó nhiều người được quốc tế tôn trọng và thừa nhận. Các nhà toán học này đã đào tạo nên hàng chục ngàn giáo viên giảng dạy toán từ bậc phổ thông đến đại học trên khắp mọi miền đất nước. Cũng các nhà toán học ấy đã góp phần đào tạo nên hàng vạn kỹ sư hoạt động trên tất cả các lĩnh vực khoa học, kỹ thuật, kinh tế, an ninh và quốc phòng. Từ chỗ chủ yếu phải gửi sinh viên và cán bộ trẻ ra nước ngoài học tập, ngày nay chúng ta đã có được một đội ngũ cán bộ nghiên cứu toán học trình độ cao, đủ sức cập nhật những tri thức toán học mới nhất, sáng tạo hàng ngàn công trình khoa học công bố trên các tạp chí toán học thế giới, hợp tác một cách bình đẳng với đồng nghiệp quốc tế. Chính thông qua nghiên cứu khoa học, chúng ta đã đủ sức tự đào tạo hàng trăm TS, hàng chục TSKH đạt tiêu chuẩn quốc tế. Nhờ những đóng góp vào những nghiên cứu mang tính đột phá trong nhiều lĩnh vực toán học khác nhau, năm 1995 Viện Toán học đã được công nhận là một trong 10 viện xuất sắc (Institute of Excellence) của Viện hàn lâm khoa học thế giới thứ ba.

Ngay từ buổi đầu phát triển, những nhà toán học Việt Nam đã rất quan tâm đến việc đưa toán học phục vụ trực tiếp cho đời sống. Có thể kể ra ở đây những ứng dụng “vang bóng một thời” của vận trù học, việc giải quyết các bài toán nước thấm, nổ mìn định hướng trong thời chiến, cũng như những ứng dụng của toán kinh tế, lý thuyết hệ thống, điều khiển học, thống kê toán học, các phương pháp ngẫu nhiên và giải tích số... trong thời bình. Những ứng dụng khác nhau của toán học vào nhiều lĩnh vực của đời sống xã hội, một mặt, đã góp phần nâng cao năng suất lao động, cải tiến công tác quản lý, hỗ trợ việc sử dụng hợp lý các nguồn lực phục vụ cho sự phát triển, mặt khác, cần phải đặc biệt nhấn mạnh, các ứng dụng ấy đã từng bước góp phần không nhỏ tạo nên những cách tư duy mới trong xã hội.

Như vậy, toán học có đóng góp quan trọng trong việc nâng cao dân trí, tạo nguồn tài nguyên chất xám cho đất nước, tạo tiền đề phát huy nội lực, tiếp thu công nghệ mới trong sự nghiệp công nghiệp hoá, hiện đại hoá đất nước. Hiện tượng nước ta có trình độ dân trí và một nền giáo dục cao hơn thực trạng nền kinh tế chính là một trong những nguyên nhân khuyến khích sự đầu tư của nước ngoài.

Có thể thấy, sở dĩ toán học nước ta có được một số thành quả như trên là do những nguyên nhân chính sau đây:

Trước tiên là sự quan tâm của Đảng, Nhà nước và xã hội đến sự phát triển toán học trong một thời gian dài. Ngay trong kháng chiến chống Pháp, Nhà nước đã cho tập hợp những giáo viên ưu tú nhất mở Trường khoa học cơ bản, trong đó chủ yếu là dạy và học toán. Hoà bình lập lại, Nhà nước đã sớm quan tâm đến việc phát hiện và bồi dưỡng tài năng toán học như cho mở các lớp phổ thông chuyên toán, trong đó học sinh được cấp học bổng và hưởng tem phiếu như cán bộ nhà nước. Với tầm nhìn chiến lược, ngay trong những năm kháng chiến chống Mỹ ác liệt, Đảng và Nhà nước ta đã mạnh dạn tuyển chọn hàng vạn thanh niên ưu tú gửi đi đào tạo ở nước ngoài, trong đó ưu tiên một số đáng kể cho khoa học cơ bản và toán học. Việc đào tạo toán học ở trong nước cũng được sự quan tâm và ủng hộ của toàn xã hội.

Nguyên nhân thứ hai là sự giúp đỡ to lớn và có hiệu quả của Liên Xô và các nước XHCN ĐôngÂu cũ.Các sinh viên, nghiên cứu sinh, thực tập sinh Việt Nam thời đó đã có may mắn được làm việc với những giáo sư, viện sỹ giỏi nhất trong các trung tâm khoa học tầm cỡ thế giới. Sự đầu tư của các nước XHCN anh em cho ta thật đáng kể. Riêng ngành toán, bạn đã đào tạo cho ta khoảng hơn 200 TS và TSKH, phần lớn trong số đó vẫn đang là những cán bộ khoa học chủ chốt tại các viện nghiên cứu và các trường đại học trong cả nước. Hiện nay, ở các nước như Syri, Liban... để đào tạo được một TS toán học tại Anh phải chi khoảng 60.000 USD. Như vậy, để có được 200 người đào tạo ở nước ngoài phải chi khoảng 12 triệu USD.

Tuy nhiên, nền toán học của ta vẫn còn rất thấp kém so với các nước phát triển

Trong khoảng 15-20 năm trở lại đây, tức là từ khi đất nước chuyển sang cơ chế kinh tế thị trường, khoa học cơ bản nói chung và toán học nói riêng gặp nhiều khó khăn, xuống cấp rõ rệt và có nguy cơ suy thoái nghiêm trọng. Đội ngũ cán bộ giảng dạy và nghiên cứu toán học ở các trường đại học và các viện giảm thiểu về số lượng, lão hoá về tuổi đời và năng lực. Một bộ phận lớn cán bộ trình độ cao có độ tuổi 50 trở lên. Số người giỏi, có triển vọng ở độ tuổi 30-40 rất hiếm. Đã thế, những người này cũng không thể tập trung hoàn toàn thời gian, sức lực cho công việc chuyên môn vì phải lo kiếm sống bằng dạy thêm hoặc làm các việc khác. Một số có học hàm học vị, được đào tạo công phu đi ra nước ngoài kiếm việc làm, hoặc bỏ hẳn nghề. Một số thanh niên giỏi tốt nghiệp ở nước ngoài thì ở lại đó xin việc làm, chưa trở về nước. Trên phương diện quốc gia thì đó là hiện tượng chảy máu chất xám, là sự lãng phí nhân tài, là tổn thất nặng nề mà cuối cùng thì nền kinh tế sẽ phải gánh chịu hậu quả. Khoa học cơ bản nói chung và toán học nói riêng không còn đủ sức hấp dẫn đối với tuổi trẻ nữa. Số sinh viên theo học chuyên ngành toán không nhiều (đã có năm, chuyên ngành toán của Khoa Toán - Đại học Tổng hợp Hà Nội chỉ có 4 sinh viên ghi tên theo học!), cũng không hẳn là những người giỏi, vì phần lớn học sinh giỏi thi vào các ngành hứa hẹn cho họ một tương lai dễ dàng hơn. Sau khi tốt nghiệp ngành toán học, những sinh viên giỏi cũng không tha thiết ở lại trường giảng dạy. Phần lớn sinh viên cao học và nghiên cứu sinh toán hiện nay có tuổi đời cao và trình độ còn hạn chế. Do đó có rất ít cơ may đào tạo họ trở thành lực lượng thay thế cho những nhà toán học giỏi nhưng tuối mỗi ngày một cao.

Từ cuối năm 1997, với sự ra đời của Nghị quyết Trung ương 2 khoá VIII, với tinh thần trách nhiệm và sự năng động của các tập thể khoa học, tình hình có được cải thiện đôi chút, nhưng những khó khăn cơ bản vẫn còn đó. Nếu không sớm có được những biện pháp hữu hiệu thì chỉ mươi năm nữa chúng ta sẽ không có đủ cán bộ thực sự có trình độ để dạy toán từ phổ thông đến đại học chứ đừng nói gì đến nghiên cứu! Sự suy thoái về toán học sẽ kéo theo sự suy thoái của các ngành khác. Hậu quả xa hơn sẽ là: Chúng ta sẽ thiếu những chuyên gia giỏi để phát triển các công nghệ mũi nhọn, sẽ không có độc lập tự chủ về công nghệ, và do đó kinh tế không phát triển nhanh được. Như vậy cũng có nghĩa là, như ai đó đã nói, trong sự phân công quốc tế, con em chúng ta - một dân tộc nghìn năm văn hiến - đành phải chịu nhận những phần việc ít đòi hỏi trí tuệ nhất giữa một thời đại mà trí tuệ mới là sức mạnh chủ yếu.

Những điểm sáng hy vọng

Trong số 41 nhà khoa học được công nhận chức danh Giáo sư năm 2005, nhà toán học trẻ 34 tuổi Ngô Bảo Châu là một trường hợp đặc biệt. Anh được đặc cách phong chức danh Giáo sư mà không cần qua Phó giáo sư và cũng không câu nệ một số tiêu chí trong quy trình xét duyệt thông thường. Đây có thể được coi là một sự đột phá đáng khích lệ trong cách đánh giá và khuyến khích tài năng khoa học. Vấn đề đáng suy nghĩ là ở chỗ chúng ta cần tạo ra một môi trường khoa học thuận lợi ở ngay trong nước để từ đó có được không chỉ một Ngô Bảo Châu, bởi chúng ta đã từng có hàng trăm học sinh đoạt giải trong các kỳ thi Olympic toán học quốc tế kia mà! Phát hiện và đào tạo được các học sinh giỏi mới chỉ là bước đầu, cái chính là phải tạo được môi trường thuận lợi cho các mầm non ấy phát triển trở thành những tài năng thật sự.

Trên đây chỉ là một vài suy nghĩ bước đầu về thực trạng cộng đồng toán học đất nước. Hy vọng sẽ có dịp trở lại với vấn đề này trong tương lai gần.

Theo TC Hoạt động KH

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  mình thấy hữu ích nên cho nó sang đây

[color=blue]Sao bác không cho một cái link, có phải là đơn giản hơn không[/color]

đây.link đây,chúc mọi người vui vẻ và học tập tốt [url=http://diendantoanhoc.net/modules.php?name=News&file=article&sid=325]Click here[/url]

tiếp một bài hay nữa
“Hội chứng ễnh ương”
“Hội chứng ễnh ương” có ở rất nhiều lĩnh vực, nhiều loại người. Nào là tiếm phong, tiếm xưng. Nào là tạo dựng mặt bằng giả, uy tín giả. Kẻ mắc bệnh “ễnh ương” giống như chiếc thùng rỗng kêu to. Lắm khi lại như quả bóng bay nếu có thêm người ngoài tiếp hơi.
Một lần, trong một cuộc họp lớn của thành phố, người giới thiệu chương trình giới thiệu một tiến sĩ thuộc viện nghiên cứu... lên nói chuyện. Đối tượng nghe là các nhà nghiên cứu, các văn nghệ sĩ. Người nói chuyện hôm ấy cũng thật hóm hỉnh khi vui vẻ nói:

- Thưa các anh, các chị... Tôi là... là... thế này... Có chuyện không thực nhưng cũng là chuyện thực là sau một đêm ngủ dậy mình được chuyển đời từ Phó tiến sĩ sang Tiến sĩ.

Cả hội trường cười thoải mái. Cái hóm hỉnh của người nói xua đi cái mặc cảm về học vị anh đang mang.

Cũng là chuyện về học hàm, học vị. Một phó giáo sư, tiến sĩ nói với bạn cùng lứa với mình, cũng là người cùng ngành nữa nhưng chưa có bằng nọ cấp kia vì anh mới chỉ có "vỡ ruột" một tấm bằng đại học rằng:

- Tôi thì ông lạ gì. So trình độ có khi ông còn hơn tôi về khoản ngoại ngữ. Tôi phải cố chạy chỗ nọ chỗ kia mới lọt đấy. Viện tôi ai cũng phải tìm cách lo cho có được cái ấy để vào diện quy hoạch. Bây giờ thì yên tâm rồi.

Người tâm sự với bạn là người chân thành không biết giấu cái đáng giấu của mình. Mà có giấu cũng chẳng giấu được bởi bây giờ những chuyện học giả bằng thật cũng đang là điều khá phổ biến...

Nước mình hôm nay nhiều người thực học, thực tài. Cũng không thiếu người học ảo, tài ảo. Người thực học, thực tài thường làm nhiều nói ít. Người học ảo, tài ảo thường làm ít nói nhiều. Kẻ đại ngôn, ngoa ngôn không bị ai đóng thuế, không phạm vào luật trật tự xã hội nên nhiều khi hành tung, hành tác ngôn ngữ như một thứ nhiễu tinh thần. Việc không nguy hiểm đến tính mạng ai nhưng lại gây khó chịu cho người được chứng kiến. Bệnh này, các nhà chuyên môn đang tìm tên gọi nhưng dân gian hiện đại thì đã chỉ danh từ lâu: Đó là “hội chứng ễnh ương”.

“Hội chứng ễnh ương” này có ở rất nhiều lĩnh vực, nhiều loại người. Nào là tiếm phong, tiếm xưng. Nào là tạo dựng mặt bằng giả, uy tín giả. Kẻ mắc bệnh “ễnh ương” giống như chiếc thùng rỗng kêu to. Lắm khi lại như quả bóng bay nếu có thêm người ngoài tiếp hơi. Họ “ễnh ương” mình bằng mọi cách. Đòn bẩy tốt nhất của cách ấy là tiền. Cố đại ca tội đồ Năm Cam từng nói: "Cái gì không mua được bằng tiền sẽ mua bằng nhiều tiền". Và trong trường hợp này tiền cũng luôn luôn là cách thứ nhất để thực thi hiệu nghiệm nhất ý đồ của kẻ “ễnh ương”.

Một buổi biểu diễn ca nhạc vào loại nhàng nhàng vậy mà những M.C được coi là xinh gái đẹp trai, duyên dáng và hoạt ngôn nhất đã sử dụng rất nhiều, rất nhiều những từ bóng bẩy: "Thật là tuyệt vời, rất thích, khá đặc biệt, thật là xuất sắc, hát rất hay... Rồi thì quá hoàn hảo, xứng đáng cái nọ cái kia...". Quả tình không có từ nào hay hơn các từ này nữa. Chỉ tiếc là người hát mới độ thanh sắc bậc thường thường nhưng người giới thiệu đã "ễnh ương" họ lên mức siêu này siêu nọ. Trong nhóa nhòa của màu sắc đêm nhạc họ tụng ca nhau hết lời mà không một lần thấy ngượng. Chỉ có người nghe và người xem lắm lúc chỉ muốn bịt tai lại vì có quá nhiều những câu xưng tụng nhau như thế.

Một số cây bút trong một đôi năm đôn đáo in hết tập nọ, tập kia với giấy đẹp cùng bìa lụa, bìa cứng. Đầu quyển sách thường có lời giới thiệu rất chung chung của một vài tên tuổi. Cuối quyển sách thường là các lời nhận xét này nọ của cánh hẩu. Ai cũng biết, để có những cuốn sách loại này, trước đó là những cuộc gặp gỡ, những bữa nhậu, những phong bì đựng kim ngân lót tay. Từ đây “hội chứng ễnh ương” xuất hiện. Rượu vào lời ra. Nào họ sẽ là những tác giả này kia của đất nước. Nào là họ sánh với ông ấy, bà ấy của nhân loại. Cảm tưởng như quanh bàn bia hơi hội tụ toàn những vĩ nhân. Tiền nhập túi chữ lên giấy. Ngôn ngữ lúc ấy có cánh.

Lời khen nhau cứ choang choang. Nào là điệu nghệ, giàu xúc cảm. Nào là tài hoa trong tu từ. Rồi thì thành thạo trong ngôn ngữ dân tộc nhưng cũng rất hiện đại và nhiều ý vị sâu xa tầm nhân loại. Họ xung phong tình nguyện tổ chức các cuộc giao lưu, đi lại. Bạn bè quen mặt thì nhiều, quen văn chương thì ít. Rồi chễm chệ một ngày nào đó, người ấy thành một nhà… có chứng chỉ. Và sau đó nhiều người kể, đi đâu cũng vậy, anh ta hay mặc một cái áo trắng, có túi ngực, trong túi lấp ló "tấm thẻ nhà" của mình. Dạng này thuộc diện mua danh. Anh ta tự "ễnh ương" mình. Thật buồn là lại có những "ễnh ương" đồng đội hùa theo, tâng lên để thành một nhóm "ễnh ương" siêu hạng.

Xưa, ta có nơi thi nói phét, nói khoác. Có hẳn những làng cười hẳn hoi. Tục này là tục dân gian nhằm giải trí thư giãn của người nông dân sau lúc vất vả ruộng đồng. Nói phét, nói khoác cho vui mà chẳng chết ai. Làm gì có chuyện quả bí to bằng cái đình, vậy mà vẫn có người hả hê nghe. Làm gì có con cá quả chắn ngang đường làng khiến cái xe cút kít chạy qua bị đổ, vậy mà vẫn có người khúc khích cười…

Bây giờ, một số người có vẻ như hơn hẳn ông bà. Nói phét có mục đích. Nói khoác có cộng hưởng. Mọi cái đều bài bản hẳn hoi chứ không dân dã cười cho vui như các cụ ngày xưa. Đã có những tiến sĩ giấy ra tòa, về vườn. Có những siêu sao sập tiệm và lộ chân tướng. Cái thực dụng của không thực tài đã bị thực tế phơi trắng bụng như đô yếu trên sàn vật.

Trời mưa rào, trên các ao bèo, ruộng muống thường vang tiếng ễnh ương, ếch nhái. Mọi con đều kêu. Mỗi ễnh ương là kêu to nhất. Nhưng cũng chính nó là con vật xương xẩu, ít thịt và kém chất nhất.

trời , chả lẽ học vị VNam có giá trị thế thôi sao ?
  ma thế thì xã hội loạn mất ,ai trả lời được có bao nhiêu tiến sĩ ,giáo sư thật không ?
                   linkgreencold  bạn nghĩ sao ? ???????????

Thật ra thì các bác cho em ý kiến ạ!!!Tại việt nam còn hiện tượng mua bán bằng giáo sư,tiến sỹ như đi chợ á!!!cho nên lúc cần sử dụng thì mới lòi ra nhiều ''ễnh ương'' đến như vậy!!!nhưng cũng có những người xứng đáng chứ ạ,điển hình là tiến sỹ-thầy Lê Bá Khánh Trình trường em đây nè!!!(khoe mà :D  :blush: ngượng wá!!)

[quote="analysis"][quote="tranthanhnam"]Năm 1966, khi đất nước ta đang còn chiến tranh ác liệt, Hội Toán học Việt Nam đã được [b]Chính phủ[/b] chính thức ra quyết định thành lập. Từ đó đến nay, Hội đã trải qua một chặng đường 40 năm - một quãng thời gian không nhỏ trong sự phát triển đầy biến động của đất nước.[/quote]
cho hỏi chính phủ nào vậy?  :?:[/quote]
  tat nhien la chinh phu Vietr Nam roi! hoi la
     ( xin loi unikey tui bi hu )

hiệp định Genève, ký vào năm 1954, chia việt nam ra làm 2 miền qua vĩ tuyến 17. hội toán học việt nam được thành lập năm 1966. tôi muốn hỏi chính phủ miền nam hay chính phủ miền bắc ra quyết định thành lập?

oh , ta nhien la chinh phu Vietnam dan chu cong hoa o mien bac roi ,  boi vi luc do mien Nam chien tranh dang ac liet ma` . Song du sao toan hoc thi khong co ranh gioi cho du chien tranh hay khong thi nhung nguoi yeu toan luon bo qua moi mac cam ve chinh tri de phat trien nen toan hoc Vietnam .
Ma trong tinh hinh mien Nam chien tranh thi cung co nhieu nha toan hoc trong Nam lam lam ! boi vi o do van co nhung nguoi chien si cong san

tranthanhnam, đây là vấn đề lịch sử. câu trả lời tôi muốn là sử liệu nói về vấn đề này chứ không phải ý kiến cá nhân hay phát ngôn bừa. không lẽ học sinh việt nam chỉ có thể nói tào lao như vậy thôi sao?  :rotfl:

theo tôi nghĩ, hội toán học việt nam rất có thể được chính phủ miền nam việt nam (việt nam cộng hòa) thành lập. tại sao? vì miền bắc lúc đó nghèo đói và lo củng cố quân sự. trong khi sài gòn miền nam có thể nói là hòn ngọc viễn đông vượt hẳn thái lan. chương trình giáo dục miền nam lúc đó đã bỏ xa miền bắc...

dù sao đây chỉ là ý kiến cá nhân của tôi, cho nên tôi cũng cần một tài liệu sử học chính xác đáng tin cậy

khó kiếm tài liệu nào như vậy lắm.mà [color=red]analysis[/color] à,chuyện chính phủ nào không quan trọng trong bài viết này mà :),đừng đả động đến politic ở đây nữa,được không.
 Thân mến


Mathcamp is over.  :noo:  :noo:  :noo:  :( why doesn't AoPS have a crying emoticon? I love all ya'll, even if I didn't sign your yearbook, and wish we were still together.

What do ya'll remember most fondly (read: paraphrase your yearbook entry)? I think I enjoyed hiking (and late-night conversations) the most.

Billy.

Shortly after arriving home, I realized that Rule 1 was the thing I missed most. 5 minutes after starting school, I'm sure I'll start pining for Rule 2 as well.

That's a vague way of saying: The thing I'll miss most is all of you!

On a more specific note, I'll miss Mao a lot. I talked a little about making a web app to play Mao and am planning to register http://www.maogame.net very soon to hold open the option, although it can never be quite the same -- how could one implement the action of joking handing Josh the whole draw pile as a penalty for "being evil."

(Hugging you is somewhere down the list. :) )

-Jimmy (Koppel)

I will not paraphrase my yearbook entry since it was hastily written at a time when I was not in the mood to say much.

[quote="Darmani"]Shortly after arriving home, I realized that Rule 1 was the thing I missed most. 5 minutes after starting school, I'm sure I'll start pining for Rule 2 as well.[/quote]

Interesting - I miss rule four the most. I now have to constantly fight the urge to divide by zero. 5 minutes after starting school, I'm sure I'll start pining for rule three since I go to a school made of fire.

I know it seems strange, but I miss the [nearly] infinite supply of water in the water jugs. That water tasted really good - much better than tap water. Maybe I should have talked about how much I miss my friends, but that water was really the first thing that came to my mind.

To everyone whose yearbook I forgot to sign (which was actually everyone), I'm sorry. If you come back next year, and assuming that I do as well, I promise to sign your yearbook if you ask me to (yay for bribery).

Eric M.

After spending much of my time at home so far sleeping, I'm tempted to say the state of sleep-deprived euphoria I was in for the second half of camp, and especially the last two days of camp.

Besides that, I would have to say Mathcampers.  And math.  And games.  And IL, Puzzle Hunt, and writing LA:RPS.  And Math until we Die.  And . . . this is going to take too long.

We should perform an inversion with respect to Mathcamp, so that camp lasts 47 weeks, and the rest of the year is 5 weeks.


function
algorithm
algebra
domain
real analysis
real analysis unsolved

Let C be the set of all continuous functions from R to R. We break C in sets of 2 functions f and -f (f(x)=0 has no pair). Find an algorithm to choose exactly a function from each such pair WITHOUT using the axiom of choice.

Looks strange and non-standard, doesn't it?

There are many equivalent statements of the Axiom of Choice. The following is one of them:

For any set X there is a function f, with domain X\(0), so that f(x) is a member of x for every nonempty x in X.

Guys.. I'm afraid I forgot the solution to this one :?

It is harder to write it formally, but my idea is something like this:
Choose all functions that have the 'first' nonzero value positive for x>=0.
For the functions that are 0 for all x>=0 choose the ones that have the 'last' nonzero value positive. :D  :D
If it is not clear what I mean I will come back with a better explanation.


[quote="zmli"]It is proven that the celebration of birthdays is healthy. Statistics show that those people who celebrate the most birthdays become the oldest.

- S. den Hartog, Ph D. Thesis Universtity of Groningen.

Happy birthday, JLI.[/quote]

Hey, great!!!, do you think it is possible to get some more information about that thesis. I mean, I always hear about someone who says that in someone else's thesis there is a thing like that, but I have never read something like that myself directly from an article or from a thesis.

[quote="zmli"]It is proven that the celebration of birthdays is healthy. Statistics show that those people who celebrate the most birthdays become the oldest.

- S. den Hartog, Ph D. Thesis Universtity of Groningen.

Happy birthday, JLI.[/quote]
but maybe its people who celebrate birthdays are richer (or have more of some quality) and therefore their lives woould be longer?
for example, rich people usually celebrate their sons/daughters birthdays and a poor bum on the street cannot celebrate their birthday

[quote="zmli"]It is proven that the celebration of birthdays is healthy. Statistics show that those people who celebrate the most birthdays become the oldest.

- S. den Hartog, Ph D. Thesis Universtity of Groningen.

Happy birthday, JLI.[/quote]

Lol, that's hilarious. i wonder how he figured it out.

[quote=zmli]It is proven that the celebration of birthdays is healthy. Statistics show that those people who celebrate the most birthdays become the oldest.

- S. den Hartog, Ph D. Thesis Universtity of Groningen.

Happy birthday, JLI.[/quote]

Ah yes, the floor here is clearly made out of floor...


Let $ (x_n)_{n\geq1}$ and $ (y_n)_{n\geq1}$ two sequences so that $ \lim_{n\to\infty}\frac{x_n}{n^3}\equal{}4$ and $ \lim_{n\to\infty}\frac{y_n}{n^4}\equal{}3$. Find $ l\equal{}\frac{(x_1^2\plus{}x_2^2\plus{}...\plus{}x_n^2)(y_1^2\plus{}y_2^2\plus{}...\plus{}y_n^2)}{(x_1y_1\plus{}x_2y_2\plus{}...\plus{}x_ny_n)^2}$

I assume you mean $ l\equal{}\lim_{n\rightarrow \infty}\frac{(x_1^2\plus{}x_2^2\plus{}...\plus{}x_n^2)(y_1^2\plus{}y_2^2\plus{}...\plus{}y_n^2)}{(x_1y_1\plus{}x_2y_2\plus{}...\plus{}x_ny_n)^2}$.

[hide="Solution"]
We divide the second limit by the first to get that $ \lim_{n\rightarrow \infty}\frac{nx_n}{y_n}\equal{}\frac{4}{3}$; hence as $ n\rightarrow \infty$, $ \frac{x_n}{y_n}$ becomes arbitrarily small. Thus, by the Cauchy-Schwarz inequality,
$ l\equal{}\lim_{n\rightarrow \infty}\frac{(x_1^2\plus{}x_2^2\plus{}...\plus{}x_n^2)(y_1^2\plus{}y_2^2\plus{}...\plus{}y_n^2)}{(x_1y_1\plus{}x_2y_2\plus{}...\plus{}x_ny_n)^2}\equal{}\boxed{\infty}$.[/hide]

For simplicity, I'll use $ l \equal{} \frac {(x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2)(y_1^2 \plus{} y_2^2 \plus{} ... \plus{} y_n^2)}{(x_1y_1 \plus{} x_2y_2 \plus{} ... \plus{} x_ny_n)^2}$, find $ \lim_{n\rightarrow\infty}l$.

[hide="Simpler"]
We are given $ x_n\sim n^3, y_n\sim n^4$. Thus, $ l\sim \frac {(\sum_{k \equal{} 1}^nk^3)(\sum_{k \equal{} 1}^nk^4)}{\sum_{k \equal{} 1}^nk^7}$. Without even bothering to simplify, we can tell the top is of degree 9, while the bottom is of degree 8, so $ \lim_{n\rightarrow\infty}l \equal{} \infty$.[/hide]


Sixty- four unit cubes are placed together to create a large cube. How many cubes with integer dimensions are in the 4 by 4 by 4?

You should post all of your questions in the same topic, and show what you tried :wink:

1x1x1 cubes: 64 of them
2x2x2 cubes: 27 of them
3x3x3 cubes: 8 of them
4x4x4 cubes: 1 of them

64+27+8+1=$ \boxed{100}$.

$ 4^3\plus{}3^3\plus{}2^3\plus{}1^3\equal{}100$. To arrive at this, we use combinatorics.

also, a good formula for finding the sum of n consecutive cubes is $ {{n^2}}$*${ (n + 1)^2}$${ /4}$


quadratics
modular arithmetic
number theory unsolved
number theory

Suppose that $a$ is a quadratic residue modulo $p$, and suppose further that $p\equiv 5 \pmod{8}$.

Now then, show that one of the values: $x=a^{\frac{p+3}{8}}$ or $x=2a\cdot(4a)^{\frac{p-5}{8}}$ is a solution to the congruence $x^{2}\equiv a \pmod{m}$.

Perhaps you should post this in the number theory section.

Did you, by $m$, mean $p$?  If so, then it's a simple problem.  The desired statement is equivalent to proving that

either $a^{\frac{p-1}{4}}\equiv 1\mod{p}$, or $(4a)^{\frac{p-1}{4}}\equiv 1\mod{p}$.  Because $2$ is a quadratic nonresidue mod $p$, $4$ is a quadratic residue, but a quartic nonresidue mod $p$.  Then exactly one of $a$ and $4a$ is a quartic residue mod $p$, which is equivalent to stating that exactly one of the above statements holds.

yes, i think i've made a typo there... i've meant to say $p$ instead of $m$.


geometry
incenter
advanced fields
advanced fields open

Hi,

I have a computer program that ends up defining three lines that intersect to form a triangle. I am now trying to find the point within this arbitary tringle which minimises the perpendicular distance from each of the three lines which make up the triangle. Is there a sensible way to do this without just using a Monte Carlo method of trial and error for random points?

Thanks for your help

James Atkinson

What about those distances are you trying to minimize? If it's the sum, one of the vertices is always minimal.

I would simply construct the perpendicular bisectors of the three sides, which will be concurrent.  I believe the intersection would be the point which gives the smallest sum of distances to the three sides of the triangle.  I will consider a proof.

I suspect that what you really meant was that if you take the incenter (the intersection of the angle bisectors), than it minimizes the maximum of the 3 distances. ;) Anyway, it isn't clear to me what exactly is the functional we have to minimize :?.


Hey,

I'm just starting to study for chem olympaids.
I ahve a few questions:

1) How hard is USAChO, compared to math oylmpiads?
2) What books should I study?
3) Any other tips?

Thanks

Hello, I've got gold medal from 41st International Chemistry Olympiad and Silver Medal from International Mathematics Olympiad the year before.Do you want me to give you suggestions?
1) USAChO is too easy compared to Math Olympiad, Math is much more complex than chemistry and if you think to compare them , you gonna on the wrong way. By the way, I think USAChO problems are verily absolutely very easy and scarcely is there hard problems, No way!.I don't wanna look down on this competition, just want you to know that it isn't too hard for you. There isn't any complexity in particular. And the most interesting part is just the last parts.

2) I think to begin, since you were born, you must learn A-Z and then know how to count a number.(Kidding!) In fact, I want to tell you that Chemistry has stayed with you since you were young. You may start with any books , if only paid an attention to it a lot. Just carry out lots of problems, gather the experience from any friends and then you are better and ready to go on olympic route.

3)Other tips. I give you 5 dollars.For your attention to Olympics.
P.S. 5 dollars but priceless because it is your golden chance.

If you are a gold medalist then can you give some advice?

I am not good at practical task. But I did theoretical task very well though not as well as China team(got 56.75 from 60 in 2007 when the exam was the most difficult).

First of all, you should study much more advance than in general chemistry. General chemistry is just fundamental but you have to figure it out properly. There are some branches such as Physical Chem ,Analytical Chem, Biochem,and Organic chem which you have to learn it in basic. In fact, the olympiad require you all to meet each other and exchange your kniwledge and experience,rather than to get a medal.

In competition, in practical task, you have to plan efficiently in doing this task because the time is limited.You have to carry out the experiment and learn some important technique most frequently as you can. And the decision is the most important thing.

In theoretical task, you have to learn actively and critically think. The exam doesn't like in the university,it is much more tricky. You have to draw a good conclusion or result from only little source of goven data.

If you want to do the exam, enter the website of ICHO. search it in google

                 Best wishes


probability
function
calculus
integration
algebra
polynomial
complex numbers

I'm really uncertain as to where to ask this kind of thing, but when I took my Statistics course in college I became interested as to if some statistics notions would become applicable with complex-valued variables.

Here are a few suspicions I have concerning the field of complex-valued statistics:

1. All averages (be it geometrical, arithmetical or harmonical) would keep their real-valued definitions

2. Standard deviations and their squares, assuming we keep their real-valued definitions, would be complex numbers

3. Z-scores would be complex-valued as well

4. Medians or other ranking methods would work only if all values were made into their absolute values

If any of you know more about complex-valued statistics than I do, please tell me so.

A few comments:

1. This doesn't seem to belong in The Round Table. I assume JBL will figure out a place to send it  - maybe the probability forum.

2. A probability space is simply a measure space (of total mass 1), a random variable is simply a (measurable) function defined on that space, and an expectation is simply an integral. Integration theory has no problem dealing with complex valued functions; this is familiar ground there.

3. It would probably be best not to let standard deviation be complex valued. The idea here is that covariance is analogous to an inner product. I'd define $ \text{Cov}\,(X,Y)=E\left[(X-E[X])\overline{(Y-E[Y])}\right]$ from which $ \text{Var}\,(X)=\text{\Cov}\,(X,X)=E\left[|X-E(X)|^2\right],$ which is nonnegative.

4. On the other hand, there is a well-developed theory of $ \mathbb{R}^n$-valued random variables; I'm not sure I see much obvious motivation for $ \mathbb{C}$-valued random variables that gets us anything we don't already have with $ \mathbb{R}^2$-valued random variables.

nice post kent.

in sort of the same direction, if you wanted to have random variables taking values in the complex plane "just so that they would be there", it would just be the same as R^2, because in terms of measurable structure, they're exactly the same.  so you would have to be interested in some interaction between probability/measure theory and a special property of the complex numbers, like analyticity or algebraic closedness or something.

off the top of my head one example, could be the relationship between complex valued brownian motion and analytic and harmonic functions, and there are some basic complex analysis results that can be proven with brownian motion.

in terms of algebraic closedness, you could consider some sort of problem where you randomly chose complex coefficients of some polynomial and wanted to investigate the distribution of its roots.


Given that $a+b\leq c+1$, and $b+c \leq a+1$, and $c+a \leq b+1$
For the numbers $a,b,c \geq 0$

Prove that $a^2+b^2+c^2\leq 2abc+1$

Have fun!  :D

[hide]Well, it's easy to find $0 \le a+b+c \le 3$...[/hide]

I'd put this in Pre-Olympiad, Jan.

I think you should look at the Pre-Olympiad forum as well, now and then. I am sure there are lots of problems there you can tackle already. Remember: the sky is the limit!

[quote]I'd put this in Pre-Olympiad, Jan. [/quote]
Ok, I put it there.

Since nobody posted a solution, here's something that may help:

Prove that $a,b,c \leq 1$

$c \leq a-b+1$ and $c \geq a+b-1$
So $b\leq 1$

Analogue for $a$ and $c$

You can also show that $a,b,c\leq 1$ by contradiction.
WLOG $a\leq b\leq c$.  If at least one of $a,b,c$ is greater than 1 it would be $c$.  However
\[b+c\leq a+1<a+c\]
\[b<a\]
contradiction.
So all are less than or equal to one.  As for the original inequality add $2ab+2bc+2ca$ to both sides
\[(a+b+c)^2\leq 2abc+2ab+2bc+2ca+1\]
Now add $2(a+b+c)+1$ to both sides
\[(a+b+c+1)^2\leq 2(a+1)(b+1)(c+1)\]
Now try it from here

I am Oni from Bangladesh. I think I have proved that a²+b²+c²≤2abc+1.

             a+b≤c+1
          →a+b-c≤1   ……….. (1)
Similarly it can be shown that
              b+c-a≤1 ………... (2)
and         c+a-b≤1 ………. (3)
equation(1)+ equation(3)
            =a+b-c+c+a-b≤2
          →2a≤2
          →a≤1 …………… (4)
Similarly it can be shown that
             b≤1 ……………… (5)
and        c≤1 …………. (6)
equation (4)+ equation(5) +equation(6)
           =a+b+c≤3
If a=0, b=0 and c=0 then
                     a²+b²+c²=0
and               2abc+1=0+1=1
So                a²+b²+c²<2abc+1
If any two of the a, b and c equal to 0 and the other one is not equal to 0 then
                    a²+b²+c²≤1
and              2abc+1=0+1=1
then             a²+b²+c²≤2abc+1
if any of the a, b and c equal to 0 and the other two is not equal to 0 then
then the sum of the other two must be lesser than or equal to 0
So              a²+b²+c²≤1
and             2abc+1=0+1=1
Therefore  a²+b²+c²≤2abc+1
Then
As               a≤1, b≤1 and c≤1
So               abc ≤ 1, ab≤, bc≤1, ca≤1

 Now          ½(a) < abc if and only if bc>½ and a≠0
Otherwise, ½(a) ≥abc
So               ½(a+b+c) > abc

Suppose      bc>½ and a≠0
[Note:         A positive number (Such as y) can be expressed as the multiplication of any
                   of the two positive numbers (Such as b and c). And if we add these two
                   positive numbers (b and c) we will find different results depending on the
                   value of these two numbers (b and c). It can be proved that we will find the
                   lowest value of (b +c) if b=√y and c=√y]

 So              b+c>√ (1/2) +√ (1/2)
               →b+c> (1/√2) + (1/√2)

 Therefore   ½(b+c)> ((1/√2) + (1√2))/2
 So              ½(b+c)> (0.707+0.707)/2
Therefore    ½(b+c)>0.707
Now if         ½(a)> (1-0.707)
Then            ½(a+b+c)>0.707+ (1-0.707)
Or                ½(a+b+c)>1
So                ½(a+b+c)>abc
So Suppose  ½(a) ≤ (1-0.707)
Or                a≤2(0.293)
So                 a≤0.586
So the minimum value of ½(a+b+c)>0.707
 And as a≤ 0.586 then the maximum value of abc=0.586
So considering all the probability we find that always ½(a+b+c)>abc
So (a+b+c)>2abc
Similarly we can prove that the difference between (a+b+c) and (2abc) is always lesser than or equal to 1
Therfore    a+b+c-2abc≤1
So             a+b+c ≤1+2abc
As             a+b+c≥a²+b²+c²
So             a²+b²+c²≤2abc+1
                                               [Proved]


Πρέπει α>0 για να είναι καλά ορισμένη
Αν α<2 συγκλίνει στην θετική ρίζα της $ x^2\minus{}x\minus{}a\equal{}0$
Αν α=2 είναι σταθερή
Αν α>2 συγκλίνει στο $ \plus{}\infty$

[quote="r_boris"]Πρέπει α>0 για να είναι καλά ορισμένη
Αν α<2 συγκλίνει στην θετική ρίζα της $ x^2 - x - a = 0$
Αν α=2 είναι σταθερή
Αν α>2 συγκλίνει στο $ + \infty$[/quote]

boris εβαλα α=6 και βρηκα οτι η ακολουθια¨.

$ x_{1}=6$

$ x{n}=\sqrt{ 6 + x_{n-1}}$ εχει οριο 3 ,εκτος και αν κανω λαθος καπου

Επισης για , a<-1/4 , εχει η ακολουθια οριο στους μιγαδικους??

[quote="Durandal"][quote="τρικλινο"]Διδεται η ακολουθια¨.
.............$ x_{1} \equal{} a$..............aεR.....................
.............$ x _{n} \equal{} \sqrt { a \plus{} x_{n \minus{} 1}}$..............$ n\geq 2$.
Να μελετηθη ως προς την συγκλιση της.[/quote]

Υποθέτω εννοείς $ a\geq 0$ έτσι?
[hide]Μελέτησε τη μονοτονία της $ a_n$ και λύσε την εξίσωση $ x \equal{} \sqrt {a \plus{} x}$.[/hide]

Cheerio,

Durandal 1707

PS: Σου είναι απολύτως απαραίτητες οι τελίτσες που βάζεις? Δυσκολεύουν κάπως το διάβασμα πρέπει να ομολογήσω........................................................[/quote]

sorry Durandal ,κακη συνηθεια δευτερα φυσις (εννοω για τις τελιτσες)

Γιατι για α=0 η ακολουθια δινει οριο 0 ενω η εξισωση $ x\equal{}\sqrt{x}$ δινει δυο λυσεις αρα η ακολουθια δεν εχει οριο

Μάλλον όχι
Για α=0 η ακολουθία είναι σταθερή
Αν θέλεις κατι γενικότερο για τις $ a_{n\plus{}!}\equal{}F(a_n)$  δες το attachment

[quote="r_boris"]Μάλλον όχι
Για α=0 η ακολουθία είναι σταθερή
Αν θέλεις κατι γενικότερο για τις $ a_{n \plus{} !} \equal{} F(a_n)$  δες το attachment[/quote]

Καταπληκτική δουλειά rudy!  :first:

Cheerio,

Durandal 1707


I came across a test that is multiple choice, not that hard but it didn't have an answer key.  So I decided to post it here considering its level of difficulty.  There are 5 pages but I'm posting one at a time so one person isnt talking about the last question while the other the first.  I'll post the second page once the first page has been complete.  Dont post a comment on a page we have already finsihed please.  Just post the letters of the answer.

Well if there is no answer key and you dont have any specific questions, then dont post this because we would then have to do the whole test when all u need are a few questions answered. So if u werent sure about a specific number or problem then ask that.

Also these questions arent from mathcounts nor do I think that they are MATHCOUNTS level so im kind of questioning whether they even belong here.

@ ajain, it sounds like you want us to do a whole test for you, without even explaining the answers...   :huh:

and also, i think most of those questions are way too easy for mathcounts-level.  try the classroom-math forum.

uhh...

c
c
d
a
e?
c
d

it would be really sad if i missed any...

I can answer all but one of them with 100% certainty, im not sure about number 5 is it d or e?, but everything else i agree with dynamo.

Thats what I got.  Here's teh second page with my answers:

8. E
9. A
10. D
11. B
12. A
13. A
14. B
15. E
16. C

Thanks!  Here is the third page:

17. E
18. B
19. D
20. D

[hide="17"]
The area of the circle is $ \pi \cdot 2^2 \equal{} 4\pi$. The area of the rectangle is $ 6 \cdot 14 \equal{} 84$. Therefore, $ \frac {4\pi}{84} \approx \boxed{0.150\ \text{(B)}}$.

[b]Edit:[/b] Somehow, I put $ 1.50$ instead of $ 0.150$! Oh well, now it's fixed :D
[/hide]

[quote="Yongyi781"][hide="17"]
The area of the circle is $ \pi \cdot 2^2 \equal{} 4\pi$. The area of the rectangle is $ 6 \cdot 14 \equal{} 84$. Therefore, $ \frac {4\pi}{84} \approx \boxed{1.50\ \text{(B)}}$.
[/hide][/quote]

Thanks for pointing my mistake out, I put in 4/84 im my calc and totally forgot teh pi sign!!!!  Anyone, no one did 18, but im positive is B!! :) ill go ahead and post the next page:

21. B
22. D
23. B
24. D
25. B
26. C
27. C
28. E
29. D-----not sure, i dont know what "outlier" means?/