Hi,

I'll be graduating as the valedictorian of my class in high school this year (at least, I am at this point and hope to keep it) so I have to make a speech. But even after I've looked into Youtube, I cannot seem to find how will I be able to write a speech. I have limit of 2-3 minutes (this is asked by many faculty so I'll probably have to keep it) so it has to be pretty short. I want to be bit funny yet has a strong message as well. How should I go about this?

Please help, any graduates or past valedictorians!

P.S. I put "please don't put crap" because recently, I've noticed some spams and yeah, don't put spam on this thread because this is really serious for me and if I see a crappy spam, I'm immediately contacting an admin (I'm not joking; I'm sensitive near the date of graduation so please don't do that).  :)

I've never given a valedictorian speech, but I have done a [i]lot[/i] of presentations, so I can help you a little with the mechanics of planning a talk.

Your speech should only have one major point.  One of my advisers told me, "In a 10-12 minute talk, you really can't drive home more than [i]one[/i] major point!  A 20-25 minute talk may permit discussion of two major points, but no more."  This has pretty much been my experience.  I tried to make three points in a 15 minute talk, and it ran to 20 minutes; five minutes is barely enough for one major point.  So, if you have two or three minutes, you should only try to make one point.

Two to three minutes is really very short, but of course it won't seem that way when you first step up to the podium!  Time runs in strange ways when you're new to public speaking.  Practice beforehand to make sure you won't exceed your allotted time.

I don't know whether you feel comfortable speaking in front of people.  If you're not, you might give that some practice, too.

I can't tell you what you should say, but whatever you do say should be a tad outrageous, and uniquely you.  Be controversial.  People only pay attention to and remember things they had strong feelings about at the time.  Be yourself.  If people like you, it's because you're you.  They want to hear what you have to say, not some general thing that anyone could have said.

Some examples might help:

When I graduated from college, there was a party for all the science students where each graduating senior was supposed to stand up and tell the younger students their best advice.  Like you, the time limit was 1-2 minutes per person.

In my speech, I told the stupid joke about the science nerd who finds a frog, and the frog promises him that if he'll kiss her, she'll turn into a princess.  The science nerd refuses to kiss her, because a girlfriend's great, but a talking frog, that's REALLY COOL!  The moral of the story:  if you ever think you're weird, look around now and see who's laughing, because you're in really good company here!

My friend, who is known for taking insane course loads, said that he was graduating with majors in math, physics, astronomy, computer science, geophysics, economics, and finance.  He also said that he'd usually take 22 hours a semester (a normal course load is 12), but one time he managed to take 27 by enrolling in two universities at once.  His advice?  "There's always time for one more class."

Someone told us later that ours were the only ones she would remember, and that we couldn't have said anything more perfectly suited to each of us.

Wow, this was long.  I hope it helps!

Be humorous and self-depricating. Nothing breaks the ice better than humor. Also talk about things that have happened over the high school years that everyone remembers. People are tied together by memories and common experiences. Avoid lecturing about the future or giving advice for the future. Teenagers get enough of that already.

Short anecdotes of funny situations are nice if you can draw a message out of them. While doing so, though, try not to come across as too cheesy, or too "canned". Also, make sure some parts of the speech very directly address your classmates.

i dunno if you feel confident enough to do so, but what i think would also help is, before the speech, sketching an outline of what you want to hit in your speech, and then ad-libbing it around those key points when you do your actual speaking. that way it'll seem more personal, more you, and you won't have to do that annoying thing where you look down every three or four seconds, lol.

[quote="Lazarus"]you won't have to do that annoying thing where you look down every three or four seconds[/quote]  The way to avoid having to read your notes is to have practiced your speech enough that you have much of it memorized.  Ad-libbed speeches are almost invariably worse than well-rehearsed ones.  Similarly, knowing in advance what you're going to say allows you to more properly change tone, emphasis, etc., based on the content of your words.

Thanks for all.

I guess since I'm getting such a good help, I'll describe my "classmates."

I'm graduating from this high school in rural area in Pennsylvania where majority of students engage in farming or work force after high school. Only about 30-40 students out 140 total go to college in my class. Sports is a big thing that is open to everyone but there are still some who did not join any sport. For instance, I know some of my good friends who play in rock band and go to shows and yet not really participate in any activity, except maybe the school band.

This is sort of reason why I'm having a hard time to pinpoint a topic. I was a college preparatory student playing three sports and doing lots of community services. I honestly do not know the details about the rock band, really sports guy who "live in the weighting room," and definitely, the "farming" people. Many boys are interested in 4-wheeling but I've never done that either. I don't know, it is hard for me because I can do my best job by talking about the school and classes but then, I'd be leaving out some kids who come to school "to see the girls."

How should I go about this? Like... as the valedictorian, my #1 goal is to reach out to everyone and add humor.

[quote="JBL"]Ad-libbed speeches are almost invariably worse than well-rehearsed ones.  Similarly, knowing in advance what you're going to say allows you to more properly change tone, emphasis, etc., based on the content of your words.[/quote]

Ah! I highly disagree. I've had to do this twice and I think a nice outline that gives you freedom to change according to audience and speak freely is always better. It is really hard to pay attention to a well-rehearsed speech, because it will sound well-rehearsed and not natural. If it sounds like you are having a conversation with the audience, then they will feel like they are in a conversation and listen.

Now, I've had to do this twice (once at high-school and once at my undergrad commencement) and they were both fully written out, but I strayed from what was written when it felt right. Do not do this if you don't feel comfortable, though. I give speeches all the time (I am a high school speech and debate coach and I coach the impromptu team at that), so it was probably easier for me than many. So the best advice I can give is sound natural!

My high school one was very un-original. It was the standard "do what you love" whatever. My undergrad one I felt much more strongly about being original. I actually posted it [url=http://hilbertthm90.wordpress.com/2008/05/16/commencementagain/]here[/url] . There are definitely two routes you can take. The safe, everyone will love it, people will not remember it route, or you can try to be a little more daring. Mine definitely isn't "daring," but it does break many conventions. I refused to use any of the buzzwords that float around these things.

You can avoid that by actually saying something you really mean to the point where you essentially have the speech memorized.

[quote="HilbertThm90"]It is really hard to pay attention to a well-rehearsed speech, because it will sound well-rehearsed and not natural.[/quote]

Natural speech sounds [i]terrible[/i] -- it's full of incomplete sentences, pauses, slurred words, "uh"s and "um"s -- there is nothing worse that listening to someone talk who isn't comfortable with what they're saying.  "Rehearsed" and "rote memorized line-by-line" are not necessarily the same, and a speaker is much more likely to be able to deviate successfully from a well-rehearsed speech than an under-rehearsed speech.

I have no content-pertinent advice for the O.P., unfortunately -- I think, Silverfalcon, that you'll probably have to come up with the content yourself, sad to say.

Like I said, if you are going to have uh's and um's and you are uncomfortable with doing a natural sounding speech, then don't do it. I explained that you should not do it impromptu. This is not the same as "extemporaneous" speaking to which I was referring. You should have an outline and have practiced it some. This does not mean it should be set in stone. Natural speech does not sound terrible. It gives the speech character. Polished is difficult to listen to.

Would you rather listen to the most polished, rigorously performed piece of music performed perfectly by a machine without emotion, or would you rather listen to someone improvise a little imperfectly, but full of emotion? Of course, these are extremes, but I prefer to err on the side of the imperfect but real sound.

regarding content

seems to me you're too concerned with making your speech fit in with your school community. Give a speech from yourself--you'll not only have more to say, it'll be more entertaining.

you can give a speech that appeals to people who are different from you without knowing all about them.

Thanks for all the advice.

I first talked about our past. Then about myself and my change. Then about how the class changed and what we learned. Then about how we'll be in future.

I thought I did a very solid job and everyone I talked to say the same thing. But thank you all for your advices. Without them, I'd been lost. Thank you.

Graduated from June 4th, 2008 as the valedictorian with GPA of 99.0/100.0.  :)


calculus
integration
logarithms
algebra
partial fractions
calculus computations

giving this a cursory glance,
couldn't we just use partial fractions?

Convert the integrand into partial fractions and then integrate:

$ \int_0^1 \frac{5}{(x\plus{}1)^2} \plus{} \frac{3}{x\plus{}1} \plus{} \frac{4x\plus{}7}{x^2\plus{}1} \ dx$

$ ( \minus{}\frac{5}{x\plus{}1} \plus{} 3 \ln (x \plus{} 1) \plus{} 2 \ln(x^2 \plus{} 1) \plus{} 7 \arctan(x) )|_0^1$

Which yields the result:

$ \frac{5}{2} \plus{} 5 \ln 2 \plus{} \frac{7}{4} \pi$

Is this correct?


Let $ S$ be denote the set of all n-tuples$ (n > 1)$ of real numbers $ (a_1;a_2;...;a_n)$ such that
$ 3\sum_{i \equal{} 1} ^n a_i^2 \equal{} 502$ (1)
a)Prove that $ \min_{1 \leq i \leq j \leq n} |a_j \minus{} a_i| \leq \sqrt {\frac {2008}{n(n^2 \minus{} 1)}}$ (2)
b)Give an example of an n-tuple $ (a_1;a_2;...;a_n)$ such that condition (1) holds and for which there is an equatily in (2)

Outline of the solution: Let $ min\{a_j\minus{}a_i\}\equal{}k$. Among many possible $ (a_1,a_2,\dots,a_n)$ satisfying $ (1)$ we shall choose the n tuple so that $ k$ is maximum and show that $ (2)$ holds.

Let $ a_1\equal{}0$, then $ a_2\equal{}k$, $ a_3\equal{}\minus{}k$, $ a_4\equal{}2k$ and so on so that all the $ a_i$'s are placed on the number line with the gap of exactly $ k$. Instead of working this method starting from $ a_1\equal{}0$ if we start from $ a_1\equal{}r,r\in \mathbb{R}$ we simply end up having
\[ k_1r^2\plus{} k_2k^2\plus{}2k_3rk\equal{}\frac{502}{3}\] which yields smaller value of $ k$ compared to $ k^2\sum (1^2\plus{}2^2\plus{}..)$.

Instead of placing all the $ a_i$'s at the gap of $ k$ if al least one of them is altered, it is easy to see that again it yields smaller $ k$.

From here it is straight forward. If $ n\equal{}2h\plus{}1$ we have,
\[ k^2\big (2(1^2\plus{}2^2\plus{}...\plus{}h^2)\plus{}(h\plus{}1)^2\big)\equal{}502/3\]
\[ 4\big(h(h\plus{}1)(2h\plus{}1)\plus{}3(h\plus{}1)^2\big)\geq 4h(h\plus{}1)(2h\plus{}1)\]
\[ \Rightarrow k\leq \sqrt{\frac{2008}{n(n^2\minus{}1)}}\]

Equality appears in the case when $ n$ is even and $ a_1\equal{}0$.

Why from $ 4\big(h(h \plus{} 1)(2h \plus{} 1) \plus{} 3(h \plus{} 1)^2\big)\geq 4h(h \plus{} 1)(2h \plus{} 1)$ we can have $ \Rightarrow k\leq \sqrt {\frac {2008}{n(n^2 \minus{} 1)}}$  ?

$ k^2\big (2(1^2\plus{}2^2\plus{}...\plus{}h^2)\plus{}(h\plus{}1)^2\big)\equal{}502/3$

$ \Rightarrow k^2\equal{} \frac{502}{4\big(h(h\plus{}1)(2h\plus{}1)\plus{}3(h\plus{}1)^2\big)}$

then

$ \Rightarrow k^2 \le \frac{502}{4h(h\plus{}1)(2h\plus{}1)}$

$ \Rightarrow k\leq \sqrt{\frac{2008}{n(n^2\minus{}1)}}$

Am I clear ?

Dont explain more,here is a problem from MYM in my country,see [hide="here"]http://www.mathlinks.ro/viewtopic.php?t=168836[/hide]
To mod lock it ,please
Toevery,đừng làm thế mất hay bạn à,


for a, you could use generating functions but that's too advanced for this forum :mad:

[quote="Bachukas"]a) in how many ways 1000 can be expressed as a sum of 17 positive integers, if each integer is smaller than 100?

b) In how many ways Santa Claus can distribute 2000 candies to 200 kids, such that each kid would have maximum 15 candies?[/quote]

Hint: Balls and urns??? Search it, and understand the concept.


Putnam
calculus
derivative
geometry
college contests

Find the minimum value of
\[(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2}\]
for $0<u<\sqrt{2}$ and $v>0$.

[hide="My Solution"]
Notice that the expression above is the square of the distance between the points $\left( u,\sqrt{2-u^{2}}\right)$ and $\left( v, \frac{9}{v}\right)$ and so the minimum value of it is achieved when the curves $y=\sqrt{2-x^{2}}$ and $y=\frac{9}{x}$ are closest to each other, as it may be deduced from the plot (or by inspection) this will occur along the line $y=x$ and hence when
\[u=\sqrt{2-u^{2}}\mbox{ and }v=\frac{9}{v}\]
that is for $(u,v)=(1,3)$ producing the minimum value of 4. Here's the plot:

[url=http://allyoucanupload.webshots.com/v/2003181487180586283][img]http://aycu13.webshots.com/image/3772/2003181487180586283_rs.jpg[/img][/url][/hide]

nice soln . i did it using partial derivatives, but your solution is much better .
@jmerry- i cant get the geometry part. can you help?

[quote="jmerry"]That's exactly the right method, but the [hide="answer is"]8[/hide]

If you don't want to see the whole solution at once, here's a [hide="hint"]Think geometrically.[/hide][/quote]

I get it... let $f(u,v)$ denote the expression that was to be minimized, then $f(1,3) = (1-3)^{2}+(1-3)^{2}= 2\cdot 2^{2}= 8$.

I really had just goofed on the arithmetic: :oops_sign: .

P.S. John Scholes has posted [url=http://www.kalva.demon.co.uk/putnam/psoln/psol848.html]his solution[/url] to this problem using the partial derivatives method, he also mantains this [url=http://www.kalva.demon.co.uk/putnam.html]Putnam problem archive[/url] with all Putnam problems from the $1^{st}$ Putnam competition in 1938 up to the 2003 competition.

P.S.$^{2}$ I wonder, would you guys look at the next problem I will post, namely [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=663995#663995]Putnam 1981 B2[/url], to see if the same type of geometric method can be applied in 4-D to solve Putnam 1981 problem B2.


The first 2007 positive integers are each written in base 3. How many of these base-3 representations are palindromes? (A palindrome is a number that reads the same forwards and backwards).

(A) 100
(B) 101
(C) 102
(D) 103
(E) 104

[hide]In general, we can have a 0, 1, or 2 as digit, with the first half digits being 1 or 2.

2007 in base 3 is 2202100.

If there is a single digit, there are 2 options:  1 or 2.

If there are two digits, there are 2 options:  1 or 2 (the other is set).

If there are three digits, there are $2*3=6$ numbers.

If there are four digits, there are 6 numbers.

If there are five digits, there are $2*3^{2}=18$ numbers.

If there are six digits, there are 18 numbers.

If there are seven digits and the first digit is 1, then there are $1*3^{3}=27$ numbers.

If there are seven digits and the first digit is 2 and the second digit is 0 or 1, there are $1*2*3^{2}=18$ numbers.

If there are seven digits and the first digit is 2 and the second digit is 2, then there are only 3 numbers.

The total is 100, or $A$.[/hide]

junggi already finished posting these problems in Intermediate.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135703


number theory
least common multiple
search

I wanted to post Prob 10 about a solar system from State 1999 exam - but I have the test on pdf file and can't figure out how to cut & paste from a pdf - problem is long, plus has important pictures, so I can't retype easily -- can someone tell me how to get a problem from pdf onto this site? Or if someone knows the problem about the solar system with 4 planets and a sun, can you explain solution? (and maybe post prob for others if you're able?) If it's easier to pm, me - feel free.

yeah, that was the problem (though you really simplified by removing all orbit nonsense) but WAIT - when aligning the planets in a line, you assumed that one planet could be in front of the sun and one behind?

Hehe, this one was a tricker, aaaand...if you click that handy-dandy search button and type in "planets," you'll get a few hundred threads with this problem :D.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=planets&t=74701[/url]

(I thought ^ was one of the better ones not only because I posted it  :D )

Ahh yes, I remember this problem from when I posted it and was told about how many times it has been posted  :D

[hide="anyway"]You just have to realize that the planets can be in line on both sides of the sun.[/hide]

Yes, that's an important realization, but I don't think the wording of the problem made that clear, did it? And if you thought it COULDN'T be like that, (as I initially thought) you'd get a different (apparently, wrong) answer! What in the problem may you assume the sun could be lined up in between?

the word "line" did. a line extends in either direction infinitely.

[quote="kheg_k"]the point is to find the LCM of 20, 36, 40, and 48 months.

then divide that by 12 and add to 2000 to get answer[/quote]

There is only one problem, I can't believe they will trick us like this, but one can have half orbits, so there.


Let I be the incenter of triangle ABC, and D,E, F the projections of I onto BC,CA,AB,respectively. The incircle of ABC meets the segments AI,BI,CI at M,N, P, respectively.Show that the Simson lines of any point on the incircle with respect to the
triangles DEF and MNP are perpendicular.

Let $s,t$ be the two Simson lines. According to Darij's post [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=636738&search_id=582266813#636738]here[/url], we have that $\measuredangle \left( s, FD \right) = \frac{\pi}2-\measuredangle XFE$ and that $\measuredangle \left( t, PM \right) = \frac{\pi}2-\measuredangle XPN$.

Hence,
\begin{eqnarray*}\measuredangle \left( s, t \right) &=& \measuredangle \left( s, FD \right)+\measuredangle \left( FD, PM \right)+\measuredangle \left( PM, t \right) \\ \ &=& \measuredangle XPN-\measuredangle XFE+\measuredangle \left( FD, PM \right) \\ \ &=& \measuredangle EFN+\measuredangle \left( FD, PM \right) .\end{eqnarray*}
It remains to prove that the last sum of angles is just $\frac{\pi}2$, but it's only angle chasing, so I'll omit it.


$3.$ Deconditionam inegalitatea:
Fie $a=xr\ b=yr\ c=zr$
Rezulta $r(x+y+z)=1$, deci $r=\frac{1}{x+y+z}$
Atunci $a=\frac{x}{x+y+z}$ $b=\frac{y}{x+y+z}$ $c=\frac{z}{x+y+z}$
Astfel, membrul stang al inegalitatii este:
$\sqrt{\frac{yz}{yz+x^{2}+xy+xz}}+\sqrt{\frac{xz}{yz+y^{2}+xy+xz}}+\sqrt{\frac{xy}{yz+z^{2}+xy+xz}}$
Avem de demonstrat ca $\sqrt{\frac{yz}{(x+z)(x+y)}}+\sqrt{\frac{xz}{(y+z)(x+y)}}+\sqrt{\frac{xy}{(x+z)(y+z)}}\leq \frac{3}{2}$
Folosind inegalitatea mediilor ($MG\leq\ MA$) pentru fiecare termen din membrul stang avem:
$\sqrt{\frac{yz}{(x+z)(x+y)}}\leq \frac{\frac{y}{x+y}+\frac{z}{x+z}}{2}$
$\sqrt{\frac{xz}{(y+z)(x+y)}}\leq \frac{\frac{x}{x+y}+\frac{z}{y+z}}{2}$
$\sqrt{\frac{xy}{(x+z)(z+y)}}\leq \frac{\frac{y}{z+y}+\frac{x}{x+z}}{2}$
Prin insumare obtinem concluzia.

o singura intrebare am pentru Mircea Lascu!!!Ai scris ca fie a,b,c >0 si dupa aceea ai zis ca abcd=1.Intrebarea mea este:pana la urma cate variabile are inegalitatea 3 sau 4??

la alea cu $abcd=1$ sunt patru variabile

urmeaza 2 brute force :)

[hide="solutie 1"]$\sum\frac{1}{a+3}\leq 1$.
aduc la acelasi numitor
$\sum_{a,b,c,d}\Bigg( (a+3)(b+3)(c+3) \Bigg) \leq (a+3)(b+3)(c+3)(d+3)$.
voi distinge sumele in $a,b,c$ de cele in $a,b,c,d$ prin ce scriu sub suma.
inegalitatea de mai sus este, dupa desfacerea parantezelor :
$LHS=\sum_{a,b,c,d}\Bigg ( abc+3\sum_{a,b,c}ab+9\sum_{a,b,c}a+27\Bigg)$.
de aici se poate exprima membrul stang doar in sume de $a,b,c,d$, deci voi elimina notatia de sub suma :
$LHS=\sum abc+6\sum ab+27\sum a+27\cdot 4$.
in membrul drept este evident ca toate sumele sunt in $a,b,c,d$.
$RHS=abcd+3\sum abc+9\sum ab+27\sum a+81$.
$LHS\leq RHS$
echivalent cu
$26\leq 3\sum ab+2\sum abc$.
evident, deoarece $abcd=1$, si apoi medii.
egalitate cand $a=b=c=d=1$.[/hide]

[hide="solutie 2"]pentru inceput, $1+a+a^{2}+a^{3}=(1+a)(1+a^{2})$.
voi aduce la acelasi numitor
inegalitatea va deveni :
$\sum \Bigg( (1+a)(1+b)(1+c)(1+a^{2})(1+b^{2})(1+c^{2}) \Bigg)\leq \prod (1+a)(1+a^{2}$.
pentru inceput, il voi ignora pe $d$ si voi calcula :
$Q=(1+a)(1+b)(1+c)(1+a^{2})(1+b^{2})(1+c^{2})$.
$Q=(1+\sum a+\sum ab+abc)(1+\sum a^{2}+\sum a^{2}b^{2}+a^{2}b^{2}c^{2})$.
parantezele se desfac frumos :)
$Q=1+\sum a+\sum ab+abc+\sum a^{2}+\sum a^{3}+\sum_{sym}a^{2}b+\sum_{sym}a^{3}b+\sum a^{2}bc+\sum a^{3}bc+\sum a^{2}b^{2}+\sum_{sym}a^{3}b^{2}+\sum a^{2}b^{2}c+\sum_{sym}a^{3}b^{2}c+\sum a^{3}b^{3}+\sum a^{3}b^{3}c+a^{2}b^{2}c^{2}+\sum a^{3}b^{2}c^{2}+\sum a^{3}b^{3}c^{2}+a^{3}b^{3}c^{3}$. (se pot numara toti termenii de tip $a^{x}b^{y}c^{z}$, ies chiar 64, deci pana acum e bine).
acum trebuie introdus si $d$.
sa zicem ca $Q$ de mai sus este $Q_{d}$.
atunci $LHS=\sum Q_{a}$.
sumele de mai sus sunt initial in $a,b,c$. dupa insumarea $Q$`urilor, ele vor fi in $a,b,c,d$, tot simetrice. important este coeficientul acestora, deoarece unii termeni pot sa apara in doua $Q$`uri diferite.
termenul $1$ apare in toate $Q$`urile, deci are coeficient $4$.
termenii intr`o singura variabila apar in trei $Q$`uri ($a$ apare in $Q_{b},Q_{c},Q_{d}$).
termenii in doua varaibile apar in doua $Q$`uri (cele cu indicii acele litere care nu denumesc cele doua variabile considerate).
termenii in trei variabile apar in un singur $Q$. (cel care are ca indice litera care denumeste cea de a patra variabila).
deci, coeficientul unei sume in trei variabile ramane $1$, coeficientul unei sume in doua variabile este $2$ si coeficientul unei sume in o variabila este $3$.
acestea fiind zise, se poate exprima membrul stang. toate sumele sunt simetrice, dar se iau in calcul doar permutarile relevante. (daca de exemplu avem $\sum a$, nu il numaram pe $a$ de patru ori, ci doar o data).
mai explicit $\sum a = a+b+c+d$, $\sum ab=ab+ac+ad+bc+bd+cd$, $\sum a^{2}b=a^{2}b+b^{2}a+a^{2}c+c^{2}a+\ldots$.
$LHS=4+3\sum a+2\sum ab+\sum abc+3\sum a^{2}+3\sum a^{3}+2\sum a^{2}b+2\sum a^{3}b+\sum_{a}^{2}bc+\sum a^{3}bc+2\sum a^{2}b^{2}+2\sum a^{3}b^{2}+\sum a^{2}b^{2}c+\sum a^{3}b^{2}c+2\sum a^{3}b^{3}+\sum a^{3}b^{3}c+\sum a^{2}b^{2}c^{2}+\sum a^{3}b^{2}c^{2}+\sum a^{3}b^{3}c^{2}+\sum a^{3}b^{3}c^{3}$.
se poate verifica din nou ca numarul de termeni este intr`adevar 256.
acesta fiind membrul stang, voi trece mai departe la membrul drept. sumele de acum sunt tot simetrice, cu aceeasi observatie, si toate sunt in $a,b,c,d$.
$RHS=\bigg( 2+\sum a+\sum ab+\sum abc\bigg)\bigg( 2+\sum a^{2}+\sum a^{2}b^{2}+\sum a^{2}b^{2}c^{2}\bigg)$.
si aici parantezele se desfac frumos :)
$RHS=4+2\sum a+2\sum ab+2\sum abc+2\sum a^{2}+\sum a^{3}+\sum a^{2}b+\sum a^{3}b+\sum a^{2}bc+\sum a^{3}bc+\sum a^{2}bcd+2\sum a^{2}b^{2}+\sum a^{3}b^{2}+\sum a^{2}b^{2}c+\sum a^{3}b^{3}+\sum a^{3}b^{2}c+\sum a^{2}b^{2}cd+\sum a^{3}b^{3}c+\sum a^{3}b^{2}cd+2\sum a^{2}b^{2}c^{2}+\sum a^{3}b^{2}c^{2}+\sum a^{2}b^{2}c^{2}d+\sum a^{3}b^{3}c^{2}+\sum a^{3}b^{2}c^{2}d+\sum a^{3}b^{3}c^{2}d+\sum a^{3}b^{3}c^{3}$.
acum, comparand $LHS$ cu $RHS$, foarte multe din sumele astea sunt aceleasi, deci se reduc. mai raman cateva :)
am redus membrii la :
$LHS=\sum a^{2}+2\sum a^{3}+\sum a^{3}b+\sum a^{3}b^{3}+\sum a^{3}b^{2}$.
$RHS=\sum a^{2}b^{2}c+\sum a^{2}b^{2}c^{2}+\sum a^{2}bc+\sum ab+2\sum abc$.
acum voi verifica : cand $a=b=c=d=1$, $LHS=42$ si $RHS=42$. deci este bine (adica n`am uitat nici o suma).
$LHS\geq RHS$ (*) o voi demonstra pas cu pas.
pentru inceput
$\sum a^{2}b^{2}c\leq \sum a^{3}b^{2}$. (muirhead)
$\sum a^{2}bc\leq \sum a^{3}b$. (muirhead)
$\sum abc\leq \sum a^{3}$. (muirhead)
(*) a devenit
$\sum a^{2}b^{2}c^{2}+\sum ab\leq \sum a^{2}+\sum a^{3}b^{3}$. (**)
insa $\sum a^{2}\geq \frac{2}{3}\sum ab$.
deci (**) devine :
$\sum a^{2}b^{2}c^{2}+\frac{1}{3}\sum ab\leq \sum a^{3}b^{3}$.
insa
$\frac{2}{3}\sum a^{3}b^{3}\geq \sum a^{2}b^{2}c^{2}$ (muirhead)
si
$\frac{1}{3}\sum a^{3}b^{3}\geq \frac{1}{3}\sum a^{2}b^{2}cd=\frac{1}{3}\sum ab$ (muirhead).
aceste ultime doua inegalitati le insumez, si obtinem chiar (***).
acest fapt inchieie "mica" mea demonstratie :)[/hide]

la problema 3 nu mai pun solutia ca e identica cu cea a lui Stormraven.

Sunt sigur ca post-ul de sub al meu va contine o solutie de cateva randuri a inegalitatii $(2)$, asa ca puteti sa scroll-down cu incredere. Partea buna a acestei solutii e ca o idee asemanatoare merge pentru a demonstra generalizarea inegalitatii $(1)$ (aia care a aparut la un TST din 1999).

[quote="Mircea Lascu"]
$\textbf{2.}$ Fie $a,b,c>0$ astfel incat $abcd=1$. Aratati ca: \[\sum\frac{1}{1+a+a^{2}+a^{3}}\geq 1.\] [/quote]

Vom folosi derivate la greu.

Fie $x = \ln a, \, y = \ln b$ etc si $f : \mathbb R \to \mathbb R$, $f(x) = \frac1{1+e^{x}+e^{2x}+e^{3x}}$.

Inegalitatea pe care trebuie s-o demonstram e echivalenta cu $f(x)+f(y)+f(z)+f(t) \geq 1$ pentru $x+y+z+t=0$.

Avem $f^{\prime \prime}(x) = \frac{e^{x}}{\left( 1+e^{x}+e^{2x}+e^{3x}\right)^{3}}\cdot \left( 9 e^{5x}+11 e^{4x}+6 e^{3x}-6 e^{2x}-3 e^{x}-1 \right)$.

Eu as vrea sa demonstrez ca $f$ are un unic punct de inflexiune, asa ca voi considera $g : \left( 0,\infty \right) \to \mathbb R \, ,$ \[g(u) = 9 u^{5}+11 u^{4}+6 u^{3}-6 u^{2}-6u-1 .\] $g^{\prime \prime \prime}(u) = 540 u^{2}+264 u+36$ are $\Delta < 0$, deci $g^{\prime \prime}(u) = 180 u^{3}+132 u^{2}+36 u-12$ e strict $\nearrow$.

$g^{\prime \prime}(0) < 0$ si $g^{\prime \prime}(1) > 0$, deci exista $u_{0}\in (0,1)$ (si e unic) astfel incat $g^{\prime \prime}\left( u_{0}\right) = 0$.

Rezulta ca $g^\prime (u) = 45 u^{4}+44 u^{3}+18 u^{2}-12 u-3$ e strict $\searrow$ pe $\left( 0 , u_{0}\right]$ si strict $\nearrow$ pe $\left[ u_{0}, \infty \right)$.

Dar $g^\prime (0) < 0$ si $g^\prime (1) > 0$, deci exista $u_{1}\in \left( u_{0}, 1 \right)$ (si e unic pe $\left( 0, \infty \right)$) astfel incat $g^\prime \left( u_{1}\right) = 0$.

Asta inseamna ca $g$ scade pe $\left( 0, u_{1}\right]$ si creste pe $\left[ u_{1}, \infty \right)$.

Cum $g(0) < 0$ si $g(1) > 0$, rezulta ca exista $v \in \left( u_{1}, 1 \right)$ astfel incat $g$ este $\leq 0$ pe $\left( 0, v \right]$ si $\geq 0$ pe $\left[ v, \infty \right)$.

Asadar, $f$ are un singur punct de inflexiune, $w = \ln v$.

$f$ este concava pe $I_{0}= \left(-\infty, w \right]$ si convexa pe $I_{1}= \left[ w, \infty \right)$.

Fie $a_{i_{1}}, \ldots, a_{i_{k}}, \, k \leq 4$, variabilele aflate in $I_{0}$. Cu ajutorul inegalitatii Karamata (sau HLP), lasam o distanta mai mare intre ele, i.e. $f(a_{i_{1}})+f(a_{i_{2}}) \geq f(w)+f \left( a_{i_{1}}+a_{i_{2}}-w \right)$. In acest mod mutam toate variabilele (mai putin una) in punctul de inflexiune. Acum consideram variabilele aflate in $I_{1}$ si le egalam cu Jensen (aici sunt incluse si variabilele pe care le-am facut $w$ adineauri). Cum toti acesti pasi au micsorat $LHS$, rezulta ca este suficient sa demonstram inegalitatea in cazul $a=b=c$, i.e. \[\frac3{1+a+a^{2}+a^{3}}+\frac1{1+\frac1{a^{3}}+\frac1{a^{6}}+\frac1{a^{9}}}\geq 1 .\] Desfacand parantezele ( :) ), inegalitatea devine $2a^{9}-a^{8}-a^{7}+a^{6}-a^{5}-a^{4}+a^{3}-a^{2}-a+2 \geq 0$, i.e. $\left( a-1 \right)^{2}(a+1) \left( a^{2}+1 \right) \left( 2+a+a^{3}+2 a^{4}\right) \geq 0$, care e evidenta.

Oribil, nu?

e corect :)
si solutia mea e lunga si calculatorie :)

[hide="Solutia 2 - Setul 6,  Problema 1"]Notam $S_{1}=\sum{\frac{1}{a+3}}$ si $S_{2}=\sum{\frac{a}{a+3}}$. In mod evident, $3S_{1}+S_{2}=4.$
Trebuie sa aratam ca $S_{1}\leq 1$ si pentru aceasta este suficient sa probam inegalitatea $S_{2}\geq 1$.
Din conditia $abcd=1$ putem scrie $a=\frac{x^{3}}{yzt}$, $b=\frac{y^{3}}{xzt}$, $c=\frac{z^{3}}{xyt}$, $d=\frac{t^{3}}{xyz}$ $(x,y,z,t>0)$ si trebuie sa demonstram ca $\sum{\frac{x^{3}}{x^{3}+3yzt}}\geq 1$.
Din inegalitatea mediilor, avem $3yzt\leq y^{3}+z^{3}+t^{3}$ si analoagele.
Atunci $\sum{\frac{x^{3}}{x^{3}+3yzt}}\geq\sum{\frac{x^{3}}{x^{3}+y^{3}+z^{3}+t^{3}}}=1$ si solutia se incheie.[/hide]

frumoasa solutie :)
cred totusi ca aveti o mica greseala de tipar, e $3yzt$ la suma, si la medii la fel :)

Chiar e frumoasa solutia asta  :D Si merge si pentru problema pe care am mentionat-o in postul meu precedent: $\forall a_{1},\ldots,a_{n}> 0, \, \prod_{i=1}^{n}a_{i}= 1$ avem $\sum_{i=1}^{n}\frac1{a+(n-1)}\leq 1$. Cum ati reusit s-o gasiti?

[quote="maky"]
cred totusi ca aveti o mica greseala de tipar, e $3yzt$ la suma, si la medii la fel :)[/quote]
Asa este. Am corectat.
[color=blue]Pentru perfect_radio:[/color] am inmultit cu 3 ambii membri ai inegalitatii initiale, apoi am adunat 4 in fiecare membru. In continuare am trecut fractiile in care apar $a,b,c,d$ in membrul drept. Tinta era sa obtin o inegalitate de forma $f(a,b,c,d)\geq 1.$

[quote="Mircea Lascu"]$\textbf{1.}$ Fie $a,b,c>0$ astfel incat $abcd=1$. Aratati ca: \[\sum\frac{1}{a+3}\leq 1.\] $\textbf{2.}$ Fie $a,b,c>0$ astfel incat $abcd=1$. Aratati ca: \[\sum\frac{1}{1+a+a^{2}+a^{3}}\geq 1.\] $\textbf{3.}$ Fie $a,b,c>0$ astfel incat $a+b+c=1$. Aratati ca: \[\sum\sqrt{\frac{bc}{bc+a}}\leq\frac{3}{2}.\][/quote]

to 1]           $\frac{1}{4}-\frac{1}{a+3}\ge \frac{1}{16}In(a)$


Hi,

I did this thing back long time ago in Getting Started.. I'm pretty sure joml88 remembers it since he also did this.  :D Anyway, this is basically talking about math in AIM chatroom. It used to be pretty "loose" thing but during this summer, I'd like to run something similar to "class" on AoPS. It will be about 90 minutes (first 45 min = AMC, second 45 min = AIME; 10 minute break between them) and for 12 weeks, there will be a math chat in each week (one day). I learned that the best way to test what you know is to review with other people and that's what I plan to do.  :D Anyway, this is not suppose to happen until summer. I'll probably do practice one in May or June so if you're at all interested in "attending" OR "helping," please post here.

Thanks.  :)

EDIT: I don't know if I can do it for sure.. I'll have to see how my time goes.

[quote="Silverfalcon"]Hi,

I did this thing back long time ago in Getting Started.. I'm pretty sure joml88 remembers it since he also did this.  :D Anyway, this is basically talking about math in AIM chatroom. It used to be pretty "loose" thing but during this summer, I'd like to run something similar to "class" on AoPS. It will be about 90 minutes (first 45 min = AMC, second 45 min = AIME; 10 minute break between them) and for 12 weeks, there will be a math chat in each week (one day). I learned that the best way to test what you know is to review with other people and that's what I plan to do.  :D Anyway, this is not suppose to happen until summer. I'll probably do practice one in May or June so if you're at all interested in "attending" OR "helping," please post here.

Thanks.  :)[/quote]

Yep, I remember those :)  ...fun times.  Anyway, I'm willing to help out a little bit.

I would probably be interested (i think you pmed me about it)
but 45+10+45 minutes=100min?
and 20 bucks is kinda cheap


Find the sum: 2/3 + 1/4 + 2/15 + 1/12 + ... + 2/10200

This is to be done without a calculator. Thanks in advance!

hello
[hide="hint"]
Observe that there are two different GP.
1.GP
first term = $ \frac{2}{3}$  ratio=$ \frac{1}{5}$
2.GP
first term = $ \frac{1}{4}$  ratio=$ \frac{1}{3}$
find the total no. of terms in both the GP , use the sum formula of GP
[/hide]
Thank u.


How many palidromic numbers are there between $ 10$ and $ 1000$?

Two kinds of palindromes need to be considered: two-digit palindromes and three-digit palindromes.

Two-digit palindromes are of the form $ \overline{xx}$ so there are of course 9 of them.

Three-digit palindromes are of the form $ \overline{xyx}$. x cannot be zero, otherwise the number would have just two digits (or even zero). Thus 9 choices for x and 10 choices for y means 90 palindromes.

Total: 99 palindromes.


Να αποδείξετε ότι ο αριθμός  $ \frac{x^{2}\plus{}1}{y^{2}\minus{}5}$  δεν είναι ακέραιος για όλους τους ακεραίους $ x,y>2$

Αν ο $ y$ ειναι περιττος τοτε $ y^2 \minus{} 5 \equal{} 0mod4$ αρα πρεπει $ 4/x^2 \plus{} 1$ αδυνατο!
Αν ο $ y$ ειναι αρτιοs τοτε $ y^2 \minus{} 5 \equal{} 3mod4$.Eστω $ p$ ενας πρωτοσ διαιρετης του $ y^2\minus{}5$ με $ p\equal{}3mod4$ tote $ p/x^2\plus{}1$ Ατοπο.(λοι οι πρωτοι διαιρετεσ του $ x^2\plus{}1$ ειναι $ \equal{}1mod4$.


Let be $ x,y,z>0$ . Then prove that :

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

$ \frac{{x^3  \plus{} y^3 }}{{x^2  \plus{} xy \plus{} y^3 }} \ge \frac{{x \plus{} y}}{3} \Leftrightarrow 3(x^3  \plus{} y^3 ) \ge x^3  \plus{} 2x^2 y \plus{} 2xy^2  \plus{} y^3  \Leftrightarrow ... \Leftrightarrow (x^2  \minus{} y^2 )(x \minus{} y) \ge 0$ si se insumeaza

"si se insumeaza" = and after that the inequalities are summing up

A few questions.

First off, is there a typo in $ x^2 \plus{} xy \plus{} y^3$? Is it $ x^2 \plus{} xy \plus{} y^2$ instead?

In addition, how would we notice that $ \frac{{x^{3}\plus{}y^{3}}}{{x^{2}\plus{}xy\plus{}y^{3}}}\ge\frac{{x\plus{}y}}{3}$ without just expanding?

[quote="Zhero"]

In addition, how would we notice that $ \frac {{x^{3} \plus{} y^{3}}}{{x^{2} \plus{} xy \plus{} y^{3}}}\ge\frac {{x \plus{} y}}{3}$ without just expanding?[/quote]
That is the cleverness at inequalities.....To notice some thinks...In this inequality for instance is not so difficult to see it.....

The inequality is false for $ x\equal{}y\equal{}z>1$, the inequality Zhero gave is true if it is $ y^2$ instead of $ y^3$, with equality if $ x\equal{}y$, but if $ y>1$ that would only make the left side less.


In year N, the 300th day of the year is a Tuesday. In year N+1, the 200th day is also a Tuesday. On what day of the week did the 100th of year N-1 occur?

[hide]
call tuesday: day $ 2\mod 7$
If there are $ 365$ days in a "non-leap-year" year then the $ 200^{th}$ day in year $ N\plus{}1$ is $ 200\plus{}65\plus{}2 \equiv 1 mod 7$ - which is a monday
This suggests that year $ N$ is a leap year and therefore year $ N\minus{}1$ is not.
So $ 2\minus{}(300\plus{}(365\minus{}100))\equiv \minus{}563 \equiv 4 \mod 7$

Therefore the $ 100^{th}$ day of year $ N\minus{}1$ is a Thursday[/hide]

[hide]365-300=65
65/7=9 remainder 2
Therefore, last day of year N is a Thursday and, consequently, first day of year N+1 is a Friday.
However, 200/7= 28 remainder 4 which implies that the 200th day of the year of N+1 is a Wednesday. Therefore, the year N+1 must be a leap year which means that the year N-1 is not a leap year.

300/7= 42 remainder 6
Therefore, the first day of the year N is a Wednesday as the 300th day is a Tuesday.
365-100=265
265/7 = 37 remainder 6
Therefore, the 100th day of the year N-1 is a Thursday as the first day of the year N is a Wednesday.[/hide]

[hide]
In year N, 200th day is Sunday, and in N+1 is Tuesday, so N+1 is leap. Now 200th in N-1 is Saturday, and 100th is Thursday.
I used that x-th days in 2 consecutive years differ by 1 or 2 if it's not leap or if it is[/hide]


integration
calculus
trigonometry
logarithms
parameterization
function
complex analysis

This integral have been calculated several times on this forum. Let I be your integral. Start by showing that

$ \int^{\pi/2}_{0}f(\sin x)\,dx = \int^{\pi/2}_{0}f(\cos x)\,dx$.

Then using that, we can write

$ 2I = \int^{\pi/2}_{0}\ln(\sin x)\,dx+\int^{\pi/2}_{0}\ln(\cos x)\,dx = \int^{\pi/2}_{0}\ln\left(\frac{\sin 2x}{2}\right)\,dx$

$ =-\frac{\pi}{2}\ln 2+\int^{\pi/2}_{0}\ln(\sin 2x)\,dx$.

In the last integral change variables according to $ 2x = \frac{\pi}{2}-t$, and so

$ \int^{\pi/2}_{0}\ln(\sin 2x)\,dx =-\frac{1}{2}\int^{-\pi/2}_{\pi/2}\ln[\sin(\pi/2-t)]\,dt$

$ = \frac{1}{2}\int^{\pi/2}_{-\pi/2}\ln(\cos t)\,dt = \int^{\pi/2}_{0}\ln(\cos t)\,dt = I$.

So:

$ 2I =-\frac{\pi}{2}\ln 2+I \Rightarrow  I =-\frac{\pi}{2}\ln 2$.

hello, it is $ \int_{0}^{\frac{\pi}{2}}\ln\left(\sin(x)\right)=-\frac{1}{2}\pi\ln(2).$
Sonnhard.

[quote="Carcul"]Integration in the complex plane can also be used.[/quote]
I read about countor integration during my summer vacation and I love it. Can you please post the solution. Or possibly the steps with details so I can learn.

Thanks. :)

Sylow_Theory wrote:

[quote]I read about countor integration during my summer vacation and I love it. Can you please post the solution. Or possibly the steps with details so I can learn.[/quote]

The way I know that uses complex analysis is a bit indirect.

(i) Show using typical contour methods that $ \int_{0}^{\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}\hspace{1 mm}\,dx =\pi \ln 2$

(ii) Substitute $ x=\tan t$

(iii) Substitute $ \theta=\frac{\pi}{2}-t$

Also, I think this integral is in Ahlfor's book, but I do not have it at home so I am not sure if he does it the same way.  Alternatively, I think you could introduce a free parameter in the logarithm and differentiate with respect to it, calculate the resulting integral with complex variables and integrate back -- but I have not verified it.

[quote="sos440"]Alternative solution: you may use the formula $ \ln |\sin x| \; = \;-\ln 2-\sum_{n=1}^{\infty}\frac{\cos 2n x}{n}$[/quote]

How is this so?

[quote="mdk"]How is this so?[/quote]

$ \ln |2 \sin x| \; = \; \ln |1-e^{2ix}| \; = \; \mathfrak{Re}\ln (1-e^{2ix}) \; = \;-\sum_{n=1}^{\infty}\frac{\cos 2nx}{n}$

where $ \mathfrak{Re}(z)$ denotes the real part of $ z$. Integrate both side from $ 0$ to $ \frac{\pi}{2}$, then the summation part vanishs and we have the answer.

[quote="Sylow_theory"]I read about countor integration during my summer vacation and I love it. Can you please post the solution. Or possibly the steps with details so I can learn.[/quote]

Consider the function $ f(z) = \frac{\ln z}{z-1}$ and the contour that encloses the region inside the circle centered at Re(z) = 1 and with radius 1, and outside the circle centered at z = 0 with radius $ \epsilon < 1$.


calculus
integration
number theory solved
number theory

Show that product of the distinct prime numbers less
than n is less than 4^n.

Didn't this have anything to do with Gauss's integral?

Anyway I think you better not post fresh problems in Solved Problems section...


linear algebra
matrix
linear algebra unsolved

Let $A$ be an $n\times n$ matrix. Given any linearly independent set $\{x_1,\cdots,x_k\}$ in $R^n$ where $k<n$. Is it always possible to find $x_{k+1}\not\in\text{span}\{x_1,\cdots,x_k\}$ such that $Ax_{k+1}\in\text{span}\{x_1,\cdots,x_k,x_{k+1}\}$?

If you're talking about matrices over $\mathbb{R},$ the answer is no, essentially becuase real eigenvalues don't have to exist.

Let $A=\begin{bmatrix}1&0&0\\0&0&-1\\0&1&0\end{bmatrix}.$ Let $k=1$ and let $x_1=\begin{bmatrix}1\\0\\0\end{bmatrix}.$

Then for us to find an $x_2$ for which $Ax_2\in\text{span}\{x_1,x_2\},$ we would have to find an eigenvector for

$\begin{bmatrix}0&-1\\1&0\end{bmatrix},$ which we can't.


calculus
integration
trigonometry
derivative
real analysis
real analysis unsolved

Evaluate $ \int\frac{dx}{  \sqrt{ a\cos x \plus{}  b\sin x }  }$

hello, what do we know about $ a$ and $ b$?
Sonnhard.

For any $ a, b$, dividing inside the square root sign by $ \sqrt {a^2 \plus{} b^2}$ yields $ c \int \frac {dx}{\sqrt {\cos(x \plus{} \phi)}}$ for some constants $ c, \phi$, and this is some elliptic integral, i.e., not elementary.

I think a,b can be assumed to be any real numbers ,.

Can it be solved in this way ?

 $ \int\frac {dx}{\sqrt {\cos(x \plus{} \phi)}} \equal{} \int{\sqrt {\sec(x \plus{} \phi) } dx}$

Put  $ \sec(x \plus{} \phi) \equal{} t^2$

So it becomes $ \int\frac {2t^{2} dt }{ t^2 (\sqrt { t ^4 \minus{} 1 ) } } \equal{} \int\frac {2dt}{\sqrt { t^4 \minus{} 1 }}$

EDITED

No. The derivative of $ \sec$ is not $ \sec^2\plus{}1$.

Sorry it was a typo it is $ \sec x \ tanx$

No, if you make that substitution, you wind up with $ \int\frac{dt}{\sqrt{t^4\minus{}1}}$

Which doesn't help.

Listen to what JBL is saying.

[quote="Kent Merryfield"]No, if you make that substitution, you wind up with $ \int\frac {dt}{\sqrt {t^4 \minus{} 1}}$

Which doesn't help.[/quote]
Or maybe it [i]does[/i] help, if that is something you can recognize as an elliptic integral.


I HAVE GOT A PROBLEM .

EASY,BUT I AM NOT GETTING IT RIGHT.

QUESTION:- COMBINED AGE IS 91.
                    I AM TWICE AS OLD AS JONES WAS WHEN
                     I WAS AS OLD AS HE IS NOW!

Hmm... this seems like more of a question to understand what the question is asking or making sense of the data.
Obviously, the combined age is 91, and you are older than Jones.
I think its saying that when you were as old as he is now, you were twice as old as he was and it is asking you how old you and Jones are now.

possibly you could set it up like this:

Jones= $ x$
You=$ y$
(both current ages)
$ x\plus{}y\equal{}91$
$ z\equal{}number years ago$
$ x\minus{}z\equal{}2(y\minus{}z)$
$ x\minus{}z\equal{}2y\minus{}2z$
$ x\equal{}2y\minus{}z$

hmm...I'm not sure where to go after that but do some simultaneous stuff.

I tried guessing and checking but the closest i got was 30 and 61. Can you have decimal ages?

x= my current age, y=his current age

x+y=91          This part is obvious.
y=91-x

x=2[y-(x-y)]
x=2(2y-x)
x=4y-2x
3x=4y

Here, we have my current age, x, equal to twice [messy thing], so that's x=2(something).
"x-y" is the amount of time since you were jone's current age. Subtracting from y gives us how old jone's was at this time, which is what the problem describes.

Substituting,

3x=4(91-x)
3x=364-4x
7x=364
x=[b]52=my age [/b]
91-52=[b]39=jone's age[/b]

We can check-
Our ages sum to 91.
52-39=13 years ago
jones was 39-13=26
which is half of my age, 52.

ok got it. but @pinkmuster the 5th line that is

x=2(2y-x)
x=4y-2x.right?but

x=2(2y-x)gives
x=4y2-2x

I think.

@eli140 no decimal

Using the distributive property,

2(2y-x) yields

2(2y)-(2)x
4y-2x


geometry
geometric transformation
rotation
combinatorics unsolved
combinatorics

Difine $ X$ is a three-dimensional space.Is there a $ f: X \mapsto X$ ,for any of two different points $ A,B$ in $ X$ ,line$ AB$ is perpendicular to line $ f(A)f(B)$ ?

Rotation by $ \pi/2$.

UPD
Sorry, the space is 3d.

Any thoughts?
Please help me if you can solve


How would you type a bar over repeating numbers, such as .123123123123... to become .123 with a bar over 123?

use [code].\overline{123}[/code]

$.\overline{123}$

Just for information, there are many ways to put stuffs on the top.

$\overline {AB}$

$\overrightarrow {AB}$

$\vec {AB}$

$\bar {AB}$

yes--originally I typed $.\bar {123}$ instead of $.\overline {123}$ :P


Solve the system of eqts for $ x,y\in \mathbb{R}$ :

$ (1) \quad (3x^2\plus{}y^2)(x^2\plus{}3y^2)\equal{}{1\over x}\plus{}{1\over 2y}\\
\\
(2) \quad 2(y^4\minus{}x^4)\equal{}{1\over x}\minus{}{1\over 2y}$

I'm deleting this post.  Either I shouldn't post solutions ideas before checking for logical errors or I should not do math right after waking up.  Perhaps I can do one or the other, just not at the same time : ).

[hide="repeat post"][url]http://www.mathlinks.ro/viewtopic.php?t=286884[/url][/hide]

[hide]although  it was  not answered there[/hide]


In triangle ABC, P is any point on BC. Now find a point Q either on AB or on AC such that the line PQ divides the triangle into 2 equal halves.

Can you actually prove that PQ divides the triangle into 2 equal halves?

??

[quote="frt"][hide="Hint"]
Draw parallel line(s), and "transfer" equal areas :)
[/hide][/quote]

Can you provide the proof in detail?

[quote="frt"]Ok, here's what I did:
[hide]
Construct line $l$ that is parallel to $PA$ and goes through $C$.
Then let $D$ be the point of intersection between $l$ and $\overrightarrow{BA}$.

Note that $\triangle APC$ and $\triangle APD$ have the same area because $AP\parallel DC$.
So $\triangle ABC=\triangle ABP+\triangle APC=\triangle ABP+\triangle APD =\triangle BPD$

Therefore, $Q$ is simply the midpoint of $BD$.
[/hide][/quote]

Therefore, $Q$ is simply the midpoint of $BD$.

how did you get $Q$ here? In your assumptions, $Q$ is not mentioned?

typo?

[quote="frt"]Aren't we trying to find the location of point $Q$ on $AB$ or $AC$? :huh:

To clarify, the answer to this problem is:
$Q$ is the midpoint of $BD$ if $AB>AC$.

(If $AB<AC$, then we simply need to draw the parallel line from $B$, and the rest is the same.)

Edit: If you meant to ask how I knew that $Q$ is the midpoint, I simply used the fact that the median in a triangle bisects the area of the triangle. Is this clear now?[/quote]

Thanks... But what I meant was that in your analysis you have not mentioned $Q$ anywhere and in the last line you just mentioned that $Q$ is the midpoint.

Now its clear.

One more doubt... I think $D$ will be outside the triangle ABC in the sense that we need to extend line to arrive at $D$, but what about $Q$? It can be on the triangle or it can be outside also?

Oh yeah, good point.

If $P$ is closer to $C$, $M$ is on segment $AB$.
However, if $P$ is closer to $B$, then this makes $M$ outside $\triangle ABC$, which implies we need to draw $l$ from $B$ instead.
So, actually, the length of $AB$ and $AC$ doesn't really matter, but the location of $P$ matters. (A part of my clarification in my previous post was wrong.)


The altitude $ \overline{BD}$ divides $ \overline{AC}$ in the ratio 3:1. If $ AC$ is 12 and $ BD$ is 5, find the length of $ \overline{BC}$. Round your answer to the nearest tenth.

[asy]draw((0,0)--(30,20)--(40,0)--cycle);
draw((30,20)--(30,0));
label("A", (0,0), SW);
label("B", (30,20), N);
label("C", (40,0), SE);
label("D", (30,0), S);[/asy]

BD=5 is given, and DC=3.  Thus, BC=sqrt(25+9)=sqrt(34)=5.8


You are given a sequence of convex pentagons. You get the next pentagon from previous by connecting midpoints of its sides. Find formula for area of the nth pentagon if you have the area of all the previous pentagons in the sequence and how many areas of pentagons in the sequence are needed to get the area of nth pentagon.

PS: I hope the formulation of problem is understandable. If something is unclear feel free to ask.

We need two pentagons (this implies the question is just well formulated for n > 2). The proof follows.

Suppose wlog the center of mass of the pentagons is O = (0, 0) (it is easy to see the mass center is invariant). Let $ a, b, c, d, e$ be the vertices of the first pentagon, in this order. We define $ ab$ to be the area of the triangle ABO. This is twice the usual vectorial product. Then the area of the pentagon is

$ S_1 \equal{} \sum_{cyc} ab$

The vertices of the second pentagon are $ \frac {a \plus{} b}{2}, ..., \frac {e \plus{} a}{2}$, so its area is

$ S_2 \equal{} \sum_{cyc} \frac {a \plus{} b}{2}\frac {b \plus{} c}{2} \equal{} \sum_{cyc} \frac {ab \plus{} bc \plus{} ac}{4} \equal{} \frac {S_1}{2} \plus{} \frac {\sum_{cyc} ac}{4}$

The third pentagon has vertices $ \frac {a \plus{} 2b \plus{} c}{4}, ..., \frac {e \plus{} 2a \plus{} b}{4}$. Thus, its area is

$ \sum_{cyc} \frac {a \plus{} 2b \plus{} c}{4}\frac {b \plus{} 2c \plus{} d}{4} \equal{} \sum_{cyc} \frac {ab \plus{} 2ac \plus{} ad \plus{} 4bc \plus{} 2bd \plus{} cb \plus{} cd}{16}$

Remembering that $ xy \plus{} yx \equal{} 0$, we can observe this is equals to

$ \sum_{cyc} \frac {5ab \plus{} 3ac}{16}$

Now just observe that $ \sum_{cyc} ac \equal{} 4S_2 \minus{} 2S_1$ to obtain

$ \frac {5}{16}S_1 \plus{} \frac{3}{4}S_2 \minus{} \frac {3}{8}S_1 \equal{} \frac{3}{4}S_2 \minus{} \frac {1}{16}S_1$

Interesting. Taking $ ABCDE$ regular, this gives another way to compute $ cos \frac {\pi}{5}$.


Having a seqence of positive reals $\{a_n\}$ with a pseudo-additive property $a_{m+n} \leq a_m+a_n$ for all $m,n \in N$ prove that the sequence $\{\frac{a_n}{n}\}$ is convergent.

It's a classic but it was a headache problem for students :).

I remember also having some headaches with this problem.  :blush:    Take $ t=inf_{\frac{a_n}{n}}$ and $\epsilon>0$. Choose $n_0$ such that $\frac{a_{n_0}}{n_0}<t+\epsilon$. For all $n>n_0$ we have $ n=q_n n_0+r_n$ and $t\leq \frac{a_n}{n}\leq\frac{ max(a_0,...,a_{n_0})+ q_n a_{n_0}}{n}$. From the last relation easily deduce that $t\leq\frac{a_n}{n}<t+2 \epsilon$ if $n$ is large. Rafal, please post this type of problems in the analysis forum.

Sorry, I know it's analysis but why not put it in the olympic section (?) since it can be (in theory only I guess ;) ) an olympic type problem.
Btw., your solution is the same that I found but I think it's too difficult for an analisys exam ( it was once few years ago at university).


A computer screen shows a $ 98 \times 98$ chessboard, colored in the usual way. One can select with a mouse any rectangle with sides on the lines of the chessboard and click the mouse button: as a result,  the colors in the selected rectangle switch (black becomes white, white becomes black). Find, with proof, the minimum number of mouse clicks needed to make the chessboard all one color.

Very original, #H34N1. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=343869#343869[/url]


If $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}3.$ prove that :

 $ \frac{a}{2\plus{}b^3}\plus{}\frac{b}{2\plus{}c^3}\plus{}\frac{c}{2\plus{}b^3}\geq\frac{1}{6}(5\plus{}abc)$.

[quote="alex2008"]If $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 3.$ prove that :

 $ \frac {a}{2 \plus{} b^3} \plus{} \frac {b}{2 \plus{} c^3} \plus{} \frac {c}{2 \plus{} b^3}\geq\frac {1}{6}(5 \plus{} abc)$.[/quote]
We have:
$ \frac{a}{2\plus{}b^3}\equal{}\frac{a}{2}\minus{}\frac{ab^3}{2(2\plus{}b^3)} \ge \frac{a}{2}\minus{}\frac{ab}{6}$   (by AM-GM)
Similar,we have:
$ LHS \ge \frac{a\plus{}b\plus{}c}{2}\minus{}\frac{ab\plus{}bc\plus{}ca}{6} \ge 1$  (1)
On the other hand:
$ abc \le 1$=> $ \frac{1}{6}(5\plus{}abc) \le 1$   (2)
Since (1) and (2) we have Q.E.D :wink:  :)

Yes, he had a small mistake, of course, this problem is not hard and his way is right.
$ ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc \leq\ 4$   :)

[quote="mathVNpro"]I think something is wrong in your solution, quykhtn-qa1???! Can you check it again :lol: ???[/quote]
Sorry :blush: , must be:
$ \frac {a}{2 \plus{} b^3} \equal{} \frac {a}{2} \minus{} \frac {ab^3}{2(2 \plus{} b^3)} \ge \frac {a}{2} \minus{} \frac {ab^2}{6}$ .......
Then ,use the inequality like nguoivn write  .Thank you:)


Hi, I need some help with the equation.

[quote]Solve in $ \mathbb{N}$ : $ 3^n \minus{} 2 \equal{} m^2$[/quote]

Reducing mod 4, we see that $ n$ must be odd, $ n \equal{} 2k \plus{} 1$. Then, the equation rewrites like : $ m^2 \minus{} 3(3^k)^2 \equal{} \minus{} 2$ which is Pell-type equation in the shape of $ u^2 \minus{} 3v^2 \equal{} \minus{} 2$

We apply Pell's equation theory and we find that $ 3^k \equal{} \frac {(1 \plus{} \sqrt {3})(2 \plus{} \sqrt {3})^{i} \minus{} (1 \minus{} \sqrt {3})(2 \minus{} \sqrt {3})^{i}}{2\sqrt {3}}$

But now, I'm wedge ... I thought reduce the writing expanding the identities but it failed ...

Thanks to help me.

Hint:
[hide]Write the equation as $ 3^n \equal{} m^2\plus{}2$ and factor.[/hide]

Yes, I tried this in $ \mathbb{Z}$${ [\sqrt {-2}]}$, but I found this exercise in a lesson on Pell equation's, so I would like to solve it with Pell's equation theory.

Can someone help me, please (using Pell's equation) ?


[quote="KMK00009"]If mn divides m^2+n^2+m
Prove that m is a perfect square[/quote]
We denote that: m^2+n^2+m=bmn, so n^2 divises by m and m(m+1) divise by n.
Because gcd(m,m+1)=1 then m or m+1 divises by n.
But if n^2 divise m so m+1 divise by n, that is not true.
Then m divise by n and n^2 divise by m.
Recall m=an, so : n^2( a^2-ab+1)+an=0 then a is divise by n.
So m divise by n^2. Then m=n^2.
Continue we have m=n=1.

[quote="Jungle"]I'm not sure whether the conclusion of mr bui's proof is correct, but there is an error in line 2:

[quote] We denote that: m^2+n^2+m=bmn, so n^2 divises by m and m(m+1) divise by n.
Because gcd(m,m+1)=1 then m or m+1 divises by n. [/quote]

This only holds true if n is a prime or prime power. Anyway, here is my proof:

Suppose $ m$ is not a perfect square i.e. there exists a prime $ p$ such that $ p^{2k \plus{} 1} \parallel{} m$ for some $ k \in \mathbb{N}$.

From the given condition we may deduce that $ m \; | \; n^2$. So $ p^{2k \plus{} 1} \; | \; n^2 \Rightarrow p^{k \plus{} 1} \; | \; n$.

Now $ p^{k \plus{} 1} \; | \; (m,n)$ so $ p^{2k \plus{} 2} \; | \; m^2 \plus{} n^2 \plus{} m$. But $ p^{2k \plus{} 2} \; | \; m^2$ and $ p^{2k \plus{} 2} \; | \; n^2$, so $ p^{2k \plus{} 2} \; | \; m$. Contradiction.

Therefore $ m$ must be a perfect square.[/quote]
Thx a lot.
That's a really nice solution :)


Consider the set A = {3n + n + 1|n is a positive integer}.
For any positive integer k, let S(k) denote the sum of the digits of k.
Let B be the set {S(m)|m is in A}.
(a) Find the minimum of B.
(b) Does 1999 belong to B ?

[hide]

Firstly, when n=8, we get 201 is in A so the minimum sum is either 1, 2 or 3. All terms in A are odd, so sum of 1 is impossible. For a sum of two the number must be of the form 1000..001. Thus 3n^2 + n = 1000..00, n(3n+1) = 10^k for some k. n and 3n+1 have opposite parities so n=2^k, 3n+1 = 5^k, but we can check that doesn't work for the first few k (and for large k 3n+1 is far too small).

So the minimum is 3.

[/hide]

Still working on part b..

Hmm ... for (b) :

A suitable example suffices to prove that 1999 belongs to B.

For example : what about the digit sum of

3(10222 - 1)2 + (10222 - 1) + 1 ?


The following problem apperead lomg time ago in an AMM. Can anyone figure out a solution? Prove that the integer part of (pi)^n is a prime for only finitely many values of n.

AMM publishes solution, doesn't it? Looks really impossible.

Just a note: there is also http://www.mathlinks.ro/Forum/viewtopic.php?t=30308 .

But for some canonical reason (  :roll:  ) this topic is probably better to discuss it.

Unfortunately, I don't think a solution appeared for this one in AMM. Actually, I strongly believe the author put it as open question there, but I don't have the issue with the actual statement and context.


Pierre Fermat's father was a wealthy leather merchant and second consul of Beaumont- de- Lomagne. Pierre had a brother and two sisters and was almost certainly brought up in the town of his birth. Although there is little evidence concerning his school education it must have been at the local Franciscan monastery.

[size=134][color=red]The rest of this post has been removed by a moderator.  Lengthy quotations are usually not appropriate on the forums; please just post a link in the future, such as the link posted by t0rajir0u below.  Thanks![/color][/size]

Not sure what the point of copy-pasting this was.  Look, it is clearly available (in more readable format) here:

http://www-history.mcs.st-andrews.ac.uk/Biographies/Fermat.html

The St. Andrews site (see the link in t0rajir0u's post, then track up and around for other mathematicians and other topics) is one of the treasures of the www, and any publicity which makes more people aware of it is a good thing. But there's no point in pasting articles from it here; use references.


Hopefully by this Saturday, I'm planning to post up Mock ARML. This Mock ARML will contain Individual Round, Team Round, and Relay Round. All rules of real ARML will apply.

There are changes, however, so please read this.

To make this real, all participants must be in groups so that they can participate in all three rounds. Requirements for this:

A. Each group will consist of 3 people in SAME STATE. This is to make it real as possible. I need name of the group (make it creative.. don't be like Team A), names of the people in the team, grade, and state. Names of the people will be used for top 3 scores (8,7, and 6) on Individual.
B. All 3 members must participate in Individual Round. Each person will send the answers by themselves, not together. For the Team Round, only one person will send me all answers. These answers should NOT be LaTeXed but written so that I can see what you're saying clearly.

For instance, if the answer was $\frac{2}{1+x^2}$, you must write as 2/(1+x^2). If you write as 2/1+x^2, this will be marked wrong.

For Relay Round, same rules as Team Round apply.
C. Problems will be up in PDF and please note that these are not my problems nor are they ARML problems. I found problems of similar levels and created this Mock ARML and please, do not try to find them. Use your brain on math, not googling.

If you have a group of three people, shoot me an email at problemsolvingsession@gmail.com

I want to have as many teams as possible. Since each team is made up of only 3 members, there should be a lot of team but let's do this! Also, please spread the word to your teammates so we can really do "reality Mock ARML."

I agree.

$\frac{5}{x^{2}+x}$ is a lot easier to read than 5/(x^2+x)

Regardless, I'm in for GA.

[quote="matt276eagles"]How are you going to set up the relay and team rounds?  Also, do we need to create teams of our own, or do we just have to say which state we're from?[/quote]

All rounds ---- Individual, Relay, and Team Rounds ---- will be up about one day. So, I'm expecting people to just do them whenever you can with honor system. That is, I don't want people working on Team Rounds for about 2 hours to get all the answers.

You need to create teams with the member of people in your state. Since you're in Lehigh, I'm pretty you can find two other people who will work with you on this. Time is an issue but well, that's something that I can't help. You also need to state name of States as well.

chess64, what's wrong with LaTeX? Well.. Only thing is that Latex makes emails heavy. If you can somehow put LaTeX images on your email, that's fine. I just don't like people sending answers in PDF documents. That's what I meant by LaTeX.

So finally, we have to put our answers as 5/(x^2+x) instead of $\frac{5}{x^2+x}$?

Yeah, or DVI maybe? Or how about if I send you a link to my PDF solutions that I post on my web server? :D

I think DVIs are the best... besides, it would be to your advantage to read and grade everything considering that it'd be much easier to see.

And how would you make notations such as sigma, product symbol, and other stuff?  :)

[quote="Boy Soprano II"]What time frame(s)?

Could we submit a .tex file containing our answers?  Those are economical.[/quote]

Eh. No.

Okay, so here it is:

If you want to just send by regular text, that's fine and make sure it's clear as 5/(x^2+x).

If you want to use $\text{\LaTeX},$ then MAKE PDF DOCUMENT WITH SOLUTION TO SEND IT. After your solution was graded, I'll delete the email so it doesn't matter anyway. Do not use DVI or others. Just use PDF.

Karth, I don't know what you mean by "how do you make math symbols.." Do you mean how do I type them or you want to know how to type them for yourself? If you don't know $\text{\LaTeX},$ there is a tutorial in this site.

[quote="Silverfalcon"][quote="Boy Soprano II"]What time frame(s)?
Karth, I don't know what you mean by "how do you make math symbols.." Do you mean how do I type them or you want to know how to type them for yourself? If you don't know $\text{\LaTeX},$ there is a tutorial in this site.[/quote][/quote]

I never said that?  :huh: I already know $\text{\LaTeX}$

By the way, if you formed a team, please shoot me an email with the name of the team, name of the members, grade, and state so I can start organizing early. Thanks!

Okay, I'm in for GA, providing nothing's happening c. 12–2 o'clock EDT.  But that's something like 9–11 o'clock in California anyhow, so I'm guessing it won't be.

how about writing the answers in latex "code", but not really in latex? ;)
For example, \frac{5}{x^2+x)
It is a uniform, systematic, unambiguous way to write answers, plus if they are [i]only[/i] answers and not complete solutions, they won't be much to read through anyway.

Please note that if you do have team(s), please send me email to the address above. I already have two teams (not telling you which state though  :D ) but we need to have more than two teams to compete on this. Thanks.

P.S. Problems will be up sometime tomorrow 6:00 EST PM to 8:00 EST PM. I'd LIKE to have the teams sending emails prior to those times.

[quote="Iversonfan2005"]Unfortunately, SC can not do it since we weren't able to get a team together :(

Oh well, I hope that you (Silverfalcon) or any one else would be able to make another one of these contests again before ARML..perhaps?[/quote]

Hmm.. Not even one team? That's sad.  :(

I might be able to get a couple people from Iowa together for a team by next weekend, I'll have to talk to them at school.

Here are problems. They're due by Tuesday morning. Make sure you give me necessary informations as well.

[quote="Boy Soprano II"]Wait.  I don't get it.  For relay, do we collaborate on them?  It would be pretty hard to do it like a real relay.[/quote]

Get two other people and do a real one. I'm putting my trust on people so if you want to do more than given time (6 minutes total and extra points if you do it in 3 minutes), whatever. There's nothing I can do about people like that.

For the Team Round, the time is extended to 1 hour and 40 minutes. This is from the ratio of 3:15 people so to make it fair, 3:15 people = 20:100 minutes.

[quote="Shogia"]er 100 minutes would be an hour and forty...[/quote]

Oops.. My bad..It's fixed now.

BTW, maybe I should extend the date to next weekend? I'm saying this because there wasn't any person who sent me answers and I forgot that Sunday is the busiest day of the week (finishing all homeworks that you didn't do while doing math :D ).

On one relay question, it should be TNYWR, not TNYWF. That was a typo.

[quote="Shogia"]If you even extended it to Wednesday morning that would be a great help for my group since our math club meets Tuesday afternoon.[/quote]

New deadline: Friday night.

So, this should give people ENOUGH time. By the way (I'm not just talking to Shogia but everyone), if you think you're going to do it, can you at least shoot me an email with people who will? It seems like Georgia, Upstate New York and several other states are interested but I only received one from Florida.

all [s]individual +  team round[/s] problems have been posted in separate threads in HSM  , I just created the related post collections

[url=https://artofproblemsolving.com/community/c3736111_2006_mock_arml_iv_individual_reality]individual round[/url] + [url=https://artofproblemsolving.com/community/c3736133_2006_mock_arml_iv_team_reality]team round[/url] + [url=https://artofproblemsolving.com/community/c3736347_2006_mock_arml_iv_relay_reality]realy round[/url]  [s](I shall post relay round soon)[/s]

enjoy  / start solving

more USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more ARML mocks [url=https://artofproblemsolving.com/community/c2442143_arml_mocks]here[/url],
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


Let $n$ be a given non-negative integer. Determine the number of solutions to
$x+2y+2z=n$
for non-negative integers $x,y,z$.

[hide="hint 1"]Use problem ''Determine the number of solutions to $x_{1}+x_{2}+...+x_{m}=n$ , $x_{i}$ are nonnegative integers''.[/hide]
If you can not, please read following hint.

 [hide="hint 2"]see cases $n$ is even, $n$ is odd.[/hide]

from the hints i got
[hide]
if
$n\equiv 0\bmod{2}$ there are $\dbinom{\frac{n}{2}+2}{2}$ solutions and if
$n+1\equiv 0\bmod{2}$ then there are just as many solutions.
[/hide]
is that correct?

Before I was trying to use generating functions. does anyone know how to solve the problem that way?

[hide="big hint"]
\[y+z = \frac{n-x}{2}\in \mathbb{Z^{*}}\]
Find all applicable $x$, then sum. [/hide]

[hide="lemma"]There are $\binom{n+m-1}{m-1}$ ordered $m-$ tuples $(x_{1},x_{2},...,x_{m})$ of nonnegative integers satisfying the equation $x_{1}+x_{2}+...+x_{m}=n$.[/hide]


Not sure if this belongs here.

What would be the most succinct way to say "the remainder when $ a$ is divided by $ m$", or equivalently "the least residue of $ a$ modulo $ m$"? Both of these are such horribly puffy and wordy and huge and inconvenient expressions for such a simple concept. I've heard people say and write things like, "Since the sum of the digits is congruent to $ n$ mod 9, and the sum of the digits is less than 9, it remains to find $ n$ mod 9". This language is perfectly understandable, but I don't believe it's conventionally correct.

Umm..
I believe that "the remainder when $ a$ is divided by $ m$" is pretty distinct, but I suppose you could instead choose to go with the more succinct "$ a$ is congruent to $ b$ mod $ m$."

Or simply "$ a$ is $ b$ mod $ m$."

If we're talking about how you'd write it in a proof or something, there is, of course, the notation $ a\equiv b\pmod m$.

I don't think there's anything conventionally wrong with "$ n$ mod 9" or "...congruent to $ n$ mod 9".

[quote="leoxnlin"]I've heard people say and write things like, "Since the sum of the digits is congruent to $ n$ mod 9, and the sum of the digits is less than 9, it remains to find $ n$ mod 9".[/quote]
$ 9$ or less.  The digital root of a positive integer is in $ [1, 9]$, not $ [0, 8]$; because it can't go to zero, you have to pick a slightly different set of residues.

A computer scientist, and to a lesser extent a mathematician, would plausibly interpret "$ a \bmod m$" as "the least non-negative residue in the equivalence class of $ a \bmod m$," so this is pretty good notation unless you want to be totally unambiguous.  It's important to distinguish residue classes from their representatives.

Thanks everyone. It appears... the notation is acceptable? That's new to me. Just making sure... would writing something like $ (a \bmod{10}) < 5$ be an acceptable way to say that the unit's digit of $ a$ is less than $ 5$?

I would use $ a\equiv x(mod 10)$  and  $ x < 5$

That doesn't really work; the equivalence notation isn't really about remainders, so you haven't put any conditions on $ x$ as far as being minimal or non-negative.

Your best bet would just be to say "the units digit of $ a$ is less than $ 5$."  Also, if you're trying to do something like that that depends strongly on a particular set of residues, you should be more specific than that.  For example, sometimes instead of working with $ \{ 0, 1, 2, ... p\minus{}1 \}$ we like to work with $ \{ 0, \pm 1, ... \pm \frac{p\minus{}1}{2} \}$; this is used, for example, in certain proofs of Gauss's lemma and to speed up modular computations.

How about,

$ a\equiv x (mod 10)$ and $ 0 \le x < 5$ :)


Suppose $N$ is an $n$ digit positive integer such that
(a) all its digits are distinct;
(b) the sum of any three consecutive digits is divisible by $5$.
Prove that $n \leq 6$. Further, show that starting with any digit, one can find a six digit number with these properties.

It's obvious $a_{k+3}$ and $a_k$ have the same value modulo 5, because $a_1 + a_2 + a_3 = a_2 + a_3 + a_4$ modulo 5, etc.

 If n>6, then $a_1, a_4, a_7$ are all the same mod 5, contradiction.

We can find 145690 or rotations, 235780 or rotations.

Sorry for reviving old post but this is my doubt:
Why $a_1, a_4, a_7$ have all the same mod 5?

Obviously ak and ak+3 have same modulo five.
But in 0-9 only 2 numbers can have same mod 5
But here a1 , a4 , a7 have same mod5 and hence n<=6

It s false their are 4 pairs which has same modulo

We observe that ak+3  - ak is 0 mod 5. Now since they are digits the difference is either 0 or 5. Now it cant be 0 as digits are distinct.  So the must be 5.  But then a7-a1 = 10 a contradiction.

Can anybody prove the next part of the problem?


[hide]
perimeter is just $x$

area is $(\frac{x}{4})^{2}$[/hide]

isn't this a little too easy for HSB?

[quote="Interval"]Your content did not reveal.[/quote]
What? It's right there.

[quote="Interval"]A wire of length x is bent into the shape of a square.

(a) Express the perimeter of the square as a function of x.

(b) Express the area of the square as a function of x.[/quote]
[hide="A"]$P=x$[/hide]
[hide="B"]$A=\left(\frac{x}{4}\right)^{2}$[/hide]

[quote="Interval"]A wire of length x is bent into the shape of a square.

(a) Express the perimeter of the square as a function of x.

(b) Express the area of the square as a function of x.[/quote]
[hide="A"] each side will be $\frac{x}{4}$ by the difinition of a square. Since there are 4 sides, multiply: $\frac{x}{4}\cdot4=x$[/hide]
[hide="B"]Like I said in A, each side is $\frac{x}{4}$, so the area of a square is $\left(\frac{x}{4}\right)^{2}=\frac{x^{2}}{16}$[/hide]

[hide]
The perimeter is all 4 sides. Therefore, it'd be just $x$?

Also, for the second question, it'd be $\left(\frac{x}{4}\right)^{2}$ Because one side is $\frac{x}{4}$.

[/hide]


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

$A$ and $B$ are both $n\times n$ REAL matrices. Let $A$ be anti-symmetric, and let $B=\mathrm{diag}\,(b_1,\cdots,b_n)$, where $b_i>0$ for $i=1,\cdots,n$. Prove that, $\det(A+B)>0$.

Use the definition of determinant and partition the permutations according to the set of fixed points; we see that the determinant takes the form of a sum (over the partitions formed above) of a product of the diagonal elements (from the fixed points) and the determinant of the skew-symmetric matrix formed by crossing out the rows and columns of the aforementioned diagonal elements (the nonfixed points); thus, as all the diagonal elements are positive, the result will follow if we can prove that the determinant of any skew-symmetric matrix is nonnegative (all terms will be nonnegative and the last one, being the product of all the diagonal elements, will be strictly positive, so the sum will be strictly positive). Thus, it remains to prove that for any skew-symmetric matrix, its determinant is nonnegative.

Complexify the matrix (i.e. regard it as a transformation over $\mathbb{C}^n$) and suppose $\lambda$ is an eigenvalue with eigenvector $x$. Using the standard sesquilinear (right terminology?) scalar product and the definition of an adjoint, we note that $\lambda \langle x, x \rangle = \langle \lambda x, x \rangle = \langle Ax, x \rangle = \langle x, A^{*}x \rangle = \langle x, \overline{A}^{\mathrm{T}}x \rangle$ $= \langle x, A^{\mathrm{T}}x \rangle$ $= \langle x, -Ax \rangle$ $=$ $- \langle x, kx \rangle$ $=$ $-\overline{\lambda} \langle x, x \rangle$. As we know that $x \ne 0$, being an eigenvector, we can divide through by $\langle x, x \rangle > 0$ to find that $\lambda = -\overline{\lambda} \Rightarrow \lambda \in i\mathbb{R}$. However, we know that the eigenvalues, other than $0$, come in pairs of additive inverses (i.e. $\pm \lambda$) from the standard form in a symplectic basis (actually, for real matrices, this is easy to prove right off the bat since we've already proven that all eigenvalues are pure imaginary but they are roots of the characteristic polynomial $\in \mathbb{R}[x]$, they come in conjugate pairs which reduces to the above). Thus, as the determinant is the product of the eigenvalues, it will have the form $\prod_i \lambda_i \cdot -\lambda_i$ (with possibly a $0$ thrown in); as all the $\lambda$s are pure imaginary, we see that the determinant is nonnegative, as desired.


cos pie/7  + cos 3pie/7 + cos 5pie/7 = 1/2  i need help with this one plz


Let $x_i$ with$i=1,2,..,n$ be real numbers such that $s > rx_j$ for every $j=1,2,..,n$ with $s=\sum_{i=1}^{n}x_i$ and $r<n$ positive real number.
Prove that

\[\sum_{i=1}^{n} \frac{x_i^{2m}}{s-rx_i} \geq \frac{s^{2m-1}}{(n-r)n^{2m-2}}\]

But it is just Jensen for $f(x)=\frac{x^{2m}}{s-rx}$.  :?

It was created by Klamkin who gave a direct proof.

However you are right Myth ;)

Can Cauchy-Schwarz be proved by Jensen ?

I think so. Apparently, it is Jensen for $x^2$. :?

[quote="manlio"]Can all inequalities  be derived by Jensen? :?[/quote]

you're probably have some difficulties with proving Karamata this way  :D
or am I wrong    :)

Actually, Karamata proved it via Jensen ;)

oh really , I dont know that - I just came up with this because its really easy to prove Jensen using Karamata and I thought  reverted procces would be hard...

You can prove Holder with Jensen, which means that you can definitely use it to prove C-S.


I want to know if Jacobs University of Bremen requires SAT II. I've already found some information on the official site but there are only a few words about SAT II.

If it is unclear from their website then probably you should contact someone responsible for admissions there and ask.

No, Jacobs doesn't require SAT II. At least they didn't require it for Class of 2012; however,  one or 2 subject tests wouldn't hurt your application.


conics parabola invariant geometry parameterization calculus derivative

someone.... prove mahematiclly that only one normal can be drawn from an outside point of the parabola and three from an inside point.

 it is what i am i discussing with all my teachers all my teachers have that what you have asked is absolutly true but they have accepted that proving it mathematically is a tough job.some also approached using cardan's method but failed....

You can't prove it because it's wrong.

[b]Counterexample:[/b] Take the parabola $ y \equal{} x^2$, and the interior point $ (x,y) \equal{} (0,1/2)$. The only normal is the point connecting $ (0, 1/2)$ to $ (0,0)$. In general, if $ y < 1/2$, then there is always only one normal, whether your point is inside or outside the parabola.

I encourage you to derive the correct conditions for the number of normals.

[b]Hint 1:[/b] [hide="Reduction of the problem without loss of generality"]
All parabolas are similar. Can you prove this? Since orthogonality is invariant under scaling, it suffices to consider the case of $ y \equal{} x^2$. After you derive the conditions for this parabola, try to use some geometric intuition to describe the conditions for a general parabola, keeping in mind that all parabolas are similar.[/hide]

[b]Hint 2:[/b] [hide="Analytic geometry approach"]
Parametrize the parabola in a way that your parametrization will "keep moving" if you vary the parameter (ie, if your parameter is t = time, then the point T(t) should never slow to a stop). Then, whenever the derivative of the distance from your chosen point P with respect to t is zero, that point T is a normal point. To think why this is the case, imagine that your point T is moving, but the distance PT is not changing. In the locality of that T, the parabola is approximately a circle with center P.[/hide]


induction
number theory unsolved
number theory

Prove by means of induction that $ n$ does not divide $ 2^n\minus{}1$.

See here : http://www.mathlinks.ro/viewtopic.php?p=849217#849217


The number 169 can be written as the sum of three positive perfect squares.  What is the greatest of these perfect squares?

Let's think about this. The nearest square is $ 144$. $ 169 \minus{} 144 \equal{} 25$. We know from the Pythagorean triple 3-4-5 that $ 9$ and $ 16$ make $ 25$, so those are our three squares -- 144, 16, and 9. The biggest square is $ \boxed{144}$.


Problem 40
This is a chem problem which i feel has lost the charm due to the objective pattern,but in an non-objective pattern...interesting

spiro[3.3.3]bromo heptane reacting with Bu-Li
(2 cyclopropyl groups attached to 2 carbons of another cyclopropane and a bromo group on the remaining C of the final cyclopropane{I think i made myself clear   })

Expecting some interesting answers

is anything mentioned about temp??

Nope I think we should consider normal temp

:idea:  :idea: I think no.1 the hydrogen of the carbon having the bromine first gets knocked off.

i was thinking that probably br- is eliminated and you get a carbocation.  and ring expansion takes place (twice) and then hydrogen is eliminated, giving an alkene. :maybe:

It comes in one step of wittig so that's why I said that.

in wittigs, hydrogen is removed as it acidic, adjacent to positively charged phosphorus. but here, the hydrogen is not all that acidic, also wudnt that lead to a carbene?  and cyclopropyl methyl carbocation is very stable.

Varunarasu,i would like to have a rough structure of ur product or the iupac name of it
But,Chemrock is wrong

how do ya draw these structures chemrock.....

Yes paint da and then don't store as bmp change it to JPEG that is supp by aops da :D  :D  :D

That is not the final answer, (i think is E2 big clue)

Then butyl substituted product eh? :huh:  :huh:  :huh:

Nope ,that is what nitish Bharadwaj replied

I was thinking of some cyclobutyl substituted productr but if its E2, the product is probably something weird like this


function
real analysis
real analysis unsolved

Could anyone help me proving this?
$ \sum^{\infty}_{n\equal{}2}[\zeta(n)\minus{}1]\equal{}1$

$ \zeta(n)$ is Riemann zeta function.

$ \sum_{n\equal{}2}^{\infty}\zeta(n)\minus{}1\equal{}\sum_{n\equal{}2}^{\infty}\sum_{k\equal{}2}^{\infty}\frac1{k^n}$

Switch the order.

[color=green]Unnecessary quote removed by moderator[/color]
Thanks!


Let $ a,b,c,d$ be non-negative real numbers such that $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$. Show that
\[ (1\minus{}ab)(1\minus{}bc)(1\minus{}cd)(1\minus{}da)\geq\frac{81}{256}.\]

[quote="pvthuan"]I have tried Cauchy in several ways for it, but eventually, it was reversed.

Have you got any idea for this?[/quote]
Sorry because of my post.But can you tell me how I can buy your book if I was in Dong Nai ?

I too found that at crucial points in the argument, the inequality I wanted was backwards!  I think the following argument works though.

First a slight notational change: let’s write the problem as $ a\plus{}b\plus{}c\plus{}d\equal{}1,$ with $ a,b,c,d$ nonnegative, and establish that $ \prod (1\minus{}\sqrt{ab}) \geq 81/16.$

All of the sums $ \sum$ and products $ \prod$ are to be taken cyclically.

Step one: By the GM/HM inequality we have $ \prod (1\minus{}\sqrt{ab}) \geq \frac {256} {\sum \frac {1} {1\minus{}\sqrt{ab}}}$.  Thus it will be sufficient to prove (upon rearranging) that $ \sum\frac {1} {1\minus{}\sqrt{ab}} \leq 16/3.$

Step two: We have $ \frac {1} {1\minus{}\sqrt{x}} \leq \frac {32x\plus{}10} {9}$ for $ 0 \leq x \leq 1/4$.  This inequality is equivalent (on the given interval) to $ \sqrt x \leq \frac {32x\plus{}1} {32x\plus{}10},$ which upon squaring and rearranging is equivalent to $ (16x\minus{}1)^2 (1/4\minus{}x) \geq 0,$ which is clearly true.  (I found this by considering the line tangent to the graph of $ f(x)\equal{}1/(1\minus{}\sqrt{x})$ at $ x\equal{}1/16.$  Note here that if $ a\plus{}b\plus{}c\plus{}d\equal{}1,$ then it is immediate that $ ab \leq 1/4$ with similar inequalities for $ bc$ and $ cd$ and $ da.$

Step three: $ ab\plus{}bc\plus{}cd\plus{}da \leq 1/4.$  This is so because $ ab\plus{}bc\plus{}cd\plus{}da\equal{}(a\plus{}c)(b\plus{}d)\equal{}(a\plus{}c)(1\minus{}(a\plus{}c)) \leq 1/4,$ the last inequality being true by the GM/AM inequality.

Step four: By steps two and three we have $ \sum \frac {1} {1\minus{}\sqrt{ab}} \leq \sum \frac {32ab\plus{}10} {9} \equal{} \frac {32} {9} \sum (ab) \plus{}\frac{40}{9} \leq \frac{32}{9} \frac 14 \plus{}\frac {40}{9} \equal{} \frac {16}{3},$ and we are done.

Oh, yeah. By GM-HM inequality,
I just want to be direct.
\[ \sqrt[4]{(1\minus{}ab)(1\minus{}bc)(1\minus{}cd)(1\minus{}da)}\geq\dfrac4{\dfrac1{1\minus{}ab}\plus{}\dfrac1{1\minus{}bc}\plus{}\dfrac1{1\minus{}cd}\plus{}\dfrac1{1\minus{}da}}\geq\frac34.\]
Your step 2 is right.

I am wondering why Cauchy does not work for it.

Let $ \ell$ be the left hand side of the desired inequality. Then notice
\[ \ell\equal{}(1\minus{}ab\minus{}cd\plus{}abcd)(1\minus{}bc\minus{}da\plus{}abcd)\equal{}\left[\frac12(a\minus{}b)^2\plus{}\frac12(c\minus{}d)^2\plus{}\frac12\plus{}abcd\right]\left[\frac12(b\minus{}c)^2\plus{}\frac12(d\minus{}a)^2\plus{}\frac12\plus{}abcd\right].\]
By Cauchy Schwarz,
\[ \ell\geq \left[\frac12|(a\minus{}b)(c\minus{}b)|\plus{}\frac12|(c\minus{}d)(a\minus{}d)|\plus{}\frac12\plus{}abcd\right]^2.\]

But how to proceed?  :roll:


There are 5 trains from Chicago to Naperville in the morning, and 4 trains returning in the afternoon.
a. How many different ways can a daily trip be made?
b. If Jill and Fred make the trip each day, what is the probability they will be on the same train on a given day?

[quote="123456789"][hide]a. There are $ 5$ ways to choose a train from Chicago to Naperville, and $ 4$ ways to choose a train from Naperville to Chicago, so there are $ 5 \cdot 4 \equal{} \boxed{20}$ different ways for a daily trip to be made.[/hide]

b. Do Jill and Fred have to be on the same train in both the morning and the afternoon?[/quote]
b.  No, they do not have to be on the same train in both morning and afternoon.  Only one

since they don't need to be on the same plane you must subtract the intersection(ability to be in the same plane)i.e1/5+1/4-(1/5)(1/4)=8/20=2/5[quote]lack of money is the root of all evil[b] :D [/b][/quote]


integration
LaTeX
trigonometry
real analysis
real analysis theorems

[b]compute $\int (x*cos(x))/(1+x^2)dx$ for $x=0$ to $x=\pi$[/b]
sorry my english is worst :P
can you give me tutorial in LateX

[quote="edigoras"]
can you give me tutorial in LateX[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=5261

[quote="FMako"]$\int_{0}^{\pi} \frac{x \cos x}{1+x^2} dx$

Can anyone solve this?[/quote]
its too much calculation,
You need to solve ODE
\[ y'' - y = \frac{1 - \cos \pi x }{x}  \] then the answer is y'(1).  :)

Edit: some typos have been fixed

I've found that ODE is somewhat difficult to solve  :?
Here how i got the ODE
\begin{eqnarray*} I(p) &=& \int_{0}^{\pi} \frac{\sin px}{x^2+1} \, dx \\ I'(p) &=& \int_{0}^{\pi} \frac{x \cos px}{x^2+1} \, dx \\ I''(p) &=& \int_{0}^{\pi} \frac{-x^2 \sin px}{x^2+1} \, dx \\ \end{eqnarray*}
So $I(p) - I''(p) = \int_{0}^{\pi} \frac{x^2 \sin px + \sin px}{x^2+1} \, dx = \int_{0}^{\pi} \sin px \, dx = \frac{1}{p} (1 - \cos \pi p)$


$\sum_{i=1}^na_i=\sum_{i=1}^na_i^2$ Ptove that
1) $\sum_{i=1}^n(a_i^3+a_i)\leq2n$, where $n<10, n\in\mathbb N$
2) Inequality of 1) not correct for $n\geq50.$  :)

[quote="altusi"]i think it's true for all n
 :huh:  isn't it?[/quote]
Give reasons, please. ;)

Taking $a_{1}= ... = a_{n-1}= \frac{1}{2}$ and $a_{n}= \frac{1+\sqrt{n}}{2}$, we find
$\sum_{i}(a_{i}^{3}+a_{i})-2n = \frac{1}{8}n^{3/2}-n+\frac{7}{8}\sqrt{n}= \frac{\sqrt{n}}{8}(\sqrt{n}-1)(\sqrt{n}-7)$ which is positive for $n \ge 50$. :D

And if I understand $n-1$ EV theorem right (and it is quite possible I don't :P :P) then this is also proof of the original inequality for $1 \le n \le 49$.

[quote="Xevarion"]
And if I understand $n-1$ EV theorem right [/quote]
I don't sure! You  were deceived!:D

:( well it seems I must go study some more uses of it so I can do it right myself :P

We may write the inequality as
$\sum_{i=1}^{n}(a_{i}-1)^{2}(a_{i}-2) \leq 0$.
If $n<10$, then all $a_{i}\leq 2$, and the inequality is true.
Indeed, if $a_{1}>2$, then $(a_{1}-1/2)^{2}>9/4 \geq n/4$, which contradicts the hypothesis $\sum_{i=1}^{n}(a_{i}-1/2)^{2}=n/4$.

OK I understand what's the problem with my thought from before. Can someone prove it for $11 \le n \le 49$? It would be sufficient to prove that the max of LHS happens when all the $a_{i}$ are equal.... but how?  :o


calculus
integration
trigonometry
calculus computations

One can write your integral as

$ I: \equal{}\frac12 \int_0^1 \frac{du}{\sqrt{(1\minus{}u^2)(1\minus{}k^2 u^2)}}$ with $ k\equal{}\cos \frac{\pi}{12}\equal{}\frac{\sqrt{6}\plus{}\sqrt{2}}{4}$, and thus $ k^2\equal{}\frac{2\plus{}\sqrt{3}}{4}$ .

And this is $ \frac12 \int_0^{\frac{\pi}{2}} \frac{dx}{\sqrt{1\minus{}k^2 \sin^2 x}}\equal{}\frac12 K(k)\equal{}\frac{3^{3/4}\cdot \Gamma^3\left(\frac13\right)}{\pi\cdot 2^{10/3}}$ . See this [url=http://www.mathlinks.ro/viewtopic.php?t=302862]link[/url].


modular arithmetic
floor function
inequalities

donbalehaye $c_{1},c_{2},\dots$ az adade tabiei mafrooz ast.baraye har do adade tabiei $m$ va $n$ ba $1\leq m \leq\sum_{i=1}^{n}c_{i}$ mitavan adade tabiei $a_{1},a_{2},\dots,a_{n}$ ra chenan ekhtiar kard ke :
$m=\sum_{i=1}^{n}\frac{c_{i}}{a_{i}}$
bishtarin meghdare $c_{i}$ ra baraye har ${i}$ moshakhas konid.
(IRAN 2000)

tamame tavabe $f: R \rightarrow R$ ra chenan taein konid ke baraye har $x,y\in R$:
$f(x^{2}-y^{2})=(x-y)(f(x)+f(y))$

The numbers 1, 2, ... , 2002 are written in order on a blackboard. Then the 1st, 4th, 7th, ... , 3k+1th, ... numbers in the list are erased. Then the 1st, 4th, 7th, ... 3k+1th numbers in the remaining list are erased (leaving 3, 5, 8, 9, 12, ... ). This process is carried out repeatedly until there are no numbers left. What is the last number to be erased?

[quote="khashi70"]
be nazaram tarafe dovvomo hatta mishe $\frac{3}{2}$ gharar dad
[/quote]
hammon $\frac{3}{2}$dorost hast man havasam nabood eshteb type kardam va ba $AM-GM$kheili rahat hal mishes :wink:va mishe:
$\frac{(x^{2}+y^{2}+z^{2})^{2}}{2(xyz)(x+y+z)}\ge\frac{(x+y+z)^{3}}{18xyz}\ge\frac{3}{2}$

[quote="sepehr ahmadzadeh"]agar $x=y$ natije mishavad $f(0)=0$ va ba gharar dadane $x=-1$ va $y=0$ natije mishavad $f(1)=-f(-1)$pas gar $x=a$ va $y=1$ ,va $x=1$va $y=-1$ darim:
$f(a^{2}-1)=(a-1)(f(a)+f(1))$
$f(a^{2}-1)=(a+1)(f(a)-f(1))$
pas az indo natije mishe $f(a)=f(1).a$ pas tamae tavebei ke dar moadeley fogh sedgh mikonand baiad be soorat $f(x)=kx$ bashad. :D[/quote]
khodaii khodetoon ke soal mizanin hallesho nagin ta vaghti ke kasi hallesho nazade vaysin yeki hal kone bad khastin halletoono bezanin dar zemn hame lotf konan soalhaye jabreshoono inja bezanan bahaltar mishee

dige vaghti soal be in tabloeiro kesi hal nakardatesh chi kar mishe kard?
 :?:

hame tavabe f,g ra biabid betori ke baraye har x,y haghighy dashte bashim:
$f(x-y)=g(f(x))g(f(y))$

farz konid $p$ yek adade aval bashad be torike $p\equiv1\pmod{4}$.meghdare $\sum_{k=1}^{p-1}(\lfloor\frac{2k^{2}}{p}\rfloor-2\lfloor\frac{k^{2}}{p}\rfloor)$ ra mohasebe konid.

Bebakhsid, man nemitavanam dar zabane farsi javabe yek masale ro khoob tozi bedaham.:(

[i]Solution:[/i] Multiplying both sides by $(a+1)(b+1)(c+1)$, we obtain
\[a(c+1)+b(a+1)+c(b+1)\geq\frac{3}{4}(a+1)(b+1)(c+1).\]
Simplifying and using the given fact $abc=1$, we have
\[\\\frac{1}{4}(ab+bc+ac+a+b+c)\geq\frac{3}{2}\\\implies ab+bc+ac+a+b+c\geq 6,\]
which is clearly true by the arithmatic mean-geometric mean inequality.

Omidvaram ke javabe ghanekonadeyi bashe.

Masoud Zargar

[quote="khashi70"]If $a,b,c>0$ prove this :
$(a+b)(a+c)\geq 2\sqrt{abc(a+b+c)}$[/quote]
Expand and simplify.  Then the conclusion follows by muirhead since $(4,0,0)\succ(2,1,1)\wedge (2,2,0)\succ(2,1,1)$.

Masoud Zargar

kheyli ghashange! Chera fekram be substitution naresid? :(

[quote="sepehr ahmadzadeh"]For $a,b,c,d,e\in\mathbb R$ prove that:
$a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\leq\frac{2}{\sqrt{3}}(ab+bc+cd+de)$[/quote]
mesal naghz:a=b=c=0,d=1,e=-1

[quote="sepehr ahmadzadeh"]For $a,b,c,d,e\in\mathbb R$ prove that:
$a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\leq\frac{2}{\sqrt{3}}(ab+bc+cd+de)$[/quote]
dorostesh ine
 $a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\geq\frac{2}{\sqrt{3}}(ab+bc+cd+de)$

ba arze poozesh az hame!! az zire dastam dar raft>> khashi merC

for all positive integers $n,m$ with $n\geq m$  prove that :
$(n^{2}+n)^{m}\geq \frac{(n+m)!}{(n-m)!}\geq m!2^{m}$

[quote="khashi70"]For $0 \leq a,b,c\leq 1$ prove that :
$\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}\leq 1$[/quote]

[b][color=blue]Rahe hale aval: [/color][/b]

$\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}\le \frac{a+bc}{2}+\frac{(1-a)+(1-b)(1-c)}{2}=\frac{2+2bc-b-c}{2}=1+bc-\frac{b+c}{2}\le 1+bc-\sqrt{bc}\le 1$

[color=blue][b]Rahe hale dovom: [/b][/color]

agar $x\in [0,1]$ darim $\sqrt x\le\sqrt[3] x$ pas:

$\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}\le \sqrt[3]{abc}+\sqrt[3]{(1-a)(1-b)(1-c)}\le \frac{a+b+c+1-a+1-b+1-c}{3}=1$
[b][color=blue]Rahe hale sevom: [/color][/b]

$\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}\le \sqrt{bc}+\sqrt{(1-b)(1-c)}\le 1$

farz konid $x,y,z>0$ va $x+y+z=xyz$

sabet konid:

$(x-1)(y-1)(z-1)\le 6\sqrt 3-10$


geometry
rectangle
superior algebra
superior algebra theorems

maybe someone could explain me this text (maybe with the outline drawing or smth.. )

[url=http://img502.imageshack.us/my.php?image=25348113ni0.jpg][img]http://img502.imageshack.us/img502/9874/25348113ni0.jpg[/img][/url]
]

[quote="BlueVelvet"]http://en.wikipedia.org/wiki/Klein_four-group[/quote]

Here you can see the example with the rectangle: http://planetmath.org/encyclopedia/Klein4Group.html


superior algebra
superior algebra theorems

http://www.amazon.com/gp/reader/0387905189/ref=sib_dp_ptu#reader-link

There is a link to the text that my question is about. My question relates to the first paragraph of the "EXAMPLE" on page 2. There is says that "For if M were a set, then either $ M \in M$ or $ M \notin M$." I am confused about that statement. Why does that follow from M being a set? They defined a set as something that is an element of some class.

this is true in common logics ... accept it. it's intuitive ;).

It is intuitive...I just don't see why they needed to say "if M were a set". If M were a proper class, is it not true that either $ M \in M$ or $ M \notin M$? That seems to be Hungerford's point...


ceiling function
inequalities proposed
inequalities

prove that for every $n\ge0$ we have:
$\lceil\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}\rceil=\lceil \sqrt{9n+8}\rceil$

Note that $9n+8$ is not a perfect square. It means that $[\sqrt{9n+8}]=[\sqrt{9n+7}]$.
We have $\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}<\sqrt{9n+9}$, moreover $\sqrt{n}+\sqrt{n+2}>\sqrt{4n+3} $ and $\sqrt{4n+3}+\sqrt{n+1}> \sqrt{9n+7}$. Hence the result.


Find the number of square units bounded by the $ x$-axis, the $ y$-axis and the line $ x\plus{} y\equal{} 6$.

If we graph this, we find the the height is 6 and the base is 6. So:

$ \frac {1}{2}bh \equal{} A$

$ \frac {1}{2} \times 6 \times 6 \equal{} A$

$ \boxed{18} \equal{} A$

EDIT: Yeah, I meant $ A$. Fixed.  :oops:

When you make this, you form a triangle.

[quote="isabella2296"]If we graph this, we find the the height is 6 and the base is 6. So:

$ \frac {1}{2}bh \equal{} A$

$ \frac {1}{2} \times 6 \times 6 \equal{} A$

$ \boxed{18} \equal{} 6$[/quote]

...18=6??

Perfect place for me to try asymptote!

[asy]size(250);
Label f;
f.p=fontsize(6);
xaxis(-16, 16, Ticks(f,8,5));
yaxis(-16, 16, Ticks(f,8,5));
draw((-10,16)--(16, -10));[/asy]

How do you make arrow tips?

As dragon's picture shows, a right triangle is formed.

[quote="isabella2296"]If we graph this, we find the the height is 6 and the base is 6. So:

$ \frac {1}{2}bh \equal{} A$

$ \frac {1}{2} \times 6 \times 6 \equal{} A$

$ \boxed{18} \equal{} A$
[/quote]

But doesn't graphing take awhile? Faster way:

Rewrite x+y=6 as
x=-y+6

and

y=-x+6

Note that if y=0 in the first one, x=6 which would be the x intercept. In the second one, the y intercept is 6, so the base is 6 and height 6. So, 36/2=18


vector
function
calculus
integration
linear algebra
matrix
real analysis

i was wondering if someone could help me with this problem:

Suppose that f is a continuous, vector-valued function and that T is a linear transformation on R^3.  Show that "T o f" (stands for T
composite f) is continuous and that INTEGRAL[T o f] = T(INTEGRAL[f]).
---------------------------------------------------------------

I think i know the general epsilon - delta proof for continuity of composite
functions.  The theorem states that "If f:R^n -> R^m is continuous at
vector "a" and T:R^m -> R^p is continuous at f(a) then Tof:R^n
->R^p is continuous at vector a."

The proof is as follows:
For any E>0, we want to show that there exists a d>0 such that |x-a|<d
=> |T(f(x)) - T(f(a)| < E.

Since T is continuous at f(a), then for E>0, there exists a d1 such that
|f(x)-f(a)| < d1 => |T(f(x))-T(f(a))| < E

For f, use d1 as its "E value".  Then there exists d2 > 0 such that
|x-a|< d2 => |f(x)-f(a)|< d1.

Let d=d2.  then:

|x-a| < d => |f(x) - f(a)| < d1

but from the continuity of T, we know |f(x)-f(a)| < d1 =>
|T(f(x))-T(f(a))| < E

so |x-a| < d => |f(x) - f(a)| < d1 => |T(f(x))-T(f(a))| < E
which proves that Tof is continuous at a.

however, this proof only works if both f and T are continuous.  going back to the original question, it states that f is continuous, but it doesn't say anything about the continuity of T.  so i need to first prove that T is continuous.  How do i prove that a linear transformation on R^3 is continuous using epsilon-delta proof?  i'm not very familiar with transformations.

also, the proof i wrote above is for continuity at a specific point a.  the question just states that f is continuous - i'm assuming it's continuous everywhere.  would i still use the same proof for this question?

I don't really know how to do the second part of the problem: show that INTEGRAL[T o f] = T(INTEGRAL[f]).  i suspect that it has something to do with the chain rule but i only know how to use the chain rule for derivatives.  how do you deal with composite functions when it comes to integrals?

thanks in advance

Every linear tranformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ is continuous, so $T$ is continuous. That's why nothing was said about its continuity: linearity implies continuity.

right - i guess linearity implying continuity is quite intuitive.  but my instructor still requires a proof for it.  i'm just not sure how to do the epsilon delta proof for it - i've never done this kind of proof with transformations before.

You would like to show the following: given any $\epsilon>0,\,\exists\,\delta>0$ such that if $||x-y||<\delta,$ then $||T(x)-T(y)||<\epsilon.$

Note that this is not merely continuity, it is uniform continuity.

As a first step note that $T(x)-T(y)=T(x-y)$ because $T$ is linear. That's the key to the uniformity. From this is sufficles to show that if $||u||<\delta$ then $||T(u)||<\epsilon.$

Now we have to ask what we mean by $||u||.$ Most familiar to you is the Euclidean distance:

$||x||_2=\sqrt{x_1^2+x_2^2+\cdots+x_n^2}.$

But when it comes to norms of operators, the Euclidean norm is annoyingly complicated to work with - and we don't need to put up with that complication in this case. Let's introduce another notion of distance:

$||x||_{\infty}=\max\left(|x_1|,|x_2|,\dots,|x_n|\right).$

You can show - I'll leave this to you - that in $\mathbb{R}^n,$ $||x||_{\infty}\le||x||_2\le\sqrt{n}||x||_{\infty}.$

Now suppose $T: \mathbb{R}^n\mapsto\mathbb{R}^m$ is linear. Then $T$ is given by an $m\times n$ matrix $A$ as follows:

$(Tx)_i=\sum_{j=1}^na_{ij}x_j.$

Let $M= \max_{1\le i\le m}\sum_{j=1}^n|a_{ij}|$ - that is, the maximum row sum of the matrix.

Then $|(Tu)_i|\le\sum_{j=1}^n|a_{ij}||u_j|\le\sum_{j=1}^n|a_{ij}|\cdot ||u||_{\infty}\le M||u||_{\infty}.$

Hence, $||T(u)||_{\infty}\le M||u||_{\infty}$

If we want that in terms of the Euclidean distance, we have proved that

$\frac1{\sqrt{m}}||T(u)||_2\le M||u||_2$

and if $||u||_2<\frac{\epsilon}{\sqrt{m}M}$, then $||T(u)||_2<\epsilon.$


What is the largest possible value of the greater of the two non zero integers for which difference of their cubes equals square of their difference?

$ x^{3} \minus{} y^{3} \equal{} (x \minus{} y)^{2}$

$ (x \minus{} y)(x^{2} \plus{} xy \plus{} y^{2}) \equal{} (x \minus{} y)(x \minus{} y)$

$ x^{2} \plus{} xy \plus{} y^{2} \equal{} x \minus{} y$

$ x^{2} \minus{} x \plus{} xy \plus{} y^2 \plus{} y \equal{} 0$

This is the equation of an ellipse which passes through the points $ (0, 0), (0, \minus{} 1), (1, \minus{} 2), (2, \minus{} 2), (2, \minus{} 1)$ and $ (1,0)$.

I would like to be able to say that the answer is $ 2$ when the two integers are $ 2$ and $ \minus{} 2$ and $ x^{3} \minus{} y^{3} \equal{} 16 \equal{} (2 \plus{} 2)^{2}$, but I'm not sure as I suspect that the major axis does not run from $ (0, 0)$ to $ (2, \minus{} 2)$. However, differentiating and substituting $ x\equal{}2, y\equal{}\minus{}2$, I do get a gradient of $ 1$ which is perpendicular to the line $ y\equal{}\minus{}x$ so this line might be the major axis.