inequalities
geometry
inequalities unsolved

Let a,b,c be the positive real number and x,y,z>0 satisfying $ xy\plus{}yz\plus{}zx\equal{}3$
Prove that $ \frac{a}{b\plus{}c}(y\plus{}z)\plus{}\frac{b}{c\plus{}a}(z\plus{}x)\plus{}\frac{c}{a\plus{}b}(x\plus{}y)$≥3

[quote="Nak"][quote="SUPERMAN2"]Let a,b,c be the positive real number and x,y,z>0 satisfying $ xy \plus{} yz \plus{} zx \equal{} 3$
Prove that $ \frac {a}{b \plus{} c}(y \plus{} z) \plus{} \frac {b}{c \plus{} a}(z \plus{} x) \plus{} \frac {c}{a \plus{} b}(x \plus{} y)$≥3[/quote]
See here http://reflections.awesomemath.org/2008_1.html[/quote]
Thanks but I can not find anything from your links.Can you give me your solution?

[quote="SUPERMAN2"]
Thanks but I can not find anything from your links.Can you give me your solution?[/quote]
OK, try this http://reflections.awesomemath.org/2008_1/unexpected_ineq.pdf page 1 theorem 1

$ \mbox{If }a,b,c,x,y,z > 0 \mbox{ and } xy + xz + yz = 3, \mbox{ then: } \sum{\frac {a}{b + c}.(y + z)}\ge 3. \ (1) \\
\mbox{Acording the inequality; } \sum{a^2x_1\ge 4F.\sqrt {\sum{x_1y_1}} \ (2)}$
$ a,b,c \mbox{ the sides and F = area of triangle. Let: } \\
a = \sqrt {y + z}, b = \sqrt {x + z}, c = \sqrt {x + y} \mbox{ and } x_1 = \frac {a}{b + c}, y_1 = \frac {b}{a + c}, z_1 = \frac {c}{a + b}, \\
\mbox{the inequality (2) obtaines: }$
${ \sum{\frac {a}{b + c}.(y + z)}}\ge 4.\sqrt {\frac {\sum{yz}}{4}}.\sqrt {\sum{\frac {ab}{(b + c)(c + a)}}} \mbox{ because: } \\
\sum{\frac {ab}{(b + c)(c + a)}}\ge\frac {3}{4}, \ (3) \mbox{ obtaines (1).}$
$ \mbox{proof the inequality (3): } \\
\sum{\frac {ab}{(b + c)(c + a)}}\ge\frac {3}{4} \Leftrightarrow 4.\sum{ab(a + b)\ge 3.(a + b)(a + c)(b + c)\Leftrightarrow}$
$ \Leftrightarrow 4.\sum{(a^2b + ab^2)}\ge 3.\prod {(S - a)} = 3.\left[S^3 - (a + b + c).S^2 + S.(\sum{ab}) - abc \right]\Leftrightarrow \\
(\mbox{where: }S = a + b + c)$
$ \Leftrightarrow 4.\sum{(a^2b + ab^2)}\ge 3.\left(\sum{ab}\right).\sum{a} - 3abc \Leftrightarrow \\
\Leftrightarrow \sum{(a^2b + ab^2)}\ge 6abc\mbox{; and aply AM - GM. obtaines: } \\
\sum{(a^2b + ab^2)}\ge6.\sqrt [6]{\prod{(a^2b.ab^2)}} = 6abc.$
--------------------------------------------------------------------------------------------------------------------------
$ \mbox{The inequality: } \ (*) \ \sum_{cyclic}{\frac {a}{(a + 1)(b + 1)}\ge \frac {3}{4}\mbox{, for }abc = 1; a,b,c > 0 \\
\mbox{let: } a = \frac {x}{y}, b = \frac {y}{z}, c = \frac {z}{x}\mbox{, obtaines: }}$
$ \sum{\frac {\frac {x}{y}}{\left(\frac {x}{y} + 1\right).\left(\frac {y}{z} + 1\right)}}} = \sum{\frac {xz}{(x + y)(y + z)}} = \sum{\left(\frac {x}{y + z}.\frac {z}{x + y}\right)\ge\frac {3}{4}$
$ \mbox{ beucause }\Leftrightarrow \sum{\left(\frac {a}{b + c}.\frac {b}{c + a}\right)\ge\frac {3}{4} \\
\mbox{ for } a\rightarrow x, b\rightarrow y, c\rightarrow z.}$

Use the following auxiliary inequality;

Lemma. Let $ a,b,c,x,y,z$ be nonnegative real numbers.
Then, $ a(y \plus{} z) \plus{} b(z \plus{} x) \plus{} c(x \plus{} y) \geq 2 \sqrt {(ab \plus{} bc \plus{} ca)(xy \plus{} yz \plus{} zx)}$

Proof) Add $ ax \plus{} by \plus{} cz$ to both sides to obtain

$ (a \plus{} b \plus{} c)(x \plus{} y \plus{} z) \geq (ax \plus{} by \plus{} cz) \plus{} 2 \sqrt {(ab \plus{} bc \plus{} ca)(xy \plus{} yz \plus{} zx)}$
By Cauchy-Schwartz Inequality, we obtain

$ (a \plus{} b \plus{} c)^2 (x \plus{} y \plus{} z)^2 \equal{} ((a^2 \plus{} b^2 \plus{} c^2 ) \plus{} 2(ab \plus{} bc \plus{} ca))((x^2 \plus{} y^2 \plus{} z^2 ) \plus{} 2(xy \plus{} yz \plus{} zx))$
$ \geq (\sqrt {(a^2 \plus{} b^2 \plus{} c^2 )(x^2 \plus{} y^2 \plus{} z^2 )} \plus{} 2 \sqrt {(ab \plus{} bc \plus{} ca)(xy \plus{} yz \plus{} zx)})^2$
$ \geq ((ax \plus{} by \plus{} cz) \plus{} 2 \sqrt {(ab \plus{} bc \plus{} ca)(xy \plus{} yz \plus{} zx)})^2$

Taking the square root of both sides, we arrive at the conclusion.

Substitute $ a \minus{}> \frac {a}{b \plus{} c}$, $ b \minus{}> \frac {b}{c \plus{} a}$, $ c  \minus{}> \frac {c}{a \plus{} b}$
Let $ xy \plus{} yz \plus{} zx \equal{} 3$. Then our auxiliary inequality is equivalent to

$ \sum \frac {a}{b \plus{} c} (y \plus{} z) \geq 2 \sqrt {( \sum \frac {a}{b \plus{} c} \frac {b}{c \plus{} a})(xy \plus{} yz \plus{} zx)} \equal{} \sqrt {12( \sum \frac {a}{b \plus{} c} \frac {b}{c \plus{} a} )}$
It is enough to show that $ \sqrt {12( \sum \frac {a}{b \plus{} c} \frac {b}{c \plus{} a} )} \geq 3$
Clearing denominators, squaring and sending all terms to the RHS yields $ 3a(b \minus{} c)^2 \plus{} 3b(c \minus{} a)^2 \plus{} 3c(a \minus{} b)^2 \geq 0$
Which is trivial :)


...something I hope I do not have, but I probably do.

Is math inherently unique, does it possess some special quality no other discipline shares? I feel so. Is math "different"? Does one process and learn it differently from other subjects?

(I am trying to get more interesting discussions going here in the round table; there seems to be a slowdown of interesting things - but maybe it's just my preferences... On second thought, I can't quite imagine this topic turning into anything very interesting, but oh well.)

Mathematics has the distinction of a priori certainty. This in itself justifies any chauvinism on the part of the mathematician (but we might like to rub it in everyone else's faces once in a while with a Mathematician's Apology of course!:rotfl:).

The problem of certainty in epistemology is best expressed even all these centuries later by Descartes considerations- what knowledge do we really have apart from what can be derived from consciousness (and its irrefutable existence)? Though Descartes goes on to an absurd ontological argument for God, the answer would simply be 'none'. Hence, mathematics constitutes knowledge whenever it is based solely on a priori considerations (almost all modern theories within it are) and does not assume a naive reality!

That is, until you get into the complex and theoretical.   :wink:

After having read through these posts a discussion I had with a couple friends popped into my head: A classmate of ours (a very talented classmate at that!) is planning on going to medical school, and we commented on how much of a waste it would be. Can you believe that? We had such inflated egos as to call becoming a doctor a waste! I thought it was a pretty good example of mathematical chauvinism  :P.

[quote="annoyingpi"]Math is different in that it does not depend upon human emotion. In the humanities, subjective emotion is commonly applied and people state their opinions without being able to prove their opinions or their reasons. In the sciences, people try to leave subjectivity behind with the scientific method, but they do not always succeed. Experimental error is always present and different people tend to interpret data differently, especially if they already have some sort of preconception of what should be true and what should not. Math has none of these problems; math is logical and it is something that everyone can agree on, since a proof is either right or wrong. Math has no fuzzy areas, so it is unique and it is processed differently (completely separately from subjective emotions and preconceptions).[/quote]

I don't know if I'd call if completely lacking in 'fuzzy areas'- considered within its own scope, the foundations of mathematics becomes a truism of sorts under Godel's incompleteness theorems. Even the Zermaelo-Frankel set theory with the axiom of choice we use today as a basis for mathematics today will have paradoxes or inconsistencies when considered at some arbitrarily complex level (which, admittedly for most mathematicians won't be a problem at all!).
(this is why I like being a fictionalist- formalism didn't work, but the spirit remains both in our attitude today and in fictionalism, and in the latter we easily sidestep the consistency vs completeness issue)

max_tm's story got me somewhat guilty though; I guess we do need to engage in the seeming 'trifles' of the world sometimes!  :oops:

What I meant by fuzzy areas are cases where something can be interpreted two different ways. In math, a proof is either right or wrong, and a statement can be in exactly one of the categories of true, false, or impossible to determine. Thus, math isn't fuzzy.

Mathematicians often need to realize that even though they would not be interested in anything except math, other people want to do other things. In a world where no one wants to be a doctor, all of the mathematicians would die despite their feelings of superiority.

[quote="annoyingpi"]What I meant by fuzzy areas are cases where something can be interpreted two different ways. In math, a proof is either right or wrong, and a statement can be in exactly one of the categories of true, false, or impossible to determine. Thus, math isn't fuzzy.
[/quote]

True, but it does round back- this is only because it assumes (and, with sufficient rigour- acknowledges) just enough to make the results so.
However, as we have seen, the proof and method and hence the results ultimately succumb to being little more than tautology when considered within its own scope.

Not that I'm disappointed with this of course- I'd rather have this with its curious (seeming) misgivings rather than something hand-wavy (eg- philosophy) or goodness-forbid- something practical! (eg- about everything else!)  :D

Now that I think about it, a long time ago I wanted to become a surgeon! Maybe that's more noble than what I plan on doing now, but I think it'll suffice that everyone acknowledges the significance of what everyone else does.

Math is shockingly fuzzy.

We'd like to think that there's always a "right" and "wrong" in math. That's what you're made to believe. And, relative to other disciplines, this is pretty much the case.

But if one looks deeper, doubts will arise. How do we define "right" and "wrong"? This is inherently the same philosophical question of what truth is. Is a mathematical truth something that has a proof? What about undecidable propositions? Can math be programmed? If not, is it still entirely objective and free of the complexities of human perception? What is a proof anyways? It seems to depend on logic. Is logic universal? If no sentient beings had ever existed, would logic exist? Mathematicians and logicians have had disputes in the past over the precise axioms of logic.

We'd like to think that math and logic are universal and unique. But as physical human beings, we are bound by our own capacity to reason. It would be impossible to know whether or not our math is completely free of these hideous creatures called 'humans'. And what exactly do we mean by that? Who knows?

So we see, mathematics treads the narrow ridges of the cliff of sanity. Let your mind wander free for even a moment and you fall off into the roaring seas of philosophy. Mathematics is not at all cut and dry.

This is where my doubts of 'math chauvinism' being justifiable arise.

It's often thought that logic is what encompasses mathematics- but I'd have thought Principia Mathematica's effort and subsequent failure would have enlightened us as to how such a claim is false. It is the other way around- mathematics can give rise to different sets of logic and base theories upon them. Regardless of the consistency vs completeness problem, if logic is all we have, then ergo- mathematics is all we have in discerning truths. Since no other discipline comes even arbitrarily close to such a level- mathematical chauvinism is justified. QED.

[quote="Aqua3993"]I can just say one sentence: Math is discovered, not created.

Some might say that science is also discovered. However, much of science is heavily dependent on math. Also, scientific laws are almost never withheld forever (i.e. Newton's gravitation vs. Einstein's general relativity). Therefore, science is created by us in a sense.[/quote]

I beg to differ- mathematics is almost certainly created and not discovered (though, should one adopt sufficiently contrieved semantics about it- one can twist the words either way). I would have thought that given how unsatisfactory Platonism can be, and how inadequate logicism has been shown to be throughout, that this would be taken for granted. Modern mathematics holds to Hilbert's formalist programme (though it has been shown to lie on the shakiest foundations), and the amount of leeway one has certainly allows for theories to be developed in any direction (this is any direction imaginable really- given an internal measure of consistency holds); while this is sometimes considered an objection to the programme, it has none the less given us many a fruitful result. At this point you can propose a tautology, but it remains true that, given this freedom, mathematics is truly a creation.

'Science' as is considered today (on the Galeilean principle), is always looking for explanations by best inference. As such, it is a continually evolving creature, which reaches the next echelon just because whatever can be simply thought to be 'discovered' alters its path. Certainly, whenever mathematics is involved in building a suitable model, the degress of freedom involved often makes it the case that a theory is at best created to suit the observational evidence at a given time, but this is again a property of mathematics as I've outlined above. The difference being that, presumably, given enough 'boundary conditions' our model might one day be made unique enough that we might safely dispense of the need for 'falsifiability' and make a 'discovery' through our efforts of inference!


There was some discussion on how to sign up if your school is clueless or reluctant.
So far I did not see much enthusiasm from neighboring schools to let students from other schools even close to their math teams/ programs/ materials etc. So I do not anticipate any response if I try to sign up a student with another school for AMC.
Is there a list of schools that have room for students from other schools and would be willing to take the other students in for a test? Or homeschools that registered and having room left?
Is it possible to sign up individually?
Thank you.

I'm not sure what you mean. The AMC is already done for this year.

I mean for next year. Just doing my homework.

The AMC won't have a list of which schools have registered for the AMC until probably around next November/December. You can start checking up with schools personally right now to see if they'll let you compete next year, though.

[quote="AMCDirector"]The preceding information from Nathan is correct and is good advice.
Throughout this year make contact with local high schools and high school math teachers and math team coaches.
Check with the AMC office in December 2006 about schools in your region registered for 2007.

Steve Dunbar
MAA Director, Americna Mathematics Competitions[/quote]

Since we have to homeschool, I have had to go through this the last two years. Last year I send email to 3 local private schools. Two got back to me - one to give us the go ahead, and the other to tell us they only offer alternating years. The school we went to last year didn't plan to offer it this year, so I sent email to another private school, and within 15 minutes I had an affirmative.  Since my son has a chance of qualifying for the USAMO, I chose among the schools who would probably have a qualifier this year.

I sign my son up as a guest, which costs $\$$32, thru the AMC. AMC sent them a copy of the test, so he does not use up one of their tests. He will not be eligible for their site honors if he does well, nor will he affect their statistics if he does poorly.

Last year we were at a local math competition. I asked someone from a local public school (not our district), if he could take it there. I got a very negative response. This school had once allowed a middle school student from another district to take the test, and they had a bad experience with the mother.  He ranted long enough for me to figure out who the family was. I will stick with the private schools!


Hello, to everyone once again!!

It was today that I learned that I can give both SAT I and SAT II in the same day. So I decided to take the SAT Reasoning Test on June and live the Subject Tests for October. In Mathematics and Writing I expect >720. But I have got problems with Critical Reading(in my practice test I get 610), because a lot of the vocabulary there, is unknown to me. I guess that the aim of this test is to check your critical thought but the thing is that I don't know many key words of the passages or questions and this makes me many times not to understand what I am asked (if the passages were at Greek I could very easily answer the questions). Isn't that test a bit unfair for non-native speakers (clearly native speakers are in an advantage)?

Do you know if the universities place emphasis on this section as concerns non-native speakers. Because in this case I better improve in the remaining time my vocabulary.

cheers mate...ok now look...for non native speakers the US unis perfectly understand the fact that it is really tough for foreign students to cope with the reading test....thus if you are able to score about 610 then you are great in my opinion...if you sum up your score it comes about 2050, which is great...you can rock the sat subjects too by scoring over 700 and that's it...plus you might need to score over 100 on the TOEFL since you are greek....good work and good luck on the test day...

and that's the way the cookey crumbles....sat's are great...


Find all functions $ f$ $ :$ $ R$ $ \longrightarrow$ $ R$ that for every $ x$,$ y$ $ \in$ $ R$ :
$ (i)$ $ f(x)$ $ \leq$ $ x$
$ (ii)$ $ f(x \plus{} y)$ $ \leq$ $ f(x)$ $ \plus{}$ $ f(y)$

$ f(0)\equal{}f(0\plus{}0)\le f(0)\plus{}f(0)$ so $ 0 \le f(0)$
Then $ 0 \le f(0)\equal{}f(x\plus{}(\minus{}x)) \le f(x)\plus{}f(\minus{}x) \le x\plus{}(\minus{}x)\equal{}0$ .
Equality holds only when $ f(x)\equal{}x$ .
Thus $ f(x)\equal{}x$ is the only solution.


modular arithmetic
Diophantine equation
algebra unsolved
algebra

Can anyone find a simple way to show that

$ A_n \equal{} \frac {1}{6}\cdot \left[\left(\frac {5 \minus{} 2\sqrt {6}}{2}\right)\cdot (49 \plus{} 20\sqrt {6})^n \plus{} \left(\frac {5 \plus{} 2\sqrt {6}}{2}\right)\cdot (49 \minus{} 20\sqrt {6})^n \plus{} 1\right]$

Is a perfect square??

With  $ n\in\mathbb{N}$

Let $ \alpha \equal{} \sqrt{2} \plus{} \sqrt{3}$, and $ \beta \equal{} 1/\alpha \equal{} \sqrt{3} \minus{} \sqrt{2}$.  Then

$ 5 \plus{} 2 \sqrt{6} \equal{} (\sqrt{2} \plus{} \sqrt{3})^2 \equal{} \alpha^2, \quad 49 \plus{} 20 \sqrt{6} \equal{} (\sqrt{2} \plus{} \sqrt{3})^4 \equal{} \alpha^4$

and similarly

$ 5 \minus{} 2 \sqrt{6} \equal{} (\sqrt{3} \minus{} \sqrt{2})^2 \equal{} \beta^2, \quad 49 \minus{} 20 \sqrt{6} \equal{} (\sqrt{3} \minus{} \sqrt{2})^4 \equal{} \beta^4$.

Thus

$ A_n \equal{} \frac{1}{12} \left( \beta^2 \alpha^{4n} \plus{} \alpha^2 \beta^{4n} \plus{} 2 \right) \equal{} \frac{1}{12} \left((\alpha^{2n\minus{}1})^2 \plus{} (\beta^{2n\minus{}1})^2 \plus{} 2\right)$
$ \equal{} \frac{1}{12} \left(\alpha^{2n\minus{}1} \plus{} \beta^{2n\minus{}1}\right)^2$.

So all that remains is to prove that

$ B_n \equal{} \frac{1}{2\sqrt{3}}\left(\alpha^{2n\minus{}1} \plus{} \beta^{2n\minus{}1}\right)$

is an integer for integer $ n$.  We observe that

$ B_n (\alpha^2\plus{}\beta^2) \equal{} \frac{1}{2\sqrt{3}}(\alpha^{2n\minus{}1} \plus{} \beta^{2n\minus{}1})(\alpha^2\plus{}\beta^2)$
$ \equal{} \frac{1}{2\sqrt{3}}\left((\alpha^{2n\plus{}1} \plus{} \beta^{2n\plus{}1}) \plus{} (\alpha\beta)^2(\alpha^{2n\minus{}3} \plus{} \beta^{2n\minus{}3})\right)$
$ \equal{} B_{n\plus{}1} \plus{} B_{n\minus{}1}$.

Since $ \alpha^2\plus{}\beta^2 \equal{} 10$, it follows that

$ B_{n\plus{}1} \equal{} 10B_n \minus{} B_{n\minus{}1}$,

and since $ B_0 \equal{} B_1 \equal{} 1$, it follows that $ B_n$ is an integer for all nonnegative integers n (and in fact, all integers n, since $ B_n \equal{} B_{1\minus{}n}$).  Therefore $ A_n \equal{} B_n^2$ is a square.

Does that count as "simple?"

Actually it is fairly simple, I even managed to understand everything :)

Very helpful, thank you for the time.

I guess another possible way is to set the recurrence formula for $ A_n$, and then use MI.

Actually the original expression was a recurrence, $ A_n\equal{}98A_{n\minus{}1}\minus{}A_{n\minus{}2}$ , which is fairly easy to demonstrate, but I was wondering what the solution using recurrence theory would look like.

ALWAYS USE PELL EQUATION FOR THIS KIND OF PROBLEM!!!
note that we have
$ 5^2\minus{}2^2.6\equal{}1,49^2\minus{}6.20^2\equal{}1$
so we get that
$ 49 \pm 20.\sqrt{6}\equal{}(5 \pm 2\sqrt{6})^2$
also note that $ (5,2)$ is the fundamental answer of the pell equation
$ x^2\minus{}6y^2\equal{}1$
and $ (49,20)$ the second answer of this equation.
in general we have the set of answers $ (x_n,y_n)$ defined as follows:

$ x_n \pm y_n \sqrt{6}\equal{}(5 \pm 2\sqrt{6})^n$

which implies that

$ x_n\equal{}\frac{(5 \plus{} 2\sqrt{6})^n\plus{}(5 \minus{} 2\sqrt{6})^n}{2}$.

and for your sequence we have

$ A_n\equal{}\frac{1}{6}(\frac{(5\plus{}2\sqrt{6})^{2n\minus{}1}\plus{}(5\minus{}2\sqrt{6})^{2n\minus{}1}}{2}\plus{}1)$

or in other words $ A_n\equal{}\frac{1}{6}(x_{2n\minus{}1}\plus{}1)$
now we take a look at the equation $ x^2\minus{}6y^2\equal{}1$ we can deduce from the reccurence formula of the answers of this pell equation:
$ x_n\equal{}5x_{n\minus{}1}\plus{}12y_{n\minus{}1}$
for all $ n$,
$ x_n$ is odd
and for odd $ n$ ,$ x_n \equiv \minus{}1 \pmod {6}$
and also for all $ n$,$ x_n \equiv \plus{}1 \pmod {4}$.
so for odd $ n$ we have
$ (x_n\plus{}1)(x_n\minus{}1)\equal{}6y^2$
but from what has been said we can conclude that
$ \gcd(\frac{x_n\plus{}1}{6},x_n\minus{}1)\equal{}1$
$ \implies \frac{x_n\plus{}1}{6}\equal{}z_n^2,x_n\minus{}1\equal{}t_n^2,t_nz_n\equal{}y_n,\gcd(t_n,z_n)\equal{}1$
so we have proven that $ A_n\equal{}\frac{x_{2n\minus{}1}\plus{}1}{6}\equal{}z_{2n\minus{}1}^2$.
we r done!


Using the formulae
\[ \cos(3x) \equal{} 4\cos^3(x)\minus{}3\cos(x)\]

and let $ \cos(20^0) \equal{}t$

Assume that t can be represented as a fractional expression such as

\[ t\equal{} \frac{p}{q} \, \mbox{with} \gcd(p,q)\equal{}1\]

Hence, we have

\[ 1\equal{}(2t)^3\minus{}6t\]

or

\[ q^3\equal{}8p^3\minus{}6pq^2\]

Moreover, due to simple explanation as shown

\[ RHS \, \vdots \, p\]

so LHS must be divided by p but (p,q) =1 then it leads us to a contradiction.

We complete our proof that $ \cos(20^0)$ is a irrational number.


inequalities
calculus
inequalities proposed

$ a,b,c>0$ and $ \sum \frac {1}{1 \plus{} a} \equal{} 1$
find min $ a\plus{}b\plus{}c$

[quote="bigbang195"]$ a,b,c > 0$ and $ \sum \frac {1}{1 \plus{} a} \equal{} 1$
find min $ a \plus{} b \plus{} c$[/quote]

$ 1\equal{}\sum \frac{1}{1\plus{}a}  \stackrel{\mbox{C.B.S}}{\ge} \frac{9}{3\plus{}a\plus{}b\plus{}c}$

$ \Longrightarrow a\plus{}b\plus{}c \ge 6$

$ Min\{a\plus{}b\plus{}c\}\equal{}6$

[quote="bigbang195"]$ a,b,c > 0$ and $ \sum \frac {1}{1 + a} = 1$
find min $ a + b + c$[/quote]
Note that from CS inequality we have
\[ \begin{aligned*} 1=\sum_{cyc}\frac 1{1+a}\geq \frac 9{3+a+b+c}\\
\therefore 3+a+b+c\geq 9 \implies a+b+c\geq 6 \end{aligned*}\]
Hence the required minimum $ = 6$; attained for $ a=b=c=2 \Box$.

hello, with the help of calculus we get $ a\plus{}b\plus{}c\geq6$, the equal sign will be attained for $ a\equal{}b\equal{}c\equal{}2$.
Sonnhard.

What about this ineq  :wink:
Let $ a,b,c \ge 0$ and $ \frac {1}{a^2 \plus{} 1} \plus{} \frac {1}{b^2 \plus{} 1} \plus{} \frac {1}{c^2 \plus{} 1} \equal{} 2$
Find min: $ a \plus{} b \plus{} c$

[quote="mathstarofvn"]What about this ineq  :wink:
Let $ a,b,c \ge 0$ and $ \frac {1}{a^2 + 1} + \frac {1}{b^2 + 1} + \frac {1}{c^2 + 1} = 2$
Find min: $ a + b + c$[/quote]
From AM-GM:
\[ \begin{aligned*} \sum_{cyc}\frac 1{1+a^2}= 2 \implies \sum_{cyc}\frac{a^2}{a^2+1}=1\\
\therefore 1=\sum_{cyc}\frac{a^2}{1+a^2}\leq \sum_{cyc}\frac {a^2}{2a} = \frac{a+b+c}{2}\\
\therefore a+b+c\geq 2\]
The required minimum is $ 2$, attained for $ a=b=1;c=0$. :lol:


calculus
integration
function
induction
algebra proposed
algebra

Find all function f : Z --> Z satisfying:
$f(m^2 +n^2) = (f(m))^2 + (f(n))^2$

$\forall m, n \in Z$

sorry if this is a stupid question :D , but does Z include 0? (sorry for the ambiguity, thats the definition of Z here. i dont think Z including negative integers would be suitable for the question :) ).

[quote="vinoth_90_2004"]sorry if this is a stupid question :D , but does Z include 0 and negative integers? (sorry for the ambiguity, thats the definition of Z here).[/quote]
You are stupid :D, because Z or N is no trouble

Why do you call it "integral"?
Anyway it's a nice problem, I'll try to solve it

I solved it. I based my solution on \[(5k + 1)^2 + 2^2 = (4k + 2)^2 + (3k - 1)^2\] and similar identities to give an induction proof.

i did a similar proof which is uses maybe more straightforward equalities as yours but is mainly the same. so all solutions are either $f(m)=0$ everywhere or $f(m)=\pm m$ where we may arbritarily choose signs except $f(m^2+n^2)=+(m^2+n^2)$. one first proves $f(m)=\pm m$, then the second condition is clear by the equation and it is also clear that all these functions satisfy the conditions.

Peter

Yeah, I remember that Arne solved this problem some time ago (on the forum). I also have a solution, but the idea is basicly the same. Let's work with $n\ge0$ for now.

We assume that $f$ isn't the identically null function. Since $f(1)\in \{0,1\}$, we assume that $f(1)=1$ (because otherwise an argument similar to the one that follows shows that $f$ is identically null). We can manually show for the first few values of $n$ that $f$ has the form: $f(n)=n$ if $n$ can be written as the sum of two squares and $f(n)=\pm n$ otherwise. We assume the function has this form for $n$ up to $k-1$ and we show that it has the same form for $k$: if $k$ can be written as the sum of two squares then we're done, and if it can't, then we can find $a,b,c<k$ s.t. $k^2+a^2=b^2+c^2$ and we are, again, done.

We may extend the function on the integers by taking $f(n)=\pm n$ for $n<0$. The functions that satisfy the equation are, thus, $(1)\ f(n)=0,\ \forall n\in \mathbb Z$, $(2) \{ \begin{array}{l}f(n)=n,\ n=a^2+b^2,\ a,b\in \mathbb Z\\f(n)=\pm n\mbox{ otherwise}\end{array}$.

I think you add an useless condition $n = a^2 + b^2$ in your conclusion

This problem appeared in our 'Concours Gnral' (which was our TST in these days) in 1994, as problem #5 (to be rigourous, it was only asked for $f$ from $N$ onto $N$ and with $f(1) > 0$, but the most difficult part is here)

Pierre.

[quote="pbornsztein"]I think you add an useless condition $n = a^2 + b^2$ in your conclusion
[/quote]

Errr... What do you mean? :D

the condition $n=a^2+b^2$ is not senseless, as i mentioned before in my post. look at the equation: we must $f(m^2+n^2)\geq 0$ since the right is sum of two squares. in all other cases the sign doesn't matter, since it is squared in the functional equation.

Peter

Hmmm..
At us we have N={1..} so I couldn't use negative numbers and 0!
That's why I spent a lot of time on this problem and my soln is so ugly!
Consier f(1)=a and f(4)=b.
After a lot of computations I get a^3 and 4a^6+1 are solns for Pell's equation x^2-2y^2=-1 and this can happen only for a=1. the use induction
So much time wasted!

If we change 2 by 3 we have a new function problem:
Find all f: Z--->Z satisfying:
$f(x^3 + y^3 + z^3)=(f(x))^3 +(f(y))^3 + (f(z))^3$
for all x, y, z in Z
---------

What is happen if we change 2,3 by n
Find all f: Z--->Z satisfying:
$f(\sum_{i=1}^n x_i^n)=\sum_{i=1}^n(f(x_i))^n$
for all $x_1, x_2, ..., x_n in Z$
Please try it

I think the problem with $3$ instead of $2$ was posted by Arne and solved by Pierre.

And how about finding all f: Z--->Z satisfying:
$f(\sum_{i=1}^n x_i^k)=\sum_{i=1}^n(f(x_i))^k$
for all $x_1, x_2, ..., x_n$ in $Z$ and $k$ in $N$.
Please try this.


[hide]We see that $10-20, 20-30$, and $30-40$ all have at least a 0,1,2,3 in it. So we can start the set from $40-50$ and see that there are only 6 numbers. There are 5 more sets like this in the 2 digit set so there are 36 numbers.[/hide]

[quote="rnwang2"]How many two-digit whole numbers do not contain the digits $0$, $1$, $2$, or $3$?[/quote]
[hide]All the ones without 0-3 are the ones that count.
$6 \cdot 6=36$[/hide]

[hide]i did the whole numbers 40-99. all the previous have 0,1,2, or 3.  then 10-4 for all the numbers in a group of ten without 0,1,2, or 3. then multiply by 6 to get 36.

wow. this solution looks a lot harder than it was. :| [/hide]


A right circular cylinder with radius 2 inches and height 1 inch has the same volume as a second right circular cylinder. The radius of the second cylinder is 1 inch. How many inches are in the height of the second cylinder?

Volumes: $ V_1\equal{}r_1^2h_1\pi \equal{} r_2^2h_2\pi\equal{}V_2$
Now it's easily seen that $ h_2\equal{}4$inches


Heres the problem "For all positive integers n, the expression n! denotes the product of the first n positive integers. When 5! is expressed as an integer, what is the ones digits?"

This problem is soooooo easy!  :lol:

the answer to the first question for me is yes, this seems easy.
the answer to the problem is [hide]0[/hide]
i think it is good for people prepping for countdown to memorize factorials up to 10.
1!=1
2!=2
3!=6
4!=24
5!=[hide]120[/hide]
6!=720
7!=5040
8!=40320
9!=362880
10!=3628800
to memorise 9 and 10, 3(628)80(0)
the 8 repeats, and 628 is about pi*200

[quote="tpulak"]
This problem is soooooo easy!  :lol:[/quote]

Please refrain from posting this in the future.  If someone for some reason can't get the answer, it'll only frustrate them more to hear someone call it almost trivial.

Plus I believe this came from a lower-level (as in chapter) round countdown, and I've seen kids easily confuse problems such as

what does x equal if $ x^[7x]\plus{}2\equal{}3$?, or how many zeros does $ 50!$ end in?...etc. (and these were very high-level kids, but since it was a very competitive area, they got mixed up doing it quickly, so that question you posted is just fine.

One comment though...what's the point of posting a question *you* believe is easy and that you already got the answer to?

-mamathbug

Usually, there are three categories of problem posters

1) People that post problems that they have attempted and couldn't solve
2) People that wish to challenge others
3) People that use AoPSers to do their homework

Usually people don't post problems that they think should be so obvious unless it's in a marathon

as well as spammers and people who wish to share something, but that's getting off-topic


So I assume you ask for the explicit formula?

If so, here are some:

1) $a_n=2n-1$
2) $a_n=2^{n-1}$
3) $a_n=\frac{1}{n}$
4) $a_n=n^2-n+1$

Perhaps i'll post some later

[quote="Jan"]So I assume you ask for the explicit formula?

If so, here are some:

1) $a_n=2n-1$ :)
2) $a_n=2^{n-1}$ :)
3) $a_n=\frac{1}{n}$ :)
4) $a_n=n^2-n+1$ :)

Perhaps i'll post some later[/quote]


analytic geometry
conics
hyperbola
calculus
integration
symmetry
AMC

1)
It is known that, for all positive integer k,
1^2 + 2^2 + 3^2 +....+ k^2 = [k(k+1)(2k+1)]/6

Find the smallest positive integer k such that 1^2 + 2^2 + 3^2 +....+ k^2 is a multiple of 200.

2)
A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola x^2 -y^2 =2000^2.

3)
Find the sum of all positive integers n for which n^2 -19n +99 is a perfect square.

4)
Let n be the smallest positive integer that is a multiple of 75 and has exactly 75 positive integral divisors, including 1 and itself. Find n/75.

I'm not sure if all of these are right, but they look pretty good.

1.
Answer: 112
Solution:
Set (k(k+1)(2k+1))/6=200*a where a is some positive integer. k(k+1)(2k+1)=1200*a=2^4*3*5^2. Since 2k+1 is always going to be odd, either k or k+1 will be a multiple of 16 because they are of opposite parity. Letting k or k+1 equal a multiple of 16, a quick check shows that k=112 is the smallest possible k.

2.
Answer: 25
Solution:
x^2-y^2=2000^2
(x+y)*(x-y)=2000^2=2^8*5^6
(x+y) and (x-y) must be a factor pair of 2000^2. Notice that we can get x=((x+y)+(x-y))/2 and y=((x+y)-(x-y))/2. These are only integers when you can factor at least one 2 out of (x+y) and (x-y). There are 9*7=63 total factors of 2000^2. Subtracting the case of 2000*2000, we are left with 62 unique factors of 2000^2, or 31 factor pairs. Consider the factors of 2000^2 that are only powers of 5 or =1, of which there are 7. The other member of the pair has all 8 of the powers of 2, thus a 2 cannot be factored out of both of them and x and y aren't integers. There are 31-7=24 factor pairs that yield lattice points (x,y) and the 2000*2000 case which adds one point for 25 total lattice points.

3.
Answer: 38
Solution:
The axis of symmetry of f(n)=n^2-19n+99 is 19/2. Thus, the two solutions for f(n) equal to any number in its range have a difference of 19. This limits us to 9 pairs of numbers since n has to be positive: (1,18),(2,17),...,(9,10). Observe that the sum of the values of n that yield perfect squares is a multiple of 19. Thus, we must only check the integers 1-9. It turns out that only 1 and 9 satisfy the conditions, thus the answer is 19*2=38.

4.
Answer: 432
Solution:
The number of factors of a positive integer is found by multiplying the degree of each prime factor plus one by each other.  This is because you have the degree+1 choices for each factor. For example, 6=2^1*3^1, 2*2=4 and 6 has 4 factors: 1,2,3,6. Thus, for n to have exactly 75 factors, the degrees of its prime factors must factor into 3*5*5.

In order to be a multiple of 75, n= 3*5^2*a, where a is any positive integer. We already have 3*5*5, 2*3=6 factors for our n. Observe that 6 does not divide 75. Thus, we must make a factor of 75 only by multiplying by more 3's and 5's. This gives us either 3^4*5^2 or 3^2*5^4. The first choice is smaller, thus we finish by multiplying by 2^4, the smallest option. This gives us n=2^4*3^4*5^2 with 5*5*3=75 factors and n/75=432.

Pretty sure these are AIME problems this thread should be moved to HS Intermedate topic. Sorry to revive but, number 2 should be 98. Since both $ (x+y)$ and $(x-y)$ are to be even, and $2000^2$ factor equals $2^8$ times $5^6$, give each a 2 first. Now its $2^6$ times $5^6$. So 7 times 7 equals 98, due to negative both possibility answer should be 98.


linear algebra
matrix
vector
linear algebra unsolved

Let $ u_1,\cdots,u_s;v_1,\cdots,v_s$ be $ 2s$ points in $ \mathbb{R}^n$ such that $ \|u_i \minus{} u_j\| \equal{} \|v_i \minus{} v_j\|,\forall 1\le i,j \le j$,prove that there is a $ x \in \mathbb{R}^n$,and orthogonal matrix $ A$,such that
\[ v_i \equal{} Au_i \plus{} x,\forall 1\le i\le s\]

Since each $ u_j, v_j$ is a vector, what is exactly the meaning of $ |u_j|$? Did you want to write $ \parallel{}u_i\minus{}u_j\parallel{} \equal{} \parallel{}v_i\minus{}v_j\parallel{}$, relative to the usual inner product?

I can get you started a bit:

* You can assume via appropriate choice of $ x$ that $ u_1 \equal{} v_1 \equal{} 0$.

* Then $ |u_i| \equal{} |v_i|$ for each i, and $ |u_i \minus{} u_j| \equal{} |v_i \minus{} v_j|$.

* Manipulate the above to obtain $ \langle u_i, u_j \rangle \equal{} \langle v_i, v_j \rangle$.

Finishing up from here should be pretty easy; for example, the $ u_i$ can be reduced to an orthonormal
basis in the same way the $ v_i$ can be, and you can use the matrix that turns one basis into the other, for
example.


calculus
integration
trigonometry
real analysis
real analysis unsolved

[size=125]Prove that the integral from 0 to pi/2 of (1/(1+(tanx)^sqrt(2))) dx = pi/4.[/size]

$ \int_0^\frac {\pi}{2} \frac {1}{1 \plus{} \tan^{\sqrt {2}} x } dx \equal{} \int_0^\frac {\pi}{2} \frac {\tan^{\sqrt {2}} x}{1 \plus{} \tan^{\sqrt {2}} x } dx$ (Let $ u\equal{}\frac{\pi}{2}\minus{}x$)

$ 2 \int_0^\frac {\pi}{2} \frac {1}{1 \plus{} \tan^{\sqrt {2}} x } dx \equal{} \int_0^\frac {\pi}{2} 1 \  dx \equal{} \frac {\pi}{2}$

Therefore $ \int_0^\frac {\pi}{2} \frac {1}{1 \plus{} \tan^{\sqrt {2}} x } dx \equal{} \frac {\pi}{4}$

Having $ \tan^nx,$ no matter what $ n$ is and the result is the same.


1'm a _________. (insert anything into the blank)

its wierd, but thats what i think it sound like.

When I see IMO I think of 3 words that have the initials IMO. Almost always International Math Olympiad, but occasionally, like 2%, I think of in my opinion.

Whenever I see FTW I think AoPS game, anyone else?

How about USAMO? AIME? AMC?

USAMO

[ ] United States of America Math Olympiad
[ ] USA Math Olympiad

AIME

[ ] American Invitational Math Examination
[ ] Aime. French.
[ ] AIM AT THE TARGET!

AMC

[ ] American Math Contests
[ ] American Mathematics Contest
[ ] American Math Competition
[ ] American Mathematics Competition
[ ] America's Movie Center.

Choose one in each category for the one you think up first.

None of them. I think of the contests as the abbreviation, so when I see USAMO, I think "the difficult math competition after AIME", and stuff like that, I don't actually think of their names.

Wow, my mind thought of AIME when I saw Klebian talk about aim.

...

Wow, international monkey orientation has more google hits than international mathematical olympiad...

Not if you put it in quotes. Then, the only hits are from this thread.

when i see ANC, i think AMV, whatever that is. i think like fan made vids or something, like "rurouni kenshin amv". man, i watch too much anime. (im not even sure it's AMV...)

idk for some weird odd reason i thought of lmao when i saw imo... that was weird :D


Given $ a, b, c \geq\ 0.$ Prove that:
$ \frac{11}{3}(a\plus{}b\plus{}c) \geq\ 8 . \sqrt[3]{abc} \plus{} 3. \sqrt[3]{\frac{a^3\plus{}b^3\plus{}c^3}{3}}$
I have said that it can be solved by Am-Gm (of course, not easy).
But I can't solve the stronger of can_hang. So, I wait for the nice proofs for it and for the stronger  :)

[quote="arqady"]The following inequality is much easier.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89953
nguoivn, what is the stronger can_hang's inequality? Thank you![/quote]
My stronger one is here: http://www.mathlinks.ro/viewtopic.php?t=228831 :)

[quote="manlio"]Very nice is your proof, nguoivn. :)[/quote]
Thanks, my friend. And I think that can_hang had a nice proof for the stronger. We'll wait for him  :)

[quote="nguoivn"]Sorry, I tried but I can't find that link  :(
I only can post my proof for my easier ineq:
But really, it's one of my proofs which I like best  :)[/quote]

Dear NGOIVN we can have the resolution in formed PDF
Thank you very much  :)

[quote="toanhocmuonmau"][quote="nguoivn"]Given $ a, b, c \geq\ 0.$ Prove that:
$ \frac {11}{3}(a \plus{} b \plus{} c) \geq\ 8 . \sqrt [3]{abc} \plus{} 3. \sqrt [3]{\frac {a^3 \plus{} b^3 \plus{} c^3}{3}}$
I have said that it can be solved by Am-Gm (of course, not easy).
But I can't solve the stronger of can_hang. So, I wait for the nice proofs for it and for the stronger  :)[/quote]
Nhocnhoc has asked me to make a pdf version of his solution. And here is it :)[/quote]

Thank you very much my Friend  :)


This was a problem from the Bay Area Math Olympiad from this year:

In a national basketball tournament with N teams, 251 teams are from California. Alcatraz is a California team, and is the unique CA champion (most games won against CA teams vs. other CA teams). However, Alcatraz is the unique loser of the tournament (it has the least wins out of any country in the nation)! What is the smallest possible value of N such that this is possible?

Us Bay Area people have had a discussion on this and are not sure what rigorous standards for this would be. This is what I think is rigorous:
1) Show that a directed graph with 251 vertices and one vertex of degree 124, one of degree 126, and 249 of degree 125 is possible.
2) Show that a directed graph with 255 vertices and one vertex of degree 126, one of degree 128, and 249 of degree 127 is possible.
3) Show that it is possible to get from a graph of type 1 to a graph of type 2, without changing any existing edges.
4) Prove that 255 is the minimum N for the problem.

However, most people only proved 3 and 4, or some other combination of them. So what exactly is rigorous on this type of problem?

The noun form is just "rigor."  You need to show that the minimum is attainable.

Ehh I was sleepy when I posted that.

So ALL you have to do is show the minimum? Not that it's attainable?

I just said you need to show that the minimum is attainable.  In other words, you need to give a construction of a scenario that gives the minimum.

Sorry, I misread what you said.

If they assumed #1 and #2 (which is pretty trivial to prove), but proved the others, would that get full credit?

I don't think so, but then again, I'm not a grader.  It sounds to me like the proofs you all wrote showed that $ N$ could not be less than $ 255$, which is not the same as showing that $ N$ can take on any value $ 255$ or greater.

Wait, but to solve for the minimum, you had to have 1 and 2. So to prove that 255 is the minimum, you need to prove 1 and 2. A lot of people's proofs did not include proving 1,2. Rather, they assumed them and then solved for the minimum. Taking that into consideration, if you construct a scenario where 255 works, it does not necessarily prove 255 is the minimum, so therefore I think you must prove all 4 of the points ubey listed to have a legitimate proof, as well as showing that 255 is possible, correct?

Yes.  Any answer to this problem needs to prove two points:

- That $ N$ cannot be less than $ 255$, and
- That $ N$ can be $ 255$.


Hi I'm a newcomer and there are loads of good stuff here, I've already presented myself.I'm gonna summarize and try to explain my situation: I'm a 20 yo student in Belgium, recently I successfully passed the admission exam to the Belgian Polytechnic Institutes (there are 4 of them in french-speaking Belgium, here's a site with hundreds of ancient questions http://www.users.skynet.be/bk337103/ go to "exercices résolus") I was ranked 25/276, but here's not the point;I'd like to graduate from a US University in aerospace eng.,I'm okay at maths I think (from sept06 through may07 I followed a special 18h week maths course to pass this exam and got 80% at the end in maths), but probably not even remotely close to the Olympiad level, I didn't even know such a thing existed until very recently,(in my country of origin there's no such thing as math competitions).But still I did loads of maths; analysis(integrals,exp,ln, a littl differential functions),trig(ranging from basic identities and eqs. to challenging triangles) algebra(linear alg,syst,complex..),geom(2D,3D analytical and synthetic,locuses..).I'm attracted to 3 specific institutions Purdue Univ.,Arizona Univ. & Georgia Inst.Tech, my questions are what are the steps I need to take in Belg. to get admitted ?
I've heard that the first thing to do is the TOEFL, but is SAT required too for foreign students ?and what else should I do ?Simply walk in the US embassy and say hi chaps I'm applying to Purdue or what ?I've heard Purdue has a large Intl student pop and how on Earth can all of them afford the astro tuition fees ?I don't have that cash, what should I do to get help?
As to my "achievements" ; got once a local maths prize at my school in 6th grade + 3 times the first excellence prize in grade 7,10,12. that's all.What can I do ? sorry for being so long, but please advise me.I'd very grateful.

PS:whatever happens beginning in september 07 I start my undergrad studies at Mons Polytech. School.

Instead of US embassy, go to the website of each school that you are interested in and look for "admissions" (for international students). For example, [url=http://admissions.tamu.edu/freshmen/intnl.aspx]this[/url] is what Texas A&M requires of international applicants.

Did I mention that Texas A&M has a highly regarded Aerospace Engineering program?

Thank you,I'll check it I promise also could anyone tell me how on Earth Intl. undergraduate students can afford the tuition ??? Are all of them sons of arab oil barons ?

Some universities even have financial aid for international students. Look on the websites of the universities you're interested in.

[quote="mlok"]You may be better off doing your undergraduate studies at a state-sponsored European university, and then going to a U.S. university for graduate studies... which will be sponsored by the astronomical tuition paid by undergraduates.  :blush:[/quote]

This usually is the path followed by international students into United States universities, for the financial reason given. There are opportunities to study in the United States as an undergraduate, but for the international student, those opportunities are more varied if you have more money to pay tuition. (It is a requirement for issuance of a student visa to show ability to pay for the school program you will attend, although that ability to pay can be shown through a school scholarship offer as well as through personal funds.)

I'm desperate  :( So unless you're an IMO winner with the right connections to the right people or super rich, there is no way a good and determined student can get into the US and receive a little finaid, this is unfair ! The European system is much more equalitarian, everyone got a chance even the not so rich. :mad:

[quote="Appolonius"]So unless you're an IMO winner with the right connections to the right people or super rich, there is no way a good and determined student can get into the US and receive a little finaid, this is unfair![/quote]

That overstates the case. Apply to a lot of colleges, and see what happens. Try to develop a list of colleges known to welcome international undergraduate students--they will have financial aid policies that can help make your studies possible, or else a list price low enough that you can afford it.

Many universities in other countries are less expensive than those in the United States. What difference in quality you are interested in may not be the same as the difference desired by other students. But some United States universities, even though expensive, are considered well worth studying at by some international students. Figure out what features of a university are important to you, and then check whether any of the hundreds of universities in the United States might offer what you are looking for at an affordable price.

In relative terms, very few foreign students in U.S. universities come from Western Europe. ([url=http://opendoors.iienetwork.org/?p=89251]Source of the data[/url].) There ought to be a reason for that.
[quote]1. India 76,503
2. China 62,582
3. Korea 58,847
4. Japan 38,712
5. Canada 28,202
6. Taiwan 27,876
7. Mexico 13,931
8. Turkey 11,622
9. Germany 8,829
10. Thailand 8,765
11. United Kingdom 8,274
12. Hong Kong 7,849
13. Indonesia 7,575
14. Brazil 7,009
15. Colombia 6,835
16. France 6,640
17. Kenya 6,559
18. Nigeria 6,192
19. Nepal 6,061
20. Pakistan 5,759[/quote]

I conjecture that *one* reason is that Western Europe has the best universities in the world, after the US.

Appolonius, the US might be unfair, but I assume that an American can't get a free education in Belgium either, so I'm not sure what makes the European system more fair (in this respect).

The reason given by the OP is important: European universities have unusually low list prices. But if there is something that United States universities offer that European universities do not (for a particular student), it might still be important to pay a little to experience the United States university.

[quote="Appolonius"](in my country of origin there's no such thing as math competitions)[/quote]

That's nonsense. How could I have participated in mathematical competitions in Belgium if they do not exist?  :D

[quote]I'm a 20 yo student in Belgium[/quote]

Ok, so where from are you then?

I'm a russian citizen, but I was born in Africa precisely in Rwanda and studied in a french-language belgian school there(actually the only foreign school in the country), this is why I chose to pursue my studies in Belgium,...Now you see ?


Prove the following

1. For real numbers $ p, x, y, z$, where $ x, y, z >\equal{}0$, prove
$ x^p(x\minus{}y)(x\minus{}z)\plus{}y^p(y\minus{}z)(y\minus{}x)\plus{}z^p(z\minus{}x)(z\minus{}y) >\equal{} 0$
2. For real numbers $ p, x, y, z$, where $ x, y, z >\equal{}0$ and $ p >\equal{} 1$, prove
$ x^p(y\plus{}z)(x\minus{}y)(x\minus{}z)\plus{}y^p(z\plus{}x)(y\minus{}z)(y\minus{}x)\plus{}z^p(x\plus{}y)(z\minus{}x)(z\minus{}y) >\equal{} 0$

Thanks! :blush:

Number 1 is Schur's Inequality...

Proof rewritten for your convenience out of the kindness of my own heart:

[hide]WOLOG $ x\geq{y}\geq{z}$.

Thus $ (x\minus{}z)\geq(y\minus{}z)$ and $ x^p\geq{y^p}$.

It follows that $ x^p(x\minus{}z)\geq{y^p(y\minus{}z)}$.  We also know $ (x\minus{}y)$ is positive, so:

$ x^p(x\minus{}y)(x\minus{}z)\geq{y^p(y\minus{}z)(x\minus{}y)}$
$ x^p(x\minus{}y)(x\minus{}z)\minus{}y^p(y\minus{}z)(x\minus{}y)\geq{0}$
$ x^p(x\minus{}y)(x\minus{}z)\plus{}y^p(y\minus{}z)(y\minus{}x)\geq{0}$

Also from the WOLOG we know $ (z\minus{}x)$ and $ (z\minus{}y)$ are both negative, so $ z^p(z\minus{}x)(z\minus{}y)$ is positive.

Using this knowledge,

$ x^p(x\minus{}y)(x\minus{}z)\plus{}y^p(y\minus{}z)(y\minus{}x)\plus{}z^p(z\minus{}x)(z\minus{}y)\geq{0}$

QED.

Part 2 can be proven similarly.[/hide]

Thanks for your answer. But in #1 it is not known that $ p >\equal{} 0$, so is $ x^p > y^p$ valid? (e.g., when $ p \equal{} \minus{} 1$ you get it reversed).

At the risk of sounding dumb, how do you prove #2? Can you give a full explanation? Thanks.  :roll:

For #2: WLOG $ x\ge y \ge z$

We factor our original equation to:
$ (x\minus{}y)[x^p(y\plus{}z)(x\minus{}z) \plus{}y^p(z\plus{}x)(z\minus{}y)] \plus{} z^p(x\plus{}y)(z\minus{}x)(z\minus{}y) \ge 0$

Since $ z^p(x\plus{}y)(z\minus{}x)(z\minus{}y) \ge 0$, all we need to prove now is that the other portion of the inequality is greater than 0.

Also, $ x\minus{}y$ is greater than zero, so we just need to show:

$ x^p(y\plus{}z)(x\minus{}z) \plus{}y^p(z\plus{}x)(z\minus{}y) \ge  y^p(y\plus{}z)(x\minus{}z) \plus{} y^p(z\plus{}x)(z\minus{}y) \ge y^p(2xz \minus{}2yz) \ge 0$

Thus, the inequality is true.

[quote="t0rajir0u"]Schur's inequality as it is commonly stated contains the premise that $ p > 0$.  See [url=http://en.wikipedia.org/wiki/Schur%27s_inequality]the wiki page[/url].[/quote]

Thanks for that link. Now, what about the case when $ p < 0$? Is it still true?

Oh yeah I see now, thanks, anonymous03.


[quote="leonariso"]Problem 2

Given a triangle with sides $\alpha, \beta, \gamma$. Prove that:

$\alpha^{2}(\beta+\gamma-\alpha)+\beta^{2}(\gamma+\alpha-\beta)+\gamma^{2}(\alpha+\beta-\gamma)\leq 3 \alpha \beta \gamma$[/quote]
We use the Ravi substitution $x=\beta+\gamma-\alpha$ and cyclically. Then $\alpha= \frac{y+z}{2}$ and cyclically. After setting it to the inequality, expanding and simplifying, we get \[\frac14\cdot\left(6xyz+\sum_{\text{sym}}x^{2}y\right)\leqslant \frac38(x+y)(y+z)(z+x).\] Let $s=\sum_{\text{sym}}x^{2}y$ and $p=xyz$. We have to prove \[s+6p\leqslant \frac32(s+2p).\] After simplifying it becomes \[6p\leqslant s\] which follows directly form $\mathrm{AM-GM}$.

$\alpha, \beta$ and $\gamma$ can take any positive real value, they don't have to be the sides of a triangle...

The given inequality is equivalent to \[\alpha(\alpha-\beta)(\alpha-\gamma)+\beta(\beta-\alpha)(\beta-\gamma)+\gamma(\gamma-\beta)(\gamma-\alpha) \geq 0\] which is Schur inequality for $f(x)=x$.

[quote="me@home"]Well, normal schur it is $r=1$
Are you saying schur holds for any $f(x)$ with a specific criteria (instead of just $a^{r}$)?[/quote]
As I remember, it holds for any increasing function $f: \mathbb{R}\rightarrow \mathbb{R}^{+}$. For further gneralisations see [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=156212#156212[/url] lemma 1.


geometry
trigonometry
circumcircle
trig identities
Law of Sines
perpendicular bisector

ABC is a triangle and M is the middle point of AB. If we know that the angles CAM + MCB = 90, prove that the triangle ABC is an isosceles triangle or a right triangle.

[hide]Let $ \angle CAM\equal{}x$ and $ \angle MBC \equal{} y$.  Then we have $ \angle MCB\equal{}90\minus{}x$.  Since $ x\plus{}y\plus{}(90\minus{}x\plus{}\angle ACM)\equal{}180$, we have $ \angle ACM\equal{}90\minus{}y$.  Applying the Law of Sines on $ \Delta ACM$, we have

$ \frac{CM}{\sin x}\equal{}\frac{AM}{\sin (90\minus{}y)}\Rightarrow \frac{\cos y}{\sin x}\equal{}\frac{AM}{CM}$

Applying the Law of Sines on $ \Delta CMB$ gives

$ \frac{CM}{\sin y}\equal{}\frac{BM}{\sin(90\minus{}x)}\Rightarrow\frac{\cos x}{\sin y}\equal{}\frac{BM}{CM}$

Since $ AM\equal{}BM$, we have $ \frac{AM}{CM}\equal{}\frac{BM}{CM}$.  Thus,

$ \frac{\cos y}{\sin x}\equal{}\frac{\cos x}{\sin y}$

$ \Rightarrow \cos y \sin y\equal{}\sin x \cos x$

$ \Rightarrow \sin 2y\equal{}\sin 2x$

Since $ 0<x,y<180$, we have either $ 2y\equal{}2x\Rightarrow y\equal{}x$ or $ 2y\equal{}180\minus{}2x\Rightarrow x\plus{}y\equal{}90$.  That is, either$ \Delta ABC$ is iscosceles or right.[/hide]

Geometric
[hide]Let $ O$ be the circumcircle of $ \triangle ABC$. Extend $ CM$ to $ D$ on circle $ O$ and connect $ AD$. Then $ \angle BAD \equal{} \angle MCB$ by Inscribed Angle Theorem. Because $ \angle CAM \plus{} \angle MCB \equal{} 90^\circ$, we have $ \angle CAD \equal{} \angle CAM \plus{} \angle BAD \equal{} 90^\circ$. Therefore, $ CD$ is a diameter, $ O$ lies on $ CD$.

If $ AB$ is also a diameter, then $ \angle ACB \equal{} 90^\circ$, so $ \triangle ABC$ is a right triangle. Suppose $ AB$ is not a diameter, then $ O$ does not lie on $ AB$. Because $ OA \equal{} OB$ and $ MA \equal{} MB$, line $ OM$ is the perpendicular bisector of $ AB$. Thus, line $ CD$ is the perpendicular bisector of $ AB$. Hence, $ AC \equal{} AB$, $ \triangle ABC$ is an isoceles triangle. QED [/hide]


real analysis
LaTeX
real analysis unsolved

Fix b>1.
a) if m, n, p, q are integers, n>0, q>0, and r=m/n=p/q, prove that
                        (b^m)^(1/n)= (b^p)^(1/q)
Hence it makes sense to define b^p =(b^m)^(1/n)

b)prove that b^(r+s) = b^r  b^s

c) if x is real, define B(x) to be the set of all numbers b^t, where t is rational and t<=x, Prove that
                              b^r =sup B(r)
when r is rational. Hence it makes sense to define
                              b^x = sup B(x)
for every real x.

any ideas??
thxx~ :)

a)'s pretty easy, but how do u solve b? I guess we have to use the conclusion of a to solve to, make r = m1/n1, s=m2/n2, where m1,m2,n1,n2 are all integers.
but then I got stuck :( ... any help?

thxx

Ah, the time-tested trick of posting questions copied directly out of Rudin in the hopes that someone will do your math homework for you ...

ohh no, I'm not looking for some1 to do my hwk. (it's not like I completely don't no wat's going on....)  I'm just stuck in one step of solution...

actually, I only wanna no if this is a valid theorem: s^n * s^m = s^(n+m), n and m are integers. or do I have to prove this?

if it's valid then im done.

What is the meaning of $ s^n$?  It means a product of $ n$ copies of $ s$.  By associativity, then, $ s^m\cdot s^n$ and $ s^{m \plus{} n}$ mean exactly the same thing.  So, that's a proof, but it's so trivial I wouldn't bother including it in your write-up.

By the way, you'll get much more constructive advice if you post the problem you're working on and the progress you've made on it so that others can give you relevant hints or suggestions where you've gone wrong.  It would also help if you learned a little [[LaTeX]] -- see that link or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]this one[/url].


I need to put a footnote on table. If I put only \footnote{} on the position that I want on my table then the note isn't written. For this reason I use minipage as:

\begin{table}[]
\begin{minipage}{\textwidth}
\begin{tabular}{}
....
\end{tabular}
\end{minipage}
\caption{...}
\end{table}

The result is good but the reference use a letter and doesn't use a number. How can I use a number as reference?

Thanks.

[quote="stevem"]Somewhere you must have redefined \thefootnote. \renewcommand{\thefootnote}{\arabic{footnote}} should put numbers back again but I would look to see where the redefinition has happened. Could you post a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal working example[/url]?[/quote]

Hi,
The problem is when I do \footnote on text, the note is shown with a number, but when I do \footnote on table, the note is shown with letter.

Minimal working example:

\begin{table}
\centering
\begin{minipage}{\textwidth}
\centering
\begin{tabular}{|l|p{10cm}|p{10cm}|}
\hline
\rowcolor[gray]{.2}
A \footnote{my foot note} & B & C \\
\hline
....
\end{tabular}
\end{minipage}
\caption{}
\label{}
\end{table}

Thanks.

[quote="stevem"]Minipages use a different footnote, mpfootnote, which defaults to using letters. So to change it back to numbers use \renewcommand{\thempfootnote}{\arabic{mpfootnote}}.

Of course mpfootnote uses a different counter to footnotes outside minipages so they won't use consecutive numbers which is why the default is to use letters.[/quote]

Thank you very much.


If a, b, c and d are four multidigit numbers and

if

a + b + c + d, a + b, a + c, a + d, b + c, b + d and c + d are all square numbers,

find minimum values of a, b, c and d.

PS: earlier i posted it in different forum, but this seems to be a better place.

[quote="xxxyyyy"]PS: earlier i posted it in different forum, but this seems to be a better place.[/quote]

I think this is rather number theory so I moved it here.

I get one solution and I think it's the smallest.
(a,b,c,d)=(1502,802,98,98)

[quote="jin"]I get one solution and I think it's the smallest.[/quote]

Can you prove it's the smallest?

[quote="jin"]$a+b+c+d=x^2$
Then $x^2$ can be expressed as the sum of two square in at least two ways.
And so $x$ can be expressed as $k(a^2+b^2)$ in at least two ways.
Then I try it from 1 to 50,and I think if $x<50$,there's no such numbers.
And for $x=50$, I got it.[/quote]

Three ways.

Apologies for not mentoning that all numbers have to be different.

"JBL" has already provided the solution here...
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=479706&highlight=#479706

Please have a look.


integration
real analysis
real analysis unsolved

Integrate the following equation: \[ (x'')^{2}-2x'x'''+1=0. \]

[hide="By solving it"]
By inspection it seems like $x'' = t + 1$ might work as a particular solution. Then $x' = t^2/2 + t + c$ and $x''' = 1$. We have $t^2 + 2t + 1 - t^2 - 2t - 2c + 1 = 0$. So $c=1$.

In the homogeneous case, we have $(x'')^2 = x'x'''$ so $\frac{x''}{x'''} = \frac{x'}{x''}$ and thus $ln(x'') = \ln(x') + k$, and we have $x'' = cx'$ and $x' = Ae^{Bt}$.

Now if we just want to integrate what we had there, it's
$(ABe^{Bt} + t + 1)^2 - 2(Ae^{Bt} + t^2/2 + t + 1)(AB^2e^{Bt} + 1) + 1 = 0$

and I am too lazy but it's easy.
[/hide]
[hide="Another way"]
Let $f' = (x'')^2$. Then we have $f' - 2f'' + 1 = 0$. We can integrate to $f - 2f' + t + c = 0$. But we don't know what is $f$. We can integrate by parts to get $f = x'x'' - \int x'x''' dt = x'x'' - f'/2$ and plug in to obtain $x'x'' - 5f'/2 + t + c = 0$, and finally $x'x'' - 5(x'')^2/2 + t + c = 0$. Hopefully I didn't make a mistake.
[/hide]

notam x'=p si obtinem $(p')^2-2pp''+1=0$. notam p'=z(p), de unde p''=z'z. ecuatia devine $\frac{2z}{z^2+1}dz=\frac{1}{p}dp$ cu solutia $z^2+1=Cp$ de unde $\frac{dp}{\sqrt{Cp-1}}=dt$ si $\frac{dp}{\sqrt{Cp-1}}=-dt$.
de aici rezulta $p=\frac{Ct^2}{4}+\frac{1}{C}$ si mai departe $x=\frac{Ct^3}{12}+\frac{t}{C}+C'$

[quote="Xevarion"]
Let $f' = (x'')^2$. Then we have $f' - 2f'' + 1 = 0$. We can integrate to $f - 2f' + t + c = 0$. But we don't know what is $f$. We can integrate by parts to get $f = x'x'' - \int x'x''' dt = x'x'' - f'/2$ and plug in to obtain $x'x'' - 5f'/2 + t + c = 0$, and finally $x'x'' - 5(x'')^2/2 + t + c = 0$. Hopefully I didn't make a mistake.
[/quote]
$f' = (x'')^2$ .
$f'' = 2x''x'''$
$f' - 2f'' + 1 = (x'')^2 - 4x''x''' + 1 \ne (x'')^2 - 2x'x''' + 1 = 0$ :oops:


this one's fun

[hide="tiny hint you should know"]what's three (an odd number) factors tell you about this number? [/hide]

[quote="joejia"]A two-digit counting number has exactly three factors, and the sum of its digits is less than 10. What is the number?[/quote]
[hide]Is it 25?[/hide]

[quote="mathmanman"]Hmm, what about 30 then? (=2*3*5/3+0=3)[/quote]
[quote="h78reg"][hide]what about 15? the factors are 1,3,5 and the sum of its digits is 6.[/hide][/quote]
[hide]YOu both are forgetting that the numbers have a factor that is the number and for mathmanman you also forgot 1 is a factor
I believe my answer is correct because my number has only 3 factors 1,5,25 i looked for a number that had only one factor other than 1 and itself.[/hide]

oops. you're right then. what a mistake that was. :wallbash:

Oh yeah, ok, I get it now!
Then you're right... (I thought it was, three factors apart from 1 and the number itself, duh!)

[hide]25, becasue it is a square of a prime number, so 5, 1, and 25 are its factors[/hide]


quadratics
algebra
quadratic formula
linear equation

Where did you get this problem from? o.o

apply the distributive property to get rid of the parentheses
then either a) multiply by a common denominator
or b) convert every fraction/number into a fraction with identical denominators
from there, collect like terms, and isolate x
i personally am also freaked out by the fractions

$ \frac {5}{17} \plus{} 8(\frac {11}{23x} \minus{} \frac {67}{93}) \plus{} 67x \minus{} \frac {13}{3}(\frac {47}{101x} \minus{} \frac {23}{41}) \equal{} 7$

first, distribute anything that can be distributed

$ \frac {5}{17} \plus{} \frac {88}{23x} \minus{} \frac {536}{93} \plus{} 67x \minus{} \frac {611}{303x} \plus{} \frac {299}{123} \equal{} 7$

multiply each side by $ 150579183x$ to get rid of all the fractions

$ 44287995x \plus{} 576129048 \minus{} 867854216x \plus{} 10088805261x^2 \minus{} 303643171 \plus{} 366042079x \equal{} 1054054281x$

simplify the expression

$ 10088805261x^2 \minus{} 457524142x \plus{} 272485877 \equal{} 1054054281x$

$ 10088805261x^2 \minus{} 1511578423x \plus{} 272485877 \equal{} 0$

use quadratic formula to solve for $ x$, which i'm not gonna do since my calc can't handle these numbers

can anyone check for mistakes, i have an odd feeling it's not supposed to be like this......

11/23x = $ \frac {11}{23}x$, not $ \frac {11}{23x}$ :D

And comment: $ b^2 \minus{} 4ac < 0$, so there are no real solutions to that ... ugly... (understatement) quadratic equation :D

OH!
the x's weren't in the denominator!
that makes everything so much easier

that's still an ugly problem though
where'd you get this question?

[quote="fishythefish"]Assuming those x's were not in the denominators,

$ \frac {5}{7} \plus{} 8\left(\frac {11}{23}x \minus{} \frac {67}{93}\right) \plus{} 67x \minus{} \frac {13}{3}\left(\frac {47}{101}x \minus{} \frac {23}{41}\right) \equal{} 7$

$ \frac {5}{7} \plus{} \frac {88}{23}x \minus{} \frac {538}{93} \plus{} 67x \minus{} \frac {611}{303}x \plus{} \frac {299}{123} \equal{} 7$

Multiplying everything by 186009579...(Vallon, I fixed some of your mistakes)

$ 132863985 \plus{} 711688824x \minus{} 1076055414 \plus{} 67x \minus{} 375088623x \plus{} 452169627 \equal{} 7$

$ \minus{} 739455146x \plus{} 585033605 \equal{} 0$

$ 739455146x \equal{} 585033605$

$ x \equal{} \frac {585033605}{739455146}\approx0.7911684815$ (That's what my calculator said...)[/quote]

srry to double post, but it's too annoying getting this equation over to my last post

anyway, your +67x shoulda been multiplied by 186009579 as well
idk about the rest of it, as i'm not gonna check everything

BTW, to make this problem easier, we coulda combined like terms before multiplying by a common denominator
so instead of working with all those numbers, we could have a simple variable with a coefficient on one side and a constant on the other
then we coulda divided both sides of the equation by the coefficient of the variable to get it much easier than finding a common denominator for all that

i'll solve the problem again when i have time
if anyone else would like to attempt, good luck!

But the idea of finding the common denominator was so that we could combine like terms. Since most of the numbers are prime, it's hardest to somehow combine like terms without finding a common denominator.

Just expanding the simplification:

$ \frac{5}{17}\plus{}\frac{88}{23}x\minus{}\frac{536}{93}\plus{}67x\minus{}\frac{611}{303}x\plus{}\frac{299}{123}\equal{}7$

Let's try this multiplication again...

$ 132863985\plus{}711688824x\minus{}1076055414\plus{}12462641793x\minus{}375088623x\plus{}452169627\equal{}7$

$ 12799241994x\minus{}491021809\equal{}0$

$ 12799241994x\equal{}491021809$

$ x\equal{}\frac{491021809}{12799241994}\approx0.0383533507$... :?:

Hmm...wonder what happened.

Check this if you want, then check what you did with the common denominators...

But I agree that this is a strange problem...

what's your common denominator?

it's kinda hard to check if you don't know what you multiplied by in the first place

I multiplied by 186009579, same as last time (but I fixed the problem)...
What did you do?

I found it easier to just get rid of denominators rather than to find a common one. It should reduce the chance of making a mistake, especially if you use a calculator for multiplication. Besides, as we're going to probably use a calculator for the final division, it won't make too much of a difference if the linear equation we get is factorable.


I have the first five Arbelos journals on order, but what about the "Special Geometry" edition? What is it? Is it still sold anywhere anymore?

I have it, it has the solutions to the cover problems, I believe. You can still get it at the MAA, I think, that's where I bought it a couple years ago.

Would that be the "Volume 6" sold on the MAA website? The website only sells 5 of the volumes (it's missing one of them), so I thought that the 6 might have been a typo.

Yeah i have the arbelos journals but they seem to be kinda random to me 0.o like the way topics lead into each other


The slope of the line $\displaystyle \frac{x}{3} + \frac{y}{2} = 1$ is

$\displaystyle \text{(A)} \ - \frac{3}{2} \qquad \text{(B)} \ - \frac{2}{3} \qquad \text{(C)} \ \frac{1}{3} \qquad \text{(D)} \ \frac{2}{3} \qquad \text{(E)} \ \frac{3}{2}$

[hide]Bring the x to the other side, and multiply by 2.

You get:

$y = \frac{-2}{3} x + 2.$

Slope is $\frac{-2}{3}$

Answer is B.[/hide]


[quote="byrdgyrl"]My friend and I disagree on this problem.

If AxB=A(B-1) what does 3x7= ?

A) 21
B) 18

If you could please explain why (I know how to do the math but don't know the answer.)  Thanks! :o[/quote]

That would be

3(7-1)
= 3(6)
=18

so B is correct for this problem but everyone in real life knows that 3x7=21.

This kind of problem confuses many students, particularly given that students are used to thinking of "3 x 7" as "three times seven."

In this problem, the symbol that is used for grade-school multiplication is defined differently.  In math we often create "operations" (things that we do with numbers) and assign symbols to them.  For instance, let
\[ a*b = 2a + 3b. \]
Here we see that
\[ 3*7 = 2(3) + 3(7) = 6 + 21 = 27. \]
We follow the recipe of the operation to get a result.

So, look back at your problem and view "x" not as the normal multiplication symbol, but as an operation that is defined in the problem.


Choose a subspace $U$ of $V$ so that $T$ is an isomorphism from $U$ to its range- you can do this by choosing a preimage for each element of a basis of the range.

Then $V$ is the sum of $U$ and the null space of $T$, each of which are finite dimensional.

[quote="§outh§tar"]If T:V->W is a linear transformation on a linear space V over a field F and rank T and nullity T are finite, does this imply dim V is finite?
My intuition cannot help me here.[/quote]

the Rank plus nullity theorem, says taht:

rank(T)+nullity(T) =dim V

If rank(T) and nullity(T) are finites, then dim (V) is finite.


trigonometry
calculus
integration
function
derivative
complex numbers
calculus computations

[quote="santosguzella"]Solve this equation: $\frac{d^{4}y}{dx^{4}}-2\frac{d^{3}y}{dx^{3}}+2\frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}+y=e^{x}+e^{2x}$[/quote]

Start by solving the associated homogeneous equation:

$\frac{d^{4}y}{dx^{4}}-2\frac{d^{3}y}{dx^{3}}+2\frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}+y=0$

which has the characteristic equation: $\lambda^{4}-2\lambda^{3}+2\lambda^{2}-2\lambda+1=0$

which factors into: $(\lambda-1)^{2}(\lambda^{2}+1)=0$ so we have $\lambda = 1,1,-i, i$

so we have the homogeneous solution is $y=A e^{x}+Bxe^{x}+C\sin x+D\cos x$. Now we need a particular solution, recall that we have

$\frac{d^{4}y}{dx^{4}}-2\frac{d^{3}y}{dx^{3}}+2\frac{d^{2}y}{dx^{2}}-2\frac{dy}{dx}+y=e^{x}+e^{2x}$

I'll use the method of undetermined coefficients: for the $e^{x}$ term we have both $Ae^{x}$ and $Bxe^{x}$ as solutions to the homogeneous equation already, so we should try $E x^{2}e^{x}$ as a particular solution for this term, I get $E=\frac{1}{4}$ after plugging this in for $y$ in the diff. eq. and solving for $E$. For the other term, namely $e^{2x}$, we use $Fe^{2x}$ as our particular solution... what do you get for $F$? Put this value into the solution below to complete the solution:

$y=A e^{x}+Bxe^{x}+C\sin x+D\cos x+\frac{1}{4}x^{2}e^{x}+Fe^{2x}$.

[quote="maokid7"]how do you use the method of undetermined coefficients?[/quote]
Let's start with a simpler case. Consider this differential equation:

$y'=e^{-2x}\sin(3x)$

(OK, it's an integration problem. I said I wanted a simpler case. And it does qualify as the type of problem for which the method is used: a constant coefficient linear differential equations, nonhomegenous but with the nonhomogenity falling within the class of functions generated by exponentials, sines and cosines, and polynomials.)

The corresponding homogeneous equations is $y'=0,$ which has the general solution $y=C.$ We now need to find a particular solution, which is where the undetermined coefficients come in. As long as we're working under "real" rules, every sine must be accompanied by a cosine and vice versa. (Each term in the nonhomogeneity is associated with a particular exponential; for that mode of thinking, I will use complex numbers and refer to the value(s) $2\pm3i.$)

We guess that there is a solution of the form

$y=Ae^{2x}\cos(3x)+Be^{2x}\sin(3x).$

What we do with this guess is to take the appropriate derivatives to put back into the differential equation; the result should be a system of linear equations for the coefficients.

$y'=Ae^{2x}(2\cos(3x)-3\sin(3x))+Be^{2x}(2\sin(3x)+3\cos(3x))$

$=(2A+3B)e^{2x}\cos(3x)+(-3A+2B)e^{2x}\sin(3x).$

In the differential equation, this becomes

$(2A+3B)e^{2x}\cos(3x)+(-3A+2B)e^{2x}\sin(3x)= e^{2x}\sin(3x).$

Equating terms, we get:
\[\begin{array}{ccc}2A+3B&=&0\\-3A+2B&=&1 \end{array}\]
We solve this to get $A=-\frac{3}{13},B=\frac{2}{13}.$ From that, and adding back in the solution to the related homogeneous equation, we get that the general solution is

$y=-\frac{3}{13}e^{2x}\cos(3x)+\frac{2}{13}e^{2x}\sin(3x)+C.$

---

It's a very structured procedure, with rules for dealing with each eventuality. In his post, benorin has explained which such rules he's using where quite clearly, albeit briefly. To understand it further, you really should consult a textbook. This is standard textbook material, and is in a dozen different textbooks with titles like "Introduction to Ordinary Differential Equations" or something similar. It may also be in some calculus textbooks.


My book(Not Morrison-Boyd  :D ) says that Carbonyl groups in 'sensitive' compounds can be reduced by first formed $ \text{thioacetal}$ groups and then using $ \text{Raney\minus{}Nickel}$
those can't be reduced either by wolff-Kishner or by Clemmensen

That's right, because thioacetal reduction occurs under neutral conditions (hydrogen and a catalyst), and so it is the method of choice for those carbonyl compounds sensitive to both acids and bases.

so it's selective reduction
say i have a compound which ahs a double bond a carbon atom to which a halogen is attached and carbonyl group and i want to specifically redue the carbonyl group  then i will proceed via the $ \text{cyclic\minus{}thioAcetal}$ group

At first sight that would be correct: as the double bond could react with HCl and the chloride bearing carbon could react with the base, then the correct method would be thioacetal reduction. However, don't forget that the thioacetal forming reaction is a nucleophilic reaction, and the sulphur atom could easily react with the chloride carbon.

Yes that's the Mozingo method right?  :)

Actually I had never heard of such a thing as "Mozingo method": all Organic Chemistry books I know simply refer to Thioacetal reduction.

That's Mozingo Reduction, that's what one more text book that I referred told me.

Which is probably a textbook from India, right?


calculus
integration
function
logarithms
calculus computations

I don't think calculus is Middle School Classroom Math lol..

Anyways, it's $ \boxed{x^x(\ln(x) \plus{} 1)}$.

No elementary antiderivative. Only interesting thing related to this integral is the so-called sophomore's dream ([url=http://en.wikipedia.org/wiki/Sophomore%27s_dream]Bernoulli's identity[/url])

There is no elementary antiderivative for this function. that is there is no primitive function which is a antiderivative of this expression.

Do you mean $ \int x^x dx$ ? Anyway, $ \int x^x dx
\equal{}\sum_{k\equal{}0}^\infty(\frac{x^{n\plus{}1}}{k!(n\plus{}1)}
 \sum_{i\equal{}0}^n (\minus{}1)^i \frac{(n)_i}{(n\plus{}1)^i} (\ln x)^{n\minus{}i})$

perhaps you mean

\[ \int_0^1 {x^x } dx \equal{} \int_0^1 {e^{x\ln x} } dx \equal{} \int_0^1 {\left\{ {\sum\limits_{n \equal{} 0}^\infty  {\frac{{x^n \ln ^n x}}
{{n!}}} } \right\}} dx \equal{}\]

\[ \sum\limits_{n \equal{} 0}^\infty  {\left\{ {\frac{1}
{{n!}}\int_0^1 {x^n \ln ^n xdx} } \right\}} \underbrace  \equal{} _{u \equal{}  \minus{} \ln x}\sum\limits_{n \equal{} 0}^\infty  {\left\{ {\frac{{\left( { \minus{} 1} \right)^n }}
{{n!}}\int_0^{ \plus{} \infty } {e^{ \minus{} \left( {n \plus{} 1} \right)u} u^n du} } \right\}}\]

\[ \underbrace  \equal{} _{z \equal{} \left( {n \plus{} 1} \right)u}\sum\limits_{n \equal{} 0}^\infty  {\left\{ {\frac{{\left( { \minus{} 1} \right)^n }}
{{n!\left( {n \plus{} 1} \right)^n }}\int_0^{ \plus{} \infty } {e^{ \minus{} z} z^n du} } \right\}}\]

\[ \underbrace  \equal{} _{z \equal{} \left( {n \plus{} 1} \right)u}\sum\limits_{n \equal{} 0}^\infty  {\left\{ {\frac{{\left( { \minus{} 1} \right)^n \Gamma \left( {n \plus{} 1} \right)}}
{{n!\left( {n \plus{} 1} \right)^n }}} \right\}}  \equal{} \sum\limits_{n \equal{} 0}^\infty  {\frac{{\left( { \minus{} 1} \right)^n }}
{{\left( {n \plus{} 1} \right)^n }}}\]

	greetings from Chile, South America :)


geometry
3D geometry
tetrahedron
circumcircle
sphere
Pythagorean Theorem
geometry proposed

Each face of a tetrahedron is a triangle with sides $a, b,$c and the tetrahedon has circumradius 1. Find $a^2 + b^2 + c^2$.

Inscribe the tetrahedron in a right parallelepiped which has edge of lengths $ p,q,r$  we see that $ a,b,c$  are the lengths of diagonals of each face of the parallelepiped. Since the circumsphere of the tetrahedron is the same sphere that circumscribed the parallelepiped,so it's enough to find the circumradius of the pararellepiped,which is half of its main diagonal. and the Pythagorean theorem gives
$ p^2 + q^2 = a^2$,      $ p^2 + r^2 = b^2$    and  $ q^2 + r^2 = c^2$
Since the half of the length of the main diagonal is ${ \frac {\sqrt{p^2 + q^2 + r^2}}{2}}$

It follows that ${ \frac {\sqrt{p^2 + q^2 + r^2}}{2}} = \sqrt {\frac {a^2 + b^2 + c^2}{8}} = 1$

Therefore, $ a^2 + b^2 + c^2 = 8$


Hey could ne1 give me a simple proof for the napoleon's theorem?
(doesn't matter if its long , it shud just be understandable)

if u have challenge and thrill of pre college maths have a look at it.

Is [url=http://planetmath.org/encyclopedia/NapoleonsTheorem.html]this[/url] simple enough?

It may be simple 2 u but i m not 2 much familiar with coordinate geometry

ne oder answer

Let $ ABC$ be the original triangle and $ P,Q,R$ be points in the exterior of $ ABC$ such that $ ABP, BCQ,$ and $ ACR$ are equilateral triangles. Let $ D$ be the intersection point of the circumcircle of $ BQC$ and $ CRA$. Then $ \angle BDC\equal{}\angle CDA\equal{}180^\circ \minus{}60^\circ \equal{} 120^\circ$. Also $ \angle ADB\equal{}360^\circ \minus{}2\cdot 120^\circ \equal{}120^\circ$. Thus $ APBD$ is cyclic so $ D$ is in the circumcircle of the triangle $ APB$. Let $ O_A,O_B,O_C$ be the circumcircles of the triangles $ APB,BQA,CRA$. As $ O_AO_B$ is a midpoint normal of $ DC$ and $ O_AO_C$ a midpoint normal of $ BD$ we have $ \angle BDC \plus{}$ $ \angle O_CO_AO_B \equal{} 60^\circ$. Similarly other two angles.

Hi, everybody,
In fact what I remember to have learned about the subject was this:

[b]Divide each of the sides BC, CA and AB of a triangle in three congruent segments and construct externally, on the middle segment of each side, an equilateral triangle; then the vertices of these triangles not belonging to the sides BC, CA and AB are the vertices of an equilateral triangle[/b]

In fact, this is absolutely similar with:

[i]The centers of the equilateral triangles constructed externally on the sides of a triangle determine an equilateral triangle[/i]

A [b][u]Proof[/u][/b] showing the equality of [i]sides[/i], although it's much easier to see the equality of angles:

Let BCD, CAE and ABF be the equilateral triangles constructed externally on the sides of ABC and K, L and M respectively, their centers.
We know that AD = BE = CF ( 1 ) [ A 60 degs rotation around C will map A to E and D to B, hence AD = BE, while a 60 degs rotation around B will map C to D and F to A, with AD = CF ].

Further a 30 degs rotation around C followed by a dilation factor sqrt(3) will map L to E and K to B, so we got BE/KL = sqrt(3). Similarly we get CF/MK = BE/ML = sqrt(3), hence BE/KL = AD/MK = CF/ML = sqrt(3) ( 2 ).
Taking into account (1) we get KM = ML = KL, that is, the triangle KLM is equilateral.

Best regards,
sunken rock

The aproach with [b]complex numbers[/b] is quite easy:
Let $ \epsilon\equal{}\frac{1\plus{}i\sqrt{3}}{2}$.
Let $ a,b,c,a_0,b_0,c_0,a_1,b_1,c_1$ the complex number asociated with the vertices of the triangles.
And we have
$ b_0\equal{}a\plus{}(c\minus{}a)\epsilon$
$ c_0\equal{}b\plus{}(a\minus{}b)\epsilon$
$ a_0\equal{}c\plus{}(b\minus{}c)\epsilon$
Therefore $ 3(a_1\minus{}c_1)\epsilon\equal{}\epsilon(2c\minus{}a\minus{}b)\plus{}(\epsilon\minus{}1)(2b\minus{}a\minus{}c)\equal{}3(b_1\minus{}c_1)$

This is a more in depth version of the proof above, which I find to be a bit dense.

Similarly to above, we define $a$, $b$, and $c$ as the vertices of $\triangle{ABC}$, $a_0$, $b_0$, and $c_0$ as the third vertices of the equilateral triangles constructed from sides $BC$, $AC$ and $AB$ respectively. Finally, let $a_1$, $b_1$, and $c_1$, in that order, be the respective centers of the previously stated triangles, and let $\epsilon=\frac{1+i\sqrt{3}}{2}$.

First, we prove that $\triangle{A_1B_1C_1}$ is equilateral.

Rotating $b$ $60^{\circ}$, counterclockwise, about $c$, we obtain

$$a_0=c+(b-c)\epsilon$$

Similarly we find that

$$a_0=c+(b-c)\epsilon$$
$$b_0=a+(c-a)\epsilon$$
$$c_0=b+(a-b)\epsilon$$

Since $a_1$, $b_1$ and $c_1$ are the centroids of their respective triangles, we compute

$$a_1=\frac{1}{3}(b+2c+(b-c)\epsilon)$$
$$b_1=\frac{1}{3}(c+2a+(c-a)\epsilon)$$
$$c_1=\frac{1}{3}(a+2b+(a-b)\epsilon)$$

It suffices to prove that segment $A_1C_1$ rotated by $60^{\circ}$ counterclockwise results in $B_1C_1$, or equivalently, that $3(a_1-c_1)\epsilon=3(b_1-c_1)$. Note that we multiply by three to get rid of the one thirds we have as a result of dealing with centroids.

$$3(a_1-c_1)\epsilon=\epsilon((2c-a-b)+\epsilon(2b-a-c))=\epsilon(2c-a-b)+\epsilon^2(2b-a-c)$$

Notice that $\epsilon^2=\frac{-1+i\sqrt{3}}{2}=\epsilon-1$. Plugging in and simplifying, we get

$$(a+c-2b)+\epsilon(b+c-2a)=3(b_1-c_1)$$
as desired.

Now that we know that $\triangle{A_1B_1C_1}$ is equilateral, we have that its center is
$$\frac{1}{3}\bigg(\frac{1}{3} \sum_{cyc} b+2c+(b-c)\epsilon\bigg)$$
Our terms with $\epsilon$ cancel however, and we are left with $$\frac{3a+3b+3c}{9}=\frac{a+b+c}{3}$$
which we recognize as the centroid of $\triangle{ABC}$ as desired. $\square$


calculus integration algebra polynomial derivative function domain

Let $Q_n$ be a real polynomial of degree $n$ such that $Q_n(-x)=(-1)^nQ_n(x)\; ,\; \forall x\in {\mathbb R}.$
If $m,n$ are non-negative integers such that $m>n$ and $m+n=$even , find \[ \int\limits_{0}^1 Q_n(x)\left((x^2-1)^m\right)^{(m)} \; dx\; .  \]

Notational question: in $((x^2-1)^m)^{(m)}$, does that last $(m)$ mean the m-th derivative? In other words, are you talking about about the Legendre polynomials? (In that case, note that the domain of integration, [0, 1], is not the "natural" domain for Legendre polynomials, which would be [-1, 1]. That will play a role in the problem, and is the reason for odd-even conditions.)

Indeed ,  $f^{(m)}$  denotes the m- $th$  derivative, and also $\; \left((x^2-1)^m\right)^{(m)}=2^m m!\cdot P_m(x)\; ,$  where $P_m$ is Legendre polynomial normalized by $P_m(1)=1$.  Of course, $P_m$ is orthogonal over [-1,1] with respect to the weight function  $w(x)=1 .$
[b]Hint:[/b]  use the fact that   $P_m(-x)=(-1)^mP_m(x) .$


Let $AC$ be the hypotenuse of $\triangle ABC$. Points $D,E,F$ are on $AC$ such that $BD$ is the median, $BE$ is the bisector of the right angle and $BF$ is the height. Show that $m \angle DBE= m \angle EBF$.

Perhaps this is better for Getting Started?

[hide]$\angle FBA=90^\circ-\angle FAB=\angle C$ and $\angle DBC=\angle DCB=\angle C$ because $DB=DC=DA$. Hence $\angle EBF=45^\circ-\angle C=\angle EBD$.[/hide]


search
number theory
number theory proposed

Let p be a prime number such that $ p^2$ divides $ 2^{p \minus{} 1} \minus{} 1$ .
Prove that for any natural number $ n$ the integer $ (p \minus{} 1)(p! \plus{} 2^n)$ has at least three distinct prime divisors

[quote="ZetaX"]Next time do not post it twice (there was exactly the same in Number Theory Unsolved Problems, I deleted it now). And use some meaningfull title, or how should one determine what the problem is about¿
And I'm sure it was posted before, but don't have the time now to search for it.[/quote]
ok

It was posted before here: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=760066011&t=88579


This is a competition ladder style. Its single elimination and the last person who wins wins.

For example lets say A,B,C,D sign up. Also When I think there are enough people signed up I'll post that

A vs. B, A wins
c vs. D C wins

B and D are out

A. vs C, A wins

A is the winner

so its like the march madness tournament

here are some spots. It doesn't mean that these are the number of people playing I'm just setting it up. sign in some spots

A.person signing up will have tehre competition in FTW
B.person signing up in B will have a 50/50 shot of moving to the next round
C
D.
E
F
G
H
I

For each match there are different competitions. For example
A vs B. C vs. D

A and B could have a competition like cding each other on FTW and the winner moves on. while  c and D could have a competition like I pick a number between 1 and 100 and they each pick a number and the closest one goes on. :D

Um what are you suppose d to do in this?

its just a competition thing. Its like AOPS survivor except different style and and lot more different competitions

i'll join. i might have to drop later though if i'm too busy :|


Prove that if the real numbers $a,b$ and $c$ satisfy $a^2+b^2+c^2=3$ then
\[\frac{a^2}{2+b+c^2}+\frac{b^2}{2+c+a^2}+\frac{c^2}{2+a+b^2}\ge\frac{(a+b+c)^2}{12}\]
When does the inequality hold?

[quote="limes123"][b]Problem 2[/b]
Prove that if the real numbers a, b and c satisfy $ a^2 + b^2 + c^2 = 3$ then
$ \sum \frac {a^2}{2 + b + c^2}\geq \frac {(a + b + c)^2}{12}$.
When does the inequality hold?[/quote]
Uhh.. I solved it the day after the competition.. A bit too late :/

Cauchy-Swartz gives:
${ (\sum \frac {a^2}{2 + b + c^2})(\sum 2 + b + c^2}) \ge (a + b + c)^2$ So we just have to prove:
$ \sum 2 + b + c^2 \le 12$ or $ a + b + c \le 3$ which is obvious when $ a^2 + b^2 + c^2 = 3$. ($ a + b + c \le \sqrt {(1+1+1)(a^2 + b^2 + c^2)} = 3$)

My solution is the same but I thought that if a,b,c are real the proof would be more complicated... Any other solutions? :)

[quote="limes123"]Aaaaa right :D  no, my first idea was to use the Engel's form of CS and I didn't try to prove it in other way.[/quote]
A little off topic, but I am a bit curious :) Did you attend to the Baltic Way for Poland?

No. I found the problems on Polish math forum.

CS Engel
\[ \frac{a^2}{2\plus{}b\plus{}c^2}\plus{}\frac{b^2}{2\plus{}c\plus{}a^2}\plus{}\frac{c^2}{2\plus{}a\plus{}b^2} \ge \frac{(a\plus{}b\plus{}c)^2}{6\plus{}a\plus{}b\plus{}c\plus{}a^2\plus{}b^2\plus{}c^2}.\]
So it suffices to prove that \[ 6\plus{}a\plus{}b\plus{}c\plus{}a^2\plus{}b^2\plus{}c^2 \le 12.\]
Note that $ a^2\plus{}b^2\plus{}c^2\equal{}3$, then we only need to prove that \[ a\plus{}b\plus{}c \le 3\].
But \[ (a\plus{}b\plus{}c)^2\equal{}a^2\plus{}b^2\plus{}c^2\plus{}2(ab\plus{}bc\plus{}ca) \le a^2\plus{}b^2\plus{}c^2\plus{}2(a^2\plus{}b^2\plus{}c^2)\equal{}3(a^2\plus{}b^2\plus{}c^2)\equal{}9.\]
Hence $ a\plus{}b\plus{}c \le 3$ which completes the proof.

I'm sorry for some mistakes... Can we use CS Engel for real numbers?

[quote="Raja Oktovin"]CS Engel
\[ \frac {a^2}{2 \plus{} b \plus{} c^2} \plus{} \frac {b^2}{2 \plus{} c \plus{} a^2} \plus{} \frac {c^2}{2 \plus{} a \plus{} b^2} \ge \frac {(a \plus{} b \plus{} c)^2}{6 \plus{} a \plus{} b \plus{} c \plus{} a^2 \plus{} b^2 \plus{} c^2}.
\]
So it suffices to prove that
\[ 6 \plus{} a \plus{} b \plus{} c \plus{} a^2 \plus{} b^2 \plus{} c^2 \le 12.
\]
Note that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$, then we only need to prove that
\[ a \plus{} b \plus{} c \le 3
\]
.
But
\[ (a \plus{} b \plus{} c)^2 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca) \le a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(a^2 \plus{} b^2 \plus{} c^2) \equal{} 3(a^2 \plus{} b^2 \plus{} c^2) \equal{} 9.
\]
Hence $ a \plus{} b \plus{} c \le 3$ which completes the proof.

I'm sorry for some mistakes... Can we use CS Engel for real numbers?[/quote]
I don't know what CS Engel is... But it looks very much like Cauchy-Swartz.. And of course you can use negative values! (But remember that the square of a negative number is a positive number, so the biggest side will _always_ be non-negative) It seems like your proof is the exact same as mine.. So take a look at mine post, limes123 question and my answer. :)

[quote="Aravind Srinivas L"]Which is actually stronger? Cauchy Engel or cauchy schwartz as such? I feel i have solved a lot of problems including this one using cs.[/quote]
What is Cauchy Engel ?

[quote="Aravind Srinivas L"]the one that raja oktavin used.[/quote]
How does that differ from normal cauchy schwartz?

I've always know cauchy schwartz as:
$ (a_1^2\plus{}...\plus{}a_n^2)(b_1^2\plus{}...\plus{}b_n^2) \ge (a_1b_1 \plus{} ... \plus{}a_nb_n)^2$. Is Cauchy Schwartz = Cauchy Engel ???

[quote="Pablo09"]Hey Mathias ,
this is Cauchy Engel  $ \sum_{i \equal{} 1}^n \frac {x_i}{y_i}\geqslant \frac { (\sum_{i \equal{} 1}^n x_i)^2}{\sum_{i \equal{} 1}^n x_iy_i}$
Very useful ![/quote]
Ok. So that's just cauchy swartz $ a_i^2 \equal{} \frac{x_i}{y_i}$ and $ b_i^2 \equal{} x_iy_i$. $ \iff$ Cauchy Swartz = Cauchy Engel. QED :P :P :P

Very similar to the above, but
[hide=Solution]
From Titu's Lemma, we have \[\sum_{cyc}\frac{a^2}{2+b+c^2}\geq \frac{(a+b+c)^2}{9+a+b+c},\] so it suffices to prove that $a+b+c\leq 3$, which is obvious from $(1+1+1)(a^2+b^2+c^2)\geq (a+b+c)^2$.
[/hide]

Using C-S It suffices to prove that $(\sum 2+b+c^2)\le 12$ or $a+b+c\le 3$ from C-S $3(a^2+b^2+c^2)\ge (a+b+c)^2$

This inequality is easy to prove we just have to apply cauchy Schwartz once then we need to prove a+b+c<= 3 now $a^2+b^2+c^2$>= $(a+b+c) ^2$\3 so result follows

Note that $\sum_{cyc} \frac{a^2}{2+b+c^2} \ge \frac{(a+b+c)^2}{9+a+b+c}$ by Titu's Lemma, so it suffices to show $12 \ge 9+a+b+c \implies 3 \ge a+b+c$. But by Cauchy we know $(1+1+1)(a^2+b^2+c^2)=9 \ge (a+b+c)^2$, so $3 \ge a+b+c$.

[quote=jasperE3]Can we use C-S since $a,b,c$ are real, not necessarily positive?[/quote]

I guessed everybody used that... Btw, is Engel's form of Cauchy Schwarz is Titu's Lemma?

[quote=jasperE3]Can we use C-S since $a,b,c$ are real, not necessarily positive?[/quote]
You could have checked this [url=https://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality]here[/url] instead of asking.
[quote=SatisfiedMagma]I guessed everybody used that... Btw, is Engel's form of Cauchy Schwarz is Titu's Lemma?[/quote]
And to answer your question [url=https://en.wikipedia.org/wiki/Sedrakyan%27s_inequality]yes[/url].

[hide=#23][url=aops.com/community/user/297593][b]phoenixfire[/b][/url] · May 4, 2021, 4:20 PM [url=aops.com/community/p21798549](view)[/url][color=transparent]helo[/color]
[quote=jasperE3]Can we use C-S since $a,b,c$ are real, not necessarily positive?[/quote]
You could have checked this [url=https://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality]here[/url] instead of asking.

-----------
[color=#5b7083][aops]x[/aops] 23[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]

But I mean C-S in Engel form.

Yes, since $a\ge -\sqrt{3}$, etc. hence the quantities $2+a+b^2$, etc. are all positive. It was also stated in #4.


Find all solution to the equation $na^2+b^2=c^2$ where $(n,a,b,c) \in \mathbb{N}^4$ and $n$ is fixed.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=87739 (this here gets a special case of the problem there after dividing through $c^2$).
Just one point left to show: we need at least one solution; but $a=0, b=1=c$ works.

I did it by myself, but i didn't know how to finish the problem. I got something like $c\frac{P(p,q)}{Q(p,q)}$ which is nothing.

Dividing through $c^2$ gives after substitution $nx^2+y^2=1$, which fits into the other thread.


For how many integer values of $ x$ does the expression $ x^{2}\plus{}2x\minus{}19$ have a negative value?

The roots are $ x\equal{}\frac{\minus{}2\pm\sqrt{4\plus{}76}}{2}\equal{}\minus{}1\pm 2\sqrt{5}$.  These values are approximately 3.47 and -5.47.  So all integers -5 through 3 yield a negative value, for a total of 9 integers.


http://math.berkeley.edu/~wu/math-majors.pdf
Interesting article on whether math is taught correctly in college.

Almost any writing by Professor Hung-hsi Wu is worth reading.

http://math.berkeley.edu/~wu/

that article is really good...and i never realized that such a problem exists in even college. I know, of course, that this problem DO exist in K-12 becuase of either the lack of time or knowledge. So after reading that article, i realize that I should find a college that will not teach thier undergrad math majors with the intent of them getting a masters in math.

anyone have any ideas what some of these colleges may be?

this also brings me to another question. I don't exactly know how to word it. but I'll do the best i can. would you recommend someone to get also a masters in math to learn much much more in the math field? or is getting a bachelor's in math enough?.....i dont think that i did a great job. The thing is that i wanted to get a bachelo's in  math then get a masters in education. But I'm wondering if i should also plan on getting a master's degree in math then get my credentials becuase i will end up learning much more after a master. Hopefully this makes sense.

Does anyone know which way BU teaches math?


Let $ ABC$ be an acute triangle, $ AD_1$ is its altitude with $ D_1$ in $ BC$, $ M$ is the midpoint of $ BC$, $ AM$ meets the circumcircle of $ ABC$ at $ A_2$, $ A_2D_2$ is the perpendicular line of $ BC$
Prove that $ |A_2D_2|\geq|A_1D_1|$

you can prove this easily in this way:
Let $ A_3$ the point where AO meet again the circumcircle of ABC and $ D_3$ the projection of $ A_3$ on BC. As well-know $ \angle A_3AB \equal{} \angle A_1AC$ and so $ \overline{A_3D_3}\equal{}\overline{A_1D_1}$. Now clearly the point $ A_2$ is in the arch $ A_1A_3$ that not contain A, so $ \overline{A_2D_2} \ge \overline{A_1D_1}$


For a,b,c are all positive real,a^2+b^2+c^2=3, prove that

$ a\plus{}b\plus{}c^2 \leq \frac{7}{2}$

I have a proof but can't find out  the equality....... :?: need help :maybe:

Rewrite it as:

\[ a^2\plus{}b^2\plus{}\frac{1}{2}\ge a\plus{}b\]

AM-GM gives:

\[ a^2\plus{}b^2\plus{}\frac{1}{2}\ge \sqrt{2(a^2\plus{}b^2)}\ge a\plus{}b\]

so it's done. Equality occurs only when $ a\equal{}b\equal{}\frac{1}{2}$ and $ c\equal{}\sqrt{\frac{5}{2}}$.

[quote="4865550150"][quote="Pain rinnegan"]Rewrite it as:
\[ a^2 \plus{} b^2 \plus{} \frac {1}{2}\ge a \plus{} b\]
AM-GM gives:
\[ a^2 \plus{} b^2 \plus{} \frac {1}{2}\ge \sqrt {2(a^2 \plus{} b^2)}\ge a \plus{} b\]
so it's done. Equality occurs only when $ a \equal{} b \equal{} \frac {1}{2}$ and $ c \equal{} \sqrt {\frac {5}{2}}$.[/quote]

thx help.

My solution is that....
AM-GM:
$ \frac {(a \plus{} b)^2}{2(a^2 \plus{} b^2)} \plus{} 2(a^2 \plus{} b^2) \ge 2(a \plus{} b)$
$ 1 \plus{} 6 \ge \frac {a \plus{} b}{2(a^2 \plus{} b^2)} \plus{} 6 \ge 2(a \plus{} b \plus{} c^2)$
so,
$ 7/2 \ge a \plus{} b \plus{} c^2$

but i can't find equality from this way[/quote]

Why?

[quote="4865550150"]
$ 1 \plus{} 6 \ge \frac {a \plus{} b}{2(a^2 \plus{} b^2)} \plus{} 6$
[/quote]

You must have here equality which means $ a\equal{}b$

[quote="4865550150"]
$ \frac {(a \plus{} b)^2}{2(a^2 \plus{} b^2)} \plus{} 2(a^2 \plus{} b^2) \ge 2(a \plus{} b)$
[/quote]

You also need here equality which setting $ a\equal{}b$ gives $ a\equal{}b\equal{}\frac{1}{2}$. Then you get the value of $ c$.

Btw, your solution is the same as mine.

[quote="Pain rinnegan"][quote="4865550150"][quote="Pain rinnegan"]Rewrite it as:
\[ a^2 \plus{} b^2 \plus{} \frac {1}{2}\ge a \plus{} b\]
AM-GM gives:
\[ a^2 \plus{} b^2 \plus{} \frac {1}{2}\ge \sqrt {2(a^2 \plus{} b^2)}\ge a \plus{} b\]
so it's done. Equality occurs only when $ a \equal{} b \equal{} \frac {1}{2}$ and $ c \equal{} \sqrt {\frac {5}{2}}$.[/quote]

thx help.

My solution is that....
AM-GM:
$ \frac {(a \plus{} b)^2}{2(a^2 \plus{} b^2)} \plus{} 2(a^2 \plus{} b^2) \ge 2(a \plus{} b)$
$ 1 \plus{} 6 \ge \frac {a \plus{} b}{2(a^2 \plus{} b^2)} \plus{} 6 \ge 2(a \plus{} b \plus{} c^2)$
so,
$ 7/2 \ge a \plus{} b \plus{} c^2$

but i can't find equality from this way[/quote]

Why?

[quote="4865550150"]
$ 1 \plus{} 6 \ge \frac {a \plus{} b}{2(a^2 \plus{} b^2)} \plus{} 6$
[/quote]

You must have here equality which means $ a \equal{} b$

[quote="4865550150"]
$ \frac {(a \plus{} b)^2}{2(a^2 \plus{} b^2)} \plus{} 2(a^2 \plus{} b^2) \ge 2(a \plus{} b)$
[/quote]

You also need here equality which setting $ a \equal{} b$ gives $ a \equal{} b \equal{} \frac {1}{2}$. Then you get the value of $ c$.

Btw, your solution is the same as mine.[/quote]

whoops! you remind me.
i understand now. :lol:

好多不等式
不過還比較基礎


Putnam
integration
logarithms
function
calculus
real analysis
real analysis unsolved

$\int_{0}^{1}\frac{\ln (1-x)}{1+x^2}dx=\sum^{\infty}_{n=0}\int_{0}^{1}\frac{x^n}{n(1+x^2)}$.I'm too lazy to make the computations,but I think this should give the answer.
(We have the right to interchange the symbols $\int$ and $\sum$ because the functions $f_n(x)=\frac{x^n}{n(1+x^2)}$ verify the hypothesis of the theorem of dominant convergence...recently posted)

Do you know how to actually calculate those integrals and evaluate their sum?
That's probably not a useful method.

I think you need (-1) in that integral,
$\ln(1-x)= (-1) \sum_{k=1} \frac{x^k}{k}$.
About the integral, i found $\int_{0}^{s} \frac{x^n}{1+x^2} \, dx$ in numerical analysis test.
could you evaluate that integral analytically?

I think that my argument doesnt suffice :oops: .Anyway ,that's what I've obtained so far..:
Let $I_n=\int\frac{x^n}{1+x^2}dx$.We have $I_{n+2}+I_n=\frac{x^{n+1}}{n+1}$ so $J_{n+2}+J_n=\frac{1}{n+1}$,where $J_n=\int_0^1\frac{x^n}{1+x^2}dx$.
Since $J_n\geq J_{n+2}$ we have $\frac{1}{2n(n-1)}\geq\frac{J_n}{n}\geq \frac{1}{2n(n+1)}$.
So $\sum_2^{\infty}\frac{1}{2n(n-1)}\geq \sum_2^{\infty}\frac{J_n}{n} \geq \sum_2^{\infty}\frac{1}{2n(n+1)}$.
${\frac12\geq \sum_2^{\infty}\frac{J_n}{n} \geq \frac{1}{4}}$.


I have a bag with marbles numbered from 1 to 6.  Mathew has a bag with marbles numbered from 1 to 12.  Mathew chooses one marble from his bag and I choose two from mine.  In how many ways can we choose the marbles (where the order of my choices does matter) such that the sum of the numbers on my marbles equals the number on his?

The two marbles chosen by me determine the marble Mathew chooses, so it's just $ \binom{6}{2}\equal{}15$, i hope...

Except...no.

What if you get 1,6. Is that different from 3,4?

Its actually simpler.

The minimum sum you can make is 3, the maximum 11.

Thus 11-3+1=9.

both of our solutions are wrong...
in your solution, you found the number of ways Mathew could choose marbles such that there exists a way to draw two marbles from your bag for which their sum is the number on his marble
on mine, I didn't see that the order matters, so it should be $ 6\cdot 5\equal{}30$?

isn't it simpler thinking that every choise of two of my marbles will sure have a sum minor than 12... so the result must be all the possibly couples taken from my bag... so 6x5=30....

[quote="5849206328x"]The two marbles chosen by me determine the marble Mathew chooses, so it's just $ \binom{6}{2} \equal{} 15$, i hope...[/quote] it isn't because of order. therefore *2!

If I draw my $2$ marbles, then Mathew will only have $1$ choice to draw his marble. So the possibilities is equal to the total amount of ways I can draw my marbles, so $6$ times $5$ = $\boxed{30}$ .

I think the question should mention that the bag with $6$ marbles should mention that the marbles are being picked without replacement

[quote=GameBot] (where the order of my choices does matter)[/quote]
[hide=.]so you choose your 2 marbles in 6*5 ways(order does matter) and Mathew can draw whatever the sum of your two marbles is, so it's just 30.

Why did I list all the possible sums and multiply that by 2 smh. Took too long.

Anyways, quicker solution:
[hide=Sol]The maximum sum of my numbers is $6+5 =11.$ So we can just choose two numbers to find a sum, and Mathew's number will be already chosen. Since order matters, there are $6 \cdot 5 = \boxed{30}$ ways that we can choose marbles.[/hide]