trigonometry
geometry
rhombus
geometry unsolved

Let $ABCD$ is a rectangle. $M, N$ be the points lie on $BC, CD$ so that $\triangle AMN$ is an equilateral triangle.
Prove that: $S(CMN)=S(ABM)+S(ADN)$
With $S(XYZ)$ is the area of triangle $XYZ$.

[b]Remark:[/b] Who can prove it without use trigonometry?

It suffices to show that the sum of the altitudes of $ABM$ and $ADN$ equals the altitude of $CMN$.
Cut off the triangles $ABM,CMN$ and $ADN$ and rearrange them so they share a common base (see Graphic2.jpg).
 Since we have $\angle BAM+\angle DAN=30^{\circ}$ it follows that the arc $BD$ has the measure $120^{\circ}$. Another simple angle chasing shows that th arc $BC$ equals $60^{\circ}$, hence $C$ is the midpoint of the arc $BD.$ We deduce that $OBCD$ is a rhombus, therefore $S$ is both the midpoint of $BD$ and the midpoint of $OC$. Finally, the sum of the altitudes of $ABM$ and $ADN$ equals twice the distance from $S$ to the base and the latter is also equal to the altitude of $CMN$ which is the distance from $C$ to the base.


$ \lim_{x\rightarrow \infty} e^{[(\minus{}1)^\frac{1}{2x}]}$

hello, $ \lim_{x\rightarrow \infty} e^{[(\minus{}1)^\frac{1}{2x}]}\equal{}e$.
Sonnhard.

No. There is no unique definition of $ (\minus{}1)^{\frac1{2x}}$, and the set of possible values is dense in a curve at most points.


trigonometry
inequalities proposed
inequalities

Let $u;v;a;b;c$ be positive real numbers satisfying   $ab+bc+ca=1$
Find min $Q=ua^{2}+vb^{2}+c^{2}$

Use $x^{2}+y^{2}+z^{2}\geq 2xy\cos{A}+2yz\cos{B}+2zx\cos{C}$, where $x,y,z>0$ and $ABC$ is a triangle.


[quote="joml88"]For this class how high on the amc does it go?  Will it cover up to number 25 difficulty and how much of the class is focused on this level of difficulty?   I am just trying to decide if I should take the class.[/quote]

This was formerly the introduction to counting class iirc. If the highest level you're looking at is AMC 10/12 right now, it's an excellent class to take.

So if I do take this course Aime/olympiad counting would be a good course to take afterwards right?  Since I couldn't find a Amc/aime counting i am assuming that the olympiad counting is the next step. Am I right?  Also, when is the next Aime/olympiad class offered?

A very, very good place is the "Online Classes" page.  It's your friend. ;-)

You can also take the "Are you ready?" and "Do you need this?" tests.

I already went to those places.  I was assuming that all the classes listed were the only classes offered (now or in the future) and I just wanted to know if that was correct.   It also doesn't say when the next Aime counting is.  (well the next one is this summer but if I take the mathcounts/amc level then I obviously wouldn't take the aime level then :wink: )

The first Counting and Probability course covers all concepts tested for on the AMC plus a few things that are not.

Both Probability and Counting courses will be offered during the school year.

Also, we will be adding courses not currently listed as we move forward.  We eventually plan to cover all topics that students would encounter in competition problem solving through even the IMO and Putnam plus some other courses.

[quote="joml88"]So if I do take this course Aime/olympiad counting would be a good course to take afterwards right?  Since I couldn't find a Amc/aime counting i am assuming that the olympiad counting is the next step. Am I right?[/quote]

Theoretically, yes. However, I think it'd be better to wait on that until you're ready to tackle problems of that difficulty level. It's one thing to say you have the "prerequisites" for the AIME/Olympiad Counting, but another to be ready for problems of that level. The "Do you need this class?" tests should help you determine whether or not you should take the class.


$ 1)$ The function $ f: Z\to Z$ satisfies $ f(n)\equal{}n\minus{}3$ if $ n \geq 1000$ and $ f(n)\equal{}f(f(n\plus{}5))$ if $ n<1000$
Find $ f(84)$
$ 2)$ A function $ f$ is defined for all real numbers and satisfies:
$ f(2\plus{}x)\equal{}f(2\minus{}x)$ and $ f(7\plus{}x)\equal{}f(7\minus{}x)$,for all real $ x$.
If $ x\equal{}0$ is a root of $ f(x)\equal{}0$ what is the least number of roots $ f(x)\equal{}0$ must have in the interval $ \minus{}1000 \leq x \leq 1000$
Have fun :)

[hide="1"]
We show $ f(1004 \minus{} 5(2m \minus{} 1)) \equal{} 998$ and $ f(1004 \minus{} 5\cdot 2m) \equal{} 997$ for any positive integers $ m$ by using induction.

1)for $ m \equal{} 1$,we have $ f(999) \equal{} f(f(1004)) \equal{} f(1001) \equal{} 998$ and $ f(994) \equal{} f(f(999)) \equal{} f(998) \equal{} f(f(1003)) \equal{} f(f(1000)) \equal{} 997$ hence the proposition is true.

2)we assume $ f(1004 \minus{} 5(2m \minus{} 1)) \equal{} 998$ and $ f(1004 \minus{} 5\cdot 2m) \equal{} 997$,then
$ f(1004 \minus{} 5(2m \plus{} 1)) \equal{} f(f(1004 \minus{} 5\cdot 2m)))$
$ \equal{} f(997) \equal{} f(f(1002)) \equal{} f(999) \equal{} 998$
and
$ f(1004 \minus{} 5(2m \plus{} 2)) \equal{} f(f(1004 \minus{} 5(2m \plus{} 1))) \equal{} f(998) \equal{} 997$,
therefore our induction complete,

especially we let $ m \equal{} 92$ we have $ f(84) \equal{} 997$.[/hide]

[quote="Farenhajt"][hide="#2"]$ f(t) \equal{} f(4 \minus{} t) \equal{} f(14 \minus{} (4 \minus{} t)) \equal{} f(t \plus{} 10)$

hence $ f$ is periodic with period $ 10$. Therefore it must have at least $ 201$ zeroes in the segment $ [ \minus{} 1000, 1000]$[/hide][/quote]

So can you find a function to prove your conclusion? :)

2)$ f(t)\equal{}f(4\minus{}t)\equal{}f(14\minus{}(4\minus{}t))\equal{}f(t\plus{}10)$, so $ f(x)$ is periodic with period $ 10$ as you say , also $ f(0)\equal{}0$ and $ f(4\minus{}0)\equal{}f(4)\equal{}0$ gives

for $ x\equal{}0,4(mod 10)$, $ f(x)\equal{}0$. So the least number of roots must be at least $ 401$.

Now we defined $ f(x)$ like this :$ \begin{cases}f(x)\equal{}0 when x\equal{}0,4(mod 10) \\
f(x)\equal{}1 when x \neq 0,4(mod 10) \end{cases}$

Obviously this satisfies the condition ($ f(t)\equal{}f(4\minus{}t)$ and $ f(t)\equal{}f(14\minus{}t)$ and $ f(0)\equal{}0$), and the number of solution to $ f(x)\equal{}0$ is $ 401$,
so the answer is $ 401$.


calculus
integration
logarithms
function
trigonometry
absolute value
calculus computations

Compute the following integral

$ \int_{0}^{1} \frac {x \minus{} 1}{\ln x} dx.$

I'm pretty sure that you [i]all[/i] are going to like this one...

Regards.

Begin looking at:

$ \int^1_0 \frac{x^n\minus{}1}{\ln(x)} \,dx \equal{} \int^1_0 \left(\int  \frac{\partial}{\partial n} \frac{x^n\minus{}1}{\ln(x)} \, dn\right)\,dx \equal{}$

$ \equal{} \int \left(\int^1_0  \frac{\partial}{\partial n} \frac{x^n\minus{}1}{\ln(x)} \, dx\right)\,dn \equal{} \int \left(\int^1_0 x^n \, dx\right)\,dn \equal{} \int \frac{dn}{n\plus{}1} \equal{} \ln(n\plus{}1)$

And now just plug in $ n\equal{}1$ to get:

$ \int^1_0 \frac{x\minus{}1}{\ln(x)} \,dx \equal{} \ln(2)$

Very nice solution!

I'm just curious, under what circumstances may we switch the order of integration and what is the motivation for looking at the problem this way?

With the substitution $ x \equal{} e^{\minus{}t}$, we have

$ \int_{0}^{1} \frac{x \minus{} 1}{\log x} \, dx \equal{} \int_{0}^{\infty} \frac{e^{\minus{}t} \minus{} e^{\minus{}2t}}{t} \, dt \equal{} \log 2 ,$

and this is a well-known form of integral called [url=http://mathworld.wolfram.com/FrullanisIntegral.html]Frullani's integral[/url].

In that form, I would use $ \frac{e^{\minus{}t}\minus{}e^{\minus{}2t}}{t}\equal{}\int_1^2 e^{\minus{}tx}\,dx$ and the same interchange idea.

[quote="jmerry"]There are two standard sufficient conditions: positivity (Tonelli's theorem) or absolute convergence (Fubini's theorem). The first works well here.[/quote]

Just out of curiosity, I went to Mathworld to see what were the two theorems you talked about, and there I found that Tonelli's theorem IS Fubini's theorem!  :maybe:

[url]http://mathworld.wolfram.com/FubiniTheorem.html[/url]

Tonelli's theorem (named after Leonida Tonelli) is a successor of Fubini's theorem. The conclusion of Tonelli's theorem is identical to that of Fubini's theorem, but the assumptions are different. Tonelli's theorem states that on the product of two $ \sigma$-finite measure spaces, a product measure integral can be evaluated by way of an iterated integral for nonnegative measurable functions, regardless of whether they have finite integral.

In fact, the existence of the first integral above (the integral of the absolute value), can be guaranteed by Tonelli's theorem.

A formal statement of Tonelli's theorem is identical to that of Fubini's theorem, except that the requirements are now that $ (X,A,\mu)$ and $ (Y,B,v)$ are $ \sigma$-finite measure spaces, while $ f$ maps $ X\times Y$ to $ [0, \infty]$.

Here's a problem that has some connections with the one posted on top of this thread:

$ \int^1_0 \frac{\sin (\pi x)}{\ln x} \, dx$

[quote="coquitao"]
Compute the following integral

$ \int_{0}^{1} \frac {x - 1}{\ln x} dx.$

I'm pretty sure that you [i]all[/i] are going to like this one...[/quote]
More generally,

\begin{eqnarray*} \int_{0}^{1}{\frac {x^{\alpha } - x^{\beta }}{\ln x}\,dx} & = & \int_{0}^{1}{\int_{\beta }^{\alpha }{x^{y}\,dy}\,dx} \\
& = & \int_{\beta }^{\alpha }{\int_{0}^{1}{x^{y}\,dx}\,dy} \\
& = & \int_{\beta }^{\alpha }{\frac {dy}{1 + y}} \\
& = & \ln \frac {\alpha + 1}{\beta + 1},\,\forall\,\alpha,\beta\ge0. \end{eqnarray*}

In particular, for $ \alpha = 1$ & $ \beta = 0$ yields the value of the original integral.

By the way, how would you guys prove the convergence?


Suppose A = 1061520150601. Determine (without calculator) the smallest rational number B for which there exists a rational number C such that AB = C.

Ooh nice problem. Since it's in intermediate I'll give others a chance to post first.

You might want to edit that to "smallest positive rational number B."

A vague hint: [hide]It has to do with triangular numbers (1 6 15 20 15 6 1)  and is related to 11 ... [/hide]

The answer is [hide]1/6 [/hide]

This is a nice problem.

Nice problem

[hide]Since we are not allowed to use calculator, we can notice that :

(1+x)6 = 1 + 6x + 15x2 + 20x3 + 15x4 + 6x5 + x6

So 1061520150601 = (100+1)6 = (101)6    [/hide]

Because any non-positive integer works as well.  For example, B = 0 gives us the nice rational number C=1.  B=-7 gives us a very nice unit fraction, as well.

If you haven't seen this problem before.. how would you think of doing that?

I think 1, 6, 15, 20, 15, 6, 1 kind of "stand out". This really is a nice problem, Arne. Where did you get it? Is it original?

Oh yes, of course I meant positive rational number. Sorry

Just some experiment. In our own national olympiad the following MC - question was given in round 1 :

The number 14641 is the product of
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 or more
prime numbers.

So this was my source of inspiration. (14641 = 114)


geometry
circumcircle
search
incenter
perpendicular bisector
radical axis
power of a point

Dear jayme:
your problem is the same to ML Contest E7,R6, P3
you can see if you know chinese  http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=43&Id=12703&page=2

Let A'  be the intersection of AA' and 0  AA' is parallel to BC
angleA'DM=angleAA'P=angleAM'P  so P,M',M,D are concyclic
we know that E,U,V,F   E,D,M,F are concyclic    so  EF,BC,PM' are concurrent at X
this completes our proof

[quote="jayme"]Dear Plane Geometry and Mathlinkers,
thank you very much for your rapidity and reference.
Sorry, I can't red your langage... but if you have some time, can you translate the problem in order to I understand the notations of your proof.Thank in advance
Sincerely
Jean-Louis[/quote]
Notice IP⊥Q’T   IR⊥RT  ID ⊥ DT   =>   I，D，P，T，R are concyclic
∠PDT=∠PIT =∠EIQ    but  ∠AIQ+∠EIQ=90  ∠IDM+∠MDB=90
PS pass through midpoint of arc BC，Q’Q⊥BC
If we can prove ∠Q’IP=∠IDM , the problem is done
notice ∠Q’IP= ∠A Q’Q+ ∠Q’QP=∠MQ’Q+∠Q’MP=∠Q‘VP  （1）
we have QQ’∥ID  so (1)  is true
done!

[quote="jayme"]Dear Plane Geometry,
"Let A' be the intersection of AA' and 0 AA' is parallel to BC
angleA'DM=angleAA'P=angleAM'P so P,M',M,D are concyclic"  Yes.
"we know that E,U,V,F are concyclic" Yes.
But how you prove that "E,D,M,F are concyclic"?
If you have some time, can you please translate the problem in order to I understand correctly the notations of your proof. Thank again in advance.
Sincerely
Jean-Louis[/quote]
reference link：
http://www.mathlinks.ro/Forum/viewtopic.php?t=214426
http://www.mathlinks.ro/viewtopic.php?search_id=73358065&t=214687

sorry,dear jayme.what I say is confused,the right order should be as followed:
1.E,U,V,F are concyclic
2.P,U,V,M' are concyclic
3.EF,BC,PM' are concurrent
4.from the problem I mentioned above, we have P,D,M,M' are concyclic
5.E,D,M,F are concyclic
sorry again  :oops:  :blush:
for your problem you can see here :http://www.mathlinks.ro/viewtopic.php?t=213098[url][/url]

[quote="jayme"]Dear Mathlinkers,
let ABC be a triangle, M be an arbitrary point on the side BC, 0 the circumcircle of ABC, M' the second meet point of AM with 0,
1 the circle tangent to MA, MB and 0,
U the point of contact of 1 with BC.
2 the circle tangent to MA, MC and 0,
V the point of contact of 2 with BC
and P the point of contact of the A-mixtilinear incircle of ABC.
Prove that U, V, M' and P are cocyclic.
Sincerely
Jean-Louis[/quote]

Dear Jayme,
This (short) proof does not use inversion anywhere (as desired) and it is completely different from the (long) one following from the topics:

http://www.artofproblemsolving.com/viewtopic.php?t=213098
http://www.artofproblemsolving.com/viewtopic.php?t=209898
http://www.artofproblemsolving.com/viewtopic.php?t=186117

Circles $ (1), (2)$ are tangent to $ MA$ at $ U', V'.$ By Thebault theorem proof, $ UU', VV'$ intersect at the incenter $ I$ of the triangle $ \triangle ABC.$ Consequently, the angle $ \angle UIV$ is right and tangents to the incircle $ (I)$ from $ U, V$ are parallel to $ MA.$ Done with Thebault theorem.

Internal angle bisectors $ AI, BI, CI$ cut $ (O)$ again at $ X, Y, Z.$ The A-mixtilinear incircle touches $ AB, AC$ at $ E, F.$ By Pascal theorem for $ AZBPCY,$ the intersections $ E \equal{} AB \cap PZ,\ F \equal{} AC \cap PY,\ I \equal{} BY \cap CZ$ are collinear, $ I$ is midpoint of $ EF.$ Let $ PI$ cut $ (O)$ again at $ X'.$ $ PI, PA$ are median and symmedian of the triangle $ \triangle PEF \Longrightarrow ZX' \equal{} AY,$ so that $ XX'$ is the perpendicular bisector of $ BC,$ the angle $ \angle IPX \equal{} \angle X'PX$ is right, and $ P$ is on a circle $ (K)$ with diameter $ IX.$ (Well known, I usually use inversion in the incircle here.)

$ (X)$ is a circle with center $ X$ and radius $ XB \equal{} XC \equal{} XI.$ The circle $ (K)$ with diameter $ IX$ is internally tangent to the circle $ (X)$ at $ I.$ Radical axes $ XP$ of the circles $ (O), (K)$ and $ BC$ of the circles $ (O), (X)$ meet at the radical center $ R$ of $ (O), (X), (K).$ Radical axis of $ (X), (K),$ their single common tangent at $ I,$ goes through $ R.$

Let $ PM'$ cut $ BC$ at $ Q$ and let $ QI$ cut the circle $ (X)$ again at $ J.$ $ N$ is the midpoint of $ IJ$ and $ NX$ is the perpendicular bisector of $ IJ.$ The angle $ \angle INX$ is right $ \Longrightarrow N$ is on the circle $ (K)$ with diameter $ IX.$ The line $ PM'$ cuts the circle $ (K)$ again at $ L$ and by Reim theorem, $ IL \parallel AM'.$

Let $ IL, NX$ intersect at $ S.$ By Pascal theorem for the cyclic pentagon $ IPXLN$ (degenerated cyclic hexagon $ IIPXLN$ with double vertex $ I$), the intersections $ Q \equal{} LP \cap IN,\ R \equal{} XP \cap II,\ S \equal{} IL \cap NX$ are collinear. Since $ Q, R$ are on $ BC,$ so is $ S.$

A circle $ (S)$ with center $ S$ and radius $ SI \equal{} SJ$ cuts $ BC$ at $ U, V,$ following on this line at the same order as $ B, C.$ The angle $ \angle UIV$ is right and tangents to the incircle $ (I)$ through $ U, V$ are parallel to $ IL,$ which is parallel to $ AM'.$ Therefore, $ BC$ is tangent to the circles $ (1), (2)$ at $ U, V.$ But of course,

$ \overline{QU} \cdot \overline{QV} \equal{} \overline{QI} \cdot \overline{QJ} \equal{} \overline{QB} \cdot \overline{QC} \equal{} \overline{QP} \cdot \overline{QM'}$

$ \Longrightarrow$ the points $ P, U, V, M'$ are concyclic.

Very nice solution! Dear yetti.
A typo: $ S \equal{} IL\cap JX$, it should be $ S \equal{} IL\cap NX$.

Thank you, I fixed the typo. I hope that's it.

Dear Vladimir,
I discover now your message...
Thank very much for your help and the wonderful synthetic proof.
This result changes completely the perspective of your three reference.
Sincerely
Jean-Louis

Dear Vladimir and Mathlinkers,
a new way for a proof of this result can be found on on my website :
http://perso.orange.fr/jl.ayme   vol. 4   A new mixtilinear incircle adventure III, p. 27
Sincerely
Jean-Louis


compute

lim sum ([(2k^2-3k-5)*3^k]/(k+2)!), k=1, n and lim when n -->oo.

really need help with this one!
bye bye!

3/2

[hide]

Use :

sum a^k / k! = e^a

sum k*a^k / k! = a*e^a

sum k*a^k / k ! = a*(a+1)*e^a

The sums are done from k=0 of course.

The coefficients are chosen to eliminate e^3 !![/hide]

i/3:

Im really bad with limits, so could you explain where you got those identities from?

[quote="teratomato"]That's just the Maclaurin series for e^x.[/quote]

Or if you so like, the definition of e^x.

hmm.. I would really need a much more detailed solution.. please!

thank you!

This is really basic manipulations of power series :

. e^x = sum x^n/n! is either a definition or a Taylor expansion

. sum n*x^n/n! = sum x^n/(n-1)! = x sum x^(n-1)/(n-1)! = xe^x

. notice that n^2 = n(n-1)+n to get sum n^2*x^n/n!

HTH

I didn't get 3/2  :oops:
i got 2e^3/9  the limit... how did you actually got there...I feel so stupid.. :( .. could you help?

[quote="i/3"]
. sum n*x^n/n! = sum x^n/(n-1)! = x sum x^(n-1)/(n-1)! = xe^x
[/quote]

This is right, but your reasoning is wrong. If you are summing from n=0, then you cannot divide n into n! to get (n-1)!

What you really want to do is say:
e^x = sum [x^n/n!]
d/dx(e^x) = sum [n*x^(n-1)/n!]
So x*e^x = sum [n*x^n/n!]

And then,
d/dx(x*e^x) = sum [n^2*x^(n-1)/n!]
So x*(x+1)*e^x = sum [n^2*x^n/n!]

To simplify the original problem, you might also want to try a change of variables from k --> y = k+2. Then I think the exponential series formula will look more familiar.


Yale college Stanford MIT Princeton geometry Harvard

I'd like to know how Yale's undergrad math compare with other universities?  I have a place there, but I understand that Yale doesn't exactly have a good reputation in the sciences.

Can anyone help?

Oh, I wouldn't say at all that "Yale doesn't exactly have a good reputation in the sciences," but it does seem that the Yale undergraduate program in math is, in the opinion of some would-be math majors, less coherent than that at Chicago, Stanford, MIT, Princeton, or Harvard. But certainly a science or math degree from Yale would be well regarded as compared to the same degree from many other colleges around the United States, because, after all, it's a Yale degree and Yale is hard to get into.

There is no specific undergraduate curriculum at most schools, including most or all of the ones listed above; there are some loose requirements such as 1-2 courses each in designated areas such as analysis, algebra, geometry, and a minimum total number of courses in and out of the subject.  The student chooses their own program, and at undergraduate level there is very little that can be done at one major university that can't be done at another.   If you want to do research in a specific area with specific people, that changes things.   Yale has no shortage of top-level faculty whose enthusiastic letters of recommendation would get anybody into grad school.

I have spent the past semester at the Budapest Semester in Mathematics program.  I am a Harvard junior and my roommate is an old friend with whom I attended middle school but with whom I have had reasonably little contact since then.  He is a junior math major at Yale.  He has a very, very low opinion of the Yale math program -- his complaints include an unfriendly faculty, a very small department and poor teachers with poor English language skills.  (On that last point, he commented during the semester that his professors in Budapest, all native Hungarians, spoke better English than his professors at Yale.)  He decided halfway through the semester to return for the spring semester (that is, to spend the entire school year there).  I would note that some of his complaints are also fair complaints about the Harvard math department, most significantly the relative unfriendliness of the faculty.  By way of comparison, I have much better relationships with two of my three math professors in Budapest than with any of my professors at Harvard.

With respect to fleeting_guest's comment that "Yale has no shortage of top-level faculty whose enthusiastic letters of recommendation would get anybody into grad school,"  I would offer that first, it assumes the existence or achievability of such enthusiastic letters of recommendation, and second, that it's a terrible reason to choose to go to a particular college.

You may consider this whole post as one long, unsolicited advertisement for the BSM program if you like :-)

I don't want to hijack this thread, but Joel,  could you tell us more about BSM? Sounds like something I'm interested in, altough I'm pretty far away from being eligible to go.

Heh, give me a new thread and a couple of specific questions and I will be happy to hold forth on my views, thoughts, experiences, etc.  The short version is that I had a truly wonderful time and I think it's a great program.

[quote="JBL"]With respect to fleeting_guest's comment that "Yale has no shortage of top-level faculty whose enthusiastic letters of recommendation would get anybody into grad school,"  I would offer that first, it assumes the existence or achievability of such enthusiastic letters of recommendation, and second, that it's a terrible reason to choose to go to a particular college. [/quote]Please speak for yourself rather than attributing assumptions.

The initial poster asked about Yale compared to other universities and whether Yale "doesn't exactly have a good reputation in the sciences".  All I [i]assume[/i] is that reputation of faculty, and the quality of one's options after graduation, are relevant to answering such a question.   The message of all that I wrote was that as far as specifically mathematical options are concerned, it doesn't much matter where one studies, unless one wants to work with specific people and is advanced enough to do so.  The differences are even more marginal when comparing only between top 5-10 universities.  What will vary is the population of fellow students, the appeal of the location, the weather, the local institutional culture, and other considerations not to do with the mathematics.

This point (my own which follows, not yours which precedes) is so not worth making that I would advise anyone who cares about anything other than my own personal stubborness to skip it.

I was not speculating on your personal assumptions, but rather on the implicit assumptions of your sentence.  To say, "Yale has no shortage of top-level faculty whose enthusiastic letters of recommendation would get anybody into grad school," when giving advice to someone about what college they should apply to or attend, implicitly supposes that those letters of recommendation would be available to the student applying.

More importantly, it's still a terrible reason to go to a school.

[quote="JBL"]I was not speculating on your personal assumptions, but rather on the implicit assumptions of your sentence.  [/quote]

There are ways of writing such speculations that cannot be misunderstood as being personal, and those ways are better in a public forum.   As to your actual point,

[quote]To say, "Yale has no shortage of top-level faculty whose enthusiastic letters of recommendation would get anybody into grad school," when giving advice to someone about what college they should apply to or attend, implicitly supposes that those letters of recommendation would be available to the student applying.  [/quote]

Unless you are hinting at reasons specific to Yale that would make it harder to get such letters as compared to another university, what is the relevance?  If I had said instead that "Yale has great math courses", I don't think it would rise to the level of a notable tacit assumption that one has to (a) attend Yale, and (b) take the math courses, in order to experience their greatness, and it would be strange to comment on those as assumptions.

[quote]More importantly, it's still a terrible reason to go to a school.[/quote]

It is equally terrible or non-terrible considered as a reason to go to Yale, Harvard, Caltech, etc, so I don't see the relevance to a discussion of Yale specifically.

The OP appeared to be a person who felt sure he could be admitted to Yale (had already been admitted in the early round?) but wasn't sure he should take that opportunity. What might be a better opportunity for a student seeking a good undergraduate experience in mathematics or physical science?

[hide="Re fleeting_guest"][quote="fleeting_guest"][quote="JBL"]I was not speculating on your personal assumptions, but rather on the implicit assumptions of your sentence.  [/quote]

There are ways of writing such speculations that cannot be misunderstood as being personal, and those ways are better in a public forum.[/quote]

I wrote, "With respect to fleeting_guest's comment ... I would offer that first, [i]it[/i] assumes..."  (emph. added).  Unless you thought the "it" was directed at you, I would describe this as fairly grammatically clear: the comment was the subject, not you.

[quote="fleeting_guest"][quote]More importantly, it's still a terrible reason to go to a school.[/quote]

It is equally terrible or non-terrible considered as a reason to go to Yale, Harvard, Caltech, etc, so I don't see the relevance to a discussion of Yale specifically.[/quote]
Heh, I initially thought you were refering to your own comment.  Having read correctly, however: I agree -- if this were strictly a discussion about Yale, my comment would have seemed like something of a non-sequitur.  However, this was a discussion about advice, particularly advice which pertained to Yale.  You offered what I consider a bad piece of advice.  I offered that I thought it was a bad piece of advice (which is in fact advice as well: "don't go to school based on that consideration.")[/hide]

Sorry, tokenadult, for continuing this.

[hide="(click here to view reply to JBL re: metadiscussion)"]
[quote="JBL"][quote]There are ways of writing such speculations that cannot be misunderstood as being personal[/quote]I wrote, "With respect to fleeting_guest's comment ... I would offer that first, [i]it[/i] assumes..."  (emph. added).  Unless you thought the "it" was directed at you, I would describe this as fairly grammatically clear: the comment was the subject, not you. [/quote]Assumptions are made by people, not by comments, so the supposedly clear grammatical nuance is questionable.  More to the point, there is a definite gradation in the different ways of writing such things, e.g. (in decreasing order of personalization),

1. "Joe is assuming X"   (direct statement about person)
2. "Joe's remark that 'Oxford is great' assumes X"  (attribution of disputable assumption to person in the room)
3. "Shakespeare's remark that 'Oxford is great' assumes ..." (attribution to person not present in the room)
4. "Saying 'Oxford is great' assumes X"  (dispute assumption without relating it to a person)
5. "X is not necessarily true for reason Z, thus Oxford may not be so great"  (refute conclusion without speculating on who assumed what in reaching it)

Your comment was also negative in various ways, which (given that it named a specific user) moves it further in the personal direction than whatever is implied by sentence structure alone.

In short, I think that you are right on a "plausible deniability" standard but not on a "commonsense English reading" standard.   I realize that opinions may differ here, and that a separate thread may be a better place for discussing matters of etiquette. [/hide]

[quote]this was a discussion about advice, particularly advice which pertained to Yale.  You offered what I consider a bad piece of advice. [/quote]  The sentence about faculty and recommendations that you quoted does not contain any apparent advice; please do not attribute to me speculations that go beyond what I actually wrote.   Anyway, it would be useful and on-topic to specify exactly what is this [i]hypothetical[/i] advice about Yale/college/etc that is objectionable, and what are the objections?

I'll return to the issue that the OP apparently has misgivings about attending Yale. Where might he go where those misgivings are less of a problem?

Let me give one more comment on my perception of mathematics at Yale, relative to mathematics at Harvard: I think that both schools have an impressive list of senior faculty and place a low emphasis on teaching students so that professors (at least in the relatively lower-level classes, such as someone is likely to reach in their first two years of college) are often unfriendly towards and uninterested in their students, as well as sometimes incompentent in a classroom or at conveying knowledge.  The major difference is that at Harvard, there is a large population of students at a high level in the math department, so the peer group is larger and more substantive than at Yale.

As for TA's question, both my (Yalie) friend and I had very strongly positive opinions of U. Chicago when we were applying, although obviously not quite strong enough to get us to go there.  One of the five Stuy math team captains was accepted to Yale early and is applying to Chicago regular.  So, there is at least one school to consider.

My Yale friend whose views I tried to represent above sent me an e-mail whose relevant parts read as follows:

[quote]Everything you said about Yale is right, except that I don't really
think there's any problem with the size of the department.  (Now that
I look at it, it's not really small.)  The professors are indeed
unfriendly.  I sent an email to one of them about a recommendation
last year--no response.  So, I went to the math department a few times
looking for him.  I was told I should try leaving a note on his door.
I left a pathetic note, telling him it was fine if he didn't want to
write me a recommendation, but could he please tell me so.  I never
got a response.  One time much later I passed him on the street, but I didn't say anything.

[ redacted by forum moderator ]

I'm sure there are bad professors at most schools.  Yale compounds
this problem by having, as an earlier poster said, an incoherent
program.  If you're already at a high level of math going into college
(i.e., you realize that math is based on proofs rather than arcane
symbols), there's no introductory math class that you can possibly
enjoy.  (If you come into Yale without a high level of math, you're
really screwed.  There are no classes there that can bring someone
from the "arcane symbol" view of math to the proof-based one.)

I think one could scrape out a very good math education at Yale, like
this.  Skip the intro. classes (though there's a new professor now who
might be better) and take analysis or algebra so you're qualified for
BSM.  Take whatever class Igor Frenkel is teaching.  Take Sun-Joo
Shin's logic classes.  Now, take some of Yale's excellent non-math
classes, and then go to Budapest for a year.[/quote]

Well, as an econ/mathematics and computer science double major (sophomore currently) at Yale, I can say that the math here really isn't that bad. I disagree with your friend JBL when he stated that there is no good "intro" class in the sense that there is an equivalent to maths23a or maths25a (i believe those are the names of the classes at harvard) that is offered here (called math230). However, the professor for that class was absolutely horrendous when I took it last year. Luckily he has since been removed since he was the 1 of the nastiest people that i've ever spoken to in my entire life and didn't care at all about his students.

As per taking classes with Igor Frenkel, THIS IS A MUST!! Igor Frenkel is by far the best professor that I've had to date in the math department. I took discrete math with him last professor and though I found the class not challenging compared to some other yale math courses, Igor himself was very good. (I'm taking Sun Joo's logic class next semester :) )

But ya, I suppose that most of the professors here are not very accessible and they are kind of mean. I remember when they swore they would get some Putnam practice sessions and they did absolutely nothing until 4 days before the actual exam. How useless.

That's my 2 cents. If you have more specific questions I would be happy to answer them.

Yoel.

Quality of community can be half of the undergraduate math experience since not every professor you get will be amazing. For instance, at MIT I've had one professor for Algebra I and II who is pretty good, one professor for Analysis who was excellent, one professor for Numerical Analysis who is excellent, and one professor for Topology who is undoubtedly the worst teacher I've had in my entire life in any subject (and I thought I had a history teacher in high school who could never be topped...).

However, the quality of community, that is people to talk to, discuss math with, work with, and overall environment has made a huge difference in how those classes turned out for me in terms of me actually enjoying and learning the material.

I agree with Hashbrownie as another engineering student at Yale. Yes the department isn't going to be at the same level as Harvard and MIT but the reputation of the bad department is the awful way the school treats the lower level students. That is were the nonenglish nonteaching professors come in. Once you get past MVC, the professors actually start getting good.

Another thing that Hashbrownie pointed out is that there really isn't much of a putnam team. That sort of has to do with a lack of too many overly talented mathematicians. That is one thing that you won't get from Yale is the community of mathematicians that you would see at MIT, etc. Hashbrownie who are you?

I got this e-mail from a mathematics student at Yale:

"I have never had a class taught by a grad student. I have only had
professors. I love the professors in the math department. I don't think I've
had a single one that I couldn't approach with questions. Yale has one of the
top math graduate programs in the nation, and the quality is passed down to the
undergrad major if you so choose to go for it. After looking at other grad
schools in a number of areas (public policy, education, mathematics), I have
really come to believe that Yale gives a higher quality education than most
other schools, coming part from professors, part from opportunities, and part
from the fellow students. I find that all of our math professors are brilliant,
but I'm not really aware of how prominent they are in their fields."

To everyone who've contributed to this thread, thanks for your help.  I am going to Yale after all (for lack of a better option, of which there are only a few), but will be taking a year off.  I'll probably take some time to work on my math too, so I can hopefully move past the intro classes and take some grad classes by junior year.

I'm gunning for grad school at HPM (theoretical math), so I guess doing several grad classes in college will help my application.  It's probably rare for a Yalie to go on to graduate school in mathematics though, but I guess there's always this board for support if the community is too small.  :)

Does anyone have an idea what kind of credentials a Yale grad would need to have a chance at said grad schools?

You know, Yale's mathematics department really isn't that bad.  Sure, it's not as good as Princeton's or Harvard's or MIT's (and I'm sure a few others), but it's still well regarded, and you could definitely get into any graduate school from Yale, although (perhaps) not as easily.

Although I have an undying affection for Yale and believe that it's perfect, you don't go to Yale for its math--you go for everything else. I'll be there next year for a proposed major of mathematics, but I'll be honest and say that Yale attracted me because of things like the residential college system and atmosphere rather than the mathematics. That being said, it's not as if Yale is bad at math, it's just not quite as well-regarded, which I don't see as much of a problem for undergraduate studies.


geometry
geometric transformation
reflection

Hi volks,
does anyone know the translation to Sanskrit of "Soul's Mirror" or "Soul's Reflection"?
Tnx
Helo

translation for soul is 'atma' and for reflection 'pratibimb'
so, applying cases, we have:
                                  'atmayah pratibimbam'
i suppose its too late. this was posted 8 months back

hey analytic, ur translation is slightly wrong. :rotfl:
The correct answer is atmanah pratibimbam. :D

Soul's Mirror = [size=200]आत्मादर्श[/size]

Soul's Reflection = [size=200]आत्मप्रतिबिंब [/size]

Webmaster-Translations:
freetranslationblog.blogspot.com


Introduction: Let internal and external angle bisectors of $ \angle AXB$ intersect $ AB$ at $ M$ and $ N$. The circle $ (MON)$ is called the Apollonius circle of $ \angle AXB$.

Problem

A circle with center $ O$ and radius $ R$ is given. Let $ P$ be an exterior point according to the circle. Draw a tangent from $ P$ to the circle at $ T$. Let $ OP$ intersect the circle at $ A$ and $ B$. If the radius of Apollonius circle of $ \angle OTP$ equals the radius of Apollonius circle of $ \angle ATB$, then find $ |OP|$ in terms of $ R$.

[hide]

Let internal and external angle bisectors of $ \angle ACB$ intersect  $ AB$ at $ M$ and $ N$. Assume $ b > a$.
By bisector theorem $ AM \equal{} \frac {bc}{a \plus{} b}$, $ MB \equal{} \frac {ac}{a \plus{} b}$, and $ BN \equal{} \frac {ac}{b \minus{} a}$.
Then diameter of the Apollonius circle is $ MN \equal{} \frac {2abc}{|b^2 \minus{} a^2|}$

Let $ B$ be closer to $ P$ than $ A$. $ BT \equal{} a$ and $ AT \equal{} b$, and $ PB \equal{} x$. $ PT \equal{} \sqrt {x(2R \plus{} x)}$.

The radius of Apollonius circle of $ \angle ATB$ is $ \frac {a.b.2R}{b^2 \minus{} a^2} \equal{} \frac {2.2R}{\frac ba \minus{} \frac ab} (..1..)$.

The radius of Apollonius circle of $ \angle OTP$ is $ \frac {R. \sqrt {x(2R \plus{} x)}.(R \plus{} x)}{x(2R \plus{} x) \minus{} R^2} (..2..)$.

$ \triangle ATP \sim \triangle TBP$. Then $ \frac {TB}{AT} \equal{} \frac {BP}{PT} \Rightarrow \frac ab \equal{} \frac {x}{\sqrt {x(x \plus{} 2R)}} \equal{} \frac {\sqrt {x}}{\sqrt {x \plus{} 2R}} (..3..)$

Put $ (..3..)$ in $ (..1..)$. Then $ (..1..) \equal{} (..2..)$. We get $ x^2 \plus{} Rx \minus{} 2R^2$. Then $ x \equal{} R$ or $ x \equal{} \minus{} 2R$. So $ OP \equal{} 2R$.[/hide]


function
algebra
functional equation
algebra proposed

This functional equation is nice and has a super short and elegant solution:

Find all functions f: [b]N[/b] --> [b]N[/b] such that

xf(y) + yf(x) = (x + y)f(x2 + y2)

for all x, y and in [b]N[/b], where [b]N[/b] = {0, 1, 2, 3, ...}.

[i]Canada 2002[/i]

Just because I know that solution I do not post anything. And when I remember my monster of solution! Indeed, the solution you speak about is absolutely extraordinary.

It's almost impossible that nobody finds that sweet solution :) Come on, try it! :)

If f(x) is not constant then we can find x, y suxh that f(x) - f(y) has least positive integer value. But then
                    f(y) < f(x^2+y^2) = (xf(y) + yf(x))/(x+y) < f(x)
And we get the contradiction. So f(x) must be constant function which in fact satisfies the equation. Do you mean this one?

Namdung


geometry
3D geometry
quadratics
algebra
algebra unsolved

Find all real numbers with the following property: the difference of its cube and
its square is equal to the square of the difference of its square and the number itself.

[hide="Solution"]a^3-a^2=(a^2-a)^2
Then a^3-a^2=a^4+a^2-2a^3
3a^3=a^4+2a^2   (1).
Dividing by a^2 we get quadratic equation a^2-3a+2=0 which solutions are 2 and 1.
The other case is when we can't divide (1) by a^2 then a=0 [/hide]

[hide=solution]$a^3 - a^2 = (a^2-a)^2$, or $a^2(a^2-3a+2) = 0$. Factoring further yields $\boxed{0, 1, 2}$ as solutions.[/hide]

[hide=Solution]Let the number be $x$. Then $x^3-x^2=(x^2-x)(x^2-x)$. Note that $x=0$ works. Now dividing by $x^2$ gives $x-1=(x-1)^2$. Note that $x=1$ works. Now dividing by $x-1$ gives $1=x-1$, so $x=2$ works. The answer is hence $\boxed{0,1,2}$.[/hide]


number theory
greatest common divisor
modular arithmetic
ratio

Let $ a$ be an odd integer.Prove that $ gcd\equal{}(a^{2^n}\plus{}2^{2^n},a^{2^m}\plus{}2^{2^m})\equal{}1$ for all integers $ n$ and $ m$ with $ n \not\equal{} m$

[hide]Suppose $ g \equal{} \gcd (a^{2^n}\plus{}2^{2^n},a^{2^m}\plus{}2^{2^m}) >1$ and wlog $ n<m$

Consider an arbitrary prime, p, dividing g.

Therefore $ a^{2^n}\plus{}2^{2^n} \equiv 0 \pmod{p}$, $ a^{2^m}\plus{}2^{2^m} \equiv 0 \pmod{p}$

Therefore $ \left ( \frac{a}{2}\right )^{2^n} \equiv \minus{}1 \pmod{p}$,  $ \minus{}1 \equiv \left ( \frac{a}{2}\right )^{2^m} \equiv \left ( \left ( \frac{a}{2}\right )^{2^n} \right )^{2^{m\minus{}n}} \equiv (\minus{}1)^{2^{m\minus{}n}} \equiv 1 \pmod{p}$

Contradiction![/hide]

[quote="SimonM"]
Therefore $ \left ( \frac {a}{2}\right )^{2^n} \equiv \minus{} 1 \pmod{p}$,  $ \minus{} 1 \equiv \left ( \frac {a}{2}\right )^{2^m} \equiv \left ( \left ( \frac {a}{2}\right )^{2^n} \right )^{2^{m \minus{} n}} \equiv ( \minus{} 1)^{2^{m \minus{} n}} \equiv 1 \pmod{p}$[/quote]

Ah, but $ \frac a2 \notin \mathbb Z$ since $ a$ is an odd integer.

Could we not remedy this by showing that p = 2 is clearly a prime factor for neither, and then instead considering $ \frac{a \plus{} p}{2}$?

[quote="Yongyi781"]Ah, but $ \frac a2 \notin \mathbb Z$ since $ a$ is an odd integer.[/quote]

$ (2,p) \equal{} 1$, so $ \frac{1}{2} \equal{} 2^{\minus{}1}$ makes sense mod p

But would that not be a fractional value mod p?

What SimonM is saying is that $ 2$ has a unique inverse $ a\equiv 2^{\minus{}1} \pmod{p}$ (that is $ 2a\equiv 1 \pmod{p}$). $ p/2$ is rather unfortunate notation though.

Yeah, and funny enough, $ a2^{\minus{}1} \equiv \frac{a \plus{} p}{2} \mbox{ mod } p$. :P

Nice solution. However, what is the standard notation for multiplicative inverse though (so as to prevent misunderstandings)?
Also @Dimitris: where did you get this nice problem?

[quote="mathwizarddude"]
Also @Dimitris: where did you get this nice problem?[/quote]
104 number theory problems from Titu Andreescu-Zuming Feng(highly recomended book) :wink:

[quote="mathwizarddude"]what is the standard notation for multiplicative inverse though[/quote]
Pretty much what SimonM wrote.  The ambiguity isn't between division in the rationals and division $ \bmod p$ but between division and the Legendre symbol, but here the Legendre symbol is clearly the wrong interpretation.

For prime $ p>3,2^{p\minus{}2}\plus{}3^{p\minus{}3}\plus{}6^{p\minus{}2}\minus{}1\equiv \frac12\plus{}\frac13\plus{}\frac16\minus{}1\pmod p$ but it just happens that the fractions sum to $ 0$, which agrees with the expression being congruent 0 mod p... is this a coincidence?

No.  That's a perfectly valid chain of reasoning.

But wouldn't it be an abuse of notation when the fractions are meant to be multiplicative inverses of the denominator instead of being meant to be ratios? Why is it valid?

It's not an abuse of notation; that notation means precisely what it says, which is multiplicative inverse.  The definition of multiplicative inverse is set up so that addition of fractions still makes sense $ \bmod p$.  This is because working $ \bmod p$ gives you a [url=http://en.wikipedia.org/wiki/Finite_field]field structure[/url] compatible with the field structure of the "localization" of $ \mathbb{Q}$ at $ p$, which consists of the set of fractions with denominator not divisible by $ p$.

In other words, try to prove that if $ \frac{a}{b} \equiv c \bmod p$ and $ \frac{d}{e} \equiv f \bmod p$ then $ \frac{ae \plus{} bd}{be} \equiv c \plus{} f \bmod p$.  (The proof goes exactly the same as the proof over $ \mathbb{Q}$.)

WLOG say $ m > n$. Now let $ p$ be a prime such that $ p| (a^{2^n} \plus{} 2^{2^n})$, $ p| (a^{2^m} \plus{} 2^{2^m})$.
We have $ a^{2^n}\equiv \minus{} 2^{2^n} \pmod p\Longrightarrow (a^{2^n})^{2^{m \minus{} n}}\equiv (2^{2^n})^{2^{m \minus{} n}} \pmod p$
$ \Longrightarrow a^{2^m}\equiv 2^{2^m}\equiv \minus{} 2^{2^m}\pmod p$, which implies $ p|(2^{2^{m} \plus{} 1})\Longrightarrow p \equal{} 2$.
But since $ a$ is odd $ \Longrightarrow a^{2^n} \plus{} 2^{2^n}$, $ a^{2^m} \plus{} 2^{2^m}$ are all odd. Hence there won't exist any prime number $ p$ such that $ p$ is the common divisor of  $ a^{2^n} \plus{} 2^{2^n},a^{2^m} \plus{} 2^{2^m}$. Thus, $ \gcd (a^{2^n} \plus{} 2^{2^n},a^{2^m} \plus{} 2^{2^m}) \equal{} 1$ $ \square$


abstract algebra
algebra
polynomial
superior algebra
superior algebra unsolved

How can we show that every $ C^*$-algebra generated by a self-adjoint element, is abelian ?

thanks

$ C^*(x)$ is abelian if and only if $ x$ is normal, i.e. $ x^* x \equal{} x x^*$. the one direction is trivial, the other is clear since the polynomials in $ x$ and $ x^*$ are a commutative, dense subalgebra of $ C^*(x)$.


superior algebra
superior algebra unsolved

Let I be an Ideal of R. Let $ M' \equal{}\{a \in M: I^ka \equal{}0\}$, where $ I^k$ may depend on $ a$. Show that M' is a submodule of M.

What part are you having trouble with?  You just verify the definition of submodule.  It is a subset.  It is closed under scalar multiplication since I^k * (ra) = (I^k*r) * a ≤ I^k * a = 0.  It is closed under addition since if I^k*a = 0 and I^j*b = 0 then I^(k+j)*(a+b)=0.

The exercise suggested that I use another exercise which is the statement I put as the title of this post and I just did not see why it was needed.

Well, is $ M'$ the union of some chain of submodules, perhaps?  Liiiike... one for each $ k$?


real analysis
limit
real analysis unsolved

Hi,

Let $ \mu$ a measure on $ [0,\plus{}\infty[$  and $ \lambda$ the Lebesgue measure,  can we have all these conditions
i) $ \forall a \geq 0 \qquad \mu(\{a\})\equal{}0$

ii) $ \mu([0,\plus{}\infty[)\equal{}1$

iii) $ \exists E\subset [0,\plus{}\infty[ \qquad \lambda(E)\equal{}0 \qquad \mu(E)\equal{}1$

iv) $ \lim\limits_{y\rightarrow 0\plus{}} \mu \left( \bigcup_{n\in \mathbb{N}} [2ny,2ny\plus{}y] \right)\equal{}0$

Thanks

For example, there are lot of sets A such that $ \mu(A)\equal{}0$.
If $ (y_k)$ is a sequence which converges to $ 0^\plus{}$, then Fatou's lemma says

\[ 0 \leq     \mu \left( \liminf\limits_{k\rightarrow \plus{}\infty} y_k \bigcup_{n\geq 0}[2n,2n\plus{}1]  \right)     \leq
\liminf\limits_{k\rightarrow \plus{}\infty} \mu \left(  y_k \bigcup_{n\geq 0}[2n,2n\plus{}1] \right)\]
\[ \Rightarrow \mu \left( \liminf\limits_{k\rightarrow \plus{}\infty} y_k \bigcup_{n\geq 0}[2n,2n\plus{}1]  \right)  \equal{}0\]

I recall that if $ A_k$ is a sequence of sets then $ \liminf A_k$ is defined by $ 1_{\liminf A_k }\equal{}\liminf 1_{A_k}$

A simple example is $ y_k\equal{}(1/3)^k$, we can show that $ \liminf\limits_{k\rightarrow \plus{}\infty}\left( y_k \bigcup_{n\geq 0}[2n,2n\plus{}1] \right)$ is $ C\plus{}\mathbb{N}$ where $ C$ is the usual triadic Cantor set of $ [0,1]$.


[quote="tanlsth"]Give $ a,b,c \in R^{ \plus{} }$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$

Prove that $ \frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} a^2}{c \plus{} a} \geq 3$[/quote]

$ \Leftrightarrow \sqrt{3(a^2\plus{}b^2\plus{}c^2)}\le \sum_{cyc}{\frac{(b\plus{}c)^2\plus{}(b\minus{}c)^2}{2(b\plus{}c)}}\Leftrightarrow$ $ \sqrt{3(a^2\plus{}b^2\plus{}c^2)}\minus{}a\minus{}b\minus{}c\le\sum_{cyc}{\frac{(b\minus{}c)^2}{2(b\plus{}c)}}\Leftrightarrow$ $ \frac{(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2}{\sqrt{3(a^2\plus{}b^2\plus{}c^2)}\plus{}a\plus{}b\plus{}c}\le\sum_{cyc}{\frac{(b\minus{}c)^2}{2(b\plus{}c)}}$

Since $ \sqrt{3(a^2\plus{}b^2\plus{}c^2)}\ge a\plus{}b\plus{}c$ it is enough to prove that $ \frac{(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2}{2(a\plus{}b\plus{}c)}\le\sum_{cyc}{\frac{(b\minus{}c)^2}{2(b\plus{}c)}}\Leftrightarrow$ $ 0\le\sum_{cyc}{\frac{a}{b\plus{}c}(b\minus{}c)^2}$

[quote="tanlsth"]Give $ a,b,c \in R^{ \plus{} }$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$

Prove that $ \frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} a^2}{c \plus{} a} \geq 3$[/quote]
$ \frac {a^2 \plus{} b^2}{a \plus{} b} \equal{} \frac {a \plus{} b}{2} \plus{} \frac {(a \minus{} b)^2}{2(a \plus{} b)} \ge \frac {a \plus{} b}{2} \plus{} \frac {(a \minus{} b)^2}{2(a \plus{} b \plus{} c)}$
So we just have to prove:
$ (a \plus{} b \plus{} c) \plus{} \frac {a^2 \plus{} b^2 \plus{} c^2 \minus{} (ab \plus{} bc \plus{} ca)}{a \plus{} b \plus{} c} \ge 3 \iff$

$ 2(a^2 \plus{} b^2 \plus{} c^2) \plus{} (ab \plus{} bc \plus{} ca) \ge 3(a \plus{} b \plus{} c) \equal{} \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}(a \plus{} b \plus{} c)\iff$

$ \frac {(a \plus{} b \plus{} c)^2 \plus{} 3(a^2 \plus{} b^2 \plus{} c^2)}{2} \ge \sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}(a \plus{} b \plus{} c)$
Which is AM-GM :)

[quote=tanlsth]Give $ a,b,c \in R^{\plus{}}$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$
Prove that $ \frac{a^2\plus{}b^2}{a\plus{}b}\plus{}\frac{b^2\plus{}c^2}{b\plus{}c}\plus{}\frac{c^2\plus{}a^2}{c\plus{}a} \geq 3$[/quote]
The inequality is [url=http://www.artofproblemsolving.com/community/c6h482778p2705140]equivalent to[/url]
Give $ a,b,c \in R^{\plus{}}$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}3 .$ Prove that $$ \frac{a^2}{a\plus{}b}\plus{}\frac{b^2}{b\plus{}c}\plus{}\frac{c^2}{c\plus{}a} \geq \frac{3}{2}.$$ Because$\frac{a^2}{a\plus{}b}\plus{}\frac{b^2}{b\plus{}c}\plus{}\frac{c^2}{c\plus{}a}=\frac{b^2}{a\plus{}b}\plus{}\frac{c^2}{b\plus{}c}\plus{}\frac{a^2}{c\plus{}a}.$
[url=http://www.artofproblemsolving.com/community/c6h507514p2851214]here[/url]
For $a, b, c>0,$ and $k\geq1,$ [url=https://artofproblemsolving.com/community/c6h416432p2345631]prove that[/url]
\[\frac{a^{k+1}}{b^k+c^k}+\frac{b^{k+1}}{c^k+a^k}+\frac{c^{k+1}}{a^k+b^k}\geq\frac{3}{2}\sqrt{\frac{a^{k+1}+b^{k+1}+c^{k+1}}{{a^{k-1}+b^{k-1}+c^{k-1}}}}\]
[url=https://artofproblemsolving.com/community/c6h86713p505856]h[/url]

[quote=tanlsth]Give $ a,b,c \in R^{\plus{}}$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$ Prove that $$ \frac{a^2\plus{}b^2}{a\plus{}b}\plus{}\frac{b^2\plus{}c^2}{b\plus{}c}\plus{}\frac{c^2\plus{}a^2}{c\plus{}a} \geq 3$$[/quote]

Let $a,b,c$ be positive real numbers. [url=https://artofproblemsolving.com/community/c125h1988017p13855198]Prove that[/url]$$\frac {a^2+b^2}{a+b}+\frac {b^2+c^2}{b+c}+\frac {c^2+a^2}{c+a}\geq a+b+c.$$
[size=50]AMSP Sample Problems[/size]

[quote=sqing]
Let $a,b,c$ be positive real numbers. [url=https://artofproblemsolving.com/community/c125h1988017p13855198]Prove that[/url]$$\frac {a^2+b^2}{a+b}+\frac {b^2+c^2}{b+c}+\frac {c^2+a^2}{c+a}\geq a+b+c.$$[/quote]
Let $a,b,c$ be positive real numbers.  Prove that$$\frac {a^2+b^2}{a+b}+\frac {b^2+c^2}{b+c}+\frac {c^2+a^2}{c+a}\leq 3\left(\frac {a^3}{b^2+bc+c^2}+\frac {b^3}{c^2+ca+a^2}+\frac {c^3}{a^2+ab+b^2}\right).$$
[size=50]p/6452782811[/size]


Putnam
linear algebra
matrix
inequalities
vector
ratio
absolute value

Let $H$ be an $n\times n$ matrix all of whose entries are $\pm1$ and whose rows are mutually orthogonal. Suppose $H$ has an $a\times b$ submatrix whose entries are all $1.$ Show that $ab\le n.$

Did anyone else get an inequality solution of sorts? (the official solutions all use dot products and stuff in some way)

I did something along the lines of looking at the maximum number of times any pair of rows could match in the remaining $n-a$ columns versus the minimum number of times they had to match.

Just so it's here, a very slick solution.

Suppose that some $a\times b$ submatrix has all entries 1. Let $v=v_1+v_2+\cdots+v_b$ be the sum of these $b$ rows. We have $\|v_b\|^2=\|v_1\|^2+\|v_2\|^2+\cdots+\|v_b\|^2=nb$ by orthogonality of rows.
We also have $\|v_b\|^2=b^2+b^2+\dots+b^2+x_{a+1}^2+\dots+x_n^2\ge ab^2$, counting by components. Therefore $nb\ge ab^2$, and $n\ge ab$.

This result is reasonably well known in discrete math and theoretical computer science.  We used to use it to show that Hadamard matrices have some pseudorandom properties.

More generally, here is Lindsey's Lemma:  In every $a \times b$ submatrix, the difference between the number of $1$'s and the number of ${-1}$'s is at most $\sqrt{abn}$.

Here's a version of the argument cast more in matrix terms than vector terms.

The orthogonality condition, along with the fact that every entry has absolute value 1, means that $HH^T=nI.$

Use $J$ to denote a matrix whose entries are all $1$'s. Suppose $H$ can be block-structured as

$H=\begin{bmatrix}J_{a\times b}&A\\B&C\end{bmatrix}.$ Then $H^T=\begin{bmatrix}J_{b\times a}&B^T\\A^T&C^T\end{bmatrix}.$

Multiply this out:

$HH^T=\begin{bmatrix}bJ_{a\times a}+AA^T&JB^T+AC^T\\BJ+CA^T&BB^T+CC^T \end{bmatrix}=\begin{bmatrix}nI_a&0\\0&nI_{n-a}\end{bmatrix}.$

So $bJ_{a\times a}+AA^T=nI_a.$

The largest eigenvalue of $bJ_{a\times a}$ is $ab.$ Adding the positive definite matrix $AA^T$ can only make that largest eigenvalue larger. But all eigenvalues of $nI_a$ are $n.$ Hence $ab\le n.$

yea, i looked at the incidence sub matrix of a by n. For any 2 rows, since sum of pairwise product = 0, the entries must be equal at n/2 positions, different at n/2 positions. Then, we just sum column pairs of (1, 1) and (-1, -1).
We also have AM>GM for the min number of column pairs.

The 'ugly' part of this solution was that i had to split into cases a even and a odd, in order to get the inequality case resolved.

total no of column pairs = $a \choose 2$ (n/2)  $\geq$ $a \choose 2$ X b + (n-b)[  \floor a\2 \choose 2)  + (ceiling a/2 \choose 2)] = min number of column pairs.

You don't need to resolve into cases. The pairwise products in any one column sum to more than -a/2, so we have
0 >= (n-b)(-a/2) +ba(a-1)/2, which gives n >= ab.
That the pairwise product in any one column sum to more than -a/2 is clear with a bit of work. Go figure.

Pardon my lack of LaTeX.

[quote="Kent Merryfield"]The largest eigenvalue of $bJ_{a\times a}$ is $ab.$ Adding the positive definite matrix $AA^T$ can only make that largest eigenvalue larger.[/quote]
I got a PM asking why this statement is true. Let me respond to that.

If $X$ is any real matrix, then $MM^T$ is symmetric and (at least) positive semidefinite. Let us look at adding any two such matrices.

Suppose $A$ is a real symmetric $n\times n$ matrix. We know that $A$ is orthogonally diagonlizable, which allows us to assume WLOG that it is diagonal. Suppose $A$ has eigenvalues $\lambda_1\le\lambda_2\le\cdots\le\lambda_n.$ Suppose that $x$ is any nonzero vector in $\mathbb{R}^n.$ Then we can show (here is where assuming that $A$ is diagonal helps) that $\lambda_1\le\frac{x^TAx}{\|x\|^2}\le \lambda_n.$ But by selecting $x$ to be the appropriate eigenvector, we can have that ratio equal to $\lambda_1$ and equal to $\lambda_n.$ We can write $\lambda_n=\max_{\|x\|=1}x^TAx.$ That's the largest eigenvalue. Label $\lambda_n$ as $r(A),$ the spectral radius of $A.$

So if $A$ is real, symmetric, and positive semidefinite, then for $\|x\|=1,$ $0\le x^TAx\le r(A)$ with the maximum achieved. If $B$ is real, symmetric, and positive semidefinite, then for$\|x\|=1,$ we have $0\le x^TBx\le r(B).$

Let $C=A+B.$ Then $C$ is also real, symmetric, and positive semidefinite. It is positive semidefinite because $x^TCx=x^TAx+x^TBx\ge 0$ for all $x.$ Now let $x$ be an eigenvector of $A$ associated to the largest eigenvalue, and scale $x$ so that $\|x\|=1.$ That gives us $x^TAx=r(A).$ It follows that \[x^TCx=x^TAx+x^TBx=r(A)+x^tBx\ge r(A).\]

Hence the spectral radius (largest eigenvalue) of $C=A+B$ is at least as large as the spectral radius (largest eigenvalue) of $A.$ Which is exactly what I was claiming above: that the largest eigenvalue of $bJ+AA^T$ was at least as large as the largest eigenvalue of $bJ.$

With the disclaimer that I am not a Putnam grader (and don't really understand Putnam grading all that well), I would hope that just quoting it the way I did would have sufficed. But I don't really know.

And now that the topic has been revived, an observation that I made at the time this Putnam was sat is that the problem setters should have presented a $m\times n$ rectangular matrix. The proof stays exactly the same, and at least now, for every $n$, there exists at least such a matrix, namely for $m=1$. The problem with $n\times n$ square matrices is that such matrices don't exist for every $n$ - we get into the realm of Hadamard matrices (for existence), and that is unneeded and irrelevant for the question at hand.

Let $v_1,\ldots,v_a\in\{\pm 1\}^n$ be the rows that are the ones you need for the $a\times b$ submatrix of all $1$s. WLOG suppose the first $b$ coordinates of all these are $1$.

Note that if $x_1,\ldots,x_a\in\{\pm 1\}$, then
\[\sum_{i<j}x_ix_j\ge \binom{\lfloor a/2\rfloor}{2} + \binom{\lceil a/2\rceil}{2} - \lfloor a/2\rfloor\lceil a/2\rceil = -\lceil(a-1)/2\rceil.\]
Thus,
\[0 = \sum_{i<j}v_i\cdot v_j \ge \binom{a}{2}b - \lceil(a-1)/2\rceil(n-b).\]
If $a$ is even, this means
\[0\ge (a-1)b - (n-b),\]
so $n\ge ab$. If $n$ is odd, then this means
\[0\ge ab - (n-b),\]
so $a(b+1)\le n$, so $ab\le n$ as well. This completes the proof.


topology
function
algebra
domain
advanced fields
advanced fields unsolved

Please can someone explain why R is homeomorphic to the y-axis in R^2: I'm afraid I can't think of an actual homeomorphism between the 2 and clearly if they both share a topological property it doesn't necessary lead to the conclusion they are homeomorphic.

Also I seem to have difficulty finding an infinite topological space X that is homeomorphic to XxX, and I doubt the above example will do since it doesn't seem strict enough..... Please point me towards a proper example and explain why they are homeomorphic.

Thank you

The $ y$-axis (which I will call $ Y$) consists of all points $ \{(0,y)\} \in \mathbb{R}^2$. The projection map $ \pi_y(x,y) = y$ from $ \mathbb{R}^2 \to \mathbb{R}$ is continuous. This restricts to a continuous function $ \varphi : Y \to \mathbb{R}$ sending $ (0,y) \mapsto y$.

Now consider the zero map $ c_0 : \mathbb{R} \to \mathbb{R}$ sending $ x \mapsto 0$ for each $ x$; this is continuous. So is the identity map $ i : \mathbb{R} \to \mathbb{R}$ sending $ x \mapsto x$. Their product $ c_0 \times i$ is therefore a continuous map sending $ x \mapsto (0,x)$. The image of this map sits entirely in $ Y$, and so provides a map $ \psi : \mathbb{R} \to Y$.

Now, check to verify that $ \varphi\psi = id_{\mathbb{R}}$, and $ \psi\varphi = \id_Y$. Therefore $ Y$ and $ \mathbb{R}$ are homeomorphic.

For the second question: you need continuous maps between $ X$ and $ X \times X$ (with some other conditions); what sort of topology makes constructing continuous maps as easy as possible?

Also for the homeomorphism between $ X$ and $ X\times X$, you gotta make sure their cardinalities are in agreement (obvious, but just saying you know)...

and with what JBL says, there is a certain topology that makes [u]everything[/u] continuous!

Thanks for your help all of you!
So is the topology that's supposed to make constructing continuous maps easy the topology generated by the open sets f^-1(U) in X where f maps X onto Y and U is an open set in Y?

It's just that in this case as R is open so RxR must be open so then the function that maps R to RxR is continuous. The same is true for its inverse function and so can we deduce R is homeomorphic to RxR? (P.S. I didn't think the question I posted earlier on was strict enough to prove this.)

However, it is known that there are no continuous injective maps from RxR to R which seems to contradict the above, so please can someone explain this? (and how my reasoning above goes wrong?

Um, no the topology I'm referring to (and probably what JBL was referring to) is the "discrete topology". If $ X$ is a set, the discrete topology over $ X$ is the topology such that each subset is open (i.e., the topology is $ P(X)$, the power set). Now, with this as the topology, why is every function with $ X$ as its domain open? Now what are the conditions for $ X$ and $ X\times X$ to have the same cardinality? (Think Cantor's diagonal argument for a basic example). Now once we get a bijection from $ X$ to $ X\times X$, what does the discrete topology make this function?

[quote="LkNsngth"](Think Cantor's diagonal argument for a basic example)[/quote]  Everything else is right, but Cantor's diagonal argument is a proof that $ |{\bf R}| > |{\bf N}|$ -- it has nothing to do with the cardinalities of $ X \times X$ and $ X$.

There is a first and a second diagonal argument. The other one shows $ |\mathbb{N}|\equal{}|\mathbb{N}\times\mathbb{N}|$.

So is the fact there is no injective map from R^2 to R something to do with (possibly different) cardinalities?

Sorry, I'm slightly confused!

No, there are such maps. In fact there are a lot of them ($ 2^{|\mathbb{R}|}$ many) and hence $ \mathbb{R}$ and $ \mathbb{R}^2$ have the same cardinality.

Sorry, I meant continuous injective maps: why are there none of these between the 2 sets?

Take $ X \equal{} \mathbb{R}^{ \mathbb{N} }$ for the second question.

As for your third question, what can the image of the circle $ x^2 \plus{} y^2 \equal{} 1$ be?


This time I will "answer" with two questions for you:
Which from the below is it true?
(a) $ \mathop{\lim}\limits_{x\rightarrow0}f(x) \equal{} f(0)$
(b) $ \mathop{\lim}\limits_{x\rightarrow1}f(x) \equal{} f(1)$
If you can answer these two, you can answer about continuouity also.

P.S. Next time for elementary questions it is more appropriate to post in "Calculus computation and tutorials" senction.

[quote="grigkost"]P.S. Next time for elementary questions it is more appropriate to post in "Calculus computation and tutorials" senction.[/quote]
Which is why I moved it.

I'll handle the $ x\equal{}0$ case:

What is the limit as $ x$ goes to 0?  Well if we're coming from the left, it's clearly 0.  Coming from the right, it's still 0 so the limit is 0.  What about $ f(0)$?  Well just look at the second line and you see that it's $ 0$.  They all agree, so it's continuous.


[quote="erbed"]Prove that for any positive real numbers $ x_{1},x_{2},...,x_{n}$ :
$ \frac {1}{x_{1}} \plus{} \frac {x_{1}}{x_{2}} \plus{} \frac {x_{1}x_{2}}{x_{3}} \plus{} \frac {x_{1}x_{2}x_{3}}{x_{4}} \plus{} ... \plus{} \frac {x_{1}x_{2}...x_{n}}{x_{n \plus{} 1}} \geq 4(1 \minus{} x_{1}x_{2}...x_{n \plus{} 1})$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168848


Find all function $ f: \mathbb{N}\rightarrow\mathbb{N}$ satisfy $ f(mn)\plus{}f(m\plus{}n)\equal{}f(m)f(n)\plus{}1$ for all natural number $ n$

[hide]Let $ f(1)\equal{}c$.  Put $ n\equal{}1$ and you have

$ f(m\plus{}1)\equal{}f(m)(c\minus{}1)\plus{}1$

for all $ m$.  Applying this again yields

$ f(m\plus{}2)\equal{}f(m)(c\minus{}1)^2\plus{}c$

If instead, you input $ n\equal{}2$ into the functional equation you get

$ f(2m)\plus{}f(m\plus{}2)\equal{}f(2)f(m)\plus{}1$

Notice from the first equation we get $ f(2)\equal{}c^2\minus{}c\plus{}1$.  Equating the two expressions for $ f(m\plus{}2)$ yields

$ f(m)(c\minus{}1)^2\plus{}c\equal{}(c^2\minus{}c\plus{}1)f(m)\minus{}f(2m)\plus{}1$

$ \Rightarrow c\minus{}1\equal{}cf(m)\minus{}f(2m)$

It is easy to see from the fact that $ f(m\plus{}1)\equal{}f(m)(c\minus{}1)\plus{}1$ that if $ c>1$, then there will be a large enought $ m$ for which $ cf(m)\ge f(2m)$ which contradicts the fact that $ c\minus{}1>0$.  Thus, we must have $ c\equal{}1$.  Which from $ f(m\plus{}1)\equal{}f(m)(c\minus{}1)\plus{}1$ shows that $ f(m)\equal{}1$ for all $ m$, and this satisfies the functional equation.
[/hide]

@cosinator: You miss the function $ f(n)\equal{}n\plus{}1$
I'll explain my ideas, just my ideas, because my solution is too long at the problem, and I'm to lazy to write the solution again :mrgreen:

By setting $ m\equal{}n\equal{}1$, $ m\equal{}1, n\equal{}2$, $ m\equal{}1, n\equal{}3$, and $ m\equal{}2, n\equal{}2$, we immediately get 4 different equation with 4 different variables, which is $ f(1), f(2), f(3), f(4)$.
From that system of equation, we get 3 possibilities from $ f(1)$, which is $ \frac12,1,2$
Since $ f: \mathbb{N}\rightarrow\mathbb{N}$, we got only 2 possibilities, which is $ 1$ or $ 2$
Setting $ m\equal{}1$,
we get:
$ f(n\plus{}1)\equal{}f(n)(f(1)\minus{}1)\plus{}1$
For $ f(1)\equal{}1$, we get $ f(n)\equal{}1$.
For $ f(1)\equal{}2$, we get $ f(n\plus{}1)\equal{}f(n)\plus{}1$, and we could easily get $ f(n)\equal{}n\plus{}1$
after checking back the solution, we could see that our 2 functions satisfies the equation.
So, $ f(n)\equal{}1$ or $ f(n)\equal{}n\plus{}1$

[b]wangsacl[/b], thank you so much for pointing out my mistake.  I did indeed neglect the case $ c\equal{}2$.  Here is how my proof should have gone after the first couple of preliminary steps:

For $ c\ge3$, we have $ f(m\plus{}2)\equal{}f(m)(c\minus{}1)^2\plus{}c$, but $ (c\minus{}1)^2\ge c$ for $ c\ge\frac{3\plus{}\sqrt{5}}{2}$ and so for $ m>2$, $ f(2m)>f(m\plus{}2)>cf(m)\Rightarrow 0>cf(m)\minus{}f(2m)\equal{}c\minus{}1$, a contradiction to the fact that $ c\ge3$.  Thus, we find $ f(1)$ is either $ 1$ or $ 2$, which as wangsacl, showed yields the two solutions $ f(m)\equal{}1,f(m)\equal{}m\plus{}1$.

$\bigstar \color{blue}{\textit{\textbf{ANS:}}}$ $f(n)=1 \quad \textrm{and} \quad f(n)=n+1$ for all natural numbers $n$.
$\spadesuit \color{red}{\textit{\textbf{Proof:}}}$ It's easy to see that these are indeed solutions to the given FE. Let $f(n)=g(n)+1$, then let $F(m,n)$ denote the assertion: \[F(m,n): g(mn)+g(m+n)=g(m)g(n)+g(m)+g(n).\] We have the following four equations: \[F(1,1): g(2)=g(1)^2+g(1)\] \[F(1,2): g(3)=g(1)g(2)+g(1)+g(2)\]\[F(1,3): g(4)=g(1)g(3)+g(1)\] \[F(2,2): 2g(4)=g(2)^2+2g(2).\] From the last two equations, we get \[g(2)^2+2g(2)=2g(1)g(3)+2g(1)\] and since $g(2)=g(1)^2+g(1), g(3)=g(1)^3+2g(1)^2+2g(1)$. By substituting these into the previous equation and hunting down $g(1)$, we should get two values $0$ or $1$. If $g(1)=0$, then $F(1,n): g(n+1)=0$ which means $g(n)=0 \implies f(n)=1 \quad \forall n \in \mathbb{N}.$ If $g(1)=1$, then $F(1,n): g(n+1)=g(n)+1$ which means $g(n)=n$ by a simple induction and $f(n)=n+1 \quad \forall n \in \mathbb{N}.$ $\quad \blacksquare$


hello every1 i have a function problem that goes like this:

Find all strictly increasing functions f:N\{0}-->N\{0}, such that :
f(n+2f(n))=3f(n), for all n from N\{0}.

danke schoen!!

I can prove that 3f(n)>=f(3n), but I'm not quite sure what that does for us.

From looking at the equation it seems the answers will f(x) = x+k.
So let g(x) = f(x)-x.

The equation transforms into
g(3n+2g(n)) = g(n) and g(n+1)>=g(n).

and we are trying to prove g is constant.

Well letting g(1)=a, put n=1 and we get g(3+2a) = a, and so g(x)=a for all x from 1 to 3+2a. We can keep doing this, put n=3+2a and we get g(9+8a) = a. Doing this an arbitrary number of times we get g(x)=a for any x. So g is constant and f(x) = x+k is the only solution, for any constant k.


number theory
prime factorization
divisibility tests

[quote="weatherbottle"]Source: Massachusetts Mathematics League December 2005

Twin Primes are primes that differ by 2. There are 12 pairs of twin primes between 1 and 180. Find the largest and smallest value of $N$ if $N-1$ and $N+1$ are twin primes less than 180 and $N$ has exactly eight factors.[/quote]

[hide]$N$ can be written as $a\cdot b\cdot c$, $a^3\cdot b$, or $a^7$, where $a$, $b$, and $c$ are the factors in the prime factorization of $N$. The smallest number is $2^3\cdot 3=24$, but $N+1$ is not prime. The second smallest is $2\cdot 3\cdot 5=30$, and $N-1$ and $N+1$ are primes. $30$ is the smallest. For the largest, $a^7$ can't be the largest because $128$ isn't the largest and all other $a$ exceed $180$. For $a^3\cdot b$, either $a$ or $b$ must be two, otherwise there would be no primes in $N+1$ or $N-1$ (they would be odd). So if $a=2$, the largest number of $b$ that has primes $N+1$ and $N-1$ is $5$. $8\cdot 5=40$. If $b=2$, there is no $a$. For $a\cdot b\cdot c$, $a$, $b$, or $c$ must be two. The largest I could find is $102$. So $102$ and $30$?
I'm pretty sure there's a better way to do this. [/hide]

[hide="there is a better way"]
All numbers with eight positive divisors have the form $p^7$, $p^3q$, or $pqr$ where $p$, $q$, and $r$ are distinct primes. Also, since any prime greater than 3 must be 1 or 5 (mod 6), $N$ has to be divisible by 6. This only leaves the numbers $2^3\cdot 3=24$ (which doesn't work), $3^3\cdot 2=54$ (also doesn't work), and $6p$ for $p>3$.

The smallest such $p$, 5, works because 29 and 31 are primes, so the smallest $N$ is 30. To find the largest $N$, start with the largest $p<30$ and use trial and error. $6*29$ has a units digit of 4, so it can't work ($N+1$ would be divisible by 5). $6*23=138$ which works because 137 and 139 are both primes (they fail the divisibility tests for 2,3,5, and 11, and doing the division shows that 7 is not a divisor either. This covers all primes up to $\sqrt{139}$.) So the largest $N$ is 138.[/hide]


prove or disprove the following statements:

1. There exists countable family of uncountable, dense, pairwise disjoint subsets of real line.

2. There exists uncountable family of uncountable, dense, pairwise disjoint subsets of real line.

(2) implies (1), right?

At first I thought this will do for 1:

$ \{(x,x\plus{}1)|x\in\mathbb{Z}\}$ is a countable family of uncountable, pairwise disjoint subsets of R, such that the union of the family is dense.

But I realise this is not what you're asking:
[quote="om256"]1. There exists countable family of uncountable, dense, pairwise disjoint subsets of real line.[/quote]So every member of the family needs to be dense in R itself. I'll have to think for a moment about this one. All we need for 1 is [i]two[/i] disjoint, uncountable, dense subsets of R, which seems easy to construct.

Consider a Hamel basis $ a_j, j \in J$ for $ \mathbb{R}$ over $ \mathbb{Q}$, where $ J$ is uncountable.

1.  Consider the translates of the span of $ \{ a_j | j \neq j_0 \}$ for some fixed $ j_0$ by the span of $ \{ a_{j_0} \}$.

2.  Partition $ J$ into two disjoint uncountable sets $ J_1, J_2$, and consider the translates of the span of $ \{ a_j | j \in J_1 \}$ by the span of $ \{ a_j | j \in J_2 \}$.


我有一个重大的发现，就是当人们发生冲突的时候，其根本原因并不在于各自主要都说了些什么。原因是什么呢？我发现是大家在说完所想要说的话之后说的最后的那句话。没有这句话，大家其实都是在很友好地交流，而一旦加上了这句话，交谈就变成了吵嘴，并且会愈演愈烈，最终局面无法收拾。问题在于，这句话的内容往往和大家要说的事情毫不相干，所以，我把它称为“多余的最后一句话”。

　　举个例子吧。

　　我那天坐公共汽车去办事，车上人不多，但也没有空位子，有几个人还站着，吊在拉手上晃来晃去。

　　一个年轻人，干干瘦瘦的，戴个眼镜，身旁有几个大包，一看就是刚从外地来的。他靠在售票员旁边，手里拿着一个地图在认真研究着，眼里不时露出迷茫的神情。估计是有点儿迷路了。

　　他犹豫了半天，很不好意思地问售票员：“去颐和园应该在哪儿下车啊？”售票员是个短头发的小姑娘，正剔着指甲缝呢。她抬头看了一眼外地小伙儿，说：“你坐错方向了，应该到对面往回坐。”要说这些话也没什么，错了小伙儿下站下车马路对面坐回去呗。但是售票员可没说完，她该说那多余的最后一句话了。

　　“拿着地图都看不明白，还看什么劲儿啊！”售票员姑娘眼皮都不抬地说。外地小伙儿可是个有涵养的人，他嘿嘿笑了一笑，把地图收起来，准备下一站下车换乘去。

　　旁边有个大爷可听不下去了。他对外地小伙儿说：“你不用往回坐，再往前坐四站换904也能到。”要是他说到这儿也就完了，那还真不错，既帮助了别人，也挽回了北京人的形象。可大爷哪儿能就这么打住呢，他一定要把那多余的最后一句话说完：

　　“现在的年轻人哪，没一个有教养的！”

　　我心想，大爷这话真是多余，车上年轻人好多呢，打击面太大了吧！可不，站在大爷旁边的一位小姐就忍不住了。

　　“大爷，不能说年轻人都没教养吧，没教养的毕竟是少数嘛。您这么一说我们都成什么了！”

　　这位小姐穿得挺时髦，两根细带子吊个小背心，脸上化着鲜艳的浓妆，头发染成火红色。可您瞧人这话，不象没教养的人吧，跟大爷还“您，您”的。可谁叫她也忍不住非要说那“多余的最后一句话”呢！

　　“就象您这样上了年纪看着挺慈祥的，一肚子坏水儿的多了去了！”

　　要没有人出来批评一下时髦的小姐是不正常的。可不，一个中年的大姐说了：

　　“你这个女孩子怎么能这么跟老人讲话呢，要有点儿礼貌嘛。你对你父母也这么说话吗？”您瞧大姐批评得多好！把女孩子爹妈一抬出来，女孩子立刻就不吭气了。

　　要说这事儿就这么结了也就算了，大家说到这儿也就完了，大家该干嘛干嘛去。可不要忘了，大姐的“多余的最后一句话”还没说呢。

　　“瞧你那样，估计你父母也管不了你。打扮得跟鸡似的！”

　　后面的事大家就可想而知了，简单地说，出人命的可能都有。这么吵着闹着车可就到站了。车门一开，售票员小姑娘说：“都别吵了，该下车的赶快下车吧，别把自己正事儿给耽误了。”当然，她没忘了把最后一句多余的话给说出来：

　　“要吵统统都给我下车吵去，不下去我车可不走了啊！烦不烦啊！”

　　烦不烦？烦！不仅她烦，所有乘客都烦了！整个车厢这可叫炸了窝了，骂售票员的，骂外地小伙儿的，骂时髦小姐的，骂中年大姐的，骂天气的，骂自个儿孩子的，真是人声鼎沸，甭提多热闹了！那个外地小伙儿一直没有说话，估计他实在受不了了，他大叫一声：

　　“大家都别吵了！都是我的错，我自个儿没看好地图，让大家跟着都生一肚子气！大家就算给我面子，都别吵了行吗？”听到他这么说，当然车上的人都不好意思再吵了，声音很快平息下来，少数人轻声嘀咕了两句，也就不说话了。但你们不要忘了，外地小伙儿的“多余的最后一句话”还没说呢：

　　“早知道北京人都是这么一群不讲理的王八蛋，我还不如不来呢！”

　　想知道事情最后的结果吗？我那天的事情没有办成。我先到派出所录了口供，然后到医院外科把头上的伤给处理了一下。我头上的伤是在混战中被售票员小姑娘用票匣子给砸的。你们可别认为我参与了他们打架，我是去劝来着。我呼吁让他们都冷静一点儿，有话好好说，有没什么大事儿，没什么必要非打个头破血流。

　　我的多余的最后一句话是这么说的：不就是售票员说话不得体吗？你们就当她是个傻X，和她计较什么！

這事是真的？？
太不可思議..


AMC
AIME
geometry
search
videos
Alcumus
AIME I

I'm really sorry guys because you've probably seen a lot of these "advice" posts
but I would really appreciate help on my individual situation   :D

sooo last year on the AMC 10A i think i got like a 121.5 (freshman)
and on the AIME i got a 5 (i solved the first five in around an hour, unfortunately i didnt solve any other problems in the next two hours  :ewpu: )

i have both Aops books but i've never completely finished the first one (i went really quickly over the geometry section in the summer, but i prob need to review)
and i think my other weakness is[i] probably [/i]counting

my goals this year are to: get a decent score on the AMC 10A and get into Aime again....and (hopefully) get into USAMO...though the chances are slim i think

any advice/help on how i should prepare? thanks!!!!

Do old AIME exams, and read through the books completely.  The more problems you do, the better off you'll be.  There is no substitute for practice.

And please use the "Search" feature.  Most "advice" threads end up being the same anyway so please look for old responses first.

To improve counting, I would suggest Alcumus. It's free, reliable, and great. If you don't understand the problems, first try really hard yourself, next see if there are any videos related to the problem, and lastly, if all else fails, ask on the Alcumus forum. You are bound to succeed.

There's already one thread open on the first page of the forum and a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=246437]FAQ page[/url]; you can also use the search function.


real analysis
geometry
real analysis unsolved

it's a well known fact (and easy to prove it..) that the set {sin(n)} is dense in [0;1] ... but what about the set {n*sin(n)} ? is it dense in R ??

Yes, the range of f(x)=x*sin(x) is R.  If you think about it, sin(x) will fluctuate between 0 and 1, and x will get larger and larger, so the product x*sin(x) will get larger and larger, except when x=pi*k, so then the product is 0.

Julie, that's not the question. Of course the range of f(x)=xsin(x) for x in R is all of R. The question was whether or not, given any real $y$ and any $\epsilon > 0$, you can find an INTEGER $n$ such that $|f(n)-y|<\epsilon$.

You are right Gorbber, I've found it (this is the following of a discussion in another topic) . But as you may see, Alekk asked it a an open question (even if it not really clear...)

Pierre.

Ok, I've read the paper in the reference from Alekk (Journal of the ENS Cachan, in french).

That the sequence $U_n = n \cdot sin(n)$ is an old problem. I've seen it in an old issue of the A.M.M. and it is probably not the first reference about it. This question appears periodically in each math forum or math journal, probably because the density of the sequence $(sin(n))$ in $[-1,1]$ is a classical exercise...and obviously, it happened in that journal.

It can be proved that if $\pi = [a_0,a_1, \cdots , a_n, \cdots]$ is the expansion of $\pi$ into continued fractions then the density of $(U_n)$ in $R$ is equivalent to $limsup a_n = + \infty$.
This last point of view has been much explorated (not only for answering to the density problem as you may guess) and this is still an open problem.

More precisely :

It can be shown that $0$ is an accumulation point of $(U_n)$ if and only if for each $e> 0$ there are an infinte number of irreducible fractions $\frac p q $ such that $| \frac p q - \pi| < \frac e {q^2}$.

This directly leads to the diophatine approximation area :

If $x$ is a real number, the irrationality degree of $x$ is the least real $k$ such that $| \frac p q - x| \leq  \frac 1 {q^k}$ holds for a finite number of irreducuble fractions $\frac p q$ (if this least real number does not exist we choose $k= + \infty$).

Thus, it can be shown that : $\pi $ has a finite irrationality degree $k$ and it is known that $k \leq 8,0161$ (M.Hata, 1993 which improves a famous result of Mahler $k \leq 42$).
It is conjectured that $k=2$. Anyway, this result would be not enough to prove the density of $(U_n)$ since we are looking for a improved result, since it deals with the approximation between $\frac 1 {q^2}$ and $\frac 1 {q^{2+ \alpha}}$ for all $\alpha > 0$.

In another hand, to be more positive, the above results leads to the following :
The sequence $(n^9 \cdot sin(n) )$ is not dense in $R$. In particular, $0$ is not an accumulation point for this sequence.
And since $\pi$ is irrational, then $k \geq 2$ (see Hardy and Wright, th. 185) which leads (using Bolzano-Weierstrass theorem)  to the fact that $(U_n)$ has an accumulation point between $- \pi $ and $ \pi $.

Pierre.

thanx Pierre .. I've not read the paper because I'm not at home and I cannot view postscript here.. that's why I ask if anyone knows the problem.. I'm sorry to have post an open problem  :(

Kent Merryfield - sorry, stupid mistake, I guess I just inferred that's what "dense in" meant.  I realize that if you are posting here you can already answer a simple question like that.  whoops, sorry


prove that in any triangle $ [1\plus{}h_a( \frac{1}{h_b}\plus{}\frac{1}{h_c})][1\plus{}a(\frac{1}{b}\plus{}\frac{1}{c})]>9$     :rotfl:

hello, setting
$ h_a\equal{}\frac{2A}{a},h_b\equal{}\frac{2A}{b},h_c\equal{}\frac{2A}{c}$ in your inequality we get
$ (1\plus{}\frac{1}{a}(b\plus{}c))(1\plus{}a(\frac{1}{b}\plus{}\frac{1}{c})\geq9$
this is equivalent with
$ \frac{a}{b}\plus{}\frac{b}{a}\plus{}\frac{c}{a}\plus{}\frac{a}{c}\plus{}\frac{c}{b}\plus{}\frac{b}{c}\geq6$
and this is true, the equal sign holds if and only if $ a\equal{}b\equal{}c$.
Sonnhard.


for$x_1,x_2 ... x_n \in N$and $1\le i \le n$prove that :
$\mid (x_1-i)(x_2-i)...(x_n-i)\mid \le x_1x_2...x_n$ :roll:

posted today in http://www.mathlinks.ro/Forum/viewtopic.php?t=60377


I'm using \mathbf{\beta} for example, but it looks just the same as a normal \beta.

Sample code:

[code]
removed
[/code]

Thanks.

edit: I found the solution here: http://www.tex.ac.uk/cgi-bin/texfaq2html?label=boldgreek

Use the [i]bm[/i] or [i]amsbsy[/i] package.

You can also make anything you wish bold by using \pmb (which, I believe, stands for "poor man's bold" and just draws the symbol together with its slight shift the right or something like that):
$ \pmb{\prod_{\beta<\mathcal B}\sqrt{1\plus{}\beta^2}}\ne
\prod_{\beta<\mathcal B}\sqrt{1\plus{}\beta^2}$
On the other hand, the image quality delivered by \pmb is inferior to that of specially designed bold characters.


if f(n)=f(n-2)-f(n-1), g(n) = (-1)^n * f(n) how to prove that g(n) satisfy the Fibonaccci relation?

OH my god it's Benjy! This is Matt from HSCCiM.

So weird to see you post :P

Anyways, cool problem :)


Whoever is the top 2 after the perfect score would be send to the tiebreaker, which is a proof round. If there's no tie on the second highest score, we'll just have to choose the next highest score. If there are more than one person in that score, they're all invited to take this. The people who will take this would be soon announced and because this is a proof round, you're given 2 weeks to do this. You must write full-written solution to this problem. Show every work and explain every step.

Question 1:

$z_1,z_2,....z_5$ are non-negative real numbers that holds the following equation true:

$z_1+z_2+z_3+z_4+z_5 = 200$.

The maximum value, which is called $\mu$, is defined by $z_1+z_2$,$z_2+z_3$,$z_3+z_4$, or$z_4+z_5$. If $\mu$ lies between $\alpha$ and $\beta$ such that $\beta - \alpha = 2$, find, with fully written proof, the value of $x+y$ if $\mu - \beta + \alpha = \frac {x}{y}$. NOTE: $\alpha$ is the smaller number that is closest to the real value of $\mu$, meaning $\mu - \alpha$ is less than one.

Question 2:

Given that $a,b,c$ represent the length of the side of the $\Delta {ABC}$ and $k$ is any positive integer. Prove that for all $a,b,c,k$, the following inequality holds true:

$\sqrt {\frac {ck^2}{a+b} + \frac {bk^2}{a+c} + \frac {ak^2}{b+c}} < \sqrt {2k^2}$

The end of the two weeks isn't the 18th, is it?

No.

The thread was posted in 3/11. So, adding two weeks give you 11+14 = 25.

March 25th is the date you have to submit the proof.

For question 2, the sides of the triangle are integral lengths, right?  :lol:

Please send me the answers by Friday 3/25/05.

When will the next COP competition start?

BTW, what does COP stand for.

After the championship round has ended, which would be approximately two weeks later.

COP stands for the Contest of Opportunity. The aim is to find the students who are excel at math and strong on avoiding the careless mistakes. That's where opportunity comes to play. Strong on catching the opportunities.

who is eligbile to compete at COP? anyone?

Well, we try to discourage people who normally browse the intermediate or above forums, but basically anyone can participate.

Actually, that's not true.

Anyone's welcome. Exception would be people like JBL, ComplexZeta, or Ravi B, and as we know, the admins :D

Even your level is high, it doesn't necessarily guarantee your chance of placing first with perfect score.

It does tell you have high chance. But it doesn't necessarily [b]PROVE[/b] the chance of your winning.

The problem one has been edited one more note to it.

There could've been two possible answers for that one.

Please submit the answers by today.

P.S. It seems that there has been major problem with 1. ;) I recently found from one of contestant that there is another answer than what I thought was. But I'll accept both of them.


probability
integration
function
algebra
domain
geometry
geometric transformation

I have the following problem, I don't know the solution, I don't even know whether a known solution exists. Here it is:
Considering a continuous random variable X(t), I know:
$ P(|X(t\plus{}h)\minus{}X(t)|\le w)\equal{}\displaystyle\int^w_{\minus{}w} f(u)\,du$
with $ f(u)$ being function of $ h$ but independent of $ t$.

What I wish to know however, is the probability of $ |X(t\plus{}h)\minus{}X(t)|$ to remain confined below or equal to $ w$, for $ h$ varying continuously (in the real domain)  from $ 0$ to $ h_0$ ?

Any ideas are welcome.

After further reflections, naming $ P_{h_{0}}$ the probability I am looking for, and $ I(h)$ the $ \displaystyle\int^w_{\minus{}w} f(u)\,du$, I think:
$ P_{h_{0}}\equal{}exp[\displaystyle\int^{h_0}_0ln(I(h))\,dh]$
Obviously I may be wrong, if so, don't hesitate to let me know.

Hi Endiku

Well I'm not sure so about your formula, the first thing to do should be to specify exactly what you mean by "$ X(t)$ is a continuous random variable". I f you mean that it is an almost sure continuous stochastic process then without any further specification you cannot say anything about the law of $ X(t\plus{}h)\minus{}X(t)$. for example if we take a brownian motion, it is a continuous stochastic process whose law is specified by its indepedent increment following a gaussian law of 0 mean and variance equal to $ t\plus{}h \minus{}t\equal{} h$.

In general you first know a familiy of consistent finite dimensional laws for $ X_{t_1},...X_{t_n}$ (or better the difference $ X_{t_1},X_{t_2}\minus{}X_{t_1},...,X_{t_n}\minus{}X_{t_{n\minus{}1}}$) and then you construct by kolmogoroff theorem a probability measure for the process $ X_t$.

The fact that in your case the increments of $ X_t$ are not independent complicate the subject very much I think. Moreover I feel like you consider the the family of law (w.r.t. h) of $ X(t\plus{}h)\minus{}X(t)$ is tight which is not implied by the only almost sure continuity of the process $ X(t)$.

Well now I think that you should specify eplicitltely the law of your process and we'll try to see if we can get something about $ P_{h_0}$.

Thanks for the precision, and here is the exact case:

I am considering a fractional Brownian Motion of index-$ \alpha$ ($ 0 < \alpha < 1$), then, with probability 1, X(t) is continuous and X(0)=0, and for every $ t\ge 0$ and $ h > 0$ the increment X(t+h)-X(t) has the normal distribution with mean 0 and variance $ h^{2\alpha}$, so that:

$ P(X(t \plus{} h) \minus{} X(t)\le x) \equal{} (2\pi)^{ \minus{} 1/2}h^{ \minus{} \alpha}\displaystyle\int^x_{ \minus{} \infty} exp(\frac { \minus{} u^2}{2h^{2\alpha}})\,du$

Hi Endiku
I understand your problem now,
my advice would be to try to look at the law of $ sup_{0\leq s \leq h}\{|X_{s \plus{} t} \minus{} X_t|\}$, I don't know if there is any derivation for this functionnal for fractionnal brownian motion but I would be surprised if none has ever studied the subject.

I'll check if I can find some reference for you.

Well after a quick search (certainly incomplete ) I could only find "Upper Bound" and "asymptotic behaviour" type results for the law of this functional
a good refernce is last chapter of Robert J. Adler "An introduction to continuity, extrema, and related topics for general gausian processes"

Thanks for the reference, I shortly looked at it, and as you say it seems to focus on the asymptotic behaviour, which I am not so much interested in right now, I got to read it in more detail though, maybe I'll get some ideas out of it.

My earlier solution indeed seems false, as it does not account for the dependence. I'll keep looking around then, thanks for your help.
And any other hint or reference are most welcome.


Any one can tell me something about the M theory,If there is something like that regarding black holes and all.

Are you referring to that stuff which was known as "Strings" ??

There are two good books, one called [u]The Elegant Universe[/u] and another called [u]The Fabric of the Cosmos[/u], both by Brian Greene.

The first book describes string theory and M-Theory in basic layman's terms; and the second book explains the general structure of the Universe, including the idea that there may be many hidden dimensions.

I highly recommend both books, though they do overlap a little bit in content. There's also a 3 hour PBS/Nova [url=http://www.pbs.org/wgbh/nova/elegant/]video[/url] called The Elegant Universe, narrated by Brian Greene, which is pretty good.

thankyou Guass,By the way ,no offence but I always disliked guasses theorem.

No offense taken, I didn't invent it. :lol:

But which Gauss's Theorem are you referring to? Do you mean the one that's also called the [url=http://mathworld.wolfram.com/DivergenceTheorem.html]Divergence Theorem[/url]? If so, I quite like that one.

[quote="11th dimension"]Look at my name.[/quote]

May I infer from your Avatar that Canada is the 11th dimension? ... I always thought so  ;)

I just read an article from Scientific American concerning Inconstant constants, earlier today in a Barnes & Noble Bookstore. It talked about how relative are the physical constants and how physicists are trying to explain the nature of physical constants (like $G$ or $c$). They believe that maybe the Unified Theory or String Theory can explain how those constants became constants or maybe there are more fundamental things that we are not aware of. Just a thought that I wanted to share to some people (sorry that non-American audience will not be able to access Scientific American unless you have an international subscription  :D ).

From the books i've read i can advise you to read the following books:

1. The Elegant Universe by Brian Greene(Very interesting and exciting, its mostly about superstring theory and m-theory)
2. The Universe in A Nutshell by Stephen Hawking (Majorly about black holes and cosmology)
3. A Brief History of Time by Stephen Hawking (black holes, elementary particles and quantum theory)

Have Fun! ;)


function
inequalities
logarithms
inequalities unsolved

Is it true for all positive real a,b,c and increasing and convex function f such that
$f(a)+f(b)+f(c)+3f(\sqrt[3]{abc}) \ge 2(f(\sqrt{ab})+f(\sqrt{bc})+f(\sqrt{ca}))$?

[quote="ondrob"]If this implication is true:

If $f(x)$ is convex, then also $f(e^x)$ is convex.

Then it's Popoviciu's inequality on function $f(e^x)$ and values $\ln a,\ln b,\ln c$[/quote]
I believe it's true.

Ok, we know that $f(x)$ is increasing and convex on positive real numbers. It means that for each positive numbers $x,y$ and real number $0<a<1$ this is true:
$af(x)+(1-a)f(y)\geq f(ax+(1-a)y)$
Now it's easy to show that:
$af(e^x)+(1-a)f(e^y)\geq f(ae^x+(1-a)e^y)\geq f(e^{ax+(1-a)y})$ (last ineq is weighted AM-GM+increasing of $f(x)$)
So $f(e^x)$ is convex.

Now using popovicu ineq we have:
$f(e^{\ln a})+f(e^{\ln b})+f(e^{\ln c})+3f(e^{\frac{\ln a+\ln b+\ln c}{3}})\geq 2(f(e^{\frac{\ln a+\ln b}{2}})+f(e^{\frac{\ln b+\ln c}{2}})+f(e^{\frac{\ln c+\ln a}{2}}))$

And it's equivalent to:
$f(a)+f(b)+f(c)+3f(\sqrt[3]{abc}) \ge 2(f(\sqrt{ab})+f(\sqrt{bc})+f(\sqrt{ca})) $

I got the same result with a different method.
Anyway, I think this result is quite interesting, as in general we also have the popoviciu inequality in product form
If f is increasing and convex and $a_i$ are positive
$f(a_1)+f(a_2)+...+f(a_n)+n(n-2)f(\sqrt[n]{a_1...a_n}) \ge (n-1)(\sum f(\sqrt[n-1]{\frac{S}{a_i}}))$
where $S=\prod a_i$.
This is one of the generalizations of vasc, i believe we can do the same thing to the other generalization of vasc as well.
I think this inequality is quite nice.  :)

Quite general thing about convexity can be obtained.
Let's have two intervals $I,J$ and function $f(x)$ from $I$ to $\mathbb R$ and $f(x)$ is convex and increasing on $I$ and convex function $g(x)$ from $J$ to $I$.

$f(x)$ is convex on $I$ => For each $x,y$ from $I$ and $0<a<1$ we have:
$af(x)+(1-a)f(y)\geq f(ax+(1-a)y)$

By putting $g(x),g(y)$ instead of $x,y$ we obtain:
$af(g(x))+(1-a)f(g(y))\geq f(ag(x)+(1-a)g(y))$
Using convexity of $g(x)$ we have:
$ag(x)+(1-a)g(y)\geq g(ax+(1-a)y)$
Because of the fact that $f(x)$ is increasing we have:
$f(ag(x)+(1-a)g(y))\geq f(g(ax+(1-a)y))$
So together we have:
$af(g(x))+(1-a)f(g(y))\geq f(ag(x)+(1-a)g(y))\geq f(g(ax+(1-a)y))$
This means that function $f(g(x))$ is convex on $J$.

I believe using this we can derive some nice inequalities from inequalities for convex functions.

Yes, it is a well-known result that increasing convex functions of convex functions are convex.


algebra
system of equations
algebra proposed

There are 2 parts to this problem :

Part One :

Find all real solutions to the system of equations

x^y = z,  y^z = x,  z^x = y.

Part Two :

Find all complex solutions.
--------------------------------------------------------------
I only did the first part, I need help with Part Two.
Here's what i did for part one :

Let's raise he first equation to the xz power, the second to the xy power, and the third to the yz power. This gives :

(1') x^xyz = z^xz
(2') y^xyz = x^xy
(3') z^xyz = y^yz

but z^xz = (z^x)^z = y^z = x,
and similarly x^xy = y, y^yz = z, so

(1") x^xyz = x
(2") y^xyz = y
(3") z^xyz = z

Put n=xyz, then
x^n = x
y^n = y
z^n = z
We can easily conclude that
x = +/-1, y = +/-1, and z = +/-1
So
x = y = z = +1
x = y = z = -1

Do you agree with my solution ? Have I missed anything ?
How do you solve part two ?

From $x^{n}=x, y^{n}=y, z^{n}=z$ you cannot conclude that |x|=|y|=|z|=1, because you can also have n=1.


A six-faced die is weighted so that the probabilities of rolling a 1, 2 and 3 are equally likely; the probabilities of rolling a 4, 5 and 6 are equally likely; and the probability of rolling a 5 is four times the probability of rolling a 3. What is the probability that a 2 or a 5 occurs on a single roll of the die? Express your answer as a common fraction.

a+a+a+b+b+b=1.
We want the probability of two exclusive events so we add P(2)+P(5)=a+b=1/3.


Δεδομενης της γεννητριας συναρτησης  της ακολουθιας $ a_0,a_1,a_2$.....
Nα βρειτε την γεννητρια συναρτηση της υπακολουθιας $ a_1,a_4,a_7$......
GOOD LUCK!!!

Θεωρούμε τις 3 ρίζες της μονάδος $ 1,a,a^2$

Έστω $ A(x)\equal{} \sum_{0}^{\plus{}\infty} a_n x^n$

Tότε:

$ \frac {A(x)\minus{}a_0} {x} \equal{} \sum_{1}^{\plus{}\infty} a_n x^{n\minus{}1}$,
$ \frac {A(ax)\minus{}a_0} {x} \equal{} \sum_{1}^{\plus{}\infty} a_n (ax)^{n\minus{}1}$,
$ \frac {A(a^2 x)\minus{}a_0} {x} \equal{} \sum_{1}^{\plus{}\infty} a_n (a^2x)^{n\minus{}1}$

Προσθέτουμε και χρησιμοποιώντας τις ιδιοτητες των ριζών της μονάδας:

$ \frac {A(x)\plus{}a^2A(ax)\plus{}aA(a^2x)}{3x}\equal{} \sum_{1}^{\plus{}\infty} a_{3n\plus{}1}  x^{3n}$

Και άρα η ζητούμενη γεννήτρια συνάρτηση είναι η:

$ B(x)\equal{}\frac {A(\sqrt[3]{x}\plus{}a^2A(a\sqrt[3]{x})\plus{}aA(\sqrt[3]{x})}{3\sqrt[3]{x}}$


search
inequalities proposed
inequalities

This is first created by me and VASC, solved by Arqady and Me by Contradiction method. You can search this forum or VI forum.

[quote="quangpbc"]$ \frac{a}{\sqrt{b^{2}\plus{}3c^{2}}}\plus{}\frac{b}{\sqrt{c^{2}\plus{}3a^{2}}}\plus{}\frac{c}{\sqrt{a^{2}\plus{}3b^{2}}}\ge\frac{3}{2}$ for any nonnegative real numbers $ a,b,c$

http://www.mathlinks.ro/Forum/viewtopic.php?p=862698#862698 :)[/quote]
Someone has erased there my proof. :wink:
Let $ \frac{c^{2}}{a^{2}\plus{}3b^{2}}\equal{}\frac{z^{2}}{4},$ $ \frac{b^{2}}{c^{2}\plus{}3a^{2}}\equal{}\frac{y^{2}}{4}$ and  $ \frac{a^{2}}{b^{2}\plus{}3c^{2}}\equal{}\frac{x^{2}}{4},$ where $ x,$ $ y$ and $ z$ are non-negative numbers.
Hence, $ 16\equal{}7x^{2}y^{2}z^{2}\plus{}3(x^{2}y^{2}\plus{}x^{2}z^{2}\plus{}y^{2}z^{2})$ and it remains to prove that $ x\plus{}y\plus{}z\geq3.$
Let $ x\plus{}y\plus{}z<3$ for any $ x,$ $ y$ and $ z.$ Then it exists $ k>1$ for him holds: $ kx\plus{}ky\plus{}kz\equal{}3.$
Let $ kx\equal{}u,$ $ ky\equal{}v$ and $ kz\equal{}v.$ Then $ u\plus{}v\plus{}w\equal{}3$ and
$ 16\equal{}\frac{7u^{2}v^{2}w^{2}}{k^{6}}\plus{}\frac{3(u^{2}v^{2}\plus{}u^{2}w^{2}\plus{}v^{2}w^{2})}{k^{4}}<7u^{2}v^{2}w^{2}\plus{}3(u^{2}v^{2}\plus{}u^{2}w^{2}\plus{}v^{2}w^{2}).$
But easy to prove that $ 16\geq7u^{2}v^{2}w^{2}\plus{}3(u^{2}v^{2}\plus{}u^{2}w^{2}\plus{}v^{2}w^{2})$ is true
for all non-negative $ u,$ $ v$ and $ w$ such that $ u\plus{}v\plus{}w\equal{}3.$


In a triangle $ ABC \angle A\equal{} 120^{\circ} |AB|\equal{}17,|AC|\equal{}b,|BC|\equal{}a.$ a and b integer.How many total a values?
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calculus
integration
geometry
3D geometry
trigonometry
parameterization
analytic geometry

Evaluate the surface integral $\iint_{S}{\mathop A^\to }.dS$

$\mathop A^\to(x,y,z) = 4xz\text{ }i\text{ }+\text{ }xyz^{2}\text{ }j\text{ }+\text{ }3z\text{ }k$

S is the part of the cone $z=\sqrt(x^{2}+y^{2})$ beneath the plane $z=4$

Thanks

Divergence theorem applies to flux integrals over [b]closed [/b]surfaces, and $S$ is not closed. You need a parametrization of $S$, such as $\vec r(x,y)=\langle x,y,\sqrt{x^{2}+y^{2}}\rangle$, where $(x,y)$ is a point in the disk $D=\{(x,y)\colon x^{2}+y^{2}\le 16\}$. The flux of $\vec A$ across $S$ (in which direction? you have not specified an orientation of $S$) can be computed as $\pm\iint_{D}\vec F(\vec r(x,y))\cdot (\vec r_{x}\times \vec r_{y})dA$. You will probably want to switch to polar coordinates at the end.

Well, I guess then you could subtract off the flux through the bottom.

Thus, the correction is $-(\iint_{A}(3z) dA) =-(3*4 * \pi * 4^{2}) =-192 \pi$.

Therefore, the correct answer should be $\frac{80 \pi}{3}$.

[quote="PTynan89"]$\iint_{S}{\vec A}\vdots d \vec S = \iiint_{V}{(\del \vdots \vec A)}dV$.

Thus, the integral becomes $\iiint_{V}{(4z+xz^{2}+3)}dV$
$= \iiint_{V}{(4z+r \cos \theta z^{2}+3)}dV = \int_{0}^{2 \pi}\int_{0}^{4}\int_{0}^{z}(4z+r \cos \theta z^{2}+3) dr dz d \theta$[/quote]

Recall that $dV = r dr dz d \theta$.


Given a rectangular $ ABCD$, let $ P,\ Q$ be the points on $ BC,\ CD$ respectively.
If $ \triangle{APQ}$ is a regular triangle, then prove that $ \triangle{ABP} \plus{} \triangle{ADQ} \equal{} \triangle{CPQ}.$

Let $ \angle PAB \equal{}\alpha$ and $ AP\equal{}PQ\equal{}QA\equal{}1$. Then $ \angle QBC\equal{}30\plus{}\alpha$ and $ \angle QAD \equal{} 30\minus{}\alpha$.

Then $ \triangle{ABP}\plus{}\triangle{ADQ}\equal{}\triangle{CPQ}$  $ \leftrightarrow$ $ \sin 2 \alpha \plus{} \sin 2(30\minus{}\alpha) \equal{} \sin 2(30\plus{}\alpha)$.

Because $ \sin 2\alpha\plus{} \sin 2(30\minus{}\alpha) \equal{} 2 \sin 30 \cos (30\minus{}2 \alpha )\equal{} \cos( 30\minus{}2 \alpha )\equal{} \sin (60\plus{} 2 \alpha)$, Our proof was finished.


Who is going to the Oklahoma State 2009 on March 7? Hope you can tell us. I already know jonathanchou711, carphead0, mr.k_74, shortstack, are going. And probably, jimmyw8888, and math888. Please tell us. Oh yeha, and I am going, and so is jackblack right?

idea: post in the oklahoma forum and ask people from oklahoma to post there.

Is there a possibility that jackblack is from University :?:

he is. u know math888, who we are talking bout. the Jack

I'm assuming that U School, Jenks, Classen, and Whittier will all be there tomorrow...am I correct?
Well, I guess I shall find out tomorrow...

[quote="jonathanchou711"]shortstack is mitchell. the short kid... in 8th grade!  :lol:[/quote]

hmm...
I don't think he would like this if he sees this....  :wink:

[quote]hmm...
I don't think he would like this if he sees this.... [/quote]

I don't think he gets on anymore. He just made an account to answer the PoW's... CHEATER!!  :mad:  But b80 found out! ;)