symmetry
inequalities
inequalities unsolved

Prove that if n real numbers $ x_1,x_2,...,x_n$ satisfying $ (n\minus{}1)(x_1^2\plus{}x_2^2\plus{}...\plus{}x_n^2)\equal{}(x_1\plus{}x_2\plus{}...\plus{}x_n)^2$ ,then all of them is ≥0,or all of them is ≤0.

[quote="arqady"][quote="SUPERMAN2"]Prove that if n real numbers $ x_1,x_2,...,x_n$ satisfying $ (n \minus{} 1)(x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2) \equal{} (x_1 \plus{} x_2 \plus{} ... \plus{} x_n)^2$ ,then all of them is ≥0,or all of them is ≤0.[/quote]
When sum of squares of reals numbers equal to $ 0$ $ ?$ :wink:[/quote]
I mean that if n real numbers $ x_1,x_2,...,x_n$ satisfying $ (n \minus{} 1)(x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2) \equal{} (x_1 \plus{} x_2 \plus{} ... \plus{} x_n)^2$,then then $ x_i x_j$≥0 for all 1≤i,j≤n

Assume to the contrary, WLOG, that $ x_1x_2 < 0$. Then we have:
$ (x_1 \plus{} \cdots \plus{} x_n)^2 > (n \minus{} 1)(x_1^2 \plus{} \cdots \plus{} x_n^2) \plus{} 2(n \minus{} 1)x_1x_2$
Or, equivalently:
$ \sum_{i \ge 3} (x_1 \plus{} x_2 \minus{} x_i)^2 \plus{} \sum_{i > j\ge 3} (x_i \minus{} x_j)^2 < 0$
which is a contradiction.

[quote="bambaman"]Assume to the contrary, WLOG, that $ x_1x_2 < 0$. Then we have:
$ (x_1 \plus{} \cdots \plus{} x_n)^2 > (n \minus{} 1)(x_1^2 \plus{} \cdots \plus{} x_n^2) \plus{} 2(n \minus{} 1)x_1x_2$
Or, equivalently:
$ \sum_{i \ge 3} (x_1 \plus{} x_2 \minus{} x_i)^2 \plus{} \sum_{i > j\ge 3} (x_i \minus{} x_j)^3 < 0$
which is a contradiction.[/quote]
Could you explain more clearly please?

Yes... As you said, we want to show that $ x_ix_j \ge 0$ for all $ i \neq j$. From symmetry it is enough for $ x_1x_2$.
$ x_1x_2 < 0 \implies 2(n \minus{} 1)x_1x_2 < 0 \implies (x_1 \plus{} \cdots \plus{} x_n)^2 \equal{} (n \minus{} 1)(x_1^2 \plus{} \cdots \plus{} x_n^2) > (n \minus{} 1)(x_1^2 \plus{} \cdots \plus{} x_n^2) \plus{} 2(n \minus{} 1)x_1x_2$
or
$ 2(n \minus{} 2)x_1x_2 \plus{} (n \minus{} 2) \sum_{i \equal{} 1}^{n} x_i^2 < 2\sum_{i > j, (i,j) \neq (2,1)} x_ix_j$ (*)

Notice that $ 2\sum_{i > j, (i,j) \neq (2,1)} x_ix_j \equal{} 2(x_1 \plus{} x_2)(x_3 \plus{} \cdots \plus{} x_n) \plus{} 2\sum_{i > j\ge3}x_ix_j$. It helps us to complete the inequality into a sum of squares, as follows:
$ \sum_{i \ge 3} (x_1 \plus{} x_2 \minus{} x_i)^2 \plus{} \sum_{i > j\ge 3} (x_i \minus{} x_j)^2 < 0$

Just expand and see that it is indeed equivalent to (*). But the LHS is non-negative, which gives a contradiction and implies that $ x_1x_2 \ge 0$.


induction
inequalities
algebra open
algebra

Can anyone prove the following by induction:

[color=green]Prove by induction that

(1/1^2)+(1/2^2)+...+(1/n^2) <= 2-(1/n)

for all integers n >= 1[/color]
Thanks very much!!! :lol:

This one is more than classic! (and should probably be moved into the intermediate forum :) )

Well, obviously, it is true for $n=1$. So now, let's assume it's true for some $n \in \mathbb{N^*}$, then :
$\frac 1{1^2} + \frac 1{2^2} + \dots + \frac 1{n^2} \le 2 - \frac 1n + \frac 1{(n+1)^2} < 2 - \frac 1{n+1}$.
Thus, it's true for $n+1$, and we're done ;)

I think the original inequality is :
$2\geq \frac{1}{1^2}+\frac{1}{2^2}+\dots+\frac{1}{n^2}$

Of course, but to prove this, you first need to show the one written by [b]suedenation[/b].

Angelov,

you are using wrong ways to display the following items.

Use 1/(1*2) and 1/(2*3) or something similarly correct, and not the decimal points you used because they mean only 1.2 and 2.3 for the denominators, respectively.  You need grouping symbols around the one times two, the two times three, and so on expressions in the denominators, because of the order of operations.

1/(n-1).n really doesn't mean anything, but if you STILL
attributed multiplication to the period, then it would equal
n/(n-1) by the order of operations; you divide first by the (n-1)
quantity, and then you would multiply by the n quantity.

1/(k-1)k does not equal 1/(k-1) - 1/k, because it actually equals
k/(k-1).  The same reasons/conditions above apply.


For all $a,b,c,d,e > 0$, and $ad^{2}=bd^{2}=cd^{2}=36$, prove that

\[a^{2}+b^{2}+c^{2}+d^{2}+e^{2}> 9e\]

I hope you like this inequality.  :P I don't think it is easy, but I could be mistaking.

[hide="The beginning of the solution."]
Consider $f : \left( 0, \infty \right) \to \mathbb R, \, f(x) = x^{2}-9x+a^{2}+b^{2}+c^{2}+d^{2}$.
We have that $\Delta_{f}= 81-4 \left( a^{2}+b^{2}+c^{2}+d^{2}\right)$.
If $\Delta_{f}$ were $>0$, then $f$ would have two real roots whose sum and product would be positive, i.e. it would have two positive roots. Between those roots $f$ would be $<0$, so the problem would be wrong.
So we must necessarilly prove that $\Delta_{f}\leq 0$.
It's not that hard, it's just one big computation.
[/hide]

[quote="Jan"]$ad^{2}=bd^{2}=cd^{2}=36$[/quote]

Is this a typo, Jan? Or is it just to make your inequality look better?

If it's like you stated it
[hide="Solution"]We know that $ad^{2}= bd^{2}= cd^{2}$, so $a=b=c$. (because $d>0$) So we want to prove that \[3a^{2}+d^{2}+e^{2}> 9e\]for all $a,d,e > 0$ with $ad^{2}= 36$. Consider it as a quadratic in the variable $e$. If we could prove that $0 > \Delta = 81-4(3a^{2}+d^{2})$, we would be done. So let's just prove it!  :P
It's equivalent to $12a^{2}+4d^{2}> 81$. By AM-GM we have \[12a^{2}+4d^{2}= 12a^{2}+2d^{2}+2d^{2}\geq 3\sqrt[3]{48(ad^{2})^{2}}= 3\sqrt[3]{48\cdot 36^{2}}= 3\sqrt[3]{2^{8}\cdot 3^{5}}\]So we just need to prove that $3\sqrt[3]{2^{8}\cdot 3^{5}}> 81$, i.e. $2^{8}\cdot 3^{8}> 3^{12}$, i.e. $2^{8}> 3^{4}$, i.e. $4 = 2^{2}> 3$, which is true.[/hide]

Ok, I must have been very tired last night, to have missed the $a=b=c$ condition. It makes the inequality very stupid.  :(

Here is what I had in mind:

\[\frac{a^{2}}{2}+\frac{a^{2}}{2}+\frac{e^{2}}{8}+\frac{e^{2}}{8}\geq ae\]

Sum up for all variables. Now, $ae+\frac{de}{6}+\frac{de}{6}\geq 3 \sqrt[3]{\frac{ad^{2}e^{3}}{36}}= 3e$

So sum up over all variables. Now, clearly the equality case is different in both inequalities, so its is strict.


linear algebra
matrix
vector
function
integration
calculus
derivative

Show that the set of equations $ \mathbf{\dot{x}} \equal{} \mathbf{Ax}$, where $ \mathbf{x}(t)$ is a column vector and $ \mathbf{A}$ is an $ n\times n$ matrix with constant elements, has solutions of the form $ \mathbf{x} \equal{} \textup{exp}(\mathbf{A}t)\mathbf{x}_0$, where $ \mathbf{x}_0$ is a constant vector and
\[ \textup{exp}(\mathbf{A}t) \equiv \mathbf{I} \plus{} \mathbf{A}t \plus{} \frac{1}{2!}\mathbf{A}^2t^2 \plus{} \frac{1}{3!}\mathbf{A}^3t^3 \plus{} \ldots .\]This method can be used to solve differential equations systems such as $ \dot{x} \equal{} 3x \plus{} 2y, \dot{y} \equal{} \minus{}5x \minus{} 3y$. Would you expect this method to work if the elements of $ \mathbf{A}$ are not constant?

[quote="aidan"]Would you expect this method to work if the elements of $ \mathbf{A}$ are not constant?[/quote]I'll answer your question with a question of my own: can you construct a matrix-valued function $ A(t)$ on an interval $ I$ such that $ \int_I A(t)\,dt$ converges (as an improper integral), but solutions of $ y'(t) \equal{} A(t)y(t)$ do not? [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=172101]Thread - no solutions posted at this time[/url].
Of course, any solution to that is obviously a case in which $ \exp\left(\int A(t)\,dt\right)\cdot y_0$ aren't solutions to the differential equation.

[Edit- silly mistakes in that last expression removed]

I have 3 questions:

(i) Does a matrix valued function mean a matrix with functions of $ t$ as its entries?

(ii) Does $ \int_{I}A(t)\,dt$ mean integrating every entry of $ A$?

(iii) Why are we looking at convergence of the integral of $ A$?

Yes, yes, and we're doing that because it's a different (more specific) question than what you asked.

[quote="jmerry"]Of course, any solution to that is obviously a case in which $ \exp\left((\int A(t)\,dt\right)\cdot y_0(t)$ aren't solutions to the differential equation.[/quote]

And at the risk of sounding stupid.. why is the above true?

In that example, $ \exp\left(\int A(t)\,dt\right)\cdot y_0$ is not a solution- because it's defined on the whole interval, and there aren't any solutions defined on the whole interval.

Of course, if you feel like something more mundane- $ \exp(A\plus{}B)\equal{}\exp(A)\exp(B)$ only if $ A$ and $ B$ commute. Consider what this does to differentiation formulas.

Then in the $ 2 \times 2$ case could such a matrix $ A$ be $ \begin{pmatrix} e^{\minus{}t} & e^{\minus{}2t} \\ e^{\minus{}2t} & e^{\minus{}t} \end{pmatrix}$ - all entries converge for $ \int_I A(t)$, for any interval $ I$?

Now, which question are you trying to answer, aidan? If it's the one I asked, stop. You're not going to get it, as evidenced by that nicely behaved matrix you just posted.

If it's the original question, where we merely want to show that $ x(t)\equal{}\exp\left(\int A(t)\,dt\right)x_0$ is not a solution to $ x'\equal{}Ax$, go simpler. You don't want to take the exponential of a matrix that already has exponentials in it. Also, the $ A$ you wrote doesn't work as an example, because everything still commutes.

[quote="jmerry"]You're not going to get it, as evidenced by that nicely behaved matrix you just posted.[/quote]
I thought so.. seems like pretty advanced stuff. I'll just try to answer my original question alright - what if $ A \equal{} \begin{pmatrix} t^{\minus{}1} \plus{} 1 & t^2 \minus{} 1 \\ \minus{}1 & 2t^2 \plus{} 1 \end{pmatrix}$? How do you differentiate $ \exp (At)$ in this case? o.0


If someone has the answers, could you post it?

I don't have the solution book.

Thanks!!  :lol:

Now a ranger  I have the answers but you just want answers not solutions right?


[quote="nayel"]$ P$ is a point on a circle with diameter $ AB$. $ M$ is the orthogonal projection of $ P$ on $ AB$. The circles with diameters $ MA$ and $ MB$ meet segments $ AP$ and $ BP$ at $ Q$ and $ R$, respectively. Prove that $ QR$ is a common tangent of the latter circles.[/quote]
[hide="Solution"]
Notice that $ \angle APM\equal{}90\minus{}\angle MPB\equal{}\angle PBA$ and $ \angle MPB\equal{}90\minus{}\angle APM\equal{}\angle PAB$.  Let $ X$ is the intersection between the circle with diameter $ AM$ and $ AP$ and let $ Y$ be the intersection with the circle with diameter $ BM$ and $ PB$.  Thus, $ MX\perp AP$ and $ MY\perp BP$.  Hence, $ \angle PXM\plus{}\angle PYM\equal{}90\plus{}90\equal{}180$.  Hence, $ PYMX$ is cyclic.  Thus, $ \angle XYM\equal{}\angle XPM\equal{}\angle APM\equal{}\angle PBA\equal{}\angle YBM$ so $ XY$ is tangent to the circle with diameter $ MB$.  Similarly, $ \angle MXY\equal{}\angle MPY\equal{}\angle PAB\equal{}\angle YAM$, so $ XY$ is tangent to the circle with diameter $ MA$.  Thus, $ XY$ is the common tangent between the two circles. [/hide]

[hide="My Solution"]It's not difficult to see that $ \angle APQ \equal{} 90^{\circ},MQ\perp AP\wedge MR\perp BP\implies PQMR$ is a rectangle. Thus $ QR \equal{} PM$. Hence $ N \equal{} QR\cap PM\implies NQ \equal{} NM \equal{} NR$. But since $ NM$ is tangent to circle $ AMQ$ (resp. $ BMR$) we conclude that $ NQ$ (resp. $ NR$) is also tangent to circle $ AMQ$ (resp. $ BMR$), since tangents from a given point to a given circle are of equal length. Thus $ QR$ is a common tangent of the two circles. [/hide]


vector
linear algebra
matrix
abstract algebra

$T: R^{2}\to R^{2}$ defined T(x,y)=(x,0) is represented in the standard ordered basis by the matrix $A = \left[{\begin{array}{*{20}c}1 & 0 \\ 0 & 0 \\ \end{array}}\right]$

$T^{2}= T$. Prove that if S is a linear operator on $R^{2}$such that $S^{2}= S$, then S = 0, S = I, or there exists an ordered basis B of $R^{2}$such that the matrix of S in B is A.

I can prove that in each condition, S is idempotent. But I can't prove that they are necessary. Perhaps someone has a different way of approaching it?

$S$ acts as the identity on its range. The range of $S$ can have dimension $0$, $1$, or $2$. The only nontrivial case is dimension $1$. Choose the first element of basis from the range of $S$, and the second from the kernel of $S$.

If the range of S is 2, how is it nontrivial? It is invertible, but I don't see easily that S is the identity.

Reread the first sentence in mlok's post.


i don't understand why it $\leq 2006^2 +2.2006$?
     can you help me ?
  i think that you guess it ?
              thank you,imortal  !

[quote="imortal"]ok that makes the problem very easy
then

$\sum \leq \sum x_i^2 + \sum x_i \leq 2.2006^2 +2.2006$

understand?:)[/quote]
I don't think that's the answer because $x_{1} \geq x_{2} \geq ...\geq x_{2006}$
but I still think this one is not hard

Hongqui,your solution is very good
 thanks very much ,Hongqui


linear algebra
matrix
vector
algebra
system of equations
superior algebra
superior algebra theorems

Hi!

Is there a way implement gaussian elimination for a system of equations in a finite field $ \mathbb{Z}_p[x]$ with Maple? I have been searching for something but to no avail.  :mad:

Thanks in advance,

Los

[quote="MeKnowsNothing"]Can't you just run normal Gaussian elimination and then take the answer mod $ p$?[/quote]

Assuming the answer provided is in terms of rational numbers, rather than floating point, yes. Bud mod $ p$ is much less computationally intensive, and has a significantly smaller memory requirement than rational numbers, which very rapidly explode in memory and start requiring BigIntegers to store the numerators and denominators.

I think there are no floating points in $ \mathbb{Z}_p[x]$. :wink: And Maple can do calculations in huge precision. I'm not sure but I guess the only limit of the number of digits comes from the available memory and hard disc space.

[quote="MeKnowsNothing"]I think there are no floating points in $ \mathbb{Z}_p[x]$. :wink: [/quote]

That's... kind of my point. You suggested doing the "normal" Gaussian elimination and then taking it mod $ p$. Assuming the "normal" one is over $ \mathbb{Q}$ or $ \mathbb{R}$, there [i] might [/i] be floating point in the answer that could cause trouble by just taking it modulo $ p$.

And, yes, presumably the only limit to the precision is RAM and (possibly) HD space, but doing lots of high-precision calculations can be slow -- if you're working on a $ 1000 \times 1000$ matrix, for example, that extra multiplicative factor is going to hurt. But if you do all the calculations mod $ p$ [i] as you go [/i], rather than just at the end after doing the calculation in $ \mathbb{Q}$, then there is never a blowup in integer size beyond $ p^2$ in any intermediate steps -- so, if your prime is (relatively) small, then the calculations will be quick and efficient.

From an algorithmic and scalability point of view, you want to be doing your computations entirely in your finite field whenever possible. For small problems, you might get away by hammering it from $ \mathbb{Q}$ into $ \mathbb{Z}_p$, but it's honestly a dirty hack at best.

so wait, if you naively perform gaussian elimination, you can expect to have as an answer a vector whose components are rational numbers. $ \mathbb{Z}_3/(x^2 \plus{} 1)$ is a field, so division is a well-defined operation. Is it possible to convert, say $ \frac {1}{2}$, to an integer?

Thanks!

Los

Since $ \mathbb{Z}_p$ is a field, convert $ a/b$ to $ b^{\minus{}1}a$ where $ b^{\minus{}1}$ is compute mod $ p$. However, I do recommend that [i] you avoid doing this if at all possible [/i], since it's dirty dirty hack :).

[quote="Nukular"]where $ b^{ \minus{} 1}$ is compute mod $ p$. [/quote]

how do you do that? So, how would one compute $ \frac {1}{2}$ in the field $ \mathbb{Z}_3/(x^2 \plus{} 1)$?

Thanks!

It would be 2, since $ 2 \times 2 \equal{} 1$.


inequalities
logarithms
number theory unsolved
number theory

prove that $ p_{n\plus{}1}\minus{}\frac{p_n}{36}>H_n$ where $ p_n$ - n-th prime, $ H_n$-harmonic numbers

The left hand side is roughly $ n\ln n$ (and trivially greater than $ 35/36 n$), the right hand side is approximately $ \ln n$. That is, the left hand side is much greater then the right hand side.

Thanks, but you proved it holds for sufficiently large $ n$, this proof (or I misunderstand something) doesn't give the $ N$ s.t. inequality holds for all $ n>N$.

How can we prove it for all $ n$ ?

\[ p_{n\plus{}1} \minus{} \frac{p_n}{36} \geq (1\plus{}p_n) \minus{} \frac{p_n}{36} \equal{} 1 \plus{} \frac{35}{36} p_n > 1 \plus{} \frac{p_n}{2} > 1 \plus{} \frac{n}{2} > 1 \plus{} \underbrace{\frac12 \plus{}\dots \plus{}\frac12}_{n\minus{}1\text{ times}}\geq 1 \plus{} \frac{1}{2} \plus{}\dots \frac{1}{n} \equal{} H_n.\]


Harvard
college
Princeton
MIT
Gauss
Ross Mathematics Program

I have talked to quite a few people about this.  It is a non-trivial decision for me.  My own personal situation is complicated, but it could be helpful to know a few things.  I am highly interested in pure math.  In high school I've done fairly well in contest math, but most of my energy is devoted toward learning other math.  My interests are actually relatively broad: hence the liberal arts schools.  So any comments on other programs may be helpful.  In particular, I like such things as philosophy, physics, public policy, diplomatic relations to China (I'm Chinese), cs.  Also, maybe if anybody happens to know what sort of alumni connections each school entails more.  I would assume it would be far more advantageous to go to Harvard if you are going to be an entrepreneur, but I also have heard that the NSA heavily favors Princeton.  Obviously, there are many more important factors to consider.  I have given up finding the perfect choice, but I certainly hope to make an informed-as-possible choice.  Why is this is so unclear!  Only got until May 1.

Thanks a lot guys!

MIT is one of a couple I ruled out mainly on the basis of two visits I made.  First time I stayed for a week at Random and determined that was too much for me.  The second time, I didn't enjoy that kind of cpw.  In general, I am less applied and more theoretical anyways.  Plus, the vibe was never one of happiness around MIT, maybe more hyperness in my eyes.  I don't mean to offend anybody, but it simply didn't feel like the right place for me.

it sounded to me that princeton doens't focus all that much on their undergrads. also, my friend who went there didn't like the social scene very much. i like MIT very much and i feel that staying at Random may have been a bad idea- not all MIT people are Randomites and you dont' have to fit in at Random to go to MIT... we also have a theoretical math track, but i digress.
i like harvard's campus alot, though i'm not quite sure about the academics there... haven't really thought about it since i didn't apply there!

Hmm.... as the three schools I am going to apply to are MIT, Harvard and Princeton, I will also have this dillema (If I don't get into MIT, and then I'd pick that  :P )

But between Harvard and Princeton, that's a tough call. I'd lean a little more towards Harvard, but it's still a difficult decision.

Sure it's a tough decision, but I mean really, can you go wrong either way?

I went to Princeton for undergraduate and worked at both Harvard and MIT for summers during my undergraduate years. Each of the schools have their own unique advantages and disadvantages.

Academically, they're all very good in terms of undergraduate math education. All of them have more opportunities at every level (casual math interest, strong math interesting, super-math-freak status) than anyone could take advantage of. There are of course differences between the departments - mostly due to size rather than quality. But all three schools graduate strong math majors who go on to do various things, and they're all about equally good in training pure mathematicians.

More than anything else, its the social environments of each of the schools that are different. Princeton's social scene is very much dominated by the Eating Clubs. That doesn't mean that there's no place for someone outside of the eating club system, but you'll be more outside of the "mainstream" student body if you don't participate.

The upside to the Eating Club system is the strong connection that you make with your fellow members and the networking opportunities that last often well after college. Princeton has one of the strongest and most active alumni associations of any school.

Harvard also has a strong alumni association, and because of its larger size, a more diverse offering of courses. It also has a very different surrounding atmosphere than Princeton (which some people love and some people hate - I personally love the Cambridge/Boston area).

But one of the stronger selling points for Harvard (and for MIT for that matter) is their proximity to each other and the ability to cross register at each school. Going to one of those schools is like getting duel citizenship to the other.

As far as the courses you said you're interested in, both Princeton and Harvard have a plethora of courses in public policy and diplomatic relations; and I'm sure both also have strong courses in philosophy and East Asian studies. In addition, I know that Princeton now has some new courses specifically focusing on entrepreneurship. But I'm sure both schools have a fairly high number of alumni entrepreneurs.

Personally if I had to do it over again, I probably would have chosen Harvard or MIT over Princeton just because I like the urban environment and the social scenes better there, and also because of the larger number of course offerings. I think I also would have liked to have more sciency/math people around (they were few and far between at Princeton).

Academically, however, I thought Princeton was great. I took lots of interesting classes (even outside of mathematics), and I can say that I learned a lot. I just wish that I had been more social and met more people while I was there. Perhaps I should have joined an eating club.

Yes, I was afraid that princeton wouldn't be nerdy enough for me, but after visiting and meeting math/cs/physics majors, I was convinced it would be alright.  As far as surroundings go, I want to go out at 3 in the morning wandering the beautiful campus at Princeton thinking about math.  I want to hear some birds, not sirens and fire trucks, and the splash of melted snow in potholes as cars run over.  Lol, I'm being a little over the top, but you get the gist of my feeling.  I don't know if I find cities such a drawing point for me.  Sorry I sound so biased towards princeton, I am reacting against a lot of pressure from people I know to go to Harvard.  I am open to what anybody has to say.  I particularily appreciated what gauss had to say.  As far as undergraduate focus is concerned, I was almost sure that princeton does a far better job considering that they don't have major graduate schools (well, math at princeton is a different story.)  and just from listening to students talk about profs.

If you're enrolled in either MIT or Harvard can you really take classes at the other one?

There's transfer courses all the time. I have a friend who's enrolling in Harvard for Fall '07, and she will be taking her math courses at MIT for some reason.

Cross-registration between Harvard and MIT is permitted quite routinely, but it can be inconvenient because the two colleges don't run on the same academic term schedules. Still, it can be done, and it makes the math departments of each school stronger, because students can take the best courses from either school to build a math program for themselves, as long as the scheduling works out.

Are intense math majors ever really "mainstream" anywhere?

Some people cite the fact that many graduate courses at Harvard and MIT are really just for undergraduates, while at Princeton graduate courses are often simply too difficult for undergrads.  I don't know how big of a deal that is.  Also, I don't like it when MIT people tell me that their classes are generally "less substantial" than at Harvard or Princeton.  That is to say, you are seriously killing yourself to take more than 5 classes at a place like that (even for people I consider to be ridiculously strong), while at MIT I know several people (undoubtedly also abnormally strong and crazy), who have taken 6, 7, or even 8 classes.

People have also pointed out that Harvard generally does a very good job with courses for non-majors.  That is to say, they don't just give a dumbed down version of the course majors might take, they build a course assuming you are smart and know nothing in that field and get to the "good stuff".

About math department exactly, I hear that Harvard has a weird system.  That their faculty is divided thanks to the way they give tenure.  That is to say that usually when they get a chance for a tenured professor they make a short list of the 5 smartest people in the world and then ask them to come and be tenured immediately.  Thus, I hear it is basically impossible to become tenured if you are actually there.  This creates a divide between professors and guess who the undergraduates get if they get a professor at all?

Another question I like to ask people is what is the purpose of a city?  Do people go in and do stuff all the time? I don't think I would considering it seems highly unlikely a single person could really fully take advantage of all the opportunities at these universities.  Or is it just exciting to be near a city, since it feels busy?  Or do people just really want to be able to get a burrito at a good restaurant at midnight?  I don't really understand the point of a city.  Maybe I'm just not outgoing or social enough.

I hear that princeton's math department is massive, like twice the size of harvard's.  But there are far less math majors at princeton.  Plus Institute of Advanced Studies!  That place has got to be the center of all pure math at this point.  I hear that 37/40 of fields medallist have gone there and there are like 4 there right now.  I know it is not at all officially connected, but wow.  It seems true that Harvard and MIT's departments complement and help a lot, but then again who wants to deal with stupid schedule differences, 30 minutes of travel to get to class and in boston's weather most of the time?

Also, happiness wise, it seemed to me people were actually far happier at princeton.  It definitely feels more nurtureful.  I have been told that the "flexibility" of the math major at Harvard has resulted in people leaving there with a degree and not really being even ready for grad school.  Harvard does a real good job of attracting a lot of olympiad math people there, maybe that should be a big draw for me.  So many smarties, but I have found that some people doing olympiad math do it for the wrong reasons: it makes them feel smarter than others, it will help them get into college, etc.  I mean, I enjoy contest math a lot, but a lot of it is repetition, memory, and is pretty darn useless too, but it does a lot to inflate our egos.  If you go to MOSP you are one of the top 50 students in the nation.  Sometimes it makes me sick.  That's why I actually feel that going Ross was far more valuable than going to MOSP.  Ross kids don't come in feeling real smart, they simply are there to learn and do stuff.  But many are actually ridiculously smart too.  It's just an attitude difference.  In my eyes, MOSP is Harvard and Ross is Princeton.  Call it gut feeling or whatever, but after speaking visiting and talking to a few students, professors, or people who generally know these kind of things, that's the vibe I got.  But I admit, both schools are probably totally stuck up and princeton probably even more so.

People actually leave Harvard, people don't generally leave princeton, their admissions officer told me that like nobody leaves, so they don't have a transfer in system.  This is comforting in knowing that I can always transfer to Harvard (I don't know why they wouldn't accept me after they already accepted me), but not the other way around.

I don't know why I started posting everything I felt about the two schools.  I'll stop now.  I think I kind of resent that Harvard seems to be the default choice by reputation, and it requires strong reasons to not go to Harvard.

I do have a question about undergraduate college in general.  How important is it in the great scheme of things?  At some point, wouldn't your other accomplishments start overshadowing the piece of paper you are given for taking some classes?  Or are people going to judge you forever for the place you go to.  I mean, some people dislike not being able to double major at princeton while they citing the ease of getting a masters at harvard at the same time or getting joint concentrations, the ease of requirements, etc.  But couldn't just take the courses and learn the stuff?  Isn't what you know, not what your degree is?  I have been told that this is simply naive thinking.

Alright, the rant is over.

It seems to me like you really like a lot of things about Princeton and that it fits you well. You should ultimately make your decision based on what [i]you[/i] feel - not on what other people think. There is no absolute right answer to these types of questions. There's a right place and a right school for everyone. Pick the place that makes you feel the most welcome and comfortable, and where you most like the people and the opportunities there. If you do that, you can't go wrong with your choice.

[quote="pdiao"]I do have a question about undergraduate college in general.  How important is it in the great scheme of things?  At some point, wouldn't your other accomplishments start overshadowing the piece of paper you are given for taking some classes?  Or are people going to judge you forever for the place you go to.  I mean, some people dislike not being able to double major at princeton while they citing the ease of getting a masters at harvard at the same time or getting joint concentrations, the ease of requirements, etc.  But couldn't just take the courses and learn the stuff?  Isn't what you know, not what your degree is?  I have been told that this is simply naive thinking. [/quote]

Each has an excellent reputation, and I think anyone would be impressed by seeing that you are a graduate of either one  ;)
The truth be told, reputation is the biggest difference between upper echelon schools and schools which, while excellent, don't have the same name recognition (like University of Michigan, for example)
The difference in reputation between Harvard and Princeton is virtually nil, and what you actually learn is far more important.

[quote="LordoftheMorons"]Hmm.... as the three schools I am going to apply to are MIT, Harvard and Princeton, I will also have this dillema[/quote]Don't jynx yourself.

MIT is great in theoretical stuff...
im very anti applied, and my interests are more toward the arts, philosophy and that kinda stuff and im going to study math at MIT!
by the way, for which univ did u decided?

[quote="Mariag"]it sounded to me that princeton doens't focus all that much on their undergrads. [/quote]

Having been an undergrad at Princeton, I strongly disagree with this statement.  I would say that one of the things that distinguishes Princeton from Harvard/Stanford/MIT is its focus on undergrads.  Princeton doesn't have a business school or a law school or a med school, etc.  Maybe this is different in the arts, but in the sciences, there was a clear focus on the undergrads (and undergraduate life, except the food, which was wretched).

[quote="pdiao"]Thanks everybody.  I'll be going to Princeton next year.  Maybe I'll see some of you there!   It's all really exciting business.[/quote]

Congrats! I saw this thread obviously too late to give any input, but it looks like you made the right decision based on what you've said.

[quote="pdiao"]MIT is one of a couple I ruled out mainly on the basis of two visits I made.  First time I stayed for a week at Random and determined that was too much for me. [/quote]

Mistake number one.

[quote="pdiao"] Plus, the vibe was never one of happiness around MIT, maybe more hyperness in my eyes. [/quote]

Sorry the school actually makes you earn a grade :P jk. See you in Boston next year!

[quote="pdiao"]
I do have a question about undergraduate college in general.  How important is it in the great scheme of things?  At some point, wouldn't your other accomplishments start overshadowing the piece of paper you are given for taking some classes?  Or are people going to judge you forever for the place you go to.  I mean, some people dislike not being able to double major at princeton while they citing the ease of getting a masters at harvard at the same time or getting joint concentrations, the ease of requirements, etc.  But couldn't just take the courses and learn the stuff?  Isn't what you know, not what your degree is?  I have been told that this is simply naive thinking. [/quote]

That's what I think.  So what if you get a 2200 or a 2400 on the sat, or however many degrees from whatever college, so what.  How will the obscure vocabulary you were luckily tested on or the good test day you had that one time make you better than someone who got an similarly impressive, although slightly lower, score?  Tests and degrees aren't a measure of intelligence.   As long as you go to a good school, do well and learn, what's the degree do except add to prestige and "bragging rights"?

I've always found the vocabulary and reading parts demeaning. I like to take my time reading; I have my own type of vocabulary; I like having my own type of style. However, it seems the high and mighty create some type of elitist system. The board was scrutinized once for creating a vocabulary system where the majority of rich people would understand what the word was; the rich partook in such extravagent outdoor activities.

I think a large vocabulary is nice for a change of words in speech; however, some of the greatest writers in time have discouraged it. This would include Strunk, and I believe Devlin had his thoughts on the matter, too.

"Now in order to do his best a man of ordinary intelligence and education should be able to express himself correctly both in speaking and writing, that is, he should be able to convey his thoughts in an intelligent manner which the simplest can understand." - [u]How to Speak and Write Correctly[/u] by Joseph Devlin

Screw the standardized tests; they can go straight to h e double hockey sticks.

The only reason people have to act with a large vocabulary is because the northeast was a very literate place throughout U.S. history. This trend doesn't seem to stop anywhere; therefore, people will act as snooty as they wish to be, as long as they wish to be.

When going to an institution you want to be able to do research and development. I'm tired of all the people that want to have bragging rights.

Pardon the expletives.

Vocabulary is learnable for anyone willing to do enough reading to be ready for a selective college. Thus a vocabulary test is a reasonable element of the admission process for such a college. Mathematics at the university level involves learning a fair amount of specialized vocabulary, so a vocabulary test makes sense for would-be math majors too.


geometry
3D geometry
pyramid
induction
graph theory
advanced fields
advanced fields unsolved

Assume that a face of a convex polyhedron $ P$ has a common edge with every other face. Show that there exists a simple closed polygon that consists of edges of $ P$ and passes through all vertices.

[i]L .Lovasz[/i]

everything starts with a pyramid: if you have the pyramid, the work is done.

if not, choose two linked vertices outside the preferred face (which i'll call the base). if you collapse them, you obtain a polyhedron satisfying the same hypothesis, with less vertices and the same number of edges for the base. now, once that you have a path on the smaller polyhedron, this will pass through the collapsed vertex, and you can lift it to a path on the original polyhedron.
by induction, you have the conclusion.

Nope, the path cannot always be lifted: the path in the collapsed graph may enter and exit the collapsed vertex via the edges incident to the same uncollapsed vertex and then the lifted path will not visit the second uncollapsed vertex.

Strange, it looks like this problem may be solved by some stupid induction.

Consider the planar graph instead polyhedron so that our specific facet is outer polygon $ A_1A_2\dots A_k$. Note that the degree of each $ A_i$ equals 3. Remove the vertices of outer face and edges passing from them. It is easy to see that the rest graph is a tree. It has some vertice $ B$ of degree 1. Let $ B$ be joined with $ A_m$ and $ A_n$. Then the broken line $ A_mBA_n$ divides our polygon onto two parts, in one of which our tree lies. Clearly, the vertex $ B$ is joined with all $ A_i$'s between $ A_i$ and $ A_j$ in the other part. Relace all these vertices, including $ A_i$, $ A_j$ and $ B$ to the new vertice and get a smaller graph with the same structure. It has Hamiltonian cycle, easy to see that hence initial graph also has. Some cases, when $ k$ becomes equal to 1 or 2 after our operation, are analagous.


You have three identical looking fair 4-sided dice with these numbers on the faces:  4, 4, 4, 9.
Your friend has three identical looking fair 4-sided dice with these numbers on the faces:  6, 6, 6, 1.

When you look down at a 4-sided die, the face not showing directly is the face used, because
they are tetrahedral shaped.
When two or more dice are thrown, the numbers on the used faces are added together for the
composite number.

You throw your one die and your friend throws one of his/her own dice, apiece,
likewise for 100 times.  Your friend beats you the majority of rolls this time.

Then you throw all of your three dice, and your friend throws his/her all three dice,
apiece, for 100 times.  Your friend beats you the majority of rolls this time.

You are going to roll your die/dice against the same number of your friend's die/dice 100 times
once more.  Is it possible to beat your friend in the majority of rolls this time by choosing from
the possibilities below?

Choose one:

A) Each throws one die of his/her own
B) Each throws two dice of his/her own
C) Each throws three dice of his/her own

This problem is based on the concept of expected value, or the average (thus, "expected") amount to be rolled in terms of weighted possibilities. In this case, the expected value when rolling one dice each for you and your friend respectively are $ \frac{3(4)\plus{}9}{4}\equal{}5.25$ and $ \frac{3(6)\plus{}1}{4}\equal{}4.75$, so you are expected to win. For some others, due to combinatorial windings and such, it could be otherwise.

And, of course, it is always possible to beat your friend the majority of the time by chance.  :P There's also the argument that identical looking doesn't necessarily mean identical.

This is false, at least for one die being rolled apiece.

You:  4, 4, 4, 9

Friend:  6, 6, 6, 1

As far as rolling one die apiece:
-----------------------------------------

You may match the possibilities of the rolls from you against the rolls
from the friend:

A 6 beats a 4 three times, and there are three sixes.  That equals nine
pairings.  There aren't any other pairings where a friend's number beats
your number in a die versus die event.  This is a 4 by 4 equals 16 pairings.
So, there are nine pairings out of 16 pairings where the friend beats you.

The friend has a $ \frac {9}{16}$ probability of beating you, *despite*

the expected values you quoted, so they must "not be in play here."

"Identical looking" was to signify that all three dice have the same sets of
numbers on their sides for a particular player.

You still have to match the possibilities of the rolls of two and
three dice against each other, respectively, to check probabilities,
or you may have an alternate method.

***This problem is not yet solved in this thread.


function
induction
algebra unsolved
algebra

Find all functions $f: \mathbb R \to \mathbb R$ satisfying \[ f(x^2\plus{}f(y)\minus{}y)\equal{}(f(x))^2\minus{}f(y)\] for all $ x,y \in \mathbb R$.

[quote="SUPERMAN2"]Find all function $ f: R \to R$ satisfying  $ f(x^2 \plus{} f(y) \minus{} y) \equal{} (f(x))^2 \minus{} f(y)$ $ \forall$ $ x,y \in R$[/quote]

Let $ P(x,y)$ be the assertion $ f(x^2\plus{}f(y)\minus{}y)\equal{}f(x)^2\minus{}f(y)$

1) Suppose $ \exists u$ such that $ f(u)\equal{}0$
$ P(u,u)$ $ \implies$ $ f(u^2\minus{}u)\equal{}0$ and $ P(\minus{}u,u)$ $ \implies$ $ f(u^2\minus{}u)\equal{}f(\minus{}u)^2$ and so $ f(\minus{}u)\equal{}0$
$ P(0,u)$ $ \implies$ $ f(\minus{}u)\equal{}f(0)^2$ and so $ f(0)\equal{}0$
$ P(x,0)$ $ \implies$ $ f(x^2)\equal{}f(x)^2$ and so $ f(x)\geq 0$ $ \forall x\geq 0$
Comparing $ P(x,y)$ and $ P(\minus{}x,y)$, we get that $ f(\minus{}x)\equal{}\pm  f(x)$

Then $ P(0,x)$ $ \implies$ $ f(f(x)\minus{}x))\equal{}\minus{}f(x)$. So $ x\geq 0$ would imply $ \minus{}f(x)\leq 0$ and so $ f(x)\minus{}x\leq 0$ and so $ x\minus{}f(x)\geq 0$ and so $ f(x\minus{}f(x))\equal{}\pm f(f(x)\minus{}x)\equal{}\pm f(x)\equal{}f(x)$ (since both $ f(x\minus{}f(x))$ and $ f(x)$ are $ \geq 0$)
So we get $ x\geq 0$ $ \implies$ $ x\minus{}f(x)\geq 0$ and $ f(x\minus{}f(x))\equal{}f(x)$
Applying this to $ x\minus{}f(x)$, we get $ x\minus{}f(x)\minus{}f(x\minus{}f(x))\equal{}x\minus{}2f(x)\geq 0$ and $ f(x\minus{}2f(x))\equal{}f(x\minus{}f(x)\minus{}f(x\minus{}f(x)))\equal{}f(x\minus{}f(x))\equal{}f(x)$ and so $ x\geq 0$ $ \implies$ $ x\minus{}2f(x)\geq 0$ and $ f(x\minus{}2f(x))\equal{}f(x)$
An immediate induction shows that $ x\geq 0$ $ \implies$ $ x\minus{}nf(x)\geq 0$ and $ f(x\minus{}nf(x))\equal{}f(x)$ $ \forall n\in\mathbb N$
And $ x\geq nf(x)$ $ \forall n$ obviously implies $ f(x)\equal{}0$ $ \forall x\geq 0$ and so (since $ f(\minus{}x)\equal{}\pm f(x)$), $ f(x)\equal{}0$ $ \forall x$

And it is immediate to check back that this function indeed is a solution.

2) Suppose $ f(x)\neq 0$ $ \forall x$
If $ \exists u$ such that $ f(\minus{}u)\equal{}\minus{}f(u)$, then Wlog say $ f(u)\leq 0$ (else swap $ u$ and $ \minus{}u$). Then $ P(\sqrt{\minus{}f(u)},u)$ $ \implies$ $ f(\minus{}u)\equal{}f(\sqrt{\minus{}f(u)})^2\minus{}f(u)$ and so $ f(\sqrt{\minus{}f(u)})\equal{}0$, which is impossible in this paragraph 2.

So $ f(\minus{}x)\equal{}f(x)$ $ \forall x$
Suppose now $ \exists u$ such that $ f(u)<0$. Then $ P(\sqrt{\minus{}f(u)},u)$ $ \implies$ $ f(\minus{}u)\equal{}f(\sqrt{\minus{}f(u)})^2\minus{}f(u)$ and so $ 2f(u)\equal{}f(\sqrt{\minus{}f(u)})^2$, which is impossible, since $ LHS<0$ while $ RHS>0$
So $ f(x)>0$ $ \forall x$
As a consequence, looking at RHS of $ P(x,y)$, we get $ f(x)^2>f(y)$ $ \forall x,y$ and so $ f(x)$ is bounded and $ \exists M$ such that $ 0<f(x)<M$ $ \forall x$
Let then $ y>M$, and so $ y>f(y)$ and $ x\geq 0$ : $ x\plus{}y\minus{}f(y)\geq 0$ and so :
$ P(\sqrt{x\plus{}y\minus{}f(y)},y)$ $ \implies$ $ f(x)\equal{}f(\sqrt{x\plus{}y\minus{}f(y)})^2\minus{}f(y)$
$ P(\sqrt{x\plus{}y\minus{}f(y)},\minus{}y)$ $ \implies$ $ f(x\plus{}2y)\equal{}f(\sqrt{x\plus{}y\minus{}f(y)})^2\minus{}f(y)$
And so $ f(x\plus{}2y)\equal{}f(x)$ $ \forall x\geq 0$, $ \forall y>M$

And so $ f(x)\equal{}a$ is a constant.
Plugging back in the original equation, we get $ a\equal{}a^2\minus{}a$ and so $ a\equal{}2$ (since $ f(x)\neq 0$ in this paragraph).

3) synthesis of solutions :
We got only two solutions :
$ f(x)\equal{}0$ $ \forall x$
$ f(x)\equal{}2$ $ \forall x$

[quote="nguyenvuthanhha"][i]Nice , actually , you have forgotten that you did that problem recently

 And you have solved one hard problem in two distinct ways [/i] :lol:[/quote]

Ohhh, you're right! you submitted this problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=286774]Here[/url].

I forgot  :lol:

And, btw, the two solutions I gave are not very different.


Hello,

Can someone here suggest an easy way to count words in a PDF file created in TeXnic Center under Windows XP?  What I could get off Google is I'd have to download ActivePearl then download a Perl script.  Or would it be easier to create a DVI file and count the words.

Is there a simpler method?

Thank you.

[quote="stevem"][url=http://www.winedt.com/]WinEdt[/url] has a word count feature. It is shareware but has a 31 day free trial.[/quote]

Hello.  Thanks for replying.  I had just found this site:

http://folk.uio.no/einarro/Services/texcount.html

I just did a "browse", found the file on my machine, pressed "submit" and it reports the document and color-codes the words it's counting and not counting.  It skips all the math code but I think there are options to control what is and isn't being counting.


Let us name the graphs of order n with the following property "almost perfect" graphs(This is just my terminology!):

All but two vertices have distinct degree,i.e.,only two vertices have the same number of degrees. Besides, multiple edges and loops are not allowed. For instance, a tree with 3 vertices, or a connected graph with just two vertices and an edge satisfy these properties. Now the question is :

Can we describe "almost perfect" graphs in such a way that they have distinctive properties and special relations?

Every "almost perfect" graph $G$ on $n+1$ vertices ($n\ge 2$) either has a vertex with degree $n$ or a vertex with degree zero. If $x\in G$ has degree zero or $n$, then $G-x$ (the graph obtained from $G$ by removing the vertex $x$ and all edges attached to it) is also almost perfect; $y$ and $z$ still have distinct degrees in $G-x$ if they had distinct degrees in $G$. Since there are no "perfect" graphs with $n$ distinct degrees on $n$ vertices, $G-x$ has no vertex of degree zero if $x$ had degree zero, and $G-x$ has no vertex of degree $n$ if $x$ had degree $n$.

It is easy to see that the complement $G^{*}$ of an almost perfect graph $G$ (each edge is removed, and each nonedge is made into an edge) is an almost perfect graph. The above argument gives us a reduction procedure; given an almost perfect graph $G_{n}$ on $n+1$ vertices, we construct a sequence of almost perfect graphs $G_{i}$ as follows:
For each $i$, $N\ge i> 1$, let $G_{i-1}=(G_{i}-x_{i})^{*}$, where $x_{i}$ has degree either $i$ or $0$ in $G_{i}$. If $G_{n}$ has a vertex of degree $n$, all of the $x_{i}$ have degree $i$, and if $G_{n}$ has a vertex of degree zero, all $x_{i}$ have degree zero.

This deterministic process is reversible: $G_{i}$ is obtained from $G_{i-1}$ by taking $(G_{i-1})^{*}$ and then adding a new vertex $x_{i}$, with all edges from $x_{i}$ if $G_{i-1}$ had a vertex of degree $i-1$, and no new edges if $G_{i-1}$ had a vertex of degree zero. Any two graphs obtained this way are isomorphic.

There are exactly two almost perfect graphs on two vertices, one of each type. By this argument, there are exactly two almost perfect graphs (up to isomorphism) on $n+1$ vertices for
each $n\ge 2$, which are complements of each other.
They're pretty easy to construct, so we can find out anything we want to about them. An explict construction:
Let the vertices be $v_{0},v_{1},\dots,v_{n}$. Connect $v_{i}$ and $v_{j}$ ($i\neq j$) with an edge whenever $\max\{i,j\}$ is even (for the complement, use odd in place of even).

Ooops, thanks for this detailed answer. I had a guess about the duality of "almost perfect graphs" on n vertices but now i have learned why it is indeed correct. By the way, i name these graphs "almost perfect" but do they have a commonly-accepted name?


calculus
integration
function
real analysis
real analysis unsolved

Anyone have any idea how to approach this one? Integration by parts leads to to complicated integrals of the error function:

$ \int_0^\infty\frac{re^\frac{\minus{}r^2}{2k^2}}{r^f\minus{}b}\,dr$  $ k,f,b$ constant reals

If $ f>0$ and $ b>0,$ then the integral diverges - note that the denominator goes to zero at a point in the domain.

Even when it converges, there is no particular reason for the answer to be nice. When $ b\equal{}0$, say, we get the Gamma-function, which is nothing but just a different name for the very same integral.

What exactly do you want from it? :?

[quote="transseries"]
$ G_{i,j}^{n,m}$ is the Meijer $ G$ function, $ Ei(a,x)$ is the exponential integral function
[/quote]
Names, names, names... :)

Maple you say. I should really learn to use that ^_^

Man those functions look ugly - these are supposed to go into a computer code, and it is going to run slow as all hell with those in it. Course, three cases probably won't do it either :P Simpson's rule + lookup table it is.

Thanks guys.

Actually, I have more to ask

Googling around found me this article from wikipedia: http://en.wikipedia.org/wiki/Weierstrass_transform#Transforms_of_some_important_functions

Now, that says that this transform will give me an analytic approximation of my function, which is fine - I don't need a precise solution as long as I can fine tune it to be arbitratily good. Unfortunately, this will approximate the function I want, but not the integral of the function I want. However, this line in the article:

[quote]If f is integrable over the whole real axis (i.e. f ∈ L1(R)), then so is its Weierstrass transform F, and if furthermore f(x) ≥ 0 for all x, then also F(x) ≥ 0 for all x and the integrals of f and F are equal. [/quote]

seems to imply that integrating the approximation i get for f will get me an approximation for its integral. Can someone with better knowledge of this transform verify this for me? Because the transform is relatively easy to integrate since it is over y instead of r.

You seem to be quite confused here. The Weierstrass transform has nothing to do with your problem.


Suppose that a,b,x are positive integers such that:
x^(a+b)=a^b.b
Show that a=x and b=x^x


$ a,b,c > 0$,find the best $ k$ such that
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\geq k\left(\frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\right)\]

[quote="Tourish"]$ a,b,c > 0$,find the best $ k$ such that
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq k\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\]
[/quote]

when k=1
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} \left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\]
$ \equal{} \frac {a^4 c^2 \plus{} a^3 (b \plus{} c) (b^2 \plus{} c^2) \plus{} a^2 b (b^3 \plus{} b^2 c \plus{} c^3) \plus{} a b^2 c^2 (b \plus{} c) \plus{} b^2 c^3 (b \plus{} c)}{abc(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
$ > 0$

[quote="kuing"][quote="Tourish"]$ a,b,c > 0$,find the best $ k$ such that
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq k\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\]
[/quote]

when k=1
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} \left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\]
$ \equal{} \frac {a^4 c^2 \plus{} a^3 (b \plus{} c) (b^2 \plus{} c^2) \plus{} a^2 b (b^3 \plus{} b^2 c \plus{} c^3) \plus{} a b^2 c^2 (b \plus{} c) \plus{} b^2 c^3 (b \plus{} c)}{abc(a \plus{} b)(b \plus{} c)(c \minus{} a)}$
$ > 0$[/quote]

$ \frac {a^4 c^2 \plus{} a^3 (b \plus{} c) (b^2 \plus{} c^2) \plus{} a^2 b (b^3 \plus{} b^2 c \plus{} c^3) \plus{} a b^2 c^2 (b \plus{} c) \plus{} b^2 c^3 (b \plus{} c)}{abc(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

$ \frac {a}{b} > \frac {a}{b \plus{} c}$

$ \frac {b}{c} > \frac {b}{c \plus{} a}$

$ \frac {c}{a} > \frac {c}{a \plus{} b}$

Therefore  $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}>\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}$

[quote="hedeng123"][quote="kuing"][quote="Tourish"]$ a,b,c > 0$,find the best $ k$ such that
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq k\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\]
[/quote]

when k=1
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} \left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)\]
$ \equal{} \frac {a^4 c^2 \plus{} a^3 (b \plus{} c) (b^2 \plus{} c^2) \plus{} a^2 b (b^3 \plus{} b^2 c \plus{} c^3) \plus{} a b^2 c^2 (b \plus{} c) \plus{} b^2 c^3 (b \plus{} c)}{abc(a \plus{} b)(b \plus{} c)(c \minus{} a)}$
$ > 0$[/quote]

$ \frac {a^4 c^2 \plus{} a^3 (b \plus{} c) (b^2 \plus{} c^2) \plus{} a^2 b (b^3 \plus{} b^2 c \plus{} c^3) \plus{} a b^2 c^2 (b \plus{} c) \plus{} b^2 c^3 (b \plus{} c)}{abc(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

$ \frac {a}{b} > \frac {a}{b \plus{} c}$

$ \frac {b}{c} > \frac {b}{c \plus{} a}$

$ \frac {c}{a} > \frac {c}{a \plus{} b}$

Therefore  $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}$[/quote]

 :D

I just uesing compute ... no think of ... hai da cuo fu hao , han ... :blush:

but max of k is 1, how to prove it ?

is need to find min of this:
$ \frac{(1\plus{}t) (1\plus{}u) (t\plus{}u) \left(u\plus{}t \left(t\plus{}u^2\right)\right)}{t u \left(1\plus{}t \left(1\plus{}t\plus{}t^2\right)\plus{}u\plus{}t (3\plus{}t) u\plus{}(1\plus{}t) u^2\plus{}u^3\right)}$
where $ t,u>0$.

You can easily check (no computer needed :D) that for small $ \epsilon>0$, it is false for $ k=1+2\epsilon$ and ${ (a,b,c)=(2,\epsilon^2},\epsilon^3)$  :lol:

[quote="Tourish"][quote="spanferkel"]You can easily check (no computer needed :D) that for small $ \epsilon > 0$, it is false for $ k = 1 + 2\epsilon$ and ${ (a,b,c) = (2,\epsilon^2},\epsilon^3)$  :lol:[/quote]

 :lol: That's exactly what I want to see.[/quote]

me too... :D

[quote="aadil"]i think k=2 is the maximum[/quote]

I think you're certainly wrong.

[quote]aadil wrote:
i think k=2 is the maximum

I think you're certainly wrong.[/quote]
please tell me why.

:rotfl: if there is a counterexample for all $ k \equal{} 1 \plus{} 2\epsilon$, $ \epsilon > 0$ small, I assure you that it is also wrong for $ \epsilon \equal{} \frac12$. By the way, the one above is still a counterexample if you put $ \epsilon \equal{} \frac12$. Try it.

Now it is almost trivial (see hedeng123's proof) that the ineq holds for $ k \equal{} 1$, i.e. $ \epsilon \equal{} 0$. So that's all.


geometry
incenter
angle bisector
geometry proposed

given the equilateral triangle XYZ is inscribed in the circle (O).Let P is an arbitrary point inside the triangle which is not on the sides .PX,PY,PZ cuts (O) at A,B,C,respectively.Let D,E,F is the center of the inscribed circle of the triangle PBC,PCA,PAB.Prove that AD,BE,CF are concurrent
The similar problem in solid geometry is also true

It was appeared in my shool TST. The solution can be found at [url=http://forum.mathscope.org/showthread.php?t=10272]here[/url]

[b]Lemma:[/b] Let $ U$ be an isodynamic point of $ \triangle ABC$ and $ I,A',B',C'$  the incenters of $ \triangle ABC,\triangle UBC,\triangle UCA$ and $ \triangle UAB.$ Then lines $ UI,AA',BB',CC'$ concur.

By isodynamic point property $ AU \cdot BC \equal{} BU \cdot CA \equal{} CU \cdot AB \ (\star).$ This is, angle bisectors of $ \angle BUC,$ $\angle CUA,$ $\angle AUB$ pass through the feet of the angle bisectors of $ \angle BAC,\angle ABC ,\angle BCA.$ Thus, let $ V$ be the common foot of the inner angle bisectors of $ \angle BAC,\angle BUC$ and let $ X \equiv UI \cap AA'.$ By Menelaus' theorem for $ \triangle UIV$ cut by the transversal $AXA',$ keeping in mind the relation $ (\star),$ we have

$ \frac {XI}{UX} \equal{} \frac {AI}{AV} \cdot \frac {VA'}{A'U} \equal{} \frac {AB \plus{} AC}{AB \plus{} BC \plus{} CA} \cdot \frac {BC}{UB \plus{} UC}  \Longrightarrow$

$\frac {XI}{UX} \equal{} \frac {AC}{UC} \cdot \frac {BC}{AB \plus{} BC \plus{} CA} \ (1)$

Similarly, if $ X' \equiv BB' \cap UI,$ by Menelaus' theorem for the triangle formed by $ U,I$ and the foot of the B-angle bisector, cut by the transversal $ BX'B',$ we get the relation

$ \frac {X'I}{UX'} \equal{} \frac {BC \plus{} AB}{AB \plus{} BC \plus{} CA} \cdot \frac {AC}{UA \plus{} UC} \equal{} \frac {AC}{UC} \cdot \frac {BC}{AB \plus{} BC \plus{} CA} \ (2)$

From $ (1)$ and $ (2)$ it follows that $ X \equiv X'$

$ \Longrightarrow$ lines $ AA',BB',UI$ concur at $ X.$ Analogously, we have $ X \in CC'.$
____________________________________________

Back to the problem, since the circumcevian triangle $ \triangle XYZ$ of  $ P$ with respect to $ \triangle ABC$ is equilateral, then $ P$ becomes an isodynamic point of $ \triangle ABC.$ From the previous lemma, we get that $ AD,BE,CF$ concur and furthermore the concurrency point lies on the line connecting $ P$ with the incenter $ I$ of $ \triangle ABC.$

Is there an inversive proof to this? Perhaps invert wrt to P which is an isodynamic point of ABC and then ABC turns into an equilateral triangle...


function
induction
modular arithmetic
number theory unsolved
number theory

Let $ f: Z--> Z $ be a fuction satisfying the conditions :
 $ f(0) =1 ; f(f(x))= x+4f(x) \forall x\in Z $ .
 Find all natural numbers $ n \geq 1 $ such that $ f_n(0) $ is divisible by $ 20^{11} $ , where $ f_1(x) = f(x) ; f_n(x)=f(f_{n-1}(x)) $ .

[quote="peeta"]i dont know if it helps anyone but i can prove:
$f_n(0)=f_{n-2}(0)+4f_{n-1}(0)=\frac{1}{2\sqrt5}\left((2+\sqrt5)^n-(2-\sqrt5)^n\right)$
with $f_0(0)=0$ and $f_1(0)=1$[/quote]
  I think this one can't help to solve this problem

This is a fairly straightforward qn requiring no special trick, but rather uses only induction repeatedly. I shall just give the outline of the proof.

For simplicity sake of my typing, we shall let $a_n = f_n(0)$. We shall tackle this problem by dividing the problem into two parts, finding out for which $n$ does $2^{22}$ divide $a_n$, and for which $n$ does $5^{11}$ divide $a_n$.

First, prove by induction that  $a_{2k} \equiv 0 \pmod{4} \forall k \in \mathbin{N}$. Next, prove by induction that $a_{2k-1} \equiv 1 \pmod{16} \forall k \in \mathbin{N}$. Using these two lemmas, prove by induction that $2^m \mid \mid k$ iff $2^{m+2} \mid \mid a_{2k}$. This implies that for $2^{22}$ to divide $a_{2k}$, we must have $2^{20} \mid k$.

For the second part of the problem, we prove by induction that if $k= 5^qr, \gcd (5, r) = 1$, then $a_{2k} \equiv 5^q \pmod{5^{q+1}}$. Here, the formula given by [i]peeta[/i] would be useful. This implies that for $5^{11}$ to divide $a_{2k}$, we must have $5^{11} \mid k$.

$\therefore 20^{11} \mid f_n (0)$ iff $2^{21}5^{11} \mid n$.

Perhaps there might be typo or calculation errors, but the main idea is there.


calculus
integration
inequalities
calculus computations

Prove that $ \int_0^1(1 \minus{} x^2)^n\ge\frac1{n \plus{} 1}.$

What's the simplest way to prove this?

Ah, I was makin' it too complicated, I thought on $ \frac1{n \plus{} 1} \equal{} \int_0^1 x^n\,dx.$

Thanks for your answer.


superior algebra
superior algebra solved

Let  $(G,\ast)$ be finite group and $S,T<G$ then $card(ST)=\frac{card(S)card(T)}{card(S\cap T)}$.
where $ST=\{ s\ast t : s\in S , t\in T \}$ and $card(A)=$ numbers of element of $A$.

$ST$ is a set. One can define a relation $x R y$ if $x^{-1}y \in S$. This relation induces a partition of $ST$, and every class $aS$ has cardinal $Card(S)$. One can define a group action  $f: T \times \frac{ST}{S} \to  \frac{ST}{S}$ on this set: $f(t,aS)=(ta)S$. Clearly there is only one orbit, and the cardinal of the orbit is $\frac{Card(T)}{Card(Stab(aS))}$ for any class $aS$. But $Stab(S)=T \cap S$.

We can count the number of repetitions of each element. Suppose $u \in S \cap T$, then $st = (su)(u^{-1}t)$, so there is at least $o(S \cap T)$ repetitions for each element. Howver, if $st = s't'$, then $s'^{-1}s = t't^{-1}$, so $s'^{-1}s = t't^{-1} = u \in S \cap T$, giving $s = s'u$ and $t = u^{-1}t'$, so there are exactly $o(S \cap T)$ repetitions for each element. Then the result follows.


number theory
relatively prime
superior algebra
superior algebra unsolved

One can show that a finite set G closed under an associative product for which both cancellation laws hold is a group. For example, one can use this to show that the integers relatively prime to n modn is a group under modn multiplication.
   However, my question is, can anyone give me an example where if only one cancellation holds then G may not be group. Clearly for such a finite G, the product defined, can not be commutative for if it were then if one cancellation law held the other would also. Is there an obvious example???
This question was taken from Herstein's Topics In Algebra.
Currently I am trying to solve every exercise in the text.


i was reading my counting and probability book preparing for the AMC's and ran across something that didnt seem right.  The book stated in a blue box or something that $ 0! \equal{} 1$ because 0 books can be arranged in 0 slots or spaces in one way, doing nothing.  This didnt make sense to me.  Could someone explain this concept or tell me that the book is just wrong.  Thanks

There's only 1 way to arrange 0 items, and that way is to do nothing.
 ;-)

so your saying that the books right and somehow $ 0! \equal{} 1$ even though $ 0 \times$ anyting is 0???

Well normally for a factorial you multiply the integers until you go down to 1... in this case 0 < 1. So you have to step away from math and use your common sense  :D

Cause with 0 books and 0 slots, there's nothing to do.  "nothing" is one way of doing "something".  :rotfl:

@ZhangPeijin - But using $ P(0,0)$ to prove $ 0!\equal{}1$ is circular reasoning, and thus invalid.

EDIT: what the guy above said

[quote="ZhangPeijin"]Yes because $ X!$ is actually $ P(X,X)$[/quote]

Peijin is correct because we use P to determine how many ways to arrange items in slots etc. As other people said, 1 way to arrange nothing into nothing.

Anyways, plug in 0 ! in a calculator. you get 1

Using this reasoning we get...
$ 0!\equal{}P(0,0)$
$ P(0,0)\equal{}\frac{0!}{(0\minus{}0)!}\equal{}\frac{0!}{0!}$
$ x^0\equal{}\frac{x}{x}\equal{}1$
Therefore,
$ \frac{0!}{0!}\equal{}1$
Since $ \frac{0}{0}\neq1$, $ 0!\neq0$
But then of course my reasoning is completely wrong because $ \frac{0}{0}$ is undefined.

Here's another way: for 0!, you aren't multiplying 0 times anything (like you would multiply 5*4*3*2*1 to get 5!), you just don't multiply anything in the first place. If you have an equation like 2x=2 and you divide by 2 to find x, on the right side the 2s cancel so you aren't multiplying anything, so that's what 0! must be, and you know it's 1 since 2/2=1. Sorry if my method makes no sense whatsoever, it might be easier to try putting 0 books on a bookshelf and see what happens. :oops:

[quote="ZhangPeijin"]Using this reasoning we get...
$ 0! \equal{} P(0,0)$
$ P(0,0) \equal{} \frac {0!}{(0 \minus{} 0)!} \equal{} \frac {0!}{0!}$
$ x^0 \equal{} \frac {x}{x} \equal{} 1$
Therefore,
$ \frac {0!}{0!} \equal{} 1$
Since $ \frac {0}{0}\neq1$, $ 0!\neq0$
But then of course my reasoning is completely wrong because $ \frac {0}{0}$ is undefined.[/quote]

For some reason, I feel like this argument uses circular reasoning

If you have nothing, you can only do nothing, because factorials often have to do with probability.  For example, if you have 0 pieces of candy, and 5 different wrappers that can go with any piece, you can do nothing, but that is an option.  That is why

0! = 1

[quote="stevenmeow"]There is a recursive formula for factorials. if u know the (n-1)th factorial, then the nth factorial is n*(n-1)!
or n!=n*(n-1)!
we could work backwards to find 0!

nor n=1, we know that 1! is 1
then substitute into the above equation...
1!=1*(1-1)!
1=1*0!
1=0![/quote]
no1 is commenting on my solution. I really hoped that this would help but its like evry1 has been skipping this solution and stating over and over that there is 1 way to arrange 0 pieces. maybe if i quote myself ppl will acknowledge it.

Jeez. Just wait a bit mr. meow. So impatient.  Yes that is most likely the best solution simply using the definition of factorials.  Same for exponents if you wanted to prove $ x^0 \equal{} 1$ you use the fact that $ x^{y \minus{} z} \equal{} \frac {x^y}{x^z}$.

But can we stop discussing this.  By pure logic, the answer has to be 1.  It's simply how many ways we can permute 0 objects.


calculus
inequalities proposed
inequalities

Let $a_i$ positive reals for $i=1,2,..,n$ and $m,p$ positive reals such that $m \geq p$. Prove that

$ (a_1^m+...+a_n^m)^p  \leq (a_1^p+...+a_n^p)^m$

A proof without calculus, please  :)

I think I found something nice. Assume that $ a_n$ is the greatest among them. Then write that $\sqrt [m]{a_1^m+...+a_n^m}=a_n\cdot\sqrt [m]{ (\frac{a_1}{a_n})^m+...+1}\leq a_n\cdot\sqrt[p]{ (\frac{a_1}{a_m})^m+...+1}\leq\ a_n\cdot\sqrt[p] {(\frac{a_1}{a_n})^p+...+1}$ which is what we wanted. I just use elementary properties of the exponential and sqrt functions.

Fantastic solution, Harazi.

Thank you very much. :)

We can assume that:
$a_1^p+a_2^p+...+a_n^p=1$
We need to prove:
$a_1^m+a_2^m+...+a_n^m \le 1$
true because $a_i \le 1$ then $a_i^m \le a_i^p$

This proof is really extraordinary short and nice as uses the homogeneity of the ineq.

Good job !!!! :)


what is the ans to this
In how many positive four-digit integers that are not multiples of 1111 do the digits form an arithmetic sequence?

We're not going to answer here so you can get your Alcumus score up.  After you answer incorrectly twice and are on the solution page, use the "Ask on Forum" button if you don't understand the solution.


modular arithmetic
geometry
3D geometry
number theory proposed
number theory

$1$.Prove that $y^{2}=x^{3}-73$ has no solutions for $x,y$ integers.
$2$.If $m$ is an integer which does not have a prime divisor in form $3(mod\ 4)$and $n$ is an arbitrary integer, Prove that the equation  $y^{2}+4m^{2}=x^{3}+(4n-1)^{3}$ does not have solution for $x,y$ integers.

[quote="doicu"]hallo,
to the first one, take $\bmod{7}$, i think .  :)[/quote]
bravo doicu

[quote="doicu"]hallo,
to the first one, take $\bmod{7}$, i think .  :)[/quote]

Dude, it's not supposed to be an easy problemo i think :wink:
do you think $\mod\ 7$ is enough?!!!

I do not understand how mod 7 yields a solution. What if $y \equiv 2 \pmod{7}$ and $x \equiv 0 \pmod{7}$. The equation becomes $4 \equiv-3 \pmod{7}$ which is not a contradiction...

Well mod 7 $y^{2}\equiv x^{3}+4$ so cubes mod 7 are 1,1,-1,1,-1,-1 which is really nice. Thus $y^{2}\equiv \pm1+4 =3,5$ neither of which are squares.

I don't know why I'm bumping a lot of topics today but oh well... it wasn't answered :wink:

Any ideas on q.2?
To make it useful, here's my proposed question:
Solve \[y^{2}=x^{3}+73\]

[quote="me@home"]Solve
\[y^{2}=x^{3}+73 \]
[/quote]

 $(3,10)$ is a solution. so the equation has solution(s)! this is all i have. :wink:

I know that why I posted it :wink: especially if you try the mod 7 trick like before, you find that not a single soluiton is screened out!! :P


Suppose that $A$ and $B$ are real matrices such that $A^t=A$, $v^tAv \geq 0$ for all $v\in\mathbb{R}^n$ and $AB+BA=0$.Show that $AB=BA=0$ and give an example where neither $A$ nor $B$ is zero

Examples are not difficult to give (choose A diagonal with nonnegative entries). For the more difficult part, notice that A is unitary diagonalisable, thus we can find unitary $U$ and $D$ with non-diagonal entries zero and all elements on the diagonal nonnegative such that $A=U^{-1}DU$.Put $ X=UBU^{-1}$. The condition can also be written $ DX+XD=0$. Thus $ (d_i+d_j)x_{ij}=0$ from where $ d_i x_{ij}=0=d_j x_{ij}$. This can be written $DX=XD=0$ and from here $ AB=BA=0$.


calculus
integration
function
calculus computations

The average value of function f(x,y,z) over solid S is defined:

$ \frac {\int \int \int f(x,y,z)dV}{\int \int \int dV} $

Find the avg. value of $ f(x,y,z) = 24xy^2z^3 $ over the solid S in the first octant bounded by the planes x = 0, y = 0 , z = 0, and 3x + y + z = 9.  Sketch the solid "S".

$\iiint_\Omega f(x,y,z)dxdydz = \int_0^\frac{1}{3}dx \int_0^{9-3x}dy \int_0^{9-3x-y} f(x,y,z)dz$.

I think the limits on that first integral should be from 0 to 3.


Let $a, b, c$ be non-negative reals. Show that for all real $k$,

\[
\sum a^k(a-b)(a-c) \ge \sum \frac{\mbox{min}(a^k,b^k)(a-b)^2}{2}
\]

(where $a, b, c \ne 0$ if $k \le 0$) and determine where equality holds for 1) $k>0$, 2) $k=0$, and 3) $k < 0$.

WLOG assume $a^k \ge b^k \ge c^k$, then we can write the inequality as
$2(a^k-b^k)(a-b)(a-c)+(b^k-c^k)(a-b)^2 \ge 0$
So it's done, because a-b and a-c are of the same sign.

Exactly. Equality holds for $k>0$ if and only if the two largest variables are equal, identically for $k=0$, and for $k<0$ if and only if the two smallest variables are equal.

Actually, I believe there is still more! Prove the related upper bound (for the same $a,b,c,k$):

$\sum \frac{\mbox{max}(a^k,b^k)(a-b)^2}{2} \ge \sum a^k(a-b)(a-c)$

When does equality hold?


James wants to make a 25 percent fruit juice drink by adding fruit juice to a 2 liter mixture of 10 percent fruit juice. How many liters of fruit juice does he he have to add?

[hide]There are $\frac1{10}\cdot2=\frac15$ liters of fruit juice to start.  Let $x$ be the amount of fruit juice added.  Thus, we have the equation:

$\frac{\frac15+x}{2+x}=\frac14$

$\frac45+4x=2+x$

$3x=\frac65$

$x=\frac25$[/hide]

[quote="jigger"]$x+\frac{1}{10}2000 = \frac{1}{4}2000$

$x+200 = 500$

$x = 300$

 So, he needs to add 300 mL or 0.3 L of Fruit Juice to make his mixture contain 25% Fruit Juice[/quote]

You have to remember that the total amount of liquid also increases when you add more fruit juice, so you will need to account for that also  :wink:


ratio geometry trigonometry algebra polynomial Vieta conics

Does anyone know what kinds of theorems come up alot on the AMCs?  for example, the angle bisector theorem, simon's favorite factoring trick, modulos are quite useful, whenever the problem asks for ratios look for similar triangels or look to see where you can draw a parallel line.  If one angle is twice another angle, a suggestion would be to bisect the bigger angle so you can use the angle besector theorem.

are there any other things I should look out for?  or anything else I should add to my list?

I am still trying to be more well-rounded, even though I'm not at all.  :blush:

The question concept isn't as hard, but how to apply it in different types of questions and when is a little bit more difficult to me.

Hmm it's mostly like:

Combinatorics: Most basic combinatorics/counting/probability stuff appears on here, sometimes slightly difficult counting problems involving casework or correspondences, also various forms of arithmetic/geometric sequences (though this is sort of algebra) and binomials appear

Geometry: Make sure you know trig, and law of sines and law of cosines come up all over the place. Other than that it's mostly similar triangles and most basic geometry facts.

Algebra: Much of math is related to algebra, know about polynomials and polynomial roots in detail (including Vieta), complex number facts are very important, manipulating equations and expressions tends to show up, also logs, basic conic sections, and graphs of functions often come up. Also maybe just maybe there might be a simple AM-GM or something, but most of the inequalities involve simple algebra.

Number Theory:  Definitely know modular arithmetic, know important prime number facts (like number of factors, inverses mod a prime, etc), and base number problems may come up.

Just keep in mind that these "tricks" you see won't really help you as you advance to higher math contests, although these are most of the concepts that often show up on the AMC.

Yeah I would definitely practice combinatorics. If you look at the solution manual to past AMC's you will notice that on probability and enumeration problems, the manual typically offers two solutions: one that is long, drawn-out, and involves overcounting and subtraction, and another (much shorter one) that uses combinatorial aruments. Being comfortable with the latter can save a significant amount of time toward the end of the test.


[quote="GameBot"]You have 50 dimes and 20 quarters. What percent of the value of your money is in quarters?[/quote]

The total amount of money, in cents, is $ 50 \times 10 \plus{} 20 \times 25 \equal{} 1000$. Of these 1000 cents, $ 20 \times 25 \equal{} 500$ is of quarters, so our answer is $ 500 \div 1000 \equal{} 50$ percent.


A convex quadrilateral $ ABCD$ is inscribed in a circle $ (O)$.Let $ P\equal{}AC\cap BD$.The circumcircles of $ \triangle ABP$ and $ \triangle CDP$ intersect again at $ Q$.Prove that if $ O,P,Q$ are $ 3$ distinct points,then $ OQ \perp PQ$.

It is a very well know fact. Let $ E\equiv AB\cap CD$ and $ F\equiv AD\cap BC$, we have that among three points $ P,E,F$, each is the pole of the line connected the two others with respect to $ (O)$. Thus, the result is followed.

Can you explain a little more? I see how each of P,E,F is the pole of the line connected to the other two but why does that show $ OP \perp OQ$? I can see $ OF \perp PQ$, $ OP \perp EF$


Hi chavez!
Please, read http://www.mathlinks.ro/viewtopic.php?p=140098 . The title "Very Nice Problem" is out of law! Try to find something more appropriate.

We are to make some little estimates.
First of all total number of points is $(2x+9)(x+4)$ and we know that $10\mid (2x+9)(x+4)$, so $x=8,16,...$. On the other hand Venezuelan got at least $\frac{x(x-1)}{2}$ points, so $\frac{x(x-1)}{2}\leq \frac{(2x+9)(x+4)}{10}$. It follows $x\leq 8$. Therefore $x=8$ and Venezuelan got 30 points and 28 of them they achived inside their group. Thus they got 2 additional points from Brazilian. It means that the maximum number of points that one Venezuelan player made is 7+2=9.


geometry
circumcircle
symmetry
incenter
geometry proposed

Let $ O$ be a point inside triangle $ ABC$. Draw the circumcircles $ (OBC),(COA),(AOB)$. Deonte D,E,F be the midpoint of the arcs $ BOC,COA,AOB$(which contain O) of the circles $ (OBC),(OCA),(OAB)$. Show that $ O,D,E,F$ are on a circle.

This problem directly follows from one well-known lemma(maybe not so well-known,but anyway,it's very useful :) ):
Lemma:
Let $ \triangle ABC$ be a triangle,and $ O$ is an arbitrary point on a plane.Also let $ P,Q,R$ be a points on circumcirlces of $ \triangle BOC,\triangle AOC,\triangle AOB$,respectively.
Then $ P,Q,R,O$ are concyclic if and only if $ \frac {BP}{PC}\cdot\frac {CQ}{QA}\cdot\frac {RA}{AB} = - 1$.
I remember,that this lemma was discussed here,but i forgot the link,but i still remember its proof... :wink:
[hide="Hint"]Inversion with center at $ O$
[/hide]

Denote by $ O_1$, $ O_2$, $ O_3$ be the centers of the circumcircles of $ \triangle BOC$, $ \triangle COA$, $ \triangle AOB$, respectively. Call $ X$, $ Y$, $ Z$ the symmetry points of $ O$ w.r.t. $ O1$, $ O_2$, $ O_3$, respectively. We have $ A$, $ Y$, $ Z$ are collinear, it's the same for $ B$, $ Z$, $ X$ and $ C$, $ X$, $ Y$. On the other hand, we have $ XD$, $ YE$, $ ZF$ are concurrent at the incenter $ I$ of triangle $ XYZ$, so we have $ \measuredangle IDO \equal{} \measuredangle IEO \equal{} \measuredangle IFO \equal{} 90^{\circ}$. Hence, points $ O$, $ D$, $ E$, $ F$ lie on the circle with diameter $ OI$.

@mr.danh: Please give your posts some meaning titles.

[quote="mr.danh"]Dear [b]Erken[/b],
You introduced one lemma which helps us to solve this problem.
Can you post its proof of lemma ?[/quote]
See a hint,and the solution will become obvious :wink:
I can't write the details right now,i'm at my TST,i'll come back at 24th of May


Two altitudes of a triangle are of length 1 and 2. If h denotes the length of the third altitude, then show that 2/3 < h < 2.

Let $ A$ be twice the area of the triangle.
Then if sides opposite to altitudes of lengths $ 1,2,h$ have lengths $ a,b,c$ respectively then:
$ a\plus{}b>c$ and $ b\plus{}c>a$ so; also $ a\equal{}\frac{S}{1}, b\equal{}\frac{S}{2}, c\equal{}\frac{S}{h}$ so get:
$ \frac{S}{1}\plus{}\frac{S}{2}>\frac{S}{h}$ and $ \frac{S}{h}\plus{}\frac{S}{2}>\frac{S}{1}$ so:
$ h>\frac{2}{3}$ and $ h<2$ and the result follows.


calculus
integration
function
trigonometry
calculus computations

Show:

\[ \int_a^b \frac{1}{\sqrt{\minus{}(x\minus{}a)(x\minus{}b)}}dx\equal{}\pi ;\quad a<b\in\mathbb{R}\]

Anyway, took me some effort to do and I though it was interesting that it's always $ \pi$; probably easy for you guys. :)

Complete the square and it's an arcsin-thingy. In the original proof of the Fatou/F. and M. Riesz theorem on peak sets of functions in the disc algebra, these integrals play a key role.

[quote="shawtoos"]Show:
\[ \int_a^b \frac {1}{\sqrt { \minus{} (x \minus{} a)(x \minus{} b)}}dx \equal{} \pi ;\quad a < b\in\mathbb{R}
\][/quote]
Substitute $ u\equal{}x\minus{}a,$ the integral becomes $ \int_{0}^{b\minus{}a}{\frac{1}{\sqrt{u(b\minus{}a\minus{}u)}}\,du}.$ Let $ u\equal{}\frac{(b\minus{}a)(1\plus{}\lambda )}{2},$ the integral becomes $ 2\int_{0}^{1}{\frac{1}{\sqrt{1\minus{}\lambda ^{2}}}\,d\lambda }\equal{}\pi .\quad\blacksquare$

I also completed the square and routed it through an inverse trig expression; pretty messy though.  Nice clean approach Black. :)

Put $ x \equal{}a \cos^2 \theta \plus{} b \sin^2 \theta$  :roll:


Hi, I need some advice on choosing a rigorous and  challenging text on introductory algebra and geometry. I've done some browsing and found these names Lial/Miller, Foerster, Dolciani, Hall and Knight, and Weeks and Adkins. If anyone has opinions on these texts, preferences, level of difficulty, or knows of other texts worth looking at, please let me know.

Thank you


I was wondering what you are in : band, orchestra, or singers.
If you are in band or orchestra, then could you tell me what instrument you play?

btw, i play trumpet!
yeah! trumpet ftw.

I currently play piano and handbells.

I used to play clarinet.

Piano, Cello, and a little bit of others.

Clarinet the instrument for awesome people.

piano, cello, violin, guitar (but fail at all)

cmon, it all string, perrcusion and wood-wind.

NO BRASS?!?!

I play piano (justed started so I stink) and guitar.

Piano, glockenspiel, any orff instruments pretty much. GO MUSIC STUDIO!

piano, snare drums, bass drums, marimba, xylophones, glockinspiel, trumpet, and many more I can't remember

i want 2 be awesome monkey girl! ! ! !
i uh play tht instrument to. . . . :) just dont ask to hear me play

Recorder.
meh
ya. that's it.
but oh well

HMMMM
oh yeah. level 10 piano...................... and i also play guitar.

I can play the recorder too, k00lperson :D

These levels are like the book level. 1-10........ 1being the lowest. and 10 being the highest. And  yes, I am telling the truth. Ive been taking piano for 7 years.

Well, its not always that accurate, I always played level 10 songs to my level 7 test  :o


[b]Let [/b]$ A,B \in M_{n}\mathbb{(C)}$ , $ k \in N^{*}$
Such that :
            $ Tr(\sum_{i = 1}^{k}{A.X^{i}}) = Tr(\sum_{i = 1}^{k}{B.X^{i}})$ ,$ \forall X \in M_{n}\mathbb{(C)}$
[b]Prove that[/b] : A=B
"[i]Do math to love math , love math must do math ,do math , love math , math is my life !!![/i]"

Let $ (E_{ij})$ be the canonical basis of $ \mathcal{M}_n(\mathbb{C})$; assume that $ Tr(A\sum_{i \equal{} 1}^{k}{X^{i}}) \equal{}0$ for all $ X$. With $ X\equal{}E_{ij}$ and $ i\not\equal{}j$ we find $ Tr(AE_{ij})\equal{}0$ (because $ E_{ij}^2\equal{}0$) or $ a_{ji}\equal{}0$. Thus $ A$ is a diagonal matrix. With $ X\equal{}E_{ii}$ we find $ kTr(AE_{ii})\equal{}0$ or $ a_{ii}\equal{}0$. Thus $ A\equal{}0$.  QED

[b]Thanks you ![/b]
$ E_{ij}$ [b]is nice problem of math !![/b]


function
real analysis
real analysis theorems

what is the use of Green functions for a linear system?


Solve system of equations
$ e_1: x \plus{} y \plus{} z \equal{} 0$
$ e_2: x^2 \plus{} y^2 \minus{} z^2 \equal{} 12,$
$ e_3: x^4 \plus{} y^4 \minus{} z^4 \equal{} 96$

[hide]Simply by substituting $ z\equal{}\minus{}x\minus{}y$ in the other two equations and after elementary rearrangements, we get $ (x,y,z)\equal{}\{(3,\minus{}2,\minus{}1),(\minus{}2,3,\minus{}1),(\minus{}3,2,1),(2,\minus{}3,1)\}$[/hide]


Let $ A$ be the set of numbers in the Fibonacci sequence.

Let $ M\equal{}\{\frac{x}{y}|x,y\in A\}$. Is the set $ M$ dense in the set of real numbers?

it is not density. Exactly: for any $ x\not \equal{} q^n,n\in Z, q \equal{} \frac {1 \plus{} \sqrt 5}{2}$ exist $ \delta (x) > 0$, suth that in $ (x \minus{} \delta (x),x \plus{} \delta (x))$ there are only finit members of M.


Se da un triunghi MNP. Fie (MS bisectoarea unghiului NMP , S apartine (NP).
 Fie A un punct al segmentului MS. Perpendiculara in A pe NA intersecteaza  MN respectiv NP in punctele B respectiv C. Daca unghiurile MBA si PAM sunt congruente sa se arate ca d(A;NP)=d(A;MP)

O idee: Aplici Teorema Sinusurilor in triunghiurile $MNP, MAP, MAN$, după care obligi $\measuredangle MNA = \measuredangle MAP-90^\circ$. Mai rămân "nişte calcule" de făcut.

O singura intrebare:
Vi se pare normal sa apara aceasta pb la olimpiada faza de sector la clasa a7-a?
perfect_radio multumesc pentru hint, dar nici nu am apucat sa predau la
clasa notiunea de sinus
Inseamna ca are si o solutie elementara, si acea solutie chiar ma macina binisor

cred ca merge sa fie facuta cu urmatoarea observatie interesanta (sper ca vad bine eu aici...)
Voi face niste sume mici de unghiuri :
$\angle CAM=\angle CAP+\angle PAM=\angle CAP+\angle BMA$.
$\angle CAM=\angle MBA+\angle BMA$ (unghi exterior)
asadar $\angle BMA=\angle CAP$.
Apoi, $\angle NAP=\angle NAC+\angle CAP=90^{\circ}+\angle BMA=90^{\circ}+\frac{\angle NMP}{2}$.
aceasta ultima relatie este ceva cunoscut, este chiar masura unghiului $NIP$, unde $I$ este centrul cercului inscris in triunghi.
Cum $A$ si $I$ ambele sunt pe bisectoare, rezulta ca ele coincid, asadar $PA$ este bisectoare, deci $d(A,NP)=d(A,MP)=r$ (raza cercului inscris).
Faptul ca intr-un triunghi $XYZ$ oarecare, cu $I$ centrul cercului inscris are loc relatia $\angle YIZ=90^{\circ}+\frac{\angle X}{2}$ iese cu adunari de unghiuri imediat :)

e bine ce`am zis eu aici ?

edit : gresisem un semn in latex


function
limit
calculus
integration
geometry
geometric transformation
reflection

I was playing around with some series and found some interesting stuff. Basically, I was just like "what will converge? Okay so let's approximate it in Maple and see if it is nice." Now, $\lim_{n \to \infty}\zeta(n) = 1$ and the convergence is very fast, so I tried subtracting one from that. And I found some stuff which I was slightly disappointed to discover were already on Mathworld, but as usual Mathworld doesn't have a proof, so I was wondering if anyone would care to try their hands at:

$\sum_{n=2}^\infty (\zeta(n)-1) = 1$

(In case you're not familiar with the notation,
$\zeta(n) = \sum_{k=1}^\infty \frac{1}{k^{n}}$
and it is known as the Riemann Zeta Function.)

There are similar ones. Equations 110-113 on
http://mathworld.wolfram.com/RiemannZetaFunction.html

[quote="Kent Merryfield"]What you've written is this:

$\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac1{k^{n}}=?$

Now, I don't know about you, but every time I see a double sum (or a double integral, or the integral of a sum, or the sum of an integral), there's something staring me in the face that I just [b]have[/b] to try.

Go ahead. Do what needs to be done.[/quote]
Using the above double summation mentioned above, what I did was bring the outer summation sign inside.

So, $\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac1{k^{n}}= \sum_{k=2}^{\infty}\sum_{n=2}^{\infty}\frac1{k^{n}}= \sum_{k=2}^{\infty}\frac{1}{k^{2}}(1+\frac{1}{k}+\frac{1}{k^{2}}+\ldots) = \sum_{k=2}^{\infty}\frac{1}{k^{2}}\cdot\frac{1}{1-\frac{1}{k}}$
$=\lim_{N\rightarrow\infty}\sum_{k=2}^{N}\frac{1}{k(k-1)}= \lim_{N\rightarrow\infty}\sum_{k=2}^{N}(\frac{1}{k-1}-\frac{1}{k}) = \lim_{N\rightarrow\infty}(1-\frac{1}{N}) = 1.$

Is this the correct way to evaluate the given sum?

[quote="blahblahblah"]Yeah, interchanging the order of summation is what he was getting at.[/quote]
So, if we have three or more summation signs, then how would we interchange the order of summation? I hope you get my question.

Well you could interchange however you like (assuming the convergence is all ok). What change you want to do will depend on the nature of the problem.

Justification? I've got a long writeup of the proof somewhere (I wrote it up for a class) but I'm not sure I want to dump all of that here right now.

For finite sums, interchanging the order of summation is a reflection of the commutativity and associativity of addition. For infinite sums, we have to deal with two limit processes. There are examples in which it doesn't work. [color=green][Edit: see Yury's example below.][/color] However, the interchange of two infinite sums is justified in either of two situations:

(1) If the terms are all nonnegative. (This allows for the possibility that the whole sum is infinite - if so, it's infinite both ways), or

(2) If the whole thing is absolutely convergent - that is, if the double sum (either way) of the absolute values is finite.

In this particular situation, the terms are nonnegative. (And, since we get a finite sum, also absolutely convergent.)

[quote="me@home"]How would you prove that it is legal to interchange summation signs (i. e. you don't need parenthesis)[/quote]

The usual hypothesis you want is that $a_{ij}$ is absolutely summable, e.g,

$\sum_{i}\sum_{j}|a_{ij}|<\infty$

and you can prove this (for example) by using power series in a clever fashion.

It is also permissible under the conditions that $a_{ij}\geq 0$ for every $i,j$ - the simplest proof here is probably measure-theoretic.

e:  Never mind the timestamp - Dr. Merryfield beat me by about a minute and the system wouldn't let me post because it said that someone had already posted.  One day I will beat him to the punch and it will be glorious!

You can take the fact that blahblahblah and I said the same thing as evidence that the theorem in question is well-known.

[quote="me@home"]How would you prove that it is legal to interchange summation signs (i. e. you don't need parenthesis)[/quote]
Consider the sequence \[a_{ij}= \begin{cases}\frac{1}{i}, &\text{if }i=j\\-\frac{1}{i}, &\text{if }j= 2i\\ 0,&\text{otherwise}\end{cases}\] Then \[\sum_{i=1}^{\infty}\sum_{j=1}^\infty a_{ij}= \sum_{i}(\frac{1}{i}-\frac{1}{i}) = 0.\] Now we interchange the order of summation \[\sum_{j=1}^\infty\sum_{i=1}^\infty a_{ij}= \sum_{j=1}^\infty \frac{(-1)^{j+1}}{j}=\ln 2.\] We conclude that  :maybe: \[0 = \ln 2.\] On a more serious note, we can claim that \[\sum_{i=1}^{\infty}\sum_{j=1}^\infty a_{ij}= \sum_{j=1}^{\infty}\sum_{i=1}^\infty a_{ij}\] if the series converges absolutely. That is \[\sum_{i=1}^{\infty}\sum_{j=1}^\infty |a_{ij}| < \infty.\] Update: Took so long to answer. Kent Merryfield and blahblahblah answered before me...  :oops:

This much spoken of "well known theorem" is Theorem 8.3 on pg. 175 of Rudin's Principles of Mathematical Analysis (afectionately refered to as "Baby Rudin") and it reads thus:

[quote="Principles of Mathematical Analysis, pg. 175"][b]Theorem 8.3[/b]  Given a double sequence $\{ a_{ij}\},\ i=1,2,,3,\ldots,\ j=1,2,,3,\ldots ,$ suppose that \[\sum_{j=1}^\infty\sum_{j=1}^\infty \left| a_{ij}\right| = b_{i}\ \ (i=1,2,,3,\ldots ),\] and $\sum b_{i}$ converges. Then \[\sum_{i=1}^{\infty}\sum_{j=1}^\infty a_{ij}= \sum_{j=1}^{\infty}\sum_{i=1}^\infty a_{ij}.\] [/quote]

The other half of the theorem is on pages 22/23 of Green Rudin :-)

question about sums and integrals

is it always ok to say that $\int\left(\sum_{x=0}^{n}{f(x)}\right)dn=\sum_{x=0}^{n}\int f(x)dx$?

or must some special conditions be met first.
since i tried it w/ $f(x)=x^{2}$ and it seemed to work.

maokid7, your notation doesn't make a lot of sense. I'll answer the question I think you meant to ask: can one justify

$\int_{E}\sum_{n=0}^{\infty}f_{n}(x)\,dx=\sum_{n=0}^{\infty}\int_{E}f_{n}(x)\,dx\ ?$

If we are operating within the Riemann theory of integration, then we would need for $E$ to be a compact interval (or otherwise a set on which the Riemann integral can be defined). A sufficient condition for interchanging sum and integral is that the series of functions converge uniformly.

That's a useful theorem, but there are many instances in which one finds the hypothesis of uniform convergence to be too restrictive. We're better off operating within the Lebesgue theory of integration. In that case, we merely need $E$ to be measurable, and the interchange of integration is justified under two circumstances:

(1) If the functions $f_{n}$ are nonnegative.

This is a corollary of the Monotone Convergence Theorem (and allows for the posibility of an infinite answer - if infinite one way, then also infinite the other way).

(2) If $\int_{E}\sum_{n=0}^{\infty}|f_{n}(x)|\,dx<\infty$

This is a consequence of the Dominated Convergence Theorem.

You'll notice that we have for this theorem the same pair of conditions that we quoted for the double-sum theorem. That pair of conditions appears yet one more time. We can justify the interchange of order of integration in a double integral if the function is nonnegative (Tonelli's Theorem) or if the double integral of the absolute value is finite (Fubini's Theorem).

In all cases, we can construct examples in the spirit of what Yury wrote above: functions or sequences that are sometimes positive and sometimes negative, too big to be absolutely convergent, and sensitive to the order of operations.

thanks for the clearification.
and yes that was the question that i meant to ask

Hmm. Well, thank you for this exposition. What are the commonly used criteria for interchanging differentiation and integration, i.e. $\frac{d}{dt}\int f(x, t)\,dx = \int \frac{\partial}{\partial t}f(x, t)\,dx$?

That's a more delicate question, but it's also about interchanging a limit and an integral.

$\frac{d}{dt}\int_{E}f(x,t)\,dx= \lim_{h\to0}\frac1h\left[ \int_{E}f(x,t+h)\,dx-\int_{E}f(x,t)\,dx\right]$

$=\lim_{h\to0}\int_{E}\frac{f(x,t+h)-f(x,t)}h\,dx$

Now assume the hypotheses of the Mean Value Theorem have been met:

$=\lim_{h\to0}\int_{E}\frac{\partial}{\partial t}f(x,\xi)\,dx$

where $\xi=\xi(x,t,h)$ lies between $t$ and $t+h.$

How to proceed next depends on what kind of estimates we have for that derivative. If $f$ is $C^{1}$ and $E$ is a compact set, then we can argue that $\frac{\partial}{\partial t}f(x,\xi) \to \frac{\partial}{\partial t}f(x,t)$ uniformly, and justify the interchange of integrals by uniform convergence.

Failing that, we'll be looking for a way to apply the Lebesgue Dominated Convergence Theorem - something which requires us to have a useful dominating function.

The common thread in all such arguments will be the use of the Mean Value Theorem to estimate the difference quotient.


Prove that for every $n$,

the equation   $x^{3}+y^{3}+z^{3}+t^{3}+v^{3}=n$  has solution.

this was JUST posted...
http://www.mathlinks.ro/Forum/viewtopic.php?t=48334


1. Selenium (Se) is similar to sulfur in its properties and francium (Fr) is an alkali metal. What is the formula for francium selenite?

(a) $ FrSeO_2$
(b) $ Fr_2SeO_4$
(c) $ Fr_2SeO_3$
(d) $ Fr_2Se_2O_3$

2. Two pure organic compounds melt at 112 degrees Celsius and 114 degrees Celsius, respectively. If equal quantities of them are mixed, at what temperature will the mixture begin to melt?

(a) below 112
(b) at 112
(c) between 112 and 114
(d) above 114
[hide]The answer is somehow (a). Does this have to do with the concept of azeotropes?[/hide]
3. Which substance boils at the highest temperature?

(a) $ CH_3Cl$
(b) $ CH_2Cl_2$
(c) $ CHCl_3$
(d) $ CCl_4$
[hide]I thought since $ CHCl_3$ was the most polar, it should have the highest bp. But apparently the answer is (d), even though carbon tetrachloride is nonpolar.[/hide]
4. Which species can act as an oxidizing agent but not as a reducing agent?

(a) $ Cl_2$
(b) $ Cl^\minus{}$
(c) $ ClO_2^{\minus{}}$
(d) $ ClO_4^{\minus{}}$

What's funny is I've been doing past Chemistry Olympiad Regional exams and I had trouble with those same questions.  Question 3 stumped me a bit as well, as I had the same thoughts as you in regards to the answer being C.  Hopefully someone can explain why it isn't.

1. C.

Sulfite = $ \mathrm{SO_{3}^{2 \minus{} }}$
Francium has a +1 charge
So, selenite probably is $ \mathrm{SeO_3^{2 \minus{} }}$
And the answer is $ \mathrm{Fr_2SeO_3}$

2. A.

I would say this could be done as a colligative properties problem...
Assume you have the two organic compounds dissolved in each other (which is likely). Then, the freezing point will be lowered. Because the freezing point is lowered, the melting point is lowered. So, the melting point will be below either temperature.

Also, you could kind of describe this as a solid state problem...
When either organic compound is solid, there is a certain amount of attraction (packing) between the individual molecules. When they are mixed (homogeneous), there are "impurities" which makes the intermolecular forces between the individual molecules weaker. So, these "bonds" are easier to break. The temperature is lowered.

3. D.

I honestly would put B as my first choice...because B has the largest vector dipole moment quantity. However, D is correct. In alkyl halides, only two properties govern the boiling point: London dispersion forces and molecular weight.
$ \mathrm{CCl_4}$ has the largest surface area (London dispersion force) and highest molecular weight by far. So it has to be D.

4. D.

An oxidizing agent is reduced (oxidation number goes down); a reducing agent is oxidized (oxidation number goes up). Chlorine's extreme oxidation numbers are -1 and +7 (refer to the periodic table). In perchlorate ion, the chlorine atom has the maximum oxidation number of +7. So, perchlorate can be reduced but not oxidized.

(keep in mind that perchlorate is an EXTREMELY stong oxidizing reagent)

I feel like I did these problems not too long ago...on some local tests...

One more:

A chemical reaction is carried out twice with the same quantity of reactants to form the same products but the pressure is different for the two experiments. Which does not change?

(a) $ K_p$
(b) heat released
(c) $ \Delta T_{surr}$
(d) work done
[hide]The answer seems to be $ K_p$ since changing the pressure just shifts the equilibrium, so does not change the equilibrium constant. But I was wondering why/how do the others change?[/hide]

Say an equilibrium exists of the form

$ \mathrm{A}_{(g)}\rightleftharpoons\mathrm{B}_{(g)}$

There is some heat of reaction $ H$ associated with this equilibrium. So, when the reaction proceeds, heat will be released to or absorbed from the surroundings. B changes. Also, C changes ($ q\equal{}mc\Delta T$).

I do not remember the sign conventions (someone could check this for me), but basically,

$ W\equal{}Q\plus{}PV$

So, a change in $ Q$, or the heat released or absorbed, results in a change in $ W$, or the amount of work done.

[quote="Jdlkem"]$ W \equal{} Q \plus{} PV$[/quote]

And from where come from that amazing relation?

We know that W=F*d, but energy is lost in this case so we have to add it on, this is Q.

So, working backwards

W=Q+PV

W=Q + (F/A) (A*s) where s is an arbitrary unit of length, A is area, and F is force

W=Q + F*s, which rounds back to my explanation in the beginning.

[quote="jdlkem"]
I would say this could be done as a colligative properties problem...
Assume you have the two organic compounds dissolved in each other (which is likely). Then, the freezing point will be lowered. [color=red]Because the freezing point is lowered, the melting point is lowered. [/color]So, the melting point will be below either temperature.
[/quote]

Wait...isn't the boiling point elevated due to colligative properties?

Yes, but in the sentence you quoted there is no reference to boiling point.

this is embarrassing... I misread the post.  Thanks guys.

isnt the melting same as the boiling point because it initiates the same link between solid and liquid, just going one way or another?

what i mean is like this.

Solid-->liquid point is same as liquid-->solid point.

Also, does anyone know what temperature you have to heat Carbon dioxide at 2psi to to get plasma? im lost. :oops:
What's tautology? i dont get it.

Boiling point is liquid-->Gas or the other way round.

Melting point is Solid-->Liquid.

[hide="Btw"]
The tautology thing is signature :rotfl: . It doesn't have any relation to the post. It appears at the bottom of every post he posts[/hide]


trigonometry
integration
function
limit
real analysis
real analysis unsolved

Suppose
\[0 < \delta < \pi \]

\[f(x) = \left\{ \begin{array}{l}1{\rm }if|x| \le \delta \\ 0{\rm }if\delta < |x| \le \pi \\ \end{array}\right. \]

\[f(x+2\pi ) = f(x){\rm }\forall x \]

(a) Compute the Fourier coefficients of f.

(b) Conclude that
\[\sum_{n = 1}^\infty{\frac{{\sin (n\delta )}}{n}= \frac{{\pi-\delta }}{2}}\]
\[0 < \pi < \delta \]

(c) Deduce from Parseval's theorem that
\[\sum_{n = 1}^\infty{\frac{{\sin^{2}(n\delta )}}{{n^{2}\delta }}= \frac{{\pi-\delta }}{2}}\]

(d) Let
\[\delta \to 0 \]
and prove that \[\int_{0}^\infty{\left({\frac{{\sin (x)}}{x}}\right)^{2}dx = \frac{\pi }{2}}\]

(e) Put
\[\delta = \frac{\pi }{2}\]
in (c). What do you get?


Here I'd like to compare two concepts which IMHO are really close to each other and perhaps related in nature: proof and algorithm.

As you know algorithm is a computational procedure (process) that transmutes and changes input to output (in its shape, content, etc.), i.e. an algorithm maps an input to specific desired output. A mathematical proof is a flow of stepwise logical reasons to finally base an "conclusion" on "premise(s)". Think of it as a construct built by bricks (one on another).

It's something that makes us to name these two concepts differently, something in them, in their purposes. For proofs we have a journey from premises to conclusion, for algorithms ...!

Why writing about such strange binary relation?!
Well, I had this idea when I was reading a book in "success literature" (wasn't it "7 habits of ...?!") when the autor stated that the journey is much important than the goal (if you want to live happy), live at present not future, and such things!

yeah, hard work pays off later, laziness pays off now!  :rotfl:
anyway i don't find the analogy particularly well-thought-of, unless the proof is of the algorithm itself.
but the idea seems quite interesting. i love analogies. what do you want to do in this thread anyway? :)