Express the product $ 15 \frac{1}{2} \times 4 \frac{1}{3}$ as a mixed number.

One way to approach this is to change the mixed numbers into improper fractions (for ease of multiplication). Our fractions become $ 15\frac{1}{2} \equal{} \frac{31}{2}$ and $ 4\frac{1}{3} \equal{} \frac{13}{3}$. When we multiply these fractions, we get $ \left(\frac{31}{2}\right)\left(\frac{13}{3}\right) \equal{} \frac{403}{6}$. Turning this into a mixed number results in an answer of $ \boxed{67\frac{1}{6}}$.


One sixth of a steel beam of a bridge is located in cement below the bed of a river. Two fifths of the beam is in the water, and 78 feet is located above the water. How long is the pole?

[hide]
$\frac{1}{6}+\frac{2}{5}=\frac{17}{30}$
so 17/30 are in the water and 13/30 are out of the water.  if 78 is 13/30 then you would do $\frac{78}{\frac{13}{30}}=\boxed{180}$ feet[/hide]


Let $ p > 2$ be an odd number and let $ n$ be a positive integer. Prove that $ p$ divides $ 1^{p^n}\plus{}2^{p^n}\plus{}\cdots\plus{}(p\minus{}1)^{p^n}$.

[hide]Well $ 1 \equiv 1 \minus{} p \pmod p$ and so on, and $ 1 \minus{} p$ is the negative of $ p \minus{} 1$, and the exponent, $ p^n$ is odd, so that expression will be 0 mod p, as the terms will keep pairing up and cancelling out.[/hide]


advanced fields
advanced fields theorems

Can anyone help...
Is it possible to deduce Cassini's Formula
Tn^2=Tn+1 x Tn-1 - 1 for n even
Tn^2=Tn+1 x Tn-1 + 1 for n odd

see here http://www.mathlinks.ro/Forum/viewtopic.php?t=136618  :wink:


inequalities calculus derivative logarithms

Your relation makes no sense.  Maybe the sum of their reciprocals is $1$?

Anyways, it looks like an application of Jensen's on $\ln(1/x-1)$ to me.

To elaborate, the second derivative of $\ln |\frac 1x-1 |$ is $\frac 1{x^2}-\frac 1{(x-1)^2}$ which is positive when $x<0$.  Because there is an even number of negative numbers here, we may replace everything by its absolute value.  Now things get more interesting

By Jensen's, ${\prod_{2000} (\frac {1}{x_i}-1)\geq (\frac {2000}{\sum x_i}-1})^{2000}$

But here we have problems - we can't attain the numbers you have unless $\sum x_i=1/2000$, which can't happen unless some of the $x_i$'s are positive, etc, etc.

Please come up with a proper formulation.

actually, we can make it work, we just need to switch the sign and we need $2000$ positive numbers which sum to $1/2000$.  My first post contains all the information you need to solve such a modified question...

[quote="limkimsrun"]Sorry for my big mistake  :blush:
the real fact is that $ a_1, a_2....,a_{2000}$ are all positive real numbers.
anyway, are there any proof for Jensen's inequality? Please post it for all.[/quote]

You can use backward induction.
First you may prove it for $n=2^k$
Next you can prove that if $n=k$ is true, $n=k-1$ is also true by putting one of the variables = the mean of others.

[quote="blahblahblah"]Are you talking with weights or without them?[/quote]
i'm talking about the one without weights. I don't know what he wants though.


calculus
integration
trigonometry
logarithms
special factorizations
algebra
partial fractions

Divide the numerator and denominator with x^2. In the numerator make the 1/x^2 = 2/x^2 by dividing both numerator and denominator with 2. Then adjust it as (1+1/x^2)-(1-1/x^2)
Now distribute the denominator, The one below the left expression should read as (x-1/x)^2+3and the one below the right one should be (x+1/x)^2 -1.

Then take the things inside the brackets in the denominators as something t or z maybe , substitute them apply the formula. And be sure to change the limits as per  the variables through substitution.

Completing the square in the denominator yields $ (x^{2}\plus{}\frac{1}{2})^{2}\plus{}\frac{3}{4}$.  Now you can pull out $ \frac{3}{4}$ to turn this into $ \frac{4}{3}\int\frac{dx}{\frac{4}{3}(x^{2}\plus{}\frac{1}{2})^{2}\plus{}1}\equal{}\frac{4}{3}\int\frac{dx}{(\frac{2}{\sqrt{3}}(x^{2}\plus{}\frac{1}{2}))^{2}\plus{}1}$.  Now make the appropriate substitutions to turn this into an easy trig integral.

Which trig integral to be precise? Are you suggesting that one should substitute x by 1/root 2 (tanA) or something like that? Believe me it's gonna be hell lot of an integral!!

hello, the right result of the integral
$ \int_{0}^{1}\frac{1}{x^4\plus{}x^2\plus{}1}\,dx$ is $ \frac{1}{4}\ln(2)\plus{}\frac{1}{12}\pi\sqrt{3}$
Sonnhard.

Thanks Sonnhard, I've edited my solution.


calculus derivative logarithms integration function algebra domain

Does anyone know why the derivative of logx is x^-1?

Well, this question is not very suitable to appear in the pre-olympiad forum... Anyway, here it goes.
$\frac{d}{dt}e^t=e^t$
Let $x=e^t$, then $t=\ln x$
$\frac{dx}{d\ln x}=x$
$\frac{d\ln x}{dx}=\frac{1}{x}$
 :D

Which definition of $log$ do you take ? There are so many equivalent definitions.

Sometimes $log$ is defined as $\log{x} = \int_1^x \frac{1}{t} dt$  which makes things easy in this case  :D

But then you would still have to prove, with that definition, $\ln (e^x) = x$, and I think it's a little easier to take that as the definition and then work towards the derivative  :D

Well, actually the derivative of $\ln{x}$ is $\frac{1}{x}$ but the derivative of $\log_ax$ is $\frac{1}{x lna}$.
Hope this was useful.

[quote="Hokkage"]
Hope this was useful.[/quote]

Unfortunately it wasn't - many of us use $\log$ where most people use $\ln$.  The reason for this is that the base 10 logarithm almost never comes up in higher mathematics (maybe in number theory and computer science, at most)

The two definitions of $\ln$ I see most commonly are its definition as a) the antiderivative of $1/x$ that takes the value 0 at point 1 (the link between the logarithm and the corresponding exponential function is then shown quite easily, as are the properties of the logarithm, using derivatives), and b) the inverse function of the Neperian exponential. The properties of the logarithm can then be found via the properties of the exponential function. The first definition seems more elegant to me (for whatever reasons). I can't really think of a definition of $\ln$ that would lead to any serious problems in proving the common results about logarithms and exponentials.

On a related note, I have several textbooks (especially older, French books) that use Log (always written with an uppercase L) to denote the "natural" logarithm ($\ln$).

Yes in many books it is stated that the logarthm is the antiderivatif of $\frac{1}{x}$. But some books take another idea (maybe using historical background).
The logratihm emerges because it help people in doing calculation in large value numbers. It usefull if we can change the multiplication to addition.
So we lead to this definition (from Hille and Salas [i]Calculus, one and several variable[/i]).
The function $L(x)$ is the function whose domain is positive real number so that \[ L(ab) = L(a)+L(b) \qquad \forall \, a, b \in \text{domain}  \]
Here we not define $L(x)$ for complex number. If we let $a= \frac{c}{d}$ and $b=d$, then  $L \left(\frac{c}{d} \right) + L(d)= L(c) \Rightarrow L \left(\frac{c}{d} \right) = L(c) - L(d)$. And if $c=d$ we have $L(1) = 0$.
About the antiderivatif of $\frac{1}{x}$. It was observed from
\begin{eqnarray*} L'(x) &=& \lim_{h \rightarrow 0} \frac{L(x+h) - L(x)}{h} \\ &=& \lim_{h \rightarrow 0} \frac{L \left(\frac{x+h}{x} \right)}{h} \\ &=& \lim_{h \rightarrow 0} \frac{L \left(\frac{x+h}{x} \right) - L(1) }{h} \\ &=& \lim_{h \rightarrow 0} \frac{L \left(1 + \frac{h}{x}\right) - L(1) }{h} \\ &=& \lim_{h \rightarrow 0} \frac{1} {x} \frac{L \left(1 + \frac{h}{x}\right) - L(1) }{\frac{h}{x}} \\ &=& \frac{1}{x} L'(1) \end{eqnarray*}
Now there is a hope for us to determine what is the antiderivatif of $\frac{1}{x}$. That is the logarithm function for which $L'(1)=1$, this function is denoted by $\ln x$ appears from naturally choice of $L'(1)$.  The other choice of $L'(1)$ will lead to other base of logarithm function.

Using a.b initio

i.e, $f'(x)=lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
you will get what you require.


Let $ a, b, c >0$. Prove that: $ \boxed{a^4\plus{}b^4\plus{}c^4\ge (a\plus{}b\plus{}c)abc}$

AM-GM ; $ a^4\plus{}a^4\plus{}b^4\plus{}c^4 \geq 4a^2bc$  :lol:

[quote="thanhnam2902"]You can post full solution?[/quote]
$ a^4\plus{}a^4\plus{}b^4\plus{}c^4 \geq 4a^2bc$
$ b^4\plus{}b^4\plus{}c^4\plus{}a^4 \geq 4ab^2c$
$ c^4\plus{}c^4\plus{}a^4\plus{}b^4 \geq 4abc^2$
$ \therefore 4(a^4\plus{}b^4\plus{}c^4) \geq 4abc(a\plus{}b\plus{}c)$   :wink:

[quote="apollo"][quote="thanhnam2902"]You can post full solution?[/quote]
$ a^4 \plus{} a^4 \plus{} b^4 \plus{} c^4 \geq 4a^2bc$
$ b^4 \plus{} b^4 \plus{} c^4 \plus{} a^4 \geq 4ab^2c$
$ c^4 \plus{} c^4 \plus{} a^4 \plus{} b^4 \geq 4abc^2$
$ \therefore 4(a^4 \plus{} b^4 \plus{} c^4) \geq 4abc(a \plus{} b \plus{} c)$   :wink:[/quote]

We can apply twice times the well-known inequality $ a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca$


calculus
derivative
function
logarithms
calculus computations

A function f(x) is defined for $ x$ $ \geq$ 2 and k is a given positive number.Prove that

1. if  $ \frac{d(xf(x)}_{dx}$ $ \leq$  $ -kf(x)$ then  $ f(x)$ $ \leq$ $ Ax^{-1-k}$

2. if  $ \frac{d(xf(x)logx)}_{dx}$ $ \leq$ $ -kf(x)$   then $ f(x)$ $ \leq$ $ Bx^{-1}.(logx)^{-1-k}$ , where A and B are independent of x.

Do or don't $ A$ and $ B$ depends on $ k$? If $ A$ and $ B$ depends on $ k$ just use the following hints
[hide=" Hints"]
1) Consider the function $ g(x)\equal{}x^{k\plus{}1}f(x)$
2) Consider the function $ g(x)\equal{}(\log x)^{k\plus{}1}xf(x)$[/hide]


I'm self studying Markov Chains and very, very confused at the same time.

How do you find the limiting matrix of an absorbing Markov chain by partitioning the transition matrix into submatrices?

Supposedly the limiting matrix should look like [url=http://img387.imageshack.us/img387/691/markovabsorb7lg.jpg]this[/url], where M = (I - NN)^-1. But what if NN or NA is not a square matrix? Then it would not be possible to end up with a matrix of the orignial dimensions....

Am I misunderstanding the definition of NN and/or NA? For example, is [url=http://img379.imageshack.us/img379/6098/absorb20ef.jpg]this[/url] correct?

Or am I just missing something extremely obvious?

Any and all help is appreciated,

sphoeniXee

I encountered this problem in statistics and I wanted to go to the bone with it

I saw it this way, a Markov matrix has a property : every column has sum 1, and all elements are positive (and therefore at most 1)
the element$A_{ij}$ shows the probability to go from state j to state i
now in a Markov chain you wanna do these things in a sequence, so yes, it makes no sense if the matrices are not square!

now these matrices do not always converge to a single state
you see, you can prove that they always have eigenvalue 1, and that all eigenvalues are at most 1 in modulus
However things get grim if there are others modulus 1 eigenvalues.
That is the condition : all none 1 eigenvalue are strictly smaller in modulus.
In that case you can find the limit matrix quite easy (I mean no analysis is involved)
Just solve the matrix equation $AX=X$
However, don't just plunge into this one, interpret!  Every colymn must be eigenvector of A with eigenvalue 1
Just finding the eigenvector and making every column of X that one, will give you the correct limit matrix

Is there a way to see whether a matrix satisfies the condition for convergence.  Yes, but is doesn't work both ways
You see, if your matrix A has a natural power with only strictly positive elements, then you can be sure.
However, I can show you matrices that do not have such a power and do converge.

I want to stress my point again that this nonconvergence problem in statistics is really only a technical detail, as you see how mild the conditions are (usually your matrix has every element strictly positive)

If you are interested, I have some proofs (copied from the net) of all this, and some examples in maple files (you don't happen to speak dutch by any chance)

You can choose between PM and this topic to ask, I'm just happy to finally answer something here

Sorry I didn't look at your images, they are way too small apparently?)


Consider the sequence of numbers: $a_1,a_2 \cdots a_n$ which is a permutation of the set $1,2 \cdots ,n$ where $n = 2004$. Prove we can find a pair $i,j$ s.t. at least one of the two following equations is true:
$|a_i - a_j| = 2002 - |i - j|$
$|a_i - a_j| = |i - j|$

Please solve this problem people!
if you don't i will sit here and start crying  :(  :(

I haven't thought about the problem so much (have no time for the moment), but if we try to consider smaller examples:
Let our numbers be 1,2,3,4. Then the following permutation: 2,4,1,3 doesn't satisfy the first condition, and neither the second if we try to generalize our second condition to: |a_i - a_j| = n-2 - |i-j|, which is (in our smaller example): |a_i - a_j| = 2 - |i-j|. So if we think that that should be the generalization for arbitrary n, then it doesn't work for our example. n-1 works though.

So if we consider 1,2,3,4,5,6, the following permutation doesn't satisfy condition 1): 2,4,6,1,3,5. So for arbitraray n, we first list even integers, then odd. Maybe if we could prove that these are the only permutations (and maybe their cyclical shifts) that doesn't satisfy condition 1) ?

That's just a fast idea, maybe you have probably tried it.

hmm, so which one are you referring to as the first condition and the second one?
about the first condition $|a_i - a_j| = |i - j|$ finding all permutations that do not satisfy this for all pairs $(i,j)$ is extremely difficult and not do-able. if you think about it [plot the points $(i,a_i)$ on a $n$ by $n$ chessboard] it's equivalant to the $N$ queens problem, which is, as far as i know, unsolved.
however about the second condition, the set of permutations not satisfying may be nice-ish.
thanks for the help anyway. any more help would be very appreciated.

i think the problem just asks if a each permutation of 2004 elements contains
some inversion or transposition.
or something like that

spx2, could you be a bit more explicit? what's an inversion/transpotation in this sense? do you have any ideas of proof?

I think it's a bit difficult. can you give the solution epitomy01? :oops:

sorry i don't have a solution ..
that's why i posted in the unsolved problems section as opposed to solved problems section.


function
number theory
real analysis
real analysis solved

does anyone heard of this???  using cesaro sum, someone on the internet claims that \[\sum_{n=1}^\infty n = -\frac{1}{12}\]

*Sigh* Riemann Zeta function... It's $\zeta(-1)$. But don't worry too much about it.

oh i see, so you stick z = -1 there??

like \[\sum\frac{1}{n^{-1}}\] ???

Yes. I don't know exactly what happens, I am not an expert on this kind of number theory (not yet? :D) but it's something about reconstructing the zeta landscape. I also know that Ramanujan discovered this on his own (I mean, without reading about advanced number theory anywhere). Pretty impressive, he sort of re-invented the whole Riemann zeta function thing.

The brief version is that the sum

$\zeta(s)=\sum^{\infty}_{n=1} \frac {1}{n^s}$

defines an analytic function for $\text{Re}(s)>1$, as a convergent $p$-series.  However, it has an analytic continuation to $\text{Re}(s)>0\cap\{1\}$, which is quite easily derived.  This means that we can speak meaningfully about $\zeta(1/2)$, even though plugging such a sum into the original series would result in a divergent series.  Next, we can, through use of the gamma function, extend it to $\mathbb{C}\setminus\{1\}$.  That means that there is a function $\beta(s)$ such that $\beta(s)=\zeta(s)$ on $\text{Re}(s)>1$, but such that $\beta$ is defined on $\mathbb{C}\setminus\{1\}$.  This is called the Riemann Zeta Function.  Now, it so happens that $\beta(-1)=1/12$, but since we identify the values of $\zeta$ with those of $\beta$, we may say (some what facetiously) that $\zeta(-1)=1/12$.


I found very interesting Miquel theorems. Most interesting for me was this one: http://www.gogeometry.com/miquel_pentagram1.htm
Inspired of that theorem using geogebra I found the configuration on the picture.

To be easier for you - take a look at the file attached.
It is given a pentagram (five rays star). The vertices of the pentagram lies on a circle. There are constructed five smaller circles.
Prove that their second intersecting points lies on a circle.

My questions are:
1. Is it a famous theorem or a problem from math competition?
2. How to prove that statement?
3. What is its level of difficulty?
4. Is it a beatiful problem?

I also don't know how to solve the problem. I think it can be solved only by using inscribed angle/quadrilateral properties and additional constructions.

If you can solve the problems in the attached pictures our initial problem is solved.

1. In the IntersectingLines.png - you should prove the all five lines intersects at a common point.
2. In the InscibedQuad.png - you should prove the quadrilateral marked is inscribed.

If you have some questions - please ask me.

[quote="borlislav_mirchev"]It is given a pentagram (five rays star). The vertices of the pentagram lies on a circle. There are constructed five smaller circles. Prove that their second intersecting points lies on a circle. [/quote]
See [i]Les pentagrammes de Miquel et de Morley[/i], Volume (4) [url=http://pagesperso-orange.fr/jl.ayme/]here[/url].

Thank you very much these materials they are very interesting ... I made mistakes in all my solutions ... it seems that the problem is very hard or incorrect. But I believe it is simply a hard problem.

I'm sorry all my attempts to solve the problems failed ... my friends also failed ... I'll give 50 euros for first correct solution with matherial used in high school.
The tallent have no price ... but it is just a bonus for invested time.
Proof that the statements are not correct is also a solution.

Dear Mathlinkers,
Here is the solution for 50 Euros   :D

See my solution here :
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=393621

This problen is from :
International Mathematical Olympiad 2009
Hong Kong Team Selection Test 1
See here:
http://gifted.hkedcity.net/Gifted/Download/0809IMOTest/IMO-Test%201-August-2008.pdf

Dear skytin
Thank you very much for very useful information and the excellent solution. The strange thing is I discovered the statement at my own.
Some of the statements I rediscovered appeared also at Sharygin MO 2010 and Romania TST 1993 other were some well known theorems but not for me :(


Let $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$. Prove that: $ \frac{a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac{b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac{c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \leq \frac{6}{5}$

[quote="D.I.Culianop"]Let $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$. Prove that: $ \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \leq \frac {6}{5}$[/quote]
The inequality is true for all $ a,b,c \ge 0$.

\[ \sum {\frac{{a\left( {b \plus{} c} \right)}}{{a^2  \plus{} \left( {b \plus{} c} \right)^2 }}}  \le \sum {\frac{{4a}}{{4a \plus{} 3b \plus{} 3c}}}\] true by AM-GM
\[ \sum {\frac{{4a}}{{4a \plus{} 3b \plus{} 3c}}}  \le \frac{6}{5}\] true by Cauchy Schwarz

[quote="D.I.Culianop"]Let $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$. Prove that: $ \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \leq \frac {6}{5}$[/quote]
We have:

 $ \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2}\equal{}\frac {a(1\minus{}a)}{a^2 \plus{} (1\minus{}a)^2} \le\ \frac{27}{25}a\plus{}\frac{1}{25}$

$ \leftrightarrow (3a\minus{}1)^2(6a\plus{}1) \ge\ 0$

$ \rightarrow LHS \le\  \frac{27}{25}(a\plus{}b\plus{}c)\plus{}\frac{3}{25}\equal{}RHS$

We have done! :)

Thank you! Your solutions are very nice! Speed!

onlylove_math  used Tangent lines
I think so

[quote="D.I.Culianop"]Continue with an ine from USA MO 2003: Prove that for any positive real numbers $ a, b, c$ the following ine holds
$ \frac {(2a \plus{} b \plus{} c)^2}{2a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {(2b \plus{} c \plus{} a)^2}{2b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {(2c \plus{} a \plus{} b)^2}{2c^2 \plus{} (a \plus{} b)^2} \leq 8$
I'm trying with a way of TiTu Andreescu teacher. But the end of this ine so hard. I can't understand![/quote]
I think it 's similar  :wink:

Assum  $ a\plus{}b\plus{}c\equal{}3$

Need prove

$ \frac {(a\plus{}3)^2}{2a^2 \plus{} (3\minus{}a)^2} \plus{} \frac {(b\plus{}3)^2}{2b^2 \plus{} (3\minus{}b)^2} \plus{} \frac {(c\plus{}3)^2}{2c^2 \plus{} (3\minus{}c)^2} \leq 8$

We have :

$ \frac {(a\plus{}3)^2}{2a^2 \plus{} (3\minus{}a)^2} \le\ \frac{4}{3}a\plus{}\frac{4}{3}$

$ \leftrightarrow (12a\plus{}9)(a\minus{}1)^2 \ge\ 0$

We have done! :)

[quote="D.I.Culianop"]Let $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$. Prove that: $ \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \leq \frac {6}{5}$[/quote]

This problem is from Japan Olymplics.(1997)
You can find it.

[quote="D.I.Culianop"][quote="onlylove_math"]
...
We have :

$ \frac {(a \plus{} 3)^2}{2a^2 \plus{} (3 \minus{} a)^2} \le\ \frac {4}{3}a \plus{} \frac {4}{3}$

...
[/quote]
^^ Can you show me the way to this step?[/quote]

$ \frac {(a \plus{} 3)^2}{2a^2 \plus{} (3 \minus{} a)^2} \le\ xa \plus{} y$

We 'll find $ x,y$

Let $ f(a) \equal{} \frac {(a \plus{} 3)^2}{2a^2 \plus{} (3 \minus{} a)^2} \minus{} xa \minus{} y$

Because equality hold on $ a \equal{} b \equal{} c \equal{} 1$. So we have $ f(1) \equal{} 0$ and $ f '(1) \equal{} 0$

So we can find $ x;y$ from $ f(1) \equal{} 0 ; f '(1) \equal{} 0$  :wink:

Please show me my solution is true or false!

[quote="D.I.Culianop"][quote="D.I.Culianop"]Let $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$. Prove that: $ \frac {a(b \plus{} c)}{a^2 \plus{} (b \plus{} c)^2} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} (c \plus{} a)^2} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} (a \plus{} b)^2} \leq \frac {6}{5}$[/quote]
This is my solution: LHS = $ \frac {\frac {b \plus{} c}{a}}{1 \plus{} (\frac {b \plus{} c}{a})^2} \plus{} \frac {\frac {c \plus{} a}{b}}{1 \plus{} (\frac {c \plus{} a}{b})^2} \plus{} \frac {\frac {a \plus{} b}{c}}{1 \plus{} (\frac {a \plus{} b}{c})^2} \equal{} \frac {x}{1 \plus{} x^2} \plus{} \frac {y}{1 \plus{} y^2} \plus{} \frac {z}{1 \plus{} z^2} \leq \frac {6}{5}$
With $ x, y, z \geq 0$ and $ x \plus{} y \plus{} z \plus{} 2 \equal{} xyz$ (It is from "Two useful substitutions - Titu Andreescu").
Assum $ x \plus{} y \plus{} z \equal{} \frac {12}{5}$. We have:
$ \sum \frac {x}{1 \plus{} x^2} \equal{} \sum x.\frac {1}{1 \plus{} x^2} \equal{} \sum x.\frac {(\frac {1}{2} \plus{} \frac {1}{2})^2}{1 \plus{} x^2} \leq \sum x(\frac {1}{4} \plus{} \frac {\frac {1}{4}}{x^2}) \equal{}$
$ \sum (\frac {x}{4} \plus{} x.\frac {\frac {1}{4}}{x^2}) \equal{} \sum (\frac {x}{4} \plus{} \frac {\frac {1}{4}}{x}) \equal{} \sum (\frac {x}{2}) \equal{} \frac {x \plus{} y \plus{} z}{2} \equal{} \frac {6}{5}$
^^! I don't know it's true or false? Hep me![/quote]
I think It has some mistake.

We can't assum $ x \plus{} y \plus{} z \equal{} \frac {12}{5}$

And why you obtains $ \sum (\frac {x}{4} \plus{} \frac {\frac {1}{4}}{x}) \equal{} \sum (\frac {x}{2})$

$ \frac{1}{\frac{4}{x}} \equal{} \frac{x}{4}$ ^^! Sometimes iam ... and i don't know What am i doing!

Why do i can assume " $ x\plus{}y\plus{}z\equal{}\frac{12}{5}$" ???

halo...
after you transform $ LHS$ to $ \frac{x}{1\plus{}x^2}\plus{}\frac{y}{1\plus{}y^2}\plus{}\frac{z}{1\plus{}z^2}$, you've just made new inequality which is not homogenous. so you can not assume $ x\plus{}y\plus{}z\equal{}\frac{12}{5}$.

sorry for my bad english...

[quote="Raja Oktovin"]halo...
after you transform $ LHS$ to $ \frac {x}{1 \plus{} x^2} \plus{} \frac {y}{1 \plus{} y^2} \plus{} \frac {z}{1 \plus{} z^2}$, you've just made new inequality which is not homogenous. so you can not assume $ x \plus{} y \plus{} z \equal{} \frac {12}{5}$.

sorry for my bad english...[/quote]
Why isn't $ \frac {x}{1 \plus{} x^2} \plus{} \frac {y}{1 \plus{} y^2} \plus{} \frac {z}{1 \plus{} z^2}$ homogenous?

It's from Viet Nam university entrance exam. And i think the solution use Cauchy-Schwarz is the best solution!


Does anybody like my constants ( at my signature )  :blush:  :?:

i find them original.  the "big" question is are they rational or irrational  :?: algebraic or not  :?:

i believe it is easy to prove the transcendality of the first and the irrationality or rationality of the second.

i really believe it is simple but i cant see it yet  :oops:  :oops:

i even expect the second constant to be algebraic of low degree if not rational.

but im just guessing here, can anybody prove their irrationality  :?:

any comments would be appreciated

greetz
tommy1729

On the second constant, what do you mean by 'written backwards'  ?    :huh:


calculus
integration
calculus computations

You might start by trying to figure out what it means- it looks like nonsense to me.


MATHCOUNTS
quadratics
algebra
quadratic formula
special factorizations

Try to derive the quadratic formula of (-b :pm:  :sqrt:(b :^2:-4ac))/2a.

Post your thoughts, work, full solution, etc.  Please use spoiler.

Yup, but why did you add [hide](B/2A):^2:[/hide]?[/hide]

Yeah, I just wanted people to know why that step was taken.

[hide](x+y)^2=x^2+2xy+y^2.

Suppose you have x^2+16x, and you want to complete the square.  16x=2xy, or y=8.  So, y^2 is 64.  Then the rest is just a generalization.[/hide]

yes...I was completing the square...that's how we learned it and how we had to do it on our quiz today...

Quote:[hide]so he could use the completing the square as rep123max said??[/hide]

he?!?!?!?!?!

Pay very close attention to the process of completing the square.  It's not simply a method that will derive the quadratic formula that you should simply forget thereafter.  Completing the square is by far the more important part of the process of solving quadratics and can shed light on many interesting math problems.  Here's an example:

Find all real ordered pairs of solutions, (x, y), to the system of simultaneous equations
x2 - 14y + 73 = 0,
y2 - 10x + 1 = 0.

Can anyone solve this?

MCrawford wrote:Find all real ordered pairs of solutions, (x, y), to the system of simultaneous equations
x2 - 14y + 73 = 0,
y2 - 10x + 1 = 0.

Can anyone solve this?

Hint:

[hide]Adding together and completing square: (x-5)2 +(y-7)2 = 0 [/hide]

Fuzzy, what are you doing answering challenge questions on this forum. You know better. *wags a finger at you*

As an extension, try using the same method on this system:

x^2 - 14y + 64 = 0

y^2 - 10x + 1 = 0

Observations (spoiler):

[hide]

1.  This made me think about the graphs both of the original systems and of the 'solutions'.

2.  Is the algebraic solution to system #2 correct or not? Why?

3.  If the method does not yield the solution to system #2, is it trustworthy for system #1?

4.  What IS the answer to system #2?  Method both with and without TI?[/hide]

Adding the equations gives [i]potential[/i] solutions.  Anything that satisfies the two original equations necessarily satisfies their sum.  However, the converse is not true.

4 = 4
6 = 6
10 = 10

But from
10 = 10
4 = 7
6 = 3
doesn't work.

It's a matter of keeping track of the meaning of each act you perform on equations.

[quote="captcha000"]so is there any other way to be sure that you find the all the right answers other than substituting??[/quote]

Once you add the equations and complete the square it should be obvious that the solutions set is very limited.  Then yes, you need to be sure that those solutions also fit the original equations.  Sometimes there might be a clever way available to do that, but don't count on it.


This was a question on my Algebra 1 math test that I took this weekend.

-- In how many zeros does 920! end.  I know what the answer is, but I have no idea how to get to the answer. Does anyone know a way to solve this besides doing 920*919*918...*3*2*1?

lol that from the Vestavia test???? did u join cuz the guy mentioned it? thats y i joined but heres the way 2 do that question.[hide]you no to have a 0 at the end you need to have a multiple of 5 and 2. Since there are more 2s than 5s, you find how many 5s. You find this by taking 920 and dividing it by 5, then 25, then 125, then 625. Then add them togethor. so it would be $184+36+7+1=228$[/hide]

You can find the number of zeroes by finding how many 10s are there in the factorization. 10 factors to 2*5, so you want to find how many 5's there are, since 2's obviously outnumber 5's, and you want to find how many pairs of 5 and 2 there are.

For 920!, 920/5=184 numbers with atleast 1 five
add one more for each 5 squared, 920/25=36
5^3, 950/125=7
5^4, 920/625=1

184+36+7+1=228

WOW THAT WAS EASY. Since there are obviously more 2's than 5's you find how many 5's since your trying to find a multiple of 10. You can do it either 1 of 2 ways.

First: [hide]Divide 920 by powers of 5 until the power of 5 is bigger than 920. so 920/5=184, 920/25=36, 920/125=7, 920/625=1. Then you add 184+36+7+1=228
[/hide]

Second: [hide]Or you can do 920/5=184/5=36/5=7/5=1 184+36+7+1=228[/hide]


well once in a while, I would want LaTeX not to pack my paragraphs too close to each other. So I use the \\ command with its parameter as [5 mm]. This will give me a reasonable line distance between the paragraphs.

This is done by trial and error, thus it's not really ideal (being a math. student I have unfortunately developed a hatred towards things that can't be use to construct or solve complex things). So I was wondering if there exist any documentation that actually gives a definition/example/whatever that shows what do 1 point, 1 mm, etc. would be like.

I usually do \vskip x in (where x can be a decimal, such as .25). That's for "vertical skip x inches". there's also \hskip.

Well, intuitively the parameters just indicate what the length of the skip is.  You can get out a ruler to see how long a skip 1 mm or 1 in is - alternatively, you could just generate a test file with two lines (using \hline, for example) and a skip between them to see how big the space is.  A pt is 1/72 of an inch, if you're wondering.

thx for the reponse :). I was being dumb, totally forgot about my ruler.

Since my article is 12pt., I think skipping one line would mean skipping 12pt. The test result is good so far :)

[quote="DPatrick"]\vspace{\baselineskip} will skip exactly one line.[/quote]
Oh, wow - that's very handy!

[quote="DPatrick"]\vspace{\baselineskip} will skip exactly one line.[/quote]

lol that's cool except the command is quite long :P. I think I will use this since it's very easy to remember (I think). Thank you for helping out.


geometry
circumcircle
inradius
inequalities
perimeter
geometric inequality
inequalities proposed

Given triangle $ABC$ circumradius $R$, inradius $r$ area $S$ prove that
$\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\ge\frac{27}{8S(4R+r)}$

[color=purple]If i have enough time and intelligent i'll solve it.But gemath can you help me ? Here is a problem of JiChen , too
    http://www.mathlinks.ro/Forum/viewtopic.php?t=140281            [/color]

Don't reply meaningless, if you can do post your solution!
And if you asked Ji Chen and he didn't want to help you with his inequality then why can I help you ? However see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=3485]the link[/url]  there is a solution of Harazi

It is one of my solution and the other way we can prove it base on the stronger inequality
$\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\ge\frac{9}{8S(4R+r)}+\frac{1}{s}(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}})\ge\frac{27}{8S(4R+r)}$

[quote="gemath"]Given $ \triangle ABC$ with circumradius $ R$, inradius $ r$ and area $ \Delta$, prove that

$ \frac {1}{a^{3}} \plus{} \frac {1}{b^{3}} \plus{} \frac {1}{c^{3}}\ge\frac {27}{8\Delta(4R \plus{} r)}$.[/quote]
Actually, the following refined inequality chain holds

$ \sum{\frac {1}{a^{3}}}\geq\frac {1}{9}\left(\sum{\frac {1}{a}}\right)^3\geq\frac {3}{8}\prod{\left(\frac {1}{b} \plus{} \frac {1}{c}\right)}\geq\frac {27}{8\Delta(4R \plus{} r)}$.

Let $ a,b,c,$ are the sides and $ s$ is the semi-perimeter, then

$ \Delta \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$,

$ R \equal{} \frac {abc}{4\sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}}$,

$ r \equal{} \sqrt {\frac {(s \minus{} a)(s \minus{} b)(s \minus{} c)}{s}}$.

Put $ s \minus{} a \equal{} x,s \minus{} b \equal{} y,s \minus{} c \equal{} z$, the last geometric inequality is equivalent to the following algebraic inequality

$ (x \plus{} y \plus{} z)(yz \plus{} zx \plus{} xy)\geq \frac {9(y \plus{} z)^2(z \plus{} x)^2(x \plus{} y)^2}{(2x \plus{} y \plus{} z)(2y \plus{} z \plus{} x)(2z \plus{} x \plus{} y)}$

for positive real numbers $ x,y,z$.

[hide="Proof."]$ (x \plus{} y \plus{} z)(yz \plus{} zx \plus{} xy)(2x \plus{} y \plus{} z)(2y \plus{} z \plus{} x)(2z \plus{} x \plus{} y) \minus{} 9(y \plus{} z)^2(z \plus{} x)^2(x \plus{} y)^2$

$ \equal{} 2\left[x^5(y \plus{} z) \plus{} y^5(z \plus{} x) \plus{} z^5(x \plus{} y)\right] \plus{} 2xyz(x^3 \plus{} y^3 \plus{} z^3)$

$ \minus{} 4(y^3z^3 \plus{} z^3x^3 \plus{} x^3y^3) \minus{} xyz\left[x^2(y \plus{} z) \plus{} y^2(z \plus{} x) \plus{} z^2(x \plus{} y)\right]$

$ \equal{} 2\sum{yz\left(y^4 \plus{} z^4 \minus{} 2y^2z^2\right)} \plus{} xyz\sum{\left(y^3 \plus{} z^3 \minus{} y^2z \minus{} yz^2\right)}$

$ \equal{} \sum{yz(y \plus{} z)(x \plus{} 2y \plus{} 2z)(y \minus{} z)^2}\geq0$.[/hide]
[hide="Exercises."]If $ a,b,c$ are the sides of a triangle $ ABC$, $ R$ is the radius of circumcircle, $ r$ is the radius of incircle and $ \Delta$ is the area, then

$ \frac {1}{a^{3}} \plus{} \frac {1}{b^{3}} \plus{} \frac {1}{c^{3}}\ge\frac {51}{8\Delta(6R \plus{} 5r)}$;

$ \frac {1}{a^{3}} \plus{} \frac {1}{b^{3}} \plus{} \frac {1}{c^{3}}\ge\frac {3\sqrt {3}}{4Rr(R \plus{} 7r)}$;

$ \frac {1}{a^{3}} \plus{} \frac {1}{b^{3}} \plus{} \frac {1}{c^{3}}\ge\frac {1}{\sqrt {3}r\left(R^2 \plus{} 4r^2\right)}$.[/hide]


trigonometry
vector
analytic geometry
geometry
3D geometry
tetrahedron

Let $ \alpha ,\ \beta ,\ \gamma$ be acute angles such that $ \cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma =1$.

Prove that $ \frac{3}{4}\pi <\alpha+\beta+\gamma <\pi$.

can we somehow use the following fact for vectores?
let $ \vec{v}\in \mathbb{R}^{3}$ be a vector which has angle $ \alpha,\beta,\gamma$ with coordinates $ x,y,z$ then we have that:
$ \cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1$
and also let $ x_{1},x_{2}\in \{\alpha,\beta,\gamma\}$ and $ x_{1}\not= x_{2}$ then $ \frac{\pi}{2}\leq x_{1}+x_{2}$
because in this case we get that:

$ \frac{\pi}{2}\leq \alpha+\beta$

$ \frac{\pi}{2}\leq \beta+\gamma$

$ \frac{\pi}{2}\leq \gamma+\alpha$

$ \Rightarrow \frac{3}{4}\pi\leq \alpha+\beta+\gamma$

[hide="Hint"]Consider the tetraedron $ OABC$ for which $ OA\perp OB\perp OC\perp OA$.
Then for the angles $ \alpha$, $ \beta$, $ \gamma$ between the planes $ (OBC)$, $ (OCA)$, $ (OAB)$ with the plane $ (ABC)$ exists the relation $ \cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma =1$.[/hide]


The rate at which a river flows is one-third the speed of a boat in still water. If that boat travels down and then back upriver for two hours, it will be 16 miles short of its starting point. What is the speed of the boat in still river?


This is really weird.

 Problems - Correct: 5    Incorrect: 0    Give Up: 0    Percentage: 100.00

AoPS Karma: 1000

I've been doing all these problems, but and rating all of them, but I don't appear to be gaining any karma.
Can any help :?:  :?:

If you're in the diagnostic phase, you don't earn karma.

I thought there wasn't going to be a diagnostic phase anymore?

Before, you just had to do 60 problems.
Is it the same?
Can anyone :help::?:

I think it is a 10 question diagnostic phase now, because it took me 10 questions before I got out of the diagnostic phase and I did like 2-3 problems before the release of the new version.
Also did you watch any videos?

You can't watch videos during the diagnostics phase, unless you mean when Alcumus was down.

So basically the solution to your problem, atleast from what we know, is just to do a few more problems.  :P

What's really weird is that I can watch videos, but I'm not gaining/losing any karma.

If you have rated the problems and watched videos evenly, the karma would appear to be the same because the karma you spent and gained is equal. Otherwise, you must finish watching a video or do a problem to lose karma. If you appear to have finished watching the video and you still stay at the same karma, its probabily a system error.

i think that is weird too but you really don't want to complain since you don't want to waste karma ok?

Well, there is a phase where you dont earn karma, and since you have done 5 (your phase is 60 and below) you wont earn or lose any.

why does it matter? you don't lose any karma points anyways and your on a pretty good streak.

Watch videos and rate/comment them
this gains you 5 karma points (watching a video = -5 pts/ rating a video = +5/ commenting a video = +5; 10-5 = +5)
Also you can report bugs (if successful, you earn +10 points i think)
Solving problems and rating them (+/- 0)

See [url]http://www.artofproblemsolving.com/Alcumus/Instructions.php#karmapoints[/url].

Also, note that there are three categories in which you can rate problems.  Thus, if you solve a problem (-5) and rate it in all 3 categories (+5*3=15) you will end up gaining 10 karma.

[quote="Butler"]Everything these guys have said doesnt work for me. I rate and comment on both the videos and problems, yet my karma keeps getting lower. :o  :o  :o[/quote]

I would conduct some tests.

1: Check your karma, solve a problem w/o rating it, and check your karma again.
2: Check your karma, solve a problem and rate in one category, and check your karma again.

With a few more tests like this, you may be able to pinpoint the problem and get help from developers/admins.

Ah, now i get. I guess the computer doesn't always update when i refresh it. :oops:


I don't get how you choose y's or x's from (x+y)s
$(x+y)^5$, etc.
like the counting part in its expansion $\binom{5}{0}x^5y$, and so on I believe

Help?

In the binomial expression $(x + y)^n$, the power of $x$ is increasing is decreasing starting from $x^n$ and the power of $y$ is increasing starting from the second term.

For example, $(x + y)^4$, using the Pascal's triangle you will have the coefficients $1,4,6,4,1$.  Then plugging in the values of $x$ and $y$, you will have $x^4 + 4x^3y + 6{x^2}y^2 +4xy^3 + y^4.$

As you can see the power of x is increasing and the power of y is decreasing.

Hope it helps!


A sequence $ \{a_n\}$ is given by:

$ a_1\equal{}1$, $ a_{n\plus{}1}\equal{}\frac{3a_{n}\minus{}1}{4a_{n}\minus{}1}$.

Find $ a_n$.

hello, what du you mean with find $ a_n$? do you mean an explicit function $ f(n)$ for that sequence?
Sonnhard.

Hi,Sonnhard.

I mean we should find the nth term of this sequence in terms of $ n$.


AMC
AIME
logarithms
LaTeX
number theory
relatively prime

$\frac{2}{\log_{4}{2000^6}}+\frac{3}{\log_{5}{2000^6}}$ can be written as $\frac{m}{n}$ where $m,n$ are relatively prime positive integers. Find $m+n$.

Answer:

[hide]Use the property of logs and take out $6$ from both sides and multiply it to the logs. Then when you bring the logs to the numerator, you switch the bases so now it would be $\log_{2000}{4}+\log_{2000}{5}$. Then you can bring the $2$ and $3$ to the exponent of $4$ and $5$ so it would be $\log_{2000}{4^2}+\log_{2000}{5^3}$ and then you can use property of logs to multiply so $\log_{2000}{4^2(5^3)}=\log_{2000}{2000}=1$, and thus the fraction would be $\frac{1}{6}$ and $m+n=7$.[/hide]

[quote="deej21"]Iverson, you don't write "of" in the middle of latex commands.  Here's how to properly write a log in base-argument format.

$\log_{base} {argument}$

code: \log_{base} {argument}

And just for clarification, the "switching the bases" Iverson refers to is the property $\log_a b = \frac 1 {\log_b a}$ for $a, b > 0$[/quote]

Sorry deej, I fixed it :)

[quote="Rushil"]AIME is basically equiv to RMos in other countries. It's this simple!!! :rotfl:  :rotfl:  :rotfl:[/quote]Just out of curiosity, what's RMO?

Regional Mathematics Olympiad. Most big countries hold this in different states/regions etc. as the first stage of the Olymopiad before the National Olympiad

What is your point???  I don't really see what is so funny.  If you knew about the AMC's then you would know that the AIME is just like a RMO, as described by you.  Also, a bunch of your recent posts are borderline spam Rushil...they're not yet spam but I thought I would just inform you that they are nearing it.

Sorry if you thought I was spamming. I'll try not to do so in the future.

Also I don't find anything funny.. just couldn't find a perfect emoticon

My point is that usually the RMOs(the one's I have seen) are subjective and tougher. I was wondering how students who get through the (as I see - I might be wrong) easier first stage are able to win almost the highest no. of golds at IMO!!!!

Rushil, why have you been bringing back a few of my really old threads? ;)


Let $ ABC$ is a triangle . $ M$ is a arbitraty point in plane . Bisectors of angles $ BMC$ , $ CMA$ , $ AMB$ intersects $ BC$ , $ CA$ and $ AB$ in points $ A_1$ , $ B_1$ , $ C_1$ accordingly . Prove that $ S_{A_1b_1C_1}
\le \frac{1}{4}S_{ABC}$.

[quote="Sergic Primazon"]Let $ ABC$ is a triangle . $ M$ is a arbitraty point in plane . Bisectors of angles $ BMC$ , $ CMA$ , $ AMB$ intersects $ BC$ , $ CA$ and $ AB$ in points $ A_1$ , $ B_1$ , $ C_1$ accordingly . Prove that $ S_{A_1b_1C_1} \le \frac {1}{4}S_{ABC}$.[/quote]
Because exists $ O$ for which $ AA_{1}\cap BB_{1}\cap CC_{1}\equal{}\{O\}.$ :wink:


AMC
USA(J)MO
USAMO
Princeton
college
number theory

a few q's
1) what score on a section (particular NT/algebra/combinatorics) is typically needed to make individual finals? how many kids usually make it? Is just making the finals very prestigious on college apps or do u need to win it (as compared to say making USAMO)? Are there trophies-prizes etc. Is a month too short to prepare? Thanks,
mathemonster

If college applications are the only thing motivating you to study for this competition, then perhaps you should rethink your motivations for entering the competition at all.  I don't mean this in a harsh way, but shouldn't math competitions be [i]fun?[/i]  :)

(But for what it's worth, this is a very young competition, so I doubt most colleges take it very seriously with the possible exception of Princeton itself.)

The competition itself is only three years old so a ranking in division A won't be quite as prestigious as making USAMO.
If you're in division B, you needed a score of 11/40 in number theory and 8/40 in combinatorics.
If Division A, you needed a 13/40 in number theory and 14/40 in combinatorics.

(But the Division A problems were harder than Division B last year)

i thought there were only 10 q's for NT and combo each so wouldn't the points be out of 10?

no, each question was worth more than one point -- the first three were each worth three points, the next four were each worth four, and the final three were worth five

Here's the link to last year's individual results in Division A (for all four tests).

[url]http://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition_2007_Division_A_results[/url]

They are changing the rules for this year, it seems though. Look here: [url]https://cgi.math.princeton.edu/mathclub/index.php/Official_Competition_Rules[/url]

[b]Each correct answer will be worth 1 point, so that a full team of ten students can gain a maximum of 160 points from the Individual Tests.[/b]

Unfortunately, someone got his/her math wrong. :P

[b]A full team will consist of 8 members.[/b]

Thus, a full team of eight students can get a maximum of 160 points, since an individual test is 10 points, one student can take two tests, and one team can have eight students. If a team had ten students, then the maximum score from the Individual Tests would be 200 points.

Just pointing that out.


There are five statements:
[b](i)[/b] Statement [b](ii)[/b] is true
[b](ii)[/b] At most one of these five statements is true
[b](iii)[/b] All of these five statements are true
[b](iv)[/b]
[b](v)[/b]

The last two statements are in invisible ink. Which of the statements are true?

[quote="236factorial"][hide]If (i) were true, then (ii) would be true. However, (ii) says that only one can be true, and it contradicts what we first proposed. Thus (i) can't be true, and neither can (ii). Since (i) and (ii) can't be true, (iii) can't be true either. Now how will I determine whether the ones in invisible ink are true? [/hide][/quote]
You almost finished 236factorial.
[hide="Check this out"]So you already said (i),(ii),(iii) are all false. If (iv),(v) are both false, (ii) would become true because no statement is true and statement (i) becomes true, contradiction. If one of (iv),(v) is true, same contradiction as above. If (iv),(v) are both true, then (i),(ii),(iii) remain false. Conclusion: (iv) and (v) are both true[/hide]

[quote="10000th User"]
You almost finished 236factorial.
[hide="Check this out"]So you already said (i),(ii),(iii) are all false. If (iv),(v) are both false, (ii) would become true because no statement is true and statement (i) becomes true, contradiction. If one of (iv),(v) is true, same contradiction as above. If (iv),(v) are both true, then (i),(ii),(iii) remain false. Conclusion: (iv) and (v) are both true[/hide][/quote]

Good job  :)


probability
logarithms
probability and stats

A certain  insurance company devised a model to predict the price change of a certain action. The resultig model was $ Ln(X_k \plus{} 1/X_K)$ and has a certain  distribution with expectation 0.01 and variance 0.0009, where X_i is the  closing price  of the action on  day  i. Assuming that  the daily price changes are independent  day  after  day and that  the closing price of the action on day 1 is 1 dollar.  Calculate the probability that the closing price of the action on day 101  is at  least  4 dollars.

 Note.- The subindice on the numerator is supposed to be k+1

Some questions to clarify:

What is L and n?
The subindex in the numerator is supposed to be k+1 means  $ \frac{X_{k\plus{}1}}{X_k}$?
The given distribution is Gaussian?

Sorry  for not making  my  post  very  clear.  Anyway  Ln =  natural  logarithm, your  guess is correct on the subindice , and for  the last question it is not  known  exactly  which distribution it has ,  just  that its expectation is .01 and its variance .0009. I think the whole Ln part  is just  a bluff. What  i think  is the solution is something  of this nature :

  Define $ Y_i$ as  the  price change  after day i. So  we want to calculate P[ Y_1+Y_2+Y_3+...+Y_100 > 3].
 We know  that  the Y_i 's are independent  and identically distributed (who knows exactly which one) we also know the expectation and variance , so  we can  apply the central limit theorem  to  approximate the probability.  I posted  this   to confirm  answers.

we want $ \sum_{k\equal{}1}^n \ln ( \frac{x_{k\plus{}1}}{x_k} ) \equal{} \ln \left( \prod_{k\equal{}1}^n \frac{x_{k\plus{}1}}{x_k} \right) \equal{} \ln \frac{x_{n\plus{}1}}{x_1} \equal{} \ln x_{n\plus{}1} > \ln 4$

$ P(S_{100}>\ln 4) \equal{} P( z > \frac{ \frac{\ln 4}{100}\minus{}.01}{\frac{.0009}{\sqrt {100}}} )$

$ P(z >  \frac{1000(\ln 4 \minus{} 1)}{9})$

same z value means same p value so i didnt look it up


inequalities
geometry
parallelogram
triangle inequality
inequalities unsolved

Given complex number $a,b,c$. Let $|a+b|=m,|a-b|=n$. Suppose $mn \neq0$. Prove that
\[\max\{|ac+b|,|a+bc|\}\geq\frac{mn}{\sqrt{m^{2}+n^{2}}}. \]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130463]Whole Paper[/url]

I don't understand your last step... and, can't the square of a complex number be negative?

Anyways, we proceed similarly -- we prove the stronger result that
\[(m^{2}+n^{2})(|ac+b|^{2}+|a+bc|^{2}) \geq 2m^{2}n^{2}. \]
We have $m^{2}+n^{2}= 2|a|^{2}+2|b|^{2}$ (it's not hard to prove by expanding out everything with conjugates, but is there a name for this?) so Cauchy and the triangle inequality yield
\[(|b|^{2}+|a|^{2})(|ac+b|^{2}+|a+bc|^{2}) \geq (|abc+b^{2}|+|a^{2}+abc|)^{2}\geq |b^{2}-a^{2}|^{2}= m^{2}n^{2}. \]
which gives us the desired inequality...

[quote="ThAzN1"]I don't understand your last step... and, can't the square of a complex number be negative?[/quote]
For complex number, the last step is
\[=|(|a|^{2}+|b|^{2})c+a\overline{b}+\overline{a}b|^{2}\geq 0.\]

[quote="ThAzN1"]
We have $m^{2}+n^{2}= 2|a|^{2}+2|b|^{2}$ (it's not hard to prove by expanding out everything with conjugates, but is there a name for this?) [/quote]

Of course, this is called (at least from where I came from :) ) a parallelogram theorem. If you look more cearfuly you can see that $m$ and $n$ are length of diagonals in parallelogram which sidelengths are $a$ and $b$ :)

Bye

You can reduce the problem to,
By the formula,
$\max \{x,y\}= \frac{1}{2}|x-y|+\frac{1}{2}(x+y)$
Whatever it is.


Consider the sequencev {$ x_{n}$} define by :
       $ x_{1}$=$ a$ $ \geq$ 1 ;$ x_{n\plus{}1}$= 2$ x_{n}^{2}$-1
   Determine the formular of $ x_{n}$.

If $ x_{n}\equal{}ch(y)$, then $ x_{n\plus{}1}\equal{}ch(2y)$.

If you don't understand Rust's solution then put $ a \equal{}\frac{1}{2}(b\plus{}1/b)(b\geq 1)$ then by induction on $ n$ we have $ x_{n}\equal{}\frac{1}{2}(b^{2^{n\minus{}1}}\plus{}b^{\minus{}2^{n\minus{}1}})$.


Counting Diagonals with Division

We use division to correct for overcounting to find the number of diagonals in an octagon and in a complex polyhedron.

how do calculate the area of the shape??  :?:

Well, there were 18 squares with side length 1, and 8 equilateral triangles with side length 1.  Thus, the surface area is $18+4\sqrt{\frac{3}{4}}=21.46$ square units.

Although I am curious how to compute the volume :D.

To calculate the volume, use complementary counting. Find the volume of the cube that that the polygon fits in and subtract the volume of the triangular prisms at the corners. There are 48 edges.

@bluecarneal: where do you live in florida? I live in florida also. send me a private message.


So in homology we have
[X = things that are boundaries of something else] $ \subseteq$ [Y = things whose boundary is 0]

But in cohomology we have
[A = things that are the coboundary of something else] $ \subseteq$ [B = things that are 0 on anything whose boundary is 0] $ \subseteq$ [C = things that are 0 on anything that is the boundary of something else]

(A,B,C,X,Y are letters I'm making up to refer to it)

So in homology we take Y/X, but in cohomology, we don't take C/B (which is sort of like Y/X), but we take C/A, which gives that extra B/A. I've been told that cohomology gives more information than homology does - is this why?

the reason is that the (singular, e.g.) cohomology groups constitute a cohomology *ring*; the multiplication is the cup product. this ring structure inherits more topological information than just the group structure.

Although the picture gets even worse, because the cohomology ring in the suspension of a space has no nontrivial products; this means its impossible to just use the cohomology ring structure to determine whether or not two spaces are stably homotopic (meaning, in the limit, their suspensions are homotopic). There are natural transformations $ H^n(\minus{}) \to H^{n\plus{}m}(\minus{})$ called cohomology operations that can tell you a bit more, especially if these operations are stable (natural w.r.t the suspension functor). To be fair, some of these operations (steenrod squares) are derived [i]from[/i] the cup product, but still... :)


Guys post ur scores as and wen u appear for BITSAT........

Unlike other exams, there will be no ambiguity in your scores as you'd get 'em as soon as you complete the test........so let's be honest and not post saying I'd get a score say between x and y or this is a probable solution or something......... :huh:

(although this is not worth mentioning, just felt a little duty bound and compelled to type it out!)

Godspeed people.....Godspeed :ninja:

theres already a thread for it in GD, started by yours truly!


I've had discussions(and fights :lol: ) with my friebds on whether to be successful,u have to be a greedy pig and have 0 mral values or be a good,kind person?

what da ya guys think? ;)

well, first of all, it depends on what you mean by successful. second of all, you don't need to be greedy or kind to have success. anywhere in between works too.

Most successful people I know also spell/typo check their memos. It's definitely a start. :)

to be successful, you either need a lot of luck or skill, one or the other usually.

Being ruthless/ amoral is definitely not a necessary condition for being successful (although I'm sure that [i]some [/i] degree of self-interest is usually required). [b]However[/b], I have observed many mediocre talents rise to positions above what they deserve by being ruthless. Such is the nature of our social institutions-- it is easier to bring other people down than to make oneself excel. If we lived in a culture that put more emphasis on fair play than just winning, perhaps things would not be this way.

Also, competing with ruthless people may put one in a situation where one must also be ruthless or get screwed over (like in the prisoners' dillemma). It is best to avoid such things altogether, in my opinion.

i agree, being ruthless won't get you anywhere because success has a lot to do with how people see you in society.

Do you mean successful in this life or the afterlife?

i think he means this life........... :?

This seems like a rather easy question

If being successful means being a good person in life, then the answer is obvious.

Everyone has their own definition of personal success. I figure that if I'm happy in life, then I'll be successful.

if i am a Christian and live it out, i have had a successful life :)


Structure two positive series such that:
(1) $\sum a_n,\sum b_n$,disverges
(2)$a_n \geq a_{n+1},b_n \geq b_{n+1}$
(3)$\sum \min\{a_n,b_n\}$ converges

Take any convergent positive series $\sum c_n$, where $c_n$ is non-increasing, and then, for an increasing sequence $(k_n)_{n\ge 1}$ consisting of positive integers, put $a_m=c_{k_{2n-1}},\ \forall m\in[k_{2n-1},k_{2n}-1]$, and $b_m=c_{k_{2n}},\ \forall m\in[k_{2n},k_{2n+1}-1]$ (these relations hold for all $n\ge 1$). It's obvious that the sequences $(a_n)_n,(b_n)_n$ are non-increasing, and, if $k_{n+1}-k_n$ is sufficiently large, the series $\sum a_n,\sum b_n$ wil be divergent.

P.S. I forgot to mention: $a_m=c_m$ for all $m\in[k_{2n},k_{2n+1}-1]$, and $b_m=c_m,\ \forall m\in[k_{2n-1},k_{2n}-1]$.

There might be some mistakes, but I'm sure you'll see what I mean if you draw a picture :).


advanced fields
advanced fields unsolved

The proof is not included in the book I found the problem in?  Hints and/or solutions?

Prove:

$ \displaystyle\sum_{n \equal{} \minus{} \infty}^{\infty}{( \minus{} 1)}^{n} {{{q}^n}^2} \equal{} \displaystyle\prod_{n \equal{} 1}^\infty\frac {1 \minus{} q^n}{1 \plus{} q^n}$.

Do you mean $ q^{n^2}$? $ (q^n)^2$ is not the same, and what you wrote doesn't look good anyway.

We have the series expansion $ \frac{1\minus{}t}{1\plus{}t}\equal{}1\minus{}2t\plus{}2t^2\minus{}2t^3\plus{}\cdots$. Multiply all of them together and this is a combinatorial identity; the number of partitions of $ n$, weighted by $ 2^j$ where $ j$ is the number of distinct terms and by $ (\minus{}1)^k$ where $ k$ is the total number of terms, is $ 2\cdot (\minus{}1)^n$ when $ n$ is a nonzero perfect square and zero otherwise.

So there's a hint: you've misclassified this, and it's really combinatorics.

hello, see here
http://mathworld.wolfram.com/EllipticThetaFunction.html
Sonnhard.

[quote="jmerry"]Do you mean $ q^{n^2}$? $ (q^n)^2$ is not the same, and what you wrote doesn't look good anyway.

We have the series expansion $ \frac {1 \minus{} t}{1 \plus{} t} \equal{} 1 \minus{} 2t \plus{} 2t^2 \minus{} 2t^3 \plus{} \cdots$. Multiply all of them together and this is a combinatorial identity; the number of partitions of $ n$, weighted by $ 2^j$ where $ j$ is the number of distinct terms and by $ ( \minus{} 1)^k$ where $ k$ is the total number of terms, is $ 2\cdot ( \minus{} 1)^n$ when $ n$ is a nonzero perfect square and zero otherwise.

So there's a hint: you've misclassified this, and it's really combinatorics.[/quote]

Oh, sorry about that.  I mean $ {q}^{n^2}$

hello, i think he meant
$ \sum_{n\equal{}\minus{}\infty}^{\infty}(\minus{}1)^nq^{n^2}$
Sonnhard.


algebra
system of equations
geometry proposed
geometry

Prove that the triangle $ABC$ is equilateral if and only if the following system has at least two solutions and in this case interpret geometrically:
\[ x+y+z=1 \] \[ bc\cdot x+ca\cdot y+ab\cdot z=2R(R+r) \] \[ \left(b^2+c^2\right)\cdot x+\left(c^2+a^2\right)\cdot y+\left(a^2+b^2\right)\cdot z=6R^2 \]

The system of equations

$\left(\begin{array}{ccc} 1&1&1\\\ bc&ca&ab\\\ b^2+c^2&a^2+c^2&b^2+a^2\end{array}\right) \cdot\left(\begin{array}{c} x&y&z\end{array}\right)=\left(\begin{array}{c} 1&2R(r+r)&6R^{2}\end{array}\right)$

has two or more solutions if $det\left(\begin{array}{ccc} 1&1&1\\\ bc&ca&ab\\\ b^2+c^2&a^2+c^2&b^2+a^2\end{array}\right) =0$

But $det\left(\begin{array}{ccc} 1&1&1\\\ bc&ca&ab\\\ b^2+c^2&a^2+c^2&b^2+a^2\end{array}\right) =0 \Longrightarrow$

$(a-b)(b-c)(c-a)(a+b+c)=0 \Longrightarrow\, a=b=c$

What do you mean by interpret geometrically ??  :)

[quote="Ramanujan"]The system of equations
$\left(\begin{array}{ccc} 1&1&1\\\ bc&ca&ab\\\ b^2+c^2&a^2+c^2&b^2+a^2\end{array}\right) \cdot\left(\begin{array}{c} x&y&z\end{array}\right)=\left(\begin{array}{c} 1&2R(r+r)&6R^{2}\end{array}\right)$

has two or more solutions if $det\left(\begin{array}{ccc} 1&1&1\\\ bc&ca&ab\\\ b^2+c^2&a^2+c^2&b^2+a^2\end{array}\right) =0$
........................................................................................
$(a-b)(b-c)(c-a)(a+b+c)=0 \Longrightarrow\, a=b=c.$[/quote]
In my opinion, you made two mistakes:
$1.\ \ (a-b)(b-c)(c-a)=0\Longrightarrow a=b$ [b]or[/b] $b=c$ [b]or[/b] $c=a$;
$2.$ the system $x+y+z=3$, $x+2x+3z=6$, $3x+4y+5z=10$ has the determinant of the coefficients equal with $0$ but the system hasn't solutions !

I know that
[quote]the system $x+y+z=3, x+2x+3z=6, 3x+4y+5z=10$ has the determinant of the coefficients equal with $0$ but the system hasn't solutions[/quote]
but we dont need that(we need that for the converse). :)
Also i know that
[quote]$(a-b)(b-c)(c-a)=0\Longrightarrow a=b$ or $b=c$ or $c=a$[/quote]
but i just wrote the idea not the complete solution because i was in hurry.  :)
Ok .I'll continue the solution.

From $(a-b)(b-c)(c-a)(a+b+c)=0\Longrightarrow a=b$ or $b=c$ or $c=a$.
We take $b=c$ WLOG.
For the triangle $ABC$ since $b=c$ we get $R=\frac{b^2}{\sqrt{2b-a}\,\sqrt{2b+a}}$ and $r=\frac{a}{2}\frac{\sqrt{2b-a}}{\sqrt{2b+a}}$.

Now the system $\left(\begin{array}{ccc} 1&1&1\\\ bc&ca&ab\\\ b^2+c^2&a^2+c^2&b^2+a^2\end{array}\right) \cdot\left(\begin{array}{c} x&y&z\end{array}\right)=\left(\begin{array}{c} 1&2R(R+r)&6R^{2}\end{array}\right)$ becomes

$\left(\begin{array}{ccc} 1&1&1\\\ b^2&ba&ab\\\ 2b^2&a^2+b^2&b^2+a^2\end{array}\right) \cdot\left(\begin{array}{c} x&y&z\end{array}\right)=\left(\begin{array}{c} 1&2R(R+r)&6R^{2}\end{array}\right)$

or equivalently after eliminating

$\left(\begin{array}{ccc} 1&1&1\\\ b-a&0&0\\\ 0&0&0\end{array}\right) \cdot\left(\begin{array}{c} x&y&z\end{array}\right)=\left(\begin{array}{c} 1&\frac{(b-a)(2b^2-a^2)}{4b^2-a^2}&0\end{array}\right)$.

But there are at least two different solutions of the equation ${(b-a)x=\frac{(b-a)(2b^2-a^2)}{4b^2-a^2}}$ therefore $b-a=0$ or $b=a$.

For the converse we assume that $a=b=c$
Then the system becomes (in this case $R=\frac{a}{\sqrt3}$ and $r=\frac{a}{2\sqrt3}$)

$\left(\begin{array}{ccc} 1&1&1\\\ a^2&a^2&a^2\\\ 2a^2&2a^2&2a^2\end{array}\right) \cdot\left(\begin{array}{c} x&y&z\end{array}\right)=\left(\begin{array}{c} 1&a^2&2a^2\end{array}\right)$

or equivalently

$\left(\begin{array}{ccc} 1&1&1\\\ 0&0&0\\\ 0&0&0\end{array}\right) \cdot\left(\begin{array}{c} x&y&z\end{array}\right)=\left(\begin{array}{c} 1&0&0\end{array}\right)$
and this system has more than one solutions.

It seems correct now  :)  :roll:


I found this problem while reading Chinese math book.
I have never seen the technique used in its solution, so I was wondering if AoPS users were familiar with it.

$ x^2\minus{}2ax\plus{}a\equal{}0$ is an equation in $ x$. One of its roots is in between the interval $ (0,1)$ and another root is greater than 2. Find all conditions that real $ a$ must satisfy.

Any solution would be appreciated. :)

[quote="t0rajir0u"]Is the technique to which you refer the intermediate value theorem for polynomials?  All we require for this to be true is that (letting $ f(x) \equal{} x^2 \minus{} 2ax \plus{} a$) $ f(0)$ and $ f(1)$ have opposite signs and $ f(2)$ is negative.  (This then requires that $ f(1)$ is negative, so $ f(0)$ is positive.)[/quote]
Yeah that was it :)


function
algebra
polynomial
combinatorics unsolved
combinatorics

How many nonegative integer solutions are the to $ x_1 \plus{} x_2 \plus{} x_3 \equal{} 25$ where $ x_1$ is even, $ x_2$ is at least 3, and $ x_3$ is at most 5?

The generating function is $ \left(1 \plus{} x \plus{} x^2 \plus{} x^3 \plus{} x^4 \plus{} x^5\right)\left(\frac {1}{1 \minus{} x^2}\right)\left(\frac {1}{1 \minus{} x} \minus{} x^2 \minus{} x \minus{} 1\right)$. Is there any way to extract the coefficient of $ x^{25}$ without doing like 3 pages of algebra?

Sure -- if you bring that all over a common denominator, it'll work out to $ \frac{\textrm{some polynomial}}{(1 \minus{} x)^2}$.  Get the series for $ \frac{1}{(1 \minus{} x)^2}$ and you're all set.


Stanford college search geometry MIT Harvard Yale

Hi all,

I'm graduating from college this semester and am looking to continue studies at graduate school either one or two years from now. I'm thinking of studying either computer science or mathematics, but will mostly likely combine the two.

How important are the USNews grad school rankings? How important is it to find a lab that you like and a professor/academic culture that you thrive in vs. the prestige of the school? Also, how do you make a decision like that based on what little you know of the school from a one-two day visit? Are there any reliable sources of information?

I would also be very grateful if any graduate students/professors/anyone in the know can give me some advice. In particular, I'm looking at schools like Stanford, UC Berkeley, UCLA, UCSD, and maybe MIT/CMU (though I'm leaning heavily toward California because of the weather =) ).

Thanks!

Put little stock in the US News rankings.  Make your choice based on, as you said, "a lab that you like and a professor/academic culture that you thrive in".

You can't see much in a 1-2 day visit, but you can talk to the current grad students there and possibly learn from their years of time at the place. Although judging by what happened when prospective grad students came to visit my school, the current grad students aren't always too helpful... when I was there, they only seemed to be able to tell the prospectives about how they play soccer and bridge and Conway football and there's tons of free food...

edit: so that I'm actually saying something in this post, the lesson you should take is that you should come to those visits prepared with questions to ask of the current grad students. If you don't ask them anything specific, they might have no idea what to tell you and just ramble about random junk.

Good luck with your grad school search, joshuatan!  I should be able to help you, because I have just gone through the grad school applications/grad school visits this fall and spring.  I'm not in math or CS, but my advice should apply to graduate school more broadly.

[quote]How important are the USNews grad school rankings?[/quote]

They tell you something about the quality of the program, but maybe not as much as you would think.  If you have your heart set on studying a particular subdiscipline of math or CS, you would do better to ask around about which programs are strong  in that subdiscipline.  You'll end up in a program which is a better fit for your interests, which means that you'll have a better experience while you're there.  It will also help you when you graduate and are looking for a job, because people don't think, "How good is that program in the USNews grad school rankings."  They think, "How good is that program in the area he did his dissertation in?"

[quote]How important is it to find a lab that you like and a professor/academic culture that you thrive in?[/quote]

This is of paramount importance for your happiness and the quality of the work you do.  Supervisors can have near-absolute power over the students they supervise.  I have seen horror stories unfold where the professor was verbally/emotionally abusing their students, and the students didn't feel they could stand up to him or speak up to the administration about it.  Make sure that your supervisor is someone who you trust and who you like as a person.  Ask the grad students at the places you visit specifically what the various professors are like to work under.

[quote]Also, how do you make a decision like that based on what little you know of the school from a one-two day visit?[/quote]

For me, it wasn't very hard.  I came with a list of questions I needed to answer to my own satisfaction, and then asked around until I had answered them to my satisfaction.  I set up a lot of appointments beforehand, so that I knew I would be able to talk with all the people I felt I needed to.  But, in practical terms, I knew within an afternoon whether I wanted to go to a particular school-- the differences in the atmosphere of the schools were so loud and clear.

[quote]Are there any reliable sources of information? [/quote]

Probably your own feelings.  If a place turns you off, but you can't put your finger on it, that's a bad sign, and you shouldn't second-guess it.  If you feel comfortable with the people in a place, then that's good, and probably more reliable information than anything anyone could tell you about the atmosphere of the program.  Grad students are usually the best sources of information about the academic experience, and the experience of day-to-day living, because they're the ones living it, and usually aren't as devoted to trying to recruit you.  You have to be careful about what people tell you, though.  Often people will start describing their school, and sweet music will play in the perfumed air and rose petals will fall from the sky...

Anyway, good luck to you!  If you need advice about what to do/ask about on a grad school visit, feel free to PM me.

I wouldn't disregard USNews rankings.  They are fairly accurate.  (But look at the score rather than the rank and apply some reasonable error bars.)  The problem is that there's really nothing else to consult.  phds.org allows you to rank institutions using your own balance of factors, but its uses the NRC rankings which are a million years old (I don't know how old the rest of the data is).  The new NRC rankings are coming out within a year; those will definitely be worth paying attention to.

Choosing a graduate program in math is very different from other sciences.  (I don't know much about CS.)  Math might be the only subject in which, "I have little idea of what I want to study and no idea who I want to worth with," is a perfectly reasonable thing to say.  If you don't know what you want to study, you should look at the program as a whole.  If you know what general field you want (e.g. algebra), you should look at groups of profs rather than individual profs.  Only if you know *exactly* what you want to study should you focus on individual professors.  (Depending on what you mean by combining math and CS, you might be in the second category or the third.)  In all cases, you should put in some level of preference for more prestigious departments.

Btw, even if you plan to do something at the intersection of math and CS, you will have to choose between them when entering a graduate program, and your decision will have a big effect on the sort of education you receive.  Of the schools you mentioned, I do know that Stanford has a Scientific Computing and Computational Mathematics Program that might interest you.

[quote="yenlee"]The problem is that there's really nothing else to consult.[/quote]  Any professors at your current institution would probably be happy to give you advice on this subject.  I found my professors extremely helpful in pointing out institutions I wouldn't have thought to look at (often because their overall rankings are not so high even though they had top-notch professors in certain fields) and particular professors to go talk to during my visits.  Of course, any professor will be most helpful in whatever area he or she is most focused in.

Hi all,

Thanks for taking the time to post and for the advice! If anyone feels like he has anything else to add please don't hesitate  :)

JBL is, of course, correct.  Get advice from your profs.  They're going to vaguely agree with US News about what the top math departments are, because everyone agrees on that.  Your profs are likely to know about the relative strengths and weaknesses of the top departments.  (Btw, weaknesses in an area doesn't mean that the profs are weak; it means that the dept has no profs in that area.)  Once you go beyond that, mathematicians tend to only know a lot about their own field, and this is where US News becomes more useful.  (The math subfield rankings in US News are totally useless, though.  They barely make any sense.)  When you're an undergrad, it's hard to appreciate how really narrow most mathematicians' research scope is.

To summarize:  If you want to do math but don't know what you want to study, then blindly following US News is a non-horrible strategy.  That's all I'm saying.  (But perhaps JBL disagrees with this also.)

One more reason to listen to your profs:  Recommendations are really important, so if you don't attend a big name school, your profs are likely to know where their recs might be taken more seriously.

[quote="yenlee"]To summarize:  If you want to do math but don't know what you want to study, then blindly following US News is a non-horrible strategy.  That's all I'm saying.  (But perhaps JBL disagrees with this also.)[/quote]

No, I don't disagree.  As far as "decent gist of overall rankings" goes, US News does a fine job.  If I were to differ at all, the only reason would be one of emphasis: there are factors that I think will make more difference in your life than the "overall strength" of your program.  Some of these factors are mathematical (is there a professor working on questions that are interesting to you, does that person take students, does that person interact with students in a way you'll appreciate, are there other options as well, is the student body strong) and some are nonmathematical (what is the size of the stipend, how's grad student housing, do you need a car to get around, is the department generally friendly, how much teaching is there, are other graduate students all miserable) and I don't think the US News ranking captures this, while current professors and current graduate students (and departmental web pages) can probably provide a lot of this information to you.  But I think yenlee and I both agree that acquiring more information is better than acquiring less for purposes of determining where to apply and where to go.

One resource that I don't think anyone has raised: another good tool to use is a list of departmental placements, i.e. where people got jobs after leaving grad school.  All of the schools I got into made such a list available to acceptees, and they might make a list available to potential applicants if you go and ask.  Such a list is a decent measure of what presige translates to in the academic job market.

Okay, I am a complete chump, because I didn't even know that schools gave out that sort of information to prospective students.  Needless to say, that is a very valuable piece of information.

Something that hasn't been mentioned yet is teaching preparation.  Most students go to grad school planning to do research, but aside from the top programs, the majority of PhDs will get jobs based on their teaching ability.  Some grad programs pay a lot of attention to teacher preparation while others don't.

Thanks for the replies.

I'm actually on scholarship all the way up to grad school from a certain RI, so I don't have to worry about stipend/financial matters at this point.

Thus the quality of the school/nature of the program/quality of life matter much more to me.

Most of it depends where you what you want to study.

Math would probably be MIT or Caltech. But Caltech right now is in more of an up-and-coming place which has a large need of well-trained mathmatechians (spelled it wrong), so it gets the slight edge. Medicine is probably John Hopkins. Law: most people think Harvard, but Stanford is probably better, and, as someone said before, don't put too much stock in the U.S. News and World Report rankings. Also, much like Caltech, Stanford is located in Palos Altos, near Silicon Valley, home of a lot of new businesses, so if you want to be a corporate lawyer, Stanford is the obvious choice. Business is probably Penn State (U.S. News got this one right).

No offense, but it seems you just named the most "prestigious," expensive, and sometimes overrated graduate schools from the top of your head. That isn't helping him much.

Also, its University of Pennsylvania that's known for it's graduate school, not Penn State.  :wink:

Academic graduate schools (not professional schools) aren't expensive -- typically, they pay Ph.D. students a stipend.  I agree with you about the quality of the comments, though.

[quote]Medicine is probably John Hopkins.[/quote]

No such place.  It's Johns Hopkins so get it right.  And we are great and plenty of things besides medicine.

[quote="JRav"]No such place.  It's Johns Hopkins so get it right.  And we are great and plenty of things besides medicine.[/quote]
My dad used to joke that the university was founded by John Hopkin and his brother, John Hopkin, and they couldn't decide how to pluralize it.

DonkeyKingKong, your other post says that you are in 8th grade, but somehow you are an authority on graduate schools?  I hope that at least you got these opinions from your parents or something.  Regardless of where they came from, they're wrong, so let me at least set some of the record straight.

Law:  Um, no.  Rightly or wrongly, "most people" think of Yale as being the most prestigious.  The comment about Harvard vs Stanford and corporate law is completely silly.  Here is a more correct (but still overly simplified) statement:  Choose Harvard over Stanford if you want to live and work on the East Coast.  Choose Stanford over Harvard if you want to live and work on the West Coast.  By the way, there's a little town called New York City that has a wee bit more business activity than Silicon Valley.

Medicine:  I'm not contradicting your assertion that Johns Hopkins is the best--Johns Hopkins has a great great medical school--but people should know that Harvard Med is still Harvard Med.  Rightly or wrongly, Harvard Med's reputation is untouchable.

Math:  I don't understand how having a "need of well-trained mathematicians" can be a plus.

Business:  What Anaxerzia said.

[quote="DonkeyKingKong"]I'm in eighth grade but I am a genius.[/quote]

Wow do you have an ego.

@all: I know this person in RL.

A) An SAT score of 2140 doesn't characterize a genius
B) An IQ of 135 doesn't characterize a genius

The only reason why you are saying Stanford is the best for everything isbecause your cousins went there.

Not a good reason.

Please stop spamming up a good forum.

Thanks.

In addition, with all of the MATHCOUNTS questions you posted a few days ago, I'm doubting your genius status.  Those were trivially easy questions.  You think you're a genius, go have a chat with Terrence Tao or Reid Barton and see how fast you get lost.  Also, if you're genius, apply for the Johns Hopkins Program for Talented Youth and see if you get in.
[quote]Medicine: I'm not contradicting your assertion that Johns Hopkins is the best--Johns Hopkins has a great great medical school--but people should know that Harvard Med is still Harvard Med. Rightly or wrongly, Harvard Med's reputation is untouchable.[/quote]

We are the best.  Basically every US med school is modeled after the Hopkins model.  Harvard is great-Hopkins is better.

Man you're so full of yourself... Be more humble dude, as I understand it you're even in the beginning of your undergrad.

[quote="JRav"]In addition, with all of the MATHCOUNTS questions you posted a few days ago, I'm doubting your genius status.  Those were trivially easy questions.  You think you're a genius, go have a chat with Terrence Tao or Reid Barton and see how fast you get lost.  Also, if you're genius, apply for the Johns Hopkins Program for Talented Youth and see if you get in.
[/quote]
He's in eight grade. Yes it's good that you and Tao find MATHCOUNTS questions easy.

Kalle I definitely don't think that is the case. JRav was simply being objective to the case of this kid who claims to be able to go to MIT next year. He isn't saying this kid is stupid, but usually kids who go to top universities when young are extremely intelligent like Terrence Tao. Terrence Tao went to the IMO when he was 12 or so (I think) for Australia. This kid is doing MathCounts. Terrence Tao learned university level mathematics when he was like 10. This kid did precalc at 13. This kid obviously shuows a good degree of intelligence, but to publically declare that ones credentials qualify for MIT at 13? I feel JRav is justified. Also, he isn't touting his own credentials either, just brining the kid down to Earth...

I think both of you guys made the mistakes to feeding this troll.  :wink:

[quote="DonkeyKingKong"]Also, much like Caltech, Stanford is located in Palos Altos, near Silicon Valley...[/quote]

It's actually called Palo Alto... and I'm not sure what you mean by "much like Caltech," but your sentence structure suggests that Caltech is also located near Silicon Valley, which I can assure you is bogus.

[quote="#H34N1"]The only reason why you are saying Stanford is the best for everything isbecause your cousins went there.[/quote]

This may be true... but Stanford is really the best :wink:.


function calculus integration email geometry

Consider a piston cylinder arrangement containing an ideal gas at conditions P1,V1 which undergoes a irreversible change to final conditions P2,V2.The atmospheric pressure is P0 and the mass of the piston is negligible.

For such a process we usually write,
work done on the sys. w1=p(ext.)[V2-V1]=P0[V2-V1]   (in this case)
now, work done by the sys.=-w1=-PO[V2-V1]
How can we justify the 2nd equality?
work done by the gas is =force*displacement
                                 =$\int.\$P*dV where P,V defines the conditions of the gas at an intermediate stage where both continually change with time,
then how could it be equal to  -P0[V2-V1]? :huh:

Keep in mind that you have an irreversible process.  Since w is NOT a state function, it is path-dependent.  The equation $w_{\text{on system}}=-P_{0}\Delta V$ IS applicable, because we know $dw = P_\text{ext}dV$ (always).

However, we cannot use your integral because it is not always true that $P_\text{gas}= P_\text{ext}$.  Also, be sure to watch your signs; I think you might have mixed them up.

yes,I understood that we cannot have the integral sign as the pressure is abruptly changing from $P_{1}$ to $P_{2}$.
but now consider the work done by the gas in a short duration i.e.$PdV$ and sum up all such works for the whole time period,will its modulus be eqUal to the modulus of $W_{on system}$ ($P_{0}(V_{2}-V_{1})$, assume $V_{2}>V_{1}$,how?

I am not sure what you mean by "modulus."  However, I think that you bring up a very good point, and touch on something that I have wondered about before too.  I know that $w_\text{on system}=-P_\text{ext}dV$ is always valid, and that $w_\text{on system}=-P dV$ only in a process where $P=P_\text{ext}$.  However, I must say that I am not sure why this is so.

Actually, I sat down and came up with the following experiment, which I think pretty much sums is what you are asking in your question (please let me know if this is not what you are asking):

I assume that in an irreversible ideal gas expansion (a) $dw_\text{on gas}=-P_\text{ext}dV$   and (b) $dw_\text{on gas}=-dw_\text{on surroundings}$.

So then consider a sealed container with a divider in the middle separating two samples of ideal gases, the left side at pressure $P_{1}$ and the right at pressure $P_{2}$ (as in the attached picture), where $P_{1}>P_{2}$.  If we let the $P_{1}$ side expand by a differential volume $dV$, then the differential work according to that side's point of view is \[dw_\text{on left side}=-P_\text{ext}dV =-P_{2}dV.\] But let's look at the right side now, which must expand by $-dV$.  Then \[dw_\text{on right side}=-P_\text{ext}(-dV) = P_{1}dV.\] But it must be that $dw_\text{on left side}=-dw_\text{on right side}$, so $-P_{2}dV =-P_{1}dV$, or $P_{1}= P_{2}$, contrary to assumption.

I sent an email to someone way smarter than I, I'll let you know if I get a resolution to this.  In the meanwhile, anyone on this board have any ideas?

Yes,that was exactly what I wanted to ask.I mentioned about the "modulus" because I wanted to tell that I'm not worried about the signs(+or-) but only about the expression and the ex. you chose exactly expresses it.

Ok, so after polling a rather large number of contacts, I finally found a source that noted the, in hindsight, apparently obvious thing that I was missing--the divider itself will acquire a kinetic energy, and this kinetic energy makes up for the apparent "missing work."

Anyway, here's my resolution, based on what others have told me/I have thought about on my own:
Let A be the cross-sectional area of the divider.

(1) The "missing work" is accounted for in the kinetic energy of the divider (which slides spontaneously, since it has a net force acting on it).  If this expansion is forced to proceed slowly, then this excess energy is transfered to whatever agent holds back the divider.

(2) Let us consider the net force on the gas of side 2.  Since it exterts a force of P2 A on the divider, by Newton's Third Law the divider exterts a force of P2 A on the gas.  Furthermore, the net force on the divider itself is (P1-P2)A in the direction of side 2, so (assuming negligible mass of the divider, which is interesting since then this expansion will happen incredibly fast.  If this mass is non-negligible, perhaps the best thing to do is to arbitrarily assign the divider to one half of the system or the other; the desired equations follow rather simply) it exerts a force of an additional (P1-P2)A on side 2.  So the net force on side 2 is indeed P2 A, which is P(ext) A.

(3)  A similar analysis shows that the force on side 1 is P1 A.  Thus this resolves the problem.

If this is not clear, or doesn't seem to work, please let me know.


limit
induction
algebra unsolved
algebra

Consider the sequence $ a_1\equal{}\frac{3}{2}, a_{n\plus{}1}\equal{}\frac{3a_n^2\plus{}4a_n\minus{}3}{4a_n^2}.$
$ (a)$ Prove that $ 1<a_n$ and $ a_{n\plus{}1}<a_n$ for all $ n$.
$ (b)$ From $ (a)$ it follows that $ \displaystyle\lim_{n\to\infty}a_n$ exists. Find this limit.
$ (c)$ Determine $ \displaystyle\lim_{n\to\infty}a_1a_2a_3...a_n$.

[quote="moldovan"]Consider the sequence $ a_1 \equal{} \frac {3}{2}, a_{n \plus{} 1} \equal{} \frac {3a_n^2 \plus{} 4a_n \minus{} 3}{4a_n^2}.$
$ (a)$ Prove that $ 1 < a_n$ and $ a_{n \plus{} 1} < a_n$ for all $ n$.
$ (b)$ From $ (a)$ it follows that $ \displaystyle\lim_{n\to\infty}a_n$ exists. Find this limit.
$ (c)$ Determine $ \displaystyle\lim_{n\to\infty}a_1a_2a_3...a_n$.[/quote]

Solution for part $ (a)$:

First will prove by induction that $ 1<a_n<3$ for all positive integers $ n$.

For $ n\equal{}1$ it's trivial. Suppose that $ 1<a_n<3$. So $ a_{n\plus{}1}>1$ is equivalent to: $ 3a_n^2\plus{}4a_n\minus{}3>4a_n^2$ or $ (a_n\minus{}1)(a_n\minus{}3)<0$, which is true. $ a_{n\plus{}1}<3$ is equivalent to $ 9a_n^2\minus{}4a_n\plus{}3>0$ discriminant of this equation is negative so it is always true.

Now we will prove that $ a_n>a_{n\plus{}1}$ denote $ a_n$ by $ x$. We need to prove that $ f(x)\equal{}4x^3\minus{}3x^2\minus{}4x\plus{}3>0$ for any real $ x$ that satisfy $ 1<x<3$. $ f'(x)\equal{}12x^2\minus{}6x\minus{}4$ both solution are smaller that $ 1$. So $ f(x)$ is increasing on $ x$ bigger than $ 1$. So $ f(x)>f(1)\equal{}0$


a question from MATHEMATICAL CIRCLES.
                                          BLASE + LBSA =BASES.
it is known that equal digits are represented by equal letters and different digits with different letters.
two giant computers came up with two different solutions to the riddle.
       is this possible or does one pf them need repair? :maybe:

A + B = C has a bunch of solutions.  doesnt mean my brain needs to be fixed

One of the computers needs repair, unless someone else can find another solution.

[hide="Answer"] Write the equation in a column form, with $ BASES$ set below $ BLASE$ and $ LBSA$.  Looking at the first and third columns, [b]E[/b] and [b]B[/b] must differ by exactly 1 or exactly 9 to have the same digit in the sum, requiring the second column to carry over a 1 unless [b]E[/b] is 9.  If by 9, [b]E[/b] cannot be 0, because $ A\equal{}S$.  It cannot be 9, because [b]S[/b] would also be 9 (see second column, even with carrying).  So they must differ by 1.

In addition, the first column can't carry.  Otherwise:

$ 2S\plus{}1\equal{}E\plus{}10$
$ E\plus{}A\equal{}S\plus{}10$
which, through manipulation, implies that:
$ E\plus{}2A\equal{}29$
which is impossible.

Thus, $ E\equal{}B\plus{}1$, $ S\geq5$, $ E\plus{}A\equal{}S$, $ 2S\equal{}E\plus{}10$, [b]E[/b] must be even, [b]A[/b] must be even by the fourth column (third doesn't carry if the first doesn't), so [b]S[/b] must be even.

Now lets try the two cases centered around [b]S[/b].

$ S\equal{}6$ does not work, as $ L\equal{}E$.

$ S\equal{}8$ does, with $ 51286\plus{}1582\equal{}52868$.

This means only one solution to the problem.
[/hide]


algebra
polynomial
limit
real analysis
real analysis unsolved

If $ P$ is a polynomial ,$ P \in \mathbb R[X]$ compute:

\[ \lim_{n\rightarrow \infty} (\sqrt[n\plus{}1]{|P(1)\cdot P(2)\cdot ..\cdot P(n\plus{}1)|}\minus{}\sqrt[n]{|P(1)\cdot P(2)\cdot ..\cdot P(n)|})\]

[i]N.Pavelescu[/i]

Obs: for $ P(n)\equal{}n$ it's the famous one:  $ \frac{1}{e}$

Something must be wrong, it seems that the limit is $ \infty$.

Let $ P(x)\equal{}x^{2}$ and let $ x_{n}$ be your sequence. Then

$ x_{n}\equal{}\left(\sqrt[n\plus{}1]{(n\plus{}1)!}\minus{}\sqrt[n]{n!}\right)\left(\sqrt[n\plus{}1]{(n\plus{}1)!}\plus{}\sqrt[n]{n!}\right)\rightarrow \infty$,

since $ \sqrt[n]{n!}\rightarrow \infty$.

Perhaps $ P$ is a polynomial of degree 1.

Nothing is wrong :D .I managed to prove that if  $ P$ is a polynomial with degree $ \geq 2$ then the limit is $ \infty$ .Otherwise is $ \left | \frac{a}{e} \right|$ , where $ P(X)\equal{}aX\plus{}b$

OK. I understand what you meant by that.
For solving the problem you need to consider, due to the end behavior of polynomials, only polynomials of the form $ P(x)\equal{}ax^{k}$.
In this case it is not hard to prove that the limit equals $ |a|/e$ if $ k\equal{}1$ and $ \infty$ if $ k\geq 2$, since all you have to do is to use Lalescu's  limit and the $ a^{k}\minus{}b^{k}$ formula.

What if $ P \equal{} x^r$ for any $ r \in (0, 2)$?

Let $ y_{n}$ be the generalized Lalescu's sequence, i.e.

$ y_{n}\equal{}(\sqrt[n\plus{}1]{(n\plus{}1)!})^{r}\minus{}(\sqrt[n]{n!})^{r}$.

Then using the Mean Value Theorem with $ f(x)\equal{}x^{r}$ we get that

$ y_{n}\equal{}L_{n}r c_{n}^{r\minus{}1}$,

where $ c_{n}\in [\sqrt[n\plus{}1]{(n\plus{}1)!},\sqrt[n]{n!}]$. (the order of the two points is not important).

Thus, it follows that the limits is $ 0$ if $ 0<r<1$,  $ \infty$ if $ r>1$ and the case when $ r\equal{}1$ the limit is $ 1/e$.


This is a propblem some of my friends in Physics B were having trouble with, so I thought beginners in physics could learn from it:

Suppose you adjust your garden hose nozzle for a hard stream of water.  You point the nozzle vertically upward at height of 1.5 m above the ground.  When you quickly move the nozzle away from the vertical, you hear the water striking the ground next to you for another 2 seconds.  What is the water speed as it leaves the nozzle?

Is the answer remotely close to
[hide]9.05 m/s[/hide]
?
I'll explain if that somehow manages to turn out right...

Unless I missed something, you are both correct.  Good job. :)

[quote="Spoon"]I think this is a bad application to explain and test a simple concept, because if the hose was on for any amount of time, the water falling back onto the hose's stream distorts it's maximum height, making our simple calculation impossible. A cannon shooting a ball up would be a better example :o.[/quote]

I agree (this same thing bothered me when I first read the problem).  The only reason this question gave anyone trouble was because their teach didn't give them the equation $x=x_0+v_0t+\frac{1}{2}at^2$, instead he gave them $x=v_0t+\frac{1}{2}at^2$ and told them that they would [b]never[/b] use the $x_0$.

[quote="Jimmy"]and told them that they would [b]never[/b] use the $x_0$.[/quote]
Well that's just silly.

[quote="Spoon"]I think this is a bad application to explain and test a simple concept, because if the hose was on for any amount of time, the water falling back onto the hose's stream distorts it's maximum height, making our simple calculation impossible. A cannon shooting a ball up would be a better example :o.[/quote]

Spoon,

What water "falling back onto the hose's stream?" Water leaving the hose would form a continuous column. The maximum height of the column is as high as any water particle gets.

[b]Bernoulli's Principle:[/b]

$\frac{1}{2}\rho v^2+\rho gh + P_{\text{gauge}} = \text{ constant}$ .


geometry
3D geometry
geometric transformation
rotation
symmetry
invariant
modular arithmetic

[quote="boxedexe"][i]Is it possible to put 100 (or 200) points on a wooden cube suck that by all rotations of the cube the points map into themselves? Justify your answer.[/i]

Masoud Zargar[/quote]
What do you mean exactly? Does $f: [0,1]^{3}\rightarrow [0,1]^{3}$ map $f(a)\rightarrow a$ or $f: A \rightarrow A$? Assuming the latter then place a figure with $\pi/2$ rotational symmetry on each side, centered at the origin. Make sure the figures are congruent though.

The image of the points is invariant upon rotation.  Moreover, it is not always possible to construct such a map. :wink:

I wonder if this works. Notice that if you place a point, you immediately have to place some number of other points which are the possible positions of the first point under rotations. There are two cases of interest:

(1) The point $p$ is on a diagonal, but not the center of the cube. Under all rotations, it takes $8$ different positions.

(2) The point $p$ is on the line segment connecting the centers of opposite faces, but not the center of the cube. Under all rotations, it takes $6$ different positions.

So if we take $11$ points of the first type and $2$ points of the second type, we should get $100$ points on the cube. For $200$, we can simply take $25$ points of the first type.

[quote="boxedexe"]Consider modulo 6. You can't do it for 100 points since you can prove that the number of points $p$ must satisfy $p\equiv 0,2\pmod 6$.[/quote]

Are you sure? If you pick a point on a space diagonal, it has $8$ different positions. Also, if you pick the center, it only has $1$.

Notice that if you have points at the center of a face,midpoint of each edge, and at vertices, in order to have points mapping to themselves, they either have to all exist, or none.  Thus, there are wither $0$ or $12$ points at the midpoints, $0$ or $8$ vertex points, an $0$or $6$ face points. All other points have 24 images, so all other points are a multiple of $6$.Also notice that $6$ divides the number of edge and face points as well. As a result the number of points $p$ satisfies $p\equiv 0,8\equiv 0,2\pmod 6$.  $(100,200)\equiv (4,2)\pmod 6$.  The conclusion follows.

[quote="Hamster1800"]Paladin was thinking that we could put points on the inside of the cube. Boxedexe thinks that we can only put points on the surface of the cube. Since the problem describes it as a wooden cube, I'd be inclined to go with boxedexe's answer.[/quote]

Oh, I see. Thanks for the clarification :roll:.


Why are copper ions often said to show anomolous stability?

I have never seen the term before, but my guess would be that $\text{Cu}^{+}$ is especially stable, compared to other transition metals, due to its complete $d$ sublevel.  That said, $\text{Cu}^{2+}$ tends to be more common, and no reason comes to mind for that to be especially stable.


suppose$ AH_1,BH_2,CH_3$ are the three altitudes,we want to find the lengths $ a,b,c$

if u consider the area of triangle $ S$,we have:

$ S\equal{}a.AH_1\equal{}b.BH_2\equal{}c.CH_3$

then the values of $ \frac{a}{b}$and $ \frac{b}{c}$and $ \frac{c}{a}$ are constant,

and u can easily find the values of $ a,b,c$ accordingly :wink:

[quote="zweig"]The foots' altitudes of a triangle are given. Construct the triangle.[/quote]

 The incenter $ I$ of the triangle with vertex the feet of the altitudes $ A_1,A_2,A_3$  , is the orthocenter of the initial triagle which we want to construct.It suffices then to  draw the perpendiculars at  $ IA_1$ ect.

 Babis

The incenter I of the triangle with vertex the feet of the altitudes A_1,A_2,A_3 , is the orthocenter of the initial triagle which we want to construct.It suffices then to draw the perpendiculars at IA_1 ect. [quote]

I think what you said doesn't give a way to find the triangle. You also need the colinearity between the vertex of the altitude that doesn't belong to the side, another vertex of another altitude that doesn't belong to the side and the vertex of the remaining altitude that belongs to the side.

Maybe I don't get the idea. Can you be more specific?[/quote]


function
limit
real analysis
real analysis unsolved

$f$:$R\rightarrow R+$ continuous, periodic with period irrational
M=max f
Compute $\lim_{n\rightarrow +\infty}\frac{f(1)f(2)...f(n)}{M^n}$

are you sure the problem is exact because if so, it is quite clear it is $0$ :
you can suppose that $f: \mathbb{R} \to  [0;1]$, and the function is not identically equal to $1$. So there is an $0<\epsilon <1$ such that the value $1- \epsilon$ is achieved. You want to find $\lim u_n=f(1)f(2)..f(n)$. It is obvious that $u_n$ is a non increasing positive sequence, and because of the irrationality of $\alpha$, there are infinitely many $n$ such that $0 \leq f(n) \leq 1-\frac{\epsilon}{2}$ ??