[hide]Yes, I have posted about KVPY in other forums. However, I am starting a new forum in order to generate more interest and spread information and to know it's significance[/hide]

[b]KVPY - Kishore Vaigyanik Protsahan Yojana (National Science Fellowship) [url=http://www.iisc.ernet.in/kvpy]KVPY[/url].
Please follow the above link for the best information regarding the scholarship etc.[/b]

The KVPY is the most competitive scholarship/exam in India. As scholarships in India are almost all governmental (private haven't yet established themselves - our system relies too much on govern.), this scholarship and its exam are as tough as , if not more than, the National Olympiads in various subjects. I say tougher for it has difficult problems in the exam and a hard interview(believe me - i am not bragging) in ALL 4 Science subjects - Math, Physics, Chem and Bio. At the camp, I found that [i]all[/i] students were either Olympiad clearants or were taking the exam next year. So there's a nice correlation with the Natioanl Olympiads - I'll leave you to find that more explicitly. Please consider all this in context of the situation in India- IIT ,etc. Though our Olympiad scores aren't the best, we're definitely up there[hide]( I know most here are the best and don't like being compared with OK Indians but just context)[/hide] At the camps , we had lectures and were later given research projects which we will pursue throughout the year. If you ant info on what was taught in the camps, etc. please post!!! For everyone clearing Class X(or equiv. grade) go to the site and register. It is THE most prestigious scholarship in India.

[b]Does anybody have any comments!!??!!??!![/b]

Please give me information on how to prepare for kvpy. Also if camp material, or any other material, last papers, or sample papers are available, please let me know
Thanks
kjh

It seems you are an India. Where do you live?
Regarding prep material, I never did any and as far as I know, they never release their papers or any syllabus. The exam is basically a true test of intelligence. The paper is very good. I fyou r concepts are right, you'll get through! Regading camp material, I don't know how it will be helpful in preparing for the exam. We were taught quite advanced things -- totally unrelated to Class XII Syllabus. If you want it just for knowing, please post and I'll see if I can get them in time-- I've got a whole notebbook!!!

could you please suggest how to prepare? for nts we had some guideline material, sample papers etc.
kjh

any idea where to get kvpy sample papers?

[quote="Rushil"]Some of the answers here are specifically for kjh:-

Since I believe the written exam is on 6th November, there's not much that can be prepared. As I said, the KVPY paper is set by IIT Prof's and is just like IIT Screening. But just because I'm saying it's IIT type DOES NOT mean that you start preparing just for KVPY by doing loads of past IIT papers etc... The exam is meant to test your intelligence and let it do just that - don't PREPARE!!! I did not and neither did most of the others I knew at the camp! - Just use your intelligence -if you have that 'scientific' thing in you , you'll get it - that's the exact purpose of KVPY after all. But if yo don't get it, don't be disheartened at all - I'm sure you are the best in your own field!!!

Regarding the connection with Olympiads, the only difference might be in the Math Section. In PCM ( regarding BIO - don't ask me!!!), the syllabus for Physics and Chemistry and even the type of questions is just like IIT> However, in Maths, since the Olympiads are much different than IIT. the Math section in KVPY is also a bit different. It had interesting questions of the type that usually appear in Math Olympiads( unlike IIT-JEE). But don't prepare!!! Just use your knowledge - nothing more than Class X and a bit of Calculus required, and solve problems logically... again is analytic ability and intelligence - the things you'd expect in a budding scientist!!!

No past papers/interview info is released(as far as I know) > Don't get into these things. Just be yourself!!!

Hope this helps. Happy to clarify further![/quote]

hello rushil i am a 11th std student in india
i am also appearing for kvpy and i have no idea about what kind of exam it is
can you give 1 or 2 sample problems of that model

Rushil,
Can you give some information about KVPY by project? of what level these projects are ? can we APPEAR FOR THIS NEXT YEAR ALSO?
you know site from where information about previous projects may be available?
tahnks

I dont have much info on projects since I did not do it. Niether do I know of any site where they have released these. They didnt even tell us about these on the camp - i.e. the actual matter in these projects. So I am unable to answer you!

if any one can comment on kvpy by project? for XI XX std?

any other exam for XI the std like kvpy?

I dont know. If you do, PLEASE TELL!!! Same for Class XII!

Can you please give some idea about the interview for KVPY ? what do you suggest for preparing for it? when is it held generally? how big is the pannel? could you please guide?

HI...This is rajarshi from Delhi.
I have qualified for KVPY in both SA and SP..
but am completely clueless abiout the interview.
Dear BHAIYYA who has cleared KVPy,
please tell me what type of ques they ask in interview???
scientific que.s or just abt ur interest in research?

Plz help..
Thanks

HI GUYS

m govinda upadhyay from bits-pilani,goa campus,doing msc(hons)phy integrated.i m selected for SB stream this year.can some body help me.i want to know how does kvpy help in future.wat i suppose to do now?????wat are the benifites of being KVPY scholar?????plz let me know
bye
kip smiling

whre is moderater....plz give reply to my question....
govi...[/quote]

[quote="govi"]whre is moderater[/quote]

The moderator is in Minnesota, the United States. I'm always very glad to see discussion of university entrance examinations and scholarships in other countries here in this College Forum, but my volunteer job description doesn't include answering every possible question that someone might ask here. I wish you well in obtaining a scholarship, and in obtaining an answer to your question--most likely from another participant who lives in India, who would be better informed about the KVPY.

This thread should be moved to the Indian national forum.

I'm happy to have general questions about college admission in any country here, but let me investigate whether or not I can copy this thread into the [url=http://www.artofproblemsolving.com/Forum/index.php?f=362]Indian national forum[/url]. I see, now that I have checked there, that that forum also has a thread about KVPY, so I'll just leave this right here, with the suggestion to look in the other forum too for possible additional information.

Hi,
My Brother cleared KVPY written test SA stream.
Can anyone give suggestions on what to prepare for the interview?And what type of questions will be asked.
Thanks,
Mounica.


superior algebra
superior algebra solved

Since I haven't found this problem on the forum yet, I'll post it :)
Prove that if in a ring exist $a,b$ s.t. $a^3+b^3=ab-ba=1$ then $1+1+1=0$.


geometry
parallelogram
projective geometry
geometry proposed

let $ ABCD$ be a paralellogram,consider points$ M$and$ N$,on segments$ AB$,$ CD$,respectively.

$ P$and$ Q$,are the intersections of$ (AN,MD)$and$ (BN,MC)$,respectively.

prove that,$ PQ$,goes through a fixed point.

It is not hard  to prove using Pappus theorem that $ PQ$ passes through the dioganals intersection point.


algebra
polynomial
geometry
3D geometry
complex numbers
calculus
calculus computations

I have been trying to look up a simple way to find the roots of a cubic equation.  There are lots of variations on the same drawn out process.  Each time I try to find the roots, I don't get the correct answers.  I don't know if I'm getting lost in the process or what.  Will someone please help?  I have two equations that I have been working.  The answers are: -2, -4, 7 for the first equation and -2, 1, 3 for the second equation but I have yet to arrive at those answers myself.
f(x)=x^3-x^2-34x-56    and   x^3-2x^2-5x+6=0

Any assistance would be MUCH appreciated.  Thanks!

The solution to a general cubic polynomial is very tedious, but if you're interested, check this out [url]http://mathforum.org/dr.math/faq/faq.cubic.equations.html[/url].

As for your two equations, use the rational roots theorem first and if you then find at least one root, you can always solve the resulting quadratic.

Also, the general process for solving a cubic will never work quite right if that cubic has three real roots- it will always be expressed in terms of cube roots of complex numbers, which can't be evaluated except by solving a cubic just like the original (or trisecting an angle).


group theory
abstract algebra
superior algebra
superior algebra solved

Let $G'$ be the commutator of the group $G$, $p$-a prime number and assume that $p$ divides $|Z|$ ($Z$-the center of $G$ ), but doesn't divide $|G' \cap Z|$. Prove that $G$ has a subgroup of index $p$.

I assume that $C\cap Z$ means $G'\cap Z$? In that case:

$G'Z/G'$ is a subgroup of $G/G'$, and it's isomorphic to $Z/Z\cap G'$, whose order is divisible by $p$. This means that the order of $G/G'$ is divisible by $p$. Now, $G/G'$ is abelian, and this means that given a divisor $d$ of its order, it has a subgroup of order $d$. Let $H$ be a subgroup of $G$ s.t. $|H/G'|=\frac{|G/G'|}p$. Then we have $|H|=\frac{|G|}p$, and we're done.


inequalities
induction
number theory
inequalities unsolved

Define the scalar product of $ m$ sequences $ \{a_{i1}, \ldots, a_{in}\}$ (where $ 1\le i \le m$) of length $ n$ to be

$ \sum_{j\equal{}1}^{n}{\prod_{i\equal{}1}^{m}{a_{ij}}}$.

Is it true that if the $ m$ sequences are similarly sorted, then their scalar product is maximal? (As compared to other permutations of the elements in each sequence)

EX: (can be solved by AM-GM for each case, but w/e)
For $ a,b,c>0$, sequences are $ \{a^3, b^3, c^3\}, \{a^2, b^2, c^2\}, \{a,b,c\}$

So $ \sum_{cyc} a^6 \ge \sum_{cyc}{a^5b}$,
$ \sum_{cyc} a^6 \ge \sum_{cyc}{a^4b^2}$,
$ \sum_{cyc} a^6 \ge \sum_{cyc}{a^3b^2c}$, etc.

It is true but it is neither a number theory question, nor an open problem. If you had posted it in the Inequalities section, where it belongs, you would get an answer immediately together with the name of this inequality as a bonus (I suspect it is still called Chebyshev's inequality but I'm not 100% sure). As to the proof, just show by induction on $ m$ that not only the entire sum is maximized by the decreasing rearrangement, but also, for each $ k\equal{}1,\dots,n$ the sum of first $ k$ terms in the decreasing rearrangements scenario is at least as the sum of any $ k$ terms in any other scenario.

This is only a rough idea, not quite rigorous.

If $ \prod_{i \equal{} 1}^{m}{a_{ij_1}}>\prod_{i \equal{} 1}^{m}{a_{ij_2}}$ but there exist an element $ a_{i_1j_1}$ which is smaller than an element $ a_{i_2j_2}$, then we can swap the two elements to obtain a bigger sum.
Therefore if the sum attains max, $ \prod_{i \equal{} 1}^{m}{a_{ij_1}}>\prod_{i \equal{} 1}^{m}{a_{ij_2}}$ implies $ a_{i_1j_1}\ge a_{i_2j_2}$ for all $ i_1,i_2$. The result follows

Oops, I'm very sorry... I definitely clicked in the wrong forum somehow :blush:

By the way, also posted at
http://www.mathlinks.ro/Forum/viewtopic.php?t=24822

  darij


[quote="Vegre nem butulok votabb"]p is a prime number ,prove that among the p+1 distinct positive integers there must be a pair of number (m,n ) and  their biggest common factor is d   thus  m/d or  n/d  is larger than  p+1[/quote]

isn't it [larger or equal], instead of just [larger]?

There are counterexample: if p+1 distinct p.integers are $ 1,2,\cdots,p\plus{}1$, any m/d or n/d cannot be larger than p+1.

If it's [larger or equal]..

---------------------------------------------------------------------------

Assume $ \forall m,n.$     $ \frac{m}{d} , \frac{n}{d} \le p$.

In p+1 distinct positive integers, there must be two positive integers M,N s.t $ M \equiv N (mod p)$

Then $ \frac{M}{d} \equiv \frac{N}{d}$ ( $ \because$ p is a prime number )

However, in our assumption, $ \frac{M}{d}, \frac{N}{d} \le p$

so it makes contradiction, because  $ \frac{M}{d}, \frac{N}{d}$ are both positive integers.

QED.


geometry
3D geometry
geometric transformation
rotation
algebra
binomial theorem
combinatorics unsolved

Hi, I've just started combinatorics (basic) and just some beginner questions I can't do, if anyone could help with them it would be very much appreciated  :lol: [Note: if it's possible please don't leave out any necessary steps or procedures I'm a beginner at these and pretty n00b haha]

1. Prove $ \left(^n_0\right)\plus{}\left(^n_1\right)\plus{}\left(^n_2\right)\plus{}...\plus{}\left(^n_n\right) \equal{} 2^n$

2. Which are there more of among the natural numbers between $ 1$ and $ 10^6$: Numbers that can be represented as a sum of a perfect square and a (positive) perfect cube or numbers that can not be?

3. Two of the squares of a $ 7 \times 7$ checkerboard are painted yellow and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane of the board. How many inequivalent color schemes are possible?

[hide="1"]By the Binomial Theorem,

$ (a \plus{} b)^n \equal{} \binom{n}{0}a^n \plus{} \binom{n}{1}a^{n\minus{}1}b \plus{} ... \plus{} \binom{n}{n\minus{}1}ab^{n\minus{}1} \plus{} \binom{n}{n}b^n$

So,

$ (1\plus{}1)^n \equal{} \binom{n}{0} \plus{} \binom{n}{1} \plus{}... \plus{} \binom{n}{n} \equal{} 2^n$.

Note: This is actually a very well known fact. It also shows why the number subsets in a set of $ n$ elements is $ 2^n$ (do you see why?) .[/hide]

[hide="2"]There are $ 10^3$ perfect squares and $ 10^2$ perfect cubes between 1 and $ 10^6$, inclusive. The number of integers which can be represented by a perfect square and a perfect cube between $ 1$ and $ 10^6$, inclusive, is less than $ 10^3\cdot10^2\equal{}10^5$, which is less than half of $ 10^6$. So there are more numbers which cannot be represented in such way.[/hide]

[hide="3"]If the two yellow squares are not symmetric about the center of the board, we can rotate the board 4 times to get equivalent color schemes. Otherwise, we can only rotate it twice. There are 24 possible positions in which the yellow squares are symmetric about the center of the board, so there are $ \frac{24}2\equal{}12$ inequivalent color schemes here. Note that there are $ \binom{49}2\equal{}1176$ possible positions of the yellow squares. So there are $ 1176\minus{}24\equal{}1152$ possible positions in which the yellow squares are not symmetric about the center of the board, so we get $ \frac{1152}4\equal{}288$ inequivalent color schemes. Thus, the total number of inequivalent color schemes is $ 12\plus{}288\equal{}300$.[/hide]


algebra
difference of squares
special factorizations

[hide="Solution"]Rewriting and shifting the index on the RHS a bit, we have to prove:

$ \sum_{k\equal{}0}^{n}(n(2n\plus{}1)\plus{}k)^2\equal{}\minus{}(2n(n\plus{}1)\plus{}1\plus{}n)^2 \plus{}\sum_{k\equal{}0}^{n}(2n(n\plus{}1)\plus{}1\plus{}k)^2$

Or, equivalently,

$ \sum_{k\equal{}0}^{n}(n(2n\plus{}1)\plus{}k)^2\minus{}(2n(n\plus{}1)\plus{}1\plus{}k)^2\equal{}\minus{}((2n\plus{}1)(n\plus{}1))^2$

Let's play around with $ (n(2n\plus{}1)\plus{}k)^2\minus{}(2n(n\plus{}1)\plus{}1\plus{}k)^2$. Difference of squares reduces it to $ \minus{}(n\plus{}1)(4n^2\plus{}3n\plus{}2k\plus{}1)$. So we have, on the LHS,

$ \minus{}(n\plus{}1)\sum_{k\equal{}0}^{n}(4n^2\plus{}3n\plus{}1)\plus{}2k\equal{}\minus{}(n\plus{}1)((4n^2\plus{}3n\plus{}1)(n\plus{}1)\plus{}n(n\plus{}1))$

$ \equal{}\minus{}(n\plus{}1)^2(4n^2\plus{}4n\plus{}1)$

$ \equal{}\minus{}(n\plus{}1)^2(2n\plus{}1)^2\equal{}\minus{}((n\plus{}1)(2n\plus{}1))^2$ $ \Box$[/hide]

[quote]

For $ n \ge 1$, show the following:

$ [n(2n \plus{} 1)]^2 \plus{} [n(2n \plus{} 1) \plus{} 1]^2 \plus{} ... \plus{} [2n(n \plus{} 1)]^2 \equal{}$

$ [2n(n \plus{} 1) \plus{} 1]^2 \plus{} [2n(n \plus{} 1) \plus{} 2]^2 \plus{} ... \plus{} [n(2n \plus{} 3)]^2$
[/quote]

I prefer grn_trtle's solution to the following, mostly because it doesn't rely
on knowing a formula for the sum of squares of integers from $ 1$ to $ n$.

Alternate solution:
[hide]

If you know the formula for the sum of the squares of the integers $ 1$ to $ n$ is equal to $ \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}$,
then that could be used here.

Left-hand side:

The sum from (where $ n \equal{} 1$ to the last term) minus the sum from
(where $ n \equal{} 1$ to the term just before the first term):

$ \sum_{k \equal{} 1}^{n} ((2k(k \plus{} 1))^2 \minus{} \sum_{k \equal{} 1}^{n}{(k(2k \plus{} 1) \minus{} 1)^2} \equal{}$

$ \frac {2k(k \plus{} 1)(2k(k \plus{} 1) \plus{} 1)(4k(k \plus{} 1) \plus{} 1)}{6} \minus{} \frac {(k(2k \plus{} 1) \minus{} 1)(k(2k \plus{} 1))((2k(2k \plus{} 1) \minus{} 2) \plus{} 1)}{6} \equal{}$

$ \frac {(2k^2 \plus{} 2k)(2k^2 \plus{} 2k \plus{} 1)(4k^2 \plus{} 4k \plus{} 1)}{6} \minus{} \frac {(2k^2 \plus{} k \minus{} 1)(2k^2 \plus{} k)(4k^2 \plus{} 2k \minus{} 1)}{6} \equal{}$

$ \frac {k(24k^4 \plus{} 60k^3 \plus{} 50k^2 \plus{} 15k \plus{} 1)}{6}$ $ (*)$

Right-hand side:

The sum from (where $ n \equal{} 1$ to the last term) minus the sum from
(where $ n \equal{} 1$ to the term just before the first term):

$ \sum_{k \equal{} 1}^{n}(k(2k \plus{} 3))^2 \minus{} \sum_{k \equal{} 1}^{n}(2k(k \plus{} 1))^2 \equal{}$

$ \frac {k(2k \plus{} 3)(k(2k \plus{} 3) \plus{} 1)(2k(2k \plus{} 3) \plus{} 1)}{6} \minus{} \frac {(2k(k \plus{} 1)(2k(k \plus{} 1) \plus{} 1)(4k(k \plus{} 1) \plus{} 1)}{6} \equal{}$

$ \frac {(2k^2 \plus{} 3k)(2k^2 \plus{} 3k \plus{} 1)(4k^2 \plus{} 6k \plus{} 1)}{6} \minus{} \frac {(2k^2 \plus{} 2k)(2k^2 \plus{} 2k \plus{} 1)(4k^2 \plus{} 4k \plus{} 1)}{6} \equal{}$

$ \frac {k(24k^4 \plus{} 60k^3 \plus{} 50k^2 \plus{} 15k \plus{} 1)}{6}$ $ (**)$

The work is not shown to arrive at $ (*)$ and $ (**)$.
[/hide]


trigonometry
logarithms
integration
calculus
function
limit
real analysis

$p<1$,
prove $\sum \frac{\cos {\ln n}}{n^p}$ disverges.

The blocks of positive and negative terms are very long here- that's why we don't get conditional convergence. The subtitution $x=\ln n$ gives you effectively $\int_0^M\cos x\cdot e^{(1-p)x}\, dx$, which should diverge for all $p\le1$.

Consider the terms between $a_n=e^{(n-\frac12)\pi}$ and $a_{n+1}=e^{(n+\frac12)\pi}$. These terms all have the same sign, so by the integral estimate, $\sum_{a_n<k<a_{n+1}}\frac{\cos\ln k}{k^p}=\int_{a_n}^{a_{n+1}}\frac{\cos\ln x}{x^p}\, dx+O(e^{-pn\pi})$
$=\int_{(n-\frac12)\pi}^{(n-\frac12)\pi}\cos t\cdot e^{(1-p)t}\, dt+O(e^{-pn\pi})$
This integral clearly tends to infinity when $p<1$, so the differences between partial sums are unbounded. We would actually need that part of the sum to tend to zero, so the fact that it tends to $2(-1)^n$ when $p=1$ shows that the sum diverges in that case as well.
The error in the integral estimate is no more than the total variation of the function multiplied by the lengths of the small intervals- that's where $O(e^{-pn\pi})$ comes from.

$a_n=E(e^{-\pi/3+2n\pi})+1$
$b_n=E(e^{\pi/3+2n\pi})$

$\sum_{a_n}^{b_n}\frac{\cos(\ln n)}{n^p}\geq1/2\frac{b_n-a_n+1}{b_n^p}=c_n$

but $c_n \geq Ke^{2\pi n (1-p)}$ wich tends to infinity (K>0)

The Cauchy criterion is not satisfy the serie is divergent

to jmerry,can you tell me how $O(e^{-pn\pi}) $ come from for more detail??
to Moubinool, your solution is nice :)
thanks to you two :)

I think i find a more general problem:
for every sequence $\{a_n\}$
if such that $n(a_{n+1}-a_n)$ is bounded and $\lim_{n \to \infty}a_n$ is not equal to 0.
prove:$\sum \frac{a_n}{n^p}$ is divergent


In $ \triangle ABC$ with area $ S$, find the side lengths such that the length of $ BC$ is minimized.

What?  Can't we make the height to BC arbitrarily long, so BC can be arbitrarily small for a given area?

[quote="CatalystOfNostalgia"]Maybe BC = min(AB,BC,CA)?[/quote]
Am I missing something?  How would that change anything?  Unless you meant max(AB,BC,CA)...

Oops I forgot the rest of the problem. Sorry. Can a mod please lock this?


function
geometry
geometric transformation
reflection
algebra
polynomial
complex analysis

This is my proof which I think is easy to understand and accessible to beginners but I'm not sure I have everything precisely correct.  I've added a little to the standard presentation.  If anyone can point out any incorrect or poorly interpreted concepts in the proof, that would be nice.  The reason I say "accessible" is I think sometimes proofs in texts are a little difficult to understand for beginners.  Just my personal opinion, that's all and hopefully I'll not offend anyone.

Theorem:
Let $ f(z)$ have an essential singularity at $ z_s$. Then $ f(z)$ comes arbitrarily close to any complex number an infinite number of times in any deleted neighborhood ($ D'(z_s,\delta)$) of $ z_s$.

Proof:
Assume the contrary: There exists a complex number $ h$ such that for arbitrary $ \delta$ and $ \epsilon$,  $ |f(z) \minus{} h|\geq \epsilon$ for all $ z\in D'(z_s,\delta)$.  This then makes the function
\[ g(z) \equal{} \frac {1}{f(z) \minus{} h}
\]
bounded in the deleted neighborhood of $ z_s$ with $ \displaystyle |g(z)| < 1/\epsilon$ for $ z\in D'(z_s,\delta)$.  If a function is bounded in a singular neighborhood, then the limit as $ z$ approaches the singularity exists. Because of the existence of this limit, any singularity imparted to $ g(z)$ by association with $ f(z)$ in the expression above is artificial or ``removable'' and reflects a deficiency in notation rather than analyticity. Consider for example the singularity of the function $ \displaystyle h_1(z) \equal{} \frac {1}{z}$ and it's ``removal'' by the expression $ \displaystyle h_2(z) \equal{} \frac {1}{h_1(z)}$.
The function $ g(z)$ then, by virtue of the assumptions made regarding $ f(z)$ and the existence of the limit above, is forced to be analytic throughout the undeleted neighborhood of $ z_s$.  This then would force
\[ f(z) \equal{} \frac {1}{g(z)} \plus{} h
\]
to be analytic if $ g(z_s)\ne 0$, or having a pole of multiplicity $ m$ if $ z_s$ is a zero of multiplicity $ m$ of $ g(z)$.
A contradiction is reached in both cases since it was assumed $ f(z)$ had an essential singularity at $ z_s$.  Thus there must exist a point $ z_1\in D'(z_s,\delta)$ for $ \delta > 0$ such that $ |f(z_1) \minus{} h| < \epsilon;\quad h\in\mathbb{C},\quad \epsilon > 0$.
Now we apply the theorem again but this time look for a point $ z_2\in D'(z_s,\delta_2)$ with $ \delta_2 \equal{} 1/2|z_1|$. Using the same analysis we conclude there is another point such that
\[ |f(z_2) \minus{} h| < \epsilon;\quad z_2\in D'(z,\delta_2).
\]
Since $ \delta$ was arbitrary, we can do this an infinite number of times with $ \delta_k \equal{} \frac {1}{2}|z_{k \minus{} 1}|$ and thus conclude in the vicinity of an essential singularity, $ f(z)$ comes arbitrarily close to any value an infinite number of times.

[edit] corrected formulas as per Extremal below

in your proof $ c \equal{} h$ ?
also comes to arbitrary to any complex number: case when such number is equal to $ \infty$ you need to consider .
I think you don't need $ \epsilon$... and e.t.c .
Idea is true.
and the name of this theoerm as I know is $ Sokhockogo$ (I don't know How it will be in english  :blush: )

[quote="shawtoos"]
A contradiction is reached in both cases since it was assumed $ f(z)$ had an essential singularity at $ z_s$. [/quote]
The proof stops right there. This is just a proof copied by a book, but completely misunderstood in the end. You have shown that under the assumption of the property not holding, $ z_s$ must be a removable singularity or a pole. But since it is an essential singularity (and not a removable singularity or a pole) there is a contradiction.

Thanks Extremal.  I corrected it.  Kalle, how do I completely mis-understand this?     Actually I thought I understood it well.  I'm surprised.   My objective is of course not to take credit for the proof but rather present it in a manner that's easy to understand by beginners with additional information I think would be helpful in that regard?  Is this not the case?

[quote]
Because of the existence of this limit, any singularity imparted to $ g(z)$ by association with $ f(z)$ in the expression above is artificial or ``removable'' and reflects a deficiency in notation rather than analyticity. Consider for example the singularity of the function $ \displaystyle h_1(z) \equal{} \frac {1}{z}$ and it's ``removal'' by the expression $ \displaystyle h_2(z) \equal{} \frac {1}{h_1(z)}$
[/quote]
Do you really think this passage clarifies anything? Kalle's point was that your text reflects mainly your personal struggle with the theorem and, if anything, just shows that you had trouble in some spots. The final part is OK, of course, once you want "infinite number of times". The first part shows only that you approach the value at least once in each punctured neighborhood. The problem is that rather then stating this claim explicitly, you started with "Assume the contrary..." and messed a few things in the very first phrase (the quantifiers on $ \varepsilon$ and $ \delta$ should be $ \exists$, not $ \forall$, etc.).

[quote="fedja"]
Do you really think this passage clarifies anything? [/quote]

I guess not Fedja.  I'll try to come up with something better in the future and I see where I need to change the language to "for some $ z$"  but can't now since my edit time has expired.

I don't mind being corrected guys, not at all.  And I accept having a struggle with the entire field.  It's quite formidable to me but wonderfully beautiful at the same time :)

The thing is, the last part shouldn't be necessary if you correctly state "Assume the given condition does not hold. This means that there is a complex number $ h$ and a $ d > 0$ such that $ |f(z) \minus{} h| > d$ in a punctured neighborhood of the singularity". Proceed to prove that this gives a contradiction.

Maybe this helps you.  There is no difference between the statements:

Then $ f(z)$ comes arbitrarily close to any complex number an infinite number of times in any deleted neighborhood ($ D'(z_s,\delta)$) of $ z_s$.

Then $ f(z)$ comes arbitrarily close to any complex number in any deleted neighborhood ($ D'(z_s,\delta)$) of $ z_s$.

The infinite number of times thing here is completely unnecessary; how does f come arbitrarily close to something without doing it an infinite number of times? This isn't at all picard's theorem that says that $ f$ takes on every value infinitely many times.

Now prove that an entire function either is a polynomial or has en essential singularity at infinity.

And honestly, this proof is very simple and easy. Maybe read some rigorous real analysis before you tacke complex

[quote="Kalle"]The thing is, the last part shouldn't be necessary if you correctly state "Assume the given condition does not hold. This means that there is a complex number $ h$ and a $ d > 0$ such that $ |f(z) \minus{} h| > d$ in a punctured neighborhood of the singularity". Proceed to prove that this gives a contradiction.

Maybe this helps you.  There is no difference between the statements:

Then $ f(z)$ comes arbitrarily close to any complex number an infinite number of times in any deleted neighborhood ($ D'(z_s,\delta)$) of $ z_s$.

Then $ f(z)$ comes arbitrarily close to any complex number in any deleted neighborhood ($ D'(z_s,\delta)$) of $ z_s$.

The infinite number of times thing here is completely unnecessary; how does f come arbitrarily close to something without doing it an infinite number of times?
 This isn't at all picard's theorem that says that $ f$ takes on every value infinitely many times.[/quote]

Thanks.  That's very helpful to me although I don't understand why you mention Picard.  This is Casorati-Weierstrass although I plan to do likewise with Picard's Theorem shortly albeit not in the formal context of a proof this time.  You are welcome to point out problems with my presentation there if you wish.

[quote]Now prove that an entire function either is a polynomial or has en essential singularity at infinity.[/quote]

I'll work on that one.

This though I consider inappropriate as this is a tutorial section for beginners:

[quote]
And honestly, this proof is very simple and easy. Maybe read some rigorous real analysis before you tacke complex[/quote]


Tell me what you think of this:
     Scientists have crazy theories that if we ship a large amount of Carbon Dioxide to Mars the planet will heat enough for the iron core to melt. This would create a magnetic shield blocking solar wind. If there are giant plant farms installed after this they will turn the Carbon Dioxide into oxygen and make Mars habitable, and Earth Global Warming Free. And even if the the core doesn't melt, if they have indoor plant farms they could live in bunker homes on Mars.(who would want to I don't know.)

Here are some questions:
1. Is this possible in any way?
2. How will water get to Mars to water the plants?
3. How will we get all this material to Mars?'
4. Could we make the rocket inhale the Carbon Dioxide it creates and take that to Mars too?
5. Will this stop Global Warming?

I'd prefer if multiple people answer to wee if we have different opinions.

There is one thing I can ensure you: you can't melt iron by global warming, especially not on mars which is even farer away from sun than earth (note that even on venus with very much glasshouse effect and nearer to the sun, we still don't get near to the >1800K we need to melt iron). I have very much doubt that any serious scientist has claimed this.
Conversely, I think the solar wind is one of the neglegtible problems in colonizing mars...
And you don't even want to try to send the carbon dioxide anywhere away at the moment as starting a rocket causes much more carbon dioxide ;)

[quote="AstroPhys"]You should post this in the Astronomy and Engineering Forum.  :)[/quote]
As in right [url=http://www.artofproblemsolving.com/Forum/index.php?f=449]here[/url]

Well if Mars really does heat up a lot, then the ice caps will melt and there will be water for the farms.

Assuming everything else works... :roll:

It probably never will. Mars is farther away from the sun, and has a somewhat cool core. So its ice caps will never melt. That is if we don't do anything to it.

Just my thoughts...

[quote="dbkarp"]Tell me what you think of this:
     Scientists have crazy theories that if we ship a large amount of Carbon Dioxide to Mars the planet will heat enough for the iron core to melt. This would create a magnetic shield blocking solar wind.[/quote]

Where would it all come from? By your logic, if there was that much carbon dioxide on Earth, Earth would be melting too.

[quote="dbkarp"]If there are giant plant farms installed after this they will turn the Carbon Dioxide into oxygen and make Mars habitable,[/quote]

They would have to be pretty giant to process such a "large amount" of carbon dioxide.

[quote="dbkarp"]and Earth Global Warming Free.[/quote]

You can't possibly ship that much carbon dioxide anyway. Do you realize how big the atmosphere is? Even though carbon dioxide only takes up about 300 parts per million of it, it's still a lot. And how would you separate it from the rest of the atmosphere?

[quote="dbkarp"]And even if the the core doesn't melt, if they have indoor plant farms they could live in bunker homes on Mars.(who would want to I don't know.)[/quote]

There are many problems involved in transporting humans on such long journeys through space (several months, in this case). For one, keeping someone in zero-gravity for that long is dangerous.

[quote="dbkarp"]Here are some questions:
1. Is this possible in any way?[/quote]

No.

[quote="dbkarp"]2. How will water get to Mars to water the plants?[/quote]

I suppose if you melted the ice caps, maybe you might get a little water from that. But certainly not enough to support "giant" plant farms.

[quote="dbkarp"]3. How will we get all this material to Mars?[/quote]

Precisely my point.

[quote="dbkarp"]4. Could we make the rocket inhale the Carbon Dioxide it creates and take that to Mars too?[/quote]

It would be "inhaling" the other exhaust gases too, many of which are highly dangerous.

[quote="dbkarp"]5. Will this stop Global Warming?[/quote]

No.

I am more optimistic on terraforming Mars.

Evidences have shown that water had flown on Mars surface before. So where has water gone?

I can think of two possibilities:

1. some water evaporated into Mars thin atmosphere and eventually escaped into space.
2. some water might be frozen under the surface of Mars.

I believe that someday we will find fair amount water (ice) underneath the Mars surface. With the water resource to tap, we are one step closer to terraforming Mars.

Another problem is the energy source. It's impractical to haul conventional fuel from Earth to Mars due to the cost-effectiveness. But once we have the controllable fusion technology, it will be feasible to carry a small amount of fusion fuel to Mars and support a Martian base in an-enclosed environment. We can setup facilities to generate CO2 that are locked in the Martian rocks. Eventually, Mars atmosphere will be thick enough to sustain life (e.g. plants first) migrated from Earth.

On a side note, I think it's even easier to setup a lunar-base, due to the fact that the Moon is much closer to Earth. Of course, there will be no hope of creating a thick enough atmosphere on the Moon, but we can setup enclosed environment. The energy source is plenty on the Moon. The solar energy is ready for us to use. Also there are helium-3 as fusion fuel in the future (once we know how to utilize it). Now, NASA engineers are busy finding water source on the Moon.

Here are some things to consider when you try to colonize Mars:

1. It will be very expensive. We're gonna have to spend, oh, A QUADRILLION dollars just to make machines that produce greenhouse gases to heat up Mars? And where are we gonna get the money to raise a ton of plants to convert the CO2 into oxygen?

2. It will take CENTURIES, maybe even a few millennia, to completely transform the Martian surface into a plant-covered one. And how long will it take to heat up the atmosphere? Will transforming Mars REALLY happen fast enough to meet our world's skyrocketing population?

3. What about Martian life? By terraforming Mars, we might destroy native Martian life (if there is any) before we learn about it. Our warmer atmosphere formed by the transformation of Mars might kill off whatever Martian microbes rely on cold to survive.

4. It's just plain dangerous. Face it--we humans are gonna die on Mars rather than survive. It's WICKED cold there and the thin atmosphere give ZERO protection against solar radiation. We won't have fuel to transport the humans back if anything goes wrong--in fact, we can't even get ROBOTS back from Mars! Say goodbye to those who volunteer for the next manned mission to Mars.

Yes life is habitable in mars :thumbup:
[hide]only for anaerobic bacteria. :rotfl: [/hide]only about 10 humans can go to mars for each launch.to transfer at least 1% of population=money of all countries and also value of all properties they have.it can also be more than that.

It really wouldn't be that expensive if we simply just brought a large greenhouse and filled it with plants and humans. Humans would produce CO2 and in return plants would make oxygen. It's extremely simple once you think about it. You would just pump a little co2 in to get it started and we're off and running!

[quote="El_Ectric"][quote="hshiems"]
4. It's just plain dangerous. Face it--we humans are gonna die on Mars rather than survive. It's WICKED cold there and the thin atmosphere give ZERO protection against solar radiation. We won't have fuel to transport the humans back if anything goes wrong--in fact, we can't even get ROBOTS back from Mars! Say goodbye to those who volunteer for the next manned mission to Mars.[/quote]
Do you seriously think you know more than the scientists planning this?[/quote]
I'm just saying that the scientists have to be EXTRA CAREFUL when planning this--this is more dangerous than a walk in the park. If they mess up, the humans on the mission are basically dead.

[quote="hshiems"]we can't even get ROBOTS back from Mars![/quote]
Any basis for thic probably completely unfounded claim¿ There is a huge difference between "not being able to" and "having no interest to do".

And as a matter of fact, getting those humans back is very cost intensive, which is the reason why some space agencies are thinking about sending rather old astronauts on a one-way trip.

It really might be a matter of effort...
In the 60's we got a man to the moon (AND BACK!) with less computational power than an ipod. That's something.
Clearly our rocketry hasn't advanced as much, maybe it's even backslided a bit. But there are some advancements such as Ion rockets and such allow us to move much much faster.

I think there is a girth of technology to draw on for such a mission, and if humanity put a similar strong effort to accomplish a mission to Mars and back we totally could, but we just aren't without a compelling reason.

As for terraforming, that would take lots of new technology... centuries long process of adding CO2 to the atmosphere. Not to mention, I think Mars gradually loses it's atmo over time because it's lighter than the earth, and doesn't hold on to water vapor (go to the terraforming mars wiki and look under effects of space weather, I can't link yet because I'm too new).

But a colony may very well be possible this century with green houses for crops and such. Still new tech, but fessible I'd think!

Well we could make the planet heat up to sustainable temperatures causing a runaway greenhouse effect after melting the dry ice caps.  Then we stick a magnet in the core  to create a magnetic field.  Or we could speed up the rotation of mars by strapping booster engines on it's equator.  Or smashing an asteroid to eed up it's rotation and make the liquid iron it does have churn more increasing it's magnetic field.  Themagnet idea reguires an extremely large magnet.  Or we could conduct a strong electrical current through mars to generate the magnetic field and protect us from the solar wind.

Mars is actually in the habitable zone of our solar system; Scientists have discovered ice that they think is water, and some people speculate that it was once inhabited by some form of life, but something happened to turn it into what it is now (I should know; according to my profile I live there ;)).

1. Yes
2. Duh. Bring it there. Besides. There are some [size=50]tiny[/size] bits of water-ice.
3. You're really asking this... Rocket
4. Why would you want to do that? Mars is mostly carbon dioxide already!
5. Heck naw. Mars is smaller than Earth. That means not everyone can go to Mars. Therefore still global warming. Besides... some global warming is not from humans.

1.improbable
2.There are water ice on mars
3.wait until nuclear fusion thrusters comes out, then turn asteroids into starships. Another option is the mars dirct plan, where you make fuel for return joerny at mars.
4. AND 5. no, but there are carbon dioxide ice on mars, so we could use that. we could use heat from the fusion plants to melt lots of ice at once.

The global warming wouldnt melt the core, but we could drill into the center and pump plasma using magnetic fields from the fusion reactors. There was a Star Trek: the Next Generation episode where they melted a freezing core of a planet by plasma. We could install giant magnets(and power it by fusion), but exposur to high levels of electromagnetic radiation could cause cancer and stuff. If we were to heat mars so that the core would melt, the surface would melt too.

[quote=herpnerp]1. watch the martian
2. watch the martian
3. watch the martian
4. watch the martian
5. watch the martian
if you would watch the martian you walnut even have these questions in the first place
btw longs it would take many years and 100s of rockets to get all the  materials to mars and money that no one is willing to spend[/quote]

1.READ the martian
2.READ the martian
3.READ the martian
4.READ the martian
5.READ the martian

Well in an astride that landed on earth they found it was martian rock and they found small microscopic organisms so maybe at one point there was life but not anymore.


In an all-aginst-all tournament, three teams are said to `beat round' each other if each of them wins exactly once when only games among those three teams are considered. At most how many beat-rounds may happen in a tournament with 23 participating teams?

Let B(n) be the answer for n teams.  We claim $B(n) = (\frac{n-1}{2})^2$ for n odd.

It is easy to get such an example:  suppose $A$ is a set of 11 people who beat $X$, and $X$ beat a set $B$ of 11 people, and everyone in $B$ beat everyone in $A$.  Then $Xba$ is a beat round for all $b \in B$ and $a \in A$.

However I cannot prove because of the variety of configurations for which this is possible.  For example, taking a simple case, with players ABCDE, and "XYZ" meaning x beats y beats z beats x:

AEB AED CEB CED

Or

AEB CED ADC CBD

----

Please post answer ?


probability
expected value
probability and stats

A software package consists of $ 12$ programs, five of which must be upgraded. If $ 4$ programs are randomly chosen for testing,

$ \textbf{(A)}$ What is the probability that at least two of them must be upgraded?
$ \textbf{(B)}$ What is the expected number of programs, out of the chosen four, that must be upgraded?

I got this problem marked wrong and would like to know how to do it properly.
Thanks.

i)Well, no. of ways 4 programs can be chosen out of 12 = 12C4

No. of favourable conditions = 5C2 X 10C2(fixing two upgradable programme beforehand so that they always occur in the selection.)

Hence probability,

5C2X10C2/12C4

I couldn't understand what the second part meant.

[quote="A K Bhattacharyya"](fixing two upgradable programme beforehand so that they always occur in the selection.)[/quote]  Which means that you overcount all those times when I choose more than two of the five special programs.  For example, there are five ways to choose four special programs; you've counted each of these six different ways.  Probably the easiest way to proceed here is to find the probability you choose zero special programs and the probability you choose exactly one special program, then take the complement.

The second question is a wonderful place to use linearity of expectation.


18. A simple pendulum of length L is constructed from a point object of mass m suspended by a massless string attached to a fixed pivot point. A small peg is placed a distance 2L/3 directly below the fixed pivot point so that the pendulum would swing as shown in the figure below. The mass is displaced 5 degrees from the vertical and released. How long does it take to return to its starting position?

(image attached)

a) $ \pi\sqrt\frac{L}{g}\left(1\plus{}\sqrt\frac{2}{3}\right)$
b) $ \pi\sqrt\frac{L}{g}\left(2\plus{}\frac{2}{\sqrt3}\right)$
c) $ \pi\sqrt\frac{L}{g}\left(1\plus{}\frac{1}{3}\right)$
d) $ \pi\sqrt\frac{L}{g}\left(1\plus{}\sqrt3\right)$
e) $ \pi\sqrt\frac{L}{g}\left(1\plus{}\frac{1}{\sqrt3}\right)$

I have no clue how to approach this.

Let me take a shot:

In case your difficulty arose from misunderstanding the question, here're some hints:
1) The mass starts from the position at the left (in that bent position)
2) The "lower" pendulum" (the "peg" of length L/3) swings to the vertical position
3) The "whole" pendulum, of length L, swings to the rightmost position
4) Reverse all of the above, such that the mass returns to the left.

[hide="Solution"]

$ T \equal{} \frac{1}{4}(2\pi \sqrt{\frac{L}{3g}})$(at vertical) $ \plus{}\frac{1}{4}(2\pi \sqrt{\frac{L}{g}})$(at right) $ \plus{}\frac{1}{4}(2\pi \sqrt{\frac{L}{g}})$(at vertical again) $ \plus{}\frac{1}{4}(2\pi \sqrt{\frac{L}{3g}})$(at left again)

$ \equal{}\pi \sqrt{\frac{L}{g}}(\frac{1}{\sqrt{3}}\plus{}1)$

[/hide]

Wow, I feel stupid. I did understand the question, but I can't believe I forgot the period of a pendulum. Thanks!


Find the interger part of:
$ A\equal{}\sqrt[n]{1\minus{}\frac{x}{2008}}\plus{}\sqrt[m]{1\plus{}\frac{x}{2008}}$ where $ x$ is a real number in $ [\minus{}2008;2008]$ and $ m;n$ are natural numbers,$ m \geq n \geq 2$

Let $ y\equal{}\frac{x}{2008}$, then $ y\in [\minus{}1,1]$. Then $ A\equal{}\sqrt[n]{1\minus{}y}\plus{}\sqrt[m]{1\plus{}y}$. Because $ 0\le 1\plus{}y,1\minus{}y\le 2$ and one of $ max(1\minus{}y,1\plus{}y)\ge 1$ we get $ 1<A<3$. If $ x\equal{}0$ we get $ [A]\equal{}A\equal{}2$ else $ A<2\plus{}y(\frac 1m \minus{}\frac 1n )$ and for $ y>0$ we get $ [A]\equal{}1.$

[quote="Rust"]Let $ y \equal{} \frac {x}{2008}$, then $ y\in [ \minus{} 1,1]$. Then $ A \equal{} \sqrt [n]{1 \minus{} y} \plus{} \sqrt [m]{1 \plus{} y}$. Because $ 0\le 1 \plus{} y,1 \minus{} y\le 2$ and one of $ max(1 \minus{} y,1 \plus{} y)\ge 1$ we get $ 1 < A < 3$. If $ x \equal{} 0$ we get $ [A] \equal{} A \equal{} 2$ else $ A < 2 \plus{} y(\frac 1m \minus{} \frac 1n )$ and for $ y > 0$ we get $ [A] \equal{} 1.$[/quote]
$ y<0$,How many cases?


This is one of the harder problems I have recently encountered, and I've actually found out that it's regarded as pretty classic now. Nevertheless, solving it can be a lot of fun (unless you're solving it at 2pm in order to turn it in next morning, trust me on this).

Ok, so here it goes:

A space station is geosinchronously orbitting the Earth, at distance $ r_0$ from its center. An astronaut floats in space at a distance $ \epsilon\ll r_0$ from the station, positioned directly between the station and the center of Earth. The astronaut throws a beer can in such a way that, as seen from the astronauts on the station, the can orbits around the station while always remaining in the plane of the station's orbit around the Earth. Find:
1) the initial velocity vector the astronaut must give to the beer can in order to achieve the described effect;
2) the equation of the beer can's orbit, as seen from the space station.

hi im a newbie to this forum
im a mathematician and not a physicist anyway but ill give this a go

a few question:
1) is the space station rotating about its axis?
2) what do you define as "orbiting?" is it constant angular momentum? constant radial distance? or merely taking one revolution around the space station?

e.g.
would a orbit of smaller radius work for teh beer can?

srry for my stupidity

The questions are by no means stupid and they're certainly welcome :)

1) No. The station can be imagined to be a point rotating around the Earth, and since it is known that it's in the geosinchronous orbit we know its angular velocity. (In fact, I used the term a bit erroneously, I should have said the orbit is "geostationary", i.e. it is the orbit in the Earth's equatorial plane.)

2) 'Orbiting' means that, as seen from the station, the can moves periodically in a closed trajectory (i.e. ellipse or circle) around the station. In general, the situation is best described by your third and final guess.


since $p(x)=x^{5}+x^{2}+1$ and it has roots $r_{1}$, $r_{2}$, ..., $r_{5}$.
and $q(x)=x^{2}-2$

$p(\sqrt{x+2})$ will have the roots $q(r_{1})$, $q(r_{2})$, ..., $q(r_{5})$

$(\sqrt{x+2})^{5}+(\sqrt{x+2})^{2}+1=0$
$(x+2)^{\frac{5}{2}}=-1-x-2$
$(x+2)^{5}=(-x-3)^{2}$

if you expand it, the highest degree of the polynomial is $5$ and the leading coefficient is $1$. After you move everything to the left side, the constant coefficient is $2^{5}-3^{2}=32-9=23$

so the product will be $\boxed{-23}$ ?

i am not sure if i am right

[quote="scorpius119"]yes, it's -23, and there is a simpler solution...

[hide="very big hint"]
$x^{2}-2=(x+\sqrt{2})(x-\sqrt{2})$, so the product of $q$'s is equal to
\[\left( \prod_{n=1}^{5}(r_{n}+\sqrt{2}) \right) \left( \prod_{n=1}^{5}(r_{n}-\sqrt{2}) \right) \]
[/hide][/quote]

woa, I guess I ODed.
sorry, I don't get what are those symbols (the one looks like pi).. maybe I am a sophomore and I haven't see it before. please tell me what are those symbols and how we use it
:-) ty

Product from n=1 to 5 applied to the terms

[quote="scorpius119"]yes, it's -23, and there is a simpler solution...

[hide="very big hint"]
$x^{2}-2=(x+\sqrt{2})(x-\sqrt{2})$, so the product of $q$'s is equal to
\[\left( \prod_{n=1}^{5}(r_{n}+\sqrt{2}) \right) \left( \prod_{n=1}^{5}(r_{n}-\sqrt{2}) \right) \]
[/hide][/quote]

that is very nice!!!

My solution from WOOT:

The key is that $q(x)=\left(x+\sqrt2\right)\left(x-\sqrt2\right)$.
$$\prod_{i=1}^5q(r_i)=\prod_{i=1}^5(r_i+\sqrt2)\cdot\prod_{i=1}^5(r_i-\sqrt2)=p\left(\sqrt2\right)p\left(-\sqrt2\right)=\boxed{-23}$$

How did you get that $\prod_{i=1}^5(r_i+\sqrt2)\cdot\prod_{i=1}^5(r_i-\sqrt2)=p\left(\sqrt2\right)p\left(-\sqrt2\right)$?

Since $p(x)=\prod_{i=1}^5(x-r_i)$, I plugged $x=\pm\sqrt2$.


analytic geometry
limit
calculus
calculus computations

A sequence $ a_1,\ a_2,\ a_3,\ \cdots$ are defined as follows.
Let $ a_1\ (a_1\neq 0,\ a_1\neq 2)$ be a given real first term. When $ a_2$ is given, take a point $ P$ with $ x$ coordinate $ a_n$ on the curve $ y \equal{} \frac {1}{x} \minus{} 1$. Let $ Q(a_{n \plus{} 1},\ 0)$ be the intersection point of the tangent line drawn at $ P$ and the $ x$ axis. Find $ \lim_{n\to\infty} a_n$.

$ g(x)\equal{}f(x_0) \plus{} f'(x_0)(x\minus{}x_0)$ is a tangent line equation.

We have $ g(a_{n\plus{}1})\equal{}0$ and $ x_0 \equal{} a_n.$

$ 0\equal{}\frac{1}{a_n}\minus{}1\minus{}\frac{1}{a_{n}^2}(a_{n\plus{}1}\minus{}a_n).$

$ a_{n\plus{}1}\equal{}2a_n \minus{} a_{n}^2.$

If $ a_1\in (0,2)$ then $ \lim_{n\to \infty}a_n \equal{} 1.$

If $ a_1 < 0$ then $ a_{n\plus{}1} < 2 a_n < 2^{n}a_1.$ Limit does not exist.

It $ a_1 > 2$ then $ a_2 < 0$ and get previous case.

[quote="kunny"]How did you solve the reccurence?[/quote]
 if  $ a_1 \equal{} 1$ then $ a_2 \equal{} a_3 \equal{} ... \equal{} a_n \equal{} 1$
if $ a_1 \in (0,2)$  we have :  $ a_{n \plus{} 1} \minus{} 1 \equal{} \minus{} (a_n \minus{} 1)^2 \equal{} \minus{} (a_{n \minus{} 1} \minus{} 1)^4 \equal{} ... \equal{} \minus{} (a_1 \minus{} 1)^{2n}$
by other hand,  $ |a_1 \minus{} 1| < 1$ Hence  $ \lim\limits_{n \to \infty} x_n \equal{} 1$


Let $n$ be odd number. Prove that $n(n^{4}-1)$ is multiple of 240.

I'll assume n isn't allowed to equal one, or else you're multiplying by zero.

For (n=3), expression = $1*240$
Our next value (n=5) is $13*240$
Our next value (n=7) is $70*240$
Our next value (n=9) is $246*240$
Our next value (n=11) is $671*240$

We can rewrite the expression for any integer.
$E \rightarrow (2k+1)((2k+1)^{4}-1)$
$E \rightarrow (2k+1)(2k)(2k+2)(4k^{2}+4k+2)$

Ignoring the first factor for now, we see that the last three factors become $2(k)(k+1)(2k^{2}+2k+1)$
This means that these three factors have an even product.
Which means that $(2k+1)(k)(k+1)(2k^{2}+2k+1)$ is a multiple of 120.
I'm stuck here. Can anyone help me?

$240=2^{4}\cdot 3\cdot 5$

$n(n^{4}-1)=n(n^{2}+1)(n+1)(n-1)$

from pigeonhole , one of $n-1,n,n+1$ is a multiple of $3$

since $n$ is odd , we must have $n+1,n-1,n^{2}+1$ all even . But one of $n+1,n-1$ have to be multiple of $4$ . So it is a multiple of $2^{4}$ .

If $n$ is a multiple of $5$ , we are done .
Else , $n^{4}-1$ will be a multiple of $5$ since $n^{4}\equiv 1\mod 5$ for $n$ which is not a multiple of 5.

[color=blue]Similar Example[/color] Chiba University entrance exam 2007

Let $n$ be odd number.Answer the following questions.

(1) Prove that $n^{2}-1$ is a multiple of 8.
(2) Prove that $n^{5}-n$ is a multiple of 3.
(3) Prove that $n^{5}-n$ is a multiple of 120.

[quote="junggi"][quote="DiscreetFourierTransform"]I'll assume n isn't allowed to equal one, or else you're multiplying by zero.
[/quote]
i dont think that this actually matters, since 0 is a multiple of any number

@kunny: er aren't those all proved in shyong's solution?[/quote]

shyong's solution is correct. :)

[quote="Altheman"]in mod 5, it is equivalent to: (n)(n-1)(n+1)(n^2+1-5)=(n-2)(n-1)(n)(n+1)(n+2)[/quote]

This is the key point. :wink:

(1) $n=2k+1 \implies n^{2}= (2k+1)^{2}-1 = 4k^{2}+4k = 4k(k+1)$
Since $2|k(k+1)$, we're done.

(2) $n^{5}-n=n(n-1)(n+1)(n^{2}+1)$. Apply pigeonhole over the first 3 factors.

(3) $n^{5}-n=n(n-1)(n+1)(n^{2}+1)$. We already know this is divisible by 3. Also by pigeonhole, one of $(n-1)$ and $(n+1)$ is divisible by 4 and the other is divisible by 2 but not 4, so the product must be divisible by 8. Left to prove is that it is divisible by 5. If either n, (n-1), or (n+1) is divisible by 5, we're done.

Otherwise, n is congruent to 2 or 3 mod 5, and $n^{2}+1 \equiv 0$ mod 5, and we're done.


inequalities
function
inequalities unsolved

for $a_1,a_2,a_3,...,a_n >0$ prove:
$(1+a_1)(1+a_2)(1+a_3)...(1+a_n) \geq (1+\sqrt[n]{a_1a_2a_3...a_n})^n$
[hide]i have posted this before in algebra section but no one was interested for it there[/hide]

its a general form of holder

---

if you want a proof, can show by induction.

n=2 is cauchy

now suppose

$\prod^n (1+a_i) \geq (1 + \sqrt[n]{\prod^n a_i})^n$.

Let $r = \prod^n a_i$ and $x > 0$.

It remains to show:

$(1 + \sqrt[n]{r})^n(1+x) \geq (1 + \sqrt[n+1]{rx})^{n+1}$

which is just the specific version of holders here:

http://mathworld.wolfram.com/HoeldersInequalities.html

it's exactly huygen's inequality
To prove: $\prod (a_i+b_i) \ge (\sqrt[n]{\prod a_i}+\sqrt[n]{\prod b_i})^n$
Proof:
As it's homogeneous we can assume all $a_i+b_i$=1, then
$\sqrt[n]{\prod a_i}+\sqrt[n]{\prod b_i} \le \frac{1}{n} (\sum a_i + \sum b_i) = 1$
Finished.

we can also just sum up all inequalities
$S_k \geq \frac{n!}{k!(n-k)!} (a_1 a_2 ... a_n)^{\frac{k}{n}}$
just AM-GM where $S_k$ represents all sums with k terms of $a_1 ... a_n$

or another way is to apply Jensen to the function $1+e^x$


logarithms
inequalities
real analysis
real analysis unsolved

construct a cantor-type set on [0,1] by removing from each interval remaining at the kth stage a sub-interval of relative length $ a_k, 0 < a_k < 1$. show that the remainder has measure 0 iff $ \sum_{k \equal{} 1}^{\infty}a_k \equal{} \infty$.

The measure of the what remains after the $ n\minus{}$th step is $ (1\minus{}a_1)\ldots(1\minus{}a_n)$.
We want this to tend to zero, i.e. $ P_N\equal{}\prod_{n\equal{}1}^N(1\minus{}a_n)\to 0$. Since all the terms are positive we may plug in the logarithm, and we get (with the use of series expansion):
$ \log(P_N)\equal{}\minus{}\sum_{n\equal{}1}^N\sum_{k\equal{}1}^{\infty}\frac{a_n^k}{k} \equal{} \minus{}\sum_{n\equal{}1}^N a_n\minus{}\sum_{n\equal{}1}^N\frac{a_n^2}{2}\minus{}\sum_{n\equal{}1}^N\frac{a_n^3}3\minus{}\ldots$.
Now, if $ \sum_{n\equal{}1}^Na_n\to\infty$ we see that $ \log(P_N)\to\minus{}\infty$, i.e. $ P_N\to 0$.
If $ \sum a_n$ converges to some $ L<\infty$:
There is some $ i_0$ so that $ a_n<\frac 12, \forall n\ge i_0$ and this implies that $ a_n<\frac1{\sqrt[k\minus{}1]k}, \forall k\ge 2$. Hence we have that $ a_n^k<\frac1k a_n,\forall n\ge i_0$.
$ |P_N|\le \sum_{n\equal{}1}^Na_n\plus{}\sum_{n\equal{}1}^N\frac{a_n^2}2\plus{}\ldots\le \left(\sum_{n\equal{}1}^Na_n\right)(1\plus{}\frac1{2^2}\plus{}\frac1{3^2}\plus{}\frac1{4^2}\plus{}\ldots)\plus{}\sum_{n\equal{}1}^{i_0\minus{}1}\sum_{k\equal{}1}^{\infty}\frac{a_n^k}{k}$
$ \le L\frac{\pi^2}6\minus{}ln(1\minus{}a_1)(1\minus{}a_2)\ldots(1\minus{}a_{i_0\minus{}1})$, hence $ P_N\not\to0$

i had a quite similar proof. as you noted the product $ (1\minus{}a_1)(1\minus{}a_2)...(1\minus{}a_n)$ is the measure of the remainder at the nth step. i used the fact that $ p_n$ converges to a non zero limit iff $ \sum_{n\equal{}1}^{\infty}a_n$ converges ( ofcourse this is only true for non negative a_n.)

( sketch of proof- suppose that the product converges.  use the inequality $ 1\plus{}x\leq e^x$ for all real x. then substituting $ \minus{}a_k$ for x, k=1,2..n and multiplying the inequalities we get that $ p_n\leq e^{\minus{}s_n}$ since we assumed $ p_n$ converges to a non zero limit and p_n's are decreasing, we see that they are bounded below by a positive constant. from the above inequality we find that $ s_n$ are bounded above and hence $ \sum_{n\equal{}1}^{\infty}a_n$ converges. the converse is also proved using a similar argument)


function
algebra
functional equation
real analysis
real analysis unsolved

If $f$ is a continuous function satisfying $\frac{f(x)+f(y)}{2}\ge f(\frac{x+y}{2})$ for all real $x$ and $y$ then it is convex. But what if $f$ is not assumed to be continuous?

Then it doesn't have to be convex.  Are you asking for an example of such a function?

Any solution to the functional equation $f(x)+f(y)=f(x+y)$ satisfies this. There are discontinuous solutions to that equation; their graphs are dense in the plane.

Since the proof of existence requires the axiom of choice, you're not going to get an explicit construction.

Jmerry is referring to [url=http://mathworld.wolfram.com/HamelBasis.html]Hamel Basis[/url]

An Ostrowski's result: if $\forall{x\not=y},\frac{f(x)+f(y)}{2}>f(\frac{x+y}{2})$ and if $f$ is bounded in a neighborhood of ONE  point then $f$ is strictly convex.


Signups start
8 ppl are needed.
This will be single elimination.

i join.
we only need 1 more person. We can do doulbe elmination then.

Me vs. Latex
Izzy vs. maybach
You guys have 1 week to finish games
This is best of three right now.

$ \text{\LaTeX}$ defeats cgyao, 2-0(4-1,4-1)...

Oh no!!

Next round!
Izzy vs. LaTeX
cgyao15 vs. maybach!
Winner of Izzy and LaTeX will get a bye next round.
Loser of cgyao15 and maybach will be eliminated.

i am eliminated.
Izzy LaTeX you have till next week.

$ \text{\LaTeX}$ defeats izzy 2-1(4-3,1-4,4-1)... I'm surprised.....  :o

[quote="cgyao15"]gratz LaTeX!
Scoreboard
LaTeX 2pts
izzy, maybach 1pt
izzy vs. maybach you have till next week.
LaTeX gets a bye![/quote]

Wait, so I play the winner of izzy vs. maybach, when its over?

Yes.
Then you have two chances to win.
:)

I PMed izzy.  Izzy's currently in a hotel and we need an extension.

1 week extension granted, more will be if needed.

Who shall win the maybach izzy game?
PM me your ideas.

PMing maybach. I PM'd him yesterday but I think it didn't go through.

i will join. highest is 1415. currently is 1250.

Sorry.  It was school.  I played against izzy and failed 0-2.  Bezerk, the game's almost over!  Read the posts!  Also, jjx1 banned you from all future tourneys, remmber? :wink:

OK, it's izzy v. LaTeX now I guess

[quote="$ LaTeX$"]Uh..... cgyao, Isn't this tourney just going to keep going in circles? I've already played izzy once before in this tourney, though I think you're aware of this.[/quote]

The last time you played izzy was for a bye(so you wouldn't have to have played me and face the .0000001% chance that you'd lose), and this game is for the championship.

I die.

How many games do we have to play? Best of 3 or 5?

[quote="$ LaTeX$"]Wait, so whoever wins, wins the whole thing, and my previous game against Izzy basically had no effect on this game...?  I forfeit....  :(  jk...... Even though I'll lose.  :([/quote]

You're advantage was a bye.  Now, you've used your advantage, and the game is even steven.

[quote="cgyao15"]3. If you survive izzy, then it'll be 5.[/quote]

Wait, so if I go up 2-1, the tourney is over, but if izzy is up 2-1, one of us has to get to 3 up?

Yes. That's correct, basically saying that you have two chances to beat izzy(PS USE THEM WELL!:))

LaTeX izzy what's happenening i felt like giving you 2 weeks almost and yet the results arne't here?  :o

how come every time i get to 1300 somebody brings me back to 1200some?

STOP SPAMMING. mods can you edit this post to be Latex wins?

LaTeX wins


Beauty and Morality are Species of Value/

It is human nature to be attracted to the mere appearance of things; the survival of many kinds of animals is dictated by the ability of the male and female to attract one another resulting from the colors and forms of eye appeal. We dress in the morning often based upon what type of trial we are facing; we gain a sense of confidence when we are confident of our appearance.

Our culture provides us little incentive to examine the common principles of our nature in such matters as morality and aesthetics. Such principles represent the very foundation for our actions. We finish our formal schooling without even rudimentary comprehension of these fundamental aspects of our nature. Not only do we finish our schooling with this fundamental ignorance but we leave schooling with a disdain and dismissive attitude of such matters.

We finish schooling with a prejudice against our self. We develop a satisfaction only when we think of our self as being surrounded by objects and laws independent of our self. We finish school unaware of the psychology which is the instrument of our speculations about these laws and principles. We aggressively dismiss the exclusively “subjective and human department of imagination and emotion…we have still to recognize in practice the truth that from these despised feelings of ours the great world of perception derives all its value, if not also is existence…had our perceptions no connection with our pleasures, we should soon close our eyes on this world”.

I think that specialization is perhaps a necessity but it is not necessary, nor is it health, for us to graduate sophomores who lack the rudimentary knowledge of fundamental human capacities and limitations? Also the self congratulatory attitude resulting from a mistaken hubris leaves us handicapped in any effort to develop a sophisticated comprehension of our problems after our school daze are over.

[b]Criticism emphasizes deliberate judgment whereas enjoyment emphasizes the instinctive and immediate.

Criticism implies judgment and aesthetes (having an affecting sensitivity to beauty) imply perception. To reach a common ground between the two we must consider perceptions that are more than passive but are critical. Also we must adjust our notion of criticism to include “those judgments of value which are instinctive and immediate, that is, to include pleasures and pain”.

If we also narrow our concept of aesthetics (pleasing in appearance) so as to exclude all perceptions which are not appreciations, i.e. which do not find value in their objects, we can reach a “sphere of critical or appreciative perception”.

Thus, aesthetics is “concerned with the perception of values”.

Self consciousness is the precursor of the possibility of worth. For the existence of ‘good’ in any form, emotional consciousness is required. “Observation will not do, appreciation is required.”

From this we can assert an axiom that is important for all moral philosophy; and science of morality should it ever come to be. “There is no value apart from some appreciation of it.”

Spinoza informs us that we desire nothing because it is good but that it is good because we desire it. We can find value in that which is not instinctively good only because it is derivative of the instinctively appreciated. “The verbal and mechanical proposition, that passes for judgment of worth, is the great cloak of ineptitude in these matters…Verbal judgments are often instruments of thought but it is not by them that worth can ultimately be determined.”[/b]

Quotes from “The Sense of Beauty” by George Santayana

This is not a space designed for the posting of manifestos.  (Perhaps you should invest in your own blog?)  See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214779 for rough posting guidelines -- in particular, you're flirting with failing 3) and [b]2.[/b]

Actually this [b]might[/b] start an interesting discussion if it were at least somewhat clear what exactly the OP wanted to discuss or if he used more "down to Earth" English to convey his points :roll:. Some passages like, for example,
[quote]
We develop a satisfaction only when we think of our self as being surrounded by objects and laws independent of our self.
[/quote]
are terribly hard to comprehend (I mean, I understand each word and the grammatical structure of this particular sentence but all my attempts to discern its meaning failed miserably :(). The general ideas seem to be that the "technological progress" without "moral progress" is a dead end (which I strongly agree with), that the "subjective emotional perception of value" is in some sense superior to the "objective critical perception of value" (which I strongly disagree with) and that the current structure of college education should be changed to include courses in "social behavior" and "psychology", (which I somewhat agree with). Did anybody see anything beyond these three points in the original post?

If you google "It is human nature to be attracted to the mere appearance of things", and then click "repeat the search with the omitted results included", you will find that this same post appears on many other internet forums. The same is true for cobert's two other posts.


geometry incenter ratio inradius circumcircle

[quote="shadysaysurspammed"]I"m feelin too lazy to say stuff in detail,so bear with me.
In this question,we know that in an equilateral triangle,the incentre,circumcentres and evry thing coincides.So the median too coincides with the altitude,since radii of incircle is ependicular to the side and so is altitude,hence ratio is 2:1(median thinge).

HEnce,radius if incentre divides altitude in the ratio ...whatever the question asks.....zzzzzzzzgoodnight[/quote]
wouldnt it be,
$\frac{2}{3}$ of the altitude? :huh:

No it is one third because it is from the center to the nearest side on the triangle not the point of one

I dont understand what you're trying to say, the radius would be from the centre to the vertex of the triangle? :huh:

[quote="Drunken_Math"][quote="236factorial"]I'm not sure how easy this proof is so mods be thoughtful:

Prove, using algebraic methods, that the length of the radius of the inscribed circle in an equilateral triangle is one-third of the length of the altitude of the triangle.[/quote]
[hide]Say each side of the equilateral triangle was equal to $x$, the height is $h$, and the inradius was $r$

$r(\frac{3x}{2})=\frac{xh}{2}$

$3xr=xh$

$3r=h$

$r=\frac{h}{3}$[/hide][/quote]
I don't think it is Algebraic. You used the area equation, which is also geometrical. I think the proper way of solving 236!'s problems is using coordinates.

[quote]I dont understand what you're trying to say, the radius would be from the centre to the vertex of the triangle?[/quote]
This inradius is the distance from the intersection of the angle bisectors to the sides, the circumradius is the distance from the intersection of the perpendicular bisectors of the sides to the vertex. This problem said Inradius.

Read my post in the follwoing forum by clicking on the following link:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=316785#p316785]www.mathlinks.ro/Forum/viewtopic.php?p=316785#p316785[/url]

Masoud Zargar


geometry
analytic geometry
geometric transformation
reflection
vector
ratio
vector geometry

Given the coordinates of a point $ P$, is there a way to find the coordinates of its reflection $ P'$ about the plane $ \pi$?

Also, is there any formula for finding the intersection between a line and a plane, given the vector form or Cartesian form of the plane and line?

1. yes there is. the trick is to find out the coordinates of the foot of the perpendicular from $ P$ on to the plane $ \pi$.
that point is nothing but the midpoint of $ P$ and $ P'$. so coordinates of $ P'$ can be found

2. i don't think any formula is necessary because, in the case of cartesian form, it just reduces to solving 3 equations with 3 unknowns, which can be easily done
:)

How do you find the coordinates of the foot of the perpendicular of the point $ P$ projected onto the plane $ \pi$ then?  :P

For the intersection of the line and plane, you just compare $ x$,  $ y$ and $ z$ coordinates of the line and plane?

have you learnt what direction ratios of a line are? if you havent, my explanations may prove to be futile. :)

I believe i've covered it in the vectors course i just completed, go ahead =P I'm just trying to strengthen my grasp of my concepts n develop techniques for routine problems, e.g. intersection of line and line, shortest distance between 2 skew lines, etc

Vector Form:
Find the line past the given point that is parallel with the plane's normal vector. Then plug it in the plane's equation (and then dot product).

I've just recently self-learned some vector geometry, so I might be wrong...

well then, in the problem, the equation of the plane as well as coordinates of point will be given.
so assume the foot of perpendicular to be (a,b,c) and point P to be (p,q,r) and eqn of plane to be Ax + By +Cz + D = 0

direction ratios(dr's) of line perpendicular to plane are A, B, C.
also the drs of the same line are p-a,q-b,r-c.

so $ \frac {A}{p\minus{}a}\equal{}\frac {B}{q\minus{}b}\equal{}\frac {C}{r\minus{}c} \equal{} k$

now, get p, q and r in terms of A,B,C,a,b,c (which are all known) and k.
Now, substitute those (p,q,r) in the equation of the plane as that point lies on the plane.
solve for k and hence get values of p, q and r in terms of knowns :)


[hide="hint"]
$\sin 2\theta = 2\sin\theta\cos\theta$
$\sin^{2}\theta+\cos^{2}\theta = 1$
[/hide]


function
algebra
domain
abstract algebra
group theory

If $ f(2002) \not \equal{}0$ and $ f(n\plus{}3)\equal{}{f(n\minus{}1)\over f(n\plus{}1)}$,then

Find $ f(2002)\times f(2008)\equal{}$?

[quote="PKMathew"]If $ f(2002) \not \equal{} 0$ and $ f(n \plus{} 3) \equal{} {f(n \minus{} 1)\over f(n \plus{} 1)}$,then

Find $ f(2002)\times f(2008) \equal{}$?[/quote]

what is the domain and the image of this function?

[quote="PKMathew"]If $ f(2002) \not \equal{} 0$ and $ f(n \plus{} 3) \equal{} {f(n \minus{} 1)\over f(n \plus{} 1)}$,then

Find $ f(2002)\times f(2008) \equal{}$?[/quote]

[quote="JBL"]Assume $ f: {\bf Z} \to {\bf R}\smallsetminus\{0\}$, for example.  (Actually, any abelian group with multiplication would make a fine range set here, and any commutative ring with unity would make an acceptable domain.)[/quote]

If it is $ f: {\bf Z} \to {\bf R}\smallsetminus\{0\}$, then there are infinitely many solutions.

Let $ f(2008)\equal{}x$, $ f(2006)\equal{}y$, then

$ f(2004)\equal{}xy$    ,     $ f(2002)\equal{}xy^2$

So $ f(2002)f(2008)\equal{}x^2y^2$ is not fixed as $ x,y$ are arbitrary.

Well, you're asked essentially for the sum of two consecutive elements of the Fibonacci sequence without being told which two elements you're looking for -- the fact that this is underdetermined has little to do with the choice of range, unless it's taken to be something very artificial (like, say, $ f: {\bf Z} \to {\bf Z}$).


calculus
integration
logarithms
calculus computations

([b]8[/b]) Evaluate the integral $ \int_0^1\ln x \ln(1\minus{}x)\ dx$.

Again this integral. Can someone provide a link?

In summary, you have to do this:

1) Write ln(1-x) as a power series in x.

2) Compute the integral $ \int_0^1x^n \ln x \,dx$

3) Express the original integral as an infinite series.

4) Compute the value of the series.

[hide="Final answer"]$ 2 \minus{} \frac{\pi^2}{6}$[/hide]

Link to a thread in which this problem is thoroughly discussed: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190457]here[/url]


inequalities
induction
function
triangle inequality
inequalities proposed

Let $x_1,x_2,\ldots,x_n$ be real numbers. Prove that
\[ \sum_{i,j=1}^n |x_i+x_j|\geq n\sum_{i=1}^n |x_i| \]

For $n=2$ it is like:
$|a+b| \geq 2(|a| + |b|)$
or (because I don't know how should I understand the sum on the left side)
$2|a+b| \geq 2(|a| + |b)$

Both of these inequalities are not true.

For $n=2$ we have $|a+a|+|a+b|+|b+a|+|b+b| \geq 2(|a|+|b|)$ - you should take all permutations.

Everything I can say is that : It's very nice problem.

Hungkhtn,I can prove the orginal inequality for $n=3,4$ and $3(\sum_{1 \leq i<j \leq n}|x_{i}+x_{j}|)\geq (n-1)\sum_{i=1}^n|x_{i}|$,but it's weaker than the orginal problem,so i can't solve this.Please Hungkhtn or Nima Ahmadi Pour,can youpost your solution?Thanks so much :?  :?

My own solution is rather complicated, but I know there are other solutions that are simpler.
Anyway here is my solution:
We use induction on $n$.
Easily we can verify that if $x_i=0$ for some $i$ then the result can be derived from case $n-1$.
Now we will show that if $x_i=-x_j$ for some $i$ and $j$ then the result can be derived from the case $n-2$.
First observer a small lemma:
Lemma : $|x-y|+|x+y|\geq |x|+|y|$
Proof : By triangle inequality we have $|x-y|+|x+y|\geq 2|x|$ and $|x-y|+|x+y|\geq 2|y|$. Adding these two and dividing by two we get the result.
Now suppose that $x_1=-x_2$. Now by lemma we have the following inequalities:
\[ |x_1+x_i|+|x_2+x_i|\geq |x_1|+|x_i| (3\leq i\leq n) \]
\[ |x_i+x_1|+|x_i+x_2|\geq |x_2|+|x_i| (3\leq i\leq n) \]
And we obviously have :
\[ |x_1+x_2|=0 , |x_2+x_1|=0 , |x_1+x_1|=2|x_1| , |x_2+x_2|=2|x_2| \]
And from the induction hypothesis for $n-2$ we have :
\[ \sum_{i,j=3}^n |x_i+x_j|\geq (n-2)\sum_{i=3}^n |x_i| \]
By adding all the above inequalities/equalities we have
\[ \sum_{i,j=1}^n |x_i+x_j|\geq n\sum_{i=1}^n |x_i| \]
which is what we wanted.
Now we may assume that no $x_i$ is zero and for no $i,j$ we have $x_i+x_j=0$.
Now take the function $f$ to be $\sum_{i,j=1}^n |x_i+x_j|-n\sum_{i=1}^n |x_i|$.
Now assume that we add some $x_i$ a small value $\epsilon$. Then some terms in $f$ will increase by $\epsilon$ and some will decrease and some do not change. (Count the term $|x_i+x_i|$ twice)
But since we have that no term in $f$ is zero, if rather than increasing $x_i$ by $\epsilon$ we decrease it by $\epsilon$ the terms in $f$ that were increasing will decrease and the ones that were decreasing will increase.
So we can either increase $x_i$ by $\epsilon$ or decrease it such that $f$ is not increased (At this point $f$ behaves like a linear function in $x_i$)
Also note that value of $\epsilon$ can be increased until one of the terms in $f$ become zero, but then value of $f$ is by the above arguments, non-negative and so we have the results.
So assume that $\epsilon$ can be increased to infinity without having any of the terms become zero.
But we have this propery for every $x_i$ so there is no $x_i$ between two others, that is there are $a$ of $x_i$s equal to $t$ and $b$ others equal to $s$. ($a+b=n$)
Supposing that $|t|<|s|$ ,just increase the $\epsilon$ for all $x_i$s equal to $t$ until we have $|t|=|s|$ or $t=0$.
But if $t=0$ then we are done. So assume $t\neq 0$.
But then $t=s$, or else we have $t=-s$ which can be derived from the case $x_i+x_j=0$.
But then the inequality becomes like this :
\[ 2n^2|t|\geq n^2|t| \]
which is obviously true.

[quote="Albanian Eagle"]can't we just use karamata ineq for the convex function |x|[/quote]
Yes, we can (But only jensen is enough)
It is one of my friends simple solutions (I told you before, others solutions are simpler).
If we make the negative $x_i$s to become equal to their arithmetic mean, by jensen the left side will decrease, and the right side obviously doesn't change.
So we do this for the negative ones, and then for the positive ones.
Then the ineqality becomes in four varianles and it can be easily solved by hand.

My idea is the following:
Use $|a_i+a_j|+|a_i-a_j|=2max(|a_i|,|a_j|)$, then we have to demonstrate
$2\sum_{i,j=1}^n max(|a_i|,|a_j|)\geq n\sum |a_i|+\sum_{i,j=1}^n |a_i-a_j|$

Divide $a_i$ in two groups: $u$ numbers are positive, $v$ numbers are negative, $u+v=n$.
Positive numbers we redenote $S^+_u\geq S^+_{u-1}\geq...\geq S^+_1\geq 0$ (arranging in order)
Negative numbers we redenote $S^-_v\geq S^-_{v-1}\geq...\geq S^-_1\geq 0$ (arranging in order by modulo rule - inversing their sign).
And also they are arranged as $S_n \geq S_{n-1}\geq ... \geq S_1\geq 0$ (full notation).
It will be clear in futher steps.

$\sum_{i,j=1}^n max(|a_i|,|a_j|)=2nS_n+(2n-2)S_{n-1}+...+2S_1$

$n\sum |a_i|=n(S_n+S_{n-1}+...+S_1)$

$\sum_{i,j=1}^n |a_i-a_j|=2v\sum_{i=1}^uS^+_i+2u\sum_{j=1}^vS^-_j+$

$+(2uS^+_u+2(u-2)S^+_{u-1}+...+(-2u)S^+_1)+(2vS^-_v+2(v-2)S^-_{v-1}+...+(-2v)S^-_1)$.

(Last step considering combinations positive with negative, positive with positive, negative with negative numbers)
Thus we have to prove:

$4(nS_n+...+S_1)\geq n(S_n+...+S_1)+((2v+2u)S^+_u+(2v+2(u-2))S^+_{u-1}+...+(2v-2u)S^+_1)+((2u+2v)S^-_v+(2u+2(v-2))S^-_{v-1}+...+(2u-2v)S^-_1)$

$4(nS_n+(n-1)S_{n-1}+...+S_1)\geq 3nS^+_u+(3n-4)S^+_{u-1}+...+(n+2v-2u)S^+_1+3nS^-_v+(3n-4)S^-_{v-1}+...+(n+2u-2v)S^-_1$

Hence we have to prove that row $a=(4n, 4(n-1),...,4)$ is greater at least than row $b=(3n,3n,3n-4,3n-4,..., min(3n+2u-2v,3n+2v-2u))$ as the worst case.

1 case: take first $2k$ from each row:
$4(n+..+n-2k+1)=4\cdot 2k\frac{n+n-2k+1}{2}=4k(2n-2k+1)$
$2(3n+(3n-4)+...+(3n-4k+4))=2k\frac{3n+3n-4k+4}{2}=2k(3n-k+1)$

and we want to prove $4k(2n-2k+1)\geq 2k(3n-2k+2)$ equaling $n\geq 2k$ - true.

2 case: take first $2k+1$ from each row:
$4(n+..+(n-2k+1)+n-2k)=4\cdot 2k\frac{n+n-2k+1}{2}+4(n-2k)=4k(2n-2k+1)+4(n-2k)=4(2kn-2k^2+n-k)$
$2(3n+(3n-4)+...+(3n-4k+4)+(3n-4k))=2k\frac{3n+3n-4k+4}{2}+2(3n-4k)=2k(3n-2k+2)+2(3n-4k)=2(3kn-2k^2+3n-2k)$

wishing to demonstrate $4(2kn-2k^2+n-k))\geq 2(3kn-2k^2+3n-2k)$
equaling $n(k-1)\geq 2k^2-k$, true because of $n\geq 2k+1$ and the problem is solved.

Here's mine. I think it's too short, maybe I'm wrong, but... maybe not
\[ \{a_1,a_2,...,a_n\}=\{b_1,b_2,...,b_r\}\cup \{-c_1,-c_2,...,-c_s\} \]
(devide set $\{a_1,a_2,...,a_n\}$ to negative and non-negative groups).
With $r+s=n,b_i \ge 0,c_j \ge 0$. Denote
\[ R=\sum_{i=1}^r b_i,S=\sum_{j=1}^s c_j \]
So the inequality can be rewrite as follow
\[ 2\sum_{i=1}^r\sum_{j=1}^s|a_i-b_j| \ge (s-r)(R-S) \]
Assume that $s \ge r$, so we can suppose that $R \ge S$. In this case, notice that
\[ \sum_{i=1}^r\sum_{j=1}^s|a_i-b_j| \ge\sum_{i=1}^r(sa_i-S)=sR-rS \]
So we need to prove
\[ 2(sR-rS ) \ge (s-r)(R-S) \]
But it's immediately true because $s \ge r$ and $R \ge S$.

[quote="Nima Ahmadi Pour"]Yes, we can (But only jensen is enough)[/quote]

How do we exactly use Jensen?

Anyway, the natural idea is your first idea, with the linear functions. I couldn't solve it for $x_i+x_j=0$, so thanks.

Hungkthn, I think it's $2\sum_{i=1}^r\sum_{j=1}^s|a_i-b_j| \ge (s-r)(R+S)$, but I may be wrong.

Sorry, I don't understand what $\sum_{i,j}^n |x_i+x_j|$ means. Help!

$\sum_{i,j=1}^n |x_i+x_j|$ means that the sum runs over all pairs $(i, j)$, where $i, j \in \{1, ..., n\}$.

[quote="perfect_radio"][quote="Nima Ahmadi Pour"]Yes, we can (But only jensen is enough)[/quote]
How do we exactly use Jensen?
[/quote]
Just note that $|x+a|$ is a convex function and also sum of some convex functions is also convex.
So if we take out the terms on the left such that they are like $|a+b|$ where $a$ is negative and $b$ is positive then these terms can be grouped like this $\sum_{x_i\in \mathbb R^-}(|x_i+a_1|+|x_i+a_2|+\ldots+|x_i+a_k|)$.
Also note that sum of other terms does not change by putting the mean of negative ones instead of them.
So now by jensen its pretty simple.

Now take the function $f$ to be $\sum_{i,j=1}^n |x_i+x_j|-n\sum_{i=1}^n |x_i|$.
Now assume that we add some $x_i$ a small value $\epsilon$. Then some terms in $f$ will increase by $\epsilon$ and some will decrease and some do not change. (Count the term $|x_i+x_i|$ twice)
But since we have that no term in $f$ is zero, if rather than increasing $x_i$ by $\epsilon$ we decrease it by $\epsilon$ the terms in $f$ that were increasing will decrease and the ones that were decreasing will increase.
So we can either increase $x_i$ by $\epsilon$ or decrease it such that $f$ is not increased (At this point $f$ behaves like a linear function in $x_i$)
Also note that value of $\epsilon$ can be increased until one of the terms in $f$ become zero, but then value of $f$ is by the above arguments, non-negative and so we have the results.
So assume that $\epsilon$ can be increased to infinity without having any of the terms become zero.
But we have this propery for every $x_i$ so there is no $x_i$ between two others, that is there are $a$ of $x_i$s equal to $t$ and $b$ others equal to $s$. ($a+b=n$)
Supposing that $|t|<|s|$ ,just increase the $\epsilon$ for all $x_i$s equal to $t$ until we have $|t|=|s|$ or $t=0$.
But if $t=0$ then we are done. So assume $t\neq 0$.
But then $t=s$, or else we have $t=-s$ which can be derived from the case $x_i+x_j=0$.
But then the inequality becomes like this :
\[ 2n^2|t|\geq n^2|t|  \]
which is obviously true.[/quote]

  Hi I have read your solution can you explain this side, I could not understand :huh: . Becaouse you have used function that I could not understand

Smth similar was give in Romanian TST. For this one.
Order the numbers $a_1<a_2...<a_n$, and assume that $a_1+a_n\ge 0$.

It can be shown easily that decreasing, all numbers $\ge -a_1$ until they become $-a_1$ the ineq eventually sharpens. Hence we can assume that $a_1+a_n=0$.

Now use an induction hypotesis for the middle $n-2$ terms.

It was exactly the same problem xirti... Actually, our problem was far worse from an aesthetic view.

[quote="xirti"]I didn't remember it very well :D ...So, another problem apearing in more than one country at TST....[/quote]
No this is not a shortlist problem. The author (Dr. Mohsen Jamali) said it was not.
I think in the post of the other similar problem (the romanian one) it is explained more.

P.S. : Abdurashidjon, I don't know where you don't understand, please explain more.

I'm sorry for writing on old topic, but I've found short and elementary proof, so I want to post it.

If all numbers have the same sign, inequality is trivial because $LHS=2n\sum |x_i| \ge n\sum |x_i|=RHS$.

Since inequality is symmetric and it doesn't change when we invert all signs, we can wlog suppose that there exist $\frac{n}{2} \le k <n$ such that $x_1 \ge... \ge x_k \ge 0>x_{k+1} \ge ... \ge x_n$. If $a$ and $b$ have the same sign, then $|a+b|=|a|+|b|$, so we have:

$\sum_{i,j=1}^{n} |x_i+x_j|= \sum_{i,j=1}^{k} |x_i+x_j| + \sum_{i,j=k+1}^{n} |x_i+x_j| + 2\sum_{i=1}^{k}\sum_{j=k+1}^{n} |x_i+x_j| \\
=2k\sum_{i=1}^{k} |x_i|+2(n-k)\sum_{i=k+1}^{n} |x_i|+ 2\sum_{i=1}^{k}\sum_{j=k+1}^{n} |x_i+x_j|$

Since $2k \ge n$, it remains to prove:

$(2k-n)\sum_{i=1}^{k} |x_i|+2\sum_{i=1}^{k}\sum_{j=k+1}^{n} |x_i+x_j| \ge (2k-n)\sum_{i=k+1}^{n}|x_i|$

But using trialngle inequality for each $j=k+1,...,n$, we have: $|x_i|+|x_j+x_i| \ge |x_j| \; \; (1)$ for some $i=1,2,...,n$.

Since there are $(2k-n)\cdot k$ elements in $(2k-n)\sum_{i=1}^{k} |x_i|$, and for each $j=k+1,...,n$ there are $2k$ summands of the type $|x_i+x_j|$ in $2\sum_{i=1}^{k}\sum_{j=k+1}^{n} |x_i+x_j|$, we can see that for every $j$ we can find $2n-k$ sums of the type $(1)$ on the LHS to prove last inequality.

Hence, inequality is proved. $\blacksquare$

Suppose $g(n)=\sum_{i=/j=1}^{n} |x_i+x_j|-n\sum_{i=1}^{n}|x_i|$ ,
note what we’ve to show is, $g(n)\geq 0$ for all $n\in\mathbb N$.
Now  $WLOG$ suppose $0\leq x_1\leq x_2\leq x_3...\leq x_p$  and $x_{p+1}=-y_{p+1}<0,....,x_n=-y_n<0$.
Also suppose $y_i\in\ [x_{f(y)},x_{f(y)+1}]$.
So for a fixed $i$ that is $y_i(2f(i)-p)+(\sum x_i)-2(x_1+x_2+...x_{f(i)})$.
So finally $g(n)= \sum [y_i(2f(i)-p)+(\sum x_i)-2(x_1+x_2+...x_{f(i)})]$.
So, $g(n)=\sum x(n-p)-p(y_{p+1}+y_{p+2}+....y_n)+2\sum (y_if(i)-(x_1+x_2+...+x_{f(i)}))$.
Now let’s keep $f(i)$’s  as well as those $x$’s constant.
Not consider  a $y_t$ and keep others fixed. Note it’s coefficient is a fixed natural number.
So depending on coefficient either $g(n)$ is increasing or not so. So certainly minimum must occur
either at $y_t=x_{f(t)}$ or $y_t=x_{f(t)+1}$ , or basically say at end points.
Now so basically we need to look the case where $y_i$’s are nothing but permutations of those positive $x_i$’s .
Now for some $y_m=x_i$, keeping other $x_i$’s fixed we get $g(n)=(2i-1-p)x_i+c$ for some constant $c$.
Now note if $i\geq \frac{p+1}{2}$ then $g(n)$ is increasing so minimum at $i=[\frac{p+1}{2}]$
similarly for other case again minimum occurs at that point.
So now we can , we need to check the case all negative quantities are equal.
But also similarly we would also had to check when all those positives are equal.
So now problem is though for $n$ variables but  among them some are $+a$ while rest are $-b$, and
this is just an easy case chase, so done.


Show that for all positive real numbers $ a, b, c$, $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \leq \left (\frac {a}{b} \right) {}^2 \plus{} \left (\frac {b}{c}\right ){}^2 \plus{} \left (\frac {c}{a}\right) {}^2$.

Also, how can one show that for any real numbers $ a, b, c$, if $ a \plus{} b \plus{} c \equal{} n$, then the maximum obtainable value of $ ab \plus{} ac \plus{} bc \equal{} \frac{n^2}{3}$?

[hide="First question"]Let $ x \equal{} \frac{a}{b}, y \equal{} \frac{b}{c}, z \equal{} \frac{c}{a}$. We know $ xyz \equal{} 1$ and we want
$ x \plus{} y \plus{} z \leq x^2 \plus{} y^2 \plus{} z^2$

Equivalently,
$ x^{4/3}y^{1/3}z^{1/3} \plus{} x^{1/3}y^{4/3}z^{1/3} \plus{} x^{1/3}y^{1/3}z^{4/3} \leq x^2 \plus{} y^2 \plus{} z^2$

By Weighted AM-GM,
$ \frac{4x^2 \plus{} y^2 \plus{} z^2}{6} \geq x^{4/3}y^{1/3}z^{1/3}$

Summing cyclically gives the desired inequality.[/hide]

[hide="Second question"]Square the given equality.
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca) \equal{} n^2$ (1)

By AM-GM, $ \frac{a^2 \plus{} b^2}{2} \geq ab$. Summing cyclically gives
$ a^2 \plus{} b^2 \plus{} c^2 \geq ab \plus{} bc \plus{}ca$

Applying this to (1)
$ n^2 \geq 3(ab\plus{}bc\plus{}ca)$

$ \frac{n^2}{3} \geq ab \plus{} bc \plus{} ca$

Equality holds when $ a \equal{} b \equal{} c \equal{} \frac{n}{3}$, so this is the maximum.[/hide]

[hide="a possible other approach"]
start off like zuton x=a/b, y=b/c, z=c/a
then say z=1/xy
so $ x\plus{}y\plus{}\frac{1}{xy}\le x^2\plus{}y^2\plus{}\frac{1}{x^2y^2}$
by AM GM $ x\plus{}y\plus{}\frac{1}{xy}\le 3$
can anyone make something magically out of this?
[/hide]

[quote="CircleSquared"]start off like zuton[/quote]
Eh? :P

$ 3 \leq x^2 \plus{} y^2 \plus{} \frac1{x^2y^2}$ is not true in general, so I don't think you'll get anywhere with AM-GMing the left.

Clever solutions, MellowMelon! =D I got the first using a mix of RMS-AM-GM (I'll post my solution below), but I really didn't get the second one using plain algebra. I'm also completely unfamiliar with weighted AM-GM.

Note that $ \sqrt{\frac{({a \over b})^2 \plus{} ({b \over c})^2 \plus{} ({c \over a})^2}{3}} \geq \frac {{a \over b} \plus{} {b \over c} \plus{} {c \over a}}{3}$. Squaring and simplifying, $ ({a \over b})^2 \plus{} ({b \over c})^2 \plus{} ({c \over a})^2 \geq \frac {({a \over b} \plus{} {b \over c} \plus{} {c \over a})^2}{3}$. Now we'll show that $ \frac {({a \over b} \plus{} {b \over c} \plus{} {c \over a})^2}{3} \geq {a \over b} \plus{} {b \over c} \plus{} {c \over a}$. Assume otherwise. Then we have that $ {a \over b} \plus{} {b \over c} \plus{} {c \over a} < 3$, i.e., $ \frac {{a \over b} \plus{} {b \over c} \plus{} {c \over a}}{3} < 1$. But by AM-GM, $ \frac {{a \over b} \plus{} {b \over c} \plus{} {c \over a}}{3} \geq 1$. Contradiction.

Minor refinement: Your "assume otherwise" step at the end isn't really necessary. You can just reduce the inequality you have to $ \frac{\frac ab \plus{} \frac bc \plus{} \frac ca}{3} \geq 1$ and say "yay it's true". No need for contradictions. ;)

Weighted AM-GM in most cases is just applying AM-GM with repeated terms. When I said

$ \frac{4x^2 \plus{} y^2 \plus{} z^2}{6} \geq x^{4/3}y^{1/3}z^{1/3}$

was true by weighted AM-GM, I was essentially using AM-GM on the following.

$ \frac{x^2 \plus{} x^2 \plus{} x^2 \plus{} x^2 \plus{} y^2 \plus{} z^2}{6} \geq x^{4/3}y^{1/3}z^{1/3}$

Weighted AM-GM is slightly stronger though. For instance, it can prove that

$ \frac{\sqrt{2}a \plus{} b}{\sqrt{2} \plus{} 1} \geq \sqrt[\sqrt{2}\plus{}1]{a^{\sqrt{2}}b}$

in one step, while proving that with AM-GM requires a limiting argument involving a sequence of rationals converging to an irrational.

(though I couldn't tell you when you'd need a hideous radical like the one on the RHS of the above)

[hide="Another solution to the first question"]Let $ x\equal{}a/b$, $ y\equal{}b/c$, and $ z\equal{}c/a$.  Substituting, the equation becomes $ x\plus{}y\plus{}z\le x^{2}\plus{}y^{2}\plus{}z^{2}$, where $ xyz\equal{}1$.

By AM-GM, we have $ \frac{x\plus{}y\plus{}z}{3}\ge 1$.  Therefore,

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

But by Rearrangement, $ xy\plus{}yz\plus{}zx\le x^{2}\plus{}y^{2}\plus{}z^{2}$, so $ x\plus{}y\plus{}z\le x^{2}\plus{}y^{2}\plus{}z^{2}$ as desired.[/hide]


Are there any South Carolinians active on the board this year?

[quote="not_trig"]Err yes Joe and Shobhit are gone. And they were like half your forum two years ago... if not like 2/3. (of course I say 2 years ago because your forum wasn't active last year)[/quote]

Yeah, I'm not really active anymore =(  I'm pretty immersed in MIT right now.  I'll keep an eye on the forum every once in a while in case there's any activity though...


$a_{1}= 2007$
For all positive integers $n$, $a_{n+1}= \frac{1}{a_{n}+\frac{1}{a_{n}+\frac{1}{a_{n}+\cdots}}}$.
Compute $\lim_{n \to \infty}{a_{n}}$.

$a_{n+1}= \frac{1}{a_{n}+a_{n+1}}$

$a_{n+1}^{2}+a_{n}a_{n+1}-1=0$

$a_{n+1}=\sqrt{\frac{a_{n}^{2}}{4}+1}-\frac{a_{n}}{2}$

If I'm not mistaken, this approaches the value $\frac{\sqrt{2}}{2}$, but I'm not sure how to prove that...

[hide="solution"]We look for a way to define the recursive seris such that it's muchh easier to compute.

$a_{n+1}=\frac{1}{a_{n}+a_{n+1}}$

$1=a_{n+1}^{2}+a_{n+1}a_{n}$

$a_{n+1}=\frac{-a_{n}+\sqrt{a_{n}^{2}+4}}{2}$

So a_2 is roughly 0, then a_3 is roughly 1, then a_4 is roughly 1 again, and again, then so on.

I think the limit is 1, even though my solution isn't that rigorous...[/hide]

[hide="Solution"]We know that if the series converges to some number $S$, then
$S = \frac{1}{S+\frac{1}{S+\frac{1}{S+\cdots}}}$
$S = \frac{1}{S+S}$
$S = \frac{1}{2S}$
$S^{2}= \frac{1}{2}$
$S = \boxed{\frac{\sqrt{2}}{2}}$[/hide]


Please post all future CTC tourney ideas here.

My method is:

Each participant starts with a rating of 0.  You must cd other players to bring your rating up.  Rating changes is based on a chart that I made, based on rating differences, the expected winner's gain, and the expected loser's gain.  Also, your rating change is affected by the score of which a player wins. (for example, rating is changed more when you win 4-0 than by winning 4-3.)

Comments/Suggestions/Ideas go in this thread.

before i vote plz put the rating methods because i didnt have time to see them

Could you post the entire rating system?  That way we could work out bugs.

ELO is the one used on AoPS.

Ladder is where everyone is put in a ladder, and the goal is to get to the top. You move up by beating people ahead of you.

I like the ladder method.
For people who don't know what ladder is here is an example:
A v B #1
C v D #2
E v F #3
G v H #4

Basically, someone starts in every 'rung'.
The winner climbs up.
The loser goes down.
Exceptions:
You are in the top game, or #1 you can not move up.
You are in the bottom game, or #4, you can not move down.
You can then easily see rankings by playing the last game.

No one likes my method, so I guess that's not the best idea.

This will not factor into the actual decision.  Another poll will.  This poll is just like a preview.

The definition of an ELO rating was slightly inaccurate. On AoPS, it was modified to suit multi-player games.

To learn about the ELO rating, please click [url=http://en.wikipedia.org/wiki/Elo_rating_system]here[/url].

[quote="herefishyfishy1"]I don't like the ladder method because then you can only play people that have a ranking right next to yours. Not that I can vote anyway, because I'm not in CTC.[/quote]

Yeah, the ladder method is very restricting.


Find all integer sets $(a,b,c,d)$ satisfying:
i)$a^2+1=bc$
ii)$b^3+1=a^2d$ :)

I saw this problem several times till now, but I don't find it easy.


quadratics
number theory proposed
number theory

The sequence $ (u_n)$ is defined by $ u_1\equal{}7, u_{n\plus{}1}\equal{}2u_n^2\minus{}1$ for every $ n\ge 1$. Prove that no term of sequence is divisible by 2003.

$ 2003|u_{n\plus{}1}\equal{}2u_{n}^2\minus{}1$
$ 2u_{n}^2\minus{}1\equiv 0\mod 2003$
$ 2u_{n}^2\equiv 1\mod 2003$

So $ 2$ is a quadratic residue mod 2003, but $ 2$ is only a quadratic residue mod primes that are $ 1$ or $ 7$ mod $ 8$. Since $ 2003\equiv 3\mod 8$, $ 2$ is not a quadratic residue.


We consider all 10-digit numbers where each of the ten digits $0, 1, \ldots, 9$ occurs exactly once. Prove that there are at least 50,000 numbers which are divisible by 11.

you mean $0,1,...,9$ right? ;)

anyway it's easy to show that $11|(a_0a_2...a_9)$ means $x_0-x_1+x_2-x_3+x_4-x_5+x_6-x_7+x_8-x_9=0 \pmod{11}$

the max value is 20, the min value -20 so we only need to check 11,0,-11. It can't be 0 because of the parity, so only 11 and -11 is left. Which means the sums of odd/even index digits are 17 and 28. So we simply need to see in how many ways 17 can be formed with 5 different digits.

If highest is 9: 4-3-1-0, 5-2-1-0
If highest is 8: 6-2-1-0, 5-3-1-0, 4-3-2-0
If highest is 7: 6-3-1-0, 5-4-1-0, 5-3-2-0, 4-3-2-1
If highest is 6: 5-4-2-0, 5-3-2-1
5 or less isn't possible, so we got a total of 9 options. Each one can be permutated in 120 ways, on both sides. So that gives 2*9*120*120 = 259200 ways. Did I miss something? Or was there an easier proof just to show the 50k bond?