$ a,\ b,\ c,\ d$ satifies the equations : $ a\plus{}b\equal{}5,\ c\plus{}d\equal{}2,\ ad\equal{}bc\equal{}1$. Find the values of $ ac\plus{}bd,\ \frac{a}{c}$.

Lets hope that the system is consistent...

$ (a\plus{}b)(c\plus{}d)\equal{}ac\plus{}ad\plus{}bc\plus{}bd\equal{}ac\plus{}bd\plus{}(ad\equal{}1)\plus{}(bc\equal{}1)\equal{}ac\plus{}bd\plus{}2\equal{}10$. $ ac\plus{}bd\equal{}8$.

Then $ c\plus{}d\equal{}2$, $ a\plus{}b\equal{}d^{\minus{}1}\plus{}c^{\minus{}1}\equal{}\frac{c\plus{}d}{cd}\equal{}\frac{2}{cd}\equal{}5$,  so $ cd\equal{}\frac{2}{5}$.

Then $ \frac{a}{c}\equal{}\frac{ad}{cd}\equal{}\frac{1}{\frac{2}{5}}\equal{}\frac{5}{2}$.

[quote="Altheman"]$ a \plus{} b \equal{} d^{ \minus{} 1} \plus{} c^{ \minus{} 1}$[/quote]
Can you explain this? :blush:

$ a \equal{} d^{\minus{}1}, b \equal{} c^{\minus{}1}$.

[quote="Altheman"]Lets hope that the system is consistent...

$ (a \plus{} b)(c \plus{} d) \equal{} ac \plus{} ad \plus{} bc \plus{} bd \equal{} ac \plus{} bd \plus{} (ad \equal{} 1) \plus{} (bc \equal{} 1) \equal{} ac \plus{} bd \plus{} 2 \equal{} 10$. $ ac \plus{} bd \equal{} 8$.

Then $ c \plus{} d \equal{} 2$, $ a \plus{} b \equal{} d^{ \minus{} 1} \plus{} c^{ \minus{} 1} \equal{} \frac {c \plus{} d}{cd} \equal{} \frac {2}{cd} \equal{} 5$,  so $ cd \equal{} \frac {2}{5}$.

Then $ \frac {a}{c} \equal{} \frac {ad}{cd} \equal{} \frac {1}{\frac {2}{5}} \equal{} \frac {5}{2}$.[/quote]

Needless to say, your answer is correct.

For latter part, my solution is as follows.

$ ad \equal{} bc\Longrightarrow \frac {a}{c} \equal{} \frac {b}{d} \equal{} k$, we have $ a \equal{} ck,\ b \equal{} dk$.
$ \therefore a \plus{} b \equal{} k(c \plus{} d)\Longleftrightarrow 5 \equal{} 2k$, yielding $ k \equal{} \frac {a}{c} \equal{} \frac {5}{2}$.


function
factorial
floor function
algebra

Find $ I_{n}$ in each of the following cases, where $ I_{n}$ is a function of the positive integer $ n$ and $ I_{1} \equal{} 1$:

(i) $ I_{n} \equal{} nI_{n\minus{}1}$;

(ii) $ I_{n} \minus{} I_{n\minus{}1} \equal{} n$;

(iii) $ I_{n} \plus{} I_{n\minus{}1} \equal{} n$.

[quote="AndrewTom"]Find $ I_{n}$ in each of the following cases, where $ I_{n}$ is a function of the positive integer $ n$ and $ I_{1} \equal{} 1$:

(i) $ I_{n} \equal{} nI_{n \minus{} 1}$;

(ii) $ I_{n} \minus{} I_{n \minus{} 1} \equal{} n$;

(iii) $ I_{n} \plus{} I_{n \minus{} 1} \equal{} n$.[/quote]

[hide="i"]
hey look it's the recursive definition of the factorial!

yes, it's $ I_n\equal{}n!$
[/hide]
[hide="ii"]
The given is equivalent to $ I_n\equal{}n\plus{}I_{n\minus{}1}$
so the first few terms are $ 1, 1\plus{}2, 1\plus{}2\plus{}3, 1\plus{}2\plus{}3\plus{}4$
or the triangular numbers exactly. it is well known that the formula is $ I_n\equal{}\frac{(n)(n\plus{}1)}{2}$
[/hide]
[hide="iii"]
The given is equivalent to $ I_n\equal{}n\minus{}I_{n\minus{}1}$
So the first few terms are $ 1, 2\minus{}1, 3\minus{}2\plus{}1, 4\minus{}3\plus{}2\minus{}1$
For odd n, working from left to right, every 2 terms cancel out to 1 except for the last. $ I_n$ would therefore be $ \frac{n\minus{}1}{2}\plus{}1\equal{}\frac{n\plus{}1}{2}$
For even n, each two terms cancel out to make 1, so $ I_n\equal{}\frac{n}{2}$
This is a piecewise defined function unless you want to be tricky with the floor function.
[/hide]

[hide="III"]Yeah, you can write it as a piecewise (as stevenmeow stated), or with the floor function:

$ I_n \equal{} \lfloor{\frac{n\plus{}1}{2} }\rfloor$[/hide]


(The original form of the question is here
http://www.geocities.com/johntyung/Q3.pdf )

Prove that

(n choose 0) + (n choose 2) + (n choose 4) +...+ (n choose n) =2^(n-1)

and

(n choose 1) + (n choose 3) +...+ (n choose n-1) = 2^(n-1)

by using a combinatorial argument.
(Hint :
a) Consider the following construction: given a subset S of {1,2,....,n}, construct a new subset T by the following recipe : if 1 is in the set S, let
T = S \ {1}
if 1 is not in the set S, let T = S union {1}
Show that this gives a bijection between the subsets of {1,....,n} of even cardinality and the subsets of {1,...,n} of odd cardinality.
b) conclude that this proves the result.

The hint is really a full solution- do you understand it?

Yes,  I am unclear as to how to construct a written answer.

Can u shed some light on this.?

By a) , the number of subsets of {1,2,3.....n} of even cardinality is the same as the number of subsets of odd cardinality.Since the number of subsets of {1,2,3,...n} is 2^n , then the number of subsets of even cardinality is 2^(n-1).


Show that if $ |s_{n\plus{}1}\minus{}s_n|<2^{\minus{}n}$ for all $ n$, then $ s_n$ is Cauchy.

You have to establish: $ |s_{n+k}-s_{n}|<\epsilon$ for all $ k\in \mathbb{N}$ and $ n\geq n_{0}$.

Just write

\begin{eqnarray*}
|s_{n+k}-s_{n}|&=&|s_{n+k}-s_{n+k-1}+s_{n+k-1}-\ldots+s_{n+1}-s_{n}|<\\
&<&|s_{n+k}-s_{n+k-1}|+\ldots+|s_{n+1}-s_{n}|
\end{eqnarray*}

and use the condition from the task.

I see that $ |s_{n \plus{} k} \minus{} s_n| < \frac {k}{2^n}$, but we want $ \frac {k}{2^n} < \epsilon$, and I can't seem to find the proper condition since $ k$ can be arbitrarily large.

[quote="aspiringchemlete"]I see that $ |s_{n + k} - s_n| < \frac {k}{2^n}$, but we want , and I can't seem to find the proper condition since can be arbitrarily large.[/quote]

You can make $ |s_{n + k} - s_n| < \frac {k}{2^n}$ stricter.

\begin{eqnarray*}
|s_{n+k}-s_{n}|&<&|s_{n+k}-s_{n+k-1}|+\ldots+|s_{n+1}-s_{n}|<\\
&<& \frac{1}{2^{n+k-1}}+\frac{1}{2^{n+k-2}}+\ldots+\frac{1}{2^{n}}=\\
&=&\frac{1-(\frac{1}{2})^{k}}{2^{n-1}}<\\
&<&\frac{1}{2^{n-1}}
\end{eqnarray*}


You can add without switching bases, you just have to carry after whatever base you are using (in this case 7) rather than 10.  Correct me if I'm wrong.

My method is usually just the same thing you do in base 10. If you go over the base $b$, then leave whatever is residue $\mod b$ and then carry the rest over.


Sorry for my poor English. I'm a beginner, I can't see the contest problems, where can I find it? Thank you for helping me.

Click on [color=blue]Resources[/color] - top right-hand-corner of screen.

Click on contest next to resources.
Is Mathlinks contest active?


group theory
superior algebra
superior algebra unsolved

Prove that there is no nonabelian finite simple group whose order is a Fibonacci number.

It occured in the attachment in this message but I couldn't find a reference or discussion on it:
http://www.mathlinks.ro/viewtopic.php?t=238048

I managed to dig up a reference: http://pphmj.com/abstract/1673.htm


geometry
rectangle
number theory
prime factorization

How do you solve this?
7. Rectangle ABCD has all integer side lengths and its interiot is divided into seven non-overlapping rectangular regions: two 8 by 10 regions, a 10 by 12 region, a 13 by 5 region, a 13 by 13 region, a 10 by 22 region and a 3 by 16 region. All linear measurments are given in feet. What is the sum of the length and width of rectangle ABCD?

[hide="Hints"][hide="Hint 1"]What is the area of ABCD?[/hide]

[hide="Hint 2"]Based on the prime factorization of the area of ABCD, what can you conclude about the side lengths.[/hide]

[hide="Hint 3"]Which side lengths don't work?[/hide][/hide]


Consider the equation $ ax\plus{}by\plus{}cz\plus{}d\equal{}0$, $ a,b,c,d\in \mathbb{Z}, abcd\not\equal{}0$.
If for every $ m\in\mathbb{Z}_\plus{}$,$ ax\plus{}by\plus{}cz\plus{}d\equal{}0(\bmod m)$ is solvable, show that
$ ax\plus{}by\plus{}cz\plus{}d\equal{}0$ is solvable on $ \mathbb{Z}$

$ (a, b, c)$ generates some ideal $ (e)$.  Letting $ m \equal{} e$, it follows that $ d \in (e)$, from which the result follows.  This generalizes to any PID.  See [url=http://en.wikipedia.org/wiki/Hasse_principle#Intuition]Hasse principle[/url].


\begin{titiepage}
.........
\end{tilepage}

and

\maketitle

Can   the   two  use  the  same  time?   For   example,  Does    the  following  code    work,  and  get  no  mistake and   warning?

Many  thanks!

 [code]\documentclass{article}

\begin{document}

\begin{titlepage}

gggggggggg
\end{titlepage}

\title{mikte}
\author{a}
\date{}
\maketitle
\begin{abstract}
fffffffffffffffffffffff
\end{abstract}

\tableofcontents

\section{ddddddd}fffffff
\section{ddddddd}fffffff

\end{document}[/code]

It makes no sense to use both at the same time. \maketitle uses a built-in style to produce a page whereas \begin{titlepage}...\end{titlepage} allows you to design your own title page.


probability
factorial
algorithm
combinatorics proposed
combinatorics

$ n$ mathematicians are seated randomly around a conference table. After they break for lunch, they seat themselves randomly again. As $ n$ gets large, what is the probability that there is some pair of mathematicians who were neighbors both before and after lunch?

Okay, here's a way to start: we'll take circular symmetries into account at the very end, so for now consider there to be $ n!$ seating arrangements.  Let $ A_i$ be the set of arrangements in which mathematician $ i$ is seated next to mathematician $ i \plus{} 1$ (taking indices modulo $ n$).  We want $ 1 \minus{} \frac {\left|\bigcup_i A_i\right|}{n!}$, and we'll proceed by inclusion-exclusion.  In order to do this, we need to compute expressions of the form $ \left| \bigcap_{i \in S} A_i\right|$ for arbitrary $ S \subseteq [n]$.  Now, we obviously need to consider each subset as a union of several "blocks" since (for example) in the case $ A_i \cap A_{i \plus{} 1}$ we have that mathematicians $ i$, $ i \plus{} 1$ and $ i \plus{} 2$ must be seated in three consecutive chairs, etc.  If $ |S| \equal{} m < n$ and $ S$ is a union of $ k$ maximal blocks, then there are $ 2^k \cdot n \cdot (n \minus{} m \plus{} k \minus{} 1)!$ permutations in $ \bigcap_{i \in S} A_i$.  (That is, each block can be oriented clockwise or counterclockwise; we can seat mathematician number 1 in any seat we like; and we then have $ n \minus{} m \plus{} k \minus{} 1$ blocks and individuals to arrange.)

Now, if I'm not mistaken, there are exactly $ \frac {n}{k} \binom{n \minus{} m \minus{} 1}{k \minus{} 1} \cdot \binom{m \minus{} 1}{k \minus{} 1}$ subsets of $ [n]$ with $ m$ elements and $ k$ blocks.  (I have not actually sat down and tried to prove this yet.)  Assuming I haven't made any monumental errors so far, this means that the sum over all $ m$-subsets of $ [n]$ contributes to our inclusion-exclusion precisely
\[ \sum_{k \equal{} 1}^m \frac {n}{k} \binom{n \minus{} m \minus{} 1}{k \minus{} 1} \cdot \binom{m \minus{} 1}{k \minus{} 1} \cdot 2^k \cdot n \cdot (n \minus{} m \plus{} k \minus{} 1)!\]
permutations ($ m \neq 0, n$) and that we now want to take the alternating sum of this expression as $ m$ goes from $ 1$ to $ n \minus{} 1$ (and in addition to take into account the terms $ m \equal{} 0, n$, which are easy to understand).  Now, this seems terribly ugly, and I have not double-checked anything I've written, so it's not clear whether this leads to a pleasant (or even reasonably usable) formula; unless it does, it's unfortunately not clear how to extract the data you've actually asked about (though the nature of PIE would let us extract at least some bounds by choosing only a fixed number of terms).

Hopefully I'll remember to return to this soon (or someone else will come along and do a better job ;)).

The [url=http://www.research.att.com/~njas/sequences]OEIS[/url] entry for this sequence ([url=http://www.research.att.com/~njas/sequences/A089222]A089222[/url]) contains what appears to be a brute-force factorial-time (!!) algorithm for computing these numbers, but also some links to resources that I have not yet looked at.

There are $ n$ bad events here, corresponding to the $ n$ pairs of neighbors from the first table that could be repeated at the second table.
Each bad event occurs with probability $ \frac{2}{n\minus{}1}$.  Since we have many rare events and they're "nearly" independent (seatings at one part of the table don't have much impact on seatings elsewhere), morally we'd expect the number of bad events which occur to be Poisson with mean $ \frac{2}{n\minus{}1} n \approx 2$, and the probability none happens to be $ \approx (1\minus{}\frac{2}{n\minus{}1})^n \approx e^{\minus{}2}$.

I believe that it's possible (but tedious) to make that intuition rigorous by showing that every moment of the distribution of the number of bad events converges to the corresponding moment of the Poisson distribution.  Here's a sketch of how I'd go about showing it.

Let the mathematicians be denoted $ (a_1, \dots, a_n)$ based on their morning seating, and let $ X_i$ be the indicator variable of the event that $ a_i$ and $ a_{i\plus{}1}$ are next to each other after lunch.  As noted before, we have $ E(X_i)\equal{}\frac{2}{n\minus{}1}$.  Not let $ k$ be fixed.  Call a tuple $ (i_1, \dots, i_k)$ [b]spread[/b] if it contains no two consecutive elements.  Then show:
[b]Lemma 1[/b]: If $ (i_1, \dots, i_k)$ is spread, then $ E(X_{i_1}X_{i_2}\dots X_{i_k})\equal{}(1\plus{}o(1))n^{\minus{}k} 2^k.$

[b]Lemma 2[/b]: Almost all tuples of size $ k$ are spread, so the contribution of the spread tuples is in the limit equal to the Poisson moment.

[b]Lemma 3[/b]: The contribution of the non-spread tuples to the moment is negligible.

Okay, good, I figured someone who knew some probability would be able to help :)

To finish my approach to an exact formula: the $ 2^k \cdot n \cdot (n \minus{} m \plus{} k \minus{} 1)!$ in my previous post should be just $ 2^k \cdot n \cdot (n \minus{} m \minus{} 1)!$.  (When we have an intersection of $ m$ of the $ A_i$ forming $ k$ blocks, these blocks include a total of $ m \plus{} k$ people; this leaves a total of $ n \minus{} (m \plus{} k) \plus{} k \equal{} n \minus{} m$ blocks and individuals to be arranged.  Each of the $ k$ blocks can go either clockwise or counterclockwise, and we may fix the position of mathematician 1, after which the remaining $ n \minus{} m \minus{} 1$ blocks and individuals must be linearly ordered.)  With this correction, the number of "good" permutations is
\[ n! \plus{} ( \minus{} 1)^n \cdot 2n \plus{} \sum_{m \equal{} 1}^{n \minus{} 1} ( \minus{} 1)^m \sum_{k \equal{} 1}^m \frac {n}{k}\binom{n \minus{} m \minus{} 1}{k \minus{} 1}\cdot\binom{m \minus{} 1}{k \minus{} 1}\cdot 2^{k}\cdot n\cdot (n \minus{} m \minus{} 1)!\]
(For reasons I haven't quite figured out and don't plan to, this is only valid for $ n \geq 3$.)  This formula is horrible and doesn't obviously simplify.

Edit: another nearly identical sequence in the OEIS: http://www.research.att.com/~njas/sequences/A078603
2nd edit: and here's one that's actually useful: http://www.research.att.com/~njas/sequences/A002816 confirms that the answer is $ e^{\minus{}2}$.

But what if we additionally demand that all mathematicians must be sitting on a different chair, compared to the first sitting arrangement?


Ross Mathematics Program
summer program
Mathcamp
Putnam

Not to be outdone by the Mathcamp list, this thread is for all who will be attending the Ross program this summer to say so.  (Though unlike Mathcamp, admission at Ross has not been closed out yet.)

I am attending as a Junior Counselor.

Im going as a Junior Counselor! As far as I know Zeke is coming back as a Junior Counselor too.

P.S. do you know if Tim is returning as a counselor?

that's great, I'm gonna be bugging him all summer to help improve my problem solving skills.  It's nice to have two counselors (Tim and Sawyer) who were Putnam HMs, that will be a great resource (especially because I wont have any NT sets to do).

And I will be returning as a lowly second year.

Dan, its awesome that you are returning!

I'm probably going to the Ross Program. I've been tearing my hair out the last few weeks deciding between Ross and Mathcamp...

well of course I would have to say go to Ross, but both programs are very good.  I cannot make any type of comparison (as I have not been to Mathcamp), but there is one kid I know of who went to Mathcamp two years ago and Ross last year.  He never really said which one he liked better but he is coming back to the Ross program as a junior counselor (and I believe he is getting his friend to come) so that may say something.

i know of at least two people from thomas jefferson that are going:

michael druker and menyoung lee

I'm actually planning on going to a math camp this summer although i know it's quite early...perheps too early...but nevertheless i wanted to know if anybody knows whether the recommandations from teachers for any math camp are the same each year. I planning to apply to a few math camps and I didnt want to throw all the recommandations on one teacher all at once..I don't exaculty think that she would enjoy that very much. ...so i thought that I could give her a few now so she won't end up with too many later...thanks


conics
parabola
analytic geometry
graphing lines
slope
calculus
derivative

Consider the parabola $ y \equal{} x^2$ in coordinate plane. tProve hat the slope of the tangent straight line (try to be intuitive here) to the parabola at the point $ (x,x^2)$ is $ 2x$.

:)

[hide="Note"]I don't expect a solution using calculus neither limits...but it is always welcome... :) [/hide]

I show the proof without differentiation as follows.

Suppose that the parabola $ y \equal{} x^2$ touches to the line $ y \equal{} ax \plus{} b$ at the point of tangency $ (t,\ t^2)$,
we have the equality $ x^2 \minus{} (ax \plus{} b)\equiv (x \minus{} t)^2\Longleftrightarrow ax \plus{} b \equal{} 2tx \minus{} t^2$, yielding the desired slope is $ 2t$.

To avoid messing up the notation, we'll consider the point $ (a,a^2)$. The equation of a line passing through that point and having the slope $ k$ is $ y\minus{}a^2\equal{}k(x\minus{}a)$. Now we set up the condition that it has only one intersection point with the parabola:

$ y\equal{}x^2$
$ y\minus{}a^2\equal{}kx\minus{}ka$

$ x^2\minus{}a^2\minus{}kx\plus{}ka\equal{}0$

The discriminant must be zero:

$ k^2\minus{}4(ka\minus{}a^2)\equal{}0$
$ k^2\minus{}4ka\plus{}4a^2\equal{}0$
$ (k\minus{}2a)^2\equal{}0$

$ k\equal{}2a$

as desired.

(Using calculus) This simply means finding the derivative of $ y\equal{}x^2$ and plugging in the x-coordinate which in this case is x so it doesn't matter. Anyways, $ \frac{d}{dx} x^2 \equal{} 2x$. So $ y' \equal{} 2x$ and at point $ (x,x^2)$ the slope of the tangent will be (by plugging in x in 2x) is clearly $ 2x$.

[quote="kunny"]ElChapin doesn't require the solution using calculus. :wink:[/quote]

But he said it's welcome, he even put a smily face lol.

My proof is similar to Farenhajt' proof.
[hide]
The equation of the straight line with slope a and through the point $ (b,b^2)$ is $ y\equal{}a(x\minus{}c)\plus{}c^2$.
Intuitively we know that $ x^2\ge a(x\minus{}c)\plus{}c^2$ or equivalently $ x^2\minus{}xa\plus{}ca\minus{}c^2\ge 0$. Then, this last expression as a cuadratic polynomial in $ x$ must have a non positive discriminant.
That is $ 0\ge a^2\minus{}4(ca\minus{}c^2)\equal{}(a\minus{}2c)^2$ but, $ (a\minus{}2c)^2\ge 0$ since $ a$ and $ c$ are reals, thus $ a\equal{}2c$.
[/hide]

:)

The equation of the tangent $ TT$ in the point $ T\left(x,x^2\right)$ at the parabola $ Y\equal{}X^2$ is $ Y\equal{}x^2\plus{}p(X\minus{}x)$ ,

where $ p$ is its slope. Since the tangent is [u]over all[/u] under parabola (?!) obtain that $ X^2\ge x^2\plus{}p(X\minus{}x)$

for any $ X\in\mathbb R$ $ \Longleftrightarrow$ $ (X\minus{}x)(X\plus{}x\minus{}p)\ge 0$ for any $ X\in\mathbb R$ , what is possibly iff $ x\equal{}p\minus{}x$ , i.e. $ p\equal{}2x$ .

In conclusion, for any $ x\in \mathbb R$ the slope $ p(x)$ of the tangent to the parabola in the point $ X(x,x^2)$ is $ p(x)\equal{}2x$ .


very nice problem..
please, give me a solution  :D  :D

Please, post your complete solution  :D Very nice problem =D

For which one?
The idea I used for the initial problem (with $\frac{\dots}{n}$) is the following. We put $a_i=i$ all $i\leq n/2$ and $a_i=i$ or $a_i=2i$ for $i>n/2$.

It wasn't a complete solution, but just an idea. Note, that LHS depends only on the count of whose $a_i$, which are equal to $2i$. Using that we should adjust RHS. (and again, it is an idea ;) )

If you complete your solution, please, post  :D

I have it for all sufficietly large $n$. I am too lazy to write it... :P

:D  :D
If you find one little solution, please, post  :)

Does anyone have a complete solution of the original problem?

[quote="Myth"][quote="Valentin Vornicu"]Find all positive integers $n$ for which there exist distinct positive integers $a_1,a_2,\ldots,a_n$ such that
\[ \frac 1{a_1} + \frac 2{a_2} + \cdots + \frac n { a_n} = \frac { a_1 + a_2 + \cdots + a_n } n. \][/quote]
I just wonder whether Valentin did it knowingly. ;) :lol:  :lol:  :lol:
I have analyzed some cases and encountered some problems with it. Indeed, first of all I invented solution for all $n$ satisfying the Pell's equation $x^2-5(n+1)^2=-4$ :). After that, the great idea come to me and I improved my construction, so I think it covers all sufficiently large $n$ (maybe $n>20$ or something like that). It is really "[i]a relatively hard problem[/i]". :lol:

[b]The right stetement is the following[/b] (thanks to Bogdan Enescu for RMC2004 book :) )
[i]Find all positive integers $n$ for which there exist distinct positive integers $a_1,a_2,\ldots,a_n$ such that
\[ \frac 1{a_1} + \frac 2{a_2} + \cdots + \frac n { a_n} = \frac { a_1 + a_2 + \cdots + a_n } 2. \][/i]

Do you see the difference? :D  :lol:  :lol:  :lol:[/quote] :oops_sign:  :oops_sign: My bad ... I did it this time :heli: Anyway it's great you solved the problem I accidentaly created :D :thumbup:


[hide]how many minutes past 3 will the minute hand coincide with the hour hand?
how many minutes past 3 will the minute hand be perpendicular with the hour hand?
express your answer as a common fraction for both.[/hide]

[hide="1"]For your first problem, you can set up an equation where $ x$ is equal to the number of minutes that elapse.
The minute hand travels at a speed of $ x$,for every minute, it advances a minute on the clock. The hour hand travels at a speed of $ x/12$, for every 12 minutes elapsed, it travels a minute on the clock. Setting up an equation,
$ x=\frac{x}{12}+15$(The 15 is there because the hour hand has a 15 minute head start (on the clock))
Multiplying by 12,
$ 12x=x+180$
$ 11x=180$
$ x=\frac{180}{11}$
It is $ \frac{180}{11}$ minutes until they coincide[/hide]
[hide="2"]You set up an equation similar to the first.
Since there will be a 15 minute difference, It is either $ x+15=\frac{x}{12}+15$, or
$ x=\frac{x}{12}+30$
Solving the first, we get $ x=0$
If this doesn't count, We solve the second, getting $ x=\frac{360}{11}$
[/hide]

Couldn't you solve this problem almost the same way as the other clock problem?

I was setting up an equation for that.. until I saw your solutions.  :(

[quote]cyrazeno, i think some one hates you, all the posts i've seen was marked spam

:spam: :spam: :spam:[/quote]

Doesn't take a genius to figure that out.
(Person's not me, if you had thought so.)


function
calculus
integration
linear algebra
matrix
vector

Question:

Apply the Gram-Schmidt orthogonalization procedure to the canonical basis ${1, x, x^{2}, x^{3}, x^{4}}$ in order to find an orthonormal basis for the space P4([0, 1]) with respect to the inner product <p(x), q(x)> =int(0,1) p(x)q(x) dx

Answer:

This is a pretty straight algorith, and when carrying out, I get:

$e1 = 1$

$e2 = x-1/2$

$e3 = x^{2}+1/6-x$

$e4 = x^{3}-1/20+3/5*x-3/2*x^{2}$

$e5 = x^{4}+1/70-2/7*x+9/7*x^{2}-2*x^{3}$

Now, I am having a real problem...how would I use this basis to find the adjoint of the linear map?

Could somebody help walk me through this please?

Well, there are 5 orthogonal functions - used to find projections of f(x) a simple way - integral int(0,1) ei(x)f(x) dx. The Gram-Schmidt orthogonalization procedure aplied to do it.

What the proplem else u have? Seems, all, that we have and need from orthogonal basis - is the shown simplicity. Problems could be in terminology.

Why you need a linear map, what spaces it connects, f(x) and what else? If you will transforme orthogonal functions by linear operator, you have to find projections on its eigen functions, orthogonal basis has here mach less meaning, besides - these questions are outside of interests of matrix algebra - we can find here some analogies only. Vector analysis and Matrix algebra uses the Gram-Schmidt orthogonalization procedure, but linear operators are more complicated objects, then matrices, to discuss the same free analogies of inversions, transpositions, etc.. In many cases it's whole abstractions. And your interest it seems look like very practical.

The similar discussion:
http://www.physicsforums.com/showthread.php?t=156953


Find the smallest positive integer $n$ divisible by 999 without 9's in the digits of $n$.

[quote="behemoth571"][hide="Possible Solution"]$(999)(117)=116883$
[/hide][/quote]

How did you get that? And actually, I don't have the answer.... ;)

[hide="partial"]
As we go up on the multiples of 999, we see that the last three digits decrease by 1 each time. for example, 999*1=999, 999*2=1998, ....etc. The first multiple of 999 that has the last three digits of non-9 is when that three digit is 888. That is when it is 111888. So it would be $999\cdot112 = 111888$
[/hide]


Find the expression and domain of $f(x)$ if $f \left( \frac{2}{x}+1 \right)=\lg{x}$.

Put $y={2\over x}+1\iff x=\frac{2}{y-1}$, giving $f(y)=\lg\frac{2}{y-1}$, which is defined for $y>1.$


Well as you read the description you know what to do. This place is for people to put what blathering nonsense and talk about online games they like.

I really enjoy the Neverwinter Nights gameplay online (though it depends alot on which server you join  :wink: ).  I only have NWN, not NWN2, so I'm not sure how the new one is in comparison.

RPG?
MMORPG?
I like Bunni: How we first met.
It's so sweet...
AND...
Runescape
I don't play it as much as I used to anymore....

Eternal Duel and eRepublik. There's a thread specifically on eRepublik for AoPS, and my clan in Eternal Duel has the same acronym as that of AoPS in case anyone from AoPS also plays, but I doubt anyone does.

I really like this one game
Forgot the name
Got it from somewhere on the games section

Sonny 2
Battalion: Arena beta (not the actual because you have to pay to get all the features)
Learn to Fly
Feudalism
Unreal Flash (both the original and Unreal Flash 2007)
Void

There are so many more... I can't bother to remember them all.

[quote="meewhee009"][quote="evilhamster"]Sonny 2
Battalion: Arena beta (not the actual because you have to pay to get all the features)
[b]Learn to Fly[/b]
Feudalism
Unreal Flash (both the original and Unreal Flash 2007)
Void

There are so many more... I can't bother to remember them all.[/quote]
OHMIGOSH!
That one flying penguin game?
I love(d) that :)[/quote]

me too!!!!!!!!!!!!!

I am addicted to the Flash gaming website [url=http://www.kongregate.com/]Kongregate[/url].  Some of my favorite games include Castlewars, Hedgehog Launch, Amorphous+, Four Second Frenzy, and Gravitee.

maplestory, penguin diner, castle wars, whatnot.  :)

FTW, Maplestory, and cubefield.   Totally awesome gamez.

Kingdom of Loathing.

Text based RPGs FTW.

I like Nitrome.com. It's doesn't have very many fans, but it's a good site.

Hmm...best games...

sonny (stuck on sinjid's shadow)
sonny 2 (corruptor beaten)
gemcraft (haven't played that much because of gemcraft 0)
gemcraft 0 (beaten 4th arcane guardian)
super crazy guitar maniac deluxe 2 (all perfects)
ffr (i'm guest15937)
amorphous+ (all awards)
and quite a few others

My online game is Mario and Mario is simply a joy to play. It's the plumber in his most pure form, running left to right, growing big and throwing fireballs, jumping huge pits and exploring underground tunnels, Swimming, Floating, Finding hidden areas.


For some reason, when I put \includegraphics{filename} in the solutions template, it puts the picture on its own page, even though I didnt create it to be so big.  Is there any way to resolve this? Thanks

Try typing or changing values in \vspace 4in or \hspace 4in right before the include.  Go here to read more: http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictMan.php#spacing
You really should check out all the stuff in the LaTeX help section.


Raffle tickets are sold at a price of three for $ \$2$. Jeff bought $ \$16$ worth of tickets, and a total of $ \$7200$ worth of tickets were sold. What is the probability that Jeff will win the raffle? Express your answer as a common fraction.

7200 * 3/2 = 10800 because you get 3 tickets per 2 dollars.

16/2 * 3 because divide the money and find tickets. 24/10800 = 12/5400 =
6/2700 = 3/1350 = 1/450


inequalities
geometry
inequalities unsolved

[quote="Lee Sang Hoon"]If $ x$, $ y$, $ z$ are three nonnegative reals, then prove that

$ \sum_{\text{cyc}}\sqrt {\left(z + x\right)\left(x + y\right)}\geq x + y + z + \sqrt3\cdot\sqrt {yz + zx + xy}$,

where the $ \sum_{\text{cyc}}$ sign means cyclic summation.[/quote]
$ \sum_{cyc}\sqrt {(z + x)(z + y)}\geq x + y + z + \sqrt {3(xy + xz + yz)}\Leftrightarrow$
$ \Leftrightarrow2\left(x + y + z - \sqrt {3(xy + xz + yz)}\right) - \sum_{cyc}\left(\sqrt {z + x} - \sqrt {z + y}\right)^{2}\geq0.$
But $ 2\left(x + y + z - \sqrt {3(xy + xz + yz)}\right) - \sum_{cyc}\left(\sqrt {z + x} - \sqrt {z + y}\right)^{2} =$
$ = \sum_{cyc}(x - y)^{2}\left(\frac {1}{x + y + z + \sqrt {3(xy + xz + yz)}} - \frac {1}{\left(\sqrt {z + x} + \sqrt {z + y}\right)^{2}\right)} =$
$ = \sum_{cyc}\frac {(x - y)^{2}\left(z + 2\sqrt {(z + x)(z + y)} - \sqrt {3(xy + xz + yz)}\right)}{\left(x + y + z + \sqrt {3(xy + xz + yz)}\right)\left(\sqrt {z + x} + \sqrt {z + y}\right)^{2}\right)} =$
$ = \sum_{cyc}\frac {(x - y)^{2}\left(z + \frac {4z^{2} + xy + xz + yz}{2\sqrt {(z + x)(z + y)} + \sqrt {3(xy + xz + yz)}}\right)}{\left(x + y + z + \sqrt {3(xy + xz + yz)}\right)\left(\sqrt {z + x} + \sqrt {z + y}\right)^{2}\right)}\geq0.$ :)

Slightly stronger:

[color=blue][b]Problem.[/b] If x, y, z are three reals such that the numbers y + z, z + x, x + y and yz + zx + xy are all nonnegative, then

$ \sum\sqrt {\left(z \plus{} x\right)\left(x \plus{} y\right)}\geq x \plus{} y \plus{} z \plus{} \sqrt3\cdot\sqrt {yz \plus{} zx \plus{} xy}$,

where the $ \sum$ sign means cyclic summation.[/color]

[i]Solution.[/i] Applying the Conway substitution theorem ([url=http://www.mathlinks.ro/Forum/viewtopic.php?p=204393#204393]http://www.mathlinks.ro/Forum/viewtopic.php?t=2958 post #3[/url]) to the reals x, y, z (in the role of u, v, w), we see that, since the numbers y + z, z + x, x + y and yz + zx + xy are all nonnegative, we can conclude that there exists a triangle ABC with sidelengths $ a \equal{} BC \equal{} \sqrt {y \plus{} z}$, $ b \equal{} CA \equal{} \sqrt {z \plus{} x}$, $ c \equal{} AB \equal{} \sqrt {x \plus{} y}$ and area $ S \equal{} \frac12\sqrt {yz \plus{} zx \plus{} xy}$.

Now,

$ \sum\sqrt {\left(z \plus{} x\right)\left(x \plus{} y\right)} \equal{} \sum\sqrt {z \plus{} x}\cdot\sqrt {x \plus{} y} \equal{} \sum b\cdot c \equal{} bc \plus{} ca \plus{} ab$;
$ x \plus{} y \plus{} z \equal{} \frac12\left(\left(y \plus{} z\right) \plus{} \left(z \plus{} x\right) \plus{} \left(x \plus{} y\right)\right)$
$ \equal{} \frac12\left(\left(\sqrt {y \plus{} z}\right)^{2} \plus{} \left(\sqrt {z \plus{} x}\right)^{2} \plus{} \left(\sqrt {x \plus{} y}\right)^{2}\right)$
$ \equal{} \frac12\left(a^{2} \plus{} b^{2} \plus{} c^{2}\right)$;
$ \sqrt3\cdot\sqrt {yz \plus{} zx \plus{} xy} \equal{} 2\sqrt3\cdot\frac12\sqrt {yz \plus{} zx \plus{} xy} \equal{} 2\sqrt3\cdot S$.

Hence, the inequality in question,

$ \sum\sqrt {\left(z \plus{} x\right)\left(x \plus{} y\right)}\geq x \plus{} y \plus{} z \plus{} \sqrt3\cdot\sqrt {yz \plus{} zx \plus{} xy}$,

becomes

$ bc \plus{} ca \plus{} ab\geq\frac12\left(a^{2} \plus{} b^{2} \plus{} c^{2}\right) \plus{} 2\sqrt3\cdot S$.

Multiplication by 2 transforms this into

$ 2\left(bc \plus{} ca \plus{} ab\right)\geq\left(a^{2} \plus{} b^{2} \plus{} c^{2}\right) \plus{} 4\sqrt3\cdot S$,

or, equivalently,

$ 2\left(bc \plus{} ca \plus{} ab\right) \minus{} \left(a^{2} \plus{} b^{2} \plus{} c^{2}\right)\geq 4\sqrt3\cdot S$.

Using the notation $ Q \equal{} \left(b \minus{} c\right)^{2} \plus{} \left(c \minus{} a\right)^{2} \plus{} \left(a \minus{} b\right)^{2}$, this rewrites as

$ \left(a^{2} \plus{} b^{2} \plus{} c^{2}\right) \minus{} Q\geq 4\sqrt3\cdot S$,

what is equivalent to

$ a^{2} \plus{} b^{2} \plus{} c^{2}\geq 4\sqrt3\cdot S \plus{} Q$.

But this is the well-known [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31598]Hadwiger-Finsler inequality[/url]. $ \blacksquare$

  Darij

Thank you  very much. Those are ingenious solutions.

By the way, the following inequality is also true.
For non-negatives $x$, $y$ and $z$ prove that:
\[2\sum_{cyc}\sqrt{(x+y)(x+z)}\leq 3(x+y+z)+\sqrt{3(yz+zx+xy)}\]

I have fixed my LaTeX mistakes.
[quote=arqady][quote="Lee Sang Hoon"]If $ x$, $ y$, $ z$ are three nonnegative reals, then prove that

$ \sum_{\text{cyc}}\sqrt {\left(z + x\right)\left(x + y\right)}\geq x + y + z + \sqrt3\cdot\sqrt {yz + zx + xy}$,

where the $ \sum_{\text{cyc}}$ sign means cyclic summation.[/quote]
$ \sum_{cyc}\sqrt {(z + x)(z + y)}\geq x + y + z + \sqrt {3(xy + xz + yz)}\Leftrightarrow$
$ \Leftrightarrow2\left(x + y + z - \sqrt {3(xy + xz + yz)}\right) - \sum_{cyc}\left(\sqrt {z + x} - \sqrt {z + y}\right)^{2}\geq0.$
But $ 2\left(x + y + z - \sqrt {3(xy + xz + yz)}\right) - \sum_{cyc}\left(\sqrt {z + x} - \sqrt {z + y}\right)^{2} =$
$ = \sum_{cyc}(x - y)^2\left(\frac {1}{x + y + z + \sqrt {3(xy + xz + yz)}}- \frac {1}{\left(\sqrt {z + x} + \sqrt {z + y}\right)^2}\right)=$
$ = \sum_{cyc}\frac {(x - y)^2\left(z + 2\sqrt {(z + x)(z + y)} - \sqrt {3(xy + xz + yz)}\right)}{\left(x + y + z + \sqrt {3(xy + xz + yz)}\right)\left(\sqrt {z + x} + \sqrt {z + y}\right)^2}=$
$ = \sum_{cyc}\frac {(x - y)^2\left(z + \frac {4z^{2} + xy + xz + yz}{2\sqrt {(z + x)(z + y)} + \sqrt {3(xy + xz + yz)}}\right)}{\left(x + y + z + \sqrt {3(xy + xz + yz)}\right)\left(\sqrt {z + x} + \sqrt {z + y}\right)^2}\geq0$ :)[/quote]

why $ S = \frac12\sqrt {yz + zx + xy}$? thanks.

[quote=arqady]For non-negatives $x$, $y$ and $z$ prove that:
\[2\sum_{cyc}\sqrt{(x+y)(x+z)}\leq 3(x+y+z)+\sqrt{3(yz+zx+xy)}\][/quote]

[img]http://s13.sinaimg.cn/middle/0018zOAxgy70UgpIkiwfc&690[/img]

[quote=Lee Sang Hoon]If $ x$, $ y$, $ z$ are three nonnegative reals, then prove that

$ \sum_{\text{cyc}}\sqrt {\left(z \plus{} x\right)\left(x \plus{} y\right)}\geq x \plus{} y \plus{} z \plus{} \sqrt3\cdot\sqrt {yz \plus{} zx \plus{} xy}$,

where the $ \sum_{\text{cyc}}$ sign means cyclic summation.[/quote]

Very beautiful problem .

[quote=Lee Sang Hoon]If $ x$, $ y$, $ z$ are three nonnegative reals, then prove that

$ \sum_{\text{cyc}}\sqrt {\left(z \plus{} x\right)\left(x \plus{} y\right)}\geq x \plus{} y \plus{} z \plus{} \sqrt3\cdot\sqrt {yz \plus{} zx \plus{} xy}$,

where the $ \sum_{\text{cyc}}$ sign means cyclic summation.[/quote]

[b]here's my final proof: [/b]
Square both sides:
$$(\sum\sqrt{x^2+3})^2\ge(\sum x+3)^2$$
Then
$$\sum x^2+9+2\sum\sqrt{(x^2+3)(y^2+3)}\ge\sum x^2+9+2\sum xy+6\sum x$$
Or
$$2\sum\sqrt{(x^2+3)(y^2+3)}\ge2(\sum xy+3\sum x)$$
Or
$$\sum\sqrt{(x^2+3)(y^2+3)}\ge 3(\sum x+1)$$
Which is equivalent to:
$$\sum\sqrt{(x^2+\sum xy)(y^2+\sum xy)}\ge 3(\sum x+1)$$
Or
$$\sum\sqrt{(x+y)(x+z)(y+x)(y+z)}\ge 3(\sum x+1)$$
Or
$$\sum (x+y)\sqrt{(x+z)(y+z)}\ge 3(\sum x+1)$$
Or
$$\sum (x+y)\sqrt{z^2+\sum xy}\ge 3(\sum x+1)$$
Which is equivalent to prove:
$$\sum (x+y)\sqrt{z^2+3}\ge 3(\sum x+1)$$
Then by C-S we know:
$$(z^2+3)(1+3)\ge(z+3)^2$$
$$(z^2+3)\ge\frac{(z+3)^2}{4}$$
Hence by replace this:
$$\sum (x+y)\sqrt{z^2+3}\ge\sum (x+y)\sqrt{\frac{(z+3)^2}{4}}$$
Or
$$\sum (x+y)\sqrt{z^2+3}\ge\frac{1}{2}\sum (x+y)(z+3)$$
Hence we need to prove:
$$\frac{1}{2}\sum (x+y)(z+3)\ge 3(\sum x+1)$$
Or
$$\sum xz+yz+3x+3y\ge6\sum x+6$$
Or
$$2\sum xy +6\sum x\ge6\sum x+6$$
Or
$$2\sum xy\ge 6$$
$$\sum xy=3\ge3$$
Which is true so we're done...!

First rewrite as
\[(\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x})^2 \ge 4(x+y+z) + 2\sqrt{3(xy+yz+zx)}.\]
Using the substitution $a = \sqrt{y+z}$, etc, we can rewrite as
\[(2ab+2bc+2ca-a^2-b^2-c^2)^2 \ge 3(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4)\]
which is just 4th degree Schur.


Let $ f(x)$ be a function in $ \mathbb{Z}_\plus{}$ such that

$ x \equal{} 1\Longrightarrow y \equal{} 2$
$ x \equal{} 2\Longrightarrow    y\equal{}  8$
$ x \equal{} 3\Longrightarrow    y\equal{} 12$
$ x \equal{} 4\Longrightarrow    y \equal{} 14$
$ x \equal{} 5\Longrightarrow    y \equal{} 18$
$ x \equal{} 6\Longrightarrow    y \equal{} 20$
$ x \equal{} 7\Longrightarrow    y \equal{} 24$
$ x \equal{} 8\Longrightarrow    y \equal{} 30$
$ x \equal{} 9\Longrightarrow    y \equal{} 32\equal{}   30 \plus{} 2$
$ x \equal{} 10\Longrightarrow  y \equal{} 38\equal{}   30 \plus{} 8$
$ x \equal{} 11\Longrightarrow  y \equal{} 42\equal{}30 \plus{} 12$
$ \ldots$

Find $ f(x)$.

Tried to figure out whats the connection between the first 8 values of x and gave up . hope some1 post the sollution fast cause im fackin nervous  :mad:

[quote="MellowMelon"]Are we supposed to give like a concise form? Assuming you're indicating a pattern, you could just give an individual definition for each residue mod 8...[/quote]

Any solution is fine  :)

[hide="My attempt at a concise form:"]
First, I did $ y\equal{}4x\minus{}2\minus{}2\lfloor\dfrac{x\minus{}1}{8}\rfloor$. The $ 4x\minus{}2$ part somewhat approximates the function, and the $ 2\lfloor\dfrac{x\minus{}1}{8}\rfloor$ part decreases $ 2$ for every $ 8$ values of $ x$, which will make the function as a whole increase by $ 30$ for every $ 8$ values of $ x$.
Now the function must be somewhat adjusted for different residues of $ x \bmod 8$.
Denote $ z_8\equal{}x\minus{}8\lfloor\dfrac{x}{8}\rfloor$ as the residue of $ x \bmod 8$. Also, denote $ z_4\equal{}x\minus{}4\lfloor\dfrac{x}{4}\rfloor$ as the residue of $ x \bmod 4$.
The values of $ z_8$ from $ 0$ to $ 7$ must map out to $ 0, 0, 2, 2, 0, 0, \minus{}2,$ and $ \minus{}2$, respectively. A "sign test" of $ \dfrac{|4.5\minus{}z_8|}{4.5\minus{}z_8}$ can be combined with a floor function on $ z_4$ to produce the desired result. $ \dfrac{|4.5\minus{}z_8|}{4.5\minus{}z_8} \cdot 2\lfloor\dfrac{z_4}{2}\rfloor$ maps out $ z_8$ as desired.
Now, combining these together, we get
\[ y\equal{}4x\minus{}2\minus{}2\lfloor\dfrac{x\minus{}1}{8}\rfloor\plus{}2\dfrac{|4.5\minus{}z_8|}{4.5\minus{}z_8} \cdot \lfloor\dfrac{z_4}{2}\rfloor\]
Replacing the values of $ x_4$ and $ x_8$ gives
\[ y\equal{}4x\minus{}2\minus{}2\lfloor\dfrac{x\minus{}1}{8}\rfloor\plus{}2\dfrac{|4.5\minus{}(x\minus{}8\lfloor\dfrac{x}{8}\rfloor)|}{4.5\minus{}(x\minus{}8\lfloor\dfrac{x}{8}\rfloor)} \cdot \lfloor\dfrac{x\minus{}4\lfloor\dfrac{x}{4}\rfloor}{2}\rfloor\]
After putting it in my graphing calculator (and typoing 3 times  :mad:), I've verified this to be true! A concise form!
[/hide]

I have a weird hunch it involves flooring.  And notice how each of the terms is even. Hmm.

Very Good job [b]Brut3Forc3[/b].

:clap2:

:D

Is there an easier way to express it? (I sure hope so...)

Suppose now that $ F(x) \equal{} \displaystyle \sum_{i \equal{} 1}^xf(i)$, where $ x$ is a positive integer. Find $ F(x)$ in terms of $ x$.


calculus
integration
number theory unsolved
number theory

Prove that:Equation $ \ \frac{x}{y}\minus{}\frac{y}{z}\plus{}\frac{z}{x}\equal{}9$ has infinite integral solutions
$ \ (x,y,z)$ and all triples $ \ (x,y,z)$ also satisfy $ \ \sqrt[3]{xyz}$  $ \  \epsilon Z$

[quote="duongptnkvn"]Prove that:Equation $ \ \frac {x}{y} \minus{} \frac {y}{z} \plus{} \frac {z}{x} \equal{} 9$ has infinite integral solutions
$ \ (x,y,z)$ and all triples $ \ (x,y,z)$ also satisfy $ \ \sqrt [3]{xyz}$  $ \ \epsilon Z$[/quote]
And more
If $ |\frac {x}{y}| \plus{}|\frac {y}{z}| \plus{}| \frac {z}{x}|$ is a integer number
then $ |x|\equal{}|y|\equal{}|z|$
It is very easy
My hint is denoting $ d\equal{}gcm(x,y,z)$.I will post my solution later,because I want see a  full proof for it,my one is not short


Ciudat mi-am adus aminte acum de ea.... ma rog am avut-o la examenul din vara si la marire in toamna si tot n-am facut-o. As fi tare recunoscator, daca cineva ar putea sa posteze o solutie urmatoarei teoreme:

  [color=blue] [b] Teorema(Poincare).[/b]

   Presupunem ca $U$ este stelata intr-un punct $x_{0}\in U$ si forma $dif \omega$ este inchisa. Atunci $\omega$ este exacta.

   [/color]

Ai putea defini un pic termenii/literele folosite? ($\omega$, $dif \omega$, exact)

(sau echivalentele lor in engleza)

Da. Spunem ca multimea $U$ este [color=blue] stelata [/color] intr-un punct $x_{0}\in U$ daca oricare ar fi $x\in U$
avem segmentul $[x_{0}, x]$ inclus in $U$.

Spunem ca $1$- forma diferentiala $\omega$ este [color=blue] exacta [/color] daca exista o functie $f: U\to\mathbb{R}$ derivabila cu derivata continua astfel incat $\omega =d f$.

  Sper ca acum este clar!

Daca nu ti-o mai amintesti, e o singura solutie: gandeste-te si tu cateva minute, si o sa vezi ca poti sa scoti demonstratia fara probleme, ca nu e mare lucru de capul ei :).

Deci ai $\omega=\sum_{i=1}^nf_idx_i$ definita pe un deschis stelat in raport cu $0$ o sa-l iau eu, ca sa nu ma incurc cu $x_0$ ala. Stii ca pentru toti $i,j\in\overline{1,n}$, ai $\frac{\partial f_i}{\partial x_j}=\frac{\partial f_j}{\partial x_i}$, so vrei sa arati ca exista o functie $f$ definita in $U$ astfel incat $\frac{\partial f}{\partial x_i}=f_i,\ \forall i\in\overline{1,n}$.

Hai sa vedem cam cum ar trebui sa arate $f$ asta (asa am gandit eu cel putin). Fixeaza-ti un punct $u=(u^1,\ldots,u^n)\in U$ diferit de zero (diferit de punctul in raport cu care $U$ e stelata), si defineste $\varphi(t)=f(tu)$. Pentru $t_0$ fixat, avem $\varphi'(t_0)=\lim_{t\to t_0}\frac{f(t_0u+(t-t_0)u)-f(t_0u)}{t-t_0}=Df(t_0u)(u)$. Noi am vrea ca aplicatia liniara de la $\mathbb R^n$ la $\mathbb R\ Df(t_0u)$ sa fie produsul scalar cu $(f_1(t_0u),\ldots,f_n(t_0u))$, deci $\varphi'(t_0)=\sum_{j=1}^nf_j(t_0u)u^j$, pentru orice $t_0$ in domeniul de definitie (care consta din acei $t$ pentru care $tu\in U$). Asta inseamna ca poti sa definesti $f(tu)=\varphi(t)=\int_0^t\left(\sum_{j=1}^nf_j(su)u^j\right)\mbox ds$. Acum pentru $t=1$ avem exact ce voiam: o expresie pentru valoarea lui $f$ intr-un punct $u$ arbitrar al lui $U: \ f(u)=\int_0^1\left(\sum_{j=1}^nf_j(su)u^j\right)\mbox ds$. Mai ramane sa verificam ca intr-adevar $\frac{\partial f}{\partial x_i}=f_i,\ \forall i\in\overline{1,n}$.

Derivam in raport cu $x_i$ si trecem derivata sub integrala. Avem deci $\frac{\partial f}{\partial x_i}(u)=\int_0^1\left(\sum_{j=1}^n\frac{\partial f_j}{\partial x_i}(su)su^j+\sum_{j=1}^nf_j(su)\frac{\mbox du^j}{\mbox dx_i}\right)\mbox ds$ ($u^j$ e privita ca functia de $u$ care il duce pe $u$ in $u^j$, adica proiectia pe coordonata $j$; asta inseamna ca $\frac{\mbox du^j}{\mbox dx^i}=1$ daca $i=j$, si $0$ altfel). Rezulta de aici ca $\frac{\partial f}{\partial x_i}(u)=\int_0^1\left(\sum_{j=1}^n\frac{\partial f_i}{\partial x_j}(su)su^j+f_i(su)\right)\mbox ds\ (*)$ (am schimbat $\frac{\partial f_j}{\partial x_i}$ cu $\frac{\partial f_i}{\partial x_j}$). Acum, daca $\psi(s)=sf_i(su)$, se observa usor ca sub integrala din $(*)$ e exact $\psi'(s)$, deci integrala o sa fie egala cu $\psi(1)-\psi(0)=f_i(u)$, si gata.

Edit:

Din explicatiile pe care le-ai dat, inteleg ca te refereai la $1$-forme, nu? Ca e si o varianta generala, care suna asa: daca $A$ e deschisa si stelata, atunci orice $k$-forma ($k\ge 1$) diferentiala inchisa in $A$ e exacta, adica pentru orice $k$-forma $\omega$ definita in $A$ care satisface $d\omega=0$, exista o $k-1$-forma $\eta$ in $A$ astfel incat $d\eta=\omega$.

Extrem de clar!!! Frumos! Vad ca ai inceput sa explici aproape ca Darij. :)  ;) Keep going!


What is the number of distinct ways of arranging the letters in the word AVERAGE?

[quote="AceOfDiamonds"]AVERAGE has 7 letters, 2 Es, 2 As   so answer is
$ \frac{7!}{2!*2!}\equal{}\frac{720}{4}\equal{}180$[/quote]
$7!$ doesn't equal $720$, it equals $5040$

So $\frac{5040}{4}=\boxed{1260}$

[quote=tangzhenzhi]I do not understand the 2! times 2! part.[/quote]

We divide by two for each of the repeats because we can switch the positions of the E's and A's and it won't matter.


Just for fun, I was wondering which ones of you had been in the Scripps National Spelling Bee program.

Rounds 5-7 were on ESPN/ESPN2, and the last 13 spellers (rounds 8-20) were on ABC primetime last year.

See: [url]http://www.spellingbee.com[/url].

[quote="kate25"]I was in the program through middle school. I was in the Kansas state finals in eighth grade.[/quote]

we never have spelling bees...idk why...i was the only fifth grade that could spell

pneumonoultramicroscopicsilicovolcanoconiosis

and

antidisestablishmentarianism

in my school

Mycroft's Brother has been, and a good number of other people I know who aren't AoPSers. But I haven't and won't myself.

*unless I become a NSB official when I grow up, but that's not likely; I even more doubt that I'll take Mr. Bailly's job ever :D

In 6th grade(I skipped one), I was amazed at myself that I got to third place, and then I dropped out. :( But the other two contestants got all of these easy words wrong, and if I stayed in, I would've won!

[quote="mathnerd314"]I got second place at the regional bee last year (8th grade), and then dropped out because of "dirndl".

[quote="7h3.D3m0n.117"]antidisestablishmentarianism[/quote]

Ooh! Ooh! I recently did a speech about antidisestablishmentarianism![/quote]

wow lol a lot of my friends know the word but they don't know the meaning  :wink:

I'm not your friend, but I don't know the meaning...

I don't think the kids' parents should force them to learn all of those hards words! That must be just plain bore! The highest I have ever taken in a spelling bee is 23rd at state, and only 1 person from WI got to go to the national one. I just did them to get out of school :P . I never expected to win because I do not like slaving long and hard to learn to spell flauccinauccinilifilipilification or something like that :blush:

[quote="kstan013"]I don't think the kids' parents should force them to learn all of those hards words! That must be just plain bore![/quote]

Some kids do it not because their parents force them, but for their own reasons. I personally find it fascinating to see where the English language comes from.

And it's floccinaucinihilipilification, by the way. You spelled the "o" with an "au", the second "cc" should only be spelled with one, and you inserted an extra "fili" before "pili". :harhar: ;)

[quote="mathnerd314"][quote="kstan013"]I don't think the kids' parents should force them to learn all of those hards words! That must be just plain bore![/quote]

Some kids do it not because their parents force them, but for their own reasons. I personally find it fascinating to see where the English language comes from.

And it's floccinaucinihilipilification, by the way. You spelled the "o" with an "au", and the second "cc" should only be spelled with one. :harhar: ;)[/quote]

floccinaucinihilipilification is the longest word without an e too lol

[quote="jhredsox"]ok, question, What's the longest word you can make using only the top line of letters on a keyboard?
-jorian[/quote]

QWERTY

i believe thats actually a word

You could have purchased it (in print) at the spelling bee store last year. I have the dictionaries for 2003-2006 plus Hexco stuff. Doing the spelling bee is not cheap!

The national office gives you the electronic versions of lots of study aids (for free) when you get to nationals. I don't think it's an issue to post them since they can't be obtained anymore.

I was in the Nationals this year.  I missed in the 5th round.

That's not bad. What word did you miss? If you can go back next year, you can still improve considerably (although it is not always easy to get better than round 5). I missed the cut for even ESPN in a monster round that took out almost 1/2 of all the spellers.

What I wonder about this year is the relative difficulty of the written round. The statistics for that are usually not released on the website but are given to spellers.

Thanks for the word banks Lawerence (right?). Yeah....

I never got past the county finals because I live in the same county as the Gunturi brothers (Sai won in like 2003 and his brother Abhiram went to the Nat'l bee each year after that until I left middle school)

I made it to Nats this year, but got out in the third round (Aargh!). I made it to the fifth round last year, though.

I'm thinking about spending time to coach students at my former middle school weekly afterschool and resurrect its dead spelling bee program. Some incoming elementary school kids have already been to regional finals several times; it would be a pity if they never competed again due to a lack of school support.

I will probably dig out some old Hexco stuff, get the Spell It!'s from the newspaper agency, and leave the administrative work to the former coordinator (i.e., paying the 99 dollars for the School Pronouncer Guide and other district fees).

What other steps do you think I should take?

Wait, is the national champion's name spelled

Evan O'Dorney or

[b]Evan O'Dorkey[/b].  I always thought it was the latter and since the n and k are near on the keyboard I thought there was a mistake.  I cant believe so many people mispelled his name.

Its [b]E-V-A-N__O'-D-O-R-K-E-Y[/b]

That's a really bad joke. It's Evan O'Dorney, and he deserves respect no matter what you think of his personality.

oh I think hes a genius and all (way way smarter than me)

but hes going to fail at life.  [b]social skills[/b]$ >$[b]mental skills[/b]+$ \infty$

A person that knows abosolutely nothing will still be better than him because hes going to get backed into a corner and wont get himself out of it.

2007 scripps national spelling bee champ

[quote="shaggy75"]i got 6th in north california regionals[/quote]
I dont get it? Are you evan o'dorney?


Let $ d(n)$ be the sum of proper divisors of $ n$. If $ d(a) \equal{} b$ and $ d(b) \equal{} a$, where $ a \ne b$, then $ a$ and $ b$ are an amicable pair and each of $ a$ and $ b$ are called amicable numbers. For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore $ d(220) \equal{} 284$. The proper divisors of 284 are 1, 2, 4, 71 and 142; so $ d(284) \equal{} 220$. Evaluate the sum of all the amicable numbers under 10000.

[hide]
There is no really good method to this one. Just search them online. :D

$ 220 \plus{} 284 \plus{} 1184 \plus{} 1210 \plus{} 2620 \plus{} 2924 \plus{} 5020 \plus{} 5564 \plus{} 6232 \plus{} 6368 \equal{} \boxed {31626}$[/hide]

In general project euler problems are meant to be solved by computers (and without help!)

Basically, you are searching for numbers such that $ d(d(a))\equal{}a$ but $ d(a)\ne a$. This is easy to run in say mathematica.

It's easy enough with a computer. I was wondering if there was a mathematical way. It seems not.


Consider the numbers $ a_1, a_2,..., a_n$. We take two numbers $ a_i$ and $ a_j$, with $ a_i \neq a_j$, and replace them by $ \frac{a_i\plus{}a_j}{2}$ and $ \frac{a_i\plus{}a_j}{2}$ (yes, the same number). Prove that after a finite number of substitutions, all the numbers will be equal.

How does a finite number of such substitutions get $ (a_1,a_2,a_3) \equal{} (0,0,2)$ to be three identical numbers?

@MellowMelon, the poster said that $ a_i \ne a_j$.

Nonetheless, when applying that function to two numbers, you will always end up with those two numbers being equal and the third having another value.  For example, you might start out with $ (0, 0, 2)$, go to $ (1, 0, 1)$, $ (0.5, 0.5, 1)$, $ (0.5, 0.75, 0.75)$ and so on, but the three numbers will never be equal.

123: $ a_i \neq a_j$ only applies to doing the operation. No such condition is applied to the sequence. Either way, $ 0,1,3$ is another counterexample.

Here's a save, though it's quite easy and may not be as hard as whatever problem the OP intended.

Let $ a_1, a_2, \ldots a_n$ be a sequence of $ n$ real numbers. For $ m \mid n$, we allow the operation of selecting $ m$ numbers from the sequence $ a_{k_1}, a_{k_2}, \ldots a_{k_m}$ with $ 1 \leq k_1 < k_2 < \ldots < k_m \leq n$, and turning all $ m$ numbers into $ \frac{a_{k_1} \plus{} a_{k_2} \plus{} \ldots \plus{} a_{k_m}}{m}$. Show that all numbers can be made the same through this operation.

Another save of the problem:

For which $ n$ does the OP's statement hold for all initial choices?


Let $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}1$. Prove that:
$ \frac{\sqrt{a^2\plus{}abc}}{ab\plus{}c}\plus{}\frac{\sqrt{b^2\plus{}abc}}{bc\plus{}a}\plus{}\frac{\sqrt{c^2\plus{}abc}}{ca\plus{}b}\leq\frac{1}{2\sqrt{abc}}$

Let a=0 , b>0, c>0  .Please try the inequality by this case .1/(b+c)>= 1/b,is it possible, e.g . when b=c = 1/2 ?

[quote="christ"]Let a=0 , b>0, c>0  [/quote]

but you can't put $ a\equal{}0$, you have $ a, b, c >0$ as an assumption.


Cono Sur Math Olympiad training list 1 (Brazil)

Altitudes $ AD$ and $ BE$ of the triangle $ ABC$ meet at $ H$. Let $ X$ and $ Y$ be midpoints of $ AB$ and $ CH$, respectively. Prove that $ XY$ is perpendicular to $ DE$

[asy]pair A=(0,0), B=(10,0), C=(7,9), D, E, X, Y, H; D=foot(A,B,C); E=foot(B,A,C); H=intersectionpoint(A--D,B--E); X=midpoint(A--B); Y=midpoint(C--H); draw(A--B--C--A..B--E..A--D..C--foot(C,A,B)); draw(Y--E--X--D--Y..Y--X..E--D); dot(A^^B^^C^^D^^E^^H^^X^^Y); label("A",A,SW); label("B",B,SE); label("C",C,NE); label("D",D,NE); label("E",E,NW); label("H",H,SW); label("X",X,SW); label("Y",Y,NW);[/asy]$ ABDE$ is cyclic, so $ X$ = midpoint of $ AB$ = circumcenter of $ ABDE$.
$ CDEH$ is cyclic, so $ Y$ = midpoint of $ CH$ = circumcenter of $ CDEH$.
$ XA=XE=XD=XB$=circumradius of $ ABDE$.
$ YC=YE=YH=YD$=circumradius of $ CDEH$.
since $ XE=XD$ and $ YE=YD$, $ XY\perp ED$.

If you know about the 9 point circle, this problem is pretty trivial.


vector
induction
graph theory
combinatorics proposed
combinatorics

Given an $n\times n$ array filled with symbols s.t. the rows are distinct, show that we can remove a column s.t. in the resulting array the rows are still distinct.

I know it's easy, but doesn't it look nice? :)

Hmmm...I don't think it is as easy as you said, but...
Anyway, It also appeared in TOT 1980 and in the course in graph theory that I wrote for the olympic french contestants (see : http://www.animath.fr)

Pierre.

I read that graph theory tutorial last year when preparing for IMO, and I can say it is really great, Pierre!!!
About the solution, I think it was something like this, but maybe I am wrong.
Consider a graph: the rows. Now take out each column $k$. Some to rows $i,j$ will be the same, so add edge $(i,j)$ to the graph!. There are n colums, so we will get n edges, so there will be a cycle!
I think this is bad, to have a cycle I mean.....

Yes, same thing.
But I don't think the graph approach is easy to think about.

Pierre.

Indeed, I spent quite many time trying a wrong approach before I found this idea with tree. And I think harazi ( :mad: ) missed then he wrote the following message http://www.mathlinks.ro/Forum/viewtopic.php?p=123333#p123333 . It should be addressed to grobber in this case.

This problem is a famous lemma of Bondy discussed on the first page of Bollobas' book on combinatorics of finite set systems.

Do you have a precise reference?

Pierre.

The title of Bollobas' book is something like "Combinatorics" or "Combinatorics of Finite Sets" or "Combinatorics of Set Systems".
It was published 10-15 years ago based on his lectures in the 1980's at Louisiana State University.
Bondy's theorem is the [i]mise en scene[/i] for the first several chapters, on set systems (i.e. families of subsets of a finite set).

(Edit: here it is.
B. Bollobas, "Combinatorics: Set Systems, Hypergraphs, Families of Vectors and Probabilistic Combinatorics", publ. 1986, Cambridge University Press)

I will solve this problem with induction:
We know that it is true for n=2.
Suppose the statement is true for n=k. it is trivial that if it is true for a  k * k  array ,then  it will be true for any  p * k array (p <= k).  now we have a (k+1) * (k+1) array.
Take out the first column. We will have at least 2 the same rows, if not the problem is solved . now we have k columns and at last k distinct rows.
By using induction we can take out a column such that the rows are distinct . then add the first column to the set.

The version in OTIS has $n = 50$, but it's pretty much the same thing...
[hide=Solution]
Assume the contrary. Then, for each column, there exists two rows which differ only by that column, otherwise we could remove that column and be fine. For each of the $50$ columnns, let $\{i_k, j_k\}$ be an unordered pair of rows for which row $i_k$ and row $j_k$ differ only at column $k$ (the $k$th place from the left). Now, take the graph $G$ on each of the $50$ rows, and connect only each $i_k$ and $j_k$ on the graph for $1\leq k\leq 50$. Then, note that $G$ has no cycles (of length $\geq 3$), since if there existed a cycle of rows $a_1, a_2, \cdots, a_k$ in $G$ for some $k\geq 3$, then $a_2$ would differ by $a_1$ in one place, $a_3$ would differ by $a_2$ in another place, so $a_3$ would differ by $a_1$ in $2$ places, $a_4$ would differ by $a_3$ in yet another place, so $a_4$ would differ by $a_1$ in $3$ places, etc., so $a_k$ would differ by $a_1$ in $k-1$ places. However, since $a_k$ is connected to $a_1$, $a_k$ differs by $a_1$ in $1$ place, but $k-1 \geq 2$, a contradiction. Thus, $G$ has no cycles, so $G$ is a forest. Thus, since $G$ is a graph on $50$ vertices, we have that $G$ has at most $50 - 1 = 49$ edges. However, $G$ actually has $50$ edges, since the only edges in $G$ are the ones between $i_k$ and $j_k$ for each $1\leq k\leq 50$, which is a contradiction. Thus, our original assumption was false and so there does exist such a column which you can remove and still have the rows be pairwise distinct, as desired. $\blacksquare$


[quote="DavidL"]Prove:

For the terms 1/n, 2/n, 3/n ... n/n,

If d | n, exactly Euler'sTotient(d) of the terms in the above list have d as their denominator when reduced to lowest terms.

I am at a loss! Help please?[/quote]

I cant understand the problem, can you give a detail example?

Say I choose 8 to be n.

then in the set {1/12, 2/12, 3/12, 4/12, 5/12, 6/12, 7/12, 8/12, 9/12, 10/12, 11/12, 12/12}, for any d such that d | n, (Let's choose d = 4) there are exactly EulerTotient(4) = 2 fractions when reduced to lowest terms have denominator 4.

These two fractions are 3/12 = 1/4, 9/12 = 3/4, and no others.

But how do you prove this?


Let $x,y,z$ be positive numbers. Prove that:
\[ \left(\frac{x}{2x+y}\right)^{3}+\left(\frac{y}{2y+z}\right)^{3}+\left(\frac{z}{2z+x}\right)^{3}\geq\frac{1}{9}. \]

[quote="Beat"]This is Classic. We will use only AM-GM:
$\left(\frac{x}{2x+y}\right)^3+\left(\frac{y}{2y+z}\right)^3+\left(\frac{z}{2z+x}\right)^3\ge \frac{3xyz}{(2x+y)(2y+z)(2z+x)}\ge \frac{3xyz}{(\frac{2x+y+2y+z+2z+x}{3})^3}=\frac{3xyz}{(x+y+z)^3}\ge\frac{1}{9}$[/quote]

the last it's wrong

[quote="libra_gold"]
with $u+v+w<1$, we have $(-u+v+w)(+u-v+w)(+u+v-w)<(1-2u)(1-2v)(1-2w)=uvw$

which leads to contradiction![/quote]

How it leads to contradiction ?  :?:

i think $uww>(u+v-w)(u-v+w)(-u+v+w)$ is true by Schur :!:

[quote="andyciup"]Is is true that $\frac x{2x+y}+\frac y{2y+z} +\frac z{2z+x} \geq 1$?Because if this is true then the problem is solved.[I think][/quote]

i don't think it true...
for example: $x\rightarrow0^+, y=1, z=1$ LHS=$\frac{5}{6}$

I am pretty sure this one can be done by mass expanding and a Murihead bash, but there's probably a nicer solution.


$A\in M_{3}(C)$ o matrice pentru care $det(A+3A^{t})=0$

Demonstrati cä $det(A+A^{t})=(-8/3)det(A)$

unde $A^{t}$ transpusa matricii A


Let $ k\geq1$.Find f:R-->R such that
$ f(x^{k}+f(y))=y+f(x^{k})$

Earlier I couldn't get anywhere with the problem actually in the first post, where the rightmost $ k$ has moved into the function. Does the problem in the first post have a reasonable solution?

Moving the $ k$, as in the actual TST problem, made it a bit easier.


A , B and  C  are acute angles such that

CosA =TanB

CosB =TanC

CosC =TanA

Prove that SinA = SinB = SinC  =2.Sin18 degrees

If we multiply across by $\cos$ and replace $\cos^2$ with $1-\sin^2$


For, #1, if we just list all the possibilities without proof, will we still receive full credit (if we cover all possibilities)?

We can't answer questions like this while the round is still in progress.


function
trigonometry
calculus
calculus computations

Hey :) not sure if this is the right forum but i couldn't find anywhere else to put it.

I'm working on a fourier series question, and im having problems determining the period. Here's a scan of the question:

[img]http://homepages.ihug.co.nz/~gem/im%20a%20geek.jpg[/img]
The question states that the function is repeated every 2pi, but does that mean 2pi is the period?
Upon closer examination i realised that the function reaches the same point at -pi/2 and pi/2, indicating a period of pi?

also, just to check:
this is an odd function, so i would obtain ao = an = 0
and bn = (4/pi*n)(-1 - (-1)^n)   where [(-1)^n = cos(n*pi)]

and another thing :) is there a quick and easy method to check these things? i've been trying to write an excel spreadsheet but i can't work out how to do it.

any help would be appreciated!
thanks
-brad

Oh and if there is a more appropriate forum for this kind of thread, would an admin please move it there? :) thanks

I get different coeffecients than you for $b_n$
(I assume $b_n$ stands for the coeffecient of $\sin (nx)$ in the series).

I get that $b_n = \frac{-8}{\pi n}$ for $n \equiv 2 \bmod 4$, (i.e. for all n that leave a remainder of 2 when divided by 4), and zero otherwise.

This means $f(x) = \sum_{n=1}^{\infty} \frac{-8}{\pi (4n-2)} \sin \big((4n-2)x\big)$

I graphed the first 5 terms of this series on a graphing calculator and it looks correct. A graphing program is a good way to see if your series is accurate.

Thanks for the reply :)

I'm not sure if i can use that mod thing because we haven't been taught it yet, so i think i should just stick with the standard formulae. Would you mind posting some working for me? i can't seem to get my head around this one :<

oh and what sort of graphing program? any in particular?


algebra
polynomial
quadratics
number theory proposed
number theory

Find the smallest odd integer $ k$ such that: for every $ 3\minus{}$degree polynomials $ f$ with integer coefficients, if there exist $ k$ integer $ n$ such that $ |f(n)|$ is a prime number, then $ f$ is irreducible in $ \mathbb{Z}[n]$.

Quickly we can see that $ k\leq 7$.

Suppose $ f$ has a linear factor in $ \mathbb{Z}[n]$. Then

$ f(n) \equal{} (n\minus{}a)g(n)$

where $ g$ is a quadratic polynomial with integer coefficients. But $ |g(n)|\equal{}1$ is possible only for four values of $ n$ and $ |n\minus{}a|\equal{}1$ only for two values of $ n$. Hence $ f(n)$ can be prime at most for $ 6$ values of $ n$. Therefore if $ |f(n)|$ is prime for $ 7$ or more values of $ n$, then $ f$ is irreducible.

There is definitely better though. I'm willing to bet $ k\equal{}5$.

$ k>3$ because
$ f(n)\equal{}(n\minus{}3)(n^2\plus{}1)$ is prime for $ n\equal{}0,2,4$ and reducible.


Hi, I have been using Matlab, but it seems that it is does not have enough precision to compute complex roots for what I am trying to do. What other software are there that have better precision? If I know how to program, is it feasible for me to implement a polynomial root solve on my own (either in Matlab or other language)?

- Thanks

1. You can compute the roots of a polynomial with any precision in Maple.
2. Yes, it is easy to write a program that finds roots with any precision.


inequalities
LaTeX
logarithms
inequalities open

I am sorry if this problem has been posted.  :?

Find the smallest positive $k$ such that, if $x,y,z>0$ and $xyz=1$, then
\[\frac{1}{(1+x)^k}+\frac{1}{(1+y)^k}+\frac{1}{(1+z)^k} \ge \frac{3}{2^k}.\]

I wonder wt the minimum value is when k=1?

I think that the minimal value verifies $2^k=3$, but I do not know how to prove. I know Vasc has a proof, but I think it is useless to ask him to post the solution. This kind of problems need a huge amount of computations and analysis.

We can solve them,not necessary ask Vasc.

The better inequality:

$\sum\frac{1}{(a+1)^k}\ge min(1,\frac{3}{2^k})$
For all $k\ge 0$

Well, then probably someone will want to post a solution. If I do not ask too much.

ok ,My result is the same as Hungkhtn and here is my approach :

We can assume that a<=b<=c ,then bc>=1
Lemma : 1/(1+b) +1/(1+c) >= 2/ (1+ :sqrt: bc) =2/(1+1/ :sqrt: a) ( easyly prove)

First case : k<=1

:Sigma: 1/(1+a)^k  >=  Sigma 1/(1+a) >= 1/(1+a) +2/(1+ 1/:sqrt: a) > 1

Second  case : k>=1

We have : 1/(1+b)^k +1/(1+c)^k >= 2 [ (1/1+b +1/1+c )/2 ]^k  >= 2 .1/(1+1/ :sqrt: a)^k =f(a)

We will prove that : f(a) + 1/(1+a)^k = g (a) >=min(1,3/2^k)

We work on g'(a) and the rest is for you

The rest is not easy.
And for this problem,the rest is the most difficult in the solution.
-------
Where are you from,Namamhams?

I agree with you .It's not easy ,and in my opinion ,it's  not so diffcult .I want somebody who know Latex will help me complete this .I don't know Latex ,and also i don't have much time .

@ : i'm from VietNam  :D And you're too :lol:

Ok, I abandon my idea of seeing a solution for this inequality... :(

Ok , if nobody post the  solution , i'll post my complete solution as soon as possible .

For particular,we can prove from :

$\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \ge 1$

If $abcd=1$.

And let $d=1$.We have result.

well,this inequality was the third problem in Chinese Team Selection Test 2st day.
I have all the problems.But they are Chinese version.

I am still waiting for the solution :?  :(  :?

I use the following version of Holder, which is similar to "CS in engel form", and has been discussed somewhere (i forget where.)

H$\ddot{o}$lder version 2: for $u, v \in N$, $\sum_{i=1}^n \frac{x_i^u}{y_i^v} \geq \frac{ (\sum_{i=1}^n x_i)^u }{ n^{u-v-1}(\sum_{i=1}^n y_i)^v}$ with equality when $x_i/y_i$ are equal for all i.

Now for the proof.  Change of variables, x to 1/x.  We claim all positive k work.

Use the holders ver2 above.  Let $s = \sum x$.  We have:

$\sum \frac{x^k}{(1+x)^k} \geq \frac{3(\sum x)^k}{(\sum 1+x)^k} = 3(\frac{s}{3+s})^k$.

We want to show $\frac{s}{3+s} \geq \frac{1}{2}$.  This is just $s \geq 3$, which is clearly always true.

So infact, it works for all positive k.

---

Edit:  oops, i dont think can use holders for non integer argument.  But still sheds light on problem: i still believe answer is any positive k.

---

Second edit:  What about jensen?

f(x) =  1/(x+1)^k

$LHS \geq 3f(\frac{x+y+z}{3}) \geq 3f(\sqrt[3]{xyz}) = \frac{3}{2^k}$

---

third edit: yea its false

Your solutions are false
for the first: you forget the condition $u\ge v+1$
for the second:$f(\frac{x+y+z}{3})\le f(\sqrt[3]{xyz})$

Yes, false.
Obviously it's not true for all positive k.

for my first soln, does it hold if u and v are not integer?

---

because, using the same theorem (holder v2):

$\sum \frac{1^{k+1}}{(1+x)^k} \geq \frac{3*3^k}{(3+s)^k}$

where the result follows as before, but we have used $u \geq v + 1$.

Can anyone prove it for [size=150][color=red]k=3[/color][/size]

isn't the case k=3 a vietnam tst question last year? It was posted on the forum for some time already.

I want you post in this forum !
I need it very much !
Thank you ! please help me !

Proving it for larger values of $k$ is easy. If $abc = 1 \Longrightarrow \frac{1}{(1+a)^k} + \frac{1}{(1+b)^k} + \frac{1}{(1+c)^k} \ge \frac{3}{2^k}$ holds for for $k=k_0$, then for $k_1 > k_0$, we can write
\[ \sum_{cyc} \frac{1}{(1+a)^{k_1}} \ge \frac{3}{2^{k_1}} \iff \left( \frac{\sum_{cyc} \left(\frac{2}{1+a}\right)^{k_1}}{3}\right)^{1/k_1} \ge 1 (*)  \]
Then power-mean shows
\[ \left(\frac{\sum_{cyc} \left(\frac{2}{1+a}\right)^{k_1}}{3}\right)^{1/k_1} \ge \left( \frac{\sum_{cyc} \left(\frac{2}{1+a}\right)^{k_0}}{3}\right)^{1/k_0} (**)  \]
Hence, if the right side of (**) is at least 1, which is equivalent to the desired for $k=k_0$, then certainly (*) holds. We have that $k = 2$ is valid because it is a special case of the well known problem
\[ \sum_{cyc} \frac{1}{(1+a)^2} \ge 1  \]
for $abcd = 1$. (Take $d=1$.)

dont we find $k\geq \frac{\ln 3}{\ln 2}$ from the InflectionPointTheorem(Myth) and RightConvexFuntionTheorem(Vasc)
($f(x)=\frac{1}{1+e^xk}$ is right convex and because $\sum lnx=0$we can apply the above theorems)


[quote="t0rajir0u"]Phone numbers.  (429) 496-73** probably belongs to some company.  Have you looked at the results more carefully?[/quote]

Well, that would probably only cause a spike for one number, though... and they're all continuous numbers, so that's ruled out...

Actually, large institutions frequently own a consecutive block of phone numbers.  (For example, my undergraduate university keeps trying to solicit donations from me, and all the calls come in the form (617) 384-55xx, but they "xx" part varies.  This is typical.

Unfortunately for this theory, area code 429 doesn't exist (at least not in the US).

[quote="nick42"]remind me why 4294967296 is such a popular number.  :?: [/quote]
It's one more than the largest number expressible using 32 bits, and IPv4 in particular uses 32 bits, so this is the total number of possible IP addresses in IPv4.

[quote="nick42"]btw, how bored are you?  :D[/quote]

Quite. I just had ECOO.


can someone please explain me how electron flows through the circuit ?
what assumptions..law...

Voltage difference and mediums through which such a difference can be noted on. Simply, a conductor. Electrons will move away from the negative charge, as likes repel, and towards the positive charge, as opposites attract.

Good ya?

can u be more specific in ur explaination and start with the field concept rather than the potential diff. concept

You can use a battery which establishes a potential difference across the circuit, as well as an associated electric field. The electric field is constant. Thus electrons experience an electrical force which leads them to the higher potential side of the battery. However, random collisions between electrons and atoms ultimately eliminate any net accelaration that the electrons experience, so they then flow through the circuit at what is known as their drift speed. The retarding force of the collisions is similar in many ways to the retarding force of friction when a projectile is traveling through air.

Suppose the field is not constant. Then electrons would travel at different speeds in different parts of the wire since they would be experiencing different accelerations (still valid even when we take into account the retarding forces). Then we would be able to measure the current in two different parts of the wire, and get two different results. But by experiment we know that the current is constant throughout (assuming the wire is identicle throughout).

but as far as i know if u measure the current through an ammeter then it's principle is based again on ohm's law whose 'proof'  again is based on the fact that the field is constant

No ohm's law is based off of experiment. It was first observed that $ I$ is proportional to $ V$ and then $ R$ is defined so that $ V=IR$.

well anyway in a normal wire the electrons always 'flow' with different speeds it is their average speed which we assume to be charctersitics of teh material, temperature....
but the current is same everywhere because the current $ I$ is directly propotional to $ v_{d}$ the drift speed of electrons


MATHCOUNTS probability geometry blogs AMC AIME ARML

Do anybody know any good mathcounts websites? I know goodly maths and MathCounts drills. Also what do I need to know? Thanks!  :lol:

By mathcounts drills, I'm assuming you mean http://mathcounts.saab.org/mc.cgi ?

Because that website is extremely helpful. I'd recommend that you do at least some of those every day.

And by what you need to know, that's sort of a vague question. I'd say that you should have a pretty good mastery of Algebra 1, know basic counting and probability, and basic geometry.

you need to be good at word problems and translating them into algebra.
You need to be really good at counting and probability and geometry.
you need to be fast
but most of all, you need to be lucky

MY 800th post :wow:  :wow:  :wow:

Make sure you give the correct units. IF they ask you for miles, don't asnwer in millimeters

I got the two minutes call for target at my latest chapter, and the 20 and 10 minutes calls at the sprint.

Last year we were allowed a timer but this time we weren't...

[quote="mathnerd314"][quote="moogra"]get really fast at solving problems. In countdown, you only need to rea dthe key words. Usually, the first few words aren't important[/quote]

Just make sure you aren't missing anything like "not" or "without replacement".

[quote="bryantacademy"]when you get the 10 minutes call (assumming you are going to MATHCOUNTS), guess and check is what you should do.[/quote]

Funny. All I've ever gotten was three-minute calls.[/quote]

Same here. I always got three minute calls. I been to the school round and chapter rounds so far this year.

We got 5 minute calls (maybe it was 10, can't remember) for sprint, and 10 second calls for target.

at nationals there was this giant digital clock with the amount of time left for all rounds

Officially, they say you get 3 minute calls, but sometimes the proctors improvise  :P .

Well, if there is a clock, look up every once in a while. Sometimes during the sprint round, your brain might need some resting up (for about 10 seconds), especially since it's 40 minutes long.

you mean that literally, right? i actually FELL ASLEEP during the school sprint round

AAAAAAHHHHHHH
NOW YOU KNOW MY NAME :nhl: :stink:

I wonder if someone can find me???  You have a 50% chance.

xxm0rg07hxx:are you karan? i don't think your homeschooled
biffandoc: i dont know if i made states, my chapter competition is saturday.

[quote="easyas3.14159..."]xxm0rg07hxx:are you karan? i don't think your homeschooled
[/quote]
Yep.  You also could have found out from the flag below my name on my posts.  Jonathan is clearly not an Indian name.  And I was in 7th grade last year.  Or you could see my blog, my name's in that.

Yes, I know your name, Tommy Gill.
MUHOHAHAHAHAHAHA :P
YOU'RE DOOMED!!!!!!!!!

AAAAHHHHH SAVE ME :nhl:

YOURE DOOMED FOR DOUBLE POSTING

are we playing the guess my name game?!?!?
fun!
try to guess mine. I think there's enough clues on this website already but you might have to take a while. No fair if you know me in real life!

You're Kate Rudolph, muahaha
bet no one could guess my name. :D

ur too fast!!!
I bet you also found a picture, and know what school I go to in what state and what grade...
why oh why did my mom give the mmy picture???

:lol: I just searched your posts, the one that said you were the only perfect scorer in Illinois sorta gave it away.

Hi, Mathletes:
Do you know these two:
http://jenksps.org/ms/Activity_n_Academics/MATHCOUNTS/Materials/Mathcounts%20prestate.doc.pdf
http://jenksps.org/ms/Activity_n_Academics/MATHCOUNTS/Materials/MATHCOUNT%20Bible%20full.pdf

I found them on Mathcounts.org Forum.
Printing them out.
Enjoy!

The Mathcounts Bible is an excellent resource; I used to study from it.  It will help you especially for Countdown, but for everything else too.  The first six pages, by the way, are the original version.  You need to know everything on the first six pages forwards and backwards except the first 100 triangular numbers and all the random fraction/decimal equivalencies.  If you know the first 10-15 triangular numbers and decimal equivalencies for everything with denominator up to 12, you'll be fine.

The last six pages of the Mathcounts Bible posted above are a lot of random stuff that you probably don't need to know.  Also, the tricks on that other page are nice to know, but they won't help you nearly as much as just learning to be fast at multiplying things, etc.  A good way to get fast at arithmetic is to learn to group things differently; for example, to multiply 35 by 16, rewrite it as 70*8, which is easy to do.

Some tips and tricks, a lot of which apply to competitions in general:
--I like the idea of having a predetermined number to guess on the Sprint round.  This saves you the temptation to go through and check your guess, or to see if it's reasonable, which takes too much time.  On the target round, though, since you've presumably at least read both problems, you should come up with a guess that sounds reasonable on the spot.
--If you're coming to the end of a round (i.e. the last minute or three) and it looks like you're not going to make any more progress, go back and check your work on other problems.
--The best way to check your work is not to go back through and check each step; it's much more reliable and efficient to start over, solve the problem again (preferably in a different way if possible) and make sure you get the same answer.
--After each round has ended, don't think too much about the problems from that round.  It's OK to discuss them a little bit with other students in between rounds, but you need to make sure that (a) you give your brain a rest, and (b) you are not distracted in future rounds by problems you couldn't solve (or answered wrong) in past rounds.
--Make sure you're answer is (a) what is asked for, and (b) sane.  If they ask for the number of feet, don't give the number of miles.  If they ask for the diameter, don't give the radius (this cost me an AIME point last year).  If they ask for an absolute value, don't write down a negative number.  If they ask for the number of ways to arrange n things given a certain constraint, your answer had better be smaller than n!, etc.  These are the kinds of things that you can check right after you finish the problem.
--On the team event, don't count a problem as checked unless two different people solved it completely independently and got the same answer.  If you're checking a problem and require help from the person that initially solved it, your check is a little bit tainted.

Also, know the prime factorization of the current year.  It doesn't come up a whole lot in Mathcounts (more in ARML), but it never hurts to know: 2006=2*17*59.

[quote="Sly Si"]Also, know the prime factorization of the current year.  It doesn't come up a whole lot in Mathcounts (more in ARML), but it never hurts to know: 2006=2*17*59.[/quote]

It also never hurts to know that the units digit of $7^{2006}=9$ and that of $3^{2006}$ is also 9.
Maybe even know that $2^{2006}$ ends with 4? $8^{2006}$ ends with 4 also?

That's y u don't give out clues as to ur true identity.
so far: biffanddoc=Tommy Gill
K8reindeer=Kate Rudolph

You can find pictures of some people at this site:
[url=http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2005-8a/05amc8perf.html]I'm so evil :D  :D[/url]


inequalities
induction
floor function
algebra
binomial theorem

For $ \{1,2,3,4...,n\}$ and each of its non-empty subsets a unique [b]alternating sum[/b] is defined as follows:Arrange the numbers in the subset in the decreasing  order and then,beginning with the largest,alternately substract and add succesive numbers.(for example, the alternating sum for $ \1,2,3\}$ is - $ 3 - 2 + 1 = 2$ and for $ \{n\}$ it is simply n. Find a formula for the sum of all the alternating sums of all the subsets.

the element $ k$ occurs in $ e_k\equal{}\binom{n\minus{}k}{0}\plus{}\binom{n\minus{}k}{2}\plus{}\cdots$ subsets as an even numbered element from the top, and in $ o_k\equal{}\binom{n\minus{}k}{1}\plus{}\binom{n\minus{}k}{3}\plus{}\cdots$ subsets as an odd numbered element from the top. but for $ k<n$, $ e_k\minus{}o_k\equal{}0$, a can be seen by expanding $ (1\minus{}1)^{n\minus{}k}$ using the binomial theorem. so the only element that contributes to the sum is $ n$, and it contributes $ n2^{n\minus{}1}$. ($ n$ occurs in $ 2^{n\minus{}1}$ subsets and is the largest element in each of them.)

Another way to think about it is that adding the number n to any subset changes all the numbers below it from positive to negative and vice versa, so they all cancel out.

from the above two posts my answer seems to be different  :o but this is wat i did...
[hide="my try"]
First let, us a consider the following set X : $ \{a,b,c,d\}$ with $ a > b > c > d$
(Note that the inequality signs are all strict as we are dealing with the set of first n natural numbers)
Let $ S_i$ be the alternative summation of all subsets of the above set with $ i$ elements
Let us go according to the elements of the subsets of set X :
[b]1 element[/b]:$ \{a\}$ $ \{b\}$ $ \{c\}$ $ \{d\}$
Therfore ,
$ S_1 = a + b + c + d ( = K)$

[b]2 element[/b]:$ \{a,b\}$ $ \{a,c\}$ $ \{a,d\}$ $ \{b,c\}$ $ \{b,d\}$ $ \{c,d\}$
Therefore ,
$ S_2 = (a - b) + (a - c) + (a - d) + (b - c) + (b - d) + (c - d) = 3a + 2b + c - b - 2c - 3d = 3a + b - c - 3d$

[b]3 element[/b]:$ \{a,b,c\}$ $ \{a,b,d\}$ $ \{a,c,d\}$ $ \{b,c,d\}$
Therefore ,
$ S_3 = (a - b + c) + (a - b + d) + (a - c + d) + (b - c + d) = 3a - b - c + 3d$

[b]4 element[/b]:$ \{a,b,c,d\}$
Therefore ,
$ S_4 = a - b + c - d$

Now add all these -
The Total Alternative Summation of Set X $ = \sum_{i = 1}^4 S_i = 8a$

For convenience sake , Let the Total Alternative Summation Of a Set X with Z elements and maximal element as $ a$ be expressed as $ F(X_a) = \sum_{i = 1}^Z S_i$
Similarly, we observe that

\begin{eqnarray*} F(X_a - \{d\}) = 4a \\
F(X_a - \{c,d\}) = 2a \\
F(X_a - \{b,c,d\}) = a \end{eqnarray*}

thus , We can conclude that -
[i]A set X with $ A$ as the maximal element and $ j$ elements has $ F(X_A) = 2^{j - 1}A$[/i]

Consider the Set X with $ A$ as the maximal element and $ j$ elements , we observe the base case for $ j = 1$ is satisfied so by induction hypothesis we have to prove that -
For a Set X' with $ A$ as the maximal element and $ j + 1$ elements  , $ F(X'_A) = 2^{j}A$
This can be proved as follows -
Let the new element added be - $ a_{j + 1}$.
Note that $ max\{j\} = A - 1$ because if $ j > A$ then A cannot be the maximal element as all the elements are taken from the set of first n natural numbers,
$ \{\underbrace{1,2,3,.....,A - 1},A,A + 1,.....,n\}$

Lets separate the new $ 2^{j + 1}$ subsets into 2 cases :
[u]Case 1[/u] : $ a_{j + 1}$ is absent.

In this case the alternative summation is the same as the previous case when $ j$ elements were there and so ,
$ F(Y_{A}) = 2^{j - 1}A$
where,
$ Y = X - a_{j + 1}$

[u]Case 2[/u] : $ a_{j + 1}$ is present

In this case,
the element $ a_{j + 1}$ is added as an extra element to every subset in the first Case.
Since the alternating summation is calculated from the LEFT it only matters to check whether the number of elements is odd or even and suitably $ a_{j + 1}$ is added or subtracted from the alternative summation of that subset.
If $ a_{j + 1}$ is added to a subset with $ 2k$ elements ,then it will be added in the [b]alternating summation of that set without $ a_{j + 1}$[/b]

If $ a_{j + 1}$ is added to a subset with $ 2k + 1$ elements ,then it will be subtracted in the [b]alternating summation of that set without $ a_{j + 1}$[/b]

Thus the alternative summation of the new set(Case 2 sets) would depend on how many times $ a_{j + 1}$ is added or subtracted.

Number of times $ a_{j + 1}$ is added
= Number of ways of choosing $ 2k$ subsets = $ \sum_{k = 1}^{\lfloor{\frac {j}{2}}\rfloor} \binom{j}{2k}$

Number of times $ a_{j + 1}$ is subtracted
= Number of ways of choosing $ 2k + 1$ subsets = $ \sum_{k = 1}^{\lfloor{\frac {j}{2}}\rfloor} \binom{j}{2k + 1}$

but we know that -
$ \sum_{k = 1}^{\lfloor{\frac {j}{2}}\rfloor} \binom{j}{2k} = \sum_{k = 1}^{\lfloor{\frac {j}{2}}\rfloor} \binom{j}{2k + 1}$
[hide="how?"]
In binomial theorem expansion of $ (x + a)^n$ put $ x = 1$ and $ a = - 1$.

Thus, the number of times $ a_{j + 1}$ is added = number of times $ a_{j + 1}$ is subtracted

Or,
the alternative summation for second case $ = 2^{j - 1}A + a_{j + 1}(\sum_{k = 1}^{\lfloor{\frac {j}{2}}\rfloor} \binom{j}{2k} - \sum_{k = 1}^{\lfloor{\frac {j}{2}}\rfloor} \binom{j}{2k + 1} = 2^{j - 1}A + a_{j + 1}(0) = 2^{j - 1}A$

Thus,
$ F(X'_A) = \sum_{i = 1}^{j + 1} S_i = 2^{j - 1} + 2^{j - 1} = 2^{j}$
and hence proved.
[/hide][/hide]

P.S. i first posted something else and finally it seems i was doing things unnecesarily and also... my answer is the same by the same method as pluritiq has posted above ... sorry..


Can 2 conjugated dienes react with each other like diels alder? If so which diene will add conjugately to the other? Please give an example if possible.

The simplest example is cyclopentadiene, which exists (when pure) as its cycloadduct. For different conjugated dienes, a simple rule is that the electron richer diene reacts with the most electron deficient double bond, or the most electron deficient diene reacts with the electron richer double bond. However, for a correct prediction, one have to consider the orbital coefficients of HOMO and LUMO of both reactants.


limit
inequalities
integration
function
Support
real analysis
real analysis solved

$f\in L^p(\mathbb{R}) (1\leq p<\infty)$.
$f_h(x) = f(x+h)$.
$r,s>0$ and $r+s=p$, Show that
$\lim_{|h|\to \infty} \|f_h^r\cdot f^s\|_1 = 0.$

A use of Hölder's Inequality gets us:

$||f_h^r\cdot f^s||_1\le \left(\int |f_h|^p\right)^{\frac{r}{p}}\left(\int |f|^p\right) ^{\frac{s}{p}} = ||f||_1^{\frac{r}{p}}||f||_1^{\frac{s}{p}}=||f||_1.$

We now need a "nice" dense subset of $L^1.$ There are many choices, but a simple one will do here: integrable functions of compact support are dense in $L^1$. That is, let $f_N=\chi_{[-N,N]}\cdot f.$ By the Dominated Convergence Theorem, $f_N\to f$ in $L^1.$

So, write $f=g+b$ where $g$ has compact support and $||b||_1<\epsilon.$

- I'll be back when I finish the details of this.

We might as well assume that all our functions are nonnegative; if not, replace them by their absolute values.

About that split of $f$ into $g+b$ where $g$ has compact support and $b$ is small: we can do that in such a way that the supports of $g$ and $b$ are non-overlapping, and then we not only have $f=g+b$, we also have $f^r=g^r+b^r.$ Thus:

$f_h^r\cdot f^s=(g_h^r+b_h^r)(g^s+b^s)= g_h^r\cdot g^s + b_h^r\cdot g^s + g_h^r\cdot b^s + b_h^r\cdot b^s.$

Now take the $L^1$ norm of $f_h^r\cdot f^s$:

$||g_h^r\cdot g^s||_1=0$ for sufficiently large $h$.

For the other pieces, we use the Hölder's inequality argument at the top of the previous post:

$||b_h^r\cdot g^s||_1\le ||b||_1^{\frac rp}||g||_1^{\frac sp} \le \epsilon^{\frac rp}||f||_1^{\frac sp}.$

$||g_h^r\cdot b^s||_1\le ||g||_1^{\frac rp}||b||_1^{\frac sp} \le \epsilon^{\frac sp}||f||_1^{\frac rp}.$

$||b_h^r\cdot b^s||_1\le ||b||_1^{\frac rp}||b||_1^{\frac sp} \le \epsilon.$

The sum of these last three pieces can be made arbitrarily small. Hence $\limsup_{h\to\infty}||f_h^r\cdot f^s||_1$ can be made arbitrarily small.

[quote="Kent Merryfield"]then we not only have $f=g+b$, we also have $f^r=g^r+b^r.$ Thus:[/quote]
Why do we have $f^r = g^r + b^r$?

[quote]Why do we have$ f^r = g^r + b^r$?[/quote]
Because $g$ was $f\cdot\chi_I$ for some interval $I$ and $b$ was $f-g.$ At every point, at least one of $g$ and $b$ is zero; hence when we raise $g+b$ to a power, we are raising only one of them to that power.[/quote]


define $\rho_{1}=\frac{1+\sqrt{5}}{2}$ and $\rho_{2}=\frac{1-\sqrt{5}}{2}$.

If $x=y$, $x\in\{\rho_{1},\rho_{2}\}$, and $x=z$ or $x+z=1=\rho_{1}+\rho_{2}$.

Now suppose $x>y$. Then if $x \geq \rho_{1}$ or if $x \leq \rho_{2}$, $y+1=x^{2}\geq x+1$ a contradiction.

If  $\rho_{1}>x>\rho_{2}$ you can find also a contradiction

Rather than "define", I'd suspect x=y=z from symmetry then plug in to get equations yielding those values.


calculus
integration
number theory
divisibility tests
puzzles

Determine the smallest  non-zero number constituted entirely by ones, twos and zeroes (but not consisting of any leading zeroes) which is divisible by 17670016.

What would be the smallest  non-zero number constituted entirely by ones, twos, threes and zeroes (but not consisting of any leading zeroes) which is divisible by 8835008?

This is a very nice question, indeed! So far, I haven't been able to come up with a 'clean' mathematical solution but here are two ideas that might be useful (N will denote the number 17670016):

N = 17670016 = 2*2*2*2*2*2*2*7*13*37*41 = (2^7)*7*13*37*41

1) First step in solving this problem should be to find some 'evidence' that there actually is SOME number satisfying the required properties and moreover, this usually also gives us some upper bound on the size of the result. In this case, this step is quite easy to perform.

Consider all the numbers consisting of just 1's (i.e. 1, 11, 111, 1111, ...) and calculate their remainders modulo N. As there are only finitely many possible remainders, sooner or later we'll find two numbers with the same remainder. Now, look at their difference and verify that it indeed satisfies all the properties of the problem statement ([b]well, except for being the smallest such integer[/b]). Obviously, we don't have to look at numbers consisting of more than [b]n[/b] digits, so we also
have some VERY weak upper bound on the solution.

In fact, in this particular case, the numbers consisting 37 ones and 7 ones give the same remainder. In anther words, a number consisting of 30 ones followed by 7 zeroes is divisible by our N. ([b]n=37[/b])

It means '1111111111 1111111111 1111111111 0000000' satisfies the question but we are far away from stating if that is the smallest.

Now that we know that we are not dealing with a problem-with-no-solution, we can try to find the solution using, for example, a brute-force approach (which still wouldn't look quick enough in this case; the bound we have obtained is still too high).

The first brute force approach may be getting all the multiples of N and checking if it satisifies the condition. We have to find out the least positive integral value of k such that 17670016*k < 1111111111 1111111111 1111111111 0000000.

The other way is to find out all the possible numbers ([b]at most 37 digits and less than 1111111111 1111111111 1111111111 0000000[/b]) consisting of only 1's, 2's and 0's and checking if they are divisible by N = 17670016.

I've written a small computer program which found out the actual solution consisting of eleven digits which is [b]22211210112[/b] = 17670016*1257.

Would let you know if I get through a 'more' mathematical answer.

I get 20223333312 for the second question, again by brute force (with some small help from knowing what the last few digits might be, using divisibility by 8 and stuff like that.).

Anyone have a nice way of solving these?

--Joshua Zucker


Let $ a,b,c$ be nonnegative real which sum $ 3$. Prove that

1/ $ \sqrt{\frac a{1\plus{}3b^2}}\plus{}\sqrt{\frac b{1\plus{}3c^2}}\plus{}\sqrt{\frac c{1\plus{}3a^2}}\ge\frac3{2}$

2/ $ \sqrt{\frac a{1\plus{}3a^2}}\plus{}\sqrt{\frac b{1\plus{}3b^2}}\plus{}\sqrt{\frac c{1\plus{}3c^2}}\le\frac3{2}$

[quote=dduclam]Let $ a,b,c$ be nonnegative real which sum $ 3$. Prove that
1/ $ \sqrt{\frac a{1\plus{}3b^2}}\plus{}\sqrt{\frac b{1\plus{}3c^2}}\plus{}\sqrt{\frac c{1\plus{}3a^2}}\ge\frac3{2}$
[/quote]
Holder work
\[\left(\sum \sqrt{ \dfrac{a}{1+3b^{2}}}\right)^2 \sum a^2(1+3b^2)(a+b+3c)^3 \geqslant \left[\sum a(a+b+3c)\right]^3.\]

[quote=dduclam]Let $ a,b,c$ be nonnegative real which sum $ 3$. Prove that

2/ $ \sqrt{\frac a{1\plus{}3a^2}}\plus{}\sqrt{\frac b{1\plus{}3b^2}}\plus{}\sqrt{\frac c{1\plus{}3c^2}}\le\frac3{2}$[/quote]
The Vasc's LCF Theorem helps here very well.


topology
real analysis
real analysis unsolved

Prove that $R$ is not a countable union of infinitely many close sets, any 2 of them having an empty intersection.

Let $F_1,F_2,\ldots$ be our closed sets. The set $\mathbb R\setminus F_1$ is open, so for every point it contains there's a neighborhood inside this set. We can find a point inside this set s.t. none of its neighborhoods are included in one of our closed sets (we easily get a contradiction otherwise). Now take such a point and a closed interval $I$ around it which is also included in $\mathbb R\setminus F_1$. Now repeat the process (look at $I\setminus F_2$, which is open in the subspace topology of $I$, choose a point s.t. none of its neighborhoods is included in just one $F_i$, take a closed interval around that point, and so on). We get a descending chain of intervals $I_n$ s.t. no point in $I_n$ belongs to any of $F_1,F_2,\ldots,F_n$. This descending chain of intervals has non-void intersection, which can't be in any of our closed sets $F_i$.

Does this work?

Here's the general version, due to Sierpinski:

Show that a compact connected topological space cannot be partitioned in countably many (but more than $1$) non-void closed sets.

I don't want to revive dead threads, but sam-n asked me if I had a proof of the theorem stated in the post above a few days ago, so I'm posting it.

First of all, let's show that a connected component $B$ of a proper closed subset $A$ of a connected compact Hausdorff space $X$ intersects the boundary of $A$.

Suppose the contrary. Then $B$ contained in the interior of $A$. For a compact space, the connected components are the same as the pseudocomponents (i.e. the connected component of $x$ equals the intersection of all the clopen sets containing $x$). We apply this to the subspace $A$ of $X$ (since $A$ itself is compact), so $B=\cap F_i$, where $F_i$ are all the clopen sets containg some $b\in B$. The interior of $A$ is an open set containing $\cap F_i$, so the border of $A$ is covered by the complements of the $F_i$'s, which means that it's covered by finitely many of them, forming a family $\mathcal F$. The intersection $F$ of the $F_i$'s which are complements of the elements of $\mathcal F$ is again clopen (in $A$, with the subspace topology inherited from $X$), and it's included in the interior of $A$. This means that $F$ is both the intersection between a closed subset of $X$ and $A$, and thus closed, and the intersection between an open subset of $X$ and the interior of $A$, and thus open. This means that $F$ is clopen in $X$, a contradiction (since $X$ is connected). This proves the assertion above.

Assume now that $X=\bigcup F_n$, where the $F_n$'s are closed sets which partition $X$ (if we assume that there are at least two non-void, then there are infinitely many, so we might as well assume that all of them are non-void). Cover $F_1,F_2$ with open sets $U,V$ respectively which do not intersect, and select a connected component $E_1$ of $\bar V$ which contains $x\in F_2\subset \bar V$. According to the above, this component intersects the border of $\bar V$, which is disjoint from $F_2$, and, since it also contains $x\in F_2$, it's a closed connected subset of $X$ which does not intersect $F_1$, partitioned into at least two closed subsets. Again, we select $E_2\subset E_1$ which does not inetrsect one of these closed sets, but intersects at least two others, and so on, thus getting $E_1\supset E_2\supset\ldots\supset E_n\supset\ldots$ s.t. $E_n\cap \left(F_1\cup\ldots\cup F_n\right)=\emptyset$. The intersection of all the $E_n$'s is non-empty and does not belong to any of the $F_n$'s, so we have a contradiction.

I think theres some problem in Grobber's sol.when you say"We get a descending chain of intervals  s.t. no point in  belongs to any of . This descending chain of intervals has non-void intersection, which can't be in any of our closed sets . "

Are you using Cantor's Intersection Theorem.


geometry
similar triangles
Pythagorean Theorem

A right triangle $ABC$ with hypotenuse $AB$ has side $AC = 15$. Altitude $CH$ divides $AB$ into segments $AH$ And $HB$, with $HB = 16$. The area of $\triangle ABC$ is:
[asy]
size(200);
defaultpen(linewidth(0.8)+fontsize(11pt));
pair A = origin, H = (5,0), B = (13,0), C = (5,6.5);
draw(C--A--B--C--H^^rightanglemark(C,H,B,16));
label("$A$",A,W);
label("$B$",B,E);
label("$C$",C,N);
label("$H$",H,S);
label("$15$",C/2,NW);
label("$16$",(H+B)/2,S);
[/asy]
$\textbf{(A) }120\qquad
\textbf{(B) }144\qquad
\textbf{(C) }150\qquad
\textbf{(D) }216\qquad
\textbf{(E) }144\sqrt5$

You are right minsoens. :)

We are to use Architas' theorem two times. :lol:

[hide="Answer"]Well, intuitively, I saw that $AH=9$, $CH=12$, $BC=20$ works... so $\frac{1}{2}\cdot 20\cdot 15=150\Rightarrow \boxed{C}$.[/hide]

YooSam how do we know that ach and chb exist as right triangles?(i really don't know geometry well).