combinatorics
IMO Shortlist
double counting

There are $10001$ students at an university. Some students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form several societies (a club may belong to different societies). There are a total of $k$ societies. Suppose that the following conditions hold:

[i]i.)[/i] Each pair of students are in exactly one club.

[i]ii.)[/i] For each student and each society, the student is in exactly one club of the society.

[i]iii.)[/i] Each club has an odd number of students. In addition, a club with ${2m+1}$ students ($m$ is a positive integer) is
in exactly $m$ societies.

Find all possible values of $k$.

[i]Proposed by Guihua Gong, Puerto Rico[/i]

Easy:

Suppose $A$ belongs to clubs $C_1,C_2...C_a$, then in these clubs A apears in $\sum|C_i|-1$ pairs, so we get this sum is equal to $10000$. and since every of these clubs is in $\frac{|C_i|-1}{2}$ societies, $A$ appears in $5000$ societies.

Now every Society contains $10001$ students, so we easily get, counting edges in the bypartite graph formed by students and societies, and an edge if a student goes in that society that $k=5000$

I think that you could probably eliminate one of conditions (1) and (2) and it would probably still be sufficient to solve the problem (Pascual's solution doesn't look like it really uses condition (2), for example).

Is there a proof that those conditions are met :?:

[hide="Bound"]Suppose there are $n$ clubs, and let club $C_i$ contain $a_i$ students and be contained in $b_i$ societies. Condition (i) gives
\[
\sum_{i=1}^n \binom{a_i}{2} = \binom{10001}{2}.
\]
Condition (ii) gives
\[
10001 \cdot k = \sum_{i=1}^n a_ib_i.
\]
Condition (iii) gives
\[
a_ib_i = a\left( \frac{a_i-1}{2} \right) = \binom{a_i}{2}.
\]
Thus the conditions together imply $1001\cdot k = \binom{10001}{2}$, or $k=5000$. [/hide]

[hide="Construction"]There is only one club containing all $10001$ students. There are $5000$ societies each containing this one club. All conditions are met. [/hide]

[hide="Comments"]In searching for a construction I tried to use condition (i) as follows: Consider $10001$ lines in a plane in general position; two lines intersect in one point. This suggests letting the intersection points be clubs and lines be students, where a club contains a student exactly when the point lies on the line. Condition (iii) says every intersection of two lines is in fact an intersection of $\ge 3$ lines (specifically, it is the intersection of an odd number of lines). I remembered a problem which stated that this implies all lines intersect in exactly one point; this is exactly the construction given above.

My question is: is there a more interesting construction? The wording of the problem suggests there should be "several" clubs, not just one. My interpretation suggests the above is the only construction, but it is by no means the only way of finding constructions... [/hide]

[hide="@Monkey"]Pascual's solution uses condition (ii) to conclude each society includes $10001$ students. If no club in some society included a particular student, this student and society would contradict condition (ii). [/hide]

Pretty straightforward problem.

Label the students $s_1, s_2, \cdots , s_{10001},$ the clubs $C_1, C_2, \cdots , C_n$ and the societies $S_1, S_2, \cdots , S_k.$ Let the number of students in club $C_i$ be $2f(C_i)+1,$ and let $g(S_i)$ denote the number of clubs in society $S_i.$

The last condition implies that club $C_i$ is in exactly $f(C_i)$ societies. Now, for any club $C_i,$ there are $\dbinom{2f(C_i)+1}{2} = f(C_i) (2f(C_i)+1)$ pairs of students. Since each pair of students is in a unique club, $\displaystyle \sum_{i=1}^{n} f(C_i) (2f(C_i)+1)$ must denote the total number of student pairs, so $\displaystyle \sum_{i=1}^{n} f(C_i) (2f(C_i)+1)   = \dbinom{10001}{2}.$ Now, each student is contained in exactly one club of a society, so a society contains exactly 10001 students. The total number of students (counted with multiplicity) in all societies is hence $10001 \times k.$ However, the sum of number of students of all societies is also equal to the sum of the number of students in each club times the number of times that club appears in all the societies, so $10001 \times k = \displaystyle \sum_{i=1}^{n} f(C_i) (2f(C_i)+1).$ Combining them, we easily get that $k=5000.$

For equality, set $f(C_1)=10001, f(C_2) = f(C_3) = \cdots = 0$ and $g(S_1) = g(S_2) = \cdots = g(S_{5000}) = 1$ where $C_1 \in S_i \quad \forall \ i.$

Standard problem. Simply double count the number of ordered triples (P,C,S) where student P is in a club C of society S. The second condition gives $T=10001k$ triples. The third condition gives $T$ is the sum over all clubs of $m(2m+1)$. To make use of the first condition, simply double count the number of triples (P1,P2,C) where students P1, P2 are in the same club C. You get: the sum over all clubs of (2m+1) choose 2 is equal to 10001 choose 2. Multiply both sides by 2 and equate with T to solve for

Easy:Count (student,club,society) such that student is in club and club is in society.

Nothing different, but I was solving this recently, so here goes :)

Let there be $t$ clubs $C_1 , C_2, \cdots C_t$.
Consider the set
$$X = \{ (s , C , S) | s \in C , C \in S, s \text{ is a student } , C \text{ is a club, and } S \text{ is a society.} \} $$

We calculate $|X|$ in two ways. First fix $s$ and $S$. From (ii), there is exactly one club $C_i$ for which $(s,C_i,S) \in X$. Thus
$$|X| = 10001k.$$

Now fix the club, say $C_i$. It has $|C_i|$ students, and it belongs to $\frac{|C_i| - 1}{2} $ societies from (iii). Thus
$$|X| = \sum_{i=1}^t \frac{|C_i|(|C_i|-1)}{2}.$$

Now consider the set
$$Y = \{( \{s_1 , s_2\} , C) | s_1 , s_2 \text{ are different students belonging to the club } C.\}$$

Again, we calculate $|Y|$ in two ways. First fix $C = C_i$. There are ${|C_i| \choose 2}$ such pairs of students in the club. Thus
$$|Y| = \sum_{i=1}^t {|C_i| \choose 2} $$

Now fix the pair $\{s_1 , s_2\}$. From (i), there is exactly one club $C$ for which $(\{s_1,s_2\} , C) \in Y$. Therefore
$$|Y| = {10001 \choose 2}$$

From all these relations, we have

$$10001k = |X| = \sum_{i=1}^t \frac{|C_i|(|C_i|-1)}{2} = \sum_{i=1}^t {|C_i| \choose 2} = |Y| = {10001 \choose 2}$$
so that
$$k = 5000.$$

To show that $k=5000$ indeed works, consider just $1$ club $C$ (that is, $t=1$) which belongs to $5000$ societies and such that all students are a member of this club. Clearly (i),(ii) and (iii) are satisfied.

Fix a student $T$ and let $C_1, C_2, \dots, C_r$ be all the clubs he is in. By the first condition, $$\sum _{i=1}^r |C_i| = 10000+r$$
From the second condition, each society has exactly all the students in the university. So $T$ appears $k$ times in all $k$ societies exactly once. Moreover, the clubs within each society is disjoint. So $C_1, C_2, \dots ,  C_r$ are in different societies.
From the third condition, observe that each $C_i$ appears in $\frac{|C_i|-1}{2}$ societies. So combining this with the previous equation,$$k = \sum _{i=1} ^r \frac{|C_i|-1}{2} = \frac{10000+r-r}{2} =5000$$

Now a construction for $k=5000$. Put all students into just one club and put this club in each of $5000$ societies. We are done.

Let us write n= 10001.Denote by T the set of ordered triples (a,C,S ), where a is a student, C a club, and S a society such that a∈C and C∈S . We shall count|T |in two different ways. Fix a student a and a society S. By (ii), there is a unique club C such that (a,C,S )∈T .Since the ordered pair(a,S ) can be chosen in nk ways,we have that|T |= nk. Now ﬁx a club C. By (iii), C is in exactly (|C|−1)/2 societies, so there are |C|(|C|−1)/2 triples from T with second coordinate C. If C is the set of all clubs, we obtain|T |= ∑C∈C |C|(|C|−1) 2 . But we also conclude from (i) that ∑ C∈C |C|(|C|−1) 2 = n(n−1) 2 . Therefore n(n−1)/2= nk, i.e., k = (n−1)/2= 5000. On the other hand, for k = (n−1)/2 there is a desired conﬁguration with only one club C that contains all students and k identical societies with only one element (the club C). It is easy to verify that (i)–(iii) hold.

[hide=Solution][quote=ISL 2004 C1]There are $10001$ students at an university. Some students join together to form several clubs (a student may belong to different clubs). Some clubs join together to form several societies (a club may belong to different societies). There are a total of $k$ societies. Suppose that the following conditions hold:

[i]i.)[/i] Each pair of students are in exactly one club.

[i]ii.)[/i] For each student and each society, the student is in exactly one club of the society.

[i]iii.)[/i] Each club has an odd number of students. In addition, a club with ${2m+1}$ students ($m$ is a positive integer) is
in exactly $m$ societies.

Find all possible values of $k$.
[/quote]
[b][color=#000]Solution:[/color][/b] Let $x_i$ denote the no. of students in Club $C_i$ and $y_i$ denote the no. of societies in which $C_i$ is contained. Thus Condition (iii) $\implies$ $x_i=2y_i +1$. Let no. of clubs be $n$. Double Counting the sets
\begin{align*} \mathcal{S} = \{ \left( C_i , \{ S_j, S_k \} \right) \} \implies \sum_{i=1}^n \binom{x_i}{2} = \binom{10001}{2} \end{align*}
\begin{align*} \mathcal{T} = \{ \left( C_i, S_j, s_k  \right) \} \implies 10001 k = \sum_{i=1}^n x_i y_i = \sum_{i=1}^n \frac{x_i (x_i -1)}{2} = \sum _{i=1}^n \binom{x_i}{2} \end{align*}
\begin{align*} \implies \binom{10001}{2} = 10001 k \implies \boxed{k= 5000} \qquad \blacksquare \end{align*}[/hide]

The value $k=5000$ is achieved if there is exactly one club, and $5000$ societies consisting of only that club. We show it's the only value.

Condition (i) implies \[ \binom{10001}{2} = \sum_{C \text{ club}} \binom{|C|}{2} \]
On the other hand, condition (ii) means that each individual society is a set of clubs whose disjoint union is all $10001$ students, so we have \[ k \cdot 10001 = \sum_{C \text{ club}} |C| \cdot \frac{|C|-1}{2} \] where $\frac{|C|-1}{2}$ comes from (iii).

From these two equations we see $k = 5000$ is the only possible value.

Let $C$ be the number of clubs and let $k_1 , k_2 , \dots , k_{C}$ be the number of students that each club has.
First, we count the number of triplets $(s_1 , s_2 , c)$ where $s_1 , s_2$ are students whose common club is $c$.
If we fix the students, there is only one way to choose $c$ so there are $\binom{10001}{2}$ triplets.
If we fix the $c$ first, then there are exactly $\binom{k_1}{2} + \binom{k_2}{2} + \dots + \binom{k_C}{2}$ triplets.
Thus, we have that $\sum_{i=1}^{C} \binom{k_i}{2} = 10001\cdot 5000$.
Now, we count the triplets $(s,c,t)$ where $s$ is a student, $c$ is student's one of the clubs, $t$ is this club's one of the societies.
If we fix the student and society, there is only one way to choose $c$ so there are $10001k$ such triplets.
If we fix the club, then there are exactly $\frac{k_1(k_1 - 1)}{2} + \dots + \frac{k_C(k_C - 1)}{2} = \sum_{i=1}^{C} \binom{k_i}{2} = 10001\cdot 5000$ such triplets.
Thus, we have $10001k = 10001\cdot 5000 \implies k=5000$.
It is easy to construct an example for $k=5000$. Let there be only one club and all the students are joined to this club, and this club is in every society. It is clear that this works and we are done. $\square$

We claim that the only possible value of $k$ is $\boxed{5000}$. This can be achieved if there is one club consisting of everybody, and that club is in $5000$ societies.

Consider a person $A$. Condition (i) implies that the set of clubs containing $A$ contain every other person exactly once. Since every club has an odd number of people, for every club containing $A$ the other students must come in pairs. However, notice that if a club containing $A$ contains $k$ other pairs of students, it will have $2k+1$ people overall and thus must be in $k$ societies.

There are $10000$ other people, and therefore $5000$ pairs of other people. Thus there are a total of $5000$ societies that the clubs containing $A$ are in. Since no society can contain two clubs with $A$ in them, all $5000$ of these societies are distinct. Furthermore, the presence of another society would imply that there is a society with no club containing $A$, a clear contradiction. The proof is complete.

ISL marabot solve

Consider a random student $P$ and let $S$ be the set of clubs $P$ is in. By the first condition, every other student is in exactly one of the clubs in $S$. Now consider some club in $S$. Suppose it has $p$ disjoint pairs of students excluding $P$. Then it must be in $p$ societies. Since there are totally $5000$ pairs, $P$ must be in $5000$ societies. The existence of another society implies that $P$ is not in it, a contradiction. So the only possible value of $k$ is $\boxed{5000}$. This is achievable if there is exactly one club that consists of everyone and it is in $5000$ societies.

success is 99% understanding the problem, 0.99% perspiration and 0.01% inspiration.

Let $S$ be the number of ways to pick a society, a club in that society, and a student in that club. Picking the society and the student, we know that there exists exactly one club that satisfies the conditions. Thus, $S=10001k.$

Now let the clubs have $c_1,c_2,\dots,c_n$ people, respectively. These clubs belong to $\frac{c_1-1}{2},\frac{c_2-1}{2},\dots,\frac{c_n-1}{2}$ societies, respectively. Thus, $S=\sum_{n}^{i=1}{\binom{c_i}{2}}.$ This is summing the number of pairs of people in any given club over all clubs. However, each pair of people is in exactly one club so this is just $\binom{10001}{2}.$ Thus, $k=5000.$ This is reachable with just one club with all the people, and this club is in all the societies.

The answer is $k=5000$. This is achievable with one club of all $10001$ people in $5000$ societies. This can easily be seen to work.
Let the clubs be $C_1,C_2, \cdots, C_k$ and the people be $P_1,P_2, \cdots , P_{10001}$. We count the number of triples $(P,C,S)$ where $S$ is a society, $C$ is a club in $S$, and $P$ is a person in $C$. Note that if we pick the club first we have that this sum is $$\sum_{C \text{ club}} |C_i| \cdot \frac{|C_i|-1}{2} = \sum_{C \text{ club}} \binom{|C_i|}{2}$$ using condition (iii). Next note that $$\sum_{C \text{ club}} \binom{|C_i|}{2} = \sum_{\text{unordered pair } (P_i, P_j) \text{ in club } C} 1 = \binom{10001}{2},$$ where the last equality follows from condition (ii). Next, if we pick the person and society first, the club is automatically chosen so the sum is $10001k$. This means $10001k=\binom{10001}{2}$, so $k=5000$.

Let $C_i,$ where $1\le i\le \ell,$ be the size of the clubs. By (i), we see $$\binom{10001}{2}=\sum_{i=1}^{\ell}\binom{C_i}{2}.$$ Notice by (ii), all students must be part of any society. Hence, if we want to choose a student, a club, and a society (such that the student is in the club is in the society), we can just choose the society and then choose any student, which we can do in $10001k$ ways (this would also determine the club by (ii)). Alternatively, we can proceed by casework on which club the student is in. For each club $C_i,$ we can choose $C_i$ students and $\frac{C_i-1}{2}$ societies by (iii). Thus, we see $$10001k=\sum_{i=1}^{\ell}C_i\cdot\frac{C_i-1}{2}.$$ This implies $k=5000.$ This works by letting all the students be in one club which is part of $5000$ societies. $\square$

Let there be $c$ clubs. For each club, let the number of members in it be $a_1,a_2,\dots,a_c$.
From the first condition, clubs contribute distinct pairs to all student pairs: $$\binom{10001}{2}=\binom{a_1}{2}+\binom{a_2}{2}+\dots+\binom{a_c}{2}.$$
From the second and third conditions, each club contributes $(2m+1)m$ student-society pairs: $$10001k=\binom{a_1}{2}+\binom{a_2}{2}+\dots+\binom{a_c}{2}.$$
We see that the two equations have equivalent right hand sides, so $k=\boxed{5000}$. This can be achieved by just having every student in $1$ club, which is in all $5000$ societies. QED.

We claim that the answer is $k=5000$. This can be achieved by having 1 club that contains all of the students and is in all 5000 societies.\\

Suppose that there are $n$ clubs, and let club $i$ have $2m_i+1$ students.\\

From condition (i), we have $${2m_1+1 \choose 2}+{2m_2+1 \choose 2}+\cdots +{2m_n+1 \choose 2}={10001 \choose 2},$$ since both sides are equal to the number of pairs of students.\\

Consider the number of triples $(a,b,c)$ such that club $b$ contains student $a$ and is in society $c$. By summing over all clubs, this is equal to $$m_1(2m_1+1)+m_2(2m_2+1)\cdots +m_n(2m_n+1).$$ However, condition (iii) tells us that this is also equal to $10001k.$ Therefore, we have $$m_1(2m_1+1)+m_2(2m_2+1)\cdots +m_n(2m_n+1)=10001k.$$ However, note that $m_i(2m_i+1)={2m_i+1\choose 2},$ so combined with our equation from condition (i), we have $${10001\choose 2}=10001k\rightarrow k=5000,$$ so we are done.

Double count pairs $(s, C, S)$ where student $s$ belongs in club $C$, which belongs in society $S$. For every pair $(s, S)$, there exists a unique $C$, so there are $10001k$ such tripltets. On the other hand, for every $C$, there exsit $m(2m+1)$ pairs for $2m+1$ students in club $C$. But then $$\sum_C {2m+1 \choose 2} = \sum_C m(2m+1) = 5000 \cdot 10001,$$ so thus we must have $k=5000$. For a construction, put all students in one club, which is part of all $5000$ societies.

This solution uses the deep fact that $\binom{2m+1}{2}=m(2m+1)$.

For a club $c$ let $s(c)$ denote the number of societies it belongs to. Then $\sum s(c)(2s(c)+1)=\sum \binom{2s(c)+1}{2}$ over all clubs $c$ will count both the number of pairs of students and the number of student-society pairs. There are $10001\cdot 5000$ of the former so we obtain $k=5000$ only. A construction is to create a single club which belongs for some reason to $5000$ societies and put every student into it.

combo is hard

The answer is $k = 5000$. Let the clubs be $C_1, C_2, C_3 ... C_i$.  Then condition i) tells us that \[  \dbinom{10001}2 = \sum \dbinom{|C_j|}2. \] We count the number of pairs $(p, q, r)$ such that student $i$ is in club $j$ with society $k$. Condition ii) and iii) together tell us that each club $C_m$ contributes $|C_m|(2|C_m|+1)$ to this number. By summing this across all clubs, we obtain \[\sum \dbinom{|C_j|}2 = 10001k \]. This implies that $k = 5000$, and a construction is to let all students be in $1$ club with $5000$ societies.

This problem made me so mad.

Number the clubs $1,2, \ldots, \ell$. Let $s_i$ and $t_i$ denote the number of students in the $i$'th club. Now, we will do the following genius calculations \[ 10001k = \sum s_it_i = \sum s_i \left ( \frac{s_i-1}{2} \right ) = \sum \binom{s_i}{2} = \binom{10001}{2} \implies k = 5000\] Construction for $k=5000$ is just one club with every student and part of $5000$ societies.

Let there be $n$ clubs such that the $i$th club has $2s_i+1$ members. We see that $\sum \dbinom{2s_i+1}2 = \dbinom{10001}2$ by the first condition. Consider the sum of the number of people across all societies. By the second condition, this is $10001k$ but by the third condition, this is also $\sum s_i(2s_i+1) = \sum \dbinom{2s_i+1}2 = \dbinom{10001}2$. Then it follows that $k=5000$.

bruh this one is so weird to wrap your mind around even though its standard double counting:

Notice that the number of pairs of students =10001 choose 2 are equal to choosing any club and taking two students, since those two students are only in this single club together, so we sum over all clubs=sum over 2m+1 choose 2=sum over m(2m+1). On the other hand, this is also equivalent to choosing a student and a society because any student-society makes up a distinct club in m(2m+1) ways. Hence 10001C2=2m+1C2=10001k, which readily implies k=5000! $\blacksquare$

Consider the set $\mathcal{A}=\{ (s , C , S_{0}) | s \in C , C \in S_{0}, s \text{ is a student } , C \text{ is a club, and } S_{0} \text{ is a society.} \}$\\

So since for each student and each society, the student is exactly in one club of the society we have $|\mathcal{A}|=10001k$. Also consider a particular club $C_{i}$ , denote $|C_{i}|$ as number of students in it.\\

So we have $|C_{i}|$ many students in a particular club $C_{i}$ , now there are $\frac{|C_{i}|-1}{2}$ societies. consider total such clubs to be be $n$ we have $\sum_{i=1}^{i=n} |C_{i}|\cdot \left(\frac{|C_{i}|-1}{2}\right)=|\mathcal{A}|$\\

Also, Notice for all clubs we have total pair of distinct students to be $\sum_{i=1}^{i=n} \binom{|C_{i}|}{2}$\\

and also clearly from $(i)$ there is exactly one club for which two distinct pair of students are in a club so we have $\binom{10001}{2}=\sum_{i=1}^{i=n} \binom{|C_{i}|}{2}$\\

so we have $10001k=\binom{10001}{2} \implies \boxed{k=5000}$. $\blacksquare$

edit: 850th post to this cool problem!  :coolspeak:

Let there be $n$ clubs, with the $i$th club having $a_i$ members. Then from the first condition, it must be that
\[ \binom{a_1}{2} + \binom{a_2}{2} + \dots + \binom{a_n}{2} = \binom{10001}{2} \]
But the second and third condition imply that
\[ \binom{a_1}{2} + \binom{a_2}{2} + \dots + \binom{a_n}{2} = 10001k \]
It immediately follows that $k = \boxed{5000}$. $\blacksquare$

Label the clubs $C_i$ and the societies $S_i$, with corresponding member counts $c_i$ and $s_i$. The first condition tells us the total number of pairings is
\[\binom{10001}{2} = \sum \binom{c_i}{2}.\]
The second and third conditions give us the total sum of member counts across all societies as
\[10001k = \sum s_i = \sum \frac{c_i-1}{2} \cdot c_i = \sum \binom{c_i}{2}.\]

Equating these, we find $k=\boxed{5000}$. $\blacksquare$

Let the clubs be $C_1, C_2, \ldots, C_j$, with club $C_i$ having $x_i$ members for some odd $x_i$. Then by the first condition, we have
$$\binom{10001}{2} = \sum_{i = 1}^j \binom{x_i}{2}.$$
By the second condition, we have
$$10001k = \sum_{i = 1}^j x_i \cdot \frac{x_i - 1}{2} = \sum_{i = 1}^j \binom{x_i}{2},$$
by considering the sum of the number of members across all $k$ clubs. Therefore $k = \tfrac{10000}{2} = 5000$. For a construction, take one and only one club which has all $10001$ students and is in $5000$ societies.


calculus
integration
trigonometry
function
logarithms
calculus computations

$ \int \frac {sin^{2}x}{1 \minus{} e^{x}}dx$OR$ \int \frac {sin^{2}x}{1 \plus{} e^{x}}dx$

Mathematica says this involves hypergeometric functions.

[quote="Timestopper_STG"]I think expand the denominator by using maclaurin series and then by parts each term should work.[/quote]

I don't think it's a good way. If we sub $ sin^{2}x\equal{}\left ( \frac{e^{ix}\minus{}e^{\minus{}ix}}{2i} \right )^{2}$,then do the intergral ,is that right ?

[quote="IndoChina"]centry57... looking at your previous posts, my guess is that your original problems are
$ \int_{ \minus{} \alpha }^{\alpha }\frac {\sin ^{2}x}{1 \plus{} e^{x}}dx$ and $ \int_{ \minus{} \alpha }^{\alpha }\frac {\sin ^{2}x}{1 \minus{} e^{x}}dx$
Right?[/quote]

I'm sorry ,I want to know $ \int\frac{sin^{2}x}{1\plus{}e^{x}}dx$

Sigh. All the people who think somehow that antiderivatives are more interesting than integrals. Look, this antiderivative exists, except at the places where the denominator is zero. We could always write that antiderivative as a variable-endpoint integral, such as $ \int_0^x\frac{\sin^2t}{1\plus{}e^t}\,dt.$

Now, as to whether that can be written in closed form as an elementary function ... I don't care all that much, but let me ask you this: Do you have some [i]a priori[/i] reason to believe that this antiderivative is an elementary function? Or did you just make it up as something you didn't immediately see a path for? Because if you just make up problems like that, then 99% of the time the answer is going to be "not an elementary function."

I say that not knowing whether that's the case here or not - but if you don't see a method, the odds are long against you. (Again, I don't even think it's all that interesting a question.)

Integrals - meaning definite integrals - make for much more interesting questions.

hello, Mathematica says
$ \int \frac {\sin(x)^2}{1 + e^x}\,dx =$
$ \frac {1}{8}(4x - ie^{ - 2ix}\text{Hypergeometric2F1}[2i,1,1 - 2i, - e^x] +$$ ie^{2ix}\text{Hypergeometric2F1}[2i,1,1 + 2i, - e^x] - 4\log[1 + e^x])$.
Sonnhard.

Which is a specific way of saying, "not an elementary function." Which is what I suspected all along.


inequalities
geometric inequality
inequalities proposed

[quote="gemath"]Given are $\triangle ABC$ prove that
$sin\frac{A}{2}+sin\frac{B}{2}+sin\frac{C}{2}+\frac{1}{2}(cos^{2}\frac{A}{2}+cos^{2}\frac{B}{2}+cos^{2}\frac{B}{2}-cos\frac{B}{2}cos\frac{C}{2}-cos\frac{C}{2}cos\frac{A}{2}-cos\frac{A}{2}cos\frac{B}{2})\le \frac{3}{2}$ :)[/quote]
Is anybody interested in?  :|  :P

an idea  ?:w00tb: or  :stretcher:  :thumbup:

I am bad at inequalities, so the stretcher applies to me :(
at least for now I will make it easier to read (at least for myself) \[\frac{3}{2}+\left(\mathbf{C}\frac{A}{2}\right)\left(\mathbf{C}\frac{B}{2}\right)+\left(\mathbf{C}\frac{B}{2}\right)\left(\mathbf{C}\frac{C}{2}\right)+\left(\mathbf{C}\frac{C}{2}\right)\left(\mathbf{C}\frac{A}{2}\right)\geq \mathbf{S}\frac{A}{2}+\mathbf{S}\frac{B}{2}+\mathbf{S}\frac{C}{2}+\frac{\mathbf{C}^{2}\frac{A}{2}+\mathbf{C}^{2}\frac{B}{2}+\mathbf{C}^{2}\frac{C}{2}}{2}\]

Thank me@home for your interest
[quote="gemath"]Given are $\triangle ABC$ prove that
$sin\frac{A}{2}+sin\frac{B}{2}+sin\frac{C}{2}+\frac{1}{2}(cos^{2}\frac{A}{2}+cos^{2}\frac{B}{2}+cos^{2}\frac{B}{2}-cos\frac{B}{2}cos\frac{C}{2}-cos\frac{C}{2}cos\frac{A}{2}-cos\frac{A}{2}cos\frac{B}{2})\le \frac{3}{2}$ :)[/quote]
Can Vasc or anyboy do it ? :D
Vasc, are you interested in geometric inequality, I have some hard problems in it, but maybe nobody is interested in !!  :(


[hide]
I used an equation.  X is the distance between home and work:
$\frac{x}{60}+\frac{x}{40}=2$
$\frac{2x}{120}+\frac{3x}{120}=2$
$\frac{5x}{120}=2$
$x=48$
So it took her $\frac{48}{40}=1.2$ hours, or $72$ minutes, to get back.[/hide]

Hope I did that right.

[quote="sharkman"]Rita drove from her home to work at an average speed of 60 miles per hour and returned home alone the same route at an average speed of 40 miles per hour.  If her total driving time for the trip was 2 hours, how many minutes did she take to drive from her job back home?[/quote]

[hide="Fast solution"]She spent $\frac{60}{40+60}$ of her time going home, so the answer is $2\cdot \frac{3}{5}= \frac{6}{5}$ hours, or $72$ minutes.[/hide]
This is more like mathcounts level...

[quote="sharkman"]Rita drove from her home to work at an average speed of 60 miles per hour and returned home alone the same route at an average speed of 40 miles per hour.  If her total driving time for the trip was 2 hours, how many minutes did she take to drive from her job back home?[/quote]

[color=white]I've seen like, 4 problems which are almost the exact same?  Including one in Classroom math or something?[/color]

[hide="My Half-Solution"]

$\frac{x}{40}+\frac{x}{60}=2$
$\frac{2x}{120}+\frac{3x}{120}=2$
$2x+3x=240$
$5x=240$
$x=48$
$\frac{48}{40}=\frac{12}{10}=1.2$
$72\mbox{ minutes}$[/hide]


[code]

a_{jk} =
\begin{cases}
M & \text{if} job r_j < k - p\\
w_j(k - d_j) & \text{if} r_j \geq k - p\\
\end{cases}
[/code]

I am using the above code (with dollar signs before and after, but I couldn't put those in here since it messed up the text), but it is giving me the errors of "Undefined Control Sequence" and "Misplaced Alignment tab character &." What am I doing wrong?
I am using Texnic Center.

It works fine here (not sure where 'job' came from) especially if you put a space after 'if':
$a_{jk} =
\begin{cases}
M & \text{if } r_j < k - p\\
w_j(k - d_j) & \text{if } r_j \geq k - p\\
\end{cases}$

In TeXnicCenter you need the amsmath package loaded to use the cases environment


Please post here ranking of Graduate school in Math of Netherlands. Thank you very much!  :)

Only Netherland or also Flanders (Belgium, next to it, speaking same language)?

And in what area? I mean the ranking is different for pure maths/applied maths/mathematical physics/... and even within each branch there's difference in which is the best (Algebra/Geometry/Combinatorics/Analysis/NumberTheory/...)

Tell me your field(s) of interest and I'll ask around which university fits you best. :)

[quote]
Only Netherland or also Flanders (Belgium, next to it, speaking same language)? [/quote]
All, if you can help me. :)

[quote="Peter"]
Tell me your field(s) of interest and I'll ask around which university fits you best. :)[/quote]
My fields: Number Theory, Algebraic Geometry, Non-commutative Geometry. Please post here websites of those schools, if you can.

Thank you very much.  :wink:

[quote="N.T.TUAN"]Number Theory[/quote]Leiden and Leuven are the two names popping to my mind here (asssuming you're leaning towards algebraic number theory).

[quote="N.T.TUAN"]Non-commutative Geometry[/quote]If you mean moufang quadrangles, generalized hexagons and alike geometry that relies heavily on non-commutative algebra, Ghent is your name.
(if that's not what you meant, I'll need more explanation)

[quote="N.T.TUAN"]Algebraic Geometry[/quote]I'm not sure about this one, I guess it is common and most universities are working on this?
(at least I know that here in Ghent we have strong people working on it, but no idea about other universities)


in spoila:

[hide] sin4x=a. cos8x=1-2sin:^2:4x, so cos8x=1-2a :^2: . [/hide]


AMC AIME MATHCOUNTS USA(J)MO USAMO Euler \/closed

under most of your names, you have some sort of conjecture, or hypothesis, or theory, and a number of question marks, do you guys just put that there? or do you have to do something to get that?

I think you need about 20 or 50 before the first level (P vs. NP) kicks in.

http://www.claymath.org/millennium/   is the website which can tell you about the specific 7 problems (and why they are important).

I have been online for only a short time. I don't know any netiquette. Can you teach me some? BTW I like the mandelbrot set.  I did a science fair project on fractals.

It is actually for my own good. I want to start having a habit of having good grammar and only posting in the My classes section.

I think, as he made it quite clear, he has a problem with your purposefully taking as your avatar an image that makes you appear to be at first glance another user of the forum.  There are hundreds of pictures of the mandelbrot fractal available, and you picked the only one that MCrawford uses.

If you want to work on your grammar and understandability, re-read everything you post here to make sure that it actually makes sense.

Ok thanyou tokenadult, and jbl. You guys just reminded me that avatars must be for a cause. Kerry must win the BIG DAY ON TUESDAY. So I'm putting a Bush poster and Kerry line.

I've heard a little joke going around that the Bush campaign should choose as its slogan "Bush-Cheney 2004:  Don't change horsemen mid-apocalypse."  ;)

What is AOPS?

[quote="tokenadult"][quote="churchilljrhigh"]instructors don't have them they have something that says instructors so i am churchill almighty i have both poincare conjecture and instrucotr under my belt.[/quote]

You really need to learn some netiquette and learn to stop trying to impersonate the AoPS instructors.[/quote]

[quote="JBL"]AoPS = "Art of Problem Solving"[/quote

How do people qualify to teach the online courses?

Haha.  You work for AoPS.  You can read about Richard and Mathew in the "about us" section of this website.

Once you start high school,  pre high school math competitions (mathcounts) still count for things?

[quote="churchilljrhigh"]OK you ace the AIME you win national mathcounts you get a p.h.d in mathematics you get an exceptional ranking in USAMO and you write a nyt bestseller math book.[/quote]

What is a conjecture?  How could they exist if they're not proven?

[quote="churchilljrhigh"]the number of posts you have you get a conjecture that is one of the seven famous not proven mathematical conjectures[/quote]

Not really; they may look good on your resume as with the aformenioned example.

A conjecture (in English) is something seemingly true, but hasn't been proven yet. They can exist because it could be a trivial fact to some people but can't/haven't been proven yet. (like 1+1=2 is trivial but I'm guessing most people can't "prove" that 1+1=2)

[quote="Mathhelp"]What is a conjecture?  How could they exist if they're not proven?[/quote]

The word "conjecture" basically means "educated guess." And in mathematics, a conjecture is a statement about what MAY BE a provable mathematical fact, for which no proof has been found. One of the most famous mathematical conjectures is Goldbach's conjecture, from a personal letter from Christian Goldbach (1690-1764) to Leonhard Euler (1707-1783), that every even number greater than or equal to 6 can be written as the sum of two odd prime numbers. BILLIONS of numerical examples have been examined by computer, witbout anyone finding any counterexample, but no one has come up with a proof to cover ALL even numbers greater than or equal to six, which is what is needed to turn this statement from being a conjecture to being a theorem.

The theorem proved by Andrew Wiles that is popularly known as Fermat's Last Theorem was really a conjecture until Wiles proved it, but because Fermat claimed to have a proof (which math historians are quite sure he didn't have), it was always informally referred to as a theorem, not as a conjecture. I would like to nudge popular usage in the direction of referring to this statement as the Fermat-Wiles Theorem, giving credit to both the person who posed the problem and the person who solved it.

[quote="ComplexZeta"]In general, the name of a theorem is the name of the poser rather than the name of the solver.[/quote]

That idea has all sorts of interesting implications.

Are you going to start posing tons of hard questions for future generations to solve? Fermat did lots of that. Euler (dis)proved most of them.


This diagram represents a thin belt stretched taut around three pulleys each 2 feet in diameter. The distances between the centers of these pulleys are 6,9, and 13 feet. Using 3.14 for $ \pi$, express the number of feet in the length of the belt to the nearest hundredth.

[asy]pair rotateleft(pair p){
    return (-p.y, p.x);
}
void drawCircle(pair center, real radius){
   draw(center+(radius,0)..center+(0,radius)..center-(radius,0)..center-(0,radius)..cycle);
}

pair normalize(pair p){
    return p/sqrt(p.x*p.x + p.y*p.y);
}

void drawTangent(pair left, pair right, real rad){
    pair delta = normalize(rotateleft(right-left))*rad;
    draw(left+delta--right+delta);
}

real rad = 1;
pair A=(0,0), B=(9,0), C=(12,-5);
drawCircle(A,rad);
drawCircle(B,rad);
drawCircle(C,rad);
draw(A--B--C--cycle,dashed);
drawTangent(A,B,rad);
drawTangent(B,C,rad);
drawTangent(C,A,rad);[/asy]

Let the vertices of the triangle be $ A$, $ B$ and $ C$.  The length of the belt is the perimeter of the triangle plus the arcs.  The total length of the arcs is $ \frac {3(360) \minus{} 6(90) \minus{} 180}{360} \equal{} 1$ times the circumference of one circle, so the arcs have length $ 2\pi$ and the perimeter of $ ABC$ is $ 6 \plus{} 9 \plus{} 13 \equal{} 28$, so the belt's length is
\[ 2\pi \plus{} 6 \plus{} 9 \plus{} 13\approx \boxed{34.28}
\]


Let $ x(t)$ be the solution of $ \frac {dx}{dt} \equal{} x^2 \plus{} t^3$ with $ x(0) \equal{} 0$. Show that for each $ 0 < t < 1$ we have $ x(t) < t^4 \plus{} t^9.$

It suffices to show is that $ y(t)\equal{}t^4$ is a "supersolution", i.e., $ y'\ge y^2\plus{}t^3$. This reduces to $ 4t^3\ge t^8\plus{}t^3$, which is trivial on $ [0,1]$. I'm pretty sure you meant something else. Perhaps, the equation was $ x'\equal{}x^2\plus{}4t^3$ (then you'll really need the $ t^9$ term though the solution remains exactly the same).


calculus
integration
calculus computations

I was tutoring today and couldn't show my student how to integrate this  :blush:

maple gives 1/4*exp(2*x)/(2*x+1) as an answer

setting exp(2*x)dx=dv will give increased powers of 1/(2*x+1) in another integral and going the other way gives ln(2*x+1) which doesn't seem to help me.

first change variable $u=2x$, and it becomes $\frac{1}{4}\int \frac{ue^u}{(u+1)^2}du$

\begin{eqnarray*}
\int \frac{ue^u}{(u+1)^2}du &=& \int \frac{ue^u}{(u+1)^2}d(u+1)\\
&=&-\int ue^ud\left(\frac{1}{u+1}\right)\\
\mbox{integration by parts}&=&-\frac{ue^u}{u+1} + \int \frac{e^u+ue^u}{u+1}du\\
&=& e^u(-\frac{u}{u+1}) + \int e^u du = e^u (1-\frac{u}{u+1})\\
&=& \frac{e^u}{u+1}
\end{eqnarray*}

well... wow

i don't understand the second step where 1/(1+u)^2 dissappears.

also this it a trick that my highschool calculus student won't understand...
do you have any simpler suggestions?  or a good way of explaining you solution.

okay..it is, in fact, changing of variable.

$d(\frac{1}{u+1}) = -\frac{1}{(u+1)^2}du$

It's just a notational thing.  The obvious thing is to integrate by parts to break off the $\frac {1}{(1+x)^2}$; do that and the indefinite integral becomes easy to find.


geometry
calculus
number theory
complex analysis

Hi, several of the threads here have me wondering what the word on the street is about scoring Olympiad-level competitions. I have read in a lot of places that the IMO explicitly is based on PREcalculus mathematics (although it presupposes a curriculum with much stronger geometry and number theory and combinatorics preparation than is typical in the United States curriculum). But if someone competing at the IMO had an extraordinary background in math, and knew how to use complex analysis (to give a fanciful example) to solve an Olympiad problem, would that be deemed a "correct" solution? Or is it only Olympiadly correct to use secondary school mathematics methods to solve the problems?

Just wondering, that's all.

You can use whatever you want to solve the problems. If you want to prove the Riemann Hypothesis on a contest and then use the Riemann Hypothesis in your solution, you'll still get full credit if it's correct.

Yeah, you can use calculus or whatever you want. There is one point bad in doing it though, in the fact that if you use a calculus solution and make a small slip, then they're quite likely to give you no marks. You don't often get part marks for something like that.

At MOP 2002, someone used Fourier Analysis to solve a problem.

I have wondered that myself on several occasions. I don't know the answer.

[quote="Fierytycoon"]Hmm, what are the qualifications of the IMO judges? What if some kid used math that was beyond the knowledge of any of the judges? :D[/quote]

[quote="JBL"]That is highly improbable.  The judges are mostly professional mathematicians.[/quote]

Interestingly, this question came up in a [url=http://groups.google.com/groups?&threadm=bmhlgi%24qfa%241%40oravannahka.helsinki.fi&rnum=1]thread on the Usenet sci.math newsgroup[/url], so it's at least a plausible concern to some mathematicians who read about Olympiad competitions in the popular press. I wouldn't lose sleep over it, though.

http://groups.google.com/groups?&threadm=bmhlgi%24qfa%241%40oravannahka.helsinki.fi&rnum=1

It is my understanding that at the IMO, scores are determined in the presence of the team leader, who can argue for points. If a judge is not aware of a certain theorem, the leader just has to present the theorem in a credible resource and they have to give credit.

Apparently this came up in 1994; one USA competitor wrote a one line proof citing a theorem identical to the problem statement which the judges were not aware of, but the US was able to find a book in which the theorem was published at a Hong Kong library, thus the competitor recieved full points. The details may be wrong, but something like that occurred.

Olympiads are designed so that all problems CAN be solved with precalculus mathematics, but they do not NEED to be; all mathematics is fair game.  I think that they try to avoid problems which would be easy with higher math but difficult without it.


conics ellipse integration vector calculus parameterization geometry

Compute [tex]\oint_c(y+z)dx + (z-x)dy + (x-y)dz [/tex] using Stoke's theorem, where c is the ellipse [tex]x(t) = asin^2t, \ y(t) = 2asintcost, z(t)  = acos^2t, 0\leq t \leq \pi [/tex]

the version of stoke's theorem I learned is:
[tex]
\int_c \overrightarrow{F} \cdot d\overrightarrow{r}
= \int_s curl \overrightarrow{F} \cdot d\overrightarrow{S}
=\iint_s curl \overrightarrow{F}\cdot \overrightarrow{n} \cdot dS
[/tex]

where S is the elliptical surface bounded by the curve c, F is a vector field and n is the unit vector pointing out at that point.

In this case, [tex]F = <y+z, z-x, x-y>[/tex], and I calculated curl F to be [tex]<-2, 0, -2>[/tex].

So we have to find

[tex]\iint_s <-2, 0, -2> \cdot \overrightarrow{n} \cdot dS [/tex]

How would I find [tex]\overrightarrow {n} [/tex] and dS, and also the bounds of integration for the double integral?

Note that the curve lies on the plane $x+z=a.$ It would be useful to rewrite the parameterization as:

$x=a\left(\frac12-\frac12\cos2t\right)$
$y=a\sin2t$
$z=a\left(\frac12+\frac12\cos2t\right)$

You have many choices for the surface, but if you have a chance to use a plane, you should, for the sake of simplicity.

This plane will be relatively easy to project to a two-dimensional integral.

If $z=g(x,y)$ we can write $\overrightarrow{dS}=\left(-\frac{\partial g}{\partial x}\overrightarrow{i}-\frac{\partial g}{\partial y}\overrightarrow{j}+\overrightarrow{k} \right)\,dx\,dy.$

(Or the negative of that, depending on the orientation of the curve.)

In our specific case, we have $\overrightarrow{dS}=(\overrightarrow{i} +\overrightarrow{k} )\,dx\,dy$

and, (using your calculation of the curl, which is correct)

$\text{curl}\overrightarrow{F}\cdot\overrightarrow{dS}=-4\,dx\,dy.$

Projected onto the $x$-$y$ plane, we have the curve $x=a\left(\frac12- \frac12\cos2t\right),y=a\sin2t.$ This is an ellipse centered at $\left(\frac a2,0\right)$ with semi-major axis $a$ and semi-minor axis $\frac a2,$ traced out clockwise in the $x$-$y$ plane. Since clockwise is "backwards," we must reverse the sign in $\overrightarrow{dS}.$ We now have

$\text{curl}\overrightarrow{F}\cdot\overrightarrow{dS}=4\,dx\,dy.$

Integrate this over the ellipse whose area in the $x$-$y$ plane is $\frac{\pi a^2}2$ and our final answer for the integral is simply $2\pi a^2.$


Θα γράψω άλλη μία άσκηση κάπως διαφορετική από την προηγούμενη.

Έστω τυχαίο τρίγωνο ΑΒΓ και το ύψος ΑΔ. Από το Δ φέρνουμε τις κάθετες ευθείες ε1 και ε2 προς τις πλευρές ΑΒ και ΑΓ αντίστοιχα. Έπειτα από το Β και το Γ κατασκευάζουμε τις ευθείες ε3 και ε4 αντίστοιχα οι οποίες είναι παράλληλες προς την ΑΔ. Οι ε1 και ε3 τέμνονται στο Ν ενώ οι ε2 και ε4 τέμνονται στο Μ. Να αποδείξετε ότι η ευθεία ΜΝ διέρχεται από το ορθόκεντρο Η του τριγώνου ΑΒΓ.

Την έλυσα με αναλυτική αλλά συνθετικά δεν έχω βρει ακόμα λύση και είναι πολύ αργά για να συνεχίσω τώρα.

Σημειωτεόν ότι δε χρησιμοποιώ ότι το Δ είναι ίχνος του ύψους γιατί είμαι 100% σίγουρος (και με αναλυτική το απέδειξα κιόλας) ότι ισχύει και στη γενικότερη περίπτωση όπου το Δ είναι οποιοδήποτε εσωτερικό σημείο της ΒΓ (οι ε3 και ε4 όχι παράλληλες στην ΑΔ αλλά κάθετες στη ΒΓ, το οποίο ισχύει φυσικά και σε αυτή την ειδική περίπτωση)

Μπορεί κάποιος να δώσει αναλυτικά μια συνθετική απόδειξη για τη γενική περίπτωση? :?:

Θα δωσω μια λυση (δυστυχως μονο στην ειδικη περιπτωση) με ομοιοτητες και Θαλη.

Φερνουμε το υψος BD' και δια του N παραλληλη προς αυτο που τεμνει την BC (την προεκταση της) στο Ν'
Τωρα ισχυει:
$ \angle{CMD} \equal{} \angle{C} \equal{} \angle{BNN'}, \angle{BND} \equal{} \angle{B}$
Αρα τα ορθογωνια τριγωνα $ BNN', DAC$ ειναι ομοια, οπως επισεις και τα $ BDN, DAB$
Αρα:
$ \frac{N'B}{AD} \equal{} \frac{NB}{DC} \Rightarrow N'B*DC \equal{} AD*NB$ (1)
και
$ \frac{NB}{BD} \equal{} \frac{DB}{AD} \Rightarrow AD*NB \equal{} BD^2$ (2)
Απο τις 2 τελευταιες σχεσεις προκειπτει:
$ N'B*DC \equal{} BD^2 \Rightarrow \frac{N'B}{BD} \equal{} \frac{BD}{DC}$, απο οπου λαμβανοντας υπ οψην οτι:
$ NN' \parallel BD' \parallel DM, NB \parallel AD \parallel MC$
προκειπτει οτι τα δυο υψη διαιρουν το ευθιγραμμο τμημα $ MN$ στον ιδιο λογο, αρα το σημειο τομης τους (δηλαδη το ορθοκεντρο) ανηκει στο τμημα αυτο.

Νομιζω οτι εχω και την γενικη περιπτωση

Αν $ K$ τυχαιο σημειο της $ BC$ και $ S, S'$ τα αντιστοιχα σημεια που οριζονται πανω στις $ {\epsilon}_3, {\epsilon}_4$, τοτε απο το $ S$ φερνουμε παραλληλη στο υψος $ BD'$ που τεμνει την $ BC$ στο $ S''$. Τοτε απο την ομοιοτητα των $ S''S'B, N'NB$ και των $ BND, BS'K$ προκειπτει ευκολα $ \frac {S''B}{BK} \equal{} \frac {N'B}{BD}$ και σε συνδιασμο με τις αναλογιες που βρηκαμε στην προηγουενη αποδειξη εχουμε τελικα: $ \frac {S''B}{BK} \equal{} \frac {BD}{DC}$, απο οπου οπως και στην ειδικη περιπτωση προκειπτει οτι τα υψη $ AD, BD'$ τεμνονται πανω στο ευθιγραμμο τμημα $ SS'$ :)

Πειτε μου αν χανει καπου η αποδειξη...

Το πρόβλημα αληθεύει επίσης και για τυχαίο σημείο $ H,$ στο εσωτερικό δοσμένου τριγώνου $ \bigtriangleup ABC,$ όπου $ (\epsilon_{1})\parallel CH,\ (\epsilon_{2}\parallel BH$ και $ (\epsilon_{3})\parallel AD\parallel (\epsilon_{4}).$

Αποδεικνύεται εύκολα με το [b][size=100]θεώρημα Θαλή,[/size][/b] ως άμεση εφαρμογή του.

$ \frac {MC}{ML} \equal{} \frac {DC}{DB} \equal{} \frac {NK}{NB}$ $ \Longrightarrow$ $ M,\ H,\ N,$ συνευθειακά σημεία και η πρόταση έχει αποδειχθεί.

Κώστας Βήττας.

Αυτά είναι... Κοντά ήμουν πάντως... Ωραίος Nick. Πολύ όμορφη η λύση. :)


function
complex analysis
complex analysis unsolved

Excuse me for being stupid, but I can't prove that the composition of two analytical functions is again analytical. Does the result remain true for entire functions? Could anyone post a proof, in case the assertions are correct? Thanks.

Doesn't it simply mean that it's complex differentiable?  So then shouldn't it be as easy to prove as the fact that the composition of two differentiable functions is differentiable? :? Or maybe you don't wnt to use the fact that it's analytic iff it's holomorphic?

By analytic, do you mean it can be represented by a power series and by entire, do you mean it is complex differentiable?  The reason I am asking is that most complex analysis books seem to consider them synonymous...

No, I do not have a simple proof, that's why I eventually realized that by analyic you actually meant the standard definition of analyticity.

[quote="harazi"]Grobber, do you have a simple proof for the fact that complex differentiable and analytic functions are the same? Sorry, it doesn't seem simple to me. Here by analytic I mean that $f$ is developable in entire series around each point from its domain.[/quote]

this is a simple consequence of Cauchy's theorem. You can find a simple proof in the book by H. Cartan: théorie élémentaire des fonctions analytiques.

Michael


Euler
limit
function
integration
logarithms
calculus
complex analysis

Prove: ${\gamma = \lim_{n \to+\infty}(n-\Gamma(\frac1n}))$.

${\lim_{n \to\infty}(n-\Gamma(\frac1n}))= \lim_{n \to\infty}\frac{1-\Gamma(1+\frac1n)}{\frac1n}=-\Gamma'(1).$

That doesn't solve the problem but it's an interesting way of restating it.

Here is a solution where I use complex analysis (after all, this is where $\Gamma$ really belongs).

First of, let $H(z) = \prod_{k=1}^\infty\left(1+\frac{z}{k}\right)e^{-z/k}$ be the canonical product associated to $-1,-2,-3,\ldots$. We know this is an entire function and that the product converges absolutely locally uniformly in $\mathbb{C}$. We have $H(0)=1$ and $H'(0) = 0$. So we get that $H(z) = 1+O(z^{2})$ and thus also $1/H(z) = 1+O(z^{2})$.

I don't know if this is well known, but in my complex analysis course we showed that $\Gamma\left(z\right) = \frac{e^{-\gamma z}}{zH(z)}$. This follows quite easily from the Stirling formula, but of course the Stirling formula is difficult to prove, so this solution might be a bit overkill.

Now $n-\Gamma\left(\frac{1}{n}\right) = n\left(1-\frac{e^{-\gamma/n}}{H\left(\frac{1}{n}\right)}\right) = n\left(1-\left(1-\frac{\gamma}{n}+O\left(\frac{1}{n^{2}}\right)\right)\right) \rightarrow \gamma$

[quote="Kalle"]Here is a solution where I use complex analysis (after all, this is where $\Gamma$ really belongs). [/quote]
Sure, and it's a nice solution. But I'll leave the topic right here rather than moving it, becuase I think there's still room for an entirely real solution.

[quote="Kent Merryfield"]becuase I think there's still room for an entirely real solution.[/quote]

I think I found such a solution.

Notice that
\[\begin{aligned}n-\Gamma\left(\frac{1}{n}\right) & = n \int_{0}^{+\infty}e^{-x}\,dx-\int_{0}^{+\infty}e^{-x}x^{\frac{1}{n}-1}\,dx \\ & = n \int_{0}^{+\infty}e^{-x}\,dx-\int_{0}^{+\infty}n x^{\frac{1}{n}}e^{-x}\,dx \\ & =-\int_{0}^{+\infty}\frac{x^{\frac{1}{n}}-1}{\frac{1}{n}}e^{-x}\,dx\end{aligned}. \]
Hence
\[\lim_{n \to+\infty}\left(n-\Gamma\left(\frac{1}{n}\right)\right) =-\lim_{n \to+\infty}\int_{0}^{+\infty}\frac{x^{\frac{1}{n}}-1}{\frac{1}{n}}e^{-x}\,dx. \]
It's obvious that
\[\lim_{n \to+\infty}\frac{x^{\frac{1}{n}}-1}{\frac{1}{n}}=\ln x. \]
Using the dominated convergence theorem, we get
\[\lim_{n \to+\infty}\int_{0}^{+\infty}\frac{x^{\frac{1}{n}}-1}{\frac{1}{n}}e^{-x}\,dx =\int_{0}^{+\infty}\lim_{n \to+\infty}\left(\frac{x^{\frac{1}{n}}-1}{\frac{1}{n}}e^{-x}\right)dx =\int_{0}^{+\infty}\left(\ln x\right) e^{-x}\,dx \] and it is well-known that this integral is equal to $-\gamma$.

(We can use $g$ as the "dominating function", where $g(x) = \sqrt{x}$ for $0 < x < 1$ and $g(x) = x^{2}$ for $x \geq 1$.)

So, we're done?

Arne I think it's not necessery to do all that in order to get $\underset{n\rightarrow \infty }{\lim }\left( n-\Gamma \left( \frac{1}{n}\right) \right) =-\int_{0}^{+\infty }\left( \ln x\right) e^{-x}dx$
Kent has proved that : $\underset{n\rightarrow \infty }{\lim }\left( n-\Gamma \left( \frac{1}{n}\right) \right) =-\Gamma^{\prime }\left( 1\right) =-\int_{0}^{+\infty }\left( \ln x\right) e^{-x}dx$
since $\left( \forall x>0\right) : \Gamma^{\prime }\left( x\right) =\int_{0}^{+\infty }\left( \ln t\right) t^{x-1}e^{-t}dt$

[quote="Arne"][quote="erdos"]$\left( \forall x>0\right) : \Gamma^{\prime }\left( x\right) =\int_{0}^{+\infty }\left( \ln t\right) t^{x-1}e^{-t}dt$[/quote]

Well, I didn't know that. Is it hard to prove?[/quote]
No, you just apply a theorem , I think in english it's called "DERIVATION UNDER THE INTEGRAL SIGN"
Try some google with it if you don't know it

The difference is really very minor. Kent uses the knowledge of a functional equation to skip a step, and if you want to apply derivation under the integral sign you still must motivate it with dominated convergence.

One can use the Weierstrass' formula :
\[\forall x > 0, \quad \frac 1{\Gamma(x)}= xe^{\gamma x}\prod_{n=1}^{+\infty}\left[\left(1+\frac{x}{n}\right)\right]\]

to prove that
\[\forall x > 0, \quad \frac{\Gamma'(x)}{\Gamma(x)}=-\gamma-\frac{1}{x}+\sum_{n=1}^{+\infty}\frac x{n(x+n)}.\]

Hence, for $x = 1$, since $\Gamma(1) = 1$,
\[\Gamma'(1) =-1-\gamma+\sum_{n=1}^{+\infty}\frac 1{n(n+1)}=-\gamma.\]


[quote="peregrinefalcon88"]I dont do any programs outside of school and mathcounts so i only have an academic recomendation and just got a second recomendation from my 7th grade teacher is that ok? I mean i think i got 5 test questions and i dont want to not go to camp just because i dont have an outside recomendation.[/quote]

Can you get a second recommendation from someone who knows you in a non-academic context?  That doesn't mean Mathcounts.  It could be a friend of the family, or any adult who knows you outside of school.  If you absolutely don't have anyone outside of school that's OK, but try to think more broadly before you resort to another teacher!

I have already gotten a recomendation from another teacher but I really dont know any of my moms friends. She has so many of them and all i really have to do when they come over is say hello and goodbye. I do have some friends from school but the only friends whose parents i know dont speak english so i dont think i can ask their parents to write me a recomendation either.

Do you guys think parents can be recommenders?

it can also be a sports coach .... if you play any sort of sports...
i think [private] music teacher might work?

Let me make this clear before you all give more suggestions

I wake up then go to school, go to mathcounts on mondays, come home, do homework and anything else my mom needs then do whatever i want if i have time. Thats my average day.

[quote="E^(pi*i)=-1"]I got my non-academic recommendation from my next-door neighbor.  :|

Out of curiosity, is your Mathcounts coach the one writing the academic recommendation?  If you must use a teacher for the non-academic letter, they would probably be the best choice.

If that fails . . . the Mathcamp staff is a pretty polyglot group  :) .  What language do your friend's parents speak?[/quote]

These are all great suggestions.  If all else fails, use the MathCounts coach as a nonacademic recommendation and make sure to give them the instructions for the nonacademic recommender!

Good luck!

My moms friend speaks chinese but i gave my nonacademic recommendation to my 7th grade math teacher and my academic recomendation to my mathcounts teacher just because my mathcounts teacher probably knows me better academicly and i already asked my 8th grade teacher for a recomendation so i dont want to bother her again but it shouldnt make a difference because there are so few people in almost every class in my school who dont talk during the lesson :P

[quote="JasmineMoMo"]What if my badminton coach happens to be my social sciences teacher? does that count? coz he wrote a bunch of stuff about my academic too[/quote]

That's fine, so long as he talks about out-of-school stuff as well.

[quote="e=mc2.kroll"]my "outside" recommendation is from my history teacher, since her period with me is probably the time of the day where i am least concerned with math^^
But i assume that since she has probably the best shot at assessing my non-math related maturity it shouldn't be that big of a deal.

If im wrong, tell me, I can get more people to send in recs.[/quote]

Our goal is more "out of school" than "non-math;" it would be nice if you could provide one from someone who's interacted with you outside of school.


linear algebra
matrix
vector
superior algebra
superior algebra solved

assume that $A,B$ be orthognals matrices and we know that $det(A)+det(B)=0$ prove or disprove that $det(A+B)=0$

The determinant of a real orthogonal matrix is either 1 or -1, so the condition that $\det A + \det B = 0$ is the same as $\det A \det B = -1$, or $\det A^TB = -1$. The eigenvalues of an orthogonal matrix are either 1, -1, or non-real numbers of magnitude 1. The last come in conjugate pairs with product 1, so any orthogonal matrix with determinant -1 must have some eigenvalue be -1.
Since $A^TB$ is orthogonal with determinant -1, there is some vector $x$ such that $A^TBx = -x$. This gives $Bx=-Ax$, so $(A+B)x=0$ and $\det(A+B)=0$.


calculus
integration
inequalities
function
Support
analytic geometry
graphing lines

f:R--->R a function $ C^{2}$
with

$ \int_{R}f^{2}, \int_{R}f''^{2}$ are finite

prove that $ \int_{R}f'^{2}\leq \sqrt{\int_{R}f^{2} \int_{R}f''^{2}}$

If $ f$ also has compact support, then we can use integration by parts $ \int_{\mathbb R} f'(x)f'(x)dx\equal{}\minus{}\int_{\mathbb R} f(x)f''(x)dx$ followed by Cauchy-Schwarz. In general we get a boundary term $ ff'\bigg|_{a}^b$, which does not necessarily tend to zero as $ a\to\minus{}\infty$ and $ b\to\infty$. However, all we need is $ \liminf_{x\to\infty} |f(x)f'(x)|\equal{}0$, which can be established as follows. If $ f(x)f'(x)\ge \epsilon>0$ for all $ x>b$, then the slope of $ f^2$ in this interval is at least $ 2\epsilon$, which contradicts the finiteness that $ \int_b^{\infty}f(x)^2\,dx$. If $ f(x)f'(x)<\minus{}\epsilon$ for all $ x>b$, then $ f^2$ must eventually become negative, a contradiction. Thus $ \liminf_{x\to\infty} |f(x)f'(x)|\equal{}0$.


geometry
circumcircle
angle bisector
geometry unsolved

Let $ D$ be the point on the base $ BC$ of an isosceles triangle $ ABC$ such that $ BD : DC \equal{} 2 : 1$, and let $ P$ be the point on the segment $ AD$ such that $ m(BAC) \equal{} m(BPD)$. Prove that $ m(DPC) \equal{} m(BAC)/2$.

[hide]Let the circumcircle of triangle $ BPD$ intersects side $ AB$ at $ E$ ($ E\ne B$) and let $ F$ be the point on side $ AC$ such that $ AE\equal{}AF$. Also let $ \angle ABC\equal{}\angle ACB\equal{}x$. Note that $ \triangle AEF$ is also an isosceles triangle, so $ \angle AEF\equal{}\angle AFE\equal{}x$. We have $ \angle APE\equal{}180^{\circ}\minus{}\angle EPD\equal{}\angle ABC\equal{}\alpha$. Therefore $ A,F,P,E$ lie on a circle. By Miquel's Theorem, $ F,C,P,D$ also lie on a circle. Thus $ \angle DPC\equal{}\angle DFC$. Let $ M$ be the midpoint of $ BC$. Note that $ \angle DEB\equal{}\angle DPB\equal{}\angle CAB$, so $ DE\parallel AC$, and hence $ AE: AB\equal{}CD: CB\equal{}1: 3$. Hence we have $ CD: CM\equal{}2: 3\equal{}CF: CA$, so $ FD\parallel AM$, which gives $ FD\perp DC$. So $ \angle DFC\equal{}90^{\circ}\minus{}x$. Hence $ \angle DPC\equal{}90^{\circ}\minus{}x\equal{}\frac{180^{\circ}\minus{}2x}2\equal{}\frac{\angle BAC}2$. Q.E.D.[/hide]

If $ AD$ intersects the circle $ (ABC)$ at $ \{ A, M \}$ and $ E$ is the midpoint of $ BM$, the triangle $ \triangle BPM$ is isosceles, $ MD$ is the angle bisector of $ \angle BMC$, hence $ MC \equal{} ME$, and $ PEMC$ is a kite, but $ PE$ is the angle bisector of $ \angle BPD$, done.

Best regards,
sunken rock

P.S. Very nice solution, [b][i]Johan[/i][/b]!

We can see that $ P$ is the intersection of $ AD$ with the circle $ (ABQ)$, $ Q$ being the intersection of the tangents to the circle $ (ABC)$. From the equality $ \angle AQP \equal{} \angle ABP \equal{} \angle PAC$ we conclude that the quadrilaterals $ AQBP$ and $ CABE$ ( $ E$ being the intersection of $ AD$ with the circle $ (ABC)$ ) are similar, hence $ \frac{AP}{PB}\equal{}\frac{CE}{BE}$, but $ \frac{CE}{BE}\equal{}\frac{CD}{BD}\equal{}\frac{1}{2}$, consequently $ BP\equal{}2\cdot AP$ ( 1 ).
If  $ M$ is the midpoint of $ BP$, then $ \triangle ABM \equal{} \triangle CAP$,  $ \measuredangle AMP\equal{}\measuredangle CPD$ ( 2 ).
But from $ AP \equal{} PM$, $ \measuredangle BAC\equal{}\measuredangle DPB \equal{} 2 \cdot \measuredangle AMP\equal{}\measuredangle CPD$, hence $ \measuredangle CPD \equal{}\frac{\measuredangle BAC}{2}$.

Best regards,
sunken rock

The [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=39978]link[/url] contains the inverse of the problem asked here.

My solution:

Let $ Q \in BP $ satisfy $ BQ=AP $ .

Since $ \angle QBA=\angle BPD-\angle BAD=\angle BAC-\angle BAD=\angle PAC $ ,
so we get $ \triangle ABQ \cong \triangle CAP $ . ... $ ( \& ) $
Since $ \text{S}[\triangle ABP]: \text{S}[\triangle ACP]=BD:DC=2:1 $ ,
so combine with $ ( \& ) $ we get $ Q $ is the midpoint of $ BP $ and $ \angle QAP=\angle PQA $ ,
hence $ \angle BAC=\angle BPD=2\angle PQA=2\angle DPC $ .

Q.E.D

Here is my solution: (I am not yet acquainted with LateX so I posted a pic of my solution, I am learning LateX)


trigonometry
geometry
angle bisector
similar triangles
projective geometry
geometry proposed

Let $ ABC$ be a right triangle with $ \angle BAC\equal{}90^{\circ}$. Let $ AD$ be the angle bisector of $ \angle BAC$, $ K$ point on $ AD$ such that $ \angle BKA\equal{}90^{\circ}$ and $ H$ the foot of $ A$-altitude. Let $ X$ be the intersection point of $ AB$ and $ HK$. Prove that the intersection point of $ DX$ and $ AC$ is the midpoint of $ AC$.

1)suppose that $ AC<AB$ by menalus theorem's in triangle $ ABC$ and line $ (XD)$ : $ (*) \frac{XA}{XB}.\frac{AB}{AC}.\frac{MC}{MA}=\frac{XA}{XB}.\frac{DB}{DC}.\frac{MC}{MA}=1$, where $ M$ is the intersection of $ AC$ and $ XD$ .
since $ ABKH$ is cyclic we have :`
$ <HXA=180-(<XAH+<XHA)=180-((180-<C)+(<ABK))$=$ <C-45$.(sine $ <ABK=45$).

now by sin law in triangle $ XHA$ ,$ \frac{XA}{sin(45)}=\frac{HA}{sin(<C-45)}$ and  we can easily to find $ XA(AB-AC)
=AB.AC$ and so

$ XA.AB=(XA+AB).AC=XB.AC$  and we get the result from ($ *$)

2)if $ AC>AB$ by the same way we find $ <HXB=<B-45$ and by sin law in triangle $ HXB$ :$ \frac{XB}{sin(45)}=\frac{HB}{sin(<B-45)}$.we continue as the previous case .

3)if $ AB=AC$ there are nothing to prove since $ H=K=D$.

Here's my proof. Assume $ AC\neq AB$. Let $ M$ be the midpoint of $ BC$ and $ P$ the intersection point of $ AM$ and $ BK$. We know that $ ABKH$ is cyclic, hence $ \angle HAK\equal{}\angle HBK$. Now using well known fact $ \angle HAD\equal{}\angle MAD$ we see that $ ABPD$ is also cyclic. Let $ E$ be the intersection point of $ PD$ and $ AC$. By angle chasing $ \angle CAH\equal{}\angle CBA\equal{}\angle BDP\equal{}\angle HDE$ hence $ ADHE$ is cyclic. Now we can write $ \angle AHE\equal{}\angle ADE\equal{}45^{\circ}$ and $ \angle BHK\equal{}\angle BAK\equal{}45^{\circ}$ so $ K,H,E$ are collinear and $ PE\perp AC$. Using Menelaus theorem $ \frac{AX\cdot BD\cdot CF}{BX\cdot CD\cdot AF}\equal{}1$ (*) hence it suffices to prove that $ \frac{AX}{BX}\equal{}\frac{AC}{AB}$. Using Menelaus again we obtain $ \frac{AX\cdot BH\cdot CE}{BX\cdot CH\cdot AE}\equal{}1$ (1). Now using similar triangles $ \frac{BH}{AH}\equal{}\frac{AB}{AC}$ and $ \frac{AH}{CH}\equal{}\frac{AB}{AC}$ hence $ \frac{BH}{CH}\equal{}(\frac{AB}{AC})^2$ (2) and $ \frac{CE}{AE}\equal{}\frac{CD}{BD}\equal{}\frac{AC}{BC}$ (3). From (1), (2), (3) we get $ \frac{AX}{BX}\equal{}\frac{AC}{AB}$ and now using it in (*) we get $ AF\equal{}CF$ :)

$ \triangle\ ABC\ \ ,\ \ \left\|\begin{array}{ccc} D\in (BC) & , & \widehat {DAB}\equiv\widehat {DAC} \\
\ K\in AD & , & BK\perp AD \\
\ H\in BC & , & AH\perp BC \\
\ X\in AB\cap HK & , & Y\in AC\cap XD\end{array}\right\|$ $ \implies\boxed {YA \equal{} YC\Longleftrightarrow A \equal{} 90^{\circ}}$ .

[color=darkblue][b][u]Proof.[/u][/b] Apply the [b]Menelaus' theorem[/b] to the transversal $ \overline {XDY}$ and $ \triangle ABC\ \ : \ \ \frac {XB}{XA}\cdot\frac {YA}{YC}\cdot\frac {DC}{DB} \equal{} 1$ .

Thus, $ \boxed {\ YA \equal{} YC\ }$ $ \Longleftrightarrow$ $ \frac {XB}{XA} \equal{} \frac {DB}{DC}$ $ \Longleftrightarrow$ $ \frac {KB}{KA}\cdot\frac {\sin\widehat {XKB}}{\sin\widehat {XKA}} \equal{} \frac cb$ $ \Longleftrightarrow$ $ \tan\frac A2\cdot \frac {\cos B}{\sin B} \equal{} \frac cb$ $ \Longleftrightarrow$

$ \tan\frac A2 \equal{} \frac {\sin C}{\cos B}$ $ \Longleftrightarrow$ $ \frac {\sin A}{1 \plus{} \cos A} \equal{} \frac {\sin C}{\cos B}$ $ \Longleftrightarrow$ $ \frac {a}{1 \plus{} \cos A} \equal{} \frac {c}{\cos B}$ $ \Longleftrightarrow$ $ a\cos B \equal{} c\cdot (1 \plus{} \cos A)$ $ \Longleftrightarrow$

$ c\cdot\cos A \plus{} (c \minus{} a\cdot\cos B) \equal{} 0$ $ \Longleftrightarrow$ $ c\cdot\cos A \plus{} b\cdot\cos A \equal{} 0$ $ \Longleftrightarrow$ $ (b \plus{} c)\cdot\cos A \equal{} 0$ $ \Longleftrightarrow$ $ \boxed {\ A \equal{} 90^{\circ}\ }$ . [/color]

[hide]We can say that the problem is also true, for an arbitrary point $ D,$ on the hypotenuse $ BC$ of the given right triangle $ ABC.$

So, the point $ X$ is the harmonic conjugate of $ E,$ with respect to the points $ A,\ B,$ where $ E\equiv AB\cap DP$ and $ P\equiv AH\cap BK.$

Hence, we conclude that the pencil $ D.XBEA$ is also in harmonic conjugation and then, its ray $ DX,$ bisects the side-segment $ AC,$ because of $ AC\parallel DE$ and the proof is completed.[/hide]

Kostas Vittas.

Thank you for your nice solutions and generalizations :)

[color=darkblue][size=117][b]Very nice the Vittasko's extension and proof ![/b][/size][/color] [color=darkblue]Here is my proof in the general case.[/color]

$ \triangle\ ABC\ \ ,\ D\in (BC)\ ,\ \left\|\begin{array}{ccc} K\in AD & , & BK\perp AD \\
\ H\in BC & , & AH\perp BC \\
\ X\in AB\cap HK & , & Y\in AC\cap XD\end{array}\right\|$ $ \implies\boxed {YA \equal{} YC\Longleftrightarrow A \equal{} 90^{\circ}}$ .

[color=darkblue][b][u]Proof.[/u][/b] Denote $ m(\widehat {BAD}) \equal{} \phi$ . Apply the [b]Menelaus' theorem[/b] to the transversal $ \overline {XDY}$ and $ \triangle ABC\ :$

$ \frac {XB}{XA}\cdot\frac {YA}{YC}\cdot\frac {DC}{DB} \equal{} 1$ . Thus, $ \boxed {\ YA \equal{} YC\ }$ $ \Longleftrightarrow$ $ \frac {XB}{XA} \equal{} \frac {DB}{DC}$ $ \Longleftrightarrow$ $ \frac {KB}{KA}\cdot\frac {\sin\widehat {XKB}}{\sin\widehat {XKA}} \equal{} \frac {AB}{AC}\cdot\frac {\sin\widehat{XDB}}{\sin\widehat {XDC}}$ $ \Longleftrightarrow$

$ \tan\phi\cdot\frac {\cos B}{\sin B} \equal{} \frac {\sin C}{\sin B}\cdot\frac {\sin\phi}{\sin (A \minus{} \phi)}$ $ \Longleftrightarrow$ $ \cos B\sin (A \minus{} \phi ) \equal{} \sin C\cos\phi$ $ \Longleftrightarrow$

$ \sin (A \plus{} B \minus{} \phi ) \minus{} \sin (B \minus{} A \plus{} \phi ) \equal{} \sin (C \plus{} \phi ) \plus{} \sin (C \minus{} \phi )$ $ \Longleftrightarrow$ $ \sin (B \minus{} A \plus{} \phi ) \plus{} \sin (C \minus{} \phi ) \equal{} 0$ $ \Longleftrightarrow$

$ \sin\frac {B \plus{} C \minus{} A}{2}\cos\frac {A \plus{} C \minus{} B \minus{} 2\phi }{2} \equal{} 0$ $ \Longleftrightarrow$ $ \cos A\sin (B \plus{} \phi ) \equal{} 0$ $ \Longleftrightarrow$ $ A \equal{} 90^{\circ}$ .

 [/color]

Let $ Y\equal{}HK\cap DX$, We can prove $ DX//AC$, Then, using pencil D.


Prove that the internal bisectors and the external angle bisector of the third angle meet the opposite sides in three collinear points.

anyone interested in this ?
it is quite easy but i want to see the approach of other people

Let $M,N$ the feet of the internal bisectors of $ B,C$ and $ L$ the foot of the external bisector of $ A.$ By angle bisector theorem we have

$\frac {BN}{NA} \equal{} \frac {a}{b} \ ,\ \frac {AM}{MC} \equal{} \frac {c}{a} \ ,\ \frac {LC}{LB} \equal{} \frac {b}{c}$

$ \frac {LC}{LB} \cdot \frac {BN}{NA} \cdot \frac {AM}{MC} \equal{} \frac {b}{c} \cdot \frac {a}{b} \cdot \frac {c}{a} \equal{} 1 \Longrightarrow$ by Menelaus' theorem $ M,N,L$ are collinear.


At JMP (see http://scitation.aip.org/dbt/dbt.jsp?KEY=JMAPAQ&Volume=46&Issue=1 for free samples) the citations are different than in math journals.

In text, they are like: "see Marriott^1" (a small upperscript index 1 appears with a link to the reference 1).

At the references, at the end of the paper it's like: "1Marriot, J., .....", where 1 is an upperscript index again.

How could I produce such citations and references in Latex? Also, I'd like that "1" from the citation be a hyperlink to "1" from the References.

Thanks,

Jean [/quote]

Use the American Institue of Physics files aip.sty, aip.bst available from [url=http://tug.ctan.org/tex-archive/biblio/bibtex/contrib/phy-bstyles/]CTAN[/url] and the hypreref package for the links.

I went to CTAN and got the files. But I still can't produce the references as they did (as I said above in the previous message). I just want to make it appear, like "Smith^1", a small 1 upperscript with a link to the references. And at the references should be an upper 1 followed by Smith. Could show me please what are the commands or where to find them?

Thanks,
Jean

There's nothing to do as aip.sty does it all for you eg
[code]\documentclass{article}
\usepackage{aip}
\begin{document}
This is a reference to a paper \cite{paper}
\begin{thebibliography}{99}
\bibitem{paper} Some paper or other
\end{thebibliography}
\end{document}[/code]produces the result you want.


[quote="Rep123max"]I just followed all the instructions on downloading the programs, but I have a problem. Im trying the basic thing, and it keeps saying bad option. Was I supposed to do something to MikTex to like activate it (I downloaded and installed it) before I downloaded-installed TexNic center?[/quote]

You'll have to be a little more clear about where you're getting the error and what it says.  I don't know what you mean by 'the basic thing'.

You should have first downloaded the MiKTeX, and installed it (to do this, just find the file you downloaded and click on it), then done the TeXnicCenter.  If you haven't finished with the MiKTeX, do that, then go back to the TeXnicCenter.

The other hello files are fine - they are generated automatically (and used for things like building indexes and doing referencing, which you don't need to worry about for now).

Try this: switch the LaTeX->DVI at the top to LaTeX->PDF.  Do Ctrl-F7 then F5.  Did that work?

Deleting them is fine - they'll be recreated every time you compile the file.

You probably won't ever be creating DVI files, but if you do, do this in TeXnicCenter:

1) Click Build.
2) Click Define Output Profiles
3) Select LaTeX => DVI on the left
4) Click the Viewer tab
5) Put C:\TEXMF\MIKTEX\BIN\YAP.EXE in the 'Path of executable' field.

That's interesting. My Adobe Reader seems to lock-read the document, and TeXNicCenter then reports a write failure...

[quote="confuted"][quote="Osiris"]If you're going to be submitting paper (as opposed to electronic) copies of your documents, DVI is fine. DVI generation is loads faster than PDF generation, and you don't have to shut down YAP every time you want rebuild the document.[/quote]
Nor do you have to shut down adobe reader every time, at least if you're using 6.0.[/quote]

This is one of the things I really like about TeXnicCenter over my old emacs and DOS solution.

I downloaded TeXnicCenter today... what a difference from DOS!


f(1),f(2),...........f(n)    is  a sequence of real numbers.

       f(1) = 2.

 [ f(n+1) ]  [ f(1) + f(2) +..................+ f(n) ]   =  n               for  all  n  (natural number)

     Prove that  f(2009)  >  0.9995

 Sir,please SEND  the solution  to me   [ nuthetishivdeep@gmail.com ] .

You already posted this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=302261]here[/url] and it is a current problem . Mods , lock this topic .


A monster buys a camera and a case worth 1/2 of the camera's cost, an ice cream worth 1/2 of the case's value, and a cookie for 1/2 of the ice cream's value, all for $ \$$105. How much did the cookie cost?

$ \textbf{(A)} \$7\qquad\qquad\textbf{(B)}
\$14\qquad\qquad\textbf{(C)} \$56\qquad\qquad\qquad\textbf{(D)}
\$57.25\qquad\qquad\qquad\textbf{(E)} \$105$

How many distinct solutions exist for the equation $ 2x^3 \minus{} 5x^2 \plus{} 4x \minus{} 1 \equal{} 0$?

$ \textbf{(A)} 0\qquad\qquad\qquad\textbf{(B)} 1\qquad\qquad\qquad\textbf{(C)} 2\qquad\qquad\qquad\textbf{(D)} 3\qquad\qquad\qquad\textbf{(E)} 4$

What is the area of a polygon with vertices at: (0,0), (3,4), (6,0), (3,-4), and (0,3)?

$ \textbf{(A)} 4.5\qquad\qquad\qquad\textbf{(B)} 19.5\qquad\qquad\textbf{(C)} 21.5\qquad\qquad\qquad\textbf{(D)} 24.0\qquad\qquad\qquad\textbf{(E)} 28.5$

What is the average of all two-digit integers that satisfy the condition that the number is equal to the sum of the ten's digit and the unit's digit and the product of the ten's digit and the unit's digit?

$ \textbf{(A)} 56\qquad\qquad\qquad\textbf{(B)} 57\qquad\qquad\qquad\textbf{(C)} 58\qquad\qquad\qquad\textbf{(D)} 59\qquad\qquad\qquad\textbf{(E)} 60$

What is the units digit of the following sum:
\[ (1!)^2 \plus{} (2!)^2 \plus{} \dots \plus{} (1337!)^2?
\]
$ \textbf{(A)} 5\qquad\qquad\qquad\textbf{(B)} 6\qquad\qquad\qquad\textbf{(C)} 7\qquad\qquad\qquad\textbf{(D)} 9\qquad\qquad\qquad\textbf{(E)} 0$

[hide="Problem 11"]This is just a lot of smaller and smaller fractions. Doing the division results in $ \boxed{\textbf{A}}$.[/hide]

[hide="Problem 12"]The quick reaction is to notice that the polynomial is to the third power so it has 3 roots. [b]However[/b], this polynomial has a double root (try root locating and see). Hence, this polynomial only has 2 distinct solutions, so the answer is $ \boxed{\textbf{C}}$.[/hide]

[hide="Problem 13"]We divide this into triangles, or we can use Pick's theorem. Either method results in $ 28.5$, or $ \boxed{\textbf{E}}$.[/hide]

[hide="Problem 14"]We add least and greatest or whatever and divide by two blah answer is $ \boxed{\textbf{D}}$.[/hide]

[hide="Problem 15"]We note that above $ 4!$, the units digit is 0. We compute the last digits of $ (1!)^2\plus{}(2!)^2\plus{}(3!)^2\plus{}(4!)^2$, and our result is $ 7\implies\boxed{\textbf{C}}$.[/hide]


AMC
AIME
USA(J)MO
USAMO
MATHCOUNTS
complementary counting

Poll above.

As for me, No for AIME, Yes for Nationals

You still have 4 years, no worries.
Anyways, yeah, no Nats for me.

qualified for both :lol:

hmm, andrew, jonathan, allison, peijin, jason, me, didn't they all go to chapter?

[quote="ZhangPeijin"]Anyways, you still have 4 years to qualify, no worries.[/quote]

Two of which will give me a higher chance of qualifying for USAMO :|

I qualified for... All three of them... 1, 2, ... oh wait. That just shows how bad I am at counting  :( . Peijin is real nice.  I should start making fun of him.

Yeah, you should complement about how awesome I am, and such.

oh yeah, you couldn't count to 5, we should definitely bow down to your awesomeness now

I meant so say complement because of the math things and such.
Like the complement of an angle?
-laughter-
Aha...

[quote="Yongyi781"][quote="ZhangPeijin"]Yup, lets see you count 5 dots on a piece of messy paper when you're stressed and you only have 1 minute left on the AMC 10B.[/quote]

When I count five dots on a piece of messy paper, I say, "1, 2, 3, 4, 5", not ... erm... how else do you count to 5?

How many ways can you count from 1 to 5 using 5 numbers from 1 to 10 at most once? (ex. 1, 9, 8, 7, 5)?

I can always conjure MATHCOUNTS style questions from thin air! :D[/quote]
No I mean as I miss one dot, and say 4, and get the question wrong.
Then I look back at my scrap and find 5 dots.
...

[quote="Yongyi781"]How many ways can you count from 1 to 5 using 5 numbers from 1 to 10 at most once? (ex. 1, 9, 8, 7, 5)?[/quote]

1 and 5 are set, so rest is 7C3 so 35.  :lol:

Well anyways, I've never done the AMC's I'm planning to next year, but I'm going to nats.

Hmm...  The first term would be 1, the last term would be 5.  Thus, we can choose from 7 terms to fill in the three.  Thus, it would be 7x6x5=$ \boxed {210}$ ways. :P Or at least I think that's right...

Aaaaaaanyways, I qualified for both, which for my state was pretty easy to accomplish.

Here's an MC-style problem to think about:
How many ways can you count from 1 to 5 using 5 numbers from 1 to 10 at most once, where integers next to each other in the sequence cannot have a positive difference of 1?  Muhahahaha!  An even harder question based on Peijin's inability to count! (no offense)

EDIT: Beaten.  Wait, never mind.  Ihatepie, order matters in this case, so it's permutations rather than combinations.  Ha!

Muhahahahahaha, my problem totally pwnages!!!!  Mwahahahahaha!  Sorry...  Hmm, anyone else wanna try? :lol:

1 and 5 are set.
The middle 3 can be any numbers right?
So its just $ 9\cdot8\cdot7\equal{}504$

I would say taking 210 and subtracting all the cases where at least one pair consecutive integers have a difference of 1... but I'm too lazy to do that now :P

Oh, and by the way,

[quote="ZhangPeijin"]1 and 5 are set.
The middle 3 can be any numbers right?
So its just $ 9\cdot8\cdot7 \equal{} 504$[/quote]

But 1 and 5 are already taken, so there are 8 options left (7 if you don't count 0). So it's $ 8\cdot7\cdot6 \equal{} 336$ or $ 7\cdot6\cdot5 \equal{} 210$. I assumed that 0 was not allowed, but you could've answered both ways.

[quote="ZhangPeijin"]1 and 5 are set.
The middle 3 can be any numbers right?
So its just $ 9\cdot8\cdot7 \equal{} 504$[/quote]
Oh, sorry, forgot to mention numbers must be distinct, or did I mention that?  Too lazy to scroll up to check...  Hmm, Yongyi, I think that's the best way to tackle it, but instead of listing so many, the number of opposite cases is less cuz you know that either the digits are all going up 1 or all going down 1.  It's impossible to have, say 1, 2, 3, 2, 5 as one cuz all digits are distinct.  That makes things easier.

Meh, if you cant use 1, 5, or 0, then it is $ 7\cdot6\cdot5$
Are you guys saying there's some other limitation?

The numbers next to each other cannot differ by 1.  Read my post, not Yongyi's for the question.

[quote="Math Geek"]where integers next to each other in the sequence cannot have a positive difference of 1?  [/quote]

well....

I made AIME, but state's on April 5th since it got rescheduled.

Oh...
Anyways, gotta do my S&D project thats due tomorrow, solve it later.

Hah, if I don't make USAMO this year that will be a major fail. I almost made it last year...

The whole of california sends only four to the nationals. Our state is very hard to qualify for nationals.

I take it Cali and Texas are two of the hardest states?
..Aren't they?

Wait... for the counting problem it said numbers from 1-10. If you can't use 1 and 5 that leaves 8, not 7.

I don't even take AMCs or AIMEs, and I never made Nats. Unfortunately, so sad :mad:  :(

Start from the back... The number before $ 5$ cannot be $ 4,6$.  There are thus $ 6$ choices for that number.

If it is $ 2$, then the number before it can be $ 4,6,7,8,9,10$

If it is $ 10$, it can be $ 2,3,4,6,7,8$

For the other $ 4$ possibilities, there are only $ 5$ other choices since they each get rid of the two numbers next to it.  This averages to $ 5.4$ =P

$ 6*5.4*?$

Depending on what the third digit was, we can find the second digit, which cannot be 2.  Err... ya xD.  I would just assume that the ? ends up being 5 since it should be an integer. =).  Then the answer is: $ 162$. lol. Probably wrong though since i'm really tired right now

Or...
Complementary counting, but it still takes too long =P


number theory
greatest common divisor
relatively prime

Hey,
I have been a lurker here for a while, but just finally registered because I cannot, for the life of me, figure out how to do this problem.  It probably is pretty easy, but I just don't see it.  Thanks for your help.

Let m and n be positive integers. Show that the smallest positive integer a such that m divides kn is k = m/gcd(m,n).

Thanks.

Do you mean $k$ instead of $a$? If so, then [hide="here is my take on the problem"]
Note that this choice of $k$ will always work, it remains only to show that there are no smaller $k$.

Let the product of all the primes that divide $m$ but do not divide $n$ be $q$, and the product of all the primes that divide both $m$ and $n$ be $r$. (repeats included on both occasions).
Now $\gcd(m,n)=r$ and $qr=m$. We know that $qr|kn$ and since $\gcd(q,n)=1$ we have $q|k$. Hence $k \geq q$. But if $k=\frac{m}{\gcd(m,n)}$ then $k=q$, so this is clearly the minimum.[/hide]

$m=db+c$
$n=fb+c$

Suppose a prime $p$ divides $b$.

Then it divides $db$ and so divides $m$ is and only if it divides $c$. Also, it divides $fb$ and so divides $n$ if and only if it divides $c$. Hence the set of primes that divide both $\gcd(m,b)$ and $\gcd(n,b)$ are the same, and by considering whether or not $p^{2}$ divides $c$ etc. we get that $\gcd(m,b)=\gcd(n,b)$.

$m=n(\mod{b}) \rightarrow \exists{x}\in \mathbb{Z}: bx+m=n$

So $\gcd{(b,n)}=\gcd{(b,bx+m)}=\gcd{(b,m)}$

Ok, I understand all the steps in that second response except the last one.  How do you get gcd(b,bx+m) = gcd (b,m)?

We know that $\gcd(a,b)=\gcd (a,b-a)$

To see why, say $d=\gcd(a,b)$ That means there is an $n$ and $m$ such that $a=dn$ and $b=dm$, and that $\gcd(n,m)=1$

So $\gcd(a,b-a)=\gcd(dn,d(m-n))=d\gcd(n,m-n)$

But can you see why $\gcd(n,m-n)=1$? Because they're relatively prime!
So we get $\gcd(a,b)=\gcd(a,b-a)$

Returning to our problem: We have
$\gcd(b,bx+m)=$
$\gcd(b,bx+m-b)=$
$\gcd(b,b(x-1)+m)=$
$\gcd(b,b(x-1)+m-b)=$
$\gcd(b,b(x-2)+m)=$
$\cdots=$
$\gcd(b,b(x-x)+m)=$
$\gcd(b,m)$

Sorry if this is confusing, its late here in Ireland and im flying to Switzerland in the morning and need to hunt for my passport. Seamus out!

Hmmm, well that still seems strange to me.  The hint in the back of the book (it's an old textbook I got for hardly anything, can't beat that) say to use the definition of GCD to show that each of them are less than or equal to the other...how would I do that?

Anyone?  This problem has been driving me crazy all night.

[quote="redfalcon"]Thanks, I see how it works now.  One more problem that I'm working on is this one:

Suppose b, m and n are positive integers. If m == n (mod b), prove that gcd(b,m) = gcd(b,n).

Any input?[/quote]

$n=b\cdot k+m$ for some integer $k$ using the mod condition

$\gcd(b,n)=\gcd(b,b\cdot k+m)=\gcd(b,m)$

(since $\gcd(a,b)=\gcd(a,b-a)$)

Ok, but then how do I prove that gcd(a,b)=gcd(a,b-a)?  Also, how is it done with the definition thing because that has me baffled?

Thanks to everyone for all your help too.

To prove that $\gcd(a,b)=\gcd(a,b-a)$ just consider all the factors of $\gcd(a,b)$.

There is an elementary rule in number theory that states that if $d|m$ and $d|n$ then $d|mx+ny$ for all integers $x,y$.

Now since $d|a$ and $d|b$ ($d=\gcd(a,b)$) $d|a-b$ (here $m=1$,$n=-1$).

Conversely, if $d|a$ and $d|b-a$ then $d|a+(b-a)=b$

Hence $\gcd(a,b)=\gcd(a,b-a)$


A solitaire game is played with an even number of discs,each coloured red on one side and green on the other side.Each disc is also numbered,and there are two of each number;i.e $\{1,1,2,2,3,3,...,N,N\}$ are the labels.The discs sre laid out in rows with each row having at least three discs.A move in this game consists of flipping over simultaneously two discs with the same label.Prove that for every initial deal or lay out there is a sequence of moves that ends with a position in which no row has only red or only green sides showing.

First we exclude inductively the case when, for some $k$, both $k$ discs are in the same row.

If they're in a 3-disc row (two $k$ discs and an $m$ disc), remove the row and place an $m$ disc in the row containing the other $m$ disc, so that they're of the same color. Solve this case inductively. Then isolate one of the $m$ discs and put it together with two $k$ discs to recreate our row. Clearly the row containing the other $m$ doesn't fail, and we can fix the new row.
If they're in a 4-disc row of the form $k, k, m, m$ we just remove the row and proceed inductively in the obvious way. If they're in a 4-disc row of the form $k, k, m, l$, place the $l$ disc in the row containing the other $l$, solve this case by the above (we'd done it for $k, k, m$) and put back the $l$. If they're in a 5-disc row, just remove both $k$ discs and proceed inductively.

So these cases are done. Now suppose no two discs with the same number are in the same row. We will prove that, if there's a non-zero number of monochromatic rows, we can eventually decrease this number; and this will be enough.

Take a monochromatic row (we'll call it an F-row, for forbidden) and switch one of its discs. If the other disc with that number is in a row that was either F or doesn't become F because of this move, we've decreased the number of F-rows. Suppose that the row we just fixed is $r_1$, and we made some other row $r_2$ F. Fix $r_2$ (not switching the same number!). If we don't mess up any new rows, we win. Otherwise, fix the new messup $r_3$, and so on. Eventually (after a finite number of moves), either we will win or we will get some $i<j$ such that $r_i=r_j$. This would imply that there's a cycle of rows, say $R_1, R_2, \ldots, R_k$ such that $R_1$ contains $1, 2$; $R_2$ contains $2, 3$; ...; $R_{k-1}$ contains $k-1$ and $k$, and $R_k$ contains $k$ and $1$, and just one of them is F. Now we'll show a way to make them all good. First switch some discs so that the $1, 2$ in $R_1$ are distinct; then switch $3$ if necessary, so that the $2, 3$ in $R_2$ are distinct; and so on. If in the end we get the $k, 1$ in $R_k$ distinct, or $R_k$ contains also some disc of the other color, we've won. If we get $R_k$ an F, then switch all $1, 2, \ldots, k$, and all $R_1, \ldots, R_k$ will be good.

The procedure will finish after a finite number of steps. We're done.


geometry
trapezoid
rotation
geometric transformation
dilation
ratio
vector

For those of you who know how to use Geometer's Sketchpad, how would you draw a trapezoid with bases $5$ and $10$ and base angles $20^{\circ}$ and $45^{\circ}$?

I'm having trouble constructing such a figure exactly. I don't want just approximations.

Draw AB=10 cm, then rotate B around A for 20 degress and construct line AB', then rotate A around B for 45 degrees and construct line BA', then construct the intersection of the two lines, let it be E, and then construct the midline of triangle ABE and you have your trapezoid.

Ok. That works.

But what if one base was not half the other? For example, what if the other base was $8$?

Then you should mark E as center, dilate A with the ratio ${8\over 10}$ — that is, with the ratio ${CD\over AB}$, whatever CD may be — and draw the parallel with AB through A' (later you can relabel it as D).

Or you can find M on AB such that AM is equal to the other base and draw parallel with AE through M. It's intersection with BE will be C.

Nice. Thanks!

Also, do you know how to make the right angle symbol (a little square)?

to the first  question,  i just use the graph thing and can draw accurate pictures usint that

i think you just have to draw them in for the right angles

There's no ready-made command for the right angle symbol, but you can visit [url]http://whistleralley.com/GSP/sketchpad.htm[/url] and download [i]Geo_Notation[/i] file. It contains tools for always-needed-but-very-cumbersome-to-produce symbols, such as right-angle symbol, ticks on congruent segments, arcs on congruent angles, vector arrows, arrows for parallel lines, crosses instead of dots for points etc.

For how to import and use GSP Tools, please consult the manual or Help, it's just too extensive to explain here.


[hide]
If $ n\equal{}2$, the result is immediate.
Otherwise, the expression on the right can be separated:
$ 2^{n \minus{} 1} \plus{} 2*3^{n \minus{} 1} \plus{} ... \plus{} (n \minus{} 1)n^{n \minus{} 1} \plus{} (n \minus{} 1)$.
By Fermat's little theorem, this expression is $ \equiv \pmod{n}$ to:
$ 1 \plus{} 2 \plus{} 3 \plus{} ... (n \minus{} 2) \plus{} 0 \plus{} (n \minus{} 1)$
$ \equiv \frac {(n \minus{} 1)n}{2}$
$ \equiv 0 \pmod{n}$.[/hide]


Can somebody explain the shape, the hybridisation and the electronic structure (orbital wise ofcourse) of this compound.

$ [Co(NH_3)_6]^{2\plus{}}$

shudn't it be square bipyramidal????????????

[quote="Chronoz"]Can somebody explain the shape, the hybridisation and the electronic structure (orbital wise ofcourse) of this compound.

$ [Co(NH_3)_6]^{2 \plus{} }$[/quote]

The geometry is $ \text{octahedral}$ and the hybridization is $ d^{2}sp^{3}$.

i hope square bipyramidal & octahedral r same

@chronoz--checkout NCERT-(ch-Coordination compound )


Gauss
algebra
polynomial
Ring Theory
number theory
prime factorization
number theory unsolved

Let $m=p_{1}^{k_{1}}...p_{s}^{k_{s}}$, then the ring \[Z/mZ=\sum_{i=1}^{s}Z/p_{i}^{k_{i}}.\]
Because $(Z/p^{k}Z)^{*}$ had even elements if p is odd or p=2,k>1, number of solutions $x^{2}=a(mod \ m)$ is $2^{s}$ if m is odd or 4|m, else (when m=2(mod 4)) number of solutions is $2^{s-1}$.

[quote="Keno20"]Assume m>2, (a,m)=1 and a is a quad residue mod m. prove that the congruence  x^2≡a (mod m) has exactly 2 solutions if and only if m has a primitive root.

unknown source[/quote]

That isn't true.

Say, $m=35$.
Now by Gauss' theorem this doesn't have primitive roots.
But it sure does have solutions to,
$x^{2}\equiv a \mod 35$ for some $\gcd(a,35)=1$
Now if $y$ is solution then $m-y\not \equiv y$ is a solution.

[quote="t0rajir0u"][quote="sylow_theory"]But it sure does have solutions to,
$x^{2}\equiv a \mod 35$ for some $\gcd(a,35)=1$[/quote]

The problem specifies [b]exactly two[/b] solutions.  There are four, I think, in $\mathbb{Z}_{35}$.[/quote]
Ah! I see now.

I think you are right about Z_35 having 4 solutions.
Because the polynomial ring Z_35[x] does not obey the rule that there are at most n=2 solutions. For Z_35 is not a field.

Prime factorization, CRT kills it, considering $\mathbb{Z}_{m}$ has primitive roots iff $m$ is a power of a prime or double a power of a prime.

I don't know much about rings or fields, so my idea is a bit less nice.

it's actually quite simple to show that no primitive roots mod m implies the residue equation has 3 or more solutions.

for the other way, just use the primitive root to express two solutions to the equation and show that they are the same which follows simply from some very basic properties of primitive roots.


Prove that for all real positive numbers $ x,y,z$, the following inequality is true :

$ \frac {x}{13x \plus{} y \plus{} z} \plus{} \frac {y}{x \plus{} 13y \plus{} z} \plus{} \frac {z}{x \plus{} y \plus{} 13z} \leq \frac {1}{5}$

write the ineq as:

$ \sum{\frac{y\plus{}z}{13x\plus{}y\plus{}z}}  \geq  \frac{2}{5}$ and then cauchy shwarz :wink:

anas, your solution is wrong since a,b,c are[b] reals[/b], not positive reals.

I was wrong again , i edited  :blush: Sorry great math .

That's alright, alex

Here is my solution

$ \frac{a}{12a\plus{}a\plus{}b\plus{}c} \le \frac{a}{25}\left(\frac{4}{3a}\plus{} \frac{1}{a\plus{}b\plus{}c}\right)$


Accordingly,Human there are 46 chromosomes. In
man,x,y.And in woman double x.Each contain a complete "[b]Genetic[/b] Material[b]"[/b] that determine the humanly inheritance.Is there any further "[b]Divisional Components[/b]"[b](Ex.Genetic Sequencing Arrangement[/b].)taht explain the Physical Make-up of each human being?

                                                    Respectfully,
                                                   A.Eddington

 Postscript:
               Example Of Physical Attributes:
             5 Feet,3 inches tall,Semi-Slender Body Built,Semi-Oblong Face,Thin Eye Brows,Expressive Eye Ball,Not So Thin Lip,
White Skin,Cutely Curve Eye Laces,Black Short Hair,Proportional
Nose Figure.

                                                            Same

Alright...

-First off, each chromosome doesn't contain a complete copy of the genetic material.  The entire set of chromosomes however, does.

-The short answer to your question is "yes"; there are other things than just the nucleotide sequences that determine the "physical make-up" of a human being; things like differential gene expression, environmental interactions, alternative splicing, mitochondrial DNA, and (as always) random chance.

-Re: your postscript.  Many of those things are partially environmentally determined, like height, semi-slender body, etc.

Just a small bit to add, as this seems to relate to the question at hand, you might wish to read more about epigenetics and quantum biology.

Offhand, I know of one video in which epigenetics is at least partially discussed (in reference to the feasibility of a cloned embryo forming 'normal' offspring):

http://mitworld.mit.edu/video/138/

It requires RealPlayer to view, but I certainly thought it was an interesting lecture.  It addresses some of the ethical and practical issues of 'cloning' and the like.


inequalities
LaTeX
inequalities proposed

Prove that for positive reals $ a,b,c$ such that $ abc\equal{}1$,
the following inequality holds-
$ \frac{a^3\plus{}b^3\plus{}c^3\plus{}3}{4} \ge \frac{b}{a\plus{}c}\plus{}\frac{c}{a\plus{}b}\plus{}\frac{a}{b\plus{}c}$

[quote="geniusbliss"]Prove that for positive reals $ a,b,c$ such that $ abc \equal{} 1$,
the following inequality holds-
$ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3}{4} \ge \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a}{b \plus{} c}$[/quote]
$ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3}{4} \ge \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a}{b \plus{} c}$
$ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc}{4abc} \ge \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a}{b \plus{} c}$
$ \frac {\sum_{cyc}a^3\plus{}3abc}{4abc} \ge \frac{\sum_{cyc}a^3\plus{}\sum_{cyc}a^2(b\plus{}c)\plus{}3abc}{\sum_{cyc}(a^2(b\plus{}c))\plus{}2abc}$
$ \frac {2(\sum_{cyc}a^3\minus{}abc)}{8abc} \ge \frac{\sum_{cyc}a^3\plus{}abc}{\sum_{cyc}(a^2(b\plus{}c))\plus{}2abc}$
because of the two inequalities
$ 2(\sum_{cyc}a^3\minus{}abc) \geq \sum_{cyc}a^3\plus{}abc$
$ \sum_{cyc}a^2(b\plus{}c)\plus{}2abc \geq 8abc$
so it has been solved.

[quote="geniusbliss"]Prove that for positive reals $ a,b,c$ such that $ abc \equal{} 1$,
the following inequality holds-
$ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3}{4} \ge \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a}{b \plus{} c}$[/quote]

your inequality is equivalent to
$ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3}{4}\minus{}\frac{3}{2} \ge  \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a}{b \plus{} c}\minus{}\frac{3}{2}$

and using the two flowing identites

$ \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {a}{b \plus{} c}\minus{}\frac{3}{2}\equal{}\sum_{cyc}\frac{(a\minus{}b)^2}{(b\plus{}c)(c\plus{}a)}$
$ \frac {a^3 \plus{} b^3 \plus{} c^3 \plus{} 3}{4}\minus{}\frac{3}{2}\equal{}\frac{a^3 \plus{} b^3 \plus{} c^3\minus{}3abc}{4}\equal{}\frac{(a\plus{}b\plus{}c)(\sum(a\minus{}b)^2}{8}$

and since $ \frac{(a\minus{}b)^2(a\plus{}b\plus{}c)}{8} \ge \frac{(a\minus{}b)^2(a\plus{}b)}{8} \ge \frac{(a\minus{}b)^2}{(b\plus{}c)(c\plus{}a)}$

because $ (a\plus{}b)(b\plus{}c)(c\plus{}a) \ge 8abc$

we are done !

[i]another soultion[/i]

we have  by CS

$ \frac {a}{b} \plus{} \frac {a}{c} \ge \frac {4a}{b \plus{} c}$

and hence it sufficent to show that :

$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge \sum_{cyc}\left(\frac {a}{b} \plus{} \frac {a}{c}\right) \equal{} \sum_{cyc}ab(a \plus{} b)$

which is schur  :)

By AM-GM:
$ \frac{a^3\plus{}b^3\plus{}c^3\plus{}3}{4}\ge\frac{a\sqrt{a}\plus{}b\sqrt{b}\plus{}c\sqrt{c}}{2}\equal{}\frac{a}{2\sqrt{bc}}\plus{}\frac{b}{2\sqrt{ac}}\plus{}\frac{c}{2\sqrt{ab}}              \ge\frac{a}{b\plus{}c}\plus{}\frac{b}{a\plus{}c}\plus{}\frac{c}{a\plus{}b}$
Easy! :)

[quote="akashram"]Just an application of c.s inequality[/quote]
well humour me with the solution then..

I am sorry I am not able to use latex properly :D

And MRgeniusbliss I have not posted any solution withot working

geniusbliss, you could use the $ PM$ to ask Akash Ram to type humorous things for you to laugh(if it really is humorous) and AkashRam, you may also use the $ PM$ to inform geniusbliss about whether your post is genuine and whether you have worked out the solution or not, and it is advisable for you to learn a but of latex. Refer to the AoPs for latex and you can click on the latex formulae of other people's posts and look at the latex commands. I think as far as inequalities are concerned, you need to know how to use
\sigma, \frac, \sqrt frequently. Rest is all formulae using  dollar symbols.

BTW, Abdek, your proof is really nice, superb!!!


A polynomial P has four roots, 1/4, 1/2, 2, 4. The product of the roots is 1, and P(1) = 1.
Find P(0).

I wonder why the original problem gives you product of the roots, when you already know the four roots.  :huh:


[quote="most-wanted"]question no.1
$2x-y = 5$
how to find that x = -3y ?
and how to draw the slope for this ?

i hope i get fast reply.

Question No.2
yeah the second eqution is
I.   $2x-y = 5$
II. $3y-x = 15$[/quote]
I don't understand what you are saying for the first. For the second I am assuming you are looking for x and y, in which case
[hide]Add y-5 to both sides to get
y=2x-5. Punching this into the second equation, we have, 6x-15-x=15, which solves to x=6. Plugging back in, we get y=7.[/hide]

[quote="most-wanted"]question no.1
$2x-y = 5$
how to find that x = -3y ?
and how to draw the slope for this ?

i hope i get fast reply.

Question No.2
yeah the second eqution is
I.   $2x-y = 5$
II. $3y-x = 15$[/quote]
[hide="1"]$2x-y=5$ is in standard form. To find the slope, you put it into slope-intercept form. $y=2x-5$
The slope is the coefficient of $x,$ which is $2$ in this case. Make sure the coefficient of $y$ is $1$ in slope-intercept form.[/hide]
[hide="2"]$2x-y = 5$
$y=2x-5$

Substitute that into the second equation.

$3(2x-5)-x=15$
$5x=30$
$x=6$

Now we can find $y=7.$ The answer is $(6,\ 7).$[/hide]


trigonometry
inequalities
inequalities unsolved

$ A,B,C$ are angles of $ \Delta ABC$,$ u,v,w > 0$are given const,find the maximum and minimum value of the following expression.
\[ u\sin{A} \plus{} v\sin{B} \plus{} w\sin{C}\]

???

Min: $ u,v,w\rightarrow 0$ max: $ u,v,w\rightarrow \infty$

So it remains to show min and max of sum of sines... That's easy (Jensen for example for one of them)

:blush: Sorry,I missed something.I mean $ u,v,w$ are given const.

:( Why no one help me....I don't think Jensen work here.

im not so sure about this.
[b]my solution:[/b]
WLOG let $ u < = v < = w$ and $ sin A < = sin B < = sin C.$
then by Chebychev's inequality,
$ u sin A + v sin B + w sin C < = (u + v + w)(sin A + sin B + sin C)/3 (*)$
then by Jensen's inequality,
$ (*) \< = (1/2)(u + v + w)\sqrt3.$
hence we have a maximum.
minimum is obviously when $ A = B = 0,$ and $ C = 180$.

[quote="Raja Oktovin"]@zool007: i think you can not assume $ u \le v \le w$ since it's given constant and while $ A,B,C$ runs all along positive real numbers satisfying $ A \plus{} B \plus{} C \equal{} 180^{\circ}$. or, may be you can work in cases?[/quote]
yeah, that's why i wasnt sure :D

[quote="Elixir10"]Since $ A,B,C \in [0,\pi]$, $ \sin{A},\sin{B},\sin{C} > 0$, so the minimum is $ 0$, attained at $ (A,B,C) = (0,0,\pi)$ or any of its permutations.

Then, we note that on $ [0,\pi], \sin{x} > 0$ so $ \frac {d^2x}{dy^2} = - \sin{x} < 0$, so $ \sin{x}$ is concave downward on this interval. Thus, by Jensen's Inequality,

$ u\sin{A} + v\sin{B} + w\sin{C}\ge(u + v + w)(\sin{\frac {A + B + C}{u + v + w}) = (u + v + w)(\sin{\frac {\pi}{u + v + w})}}$[/quote]

I believe that by Jensen's, we get $ LHS\le (u+v+w)\sin \frac{uA+vB+wC}{u+v+w}.$

[quote="earldbest"][quote="Elixir10"]Since $ A,B,C \in [0,\pi]$, $ \sin{A},\sin{B},\sin{C} > 0$, so the minimum is $ 0$, attained at $ (A,B,C) = (0,0,\pi)$ or any of its permutations.

Then, we note that on $ [0,\pi], \sin{x} > 0$ so $ \frac {d^2x}{dy^2} = - \sin{x} < 0$, so $ \sin{x}$ is concave downward on this interval. Thus, by Jensen's Inequality,

$ u\sin{A} + v\sin{B} + w\sin{C}\ge(u + v + w)(\sin{\frac {A + B + C}{u + v + w}) = (u + v + w)(\sin{\frac {\pi}{u + v + w})}}$[/quote]

I believe that by Jensen's, we get $ LHS\le (u + v + w)\sin \frac {uA + vB + wC}{u + v + w}.$[/quote]

Then,what's the maximum of $ (u + v + w)\sin \frac {uA + vB + wC}{u + v + w}$???

My apologies; it appears that earldbest is correct and I am wrong. Unfortunately, this brings us no nearer to solving the problem. :|

I think zool007's argument might work, considering that $ A,B,C$ are pretty much flexible. The thing is, will equality always be possible for any values of $ u,v,w$?

By Lagrange multipliers, it is easy to see that for the maximum, $ u\cos A \equal{} v\cos B \equal{} w \cos C$. So we are looking for a triangle such that $ cosA \equal{} \frac ku$ etc. with a certain $ k > 0$.

By the identity for any triangle $ \sum cos^2A \plus{} 2cosAcosBcosC \equal{} 1$, we obtain the equation $ k^3\frac {2}{uvw} \plus{} k^2\sum\frac {1}{u^2} \minus{} 1 \equal{} 0$. By Cardano's formula, this may yield a somewhat lengthy formula for $ k$ and then an even longer one for the maximum. I wonder if some ugly terms would eventually cancel out. :(  :blush: