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Well, to be more precise the question is:
If modulus of a and modulus of b are both less than one, show that the function
$f(z)=(\frac{z-a}{1-\overline{a}z})(\frac{1-\overline{b}z}{z-b})$ for mod of z less than one,
takes every value $w_{0}$ such that modulus of $w_{0}\neq 1$, exactly once.

by the way, whats latex symbol for modulus?? :)

Vertical bars | | will do (as in $|w_{0}|$). Connoisseurs use \lvert and \rvert (as in $\lvert w_{0}\rvert$).

You forgot to mention that $a\ne b$. Substitution $z'=\frac{z-b}{1-\bar b z}$ reduces the problem to the case $b=0$, but I don't know what to do next. :|

Hmm.. i think i made some progress on this so let me update it:
i think i should use rouche's theorem:
recall, rouche's theorem states that if you have a simple closed curve call it $C$ and if $f$ and $g$ are analytic (holomorphic) inside $C$ and such that $\lvert f(z) \rvert < \lvert g(z) \rvert$ for $z$ on $C$, then $f$ and $f+g$ have same number of zeros, or more generally have same $N-P$ value where $N$ is the number of zeros and $P$ is the number of poles.
Okay, so now, consider the following:
$(\frac{z-a}{1-\overline{a}z})(\frac{1-\overline{b}z}{z-b})-w_{0}=0$
if we let $g(z)$ to be $-w_{0}$ then above equation is $f(z)+g(z)=0$.
number of solutions to above equation is of course 'zero' for $f(z)+g(z)$ and by Rouche is same as the number of zeros of $f(z)$. (Our discussion hasn't incorporated poles yet). so the number of zeros of $f(z)$ is, i guess, $z=a$ from the inspection.(for a fixed $w_{0}$. As $w_{0}$ varies, there is precisely one zero namely $z=a$ for some $a$ in the disk of radius one. I think this shows that function takes every value $w_{0}$ inside disk once. Now for $w_{0}$ outside the disk:
above equation can be rewritten as:
$\frac{(\frac{z-a}{1-\overline{a}z})}{(\frac{z-b}{1-\overline{b}z})}-w_{0}=0$.
i mean the reason i did this is because i know that mapping $\frac{1}{z}$ maps points inside the disk to outside the disk...
hmm... but i'm not quite sure here because if we use same reasoning as above, i.e. $z=b$ this makes denominator zero and hence becomes pole of the equation above?? hmm... plz help here... i think i'm on the right track though :)

your statement of rouches theorem is wrong.according to your hypothesis, $g$ and $f+g$ have the same number of zeros inside C, you probably got them mixed up.

oops sorry i think i was meant to say:
$\lvert f(x) \rvert > \lvert g(x) \rvert$
which means $f(x)$ and $f(x)+g(x)$ have same number of $N-P$ value...
anyway, i think people can't be bothered to read all the stuff i wrote so never mind... ill try figure it out myself  :maybe:


Find all pair of positive integers $ (x,\ y)$ such that $ 2x^2y^2 \plus{} y^2 \equal{} 26x^2 \plus{} 1201.$

$ (2x^2\plus{}1)(y^2\minus{}13) \equal{} 1178 \equal{} 2\cdot 19\cdot 31$

$ 2x^2\plus{}1$ is odd, and $ 2(3)^2\plus{}1\equal{}19$ is the only solution

Leaving $ y^2\minus{}13\equal{}62 \Rightarrow y\equal{}7$

So $ (3,7)$ is the only positive integer solution

Rewrite the equation as $ (2x^2\plus{}1)(y^2\minus{}13)\equal{}2^2 \times 3^3 \times 11$

$ (2x^2\plus{}1)| 3^3 \times 11$

By trial we get $ (x,y)\equal{}(4,7),(7,5)$ as the only solution.


If you buy two squiggles at the regular price, you get the third one for a penny. If I paid $ \$2.02$ for 22 squiggles, what is the regular price, in cents, of one squiggle.

Solve a linear system.

R=regular price
D=discounted

R=2D
R+D=21 (notice I use 21 because we find the lower multiple of 3)

R=14
D=$ 7$
But notice, we took one out of the 22, so R now =$ 15$.
Subtracting $ \$0.07$ from $ \$2.02$ ($ 7$ discounted squiggles sum to $ \$0.07$), we get $ \$1.95$.
$ \$1.95/15\equal{}\$\fbox{0.13}$ per regular squiggle.

I'm sorry, I don't understand your solution...

Let S be the regular price. Then, the price of 3 is $ 2S\plus{}1$.

The price of 21 is then $ 14S\plus{}7$.

$ 14S\plus{}7\plus{}S\equal{}15S\plus{}7\equal{}202$
$ 15S\equal{}195$
$ S\equal{}13$.


algebra
system of equations
number theory solved
number theory

Solve the system of equations

(1+4 2x-y)*51-2x+y = 1 + 22x-y+1
y3 + 4x + 1 + ln(y2 + 2x) = 0.


Let $A=\{1,2,\ldots,n\}$ and $B_1,B_2,\ldots,B_b\in A$ such that $|B_i|=k$,
if every two element of $A$ are at $\lambda$ from $B_i$ together .Prove that exist a constant $r$ that every element of $A$ is in each of $B_i$.

For each $x\in A$, let $r_x$ be the number of $B_i$'s it belongs to. Now fix $x\in A$ count the pairs $(y,B)$ such that $x\ne y\in A$, and $B$ is one of the $B_i$'s which contains both $x$ and $y$. We can do this in two ways: first coun the $y$'s (there are $n-1$ of them), and for each $y$ coun the $B$'s which contain both $x$ and $y$ (there are $\lambda$ of them). We thus get $\lambda(n-1)\ (*)$. On the other hand, we can ount the $B$'s which contain $x$ first (there are $r_x-1$ of them), and then, for each $B$, we can count the $y\ne x$ which belong to $B$ (there are $k-1$) of them. We get $r_x(k-1)$. This is equal to $(*)$, and this holds for every $x$, but $(*)$ does not depend o $x$, so $r_x$ must be the same for all $x$.

This is like the first thing one proves in [url=http://mathworld.wolfram.com/BlockDesign.html]design theory[/url], I think :).


well 1 + 1/2 + 1/3 + ... + 1/n-1 + 1/n diverges and does not equal an integer for any value of n greater than 1.  This is so because the denominators are always changing.  So the above statement is
1 + (n/2)/n + (n/3)/n + ... + (n/(n-1))/n + 1/n

[quote="demji"] This is so because the denominators are always changing.  So the above statement is
1 + (n/2)/n + (n/3)/n + ... + (n/(n-1))/n + 1/n[/quote]

So...?

Changing denominators don't necessarily mean
that the sum can not be an integer.

One well known proof:

[hide]Let  k be the   largest power or 2 less than or equal to  n)

Now if sum is  p, multiply both side by  (k/2)*(product of all odd numbers up to n )  then except for  the term corresponding to (1/k) every other term would be an integer  ==> contradiction.[/hide]

[quote="Gyan"]One well known proof:
[/quote]

Yes...that's a proof....

When I was doing this exercise I came up with a
different argument...

Here it goes...:

Every harmonic number (except for the first one)
can be written as:

              odd number / even number   (*)

Thus an harmonic number can not be an integer.

A simple inductive proof is enough to prove (*)

[quote]A simple inductive proof is enough to prove (*)[/quote]
What is the simple inductive proof when     n in the next term (1/n)  is odd?

[quote="Gyan"]What is the simple inductive proof when     n in the next term (1/n)  is odd?[/quote]

Upsss!!!!..I guess the inductive step was not
as easy as I thought it was....

I admit it...I missed that case..... :blush:

I think this exact problem was in a previous topic...

[quote="Treething"]It's simple to prove that 1/1 + 1/2 + 1/3 + 1/4 + ... + 1/n as n approaches infinity diverges (harmonic series), but a diverging sequence could include an integer. Take

2 + 4 + 8 + 16 + ... + 2^n as n approaches infinity. This sequence obviously diverges, but for any value n, the sum of the series is an integer.[/quote]

The harmonic series is an example of a diverging series which does NOT
include any integer.

Yeah, the key is to consider the largest power of 2 in the numerator and denominator, respectively.

[quote="josejacobi"]Prove that 1+1/2+1/3+......+1/n can NEVER be an integer for any given n>1[/quote]
what if it doesnt start with 1?

[quote="josejacobi"][quote="Scrambled"]what if it doesnt start with 1?[/quote]

If the sequence does not start with one, the sum will
be always an odd numerator over an even denominator.
Thus, the sum can never be an integer.[/quote]
Scrambled means sum $S=\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{m}$.

Anyway, if $n>m/2$ then $S<1$, and if $n\leq m/2$ then there is a power of 2 among numbers $n$, $n+1$, ..., $m$ and we apply previous argumentation.


Evaluate the sum of the series

$\sum_{k=1}^{\infty}{\frac{1}{k(k+1)(k+2)}}$

Thanks :wink: Please help me to solve it

[quote="miyomiyo"][quote="miyomiyo"][hide="Solution"]Use partial fraction decomposition to get $\frac{1}{k(k+1)(k+2)}= \frac{\frac{1}{2}}{k}+\frac{-1}{k+1}+\frac{\frac{1}{2}}{x+2}$.
Because you get $\frac{1}{2}+\frac{1}{2}-1 = 0$ for every denominator $3$ and above, we just need to worry about $1$ and $2$.
For $1$, we only have $\frac{\frac{1}{2}}{1}= \frac{1}{2}$.
For $2$, we have $\frac{-1}{2}+\frac{\frac{1}{2}}{2}= \frac{-\frac{1}{2}}{2}=-\frac{1}{4}$.
Finally, we get $\frac{1}{2}-\frac{1}{4}= \boxed{\frac{1}{4}}$.[/hide][/quote]
I did. :wink:[/quote]

hmm spent a couple of hrs on this qn  :blush:

nicely done. Do you know of any other way of solving other than partial sum

I only ask because i Found this question in a 2 unit book and partial sum is 4 unit method

In Aust. maths is classed accoring to 4 levels

general
2unit
3unit
4unit

[quote="Ravi B"]A slightly different approach is to notice that
[hide]
\[\frac{1}{k(k+1)(k+2)}= \frac{1}{2}\left( \frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}\right) \, . \]
[/hide][/quote]Wow, nice.  But how do you "notice" this?

I'm not exactly sure.  I knew the decomposition
\[\frac{1}{k(k+2)}= \frac{1}{2}\left( \frac{1}{k}-\frac{1}{k+2}\right). \]
Then I divided both sides by $k+1$.


Let $ a,b,c$ positive reals such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that

$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\geq\frac {1}{3}(7 \plus{} \frac {1}{a \plus{} b \plus{} c})$

[quote="Ligouras"]Let $ a,b,c$ positive reals such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that

$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\geq\frac {7}{3} \plus{} \frac {1}{a \plus{} b \plus{} c}$[/quote]
I believe that this inequality is not valid.

Take $ a\equal{}1, b\equal{}1, c\rightarrow 0^\plus{}$.

[quote="FantasyLover"][quote="Ligouras"]Let $ a,b,c$ positive reals such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Prove that

$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\geq\frac {7}{3} \plus{} \frac {1}{a \plus{} b \plus{} c}$[/quote]
I believe that this inequality is not valid.

Take $ a \equal{} 1, b \equal{} 1, c\rightarrow 0^ \plus{}$.[/quote]

the correct version is:

$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\geq\frac {1}{3}(7 \plus{} \frac {1}{a \plus{} b \plus{} c})$

Sorry......  :(


probability
analytic geometry
integration
calculus
induction
pigeonhole principle
combinatorics unsolved

Provided a square lattice with N small boxes. Let N $\geq 3$ be an odd number  . The caterpillar 'Number-fullly' sits in the small box exactly in the center of the lattice. Everyone of the remaining small boxes contains a positive whole number. Over the distribution of the numbers it is only well-known that in no two fields the same number is. Number-fullly wants to leave the lattice square. But Number-fullly must keep to the following rules. It only can move to neighbouring boxes which means the neighbouring box must have a common side with the current box. And Number-fullly has to 'eat' each number of the squares it passes though. Each number n has a weight of $\frac{1}{n}$ kg. And Number-fullly can eat no more than 2 kg numbers.

Analyse for

a) for N = 2003,
b) for all odd numbers of $N \geq 3$,

whether the numbers can be distributed in the lattice so unfavorably that Number-fullly cannot find a way to leave the square lattice where there are at most 2 kg numbers !

Note that the problem was posed at the DeMo2003.

Orl, do you mean that each number has the same weight ?
Or have all numbers different weights, for example, is the weight of 5 equal to 1/5 ?

Yeah, the second statement is correct. No two fields include the same number. So following from the fact that each number has a weight of 1/n kg all the weights are different.

But I don't see how to solve it. Any hints ?

My supervisor solved this problems with help of probability theory. But I don't want to write his solution, because there are specific terminology.

Anyway since problem were presented on DeMo2003 it must have comprehensible simple solution. Am I right?

I have a solution using elementary estimates. So you are right, Myth ! So if there isn't anybody who wants to provide his own solution I will write it down in due course.

a.)

The answer is no. The caterpillar Numberfullly can always find an opportunity to leave the square
lattice where there at most 2 kg numbers.
For the square lattice we use integer coordinates x,y with -1001 \leq x \leq 1001 and
-1001 \leq y \leq 1001. At first Numberfullly sits on the square (0,0). Now we will show that one
of the following moves will solve the problem:

1.)   (0,1)  -> (0,2)   -> (0,3)   -> (0,4)   -> ... -> (0,1001)
2.)   (0,1)  -> (1,1)   -> (1,2)   -> (1,3)   -> ... -> (1,1001)
3.)   (1,0)  -> (1,1)   -> (2,1)   -> (3,1)   -> ... -> (1001,1)
4.)   (1,0)  -> (2,0)   -> (3,0)   -> (4,0)   -> ... -> (1001,0)
5.)   (1,0)  -> (1,-1)  -> (2,-1)  -> (3,-1)  -> ... -> (1001,-1)
6.)   (0,-1) -> (1,-1)  -> (1,-2)  -> (1,-3)  -> ... -> (1,-1001)
7.)   (0,-1) -> (0,-2)  -> (0,-3)  -> (0,-4)  -> ... -> (0,-1001)
8.)   (0,-1) -> (-1,-1) -> (-1,-2) -> (-1,-3) -> ... -> (-1,-1001)
9.)   (-1,0) -> (-1,-1) -> (-2,-1) -> (-3,-1) -> ... -> (-1001,-1)
10.)  (-1,0) -> (-2,0)  -> (-3,0)  -> (-4,0)  -> ... -> (-1001,0)
11.)  (-1,0) -> (-1,1)  -> (-2,1)  -> (-3,1)  -> ... -> (-1001,1)
12.)  (0,1)  -> (-1,1)  -> (-1,2)  -> (-1,3)  -> ... -> (-1,1001)

These ways include 12008 squares. Eight squares are involved in at least two ways, within these 8 squares
four of them are involved in three ways. We denote the squares which are used for three ways with a_1, a_2,
a_3 and a_4, those which are used twice (but not three times) with a a_5, a_6, a_7 and a_8 and the other
squares only being used for one way with a_9, a_10,...,a_12008. All these 12 ways include altogether

m = \sum^12008_(k=1) 1/a_k + \sum^8_(k=1) 1/a_k + \sum^4_(k=1) 1/a_k

kg numbers. The sums on the rhs. will take their greatest values (under the problem's conditions) when we
sum up the reciprocal terms of the first 12008, eight and four positive integers respectively. Hence we get:

m \leq M = \sum^12008_(k=1) 1/k + \sum^8_(k=1) 1/k + \sum^4_(k=1) 1/k

Within these 12 ways Numberfully will choose the way with the least amount of numbers and has to eat at most
M/12 kg numbers. So we have to prove M < 24. I provide two possibilities:

(i) \sum^n_(k=1) 1/k  <  \int^(n+1/2)_(1/2) 1/x dx = ln(2n+1).

Proof: f(x) = 1/x is concave implies 1/2 (f(k-t)+f(k+t)) > f(k). Integration over -0.5 \leq t \leq 0.5 and
summing over the interval gives the assumtion. With the help of a calculator one gets the estimate:

M < ln(24017) +  \sum^8_(k=1) 1/k + \sum^4_(k=1) 1/k < 10.09 + 2.72 + 2.09 = 14.90.

(ii) Without a calculator (which is forbidden at the DeMo) one can obtain M < 24 like that: There 2^s integers
with 2^s \leq k \leq 2^(s+1) so we get:

\sum^(2^(s+1)-1)_(k=2^s) 1/k < \sum^(2^(s+1)-1)_(k=2^s) 1/2^s = 1 which implies
\sum^(2^(s+1)-1)_(k=2^s) 1/k < M

Hence we get because of 16383 = 2^14 - 1

M < \sum^16383_(k=1) 1/k + \sum^15_(k=1) 1/k + \sum^7_(k=1) 1/k < 14 + 4 + 3 = 21.

b.)

For all odd N Numberfully can find a way in order to leave the square lattice.

Denote the length of a way as the number of eaten numbers or the crossed squares on this way respectively. The
original distance will be denoted by the shortest way from the starting square to the current square. Let a
regular way denoted by a way whose starting point is the middle square and whose original distance increases at
each step.
Let us consider composition of ways which will be called collections. A collection includes one or more ways and
each way may occur more than once. (That's exactly the difference between a collection and a non-empty set of ways.
Every set has the property that each element is only included once.) In the collection the multiple occurence
a certain is the number how often a certain way is included in the collection. Ways belonging to a collection will
be called elements.
The N x N square lattice will be included in the middle of a bigger (2N-1) x (2N-1) square lattice where there are
further numbers like in the small square lattice. Consider regular ways of length N-1. Each way will make
Numberfully leave the N x N square lattice.

We show that for each positive integer n there is a collection C_n of ways with the following property:
The collection C_n includes finite many ways. Each way of C_n is regular and has lenth n. Consider two arbitrary
squares with the same original distance. Then both have the same number of ways from C_n (all multiple occurences
included). The proof is by induction on n:

n = 1:

All the four ways with lenth 1 beginning at the starting square to its four neighbour squares generate a collection C_n.

n-1 -> n:

Assume there is a collection C_k of ways with length which satisfy the required property. Based on this collection
we construct further collections C_k,1 , C_k,2 , ... , C_k,(k+1) whose elements are regular ways with length k+1. Each
of these collections contains as many elements as C_k. Each of the collections C_k,i will generated from C_k like that:
Within 4k+4 squares with original distance k+1 four squares are selected whose mid(dle) points generate a square. There
k+1 possibilities to do that. Each of the possibilities is denoted by a collection C_k,i. Each way from C_k will be
extented to a square with original distance k+1 so that the four chosen destination goals will be not reached. The
extended ways generate the collections C_k,i. So the collection C_(k+1) will be generating by putting together all the
collection C_k,i. Ways being elements of several C_k,i will be added according to its multiple occurence.

Let C := C_(N-1) be a finite collection of regular ways with length N-1 and with the property that squares whose
original distance equals will appear in the same number of ways. The number of elements of C is denoted by c > 0.
The idea is to find an estimate for the average weight of those ways of C. And then one applies the pigeonhole
principle in order to show there is at least one way which is not greater than the average weight.

If a square has the distance k from the starting point and the square is assigned with the number n, then this square
will be crossed by c/(4k) ways of C. So it 'contributes' c/(4kn) to the average weight. G_k denotes the amount of
numbers of all squares with distance k to the average weight. So we get:

G_k = 1/(4k) * (1/n_k,1 + 1/n_k,2 + ... + 1/n_k,(4k)).

The average weight G will be generated by summing up G_1, G_2,... , G_(N-1). G takes the biggest value if we assume
the following situation: n_1,1,...,n_1,4 \in {1,...,4}, n_2,1,... ,n_2,8 \in {5,...,12} etc. So G may be estimated:

G \leq 1/4 * (1 + 1/2 + 1/3 + 1/4) + 1/8 * (1/5 + ... + 1/12) +

       ... + 1/[4(N-1)] * (1/[4*1 + ... + 4(N-2) + 1] + ... + 1/[4*1 + ... + 4(N-2) + 4(N-1)])

  \leq (1/4)*4*1 + (1/8)*8*(1/5) + ... + (1/[4N-4])*4N-4*(1/[4*1 + ... + 4(N-2) + 1])

  \leq 1 + 1/5 + ... + (1/[4*1 + ... + 4(N-2) + 1])

  \leq 1 + 1/4 + ... + (1/[4*1 + ... + 4(N-2)])

  = 1 + (1/4) * \sum^(N-2)_(i=1) 2/[i(i+1)]

  = 1 + 1/2 * (\sum^(N-2)_(i=1) 1/i - \sum^(N-2)_(i=1) 1/(i+1))

  \leq 1 + (1/2)*1

  < 2.

As the average weight which is connected with the ways of the collection is less than 2, we may conclude that within these ways there must be a single regular way where there only numbers whose whole weight is less than 2 kg.  Consequently Numberfully finds a way to leave the square lattice. The proof also shows that one also might assume a weight of 1.5 kg.

Wow orl, is this the official solution ?

This problem was the most challenging one at the DeMo2003. Even several IMO2003 participants did not too well at it. Most solutions were only directed to the special case N = 2003 and often could not be extended in order to cover part b.). The IMO participant Christian Reiher received the highest score of 5/7 at this problem. But also his solution included several small mistakes. The original idea of the problem is the fact that the sum 1/4 + 1/9 + ... is finite. So one might get the idea to distribute these numbers in the plane in a certain way. And moving along the axes you will always get a finite sum. The problem may be easily formulated by eating numbers and after that the caterpillar Numberfullly was included in the problem very quickly as well. The problem was created by the German student Ben Liese.

After this BIG terrible (without any offence, Orlando) solution I MUST write my supervisor's "probabilistic" solution.

We consider only ways of caterpillar which hold away the caterpillar from cell [0,0].

It is clear Probability(caterpillar will visit cell [0,0])=1.

Let us construct way of caterpillar in such way that the following holds:
P(caterpillar will visit cell [x,y])=1/4(|x|+|y|).

It is easy to do by induction on n=|x|+|y|. For example, we can move from [0,0] with probability 1/4 to any of four cells: P([1,0])=P([0,1])=P([0,-1])=P([-1,0])=1/4. From [1,0] we move with probability 1/8 to [2,0] and with probability 1/16 to [1,1] and [-1,1] (1/8+1/16+1/16=1/4). The same we do with [1,0],...  Further we move from [2,0] to [3,0] with probability 1/12 and to [2,1] and [2,-1] with probability 1/48, from [1,1] to [2,1] and [1,2] with probability 1/16 and so on. You may draw figure to understand this.

Let us calculate expected weight of numbers on caterpillar way \sum Weight[x,y]*P([x,y]). If this value is less than 2 then there are way with weight<2 and answer will be NO.

\sum Weight[x,y]*P([x,y]) = \sum {s=1,2,3,...} (1/4s * \sum {|x|+|y|=s} Weight[x,y] ) \leq 1/4(1+1/2+1/3+1/4) + 1/8(1/5+1/6+...+1/12) + 1/12(1/13+...+1/24)+...<2 !!!!

[quote="Myth"][quote="orl"]Prove part b.) for at most 1 kg.[/quote]
That is actually what I did!
Indeed, in my last estimate:
\[\textstyle \frac{1}{4}\cdot \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) + \frac{1}{8}\cdot \left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{12}\right) + \frac{1}{12}\cdot \left(\frac{1}{13}+...+\frac{1}{24}\right)+...\approx 0.847<1.\]

That's all.

I hope we can move it to Solved Section now. :)[/quote]

I did not look at your proof so I assumed it was just proven what was requested. Excellent performance, Myth. I think grobber must be aware that he get chased by you in the number of solved problems. Congrats for your 2000th post and keep up the superior work.  :)

[quote="Myth"]:blush:  :blush:  :blush:
I am sure that Grobber and Pierre are superior solvers.[/quote]

You are wrong, Myth.  :P  You are [b]ALL[/b] superior problem-solvers. And thanks your efforts all the old problems in the forums get cleared up.  ;)

Oh no !

please don't move the problem to the Solved Section. There is a whole universe of questions around it. Here is one which probably will never have a final answer.

Let N be a fixed odd number. For a given distribution D of the numbers 1,...,N^2 on the square we denote by  F(D,P) the weight of the numbers which the caterpillar has eaten on its path P. Find good upper and lower estimates of

S(N) := Max_D Min_P F(D,P)

Myth has shown that S(N) < 0,8466016616 for all N. What about lower estimates? Are there ``good'' strategies for distributing the numbers such that Min_P F(D,P) becomes large?

And what happens if the weight of the number n is not 1/n but f(n), for example f(n)=1/n^p?

So many problems ...

Dear Elias,

if you want to have your additional questions answered I request you to make a new post for it. Thanks.

BTW did you notify Dieter Gronau about the post with the past IMO problem authors ?


How many different 4 digit numbers can be created by rearranging the digits in 2005?

[quote="math92"][hide=" :D  :rotfl: "]
I think you would do 4! (4*3*2*1)  That gives you 24.[/hide][/quote]

TRY AGAIN![hide="hint"]dont forget there are 2 0's![/hide]

Hey there,

do numbers with first digit 0 count? If not, there are 6 different ways:

[hide]2005
2050
2500
5002
5020
5200[/hide]

[quote="john_nash"]Hey there,

do numbers with first digit 0 count?[hide] If not, there are 6 different ways:

2005
2050
2500
5002
5020
5200[/hide][/quote]

You are correct! :) No, numbers that begin with 0 do not count because then it would not be a 4 digit number!

[hide]
Numbers with 0s in the front aren't 4 digits, so it's:
4!/2!=12.
Since this includes zeros, you have to subtract  and  for each one with 0, and there are 6 of them.
So you have 6[/hide]

[quote="pascal_1623"]Very tricky
[hide]
Numbers with 0s in the front aren't 4 digits, so it's:
4!/2!=12.
Since this includes zeros, you have to subtract  and  for each one with 0, but since there can be 0 and 0 consecutive and have 25 or 52, you subtract one 1.
So you have 5[/hide][/quote]

Actually there are 6 and I can't figure out what exactly you did wrong.

Thats easy.
[hide]$4!/2!=12$ Since there are 2 0s you subtract 2 from 12 and get 10, then you subtract 4 from 10 since there are 4 digits in 2005. There are 6 ways.[/hide]


inequalities
Cauchy Inequality
inequalities unsolved

Let $ a,b,c > 0$ .Prove that :
$ \frac {a^4}{ab^2 \plus{} 1} \plus{} \frac {b^4}{bc^2 \plus{} 1} \plus{} \frac {c^4}{ca^2 \plus{} 1} \geq \frac {abc(a \plus{} b \plus{} c)}{1 \plus{} abc}$

[quote="M.A"]first i used T2's lemma after that i multipled. and the problem solved. :wink:[/quote]

T2's lemma?

T2's lemma=cauchy lemma=another format of cauchy inequality

here is the solution: (so sorry i dun know how to type the summation sign,all summation are cyclic here)

summation{a^4/(ab^2+1)}>=abc(a+b+c)/(1+abc)
by cauchy lemma,

<=>(summation{a^2})^2/(summation{ab^2+3}=3(summation{a^2})^2/3(summation{ab^2+3}>=(summation{a^2})(a+b+c)^2/3(summation{ab^2+3})>=abc(a+b+c)/(1+abc)

<=>(summation{a^2})(a+b+c)^2/3(summation{ab^2+3})>=abc(a+b+c)/(1+abc)---(*)

consider summation{a(b^4)c+a^3(b^2)c}>=summation{2a^2(b^3)c}---(1)
             summation{a^3}+summation{ab^2+ac^2}>=9abc----(2)

(1)+(2),it yields (*), equal sign holds as a=b=c.

BTW, i wonder if there is any other solution, esp SOS. it's reali hard to make it homogenous.....

[quote="conan_naruto236"]Let $ a,b,c > 0$ .Prove that :
$ \frac {a^4}{ab^2 + 1} + \frac {b^4}{bc^2 + 1} + \frac {c^4}{ca^2 + 1} \geq \frac {abc(a + b + c)}{1 + abc}$[/quote]
This is my proof :
Let $ abc = k$ , we have :
$ \frac {a^4}{ab^2 + 1} + \frac {b^4}{bc^2 + 1} + \frac {c^4}{ca^2 + 1} \geq \frac {abc(a + b + c)}{1 + abc}$
$ \Longleftrightarrow \frac {a^4}{ab^2 + \frac {abc}{k}} + \frac {b^4}{bc^2 + \frac {abc}{k}} + \frac {c^4}{ca^2 + \frac {abc}{k}} \geq \frac {a + b + c}{1 + \frac {1}{k}}$
$ \Longleftrightarrow \frac {a^3}{b(b + \frac {c}{k})} + \frac {b^3}{c(c + \frac {a}{k})} + \frac {c^3}{a(a + \frac {b}{t})}\geq \frac {a + b + c}{1 + \frac {1}{k}}$
That inequality is right because :
By Holder inequality ,we have :
${ (\frac {a^3}{b(b + \frac {c}{k})} + \frac {b^3}{c(c + \frac {a}{k})} + \frac {c^3}{a(a + \frac {b}{k})})(b + c + a)(b + \frac {c}{k} + c + \frac {a}{k} + a + \frac {b}{k}})\geq(a + b + c)^3$
or $ \frac {a^3}{b(b + \frac {c}{k})} + \frac {b^3}{c(c + \frac {a}{k})} + \frac {c^3}{a(a + \frac {b}{k})}\geq \frac {a + b + c}{1 + \frac {1}{k}}$


geometry
algebra
area
rectangle
equation
IMO Shortlist

Ali Barber, the carpet merchant, has a rectangular piece of carpet whose dimensions are unknown. Unfortunately, his tape measure is broken and he has no other measuring instruments. However, he finds that if he lays it flat on the floor of either of his storerooms, then each corner of the carpet touches a different wall of that room. If the two rooms have dimensions of 38 feet by 55 feet and 50 feet by 55 feet, what are the carpet dimensions?

   Sorry to revive.
   Answer:$25*50$.
   Hint:In fact the length of the room that is given to be $55$ is no use.
   We can solve this by solving a diophantine $1056k^2-48kq+1056=0$.($q$ is the lenghth of the room and $k$ is the similarity ratio.


How many ordered triples of positive integers $ (a, b, c)$ exist such that all of the following are true?\\
$ a > b$\\
$ b > c$\\
$ a \div c \equal{} b$\\
$ a < 10$

Notice that $ a$ must have at least $ 3$ proper factors. If it has $ 2$ proper factors, it is a square and $ b\equal{}c$, for example $ (9,3,3)$. If it has $ 1$ proper factor, it is prime and $ a\equal{}b$, for example $ (7,7,1)$. These cases are not allowed as the rules dictate that the three variables are distinct integers.

There are two numbers below $ 10$ that have $ 3$ proper factors, and none that have more. These two numbers are $ 6$ and $ 8$, which clearly bring about the corresponding triples $ (6,3,2)$ and $ (8,4,2)$. Thus, there are $ \boxed{2}$ triples of positive integers that satisfy all four of the given conditions.


geometry
symmetry
calculus
LaTeX
induction
number theory
modular arithmetic

Hi Friends,

I am a newbie on the forum from India preparing for IMO 2010

Please help me out with the following problems.

[u][b]Number Theory
[/b]
[/u]

1). Find, showing your method, a six-digit integer n with the following  properties:   (i)  n  is  a  perfect  square,  (ii)  the number formed by the last three digits of  n is exactly one greater  than  the  number  formed  by  the  ﬁrst  three  digits of n.  ( for example n might look like 123124 etc)

[u][b]Difference Equations
[/b][/u]

2).  For each positive integer c, the sequence un  of integers is
deﬁned by
$ u1 \equal{} 1$, $ u2 \equal{} c$,    $ u_n \equal{} (2n \plus{} 1)u_{n \minus{} 1} \minus{} (n^2 \minus{} 1)u_{n \minus{} 2}$ , (n ≥ 3).
For which values of c does this sequence have the property that $ u{i}$  divides $ u{j}$  whenever i ≤ j?

What should be the general approach for these kind of number theory problems

[u][b]Geometry
[/b][/u]

3).  A square piece of toast ABCD of side length 1 and centre O is cut in half to form two equal pieces ABC and CDA. If the riangle ABC has to be cut into two parts of equal area, one would usually cut along the line of symmetry BO. However, there are other ways of doing this.  Find, with justiﬁcation, the length and location of the shortest straight cut which divides the triangle ABC into two parts of equal area.

The only approach I can think is setting up a co-ordiante system on the square and trying to minimize cut length through calculus any direct geometrical approach will be really helpful

Thanks in advance  :D [/b]

[b]1)[/b] There are exactly $ 4$ solutions. Make the Ansatz $ N \equal{} X^2 \equal{} 1000Y \plus{} (Y \plus{} 1)$ with $ 316 < X < 1000$ and $ 100\leq Y < 999$ and $ Y \plus{} 1$ describing the number corresponding to the last $ 3$ digits. This yields $ (X \plus{} 1)(X \minus{} 1) \equal{} 7\cdot 11\cdot 13Y$ and $ X\equiv\pm 1\bmod{p}$ for $ p\in\{7,11,13\}$. Taking into account the boundaries for $ X$ we can assume $ X \equal{} 325 \plus{} S \plus{} 13T$ with $ S\in\{\pm 1\},0\leq T\leq 51$. Working modulo $ 11$ resp. $ 7$ we get $ \pm 1\equiv 6 \plus{} S \plus{} 2T\bmod{11}\Longrightarrow T\equiv \pm 6\minus{}3\minus{}6S\bmod{11}$ resp. $ \pm 1\equiv 3 \plus{} S \plus{} 6T\bmod{7}\Longrightarrow T\equiv 3\plus{}S\mp 1\bmod{7}$. This yields $ T\equiv 7,8,9\bmod{11}$ and $ T\equiv 1,3,5\bmod{7}$. The first congruence gives $ T \equal{} 8 \plus{} R \plus{} 11U$ with $ R\in\{0,\pm 1\},0\leq U\leq 4$. From $ T\equiv 1 \plus{} R \plus{} 4U\in\{1,3,5\}\bmod{7}$ we get as the only candidates $ (U,R)\in\{(0,0),(1,0),(2,\minus{}1),(2,1),(3,\minus{}1),(4,0)\}$ $ \Longrightarrow T\in\{8,19,29,31,40,52\}\;$$ \Longrightarrow X\minus{}S\in \{429,572,702,728,845,1001\}$. A final check of residues and boundaries of $ X$ yields $ X\in\{428,573,727,846\}$ with corresponding numbers $ N\in\{183184,328329,528529,715716\}$.

[b]As for question 2)[/b] pls clarify your problem statement, e.g., I could interpret your recursion as $ u_n \equal{} (2n \plus{} 1)u_{n \minus{} 1} \minus{} (2n \minus{} 1)u_{n \minus{} 2}$ or as $ u_n \equal{} (2n \plus{} 1)u_{n \minus{} 1} \minus{} (n^2 \minus{} 1)u_{n \minus{} 2}$ etc. Pls use Latex for clarification.

[quote="hjbrasch"][b]1)[/b] There are exactly $ 4$ solutions. Make the Ansatz $ N \equal{} X^2 \equal{} 1000Y \plus{} (Y \plus{} 1)$ with $ 316 < X < 1000$ and $ 100\leq Y < 999$ and $ Y \plus{} 1$ describing the number corresponding to the last $ 3$ digits. This yields $ (X \plus{} 1)(X \minus{} 1) \equal{} 7\cdot 11\cdot 13Y$ and $ X\equiv\pm 1\bmod{p}$ for $ p\in\{7,11,13\}$. Taking into account the boundaries for $ X$ we can assume $ X \equal{} 325 \plus{} S \plus{} 13T$ with $ S\in\{\pm 1\},0\leq T\leq 51$. Working modulo $ 11$ resp. $ 7$ we get $ \pm 1\equiv 6 \plus{} S \plus{} 2T\bmod{11}\Longrightarrow T\equiv \pm 6 \minus{} 3 \minus{} 6S\bmod{11}$ resp. $ \pm 1\equiv 3 \plus{} S \plus{} 6T\bmod{7}\Longrightarrow T\equiv 3 \plus{} S\mp 1\bmod{7}$. This yields $ T\equiv 7,8,9\bmod{11}$ and $ T\equiv 1,3,5\bmod{7}$. The first congruence gives $ T \equal{} 8 \plus{} R \plus{} 11U$ with $ R\in\{0,\pm 1\},0\leq U\leq 4$. From $ T\equiv 1 \plus{} R \plus{} 4U\in\{1,3,5\}\bmod{7}$ we get as the only candidates $ (U,R)\in\{(0,0),(1,0),(2, \minus{} 1),(2,1),(3, \minus{} 1),(4,0)\}$ $ \Longrightarrow T\in\{8,19,29,31,40,52\}\;$$ \Longrightarrow X \minus{} S\in \{429,572,702,728,845,1001\}$. A final check of residues and boundaries of $ X$ yields $ X\in\{428,573,727,846\}$ with corresponding numbers $ N\in\{183184,328329,528529,715716\}$.[/quote]

Dear Sir, thank you for the beautiful solution to this problem, I am a beginner to this kind of advanced modular arithmetic and numerical analysis, could you provide me with some good resources or books to learn more about the same

[quote="hjbrasch"][b]As for question 2)[/b] pls clarify your problem statement, e.g., I could interpret your recursion as $ u_n \equal{} (2n \plus{} 1)u_{n \minus{} 1} \minus{} (2n \minus{} 1)u_{n \minus{} 2}$ or as $ u_n \equal{} (2n \plus{} 1)u_{n \minus{} 1} \minus{} (n^2 \minus{} 1)u_{n \minus{} 2}$ etc. Pls use Latex for clarification.[/quote]

Sir it is the latter $ u_n \equal{} (2n \plus{} 1)u_{n \minus{} 1} \minus{} (n^2 \minus{} 1)u_{n \minus{} 2}$

2) The only solutions are $ c\equal{}2,4$. First, calculate $ u_3\equal{}7c\minus{}8\equal{},u_4\equal{}48c\minus{}72,u_5\equal{}360c\minus{}600, u_6\equal{} 3000c\minus{}5280$. Checking the divisibility property observe that $ u_2\equal{}c|u_3\equal{}7c\minus{}8$ implies $ c\in\{1,2,4,8\}$. If $ c\equal{}8$, we get the contradiction $ u_3\equal{}48|u_4\equal{}312$. The case $ c\equal{}1$ yields the contradiction $ u_5\equal{}\minus{}240|u_6\equal{}\minus{}2280$.
For $ c\equal{}4$ we have $ u_1\equal{}1,u_2\equal{}4,\forall n\geq 3: u_n\equal{}\frac{(n\plus{}2)!}{3!}$ and for $ c\equal{}2$ we have $ u_n\equal{}n!$ (both assertions are verified by straightforward induction on $ n$). These series obviously have the divisibility property, since $ \forall n: \,u_n|u_{n\plus{}1}$ completing the proof.

P.S. Note, that for general $ c$ the series has many nice properties, e.g., $ \forall n\geq 3: u_n\equal{}(n\plus{}2)u_{n\minus{}1}\plus{}(n\minus{}1)!(c\minus{}4)$ holds.


find all$ x,y \in N$ that :
\[ 2x^2y^2 \plus{} y^2 \equal{} 26x^2 \plus{} 1201
\]

[hide]
$ (2x^2 \plus{} 1)(y^2 \minus{} 13) \equal{} (2)^2(3)^3(11)$

$ 2x^2 \plus{} 1$ is odd so it can't have any factors of $ 2$
$ 2x^2 \plus{} 1 \equal{} {1,3,9,11, 27, 33, 99, 297}$
Taking intiger solutions
$ x \equal{} \pm({1,2,4,7})$

$ \Rightarrow y^2 \minus{} 13 \equal{} {396,132,36,12}$
Taking intiger solutions
$ y \equal{} \pm({7, 5}) \longrightarrow x \equal{} \pm({4,7})$

So the only solutions are $ (\pm 4, \pm 7)$ and $ (\pm{7},\pm{5})$[/hide]


[b]Mods, could you move this? Sorry! [/b]So all of you love Mathcounts..... but does anyone like georaphy, especially the National Geographic Bee? Its very demanding (I participated in it!), but is pretty cool, giving off a Mathcounts-like vibe. Sign up by October 15th!

P.S. Here's a sie to look into: geobeecentral.com.I'm samus there, too. Its comprehensive and still growing. Math geniuses in AoPS, geography geniuses in Geobeecentral (Really, look at some of the members).

P.P.S. Go to the message board on the bottom and sign up. You'll drop math and go to geography in a day.

I'm not advertising, I don't own the site.... just to let you know.

My junior high school participated in the national geography bee when I was in seventh and eighth grade. I won my school competition when I was in seventh grade, but I wasn't very interested in competitions back then, so I didn't try very hard on the round to get to the state level. (To be honest, one of my teachers was giving an interesting lecture that I didn't want to miss, so I rushed through it.) Therefore I ended there. I then completely messed up in eighth grade and missed a question in the class round and was eliminated right then.

Incidentally, Peter VDD, if you want to say that writing in big letters is not a good idea, then it might be wise for you not to do the same thing. I think you have already made quite a few moderators deaf to your requests.

[quote="ComplexZeta"]Incidentally, Peter VDD, if you want to say that writing in big letters is not a good idea, then it might be wise for you not to do the same thing.[/quote]

ugh? :shock:

Can you please show me one post of mine where I wrote in big letters or shouted ay anyone??  :|

[quote="Peter VDD"]No need to [size=200][b]Shout![/b][/size] at the mods like that, you'll make them deaf so they don't hear you next time :D[/quote]

I got to the state bee once on a total fluke - I hadn't studied at all for the local round. I was able to get to states, but then I got nervous and got only 3 questions right in the preliminary round. My brother has gotten to states the past two years. This year he is really studying geography hard because he wants to make it at least to the final round of states. I help him some, though I'm not terribly interested in geography. My brother is though; he has said he may even major in geography.

I'm in 8th grade now and almost won the school bee last year - wth 6 out of 7. Geography's not a thing I take too competitively like I do with math and writing, so I didn't study, and I *forgot* on my 7TH QUESTION that Wellington and Aukland were in New Zealand. Oops. Anyway, that's not as bad as another sixer last year who screwed up on what country Vienna was in. (Think Mozart, you uneducated turd!)
So I want to do better this year.

[color=green]Oops. :blush: Anyways, congrats to you guys. With geography, I won't ever forget trying to correctly spell all the countries and capitals and stuff in the Middle East and Africa. ;) One would have to admit that it's pretty hard trying to spell something correctly when you've only heard of it once or twice in your lifetime. :?[/color]

Ok    3 Challenge questions in Georgraphy.

See how you do  without any references (no googling) and then check yourself. If you know the answer please don't answer it here (or put it in spoilers) just tell us how well did you do.

Q1: What is the north-most, south-most, east-most and west-most state in United states of America.

Q2. Name a 5 letter capital which  does not contain (a,e,i,o or u),  and will get you a scrabble score of 20+

Q3.  What is the highest peak in India if Himalaya mountains  are excluded.

In seventh grade, people at my school weren't allowed to compete for some reason.

In eighth grade, I won the school bee.  I did what was actually worse than I could have on the written test, but got to state anyway.  There, I got the strangest questions I've ever heard, and thus was eliminated quickly even though I had studied a lot

Oh well.  I'll never forget that 'truck painting' is common in Pakistan!

[hide]

1)  North-most:  Alaska

     South-most:  Hawai'i (?)

     West-most:  Alaska

[/hide]

Okay - A hint for:
[quote]Q2. Name a 5 letter capital which does not contain (a,e,i,o or u), and will get you a scrabble score of 20+ [/quote]
Here is where an obelisk stands to mark "the center of Asia"
Can  you name the city?
  [img]http://www.fotuva.org/travel/images/centre_small.jpg[/img]

Okay another hint: The cute  four year native  from the said place is  being friendly with   another native in upsatate  Ney York,
[img]http://www.petoftheday.com/archive/2000/June/15.jpg[/img]

Are you reffering to a national capital?

Well   it used to be an independent  nation, <rest in spoiler> [hide] it was a part of former USSR but now know as (If I am not wrong) Republic of Tuva and it is a member of Russiona fedration. For the name of the capital see the above picture, it is spelled  there [/hide]


Given $ n>0$ be an integer.Let $ A$ be the set of all integers $ m$ such that $ |m|<2n$.Prove that whenever $ 2n\plus{}1$ elements are chosen from $ S$,there exist $ 3$ elements whose sum is $ 0$.


Given $ \frac{\log a}{p} \equal{} \frac{\log b}{q} \equal{} \frac{\log c}{r} \equal{} \log x$, all logarithms to the same base and $ x \neq 1$. If $ \frac{b^2}{ac} \equal{} x^y$, then $ y$ is:

$ \textbf{(A)} \frac{q^2}{p\plus{}r} \qquad \textbf{(B)} \frac{p\plus{}r}{2q} \qquad \textbf{(C)} 2q \minus{} p \minus{} r \qquad \textbf{(D)} 2q \minus{} pr \qquad \textbf{(E)} q^2 \minus{} pr$

From the given we have $ x^p\equal{}a$, $ x^q\equal{}b$, and $ x^r\equal{}c$. $ \frac{b^2}{ac}\equal{}\frac{x^{2q}}{x^{p\plus{}r}}\implies{y\equal{}2q\minus{}p\minus{}r} (C)$


A student spins the spinner twice and records the number that
the spinner lands on each time. What is the probability that the
sum of the two recorded numbers is less than 4, if the
probability of spinning each number is $ \frac 14$? Express your answer as a common fraction.

[asy]size(100);
draw(Circle((0,0),20));
draw((20,0)--(-20,0));
draw((0,20)--(0,-20));
label("1",(-8,11),S);
label("3",(-8,-11),N);
label("2",(8,-11),N);
label("4",(8,11),S);
draw((0,0)--(5,11),Arrow);[/asy]

You get a sum of less than 4 when you spin 1-1, 1-2, and 2-1. There is a total of 4*4, or 16 possibilities, so our answer is 3/16.


inequalities
limit
inequalities unsolved

what will be minimum
?? and when achived equality?

[quote="Zamfirmihai"]I have a little challange for you:
prove that if $a,b,c$ are the sides of a triangle then:

$\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}}>\frac{4}{3}$[/quote]
You are wrong.
I find it!
after transformation of this inequality you will get
$\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+\frac{1}{x^{2}y^{2}}}}$
now let $y^{2}=\alpha^{6}$ and $x^{2}=\frac{1}{\alpha^{4}}$  :D . then
we have
$\frac{1}{\sqrt{1+\frac{1}{\alpha^{4}}}}+\frac{1}{\sqrt{1+\alpha^{6}}}+\frac{1}{\sqrt{1+\frac{1}{\alpha^{2}}}}$
then $\lim_{\alpha\to 0}\frac{1}{\sqrt{1+\frac{1}{\alpha^{4}}}}+\frac{1}{\sqrt{1+\alpha^{6}}}+\frac{1}{\sqrt{1+\frac{1}{\alpha^{2}}}}=1$

[quote="Zamfirmihai"]what are x,y?? and the inequality can`t hold because I put $>$, not $\geq$[/quote]
Yes i know you wrote $>$ but it's false.
$x=\frac{b}{a}$ and $y=\frac{c}{b}$

[quote="arqady"][quote="Twins"]$a,b,c \in R^{+}$, prove that:
$1<\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}}\leq \frac{3\sqrt{2}}{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17549[/quote]
And here : http://www.mathlinks.ro/Forum/viewtopic.php?t=148356

[quote="HTA"][quote="Extremal"][quote="Zamfirmihai"]$2\sum_{cyc}\frac{a^{2}}{2a^{2}+b^{2}}\geq2\frac{(\sum a)^{2}}{3\sum a^{2}}>\frac{2}{3}*2=\frac{4}{3}$ because $a<b+c \Rightarrow a^{2}<ab+ac \Rightarrow a^{2}+b^{2}+c^{2}<2ab+2ac+2bc$[/quote]
when you use this:
$a<b+c \Rightarrow a^{2}<ab+ac \Rightarrow a^{2}+b^{2}+c^{2}<2ab+2ac+2bc$
it means :
$a<b+c$
$b<a+c$
$c<a+b$
but we don't know it.
you must prove it for all real $a,b,c >0$ :wink:
or in other words:
$a^{2}+b^{2}+c^{2}<2ab+2ac+2bc$
if put $a\to 0$ then  :
$b^{2}+c^{2}<2bc$
holds?[/quote]
i think we already have a,b,c are three sides of a triagle  :roll:[/quote]

[quote="Twins"]$a,b,c \in R^{+}$, prove that:
$1<\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}}\leq \frac{3\sqrt{2}}{2}$[/quote]

[quote="Extremal"][quote="Zamfirmihai"]$2\sum_{cyc}\frac{a^{2}}{2a^{2}+b^{2}}\geq2\frac{(\sum a)^{2}}{3\sum a^{2}}>\frac{2}{3}*2=\frac{4}{3}$ because $a<b+c \Rightarrow a^{2}<ab+ac \Rightarrow a^{2}+b^{2}+c^{2}<2ab+2ac+2bc$[/quote]
when you use this:
$a<b+c \Rightarrow a^{2}<ab+ac \Rightarrow a^{2}+b^{2}+c^{2}<2ab+2ac+2bc$
it means :
$a<b+c$
$b<a+c$
$c<a+b$
but we don't know it.
you must prove it for all real $a,b,c >0$ :wink:
or in other words:
$a^{2}+b^{2}+c^{2}<2ab+2ac+2bc$
if put $a\to 0$ then  :
$b^{2}+c^{2}<2bc$
holds?[/quote]
I said that $a,b,c$ are the sides of a triangle


function
vector
algebra unsolved
algebra

Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfying
$f(x)+f(y)^2=f(x+y^2) \quad \forall x, y\in \mathbb{R}$

First, $x=y=0$ gives $f(0)=0$. Then, $x=0$, $y=1$ gives $f(1)=f(1)^2$ so $f(1)=1$. (Otherwise there's the $f(x)\equiv0$ solution).

$y=1$ then gives $f(x+1)=f(x)+1$ for all $x$. If $x=n$ (positive integer) we then have

$f(n)=1+f(n-1)=2+f(n-2)=\cdots=n+f(0)=n$.

In addition $f(-1)=f(0)-1=-1$ and so on, giving $f(n)=n$ for all integers $n$.

Now we have to extend this to all reals. Let $z\in\mathbb{R}$ and $y$ be any integer and define $x=z-y^2$. Then $f(z)=f(x+y^2)=f(x)+f(y)^2=f(x)+y^2=f(x)+z-x$ giving

$f(z)-z=f(x)-x$ for all $x,z\in\mathbb{R}$ which differ by a square.

This seems to go towards $f(x)=x$ for all $x$ but where to go next?

Given $x=y=0$we have:$f(0)=0$
Given$x=0$we have:$f(y^2)=(f(y))^2$
$\Rightarrow f(x+y)=f(x)+f(y) \forall x\in R,y\geq 0$
Given $x=-y,y>0$,we have:$f(-y)=-f(y)\forall y$
$\Rightarrow f(x+y)=f(x)+f(y) \forall x,y$
But $f(x)=(f(\sqrt{x}))^2 \Rightarrow f(x)\geq 0 \forall x\geq 0$
$\Rightarrow f(y)=f(x)+f(y-x)\geq f(x) \forall y>x$
$\Rightarrow f$ is monotonic$\Rightarrow f(x)=kf(x)$
$f(x^2)=(f(x))^2\Rightarrow k=0$or$1\Rightarrow f(x)\equiv 0$or$f(x)=x$

[quote="dondigo"][b]Vietnamesegauss89[/b]. How did you know, that if a function is monotonic, then $f(x)=kf(x)$.[/quote]
Because $f(x+y)=f(x)+f(y) \forall x,y \in R$and f is monotonic
Can you understand [b]dondigo[/b]? :lol:

[quote="dondigo"][b]Vietnamesegauss89[/b]. How did you know, that if a function is monotonic, then $f(x)=kf(x)$.[/quote]

It's a classical result, proven by Cauchy :)

Are you sure this result is true ?
I know it is true for continuous function, or continuous at the origin function.
But I neved heard it to be true for monotonic functions. Can you give some link ?

It's true for any function that isn't to ugly.  ;)
More exactly, every solution to $f(x+y)=f(x)+f(y)$ different from $f(x)=kx$ is dense in the plane, which is ruled out be all things mentioned above.
Proof that it is indeed dense:
Assume $f(z) \neq f(1) \cdot z$, or more generaly (indeed it's no difference) $f(u) \neq \frac{u}{v} f(v)$.
By the known methods for this equation we have for any rational $r$: $f(rx)=rf(x)$.
Now it suffices to show that when given two vectors $u,v$, non lineary dependent as real vectors, that the sums $su+tv$ with rational $s,t$ are dense in the plane.
Since $u,v$ build a base, we can at first find real $s',t'$ such that $s'u+t'v$ is a given point. Now just using that the rationals are dense in the reals, we get our 'rational point' $(s,t)$ (seen in respect to this base) as close as we want (when $\epsilon >0$ is given, take some rationals $s,t$ with $|u| \cdot |s'-s| , |v| \cdot |t'-t| < \frac{\epsilon}{2}$, then $|(su+tv)-(s'u+t'u)| \leq |(s-s')u| + |(t-t')v| < \epsilon$).


Let a,b,c,d be real of numbers,a$\neq$0.Prove that:
   there exist x,y are real numbers satisfy that f(x)<0; f(y)>0

The equation has the form $ax^3+bx^2+cx+d$.
We know that every third degree equation has at least one solution in $R$.

Let  $r$ the root of the equation and let  $r\in\left (k,l\right )$.

Then by Bolzano's theorem we have the desired result choosing the $x,y$

$\lim_{x \to \infty} f(x) = \pm \infty$ and $\lim_{x \to \infty} f(x) = \mp \infty$.  :D

Can you post the detail solution,silouan???
  I don't know Bolzano's theorem
 Anyone has the solution???HELP ME


algebra
polynomial
superior algebra
superior algebra unsolved

How would one create a field of 8 elements? How about 14 elements? What's the general process?

The only possible orders of finite fields are of the form $ p^k$ where $ p$ is a prime.  One creates such a field by creating the field with $ p$ elements $ \mathbb{F}_p$ and quotienting by an irreducible polynomial of degree $ k$.  For example, one can construct the field $ \mathbb{F}_8$ by finding an irreducible cubic over $ \mathbb{F}_2$.  A field of order $ 14$ does not exist.

So $ \mathbb{Z}_{2}[x]/(x^{3}\plus{}x\plus{}1)$ is a field with 8 elements? Although how do I prove that all elements in $ \mathbb{Z}_{2}[x]/(x^{3}\plus{}x\plus{}1)$ are of the form $ ax^{2}\plus{}bx\plus{}c$ where $ a, b, c \in \mathbb{Z}_{2}$?

[quote="1/(ln x)"]Although how do I prove that all elements in $ \mathbb{Z}_{2}[x]/(x^{3} \plus{} x \plus{} 1)$ are of the form $ ax^{2} \plus{} bx \plus{} c$ where $ a, b, c \in \mathbb{Z}_{2}$?[/quote]
Since you can replace $ \overline{x}^3$ with $ \minus{}\overline{x}\minus{}1$, you can keep reducing the highest power of $ \overline{x}$ in a polynomial expression until it is no more than 2.


geometry
circumcircle
vector
geometry proposed

We consider an acute triangle $ \ ABC$ cu $ \ I$ the incenter.$ \ AI$ meets $ \ (BC)$ at $ \ D$, $ \ BI$ meets $ \ (CA)$ at $ \ E$ and also $ \ CI$ meets $ \ (AB)$ at $ \ F$.We take the points $ \ I1,I2,I3$ so that $ \ D$-midpoint of $ \ (II1)$,$ \ E$-midpoint of $ \ II2$ and $ \ F$-midpoint of $ \ II3$.If $ \ I$ is the circumcenter of traingle $ \ I1I2I3$ show that $ \ ABC$ is equilateral.

[i]Stancioiu Sorin,, Romania[/i]

[hide]The problem is trivial if we use vectors but this pb I think it's a 7-th grade problem because I found a solution without vectors and only with some known ineq.[/hide]

II1=II2=II3->ID=IE=IF
 Define that incircle meets triangle at A1,B1,C1.
 (IA1D)==(IB1E)==(IC1F) is trivial. And make AB=a+b, BC=b+c, CA=c+a, DA1=x

 a+b:b+c=a-x:c+x
 a+c:a+b=c+x:b-x
 a+c:b+c=a+x:b-x

  calculate it and we get x=0
 Therefore incircle=circumcircle-> It is equilateral.


[quote="Jumbler"]Let $a,b,c\in R^{+}$, $a+b+c=2$
Prove $\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca}\geq \frac{27}{13}$[/quote]
$\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca}\geq\frac{9}{3+ab+ac+bc}.$
And remain to prove that $\frac{9}{3+ab+ac+bc}\geq\frac{27}{13},$ which trivial.


function
calculus
integration
algebra proposed
algebra

let $ x_k$ denote a solution to $ \frac {f(x)}{k^n} \equal{} k \minus{} x$. it's clear that $ x_k\in [0,k]$, and in fact, as shown below, for $ k\geq 2$, $ x_k\in [1,k \minus{} 1]$.

it's easy to check that $ f(x) \equal{} (x \plus{} 1)^n$ satisfies the conditions: it's strictly increasing, and we may take $ x_k \equal{} k \minus{} 1$. now we'll show that this is the only function.

by considering $ k \equal{} 1$, there's an $ x$ s.t. $ f(x) \equal{} 1 \minus{} x$. we can't have $ x \equal{} 1$ or else $ f(1) \equal{} 0$, in which case $ f(0) < f(1)$ is impossible. thus $ x \equal{} 0$, and $ f(0) \equal{} 1$. this is our base case.

now assume that $ f(x) \equal{} (x \plus{} 1)^n$ for $ x \equal{} 0, 1, ..., r$, $ r\geq 0$.

note that for $ k\geq 2$, the RHS of the equation $ \frac {f(x)}{k^n} \equal{} k \minus{} x$ is at least $ k \minus{} (k \minus{} 1) \equal{} 1$ and at most $ k \minus{} 1$. this is because we cannot have $ x \equal{} k$ (or else $ f(k) \equal{} 0$, which, as above, is impossible) and we cannot have $ x \equal{} 0$ (or else $ f(0) \equal{} k^{n \plus{} 1}$>1, contradicting $ f(0) \equal{} 1$).

it then follows that for every $ k\geq 2$, $ k^n\leq f(x_k) < k^{n \plus{} 1}$, where $ x_k\in [1,k \minus{} 1]$ (*)

now, by (*), $ (r \plus{} 2)^n\leq f(x_{r \plus{} 2}) < (r \plus{} 2)^{n \plus{} 1}$ and $ x_{r \plus{} 2}\in [1,r \plus{} 1]$. by the inductive hypothesis, $ f(0), ..., f(r)$ are all $ \leq (r \plus{} 1)^n < (r \plus{} 2)^n$. thus $ x_{r \plus{} 2} \equal{} r \plus{} 1$, from which we get $ \frac {f(r \plus{} 1)}{(r \plus{} 2)^n} \equal{} r \plus{} 2 \minus{} (r \plus{} 1) \equal{} 1$, i.e. $ f(r \plus{} 1) \equal{} (r \plus{} 2)^n$, completing the induction.

[quote="x164"]Let $ n \in \mathbb{Z}_{ > 0}$ be a positive integer.

Find all strictly increasing functions $ f: \mathbb{Z}_{\geq 0} \rightarrow \mathbb{Z}_{\geq 0}$ such that the equation
\[ \frac {f(x)}{k^n} \equal{} k \minus{} x\]
has an integral solution $ x$ for all $ k \in \mathbb{Z}_{ > 0}$.[/quote]
For every $ k$ let $ h : \mathbb{N} \to \mathbb{N}_0$ be the solution such that:
$ \frac {f(h(k))}{k^n} \equal{} k \minus{} h(k)$
Let $ g(k) \equal{} \frac {f(h(k))}{k^n}$. Obviously $ g(k) \in \mathbb{N}_0$.
We know that $ h(k) \le k$, and it is easy to prove that $ h(k)$ is injective. (If $ h(k_1) \equal{} h(k_2)$ then $ k_1 \minus{} \frac {f(h(k_1))}{k_1^n} \equal{} k_2 \minus{} \frac {f(h(k_1))}{k_2^n}$)
If $ h(n) \equal{} n$ for some $ n$ then $ f(n) \equal{} 0$, but $ n \ge 1$ and $ f(n) > f(0)$ is impossible. Hence $ h(k) \le k \minus{} 1$. Since $ h$ is injective we easily find $ h(k) \equal{} k \minus{} 1 \forall k \ge 1$ and therefore $ f(k \minus{} 1) \equal{} k^n \forall k \ge 1 \iff f(x) \equal{} (x \plus{} 1)^n \forall x \ge 0$.
So the only solution is $ f(x) \equal{} (x \plus{} 1)^n \forall x \ge 0$ which is easily seen to be a solution.

We can replace the "strictly increasing function $ f : \mathbb{N}_0 \to \mathbb{N}_0$" condition with "any function $ f : \mathbb{N}_0 \to \mathbb{N}$".

[quote="voong"][quote="pco"][quote="Mathias_DK"] ...We know that $ h(k) \le k$, and it is easy to prove that $ h(k)$ is injective. (If $ h(k_1) \equal{} h(k_2)$ then $ k_1 \minus{} \frac {f(h(k_1))}{k_1^n} \equal{} k_2 \minus{} \frac {f(h(k_1))}{k_2^n}$)
[/quote]

And why would "$ k_1 \minus{} \frac {f(h(k_1))}{k_1^n} \equal{} k_2 \minus{} \frac {f(h(k_1))}{k_2^n}$" be impossible ?[/quote]

because $ p(x) \equal{} x \minus{} \frac {k}{x^n}$ is strictly increasing in our condition.

( $ h(k_1) \equal{} h(k_2)$ , then $ f(h(k_1)) \equal{} f(h(k_2)) \equal{} k$)[/quote]
Ohh yes. I forgot that $ f>0$. Sorry for my stupid question and thanks for your answer.

For $k = 1$, we have
$$f(a) + a = 1$$
 for some non-negative integer $a$. $a$ here is of course less than $2$. $a = 1$ would imply $f(1) = 0$, which would mean $f(0) < 0$, which is not possible. Therefore, we get $a = 0$ and $f(0) = 1$.
Now we prove that $f(j-1) = j^n$ for all natural numbers $j$ using strong induction on $j$.
The base case $j = 1$ is done.
Assume that all natural $j$ less than $m$ satisfy our induction hypothesis.
Then, considering $k = m+1$ we get
$$f(b) = (m+1)^n(m+1-b)$$
for some natural number $b$. As $f(0) > 0$, $f(b) > 0$. As $(m+1)^n$ must divide $f(b)$, $f(b) \geq (m+1)^n$. This means $b > (m-1)$ from our induction hypothesis. Also, as $(m+1-b)$ needs to be positive, $b < (m+1)$,  which means $b = m$, which shows that
$$f(m) = (m+1)^n$$
as required.
It's easy to check that this satisfies the given conditions.


let $ x,y,z\geq 0$ prove that $ \sum \frac{x}{\sqrt{{y}^{2}\plus{}{z}^{2}}}\geq 2$

Let  $ T \equal{} \sum_{cyc}\frac {x}{\sqrt {y^2 \plus{} z^2}}$. Then by the Holders inequality we get
\[ T^2\cdot\bigl(x(y^2 \plus{} z^2) \plus{} y(z^2 \plus{} x^2) \plus{} z(x^2 \plus{} y^2)\bigr)\geq \bigl(x \plus{} y \plus{} z\bigr)^3 .
\]
So it is enough to prove that    $ \frac {(x \plus{} y \plus{} z)^3}{x(y^2 \plus{} z^2) \plus{} y(z^2 \plus{} x^2) \plus{} z(x^2 \plus{} y^2)}\geq 4$ ,
which is equivalent the following inequality
\[ x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz\geq x(y^2 \plus{} z^2) \plus{} y(z^2 \plus{} x^2) \plus{} z(x^2 \plus{} y^2).
\]
Using Schur's inequality we have:
\[ \displaystyle\sum_{cyc}x(x \minus{} y)(x \minus{} z)\geq 0 ,
\]
or
\[ x(y^2 \plus{} z^2) \plus{} y(z^2 \plus{} x^2) \plus{} z(x^2 \plus{} y^2)\leq x^3 \plus{} y^3 \plus{} z^3 \plus{} 3xyz .
\]
So  $ x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz \geq x^3 \plus{} y^3 \plus{} z^3 \plus{} 3xyz\geq x(y^2 \plus{} z^2) \plus{} y(z^2 \plus{} x^2) \plus{} z(x^2 \plus{} y^2)$ and we are done.

Equality holds iff $ x\equal{}0, y\equal{}z>0$ or it's permutations.

$ \mbox{If: } x^2=a, y^2=b,z^2=c\mbox{, then: }\\
\sum_{cyclic}{\frac{x}{\sqrt{y^2+z^2}}}=\sum_{cyclic}\sqrt{\frac{x^2}{y^2+z^2}}=\sum_{cyclic}{\sqrt{\frac{a}{b+c}}}=\sum_{cyclic}{\sqrt{\frac{a^2}{a.(b+c)}}}=\\
=\sum_{cyclic}{\frac{a}{\sqrt{a.(b+c)}}}\ge \sum_{cyclic}{\frac{a}{\frac{a+b+c}{2}}=2.\sum_{cyclic}{\frac{a}{a+b+c}}=2.}$

$ LHS^2(\sum x(y^2 \plus{} z^2) \ge (x \plus{} y \plus{} z)^3$
$ \plus{} )(x \plus{} y \plus{} z)^3 \ge 4 \sum x(y^2 \plus{} z^2)$
$ < \equal{} > x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz \ge \sum x(y^2 \plus{} z^2)$
Which is true by shur :)

This is shortly I00's proof.
However, I like these proofs very much :lol:


calculus
derivative
geometry
trigonometry
absolute value
calculus computations

The sides a,b,c of a triangle are measured with a possible % error of e and the area calculated. Prove that the possible % error in the area is approximately 2e or 2ecotBcotC according as the triangle is acute-angled or obtuse-angled at A.

I can only show the acute case.

Since you claim that you can do the acute case, I assume that you can show that
\[ \frac{dS}S\equal{}\frac{\cos A}{\sin B\sin C}\frac{da}a
\plus{}\frac{\cos B}{\sin A\sin C}\frac{db}b
\plus{}\frac{\cos C}{\sin A\sin B}\frac{dc}c\]
($ S$ is the area).
Now, when you have three acute angles, all three coefficients are positive, so the sum of their absolute value is the same as their sum, which is $ 2$. If $ A$ is obtuse, the first coefficient is negative, so you need to show that
\[ \minus{}\frac{\cos A}{\sin B\sin C}
\plus{}\frac{\cos B}{\sin A\sin C}
\plus{}\frac{\cos C}{\sin A\sin B}\equal{}2\cot B\cot C\]
which is a straightforward exercise in basic trigonometry.

Unfortunately I started off (perhaps incorrectly I am now thinking) by using the cosine rule to get:

$ a.da\equal{}b.db\plus{}c.dc\minus{}b.cosA.dc\minus{}c.cosA.db\plus{}2.b.c.sinA.dA$

so that

$ dA\equal{}\epsilon\frac{(a^2\minus{}b^2\minus{}c^2\plus{}2.b.c.cosA)}{b.c.sinA}\equal{}0$

by the cosine rule (the error term $ \epsilon\equal{}\frac{da}{a}$ etc).

I then just used $ S\equal{}\frac{1}{2}.b.c.sinA$ to get $ \frac{dS}{S}\equal{}2.\epsilon$ whether A is acute or obtuse.

I am still uncertain why this does not work.

I see that your formula for $ \frac{dS}{S}$ is derived from

$ \frac{dS}{S}\equal{}\frac{db}{b}\plus{}\frac{dc}{c}\plus{}\frac{cosA}{sinA}.dA$

and similarly for the other angles. Then using

$ S\equal{}\frac{1}{2}.a^2.\frac{sinB.sinC}{sinA}$ to get $ \frac{dS}{S}$

and substituting in values for $ dB$ and $ dC$ to get the required formula.

Thanks for the help. :lol:

[quote="apatch"]
I am still uncertain why this does not work.
[/quote]
It does but you should understand that to change the area as much as possible, you should increase $ b$ and $ c$ and [b]decrease[/b] $ a$ in the obtuse case. ;)


inequalities
function
calculus
derivative
integration
geometry
real analysis

Let $ f: [0,1]\to\mathbb R$ be a derivable function, with a continuous derivative $ f'$ on $ [0,1]$. Prove that if $ f\left( \frac 12\right) \equal{} 0$, then
\[ \int^1_0 \left( f'(x) \right)^2 dx \geq 12 \left( \int^1_0 f(x) dx \right)^2.\]

lemma: if $ f(x)\in C^{1}([a,b],R)$ and $ f(a)\equal{} 0 or f(b)\equal{}0$ then we have
$ \left( \int^a_b f(x) dx \right)^2\leq \frac {(a \minus{} b)^2}{2}\int^a_b \left( f'(x) \right)^2 dx$. apply this lemma 2 times for $ (a,b) \equal{} (0,1/2)$ and $ (1/2,1)$

Nice. Where is this lemma coming from though?

my mistake. in fact it is $ \left( \int^a_b |f(x)|^{2} dx \right)\leq \frac {(a \minus{} b)^2}{2}\int^a_b \left( f'(x) \right)^2 dx$ .doesnt matter cos it works anyway.
we can prove it by using Cauchy-Schwartz . supposing $ f(a)\equal{}0$ then $ f(x)^{2}\equal{}(\left( \int^a_t f(x)' dx \right)^2\leq\frac{t\minus{}a}{2}\left( \int^a_b |f(x)'|^{2} dx \right)$ integral it from $ a$ to $ b$ and we have $ \left( \int^a_b |f(x)|^{2} dx \right)\leq \frac {(a \minus{} b)^2}{2}\int^a_b \left( f'(x) \right)^2 dx$

same way with $ f(b)\equal{}0$

i find it quite strange because $ \frac{1}{12}$ is replace by $ \frac{1}{32}$ maybe there r faults somewhere  :)

The best possible constant comes from the $ 2$-norm of $ \frac12\minus{}x$. Since $ \int_0^1 \left(\frac12\minus{}x\right)^2\equal{}\frac1{12}$, this constant is indeed 12. We get equality if and only if $ f'(x)$ is a real multiple of $ \frac12\minus{}x$.

First, write $ f(t)\equal{}\int_{\frac12}^t f'(x)\,dx$. Define $ g(x)\equal{}f'(x)$. The LHS is $ \int_0^1 (g(x))^2\,dx$, and the RHS becomes $ \left(\int_0^1 \int_{\frac12}^t g(x)\,dx\,dt\right)^2\equal{} \left(\int_0^1 \int_x^{\frac12}g(x)\,dt\,dx\right)^2$
$ \equal{}\left(\int_0^1 \left(\frac12\minus{}x\right)g(x)\,dx\right)^2$

In the language of inner products, we want to show that $ \|g\|^2\ge 12(g\cdot h)^2$, where $ h(x)\equal{}\frac12\minus{}x$. The standard C-S inequality gives $ (g\cdot h)^2\le \|g\|^2\|h\|^2\equal{}\frac1{12}\|g\|^2$, and we are done.

Sorry
May be I didn't understand something, but you think that
\[ \left(\int_0^1 \int_{\frac12}^t g(x)\,dx\,dt\right)^2\equal{} \left(\int_0^1 \int_x^{\frac12}g(x)\,dt\,dx\right)^2\]
is true for all $ g(x)$ or a something misunderstood?

But if we put $ g(x) \equal{} x$, first integral will be equal 1/576, and second 1/144.

You can drop the squares- it's about interchanging the order of integration, and that's really $ g(x,t)$.

If the 2-dimensional region were traced positively, there would be no problem. As it is- I didn't account properly for the interchange.

[hide="Erroneous solution hidden"]A corrected version: $ \int_0^1\int_{\frac12}^t g(x,t)\,dx\,dt$
$ \equal{} \int_0^1\int_x^{\frac12}g(x,t)\,dt\,dx \plus{} \int_0^{\frac12}\int_{\frac12}^0 g(x,t)\,dt\,dx \plus{} \int_0^{\frac12}\int_{\frac12}^1 g(x,t)\,dt\,dx$

For $ g$ a function of $ x$ alone, this simplifies to $ \int_0^1 (x \minus{} \frac12)g(x)\,dx \plus{} \int_0^1 \frac12g(x)\text{sgn}(x \minus{} \frac12)\,dx$ (That sgn is $ 1$ for $ x > \frac12$ and $ \minus{} 1$ for $ x < \frac12$).

How does this change the result? We're looking at $ \langle g(x),(x \minus{} \frac12) \plus{} \frac12\text{sgn}(x \minus{} \frac12)\rangle$. $ \int_0^1 \left((x \minus{} \frac12) \plus{} \frac12\text{sgn}(x \minus{} \frac12)\right)^2\,dx \equal{} 2\int_{\frac12}^1 \left((x \minus{} \frac12) \plus{} \frac12\right)^2\,dx$
$ \equal{} \int_{\frac12}^1 x^2\,dx \equal{} \frac14$

That means we only get $ \int_0^1 f(x)^2\,dx\le 4\left(\int_0^1 f'(x)\,dx\right)^2$. Equality is impossible, but we can approach arbitrarily closely by making $ f'(x)$ close to a real multiple of $ h(x) \equal{} \begin{cases}x & x > \frac12 \\
x \minus{} 1 & x < \frac12\end{cases}$. In this case, $ f(x)$ is approximately a multiple of $ \begin{cases}\frac {x^2}{2} \minus{} \frac18 & x > \frac12 \\
\frac {x^2}{2} \minus{} x \plus{} \frac38 & x < \frac12\end{cases}$.

It's likely the proposer made a similar mistake to the one I did.[/hide]

I wonder where, I can find official solution of Romanian national olimpiad 2008, most of all I am interested in author's  solution of this problem.
Jmerry you gave a contrexample or some thougth?
\[ \begin{cases}\frac{x^2}{2}\minus{}\frac18&x>\frac12\\
\frac{x^2}{2}\minus{}x\plus{}\frac38&x<\frac12\end{cases}\]
Is not good enough, it hasn't derivative in $ \frac12$

I made multiple mistakes there, both in the calculus and in the geometry of the interchange. On the other hand, your objection isn't one of them; I said we can approach arbitrarily closely, not that equality actually happens.
A fixed version:

We want to find $ C$ so that $ C^2\int_0^1 (f'(x))^2\,dx\ge \left(\int_0^1 f(x)\,dx\right)^2$ for all $ C^1$ $ f$ with $ f(\frac12)\equal{}0$. The problem suggests $ C^2\equal{}\frac1{12}$.

First, write $ f'(x)\equal{}g(x)$ and $ f(x)\equal{}\int_{\frac12}^{x} g(y)\,dy$. The RHS becomes $ \left(\int_0^1\int_{\frac12}^x g(y)\,dy\,dx\right)^2$; we want to interchange the order of integration. The region we integrate over is the triangle between $ x\equal{}0,y\equal{}\frac12,y\equal{}x$ traced negatively, together with the triangle between $ y\equal{}\frac12,x\equal{}1,y\equal{}x$ traced positively. Interchanging the order, we get $ \left(\minus{}\int_0^{\frac12}\int_0^y g(y)\,dx\,dy\plus{} \int_{\frac12}^1 \int_y^1 g(y)\,dx\,dy\right)^2$
We evaluate the inner integral, and this becomes $ \left(\int_0^{\frac12}\minus{}yg(y)\,dy\plus{}\int_{\frac12}^1 (1\minus{}y)g(y)\,dy\right)^2 \equal{}\left(\int_0^1 g(y)h(y)\,dy\right)^2$, where $ h(y)\equal{}\begin{cases}\minus{}y&y<\frac12\\1\minus{}y&y>\frac12\end{cases}\equal{} (\frac12\minus{}y)\plus{}\frac12\text{sgn}(y\minus{}\frac12)$.
(Note that I had the sign of the $ \frac12\minus{}y$ wrong last time)

By C-S, we have $ \|h\|^2\|g\|^2\equal{}\int_0^1 (h(y))^2\,dy\int_0^1 (g(x))^2\,dx\ge \left(\int_0^1 g(y)h(y)\,dy\right)^2\equal{}\langle g,h\rangle^2$. A simple calculation shows that $ \|h\|^2\equal{}\frac12$, and $ C\equal{}\frac12$. Equality holds if $ g$ is a real multiple of $ h$, which is impossible by continuity of $ g$. We can approach arbitrarily close to equality by approximating $ h$ in the $ L^2$ sense by continuous functions.


Let a funtion f in positiveintegers and a positive integer such that
f(a)=f(1995), f(a+1)=f(1996), f(a+2)=f(1997)
additionally f(n+a)=$\frac{f(n)-1}{f(n)+1} $ for each positive integer n.
Show that f(n+4a)=f(n)
and find the min positive price of a

No it's not.

\[f(n + 2a) = \frac{f(n + a) - 1}{f(n + a) + 1} = \frac{\frac{f(n) - 1}{f(n) + 1} - 1}{\frac{f(n) - 1}{f(n) + 1} + 1} = -\frac{1}{f(n)}\]

and hence \[f(n + 4a) = -\frac{1}{f(n + 2a)} = f(a).\]

And I guess $a$ can be equal to 1... (depends on how you choose the values of $f(1995)$, $f(1996)$ and $f(1997)$, I am not sure about the way you are stating the problem)


I need help in understanding the order of a chemical reaction, can any body explain elaborately or give links to useful articles?

Why is the slowest step the rate determining step of a reaction?
Why does order of a complex reaction depend upon the rate determining step?
What is the relation between order of a reaction and its molecularity?
Why do reactions generally not have order greater than 3?

Also I would like suggestions about good physical chemistry books which clearly explain the above said queries and concepts.

Please I need help urgently,

Thanx in advance!

you can read from Skoog or P.W.Atkins

and for the link .
this one's good I believe [url]http://en.wikipedia.org/wiki/Order_of_reaction[/url]


In the diagram below,  $\odot O$, $\odot P$, and $\odot Q$ are mutually tangent. $\odot Q$ is also tangent to $CD$. $AB$ is a diameter of $\odot O$ and $AC$ is a diameter of $\odot P$. $\triangle CDB$ is isosceles with legs $CD$ and $BD$. Prove $QC \perp AC$.

is this "the" fuzzylogic?  If so, welcome back!

Huh, what is this problem looking for in the "Pre-Olympiad" section?

Anyway, it was solved here: http://www.mathlinks.ro/Forum/viewtopic.php?t=18266

  Darij

[quote]Huh, what is this problem looking for in the "Pre-Olympiad" section?

Anyway, it was solved here: http://www.mathlinks.ro/Forum/viewtopic.php?t=18266 [/quote]
oops, sorry

It is a Japanese temple problem:
[url]http://www2.gol.com/users/coynerhm/0598rothmanbox1a.html[/url]

Dear Mathlinkers,
for a synthetic proof, you can see

http://jl.ayme.pagesperso-orange.fr/  vol. 4 La fameuse san Gaku

Sincerely
jean-Louis


Suppose f belongs in BV[0,1] and is continuous at 0.Prove that for every e>0 there is a d >0 s.t.
the total variation of f on [0,d] is less than e.


Find all solutions of x^(n+1)-(x+1)^n=2001 in positive integers x and n.

$ x^{n+1}-(x+1)^{n}\equiv-1\equiv 2001\mod x\implies x|2002$.

$ 2002=2*7*11*13$. It is just a matter of testing now...Perhaps there is a faster way to reduce the cases...


geometry
parallelogram
ratio
area of a triangle

The diagonals of a parallelogram divide it into $4$ triangles, one of which has $G$ as its centroid.  If the area of the parallelogram is $180$, find the largest possible area of a triangle with vertex $G$ and one side which is also a side of the parallelogram.

I'm not sure if this is right, but here goes. Hope it's not too hard to follow.  :|

[hide="Solution"]Let the parallelogram have vertices $A, B, C, D$ and intersection of diagonals be $E.$

WLOG, say $G$ is the centroid of triangle $ADE.$ Draw the median through $G$ and $E$ and let it meet $AD$ and $BC$ at $P$ and $Q,$ respectively. If follows that $PQ\parallel AB\parallel CD$ and $\triangle APE\sim \triangle ACD$ by a ratio of $1: 2.$

Hence, the areas of triangle $CGD$ and $ABG$ are equal, and the sum of their areas is half the area of the parallelogram, or $90.$ Um, each has area $45.$

Now, there's "$90$ more area" to distribute among triangles $AGD$ and $BGC.$ Recall that the medians of a triangle cut the triangle into six equal areas. Hence, the area of triangle $AGD$ is one third that of triangle $ADE,$ or 1/12 that of the parallelogram $ABCD.$ Now, triangle $BGC$ has $5/12$ of the parallelogram's area, or $75.$

So the largest of the four areas (15, 45, 45, 75) is $75.$

[/hide]


I do believe this has a nice idea? Do you? I'm searching for other solutions.

I'm sorry for picking up so old a post
But we actually get a solution of it.

We just need to prove that for all $ x_1,x_2,\ldots,x_n \in \mathbb{R_\plus{}}$ there is
$ \sum_{i\equal{}1}^n \frac{x_1}{\sum_{j \neq i} x_j} \sum_{1 \le i<j \le n}x_ix_j \le \frac{n}{2}\sum_{i\equal{}1}^n x_i^2$
Rewrite it into
$ \left( \sum_{i\equal{}1}^n \frac{x_i}{S\minus{}x_i} \right) \left(S^2\minus{} \sum_{i\equal{}1}^n x_i^2 \right) \le n \sum_{i\equal{}1}^n x_i^2$
where $ S\equal{}\sum_{i\equal{}1}^n x_i$. It's equivalent to
$ S^2 \sum_{i\equal{}1}^n \frac{x_i}{S\minus{}x_i} \le \sum_{i\equal{}1}^n x_i^2 \sum_{i\equal{}1}^n \left( \frac{x_i}{S\minus{}x_i}\plus{}1 \right)\\
\Longleftrightarrow S \sum_{i\equal{}1}^n \frac{x_i}{S\minus{}x_i} \le \sum_{i\equal{}1}^n x_i^2 \sum_{i\equal{}1}^n \frac{1}{S\minus{}x_i}\\
\Longleftrightarrow S \le \sum_{i\equal{}1}^n \frac{1}{S\minus{}x_i} \sum_{j \neq i} x_j^2$
Use Cauchy-Schwarz we get it.


I just got $\LaTeX$ and I'm trying to make my first document. But there's a problem:

Here's my document so far:

\documentclass[11pt]{article}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{amsmath}

\title{Blah}
\author{Drunken_Math}
\begin{document}
Blah
\end{document}

But when I press Ctrt+F7 and then F5.. the document only shows Blah, and not the Title and Author.

Can you tell me what's going on?

You have to have a \maketitle command somewhere.  \title, \author, etc. just define the title, author, etc. -- \maketitle actually prints it.


trigonometry
logarithms
inequalities
function
limit
inequalities proposed

Prove that $ (\cos x)^{\cos x} > (\sin x)^{\sin x}$ for all real numbers $ x \in (0,\pi/4)$.

Take natural logarithm on both sides. It is equivalent to prove:
$ (\cos x)\ln{\cos x} > (\sin x)\ln{\sin x}$

For $ x \in (0,\pi/4)$, from graph we get, $ \cos x > \sin x$
Since, $ \ln x$ is increasing, the other ineq holds as well for the given interval. Isn't it?
 :)

nope, your conclusion is false because $ 0>\ln(\cos x)>\ln(\sin x)$, so we can't multiply the inequalities...

[hide="{my approach}"]
i tried the following, let $ \cos x=\sqrt{b}, \sin x=\sqrt{1-b}$, with $ 1>b>\frac{1}{2}$. then the inequality is equivalent to

$ \sqrt{b}\ln\sqrt{b}>\sqrt{1-b}\ln\sqrt{1-b}$, or equivalently, $ -\sqrt{1-b}\ln(1-b)>-\sqrt{b}\lnb$

note that both terms in the inequality are positive so we can square it. we would have to prove that $ f(b)=(1-b)\ln^2(1-b)-b\ln^2b>0$, for all $ b\in\left(\frac{1}{2},1\right)$

we have that
$ f''(b)=\dfrac{2\ln(1-b)}{1-b}-\dfrac{2\ln b}{b}+\dfrac{2}{1-b}-\dfrac{2}{b}$
and
$ f'''(b)=\dfrac{2\ln(1-b)}{(1-b)^2}+\dfrac{2\ln b}{b^2}<0$

so $ f''(b)$ is strictly increasing, and it follows that $ f''(b)<f'''\left(\frac{1}{2}\right)=0$, so $ f$ is a concave function... then $ f$ takes its minimum at its endpoints...

and this is where i'm kind of stucked... can we try this? let $ g$ a function on $ \left[\frac{1}{2},1\right]$ defined by $ g(b)=f(b)$ for $ b\in\left(\frac{1}{2},1\right)$ and $ g\left(\frac{1}{2}\right)=g(1)=0$... we have that $ \lim_{x\rightarrow\frac{1}{2}^{-}}g(x)=0$ and $ \lim_{x\rightarrow 0^{+}}g(x)=0$, so $ g$ is continous... is it valid from this to conclude that $ f(b)>0$ for all $ b\in\left(\frac{1}{2},1\right)$?
[/hide]

Yes, you are right sir. I missed that part, and that's why the problem seemed so easy to me! Sorry! :blush:

Anyway, your approach is very good sir. At the last I failed to understand one thing sir, which is:
how are we concluding the graph of $ f(x)$ to be over-tangent within $ \left[ \frac{1}{2},1\right]$?

I was thinking of a different approach:

Let $ u\equal{}(\cos x)^{\cos x}$
thus, $ \ln u\equal{}(\cos x)\ln {\cos x}$
$ \Rightarrow \frac{u'}{u} \equal{} \minus{}{\sin x}(1\plus{}\ln {\cos x})$
$ \Rightarrow u'\equal{}(\cos x)^{\cos x}[\minus{}{\sin x}(1\plus{}\ln {\cos x})]$

similarly considering $ v\equal{}(\sin x)^{\sin x}$, we get;
$ v'\equal{}(\sin x)^{\sin x}[{\cos x}(1\plus{}\ln {\sin x})]$

now taking $ f(x)\equal{}u\minus{}v$ we are to show if $ (u'\minus{}v')>0$ for $ x \in (0, \frac{\pi}{4})$.

i think that your approach would not lead us to a solution, since $ u' \minus{} v' > 0$ implies that $ f$ is increasing and we have "kind of" equality cases at $ 0$ and $ \pi/4$...

what i wanted to say in my solution was that since $ g$ is continous (and $ f$ convex) then $ g$ is convex, from where we conclude that it takes its minimum at its endpoints, this is $ g(x)\geq \min\left\{g\left(\frac {1}{2}\right),g(1)\right\} \equal{} 0$, and we're done...

i don't know if it's necessary to show that no other value $ x$, distinct from $ \frac {1}{2},1$, satisfies that $ f(x) \equal{} 0$, but i think that this is easy to prove..

*please don't call me sir, i'm just 19  :D


trigonometry
logarithms
integration
calculus
calculus computations

Solve:

$ y^{\prime\prime\prime} \minus{} y^{\prime\prime} \plus{} y^{\prime} \minus{} y \equal{} \sec x$

 :)

$ \begin{aligned}y(x) \equal{} &\textbf{C}_1 e^x \plus{} \textbf{C}_2 \sin x \plus{} \textbf{C}_3 \cos x \plus{} \\& \frac {1}{2}\left((\sin x \minus{} \cos x)\ln(\cos x) \minus{} x(\sin x \plus{} \cos x) \plus{} e^x \int_{0}^{x}{\frac {e^{ \minus{} t}}{\cos t}\textbf dt}\right)\end{aligned}$
:D

Your answer is correct [b]Yuriy Solovyov[/b]  :)


modular arithmetic
superior algebra
superior algebra solved

First of all, $m|n$ (in order for $n\mathbb Z$ to be a factor ring of $m\mathbb Z$). A complete set of classes $\pmod{n\mathbb Z}$ in $m\mathbb Z$ is $\widehat{m\cdot 0},\widehat{m\cdot 1},\ldots,\widehat{m\cdot(k-1)}$ where $k=\frac nm$. Since we can't have divisors of $0$, we find the fact that $k$ is a prime to be a necessary condition. The existence of a multiplicative identity element translates to the existence of some $i_0\in\overline{0,k-1}$ s.t. $n=mk|(mi_0-1)\cdot mi,\ \forall i\in\overline{0,k-1}$. It's clear that this is equivalent to $(m,k)=1$, so a necessary condition is: $(m,n)=(m,pm)$ with $p$ prime and $p\not |\ m$.

Conversely, the axioms for a field are easy to check for such a pair (too lazy to do it now :)), so I guess this is a sufficient condition as well.


hi guys
i am new to this forum so hope to interact with you guys more. i am doing an undergraduate course in singapore but i am from india. prepared hard for iit but bad luck ne ways have fun and also check out this site http://www.alchemy-education.com and gimme your suggestions and comments on da site.
thanks[/i]

I like it! Well...I did not see the whole of it but the maths section looked great!

It seems the site is build for adsense  :D

you have blended the advertisements very well dude

cheers !!!
http://www.educationtrak.com


Prove that every integer can be expressed in the form $x^{2}+y^{2}-5z^{2}$, where $x, y, z$ are integers.

You might as well post every problem from the red/green books :P

Actually, from your posts I decided to buy these books when I saw them at Powell's, there are very nice but the lack of latex really takes away from the experience...

Hmm.  Can similar reason as http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129978 be applied here?


LaTeX
function
calculus
derivative
combinatorics unsolved
combinatorics

Let $ p\left(n\right)$ be the $ n$-th partition number for any $ n\in\mathbb{N}$. Then,
$ np\left(n\right)\equal{}\sum_{k\equal{}1}^nd\left(k\right)p\left(n\minus{}k\right)$,
where $ d\left(n\right)$ means the sum of the divisors of $ n$ for any $ n\in\mathbb{N}$.

Please see http://www.artofproblemsolving.com/Forum/index.php?t=123690 and [[LaTeX]] to learn about making your posts more legible, and proofread the problems when you post them (here $ i$ and $ k$ should be the same letter). [Mod edit: I fixed it for him.] Also, please consider choosing more informative titles for your posts -- for example, "Partition identity" would be an improvement on the title of this thread.  Finally, the standard notation for the sum of divisors is $ \sigma$, not $ d$ (which is usually used for the [i]number[/i] of divisors).

The generating function approach is quite natural here: we have $ P(x) : = \sum_{n \geq 0} p(n)x^n = \prod_{i \geq 1} \frac {1}{1 - x^i}$ and so we may take a derivative of both sides and multiply through by $ x$ to get that
\[ \sum_{n \geq 0} np(n) x^{n} = \sum_{j \geq 1} \frac {j x^{j}}{1 - x^j} \cdot \prod_{i \geq 1} \frac {1}{1 - x^i} = P(x) \cdot \sum_{j \geq 1} \frac {j x^{j}}{1 - x^j}. \hspace{2cm} (1)
\]
The coefficient of $ x^{n}$ in the left side of this equation is the $ np(n)$ that we're looking for.  We also have that
\begin{align*} \sum_{j \geq 1} \frac {j x^{j}}{1 - x^j} & = \sum_{j \geq 1} j x^{j - 1} \sum_{i \geq 0} x^{ij} \\
& = \sum_{i \geq 0, j \geq 1} jx^{j (i + 1)} \\
& = \sum_{m \geq 1} \sum_{j \mid m} j x^m \\
& = \sum_{m \geq 1} \sigma(m) x^m. \end{align*}
Thus the coefficient of $ x^n$ on the right side of $ (1)$ is exactly $ \sum_{k \leq n} \sigma(k) p(n - k)$, as desired.

How about a bijective proof?  $ n p(n)$ is the number of ways to draw the Young diagram (Ferrers diagram) of a partition of $ n$ and fill in one square.  Why is $ \sum_{k \leq n} \sigma(k) p(n - k)$ the same thing?


linear algebra
matrix
linear algebra unsolved

let A and B be nxn mxes, adj AB = adj B adj A. Use this to show That |adj A| = |A|^(n-1).

Denote by adj A the transposed matrix of cofactors of A.

Thanks!

http://www.mathlinks.ro/Forum/viewtopic.php?t=39567

Follow this topic, the identity $A Adj(A)=|A| I$ is better suited, my post was about the singular cases, your formula has been asked by viking


algebra
function
domain
calculus
derivative
integration
real analysis

Let $ \Omega \subset \mathbb R^N$ be an open domain with smooth boundary. We recall the notation $ C_0^\infty \left( \Omega \right)$ to be the set of all smooth function $ \varphi : \Omega \to \mathbb R$ with compact support. Can you derivative the following
\[ I\left( u \right) \equal{} \int\limits_\Omega {\left| {\nabla u} \right|^2 dx}
\]
provided $ u \in C_0^\infty \left( \Omega \right)$. This is a simple question coming from calculus of variation.

Let's compute the directional derivative in the direction $ h\in C_0^{\infty}(\Omega)$. Since
$ |\nabla (u\plus{}th)|^2\equal{}|\nabla u|^2\plus{}2t\langle \nabla u,\nabla h\rangle\plus{}t^2|\nabla h|^2$, the derivative is $ 2\int_{\Omega}\langle \nabla u,\nabla h\rangle dx$, which is equal (by integration by parts) to $ \minus{}2\int_{\Omega}(\Delta u) h\, dx$. If $ u$ is a stationary point for $ I$, then it follows that $ \Delta u\equal{}0$, i.e. $ u$ is harmonic.


If a$b$c=b/ac, what will a$b$c$d$...$x$y$z? (The $ operation only works if there are three numbers, a, b, and c. For example, 1$3 is undefined but 2$3$6=1/4)

If you like this one I'll think of more.

The operation isn't associative. You need to put parentheses in.

Mmm...OK, this might be better: a$b$c*b$c$d*c$d$e*...*x$y$z

I'd still like to see parentheses, but I now assume you mean (a$b$c)*(b$c$d)*(c$d$e$)*...*(x$y$z). Is that correct?

thats not much of a MC problem...the answer would be a pretty long string of variables that u wouldn't like to copy...

Actually, it's quite simple. Just try it out.

....these are some pretty dumb questions.

yup, I agree, yet there are a hidden dragon in each one of these seemingly calm and unthreatning crouching tarepandas... :lol:

dude...the problems r ridiculously pointless...

they're basic rules of math...

all we need to do is write out this long chain of fractions and cancel out and stuff...

...*sigh* then please post your answer and the trap in this problem por favor.

b/c personally i'd like to spend time on something worth thinking about...

simon, you probly wouldnt want do write out the answer either...as ridiculously simple as it is...

If someone (read: Tare) wants me to post the answer here, then I will.

no, Simon, please don't. Miths, at least try the problem. I assure you this (usually in a slightly different form) does come up in mathcounts.

Mystic, as far as the answer goes...

all we need to do is multiply (b/ac)(c/bd)(d/ce)...most of the terms cancel out but we're still left with a long string in the denominator

Thank you Mystic. At least give it a chance, pleeeeeeease? *bambi eyes* :shock:

...wait,  :shock: isn't exactly a "bambi eye", isn't it... :lol:

You're very close...in fact if you move the operations around you'd get it right :)


Separate solutions are made of these substances by dissolving 10 g in 1000 g of benzene. Which will start to freeze at the lowest temperature?
a. CH2Cl2
b. CHCl3
c. CCl4
d. all will freeze at the same temperature

hint:
[hide] compare molality ^^[/hide]

[hide]We want to find the solution with the highest molality. Looking at the choices:

a. 10/84=0.119
b. 10/119.5=0.084
c. 10/152=0.066.

And a is the correct answer. [/hide]


logarithms
function
limit
trigonometry
calculus
calculus computations

I figured out that this sequence converges, but I haven't been able to figure out to what?

An= [n^2/(2n+1)] *sin(1/n)

and also, if it's not too much trouble, does the sequence  An=(1/ln(n)) diverge?

[quote="Keno20"]
and also, if it's not too much trouble, does the sequence  An=(1/ln(n)) diverge?[/quote]
$ \frac{1}{\ln n}>\frac{1}{n},\,\forall \,n\ge 2,$ hence, diverges.

if you want to find where is converges you can take a function f(x)=[code]1/(2*x+x^2)*sinx[/code] for x>0
and f(x)=1/2 for x=0 then show with de hospital it is continiues and that means for the sequence 1/n ->0 f(1/n)->1/2

Knowing already that a_n converges, then its limit is

$ \lim a_n \equal{} \lim \frac {n^2}{2n \plus{} 1} \sin \frac1n \equal{} \lim \frac {n}{2n \plus{} 1} \cdot \frac {\sin \frac1n}{\frac1n} \equal{} \frac12 \cdot 1 \equal{} \frac12$.

[quote="Bipshula"][quote="Keno20"]
and also, if it's not too much trouble, does the sequence  An=(1/ln(n)) diverge?[/quote]
$ \frac {1}{\ln n} > \frac {1}{n},\,\forall \,n\ge 2,$ hence, diverges.[/quote]

the sequence $ a_n\triangleq 1/n$ converges to $ 0$...

Oops, I misread the problem, I thought it was a series!

Well, it isn't surprising that you thought it was the series (reason no. 1: it's pretty obvious in case of sequences, reason no. 2: see the thread title).


function
real analysis
real analysis unsolved

Let $ f,g: \mathbb{R}\minus{}>\mathbb{R}$ be continuous bijective functions such that
\[ f\left( g^{\minus{}1}(x)\right)\plus{}g\left( f^{\minus{}1}(x)\right)\equal{}2x ,\forall x \in \mathbb{R}\]
Prove that if $ f$ and $ g$ take the same value in a point, then they coincide.


Give $ a,b,c>0$ and $ ab\plus{}bc\plus{}ca\equal{}3$. Prove that:

$ \frac{1}{\sqrt{1\plus{}(a\plus{}b)^3}\minus{}1}\plus{}\frac{1}{\sqrt{1\plus{}(b\plus{}c)^3}\minus{}1}\plus{}\frac{1}{\sqrt{1\plus{}(a\plus{}c)^3}\minus{}1} \geq \frac{3}{2}$

I thinks this is easy problem.
$ 1\plus{}x^3\equal{}(1\plus{}x)(1\minus{}x\plus{}x^2) \le (1\plus{} \frac{x^2}{2} )^2$
let x=a+b,x=b+c,x=c+a,we get:
 $ \sqrt{1\plus{}(a\plus{}b)^3} \minus{}1 \le 1\plus{} \frac{(a\plus{}b)^2}{2} \minus{}1 \equal{} \frac{(a\plus{}b)^2}{2}$
And finally,we need to prove :
$ \frac{1}{(a\plus{}b)^2} \plus{} \frac{1}{(b\plus{}c)^2} \plus{} \frac{1}{(c\plus{}a)^2} \ge \frac{3}{4}$
or $ \frac{1}{(a\plus{}b)^2} \plus{} \frac{1}{(b\plus{}c)^2} \plus{} \frac{1}{(c\plus{}a)^2} \ge \frac{9}{4(ab\plus{}bc\plus{}ca)}$
This is Iran TST96

Nice proof, Ngoc Anh. And nice hint to creat, vipCD_A1  :)


articles
USAMTS
probability
expected value

This problem is really simple if you read the article in USAMTS site..  :D

A company is making bulbs for flashlights. They know that the probability of a bulb being defective is $\frac{1}{100}$. The company makes a profit of 0.80 dollars for every non-defective bulb it sells. However, each defective bulb costs the company 8 dollars in replacement costs. How much money does the company really make per bulb?

[hide]well, out of 100 bulbs, 99 will work and 1 won't, leaving the total profit at $71.20 for 100 bulbs, or $0.712 per bulb. although i think i made an error somewhere.... :? [/hide]

Correct.

We have 1/100 chance that it won't work and 99/100 chance that it will work.

So, the expected value is 99/100*0.80 + 1/100*(-8) which gives 0.712. Since we are dealing money here, the final answer is 0.71 dollars.  :)

was this problem on the USAMTS? it doesn't really require proof.........

Nope.

This one was from internet. It's easy simple concept one.


As figure shown, $ AB$ is diameter.$ \widehat{AE}\equal{}\widehat{ED}\equal{}\frac{1}{2}\widehat{BC}$

If $ \overline{BC}\equal{}6,\overline{BE}\equal{}4$, then $ \overline{AB}\equal{}?$

Am I missing something here?

$ AE \equal{} \frac{1}{2} BC \equal{} 3$. Triangle $ ABE$ is a right-angled at $ E$, because the angle in a semi-circlle is a right-angle.

$ BE \equal{} 4$. Therefore, by Pythagoras, $ AB \equal{} 5$.

But then $ BC$ is a chord whose length is more than that of the diameter, which cannot be. So the question is flawed.

Am I wrong?

[quote="AndrewTom"]Am I missing something here?

$ AE \equal{} \frac {1}{2} BC \equal{} 3$. Triangle $ ABE$ is a right-angled at $ E$, because the angle in a semi-circlle is a right-angle.

$ BE \equal{} 4$. Therefore, by Pythagoras, $ AB \equal{} 5$.

But then $ BC$ is a chord whose length is more than that of the diameter, which cannot be. So the question is flawed.

Am I wrong?[/quote]Yep you are wrong here:

[quote="AndrewTom"]Am I missing something here?

$ AE \equal{} \frac {1}{2} BC \equal{} 3$.[/quote]Take an arc $ AB$ and let $ M$ be midpoint of this arc. Draw chords $ AM,BM$. So, is your claim still true?

[hide="my attempt"]ACBD is a rectangle.
AEB and FEB are congruent
AD=6, BE=4 (these are altitudes of ABF with different bases).

The bases therefore must be in the ratio 2:3

Let BF=2a, AF=3a

$ BE \equal{} \sqrt {(2a)^2 \minus{} (1.5a)^2} \equal{} \sqrt {\frac {7}{4}}a \equal{} 4$

$ a \equal{} \frac {8}{\sqrt {7}}$

$ AB \equal{} BF \equal{} \frac {16\sqrt {7}}{7}$[/hide]

Thanks. I misread the question, not realising that the first line refers to lengths of arc and not chords.


Determine all numbers $ \overline{abcd}$ such that $ \overline{abcd}\equal{}11(a\plus{}b\plus{}c\plus{}d)^2$.

$ 1000 \le \overline{abcd} < 10000$
$ 91 \le (a \plus{} b \plus{} c \plus{} d)^2 \le 909$
$ (a \plus{} b \plus{} c \plus{} d) \in \{10,11,\cdots, 30\}$

On the other hand, $ x \equiv \overline{abcd} \equiv a \plus{} b \plus{} c \plus{} d (9) \Rightarrow x \equiv 2 x^2  (9) \Rightarrow x \equiv 0 (9)$ or $ x \equiv 5 (9)$

Combining both implies $ (a \plus{} b \plus{} c \plus{} d) \in \{14,18,23,27\}$

obtaining the following solutions for $ \overline{abcd} \in \{2156,3564,5819\}$


The limit of a sum of an infinite number of terms in a geometric sequence is a/(1-r) where a denotes the first term and -1 :ang: r :ang: 1 denotes the common ratio. The limit of the sum of their squares is _________.  (In terms of a and r)

[hide]When you square everything I believe that you are still left with a geom. series.  The only difference is that the first term is squared and the ratio from each consecutive term is squared.  So the sum is [tex]\frac{a_1^2}{1-r^2}[/tex][/hide]

I thought (1-r) :^2: =1-2r+r :^2:
Maybe I am just not following your logic.

Hes not squaring [tex]\frac{a}{1-r}[/tex]. Hes squaring the first term and the ratio of change, getting. [tex]\frac{a^2}{1-r^2}[/tex]


In a library that contains N books every day we choose one of the books at random for consultation, and return it to the shelf. What is the expected number of days it will take us to consult all the books in the library at least once?

This has been known commonly as the "coupon collector" problem.  For anyone with "The Book" cf. p.158.

I present work along the same lines as "The Book."  If $ k$ books have already been picked, the probability that it will take $ s$ more days to pick a new book is $ \left (\frac{k}{N}\right )^{s-1}(1-\frac{k}{N})$.  Hence the expected number of drawings for the next book is

$  \sum_{s=1}^{\infty}\left (\frac{k}{N}\right )^{s-1}(1-\frac{k}{N})s=\sum_{s=0}^{\infty}\left (\frac{k}{N}\right )^{s}(s+1)-\sum_{s=0}^{\infty}\left (\frac{k}{N}\right )^{s}s=\frac{1}{1-\frac{k}{N}}$.

Hence the expected number of days is

$ \sum_{i=0}^{N-1}\frac{1}{1-\frac{k}{N}}= \frac{N}{N}+\frac{N}{N-1}+\cdots+\frac{N}{2}+\frac{N}{1}=N\cdot H_{N}$

which is approximately $ N$ log $ N$.


geometry
3D geometry
sphere
calculus
derivative
function
percent

By approximately what percentage does the volume of a sphere increase if the racius increases from 10 to 11 units?

All the examples and such I've seen its either completely in percentages or in unites... how do I get from the rate of change to the approx. percentage

I'm not sure if this is right, but i don't think this problem requires derivatives at all. Well, first of all, recall that the volume V of a sphere as a function of radius r is:

V(r) = (4/3)pi(r)^3

the percent increase from r = 10 to r = 11 is just V(11) divided by V(10)

the 4/3(pi) cancels out so you just have (11)^3/(10)^3. I didn't have a calculator so if i did my arithmetic right, it comes out to be about 1.221, which corresponds to a 22.1% increase in volume

The point is to use derivatives to approximate the change in volume. From

$V = \frac{4}{3}\pi r^{3}$

we get

$dV = 4 \pi r^{2}dr$

so if $r = 10$ and $dr = 1$, we have

$dV = 4 \pi (10)^{2}\cdot 1 = 400 \pi$.

So the percentage change is about

$\frac{dV}{V}= \frac{400 \pi}{\frac{4}{3}\pi 10^{3}}= \frac{12}{40}= 30\%$.

On a side note, does anyone know if using $dV$ and $dr$ here is inappropriate? Would it be more correct to use $\Delta V$ and $\Delta r$?


calculus trigonometry function integration

The number of hours of daylight, d, in Hartsville can be modeled by

d= 35/3 + (7/3)sin((2pi/365)*t)

where t is the number of days after March 21. The day with the greatest number of hours of daylight has how many more daylight hours than May 1? (March and May have 31 days each. April and June have 30 days each.)

I dont know how to: a) find the day with the greatest number of hours of daylight b) calculate all this without a calculator and c) Im not sure if 2pi/365 = 360/365 ?

Please explain, thanks.

Move this to the calculus forum.  To find the day with the maximum daylight, it is optimal (in my opinion at least) to use calculus.

three things:
1. Why did you post it in two places (HSB and here) ?
2. What book did you get this problem from?
3. you probably can't do the sine function without integration

This is not calculus, or at least its intended solution does not require calculus. It's typical in pre-calculus classes.

a) You're not supposed to find the day with the greatest number of hours of daylight.
b) If you're not allowed to use a calculator, the angle inside the $ \sin$ will most likely work out to a nice number, like $ \frac{\pi}{2},\frac{\pi}{3},\frac{\pi}{6}$, etc. If you haven't memorized the sin, cos, and tan of these and other special angles, you definitely should invest the time. Calculus is hard enough without having to repeatedly worry about this simple trig stuff.
c) Calculus is done in radians. While it is true that $ \frac{2\pi}{365}$ radians is equal to $ \frac{360}{365}$ degrees, it is for the most part irrelevant to the question and you should get used to thinking in terms of radians.

[hide]You only need to know how many hours of daylight that day has, that is, what is the maximum value of $ d$. So the question becomes: What is the maximum value of $ \sin\theta$? You should know that this maximum is 1. Then the maximum of $ d$ is $ \frac{35}{3}\plus{}\frac73\cdot 1\equal{}14$. Plug in the appropriate value of $ t$ to find the number of hours of daylight May 1 has, and subtract this number from 14 for the final answer.[/hide]

Poincare, I posted this in HSB because JRav suggested that I post it in calculus forum instead of here, but I couldn't find it, so..

Bictor, thanks for the solution, is there any other way of doing it besides the one that requires me to know that max of sin(theta) is 1?  :) It mentions the number of days of certain months, so I guess that intended solution was different.

I wanted to know if its okay to write 360/365 instead of 2pi/365 when its not mentioned that 365 represents degrees? Thanks.

If you decide to post it in the calc forum (if you can find it), I'll see if I can work out a solution for you.  Instead of being required to know that the maximum value of $ \sin$ is 1, you instead can just find where the maximum value occurs, and then plug it in.


The following inequality is from my friend, Nguyen Van Thach. The problem stated

If $ a,b,c$ are positive real numbers, then show that
$ \sqrt{\frac{a^3}{a^3\plus{}7abc\plus{}b^3}}\plus{}\sqrt{\frac{b^3}{b^3\plus{}7abc\plus{}c^3}}\plus{}\sqrt{\frac{c^3}{c^3\plus{}7abc\plus{}a^3}} \ge 1.$

I wait for some nice proofs.

I am sorry, but I dont think this way works, my friend. Please check it again.

I think  we can solve it by Holder (hope I didn't   made any mistake)

by holder inequality it follows that :
\[ (LHS)^2( \sum{\frac{a^3+7abc+b^3}{(b+c)^3}}) \geq  (\sum{\frac{a}{b+c}})^3\]

all we have to prove is :

 \[ (\sum{\frac{a}{b+c}})^3 \geq  \sum{\frac{a^3+7abc+b^3}{(b+c)^3}}\]

but:
$ RHS= \sum{\frac{a^3}{(b+c)^3}} +3-3\sum{\frac{bc}{(b+c)^2}} +7\sum{\frac{abc}{(b+c)^3}} - \sum{\frac{c^3}{(b+c)^3}}\leq \sum{\frac{a^3}{(b+c)^3}}+ 3-\frac{3}{8}-3\sum{\frac{bc}{(b+c)^2}} +\frac{7}{4}\sum{\frac{a}{b+c}}= \sum{\frac{a^3}{(b+c)^3}} +\frac{7}{4}\sum{\frac{a}{b+c}}-3\sum{\frac{bc}{(b+c)^2}}+\frac{21}{8}$

$ LHS= \sum{\frac{a^3}{(b+c)^3}} +3(\sum{\frac{a}{b+c}})( \sum{\frac{ab}{(b+c)(a+c)}}) -3\frac{abc}{(a+b)(b+c)(c+a)}$

hence the ineq is equivalent :
\[ (\sum{\frac{a}{b+c}})(\frac{3\sum{ab(a+b)}}{\prod{(a+b)}}-\frac{7}{4}) + 3\sum{\frac{ab}{(a+b)^2}} \geq \frac{21}{8}\]
by cauchy :
\[ \sum{\frac{a}{b+c}} \geq \frac{ (a+b+c)^2}{2(ab+ac+bc)}\]
we will try to fin a lower bound to : $ S= \sum{\frac{ab}{(a+b)^2}}$
again by cauchy :
\[ (a+b)(b+c)(c+a)S = \sum{\frac{ab(a+c)(b+c)}{(a+b)} = 3abc\sum{\frac{a}{b+c}} + \sum{\frac{(ab)^2}{a+b}} \geq 3abc\frac{(a+b+c)^2}{2(ab+ac+bc)} + \frac{ (ab+ac+bc)^2}{\sum{ab(a+b)}}}\]
then pqr finishes the job ( only easy computations ) I haven't check because I don't have pen and paper righ now
Good night/morning ( 3am in   my country
 :sleeping: )

[quote="anas"]I think  we can solve it by Holder (hope I didn't   made any mistake)

by holder inequality it follows that :
\[ (LHS)^2( \sum{\frac {a^3 + 7abc + b^3}{(b + c)^3}}) \geq (\sum{\frac {a}{b + c}})^3\]
all we have to prove is :
\[ (\sum{\frac {a}{b + c}})^3 \geq \sum{\frac {a^3 + 7abc + b^3}{(b + c)^3}}\]
but:
$ RHS = \sum{\frac {a^3}{(b + c)^3}} + 3 - 3\sum{\frac {bc}{(b + c)^2}} + 7\sum{\frac {abc}{(b + c)^3}} - \sum{\frac {c^3}{(b + c)^3}}\leq \sum{\frac {a^3}{(b + c)^3}} + 3 - \frac {3}{8} - 3\sum{\frac {bc}{(b + c)^2}} + \frac {7}{4}\sum{\frac {a}{b + c}} = \sum{\frac {a^3}{(b + c)^3}} + \frac {7}{4}\sum{\frac {a}{b + c}} - 3\sum{\frac {bc}{(b + c)^2}} + \frac {21}{8}$

$ LHS = \sum{\frac {a^3}{(b + c)^3}} + 3(\sum{\frac {a}{b + c}})( \sum{\frac {ab}{(b + c)(a + c)}}) - 3\frac {abc}{(a + b)(b + c)(c + a)}$

hence the ineq is equivalent :
\[ (\sum{\frac {a}{b + c}})(\frac {3\sum{ab(a + b)}}{\prod{(a + b)}} - \frac {7}{4}) + 3\sum{\frac {ab}{(a + b)^2}} \geq \frac {21}{8}\]
by cauchy :
\[ \sum{\frac {a}{b + c}} \geq \frac { (a + b + c)^2}{2(ab + ac + bc)}\]
we will try to fin a lower bound to : $ S = \sum{\frac {ab}{(a + b)^2}}$
again by cauchy :
\[ (a + b)(b + c)(c + a)S = \sum{\frac {ab(a + c)(b + c)}{(a + b)} = 3abc\sum{\frac {a}{b + c}} + \sum{\frac {(ab)^2}{a + b}} \geq 3abc\frac {(a + b + c)^2}{2(ab + ac + bc)} + \frac { (ab + ac + bc)^2}{\sum{ab(a + b)}}}\]
then pqr finishes the job ( only easy computations ) I haven't check because I don't have pen and paper righ now
Good night/morning ( 3am in   my country
 :sleeping: )[/quote]
Thanks, anas. I will check carefully your solution but would you please try a solution with less calculations and without pqr?

[quote="anas"]hence the ineq is equivalent :
\[ (\sum{\frac {a}{b \plus{} c}})(\frac {3\sum{ab(a \plus{} b)}}{\prod{(a \plus{} b)}} \minus{} \frac {7}{4}) \plus{} 3\sum{\frac {ab}{(a \plus{} b)^2}} \geq \frac {21}{8}\]
[/quote]
I have checked your idea, and I think this inequality must be
$ (\sum{\frac {a}{b \plus{} c}})(\frac {3\sum{ab(a \plus{} b)}}{\prod{(a \plus{} b)}} \minus{} \frac {7}{4}) \plus{} 3\sum{\frac {ab}{(a \plus{} b)^2}} \geq \frac {21}{8}\plus{}\frac{3abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}.$
If my calculations are not wrong, this inequality is true. But do you have any simple proofs for it, anas?