Given real $A$, find all solutions $(x_1, x_2, ... , x_n)$ to the $n$ equations ($i = 1, 2, ... , n$):
$x_i\cdot |x_i|- (x_i - A)\cdot |x_i - A| = x_{i+1}|x_{i+1}|$, where we take $x_{n+1}$ to mean $x_1$.

[i]Edited by Myth[/i]

One can suppose that $A\geq 0$ (if $A\leq 0$, then $x_{i} \rightarrow -x_{i}$).
Consider
$f(x)=x|x|-(x-A)|x-A|$.
\[\forall x\leq 0, f(x)=-2Ax+A^{2}\]
\[\forall x\in [0,A], f(x)=x^{2}+(x-A)^{2}\]
\[\forall x\geq A, f(x)=2Ax-A^{2}\]
(a graph might help...)
One notices that $f\geq 0$, and that $f(x)\leq A^{2}$ iff $0\leq x\leq A$. As we have $f(x_{n})=x_{n+1}|x_{n+1}|$, we know that all the $x_{i}$ are $\geq 0$.
Suppose that $x_{1}>A$. This implies that $f(x_{1})>A^{2}$, so $x_{2}^{2}>A^{2}$ and $x_{2}>A$. With an easy induction we show that all the $x_{i}$ are $>A$. But we have:
\[x_{1}^{2}-x_{2}^{2}=(x_{1}-A)^{2}\]
\[x_{2}^{2}-x_{3}^{2}=(x_{2}-A)^{2}\]
\[\dots\]
\[x_{n}^{2}-x_{1}^{2}=(x_{n}-A)^{2}\]
By summing all these identities we get
\[0=\sum_{i=1}^{n}{  (x_{i}-A)^{2} }\ (*)\]
This cannot be true: all the terms in the sum are strictly positive. The same argument works of $x_{1}<A$. This implies that all the $x_{i}$ are $<A$, and (*) gives the contradiction.
We therefore have necessarily: $x_{1}=A$. This implies that all the $x_{i}$ equal A. One checks easily that $x_{i}=A$ is the only solution.


inequalities
induction
calculus
inequalities unsolved

[hide]
replace "2003" with 2m + 1, and induct on m. Induct on m,
of course you will have to show the 3 variable case before....
[/hide]

toodles,

i'm not sure how induction will work especially because the condition in the question will not hold if you try to use the IH... unless you show the 3-variable case more generally.

Well actually...I sorta meant that and sorta didnt :( . Anyway, here is a newer solution I have,and it requires calculus, and please o please tell me if its wrong, cause i keep on making errors that are pointed out much later . (im not much of a good proof readers, bieng fairly new the inequalities and stuff).

Okay so new way:

Assume results to hold for the n = 2001: ie x3>=x4>=x5...>=x2003 (you can generalize this to 2k+1, im just giving example, the case with 1 variable is  trivial to prove)

so now consider when we append x >= x2 in front of this sequence to get x>=x2>=x3..>=x2003 to get
(treat x as a variable, rest as constants)
(x)^n - (x2)^n + (x3)^n....-(x2002)^n + (x2003)^n >= (x - x2 + x3 + x4 - x5...+x2003)^n

now let f(x) = x^n. So we have

f(x) - f(x2) + f(x3) ...+f(x2003) >= f(x-x2+x3-x4....+x2003)

Now we check to see if this holds when x = x2, and it indeed does, since it reduces to the 2001 case which we assumed true.

now if we show that f'(x) - f'(x - x2 + x3 - x4...+x2003) >= we are done.

so n(x)^n-1 >= n(x-x2+x3-x4...+x2003)^n-1.

Divide both sides by n, and take the n-1th root

so we get x >= x - x2 + x3 - x4 + x5 - x6...

So you get x2 + x4 + x6..+ x2002 >= x3 + x5 + x7 ..+ x2003

which, by the condition of the problem holds true.
I think this is correct. :?

Toodles,


solve system equation :
$x^3(3y+1)=8$and$x(y^3-1)=6$

cubing the 2nd equation and dividing it by the first one, we get:
$27y^6-81y+81y^3-3y-28=0$
And i m not good at factoring or solving 6th degree polynomials

I solved for $y$ in each equation and got:
$\frac{8}{3x^3}-\frac{1}{3}=y=\sqrt[3]{\frac{(x+6)}{x}}$

(There are two other possible expression for the right side.)

I used Mathcad to solve and obtained:
$x=\sqrt[3]{4(i\sqrt{3}-1)}+\frac{4}{\sqrt[3]{4(i\sqrt{3}-1)}}-2\approx1.064$

I believe that there is another solution.


[hide]
$Q_n \equiv 12^n+43^n+(-43)^n+(-12)^n \pmod{1993}$

If $n$ is odd, we clearly have $12^n+(-12)^n \equiv 0 \pmod{1993}$ and $43^n+(-43)^n \equiv 0 \pmod{1993}$ so it is divisible by $1993$.

If $n = 2k$ with $k$ odd, we have $Q_n \equiv 2 \cdot 144^k+2 \cdot 1849^k \equiv 2(144^k+(-144)^k) \equiv 0 \pmod{1993}$ so it is also divisible by $1993$.

If $n = 2k$ with $k$ even, we have $Q_n \equiv 4 \cdot 144^k \pmod{1993}$, which is never $0$, so it is not divisible by $1993$.

Hence $Q_n$ is divisible by $1993$ iff $n$ is odd or $n = 2k$ for $k$ odd.[/hide]

This doesn't have anything to do with complex numbers and geometry, right? :D

Maybe it can be somehow solved by Complex Numbers.  :)


Hello,
At the beginning I would like to excuse if this is not a proper section of this forum.
I have a problem with converting integers to negabinary. According to http://en.wikipedia.org/wiki/Negabinary we just simply repeat dividing by -2 and each number in negabinary is a rest from this division. But I have a problem with the example from wikipedia:
[quote]
13 / -2 = -6, remainder 1
-6 / -2 =  3, remainder 0
 3 / -2 = -1, remainder 1
-1 / -2 =  1, remainder 1
 1 / -2 =  0, remainder 1[/quote]
Exactly with the second line from the bottom. Why -1/-2 = 1 and the rest is 1? For my it's just 0.5 and 0 in Z. Can anyone explain me that ?
Regards,
apc.

The remainder is taken to be non-negative, so it's either $0$ or $1$. From that it should be obvious that we must write $-1 = 1 \cdot (-2)+1$.

-1/-2 = 1/2 = 0, remainder 1
What's wrong with it?


Hello, i got all these from a math puzzle book...
I dont know what category they should belong in since they're so random...

ALL ANSWER ARE BETWEEN 1 AND 9.

The dimensions of A Square's imaginary land.

The minimum number of incongruent square that tile a (non-square) rectangle.

The order of the symmetry group of the square.

The largest interger n such that S{subscript n} is a solvable group.

The minimum number of edges in any non-planar graph.

(pi^2)/(2 sigma k^-2)

I couldn't solve these ones.
Any help would be greatly appreciated!

The first one is $ 2$.  It's from the book Flatland.

For the order of the symmetry group of the square:

[hide] this is the dihedral group with order 8, so the answer is 8.[/hide]

For smallest number of edges in non-planar graph:

[hide]Kuratowski's Theorem states that every non-planar graph contains or has a subdivision of K_5 or K_3,3. K_5 has 10 edges and K_3,3 has 9 edges, so the minimal number is 9. (Subdivisions can only have more edges, not less edges.)[/hide]

For the last one, I'm assuming that you're asking
\[ \frac{\pi^2}{2 \sum_{k\equal{}1}^{\infty} \frac{1}{k^2}}?\]In that case, it equals \[ \frac{\pi^2}{2 \cdot \frac{\pi^2}{6}}\equal{}\boxed{3}.\]

[quote="123456789"]For the last one, I'm assuming that you're asking
\[ \frac {\pi^2}{2 \sum_{k \equal{} 1}^{\infty} \frac {1}{k^2}}?
\]
In that case, it equals
\[ \frac {\pi^2}{2 \cdot \frac {\pi^2}{6}} \equal{} \boxed{3}.
\]
[/quote]

that's what i didn't get. There no numbers above or under the sigma symbol, so i didn't know where to start and where to end.

Well if it is asking you K^-2 then you should immediately get that it is undefined for 0. Then, you should also get that if it would want you to start from a negative number, it must also include 0 which is what it can not, so you must start with k=1. I'm not that familiar with sigma but if it gives you nothing above the symbol, then it means "to infinite" or something?

[quote="iamhe"]The largest interger n such that S{subscript n} is a solvable group.[/quote]
This isn't really a fair question.  The answer is [url=http://en.wikipedia.org/wiki/Abel-Ruffini_theorem]4[/url], and this turns out to be the reason quintic equations don't have a general solution, but this is more or less trivia.

[quote="iamhe"]
The minimum number of edges in any non-planar graph.[/quote]

Non-planar graphs contain $ K_5$ or $ K_{3, 3}$; the former has 10 edges, while the latter has 9, so the answer is $ \boxed 9$.


search
inequalities unsolved
inequalities

Let $a,b,c$ be positive numbers.Prove that
$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}>2$

This problem is maybe.. for the 100 time in this forum
Use: Search
$\sqrt{\frac{a}{b+c}}\geq \frac{2a}{a+b+c}$


function
floor function
logarithms
number theory proposed
number theory

Let $ x$ be a positive integer, $ a \equal{} \lfloor \log_{10}x\rfloor$, $ b \equal{} \left\lfloor \log_{10}\frac {100}x\right\rfloor$. Here $ \lfloor c\rfloor$ denotes the greatest integer less than or equal to $ c$. Find the greatest possible value of $ 3a^2 \minus{} 4b^2$.

If $ a \in \mathbb{N}$ then $ \max\{3a^2 \minus{} 4b^2\} \equal{} \minus{} \min\{a^2 \minus{} 16a \plus{} 16\} \equal{} 48$.Otherwise $ a\in \mathbb{R}\setminus \mathbb{N}$ and $ \max\{3a^2 \minus{} 4b^2\} \equal{} \max\{3a^2 \minus{} 4(2 \minus{} (a \plus{} 1))^2\} \equal{} 12$.[]


[quote="archimedes1"][quote="kohjhsd"][quote="archimedes1"]Two handymen, Peter and Paul, agree to paint the interior of a house and lay hardwood floor.  The total area of the walls to be painted is $4000$ square feet and the total floor area is $1500$ square feet.  In an hour of hard work, Peter can either paint $40$ square feet or lay $15$ square feet of hardwood, while Paul, a better hardwood layer, can either paint $20$ square feet in an hour or lay $30$ square feet of hardwood.  What is the minimum time (in hours) in which Peter and Paul, working together, can finish the job?[/quote]

[hide="somehow not very sure :| "]Because Peter is faster in painting and Paul is faster at laying hardwood, we first let one of them finish their specialty: Peter will need 100 hours and Paul will need 50 hours. Let Paul finish laying the floor in first $50$ hours. Then Peter would have done the 2000 sq.ft. of painting. Both now finish painting the leftover 2000 sq.ft. in 60 sq. ft. per hour, which should take $33 \frac{1}{3}$ hours. In total, they need $50+33 \frac{1}{3}=\boxed{83 \frac{1}{3}}$ hours in minimum.[/hide][/quote]
That's what I got, but it's wrong.
EDIT:  Never mind, I miswrote the answer on the answer sheet I turned in but had written your answer on the question paper that I got to keep.  So it might be right.[/quote]
If it's not right by any chance, check if they should paint ceilings or floors too... It says only "interior" so it was somewhat confusing for me to solve...


function
trigonometry
complex numbers
algebra unsolved
algebra

Find all functions pairs (f,g), where each of f and g is a function defined on the integers and with values, such that, for all integers a and b.
     f(a+b)=f(a)g(b)+f(b)g(a)
     g(a+b)=g(a)g(b)-f(a)f(b)
 :) Very stange that i have tried to solve it for real numbers. And only now saw it that it is for integer values

Let $ h(n)\equal{}f(n)\plus{}g(n)\sqrt{\minus{}1}$ for every integer $ n$, then from relation about $ f$ and $ g$ we have $ h(a\plus{}b)\equal{}h(a)h(b)$. So we have $ h(0)\equal{}0$ or $ h(0)\equal{}1$.

If $ h(0)\equal{}0$ then $ h(a)\equal{}0$ for every integer $ a$.

If $ h(0)\equal{}1$ then $ h(a)\equal{}h(1)^a$  for every integer $ a$.

It seems $ f(x)\equal{}\sin x$ and $ g(x)\equal{}\cos x$.

Also I found that f(0)=0 and g(0)=1

***But do sin(a) and cos(a) have the integer values for integer a?

Sorry...

I found a serious mistake on #2. I will rewrite here.

Let $ h(a) \equal{} g(a) \plus{} f(a)\sqrt { \minus{} 1}$ for every integer $ a$.

Then for every integers $ a,b$ ,

$ h(a)h(b)$
$ \equal{} (g(a) \plus{} f(a)\sqrt { \minus{} 1})(g(b) \plus{} f(b)\sqrt { \minus{} 1})$
$ \equal{} g(a)g(b) \minus{} f(a)f(b) \plus{} (g(a)f(b) \plus{} f(a)g(b))\sqrt { \minus{} 1}$
$ \equal{} g(a \plus{} b) \plus{} f(a \plus{} b)\sqrt { \minus{} 1} \equal{} h(a \plus{} b)$

So let $ a \equal{} b \equal{} 0$ , then $ h(0)^2 \equal{} h(0)$, and $ h(0) \equal{} 0$ or $ h(0) \equal{} 1$.

(case 1) If $ h(0) \equal{} 0$ , then for every integer $ a$,
$ g(a) \plus{} f(a)\sqrt { \minus{} 1} \equal{} h(a) \equal{} h(a \plus{} 0) \equal{} h(a)h(0) \equal{} 0$, and $ g(a) \equal{} f(a) \equal{} 0$.

(case 2) If $ h(0) \equal{} 1$ , then for every positive integer $ a$ , $ h(a) \equal{} h((a \minus{} 1) \plus{} 1) \equal{} h(a \minus{} 1)h(1)$ .
So using MI , $ g(a) \plus{} f(a)\sqrt { \minus{} 1} \equal{} h(a) \equal{} h(1)^a \equal{} (g(1) \plus{} f(1)\sqrt { \minus{} 1})^a$, and $ h(1)h( \minus{} 1) \equal{} h(0) \equal{} 1$,
that says $ h(1) \ne 0$.

So for every integer $ a$

$ g(a) \equal{} \frac {(g(1) \plus{} f(1)\sqrt { \minus{} 1})^a \plus{} (g(1) \minus{} f(1)\sqrt { \minus{} 1})^a}{2}$

and

$ f(a) \equal{} \frac {(g(1) \plus{} f(1)\sqrt { \minus{} 1})^a \minus{} (g(1) \minus{} f(1)\sqrt { \minus{} 1})^a}{2\sqrt { \minus{} 1}}$.

But $ f(\minus{}1)$ and $ g(\minus{}1)$ must be integers. So $ \frac{f(1)}{f(1)^2\plus{}g(1)^2}$ and $ \frac{g(1)}{f(1)^2\plus{}g(1)^2}$ must be integers.

And from $ |f(1)| \le f(1)^2 \le f(1)^2\plus{}g(1)^2$ and $ |g(1)| \le g(1)^2 \le f(1)^2\plus{}g(1)^2$,

we obtain $ |f(1)|\equal{}1$ and $ |g(1)|\equal{}0$ , or $ |f(1)|\equal{}0$ and $ |g(1)|\equal{}1$.

1) If $ f(1)\equal{}1$ and $ g(1)\equal{}0$, for every integer $ a$ we have $ f(a)\equal{}1$ and $ g(a)\equal{}0$ .

2) If $ f(1)\equal{}\minus{}1$ and $ g(1)\equal{}0$, for every integer $ a$ we have $ f(a)\equal{}(\minus{}1)^a$ and $ g(a)\equal{}0$ .

3) If $ f(1)\equal{}0$ and $ g(1)\equal{}1$, for every integer $ a$ we have
       $ f(4a)\equal{}1$ and $ g(4a)\equal{}0$ ,
       $ f(4a\plus{}1)\equal{}0$ and $ g(4a\plus{}1)\equal{}1$ ,
       $ f(4a\plus{}2)\equal{}\minus{}1$ and $ g(4a\plus{}2)\equal{}0$ and
       $ f(4a\plus{}3)\equal{}0$ and $ g(4a\plus{}3)\equal{}\minus{}1$ .

4) If $ f(1)\equal{}0$ and $ g(1)\equal{}\minus{}1$, for every integer $ a$ we have
       $ f(4a)\equal{}1$ and $ g(4a)\equal{}0$ ,
       $ f(4a\plus{}1)\equal{}0$ and $ g(4a\plus{}1)\equal{}\minus{}1$ ,
       $ f(4a\plus{}2)\equal{}\minus{}1$ and $ g(4a\plus{}2)\equal{}0$ and
       $ f(4a\plus{}3)\equal{}0$ and $ g(4a\plus{}3)\equal{}1$ .

THank you Makoto Kohno.
But what about all pair of functions f and g?

My idea is based on complex numbers and $ e^{x\sqrt { \minus{} 1}} \equal{} \cos{x} \plus{} \sqrt { \minus{} 1} \sin{x}$.

And I thought $ h(a) \Leftrightarrow (f(a),g(a))$ is unique because $ \{1,\sqrt { \minus{} 1} \}$ is independent on $ \mathbb{Z} \subset \mathbb{R}$.

So if $ h(0)$ and $ h(1)$ can be determined , $ h(a)$ can be determined for all $ a\in\mathbb{Z}$ , and then $ f(a)$ and $ g(a)$ can be determined for all $ a\in\mathbb{Z}$


Hi, if anybody could solve this problem, I would be very greatful.
Thank You in Advance
See attachment for the problem

There are at most $ 1990^2$ pairs that appear in some element of $ \mathcal{P}$, and $ \binom{100}{2}$ pairs appear in each element of $ \mathcal{P}$, so we have $ |\mathcal{P}| \cdot \binom{100}{2} \leq 1990^2$ and $ |\mathcal{P}| \cdot \leq \frac {1990^2}{\binom{100}{2}} \equal{} 800.020202\ldots$.  (In fact, the problem is extra silly because actually we can replace $ 1990^2$ with $ 1990 \cdot 1989$, as none of the pairs $ (x, x)$ appear in any element of $ \mathcal{P}$, and this reduces our upper bound to 799.)


Consider the Following numbers as PERIODIC --
0000 , 0101 , 110110
that is those numbers which can be formed by a smaller number [ two or more occurences ] .

So , numbers like - -

110 , 1001 etc.

are aperiodic numbers.

Number given the numbers of 1's [Assume it to be a ] and 0's[Assume it to be b ] ,  we need to tell the number of aperiodic numbers with exactly a 1's and b 0's.
For Example --

if a = 2 , b = 2
then total aperiodic numbers are -- 4.

I'm not 100% sure, but I think [url=http://www.research.att.com/~njas/sequences/A007727]A007727[/url] is the associated sequence in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url] in the case $ a \equal{} b$.  In general, though, this should be quite complicated: it's $ \binom{a\plus{} b}{a}$ whenever $ (a, b) \equal{} 1$ but smaller otherwise.

Let $ s(a, b)$ denote the number of aperiodic strings with $ a$ ones and $ b$ zeroes.  The set of all strings with $ a$ ones and $ b$ zeroes can be partitioned according to their maximum period $ d$ (which must divide both $ a$ and $ b$), which gives

$ {a \plus{} b \choose a} \equal{} \sum_{d | \gcd(a, b)} s \left( \frac{a}{d}, \frac{b}{d} \right).$

Note that if we write $ s(a, b) \equal{} s(n, a_1, b_1)$ where $ n \equal{} \gcd(a, b), a_1 \equal{} \frac{a}{n}, b_1 \equal{} \frac{b}{n}$ then among all the terms above $ a_1, b_1$ are fixed, so Mobius inversion applies and we get

$ s(n, a_1, b_1) \equal{} \sum_{d | n} {d(a_1 \plus{} b_1) \choose da_1} \mu \left( \frac{n}{d} \right) \Leftrightarrow$
$ s(a, b) \equal{} \sum_{d | \gcd(a, b)} { \frac{a \plus{} b}{d} \choose \frac{a}{d} } \mu(d).$


在公差为d(d>0)的正数等差数列$a_1,a_2,a_3,...a_{3n}$中,任取$n+2$个数,证明其中必存在两个数$a_i,a_j (i>j)$,满足不等式$1<\frac{a_i-a_j}{nd}<2$

只要很簡單的用一次抽屜原理就能證出, 不是嗎?
[hide]在那n+2個數中, $\exists i>j,\  \max_{i>j}(a_i-a_j)\leq2(n-1)d$[/hide]

[quote="dingdongdog"][quote="ifai"]只要很簡單的用一次抽屜原理就能證出, 不是嗎?
[hide]在那n+2個數中, $\forall i>j,\  \max_{i,j}(a_i-a_j)\leq2(n-1)d$[/hide][/quote]
Let i=1 and j=3n.And then it turns wrong.[/quote]
真不好意思──我似乎把"$\exists$"打成"$\forall$"(上面已作出了修改).  :blush:
此外, i,j 是指那n+2個數中其中兩個的序號.

还是看不懂,可以说明白一些吗?

a complete solution:
Let $A=\{a_i|1\le i\le n+1\}$. Note that for each element of A there is exactly n-1 $a_i$'s which is not a element of A such that $1<\frac{a_i-a_j}{nd}<2$(i+n+1,...i+2n-1). Define $B_i=\{a_{i+n+1},...,a_{i+2n-1}\}$ for $1\le i\le n+1$.
Suppose k of the chosen n+2 $a_i$'s are elements of A. Then there are n+2-k $a_i$'s not element of A. Note that the union of the corresponding $B_i$ of the k elements of A has at least k+n-2 elements. Since $(n+2-k)+(k+n-2)>3n-(n+1)$, the result follows.

[quote="jerry_science"]
所以b(n+2)-b(i+1)<=n,b(n+2)-bi>=2n...........(2)
[/quote]
I think there must be something wrong here.

用抽屉原则可以阿，但要讨论an+2的大小。


y^2=x^3-p^2 x . find all integer ordered pairs (x,y) where p is a prime such that p=3(mod 4)

What is IMOTC? (Asked in order figure out the best location for this topic.)

IMOTC is the International Mathematics Olympiad Training Camp, this post must be in the olympiad section, number theory forum.

I meant the IMO training camp that is conducted in India. For details see http://en.wikipedia.org/wiki/IMOTC


I'm not sure about the level of this problem. So, I'd be glad if moderators don't mind to move it.

[quote] A new company with just two employees, Andri and Budi, rents a floor of a building with
12 offices. How many ways are there to assign different offices to these employees?
[/quote]
What does this problem means? 1 office for each employees or all office must be given to the employees?

EDIT: to obey the etiquette of mathlinks

Im not exactly sure how you want this.

[hide]
Assuming both workers get disjont tasks but their union is all of them.
Let $ A$ and $ B$ be sets. And we want $ |A\cup B| =12$ with $ A\cap B = \{ \}$.
Draw a Venn diagramm. There are three sections. $ A$ only. $ B$ only. The intersection.
We want to put 12 objects into 2 sections because the intersection is empty.
That is $ {{12+2-1}\choose 2}$.
[/hide]

Common sense dictates that it should be one office per employee.  Mathematical curiosity dictates that you should solve both problems anyhow.  The etiquette guidelines of this forum dictate that you should not post several questions on the same thread and certainly that you should not place your questions under "hide" tags.

Oh, I'm sorry. Actually, I'm not too familiar with this forum before.  :blush: Thanks!
I wll edit it soon.
So, I suppose to write 2 solution for this problem?  :o


inequalities
induction
number theory proposed
number theory

[quote="moldovan"]Find all the positive integers $ a$ such that $ \frac {5^a \plus{} 1}{3^a}$ is a positive integer.[/quote]
This problem is not very hard . Let's solve it ! . Suppose $ a$ satisfy condition $ 3^a|5^a\plus{}1$ ,then it is not hard to show that a is an odd number . By the lifting exponent lemma we have:  $ v_3(5^a\plus{}1)\equal{}v_3(1\minus{}(\minus{}5)^a)\equal{}v_3(6)\plus{}v_3(a)\equal{}v_3(a)\plus{}1$ . Because $ 3^a|5^a\plus{}1$ so $ v_3(a)\plus{}1\geq a$ . The rest is easy by some inequalities and $ a\equal{}1$ is the unique solution of this problem .
Note :Lifting exponent lemma said that if $ p|a\minus{}b$ and $ \gcd(a,b)\equal{}1$ then $ v_p(a^n\minus{}b^n)\equal{}v_p(a\minus{}b)\plus{}v_p(n)$ .This lemma can be proven by induction .

[quote="TTsphn"]
Note :Lifting exponent lemma said that if $ p|a \minus{} b$ and $ \gcd(a,b) \equal{} 1$ then $ v_p(a^n \minus{} b^n) \equal{} v_p(a \minus{} b) \plus{} v_p(n)$ .This lemma can be proven by induction .[/quote]
Er... What does $ v_p(n)$ mean?

$ v_p(n)$ is the largest integer k such that $ p^k|n$  :)

[quote="TTsphn"]The rest is easy by some inequalities and $ a \equal{} 1$ is the unique solution of this problem . [/quote]

I still don't get this part  :D
Could you fill in the details? Thanks :)

Let $ n\equal{}3^t.k$ where $ t\geq 0$ and $ \gcd(3,k)\equal{}1$ ,by the condition we have $ t\plus{}1\geq 3^tk\geq 3^t$ ,this inequality hold for only $ t\equal{}0$ and $ k\equal{}1$  and $ a\equal{}1$ follows from this fact .

Is it necesarry for p to be prime?
P.S Yes, it is. I found it while surfing the net :blush:


inequalities
induction
algebra unsolved
algebra

for all elements $ a_m$ and $ a_n$ of the sequence $ a_1$ , $ a_2$, $ a_3$, ...
the following inequality holds:

$ | a_m \plus{} a_n \minus{} a_{m \plus{} n} | \leq \frac {1}{m \plus{} n}$

Show that the sequence is an arithmetic progression.

regards

P.S.:
The problem was proposed by arthur engel during the 1997 German TST if I remember correctly

We had $ |a_{2n}-2a_n|\le \frac{1}{2n}, \ \frac{1}{3n}\ge |a_{3n}-a_{2n}-a_n|\le |a_{3n}-3a_n-(a_{2n}-2a_n)|$.
It give (by induction) $ |a_{mn}-ma_n|\le \frac 1n (\frac 12 +\frac 13+...+\frac 1m)=\frac 1n (H_m-1)$.
Therefore $ |ma_n-na_m|\le \frac 1n (H_m-1)+\frac 1m (H_n-1)$. It equavalent to
$ |\frac{a_n}{n}-\frac{a_m}{m}|\le \frac{1}{n^2}g(m)+\frac{1}{m^2}g(n)$, were $ g(k)=\frac{1}{k}(\frac 12+\frac 13+...+\frac 1k)<\frac{ln k}{k}\to 0$, when ${ k\to \infty}$.
It give $ |\frac{a_n}{n}-\frac{a_m}{m}|\le |\frac{a_n}{n}-\frac{a_k}{k}|+|\frac{a_k}{k}-\frac{a_m}{m}|\le (\frac{1}{n^2}+\frac{1}{m^2})g(k)+(g(m)+g(n))\frac{1}{k^2}\to 0$, when $ k\to \infty$. Therefore $ a_n=nconst.$


Putnam inequalities probability LaTeX function combinatorics logarithms

I can only repeat solution (maybe blahblahblah ?), which was posted before crash.
We can rewrite it as
\[C_{m+n}^nm^mn^n< (m+n)^{m+n}=\sum_{k=0}^{m+n}C_{m+n}^km^kn^{m+n-k}=...+C_{m+n}^nm^mn^n+...\]
That's all.

(I'm new to this forum, so I don't know how to write the equations fancily yet.)

The solution I came up with was similar to the above one, but a slightly different flavor:

The inequality rearranges to become:

choose(m + n, m) * (m / (m+n))^m * (n / (m+n))^n < 1

Notice that the left hand side of this inequality is simply the probability of obtaining success exactly m times out of m+n tries, given that the probability of a single success is m / (m + n). Since this is a probability, it obviously must be  <= 1. Also, it can only be equal to 1 if either m / (m+n) = 1 or n / (m+n) = 1, i.e. either m = 0 or n = 0 (here 0^0 is defined as 1). But since m and n must be positive integers, the strict inequality holds.

Certainly. That is algebraically equivalent to the solution already posted.

A pretty version of the inequality in your post:
${m+n\choose m}\cdot \left(\frac{m}{m+n}\right)^m\cdot \left(\frac{n}{m+n}\right)^n<1$

The "fancy" equations here are $\LaTeX$; enclose the code in dollar signs to use it. You can see the code by hovering over or copying what you see in the posts.

[quote="jmerry"]Let $m$ and $n$ be positive integers. Show that
$\frac{(m+n)!}{(m+n)^{m+n}} < \frac{m!}{m^m}\cdot\frac{n!}{n^n}$[/quote]
I discovered this result  months ago:

http://www.ddtoanhoc.net/forum/index.php?showtopic=150

Let $m_i$ are non-negative integer numbers. $m = \sum_{i=1}^n m_i$.

Then we have:

\[\frac{m!}{m_1!m_2!\cdots m_n!} \leq 2^{H(m_1/m,m_2/m,...,m_n/m)}\]

where $H(p_1,p_2,...,p_n) \stackrel{\text{def}}{=} -\sum p_i \log p_i$ is the( Shanon) entropy function.

Simplest solution is:

for $p_i\geq 0, \sum p_i = 1$, we have:

\[ \sum{}_{m{}_1,m_2,...,{}_m{}_n} {m \choose m_1,m_2,...,m_n} p_1^{m_1}p_2^{m_2}...p_n^{m_n} = 1\]

Hence

\[{m \choose m_1,m_2,...,m_n} p_1^{m_1}p_2^{m_2}...p_n^{m_n} \leq 1 \]

holds for any $p_i$.

Putting $p_i = m_i/m$ terminates the proof.

The L.H.S is the probability of exactly $m_i$ times of face $i$ for a biased n-coin with $p_i = m_i/m$

[quote]

Với xác suất $ p = {p_1, p_2 ... p_n}$ định nghĩa entropy của p như sau$H(p) =-\sum p_k \log_2 p_k$
.
0 log 0 được định nghĩa là 0.
Chú ý là $\sum p_k = 1$  nên entropy luôn không âm.

Chứng minh rằng với $m_i$ nguyên không âm và $\sum m_i = m$  thì

\[\frac{m!}{m_1!m_2!\cdots m_n!} \leq 2^{H(m_1/m,m_2/m,...,m_n/m)}\]

Bài này khá là cơ bản trong Information Theory nhưng định lí trên lại có một số cách giải sơ cấp. Mình tìm ra 2 cách giải bằng quy nạp, một bằng cách dùng bất đẳng thức Stirling. Các bạn tham khảo nhé.

Lớp 11+ có thể giải được bài này. Have fun!
[/quote]

[quote="jmerry"]Let $m$ and $n$ be positive integers. Show that
$\frac{(m+n)!}{(m+n)^{m+n}}< \frac{m!}{m^{m}}\cdot\frac{n!}{n^{n}}$[/quote]

Can the beta function be used to prove this problem? We have: \[B(m+1,n+1)=\int_{0}^{1}x^{m}(1-x)^{n}\ dx = \frac{\Gamma (m+1)\Gamma (n+1)}{\Gamma (m+n+2)}= \frac{1}{m+n+1}\frac{m!n!}{(m+n)!}\] and since the problem is equivalent to proving that \[\frac{m^{m}n^{n}}{(m+n)^{m+n}}< \frac{m!n!}{(m+n)!}= (m+n+1)\int_{0}^{1}x^{m}(1-x)^{n}\ dx\] for $0\leq x\leq 1,$ consider the function $f(x)=x^{m}(1-x)^{n}$ so that \[f^{\prime}(x)=mx^{m-1}(1-x)^{n}-nx^{m}(1-x)^{n-1}\]  \[=x^{m-1}(1-x)^{n-1}[m-(m+n)x]=0\]  \[\Rightarrow x=0,\frac{m}{m+n},1\] are the critical points of $f(x)$ where \[f\left( \frac{m}{m+n}\right) = \frac{m^{m}n^{n}}{(m+n)^{m+n}}\] is the global maximum of $f(x)$ over the interval [0,1] for all $m,n=1,2,\ldots$ (cf. [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110120]this post[/url]) and since $f(x)$ is: (i) positive for $x\in (0,1)$; (ii) continuous for $x\in [0,1]$; (iii) $f(0)=f(1)=0$ and $f(x)$ has no other zeros; (iv)  $f(x)$ is concave on (0,1) (since f is continuous+twice differentiable this means that  $f^{\prime\prime}\left( x\right) <0$ on (0,1)); and (i)->(iv) imply that (v) the triangle with verticies at the points $\left\{ (0,0), \left( \frac{m}{m+n}, \frac{m^{m}n^{n}}{(m+n)^{m+n}}\right) , (1,0) \right\}$ lies entirely under the (closed) graph of $f(x)$ and hence the area of the triangle is less than the integral which is less than the maximum of the function, that is \[\frac{1}{2}\ \frac{m^{m}n^{n}}{(m+n)^{m+n}}\leq \int_{0}^{1}x^{m}(1-x)^{n}\ dx\leq \frac{m^{m}n^{n}}{(m+n)^{m+n}}\] which implies that \[\frac{(m+n+1)}{2}\ \frac{m^{m}n^{n}}{(m+n)^{m+n}}\leq \frac{m!n!}{(m+n)!}= (m+n+1)\int_{0}^{1}x^{m}(1-x)^{n}\ dx \leq (m+n+1)\frac{m^{m}n^{n}}{(m+n)^{m+n}}\] and since $\frac{(m+n+1)}{2}>1$ for all $m,n=1,2,\ldots$ the proof has been shown (need to slim it down though).

I assigned the 2004 Putnam to my Putnam preparation class. Patterns_34, who is in that class, gave the start of an argument but didn't finish it. I'm here to provide the finish for him. The beginning of this argument is generally the same as benorin's post above, but the finish is a little different.

We'll make use of two facts:

[b]Beta integral:[/b] for $n,m$ nonnegative integers, $\int_0^1x^n(1-x)^m,dx=\frac{n!m!}{(n+m+1)!}.$

[b]Bernoulli's inequality:[/b] For $n\in\mathbb{N}$ and $x\ge-1,$ $(1+x)^n\ge 1+nx.$

Both of these are, or can be, classic exercises in induction. In particular, one can use integration by parts to find a recursion for the Beta integral for integer coefficients, which is all we need here. (There are other arguments for non-integer coefficients, but we don't need them.)

So we are trying to prove that:
\[\frac{n^nm^m}{(n+m)^{n+m}}<\frac{n!m!}{(n+m)!}=(n+m+1)\int_0^1x^n(1-x)^m\,dx.\]
Now, multiply through by $\frac{(n+m)^{n+m}}{n^nm^m}$ and rearrange a little. We want to prove that
\[(n+m+1)\int_0^1\left(\frac{(n+m)x}n\right)^n\left(\frac{(n+m)(1-x)}m\right)^m\,dx>1\]
\[\int_0^1\left(\frac{(n+m)x}n\right)^n\left(\frac{(n+m)(1-x)}m\right)^m\,dx>\frac1{n+m+1}\]
The integrand has a maximum value of $1$ at $x=\frac{n}{n+m}.$ (We don't technically need to prove that, but we need it for motivation.) Now we make the substitution $x=\frac{n}{n+m}+u.$ The changes the interval of integration to $u\in\left[-\frac{n}{n+m},\frac{m}{m+n}\right]$ and it changes the integrand into
\[\left(1+\frac{n+m}n\,u\right)^n\left(1-\frac{n+m}m\,u\right)^m.\]
By Bernoulli's inequality, for an appropriate range of $u$ (which we will verify later),
\begin{align*}\left(1+\frac{n+m}n\,u\right)^n\left(1-\frac{n+m}m\,u\right)^m&\ge(1+(n+m)u)(1-(n+m)u)\\&=1-(n+m)^2u^2\end{align*}
We will only invoke this inequality for $u\in\left[-\frac1{n+m},\frac1{n+m}\right],$ which is contained in the interval of integration. Here is the finishing computation:
\begin{align*}\int_0^1\left(\frac{(n+m)x}n\right)^n\left(\frac{(n+m)(1-x)}m\right)^m\,dx&=\int_{-\frac{n}{n+m}}^{\frac{m}{n+m}}\left(1+\frac{n+m}n\,u\right)^n\left(1-\frac{n+m}m\,u\right)^m\,du\\
&\ge\int_{-\frac{1}{n+m}}^{\frac{1}{n+m}}\left(1+\frac{n+m}n\,u\right)^n\left(1-\frac{n+m}m\,u\right)^m\,du\\
&\ge\int_{-\frac{1}{n+m}}^{\frac{1}{n+m}}1-(n+m)^2u^2\,du\\
&=\frac43\cdot\frac1{n+m}\\
&>\frac1{n+m}>\frac1{n+m+1}\end{align*}
which is what we wanted to prove.

Alternatively, an induction argument works, although it is a tad bit messy.

Note that the inequality can be rearranged to \[\dbinom{m+n}m < \dfrac{(m+n)^{m+n}}{m^mn^n}.\] It suffices to prove that this inequality is true.  To proceed, use strong induction on $m+n$.  Base cases are easy, so I will ignore them.  For the inductive step, assume $m\geq 2$ and $n\geq 2$, and remark that by Pascal's Identity \begin{align*}\binom{m+n}n &= \binom{m+n-1}{n-1} + \binom{m+n-1}n\\&<\dfrac{(m+n-1)^{m+n-1}}{m^m(n-1)^{n-1}} + \dfrac{(m+n-1)^{m+n-1}}{(m-1)^{m-1}n^n}.\end{align*} Proving that this is less than $(m+n)^{m+n}/(m^mn^n)$ is equivalent through rearrangement to showing that \[\dfrac{m^m}{(m-1)^{m-1}} + \dfrac{n^n}{(n-1)^{n-1}} < \dfrac{(m+n)^{m+n}}{(m+n-1)^{m+n-1}}.\] Consider the function \[f(x) = \left(1 + \dfrac{1}{x}\right)^x.\] Note that through standard procedures one may conclude that $f(x)$ is increasing along the positive real numbers.  This in turn means that \[\left(\dfrac{m}{m-1}\right)^{m-1} < \left(\dfrac{m+n}{m+n-1}\right)^{m+n-1}\qquad\text{and}\qquad \left(\dfrac{n}{n-1}\right)^{n-1} < \left(\dfrac{m+n}{m+n-1}\right)^{m+n-1}.\] Thus \begin{align*}m\left(\dfrac{m}{m-1}\right)^{m-1}+n\left(\dfrac{n}{n-1}\right)^{n-1} &< m\left(\dfrac{m+n}{m+n-1}\right)^{m+n-1} + n\left(\dfrac{m+n}{m+n-1}\right)^{m+n-1} \\&= (m+n)\left(\dfrac{m+n}{m+n-1}\right)^{m+n-1},\end{align*} or \[\dfrac{m^m}{(m-1)^{m-1}} + \dfrac{n^n}{(n-1)^{n-1}} < \dfrac{(m+n)^{m+n}}{(m+n-1)^{m+n-1}},\] as desired.

[quote=jmerry]Let $m$ and $n$ be positive integers. Show that
$\frac{(m+n)!}{(m+n)^{m+n}} < \frac{m!}{m^m}\cdot\frac{n!}{n^n}$[/quote]

Proof
Suppose not,
Then there exists $m,n \in \mathbb{N}$ such that $\frac{(m+n)!}{m!n!}*m^m*n^n > (m+n)^{m+n}$
Now, consider the following:
Suppose we have $m+n$ balls all of different sizes and $m$ black boxes all with different labels and $n$ white boxes all with different labels.
Now, we choose $n$ balls out of the $m+n$ balls and paint them white colour, and we paint the remaining $m$ balls with black colour.
See this can be done in number of ways = $\frac{(m+n)!}{m!n!}$
Now, we put the balls in the boxes such that no black box contains white ball and no white box contains black ball.
Then see that each black ball have $m$ choices (all boxes were distinct) and each white ball have $n$ choices.
Further number of black balls = $m$ and number of white balls =$n$.
So total number of ways to do this = $\frac{(m+n)!}{m!n!}*m^m*n^n$
Now, consider a different activity
We are given $m+n$ boxes out of which $m$ boxes are black colour and contain black paint and the $n$ boxes are white colour and contain white paint.
Now,
further we are given $m+n$ balls all different sizes.
We place each ball in a black box or a white box and assign black colour to the ball if t goes to a black box and assign white colour to the ball if it goes to the white box.
Note that if, $A$ be the set of outcomes in the previous activity and $B$ be the set of outcomes in the second activity then, $A$ is a subset of $B$ and $B \ne A$ (because $n>0$  and  $m>0$)
Therefore, number of ways to complete activity $B$ is greater than the number of ways to complete activity $A$.
Now, number of ways to complete activity $B$ = $(m+n)^{m+n}$
The,  number of ways to complete activity $A$ = $\frac{(m+n)!}{m!n!}*m^m*n^n$

So this gives us a contradiction to our assumption

$\frac{(m+n)!}{m!n!}*m^m*n^n > (m+n)^{m+n}$

[b][u]Hence Proved[/u][/b]

Note that the given inequality is equivalent to
\begin{align*}
\frac{(m+n)!}{m!n!} = \binom{m+n}{m} &< \frac{(m+n)^{m+n}}{m^mn^n}  \\
&= \frac{1}{m^mn^n} \sum_{k=0}^{m+n} \binom{m+n}{k} m^k n^{m+n-k} \\
&=  \sum_{k=0}^{m+n} \binom{m+n}{k} \left( \frac{n}{m}\right)^{m-k},\\
\end{align*}
which is obvious since the $k = m$ term of the sum is exactly $\binom{m+n}{m} = \frac{(m+n)!}{m!n!}$.


I got answer A. But my friend got D. We are trying to resolve this.
Can someone help us?
Thanks.

[hide]
It's D based on an answer key someone sent me.
[/hide]

Perhaps posting the problem and the proposed solutions might help clarify this?  :)

I got D, and so did one of my friends, but I'm not sure.

Because the TV screen is made up of two $ 3: 4: 5$ triangles, and we know that the hypotenuse is $ 27$, then the width is $ 27 \cdot \frac {4}{5} \equal{} \frac {108}{5}$. The height is $ 27 \cdot \frac {3}{5} \equal{} \frac {81}{5}$.

Then, the white part of the screen has a width of $ \frac {108}{5}$, and a height of $ \frac {1}{2} \cdot \frac {108}{5} \equal{} \frac {54}{5}$.

Then, the height of each darkened strip is simply $ \frac {\frac {81}{5} \minus{} \frac {54}{5}}{2} \equal{} \frac {\frac {27}{5}}{2} \equal{} \frac {27}{10} \equal{} 2.7 \Rightarrow \boxed{D}$

Hmmmm... thats #9 on the AMC 12A... and yes, the answer is 2.7\

Oh, and an easier way would be to pretend its actually a 3-4-5 triangle. Then to get the aspect ratio to 2:1, you would have to take 1 off of the height, or .5 from the top and bottom.  Multiply by 27/5 to scale it back up, and you get 2.7


number theory
prime numbers
Additive Number Theory

Show that [list=a][*] infinitely many perfect squares are a sum of a perfect square and a prime number, [*] infinitely many perfect squares are not a sum of a perfect square and a prime number. [/list]

Take $n \in \mathbb{N}$.
Now it is obviously, that every square $(n+1)^2$ with is the sum of a square, i.e. $n^2$ and an odd number, i.e. $2n+1$.
Since every odd number $2n+1$ is either a prime number or not a prime number (and, of course, there exist infinitely many odd primes and infinitely many odd composite numbers), there are infinitely many odd prime numbers as well as infinitely many odd composite numbers, so there are infinitely many squares $(n+1)^2$ which are a sum of a square and a prime number as well as a sum of a square and an odd composite number, which is not prime.
This solves a.
The solution for b also needs, that for all $a$ with $a<n$,  $(n+1)^2-a^2=(n+a+1)(n-a+1)$ is an odd composite number and hence, $(n+1)^2$ is not a sum of an ordinary (is it okay like this, Xorg? ;) ) square and a prime number.

Hallo,

for b it only suffices to prove that  (n +1)^2 - n^2 is composite for infinitely many n,
because the difference between 2 squares is surely composite by (a-b) (a+b) ;except for b equals a-1 .
For n = 2^(2a+1) - 1, we have 2n + 1 = 2^(2a+2) - 1 which factors for infinitely many a and is therefore not a prime .


probability
combinatorics theorems
combinatorics

I'm preety bad in Combinatorics, what books, and other literature do you recommend that can help me improve my level?

Thanks

yeees i have the same question
i can't understand combinatorics  :(
a good book maybe will help me
please post some links or atach one =D
plisssss
xD  :D  :D

look at this link [url]http://www.openmathtext.org/downloads.html[/url] the probability notes.
There you can find the basics of combinatorics :wink:


geometry
calculus
integration
MATHCOUNTS
absolute value

Hi,

I have a few problems that I don't really understand. Could someone please explain them? Thanks.

1. In the first two-mile race, Kendi beat Jerri by 1/3-mile, and in the second two-mile race, Jerri beat Micah by 1/2-mile. If Kendi and Micah run a third two-mile race, and all three runners run at the same speed in every race, by how many miles will Kendi beat Micah? Express your answer as a common fraction.

2. What is the number of square units in the area of the region bounded by the graphs of y = -|x| + 2 and y = |x| - 2?

Have you sketched the graphs for 2? Upon sketching, I think you will have to find some limits and integrate.

EDIT: This[url=http://archives.math.utk.edu/visual.calculus/5/area2curves.3/index.html]link[/url]might be useful.

[hide="maybe"]The limits I think come out as 2 and -2, so the integral might be: $ \int^2_{\minus{}2} (|x|\minus{}2)\minus{}(\minus{}|x|\plus{}2)dx \equal{} \int^2_{\minus{}2} 2|x|\minus{}4dx$[/hide]

Just find the area of the region bounded by the x-axis and the graph of the first equation and multiply by two.

[quote="the future"][quote="beans"]Have you sketched the graphs for 2? Upon sketching, I think you will have to find some limits and integrate.[/quote]

Sorry but mathcounts doesnt use calculus.  ALthough just by sketching it (your first hint), you dont need  to integrate, just find the areas of the geometric parts[/quote] Ah sorry. (Noob moment - I just clicked on unanswered posts!) :oops:

Reve, please ignore my edit.

[quote="alanchou"]Actually, in sketching the graph, you get a square with vertices at (2,0),(0,2),(-2,0), and (0,-2).

Graphing a linear absolute value equation results in straight lines, if I'm not mistaken.[/quote]

yea thats right so its just $ (\sqrt {8})^{2} \equal{} 8$.

This is alot better solution then calculus, easier and simpler to solve.

Infact, most calulus solutions of this type with absolute values are better off solving the elementary way with regular geometric shapes.


You have to guess the names of the players and their positions. You are allowed to ask me.

Who's on first, What's on second, I Don't Know is on third  :rotfl:  :rotfl:  :rotfl:

The bases are not the only positions, there is also an outfield, and all that stuff, but you know the most important part.


Because Sally left out one eighth of her life when she responded to Silly Sod
[hide] $21: 7/8=24$
  => Sally is 24 years old[/hide]

Well, now there're 5 answers, all the same, so, as in other posts in this section, don't continue posting the answer, it's clear... That's one problem in this section: 10 consecutive and equal solutions... I think the answer, now, is well-known, so it's useless to post again the same.

[quote="Arvind_sn"]Sally was asked her age by Silly Sob. Sally said that her age was 21, when her age was some number. Assuming that Sally left out one eighth of her life when she responded to Silly Sob, how old was Sally?[/quote]

[hide]7x/8=21
7x=168
x=168/7
x=24[/hide]

[quote="Arvind_sn"]Sally was asked her age by Silly Sob. Sally said that her age was 21, when her age was some number. Assuming that Sally left out one eighth of her life when she responded to Silly Sob, how old was Sally?[/quote]

[hide]
$7x/8=21$
$x=24$[/hide]

[hide][b]24 years old[/b]

Let x=her real age

(8/7)21=(7/8)x(8/7)

24=x

x=24[/hide]

FINALLY I M A RIEMAN HYPOTHESIS RANK!!!

I haven t been one for like 5 months.

[quote="Arvind_sn"]Sally was asked her age by Silly Sob. Sally said that her age was 21, when her age was some number. Assuming that Sally left out one eighth of her life when she responded to Silly Sob, how old was Sally?[/quote]

[hide] 21=7/8 of x
          x=24 [/hide]

Sally is actually 24....

21 * 7/8 =24

-soccerstar123

REVIVE OLD TOPICS??? THAT'S NOT OLD THAT'S THE WHOLE PART OF THIS FORUM..

-SOCCERSTAR123


Of course, I am math-oriented, so I want to master LaTeX and Matlab. I also know Java very well, and I am in the process of learning C++ and I want to learn Python.
I was wondering what other people thought were the most important programming languages to learn for the future?

Depends what you want to do.  If you see yourself going into, say, quantitative finance, then I think C++ and Java are the commonly used languages today.  In engineering, Matlab might be more common.  In scientific computing (or at least in applied math), however, it seems that by and large Fortran 95 (basically the same Fortran that's been in use for over half a century) is still the standard.  Matlab is almost never used when efficiency is an issue -- personally I use it only to make quick proof-of-concept type computations.  For example if I have a big numerical simulation which has no hope of running on my desktop, I might write up a Matlab code to simulate a stripped down bare-bones version of the same model to demonstrate some qualitative behavior.

High performance and parallel computing is finding increasingly more use in science, and this trend seems likely to continue.  Most grad students that work on biophysical models in my department, for example, typically use a computing cluster to do all their numerical simulations.  The major difficulty is developing parallelizable code, which is something Fortran is somewhat better suited to (though C is used too).  But in the future it seems almost inevitable that specialty languages that streamline parallel computing will become the norm; a couple exist now, but they've got their problems, and really aren't too popular, e.g. Haskell, and SISAL.  In the last few years there's been a lot of DARPA funding toward the development of a high performance computing language, with all the big players (IBM, Cray, Sun), developing their own.  I guess the future will tell which, if any, winds up being the standard.

So if you think you might wind up doing some high performance computing one day, it wouldn't hurt to keep up with these developments, and learn some of the theory behind parallel computing in general.

Thank you for your explanation; I will look into languages like Fortran and some of the other ones you mentioned...

AS JoeBlow said, it really does depend...

If you wanna go for scripting servers and networking, SQL, Java, and HTML are all must knows.

For general purpose programming, I find Ruby really nice to use. It's simpler than C++ and is just as functional. I know someone who works with Ruby in scripting disaster-predicting software.

Just, whatever you do, DO NOT LEARN BASIC!
It will nearly permanently mess up you're conceptions of coding, and beyond basic ( no pun intended :p) stuff, it really just doesn't make the cut.

Hope you find what works for you!

if you really want to go to the bricks and mortar try writing some specifications in CafeObj

C++ has a limitless potential, and Python is a language that greatly simplifies mathematical calculations. For example, 20 lines of code in C++ may only be like 3-5 lines in Python.

Python is becoming increasingly popular among scientists (at least in my maths department): developing time is greatly reduced (compared to c++, say) and graphical representation of data are very easily implemented. These days, I always code in Python for small simulations. (some banks use Python (eg: Barclays) and I know that the NASA used it for several projects - this is quite robust )

The language R is also very popular for statisticians.

Thank you for your suggestions... I think I will look at C++ and Python first and see where I can go from there.


Let $ x,y,z>0$ such that $ (x\plus{}y)(y\plus{}z)(z\plus{}x)\equal{}1$ prove that
$ \sqrt{xy(x\plus{}y)}\plus{}\sqrt{yz(y\plus{}z)}\plus{}\sqrt{xz(x\plus{}z)}>2\sqrt{\frac{xy\plus{}yz\plus{}zx}{3}}$

[quote="hoangvn.fix"]Let $ x,y,z > 0$ such that $ (x \plus{} y)(y \plus{} z)(z \plus{} x) \equal{} 1$ prove that
$ \sqrt {xy(x \plus{} y)} \plus{} \sqrt {yz(y \plus{} z)} \plus{} \sqrt {xz(x \plus{} z)} > 2\sqrt {\frac {xy \plus{} yz \plus{} zx}{3}}$[/quote]
The following stronger inequality is also true.
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ (x \plus{} y)(y \plus{} z)(z \plus{} x) \equal{} \frac {512}{729}.$ Prove that:
\[ \sqrt {xy(x \plus{} y)} \plus{} \sqrt {yz(y \plus{} z)} \plus{} \sqrt {xz(x \plus{} z)}\geq 2\sqrt {\frac {xy \plus{} yz \plus{} zx}{3}}\]

[quote="hoangvn.fix"]i would like to see your solution for this problem! Dear mathlinkers[/quote]
Will your way work for the stronger ?  :)
$ \sqrt{xy(x\plus{}y)}\plus{}\sqrt{yz(y\plus{}z)}\plus{}\sqrt{zx(z\plus{}x)} \ge \sqrt[3]{2}$ $ .\sqrt{ab\plus{}bc\plus{}ca}.\sqrt[6]{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$

$ k\equal{}\sqrt[3]{2}$ is also the best value for this ineq still trues:

$ \sqrt{xy(x\plus{}y)}\plus{}\sqrt{yz(y\plus{}z)}\plus{}\sqrt{zx(z\plus{}x)} \ge k.$ $ \sqrt{ab\plus{}bc\plus{}ca}.\sqrt[6]{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$
It exists a nice proof by only AM-GM  :lol:

[quote="nguoivn"]

$ \sqrt {ab(a \plus{} b)} \plus{} \sqrt {bc(b \plus{} c)} \plus{} \sqrt {ca(c \plus{} a)} \ge \sqrt [3]{2}$ $ .\sqrt {ab \plus{} bc \plus{} ca}.\sqrt [6]{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

It exists a nice proof by only AM-GM  :lol:[/quote]

Besides, I think the following also trues, too (I still haven't try it):
$ \sqrt {ab(a \plus{} b)} \plus{} \sqrt {bc(b \plus{} c)} \plus{} \sqrt {ca(c \plus{} a)} \le \sqrt []{3}$ $ .\sqrt {ab \plus{} bc \plus{} ca}.\sqrt [6]{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$  :maybe:

[quote="nguoivn"][quote="hoangvn.fix"]i would like to see your solution for this problem! Dear mathlinkers[/quote]
Will your way work for the stronger ?  :)
$ \sqrt {ab(a\plus{} b)} \plus{} \sqrt {bc(b \plus{} c)} \plus{} \sqrt {ca(c \plus{} a)} \ge \sqrt [3]{2}$ $ .\sqrt {ab \plus{} bc \plus{} ca}.\sqrt [6]{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

$ k \equal{} \sqrt [3]{2}$ is also the best value for this ineq still trues:

$ \sqrt {ab(a \plus{} b)} \plus{} \sqrt {bc(b \plus{} c)} \plus{} \sqrt {ca(c \plus{} a)} \ge k.$ $ \sqrt {ab \plus{} bc \plus{} ca}.\sqrt [6]{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
It exists a nice proof by only AM-GM  :lol:[/quote]
See here: http://can-hang2007.blogspot.com/2009/12/inequality-8-t-q-anh.html


vector
real analysis
real analysis unsolved

Show that a real normed vector space is finite-dimensional iff the closed unit ball is compact.

Let's suppose the unit ball B $||x||\leq1$ is compact, thus there must be finite sequence of points $a_i$ which can approximate the unit ball with desired accuracy, let it be ball with radius $\frac{1}{2}$,i.e. $B$ is contained in the union of finite number of ball with centers at $a_i$ and radiuses $\frac{1}{2}$. Denote by $V$ the finite-dimensional space, generated by the point $a_i$. We'll prove that $V=E$ (by $E$ we've denoted out given normed space):If it is not there must be point $x\in E\V$. Since $V$ is closed (as any finite-dimensional subspace),denote  $d(x,V)=\alpha > 0$. From the definition of the distance there must be point $y\in V$ such that $\alpha \leq ||x-y|| \leq \frac{3\alpha}{2}$ and denote $z=\frac{(x-y)}{||x-y||}$ (1), since $z$ belongs to $B$ there must be an index $i$ s.t. $||z-a_i|| \leq \frac{1}{2}$: And finally it is left to write that
$x=y+||x-y||z=y+||x-y||a_i+||x-y||(z-a_i)$ (*) and to notice that $y+||x-y||a_i \in V$. Therefore by the denotion $d(x,V)=\alpha > 0$ we must have $||x-y||\cdot ||z-a_i|| \geq \alpha$, from where according to (1) we get a contradiction with $||x-y|| \geq 2\alpha$


AMC
USA(J)MO
USAMO
AIME
AMC 12
AMC 10
AMC 12 A

For freshmen and sophomores who want to qualify for the USAMO, does the score you get on the AMC-12 or AMC-10 matter or just your AIME score?

Just your AIME score matters. As long as that is higher than the floor value, you're in.

don't you also automatically get in if your AMC 12 score is 150?
or was that a rumor i heard...

No.  I remeber that a lot of the 150s last year didn't get to take the USAMO.

[quote="JesusFreak197"]I thought no one got a 150 on the AMC-12, just the AMC-10. Or was that only in California?[/quote]

That must've been only in California :P   Zach Abel got a 150 on the AMC 12 A and a bunch of people got a 150 on the AMC 12 B.

also Adam from WA got a 150 on AMC12 (he tied Zach able with 150 in AMC and 14 in AIME)

Does anyone know what the floor score was last year for freshmen and sophomores?

Last year as a sophomore I got 8 right on the AIME I (from what I marked on the booklet next to the questions; I could have always mis-bubbled), and I didn't qualify.

Did I make some mistakes bubbling, or did I just miss the cut to USAMO?

[quote="white_horse_king88"]It was a record high cutoff also... I suspect it will be 9 again? Maybe this year.[/quote]

i heard that was not going to happen this upcoming year, because the AMC has been preparing the AIME 2006 for a long time, after noting that 2005's was too easy. This I saw from one of the other posts on this forum.


the three sides of $\triangle{ABC}$ are a,b,c.then we make a $\triangle{A_1B_1C_1}$,with A located on $A_1B_1$,B located on $C_1B_1$,C located on $A_1C_1$,and $\frac{AA_1}{b^2c}=\frac{BB_1}{c^2a}=\frac{CC_1}{a^2b}$.
Prove that  $\triangle{ABC}$and $\triangle{A_1B_1C_1}$ are similar.

[quote="ehsan2004"]I think you meant was $a= BC , b= CA$ and $c= AB$?? Am I wrong  ?? :huh:[/quote]
You're right.

Dingdongdog, are you sure about the statement?
An olympiad 2002 problem reads as follows:
Let $ABC$ be a triangle. Points $A_1,B_1,C_1$ are chosen on the lines $AC,AB,BC$, respectively, such that $A\in (CA_1),B\in (AB_1),C\in (BC_1)$ and $\frac {AA_1}{bc^2}=\frac {BB_1}{ca^2}=\frac {CC_1}{ab^2}$. Prove that triangles $ABC$ and $A_1B_1C_1$ are similar.

Still no one?To prove either mine or enescu's is OK.

For the problem I stated:
suppose $\frac{AA_1}{bc^2}=\frac{BB_1}{ca^2}=\frac{CC_1}{ab^2}=x$. The cosine law in triangle $A_1AB_1$ gives
\[A_1B_1^2=(xbc^2)^2+(c+xca^2)^2-2xbc^2(c+xca^2)\cos {\angle {A_1AB_1}}.
\] But $\angle {A_1AB_1}=180^{\circ}-\angle {BAC}$, hence $ \cos {\angle {A_1AB_1}}=-\frac {b^2+c^2-a^2}{2bc}$. Replacing in the equality above yields
\[\frac{A_1B_1^2}{c^2}=1+x(a^2+b^2+c^2)+x^2(a^2b^2+b^2c^2+c^2a^2).
\]The conclusion is obvious.


Fie doua triunghiuri $ABC,\ A'B'C'$.

Sa se arate ca urmatoarele patru afirmatii sunt echivalente:

$\blacksquare\ 1.\ ABC\equiv A'B'C';$

$\blacksquare\ 2.\ aa'+bb'+cc'=9RR';$

$\blacksquare\ 3.\ 4(m_am'_a+m_bm'_b+m_cm'_c)=9RR'.$

$\blacksquare\ 4.\ aa'+bb'+cc'=4(m_am'_a+m_bm'_b+m_cm'_c).$

2=>1. Dacă se foloseşte inegalitatea Cauchy-Buniakowski-Schwarz şi formula
       $OH^2 = 9R^2 - (a^2 + b^2 + c^2)$
   din care rezultă $a^2 + b^2 + c^2 \leq 9R^2$
   iese că triunghiurile ABC şi A'B'C' trebuie să fie echilaterale, nicidecum congruente.

Doua triunghiuri echilaterale sunt congruente.
Ai dreptate totusi din alt punct de vedere. Voi reveni cu o usoara corectura.
Trebuie dovedit care dintre ele sunt echivalente cu "$ABC\sim A'B'C'$",
care sunt echivalente cu "$ABC\equiv A'B'C'$" si care sunt echivalente
(daca exista) cu "$ABC,\ A'B'C'$ sunt simultan echilaterale".
Astfel unele afirmatii ar putea fi slabite prin echivalenta cu cele doua triunghiuri asemenea si nu neaparat congruente ! Deja m-a apucat ameteala "logica". Stop.


Let $ a,b,c$ be positive real numbers .

Prove that :

$ 2(a^5 \plus{} b^5 \plus{} c^5) \plus{} 9(a^3b^2 \plus{} b^3c^2 \plus{} c^3a^2) \geq 11(a^2b^3 \plus{} b^2c^3 \plus{} c^2a^3)$

[quote="IW@IT"]I think there is a typo in $ RHS$,[/quote]

Oh , sr , i edited :)

I think Materazzi wants to use

$ c\equal{} \min\{a,b,c\}$ then $ a\equal{}x\plus{}c$ and $ b\equal{}y\plus{}c$ ($ x,y \ge 0$)

After that, expand the whole inequality and we have the desired result.

[quote="great math"]I think Materazzi wants to use

$ c \equal{} \min\{a,b,c\}$ then $ a \equal{} x \plus{} c$ and $ b \equal{} y \plus{} c$ ($ x,y \ge 0$)

After that, expand the whole inequality and we have the desired result.[/quote]

No , i want to see another solution for this :)

Thanks you :)

It's an old problem of Nguyen Anh Cuong. And I think it's not easy to find a nice proof for it. But of course, I'll try  :oops:

[quote="Stephen"][quote="great math"]I think Materazzi wants to use

$ c \equal{} \min\{a,b,c\}$ then $ a \equal{} x \plus{} c$ and $ b \equal{} y \plus{} c$ ($ x,y \ge 0$)

After that, expand the whole inequality and we have the desired result.[/quote]

That method works, but it is very messy and takes a lot of time :rotfl:[/quote]

Maybe ... but not easy :)


can someone prove the equivalence of these two definitions of boiling point that it is the temp. at which vapour pressure  of the liquid equals the atmospheric pressure and it is the temp. at which the liquid vapourizes

The boiling of a liquid - when the molecules have enough energy to overcome the intermolecular forces existing in the liquid and transform in vapour - occurs not only at the surface of it but also (and predominantly) inside it. In the surface the molecules simply go directly to vapour after overcoming the forces that were exerted on them. Inside the liquid, a bubble of vapour has to form (and not colapse) and, because it is less dense than the liquid it rises to the surface and then breaks, with its vapour mixing with the already present vapour (and air) at the surface. However, for a recently formed bubble of vapour to survive inside the liquid, the pressure exerted by the vapour on the "walls" of the bubble must be greater or equal to the outside pressure: otherwise, it will colapse. And what is that outside pressure? From hydrostatics we know that in a given point of a liquid the total pressure is the sum of the atmosferic pressure with the hydrostatic pressure (the pressure due only to the liquid). As this last one is always neglegible, then for a bubble to be able to form and survive its vapour pressure (the pressure exerted by the vapour, not only at the surface of the liquid but also inside a bubble: the pressure exerted by the vapour form of the liquid) must be greater or equal to the atmosferic pressure. Of course, when this happens then the liquid is able to vapourize: its molecules (inside and at the surface) go to the vapour state.

Another statement is the following: the boiling point of a substance is the maximum temperature for which the liquid state of that substance can exist.

also when u say the pressure inside the bubble must be graeater than or equal to the atmospheric pressure .then why isn't the vapour pressure due to already present vapour added here.i mean there are already some vapours so pressure at the top of the liquid equals this pressure and the atomspheric pressure(by the way i really don't in some sense understand why the atmospheric pressure gets added here. because i f u have to have vapours then u need to carry out your experiment in a close dcontainer otherwise these vapour molecules would escape.if it is close dthen what about the atm. pressure

ok i f i  am not wrong is this one of the assumptions in ur proof that the pressure due to the molecules inside the bubble is same as the vapour pressure

[quote="Pardesi"]how do u imagine a 'bubble of vapour'[/quote]

A bubble of vapour is simple an aggregate of molecules with enough energy to be in the gaseous state, inside the liquid. The form of the bubble is spherical because the pressure inside a fluid acts equally in all directions.

[quote="Pardesi"]then why isn't the vapour pressure due to already present vapour added here.i mean there are already some vapours so pressure at the top of the liquid equals this pressure and the atomspheric pressure[/quote]

Because that vapour pressure contributes very little to the total pressure exerted on the (surface of the) liquid: that vapour pressure is a partial pressure (because the vapour mixes in the atmosphere), and because $ P_{i}= x_{i}P$, that partial pressure is always neglegible, as the atmosphere is such a "big solution".

[quote="Pardesi"]if it is close then what about the atm. pressure[/quote]

If a container is closed then the liquid will boil with greater difficulty because precisely the vapour accumulates above the liquid and now the total pressure to overcome is increasingly higher as the liquid is heated. That's why cooking food in a closed vessel (the appropriately called [i]pressure cooker[/i]) is much faster than in an open saucepan: the water can attain higher temperatures without boiling.


factorial
calculus
integration
real analysis
real analysis unsolved

Determine the volume of the region in $ \mathbb{R}^{2n}$=$ \mathbb{R}^{2}$x$ \mathbb{R}^{2}$x....x$ \mathbb{R}^{2}$ ,$ n \in \mathbb{N}^{*}$ defined by :
                  $ |z_1| \plus{} |z_2| \plus{} ... \plus{} |z_n| \leq r$
 Where $ z_i \equal{} (x_i,y_i)$ and $ |z_i| \equal{} \sqrt {x_i^2 \plus{} y_i^2},i \in \{1,2,..,n\}$ is the euclidean norm in $ \mathbb{R}^2$

Again using $ r \equal{} 1$ and $ z_i \equal{} r_i e^{\pi i \theta_i}$ and $ 0\leq r_i \leq 1$ and $ 0\leq \theta _i \leq 2 \pi$
The Region is similar to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=323138]other one[/url]. But the jacobian is $ \prod _{i \equal{} 1}^n r_i$ (easly calculated)
This will make the change $ n$ to $ 2 n$ in the factorial
The integral is independent of theta.
So the volume is $ \frac {(2 \pi )^n r^n}{(2 n)!}$


geometry
3D geometry
pigeonhole principle

In the interior of a cube of side $6$ there are $1001$ unit cubes with the faces parallel to the faces of the given cube. Prove that there are $2$ unit cubes with the property that the center of one of them lies in the interior or on one of the faces of the other cube.

[i]Dinu Serbanescu[/i]

[hide]
Let $S$ be the set of the centers of the $1001$ unit cubes. Since each unit cube must be completely contained within the larger cube, all points in $S$ must be contained in a $5 \times 5 \times 5$ unit cube with the same center as the large cube. Partition the $5 \times 5 \times 5$ cube into $1000$ separate $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$ cubes. By Pigeonhole, two of the $1001$ cubes' centers must lie in the same $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$ cube. But if two centers are in the same $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$, they are within $\frac{1}{2}$ unit of each other in each direction; therefore each must be contained in the unit cube centered at the other (or on the face).[/hide]


If we decide to go back to AMP, will we have to fill out a new application, including the admission test and recommendation letters, or are we accepted automatically?

[quote="CatalystOfNostalgia"][quote="Altheman"]Thats a good question...In my opinion (which doesn't matter), it would be a good idea to do the problems just for the sake of doing problems. I emailed one of the directors who might be able to give an official answer.[/quote]

I was planning on doing the problems, but I was mainly concerned about looking for more people to write recommendation letters.[/quote]
I think I heard somewhere that you don't have to reapply, unless you created problems, in which case it's unlikely they'd accept you again in the first place.

If you're a returning applicant, you don't need to solicit new letters of recommendation or pay the application fee.

We do need you to resubmit the personal information form. Also, we ask that you take an admission test so that we can track your progress and so that you get even more practice formally writing up solutions. Don't worry--if you've been to AMSP before, we won't scrutinize your performance on the test.

The optional parents' statement will not affect the admission of returning applicants; we know you well enough already. Parents should submit it only if there's new information about the student that we should know that wouldn't show up elsewhere in the application.

To summarize:
[b]For applicants who have been to AMSP before...[/b]
[b]Required: [/b] Personal Information Form, Admission Test
[b]Not Required: [/b]Letters of Recommendation, Application Fee, Parents' Statement
Hope this clears things up.

-Zac


Let $ a,$ $ b$ and $ c$ are natural numbers. Prove that the following number is natural too.
\[ \frac {(a^2 \plus{} bc)(b^2 \plus{} ac)}{(a \plus{} c)(b \plus{} c)} \plus{} \frac {(a^2 \plus{} bc)(c^2 \plus{} ab)}{(a \plus{} b)(b \plus{} c)} \plus{} \frac {(b^2 \plus{} ac)(c^2 \plus{} ab)}{(a \plus{} b)(a \plus{} c)}
\]

hello arqady, your term is equivalent to $ a^2\plus{}b^2\plus{}c^2$ and this is always a natural number if $ a,b,c$ are natural numbers.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello arqady, your term is equivalent to $ a^2 \plus{} b^2 \plus{} c^2$ and this is always a natural number if $ a,b,c$ are natural numbers.
Sonnhard.[/quote]

How to simplify the expression?

[hide]
Note that $ (a + b)(a + c) = a^2 + ab + ac + bc$

Therefore we have

$ \sum \frac {[(a + b)(a + c) - a(b + c)]\cdot[(b + c)(b + a) - b(a + c)]}{(a + c)(b + c)}$

$ = \sum \left[ (a + b) - a\cdot\frac {b + c}{a + c}\right]\cdot\left[ (b + a) - b\cdot\frac {a + c}{b + c}\right]$

$ = \sum (a + b)^2 - (a + b)\left[{a\cdot\frac {b + c}{a + c} + b\cdot\frac {a + c}{b + c}\right] + ab}$

Now since $ (a + b)^2$ and $ ab$ are both in $ \mathbb{N}$ we can leave them out

$ \sum (a + b)\left[{a\cdot\frac {b + c}{a + c} + b\cdot\frac {a + c}{b + c}\right]}$

$ \sum \frac {a(b + a)(b + c)}{a + c} + \frac{c(b + c)(b + a)}{c + a}$

$ \sum (b + c)(b + a)$

And this is in $ \mathbb{N}$

Since natural numbers are closed under addition, the above expression must be a natural number  :) [/hide]

[hide="I think this works?"]
By symmetry, since the degree of each term in the numerator (after combining into one fraction) is $ 5$ and $ 3$ in the denominator, the resulting expression must have each term that's not in the remainder of degree $ 2$ (since each time we take something out of the numerator by division, we subtract something of degree $ 5$ divisible by the denominator of degree $ 3$, the degree of the numerator stays the same (that is, either every term stays at degree $ 5$ or the remainder is $ 0$) and what we've taken out is of degree $ 2$). Note that the remainder must be $ 0$ from the substitutions $ a \equal{} \minus{} b$, $ b \equal{} \minus{} c$, and $ c \equal{} \minus{} a$ (for

$ a \equal{} \minus{} b$, we get $ (a^2 \plus{} bc)(b^2 \plus{} ca)0 \plus{} (c^2 \minus{} a^2)(a^2 \minus{} ac)(c \plus{} a) \plus{} (a^2 \plus{} ac)(c^2 \minus{} a^2)(c \minus{} a)$

$ \equal{} (c^2 \minus{} a^2)(a(a \minus{} c)(a \plus{} c) \plus{} a(a \plus{} c)(c \minus{} a)) \equal{} (c^2 \minus{} a^2)0 \equal{} 0$, and by symmetry it follows for the other substitutions as well),

so the numerator is divisible by $ (a \plus{} b)(b \plus{} c)(c \plus{} a)$. Now, to find the expression,

$ \frac {(a^2 \plus{} bc)(b^2 \plus{} ca)(a \plus{} b) \plus{} (c^2 \plus{} ab)(a^2 \plus{} bc)(c \plus{} a) \plus{} (b^2 \plus{} ca)(c^2 \plus{} ab)(b \plus{} c)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

$ \equal{} x(a^2 \plus{} b^2 \plus{} c^2) \plus{} y(ab \plus{} bc \plus{} ca)$.

Substituting $ (a,b,c) \equal{} (1,1,0)$ gives $ 2x \plus{} y \equal{} 2$;

$ (a,b,c) \equal{} (1,1,1)$ gives $ 3x \plus{} 3y \equal{} 3\implies x \plus{} y \equal{} 1$.

Solving yields

$ (x,y) \equal{} (1,0,0)\implies \frac {(a^2 \plus{} bc)(b^2 \plus{} ca)(a \plus{} b) \plus{} (c^2 \plus{} ab)(a^2 \plus{} bc)(c \plus{} a) \plus{} (b^2 \plus{} ca)(c^2 \plus{} ab)(b \plus{} c)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

$ \equal{} a^2 \plus{} b^2 \plus{} c^2\in\mathbb N$.
[/hide]

[b]Edit:[/b] I hope this works. :blush:

[quote="math154"][hide="I think this works?"]
By symmetry, since the degree of each term in the numerator (after combining into one fraction) is $ 5$ and $ 3$ in the denominator, the resulting expression must have each term that's not in the remainder of degree $ 2$ (since each time we take something out of the numerator by division, we subtract something of degree $ 5$ divisible by the denominator of degree $ 3$, the degree of the numerator stays the same and what we've taken out is of degree $ 2$), so we have

$ \frac {(a^2 \plus{} bc)(b^2 \plus{} ca)(a \plus{} b) \plus{} (c^2 \plus{} ab)(a^2 \plus{} bc)(c \plus{} a) \plus{} (b^2 \plus{} ca)(c^2 \plus{} ab)(b \plus{} c)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

$ \equal{} x(a^2 \plus{} b^2 \plus{} c^2) \plus{} y(ab \plus{} bc \plus{} ca) \plus{} \frac z{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$.

Substituting $ (a,b,c) \equal{} (1,1,0)$ gives $ 2x \plus{} y \plus{} \frac z2 \equal{} 2\implies4x \plus{} 2y \plus{} z \equal{} 4$;

$ (a,b,c) \equal{} (1,1,1)$ gives $ 3x \plus{} 3y \plus{} \frac z8 \equal{} 3\implies24x \plus{} 24y \plus{} z \equal{} 24$;

$ (a,b,c) \equal{} (2,1,0)$ gives $ 5x \plus{} 2y \plus{} \frac z6 \equal{} 5\implies30x \plus{} 12y \plus{} z \equal{} 30$.

Solving (one quick way to see the solution is to note that the coefficients of $ x$ are all equal to the RHS of the equations) yields

$ (x,y,z) \equal{} (1,0,0)\implies \frac {(a^2 \plus{} bc)(b^2 \plus{} ca)(a \plus{} b) \plus{} (c^2 \plus{} ab)(a^2 \plus{} bc)(c \plus{} a) \plus{} (b^2 \plus{} ca)(c^2 \plus{} ab)(b \plus{} c)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$

$ \equal{} a^2 \plus{} b^2 \plus{} c^2\in\mathbb N$.
[/hide][/quote]

Very Nice! But why is $ z$ necessarily an integer? couldn't it be a symetric expression with $ a,b,c$...

You're correct; it doesn't have to be an integer (in fact it can't if it's not $ 0$). I think I fixed it.

Also, it seems that if there is a symmetric "remainder" then there isn't really any point in the division in the first place because the "remainder" is still of the degree of the dividend, and you haven't actually done anything nor is the "remainder" actually a remainder. :|


Let $a,b$ and $c$ be positive real numbers such that $ab+bc+ca=3$. Prove that
\[ \frac{a(b^2+c^2)}{a^2+bc}+\frac{b(c^2+a^2)}{b^2+ca}+\frac{c(a^2+b^2)}{c^2+ab}\ge 3 \]

I am trying to prove:

$12\geq a^3+b^3+c^3+3abc$

Davron

[quote="Vasc"]Nice inequality!
I think it suffices to prove that
$\sum\frac{bc(a^2+b^2)(a^2+c^2)}{(b^2+ca)(c^2+ab)} \geq \sum{bc}$.[/quote]

Yes!! In fact, we have
\[ \sum_{cycl}\frac{a(b^2+c^2)}{a^2+bc} \ge \sqrt{3(ab+bc+ca)} \]

[quote="manlio"][quote="Vasc"]Nice inequality!
I think it suffices to prove that
$ \sum\frac{bc(a^{2}\plus{}b^{2})(a^{2}\plus{}c^{2})}{(b^{2}\plus{}ca)(c^{2}\plus{}ab)}\geq\sum{bc}$.[/quote]

How to prove it?

Thank you very much.[/quote]
$ \sum_{cyc}\frac{bc(a^{2}\plus{}b^{2})(a^{2}\plus{}c^{2})}{(b^{2}\plus{}ca)(c^{2}\plus{}ab)}\geq\sum_{cyc}{bc}\Leftrightarrow\sum_{cyc}(a^{5}\minus{}a^{4}b\minus{}a^{4}c\plus{}2a^{3}bc\minus{}a^{2}b^{2}c)\geq0,$
which true by Schur and Miurhead.


Let $p$ be a prime greater than 2. Prove that $\frac{2}{p}$ can be expressed in eactly on e way in the form $\frac{1}{x} + \frac{1}{y}$ where x and y are positive integers with $x > y$.

$\frac{1}{x} = \frac{2y-p}{py}$
$2y > p$
$2y = p + k$ for some $k \in \mathbb{N}$
$\frac{1}{x}= \frac{k}{py}$
so $k = p$ or $k | y$
Let $k | y$
then:
$p = (2n-1)k$ for some natural $n$
so $k = 1$ or $n = 1$ (and $k = p$)
But if $k = p$ we have got:
$\frac{1}{x}=\frac{1}{y}$ which can't be.
So $k = 1$
$y=\frac{p+1}{2}$
$x=\frac{(p+1)p}{2}$

$\frac{2}{p} = \frac{1}{x} + \frac{1}{y}$

$y \cdot 2 \cdot x = p(x+y)$

WLOG let x = kp.

$y \cdot 2 \cdot k = kp + y$

Now y|kp $\implies$ k = my or y = np, n|k.

For the first case, assuming k = my,

$y \cdot 2 \cdot my = y(mp+1)$

$2my = mp+1$

$m(2y-p) = 1$

This implies that m = 2y - p = 1.  Thus y = $\frac{p+1}{2}$.  Also, k = y $\implies$ x = yp $\implies \frac{1}{x} + \frac{1}{y} = \frac{2}{(p+1)p} + \frac{2}{p+1} = \frac{2}{p}$.  Noting that p+1 is even, this does indeed work.

Now consider the second case, where y = np.  Then we have

$y \cdot 2 \cdot k = kp + y$

$2npk = pk + np$

$2nk = k+n$

$(2n-1)(2k-1) = 1$

We then have n = k = 1 $\implies$ y = x = p, which is not permitted in the problem.

[quote="ZetaX"]When solving this, try the slighty more general question:
http://www.mathlinks.ro/Forum/viewtopic.php?t=28948[/quote]

but my questions specifies $p$ as being a prime number. the problem from the link does not.

Yes, but why not trying the other one (it's not that much harder)  :)


geometry
MATHCOUNTS
area of a triangle
Heron\u0027s formula

What is the area of a triangle with legs 14 feet and 13 feet and a base of 15 feet?

I dont get how to find the height.

Somebody Help!

[hide]Use Heron's Formula

$\sqrt{s\cdot(s-a)\cdot(s-b)\cdot(s-c)}$

Where s=the semiperimeter of the triangle (all the sides added together and then divided by 2) And A, B, and C are the three sides of the triangle.

Plugging in your numbers we get:
$\sqrt{21\cdot7\cdot6\cdot8}=84$[/hide]

thanks, i think thatll come i handy.  i got to remember that.  i read about it, but i forgot when and how to use it.

[quote="magixter"]It also helps to know Brahmagupta's formula (in other words, Heron's for quadrilaterals).
\[\sqrt{(s-a)(s-b)(s-c)(s-d)}\]
[/quote]
Keep in mind that that's only for cyclic quads
[hide="to derive heron's"]
Drop an altitude to a side with length z.
Call the altitude length x.
Call the parts of the side the altitude splits r and z-r.
Use pythag.
Solve for x.[/hide]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111675[/url]

way of doing it without heron's

I'm not sure I have ever seen a problem in MATHCOUNTS in which Brahmagupta's formula provides an advantage over a simpler method.  But the formula isn't really any harder than Heron's, provided you remember how to use it.

Heron's Formula, on the other hand, does come in handy once in a while.

[hide="solution without Heron"]
A $13-14-15$ triangle is a common triangle you will find on math contests throughout high school.  I suggest you learn about this triangle.

The $13-14-15$ triangle can be split into two right triangles.  A $5-12-13$ right triangle and a $9-12-15$ right triangle, where $12$ is the altitude to the side of length $14.$  So, $A=\frac12bh,$ where $b=14,$ and $h=12.$  $A=\frac12\times 14\times 12=84.$[/hide]

thanks for helping me sort that out, rcv and everyone else.  i woulda been lost.  so, lemme get this straight.  i should know about herons formula, but not worry about the other formula?  gotcha, ill study up on that right now.  you know, 1 month from today is the first day of nats in tx


geometry
calculus
integration
calculus computations

Can anyone help me solve this word problem.
Find the area of the region bounded on the left by y=square root of x, on the right by y=6-x, and below by y=1
Thanks .

First, draw a picture.

Then solve $ y \equal{} \sqrt {x}$ as $ x \equal{} y^2$ and $ y \equal{} 6 \minus{} x$ as $ x \equal{} 6 \minus{} y$

We can take slices of this parallel to the $ x$ axis and add those up to get the area we should get:
\[ A \equal{} \int_1^?(6 \minus{} y) \minus{} y^2\,dy
\]
Now you have one problem to solve: what should that question mark I wrote be, and how are you going to figure that out?

Thanks i got it .
 A= Integral of (-y+6-y^2) with limit 2upper and 1 lower
if I equal this too equation y^2 and -y+6 to each other then solve to get the limits


algorithm
number theory
greatest common divisor
number theory proposed

Find all m,n which are positive number such that
$ n^5\plus{}n^4\plus{}1\equal{}7^m$
 :lol:

{5,4,0} is a complete residue class modulo 3, so $ n^2\plus{}n\plus{}1|n^5\plus{}n^4\plus{}1$. long division yields $ n^5\plus{}n^4\plus{}1\equal{}(n^2\plus{}n\plus{}1)(n^3\minus{}n\plus{}1)$
we need to solve the system $ n^2\plus{}n\plus{}1\equal{}7^{k_1}, n^3\minus{}n\plus{}1\equal{}7^{k_2}$ ($ n\ge 1$, hence $ k_1, k_2 >0$. for $ n>2: k_2>k_1$)

we will use Euclid's algorithm:
$ (n^3\minus{}n\plus{}1,n^2\plus{}n\plus{}1)\equal{}(n^3\minus{}n\plus{}1\minus{}n(n^2\plus{}n\plus{}1), n^2\plus{}n\plus{}1)\equal{}(\minus{}n^2\minus{}2n\plus{}1,n^2\plus{}n\plus{}1)\equal{}(2\minus{}n,n^2\plus{}n\plus{}1)\equal{}(2\minus{}n,5n\minus{}3)\equal{}(7,n\minus{}2)$
Because $ 7|gcd$, we must have gcd=7, and $ k_1\equal{}1$, which implies that n=2. we can find the m: $ 2^5\plus{}2^4\plus{}1\equal{}49\equal{}7^2$, so (2,2) is a solution. it's easy to check that n=1 doesn't give a solution so this is the only one.

[quote="Tomaths"][quote="bambaman"]{5,4,0} is a complete residue class modulo 3, so $ n^2 \plus{} n \plus{} 1|n^5 \plus{} n^4 \plus{} 1$.[/quote]

Can you explain? I'm not sure if I follow.[/quote]

Its an algebraic result. Suppose $ \omega$ is a root of $ n^2\plus{}n\plus{}1$, i.e. $ \omega^3\equal{}1$. Then $ \omega^a\plus{}\omega^b\plus{}\omega^c\equal{}0$ whenever $ \{a,b,c\}$ are distinct mod $ 3$... because you mod out powers of $ \omega^3$. It's just your typical roots of unity number theory


algebra
function
domain
real analysis
complex analysis
complex analysis unsolved

How could i show that there exists an entire map bounded every where except on $ 0 \leq Re(z)\leq 1$ ?
i did not succeed in constructing one .

Consider $ e^{e^z}$. Notice that it tends to $ 0$ fairly fast in the domain $ \Omega$:
[asy]size(150);
path f(real r)
{
return (10,r)--CR((0,0),r,90,270)--(10,-r);
}
fill(f(4)--reverse(f(2))--cycle,yellow);
fill(f(2)--cycle,rgb(1,1,0.7));
DPA(f(4)^^f(2));
D("0",7,(0,0));
MP("\frac{2\pi}3",(0,2),7,W);
MP("\frac{4\pi}3",(0,4),7,W);
MP("\Omega",(5,3),(0,0));
MP("\omega",(5,0));
DPA((0,0)--(10,0)^^(0,-4.5)--(0,4.5),black);[/asy]
Now choose smooth $ g(z)$ that is identically $ 1$ in $ \omega$, and descends to $ 0$ within $ \Omega$. The function $ G(z) \equal{} g(z)e^{e^z}$ is identically $ 0$ outside $ |\Im z| < 2\pi$, say, and $ \bar\partial G$ is not large and decaying fast. Now it remains to solve $ \bar\partial H \equal{} \minus{} \bar\partial G$ by usual convolution with $ \frac 1{\pi z}$ in the class of bounded functions, put $ f(z) \equal{} G(z) \plus{} H(z)$ and make a linear change of variable.


algebra
polynomial
advanced fields
advanced fields unsolved

Let $ P(z)$ be a polynomial of degree $ n$ with complex coefficients, \[ P(0)\equal{}1, \;\textrm{and}\  \;|P(z)|\leq M\ \;\textrm{for}\  \;|z| \leq 1\ .\] Prove that every root of $ P(z)$ in the closed unit disc has multiplicity at most $ c\sqrt{n}$, where $ c\equal{}c(M) >0$ is a constant depending only on $ M$.

[i]G. Halasz[/i]

here are really nice solutions by iura and harazi

http://www.mathlinks.ro/viewtopic.php?t=118727


I am majoring in applied math at a top university in which most people have the option to take a course pass/fail every term (most people p/f humanities).  I am looking to find a job as a quant in the future, so I was wondering if future employers would mind if on my transcript, I had 5 courses every semester, with one being p/f (usually a humanities course or a biology course), provided that I got good grades in my math/science courses. Would the people looking at my transcript think I am lazy, or would they not care about the p/f's?

Thanks

I am at a similar university and I am taking one course this semester P/F.  That said, it will probably be the only course I take this way.  I would not take a course every semester this way, despite what you get in your other classes.  That's just my opinion.

By the way, where do you go? (if I had to guess, I would guess Brown or Hopkins)

Carnegie Mellon

My philosophy is that I'd rather spend mor etime and excel in the courses I'm interested in and not worry about biology and humanities courses.

You should still take them for grades.  I will be taking 2 math courses next semester, an econ course, and a humanities course.  I'm still taking them for a grade.


At a program such as RSI, if you're doing research in a math field, what is the average day like?

Great question!  Here's a rundown of my average day at RSI doing math research on a day with no guest lecture.

11:30:  Wake up.
11:32:  Shower.
11:40:  Eat breakfast.
12:00:  Mull over some math.
1:00:  Meet up with mentor.
2:30:  Lunch or go to library to look up stuff.
3:00:  Go to computer lab, do some calculations, waste time on AoPS.
6:00:  Dinner.
6:30:  Go to dorm, read some papers, waste time on AoPS.
11:00:  Frisbee.
12:00:  Grab a Vault, go to computer lab, do some algebra, waste time on AoPS.
(Random:  Random activity, planned or otherwise.)
1:30 - 4:00:  Fall asleep.

:D

I got a little more work done in the last two weeks, but as you can see, math research at RSI is somewhat laid-back.  (The lab science people, who have to get up early to get lab time, are generally fairly jealous.)


function
integration
calculus
calculus computations

If $ f(x \plus{} T) \equal{} \minus{} f(x)$, can I call $ f$ a quasi-periodic function?

Or please tell me other words to describe the above property of $ f$!

Thank you.

Happy New Year 2009 to All :lol:

I'd call it antiperiodic.

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

Note that such a function is periodic of period $ 2T.$

Suppose that $ f$ is integrable on each period. Define the Fourier coefficients

$ \widehat{f}(k) \equal{} \frac1{2T}\int_0^{2T}f(x)e^{\minus{}\pi ikx/T}\,dx.$

The stated algebraic condition on $ f$ would imply a certain algebraic condition on $ \widehat{f}.$ Can you derive that condition?


There is a class of three digit numbers that can be factored into $ x\cdot y\cdot z$, where $ x$, $ y$, and $ z$ are primes. If the hundreds digit of such a number is $ a$, the tens digit $ b$, and the ones digit $ c$, find all numbers such that

$ x \equal{} a\plus{}b\plus{}c$
$ y \equal{} a\minus{}b\plus{}c$
$ z \equal{} c$


calculus
integration
logarithms
calculus computations

Hi :) !!!

$ \int (t \minus{} 2)ln(t) dt \equal{} \int t ln(t) dt \minus{} 2\int ln(t)dt$

$ \int t ln(t)dt \equal{} \frac {1}{2}t^2 ln(t) \minus{} \frac {1}{4}t^2 \plus{} c_2$

and $ \int ln(t)dt \equal{} tln(t) \minus{} t \plus{} c_1$

$ \int (t \minus{} 2)ln(t) dt \equal{} \int t ln(t) dt \minus{} 2\int ln(t)dt \equal{} ln(t) \left( \frac {t^2}{2} \minus{} 2t \right) \plus{} 2t \minus{} \frac {t^2}{4} \plus{} cte$

and thx :)

Very Easy!

$ \frac {d}{dt}\left(\frac {t^2}{2} \minus{} 2t\right)\ln t \equal{} (t \minus{} 2)\ln t \plus{} \frac {t}{2} \minus{} 2$, yielding the integral is $ \left(\frac {t^2}{2} \minus{} 2t\right)\ln t \minus{} \frac {t^2}{4} \plus{} 2t \plus{} Const.$


Given $ a,b,c,d$ positive integers satisfying

$ \boxed{1}  d > c > b > a > 0$

$ \boxed{2}  a,b,c,d$ are all odd integers

$ \boxed{3} ad \equal{} bc$

$ \boxed{4} a \plus{} d \equal{} 2^{k}$

$ \boxed{5} b \plus{} c \equal{} 2^{m}$

$ k, m     \in  \mathbb{Z}$

Prove that $ a \equal{} 1$

er ....guys I am getting a really stupid contradiction. I should be wrong .
Like from the given  condition .....
Assume $ a \neq 1$
set $ a \equal{} 2a_1 \plus{} 1$  and $ d \equal{} 2d_1 \plus{} 1$
adding  the above we  get $ 2a_1 \plus{} 2d_1 \plus{} 2 \equal{} 2^{k}$
$ \implies a_1 \plus{} d_1 \equal{} 2^{k \minus{} 1} \minus{} 1$
$ \implies a_1 \plus{} d_1$ is odd
Let $ a_1$ be even and $ d_1$ be odd
$ a_1 \equal{} 2a_2$ and $ d_1 \equal{} 2d_2 \plus{} 1$
$ a_1 \plus{} d_1 \equal{} 2a_2 \plus{} 2d_2 \plus{}1 \equal{} 2^{k\minus{}1} \minus{} 1$
$ a_2 \plus{} d_2 \equal{} 2^{k\minus{}2} \minus{} 1$
proceeding this way taking $ a_i$ to be even and  $ d_i$ to be odd where $ a_{i}\equal{}2a_{i\plus{}1}$ and $ d_{i} \equal{} 2d_{i\plus{}1} \plus{} 1$  where $ a_{i}$ and $ d_{i}$  are positive integers
we get a system of equations
                   .
                   .
                   .
                   .
                   .
                   .
and we get something sort of a infinite descent
But now consider the (k-1)th equation
$ a_{k\minus{}1} \plus{} d_{k\minus{}1} \equal{} 1$
$ a_{k} \plus{} d_{k} \equal{} \minus{}1$
$ \implies$ at least one of $ a_k$ or $ d_k$ is negative but this implies $ a_{k\minus{}1}$ or $ d_{k\minus{}1}$ should be negative and arguing this way we might get $ a$ or $ d$ to be negative.You also have $ a\equal{}2a_1 \equal{}4a_2\equal{}......................\equal{}2^{n}a_n$ This means that if $ a_n$ is a positive integer we have
$ a > 2^{n}$  ....But these  both  cannot happen so $ a_1$ or $ d_1$ should be zero
$ \implies a$ or $ d \equal{}1$
Similarly $ b$ or $ c$ must be 1.
If $ d$ is the greatest it cannot be $ 1$
Since $ c$ is greater than $ b$ ....$ c \neq 1$
$ \implies b\equal{}1$
$ \implies b\equal{}a$
which is not possible isnt it  :(
Oh i am getting somethin stupid  :rotfl:

[quote="srinath.r"]
But now consider the (k-1)th equation
$ a_{k \minus{} 1} \plus{} d_{k \minus{} 1} \equal{} 1$
$ a_{k} \plus{} d_{k} \equal{} \minus{} 1$[/quote]

Shouldn't it be
$ a_{k} \plus{} d_{k} \equal{} 0$ ?

ya you are right ....I was a stupid to substitute $ 2^{0} \equal{} 0$   :wallbash: .
When $ a_2 \plus{} d_2 \equal{} 2^{k \minus{} 2} \minus{} 1$
then $ a_k \plus{} d_k \equal{} 2^{k \minus{} k} \minus{} 1$
$ \implies a_k \plus{} d_k \equal{} 0$
But still has it set things right for us .... I dont think so?
This means that  one of the numbers in the last equation  should be negative like $ a_k \equal{} \minus{} d_k$
or both of them should be zero .
With these in our hand and we go back in the system of equations we still get a contradiction o ,dont we ?....

er..........hello guys......can anyone tell me the mistake ..please.

I am proceeding that way like setting $ a_{i}$ even . Even if  I change it into odd in the middle I dont think it affects the contradiction.


AMC
AIME
geometry
trigonometry
USA(J)MO
USAMO
AIME I

Which chapters of aops 2 are the most important to know for the aime? I am going to try to learn all of it, but I want to know what order I should do it in.

Just do it in the regular order, I say.

If you really must know, I suppose limits(8), inequalities(14), and Concurrency/Collinearity Geometry (21+22) are not really needed. Chapters 10-12 are also pretty rare on the AIME.

Umm, the AoPS vol. 2 is WAY too confusing for me. I have master vol. 1, but vol. 2 leaves me befuddled...

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10806&start=20[/url]

This is a really cool thread that explains everything.  :D

Yes, math books can be hard to read sometimes.

I feel your pain bpms...

I find that alot of books (not dissing the writing ability of AoPS), not just AoPS books, all math books, introduce a new concept and immediately go to a very advanced proof using this concept. I have particularly seen this in Number Theory by Andrews and ACoPS. I guess reading these kinds of books requires alot of experience.

I have a lot of trouble with AoPS2 also. What I do is, read stuff a bunch of times, then have someone else explain the concept to me, then read it again. Sometimes, after each sentence, I will rephrase the sentence in my own words, to make sure that I understand it. While doing lots of practice problems, and posting the occassional question here, I have gotten through a few chapters. :rotfl:

I am finding that learning some chapters makes it easier to learn some other chapters, and so on.... The best thing to have with a book that is too hard for you is someone that is really smart to explain the stuff that you don't get too you.

Huh that is really weird. I had no trouble at all going through AoPS volume 2, and I was in 8th-9th grade. Everything is explained so simply... I really don't understand why you guys are having difficulty.

@13375P34K43V312: NT by Andrews is different from AoPS. It is not intended for high schoolers but rather an undergraduate audience. This is why it is written with less explanation than AoPS for example. But AoPS is written in a very readable, conversational tone.

It is true however that you will not understand everything the first time you read it. You have to read it multiple times, as Saiem just said.

[quote="Pakman2012"]This is in accordance with Saiem's Let It Simmer theory.

Sometimes you just gotta read it once, try really hard to understand it, and then leave it alone.  Often understanding comes with time.  Trying to explain it to someone else also helps.

I didn't really start to understand AoPS until I started reading it again this summer.  Not everything takes continual repetition, some things just take time to understand.[/quote]
He's right, I finally understand the chapter on trig in vol. 2. I am happy.

I like the term "active reading," coined by Paul Zeitz in ACoPS. I find having paper and pencil in hand and copying the proofs, while filling in skipped steps is extremely helpful.

[quote="chess64"][quote="Elemennop"][quote="chess64"]Huh that is really weird. I had no trouble at all going through AoPS volume 2, and I was in 8th-9th grade. Everything is explained so simply... I really don't understand why you guys are having difficulty.[/quote]

Maybe not everyone is a math genius and things dont work out instantly in their head?

I dunno, I'm probably COMPLETELY WRONG>[/quote]

Maybe I'm not a math genius either and things dont work out instantly in my head?

I dunno, I'm probably COMPLETELY WRONG>

hey wait a minute elemennop you're the math genius wtf are you talking about?????[/quote]

Um, yea. Well, we should probably either shut down this thread or start acting correctly. But it is true that AoPS is hard to understand if you're starting from school math only. (People in my school think it's hard.)

[quote="Karth"][/quote][quote="chess64"][quote="Elemennop"][quote="chess64"]Huh that is really weird. I had no trouble at all going through AoPS volume 2, and I was in 8th-9th grade. Everything is explained so simply... I really don't understand why you guys are having difficulty.[/quote]

Maybe not everyone is a math genius and things dont work out instantly in their head?

I dunno, I'm probably COMPLETELY WRONG>[/quote]

Maybe I'm not a math genius either and things dont work out instantly in my head?

I dunno, I'm probably COMPLETELY WRONG>

hey wait a minute elemennop you're the math genius wtf are you talking about?????[/quote][quote="Karth"]

Um, yea. Well, we should probably either shut down this thread or start acting correctly. But it is true that AoPS is hard to understand if you're starting from school math only. (People in my school think it's hard.)[/quote]

who's acting incorrectly :mad: :police:

anyway i started from school math only myself

i think you mean it's hard to understand if you don't know school math well

I think there is a difference of expectations here.  With math books (real math books, not average high school textbooks), you don't "get" everything right away; you have to think about it.  This isn't hard; this is normal.  If it reads like a novel (even a difficult novel), then it's really easy for you.  I mean, really, in math, if things seem easy, then you're not challenging yourself.  My mom (who is a mathematician) once even said to me, "If you don't feel stupid, you're not really doing math."

I think that some people may "understand" better than others. However, "understand" doesn't mean anything if you can't "apply" it on real problems.

I like Saiem's theory because that's what I did too. Try hard to understand it but if it doesn't work out, just come back to it later when you had more practices.

As for "active reading," it's funny how someone mentioned that. In any reading, there are two types of reading ----- passive reading and active reading. In most reading you do with regular books, you can be passive, that is just reading the word and simply understand it, and still be OK. But in active reading, you really have to think as you read it, and this applies to books like ones by Charles Dickens and other classics. AoPS 2's wording is passive ----- its whole point was to create the book that was user-friendly. I don't think anyone should argue that because honestly, it's pretty easy to read the book. But in some ways, it also requires active reading ----- or should I say active math ----- because you can't really understand it until you tried the problems. If you're having trouble on this "active math" part, use Mu Alpha Theta problems. By that, I mean that if trig confuses you, go to Mu Alpha Theta and work on bunch of trig problems. They're pretty easy but build you solid base.

[quote="Silverfalcon"]However, "understand" doesn't mean anything if you can't "apply" it on real problems.[/quote]

Exactly! This is why I suck at math.

It said that 120 students from grades 7th -10th qualify with an AIME score of 8 or above on the list of USAMO qualifiers. How do sixth graders qualify when they have that much ability? Can they?

I think the 7th-10th grade indicates that there wasn't a sixth grader that qualified, aka not that much ability?

[quote="Boy Soprano II"]Do you mean how would a hypothetical sixth-grader who had the ability to qualify for USAMO qualify?

Those cutoffs were just for this particular year.  If there were a sixth-grader with sufficient ability, (s)he would be able to qualify through the floor score, like all other students in tenth grade or below.  There is no lower bound on ages for participation in any of the AMC's competitions.[/quote]
That's what I meant. I think that Reid Barton probably qualified in 6th grade. He wrote 3rd on it in 9th grade...

[quote="Karth"][quote="Boy Soprano II"]Do you mean how would a hypothetical sixth-grader who had the ability to qualify for USAMO qualify?

Those cutoffs were just for this particular year.  If there were a sixth-grader with sufficient ability, (s)he would be able to qualify through the floor score, like all other students in tenth grade or below.  There is no lower bound on ages for participation in any of the AMC's competitions.[/quote]

So technically, a five-year-old could enter the AMCs, go to USAMO, enter MOSP and pwn Arnav  :D[/quote]
No. It just wouldn't happen. A fifth grader who is ridiculously talented could do it, but no way a five year old. :P

[quote="probability1.01"][quote="MysticTerminator"]dude I must disagree with Karth in this case no 5-year-old people pwning me please[/quote]

Yeah that would just be quite a turkey, especially if $5 = p^{2}+k$, with k odd (but not necessair greater than one) :maybe:[/quote]

?

@

what are you talking about?


function
calculus
derivative
trigonometry
real analysis
real analysis solved

Find all real functions f with Darboux property such that f(x+y)=f(x+f(y)).

Here's a question related to this: I know that if $f,g$ have Darboux property, then it doesn't follow that $f+g$ does too, but what if $g$ is continuous? So the question is: if $f$ has DP and $g$ is conbtinuous, does it follow that $f+g$ has DP?

I don't even need that much. I only need an affirmative answer in the case $g(x)=x,\ \forall x\in \mathbb R$.

For any $a,b$ s.t. $f(a)\ne f(b)$ and $\lambda$ strictly between $f(a)$ and $f(b)$, there is $c$ strictly between $a,b$ s.t. $f(c)=\lambda$.

I think this looks harder than it actually is (unless I'm making some mistake :)). That question I asked is an interesting one, but we don't need to answerit in order to solve this problem.

We can see that $f(x)=f(x+f(y)-y),\ \forall x,y\in \mathbb R$. Let $I=f(\mathbb R)$ and let $F$ be the set of fixed points of $f$. If $I$ is a degenerate interval, then we're done: $f$ is constant, and it's obvious that all such functions verify the equality. If $I$ is non-degenerate, assume it's bounded from above (it's exactly the same if we assume it's bounded from below). Then We have $f(x)=x,\ \forall x\in I,\ f(x)<x,\ \forall x>t$, where $t=\sup I\Rightarrow t\in \mathbb R$. Take a sequence $y_n$ which tends to $t$ from above. Then, from Darboux's property, we can find a sequence $z_n\in (t,y_n)$ s.t. $f(z_n)$ is "really close" to $t$. Since $y_n\to t\Rightarrow z_n\to t$, it means that $f(z_n)-z_n\to 0$. We have $f(x)=f(x+f(z_n)-z_n),\ \forall x\in \mathbb R,\ n\in \mathbb N$. If $n$ is large enough, we will be able to choose $x$ s.t. $x,x+f(z_n)-z_n\in I$, and we have a contradiction, because $f$ is injective on $I$.

All of this shows that $I$ must either be degenerate, in which case $f$ is constant, or $I=\mathbb R$, in which case $f=1_{\mathbb R}$.

anyway, l think that to satisfate the darboux property, a function must be monotonic or continuous, exept if $f(a)$ and $f(b)$ are very close.

That's wrong, man.. The Darboux property is much weaker than continuity. You do know, however, that all continuous functions have Darboux property. Even more than that: all derivatives have Darboux property.

Here's a non-monotonic and non-continuous function: $f(x)=\sin \frac 1x,\ x\ne 0,f(0)=0$. This only has one discontinuity point, but nastier examples can be given.

that's what l thought, but l don't have formulated it good  :D .

grobber, with the inverse function $f(x)=\frac1x$, $f(0)$ isn't definite (the division by $0$ is forbidden).

Sorry, I meant $\sin \frac 1x$. I've edited it. :D


Please read the foll and respond soon  :surrender:
Can you make Pardesi a moderator of the Indian Forum.
The forum seems to be inactive as the old moderator seems to have not replied for quite some time.
Also can we start a subforum in India named IIT and Academics?
We have the exam in April so we need it fast  :idea:
Please reply  :surrender:
Thank you
 :)

or atleast we could have a renaming of that so that indians out here feel it's relevance...anyway anything with IIT on it is seemingly quite relevant to anybody in India  :D .
many people r joining especially for this and since chemistry Forum is probably the only active non- Mathematical Subject Form we had to continue our discussions about this in a thread there irrelevantly.  :rotfl:


Sorin, te las pe tine sa demonstrezi ca are loc urmatoare egalitate:

                               \[ \sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}=\frac{r}{4R}.  \]

Eu ma gandisem sa o rezolv altcumva. ;) Dar cred ca pot demonstra si egalitatea ta, Caesar.Nu m-am gandit mult, dar cred ca iese folosind Teorema Sinusurilor si formulele arie unui triunghi: S=pr si S=abc/4R.Acum sa vezi cum m-am gandit eu.Tot in cartea <ta> am vazut inegalitatea sinA/2=<a/(b+c).Continuam si pentru sinB/2,sinC/2 si trebuie, in final, sa aratam ca (a+b)(b+c)(c+a)>=8abc, care e adevarata, fiind inegalitatea lui Cauchy pentru medii.Mai tarziu voi posta solutia egalitatii tale.Acum sunt obosit si in plus nu prea mai am asa mult timp ca am concursul Pitagora la Valcea si trebuie sa ma pregatesc. ;) Ca sa particip la Titeica trebuie sa iau aici in primii trei.
Mai vorbim.

Sorin, daca esti pasionat de trigonometrie vezi [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=112783]Identitati trigonometrice[/url]. Un "old topic" foarte folositor in acest sens. Acolo ai sa gaseshti multe identitati interesante si unlele rezolvate ;)

Inca o data iti multumesc Slizzel.Majoritatea identitatilor le stiam, dar acum stiu si demonstratia lor(la unele doar).


[quote="pkerichang"][quote="bpms"]Where can I find Olympiad problems about the same difficulty as round 2 of USAMTS?

Thanks,
bpms[/quote]

.....almost none??
Well, I guess a few number 1's and 2's are about the same difficulty, but I doubt they'll be that easy....But again I haven't done a lot of Olympiads, so I don't know (Somehow I find most of the British and Hugarian Olympiad easier than others tho, but that maybe just be myself...)[/quote]
Not necessarily Olympiad problems, I was just referring to proof questions of that difficulty.

[quote="mustafa"]The USAMTS website has past problems on it under the problems section.[/quote]
Well yes, but I was referring to problems that are specifically that difficulty. From what I can gauge this was one of the easiest USAMTS rounds ever...

USAMTS is one of, if not the easiest proof contests. In fact, since the only other proof contests I can think of are olympiads (barring Putnam of course), it probably is the easist.

but it's a great way to start of doing proofs, hopefully.


A circle of diameter 1 is inscribed in a square, dividing it into five regions: the region inside the circle, and a region at each corner. Find the diameter of a circle inscribed in one of the corner regions.

EDIT: sorry about that. fixed.
EDIT 2: Added a picture for clarification

[quote="pianoforte"]A circle of diameter 1 is inscribed in a circle, dividing it into five regions: the region inside the circle, and a region at each corner. Find the diameter of one of the circles inscribed in one of the corner regions.[/quote]

A circle in a circle? Huh?

Square, obviously.
That's not that hard to figure out, you should just assume what should be obvious and work from there...

i don't really understand the problem in the first place.

 :(  :trampoline:

Big circle has radius 0.5. The diagonal is $\sqrt{2}*0.5$ long (from the center to the corner of the square). Let r be the radius of the small circle.
$0.5+(\sqrt{2}+1)r = \sqrt{2}*0.5$
$r ~ 0.086$