$ x,y,z >0$ and $ x\plus{}y\plus{}z \equal{} 1$. show that
$ \sum\limits_{x,y,z}  \frac{x(1\minus{}x)^2}{1\plus{}x}  \geq  \sqrt{3xyz}$

HARD!

[quote="pavarotti"]$ x,y,z > 0$ and $ x \plus{} y \plus{} z \equal{} 1$. show that
$ \sum\limits_{x,y,z} \frac {x(1 \minus{} x)^2}{1 \plus{} x} \geq \sqrt {3xyz}$

HARD![/quote]
It's not hard, pavarotti! See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169390

[b]arqady[/b], i don't understand! These problems are distinc! And the problem at the page http://www.mathlinks.ro/Forum/viewtopic.php?t=167965 is a problem of trigonometric. I haven't the solution for it!

I think, pavarotti, the following problems:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167653
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167965
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169390
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168941
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169626
are the same problems. :wink:

Oh, i see. But... i don't any solution for one of them! :(


The 18th(2004-2005) Korean MO Round 2
7th November, 2004
Original test papers are here. (only in Korean not English)

Problem 1 of high school paper and Problem 5 of middle school paper are the same.

I have solutions collected from several students,
(the KMO committee does not support official solutions :( )
but not done texing yet.

If someone can translate this, please post it to National Olympiads forum.

-----
(May 10, 2005)
I have done texing the solutions.
The pdf file is attached.
(only in Korean not English, sorry)

[quote="chavez"]where?[/quote]

The attached files are those.

[quote="Omid Hatami"]
Can you please post the problems in PDF
[/quote]

they are.


If I have 18 dollars to spend at a restaraunt, and I have to pay a 5% tax on the food and a 15% tip on the food, what is the largest amount of money I can spend on food?

For example, if I spend 10 dollars, I will have to pay a 50 cent tax and a 1.50 dollar tip.

You could make an equation: lets say you spent x dollars on the food

x+(0.05x)+(0.15x)=18

Then you have 1.2x=18

Therefore, x=18/1.2=15

and you spend 15 dollars on the food

[hide][b]I'm not sure if I'm right but I would say  $\$$14.40

He has $\$$18.00
5% of that is  $\$$ 0.90
15% is  $\$$ 2.70
Even easier u could do it like this: 5 and 15 equal 20-20% is  $\$$ 3.60[/hide]

P.S. How do you do that Reveal hidden content and how do u make the diagrams?

 [/b]

[color=darkred][size=75][Hide tags added][/size][/color]

you click the hide button on the top and when your done you click it again

[hide]x + .05x + .15x = 18
1.2x = 18
x = 15[/hide]


calculus
integration
geometry
3D geometry
number theory
prime numbers

The number of positive integers $\leq 100$ having exactly four (distinct positive integral) divisor is:

(a) < 10
(b) > 10 and < 20
(c) 26
(d) 32
(e) none of above

I get:

[hide="Expressed as product"]
32 because:

2*3
2*5
2*7
2*13
2*17
2*19
2*23
2*31
2*37
2*41
2*43
2*47
3*5
3*7
3*11
3*13
3*17
3*19
3*23
3*29
3*31
5*7
5*11
5*13
5*17
5*19
7*11
7*13  [/hide]

What do you think and make sure do not look at my answer until you tried. :D

There are $28$ numbers that are a product of two primes. However, there are also $2$ numbers that are cubes of a prime number. These also work, so the answer is $30$, or $(E)$.

Whoa, I'm confused about this. Can someone give a really good explanation please  :oops:

The numbers can be written as m^n x p^q, where m and n are prime numbers.  the number of factors a number has is always (n+1) x (q+1), so n and q have to be 1.  If you let m=2, then p can be all the prime numbers from 3 to 47...14 numbers.  Let m=3 and p will go from 5 to 31...9 numbers.  Let m=5 and p goes from 7 to 19 (5 numbers.  Finally let m=7 , then p can only be 11 and 13. [size=150]So: 14+9+5+2=30 numbers[/size][/code][/list]


An element of a current-carrying wire can apply a force or torque on another element of the [b]same[/b] wire when :-

a) When the wire is straight

b) When the wire is of any other shape except straight

c) Every time

d) Never

[quote="yugos"]It is correct.
Explain the answer to me.[/quote]

$ \text{Consider two elements of the wire carrying currents }i_a \text{ and } i_b. \\
 \\
\;\text{Use,} \\
 \\
\vec{F} \equal{} i_b\vec{L} \times \vec{B}_a \equal{} i_b|\vec{L}\parallel{}\vec{B}_a|\sin\theta \\
 \\
B_a \equal{} \frac {\mu_0i_a}{2\pi d} \\
 \\
\;\theta \equal{} \text{Angle between }\vec{B} \text{ and }\vec{L} \equal{} 0 \text{ for current through same, straight wire} , \; \neq 0 \text{ otherwise.}$


algebra
polynomial
function
inequalities
real analysis
real analysis unsolved

Consider the polynomial:
\[ g_n(a,b)\equal{}\sum\limits_{i\equal{}0}^{n}\frac{(a\plus{}1)_i(b)_{n\minus{}i}}{(c\plus{}1)_i(c)_{n\minus{}i}i!(n\minus{}i)!}
\]
with positive $ a$, $ b$ and $ c$. Here $ (z)_m\equal{}z(z\plus{}1)\cdots(z\plus{}m\minus{}1)$ is shifted factorial.

1. Prove that $ g_n(a,b)>g_n(b,a)$ if $ b>a$.  I have a proof based on my previous post with $ \delta\equal{}1$, but may be another proof can be found which extends to $ \delta\ne{1}$.

2. Since $ b\minus{}a$ clearly divides $ g_n(a,b)\minus{}g_n(b,a)$ the function $ P_{n}(a,b)$ below is a polynomial. Prove that all coefficients of the polynomial
\[ P_{n}(a,b)\equal{}\frac{g_n(a,b)\minus{}g_n(b,a)}{b\minus{}a}\equal{}P_{n}(b,a),
\]
are positive.  This would imply 1. of course.

[quote="dimono"]
1. Prove that $ g_n(a,b) > g_n(b,a)$ if $ b > a$.  I have a proof based on my previous post with $ \delta \equal{} 1$, [color=red]but may be another proof can be found which extends to $ \delta\ne{1}$[/color].
[/quote]
Erm, where is $ \delta$ in your new problem? :?

I meant that taking $ \delta \equal{} 1$ in my previous post provides a proof for item 1. in this post through some manupulations.

I am searching for another proof in order to extend it to the following situation.  Put
\[ g_n(a,b;\delta) \equal{} \sum\limits_{k \equal{} 0}^{n}\frac {(a \plus{} \delta)_k(b)_{n \minus{} k}}{(c)_k(c)_{n \minus{} k}k!(n \minus{} k)!}
\]
then  $ g_n(a,b;\delta) > g_n(b,a;\delta)$ when $ b > a > 0$, $ \delta > 0$.  It is sufficient to show it for $ 0 < \delta < 1$ then I know how to extend to all other positive $ \delta$.

Should one of $ c$'s in the denominator be $ c\plus{}\delta$? I asked because it makes more sense to look at the problem you cannot solve directly than to try to find an alternative solution to the problem you know how to solve in hope that it will generalize...

The answer about $ c$ is no - so the form given in the post is correct.

Nevertheless, the idea to transfrom the expression in a way that shift is in $ c$ while the numerators are the same is very good.  This is in fact the method that leads to solution in case $ \delta\equal{}1$.  I do not see how to do it however...

Well, it's true that I have a solution for the item 1. ($ g_n(a,b)>g_n(b,a)$), but I do not know how to prove item 2 (P_n(a,b) has positive coefficients).  Might it be some algebraic agrument that leads to it?

Ok, this time the assertion is, indeed, true and quite easy. First of all, assume that $ k\le n/2$ and $ \underbrace{(a \plus{} \delta)_k(b)_{n \minus{} k}}_{ \equal{} r} \plus{} \underbrace{(a \plus{} \delta)_{n \minus{} k}(b)_{k}}_{ \equal{} s}\le \underbrace{(a)_k(b \plus{} \delta)_{n \minus{} k}}_{ \equal{} v} \plus{} \underbrace{(a)_{n \minus{} k}(b \plus{} \delta)_{k}}_{ \equal{} u}$
Our aim is to show that then the same inequality is true for $ k \minus{} 1$.

Note that since $ (a \plus{} \delta)_\ell(b)_\ell\ge (a)_\ell(b \plus{} \delta)_\ell$ for all $ \ell$, we have $ rs\ge uv$. Also, clearly $ v\ge u$. Together with $ v \plus{} u\ge r \plus{} s$, this yields $ v\ge r,s\ge u$ (this is the only chance to have bigger sum and smaller product). Now let us look at what we need to show for $ k \minus{} 1$. It is
\[ D(\delta) \equal{} v\frac {b \plus{} \delta \plus{} n \minus{} k}{a \plus{} k \minus{} 1} \plus{} u\frac {a \plus{} n \minus{} k}{b \plus{} \delta \plus{} k \minus{} 1} \minus{} r\frac {b \plus{} n \minus{} k}{a \plus{} \delta \plus{} k \minus{} 1} \minus{} s\frac {a \plus{} \delta \plus{} n \minus{} k}{b \plus{} k \minus{} 1}\ge 0
\]
For $ \delta \equal{} 0$, we clearly have $ \frac {b \plus{} n \minus{} k}{a \plus{} k \minus{} 1}\ge \frac {a \plus{} n\minus{} k}{b \plus{} k \minus{} 1}$, so the desired inequality is just a combination of $ v\ge r$ and $ v \plus{} u\ge r \plus{} s$ with positive coefficients (here I treat $ u,v,r,s$ as constants, not as functions of $ \delta$!). Differentiating, we get
\[ D'(\delta) \equal{} v\frac {1}{a \plus{} k \minus{} 1} \minus{} u\frac {a \plus{} n \minus{} k}{(b \plus{} \delta \plus{} k \minus{} 1)^2} \plus{} r\frac {b \plus{} n \minus{} k}{(a \plus{} \delta \plus{} k \minus{} 1)^2} \minus{} s\frac {1}{b \plus{} k \minus{} 1}
\]
which is obviously non-negative since $ v\ge s$ and $ r\ge u$.

Now, we see that our inequality is equivalent to
\[ \sum_{k\le n/2}\frac 1{(c)_k(c)_{n \minus{} k}}M_k\ge 0
\]
where $ M_k \equal{} \frac {1}{k!(n \minus{} k)!}[(a \plus{} \delta)_k(b)_{n \minus{} k} \plus{} (a \plus{} \delta)_{n \minus{} k}(b)_{k} \minus{} (a)_k(b \plus{} \delta)_{n \minus{} k} \minus{} (a)_{n \minus{} k}(b \plus{} \delta)_{k}]$ enjoy the property that if $ M_k\le 0$, then all previous $ M_j\le 0$ as well. Since the $ c$ factors increase, it is enough to prove that $ \sum_{k\le n/2}M_k\ge 0$, i.e., that the inequality is true without $ c$-factors in the definition of $ g(a,b;\delta)$. But then it is an identity, because both sides are just the $ n$-th coefficient of $ (1 \minus{} x)^{ \minus{} a \minus{} b \minus{} \delta} \equal{} (1 \minus{} x)^{ \minus{} a \minus{} \delta}(1 \minus{} x)^{ \minus{} b} \equal{} (1 \minus{} x)^{ \minus{} a}(1 \minus{} x)^{ \minus{} b \minus{} \delta}$.

That's all. Unfortunately, I'm not any good in algebra, so I'll leave the problem about the positivity of coefficients to somebody else :P.

This is exactly the sort of proof I was looking for but was unable to complete.  Thanks, Fedja!!!

Please also check your private messages.


function
algebra
domain
polynomial
real analysis
real analysis unsolved

Hi, I have a problem following:
Let f be a function from $ R$ to $ R$ such that $ f(f(x))\equal{}\minus{}x$. Prove that f has infinity discontinuous points.

First note that f is unbounded, since is codomain is a subset of codomain of $ \minus{} x$, which is $ \mathbb R$ itself.
Moreover the function's domain is $ \mathbb R$, so f is bounded in every bounded subset of $ \mathbb R$. So $ \mid f(x) \mid \to \infty \Leftrightarrow \mid x \mid \to \infty$.
Now, let $ f(0) \equal{} c$. Then $ f(x) \equal{} 0 \Rightarrow \minus{} x \equal{} f(f(x)) \equal{} f(0) \equal{} c$: then $ \minus{} c$ is the only element such that $ f(x) \equal{} 0$ (i.).
Suppose points of discontinuity are finite: so exists $ t > 0$ such that $ t > \mid c \mid \wedge \forall x \in \mathbb R, \mid x \mid > t \Rightarrow f$ is continous in x: because of (i.) f is definitely located in an only quadrant, but in $ \displaystyle (x,f(x),f(f(x)))$ there are surely only one variation and one permanence in sign, hence (for x>0) exactly one of the points (x, f(x)), (f(x),-x) belongs to the first quadrant and one belongs to the fourth: but this is a contradiction, since $ \mid f(x)\mid$, $ \mid x\mid \to \infty$ and f is continous for $ \mid x \mid > t$ by assumption.

I'm curious anyway to know if is there a solution using associated polynomials and matrices, since variation and permanence remind me that :)


Let $ f(x)$ be a function defined on [0;1] such that the following properties hold :
i) $ f(1)\equal{}1$
ii) $ f(x)\equal{}\frac{1}{3}(f(\frac{x}{3}\plus{}f(\frac{x\plus{}1}{3}\plus{}f(\frac{x\plus{}2}{3})$ for all $ x\in [0;1]$
iii) for every a possitive but can be arbitraily small,there exists a possitive number b(b depends on a) such that:For all $ x,y\in[0;1]$ such that $ |x\minus{}y|\leq b$,we have $ |f(x)\minus{}f(y)|\leq a$.Prove that $ f(x)\equal{}1$ for all $ x\in[0;1]$

we take $ a > 0$
let $ E_a = \{b > 0|\ \forall x,y\in[0,1]: \ |x - y|\le b\Rightarrow |f(x) - f(y)|\le a\}$
we have $ E_a\neq \emptyset$ and $ c\in E_a \Rightarrow ]0,c]\subset E_a$
let $ c\in E_a$
$ \forall x,y\in [0,1]$ such that $ |x - y|\le 3c$
we have $ \forall i\in\{0,1,2\}: \ |\frac{x+i}{3}-\frac{y+i}{3} |\le c$
$ \Rightarrow |f(\frac{x+i}{3})-f(\frac{y+i}{3}) |\le a$
then $ |f(x) - f(y)|\le \frac {1}{3}\sum_{i = 0}^{2}|f(\frac {x + i}{3}) - f(\frac {y + i}{3}) |\le a$
then $ 3c\in E_a$
and we get $ \forall n\in\mathbb{N}: \ 3^nc\in E$ $ \Rightarrow$ $ \exists N\in\mathbb{N}: \ 1\in ]0,3^{N}c]\subset E_a$
forall $ x,y\in [0,1]: \ |x - y|\le 1\Rightarrow |f(x) - f(y)|\le a$
then $ {\forall a > 0,\forall x,y\in [0,1]: \ |f(x) - f(y)|\le a\Rightarrow \forall x,y\in [0,1]: f(x) = f(y) = f(1) = 1}$


linear algebra
matrix
limit
function
absolute value
linear algebra unsolved

Hi,

let $ A\in \mathbb{R}^{n\times n}_{\geq 0}$ be a symmetric real matrix.
It is well-known (Gelfand's formula; valid in a much more general
setting as well) that when $ \lVert \rVert$ is any consistent matrix
norm, then
$ \lim_{k\rightarrow} \lVert A^k \rVert^{\frac{1}{k}} \equal{} \rho(A)$
where $ \rho(A)$ is the spectral radius of $ A$, i.e. the largest
modulus of a eigenvalue of $ A$.

I  (by numerical experiments)  stumbled upon the fact (?) that
when as a norm (?) I take $ A\mapsto \underbrace{[1,\dotsc, 1]}_{n} A \underbrace{[1,\dotsc, 1]^T}_{n}$
the same convergence takes place. However, I do not know
whether this is a consistent matrix norm. I do not know a name for it.
It is equal to the square of the Frobenius norm of $ A$.

Does anyone of you have an explanation for this?

Salut,
Sébastien

No, it does not equal the square of the
Frobenius norm, sorry. But the formula
I gave is the one I mean.

So does anyone of you know whether
the sum of all entries (since this is what I
defined) of a non-negative
real matrix is a consistent matrix norm
and whether it has a name?

The function $ \gamma : \,\, \mathbb{C} ^{m \times n} \longrightarrow \,\, \mathbb{R}_0^ +$ defined by $ \gamma (A) = \sum_{i,j} |a_{ij}|$ can be shown to be a matrix norm. In your case, since all elements of your matrices A are already non negative, you have $ \gamma (A) = \sum_{i,j} |a_{ij}| = \sum_{i,j} a_{ij}$, and as Gelfand's formula is valid for any matrix norm, it is no surprise you find that

$ \lim_{k \to + \infty} [\gamma(A^k)]^{1/k} = \rho (A)$. The norm $ \gamma (A)$ is sometimes referred to as the [i]the absolute value sum norm[/i] and denoted $ \parallel{}A\parallel{}_1$, since it is an Hölder norm with p = 1.

I'd be very hesitant to use the notation $ \|A\|_1$ for this, for fear of confusing it with the $ \ell^1$ operator norm, which is the maximum column sum.

Yes, that notation makes confusion with the induced norms (however, some texts use it), that's why I prefer to use $ \gamma (A)$.


In a polynomial like $ S^3 \minus{} 21S \plus{} 20 \equal{} 0$, since the sum of the coefficients is $ 0$, $ 1$ must be a root.  Are there any other rules similar to this one?

Thanks!

Yes.  If $ P(r) \equal{} 0$, then $ r$ must be a root.   :D

(In all seriousness, the only other really nice case of this rule occurs for $ r \equal{} \minus{}1$.)

[quote="t0rajir0u"]Yes.  If $ P(r) \equal{} 0$, then $ r$ must be a root.   :D

(In all seriousness, the only other really nice case of this rule occurs for $ r \equal{} \minus{} 1$.)[/quote]

Is there any connection with Bezu Theorem ?


A weighted coin lands heads with probability p=2/3.  Find the probability that an even number of heads come up when the coin is tossed 50 times.

I noticed a weird pattern.  Whenever x (the number of tosses) is even, the probability is one half the denominator of the fraction rounded up.  When x (the number of tosses) is odd, the probability is one half the denominator of the fraction rounded down.  Of course, the number in the denominator is 3^x.  Therefore, since 50 is even, we have [(3^50)/2+0.5]/3^50.  This is close to 1/2, though it's slightly more.  I can't prove this pattern and I don't know even if it works.  I only investigated the first 5 cases.

[hide]Adding up all cases (0,2,4,...50 heads) gives $P=\sum_{n=0}^{25} \binom{50}{2n}(2/3)^{2n}(1/3)^{50-2n} =3^{-50}\sum_{n=0}^{25} \binom{50}{2n} 2^{2n}$
Using Pascal's Identity, $P=3^{-50}\left( 1+ \sum_{n=1}^{25} \left(\binom{49}{2n}+\binom{49}{2n-1} \right) 2^{2n} \right)$
Use binomial theorem to clean up most of terms, $P=3^{-50}\left( 3^{49}+\sum_{n=1}^{25} \binom{49}{2n-1}2^{2n-1} \right)$
Repeating the process gives $P=3^{-50}\left(3^{49}+3^{48}+\sum_{n=1}^{24} \binom{48}{2n} 2^{2n}\right)$
Repeating again and again until remainder drops out completely, $P=3^{-50}\left( 1+\sum_{n=0}^{49} 3^n\right) =3^{-50}\frac{3^{50}+1}{2}=\frac{3^{50}+1}{2*3^{50}}$[/hide]
(I fixed the error. This answer makes more sense...)

When you cleaned up, did you do:
$P= \left( 1+ \sum_{n=1}^{25} \left(\binom{49}{2n}+\binom{49}{2n-1} \right) 2^{2n} \right) \\ \Rightarrow 1 + \sum_{n=1}^{25} \left(\binom{49}{2n}(2^{2n}) + 2\binom{49}{2n-1}(2^{2n-1}) \right) \\ \Rightarrow 1 + \sum_{n=1}^{25} \left(\binom{49}{2n}(2^{2n})\right) + 2\sum_{n=1}^{25} \left(\binom{49}{2n-1}(2^{2n-1})\right)$, because then I think you left out a part.

o i see. That last part remaining should be not $\sum_{n=1}^{25} \binom{49}{2n-1}$ but $\sum_{n=1}^{25} \binom{49}{2n-1} 2^{2n-1}$.

FWIW - An intutive way (this problem may have been asked here before).. The intutive answer is closer to $1/2$ so ..
[hide]Let the probablity(of even number of heads)  after  $n$ throws  be  $0.5 + x_n$  (Probality that odd number of heads = $0.5-x_n$)

Now for $n+1$ throws we have  ( p(even number of heads before , tail now )+ p(odd number of heds before  and head now) )

$0.5+x_{n+1} = (0.5+x_n)(1/3)+(0.5-x_n)(2/3) =0.5+(-1/3)x_n$
or $x_{n+1} = - (1/3)x_n$  obviously $x_1 = (1/3-1/2)=-1/6$  so the required probablity = $0.5+(1/2) (-1/3)^n$ etc...[/hide]


trigonometry
AMC
AIME
logarithms
number theory

umm I plan on using the AoPS textbooks as my main source of study during the summer, but i'm not sure of the pace i should be going at. Do you think it's possible to cover the entire first textbook and possibly master it within summer break?  :blush:

You could cover it.

It takes a long time to truly master it.

I guess you could do the problems in the book up to maybe the final few in a chapter and save those for later if you can't do them.

You dont have to finish AoPS 1 before you start 2. 2 has some important things such as the trig theorems and identities, number theory, sums and sequence etc. You could easily work through them together.

[quote="sapphyre571"]also
don't try to do it all in one week
i read the entire first one the week before AIME
and completely forgot everything i learned about logarithms and summation notation
so take it slow
a chapter a day maybe?[/quote]

Well, it helps to read things over and over again.  That way you won't forget as easily.

And a chapter a day seems too much.  Try a chapter a week :)


calculus
integration
analytic geometry
function
limit
real analysis
real analysis unsolved

I can solve this problem easily if they had general equation boundaries, but the infinite boundaries are confusing me. I would appreciate it if someone could walk me through it. The problem is number 105 in the homework packet that I've linked below. I am very sorry, but I don't know how to input all the calculus symbols.

http://www.math.ufl.edu/%7Ehuang/calc3/fall2007.pdf

Famous problem:

Let $ I\equal{}\int_{\minus{}\infty}^{\infty}e^{\minus{}x^{2}}\ dx$.  Then extending it to another dimension,
$ I^{2}\equal{}\iint_{S}e^{\minus{}x^{2}\minus{}y^{2}}\ dy\ dx$ where $ S$ is the disc indicated.
Now we should perform a change of variables, switching into polar coordinates:
$ I^{2}\equal{}\iint_{S}re^{\minus{}r^{2}}\ dr\ d\theta$
This is a circle so obviously $ \theta$ goes from $ 0$ to $ 2\pi$ and it has an infinite radius, so $ r$ goes from $ 0$ to $ \infty$.  Now plug in:
$ I^{2}\equal{}\int_{0}^{2\pi}\int_{0}^{\infty}re^{\minus{}r^{2}}\ dr\ d\theta\equal{}\int_{0}^{2\pi}\left(\minus{}\frac{1}{2e^{r^{2}}}\right)_{0}^{\infty}\ d\theta\equal{}\int_{0}^{2\pi}\frac{1}{2}\ d\theta\equal{}\pi$.
So we have shown that $ I^{2}\equal{}\pi$, and therefore, $ I\equal{}\sqrt{\pi}$, which is the ultimate goal of the problem.

This particular trick [b]only[/b] works for the whole plane, and only for this function. You couldn't do it for other regions, because the function doesn't have an elementary antiderivative with respect to either $ x$ or $ y$. We need a region that is particularly nice in both rectangular and polar coordinates, and that must be the entire plane (or a quadrant).

Wait I'm still a bit confused. How did you go from I to I^2? How did you change e^(-x^2) to e^(-x^2-y^2)?

This particular trick relies on the fact that the value of a definite integral is independent of the name of the variable used to integrate it.  Thus

$ I \equal{} \int_{ \minus{} \infty}^{\infty} e^{ \minus{} x^2} \, dx \equal{} \int_{ \minus{} \infty}^{\infty} e^{ \minus{} y^2} \, dy$.

We can then write (using repeatedly the fact that integration is a linear operator, i.e. $ c \int f(x) \, dx \equal{} \int cf(x) \, dx$):

$ I^2 \equal{} I \int_{ \minus{} \infty}^{\infty} e^{ \minus{} x^2} \, dx \equal{} \int_{ \minus{} \infty}^{\infty} Ie^{ \minus{} x^2} \, dx \equal{} \int_{ \minus{} \infty}^{\infty} \left( \int_{ \minus{} \infty}^{\infty} e^{ \minus{} y^2} \, dy \right) e^{ \minus{} x^2} \, dx \equal{} \iint_{S} e^{ \minus{} y^2 \minus{} x^2} \, dy \, dx$.

The value of this strange rewrite is the coordinate substitution that follows.  Note that $ S$ actually refers to the entire $ xy$ plane.

I think that it would be nice to note that there is another method to solve this problem:

Consider the two functions $ f(x) \equal{} \int_{0}^xe^{ \minus{} t^2}\ dt$ and $ g(x) \equal{} \int_0^1\frac {e^{ \minus{} x^2(1 \plus{} t^2)}}{1 \plus{} t^2}\ dt$.  Then note that $ 2f(x)f'(x) \plus{} g'(x) \equal{} 0$, that is, $ [f(x)]^2 \plus{} g(x)$ is a constant.  Put $ x \equal{} 0$ to obtain $ [\lim_{x\rightarrow\infty}f(x)]^2 \equal{} \frac {\pi}{4}\implies \int_{ \minus{} \infty}^{\infty}e^{ \minus{} x^2}\ dx \equal{} \sqrt {\pi}$.


What would happen if ammonia solution is added to calcium nitrate solution??
Any precipitate Ca(OH)2(s)?

Thanks!!

Is it possible that [Ca(NH3)4]2+ is formed?

[quote="Kent Merryfield"][quote="BanishedTraitor"]I would say no because $Ca(OH)_{2}$ is really quite soluble.[/quote]
Sure. It's soluble enough that a calcium hydroxide solution ("lime water") can have enough calcium in solution that you get a nice visible precipitate of calcium carbonate when you add carbon dioxide.

But it's less soluble than calcium nitrate. This is a case for which there's no substitute for quantitation - you can't just say "a calcium nitrate solution" and "an ammonia solution." You need to specify concentrations, amounts, pH, the solubility of calcium hydroxide, the pKa of the ammonium ion - all of it. Chemistry is a quantitative science.[/quote]

That's true.

Looking at the electronic distribution, I wouldn't say Calcium had form complex ions.

There is whole serie of Ca/NH_3 complexes, up to Ca(NH_3)_6^{2+}. But the complex creation constants are rather small - like 5x10^5 for this reaction:

Ca(NH_3)_6^{2+} -> Ca^{2+} + 6NH_3


How many people are in your club now?

there are about 15-20 people in my club, a drop from last year, when it was 45-55.

Nobody seems to like math at my school :( because there are only about 10 people in the club.

Our school's MC is weird. 2 years ago, we had 10. last year, we had 15. Now we have 17 or 18.  :huh:

Woah! That's amazing! The most my team ever had was like 8!

I always thought there were more people in Mathcounts in other schools in bigger cities...

we have about 20 kids but 10 are stupid 5 are ok and 5 are smart so the top five are competing for getting in the four spots for the main smart team

Small, private gifted school -- 140 5th-8th, we have about 15, including 2 4th's and a 5th

My school had enough for a main team and an alternate, and that was it. I don't know how many people knew about it, though; basically, our advanced math teacher picked the best people. :P

None. We don't have one. My coach is really dumb because around December, January, all of the 6-8 adv students take the school sprint round and the top 12 on to the next round and then take the target and add up scores to make an "a" and "b" team.

Ha ha

my school stinks more!

we don't have a mathcount club
we don't do mathcount

we dno't even have a normal math club
this is something I've been trying to work on.

We split up our club. We have 6th and 7th in one room and 8th in another. We can't fit us all in one room... I think we have about 45.

That was the case 2 yrs. ago at my school. 6th grade in one room, 7th and 8th in another. About 30 something people out of 350 or so students. Sadly, it dropped last year to about 14, three 7th, no 8th and the rest 6th out of like 400, though we did do better in chapter that year, but it probably was a fluke. This year we have about the same, (out of around 500 people). I'm the only 8th grader :(

I believe around 24...

we don't have any insanely smart people anymore... even I'm stupid now

[quote="Treething"]...even I'm stupid now[/quote]

That has definitely affected your posting habits on AoPS as well :P When was the last time you posted a solution to a problem?

Half our mathcounts attendants are dumb 6th grade girls. When voting for our team t-shirt, they teammed up and voted on a shirt with a Pikachu on it  :(  It probably will cost a ton too.

Mathcounts doesn't sponsor the team T-shirt? They're not affiliated with them?

No wonder my team never had one.

500th post!!!

We have 130 in the class and about 80 that attend evening meetings on a regulatr basis.  Breakdown...40 6th graders and 45 7th and 8th graders.[/quote]

Are you serious??? What school is that???? Wow...

JBoyd is the coach of (I think) Fort Settlement Middle in Texas; he had 3 people go to nationals last year. (Texas was the top team; they also had some people in the Countdown Round).

Oh, I didn't realize that was him :blush:


If $A+B+C=\pi$, prove that $sinA+sinB+sinC=4cos(\frac{A}{2}).cos(\frac{B}{2}).cos(\frac{C}{2})$


I am a freshmen at my high school trying to figure out what courses I should take.
How should I go about asking successful students at my high school who  don't know well for advice on course selection and other questions in general?
[hide="What I am trying to find out"]
[b]In general, I'm trying to find a "right" order of classes for me to take at my high school. I understand that my preferences will change and that nothing I decide now is final, but I still want to get a good feel for what's is coming and what I should pursue.[/b]

[b]Things I'm Trying to "Get a good feel for"[/b]
- how I should balance out my classes so that I don't overload myself or give myself too easy a time in high school
- which classes which interest me personally and are beneficial for me to take academically
- how I can prepare for extracurricular competitions such as the AMC
- how difficult certain classes are and which will help me learn the most
[/hide][hide="Who I am Asking and Why"]
[b]Who I want to ask:[/b]
-students in my high school who have been most successful in academics and math, science, or technology related extracurricular activities and clubs.
-students who have done well in terms of course difficulty, grades, and standardized test scores.

[b]Who I don't want to ask:[/b]
-teachers who only promote themselves or don't want to say things that dissuade me from taking their classes.
-counselors who have not taken the classes at my high school themselves or were not successful in them.
-parents of students in my high school because they also don't have first-hand accounts of classes at my high school.
- students who are having extreme difficulty getting through my high school and are not doing well (people with Cs and lower... or minimalists who don't care about their education)
[/hide][hide="What I am having trouble with"]
[b]Ideally, I want to ask students questions through email because it is difficult to get to them for a while in school and even more so to ask many questions.

Here are some of the difficulties I'm having:[/b]
- determining who to ask (I usually look for the officers of the successful clubs at my school)
- finding the people I'm interested in! My school posts pictures of them on the intranet but I can't always recognize them
- asking many questions in person, who are often busy
- [b]asking many questions without seeming rude and trying to get honest, not-so-fluffy answers...[/b]
[/hide][hide=" Some Specific Questions I want to ask"]
In your experience,

[b]What have been the more challenging or time consuming classes?

Which classes that you taken do you think are the less challenging courses that you would recommend for someone to balance out harder classes with?

What order science and math classes did you take and why did you choose it? If you could have changed some things about your scheduling, what would you change and why?[/b]
[/hide]..............................................................................>>
[b]If you wanted to ask many questions, felt uncertain about what classes you wanted to take in high school, and wanted to ask the students you thought were the hot-shots, the smart people, the geniuses,... in addition to the average students... how would go about doing it?[/b]

[hide="Why I put my topic here:"]This is a question about college because I need to start preparing to try to get into college in high school.[/hide]

This is a somewhat unusual request.  It's been my experience that most high-achieving students take  similar courses (the most challenging core classes in each subject they can find) and don't spend much time deciding.  It's generally fairly obvious from the course catalog.

If you want some advice specific to your school, though, I'd say your best bet would be to join clubs.  As you become a part of the social environment of that club, you'll find the senior members more willing to dispense honest advice.  Some of them will tell you not to worry about these things so early in high school, and you should give that some thought, too.  Freshman year of high school is a time of exploration - academically, socially, [i]and[/i] personally.  Don't stress yourself out worrying about college now.  Just do what you have a passion for and the rest will come naturally.

Additionally, this is what your school counselors are for, but you haven't mentioned them.  Do you not trust them?

Other people have also encouraged me to explore and not to stress myself, and I completely agree with them.

However, in my past experience, I've never been excited or interested in most of the school classes I take although I realize that this might change in high school. Extracurricular competitions and preparation for them have always interested me and taught me more than school classes. This is especially true for math and I still feel that this is true. This is why I want to arrange a set of courses that lets me follow my interests outside the constraints of school classes while still challenging myself academically.

To do this, I want to know what many of the classes are like in my school terms of difficulty and time commitment. Something that I want to do that I didn't mention in the above post was that I want to scout for people who have taken classes that I'm interested in and ask them about the classes.

I've joined clubs. There are smart people there like you said. As I become a part of the social environment? Honestly, I'm timid talking to and asking the successful senior members questions and there is rarely time to ask them for more than a few minutes or two at a time. Imagine you're a freshmen... are you going to approach a group of really smart seniors, pick one, and start asking him questions?

I briefly mentioned my counselor under "Who I am Asking and Why." I'm looking for people who have actually taken classes at my school, not adults who can only look at the course descriptions and get second hand accounts of student's experiences with them.

[quote="aufha"]However, in my past experience, I've never been excited or interested in most of the school classes I take although I realize that this might change in high school.[/quote]

You're likely to continue to not be interested if you're going to a public school.  I suggest that you look for intellectual stimulation outside the classroom directly and not worry too much about finding the most stimulating courses.  College is where the stimulating courses are.

[quote="aufha"]Honestly, I'm timid talking to and asking the successful senior members questions and there is rarely time to ask them for more than a few minutes or two at a time. Imagine you're a freshmen... are you going to approach a group of really smart seniors, pick one, and start asking him questions?[/quote]

Why not?  I'm sure they'd be flattered you're asking them for advice.  In any case, if they're rude or dismissive, they may not be the kind of people you want to emulate anyway.

[quote="aufha"]I briefly mentioned my counselor under "Who I am Asking and Why." I'm looking for people who have actually taken classes at my school, not adults who can only look at the course descriptions and get second hand accounts of student's experiences with them.[/quote]

The value of (experienced) counselors is that they get second-hand accounts of a lot of [i]different[/i] students' experiences and can condense them to give you the most relevant advice.  While any given senior you ask may not have a situation that fits yours, counselors are (or rather, should be) equipped to consider your situation and adapt their advice appropriately.

I'd like to mention that you should not assume that classes cannot possibly be interesting just because it's school and school is boring. My middle school classes were abysmally boring. I carried around a notebook, and during my classes I created a science fiction adventure game and played myself. It involved drawing lots of silly pictures on graph paper and doing a TON of arithmetic. But when I got to high school, I actually had interesting classes and teachers. Especially the AP sciences are good, if you have a decent teacher (and the class isn't just "prepare for AP test" but actually "learn the subject"), and languages can be good once you are far enough along to read literature.

What would be appropriate to ask a counselor and which would be better to ask a student about?
Do you think that asking your counselor a lot about a specific class is okay?

If you were a counselor, what kind of questions would you not want to hear? Which would you want to hear?

Anyway, thank you guys and Hamster1800, who private messaged me a long note, for responding to my post. I appreciate your time and advice.


Αποδειξτε οτι για καθε φυσικο $ n$ υπαρχει τριαδα ακεραιων $ (x,y,z)$ ωστε vα ισχυει η παρακατω συνθηκη:
$ n\equal{}x(x\plus{}1)\plus{}y(y\plus{}1)\minus{}z(z\plus{}1)$

Καταπληκτική άσκηση. Έχω την εντύπωση ότι βγαίνει με επαγωγή. Θα πω σίγουρα αφού προσπαθήσω.

Μανιέ μιας και είπες επαγωγή ίσως βγαίνει με ισχυρή επαγώγη αλλά δεν είμαι σίγουρος. Πάντως νομίζω ότι βγαίνει και με αλλο τροπο αλλα δεν ειμαι σιγουρος θα τη δω .Πολυ ωραια άσκηση.

Sorry guys, αλλά εμένα γιατί μου φαίνεται ότι υπάρχει λάθος στην εκφώνηση? Ο αριθμός της RHS είναι πάντα άρτιος + άρτιος - άρτιος = άρτιος, όπερ άτοπο αν $ n\equal{}$ ο αγαπημένος σας περιττός.

:huh:  :?:

Cheerio,

Durandal 1707

Η άσκηση είναι όντως λάθος. Όταν το n είναι άρτιος, υπάρχει μια ωραία λύση που χρησιμοποιεί μόνο απλή επαγωγή. Μπορείτε να τη βρείτε;

Μα αν το n ειναι αρτιος ισχυεί για ολα τα χ, y , z για οτιδηποτε x , y, z   γιατι οτι και να ειναι το χ, y, z  εχουμε αρτιο+αρτιο-αρτιο=αρτιος

Αυτό ακριβώς λέμε. Η άσκηση αντιφάσκει μόνο όταν το n δεν είναι άρτιος. Αλλά μην ξεχνάς ότι τα ζητούμενα είναι τα x, y, z και όχι το n.

Συμφωνώ αλλά απάντησα στη δική σου ερώτηση για ν αρτιος

Για n άρτιο η άσκηση γίνεται:
Αποδειξτε οτι για καθε φυσικο $ n$ υπαρχει τριαδα ακεραιων $ (x,y,z)$ ωστε vα ισχυει η παρακατω συνθηκη:
$ 2n \equal{} x(x \plus{} 1) \plus{} y(y \plus{} 1) \minus{} z(z \plus{} 1)$

Η λύση σου αποδεικνύει την ύπαρξη του n, όχι της τριάδας.

Φυσικα δν ισχυει για περιττο.Συγνωμη για τν παραληψη.Μονο για καθε αρτιο.....

Εαν $ n$ άρτιος τότε $ n\equal{}2k$ για κάποιο ακέραιο $ k$. Αρκεί τώρα να παρατηρήσουμε ότι

$ k(k\plus{}1)\minus{}(k\minus{}1)k\equal{}2k$ οπότε για κάθε $ n \equal{} 2k$ παίρνουμε την τριάδα $ (x,y,z)\equal{}(0,k,k\minus{}1)$ και έχουμε ολοκληρώσει την άσκηση.

Αλέξανδρος

Σωστα Αλεξ.Και αν οι $ x.y,z$ ειναι μη μηδενικοι?

Αρκεί η βοηθητική πρόταση:
Αποδειξτε οτι για καθε φυσικο $ n$ υπάρχουν $ x, z \equal{} x \minus{} 1$ ωστε vα ισχυει η παρακατω συνθηκη:
$ 2n \equal{} x(x \plus{} 1) \minus{} z(z \plus{} 1)$

Απόδειξη:
[hide]
Για $ n \equal{} 2$ υπάρχει το ζεύγος $ (1,0)$
Έστω ότι υπάρχει ζεύγος $ (x, z)$ για $ n \equal{} k$ τέτοιο ώστε $ 2n \equal{} x(x \plus{} 1) \minus{} z(z \plus{} 1)$.
Το ζεύγος (x+1, z+1) ικανοποιεί τη συνθήκη για $ n \equal{} k \plus{} 1$, αφού:
$ n(x \plus{} 1, z \plus{} 1) \minus{} n(x,z) \equal{} (x \plus{} 1)(x \plus{} 2) \minus{} (z \plus{} 1)(z \plus{} 2) \minus{} ( x(x \plus{} 1) \minus{} z(z \plus{} 1) ) \equal{} 2(x \plus{} 1) \minus{} 2(z \plus{} 1) \equal{}$
$ \equal{} 2(x \minus{} z) \equal{} 2$
[/hide]

Άρα δυο λύσεις του αρχικού προβλήματος είναι οι $ (x, 0, x \minus{} 1)$, $ (x, \minus{}1, x \minus{} 1)$ (στη δεύτερη περίπτωση οι αριθμοί αρχίζουν από μεγαλύτερους του 2, το οποίο εξετάζεται σαν ειδική περίπτωση).

Όμοια βγαίνει ότι αρκεί $ x \plus{} y \minus{} z \equal{} 1$ για τη σταθερή μεταβολή του n κατά δύο όταν οι μεταβλητές αυξάνονται κατά 1.

Κυριοι οπως ειπα δν θελω κατι τετοιο.....Θελω να το ερευνησεται και οταν κανενας απο τους $ x(x\plus{}1)$,$ y(y\plus{}1)$,$ z(z\plus{}1)$  δν μηδενιζεται.....Εξαλλου ετσι γινεται και πιο ενδιαφερον!Ευχαριστω για την κατανοηση σας.

Όμοια με την προηγούμενή μου απόδειξη βγαίνει ότι αρκεί $ x \plus{} y \minus{} z \equal{} 1$ για τη σταθερή μεταβολή του n κατά δύο όταν οι μεταβλητές αυξάνονται κατά 1. (Η περίπτωση για 2 μπορεί να γίνει με πολλούς μη μηδενικούς τρόπους.)

Πολυ δυναμικο επισης... :)Και αν σε ρωτουσα να μου βρεις τους μη μηδενικους $ x,y,z$ οταν $ n\equal{}2008$....?

Καλησπέρα και πάλι,

Γράφεις τον $ n$ στη μορφή $ n \equal{} 2(k \plus{} 2)$ (επιλέγοντας κατάλληλα το $ k$) και παρατηρείς ότι το ζεύγος

$ (x,y,z) \equal{} (k \plus{} 1,1,k)$ είναι ok.

Στην περίπτωση που $ k \equal{} \minus{} 1$ τότε $ n \equal{} 2$ οπότε επιλέγω $ (x,y,z) \equal{} (1,t,t)$ όπου t τυχαίο με $ t \neq \minus{} 1,0$, ενώ εαν $ k \equal{} 0$ τότε $ n \equal{} 4$ οπότε επιλέγω $ (x,y,z) \equal{} (3,3,4)$.
Τέλος εαν $ k \plus{} 2 \equal{} 0$ δηλαδή $ k \equal{} \minus{} 2$ τότε $ n \equal{} 0$ οπότε επιλέγω $ (x,y,z) \equal{} (2,2,3)$

Έτσι έχουμε πάντα λύση ακόμη και στην περίπτωση όπου $ x(x \plus{} 1),y(y \plus{} 1),z(z \plus{} 1) \neq 0$.

Είχες μήπως κάποια άλλη υπαρξιακή απόδειξη στο μυαλό σου ? Εαν ναι γράψε την γιατί πιστεύω ότι θα έχει μεγάλο ενδιαφέρον η σκέψη που έκανες για να την κατασκευάσεις. Αυτή μετράει και όχι αυτή καθεαυτή η άσκηση και η ευκολία ή δυσκολία της. Έχει καθόλου σχέση με τρίγωνους αριθμούς ή κάτι παρόμοιο ή λύση κάποιων ισοτιμιών και κλασσική Θεωρία Αριθμών?

Αλέξανδρος

Η λυση μου βασιζεται στο λημμα οτι καθε ακεραιος μπορει να παρασταθει με τν μορφη $ x^2\plus{}y^2\minus{}z^2$ με απειρους τροπους και με τους x,y,z θετικους ακεραιους.... Οχι δν χρησιμοποιω τριγωνους αριθμους, απλες τεχνικες....

Το θέμα γίνεται πιο ενδιαφέρον αν ζητούνταν το πλήθος των τριάδων που μπορούν να δημιουργηθούν.
Είναι άπειρες, στηρίζοντας την απόδειξη στο ίδιο λήμμα.

Παιδιά αν διαιρέσουμε με 2(αφού έχουμε n=2k ) τότε σχηματίζονται τριγωνικοί αριθμοι και από εδώ βγαίνει ωραία λύση

Ακυρο το τελευταιο(εννοω τριγωνους) δεν ειχα διαβασει τα ποιο πάνω post sry :blush:

Μετά όμως πρέπει να αποκλείσεις όλους τους υπόλοιπους ακεραίους εκτός τα μισά


Find the positive difference between the solutions for $ |x \plus{} 3| \equal{} 6$.

There are two cases - $ x \plus{} 3 \equal{} 6$ and $ x \minus{} 3 \equal{} \minus{} 6$. The first case gives us a solution of $ 3$ and the second $ \minus{}3$, so the final answer is $ \boxed{6}$.

The positive difference between the solutions for $ |x\plus{}h|\equal{}c$ is $ 2c$, so the positive difference here is $ 2(6)\equal{}\fbox{12}$.


replace 2004 with 2k
Prove: $\frac{{a^{2k}}}{{b^{k}}}+\frac{{c^{2k}}}{{d^{k}}}= \frac{2}{{(b+d)^{k}}}$
Are you sure it's not $a^{2}+c^{2}=1$? Because then by Cauchy we have:
$\frac{{a^{4}}}{b}+\frac{{c^{4}}}{d}\geq  \frac{(a^{2}+c^{2})^{2}}{{b+d}}$
But if $a^{2}+c^{2}=1$ then we have equality; so $b: a^{2}=c: d^{2}$
Notice $\sqrt[k]{\left(\frac{a^{2}}{b}\right)^{k}+\left(\frac{c^{2}}{d}\right)^{k}}\geq 2^{\frac{1}{k}-1}\left(\frac{a^{2}}{b}+\frac{c^{2}}{d}\right)$ from AM-GM (actually powermean or generalization whatever its called)
but in fact, we have equality (we established $b: a^{2}=c: d^{2}$ from Cauchy) and that is the equality condition we need for powermean
So $LHS=\left(2^{\frac{1}{k}-1}\left(\frac{a^{2}}{b}+\frac{c^{2}}{d}\right)\right)^{k}=2\frac{\left(\frac{a^{2}}{b}+\frac{c^{2}}{d}\right)^{k}}{2^{k}}\geq 2\frac{\left(2\frac{a^{2}+c^{2}}{b+c}\right)^{k}}{2^{k}}=\frac{2}{(b+d)^{k}}$
and I'm pretty sure we have equality in the last inequality (i don't know too much about the last inequality but I've heard about it)
so $\frac{{a^{2k}}}{{b^{k}}}+\frac{{c^{2k}}}{{d^{k}}}= \frac{2}{{(b+d)^{k}}}$ as desired


function
number theory
least common multiple
trigonometry
limit
calculus
derivative

Can we express the function $f(x)=2^x+3^x+9^x$ as the sum of finite periodic functions?

The coefficients of the corresponding Fourier Series don't ever become zero, so... no?

Can you give some proof that only use high school knowledge?

Okay.  Uh... continuous periodic functions must be bounded, so a sum of finite continuous periodic functions is also bounded, and since $f(x)$ is not bounded it cannot be represented as such?

Guys, nobody mentions anything about continuous functions in the statement. ;)

I agree, there is no statement about continuous functions.

[quote="JohnConstantine"]But the sum of periodic function isn't periodic with period equal to the $lcm$ of the periods of the single functions? $2^x+3^x+3^{2x}$ doesn't seem periodic[/quote]

I considered this yesterday, but if the periods are incommeasurable then this is not so.  For example, consider $g(x) = \sin x + \sin \pi x$.  :)

Completing my earlier proof:  It is necessary that at least one of the periodic functions involved be unbounded because $f(x) = 2^x + 3^x + 9^x$ is unbounded.  If a periodic function $P(x) = P(x+p)$ is unbounded, then there must be some $x_0$ such that $\lim_{x \to x_0} P(x_0) = \pm \infty$, and because the function is periodic this must occur at an infinite number of points $P(x_0 + kp)$ for $k \in \mathbb{Z}$ - hence, the function is discontinuous.

If any of the periodic functions are discontinuous then the sum of the functions must be discontinuous - however, $f(x) = 2^x + 3^x + 9^x$ is not discontinuous.   Q.E.D.

It still seems an odd question to ask.  By what principle would you even suspect $f(x) = 2^x + 3^x + 9^x$ of being expressible as the sum of a finite number of periodic functions?

[quote="jin"]I've got a question,why do you say "If any of the periodic functions are discontinuous then the sum of the functions must be discontinuous "?
The problem is a mathematic olymiad training problem in our school and we've got some idea but no proof.[/quote]

Err... because it's true.  For example, $f(x) = \sin x + \tan x$ is discontinuous at $x = \frac{ \pi }{2}$ because $\tan x$ is not defined at the point where it is unbounded.  This is necessary in order for $\tan x$ to be both unbounded and periodic, as I have shown.

Before I hear objections, we next consider $f(x) = -\tan x + \tan x$.  This function is [i]still discontinuous at $x = \frac{ \pi }{2}$ even though $\lim_{x \to \frac{\pi}{2} } f(x)$ exists.[/i]  That is because actually plugging the number in and evaluating still results in an undefined value.  And even if you consider it a removable discontinuity, the sum is now bounded again, and the previous argument applies.

(Seriously, I don't understand why this problem isn't trivial.)

suppose $f(x)=\tan x(x\neq\frac{k\pi}{2}), 0(x=\frac{k\pi}{2})$ $g(x)=2-f(x)$,then both of $f(x)$ and $g(x)$ is discontinuous but $f(x)+g(x)$ is continuous.

I think t0rajir0u means undefined instead of discontinuous in some instances.

I think you are right, jin. The problem doesn't seem that easy to be dealt by discussing continuity.

I have a solution here, hope it is right :)
    Assume that the periodic functions $f_1,f_2,...,f_n$ have periods $t_1,t_2...,t_n$ and $2^x+3^x+9^x=f=f_1+f_2+...+f_n$
    Let $m$ be an arbitrary positive integer. For every $k=0,1,...,n-1$, let $S_k=\{(a_1,a_2,...,a_n)\; ; \;a_1,a_2,...,a_n\in\{0,1,...,n-1\}\text{ and }n|a_1+a_2...+a_n-k\}$. And let
\[ U_k=\displaystyle\sum_{(a_1,a_2,...,a_n)\in S_k}f(ma_1t_1+ma_2t_2+...+ma_nt_n)  \] Then we have $U_0=U_1=...=U_{n-1}=\frac{1}{n}\displaystyle\sum_{i=1}^n\sum_{(a_1,a_2,...,a_n)\in S}f_i(ma_1t_1+ma_2t_2+...+ma_nt_n)$, where $S=S_0\cup S_1\cup...\cup S_{n-1}$.
However, when $m$ is large enough, $U_0$ contains $f(m(n-1)(t_1+...+t_n))$ and must be larger than other $U_i$'s, contradiction!

But what's wrong with the argument that a sum of finite periodic functions is either bounded or undefined at an infinite number of points, neither of which can satisfy the properties of $f(x)$?

Well, it is just as jin said, the sum of two discontinuous function may be continuous, and the sum of two unbounded function may be bounded.

I don't really follow that argument- here's another.

Suppose $f=f_1+f_2+\dots+f_n$, where $f_i$ has period $t_i>0$.
Consider $S=\sum_{\epsilon_j\in\{0,1\}}(-1)^{n-\sum_j\epsilon_j} f(\sum_j\epsilon_jt_j)$

We have $S=\prod_j\left(2^{t_j}-1\right)+\prod_j \left(3^{t_j}-1\right)+\prod_j \left(9^{t_j}-1\right)>0$ by direct calculation. On the other hand, by periodicity of $f_i$,
$\sum_{\epsilon_j\in\{0,1\}}(-1)^{n-\sum_j\epsilon_j} f_i(\sum_j\epsilon_jt_j)$
$=\sum_{\epsilon_j\in\{0,1\},j\neq i}(-1)^{n-\sum_{j\neq i}\epsilon_j} (f_i(\sum_{j\neq i}\epsilon_jt_j)-f_i(\sum_{j\neq i}\epsilon_jt_j+t_i)=0$, so $S=0$, a contradiction.

This actually relies on the fact that all derivatives of $f$ are strictly positive.

Check out [url=http://mathworld.wolfram.com/FourierSeries.html]Fourier Series[/url]

That's right, but I hope we can prove a stronger result, which does not even require continuity.

[quote="Darkseer"]I think your solution is quite good, jmerry.
[quote="jmerry"]This actually relies on the fact that all derivatives of $f$ are strictly positive.[/quote]Thanks for this observation!

Then I suppose all the polynomials can be expressed as the sum of finitely many periodic function? I already see that $f(x)=x$ can...

But, what kind of functions can be expressed as
$\displaystyle\sum_{i=1}^{\infty} f_i$, where each $f_i$ is periodic?[/quote]

I don't see how $f(x) = x$ can be represented as the sum of [i]finitely[/i] many periodic functions  :huh:  I would be interested in seeing this representation...

As to your other question, minimally, functions that are discontinuous at only a finite number of points are expressible as the sum of an infinite number of periodic functions.

By Zorn's Lemma, being a vector space over $Q$, $R$ has a basis $B\subset R$. And every real $x$ can be uniquely expressed as
$r_1b_1+r_2b_2+...+r_kb_k$, where $r_i\in Q$ and $b_i\in B$ (and $k$ is finite).

Choose $b_0\in B$. For every real $x=r_0b_0+r_1b_1+r_2b_2+...+r_kb_k$ we may define
$f_1(r_0b_0+r_1b_1+r_2b_2+...+r_kb_k)=r_0b_0$, which has period $b'$ for any $b_0\not=b'\in B$
and
$f_2(r_0b_0+r_1b_1+r_2b_2+...+r_kb_k)=r_1b_1+r_2b_2+...+r_kb_k$, which has period $b_0$
and thus $x=f_1(x)+f_2(x)$.

:omighty:

Needless to say, I now greatly desire a graph of these two functions :)

A graph? Hamel bases do not lend themselves to such ideas. The graph of any of these functions is dense in the plane.

Extending this idea, here's how to write $x^n$ as a sum of $n+1$ periodic functions, from which we could construct any polynomial.

Let $b_1,b_2,\dots,b_{n+1}$ be $n+1$ real numbers which are linearly independent over the rationals. Let $B$ be a Hamel basis for $\mathbb{R}$ containing these basis vectors.

For each rational number, write $x=x_1+x_2+\dots+x_{n+1}=\sum_{j=1}^n a_jb_j+x_{n+1}$, where the $a_j$ are the coefficients of $x$ with respect to $B$ at $b_j$.

Write $x^n=(x_1+x_2+\dots+x_{n+1})^n$ as a sum of monomials in the $x_j$.
Let $f_1(x)$ be the sum of the terms which are not divisible by $x_1$.
Let $f_2(x)$ be the sum of the terms which are divisible by $x_1$ but not $x_2$.
Let $f_3(x)$ be the sum of the terms which are divisible by $x_1x_2$ but not $x_3$.
...
Let $f_{n+1}(x)$ be the sum of the terms which are divisible by $x_1x_2\cdots x_n$.

Since no term is divisible by $x_1x_2\cdots x_nx_{n+1}$, $\sum_{j=1}^{n+1}f_j(x)=x^n.$ Also, each $f_j$ has period $b_j$.


[quote="Art of Owna"]i know theres a way to do problmes like $\cos^{-1}{x}=\sin^{-1}{x}$

but is there a way to do problmes like $\cos{x}=\sin{x}$?[/quote]
Hmm maybe I'm crazy but

$\cos{x}=\sin\left({\frac{\pi}{2}-x}\right)$.

 :maybe:

Yes, that's a good way to do it.  If you don't like reasoning about angles, you could always do it [hide="the algebraic way"] $\cos x-\sin x = \sqrt{2}\cos (x+\frac{\pi}{4}) = 0$

The general formula is given by $A \cos x+B \sin x = \left< A, B \right> \cdot \left< \cos x, \sin x \right> = \sqrt{A^{2}+B^{2}}\cos \left( x-\arctan \frac{B}{A}\right)$. [/hide]

$\cos x = \sin x$ is only true when x has a value of $(45+180n)^{\circ}$


function
combinatorics proposed
combinatorics

Suppose that $p(n)$ is the number of ways to express $n$ as a sum of some naturall numbers (the two representations $4=1+1+2$ and $4=1+2+1$ are considered the same). Prove that for an infinite number of $n$'s $p(n)$ is even and for an infinite number of $n$'s $p(n)$ is odd.

The following one also holds:

Prove that the number of ways in which $ n$ can be expressed as a sum of different positive integers is an equal to the ways in which $ n$ can be expressed as a sum of odd positive integers.

Well, this one, [b]delegat[/b], is really well-known, and thus has surely been discussed before on ML.

That said, it's quickly proved with generating functions, so I'll write it here :
let $ D$ be the number of ways in which $ n$ can be expressed as a sum of different positive integers and let $ O$ be the number of ways in which $ n$ can be expressed as a sum of odd positive integers,
then $ D = (1 + x)(1 + x^2)\cdots = \frac {1 - x^2^}{1 - x} \cdot \frac {1 - x^4}{1 - x^2^} \cdots = \frac 1{1 - x} \cdot \frac 1{1 - x^3^} \cdots = O.$
(of course, I can't help but recommend [url=http://www.math.upenn.edu/~wilf/gfology2.pdf]generatingfunctionology[/url] by Herbert Wilf)

That said, there is a bijective proof of this fact, which I invite you to try and find.
For more of those, see also [url=http://www.mathlinks.ro/viewtopic.php?t=26379]# of partitions of type 1 = # of partitions of type 2[/url]

Now, as for the original problem of this thread, it should be mentioned that the parity of $ p(n)$ is in general a very difficult problem. (this question was raised by Ramanujan)
This particular problem was solved by Kolberg in the 60's, and his proof is based on the identities :
$ \sum_{n \ge 0} p(n) x^n = \prod_{n \ge 1} \frac 1{1 - x^n}$
and
$ \prod_{n \ge 1} (1 - x^n) = \sum_{n \in \mathbb{Z}} ( - 1)^nx^{(3n^2 + n)/2}.$ (namely, the [url=http://mathworld.wolfram.com/PentagonalNumberTheorem.html]Pentagonal Number Theorem[/url])

Actually, a pretty hard conjecture is that, (roughly,) the partition function takes as many even values as odd values.

N.A.P., I recommend the [i]excellent[/i] book(let)

[i]INTEGER PARTITIONS[/i] by G.E.Andrews and K.Eriksson, Cambridge Univ. Press, 2004.

On page 125 the following "dramatic discovery" about divisibility properties of p(n) by Ken Ono is reported:

|{m: m \leq n and p(m) is even}|/n  --> 1/2, as n goes to infinity.

That is, approximately half the values of p(n) are even!

(Ken Ono's paper [i]Distribution of the partition function modulo m[/i] appeared in the [i]Annals of Mathematics[/i], vol. [b]151[/b] (2000), 293-307.)

******
EDIT. Sorry for the confusion.
You are absolutely [b]right[/b], mathmanman.
I better had gone to bed yesterday ...

But, indeed,  the following "very old" reference is of interest for the problem at hand:
Kolberg, O. [i]Note on the Parity of the Partition Function.[/i] Math. Scand. [b]7[/b] (1959), 377-378
http://www.mscand.dk/article.php?id=1555
There it is proven that there are infinitely many even and odd values of p(n).

Check also the following Mathworld entry -> http://mathworld.wolfram.com/PartitionFunctionPCongruences.html
and follow the links given therein ...

Sure about that?
I believe this conjecture is still wide open.

Actually, I think Ono's work was about finding new congruences for the partition function modulo prime $ m,$ and that his results "only" yield something like $ \frac {|{m: m \leq n \text{ and } p(m) \text{ is even }}|}n \gg \sqrt {n},$ which is still quite far from the mentioned conjecture.

EDIT : indeed, after checking [url=http://books.google.com/books?id=_5gsjNr0lWAC&printsec=frontcover]that book[/url] on google books one can see that on page 125, it is written that this result is still not known.
EDIT 2 : and I see that the mentioned paper from K. Ono is [url=http://www.emis.de/journals/Annals/151_1/ono.pdf]available online[/url] and confirms what I've written above.


A hexagon with consecutive sides of lengths 2, 2, 7, 7, 11, and 11 is inscribed in a circle.  Find the radius of the circle.

Radius is 7, see http://mathforum.org/kb/thread.jspa?forumID=129&threadID=356037&messageID=1089504


geometry
circumcircle
geometric transformation
homothety
geometry proposed

Two circles $ \gamma_{1}$ and $ \gamma_{2}$ meet at points $ X$ and $ Y$. Consider the parallel through $ Y$ to the nearest common tangent of the circles. This parallel meets again $ \gamma_{1}$ and $ \gamma_{2}$ at $ A$, and $ B$ respectively. Let $ O$ be the center of the circle tangent to $ \gamma_{1},\gamma_{2}$ and the circle $ AXB$, situated outside $ \gamma_{1}$ and $ \gamma_{2}$ and inside the circle $ AXB.$ Prove that $ XO$ is the bisector line of the angle $ \angle{AXB}.$

[b]Radu Gologan[/b]

Inversion with center $ X$ and power $ XY^2$ takes $ A, B$ into $ A', B',$ the line $ p \equiv AYB$ into circumcircle $ (P')$ of the $ \triangle A'YB',$ the circles $ \gamma_1, \gamma_2, \odot (AXB)$ into the lines $ YA', YB', A'B',$ the circle $ (O)$ tangent to $ \gamma_1, \gamma_2, \odot (AXB)$ in the specified manner into the incircle $ (O')$ of the $ \triangle A'Y'B'.$ $ X, O, O'$ are collinear, the inversion center $ X$ being the homothety center of the circle $ (O)$ and its image $ (O').$ The common tangent $ t$ of $ \gamma_1, \gamma_2$ closer to $ Y$ and parallel to $ p$ goes to a circle tangent to the linest $ YA', YB'$ and internally tangent to $ (P')$ at $ X,$ i.e., to the mixtilinear Y-incircle $ (T')$ of the $ \triangle YA'B'.$ If $ YO'$ cuts $ (P')$ again at $ M,$ then the angle $ \angle MXO'$ is right $ \Longrightarrow$ $ XO' \equiv XO$ bisects the angle $ \angle A'XB' \equiv \angle AXB.$ For example, see [url]http://www.mathlinks.ro/viewtopic.php?t=233986[/url] and elsewhere.

Inversion with Pole and Polar are very very 2 strong tools, Inversion makes your solution of a geometry problem smooth and clearly, especially this kind of problem, but Inversion with Pole and Polar are not easy to feel it when we first approach it. :blush:


linear algebra
matrix
geometry
algebra
difference of squares
special factorizations

This forum is a little more advanced then this, I know, but I have a Math Exam on Wednesday, and don't understand enough to pass. I was wondering if anyone on here would explain any of this to me, I'll put down all the questions I didn't understand and if anyone could explain how to do any of them, it'd be greatly appreciated :)

Part E.   Add or Subtract

(8x + 3) + (-2x + 2)

3x + 4y - 6x - y + 3

(4x + 8) - (x + 4)

5x - 3x + 2y + y

(3xsquared + 2y) + (2xy - xsquared)

2xy + ysquared - 4xy - 6ysquared

Part F.  Matrix Problems

a.What are the dimensions?
b.What would the new matrix be if it were multiplied by a factor of 3?

| 2   3   -2   6 |
| 8   4    8   5 |
|10  1   -5  -2 |

Add the following matrices.

3   1   5
8   6   4       <--- A
2   5   4

3    -5     5
-1   -2   -10     <--- B
-8    6    -1

Part H.    Multiply or Divide

3(2x+5)

-27x divided by 3

(x+3)(x+5)

(x+y) (x-y)

45ysqaured + 9xsquared divided by 9

Sorry for long post, just need help with any of these, thank you :):)

my advice to you, since you have until wednesday... is to go ask your teacher for help

[quote=".VicE"]This forum is a little more advanced then this, I know, but I have a Math Exam on Wednesday, and don't understand enough to pass. I was wondering if anyone on here would explain any of this to me, I'll put down all the questions I didn't understand and if anyone could explain how to do any of them, it'd be greatly appreciated :)

Sorry for long post, just need help with any of these, thank you :):)[/quote]It would be wize next time to post in the right place. This is the Olympiad Section as you can see, and you are talking about stuff that should be in the highschool -> Getting Started. I've relocated this topic there.

Ask your teacher... but here's a start: (I'm not gonna do all of them incase this is homework or something)
Part E. Add or Subtract

[quote](8x + 3) + (-2x + 2) [/quote]
If there is a plus inbetween pemdas (order of operations) says you can just take it away.  So (8x + 3 + (-2x) + 2)
That's 8x-2x+3+2 or [b]6x+5[/b]

[quote]3x + 4y - 6x - y + 3 [/quote]
Just regular adding.  Try regrouping.  3x-6x+4y-y+2 which is [b]-3x+3y+2[/b]

[quote](4x + 8) - (x + 4)[/quote]
times the subtraction through.  4x+8[b]-[/b]x[b]-[/b]4 you can do it from there

[quote](3xsquared + 2y) + (2xy - xsquared) [/quote]
Group similar things

[quote]a.What are the dimensions?[/quote]
Length Width and Depth (time, etc. etc.)
[quote]b.What would the new matrix be if it were multiplied by a factor of 3?

| 2 3 -2 6 |
| 8 4 8 5 |
|10 1 -5 -2 | [/quote]

 Just multiply every number inside by three

[quote]Add the following matrices.

3 1 5
8 6 4 <--- A
2 5 4

3 -5 5
-1 -2 -10 <--- B
-8 6 -1 [/quote]
add corresponding numbers (top-left, top middle, top right,...)

[quote]3(2x+5) [/quote]Multiply 2x by three and 5 by three (distributive property)

[quote]-27x divided by 3 [/quote]
Divide -27 by 3 and times it by x again.

[quote](x+3)(x+5) [/quote]
FOIL
[quote](x+y) (x-y) [/quote]
Difference of squares or just FOIL

[quote]45ysqaured + 9xsquared divided by 9 [/quote]
divide each term by nine

Thanks a lot :):) It's not homework, it's exam review that doesnt need to be passed in, and I can't ask my teacher because there's a %100 chance of a 50cm blizzard tommorow, so I came here. Sorry for posting in the wrong place, and thanks for the help :)

Good job somedude, but the problem is... these are the basics of the basics. The best advise for .VicE is that you MUST consult your teacher for help. You will have to work hard to understand and get past manipulation of algebraic expressions. Since your exam is Wednesday, you are running out of time and so asking the teacher for help is the best help for you at this moment.

PS: Valentin, topic is not even High School, it's more like Middle School Algebra.

Well, I learned to do basic stuff with matrices, ie. add, multiply, find 2x2 determinant, in middle school, but I wasn't actually taught it to me.  The algebra part I did in like 5th or 6th grade as part of the normal curriculum...

[quote="shinwoo"]What's a matrix?

( I need help.. )[/quote]If it helps you with anything it's not a A.I.-controlled virtual environment designed to keep people at sleep happy so that nasty robots harvest their energy out to feed upon it :)

Seriously it's a table, like this one: \[ \left( \begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{array} \right) \]

[quote="shinwoo"]oh...

I never heard of that word before, except from the movie Matrix.

i guess it's alittle complicated..matrix..sounds cool though[/quote]

lol :P .

Matrices can be used for solving equations (you hafta get it into an x by x+1 matrix and get it into reduced row-echelon form), or finding the cross product of vectors. I'm sure there are many more uses that I can't think of right now :P .

[quote="Valentin Vornicu"][quote="JBL"]I don't know anyone who learned matrices in middle school.  Most middle schools in the U.S. don't teach algebra, either.[/quote]That's what I thought also.

PS FYI I learned matrices in middle school :) (I mean that I've taught myself matrices in middle school :D).[/quote]
Most middle schools but not all. My niece was taught Algebra I in 7th grade and learned matrix addition and multiplication. Then she took Geometry in 8th grade. She attended a typical middle school in New Jersey. So, it depends on the school that offers it, I believe. And no, she is no math whiz or anything, just an above-average high school girl now.  :P

PS FYI I taught myself matrices in middle school too!!! Not a hard concept, you know what I mean, Valentin  ;) ? :lol:  :D

That's interesting - my middle school algebra teacher insisted that the class learn matrices, so we did. It was interesting - so I looked it up in Volume II and studied it further.

Well, okay, apparently I do know a couple of such people, then.  I'd never heard of matrices as part of an algebra 1 curriculum, but algebra 1 isn't a typical middle school course anyhow.  (It is typical for advanced students, I suppose.)  I didn't learn matrices until precalculus, and I didn't [i]really[/i] learn matrices until Linear Algebra in college.


If $ n$ is an integer greater than 1 then prove that $ n$ doesn't divide $ 3^{n}\minus{}2^{n}$

Posted many time :
If n is a natural satisfy $ n|(a\plus{}1)^n\minus{}a^n$ then $ n\equal{}1$
Solution : Suppose $ n>1$
Call p is least prime of $ n$ then we have :
$ p|\gcd((a\plus{}1)^n\minus{}a^n,(a\plus{}1)^{p\minus{}1}\minus{}a^{p\minus{}1})\equal{}(a\plus{}1)^{\gcd(n,p\minus{}1)}\minus{}a^{\gcd(n,p\minus{}1)}\equal{}1$
So $ p|1$
Tradition!
Imply that $ n\equal{}1$


A special rubber ball is dropped from the top of a wall that is sixteen feet high. Each time the ball hits the ground it bounces back only half as high as the distance it fell. The ball is caught when it bounces back to a high point of one foot. How many times does the ball hit the ground?


logarithms
calculus
calculus computations

How to determine whether series $ \sum_{n\equal{}2}^{\infty}\frac{\sqrt[n]{n}\minus{}1}{\ln n}$ converges or diverges?

No it isn't. The serie diverges because
\[ n^{1/n} \equal{} \exp((\ln n) /n) \equal{} 1 \plus{} (\ln n)/n \plus{} O((\ln n)^2/n^2\]
then
the egeneral term of the series is
\[ 1/n \plus{} O((\ln n)/n^2\]
. Your application of Cauchy--condensations test (I suppose) is wrong because the series converges if and only if
\[ \sum_{n \equal{} 1} 2^n f(2^n)\]
converges. Moreover the general term shuold decrease monotonically and this is true.

[quote="Rafikafi"]Since each Term is positive, we can use the Cauchy-Verdichtungssatz,.....[/quote]

hopefully this time it's correct

\[ \sum_{n\equal{}2}^{\infty}\frac{n^{1/n}\minus{}1}{\ln(n)}\equal{}\sum_{n\equal{}2}^{\infty}\frac{\frac{n^{1\plus{}1/n}}{n}\minus{}1}{\ln(n)}
\ge \sum_{n\equal{}2}^{\infty}\frac{\frac{1\plus{}(n\minus{}1)(1\plus{}1/n)}{n}\minus{}1}{\ln(n)}\equal{}\sum_{n\equal{}2}^{\infty}\frac{n\minus{}1}{n^2\ln(n)}
\equal{}\sum_{n\equal{}2}^{\infty}\frac{1}{n\ln(n)}\minus{}\sum_{n\equal{}2}^{\infty}\frac{1}{n^2\ln(n)}>
\sum_{n\equal{}2}^{\infty}\frac{1}{n\ln(n)}\minus{}1 \minus{}\sum_{n\equal{}1}^{\infty}\frac{1}{n^2}
\equal{}\sum_{n\equal{}2}^{\infty}\frac{1}{n\ln(n)}\minus{}1\minus{}\frac{\pi^2}{6}\]

and $ \displaystyle \sum_{n\equal{}2}^{\infty}\frac{1}{n\ln(n)}$ diverges, if you apply Cauchy verdichtungssatz correctly this time


limit
real analysis
calculus
calculus computations

we have $f: \left[a,+\infty\right)\rightarrow\Re$ continuous function.If $\lim_{x\rightarrow+\infty}=\ell\in\Re$ then prove that $f$ is uniformly continuous in $\left[a,+\infty\right)$.

$f$ is uniformly continous on every compact set $[a,b]$, so just fix it up by requiring b large enough that we don't get a large variation to the right of say $b-1$.

Could you explain your opinion because I don't speak so good english...??? :blush:

Choose $\epsilon$. Find $M$ so that $|f(x)-L|<\frac{\epsilon}2$ when $x>M$.
Now, $|f(x)-f(y)|<\epsilon$ whenever both $x$ and $y$ are greater than $M$. Use uniform continuity on compact intervals (say, $[a,M+1]$) to finish; we have two estimates, and we want to make sure at least one always applies.


inequalities
geometry
circumcircle
inradius
trigonometry
inequalities proposed

Given triangle $ABC$ semiperimter $s$ ,circumradius $R$, inradius $r$ prove that
$\frac{s^{2}}{r}\ge 4R+r\ge\sqrt{3}s$
 :)

[b]Problem.[/b]Given triangle $ABC$ semiperimter $s$ ,circumradius $R$, inradius $r$ prove that
$4R+r\ge\sqrt{3}s$.

[b]Solution.[/b] We have $r=4R\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}$ and $s=\frac{a+b+c}{2}=R(\sin{A}+\sin{B}+\sin{C})=4R\cos{\frac{A}{2}}\cos{\frac{B}{2}}\cos{\frac{C}{2}}$.

Therefore we need prove $1+\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}\geq \sqrt{3}cos{\frac{A}{2}}\cos{\frac{B}{2}}\cos{\frac{C}{2}}$. This is easy and we are know it .

[quote="gemath"]Given triangle $ABC$ semiperimter $s$ ,circumradius $R$, inradius $r$ prove that
$\frac{s^{2}}{r}\ge 4R+r$[/quote]
Are you sure? :D

:D  :D Why can't I be sure ? and by the way maybe your solution is too long but it is true!  :D

Because We have $s^{2}\geq 16Rr-5r^{2}$.

Oh that's right it should be $\frac{s^{2}}{3r}\ge r(4R+r)\ge\sqrt{3}r$
and your solution with the inequality $s^{2}\ge 16Rr-5r^{2}$ is right.
However my inequality isn't too trivial it is corollary of the following if you want can try it
$\frac{(xa+yb+zc)^{2}}{4(xy+yz+zx)}\ge r(4R+r),\forall x,y,z>0$
And the inequality $s^{2}\ge 16Rr-5r^{2}$ isn't trivial as $IG^{2}\ge 0$ try it [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=738011#738011]here[/url]
 :)

What is your solution? My solution is long!
[quote="N.T.TUAN"][b]Problem.[/b]Given triangle $ABC$ semiperimter $s$ ,circumradius $R$, inradius $r$ prove that
$4R+r\ge\sqrt{3}s$.[/quote]

My solution is simple, it is base on the equality
$\frac{4R+r}{s}=\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}\ge \sqrt{3}$ :D


So, how did you all do? I thought it was a bit easier than expected. Any other opinions?

It was considerably different than years past- especially the math on the multiple choice.  They didn't have the answers like this $\frac{(1.2)(760)(273)}{(500)(298)}$ anymore; they were just numbers.  I know a lot of people in my class had issues with that.  There was a lot more lab skills and descriptive chem too, compared to previous years.  But I think I did alright anyway.

I bombed it.

I flew over the MC in 30-40 minutes but it was hell double-and triple-checking every single answer for the rest of the 90 min... Oh, wasn't that FR #2 just ugly!! and I never expected to get anything right on #4. (sigh) I loved number 8, though...

I'm just glad it's over. ;)

i dont know why i took so long on the MC.... but i guess i did okay on it...

yeah i saw number 2 and freaked so i chose number 3... which wasnt all that better of a problem. geez, i rushed through number 8 with 4 minutes left.... that whole section was a blur to me. man i hope i at least got some credit on that... lol

i didnt know the math portion of it used to do things like that, inteluser. so to me, it wasn't a surprise. although it was somewhat time consuming >.<

oh well, let's just all hope for 5's!

The Mock AP exam I took boosted my confidence so much...
then today just destroyed all of it.

This was the first time I actually looked at a mandatory FRQ and just DID NOT KNOW how to do it. It was definitely not good for me. But CLASS IS OVER!!!

yeah true. although we have a test over electrochemistry this friday!  :(

what are all of yall doing in chem class now the AP is over? we're gonna make tyedye (sp?) shirts, b/c it is apparently an application of organic chem?? and make food and stuff  :lol:

i would say it was okay,
if only i had 5 more minutes, i would've gotten a 100 on the MC :D (not really!!)
but the test was OK, i missed a couple parts in the free response, but it was good.

Tie-dye (I think it's spelled this way) shirts and food? I wish.

My teacher is insisting that we have a final in June with all the other classes. Plus, she says she's still gotta put some grades in--meaning even more work.. red light!! What's the use? I was hoping she'd be nice and let us play after the entire class bombed the test, but apparently that's not happening.  :(

[quote="pkothari13"]yikes that stinks...  :( in our district, juniors and seniors get exemptions if their semester grades are above an 85 and 80 respectively... our school ends may 26 or 27 so i guess thats why we're not working anymore...[/quote]
wow...that's weired.... :lol: our schools ends at june 26th ;)
we're not gonna have a final, but we still have work to do. like we have labs and other stuff.
but our Chem teacher, has said that if anyof juniors get a 5 on the Ap test, their grade would atomaticaly be an A :)

for me, the MC was OK. it was quite expected since my teacher gave us practice MCs for every test and some more for practice over spring break (over preparation mayb?)
also, we had a policy of doing the net ionic equation from one test every day for the last 3 weeks or so, so the net ionics were super easy for me :)
the other FRQs were meh. i messed up at least one part in each one of them, and completely bombed the lab question (#5) because we just didn't do that many labs in school  :blush:

[quote="pkothari13"]well we start school in august, so we end in may

man, we still have a test over electrochem this friday. i would much rather have the 5 = A policy... our teacher was TERRIBLE this year... she didnt go over about 6-7 of the chapters we need to know for the AP. her exact words upon me asking about her lack of teaching? "oh honey pie! dont worry! we are ALL screwed for the AP! dont act like you're the only one!"
hahaha, how great.  :? but then i was oh so comforted by her subsequent statement... "but i am confident that with MY teaching, all my students will be guaranteed a TWO or higher on that test, so dont worry!"

woo woo.[/quote]
that gave me a good laugh... :rotfl:

hahaha yeah.... it was comforting to know that when i was taking the ap... i knew i wouldnt get a grade of 1.  :) *sarcasm* haha

[quote="pkothari13"] "oh honey pie! dont worry! we are ALL screwed for the AP! dont act like you're the only one!"
hahaha, how great.  :? but then i was oh so comforted by her subsequent statement... "but i am confident that with MY teaching, all my students will be guaranteed a TWO or higher on that test, so dont worry!"

woo woo.[/quote]

It arouses oh so much empathy to hear that someone else had a fabulous teacher for AP Chemistry, too.

It's her first year teaching AP (we're guinea pigs) and I seem to remember what she said at the beginning of the year--"Last year we had 3 5's. Good, but nothing to go home and brag about." Post-test discussion tells me that we'll be lucky to even have one 5 this year... Heh. That's almost on par with the terrible disaster that happened last year in AP Bio at my school--only one 5 and a vast majority of  1-2's. I wish I went to a more science- (and for that matter, math-) oriented school, but there aren't too many options where I live :(

Back on topic. Never thought those oxygen and hydrogen tests in seventh grade (another school) would be so helpful... Took a random guess on nitrogen and said the splint would just go out. 6 wasn't that great (I put 'square pyramidal' instead of 'triangular pyramidal'.. oops) but 8 was lovely. Might even get a perfect score on that one.. which will unfortunately be offset by 1 and 2. I seem to mess up every calculator-allowed test (or section) I take. Takes me too long to punch in the numbers, etc...

Anyway, gotta enjoy these two wonderful months before scores arrive! Woohoo!

yeah it was this lady's first year teaching as well

i guessed the same thing on nitrogen also, lol oh well. we dont even have an AP bio or AP physics course in our school, and we might not have a stats teacher next year too. woo hoo

[quote="frankthetank"][quote="sky9073"]

Back on topic. Never thought those oxygen and hydrogen tests in seventh grade (another school) would be so helpful... Took a random guess on nitrogen and said the splint would just go out. 6 wasn't that great (I put 'square pyramidal' instead of 'triangular pyramidal'.. oops) but 8 was lovely. Might even get a perfect score on that one.. which will unfortunately be offset by 1 and 2. I seem to mess up every calculator-allowed test (or section) I take. Takes me too long to punch in the numbers, etc...

Anyway, gotta enjoy these two wonderful months before scores arrive! Woohoo![/quote]

What the heck is "triangular pyramidal?"  Isn't that tetrahedral?  And I think I did put "square pyramidal" for one of the questions.  I don't remember which one, though.[/quote]
they are basically the same thing, an example of tetrahedral would be $NH_3$

[quote="amirhtlusa"][quote="frankthetank"]What the heck is "triangular pyramidal?"  Isn't that tetrahedral?  And I think I did put "square pyramidal" for one of the questions.  I don't remember which one, though.[/quote]
they are basically the same thing, an example of tetrahedral would be $NH_3$[/quote]

Oops. Wrong word. It's trigonal pyramidal, not triangular.

Trigonal pyramidal is not necessarily tetrahedral.
For example, CH4 is tetrahedral, but NH3 is not (lone pairs exert more repulsive forces.. VSEPR theory). NH3 would be trigonal pyramidal. Check the bond angles: the bond angle for CH4 is 109.5 but for NH3 it's only 107.3.

The compound in question was probably SF4 on number 6. The central S was bonded to 4 F's and has an extra electron pair... Wouldn't that make a PCl5-like structure (trigonal bipyramidal) except missing an atom on one side (and therefore trigonal pyramidal)?

Bleargh.

$Pcl_5$ is simply [b]trigonal BI-pyrimidal[/b] means that there are 3 CL's that are in a plane with angle of 120 from each other. and the other two Cl's are in the space with angle of 180 from each other and 90 from other three Cl's.
yes, $SF_4$ will form a similar shape(the shape is called seasaw).
i thought that was easy... ;)

i put irregular tetrahedral instead of seasaw, but im sure they accept both

does anyone know anywhere on the web, that has the solutions to this years problem??

I believe the rubrics come out in several months (september or so?).

Ag2S as one of the answer for #5

agree?

our physics teacher wrote his answers, so we get to compare. Im sure you can get your chem teacher to do the same

I HATED THAT QUESTION...
i misread sulfide as, sulfate, so i answered the whole thing wrong... :(  but hopefully it was only one part of a question. so it doesn't matter
and yes $Ag_2S$ was the correct answer.

i personally thought this exam was easy...

i knew how to do every single MC except for one

lol, and the question w/ "burning stick... oxygen, nitrogen, and hydrogen" was veryyyyy vague

i put down "hmm.. well in the presence of ONLY hydrogen and nitron (just as you have stated)... the flame will go out... however w/ a little oxygen present, the hydrogen would go BOOM! and the nitrogen would still go out"

lol, the wording was very vague...

i got my results back today (August 11)!!!
i got 5 on chem.
just wondering how did every one do?

Well since I got owned on the last FRQ and didn't study much at all for the exam in general, I got a 3. Yikes.


[b][color=blue]Pregunta 04:[/color]
Sabado 05 de mayo[/b]

En un tablero $2007 \times 2007$ casillas, colorean de negro $P$
casillas de modo que se cumple la siguiente condicion:\\
$\emph{No hay 3 casillas negras que formen un L-trinomio en cualquiera de sus 4 orientaciones.}$

Encuentre el menor valor de $P$ para que al pintar una casilla
mas, ya no se cumpla la condicion dada.

$\text{\LaTeX}{}ed$ by carlosbr

Creo que este problema es bien conocido, vean que habia sido propuesto en un reto de la semana muy viejo, generalizado. Yo hice una solución y creo que en el momento se la mande a pascual, pensaba que la habia pegado ahi pero ahora que encontre el link me doy cuenta que nunca la puse.  La idea es ver que en un tablero de $2\times 2$ debe haber máximo 2 casillas y con esa idea poner cotas y una construcción. Si no me acuerdo mal la construcción es colorear por columnas una si y una no de manera que queden coloreados las casillas de los bordes, que no se pueden colorear mas es simplemente casillas.  Ahi esta el link.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=59768[/url]

Creo que el problema es parecido pero no el mismo, ya que en el del reto de la semana si m=n=2007, la rpta seria $2*2007^{2}/3$, ya que en el de peru la rpta es 2007*669, sim embargo si coloreo las casillas de las filas 2, 4, 6, ....., 2006; cumple la condicion y solo son 2007*1003 casillas.En este link hay una solucion bastante interesante:

http://es.geocities.com/selectivos_peru/selimo2007.pdf

Perdon es que pense que era el menor numero para el que no importara la forma en que se pusieran siempre si se ponia una mas el se formaba una ficha. Porque claramente con $\frac{2007^{2}}{3}$ fichas puedo lograr una configuracion tal que si existan casillas que se pueden llenar de manera que todavia se cumpla la condición (simplemente con la construcción que yo dije alternando una columna si y otra no). Por otro lado creo que tu respuesta al caso particular del reto esta mal. Porque si coloreas $2007 \times 1003$ columnas alternadas funciona y $2007\times 1003 < \frac{2\times 2007^{2}}{3}$  :D   :blush:


inequalities geometry USAMTS AMC USA(J)MO USAMO

hey guys

first of all happy new yr to u all

 :)

now plz post some nice problems for discussion under this very thread

i am waiting for reply from gauravpatil, akashnil.....

so plz reply and guys and post some real nice problems  :wink:  :P  :help:

There are many unsolved problems here
http://www.mathlinks.ro/viewtopic.php?t=233160

i'm unable to grasp them

Well Rijul Saini, most of Apratim's inequality problems are straight forward trigonometric inequalties if u sybstitute them with the well known formulae.
Work on the combi its just logic.

Hello jaswinder.

Please try out the problems in the unsolved sections in geometry and combinatorics and other such topics. The problems are nice and challenging and every day a lot of new problems are posted. Moreover many people from many states have posted their respective INMOTC problem sheets. Please try them.

@ jaswinder what reply are u expecting from me

@gauravpatil

i am waiting for u to post some real nice questions under this very thread

[hide]and of course

[/hide][hide]u should never forget that

[/hide][hide]its must for all of u

[/hide][hide]plz remember that

[/hide][hide]don't post any hints.......... :P  :wink: [/hide]

and what makes u thinkthat i know questions that you will call nice? :?:

[quote="Jaswinder"]@gauravpatil

i am waiting for u to post some real nice questions under this very thread

[hide]and of course[/hide]
[hide]u should never forget that[/hide]
[hide]its must for all of u[/hide]
[hide]plz remember that[/hide]
[hide]don't post any hints.......... :P  :wink: [/hide][/quote]
Hi Jaswinder. Gaurav and others are right. It's no use asking people up for nice problems. This site is not only the Indian forum. In fact Indian forum is more for discussions on some indian things (that is about different indian exams), some humour and stuff etc. than for purely academic discussions on math olympiad. That is primarily because, there is a much bigger and global pool of users in the non-local forums, for example, the [url=http://www.mathlinks.ro/index.php?f=217]Olympiad section[/url]. In there, you need not ask for problems, most of the time there are problems waiting to be solved. :)

[quote="Aravind Srinivas L"]Befuddlers did you get the Pompeiu's Theorem Generalization?
I know that u succeeded in getting it for a squre. What about Regular n gons?
And you too post the USAMTS USAMO problems da.[/quote]

Yep i finished the proof for regular n gons. :lol:

If u want i can post maybe the outline of my proof when i am free(Not in the next 2 months for sure) :rotfl:

coz even im a bit slow in typing and this project is a bit long for typing.

EDIT: The USAMO and USAMTS problems are in the resource section Aravind.

hey guys if sometimes in an olympiad if we are not able to get an inequality

then can we assume its converse to be true and attain a contradiction :P

i wanted to know whether this is acceptable at an olympiad or not

and if yes then will i get the same credit as the person who has proved it true will get as i have just assumed its converse and attained a contradiction...

plz reply and tell me guys :wink:

Yes you can assume the converse. I believe same credit if both solutions are flawless.

But be careful about what exactly converse means. When we are going to prove an ineq of $ a,b,c$, we have to prove that the ineq is true for all triples $ (a,b,c)$ in the range given. But when we assume the contrary, we should assume that there exists some triple $ (a_0,b_0,c_0)$ for which the ineq doesn't hold. And then prove contradiction. But if we assume that for all triples $ (a,b,c)$ the ineq doesn't hold, and then draw contradiction, we'll be wrong.

and if you have a contradiction to some end of a converse you can  backtrace it to get a solution which is direct so there isnt much of a difference


[«Tre gentiluomini giocano insieme: nella prima partita il primo perde, a favore degli
altri due, tanto denaro quanto ne possiede ciascuno di loro. Nella successiva, il secondo gentiluomo perde a favore di ciascuno degli altri due tanto denaro quanto essi già ne possiedono. Da ultimo, nella terza partita, il primo e il secondo guadagnano ciascuno dal terzo gentiluomo tanto denaro quanto ne avevano prima. A questo punto smettono e trovano che ciascuno ha la stessa somma, cioè 24 luigi. Si domanda con quanto denaro ciascuno si sedette a giocare».]

Three man are playing cards, in the first game the first man loses, to credit of the other two men, as much money as each of them own; during the second game, the second man loses to credit of the other men as much money as they already own. In the end, during the third game, the first man and the second one gain each, from the third man, as much money as they owned before. They stop playing and each one own 24 luigi, so how much money each of them owned when he started playing?

I'm not sure if I interpreted this problem correctly, but here is how I understand it:

[hide]Let's work backwards.
At the end, the "scores" are:
$ \$24\quad \$24\quad \$24$

The third man must have given the first two $ \$12$ each, so the scores at the end of the second round were:
$ \$12\quad \$12\quad \$48$

Repeating this process,

$ \$6\quad \$42\quad \$24 \textrm{ (scores at end of first round)}$

$ \$39\quad \$21\quad \$12 \textrm{ (scores at beginning)}$.

Thus,
$ \textrm{First man had }\$39$
$ \textrm{Second man had }\$21$
$ \textrm{Third man had }\$12$[/hide]


If two items from the list $ 4\%$, $ 0.4$, $ \frac{1}{25}$, $ \frac14$, $ \frac{40}{1000}$ are selected at random without replacement, what is the probability that their values are equal? Express your answer as a common fraction.

Put them in decimal form:

0.04, 0.4, 0.04, 0.25, 0.004

There is one match out of 5C2=10 total, so $ \frac{1}{10}$ is the answer.

[quote="AIME15"]Put them in decimal form:

0.04, 0.4, 0.04, 0.25, 0.004

There is one match out of 5C2=10 total, so $ \frac {1}{10}$ is the answer.[/quote]

What?

$ 4\%\equal{}\frac{1}{25}\equal{}\frac{40}{1000}\equal{}\frac{1}{25}$.

Thus, $ \frac{3}{10}$.

[quote="AIME15"]Put them in decimal form:

0.04, 0.4, 0.04, 0.25, [b]0.004[/b]

There is one match out of 5C2=10 total, so $ \frac {1}{10}$ is the answer.[/quote]

40/1000=0.04 :wink:


[b][i][u]a,b and c are réels numbers.Find all value possible of M such that:
a²+b²+c²-ab-bc-ca>=M(a-b)(b-c) [/u][/i][/b]

Since $(a-c)^{2}$ and $(a+c-2b)^{2}$ are nonnegative, $M=-1$ and $M=3$ work. By linear combinations of the two, we can get any $M$ in the interval $[-1,3]$. Finally, we know we can't do better because $a=c=1,\ b=0$ requires $M\geq-1$ and $a=2,\ b=1,\ c=0$ requires $M\leq 3$.


Let $ P(x) \in Z[x]$ and irreducible on $ Q[x]$ .Prove that $ P(x)$($ x$=1;2;3;... )has infinity prime divisors .

i don't think we need that $ f(x)$ is irreducible...

assume first that $ f(0)\equal{}1$. for $ n$ larger than some constant $ N$, $ f(n!)\neq 1$, so that $ f(n!)$ has prime divisors. now note that any prime divisor $ p$ of $ f(n!)$ must be bigger than $ n$; otherwise, $ p|n!\equal{}n!\minus{}0|f(n!)\minus{}f(0)\equal{}f(n!)\minus{}1$, which would imply that $ p|1$. so we can find arbitrarily large primes dividing the values of $ f$. if $ f(0)\neq 0$, we can set $ g(x)\equal{}\frac{f(f(0)x)}{f(0)}$. then $ g(0)\equal{}1$ and the preceding argument shows that arbitrarily large primes divide $ g$. it's then clear that arbitrarily large primes divide $ f$ as well (because $ f(0)$ has finitely many prime divisors).

otherwise, $ f(0)\equal{}0$. but in that case, $ p|f(p)$ for all primes $ p$. so all cases are covered.


http://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem

http://en.wikipedia.org/wiki/Cauchy_sequence

How  to  use  the  Cauchy sequence  to  prove  that     every open cover of     closed interval  $ [a,b]$       has a finite subcover?
thank!

what definition of compact are you using? the usual definition *is* "every open cover has a finite subcover" ...


logarithms
number theory
relatively prime

The number \[ \frac 2{\log_4{2000^6}}+\frac 3{\log_5{2000^6}} \] can be written as $\frac mn$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

yeah i solved this the same way as jesusfreak.
no wonder this was #1. it took [u]even me[/u] less than 2 min. to solve it.  :P

[hide]Because we have two similar logarithms of different bases, it seems logical to convert them to the same base.

$\frac 2{\log_4{2000^6}}+\frac 3{\log_5{2000^6}}$ = $\frac{2}{\frac{6\log2000}{\log4}} + \frac{3}{\frac{6\log2000}{\log5}}$ =$\frac{\log2000}{6\log2000}$ = $\frac{1}{6}$.

Therefore, our answer is $1+6=\boxed{7}$.[/hide]

[hide=An alternate solution]First, we note that $\dfrac{2}{\log_{4}2000^6}=\dfrac{\log_{4}16}{\log_{4}2000^6}$, so the first term becomes $\log_{2000^{6}}16$.

Similarly, the second term becomes $\log_{2000^6}125$, so our answer is $\log_{2000^6}2000$, or basically $\dfrac{1}{6}$. Thus, $m + n = \boxed{007}$. [/hide]


[quote="Johan Gunardi"]What is the pattern for the denominator? I can't find it in the [url=http://www.research.att.com/~njas/sequences/Seis.html]sequence encyclopedia[/url].[/quote]

The pattern of the sequence is $ \frac {n^3}{(\frac {{n}\times(n \plus{} 1)}{2})^2\times(\frac {(n \minus{} 1)\times n}{2})^2}$

So from $ n \equal{} 2$, we get the first number
The $ 100^{th}$ number is $ \frac {101^3}{(\frac {101\times102}{2})^2\times(\frac {100\times101}{2})^2}$

I also couldn't find the pattern.. :P
I get the pattern from the solution

EDIT: oops, the $ 100^{th}$ should be like hsiljak's
I've edited this post :blush:

How about telescoping?

[hide]It's easy to find $ \frac {4}{(n \minus{} 1)^2} \plus{} \frac {16}{n} \minus{} \frac {4}{(n \plus{} 1)^2} \minus{} \frac {8}{n \minus{} 1} \minus{} \frac {8}{n \plus{} 1}$ after you simplify your general term to $ \frac {16}{n(n^2 \minus{} 1)^2}$.

Now, most of the terms cancel out. Sum what's left and you'll get

[hide]$ \frac {k^4 \plus{} 2k^3 \plus{} k^2 \minus{} 4}{k^2(k \plus{} 1)^2}$ where k is the last term (in your case, $ k \equal{} 100$). Of course, I might have made a silly mistake somewhere in the calculations.

BTW, answer should be $ \frac {25 502 499}{25 502 500}$.[/hide][/hide]

EDIT: I've noticed that n=100 doesn't generate the 100th, but 99th number. So the number above isn't correct, the correct one is

[hide]$ \frac {26 532 800}{26 532 801}$[/hide]

[quote="IW@IT"]Evaluate
$ \frac {2^3}{3^2} \plus{} \frac {3^3}{18^2} \plus{} \frac {4^3}{60^2} \plus{} \frac {5^3}{600^2} \plus{} \dots \plus{} n$

$ n$ is the $ 100^{th}$ number from the sequence

The pattern of the sequence is $ \frac {n^3}{(\frac {{n}\times(n \plus{} 1)}{2})^2\times(\frac {(n \minus{} 1)\times n}{2})^2}$

I also couldn't find the pattern.. :P
I get the pattern from the solution[/quote]

I think, $ 0$, $ 3$, $ 18$, $ 60$, $ \color{red}{150}$, $ 315$, $ 588$, $ 1008$, $ 1620$, $ 2475$, $ \cdots$

Oh, thx for the correction
I didn't check whether the pattern is suitable with the problem or not...
I just copied that problem from the book. :blush:

So, it should be $ 150^2$, not $ 600^2$


Incredibly, the following has a completely elementary solution: the least primitive root mod $p$ is $O(p^{1/2+\epsilon})$ for all $\epsilon>0$.

the only similar result I've seen is $g_{p}>C log p$ for a constant C and infinitely many primes p (it doesn't have an elementary solution)

I'm guessing that OP means [quote]For every $\epsilon>0$ there is an $N$ such that for all $p>N$, the least primitive root modulo $p$ is at most $p^{1/2+\epsilon}$[/quote]


Q: What's the approximate AoPS user distribution? How many live in the West, and the East, and international? Thanks, I wasn't sure if this was too trivial for the Questions and Suggestions Forum.


function
real analysis
real analysis unsolved

try to consider derivetive of this function.
 then you can see that
when
$x\in (0,c)$ then $f'(x)>0$ and when $x \in (c,1)$ then $f'(x)<0$
...
after you see that the function his min value achieved  in  point $0,1$
if I have not  mistake it's not easy inequality.

no, it is not true , with this conditions you said $f$ has maximum on $[0,1]$ not minimum. :)

Yes I agree with you : the function achieved his max on $[0,1]$, but I said
the function  take min in point $0$ and $1$. you don't understood what I said  :P

[quote="martin1985"]problem: m,n-naturals, x in[0,1]. prove that (1-x^n)^m +(1-(1-x)^m)^n >=1[/quote]
This problem posted before, very old. I remember that in that topic has a solution use probably. I try find help you , But i cann't. If grobber, harazi or Myth see this topic then they will help you  :wink:


Let $ a,b,c$ be non-negative real numbers whose $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Find the maximum value of
$ P\equal{}ab^2\plus{}bc^2\plus{}ca^2\minus{}abc$

[quote="ricardokaka"]Let $ a,b,c$ be non-negative real numbers whose $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Find the maximum value of
$ P \equal{} ab^2 \plus{} bc^2 \plus{} ca^2 \minus{} abc$[/quote]
Assume $ (a\minus{}b)(a\minus{}c) \le 0$, then
$ P\equal{} ab^2 \plus{} bc^2 \plus{} ca^2 \minus{} abc \le ab^2\plus{}abc\plus{}ac^2\minus{}abc \equal{}a(b^2\plus{}c^2)\equal{}a(3\minus{}a^2) \le 2$.


integration
calculus
probability and stats

given: $ f_{X,Y}(x,y) \equal{} 8xy$ when $ 0\le x \le y \le 1$

The solutions say:

$ f_x(t) \equal{} \int_t^1 f(x,y)\,dy.$

$ f_y(t) \equal{} \int_0^t f(x,y)\,dx.$

Can someone please explain the limits of the integration in each case? Why is it between t and 1 in the x case, and between 0 and t in the y case?

When I have this kind of problems how do I figure out what the limits of the integration should be ?

Thanks.

Suppose instead we had written

$ f_{X,Y}(x,y) \equal{} \left\{ \begin{array}{ll} 8xy & 0 \le x \le y \le 1 \\ 0 & \textrm{otherwise} \\ \end{array} \right.$

Then we would have, for instance

$ f_X(t) \equal{} \int_{y\equal{}0}^1 f_{X,Y}(t,y) \, dy$.

But how does this simplify according to the above definition of the joint density?  The integrand is zero whenever $ y < t$.  Thus it is natural to partition the interval of integration into $ [0,t) \cup [t,1]$ and discard the first part, leaving you with the claimed expression.  A similar principle applies to the marginal density of Y.


analytic geometry
conics
parabola
calculus

For a given positive real number $k$, let $S$ be the set of points in the coordinate plane equidistant from $(0,k)$ and the line $y=-k$. A circle intersects the graph of $S$ in only one point, the origin, and the rest of the circle is above the graph of $S$. In terms of $k$, what is the largest radius this circle could have?

I found that the max radius is $2k$. I found that S is the parabola $y=\frac{x^2}{4k}$ (using algebra). I then used calculus, the max radius here is the radius of curvature at $x=0$. The curvature at a point is \[ \frac{d(\arctan{\frac{dy}{dx}})}{ds}  \] where s is the arclength. I then calculated this, evaluated it for $x=0$, and found the reciprocal for the radius of curvature, which was $2k$.

I'm guessing there's a non calculus method as one's not supposed to post calculus problems in here, but that's how I did it...

Yes, it would be better to have a solution that doesn't use calculus.

[hide="hint"]Once you figure out the graph of $S$, note that the origin must be the lowest point on the circle. (Why?)[/hide]


Given $y_{1}$ and $y_{2}$ are solutions of:
$y''+6y'+5y = x$ then for any constants $c_{1}$ and $c_{2}$, $c_{1}y_{1}+c_{2}y_{2}$ is also a solution.

Is this true?
I was thinking that it is by the theorems of the Superpostion principles of homogeneous and non-homogeneous DE's.

Well, let's check.  $y_{1}= \frac{5x-6}{25}$ is a solution, and so is $y_{2}= \frac{5x-6}{25}$, and $c_{1}=-1$ and $c_{2}= 1$ are constants.  Is $-1\cdot y_{1}+1\cdot y_{2}= 0$ a solution?

The obvious way to test this (and find constraints on the $y_{i}$ or $c_{i}$ to actually make some version of it true) would be to plug $c_{1}y_{1}+c_{2}y_{2}$ into the general expression and solve the result, using the fact that $y_{1}$ and $y_{2}$ are assumed to be solutions.

:Edit: Um, right, I can differentiate just fine, thanks.

Actually, $y_{1}=\frac{x}{5}$ is not a solution.

But JBL made a good point: plug $c_{1}y_{1}+c_{2}y_{2}$ into the equation and you'll get $(c_{1}+c_{2})x$ on the righthand side. Which is not what you want unless $c_{1}+c_{2}=1$.