calculus
integration
calculus computations

Hello ;) !!

$ Let$  $ t\equal{}arctan(x)$

than:

$ H\equal{} \int \frac{e^{\alpha arctan(x)}}{(1\plus{}x^2) \sqrt{1\plus{}x^2}}dx \equal{} \int cos(t)e^{\alpha t}dt$

1) $ if$ $ \alpha \equal{} 0$ than $ H\equal{}\int cos(t)dt \equal{}sin(t)\plus{}cte \equal{}sin(arctan(x))\plus{}Cte\equal{} \frac{x}{\sqrt{1\plus{}x^2}}\plus{}Cte$

2) $ if$ $ \alpha \neq 0$:
by double integration by partie we find:

$ \int cos(t)e^{\alpha t}dt \equal{} e^{\alpha t} \frac{sin(t)\plus{} \alpha cos(t)}{1 \plus{} \alpha^2}\plus{}Cte$

than:

$ \int \frac{e^{\alpha arctan(x)}}{(1\plus{}x^2)^\frac{3}{2}}dx \equal{} e^{\alpha arctan(x)} \frac{x\plus{}\alpha}{(1\plus{}\alpha^2) \sqrt{1\plus{}x^2}} \plus{} Cte$
_____________________________________________________
lahoucine

Wrong. Are you saying that $ \sqrt{A^2}\equal{}A$ for $ A\in\mathbb{R}?$

[quote="kunny"]Wrong. Are you saying that $ \sqrt {A^2} \equal{} A$ for $ A\in\mathbb{R}?$[/quote]

NO, but we put that $ t\equal{}arctan(x)$ than $ |t| \leq \frac{\pi}{2}$ that's $ |cos(t)|\equal{}cos(t) \geq 0$

_______________________________________________
lahoucine


inequalities
function
inequalities unsolved

Given $n$ real numbers $a_1$, $a_2$ $\ldots$ $a_n$. ($n\geq 1$).  Prove that there exists real numbers $b_1$, $b_2$ $\ldots$ $b_n$ satisfying:
(a) For any $1 \leq i \leq n$, $a_i - b_i$ is a positive integer.

(b)$\sum_{1 \leq i < j \leq n} (b_i - b_j)^2 \leq \frac{n^2-1}{12}$

Lemma 1:

Assume that we have $a_1\geq a_2\geq \ldots a_n\geq 0, 0 \leq b_1\leq b_2\leq \ldots b_n$ with $a_1+a_2+\ldots+a_k\leq b_k$ for $1\leq k \leq n$. Then $a_1^2+\ldots+a_n^2\leq b_1^2+(b_2-b_1)^2+\ldots+(b_n-b_{n-1})^2$.

Proof: if suffices to apply Karamata for the convex and [i] increasing [/i] function $x^2$.

Lemma 2: The sum of all numbers less than $n$ and of the same parity with $n-1$ is $\frac{n(n^2-1)}6$.

This is proven by induction.

Let $S=\sum_{1\leq i<j\leq n} (b_i-b_j)^2=n(b_1^2+\ldots+b_n^2)-(b_1+\ldots+b_n)^2$

Assume that $b_1,b_2,\ldots,b_n$ realize the minimum. It's clear that we can replace all $b_i$ by $b_i-t$ invariating $S$, so we can assume that $b_1+\ldots+b_n=0$. Now let $k$ of the numbers $b_i$ be positive and $l=n-k$ of them be negative, with $k \geq \frac n2$. Assume that we subtract $1$ from  $p$ of the positive numbers. $S$ then will change by $np-p^2-2ns$ where $s$ is the sum of the $p$ numbers. Since $S$ is minimal, this value is non-negative. Therefore we deduce that the sum of any $p$ positive numbers is at most $\frac{p(n-p)}{2}$. By letting $p=k$ we deduce that the sum of the all positve numbers is at most $\frac{k(n-k)}2$. So the sum of any $p$ positive numbers is at most ${min \{\frac{k(n-k)}2, \frac{p(n-p)}2}\}$. Now applying Lemma 1 for these numbers we get that the sum of squares of the positive numbers is at most:

$\sum_{i=1}^{n-k} (\frac{i(n-i)}{2n}-\frac{(i-1)(n-i+1)}2n)^2=\sum_{i=1}^{n-k} \frac{(n-2i+1)^2}{4n^2} \leq \frac 1{4n^2}\sum i^2)$, where the sum is taken over all $i$ less than $n$ of the same parity with $n-1$. Applying Lemma 2, this sum is at most $\frac{n(n^2-1)}{24n^2}=\frac{n(n^2-1)}{24n}$.

Reasoning analogously for the negative $b_i$ (just now add $1$ instead of substracting), we deduce that the sum of squares of the negative numbers is also at most $\frac{n(n^2-1)}{24n}$.

Hence $S=n(b_1^2+\ldots+b_n^2)\leq n(\frac{n(n^2-1)}{24n}+\frac{n(n^2-1)}{24n})=\frac{n^2-1}{12}$. QED

Happy Easter!

why can you $b_1+b_2+\cdots+b_n=0$
that means $a_1+a_2+\cdots+a_n$ is integer. :?:

[quote="shobber"]Given $ n$ real numbers $ a_1$, $ a_2$ $ \ldots$ $ a_n$. ($ n\geq 1$).  Prove that there exists real numbers $ b_1$, $ b_2$ $ \ldots$ $ b_n$ satisfying:
(a) For any $ 1 \leq i \leq n$, $ a_i \minus{} b_i$ is a positive integer.

(b)$ \sum_{1 \leq i < j \leq n} (b_i \minus{} b_j)^2 \leq \frac {n^2 \minus{} 1}{12}$[/quote]
It is obvious that if there exists reals $ b_i$ such that $ a_i \minus{} b_i$ is an integer, then there exists reals $ b_i$ such that $ a_i \minus{} b_i$ is a positive integer with the same sum of squares of difference - thus not changing (b). So replace (a) with (c) For any $ 1 \leq i \leq n$, $ a_i \minus{} b_i$ is an integer.

Let a sequence $ \underline{c} \equal{} (c_1,c_2,..,c_n)$ be "valid" if there exists $ b_1,b_2,...,b_n$ such that $ c_i \equal{} nb_i \minus{} \sum_{i \equal{} 1}^n b_i$, and $ b_i \minus{} a_i$ is an integer.
Define the transformation $ \Delta_i$ as:
$ \underline{c} \equal{} (c_1,c_2,c_3, \cdots, c_n)$ $ \to (c_1 \minus{} 1,c_2 \minus{} 2, \cdots, c_{i \minus{} 1} \minus{} 1, c_i \plus{} n \minus{} 1, c_{i \plus{} 1} \minus{} 1, \cdots, c_n \minus{} 1) \equal{} \underline{c'}$
Then $ \underline{c'}$ is obviously a valid sequence if $ \underline{c}$ is.

Let $ \underline{d} \equal{} (d_1,..,d_n), d_i \equal{} na_i \minus{} \sum_{j \equal{} 1}^n a_j$.
Define $ f(\underline{x}) \equal{} \sum_{i \equal{} 1}^n x_i^2$, and let $ f$ be defined on all sequences $ \underline{x}$ which can be obtained from $ \underline{d}$ with a finite number of $ \Delta_i$ operations.

First I'll show that $ f(\underline{x})$ has a global minimum. Let $ \underline{y} \equal{} (y_1, \cdots, y_n)$ be an permutation of $ \underline{x}$ such that $ y_1 \le y_2 \le \cdots \le y_n$. If $ y_n \minus{} y_1 > n^2$ there must exists an $ k$ such that $ y_{k \plus{} 1} \minus{} y_k > n$. Then consider $ \underline{x'}$, which is $ \underline{x}$ after the operations: $ \Delta_1\Delta_2\cdots\Delta_k$.
As $ \sum_{i \equal{} 1}^n x_i \equal{} 0$ we find that: $ f(\underline{x}) \equal{} \sum_{i \equal{} 1}^n x_i^2 \equal{} \frac {1}{n} \sum_{1 \le i < j \le n} (x_i \minus{} x_j)^2$.
From that and the fact that $ \underline{x'}$ is a "contraction" of $ \underline{x}$ it is easy to see that $ f(\underline{x'}) < f(\underline{x})$. Furtermore the difference between the biggest and smallest element in the two sequences $ \underline{x}, \underline{x'}$ differ by exactly $ n$. So we can form a sequence where we can get from $ \underline{x_i}$ to $ \underline{x_{i \plus{} 1}}$ with the operations $ \Delta_1 \cdots \Delta_{k_i}$ for some $ k_i$, such that the difference between the greatest and smallest element in $ \underline{x_n}$ is at most $ n^2$:
$ \underline{x} \equal{} \underline{x_0} \to \underline{x_1} \to \cdots \to \underline{x_n} \equal{} \underline{x''}$
such that the difference between the biggest and the smallest element of $ \underline{x''}$.
Since $ \underline{x''} \equal{} (x''_1, \cdots, x''_n)$. Since $ \sum_{i \equal{} 1}^n x''_i \equal{} 0$, we see $ |x''_i| \le n^2$. And since the fractional part of $ x''_i$ is given by $ a_i$, $ \underline{x''}$ belongs to a finite set. There is clearly a global minimum in this finite set, and therefore $ f(\underline{x})$ has a global minimum.

Let $ \underline{m}$ be the global minimum of $ f$. I'll prove that $ f(\underline{m}) \le \frac {n(n^2 \minus{} 1)}{12}$.
$ \underline{m} \equal{} (m_1,m_2,\cdots,m_n)$
Then $ \underline{m'} \equal{} \Delta_1\Delta_2\cdots\Delta_k\underline{m} \equal{}$
$ (m_1 \plus{} (n \minus{} k),m_2 \plus{} (n \minus{} k), \cdots, m_k \plus{} (n \minus{} k), m_{k \plus{} 1} \minus{} k, m_{k \plus{} 2} \minus{} k, \cdots, m_n \minus{} k)$

Then $ f(\underline{m'}) \minus{} f(\underline{m}) \equal{}$
$ \sum_{i \equal{} 1}^k 2(n \minus{} k)m_i \plus{} (n \minus{} k)^2 \plus{} \sum_{i \equal{} k \plus{} 1}^n \minus{} 2km_i \plus{} k^2 \equal{}$
$ 2n\sum_{i \equal{} 1}^k m_i \plus{} k(n \minus{} k)n \ge 0 \iff$
$ \sum_{i \equal{} 1}^k m_i \ge \minus{} \frac {k(n \minus{} k)}{2} \iff$
$ \sum_{i \equal{} k \plus{} 1}^n m_i \le \frac {k(n \minus{} k)}{2} \equal{} \frac {(n \minus{} k)(n \minus{} (n \minus{} k))}{2}$.
Since we could have changed $ \Delta_1\cdots\Delta_k$ by $ \Delta_{i_1}\cdots\Delta_{i_k}$ for any abritary different indices $ i_1, \cdots, i_k$, we find:
$ \left | \sum_{j \equal{} 1}^k m_{i_j} \right | \le \frac {k(n \minus{} k)}{2}$
Consider the sequence $ \underline{M} \equal{} (M_1, \cdots, M_n), M_i \equal{} \frac {n \plus{} 1}{2} \minus{} \frac {i}{n}$.
Then $ \sum_{i \equal{} 1}^k M_i \equal{} \frac {k(n \plus{} 1)}{2} \minus{} \frac {k(k \plus{} 1)}{2} \equal{} \frac {k(n \minus{} k)}{2}$, and therefore $ (M_1,M_2, \cdots, M_n) \succ (m_1,m_2, \cdots, m_n)$. (Wlog $ \underline{m}$ is sorted decreasingly)
Therefore by Karamata and the fact that $ x \to x^2$ is convex:
$ f(\underline{m}) \le \sum_{i \equal{} 1}^n M_i^2 \equal{} \frac {n(n^2 \minus{} 1)}{12}$!

Since $ \underline{m}$ is obtained by $ \Delta_i$ transformations from $ \underline{d}$, which is clearly valid, $ \underline{m}$ is valid. Hence there exists reals $ b_1,b_2,..,b_n$ such that $ b_i \minus{} a_i$ is an integer and:

Let $ m_i \equal{} nb_i \minus{} \sum_{i \equal{} 1}^n b_i$. Then:
$ n^2\sum_{1 \le i < j \le n} (b_i \minus{} b_j)^2 \equal{} \sum_{1 \le i < j \le n} (m_i \minus{} m_j)^2 \equal{}$
$ n\sum_{i \equal{} 1}^n m_i^2 \minus{} \left ( \sum_{i \equal{} 1}^n m_i^2 \plus{} 2\sum_{1 \le i < j \le n} m_im_j \right ) \equal{} n \sum_{i \equal{} 1}^n m_i^2 \minus{} \left ( \sum_{i \equal{} 1}^n m_i \right )^2 \equal{}$
$ n \sum_{i \equal{} 1}^n m_i^2 \le n\frac {n(n^2 \minus{} 1)}{12} \equal{} n^2 \frac {n^2 \minus{} 1}{12} \iff$.
$ \sum_{1 \le i < j \le n} (b_i \minus{} b_j)^2 \le \frac {n^2 \minus{} 1}{12}$

So $ b_1,b_2,..,b_n$ fullfilles (b) and (c). Hence for some $ k \in \mathbb{Z}$ $ b_1 \plus{} k,b_2 \plus{} k, ..., b_n \plus{} k$ satisfies both (a) and (b), which is the conclusion :)


For the first one I get tangent 4y+x-6=0 and normal y-4x+7=0. Can you show your working so we can see what you did wrong?

Repmax, I count at least a couple of errors, either typos or arithmetic, in your solution.  Look it over.

Hmm.. I think I see the problem for the first one  :shock: Maybe you miscopied the question, as (2,1) isn't on the circle..

Did it mean "through," not "at" the point (2, 1)?  That might work.  Either way, watch for simple copying errors, Repmax . . . you have at least a couple of switched signs and so on between your write-up of the question and solution.


geometry
ratio
analytic geometry
circumcircle
graphing lines
slope
geometric transformation

[color=darkblue]Let $ A, B$ are fixeds point and $ M$ is a change point such that $ MA \equal{} 3 MB$. Let $ H$ is orthocenter of $ MAB$ triangle. Find locus of point $ H$.[/color]

I simulated the figure and the shape of the locus of $ H$ seems like a cubic. See my figure below
I did not use exactly the same ratio  but it is very approximate  The curve does not change so much because of that.
Move $ M$ on the Apolonius circle , forget about the notation in the figure, just see the trajectory of $ H$ . Is it interesting ?

[geogebra]ae1e86f862223b970301fc9b413d9a795e094567[/geogebra]

[color=darkblue]Thank you! I find locus of point $ M$ by use Geometer's Sketchpad as your post. But I want understand that, why we get it so.[/color]

The analytic solution is pretty straight forward, but I have no idea what the geometric structure is other than it looks like a chromosome lol.
[hide]
WLOG, give $ A$ and $ B$ coordinates of $ (1,0)$ and $ (9,0)$.  Then the apollonius circle, the locus of $ M$, is described by the equation $ x^2 \plus{} y^2 \equal{} 9$.  Thus, if $ M \equal{} (p,q)$, then $ p^2 \plus{} q^2 \equal{} 9$.  Let $ O$ be circumcenter, $ G$ the centroid and $ H$ the orthocenter.  Obviously $ G \equal{} 1/3(A \plus{} B \plus{} M) \equal{} 1/3(10 \plus{} p,q)$.
Let $ F \equal{} 1/2(A \plus{} M) \equal{} 1/2(p \plus{} 1,q)$.  Then the slope of line $ FO$ is $ \frac {1 \minus{} p}{q}$, and following a simple calculation involving using $ p^2 \plus{} q^2 \equal{} 9$ we get that the equation of the line $ FO$ is $ y \equal{} ((1 \minus{} p)/q)x \plus{} 4/q$.  Obviously $ O$ lies on perp. bisector of $ AB$, or $ x \equal{} 5$, so its y-coordinate is $ 5((1 \minus{} p)/q) \plus{} 4/q \equal{} \frac {9 \minus{} 5p}q$.
Since a homothety of -1/2 centered at G sends H to O, $ H \equal{} \minus{} 2O \plus{} 3G \equal{} (p,\frac { (p \minus{} 9)(p \minus{} 1)}{\pm\sqrt {9 \minus{} p^2}})$.  Therefore the equation of the locus is
\[ y \equal{} \frac { (x \minus{} 9)(x \minus{} 1)}{\pm\sqrt {9 \minus{} x^2}}\]
for $ x\in[ \minus{} 3,3]$.[/hide]


Let an be a sequence of natural numbers s.t
a_k<a_(k+1)<a_k+2001 for every k natural.
Prove that the set of prime divisors of the numbers ak is infinite.

Suppose that set is finite and it consists of p_1,...p_r. Then we obviously have  \sum (n \geq 1) 1/a_n \leq  \sum (k_1,...,k_r \geq 0) 1/[(p_1)^k_1*...*(p_r)^k_r)]=p_1*...*p_r/(p_1-1)...(p_r-1). Thus  \sum 1/a_n<infinity. This is a contradiction, since a_n \leq a_1+2001(n-1) and so  \sum 1/a_n \geq  \sum 1/(a_1+2001(n-1))=infinity.


Given that $ 5^{x\plus{}1}\equal{}30$, what is the value of $ 5^{3x\plus{}1}$?

$ 5^{3x+1}=5^{3x+3}: 5^2=1080$
The ratio sign is the same thing as division, and 3x+3-2 is equal to 3x+1.


Could you help me with this?
Find the solution of this diferential equation
$ (x\minus{}y)d_x \plus{}(x\plus{}2y)d_y\equal{}0$
Thanks for your time!

hello, writing your equation in the form $ \frac {dy}{dx} = \frac {y - x}{x + 2y} = \frac {\frac {y}{x} - 1}{1 + \frac {2y}{x}}$ and by substituting $ y = u\cdot x$ you will get $ u^'x + u = \frac {u - 1}{1 + 2u}$.
I hope this is helpfuly.
Sonnhard.


Ok, let's give other LMO 1965/1984 problems, but those I know how to solve  :D

There are $n$ glasses each big enough to hold all the water. Initially, all glasses contain the same amount of water. It is allowed to pour from any glass to any other glass as much water as in the second glass. For which $n$ is it possible to collect all water into one glass?

Pierre.

As Myth said : 'Trust, but check...'  :P

Pierre.

If you assume each glass has capacity $n$ and initially, each glass contains a quantity of $1$, then in any given moment, each glass can contain only a power of $2$, so, in the end, when a single glass contains $n$, we find that $n$ must be a power of $2$.

Conversely, it's really easy to put everything in a single glass when $n=2^k$.

[quote="pbornsztein"]As Myth said : 'Trust, but check...'  :P [/quote]
 :P  :P  :P


$A,B\in M_{3}(Z)$ such that $det(A)=det(B)=0, \; AB=BA$
Let $q\in N-(0)$ fixed

Prove that there exist $m,n$ integers such that $det(A^q+B^q)=m^q +n^q$

Cute problem :).

Write $x^q+1$ as $(x-x_1)(x-x_2)\ldots (x-x_q)$. We thus have $\det(A^q+B^q)=\det(A-x_1B)\det(A-x_2B)\ldots\det(A-x_qB)$. Put $P(x)=\det(A-xB)$. Now, $P(x)$ has second degree (because the dominant coefficient is $0$), and $P(0)=0$. It means that $P(x)=x(-ax+b)$ for some integers $a,b$. We thus get $\det(A^q+B^q)=\prod_{i=1}^q x_i\cdot\prod_{i=1}^q (-ax_i+b)=\prod_{i=1}^q x_i\cdot (a^q+b^q)$, and it's clear from here (if $q$ is even, then there's nothing to do, and if it's odd, we replace $a,b$ with $-a,-b$ respectively).

[quote="grobber"]Cute problem :).

[/quote]

Thank you


geometry
3D geometry
algebra
factorization
sum of cubes
number theory unsolved
number theory

Suppose that the sum of cubes of three consecutive positive integers is a perfect cube.Prove among the there integers,the middle one is divisible by 4

So  we have
$(a-1)^{3}+a^{3}+(a+1)^{3}=b^{3}$
$3a^{3}+6a=b^{3}$, 3|b => b=3c,
$3.a(a^{2}+2)=27c^{3}$,
(1) $a(a^{2}+2)=9c^{3}$.

Let a be odd. Then a and (a^2+2) are coprime. So we have
$a=t^{3}$,  $a^{2}+2=9u^{3}$ or
$a=9t^{3}$,  $a^{2}+2=u^{3}$.
This possibilities lead to equatons
$t^{6}+2=9u^{3}$ and
$81t^{6}+2=u^{3}$.
Both are contradictions mod 9, because cubic residues mod 9 ar 0,1 and -1.

Let a be even. Then RHS in (1) is even and therefore divisible by 8. Because $a^{2}+2\equiv 2 (mod 4)$ is not divisible by 4, a must be divisible by 4, QED.

Do you know a nice solution of the equation? I think only
$3^{3}+4^{3}+5^{3}=6^{3}$ works.


[hide]If Rocky thows a 6, Bullwinkle must throw a 6, and any number 1-6 with his other die. This repeats for all the other numbers. $\frac 16 \cdot \frac 16 \cdot \frac 16 \cdot  21 \cdot 2 = \frac {14}{72}$.[/hide]

i agree with tarquin's solution, but not completely, i don't know why he multiplied the whole thing by $2$.
i think the answer is:
 [hide="answer"] $\frac{7}{72}$[/hide]

er...not sure that i am right, but i agree with tarquin, he multiplies by 2 because of Bullwinkle having two dice, so he can get the same number as Rocky on either his first or second dice

but i don't think that order matters... ;)
if it does, then i agree too...

[quote="MCrawford"]Rocky and Bullwinkle are playing Risk.  Rocky rolls one six-sided die, while Bullwinkle rolls two of them.  What is the probability that Rocky's roll is as high as Bullwinkle's largest number?[/quote]

If Rocky throws a 6, then it is guarunteed that his roll is as high as Bullwinkle's largest number. (1/6)

If Rocky throws a 5, then Bullwinkle must throw a 1-5 for both of his rolls. (1/6*5/6*5/6)

(1/6*4/6*4/6)
(1/6*3/6*3/6)
(1/6*2/6*2/6)
(1/6*1/6*1/6)

The sum should account for the probability of Rocky number being as high as Bullwinkle's highest number.

I'm assuming that order does matter.

So $91/216$

[quote="Treething"][hide="Solution"]
(1 + 4 + 9 + 16 + 25 + 36)/216

[b]91/216[/b]

Funny, I got a very different answer...
[/hide][/quote]

Yea, and if you visualize a 6 by 6 grid, then this is another way to get to the solution. Very nice Treething.

[quote="bob123"][hide]If Rocky throws a 1, then Bullwinkle must throw 2 1s. If Rockey throws a 2, then Bullwinkle must throw either 1,2, 2,1, or 2,2 (order matters, I think/hope). Then you do (1+3+5+7+9+11)/216=1/6... I'm probably wrong, but that's ok :D[/hide][/quote]

The reason we are all getting such different answers is that some of us are assuming MCrawford to mean [i]at least as large as[/i] and some of use are assuming the MCrawford to mean [i]exactly as large as[/i]. Treething and I assumed the first case, and [i]I think[/i] tarquin (and I know bob123) assumed the second case.

I don't quite understand tarquin's solution. Could you explain it?


geometry
circumcircle
incenter
angle bisector
geometry unsolved

a)Given a triangle ABC,consider the altitude and the angle bisector at each vertex. Let $P_{A}$ be the intersection of the altitude from B and the bisector at C, and $Q_{A}$ the intersection of the bisector at B and the altitude at C. These determine a line $P_{A}Q_{A}$.The line $P_{B}Q_{B},P_{C}Q_{C}$ are analogously defined. Show that these three lines are concurrent at a point on the line joining the circumcenter and the incenter of triangle ABC.Characterize this point more precisely
b)Given a triangle ABC,consider the median and the angle bisector at each vertex. Let $P_{A}$ be the intersection of the median from B and the bisector at C, and $Q_{A}$ the intersection of the bisector at B and the median at C. These determine a line $P_{A}Q_{A}$.The line $P_{B}Q_{B},P_{C}Q_{C}$ are analogously defined. Show that these three lines are concurrent at a point on the line joining the circumcenter and the incenter of triangle ABC.Characterize this point more precisely

Hi $James$.
See my own problem this[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=67271]www.mathlinks.ro/Forum/viewtopic.php?t=67271[/url] ;)


Find the limit of the sequence $ (x_n)_{n \ge 1}$ where $ x_n\equal{}\sum_{k\equal{}1}^{n}\frac{1}{n\plus{}\sqrt[3]{(k\plus{}1)^2(k^2\plus{}1)^2}}, \forall n \ge 1.$

note that $ ((k\plus{}1)^{2}(k^{2}\plus{}1)^{2})^{1/3}\equal{}k^{2} \plus{} O(k)$
thus
$ \sum \frac{1}{n\plus{}k^{2}\plus{}O(k)} \equal{} \sum \left( \frac{1}{n\plus{}k^{2}} \plus{} O(\frac{1}{n^{2}})  \right) \equal{} \sum \frac{1}{n\plus{}k^{2}} \plus{} O(\frac{1}{n})$
thus there is no big difference between your sum and this one :
$ \sum \frac{1}{n\plus{}k^{2}}$ and this one you can calculate yourself.


Let $ x$ be a real number such that:  $ \tan^5x \plus{} \tan x \plus{} 1 \equal{} 0$,

calculate:  $ \tan x \plus{} \csc^2x$.

Greetings  :D

$ \tan^5x \plus{} \tan x \plus{} 1 \equal{} (\tan^2 x \plus{} \tan x \plus{} 1) (\tan^3 x \minus{} \tan^2 x \plus{} 1) \equal{} 0$

$ \Rightarrow \tan^3 x \minus{} \tan^2 x \plus{} 1 \equal{} 0 \ (\because \ \tan^2x \plus{} \tan x \plus{} 1 \ne 0 \ \forall \ x \in \mathbb{R})$

$ \therefore \tan x \plus{} \csc^2 x \equal{} \tan x \plus{} \frac {1 \plus{} \tan^2x}{\tan^2 x} \equal{} \frac {\tan^3 x \plus{} \tan^2 x \plus{} 1}{\tan^2x}$

$ \equal{} \frac {2 \tan^2 x}{\tan^2 x} \equal{} 2$

Very nice thanks hsbhatt.

Greetings  :D

This problem comes from a current olympiad in Peru, this is not the time to post the problems


is there any restriction for $x,y$ ?because there seems to have many answer i.e $x=y=0$

Yes, I think that there are infinitely many solutions. If we solve $3\sin^{2}{x}+2\sin^{2}{y}=2\sin{x}$ as a quadratic in terms of $sin{x}$, we get:
$sin{x}=\frac{1+-\sqrt{1-6cos^{2}{y}}}{3}$
And then it is easy to verify that $cos^{2}{x}+cos^{2}{y}$ depends only on $cos{y}$ and can take infinitely many values.


I am used to typesetting presentation slides using the beamer class. One thing that I like about this class is the way that it redifines the amsthm environments to be displayed in colored boxes. I'd like to figure out how to reproduce this effect with other document classes, but I haven't been successful in my efforts. Does anybody know how it would be done? Thanks.

The only way to find out how it's done (short of emailing the author) is to closely examine the files. Meanwhile, would \colorbox or \fcolorbox with the color package be a useful alternative? See page 6 of [url=ftp://tug.ctan.org/pub/tex-archive/macros/latex/required/graphics/grfguide.pdf]Packages in the ‘graphics’ bundle[/url]


hello, we get $ (y\minus{}x\plus{}1)^3\equal{}x^3\plus{}1$ expanding we have
$ y^3\minus{}3y^2x\plus{}3y^2\plus{}3yx^2\minus{}6yx\plus{}3y\minus{}2x^3\plus{}3x^2\plus{}3x\equal{}0$
and now you must solve this equation with respect to $ x$.
Sonnhard.


Find all primes $ p$ and $ q$ such that $ p$ divides $ q\plus{}6$ and $ q$ divides $ p\plus{}7$.

Let $ (q\plus{}6)/p\equal{}a$, where $ a$ is a positive integer. Thus $ q\equal{}ap\minus{}6$. We have $ (ap\minus{}6)|(p\plus{}7)$. Hence $ p\plus{}7\ge ap\minus{}6$, i.e., $ (a\minus{}1)p\le 13$. Now, it remains to check the cases: (i) $ a\ge3$; (ii) $ a\equal{}2$, (iii) $ a\equal{}1$. I think it's easy now.

By the way, what is IrMO?


How many times does the factor 5 occur in the prime factorization of (65!)?

Every 5th number has a factor of 5, so there are $ \left\lfloor\frac{65}{5}\right\rfloor \equal{}13$ numbers with at least one factor of 5.

In addition, every 25th number has another factor of 5.  We haven't counted these extra 5s yet, so we need to add $ \left\lfloor\frac{65}{25}\right\rfloor \equal{} 2$.

There are clearly no numbers with more factors of 5, so in the factorization of $ 65!$ there are $ 13\plus{}2\equal{}\boxed{15}$ times 5 occurs


algebra
polynomial
induction
strong induction

How would I prove that a polynomial with degree n has "at most" n roots?

I have somewhat of an attempt, please look it over:
(1) if a polynomial f(x) = (x+/-r)q(x) such that f(r/-r) = 0, then the degree of q(x)=degree of f(x) - 1. (not sure if I have to prove this..)

With induction:
ax+b = (x+b/a)a , ax+b is of degree 1 and has one root.

take an arbitrary polynomial of degree k that has k roots, and call that polynomial q(x).
q(x)(x-/+r) = f(x) : f(x) is of degree k+1 and has k+1 roots.

so k -> k+1 is established, therefore any polynomial of n degree >= 1 has n roots.
Then I guess if there are no such binomial factors such that f(x) = (x+/-r)q(x) = 0, then there are no roots. But I'm also confused about the AT MOST n roots part, how would I establish that a polynomial of degree n has AT MOST n roots?

thanks

[quote="darkdieuguerre"]The [url=http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra]Fundamental Theorem of Algebra[/url] renders this trivial.[/quote]

 Hi, I don't think I'm supposed to use the fundamental theorem of algebra - I can just prove it by induction right?

[quote="darkdieuguerre"]I'm not sure you can prove it by induction. An induction proof would have to assume that whatever polynomial you're working with does have a solution, but how would you prove that in general for a generic polynomial?[/quote]

 Well, I'm saying that if the polynomial has solutions, it would have n solutions and so n roots would be the largest number of roots since if there are no solutions, there are no roots?

You want a theorem of elementary algebra here: a root is a factor, and a factor is a root. The FTA is irrelevant, as this result works over any field, not just $ \mathbb{C}$.

A proof, which uses the ideas of the OP cleaned up:

We prove this by induction on the degree of the polynomial. Clearly, a nonzero polynomial of degree zero has no roots; that's our base case. Suppose as the induction hypothesis that all nonzero polynomials of degree at most $ n$ have at most $ n$ roots.
Choose an arbitrary polynomial $ p(x)\equal{}a_{n\plus{}1}x^n\plus{}a_nx^n\plus{}\cdots\plus{}a_0$ of degree $ n\plus{}1$. The leading coefficient $ a_{n\plus{}1}$ is nonzero, so $ p$ is not the zero polynomial. We split into two cases:
Case 1: $ p$ has no roots. Then we're done, as clearly $ 0\le n\plus{}1$.
Case 2: $ p$ has a root $ r$. Then, by the theorem I mentioned above, $ p(x)\equal{}(x\minus{}r)\cdot q(x)$ for some polynomial $ q(x)\equal{}a_{n\plus{}1}x^n\plus{}b_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots\plus{}b_0$ of degree $ n$. By the induction hypothesis, $ q$ has at most $ n$ roots. The roots of $ p$ are the roots of $ q$ together with the new root $ r$, and there are at most $ n\plus{}1$ of these.

I don't know if this proof is correct.

[hide="Alternative Proof"]We first assume without loss of generality that the zero polynomial is not considered.

Let $ S$ contain the set of all positive integers for which there exists a polynomial of degree $ n$ with more than $ n$ roots. If $ S$ is empty, then the proof is done. If $ S$ is nonempty, then by the well-ordering property, there exists a smallest integer $ k$ in the set. We note that $ k > 1$ because two lines can only intersect at one point or at infinitely many points, and we automatically ignore the second option as per our original assumption.

So, consider the polynomial $ f(x)$ of degree $ k$. Since it has more than $ k$ roots, consider an arbitrary root $ r_1$, and compute $ \displaystyle\frac{f(x)}{x\minus{}r_1}$. This is a polynomial of degree $ k\minus{}1$ with more than $ k\minus{}1$ roots, but this implies that $ k\minus{}1$ is in $ S$, which contradicts the original assumption that $ k$ was the smallest integer in $ S$. Therefore, $ S$ must be nonempty, so the claim is proven.[/hide]

[quote="jmerry"]You want a theorem of elementary algebra here: a root is a factor, and a factor is a root. The FTA is irrelevant, as this result works over any field, not just $ \mathbb{C}$.

A proof, which uses the ideas of the OP cleaned up:

We prove this by induction on the degree of the polynomial. Clearly, a nonzero polynomial of degree zero has no roots; that's our base case. Suppose as the induction hypothesis that all nonzero polynomials of degree at most $ n$ have at most $ n$ roots.
Choose an arbitrary polynomial $ p(x) \equal{} a_{n \plus{} 1}x^n \plus{} a_nx^n \plus{} \cdots \plus{} a_0$ of degree $ n \plus{} 1$. The leading coefficient $ a_{n \plus{} 1}$ is nonzero, so $ p$ is not the zero polynomial. We split into two cases:
Case 1: $ p$ has no roots. Then we're done, as clearly $ 0\le n \plus{} 1$.
Case 2: $ p$ has a root $ r$. Then, by the theorem I mentioned above, $ p(x) \equal{} (x \minus{} r)\cdot q(x)$ for some polynomial $ q(x) \equal{} a_{n \plus{} 1}x^n \plus{} b_{n \minus{} 1}x^{n \minus{} 1} \plus{} \cdots \plus{} b_0$ of degree $ n$. By the induction hypothesis, $ q$ has at most $ n$ roots. The roots of $ p$ are the roots of $ q$ together with the new root $ r$, and there are at most $ n \plus{} 1$ of these.[/quote]

  Hi, thanks for the post - this was the idea I was looking for.
do I have to prove that n+1 roots would be the MOST or that a polynomial could either have no roots or n+1 roots?
and is this very different from my original post? I don't think so

You simply have to prove that a polynomial with degree $ n\plus{}1$ can have at most $ n\plus{}1$ roots. That's how all the non-strong induction proofs work; you assume that the claim is true for a value $ k$, and then use that assumption to prove the claim true for $ k\plus{}1$.

[quote="darkdieuguerre"]You simply have to prove that a polynomial with degree $ n \plus{} 1$ can have at most $ n \plus{} 1$ roots. That's how all the non-strong induction proofs work; you assume that the claim is true for a value $ k$, and then use that assumption to prove the claim true for $ k \plus{} 1$.[/quote]

 I know, but is it okay to say without proof that a polynomial either has n roots or no roots? also, induction shows that a polynomial of degree n will have n roots, but can a polynomial of degree n have n+1 roots? (I know it won't but is it necessary to show this?)

I don't think so. You have proved that such a polynomial could exist by setting an upper bound on the number of roots, but that isn't a sufficient condition to state that such polynomials with an intermediate number of roots may exist; it could exist in theory but from the proof you can draw no such conclusion.

You don't need to show that; the induction proof holds without that information.

I never said that it has the full number of roots if it has one; there's an "at most" in that part as well. For all we know, $ q$ could have no roots.
There's also the question of whether you're counting with multiplicity; does $ (x\minus{}1)^3$ have one root or three?

[quote="jmerry"]I never said that it has the full number of roots if it has one; there's an "at most" in that part as well. For all we know, $ q$ could have no roots.
There's also the question of whether you're counting with multiplicity; does $ (x \minus{} 1)^3$ have one root or three?[/quote]

 yes, but we take 3 roots to the most amount of roots a polynomial of degree 3 could have right? and from our induction method, we've shown that a polynomial degree n has n roots, but is this the most amount of roots? I don't think a case of 1 root coming from a polynomial of degree 3 is of importance, since we want to find the case when it has the most roots

[quote="darkdieuguerre"]My above proof actually is a proof that there exists no polynomial (other than the zero polynomial) that has more solutions than its degree.[/quote]
 ah, excellent. Thank you, I will take a better look at it

[quote="darkdieuguerre"]The [url=http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra]Fundamental Theorem of Algebra[/url] renders this trivial.[/quote]

This works.

We know that the FTA makes such a statement true, but a proof was desired that used elementary algebra, since the FTA does not, as of yet, have a purely algebraic proof.


geometry
vector
trigonometry
perimeter
analytic geometry
AMC

A circle of radius 2 is centered at O. Square OABC has side length 1. Sides AB and CB are extended past B to meet the circle at D and E, respectively. What is the area of the shaded region in the figure, which is bounded by BD, BE, and the minor arc connecting D and E?

$ \mathrm{(A) \ } \frac{\pi}{3}\plus{}1\minus{}\sqrt{3}\qquad \mathrm{(B) \ } \frac{\pi}{2}(2\minus{}\sqrt{3}) \qquad \mathrm{(C) \ } \pi(2\minus{}\sqrt{3})\qquad \mathrm{(D) \ } \frac{\pi}{6}\plus{}\frac{\sqrt{3}\plus{}1}{2}\qquad \mathrm{(E) \ } \frac{\pi}{3}\minus{}1\plus{}\sqrt{3}$

[img]http://www.artofproblemsolving.com/Wiki/images/a/a4/2006amc10b19.gif[/img]

***I drew a circle with DB and BE as the radii and with arc DE.  I found that DB and DE were $ \sqrt {3} \minus{} 1$, so I used this dimension to find the area of circle B.  Knowing that Angle DBE is a right angle, I took 1/4 of the area of circle B to get the answer $ \approx0.4208936072$.

No need for decimals. By 30-60-90 triangles, $ AD\equal{}CE\equal{}\sqrt{3}$, so $ BD\equal{}BE\equal{}\sqrt{3}\minus{}1$. As $ DBE$ is a right angle, the area is simply $ \frac{\pi(\sqrt{3}\minus{}1)^2}{4}\equal{}\frac{\pi}{2}(2\minus{}\sqrt{3})$

[quote="davidyko"]No need for decimals. By 30-60-90 triangles, $ AD \equal{} CE \equal{} \sqrt {3}$, so $ BD \equal{} BE \equal{} \sqrt {3} \minus{} 1$. As $ DBE$ is a right angle, the area is simply $ \frac {\pi(\sqrt {3} \minus{} 1)^2}{4} \equal{} \frac {\pi}{2}(2 \minus{} \sqrt {3})$[/quote]

davidyko- Glad to hear that you agree with my solution, but the answer key says that the correct answer is $ A$.  http://www.artofproblemsolving.com/Wiki/index.php/2006_AMC_10B_Problems/Problem_19

[hide="hint"]Draw OD and OE. Notice that if u drpo perpendiculars to  OA from OE you get a 30-60-90 degree triangle. Now subtract the area of those sectors by seeing that they are 30-degree sectors from the larger sector and subtract the areas of OBE and OBD  by simply observing that the base is $ \sqrt {3} \minus{} 1$ and the height is 1. [/hide]

The problem is pretty easy with analytic geometry and vectors... let the centre of the circle be the origin, and you have $ x^2 \plus{} y^2 \equal{} 4$ for the circle. Then, it's easy to get that $ D(1, \sqrt {3}), E(\sqrt {3}, 1)$ since you use the lines $ x \equal{} 1, y \equal{} 1.$ Then, use the dot product on the vectors $ \vec{OA} \cdot \vec{OB} \equal{} \|\vec{OA}\|\|\vec{OB}\|\cos\theta \implies \theta \equal{} \frac {\pi}{6}.$ So, if we have triangle $ ODE,$ we have the outer "crust" of sector ODE to be : sector ODE - [ODE] = $ \frac {1}{12}\cdot \pi \cdot 2^2 \minus{} \frac {1}{2} \cdot 2 \cdot 2 \cdot \cos 60 \equal{} \frac {\pi}{3} \minus{} 1.$ Now, since $ BE \equal{} BD \equal{} \sqrt {3} \minus{} 1,$ $ \Delta BED$ is a half square, so the entire total area is a grand $ \frac {\pi}{3} \minus{} 1 \plus{} \frac {(\sqrt {3} \minus{} 1)^2}{2} \equal{} \frac {\pi}{3} \minus{} 1 \plus{} 2 \minus{} \sqrt {3} \equal{} \frac {\pi}{3} \plus{} 1 \minus{} \sqrt {3}.$ $ \blacksquare$

[quote="diophantient"][quote="Twiz"][quote="davidyko"]No need for decimals. By 30-60-90 triangles, $ AD \equal{} CE \equal{} \sqrt {3}$, so $ BD \equal{} BE \equal{} \sqrt {3} \minus{} 1$. As $ DBE$ is a right angle, the area is simply $ \frac {\pi(\sqrt {3} \minus{} 1)^2}{4} \equal{} \frac {\pi}{2}(2 \minus{} \sqrt {3})$[/quote]
The problem is that region $ BDE$ is not a quarter of a circle.[/quote]wait why not?[/quote]

$ BE\equal{}\sqrt{3}\minus{}1$ and if you draw a line through $ OB$ and lable the point where it interesects the circle as $ Q$  then $ BQ\equal{}2\minus{}\sqrt{2}$

[quote="diophantient"][quote="Twiz"][quote="davidyko"]No need for decimals. By 30-60-90 triangles, $ AD \equal{} CE \equal{} \sqrt {3}$, so $ BD \equal{} BE \equal{} \sqrt {3} \minus{} 1$. As $ DBE$ is a right angle, the area is simply $ \frac {\pi(\sqrt {3} \minus{} 1)^2}{4} \equal{} \frac {\pi}{2}(2 \minus{} \sqrt {3})$[/quote]
The problem is that region $ BDE$ is not a quarter of a circle.[/quote]wait why not?[/quote]

Assume for the sake of contradiction that $ BDE$ is a quarter of a circle. Since $ \angle DBE \equal{} \pi/2,$ this implies that point $ B$ is the centre of the circle whose perimeter contains arc $ DE.$ But we already know that $ O$ is the centre of the very same circle. This is a contradiction, since the two points are not at the same coordinate and a circle can only have one centre. $ \blacksquare$

[quote="Tiger3"]For this question, if I haven't learned about vectors yet, is there another way to do this problem?[/quote]
check the hint posts by sonny in this topic, they lead to a non-vector solution.  (I think both solutions are the same difficulty.)

Ugh, I feel so stupid, screwing up on an AMC 10 problem.  :blush:

oh yeah,
i remember this problem...

i actually posted it a few months ago,
but i know how to get it now...

AoPS is such a great and helpful math society!!!!

 :gathering:


i got 2nd place LA county 26th-ish in California would this help me at all?

I would probably say not, but that's just me. :D

[quote="mathgeniuse^ln(x)"][quote="Drunken_Math"]I'd say not, since universities don't usually look at middle school accomplishments too much. Unless you got in countdown at Nationals...[/quote]

Or close to countdown at nationals.[/quote]

such as...i don't know...say 14th?
[hide]just kidding[/hide]

[quote="mathgeniuse^ln(x)"][quote="Drunken_Math"]I'd say not, since universities don't usually look at middle school accomplishments too much. Unless you got in countdown at Nationals...[/quote]

Or close to countdown at nationals.[/quote]

or won nationals :D

[quote="Treething"][quote="mathgeniuse^ln(x)"]Erg.

I will beat you something mathy, Treething.

Though I probably already have.

No Pressure, No Pressure Treething.

[color=white]But you are going down[/color][/quote]

eh you know I suck at AIME, but let's see who does better next year...

[color=white]Haha... let's see[/color][/quote]

And you also hopefully will go down in Math League and AMC.

[color=yellow]I didn't mean hopefully as in I want you to lose, but yeah.[/color]

Colorado School of Mines and University of Denver give out half-scholarships to the top four at state each year. What my problem is is that I was in top four twice, but I can't get a half-scholarship twice.

2005 - 4th (6th written)
2006 - 1st (1st written)

I think I'll be able to get into either of those places.

Just put MATHCOUNTS on your college resume. It'll help some way or another.

[quote="Ignite168"][quote="robinhe"]Maybe I don't have any outstanding math achievements [hide]beating Neil in 7th grade Kumon math challenge and taking 1st place[/hide], but I have other stuff.

We will just have to all wait until high school is over to talk about this I guess. :P[/quote]

Kumon is a scam and a joke[/quote]
I don't have much experience, but I wll say that Kumon is pretty useless.

[quote="robinhe"]No it is not useless.
No it is not a joke.
No it is not a scam.
Yes the problems are legitimate.
Yes the grand prize is $\$1500$
Yes it is VERY HELPFUL!!!!!![/quote]

Is it a competition sponsored by Kumon or run by Kumon? I'm not saying the competition is a scam or joke, but Kumon definitely is.

[quote="robinhe"]It is run by Kumon.

It is called THE KUMON MATH CHALLENGE (sorry for the caps)

The Kumon program helps enhance your math skills a lot.[/quote]
I have seen the ads, and to my opinion it is meant for kids that aren't good at math, reading and other stuff.
But I probably have no idea what I'm talking about.
I never do.

[quote="SplashD"]speaking of which, what was your AIME score?[/quote]

My AIME was a 9/15, Treething's was an 8/15. :lol:

My AMC 10 was a 146.5/150, Treething's was a 140.5/150. :lol:

My Math League was a 40/40, Treething's was a 39/40. :lol:

I think that clarifies it.

I got 6/15 on my AIME. 138 on my AMC10. and I didn't take the other test. So I'm worse than both of them.
 :D

Most college applications specifically ask for achievements and awards AFTER EIGHTH GRADE, so you can consider MATHCOUNTS as practice for what you will do later. The applications for MIT and Caltech specifically ask for AMC 10/12 and AIME scores.

Cool...they value math competitions...

Math League before High School shouldn't count as a win if you beat somebody...only reason why treething would miss a question is because of a stupid mistake

Darn! I got a 99.5 and 92.5 on AMC 12 A\B ! Well, at least I was 10 years old then......

MIT admits people every year who either

a) didn't take AMC tests at all

or

b) didn't do particularly well on the AMC tests they did take.

But because MIT is an engineering and hard science college, mostly, it never hurts to show some evidence of strong math ability when applying to MIT.

Any other kind of achievement is also good to show when applying to MIT. Learning Chinese as a second language at high school age, if you can demonstrate that somehow, is one sign of being smart and diligent. See

http://ben.mitblogs.com/archives/2005/10/many_ways_to_de.html

for a statement about MIT admission from one of the MIT admission officers, which I find quite interesting.

[quote="myc"]I'm an incomming sophomore,  AMC 10 score last year 96.  my goal is to get at least a 120 this year.  advice, anyone?  or in other words, help!!!  Can you get into a really good  college, say MIT, even if you don't have exceptional AMC scores?  I'm pretty good at chinese.  I can read the chinese newspaper.  How much does that help towards college apps?  n english is my first language.[/quote]

Take AP Chinese next May! Did you just learn Chinese from high school age? and you can actually read newspaper (e.g. Renmin Daily)? If so, that's awesome. If AP is not enough to demonstrate your frequency in Chinese, you can actually go to China, and take one of the entrance examinations (e.g. Zhongkao or Gaokao).

aznosamaboy, check your private messages.

I can read the newspaper if its in simplified chinese.  I'm horrible at tradition.  Can you take the zhongkao in the U.S?  Do you have to go to China to take it?  Have you taken it before, is it hard?

thanks :)

Slightly off-topic, but what Chinese newspapers do you usually read?


probability
combinatorics unsolved
combinatorics

[quote="limonka"]Hi, I have a question about probabilities.
John tosses consecutively a coin. If appears a head he wins a dollar, if it appers a tail he loses a dollar. Game is over when John loses all his money or earns [b]y[/b] dollars (y has previously been defined). At the beginning John has [b]x[/b] dollars. What's the probability that John will lose?
pls help! :roll:[/quote]

It is a very well known problem.
Let $a_{i}$ be the probability of loosing when having $i\leq y$ dollars. We have the following equations :
$a_{0}=1$, $a_{i}=\frac{1}{2}a_{i+1}+\frac{1}{2}a_{i-1}$ and $a_{y}=1$
So $a_{i+1}-a_{i}=a_{i}-a_{i-1}$ and $a_{i}=1-i(1-a_{1})$ hence $1-a_{1}=\frac{1}{y}$

So the answer is $1-\frac{x}{y}$

--
Patrick


Let $ a$, $ b$, $ c$ be real numbers satisfying $ 0 \leq a \leq 1$, $ 0 \leq b \leq 1$, $ 0 \leq c \leq 1$ and $ a \plus{} b \plus{} c \equal{} \frac {3}{2}$.
Given a number $ n$.
Find the greatest possible value of the expression $ P \equal{} a^{n} \plus{} b^{n} \plus{} c^{n}$.

max(1+(1/2)^n,3/2^n)  I think.
n is real
just use karamata (when x^n is convex)
jensen(while x^n is concave)

[quote="GacKiem"]The problem is obvious if $n\in[0,1]$. It's wrong if $n<0$. I don't think it's hard if $n\ge 1$. It easy be convex function. The maximum only attain for $(a,b,c)=(1,1/2,0)$.[/quote]
Right. with Karamata Inequality it's easy for $n\geq 1$


A 36-year-old father has a 12-year-old son. In how many years will the father be exactly twice the age of the son?

There will always be a $ 24$ year gap in between their ages. The son must be $ 24$ years old to be half the age of the father, so the answer is $ 24\minus{}12\equal{}\boxed{12}$.

[hide="answer"]their age differ. is 24.

since we're looking for the age of the sun we get the equation

24-12=12

so the answer is $\boxed{12}$[/hide]

[b][i][u]please rate my post thanks!1[/u][/i][/b]

please be fair though thanks [=

 :bye:  :bye:  :bye:  :bye:

[hide]The age different is 24  24-12=12 answer 12[/hide]
Problem and solution above.


AMC
AIME
vector
trigonometry
trig identities
Law of Cosines

A farmer and his son are removing a large boulder from a well. They are pulling on ropes attached to the boulder. The ropes make a $ 45$ degree angle with each other. If the farmer is pulling with a force of $ 115$ Newtons and his son is pulling with a force of $ 105$ Newtons, then their combined force can be expressed to the nearest integer is $ ABC$ Newtons. Find $ A \plus{} B \plus{} C$

You can just simply use vectors for this problem.

Let the force vector of the son be $ <105,0>$. Since the forces are 45 degrees apart, the farmer's force vector can be expressed as $ <\frac{115}{\sqrt{2}},\frac{115}{\sqrt{2}}>$. Their combined force can be expressed with the sum of the two vectors, which is

$ <\frac{115}{\sqrt{2}}\plus{}105,\frac{115}{\sqrt{2}}>$.

The force (in Newtons) is the magnitude of this vector. Calculating it gives, to the nearest integer, 203 Newtons.

$ 2\plus{}0\plus{}3\equal{}\boxed{5}$, giving our desired answer.

Yeah, I used vectors to solve it, but I also applied the Law of Cosines
but anyway, I got the same answer, so it doesn't matter


If x(n)=a(n)a(n)-a(n-1)a(n+1) and y(n)=a(n)[a(n-1)+a(n+1)] are convergent to zero, is a(n) convergent to zero? Justify or give a counterexample. (Cristinel Mortici)

I think a_n converges to 0.

Let's assume it doesn't. Then there is an a>0 and a subsequence a k_n s.t. |a k_n|>=a for all n in N*. This means that the sequence a k_n-1 + a k_n+1 ->0 as n->00 (*).

For every   \varepsilon >0 there is an n_ \varepsilon s.t.
|a k_n 2 -a k_n-1 * a k_n+1|<= \varepsilon for all n>=n_ \varepsilon , so
a k_n-1 * a k_n+1 >= a 2 -  \varepsilon for all n>=n_ \varepsilon. If we take  \varepsilon < a 2 it means that a k_n-1 and a k_n+1 have the same sign for all n>=n_ \varepsilon , so from (*) we get a k_n-1 and a k_n+1 both converge to 0 as n-.0, so from x_n-.0 we get a k_n 2 ->0 as n->00, but this is a contradiction because we had assumed a k_n 2 >= a 2 for all n in N*.

The contradiction shows us that the assumption that a_n doesn't converge to 0 was wrong, so a_n->0 as n->00.

Is it right?


factorial
AMC
AIME
geometry
3D geometry
number theory
relatively prime

(Me) Let $ x \equal{} \sqrt [3] {6 \plus{} 5\sqrt {4}} \plus{} \sqrt [3] {6 \minus{} 5 \sqrt {4}}$. If $ \frac {1}{x[(x \plus{} 3)(x \plus{} 4) \minus{} 7x]} \equal{} \frac {k}{m}$ where $ k,m$ are relatively prime and positive. Find the product of last two digits of factorial $ (m \minus{} k)!$.

Level: Second Round... Very similar to AIME-style problem

[hide="Solution"]
Let $ a \equal{} \sqrt [3]{16}, b \equal{} \sqrt [3]{ \minus{} 4}$; then $ a^3 \plus{} b^3 \equal{} 12,\, a \plus{} b \equal{} x,\, a^3b^3 \equal{} \minus{} 64 \Leftrightarrow ab \equal{} \minus{} 4$. Factorizing,
\[ (a \plus{} b)^3 \equal{} x^3 \equal{} a^3 \plus{} b^3 \plus{} 3ab(a \plus{} b) \equal{} 12 \minus{} 12x \Longleftrightarrow x^3 \plus{} 12x \equal{} 12
\]
Thus $ \frac km \equal{} \frac {1}{x^3 \plus{} 12x} \equal{} \frac {1}{12}$, so $ (m \minus{} k)! \equal{} 11!$ .. which ends in two zeros?
[/hide]

Wow, I was stupid. I was sitting here forever, because I didn't realize I could simplify $ \sqrt{4}$ to $ 2$.

Yes that's correct.  :)

What did you think of this problem? The idea is pretty classic and there isn't really "mystery" to it once you get to cube it.

$ x \equal{} \sqrt [3] {6 \plus{} 5\sqrt 4} \plus{} \sqrt [3] {6 \minus{} 5\sqrt 4}$

Hence $ x \minus{} \sqrt [3] {6 \plus{} 5\sqrt 4} \minus{} \sqrt [3] {6 \minus{} 5\sqrt 4} \equal{} 0$

$ \Rightarrow x^3 \minus{} 12 \equal{} 3 x \sqrt [3] {6 \plus{} 5\sqrt 4} \ \sqrt [3] {6 \minus{} 5\sqrt 4} \equal{} \minus{} 12x$

Hence $ x^3 \plus{} 12x \equal{} 12$

$ \frac {1} {x[(x \plus{} 3)(x \plus{} 4) \minus{} 7x]} \equal{} \frac {1}{12}$

Hence k =1 and m = 12

So (m-k)! = 11! which ends in a zero and so the product of the last digits is $ \boxed{0}$

What's with the calculation at the end with $ k$ and $ m$? It seems simpler to just ask for $ k\plus{}m$. Even taking the factorials from $ 1$ to $ 5$, you get $ 1,2,6,24,120$, which give answers $ 1,2,6,8,0$. These 5 choices are much more limited than 000 through 999.

[quote="123456789"]Wow, I was stupid. I was sitting here forever, because I didn't realize I could simplify $ \sqrt {4}$ to $ 2$.[/quote]

I didn't realize that either until I finished typing  :lol: . But it doesn't actually help anyway.

[quote="Brut3Forc3"]What's with the calculation at the end with $ k$ and $ m$? It seems simpler to just ask for $ k \plus{} m$. Even taking the factorials from $ 1$ to $ 5$, you get $ 1,2,6,24,120$, which give answers $ 1,2,6,8,0$. These 5 choices are much more limited than 000 through 999.[/quote]

True and really, why did I add that part? Oh, just for fun of it haha... It's actually not that easy for people who are not familiar with factorials ending with zero because you need to know to count # of 5's.

Another thing was... I get tired of just adding $ k\plus{}m$ now and then.  :lol:

@hsbhatt: How did you get to:
$ \Rightarrow x^3 \minus{} 12 \equal{} 3 x \sqrt [3] {6 \plus{} 5\sqrt 4} \ \sqrt [3] {6 \minus{} 5\sqrt 4} \equal{} \minus{} 12x$

$ \sqrt [3] {6 + 5\sqrt 4} \ \sqrt [3] {6 - 5\sqrt 4} = \sqrt [3]{(16)( - 4)} = - 4$. Remember that $ \sqrt {4} = 2$. Also:

\begin{align*} x^{3} & = \left[\left(\sqrt [3] {6 + 5\sqrt 4}\right) + \left(\sqrt [3] {6 - 5\sqrt 4}\right)\right]^{3} \\
& = 12 - 12 \left[\sqrt [3] {16} + \sqrt [3] { - 4}\right] \\
& = 12 - 12x \end{align*}

If you still don't get it, make the substitution $ x = a + b$, expand $ (a + b)^{3}$ using the binomial theorem, and then make the substitutions $ a = \sqrt [3]{16}, \ b = \sqrt [3]{ - 4}$.

Its a direct result of the identity that if $ a\plus{}b\plus{}c \equal{} 0$ then $ a^3\plus{}b^3\plus{}c^3 \equal{} 3abc$

Very true, though I didn't actually see that it applied until now. XD

wow, i read the problem, then 123456789's comment about how he didnt see how sqrt4 simplified, then worked on the problem for a few minutes, and then realized that I also did not see that sqrt4 simplified.........


$ \int sin^{3/2}2\theta sin \theta d \theta \equal{} 2 \sqrt{2} \int sin^{5/2} \theta cos ^{3/2} \theta d \theta$

After that ...

hello
[hide="Probably the worst way but nothing else is striking in my mind at this moment.Even then click on this to see a stupid way "]
$ \int sin^{5/2} \theta cos ^{3/2} \theta  d \theta$
=$ \int \frac {cos^{3/2}\theta}{sin^{3/2}\theta sin^{ \minus{} 4}\theta}d\theta$
=$ \int cot^{3/2}\theta cosec^{4}\theta d\theta$
now put $ cot\theta \equal{} t$ ,$ cosec^2\theta d\theta \equal{} \minus{} dt$
u will get the result.[/hide]
Thank u.

[quote="Dr Sonnhard Graubner"]hello, i don't think that you can express your integral with the known elementary functons.
Sonnhard.[/quote]
But Dr. Graubner,
 As Kabi has solved it  [hide=" by following method"]$ given \equal{} 2\sqrt{2} \int (cot \theta)^{3/2} (1\plus{}(cot \theta)^{2})* {cosec^{2}\theta} d\theta$ [/hide]
it can be done. :?:  :?:

Hello
What's going on guys?
Is there any prob Dr.?Kindly tell my mistake.
Thank u.

hello, i think, there is a little problem with your integral, because you have
$ 2\sqrt{2}\int (\cot(x))^{3/2}(cosec(x))^4\,dx$, then you put $ t\equal{}cot(x)$ and we have $ \minus{}dt\equal{}(cosec(x))^2\,dx$, but in your integral is $ (cosec(x))^4$?
What would be the result of your integral?
Sonnhard.

hello
Well Dr.,I have done such question in my first year of high school many times and this way have always gave me the right answer.
I just did this::
$ cosec^{4}x$=$ (cosec^2x)^2$
Thank u.

hello kabi, that's clear for me, but the problem comes after that. Would you write down your whole solution please?
Sonnhard.

actually,dr.
we write one $ cosec^2x$  as  $ 1\plus{}cot^2x$ and leave the other for being dt.

hello
[hide="solution continue"]
    $ \int cot^{3/2}\theta cosec^{4}\theta d\theta$
after substituting
$ cot\theta\equal{}t$ and $ cosec^2\theta d\theta\equal{}\minus{}dt$
=$ \minus{}\int t^{3/2}(1\plus{}t^2)dt$
=$ \minus{}\int (t^{3/2}\plus{}t^{7/2})dt$
=$ \minus{}\frac{2}{5}t^{5/2} \minus{} \frac{2}{9}t^{9/2}$
now put t=$ cot\theta$ again
Tell me if u feel any step is wrong.
[/hide]
Thank u.

Kabi, that is the one I got as a solution.

Thanx

hello
Thank god .Thank to u dude for telling me that d answer is right.
Thank u

hello kabi, your primitive is ok, substituting $ t \equal{} \cot(x)$ we get $ f(x) \equal{} \frac { \minus{} 2}{5}\cot(x)^{\frac {5}{2}} \minus{} \frac {2}{9}\cot(x)^{\frac {9}{2}}$, differentiating $ f(x)$ with respect to $ x$ we get $ f'(x) \equal{} \frac {(\cot(x))^{\frac {3}{2}}}{(\sin(x))^4}$ and that's ok. It was my error.
Sonnhard.


I have:

[code]\documentclass[11pt]{article}

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}

\begin{document}

\end{document}[/code]

Compilation worked out, but when I tried to see the PDF, the following message appears:

[img]http://img150.imageshack.us/img150/7198/errorvu2.png[/img]

After that I went to the Output Profiles and I verified LaTeX => PDF's viewer, this appears:

[img]http://img212.imageshack.us/img212/6519/viewerny3.png[/img]

I don't know what could be wrong.

Hi, I think this might be because you haven't actually typed anything for your document.

I did what you did and the same error message appeared, however when I wrote a sentence it generated the pdf file.


Find a combinatorial technique in order to calculate the sum of all $n\leq 10$ digit integers with $n$ distint digits.

Masoud Zargar

If $n=2$ then the required number is $99*45$
For $n\geq3$:
Let's write each number as $a_1*10^{n-1}+a_2*10^{n-2}...+a_n$.
For each digit $a_2,..., a_n$ there are $9*9*8*7...(12-n)$ numbers that contain that digit in that exact position and have all distinct digits.
We will add all the $a_i$ together for every $i$ and then add together the n sets we will get.
Every set will have value of:
$(1+2+...+9)*10^{n-i}*9*9*8...*(12-n)$
Adding the sets together, the result will be:
$45*(11...1)*(9*9*8*...(12-n)$, where the number $11...1$ has $(n-1)$ digits.

Well, for n = 2, we have a different case because the number $9*9*...(12-n)$ by which we multiply the result will actually be $9$ instead of $12-n=10$ for $n=2$. Therefore we have a different case. But the formula will be relatively the same: $45*11*9$. ;)

Well Rem, show your solution for $n=3$ with the same reasoning you have derived the formula.

Masoud Zargar

For $n=2$, shouldn't it be $98 \times 45$?

Since for any non-zero number, when in the tens digit, it occurs $9$ times since their are $9$ other choices for the units digit. So that sum is $9\times10$ times the number.
But when in the units digit, it only occurs $8$ times since $0$ cannot be in the tens digit. So their sum is $98$.

There are 10 digits!!! :!:
So, for the ones part, for last digut we have $9$ possible numbers so my answer should be right.

Perhaps you misunderstood my problem.  When I say find the sum..., I mean: find the sum of all three digit integers with distinct digits; in general, with $n$ digits.

Masoud Zargar

I posted the solution for that yestterday, it is half a page above!

When you all figure out who did or didn't solve what when and how, consider the generalization of the problem to base $b$.

[quote]
WTF?
There are 10 digits!!!
So, for the ones part, for last digit we have 9 possible numbers so my answer should be right [/quote]

no, there are 8 possible numbers: no leading zeros.

We have $1k,2k,3k,4k,5k,6k,7k,8k,9k$ that's $9$ numbers

but if k=1, 1k doesnt count.
if k=2, 2k doesnt count, and so on

:wallbash_red:
totally forgot about the distinct digits rule

[quote]rem
If n=2 then the required number is 99*45
For $n\geq3$:
Let's write each number as $a_1*10^{n-1}+a_2*10^{n-2}...+a_n$.
For each digit $a_2,..., a_n$ there are $9*9*8*7...(12-n)$ numbers that contain that digit in that exact position and have all distinct digits.
We will add all the $a_i$ together for every $i$ and then add together the n sets we will get.
Every set will have value of:
$(1+2+...+9)*10^{n-i}*9*9*8...*(12-n)$
Adding the sets together, the result will be:
$45*(11...1)*(9*9*8*...(12-n)$, where the number 11...1 has (n-1) digits.[/quote]
This solution is actually correct except for the fact that it treats the most significant digit, $a_1$, just like any other digit. This should not be so, since there are no leading zeros. Therefore instead of $9*9*8*7...(12-n)$, it should be $8*(8*7...(11-n))=8*\frac{8!}{(10-n)!}$. Another special case occurs when we actually add all the most significant digits. Instead of $8*\frac{8!}{(10-n)!}$ it should be $9*\frac{8!}{(10-n)!}$. The final solution is then
$45*8*(11...1)*\frac{8!}{(10-n)!}+45*9*10^{n-1}*\frac{8!}{(10-n)!}$
where $(11...1)$ has $n-1$ digits.
For base b, it is:
$\frac{b(b+1)}{2}*(b-2)*(11...1)*\frac{(b-2)!}{(b-n)!}+\frac{b(b+1)}{2}*(b-1)*b^{n-1}*\frac{(b-2)!}{(b-n)!}$
where $(11...1)$ has n-1 digits.

rem, calm down with the acronyms.  In general, if what an acronym stands for is unacceptable for the boards, then the use of the acronym is also discouraged.

Sorry. :blush:
By the way, I just noticed that the author of the first post wanted just the combinatorial technique and not the answer. ;)


function
number theory
least common multiple
combinatorics unsolved
combinatorics

Determine the number of bijective functions $f: \{1,2,..,n\}\to\{1,2,...,n\}$ such that $\underbrace{f\circ f \circ ... \circ f}_{k}(x) =x,\forall x\in\{1,2,...,n\}$.

well.... given a bijective $f$, $f$ partitions the set $\{1, ..., n\}$ into orbits. (the orbit of an element $m$ is the set of all values of $f^{(k)}(m)$ as $k$ varies.  if the $p$ orbits of $f$ number $k_1, ..., k_p$ elements, respectively, then $f^{(k)}(x) = x$ for all $x$ iff $\mbox{lcm}(k_1,...,k_p) | k$.

so the problem has to do with finding numbers of partitions $k_1+...+k_p = n$ such that $\mbox{lcm}(k_1,...,k_p) | k$... and this isn't an easy problem as far as the people i know know


A map is to be colored using exactly four colors. If seven different colors are available, in how many ways can the four colors be chosen?

[hide] We have $ \binom{7}{4}\equal{}\frac{7!}{4!3!}\equal{}\boxed{35}$.[/hide]


algebra
polynomial
function
integration
inequalities
triangle inequality
real analysis

Let $ \text{Lip}_{K}[a,b]$ be the set of Lipschitz functions on $ [a,b]$ with Lipschitz constant at most $ K$.  Prove that if $ f\in\text{Lip}_{K}[a,b]$, then $ f$ can be uniformly approximated by a polynomial in $ \text{Lip}_{K}[a,b]$.

I guess this is equivalent to saying that the Lipschitz polynomials with constant up to $ K$ on $ [a,b]$ are dense in $ \text{Lip}_{K}[a,b]$.  But how to prove that?

We know $ \text{Lip}_{K}[a,b]\subset C[a,b]$, which is compact, and we know these Lipschitz functions form a subalgebra.  I also know Lipschitz functions separate points and vanishes at no points.  So I was thinking that Stone-Weierstrass could help, but how do I apply it to this situation, or am I completely off?

If $ X$ is compact, which we have, then $ \text{Lip}_{K}(X)$ forms a subalgebra.

$ f(x)\equal{}x$ is $ 1$-Lipschitz on $ K\equal{}[0;1]$. But is $ f^2(x)\equal{}x^2$ also a $ 1$-lipschitz function ?

Someone suggested to me that since $ [a,b]$ is compact, and polynomials in $ \text{Lip}_{K}[a,b]$ satisfy the conditions of Stone-Weierstrass, the Lipschitz polynomials are dense in $ C[a,b]$, which definitely makes sense.  But does this necessarily imply that they are also dense in $ \text{Lip}_{K}[a,b]$?

Actually, Lipschitz functions with a fixed constant don't satisfy those conditions- they're not an algebra, as neither addition nor multiplication work. We need different techniques.

I would do a two-step approximation.
First, we approximate $ f$ uniformly with a $ C^1$ function $ g$ such that $ |g'(x)|\le K$ everywhere. Choosing $ g(x)\equal{}\frac1{\delta_1}\int_x^{x\plus{}\delta_1}f(t)\,dt$ works here, as $ |g'(x)|\equal{}\left|\frac{f(x\plus{}\delta_1)\minus{}f(x)}{\delta_1}\right|\le K$ and $ |g(x)\minus{}f(x)|\equal{}\left|\frac1{\delta_1}\int_x^{x\plus{}\delta_1}f(t)\minus{}f(x)\,dt\right|\le \frac1{\delta_1}\int_x^{x\plus{}\delta_1}K(t\minus{}x)\,dt\equal{}\frac{K\delta_1}{2}$.
Second, approximate $ g'$ uniformly to within $ \delta_2$ with a polynomial $ p'$ that doesn't exceed $ K$ on $ [a,b]$. We can use the Weierstrass theorem, and then multiply the approximation by something slightly less than 1 if necessary. Define $ p(x)\equal{}g(a)\plus{}\int_a^x p'(t)\,dt$; we get $ |p(x)\minus{}g(x)|\equal{}\left|\int_a^x p'(t)\minus{}g'(t)\,dt\right|\le \delta_2(x\minus{}a)\le \delta_2(b\minus{}a)$.
Combining the two estimates, $ |p(x)\minus{}f(x)|\le \frac{K\delta_1}{2}\plus{}\delta_2(b\minus{}a)$. If we choose $ \delta_1\equal{}\frac{\epsilon}{K}$ and $ \delta_2\equal{}\frac{\epsilon}{2(b\minus{}a)}$, this $ p$ is an approximation to within $ \epsilon$.

[quote="JRav"]I don't see how this is pertinent to a solution.[/quote]
Please see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=232342]this thread[/url].  alekk was trying to get you to realize that the conditions for Stone-Weierstrass don't hold.

[quote="jmerry"]I would do a two-step approximation. [/quote]
I appreciate the response.  Today I came up with the following; does it work?

Let $ (p_{n})\rightarrow f$, where $ p_{n}\in\text{Lip}_{K}[a,b]$ for each $ n$.  Then for any $ x\in [a,b]$, $ n\geq 1$
$ \left|p_{n}(x)\minus{}p_{n}(y)\right|\leq Kd(x,y)$ and $ \left\|p_{n}\minus{}f\right\|<\varepsilon$.  Then by the triangle inequality,
$ \left| f(x)\minus{}f(y)\right|\leq \left| f(x)\minus{}p_{n}(x)\right|\plus{}\left|p_{n}(x)\minus{}p_{n}(y)\right|\plus{}\left| p_{n}(y)\minus{}f(y)\right|$
$ <\varepsilon\plus{}Kd(x,y)\plus{}\varepsilon'$.

As we take $ n\rightarrow\infty$, $ \varepsilon,\varepsilon'\rightarrow 0$ so $ \left| f(x)\minus{}f(y)\right|\leq K d(x,y)$ and hence $ f\in\text{Lip}_{K}[a,b]$.  Does this necessarily work?

That looks like the converse: if you can approximate it uniformly by equi-Lipschitz functions, it's Lipschitz.

So then that doesn't necessarily imply my problem.  So your method requires us to assume that $ f$ is already Lipschitz and then show that we can approximate it, which makes sense.


function
inequalities
abstract algebra
algebra
polynomial
real analysis
real analysis unsolved

Suppose that $B(z) = b_{0}+b_{1}z+b_{2}z^{2}+\cdots$ is a square summable power series, i.e.
$\sum_{n=0}^\infty |b_{n}|^{2}< \infty$.

Suppose further that $B(z), zB(z), z^{2}B(z), \ldots$ form an orthonormal sequence in the space of square summable power series, i.e.
$\sum_{n=0}^\infty |b_{n}|^{2}= 1$
and
$\langle z^{j}B(z), z^{k}B(z)\rangle = 0$ iff $j \ne k$

(The inner product is what you'd expect: $\langle A(z), B(z) \rangle = \sum_{n=0}^\infty a_{n}\overline{b_{n}}$.)

Prove that $|B(w)| \le 1$ for all $w \in \mathbb{C}, |w| < 1$. Additionally, the inequality is strict if $B(z)$ is not a constant.

Consider the Szego kernel $k_{w}(z)=\frac{1}{1-z\bar w}$: it has the property that $\langle f,k_{w}\rangle=f(w)$ for any $f\in H^{2}$. (This follows from $k_{w}(z)=\sum_{n\ge 0}\bar w^{n}z^{n}$.) Thus, for $n\ge 0$ we have $\langle z^{n}B(z),k_{w}\rangle=w^{n}B(w)$. Now square-sum this:
\[\sum_{n\ge 0}|\langle z^{n}B(z),k_{w}\rangle|^{2}=|B(w)|^{2}\sum_{n\ge 0}|w|^{2n}. \]
The orthonormality assumption implies that the sum on the left is at most $\parallel k_{w}\parallel^{2}=\sum_{n\ge 0}|w|^{2n}$. Hence $|B(w)|^{2}\le 1$. The statement about strict inequality follows from the maximum principle. In addition, the proof shows that $\{z^{n}B(z)\}$ is an orthonormal [b]basis[/b] iff $B$ is a unimodular constant.

At first I unsuccessfully tried to prove that multiplication by $B$ is a contraction on $H^{2}$. Could this approach possibly work?

[quote="mlok"]
At first I unsuccessfully tried to prove that multiplication by $B$ is a contraction on $H^{2}$ [/quote]
Either I misunderstand something, or it is next to obvious: just expand any function $f\in H^{2}$ into its Taylor series $f(z)=\sum_{k}a_{k}z^{k}$, multiply by $B$ and use the orthogonality relations and the condition $\|B\|_{H^{2}}=1$ to get $\|Bf\|=\sum_{k}|a_{k}|^{2}=\|f\|$ (to be formally correct, you have to apply this argument to polynomials first and use some density tricks, of course, but I do not see any principal difficulty on this way) :?

Well, it was late when I posted my answer. :wacko:
The contraction proof also gives $\sup_{|w|<1}|B(w)|=1$, which is not so obvious from the reproducing kernel proof.


floor function
geometry
3D geometry
sphere
quadratics
logarithms
combinatorics unsolved

A hypercube $ Q(n,k)$ is defined to be the set $ \{1,2,...,k\}^n$.  If we want to cover this hypercube with the surfaces of $ m$ ($ n$-dimensional) hyperspheres.  What is the least possible value of $ m$?

I did it by a simple Combinatorial Nullstellensatz argument and showed that if $ m(n,k)$ is the minimum possible $ m$, then
\[ 1 + n\lfloor{\frac {k - 1}{2}}\rfloor \leq m(n,k) \leq 1 + n\lfloor{(\frac {k - 1}{2}})^2\rfloor \,.
\]
However, this does not work for large $ k$.  The lower bound is exact for $ k = 1$, $ 2$, $ 3$, and $ 4$ and a special case with $ n = 2$ and $ k = 5$ (as shown in the attachment).  It is trivially true when $ n = 1$.

Can anyone help improve this bound?

P.S.  It does not have anything to do with any paper or a thesis.  I'm just curious.

When $ n\ge 5$ and $ k\gg n$, you can try to show that each sphere contains at most $ O(k^{n\minus{}2})$ points in your hypercube (it is certainly true if the radii are $ O(k)$ but I'm not sure how to do it for larger radii). This will show that the quadratic growth in $ k$ is inevitable. In dimension $ 2$, for large $ k$, you may have spheres with much more than 1 point on them (almost $ 2^{\log n/\log\log n}$ points, if I remember it right). Using them, you can get a slightly slower than quadratic growth though in the power scale it'll still be $ \ge c_\delta  k^{2\minus{}\delta}$ for all $ \delta>0$.


A spring has a natural length of 20 cm. If a 25-N force is required to keep it stretched to a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm?

Here is what I did.

30cm - 20cm=10cm=.1m  .1mk=25N , 25N/.1m = 250 N

w= x=.20m to x=.25m 250x= 125x^2

125(.25^2 - .2^2)= 45/16 J or N-meters

You can lay out your problem like this.

Extension in spring, x = 0.1m
Tension in spring, F = 25N
Spring Constant, k = F/x = 25/.1 = 250

k = 250 N/m
========

Work done during spring extension, W = (1/2)kx²

When spring extension, x = 5 cm (=0.05m), then

W = (1/2)*250*(0.05)²
W = 125*0.0025
W = 0.3125 Nm
==========

Here is an alternative method.

The relationship between spring force and extension is linear! So, ...

25 N extends spring by 10 cm
Therefore, 12.5N extends spring by 5 cm
Spring force is linear, then average force f = 12.5/2 = 6.25N
Distance moved by (average) force is s = 0.05 m

Work done = (average) force * distance moved by (average) force = fs

W = 6.25 * 0.05 = 0.3125
W = 0.3125N
========


calculus
derivative
limit
algebra
binomial theorem
calculus computations

Hello,
How do you prove that the $ (x^{n})^{\prime}\equal{}nx^{n\minus{}1}$ by using the definition of derivatives, as in in terms of limits?

$ (x^{n})^{\prime}\equal{}\lim_{x\to0}\frac{(x\plus{}h)^{n}\minus{}x^{n}}{h}$

I think about using the binomial theorem, but I'm not sure how to use it.

Can somebody help please?

Thank you very much.

Yes, it's the binomial theorem. Expand out $ (x\plus{}h)^n$, and note that the terms with higher powers of $ h$ will go to zero.

But what happens after that?

OK,
I have $ \frac {x^{n}\plus{}x^{n\minus{}1}h\plus{}x^{n\minus{}2}h^{2}...\plus{}x^{2}h^{n\minus{}2}\plus{}xh^{n\minus{}1}\plus{}h^{n}} {h}$ which becomes $ x^{n\minus{}1}$.

I don't see where the $ n$ comes in.

That's not the binomial theorem.

Let's start a little less ambitiously: try the very specific case $ n\equal{}2.$ Then when you've done that, try the specific case $ n\equal{}3.$

By the way, the Binomial Theorem is $ \sum_{r\equal{}0}^n  {\binom{n}{r}x^{n\minus{}r}y^r}$  where $ \binom{n}{r} \equal{} \frac{n!}{r!(n\minus{}r)!}$

[quote="Kent Merryfield"]Let's start a little less ambitiously: try the very specific case $ n = 2.$ Then when you've done that, try the specific case $ n = 3.$[/quote]
Following up on this, we start with some expansions:

\begin{eqnarray*} (x + h)^{2} & = & x^{2} + 2xh + h^{2} \\
(x + h)^{3} & = & x^{3} + 3x^{2}h + 3xh^{2} + h^{3} \\
(x + h)^{4} & = & x^{4} + 4x^{3}h + 6x^{2}h^{2} + 4xh^{3} + h^{4} \\
& \vdots & \\
(x + h)^{n} & = & x^{n} + nx^{n - 1}h + \text{terms containing }h^2. \end{eqnarray*}

Hence $ \frac {(x + h)^{n} - x^{n}}{h} = nx^{n - 1} + \text{terms having at least one }h.$ Now as $ h\to0$ we have the expected result.

Healker, you of course meant
\[ (x + h)^{n} = x^{n} + nx^{n - 1}h + \text{terms containing }h^2.
\]

Of course. It's been edited now, thanks.

oh my bad  :blush:

I should have been more careful...

That explains it.


I have a monotonic sequence $ \{a_{n}\}$ such that $ \{a_{n\plus{}1}\}>\{a_{n}\}$ and such that $ \{a_{n}\}<M$ for all n.

I am trying to prove that this sequence converges without using the least upper bound property of real numbers.

Any hints would be appreciated.

Hint: the property "every bounded increasing sequence has a limit" is equivalent to the least upper bound property.

I'm not sure I understand.. are you saying I should prove that this sequence has a 'least upper bound' property?

I did not mention this before, but the elements of the sequence and M are rational numbers, and I can only use the rational numbers to show convergence. So I can't use the fact that it converges to a real number.

What mlok is saying is that this is equivalent to the least upper bound property. You [b]need[/b] one of the completeness axioms, such as the least upper bound property, to prove this.
In particular, it's not true if you don't have that property, such as if all our numbers are rational.

I see. Well I'm actually trying to use the above theorem to prove something larger .. that the sequence is a Cauchy sequence, where a sequence $ \{a_{n}\}$ is Cauchy if for all epsilon>0, there's a natural N such that $ |a_{n}\minus{}a_{m}|$, where m,n>N.

Now I'm under the impression that my first method, proving convergence, will not be the best way to prove this.

That's weaker than convergence, and you can prove that it's Cauchy without invoking completeness. I'd prove the contrapositive here: if it's increasing and not Cauchy, it's unbounded (there's some $ \epsilon$ that we can always step up by).

This does require one property of the reals/rationals that's stronger than the ordered field axioms: the Archimedean property that the sequence $ c,2c,3c,\cdots$ is unbounded for any $ c>0$.


Wait, how could AB trisect MN? There's a theorem that says two lines can't intersect at two points...
Do you mean square ABCD trisects MN?

lols
im not sure either
perhaps it means it intersects it at one of hte trisection points

Is this possible?  I can't draw a diagram that satisfies all of the given conditions...

Generally, you need two points to trisect a segment.  However, when only one point (the intersection of the two lines) exists, but it is in such a position that it divides a segment into a ratio of 2:1, it can be called a trisector as well.  Thus, for example, "the centroid of any triangle bisects the three medians."  So, it just says that segment AB intersects MN in a point that divides MN into two segments, one of which is longer than the other.  I do believe the question leaves it open as to which pieces is longer, however.

[quote="darkquantum"]Vertex A of square ABCD is placed at the center of square MNPQ, and AB trisects MN. If AB=MN=1, find the area of hte overlap of ABCD and MNPQ.[/quote]

I don't think it really matters exactly how MNPQ is oriented in terms of the angle it forms with ABCD. Basically, if X and Y are the two intersection points of MNPQ and ABCD and N is the vertex of MNPQ that is inside ABCD the overlap will be [ANX]+[ANY]=[ANX]+[AMX](sim. triangles)=[AMN]=1/4[MNPQ]=1/4. Is my reasoning wrong somewhere?

You are quite correct.  Just look at the areas swept out by each of the two edges of the square as it rotates.


calculus
derivative
linear algebra
linear algebra unsolved

find all linear maps of $M_n(K)$ that satisfy:
$f(AB)=f(A)B+Af(B)$.  (like a derivative ..)

You already posted this one, Alekk. It is from Moubi's proof that I found the useful idea about $Vec f(x)= K Vec (X)$.  ;)


In $\triangle ABC$, three points are placed $R$, $S$, and $T$, on the segments $BC$, $AC$, and $AB$ such that $BR=RC$, $CS=3SA$, and $\frac{AT}{TB}= \frac{p}{q}$. If the area of $\triangle RST$ is twice the area of $\triangle TBR$, find $\frac{p}{q}$.

EDIT: Thanks for pointing that out.

um...am I just confused? because if R, S, T are on ABC, I'm pretty sure that the area of RST can't be greater than that of ABC. Is it supposed to be vice versa?

[hide] Say the area of Triange ABC is 8x.  Now, draw AR,RT, RS ST, and SB.  Since AR bistects BC, we have [ABR]=[ARC]=8x/2=4x.  Now, since 3SA=SC, we have that 3[RAS]=[SRC], and [RAS]+[SRC]=4x.  This implies that [SRC]=3x.  Now, since AT/BT=p/q, we have that [ATR]/[TBR]=p/q.  We also know that [ATR]+[TBR]=[ABR]=8x/2=4x.  This means that [TBR]=(4xq)/(p+q).  Take a look at Triangle ABS.  Since AS=AC/4, we have that [ABS]=8x/4=2x.  Therefore, because AT/TB=p/q, [AST]=2xp/(p+q).  Adding up the areas of Trianges AST, TBR, and SRC, we get that [AST]+[TBR]+[SRC]=4xq/(p+q)+2xp/(p+q)+3x=[ABC]-[RST]=8x-2[TBR]=8x-8xq/(p+q).  Now, we can divide everything by x to get 4q/(p+q)+2p/(p+q)+3=8-8q/(p+q).  Therefore, 12q/(p+q)+2p/(p+q)=5.  Thus, 12q+2p=5(p+q).  So, 12q+2p=5p+5q; 7q=3p.  This means that p/q=7/3.  [/hide]

I am pretty sure that the answer is $\frac{7}{3}$. I would like someone to prove it. My method is long and ugly, with the area formula $A=\frac{ab\sin{\theta}}{2}$ and the law of sines. Is there a more elegant way?

[quote="i_like_pie"]It's supposed to be $[ABC]=2[RST]$.[/quote]

Are you sure, since apparently it is something else.


Is $ 0.999\ldots \equal{} 1$ true?

 :huh:

Sort of.
In math last year, we had this debate about pi and 0.99999... and half of us thought it was indeed 1, but the other half thought it was just 0.99999... forever.

EDIT: SORRY I forgot to add that 0.9999 is, in fact, 1. (pressed submit button too soon!!)

$ 0.\overline{9}\equal{}\frac{9}{10}\plus{}\frac{9}{100}\plus{}\frac{9}{1000}\plus{}\dots\equal{}\frac{\frac{9}{10}}{\frac{9}{10}}\equal{}\boxed{1}$.

Ducky, you don't need to spam, especially if you're not helping the OP.

I have a question... How did you get
$ \frac{\frac{9}{10}}{\frac{9}{10}}?$

Read the bottom of [url=http://www.homeschoolmath.net/teaching/rational_numbers.php]this[/url] page to see where I first learned about this:

$ x\equal{}.\overline{9}$

$ 10x\equal{}9.\overline9$

$ 10x\minus{}x\equal{}9.\overline9 \minus{} .\overline 9$

$ 9x\equal{}9$

$ x\equal{}1$

It is true that $ .\overline 9\equal{}1$.

[quote="SuperNerd123"]I have a question... How did you get
$ \frac {\frac {9}{10}}{\frac {9}{10}}?$[/quote]

Formula for sum of infinite geometric series.

Hmmm...PowerofPi's explanation seems right, as does AIME's.
I think I'll go with $ 0.999\ldots \equal{} 1$ :)

EDIT: I tried PowerofPi's explanation with $ 1.\overline9$ and here's what I got:
$ x\equal{}1.\overline9$

$ 10x\minus{}x\equal{}19.\overline9\minus{}1.\overline9$

$ 9x\equal{}18$

$ x\equal{}2$

Yes, I know that you could just say that $ 1.\overline9\equal{}1\plus{}0.\overline9$ but I just wanted to test this out. :)
It works for all numbers ending with .999... because at the end it always cancels out.

Isn't there some kind of rule for all repeating decimals? Like with 0.11111.... 0.22222, etc.

As it said in my post's link, you can do something similar to the following:

$ x\equal{}.a\overline{bc}$

$ 100x\equal{}ab.c\overline{bc}$

$ 99x\equal{}ab.c$

$ x\equal{}\frac{abc}{990}$

You can do similar things with more or less repeating parts (b and c), and more or less terminating parts (a).

This is a little unrelated, but consider this:

If $ a$ is positive, the quotient $ \frac{a}{b}$ increases as $ b$ decreases. If $ b$ was equal to $ 0.000...0001$ (infinite zeroes), the quotient $ \frac{a}{b}$ would be equal to positive infinity. $ 0.000...0001 \equal{} 1 \minus{} 0.999...$ Since $ 0.999... \equal{} 1$, $ 0.000...0001 \equal{} 0$. Thus, dividing by $ 0$ appears to yield infinity for any nonzero dividend.

Hmm...I thought dividing by 0 is undefined...but you make an interesting point... :huh:

You can't put a 1 after an infinite number of 0's... You'll never get to the 1 so all the digits after the decimal point are 0 so the number is 0. Therefore $ 1\minus{}0.\overline{9}\equal{}0,$ so $ 1\equal{}0.\overline{9}.$

Don't divide by zero (except under admin supervision)!

As for the question: We know .1111111111... is equal to 1/9, right? And .9999999999.... is clearly equal to 9 times .11111111111.... Therefore, since .111111.... is 1/9, 9 times 1/9 is just 1. Therefore, .999999999....=1.

Well, if $ x\equal{}1$, then $ 10x\equal{}10$, so $ 9.\overline9$ is $ 10$. And $ 10\minus{}1\equal{}9$, not $ 8.9$. You are contradicting yourself.

[quote="SuperNerd123"][quote="PowerOfPi"]Read the bottom of [url=http://www.homeschoolmath.net/teaching/rational_numbers.php]this[/url] page to see where I first learned about this:

$ x = .\overline{9}$

$ 10x = 9.\overline9$

$ 10x - x = 9.\overline9 - .\overline 9$

$ 9x = 9$

$ x = 1$

It is true that $ .\overline 9 = 1$.[/quote]
I don't think this is correct because then we could do this:
$ x = .\overline{9}$

$ 10x = 9.\overline9$

Then if we minus x from this it will be like minusing 1 because $ .\overline9 = 1$ So

$ 9x = 8.\overline9$
That meanse
$ x = .\overline9$[/quote]

In your 3rd step, you assumed that x=1 (which is the same as .999999...=1) in order to get x=.99999999999 so really theres nothing wrong with PoP's or your answer. You even said in the middle that .999999=1, so you can change the last step too to make it x=1

The least controversial way to do this is to do the sum of an infinite geometric sequence, like AIME did

Just FYI, I wrote a bit about repeating decimals [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=282644]here[/url].

Yup, in AIME15's link, it says that
$ 0.\overline{x} \equal{} \dfrac{x}{9}$ so $ 0.\overline{9} \equal{} \dfrac{9}{9} \equal{} 1$.

And thanks everyone for answering! :D

EDIT: I just realized that AIME's $ 0.\overline{x}\equal{}\dfrac{x}{9}$ is similar to solafidefarms' solution. :)


topology
vector
Functional Analysis
advanced fields
advanced fields unsolved

Could anyone of you please have a quick look at the following problem?

"Let $ X$ be an infinite-dimensional (Hausdorff) Fréchet space. Prove that the dual space $ X^*$, with its weak*-topology, is of the first category in itself."

And yet another question: Does this statement remain true, if $ X$ is not complete (thus only a Hausdorff, metrizable, locally convex, $ \infty \minus{} \dim$ topological vector space)?

Thank you very much in advance!

In a Banach space it is enough to write $ X^*$ as the union of closed balls centered around zero. In the weak*-topology any open set is unbounded, so clearly a closed ball is nowhere dense. I don't know anything about Frechet spaces, but I'm sure you can come up with something.

As for the second question the dual of $ X$ is the same as the dual of the completion of $ X$.

Thank you for your answer, that was my intuition, too. I had some problems with the details in Frechét spaces first, but I found a way around it that is basically the same as for Banach spaces.

[hide="If anyone is interested:"]
Just write $ X' \equal{} \bigcup_{n\in\mathbb{N}}K_{B_{\frac{1}{n}}(0)}$, where $ K_{B_{\frac{1}{n}}(0)} \equal{} \{T\in X' \big\vert |Tx| \leq 1 \,\forall x\in {B_{\frac{1}{n}}(0)}\}$ is the polar of $ B_{\frac{1}{n}}(0) \subset X$. All those are weak*-compact due to Banach-Alaoglu, thus bounded. Since $ \dim X' \equal{} \infty$ (and this is fact all we need aside from metrizability - we could use local convexity of $ X$ and Hahn-Banach here for example -) bounded sets are nowhere dense in $ X'$ since any weak*-open set in $ X'$ contains an infinite-dimensional subspace.
[/hide]

This is true in every metrisable LCTVS X if dim(X)=oo.The union of the Polars of a countabel local base in X is X'.But a Polar of a 0-neighbourhood is weak*-kompact (by Alaoglu-Bourbaki) a fortiori weak*-boundet.
Patrick

Well, that's exactly what I wrote in my post above yours. Thank you anyway.

oh,sorry :-)If E is a normed space on can say more:(E',weak*) barrelled <=> dim(E)<oo.If E is a banach space we have a nice analog:(E,weak) barrelled <=> dim(E)<oo.
Patrick


What's the most elegant solution of this problem:

Triangle ABC has sides AC, BC and AB measuring 18, 24 and 30 units, respectively.  If D is the midpoint of segment AB, what is the length of segment CD?

$ \triangle{ABC}$ is a special triangle. $ 18: 24: 30=...$

The previous posters were hinting at the fact that your 18:24:30 triangle is just an enlarged 3:4:5 right triangle.  Sketch a right triangle with the vertices labeled as shown ... connecting C to D.

It is helpful to know a few things ... in a right triangle, the midpoint of the hypotenuse is the center of the circumscribed circle.  If you drop a perpendicular from D to BC and from D to AC, you have divided the triangle into four congruent triangles, all similar to the original ... both of these observations are relatively easy to prove ... play with it a little and I am sure you will see what I am pointing out.

[quote="jbatterson"]

It is helpful to know a few things ... in a right triangle, the midpoint of the hypotenuse is the center of the circumscribed circle.  If you drop a perpendicular from D to BC and from D to AC, you have divided the triangle into four congruent triangles, all similar to the original ... both of these observations are relatively easy to prove ... play with it a little and I am sure you will see what I am pointing out.[/quote]

Very Good! :lol:

[quote="vliu"]It would just be easier to calculate the area of the triangle using herons. Once the area is calculated u could find CD because u know that it is the altitude and the base would be line AB. Once the area is found multiply the are by 2 and then divide by line AB to get altitude or CD[/quote]
[quote="Peterhi"]If D is the midpoint of segment AB[/quote]
CD isn't the altitude to AB, it is the median.

funny, i was reading a book about medians when i noticed you post :P


analytic geometry
geometry
parallelogram
power of a point
radical axis
geometry proposed

Let a quadrilateral $ABCD$ such that:
$\angle{D}=\angle{B}=90^{o}$ and $AD=AB$. Take two points $E,Z$ on the segments $AD,AB$ respectively.Name $M$ the midpoint of $EZ$ and $K$ the intersection of $BD$ with the perpendicular line from $C$ to $EZ$.
Prove that the points $K,M,A$ are collinear.

Dear Nick.

Thank you for a beautiful problem, reminds to me another old well known ( at least to me) Lemma, based on which, we can have a solution of the one that you presented.

[b][size=109]LEMMA.[/size][/b] (I continue your wording)

An isosceles triangle $AB'D'$ $($ $AB' = AD'$ $),$ is given and let $K'$ be, an arbitrary point on $B'D'.$

Through $K',$ we draw two lines parallel to $AB',$ $AD'$ and we denote as $E,$ $Z,$ their intersection points with $AD',$ $AB',$ respectively.

Prove that the midperpendicular of the segment $EZ,$ passes through a constant point.

Best regards, Kostas Vittas.

PS. I will post here next week, the solution I have in mind, based on the above Lemma.

$($ In your configuration, we denote as $K',$ the intersection point of the lines through $E,$ $Z$ and parallel to $AB,$ $AD,$ respectively. Also the points $B',$ $D',$ are the intersection points of $AB,$ $AD$ respectively, from the line through $K'$ and parallel to $BD$ $).$

Because $AB=AD$ and $\angle ABC=\angle ADC=90$ so $BC=CD$ So that $\angle CBD=\angle CDB$
Let the line parallel with $EZ$ through $K$ meet $AD,AB$ at $F,N$. As a result, $CK \bot FN$ $(1)$
Otherwise $\angle CBN=\angle CDF=90$ so $C,K,F,D$ and $C,K,B,N$ are cyclic
Thus $\angle CFK=\angle CDB=\angle CBD=\angle CNK$
so triangle $NCF$ is isosceles
and from (1) we have $K$ is the midpoint of $FN$
This means $A,K,M$ are collinear $q.e.d$

[quote="silouan"]Huyền Vũ congrats very nice solution . :)
But I am sure that Nick has another solution  :wink:
Could you post it Nick?[/quote]

No Silouan!!I have no time at all!!!I have to prepare the lamb.... :P

So I wish you a nice <<eating>>  :P .
I found also a solution using coordinate geometry .
The little trick is to take the (0,0) at the midpoint of CD and
(-d,0) to B and (d,0) to D , (0,a) to A and (0,c) to C . :wink:

Through $K,$ we draw two lines parallel to $AB,$ $AD$ and we denote as $E',$ $Z',$ their intersection points, from the sidelines $AD,$ $AB,$ respectively.

Because of the parallelogram $AE'KZ',$ the segment line $AK,$ passes through the midpoint of the segment $E'Z'$ and it is enough to prove that $E'Z'\parallel EZ.$

We draw the circles $(Z'),$ $(E'),$ centered at $Z',$ $E'$ respectively, taken as their cords the bases of the isosceles triangles $\bigtriangleup Z'BK,$ $\bigtriangleup E'DK$, respectively, and we denote and we denote as $L,$ their second common point, the other than $K$.

The segment lines $CB,$ $CD,$ are tangent lines of $(Z'),$ $(E')$ respectively, because $CB\perp EB,$ $CD\perp ED.$

Because of now, $CB = CD,$ we conclude that the point $C,$ lies on the segment line $KL,$ as the radical axis of $(E'),$ $(Z').$

Hence, we have $CK\equiv KL$ $\Longrightarrow$ $CK\perp E'Z'$ $\Longrightarrow$ $E'Z'\parallel EZ$ and the proof is completed.

Kostas vittas.

Well, vittasko's solution is really very nice


algebra
function
domain
modular arithmetic
superior algebra
superior algebra unsolved

Heres a routine excercise:
Show that R[X]/(X^2 +1) is isomorphic to C.

Are you sure? It seems to me that it is not even integrity domain! Probably you wanted $ x^2+1$. Then indeed it is routine.

Oops very very sorry I will edit the post. :blush:

Denote the ideal $(x^2+1)$ by $\mathfrak a$.

For any $f(x)\in \mathbb{R}[x]$, $f(x)=q(x)(x^2+1)+ax+b$, where $a,b\in \mathbb{R}$, i.e., $f(x)\equiv ax+b\pmod{\mathfrak a})$

So for any linear form $ax+b$ and $cx+d$, $ax+b\equiv cx+d\pmod{\mathfrak a}$ if and only if $a=c$ and $b=d$.

Hence $\mathbb{R}[x]/(x^2+1)=\{\overline{ax+b}|a,b\in\mathbb{R}\}\cong \mathbb{C}$.


analytic geometry
geometry unsolved
geometry

[quote="krovacek"]I know how to compute barycentric coordinates of a point in a triangle:

P = u*A+v*B+w*C
u+v+w = 1

P.. point
A,B,C .. triangle
u,v,w .. barycentric coordinates

But how do I compute barycentric coordinates of a point in square? (or regular n-angle)[/quote]

You can do it the same way :)
Even the polygon could be nonregular :)
And coordinates could be simply proportional to the distances from P to respective sides of polygon (instead of $\sum X_{i}=1$).

(Then the equation of a line is $\sum_{i=1}^{n} X_{i}k_{i}=0$


Hi... Well, I'm writing this because (AS YOU CAN SEE) the flag of my country doesn't appear  :( It isn't so important but I feel a bit discriminated  :? Just if you want to know, I live in El Salvador, Central America. Thank you...

your flag is not the only problem, but many other countries have problems with their flags as well. hmm lets give our greatest admin Valentin some time to fix it :D


Let $ a$ be a positive integer. How many non-negative integer solutions $ x$ does the equation \[ \left[\frac{x}{a} \right] \equal{} \left[\frac{x}{a\plus{}1} \right]?\] Note that for each $ z \in \mathbb{R},$ $ [z]$ denotes the greatest integer which is not greater than $ z.$

$ {[x/a]\equal{}[x/(a\plus{}1)]\iff \exists p,r,r'\in\mathbb{N}\equal{}\{0,1,2,...\}: \ x\equal{}pa\plus{}r\equal{}p(a\plus{}1)\plus{}r': \ (r<a,r'<a\plus{}1)}$
we have $ p\plus{}r'\equal{}r<a$ (*)
so it's easy to see that $ |\{x\in\mathbb{N}: \ [x/a]\equal{}[x/(a\plus{}1)]\}|\equal{}|\{(r',p)\in\mathbb{N}: \ r'\plus{}p<a,\ r'<a\plus{}1\}|$
it's $ \sum_{r'\equal{}0}^{a\minus{}1}(a\minus{}r')\equal{}\sum_{k\equal{}1}^{a}k\equal{}\frac{a(a\plus{}1)}{2}$

for exemple if $ a\equal{}2$: $ [x/2]\equal{}[x/3]\iff x\in\{0,1,3\}$: $ 3$ solutions$ \equal{}\frac{2.3}{2}$
r=p+r'

$ x \equal{} ma \plus{} n$, where $ n\in [0,\; a \minus{} 1]$. Then

$ m \equal{} \left[m\cdot\frac {a}{a \plus{} 1} \plus{} \frac {n}{a \plus{} 1}\right]$

If $ m \equal{} n$ we have $ a$ solutions $ x_m \equal{} m(a \plus{} 1)\;\forall m \in [0,\; a \minus{} 1]$.

Then $ m \equal{} m\cdot\frac {a}{a \plus{} 1} \plus{} \frac {n}{a \plus{} 1} \minus{} \left\{m\cdot\frac {a}{a \plus{} 1} \plus{} \frac {n}{a \plus{} 1}\right\}$

$ 1 > \left\{m\cdot\frac {a}{a \plus{} 1} \plus{} \frac {n}{a \plus{} 1}\right\} \equal{} \frac {n \minus{} m}{a \plus{} 1}\geq 0.$

So, $ m\leq n\leq a \minus{} 1$

If $ n > m$ then should be

$ m \plus{} 1 > m\cdot\frac {a}{a \plus{} 1} \plus{} \frac {n}{a \plus{} 1} > m$

or

$ m \plus{} a \plus{} 1 > n$ but it is true for all our $ n$.

So, finally there exist $ \frac {a(a \plus{} 1)}{2}$ solutions


"The number of cents pre kilometer it cost to drive a car depends on how gast you drive it. At low speeds the cost is high because the engine operates inefficiently, while at high speeds the cost is high because the engine must overcome wind resistence. At moderate sppes the cost reaches a minimum. Assume, therefore, that the number of cents per kilometer varies quadratically with the number of kilometer per hour (kph). "

then it asked: "Supppse that it costs 28, 21, and 16 cents pre kilometer to drive at 10, 20, 30 kph, respectively. Write the function"
how would you get the function?

It's rather impolite to post the same question over and over again, even when you tweak it slightly in between posts, and especially when someone has already been kind enough to offer you some help (despite my constant disparagement) on another forum.  Perhaps this is a question you should be asking your teacher.

I think this post was in other forum also so I'll lock this.


Nice inequality, Mathboy !
Actually, the stronger inequality holds

$x+y+z >2\sqrt{a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}$.

[quote="mathboy_cntHDVN"]$ax+by+cz=xyz$
prove that :    $x+y+z\geq \sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c}$
   $a;b;c;x;y;z>0$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1899


Four triangles are circumscribed around a circle with radius 1 such that their 12 vertices form a regular dodecagon. WHat is the area of the dodecagon?

[hide="A bit long, but it does not use trig."]It stands to reason that the four triangles are equilateral. Now, we can determine that the distance between the center of the circle and any of the vertices of each circumscribed triangle is 2 units. Hence we can conclude that the regular dodecagon is inscribed in a circle of radius 2.

Then, this dodecagon can be divided into six congruent kites. Let $ ABOC$ be one of the kites where $ O$ is the center of the circle, $ AO \equal{} BO \equal{} CO \equal{} 2$.

We note that $ AO \equal{} CO$ and $ \angle AOC \equal{} 60^{\circ}$, so $ AC \equal{} AO \equal{} CO \equal{} 2$. Since $ ABOC$ is a kite, its area is given by $ A_{ABOC} \equal{} \frac {1}{2}d_1d_2 \equal{} \frac {1}{2}(2)(2) \equal{} 2.$

Hence the area of the dodecagon is $ 2 \times 6 \equal{} 12$.[/hide]


induction
number theory unsolved
number theory

The sequence $ (x_n)$ is defined as; $ x_1\equal{}a$, $ x_2\equal{}b$ and for all positive integer $ n$, $ x_{n\plus{}2}\equal{}2008x_{n\plus{}1}\minus{}x_n$. Prove that there are some positive integers $ a,b$ such that $ 1\plus{}2006x_{n\plus{}1}x_n$ is a perfect square for all positive integer $ n$.

$ 1 \plus{} 2006x_1x_2$ is supposed to be square. This inspire us to choose $ x_1x_2 \equal{} 2006 \plus{} 2$, in other words, $ a \equal{} 1$ and $ b \equal{} 2008$.
So $ x_0 \equal{} 0$, $ x_1 \equal{} 1$ and $ x_2 \equal{} 2008$, with recurence relation $ x_{n \plus{} 1} \equal{} 2008x_n \minus{} x_{n \minus{} 1}$. We we'll prove by induction that $ 1 \plus{} 2008x_nx_n \plus{} 1 \equal{} x_n^2 \plus{} x_{n \plus{} 1}^2 \Leftrightarrow 1 \plus{} 2006x_nx_{n \plus{} 1} \equal{} (x_{n \plus{} 1} \minus{} x_n)^2$.

For $ n \equal{} 0$ this is obvious. $ 1 \plus{} 2008x_{n \plus{} 2}x_{n \plus{} 1} \equal{} 1 \plus{} 2008(2008x_{n \plus{} 1} \minus{} x_n)x_{n \plus{} 1} \equal{} 1 \plus{} 2008^2x_{n \plus{} 1}^2 \minus{} 2008x_nx_{n \plus{} 1} \equal{} x_n^2 \plus{} x_{n \plus{} 1}^2 \minus{} 2008x_nx_{n \plus{} 1} \plus{} 2008^2x_{n \plus{} 1}^2 \minus{} 2008x_nx_{n \plus{} 1} \equal{} (2008x_{n \plus{} 1} \minus{} x_n)^2 \plus{} x_{n \plus{} 1}^2 \equal{} x_{n \plus{} 2}^2 \plus{} x_{n \plus{} 1}^2$ and that's it :)

Bye

PS I don't know why does it format like this, if you can fix it, feel free to do it :)

We choose $ a \equal{} 1$ and $ b \equal{} 2008$  . Then by induction or other method can easily prove that

$ (x_n \minus{} x_{n \minus{} 1})^2 \equal{} 1 \plus{} 2006x_nx_{n \minus{} 1}$

For all these the basic idea was to find a,b such that $ (b \minus{} a)^2 \equal{} 2006ab \plus{} 1$

Edit : Sponge Bob was faster  :P

In the same method I found that :
Prove that for $ m\in N$ and $ m\geq 3$ . Consider the sequence :
$ x_1 \equal{} a,x_2 \equal{} b$
$ x_{n \plus{} 2} \equal{} mx_{n \plus{} 1} \minus{} x_n$
Prove that for every k there exist $ a,b\in N$ such that $ k^2 \plus{} (m \minus{} 2)x_{n}x_{n \plus{} 1}$ is a perfect square for all n.


Is there anybody here has any pdf file for Thesis paper in any subject of Abstract algebra or Topology.?

I'd like to see one.  :wink:

Typically, universities will post on their website the theses of recent graduates.  From my school, here are some recent ones.  Since I don't know what any of these things are, you'll have to determine if any of them are applicable.

http://www.mathematics.jhu.edu/new/grad/recentphd.htm

[quote="JRav"]I'll take that to mean it was a help?  No problem.

I did a search on that other CJ Ash paper you wanted but unfortunately, I can't seem to find it.  What is the full name of that journal?[/quote]

Yes it was really Helpful . I'm so thankful

For :

Ash, C. J.
A consequence of the axiom of choice.
J. Austral. Math. Soc. 19 (1975), 306--308.

it could be  [b]Journal of the Australian Mathematical Society[/b]

Thank u again dude.

 :)

I also looked it up, but our library only goes back to 1980 for that journal.

I thank u HilbertThm90.

I still wait , I wish someone can find it...

 :)


Sorry, I couldn't really word my title correctly...
Anyways, I recall an easy way to find "ex) how many powers of 2 are in 58! ?"
I would divide 58 by 2 and then take out the remainder and divide by two again and later add all the quotients or something like that.
Could anyone possibly "clearly" remember/know this?  Thanks

If you want to find how many times $ n$ appears in $ m!,$ continually divide $ m$ by powers of $ n$ starting from $ n^1$ and flooring it each time until you get $ 0$ as the result. Then add them up. For your example, \[ \left\lfloor\frac{58}{2^1}\right\rfloor\plus{}\left\lfloor\frac{58}{2^2}\right\rfloor\plus{}\left\lfloor\frac{58}{2^3}\right\rfloor\plus{}\left\lfloor\frac{58}{2^4}\right\rfloor\plus{}\left\lfloor\frac{58}{2^5}\right\rfloor\plus{}\left\lfloor\frac{58}{2^6}\right\rfloor\equal{}29\plus{}14\plus{}7\plus{}3\plus{}1\plus{}0\equal{}54.\]


geometry
incenter
Euler
function
perimeter
circumcircle
number theory

The questions can be downloaded as a pdf file [url=http://www.mathlinks.ro/Forum/download.php?id=901]HERE[/url].

[b]First Day [/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18413]Problem 1[/url]
Find all integers $n$, such that the following number is a perfect square \[N= n^4 + 6n^3 + 11n^2 +3n+31. \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=124052#124052]Problem 2[/url]
Let $ABCD$ be a convex quadrilateral, $I_1$ and $I_2$ be the incenters of triangles $ABC$ and $DBC$ respectively. The line $I_1I_2$ intersects the lines $AB$ and $DC$ at points $E$ and $F$ respectively. Given that $AB$ and $CD$ intersect in $P$, and $PE=PF$, prove that the points $A$, $B$, $C$, $D$ lie on a circle.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=124057#124057]Problem 3[/url]
Find all reals $k$ such that \[ a^3+ b^3+c^3 + d^3 \geq k(a+b+c+d)  \] holds for all $a,b,c,d \geq -1$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=124059#124059]Problem 4[/url]
Let $\mathbb{N}$ be the set of positive integers. Let $n\in \mathbb{N}$ and let $d(n)$ be the number of divisors of $n$. Let $\varphi(n)$ be the Euler-totient function (the number of co-prime positive integers with $n$, smaller than $n$).

Find all non-negative integers $c$ such that there exists $n\in\mathbb{N}$ such that \[ d(n) + \varphi(n) = n+c , \] and for such $c$ find all values of $n$ satisfying the above relationship.

[b]Second Day[/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=124062#124062]Problem 5[/url]
The sequence $\{a_n\}_{n}$ satisfies the relations $a_1=a_2=1$ and for all positive integers $n$,
\[ a_{n+2} = \frac 1{a_{n+1}} + a_n . \]
Find $a_{2004}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=124061#124061]Problem 6[/url]
All the grids of a $m\times n$ chess board ($m,n\geq 3$), are colored either with red or with blue. Two adjacent grids (having a common side) are called a "good couple" if they have different colors. Suppose there are $S$ "good  couples". Explain how to determine whether $S$ is odd or even. Is it prescribed by
some specific color grids? Justify your answers.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=124063#124063]Problem 7[/url]
Let $\ell$ be the perimeter of an acute-angled triangle $ABC$ which is not an equilateral triangle. Let $P$ be a variable points inside the triangle $ABC$, and let $D,E,F$ be the projections of $P$ on the sides $BC,CA,AB$ respectively. Prove
that \[ 2(AF+BD+CE ) = \ell \] if and only if $P$ is collinear with the incenter and the circumcenter of the triangle $ABC$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=124046#124046]Problem 8[/url]
Let $a,b,c$ be three positive real numbers. Prove that
\[ 1 < \frac a{\sqrt { a^2+b^2} }+ \frac b{\sqrt { b^2+c^2}} +
\frac c{\sqrt { c^2+a^2}} \leq \frac {3\sqrt 2}2 . \]

I have the set of problems, but in Chinese, does anyone interested?
Or does anyone can translate it to English?

Yes. Problem 7 was absolutely clear for me, but I didn't write solution in this topic, because I thought there is separate topic for this problem.
I see some more problems in this set which are quite easy.

I noticed that the first problem looks very similar to the first problem of the same contest 2 years ago (as posted in the same forum)

I have put each problem into a separate topic. Also a pdf file with the problems is available for download.

Hmm.. I am a bit dissapointed..
I think this olympiad is a rather easy one, At least, it seems easier than other olympiads and MUCH easier than CMO.
Russian Regional Round is also harder.

But iura, this is a regional olympiad. As far as I understood, the winners of this olympiad advance to China Math Olympiad. So it's difficulty level is ok.

In my opinion, pr.8 and 2 are not so easy.
However, pr.1, 3-7 are quite simple.

I think only problem 8 is difficult, others are quite easy...

[quote="zhaoli"]CWMO is hold to encourage the students in western China to increase interest in math, and the best 1~2 students was qualified for CMO.[/quote]

Hong Kong is invited to join CWMO these 3 years too.

What is Western China? Tibet and Xinjiang, or Shantong? east west confusion :D

greets,
psykyk