vector
linear algebra
matrix
linear algebra unsolved

Let T:V---> W be a linear transformation. Show that if X1,...,Xr are vectors in V such that T(X1),...,T(Xr) are linearly independent, then so are X1,...,Xr.

thanks!

Try proving the contrapositive: if $x_1,x_2,\dots,x_n$ are linearly dependent, then $T(x_1),T(x_2),\dots,T(x_n)$ are also linearly dependent.

We can prove it the contrapositive as follows:

If $x_1,\ldots,x_r$ are not linearly independent, it follows that we can write

$\sum_i\alpha_ix_i=0$ for some $\alpha_i\in F$, where not all the $\alpha_i$'s are zero.  Then

$T\left(\sum_i\alpha_ix_i\right)=\sum_i \alpha_i T(x_i)=0$.  But this means that $T(x_1),\ldots,T(x_n)$ are not linearly independent.

I started this post before I saw jmerry's post, but I switched windows while composing it and hence got posted under.

How can i prove x1,...,xr are linearly independent? I can prove T(X1),...,T(Xr) are linearly independent if I know  x1,...,xr are linearly independent, but the question didn't say x1,...,xr are linearly independent. Do you guys have any idea to prove it? Thanks!

The proof blahblahblah gave is correct!

Suppose that the vectors $x_1,\dots,x_r$ are NOT linearly independent.
Then there exist scalars $a_1,\dots,a_r$ (in some field), NOT ALL $0$, such that $a_1x_1 + \dots + a_rx_r = 0$. Now, since $T(0) = 0$, we get \[ 0 = T(0) = T(a_1x_1 + \dots a_rx_r) = a_1T(x_1) + \dots + a_rT(x_r).  \] Hence there exists some non-trivial combination of $T(x_1),\dots,T(x_r)$ which is $0$.
Hence $T(x_1),\dots,T(x_r)$ are linearly dependent. Contradiction.
Hence $x_1,\dots,x_r$ are linearly independent and we are done.

Of course this is just blahblahblah's proof. I only tried to make it "clearer" for you.

I don't see any $a_0$ appear in his proof. And I'm telling you again - the proof is correct!!

That doesn't matter, the $\alpha_i$ are just "coefficients" in some field $F$!

Mr.Arne, If you said a_1,\dots,a_r in some field, Not All 0, how can i get  a_1x_1 + \dots + a_rx_r = 0 .  Also, dose  \alpha_i have to be in V. Thanks!

[quote="xyidan"] Not All 0, how can i get  a_1x_1 + \dots + a_rx_r = 0.[/quote]

I'm assuming that the vectors $x_1,\dots,x_r$ are linearly dependent.

In that case, there must exist $a_i$, not all $0$, such that $a_1x_1 + \dots + a_rx_r = 0$.

(Check the definition of "linearly dependent vectors")

[quote="xyidan"]Also, dose  \alpha_i have to be in V. Thanks![/quote]

No, the scalars $a_i$ need to belong to some field $F$. The $a_i$ are NOT vectors themselves.

(Check the definition of a vector space)

[quote="xyidan"]Also, blahblahblah, you said a0 belong to F, what is in your proving? Thanks![/quote]

The vector spaces $V,W$ are both over some common field (that is, I mean the scalars in $V,W$ both belong to some common field) which is typically denoted in the literature by $F$.

By the way, if you are not familiar with it, you can prove that the truth of statement $A$ implies the truth of statement $B$ by proving that if statement $B$ is false, then statement $A$ will also be false.  This is called the contrapositive, and it turns out that this is often easier to prove than the original statement.

Because you prove it is linearly Dependent for T(x1),...,T(x2) but the question said T(x1),...,T(x2) is linearly Independent, the statement is fasle,right?

I assumed that $x_1,\ldots,x_r$ were linearly dependent, and I proved that this implies that the $T(x_i)$'s are linearly dependent (the intuitive view - a linear transformation cannot add any more invertibility to a matrix then there already is).  This is the same as proving that if the $T(x_i)'s$ are linearly independent, then so are the $x_i$'s.


zhaobin  When does equality hold? I think there is no equality sign.
 Since a,b,c >0  \[ a^2+ab+b^2> \frac{9}{16}* (a+b)^2\].

[quote="kueh"] hmm there should be same number of terms on each side...[/quote]

What do you mean? I think that there is the same number of terms on each side.

OA=a,OB=b,OC=c,[tex]\hat {AOB}=120^0[/tex],...
Now we have to prove:[tex][ABC] \le \frac{3\sqrt{3}}{4}R^2[/tex]
But this is easy!

i'm probably missing something very simple, but why can't you just use AM-GM?
LHS = cyc-prod(a^2+ab+b^2) >= cyc-prod(3ab) >= [1/3*cyc-sum(3ab)]^3 = (ab+bc+ca)^3 = RHS...

I suppose it is just holder?
[ab a^2 b^2]
[a^2 ac c^2]
[b^2 c^2 bc]
finished. :)


analytic geometry
geometry
geometric transformation
reflection
dilation

(1) Point A(6,3) is reflected in the x-axis. Find the coordinates of A', its image.

(2) In a plane, what is the locus of points 8 centimeters from a given line?

(A) two paralel lines (C) a line
(B) two intersecting lines (D) a circle

(3) Point P lies on line m. How many points are both 5 units from line m and 6 units from point P.

(4) What equation describes the locus of points 5 units from point (3,-4)?

(A) (x+3)squared+(y-4)squared=25 (C) (x-3)squared+(y+4)squared=5
(B) (x+3)squared+(y-4)squared=5 (D) (x-3)squared+(y+4)squared=25

(5) In the accompanying diagram of triangle ABC, side AB is extended to D. If m<ACB=(x+30)degrees, m<CAB=(2x+10)degrees, and m<CBD=(4x+30)degrees, what is the value of x?

(6) A translation maps (1,4) onto (7,-3). Write the image of (5,10) under the same translation.

(7)Point P(-3,6) is reflected in the line whose equation is y=x. In which quadrant does the image P' lie?

(Cool What is the image of (-2,5) after a reflection in the origin?

(9) If a translation maps point A(-3,1) to point A'(5,5), the translation can be represented by

(A) (x+8, y+6) (C) (x+8, y+4)
(B) (x+2, y+4) (D) (x+2, y+6)

(10) If the lengths of the legs of a right triangle are 5 and 12, what is the length of the hypotenuse?

(A) rad 17 (C) 17
(B) rad 119 (D) 13

(11) The coordinates of two points are A(1,4) and B(3,4). What is an equation of the locus of points equidistant from A and B?

(A) y=2 (C) x=2
(B) y=4 (D) x=4

(12) The equation of the locus of points 5 units from the origin is

(A) y=5 (C) x=5
(B) x squared+y squared=5 (D) x squared+y squared=25

(13) What are the coordinates of N', the image of N(5,-3) under a reflection in the y-axis?

(14) How many points are 2 centimeters from a given line and 2 centimeters from a given point on that line?

(A) 2 (C) 3
(B) 0 (D) 1

(15) Find the image of (3,-2) under the dilation D2.

(16) The line whose equation is y=-3 is reflected in the x-axis. Write an equation of the image of that line.

(17) Triangle ABC has coordinates A(1,2), B(0,5), and C(5,4).

(PART A) State the coordinates of triangle of A'B'C', the image of triangle ABC after the translation which maps (x,y) to (x-6, y+3).

(PART B) State the coordinates of traingle A''B''C'' the image of triangle of A'B'C' after a reflection in the x-axis.

(PART C) State the coordinates of triangle A'''B'''C''', the image of triangle A''B''C'' after a reflection in the origin.

This sort of seems like homework:(......I don't like homework :D

I don't think we are supposed to put homework on AoPS.

Yes, these look like homework, and that's not the point of this forum.  These forums are for students to learn the math themselves, one problem at a time.  These are also not appropriate for the Classroom Math forum.  I'm going to lock and move this thread.


Gauss
quadratics
number theory proposed
number theory

Find all pairs of positive integers $ (x,n)$ satisfying $ x^3 \plus{} 2x \plus{} 1 \equal{} 2^n$

first with fermat we get $ x^3\plus{}2x\plus{}1\equal{}1(mod3)$ that means $ n$ is even. ignore the trivial case with $ n\equal{}2$ and suppose $ n>2$. we easily get $ x\equal{}\minus{}3(mod8)$. now add both sides $ 2$ and we get $ x^3\plus{}2x\plus{}3\equal{}(x\plus{}1)(x^2\minus{}x\plus{}3)\equal{}2^n\plus{}2$. this means that $ \minus{}2$ is a square residue for every prime $ p$ dividing $ (x\plus{}1)(x^2\minus{}x\plus{}3)$. with gauss criteria we easily check that $ \minus{}2$ is a quadratic residue $ (modp)$ if and only if $ p\equal{}1or3(mod8)$. but $ x^2\minus{}x\plus{}3\equal{}\minus{}1(mod8)$ and cannot have all primes in it's factorization of that form (obviously since $ 3^k$ cannot be of the form $ 8k\minus{}1$). so we obtained a contradiction and the only solution left is the trivial one $ (x,n)\equal{}(1,2)$. actually this is a standard technique for these type of equations but this was quite nice.


Hello

How do I solve this problem:

If n is integer:  2^n = 2n + 2

How do the two sides grow as $ n$ gets large?

hello, write your equation in the form $ 2^{n\minus{}1}\equal{}n\plus{}1$ and begin with $ n\equal{}1$.
Sonnhard.

I know that one solution is 3 but how can I prove that there is no more solutions ? (without creating some function and prove that it is increasing)

hello, you can prove $ 2^{n\minus{}1}>n\plus{}1$ for $ n>3$ by induction.
Sonnhard.


algebra
polynomial
Diophantine equation
parametric equation
roots
IMO Shortlist
IMO Longlist

The equation
\[x^5 + 5 \lambda x^4 - x^3 + (\lambda \alpha - 4)x^2 - (8 \lambda + 3)x + \lambda \alpha - 2 = 0\]
is given. Determine $\alpha$ so that the given equation has exactly (i) one root or (ii) two roots, respectively, independent from $\lambda.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Are we talking about not-necessarily-different real roots?  If so, part (ii) is impossible, since an odd-degree polynomial will have always an odd number of real roots (some of them possibly equal).

Are we talking about not-necessarily-real different roots?  If so, part (i) is impossible, since the polynomial should be of the form $ (x \plus{} r)^5$, yielding $ \lambda \equal{} r$ and $ \minus{} 1 \equal{} 10r^2$, obviously impossible for arbitrary $ \lambda$.

Finally, if we write the polynomial as $ \left(x^5 \minus{} x^3 \minus{} 4x^2 \minus{} 3x \minus{} 2\right) \plus{} \lambda(5x^4 \plus{} \alpha x^2 \minus{} 8x \plus{} \alpha) \equal{} p(x) \plus{} \lambda q(x)$, obviously any root that is independent of $ \lambda$ must satisfy $ p(x) \equal{} q(x) \equal{} 0$, but frankly I do not know what to look for now because as I stated above I do not understand what is meant by exactly one root and exactly two roots...

Edit: maybe there is an erratum in the problem, since "suspiciously enough", $ \alpha\equal{}\minus{}3$ makes $ q(x)\equal{}p'(x)$, so any $ \lambda$-independent (possibly multiple) root would be a double root of $ p(x)$, but $ p(x)$ does not have a double real root...

Edit 2: OK, silly me!  The one or two roots are obviously $ \lambda$-independent roots, regardless of how may real or complex roots the polynomial has!!!

Continuing with the idea above that any $ \lambda$-independent root must satisfy $ p(x)\equal{}q(x)\equal{}0$, note that $ p(x)\equal{}(x\minus{}2)(x^2\plus{}x\plus{}1)^2$ has the only real root $ x\equal{}2$.  Now, any $ \lambda$-independent root will also be a root of $ r(x)\equal{}5p(x)\minus{}xq(x)\equal{}(\alpha\plus{}5)x^3\plus{}12x^2\plus{}(\alpha\plus{}15)x\plus{}10$.  Then...

(i) in order to have exactly one root, it must be real, since complex roots come in conjugate pairs both for $ p(x)$ and $ q(x)$ because their coefficients are real (unless stated otherwise I am assuming $ \alpha$ and $ \lambda$ to be real), and it must be the only real root of $ p(x)$, so $ 0\equal{}8(\alpha\plus{}5)\plus{}48\plus{}2(\alpha\plus{}15)\plus{}10$ yielding $ \alpha\equal{}\minus{}\frac{64}{5}$.

(ii) in order to have exactly two roots, they must be complex, and roots of both $ (x^2\plus{}x\plus{}1)^2$ (ie, of $ x^2\plus{}x\plus{}1$) and of $ q(x)$.  Then, since $ q(x)\equal{}(5x^2\minus{}5x\plus{}\alpha)(x^2\plus{}x\plus{}1)\minus{}3x\minus{}\alpha x$, we must have $ \alpha\equal{}\minus{}3$, in which case we find $ q(x)\equal{}p'(x)$, and the complex double roots of $ p(x)$ are also roots of $ q(x)$, or $ \lambda$-independent roots of the given polynomial.


logarithms
ratio
inequalities
real analysis
real analysis unsolved

Does series $ \sum_{n\equal{}1}^\infty\frac{x^n}{(1\plus{}x^n)^n}$ converge uniformly on $ [0,1]$?

$ \frac{x}{1\plus{}x^n}$ reaches its maximum value when $ x\equal{}(n\minus{}1)^{\minus{}1/n}.$ Thus the maximum value of the $ n$th term is $ \frac{1}{(n\minus{}1)\left(1\plus{}\frac1{n\minus{}1}\right)^n},$ which is asymptotic to $ \frac1{ne}.$ So we're not going to get this as cheaply as the Weierstrass M-test.

We also have that the terms are less than $ x^n,$ so we have uniform convergence on $ [0,r]$ for each $ r<1.$ That's not good enough either.

And we have pointwise convergence for all $ x\in[0,1],$ since at $ x\equal{}1$ it's just the series $ \sum 2^{\minus{}n}.$ Still not good enough.

Right now, I don't have the answer - still thinking about it.

just a remark: Dini criterion shows that it suffices to show that $ f(x)\equal{}\sum_{1}^{\infty} \frac{x^n}{(1\plus{}x^n)^n}$ is continuous, and the only tricky thing is to show that $ f$ is continuous at $ x\equal{}1$. To do that, one has to show (the other part is immediate) that for any $ \epsilon>0$ there exists $ \eta$ such that $ x > 1\minus{}\eta$ implies $ f(x)<1\plus{}\epsilon$.

It is simple but not "too simple". While the usual Weierstrass test does not apply, the modified one does. The criterion is that a series $ \sum_n f_n(x)$ converges uniformly if for every $ \varepsilon > 0$, there exist positive majorants $ m_n$, $ M_n$ and sets of indices $ S(x)$ such that $ \sum_n m_n < \plus{} \infty$, $ \sum_{n\in S(x)}M_n < \varepsilon$, and $ |f_n(x)|\le m_n$ for $ n\notin S(x)$, $ |f_n(x)|\le M_n$ for all $ n$.

Now just define $ M_n \equal{} \frac 1n$, $ m_n \equal{} \frac {\log^2 N(\varepsilon)}{n\log^2 n}$, $ S(x) \equal{} \left\{n: n > N(\varepsilon), \frac 1{n\log^2 n} < x^n < \frac {2\log n}n\right\}$ where $ N(\varepsilon)$ is some large number to be chosen later. We need to show that $ \sum_{n\in S(x)}\frac 1n$ is small. To this end, it suffices to show that for any two indices $ m,n\in S(x)$, the ratio $ \frac mn$ is close to $ 1$. If $ S(x)$ contains two indices $ m > n$, we get
\[ \left[\frac {2\log n}n\right]^{m/n} > (x^n)^{m/n} \equal{} x^m > \frac 1{m\log^2 m}\]
whence, since both $ m,n$ are large, $ n^{(1 \minus{} \varepsilon)\frac mn}\le m^{1 \plus{} \varepsilon}$ and $ n^{(1 \minus{} 2\varepsilon)\frac mn}\le m$. If $ N(\varepsilon) > e^2$, this implies the inequality  $ e^{2[(1 \minus{} 2\varepsilon)(\frac mn \minus{} 1) \minus{} 2\varepsilon]} < n^{(1 \minus{} 2\varepsilon)(\frac mn \minus{} 1) \minus{} 2\varepsilon}\le \frac mn$ and, thereby, the estimate
$ 2[(1\minus{}2\varepsilon)(\xi \minus{} 1) \minus{} 2\varepsilon]\le\log\xi\le\xi \minus{} 1$ for the ratio $ \xi \equal{} \frac mn$. But then $ \xi \minus{} 1 < 5\varepsilon$, and we are done.

$ \sum_{n \equal{} 1}^\infty\frac {x^n}{(1 \plus{} x^n)^n}\equal{}\sum_{n\equal{}1}^{\infty}\left\{\frac{1}{(1\plus{}x^n)^{n\minus{}1}}\minus{}\frac{1}{(1\plus{}x^n)^{n}}\right\}$


In the figure below, the largest square has a side length of 80 cm. What is the area of the smallest circle? Express your answer in terms of $\pi$.

[img]http://mathcounts.saab.org/mcgif/p553_3.gif[/img]

[hide="solution"]
The answer is $400\pi \ cm^{2}$ (how do you do superscript?)
The outermost square has a side length of $80$ cm. The next inner square has a sidelength of $40\sqrt{2}$ cm. The innermost square has a side length of $20\sqrt{2}\sqrt{2}$ cm. That is $40$ cm. So the Radius of the smallest circle is $20$ cm, and the radius squared is $400$.
[/hide]

[quote="ProtestanT"][hide="solution"]
The answer is $400\pi cm2$ (how do you do superscript?)
The outermost square has a side length of $80$ cm. The next inner square has a sidelength of $40\sqrt{2}$ cm. The innermost square has a side length of $20\sqrt{2}\sqrt{2}$ cm. That is $40$ cm. So the Radius of the smallest circle is $20$ cm, and the radius squared is $400$.
[/hide][/quote]
Correct.  :)
To do superscripts, use ^. For example, 5^7 produces $5^{7}$. However, if you need something like 5^50 (where the exponent is more than one digit), you need to do 5^{50}. 5^50 produces $5^{5}0$, while 5^{50} produces $5^{50}$.
Also, to display "cm" in LaTeX, use \text{cm} or \mbox{cm}, which will both produce $\mbox{cm}$. However, you should try to avoid using text in LaTeX because it makes posts take much longer to load.
But you usually don't include units in MathCounts answers anyway.  :P

[quote="lotrgreengrapes7926"]Generalization: If the largest circle is $O_{1}$, the next largest $O_{2}$, etc, find the area of the circle $O_{n}$.[/quote]
[hide="Hint"] The radius of each circle is $\frac{1}{\sqrt2}$ times less than the preceding circle. [/hide]


I have a problem and i can't solve it
2^45 * sin1° sin3° ... sin89°= sqrt (2). HElp me

See the last problem in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=43118]10 Trigonometry Questions[/url].


function
logarithms
limit
calculus
calculus computations

[b][u]without[/u][/b] reducing the LHS expression to a relation involving the zeta functions,
show otherwise :D

$ \sum_{j=2}^\infty\;\;\sum_{i=2}^\infty\;\frac{1}{i\cdot j^{i}}\;=\;\boxed{1-\gamma}$

[hide="*** Click to reveal the hidden content ***"]
\begin{eqnarray*}\sum_{j=2}^{\infty}\sum_{i=2}^{\infty}\frac{1}{i \; j^{i}}& = & \sum_{j=2}^{\infty}\left(-\frac{1}{j}+\sum_{i=1}^{\infty}\frac{1}{i \; j^{i}}\right)\\ & = & \sum_{j=2}^{\infty}\left(-\frac{1}{j}-\ln \left( 1-\frac{1}{j}\right) \right)\\ & = &-\sum_{j=2}^{\infty}\left( \frac{1}{j}-\ln \left( \frac{j}{j-1}\right) \right)\\ & = &-\lim_{n\to\infty}( H_{n}-1-\ln n )\\ & = & 1-\gamma\\ \end{eqnarray*}
[/hide]


2005 China High School Math League was held last(this?) Sunday.

See

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=56580

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=56575

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=56579


Show that $n$ does not divide $2^{n}-1$ for $n\ge2$.

Suppose that there exists such $n>1$ and is the smallest with this property.
Let $p$ be the biggest prime wich divides $n$.
$n=m.p^{\alpha}$ with  $m, \alpha$ natural numbers and $p, m$ coprimes.
Note that $p^{\alpha}$ not divides $\phi(n)$ and remenber that $2^{\phi(n)}\cong 1 mod(n)$.
Then, $(n,\phi(n))$ divides $d=m.p^{\alpha-1}$ and $2^{d}\cong 1 mod(n)$.
In particular, we have $2^{d}\cong 1 mod(d)$ wich contradicts the minimality of $n$.

[quote="jonfv"]Then, $(n,\phi(n))$ divides $d=m.p^{\alpha-1}$ and $2^{d}\cong 1 mod(n)$.
[/quote]
It is not true.
Consider minimal prime divisor p.  Let m minimal degree k, suth that $p|2^{k}-1$. We have $1<m<p$, because $2^{1}-1=1$, therefore minimal prime divisor $p_{1}$ number m less than p and $p_{1}|n$ give contradition.


limit
induction
ratio
floor function
function
real analysis
real analysis unsolved

Let $(x_{n})_{n\geq 1}$ a sequence of strictly pozitive numbers, with the following property $\lim_{n\rightarrow\infty}{\frac{x_{n+2}}{x_{n}}}=2$. Show that $\lim_{n\rightarrow\infty}{x_{n}}=+\infty$

$a_n=x_{2n}>0,b_n=x_{2n+1}>0$.
Thus $\exists N \forall n>N \frac{a_{n+1}}{a_n}>1,5$ then $\forall n>N \; a_n > (1,5)^{n-N}a_N \to \infty,\; n\to \infty$. The same for $b_n.$
So $a_n,b_n\to \infty.$
Thus $x_n\to \infty$

I see the ideea, but i don't understand the way you wrote the solution.  :?

All that 1.5 is doing in that solution (if I may use the American 1.5 rather than the European 1,5) is that it's some number between 1 and 2.

Here's the same proof, repeated; maybe it will make sense the second time.

Suppose $x_n$ is a sequence of positive numbers and that $\liminf_{n\to\infty}\frac{x_{n+1}}{x_n}=r,$ where $r>1.$

Since ${\frac{x_{n+1}}{x_n}}$ is eventually arbitrarily close to $r,$ or larger, then it is eventually greater than $s.$

Choose $N$ such that for all $n\ge N,\,\frac{x_{n+1}}{x_n}\ge s.$

We can prove by induction that for $n>N,\,x_n\ge x_Ns^{n-N}.$ Hence, $x_n$ is eventually greater than a sequence which tends to infinity, and thus tends to infinity itself.

This is very closely related to the proof of the ratio test for the convergence of series.

But note that $\lim_{n\to\infty}\frac{x_{n+1}}{x_n}=1,$ even if we have $x_{n+1}>x_n$ for all $n,$ would not be sufficient information for us to conclude anything. We really need the room in between $1$ and $r$ to slip in that other number, $s.$

We can make generalization.
If $\exists \; \alpha > 1,\; m \in {\mathbb N}\; : \lim_{n\to\infty} \frac{x_{m+n}}{x_n} = \alpha$ then $x_n \to \infty$

[quote="anik_andrew"]It's some more intresting, when $m$ is depend of $n$. For example
$\lim_{n\to\infty} \frac{x_{n+n}}{x_n} =2$. Is it follows that $x_n \to \infty$?
[hide="hint"]
$x_n=2^p$, where $n=2^pq$,where $q$ is odd.
[/hide][/quote]
No. For $n = 2^k m$ with odd $m$, define $x_n = 2^{k-m}$. Then $\frac{x_{2n}}{x_n} = 2$ for all $n$, but $\lim_{k \to \infty} x_{2k+1} = 0$.

hi,what about \[ \lim_{n\to +\infty}\frac{x_{n+\lfloor \sqrt{n} \rfloor}}{x_n}=2  \]
then \[ \lim_{n \to \infty} x_{n} = +\infty  \]

[quote="fedja"]Seems that the discussion of this topic won't stop until somebody answers the general question: for which functions $f: \mathbb N\to\mathbb N$ satisfying $f(n)>n$ for all $n$ the condition $\lim_{n\to\infty}\frac{x_{f(n)}}{x_n}=2$ implies that $x_n\to+\infty$ (as before, it is assumed that all $x_n>0$). :P[/quote]
Within the smaller class of functions $f$ that are strictly increasing, this conclusion is correct iff $\lim_{j\to \infty}\left(f(j)-j)\right) < \infty$. Note that $k(j) = f(j)-j$ is non-decreasing in this case, thus this is equivalent to the requirement that $f(j)-j$ is eventually constant.

Proof: Suppose $k(j)$ becomes eventually constant, $k(j) = K$, and $\lim_{n\to\infty}\frac{x_{f(n)}}{x_n}=2$.  After throwing away an initial segment of the sequence, we may assume that $f(n) = n+K$ for all $n$ and $\frac{x_{n+K}}{x_n}\ge \frac{3}{2}$ for all $n$.

For $j = 1, 2, \dots$, set $m_j = \min \{x_{(j-1)K+1}, \dots, x_{jK}\}$. Then $m_{j+1} = x_{n+K}$ with $n$ between $(j-1)K+1$ and $jK$, and thus $m_{j+1} = x_{n+K}\ge \frac{3}{2}x_n \ge \frac{3}{2}m_j$. Therefore $m_j \to \infty$ and also $x_n \to \infty$.

Suppose now $\lim_{j\to \infty}k(j) = \infty$. Then there are sequences with  $\frac{x_{f(n)}}{x_n}=2$ for all $n$, yet $\liminf_{n\to \infty} x_n = 0$. Here's how to construct one:
Define disjoint subsets $A_j \subset \mathbb{N}$ and numbers $a_j \in \mathbb{N}$ as follows: $A_1 = \{1, f(1), f(f(1)), \dots\}$, $a_1 =\min A_1 = 1$.
Suppose $A_1, \dots, A_m$ have already been defined together with their minima $a_j$.

I claim that that $\cup_j A_j \neq \mathbb{N}$. Indeed, let $n_0$ be so large that $f(n) \ge n+2m$ for $n \ge n_0$. Consider the set $B=\{n_0+1, n_0+2, \dots, n_0+m\}$. Then $B$ has at most one element in common with each $A_j$, for otherwise we would have a number $n \ge n_0$ from $A_j$ whose successor in  $A_j$ is also in B, $f(n) \le n_0 + m$. But that not possible by the choice of $n_0$.

If $B \cap \left(\cup_j A_j\right) \ne \emptyset$, we have found a number not in $\cup_j A_j$. Otherwise, each number $n \in B$ appears in exactly one $A_j$, but then $\{n_0+m+1, \dots, n_0+2m\} \cap A_j = \emptyset$ for all $j$, since $f(n) \ge n_0+1+2m$ for all $n \in B$.

Therefore we can find $a_{m+1} = \min \{n \in \mathbb{N} \, | \, n \notin \cup_j A_j\}$. Define $A_{m+1} = \{a_{m+1}, f(a_{m+1}), f(f(a_{m+1})), \dots\}$. Clearly this set has empty intersection with each $A_j, \, j \le m$, since $f$ is 1-1.

We end up with a decomposition $\mathbb{N} = \cup_{j=1}^\infty A_j$, and all $A_j$ are disjoint and infinite.

Now define the sequence $x_n = 2^{k-j}$, if $n \in A_j$ and $n = f\circ \dots \circ f(a_j)$ ($k$-fold composition. One checks that $\frac{x_{f(n)}}{x_n} = 2$ for all $n$ and $x_{a_j} = 2^{-j}$. :)


Let ABC be a triangle.
  Prove that:
    $ab+bc+ca \ge 18Rr.$

Yes,not hard.$Rr=\frac{abc}{a+b+c}$
and $(ab+bc+ca)(a+b+C)\geq 9abc$is obvious.

[quote="Cezar Lupu"]Try to prove this easy but stronger inequality \[a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}\geq 18Rr.\] [/quote]

But still AM \GE GM kills it.....

[quote="Cezar Lupu"]Try to prove this easy but stronger inequality \[a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab}\geq 18Rr.\] [/quote]

$R= \frac{abc}{4S}\Rightarrow bc= \frac{R \cdot 4S}{a}$
$\sum \sqrt{a^{2}\cdot bc}$$= \sum \sqrt{R \cdot 4S \cdot a}\ge 18Rr$
$\Leftrightarrow \sum \sqrt{Sa}\ge9r \sqrt{R}$
$\Leftrightarrow \sum \sqrt{a}\ge 9r \cdot \frac{\sqrt{abc}}{2S}= 9 \cdot \frac{\sqrt{abc}}{a+b+c}$
$\Leftrightarrow (a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c}) \ge 9\sqrt{abc}$
$by AM-GM \Rightarrow abc \sqrt{abc}\ge abc \sqrt{abc}$  $true.$


If the volume of a regular hexadecimalhedron space polyhedron is $ \frac{32\pi^{2}\sqrt{6}}{225}$, what is the side length and the surface area?

Thanks!

1. When you say hexadecimalhedron, do you mean a polytope in 16 dimensions?
2. When you talk about its volume, do you mean its 16-dimensional volume?
3. When you talk about its surface area, do you mean the combined 15-dimensional volume of its 15-dimensional faces?

Umm...

I think a hexadecimalhedron is a 16-faced polyhedron.
Hope that helps. Maybe I'm wrong, but...

I originally thought that too, but there is no regular 16-faced polyhedron.

Well, there is a polyhedron with 16 congruent faces: the [url=http://mathworld.wolfram.com/GyroelongatedSquareDipyramid.html]gyroelongated square dipyramid[/url].  :)


Has anyone gotten the email/paper invitation yet?

NO, not yet. I hope I should have gotten an email last week, but it didn't happen. I think they mail the letters first, because it takes more time to arrive. Then email us which takes little time to arrive. :!:

Ummm...still haven't gotten it yet.  My name has always been on the list though, so...

since they just added 57 more students onto the list, i presume they need another day or two to process the emails in database. yes! we have 2 qualifier from my school now.

are the student numbers random or is it ranked some way...like your place or something?

See, trust us, we do what we say, given time to do our job properly.  Some public thanks to our staff, particularly Marsha and Rachelle for getting this done while the USAMO ground was shifting beneath them, speaking metaphorically.

The numbers are random.  Reading any other meaning into them is numerology.

Steve Dunbar
AMC Director

[quote="JSteinhardt"]I just noticed that Rachelle's last name is Jensen. That's pretty pwnage.[/quote]

Yeah, there was also a Jensen on the list of qualifiers.

And, of course, Jason Trigg.

[quote="beta"][quote="Philip_Leszczynski"]Are you not supposed to reveal your student number?[/quote]

I don't see why you would want to do that... (unless you want everyone to know your score)[/quote]

See other topic  ;)

I got the email at 12:44, and the letter today during school (around 9:00 AM.)

>I got the letter at school,but not the email. What should I do?

Read the letter, and follow the directions.  The letter and email say the same thing.

Steve Dunbar
AMC Director

I think I'm using the wrong term here.  By letter I meant the S-III form. I have not received my USAMO number or any other information.

Your teacher should know if they received the letter from amc...

It's already April. 10th. Why haven't I received the official mail yet?
I did receive their email, but I want to see a paper, more official one.
In the email, they say, I will receive a letter very soon, but....
I just want to know, has everybody also received a mail besides the email?
Thank you. :)

Yes, I just received a letter in the mail today.  It had my USAMO number and three papers (S-I, S-II, and S-III)

I got my paper letter today, but I already submitted my forms... what a waste of trees.

[quote="First Page of Letter"][b][size=200]Congratulations USAMO Qualifier![/size][/b][/quote]
 :D

EDIT: "First Page of Letter [i]wrote[/i]"  :rotfl:

:lol:  got it today. No worry about that.

my brother received it a long long time ago through the mail


conics
ellipse
LaTeX
analytic geometry
geometry
geometric transformation
rotation

Hi ,
i wanted to find the cartesian equation of an ellipse from the wannabe 'definition ' of it as the set of plan points that fit : MA + MB = k , k>0 , but i failed .
i want to know what is the primary definition of an ellipse , the real definition , is it " the set of pts such that MA + MB =  k"and then all other formulas come from it? :huh:

i can give u the definition:
An ellipse is the locus of a point which moves so that its distance from focus bears to its distance from a fised straight line(directrix) a constant ratio(eccentricity) which is less than unity.
Ellipse is also defined as the locus of a point the sum of whose distances from two points is K and greater than the distance between the two fixed points.

(x-x`)(x-x`)/ a.a + (y-y`)(y-y`)/b.b=1

To rephrase the equation in LaTeX:
$ \frac{(x\minus{}x_0)^2}{X^2} \plus{} \frac{(y\minus{}y_0)^2}{Y^2} \equal{} 1$
X is the length of the axis parallel to the x-axis. Y is the length of the axis parallel to the y-axis. Usually, we use a for the one that's longer. $ (x_0,y_0)$ is the center.
The focal points (the points from which the lengths are found that are supposed to sum to a constant) are $ \sqrt{a^2\minus{}b^2}$ away from the center if a is the longer axis's length and b is the other one. All this only applies to ellipses whose axises are indeed horizontal and vertical. If you want slanted ones, I forget my conic sections, but it has something to do with $ ax^2 \plus{} bxy \plus{} cy^2 \plus{} dx \plus{} ey \plus{} f$, the equation for a general conic section, with some restrictions on $ b^2 \minus{} 4ac$. Less than 1 or less than 0? What's the eccentricity formula? Um.

[url=http://mathworld.wolfram.com/Ellipse.html]This MathWorld page[/url] contains the algebraic derivation of the equation of an ellipse given the definition, under the assumptions that the axes are parallel to the coordinate axes and that the center of the ellipse is at the origin.

To remove the first assumption, we can apply a rotation matrix.
To remove the second assumption, we can just replace $ x$ with $ x \minus{} h$ and $ y$ with $ y \minus{} k$ (ie, apply a translation).


Is $ \sqrt{2}$ the limit of a sequence of numbers of the form $ n^{1/3}-m^{1/3}$ $ (n,m = 0,1,2,...)$? Why or why not?

No, becuse the equation $ \sqrt[3]{n}-\sqrt[3]{m}=\sqrt{2}$ don`t have integer roots

Sorry, you can.  This problem is A2 on the 1990 Putnam.  It has been posted on the forum before:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1175361169&t=106692


combinatorial geometry
combinatorics unsolved
combinatorics

Let $ S$ be a set of $ 6n$ points in a line. Choose randomly $ 4n$ of these points and paint them blue; the other $ 2n$ points are painted green. Prove that there exists a line segment that contains exactly $ 3n$ points from $ S$, $ 2n$ of them blue and $ n$ of them green.

This is just a discrete intermediate value theorem :) -- if the interval covering the leftmost $ 3n$ points covers fewer than $ 2n$ blue points then the interval covering the rightmost $ 3n$ points covers more than $ 2n$ blue points, so as we increment point-by-point (i.e., take a new interval that includes all but one point of the previous interval and one new point) some interval covering $ 3n$ points must cover exactly $ 2n$ blue points.


[quote="Klebian"][quote="vishalarul"]The angle x intercepts an arc AB. By definition, x = measure (AB). Also, y = measure (AB)/2 (this is a theorem). So x=2y
Simple.  :P
If you wanted me to prove the theorem...[/quote]
Sorry, that's what I meant. Prove the theorem.[/quote]
Connect $O$ and the point on the circle. Then you have two isosceles triangles. Then use the fact that the exterior angle is equal to the sum of the other two angles of the triangle.

When you connect the center to the point on the circle, you don't necessarily have to get 2 isosceles triangles.

The distanse from the point on the circle to the circle centre is a constant. That's why there are isosceles triangles.

Ok, but your proof only covers 1 case. Suppose the point were moved to the far right or to the far left. What would happen then? You need to show that no matter where the point on the circle is placed, x=2y.

[hide="Proof"]$m+n=y$

$\pi-2m+\pi-2n=2\pi-x$

$2\pi-2\left(m+n\right)=2\pi-2y=2\pi-x\implies\boxed{x=2y}$[/hide]

[hide]Connect the center  O to the point. We then have 2 isosceles triangles, where the obtuse angles are 180 - y, and the congruent angles are y/2. 2(180-y) + x = 360.
x-2y = 0
x = 2y
[/hide][/hide]

That is just a specific case, but the idea is there.

[quote="me@home"]By definition if two angles intercept the same arc, they are equal. So for any drawing (placement of the third point) you can move around the point and it becomes the diagram.[/quote]
He was asking you to show that theorem. So you cannot use it here.

[quote="me@home"][quote="Klebian"]Idea from AOPS vol.1, Prove that x = 2y[/quote]
I think that's all he wants to prove...[/quote]
By definition is a pretty lame proof, I actually wanted you to prove the theorem/definition/whatever.

[hide]If y is to the far left, draw a line from O to y's vertex. This creates two isoceles triangles. Label the angle created from y's left leg and x's left leg b, and the angle created from y's right leg and x's left leg a. A quick look at the diagram tells you a+y=b, and x=(180-2a)-(180-2b) which simplifies to x=-2a+b. Substituting in a+y=b gives you x=-2a+2(a+y), which simplifies to x=2y.[/hide]


calculus
integration
inequalities
function
real analysis
real analysis unsolved

How to show that if $\mu(X)=1$ and $f: X\to ]0,\infty[$ is measurable, then
\[\int_{X}f^{-1}d\mu\leq \int_{X}f\,d\mu?\]

MeKnowsNothing: The words "measurable function" or "$L^{p}$ space" (or anything similar) do not belong in the "Calculus Computations and Tutorials" forum. Post them here instead.

As for the problem: what do you mean by $f^{-1}?$ If you mean reciprocal, then this is obviously false, as can be shown by silly examples, such as $f(x)=\frac12.$ If you mean inverse function, then $f^{-1}$ would no longer be defined on $X$ and the problem makes no sense.

What did you mean to say?

Sorry. I meant
\[\left (\int_{X}\frac{1}{f}\,d\mu\right )^{-1}\leq\int_{X}f\,d\mu. \]

This is a special case of Jensen's inequality. Suppose $\mu(X)=1$ and $\varphi$ is convex on the range of $f.$ Then

$\varphi\left(\int_{X}f\,d\mu\right)\le\int_{X}\varphi(f)\,d\mu$

In this case, $\varphi(x)=\frac1x$ for $x\in(0,\infty).$


Most of the items I write I like to have  1" margins around the document, so I've been using the fullpage package in the preamble.

[code]\usepackage{fullpage}[/code]

That works pretty well.  Unfortunately another package that I am using a lot is the fancy header package, for example I might have something like this

[code]\usepackage{fancyhdr}   \pagestyle{fancy}
\lhead{My name\\ some header text here}
\rhead{January 19, 2007}[/code]

It looks great, except that the document text is running into the header text at the top of the page and that's no good.

[img]http://www.orangesynergy.com/pics/header.jpg[/img]

So the only way I've found to get around this is to have a some 'filler' text right after the \begin{document} command such as

[code]\begin{document}
`
\newline
`
\newline

 Now I start my text here...[/code]

The only problem is that it's not the most elegant solution and it also puts those apostrophes or whatever in the document.  I'm curious, is there a way to make text invisible... Or another method to have 1 inch margins for a paper, but also have running header?  I checked out [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=header+margins&t=115787]a[/url] [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=header+margins&t=14311]few[/url][url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=header+margins&t=12783]threads[/url] but I don't think I saw much pertaining to this...

Thanks in advance

The easiest way to change margins is to use the Geometry package. To get 1 inch margins all you need to use is \usepackage[margin=1in]{geometry} in the preamble. Full instructions are [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/manual.pdf]here[/url]. It should work well with the fancyhdr package.

Wow.  Thanks for the speedy response.  Using the geometry package seems to work and doesn't interfere with the header.   :)


this is probably a stupid question but is $\Im(a+bi)$ $b$ or $bi$?

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

One reason this definition is more useful is that when considering functions $f(x+yi) : \mathbb{C}\to \mathbb{C}$ we might like to decompose them into the form $u(x, y)+i v(x, y)$ where $u, v : \mathbb{R}^{2}\to \mathbb{R}$.


This is probably a dumb question but what does "how many standard deviations this differs from zero" mean?

ex how many standard deviations does 0.265960357 differ from zero. Search engines weren't useful

Of course they weren't useful because you have no data.  Standard deviations aren't arbitrary numbers, they're based on a formula which needs your data.  One standard deviation corresponds to a certain percentile (like 19 or something) and the more standard deviations away from the mean something is, the greater the percentile it's in.  Think of IQ scores.  The mean is about 100.  If you are 1 SD from the mean to the right, you're in the 68th percentile (or something like that).  Another SD is an even higher percentile.


integration
real analysis
real analysis unsolved

How can we prove this?
$\sum_{n\geq0}(\frac{1}{3n+1}-\frac{1}{3n+2}) = \frac{\pi}{3\sqrt{3}}$


Let $ a,\ b,\ c$ be postive real numbers such that $ \frac{a^2\plus{}b^2}{(a\plus{}b)^3}\plus{}\frac{b^2\plus{}c^2}{(b\plus{}c)^3}\plus{}\frac{c^2\plus{}a^2}{(c\plus{}a)^3}\equal{}\frac{3}{2}$.

Prove that $ a^2\plus{}b^2\plus{}c^2\geq \frac{3}{4}$.

[quote="kunny"]Let $ a,\ b,\ c$ be postive real numbers such that $ \frac {a^2 \plus{} b^2}{(a \plus{} b)^3} \plus{} \frac {b^2 \plus{} c^2}{(b \plus{} c)^3} \plus{} \frac {c^2 \plus{} a^2}{(c \plus{} a)^3} \equal{} \frac {3}{2}$.

Prove that $ a^2 \plus{} b^2 \plus{} c^2\geq \frac {3}{4}$.[/quote]

We have:
$ \frac {3}{2}\equal{}\frac {a^2 \plus{} b^2}{(a \plus{} b)^3} \plus{} \frac {b^2 \plus{} c^2}{(b \plus{} c)^3} \plus{} \frac {c^2 \plus{} a^2}{(c \plus{} a)^3} \ge \sum \frac{(a\plus{}b)^2}{2(a\plus{}b)^3} \equal{} \sum \frac{1}{2(a\plus{}b)}$

By Cauchy-Schwarz inequality:
$ \sum \frac{1}{2(a\plus{}b)} \ge \frac{9}{4(a\plus{}b\plus{}c)}$

It follows that:

$ \frac{9}{4(a\plus{}b\plus{}c)} \le \frac{3}{2}$ so $ a\plus{}b\plus{}c \ge \frac{3}{2}$

So $ a^2\plus{}b^2\plus{}c^2 \ge \frac{(a\plus{}b\plus{}c)^2}{3} \ge \frac{3}{4}$

Oh, the problem seemed to be easy such that you could solve so fast. :lol:
You could prove it in basic way. My solution seemes to be overdone.

Here is my solution.

For $ a>0,\ b>0$, $ \frac{a^2\plus{}b^2}{2}\geq \left(\frac{a\plus{}b}{2}\right)^2,\ (a\plus{}b)^3\geq 3(a\plus{}b)\minus{}2$.

$ \therefore (a^2\plus{}b^2)\{(a\plus{}b)^3\plus{}2\}\geq \frac{3}{2}(a\plus{}b)^3\Longleftrightarrow (a^2\plus{}b^2)\left\{1\plus{}\frac{2}{(a\plus{}b)^3}\right\}\geq \frac{3}{2}$.

Similarly $ (b^2\plus{}c^2)\left\{1\plus{}\frac{2}{(b\plus{}c)^3}\right\}\geq \frac{3}{2},\ (c^2\plus{}a^2)\left\{1\plus{}\frac{2}{(c\plus{}a)^3}\right\}\geq \frac{3}{2}$.

Adding these inequalities and dividing this by 2, we have $ a^2\plus{}b^2\plus{}c^2\plus{}\frac{a^2\plus{}b^2}{(a\plus{}b)^3}\plus{}\frac{b^2\plus{}c^2}{(b\plus{}c)^3}\plus{}\frac{c^2\plus{}a^2}{(c\plus{}a)^3}\geq \frac{9}{4}$, yielding $ a^2\plus{}b^2\plus{}c^2\geq \frac{3}{4}$.


algebra
polynomial
function
algebra proposed

Find all polynomials $P$ with real coefficients such that \[ P(x)P(2x^2\minus{}1)\equal{}P(x^2)P(2x\minus{}1)\] for all $x \in \mathbb R$.

Every constant polynomial satisfies given condition. Further we suppose P is not constant.

Let $ f(x) \equal{} \frac {P(2x \minus{} 1)}{P(x)}$. Let $ t\not\in\{ \minus{} 1,1\}$ satisfy $ P(t^n)\neq 0$ for all positive integers n (such t clearly exists). We have $ f(t) \equal{} f(t^2) \equal{} f(t^4) \equal{} \cdots \equal{} y$ for some y, thus $ P(2t^{2^n} \minus{} 1) \equal{} yP(t^{2^n})$ for all positive integers n. This is polynomial equation and has infinitely many roots, therefore for all real x
$ P(2x \minus{} 1) \equal{} yP(x)$. Now let P(x)=Q(x-1), so
$ Q(2x \minus{} 2) \equal{} yQ(x \minus{} 1)$.
If r is (complex) root of Q, so is 2r, 4r, etc. which means the only root of Q is 0, $ Q \equal{} ax^m$, $ P \equal{} a(x \minus{} 1)^m$.

Condition $ P \equal{} a(x \minus{} 1)^m$ is also sufficient and for m=0 includes the constant functions excluded in the begining.

[quote="MJ GEO"]thanks.my solution is exactly like you[/quote]

Cool, and could you, please, give us the solutions you promised us one week ago ?


LaTeX
limit
calculus
calculus computations

\[ First /time /i /used /latex / : P\]
\[ a_n=(3^n+2.5^n)^({}^1{}^/{}^n^)\]
\[ limit=?\]
how can we will make space  :oops:
Thanks and best regards!

[hide]$ 3 < \sqrt [n]{3^n \plus{} 2.5^n} < \sqrt [n]{2 \times 3^n} < 3\sqrt [n]{2}$

Send $ n \rightarrow \infty$, $ \sqrt [n]{2} \rightarrow 1$

Therefore the limit is 3[/hide]

Alter for desired rigour.

LaTeX code is in blue. Here are some things that might have been relevant to your post:
[color=darkblue]
\text{This is how you incorporate plain-text}[/color] produces $ \text{This is how you incorporate plain \minus{} text}$

[color=darkblue]\; [/color]produces a single space

[color=darkblue]\qquad[/color] produces a larger space

[color=darkblue]\frac{a}{b}[/color] produces $ \frac {a}{b}$

[color=darkblue]\sqrt[n]{k}[/color] produces $ \sqrt[n]{k}$
[color=darkblue]
\lim_{a \to \infty} c[/color] produces $ \lim_{a \to \infty} c$


algebra
polynomial
rotation
geometry
geometric transformation
algebra unsolved

Let $a_{1},a_{2},...,a_{n}$ be $n$ complex numbers. Define the complex polynomial $P(x)=(x-a_{1})(x-a_{2})...(x-a_{n})$
  Prove that there exists a complex number  $\epsilon$ such that $|\epsilon|=1$, and
   $|P(\epsilon)| > \frac{(1+|a_{1}|)(1+|a_{2}|)...(1+|a_{n}|)}{3^{n}}$

No reply after so long? Could somebody solve this?

Consider the values of $ P$ at $ e^{i\frac {2k\pi}{m}}$, where $ k = 1,2,...,m$ (the $ m$th roots of unity).
\[ \prod |P(e^{i\frac {2k\pi}{m}})| = \prod |a_{i}^m - 1|
\]
It suffices to show that $ \prod |a_{i}^m - 1| > \prod(\frac {1 + |a_i|}{3})^m$ for some $ m$. We partition $ a_i$ into the following three categories:

1. $ |a_i| > 1$
We have $ |a_{i}^m - 1| - (\frac {1 + |a_i|}{3})^m > (1 - \frac {1}{2}(\frac {2}{3})^m)|a_i|^m - (1 + \frac {1}{2}(\frac {2}{3})^m)$, which is positive for sufficiently large $ m$.
2. $ |a_i| < 1$
This case is a bit simpler. $ |a_{i}^m - 1| - (\frac {1 + |a_i|}{3})^m\ge1 - |a_i|^m - (\frac {1 + |a_i|}{3})^m$, which converges to 1  as $ m\longrightarrow\infty$.
3. $ |a_i| = 1$
Let $ S_m = \{ a_{1}^m,a_{2}^m,...a_{k}^m\}$, where $ a_i$ are the points on the unit circle. Let $ AB$ be the longest arc on the circle which does not contain any points in $ S_m$ (other than $ A$ and $ B$). For sufficiently large $ m$, we would have $ AC,BC>(\frac {2}{3})^m$, where $ C$ is the midpoint of arc $ AB$. Now rotate the complex plane $ \mathbb{C}$ about the origin by a suitable angle such that $ C$ is mapped to 1. In such a way, $ |a_{i}^m - 1|$, which represents the distance between $ a_{i}^m$ and 1, must exceed $ (\frac {2}{3})^m$.

Note that the arguments used in the previous two cases are not affected by the rotation as it is just a multiplication of $ e^{i\theta}$, of which the norm is 1.


Let be $ p,q\in \mathbb{N}^*\ ,\ 1\le q\le p$ and let $ a\equal{}\left(\sqrt{p^2\plus{}q}\plus{}p\right)^2$ . Prove that $ a\in \mathbb{R}\backslash \mathbb{Q}$ and $ \{a\}> \frac{3}{4}$ .

I could  just prove that $ a$ is irational number.

$ a\equal{}(\sqrt{p^2 \plus{}q} \plus{}p)^2 \equal{}2p^2 \plus{}q \plus{}2p\sqrt{p^2 \plus{}q}$

Supose that number $ \sqrt {p^2 \plus{}q}$ is integer then $ p^2\plus{}q$ is perfect square,$ {p^2 \plus{}q}\equal{}k^2 \implies q\equal{}(k\minus{}p)(k\plus{}p)$ that is imposible since $ q\leq p$
So $ p^2 \plus{}q$ is not perfect square $ \implies a$ is irational.

I have no idea about $ \{a\} >\frac{3}{4}$

:blush:

[quote="alex2008"]Let be $ p,q\in \mathbb{N}^*\ ,\ 1\le q\le p$ and let $ a \equal{} \left(\sqrt {p^2 \plus{} q} \plus{} p\right)^2$ . Prove that $ a\in \mathbb{R}\backslash \mathbb{Q}$ and $ \{a\} > \frac {3}{4}$ .[/quote]

Clearly $ \{a\}\equal{}\{\sqrt{4p^4\plus{}4p^2q}\}$. Since $ (2p^2\plus{}q\minus{}1)^2<4p^4\plus{}4p^2q<(2p^2\plus{}q)^2$, we see that in fact
\[ \{a\}\equal{} \sqrt{4p^4\plus{}4p^2q}\minus{}(2p^2\plus{}q\minus{}1) \]
holds. Then the desired inequality $ \{a\} > 3/4$ is equivalent to
\[ \sqrt{4p^4\plus{}4p^2q} > 2p^2\plus{}q\minus{}\frac14, \]
which in turn is equivalent to
\[ 2p^2\plus{}q > 2q^2\plus{}\frac18,\]
which is obviously true.


Prove  that for any prime p$ \geq$5
$ \sum_{0<k<\frac{2p}{3}}(_{k}^{p})$ is divisible by $ p^{2}$

http://www.mathlinks.ro/viewtopic.php?t=150392

  darij


AMC AIME factorial calculus integration function algebra

From 1992 AIME, #15

For how many integers n in {1, 2, 3, ... , 1992} is it true that no number m! (for integral m) ends in n zeros?

(Paraphrased)

This one seems awfully easy for being #15...

Yeah, it seems fairly easy to me, too:

[hide]It's clear that the number of zeros is going to be determined by the number of factors of 5.  Let fl(x) be the greatest integer function: e.g. f(pi)=3, f(17)=17.

Then we define g(x) = fl(x/5) + fl(x/25) + fl(x/125) + ...

Then if g(x) = k, then x! has k zeros at the end.

We need to find x such that g(x)=1992.  We take as a rough guess 4*1992 = 7968.

(The 4 comes from the reciprocal of 1/5 + 1/25+1/125+1/625+...)  This would be exactly the correct answer except for the fl( )'s in the definition.

We compute g(7968)=1988.  We eventually find that g(7980)=1992.

Note that g(m)>g(n) iff fl(m/5) > fl(n/5).  So g(x) takes on only fl(7980/5) positive values less than or equal to 1992.  So the answer is 1992-fl(7980/5) = 396.

[/hide]

--Dan


extends to the radius OA of a circle of a length equal to AB same to OA; take to a tangent to the circle, on which there is the perpendicular NA and BC.
Assuming that the angle <OAC is equal to 126º, calculate the value of <ACB

[hide="Solution"]Note that since $ OA \equal{} OB$ and $ OD \parallel AN \parallel BC$, $ CN \equal{} ND$ and $ \angle ACB \equal{} \angle NAC \equal{} \angle DAN \equal{} \angle ODA \equal{} \angle OAD$, since $ OA \equal{} OD$. Thus $ \angle OAC \equal{} 3 \angle ACB$, so $ \angle ACB \equal{} 42$[/hide]

But the question is worded in a confusing way.
For others, the tangent is any random tangent at any point on the circle, and AN and BC are the perpendiculars to it.

could you post the image about the question.??It´s sort of confused, I think

I have it like this.
[geogebra]10bef0401e3b10298872298fe2ab1e6dbb894091[/geogebra]


linear algebra
matrix
algebra unsolved
algebra

hi

 i have a set of undetermined linear equations with unknowns 13 and set of equation=3
my equation formulates as

Ax=min

with the constraint that the sum of unknowns equals the number of unknowns in my case x1+.....x13=13

is it possible in matlab
       Regards,

Could you post the complete set of equations?

According to theorems of linear algebra, you can set $ g_5, \ldots g_{13}$ equal to $ 0$, $ g_4$ to any value, and then use Gaussian elimination to find values for $ g_1, g_2, g_3$ that make all three of those equations equal to $ 0$. You can then get any value for $ g_1 \plus{} g_2 \plus{} \ldots \plus{} g_{13}$ by multiplying $ g_1, g_2, g_3, g_4$ (the only nonzero terms in that sum) by any real coefficient. This is regardless of your original matrix.

i couldnt follow ur answer exactly can u please eloborate more .......i actually want only non zero wieghts ..........
Thankyou
regards,


A student read for his exams for 37 days such that:
Every day he reads at least one hour
Every day he reads integer number of hours
And he must read exactly 60 hours
Prove that there are two successive days where he read 13 hours

Correct formulation of the problem (19th Hellenic Mathematical Olympiad 2002, seniors, problem 2):

[color=blue][b]Problem.[/b] A student of the National Technical University was reading advanced mathematics last summer for 37 days according to the following rules:
[b](i)[/b] He was reading at least one hour every day;
[b](ii)[/b] He was reading an integer number of hours, but not more than 12, each day;
[b](iii)[/b] He had to read at most 60 hours in total.
Prove that there were some successive days during which the student was reading exactly 13 hours in total.[/color]

  Darij

Standard pigeonhole. I am sure you can do this Silouan.

Hint

[hide]Let $a_i$ be the number of hours he read on day $i$, we then need to prove that for some $j$ and $k$, $a_j + a_{j + 1} + ... + a_k = 13$... So, consider all sums of the form $a_1 + ... + a_k$.[/hide]


A fishing rod 16 feet long is made of two parts. In order to function properly, $ \frac35$ of the longer part must equal the length of the shorter part. How many feet are in the length of the longer part of the pole?

We just find and solve an equation, where our variable, $ x$, represents the longer part:

$ \frac {3}{5}x \plus{} x \equal{} 16$

$ \frac {8}{5}x \equal{} 16$
$ x \equal{} \fbox{10}$


Let $a,b,c$ positive numbers with $a+b+c=1$. Prove that:

$\frac{a^{2}+b}{b+c}+\frac{b^{2}+c}{c+a}+\frac{c^{2}+a}{a+b}\geq 2$

Nice problem :) Here is the solution that came to my mind:
Observe that \[a^{2}+b=a^{2}+b(a+b+c)=a^{2}+ab+b(b+c)\] ,which means that our ineq rewrites as: \[\sum_{cyc}\frac{a^{2}+ab}{b+c}+b\geq 2\] ,and because we have a+b+c=1,our ineq is equivalent to: \[\frac{a^{2}+ab}{b+c}+\frac{b^{2}+bc}{c+a}+\frac{c^{2}+ac}{a+b}\geq 1\] ,which is equivalent to : \[\sum_{cyc}a(a+b)^{2}(a+c)\geq(a+b)(b+c)(c+a)\] Now because a+b+c=1,we have $(a+b)(a+c)=a+bc$,therefore our ineq is equivalent to \[\boxed{\sum_{cyc}a(a+b)(a+bc)\geq(a+b)(b+c)(c+a)}\] .First of all we take care of the left term:
$a(a+b)(a+bc)=a^{3}+abc(a+c)+a^{2}b$,which means that the left side of the inequality is equal to \[a^{3}+b^{3}+c^{3}+2abc+a^{2}b+b^{2}c+c^{2}a\] Now we take care of the right side of the inequality:
By a well-known identity we have \[(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ac)-abc\] Judging by our last three relations,we have to prove \[a^{3}+b^{3}+c^{3}+3abc+a^{2}b+b^{2}c+c^{2}a\geq ab+bc+ac\] Now we are really close to proving the ineq..we have just one more step:
Observe that \[abc+a^{2}b-ab=ab(a+c-1)=-ab^{2}\] ,therefore we only ned to prove \[a^{3}+b^{3}+c^{3}-(ab^{2}+bc^{2}+ca^{2})\geq0\] ,which is equivalent to: \[a^{3}+b^{3}+c^{3}\geq a^{2}b+b^{2}c+c^{2}a\] :D  :D

$\{\begin{array}{c}b+c=x\\\\ c+a=y\\\\ a+b=z\\\\ a+b+c=1\\\\ \{x,y,z\}\subset (0,1)\end{array}$ $\Longrightarrow$ $\{\begin{array}{c}a=1-x\\\\ b=1-y\\\\ c=1-z\\\\ x+y+z=2\end{array}$ $\Longrightarrow$ $\{\begin{array}{c}\frac{a^{2}+b}{b+c}=\frac{x^{2}-x+z}{x}\\\\ \sum\frac{a^{2}+b}{b+c}=-1+\sum \frac{z}{x}\\\\ \sum\frac{z}{x}\ge 3\cdot\sqrt [3]{\frac{z}{x}\cdot\frac{x}{y}\cdot\frac{y}{z}}=3\end{array}$ $\Longrightarrow$ $\boxed{\ \sum \frac{a^{2}+b}{b+c}\ge 2\ }\ .$

very nice solutions! now i`ll post mine, which is, i think, uglyer than yours :)

so we have to prove that:

$\frac{a^{2}+b}{b+c}+\frac{b^{2}+c}{c+a}+\frac{c^{2}+a}{a+b}\geq 2$ with $a+b+c=1$

i`ve started thinking about cauchy, so our ineq is equivalent with:

$\sum{\frac{(a^{2}+b)^{}}{(a^{2}+b)(b+c)}}\geq 2$

now let`s make the notation:

$\sum{\frac{(a^{2}+b)^{}}{(a^{2}+b)(b+c)}}= LHS$

by cauchy we  have:

$LHS \geq \frac{(\sum{a^{2}}+1)^{2}}{\sum{a^{2}b}+\sum{ab^{2}}+\sum{a^{2}}+\sum{ab}}$

so it remains to prove that:

$\frac{(\sum{a^{2}}+1)^{2}}{\sum{a^{2}b}+\sum{ab^{2}}+\sum{a^{2}}+\sum{ab}}\geq 2$

$\Rightarrow (\sum{a^{2}})^{2}+2\sum{a^{2}}+1 \geq 2\sum{a^{2}b}+2\sum{ab^{2}}+2\sum{a^{2}}+2\sum{ab}$

$\Rightarrow (\sum{a^{2}})^{2}+1 \geq 2\sum{a^{2}b}+2\sum{ab^{2}}+2\sum{ab}$

now we observe that:

$2\sum{a^{2}b}+2\sum{ab^{2}}= 2\sum{a^{2}}-2\sum{a^{3}}$, the proof of this thing is trivial, because we have $a+b+c=1$

now we replace it in our ineq:

$(\sum{a^{2}})^{2}+1 \geq 2\sum{a^{2}}-2\sum{a^{3}}+2\sum{ab}$

$\Rightarrow (\sum{a^{2}})^{2}+1+2\sum{a^{3}}\geq 2\sum{a^{2}}+2\sum{ab}$

but by chebyshev we have that :

$3\sum{a^{3}}\geq (\sum{a})(\sum{a^{2}})$, but $a+b+c=1$

$\Rightarrow \sum{a^{3}}\geq \frac{\sum{a^{2}}}{3}$

by replacing what we obtained with chebyshev in our ineq we have:

$(\sum{a^{2}})^{2}+1+2\cdot\frac{\sum{a^{2}}}{3}\geq 2\sum{a^{2}}+2\sum{ab}$

but $\sum{a^{2}}+2\sum{ab}=1$, because $(a+b+c)^{2}=1$, so we have:

$(\sum{a^{2}})^{2}+1+2\cdot\frac{\sum{a^{2}}}{3}\geq \sum{a^{2}}+1$

$\Rightarrow (\sum{a^{2}})^{2}+2\cdot\frac{\sum{a^{2}}}{3}\geq \sum{a^{2}}$

$\Rightarrow \sum{a^{2}}+\frac{2}{3}\geq 1$

$\Rightarrow \sum{a^{2}}\geq 1/3$

but we have that $(\sum{a})^{2}=1$, so our ineq is equivalent with:

$3\sum{a^{2}}\geq (\sum{a})^{2}$ $\Rightarrow \sum{a^{2}}\geq \sum{ab}$, which is true :)

and now we can say that our proof is [b]done[/b].

It's problem [i]23[/i] from [b]Old and New Inequalities[/b]. pohoatza's solution is the one given there.
But there's no source mentioned. :(

I've got another solution, but I'm not sure if it's right.
I'm very bad in solving inequalities but that's my solution:

$\frac{a^{2}+b}{b+c}+\frac{b^{2}+c}{c+a}+\frac{c^{2}+a}{a+b}\geq 2$

$\frac{a^{2}+b}{1-a}+\frac{b^{2}+c}{1-b}+\frac{c^{2}+a}{1-c}\geq 2$

Edit: oh, I saw my mistake. I try to correct it

[b]An easy extension :[/b] $\{\begin{array}{c}k>0\\\ \{a,b,c\}\subset (0,1)\\\ a+b+c=1\end{array}\Longrightarrow \sum_{\mathrm{cyc}}\frac{a^{2}+kb+k-1}{b+c}\ge 2(3k-2)\ .$


vector
linear algebra
matrix
analytic geometry

A vector is a sort of one dimensional array of numbers and a matrix is two dimensional. Is there any terms for three dimensional or higher arrays of numbers?

The word you're looking for is [url=http://en.wikipedia.org/wiki/Tensor]tensor[/url].

A vector is a geometric object with magnitude and direction (this means that a vector can be multi-dimensional). For example, the vector $ AB$ can be expressed by the points $ A\equal{}(1,0)$ and $ B\equal{}(0,1)$. However this is what is known as a free vector, which is a vector with an initial point other than the origin. When you see a vector with only one coordinate, it generally implies that is is a bounded vector with the origin as its initial point. Given a vector in $ R^n$, it endpoint can be arranged as a row or column matrix.


AMC
USA(J)MO
USAMO
AIME
AwesomeMath
summer program
PROMYS

Hello. I am aiming to make the USAMO and maybe get a few problems right. What camps would you guys recommend (for me for next year)? To give you an idea of my level, I cannot solve USAMO questions in general, occasionally can get an AIME question which is over #10, however most of the time I can hopefully get over a 5 on practice AIMEs. I have only worked through up to chapter 20 in AOPS 1 and will hopefully by the end of the year be done AOPS 2. I also hope to eventually work through Mathematical Circles and ACOPS. I am going into Grade 10. I am willing to work hard.

Any help is greatly appreciated.

(PS, here is a weird question: if you guys do not get enough sleep do you find that you cannot solve as well? Sorry to ask, but this idea has really been bothering me).

Thanks.

the ones i can think of off the top of my head are awesomemath and mathzoom. i'm not sure if some of the other math camps are for preparing specifically for contests or just teach math in general (such as promys, ross, mathcamp. someone else who knows more than i do can answer this).

and for your question... for me, i'd have to say yes, i need to be well rested to solve problems.

PROMYS, ROSS, and mathcamp are not for preparing for contests. They may do a bit of prep at mathcamp, but definately not at promys/ross. Its too late to do any camps this summer anyways.


[quote="myc"]If five standard fair dice are tossed simultaneously, what is the probability that the outcome has a sum greater than 28? (mathcounts 1989)

  following the example in aops book I pg. 319, I considered the first and second dice as a unit, third and fourth dice as a unit, and the fifth dice by itself.  The configuration which yeild more than 28 are:

12   12    5    -> 29    probability= $\frac{1}{36}*\frac{1}{36}*\frac{1}{6}$
11   12    6    -> 29    probability= $\frac{2}{36}*\frac{1}{36}*\frac{1}{6}$
12   12    6    -> 29     probability= $\frac{1}{36}*\frac{1}{36}*\frac{1}{6}$

if I  add the three probabilities together, I get $\frac{4}{6^{5}}$ .  The answer however is $\frac{1}{6^{4}}$

There are $\frac{5!}{2!3!}$ ways to pair up the first two dice. There are $\frac{3!}{2!1!}$ ways to pair up the 3rd n 4th dice.
the example in the book doesn't do this.  Is it because the dice are not distinct?

Thank you :)[/quote]
It's because you forgot a case:
$(12, 11, 6) = 29$
Probability = $\frac{1}{36}\cdot \frac{2}{36}\cdot \frac{1}{6}= \frac{2}{6^{5}}$.
If you add this to the total you had, your answer would then be $\frac{2}{6^{5}}+\frac{4}{6^{5}}=\frac{1}{6^{4}}$, the correct answer.

Here's another way to look at the problem:

There are five ways of getting 29:
5,6,6,6,6
6,5,6,6,6
...
6,6,6,6,5

Each of these has probability $\frac{1}{6^{5}}$, so the probability of getting 29 is $\frac{5}{6^{5}}$.

There is one way to get 30:
6,6,6,6,6

Thus the probability of getting 30 is $\frac{1}{6^{5}}$.

Adding the two up, we get $\frac{1}{6^{4}}$.


If the three points of contact of a circle inscribed in a triangle are joined, the angles of the resulting triangle:

$ \textbf{(A)}\ \text{are always equal to }60^\circ \\
\textbf{(B)}\ \text{are always one obtuse angle and two unequal acute angles} \\
\textbf{(C)}\ \text{are always one obtuse angle and two equal acute angles} \\
\textbf{(D)}\ \text{are always acute angles} \\
\textbf{(E)}\  \text{are always unequal to each other}$

[hide="Click for solution"]
[img]http://i43.tinypic.com/2rps0p1.jpg[/img]
$ \angle A'\equal{}\frac{a}{2}$, $ \angle A\equal{}180^\circ\minus{}a \implies a\equal{}180^\circ\minus{}\angle A$.  $ \angle A'\equal{}\frac{a}{2}\equal{}90^\circ\minus{}\frac{\angle A}{2} \implies \angle A' < 90^\circ$.

Similarly we can show $ \angle B'<90^\circ$ and $ \angle C' < 90^\circ$.  The answer is thus $ \boxed{\textbf{(D)}}$.
[/hide]


AMC AIME geometry trigonometry algebra polynomial ARML

What are you going to do with only 1 week left until AIME.  For me, Im going to do lots of geoemtry because geometry occurs the most on AIMEs

I'm just gonna chill, cause I had the whole year to prepare, and if I didn't do it well enough I'm not gonna learn everything in this last week.

I will just do 1 or 2 exams just so I don't lose my cool (which causes me to be dumber). Then, I'll either go on a date 'cause that's the way to go!  :D

[quote="tjhance"]I will do a few more tests for practice, and then on Sunday and Monday I will probably do nothing except rest and make sure I have enough paper and pencils and stuff. I'm even doing Monday's math homework ahead of time just so I won't tire my brain out with extra math.[/quote]

i think that's a little overkill.

For me, i just treat it routinely, since my body is already used to certain things happening at certain times. I will try to get some more sleep, but at States it failed. I was up until 12:30 the nigth before States cuz i couldnlt' fall asleep. I'll try to not do too much on Monday, but i'll go through SUnday on full throttle.

Actually getting ahead on homework is routine for me anyway.

the best thing to do now is to rest and take it easy. Also, I like to get PUMPED the night before any math comp. by listening to some "pumping" music. Lastly, green tea the morning of to clear the mind.[/quote]

===THE LAST WEEK OF===

Probably studying chemistry

===THE DAY OF===

I will do the first few problems on an old AIME (say 1-3) the day of the AIME to get my mind warmed up, so I'm not like, spacing out at the beginning of the test.

Also, I will have a minimum of 5 pencils sharpened, a good eraser, and my lucky holographic Charizard card in my pencil case, ready to go.

I will get to the science lecture hall 5 minutes early so I can find a place where three desks in a row have no gum on the bottom and properly latch when raised up.

Over this week, do like 5 problems a day on top of my usual stressful school week (plus concerts all day on friday will be fun).  Weekend: Relax mostly, but do a practice AIME (maybe the 2006 II, never done that).  Monday: Relax.

I've actually kept trying to convince my friends who also take AMC/AIME etc. to go for a short jog before the test to clear my head...but they disagree...

I'm gonna do a bit of math (AIMEs) over the weekend, not much, just enough to keep my brain going. Maybe read through some of the stuff I learned in AoPS II...
All in all, I'll take it easy.

i don't really have a very good understanding of higher level math (first year doing AIME), do u think it would be a better idea to spend a lot of time on trig or geometry?

Yeesh, you need both, I should think. Your call, really, this close to the exam.

Okay, so you're taking the AIME II. You still should brush up on both. Trig identities (well, they can all be derived from Euler's formula, the complex number trig one), as well as some geometry stuff, Stewart's Theorem, and all the good stuff about triangles. Reviewing some polynomial stuff would be good too.

[quote="h_s_potter2002"]Take the Math part of the Ohio Graduation Test, while everyone else at my school gets to sleep in.[/quote]

 :Chuckle: I don't get to sleep in that day, either, but that's because I'll be taking the AIME while the sophomores are taking [i]their[/i] OGT's. :D

[quote="NeverOddOrEven"][quote="paladin8"]I'm going to eat old AIMEs for breakfast so I can absorb all the knowledge.[/quote]

mmm yum. that sounds good.[/quote]
stuff meat or bread in the printer and then print

I'll review some of the recent aimes (06,05,04...), and take it easy.

Interestingly, the coach of my ARML team gave everyone the AIME from 1995.

[quote="davidyko"]You still should brush up on both. Trig identities (well, they can all be derived from Euler's formula, the complex number trig one), as well as some geometry stuff, Stewart's Theorem, and all the good stuff about triangles. Reviewing some polynomial stuff would be good too.[/quote]

to "review" and "brush up", you have to know. i have a lot of learning to do... :weightlift:


rotation
analytic geometry
geometry proposed
geometry

Let ABCD is a square.Proove that point M is on the line BD <=>MD^2+MB^2=2MC^2.

[hide="Hint"]Rotate the whole square around the point $A$ and then apply Pythagoras' Theorem. ;) [/hide]

Put ABCD into a coordinate plane. Then use the distance formula $d^2=(x_1-x_2)^2+(y_1-y_2)^2$, we can prove that for any point M in ABCD, we will have: $MA^2+MC^2=MB^2+MD^2$, thus we will have $MA=MB$, which gives $M$ is on $BD$.

[hide="Here is a easy generalization - the Shobber - idea."]
Let $ABCD$ be a rectangle. Prove that $\left(\forall\right)M,\ MA=MC$ $\Longleftrightarrow$ $MD^2+MB^2=2MC^2\ \ (1)$. Indeed, from the wellknown relation $MD^2+MB^2=MA^2+MC^2$ results that the relation $(1)$ is equivalently with the relation $MA=MC.$[/hide]


:first:  hi everyone don't you just love aops emoticons... i think there great....  :pilot:  :spam:  i mean i've seen alot of emoticons from yahoo, aol, hotmail... but aops smiles Rule...


Show that
 $ \frac {1} {\cot9^\circ\minus{}3\tan9^\circ} \plus{} \frac {3} {\cot27^\circ\minus{}3\tan27^\circ} \plus{} \frac {9} {\cot81^\circ\minus{}3\tan81^\circ} \plus{} \frac {27} {\cot243^\circ\minus{}3\tan243^\circ} \equal{} 10\tan9^\circ$.

hello, use that
$ \tan\left(\frac{\pi}{20}\right)\equal{}1\plus{}\sqrt{5}\minus{}\sqrt{5\plus{}2\sqrt{5}}$
Sonnhard.

Thank you, but using the value for tan9° means that I have to do all the calculation. I can't even imagine that anyone could solve this problem like that, especially because this is a problem proposed for a contest.

Dr. Graubner, could you please demonstrate that result?

this is a geometry forum, do you have a geometric interpretation of this aside from trigonometric manipulation?

A friend of mine gave me the formula  $ \frac{1}{ \cot x\minus{}3 \tan x}\equal{} \frac{1}{8}(3\tan3x \minus{} \tan x).$ This solves the problem.

The problem just changed its face. Demonstrate the above result!

I am quite intrigged with Dr. Graubner's result. Could it be demonstrated here either?

Use $ \tan 3x \equal{} \frac{\sin3x}{\cos3x} \equal{} \frac{3\sin x\minus{}4 \sin^3 x}{4 \cos^3 x\minus{}3 \cos x}\equal{}\frac{\tan x (3\minus{}\tan^2 x)}{1\minus{}3\tan^2 x}$ and $ \cot x \equal{}\frac{1}{\tan x}.$

Thank You! Dr. Graubner's resul still missing...

You can start with \[ sin 5x \equal{} 16{sin}^5 x\minus{} 20 {sin}^3 x \plus{} 5  sin x\]
(Use often $ 6\plus{}2\sqrt{5} \equal{} {(\sqrt{5} \plus{} 1)}^2$).


pigeonhole principle
ceiling function
articles

I really need some help with the Pigeonhole Principle. I understand the theorem but i don't really understand how to apply it to problems.

-----------------------
thanks

[quote="mewto55555"]Say you have 3 gumballs in a gumball machine of each color:red,blue,and green. How many times must you get a gumball to be [b]sure[/b] you have 2 of the same color?[/quote]

To be sure, you have to think of the worst possible solution.  Lets say first you got a red.  Second you got a blue.  Third you got a green.  You know have one of each color with three times.  Which ever color you draw next, it will add to the 1 already that you and make two.  So we will have to choose 4 times to be sure we get 2 of each color.

another one:

You have 15 rooms, and 164 people.
what is the least amount of people you will find in the most crouded room?

I believe this is $ \boxed{11}$.

I did $ \lceil 164 \div 15 \rceil$.

For some excellent pigeon hole principle problems and solutions/explanations:
http://www.maths.manchester.ac.uk/~avb/pigeon.html

Why isn't the answer to that room problem 12?
I figure, you throw 10 people in each room, leaving 14 people to be sorted.
Put 1 person into each of 12 rooms, we have 2 people left. Put those 2 into one room (a room with 10 people)
Now we have 2 rooms with 10 people, 12 rooms with 11 people, 1 room with 12 people.
So, the most crowded room has 12 people.

see this good article:[url]http://www.math.ust.hk/~mabfchen/Math232/Pigeonhole.pdf[/url]
the exaple 1.3 of this article is a tipical applycation of the pigeonhole principle... :wink:

anyway the secret in these problems is the consider correctly wich of tha data are the pigonholes and wich the pigeons... :wink:  (some times this may not be easy...)

prove if you put 4 points in a side 3 equilateral triangle then 2 of them will be within 1.5

I don't understand what your problem is asking mathemonster.  Can someone explain what it means?

I believe he is asking that if you place 4 dots on the side of an equilateral triangle with length 3, prove that two of the dots must be 1.5 away from each other.

Here's another one:

Given a 2 X 2 square, show that if 5 points are placed in the square, there will be a pair of points that are, at most, $ \sqrt{2}$ units apart.


I'll give it a try:

applying AM-HM will yield
$\frac{a_1^3}{S-a_1}+\frac{a_2^3}{S-a_2}+\cdots +\frac{a_n^3}{S-a_n}\geq \frac{n^2}{s(\frac{1}{a_1^3}+\frac{1}{a_2^3}+....+\frac{1}{a_n^3}) - (\frac{1}{a_1^2} + ......\frac{1}{a_n^2})}$

For the denominator, substitute $a_1+a_2+\cdots a_n=S$ and expanding, will get $a_1(\frac{1}{a_2^3}+\frac{1}{a_3^3}+...\frac{1}{a_n^3}) + a_2(\frac{1}{a_1^3}+\frac{1}{a_3^3}+...\frac{1}{a_n^3}) +..... + a_n(\frac{1}{a_1^3}+\frac{1}{a_2^3}+...\frac{1}{a_{n-1}^3})$

applying AM-GM by pairing the values up, will yield
$(n-1)(\frac{1}{a_1a_2} + \frac{1}{a_2a_3}+\frac{1}{a_3a_4}+...+ \frac{1}{a_na_1})$

apply AM-HM again to get $(n-1)(\frac{n^2}{1})$

so the whole thing becomes $\frac{a_1^3}{S-a_1}+\frac{a_2^3}{S-a_2}+\cdots +\frac{a_n^3}{S-a_n}\geq \frac{n^2}{(n^2)(n-1)} = \frac{1}{n-1}$

phew....finally finished. i hope it's correct and clear. correct me if i'm wrong....  :|

[quote="violetcraze"]applying AM-GM by pairing the values up, will yield
$(n-1)(\frac{1}{a_1a_2} + \frac{1}{a_2a_3}+\frac{1}{a_3a_4}+...+ \frac{1}{a_na_1})$
[/quote]

Sorry , but I cant see how this steps can be done by AM-GM  :) .Explain ?

It looks to me like to get the $\frac{1}{a_ia_j}$ term, she paired the $\frac{a_i}{a_j^3}$ and $\frac{a_j}{a_i^3}$ term, but there are some problems with it:
1.  That gives you a symmetric inequality, not a cyclic one.  So it also contains things like $\frac1{a_1a_3}$ which you don't really want.
2.  It's an application of AM-GM, but it does in the wrong direction.  That makes the denominator smaller, so it makes the whole small side larger, so even if it did give the right denominator, it wouldn't do you any good.

[quote="luimichael"][b]Step 1[/b]
Apply Chebyschev inequality:
$\frac{a_1^3}{S-a_1}+\frac{a_2^3}{S-a_2}+...+\frac{a_n^3}{S-a_n}\\ \geq {\frac 1 n}(a_1^2+a_2^2+...+a_n^2)\cdot (\frac{a_1}{S-a_1} +\frac{a_2}{S-a_2}+...+\frac{a_n}{S-a_n})$
$\geq \frac 1 n (a_1a_1+a_2a_3+...+a_na_1)\cdot (\frac{a_1}{S-a_1} +\frac{a_2}{S-a_2}+...+\frac{a_n}{S-a_n})$ by Re-arrangement inequality
$=\frac 1 n \cdot (\frac{a_1}{S-a_1} +\frac{a_2}{S-a_2}+...+\frac{a_n}{S-a_n})$
[b]
Step 2[/b]
$\frac{a_1}{S-a_1} +\frac{a_2}{S-a_2}+...+\frac{a_n}{S-a_n} \\ =\frac S {S-a_1} +\frac S {S-a_2}+...+\frac S {S-a_n}-n \\ \geq \frac{n^2}{n-1}-n \\ =\frac n {n-1}$

Combining Step 1 and Step 2, we got the result.

[b]Remark[/b]
By AM-HM inequality:
$[(S-a_1)+(S-a_2)+...+(S-a_n)](\frac 1 {S-a_1}+\frac 1 {S-a_2}+...+\frac 1 {S-a_n})\geq n^2\\ \implies (n-1)S(\frac 1 {S-a_1}+\frac 1 {S-a_2}+...+\frac 1 {S-a_n})\geq n^2 \\ \implies \frac S {S-a_1} +\frac S {S-a_2}+...+\frac S {S-a_n}\geq \frac{n^2}{n-1}$[/quote]

Bravo  :D !

How i did is by Cauchy in engel form then use
(i)$(n-1)\sum a_i^2\geq 2\sum a_ia_j$
(ii) $(a_1^2+a_2^2+\cdots a_n^2)(a_2^2+a_3^2+\cdots +a_1^2)\geq (a_1a_2+a_2a_3+\cdots a_na_1)^2$
    which is $\sum a_i^2\geq 1$

and complete the prove  :)

[quote="luimichael"][quote="Lovasz"]All are unnecessary. Just applying Cauchy-Schwarz in Engel form.[/quote]

I have thought about that before but I don't know how to deal with the powers and also there is no clue on how to relate the given condtion. :blush:

Do you mind showing us your proof? :)[/quote]
[quote="shyong"]For positive real $a_i$ such that $a_1a_2+a_2a_3+a_3a_4+\cdots a_{n-1}a_n+a_na_1=1$ .If $a_1+a_2+\cdots a_n=S$ , prove that

$\frac{a_1^3}{S-a_1}+\frac{a_2^3}{S-a_2}+\cdots +\frac{a_n^3}{S-a_n}\geq \frac{1}{n-1}$

 :)[/quote]
$\frac{a_1^3}{S-a_1}+\frac{a_2^3}{S-a_2}+\cdots +\frac{a_n^3}{S-a_n}$
$\ge \frac{(a_1^2+a_2^2+\cdot+a_n^2)^2}{(n-1)(a_1a_2+a_2a_3+\cdots+a_na_1)}$
We know the famous ineq:
$a_1^2+a_2^2+\cdots+a_n^2 \ge (a_1+a_2+\cdots+a_n)^2$
$\cdots$

Cauchy-Schwarz's in Engel form:

Let $a_i$ and $b_i,\ i=\overline{1,n}$ be positive reals, then we have:
\[ \frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots+\frac{a_n^2}{b_n} \ge \frac{\left(a_1+a_2+\cdots+a_n\right)^2}{b_1+b_2+\cdots+b_n}  \]

[quote="luimichael"][quote="Lovasz"]

We know the famous ineq:
$a_1^2+a_2^2+\cdots+a_n^2 \ge (a_1+a_2+\cdots+a_n)^2$
$\cdots$[/quote]

This inequality is incorrect in general.  :!:[/quote]

I'm pretty sure he meant $a_1^2+a_2^2+\cdots+a_n^2 \ge a_1a_2+a_2a_3+\cdots+a_na_1$, which would give the same result as shown above.

Yes, you're right, paladin8. I meant $a_1^2+a_2^2+\cdots+a_n^2 \ge a_1a_2+a_2a_3+\cdots+a_na_1$.
[hide="Sorry for the mistake."]I typed it while I hadn't square time. [/hide]

[quote="shyong"]

Sorry , but I cant see how this steps can be done by AM-GM  :) .Explain ?[/quote]

oops....sorry. it's really in the wrong direction.
 :blush:

[quote="M4RI0"]How did yo get: $\frac{a_1^3}{S-a_1}+\frac{a_2^3}{S-a_2}+\cdots +\frac{a_n^3}{S-a_n}$ $\ge \frac{(a_1^2+a_2^2+\cdot+a_n^2)^2}{(n-1)(a_1a_2+a_2a_3+\cdots+a_na_1)}$ by using Cauchy :?:[/quote]

$\frac{a_1^3}{S-a_1} = \frac{a_1^4}{a_1(S-a_1)}$. Do that to each term; then apply Cauchy in Engel form.

EDIT: On second thought, the denominator isn't exactly that (it should be more like $\sum_{i \neq j} a_ia_j$). The proof still works, I think, as

$(n-1)(a_1^2+a_2^2+\cdots+a_n^2) \ge \sum_{i \neq j} a_ia_j$.


For a given positive integer $ m$, find all triples $ (n,x,y)$ of positive integers such that $ (x^2\plus{}y^2)^m\equal{}(xy)^n$, where $ m,n$ are coprime.

let $ xy\equal{}z^m$ and $ x^2\plus{}y^2\equal{}z^n$
n>m obviously!
now consider a prime p and its exponent in the prime factorizations of x,y :wink:


why does y = |x+4| shifts to the left and vice versa?

Do you mean why adding $ 4$ to $ x$ in the expression $ y \equal{} |x|$ shift to the left?

Think of the $ x$-coordinate as a time.  When we add $ 4$ to it, everything happens $ 4$ units earlier, so everything you see on the graph you'll see $ 4$ units earlier than usual - that is, $ 4$ units to the left of where it used to be.  Does that help?

Since you're going to increase x by 4, that means that x has the oppurtunity of being 4 less than normal. So, now everything can be 4 less than what it is in the parent function. Similarly, if you subtract 4 from x, then it shifts to the right, since x has to be 4 greater to do what it did before. I hope that makes some sort of sense...

Another way to think of it, perhaps more mechanical, is this:

When you have "$ \plus{}4$" with $ x$ coordinate, what you have to do is to move THE ORIGIN ALONG X AXIS 4 units to the RIGHT (in the positive direction). That would result in the GRAPH being moved to the LEFT by the same amount.

In the similar manner, if you have $ y \equal{} |2x|$, then you have to STRETCH THE X AXIS twice, preserving the origin, which would result in the GRAPH BEING SQUEEZED horizontally.

Therefore, with some combined expression, such as $ y \equal{} |2x\plus{}1|\minus{}2$, first you write it thus: $ y\plus{}2 \equal{} |2x\plus{}1|$, and then you: 1) stretch the x axis twice, 2) move the origin along new x axis by 1 to the right, 3) move the origin along y axis up by 2. After all those operations, you can redraw the graph into a coordinate system similar to the initial one, if you want.

Alternatively, you can make a table of original $ x$ and $ y$, then another with the $ x$ and $ y \equal{} |x\plus{}4|$. Graph both of these by hand and see why it is shifting the way it is.

I always thought of it in this way:

For example in y = x + 4,

y = x is moved [u]up[/u] 4 units,

this is because your x-ordinate is still just x, but your y-ordinate is x + 4, 4 units higher along the y axis, which is what y= x + 4 is really saying. This just makes the graph look like you've moved it 4 units to the left when really this is an effect of moving it up 4 units.

Hmm, I hope I explained that well enough...

That interpretation is only valid for lines.  I believe the OP is talking about the general transformation from the graph of $ y \equal{} f(x)$ to the graph of $ y \equal{} f(x\plus{}4)$.  No "up" movement explains the behavior of this transformation in the general case (consider, for example, $ f(x) \equal{}\sin x$).


Show that R*R with dictionary order topology is metrizable.

Apply Urysohn metrization theorem. It is easy to verify that $ \mathbb{R}^2$ with dictionary topology is second-countable, Hausdorff and regular.


Each of the six members from the Skateer University hockey
team shakes hands with each of the six members of the Iceburg
Tech team. How many handshakes take place?

Each member from each team shakes hand once with every player from the other team, so we have $ 6\times{6}\equal{}\boxed{36}$.


inequalities
calculus
algebra
binomial theorem
inequalities unsolved

According to Bernoulli inequality, if $ x$ be a real number greater than $ \minus{}1$ and $ r$ be a positive real greater than or
equal to $ 1$, then $ (1\plus{}x)^r\geq1\plus{}rx$.
Is there a proof for this inequality without using derivation or Mean Value theorem? (or an elementary proof)

This is equivalent to proving
\[ r\minus{}1\plus{}a^r\geq ra\]
It can be proven using only the Arithmetic-Geometric mean inequality.
Try it first for when $ r$ is an integer. Next prove it for $ r\in \mathbb{Q}$. And then finish it off by using continuity to extend over the real case  :)

[quote="Albanian Eagle"]This is equivalent to proving
\[ r \minus{} 1 \plus{} a^r\geq ra\]
It can be proven using only the Arithmetic-Geometric mean inequality.
Try it first for when $ r$ is an integer. Next prove it for $ r\in \mathbb{Q}$. And then finish it off by using continuity to extend over the real case  :)[/quote]
OK, but is there a way to prove this inequality directly for irrational values of $ r$ without using continuity?

[quote="Dr Sonnhard Graubner"]hello, see here
http://mathworld.wolfram.com/BernoulliInequality.html
Sonnhard.[/quote]
But this argument hold only for positive integers $ r$, not all positive real $ r\geq1$.

[quote="hana1122"][quote="Albanian Eagle"]Well, let me ask you how do you define a number like $ a^r$ for an irrational $ r$?[/quote]
I understand your mean. but I think something like Newton's generalized binomial theorem can solve it. what is your opinion?[/quote]
Well, a real power can only be defined by countinuity from rational powers. Since I gave you the outline of a proof which avoids calculus for the rational case you can see that it extends naturally to the real case. Avoiding continuity or limits means not knowing what $ a^r$ is when $ r$ is not rational  :wink:

[quote="Dr Sonnhard Graubner"]hello, i think this is what you are searching for
http://www.dma.ulpgc.es/~aplaza/ficheros/investigacion/ficheros/pww_feb09.pdf
Sonnhard.[/quote]
Nice and simple proof. Thanks a lot.


Câu 1

  cho dãy số {$ a_n$} thỏa mãn:
           $ a_1 = 2$

         $ a_1 + a_2 +............+ a_n = n^2a_n$   :forall n>1

       tìm $ \lim_{n\to\infty}$ $ n^2 a_n$

Câu 2

Tính tích phân $ I_n$ = $ \int_0^{\pi}$ $ \frac{sin nx}{sin x}$   :forall n :in N

Câu 3

cho hàm số f(x) liên tục trên [0,1] và  $ \int_0^1 f(x)\,dx$<$ \frac{1}{2008}$

chứng minh rằng phương trình f(x)=$ x^2007$ có í nhất 1 nghiệm thuộc khoảng (0,1)

Câu 4

 cho hàm số f(x) liên tục trên đoạn [0,1] và thỏa mãn f(0)=0 , f(1)=1.
chứng minh rằng tồn tại 2 số a,b thuộc khoảng (0,1) , a  :neq  b sao cho $ f^'(a).f^'(b)$ =1

Câu 5

Cho hàm số f: [a,b] -> [a,b] ,thỏa mãn

    |f(x)-f(y)|<|x-y| ;  :forall   x,y  :in  [a,b] , x :neq  y

chứng minh rằng phương trình f(x) = x có duy nhất nghiệm trên [a,b]

Câu 6

cho IK là đoạn vuông góc chung của 2 đường thẳng chéo nhau a,b (I :in  a , K :in  b), M và N là hai điểm bất kì lần luợt thuộc a và b sao cho IM+KN = MN . Trong số các điểm cách đều các đường thẳng a,b và MN, hãy tìm điểm có khoảng cách đến mỗi đường nói trên là ngắn nhất

[quote="L_Euler"]Câu 1

  cho dãy số {$ a_n$} thỏa mãn:
           $ a_1 = 2$

         $ a_1 + a_2 + ............ + a_n = n^2a_n$   :forall n>1

       tìm $ \lim_{n\to\infty}$ $ n^2 a_n$

Câu 2

Tính tích phân $ I_n$ = $ \int_0^{\pi}$ $ \frac {sin nx}{sin x}$   :forall n :in N

Câu 3

cho hàm số f(x) liên tục trên [0,1] và  $ \int_0^1 f(x)\,dx$<$ \frac {1}{2008}$

chứng minh rằng phương trình f(x)=$ x^2007$ có í nhất 1 nghiệm thuộc khoảng (0,1)

Câu 4

 cho hàm số f(x) liên tục trên đoạn [0,1] và thỏa mãn f(0)=0 , f(1)=1.
chứng minh rằng tồn tại 2 số a,b thuộc khoảng (0,1) , a  :neq  b sao cho $ f^'(a).f^'(b)$ =1

Câu 5

Cho hàm số f: [a,b] -> [a,b] ,thỏa mãn

    |f(x)-f(y)|<|x-y| ;  :forall   x,y  :in  [a,b] , x :neq  y

chứng minh rằng phương trình f(x) = x có duy nhất nghiệm trên [a,b]

Câu 6

cho IK là đoạn vuông góc chung của 2 đường thẳng chéo nhau a,b (I :in  a , K :in  b), M và N là hai điểm bất kì lần luợt thuộc a và b sao cho IM+KN = MN . Trong số các điểm cách đều các đường thẳng a,b và MN, hãy tìm điểm có khoảng cách đến mỗi đường nói trên là ngắn nhất[/quote]

Đề thế này thì chắc mình dân chuyên mới vào nổi  :(  :(

cũng ko khó đâu.nó chỉ quanh quẩn mấy loại thôi


Putnam
integration
linear algebra
matrix
algebra
polynomial
calculus

Let $S_n$ denote the set of all permutations of the numbers $1,2,\dots,n.$ For $\pi\in S_n,$ let $\sigma(\pi)=1$ if $\pi$ is an even permutation and $\sigma(\pi)=-1$ if $\pi$ is an odd permutation. Also, let $v(\pi)$ denote the number of fixed points of $\pi.$ Show that
\[ \sum_{\pi\in S_n}\frac{\sigma(\pi)}{v(\pi)+1}=(-1)^{n+1}\frac{n}{n+1}.  \]

First,
\[ \sum_{\pi\in S_n}\frac{\sigma(\pi)}{v(\pi)+1} = \sum_{\pi\in S_n}\int_0^1\sigma(\pi)x^{v(\pi)}dx = \int_0^1\sum_{\pi\in S_n}\sigma(\pi)x^{v(\pi)}dx  \] Now we note that $\sum_{\pi\in S_n}\sigma(\pi)x^{v(\pi)}$ is the determinant of the $n\times n$ matrix with all diagonal entries equal to $x$ and all other entries equal to $1.$ By subtracting the first row from all others, we reduce it to a matrix where only the entries in the first row, first column, and the diagonal are nonzero. The rest is a straightforward computation.

Note:
 sum sigma(pi)/v(pi) +2 = (-1)^n /n+2. for all n.
Using this it is asy to prove the problem.

That is a beautiful solution, lego. One refinement: if $J$ is the $n\times n$ matrix with all entries equal to 1, it has rank 1 and trace $n$, so its characteristic polynomial is $\lambda^{n-1}(\lambda-n)$ and $\det(J-(1-x)I)=(-1)^n(1-x)^{n-1}(1-x-n)$. The integral is then easy to evaluate by the substitution $t=1-x$.

I can't top that, but I do have a solution that's probably easier to find; it was the first method I tried, and it felt very routine.

For each set $A\subset{1,2,\dots,n\}}$, let $O(A)$ be the number of odd permutations fixing $A$, $E(A)$ be the number of even permutations fixing $A$, $P(A)$ be the number of odd permutations fixing exactly $A$, and $F(A)$ be the number of even permutations fixing exactly $A$.
If $|A|\le n-2$, $O(A)=E(A)=\frac{(n-|A|)!}{2}$. If $|A|=n-1$ or $|A|=n$, $O(A)=0$ and $E(A)=1$.
By the inclusion-exclusion principle, $P(A)=\sum_{B\supseteq A} O(B)\cdot(-1)^{|B|-|A|}$ and $F(A)=\sum_{B\supseteq A} E(B)\cdot(-1)^{|B|-|A|}$.
We have $\sum_{\pi\in S_n}\frac{\sigma{\pi}}{v(\pi)+1}=\sum_{A}\frac{F(A)-P(A)}{1+|A|}$, sorting by the fixed set of each permutation. Obviously, we need to calculate $F(A)-P(A)$.
$F(A)-P(A)=\sum_{B\supseteq A} (E(B)-O(B))(-1)^{|B|-|A|}$. All terms of this sum are zero except for the sets $B$ with at least $n-1$ elements, each of which have $E(B)-O(B)=1$. There is one set with $n$ elements, and there are $n-|A|$ sets with $n-1$ elements containing $A$, so $F(A)-P(A)=(-1)^{n-|A|}+(-1)^{n-|A|-1}(n-|A|)=(-1)^{n+1-|A|}(n-|A|-1).$

We now have
$\sum_{\pi\in S_n}\frac{\sigma{\pi}}{v(\pi)+1}=\sum_{A}\frac{(-1)^{n+1-|A|}(n-|A|-1)}{|A|+1}= (-1)^{n+1}\sum_{k=0}^n (-1)^k\binom{n}{k} \frac{n-k-1}{k+1}$
Since $\frac{n-k}{k+1}\cdot\binom{n}{k}=\binom{n}{k+1}$ and $\frac1{k+1}\binom{n}{k}=\frac1{n+1}\cdot\binom{n+1}{k+1}$, we have
$\sum_{k=0}^n (-1)^k\binom{n}{k} \frac{n-k-1}{k+1}= \sum_{k=0}^n (-1)^k\cdot\binom{n}{k+1}-\frac{(-1)^k}{n+1}\cdot\binom{n+1}{k+1}.$
By the binomial theorem (evaluating at $-1$), we have $\sum_{\pi\in S_n}\frac{\sigma{\pi}}{v(\pi)+1}= (-1)^{n+1}\left((1-(1+(-1))^n)-\frac1{n+1}(1-(1+(-1))^{n+1})\right)$
$= (-1)^{n+1}\left(1-\frac1{n+1}\right)=\frac{(-1)^{n+1}n}{n+1}$

After I presented lego's solution to my class, I speculated that the problem might have been "reverse engineered" from this argument.

Here's the solution I came up with last week (looking at the Putnam for the first time...)

Let f_n(x) = sum_{pi in S_n} x^{v(x)} pi(x).  By considering the size of the cycle of sigma containing n, I got the recurrence relation f_n(x) = x f_{n-1}(x) - (n-1) f_{n-2}(x) + (n-1)(n-2) f_{n-3}(x) - ... +- (n-1)! f_1(x) -+ (n-1)!.  By considering f_n(x) - x f_{n-1}(x) this simplifies to f_n(x) - x f_{n-1}(x) = -(n-1) f_{n-2}(x) - (n-1) (f_{n-1}(x) - x f_{n-2}(x)).  It is then straightforward to check that f_n(x) = (x-1)^{n-1} (x+n-1) satisfies this recurrence and gives the right answer for n=1,2.  Then you integrate f_n(x) from 0 to 1 to get the given formula.

After the 3 simulated hours, I found the following refinement: if you see what's involved in calculating exp(xy - x^2/2 + x^3/3 - ...), you see the coefficient of x^n y^m is exactly 1/n! sum_{pi in S_n, v(pi)=m} sigma(pi), so it's equal to sum_{n=0}^infty f_n(x) y^n/n!.  (This follows from the fact that there are n!/(1^s_1 s_1! 2^s_2 s_2! ...) permutations with s_1 fixed points, s_2 2-cycles, etc., and the sign of such a permutation is (-1)^{s_2+s_4+...}.)  On the other hand, it's also equal to exp(ln(1+x) + x(y-1)) = (1+x) exp(x (y-1)).  By expanding this appropriately, you recover the formula f_n(x) = (x-1)^{n-1} (x-n+1).

Here's a thought:  $ \frac{1}{\nu(\pi) \plus{} 1}$ is a class function, which means it should be a linear combination of the irreducible characters of $ S_n$.  The problem asks us to compute $ \left< \frac{1}{\nu(\pi) \plus{} 1}, \sigma(\pi) \right>$, or the coefficient of $ \sigma(\pi)$ (which is an irreducible character) in the expansion of $ \frac{1}{\nu(\pi) \plus{} 1}$.  Are there any general techniques for arguments like this?

Here's another thought:  jmerry's probabilistic argument [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=262523]here[/url] has convinced me that every integration argument using generating functions can be converted into a probabilistic one.  In this case, $ \frac{1}{\nu(\pi) \plus{} 1}$ is the probability that, given a permutation in $ S_{n\plus{}1}$ that fixes $ 1$, a randomly chosen fixed point is $ 1$.  I'd be very interested if anyone could complete the argument this way.

The same idea as in the link gives that the average of $ \frac1{\nu(\pi)+1}$ is $ \frac1{n+1}\cdot\frac{(n+1)!}{(n+1)!-D_{n+1}}$, where $ D_{n+1}$ is the number of derangements $ (n+1)!(1-1+\frac12-\frac16+\cdots+(-1)^{n+1}\frac1{(n+1)!})$. This is approximately $ \frac{e}{e-1}\cdot\frac1{n+1}$, and it's also the wrong question.

OK, we need to incorporate the sign of the permutation, and the probabilistic argument doesn't do that naturally. We calculate two probabilities: $ Q_0$ is the probability that given an even permutation in $ S_{n+1}$ that fixes $ 1$, a randomly chosen fixed point is $ 1,$ and $ Q_1$ is the probability that given an odd permutation in $ S_{n+1}$ that fixes $ 1,$ a randomly chosen fixed point is $ 1$. The desired sum is $ \frac{n!}{2}(Q_0-Q_1)$ (assuming $ n\ge 2$), since $ \frac{n!}{2}$ of each type of permutation fix $ 1$.
Now, we count even derangements $ E_{n+1}$ and odd derangements $ O_{n+1}$. This is of course an inclusion-exclusion count; we use notations $ O(A),E(A),P(A),F(A)$ as in my previous post, only referring to permutations in $ S_{n+1}$.
We get $ O_{n+1}=P(\{\})=\sum_{B}E(B)\cdot (-1)^{|B|}=\sum_{k=0}^{n+1}(-1)^k\lfloor\frac{(n+1-k)!}{2}\rfloor\cdot\binom{n+1}{k}$
$ =-\frac12(n+1)(-1)^n-\frac12(-1)^{n+1}+\frac12\sum_{k=0}^{n+1}(-1)^k(n+1-k)!\cdot\binom{n+1}{k}$
$ =-\frac{(-1)^n n}{2}+\frac12\sum_{k=0}^{n+1}(-1)^k\frac{(n+1)!}{k!}=\frac12(D_{n+1}-(-1)^n\cdot n)$
The even derangements are almost the same, reversing the sign of the "error" term.

If we choose an odd non-derangement permutation at random and then choose a fixed point randomly, the probability that we choose $ 1$ is $ \frac1{n+1}$ by symmetry. There are $ (n+1)!-O_{n+1}$ non-derangement permutations, of which $ \frac{n!}{2}$ fix $ 1$. This makes $ Q_1=\frac1{n+1}(\cdot\frac{(n+1)!-O_{n+1}}{n!/2}$ and similarly $ Q_0=\frac1{n+1}\cdot\frac{(n+1)!-E_{n+1}}{n!/2}$. Our sum $ \frac{n!}{2}(Q_0-Q_1)=\frac{n!}{2}(\frac1{n+1}\cdot\frac{(n+1)!-E_{n+1}}{n!/2}-\frac1{n+1}\cdot\frac{(n+1)!-O_{n+1}}{n!/2})$
$ =\frac1{n+1}((n+1)!-E_{n+1}-(n+1)!+O_{n+1})$
$ =\frac1{n+1}(-\frac{D_{n+1}}{2}-\frac{(-1)^n n}{2}+\frac{D_{n+1}}{2}-\frac{(-1)^n n}{2})=\frac{(-1)^{n+1}n}{n+1}$

The structure is really very similar to my routine inclusion/exclusion bash, and has no obvious relation to the clever integration of a determinant.

[quote=ThE-dArK-lOrD][quote=lego]First,
\[ \sum_{\pi\in S_n}\frac{\sigma(\pi)}{v(\pi)+1} = \sum_{\pi\in S_n}\int_0^1\sigma(\pi)x^{v(\pi)}dx = \int_0^1\sum_{\pi\in S_n}\sigma(\pi)x^{v(\pi)}dx  \][/quote]

[hide=A less ingenious way to finish]Denote $\sum_{\pi \in S_n}{\sigma (\pi) x^{v(\pi )}}$ by $F_n(x)$. We have
$$F_{n+1}(x)=xF_n(x) +\sum_{k=0}^{n-1}{F_k(x)(-1)^{n-k}\binom{n}{n-k}(n-k)!}.$$
Setting $S_n(x)=F_n(x)/n!$, we get
\begin{align*}
(n+1)S_{n+1}(x) & =xS_n(x)+\sum_{k=0}^{n-1}{(-1)^{n-k}S_k(x)}\\
(n+2)S_{n+2}(x) & =xS_{n+1}(x)+\sum_{k=0}^{n}{(-1)^{n+1-k}S_k(x)}.
\end{align*}
So, $$(n+2)S_{n+2}(x)+(n+1)S_{n+1}(x)=xS_{n+1}(x)+xS_n(x)-S_n(x).$$
Let $K(y)=\sum_{k=0}^{\infty}{S_k(x)y^k}$ be the generating function of $\{ S_n(x)\}$.
Note that $S_0(x)=1,S_1(x)=x$. From the recurrence, we get
\begin{align*}
& \left( \frac{K'(y)-x}{y} \right) +K'(y) = x\left( \frac{K(y)-1}{y}\right) +xK(y)-K(y)\\
\implies & \frac{K'(y)}{K(y)} = \frac{xy-y+x}{y+1}\\
\implies & K(y)=(y+1)e^{(x-1)(y+1)+C_y}.
\end{align*}
Since $K(0)=1$, we get $C_y=-(x-1)$ and $K(y)=(y+1)e^{(x-1)y}$.
Finally,
$$ \sum_{k=0}^{\infty}\left( \int_0^1F_k(x) dx\right) \frac{y^k}{k!}  =\int_0^1K(y) dx  =\frac{y+1}{y}(1-e^{-y}) = \sum_{k=0}^{\infty}{\left( \frac{(-1)^{k+1}k}{k+1}\right) \frac{y^k}{k!}}.$$[/hide][/quote]

nice finish!

Let $f(n)$ denote the number of even derangements minus number of odd derangements of $\{1,\cdots,n\}$

Main claim: $f(n)=(-1)^{n-1}(n-1)$ for all $n\ge 1$

Proof: We use strong induction on $n$, with base cases, $n=1,2$ being obvious.

Inductive step: I claim $\sum\limits_{i=0}^n \binom ni f(i)=0$ for all $n\ge 3$. Let $A$ denote the set of fixed points of a permutation of $\{1,\cdots,n\}$. Then, we get a derangement of $n-|A|$ elements. There are $\binom ni$ ways to choose $A$ with $i$ elements. Therefore, this counts the number of even permutations of $\{1,\cdots,n\}$ minus number of odd permutations of $\{1,\cdots,n\}$.

With this in mind, the conclusion holds by binomial theorem (multiplied by a constant)

Now, we want $\sum_{\pi \in S_n} \frac{\sigma(\pi)}{\nu(\pi)+1}=\sum\limits_{k=0}^n \frac{1}{k+1} \binom nk f(n-k) $

$=\sum\limits_{k=0}^n \frac{1}{k+1} \binom nk (-1)^{n-k-1}(n-k-1)=\sum\limits_{k=0}^n (\frac{1}{k+1} \binom nk (-1)^{n-k-1}n)-\binom nk (-1)^{n-k-1}$

Note $\sum\limits_{k=0}^n \binom nk (-1)^{n-k-1}$ vanishes.

Therefore, we sum $n\sum\limits_{k=0}^n \frac{1}{k+1} \binom nk (-1)^{n-k-1}$

Note $\frac{1}{k+1} \binom nk = \frac{1}{n+1}\binom{n+1}{k+1}$

Thus we have $\frac{n}{n+1} \sum\limits_{k=0}^n  \binom {n+1}{k+1} (-1)^{n-k-1}$

Again, $ \sum\limits_{k=-1}^n  \binom {n+1}{k+1} (-1)^{n-k-1}$ vanishes.

Therefore, $\sum\limits_{k=0}^n  \binom {n+1}{k+1} (-1)^{n-k-1}=-(-1)^{n}$, as needed.


algebra
functional equation
linear algebra
linear algebra unsolved

Hi,

Find all applications from $ \mathcal{M}_n(\mathbb{R})$ to $ \mathcal{M}_n(\mathbb{R})$

such that
\[ f(A \plus{} B) \equal{} f(A) \plus{} f(B)
\]
And
\[ f(AB) \equal{} f(A)f(B)
\]

At the end of http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45938 you'll find a generalization of your question, but similar steps might be applied here.

Hello JC_math, this is not a generalization of my question, but of course this is connected to our problem . Indeed the maps considered above are not supposed to be linear.

For instance if you are in dimension 1, you get the following functional equation:
$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ and $ f(xy)\equal{}f(x)f(y)$ which can be solved using properties of the field $ \mathbb{R}$
In fact if the field is $ \mathbb{C}$ there are a lot of discontinuous solution satisfying the functional equation, and it is possible to construct discontinuous mapping for higher dimensions also


The entire Putnam exam is available on the web at
http://www.unl.edu/amc/a-activities/a7-problems/putnam/
for anyone interested.

Juet letting you all know that the solutions have been put up for the exam. Some of them are really quite elegant!

B6 was tough!  The closest I got was with an extra factor of 1/2 on the right hand side.

B6 may be the toughest Putnam question I have seen to date.

Professor Noam Elkies presented 3 of the solutions to the questions today -- B5 he did essentially the same way I did, which made me happy.  He did A6 with generating functions, which I thought was cool, since it avoided the "how would you ever think of this" element.  For B4, he used Galois theory.  I thought it was very funny -- very overkill.

Yeah, there is a Galois Theory solution to B4 on the AMC website.


trigonometry
linear algebra
matrix
real analysis
real analysis unsolved

y''' + 25y' = csc(5x)

i got the y (complimentary) = C1 + C2cos5x + C3cos5x. I'm just having minor difficulties getting the y (particular).

Of course, you meant to say

$y_{c}(x)=c_{1}+c_{2}\cos 5x+c_{3}\sin 5x.$ The Wronskian matrix is

$\begin{bmatrix}1&\cos 5x&\sin 5x\\ 0&-5\sin 5x&5\cos5x\\ 0&-25\cos 5x&-25\sin 5x\end{bmatrix}.$

I assume you're using the method of variation of parameters. Can you show what you've done so far?

so far i got
W = 125
W1 = 5csc5x
W2 = -5cos5xsin5x=-5(cos5x)^2 . tan5x
W3 =-5

therefor

U'1 = csc5x/25
U'2 = -cos5xsin5x/25
U'3 = -1/25

y=C1 + C2cos5x + C3sin5x + (1/5)Intan(5x/2) + (1/250)(cos5x)^3 - (1/25)xsin5x

i got that answer. can anyone verify that for me.[/code]


Ross Mathematics Program
summer program
PROMYS

After being denied from Promys..... ....  ....        ...............

I am going to fill out an app to Ross, so Im curious.
It says deadline = June 1, but is it rolling?

I don't think its rolling, just try to send it in as soon as possible--although there was a camper last summer who managed to get his application in the day before the program started and was accepted, so you never know.  If you get accepted and plan on going PM me, I am gonna be a Junior Counselor at Ross this summer.

askldj al;dkjalkdjalskdj klasdj alksdj alksdj adkja

My fickle self has decided now not to do the ross app. Im sitting here at 8:18 typing this with a giant headache, tests to study for tomorrow (well, just chemistry... i was gone for all but 1 day of the lesson), a bed that calls to me, ap tests just around the corner, and all i can think about is how horrible this day has been....

not only did i not get accpeted to promys, i also found out that my school isnt going to offer stat or physics ap next year, so now i have to do that online, plus find more courses. I have decided instead of going to a camp to just go to mao nationals and learn about the rest of everything (i.e. olympiad math, applied math (my project), and the rest of the sciences) just as if "I stood on the shoulders of giants".

Bah... this is just my 4 cents about my horrible day, didnt mean to amke it into a journal, but alas, if that is what it has come to, so be it.


Find all triples $(x,y,z)$ of real numbers such that

$x^{2}+2yz+5x=y^{2}+2zx+5y=z^{2}+2xy+5z=2$

from the first two equation we got:
$x^{2}-y^{2}+5(x-y)=2z(x-y)=0 \implies x=y~ or~ z=0$
If x=y from the last two equations we got:
$y^{2}-z^{2}+5(y-z)=2x(y-z)=0 \implies z=y~or~ x=0$ But x cannot be 0 otherwise y=0 and the firs equation cannot be equall with 2. so y=z .
x=y=z then $x^{2}+2x^{2}+5x=2\implies 3x^{2}+5x-2=0\implies x=y=z=-2~or~x=y=z=\frac{1}{3}$

If z=0 then $x^{2}+5x-2=0,y^{2}+5y-2=0,xy=1$ wich is never satisfied.

[quote="cckek"]from the first two equation we got:
$x^{2}-y^{2}+5(x-y)=2z(x-y)=0$[/quote]

Why they both are to $0$?

Here is my solution:

Subtracting equations $2nd$ from $1st$, $3rd$ from $2nd$,$1st$ from $3rd$ we get:

$(x-y)(x+y-2z+5)=0$
$(y-z)(y+z-2x+5)=0$
$(z-x)(z+x-2y+5)=0$

Let's say $y \neq z$ and $z \neq x$ so we have $y+z-2x+5=0=z+x-2y+5$. From there we get $x=y$. Then $z=x-5$. Put this into first given equation and we have $3x^{2}-5x-2=0$ and roots are $x=2$ and $x=-\frac{1}{3}$.

So we get solutions: $(x,y,z)=(2,2,-3),(-\frac{1}{3},-\frac{1}{3},-\frac{16}{3})$

In the case when $x=y=z$ we get $3x^{2}+5-2=0$ we get solutions $x=-2$ and $x=\frac{1}{3}$.

So all solutions are: $(x,y,z)=(-2,-2,-2), (\frac{1}{3},\frac{1}{3},\frac{1}{3}), (2,2,-3), (-\frac{1}{3},-\frac{1}{3},-\frac{16}{3})$ and all permutations.


The given relation is equivalent to :
$log_{3}2=log_{11-2^{x}}(7-3^{x})$
Let $a,b\in \mathbb R$ such that $11-2^{x}=3^{a},(1)$ and $7-3^{x}=2^{b},(2)$ using those notations we obtain that $a=b$ and from (1) and (2) $\rightarrow$ $log_{3}(11-2^{x})=log_{2}(7-3^{x})$ Let $f: D\to  \mathbb R$
$f(x)=log_{3}(11-2^{x})-log_{2}(7-3^{x})$ , $\rightarrow$ $f'(x)>0$ $\rightarrow$ f is increasing $\rightarrow$ $x=1$ is unique root of the equiation $f(x)=0$ and we are done :D