probability
combinatorics unsolved
combinatorics

Eight players $ P_1,P_2,........,P_8$take part in a knockout competition. $ P_i$ beats $ P_j$ iff $ i>j$
Find probability that $ P_4$ reaches the finals ?

I'm not sure, but is it...
[hide]Find the probability that $ P_4$ wins 3 games. Thus $ P_4$ have to play only $ P_1$, $ P_2$, and $ P_3$. Since there are 7 other players, the different combination of players $ P_4$ can play against is $ \binom{7}{3} \equal{} 35$, so the probability is $ \frac{1}{35}$?

I apologize if I am incorrect![/hide]

I think something different was meant...but we got the same answer, oddly enough.

[hide]Is a knockout tournament a single elimination tournament?  If so, then that mean $ P_4$ must play against two of $ P_1, P_2, P_3$.  That means that in the first round $ P_4$ plays against one of those three, and the other two play against each other.  Then $ P_4$ must play against the winning one of those.  From here it's easy: the probability that the first round is as we want it is $ \frac{3*3}{7*5*3*1}$, and then the probability that the second round is as we want it is $ \frac{1}{3}$: thus, our final probability is $ \frac{1}{35}$.[/hide]

[quote="K81o7"]

 From here it's easy: the probability that the first round is as we want it is $ \frac {3*3}{7*5*3*1}$, [/quote]

Could you plz explain that in a bit more detail--  $ \frac {3}{7}$ =prob for $ P_4$ playing 1,2,3
Prob that the other two play each other is $ \frac{1}{^6C_2}$

Did you use some other formula or trick
because i see $ \Pi (2n \minus{} 1)$ in the $ D^r$ ?
Thx a lot

[quote="|nSan|ty"][quote="K81o7"]

 From here it's easy: the probability that the first round is as we want it is $ \frac {3*3}{7*5*3*1}$, [/quote]

Could you plz explain that in a bit more detail--  $ \frac {3}{7}$ =prob for $ P_4$ playing 1,2,3
Prob that the other two play each other is $ \frac {1}{^6C_2}$

Did you use some other formula or trick
because i see $ \Pi (2n \minus{} 1)$ in the $ D^r$ ?
Thx a lot[/quote]

The total number of different ways to match up $ 8$ people into $ 4$ pairs is $ 7*5*3*1$: the way you do this is as follows.  Take the first person, and choose someone for that person to go against.  Now pick the next person who hasn't already been pair.  Repeat.  The total number of ways to pair up the people the way we want is $ (3*1)*(3*1)$

Another way to think about it (or more accurately, a rewording of some previous thoughts):

Suppose we want to find the odds $ i$ makes the finals.

We can think of the tournament as happening in two stages:

Stage 1: We assign 3 teams to be in the same half of the bracket as $ i$.
Stage 2:  $ i$ wins his half (and makes the final) if all three assigned teams were chosen from $ \{1, \dots i \minus{} 1 \}$.

There are $ \binom{7}{3} \equal{} 35$ ways to choose the three teams, and $ \binom{i \minus{} 1}{3}$ ways to choose three teams so that $ i$ wins, so the odds are $ \frac {\binom{i \minus{} 1}{3}}{35}$.

Taking $ i \equal{} 4$ gives the answer to this problem


function
geometry
geometric transformation

I need to find if the following equation determines y as a function of x.  Any help would be great.

x=(x+y/(x-y)

Thanks.

we have \[x=\frac{x+y}{x-y}\]
from componendo-dividendo (i don't know if my english is correct)\[\frac{x+1}{x-1}=\frac{x}{y}\]so \[y=\frac{x(x-1)}{x+1}\]

Thanks.
Can you explain how this componendo-dividendo  works?  That's new to me.

So y IS a function of x because it results in one solution.
Thanks for your help.

[quote="nayel"]if $\frac{a}{b}=\frac{x}{y}$ then $\frac{a+b}{a-b}=\frac{x+y}{x-y}$ this is called componendo-dividendo. :D[/quote]

Challenge: Anyone care to prove it?  Another one in this vein:

If $\frac{a}{b}=\frac{x}{y}$, then $\frac{a}{b}=\frac{x}{y}=\frac{a+cx}{b+cy}$ for any constant $c$.

hmm, is there an English translation to these kinds of properties? Anyways, here's the proof:

If $\frac{x}{y}=\frac{a}{b}$ then $\frac{x+y}{y}=\frac{a+b}{b}$.
Similarly $\frac{x-y}{y}=\frac{a-b}{b}$ or $\frac{y}{x-y}=\frac{b}{a-b}$.

Multiply both these identities together:
$\frac{x+y}{y}\times\frac{y}{x-y}=\frac{a+b}{b}\times\frac{b}{a-b}$
$\frac{x+y}{x-y}=\frac{a+b}{a-b}$

[quote="rrusczyk"]Challenge: Anyone care to prove it?  Another one in this vein:

If $\frac{a}{b}=\frac{x}{y}$, then $\frac{a}{b}=\frac{x}{y}=\frac{a+cx}{b+cy}$ for any constant $c$.[/quote]

just let $a=kb$ and $x=ky$ and it immediately follows. :D

or you can just expand it out...

if c=0, it is obvious...else
$\frac{a}{b}=\frac{a+xc}{b+yc}\iff ab+acy=ab+bcx\iff acy=bcx\iff \frac{a}{b}=\frac{x}{y}$

Grats! Alex just got 3500 posts!
And yes, this is something that comes in handy on systems of equations.
It's a nice formula to have memorized too.


trigonometry
limit
function
real analysis
real analysis unsolved

[quote="Laura S"]Define $f(x)=x^3 -\sin^2{x}\tan{x}$.

Is $f(x)$ positive or negative on $(0,\pi/2)$, or  does it change its sign?

Thanks a lot for any help.
-Laura[/quote]
It's always negative. To see this, it's sufficient to show that $\sin^3 x > x^3 \cos x$ for $0 < x < \pi/2$ or $\left(\frac{\sin x}{x}\right)^3 > \cos x$ for $0 < x < \pi/2$.

Now from the Taylor series for the sine and cosine, $\sin x > x - \frac{x^3}{6}$ and $\cos x < 1 - \frac{x^2}{2} + \frac{x^4}{24}$ for all positive $x$ ([i]use the fact that both series are alternating for positive $x$[/i]).

Therefore
$\left(\frac{\sin x}{x} \right)^3 > 1 - 3\frac{x^2}{6} + 3\frac{x^4}{36} - \frac{x^3}{216} = 1 - \frac{x^2}{2} + \frac{x^4}{12} - \frac{x^3}{216}> 1 - \frac{x^2}{2} + \frac{x^4}{24} > \cos x$
as long as $\frac{x^4}{12} - \frac{x^6}{216} > \frac{x^4}{24}$, i.e. certainly for $0 < x < 3$.

I waited too long to post and hpe beat me to it. But anyway ...

Fourier series seem unlikely to be informative. But let's look at power series:

$\sin^2x=\frac12(1-\cos 2x)=x^2-\frac13x^4+\frac2{45}x^6+O(x^8)$

$\tan x=x^3+\frac13x^5+\frac2{15}x^5+O(x^7)$

Multiply those together to get

$\sin^2x\tan x=x^3+\frac1{15}x^7+O(x^9)$

That's not definative proof, but we know know the following:

There is an interval of the form $(0,a)$ on which $x^3-\sin^2x\tan x<0.$

Since $\lim_{x\to\frac{\pi}2^2}x^3-\sin^2x\tan x=-\infty$, there is also an interval of the form $\left(b,\frac{\pi}2\right)$ on which it is negative.

So far it seems likely that this function is negative on the whole interval, and that's what numerical work shows.


There is an eigenvector $ V\neq0$ such that $A^2BV = -V$, because   $-1$ is an eigenvalue of$ A^2B $.
$zABV-tA^2BV=(t-z)V \Longrightarrow zABV+tV = (t-z)V $ $\Longrightarrow  ABV=-V  \Longrightarrow AV=V$
$xBAV-(y+x)ABV=yV \Longrightarrow xBAV+(y+x)V=yV $ $\Longrightarrow BAV=-V \Longrightarrow BV=-V$
Finally :  $(AB-BA)V=0 \Longrightarrow det(AB-BA)=0$ , and, $(A+B)V=0 \Longrightarrow det(A+B)=0$

 :cool:

[quote="Moubinool"]$A,B\in M_n(R)$ such that $-1$ is an eigenvalue of $A^2B$
suppose
there exist $z>t>0$ with $zAB-tA^2B=(t-z)I_n$
there exist $x>y>0$ with $xBA-(y+x)AB=yI_n$

Prove that $det(AB-BA)=0$[/quote]

From the first equality we have $z(AB+I_n)=t(A^2B+I_n)$, from where we get $det(AB+I_n)=0$ since $det(A^2B+I_n)=0.$ From the second equality we have $x(BA-AB)=y(AB+I_n)$ which means $det(AB-BA)=0$

Diogene you get $det(A+B)=0$ I did not notice this when I wrote this problem
thank's again.


trigonometry
algebra unsolved
algebra
144

Easy $x=2cos(2z)$, then equation equavalent to $cos(6z)=cosz (cosz>0)$.


Let $n$ be a integer positive.Find the number of ordered triples of sets $(A,B,C)$ such that $A\cup B\cup C=\{1,2,...,n\}$ and $A\cap B\cap C=\emptyset$.

Hmm... $6^n$?

Each element from $\{1,2,\ldots,n\}$ has $6$ possibilities: belonging only to $A$; belonging only to $B$; belonging only to $C$; belonging only to $A$ and $B$; belonging only to $B$ and $C$; belonging only to $C$ and $A$. Note that we only excluded the possibilities "belonging to none of $A$, $B$ and $C$" (because $A\cup B\cup C = \{1,2,\ldots,n\}$, so any element from the latter set must belong to some of the three former sets) and "belonging to all sets $A$, $B$, $C$" (because $A\cap B\cap C = \emptyset$).


blogs
geometry
circumcircle
geometry proposed

[color=darkred][b]Let $ ABC$ be an acute triangle with  the incircle $ w$ and the circumcircle $ \Omega (O,R)$.
The circle $ w_{a}(P_{a},p_{a})$ is tangent internally to the circle $ \Omega$ at the point $ A$ and tangent externally to $ w$.
The circle $ \Omega_{a}(Q_{a},q_{a})$ is tangent internally to the circle $ \Omega$ at the point $ A$ and tangent internally to $ w$.
Define analogously the pair of circles $ \left\{w_{b}\ ,\ \Omega_{b}\right\}$ and $ \left\{w_{c}\ ,\ \Omega_{c}\right\}$.
Prove that $ \boxed{8\cdot P_{a}Q_{a}\cdot P_{b}Q_{b}\cdot P_{c}Q_{c}\le R^{3}}$ with equality if and only if the triangle $ ABC$ is equilateral.[/b][/color]
[hide="Hint."][color=darkblue]I proved that $ \frac{p_{a}}{p\minus{}a}\equal{}\frac{q_{a}}{p}\equal{}\frac{R(p\minus{}a)}{bc}\equal{}\frac{P_{a}Q_{a}}{a}$ because $ P_{a}Q_{a}\equal{} q_{a}\minus{}p_{a}$ $ \implies$$ P_{a}Q_{a}\equal{}\frac{a^{2}}{4r_{a}}$ a.s.o. $ \implies$ $ 8\cdot\prod P_{a}Q_{a}\equal{} 2rR^{2}\le R^{3}$.

[b]Notation.[/b] $ r_{a}$ is the $ A$- exinradius of the triangle $ ABC$ a.s.o. I used the relations $ abc \equal{} 4Rpr$, $ r_{a}r_{b}r_{c}\equal{} p^{2}r$ and $ 2r\le R$.[/color][/hide]


Determine all positive integers for which $d(n)=\frac{n}{3}$ holds.

[quote="Robert Gerbicz"]$ d(n)\leq \frac{n}{6}+5$ is true, because up to $ \frac{n}{6}$ there are at most $ \frac{n}{6}$ divisors of n, and if $ d>\frac{n}{6}$ is a divisor of n then we can write: $ n=k*d$ where $ 1\leq k <6$ so there can be at most 5 different $ d$ values.

From equation: $ \frac{n}{6}+5\geq d(n)=\frac{n}{3}$ from this inequality: $ \frac{n}{6}+5\geq \frac{n}{3}$ so $ n\leq 30$. Checking all positive integers up to 30 we get the solutions: \[ n=9,\; 18,\; 24\][/quote]


Prove the inequality: $ \sum_{i \equal{} 1}^{n}{\frac {a_i}{a_{i \plus{} 1} \plus{} a_{i \plus{} 2}}}\ge 0,989\cdot{\frac {n}{2}}$ for all possitive real number $ a_1$ ,$ a_2$,......,$ a_n$

NOTE: $ a_{n \plus{} 1} \equal{} a_1$  and $ a_{n \plus{} 2} \equal{} a_2$

Thanks, _tequilla: can you post solution?

I don,t remember this solution
http://mathworld.wolfram.com/ShapirosCyclicSumConstant.html


[hide]Multiple choice can be helpful.  I quickly figured out that where:
_._ _ _
_._ _ _
those two numbers are timesed together, the 4 position on the 1st line times the 1st position on the 2nd line equals the 3rd number after the decimal point.  I then proceded to find out that number on the answers, concluding that [b][i]B[/i][/b] was the answer.[/hide]

This is a pretty nasty process but you can also use binomial expansion.

As I said, it's not pretty.  :)

Not too hard.

[hide="Just square each of the options"]$\boxed{B}$[/hide]

[hide]for these type of problems, it's really, really easy to solve using calculators
seriously, even for the SAT, they let you use a calculator

but like jesus said, you can square all the options and see that is B[/hide]

I'm guessing by some of the problems I've seen that calculators aren't allowed. I do things like multiply by hand 15 digits of $\pi$ by 13 digits of $e$, so it's really no problem for anyone like that. :P

Or we could just notice that $1.0045^2 = (1+0.0045)^2 = 1 + 0.009 + 0.00\textrm{(somethingsomething)}$ is really close to 1.009. Evaluating the second largest number, i.e. $0.003$, makes us sure of answer B.

For anyone who has taken calculus, I have another solution. I'm not sure a solution involving Calculus is "allowed" in Getting Started, but I'm only posting this for anyone's edification. Please don't hurt me :).

[hide="Solution"]
Linear approximation:
$f(x+\Delta x)\approx f(x)+f'(x)\Delta x$

Our $f(x+\Delta x)$ is $\sqrt{x+\Delta x}$, with $x=1$, $\Delta x=0.009$.
$f'(x)=\frac{1}{2\sqrt{x}}$

So, we have $\sqrt{1.009}\approx \sqrt{1}+\frac{1}{2\sqrt{1}}(0.009)$
$=1+0.0045=1.0045$
[/hide]

When  $x$ is small, $(1+x)^n$ is about $1+nx$ [hide](If you forget  $x^2$ and heigher terms) .. here n=1/2, so  the answer is about  $1+ .009/2$  or $1.0045$[/hide]

gyan, thing actually turns out to be the same as the linear approximation method...

[hide="Gyans case"]
$f(1+x)\approx f(1)+f'(1)x$ for small $x$.
$(1+x)^n\approx 1^n+n(1)^{n-1}(x)\\ \approx 1+nx$
[/hide]

So yeah, same thing :P. But that's a nice rule to know if you don't have calculus handy.

-BLt


what is the difference between the sum of the first 20 positive even integers and the sum of the first 20 odd integers?

Here is a longer solution
[hide] The sum of the first 20 odd integers is $20^{2}=400$ and the sum of the first 20 even integers is $2+4+6...+40=2(1+2+3...+20)=\frac{(2)(20+1)(20)}{2}=21(20)=420$.  $420-400=20$, therefore, the answer is $20$. [/hide]
Here is a shorter solution
[hide]  What we are looking for is $(2+4+6+8...+40)-(1+3+5+7...+39)=(2-1)+(4-3)+(6-5)...+(40-39)=1+1+1...+1$ for 20 ones.  This is just $20*1=20$, which is our answer.  [/hide]

[hide]The sum of the first 20 even integers is $n^{2}+n$, or 420. The sum of the first 20 odds is $n^{2}$ or 400. So the difference would be 20.[/hide]

[hide="Fast Solution"]Looking at each pair (one odd and one even), we see that the difference is always $1$. There are $20$ pairs, so  $x=1\cdot20=\boxed{20}$.[/hide]


[quote="RohanSingh"]I have two independent computers. One is running with WINXP and other LINUX. I have broadband connection in my WINDOWS machine. I want internet facility in both the computers. Help needed.[/quote]

Are you serious? As you are Linux user??

You may know some device called "Router" that can help you.

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


[quote="Soarer"]It seems that you really didn't receive my message, anyway..

Put $x_i = \frac{1}{1+a_i}$, and then it suffices to prove
$\sum \sqrt{\frac{1-x_i}{x_i}} \ge (n-1) \sum \sqrt{\frac{x_i}{1-x_i}}$
$\Leftrightarrow \frac{1-nx_1}{x_1(1-x_1)} \ge 0$
$\Leftrightarrow \frac{(x_2-x_1)+...+(x_n-x_1)}{x_1(1-x_1)} \ge 0$
$\Leftrightarrow \sum_{i > j} (x_i - x_j) \left(\frac{1}{\sqrt{x_j(1-x_j)}} - \frac{1}{\sqrt{x_i(1-x_i)}}\right) \ge 0$
$\Leftrightarrow \sum_{i > j} (x_i - x_j) \left(\frac{\sqrt{x_i(1-x_i)}-\sqrt{x_j(1-x_j)}}{\sqrt{x_ix_j(1-x_i)(1-x_j)}}\right) \ge 0$
$\Leftrightarrow \sum_{i > j} (x_i - x_j) \left(\frac{x_i(1-x_i)-x_j(1-x_j)}{(\sqrt{x_i(1-x_i)}+\sqrt{x_j(1-x_j)})\sqrt{x_ix_j(1-x_i)(1-x_j)}}\right) \ge 0$
$\Leftrightarrow \sum_{i > j} (x_i - x_j)^2 \left(\frac{1-x_i-x_j}{(\sqrt{x_i(1-x_i)}+\sqrt{x_j(1-x_j)})\sqrt{x_ix_j(1-x_i)(1-x_j)}}\right) \ge 0$[/quote]

Thank you my brother you the one see you !!! :)


Go to Google Maps, click "Get Directions," and get directions from Boston, MA to London, England.

 :rotfl:

LOL
swimming across the Atlantic Ocean is gas-efficient, though

That's not as weird as this:

If you ask Google Maps to get directions from New York to London,
it actually says "Swim across the Atlantic Ocean.  3,426 miles"  It's item #24 on the list.

 :rotfl:

That would be what he said in the first post...

Well he forgot to mention that it actually says "swim across the atlantic ocean"
which is the weird part.

I'm pretty sure that was the reason he had a " :rotfl: " smiley....

lol
i think they might have done that on purpose -_-


Slove for $ x$
$ \sqrt{x\plus{}\sqrt{x^2\minus{}1}}\equal{}\frac{9\sqrt{2}}{4}(x\minus{}1).\sqrt{x\minus{}1}$

Note that the left side is equivalent to
$ \frac {\sqrt {x \plus{} 1} \plus{} \sqrt {x \minus{} 1}}{\sqrt {2}}$
So $ \sqrt {x \plus{} 1} \plus{} \sqrt {x \minus{} 1} \equal{} \frac {9}{2}(x \minus{} 1)\sqrt {x \minus{} 1}$
$ 2\sqrt {x \plus{} 1} \equal{} (9x \minus{} 11)\sqrt {x \minus{} 1}$
Squaring, this is equivalent to the cubic
$ 81x^3 \minus{} 279x^2 \plus{} 315x \minus{} 125 \equal{} 0$
$ (3x \minus{} 5)(27x^2 \minus{} 48x \plus{} 25) \equal{} 0$
Which has roots at $ x \equal{} \frac {5}{3}, \frac {8 \pm \sqrt {11}i}{9}$

By the way, why does everyone write slove instead of solve?

@mods: Move this to temp, please (MY active problem).


Let X be a compact Hausdoff space. Let f: X -> R. The graph of f is defined as the set G(f) = { (x, f(x)) : x in X} included in X x R. Prove that f is continuous if and only if     G(f) is compact.

[quote="1234567a"]how can $ G(f)$ be compact in $ X\times \mathbb{R}$?  :maybe:
example $ id : \mathbb{R}\to\mathbb{R}$ then $ G(f) \equal{} \{(x,x) : x\in\mathbb{R}\}$ is not compact on $ \mathbb{R}\times\mathbb{R} \equal{} \mathbb{R}^2$.[/quote]

That example does not work because X has to be compact in the original problem. R is not compact.

$ f$ is continuous so $ G(f)$ is closed in $ X\times f(X)$ which is compact so it must be compact.
now suppose that $ G(f)$ is compact then it is sequentially compact,  let $ x_{n}\to x\in X$  then we have $ z_n \equal{} (x_{n},f(x_{n}))$ has a subsequence $ z_{n_{k}}\to z \equal{} (a, f(a))\in G(f)$  so we have $ x \equal{} a$ and $ f(x_{n})\to f(x)$ so $ f$ is continuous ! hope that I am right this time :lol:

Why is it sufficient to show that $ f$ is continuous in terms of sequences?

Proof of one implication: Assume that $ G(f)$ is compact. The  projection $ p: G(f)\to X$ onto the first coordinate is a continuous bijection. Since $ G(f)$ is compact and $ X$ Hausdorff, $ p$ is a homeomorphism. Then $ f\equal{}q\circ p^{\minus{}1}$ is continuous as composition of continuous mappings with the projection $ q: G(f)\to \mathbb R$ onto the second coordinate.

[quote="Third Edition"]Why is it sufficient to show that $ f$ is continuous in terms of sequences?

Proof of one implication: Assume that $ G(f)$ is compact. The  projection $ p: G(f)\to X$ onto the first coordinate is a continuous bijection. Since $ G(f)$ is compact and $ X$ Hausdorff, $ p$ is a homeomorphism. Then $ f \equal{} q\circ p^{ \minus{} 1}$ is continuous as composition of continuous mappings with the projection $ q: G(f)\to \mathbb R$ onto the second coordinate.[/quote]

I'm sorry. But why do p and q have to be continuous?

[quote="Third Edition"]Show that, given topological spaces $ A$ and $ B$, then, say, the projection $ P: A\times B\to A$, $ (a,b)\mapsto a$ is continuous. This implies that the  restriction of $ P$ to every subspace of $ A\times B$ is continuous.[/quote]

Thanks a lot. That makes sense.


Let $ n$ denote the smallest positive integer that is divisible by both $ 4$ and $ 9$, and whose base-$ 10$ representation consists of only $ 4$'s and $ 9$'s, with at least one of each. What are the last four digits of $ n$?

$ \textbf{(A)}\ 4444\qquad
\textbf{(B)}\ 4494\qquad
\textbf{(C)}\ 4944\qquad
\textbf{(D)}\ 9444\qquad
\textbf{(E)}\ 9944$

[hide]In order for number to be divisible by 9, sum of it's digits must be divisible by 9 so there are at least nine 4's. Since we have at least one digit 9, we are looking for a number with nine 4's and one 9 such that 9 has least possible value (in form $ 9*10^{k\minus{}1}$ where $ k$ is position of the 9 looking from right to left). It's enough to test last 4 digits to see if the number is divisible by 4 since $ \displaystyle\sum_{i \equal{} 4}^{9}{4*10^i}$ is divisible by 4. Let's consider different cases for the last digits:

1) $ 4449$ is not divisible by 4
2) $ 4494$ is not divisible by 4
3) $ 4944$ IS divisible by 4 so this is the solution we are looking for.

The answer is $ \textbf{(C)}\ 4944$.[/hide]

[hide]
If we add up all the digits then they must be divisible by $ 9$. Therefore $ 4k \plus{} 9j \equiv 0 \mod 9$
So the small est number will have $ \Rightarrow k \equal{} 9, j \equal{} 1$

Also from a number to be divisible by $ 4$ the last to digits must be divisible by $ 4$
$ \Rightarrow 4444444944$ is the smallest number $ \longrightarrow \boxed{C}$[/hide]

Both of you have the same solution I think, but just worded differently.


geometry
3D geometry
congruent triangles

[quote="236factorial"]A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diamter 10 and altitude 12, and the axes of the cylinder and the cone coincide. Find the radius of the cylinder, and express your answer as a common fraction.

[hide="little hint"]Try to look at half the cone in a 2-D view. [/hide][/quote]

[hide]
Looking at the cone and cylinder as a square inscribed in an isosceles triangle:
$x$ is the side length of the square

The area of the triangle is 60.

The area of the square is $x^2$
The combined area of the 2 congruent triangles having sides on the side w/length 10 is $\frac{10x-x^2}{2}$.
The area of the triangle on top is $\frac{12x-x^2}{2}$

Adding the 3 up, you get $11x$.  Since this equals the area of the big triangle, $x=\frac{60}{11}$.  We want to find the radius, so we divide by 2 to get $\frac{30}{11}$.
[/hide]


edit: tried to delete it but the BB won't let me (this was not a problem taking from usamts. in fact, I don't know what that is).

this is like the usamts problem 3...it is not that close so i don't blame anybody, but i think it would be better if you held back on this for about a week


Evaluate the infinite sum:
1/n(n+2) as n goes from 1 to infinity.

note that $ \frac{1}{n(n\plus{}2)}\equal{}\frac{1}{2}(\frac{1}{n}\minus{}\frac{1}{n\plus{}2})$

so the sum you want equals half of $ \frac{1}{1}\minus{}\frac{1}{3}\plus{}\frac{1}{2}\minus{}\frac{1}{4}\plus{}\frac{1}{3}\minus{}\frac{1}{5}\plus{}\frac{1}{4}\minus{}\frac{1}{6}\plus{}....$ which is $ \frac{3}{4}$


Prove that $ \sum_{i\equal{}0}^{n} {n \choose k} \equal{} 2^n$.

[hide]Apply [b]Binomial Theorem[/b]: $ \boxed{(a\plus{}b)^n \equal{} \sum_{i\equal{}0}^{n}{n\choose k} a^k b^{n\minus{}k}}$
for $ a\equal{}b\equal{}1$[/hide]

Suppose that we have n patients (ordered from 1 to n) which can be ill or not. How many possibilities we have?
Well, it's possible that nobody is ill (1 case).
But it's also possible that there is exactly 1 sick patient and we have obviously as may cases as patients : $ Cases \equal{} n \equal{} {n \choose 1}$.
And we may have two ills: we have n possibilities for the 1st patient and (n - 1) for the second, so $ Cases \equal{} n(n \minus{} 1) \equal{} {n \choose 2}$.
With three sick patients,, we can have... [i]et coetera.[/i]
The last possibility is that everybody is ill: 1 case, since $ 1 \equal{} {n \choose n}$.
Phew. So we have a total of $ \sum_{i \equal{} 0}^{n} {n \choose k}$ cases. But wait! Suddenly, I realize that there is a very more straightforward way to compute all the possible cases, because every patient can either be sick or sound  (2 possibilities for everyone) and we have clearly $ 2^n$ cases. This demonstrates the equality.

Notify me any mathematical mistake.
For any grammar mistake, see my signature :)

[quote="pontios"][hide]Apply [b]Binomial Theorem[/b]: $ \boxed{(a \plus{} b)^n \equal{} \sum_{i \equal{} 0}^{n}{n\choose k} a^k b^{n \minus{} k}}$
for $ a \equal{} b \equal{} 1$[/hide][/quote]

Okay.  Prove the binomial theorem :)

Another interpretation:  the number of subsets of a set with $ n$ elements ($ 2^n$) is the number of subsets of size $ k, k \equal{} 0, 1, 2, 3, ... n$.

[quote="t0rajir0u"][quote="pontios"][hide]Apply [b]Binomial Theorem[/b]: $ \boxed{(a + b)^n = \sum_{i = 0}^{n}{n\choose k} a^k b^{n - k}}$
for $ a = b = 1$[/hide][/quote]Okay.  Prove the binomial theorem :)[/quote]
Why we are not allowd to use it?

Ok, it's not that hard!  :)
[hide]For $ n = 1$ we have $ (a + b)^1 = \sum_{k = 0}^{1}{1\choose k} a^k b^{1 - k} = a + b$

$ (a + b)^{n + 1} = (a + b)\sum_{k = 0}^{n}{n\choose k} a^k b^{n - k}$

$ = \sum_{k = 0}^{n}{n\choose k} a^{k + 1} b^{n - k} + \sum_{k = 0}^{n}{n\choose k} a^k b^{n - k + 1}$

$ = \sum_{k = 1}^{n + 1}{n\choose k - 1} a^{k} b^{n - k + 1} + \sum_{k = 0}^{n}{n\choose k} a^k b^{n - k + 1}$

${ = b^{n + 1} + \sum_{k = 1}^{n}{[\binom{n}{k - 1}} + \binom{n}{k}]} a^{k} b^{n - k + 1} + a^{n + 1}$

$ = \sum_{k = 0}^{n + 1}{n + 1\choose k} a^k b^{n + 1 - k}$

and the proof for $ \boxed{\boxed{\sum_{k = 0}^{n} {n \choose k} = 2^n}}$ is now complete

Should I also prove the formula ${ \binom{n}{k - 1}} + \binom{n}{k} = \binom{n + 1}{k}$  or it's ok :?:
----------------------------------------------------------------------------------------------------------------------------- [/hide]

But let's see another way, similar to GioMott's

There are $ \binom{n}{k}$ n-digit binary numbers with $ k$ ones

There are $ S = \sum_{k = 0}^n{n \choose k}$ numbers containing $ 0,1,2,...$ or $ n$ ones

But there are in total $ 2^n$ binary numbers with $ n$ digits, so $ S = 2^n$


True/False:
1. Is every countable first countable $ T_3$-space metrizable?
2. Is every separable first countable $ T_3$ space metrizable?

Attempt
1. countable + first countable $ \Rightarrow$ second countable. second countable + $ T_3$ $ \Leftrightarrow T_4$. So by Urysohn Theorem, since this is second countable + $ T_4$, this is metrizable. But this argument doesn't work b/c $ T_3 \not \Rightarrow T_4$ in general. So is this false? What is the counterexample?
2. separable + first countable $ \Rightarrow$ second countable. Now, the argument is the same as above.
Thanks.

What's your definition of T3? Normally it does not include the Hausdorff property, and both statements are wrong without assuming this(take the indiscrete topology on a set with two elements).


Let $ P$ be an internal point of triangle $ ABC$. The line through $ P$
parallel to $ AB$ meets $ BC$ at $ L$, the line through $ P$ parallel to $ BC$
meets $ CA$ at $ M$, and the line through $ P$ parallel to $ CA$ meets $ AB$ at
$ N$. Prove that

$ \frac{BL}{LC}\times\frac{CM}{MA}\times\frac{AN}{NB}\le\frac{1}{8}$

and locate the position of $ P$ in triangle $ ABC$ when equality holds.

let $ L'N, M'L, N'M$ meet at $ P$, where $ L, L'\in BC, M, M'\in AC, N, N' \in AB$. denote by $ x, y, z$ the areas of $ PNN', PMM', PLL'$, respectively. let $ BL \equal{} k_1a, LL' \equal{} k_3a, and L'C \equal{} k_2a$, where $ a$ is the length of $ BC$. then $ x \equal{} k_1^2K, y \equal{} k_2^2K, z \equal{} k_3^2K$, where $ K$ is the area of $ ABC$. Then
\[ \frac {BL}{LC} \equal{} \frac {k_1}{k_2 \plus{} k_3} \equal{} \frac {\frac {\sqrt {x}}{\sqrt {K}}}{\frac {\sqrt {y}}{\sqrt {K}} \plus{} \frac {\sqrt {z}}{\sqrt {K}}} \equal{} \frac {\sqrt {x}}{\sqrt {y} \plus{} \sqrt {z}}\]
Similarly,
\[ \frac {CM}{MA} \equal{} \frac {\sqrt {z}}{\sqrt {x} \plus{} \sqrt {y}}, \frac {AN}{NB} \equal{} \frac {\sqrt {y}}{\sqrt {z} \plus{} \sqrt {x}}\]
So the inequality to prove becomes
\[ (\sqrt {x} \plus{} \sqrt {y})(\sqrt {y} \plus{} \sqrt {z})(\sqrt {z} \plus{} \sqrt {x})\geq 8\sqrt {xyz},\]
which is easily proven by AM-GM. equality occurs iff $ x \equal{} y \equal{} z$, and it's easy to see that this happens iff $ P$ is the centroid of $ ABC$.

We can also get the same inequality as follows:

Let $ X,Y,Z$ be the intersection of lines $ AP, BP, CP$ with $ BC, AC, AB$

$ \frac{BL}{LC}\equal{}\frac{BL\plus{}LC}{LC}\minus{}1$
$ \equal{}\frac{BC}{LC}\minus{}1$
$ \equal{}\frac{CZ}{CP}\minus{}1$ by similar triangles
$ \equal{}\frac{[ABP]}{[ABC]}\minus{}1$ since the triangles have the same bases

Letting $ [ABP]\equal{}x, [BCP]\equal{}y, [CAP]\equal{}z, [ABC]\equal{}x\plus{}y\plus{}z$ and clearing denominators we get the same inequality as above.


Using Am-Gm, we only need to prove:$ a^2b\plus{}b^2c\plus{}c^2a\plus{}abc \leq\ 4$ (always trues)
@PhongVan: I like the names of your topic  :lol:

[quote="PHONGVAN"]Let $ a,b,c \geq 0$ ,Prove that
 $ \sqrt{\frac{a^2}{2(a\plus{}b)^2\plus{}c^2}}\plus{}\sqrt{\frac{b^2}{2(b\plus{}c)^2\plus{}a^2}}\plus{}\sqrt{\frac{c^2}{2(c\plus{}a)^2\plus{}b^2}}\leq\ 1$[/quote]
By Cauchy-Schwarz we obtain:
$\left(\sum_{cyc}\sqrt{\frac{a^2}{2(a\plus{}b)^2\plus{}c^2}}\right)^2\leq(a+b+c)\sum_{cyc}\frac{a}{2(a\plus{}b)^2\plus{}c^2}\leq1$,
where the last inequality is equivalent to
$\sum_{cyc}(2a^6+2a^5b-3a^4b^2+2a^4c^2+2a^4bc+a^3b^3-7a^3b^2c+a^2b^2c^2)\geq0$,
which is obvious.


How high do people score at state mathcounts?  I'm interested to see if my brother can make nationals either this year or next (he's a 7th grader right now and can get about 20/30 on the sprint tests)

I guess 40+ would get u into countdown...I didn't go to nationals myself, but w/e.

Are scores weighed?  There are 30 problems on the sprint test, and 8 on the target test, so aren't the individual scores out of 38?  Sorry about my ignorance, I didn't do much mathcounts when I was in middle school

Each target question is worth two points. That has always been the standard.

Oh I see.  How are the team scores calculated?  Are the questions on the team test also 2 points each?

Team scores are the average of the individual scores (but still divided by 4 if you have less than 4 people) plus team score (each question is 2 points)

[quote="samath"]Jonathan, what score did you get at state last year?
You got third place, which clinches a spot on the national team.

Top score at state in 2005 was 44; top score in 2006 was 43. I think.

Usually 40 would get you into the countdown round.
I'm not sure if you could get in with a sprint score of 20 though ... because that means a maximum score of 36.[/quote]

Eh. I got 39, I think 25/14? Not very good, but then again, I wasn't very good... :whistling:

So considering that score, 40 might be a little high...

Not me, but it would be fun to grade those papers and see participants compete. Where do you get those volunteer emails?

[quote="samath"]I'll probably be helping coach the people on our team ...  :)[/quote]

Let me guess you are the kid who got 44 in 2005 and 43 in 2004.

That is mucho good.

Sorry for coming into this forum, but I couldn't help over hear.

Good Job on getting 6th Place at MathCounts.

[quote="jmerry"]I just got the volunteer recruiting email- I'll be at the Seattle chapter (on the UW campus) Feb. 10. That means I'll probably be grading, of course.

Will any of you be there as a competitor?[/quote]

I'll be there as a volunteer too! I just emailed Dan Cory about volunteering, so that's how I became one. I'm probably going to do the countdown round instead of grading.

Not impossible if you get a perfect on target, but I doubt it would happen on written alone (I got 41 at states last year and 2nd place written)..., but you would probably get into countdown.


Hi all,

I want to know the method for find all value of x such that a expression be a perfect square.

For example, I want to find all value of x, x € N, such that (3x - 19)(3x + 5) be a perfect square.

Thanks ^^


What is thee maximum number of distinct lines that can be drawn through ten distinct points if no three of the points are collinear?


Does anyone know where to find past USAMO index cutoff's and floor score cutoff's on the AIME for past years?

Please beware that this topic was discussed many many times and hence, here are couple links that I easily found by Search button.

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Index+floor&t=35905
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27485

[quote="beta"]What's usually the MOP index cutoff for Blue Group (in the years where they used MOP index)?[/quote]

Kind of like asking what's the weather usually like on July 4th? I mean, you can get a general idea from looking at past years but you'll never really know because it changes every year.

Factors for cutoff

1. Difficulty of test
2. Overall skill of competitors (Would the Winner/Blue cutoff have been 29/22 if the 2005 USAMO was given in 2004? No way ... it would have been like 37/29)


AMC
AIME
analytic geometry
graphing lines
slope

Hi,

I don't understand how the coordinates of $ S$ was derived from just knowing the slope of line PR in solution #2. Don't we need to know at least one of the points of point $ S?$

Here's the link: http://www.artofproblemsolving.com/Wiki/index.php/1990_AIME_Problems/Problem_7

Thank you!

Hi limac.

Since the slope of $ PRS$ is $ \frac{\minus{}4}{3}$, if we go straight across $ 3$ units to the right from $ P$ to $ T$, say, and then down $ 4$ units we will meet the line $ PRS$ at $ U$, say, and have completed a $ 3\minus{}4\minus{}5$ right-angled triangle $ PTU$.

Since $ PS\equal{}25$, we can go $ 15$ to the right to $ V$ say, and then down $ 20$ to get a triangle $ PVS$, similar to triangle $ PTU$.

As we started from $ (\minus{}8, 5)$, the coordinates of $ S$ must be $ (\minus{}8\plus{}15, 5\minus{}20) \equal{} (7, \minus{}15)$.

I hope this makes sense.

Oh ok. Thank you very much andrewtom. :)


algebra
polynomial
linear algebra
matrix
Gauss
group theory
abstract algebra

Let $ M\in \mathcal{M}_n(\mathbb{R})$ such that $ \chi_M(X) \equal{} \det(M \minus{} XI_n)\in \mathbb{Z}[X]$.
Find all monic $ P\in \mathbb{Z}[X]$ s.t. $ \chi_{P(M)}(X)\in \mathbb{Z}[X]$.

This is a really nice one !!! :-)

Let $ \lambda_1,\ldots,\lambda_n$ be the eigenvalues of $ M$ in $ \mathbb C$, counted with multiplicities. They are integers in a number field $ K$. We can choose $ K$ to be the splitting field of $ \chi_M(X)$, say. Let $ G$ denote the Galois group $ Gal(K: \mathbb Q)$. We have
\[ \chi_{M}(X) \equal{} \prod_{i \equal{} 1}^n(X \minus{} \lambda_i)
\]
Then $ P(\lambda_1),\ldots,P(\lambda_n)$ are the eigenvalues of $ P(M)$: By Jordan, there are a diagonal matrix $ D$ with the $ \lambda_i$'s on the diagonal and a nilpotent matrix $ N$ (both over the complex numbers) $ DN \equal{} ND$ such that $ M$ is similar to $ D \plus{} N$. From $ P(D \plus{} N) \equal{} P(D) \plus{} N'$ with a nilpotent matrix $ N'$ we can read off that the eigenvalues of $ P(D \plus{} N)$ - and equivalently of $ P(M)$ - are $ P(\lambda_1),\ldots,P(\lambda_n)$, counted with multiplicities. Thus
\[ \chi_{P(M)}(X) \equal{} \prod_{i \equal{} 1}^n(X \minus{} P(\lambda_i))
\]
Let $ f(X)\in\mathbb Q[X]$ be an irreducible factor of $ \chi_M(X)$.  Then $ f(X)\in\mathbb Z[X]$, by Gauss lemma, and it splits over $ K$ with
\[ f(X) \equal{} \prod_{\sigma\in G/G_\lambda}(X \minus{} \sigma\lambda)
\]
for some $ \lambda \equal{} \lambda_{i_0}$ where $ G_\lambda\subset G$ denotes the subgroup of elements fixing $ \lambda$. Now it suffices to show that
\[ f_P(X): \equal{} \prod_{\sigma\in G/G_\lambda}(X \minus{} P(\sigma\lambda))\in\mathbb Z[X]
\]
since $ \chi_{P(M)}(X)$ is the product of the polynomials $ f_P(X)$ where $ f(X)$ runs through the irreducible factors of $ \chi_M(X)$, counted with multiplicities. For every $ \tau\in G$ define
\[ f_P^\tau(X) : \equal{} \prod_{\sigma\in G/G_\lambda}(X \minus{} \tau(P(\sigma\lambda)))
\]
Since $ P(\sigma\lambda) \equal{} \sigma(P(\lambda))$ and $ \tau$ induces a permutation on $ G/G_\lambda$, we get $ f_P^\tau(X) \equal{} f_P(X)$. This implies that the coefficients of $ f$ are in the fixed field of $ G$, i.e. they are rational. Since $ f(X)$ is the minimal polynomial of the eigen value $ \lambda$ which is is an integer of $ K$, we have $ f(X)\in\mathbb Z[X]$. ... Done  :P

It looks surprising somehow that every $ P[X]\in\mathbb Z[X]$ fulfils the condition.  :maybe:


[quote="sen"]Prove that a set of n integers has a non-empty subset whose sum is divisible by n.[/quote]
[hide]Let the $n$ integers be $a_{1}, a_{2},...,a_{n}.$
Consider the sums
$S_{1}=a_{1}$
$S_{2}=a_{1}+a_{2}$
...
$S_{n}=a_{1}+a_{2}+...+a_{n}$

Now consider $S_{i}(\mod n)$ and use pigeonhole.[/hide]

This problem is a classic...
http://www.mathlinks.ro/Forum/viewtopic.php?t=79936

[quote="gotztahbeazn"]do you also have to consider the sums $a_{2}+a_{3}$, $a_{2}+a_{4}$ etc?[/quote]
Not really. You just need to consider the following $n$ sums: $a_{1},a_{1}+a_{2},a_{1}+a_{2}+a_{3},a_{1}+a_{2}+a_{3}+a_{4},...,a_{1}+a_{2}+...+a_{n}.$

Now, think what would you get if each of those sums is divided by $n$. What remainders would you get?

[hide]Assume that some set of $n$ integers does not have a non-empty subset divisible by $n$.  If any two of the sums $S_{n}$ and $S_{m}$, with $n<m$, have the same remainder $\pmod{n}$, then $a_{n+1}+a_{n+2}+\cdots+a_{m}\equiv 0\pmod{n}$.  Therefore, no two of the sum are equivalent $\pmod{n}$.  Then some sum is equivalent to $0\pmod{n}$, and this sum represents a subset divisible by $n$.[/hide]


Let's denote the space of integrable functions by L1 space, and the space of square integrable functions by L2 space. Can you give a function f as an example for each of the following?

1. f belongs to L1 but not L2
2. f belongs to L2 but not L1
3. f belongs to both L1 and L2

Thanks

let try with $ f\equal{}\frac{1}{x}$ and some modifications


algorithm
number theory unsolved
number theory

[quote="Tanequil"]I can see how the continued fractions algorithm for sqrt(N) gives a solution to x^2-Ny^2=1, take the (n+1)th convergent and put the denominator as y, the numerator as x, or failing that (if that gives -1), wait another n indices and take that convergent. But how would you prove that this algorithm goes through all the solutions to the equation, and doesn't miss any out?[/quote]

I am not sure what you are asking.

[b]Theorem:[/b] Let $n$ be a [b]non-square[/b] then if $x_{0},y_{0}$ is a non-trivial solution to equation $x^{2}-ny^{2}=1$ then $x_{0}/y_{0}$ is a convergent of the countinued fraction for $\sqrt{n}$.

The proof is elegant, it uses the fact about diophantine approximation, i.e. how "good" an approximation in integers can be. When the approximation it too "good" then it must be a convergent of the continued fraction. That is the general idea.

[b]Proof:[/b] Thus $(x_{0}-y_{0}\sqrt{n})(x_{0}+y_{0}\sqrt{n})=1$.
Thus, $\frac{x_{0}}{y_{0}}-\sqrt{n}= \frac{1}{y_{0}(x_{0}+y_{0}\sqrt{n})}$
Thus, $0< \frac{x_{0}}{y_{0}}-\sqrt{n}< \frac{\sqrt{n}}{y_{0}(y_{0}\sqrt{n}+q\sqrt{n})}= \frac{\sqrt{n}}{2y_{0}^{2}\sqrt{n}}= \frac{1}{2y_{0}^{2}}$
Of course,
$0<\left| \frac{x_{0}}{y_{0}}-\sqrt{n}\right| < \frac{1}{2y_{0}^{2}}$
So it must be a convergent of the the continued fraction.
Because its diophantine appoximation is too "good".

I hope you understand what I mean by "good".


Given a fixed angle $\angle XYZ$ ($X,Y,Z$ are fixed points) and some arbitrary ray $p$ starting from $X$ and intersecting $YZ$ at some variable point $K$. Is the locus of the excentre $I_{X}$ a straight line. I want to know?

Your problem is not clearly enough. You want to find the locus of the excentre of which side of which triangle?

XYK is the triangle I'm talking about. Sorry!

Since triangle $XYZ$ is fixed, so $I_{X}$ always lies on the exterior angle bisector of $\angle{XYZ}$.


function
calculus
derivative
calculus computations

Find and sketch the solution of
\[ \ddot{y} \plus{} y \equal{} H(t \minus{} \pi) \minus{} H(t \minus{} 2\pi),\]where $ H$ is the Heaviside step function, subject to
\[ y(0) \equal{} \dot{y}(0) \equal{} 0,\]and
\[ y(t), \dot{y}(t) \textup{  continuous at  } t \equal{} \pi, 2\pi.\]

The Laplace transform isn't in the syllabus of my course; is there any other way to solve it?

You can solve it piecewise. You really have three separate problems.

Problem 1: $ \ddot{y}\plus{}y\equal{}0;\ y(0)\equal{}0,\dot{y}(0)\equal{}0.$ Answer valid on $ [0,\pi].$

Problem 2: $ \ddot{y}\plus{}y\equal{}1;\ y(\pi)$ and $ \dot{y}(\pi)$ determined by solution to problem 1. Answer valid on $ [\pi,2\pi].$

Problem 3: $ \ddot{y}\plus{}y\equal{}0;\ y(2\pi)$ and $ \dot{y}(2\pi)$ determined by solution to problem 2. Answer valid on $ [2\pi,\infty).$

Can you do that?

I get why it can be split into three separate problems - but is $ H(0)$ assumed to be 1?

It makes no difference what you assume $ H(0)$ to be. Solve the differential equation on the interior of each of those intervals, and then extend the function (and its first derivative) by continuity to the endpoints of each of those intervals.

If you want to get more theoretical and pedantic: the equation does not have a classical solution, as we cannot find $ y\in C^2$ that satisfies it. The second derivative won't exist everywhere. So what are we asking for here? We're asking for a weak solution - a solution in the sense of distributions. Considered as a distribution, $ H$ acts as a locally integrable function. As a locally integrable function, it doesn't matter what its value is considered to be at any single point, or for that matter on any set of measure zero.


search
combinatorics unsolved
combinatorics

There is a $ n*n$ form.We print these grids with black and white. Suppose every two neighboring grids have at least one black grid. Find the number of ways that we can print the form.

[quote="pluristiq"]"every two neighboring grids have at least one black grid."

what does that mean? for every pair of neighboring cells, at least one of them is black??[/quote]
Yes,you're right.
I'm so sorry that I didn't explain the problem clearly as a Chinese.

It's sequence [url=http://www.research.att.com/~njas/sequences/A006506]A006506[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  It's been discussed at least twice before on the forum:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=599928754&t=226209
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=599928754&t=311192


how do i factor this?

36a^2b+6ab

would it be 6ab(6a)?? i think i am missing a b then?

Yup your so close,
its $6ab(6a+1)$ because when u multiply it you get
$36a^{2}b+6ab$


If x and y are real numbers such that $\frac{{x-y}}{5}$ and $\frac{{x y}}{5}$ are integers, are x and y necessarily so (that is, integers) ?

My Solution

[hide]A factor of one of the numbers must be 5 because (xy)/5 is an integer. Lets call this one 5n. The difference between 5n and the other number must be a mutiple of 5, so the other number can be witten as 5n+5p. A five can be factored out making the second number 5(n+p), therefore both numbers are mutiples of 5 and (xy)/25 is always an integer.[/hide]

Correct, piac31 !     :)

Another approach:

[hide]

$x-y = 5M\,\,\, \Rightarrow \,\,x = y+5M$  then with $xy = 5N\,\,\,$ we get $\,\,\,\left({y+5M}\right)y = 5N \,\,\,\,\,\, (M, N$ integers$)$

From the fact that 5 is a prime number, integer y is necessarily divisible by 5 (otherwise nor $y+5M$ nor $y$ would have 5 as a factor and, therefore, their product would not be $5N$ where $N$ is an integer!) ; now things are done: $x$ is also divisible by 5 (from $x = y+5M$) and $xy$ "has (at least) two 5´s".  [/hide]

[quote="fskilnik"]Good, DiscreetFourierTransform!

You were promoted to the "2nd Phase" of the Question...    ;)

Proof or give a counter-example:

If x and y are integers such that $\frac{{x-y}}{5}$ and $\frac{{xy}}{5}$ ARE integers, is necessarily $\frac{{xy}}{25}$ an integer?[/quote]

this is nice, note that x,y are same partity mod 5 by the first condition, and since one of them is divisible by 5 (second condition and 5 is prime), hence both are divisible by 5, showing that the third expression is also an integer

I´m glad you like it, Altheman!   :)

Good argument, by the way. (I tried to avoid "mod´s" explicitly but... sure, it´s a very good insight on what is "going on" there!)
For the interested readers, have a look at "Modular arithmetic" here:  http://en.wikipedia.org/wiki/Modular_arithmetic

Best regards,
Fábio.


To find that $ (431^2\plus{}23)\mod{703}\equal{}192$, do you have to calculate 431^2, add to 23, then find some integer $ n$ such that $ (431^2\plus{}23\minus{}n)/703\equal{}k, k$ an integer?

The easiest thing to do is to factor $ 703 \equal{} 19 \cdot 37$, then calculate $ 431^2 \plus{} 23 \bmod 19$ and $ 431^2 \plus{} 23 \bmod 37$ (both of which are small), then apply the [url=http://mathworld.wolfram.com/ChineseRemainderTheorem.html]Chinese Remainder Theorem[/url].  (This is assuming we do not already know its value in advance.  If we do, then it is [hide="even easier!"] Note that $ 192 \minus{} 23 \equal{} 169 \equal{} 13^2$.  The desired equivalence is equivalent to

$ 431^2 \minus{} 13^2 \equiv 0 \bmod 703 \Leftrightarrow$
$ 444 \cdot 418 \equiv 0 \bmod 703$.

Because $ 37 | 444$ and $ 19 | 418$, we are done. [/hide]

Note that in calculations $ \bmod m$ at any point we are free to take the remainder upon division by $ m$, which speeds up calculations significantly.


[quote="quykhtn-qa1"]let a,b,c be positive real numbers such that a+b+c=3;prove that
12(ab+bc+ca) ≤(abc)^2 +8abc +27         :lol:[/quote]
let $ a\leq\ b\leq\ c$
$ f(a,b,c) \equal{} a^2 b^2 c^2  \plus{} 8abc \plus{} 27 \minus{} 12(ab \plus{} bc \plus{} ac)$
$ f(a,b,c) \minus{} f(a,\frac{{b \plus{} c}}{2},\frac{{b \plus{} c}}{2}) \equal{} (b \minus{} c)^2 (3 \minus{} 2a \minus{} \frac{{a^2 (bc \plus{} (\frac{{b \plus{} c}}{2})^2 )}}{4})
\ge 0$
The remain is obvious. :)


Consider the distinct natural numbers $ a_1,a_2,...,a_{12}$ that have the following property:
all the positive differences between any two naturals $ a_i$ and $ a_j$ form a set of $ 20$ distinct naturals.
$ (a)$ Prove that $ \max \{ a_1,a_2,...,a_{12} \}\minus{}\min\{ a_1,a_2,...,a_{12} \}\equal{}20$.
$ (b)$ Give an example of such $ 12$ naturals.

[quote="Inequalities Master"]Consider the distinct natural numbers $ a_1,a_2,...,a_{12}$ that have the following property:
all the positive differences between any two naturals $ a_i$ and $ a_j$ form a set of $ 20$ distinct naturals.
$ (a)$ Prove that $ \max \{ a_1,a_2,...,a_{12} \} \minus{} \min\{ a_1,a_2,...,a_{12} \} \equal{} 20$.
$ (b)$ Give an example of such $ 12$ naturals.[/quote]

$ (a)$ is false,
we suppose that $ \{a_1,a_2,...,a_{12}\}$ is solution
we can see that also $ \{2a_1,2a_2,....,2a_{12}\}$ is solution.
and we have $ 2(max(a_i)\minus{}min(a_i))\equal{}max(2a_i)\minus{}min(2a_i)$ then $ max(a_i)\minus{}min(a_i)\neq 20$ or $ max(a_i)\minus{}min(a_i)\neq 20$
i think you want prove that $ \max \{ a_1,a_2,...,a_{12} \} \minus{} \min\{ a_1,a_2,...,a_{12} \}\ge 20$.
because $ \exists i,j: \ a_i\minus{}a_j\ge 20$

$ (b)$
$ \{k|\ k\equal{}1,2,...,11\}\cup \{21\}$ is solution we we have $ max\minus{}min\equal{}20$
$ \{2k|\ k\equal{}1,2,...,11\}\cup \{42\}$ is solution we we have $ max\minus{}min\equal{}40\ge 20$

I'm very sorry.
Here is the correct enunciation of the given problem.
Consider the distinct natural numbers $ a_1,a_2,...,a_{12}$ that have the following property:
all the positive differences between any two naturals $ a_i$ and $ a_j$ form a set of $ 20$ consecutive naturals.
$ (a)$ Prove that $ \max \{ a_1,a_2,...,a_{12} \} \minus{} \min\{ a_1,a_2,...,a_{12} \} \equal{} 20$
$ (b)$ Give an example of such $ 12$ naturals.


I found this problem at http://mathcircle.berkeley.edu/poly.pdf . I almost have the solution, except I am having a small problem at the end. I shall note where I have  trouble in <...>

Problem:

Let r, s, t be the roots of the complex polynomial x^3 + ax^2 + bx + c. Find (in terms of a, b, c) a cubic polynomial whose roots are rs + t, st + r, tr + s.

Solution:

From Viete's formula, we can see that

	-a = r + s + t
	 b = rs + st + tr
	-c = rst.

If we let the cubic polynomial with roots rs + t, st + r, and tr + s be p_3 x^3 + a_2 x^2 + p_1 x + p_0, we can, by applying Viete's formulas again, see that

	-p_2/p_3 = (rs + t) + (st + r) + (tr + s)
                              = b - a

	 p_1/p_3 = (rs + t)(st + r) + (st + r)(tr + s) + (tr + s)(rs + t)
	             = (rts^2 + sr^2 + st^2 + rt) + (srt^2 + ts^2 + tr^2 + rs) + (str^2 + tr^2 + rs^2 + st)
	             = rst(r + s + t) + [rs(r + s) + st(s + t) + tr(t + r)] + (rs + st + tr)  <issues on this line>

	-p_0/p_3 = (rs + t)(st + r)(tr + s)
		 = (rts^2 + sr^2 + st^2 + rt)(tr + s)
		 = (rst)^2 + str^3 + srt^3 + (rt)^2 + rts^3 + (rs)^2 + (st)^2 + rst
		 = (rst)^2 + rst(r + s + t) + [(rs)^2 + (st)^2 + (tr)^2] + rst
		 = c^2 + ac + (b^2 - 2ac) - c
		 = c(c - a - 1) + b^2

As you can see, I am having difficulty expressing rs(r + s) + st(s + t) + tr(t + r) in terms of a, b, and c. Anyone have any suggestions?

Fierytycoon

oh, silly me. musta been tired or something. that was simple. thx

Fierytycoon


Find the coefficient of x in the expansion of $(2x^{2}+3x-1)(5x-7)$

[quote="mathgeniuse^ln(x)"]Word of Warning:
You would have gotten that question wrong, since the coefficient is not $-26x$ but $-26$[/quote]

 :P

I was going to type another sentence that said the coefficient was -26, but I thought it was kind of obvious.


[quote="Yimin Ge"]Prove for all positive reals $x,y,z$:

\[ \displaystyle\sum_{cyclic}\frac{x^3}{x^2+xy+y^2} \geq \frac{x+y+z}{3}  \][/quote]

  It is tooooooo old.

$\frac{x^3}{x^2+xy+y^2} \geq \frac{2x-y}{3}$

[quote="Ilthigore"]Why? Sorry, I'm not able to work out the argument... if you could elaborate a little that would be helpful.[/quote]

Yeah, if one knows Gibbenergys inequality, it's trivial
$\frac{x^3}{x^2+xy+y^2} \geq \frac{2x-y}{3}$
$\Leftrightarrow (x-y)^2(x+y)\geq 0$

Then doing this for all other fractions as well yields to the inequality we want to prove.

I didn't know that inequality before (or at least I didn't thought of it), my proof is much longer (I won't post it because I feel embarrassed for it  :blush: )

[b]Saudi Arabia 2013:[/b]
1.Let $x,y \in[-1,1]$ ,prove that \[2\sqrt{(1-x^2)(1-y^2)}\le 2(1-x)(1-y)+1.\]
2..Let $a,b,c$ be positive real numbers such that $bc+ca+ab=1$ . Prove that\[a\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}+\sqrt{a^2+ab+b^2}\ge \sqrt{3}.\]
3.Let $a,b,c$ be positive real numbers  . Prove that\[\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\ge \frac{a+b+c}{3}.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=54&t=3533&hilit=Book%3AInegalitati%3B+Authors%3AL.Panaitopol%2CV.Bandila%2CM.Lascu]Book: Inegalitati; Authors:L.Panaitopol,V. Bandila,M.Lascu:[/url]
Prove the inequality:
$ \frac{x_{1}^3}{x_{1}^2+x_{1}x_{2}+x_{2}^2}+ \frac{x_{2}^3}{x_{2}^2+x_{2}x_{3}+x_{3}^3} + \ldots + \frac{x_{n}^3}{x_{n}^2+x_{n}x_{1}+x_{1}^2} \geq \frac{1}{3}(x_{1}+x_{2}+ \ldots +x_{n})$  $ \forall x_{1},x_{2}, \ldots ,x_{n}>0$.

[quote="Yimin Ge"]Prove for all positive reals $x,y,z$:

\[ \displaystyle\sum_{cyclic}\frac{x^3}{x^2+xy+y^2} \geq \frac{x+y+z}{3} \][/quote]
It's also from [i]Tournament of the Towns 1998[/i].

[quote="sqing"]
2..Let $a,b,c$ be positive real numbers such that $bc+ca+ab=1$ . Prove that\[a\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}+\sqrt{a^2+ab+b^2}\ge \sqrt{3}.\]
[/quote]
$\sum_{cyc}a\sqrt{b^2+bc+c^2}\geq\sum_{cyc}\frac{\sqrt3a(b+c)}{2}=\sqrt3$.

[quote="Yimin Ge"]Prove for all positive reals $x,y,z$:

\[ \sum_{cyc}\frac{x^3}{x^2+xy+y^2} \geq \frac{x+y+z}{3} \][/quote]
\begin{align*}\frac{x^3}{x^2+xy+y^2} \geq \frac{2x-y}{3} \iff 3x^3 \geq (2x-y)\left(x^2 + xy + y^2 \right) \iff x^3 + y^3 \geq x^2y + xy^2, \end{align*}which is true by AM-GM, Rearrangement, Muirhead, or any other justification. Summing three similar inequalites, we have \[\sum_{cyc}\frac{x^3}{x^2+xy+y^2} \geq \frac{x+y+z}{3},\]as required.


function
inequalities
real analysis
real analysis unsolved

1)Find all continuous functions $f: [0..1]\to R$ which verify:
$f(x)=\sum_{n=1}^\infty \frac{f(x^n)}{3^n}$
2)Find all continuous functions $f: [0..1]\to R$ which verify:
$f(x)=\sum_{n=1}^\infty \frac{f(x^n)}{2^n}$

(a) By choosing $x$ s.t. $f(x)$ is the maximum/minimum of $f$ on $[0,1]$ we reach the conclusion that both the maximum and the minimum are $0$, so $f$ must be identically null.

(b) Suppose we can find $x\in(0,1)$ s.t. $f(x)<f(0)$. Then, the sequence $f(x),f(x^2),\ldots$ will have a minimum at some $f(x^m)$, and this minimum is $<f(0)$. It's clear now that we can't possibly have $f(x^m)=\sum_{n\ge 1}\frac{f(x^{mn})}{2^n}$, as we should have (because $f(x^m)\le f(x^{mn}),\ \forall n\ge 1$, and the inequality is strict for large enough $n$), so we get a contradiction. In the same way, we can't find any $x\in(0,1)$ with $f(x)>f(0)$, so $f$ is constant (and conversely, every constant function works).

[quote="Diogene"]Actually, you can put "bounded functions" in place of  "continous functions"  :D
 :cool:[/quote]

Of course, only for point 1° .

 :cool:


inequalities
Functional Analysis
real analysis
real analysis unsolved

$ X, Y$ are Banach spaces and let $ T: X\rightarrow Y$ be a linear map. Suppose that for all $ f\in Y^*, f\circ T$ is in $ X^*$.
Show that $ T$ is continuous.

I know somehow I need to use the Principle of Uniform Boundedness, and alos I was told the hint of letting $ A\equal{}\{f\circ T| \|f\|\equal{}1\}$, but I am still confused. Please help.

i think i remember doing this problem a few years ago.  take $ f\in Y^*$ with $ \|f\| \equal{} 1$ and $ f(x) \equal{} \|x\|$ (such $ f$ exists by the Hahn-Banach theorem).  Now $ (f\circ T)(x) \equal{} \|T(x)\|$ for each $ x$.

On one hand, since $ f\circ T\in X^*$, if $ x_n\to x$, then $ (f\circ T)(x_n \minus{} x)\to 0$

On the other hand, by the above, $ (f\circ T)(x_n \minus{} x) \equal{} \|T(x_n \minus{} x)\| \equal{} \|T(x_n) \minus{} T(x)\|\to 0$, so $ T(x_n)\to T(x)$.

I didn't use the uniform boundedness principle -- where's my mistake?

Your first line doesn't work- you can choose such an $ f$ for any particular $ x$, but you can't do it for all $ x$ at once. That means you don't get to choose $ (f\circ T)(x)$; after all, $ \|T(x)\|$ is not a linear map.

[quote="jmerry"]after all, $ \|T(x)\|$ is not a linear map.[/quote]

you don't say!  i hate functional analysis.

how's the following?

Let $ A$ be as in the hint.  From the inequality $ |(f\circ T)(x)|\leq \|f\|\|Tx\|$, we see that for each $ x\in X$, $ \sup_{g\in A} |g(x)| \leq \sup_{g\in A} \|g\|\|Tx\| \equal{} \|Tx\| < \infty$.

The uniform boundedness principle then implies that $ \sup_{f\in A} \|f\circ T\| < \infty$.  By definition, this is $ \sup_{f\in A}\sup_{\|x\| \equal{} 1} |(f\circ T)(x)| \equal{} \sup_{|x\| \equal{} 1}\sup_{f\in A} |f(Tx)|$

Now, for $ x\not \equal{} 0$, we can find $ f\in X^*$ with $ \|f\| \equal{} 1$ and $ |f(x)| \equal{} \|x\|$ (this is the correct statement of what i botched earlier), so $ \sup_{f\in A} |f(Tx)| \geq \|Tx\|$, and, in particular, we see that $ \|T\|\triangleq \sup_{\|x\| \equal{} 1} \|Tx\| \leq \sup_{f\in A} \|f\circ T\| < \infty$.  But this is precisely equivalent to boundedness (hence continuity) of $ T$.


geometry
trapezoid
angle bisector
geometry unsolved

Let be given two circle $ (O_{1})$, $ (O_{2})$ with centers $ O_{1}$, $ O_{2}$, with distinct radii, externally touching each other at a point $ T$. Let $ O_{1}A$ be a tangent to $ (O_{2})$ at a point $ A$, let $ O_{2}B$ be a tangent to $ (O_{1})$ at a point B so that the points $ A$, $ B$ are on the same side with respec to the line $ O_{1}O_{2}$. Let $ H$ be the point on $ O_{1}A$, $ K$ be the point on $ O_{2}B$ so that the lines $ BH$, $ AK$ are perpendicular to $ O_{1}O_{2}$. The line $ TH$ cuts $ (O_{1})$ again at $ E$, the line $ TK$ cuts $ (O_{2})$ again at $ F$. The line $ EF$ cuts $ AB$ at $ S$. Prove that the lines $ O_{1}A$, $ O_{2}B$ and $ TS$ are concurrent.

The solution of this problem is hidden in its configuratuion.

First of all we must prove that $ O_{1}E\parallel O_{2}A$ and $ O_{2}F\parallel O_{1}B$ (not difficult).

So, we conclude that $ BE\parallel AF$ and then the line connecting their midpoints $ M,$ $ M',$ passes through the point $ S\equiv AB\cap EF$

$ ($ from the trapezium $ AEBF$ $ ).$

We can easy to prove that $ \frac{SM}{SM'}= \frac{TO_{1}}{TO_{2}}$ and that $ TS'\parallel O_{1}M\parallel O_{2}M'$ where $ S'\equiv O_{1}A\cap O_{2}B.$

Hence, the line segment $ TS',$ passes through the point $ S$ and complete the solution.

Kostas VIttas.

PS. I will post here the details of the solution I have in mind, after 10 days when will come back from the vacations in [b][size=100]Leros island[/size][/b]. :lol:

Denote by $ R$ the intersection of $ O_{1}A$ and $ O_{2}B$.
We have $ \angle O_{1}BO_{2}=\angle O_{1}AO_{2}=90^\circ$, so the quadrilateral $ O_{1}O_{2}AB$ is cyclic. It follows that $ \triangle O_{1}BR\sim\triangle O_{2}AR\Rightarrow\frac{O_{1}R}{O_{2}R}=\frac{O_{1}B}{O_{2}A}=\frac{O_{1}T}{O_{2}T}$. Hence $ RT$ is angle bisector of $ \angle O_{1}RO_{2}$ $ (1)$.
We'll prove that $ SR$ is angle bisector of $ \angle ARB$.

We have $ \angle O_{1}BH=\angle O_{1}O_{2}B=\angle O_{1}AB\Rightarrow\triangle O_{1}BH\sim\triangle O_{1}AB$
$ \Rightarrow O_{1}B^{2}=O_{1}H\cdot O_{1}A\Rightarrow O_{1}T^{2}=O_{1}H\cdot O_{1}A\Rightarrow\triangle O_{1}TH\sim\triangle O_{1}AT$
It follows that $ \angle O_{1}AT=O_{1}TH=\angle O_{1}ET$ and then the quadrilateral $ O_{1}TAE$ is cyclic.
Thus $ \angle EO_{1}A=\angle ETA=180^\circ-\angle ETO_{1}-\angle ATO_{2}$ $ =180^\circ-\angle TAO_{1}-\angle O_{2}AT=90^\circ$.
So $ EO_{1}\perp O_{1}A\Rightarrow EO_{1}\parallel AO_{2}$. Similarly, we also have $ O_{1}B\parallel O_{2}F$.

On the otherhand, triangles $ O_{1}EB$ and $ O_{2}AF$ are isosceles, and $ \angle EO_{1}B=90^\circ-\angle BO_{1}A=90^\circ-\angle BO_{2}A=\angle AO_{2}F$.
So $ BE\parallel FA$, it follows that $ \frac{SB}{SA}=\frac{BE}{AF}=\frac{O_{1}B}{O_{2}A}=\frac{RB}{RA}$. Hence $ RS$ is the angle bisector of $ \angle ARB$ $ (2)$

From $ (1)$ and $ (2)$ we get $ T,\,R$ and $ S$ are collinear and so $ TS,\,O_{1}A$ and $ O_{2}B$ are concurrent as we wanted.


inequalities function calculus derivative

For all $ a,b,c\in\mathbb{R}^ \plus{} , a \plus{} b \plus{} c \equal{} 1$, prove that:
$ \sum_{cyc}\frac {27a^4 \plus{} 1}{a^2 \minus{} 2a \plus{} 1}\geq9$
and determine when the equality holds.

PS: Anyone interested to solve it using Jensen's Inequality?  :wink:

By Jensen

$ f(x) \equal{} \dfrac{1}{(x \minus{} 1)^2} \longrightarrow f''(x) \equal{} \frac {6}{(x \minus{} 1)^4} > 0, x \in\mathbb{R}^ \plus{}$

$ \Rightarrow \sum_{cyc} \frac {1}{(a \minus{} 1)^2} \ge 3\cdot \dfrac{1}{\left(\frac {a \plus{} b \plus{} c}{3} \minus{} 1\right)^2} \equal{} \frac {27}{4}$

$ f(x) \equal{} \dfrac{x^4}{(x \minus{} 1)^2} \longrightarrow f''(x) \equal{} \frac {2x^2(x^2 \minus{} 4x \plus{} 6)}{(x \minus{} 1)^4} > 0, x \in \mathbb{R}^ \plus{}$

$ \Rightarrow \sum_{cyc} \frac {a^4}{(a \minus{} 1)^2} \ge 3\cdot \dfrac{\left(\frac {a \plus{} b \plus{} c}{3}\right)^4}{\left(\frac {a \plus{} b \plus{} c}{3} \minus{} 1\right)^2} \equal{} \frac {1}{12}$

$ \Rightarrow \sum_{cyc} \frac {27a^4 \plus{} 1}{(a \minus{} 1)^2} \ge 27\cdot\frac {1}{12} \plus{} \frac {27}{4} \equal{} 9$

QED

Sorry, I though splitting the function up would make the differentiation easier but i was wrong...

$ f(x) \equal{} \dfrac{27x^4 \plus{} 1}{(x \minus{} 1)^2} \longrightarrow f'(x) \equal{} 12x^3 \longrightarrow f''(x) \equal{} 36x^2 > 0$

$ \Rightarrow \sum_{cyc} \frac {27a^4 \plus{} 1}{(a \minus{} 1)^2} \ge 3 \cot \frac {27\left(\frac {a \plus{} b \plus{} c}{3}\right)^4 \plus{} 1}{(\frac {a \plus{} b \plus{} c}{3} \minus{} 1)^2} \equal{} 9$

QED

EDIT: sorry again, major screw up in differentiation!

Is the differentiation wrong? I'm just not sure with your results, ocha.
$ f'(x)\equal{}\frac{54a^4\minus{}108a^3\minus{}2}{(a\minus{}1)^3}$

If we show that $ \frac {27a^4 \plus{} 1}{a^2 \minus{} 2a \plus{} 1} \ge 18a \minus{} 3$ then we are done by cyclically summing.

$ 27a^4 \plus{} 1 \ge (18a \minus{} 3)(a^2 \minus{} 2a \plus{} 1)$

$ 27a^4 \plus{} 1 \ge 18a^3 \minus{} 39a^2 \plus{} 24a \minus{} 3$

$ 27a^4 \minus{} 18a^3 \plus{} 39a^2 \minus{} 24a \plus{} 4\ge 0$

$ (3a \minus{} 1)^2(3a^2 \plus{} 4) \ge 0$.

Q.E.D.


Here's the problem I am trying to do.

The population of the city grew 12% for each of 3 consecutive years. At the end of that time, the population was 50,577. Find the initial population at the start of the 3 years.

The book I am using does not tell me the answer nor offers an explanation. Heres the two ways me and a freind attempted the problem.

First, I said take .12 percent multiply that by the given 50,577. And subtract that from 50,577. That reduces it to what it was the previous year. Then take the newly found number of the previous year and do the same thing. This then did not seem logical in that I am removing 12 percent of the current population not what it was originally.

My freind tried it this way. He wanted to multiply  the 12 percent by 3 making it 36 percent and deduct 36 percent from the 50,577. This to me didn't seem right either because it is again taking the percentage of the current value.

Unfortunatly im no longer in school and trying to get it back so I don't have access to a teacher I hope someone on this forum can help me. :(

How about you work forward from the given information and build upon an equation like lego blocks.  ;)

Each of the three years, the city grew 12%.

This means that at the end of each year, the total for that year was multiplied by $1.12$.

The beginning total is $x$.

So, we have $x * 1.12$ is the first year total.

Then $x * 1.12^2$ is the second year total.

Finally, $x * 1.12^3$ is the total we have.

So the equation is $x * 1.12^3 = 50577$. I hope this helps and is not overkill.

Use a calculator and do $\frac{50577}{1.12^3}$.

Try to use the information given and see what might have led up to the answer. This is often important in proofs. IT's called the penultimate step of solving a problem.


number theory
greatest common divisor
number theory proposed

[b](a)[/b] Let $ a,b,c$ positive integer such that $ a \plus{} b \plus{} c \mid a^2 \plus{} b^2 \plus{} c^2$. Show that $ a \plus{} b \plus{} c \mid a^n \plus{} b^n \plus{} c^n$ for infinitely many positive integer $ n$. [i](Laurentiu Panaitopol)[/i]

[b](b) [/b]Show that there exist infinitely many triple of positive integers $ (a,b,c)$ such that $ \text{gcd}(a,b,c) \equal{} 1$ and $ a \plus{} b \plus{} c \mid a^2 \plus{} b^2 \plus{} c^2$ and $ a \plus{} b \plus{} c \nmid a^n \plus{} b^n \plus{} c^n$ for infinitely many positive integer $ n$. [i](Paolo Leonetti)[/i]

for the first part see the solution in PEN.
let's suppose that we have found such integers $ a,b,c$ with $ 6 < a \plus{} b \plus{} c |a^2 \plus{} b^2 \plus{} c^2$.
then if there exists a prime dividing $ a \plus{} b \plus{} c$ which is greater than $ 3$ then we put $ n \equal{} k(p \minus{} 1)$ and in this way we had constructed infinitely many integers $ n$ with the property mentioned.
so all primes dividing this number must be either $ 3$ or $ 2$.
the exponent of at least one of them is greater than $ 1$.
it can be $ 3$ i mean $ 9 |a \plus{} b \plus{} c$,for this case just see that we have to put $ n \equal{} 6k$ and we are done.
it can also be $ 2$.  in this case we must have $ 4| a \plus{} b \plus{} c |a^2 \plus{} b^2 \plus{} c^2$.one can easily check that it can be true if $ a,b,c$ are even which contradicts $ \gcd(a,b,c) \equal{} 1$.
so all that we must do is to find infinitely many such integers $ a,b,c$ with $ \gcd(a,b,c) \equal{} 1$ and $ a \plus{} b \plus{} c | a^2 \plus{} b^2 \plus{} c^2$.
for this just put $ a \equal{} 1$ then we have only to find infinitely many integers   $ b,c$ such that $ 1 \plus{} b \plus{} c | 2bc \minus{} 2$.
now put $ c \equal{} b^2$.we r done!
so just by putting $ (a,b,c) \equal{} (1,s,s^2),s>1$ we have found infinitely many triples in such a way that they satisfy the condition mentioned above.

Where it is in PEN? I cannot find it..However your solution seems to work.  :lol: You can see mine [url=http://bboyjordan.wordpress.com/2009/10/22/136/]here[/url]


conics
projective geometry
geometry proposed
geometry

Hi, everyone in mathlinks.
For two triangles ABC and A'B'C' on a circle , prove that six segments of them touch a conic out of the conic.
May I ask you a favor? this problem really pisses me off.  :(
Thanks.

Define intersections:

$ U \equal{} AB \cap C'A'$
$ V \equal{} BC \cap C'A'$
$ W \equal{} BC \cap A'B'$

$ X \equal{} CA \cap A'B'$
$ Y \equal{} CA \cap B'C'$
$ Z \equal{} AB \cap B'C'$

$ P \equal{} CC' \cap BB'$
$ Q \equal{} B'V \cap CZ$
$ R \equal{} VY \cap WZ$

By Pascal theorem for cyclic hexagon $ AC'BA'CB',$ intersections $ U, P, X$ are collinear,
By Pappus theorem for lines $ BVC$ and $ C'ZB',$ intersections $ U, P, Q$ are collinear $ \Longrightarrow$ points $ U, P, Q, X$ are collinear.
By Pappus theorem for lines $ VWC$ and $ ZYB',$ intersections $ R, Q, X$ are collinear $ \Longrightarrow$ points $ U, P, Q, R, X$ are collinear.
It follows that $ R \equal{} VY \cap WZ$ is on $UX$ $ \Longrightarrow$ main diagonals $ UX, VY, WZ$ of hexagon $ UVWXYZ$ are concurrent at $ R.$
By Brianchon theorem, hexagon $ UVWXYZ$ is tangential to a conic.

Thanks , yetti
Despite my poverty at math ,  I have got the same solution as yours , yetti .
So I am so happy  :)


For the coming school year, I really want to do a mathematics project, but all of the math projects I have seen involve areas of math I am not familiar with. For those of you who have done math projects, where did you find the source of your topics, and how did you go about finding out more about? What are some areas of math you would advise me to try?

Thanks!

Are you trying to create some novel project with a new proof of a theorem or some new conjecture?  Or are you basically just doing a math project?  I did a math fair project in HS on analytic number theory, Riemann-Zeta, and Goldbach's Conjecture.  I basically just did some reading, contacted some professors, and learned what I could.

Thanks for the reply.

I'm mostly looking for an interesting math-related project to work on; it doesn't have to be groundbreaking or anything. However, I can't seem to find a topic that seems do-able for me, as most of the unknown things in current math goes way over my head. When you contacted profs, did you already have your research problem posed, or did you just have some general theorem you thought was interesting and wanted to expand on it?

Thanks!

After I read a book on Fermat's Last Theorem, I became interested in the theory of elliptic curves and modular forms.  I knew I would be visiting colleges so I can contacted a professor at a college I would be visiting, just for some general information.  He steered me away from those areas since they are a bit advanced, but showed me some stuff on Riemann-Zeta instead, which I found interesting.  I then tried to find more information on it and went from there.

I looked up the Riemann-Zeta function on Wikipedia and it went way over my head, as I was not familiar with most of the concepts described. Do you have any recommendations for books that would introduce me to a few of the current topics of interest in math that explains some of the concepts involved? Also, do you still have a soft-copy of your project that I could look at?

I think my problem is mostly that the current topics of interest are usually too hard for me to understand, and the topics that I do understand, I can't really find something to work on that someone else hasn't already done.   :(

Thanks so much for all of your help!

You're in high school, yes?  You're not going to find something that hasn't already been done on your level.  Unless this is a project for Siemens, Intel, ISEF,... you really shouldn't need such a groundbreaking project.

I chose Riemann-Zeta because it interested me.  That doesn't mean it will interest you.  I attached the paper I wrote.  Some of the $ \text{\LaTeX}$ is messed up but it should give you a nice idea.

Thanks! I was trying to find a math-topic for science fair. I guess I should have been more specific about that.

I'll look through your project, and hopefully I'll get a better idea about what I'm looking for/planning to do.  :)


MATHCOUNTS
Vieta
AMC
AIME
geometry
AMC 12
HMMT

I think I'm going to buy the most two popular books - I'm new to AMC and I'm wondering what I should study...

If you need my ability level to help me, well, I couldn't attend the mathcounts state, but my scores were 38 to 42 for mathcounts... The problem is, AMC is different formats, strategies, and level...

If you're coming out of middle school, the best approach is to go for something more general, such as AoPS 2. Much of it will be helpful.

As you go through high school, you will begin to find your relative strengths and weaknesses, at which time it makes sense to buy specialized books. Don't rush deciding what your relative strengths and weaknesses are, though; something that might seem to be a weakness might easily be fixed by a few posts on AoPS.

The good thing about the AMC's is that, unlike Mathcounts, there is one consistent format. The best way to get used to that is by practicing under test-like conditions (that means stopping at 75:00 and not 75:12, and it might mean asking a teacher for a few bubble answer-sheets).

I see, that like 8 others so far, that I prefer AoPS vol. II. Why? Because it's a really excellent resources in general; anything from logs to Menelaus' Theorem to Vieta's is in there. And as Zuton said, you can always find out your weaknesses later.

Thank you guys.

But as for Volume II, will I actually be able to use it on AMC 12? and AIME?

Because I remember my friend warned me from buying Volume II for mathcounts due to it not being used at all. I will be able to apply Volume II material on AMC and AIME, right?

(If so, volume 2= one of my books targeted)

Seeing as you've basically decided Volume 2 as one of your books, I put geometry since you see a lot of that on AMC12

AoPS 2 will help a ton on the AIME, and probably a bit on the AMC 12, too (the last few questions)
So yeah, definitely get that first, and like the others have said, you'll find your strengths and weaknesses later on.

I personally would recommend intro to number theory, because a good foundation in number theory can prove invaluable later on, in speeding up calculations, avoiding careless mistakes, and using the integer answer criterion of the AIME cleverly.

I suggest AoPS Volume 2. It's especially designed for high school math contests like Mandelbrot, HMMT, ARML, and AMC competitions. Also, it's a useful textbook for Olympiad contestants.


[code]
class StringModify {
  public static void main(String[] args){

  String String= "This is the message!!";
  String K="v";

  byte[]p = K.getBytes();
  byte[]s = String.getBytes();
  int d = s.length;
  byte[]z = new byte [d + d/5];
  int j = d+ d/5;
  int h = 0;
  int y = 1;

   for(y=1;y<=d;y++) {
   for(h=0;h<=j;h++) {

   if (h%5 != 0){
   z[h] = s[y];

}
   else if (h%5 == 0) {
   z[h]=p[0];
   y--;

}

  }

}
  for (h=0;h<=j;h++) {
  z[h] = ++z[h];
  h++;

}

  for (h=0;h<=j;h++) {
  if (h%3==0) {
  z[h] = (++z[h]);

}
  }
  String strA = new String(z);
  System.out.println(strA);
}
 }

[/code]

I can compile this program just fine, but whenever I try to run it in my command shell I get the ArrayOutofBoundsException error with -1 specifically listed at line 20.  I think I know what the problem is, but it still won't fix!
Since 0%5=5, the second part of the first if/then statement will be satisfied so y will be decreased.  After noticing that, I took precautions against this and tried setting y=1.  It still gives me the same error!

Any help is appreciated, and please excuse this if it is a really noob mistake. I only started java 2 days ago so go easy on me. :)
Thanks in advance.

What is your program even supposed to do?

[hide="Commentary"]
class StringModify {
  public static void main(String[] args){

  String String= "This is the message!!"; //Ew, no. Do not declare variable names as reserved words. I am surprised that this compiled at all.
  String K="v";

  byte[]p = K.getBytes();
  byte[]s = String.getBytes();
  int d = s.length;
  byte[]z = new byte [d + d/5];
  int j = d+ d/5;  //Minor nitpick: You can declare j first and then declare z as new byte[j] to save the computer some calculations
  int h = 0;
  int y = 1;

   for(y=1;y<=d;y++) { //java array indices go from 0 to size-1, not 1 to size. So <= size will cause the loop counter to eventually try to access array[size] which does not exist.
   for(h=0;h<=j;h++) { //Same error here.

   if (h%5 != 0){
   z[h] = s[y];

}
   else if (h%5 == 0) { //new if statement not necessary, just use else.
   z[h]=p[0];
   y--; //Decrementing y? Why? This is what apparently caused the -1 array index error, (And never change the loop index like that in a nested for loop)

}

  }

}
  for (h=0;h<=j;h++) {
  z[h] = ++z[h]; //Bad coding style (for Java) but acceptable
  h++;

}

  for (h=0;h<=j;h++) {
  if (h%3==0) {
  z[h] = (++z[h]); //Bad coding style (for Java) but acceptable

}
  }
  String strA = new String(z);
  System.out.println(strA);
}
 }
[/hide]

I revised the code a bit just to get rid of all the bugs, but I am not sure if it does what you intend for it to do.

[code]class StringModify {
	public static void main(String[] args){

	String msg = "This is the message!!";
	String K="v";

	byte[]p = K.getBytes();
	byte[]s = msg.getBytes();
	int d = s.length;
	int j = d + d/5;
	byte[]z = new byte [j];
	int h = 0;
	int y = 1;

	while(h<j)
	{
		if (h%5 != 0)
		{
			z[h] = s[y];
			y++;
		}
		else
		{
			z[h]=p[0];
		}
		h++;
	}

	for (h=0; h<j-1; h += 2)
	{
		z[h] = ++z[h];
	}

	for (h=0;h<=j;h++)
	{
		if (h%3==0)
			z[h] = ++z[h];
	}

	String strA = new String(z);
	System.out.println(strA);
	}
}[/code]

It would be nice if you used [url=http://stackoverflow.com/questions/209015/self-documenting-code]self documenting variables[/url].

I will certainly look into the problems you have showed me.  The purpose of the program was to encrypt the string, sorry if I did not make that clear.
The goals of the program were to take the String "This is the message!!" and
1. Add a v at every 5th spot
2. Increase each ascii value by one.
3. Increase every 3rd ascii value once more.

The reason I put y as decrementing was because the number of elements of z ( the product string ) was greater than the elements of s ("This is the message!!") and therefore the array was out of bounds when I tried to bring the elements of s onto z.

Once again, thanks for your time and your comments.

Then this should work:

[code]class StringModify {
   public static void main(String[] args){

   String msg = "This is the message!!";
   String K="v";

   byte[]p = K.getBytes();
   byte[]s = msg.getBytes();
   int d = s.length;
   int j = d + d/5;
   byte[]z = new byte [j];
   int h = 0;
   int y = 0;

   while(h<j)
   {
      if (h%5 != 0)
      {
         z[h] = s[y];
         y++;
      }
      else
      {
         z[h]=p[0];
      }
      h++;
   }

   for (h=0; h<j-1; h++)
   {
      z[h]++;

      if (h%3==0)
         z[h]++;
   }

   System.out.println(new String(z));
   }
}[/code]

The code worked perfectly.  If you mind entertaining one more of my questions, can you please tell me what I coded wrong? This is my first attempt to learn a computer language and a lot of things are not clear to me.  When I look at your code its a lot more efficient.  Was using the nested for loop a bad idea when I could just use the while loop like you did?  And does the line while(h<j) imply that h increases?
Thank you for the code and your time.

EDIT:
Never mind the last question, I see the h++ located near the end of the while loop.

Nesting loops means that the inner loop loops once for every single increment in the outer loop. You wanted h and y to simultaneously change, so a while loop makes more sense (just increment both of them in the loop under a conditional)

The way your loops were nested, y would start at 1, and h would go all the way from 0 to j before the inner loop ends. Then, y goes to 2, and h starts at 0 again, obviously not what you want.

(Indenting code could help you in understanding this a little better.)

Ahh ok.  Thanks for being so helpful.
Say do you know of a site called the Euler Project? If you don't check it out, it's really tight.


[b][i][size=150][color=green]If there exists positive integers $ x,y,z$ satisfying $ 28x\plus{}30y\plus{}31z\equal{}365$.Then find the value of $ z\minus{}2x$ for some such triplet.[/color][/size][/i][/b]

[hide="solution"]Since $ 28x\plus{}30y$ is even, and 365 is odd, $ 31z$ must be odd, and therefore $ z$ is odd. Let $ z\equal{}2z_1\plus{}1$ for some non-negative integer $ z_1$. Therefore we have $ 28x\plus{}30y\plus{}62z_1\plus{}31\equal{}365$, or $ 14x\plus{}15y\plus{}31z_1\equal{}167$.
Therefore $ z_1\in \{ 0,1,2,3,4,5\}$. Trying each and every value for $ z_1$, we see that $ z_1\equal{}3$, $ x\equal{}1$, and $ y\equal{}4$. Therefore $ z\equal{}7$, and $ z\minus{}2x\equal{}5$.[/hide]

[hide="Easier"]
I think apratimdefermat intended us to do this...

$ 28x\plus{}30y\plus{}31z\equal{}30(x\plus{}y\plus{}z)\plus{}z\minus{}2x$

The only possible value for $ x\plus{}y\plus{}z$ is 12.

Thus, the value of $ z\minus{}2x$ is $ 360\minus{}12*30\equal{}5$

Answer: 5[/hide]

[quote="ffao"][hide="If we do not require proof"]
February has 28 days.
January, March, May, July, August, October, December have 31 days.

That makes $ z \minus{} 2x \equal{} 7 \minus{} 2 \equal{} 5$
[/hide][/quote]

hahahaah wow thats really neat


geometry
3D geometry
pyramid
sphere
vector
geometry unsolved

The base of a regular pyramid is a regular $2n$-gon $A_1A_2...A_{2n}$. A sphere passing through the top vertex $S$ of the pyramid cuts the edge $SA_i$ at $B_i$ (for $i = 1, 2, ... , 2n$). Show that $\sum\limits_{i=1}^n SB_{2i-1} = \sum\limits_{i=1}^n SB_{2i}$.

This is an easy exercise on scalar products. For the sake of simplicity, suppose that we deal with an hexagonal pyramid with vertices $A_1A_2A_3A_4A_5A_6$ forming a regular hexagon of center $C$ and that $S$ is the vertex of the pyramid and $O$ the center of the sphere passing through $S$. We will denote the vector $SC$ as $x$, the vector $SO$ as $y$ and the vectors $CA_i$ as $u_i$, then the left hand sum equals
\[ \frac 2{\|x+u_1\|}\left((x+u_1)\cdot y+(x+u_3)\cdot y+(x+u_5)\cdot y\right)= \frac 2{\|x+u_1\|} (3x+u_1+u_3+u_5)\cdot y=\frac 6{\|x+u_1\|}x\cdot y  \]
and now it is evident that the left hand sum gives the same value.


Find a complete metric space (X,p) and two descending sequences of non-empty compact subsets of X,say{C'(n)} and {C''(n)},such that ∩C'(n)=Ø，and C''(n)≠ Ø,but diam(∩C'(n)) <lim diam (C''(n))(for n →∞).

{C(n)}: n is the sub. from 1 to ∞.

 :)

Check what you've written. It doesn't seem to make any sense... :? The best I can make out of it is:

Suppose that $ X$ is a metric space and $ C_n$ is a decreasing sequence of compact subsets of $ X$. Is it possible that $ \text{diam\,}(\cap_n C_n)<\inf _{n}\text{diam\,}C_n$?

Is that what you meant?

No.
Sorry for the bad writing.
Here is what it mean.

[quote="fedja"]This is different from what I wrote only in one respect: it asks to find a decreasing sequence $ C_n'$ of non-empty compact sets with empty intersection, which is, clearly, impossible, in addition to the sequence $ C''_n$ from which one asks exactly what I wrote ($ \lim$ or $ \inf$, there is no difference). Check again! (and try learning basic $ \text{\LaTeX}$, it is easy :)).[/quote]

Yeap.Thanks. :)
That is what I mean.
I am learning LaTeX.


Suppose $ a,b,c\in R^{\plus{}}$,for $ n\in N^{\plus{}}$,let $ F_n\equal{}\frac{3(a^{n}\plus{}b^{n}\plus{}c^{n})}{a\plus{}b\plus{}c} \minus{} \frac{b^{n}\plus{}c^{n}}{b\plus{}c}\minus{} \frac{a^{n}\plus{}c^{n}}{a\plus{}c} \minus{} \frac{a^{n}\plus{}b^{n}}{a\plus{}b}$.
Prove that: $ F_n\geq 0$ ,$ n\equal{}1,2...$


function
vector
algebra unsolved
algebra

Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $ x,y$:

$ x f(x)\minus{}yf(y)\equal{}(x\minus{}y)f(x\plus{}y)$.

[quote="moldovan"]Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $ x,y$:

$ x f(x) \minus{} yf(y) \equal{} (x \minus{} y)f(x \plus{} y)$.[/quote]

Let $ P(x,y)$ the assertion $ xf(x)\minus{}yf(y)\equal{}(x\minus{}y)f(x\plus{}y)$

1) Obviously, the solutions set is a $ \mathbb R$-vector space ($ f$ solution implies $ \lambda f$ solution and $ f,g$ solutions imply $ f\plus{}g$ solution)

2) Dimension of solution vector-space is at most 2:
If three independant solutions $ h_1,h_2$ and $ h_3$ exist, it's easy to build a non all-zero solution $ f\equal{}\alpha h_1\plus{}\beta h_2\plus{}\gamma h_3$ which has at least two zeros.

Then, if $ f(u)\equal{}f(v)\equal{}0$ for some $ u\neq v$ :

$ P(x,u\minus{}x)$ $ \implies$ $ xf(x)\equal{}(u\minus{}x)f(u\minus{}x)$
$ P(x\plus{}v\minus{}u,u\minus{}x)$ $ \implies$ $ (x\plus{}v\minus{}u)f(x\plus{}v\minus{}u)\equal{}(u\minus{}x)f(u\minus{}x)$

And so $ Q(x)$ : $ xf(x)\equal{}(x\plus{}v\minus{}u)f(x\plus{}v\minus{}u)$

Then $ P(x\plus{}v\minus{}u,y)$ $ \implies$ $ (x\plus{}v\minus{}u)f(x\plus{}v\minus{}u)\minus{}yf(y)\equal{}(x\plus{}v\minus{}u\minus{}y)f(x\plus{}v\minus{}u\plus{}y)$ and so $ xf(x)\minus{}yf(y)\equal{}(x\plus{}v\minus{}u\minus{}y)f(x\plus{}v\minus{}u\plus{}y)$ and so :
$ (x\minus{}y)f(x\plus{}y)\equal{}(x\plus{}v\minus{}u\minus{}y)f(x\plus{}v\minus{}u\plus{}y)$
But $ Q(x\plus{}y)$ $ \implies$ $ (x\plus{}y)f(x\plus{}y)\equal{}(x\plus{}y\plus{}v\minus{}u)f(x\plus{}y\plus{}v\minus{}u)$
Subtracting these two equalities, we get $ 2yf(x\plus{}y)\equal{}2yf(x\plus{}y\plus{}v\minus{}u)$ and so : $ f(x\plus{}v\minus{}u)\equal{}f(x)$

Then $ Q(x)$ becomes $ xf(x)\equal{}(x\plus{}v\minus{}u)f(x\plus{}v\minus{}u)$ $ \equal{}(x\plus{}v\minus{}u)f(x)$ and so $ f(x)\equal{}0$
Q.E.D.

3) All solutions are $ f(x)\equal{}ax\plus{}b$
$ f(x)\equal{}1$ is obviously a solution
$ f(x)\equal{}x$ is obviously a solution
So we have a two-vectors basis of our vector space.
Hence the result.

An easier solution :
$ (x\equal{}1,y\equal{}\minus{}1) \Longrightarrow f_{(1)}\plus{}f_{(\minus{}1)}\equal{}2f_{(0)}$
$ (x\equal{} X\plus{}1,y\equal{}\minus{}1)\Longrightarrow (X\plus{}1)f_{(X\plus{}1)}\plus{}f_{(\minus{}1)}\equal{}(X\plus{}2)f_{(X)}$
$ (x\equal{}X,y\equal{}1) \Longrightarrow Xf_{(X)}\minus{}f_{(1)}\equal{}(X\minus{}1)f_{(X\plus{}1)}$
Finally, it's easy to obtain $ f_{(X)}\equal{} (f_{(1)}\minus{}f_{(0)})X \plus{} f_{(0)}$
 :cool:

Let $ x \neq0$ and $ y\equal{}0$. Then $ x f(x)\equal{}x^2$, and $ f(x)\equal{}x$ for all nonzero x. We can let $ f(0)$ be anything.

Am I missing something here?

@Bacteria : Yes. You're missing something : $ y\equal{}0$ $ \implies$ $ xf(x)\equal{}xf(x)$ and not $ xf(x)\equal{}x^2$

@Diogene : nice and quick! Congrats !

Oh wow, I saw $ (x\plus{}y)$ not $ f(x\plus{}y)$  :oops: .


Hey guys, I would really appreciate if someone could lead me in the right direction with this.  This is a possible exam question for me and I really dont know how to do it.. Thanks

Given that (x1, x2,......,xk) is linearly independent.  Show that (x1, x2, ....., xk, Y) is dependent if and only if Y is in the span (x1, x2,......, xk).  Thanks again

Elementary linear algebra may be the first place a typical student encounters a substantial amount of mathematical vocabulary for which the exact definitions matter.

First step: write out for yourself the exact, formal definition of a set of vectors being linearly independent. (And note that "dependent" is the simple negation of "independent". And that you have to be careful about your $ \exists$ and $ \forall$ quantifiers when doing a negation.)

Everything you do here is going to be keyed to that definition.

A set of vectors is said to be linearly independent provided that   t1(x1) + t2(x2) +.....+ tk(xk)=0 where t1=t2.....=tk=0 .     So this means a set that is dependent has t values that do not equal 0.

Let's suppose this is an exam in my class. I structured the question like this:

1. (a) Carefully state the definition of the [i]linear indepenence[/i] of a set of vectors.

(b) Carefully state what it means for $ y$ to be in the [i]span[/i] of $ \{x_1,x_2,\dots,x_k\}.$

(c) Suppose $ \{x_1,x_2,\dots,x_k\}$ is linearly independent. Show that $ \{x_1,x_2,\dots,x_k,y\}$ is dependent if and only if $ y$ is in the span of $ \{x_1,x_2,\dots,x_k\}.$

Now I come to your paper, and the first thing I read is

[quote]A set of vectors is said to be linearly independent provided that t1(x1) + t2(x2) +.....+ tk(xk)=0 where t1=t2.....=tk=0 .[/quote]

That's clearly intended to be your answer to 1(a). And in my grading, I count it as wrong.  I don't even give you any partial credit.

Why not? Why do I consider that to be wrong? Because [b]quantifiers matter![/b]

Here's the correct statement:

$ \{x_1,x_2,\dots,x_k\}$ is independent iff $ t_1x_1 \plus{} t_2x_2 \plus{} \cdots t_kx_k \equal{} 0$ implies that $ t_1 \equal{} t_2 \equal{} \cdots \equal{} t_k \equal{} 0.$

Or you could say this:

If $ t_1x_1 \plus{} t_2x_2 \plus{} \cdots t_kx_k \equal{} 0$, then $ t_1 \equal{} t_2 \equal{} \cdots \equal{} t_k \equal{} 0.$

Or you could say this:

$ t_1 \equal{} t_2 \equal{} \cdots \equal{} t_k \equal{} 0$ for all cases in which $ t_1x_1 \plus{} t_2x_2 \plus{} \cdots t_kx_k \equal{} 0$.

But the lesson is that the word "where" in your quote doesn't work, logically, and the word "provided" is in the wrong place to mean anything. And because it doesn't mean what it should mean, I have to count your answer as incorrect. This is a very common mistake among students taking a course like this and being exposed to serious mathematical language for the first time. Students tend to think that the meaning is in the symbols - the variables, the $ \equal{}$ signs, and so on - and overlook the fact that the precise meaning is carried by the little words that glue the symbols together. We can't sacrifice those little words.

Now, here's the logical exercise: negate this! That is, define $ \{x_1,x_2,\dots,x_k\}$ is linearly [i]dependent[/i] iff ___ ?

Note that we used either an "if ... then" implication statement or a "for all" quantifier in our definition of independence. Our negation of that should have the opposite quantifier in there somewhere - that is, words like "there exist" or "for some" should appear somewhere.

[Note: I'm moving the topic over to the "computations and tutorials" forum for linear algebra; it's a better fit over there.]

There's a cute approach to thinking about quantifiers that I think might help here, which is to think of the "for all" quantifier, $ \forall$, as an adversary (see how it looks like a little devil?) and to think of quantified statements as a game.  That is, the statement

[i]for all $ t_1, ... t_k$ such that $ t_1 x_1 \plus{} ... \plus{} t_k x_k \equal{} 0$, it is true that $ t_1 \equal{} ... \equal{} t_k \equal{} 0$[/i]

can be thought of as a game in which you have your vectors $ x_1, ... x_k$ and your adversary $ \forall$ is trying to make them sum to zero.  So he tries to send you nonzero values of $ t_1, ... t_k$ such that this is possible.  But if your vectors are linearly independent, what the above statement means is that there's no way to do this and he can't win!

So:  what happens if he [i]can[/i] win?

I do realize that your saying that my definition was wrong and I understand how yours is different than mine, but I still dont understand what step to take next.


limit
Putnam
integration
real analysis
real analysis unsolved

Find $\lim_{t\rightarrow 1^-} (1-t) \sum_{n=1}^{\infty}\frac{t^n}{1+t^n}$.

the way you stated it ti's trivial, $\sum\geq\sum\frac{t^n}{2}=\frac{t}{2(1-t)}\to+\infty$

what is $\lim_{h\to 0}h\sum_{n=1}^{\infty}f(hn)$

Much like a problem on the 2004 Putnam, we can treat this as a Riemann sum using an improper geometric partition. Consider $\int_0^1\frac{1}{1+x}\,dx$, and look at the Riemann sum using the partition $x_n=t^n$.


$a,$ $b,$ $c$ and $d$ are non-negative numbers such that $3(a^{2}+b^{2}+c^{2}+d^{2})+4abcd=16.$
Prove that $a+b+c+d\leq4.$

First we are to prove $a+b+c+d-abcd\leq 3$ for $a,\ b,\ c,\ d\geq 0.$ Then we have $16-3(a^{2}+b^{2}+c^{2}+d^{2})\geq 4(a+b+c+d)-12,$ yielding $a+b+c+d\leq 4.$

[quote="kunny"]First we are to prove $a+b+c+d-abcd\leq 3$ for $a,\ b,\ c,\ d\geq 0.$ [/quote]
It's wrong: $a=b=c=\frac{4}{3}$ and $d=0.$

The following more general statement is valid:
If $x_{1},x_{2},...,x_{n}$ are non-negative numbers satisfying $x_{1}+x_{2}+...+x_{n}=n$,  $k$ is an integer such that $2 \leq k \leq n+2$,
and $r=(\frac n{n-1})^{k-1}-1$, then
$x_{1}^{k}+x_{2}^{k}+...+x_{n}^{k}-n \geq nr(1-x_{1}x_{2}...x_{n})$.


Let $f$ be twice continuously differentiable on $(0,\infty)$ such that
$\lim_{x\to 0+}f'(x) =-\infty , \lim_{x\to 0+}f''(x) =+\infty$

Show that $\lim_{x\to 0+}\frac{f(x)}{f'(x)}= 0$

I got some kinda solution by using the L'Hopital's rule and mean-value theorem. but it just doesn't seem right. can someone help me out?

Seems that you like posting IMC problems, violetcraze  :wink:

I'll post my solution:

Let $\lim_{x \to 0+}f(x) = \infty$. Otherwise the problem is trivial. Then there exists an intervall $(0, b)$, such that $f(x) > 0, f'(x) < 0, f''(x) > 0$, if $x \in (0, b)$. By the mean value theorem

$\forall 0<x_{0}<x<a: \exists \xi \in (x_{0},x): f'(\xi ) \cdot (x-x_{0}) = f(x)-f(x_{0})$

Because $f'$ is increasing,

$f'(x) \cdot (x-x_{0}) > f(x)-f(x_{0}) \Rightarrow x-x_{0}< \frac{f(x)-f(x_{0})}{f'(x)}< 0$

This implies

$-x_{0}\leq \liminf_{x \to 0+}\frac{f(x)-f(x_{0})}{f'(x_{0})}\leq \limsup_{x \to 0+}\frac{f(x)-f(x_{0})}{f'(x)}\leq 0$
$\Rightarrow-x_{0}\leq \liminf_{x \to 0+}\frac{f(x)}{f'(x)}\leq \limsup_{x \to 0+}\frac{f(x)}{f'(x)}\leq 0$
Because this is true for all $x_{0}\in (0,b)$, we get the fact that

$0 \leq \liminf_{x \to 0+}\frac{f(x)}{f'(x)}\leq \limsup_{x \to 0+}\frac{f(x)}{f'(x)}\leq 0$

and hence $\lim_{x \to 0}\frac{f(x)}{f'(x)}= 0$


Twelve colored balls are placed in a bag. Six of the balls are red, four are white, and two are green. If two balls are drawn from the bag without replacement, what is the probability that at least one of them is white? Express your answer as a common fraction.

What's the probability we get at least 1 white? It's the same as  1-(the probability that we get no whites). This is equal to
\begin{align*}1 - \left(\frac {8}{12}\right)\left(\frac {7}{11}\right) & = 1 - \frac {2}{3}\cdot\frac {7}{11} \\
& = 1-\frac{14}{33}\\&=\boxed{\frac {19}{33}}\end{align*}


linear algebra
matrix
function
quadratics
calculus
vector
real analysis

A friend gave me this problem yesterday, and I am almost certain it involves using the Hessian matrix ([url]http://en.wikipedia.org/wiki/Hessian_matrix[/url]).

The question asks:

Given the function $ f(x,y)\equal{}ax^2\plus{}2bxy\plus{}cy^2$ and that the eigenvalues of the matrix

\[ \left(
\begin{array}{cc}
a & b  \\
b & c  \\
\end{array}
\right)\]

are both positive, must the origin be a minimum of $ f?$

I realize the Hessian of this equation is the above matrix times 2. And since the matrix is positive definite, I would think that the origin (which is a critical point) would be a minimum. However, the Hessian is just constants, and has no dependence on $ x$ or $ y$ at all, which worries me. I am also a bit skeptical because the question asks if the origin [b]MUST [/b]be a minimum.

Does anyone have any insight into this problem? Thanks!

What the Hessian is about:

For a smooth function in a neighborhood of a critical point, the behavior of the function (minus the constant which is its value at that point) is dominated by the second order terms of the Taylor expansion, which are a quadratic form. Classifying that quadratic form will in most cases tell us what is going on. (There are a few cases in which higher order terms matter in some directions.) But the essence is that, through calculus, we are using a quadratic form to approximate the function.

In this particular case, the function [b]is[/b] a quadratic form. Hence we can drop the calculus language altogether - we are not talking about an approximation.

Suppose $ A$ is a real symmetric $ n\times n$ matrix such that all of its eigenvalues are positive. Suppose $ x$ is a column vector in $ \mathbb{R}^n.$ Then the function $ Q(x)\equal{}x^TAx$ is strictly positive for all $ x\ne 0.$ Since $ Q(0)\equal{}0,$ that means $ Q$ has a minimum at $ 0.$

To prove that, make use of the fact that there exists $ U$ such that $ U^TU\equal{}I$ and $ U^TAU\equal{}D,$ where $ D$ is the diagonal matrix whose diagonal entries are the eigenvalues of $ A.$


Fie ABCD un patrulater. Sa se afle locul geom al punctelor M din planul patrulaterului cu prop ca MA*MA+MC*MC=MB*MB+MD*MD.
discutie.

daca 2 laturi opuse sunt congruente e evident ca L.g. e mediatoarea segmentului ce uneste mijloacele laturilor congruente.
dar in general?
analitic iese o dreapta
dar cu ce proprietati?

nu am stat sa termin calculele, dar here goes :) (partea esentiala e aici)
[hide="solutie"]fie $P$ mijlocul lui $AC$ si $Q$ mijlocul lui $BD$
atunci, exprim $MP$ ca mediana in triunghiul $AMC$ (eventual degenerat) :
$MP^{2}=\frac{MA^{2}+MC^{2}}{2}-\frac{AC^{2}}{4}$.
analog
$MQ^{2}=\frac{MB^{2}+MD^{2}}{2}-\frac{BD^{2}}{4}$.
deci $MP^{2}-MQ^{2}=k=\mbox{constant}$. (*)
am redus problema la una mai simpla.
acum analizez doar segmentul $PQ$ si fie $X$ proiectia lui $M$ pe dreapta $PQ$.
din (*) se obtine (cu pitagora) $XP^{2}-XQ^{2}=k$.
din teorema cosinusului obtin
$2\cdot MP\cdot PQ\cdot \cos \angle MPQ=PQ^{2}+MP^{2}-MQ^{2}$.
membrul drept este fix, deci semnul cosinusului nu depinde de $M$, adica $M$ se afla intotdeauna in unul (acelasi pt orice $M$) din cele doua semiplane determinate de perpendiculara in $P$ pe $PQ$.
analog, $M$ se afla intotdeauna in unul (acelasi pt orice $M$) din cele doua semiplane determinate de perpendiculara in $Q$ pe $PQ$.
fie $X$ proiectia lui $M$ pe $PQ$. $X$ poate fi in trei "zone" pe dreapta $PQ$ : intre $P$ si $Q$, pe semidreapta $[PQ$ dar nu pe segmentul $[PQ]$, sau pe semidreapta $[QP$ dar nu pe segmentul $[PQ]$.
apoi, cu pitagora, $k=MP^{2}-MQ^{2}=XP^{2}-XQ^{2}=(XP-XQ)(XP+XQ)$. in functie de pozitia lui $X$ (stabilita mai sus), $XP-XQ$ sau $XP+XQ$ sunt egale in modul cu $PQ$. deci $XP-XQ$ sau $XP+XQ$ este constant (si este aceeasi varianta pentru toti $X$), adica $X$ este independent de $M$.
rezulta ca locul geometric este o dreapta perpendiculara pe $PQ$, ridicata intr`un punct care se poate calcula cu usurinta
nu am stat sa termin calculele sa vad daca acesta e tot timpul in interior, sau sa exprim complet raportul :)
cred ca se poate verifica imediat si analitic acest fapt.[/hide]