Find all functions $ f,g: \mathbb Q \to \mathbb Q$ such that for all rational numbers $ x,y$ we have
\[ f(f(x) \plus{} g(y) ) \equal{} g(f(x)) \plus{} y .
\]

Fixing $ x$ and varying $ y$ we see that $ f$ is surjective, so the equation can be re-written as
\[ f(x \plus{} g(y)) \equal{} g(x) \plus{} y
\]
Now, setting $ g(0) \equal{} k$ we see, setting $ y \equal{} 0$ that $ f(x \plus{} k) \equal{} g(x)$. Note in particular that $ f(k) \equal{} k$.

Now set $ h(x) \equal{} g(x) \minus{} k \equal{} f(x \plus{} k) \minus{} k$

The functional equation is then transformed into $ f(x \plus{} h(y) \plus{} k) \equal{} h(x) \plus{} k \plus{} y$ which then becomes
\[ h(x \plus{} h(y)) \equal{} h(x) \plus{} y
\]
.

Setting $ x \equal{} 0$ and using $ h(0) \equal{} g(0) \minus{} k \equal{} 0$, we get $ h(h(y)) \equal{} y$: $ h$ is an involution.
Setting $ y \equal{} h(z)$ we get (using the fact $ h$ is an involution) $ h(x \plus{} z) \equal{} h(x) \plus{} h(z)$.

This is Cauchy's equation, and on the rationals has the unique solution set $ h(x) \equal{} cx$ with $ c$ constant. This corresponds to $ f(x) \equal{} c(x \minus{} k) \plus{} k, g(x) \equal{} cx \plus{} k$. Substituting back into our top equation,
\[ c(x \plus{} cy) \plus{} k \equiv_{x,y} cx \plus{} k \plus{} y \Leftrightarrow c^2 \equal{} 1
\]
It is a simple check to see that $ f(x) \equal{} \pm(x \minus{} k) \plus{} k, g(x) \equal{} \pm x \plus{} k$ are indeed solutions (both sides evaluate to $ x\plus{}y\pm k$), so they are precisely the set of solutions.

\[ f(f(x)+g(y))=g(f(x))+y.\quad (1)\]

First, fix $ x$. As $ y$ takes on all rational values, the right hand side takes on all rational values. Thus, $ f(f(x)+g(y))$ can take on any rational value, so $ f$ is surjective.

Now set $ y=0$, and for any value $ a$, let $ x$ be such that $ f(x)=a$ (such $ x$ exists because of surjectivity). Plug this into equation (1).

\[ f(a+g(0))=g(a).\quad (2)\]

Thus, the function $ g$ is just a shift of function $ f$. Set $ z=g(0)$, and we have $ f(a+z)=g(z)$. Plug this into equation $ 1$.

\[ f(f(x)+f(y+z))=f(f(x)+z)+y.\quad (3)\]

Now I will show that $ f$ is injective. Suppose $ f(y_1)=f(y_2)$. Then by equation (3),

\begin{align*} f(f(x)+z)+y_1-z&=f(f(x)+f(y_1-z+z))=f(f(x)+f(y_1))=f(f(x)+f(y_2))\\
f(f(x)+z)+y_2-z&=f(f(x)+f(y_2-z+z))=f(f(x)+f(y_2))\end{align*}

\begin{align*} f(f(x)+z)+y_1-z&=f(f(x)+z)+y_2-z\\
y_1&=y_2\end{align*}

$ f(y_1)=f(y_2)\implies y_1=y_2$, thus $ f$ is injective.

Now, let $ x$ be such that $ f(x)=0$, and plug this into (3).

\[ f(f(y+z))=f(z)+y.\quad (4)\]

Now, for any real $ b$, set $ y=b-z$.

\begin{align*} f(f(b-z+z))&=f(z)+b-z\\
f(f(b))&=f(z)+b-z.\quad (5)\end{align*}

This is valid for any value of $ b$. Plug in $ b=z$.

\begin{align*} f(f(z))&=f(z)+z-z\\
f(f(z))&=f(z)\end{align*}

By injectivity, we thus obtain that $ f(z)=z$. Now return to equation (5). We can now reduce it to

\begin{align*} f(f(b))&=f(z)+b-z\\
f(f(b))&=z+b-z\\
f(f(b))&=b.\quad (6)\end{align*}

Consider arbitrary $ d$. Then set $ x=f(f(d)-z)$ in equation (3) and simplify it much with equation (6).

\begin{align*} f(f(x)+f(y+z))&=f(f(x)+z)+y\\
f( f(f(f(d)-z)) + f(y+z) )&=f(f(f(f(d)-z))+z)+y\\
f( f(d)-z + f(y+z))&=f(f(d)-z+z)+y\\
f( f(d)-z + f(y+z))&=f(f(d))+y\\
f( f(d)-z + f(y+z))&=d+y\\
f(f( f(d)-z + f(y+z)))&=f(d+y)\\
f(d)-z+f(y+z)&=f(d+y)\\
f(d+y)+z&=f(d)+f(y+z)\end{align*}

Set $ h(x)=f(x+z)-z$, or equivalently $ f(x)=h(x-z)+z$. Furthermore, set $ d=e+z$ for arbitrary $ e$.

\begin{align*} f(d+y)+z&=f(d)+f(y+z)\\
h(d+y-z)+z+z &= h(d-z)+z+h(y)+z\\
h(d+y-z) &= h(d-z)+h(y)\\
h(e+y) &= h(e) + h(y)\end{align*}

This is Cauchy's Functional Equation. Over the rationals, we know that $ h(x)=mx$ for some constant $ m$. So $ f(x)=h(x-z)+z=m(x-z)+z.$ Let's plug this back into (3).

\begin{align*} f(f(x)+f(y+z))&=f(f(x)+z)+y\\
f( m(x-z)+z + m(y+z-z)+z )&=f(m(x-z)+z+z)+y\\
f( m(x+y-z) + 2z) &= f(m(x-z)+2z)+y\\
m(m(x+y-z)+2z-z)+z &= m(m(x-z)+2z-z)+z+y\\
m^2x+m^2y-m^2z+mz+z &= m^2x-m^2z+mz+z+y\\
m^2y &= y\end{align*}

Plug in $ y=1$ to obtain $ m^2=1\implies m=\pm 1$.

Case 1: $ m=-1$. We have the infinite set of solutions

\[ f(x)=-(x-z)+z=-x+2z\]
\[ g(x)=f(x+z)=-(x+z)+2z=-x+z\]

for any rational constant $ z$.

We can check that this is a valid solution. For any $ x,y$,

\[ f(f(x)+g(y))=f(-x+2z-y+z)=f(-x-y+3z)=x+y-3z+2z=x+y-z\]
\[ g(f(x))+y=g(-x+2z)+y=x-2z+z+y=x+y-z\]
\[ f(f(x)+g(y))=g(f(x))+y\]

so these solutions satisfy the given equation.

Case 2: $ m=1$. We have the infinite set of solutions

\[ f(x)=m(x-z)+z=x-z+z=x\]
\[ g(x)=f(x+z)=x+z\]

for any rational constant $ z$. We can check that these are also valid. For any $ x,y$,

\[ f(f(x)+g(y))=f(x+y+z)=x+y+z=(x+z)+y=g(x)+y=g(f(x))+y.\]

Thus our final solutions are:

\[ f(x)=-x+2z,g(x)=-x+z\]

or

\[ f(x)=x,g(x)=x+z\]

for any rational constant $ z$.

Putting $ x\equal{}0$ in the given equation gives $ f(f(0)\plus{}g(y))\equal{}g(f(0))\plus{}y$
which means that $ g$ is ''1-1'' function and $ f$ takes all values at $ Q$ .
so there exist a $ u$ so that $ f(u)\equal{}0$ . Putting $ y\equal{}u$ at the equation gives :
$ f(f(x)\plus{}g(u))\minus{}g(f(x))\equal{}u$ .Now since $ f$ is onto function (takes all values at $ Q$)
we have that for each $ m\in Q$ we have $ f(m\plus{}g(u))\minus{}g(m)\equal{}u$ (1). From the property of $ f$ that takes all values at $ Q$ we also have that there exist $ n$ such that $ f(n)\equal{}u$ and
putting at (1) where $ m$ the $ n\minus{}g(u)$ we have that $ f(n)\minus{}g(n\minus{}g(u))\equal{}u$ which means that
$ g(n\minus{}g(u))\equal{}0$ or $ g(s)\equal{}0$ for an $ s\in Q$ . Putting $ y\equal{}s$ at the initial gives
$ f(f(x))\equal{}g(f(x))\plus{}s$ which means that for each $ m\in Q$ we have $ f(m)\minus{}g(m)\equal{}s$ or
$ f(x)\minus{}g(x)\equal{}s$ (2). Putting this one in the initial gives $ g(g(x)\plus{}s\plus{}g(y))\equal{}g(g(x)\plus{}s)\plus{}y\minus{}s$ (3)
and also (2) gives us that $ g$ takes also all values at $ Q$ since $ s$ is just a constant
Putting $ x\equal{}u$ at the initial we have that $ f(g(y))\equal{}g(0)\plus{}y$ and with the help of (2) we have that $ g(g(y))\equal{}g(0)\plus{}y\minus{}s\equal{}k\plus{}y\minus{}s$ so (3) becomes
$ g(g(x)\plus{}s\plus{}g(y))\equal{}g(g(x)\plus{}s)\plus{}g(g(y))\minus{}k$ and since $ g$ takes all values at $ Q$ we will have that for each $ a,b\in Q$
$ g(a\plus{}b)\equal{}g(a)\plus{}g(b)\minus{}k$ and letting $ h\equal{}g(a)\minus{}k$ gives $ h(a\plus{}b)\equal{}h(a)\plus{}h(b)$ so from the cauchy
equation we have that $ h(a)\equal{}ca$ for each $ a\in Q$ so $ g(a)\equal{}ca\plus{}k$ or nicer
$ g(x)\equal{}cx\plus{}k$ so $ f(x)\equal{}cx\plus{}k\plus{}s$ . Now the verification :
$ c(cx\plus{}cy\plus{}2k\plus{}s)\plus{}k\plus{}s\equal{}c(cx\plus{}k\plus{}s)\plus{}k\plus{}y$ or $ c^2y\plus{}ck\plus{}k\plus{}s\equal{}k\plus{}y$
The final one holds for each $ y$ so $ c^2\equal{}1$ .
For $ c\equal{}1$ we have $ k\equal{}\minus{}s$ so the functions are $ f(x)\equal{}x$ and $ g(x)\equal{}x\plus{}k$
For $ c\minus{}1$ we have $ k\equal{}s$ so the functions are  $ f(x)\equal{}\minus{}x\plus{}2k$ and $ g(x)\equal{}\minus{}x\plus{}k$
and we are done

We know that :
$ f,g: Q\minus{}>Q$ such that. $ f(f(x)\plus{}g(y))\equal{}g(f(x))\plus{}y$ for all $ x,y \in Q$
If fixed $ x$, >Then in the right side, we have that $ y$ varied on all the rational . THEN $ f$ Varied on all the rational.then ,if $ z \in Q$ , -> Exist $ x \in Q$, such that $ f(x)\equal{}z$,
->$ f(x\plus{}g(y))\equal{}g(x)\plus{}y.$ for all $ x,y \in Q$...(1),
Now, if fixed $ y$, we have that $ f(x\plus{}g(y))$ varied on all the rational. Then $ g$ varied on all the rational.
Non, exist $ c \in Q$ , such that $ g(c)\equal{}0$,
Then, $ f(x)\equal{}g(x)\plus{}c$ for all $ x \in Q$, replacing in (1).
$ f(x\plus{}g(y))\equal{}g(x\plus{}g(y))\plus{}c\equal{}g(x)\plus{}y$, If $ g(0)\equal{}b$.
Then if $ x\equal{}0, y\equal{}0$.->$ g(b)\plus{}c\equal{}b$->$ g(b)\equal{}b\minus{}c$
$ x\equal{}c, y\equal{}b$, $ g(c\plus{}g(b))\equal{}g(c\plus{}b\minus{}c)\equal{}g(b)\equal{}b\minus{}c\equal{}g(c)\plus{}b\equal{}b$,then $ b\equal{}b\minus{}c$ , then $ c\equal{}0$.
Then $ g\equal{}f$
->$ g(x\plus{}g(y))\equal{}g(x)\plus{}y$,for all $ x,y \in Q$.
$ x\equal{}0$,->$ g(g(y))\equal{}y$, for all $ y \in Q$
if $ y\equal{}g(z)$. for all $ z \in Q$.
$ g(x\plus{}g(g(z))\equal{}g(x\plus{}z)\equal{}g(x)\plus{}g(z).$ for all $ x,z \in Q$,
But by the cauchy ecuations, $ g(x)\equal{}ax$,for all $ x \in Q$
Moreover, $ g(g(x))\equal{}ag(x)\equal{}a^2x\equal{}x$->$ a\equal{}1$ or $ a\equal{}\minus{}1$
Then.
$ g(x)\equal{}f(x)\equal{}\minus{}x$, is one solution, the other solution is $ g(x)\equal{}f(x)\equal{}x$.


function
analytic geometry
parameterization

I want to make a rating system for a (rather small) group of players for a certain game. I don't want to go into details but imagine it's chess.

I've read about the USCF rating system and the ISC (scrabble) rating system. However, the USCF rating system is extremely complicated, and the ISC one is based on score difference as well as current rating, so I can't really use either of those.

Anyway, basically, what I want is a function
$f: \mathbb{R}^{2}\times \{0, 1\}\rightarrow \mathbb{R}^{2}$
with teh following properties:
* $f(m, n, 0) = f(n, m, 1)$
* If $f(m, n, 0) = (m', n')$ then $m' < m$ and $n' > n$
* If $f(m, n, 0) = (m', n')$ then $m'+n' = m+n$
* If $f(m, n, 0) = (m', n')$ then $m-m'$ and $n'-n$ should converge to $k > 0$ as $m-n$ gets large.
* If $f(m, n, 1) = (m', n')$ then $m-m'$ and $n'-n$ should converge to $0$ as $m-n$ gets large.

Does anyone have any suggestions for this? Do those conditions seem like enough, and all the rest will be scaling? Or is there some other important factor that I'm overlooking?

What kind of formula is common for this? As far as I can tell, it seems like two rational functions would work -- one for "stronger player wins" and the other for "weaker player wins" so that one can converge to 0 and the other to some $k$ as the difference between the original ratings grows.

Does anyone have any suggestions about how large should be
$r = \frac{f(m, m, 1)}{f(m, n, 1)}$
where $n$ is much greater than $m$?

When you said $f(m, n, 0) = f(n, m, 1)$, I think you meant $f(m, n, 0)$ is the reverse (swapped coordinates) of $f(n, m, 1)$.

The USCF (Elo) system basically uses the following formula: \[f(m, n, w) = (m-\frac{k}{e^{n-m}+1}+kw, n+\frac{k}{e^{n-m}+1}-kw) \, .\]

[quote="Xevarion"]Thanks! That actually seems fine for my purposes.

I've noticed that a lot of leagues with ratings tend to have a special provisional system until players have enough games. However, most of the time those don't rate games between two players with provisional rankings. This presents a problem when the league is just starting out: do you think it's okay to just give people all the same arbitrary ranking at the start and use the regular system from teh start?[/quote]
Yes, as long as you make your system so that two even players playing each other will not result in a score change of zero.

I haven't thought too much about provisional ratings.

Going back to my previous post, in the two places I wrote $e^{n-m}$, you can use other exponentials instead.  For example, I believe that the USCF system basically uses $10^{(n-m)/400}$.  Also, the USCF value of $k$ is basically $32$.  Those might be good parameters to choose to keep the ratings similar to chess ratings.

I decided to multiply rating changes by a factor of $10^{1/(k^{2}+2)}$ where $k$ is the number of games played. It's crude, but it does weight the first few games more strongly and hopefully makes the scores a bit better distributed after the first few rounds. Also it goes to 1 pretty quickly.

What do you think of that? I can still change it.


[hide]
$\binom{n}{k}^{2}$ can be rewritten as $\binom{n}{k}\binom{n}{n-k}$.  This can be combinatorially interpreted as choosing $k$ objects from a set of $n$, and then $n-k$ objects from another set of $n$, for a total of $n$ objects.  Summed over all possible $k$, this is just the number of ways to choose $n$ objects from the two sets, or $\binom{2n}{n}$.
[/hide]

$\{m,n\}\subset\mathcal N\Longrightarrow \sum_{\begin{array}{c}0\le u\le m\\ 0\le v\le n\\ u+v=p\end{array}}\left(\begin{array}{c}m\\\ u\end{array}\right)\left(\begin{array}{c}n\\\ v\end{array}\right)=\left(\begin{array}{c}m+n\\\ p\end{array}\right)$.
[b][u]Proof.[/u][/b] You"ll find the coeficient of $x^{p}$ (left and right) from $(1+x)^{m}\cdot (1+x)^{n}=(1+x)^{m+n}$ !!!

The identity Virgil Nicula referred to is known as Vandermonde's Identity.  You can either use a very simple combinatorial argument or generating functions (as mentioned above) to prove it.  Virgil Nicula, instead of using arrays under summation and as binomial coefficients, you can use \atop and \binom, respectively, e.g.
\[\{m,n\}\subset\mathbb{N}\Longrightarrow \sum_{{0\le u\le m\atop 0\le v\le n}\atop u+v=p}\binom{m}{u}\binom{n}{v}=\binom{m+n}{p}. \]


function
combinatorics unsolved
combinatorics

In each square of an $n\times{n}$ grid there is a lamp. If the lamp is touched it changes its state every lamp in the same row and every lamp in the same column (the one that are on are turned off and viceversa). At the begin, all the lamps are off. Show that  lways is possible, with an appropriated sequence of touches, that all the the lamps on the board end on and find, in function of $n$ the minimal number of touches that are necessary to turn on every lamp.

Ciao.

[b]Fact 1[/b] [i]n moves are necessary.[/i]
With less than n moves, there exist at least a row and a column without touched lamps. The lamp at the crossing is still off.

[b]Fact 2[/b] [i]If n is odd, n moves are sufficient.[/i]
Just touch all the lamps in a row, to lit them all.

[b]Fact 3.0[/b] [i]It is useless to touch a lamp twice.[/i]
Obvious.
[b]Fact 3.1[/b] [i]If there is a solution, then n^2 moves are sufficient.[/i]

From now on, WLOG, I suppose no lamp is touched twice. Changing the order of moves doesn't change the result, so if X is the set of subsets of lamps, I can define f: X --> X, s.t. given a subset A, f(A) is the set of lamps lit after touching exactly once the lamps in A.

X is finite. For the case of n even, I will prove f is injective by proving

[b]Fact 4[/b] [i]If n is even, f is surjective.[/i]
Infact, choose a lamp L. Touch exactly the lamps in the same column or row of L. The result is that you lit L, and no other lamps.
Now, choose a set of lamps A. You can lit exactly the lamps of A by repeating the procedure described for every lamp in A, so every configuration is attainable and f is surjective.

[b]Fact 5[/b] [i]If n is even, n^2 moves are necessary.[/i]
f is bijective, so there exists a unique solution. But touching all the lamps IS a solution. []

As a remark, in the case of n even, you can see that f(f(A)) = A, so there is a very simple way to calculate the minimal sequence of moves for every desired configuration.

M.

Generalization:

Consider an array of size $m\times{n}$. If both $m$, $n$ are odd then minimum number of steps = $min(m,n)$ obtained by touching each of the buttons along the side containing the $min(m,n)$ buttons.

If only one of the sides is odd say $m$, then minimum number of steps = $m$, obtained by touching each of the buttons along the side containing m buttons.

If both $m$ and $n$ are even, then one way to alter all lights to all-ON state is to touch each of the $m.n$ buttons EXACTLY ONCE sequentially going from one row/column to the next or in any other order.

We shall prove that minimum number of steps = $m.n$ when both $m$, $n$ are even.

To prove the above claim it suffices to prove that no button can be left untouched.

Assume that a button is left untouched, then this button will be turned to ON state only if there are a total of Odd number of touches along the row and column containing this button.

So either the row or column (say row) containing this button must have odd number of touches and the other (say column) must have an even number of touches. Let us label this row as $R$ and column as $C$.

Now, each button in $C$ needs to have Odd number of touches in total, so each row of the matrix obtained after excluding $C$ and $R$ must have an Odd number of touches.

This matrix has the dimensions of $m-1\times{n-1}$, so total number of touches in this matrix must be odd $-- (1)$

Now, each button in $R$ needs to have Odd number of touches in total, so each column of the matrix obtained after excluding $C$ and $R$ must have an even number of touches, so total number of touches in this matrix must be even $-- (2)$

From $(1)$ and $(2)$ we have a contradiction.


algebra
polynomial
inequalities
algebra unsolved

Find $ a,b,c,d,e,f\in\mathbb{Z}$ such that
$ \left|\frac{aR^2\plus{}bR\plus{}c}{dR^2\plus{}eR\plus{}f}\minus{}\sqrt[3]{2}\right|<\left|R\minus{}\sqrt[3]{2}\right|$ for all $ R\ge0\in\mathbb{Q}$

Make $ R \rightarrow \sqrt [3]{2}$! The RHS tends to $ 0$, so the LHS tends to $ 0$ too.

$ \frac {at^2 + bt + c}{dt^2 + et + f} = t$,

where $ t = \sqrt [3]{2}$. We conclude $ \frac {c}{d} = 2$ and $ a = b = e = f = 0$ (the minimal polynomial of $ t$ is $ x^3 - 2$). Then we get

$ | \frac {2}{R^2} - t | < |R - t|$

But this is not true when $ R \rightarrow 0$, so there are no solutions.

As feliz, we get
$ \frac{at^2\plus{}bt\plus{}c}{dt^2\plus{}et\plus{}f}\equal{}t$ with $ t\equal{}\sqrt[3]{2}$.
From here, we get $ a\equal{}e$, $ b\equal{}f$ and $ c\equal{}2d$.
It is easy to check $ d\ne0$. We define $ A\equal{}\frac ad$ and $ B\equal{}\frac bd$. $ A,B$ are rationals.
Then the inequality becomes
$ |\frac{(A\minus{}t)R\plus{}B\minus{}t^2}{R^2\plus{}AR\plus{}B}|<1$, or
$ (R^2\plus{}AR\plus{}B)^2\minus{}((A\minus{}t)R\plus{}B\minus{}t^2)^2>0$, or
$ (R^2\plus{}tR\plus{}t^2)(R^2\plus{}(2A\minus{}t)R\plus{}2B\minus{}t^2)>0$. or
$ R^2\plus{}(2A\minus{}t)R\plus{}2B\minus{}t^2>0$.
Hence, we have either
$ 2A<t$ and $ 8B\minus{}4t^2\ge(2A\minus{}t)^2$
or
$ 2A>t$ and $ 2B\minus{}t^2>0$.

The rest should be trivial now. Actually there are many solutions.


function
geometry
geometric transformation
reflection

[quote="Interval"]Graph each function using the techniques of shifting, compressing, stretching, and/or reflecting. Start with the graph of y = x^2 and show all stages.

Before showing me how to graph the functions below, I need a clear definition of the words:

*shifting
*compressing
*stretching
*reflecting

What do the words above mean in math?

QUESTIONS:

(1) h(x) = int(-x)

(2) h(x) = sqrt({-x}) - 2[/quote]

[hide]shifting is like moving the whole graph.  It can also be interpreted be shifting the x-y axes and leaving the graph still.
so if the graph is $y=x^{2}$ than $y-2=(x+5)^{2}$ would be $y=x^{2}$ but moved 5 units to the left and 2 units up.

compressing is the same as stretching just like multiplication is the same as division.  $y=3x^{2}$ is an example.

reflecting is like...umm...reflecting.  if you reflect a graph across the y-axis, then the graph would be $y=(-x)^{2}=x^{2}$. so basically replace the x's by -x.  if is reflected across the x-axis, change all the y's to -y's.[/hide]

btw, are you studying from a book?

Why does everyone ask the same question?

I am studying from a state exam book.

I am preparing for an upcoming state exam in June 2007.  I am trying to get state certified and need to pass the Content Specialty Test in math for grades 9-12.

Is this clear?


rotation
geometry
geometric transformation
combinatorics solved
combinatorics

[quote="phuchung"]We have a regular 100-gon. Color 10 vertices red and 10 other vertices blue. Prove that there exist 4 points, 2 of them are red and the others are blue, which the distance between the red points is equal to the distance between the blue points.[/quote]

Pleaase  explain the question .  :lol:

[hide]We rotate, say, red vertices for one step, there are 99 rotations. In any rotation, at most one blue vertex can be covered with red vertex, otherwise there would be two pairs of red and blue vertices at the same distance. After all the rotations, each pair of blue and red vertex must be matched in at least one rotation. There are 10x10 = 100 such pairs, and 99 rotations, so there must be a rotation where two blue vertices are covered by two red vertices#[/hide]


calculus
derivative
logarithms
function
calculus computations

What rule(s) would you use to differentiate this?

$3^{2\sqrt{x}}$

I found that in my textbook too, but now this is even more weird. I don't know if i type it wrong on my calculator, but if we say $x=2$:

$a^{2\sqrt{2}}=22.36$

$e^{2\cdot ln(3)}=9$

What i noticed before i started this thread was that if i wanted to find the inclination in $x=$ i got different results when i did the calculation by hand compared with what my calculator told me.

What am i doing wrong here? Don't worry, won't say the book and you are lying :D

$y=3^{2\sqrt{x}}$

$\ln y = 2\sqrt{x}\ln 3$

$\frac{dy}{dx} \cdot \frac{1}{y} = \frac{\ln 3}{\sqrt{x}}$

$\frac{dy}{dx} = \frac{3^{2\sqrt{x}}\ln 3}{\sqrt{x}}$

Also, with the identity $a^x = e^{x\ln a}$, say $a=3$ and $x=2$, then we have $3^2 = e^{2\ln 3} = 9$. (Do you see your mistake?)

You could also differentiate using:

$3^{2\sqrt{x}} = e^{2\sqrt{x}\ln 3}$

I have to say i just do don't get you. The way of doing these things are different in Europe (or Denmark at least) compared with the US. Could you do this problem using the ' everytime you differentiate? If you know that system. I have never seen $y$ get differentiated. In Denmark, this is the system we use:

$(a^x)'=ln(a)\cdot a^x$

But let's say we differentiate the left side with the $y$.

Actually i think i got it, but it is just that it doesn't make much sence when i write it down. Can you tell me what i do wrong here:

$h(x)=3^{2\sqrt{x}}$

$ln'(h(x))=(2\sqrt{x}\cdot \ln(3))'$

$\frac{1}{h(x)}=\frac{\ln(3)}{\sqrt{x}}$

$1=\frac{3^{2\sqrt{x}}\ln 3}{\sqrt{x}}$

I guess you can see what i am trying to tell you. Though this method works it gives a weird solution :(

Okay, so

$y=3^{2\sqrt{x}}$

Taking logarithms

$\ln y = \sqrt{x}2\ln3$

Differentiate both sides with respect to $x$. Remember to use the [b]chain rule on the left[/b]!!

$\frac{1}{y}\frac{dy}{dx} = \frac{\ln3}{\sqrt{x}}$

(can you now see your mistake?)

Fix and fiddle to get

$y'(x)=\frac{dy}{dx}=\frac{\ln3}{\sqrt{x}}\cdot y = \frac{3^{2\sqrt{x}}\ln 3}{\sqrt{x}}$

The other way is what I suggested eariler

$y=3^{2\sqrt{x}}=e^{\sqrt{x}2\ln3}$

Differentiating we have to apply the chain rule a few times

$\frac{dy}{dx}=e^{\sqrt{x}2\ln3}\cdot \frac{d}{dx}\left(\sqrt{x}2\ln3\right)=3^{2\sqrt{x}}\frac{\ln3}{\sqrt{x}}$

Same answer :)

Still using $\frac{dy}{dx}$. Confusing system in my universe :) But i managed afterall using marks ( ' ) only.

What confused me was probably this rule:

$(a^x)'=k\cdot a^x$

But that one only goes for functions where $x$ stands alone and not like $a^\sqrt{x}$. Then we have to get help from this rule:

$a^x=e^{x\cdot \ln(a)}$

Or more like this:

$(a^x)'=a^x\cdot (x)'$

And $x$ must be something different from $x^1$ or it must involve plus, minus, multiply or divide, though not plus or minus with $0$ nor multiply or divide with $1$.

Is this how it is?


function
algebra
polynomial
arithmetic sequence
Pascal\u0027s Triangle

[hide]$\left[\begin{tabular}{ccc|c}1 & 1 & 1 & 0 \\ 4 & 2 & 1 & 5 \\ 9 & 3 & 1 & 13\end{tabular}\right]\implies \left(a,b,c\right)=\boxed{\left(\frac{3}{2},\frac{1}{2},-2\right)}$[/hide]

Oh, so I guess the easiest way to do it would be by solving a set of equations.  Okay.

[hide="Generalization"]
Suppose a polynomial $P(x)$ of known degree $n$ takes the values

$P(x_0) = p_0$
$P(x_1) = p_1$
...
$P(x_n) = p_n$

Then $P(x)$ is uniquely determined. [/hide]

[hide]Plugging in those values, we get
$a+b+c=0$
$4a+2b+c=5$
$9a+3b+c=13$
and we just need to solve for a,b, and c.

$a=3/2$ $b=1/2$, and $c=-2$. $(a,b,c)=(3/2,1/2,-2)$.[/hide]

[quote="cincodemayo5590"]how else would you have done it?[/quote]

the only faster way that i can see is to use the fact that 1,2,3 is an arithmetic sequence, so

$\frac{ f(1)+f(3) }{2}-f(2)=a=\frac{3}{2}$ since other 2 terms form an arithmetic progression, and the coefficient of a is $\frac{(x-1)^2+(x+1)^2}{2}-x^2=1$

but i suppose that you can argue that it is just easier to solve it outright, this is only slightly faster if you recognize it

I was going to ask what Lagrange Interpolation was, but [url=http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html]MathWorld[/url] helped me figure that out. This is actually pretty cool :lol:

[quote="lingomaniac88"]A function $f$ can be defined as follows:
$f(1)=0$
$f(2)=5$
$f(3)=13$
The function $f$ can be summarized as $f(x)=ax^2+bx+c$.  Compute the ordered triple $(a,b,c)$.[/quote]

Another way is to note that 1 is a root of $f(x)$.

thus: $f(x) = g(x)(x-1)$ where g(x) is a binomial.

plugging 2 and 3 in for x we get:

$5 = g(2)(1)$ and $13 = g(3)(2)$
$5 = g(2)$ and $\frac{13}{2} = g(3)$

Since $g(x) = mx + B$ (not the b in $f(x)$), we can solve it with two points

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{13}{2}-5}{1} = \frac 32$

$5 = \frac 32(2) + B$
$5 = 3 + B$
$2 = B$

Thus, we have $f(x) = g(x)(x-1) = (\frac{3}{2}x + 2)(x-1) = \frac{3}{2}x^2 + \frac 12x - 2$

So the ordered triple $(a, b, c)$ is $(\frac{3}{2}, \frac{1}{2}, -2)$.

simplifying the first three equations, we have:
$a + b + c = 0$
$4a + 2b + c = 5$
$9a + 3b + c = 13$
Subtracting the first from the second and the second from the third:
$3a + b = 5$
$5a + b = 8$
Subtracting the first from the second:
$2a = 3$
$a = \frac{3}{2}$
Plugging in $a$:
$3(\frac{3}{2}) + b = 5$
$b = \frac{1}{2}$
then substituting into the very first equation:
$\frac{3}{2} + \frac{1}{2} + c = 0$
$c = -2$

So our ordered triple is:
$\boxed{\frac{3}{2}, \frac{1}{2}, -2}$

[quote="kunny"]You can use Laglange's interpolation.[/quote]Hi kunny, why not use a simplified version of Lagrange's, aka Newton's?

From Newton's we can tabulate a 'mini-reversed Pascal triangle of subtraction'

Since we have values for $x=1,2,3$, make a small reversed Pascal triangle with [u]right minus left[/u] to get bottom entries:

\[ \begin{array}{ccccc} {\bf 0}&&5&&13\\ &{\bf 5}&&8&\\ &&{\bf 3}&& \end{array}  \]

Pick the first slanted diagonal from left to right (diagonal like the backslash \ ), they are in bold font

Polynomial is

$p(x)=\frac0{0!}+\frac5{1!}(x-1)+\frac3{2!}(x-1)(x-2)$


real analysis
function
advanced fields
advanced fields unsolved

I'd like to prove that $ f*f\equal{}f$, $ f\in L^1$ can only have solution f=0. Do I begin by taking fourier transforms and letting $ \xi \to \infty$? I'm stuck, any help would be great!

You need an $ L^1$ of what in there. After all, there are a number of perfectly valid solutions of $ f*f\equal{}f$ in $ L^1(\mathbb{T}).$

So I'll assume you're talking about $ L^1(\mathbb{R}),$ or perhaps $ L^1(\mathbb{R}^n).$

You're right to take Fourier transforms. What does that tell you about the possible values of $ \widehat{f}(\xi),$ pointwise?

Now, here's the theorem you need (which contains the Riemann-Lebesgue Lemma): if $ f\in L^1(\mathbb{R}),$ then $ \widehat{f}\in C_0(\mathbb{R}).$ That is, $ \widehat{f}(\xi)$ is continuous and tends to zero as $ |\xi|\to\infty.$ Note that both parts - both the continuity and the limit at infinity - matter here.

Why is $ \mathbb{T}$ different? The same theorem is true there, but "continuous" is a completely toothless requirement for a function (like $ \widehat{f}$) on $ \mathbb{Z}.$

[quote="Kent Merryfield"]You need an $ L^1$ of what in there. After all, there are a number of perfectly valid solutions of $ f*f \equal{} f$ in $ L^1(\mathbb{T}).$

So I'll assume you're talking about $ L^1(\mathbb{R}),$ or perhaps $ L^1(\mathbb{R}^n).$

You're right to take Fourier transforms. What does that tell you about the possible values of $ \widehat{f}(\xi),$ pointwise?

Now, here's the theorem you need (which contains the Riemann-Lebesgue Lemma): if $ f\in L^1(\mathbb{R}),$ then $ \widehat{f}\in C_0(\mathbb{R}).$ That is, $ \widehat{f}(\xi)$ is continuous and tends to zero as $ |\xi|\to\infty.$ Note that both parts - both the continuity and the limit at infinity - matter here.

Why is $ \mathbb{T}$ different? The same theorem is true there, but "continuous" is a completely toothless requirement for a function (like $ \widehat{f}$) on $ \mathbb{Z}.$[/quote]

So pointwise, the possible values of  $ \widehat{f}(\xi)$ would be 0 or 1, if I'm not mistaken. But, using the L-R lemma, it must mean that $ \hat{f}$ has to be 0, and so we conclude the only possible value for $ f \in L^1(\mathbb{R})$ (I shouldn't have been so sloppy, it is indeed $ L^1(\mathbb{R})$) is 0?

That, and you do need to say that a continuous function on $ \mathbb{R}$ (or any connected set) whose values lie in $ \{0,1\}$ must in fact be constant.

(And the reason $ \mathbb{T}$ is different is that $ \widehat{\mathbb{T}}\equal{}\mathbb{Z}$ is not connected.)

[quote="Kent Merryfield"]That, and you do need to say that a continuous function on $ \mathbb{R}$ (or any connected set) whose values lie in $ \{0,1\}$ must in fact be constant.

(And the reason $ \mathbb{T}$ is different is that $ \widehat{\mathbb{T}} \equal{} \mathbb{Z}$ is not connected.)[/quote]

Ah yes, the theorem that a bounded continuous function is constant. So, in fact in $ L^1(\mathbb{R})$ the fourier transform is constant - namely it is 0 as $ \xi \to \infty$ - and therefore f must be 0?


When $ N$ is divided by 8 the result is greater than 5. What is the least possible integer value of $ N$?


analytic geometry
geometry
3D geometry
octahedron
geometry unsolved

A convex polyhedron with $ n$ faces is given such that coordinates of all its vertices are integers and there are no other integer points inside the polyhedron or at its faces. Find the maximal possible $ n.$

[hide="an idea"]
Clearly the number of vertices is at most $ 8$, because otherwise there are two vertices with all coordinates having the same parity, so then the midpoint of the segment they define is a lattice point.  The example of $ n\equal{}8$, an octahedron, has $ 6$ vertices.  I don't know if there are any larger examples...and I have to go now...[/hide]


function
search
algebra unsolved
algebra

For which integers $ k$ does there exist a function $ f: \mathbb{N} \rightarrow \mathbb{Z}$ such that:

$ (a)$ $ f(1995)\equal{}1996,$ and
$ (b)$ $ f(xy)\equal{}f(x)\plus{}f(y)\plus{}kf(\gcd(x,y))$ for all $ x,y \in \mathbb{N}?$

[quote="moldovan"]For which integers $ k$ does there exist a function $ f: \mathbb{N} \rightarrow \mathbb{Z}$ such that:

$ (a)$ $ f(1995) \equal{} 1996,$ and
$ (b)$ $ f(xy) \equal{} f(x) \plus{} f(y) \plus{} kf(\gcd(x,y))$ for all $ x,y \in \mathbb{N}?$[/quote]

see [url]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=943898879&t=277446[/url]


geometry
parallelogram
geometry proposed

Let a parallelogramm $ABCD$. A circle with diameter $AB$ meets side $AD$ at $E$ and side $CD$ at$P,F$ such that points $D,P,F,C$ are in this order.Segment $AF$ meets $BE$ at $H$.A line through $E$ parallel to $AB$ meets $BF$ at $G$.Line $DH$ meets line $CB$ at $I$.Prove that $FI \perp GH$.

[b] Note[/b]

 [i]I have this problem with a hint in a bad paper, but I have not noted there the source. It seems IMO level, probably some ISL.[/i]

Stergiu, in the my opinion, the point $C$ is ... moreover ! The your problem made dizzy me.

[color=darkred][b]An equivalent enunciation.[/b] Let $ABC$ be a triangle with the orthocentre $H$. Define the points :

$E\in BH\cap AC$ $\mathrm{\ \ ;\ \ }$ $F\in AH\cap BC\mathrm{\ \ ;\ \ }D\in AC\ ,\ FD\parallel AB$ $\mathrm{\ \ ;\ \ }$ $G\in BC\ ,\ EG\parallel AB$ $\mathrm{\ \ ;\ \ }$ $I\in DH\ ,\ BI\parallel AC$. Prove that $FI\perp HG\ .$[/color]

Virgil sorry !
I suspected that the problem with cause troumles.As I have written, I have not the source, so I can not see again the enunciation.It seems that I saw the figure in the solution.
 According to my figure the circle meets side $DC$ at two points, so PLEASE read the problem again. I have edided a little at this point.

[hide] In the solution, the solver drops $IK \perp AF$ , proves that $\frac{KF}{KJ}= \frac{FG}{FB}$ ,whhere $J$ is the point that line $AF$ meets line $BC$,  $\triangle IJK$ similar to   $\triangle HBF$ , and finally using the relations we take that   $\triangle KFI$ is similar to   $\triangle GFH$ ect.[/hide]

 I simply wanted to see if we can have an easier solution.
[u]   Babis[/u]


Hello,
    This is a fairly good result I stumbled upon in the internet and set out to prove and finally presented a paper at Conference of Association of Mathematics Teachers of India (AMTI) 2003. It was also published in part in their journal Junior Mathematician. Actually the story is quite long. This was first discovered by one Prof Marion Walter of Oregon State University when she was experimenting with Geometer's Sketch Pad (You see she uses computers, while I absolutely avoid them in my Mathematics).
The result is :
In any triangle trisect each side and join the points of trisection to the opposite vertex. A small hexagon is obtained in the middle. The ratio of the area of the hexagon to the triangle is surprisingly a constant 1:10. I leave the details of the proof to readers. Of course, it is very easy to show that the hexagon can never be regular and so on. And a hint, the proof involves nothing but the idea of similar triangles.
I made a few extensions on similar lines and even contacted Prof Marion Walter. Now she is very old in her 80s and finds it difficult to keep replying. Nevertheless she replied to my mails quite a few times. Some extensions:
Obtain a similar ratio for n-sections of the sides. And by the way, one person at the conference made a passing remark that one might do it for angular n-sectors also. But, all my attempts proved futile  and now I am more or less convinced that there is no such result for angular sectors.
Any comments would be highly welome.
Sathej

[quote="Sathej"]
In any triangle trisect each side and join the points of trisection to the opposite vertex. A small hexagon is obtained in the middle. The ratio of the area of the hexagon to the triangle is surprisingly a constant 1:10.[/quote]

This can be easily generalised. The general version was on one of the training sheets for the South African IMO team this year.


geometry
AMC
AIME
AMC 12
USA(J)MO
USAMO
\/closed

I know there isn't going to be a intermediate geo book... but there is a course.

Will AoPS ever publish a book at that level?

Any suggestions for a book of intermediate geo?

Maybe, but Ruscyk said before that it will be a long time (at least till 2011)

It will be a long while before I write another geometry book.  In our bookstore, in the Olympiad section, there's a book from the UKMT that would probably be a good fit for you.

What is the difficulty of the problems in that book? (Medium AIME, Hard AMC12, Easy USAMO, etc.)

[quote="rrusczyk"]It will be a long while before I write another geometry book.  [/quote]

What do you mean by long? A few years?

[quote="rrusczyk"]All three of those.  It covers a pretty wide range with its examples.[/quote]

So, would I be able to answer most of the geometry AIME problems if I get [i]Plane Euclidean Geometry[/i] and thoroughly go through it?

I don't know.  I haven't worked with the book enough to know how well it correlates to AIME.  It would very likely expand your geometric horizons.

isabella--yes, at least a few years.

Is there any chance of an Intermediate Geometry book being written within the next 3 years?

No, we've never had an Intermediate Geometry course.

[quote="ssilwa"]There used to be an intermediate course?[/quote]
There is an [url=http://www.artofproblemsolving.com/School/courseinfo.php?course_id=olympiad:geometry]Olympiad Geometry[/url] course, if that is what you were referring to.

Is it fine to take Olympiad Geometry directly after taking a geometry course in school or AoPS Introduction to Geometry?

[quote="rrusczyk"]That really depends on your problem-solving experience.  Most students need a little more experience with intermediate-level problem solving (not necessarily in geometry) before taking the leap in difficulty from an honors geometry course (or our Intro Geometry course) to our Olympiad Geometry course.[/quote]
What difficulty of problems does Olympiad Geometry deal with (eg. easy/intermediate/hard USAMO). If you're familiar with the Canadian MO or Asia-Pacific MO, could you tell me how hard the problems in the course are in comparison to those olympiads?

As well, the class covers a couple of interesting techniques like inversion. Do these tend to come up often on Olympiads?

It covers a fairly wide range, from beginning olympiads to tough IMO problems.

There are problems on which inversion can be helpful, but it's rare that you absolutely need such fancy techniques to solve olympiad problems.

Sorry to bump this thread but is it still a possibility that the Intermediate Geometry book will be written?

We have no current plans to write an Intermediate Geometry book.

Please write the book asap ....  I beg you! x3 All of us here are dying for it and AoPS is the best! X3

As other countless other users have mentioned, an intermediate geometry book would be highly appreciated! It is too bad that aops isn't going to write one... (especially since it is referred to in the intro book)


hello mathlinkers this is my first post,it is a problem from IMO longlist.
We are given a circle K and a point P lying on a line g.
Construct a circle that passes through P and touches K and g.

drawing a perpendicular to g trough P.Drawing a circle with center on his perpendicular and tangent to g at P with any radii.This circle cut the K at two points(A,B).AB cut g at C.The circle trough C and the center of K cut K at points Q and S.Center of K and Q cut the perpendicular at P in the desired center of the circle.


geometry
circumcircle
cyclic quadrilateral
geometry proposed

Let $ ABCD$ be a cyclic quadrilateral with circumcircle $ \left(O,R\right)$, and let $ P$ be the point of intersection of the lines $ AC$ and $ BD$. The line $ OP$ cuts the third diagonal of the quadrilateral $ ABCD$ at a point $ Q$. Prove that $ R^{2} \equal{} OP\cdot OQ$.

[i]Remark.[/i] The [i]third diagonal[/i] of the quadrilateral $ ABCD$ is defined as the line $ EF$, where $ E$ is the point of intersection of the lines $ AB$ and $ CD$, and $ F$ is the point of intersection of the lines $ BC$ and $ DA$.

Sorry man, but since it's been unsolved here for quite some time, I'll go ahead and post the projective solution :):

The polar of $P$ is $EF$ ($E=AB\cap CD,\ F=AD\cap BC$), which passes through $Q$, so the polar of $Q$ passes through $P$. This polar is also perpendicular to $OP=OQ$, so if $T$ is one of the points where the polar of $Q$ cuts the circle, then in the right triangle $OTQ$ ($\angle OTQ=90^{\circ}$), $TP$ is the altitude, so $OP\cdot OQ=OT^2=R^2$.

I know a synthetic proof is a lot harder to find, and in the end it might be quite tricky and nice, but I don't see why one would want to look for one :).

It is only that I am not an expert of projective geometry.
Maybe someone is interested in the next proof:
We know that $ O$ is the orthocenter of $ EPF$ (the proof without poles and polars in pdf file) so $ P$ is the orthocenter of $ EOF$, draw $ EK$ and $ FJ$ altitudes in $ EOF$ (See Fig.1). Produced $ EK$ cut $ AD$ at $ G$,then $ A$ and $ D$ are harmonic conjugates wrt $ F$ and $ G$, whith $ (O_2)$ circumcircle of $ FKG$ right
triangle($ O_2$ is midpoint of diameter $ FG$), then  $ (O)$ and $ (O_2)$ are orthogonals. Now: $ OK.OF\equal{}R^2$, but also $ OK.OF\equal{}OP.OQ$ hence $ OP.OQ\equal{}R^2$. QED.
In other words $ (O)$ and $ (O_1)$ are orthogonals, where $ (O_1)$ is the circle with diameter $ EF$.


Να βρεθεί αν η παρακάτω σειρά συγκλίνει η όχι:

$\sum^{\infty}_{n=1}(-1)^{n+1}\left(\frac{1+\frac{1}{2}+...+\frac{1}{n}}{n}\right)$

Έχω αποδείξει ότι δεν συγκλινεί απόλυτα αλλά δεν μπορώ να βρω κάτι με το (-1)... :wink:

na me sympathate kiola gia ti noobia poy me dernei alla ti tha pei sygglinei?

Ότι πλησιάζει σε κάποιο όριο. Δες εδώ:

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

Vasika tora pu to ksanaskeftome.... :wallbash:  :wallbash_red: akrivos leibniz ine....
ine mia enalasusa sira....
+i akolu8ia ton apoliton timon ine f8inusa +mideniki..
mideniki gt otan to n tini sto apiro,o paranomastis tini poli pio grigora apo ton ari8miti..
+f8inusa gt
$\frac{1+\frac{1}{2}+...+\frac{1}{n}+\frac{1}{n+1}}{n+1}<\frac{1+\frac{1}{2}+...+\frac{1}{n}}{n}$
 diladi$\frac{n}{n+1}<1+\frac{1}{2}+...+\frac{1}{n}$ pu isxii..

 :D

Ωραία Ράλλη.... :)  και ωραί άσκηση bouzoukman....
Ράλλη αν θες να γράψεις το γνωστό , π ,  γράψε  \pi
kai ιδού     $\pi$

Άμα σου είναι εύκολο RAKAR ή κάποιος άλλος να δώσει μια ολοκληρωμένη λύση πως βγάζουμε ότι συγκλίνει..
Ή πως το $lim_{n\rightarrow\infty}\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}=0$ οπότε μπορούμε να χρησιμοποιήσουμε το κριτήριο του Leibniz..

[quote="kostas"][quote="bouzoukman"]Ή πως το $lim_{n\rightarrow\infty}\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}=0$ [/quote]
απο θεωρημα Stolz- Cesaro
http://planetmath.org/encyclopedia/StolzCesaroTheorem.html[/quote]
sosta kosta i apo to 8eorima stolz i pio apla epidi to
$lim_{n\rightarrow\infty}1+\frac{1}{2}+\cdots+\frac{1}{n}$
ine iso me $Inn$
to zitumeno orio ine iso me to $lim_{n\rightarrow\infty}\frac{Inn}{n}$ (1)
Opote metavenontas apo tin diakriti sti sinexi metavliti $x$ me $f(x)=\frac{Inx}{x}$ (2)kai pernontas L'Hopital vlepume oti to orio ine $0$ kai epidi i (1) ine periorismos tis (2),8a exi +ayti to idio orio..diladi to $0$... :wink:

Thanks siluan gia to ...$\pi$..

 :D

[quote][b]Rakar wrote[/b]:sosta kosta i apo to 8eorima stolz i pio apla epidi to
$lim_{n\rightarrow\infty}1+\frac{1}{2}+\cdots+\frac{1}{n}$
ine iso me $Inn$[/quote]
Xarin akribeias, den einai akribos isa iparxei mia diafora metaxi tous kathos
teinoun sto $\infty$

Sigekrimena isxiei $\gamma=\lim_{n\to \infty}\left (\sum_{i=1}^{n}\frac{1}{n}-\ln n\right )$
opou gamma i stathera tou Euler.

Kapoios mporei na deixei xoris sxetiki diskolia oti to orio $\lim_{n\to \infty}\left (\sum_{i=1}^{n}\frac{1}{n}-\ln \left (n+1\right ) \right )$ iparxei
deixnontas oti i akolouthia $a_{n}=\sum_{i=1}^{n}\frac{1}{n}-\ln \left (n+1\right )$ einai anw
fragmeni kai gnisios auxousa, epomenos siglinei. O arithmos
aytos einai h stahera $\gamma$

mia efarmogi tis am-gm

gia $n>1$ isxiei $n \left (\sqrt[n]{n+1}-1\right )<1+\frac{1}{2}+\cdots+\frac{1}{n}< n \left (1-\frac{1}{\sqrt[n]{n+1}}+\frac{1}{n+1}\right )$


USAMO
geometry
combinatorics
combinatorial geometry
Soddy

A square grid on the Euclidean plane consists of all points $(m,n)$, where $m$ and $n$ are integers. Is it possible to cover all grid points by an infinite family of discs with non-overlapping interiors if each disc in the family has radius at least $5$?

Here's a proof of why you can always fit a circle of radius $ \frac{\sqrt 2}{2}$, inside the region formed by $ 3$ circles, each of which has radius at least $ 5$.

1: First we note it's sufficient to prove it in the case where all the circles are tangent. Consider three circles $ C_{1}, C_{2}, C_{3}$, with centres $ O_{1}, O_{2}, O_{3}$.  Let $ D$ be the inner Soddy circle of $ O_{1} O_{2} O_{3}$. Clearly $ D$ lies strictly inside the triangle $ O_{1} O_{2} O_{3}$, and $ D$ is the circle externally tangent to three other circles, each of which has radius at least $ 5$. Thus if we can prove the result for these $ 3$ externally tangent circles, that it's sufficient.

2: My proof for $ 3$ externally tangent circles is just brute force. If the radii of the $ 3$ circles are $ a, b, c$, then by Soddy circle results, the radius of the Inner Soddy circle, which lies inside the region formed by these $ 3$ circles, is  $ f(a,b,c) \equal{} \frac {abc} {ab\plus{}bc\plus{}ac \plus{} 2 \sqrt{abc(a\plus{}b\plus{}c)}}$. By differentiation, it's clear that $ f(a,b,c)$ is increasing in each of it's variables, so $ f(a,b,c) \geq f(5,5,5) > \frac {1}{\sqrt{2}}$, so we can find a circle inside that has radius at least $ \frac {1}{\sqrt{2}}$, and contains a grid point, as required.

3: It's intuitely obvious that this solves the problem, but here's a formal argument: Label the circles in our infinite family $ C_{i}$. Consider any circle in this square grid, $ C_{1}$. Draw a circle tangent to $ C_{1}$, centre $ O_{1}$, with radius $ \frac {1}{\sqrt{2}}$. This circle contains a grid point, so another circle $ C_{2}$, centre $ O_{2}$ intersects this circle. Now consider any circle $ C_{n}$. As described in Step 1, we can find a circle of radius at least $ \frac {1}{\sqrt {2}}$ 'inside' these three circles. So another circle $ C_{m}$ must intersect this circle. Now repeat the argument for $ C_{1}, C_{2}$ and $ C_{m}$. Continue in this fashion, and after finitely many steps we will reach a disc of radius at least $ \frac {1}{\sqrt {2}}$, that lies outside our infinite family, which is a contradiction.

[quote="me@home"]Alternately, the smallest possible area of the gap between three circles is \[ 25\left(\sqrt{3}\minus{}\frac{\pi}{2}\right)\equal{}4.031...>4\] hence by Minkowski's theorem it contains a lattice point.[/quote]

I hate to bump an old topic, but Minkowski's theorem only applies to convex sets symmetric to the origin, so I don't know if this argument is valid.

[quote=t0rajir0u]No.  Between any three non-overlapping circles of radius greater than $\frac{ 2 \sqrt{6}+3 \sqrt{2}}{2}= 4.57...$ we can insert a circle of radius greater than $\frac{1}{ \sqrt{2}}$ in which there must be a lattice point.[/quote]

Sorry, but how do you know that the circle of radius greater than $\frac{1}{ \sqrt{2}}$ which touches three circles doesn't intersect/touch a fourth circle? Thanks!

Does anyone know why this can't happen? Thank you very much!

[hide="my solution"]
This problem is a very weird problem that seems very obvious but has many technicalities that need to be addressed, which I will hopefully do rigorously.

I claim that any space has enough room to fit a circle of radius $>\dfrac{\sqrt{2}}{2}$. This is enough to solve the problem because any circle with radius $>\dfrac{\sqrt{2}}{2}$ can contain a square with sides parallel to the axis and with side length $>1$; this means that this square covers at least one horizontal lattice grid line and at least one vertical lattice grid line, implying it covers at least one lattice point, which means that the circle also does so.

Now for the proof. Take any space, and put a circle of radius $0$ in it. Now continuously blow the circle bigger, until it is tangent with at least three circles (but they must be tangent, NOT intersected). These three circles are thus the circles that determine the limit to how big a circle in the given space can be, so we just need to focus on these circles only.

Note that we may assume these three circles are tangent, because if not we can blow them up until they are tangent and the remaining white space will be strictly smaller. If any of these circles have radius $>5$, it is easy to see that reducing the radius of that circle until it is $=5$ will make the white space strictly smaller; thus, the point at which the white space is smallest is when all three circles have radius $=5$.

Now what we just need to do is calculate the radius of the largest circle that can fit in this space, which can be easily found to be $\dfrac{10\sqrt{3}}{3}-5 > 0.77$ using the approximation $\sqrt{3}\ge 1.732$ which can be confirmed by squaring. However, note that $0.77 > 0.71 > \dfrac{\sqrt{2}}{2}$ using the approximation $1.42 > \sqrt{2}$ (which again can be confirmed by squaring) so this circle is bigger than the smallest necessary size, and so we're done.
[/hide]

EDIT: I think this solution was not rigorous enough :(

[quote=me@home]Alternately, the smallest possible area of the gap between three circles is \[ 25\left(\sqrt{3}\minus{}\frac{\pi}{2}\right)\equal{}4.031...>4\] hence by [b]Minkowski's theorem[/b] it contains a lattice point.[/quote]

[quote=Wikipedia]In mathematics, Minkowski's theorem is the statement that any convex set in $\mathbb{R}^n$ which is symmetric with respect to the origin and with volume greater than $2^n d(L)$ contains a non-zero lattice point. The theorem was proved by Hermann Minkowski in 1889 and became the foundation of the branch of number theory called the geometry of numbers.[/quote]

Is the region convex and symmetric? (If so, then everything I learned in Mathcounts about arcs and "weird areas" is false :P)

Anyway,
[hide=Solution]
Lemma. Given three non-overlapping disks in the plane, there exists a fourth circle externally tangent to all three disks, with radius at least $\frac{\sqrt{2}}{2}$.

Proof. Let the three disks have radii $r_1, r_2, r_3$, and let them be centered at points $1, 2, 3$. Let the fourth circle have radius $r_4$, and let it be centered at point $4$. (This point must exist because it is the intersection of two hyperbolas, one of them with foci at points $1$ and $2$, and the other with foci at points $1$ and $3$.)

Without loss of generality, let $\angle 134 \le 120^\circ$. Then by Law of Cosines on triangle 134, we have
$$(r_1 + r_3)^2 \le \overline{13}^2 = (r_1 + r)^2 + (r_3 + r)^2 - 2(r_1 + r)(r_3 + r) \cos \angle 134 \le (r_1 + r)^2 + (r_3 + r)^2 + (r_1 + r)(r_3 + r).$$Simplify and rearrange:
$$3r^2 + 3r (r_1 + r_3) \ge r_1 r_3$$$$3r \cdot \left( \frac{r}{r_1 r_3} + \frac{1}{r_1} + \frac{1}{r_3} \right) \ge 1.$$If $r \le \frac{\sqrt{2}}{2}$, then since $r_1, r_3 \ge 5$, we have
$$LHS \le \frac{3 \sqrt{2}}{2} \cdot \left( \frac{\sqrt{2}}{50} + \frac{2}{5} \right) = \frac{3}{50} + \frac{3 \sqrt{2}}{5}$$$$= \frac{3 + 30 \sqrt{2}}{50} < \frac{3 + 47}{50} = 1 = RHS,$$contradiction. Thus in fact $r_4 > \frac{\sqrt{2}}{2}$, as desired.

Now suppose a tiling by discs does exist. Choose a point $P$ not in any circle (this exists because circles of radius $\ge 5$ cannot tesselate the plane) and consider the three circles with least distance to $P$, where distance is measured from the closest point to $P$ on the circle. By Lemma, there exists a circle externally tangent to all three circles with radius at least $\frac{\sqrt{2}}{2}$. We can inscribe a unit square in this new circle with sides parallel to the coordinate axes. It must contain at least one grid point, which is not covered by any other circle, contradiction. Thus a valid tiling by discs does not exist.
[/hide]

Draw a small circle whose interior is disjoint from all the disks. Now, continuously expand and translate this circle until its interior has to intersect some disk. This circle is tangent now two three disks. Suppose the centers of these disks are at $A,B,C$, and that they have radii $a,b,c$. Also suppose that the circle is centered at $O$ and it has radius $r$. WLOG, suppose $\angle AOB\le 120^\circ$ (if not, permute the labels of the disks until its true). We then see that $AB^2=OA^2+OB^2-2\cdot OA\cdot OB\cos\angle AOB$. Note that $\cos\angle AOB\ge -1/2$, and $AB\ge a+b$, so
\[(r+a)^2+(r+b)^2-2(r+a)(r+b)(-1/2)\ge (a+b)^2,\]
or
\[3r^2\ge ab-3ra-3rb,\]
or
\[12r^2\ge (a-3r)(b-3r).\]
Thus, $12r^2\ge (5-3r)^2$, so in particular $r>\sqrt{2}/2$. Thus, a unit square can be inscribed in the circle, which means a lattice point is inside the circle, so its not covered by the disks. Thus the claimed situation can't be achieved.

No. Suppose it was possible. The main idea is to show that we can get a circle with radius at least $1/\sqrt2$ that does not intersect any disk; this circle then must contain a $1\times 1$ square; which contains a lattice point.

Start with a point $O$ and a red circle centered at $O$ that doesn't intersect any circle.
[list]
[*] Take a homothety at $O$ till the new circle (color it blue) is tangent to some disk $\odot(O_1)$ at a point $P$.      [*] Take a homothety at $P$ till the new circle (color it green) is also tangent to some disk $\odot(O_2)$.       [*] Expand the green circle while keeping it tangent to $\odot(O_1)$ and $\odot(O_2)$ till it is tangent to a disk $\odot(O_3)$. [/list]
[asy] size(10cm);

pair O_1 = (2,4); pair O_2 = (-5,-5); pair O_3 = (10,-3);

draw(circle(O_1,2.5)); draw(circle(O_2,3.5)); draw(circle(O_3,2));

dot(O_1); dot(O_2); dot(O_3);

dot((2,1.5)); label("$P$",(2,1.5),N); //dot((-1.8,-3.5));

label("$O_1$",O_1,N); label("$O_2$",O_2,S); label("$O_3$",O_3,S);

draw(circle((2,0),0.5),lightred); draw(circle((2,0),1.5),lightblue); draw(circle((2,-2.41),3.91),green+linewidth(0.15mm)); draw(circumcircle((2,1.5),(-1.63,-3.5),(8,-3)), darkgreen+linewidth(0.1mm)); [/asy]
[color=#00f][b]Claim:[/color][/b]  The final circle $\omega$ has radius at least $1/\sqrt2$.

[color=#00f][i]Proof:[/i][/color]  At each vertex of $\triangle O_1O_2O_3$, there is a circle at $O_i$ of radius $r_i$. If the three circles are not pairwise tangent, then you can increase one of the radii a bit, which decreases the radius of $\omega$. So WLOG assume the $\odot(O_i)$ are pairwise tangent.

WLOG $\angle O_1OO_2 \le 120^\circ$. By LoC in $\triangle OO_1O_2$, we have \begin{align*}     &-\frac12 \le \cos \angle O_1OO_2 = \frac{(r+r_1)^2+(r+r_2)^2-(r_1+r_2)^2}{2(r+r_1)(r+r_2)} \\     \implies& -(r+r_1)(r+r_2) \le (r+r_1)^2+(r+r_2)^2-(r_1+r_2)^2 \\     \implies& 3r^2\ge r_1r_2-3r(r_1+r_2) = (r_1-3r)(r_2-3r) - 9r^2 \ge (5-3r)^2-9r^2 \\     \implies& 12r^2 \ge (5-3r)^2 \\     \implies& r \ge \tfrac13(10\sqrt3-15) > 1/\sqrt2. \end{align*} This proves the claim.  $\blacksquare$

[hide=Motivation]The problem seems very unapproachable at first. To show a lattice point exists, it suffices to show a $1\times 1$ square exists. The geometry of a square doesn't work well with all the circles, so instead I tried to show a circle with radius $1/\sqrt2$ exists. The number 5 can be understood by taking all equi-radial and tiled out circles so the centers make equaliteral triangles; a quick calculation confirms that 5 is the minimum needed.

My original idea was to connect the centers of all the disks to get a ``web'', and then show that there exists a triangle in this web for which when you cut out the three sectors, the remaining area has the desired circle. This idea only works if the area in any triangle of the web has only the three sectors, and not any other circles; but this is false if the three circles are very far apart and a fourth circle comes inside the area.

To modify this idea, I instead considered expanding the circle till it hit [i]some[/i] three disks, instead of fixing them like I did in the last approach. The process ends here since tangent to three circles fixes all information. This works well, and reduces the problem to geometry.[/hide]
[hide=Remark]The [b]process[/b] can be rephrased with [b]extremal[/b] principle, as is often the case with [b]local[/b] problems. Instead show that the largest circle that does not intersect any disk has radius at least $1/\sqrt2$. The monovariant in the process is the radius, and the extremal rephrasing takes care of this.[/hide]

Quite an interesting problem - focusing [b]locally[/b] on a set of three circles is relatively easily motivated making it slightly underwhelming in my opinion.

We will prove that there is a circle of radius $\frac{1}{\sqrt{2}}$ that intersects no other circles in the plane which suffices as we get a $1$ by $1$ square in any such circle which will ultimately contain a lattice point. \\

Take a circle $\omega_1$ and the closest circle to it, $\omega_2$, we can assume that $\omega_1$ and $\omega_2$ are tangent simply by increasing the radius of $\omega_1$ by $dist(\omega_1,\omega_2)$. \\
Now, take one of the two circles tangent to both $\omega_1$ and $\omega_2$ on the exterior with radius $\frac{10}{\sqrt{3}}-5 > \frac{1}{\sqrt{2}}$, say $\Gamma$. \\

$\textbf{Claim:}$ $\Gamma$ doesn't have two distinct intersections with any of the other circles in our tiling. \\
$\textbf{Proof)}$ \\
Consider a third circle $\omega_3$ that is, without loss of generality, tangent $\omega_1, \omega_2, \Gamma$, we shall prove that the radius of $\omega_3 \geq 5$ in which case we get that the two circles can't have two intersections. \\

[hide=Remark]
In the previous argument, we can say without loss of generality that they are tangent because that minimizes the distance between $\Gamma$ and the intersection of $\omega_1$ and $\omega_2$ \\
[/hide]

Now, notice that whenever we increase the radius of $\omega_3$ ($\omega_1, \omega_2. \omega_3$ now have symmetric roles), any circle that was contained with the region created by $\omega_1, \omega_2, \omega_3$ with the former radii is still contained in the aforementioned region. \\
We can consequently assume that the three circles all have radii $5$. \\
Then, introducing the centers of the three circles, $O_1, O_2, O_3$ and computing the circumradius of $\triangle O_1O_2O_3$ which is equilateral (sidelength $10$), we have that the circle tangent to the three circles on the exterior has radius $\frac{10}{\sqrt{3}} - 5$ completing the proof. $\blacksquare$ \\

Note that for $0 \le i \le 2$, if we have a circle tangent to $i$ circles we can
increase the radius of the circle, potentially moving it, to maintain the tangencies.
Call this expanding.

Thus, if we take a point not in the circles and expand, we eventually end up with
a circle tangent to three circles in the family. We claim that this circle
has radius at least $\frac{1}{\sqrt{2}}$, which implies a lattice point within the circle.

Restrict our view to these three circles. By expanding these circles we can assume that they are
pairwise tangent.

Then, decreasing the radius of one of the circle and fixing the rest moves the tangency
points with the other circles inward, as well as increasing the curvature of the arc.

Thus, we can repeat until all three circles have radius $5$, at which point the result
directly follows.

This is a very boring problem. The answer is no; most of the diffculty is dealing with the details.

It suffices to show that we can fit a circle of radius $\frac 1{\sqrt 2}$ somewhere between the circles, as then we will be able to fit a square of side length $1$. Consider any three circles in the plane; I will show that we will be able to fit such a circle between the three.

Let $O_1, O_2, O_3$ be the centers, and let $O$ be the circumcenter of $O_1O_2O_3$. By Descartes' circle theorem, the circle with radius $\frac{10\sqrt 3 - 15}3 > \frac 1{\sqrt 2}$ centered at $O$ will not intersect any of the given circles, as the distance from a point on its rim to one of the $O_i$ is at least $5$. This is our desired circle.


For the real numbers $ x,y,z$ we have :

 $ x^2\plus{}y^2\plus{}z^2\plus{}2x\plus{}4y\plus{}6z\plus{}13 \equal{} 0$

 Find the greatest value for the  positive $ m$  , such that

  $ x\plus{}y\plus{}z \plus{} m \leq 0$.

Babis

[quote="stergiu"]For the real numbers $ x,y,z$ we have :

 $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2x \plus{} 4y \plus{} 6z \plus{} 13 \equal{} 0$

 Find the greatest value for the  positive $ m$  , such that

  $ x \plus{} y \plus{} z \plus{} m \leq 0$.

Babis[/quote]

The first equality is $ (x\plus{}1)^2\plus{}(y\plus{}2)^2\plus{}(z\plus{}3)^2\equal{}1$ and so $ X^2\plus{}Y^2\plus{}Z^2\equal{}1$.
So, the minimum value for $ X\plus{}Y\plus{}Z$ is obtained when $ X\equal{}Y\equal{}Z\equal{}\minus{}\frac{1}{\sqrt{3}}$

So the minimum value of $ x\plus{}y\plus{}z$ is obtained when $ x\equal{}\minus{}1\minus{}\frac{1}{\sqrt{3}}$, $ y\equal{}\minus{}2\minus{}\frac{1}{\sqrt{3}}$ and $ z\equal{}\minus{}3\minus{}\frac{1}{\sqrt{3}}$.

And this minimum value is $ \minus{}6\minus{}\sqrt{3}$

And $ m\equal{}6\plus{}\sqrt{3}$

We have: $ (x \plus{} 1)^2 \plus{} (y \plus{} 2)^2 \plus{} (z \plus{} 3)^2 \equal{} 1$
Then $ (x \plus{} y \plus{} z \plus{} 6)^2\leq 3[(x \plus{} 1)^2 \plus{} (y \plus{} 2)^2 \plus{} (z \plus{} 3)^2] \equal{} 3\Leftrightarrow \minus{} \sqrt {3} \minus{} 6\leq x \plus{} y \plus{} z\leq \sqrt {3} \minus{} 6$
Then $ m \equal{} 6\minus{} \sqrt {3}$

[quote="tdl"]We have: $ (x \plus{} 1)^2 \plus{} (y \plus{} 2)^2 \plus{} (z \plus{} 3)^2 \equal{} 1$
Then $ (x \plus{} y \plus{} z \plus{} 6)^2\leq 3[(x \plus{} 1)^2 \plus{} (y \plus{} 2)^2 \plus{} (z \plus{} 3)^2]$[/quote]

Which is that inequality??

$ (\sum a)^{2} \leq 3(\sum a^{2})$
because $ 3(\sum a^{2}) \minus{} (\sum a)^{2} \equal{} \sum (a \minus{} b)^{2} \geq 0$
where all sums are cyclic


ratio
geometry
geometric transformation
invariant
geometry solved

Let C1,C2,C3 be circles such that
C1 and C2 meet at distinct points A and D,
C2 and C3 meet at distinct points B and E,
C3 and C1 meet at distinct points C and F.
Let P1 be an arbitrary point on C1 and define
point P2,...,P7 as follows:
P2 is the 2nd intersection of line P1A with C2
P3 is the 2nd intersection of line P2B with C3
P4 is the 2nd intersection of line P3C with C1
   ...
P7 is the 2nd intersection of line P6F with C1
-> prove that P7=P1

(I've a nice sol ;-)

hi,

here my sol using similarity:

let O1,O2,O3 be the centers of (C1),(C2),(C3) respectively.
It's easy to show that angle(P1DP2)=angle(O1DO2) and that
EP2/EP1=EO2/EO1.
then there exist a similarity S1 of angle Angle1=(O1DO2) and ratio r1=EO2/EO1
which transform P1 to P2.
In the same way,we define the similarities S2,S3,..,S6.

Moreover we notice that
Angle1=-Angle4,Angle2=-Angle5,Angle3=-Angle6
and that
r1*r2*r3*r4*r5*r6=1.

Hence the similarity S=(S6)o(S5)o(S4)o(S3)o(S2)o(S1)
is a similarity of angle 0 and ratio 1.
Then S is either a translation or the identity.
But it can't be a translation because (C1) is invariant by S.
Then S is the identity and P7=S(P1)=P1.

Bye.

[quote="alekk"]
It's easy to show that angle(P1DP2)=angle(O1DO2) and that
[color=#FF0000]EP2/EP1=EO2/EO1[/color].
then there exist a similarity S1 of angle Angle1=(O1DO2) and ratio r1=[color=#FF0000]EO2/EO1[/color][/quote]
You probably think: DP2/DP1=DO2/DO1 and r1=DO2/DO1.

Btw, very nice solution.


logarithms
probability
videos
parameterization
expected value
geometric series

Expected Value Coin Flipping Paradox

We discover a paradox when finding the expected value of a coin flipping game.

Oh, snap.

This one's new to me. I guess the paradox just arises from the whole "infinite expected value" thing, which just seems weird. There's nothing wrong with it, though; it's a really good game for the player, but the odds of ever getting paid off are pretty low.

I think it has something to do with the fact that the game is guaranteed to end sometime (1/2+1/4+1/8+.....=1). But if the game is guanteed to end, then the expected value can't be infinite.... :huh:

Well, there isn't infinite money in the world, and there will always be a finite amount of times someone flips. So you could calculate the amount of money by how many times the game was played (I lost  :wink: ). For example, if 16 people played, each once, then the expected value is $ \log_2 16\equal{}4; 4*\frac{1}{2}\equal{}2$, so you should charge $ \$$2.
Just a note: Is this why price of services/goods go up with the size of a company?

Is this right?
[hide]
The expected number of flips that you take before you stop (all flips are heads except last one, which is tails) $ \sum_{n\equal{}1}^{\infty} {\frac{a}{2^a}}\equal{}2$.

So, you're expected to last 2 flips.
So, you're expected to win once, which means that the expected money you get is $ \$2$.
[/hide]

The paradox confuses you by making you think that the infinite geometric series has ("logically") an infinite sum. This is a case where an infinite geometric series actually has a finite sum.

The infinite geometric series that you are referring to is $ \frac{1}{2}\plus{}\frac{1}{4}\plus{}\frac{1}{8}\plus{}\frac{1}{16}\plus{}\cdots$ which does indeed sum to a finite value of 1. This is expected, as the total probability of something is always one. However, the [i]expected value[/i], as calculated in the video, is clearly infinite... :maybe:

[quote="Arvind_sn"][quote="nsreenivas"]This probably has to do with the fact that since an infinate amount of money [i]can[/i] be won, it is impossible to calculate the expected value.[/quote]

Not true.

If you do it out, I think it comes out to a nice telescoping sum. I THINK. Emphasizing the THINK.

Or a nice summation. n^2/n! perhaps? I rmr doing this...I got something like that.[/quote]

I guess. But can your really multiply infinity and 1 divided by infinity?

No, you can't. Infinity is a concept; just like $ \sqrt{\minus{}i}$.

But $ \sqrt{\minus{}i}\equal{}\frac{\minus{}\sqrt{2}}{2}\plus{}\frac{\sqrt{2}}{2}i$ or $ \frac{\sqrt{2}}{2}\minus{}\frac{\sqrt{2}}{2}i$. :wink:

Although it is true you cannot multiply or divide by infinity, at least in the real number system.  I don't know much about this topic though.

I'm not quite sure why we are discussing operations involving infinity. :maybe:  I thought that the infinite sum involved was just
\[ 0.50 \plus{} 0.50 \plus{} 0.50 \plus{} 0.50 \plus{} \cdots.
\]
Am I mistaken?

no, you are correct

but they're trying to put this into a nice summation or other formula, and that, i believe, involves infinity

I think the problem arises from the number of times one is actually able to play the game.  For example, the probability of making sixteen or more is only 1/16.  However, as the number of games played aproaches infinity, the probability of making more than 2^900,000,000 dollars approaches 1!  The expected value therefore approaches infinity as the number of games aproaches infinity.  This also implies that you should be willing to pay any amount of money for each game if you are allowed to play as many times as you want.  However, for large numbers, the age of the universe would not be enough time to come out ahead.  There is also no guarantee that you will come out ahead in any [i]finite[/i] number of games.

Okay, I'm going to going to summarize what I think, If your head hurts from what I type in the selection, please wait for another hour before you read it again, or you'll think it's wrong.

 [hide="are you ready?"] Think about this, 0.9999999999999999999999999999...
practically equals 1, but it will never reach one, though it will infinitly close the gap.  Are you still with me?

Ok, now [hide="click this if your ready"] Now think about 0.3333333333... It will never reach 0.4 cause the hundreths place will never become 4, and so on, still with me?

[hide="last one"] Think about the equation that Richard wrote: 1/2*1/2+1/4...  it will probalby never reach one, much less infinity, cause 1/2+1/4+1/8... will equal less than 0.6, as it will be 0.5......etc.  please tell me if i'm wrong and why.[/hide][/hide][/hide]

[hide="Warning! Spoiler on the problem"]
[quote="rrusczyk"]
From: rrusczyk
To: yunhua98
Posted: Thu Jul 23, 2009 6:34 am
Subject: Re: Unexpected Coin value paradox video
I don't think I follow -- to what are you referring? If you're referring to the St. Pete's Paradox (getting paid based on how many heads in a row there are), then the answer is infinite, not 1.
[/quote][/hide]

I guess I was wrong on the above post.

OMG I'm totally clueless about this paradox and its been bugging me like heck!

Yes i do think that it has something to do with the summations.

Because if you multiply the expected time someone's going to last in the game and the average of the prizes that that person could get, you should get the expected winnings

But here the expected time soneone' going to last is (1/2+1/4+1/8+1/16...)/infinity = 0, and the average of the prizes should be infinite. So the expected value should be 0*infinity, which doesnt make sense. And somehow he then translated this to 1/2+1/2+1/2...

It's sort of like how some infinite geometric series have finite sums. The amounts become totally negligible at the end.

[quote="yunhua98"]Okay, I'm going to going to summarize what I think, If your head hurts from what I type in the selection, please wait for another hour before you read it again, or you'll think it's wrong.

Think about this, 0.9999999999999999999999999999...
practically equals 1, but it will never reach one, though it will infinitly close the gap.  Are you still with me?

Now think about 0.3333333333... It will never reach 0.4 cause the hundreths place will never become 4, and so on, still with me?

Think about the equation that Richard wrote: 1/2*1/2+1/4...  it will probalby never reach one, much less infinity, cause 1/2+1/4+1/8... will equal less than 0.6, as it will be 0.5......etc.  please tell me if i'm wrong and why.[/quote]

0.999999999... does equal one (if it goes on infinitely); Mr. Rusczyk's equation is 1/2+1/2+1/2+1/2..., not 1/2+1/4+1/8+1/16..., I think.

This is a cool video, and it really makes you have to think about math more.  :lol:

Infinite. That is sad. I would basically sell the universe to play this game.

I think I know the solution to the paradox:
[hide]
Mathematically the expectation is indeed infinite. Whatever you loose while playing, at some point in the future you will be lucky enough to win back, but it may take you a billion years to be this lucky. Expectation shows the "average" amount you will make per game if you play forever. Before this lucky round comes up you may be in billions of \$ in debt.

In the real world, you can't keep playing forever. You can play for at most ~100 years before you die, and anyway you have finite money to loose even if you are Bill Gates.

If the game has an upper limit on the rounds that you are willing to play, or the debt you are willing to sustain, then the expectation isn't the right parameter to use. I think the correct way is to make the game more complicated, consider the rounds of the as part of a larger "game"  and add an ending condition (such as max rounds or max debt), and then calculate the expectation of this larger game which will not be infinite.

For example, consider the game costs 50\$ to play, and you can play it "just" a million times. Half of these times (~500K times) you will earn just 1$ (thus loose 49\$ or total of 49*500K = ~20Million \$ of debt), in just about one of the games you will flip heads 20 times and win ~million \$, and in about two of the games you will flip 19 and win ~500K, so you will probably regain a few millions back. There is a tiny chance you will flip heads 100 times and win big time (actually win more money than particles in the universe), but its not likely to happen in any of the million rounds.

[/hide]


function
trigonometry
algebra
domain
limit
real analysis
real analysis solved

Let $f:[0,1)\to (0,1)$ a continous onto (surjective) function.

a) Prove that, for all $a\in(0,1)$, the function $f_a:(a,1)\to (0,1)$, given by $f_a(x) = f(x)$, for all $x\in(a,1)$ is onto;
b) Give an example of such a function.

Another example of function would be:

q:[0,1)->(0,1), q = v O u O f O w, where

w:[0,1)->[0, inf), w(x) = x/(1-x) - bijective
f:[0, inf)->R, f(x) = x*sin(x) - surjective
u:R->(-pi/2, pi/2), u(x) = arctang(x) - bijective
v: ( -pi/2, pi/2)->(0,1), v(x) = x/pi + 1/2 - bijective

[quote="Kent Merryfield"](a) The image of $[0,a]$ must be compact, so $\exists\,\delta>0$ such that $(0,\delta)$ and $(1-\delta,1)$ lie outside of $f([0,a])$ and thus must be contained in $f((a,1)).$ By the Intermediate Value Theorem, the image of $(a,1)$ must then also contain all of the points in between and is thus all of $(0,1).$

(b) $f(x)=\frac12 + \frac{x}2\,\sin\left(\frac1{1-x}\right).$[/quote]

Kent, how did you construct this function? the author proposed the same example... so I guess there must be some logic which i can't understand now

btw, mareleG, how did you think of your function?

The function we want must be continuous on $[0,1).$ We can't do anything with it at the zero end except just have it harmlessly take on some value in the middle of the range.

The function can't have a finite number of maxima an minima. If it did, we could have it tend towards 1 or towards 0 at the open end of the domain, but not both. Suppose we did have a finite number of maxima and minima, and that $\lim_{x\to1^-}f(x)=1.$ But then, the function would have an absolute minimum somewhere, and the range of the function would have the form $[c,1).$ We couldn't get the range to be all of $(0,1)$ that way.

So we need infinitely many maxima and minima in a finite interval. That leads us to the $\sin \frac1x$ family of functions - except that the "action" has to be a 1, not zero, so make that $\sin\frac1{x-1}.$

Now, what do we want to do with the "envelope" of this curve? No one of those infinitely many maxima can be as high as 1 and no one of the infinitely many minima can be as low as 0. So we center the whole thing in the middle, at $\frac12,$ and we find some "envelope" function that expands us out to the full 0 to 1 width as $x\to1^-.$

The choice of $\frac x2$ for that "envelope" function was just a convenience - it's the simplest function I could think of with the right behavior. I can see how another person working this with the same general thought processes might arrive at the exact same function, even though there are infinitely many choices.

mareleG's example, while not identical, is in fact quite similar, and stems from the same need to construct a function with infinitely many local maxima and minima.

First built a function from an 'open' interval like [0, ∞) to R and then switching from [0, 1) to [0, ∞) and from R to (0, 1) is not hard.
The x*sinx function it's easy to build, it starts from 0 and it has alternate minimum relative and maximum relative points, such that the function covers the R set properly.
You can draw its graphical image easily-by hand and then using Matlab for example build the image for the final function. For these types of 'ugly' graphics Matlab it is quite useful and powerful. It helps you a lot to experiment if you do not have time to do it by hand  or just to see for yourself the whole picture more clearly. If anyone is interested i can post the Matlab script lines...

a) Note that $[0,a]$ is compact, so there exist $x_1 ,x_2$ with $0< x_1= \min_{x\in [0,a]} f(x), 1>x_2 =\max_{x\in [a,b]} f(x).$ Let $y\in (0,1)$, then we can choose $0<y_1 < \min(x_1 , y)$ and $1>y_2 > \max(x_2 ,y)$ and by the surjectivity of $f$, it must attain the values $y_1, y_2$ in the interval $(a,1)$, and so by the IIntermediate Value Theorem also the value $y.$

b) The following piecewise-linear function should also work. Define
\begin{align*}
f\colon [0,1) &\rightarrow (-1,1)\\
x &\mapsto \begin{cases} \frac{(-1)^{n}(1-\frac{1}{n})}{2^{-n+1}}(x-a_{n-1}) \;\; \text{if}\, x\in [a_{n-1}, a_{n-1}+ 2^{-n+1}]\\
(-1)^{n}(1-\frac{1}{n})- \frac{(-1)^{n}(1-\frac{1}{n})}{2^{-n+1}}(x-a_{n-1}-2^{-n+1}) \;\; \text{if}\, x\in [a_{n-1}+ 2^{-n+1}, a_{n}],
\end{cases}
\end{align*}
where $a_0=0, a_k =\sum_{i=1}^{k} 2^{-i}$ for $k>1.$ Since $\bigcup_{n\geq 1} [a_{n-1} , a_n]=[0,1)$, this function has the correct domain. It takes the value $(-1)^{n}(1-\frac{1}{n})$ at $x= a_{n-1}+ 2^{-n+1}$ and is hence surjective. It also takes the value $0$ at $x=a_n$ and so we see that $f$ is actually continuous. Therefore, the function $g\colon [0,1) \rightarrow (0,1), x\mapsto \frac{1}{2}(f(x)+1)$ has all desired properties.


A peep increased by $ 25\%$ is a pop. A pop decreased by $ 40\%$
is a slug, and a slug increased by $ 100\%$ is a slap. What
percent of a peep is a slap?

Let peep = p for abbrev.
A pop would then be 1.25p
A slug is 0.75p
Then two times that is 1.5p, or 150% of a peep

Another way is to let $ p$ be a nice simple number, say 100. So if $ p$ is 100, then a pop is $ 125$. That means a slug is 75, and a slap is 150. Thus our answer is $ \boxed{150}$.

I used $4$ because it is divisible by 4 (25%=1/4) and it worked out nicely. Got 4 to 6=$150\%$.


Determine with proof all those integers $a$ for which the equation $x^{2}+axy+y^{2}= 1$ has infinetely many distinct integer solutions $x, y$.

Rewrite it as $(2x+ay)^{2}+(4-a^{2})y^{2}=4$.
By this $d=a^{2}-4$ has necessarily to be positive and not a perfect square. But by Pell's equation, this also suffices:
Let $s^{2}-(a^{2}-a)t^{2}=4$, which has infinetely many solutions by Pell. Then looking $\mod 2$ yields $s \equiv at \mod 2$, thus we can find an integer $x$ such that $2x+at=s$. Then also taking $y=t$ solves our problem.
So it has $\infty$ many solutions iff $a \neq 0, \pm 1$ (see $a=\pm 2$ below; for all greater $a$, $a^{2}-4$ is never a square cause of the next square is too far away).

Edit: spam eliminated.

I don't get you. If $a = \pm 2$, we can set $(x \pm y)^{2}= 1$ and $x = 1 \mp y$, which has $\infty$ many solutions.  :maybe:

I think u're wrong...
U write $(2x+ay)^{2}+(4-a^{2})y^{2}= 4$
But then u claim that this is Pell's equation when u write $s^{2}-(4-a^{2})t^{2}= 4$
They have different sign before $(4-a^{2})$

Umm right, $a= \pm 2$ works also...

@RDeepMath91: That sign should be reverted.


trigonometry
logarithms
LaTeX
integration
calculus
function
real analysis

Find an equivalence when $x$ tends to $0^{+}$ for

1/ $\sum_{n\geq 2} \frac{\cos (2\pi nx)}{n\ln n}$

2/$\sum_{n\geq 2} \frac{ \ln n \sin(2\pi nx )}{n}$

3/ $\sum_{n\geq 2} \frac{ \sin(2\pi nx )}{\ln n}$

I'll be able to help a bit if u tell me how to write formulas here... ;)

Hi, now I've became a bit more familiar with LaTex :) Well, I see that nobody answered to yo post, so just a few ideas. Hope they will help (I studied this stuff 3 years ago, so could forget smth).

Lets's consider the following situation:

$\sum_{n\geq 1} {\cos(nx)f(x)}$,

where $f(x) \geq 0$ and $f(x)$ is decreasing and $\int_{1}^{\infty}{f(x)dx} < \infty$.

In this case it's easy (if u think it's not, let me know), that

$\sum_{k\geq n}{f(k)} = \int_{n}^{\infty}{f(x)dx} + O(f(n))$

How this can be helpful in our case? Well, you know that

$\cos(nx) = \frac{x}{2\sin(\frac {x}{2})} \int_{n-1/2}^{n+1/2} (\cos(xt))dt$

Thus,

$\sum_{n\geq 1} {\cos(nx)}{f(n)} = \frac{x}{2\sin(\frac {x}{2})} \int_{1/2}^{\infty} \cos(xt)f(t)dt + O(1))$

Now you get 1 for the multiple before integral (when $x \rightarrow 0$) and need to make substitution inside the integral. For example, when $f(x) = x^\epsilon$, then after this will get for the right side:

$\frac{x}{2\sin(\frac {x}{2})} \int_{1/2}^{\infty} \frac {\cos(xt)}{t^\epsilon}dt + O(1) = x^(\epsilon - 1)\int_{0}^{\infty} \frac {\cos(u)}{u^\epsilon}du + O(1)$. The integral is a positive constant and thus we get the required asimptotic.

The same stuff for sin(nx). Now you'll need to analyze yo functions, like $\frac{\ln(x)}{x}$, etc. If you come up with a solution let me know. Else I'll take a look during the weekend: don't have time now.


A circle centered at O and inscribed in triangle ABC meets sides AC, AB, BC at K, M, N, respectively. The median BY of the triangle meets MN at D. Show that O, D, K are collinear.

Could anyone give me some hints on this question? Thanks a lot.

Theorem 1 from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=124774#124774]http://www.mathlinks.ro/Forum/viewtopic.php?t=5830 post #4[/url], viewed from a slightly different direction.

  darij

Dear Mathlinkers,
another point of wiew can be seen on
http://perso.orange.fr/jl.ayme    vol. 3 The cross-cevian point    p. 17
Sincerely
Jean-Louis


The complex function f(z) = z^m is defined on the unit circle S ={z : |z| = 1}, where m > 1 is an integer. Determine the number N (in terms of m) of periodic points of least period 1989.

I'll just tell you what I got. If you say it's Ok, I'll post my soln (I want to make sure this is it, becausee I might have misunderstood the problem):

1989=3^2*13*17. Let x(n) be m^n-1.

I found N to be: x(1989) - [x(1989/17)+x(1989/13)+x(1989/3)] + [x(1989/(13*17))+x(1989/(3*17))+x(1989/(3*13))] - x(1989/(3*13*17)).

The problem looks more like NT to me, since all I did was counting some multiples of certain numbers of the form x(1989)/x(d), with d | 1989.

Hi grobber,

your approach might work. The answer is N = m^1989-m^663-m^153-m^117+m^51+m^39+m^9-m^3. So please post your solution. BTW the problem was used in CIST1989.


[quote="pvthuan"]1) Let $ x,y,z$ be non-negative real numbers. Prove that
\[ (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2)\geq\sqrt {3(x^2 \plus{} xy \plus{} y^2)(y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2)}.
\]
2) Let $ x,y,z$ be non-negative real numbers. Prove that
\[ (x^2 \plus{} xy \plus{} y^2)(y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2)\leq (xy \plus{} yz \plus{} zx)(x^2 \plus{} y^2 \plus{} z^2)^2.
\]
3) Let $ x,y,z$ be non-negative real numbers. Prove that
\[ (x^2 \plus{} xy \plus{} y^2)(y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2)\geq (xy \plus{} yz \plus{} zx)^2(x^2 \plus{} y^2 \plus{} z^2).
\]
I have not tried to solve. I guess it should be simple enough.[/quote]

Problem 3 is wrong.
Let $ x\equal{}y\equal{}1, z\equal{}2, LHS\minus{}RHS \equal{}\minus{}3$ (!!!!) :|

Problem 2 is not hard but I need nice solution

[quote="langthang123"]
Problem 2 is not hard but I need nice solution[/quote]

What did you use for it, [b]langthang123[/b]?

[quote="Albanian Eagle"]well, problem 2 is immediate with majorization  :wink:[/quote]

That is why I asked, does it qualify for a "nice" solution. I mean it is almost brute force general trick , isnt it?

Thanks   :D

[quote="langthang123"]Problem 2 is not hard but I need nice solution[/quote]

We can assume : $ x \plus{} y \plus{} z \equal{} 1$ , $ q \equal{} xy \plus{} yz \plus{} zx \leq\ \frac {1}{3}$ , $ r \equal{} xyz$ and notice that :

$ x^2 \plus{} xy \plus{} y^2 \equal{} (x \plus{} y)^2 \minus{} xy \equal{} (1 \minus{} z)^2 \minus{} xy \equal{} 1 \minus{} z \minus{} z(1 \minus{} z) \minus{} xy \equal{} 1 \minus{} z \minus{} (xy \plus{} yz \plus{} zx) \equal{} 1 \minus{} z \minus{} q$

And :

$ (1 \minus{} q \minus{} x)(1 \minus{} q \minus{} y)(1 \minus{} q \minus{} z) \equal{} a^{3} \minus{} a^{2}(x \plus{} y \plus{} z) \plus{} a(xy \plus{} yz \plus{} zx) \minus{} xyz \equal{} q^{2}(1 \minus{} q) \minus{} r$

We rewrite the desired inequality as :

$ r \plus{} q(5q^{2} \minus{} 5q \plus{} 1) \geq\ 0$

If $ q \geq\ \frac {1}{4}$ Apply Schur we have $ r \geq\ \frac {4q \minus{} 1}{9}$

So we need to prove that : $ (3q \minus{} 1)(15q^{2} \minus{} 10q \plus{} 1) \geq\ 0$

But this is true  :D

If $ q \leq\ \frac {1}{4}$ then
$ q(5q^{2} \minus{} 5q \plus{} 1) \geq\ 0$
We are done :)

And the following is true  :wink:
3) Let $ x,y,z$ be non-negative real numbers. Prove that
\[ \large (x^2 \plus{} xy \plus{} y^2)(y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2)\geq \frac{(x\plus{}y\plus{}z)^{2}}{3}(xy \plus{} yz \plus{} zx)^2\]


How about we start a topic that depicts theories and proofs made up by us.

I'll start with one of my own

[b]rd5493's Square Number Theory[/b]
$x^{2}=y$
$(x+1)(x-1)=y-1$

Proof:

$(x+1)(x-1)=y-1$
$x^{2}+x-x-1=y-1$
$x^{2}-1=y-1$
$x^{2}=y$

The proof above is good for a law that I came up with.  This law states that for every number $x$, $x^{2}$ equals the product $y$.  This product, in turn is one greater than the product of the two numbers $x-1$ and $x+1$.  I used this quite a lot to figure out number patterns and factoring.
Now someone else post one

[quote="chickendude"]That little thing you came up with can be generalized to find a quick way of squaring numbers in your head

Basically, if you have a number $x$ and you want to square it

$x^{2}$

$x^{2}-b^{2}+b^{2}$

$(x-b)(x+b)+b^{2}$

now, you select a number $b$ that makes this arithmetic very simple

For example:

$18^{2}$
$(18-2)(18+2)+2^{2}$
$(16)(20)+4$
$324$

I use it a lot to quickly square numbers that I don't have memorized in my head.[/quote]

In other words, for $x$, you choose the smallest positive $b$ such that $b\equiv-x\pmod{10}$?

This really shouldn't be called theories, it should be called "Arithmetic Tricks", if that is what is intended.

Is there a quick way to do the triangular numbers $\frac{x(x+1)}{2}$

Here's my next theory:

rd5493's Multiplication Theory

$x(x-1)=y$
$(x+1)(x-2)=y-2$

Proof:

$(x+1)(x-2)=y-2$
$x^{2}-x-2=y-2$
$x^{2}-x=y$

When factored...
$x(x-1)=y$

This one is helpful for finding the greatest product that can be made with 2 numbers with sum $n$.

not always vishalarul

For example

$89^{2}$

I would select b = 11

(78)(100) + 11^2

7921

[quote="vishalarul"]
This really shouldn't be called theories, it should be called "Arithmetic Tricks", if that is what is intended.
[/quote]

I agree... then, please, rd5493, explain a little clearer your tricks: I really have difficulties to follow your reasoning...  :maybe:
Speaking, for example, about your Multiplication "Theory", what do you want to demonstrate? What are your initial assumptions? If your claim is just that $x(x-1)=(x+1)(x-2)+2$ well, that's a bit trivial! But perhaps I didn't understand the usefulness of that trick...  :blush:

By the way, I'll contribute with a useful trick...

[b]Trick[/b]: To get the square of a number $\equiv 5\pmod{10}$ you have to write the product of the tens' digit and the consecutive integer, followed by the digits 25.

[b]Proof[/b]: Let the number we have to square be $n=10k+5$. Then $n^{2}=100k^{2}+100k+25=100k(k+1)+25$ and we are done!

[i]Example[/i]: $25^{2}=100(2\cdot3)+25=625$ and $1225^{2}=100(12\cdot 13)+25=15625$.

P.S.: This is of course a trivial result, but I think it's nice...


[hide]$ (1 \plus{} x)^n(1 \plus{} x)^n \equal{} (1 \plus{} x)^{2n}$
ανεπτυξε και εξισωσε τους συντελεστες του x^n...  :)
ή ενναλακτικα σκεψου ενα συνολο Χ 2n στοιχειων και X1,X2 ν στοιχειων υποσυνολα του Χ και ξενα μεταξυ τους.
Το αριστερο μελος εκφραζει τον συνδιασμο των 2n στοιχειων του συνολου ανα n.
Ομως οι συνδιασμοι αυτων των στοιχειων ανα n μπορουν να χωριστουν σε n ξενα μεταξυ τους υποσυνολα οπου το k-οστο θα εχει τους συνδιασμους οι οποιοι χρησιμοποιουν ακριβως κ στοιχεια απο το Χ1 και n-k απο το Χ2 Το πληθος των στοιχειων(συνδιασιμων) του συγκεριμενου υποσυνολου ειναι $ \binom{n}{k}^2$. αρχη αθροισματος και το ζητουμενο..[/hide]

Δείτε και εδώ http://www.artofproblemsolving.com/Forum/viewtopic.php?t=270946


I think you're done with all the problems. Then you redo the problems that you do wrong, so maybe you got that wrong and it was the last problem, so you did it again. Or not

Or there is a mistake and two problems are the exact same. What was it about?

The Alcumus AI gives you a (semi) random problem that it thinks you are ready for.  The random aspect of this is why you got the same problem twice in a row (I assume you missed it the first time).

This happened to me multiple times.


induction
limit
logarithms
function
calculus
calculus computations

Why is $1^\infty = ?$.
We're multiplying 1 with 1, forever. Aritcmetic induction would say the result is $1$. But when considered this way, $\lim{x \to \infty}1^{x}= y \Rightarrow x \ln 1 = \ln y$, a $0 \times \infty$ comes.

Can someone give a reasonable interpretion, how can something other than 1 comes out if we multiply 1 with itself many times.

Let me clarify these for you; You should first say what is meant by $1^\infty$
1. Induction merely states that $\forall n\in \mathbb{N},\; 1^{n}=1$. Note that $\infty$DOES NOT belong to$\mathbb{N}$.
2. If you want to calculate $\lim_{x \rightarrow \infty}1^{x}$, it follows easily from the definition of limit that the result is $1$.

Definition of limit? Can you explicitly do that:
$\lim_{x \to \infty}1^{x}$

And under what circumstances can it yield something other than 1?

The problem arises when you have an expression like
$\lim_{x \to \infty}f(x)^{g(x)}$
where
$\lim_{x\to \infty}f(x) = 1$
$\lim_{x \to \infty}g(x) =+\infty$

In that case the natural log trick should be the first thing you try. You can get it into an $\frac{0}{0}$ form and use L'Hopital.

[quote="Xevarion"]In that case the natural log trick should be the first thing you try. You can get it into an $\frac{0}{0}$ form and use L'Hopital.[/quote]

How?
$\lim_{x \to \infty}1^\infty = y$
$x \ln 1 = \ln y$
$0 = \frac{\ln y}{x}$

You never want to actually do operations on $1$ and $\infty$ here. You want to do your operations on functions of $x$. Otherwise what you're writing doesn't make sense. Here's what I meant (I use $\log$ for natural log):

$L = \lim_{x \to \infty}f(x)^{g(x)}$

$\log(L) = \log(\lim_{x\to \infty}f(x)^{g(x)})$

$\log(L) = \lim_{x \to \infty}g(x) \log(f(x))$

$\log(L) = \lim_{x \to \infty}\frac{\log(f(x))}{g(x)^{-1}}$

Now you have a $\frac{0}{0}$ form. :)
[Note that I am making some implicit assumptions here, like that $L$ exists and is not 0, and that $f$ and $g$ are reasonably nice funtions so I can do the log of $f(x)^{g(x)}$.]


geometry
rectangle
combinatorics open
combinatorics

Consider a rectangle that can be tiled by a finite combination of (1 x m) rectangles and (n x 1) rectangles, where m, n are positive integers.
Prove that it is possible to tile this rectangle using only (1 x m) rectangles or (n x 1) rectangles.

Once was enough:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265223


group theory
abstract algebra
algebra
polynomial
induction
Galois Theory
superior algebra

I have these two problem:

1. Let $ G$ be a finite group of order $ p^n$ where $ p$ is prime. Show that $ G$ has a normal subgroup of order $ p$ and $ p^{n \minus{} 1}$.
2. Let $ L$ be a Galois extension of $ K$ where $ Gal(L/K)$ is isomorphic to $ A_5$. Let $ f$ be an irreducible polynomial in $ K[x]$ where $ 1 < deg(f) < 5$. Show that $ f$ cannot split in $ L$. What can you say when $ deg(f) > 5$?

I thought about Sylow Theorems for the first one but am not going anywhere.

Thanks in advance,

The Sylow theorems are useless for #1.  Think about it. :)  They just tell you that the group exists and is conjugate to itself...

1) This is a classic problem. Show that there is a normal subgroup of order $ p$ using the class equation.  For the other part, use induction plus the lattice isomorphism theorem.

2) I assume you mean that $ f$ is irreducible.  Anyway, what can  you say about the Galois group of the splitting field of $ f$?  Do you know the theorem which says that $ A_5$ has no normal subgroups?

Yes I missed the part that f is irreducible. Can you provide more hint on the second one please? I'm really lost on this part.

Thanks

Work backwards: you want the Galois correspondence to produce a normal subgroup for you, what's a good choice of an intermediate field?


Consider a triangle $ ABC$, and let $ D,E$ and $ F$ be the points where the incircle of the given triangle touches the sides $ BC, CA$ and $ AB$, respectively. Consider an arbitrarly point $ P$ and denote by $ d_1,d_2$ and $ d_3$ the distances from $ P$ to the sides of the triangle $ ABC$. Prove that:

$ d_1^2\plus{}d_2^2\plus{}d_3^2\equal{}PD\cdot PE\plus{}PE\cdot PF\plus{}PF\cdot PD$


modular arithmetic
algebra unsolved
algebra

$ N$ is a four-digit integer, not ending in zero, and $ R(N)$ is the four-digit integer obtained by reversing the digits of $ N$; for example, $ R(3275)\equal{}5723$.

Determine all such integers $ N$ for which $ R(N)\equal{}4N\plus{}3$.

[quote="AndrewTom"]$ N$ is a four-digit integer, not ending in zero, and $ R(N)$ is the four-digit integer obtained by reversing the digits of $ N$; for example, $ R(3275) \equal{} 5723$.

Determine all such integers $ N$ for which $ R(N) \equal{} 4N \plus{} 3$.[/quote]
I consider we're speaking about [u]positive [/u]integers.

Let $ N \equal{} \overline{abcd}$. The equation becomes $ 1000d \plus{} 100c \plus{} 10b \plus{} a \equal{} 4(1000a \plus{} 100b \plus{} 10c \plus{} d) \plus{} 3$ and so $ 1333a \plus{} 130b \plus{} 1 \equal{} 332d \plus{} 20c$

So $ a$ is odd and, since $ RHS\le 332\cdot 9 \plus{} 20\cdot 9 \equal{} 3168$, we get $ a < \frac {3168}{1333} < 3$ So $ a \equal{} 1$ and the equation becomes : $ 667 \plus{} 65b \equal{} 166d \plus{} 10c$

So $ 166d \equal{} 667 \plus{} 65b \minus{} 10c \ge 577$ and $ d\ge 4$

Taking $ 667 \plus{} 65b \equal{} 166d \plus{} 10c$ mod 5, we get $ d \equal{} 2\pmod 5$ and, since $ d\ge 4$, we get $ d \equal{} 7$ and the equation becomes : $ 13b \equal{} 99 \plus{} 2c$

Taking this equality mod $ 13$, we get $ c \equal{} 9\pmod {13}$ and so $ c \equal{} 9$ and $ b \equal{} 9$ hence the unique solution : $ \boxed{R(1997) \equal{} 7991 \equal{} 4\times 1997 \plus{} 3}$

IMHO, these problems are quite simple and you should spend more time on them before posting for solutions.  :oops:


Let $ f(x) =x^{n}+ \cdots + x + 1$ and let $ g(x) = f(x^{n+1})$. Find the remainder when $ g(x)$ is divided by $ f(x)$.

I have a solution, but I must admit that I have seen a similar problem before..
$ f(x)=\sum_{i=0}^{n}x^{i}\implies f(x^{n+1})=\sum_{i=0}^{n}x^{(n+1)i}=n+1+\sum_{i=0}^{n}\left(x^{(n+1)i}-1\right)$.
\begin{eqnarray*}x^{(n+1)i}-1&=&\left(x^{n+1}-1\right)\left(x^{(n+1)(i-1)}+x^{(n+1)(i-2)}+\cdots+x^{n+1}+1\right)\\ &=&\left(x-1\right)\left(x^{n}+x^{n-1}+\cdots+x+1\right)\left(x^{(n+1)(i-1)}+x^{(n+1)(i-2)}+\cdots+x^{n+1}+1\right)\\ &=&\left(x-1\right)f(x)\left(x^{(n+1)(i-1)}+x^{(n+1)(i-2)}+\cdots+x^{n+1}+1\right)\end{eqnarray*}
So all of $ x^{(n+1)i}-1$ in $ n+1+\sum_{i=0}^{n}\left(x^{(n+1)i}-1\right)$ are divisible by $ f(x)$. Therefore the remainder is $ \boxed{n+1}$.
Have I made any mistake?  :maybe:

[quote="minsoens"] Therefore the remainder is $ \boxed{n\plus{}1}$.
Have I made any mistake?  :maybe:[/quote]
I forgot what the answer is  :P , but it looks right (to lazy to check).
But, I have a really short solution.  :ninja:

[quote="sylow_theory"]
But, I have a really short solution.  :ninja:[/quote]

Would this be it, by any chance?  :D

[hide]$ f(x)|x^{n\plus{}1}\minus{}1|f(x^{n\plus{}1})\minus{}f(1) \equal{} g(x)\minus{}f(1)\implies g(x)\equiv f(1)\equiv n\plus{}1\mod{f(x)}$[/hide]

[quote="sylow_theory"]I was doing a problem, and I generalized it. Its not so hard, but its a nice result.

Let $ f(x) \equal{} x^{n}\plus{}...\plus{}x\plus{}1$ and let $ g(x) \equal{} f(x^{n\plus{}1})$. Find the remainder when $ g(x)$ is divided by $ f(x)$.[/quote]

[hide="Short solution using complex numbers"]
Let $ z_{k}\equal{}e^{\frac{2ki\pi}{n\plus{}1}}$ for $ k\in\{1,2,\ldots,n\}$. We have $ z_{k}^{n\plus{}1}\equal{}1$ and $ f(z_{k})\equal{}0$ $ \forall k\in\{1,2,\ldots,n\}$

We have $ g(x)\equal{}f(x)q(x)\plus{}r(x)$ with degree of $ r(x)$ $ <n$. So $ g(z_{k})\equal{}f(z_{k}^{n\plus{}1})\equal{}f(1)\equal{}n\plus{}1\equal{}r(z_{k})$

So $ r(x)$ is a polynomial of degree $ \leq n\minus{}1$ and taking the same value for $ n$ different values. So $ r(x)$ is a constant.

And $ r(x)\equal{}n\plus{}1$[/hide]

Good solutions guys :)

But I did it like Patrick. Write it like $ g(x)\equal{}q(x)f(x)\plus{}r(x)$ and work with $ (n\plus{}1)$-th roots of unity. Substitute those in and claim $ r(x)$ is a constant polynomial.


1> Find the remainder when 10 ^10 + 10 ^100 + 10 ^1000  +…….. + 10^10000000000   is divided by 7

2> Three circles of radii 3,4, 5 resp. touch each other externally.The tangents drawn at the pts. of tangency meet at P.
Find the distance of P from any of the 3 pts of tangency.

3> The probabilities that a office-goer travels by bus,car,bike or local train are 3/7,1/7,2/7 and 1/7 resp.
The corres. probabilities that he reaches his office late while commuting by the above means are 1/9,2/9,4/9 and 1/9
Find the probability that the man has travelled by car if he reaches his office ON time .

Source of the last two problems : This year's IIT-JEE mains (IIT-JEE - Indian Institute of Technologies, Joint Entrance Exam)

[quote="djshowdown2"]3.
[hide]Conditional probability.  Caring only about occurences where he has made it on time, divide the probability of him getting to work on time by car by the total probability that he got to work on time by any means.

$\frac{\frac{2}{63}}{\frac{24}{63}+\frac{2}{63}+\frac{8}{63}+\frac{1}{63}} = \frac{2}{35}$[/hide][/quote]

Ans. is different ...

[hide]Hint : Try applying Baye's theorem[/hide]

dj, your numerator isn't what you want it to be.  Otherwise, your solution is good.

oops, I started off remembering that those represented late times, not on times, but forgot partway through the problem  :P

[hide]So $\displaystyle \frac{\frac{7}{63}}{\frac{24}{63}+\frac{7}{63}+\frac{10}{63}+\frac{8}{63}} = \frac{7}{49}=\frac{1}{7}$ , verdad?[/hide]

[hide="Number 1"]The numbers $10^x$ mod 7 with $x \ge 1$ and $x$ an integer  are 3,2,6,4,5,1,3,2,6,4,5,1,3,2.... etc.
So setting each exponent in mod 6, we get $10^4+10^4+10^4+10^4...10^4=10\cdot 10^4=10^5$.  Once we divide $10^5$ by 7, we get the remainder to be 5.

Thus, the answer is 5.[/hide]

2
[hide]Connecting the center's of the circle we form a triangle with sides lengths of 7, 8, and 9. Let $x$ denote the distance from the point P to any of the point of tangencies.  Since the sides of the triangle go directly through the points of tangencies, $x$ is perpendicular to the sides. We can find the area of the triangle in terms of $x$.

$4x + \frac {7x}{2} + \frac {9x}{2} = 12x$

By Heron's, the area is also

$\sqrt {12(5)(3)(4)} = 12 \sqrt 5$

Setting the two equal

$12x = 12 \sqrt 5 \Rightarrow x = \sqrt 5$[/hide]

[quote="djshowdown2"][quote="Chigr"]dj, shouldnt it be [hide]1/7=9/63 not 7/63 in the numerator?[/hide][/quote]

The numerator is the probability that he drives a car [i]and[/i] is on time.  So it is (1/7)(1-2/9)=7/63[/quote]

bingo !!! 1/7 is the correct answer


muon timloi giai cua trung quoc tst tren mang
ai biet tim thi thong bao ho tui voi
tui dang rat can
cag nhieu cang tot
thankssss!!! :lol:

[quote="N.T.TUAN"]In forum this  :wink:[/quote]
toi ko hieu y cua ban ???

[quote="hunter_cvp_ic"]OH ko tui muon loi` giai
ccon` de` thi` tui co' roi`
 lam` dc tam` tam nhung con` nhieu` bai` nchua giai dc
  :)[/quote]

Chưa giải được thì post ra ngoài, bên này họ giải cho. Bí quá cậu PM cho mấy bạn Trung Quốc ấy  :wink: