Let $A=\{x|2^{1+x}+2^{1-x}=a\}, \ B=\{\sin\theta|\theta\in\mathbb R\}$, and $A\cap B$ be a singleton. Find the range of $a$.

Obv. $B=[-1,1]$.
The equation
(1)   $2^{1+x}+2^{1-X}=a$
should have a single solution in the interval $[-1,1]$.
If $x_0$ is a solution in $[-1,1]$, then $-x_0$ is also a solution in $[-1,1]$, since
$2^{1+x_0}+2^{1-x_0}=a \mbox { \Rightarrow }2^{1-x_0}+2^{1+x_0}=a$
and $x_0 \in [-1,1] \mbox { \Rightarrow }-x_0\in [-1,1]$.
Hence, the only case in which equation (1) has a single solution $x_0$in $[-1,1]$ is that when $x_0=0$.
Subsituting $x=x_0=0$ in $(1)$ we get $a=4$ and this is also sufficient for equation $(1)$ to have a single solution.


Would you rather have a 50 percent chance of winning 1,000,000 dollars or a 20 percent chance of winning 3,000,000?

[hide="I highly doubt that either situation will occur any time soon"]The second choice, because it has an expected value of $ \$$600,000 as opposed to $ \$$500,000 for the first one[/hide]

[quote="t0rajir0u"]would you ever actually play it if you had to pay, say, a dollar every time?[/quote]

Only if I got to play a hell of a lot of times :)

[quote="t0rajir0u"]Personally, I'd rather take the first one.  Expected value is only a meaningful comparison if the event is going to happen a large number of times relative to the probabilities involved, which this one isn't.

Example:

I'll let you toss a coin until you get tails.  If you get $ n$ heads, I will give you $ 2^{n}$ pennies.

This game has [i]infinite[/i] expected value, but would you ever actually play it if you had to pay, say, a dollar every time?[/quote]

The people who wrote the book probably aren't expecting you to actually think about that. It's just supposed to be an exercise in expected value, and you're supposed to just plug in the numbers and take that answer. This has been mathnerd314's cynical ranting. Thank you.


The line $ y\equal{}3\minus{}2x$ contains points in how many quadrants of
the Cartesian coordinate plane?

you dont have to graph it...

a line will pass through 3 quadrants unless the slope is 0 or undefined or b=0 where $ y\equal{}ax\plus{}b$

Don't forget the case where the line passes through the origin, it then only passes through two quadrants.


Each of the players in a tennis tournament played one match against each of the others. If every player won at least one match, show that there are three players $A$, $B$, and $C$ such that $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$. Such a triple of player is called a [b]cycle[/b]. Determine the number of maximum [b]cycles [/b]such a tournament can have.

The first half is equivalent to the statement, "Every acyclic tournament is totally ordered, i.e. there is a player who wins all matches, a player who loses only once, etc., and in particular a player who loses all matches."  This is classical.  I suspect the second half has been on the forum before, and recommend a search.


This is an SAT question and I'm not sure if it goes here...

A function f(c) is defined as the area of the circle with circumference c. For what value of z is f(z) equal to f(3)+f(4)?

This was a grid in so there is no answer choices any help is great. Thanks.

[hide="Hint"]$ c\equal{}2\pi r \Rightarrow r\equal{} \frac{c}{2\pi}$. So $ f(c) \equal{} \pi\left(\frac{c}{2\pi}\right)^2 \equal{} \frac{c^2}{4\pi}$[/hide]
[hide="Solution"]$ f(3)\plus{}f(4) \equal{} \frac{3^2\plus{}4^2}{4 \pi} \equal{} \frac{5^2}{4 \pi} \equal{} f(5)$. So $ z\equal{}5$.[/hide]

[hide="Solution"]
The area of a circle is $ r^2\pi$, the circumference of a circle is $ 2\pi r$ where $ r$ is the radius.

So if $ f(c)$ is the area of a circle with circumference $ c$, then $ f(c) \equal{} \dfrac{c^2}{4\pi}$

Next we want to find a $ z$ such that $ f(z) \equal{} f(3) \plus{} f(4)$, so we calculate the R.H.S:

$ f(3) \plus{} f(4) \equal{} \dfrac{3^2}{4\pi} \plus{} \dfrac{4^2}{4\pi} \equal{} \dfrac{25}{4\pi} \equal{} \dfrac{5^2}{4\pi} \equal{} f(5)$.

A side note: If $ f(a)\plus{}f(b)\equal{}f(c)$ then triangle $ abc$ is a right triangle.
[/hide]

Edit: Beaten..

Thanks for the help guys I thought I got 5 too but I didn't think it was right


At the request of several members, we've built a tool that will allow you to include a little window into the Forum from your own site.  As you'll see, it's highly customizable, allowing you to choose which forums you can see, window sizes, colors, fonts, font styles, etc.

To learn how to use it click here:

[url]http://www.artofproblemsolving.com/Forum/marquee.php?help=1[/url]

Please let me know if you encounter any problems with it (but please read the instructions closely first if you have a problem!)

(Special thanks to Tim Sanders of [url=http://www.mathleague.org]mathleague.org[/url] for making a suggestion that led to this being built).

Nifty! Also good for advertising AoPS on a banner ad.

Awesome! Lots of good uses for this. Now it's easier for me to monitor topics in different forums. Thanks.  :first:

Nice thing... but I think you should at least check permissions. Now anyone can check moderator forum, parents forum, etc... ;)

Have you confirmed that you can look into these forums?  Please show me a sample if so - I thought I had cut that off.

I logged out and could still view forum 373. (just replacing the forums with 373 in the example link)

Maybe it's just one forum forgotten on the list if you made sort of a blocklist :)

Try it from a browser in which you are not logged in to the site - you have permission to see that forum, so you can get it in the marquee.  (Let me know if that fails.)

No, I don't think you can. I tried 373 from a friends account and 301 from mine, and couldn't get it. Though if those are the only forums listed on your account, it would be nice if it had a Permission denied message instead of blankness.

Which websites can this work in?

It doesn't work for my xanga, freewebs, or myspace.

I'm guessing they don't allow the 'iframe' tag, which would make it harder for them to control what sort of content appears on their site.

Um....there's a bug. A topic is shown that it is authored by a certain person, but it actually shows the person who last responded to it as the author. :huh:

great!!!!!! this is good for Internet explorer  :lol:


does any one know how to apply for aptitude test which is for B.Arch
i filled my form online and there was no such option

You can submit it only after 26th May 09 when JEE results are out
See this link
http://jee.iitd.ac.in/aptitude.htm

can anyone suggest some books for the aptitude test for barch that i should go through


Each point in a plane is colored either red or blue.  Show that there exists a rectangle whose all vertices are of the same color.

[hide="An Approach"]
Consider a rectangular array with 3 rows and 9 columns.

Each column has 3 points, thus there are 2^3 = 8 possible different columns. Thus if there are 9 columns in the array, two must be the same.

Also, since there are 3 points in each column, at least two must be of the same color.

Taking two points of like color from each of two identical columns will result in a rectangle with similarly colored vertexes
[/hide]


inequalities
geometry
perimeter
inradius
circumcircle
inequalities unsolved

[b]Q.1.[/b] For triangle ABC,

$ (a^2 \plus{} b^2 \plus{} c^2) \geq \frac{36}{35}(s^2 \plus{} \frac{abc}{s})$

[b]Q. 2.[/b] For an acute angled triangle ABC prove that

$ s^2 \leq \frac{27R^2}{27R^2\minus{}8r^2} (2R \plus{} r)^2$

  s – semi-perimeter, r – inradius, R - circumradius

[color=red][b]Anybody got a clue ?[/b][/color]

[color=blue][i].......[/i][/color]

Both the inequalities are easily solvable using the substitutions $ a\equal{}y\plus{}z, b\equal{}z\plus{}x, c\equal{}x\plus{}y$ where $ x,y,z$ are positive reals.
THe first one is really simple. Just keep collecting like terms and prove $ LHS\minus{}RHS \ge 0$.
As for the second one, use $ R\ge 2r$. $ R\equal{}\frac{(x\plus{}y)(y\plus{}z)(z\plus{}x)}{4\sqrt{xyz(x\plus{}y\plus{}z)}}$. $ r\equal{}\sqrt{\frac{xyz}{x\plus{}y\plus{}z}}$

hello, for the first one we have
$ {a^2+b^2+c^2\geq\frac{4}{3}s^2=\frac{36}{35}\left(s^2+(\frac{2}{3}s)^3\cdot\frac{1}{s}\right)=\frac{36}{35}\left(s^2+(\frac{a+b+c}{3})^3\cdot\frac{1}{s}\right)\geq\frac{36}{35}\left(s^2+\frac{abc}{s}\right)}$,
we have used that
$ a^2+b^2+c^2\geq\frac{1}{3}(a+b+c)^2$
and
$ \left(\frac{1}{3}(a+b+c)\right)^3 \geq abc$.
Sonnhard.


Find all positive integers $ n > 1$ such that for any two co prime factors
$ a, b$ of $ n$ the number $ a \plus{} b \minus{} 1$ is also a factor of $ n$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=267713]click[/url]

Obviously $ n\equal{}p^s$ satisfies for any prime $ p$ and any natural $ s$.
Now assume $ n$ has at least two different prime factor and $ p$ is the smallest.
We write $ n\equal{}p^sN$, where $ (N,p)\equal{}1$ and $ N$ has only greater-than-p factor.

Case 1) $ p>2$.
Since $ N\plus{}p\minus{}1|n$, $ (N\plus{}p\minus{}1,p)>1$ and/or $ (N\plus{}p\minus{}1,N)>1$.
$ (N\plus{}p\minus{}1,N)\equal{}(p\minus{}1,N)<p$. But $ N$ has only greater-than-p factor. So $ (N\plus{}p\minus{}1,N)\equal{}1$
Now $ (N\plus{}p\minus{}1,p)>1$ implies
$ N\plus{}p\minus{}1\equal{}p^t$ with $ 2\le t\le s$. (*)
Since $ s\ge2$, we have $ N\plus{}p^2\minus{}1|n$.
Since $ p^2\minus{}1\equal{}(p\plus{}1)(p\minus{}1)$ has no greater-than-p factor, $ (N\plus{}p^2\minus{}1,N)\equal{}1$.
So $ N\plus{}p^2\minus{}1\equal{}p^u$ with $ 2\le u\le s$. (**)
Now (*) and (**) give us
$ p^2\minus{}p\equal{}p^u\minus{}p^t\equal{}p^t(p^{u\minus{}t}\minus{}1)\Rightarrow t\equal{}1,u\equal{}2$.
Then $ N\equal{}1$. Contradiction!

Case 2) $ p\equal{}2$.
$ (N\plus{}1,N)\equal{}1$ implies $ N\plus{}1\equal{}2^w$ with $ 2\le w\le s$
Since $ s\ge2$, $ N\plus{}2^2\minus{}1\equal{}N\plus{}3|n$.
$ (N\plus{}3,N)\equal{}(3,N)$ implies $ N\plus{}3\equal{}3^u2^v$ with $ 1\le v\le s$.

$ 3^u2^v\minus{}2^w\equal{}2$ implies $ v\equal{}1$ and thus
$ 3^u\minus{}2^{w\minus{}1}\equal{}1$ (***)

The only possible solution for (***) is $ (u,w)\equal{}(1,2)$ or $ (u,w)\equal{}(2,4)$
Then $ N\equal{}2^w\minus{}1$ is either $ 3$ or $ 15$.
In other words, $ n\equal{}2^s*3$ or $ n\equal{}2^s*15$, where $ s\ge2$.
For $ n\equal{}2^s*3$, if $ s\ge3$, $ 8\plus{}3\minus{}1\equal{}10|n$. Contradiction!
For $ n\equal{}2^s*15$, $ 4\plus{}15\minus{}1\equal{}18|n$. Contradiction!

So $ n\equal{}12$ is the only solution other than power of primes.


The diagonal of a cube is represented by 2x. Find the total surface of the cube in terms of x.

if its space diagonal, with the side of the cube s, then 2x=Sqrt(3)*s. then s=2x/sqrt(3), and 6s^2=8x^2.
so the answer is 8x^2.

first of all, you posted the same problem twice.
second, maybe you should post all of your questions under one thread.
third, the diagonal of a cube with side $ s$ is always $ s\sqrt{3}$. therefore, the side length of the cube is $ \frac{2x}{\sqrt{3}}$. the surface area of a cube with side $ s$ is always $ 6s^2$, so the surface area is $ 6(\frac{2x}{\sqrt{3}})^2\equal{}6(\frac{4x^2}{3})\equal{}8x^2$

We can use the Pythag. Theorem to figure this out.
$ a^2\plus{}b^2\equal{}c^2$
$ a^2\plus{}b^2\equal{}2x$
In this case, $ a\equal{}b$ because we have a cube in which all sides are equal length.
$ a^2\plus{}a^2\equal{}2x$
$ 2(a^2)\equal{}2x$
We divde both sides by 2...
$ a^2\equal{}x$
We "sqaure root-ify" both sides
$ \sqrt{a^2}\equal{}\sqrt{x}$
$ a\equal{}\sqrt{x}$
The surface area of a cube is Perimeter times height
$ [4(\sqrt{x})][\sqrt{x}]$
Which is $ 5\sqrt{x}$
[b]I feel like I made a mistake along the way but I can't spot it... What did I do?![/b]

He said the space diagonal. The space diagonal is a line that starts from one corner that runs to the opposite corner running through and inside the cube.

Oh... Hah... my bad.... But still, If you were to use it, what would you find?


Solomon Golomb's selfdescribing sequence $ < f(1), f(2), \dots>$ is the only nondecreasing sequence of positive integers with the property that it contains exactly $ f(k)$ occurrences of $ k$ for each $ k$. Find the sequence.

We want $ f(k)$ to be 1 more than $ f(k \minus{} 1)$ if and only if $ f(k \minus{} 1)$ has already appeared $ f(f(k \minus{} 1))$ times in the sequence.  Another way to word this is, we want $ f(k)$ to be 1 more than $ f(k \minus{} f(f(k \minus{} 1)))$:
\[ \begin{cases} f(k) \equal{} 1 \plus{} f(k \minus{} f(f(k \minus{} 1))) \\
f(1) \equal{} 1 \end{cases}
\]
I don't know if I could find a closed form for this, but at first glance it has me stumped.  Though you could probably get some asymptotic bounds.  Any ideas?

isn't it kind of obvious?
let f(1)=1.
then the minimum value of f(2)=2, since it cannot be 1.
the minimum value of f(3)=2
then f(4)=3 and f(5)=3
then f(6)=f(7)=f(8)=4 and f(9)=f(10)=f(11)=5
and keep constructing the sequence like so.

Right, which is the recurrence I posted above.  But I imagine the OP was looking for a closed form for $ f(k)$ as a function of $ k$, or at least for asymptotic bounds, i.e. a function $ \phi(k)$ so that $ \frac {f(k)}{\phi(k)}\to 1$ as $ k\to \infty$.

There is a closed form which involves exponential function or logarithm, but I can't remember......

Well I managed to prove $ \frac{f(k)}{\phi(k)}\to 1$ as $ k\to \infty$ where
\[ \phi(k)\equal{}\left(\frac{1\plus{}\sqrt{5}}{2}\right)^\frac{3\minus{}\sqrt{5}}{2}k^{\frac{\sqrt{5}\minus{}1}{2}}\]
Can't say too much about the remainder though, except that it decays pretty slowly.


geometry
circumcircle
trigonometry
angle bisector

[quote="April"]An acute triangle $ ABC$ with $ AB \neq AC$ is given. Let $ V$ and $ D$ be the feet of the altitude and angle bisector from $ A$, and let $ E$ and $ F$ be the intersection points of the circumcircle of $ \triangle AVD$ with sides $ AC$ and $ AB$, respectively. Prove that $ AD$, $ BE$ and $ CF$ have a common point.[/quote]
An easy problem .
We have $ DE \perp AC,DF\perp AB$ and $ AD$ is bisector of $ \angle {ABC}$  so
$ AE \equal{} AF$
Other $ CF \equal{} CD.\cos {\angle {ACB}},BE \equal{} BD.\cos {\angle {ABC}}$
 $ VB \equal{} AB.\cos {\angle {ABC}},VC \equal{} AC.\cos {\angle {ACB}}$
Follow that :
$ \frac {VB}{VC}.\frac {FC}{FA}.\frac {EA}{EB} \equal{} \frac {AB}{AC}.\frac {CD}{BD} \equal{} 1$
So $ AV,BE,CF$ concurrent (Ceva's theorem )

[quote="April"][color=darkred] An acute triangle $ ABC$ with $ AB \neq AC$ is given. Let $ V$ and $ D$ be the feet of the altitude and angle bisector from $ A$, and let $ E$ and $ F$ be the intersection points of the circumcircle of $ \triangle AVD$ with sides $ AC$ and $ AB$, respectively. Prove that $ AV$, $ BE$ and $ CF$ have a common point. [/color][/quote]

[color=darkblue]Here is an easy extension which generates and another interesting problem ![/color]

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be an acute triangle with $ c < b$ and let $ M\in (BC)$ , $ N\in (MC)$ so that $ \{\begin{array}{c} x = m(\widehat {AMC}) \\
\ y = m(\widehat {ANB})\end{array}$ .

The circumcircle of the triangle $ AMN$ cut again the sides $ [AB]$ , $ [AC]$ in the points $ F$ , $ E$ respectively.

Then the lines $ BE$ , $ CF$ , $ AM$ are concurrently $ \Longleftrightarrow$ $ \cos (C - B + x - 2y) + \cos A\cos x = 0$ .[/color][/quote]

[color=darkblue][b][u]Particular cases.[/u] [/b]

$ 1\blacktriangleright\ x: = 90^{\circ}$ , i.e. $ AM\perp BC$  . In this case, $ \{\begin{array}{c} C - B + 90^{\circ} - 2y = \pm 90^{\circ} \\
\ (\ C\ < \ B\ )\end{array}$ $ \Longleftrightarrow$ $ y = C + \frac A2$ $ \Longleftrightarrow$ $ \widehat {NAB}\equiv\widehat {NAC}$ .

$ 2\blacktriangleright\ A: = 90^{\circ}$ . In this case obtain the following nice problem :[/color]

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a $ A$ - right triangle, i.e. $ AB\perp AC$ with $ c < b$ . Choose two points  $ \{\begin{array}{c} M\in (BC) \\
\ N\in (MC)\end{array}$ .

The circumcircle $ w$ of the triangle $ AMN$ cut again the sides $ [AB]$ , $ [AC]$ in the points $ F$ , $ E$ respectively.

Then the lines $ BE$ , $ CF$ , $ AM$ are concurently $ \Longleftrightarrow$ $ m(\widehat {AMC}) = 2\cdot m(\widehat {ANC})$ .[/color][/quote]

[color=darkblue]$ 3\blacktriangleright$ Here is a very easy problem which, for $ A = 90^{\circ}$, is equivalently with the previous problem :[/color]

[quote="Virgil Nicula"][color=darkred]Let $ ABC$ be a triangle. For a point  $ M\in (BC)$ denote the points

$ \{\begin{array}{c} F\in (AB)\ ,\ \widehat {FMA}\equiv\widehat {FMB} \\
 \\
E\in (AC)\ ,\ \widehat {EMA}\equiv\widehat {EMC}\end{array}$ . Then the lines $ BE$ , $ CF$ , $ AM$ are concurently.[/color][/quote]

We note that

$AE=AF$

$CE=DCcosC$

$BF=BDcosB$

So by converse of Ceva,done.......
But can anyone give a solution using the fact that the radical axes of three circles taken pairwise are concurrent....

Maths is the doctor of science

Sayantan

$\angle AVF = \angle AEF = \angle AFE = \angle AVE \implies (VA, VB; VE, VF) = -1$.

well an easy problem and as everyone has pointed,it can be easily done by Ceva's converse.
Meanwhile there is a typo in the main problem it is written that prove $AD,BE and CF$ are concurrent.I think it would be $AV,BE and CF$.

Since $ AF=AE $ we need to prove that $ \frac{BV}{VC}\frac{CE}{BF}=1 $ By the power of a point theorem $ BV=\frac{BF}{BD}BA $ so it remains to prove that $ \frac{CE}{CV}\frac{BA}{BD} =1 $ Note that $ \frac{CE}{CV}=\frac{CD}{CA} $ and now $ \frac{CD}{CA}\frac{BA}{BD}=\frac{sin \angle DAC}{sin \angle ADC} \frac{sin \angle BDA}{sin \angle BAD}=1 $

Very easy using Ceva! :D
$\angle FAD = \angle EAD \wedge DF \perp AF, DE \per AE \wedge AD$ common side $\implies \triangle AFD \cong \triangle AED$.
$\therefore ED = FD$ and $AF = AE$.
Now, $\frac{AF}{FB} \frac{BV}{VC} \frac{CE}{EA} = \frac{BV}{VC} \frac{ED/ tan \angle C}{FD/ tan \angle B} = \frac{BV}{VC} \frac{AV/ tan \angle C}{AV/ tan \angle B}=1$.
$\therefore AV, BE, CF$ concur!

Dear Mathlinkers,
a proof with harmonic division is also possible...
Sincerely
Jean-Louis

[hide="Solution using harmonic division!"]

Let $BC \cap EF = P$ and $AV \cap EF = Q$.
$AE=AF$ and $AFVE$ cyclic. So we must have $VA$ is the angle bisector of $\angle FVE$ and $AV \perp BC$. So $PFQE$ is a harmonic division.
$\therefore (P, V; B, C) \stackrel{A}{=} (P, Q; F, E) = -1, i.e.,$ it's also harmonic!
Thus $AV, BE, CF$ must concur.[/hide]

[url=https://artofproblemsolving.com/community/c6h1612634p10075334]Korea National Olympiad 1996 P8[/url]


If $ \frac{a}{b}\equal{}\frac{3}{4}, \frac{b}{c}\equal{}\frac{8}{9}$ and $ \frac{c}{d}\equal{}\frac{2}{3}$, what is the value of $ \frac{ad}{b^2}$   Express your answer as a common fraction.

ok, let's make each of the letters the same number

the third fraction, $ \frac {2} {3} \equal{} \frac {c} {d}$ and the second fraction, $ \frac {8} {9} \equal{} \frac {b} {c}$, have a c in common. let's make the two c's the same value
we multiply the numerator and denominator of the third fraction by 9, and the numerator and denominator of the second fraction by 2, to make both c's the same

$ \frac {2 \times 9} {3 \times 9} \equal{} \frac {18} {27}$

$ \frac {8 \times 2} {9 \times 2} \equal{} \frac {16} {18}$

now let's make the two b values the same. the b value in the second fraction is now equal to 16, so we multiply the numerator and denominator of the first fraction by 4

$ \frac {3 \times 4} {4 \times 4} \equal{} \frac {12} {16}$

so now we have the values,
$ \frac {a} {b} \equal{} \frac {12} {16}, \frac {b} {c} \equal{} \frac {16} {18}, \frac {c} {d} \equal{} \frac {18} {27}$

we have the values for a, b, c, and d now, so we can solve our problem

$ a \equal{} 12, b \equal{} 16, c \equal{} 18, d \equal{} 27$

we plug it into our equation
$ \frac {ad} {b^2} \equal{} \frac {12 \times 27} {16^2} \equal{} \frac {324} {256} \equal{} \frac {81} {64}$

therefore, our final answer is $ \frac {81} {64}$

if anyone sees any mistakes in that, plz tell me

Vallon put the ratios so that they are in relation with each other.

$ A : B : C : D \equal{} 12 : 16 : 18 : 27$.

This may count as spam.  Not sure.


logarithms
trigonometry
calculus
calculus computations

According to Mathematica: No, it doesn't converge. Changing the exponent 2 for e and pi doesn't affect that fact. But it is interesting to look up the series where 2 is substituted with n (the same number which multiplies e and divides pi). That one converges.

this sum $ \sum_{1}^{N} ...$  is like something $ \approx A\plus{}B\ln N$

sorry... fixed !  :(  difficult to concentrate with little kids running around you know.  :oops:

With the formula in this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=200404]topic[/url] we can derive the two formulas

$ \sum_{k\in\Bbb{Z}} \frac{\frac{\alpha^2}{2}}{k^4\plus{}\left(\frac{\alpha^2}{2}\right)^2}\equal{}\frac{\pi}{\alpha}\, \frac{\sinh\alpha\pi\plus{}\sin\alpha\pi}{\cosh\alpha\pi\minus{}\cos\alpha \pi}$

and

$ \sum_{k\in\Bbb{Z}} \frac{k^2}{k^4\plus{}\left(\frac{\alpha^2}{2}\right)^2}\equal{}\frac{\pi}{\alpha}\, \frac{\sinh\alpha\pi\minus{}\sin\alpha\pi}{\cosh\alpha\pi\minus{}\cos\alpha \pi}$ .

Your sum $ S\equal{}\sum_{k\equal{}1}^\infty \frac{1}{k^2 e^2\plus{}\frac{\pi^2}{k^2}}$ equals

$ \frac12 \sum_{k\in\Bbb{Z}} \frac{k^2}{k^4 e^2\plus{}\pi^2}\equal{}\frac{1}{2e^2} \sum_{k\in\Bbb{Z}} \frac{k^2}{k^4\plus{}\left(\frac{\sqrt{\frac{2\pi}{e}}^2}{2}\right)^2}$ .

So $ S\equal{}\frac{1}{2e^2}\, \frac{\pi}{\sqrt{\frac{2\pi}{e}}}\,
 \frac{\sinh \sqrt{\frac{2\pi^3}{e}}\minus{}\sin \sqrt{\frac{2\pi^3}{e}}}{\cosh \sqrt{\frac{2\pi^3}{e}}\minus{}\cos \sqrt{\frac{2\pi^3}{e}}}$ .


vector
geometry
3D geometry
inequalities unsolved
inequalities

Let vi, 1 \leq i \leq 95, be vectors in 3D space, all contained in the cube
C = {(x, y, z) \in R3 || |x|, |y|, |z| \leq 1}.
For all vectors of the vorm
\sum1 \leqk \leq 95skvk
where vi \in {-1, 1}, for 1 \leq i \leq 95,
find the smallest d such that there MUST exist a vector (a, b, c) of that form satisfying
a2 + b2 + c2 \leq d.

AUSTRIA _ POLAND 1995

[quote="Arne"]Let $v_i$, $1 \leq i \leq 95$, be vectors in 3D space, all contained in the cube
$C = {(x, y, z) \in R^3 || |x|, |y|, |z| \leq 1}$.
For all vectors of the form
$\sum_{1 \leqq \leq 95}{s_k v_k}$
where $v_i \in {-1, 1}$, for $1 \leq i \leq 95$,
find the smallest $d$ such that there MUST exist a vector $(a, b, c)$ of that form satisfying
$a^2 + b^2 + c^2 \leq d$.

AUSTRIA _ POLAND 1995[/quote]

I see this problem in other place and the original says that $d=48$, anyone can help me solving this?, thank you.


Let $\alpha$ and $\beta$ be two differrent values of $x$ satisfying the equation:
$acosx+bsinx=c$ with $a^{2}+b^{2}\neq 0$
Prove that:
1) $4cos^{2}\frac{\alpha}{2}cos^{2}\frac{\beta}{2}=\frac{(a+c)^{2}}{a^{2}+b^{2}}$
2) $sin(\alpha+\beta)=\frac{2ab}{a^{2}+b^{2}}$
3) $cos^{2}\frac{\alpha-\beta}{2}=\frac{c^{2}}{a^{2}+b^{2}}$

When i have posted this problem, i have idea to solve part a of problem:
From hyppothesis of it, we have:
$acosx+bsinx=c\Leftrightarrow acosx-c=-bsinx \Leftrightarrow( acosx-c)^{2}=(-bsinx)^{2}\Leftrightarrow (a^{2}+b^{2})cos^{2}x-2acosx+c^{2}-b^{2}=0$
Infer $cos\alpha$ and $cos\beta$ are two roots of the equation with variable $t$:
$(a^{2}+b^{2})t^{2}-2act+c^{2}-b^{2}=0$
Follow Vieta' theorem: $cos\alpha+cos\beta=\frac{2ac}{a^{2}+b^{2}}$ and $cos\alpha.cos\beta=\frac{c^{2}-b^{2}}{a^{2}+b^{2}}$
Hence:
$4cos^{2}\frac{\alpha}{2}.cos^{2}\frac{\beta}{2}=(1+cos\alpha)(1+cos\beta)=1+(cos\alpha+cos\beta)+cos\alpha.cos\beta=1+\frac{2ac}{a^{2}+b^{2}}+\frac{c^{2}-b^{2}}{a^{2}+b^{2}}=\frac{(a+c)^{2}}{a^{2}+b^{2}}$

Hien, the correct way to make the trig expressions is "\cos" not "cos". Like this : $\cos$ versus $cos$. The first is much neater!

I will prove part 3 of problem:
By applying similarly the way of part 1 we have: $\sin\alpha.\sin\beta=\frac{c^{2}-a^{2}}{a^{2}+b^{2}}$
Hence:
$\cos^{2}\frac{\alpha-\beta}{2}=\frac{1}{2}(1+\cos(\alpha-\beta))$
$=\frac{1}{2}(1+\cos\alpha.\cos\beta-\sin\alpha.\sin\beta)$
$=\frac{1}{2}\left(1+\frac{c^{2}-b^{2}}{a^{2}+b^{2}}+\frac{c^{2}-a^{2}}{a^{2}+b^{2}}\right)=\frac{c^{2}}{a^{2}+b^{2}}$

Who can help me solve remain part of one? Part 2!

Dear [b]felipesa[/b]
It is very simple, however i don't think that. Thank a lot!


Define $ x\otimes y \equal{} x^3 \minus{} y$. What is $ h\otimes (h\otimes h)$?

$ \textbf{(A) } \minus{} h\qquad \textbf{(B) } 0\qquad \textbf{(C) } h\qquad \textbf{(D) } 2h\qquad \textbf{(E) } h^3$

[hide]
$h \otimes (h \otimes h)$
$= h \otimes (h^3 - h)$
$= h^3 - (h^3 - h)$
$= h$
$\boxed{C}$
[/hide]


[b]Αγαπητοί φίλοι , προσοχή !!![/b]

 Όταν στέλνετε links για καλά βιβλία , [color=red]ΠΟΤΕ ΝΑ ΜΗΝ ΓΡΑΦΕΤΕ ΤΟΝ ΑΓΓΛΙΚΟ ΤΙΤΛΟ [/color]στην επικεφαλίδα του μηνύματος , ούτε στο κυρίως κείμενο !
 Τον βλέπουν και οι κατάσκοποι των εκδοτών και απαιτούν από τον Vornucu  ή από τους moderators να τα διαγράψουν.
 Χθες δεν μπήκα στο ελληνικό φόρουμ , αλλά μου είπε φίλος ότι [color=red]απαίτησαν από μένα [/color]να  διαγράψω το προηγούμενο μήνυμα του Michailk. Επειδή δεν  μπήκα, δεν είδα αυτή την προτροπή και αφού δεν διέγραψα εγώ το μήνυμα , το έκαναν άλλοι από την αμερική.

 Λοιπόν , να ακολουθήσουμε όλοι μας αυτόν τον κανόνα , ώστε να μην μας διαγράφουν τους συνδέσμους για τα καλά ηλεκτρονικά βιβλία.

 [u]Μπάμπης[/u]

Kυριε Μπαμπη,εχετε απολυτο δικιο.
πρέπει να βρούμε ένα τρόπο να αποκρύψουμε αυτές τις πληροφορίες
απο τους ''μπαμπούλες'' του είδους,γιατι ειναι
κρίμα βιβλία όπως το Compendiu*,να διαγράφονται!

Κύριε Στεργίου πολύ καλή η επισήμανση σας.
Θα ήθελα να ρωτήσω κι εγώ κάτι επί του θέματος των ηλεκτρονικών books.

Στο δίσκο του υπολογιστή μου έχω συγκεντρώσει κάπου στα 2000 βιβλία σε μορφή pdf για τους κλάδους της Ανάλυσης,Γεωμετρίας,Άλγεβρας,Μαθηματικής Φυσικής,Συνδυαστικής,Τοπολογίας,Μαθηματικής Λογικής,θεωρίας αριθμών,Κωδικοποίησης,Θεωριας Χάους,Παιγνίων αλλά και Εφαρμοσμένων Μαθηματικών.
Το συνολικό τους μέγεθος ξεπερνά τα 5 GB.

Το πρόβλημα είναι ότι προφανώς δεν μπορούν να διαβαστούν όλα απο εμένα,αλλά δεν γνωρίζω και αν μπορώ να τα διαθεσω στον κόσμο.
Δεν έχω ελεγξει Copyrights και σκέφτομαι ότι ίσως μια τέτοια διάθεση (ανέβασμα στον Rapidshare για παράδειγμα) θα ήταν παράνομη.

Μπορεί κάποιος να με διαφωτίσει επί του θέματος;  :|

Νέε μου φίλε, καλώς όρισες στην παρέα!
 Λοιπόν , ως προς το παρόν φύλλαξε καλά αυτά τα αρχεία. Σιγά σιγά θα βρούμε τρόπο να τα αποθηκεύσουμε κάπου με σχετική ασφάλεια , ώστε να είναι διαθέσιμα σε όλους.

 Σε αυτό το χώρο ίσως δημιουργήσω αργότερα ένα υποφόρουμ με φορτωμένα τέτοια βιβλία. Θα βρούμε λύση. Αρκεί να έχουμε τα βιβλία !

 Μπάμπης


modular arithmetic
induction
number theory proposed
number theory

For any positive integer n, let $f(n)$ be the number of positive divisors of $n$ which equal ±1 mod 10, and let g(n) be the number of positive divisors of n which equal $\pm 3$  mod 10. Show that $f(n) \geq g(n)$.

see http://www.mathlinks.ro/Forum/viewtopic.php?t=134074
All problem from VMO 1992 I posted :D

$\pm 1,\pm 3$ are all the coprime residues modulo $10$,
and for $n=n_{1}n_{2}$ with $\gcd (n_{1},n_{2})=1$ we have
$\{d\in\mathbb{N}: d|n\}=\{d_{1}d_{2}: d_{1},d_{2}\in\mathbb{N},d_{1}|n_{1},d_{2}|n_{2}\}$
and $(\pm 1)\cdot(\pm 1)\equiv (\pm 3)\cdot(\pm 3)\equiv \pm 1 \pmod{10}$ and $(\pm 1)\cdot(\pm 3)\equiv (\pm 3)\cdot(\pm 1)\equiv \pm 3 \pmod{10}$.
So $f(n)=f(n_{1})f(n_{2})+g(n_{1})g(n_{2})$ and $g(n)=f(n_{1})g(n_{2})+g(n_{1})f(n_{2})$
and $f(n)-g(n)=(f(n_{1})-g(n_{1}))(f(n_{2})-g(n_{2}))$.
So by induction it's enough to show $f(n)\geq g(n)$ for $n=p^{k}$ with $p$ prime and $k\in\mathbb{N}_{0}$.
But $f(p^{k})>0=g(p^{k})$ for $p=2,5$ or $p\equiv \pm 1\pmod{10}$
and $f(p^{k})=\left[{k+2\over 2}\right]\geq \left[{k+1\over 2}\right]=g(p^{k})$ for $p\equiv \pm 3\pmod{10}$.

Also $f(n)=g(n)\Leftrightarrow n$ has some prime factor $p\equiv\pm 3\pmod{10}$ with odd multiplicity
and $f(n)=g(n)+1\Leftrightarrow n$ has no prime factor $p\equiv\pm 1\pmod{10}$ and all its prime factors $p\equiv\pm 3\pmod{10}$ with even multiplicity.


When I first registered my username, rofler, I didn't realize how much I would actually go on the forum.
I hate that my name isn't capitalized, could any admin just change it to Rofler?
I know it is alot to ask, and I am not expecting much, but it would be nice :)

Thanks in advance,

rofler


Ring Theory
superior algebra
superior algebra solved

Could anyone explain why in $R=Z[x]$ that $(2,x)$ is not a princial ideal?

I will do an attempt

you do mean $R=\mathbb{Z}[x]$ right?
well note that the only units are 1 and -1

so suppose $(2,x)$ is a principal ideal, generated by $a$

well $a$ must divide $2$ and $x$ then
it is thus up to sign $1$ or $2$
however the latter does not divide $x$ , to see this explicitly, if $x=2 p(x)$ you immediately see a problem with highest power coefficient

so it is 1 , but that means the ideal is the full ring, and thus it also contains 1

$2 p(x) + q(x) x =1$ is not possible either however, as the left hand side is even for evalution with $x=0$ and the right hand side is not

I posted more general result in Solved Problems Section


Apparently I finished all the problems in the database and have 0 wrong and all that, but in the Report, I see there are topics I haven't gotten any problems from at all, yet it says "No more problems available".

A-ha! Now I remember why we originally disabled the view of any topics with 0 correct, 0 incorrect, and 0 given up on. Sadly, this produced a problem which is explained by MathMom in another post.

There are no questions in those subjects. They are merely placeholder subjects and parent subjects of other subjects.

I'll have to put in some kind of ugly hack to not display these.

Wow isabella 2296, you finished all the problems!!!! I'm not even close.....


There are n*n complex number $ a_{ij} (i \equal{} 1,2,3,...,n)(j \equal{} 1,2,3,...,n)$,and there are n complex number $ k_{j} (j \equal{} 1,2,3,...,n)$ ,and $ \sum_{j \equal{} 1}^n|k_{j}|! \equal{} 0$,and $ \sum_{i \equal{} 1}^n|\sum_{j \equal{} 1}^nk_{j}*a_{ij}| \equal{} 0$
Prove:there are n complex number $ l_{i} (i \equal{} 1,2,3,...,n)$ and $ \sum_{i \equal{} 1}^n|l_{i}|! \equal{} 0$ to make $ \sum_{j \equal{} 1}^n|\sum_{i \equal{} 1}^nl_{i}a_{ij}| \equal{} 0$

\cdot for a multiplication dot, \neq for the not-equals sign.  Anyhow, this is a linear algebra problem dressed up in silly clothes: it asserts that if a (complex) square matrix has row space of less-than-full dimension then it also has column space of less-than-full dimension, which is substantially weaker than the standard result that the dimensions of the row and column spaces are actually equal.

But can anybody prove it without the base of the linear algebra?


[quote="Demetres"]Μιας και πιάσαμε σύνολα σημείων στο επίπεδο

Να δειχθεί ότι κάθε αριθμήσιμο σύνολο σημείων Α του επιπέδου μπορεί να γραφτεί ως ένωση δύο συνόλων Β και Γ ώστε
α) Κάθε οριζόντια γραμμή του επιπέδου τέμνει το Β σε πεπερασμένο αριθμό σημείων
β) Κάθε κάθετος γραμμή του επιπέδου τέμνει το Γ σε πεπερασμένο αριθμό σημείων

Τι συμβαίνει αν το Α δεν είναι αριθμήσιμο;[/quote]

Πάρα πολύ ωραία άσκηση.

Ας κάνουμε πρώτα μιά ειδική περίπτωση, που περιέχει την ίδια δυσκολία.

Αν τα αριθμήσιμα σημεία που μας έδιναν είναι όλο τα ακέραια σημεία (m, n) του επιπέδου τότε:

Εξετάζουμε δύο ευθείες όπως την y = x και την y = -x.

Στο Β βάζουμε όλα τα σημεία που είναι πάνω και από τις δύο ευθείες ή που είναι κάτω και από τις δύο, δηλαδή
Β = {(m, n) / |m| < = |n| } , και C τα υπόλοιπα. Μιά ζωγραφιά δείχνει ότι ισχύει το ζητούμενο.

Γενικά τώρα. Τα τα σημεία βρίσκονται σε αριθμήσιμο πλήθος οριζόντιων ευθειών και, επίσης, σε αριθμήσιμο πλήθος κάθετων. Δηλαδή τα σημεία μας είναι οι κόμβοι σε ένα πλεγμα που "μοιάζει"  με το παραπάνω των ακέραιων σημείων. Και λοιπά, ακολουθώντας την ίδια ιδέα.

Φιλικά.

Μιχάλης Λάμπρου

Δημήτρη: τώρα συνειδητοποιώ ότι είσαι ο ίδιος Demetres που γράφει τόσο εξαιρετικά στο http://www.mathematica.gr.
Ο ίδιος γράφω εκεί, οπότε στο εδώ ΚΑΤΑΠΛΗΚΤΙΚΟ φόρουμ μπαίνω σπάνια. Δυστυχώς δεν τα προλαβαίνω όλα!

[quote="Demetres"]Μιας και πιάσαμε σύνολα σημείων στο επίπεδο

Να δειχθεί ότι κάθε αριθμήσιμο σύνολο σημείων Α του επιπέδου μπορεί να γραφτεί ως ένωση δύο συνόλων Β και Γ ώστε
α) Κάθε οριζόντια γραμμή του επιπέδου τέμνει το Β σε πεπερασμένο αριθμό σημείων
β) Κάθε κάθετος γραμμή του επιπέδου τέμνει το Γ σε πεπερασμένο αριθμό σημείων

Τι συμβαίνει αν το Α δεν είναι αριθμήσιμο;[/quote]

Είχα αφήσει αναπάντητο το ερώτημα για μη αριθμήσιμα.

Αποδεικνύεται ότι δεν γίνεται πάντα η παραπάνω διαμέριση.

Π.χ. στο [0, 1]x[0, 1] η χαρακτηριστική του συνάρτηση f(., .) έχει διπλό ολοκλήρωμα 1 (= το εμβαδόν).

Η παραπάνω υπόθεση α) μεταφράζεται στο ότι για κάθε σταθερό $ y_0$ η χαρακτηριστική g του Β έχει $ g(x , y_0)$  = 0 εκτός από πεπερασμένο πλήθος τιμών του x. Άρα έχει ολοκλήρωμα 0. Αν το κάνουμε αυτό για όλα τα $ y_0$ και ολοκληρώσουμε, ερχόμαστε σε αντίφαση με το θεώρημα Fubini (εναλλαγή σειράς ολοκλήρωσης δίνει το ίδιο αποτέλεσμα, και μάλιστα όσο το αρχικό διπλό ολοκλήρωμα).
Η ίδια τεχνική δείχνει ότι δεν μπορούμε ούτε αριθμήσιμο (αντί απλά πεπερασμένο) πλήθος στοιχείων σε κάθε οριζόντια γραμμή.

Δημήτρη, προσπάθησα να λύσω την άσκηση χρησιμοποιώντας μόνο πληθάριθμους, χωρίς Θεωρία Μέτρου. Δεν τα κατάφερα. Έχεις τέτοιο τρόπο; Θα χαρώ να τον ακούσω.

Φιλικά,

Μιχάλης Λάμπρου

Κε Λάμπρου, ευχαριστώ πολύ για τα καλά σας λόγια.

Η λύση μου στο (α) είναι ακριώς η ίδια με την δική σας. Το (β) είναι σαφώς πιο δύσκολο ερώτημα. Θα βάλω την λύση μου αργότερα (πρέπει να θυμηθώ κάποιες λεπτομέρειες.) Μπορείς πάντως να αποφύγεις την θεωρία μέτρου.

Κάτι όμως με έχει μπερδέψει στην λύση σας. Ενώ μου φαίνεται απόλυτα ορθή, αν αλλάξουμε το πεπερασμένο σε αριθμήσιμο η απάντηση αλλάζει! (Για να ακριβολογώ, για αριθμήσιμα μπορεί να γίνει αν και μόνο αν ισχύει η continuum hypothesis.)

[quote="Demetres"][quote="Lambrou"]
Δημήτρη, προσπάθησα να λύσω την άσκηση χρησιμοποιώντας μόνο πληθάριθμους, χωρίς Θεωρία Μέτρου. Δεν τα κατάφερα. Έχεις τέτοιο τρόπο; Θα χαρώ να τον ακούσω.
[/quote]

Υποθέτω ότι υπάρχει τέτοιος διαμερισμός του $ \mathbb{R}^2$.

Ας είναι $ A_n \equal{} \{x: (n,x)\in B\}$. Τότε $ \bigcup_nA_n$ είναι αριθμήσιμο. Παίρνω $ x \notin \bigcup_nA_n$. Τότε $ (n,x) \in C$ για κάθε n, άτοπο.

Κάτι πάντως χάνω στην απόδειξή σας. Δουλεύει η ίδια τεχνική αν αλλάξουμε το πεπερασμένο σε αριθμήσιμο και στις οριζοντιες και στις κάθετες γραμμές και αν όχι γιατί όχι;[/quote]

 Ναι, δουλεύει για αριθμήσιμα η τεχνική: Έστω για κάθε σταθερό $ y_0$ η χαρακτηριστική g του Β έχει $ g(x, y_0)$ = 0 εκτός από αριθμήσιμο πλήθος τιμών του x. Τότε το ολοκλήρωμα ως προς x, για σταθερό $ y_0$, είναι 0.
Δεν χρειάζεται η υπόθεση του συνεχούς. Η στάνταρ απόδειξη είναι ότι, για κάθε ε>0 σταθερό, καλύπτουμε το αριθμήσιμο $ (a_n)$ με διαστήματα κέντρου $ a_n$ μήκους  $ \frac {\epsilon}{2^n}$. Το συνολικό μήκος είναι ε και λοιπά.

Έχω και εγώ μία ερώτηση: Σίγουρα το $ A_n \equal{} \{x: (n,x)\in B\}$ είναι αριθμήσιμο (ή πεπερασμένο) ;  Ίσως δεν το βλέπω γιατί είναι αργά. Θα το σκεφτώ...

Καληνύχτα.

Μιχάλης Λάμπρου

[quote="Lambrou"]
Ναι, δουλεύει για αριθμήσιμα η τεχνική: Έστω για κάθε σταθερό $ y_0$ η χαρακτηριστική g του Β έχει $ g(x, y_0)$ = 0 εκτός από αριθμήσιμο πλήθος τιμών του x. Τότε το ολοκλήρωμα ως προς x, για σταθερό $ y_0$, είναι 0.
Δεν χρειάζεται η υπόθεση του συνεχούς. Η στάνταρ απόδειξη είναι ότι, για κάθε ε>0 σταθερό, καλύπτουμε το αριθμήσιμο $ (a_n)$ με διαστήματα κέντρου $ a_n$ μήκους  $ \frac {\epsilon}{2^n}$. Το συνολικό μήκος είναι ε και λοιπά.
[/quote]

Αυτό καταλάβαινα και εγώ. Αλλά εδώ αρχίζει το πρόβλημα μου. Πως το εξηγήτε αυτό

Sierpinski Decomposition: Το επίπεδο είναι ένωση δύο συνόλων Α και Β ώστε το Α τέμνει όλες τις οριζόντεις και το Β όλες τις κάθετες γραμμές σε αριθμήσιμα σύνολα.

Η απόδειξη χρησιμοποιεί Axiom of choice και Continuum Hypothesis. Τα αριθμήσιμα σύνολα που τέμνουν οι οριζόντες/κάθετες γραμμές φτιάχνονται με αξίωμα επιλογής.

Που βρίσκεται λοιπόν το λάθος; Το μόνο που μπορώ να σκεφτώ είναι ότι οι χαρακτηριστικές συναρτήσεις των Α και Β (ας περιοριστούμε π.χ. στο [0,1]^2) δεν είναι ολοκληρώσιμες. Αλλά γιατι; Δεν ισχύει ότι κάθε αριθμήσιμο σύνολο έχει μέτρο 0; Ακόμη και με αξίωμα επιλογής να το φτιάξω, με το ίδιο αξίωμα δείχνω ότι έχει μέτρο 0. Πρόβλημα στην χρήση του Fubini, δεν βλέπω. Ειλικρινά δεν ξέρω τι συμβαίνει.  :?:

[quote="Lambrou"]
Έχω και εγώ μία ερώτηση: Σίγουρα το $ A_n \equal{} \{x: (n,x)\in B\}$ είναι αριθμήσιμο (ή πεπερασμένο) ;  Ίσως δεν το βλέπω γιατί είναι αργά. Θα το σκεφτώ...
[/quote]

Η απάντηση είναι απλή. Παρόλο που το διπλοέλεγξα διότι ήμουν σίγουρος πως θα έκανα λάθος μπέρδεψα τις οριζόντιες με τις κάθετες γραμμές. :wallbash:


set theory
superior algebra
superior algebra solved

hello
sorry if this question is not really algebra, it's more set theory

suppose A is a set with subset B
suppose C contains D

let there be bijections between B and C, and between A and G
so roughly : A kinda contains C and C kinda contains A

is there then a bijection between A and C

sounds logic but not quite, it would allow me to prove that there is a bijection between
[0,1] and ]0,1[

fred

There are some ambiguities here, because you use $G$ once and $D$ once, and I think they are one and the same. Anyway, the well known Cantor-Bernstein Theorem states that if you have an injection from $A$ to $B$ and another injection from $B$ to $A$, then there is a bijection between them, so we can define an order relation on the cardinal numbers.

It is not so easy to define relation between cardinal numbers. ;)

[quote="fredbel6"]sounds logic but not quite, it would allow me to prove that there is a bijection between
[0,1] and ]0,1[[/quote]
Ha! You can easily construct bijection without Cantor-Bernstein!

Sure, we can, for example, construct a bijection between the rationals of the two sets (I won't construct this one :)), and put $f(x)=x$ if $x$ is irrational.

I am even sure it is a too complicated construction!
Explicit bijection:
$0\to 1/3$
$1/3\to 1/9$
...
$1\to 1/2$
$1/2\to 1/4$
...
$x\to x$ for all remaining $x$.  :D


limit
function
real analysis
real analysis unsolved

If $\lim_{n\to \infty}f_{n}= f$ and the functions $f_{n}$ are all monotone increasing, must $f$ be monotone increasing?
What happens if $f_{n}$ are all strictly increasing?

For the first part: just use the fact that if $a_{n}\le b_{n}$ for all $n,$ and the sequences have limits, then $\lim a_{n}\le \lim b_{n}.$

For the second part: if you're hoping for a conclusion that includes "strictly increasing," then you can't have that. Just look at $f_{n}(x)=\frac{x}{n}.$


algebra
polynomial
number theory open
number theory

Let r(n) be the number of distinct prime divisors of n
r(50)=2 (as 50=2*5*5 and has 2 distinct prime factors 2 and 5)

Then prove or disprove:

There exists a polynomial f(n) such that following holds good.
The minimum k (integer) satisfying
r(f(k))=i (i >=2 is an integer)
is a prime number.

this is not true as the number of elements in the set $ \{f(k)\}$ as k goes from 1 to n is approximately n. :wink:


Which book  should I probably read first? The Aops books or the Art and Craft of Promblem solving by Paul Zeitz for those of you that have read it?

This issue has been discussed here several times before.  I recommend you do a seach of the forum for "Art and Craft" or "Zeitz" or some such thing using the search function near the top of the page.  Also, this really isn't a getting started topic -- questions like this should go in one on the discussion forums, not one of the problem solving forums.

Well, it's definitely a newbie question. Part of the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7254]owner-designated scope of this forum[/url] is asking newbie questions about problem-solving. JBL makes the good point that now this site has an extensive database of questions that have been asked and answered before, and it is an excellent idea to use the site's search function to look for answers to what are probably Frequently Asked Questions.

The short answer is that the prerequisites assumed by AoPS volume 1 are MUCH simpler than those assumed by AoPS volume 2, which in turn are (mostly) simpler than those assumed by the Zeitz book. I own all of those books, and I would work through them in the order indicated by the previous sentence. Do the search suggested by JBL for a more detailed rationale.


number theory
Diophantine equation
number theory proposed

For fixed k,n positive numbers, with (k,n) = 1, and they are both greater or equal 1.Find the number of solutions of equation : 1/x + 1/y = k/n.

Note that, equation has solutions, x,y >=1 iff exits a,b .=1 such that:
a | n, b | n, (a,b) = 1 and k | (a+b).

When k = 2, the problem is not hard. But can you generalize this to other k?[/code]

You may write the given equation as $(n-kx)(n-ky) = n^2$.
The problem is not find the number of divisors of $n^2$ which are equals to $n$ mod[k].

Pierre.

positive numbers = positive integers?
and you want integer solutions right ?

anyway simply working out gives $n\cdot (x+y) = k\cdot xy$

go on and you get $y = \frac{n\cdot x}{k\cdot x-n}$ integer. Which is a well-known diophantine equation I think?


Prove that there exist no polyniomials $f(x)$ and $g(x)$ with integer coefficients, such that $f(x)g(x) = (x-a_1)^2(x-a_2)^2.....(x-a_n)^2 +1.$
   Ashwath

My solution is
Let $P(x) = (x-a_1)^2(x-a_2)^2.....(x-a_n)^2 +1.$. Then $P(a_k)=1=f(x)g(x)$ so

$f(a_k)=g(a_k)=1$ or $f(a_k)=g(a_k)=-1$ Also degree$[P(x)]=2n=$degree$[f(x)]$+degree$[g(x)]$.If one of the degrees of g(x) and f(x) is smaller than $n$ then the respect polynomial will be  $1$ or $-1$ ,contradiction Now we suppose that degree$[f(x)]$=degree$[g(x)]=n$. The coefficients of the biggest terms must have prod$1$. So degree$[f(x)-g(x)]\leq n-1$and because$f(a_k)-g(a_k)=0$we have that$f(x)= g(x)$. So $(x-a_1)^2(x-a_2)^2.....(x-a_n)^2 +1=[f(x)]^2$ and if $z$ integer then

$(z-a_1)^2(z-a_2)^2.....(z-a_n)^2 +1=[f(z)]^2$ so $[f(z)+(z-a_1)....(z-a_n)][f(z)-(z- a_1)....(z-a_n)]=1$  and $(z-a_1)....(z-a_n)=0$, $f(z)=\pm 1$ for each $z\in Z$, contradiction :)

Just one word about the terms of the problem : the $a_i$ must be all different. It was used by silouan to prove $f(x)= g(x)$.

Furthermore $(x-1)^2 \times (x-1)^2 \times (x-1)^2 + 1$ is divisible by $(x-1)^2 + 1$


inequalities
geometry
trigonometry
inequalities unsolved

Let $A_{i}B_{i}C_{i},\ i=1,2$ be two triangles of sides $(a_{i},b_{i},c_{i})_{i=1,2}$, areas $S_{1,2}$ and of radii $(R_{i},r_{i})_{i=1,2}$ where $R_{i},r_{i}$ denote the lenghts of the radii of the circumscribed and inscribed circles for $A_{i}B_{i}C_{i}$, respectively. Then we have \[8R_{1}R_{2}+4r_{1}r_{2}\geq a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}\geq 36 r_{1}r_{2}.\]

This looks like an old AMM problem. Is it?

Here is my solution to this cool inequality

[i]Lemma [/i] Given triangle $ABC$. Then $a^{2}+b^{2}+c^{2}\leq 8R^{2}+4r^{2}$, or equivalently $sin^{2}A+sin^{2}B+sin^{2}C \leq 2+16\prod sin^{2}\frac{A}{2}$

[i]Proof[/i] Using the well-known identity $a^{2}+b^{2}+c^{2}= 2(p^{2}-r^{2}-4Rr)$ and $cosA+cosB+cosC = 1+\frac{r}{R}$, we obtain that the inequality is equivalent to $sinAsinB+sinBsinC+sinCsinA \leq (cosA+cosB+cosC)^{2}$.

By replacing $sinAsinB = cosC+cosAcosB$ and $\sum cos^{2}A = 1-2cosAcosBcosC$, we obtain that it is sufficient to show $(1-cosA)(1-cosB)(1-cosC) \geq cosAcosBcosC$. If $ABC$ is obtuse, then it is obvious. If not, then since $tgAtgBtgC \geq cotg{\frac{A}{2}}cotg{\frac{B}{2}}cotg{\frac{C}{2}}$, the inequality is true. Thus the lemma is completely proved.

Notice that this lemma is a special case of the inequality we are trying to prove when two triangles are congruent.

Before proceeding to show the inequality, I will rewrite the problem a little bit.

[b]Inequality[/b] Given two triangles $ABC$ and $DEF$. Show that

$36rr' \leq ad+be+cf \leq 8RR'+4rr'$

[b]Solution[/b]

The left most inequality is quite straightforward, followed immediately from the fact $abc \geq 6\sqrt{3}r^{3}$ and $def \geq 6\sqrt{3}r^{3}$

The rightmost inequality is much harder. According to sine law, it follows that we need to show $\sum sinAsinD \leq 2+16 \prod sin \frac{A}{2}sin \frac{D}{2}$. By replacing $sinAsinD = sin^{2}{\frac{A+D}{2}}-sin^{2}{\frac{A-D}{2}}$, it follows that

$LHS = \sum (sin^{2}{\frac{A+D}{2}}-sin^{2}{\frac{A-D}{2}})$

Let $X = \frac{A+D}{2}, Y = \frac{B+E}{2}, Z = \frac{C+F}{2}$. Then $XYZ$ is a triangle. By applying the lemma, we obtain that

$LHS = \sum sin^{2}X-\sum sin^{2}{\frac{A-D}{2}}\geq 2+16 \prod sin^{2}\frac{X}{2}-\sum sin^{2}{\frac{A-D}{2}}$

$= 2+16 \prod sin^{2}\frac{A+D}{4}-\sum sin^{2}{\frac{A-D}{2}}$

We shall now show that

$16 \prod sin^{2}\frac{A+D}{4}-\sum sin^{2}{\frac{A-D}{2}}\leq 16 \prod sin{\frac{A}{2}}sin{\frac{D}{2}}$.

Replacing $sin{\frac{A}{2}}sin{\frac{D}{2}}= sin^{2}{\frac{A+D}{4}}-sin^{2}{\frac{A-D}{4}}, ...$, the inequality is equivalent to

$16 \prod sin^{2}\frac{A+D}{4}-\sum sin^{2}{\frac{A-D}{2}}\leq 16\prod (sin^{2}{\frac{A+D}{4}}-sin^{2}{\frac{A-D}{4}})$

or equivalently

$\sum sin^{2}\frac{A-D}{2}+16\sum sin^{2}\frac{A+D}{4}sin^{2}\frac{B-E}{4}sin^{2}\frac{C-F}{4}-16\sum sin^{2}\frac{A-D}{4}sin^{2}\frac{B+E}{4}sin^{2}\frac{C+F}{4}-16\prod sin^{2}{\frac{A-D}{4}}\geq 0 (*)$

It looks terrible, but is in fact not hard.  :)

Notice easily that $16\sum sin^{2}\frac{A+D}{4}sin^{2}\frac{B-E}{4}sin^{2}\frac{C-F}{4}\geq 16\prod sin^{2}{\frac{A-D}{4}}$.

We will now show that $sin^{2}\frac{A-D}{2}\geq 16sin^{2}\frac{A-D}{4}sin^{2}\frac{B+E}{4}sin^{2}\frac{C+F}{4}$. If we can prove it, then the whole thing is done.

We have $cotg{\frac{Y}{2}}+cotg{\frac{Z}{2}}\geq 2cotg{\frac{Y+Z}{4}}\geq 2cotg45^{0}= 2$. Hence $cos{\frac{X}{2}}\geq 2sin{\frac{Y}{2}}sin{\frac{Z}{2}}$. This implies that

$sin^{2}{\frac{A-D}{2}}= 4sin^{2}{\frac{A-D}{4}}cos^{2}{\frac{A-D}{4}}\geq 4sin^{2}{\frac{A-D}{4}}cos^{2}{\frac{A+D}{4}}\geq 16sin^{2}{\frac{A-D}{4}}sin^{2}{\frac{B+E}{4}}sin^{2}{\frac{C+F}{4}}$.

Therefore the inequality $(*)$ is proved, and we are done.


conics
geometry
3D geometry
sphere
ellipse
ratio
geometric series

Suppose we have an (infinite) cone $ \mathcal C$ with apex $ A$ and a plane $ \pi$. The intersection of $ \pi$ and $ \mathcal C$ is an ellipse $ \mathcal E$ with major axis $ BC$, such that $ B$ is closer to $ A$ than $ C$, and $ BC \equal{} 4$, $ AC \equal{} 5$, $ AB \equal{} 3$. Suppose we inscribe a sphere in each part of $ \mathcal C$ cut up by $ \mathcal E$ with both spheres tangent to $ \mathcal E$. What is the ratio of the radii of the spheres (smaller to larger)?

[hide="Diagram for Solution"](may take a moment)
[img]http://img.photobucket.com/albums/v194/mukmasterxps/hmmtguts12.png[/img][/hide]
[hide="Solution"][b]Inspiration:[/b] [i]AMC 10A 2006, number 16[/i]

Consider the projection of the diagram onto the plane containing points A, B, and C.

It is easy to see that ABC is a 3-4-5 right triangle, where in this projection, the ellipse degenerates into segment BC. The projection of the sphere in the "inside" part of the cone is the incircle of triangle ABC, and has radius 1, and thus diameter 2.

The larger sphere is tangent to BC as well, but on the other side.

Note that if we cut the cone with another plane parallel to $ \pi$ which is tangent to the other side of the insphere of the "inner" cone part, and construct a smaller sphere in this new apical piece, we get a scaled-down version of the same problem. If we iterate this infinitely, we get a geometric series.

Consider the projections of these spheres. Considering the diameters of each circle which are parallel to side BA, it is easy to see that the sum of the diameters of this geometric series of circles is AB=3.

Since the radius of the given circle is $ 2$, if we let the common ratio be $ r$, we have $ \frac {2}{1 \minus{} \frac {1}{r}} \equal{} 3 \implies r \equal{} \frac {1}{3}$.

Back-extrapolating this pattern to the large circle in question, we see that the ratio here from smaller to larger is $ \boxed{r \equal{} \frac {1}{3}}$[/hide]

Of course the cross section is the most important first step. Once you get that you can note that the ratio is $ \frac {s \minus{} a}{s} \equal{} \frac {6 \minus{} 4}{6} \equal{} \frac {1}{3}$...


Determine all non-negative integer solutions to
$ \ n^{4}_1 \plus{} n^{4}_2 \plus{} n^{4}_3 \plus{} .... \plus{} n^4_{14} \equal{} 1599$
I don't get how to solve this.
Solutions ne1?

Hint: Consider a mod based approach

[hide="Bigger hint"]
Take it all mod 16
[/hide]

[hide="Full solution"]
Taking both sides modulo $ 16,$ we get \[ n_1^4\plus{}n_2^4\plus{}n_3^4\plus{}\dots\plus{}n_{14}^4\equiv15\pmod{16}.\] $ 4$th powers modulo $ 16$ are all either $ 0,1,$ so the maximum sum is $ 14,$ thus we can't reach $ 15$ and there are no solutions.[/hide]

Which year did this come from?

Don't know what year, but it's probably older than you and me combined.

1979 (Problem 1).  I took the USAMO that year.  Back then, I think I solved this problem with a lot of casework.  :oops:


[quote="Hong Quy"]for$a_{1}, ..a_{n}$ is real positive numbers such that :
$(a_{1}^{2}-1)(a_{2}^{2}-1)..(a_{n}^{2}-1)=3^{n}.$
disprove or prove that : $\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{n}}\geq \frac{n}{2}$
 i sure it true with$n =3$ and $n=2$[/quote]
For $n=2$ it's true and nice enough.
For $n=3$ it's wrong: try $a=\sqrt7,$ $b=\sqrt{\frac{157}{7}}$ and $c=1.1.$ :wink:


limit
absolute value
real analysis
real analysis unsolved

Prove that by a suitable rearrangement of terms, a conditionally convergent series may be made to converge to any desired value, or to diverge.

Prove that no matter how the elements of an absolutely convergent series are ordered, the sum stays the same.

[quote="anik_andrew"]Is this a Riemann theorem?[/quote]

Yes it is,

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

maybe knowing the name will help you in your search.

I posted this question some time ago, and got no real response. I think it's because I ask so many questions that the best people here get tired of me  :P

Anyway, I have thought out a proof for a positive series that I think is right. Let the "original" positive series have partial sums $S_n$. Let a rearrangement of this series have partial sums $M_n$. Choose a $k$. By what we must mean by a rearrangement there exists a $k' > k$ such that every term in $S_k$ is also in $M_{k'}$. Since every term is positive we get that $S_k \le M_{k'}$ . However, in the exact same way there will exist a $k'' > k' > k$ so that $S_k \le M_{k'} \le S_{k''}$. By reasoning similar to the proof of the "squeeze theorem" this shows that $\lim_{k\to\infty}M_{k} = \lim_{k\to\infty}S_{k}$.

I have not yet really thought about how this extends to proving the same thing for absolutely convergent series.

Though this theorem looks like trivial, but the proof isn't too obvious.  :P

[quote="Kalle"]I have not yet really thought about how this extends to proving the same thing for absolutely convergent series.[/quote]

Absolutely convergent means that $\sum |a_n|$ is convergent. You have (nicely) proved the result for nonnegative series. Isn't that enough, or I'm missing something?

Fix $\varepsilon>0$. Since $\sum |a_n|$ converges, there is some $N$ such that $\sum_{i=m}^{m+k}|a_i|<\varepsilon$, for all $m\ge N$ and all $k\ge 0$. This means that for every finite set of naturals $\mathcal F$ contained in $\{N,N+1,\ldots\}$, we have $\sum_{i\in\mathcal F}|a_i|<\frac\varepsilon 2$.

Now let $\sigma$ be a permutation of the set of natural numbers, and take $M\ge N$ so large that $\{\sigma(1),\ldots,\sigma(M)\}$ contains $1,2,\ldots,N$. Then, for all $m\ge M$, you have $\left|\sum_{i=1}^ma_i-\sum_{j=1}^ma_{\sigma(j)}\right|=\left|\sum_{i\in\mathcal F}a_i-\sum_{j\in\mathcal F'}a_j\right|\ (*)$, where $\mathcal F,\mathcal F'$ are the finite sets $\{1,2,\ldots,m\}\setminus\{1,2,\ldots,N-1\}$ and $\{\sigma(1),\ldots,\sigma(m)\}\setminus\{1,2,\ldots,N-1\}$. The RHS of $(*)$ is now $\le\sum_{i\in\mathcal F}|a_i|+\sum_{j\in\mathcal F'}|a_j|<\varepsilon$.

The above says that given $\varepsilon$, we can find a natural $M$ such that $\left|\sum_{i=1}^ma_i-\sum_{j=1}^ma_{\sigma(j)}\right|<\varepsilon$ for all $m\ge M$, and this finishes it.

[quote="perfect_radio"][quote="Kalle"]Well, no, not that I see. We have shown that if $\sum b_n$ is a rearrangement of $\sum a_n$ then $\sum |b_n| = \sum |a_n|$. I don't see how it follows that $\sum b_n = \sum a_n$[/quote]

I haven't thought about this at first, but I think this works \[ \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} (a_n+|a_n|) - \sum_{n=1}^{\infty} |a_n| .  \]

Well, does it?[/quote]
I would like to believe that yes, this equality does indeed complete the solution, since both series in that are positive. If my original proof is correct, then this is probably correct too. It is always easy to overlook minor details, though. See for example all the time I wasted in one of anik_andrew's thread. So, I'm just going to say that it would be nice if my simple proof combined with your equality would indeed work as a proof.

Grobber: Thanks!

[quote="perfect_radio"]Prove that by a suitable rearrangement of terms, a conditionally convergent series may be made to converge to any desired value, or to diverge.[/quote]

Split the set of non-zero terms in two sequences, $A=\{a_n>0\ |\ n\in\mathbb N^*\},\ B=\{b_m<0\ |\ m\in\mathbb N^*\}$, such that $A$ and $B$ are ordered in the decreasing order of the absolute values of the $a_n$'s and $b_m$'s respectively. Now, you can prove just as we proved that other theorem you quoted that $S_A=\sum a_n,\ S_B=\sum b_m$ do not depend on the order in which the summation is taken. The important thing here is that because our series is conditionally convergent, $S_A=\infty,S_B=-\infty$ (a simple exercise). Now let's find a rearrangement of $A\cup B$ whose series converges to $0$. An entirely analogous process works for every other real.

Let $T=\max(\max |a_n|,\max |b_m|)$. We start adding terms to our sequence as follows. We add $a_1,\ldots,a_{t_1}$ such that $a_1+\ldots+a_{t_1}\le T,\ a_1+\ldots+a_{t_1+1}>T$. Then we add $b_1,\ldots,b_{s_1}$ such that $a_1+\ldots+a_{t_1}+b_1,\ldots,b_{s_1}\ge -T$, but $a_1+\ldots+a_{t_1}+b_1+\ldots+b_{s_1}+b_{s_1+1}<-T$. Then we add $a_{t_1+1},\ldots+a_{t_2}$ such that the sum of the terms which we have up until now is $\le T$, but if we add one more $a$ we get a sum $>T$, and so on, until the maximum of the absolute value of all the $a$'s and $b$'s that are left is $\le\frac T2$. Then, we repeat the procedure, with $\frac T2$ instead of $T$, then with $\frac T3$ instead of $\frac T2$, and so on. For every $n$, there is a large enough index $N_n$ such that all partial sums of the series we're constructing with index larger than $N_n$ lie in the interval $\left[-\frac Tn,\frac Tn\right]$, so the series converges to $0$.

If you look carefully, you'll see that we did use the fact that $S_A=\infty,S_B=-\infty$ in order to conclude that we always reach both sides of $0$. Otherwise the process described above could yield a series going to $\infty$ if $S_B>-\infty$ or $-\infty$ if $S_A<\infty$.


I'm goinh to 9th grade in September, so I'm thinking about getting an AoPS book to prepare for AMC. Does anyone have a suggeation as to which book I should get? Thanks.
(I have taken AMC 8 this year, other than that, I have also worked on old AMC problems. But I don't really feel that this is enough. I am looking for a book that would explain several topics rather than just one topic.)

AMC? AoPS should be enough...

later, you'll want Engel's Problem Solving Strategies and Zeitz's Art and Craft of Problem Solving (this is what I glean from these "which book?" topics)

Sorry if there are other similar posts around, I couldn't find any.
The problem with AoPS(actually it's my problem) is that I might not have access to a computer this summer.
Just wondering, if anyone has [i]The Art of Problem solving, Volume 1: the Basics[/i], how helpful would it be for AMC?

[quote="Phelpedo"]When he said AoPS, he meant the book  :wink:  For the AMC, you'll want volume 2; volume 1 is a little too easy, although I guess the material is enough for the most part.[/quote]

I don't think this assessment is quite accurate. From what I remember, especially from writers of the book (Mr. Rusczyk), volume I is quite sufficient for problems on the AMC. Mastering both volumes of material is essentially a large portion of what you need in order to do well on the AIME.

[quote="Phelpedo"]When he said AoPS, he meant the book  :wink:  For the AMC, you'll want volume 2; volume 1 is a little too easy, although I guess the material is enough for the most part.[/quote]
Oops :oops: . I am not very good at communicating so this is defanitely not the first time I had misunderstand something like this. Thanks for the suggestion though. By the way, is there a connection between volume 1 and volume 2 (more specificly, can you understand volume 2 without doing volume 1 first?) I think my Dad will only allow me to buy one book unless I somehow convince him...

I read volume two without first reading volume one, and it has worked well for the most part (my AMC score jumped by over 15 points).

Yeah, they sometimes refer to "we did blah in Vol. I" but it's not necessary, really.
There is some good stuff in Vol. I, though, don't ignore it.

Thanks guys!!! Looks like I'm getting a Volume 2.

Disclaimer: Do not get volume 2 three weeks before the test and try to get through it. Believe me. I know from experience; it doesn't work, you just end up burning out.

Ahem, by looking at the table of contents??
Sheesh. I mean, the practice problems in the later chapters...some of them were actually challenging.
Meh, can't give an example though, I lost my copy.  :oops:

[quote="worthawholebean"][quote="rrusczyk"]Unless you're routinely scoring in the 120s or higher on the AMC 10/12, I'd recommend starting with Volume 1 before going on to Volume 2.[/quote]
I really didn't see that much in volume 1 (from looking at the table of contents) that I did not learn from either MathCounts or my standard Alg 2 class (my school's curriculum is pretty advanced).[/quote]

And you're consistently getting over 120s on the AMC10.

I don't get it. Everyone always says that AoPS volume 1 is extraordinarily easy.. but I had decent (not great, but  not bad) AMC scores, did ok in mathcounts, learned a lot last year-- and it still takes me time to get through the geo chapters of vol 1.

[quote="rrusczyk"][quote="cat810930"]I've never done AMC 10/12 before (my school doesn't participate in them), so I guess maybe I'll try a few old AMC 10/12 contests first.
Another question - would it be easy to master the topics covered in volume 1 by simply studying from the Internet and library books? I don't think my Dad would buy me volume 2 if he does buy me volume 1. Even though I might not be good enough for vol. 2 at this moment, I am pretty sure (or at least I hope :) ) that I would be a year from now.[/quote]

Here's a thought: talk to your library about their getting the books.   Explain how top students and homeschoolers use these books, etc.  As I understand, some libraries are particularly responsive to the needs of homeschoolers, and are definitely interested in getting more non-adults using libraries.[/quote]
Thanks, I might give it a try. Though I don't see how it will happen since my library is mostly used by people who need help with English rather than people who are interested in areas like math... sometimes I feel as though I'm the only one visting the math section (which, even though is large, focuses mainly on the basics).
Luckily, the high school I'm going to specializes in math and science, so I would probably have more access to valuable resources this September. :)

Another idea: After you get AoPS 1, solve as many problems as you can, collecting your solutions in a notebook.  When you are done, with presumably more than 100 pages of solutions, show your work to your dad, and tell him how much you would like the second volume.  As a father myself, I would never say no in such a circumstance.

That's actually a great idea since my dad likes to say' "If I buy you the book, you not even going to do it" (which is not true) as a reason for not getting me the book.
Thanks, Ravi B! :lol:


[quote="the future"]suppose a set exists in which it has integers that contain the digits 1-9 exactly once and all elements of S are prime.  For example, (7,83,421,569) would be considered one of these sets as 1-9 are all used exatly once and all 4 of the numbers are prime.

Find the smallest possible total sum for all the elements for all sets that fit this rule.

For example, the sum of (7,83,421,569) is 7+83+421+569.[/quote]
Since two digit numbers will always be better than three digit numbers, the set will require two digit and one digit numbers, Also, since even digits must be in the tens digits and odds must be in the ones, then the smallest set would have a sum of:
$ 10(2\plus{}4\plus{}6\plus{}8)\plus{}1\plus{}3\plus{}5\plus{}7\plus{}9\equal{}225$

[hide="hint"]Maximize the number of one digit numbers. You can beat ZhangPeijin's number.[/hide]

[quote="ZhangPeijin"][quote="the future"]suppose a set exists in which it has integers that contain the digits 1-9 exactly once and all elements of S are prime.  For example, (7,83,421,569) would be considered one of these sets as 1-9 are all used exatly once and all 4 of the numbers are prime.

Find the smallest possible total sum for all the elements for all sets that fit this rule.

For example, the sum of (7,83,421,569) is 7+83+421+569.[/quote]
Since two digit numbers will always be better than three digit numbers, the set will require two digit and one digit numbers, Also, since even digits must be in the tens digits and odds must be in the ones, then the smallest set would have a sum of:
$ 10(2 \plus{} 4 \plus{} 6 \plus{} 8) \plus{} 1 \plus{} 3 \plus{} 5 \plus{} 7 \plus{} 9 \equal{} 225$[/quote]

Thats what I got at first too, but its the second smallest, not the smallest.

ANd you can use 3 digit numbers, but its not a good idea, the sum will be very large.

[quote="DonkeyKong"][quote="ZhangPeijin"][quote="DonkeyKong"]I am not sure how to attack this, but I got 5,89,241,367, which is 702.

I left 2,3,5,7 in the front, and left the even numbers in even slot, like tenths and thousandth.[/quote]
Question is, why would you use three digit numbers when you can use only two digit numbers?[/quote]
I think I misunderstood the question.

My second try gives me with the set(2,5,7,89,61,43) with the sum of [b]207[/b].[/quote]

ya thats correct (there are actually 3 sets but they all give the sum 207), any reason why you came to that conclusion?

DO you always minimize the number of digits?

Oh..
I forgot
Dangit!
Two is prime...
So it becomes:
$ 10(4\plus{}6\plus{}8)\plus{}1\plus{}2\plus{}3\plus{}5\plus{}7\plus{}9$


[quote="farshid_lk_park"]find all of the answers of p^k + 1 =q^m that :
k,p,q>1 and p is a coprime number[/quote]
Look up Catalan-Mihailescu


If f absolutely continuous, then it is of bounded variation on [a,b].

let the set of pairwise disjoint subintervals be just the interval [a, b] then l(a,b)= |a-b|?

This seems like it should be simple but I don't know where to start? Maybe start by saying that if it's not b.v.it can't possibly be a.c.?

Don't tell me the answer yet!

Hint: If the variation is infinite, then for all $ \epsilon > 0$, there is a subinterval of length $ < \epsilon$ on which the variation is $ >1$.


linear algebra
matrix
algebra
polynomial
complex numbers

a) Show that $\forall n \in \mathbb{N}_0, \exists A \in \mathbb{R}^{n\times n}: A^3=A+I$.
b) Show that $\det(A)>0, \forall A$ fulfilling the above condition.

a) Take $x_1$ the real root of $x^3 -x - 1=0$ and the matrix $A_{nxn}=x_1I_{nxn}$
 this matrix satisfies $A^3 - A -I = 0$ for all n natural.... right?

b) It´s enough to prove that $x_1>0$ because $Det(A)=$ product of eigenvalues and all of them satisfies $x^3 -x - 1=0$ (just multiply by a eigenvector), and the product of all complex eigenvalues is positive. So the signal of $Det(A)$ depends only of $x_1$. But the real root $x_1>0$(bolzano).

peter for a you used m and n
should  m=n?

Yes, indeed. There is no other m in the problem anyways :)

For part $b)$, suppose that $v \in N(A)$, obviously $v \in N(A^{3})$, so we have $A^{3}v=0 \Leftrightarrow (A+ I)v=0 \Leftrightarrow Av+v=0 \Leftrightarrow v=0$,  $Q.E.D$.

Usetobe, that only proves that $\det(A) \neq 0$.

For part (a), take $A = \ddag \{L_1, ..., L_{\lfloor {\frac{n}{3}} \rfloor}, O_{n-3 \lfloor {\frac{n}{3}} \rfloor} \}$, where each $L_i$ is the companion matrix of polynomial $p(x) = x^3-x-1$:

$L_i = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$.

For part (b), observe that p(x) has one real root $\alpha > 0$, and two complex conjugate roots $\beta, \bar{\beta}$. Therefore, det(A) is of the form $\det(A) = \alpha^k (|\beta|^2)^{(n-k)/2} > 0$.

i dont think the soln for part (a) in post #6 is correct. because A contains a zero matrix as a submatrix means that A is singular. but A is invertible.

another approach for part (b)
$A(A^2-I)=I$
$A^3=A+I$
$A=(A-I)(A^2+A+I)$
$det(A-I)det(A+I)=det(A^2-I) \sim det(A) \sim det(A^3) \sim det(A+I)$ so $det(A-I)>0$

$det(A) , det(A^2-I) , det(A+I) , det(A^2+A+I)$ all have the same signs (either positive or negative).
since $det(A^2+A+I)>0$ , thus $det(A)>0$.

in post #8, i dont understand why $det(A+I_n)>0$

i) moubi: Write $A=A^3-I$.
ii) $e^A\approx 1.1766433548295131458I_n+0.55184482841941241016A^2+1.2199282716787921209A$.
iii) More generally.
Let $n\geq 2$, $A\in M_n(\mathbb{R})$ and  $P\in \mathbb{R}[x]$ s.t. $P(A)=0$ and $P(0)\not= 0$.
Then $\det(A)\not= 0$ but the signum of $\det(A)$ is not fixed, except when all the real roots of $P$ are positive; in this last case, $\det(A)$ is always positive.

I have same query why in post 8, $det(A+I)>0$


Hello I was wondering how you would evaluate:

$ \sum_{x=3}^{\infty} x (\frac{5}{6})^{x-3}$

I'm only familiar with geometric and arithmetic sequences, and this is neither.

I went from 3 to 200 and got 48. Did the same from 3 to 400 and got 48. Also, I needed to get this to equal 48 exactly for my answer to a problem being correct. So, I'm pretty sure $ \displaystyle\sum_{x \equal{} 3}^{\infty} x \left( \frac {5}{6} \right)^{x \minus{} 3}\equal{}48$, I just don't know why.

Let $ S \equal{} \sum^\infty_{x\equal{}3}x\left(\frac56\right)^{x\minus{}3}$.
What is $ \frac56 S$? Match up the terms of $ S$ and $ \frac56 S$ by the powers and subtract to get a familiar series.

Consider the function $ G(x)\equal{}\sum_{k\equal{}0}^\infty (k\plus{}3)x^k\equal{}3\plus{}4x\plus{}5x^2\plus{}6x^3\plus{}\dots$ We have $ xG(x)\equal{}3x\plus{}4x^2\plus{}5x^3\plus{}\dots$, so $ (1\minus{}x)G(x)\equal{}G(x)\minus{}xG(x)\equal{}3\plus{}x\plus{}x^2\plus{}x^3\plus{}\dots\equal{}3\plus{}\frac x{1\minus{}x}$, so $ G(x)\equal{}\frac 3{1\minus{}x}\plus{}\frac x{(1\minus{}x)^2}$. Finally, plugging in $ x\equal{}\frac 56$ yields $ 3\cdot 6\plus{}\frac 56 \cdot 36\equal{}18\plus{}30\equal{}\boxed{48}$.

well here's a "strong-inductive" way to find a closed-form formula, but im afraid i dont know how to "automatize" it:
if you define \[ f_n=\sum_{x=0}^\infty x^nr^x\]then \[ f_{n+1}=\sum_{x=0}^\infty x^{n+1}r^x=\sum_{x=0}^\infty (x+1)^{n+1}r^{x+1}\]so we have\begin{eqnarray*}f_{n+1}-rf_{n+1}&=&\sum_{x=0}^\infty r^{x+1}((x+1)^{n+1}-x^{n+1})\\
&=&\sum_{x=0}^\infty r^{x+1}\sum_{k=0}^n \binom{n+1}{k}x^k\\
&=&r\sum_{k=0}^n \binom{n+1}{k}\sum_{x=0}^\infty x^kr^x\\
&=&r\sum_{k=0}^n \binom{n+1}{k}f_k\\
f_{n+1}&=&\frac{r}{r-1}\sum_{k=0}^n \binom{n+1}{k}f_k\end{eqnarray*}

Idk why he just reposted a previous post to advertise..


Actually not an AMC problem, but similar:

(from VWO (Flanders Math Olympiad) 2000 #24):

9 white and 18 black cubes with edge 1 are used to form one big cube with edge 3.  The partial area of the cube that is white is at most:

(A) 1/2
(B) 13/27
(C) 25/54
(D) 4/9
(E) 1/3

[hide]If we want to get the largest possible proportion of white on the surface of the cube, we want to use as many white cubes as possible for corners, and then use the rest for edges. Note that there are 8 corner cubes (3 faces showing), 12 edge cubes (2 faces showing), 6 face cubes (1 face showing), and 1 invisible center cube (0 faces showing). Thus we would be best off putting 8 white cubes in the corners and 1 white cube on an edge. That comes out to 8*3+1*2=26 white faces showing. On the entire cube, there are 9 faces showing per face of the large cube * 6 faces of the big cube = 54 faces. Thus we can get a maximum of 26/54=13/27 partial area being white.[/hide]

Right! Very nice, easy to understand answer.

This was the only one out of the 3 that I could solve...

[hide]Practically the problem asks for the ratio of whites on the surface of the cube at most, so you have to "show" as many whites on the outside as possible.

The corners will be the best (3 sides showing), so place 8 white cubes on all the corners and there will be one cube left.

The next to best place will be the edges (2 sides showing) so place the remaining white cube there, and fill the rest with black cubes.

Thus (3*8) white faces show on the corners (24) and two on the edge, showing the total of 26 white faces. The total amount of faces showing would be (3*3*6=54), so the answer is 26/54 or 13/27 (B).[/hide]

Quote:9 white and 18 black cubes with edge 1 are used to form one big cube with edge 3. The partial area of the cube that is white is at most:

(A) 1/2
(B) 13/27
(C) 25/54
(D) 4/9
(E) 1/3

[hide]well, i was about to put E,  1/3. but then i reread the question as i usually do, and i saw that it said "area". so then i had to do the problem again. but the 1/3 answer barely took a few seconds. so no biggie.

well. to get the most area out of the 9 white cubes. i put them at the 8 corners of the cube. at each corner, a cube gets 3 square units of surface are. so, that's already 24 square units, but we still have one more white cube to deal with. if you put the cube in the center of a side of a square, then you get 25/54. but there is another option. to put it on the middle left, top right, middle right, or middle bottom of a side of a square. this way, the white cube gets 2 square units of surface area. adding 2/54+24/54=26/54=13/27. so my answer is choice B[/hide]

if i helped a questioning aopser just a bit, my job is done

[hide]

Well to maximize the surface area, you want one cube to take up as much area as possible. The most area a partial cube can take in a bigger one is 3, being the 8 corners. 8*3 = 24 and 1 more cube needs to be used.  The next biggest area that can be covered is 2, on one of the edges. 24+2=26 over the total surface area which is 54. So therefore I got 13/27.

[/hide]

[quote="Syntax Error"]why "ElectricSnowman" would post an answer. oh the irony.[/quote]
Think about it.
Two equations:
mathfanatic not equal answer (he made the question)
ElectricSnowman equal answer
...?

spoiler. although, so many people already solved it...

[hide]so there are 9 white cubes... 3 of your faces count into the total SA of the cube when you're at the corner, and there are 8 corners, so 8x1^2x3 = 24... and the last one you fit somewhere along the edges, so you get 1^2 x2 = 2... so then you get 24+2=26 (duh) so the answer is 26/3^2x6=26/54=13/27, so B is my answer[/hide]


isn't toefl that test that has to do with grammar for foreigners?

I've taken it before  :P

[quote="Xevarion"][quote="Tiks"]Hi guys!!

Does anybody know the test dates of TOEFL iBT  in the 2007????

Thank you for your time!!! :)[/quote]
I don't see them on the TOEFL website, but I found the 2006 dates [url=http://www.ets.org/portal/site/ets/menuitem.1488512ecfd5b8849a77b13bc3921509/?vgnextoid=f368af5e44df4010VgnVCM10000022f95190RCRD&vgnextchannel=e207197a484f4010VgnVCM10000022f95190RCRD]here[/url], maybe they'll update soon.[/quote]
Yes,Xevarion,I wasn't able to find the test dates in their site too.
I am just afraid too be late with taking that test for my applications.You see I am not sure that there will be a test date in January and January is the last month to take standartized tests.

Is it true that taking TOEFL is benefiting American-born students from recent immigrant families?
After all, even though their parents might be successfull professionals speaking English at work etc., these kids can't compete in English with non-recent immigrant families (all Americans are immigrants except Native Americans, of course). Because their first language is not English, parents do not speak English at home (normally), their English scores interpreted in this context would look better?
Of course, taking some tests in your parent's native language probably makes sense too, to show your foreign language proficiency, especially since some Math programs (like Princeton) have Russian (French/German) requirements?

Yeah,guys the ETS organization at last replyed me and sent me all the test dates in the 2007  for Armenia :P (the test dates mostly the same all around the world :wink: ).The first test date in the next year is in 19 of January(in fact it is quit comfortable goes with my colleg application deadlines :lol: ).If someone wish I can send him(or her) the other test dates.Just PM me :wink: .


modular arithmetic
function
number theory
least common multiple
totient function

Are you sure that's right? It doesn't look like it.

[quote="lotrgreengrapes7926"][quote="janieluvsmusic"]Find the minimum value of c where:

$10^{c}\equiv 4\ mod\ 39$[/quote]

[hide] $10^{38}\equiv1\pmod{39}$ by Fermat's Theorem.

$10^{38}\equiv40\pmod{39}\Longrightarrow10^{37}\equiv4\pmod{39}$.

$\boxed{37}$. [/hide][/quote]

that's invalid as 39 isn't prime

[quote="janieluvsmusic"]Find the minimum value of c where:

$10^{c}\equiv 4\ mod\ 39$[/quote]
[hide]
$10^{\phi(39)}\equiv 1\mod 39$ since $\gcd(10,39)=1$

$10^{24}\equiv 1\mod 39$

$10^{23}\equiv 4\mod 39$

$\boxed{c=23}$[/hide]

23 works but it's not the minimum value.

It's necessary to take remainders modulo prime factors of $39$...

I think we actually will now have to look for a pattern.

But I won't work for now.

I doubt any theorems work now.

Euler's totient function (and as a special case, Fermat's Little Theorem) tell you an exponent that works.  They don't tell you it's the least such exponent.

Being nit-picky: you should ask for the least positive integer $c$ such that etcetcetc.

[quote="D.P.L"][quote="janieluvsmusic"]Find the minimum value of c where:

$10^{c}\equiv 4\ mod\ 39$[/quote]

is the answer 5?[/quote]

Yes -- why don't you give a solution?

[quote="JBL"]

Being nit-picky: you should ask for the least positive integer $c$ such that etcetcetc.[quote]

I was going to, but that would take more time to type. All I need is the solution now.

If you knew  $40= 39+1 = 4*10$  and   $39+10=49=7^{2}$ the answer would be a millisecond away. :)

Yes the answer is 5 but if you want to  dot the "i's and cross the "t's" ..

[hide="click here"]First we notice $(10)^{-1}\equiv 4 \bmod 39$
[hide="becuase"] after all $10 \cdot 4 = 40 \equiv 1 \bmod 39$ [/hide]
$-1$ is nice but you want a positive integer so .. we add 6 to it

[hide="becuase"]Having known how to factor  we know $39=3 \cdot 13$
Feramat's   uvacuh:  $7^{12}\equiv 1 \bmod 13$
He also says             $7^{2}\equiv 1 \bmod 3$
Showing of skills to take LCM of 2 and 12 we can say.
$7^{12 }\equiv 1 \bmod 39$
but $7^{2}\equiv 10 \bmod 39$ so we know $10^{6}\equiv 1 \bmod 39$

(we can quickly check  for factors of 6, eg $10^{2}$ and  $10^{3}$ are not  $1 \bmod 39$ so 6 is the  index or the smallest number such that $10^{this number}$ is 1.
QED

PS "uvach" is what you use when you want to show off your knowledge of
Sanskrit :)

[/hide][/hide]


Hi

I am looking for an exact statement on expanding a determinant by using one or more rows
It is well known how to calculate determinants by picking one row and then working with cofactors

But what if you, for instance, pick two rows?  I know it comes down then to calculating determinants of 2*2 matrices, and (n-2)*(n-2) matrices, and adding up their products, but what about the signs?

Can anyone help me?  Or give me a url or something?

Let $A=(a_{ij})_{i,j=\overline{1,n}}$ be a $n\times n$ matrix. Let $0<k<n$.

For arbitrary $1\leq i_i<i_2<\ldots<i_k\leq n$ and $1\leq j_i<j_2<\ldots<j_k\leq n$, define
\[ d_{(i_1,\ldots,i_k)}^{(j_1,\ldots,j_k)}=\det(a_{i_s,j_t})_{s,t=\overline{1,k}}  \]
and
\[ c_{(i_1,\ldots,i_k)}^{(j_1,\ldots,j_k)}=\det(a_{i'_s,j'_t})_{s,t=\overline{1,n-k}}  \], where $i'$s and $j'$s are chosen such that $\{i_1,\ldots,i_k,i'_1,\ldots,i'_{n-k}\}=\{1,2,\ldots,n\}$ and $\{j_1,\ldots,j_k,j'_1,\ldots,j'_{n-k}\}=\{1,2,\ldots,n\}$.

Fix $1\leq i_i<i_2<\ldots<i_k\leq n$.

[b]Laplace's Formula[/b]:
\[ \det A=\sum_{1\leq j_1<\ldots<j_k\leq n}(-1)^{i_1+\ldots+i_k+j_1+\ldots+j_k}\cdot d_{(i_1,\ldots,i_k)}^{(j_1,\ldots,j_k)}\cdot c_{(i_1,\ldots,i_k)}^{(j_1,\ldots,j_k)}.  \]


geometry
geometric transformation
reflection
ratio
cyclic quadrilateral
geometry unsolved

Let $X,Y$ is two point in circle $(O)$ , let $H$ is arbitrary point in line $XY$ .One line $(l)$ arbitrary pass through $H$ and meet $(O)$ at $M,N$ . Let $MX,NY$ meet $(O)$ at $P,Q$ respectively. Prove that $PQ$ pass through a fix point when $(l)$ change

Let $\mathcal I_{H}$ be an inversion with center H, which carries the circle (O) into itself. If H is outside of (O), then (O) has to be perpendicular to the inversion circle and the inversion has positive power $r^{2}$ (square of the inversion circle radius). If H is inside of (O), then (O) has to cut the inversion circle in diametrically opposite points and the inversion has negative power $-r^{2},$ which is equivalent to inversion with a positive power $r^{2}$ followed by a reflection in the inversion center H.

Let the line $(k) \equiv XY$ meet the circle (O) at A, B. The sides MP, NQ and the diagonal MN of the cyclic quadrilateral MPNQ meet the line (k) at X, Y, H. Let the diagonal PQ meet (k) at G. Let $\mathcal I_{X},\ \mathcal I_{Y},\ \mathcal I_{H},\ \mathcal I_{G}$ be inversions with centeres X, Y, H, G, which carry the circle (O) into itself. Inversion $\mathcal I_{X}$ carries M into P, inversion $\mathcal I_{G}$ carries P into Q, inversion $\mathcal I_{Y}$ carries Q into N, and inversion $\mathcal I_{H}$ carries N into M. Thus the Moebius transformation $\mathcal M = \mathcal I_{H}\circ \mathcal I_{Y}\circ \mathcal I_{G}\circ \mathcal I_{X}$ carries the point M to itself. In addition, $\mathcal M$ obviously carries both A and B into themselves, hence it has 3 fixed points. But a Moebius transformation with 3 fixed points is an identity transformation. For example, the general form of a Moebius transformation in complex plane is $z' = \frac{az+b}{cz+d},$ where $ad-bc \neq 0.$ For a fixed point z' = z, we have $cz^{2}+(d-a)z-b = 0.$ If this has more than 2 roots, then $c = 0,\ d = a \neq 0,\ b = 0$ and z' = z for all z.

Let a line (l') through H different from (l) and (k) meet the circle (O) at M', N'. The sides M'P', N'Q' and the diagonal M'N' of the cyclic quadrilateral M'P'N'Q' meet (k) at the fixed points X, Y, H. Let the diagonal P'Q' meet (k) at G'. In exactly the same way as above, we find that $\mathcal M' = \mathcal I_{H}\circ \mathcal I_{Y}\circ \mathcal I_{G'}\circ \mathcal I_{X}$ is also an identity transformation. It follows that $\mathcal I_{G'}= \mathcal I_{G}$ and $G' \equiv G,$ i.e., G is the fixed point of PQ.

Let's prove a stronger assertion:

[color=blue][b]Extended Problem.[/b] Let X, Y, H be three collinear points and k be a circle. For any point M on the circle k, denote by N the point of intersection of the line MH with the circle k (apart from M), and denote by P and Q the points of intersection of the lines MX and NY with the circle k (apart from M and N), respectively. Prove that the line PQ intersects the line XY at a point independent on the position of the point M on the circle k.[/color]

[i]Solution.[/i] We have to prove that for any two positions of the point M on the circle k, the corresponding lines PQ intersect the line XY at the same point.

In other words: Let $M_{1}$ and $M_{2}$ be two points on the circle k. Denote by $N_{1}$ the point of intersection of the line $M_{1}H$ with the circle k (apart from $M_{1}$), and denote by $P_{1}$ and $Q_{1}$ the points of intersection of the lines $M_{1}X$ and $N_{1}Y$ with the circle k (apart from $M_{1}$ and $N_{1}$), respectively. Similarly, denote by $N_{2}$ the point of intersection of the line $M_{2}H$ with the circle k (apart from $M_{2}$), and denote by $P_{2}$ and $Q_{2}$ the points of intersection of the lines $M_{2}X$ and $N_{2}Y$ with the circle k (apart from $M_{2}$ and $N_{2}$), respectively. We have to prove that the lines $P_{1}Q_{1}$ and $P_{2}Q_{2}$ intersect the line XY at the same point.

Applying the Pascal theorem to the cyclic hexagon $Q_{1}M_{2}N_{2}Q_{2}M_{1}N_{1}$, we see that the points $Q_{1}M_{2}\cap Q_{2}M_{1}$, $M_{2}N_{2}\cap M_{1}N_{1}$, $N_{2}Q_{2}\cap N_{1}Q_{1}$ are collinear. But the points $M_{2}N_{2}\cap M_{1}N_{1}=H$ and $N_{2}Q_{2}\cap N_{1}Q_{1}=Y$ both lie on the line XY; hence, the point $Q_{1}M_{2}\cap Q_{2}M_{1}$ must also lie on the line XY.

Now, apply the Pascal theorem to the cyclic hexagon $M_{1}P_{1}Q_{1}M_{2}P_{2}Q_{2}$. We obtain that the points $M_{1}P_{1}\cap M_{2}P_{2}$, $P_{1}Q_{1}\cap P_{2}Q_{2}$, $Q_{1}M_{2}\cap Q_{2}M_{1}$ are collinear. Since the points $M_{1}P_{1}\cap M_{2}P_{2}=X$ and $Q_{1}M_{2}\cap Q_{2}M_{1}$ both lie on the line XY, it thus follows that the point $P_{1}Q_{1}\cap P_{2}Q_{2}$ must also lie on the line XY. In other words, the lines $P_{1}Q_{1}$ and $P_{2}Q_{2}$ intersect the line XY at the same point. Proof complete.

Yetti's solution is really beautiful! I just posted mine to show that there is a more elementary way. Another solution of the Extended Problem is easy using cross-ratios on circles: Denote by U and V the (real or complex!) points of intersection of the line XY with the circle k. Let G be the point on the line XY such that (U; V; X; H) = (U; V; Y; G). Show that this point G lies on the line PQ for every position of the point M.

  darij


function
probability
integration
calculus
absolute value
probability and stats

Let $ X$ be a random variable having the probability density $ f(x)\equal{}\frac12 e^{\minus{}|x|}.$ Find the moment-generating function of $ X$.

$ m(t)\equal{}\int_{\minus{}\infty}^{\infty}f(x)e^{tx}\,dx$

$ \equal{}\int_{\minus{}\infty}^0\frac12e^xe^{tx}\,dx\plus{}\int_0^{\infty}\frac12e^{\minus{}x}e^{tx}\,dx$

$ \equal{}\frac1{2(1\plus{}t)}\plus{}\frac1{2(1\minus{}t)}\equal{}\frac1{1\minus{}t^2}$

The characteristic function would be $ \chi(t)\equal{}\frac1{1\plus{}t^2},$ by a very similar computation.

$ m(t)\equal{}\frac1{1\minus{}t^2}\equal{}1\plus{}t^2\plus{}t^4\plus{}t^6\plus{}\cdots,$ so

$ E(X^n)\equal{}\left(\frac{d}{dt}\right)^nm(0)\equal{}\begin{cases}n!&t\text{ even}\\0&t\text{ odd}\end{cases}.$

[quote="Kent Merryfield"]$ m(t) \equal{} \int_{ \minus{} \infty}^{\infty}f(x)e^{tx}\,dx$

$ \equal{} \int_{ \minus{} \infty}^0\frac12e^xe^{tx}\,dx \plus{} \int_0^{\infty}\frac12e^{ \minus{} x}e^{tx}\,dx$

...
[/quote]
Thank you, Prof. Kent Merryfield.
In fact, the question I would like to ask is: How to calculate the integral $ \int_0^{\infty} e^{(t\minus{}1)x}\,dx$?
And, is it true for every real $ t$?

Not for every real $ t.$ This particular distribution's moment generating function is only defined for $ \minus{}1<t<1.$ The integral you wrote diverges for $ t\ge 1,$ while the integral on the other side diverges for $ t\le \minus{}1.$

That's one of the theoretical advantages of the characteristic function, which is defined as $ \chi(t)\equal{}E(e^{itX}),$ which would equal $ \int_{\minus{}\infty}^{\infty}f(x)e^{itx}\,dx$ for a distribution with a density. The characteristic function is always defined and finite (in fact, no larger than $ 1$ in absolute value) for all real $ t.$ The moment generating function might only be defined on some interval in $ t$ as in this example, or in the worst case might only be defined for $ t\equal{}0$ and nowhere else.

But when the moment generating function is defined on some neighborhood of zero, it and the characteristic function each contain essentially the same information.

The moment generating function is a Laplace transform (with some sign changes); the characteristic function is a Fourier transform.


f derivabile neconstant on $ [0.1]$ s.t f(0)=0 .
Prove that exist $ c \in [0,1]$ s.t $ f(c) <f'(c)$

solve the differential equation $ f'(x)\minus{}f(x)\equal{}\minus{}h(x)$ where $ h(x)>0$


x1^{2}+d*y1^{2}=p,x2^{2}+d*y2^{2}=p*a,where p is a prime number
prove there exist x3,y3 that x3^{2}+d*y3^{2}=a


The following people have passed to the COP Round 2.

Treething
LynnelleYe
whitehorseking
RC-7th
alan
sam 3.14
h_s_potter 2002

The round 2 questions will be available very soon and the answers to the round one will be available also. (On the COP main website).

Congratulation!

Awesome... I look forward to the next round.

Yay. Time to make hundreds of careless errors :P .


function
real analysis
advanced fields
advanced fields unsolved

Let p_t(x) be the projection of x on the t direction.t lies on [0,pi) (x lies on the plane).Let K be some compact subset of the plane
How can we show that the function t-- >l(p_t(K))(l the lebesgue measure) is Borel for t in [0,pi)?

The set of points where our function is less than $M$ is open for every $M$. To see it, just note that the projection to every direction is again compact set. If, for some particular direction $t$, its measure is less than $M$, then there exists a finite union of open intervals $I_j$ of total length less than $M$ containing our projection. Therefore, the set $K$ itself is contained in finitely many rectangular boxes $Q_j$ with bases $I_j$. But the projection of each box $Q_j$ to the directions close to $t$ is only slightly larger than $|I_j|$.


What are the positive integer values of $ n$ that make $ 2^{n} \minus{}1$ divisible by $ 7$?

We consider three cases:

   [u][b]Case 1[/b][/u] : $ n\equal{}3k$ for k integer.

The we can see that: $ 2^n\minus{}1\equal{}(2^3)^k\minus{}1\equal{}8^k\minus{}1\equal{}7m$ where m is integer. Hence all the numbers of the form $ n\equal{}3k$ satisfy
the condition of the problem.

   [u][b]Case 2[/b][/u] : $ n\equal{}3k\plus{}1$

Then we can write:

     $ 2^n\minus{}1\equal{}2^{3k\plus{}1}\minus{}1\equal{}2(2^3)^k\minus{}2\plus{}1\equal{}2(7m)\plus{}1\equal{}7(2m)\plus{}1$ which is not a multiple of 7,hence this case is rejected.

    [u][b]Case 3[/b][/u]: $ n\equal{}3k\plus{}2$
Then we have:

     $ 2^n\minus{}1\equal{}2^{3k\plus{}2}\minus{}1\equal{}4(2^3)^k\minus{}4\plus{}3\equal{}4(7m)\plus{}3\equal{}7(4m)\plus{}3$ which again is not a multiple of 7.

Therefore the number $ 2^n\minus{}1$ is divisible by $ 7$ if and only if $ n\equal{}3k$ for $ k$ integer.

[hide="Much faster"]List out powers of 2 mod 7:

$ 2^0\equiv 1$
$ 2^1\equiv 2$
$ 2^2\equiv 4$
$ 2^3\equiv 1$

It clearly repeats after that, so we need the power to be $ 0\bmod 3$ to get $ 1\minus{}1\equiv 0\bmod 7$.

So any integer divisible by 3, or $ \boxed{3k,\ k\in\mathbb{Z}_{\geq0}}$.[/hide]

[hide="Just as Fast"]
$ 2^n\minus{}1 is 0 mod 7$
$ 2^n\equal{}1 mod 7$
By Fermat's Little Theorem, every multiple of 6 automatically works.

In addition, we notice that 2^3=8=1 mod 7, so we also include every multiple of 3.

Then, we're done.

[/hide]

[quote="zapi2007"][hide="Solution with fixed LaTeX"]
$ 2^n \minus{} 1 \equiv 0 \bmod 7$
$ 2^n \equiv 1 \bmod 7$

By Fermat's Little Theorem, every multiple of 6 automatically works.

In addition, we notice that $ 2^3\equiv 8\equiv1 \bmod 7$, so we also include every multiple of 3.

Then, we're done.

[/hide][/quote]

Fixed your $ \LaTeX$. You'll want to use \equiv for the equivalence symbol, and \bmod {m} or \pmod {m} for the mod part (\pmod adds parentheses).

In my opinion, my solution is a little more straightforward than yours :wink:


Putnam Harvard college Princeton MIT Stanford Duke

I just got my Putnam results envelope.

CSULB Team Rank = 42. Yay!!

I've got a class to teach right now. I'll be back to post more results later this afternoon or evening.

Does Caltech have any Putnam fellows?  :lol: (maybe someone whose surname is Lawrence?? :D )

The short answer to Albanian Eagle's questions: Yes, and Yes.

Team ranks:

1. Harvard
2. Princeton
3. MIT
4. Stanford
5. Duke

Honorable mention teams (in alphabetical order):

Caltech
Carnegie Mellon
Harvey Mudd
Toronto
Waterloo

Putnam Fellows:

Jason Bland, Caltech
Brian Lawrence, Caltech
Aaron Pixton, Princeton
Qingchun Ren, MIT
Xuancheng Shao, MIT
Arnav Tripathy, Harvard

Group N1:

Justin Bae, Harvard
Richard Biggs, Carnegie Mellon
Oleg Golberg, MIT
Yungcheng Lin, MIT
Tiankai Liu, Harvard
Konstantin Matveev, Toronto
Alison Miller, Harvard (Elizabeth Lowell Putnam Prize)
Andrei Negut, Princeton
Lingren Zhang, Duke
Yufei Zhao, MIT

Group N2:

Zachary Abel, Harvard
Lurie Boreico, Harvard
Anand Deopurkar, MIT
Anjie Guo, Waterloo
Junehyuk Jung, Chicago
Anders Kaseorg, MIT
János Kramár, Toronto
Shrenik Shah, Harvard

There are about 50 Honorable Mention names: right now I don't have time to type all of those names.

My institution (CSULB) placed one student (Joshua Lampkins) in the top 200, and two others in the top 1000, accounting for our best-ever team rank.

Nice!!

Congratulations to every one on that list of course!

The Putnam Fellows scored: 110, 97, 91, 90, 82, 82.

(The question of which Fellow got which score is not, and will not be, public information.)

The minimum score needed for group N1 was 71.

The minimum score needed for group N2 was 64.

The minimum score needed for Honorable Mention was 45.

The minimum score needed for Group I was 42. (rank $ \le 94$)

The minimum score needed for Group II was $ 30.3.$ (rank $ \le 205$)

The minimum score needed for the top scores mailing was $ 21.$ (rank $ \le 431.5$)

I see the MIT curse of picking the "wrong" team is alive and well.
Caltech was clearly hurt by the lack of a strong third member.
No word yet at UW.

[quote="jmerry"]
Caltech was clearly hurt by the lack of a strong third member.
[/quote]

Actually, I don't think Jason Bland was on the Caltech team. (He's a freshman, and I heard Caltech's team was just Brian Lawrence (duh) and the two top scores from previous years.) So they were hurt by the lack of strong second [i]and[/i] third members.

[quote="Kent Merryfield"]Lurie Boreico, Harvard[/quote]

Am I safe to assume this is [b]Iurie[/b] Boreico (Iura)?

  darij

I guess so darij ;)

Yup jason was not on the Caltech team :) too bad...

The Caltech team was Lozan Ivanov, Brian Lawrence, and Doo Sung Park.

The official mailing spelled it "Lurie." I assume they just misread it.

Who are some people with many, many AoPS/Mathlinks posts? I see Xevarion in Group II (Rank 103-205). Sly Si is also Group II.

Highest ranking Southern California ARML alum is Tedrick Leung in Group I (rank 78-94).

Honorable Mention by school:

Harvard:

Zhou Fan
Rishi Gupta

MIT:

Aleksandr Arkhipov
Thomas Belulovich
Gabriel Bujokas
Hansheng Diao
Nate Ince
Hyun Soo Kim
Sungyoon Kim
Thomas Mildorf
Kevin Modzelewski
Thanasin Nampaisarn
Eric Price
Jason Trigg
Alexandr Zamorzaev
Bohua Zhan

Yale: Dimiter Ostrev

Princeton:

Peter Diao
John Pardon
Aaron Potechin
Andrei Ungureanu
Adrian Zahariuc

Columbia: Dmytro Karabash

Cornell: Wei Quan Julius Poh

St. Joseph's (Philadelphia):  Dmytro Yeroshkin

U. Maryland: Henry Scher

Johns Hopkins: Kihyuk Hong

U. Virginia: Wuttisak Trongsiriwat

Duke:

Tirasan Khandawit
Aaron Pollack

Grinnell: Rolf Hoyer

Carleton: Jonah Ostroff

U. Chicago:

Harry Altman
Zhiwei Calvin Lin

North Texas: David Simmons

Caltech:

Evan Dummit
Timothy Black
Doo Sung Park
Gjergji Zaimi (Albanian Eagle)

Harvey Mudd:

Gregory Minton
Theodore Spaide

Stanford:

Serin Hong
Young Hun Jung
Nathan Pflueger
Jeffrey Wang (Paladin8)

UC Berkeley: Sean Eberhard

Waterloo:

Michael Lipnowski
Yin Zhao

U. Alberta: Steven Taschuk

U. British Columbia: Cedric Lin

A comparison to other recent Putnam results. It seems to stack up a a middling-difficulty exam. One influence on the percentage of zeros: it seems to have been fairly easy to get 1 or 2 points on A1. In fact, I think some credit might have been given for answers that failed to find all four roots.

$ \begin{array}{cccccc}\text{Year}&\text{No. taking}&\text{Mean}&\text{Median}&\text{pct zero}&\text{Top 500}\\ 1998&2573&14.8&10&30.3&28\\ 1999&2900&6.3&0&60.2&11\\ 2000&2818&5.3&0&57.7&11\\ 2001&2954&8.9&1&44.9&20\\ 2002&3349&11.0&3&34.7&24\\ 2003&3615&7.1&1&27.8&18\\ 2004&3733&8.4&0&53.6&22\\ 2005&3545&7.9&1&46.7&20\\ 2006&3640&6.2&0&62.6&14\\ 2007&3753&7.7&1&42.5&21\end{array}$

Yay Aaron Pollack at Duke! He is known at my school, since his dad is a professor there. He kind of grew up in the department.

does anyone know who the harvard team was? probably Tiankai, Alison, and ?

congratulations to all students

We are not in the same high school, but we are in the same county and we've went out for ARML together, during which we talked a lot about math, music and different kind of language. I feel that he is a nice guy so I have some concern with his progress. Anyway, it's really nothing important for me if I don't know that (as you say, it is not worth discussing).  :lol:

[quote="jimhu"]anyone know how well UCSD (San diego) or berkeley did? probably quite sucky[/quote]
What I know and what I don't know:

UC Berkeley had an H, two II's and two more in the top 500. UCSD had no one in the top 500.

Here's the entire list for California public universities (UC's and CSU's both):

H: UC Berkeley (1)

I: -

II: UC Berkeley (2), CSULB (1)

other top 455: UCI (2), UC Berkeley (2), UCSC (1), SJSU (2), CSU Chico (1)

I don't know anyone's team rank, other than the top 10, my own school, and two others where I've specifically been told. But UCSD is almost certainly behind our #42 placement. As for Berkeley: their test supervisor plays the identity of the team pretty close to the vest. The H (honorable mention) is a current high school student and almost certainly not on the Berkeley team; pkerichang, who is one of the II's, doesn't know whether or not he was on the team.

What is a "II"?

Also, how did you find out the results for UC Berkeley?  Is there a list posted somewhere?

A "II" is a score in the range $ [30.2,41],$ with a rank in the range $ [103,205].$ Roughly speaking, top 200 but not top 100.

My information comes from a list mailed to all Putnam supervisors (and probably a few other people) entitled "Top Participants in the 68th Annual William Lowell Putnam Mathematical Competition." I've been treating this list as public information (except for the addresses printed on it) but I don't think that it's posted on the Web anywhere.

The one person who would have more detail than that regarding UC Berkeley scores would be Prof. Kahan.

Can someone tell me how DePaul did or where I can find the team/individual scores?

Information about teams outside the top 10 is not publicly released- we can tell you about any individuals in the top 500, but nothing more. For any more details, you'll have to ask the sponsor at DePaul.

Then were there any individuals from DePaul in the top 500?

No, DePaul had no one in the top "500". For the Chicago metropolitan area, U. Chicago had an N2, two H, one II, and three top 500; Northwestern had two top 500; IIT had two top 500.

[quote="Kent Merryfield"]Rice? Have I mentioned that I was a Rice undergraduate myself?

Rice had two people in category I, which is to say rank 78-94:

Ho Chung Siu (from Hong Kong - is that you, Soarer?)
Shaui Ye

And two more Rice students in the rank 221-431.5 category.

----

The Harvard team was Tiankai, Alison, and Zachary Abel. So Harvard didn't pick the best team (yes, it would have been stronger with you) but won anyway.[/quote]

Sorry,I must point out the mistake...The name is Shuai Ye
I know him.He is a freshman at Rice.

i am interested in who from UC Berkeley is in the top 500.

could someone with the list of top 500 participants be kind enough to post this information please?  thanks!

UC Berkeley:

Honorable mention: Sean Eberhard (high school student)

Group II: Eric Chang (SoCal ARML alum), Joon Seok Lee

Other top 500: Roman Vaisberg, Lawrence Wang

[quote="galoisj"]I wanna know who ranks the first?(score 110)[/quote]
I don't know that. The person who got the 110 doesn't know that. The professors at the university of that person don't know that. You may "wanna know" but you're not going to find out.

Wow, so the fellows do not get to know their own scores? Doesn't that have some data protection issues associated with it?

I wanna know the results of U of Minnesota


algebra
function
domain
group theory
abstract algebra
topology
advanced fields

It is well-known that the torus can be constructed by using action groups; namely, $ Z^2$ acting on $ R^2$.    I know the classical model to construct the Klein Bottle, i.e., by identifying the edges, but I am figuring out if I can find some group which acts on $ R^2$ such that the equivalence classes that we get form a model as the classical one.

I hope you can understand me and if not please let me know it and I can try to explain it again.

Regards,

Adriana

I think you're looking for the group generated by the following two transformations of $ \mathbb{R}^2$:

$ f(x,y) \equal{} (x\plus{}1,y)$

$ s(x,y) \equal{} (1\minus{}x, y\plus{}1)$

Consider the action of these on the unit square, and notice how they identify the edges. You can conclude that unit square is a fundamental domain and the quotient space is the Klein bottle.

You can further analyze the group by studying the relations between $ f$ and $ s$. You'll find that

$ f^{\minus{}1}(x,y) \equal{} (x\minus{}1,y)$

$ s^{\minus{}1}(x,y) \equal{} (1\minus{}x,y\minus{}1)$

so

$ sfs^{\minus{}1}(x,y) \equal{} sf(1\minus{}x,y\minus{}1) \equal{} s(2\minus{}x,y\minus{}1) \equal{} (x\minus{}1,y) \equal{} f^{\minus{}1}(x,y)$

Therefore the fundamental group $ G \equal{} \pi_1(K)$ is a nonabliean group having a normal subgroup $ H \equal{} \langle f \rangle \cong \mathbb{Z}$ with $ G/H$ also isomorphic to $ \mathbb{Z}$. Another way to say this is that $ G$ is a semidirect product of $ \mathbb{Z} \equal{} \langle f \rangle$ by $ \mathbb{Z} \equal{} \langle s \rangle$ with a non-trivial action of $ s$ on $ f$ given by $ sfs^{\minus{}1} \equal{} f^{\minus{}1}$. This is sometimes called the infinite dihedral group (the finite dihedral group is what you get if you replace $ f$ by an element of finite order).