topology
number theory
least common multiple
Euler
ratio
geometry
algebra

Prove that there are an infinite number of primes.

[hide="Method 1"] (classic :D) Let us assume there are only a finite number of primes which range from $ p_{1}$ to $ p_{n}$. Let us define the number P which can be expressed as $ P \equal{} p_{1}p_{2}...p_{n} \plus{} 1$ Clearly no prime $ p_{i}$ can divide $ P$ without leaving $ 1$ as a remainder. Thus P itself must be prime, which leads a contradiction to our initial assumption that there are only a finite number of primes. [/hide]

A proof due to Furstenberg (and wikipedia for the details):

Define a topology on the integers by defining  $ U\subseteq\mathbb{Z}$ open if and only if it is the empty set, ∅, or a union of arithmetic sequences $ S(a, b)$, where $ S(a, b)\equal{}\{ a n \plus{} b | n \in \mathbb{Z} \} \equal{} a \mathbb{Z} \plus{} b$.  This is a topology because:
    * By definition, ∅ is open; $ \mathbb{Z}$ is just the sequence $ S(1, 0)$, and so is open as well.
    * Any union of open sets is open: for any collection of open sets $ U_i$ and $ x$ in their union $ U$, any of the numbers $ a_i$ for which $ S(a_i, x)\subseteq U_i$ also shows that $ S(a_i, x)\subseteq U$.
    * The intersection of two (and hence finitely many) open sets is open: let $ U_1$ and $ U_2$ be open sets and let $ x\in U_1\cap U_2$.   Let $ a$ be the lcm of $ a_1$ and $ a_2$. Then $ S(a, x)\subseteq S(a_i, x)\subseteq U_1\cap U_2$.

Note that since any non-empty open set contains an infinite sequence, no finite set can be open (or the the complement of a finite set can't be closed).Also, the basis sets $ S(a, b)$ are both open and closed: they are open by definition, and we can write $ S(a, b) \equal{} \mathbb{Z} \setminus \bigcup_{j \equal{} 1}^{a \minus{} 1} S(a, b \plus{} j)$ as the complement of an open set.  The only integers that are not integer multiples of prime numbers are $ \minus{}1$ and $ \plus{}1$, i.e. $ \mathbb{Z} \setminus \{ \minus{}1, \plus{} 1 \} \equal{} \bigcup_{p \mathrm{\, prime}} S(p, 0)$.  By the first property, the set on the left isn't  closed. By the second property, the sets $ S(p, 0)$ are closed. So, if there were only finitely many prime numbers, then the set on the right-hand side would be a finite union of closed sets, and hence closed. This would be a contradiction, so there must be infinitely many prime numbers.

According to

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

Ribenboim (1989) gives nine and a half proofs.

For $ n>1$, any two consecutive numbers are coprime, so $ n(n\plus{}1)$ has at least 2 different prime factors.
Then $ n(n\plus{}1)$ and $ n(n\plus{}1)\plus{}1$ are coprime, so they have 2 different prime factors between them, but $ n$ and $ n\plus{}1$ had 2 different prime factors, so $ n(n\plus{}1)(n(n\plus{}1)\plus{}1)$ has at least 3 different prime factors. This continues indefinitely.

When you say this is a marathon I feel obligated to post an original proof, however impractical it may be.

Assume there is a finite number of primes.

Euler gives the identity $ \zeta(s)\equal{}\sum_{k\equal{}1}^{\infty}\frac{1}{k^s}\equal{}\prod_{p\equal{}primes}\frac{1}{1\minus{}\frac{1}{p^s}}$.  Although we could consider the zeta of any even natural number, consider $ \zeta(2)\equal{}\frac{\pi^2}{6}$.  By the above identity, $ \frac{\pi^2}{6}\equal{}\left(\frac{1}{1\minus{}\frac{1}{p_1^2}}\right)\left(\frac{1}{1\minus{}\frac{1}{p_2^2}}\right)\left(\frac{1}{1\minus{}\frac{1}{p_3^2}}\right)...\left(\frac{1}{1\minus{}\frac{1}{p_k^2}}\right)$.  Of course, k is finite by our original assumption.  Since it can be expressed as a product of ratios of integers, it is also a ratio of integers itself and is therefore rational.  HOWEVER, $ \pi$ is not a rational, so neither is $ \frac{\pi^2}{6}$.  This is a direct contradiction.  $ Q.E.D.$

I'm not sure how great of a marathon this would be, but "collection" might be a good description. :)

[quote="Stank"]

 HOWEVER, $ \pi$ is not a rational, so neither is $ \frac {\pi^2}{6}$.
This is a direct contradiction.  $ Q.E.D.$[/quote]

This does not follow.  Just because $ \pi$ is not rational, you may not
state that $ \frac {\pi^2}{6}$ is also not rational.  An irrational number
squared and then divided by six *can* be rational.

That is correct; the mere irrationality of $ \pi$ is not, by itself, sufficient. You also have to show that $ \pi^2$ is irrational, which is a slightly different question.

As an aside: This has echos in the ancient problem of "squaring the circle", which is to say, the question of whether one can construct using compass and straightedge, a square with the same area as a given circle. If $ \pi^2$ were rational, this would be possible, but it's not possible.

A more modern way of dealing with this: $ \pi$ is known (and has been known since about the 1890's) to be not just irrational but [i]transcendental.[/i] That is, it is not the root of any polynomial equation with rational coefficients.

If $ \frac{\pi^2}6\equal{}\frac{p}{q},$ then we would have $ q\pi^2\minus{}6p\equal{}0,$ which would be an integer-coefficient polynomial equation sastisfied by $ \pi.$ We know this is not possible, so $ \frac{\pi^2}6$ is irrational.

To return to Stank's post which started this: there's a more direct and sastisfying way of using Euler's idea, namely this:
\[ \sum_{k\equal{}1}^{\infty}\frac1k\equal{}\prod_{p\equal{}\text{primes}}\frac{1}{1\minus{}\frac1p}.\]
If there were finitely many primes, the product on the right would be a finite product of real numbers and hence a finite real number. But the harmonic series is known to diverge. Hence there cannot be finitely many primes.

(Of course, this version is not remotely close to being original. And in any case, it's hard to improve on Euclid's proof of the infinity of the primes - that's a proof that has stood the test of time.)

[quote="Kent Merryfield"]You also have to show that $ \pi^2$ is irrational, which is a slightly different question.[/quote]
Spivak's book has a short chapter devoted to this, in case anyone wants a reference.

Yes, you're right that it is not sufficient for $ \pi$ to be irrational (for example, $ \sqrt{2}$ is irrational, but $ \frac{(\sqrt{2})^2}{6}$ is not).  I've since realized this but neglected to correct the post.


If the roots of $ x^2 \minus{} bx \plus{} c \equal{} 0$ are
\[ sin \frac {\pi}{7}
\]
and
\[ cos\frac {\pi}{7}
\]
then $ b^2$ equals what?

EDIT: THE ANSWER IS IN TERMS OF C

[quote="DaReaper"]If the roots of $ x^2 \minus{} bx \plus{} c \equal{} 0$ are
\[ sin \frac {\pi}{7}
\]
and
\[ cos\frac {\pi}{7}
\]
then $ b^2$ equals what?[/quote]
Hint[hide]
Because those two roots are the ONLY two roots of the equation, we can use Vieta's relations to show that $ \minus{}b$ is the sum of the two roots. $ b^2$ soon follows.[/hide]

[hide="Solution"]
By Vieta, we have

$ \sin{\frac {\pi}{7}} \plus{} \cos{\frac {\pi}{7}} \equal{} b \implies \boxed{1 \plus{} \sin{\frac {2\pi}{7}}} \equal{} b^2$

Sorry for doing a one step solution, I am too lazy at the moment.
[/hide]

[hide="Solution"]
Like how cipos2 said, we know that $ \sin \dfrac{\pi}{7} \plus{} \cos \dfrac{\pi}{7} \equal{} b$
Then $ b^2 \equal{} \sin^2 \dfrac{\pi}{7} \plus{} 2\cdot (\sin \dfrac{\pi}{7})(\cos \dfrac{\pi}{7}) \plus{} \cos^2 \dfrac{\pi}{7}\equal{} 1 \plus{} 2\cdot (\sin \dfrac{\pi}{7})(\cos \dfrac{\pi}{7}) \equal{} 1 \plus{} \sin \dfrac{2\pi}{7}$
[/hide]

[quote="DaReaper"]If the roots of $ x^2 \minus{} bx \plus{} c \equal{} 0$ are
\[ sin \frac {\pi}{7}
\]
and
\[ cos\frac {\pi}{7}
\]
then $ b^2$ equals what?

[b]EDIT: THE ANSWER IS IN TERMS OF C[/b][/quote]

$ (\sin \frac{\pi}{7} \plus{} \cos \frac{\pi}{7})^2\equal{}1\plus{}2 \cdot \sin \frac{\pi}{7} \cdot \cos \frac{\pi}{7}\equal{}1\plus{}2c$.


function
algebra
domain
floor function
trigonometry
functional equation

$f(x+1)-f(x) = 1$

how can i get all function $f$ that have this condiction?!

for example, the Floor Function has this property, and a lot of others

if we use methods for solving this we get  $f(x) = x+a$

but this is not a general solution

so, in recursive equations does not have a general solution?! is impossible to get then!?

because in differential equations we have general solutions

hm...

i need to solve this problem with fourier!?

it's like kill a ant with nuclear weapon

I did not say anything about Fourier series; I just mentioned a periodic function (with period 1). This does not have to be nice in any way, so it might not even have Fourier coefficients.

if  f is periodic. so. we can express f in terms of fourier series

but it does not give, necessarily , functions like Floor.

because floor is not periodic

Define the function however you want on $[0,1)$ and it extends uniquely to a function on the real line.

For example, if you take the function $f(x)=0$ on $[0,1)$, you get the floor function.

$f$ is definied in all possible $x$

$x$ is a complex number.

So we need a assumption for complex values

with this we get the functional equation $f(x+1)-f(x) =1$

and Floor is not periodic

see

http://functions.wolfram.com/IntegerFunctions/Floor/

If $x$ is complex define $f$ to be ANY function on $0\leq \text{Re}(z) < 1$, and put $f(z+n)=f(z)+n$ for every integer $n$.  This is the general solution.

If your domain is $\mathbb{C}$, the floor function doesn't exist. You're being inconsistent.
On $\mathbb{R}$, the values in an interval of length 1 determine the values everywhere- and you can have [i]any[/i] function in that interval. My expression $g(x)+x$ with $g$ periodic allows you to write nice solutions as sums of nice pieces, while still allowing the full range of ugly solutions- $g$ could be any function on $[0,1)$, extended to be periodic. For the floor function $\lfloor x\rfloor$, this is $g(x)=-x$ on $[0,1)$, $g(x)=n-x$ on $[n,n+1)$; that has jump discontinuities at every integer.

[The complex case is addressed in blahblahblah's post]

floor function is definied on complexes too!!

Se Wolfram's link!!!

http://functions.wolfram.com/IntegerFunctions/Floor/02/

[quote="jmerry"]Take any periodic function and add $x$.[/quote]

You mean like $f(x)=\sin(\frac{x}{2\pi})+1$?

This topic is a waste of time. The condition $f(x+1)-f(x)=1$ means exactly "$f(x)-x$ is periodic with period 1" (whenever it's domain $D$ satisfies requied condition $x\in D\Rightarrow x\pm 1\in D$, but  $\mathbb{C}$ obviously satisfies). Formally $f$ satisfies the given condition iff $f(x)-x$ is periodic with period 1 (if $f$ satisfies this, $f(x)-x$ is periodic with period 1, and if $f(x)-x$ is periodic with period 1, $f$ satisfies this condition).

[edit]
Of course, there should be $f(x)-x$, sorry.

No, it means $f(x)-x$ is periodic with period 1.  But close enough.


For $m_a,m_b,m_c$ medians of a triangle of sides $a,b,c$ prove the inequalities

$\sum \frac {m_a}{a} \geq \frac{3\sqrt{3}}{2} $

$\sum \frac {a^2}{m_bm_c} \geq 4 $

The second one is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15852]this[/url] one, after you divide it by $m_am_bm_c$.

Thr first one follows from:

http://www.mathlinks.ro/Forum/viewtopic.php?t=3171

and $(x+y+z)^2\geq 3(xy+yz+zx)$.

[quote="treegoner"]More problem here
Prove that
    $\sum \frac{ab}{m_am_b}\geq 4$[/quote]

same as this: http://www.mathlinks.ro/Forum/viewtopic.php?t=15897


inequalities
algebra
polynomial
inequalities proposed

[quote="yixinshen"]Let a,b,c,d>0 such that a+b+c+d=4.
Prove that a/(b^2+1)+b/(c^2+1)+c/(d^2+1)+d/(a^2+1)>=2[/quote]

Really?You are shenyixin?But you should post it in Inequalities Unsolved Problems :)

You can see the proof here:http://aoshoo.com/bbs1/dispbbs.asp?boardid=48&Id=13119  :lol:

[quote="yixinshen"]Thank you.
I know you are hanjingjun,but how do you know me?[/quote]

This year,in Suzhou.  :lol:

[quote="Sunkern_sunflora"]Hello shfdfzhjj(or hanjingjun?), sorry to say but ur link is showing a page that has expired!
Could u post the solution here please! :)[/quote]
Ok,just notice a/(1+b^2)=a-ab^2/(1+b^2)
and 1+b^2>=2b :lol:

PS:I think Beijing gives us a wonderful performance!Great nation!

[quote="yixinshen"]Let a,b,c,d>0 such that $ a\plus{}b\plus{}c\plus{}d\equal{}4$.
Prove that $ \frac{a}{b^2\plus{}1}\plus{}\frac{b}{c^2\plus{}1}\plus{}\frac{c}{d^2\plus{}1}\plus{}\frac{d}{a^2\plus{}1}\ge \ 2$[/quote]

[quote="Marius Mainea"][quote="yixinshen"]Let a,b,c,d>0 such that $ a \plus{} b \plus{} c \plus{} d \equal{} 4$.
Prove that $ \frac {a}{b^2 \plus{} 1} \plus{} \frac {b}{c^2 \plus{} 1} \plus{} \frac {c}{d^2 \plus{} 1} \plus{} \frac {d}{a^2 \plus{} 1}\ge \ 2$[/quote][/quote]

Yes,any problem? :maybe:

Just a second, can u tell me whether the following ineq holds or not?
For $ a\plus{}b\equal{}2$, prove that \[ \frac{a}{b^2\plus{}1}\plus{}\frac{b}{a^2\plus{}1} \geq {1}\] :maybe:

hello, is $ a$ and $ b$ positive?
Sonnhard.

hello, your inequality $ \frac {a}{b^2 \plus{} 1} \plus{} \frac {b}{a^2 \plus{} 1}\geq1$ for all positve $ a,b$ with $ a \plus{} b \equal{} 2$ is true. Proof: Let us consider $ \frac {a}{b^2 \plus{} 1} \plus{} \frac {b}{a^2 \plus{} 1} \minus{} 1 \equal{} \frac {a}{(2 \minus{} a)^2 \plus{} 1} \plus{} \frac {2 \minus{} a}{a^2 \plus{} 1} \minus{} 1 \equal{} \minus{} \frac {(a^2 \minus{} 2a \minus{} 5)(a \minus{} 1)^2}{(a^2 \minus{} 4a \plus{} 5)(a^2 \plus{} 1)}$.It is clear that $ (a \minus{} 1)^2\geq0$ and $ a^2 \plus{} 1\geq0$. Further it is $ a^2 \minus{} 4a \plus{} 5 \equal{} (a \minus{} 2)^2 \plus{} 1\geq1 > 0$ and $ \minus{} (a^2 \minus{} 2a \minus{} 5) \equal{} \minus{} (a \minus{} 1)^2 \plus{} 6 \equal{} 6 \minus{} (a \minus{} 1)^2\geq0$ for all $ a$ with $ 0\le a \le2$.
Hence the inequality is proven.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, your inequality $ \frac {a}{b^2 \plus{} 1} \plus{} \frac {b}{a^2 \plus{} 1}\geq2$ for all positve $ a,b$ with $ a \plus{} b \equal{} 2$ is true. [/quote]
It should be $ \frac {a}{b^2 \plus{} 1} \plus{} \frac {b}{a^2 \plus{} 1}\geq1$
Why you don't use Cauchy?  :D

hello, thank you for your hint, it was a typo i have corrected it.
Sonnhard.

[quote="Ji Chen"]If $ a,b,t$ be nonnegative numbers such that $ a \plus{} b \equal{} 2$, then

$ \frac {a}{b^2 \plus{} t} \plus{} \frac {b}{a^2 \plus{} t}\geq\frac {2}{1 \plus{} t},$

with equality if and only if $ a \equal{} b \equal{} 1$.

If $ a,b,c$ be nonnegative numbers such that $ a \plus{} b \plus{} c \equal{} 3$ and $ 0\leq t\leq 3 \plus{} 2\sqrt {3}$, then

$ \frac {a}{b^2 \plus{} t} \plus{} \frac {b}{c^2 \plus{} t} \plus{} \frac {c}{a^2 \plus{} t}\geq\frac {3}{1 \plus{} t},$

with equality if $ t \equal{} 3 \plus{} 2\sqrt {3},a \equal{} 0,b \equal{} 3 \minus{} \sqrt {3},c \equal{} \sqrt {3}$.

If $ a,b,c,d$ be nonnegative numbers such that $ a \plus{} b \plus{} c \plus{} d \equal{} 4$ and $ 0\leq t\leq t_{0} \equal{} 1.6351575\cdots$, then

$ \frac {a}{b^2 \plus{} t} \plus{} \frac {b}{c^2 \plus{} t} \plus{} \frac {c}{d^2 \plus{} t} \plus{} \frac {d}{a^2 \plus{} t}\geq\frac {4}{1 \plus{} t},$

with equality if $ a \equal{} 0,b \equal{} 0.70135727\cdots$ is a root of the irreducible polynomial

[/quote]
Oh,Chen ji you always gives us beautiful result.I have a proof for $ \frac {a}{b^2 \plus{} t} \plus{} \frac {b}{c^2 \plus{} t} \plus{} \frac {c}{a^2 \plus{} t}\geq\frac {3}{1 \plus{} t},$
But I think the second is to hard for me.
If $ a,b,c,d$ be nonnegative numbers such that $ a \plus{} b \plus{} c \plus{} d \equal{} 4$ and $ 0\leq t\leq t_{0} \equal{} 1.6351575\cdots$, then

$ \frac {a}{b^2 \plus{} t} \plus{} \frac {b}{c^2 \plus{} t} \plus{} \frac {c}{d^2 \plus{} t} \plus{} \frac {d}{a^2 \plus{} t}\geq\frac {4}{1 \plus{} t},$ :(

Thank u Dr Sonnhard Graubner for your nice proof.
 :)
I think now we can upgrade the ineq posted by yixinshen for $ n$ variables. Here it goes:

Let $ a_{1}\plus{}a_{2}\plus{}...\plus{}a_{n}\equal{}n$ then prove that for all $ a_{i}>0$
$ \sum_{cyc} \frac{a_{1}}{a_{2}^2\plus{}1} \geq {\frac{n}{2}}$

[quote="shfdfzhjj"]I have a proof for $ \frac {a}{b^2 \plus{} t} \plus{} \frac {b}{c^2 \plus{} t} \plus{} \frac {c}{a^2 \plus{} t}\geq\frac {3}{1 \plus{} t}$.[/quote]
See: http://www.mathlinks.ro/Forum/viewtopic.php?t=39448


$ \begin {cases} x(y\plus{}z)\equal{}35\\y(x\plus{}z)\equal{}32\\z(x\plus{}y)\equal{}27 \end {cases}$

[quote="thanhnam2902"]$ \begin {cases} x(y \plus{} z) \equal{} 35 \\
y(x \plus{} z) \equal{} 32 \\
z(x \plus{} y) \equal{} 27 \end {cases}$[/quote]

$ xy\plus{}xz\equal{}35$
$ xy\plus{}yz\equal{}32$
$ zx\plus{}zy\equal{}27$
$ 2(xy\plus{}yz\plus{}zx)\equal{}35\plus{}32\plus{}27\equal{}94$ ----> $ xy\plus{}yz\plus{}zx\equal{}47$ ---> $ yz\equal{}12$;$ zx\equal{}15$;$ xy\equal{}20$ --> $ (xyz)^{2}\equal{}3600$ --->

$ xyz\equal{}\minus{}60$ ---> $ x\equal{}\minus{}5,y\equal{}\minus{}4,z\equal{}\minus{}3$
$ xyz\equal{}60$  --->  $ x\equal{}5.y\equal{}4,z\equal{}3$


1.  Maximize the function $ f(x,y,z)\equal{}xyz$ subject to the constraints $ h(x,y,z)\equal{}x\plus{}y\plus{}z\equal{}1$   and  $ g(x,y,z)\equal{}x\plus{}y\minus{}z\equal{}0$.

I got to...
$ yz\equal{} a \plus{} b$
$ xz\equal{} a \plus{} b$
$ xy \equal{} a \minus{} b$
$ x\plus{}y\plus{}z \equal{} 1$
$ x\plus{}y\minus{}z\equal{}0$

I don't know how to continue from here... Please help...

(Let a be the legrange multiplier 1 and b the the legrange multiplier 2.)

Note that the first two equations give you that $ yz\equal{}xz.$ From this, either $ z\equal{}0,$ which would not result in a maximum, or $ x\equal{}y.$ If $ x\equal{}y,$ then $ z\equal{}2z$ and we get that the only critical point in the interior of the first octant is $ \left(\frac14,\frac14,\frac12\right).$

(You didn't say whether or not you were restricting to that octant; if not I'll let you think about the consequences.)


A sequence f(n) is defined as follows :

f(0) = 0 ;  f(n) = n - f(f(n-1))

Determine the value of  f(123456789).

I agree with Geehoon, and I suspect I know his method, but I wonder if he has a proof of it. :-)

[quote="Shamik Banerjee"][b]f(8) = 8 - f(4) = 8 - 4 = 4[/b]

... I would have fot the result of this sequence. ...[/quote]

The bolded line is in error.  I don't know what the word "fot" means.  And I think you'll find the pattern less obvious than you might have anticipated.


inequalities
trigonometry
complex numbers

Hi.  I need help on the following problem:

If  $w$ and  $z$ are complex numbers, prove that:

$2|w||z||w-z|\ge\left(|w|+|z|\right)\big|\,w|z|-z|w||$

Thanks.

let $w=a+bi$ and $z=c+di$
and $|w|=\sqrt{a^2+b^2}$
and then use some algebra and see where that gets you.

Yeah.. that gets slightly complicated.  Is there any way manipulating it and then plugging in a+bi and c+di?

Mr.Ocax

Try square the two members and then use the formula :

  $|z|^2 = z \overline z$

  I' ll try it in this way too. Of course we have to do good operation and simplifications.

 [u] Babis[/u]

If we write $w=We^{i\alpha}$ and $z=Ze^{i\beta}$ where $W,Z\geqslant 0$, then the given inequality simplifies to

$\begin{eqnarray*} 2|We^{i\alpha}-Ze^{i\beta}|\geqslant (W+Z) |e^{i\alpha}-e^{i\beta}|$

From there we have a chain of equivalent inequalities:

$2(W^2+Z^2-2WZ\cos(\alpha-\beta))\geqslant W^2+Z^2+2WZ-(W^2+Z^2+2WZ)\cos(\alpha-\beta)$

$W^2+Z^2-2WZ\cos(\alpha-\beta)\geqslant 2WZ-W^2\cos(\alpha-\beta)-Z^2\cos(\alpha-\beta)$

$W^2+Z^2-2WZ\geqslant -\cos(\alpha-\beta)(W^2+Z^2-2WZ)$

$\cos(\alpha-\beta)\geqslant -1$

and the last line is obviously true.


Find all functions $ f: \mathbb R\to\mathbb R$ such that
\[ f(x\plus{}f(y\minus{}x))\equal{}f(x)\minus{}f(y)\plus{}x\]
for all $ x,y\in\mathbb R$.

[quote="pco"][quote="nayel"]Find all functions $ f: \mathbb R\to\mathbb R$ such that
\[ f(x \plus{} f(y \minus{} x)) \equal{} f(x) \minus{} f(y) \plus{} x
\]
for all $ x,y\in\mathbb R$.[/quote]

$ x \equal{} y$ implies $ f(x \plus{} f(0)) \equal{} x$ and so $ f(x) \equal{} x \minus{} f(0)$ and so ($ x \equal{} 0$) $ f(0) \equal{} \minus{} f(0)$ and $ f(0) \equal{} 0$.

And so the unique solution would be $ f(x) \equal{} x$ but it is easy to check that this necessary condition does not work.

And so no solution for this equation.[/quote]

 Nice ! But if we had :
\[ f(x \plus{} f(y \minus{} x)) \equal{} f(x) \plus{} f(y) \minus{} x
\]
for all $ x,y\in\mathbb R$

   what should we do  to provw that $ f(x) \equal{} x\plus{}c$ ?

 Babis

[quote="stergiu"][quote="pco"][quote="nayel"]Find all functions $ f: \mathbb R\to\mathbb R$ such that
\[ f(x \plus{} f(y \minus{} x)) \equal{} f(x) \minus{} f(y) \plus{} x
\]
for all $ x,y\in\mathbb R$.[/quote]

$ x \equal{} y$ implies $ f(x \plus{} f(0)) \equal{} x$ and so $ f(x) \equal{} x \minus{} f(0)$ and so ($ x \equal{} 0$) $ f(0) \equal{} \minus{} f(0)$ and $ f(0) \equal{} 0$.

And so the unique solution would be $ f(x) \equal{} x$ but it is easy to check that this necessary condition does not work.

And so no solution for this equation.[/quote]

 Nice ! But if we had :
\[ f(x \plus{} f(y \minus{} x)) \equal{} f(x) \plus{} f(y) \minus{} x
\]
for all $ x,y\in\mathbb R$

   what should we do  to provw that $ f(x) \equal{} x$ ?

 Babis[/quote]

If $ f(a) \equal{} f(a\plus{}c)$, put $ y \equal{} x \plus{} a$ and $ y \equal{} x \plus{} a \plus{} c$, we get $ f(x\plus{}c) \equal{} f(x)$ for all x.
Put $ x \equal{} x \plus{} c$, we get $ c \equal{} 0$. So $ f$ is injective.
Put $ x \equal{} f(y)$, we get $ f(y\minus{}f(y)) \equal{} 0$. $ y\minus{}f(y)$ must be a constant because $ f$ is injective.
It is easy to check that $ f(x) \equal{} x \plus{} c$ satisfy the condition.

[quote="ringos"]
.........
If $ f(a) \equal{} f(a \plus{} c)$, put $ y \equal{} x \plus{} a$ and $ y \equal{} x \plus{} a \plus{} c$, we get $ f(x \plus{} c) \equal{} f(x)$ for all x.
Put $ x \equal{} x \plus{} c$, we get $ c \equal{} 0$. So $ f$ is injective.
Put $ x \equal{} f(y)$, we get $ f(y \minus{} f(y)) \equal{} 0$. $ y \minus{} f(y)$ must be a constant because $ f$ is injective.
It is easy to check that $ f(x) \equal{} x \plus{} c$ satisfy the condition.[/quote]

 Please , can you explain the last step ?

 '' Put $ x \equal{} x \plus{} c$, we get $ c \equal{} 0$. So $ f$ is injective. ''

 Where (to which relation)do you put this $ x\equal{}x\plus{}c$ and how do  reach that c=0?
 Thanks in advance - Babis

[quote="stergiu"][quote="ringos"]
.........
If $ f(a) \equal{} f(a \plus{} c)$, put $ y \equal{} x \plus{} a$ and $ y \equal{} x \plus{} a \plus{} c$, we get $ f(x \plus{} c) \equal{} f(x)$ for all x.
Put $ x \equal{} x \plus{} c$, we get $ c \equal{} 0$. So $ f$ is injective.
Put $ x \equal{} f(y)$, we get $ f(y \minus{} f(y)) \equal{} 0$. $ y \minus{} f(y)$ must be a constant because $ f$ is injective.
It is easy to check that $ f(x) \equal{} x \plus{} c$ satisfy the condition.[/quote]

 Please , can you explain the last step ?

 '' Put $ x \equal{} x \plus{} c$, we get $ c \equal{} 0$. So $ f$ is injective. ''

 Where (to which relation)do you put this $ x \equal{} x \plus{} c$ and how do  reach that c=0?
 Thanks in advance - Babis[/quote]

Explanation :

Let $ P(x,y)$ be the property $ f(x\plus{}f(y\minus{}x))\equal{}f(x)\plus{}f(y)\minus{}x$

Suppose $ f(a)\equal{}f(b)$. Then :

$ P(x,x\plus{}a)$ gives $ f(x\plus{}f(a))\equal{}f(x)\plus{}f(x\plus{}a)\minus{}x$
$ P(x,x\plus{}b)$ gives $ f(x\plus{}f(b))\equal{}f(x)\plus{}f(x\plus{}b)\minus{}x$

And, since $ f(a)\equal{}f(b)$, the two LHS are identical, so the two RHS are also identical,  and we have the property $ Q(x)$ : $ f(x\plus{}a)\equal{}f(x\plus{}b)$ $ \forall x$

Let then $ P(a,x)$ : $ f(a\plus{}f(x\minus{}a))\equal{}f(a)\plus{}f(x)\minus{}a$
$ Q(x\minus{}b\minus{}a)$ implies $ f(x\minus{}b)\equal{}f(x\minus{}a)$ so LHS is equal to $ f(a\plus{}f(x\minus{}b))$
$ Q(f(x\minus{}b))$ implies $ f(a\plus{}f(x\minus{}b))\equal{}f(b\plus{}f(x\minus{}b))$ so LHS is equal to $ f(b\plus{}f(x\minus{}b))$
And, since $ f(a)\equal{}f(b)$, the equality becomes $ f(b\plus{}f(x\minus{}b))\equal{}f(b)\plus{}f(x)\minus{}a$

But $ P(b,x)$ implies $ f(b\plus{}f(x\minus{}b))\equal{}f(b)\plus{}f(x)\minus{}b$

And so $ f(b)\plus{}f(x)\minus{}a\equal{}f(b)\plus{}f(x)\minus{}b$

So $ f(a)\equal{}f(b)$ implies $ a\equal{}b$ and $ f(x)$ is injective.

Then $ P(f(x),x)$ gives $ f(f(x)\plus{}f(x\minus{}f(x)))\equal{}f(f(x))$ and so injectivity implies $ f(x)\plus{}f(x\minus{}f(x))\equal{}f(x)$ and so $ f(x\minus{}f(x))\equal{}0$ and so, with injectivity again : $ f(x)\equal{}x\plus{}$constant.

[quote="pco"][quote="stergiu"][quote="ringos"]
.........
If $ f(a) \equal{} f(a \plus{} c)$, put $ y \equal{} x \plus{} a$ and $ y \equal{} x \plus{} a \plus{} c$, we get $ f(x \plus{} c) \equal{} f(x)$ for all x.
Put $ x \equal{} x \plus{} c$, we get $ c \equal{} 0$. So $ f$ is injective.
Put $ x \equal{} f(y)$, we get $ f(y \minus{} f(y)) \equal{} 0$. $ y \minus{} f(y)$ must be a constant because $ f$ is injective.
It is easy to check that $ f(x) \equal{} x \plus{} c$ satisfy the condition.[/quote]

 Please , can you explain the last step ?

 '' Put $ x \equal{} x \plus{} c$, we get $ c \equal{} 0$. So $ f$ is injective. ''

 Where (to which relation)do you put this $ x \equal{} x \plus{} c$ and how do  reach that c=0?
 Thanks in advance - Babis[/quote]

Explanation :

Let $ P(x,y)$ be the property $ f(x \plus{} f(y \minus{} x)) \equal{} f(x) \plus{} f(y) \minus{} x$

Suppose $ f(a) \equal{} f(b)$. Then :

$ P(x,x \plus{} a)$ gives $ f(x \plus{} f(a)) \equal{} f(x) \plus{} f(x \plus{} a) \minus{} x$
$ P(x,x \plus{} b)$ gives $ f(x \plus{} f(b)) \equal{} f(x) \plus{} f(x \plus{} b) \minus{} x$

And, since $ f(a) \equal{} f(b)$, the two LHS are identical, so the two RHS are also identical,  and we have the property $ Q(x)$ : $ f(x \plus{} a) \equal{} f(x \plus{} b)$ $ \forall x$

Let then $ P(a,x)$ : $ f(a \plus{} f(x \minus{} a)) \equal{} f(a) \plus{} f(x) \minus{} a$
$ Q(x \minus{} b \minus{} a)$ implies $ f(x \minus{} b) \equal{} f(x \minus{} a)$ so LHS is equal to $ f(a \plus{} f(x \minus{} b))$
$ Q(f(x \minus{} b))$ implies $ f(a \plus{} f(x \minus{} b)) \equal{} f(b \plus{} f(x \minus{} b))$ so LHS is equal to $ f(b \plus{} f(x \minus{} b))$
And, since $ f(a) \equal{} f(b)$, the equality becomes $ f(b \plus{} f(x \minus{} b)) \equal{} f(b) \plus{} f(x) \minus{} a$

But $ P(b,x)$ implies $ f(b \plus{} f(x \minus{} b)) \equal{} f(b) \plus{} f(x) \minus{} b$

And so $ f(b) \plus{} f(x) \minus{} a \equal{} f(b) \plus{} f(x) \minus{} b$

So $ f(a) \equal{} f(b)$ implies $ a \equal{} b$ and $ f(x)$ is injective.

Then $ P(f(x),x)$ gives $ f(f(x) \plus{} f(x \minus{} f(x))) \equal{} f(f(x))$ and so injectivity implies $ f(x) \plus{} f(x \minus{} f(x)) \equal{} f(x)$ and so $ f(x \minus{} f(x)) \equal{} 0$ and so, with injectivity again : $ f(x) \equal{} x \plus{}$constant.[/quote]

  Many thinks  :lol:

Michael Lambrou , Prof. at University of Crete - Greece gave this solution :

 Let  $ f(0) \equal{} c$ For $ x \equal{} y \equal{} \minus{}c$ the initial gives

                                           $ f(\minus{}c \plus{} c) \equal{} f(\minus{}c) \plus{} f(\minus{}c) \plus{} c$,
that is
               $ c \equal{} 2f(\minus{}c) \plus{} c$,
 hence $ f(\minus{}c) \equal{} 0$.

For  $ x \equal{} t \plus{} c, y \equal{} t$  the initial relation gives

                                              $ f( t \plus{} c \plus{} 0) \equal{} f(t \plus{} c) \plus{} f(t)\minus{}t\minus{}c$
Thus $ f(t) \equal{} t \plus{} c$ , which is acceptable .

 [color=blue][i]I must thank prof Lambou and  all solvers of this problem and of course Nayel who gave the basic idea to create the problem.[/i][/color]

 BAbis


Going at 50 mph in one direction, George traveled 70 kilometers in t_1 seconds. He took t_2 seconds to travel the same distance back, but instead went at 100 mph.  What was his average speed in miles per hour?

Hint:

[hide]Beware of extraneous infomation![/hide]

[quote]Going at 50 mph in one direction, George traveled 70 kilometers in $t_1$  seconds.
He took $t_2$ seconds to travel the same distance back, but instead went at 100 mph.
What was his average speed in miles per hour?[/quote][hide]
Also beware of the units: miles, kilometers, hours, seconds ...!

He drove $70 + 70 = 140$ kilometers in $t_1+t_2$ seconds.

His average speed is: $\frac{140}{t_1+t_2}\text{ km/sec} $

Then:  $\frac{140\text{ km}}{t_1+t_2\text{ sec}} \times \frac{1.609\text{ miles}}{1\text{ km}} \times \frac{3600\text{sec}}{1\text{hr}} \;= \;\frac{810,936}{t_1+t_2}\text{ mph}$[/hide]

Soroban -
[hide]Can you try to get the answer without using  t1 and t2 .. Hint:  it is possible to express the final answer in  just a  number..[/hide]

Old proverb: : Airthmetic  means better than guessing.. but Harmonic means  better than airthmetic

[quote="Gyan"]

Old proverb: Arithmetic means better than guessing.. but Harmonic means  better than arithmetic[/quote]

^the elders knew what they were talking about.. *nods head* haha, rmr what the hint said... [hide]"[size=150]beware of extraneous information[/size]"[/hide]

I HATE these kind of problems almost as much as the Patsies!
They were the worst type of Mathcounts problems ever created.

I don't think mathcounts would have this kind of problem, they generally only require you to know american units of measurements.

[quote] they generally only require you to know american units of measurements.[/quote]
Don't let such matter confuse anyone.  [hide]Do not want to make fun of anyone but just  to give a funny example here   If  someone asks, [quote]Kumar has $5$ $ ringgit $  and Hakopian  has $3$   $ringgit$   How many $ringgits$ they have  togather/?  [/quote]
instead of [quote]Joe has  $5$ $dollars$  and  Jim    has $3$ $dollars$ ....[/quote]

The answer  is no more difficult (or easy) .. One need not be familier with Malaysian money (ringgit)  or    familier with  Sanskir or Armenian names  (Kumar or Hakopian) to "solve"  the problem.  In short don't worry about the units as long as  you don't need the conversion factors.. Also as pkothari said, "beware of extraneous information" .. don't even worry about the information which is not needed to solve the problem.  (For example you don't have to know how many ringgits in a dollar,  etc to "solve" the above problem .. so even if  problem had additonal info eg " 1 dollar = 3.8 ringgits" etc .. it  could be safely ignored.
BTW  Let me rephrase the  original problem:
[quote] A person  drove with tha speed of 100 mph  one way and 50 mph  on the return journey (same distance travelled) . What is the average speed (in mph) for the whole trip" [/quote]

The answer would be same. Try it out ... (Does not matter  if the distance was 70 Km or   0.000005 parsecs .. as long as  the return journey is the same distance)

Further Hints:
[hide]If the  person drove for $100$  mph and $ 50$  mph for the [b]same ammount of time[/b], the average speed wold be       $ 75$  but here the distance travelled is the same, and thus the average would be less than $75$ .  Why? because the person  drove at slower speed of $50$ twice as long as speed of $100$.

A simple resoning along this line will show that the average speed would come out to be
$  \frac  { 2 * 50 * 100 } {(50 + 100) }  =  66 \frac {2} {3} $   $(mph) $

IOW [b] instead [/b] of taking  airthmetic  mean of  50, and 100 we have to take the [b]harmonic [/b]mean. Harmonic mean is just the reciprocal of mean of the reciprocals...[/hide]

Hope this helps.[/hide]
Hope this is useful.

Guys and Gals:

Please read the above hint ... (My previous post) if need be.. The answer is fairly straight forward..

BTW - This type of  question (with just slightly different numerical values) [b] has appeard in MANY mathcounts tests[/b] in the past  (and other tests too!) so it may be a good idea to make sure to read the hint if you have too..

Gyan gives a great explanation.... and just to show he's correct...

[hide]write the "distance = rate(time)" as

70km = t_1(50mph)  AND

70km = t_2(100mph)

then solve for the question WITHOUT canceling or converting any units... but as gyan said.. the harmonic mean is the key to these type of problems....[/hide]


function
limit
integration
calculus
real analysis
complex analysis
advanced fields

Take $\phi$ to be a test function $C^\infty_{c}(\mathbb{R})$.
There is this distribution
$\frac{1}{(\cdot)}(\phi) : = \lim_{\epsilon \to 0}\int_{|x|\ge \epsilon}\frac{\phi(x)}{x}dx$
The authors call it the "Cauchy finite part of the integral $\int_\mathbb{R}\frac{1}{(\cdot)}$.
Can someone please explain to me what it does?  :huh:  :|  :huh:

I've usually seen that as a "principal value". The key is that we take a symmetric limit, so the integral converges ($\phi(x)=a+bx+o(1)$ near zero).

This sort of integral shows up quite often in complex analysis, but obviously test functions aren't analytic.

This distribution, which I would write as $\text{p.v.}\frac1x,$ can also be considered a tempered distribution. After all, there's no problem for large $x.$

Convolution by (a scalar multiple of) this distribution generates a linear operator known as the Hilbert transform. Either fedja or I could tell you more properties of the Hilbert transform than you asked for.

But one simple excercise: consider $\text{p.v.}\frac1x$ as a tempered distribution. What is its Fourier transform?

Well it is just $\lim_{\epsilon \to 0^{+}}\int_{|x| > \epsilon}\frac{e^{-2\pi ix\xi}}{x}dx$
which we can rewrite as
$\lim_{\epsilon \to 0}\int \frac{-2i \sin(2\pi x\xi)}{x}dx$
$\lim_{\epsilon \to 0}-2i(-Si(2\pi \xi \epsilon)+\frac{\pi}{2}\text{sgn}(2\pi \epsilon \xi))$
$\boxed{-i \pi \text{sgn}(\xi)}$

I hope it's right. :)
So the fourier transform of the Hilbert transform $F^{-1}HF$ is multiplication by that? It looks to switch the parity of a function. So, I am afraid I still can't really understand what is happening on the convolution side.  :( It seems quite interesting though!

Well you are of course not obligated to do anything, but I am quite happy to hear whatever you have to say about the Hilbert transform (especially easy properties that I can prove ;))

[quote]So the fourier transform of the Hilbert transform $F^{-1}HF$ is multiplication by that?[/quote]

Yes. To make things nicer, let's divide by $\pi$ in the definition of the Hilbert transform. In other words the Hilbert transform (at least for nice enough functions) is given by convolution with $\text{p.v.}\frac1{\pi x}.$

A few items:

1. $\|Hf\|_{2}=\|f\|_{2}$ - it's an isometry on $L^{2}.$ You've got the reason why sitting right in front of you.

2. Suppose $f$ is a (sufficiently nice) real-valued function on $\mathbb{R}.$ Let $u(x,y)$ be harmonic on the upper half plane $\{(x,y): y>0\},$ $u(x,y)\to 0$ as $y\to\infty,$ and $u$ has $f$ as boundary values. In other words, $u$ is a solution to the Dirichlet problem in the upper half plane with boundary values $f.$ Then find $v$ such that $u+iv$ is holomorphic (and have $v\to0$ as $y\to\infty$ as well.)

Show that the boundary values of $v$ are $Hf,$ the Hilbert transform of $f.$

(When I say "sufficiently nice" I'm saying put whatever smoothness or rate of vanishing at $\infty$ conditions on $f$ that would be convenient for your argument.)

3. Suppose $a$ is a $C^{1}$ function supported on an inteval $I$ centered at point $x_{0}$ and that $\|a\|_{\infty}\le\frac1{|I|}$ and $\int_{I}a=0.$ Prove that $\|Ha\|_{1}\le C$ for some constant $C$ independent of $a.$

(Hint: estimate how rapidly $Hf$ decays away for points well away from $I.$)


To compete, simply post your Alcumus Points (Not karama or Alcumus score) and your FTW rating.

There will be two rounds:

[size=200]Relay[/size]

[size=150]RULES:[/size]

Using the info above, I will put you into teams. Each team will assign number 1~4, in any order they would like. I will pm the first question to the two people who are first. Those people will answer the question and pm their answer to me. The next person's answer will depend on the first person's answer. When you see TATPGBM in a problem, you can decipher it to be: The answer the person got before me. I will pm the 2nd people the first person's answer and the 2nd question. This cycle will continue on to the 4th person. There is to be no disscussion out of thread and the whole group should finish in two weeks.

[size=150]SCORING:[/size]

The first person's answer is worth 1, the second's is 2, and so on until the fourth person, for a full score of 10.

[size=200]Pressure Round[/size]

[size=150]RULES/SCORING:[/size]

There will be 5 questions. You have 2 days to work on each question as a team. The first answer you submit, no matter what question it is, will be worth 1 point. The next will be 2 and so on for all 5, for a total of 25. If you have no answer within 2 days, you will recieve 0 for that question. You may only submit each question once.

Signups:
1.
2.
3.
4.
5.
6.
7.
8.

Signs up.

ALCUMUS POINTS: 905 (yeah I know)

FTW RATING: 1857

Alcumus points is [b]not[/b] a good indicator of skill; in fact, higher points actually denotes a smaller percentage of problems answered correctly.  This is because after one gets an incorrect problem on Alcumus, they get the opportunity to later repeat it, so they get double the points.

Done complaining.

1654 Alcumus points, 2043 rating.

[quote="AIME15"]Alcumus points is [b]not[/b] a good indicator of skill; in fact, higher points actually denotes a smaller percentage of problems answered correctly.  This is because after one gets an incorrect problem on Alcumus, they get the opportunity to later repeat it, so they get double the points.
[/quote]

I know, just doing this cuz I can.

Alcumus: 4359 of 5074 (haven't really used it much)
FTW: 1169

Current signups:

PCS = pointless combined score
Signups:
1. BOGTRO - PCS = 2762
2. AIME15 - PCS = 3697
3.
4.
5.
6.
7.
8.

modular: I need your score, not ranking.

Do you have to have an Alcumus score and an FTW rating to sign up? I don’t play either of those.

[quote="rd5493"]Alcumus: 1605

FTW: 1196

Uh, modularmarc, I'm pretty sure ur alcumus score is wrong because Jongao is the number one overall player, and his is only 2340 or so...can i sign up now or what?[/quote]

I'm not sure what I have to put.

Karma
Yours: 977 of 5075
cricketk: 906 of 5075

Points
Yours: 2641 of 5075
cricketk: 597 of 5075

Alcumus Score [?]
Yours: 4456 of 5075
cricketk: 3958 of 5075

This was all I found.

no, they don't care about your ranking out of the 5 thousand + people, they just want how many points you earned, not where u stand with those points...

PS, if i want to join can i pm you my fail scores instead of posting them on here? i want to save myself the embarrassment...

...

Go to report, and it will say like
Correct: x Incorrect: y
AoPS karma: z
Points:

You want the points.

I sign up.

FTW Rating: 1872
Alcumus Points: 1002

i join

Alcumus: 226 (pitiful)

FTW: 1494

silverstien, I'm not sure if you've enrolled in this beforehand.

[quote="Dojo"]TEAM B: 4 points (somehow fantasy got the last one right)
[/quote]
Heh, pwnt.

#4 didn't need #3's answer.

Wow, everyone else in my team got their problem wrong...

I am also here.

OMG... We're tied for second???? OMG.

And I think I'm here too.  :lol:  :maybe:

yeah, the person before me got his problem wrong, so i couldn't do mine.

Well I'll just assume that everyone is here.

Remember, there are the rules:

[quote]There will be 5 questions. You have 2 days to work on each question as a team. The first answer you submit, no matter what question it is, will be worth 1 point. The next will be 2 and so on for all 5, for a total of 25. If you have no answer within 2 days, you will recieve 0 for that question. You may only submit each question once.[/quote]

1. The following table relates weight on the moon and weight on Earth mesured in pounds. What is Joe's weight on the Moon if his weight on Earth is 120 lb?

\begin{tabular}{|c|c|c|c|c|c|c|}
\hline Earth (lb) & 1 & 2 & 3 & 4 & \cdots & 100 \\
\hline Moon (lb) & 0.16 & 0.32 & 0.48 & 0.64 & \cdots & 16 \\
\hline
\end{tabular}

2. A centipede has 42 legts. If all feet down on a scale register a weight of 24 grams, how much registers when the centipede lifts 14 legs into the air?

3. If $ \alpha$ and $ \beta$ are solutions of $ x^2-4x-3=0$, what is the value of $ \dfrac{1}{\alpha} + \dfrac{1}{\beta}$?

4. Five coins look the same, but one is a counterfeit coin with a different weight than each of the four genuine coints. Using a balance scale, what is the least number of weighings need to ensure that, in every case, the counterfeit coin is shown to be heavier or lighter?

5. For positive integer $ x,y,z$, find the number of solutions that satisfy $ x^2+y^2+z^2=49$

Go go go!!!

[quote="FantasyLover"]Dojo, results?[/quote]

Did you submit?  I submitted #1 to him.

Yeah. Team Fantasy has already submitted all 5...
I'll give results at the end of the round.

Uhh when's the 'end of the round'?

It's already been 4 days... :maybe:


geometry
perimeter
inequalities
congruent triangles
triangle inequality

In centimeters, the length of each side of a triangle is a whole number.  The perimeter of the triangle is 8 centimeters.  How many different (non-congruent) triangles meet this criteria?

A.None	B.One	C.Two	D.Three    E.More than three

[hide="Solution"]We want triples of positive integers that sum to eight, where the largest integer is less than four, by the triangle inequality.

We have $ (3, 3, 2)$. This is the only valid possibility. Therefore, the answer is $ \boxed{B}$.

[/hide]

[hide] Well, you can only have one way, using the triangle inequality fomula, you figure it out.
btw, the triangle inequality is where 2 sides added together must be greater than the 3rd side.
So, the only way you can use is 3,3,2 , so the answer would be B.[/list][/hide]


Let an $ n$ order Venn Space be a set of simple closed regions $ V \equal{} \{R_1, R_2, \ldots R_n\}$ such that for every subset $ S \subseteq V$, the intersection of all regions $ \in S$ with the complement of all regions $ \in V \setminus S$ is a non-empty connected region.  Find, with proof, the largest order of which a Venn Space may be constructed, or prove that Venn Spaces of arbitrarily large order may be constructed and provide an algorithm for constructing such Spaces.

We provide a construction via induction, let $ R_1$ be a simple closed region. If a venn space of order $ n$ is constructible, we claim a venn space of order $ n \plus{} 1$ is constructible.

In the venn space of order $ n$, we claim that we can draw a closed loop that passes through all $ 2^n$ subsets [eg, intersection of the regions of the subset] of the venn space exactly once [eg, the loop enters each region and then exits it once]. Let the region that consists of the loop and its interior be $ R_{n \plus{} 1}$. Then $ V_{n \plus{} 1} \equal{} V_n \cup R_{n \plus{} 1}$ is a venn space, as every subset $ S \subseteq V_n$ is partitioned into two non-empty regions $ S$ and $ S \cap R_{n \plus{} 1}$, and further note that each of these partitions are connected.

We show the existence of the loop via induction. Consider the graph where each node denotes a subset of $ V_n$, and two nodes are connected if the regions denoted by their subsets are adjacent. It suffices to find a Hamiltonian cycle $ C_n$ in this graph; one then draws the loop in the subsets in the order given by the cycle.

If the graph for $ V_{n \minus{} 1}$ contained a Hamiltonian cycle $ C_{n \minus{} 1}$, consider two adjacent subsets $ S_1, S_2$ in the cycle. Verify that after $ R_n$ is added to $ V_{n \minus{} 1}$, the subsets $ S_1, S_2$ remain adjacent, as well as the subsets $ S_1 \cap R_{n}, S_2 \cap R_n$. Then $ C_{n \minus{} 1}$ can still be constructed in the graph of $ V_n$, as well as a second cycle $ C_{n \minus{} 1}^*$ such that if $ S_1, S_2$ are adjacent in $ C_{n \minus{} 1}$, then $ S_1 \cap R_n, S_2 \cap R_n$ are adjacent in $ C_{n \minus{} 1}^*$. Also note that $ \emptyset$ is adjacent to $ R_{n}$, and that if $ S \in V_{n \minus{} 1}$ is adjacent to $ O$, then $ R_{n} \cap S$ is adjacent to $ S$. We can break the links between $ \emptyset$ and $ S$ in $ C_{n \minus{} 1}$ and $ R_{n}$ and $ R_{n} \cap S$ in $ C_{n \minus{} 1}^*$, and link $ \emptyset, R_n$ and $ S, R_n \cap S$ to merge the two cycles into one Hamiltonian cycle $ C_n$.


search
function
number theory unsolved
number theory

Represent the number $ 989 . 1001 . 1007 \plus{} 320$ as a product of primes.

Please use the search function:

http://www.mathlinks.ro/viewtopic.php?t=26606
http://www.mathlinks.ro/viewtopic.php?t=150533


algebra
polynomial
functional equation
Iteration
IMO Shortlist
IMO Longlist

Let $ f(x)$ be a polynomial with rational coefficients and $ \alpha$ be a real number such that \[ \alpha^3 \minus{} \alpha \equal{} [f(\alpha)]^3 \minus{} f(\alpha) \equal{} 33^{1992}.\] Prove that for each $ n \geq 1,$  \[ \left [ f^{n}(\alpha) \right]^3 \minus{} f^{n}(\alpha) \equal{} 33^{1992},\] where $ f^{n}(x) \equal{} f(f(\cdots f(x))),$ and $ n$ is a positive integer.

[quote="orl"]Let $ f(x)$ be a polynomial with rational coefficients and $ \alpha$ be a real number such that
\[ \alpha^3 \minus{} \alpha \equal{} [f(\alpha)]^3 \minus{} f(\alpha) \equal{} 33^{1992}.
\]
Prove that for each $ n \geq 1,$
\[ \left [ f^{n}(\alpha) \right]^3 \minus{} f^{n}(\alpha) \equal{} 33^{1992},
\]
where $ f^{n}(x) \equal{} f(f(\cdots f(x))),$ and $ n$ is a positive integer.[/quote]

Since the equation $ x^3 \minus{}x \minus{} c \equal{} 0$ has only one real root for every $ c >
\frac{2}{3\sqrt{3}}$ , $ \alpha$ is the unique real root of $ x^3\minus{}x\minus{}33^{1992} \equal{} 0$. Hence $ f^n(\alpha) \equal{}
f(\alpha) \equal{} \alpha$ ,as desired


A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_{n}$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10}=.4$ and $a_{n}\le .4$ for all $n$ such that $1\le n \le 9$ is given to be $p^{a}q^{b}r/(s^{c}),$ where $p,$ $q,$ $r,$ and $s$ are primes, and $a,$ $b,$ and $c$ are positive integers. Find $(p+q+r+s)(a+b+c).$

[quote="sapphyre571"][hide="el solution"]

cannot make...his ninth shot
[/hide]

4everwise, seeing as this problem has an AIME-format answer, do you know what year and number problem this is?[/quote]
Why can't he make his ninth shot :?:  :ninja:

[hide="My solution"]He must miss his 1st and 2nd shot and hit his 10th shot. So then he must hit either his 8th or 9th shot (or both), and hit one of his 5th, 6th, or 7th shots and hit his third or fourth.
If he shoots his first one in in the 3rd or 4th shot, then he has to miss the other, so there are two ways to do that. Then he can either shoot both in on the 8th and 9th or shoot one in (8,9) shots and one in (5,6,7) shots so that probability is $\frac{12}{25}(\frac{27\cdot 8\cdot 7}{125^{2}})$.
If he doesn't get it in on his (3,4) shots, he either gets two in on the (5,6,7) shots or one in the (5,6,7) and two in the (8.9) shots for a probability of $(.6)^{2}(3(.4)^{2}(.6)\cdot 2(.6)(.4)+3(.6)^{2}(.4)(.4)^{2})=\frac{2^{3}\cdot 3^{6}}{5^{7}}$. Adding those two probabilities we get $\frac{2^{3}\cdot 3^{4}\cdot 37}{5^{7}}$ and multiplying by $\frac{3^{2}\cdot 2}{5^{3}}$ for the first, second, and last shot we get $\frac{3^{6}\cdot 2^{4}\cdot 37}{5^{10}}$ so the answer is $(3+2+37+5)(6+4+10)=\boxed{940}$[/hide]

my bad - I had a screwup and thought $\frac{3}{9}>.4$

I got a different answer for this, and I'm pretty positive that it's right...

[hide]The probability of any set of shots in which the basketball player makes exactly 4 out of the 10 is going to be $3^{6}*2^{4}/5^{1}0$ times the number of ways possible. For there to be a probability of 40% or less for each shot, then the first and second shots must both be missed, there can be one shot made out of 3 and 4, there can be a total of 2 after 5, 6, and 7, there can be a total of 3 after 8 and 9, and the 10th shot must be made. I started counting cases here. There are 12 ways for the shooter to make one of shots 3 and 4, one of shots 5-7, and one of shots 8-9. There are 2 ways for the shooter to make one of shots 3 and 4, and both of shots 8 and 9. There are 6 ways for the shooter to make two of shots 5-7 and one of 8-9, and there are 3 ways for the shooter to make one of shots 5-7 and both of shots 8 and 9. Adding these up, there are $12+2+6+3=23$ ways for the ballplayer to make 4 out of 10 shots the right way, so the final fraction is $3^{6}2^{4}23/5^{1}0$ which gives a final answer of $660$[/hide]

Btw, this is from the 2002 AIME II

[quote="ch1n353ch3s54a1l"][quote="sapphyre571"][hide="el solution"]

cannot make...his ninth shot
[/hide]

4everwise, seeing as this problem has an AIME-format answer, do you know what year and number problem this is?[/quote]
Why can't he make his ninth shot :?:  :ninja:

[hide="My solution"]He must miss his 1st and 2nd shot and hit his 10th shot. So then he must hit either his 8th or 9th shot (or both), and hit one of his 5th, 6th, or 7th shots and hit his third or fourth.
If he shoots his first one in in the 3rd or 4th shot, then he has to miss the other, so there are two ways to do that. Then he can either shoot both in on the 8th and 9th or shoot one in (8,9) shots and one in (5,6,7) shots so that probability is $ \frac {12}{25}(\frac {27\cdot 8\cdot 7}{125^{2}})$.
If he doesn't get it in on his (3,4) shots, he either gets two in on the (5,6,7) shots or one in the (5,6,7) and two in the (8.9) shots for a probability of $ (.6)^{2}(3(.4)^{2}(.6)\cdot 2(.6)(.4) \plus{} 3(.6)^{2}(.4)(.4)^{2}) \equal{} \frac {2^{3}\cdot 3^{6}}{5^{7}}$. Adding those two probabilities we get $ \frac {2^{3}\cdot 3^{4}\cdot 37}{5^{7}}$ and multiplying by $ \frac {3^{2}\cdot 2}{5^{3}}$ for the first, second, and last shot we get $ \frac {3^{6}\cdot 2^{4}\cdot 37}{5^{10}}$ so the answer is $ (3 \plus{} 2 \plus{} 37 \plus{} 5)(6 \plus{} 4 \plus{} 10) \equal{} \boxed{940}$[/hide][/quote]

Would you please clarify how did you get $ \frac {12}{25}(\frac {27\cdot 8\cdot 7}{125^{2}})$. ?

answer is $660$.
So basically he has to make a shot on his tenth try. Then we break it up into cases based on whether he makes a shot on his eight try, his ninth try, or both.  The probability is $\frac{(23)(2^4)(3^6)}{5^{10}}$, and hence our answer!

I would like to note you can invoke PIE very nicely in this problem:

[hide="PIE"]

We want to figure out the number of ways to get sequences of 7 that have 3 elements in them, and satisfy the circumstances.

Remember, we are at the 3rd to 9th elements since the 1, 2 and 10th turns are already predetermined. We have to miss first 2 and make the 10th turn due to ratio restrictions. So when I say first 2, I am referring to turn 3 and 4.

However, I know that I cannot have more than 2 inside of the first 2 because of restrictions on turns 3 and 4, if I had 2 there then I would have more than 2 which is restricted. Let this be conditon A.

Also, I know that I cannot have more than 3 in the first 5, because if I have that then I will automatically break the x<2 rule. The bigger picture though is fine because I will only have 3, so there will be no restrictions on first 7. Let this be condition B.

Invoking complimentary, I have $\binom{7}{3} = 35$ ways to have 3 O's contained in the 7, and I want to figure out how many don't satisfy.

I want to know P(AUB), how many break condition A, condition B, or both and subtract from 35.

So I can say that it is P(A)+P(B)-P(A and B).

Well P(A) is 5 because I must choose the first 2 and then the last one doesn't matter, P(B) is $\binom{5}{3}$ because I need to choose 3 in the first 5 to break this rule, and P(A and B) is going to be if I have the first 2 and then one in either turn 5, 6 or 7. So there are 3 here.

This gives us 5+10-3=12, or 25-12 = 23.

The rest is just probability. I know that the chances of winning 4 and losing 6 is:

$(0.4)^{4} (0.6)^{6} *23$ because there are 23 orderings and all of them are equally likely.

[/hide]

I found a quick straightforward solution. Remove the the first two tries and the last. We are simply left with putting exactly 3 wins in 7 attempts. Doing casework based on the position of the first attemp ( There are only 3 cases) . We have $2(3*2+1)+3*2+3=23$

[quote=sapphyre571][hide="el solution"]

given that $a_{1}$ through $a_{9}$ are all less than .4, we know the player cannot make either of his first or second shots or his ninth shot, but he has to make his tenth shot.
so we have to distribute 3 made shots over shot attempts 3 through 8.
we know that because of the .4 probability rule, up to 1 of the third, fourth, or fifth shot can be made. since $\frac{3}{7}>.4$, only 2 of the first seven shots can be made, but we need three makes by shot #8, so shot #8 must be made.

using this, made shot combinations that work for the problem are
3-6-8-10
4-6-8-10
5-6-8-10
3-7-8-10
4-7-8-10
5-7-8-10
6-7-8-10

so we have 7 made shot combinations that work. the probability of any one of them occurring is $.4^{4}\times.6^{6}$ or $\frac{2}{5}^{4}\times\frac{3}{5}^{6}=\frac{2^{4}3^{6}}{5^{10}}$
but since we have seven ways to do this, the target probability is
$\frac{2^{4}3^{6}7}{5^{10}}$
so the answer is (2+3+5+7)(4+6+10)=(17)(34)=578
[/hide]

4everwise, seeing as this problem has an AIME-format answer, do you know what year and number problem this is?[/quote]

[url]https://artofproblemsolving.com/wiki/index.php/2002_AIME_II_Problems/Problem_12[/url]

Um the point was [b]not[/b] to post the link, but whatever.

[b]Solution:[/b] Notice that the first two shots have to be misses. Futhermore, the first shot can be made at the third, fourth, fifth, sixth, or even seventh attempt. Doing casework on these possibilities while noting that the third  shot must be made on the eighth/ninth attempt (and the second shot must be obviously before that), we get $23$ possible ways to do this. This gives us an answer of $23(0.4)^4(0.6)^6 = \frac{2^4 \cdot 3^6 \cdot 23}{5^{10}} \Rightarrow \boxed{660}$.


Let $x,y,z,t$ be real numbers satisfying:
$(x+y)(z+t)+xy+88=0$
Find  min of: $A= x^{2}+9y^{2}+6z^{2}+24t^{2}$


Να βρεθεί για γωνίες τριγώνου η ελάχιστη τιμή της παράστασης:

$ \sum\frac {sin^2Asin^2B}{sin^2C}$

(κυκλικά όχι συμμετρικά)

Το μόνο hint που δίνω είναι ότι ο Περσικός πολιτισμός άνθησε στις όχθες της Μεσοποταμίας πολύ νωρίτερα από τον Αρχαιοελληνικό και ότι οι κρεμαστοί κήποι της Βαβυλώνας είναι ένα από τα 7 θαύματα της ανθρωπότητας...

(αν βρείτε ποια είναι η ελάχιστη τιμή το hint αυτό θα σας βοηθήσει να αποδείξετε ότι είναι όντως ελάχιστη) :wink:

 :)

Μανία με αυτή την Iran ρε παιδί μου ... Λοιπόν θέστε όπου $ x\equal{}Cot(A)$ στην Iran και θα πάρετε την ανισότητα του Κώστα . So the min is $ \frac{9}{4}$

Αν κάποιος πάει με αντικαταστάσεις και το δυικό μετασχηματισμό του βγαίνει η ανισότητα
$ \sum\frac{1}{(x\plus{}y)^4}\geq\frac{9}{16xyz(x\plus{}y\plus{}z)}$

[quote="silouan"][...]Αν κάποιος πάει με αντικαταστάσεις και το [b]δυικό[/b] μετασχηματισμό του βγαίνει η ανισότητα[...][/quote]

Όπου εγώ διάβασα τον [i]δυ[b]τ[/b]ικό[/i] μετασχηματισμό και λέω.... ααα το χάσαμε τελείως με τις πολιτισμικές συγκρίσεις ανάμεσα σε δύση και ανατολή :rotfl:

Cheerio,

Durandal 1707


modular arithmetic
number theory unsolved
number theory

Is there any number in the Fibonacci sequence whose six last digits (in decimal system) are all 9?

This problem can be greatly generalised to problems like whether there exist a number in the Fibonacci sequence whose last $k$ digits (in decimal system) are all $9$, or all $0$, or just a string of digits for that matter... For this question, the trick is to consider the Fibonacci sequence $\mod 10^6$, the trick needs just a small alteration for other generalised qns.

Considering the Fibonacci sequence $\mod 10^6$, we want to prove that $-1$ occurs in the sequence $\mod 10^6$. Take the first $10^6+1$ numbers $F_1, F_2, F_3, ...$ of the Fibonacci sequence $\mod 10^6$. They form $10^6$ pairs $(F_1, F_2), (F_2, F_3), (F_3, F_4), \ldots$. Note that any $2$ consecutive terms of the sequence will determine all succeeding and preceeding terms, thus the pair $(0,0)$ will never occur, and there are only $\leq 10^6-1$ possible pairs, and by Pigeon-hole principle, at least one of the pairs will repeat. It's easy to see that the Fibonacci sequence is periodic $\mod 10^6$ (or $\mod 10^k$ for that matter), and the pair $(1, 1)$ will be the first pair to repeat. Letting $d$ be the length of the period, we have $F_{d+1} \equiv F_1 \equiv F_{d+2} \equiv F_2 \equiv 1 \pmod{10^6}$. We finally see that $F_{d-2} \equiv -1 \pmod{10^6}$ and so is our required number. In fact, there are infinitely many such Fibonacci numbers ending with six $9$s.  :)

it's true for all $n$ where $n$ is the number of last digits equaling $9$ because
$\{r_n\}$ such that $r_{n+2} \equiv r_{n+1}+r_{n} (\mod 10^k)$ and $r_0 \equiv 0 (\mod 10^k)$, $r_1 \equiv 1 (\mod 10^k)$ is cyclic and in addition $r_{-2} \equiv -1 (\mod 10^k)$.

Could you tell me why $(1,1)$ must be the first pair to repeat?

Because if another pair is the first repeated pair (call it (a,b)) we have the pair (b-a,a) before, which is also repeated and that contradicts that the first repeated pair is (a,b).

[quote="ZetaX"]The harder question is: for which $n$ exists a Fibonacci number $F_k$ ending in exactly $n$ times $9$.[/quote]
dblues has shown it for all $n$, hasn't he?

No. I've only shown that $\forall k \in \mathbb{N} \ \exists$ a Fibonacci number that ends with $k$ nines, but not exactly $k$ nines. Yes, the question on exactly $k$ nines is indeed a harder qn.

Ah, sorry for my carelessness.

Well, since it seems I'm not going to come up with a solution anyway, could you please post your solution, Zetax?


As part of a solution of a problem in the book the following polynomial was factored :
x^3 - 3x^2 - 3x - 4 = (x - 4)(x^2 + x + 1)
Could anyone explain how was it done? I assume it is somehow an easy process since the book omited any explanations on the factoring.
Thank you![/code]

My method personally is to:
1)use Descartes's Rule of Sines to determine which roots are positive, negative, and complex
2)Looking at the last nuber (constant) and finding all the roots by taking that number and dividng by the constant of the first term and its postive and negative
3)Start plugging in psssibilities, until you find a root
4)Use synthetic division to remove that root
5)Repeat until all rational roots are found

x^3 - 3x^2 - 3x - 4

You want to break up -3x^2 and -3x into pieces that will create an a,b,a,b,a,b pattern in the coefficients/constant.

Since the last term is -4, b=-4. First coefficient is 1, a=1

1,-4,1,-4,1,-4

x^3 - 3x^2 - 3x - 4
=x^3 - 4x^2 +x^2 - 4x + x - 4
=x^2(x-4) + x(x-4) + 1(x-4)
= (x^2+x+1)(x-4)


$13$ coins are arranged all on the circumference of a circle as shown in the attached diagram.

Two players take turns either to remove any one coin or two touching coins from the circle – the winner being the player to take the last coin.

Formulate a strategy that will ensure a victory to the second player in any game.

I think I have it, no matter what the other person picks, you start off by picking one, they pick how many they want, you pick two, they pick, you pick one, so this the number you pick each turn, 1, 2, 1, 2, 1....

umm i dont think that works :?
counter example:
they pick 1, you pick 1,
they pick 2, you pick 2
they pick 1, you pick 1
they pick 2, you pick 2,
they win

I think the winner is the man who pick first.Because If you pick first then you just only need pick one coin and later they pick how many coins you pick the same and you'll be person who pick the last coin.:D

Math is my life :!:

When you pick a coin, does the circle close (to create more touching coins)?

I don't think it really matters if the circle closes

Oh I know now. Here's the answer
[hide]To startyou have to make sure that when it comes to the last 4 or 5 coins, depending on what you picked, you're picking, if it's 4 coins, pick 1 coin and you win from there. If it's five coins, pcik two and you win from there. Basically, when it goes down to 3 coins, make sure that it's their pick.[/hide]

[quote="Jiang"][hide]If the first player picks 1 coin, pick two coins from the opposite end to form two equal sections left and right. Whatever the other player does to the either side, just do the same for the other. You will eventually win and the same if they picked 2 coins in the beginning.[/hide][/quote]

Thanks ! Nice solution...


analytic geometry
radical axis
geometry unsolved
geometry

$K$ is the radical center of the circles $w_A,w_B,w_C$

$w_A=(A,r_A)$
$w_B=(B,r_B)$
$w_C=(C,r_C)$

Given the coordinates of three points $A(x_A,y_A),B(x_B,y_B),C(x_C,y_C)$

is there a formula which can give us the coordinates of $K$ ?

Of course there is a formula!

The degree of R to the three circles are

(xr-xa)^2 +(yr-ya)^2 + ra^2    (1)
=
(xr-xb)^2 +(yr-yb)^2 + rb^2    (2)
=
(xr-xc)^2 +(yr-yc)^2 + rc^2      (3)

When you substract (1) from (2) and (2) from (3) you get linear system for xr and yr with two equations.

From here it is easy to find the formula

Thank you!

But the solution (using determinants) looks very complicated

$D = 4\left| \begin{array}{cc} x_A - x_B & y_A-y_B \\ x_A - x_C & y_A-y_C \end{array} \right |$

$D_x = \left| \begin{array}{cc} p_A -p_B & 2(y_A-y_B) \\ p_A -p_C & 2(y_A-y_C) \end{array} \right |$

$D_y = \left| \begin{array}{cc} 2(x_A - x_B) & p_A -p_B \\ 2(x_A - x_C) & p_A -p_C \end{array} \right |$

and $(x,y)=(\frac{D_x}{D},\frac{D_y}{D})$

where $p_A = OA^2 - r_A^2$ is the power of the point $O(0,0)$ with respect to the circle $(A,r_A)$, similarly for $p_B,p_C$

My problem is that I want to find if the point $K$ belongs to the circle $(A,r_A)$ and the above formula is not convenient (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64502]www.mathlinks.ro/Forum/viewtopic.php?t=64502 post#8[/url]).

Is there any other (simpler) expression for the coordinates of $K$ ?


geometry
search
combinatorics proposed
combinatorics

Ten students stand on a flat surface, with the distances between them all distinct. At twelve o’clock, when the church bells start chiming, each of them throws a water balloon at their closest neighbour. What is the minimum number of students who will be hit by water?

Hi ! You can search before post :wink: .
[url]http://www.mathlinks.ro/viewtopic.php?search_id=819275322&t=219759[/url]


inequalities
limit
logarithms
probability and stats

suppose that $ X_n: \Omega \to \mathbb{R}$ is a sequence of i.i.d random variables such that $ E(X)\equal{}0$, $ Var(X)\equal{}\sigma$. So one knows that $ M_n \equal{} \frac{X_1\plus{}X_2\plus{} \ldots \plus{} X_n}{n}$ tends towards $ 0$ almost surely. I would like to know (for example) how fast the quantity $ P(M_n \in [1;2])$ decreases. I guess it decreases exponentially (it is the case if the $ X_n$ are bounded in $ L^{\infty}$ thanks to the Hoeffding inequality), but how can one relax this boundedness assumption ?

If I understand correctly, you are looking for an error estimate for the Central Limit Theorem. [url=http://en.wikipedia.org/wiki/Berry%E2%80%93Ess%C3%A9en_theorem]Berry–Esséen theorem[/url] requires only a finite third moment.

This problem is solved in large deviations theory, in particular if you know $ Ee^{tX}<\infty \ \forall t$, then for $ a>EX$
$ \lim_{n\to\infty}\frac1n \log{P(\frac{S_n}{n}\geq a)}\to \minus{}I(a)$, where $ I(z)\equal{}\sup_t [tz\minus{}\log{Ee^{tX}}]$. This is the basic and most famous fact in large deviations; i'm not familiar with the theory, but there must be some results which do not require finitness of moment generating function.


Is the "Denominator of $ \frac {1998!}{6^998}$ reduced to lowest terms" part a separate question?

Do you just want us to calculate this? $ \frac {1}{\frac {1}{1980} \plus{} \frac {1}{1981} \plus{} ... \plus{} \frac {1}{1991}}$

In that case, the answer is $ \frac{2985138929317096278796336970560}{18041689970042472430834558547}$

yes
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You can find the greatest power of two(and three) that divides $ 1998!$, then you just have to simplify.
Note that the integer part of $ \frac{n}{a}$, where $ n$ and $ a$ are positive integers, is the numbers of multiples of $ a$ less or equal than $ n$.


function
topology
advanced fields
advanced fields theorems

I'm doing some basic differential geometry. So they define an [b]immersion[/b] as an infinitely differentiable function $ x$ from an open $ U$ from $ \mathbb{R}^2 \to \mathbb{E}^3$ (euclidean space) with
\[ x: U \subset \mathbb{R}^2 \to \mathbb{E}^3: (u,v) \mapsto (x_1(u,v),x_2(u,v),x_3(u,v))
\]
such that

$ (\partial x_1/\partial u \ \partial x_2 / \partial u \ \partial x_3 / \partial u)$
$ (\partial x_1/\partial v \ \partial x_2 / \partial v \ \partial x_3 / \partial v)$

has rang $ 2$ for every $ (u,v)$. What does this actually "mean" when drawn. What happens if the rang is $ 1$, do we get some line? One only speaks of directional derivatives...  :?:

I guess it supposeably preserve some differentiable structure of a space (as a homeomorphism preserves "continuity").

It means that the induced map on the tangentspaces $ x_* : \mathbb{R}^2_p\to\mathbb{E}^3_{x(p)}$  is injective.. As you say, that it in some way preserves differentiable structure, by mapping a basis for the tangent space at $ p$ to a basis for the tangent space of $ x(\mathbb{R}^2)$ at $ x(p)$. If the rank was one, one would have the situation that the tangentplane would degenerate to a line.

[quote="Halmos"]What's an example of some $ 2$ dimensional subspace $ x(U)$ of $ E^3$ that has a $ 1$-dimensional tangent space. I can't think of it and I don't have computer algebra software to plot it

I suppose (x,y) -> (xy,xy,xy) would be an example. However, I don't quiet see how the graph looks like.[/quote]

Yes, the image is the line $ \{(t,t,t) | t\in \mathbb{R}\}$. Actually, this map has rank 1 except in the point (0,0,0) where it is zero.
Another example would be the linear projection $ (x,y)\to (x,0,0)$ which has rank 1 everywhere.


[quote="Interval"]What is the area (to the nearest square centimeter) of the largest
circle that can be cut from a square piece of sheet metal 73cm.
on each side?  Explain how you determined this.[/quote]
[hide]The diameter is 73cm and then the radius is 36.5. $\pi r^{2}=\pi(36.5)^{2}\approx4185.3868...\approx4185\text{cm}^{2}$[/hide]

[quote="Interval"]What wording in the question lead to know a diameter is part of the question?

What lead you to use A = pi(r^2)?[/quote]
Think about it. You asked for the largest circle in the square with side length 73 cm. That means the circle would be touching all sides of the square inside. So the diameter would be 73.

[quote="Interval"]What lead you to A=pi(r^2)?[/quote]
That is the area of the circle is it not? You asked for the area of the circle right?

you were asked for the area of a circle, so you use the formula A = pi(r^2). You don't know the radius, but if you knew the diameter you could find the radius...
find diameter as described above.

working backwards is sometimes useful

[quote="Interval"]So, the circle lies INSIDE the square, right?

If so, then the diameter is 73cm.[/quote]
Of course it is inside. How could it be outside? The circle was cut from the sheet of paper meaning it has to be smaller than the square. So if the circle was outside the square, that wouldn't make any sense. So yes it is inside.


If $n$ is a positive integer, prove that $\frac{2(3n)!}{n!(n+1)!(n+2)!}$ is also a positive integer

In fact, the denominator of that expression grows quite a bit faster than the numerator, giving us values between 0 and 1; you mean to ask us to show that the reciprocal of that number is always a positive integer.

Methinks induction would make short work of this, but I'm leaving my house for something I was supposed to be at 15 minutes ago.


A point on the rim of a wheel has a linear speed of 18 cm/s.  If the radius of the wheel is 20 cm, what is the angular speed of the wheel in radians per second?

Use the formula $ v\equal{}\omega r$, where $ v$ is linear speed, $ \omega$ is angular speed, and $ r$ is radius.  So we have $ 18\mathrm{{cm}/{s}}\equal{}\omega \cdot 20\mathrm{cm}\Rightarrow \omega\equal{}.9 \mathrm{s}^{\minus{}1}$.  Technically, angular speed is dimensionless so we don't really need the radians part.


This is my last problem for my presentation.  I apologize for posting here, but no one has tried in the Getting Started section:

A V-gauge is used to find the diameters of pipes. The advantage of such a device is that it is accurate and rugged, and has no moving parts to break down. The measure of angle AVB is 54°. A pipe is placed in the V-shaped slot and the distance VP is used to predict the diameter. The line VP is calibrated by listing the corresponding diameters as its units. This, in effect, establishes a function between VP and d.

a. Suppose that the diameter of a pipe is 2 cm. What is the distance VP?
b. Suppose that the distance VP is 3.93 cm. What is the diameter of the pipe?
c. Find the formula for d in terms of VP.
d. Find a formula for VP in terms of d.

Please see attachment.  P in the attachment, is shaded, and is between B and V.

Please show all detailed work. Thank you.

The attachment did not work, so posting here.

Thank you.

[quote="Bictor717"]The reason no one is responding is because you will copy their work and pass it off as your own.
How about showing some of what you have done with the problem? I'm sure people will be more helpful if they know they're not doing your homework for you.[/quote]

I already explained that.  In all presentations I state all references, APA Style.

Nevertheless, I appreciate the time the board has given.

I have the following work done, but I do not have a way to verify it.  May I get someone's opinion on my work?  I do not know how to convert into the LaTeX.

We know that the radius is CP and the diameter d=2CP. Moreover, we know that is a right triangle with.  In this triangle, we have

tan PVC = CP/VP

Since AVB = 54 degrees, we have    PVC = ½ AVB = 27 degrees.

So, by (1), we get

              tan 27 degrees = CP/VP

             CP = VP * tan 27 degrees

a.	Suppose that the diameter of a pipe is 2 cm.  What is the distance VP?

We have

                1cm = VP * tan 27 degrees

So,

VP = 1 / tan 27 degrees = 1/0.5095 = 1.96(cm)

b. Suppose that the distance VP is 3.93 cm. What is the diameter of the pipe?

We have

CP = VP * tan 27 degrees

So,

CP = 3.93 * tan 27 degrees = 2.00cm

So, the diameter of the pipe is

2CP = 4.00cm

c. Find the formula for d in terms of VP.

We have

CP = VP * tan 27 degrees

So,

d/2 = VP * tan 27 degrees

So,

d = 2 * VP * tan 27 degrees = 2 tan 27 degrees * VP = 1.02VP

d. Find a formula for VP in terms of d.

We have

CP = VP * tan 27 degrees

So,

d/2 = VP * tan 27 degrees

So,

VP = d/2 tan 27 degrees = 0.981d

It all looks good to me except that in part d it should be $VP=\frac{\frac{d}{2}}{\tan27^\circ}=\frac{d}{2\tan27^\circ}$

I think that's what he did.
[url=http://www.aw-bc.com/scp/beecher_penna_bittinger/downloads/ch05pgs411-506.pdf]Source: go down to page 23 of 96[/url]

Oh, I just got confused without the $LaTeX$...

I thought he meant $(\frac{d}{2})\tan 27$...

Gentlemen,

Thank you for your good time.  It is greatly appreciated.  Thank you for verifying my work.

Best


[quote="zanttrang"]My math teacher recently introduced me to a really neat app on the Mac called the LaTeX equation editor: http://evolve.lse.ac.uk/software/EquationEditor/
It allows small snippets of compiled LaTeX code to be exported directly as TIFF images, saving the trouble of compiling a document as a DVI or PDF, resizing it, and resaving it as an image.  Does anybody know of a similar app for Windows?  I currently use MiKTeX and TeXnicCenter (the tools recommended by the AOPS website), and they're great for compiling PDF documents, but I would greatly value the additional functionality of an app like the equation editor.[/quote]

Well, a lot less powerful than the descrition of the Equation Editor, you have the [url=http://www.dessci.com/en/products/TeXaide/]TeXaide[/url] for windows.

May be you can find a group that is disposed to program one more powerful as a Freeware project  :D .

Best,


geometry
geometric transformation
function
real analysis
real analysis solved

find all real p, such that x^p is uniformly continuous on (0,\infty).
---------
here's my solution.

p=1 is trivial, x^p is uniformly continuous.

now assume that 0<p<1.

for given  \varepsilon \in (0,1), let \delta = \varepsilon^{1/p}.
then we can prove |(x+\delta)^p - x^p| \leq \varepsilon.
let g(x) = (x+\delta)^p - x^p, thus
g'(x) = p(x+\delta)^{p-1} - px^{p-1} < 0
so g(x) decreases on (0,\infty), hence |(x+\delta)^p - x^p| \leq \delta^p = \varepsilon.

hence, x^p is uniformly continuous on (0,\infty) when 0<p<1.

a similar discussion tells us that x^p is not uniformly continuous on (0,\infty) when p>1.

--------

well, i'd like to know if there's a solution without using derivatives.

[quote="liyi"]... consistent continuous... [/quote]
What is the definition of consistent continuous ?

Sorry for my bad translation, i don't know how to translate it really.

If for any given  \varepsilon >0, there exists \delta(\varepsilon)>0, such that
|x'-x''| < \delta  ===> |f(x')-f(x'')| < \varepsilon
where x', x'' are in an interval X.

Then we say f is uniformly continuous on the interval X.

for example, if f satisfies Lipschitz condition then it is uniformly continuous.

[quote="liyi"]

If for any given  \varepsilon >0, there exists \delta(\varepsilon)>0, such that
|x'-x''| < \delta  ===> |f(x')-f(x'')| < \varepsilon
where x', x'' are in an interval X.

Then we say f is consistent continuous on the interval X.

for example, if f satisfies Lipchitz then it is consistent continuous.[/quote]
I see in french we say"uniformment continu" maybe in english
uniformly continuous.I don't know yet how to translate in romanian

Prove that f(x)=sin(x) is uniformly continuous
Is g(x) = sin(x^2) uniformly cotinuous ? Justify your answer.

it's the same in Romanian: uniform continuity.

Yes. the correct saying is 'uniformly continuous'.
I've modified the posts.

[quote="Moubinool"]
Prove that f(x)=sin(x) is uniformly continuous
Is g(x) = sin(x^2) uniformly cotinuous ? Justify your answer.[/quote]

It is well-known that sin x is uniformly continuous.
let \delta = \varepsilon.
|sin x - sin y| = |2 cos(x+y)/2 sin (x-y)/2| \leq 2|sin(x-y)/2| \leq 2 * (x-y)/2 = |x-y| \leq \delta = \varepsilon

and we can also use mean-value theorem to prove
|sin x - sin y| \leq |x-y|.
in fact, |sin x - sin y| = |cos(\ksi)| |x-y| \leq |x-y|.

But sin(x^2) is not uniformly continuous.
Let  \varepsilon =1 and \delta > 0. We are looking for |x-y|<\delta but |f(x)-f(y)|\geq 1=\varepsilon. There exists integer n(big enough) such that
|\sqrt(2n \pi+\1/2) - \sqrt(2n \pi-1/2)| = 1/(\sqrt(2n\pi+\1/2) + \sqrt(2n \pi-1/2)) < \delta. Now let x=\sqrt(2n \pi+\1/2) and y=\sqrt(2n \pi-\1/2) thus |x-y| < \delta but
|sin(x^2) - sin(y^2)| = |sin (2n \pi+1/2) - sin(2n \pi-1/2)| = 2 > 1=\varepsilon.

Good solution liyi, another one.

Let f:R--->R+\(0)=]0;+oo[ uniformly continuous.
Study the convergence of  the serie  \sum_{n\in N} 1/|f(n)|

I think it's divergent.

(a) If  \sum 1/f(n) is convergent then a_n=f(n+1)-f(n) can't have an upper bound.
(b) If f is uniformly continuous then a_n must have an upper bound.

We combine these 2 and we're done.

Proof for (a): Assume a_n is bounded, then f(n+1)<=M+f(n), so  \sum 1/f(n) >=  \sum 1/(f(1)+nM) = 00 because  \sum 1/n = 00. If we assume
 \sum 1/f(n) to be finite then a_n can't have an upper bound.

Proof for (b): For a certain d>0 there is an  \varepsilon>0 s.t. for all |x-y|<  \varepsilon we have |f(x)-f(y)|<d. Choose a natural k s.t. 1/k<= \varepsilon . This means f(n+1/k)-f(n)< \varepsilon ; f(n+2/k)-f(n+1/k)< \varepsilon ; ..; f(n+1)-f(n+(k-1)/k)< \varepsilon . We add all of these up to get f(n+1)-f(n)<k* \varepsilon , so a_n=f(n+1)-f(n) has an upper bound.

Yes, nice solution Grobber \sum 1/|f(n)| is divergent,
you can also use the
Lemma:  If  f:R--->R+\(0)=]0;+oo[
 uniformly continuous, then there exist A>0,B>0 such that for any x\in R, |f(x)| \leq A+B|x|.

Another one
f is continuous f:[0;1]->R prove that
for any d>0, there exist k>0, for any x,y \in [0;1],
 |f(x)-f(y)| \leq d+k(x-y)^2

Are you trying to say f is uniformly continuous? (instead of just "continuous", as you have written).

[quote="Moubinool"]
f is continuous f:[0;1]->R prove that
for any d>0, there exist k>0, for any x,y \in [0;1],
 |f(x)-f(y)| \leq d+k(x-y)^2[/quote]

f is only continuous on the compact [0,1], then f is uniformly continous (theorem of Heine)

There exist d>0 s.t. for any u,v \in [0,1] |u-v| \leq d =>|f(u)-f(v)| \leq e

Let k =  (2||f|| + 1)/d^2 > 0 where ||f|| = max|f|

for x,y \in [0,1]

First case: |x-y| \leq d then |f(x)-f(y)| \leq e \leq e+k(x-y)^2

Second case: |x-y| \geq d then

|f(x) - f(y)|  \leq 2.||f||  \leq 2.||f|| + 1  =k.d^2  \leq k(x-y)^2  \leq e + k(x-y)^2

Oh, I missed the fact that f is defined on [0,1]. I am aware of the theorem which states that f:[a,b]->R continuous => f is uniformly continuous, but I'm not aware of any name of this thm.

"A continuous function on a closed interval is uniformly continuous"
This theorem in China, at least in most textbooks in China, is called 'Cantor Theorem'.

but in fact, it is called 'Heine-Cantor theorem'.
"Let X, Y be uniform spaces and f:X->Y is continuous. If X is compact, then f is uniformly continuous."

http://planetmath.org/encyclopedia/HeineCantorTheorem.html

[quote="Moubinool"]Good solution liyi, another one.
Let f:R-R+\(0) uniformly continuous.
[/quote]

what does R-R+\(0) mean?
from [b]R[/b] to [b]R[/b]\{0} ?

In Romania, by R <sub>+</sub> * we denote the set of all positive numbers. I assumed this was what Moubinool meant... That's why I didn't use |f(n)| in my proof, but only f(n) (I assumed the function takes only positive values).

[quote="Moubinool"]
 Let f:R-R+\(0) uniformly continuous....[/quote]
for liyi
oooopps f:R-->R+\(0), f is a function from R to R+\(0)=]0;+oo[

what are the mathematical notations in France?

In China, we denote (0,\infty) by R<sup>+</sup> which does not contain the origin. I guess R+ in France is [0,\infty)?

We have exactly the same notations here in Romania.

in China, most notations are in Russian style, which is a little different from other European countries.

Strange... I always thought Romanian notations copied Russian notations, but now you're saying those are different. What are the Russian notations? (just curious).

Well, i don't know clearly.

In the past, tangent and cotangent are 'tg' and 'ctg', but now, it has been changed into 'tan' and 'cot'. and the natural number didn't contain zero in the past and now it do.

some other notations are now introduced, for example, the complement of a set. The complement of A was denoted by \bar{A} (a short line above A), but now it becomes CU A (A\subset U) and this notation is \complement in ams latex.

We have 'tg' and 'ctg' and I'm really not sure about the complement of a set, but I think both those notations are used in Romania. It looks like we're going off the track here  :D This topic was about something completely different, remember? :D


inequalities
induction
calculus
calculus computations

Prove for all $ n>1$

$ \sqrt{1\plus{}\sqrt{n\plus{}\sqrt{n^2\plus{}...\plus{}\sqrt{n^n}}}} < n$

Sorry if I have post a mistake solution!

With $ n \equal{} 2$ then it is OK!
With $ n > 2$ we have:
$ LHS < \sqrt {n \plus{} \sqrt {n^2 \plus{} \sqrt {n^{2^2}... \plus{} \sqrt {n^{2^n}}}}} < \sqrt {n}\sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt {1 \plus{} ... \plus{} \sqrt {1}}}}$
It is very well-known to check that: $ \sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt {1 \plus{} ... \plus{} \sqrt {1}}}} < \frac {\sqrt {5} \plus{} 1}{2}$ (Can prove it by induction)
Now this problem is OK!


A data set consists of the 151 digits used to write all the natural numbers from 1 to 80 inclusive. What is the arithmetic mean of the digits in the data set? Express your anwer as a decimal rounded to the nearest tenth.

There are...
18 ones
18 twos
18 threes
18 fours
18 fives
(When is this ever going to stop?)
18 sixes
18 sevens
9 eights
8 nines
8 zeroes

18(1+2+3+4+5+6+7)+9(8)+8(9+0)=18(28)+144=648
$ \frac{648}{151}\approx 4.3$

You have to find out how many of each number is used. There is a easy to figure out how many of each number though. We know that there will be the same number 1,2,3,4,5,6 and7 for obvious reasons. Then counting 8,9,0 are easy. Then you can use some tricks to add them up like, 18(1+2+3+4+6+7)=18(7(8)/2)


trigonometry
geometry
perpendicular bisector
geometry unsolved

Let be $ABCD$ a convex quadrilateral such that $\angle ABD=42\textdegree$, $\angle DBC=30\textdegree$, $\angle BCA=72\textdegree$, $\angle ACD=12\textdegree$. Find $\angle DAC$.

I know the answer, but I don't know how to prove.

The triangle $\triangle ABC$ is isosceles witth $\angle BCA = 72^\circ$ and $\angle ABC = \angle ABD + \angle DBA = 42^\circ + 30^\circ = 72^\circ$, hence, AB = AC = a and $BC = a \cos 72^\circ$. Let E be the vertex of an equilateral triangle $\triangle ABE$ on the same side of the line AB as the points C, D, AB = AE = BE = a. We will show that the triangle $\triangle CDE$ is isosceles with EC = ED. The triangle $\triangle ACE$ is isosceles with AC = AE and $\angle CAE = \angle BAE - \angle BAC = 60^\circ - 36^\circ = 24^\circ$, hence, $CE = 2a \sin 12^\circ$. Using the sine theorem for the triangle $\triangle BCD$ with the angles $\angle DBC = 30^\circ$ and $\angle CDB = 180^\circ - (\angle DBC + \angle BCA + \angle ACD) = 66^\circ$,

$\frac{CD}{BC} = \frac{\sin 30^\circ}{\sin 66^\circ} = \frac{1}{2 \cos 24^\circ}$

$CD = \frac{BC}{2 \cos 24^\circ} = a\ \frac{\cos 72^\circ}{\cos 24^\circ}$

Imagine a circle (A) with the center A and radius a = AB = AC = AE. The angle $\angle DCE$ of the triangle $\triangle CDE$ is equal to

$\angle DCE = \angle BCE - \angle (BAC + \angle ACD) =$

$= 180^\circ - \frac{\angle BAE}{2} - (\angle BAC + \angle ACD) = 180^\circ - 30^\circ - (72^\circ + 12^\circ) = 66^\circ$

Using the cosine theorem for the triangle $\triangle CDE$,

$DE^2 = CE^2 + CD^2 - 2CE \cdot CD \cos \widehat {DCE} =$

$= a^2 \left(4 \sin^2 12^\circ + \frac{\cos^2 72^\circ}{\cos^2 24^\circ} - 4 \sin 12^\circ\ \frac{\cos 72^\circ}{\cos 24^\circ}\ \cos 66^\circ \right)$

We will show that

$(?)\ \ \frac{\cos^2 72^\circ}{\cos^2 24^\circ} - 4 \sin 12^\circ\ \frac{\cos 72^\circ}{\cos 24^\circ}\ \cos 66^\circ = 0$

$(?)\ \ \cos 72^\circ - 4 \sin 12^\circ \cos 24^\circ \sin 24^\circ = \cos 72^\circ - 2 \sin 12^\circ \sin 48^\circ = 0$

$(?)\ \ 2 \cos^2 36^\circ - 1 - (\cos 36^\circ - \cos 60^\circ) = 2 \cos^2 36^\circ - \cos 36^\circ - \frac 1 2 = 0$

$(?)\ \ \cos 36^\circ = \frac{1 + \sqrt 5}{4} > 0$

This is exactly the value of $\cos 36^\circ$. As a result, CD = DE and $\angle DCE = \angle CDE = 66^\circ$. But then $\angle BDE = \angle BDC + \angle CDE = 66^\circ + 66^\circ = 132^\circ$. Let the line ED meet the line AB at a point F. The angle $\angle BDF = 180^\circ - \angle BDE = 48^\circ$ and the angle $\angle BFD = 180^\circ - (\angle ABD + \angle BDF) = 180^\circ - (42^\circ + 48^\circ) = 90^\circ$. Consequently, the line $ED \equiv EF \perp AB$ passing through the vertex E of the equilateral triangle $\triangle ABE$ is the perpendicular bisector of the segment AB. As a result, the triangle $\triangle ABD$ is isosceles with AD = BD and $\angle DAB = \angle DBA = 42^\circ$, which implies that the angle $\angle DAC = \angle DAB - \angle CAB = 42^\circ - 36^\circ = 6^\circ$.


calculus
integration
function
graph theory
calculus computations

If you are talking about this: $\int 0 dx $ then it's not a number it's a set of functions, the constant functions (you can consider 0 as being constant, you obtain the same result). If you are talking about $\int_a^b 0 dx$ this is obviously 0. I fail to see how this is related to hamiltonian graphs  :?

[quote="mtbmars"]OK, I'm having one of those moments where I'm split betwwen what seems like two equally valid assumptions.  When taking the integral of zero, is zero a constant?  Or does some trick work out that the integral is zero.  the second makes the hamiltonian I'm trying to work out make sense, but it seems fishy.

Any help?  I'm doubting myself and don't trust my intuition.

TKS,
BoB[/quote]
Sorry to say this but I'm completely 100% confused from what you just said.

"Where I'm split betwwen (between) what seems like two equally valid assumptions":What assumptions are you talking about?

"When taking the integral of zero, is zero a constant?": Zero is a scalar constant, a number. Indefinite integral of zero is a constant function. This is the simplest integration that you can find in calculus. Review your basics, please.

"Or does some trick work out that the integral is zero (?)": That the value of the integral is zero? "Integral of zero" and "integral is zero" are two different things.

"the second makes the hamiltonian I'm trying to work out make sense, but it seems fishy": OK this sentence really blew off my mind. If you need help with "integral of zero" or "integral is zero", "hamiltonian" is a more difficult and technical term than "integral". We know you are trying to work out and trying to make sense out of it but you will need to tell us more. What "hamiltonian" are you referring to? What subject? Physics? Graph Theory? Differential Equations? Sorry but I don't see any "hamiltonian" in real analysis. And yes, your discussion is "fishy".

"I'm doubting myself and don't trust my intuition": You bet on it. I see that your confusion is overwhelming.

I'm not particularly nit-picking on you. Your exposition needs to be clearer so that other readers can help you. No one can read your mind unless you can express your opinions/ideas/questions clearly and simply. And I apologize if you felt uneasy by this post or that I have wasted your time.

Peace, 3X.lich

Sounds like you're doing a variational calculus problem.  Can you post the actual problem?

Never mind.

I'll have someone at school straighten me out.

Think of it like this $\displaystyle \int a dx = ax + C$.  So, if $a=0$, then $ax=0x + C=C$.  $C$ is just the integral constant.

[quote="Valentin Vornicu"]I fail to see how this is related to hamiltonian graphs  :?[/quote]

A Hamiltonian means something different than a Hamiltonian graph.  It's a term from physics -- I don't remember what its equivalent in mathematics is.


Suppose there are two charges +q1 and -q2   placed at A and B resply separated by a distance a.
A line of force that emanates from the +ve charge is inclined at an angle alpha to the line AB, then obtain an expression for the angle which the line of force makes with AB even as it enters the second charge.

why did u name the problem :?: ...Problems are named only from IIT JEE thread
look at my solution [url=http://www.mathlinks.ro/viewtopic.php?t=136051]here[/url]


[hide]$\frac{1}{2}$?Since there are 4 different combinations, and you need 2 heads or 2 tails, i think the answer would be 1/2[/hide]

[quote="G-UNIT"]John flips a fair coin 3 times. Find the probability that exactly 2 of the flips come up the same.[/quote]

[hide=" Answer"]I get 3/4[/hide]

[hide]i see now, you're right the answer is 3/4. Since there are 8 combinations and TTT and HHH don't work the probability would be 6/8 or [b]3/4[/b][/hide]

[quote="MCrawford"]How might we work this problem by looking for the probably that all three flips are the same?[/quote]
[hide]1/4  We are looking for TTT or HHH and there are eight combinations so that means the probability is 2/8 or 1/4.[/hide]

[quote="MCrawford"]Nice work.  This is an example of complementary probability, a topic DPatrick will be covering thoroughly in his Introduction to Counting & Probability book that AoPS will publish in a few months.[/quote]
cool, i might get that book

[hide="wut i got..."]i got $\frac{3}{4}$[/hide]

[quote="G-UNIT"]John flips a fair coin 3 times. Find the probability that exactly 2 of the flips come up the same.[/quote]

Let us say the probability of heads = h, and the probability of tails = t then

$h + t = 1$

The three flips are independent of each other so that

$(h + t)^3 = 1^3 = 1$

The binomial expansion of the left hand side is

$(h + t)^3= h^3 +3h^2t +3ht^2 +t^3$

The terms in this expression are the probabilities that one gets 3 heads, 2 heads and 1 tail, 2 tails and 1 head, and 3 tails respectively from left to right.

The probability that exactly two flips come up the same is:

$P(2 \text{heads \& } 1\text{ tail} \cup 2 \text{ tails \& } 1\text{ head})$

$P(2 \text{heads \& } 1\text{ tail} \cup 2 \text{ tails \& } 1\text{ head}) = P(2 \text{heads \& } 1\text{ tail}) +  P(2 \text{ tails \& } 1\text{ head}) $

We know from the above that

$P(2 \text{heads \& } 1\text{ tail}) = 3h^2t = 3(\frac{1}{2})^2\frac{1}{2}$

and likewise

$P(2 \text{ tails \& } 1\text{ head})= 3ht^2 = 3(\frac{1}{2})(\frac{1}{2})^2$

$\therefore  P(2 \text{heads \& } 1\text{ tail} \cup 2 \text{ tails \& } 1\text{ head}) = \frac{3}{8} +\frac{3}{8} = \frac{3}{4}$


Prove that there exist infinitely many prime
numbers p that divide $u_{p-1}$.
this is an IMO longlisted problem

the p-1 term of the Fibonacci sequence, where $u_{0}=0 u_{1}=1$


In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If $AG=2$, $GF=13$, $FC=1$, and $HJ=7$, then $DE$ equals

[asy]
size(200);
defaultpen(fontsize(10));
real r=sqrt(22);
pair B=origin, A=16*dir(60), C=(16,0), D=(10-r,0), E=(10+r,0), F=C+1*dir(120), G=C+14*dir(120), H=13*dir(60), J=6*dir(60), O=circumcenter(G,H,J);
dot(A^^B^^C^^D^^E^^F^^G^^H^^J);
draw(Circle(O, abs(O-D))^^A--B--C--cycle, linewidth(0.7));
label("$A$", A, N);
label("$B$", B, dir(210));
label("$C$", C, dir(330));
label("$D$", D, SW);
label("$E$", E, SE);
label("$F$", F, dir(170));
label("$G$", G, dir(250));
label("$H$", H, SE);
label("$J$", J, dir(0));
label("2", A--G, dir(30));
label("13", F--G, dir(180+30));
label("1", F--C, dir(30));
label("7", H--J, dir(-30));[/asy]

$\textbf {(A) } 2\sqrt{22} \qquad \textbf {(B) } 7\sqrt{3} \qquad \textbf {(C) } 9 \qquad \textbf {(D) } 10 \qquad \textbf {(E) } 13$

[hide="Hint"]You have secants going through the circle. What property can be applied?  :) [/hide]

[hide="Giving it away......"]Power of Point :P [/hide]

[quote="Hokkage"][hide="Lazy but working way"]Using power of a point, $\overline{AH} = \frac{26}{7}$ and by adding the whole side to achieve 16 (equilateral triangle) $\overline{BJ} = \frac{37}{7}$. Then set up the other power of a point equations and the sum to get $cd=37$, $de=13$, and $c+d+e=16$. At this point I got lazy and, because it's multiple choice, plugged in the given solutions (started with the integer ones) and came out with 10 as a working answer. So D) 10.[/hide][/quote]Not quite.

A few letters off... :P

Whoops. Adding together actually gives 15, not 16.  :oops: (PS. I can add... I think)

[hide="Your error"]
Well, $AG.GF\neq AH.HJ$
[/hide]

[hide]Power of a Point tells us that AH=3 and we know that JB=6 because the triangle is equilateral.  Let a=BD, b=CE, and c=DE.  We have a system of equations.

$a(a+c)=78$
$b(b+c)=14$
$a+b+c=16$

I cant find a way to solve this quickly...[/hide]

[hide="Solution"]Answer: $\text{(A)}$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=497&sid=9716ef06b2cbcc8447c551a328d9e117[/img]

First, we apply Power of a Point to find $AH$: $(AH)(AH+7)=(2)(15)$, and $AH=3$. Now, we also know the length of $BJ$ because the triangle is equilateral. $BJ=16-7-3=6$.

We can now simultaneously solve for $BD$, $DE$, and $EC$ with Power of a Point. Let $BD$, $DE$, and $EC$ be $x$, $y$, and $z$, respectively. We get three equations.

$x+y+z=16$
$x(x+y)=6(13)$
$z(y+z)=14$.

Solving by substitution gives $y=DE=2\sqrt{22}$. [/hide]

[quote="matt276eagles"]Can you show how you solved that system?  I couldnt find a quick way. :blush:[/quote]I didn't have a quick way either.  :blush:

[hide="Ok, here is a non-quick solution of the system"]
[quote="4everwise"]
$x+y+z=16$
$x(x+y)=6(13)$
$z(y+z)=14$.

Solving by substitution gives $y=DE=2\sqrt{22}$. [/quote]

From $x+y+z=16$ we substitute to the other equations

$x(16-z) = 78$
$z(16-x) = 14$

$16x-xz = 78$
$16z-xz = 14$

substracting we get $x=z+4 \Rightarrow x+z=2z+4$

We express all the variables in terms of $y:$

$2z+4 = 16-y \Rightarrow 2z = 12-y \Rightarrow$

$z=6-\frac{y}{2}$ and $x=z+4 = 10-\frac{y}{2}$

we substitute again to the second (initial) equation

$x(x+y)=78 \Rightarrow (10-\frac{y}{2})(10+\frac{y}{2})=78$

This equation have the solution $(\frac{y}{2})^2 = 22 \Rightarrow y = +2\sqrt{22}$

[hide="(of course, if we substitute to the third initial equation we get the same thing)"]

$z(y+z)=14 \Rightarrow (6-\frac{y}{2})(6+\frac{y}{2})=14$

which have the same solution $y = 2\sqrt{22}$[/hide][/hide]

Thanks... That's what I started to do, but then I got lazy and assumed there was a faster way. :lol:

[quote="Math scientist"]what  is Power of a point?
By the way,I am a newby.[/quote]

One of the most important geometric theorems. Here's some info: http://en.wikipedia.org/wiki/Power_of_a_point

Oh my God,I didn´t learned that at school.


What is the units digit when x^16 is expressed in base 16, if x is 21 base 10?

I'm pretty sure that it doesn't work that way...

[hide="my solution"]$ \gcd (21,16)\equal{}1$, so $ 21^{\phi(16)}\equiv 1\bmod{16}$, or $ 21^8\equiv 1\bmod{16}$. Therefore $ 21^{16}\equiv 1\bmod{16}$. The units digit of a base 16 number is the remainder when that particular number is divided by 16, so the units digit of the base 16 representation of $ 21^{16}$ is 1.[/hide]

Uh, what? You mean you don't think $ a^n\bmod m\equiv (a')^n\bmod m$ where $ a\equiv a'\bmod m$?

[hide="If so..."]
$ a\equiv a'\bmod m\implies a\equal{}mk\plus{}a'$

$ a^n\equal{}(mk\plus{}a')^n\equal{}\sum_{i\equal{}0}^{n}\binom{n}{i}(mk)^{n\minus{}i}(a')^{i}$

In the RHS, terms where $ n\minus{}i>1$ (equivalently, $ i<n\minus{}1$) are divisible by $ m$, so the only term remaining if we take this mod $ m$ would be the term where $ i\geq n\minus{}1\implies i\equal{}n$:

$ a^n\equiv \binom{n}{n}(mk)^{n\minus{}n}(a')^{n}\equal{}(a')^n\bmod c$ $ \Box$
[/hide]

Euler's theorem on this one is overkill.

But we're still working in base 10. Otherwise we'd have to work in $ \mod 10_{16}$. If we're using $ 15_{16}$ we can't be working in $ \mod 16_{10}$.

[quote="PowerOfPi"][quote="MathWise"][hide="@grn_trtle"]I think what 1=2 is trying to say is that you should be trying to find the residue mod 16 of $ 21^{16}$, not $ 15^{16}$. In other words, you are trying to find $ 15_{16}^{16}\pmod{16}$, not $ 15^{16}\pmod{16}$, so the fact that $ 15^{16}\equiv1\pmod{16}$ does not show that the units digit of $ 21^{16}$ is 1 when expressed in base 16.[/hide][/quote]
But then wouldn't $ 16$ (the exponent) be in base 10? Which would be $ 10$ in base 16?[/quote]
When expressing $ x\equal{}21^{16}$ in base 16, the last digit is the residue of $ x$ modulo 16. We can work in base 10 throughout the whole problem, as long as we find $ 21^{16}\pmod{16}$. We don't actually need to convert anything (including the exponent) to base 16 when solving the problem this way, since we are not using any properties of the number's expression in base 10. Hopefully this makes it clearer!


suppose we have a sequence of real numbers $a_n$

if we have that $\left|\frac{a_n}{a_{n+1}}\right| >1$ for large n, is the series convergent? how would i reference this?

The series?

Here's a simple example. Let $a_n=\frac1n.$ Then $\frac{a_n}{a_{n+1}}=\frac{n+1}n>1$ for all $n$ but

$\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\frac1n$ diverges.

Perhaps you mean the sequence? We've got some tricks we could play.

Let $a_n=(-1)^n\left(1+\frac1n\right).$

Then $\left|\frac{a_n}{a_{n+1}}\right|=\frac{n^2+2n+1}{n^2+2n}>1.$

But the sequence $(a_n)$ does not converge; it has subsequences converging to $1$ and to $-1.$

Change that to $\liminf\left|\frac{a_n}{a_{n+1}}\right|>1$ and you have a different situation. Look up the words "ratio test."

woops, i meant [b]sequence[/b]  :blush: , i am not too familiar with the calculus vocabulary yet

and what does $\liminf$ mean?

also, i looked at ratio test, but it seems that is for series, but could i prove convergence with

$a_n>0$ and $\frac{a_n}{a_{n+1}}>1$

[quote="Kent Merryfield"][quote]but could i prove convergence with

$a_n>0$ and $\frac{a_n}{a_{n+1}}>1$[/quote]
Yes. Note that you might as well just write that as $a_{n+1}<a_n.$ In other words, we have a decreasing sequence. The Monotone Sequence Theorem (which is equivalent to the completeness of the real numbers) says that a monotone bounded sequence converges; we have a monotone sequence, and we have that it is bounded, since $0$ is a lower bound.

Note that the limit could be zero, or the limit could be positive; we could produce examples of each.[/quote]

thanks for the help  :) , i am very shaky on my calculus theorems  :(


projective geometry
geometry proposed
geometry

Let $ ABCD$ be a convex quadrilateral with opposite side not parallel. The line through $ A$ parallel to $ BD$ intersect line $ CD$ in $ F$, but parallel through $ D$ to $ AC$ intersect line $ AB$ at $ E$. Denote by $ M,N,P,Q$ midpoints of the segments $ AC,BD,AF,DE$. Prove that lines $ MN,PQ$ and $ AD$ are concurrent.

It is easy to show that NQ,MP both pass through the midpoint L of segment AD.
We make a use of Menelaus theorem and Thales's therem, the problem is solved.

The conclusion follows immediately from Desargues Theorem applied for $ RPQ$ and $ OMN$,where $ {R}\equal{}AF\cap DE$,and $ {O}\equal{}AC\cap BD$.


If P(x + 2) = x^3 + 9x^2 + 26x +23 find P(x - 2) as a polynomial in x.

It is easy to see that the first term of
$ P(x)$ has to be x^3
$ (x\plus{}2)^3\equal{}x^3\plus{}6x^2\plus{}12x\plus{}8$
subtracting from original equation, we get
$ 3x^2\plus{}14x\plus{}15$
Next term shall be then 3x^2
$ 3(x\plus{}2)^2\equal{}3x^2\plus{}12x\plus{}12$
Subtrating, we get: $ 2x\plus{}3$
Next term shall be $ 2x$
in which $ 2(x\plus{}2)\equal{}2x\plus{}4$
subtracting, we get -1

So $ P(x)\equal{}x^3\plus{}3x^2\plus{}2x\minus{}1$
And then put $ x\minus{}2$ in place of $ x$
I obtain: $ P(x\minus{}2)\equal{}x^3\minus{}3x^2\plus{}2x\minus{}1$

I wonder if there is an easier solution

[hide="Much longer solution"]Using synthetic division, we find that

$ P(x \plus{} 2) \equal{} (x \plus{} 2)(x^2 \plus{} 7x \plus{} 12) \minus{} 1$

Using synthetic division again,

$ P(x \plus{} 2) \equal{} (x \plus{} 2)[(x \plus{} 2)(x \plus{} 5) \plus{} 2] \minus{} 1$

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

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

So we have that

$ P(z) \equal{} z^3 \plus{} 3z^2 \plus{} 2z \minus{} 1$

and

$ P(x \minus{} 2) \equal{} (x \minus{} 2)^3 \plus{} 3(x \minus{} 2)^2 \plus{} 2(x \minus{} 2) \minus{} 1$

$ \equal{} x^3 \minus{} 3x^2 \plus{} 2x \minus{} 1$[/hide]