I don't know if there is an easier way, but wouldn't you use just the summation formula for e.

The Taylor series for $ e^{x}$ takes the form $ \sum_{k=0}^{\infty}\frac{x^{k}}{k!}$, where it is the exponent that is incrementing.  Here instead we have a summation of the form $ \sum_{k=1}^{\infty}\frac{k^{3}}{k!}$, where it's the base that's incrementing.

Index shifting...

[hide]Let $ S(a,b)=\sum_{x=a}^{\infty}\frac{x^{b}}{x!}$ to save me some typing.

We want to find S(1,3). Note that we can conveniently cancel an x from the numerator and denominator, to get x^2 in the numerator and (x-1)! in the denominator. Let's shift the index, so that we can write it as S(0,2)+2S(0,1)+S(0,0).

S(0,0) is equal to e. Since 0^2/0! = 0 and 0/0!, these is actually equal to S(1,2) and 2S(1,1), respectively.

So, S(1,3)=S(1,2)+2S(1,1)+e

In S(1,2), we can again cancel an x and perform an index slide to get S(1,1)+S(0,0)=S(1,1)+e

Then, S(1,3)=3S(1,1)+2e

Left to calculate is S(1,1). Again, we can go ahead an cancel an x and it becomes S(0,0)

Hence, S(1,3) = 3e+2e = 5e[/hide]

[b]Extension:[/b] Find a recursive formula for $ \frac{1}{e}\sum_{x=1}^{\infty}\frac{x^{n}}{x!}$. Can we find a closed form of this?

[quote="The Zuton Force"][b]Extension:[/b] Find a recursive formula for $ \frac{1}{e}\sum_{x=1}^{\infty}\frac{x^{n}}{x!}$. Can we find a closed form of this?[/quote]

[hide="Answer from the OEIS"][url]http://www.research.att.com/~njas/sequences/A000110[/url][/hide]
Interestingly enough, the comment that this is the desired sequence was only added to the OEIS recently (February 2007).  There is no proof in the OEIS, but it's not too hard with [hide="generating functions."]
\begin{eqnarray*}E(x) & = & \sum_{k \geq 0}\frac{x^{k}}{k!}\sum_{n \geq 0}\frac{n^{k}}{n!}\\ & = & \sum_{n \geq 0}\frac{1}{n!}\sum_{k \geq 0}\frac{x^{k}n^{k}}{k!}\\ & = & \sum_{n \geq 0}\frac{1}{n!}e^{xn}\\ & = & e^{e^{x}}, \end{eqnarray*}
and that is exactly the generating function for the Bell numbers.[/hide]
A combinatorial proof would be interesting, of course, although the $ \frac{1}{e}$ doesn't bode well for such a thing.


quadratics
number theory unsolved
number theory

Positive integers $k, m, n$ satisfy $mn=k^2+k+3$, prove that at least one of the equations $x^2+11y^2=4m$ and $x^2+11y^2=4n$ has an odd solution.

Shall we prove the following:
When $mn=k^2+k+3$, then at least one of the equations $x^2+11y^2=4m$ or $x^2+11y^2=4n$ has a solution $(x,y)$ with $xy$ being odd¿

Are the equations "$x^2+11y^2=4m$" and "$x^2+11y^2=4n$" independent?If there are simultaneous equations, it shows $m=n$
(Do I ask a silly question? :| )

[quote="ZetaX"]Shall we prove the following:
When $mn=k^2+k+3$, then at least one of the equations $x^2+11y^2=4m$ or $x^2+11y^2=4n$ has a solution $(x,y)$ with $x,y$ being odd¿[/quote]
Yes. I'll edit my problem.

Guys I think that for solve this problem we must looke at the [b]complex nubers [/b]of $a+\sqrt{11}bi$ form wher $a,b\in{Z}$ and $i^2=-1$. :?

This is the most difficult problem in CMO 2006. Only one student got the full mark 21 points in this problem. The total score of this problem recieved by the first prize students is only 21+9+9=39.

Here's a way to prove it:

Lemma 1: Prove all fctors of k^2+k+3 are expressible as form x^2+xy+3y^2. This is true as all binary quadratic forms of discrimant -11 are mutually equivalent.

Lemma 2: Use mod- bash and completion of the square to finish off problem.

Bomb

Lemma 1 p be a prime such that$\exists x\in Z+: x^2+11\vdots p$then $p=m^2+11n^2$or$4p=k^2+11l^2$(kl is odd)
Lemma 2 If m,n are odd and$4m=x_1^2+11y_1^2,4n=x_2^2+11y_2^2$Where$x_1,x_2,y_1,y_2$ are positive odd integer)then exists positive odd integer k,l such that$4mn=k^2+11l^2$
Lemma 3 if p is prime and $p=x^2+11y^2$and positive integer m such that $4mp=z^2+11t^2$(where z,t are odd integer)then there exists 2 odd integers kand l such that $4m=k^2+11l^2$By using 3 lemmas,we can easily prove the problem

One may remember the very beautiful fact tht if $xy=z^2+1$ then there are integers $a,b,c,d$ with $x=a^2+b^2,y=c^2+d^2,z=ac+bd$ (and of course $|ad-bc|=1$), connected close to Lagrange Identity $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$, and also remeber its proof by Fermat' Method of Finite Descent.

Now we try to build a similar result for our problem, trying to use Lagrange Identity $(a^2+11b^2)(c^2+11d^2)=(ac+11bd)^2+11(ab-bd)^2$. After several trials, I have come to the following Lemma which implies the problem as a consequence:

Let $k,m,n$ be positive integers with $k^2+11=4mn$ (of course $k$ is odd). Then there are positive integers $a,b,c,d$, not all even,  with $4m=a^2+11b^2,4n=c^2+11d^2,2k=ac+11bd$ (and $2=|ad-bc|$).

We use Fermat's Method of Finite Descent.

Pick up a pair $(m,n,k)$ with $m+n+k$ minimal that fails our condition.

If $k<2m,k<2n$ then $k^2+11=4mn\geq (k+1)^2$ then $k\leq 3$.
If $k=1$ then the only possibility is $\{m,n\}=\{1,3\}$ and we can take $a,b,c,d$ to be $1,1,2,0$.
If $k=3$ the only possibility is $\{m,n\}=\{1,5\}$ and we can take $a,b,c,d$ be $3,1,2,0$.

If, for example $k>2n$, then $(k-2m)^2+11=4m(n-k+m)$ and this is a triple with smaller sum. so, there are positive integers $a,b,c,d$ with $4m=a^2+11b^2,4(n-k+m)=c^2+11d^2,k-2m=ac+11bd$.

The simple computations show that $4m=a^2+11b^2, 4n=(a+c)^2+(b+d)^2. 2k=a(a+c)+11b(b+d)$ and so the conditions holds also for $m,n,k$, as if $a,b,c,d$ are not all even, the neither are $a,b,a+c,b+d$.

We can now apply out lemma to the initial condition $4mn=4(k^2+k+3)=(2k+1)^2+11$

[quote="iura"]One may remember the very beautiful fact tht if $ xy \equal{} z^2 \plus{} 1$ then there are integers $ a,b,c,d$ with $ x \equal{} a^2 \plus{} b^2,y \equal{} c^2 \plus{} d^2,z \equal{} ac \plus{} bd$ (and of course $ |ad \minus{} bc| \equal{} 1$), connected close to Lagrange Identity $ (a^2 \plus{} b^2)(c^2 \plus{} d^2) \equal{} (ac \plus{} bd)^2 \plus{} (ad \minus{} bc)^2$, and also remeber its proof by Fermat' Method of Finite Descent.

Now we try to build a similar result for our problem, trying to use Lagrange Identity $ (a^2 \plus{} 11b^2)(c^2 \plus{} 11d^2) \equal{} (ac \plus{} 11bd)^2 \plus{} 11(ab \minus{} bd)^2$. After several trials, I have come to the following Lemma which implies the problem as a consequence:

Let $ k,m,n$ be positive integers with $ k^2 \plus{} 11 \equal{} 4mn$ (of course $ k$ is odd). Then there are positive integers $ a,b,c,d$, not all even,  with $ 4m \equal{} a^2 \plus{} 11b^2,4n \equal{} c^2 \plus{} 11d^2,2k \equal{} ac \plus{} 11bd$ (and $ 2 \equal{} |ad \minus{} bc|$).

We use Fermat's Method of Finite Descent.

Pick up a pair $ (m,n,k)$ with $ m \plus{} n \plus{} k$ minimal that fails our condition.

If $ k < 2m,k < 2n$ then $ k^2 \plus{} 11 \equal{} 4mn\geq (k \plus{} 1)^2$ then $ k\leq 3$.
If $ k \equal{} 1$ then the only possibility is $ \{m,n\} \equal{} \{1,3\}$ and we can take $ a,b,c,d$ to be $ 1,1,2,0$.
If $ k \equal{} 3$ the only possibility is $ \{m,n\} \equal{} \{1,5\}$ and we can take $ a,b,c,d$ be $ 3,1,2,0$.

If, for example $ k > 2n$, then $ (k \minus{} 2m)^2 \plus{} 11 \equal{} 4m(n \minus{} k \plus{} m)$ and this is a triple with smaller sum. so, there are positive integers $ a,b,c,d$ with $ 4m \equal{} a^2 \plus{} 11b^2,4(n \minus{} k \plus{} m) \equal{} c^2 \plus{} 11d^2,k \minus{} 2m \equal{} ac \plus{} 11bd$.

The simple computations show that $ 4m \equal{} a^2 \plus{} 11b^2, 4n \equal{} (a \plus{} c)^2 \plus{} (b \plus{} d)^2. 2k \equal{} a(a \plus{} c) \plus{} 11b(b \plus{} d)$ and so the conditions holds also for $ m,n,k$, as if $ a,b,c,d$ are not all even, the neither are $ a,b,a \plus{} c,b \plus{} d$.

We can now apply out lemma to the initial condition $ 4mn \equal{} 4(k^2 \plus{} k \plus{} 3) \equal{} (2k \plus{} 1)^2 \plus{} 11$[/quote]

[i]  You are really intelligent , Iura . Very nice , solution . :lol: [/i]

 [i]But i think , there is a small mistake in your solution

Note that : $ a ; b ; c ; d$ are just  normal integer , they can be negative

  For example $ (m;k;n) \equal{} ( 3 ; 5 ; 3 )$ satisfies the equation $ k^2 \plus{} 11 \equal{} 4mn$

 But you can't find nonnegative integer $ a ; b ; c ; d$ such that

 $ 4m \equal{} a^2 \plus{} 11b^2 ; 4n \equal{} c^2 \plus{} 11d^2 ; 2k \equal{} ac \plus{} 11bd$ :) [/i] $ a ; b ;c ;d$ not all even

  [i]Otherwise , Your solution is very excellent . :lol: [/i]

Is it technically correct if I show that for every integer of the form $k^2+k+3$ we can find $(m,n)$ such that $mn=k^2+k+3$?

Here is a less clever solution.

Call a prime [b] tangfastic [/b] if it is congruent to one of $1, 3, 4, 5, 9$ modulo $11.$ Call a positive integer [b] tasty [/b] if it is expressible as $a^2 + 11b^2$ for $a, b \in \mathbb{Z}.$ In other words, a prime $p$ is tangfastic if and only if it is not $11$ and it is a square modulo  We will start by proving the following useful lemma.

[b] Lemma. [/b] If $p$ is tangfastic, then $4p$ is tasty.

[b] Proof. [/b] If $p = 3, 5$, the lemma is easily checked. Otherwise, suppose that $p \ge 23.$ Note that since $p$ is tangfastic, it's easily obtained by the Quadratic Reciprocity Theorem that $\left( \frac{-11}{p} \right) = 1.$ Therefore, Thue's Lemma implies that there exist integers $x, y \in [0, \sqrt{p}]$ so that $p|x^2 + 11y^2.$ Since $x^2 + 11y^2$ is a square modulo $p$, it must be one of $p, 3p, 4p, 5p, 9p$. If $x^2 + 11y^2 = p$ or $4p$, then we are clearly done. Otherwise, suppose that $x^2 + 11y^2 \in \{3p, 5p, 9p\}.$

First we'll show that for an integer $k$ if $3k$ can be expressed as $a^2 + 11b^2$ for integers $a$ and $b$ not divisible by $3$, then $4k$ is tasty. Indeed, observe that $36k = (a\pm11b)^2 + 11(a \mp b)^2.$ Now just pick the choice of signs so that $\frac{a \pm 11b}{3} \in \mathbb{Z}$, and we're done.

Now, observe that if $x^2 + 11y^2=3p$, then since $9 \nmid 3p$, we have $3 \nmid x, y$ and so $4p$ is tasty and we're done. If $x^2 + 11y^2 = 9p$, then if $3 | x, y$, we have $4p = \left(\frac{2x}{3}\right)^2 + 11 \left(\frac{2y}{3}\right)^2$ is tasty and we're done. Otherwise, we have that $3 \nmid x, y$, which yields that $12p$ is tasty. As $9 \nmid 12p$, we in turn have that $16p$ is tasty, so let $16p = a^2 + 11b^2$ for $a, b \in \mathbb{Z}$. By modulo $8$, it's clear that $2 |a, b$, and so hence $4p = \left(\frac{a}{2}\right)^2 + 11 \left(\frac{b}{2}\right)^2$ is tasty.

We can similarly show that for an integer $k$ so that $5k = a^2 + 11b^2$ with $5 \nmid a, b$, $4k$ is tasty. Indeed, just consider $(3a \pm 11y, a \mp 3y)$ and divide both entries of one of them by $5.$

Hence, if $x^2 + 11y^2 = 5p$, then $25 \nmid 5p$ implies that $5 \nmid x, y$. Therefore, $4p$ is tasty.

As we've exhausted all cases, the lemma is proven.

$\blacksquare$

Now, let us return to the problem. The condition of the problem implies that $4mn = (2k+1)^2 + 11$. By modulo $8$, we have that $m, n$ are both odd. Now, if $m = 1$, then $x^2+11y^2 = 4n$ has the desired odd solution (simply $(2k+1, 1)$). Otherwise, suppose that $m >1$, and analogously $n>1.$ Suppose that there was a prime divisor $p|m$ which was tangfastic but not tasty. Then, let's write $4p = a^2 + 11b^2.$ Since $p$ is not tasty, we can't have $2|a,b$, and so therefore $a, b$ are both odd. For $k \in \mathbb{Z}$, let $4k$ be called [b] Muriatic [/b] if it can be written in the form $a^2 + 11b^2$, where $a, b$ are both odd. In particular, $4p$ is Muriatic.

[b] Observation. [/b] If $t$ is an odd positive integer so that $4t = a^2+11b^2$ for odd $a, b$, then for any tangfastic odd prime $q$, $4tq$ is Muriatic.

[b] Proof. [/b] Then $4q = c^2 + 11d^2$ for some $c, d \in \mathbb{Z}$, by the lemma. We consider two cases.

[b] Case 1. [/b] $c, d$ are both even.

Then, we've that $q = c'^2 + 11d'^2$, where $c', d' \in \mathbb{Z}$ are just $c, d$ divided by two. Then, observe that exactly one of $c', d'$ is odd, and so therefore $ac' + 11bd', ad'-bc'$ are both odd. We are done since $4tq = (ac' + 11bd')^2 + 11(ad'-bc')^2$.

[b] Case 2. [/b] $c, d$ are both odd.

Then, we have that $16tq = (ac\pm 11bd)^2 + 11(ad \mp bc)^2.$ Since $a, b, c, d$ are all odd, it's clear that we can pick the signs so that $ac \pm 11bd, ad \mp bc$ are both congruent to $2$ mod $4$. From here, we can just divide two to imply the observation.

$\blacksquare$

In view of the observation, let $m = p_0p_1 \cdots p_t$, where $p_0 = p$, and $p_0, p_1, \cdots, p_t$ are (not necessarily distinct) primes. From the Observation, it can easily be shown by induction that $4p_0 p_1 p_2 \cdots p_i$ is Muriatic, for each $0 \le i \le t$. This means that $x^2 + 11y^2 = 4m$ has odd solutions, as desired. Observe that the previous argument works as long as there is a $p|mn$ so that $4p$ is Muriatic.

Now, we will analyze the remaining case (or lack thereof). Namely, we will analyze the scenario when for each prime $p|mn$, $4p$ is not Muriatic, but is expressible as $a^2 + 11b^2$ where $2|a, b.$ The following claim will help to solve the problem.

[b] Claim. [/b] If $p, q$ are positive integers so that $p$ is an odd tangfastic prime, $4p$ is not Muriatic, and $4pq$ is Muriatic, then $4q$ is also Muriatic.

[b] Proof. [/b] Suppose that $4pq = a^2 + 11b^2$ for odd positive integers $a, b.$ If $p = 11$, then $11 | a$, and so $4q = b^2 + 11 \left(\frac{a}{11}\right)^2$ implies $4q$ is Muriatic. Hence, assume that $p \neq 11$ henceforth, and write $4p = c^2 + 11d^2$. Since $4p$ is not Muriatic, we have that $2|c, d$, and therefore $p = c'^2 + 11d'^2$, where $c', d' \in \mathbb{Z}$ are $c, d$ divided by two. Now, notice that $4p^2q = (a^2 + 11b^2) (c'^2 + 11d'^2) = (ac' \pm 11bd')^2 + 11 (ad' \mp bc')^2.$ By looking modulo $p$, it's easy to see that we can select the signs so that $p| ac' \pm 11bd', ad' \mp bc'.$ From here, it remains only to note that $ac' \pm 11bd', ad' \mp bc'$ are odd, and both are divisible by $p$. These clearly imply that $4q$ is Muriatic. The claim is thus proven.

$\square$

Now, we will use the claim to derive a contradiction in this case. Let $mn = p_1p_2 \cdots p_t$, where $p_1, p_2, \cdots, p_t$ are not necessarily distinct tangfastic primes, so that $4p_i$ is not Muriatic for each $1 \le i \le t.$ It's known that $4mn = (2k+1)^2 + 11 \cdot 1^2$ is Muriatic. By the claim and downwards induction on $i$, it's easy to show that $4p_1p_2 \cdots p_i$ is not Muriatic for each $t \ge i \ge 1.$ However, this implies that $4p_1$ is Muriatic, which is clearly a contradiction.

As we've exhausted all cases, we know that one of $4m$, $4n$ is Muriatic, as desired.

$\square$

Here is a solution that uses the fact that $\mathbb{Z}[(-1+\sqrt{-11})/2]$ is a UFD. The above solutions more or less prove this implicitly.

Consider $\alpha = ((2k+1)+\sqrt{-11})/2$. Clearly $N(\alpha)=k^2+k+3=mn$. By factorizing we know that there exist $\beta,\gamma\in \mathbb{Z}[(-1+\sqrt{-11})/2]$ such that $\alpha=\beta\gamma$ and $N(\beta)=m, N(\gamma)=n$. Since $\alpha\not\in \mathbb{Z}[\sqrt{-11}]$, we can WLOG assume that $\beta\not\in\mathbb{Z}[\sqrt{-11}]$, showing that $\beta=(x+y\sqrt{-11})/2$ for odd $x,y$. The statement $N(\beta)=m$ now becomes $x^2+11y^2=4m$. This is the desired statement as $x,y$ are both odd.

The most beautiful solution I've ever seen.
We need this theorem below:
$f(x,y)=Ax^2+2Bxy+Cy^2$, if $D=AC-B^2>0$
then there exists a pair of $(u,v)\in\mathbb{Z}^2$, $(u,v)\neq (0,0) $ such that $|f(u,v)|\leq (\frac{4}{3}D)^{\frac{1}{2}}$
--------
Let $f(x,y)=mx^2+(2k+1)xy+ny^2$ and $D=mn-\frac{(2k+1)^2}{4}=\frac{11}{4}$
With the theorem above, there exists a pair of $(u,v)\in\mathbb{Z}^2$, $(u,v)\neq (0,0) $ such that $|f(u,v)|\leq\sqrt{\frac{11}{3}}<2$
$D>0\Rightarrow f(x,y)>0\Rightarrow f(u,v)=1$, that's $mu^2+(2k+1)uv+nv^2=1$
$m,n$ are both odd $\Rightarrow$ At least one of $u,v$ is odd.
WLOG $u$ is odd, with $4mnu^2+4u(2k+1)uv+4u^2v^2=4n$
Let $l=2k+1$ is odd, then $l^2=4mn-11$
Then $(2nv+lu)^2+11u^2=4n$, so $(x,y)=(2nv+lu,u)$ is the desired solution.


The number 4641 can be expressed as the product of two 2-digit whole numbers. What is the sum of these two numbers?

First we factor 4641 into $ 3 \cdot 7 \cdot 13 \cdot 17$. The two-digit numbers we can make with these numbers are $ 51$ and $ 91$, which give us a sum of $ \boxed{142}$.


topology
logarithms
limit
advanced fields
advanced fields unsolved

I've got an interesting idea (in fact an interesting article about this idea, but that's not the point). I am almost sure (or if you prefer: I've almost sure that my proof is correct) that there exist a construction of:

1. Set A in R^3, such that the capacity dimension of A is e.
2. Set B in R^4, such that the capacity dimension of B is Pi.

I found the way to solve this problem, I've been trying to find anything in topological books, but I failed...

Do you see any formal ways to prove it?

What is capacity dimension? Is that the same as Hausdorff dimension?

I think so: [url]http://mathworld.wolfram.com/CapacityDimension.html[/url]

It's almost the same - let's say, that in R^n they are the same. Which of course makes the problem easier...

I think you can fairly easily construct sequences of sets in $\mathbb{R}^n$ with Hausdorff dimension $\frac{\log a}{\log b}$ for integers $a$ and $b$, so shouldn't we be able to take the limit of a bunch of these?

Well: this formula is a good start! It's definetely better than considering the capacity dimension formula given in:

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

And now of course we should ask ComplexZeta's question:

[quote]shouldn't we be able to take the limit of a bunch of these?[/quote]

We should - but it's not as easy as it seems to be - a few assumptions - and a construction model - that's what we need now. We should answer the question:

Can the $\frac{\log (a)}{\log (b)}$ formula be generalized?

Do you need an actual set with this dimension or only proof that one exists? The former seems to me to be harder. But I think if we can come up with a sequence of pairs $(a_n,b_n)$ of integers so that $\lim_{n\to\infty} \frac{\log a_n}{\log b_n}=e$ (or $\pi$) then we can perform a modification of the standard procedure to construct sets of dimension $\frac{\log a}{\log b}$. (I talked to a professor at my university about a sort of similar problem once and he showed me how this sort of thing is done.)


This thread is about the doctrine of h*ll.  If you wish to discuss something other than the doctrine of h*ll, this is not the thread for that discussion.  I've seen enough religion threads get derailed because people would rather trash their opponents' arguments by bringing in non sequiturs than actually respond to their points.  Please do not do that.  Thank you.

In a couple of other threads, the claim has been made that the God's consigning of people to h*ll would not be just, but would instead be evil.  Loaded words like "torture," "sadism," and so on seem to be par for the course in such accusations, but I've picked what seems to me to be the objection which is most grounded in reason (rather than emotional appeal) to address: the claim that [i]H*ll, as the infliction of infinite punishment for finite sins, is inherently unjust[/i].  I want to try to briefly address this line of argument (which is, at least [i]a priori[/i], a fairly compelling one), in hopes of getting a [b]rational[/b] discussion going.

There are several ways one might object to this argument.  One might claim that:
1: H*ll is not infinite,
2: H*ll is infinite, but the pains associated with it are not infinite,
3: The pains of h*ll are infinite, but they are not punishment,
4: The pains of h*ll are infinite and punishment, but they are not inflicted,
5: Some sins (those leading to h*ll) are not finite,
6: Sins are by nature not finite,
7: It is not unjust to have infinite punishment for finite sins,
or others. As an orthodox Christian, I reject (1), (2), and (3) as options, and I think that all sins, without the action of God's grace, lead to h*ll, so (5) as distinct from (6) is meaningless.  Thus, of the objections to the argument that I have listed, only (4), (6), and (7) are open.  While I think that (7) is possibly a reasonable objection (and I'd like to see what those who use the argument have in answer to this), I'm not going to claim (7).  I am going to claim both (4) and (6), and attempt to explain why the argument cannot hold against such a doctrine of h*ll.

I will begin with (6).  How can I say that a sin is not finite, when it certainly takes only a finite amount of time, space, and so on to conceive of and act on the sin?  My answer is that time and space are not the measure of the sin.  Indeed, most people do not measure sins this way at all.  It takes a greater time to (say) successfully embezzle than to pull out a gun and murder someone in cold blood, but most people would consider the latter to be the greater sin or crime.  What, then, is the measure of a sin?  The measure, to a Christian at least, is the degree to which the sinful act falls short of the goodness of God.  But since God's goodness is perfect and infinite, anything which is in the least evil is already infinitely far from God.  Thus, though not all alike, all sins are infinite, and "the wages of sin is death."

Now, for (4).  You need not be afraid that I am going to water down the doctrine of h*ll and eliminate the very thing you are arguing against.  I hold that the pain (I use pain in more than the [i]physical[/i] sense, of course) of h*ll is such that any sane person who compared them would prefer the worst tortures on Earth to that miserable condition.

That said, I claim that it cannot be rightly said that God inflicts these pains.  Rather, the horrible pain of h*ll is the natural outcome of the sufferer's state of being.  The d*mned man is, having sinnned, infinitely far from God's perfection, and having rejected God's offer to forgive him and remake him in perfection, is doomed to remain in his own evil and apart from God.  This is what causes the great pain: God is the sole good, and separate from God there is only pain and horror.  H*ll is not [i]merely[/i] a separation from God, nor [i]merely[/i] a state of mind.  The pain is real.  But the pain of h*ll is not the pain of torture imposed, but the pain of being something which has willfully lost its only good; the punishment is for sin, not inflicted but naturally flowing from being sinful and unredeemed.  The pain of h*ll is the pain of eternal emptyness.

A doctrine of h*ll which holds what I have outlined above (or something similar) does not fall to this accusation of injustice.  H*ll, by this account, is not infinite torture imposed for a finite crime, so the argument fails to apply.  The doctrine of h*ll is by no means pleasant -- none of us wish to think of ourselves or others suffering this infinite pain -- but it is eminently just.  As C.S. Lewis put it in his [i]The Problem of Pain[/i], "In the long run, the answer to all those who object to the doctrine of h*ll, is itself a question: 'What are you asking God to do?' To wipe out their past sins and, at all costs, to give them a fresh start, smoothing every difficulting and offering every miraculous help?  But He has done so, on Calvary.  To forgive them?  They will not be forgiven.  To leave them alone?  Alas, I am afraid that is what He does."

God made hell, so anything done by hell comes indirectly from god.

But it IS indirect, and that statement didnt really provide any counter argument.  Setofallsets already said that, anyway. Also, the punishment is also coming from you. You bought your time in h*ll..

I personally do not believe in [img]http://img.photobucket.com/albums/v259/drspyder/hell.gif[/img], but if it did exist, I think it'd be a little harsh, I mean, come on, either you go to heaven or [img]http://img.photobucket.com/albums/v259/drspyder/hell.gif[/img]?

[quote="chess64"]H*ll is not infinite. It's like a hospital for those "spiritually sick." After you are cured of your "sickness", you get out of it. ;)[/quote]

Where did you hear that? So isn't that bad?  :P

[quote="236factorial"]Where did you hear that?[/quote]

It's my (and most peoples') religion's belief.

[quote="236factorial"] So isn't that bad?[/quote]

:? It's meant to be...

[quote="chess64"][quote="236factorial"] So isn't that bad?[/quote]

:? It's meant to be...[/quote]

i've heard you're hanged over a fire, but once you're dead, would you feel it?

Well, actually, according to my religion, h*ll isnt infinite, but atleast 9 months of torture beyond torture, which is why I still call it 'infinite punishment'. Either way, as to its 'either heaven or h*ll' yeah i think, but i think its not THAT hard to get into heaven.

From a religious perspective, all of the arguments proposed might seem valid. But from my perspective, they are just a way to cover up the inherent "evil" in the system.

Regarding your analysis of (6): if God's goodness is infinite, how in his right mind could he send people off to eternal suffering? I mean, even I don't do stuff that cruel, and I'm a mere [i]mortal[/i].

And (4): How is it not imposed? That doesn't make sense. Pain is inflicted, whether it be by accident or on purpose. God created everything, did he not? God created H*ll, too, with the intention of causing suffering. And that's not cool.

Actually, everyone goes to h3ll. Since the expel of Adam and Eve from the Garden of Eden, everyone is born sinful. The only way you're not going to h3ll is to believe in God, since you'll be saved from sin (and thus go to heaven and have eternal life after you die) this way.

[quote="jhollenbeck"]God cannot tolerate evil, so anything short of being holy is not acceptable.[/quote]
That sounds a tad elitest, don't you think?

[quote="jhollenbeck"]God cannot tolerate evil, so anything short of being holy is not acceptable.[/quote]

Then... why didn't he make everyone good in the first place? The only explanation I can think of is that he's playing a game with us, designing our minds to naturally think sinfully and then punishing us for it. The only way to escape this is to fight ourselves; our weapon is in the form of religion. But in the end, the whole idea of God is created by man, because we fear death, we fear nothingness. So by being religious you are fighting a pointless battle against yourself.

You say this with the belief that there is no God, but even if the 'concept' of God were man-made, which I dont believe ( i believe in revelation), that does not rule out his actual existence.  And putting a natural desire to sin only makes it so we have a challenge, if we were all naturally good there would be no point to reward..

[quote="paladin8"] Regarding your analysis of (6): if God's goodness is infinite, how in his right mind could he send people off to eternal suffering? I mean, even I don't do stuff that cruel, and I'm a mere mortal.

And (4): How is it not imposed? That doesn't make sense. Pain is inflicted, whether it be by accident or on purpose. God created everything, did he not? God created H*ll, too, with the intention of causing suffering. And that's not cool.
[/quote]
The point, though, is that the pain of h*ll is not something external to the sufferer.  It arises as a necessary consequence of his being what he is: a sinful person, separated from God.  H*ll is not so much a "place" to which God "sends" the sinful as it is the supreme state of suffering.  H*ll is not something which God could have made or not made; it is inherent in the creation of wills that are different from God.  If they may choose their good and submit to God, they must also be able to choose to deny Him.  To deny God -- to deny the One who is Himself the only good we can ever have -- is to bind oneself in h*ll.  What are you asking God to do?  To take an evil person (and, unredeemed, that is what we have all made ourselves) into heaven?  But heaven consists in the presence of God and in submission to Him, and this is precisely what this person has rejected.  To give this person some good that is separate from God?  But there is no such good.  The only option is to leave him to himself, and this is what God does.  This is h*ll, and unbearable torment, but it is the only alternative.  In the end, there are only two alternatives.  Complete and total submission to God (by His miraculous aid, to be sure), or complete and utter misery.  This is not arbitrary decree, it is the nature of things.  God is not withholding good from those in h*ll; the damned have chosen to reject the only good that there is.  This is why I say that the pain of h*ll is not inflicted.

Only humans could have created such a contrived system for the universe to run on. So I still believe it's all made-up and all, but I'll argue with it anyway.

I'm not saying that God should let sinners into heaven; of course not, I understand that part. The point is that maybe he should make it such that there are no sinners in the first place. Reveal his existence (see "Existence of God" thread for the debate) and thus prove that there is indeed a God. If he showed me that, I'd consent to believing because it'd be logical. At this stage, I have no tangible proof of such a God so I am reluctant to believe. If such is evil and sinful in the eyes of God, I'd rather not eternally be in his presence.

And the whole absoluteness of the scenario bothers me as well. Why can't there exist another world similar to ours? Or why can't sinners be reborn in a new life to prove their faith? That way everyone could end up in heaven; wouldn't that be the ideal situation? It seems that indeed the entire concept of H*ll should not exist in the first place. In a logical creation of the world, it wouldn't. But in an artificial creation out of fear, it does.

[quote="wdragyn"][quote="Setofallsets"]Complete and total submission to God (by His miraculous aid, to be sure), or complete and utter misery.  This is not arbitrary decree, it is the nature of things.  God is not withholding good from those in h*ll; the damned have chosen to reject the only good that there is.  This is why I say that the pain of h*ll is not inflicted.[/quote]

Set-
Hope your 20-hour trip home went well and you are relaxing in peace without loud music bleeding into your bedroom.

You probably already know that I'm an agnostic, leaning towards atheist.  I have been offered Christianity from birth, and rejected this path as a teenager, which I felt closed my mind, creating arbitrary rules and doctrines that people follow based on fear of an afterlife gone awry, an opiate of the masses.  I chose not to worry about God's existence, and instead try live a logical moral life based on the golden rule, because I want the world to be a better place, not because I fear h*ell.

Assume God exists.  If I make an effort to better myself in life and be good to others, but never acknowledge his existence in life, what reason would he have to d*amn me to h*ell?  I see a clear separation two parts of Christianity:  believing in God and moral values (not the hypocritical Republican kind where you hate gay people, but true moral values--which is love and acceptance towards your fellow man).  If no one believed in God, but everyone believed in living a moral life, wouldn't the world be a much better place?

What you said in your last post was pretty concrete, so why would God d*amn good people to h*ell?

Also, I realize as an employee of this company I should probably know this...but why is h*ell censored?  We can talk about abortion but we can't say h*ell??[/quote]
For the same reason GTA can get away with random killing, but as soon as you see a boob, that thing gets pulled off the shelf.

People need to learn that the human body is nothing to be scared of.

Yeah...that was kinda unrelated to my question to Set but that's just Hilary pandering to the right so she can get elected president.

[quote="paladin8"]Only humans could have created such a contrived system for the universe to run on.[/quote]I'm honestly not sure what you find so contrived about it.  If you'd explain how you find it so, perhaps I could answer you better.  How is what I've outlined above particularly contrived?

[quote="paladin8"]The point is that maybe he should make it such that there are no sinners in the first place.[/quote]I submit that the possibility of sin is inherent in the nature of a will that is independent of God's will.  If there can be a choice to submit, there must also be a choice to reject.  That's what choice is.
I'm aware, of course, that this still doesn't answer your question.  God knows (being omniscient) who among His creation will eventually choose to submit to Him; you want to know why he didn't only create those people.  Now I, not being omniscient, don't know precisely why -- but I will ask this: for whom would it be better if those people did not exist?  Certainly not for them -- if they didn't exist, there could be no good for them at all.  How could it be worse for such people to be given the choice and choose to reject God than for them to not be given a choice (or even an existence) at all?

[quote="paladin8"]Reveal his existence (see "Existence of God" thread for the debate) and thus prove that there is indeed a God. If he showed me that, I'd consent to believing because it'd be logical. At this stage, I have no tangible proof of such a God so I am reluctant to believe.[/quote]This is a little off topic, but what would constitute tangible proof?  And (think about this one) -- why is "tangible proof" necessary?

[quote="paladin8"]And the whole absoluteness of the scenario bothers me as well. Why can't there exist another world similar to ours? Or why can't sinners be reborn in a new life to prove their faith? That way everyone could end up in heaven; wouldn't that be the ideal situation?[/quote]Why would it be the case that that more lives or more time would result in more people choosing to submit?  After all, there is an action of the will involved; the scenario is not merely one of waiting to be in the right place at the right time.  To quote C.S. Lewis (who says this much better than I can):[quote="C.S. Lewis"]I believe that if a million chances were likely to do good, they would be given. But a master often knows, when boys and parents do not, that it is really useless to send a boy in for a certain examination again. Finality must come sometime, and it does not require a very robust faith to believe that omniscience knows when.[/quote]

[quote="wdragyn"]If no one believed in God, but everyone believed in living a moral life, wouldn't the world be a much better place?[/quote]Granting your scenario (that people would succeed in living morally, even if only by modern standards), I'd agree that the world would be probably more pleasant.  But if the ultimate goal is not merely a pleasant 70 years on Earth, then this is a bit of a moot point.  Of course, I'd also contend that your scenario is not possible, but that's not what you were asking.

[quote="wdragyn"]Assume God exists. If I make an effort to better myself in life and be good to others, but never acknowledge his existence in life, what reason would he have to d*amn me to h*ell?[/quote][quote="wdragyn"]so why would God d*amn good people to h*ell?[/quote]I'm sure you've heard this plenty of times before, but it's worth saying again: [i]there are no "good" people[/i].  It's not a question of "You were good, but you didn't believe the right thing; off to h*ll with you."  At the risk of sounding like I'm evangelizing -- God isn't looking for ways to send people to h*ll, He seeks the highest possible good for every person.  But God Himself [i]is[/i] the highest (indeed the only) good for any of us.  If we reject the only thing that is good for us, there will be nothing left but h*ll.  To be sure, mundane morality (that is, not explicitly tied to submission to and worship of God) is still important; to break those laws is still against God's will.  But those sins are forgivable; God has offered to forgive everything and mend the evil in us, if we but will accept His offer and submit.  But there is a reason they talk about the "unforgivable sin":  God cannot forgive a final rejection of Him, because in rejecting Him one rejects His forgiveness.

By the way, let me know if any of the things I'm saying are unclear.  I'm using religious language and terminology to some extent because those words best convey the essence of what I'm trying to say, but the superior utility of this kind of language is to no avail if the people with whom I'm talking don't understand what it means.

[quote="wdragyn"]Also, I realize as an employee of this company I should probably know this...but why is h*ell censored? We can talk about abortion but we can't say h*ell??[/quote]My best guess is that AoPS installed general filtering softwear which censors pretty much all "four-letter words".  It's just our bad luck that we happen to want to use these words legitimately.  I guess you could ask about it at the office or something...

What's contrived about it? The entire idea of H*ll is. Eternal suffering? Why would something like that exist at all? It almost seems like a myth to scare children. "Better believe in God, or you'll get sent to H*ll." If God was so good (if he was the ONLY good), H*ll wouldn't exist.

Would it be better for people who will inevitably sent to H*ll to exist? Personally, IF I thought H*ll existed, I'd rather not exist in the first place; eternal suffering sounds awfully bad. So to me it seems like God's thinking: "Hmm, I'm going to make these people, and they're going to sin without knowing it because I'm not going to tell them. Then they will suffer eternally."

Tangible proof; something that can't be argued against. Like the "proof" there is right now isn't complete; it's so easy to argue against it logically. That's why these threads exist. And is it necessary? I suppose no. But to me, like I said, I'm reluctant to devote myself to a belief that doesn't have proof. It's against what I believe as of now. Especially something of that magnitude.

OK, so maybe more lives won't help. But how about like a middle path? Something away from God, but not suffering. Or is that H*ll? Maybe H*ll is only considered eternal suffering to Christians since it is away from God and that would constitute eternal suffering for them. I'd be fine with that; as long as it's not TOO painful to live (although this life is already reasonably painful) the idea seems rational. So I guess we'll see eventually, though. After I die, I'll be very curious to find out what happens (or just be wiped from existence, in which case it doesn't matter since my memory and consciousness will vanish). I guess there's not much point in arguing this since I find it very unlikely that I will convert so this is purely for the sake of disagreement and curiosity.

Can someone illuminate me ?
Why do you write h3ll, or h*ll, and not plainly hell ?

[EDIT: Ahh sorry I just understood :), this censorship stuff is so funny :) ]

[quote="paladin8"]
Tangible proof; something that can't be argued against. Like the "proof" there is right now isn't complete; it's so easy to argue against it logically. That's why these threads exist. And is it necessary? I suppose no. But to me, like I said, I'm reluctant to devote myself to a belief that doesn't have proof. It's against what I believe as of now. Especially something of that magnitude.[/quote]

Out of curiosity, what exactly DO you believe as of now?

[quote="tetrahedr0n"]Without getting too off topic, my question is as follows:

Are the standards (or whatever decisions God uses to pick between who is saved and who isn't) the same for every individual? Or are there differences?[/quote]
God gives the same choice and opportunity to every individual. That is, Jesus died to pardon the sins of everyone, and everyone is then faced with the decision to believe it and accept forgiveness or not to. If you confess that Jesus is God and Lord of all and believe that he died and rose from the dead, then you are saved.

To me, its a nice collection of nonsense.  How can the "benevolent God" offer death, destruction, plague, "The Revelations", eternal punishment and do it all without mercy (there is no redemption in h3ll if it goes on forever)?  What sort of an example are we meant to be following, if we bury our enemies, or those that do not believe us, in eternal damnation?  Is this what we do when we "love our enemies"?

The hypocrisy in the concept of h3ll is remarkable.  My only acceptable view of it is that the concept of h3ll is man-made, and what exists in the "afterlife" (if it exists) is a surprise.

[quote="ZennyK"]God gives the same choice and opportunity to every individual. That is, Jesus died to pardon the sins of everyone, and everyone is then faced with the decision to believe it and accept forgiveness or not to. If you confess that Jesus is God and Lord of all and believe that he died and rose from the dead, then you are saved.[/quote]

All right, but what if someone is completely ignorant of Christianity? What if a human being simply is never exposed to the religion? He may be in fact following Christian doctrines, but as unaward of Jesus and the Christian God, he will be rewarded with hell?

This is my problem with hell. If it exists, who goes there? If it relies completely on belief in Jesus, then clearly there would be millions upon millions of people in hell who, from my perspective, have done nothing deserving of it.

"The wrath of God is revealed from heaven against all ungodliness and unrighteousness of men, who by their unrighteousness suppress the truth. For what can be known about God is plain to them, because God has shown it to them. For his invisible attributes, namely, his eternal power and divine nature, have been clearly perceived, ever since the creation of the world, in the things that have been made. So they are without excuse. For although they knew God, they did not honor him as God or give thanks to him, but they became futile in their thanksgiving, and their foolish hearts were darkened."

“It is not the hearers of the law who are righteous before God, but the doers of the law who will be justified. For when Gentiles, who do not have the law, by nature do what the law requires, they are a law to themselves, even though they do not have the law. They show that the work of the law is written on their hearts, while their conscience also bears witness, and their conflicting thoughts accuse or even excuse them on that day when, according to the gospel, God judges the secrets of men by Jesus Christ.”

It is not necessarily about being "exposed to the religion of Christianity." Every single human knows the law of God by nature; it is "built in" to everyone. God also reveals himself to every single person through his creation, though many refuse to acknowledge it. Because of this, no person can rightly plead ignorance.

There is also the point of "original sin": once Adam and Eve sinned, all of creation was plunged into desolation and sin (the Fall). Because of this "original sin," all people are conceived into a state of sin, so that no one does not deserve punishment. Obviously this doesn't seem quite fair, but who makes the rules in the first place? Again, it is not up to us to question the judgement of God. He is perfect, we are not. He is infinitely wiser and more just than we are. He is also perfectly graceful . . . which is why he let sinful humans kill his only son, so that we would not have to suffer the terrible punishment for our sin if we only believe it and acknowledge God as Lord.

[quote="ZennyK "]It is not necessarily about being "exposed to the religion of Christianity." Every single human knows the law of God by nature; it is "built in" to everyone. God also reveals himself to every single person through his creation, though many refuse to acknowledge it. Because of this, no person can rightly plead ignorance. [/quote]

The law of God, perhaps, but what about Jesus himself? How is one to know of Jesus without having ever been exposed to testimony of him? Is Jesus' name crucial to being saved? Or just the existence of the son of God? Or, based on your passage, is just a belief in a very general God, whether God, or Allah, or Brahma, or whoever else?

My problem with hell and the God debates is that the religious side can always answer any argument simply with the "we are human, don't know and can't understand the complexity of God" while the non-religious must use very human logic.

Next question then - what about an infant who dies before being truly conscious of religion? Since he is guilty of original sin, and he can't confess his belief in Jesus or God, how can he enter the Kingdom of God?

[quote="jmadsen"][quote="ZennyK"]There is also the point of "original sin": once Adam and Eve sinned, all of creation was plunged into desolation and sin (the Fall). Because of this "original sin," all people are conceived into a state of sin, so that no one does not deserve punishment. Obviously this doesn't seem quite fair, but who makes the rules in the first place? Again, it is not up to us to question the judgement of God. He is perfect, we are not. He is infinitely wiser and more just than we are. He is also perfectly graceful . . . which is why he let sinful humans kill his only son, so that we would not have to suffer the terrible punishment for our sin if we only believe it and acknowledge God as Lord.[/quote]
I strongly object to the idea of being punished for the sin that Adam and Eve committed thousands of years ago. Why was the sin so bad that all their descendants are to be punished for it for all of eternity? Your answer for this is "who makes the rules in the first place?" This makes God sound like a tyrant. He makes rules and enforces them, doesn't give anyone else any say, and doesn't even tell us why he made the rules? That doesn't seem like something God should do.[/quote]
And yet he gives us a way OUT of punishment, offering to replace eternal misery in h*ll with eternal joy in heaven!

[quote="tetrahedr0n"]This is my problem with -. If it exists, who goes there? If it relies completely on belief in Jesus, then clearly there would be millions upon millions of people in - who, from my perspective, have done nothing deserving of it.[/quote]

Everyone DESERVES to go there. Yes, this sounds incredibly harsh from our perspective. I mean, come on . . . I'm not too bad, and I'm sure you're not all that terrible either -- which is true from a human perspective relative to, say, a serial killer or some mass-murdering terrorist. However, God's standard is perfection -- a standard you and me and everyone else fail to satisfy. In God's eyes, you are either perfect or you are bad. If you are bad, you go to h*ll. This sounds very harsh indeed!

But, as I have already said, God loves us so much that he killed his son and took out all his wrath on him -- in h*ll -- instead of us, so that we could go to heaven.

[quote="tetrahedr0n"][quote="ZennyK "]It is not necessarily about being "exposed to the religion of Christianity." Every single human knows the law of God by nature; it is "built in" to everyone. God also reveals himself to every single person through his creation, though many refuse to acknowledge it. Because of this, no person can rightly plead ignorance. [/quote]

The law of God, perhaps, but what about Jesus himself? How is one to know of Jesus without having ever been exposed to testimony of him? Is Jesus' name crucial to being saved? Or just the existence of the son of God? Or, based on your passage, is just a belief in a very general God, whether God, or Allah, or Brahma, or whoever else?

My problem with - and the God debates is that the religious side can always answer any argument simply with the "we are human, don't know and can't understand the complexity of God" while the non-religious must use very human logic.

Next question then - what about an infant who dies before being truly conscious of religion? Since he is guilty of original sin, and he can't confess his belief in Jesus or God, how can he enter the Kingdom of God?[/quote]
And my answer is: only God knows and only God ultimately decides.

Here's the issue, though. The question seems to always be about someone else: "What about the native  African out in the middle of nowhere?" "What about some tribe in the Amazon?" "What about an unknowing infant?" -- and the list goes on. The question should not be a matter of anyone else. What about YOU? What will happen to YOU after you die? Will you end up in heaven? In h*ll? Or will you merely lose consciousness and cease to exist? It doesn't really matter for your sake or for my sake whether someone else goes to h*ll or not. It isn't even our business to know, since we don't have any say in the issue anyway.


[color=violet][b]Number of quadrilaterals which can be constructed by joining the vertices of a convex polygon of 20 sides, if none of the side of the polygon is also the side of the quadrilateral is :[/b]
[/color]
[color=blue](a) 17C4 – 15C2
(b) 15C3 .20/4

(c) 2275
(d) 2125[/color]

There are $ C^{4}_{20}$ ways to choose any 4 vertices.
Now let's count number of ways to choose 4 vertices, where at least 2 are adjacent.
[b]1) 2 vertices are adjacent and the other 2 are not.[/b]
There are 20 ways to choose 2 adjacent vertices.
...0xx0...
There are 16 left places for the other 2.
We can select them in $ C^{2}_{16}$ ways.
15 ways are to choose 2 adjacent vertices, so there are total $ 20(C^{2}_{16}\minus{}15)\equal{}2100$ ways in this case.
[b]2) 2 pairs of vertices are adjacent.[/b]
There are 20 ways to choose first pair of adjacent vertices.
...0xx0...
There are 16 left places for the other 2 adjacent vertices.
We can select them in 15 ways.
So there are total $ \frac{20\cdot15}{2}\equal{}150$ ways in this case. (We divided by 2 because we can get the same 2 pairs of adjacent vertices in 2 different ways).
[b]3) 3 vertices are adjacent.[/b]
There are 20 ways to choose 3 adjacent vertices.
...0xxx0...
There are 15 left places for the 4'th vertice, so there are total $ 20\cdot15\equal{}300$ ways in this case.
[b]4)  4 vertices are adjacent.[/b]
20 ways in this case.

So the answer is $ C^{4}_{20}\minus{}(2100\plus{}150\plus{}300\plus{}20)\equal{}2275$
 [size=150](c)[/size]


actually, im surprised that most of the calc book i have looked at do not provide a proof for this problem...im sure u can find a proof on this forum somewhere thoug.

[quote="Hokkage"]It has to do with factoring the infinite Taylor series of sine, and then matching up the x^2 coefficients.[/quote]

A rigorous solution requires far more than that

Lecture notes about this series( many proofs included):

[url=http://www.secam.ex.ac.uk/~rjc/etc/zeta2.pdf]www.secam.ex.ac.uk/~rjc/etc/zeta2.pdf[/url]

do u know where i can find a rigerous proof since the olnly one i ever see is the one by Euler. which even though it is right is not concidered rigerous.

It's likely that many AoPSers have posted rigorous proofs of this identity.  I myself posted a proof by Fourier series some time ago, and the only detail I really forgot was proving the convergence of the series, which others filled in for me.

In any case, a rigorous proof would not be elementary, and would certainly not be Pre-Olympiad level.

this question may be in the wrong forum...

anyway, if you promise not to hurt yourselves reading this, here's a link to not one, but [i]six[/i] proofs of Euler's crazy summation :huh:
Enjoy :rotfl:

[url=http://matharticles.com/ma_calc8e.html]matharticles.com/ma_calc8e.html[/url]

btw, pick "Six ways to sum a series"
ps:i fixed the link problem

http://www.american.edu/cas/mathstat/People/kalman/pdffiles/Sixways.pdf

The other link seems to have been copy-pasted from the wrong window.

Doublepost: http://www.mathlinks.ro/Forum/viewtopic.php?t=81443

Nonetheless, see http://www.mathlinks.ro/Forum/viewtopic.php?t=52359 too.

Yes  the question has been posted here many times before. From that 14 proofs of this problem
They can be found [url=http://www.maths.ex.ac.uk/~rjc/etc/zeta2.pdf]here[/url]

One of the " intutive" way (posted mutlitple times before - due to Euler ) is to look at the  function
$\frac{\sin x} x = 1 - \frac{x^2} 6 + \dots$  and consider it as a function of $x^2$
 This is zero when $x^2 = \pi^2$ or $x^2 = (2\pi)^2$ or $x^2 = (3\pi)^2$  etc ...
(Recall sum of reciprocal of roots of $1+ax+bx^2 \dots$ is given by  $a$) hence
$\frac 1 6 = (\frac 1 {\pi^2}) ( 1 + \frac 1 {2^2} + \frac 1 {3^2} + \dots$

(Details can be seen in many of the references given elsewhere)
Hope this helps.


limit
number theory proposed
number theory

A positive integer $c$ is given. The sequence $\{p_{k}\}$ is constructed by the following rule: $p_{1}$ is arbitrary prime and for $k\geq 1$ the number $p_{k+1}$ is any prime divisor of $p_{k}+c$ not present among the numbers $p_{1}$, $p_{2}$, $\dots$, $p_{k}$. Prove that the sequence $\{p_{k}\}$ cannot be infinite.

[i]Proposed by A. Golovanov[/i]

Maybe not hard,This is my solution:
$p_{k+1}\leq p_{k}+c$,
but $\lim_{x\to \infty}\frac{\pi (x)}{x}=0$
contraduction.
And there is another proof without the non-elemetry result:
$p_{k+1}\leq p_{k}+c$,suppose that $p_{k+1}=\frac{p_{k}+c}{d_{k}}$
if there exist $l$such that $d_{l},d_{l+1},....d_{l+c+2}=1$
then $p_{i+1}=p_{i}+c ,\forall i=l,\dots,l+c+2$ but there must exist $i\in[l+1,l+c+1]$ such that $p_{i}=p_{l}+(l-i)c$is divisible by $c+1$
then $p_{i}$is not a prime ,contraduction.
so there do not exist $c+2$ consecutive $p_{k}$ such that $p_{k+1}=p_{k}+c$ so $p_{k+c+2}\leq \frac{1}{2}p_{k}+c(c+1)+\frac{c}{2}$
thus $p_{k+m}\leq \frac{1}{2}p_{k}+A$
where $m,A$be constant.then$p_{k+lm}-2A \leq \frac{1}{2^{l}}(p_{k}-2A)$
when $l-\to \infty$ $p_{k+lm}\leq 0$ contraduction.

Yes, that isn't hard  :) (in contrary to some other Tuymaada's problems)

Can you post some other Tuymaada's problems?I haven't heard such a contest before...

http://www.guas.info/competit/tuyme.htm - here you can find problems from Tuymaada 2000-2006


function
real analysis
real analysis unsolved

show there is no function from R to R s.t.{x/f cont. at x}=Q(rationals)..

You could restrict f to the interval $ [0,1]$.

First show that the set of continuity points of a function $ f$ is a $ G_{\delta}$ set and then observe that the set of rationals in the interval $ [0,1]$ is not a $ G_{\delta}$ set and the problem is solved.


analytic geometry
geometry
angle bisector
geometry open

I'm wondering how to solve the following difficult problem:

[b]It is given circle k(O, r) and a segment AB. Find (the locus of) all points M from k, such that the line OM is angle bisector of the angle AMB.[/b]

I know it is not constructible by using compass and ruller - so I want the equation (if it is possible) for the locus (you are free to use analytical geometry).
I have the following responses:
1. http://www.mathlinks.ro/viewtopic.php?t=162527
2. http://lib.mexmat.ru/forum/viewtopic.php?t=9555&highlight=
I think they may be useful.

what is the significance of the circle k

In analytic terms you may think - there are given the coordinates of O and the measure of r (or equality for the triangle).
And there are given coordinates of A and B.

i got up with a solution mind
invert with circle k
the locus is a unique point

Are you sure? May you write down? I don't believe. I'm sorry.

after inversion assume there exist atleast 2 points and observe the angle of  the 2 points one reduces and other increases  :)

so contradiction
is this good enough

I cannot use inversion (I don't know anything about it) but I'll give it to a friend of mine.
How do you eliminate the second point?

one angle is greater than the other bcoz of the first point

I'll give you opinion after I learn inversion or after a friend of mine check carefully your words. Thank you for the time spent with that problem.

i,m so sorry i also have a analytical solution giving me $ \ge$ 4 points
iwill see to it

your analytical solution probably is wrong. the points are less than 4. Are you now understanding the sense of carefully written solutions?

such things are not new to me
i was at the imo training camp of india this year but didnt get selected for one of the main reasons being this still i dont mind, :)
my inversion solution was correct IF YOU WANT I will ellaborate :lol:  :D
and analytic solution was giving $ \le$ 4
sorry im troubling you a lot
a typo edited

Yes, it will be interesting to see your both solutions - analytical and inversion. I think this problem is not a trivial.

could u tell me the FREEware for which i can draw and save and post geometric images

You may use Geonext...
http://geonext.uni-bayreuth.de/index.php?id=2453

This problem still have no clear and complete solution.
The things I know:
1. It is impossible to construct M using only compass and ruller.
2. There are not more than 4 such points.
3. Most of the people attack the problem with Analytical Geometry and Inversion.

I'm interested in this problem because it seems interesting and definitely it is not very easy, it
have practical application and a former colleague of me said he is composed it at his own.


Given positive integers $a_1,...,a_n$. Find $k$ such that $\sum_{i=1}^n|a_i-k|$ is minimum.

I'll leave the proof for someone else, as it's good practice.


How do I compute in other bases?

I want to learn how to add, subtract, divide and multiply in other bases.

Can someone, who is not uptight about my questions, help me here?

Samples:

(1) Compute 24 given in base 5 + 14 also in base 5.

(2) Compute 32 given in base 5 - 14 also in base 5.

(3) Compute 44 given in base 5 times 22 also in base 5.

(4) Cumpute 1442 given in base 5 by 4 also in base 5.

Thanks

[quote="bpms"]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=132902[/quote]

LOL  :rotfl:

Anyway, I'm only going to delete your other topics about bases.  One topic should be plenty.

[hide="bases!!!"]Consider a number $(d_{1}d_{2}d_{3}\cdots d_{n})_{b}$, where each $d_{k}$ is a digit in base $b$.

First, convert the number to base to, by expanding the number as follows:

$b^{n-1}\cdot d_{1}+b^{n-2}\cdot d_{2}+\cdots+b^{0}\cdot d_{n}$

Then, convert it back to the new base by the reverse proccess.[/hide]

[hide="example"]Convert $123_{6}$ to base 10.

$123_{6}=1\cdot6^{2}+2\cdot6+3=36+12+3=51$.[/hide]

B-flat, I thought we were boycotting interval's posts not in the sticky?

I thought we'd give him a warning (notice how the others mysteriously disappeared  :ninja: )  :D


Prove:
$ \frac {a\plus{}b}{c\minus{}d}$ = $ \frac {c\plus{}d}{c\plus{}d}$ [b]if[/b] $ \frac {a}{b}$ = $ \frac {c}{d}$


algebra
polynomial
linear algebra
matrix

hi friends,

Is it possible to represent the roots of a polynomial in matrix form?

I mean let the roots of the polynomial \[ a_0x^n+a_1x^{n-1}%Error. "hodts" is a bad command.
+a_{n}=0 \] be \[ x_0,x_1,\hdots,x_n \]. Is it possible to get the roots of above equation using matrix notation?

thank you

srihari

What do you mean?

Are you asking for a formula to solve polynomial equations?

That doesn't exist, it is a well known result of Galois theory.

Are you perhaps looking for the companion matrix?
http://mathworld.wolfram.com/CompanionMatrix.html


Harvard college MIT Stanford AMC USA(J)MO USAMO

I know that Pdiao got in

Try PM-ing him

[quote="daveed"]I know that Pdiao got in

Try PM-ing him[/quote]

Done. Thanks for the info.

TJ got 6 people that I know of, half of em math guys  :) .

[quote="yunorman"]I'm so proud of the kids at my school, 14 people got into Harvard early, and 2 into MIT early :) , 2 into Stanford too, which was particularly suprising considering our school is in Boston MA, and is a public school.[/quote]

my goodness

Each famous selective school seems to take special care to find all the really bright students in its home town. Thus, for example, I would not expect a school in New Haven, CT to fare nearly as well in Harvard admission--but it would probably do well in getting students into Yale.

Corresponding to this, each famous selective school likes to get students far from its "usual" region. And that probably explains people from a Boston public school getting into Stanford (they add geographical background diversity to the entering class) and definitely explains why there is someone from Wyoming in the entering class of almost every famous selective school each year.

[quote="MithsApprentice"]Has anyone here gotten results from Harvard's EA?

If you have and got in, mind sharing the outline of your application? A lot of my friends who, in my opinion, are very qualified applicants, were deferred. So I'm just wondering what it takes to get in this year.

Thanks for sharing, btw.[/quote]

I knew a person who also got deferred). This person is a 3-time USAMO qualifier (miss this year’s MOP by 1 point/ got 20 on 2005-USAMO), female, 2400 SAT I (first try), 800s SAT II in math IIC, physics and biology and 790 in history, won gold key in writing, 240 PSAT, straight As, ranked #1 out of nearly 800 graduating seniors in her school, playing violin at almost professional level......

The level of her achievements would put her in the top 20 among the high school graduates in the entire country by any standard.

Any comments? Injustice?

[quote="qvc122"][quote="MithsApprentice"]Has anyone here gotten results from Harvard's EA?

If you have and got in, mind sharing the outline of your application? A lot of my friends who, in my opinion, are very qualified applicants, were deferred. So I'm just wondering what it takes to get in this year.

Thanks for sharing, btw.[/quote]

I quote the following from somone's Xanga (who also got deferred). The person I quoted is a 3-time USAMO qualifier (miss this year’s MOP by 1 point/ got 20 on 2005-USAMO), female, 2400 SAT I (first try), 800s SAT II in math IIC, physics and biology and 790 in history, won gold key in writing, 240 PSAT, straight As, ranked #1 out of nearly 800 graduating seniors in her school, playing violin at almost professional level......

The level of her achievements would put her in the top 20 among the high school graduates in the entire country by any standard.

Any comments? Injustice?

[quote]First off: It looks like math/science - including an IMO medalist, other USAMOers and ISEF/Siemens/RSI people - were almost all deferred.[/quote][/quote]

is that person by any chance asian? asians are at a severe disadvantage when applying

It's hard to measure who the top 20 high school graduates in the country actually are.  How do you rate international olympiad medalists against people who have done original research against people who have spent their time in high school making meaningful contributions against poverty against precocious writers against speech and debate champions against people who are absolutely all around brilliant, but don't fall in to any of the earlier categories?  I tend to think that there's more of a top 200, of which said person undoubtedly falls in.

There are a couple of factors, of course.  Elite schools like to pick up some of their legacy and minority (more asians were admitted than normal in the early round this year, c.f [url=http://www.news.harvard.edu/gazette/2005/12.15/01-early.html]here[/url]) admits in the first round of admission, but even with all of that in mind, I think that that was probably just some weird nondeterministic sort of thing that crops up in college admissions.

I think, pretty much regardless of your 'stats', unless you are an underrepresented minority, you always need to nail the essays, and you need to show that you want to go there.  That's why I've found that most students who get accepted into elite schools get accepted into a lot of elite schools.  There are always a couple who get rejected across the board and get into harvard, but those are comparatively rare.

By the way, you should be careful about posting from people's xangas without asking beforehand - I am pretty sure that I know who that person is, and I don't even follow (non-Canadian) high school contests much anymore.  Anyways, I suspect that she will get into everywhere else she applies, and into harvard rd.

[quote="blahblahblah"]It's hard to measure who the top 20 high school graduates in the country actually are.  How do you rate international olympiad medalists against people who have done original research against people who have spent their time in high school making meaningful contributions against poverty against precocious writers against speech and debate champions against people who are absolutely all around brilliant, but don't fall in to any of the earlier categories?  I tend to think that there's more of a top 200, of which said person undoubtedly falls in.[/quote]

You are right. It is very difficult to judge the top 20. She would fall in the top 200 academically. I hope you agree on that.

Yes, I do.  My sample size is small - I know 4 people who got into Harvard and 3 who are attending, and none of them impressed me that much.  Conversely, I know a half dozen people who were rejected by Harvard and are at Stanford, MIT, Princeton, etc, all of who I found more impressive than the aforementioned people.  Of course Harvard makes a few 'mistakes' in the admissions process - you can't avoid it when you're admitting 1 out of 10 people - but we don't see the applications and we don't know exactly what class compositions they're aiming for.

I will say that the aforementioned applicant will likely be successful wherever she goes.

[quote="rain"][quote="yunorman"]I'm so proud of the kids at my school, 14 people got into Harvard early, and 2 into MIT early :) , 2 into Stanford too, which was particularly suprising considering our school is in Boston MA, and is a public school.[/quote]

yunorman, what school do u go to? 14 people to harvard? thats unheard of in my region. We have most 2 and 1 is always a hispanic minority. thats crazy[/quote]

I go to Boston Latin School. 25+ into Harvard every year.

One thing that is important for applicants to privately operated colleges in the United States to understand is that NOBODY "deserves" to go to those colleges. The colleges set their own admission standards for their own reasons, and in the case of the most famous colleges there are many more applicants than spaces in each year's entering class. The college decides whom to offer admission to based on building an overall college class, NOT on the basis of a strict rank-ordering of each applicant's excellence as a future college student. This way of doing things is particularly puzzling to students whose parents come from Asia, where the system works differently, but that is the system for private college admission in the United States. Each college decides, and any privately operated college can accept or reject whomever it pleases, within only the broadest legal limits.

Good luck to this year's Harvard applicants; Harvard is definitely a fine college at which to study math. Just be sure to apply somewhere else too, especially to some suitable "safety" school where you have a sure chance of admission based on known past behavior of that college's admissions office.

[quote="MithsApprentice"]Do colleges ask for your specific PSAT score? I've never found a clear answer to this, but the general response has been a no.[/quote]

No. In fact, colleges cannot see your PSAT score.

[quote="mel"][quote="mna851"]Why is it that colleges take so many students from the state they are located in?  Why do they like to have so much representation from their home state?[/quote]

Probably has something to do with being a good neighbour too.[/quote]

I think this reason is the main reason.

I think that yield is a far more cynical but accurate reason.  Higher yield means lower admit rates means higher selectivity means higher rankings.

I don't come on this forum much, but I would like to mention that not all math/science stars were deferred by Harvard EA. In fact, two friends of mine who were top 12 in the USAMO last year were both accepted, despite being (almost) entirely math/science focused. Harvard still admits people primarily for math ability; I'm beginning to suspect that there are concrete reasons for those who aren't accepted (in particular, bad essays).

[quote="mna851"]I may be ignorant and unexperienced on the matter, but I think the essays are probably the most important thing for the average applicant.  It's a good way of distinguishing the top students in the nation, and determining which ones they would like to attend their university.[/quote]
I agree. Essays show your ability to communicate with others, which is important in life. It doesn't matter if you're amazingly smart but can't deal with people. I think that good colleges want students who can one day be influential people, and that's what they look for in the essays.

[quote="qvc122"]This person is a 3-time USAMO qualifier (miss this year’s MOP by 1 point/ got 20 on 2005-USAMO), female, 2400 SAT I (first try), 800s SAT II in math IIC, physics and biology and 790 in history, won gold key in writing, 240 PSAT, straight As, ranked #1 out of nearly 800 graduating seniors in her school, playing violin at almost professional level......

The level of her achievements would put her in the top 20 among the high school graduates in the entire country by any standard.  [/quote]

Obviously she is very accomplished by high-school graduate standards, but most of those credentials are redundant.  Any 3-time USAMO is expected to have perfect math/science SAT and to easily be in the top 5-10 of a typical high school, for example.    What is written above makes it sound like that woman is (a) very good in math, but below the top level of USAMO/IMO/Westinghouse winners; (b) solid, conscientious and competitive student (there are thousands of people each year who are #1 in a large high school), and (c) started music at an early age and is also conscientious/competitive at that (not uncommon among high-achieving Asians and other competitive populations).   Basically, a good and hard-working analytical mind with aptitude for math and music.  This is not necessarily top 20 material without further information.

One problem with top high school rank is that there are too many high schools in the United States. One figure I have seen asserts that there are 24,000 high schools in the United States. Another says there are 38,000. When high schools number in the tens of thousands, valedictorians are too numerous to be impressive to the admission officers at the very top echelon of colleges.

i got [url=http://www.nytimes.com/imagepages/2006/01/08/education/edlife/data.1.graphic.html]this[/url] from the my high school website, supposedly it is information volunteered by admissions officers at those schools!

Hi, there. I'm a Romanian student who applied EA to Harvard, and got deferred, also. I don't know exactly why, I got an 800 at the Critical Reading section of the SAT, and have participated in many Math Olympiads and in a bunch of extracurriculars. Rumor has it that Harvard does not accept international students under the Early Action Program? Does anybody have any info on that?

Were you an IMO contestant?  If you weren't, then, given the difficulty of being admitted as an international applicant, that is not especially surprising.

[quote="vlad"]Rumor has it that Harvard does not accept international students under the Early Action Program? Does anybody have any info on that?[/quote]

I talked to the Dean of Admissions at Chicago, and he told me that they usually defer international applicants under the Early program. I'm assuming that this is probably the case for other top colleges, Harvard included.

The elite schools accept a fair number of Canadians in ED/EA, although I'm well aware that Canadians are slightly 'less' international than most international applicants.

Here are some EA news:

Caltech: http://admissions.caltech.edu/news
Stanford: http://news-service.stanford.edu/news/2006/january11/admission-011106.html
Princeton: http://www.princeton.edu/main/news/archive/S13/90/15K38/index.xml?section=topstories
Harvard: http://www.news.harvard.edu/gazette/daily/2006/02/09-admissions.html

Harvard is rather straightforward in that it looks for well rounded people. Whereas MIT is much mroe likely to take USAMO qualifiers just based on that fact, Harvard looks for well rounded people. That said, however, if you are especially accomplished in math or science, you have a very good shot of getting in. But you do realize, most people who get into Harvard aren't total dorky geeks who are brilliant in math or science....Like me...

How I got in, i'll never know... :oops:

I have seen decisions handed down that are entirely beyond my comprehension. It's unfortunate and it happens; I am not alone in this regard.

There is always random chance.  I feel lucky to have gotten in, but I also feel that I played the college admissions game a little more carefully than some.  It is all a game, just like everything else.  Learn to play it!  Essays, interviews, etc.  Everything is a factor.  I feel almost that your goal is to charm your admissions officer through your application and make them feel that you are somehow a "good person".  Not just some facts on a page.  This game is tricky as is anything actually involving people and not just math.  I mean you have to make them feel good too.  Let them know why you want to go there, show that you are going to be a solid alumni for them some day.  After all, these are smart guys, they probably make a system that is pretty good for their purposes.  That purpose may or may not be to admit the academically strong, or whatever other criteria we feel is important.  Then again, I could totally be wrong, but it certainly is the feeling that I get.


How many different 3-letter strings can be formed from the letters of MATHEMATICS (no letter can be used in a given string more times than it appears in the word)?

[hide="Hint"]
Mississippi Theorem
[/hide]

Yeah, I've tried searching everywhere with anything I could, because I know I've seen (and done) similar problems a few years back. Sadly, I couldn't find anything. Could you quote the "Mississippi Theorem"?

[quote="miyomiyo"][hide="Mississippi Theorem"]
Rearrangements of a word with $n$ letters and no repetitions is $n!$.  If one letter is repeated $p$ times, another $q$ times, and so on, the number of arrangements is $\frac{n!}{p!q!r!...}$.
[/hide]

The rest is pretty easy....[/quote]
Doesn't the problem say [b]3-letter[/b] strings?

Oops, I misread the question.

MATHEMATICS

I'd split it up into casework...
[hide="Painful casework"]
Case 1:  A repeated letter is NOT chosen (M, A, T)
$5 \cdot 4 \cdot 3 = 60$
Case 2:  One repeated letter is chosen
$3 \times 5 \times 4 = 60 \times 3 (\text{arrangements}) = 180$
Case 3:  Two of the same repeated letter is chosen, third is not a repeated letter
$3 \times 1 \times 5 = 21 \times 3 = 62$
Case 4:  Two of the same repeated letter, third is a different repeated letter
$3 \times 1 \times 2 = 6 \times 3 = 18$
Case 5:  Two distinct repeated letters, third is not repeated
$3 \times 2 \times 5 = 30 \times 3 = 90$
Case 6:  All three repeated letters are chosen
$3 \times 2 \times 1 = 6 \times 6 = 36$
$60+180+62+18+90+36 = \boxed{446}$
[/hide]I probably went wrong somewhere...

Hmm, I get 399

If we just go off of miyomiyo's cases we have

Case 1:60
Case 2:180
Case 3:45 (not 62-you multiplied wrong)
Case 4: 18
Case 5: 90
Case 6: 6 (wehre did 6*6 come from)

But you dont have to have all those cases-there are two generic cases: three differnt letters or two letters the same with one different

EDIT: easier solution
[hide]For case my case 1: 8*7*6 (all three are differnet, 8 spots for frst, 7 for second...)
Case 2: 3*7*3 (3 choices for two letters the same, 7 for the single letter choice-note they can come from ones with doubles but you just chose only one of them-then 3 for rearrangement)

Add them and you get 399[/hide]

399 is the correct answer, could you post your casework?

Thanks to all!

yea, you can either go off of mimyomo's solution (but thats a little complicated-he/she did it right-just added and multipled worng) or i have another solution which I added to my previous post


Write in a more restrained mode the sum S_n=(2n+1,1)+(2n+1,3)*2^3+(2n+1,5)*2^6+..+(2n+1,2n+1)*2^3n, n>=1 where (n,k)= n!/[k!*(n-k)!].

then prove that the sum S_n is not divizible with 5. what about 3?
thank you very much!
Tiger Sandre,
Germany.

This does not belong on the intermediate forum.

This is an old IMO problem and it's not that intermediate :)

Solution at http://www.kalva.demon.co.uk/imo/isoln/isoln743.html

oh my.. I am sorry I didn't know.. my teacher gave it to me and he said it was simple.. IMO.. not that simple! :D

bye bye


symmetry
AMC
USA(J)MO
USAMO
geometry
perpendicular bisector
geometry proposed

Let a triangle $ ABC$ with $ M,N,P$ are midpoints of $ BC,CA,AB$ respectively. Call $ m,n$ are symmetry line of $ AM$ over perpendicular bisector of $ AB,AC$. Denote $ m\cap n \equal{} K$. Prove that: $ ANKP$ is a cyclic quadrangle.

see USAMO 2008 P2
the resault follows in a second


Let $ H$ be a set of real numbers that does not consist of $ 0$ alone and is closed under addition. Further, let $ f(x)$ be a
real-valued function defined on $ H$ and satisfying the following conditions: \[ \;f(x)\leq f(y)\ \mathrm{if} \;x \leq y\] and \[ f(x\plus{}y)\equal{}f(x)\plus{}f(y) \;(x,y \in H)\ .\] Prove that $ f(x)\equal{}cx$ on $ H$, where $ c$ is a nonnegative number. [M. Hosszu, R. Borges]

Let $ x \in H$ be non-zero.
Then $ f(kx) \equal{} kf(x) \equal{} c(kx)$ with $ c\equal{}\frac{f(x)}{x}$, ($ n \in \mathbb{Z})$.
If $ \{kx| k \in \mathbb{Z}\} \equal{} H$ we are ready.
Else let $ y \in H$ be a element not in this set.
We are ready, when we can show that $ f(y)\equal{}cy$.
$ f(ky) \equal{} c'(ky)$ for a certain $ c'$, and
$ f(k(x\plus{}y)) \equal{} c''k(x\plus{}y)$.
Now $ c''(x\plus{}y) \equal{} f(x\plus{}y) \equal{} f(x) \plus{} f(y) \equal{} cx \plus{} c'y$
$ (c''\minus{}c)x \equal{} (c'\minus{}c'')y$.
Thus $ c'' \equal{} c' \equal{} c$, and $ f(y)\equal{}cy$.

Instead of integers n do you mean natural numbers n, in case H is natural numbers? And please say how the last statement follows.

You should show that c is nonnegative. :thumbup:

You have to add a few lines:
In the case of at least 2 "independent" numbers $ x$ and $ y$, look at thee three numbers
$ x$, $ y$ and $ x \plus{} y$. Two of them have the same sign.
Call them $ x'$ and $ y'$.
Then ... $ (c'' \minus{} c)x' \equal{} (c' \minus{} c'')y'$.
Then $ c'' \minus{} c$ and $ c' \minus{} c''$ must be 0, because $ x'$ and $ y'$ are "independent".

[quote="PeterP"]Let $ x \in H$ be non-zero.
Then $ f(kx) \equal{} kf(x) \equal{} c(kx)$ with $ c \equal{} \frac {f(x)}{x}$, ($ n \in \mathbb{Z})$.
If $ \{kx| k \in \mathbb{Z}\} \equal{} H$ we are ready.
Else let $ y \in H$ be a element not in this set.
We are ready, when we can show that $ f(y) \equal{} cy$.
$ f(ky) \equal{} c'(ky)$ for a certain $ c'$, and
$ f(k(x \plus{} y)) \equal{} c''k(x \plus{} y)$.
Now $ c''(x \plus{} y) \equal{} f(x \plus{} y) \equal{} f(x) \plus{} f(y) \equal{} cx \plus{} c'y$
$ (c'' \minus{} c)x \equal{} (c' \minus{} c'')y$.
Thus $ c'' \equal{} c' \equal{} c$, and $ f(y) \equal{} cy$.[/quote]What do you do when $ H\equal{}\mathbb N$, $ x\equal{}4$ and $ y\equal{}2$? I am wondering if you can give a proof without invoking the taming condition on monotonicity - otherwise we could take some wild function constructed with a Hamel basis.

[quote="Third Edition"]What do you do when $ H \equal{} \mathbb N$, $ x \equal{} 4$ and $ y \equal{} 2$? I am wondering if you can give a proof without invoking the taming condition on monotonicity - otherwise we could take some wild function constructed with a Hamel basis.[/quote]

Not possible. 4 and 2 are "dependent", because then x=2y
I think the proof is correct with the extra changes.
Do you agree?

[quote="PeterP"][quote="Third Edition"]What do you do when $ H \equal{} \mathbb N$, $ x \equal{} 4$ and $ y \equal{} 2$? I am wondering if you can give a proof without invoking the taming condition on monotonicity - otherwise we could take some wild function constructed with a Hamel basis.[/quote]

Not possible. 4 and 2 are "dependent", because then x=2y
I have to make a few changes. :mad:[/quote]
Hmm, maybe I was not too clearly. Let us suppose that $ H\equal{}\mathbb N$ and let us run through the steps of your proof. Then pick some element, say $ x\equal{}4$. You did not specify any special procedure for picking $ x$, so $ x$ is simply 4. Then, by assumption, $ 4\mathbb N\subset H$ (NB: We cannot say that $ 4\mathbb Z\subset H$ - but this is not a crucial point here.) Now we pick $ y\equal{}2$ which is an element in $ H$, but not in $ 4\mathbb N$. Now how do you show that $ c_2\equal{}c_4$ without invoking any other assumption here? Maybe you are implicitly using the monotonicity of $ f$ here? ;)

[quote="PeterP"]I think the proof is correct with the extra changes.
Do you agree?[/quote]
Yes, sure. :-) The idea is pretty neat - and with some really tiny amendment you're winning the coup.  :first:

BTW: There was a real cross-posting going on in the last minutes.  :P

Let $ x \in H$ be non-zero and $ f(x)\ne 0$.
Then $ f(kx) \equal{} kf(x) \equal{} c(kx)$ with $ c \equal{} \frac {f(x)}{x}$, ($ k \in \mathbb{N})$.
If $ \{kx| k \in \mathbb{N}\} \equal{} H$ we are ready.
Else, let $ y \in H (y\ne 0)$ be a element not in this set.
We assume that $ f(y) \equal{} c'y$ with $ c'\ne c$.
If there is no such element, we are ready.
Statement:
If $ xy > 0$, then $ x\mathbb{Z} \cap y\mathbb{Z} \equal{} \emptyset$
Proof:
Suppose $ kx \equal{} my$, then (wlog) $ k > 0, m > 0$ and $ ckx \equal{} f(kx) \equal{} f(my) \equal{} c'my \equal{} c'kx$.
But $ c'\ne c$.

The requirement $ xy > 0$ can always be fullfilled. Proof:
Suppose that $ xy < 0$, then in the set $ \{x,y,x \plus{} y\}$ there are always two elements with the same sign.
If (e.g.) the two elements are $ x$ and $ x \plus{} y$, then $ cx \plus{} c'y \equal{} f(x \plus{} y) \ne c(x \plus{} y)$.
Call these two numbers $ x$ and $ y$ again.

$ c'\ne 0$, because $ kx < my < nx$ implies $ ckx < 0 < cnx$ (impossible).
Suppose wlog that $ c > c' > 0$ and $ x > 0$ and $ y > 0$.
If $ kx \le my$ then $ ckx \le c'my$.
Or: $ 1 \le \frac {m}{k}\frac {y}{x} \rightarrow \frac {c}{c'} \le \frac {m}{k}\frac {y}{x}$

We can alway find $ \frac {m}{k} \in \mathbb{Q}$ so that $ 1 < \frac {m}{k}\frac {y}{x} < \frac {c}{c'}$.
Conflict. Ready.


trigonometry
number theory proposed
number theory

Let $\theta,\phi \in (0,\frac{\pi}{2})$, in that way that $tg\theta, tg\phi \in \mathbb{Q}$. Solve the equation $sin^{2}\theta cos^{2}\theta =ctg^{2}\phi$.

$\sin \theta \cos \theta = \frac{1}{2}\sin 2\theta = \frac{\tan \theta}{1+\tan^{2}\theta}$

Hence let $\tan \theta = p, \tan \phi = q$.  Then we wish to solve

$\frac{p}{1+p^{2}}= \pm q \implies$
$q = \pm \frac{1+p^{2}}{p}$

For $\theta, \phi \in (0, \frac{\pi}{2})$ the range of $\tan$ is in the positive reals, so for every $p$ this gives a corresponding $q$ (the positive root).


How many ways can the word $BASKETBALL$ be arranged?

How many ways can the word be arranged with atleast one $A$ in the first slot?

How many ways can the word be arranged with both $A$s in the first two slots?

There are 10 letters.  That means there are 10 spots to fill.  You're picking a spot for each letter.  Dealing with the B's first, since you have 2 of them, there are ${10\choose2}$ ways to place them, assuming they are indistinct.  With two spots taken, you have 8 left.  So for the A's, there are ${8\choose2}$ ways to place those.  Next, there are ${6\choose2}$ ways to place the L's.  Now there are 4 spots left with 4 different letters.  So there are $4!$ ways for these.  You could also do them individually and get the same answer:

${4\choose1} \cdot {3\choose1} \cdot {2\choose1} \cdot {1\choose1} = \frac{4!}{3!} \cdot \frac{3!}{2!} \cdot \frac{2!}{1!} \cdot \frac{1!}{1!} = 4!$

The total number of combinations:

${10\choose2} \cdot {8\choose2} \cdot {6\choose2} \cdot 4! = \frac{10!}{2 \cdot 8!} \cdot \frac{8!}{2 \cdot 6!} \cdot \frac{6!}{2 \cdot 4!} \cdot 4! = \frac{10!}{8} = 453600$

I think you can figure out the other ones on your own.   ;)

Hmm...
Is it incorrect looking at it this way?
You have 10 letters.  Therefore, there are $10!$ ways to arrange them.  However, you have 2 A's, 2 B's, and 2 L's.  Thus, each possible permutation is counted 8 times, twice because of the B's, twice because of the L's, and twice because of the A's.  This makes it $\frac{10!}{8}$.  Something I'm missing?
Oh no wait...just realized that we get the same answer, you wrote $\frac{10!}{6}$ instead of $\frac{10!}{8}$,  :D ...remember that 2*2*2=8 instead of 6...


product A has a ratio of 1:2 to product B. product B has a ratio of 3:2 to product C. product C has a ratio of 4:7 to Product E. product E has a ratio of 2:5 to product F. product F has a ratio of 11:9 to product D. find the ratio of product A to product F.

Your two ratios between product E and F make no sense

[quote="mathgeniuse^ln(x)"][quote="Captain Sugar"]that's just a trick to confuse people. ;)[/quote]

[hide]Then I guess the answer is "no solution" since the ratios between E and F are completely messed up and you can't find the ratio between A and F.[/hide][/quote]

oops, there was a typo. i'll fix it :oops:

[quote="SplashD"]After you fixed the problem, the answer is:
[hide]$(1/2)(3/2)(4/7)(2/5)=\boxed{\frac{6}{35}}$[/hide][/quote]

Yeah, thats what I got.  I guess we got it right.


advanced fields
advanced fields unsolved

Let $ X$ and $ Y$ be topological spaces and let $ A \subset X$ and $ B \subset Y$. Is it true that the quotient spaces $ (X \times Y)/(A \times B)$ and $ X/A \times Y/B$ are equal?

Nope. In fact, it's very far from being true. To take a simple example, what happens if $ X \equal{} Y \equal{} [0,1]$ and $ A \equal{} B \equal{} \{0,1\}$ are the boundary points?

What do you mean by a quotient space when $ A$ is a subset of $ X$?  The natural definition of quotient space is quotienting by an equivalence relation; here $ X/A$ means the set-theoretic difference (at least, as I and probably randomgraph interpret it), which doesn't behave much like a quotient at all.

I much prefer seeing quotient operations defined in terms of equivalence relations, but it's fairly common to define $ X/A$ as $ X/\approx$ where $ \approx$ is just the equivalence relation which identifies all the points of $ A$ and leaves the other points untouched. In other words, it just collapses $ A$ to a point.


Let $a,b,c >0, ab+bc+ca=1$. Prove that:
$\frac{1+ab}{a+bc}+\frac{1+bc}{b+ca}+\frac{1+ca}{c+ab}\le \frac{12}{\sqrt{3}+1}$

[quote="manlio"]Can you please post your solution, NguyenDungTN ?

Thank you very much.[/quote]
Sorry manlio, I can't solve it. I'm improve it from following inequality:( I can't solve it, too  :oops: ) :
Let $ a,b,c>0, ab\plus{}bc\plus{}ca\equal{}1$. Prove that:
$ \frac{ab\plus{}c}{1\plus{}bc}\plus{}\frac{bc\plus{}a}{1\plus{}ca}\plus{}\frac{ca\plus{}b}{1\plus{}ab} \ge \frac{3(\sqrt{3}\plus{}1)}{4}$
And:
$ \frac{a\plus{}bc}{1\plus{}ab}\plus{}\frac{b\plus{}ca}{1\plus{}bc}\plus{}\frac{c\plus{}ab}{1\plus{}ca} \ge \frac{3(\sqrt{3}\plus{}1)}{4}$
Can Argady, Ji Chen,... help me?

or if we can just proof that : 9*(abc)^1/3 >= V3 +1  :oops:

[quote="NguyenDungTN"]Let $ a,b,c > 0, bc \plus{} ca \plus{} ab \equal{} 1$. Prove or disprove that

$ \frac {1 \plus{} ab}{a \plus{} bc} \plus{} \frac {1 \plus{} bc}{b \plus{} ca} \plus{} \frac {1 \plus{} ca}{c \plus{} ab}\le \frac {12}{\sqrt {3} \plus{} 1}$.[/quote]
Put $ a \equal{} t^2,b \equal{} t,c \equal{} \frac {1 \minus{} t^3}{t \plus{} t^2}(0 < t < 1)$, we have

$ \frac {1 \plus{} ab}{a \plus{} bc} \plus{} \frac {1 \plus{} bc}{b \plus{} ca} \plus{} \frac {1 \plus{} ca}{c \plus{} ab} \equal{} \frac {(1 \plus{} t)\left(1 \plus{} 3t^2 \minus{} 2t^3 \plus{} 4t^4 \minus{} t^5 \plus{} 3t^6 \minus{} 2t^7 \plus{} t^8 \plus{} t^9\right)}{t\left(1 \plus{} t^2\right)\left(1 \minus{} t^3 \plus{} t^4 \plus{} t^5\right)}$

$ \to \plus{} \infty(t\to 0^ \plus{} )$.

By the way, if $ a,b,c$ are sides of a triangle with $ bc \plus{} ca \plus{} ab \equal{} 1$, then the inequality holds.

[quote="NguyenDungTN"]Let $ a,b,c > 0, ab \plus{} bc \plus{} ca \equal{} 1$. Prove that:
$ \frac {1 \plus{} ab}{a \plus{} bc} \plus{} \frac {1 \plus{} bc}{b \plus{} ca} \plus{} \frac {1 \plus{} ca}{c \plus{} ab}\le \frac {12}{\sqrt {3} \plus{} 1}$[/quote]
I think it can is solve by CBS
we must prove
$ \sqrt{((1\plus{}ab)^2\plus{}(1\plus{}bc)^2\plus{}(1\plus{}ac)^2)(\sum \frac {1}{(a\plus{}bc)^2})} \geq \frac {12}{\sqrt{3}\plus{}1}$
Now, we can use $ p,q,r$
ok!
I hope that CBS can solve it because it not difficully

Ji Chen means that: the first inequality holds if $ a,b,c$ are sides of a triangle with $ ab \plus{} bc \plus{} ca \equal{} 1$.
This is the reason why I think vipcd_A1's inequaliy is wrong, I think it is only correct when $ a,b,c$ are sides of a triangle such that $ ab \plus{} bc \plus{} ca \equal{} 1$. Of course I have not try it too  :D
Anyway, they are really interesting. Note that the equality holds when $ a \equal{} b \equal{} c \equal{} \frac {1}{\sqrt {3}}$ so it is hard to using Mixing Variable method because if you want to using MV method, you will prove that:
\[ f(a,b,c)\le f(a, \minus{} a \plus{} \sqrt {a^2 \plus{} 1}, \minus{} a \minus{}\sqrt {a^2 \plus{} 1})
\]
So I think it is not the best way ... I will try another way.

Ji Chen, I'm not sure whether your sustitution is correct, as for any two values of $ a$ and $ b$ we can assign a value for $ c$, hence $ a$ and $ b$ don't depend on the one variable.


college AMC USA(J)MO USAMO calculus AIME MIT

Okay, im a senior in hs, hoping on transfering to uc berkely as a freshmen in college
i took calc and got a five junior year
pretty confident that i'll make the usamo this year, been scoring high on practice tests for amc.  maybe Mosp
710Math SAT
A's in math classes
wrote math research papers- love the subject
self studied calculus, number theory and been doing good on standardized testing

here comes my verbal and other stuff profile
I get C's in english
my gpa is slightly over a 3.8 weighted
i have a 430 verbal SAT (1140 combined)
1 and and 2 in psychology and history ap scores
low standardized scores.

what are my chances into a good school.  prolly nothin
thanks

[quote="tongos"]pretty confident that i'll make the usamo this year, been scoring high on practice tests for amc.  maybe Mosp
710Math SAT [/quote]

So....I'm going to be honest.  There's a little problem here.  The AMC is far, far, far, far easier than either the AIME or USAMO.  Scoring highly on the AMC means...little.  Well, it means something.  It does not mean USAMO and certainly does not mean Mosp.

Also, I'm probably going to be flamed for this statement, a lot, but I don't imagine that many people who take the USAMO got a 710 math SAT.

You noted that wanted to transfer.  That is possible.  You likely won't be accepted anywhere exceptional right now.  However, if you do well in your classes as a freshman or as a sophomore you could transfer up.  I don't think colleges even look at your SAT scores if you are a transfer student.

thanks for the feedback,
no though, your probably right, the SATmath score of 710 is probably not the common score among usamo participants.  But i've been studying my pants off for these contests and during school hours, and that 710 in math, that was a long time ago before i became interested in mathematics.

Your right, amc contests are a lot easier than that american invitational .  those questions are a little tough, i seem to be getting 7-8 right on those free response tests.  As for the usamo, i probably stand no chance for the mosp.  I can't guarentee that i'll make it to usamo, but i can say that i have a good chance, as my practice results show.

[quote="tongos"]and that 710 in math, that was a long time ago before i became interested in mathematics.  [/quote]

In that case, I entirely understand your sentiment, and apologize.  I've done the exact same thing.  Good luck.  Though I would retain the last paragraph.

Those C's in English and the 430 Verbal aren't exactly very impressive for a good school. Neither is the 1 or 2 on the APs. Most exceptional schools look at all aspects of a student's profile. Math is important, but it is still only a part of the entire package.

I think right now, you really should focus on the English. Doing relatively well in all subjects is better than doing exceptionally well in one and flunking all others.

Good luck!  =]

[quote="Magnara"]You noted that wanted to transfer.  That is possible.  You likely won't be accepted anywhere exceptional right now.  However, if you do well in your classes as a freshman or as a sophomore you could transfer up.  I don't think colleges even look at your SAT scores if you are a transfer student.[/quote]Depending on the school, they'll look at not just your SATs, but also your HS transcript even if your a sophomore in college looking into transferring.

One of the ironic things about math-centered schools is that because so many people apply with good math scores, they'll look at humanities a lot.

What mirebeka said squares with what I have heard about MIT admissions from a local parent who contacted MIT's admission office about what level of preparation is good for that school. EVERYONE applying to MIT has strong math scores, so the way to stand out in that applicant pool is to have strong verbal scores to outrank other applicants who are at your level of math scores. MIT also asks that at least one letter of recommendation come from a teacher of humanities.

If you're a senior, then it's a bit late to pull up your other scores. But if you could make the USAMO, I think that would have much impact. Either way, good luck!


In problem 4 of round 2 (the duck, duck, goose problem), it just says find all of the values of n. Do you want us to list the possible values of n, e.g. (2,5,13,47...) or write a statement that says what n can be e.g. (n>2 but divisible by 7 and n<49)?
Thanks

uh..... I believe you have to figure it out yourself....

Just decide for yourself whether you think it's more effective to list possible values or state a pattern or what. If you get the right solution I doubt you would get counted off for the way you wrote the answer.

This is the kind of question we cannot answer while a round is in progress in fairness to all participants.


The pressure $ (P)$ of wind on a sail varies jointly as the area $ (A)$ of the sail and the square of the velocity $ (V)$ of the wind. The pressure on a square foot is $ 1$ pound when the velocity is $ 16$ miles per hour. The velocity of the wind when the pressure on a square yard is $ 36$ pounds is:

$ \textbf{(A)}\ 10\frac {2}{3} \text{ mph} \qquad\textbf{(B)}\ 96 \text{ mph} \qquad\textbf{(C)}\ 32\text{ mph} \qquad\textbf{(D)}\ 1\frac {2}{3} \text{ mph} \qquad\textbf{(E)}\ 16 \text{ mph}$

[hide="Click for solution"]
$ P\equal{}kAV^2, 1\equal{}k \cdot 1 \cdot 16^2 \implies k\equal{}\frac{1}{16^2}$
$ \frac{36}{9}\equal{}\frac{1}{16^2} \cdot 1 \cdot V^2 \implies V\equal{}32$, or $ \boxed{\textbf{(C)}}$.
[/hide]


algebra
polynomial
calculus
derivative
number theory
greatest common divisor
algorithm

For which $ n\ge{1}$ does the polynomial $ f(x) \equal{} (x \plus{} 1)^n \minus{} x^n \minus{} 1$ have a multiple root?

Since this is the Real Analysis forum, I assume $ x$ is real. Since the case $ n\equal{}1$ is trivial, I assume $ n\ge 2$. The derivative $ f'(x)\equal{}n((x\plus{}1)^{n\minus{}1}\minus{}x^{n\minus{}1})$ vanishes only at $ x\equal{}\minus{}1/2$, and only if $ n$ is odd. But for such $ n$ we have $ f(\minus{}1/2)\equal{}2^{1\minus{}n}\minus{}1$ which is not zero.

how did you calculate the gcd? it looks like the euclidian algorithm but im not able to see what did. sorry.  :(

Let $ f \equal{} f(x), f' \equal{} \frac{f'(x)}{n}$.  Then

$ \gcd(f, f') \equal{} \gcd(f \minus{} (x \plus{} 1) f', f') \equal{} \gcd(f \minus{} (x \plus{} 1) f', f \minus{} x f')$


In order to crack nuts bare-fisted, one usually takes two nuts into one's hand at the same time. Why is it easier to crack a nut with this method?

hmm I'm not sure that's a good question... beacuse P=F/A and A is smaller when you put 2 nuts together, so the effective pressure is larger and overcomes the nuts surface resistance?

Cracking the shell is done by stress: uneven compression and tension on it.  This is difficult to do between bare hands which act like pillows, spreading and evening the forces.  Having each nut provide a hard and small point of contact to at least one side of the other nut allows the forces to be concentrated and applied unevenly to both nuts.  The weakest cracks first.  Similar methods are used to split wood, deliver karate blows, and penetrate surfaces with arrows and bullets.

Yep...I don't think this really needs physics analysis...just because the two nuts kill each other.


A loader has a waggon and a little cart. The waggon can carry up to 1000 kg, and the cart can carry only up to 1 kg.  A finite number of sacks with sand lie in a storehouse. It is known that their total weight is more than 1001 kg, while each sack weighs not more than 1 kg. What maximum weight of sand can the loader carry in the waggon and the cart, regardless of particular weights of sacks?

[i]Author: M.Ivanov, D.Rostovsky, V.Frank[/i]

I think that the number is $ 1000\plus{}k$ with $ k>0$ and $ k\to0$
The clue is that the wagon carries a weight $ >999$ Then you can see that the maximum is $ >1000$ And then if all sacks weigh $ 0.5\plus{}k$
 $ k>0$ and $ k\to0$ you reach the number
I don't know if the idea is correct but seems interesting

Daniel