Regrettably your answer is incorrect. :(

You should edit your question to find the value"s" i think....

"Politics/Economics"? Whoa, these Japanese University entrance exams are hard!

[hide]perhaps something like $4\cos 72k^o$  where  k=0,1,2
[hide]Of course,  $4 cos 72^o$ is somethng like $\sqrt 5 - 1$  so you could also say the other two values are $-1 \pm \sqrt 5$
[/hide]
[/hide]

[quote="xxreddevilzxx"]Well the way i did it was by very ugly algebra and mass calculation, then using the fact tha $x^5=1$ into reducing powers of remaining $x$ and then finally using $x^5-1=0$ as a way to chomp down a lot of factored terms. I will show the work at a later time..... but here is my answer:

[hide]1[/hide][/quote]
FWIW There may be (and is)  elegant methods  but  even "ugly algebra"  and brute force is not too hard,  if one remembers that:
[hide]$x^5= 1$  and  either $x=1$ or $1+x+x^2+x^3+x^4=0$  and the second equation is same as $y^2+y-1=0$ (where y=x+1/x) and final answer gets to be 2y :)  [/hide]

Hello, guys! Thank you for your replies. :)

I will give a hint. Remark that $1+4=2+3=5.$

Good luck!

kunny

[quote="Gyan"]perhaps something like $4\cos 72k^o$  where  k=0,1,2[/quote]

How?

[quote]How?[/quote] Just a fancy way to say $2(x+1/x)$  :)

[hide]Doesn't matter, if you use  [hide="trig or algebra"]and write $x=cos a + i sin a$ where a=k 72^o [/hide]  ,   its not too  hard, and  if you are lucky you may even notice:
[hide="this"]$\frac 1 {1+x} + \frac{x^3}{1+x^4} = \frac x {1+x^2} + \frac{x^2}{1+x^3} = \frac 1 x = x^4$ [/hide] right away .
[hide]$\frac 1 {1+x} + \frac{x^3}{1+x^4} = \frac{x^5} {1+x} + \frac{x^4}{x+x^5} = \frac{x^5+x^4}{1+x} = x^4$ [/hide]

[/hide]


Find all polynomials [tex]f(x) = x^n + a_{n-1}x^{n-1}+...+a_{1}x+a_{0} , a_{0} \neq 0[/tex] with integer coefficients such that
[tex]f(a_i)=0, i=0,1,...,n-1[/tex]

using the fact that if $ \frac {p}{q} $ is a root of an eqn, then $p$ divides $a_0$ and $q$ divides $a_n$ Using this we get that all of $ a_i \ i=0,1,..... n-1 $ divide $a_0$

Also the factorisation of the eqn is $ \prod _{i=0} ^n ( x - a_i) $

Also we have $ a_0 a_1 a_2 .... a_{n-1} = (-1)^n a_0 $

Now if $a_0 $ is zero , it is trivial for all nos. will divide zero (obviously)

thus we get that if $ a_0 \not= 0 $,

$ (a_1 a_2 .... a_{n-1} - (-1)^n ) = 0 $
Thus if n is even, we have that an even no. of $ a_i $ can be $-1$ and the rest should be $1$ , since $a_i$ are integers

Also, if n is odd, the situation is the opposite of the case for n = even

Is this right????? I believe that's the best we can get!!!

we know that a[0] is not 0, so for all i in{1,..,n-1} a[i] is 1 or -1,now if a[0] is 1 or -1 ,let p the number of 1 and q is the number of -1 between the root of our polynom, with p>q(because a [n-1] is not 0) we get that p-q=1  and  p(p-1)/2 + q(q-1)/2 -pq=-1 ( by the formuls between roots and coefficients) we get (p,q)=(2,1) the symetric case p<q gaves(p,q)=(1,2) so the solutions are P=(X-1)²(X+1) and P=(X-1)(X+1)²
  but if |a[0]|>1 then using 0=|P(a[0])|> |(a[0])^n|-....-|a[1]|-|-|a[0]|-1>-2 we get
|a[2]|=2 and that a[i]=(-1)^(i+1) for i in {1,...,n-1} so a[2]=-2(using the rull of signe), if n>2 then a[2]=-1 but then
P(-1) will not 0 so n is less than 2 , we get the solution P=(X-1)(X+2)


If $a_{1}, a_{2}\cdots , a_{n}$ are distinct integers, prove that the polynomial $(x-a_{1})(x-a_{2})\cdots (x-a_{n})-1$ is irreducible over the integers.

[hide]Suppose it can be factored as , where . Then  for all . Since  and  are factorizations over the integers, then  and  are both integers, and thus they must be either 1 or -1. In fact, for all  we have . Since  is zero at  different points, it must have at least  roots. However, it is well known that if , then , but then , but  is at least degree , so we have a contradiction, and the polynomial cannot be reduced over the integers.[/hide]


I need help solving this problem..

the colored part and the combined circle ( the middle part ) part have the same areas....

what's the length between the point A nad point B?

( it was from a korean worksheet, so I did my best to translate it.. :D  )

Wait, is it the sum of both colored parts equals the middle part, or one of the colored part equals the middle part?

I'm pretty sure it's both, but the book doesn't say it clearly....

that's one of the reasons y I asked AoPS

[quote="GoBraves"]Ok if its both combined, then I got 2 :pi:  as the answer. Ok here's how I did it-

Lets assume AB as x. The area of the rectangular part in the middle is 8x. The area of half of each circle is
8:pi: . So we can write the quation-
8x=8:pi: +8:pi: (We dont need to add the colored part because the area of the comined part equals the colored part)

Therefore x=2:pi:[/quote]

Wait, but if the area of the rectangular part is 8x, and you're saying 8x=16pi, doesn't 16pi only represent the area of the two half circles? And don't they overlap? And what about the shaded portion?

Oh wait. I think I get it. The overlap is the same as the shaded portion! Right?  :blush:


Prove that the polynomial $ P(x)\equal{}x^{n\plus{}1}\minus{}(x^{n\minus{}1}\plus{}x^{n\minus{}2}\plus{}...\plus{}x\plus{}1)$ has only one positive root, that we call it $ a_n$ Prove that the sequence has the limit $ \frac{1\plus{}\sqrt{5}}{2}$

[quote="TZF"][hide="Hint"]$ (x \minus{} 1) P$[/hide][/quote]

Am-Gm of bracketed term: =a(r^n-1)/r-1=(x^n-1)/x-1

Therefore, P(x)=x^n+1-(x^n-1)/x-1

(x-1)P(x)=x^n+1(x-1)-(x^n-1)

(x-1)P(x)=x^n+2-x^n+1-x^n+1

Thats all i got... can someone else work off that maybe?


Show that every positive integer can be written as a sum of no more than $ 53$ fourth powers of integers.

Before someone starts to post the proof with densities ;)
Note that $6\cdot (x_1^2+x_2^2+x_3^2+x_4^2)^2 = \sum_{1 \leq m < n \leq 4} (x_m+x_n)^4 + (x_m-x_n)^4$.
Let $n$ be our integer. Write $n=6q+r$ ($0 \leq r <6$) and then $q=a_1^2+a_2^2+a_3^2+a_4^2$ (the last one by the Four-Square-Theorem).
Every term of type $6a_i^2$ can be written as $6a_i^2=6(b_{i,1}^2+b_{i,2}^2+b_{i,3}^2+b_{i,4}^2)^2$ after writing $a_i$ as sum of four squares.
Thus, by the above, $6a_i^2$ can be written as sum of $12$ fourth powers. Now $6q=6a_1^2+6a_2^2+6a_3^2+6a_4^2$ is a sum of $48$ fourth powers, and $r$ is the sum of at most $5$ ones, thus in total $n$ is a sum of $53$ fourth powers.

Note that by such identities one gets a nice proof that when every integer can be written as sum of enough $n$-th powers, then every integer can be written as sum of enough $2n$-th powers.


abstract algebra
superior algebra
superior algebra solved

find all [b]idean[/b] of rings $\mathbb{Z}*\mathbb{Z}$.
 :(

Do you mean ideals of the ring $\mathbb{Z}* \mathbb{Z}$? How is this ring defined?

I'm sorry.I'm wrong that is [b]ideals[/b] not [b]ideans[/b] :lol:

Let $I$ be an ideal of $\mathbb Z^{2}$.
Let $n\mathbb Z$ and $m\mathbb Z$ be the projections of $I$ on the two components.
There exists in $I$ an element of the form $(n,u)$ and an element $(v,m)$ for some $u,v\in\mathbb Z$.
Since $I$ is an ideal, $I$ contains $(n,u)\times(1,0)=(n,0)$ and similarly $I$ contains $(0,m)$.
It is now obvious that $I=n\mathbb Z\times m\mathbb Z$.

Ideals in product rings are very easily described!

Notice that AxB = A+B (direct product = direct sum)

Let s1 : A -> A+B and s2 : B -> A+B
     p1 : AxB -> A and p2 : AxB -> B

Then we have id(A) = p1*s1 and id(B) = p2*s2

Also id(A+B) = s1*p1 + s2*p2

Let I be any ideal in A+B and a = p1(I) and b = p2(I) the ideals in A and B respectively. We compute I = id(I) = (s1*p1 + s2*p2)(I) = (s1*p1)(I) + (s2*p2)(I) = s1(a) + s2(b) = a+b (direct) = axb (direct).

We conclude that all ideals in A+B are of the form a+b with a and b ideals in A and B respectively.

Sorry, I don't know latex.

That's right :). Of course, it can be generalized to arbitrary finite products.

A counterexample for infinite products:

$R : = \prod_{n \in \mathbb{N}}\mathbb{Z}/2 ,~ I : = \{r \in R : \{n \in \mathbb{N}: r_{n}= 1\}\text{~ is finite~}\}$

This comes from the correspondence between ideals in products of fields and the filters of the index set; here one uses the frechet-filter, which is not generated by any subset.

Is the conclusion really correct? There is something wrong with the proof, no?

Let f and g be homomorphisms A -> B of commutative rings with 1. When is it true that (f+g)(I) = f(I) + g(I) for any ideal I?

Yes the proof has the same error as discussed in the other thread. But you can easily correct the proof.

Here's a corrected proof to clean up this mess:

We want to prove that the only ideals in AxB are axb where a and b are ideals in A and B respectively.

Let I be any ideal in AxB and put a = p1(I), b = p2(I). Suppose (x,y) is in I. Then so is (x,y)(1,0) = (x,0). Similarly, if (x',y') is in I, so is (0,y'). Therefore (x,0) + (0,y') = (x,y') is in I. But then axb < I. The converse is obvious so I = axb.

This proof also shows why the result doesn't carry over to abelian groups.  :)

In the infinite case direct product and direct sum do not coincide and ideals in product(Ai) taken over an infinite index set are not determined by the projection ideals in the components. If I is an ideal with components ai, there are at least two ideals with these components: product(ai) and sum(ai). These are different.

Yes, that's right :-). One also has to prove, that a,b are ideals (images of ideals are not always ideals), but this is no problem. Besides, the characterization also works for left- or rightideals.

The projections are surjective so ideals are taken to ideals.

You are right, commutativity is not required. But in the proof given it is essential to have 1.


Are there any home school MathCounts teams in the NorthEast/ MetroWest area?

We are looking to join one or form one if there is enough interest in the Lexington/ Bedford area.

Thanks.

I know Voyagers has one, but I don't really know much about it.

I would like to mention that if one lives in Lexington then we would be extremely supportive of any decision to abandon the (relative) freedom, enjoyment, and overall excellent education of home-schooling and attend an overcrowded public school with ridiculous class sizes, stupid classes, and my personal favorite, the smoking corner. Just because we have an awesome math team.

 :)

On a slightly more relevant note, Voyagers is actually based in Acton, not Lexington/Bedford; however they do in fact send a team which competes in the Metrowest chapter. However you probably have to join them. Mr. J.S. Trabucco (perfect628) and I, way back in 1974 (ok maybe it was 2006) went to nationals with someone who competed with them...shoot me a PM and I'll see if I can find her e-mail address.

also I noticed that Mr. J. S. Trabucco has the same first two initials as Mr. J. S. Bach


geometry
3D geometry
octahedron
probability
symmetry
rotation
counting

Six ants simultaneously stand on the six vertices of a regular octahedron, with each ant at a different vertex. Simultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each with equal probability. What is the probability that no two ants arrive at the same vertex?

$ \textbf{(A)}\ \frac {5}{256} \qquad
\textbf{(B)}\ \frac {21}{1024} \qquad
\textbf{(C)}\ \frac {11}{512} \qquad
\textbf{(D)}\ \frac {23}{1024} \qquad
\textbf{(E)}\ \frac {3}{128}$

We approach this problem by counting the number of ways ants can do their desired migration, and then multiple this number by the probability that each case occurs.

Let the octahedron be ABCDEF, with points B,C,D,E coplanar. Then the ant from A and the ant from F must move to plane BCDE. Suppose, without loss of generality, that the ant from A moved to point B. Then, we must consider three cases.

Case 1: Ant from point F moved to point C
On the plane, points B and C are taken. The ant that moves to D can come from either E or C. The ant that moves to C can come from either B or D. Once these two ants are fixed, the other two ants must migrate to the "poles" of the octahedron, points A and F. Thus, there are two degrees of freedom in deciding which ant moves to D, two degrees of freedom in deciding which ant moves to C, and two degrees of freedom in deciding which ant moves to A. Hence, there are 2*2*2=8 ways the ants can move to different points.

Case 2: Ant from point F moved to point D
On the plane, points B and D are taken. The ant that moves to C must be from B or D, but the ant that moves to E must also be from B or D. The other two ants, originating from points C and E, must move to the poles. Therefore, there are two degrees of freedom in deciding which ant moves to C and two degrees of freedom in choosing which ant moves to A. Hence, there are 2*2=4 ways the ants can move to different points.

Case 3: Ant from point F moved to point E
By symmetry to Case 1, there are 8 ways the ants can move to different points.

Given a point B, there is a total of 8+4+8=20 ways the ants can move to different points. We oriented the square so that point B was defined as the point to which the ant from point A moved. Since the ant from point A can actually move to four different points, there is a total of 4*20=80 ways the ants can move to different points.

Each ant acts independently, having four different points to choose from. Hence, each ant has probability 1/4 of moving to the desired location. Since there are six ants, the probability of each case occuring is (1/4)(1/4)(1/4)(1/4)(1/4)(1/4)=1/4096. Thus, the desired answer is 80(1/4096)=5/256.

A.

[hide="simple solution"]There are of course, $4^6$ possible movements.

Each movement in which ants from point move to adjacent point can be generalized with a pattern.

Each movement will be like a continuous line of arrows. Starting from a tip, going down to the $4$ point area, there are $4$ options. Then $2$, choosing the direction of flow. Then finally, $2$ more for each of these variations as it can be from bottom to top or top to bottom. Therefore, each possible movement has $4*2*2 = 16$ variations.

$\frac{16n}{4096}$ is the form that probability must be in, where n is number of possible movements.

Thus $P$ of the event must be in the form: $\frac{n}{256} = \boxed{A}$[/hide]

Breez, I don't think your solution ensures that the answer is A.  If $n$ was even, then the fraction could be further reduced, so E could also be a possible answer.

Let the points $ C,D,E,F$ form a square, and point $ A$ above the square and point $ B$ underneath the square. So the six points form a octahedron.

After some observations, we see that the ants can be seprate into two groups with each group forming a triangle opposite the other triangle (ex. triangle $ ACD$ vs triangle $ BEF)$
There are $ 4$ such pairs of triangles, and since on each triangle the ants can either rotates clockwise or anticlockwise there are a total of $ 4*2*2 \equal{} 4^2$ possible ways.

Another way is to have four points forming a diamond and the rest two points forming a line segment (ex. $ ADBC$ vs $ EF$). There are $ 4$ such pairs and again the ants can either choose to rotate clockwise or anticlockwise in the diamond and can either choose to moves from left to right or right to left on the line segment, there are a total of $ 4*2*2 \equal{} 4^2$ ways.

The last way is to have a single path starting from any point and ending at that point again while traveling through all the vertices only once. An example of such path is $ A \minus{} D \minus{} B \minus{} C \minus{} F \minus{} E \minus{} A$. There are $ 4*6$ such paths since they can start at any of the six vertices. But again since this is symmetric each path can either be traveling in right or left directions which gives a total of $ 4*6*2 \equal{} 3*4^2$ ways.

This is out of $ 4^6$ so the probability is $ \frac {5*4^2}{4^6} \equal{} \boxed{\frac5{256}}.$

First note that there are $ 4^6$ possible ways the ants can move

Now we consider their final destination.
Consider an octahedron $ ABCDEF$ with opposite vetecies; $ AB, CD, EF$
An ant on $ A$ or $ B$ can move to $ C,D,E,F$,
Similarly an ant of $ C$ or $ D$ can move to $ A,B,E,F$
Ants on $ E$ or $ F$ can move to $ A,B,C,D$

So all we need to ask is;
how many ways can we pick two elements from each of $ \{A,B,C,D\},\{C,D,E,F\},\{A,B,E,F\}$ in order to make $ \{A,B,C,D,E,F\}$

Well every elements can come from one of two sets so we have $ 2^6$ ways, but we cannot have $ 0,1, 3$ or $ 4$ elements comming from the same set so,(from symetry) we have over counted by $ 3\cdot 4 \plus{} 4\cdot 3 \equal{} 24$

So there are a total of $ 2^6 \minus{} 24$ ways to pick $ A,B,C,D,E,F$, but each one could have come from either of two opposite vertecies so there are $ 2(2^6 \minus{} 24)$ ways for the ants to move

$ \Rightarrow \frac {2^7 \minus{} 48}{4^6} \equal{} \frac {2^4(8 \minus{} 3)}{2^{12}} \equal{} \frac {5}{256} \longrightarrow \boxed{A}$

I don't understand this part of ocha's solution: "but we cannot have 0,1, 3 or 4 elements comming from the same set so,(from symetry) we have over counted by $3\cdot 4 + 4\cdot 3 = 24$." Can someone help clarify? Thanks.

@cheesyicecream
Please do not revive old threads, start new ones with a link to the old one.

Best to think of this problem in terms of closed loops in the octahedron, ensuring that whenever an ant moves to a new vertex, that vertex's occupant has also left.

Case 1: There is 1 closed loop. For a point you choose, the path starts out and ends at the same point without passing vertices twice. If you count them out, there are $4\times(2+2\times3)=32$ of them.

Case 2: There are 2 closed loops. These can either be along a single edge and a quadrilateral, or on two triangles. There are 12 edges on which to create a closed path and they all go 1 way, while the resulting quadrilateral can be either "clockwise" or "counterclockwise," Making the total $12\times2=24$. For two triangles, there must be one on one of the top faces of the octahedron and one on one of the bottom faces, making 4 pairs of triangles that can all go clockwise or counterclockwise: So $4\times2\times2=16$ paths. For 2 closed loops, there are $24+16=40$ paths.

Case 3: There are 3 closed loops. These can only occur in 3 edges of the octahedron. If you pick one point, there are 4 edges coming out of that point, and the other 2 edges from remaining unused vertices can be arranged 2 ways. For 3 closed loops, there are $4\times2=8$ Paths.

In Total there are $32+40+8=80$ different combinations; $\frac{80}{4096}=\frac{5}{256}$

Would it be possible to solve this problem with some graph theory? Because if we can represent the octahedron as a graph, then looops and stuff would be much easier. I'm not very experienced in this area, anyone have an idea?

ocha could you explain the part on how to remove the overcounted part? I don't understand how you arrived at 3x4 + 4x3 = 24.
thanks!

also, its ok to revive old threads as long as it isnt spam.

[quote="dgreenb801"]I think you can biject the total number of possible paths to the number of ways the ants can be on the six faces of a cube and each travel to a different face. There are two different ways to unfold a cube in a straight path, one is an "s", the other is a zigzag shape, so these are the ways the ants can move in one path without returning to a previous vertice. Can anyone finish this up?[/quote]
is this related to duality?

Sorry for the uber revive but for the solution of considering closed loops: is the case of 3 "mutual swaps" applicable? For example, A -> B, and B -> A; C -> D, and D -> C; E -> F, and F -> E?

I think I have a new solution:
Because the ants must end up at a unique position different from their start point, we can consider using derangement.
But one ant can only take one move, so we have to use PIE to account for the impossible configuration:
!6-(3*!4-3*1+1) = 240
We did not account for orientation when using derangement, so we have to divide by the number of rotations (which is 3)
So we have (240/3)/4^6 = 5/256

One can also use Rook polynomials. If $A, B, C$ are the vertices with the opposite vertices $A',B', C'$ respectively, then consider the $6 \times 6$ grid with $A, A', B, B', C, C'$ and we need to count the number of placements of 6 rooks in the grid such that the rook corresponding to any vertex is not placed in its column or its opposite vertex column.

Hello,

[hide]Casework is very simple for this problem. Notice that the 4 vertices on the base, 2 of them must move into the top/bottom vertices.

We have:
Adjacent ants move: that means there are 4 ways to choose the ants. There are also 2 ways to choose which one goes up and down. Finally, when we chose those, the 2 ants left at the base, there are 3 ways they can go. This is easy to count out. Finally, wherever those move, we have 2 vertices at the base where the ants that were previously on the top and bottom can go.
Diagonal ants move: There are 2 ways to chose these ants, and 2 ways to chose which ones go up and down. Also, there are 4 ways to choose where the ones at the base go after that, and 2 ways left for the ants previously on the top and bottom.

Adding these up you get 4*2*3*2+2*2*4*2=48+32=80, which means that the answer is A[/hide]


vector
analytic geometry
parameterization
function
geometry
3D geometry
combinatorial geometry

Suppose that $ K_i$ ($ i\equal{}1,\dots,n$) are compact connected sets in $ \mathbb R^d$ such that each set is contained in the ball of radius $ \delta$ centered at the origin and the sum $ K_1\plus{}\dots\plus{}K_n\equal{}\{y_1\plus{}\dots\plus{}y_n: y_i\in K_i\}$ is not contained in any half-space $ \{y: (y,e)\le 2\}$ where $ e$ is any unit vector and $ (\cdot,\cdot)$ is the usual scalar product. Show that if $ \delta>0$ is small enough, then this sum must contain the unit ball centered at the origin.

Well, if $ \delta<\frac2n$, there are no such sets satisfying the condition and it's trivially true. This probably isn't really what you were thinking of; I'll assume that $ \delta$ is chosen before $ n$, to make small $ \delta$ more meaningful.

It seems the core here is that the sum of many small connected pieces is almost a convex set. Easy to see intuitively, hard to quantify.

[quote="jmerry"] I'll assume that $ \delta$ is chosen before $ n$, to make small $ \delta$ more meaningful.[/quote]
Yes, of course.

I also tried to post it on [url=http://mathoverflow.net/]Math Overflow[/url]. Here are two observations that may be useful:

1) If $ a\equal{}\sum_ i x_ i$ and $ b\equal{}\sum_ i y_ i$ are in the sum, then the vectors $ v_ i\equal{}y_ i\minus{}x_ i\minus{}(b\minus{}a)/n$ are small and add up to $ 0$. There is a cute result that then you can rearrange them in such order that all partial sums are small (just constant times larger than the vectors themselves). If you switch from $ x_ i$ to $ y_ i$ along $ K_i$ in this order, you get a curve that travels from $ a$ to $ b$ within a small neighborhood of the segment $ [a,b]$.

2) The sum is $ 2d\delta$ dense in its convex hull. Indeed, if $ a\equal{}\sum_ i v_ i$ and $ v_ i$ are in the convex hull of $ K_ i$, then we can start moving $ v_ i$ until their representations as convex combinations of points in $ K_ i$ get shorter and we can do it as long as there are at least $ d\plus{}1$ vectors $ v_ i$ that do not belong to $ K_ i$ themselves (any $ d\plus{}1$ vectors in $ \mathbb R^d$ are linearly dependent, so once you are given directions in which you can move $ v_i$, you can always choose speeds that keep the sum in place). Thus, in the representation of every point in the convex hull as a sum, we need only $ d$ vectors from convex hulls and the rest may be taken in $ K_ i$. Replacing these $ d$ vectors by arbitrary points in the corresponding $ K_i$, we get a close point in the sum itself.

I feel that these two observations put together should be enough and I just do not see how to add 2 and 2 here. :wallbash:

All right, I found the missing piece (though not entirely by myself). We shall prove that if the diameters of $ K_j$ are less than $ \delta$ and the convex hull of their sum contains the unit ball, then the sum contains the origin.

[b]Step 1[/b] Without loss of generality we may assume that $ K_j$ are contained in the $ \delta$-neighborhood of the origin. Indeed, since $ 0$ is in the convex hull of the sum, we can represent it as $ \sum_j v_j$ with $ v_j\in\text{conv\,}K_j$. Now just replace $ K_j$ with $ K_j\minus{}v_j$.

[b]Step 2[/b] Since the coordinate vectors $ \pm e_k$ are in the convex hull, we can find close vectors $ f_k^{\pm}$ in the sum. Now write their representations $ f_k^\pm\equal{}\sum_j v_k^{\pm}(j)$, put $ w_k^{\pm}(j)\equal{}v_k^{\pm}(j)\minus{}\frac 1n f_k^{\pm}$ and change the order of summation if necessary so that all partial sums of $ w_k^{\pm}(j)$ are small. Now split the sum of $ K_j$ into $ d$ equal pieces (from $ 1$ to $ \frac nd$, from $ \frac nd$ to $ 2\frac nd$ and so on). The corresponding sums of $ v_k^{\pm}(j)$ will be about $ \pm \frac 1d e_k$.

[b]Step 3[/b] In the $ k$-th sum of $ K_j$, choose a continuous path that goes from approximately $ \minus{}\frac 1d e_k$ to $ \frac 1d e_k$ staying close to the straight segment (to be honest, this requires path connectedness rather than connectedness but we can always switch to small open neighborhoods and then use the routine limit procedure to take care of this minor issue). Call this path $ F_k(t)$, $ t\in[\minus{}1,1]$ (I do not care about the exact parameterization as long as $ F_k(\pm 1)$ is close to $ \pm\frac 1d e_k$).

[b]Step 4[/b] Let $ G_k(t)\equal{}\frac td e_k$ be the straight path. Now let $ \psi: [\minus{}1,1]^d\to [0,1]$ be any continuous function such that $ \psi\equal{}0$ on the boundary of the cube $ [\minus{}1,1]^d$ and $ 1$ on the slightly smaller cube $ [\minus{}1\plus{}\varepsilon,1\minus{}\varepsilon]^d$. For $ \bar t\equal{}(t_1,\dots,t_d)\in[\minus{}1,1]^d$, define
\[ F(\bar t)\equal{}\sum_{k\equal{}1}^d [(1\minus{}\psi(\bar t))G_k(t_k)\plus{}\psi(\bar t)F_k(t_k)]\,.\]
Then $ F$ maps the cube $ [\minus{}1,1]^d$ to $ \mathbb R^d$ continuously so that the boundary is mapped to the boundary of the cube $ [\minus{}1/d,1/d]^d$ in the canonical way. But then the image of $ F$ must contain the origin.

[b]Step 5[/b] It remains to note that the origin cannot be obtained when one of the coordinates $ t_k$ of $ \bar t$ is close to $ \pm 1$ because then $ \langle F(\bar t),e_k\rangle$ is close to $ \pm\frac 1d$, so the origin is attained deep inside the cube $ [\minus{}1,1]^d$ where $ \psi(\bar t)\equal{}1$. Thus, $ \sum_k F_k(t_k)\equal{}0$ but this sum is contained in the sum of $ K_j$.

So, jmerry was right: the sum of small connected compact pieces is almost convex, indeed. Fortunately, it turned out not so "hard to quantify" after all! :jump:.


analytic geometry
graphing lines
slope
exterior angle

How can you solve this...
Two circles intersect at points P and Q. A line l that intersects PQ intersect the two circles at the points A, B,C, and D( in that order along the line). It is known that <APB= 27 . Find the degree measure of <CQD.

I believe drawing segments AQ and PD will help.

I don't think there's enough information. As it stands, you could draw a degenerate case where B = C = Q, then you could choose the location of A to make APB be 27, and CQD would always be zero.

I think the points have to be distinct.

[img]http://www.initdotd.com/~nikita/Pictures/Circles.jpg[/img]

But clearly as you "move" B and C down to Q, angle CQD gets closer and closer to 0, instead of always being a single value. That's why there's not enough information in this problem.

[quote="TheAmazingOne"]clearly[/quote]

I wouldn't be so sure.  You've neglected the slope of each circle as $B$ and $C$ approach $Q$, which will determine $\angle CQD$.

[hide]
We have $\angle PAB=\angle PQC$ because both intercept the same arc, and similarly $\angle PBD=\angle PQD$. But $\angle PBD=\angle PAB+\angle APB$ (exterior angle) and $\angle PQD=\angle PQC+\angle CQD$. So we get $\angle CQD=\angle APB$, which in this case is $\boxed{27^\circ}$.
[/hide]


http://www.cnn.com/2007/US/11/13/southern.drought.ap/index.html

What would they do next if it still doesn't rain? Burning or slaying somebody as the sacrifice to the God?

[quote="solafidefarms"]Why do you mind if someone believes that words may change the weather ... ?[/quote]

Because it's obviously delusional thinking, and I'd rather not have clearly delusional people in control of important branches of government.

Incidentally, it turns out that [url=http://www.nytimes.com/2007/10/23/sports/baseball/23rockies.html?_r=1&oref=slogin]God doesn't root for the Colorado Rockies[/url], after all.

as much as i disagree with georgia people just on principle alone, i'm gonna go along with solafide here and ask whats wrong with this? he believes in religion and part of that religion is being allowed to ask your god for stuff. i don't see how it's delusional at all. and no, i'm not religious, but i don't think because someone else believes in a god means they deserve to be called delusional for that

[quote="solafidefarms"][quote="JBL"]It turns out that [url=http://www.nytimes.com/2007/10/23/sports/baseball/23rockies.html?_r=1&oref=slogin]God doesn't root for the Colorado Rockies[/url], after all.[/quote]

Also part of the Christian religion is that God is all-knowing and that if he doesn't answer prayer it is because he has some grander plan that you can't see. Who knows, God's plan may have been to cause them to lose so that next year they will be more driven and win when they deserve to win. But I don't pretend to know the mind of God, so that is idle speculation and has no bearing on reality, it's what I am biased to think.

If belief in God is delusional, and Sonny is delusional in this one area, this does not mean that Sonny is delusional in other areas. He is, actually, being a supporter of farm subsidies, but that's beside the point; calling him delusional is an ad-hominem attack on one area and does not affect the rest of his ability to govern.[/quote]

Yes that is all true. However, what about other religions? I am Jewish, and I would want there to be Jewish prayers as well. You cannot please all religions at the same time. That is why we have separation between church and state, and so we should not have any religion present in our politics.

On the other hand, if the governor and his officials want to pray for rain, they are perfectly allowed to; it would only be a problem if this prayer became official.

i dont see how they aren't seperating church and state

[quote="Buffalo"]Burning or slaying somebody as the sacrifice to the God?[/quote]I really don't see how you're making a connection between prayer and human sacrifice.   You have some extremely distorted views about religion.
[quote="JBL"]Because it's obviously delusional thinking, and I'd rather not have clearly delusional people in control of important branches of government.[/quote]I think it's pretty pathetic that just because you disagree with someone, you think their beliefs are "delusional."
Edit: It's raining right now here in Georgia, so obviously something's working out. :wink:

Seems like your God has done his job well.
Very good.
Why do you need a government then? The churches and the priests could take the place. Pray for more, like for the peace of the world, or at least let the God tell you where Bin Laden is.

[quote="solafidefarms"]You know, down here we are devout and we trust in (the Christian) God. As leader of the state, it is his responsibility to do everything he can for the state's well-being, and part of that is praying for rain because he and I both believe that God has power over the weather. Human sacrifice is not part of the Christian religion, nor is animal sacrifice.
[/quote]

If I am not wrong, there are human and animal sacrifices in the Bible.

"And he said, Take now thy son, thine only son Isaac, whom thou lovest, and get thee into the land of Moriah; and offer him there for a burnt offering upon one of the mountains which I will tell thee of."
"And Isaac spake unto Abraham his father, and said, My father: and he said, Here am I, my son. And he said, Behold the fire and the wood: but where is the lamb for a burnt offering?"

You can say that it is merely a testment of God. But since neither Abraham nor Isaac had doubted this request, it implies that human and animal sacrifice is a legitimate and normal request from the God, otherwise they would think the one who tempted Abraham is not God, but Satan or some other demons.

[quote="solafidefarms"]You know, down here we are devout and we trust in (the Christian) God. As leader of the state, it is his responsibility to do everything he can for the state's well-being, and part of that is praying for rain because he and I both believe that God has power over the weather. Human sacrifice is not part of the Christian religion, nor is animal sacrifice.

And I'd like to know how this is a sign of what's wrong with government. Why do you mind if someone believes that words may change the weather, and because they care about you they try to influence the weather through words? Should not you feel honored that they care about you that much, even if you don't believe that this has any effect?[/quote]
To pray is the duty of the priests and those believers(no matter which religions they are practicing).
As for the government, they should prepare for artificial rainfall, or import water from the other states, make sure that the people can live well. If the drought last long, they should consider importing food as well.
What had the governor done is simply a show, a dangerous political/religion-al show.

[quote]To pray is the duty of the priests and those believers[/quote]
Georgia's governor is a believer. So he prays. So what?

[quote]they should prepare for artificial rainfall, or import water from the other states, make sure that the people can live well.[/quote]
Do you have evidence of Georgia's governor neglecting his duties? Some of the measures he has taken are mentioned in the same CNN story.

[quote]What is wrong with the goverment and the people?[/quote]
Should we ask instead: what is wrong with intolerant atheists?

[quote="Twiz"]
[quote="JBL"]Because it's obviously delusional thinking, and I'd rather not have clearly delusional people in control of important branches of government.[/quote]I think it's pretty pathetic that just because you disagree with someone, you think their beliefs are "delusional."
Edit: It's raining right now here in Georgia, so obviously something's working out. :wink:[/quote]

Joel is not saying that he is delusional just because they disagree. It is not the idea that the governor believes in God or that he prays that is necessarily bothering him. It is the fact that he is relying on prayer to deal with a drought, when he should be spending his time figuring out how to legitimately handle the situation. He should understand that the weather is beyond the government's control, and focus all of the government's attention on helping the people cope with the drought. What is delusional is focusing the government's and the people's attention on what is out of their control rather than what they can do about it.

I don't usually like being spoken for, but yjneb has done a very nice job.  There've been a number of other comments sent in my general direction, and here's me addressing them in totally haphazard order:

mlok:
I'm not so sure what's intolerant about my views: if you go and find a dictionary definition of "delusion," it's quite clear that the beliefs and activities of Sonny Perdue fit the bill.  The fact that he (and soldafide and many other people) believe quite strongly in the same delusion does not make it less fictional.

The evidence that the Governor of Georgia has neglected his duties is that he waited until the city of Atlanta had 100 days of water left before doing something about it.

I think the fact that soldafide [i]took seriously[/i] my comment about the Rockies should further convince anyone not already deeply involved in this delusion that it is, in fact, totally crazy.  We have here people who believe that, as part of Its duties running the universe, captial-G God spends its time [i]picking winners of baseball games[/i].  This sounds to me about as plausible as the notion that Zeus spends his time picking up mortal chicks for fun.

soldafide:
Calling someone delusional is not an ad-hominem: it goes directly to the core of their behavior.  I didn't say, "Sonny Perdue is a Republican, so that means he's delusional."  I said something to the affect of, "Sonny Perdue has a delusion -- that makes him delusional."  I also said that I would prefer not to have delusional people running the government.  (See, e.g., current president of the United States who believes God speaks directly to him -- a belief that would have him institutionalized many places if he replaced "God" with any other nonexistent entity.)  Neither of these things is an ad-hominem.  More generally, the technique of declaring any criticism of someone void as an ad-hominem is itself an annoying behavior and fallacious argument.

Bouncing of calc rulz & "intolerance":
I certainly don't think the governor of Georgia should be [i]forbidden[/i] from praying all he wants.  Having secular authorities organize prayer rallies makes me squeamish, but also shouldn't be banned.  These activities are still pointless wastes of time.\

Twiz: That was an insult and a fallacy.  As for your response to Buffalo, lots of religions believe that animal sacrifice will lead to god(s) doing what they want.  What's particularly different about believing that a prayer rally will lead to god(s) doing what you want?  The answer: one more living chicken, but exactly the same amount of delusion.

[quote="JBL"]
Because it's obviously delusional thinking, and I'd rather not have clearly delusional people in control of important branches of government.
[/quote]
Not quite :) It is a [b]widespread[/b] delusional thinking and a widespread delusion has always been as much (actually, even much more) respected than any facts. Besides, I just do not know any completely delusion-free people, so the only way to achieve what JBL is asking for is to be governed by machines but it is a separate story.

As to the Christian religion, the main thing I hold against it is Spanish Incvisition. The intolerance of atheism is a mere response to the past (and, alas, present, though it much less concerns Christianity, than, say Islam) religious intolerance just as the modern intolerance to racism is the reaction to the past atrocities of Nazi.

As to governor's praying for rain in the Christian manner, I suspect that the governor is delusional in this particular respect not any more than JBL, just the goal of that prayer was not to get it rain but to get reelected ;)

[quote="fedja"]I just do not know any completely delusion-free people, so the only way to achieve what JBL is asking for is to be governed by machines but it is a separate story.
...
As to governor's praying for rain in the Christian manner, I suspect that the governor is delusional in this particular respect not any more than JBL, just the goal of that prayer was not to get it rain but to get reelected ;)[/quote]

I don't doubt that everyone has their personal delusions.  (And machines have their own problems, too :).)  But most of us do not specifically claim our delusions as central to our thinking about every subject, from morality to meteorology. In the case of this particular delusion, the deluded go so far as to state that their imaginary friends carry all responsibility for everything, a proposition I find quite alarming.

Buffalo: 1. that was the Old Testament; since Jesus was sacrificed for our sins, there is no longer a need for the shedding of blood to take away your sins, Jesus covers that. 2. Abraham had faith in God, and God was testing his faith [and Isaac's.]

yjneb: He's done a lot about it; he's even sued the Corps to keep them from releasing water from the lake, thus preventing them from using it.

jbl: Define: delusion "A false idea typically originating from a misinterpretation but firmly believed and strongly maintained in spite of contradictory proof or evidence." I see no contradictory proof or evidence.

Nobody knew how little water was left until there were 80 days left.

God doesn't pick winners, but he can influence events for the eventual better. "All things work together for good for those who love the Lord".

Everyone is somewhat delusional. I think you are delusional for believing in evolution. If you're scared of spiders, you're delusional. Should we then be governed by non-humans? [Apologies for being annoying.]

Ok, if prayer is just words to make some people feel good, why do you mind? If a majority feel those words are offensive, the joy of our *democracy* is that you can vote them out next election. At the same time, the *joy* of our democracy is that we *can* vote for delusional people to lead us if we so feel, and as long as they don't violate the Constitution I'm not complaining. [Did I complain when an evolutionist was elected to office? No? That's because belief in evolution doesn't violate the Constitution.]

As a meta comment, I think it's interesting to note that Buffalo began this thread by [i]laughing at[/i] Perdue, but you have refered in every instance to this as "minding" (syn: being bothered by) his behavior.  The two are rather different, and I wonder why you've interpreted one as the other.

Contradictory proof or evidence: obviously, if I could provide sufficient evidence to dissuade you of your delusion, I would have done so already.  There are lots of reasons that I can't, most of which have nothing to do with me.  However; in the case of a claim that massive supernatural forces exist that regularly interfere with the natural order of things, it seems to me perfectly reasonable to take the fact everything hums along all the time according to natural laws we have the capacity to understand as evidence of lack.  I.e., absence of evidence does not [i]in general[/i] provide evidence of absence, but when your claim is strong enough, it does.

[quote]Nobody knew how little water was left until there were 80 days left. [/quote]
This is perhaps technically true, but if so should be leading you to condemn the state government for gross incompetence: the drought's been going on for over a year at this point, and has been level D4 in parts of Georgia and neighboring states at least since June.  The first NY Times article about it that I find was July 4, and at that point it was already being viewed as a disaster.  But outdoor water use wasn't curtailed until September!

Finally, wandering ever-further off-topic: there are a very, very large number of things that a constitutional but that are also terrible ideas.  Would you really have me believe that you don't care [i]at all[/i] about these things?  That you only [i]ever[/i] complain when governmental agencies do things that are unconstitutional, never when they do things that are terrible policy, misdirected, outrageous, laughable, expensive, pointless, destructive, or self-serving?  (Not that all of these adjectives apply in this instance, of course!)

[quote="JBL"]That was an insult and a fallacy.  As for your response to Buffalo, lots of religions believe that animal sacrifice will lead to god(s) doing what they want.  What's particularly different about believing that a prayer rally will lead to god(s) doing what you want?  The answer: one more living chicken, but exactly the same amount of delusion.[/quote]
Yes, it was an insult, thank you for pointing that out. It was an insult in response to your insult of my religion. I didn't start this argument. And Buffalo wasn't referring to animal sacrifice, he was talking about human sacrifice (he used the word "somebody"), and the difference between prayer and murder should be pretty clear...

[quote="JBL"]As a meta comment, I think it's interesting to note that Buffalo began this thread by [i]laughing at[/i] Perdue, but you have refered in every instance to this as "minding" (syn: being bothered by) his behavior.  The two are rather different, and I wonder why you've interpreted one as the other.[/quote]
Really, a semantics argument? Have you perhaps run out of good arguments? Fine, I'll sink down to that level. Laughing at someone implies that you are bothered by
their behavior. Why would you laugh at someone if you approved of their behavior?

Anyways, I think the major point here is that your the aggressor here. Your trying to convince people that their beliefs are wrong. I'm just trying to argue that everyone is entitled to their own beliefs.

[quote="Twiz"]And Buffalo wasn't referring to animal sacrifice, he was talking about human sacrifice (he used the word "somebody"), and the difference between prayer and murder should be pretty clear... [/quote]
You asked what the connection between prayer and sacrifice is.  I explained it to you: both are believed by some people to get god(s) to do what they want.  Then I asked you what is fundamentally different about believing that prayer gets god(s) to do things and that sacrifice gets god(s) to do things.  Your answer was ... that killing people is bad.  This does not even remotely address my question.  [i]My[/i] answer to my question is that there is [i]nothing whatsoever[/i] different between believing that prayer gets god(s) to do things and believing that sacrifices (human, animal or otherwise) gets god(s) to do things -- both beliefs make equally little sense and have equally little evidence to support them.

[quote="Twiz"]Really, a semantics argument? Have you perhaps run out of good arguments? Fine, I'll sink down to that level. Laughing at someone implies that you are bothered by their behavior. Why would you laugh at someone if you approved of their behavior?[/quote]
Do you know what the words "meta comment" mean?  I was making an observation and an inquiry about the behavior of people (actually, one person: soldafide) in this discussion.  As another example, seem extremely irate that I called your religion a delusion.  Note that this says nothing whatsoever about the validity or lack thereof of your comments.
Continuing in a somewhat meta vein, you haven't yet offered anything to the discussion to suggest that my statement (that your religious beliefs are delusional) is wrong (except for one obviously fallacious comment), nor have you offered anything to the discussion to suggest that it was improper of my to say or believe it.  Moreover, your comment that follows "sinking down to my level" is the first time in your comment that you actually address the subject at hand.  My answer is: laughing at someone does not, generally, imply that you are bothered by their behavior.  Often, one laughs at someone else because their behavior is absurd.  That's the extent to which I can speak for Buffalo.  As for myself, I am (mildly) bothered by Perdue's behavior because I think it would be nice if people in positions of power did not spend their time spreading superstition and delusion when they could be doing something useful.

[quote]Your trying to convince people that their beliefs are wrong. I'm just trying to argue that everyone is entitled to their own beliefs.[/quote]
Your final sentence is completely untrue: you have made no discernable effort to argue that everyone is entitled to their own beliefs.  In addition, if you were making such an argument, it would be a straw man in this discussion: I have certainly not argued that Perdue, soldafide, and yourself are not entitled to have whatever wacky beliefs you have.  We are both in full agreement that you are entitled to believe that the big man in the sky listens extra-carefully when the Christians of Georgia have a prayer vigil -- it's just that I think this belief is completely delusional.

[quote="JBL"]You asked what the connection between prayer and sacrifice is.  I explained it to you: both are believed by some people to get god(s) to do what they want.  Then I asked you what is fundamentally different about believing that prayer gets god(s) to do things and that sacrifice gets god(s) to do things.  Your answer was ... that killing people is bad.  This does not even remotely address my question.  [i]My[/i] answer to my question is that there is [i]nothing whatsoever[/i] different between believing that prayer gets god(s) to do things and believing that sacrifices (human, animal or otherwise) gets god(s) to do things -- both beliefs make equally little sense and have equally little evidence to support them.[/quote]
Okay, analogy: if you want your boss to give you a raise, asking him politely may get you what you want. Stabbing your co-worker and giving the dead body to your boss is not likely to get you what you want. The idea of killing someone to get God to do something [i]is[/i] absurd because God doesn't want people to kill each other. It's a commandment.

I don't think time is the issue here. It doesn't take much time to hold prayer services. He has plenty of time to act politically (and is doing a fine job).

I argued that everyone had the right to their own beliefs when I told you that it is not right to call other people's beliefs delusional just because you do not believe them. By calling them "wacky" and "delusional" you are surely trying to convince people that they are wrong about their religious beliefs.

I don't view God as he may be pictured in some TV shows and movies. I don't think there is a big man standing on top of some clouds chilling out with Jesus and the holy spirit. I believe in a supernatural force, something bigger than man that cannot be pictured in human terms.

I [i]am[/i] sorry about personally insulting you. I don't know you and have no right to do so. I am actually probably not going to respond to your next response because I have said what I want to say, and it's pretty clear that we're not going to agree.

[quote="Twiz"]Okay, analogy: if you want your boss to give you a raise, asking him politely may get you what you want. Stabbing your co-worker and giving the dead body to your boss is not likely to get you what you want. The idea of killing someone to get God to do something [i]is[/i] absurd because God doesn't want people to kill each other. It's a commandment.[/quote]
That still doesn't answer JBL's question. You are explaining why your religion distinguishes between the two methods of getting a god to do what you want. JBL didn't ask anything about a particular religion. He was pointing out that [i]sacrifice and prayer are different manifestations of the same fundamental concept[/i] and that from the perspective of someone who does not buy into the fundamental concept, they are equally absurd. Note that there's no moral question here: presumably JBL would also agree that sacrifice is morally "worse" because you kill something (or someone!) whereas prayer is just a waste of time (in JBL's worldview).

[quote="Twiz"]I argued that everyone had the right to their own beliefs when I told you that it is not right to call other people's beliefs delusional just because you do not believe them. By calling them "wacky" and "delusional" you are surely trying to convince people that they are wrong about their religious beliefs.[/quote]
Your claim that JBL should not call beliefs other than his own delusional is NOT a defense that everyone has the right to their own beliefs. At best, you are saying that everyone has a right to be respected for their beliefs (which borders on absurd). At worst, you are attacking people's rights to their own beliefs by saying that JBL should not be allowed to hold the belief that religion is delusional! In fact, JBL's belief that religion is delusional in no way interferes with other people's beliefs about religion.

Now, I didn't read the original article and I'm not too familiar with the drought situation, so I'm not going to make any specific comments about that. However, I would like to point out that in general, although it is perfectly fine for a political leader to maintain religious beliefs and to take actions based on those beliefs [i]in private[/i], it is improper for a political leader to allow his or her religious beliefs to interfere with his or her ability to discharge official duties.

On the other hand, due to the large number of religious people in this country, it is not unreasonable to defend a political leader's religious activities by stating that it really does make a significant portion of his or her constituency [i]feel better[/i]. A political leader's job isn't just to make things function. If people want to see their governor or president or Grand Moff praying, it is quite legitimate for that person to pray in public.

Can Christians pray and ask something from the God?

In ancient China, the emperors claimed themselves to be son of the Heaven, the chosen ruler, the disasters like flood, drought, earthquake, etc. were considered the Heaven's warning/punishment to their misbehaviour, a "good" emperor should try to find out what they had done wrong, confess to the Heaven, blame himself for his guilt to the public, pray to the Heaven, saying things like "It is all my fault, please punish me only, spare the innocent people."

What would the Christian(or other religion) priests/believers think/do in such cases?


Bob bought a car for $5600. He sold it to Fred for 5/6 the price he paid for it. Fred sold it to Jim for 1/5 less than he paid for it. Jim sold it to Dan for 3/4 what he paid. What did Dan pay for the car?

The problem says that Fred sold it to Jim for 1/5 [b]less[/b] that he paid, so he paid 4/5 of the price of it and the answer is \[ 5600\cdot\frac{5}{6}\cdot\frac{4}{5}\cdot\frac{3}{4}\equal{}\boxed{2800}\]

Or did I misread too...

Well, yes, unless I confused the question


Solve each equation in the complex number system.

What exactly is the complex number system anyway?

Here are the two questions:

[1] 10x^2 + 6x + 1 = 0

[2] x^4 = 1

The complex number system is simply the set of all real, imaginary, and complex numbers. In case you don't know what the difference between imaginary numbers and complex numbers is, imaginary numbers are in the form $bi$ while complex numbers take the form of $a+bi$, where $a$ and $b$ are real in both cases.

[hide="Solutions"]1) Quadratic formula.

$x = \frac{-6 \pm \sqrt{6^{2}-4*10*1}}{2*10}= \frac{-6 \pm \sqrt{-4}}{20}=-\frac{3}{10}\pm \frac{1}{10}i$.

2. Factor.

$x^{4}-1 = 0$
$(x^{2}-1)(x^{2}+1) = 0$

The first factor obviously yields $x = \pm 1$. The second:

$x^{2}=-1$
$x = \pm \sqrt{-1}= \pm i$.

So $x = \pm 1$ or $x = \pm i$.[/hide]

What a great reply.

I will study your notes and tackle similar questions in my study book.


AMC USA(J)MO USAMO quadratics AIME algebra

Find all nonzero integers $ (a,b)$ so $ (a\plus{}b^2)(a^2\plus{}b)\equal{}(a\minus{}b)^3$

[hide]
Expanding and simplifying, we want all solutions to:

$ 2b^2\plus{}(a^2\minus{}3a)b\plus{}3a^2\plus{}a\equal{}0$

This has integer solutions iff its discriminant is a perfect square:

$ a^4\minus{}6a^3\minus{}15a^2\minus{}8a\equal{}n^2$ for some positive integer $ n$.

The LHS factors as $ a(a\minus{}8)(a\plus{}1)^2$, so we only need $ a(a\minus{}8)$ to be a perfect square, say $ m^2$.

However, $ a(a\minus{}8) \equal{} (a\minus{}4)^2 \minus{} 16$, so $ m^2$ is also 16 less than a perfect square. Thus the only possibility is $ a \minus{} 4 \equal{} 5$ so that $ (a\minus{}4)^2 \minus{} 16 \equal{} 9$.

From here, we have $ a \equal{} 9$, which gives the solutions $ (9, \minus{}6)$ and $ (9, \minus{}21)$, which are easily checked.

Before we end, however, we note $ (\minus{}1, \minus{}1)$ is another solution. Finding the hole in our logic is left as an exercise.  :D [/hide]

[hide="@alkjash"]
The hole is here:
[quote="alkjash"]
The LHS factors as $ a(a \minus{} 8)(a \plus{} 1)^2$, so we only need $ a(a \minus{} 8)$ to be a perfect square, say $ m^2$.
[/quote]
Sure, if $ a(a \minus{} 8)$ is a perfect square then so is $ a(a \minus{} 8)(a \plus{} 1)^2$, but that doesn't quite work in reverse. What if $ (a\plus{}1)^2\equal{}0$? That's where the $ a\equal{}\minus{}1$ solution comes from.

Now see if you can find your other hole and your other missing solution. :wink: [/hide]

I'm guessing it's when you write the quadratic in terms of $ a$, and after simplifying the discriminant, you have

$ (9b \plus{} 23)(b \minus{} 1) \equal{} y^2$

But I don't know how to find solutions for this one  :blush: .

EDIT: I looked at $ a(a\minus{}8)(a\plus{}1)^2 \equal{} n^2$, so I tried $ a\equal{}8$, which gave me the solution $ (8,\minus{}10)$.

Why would that be a hole? Yes that's something else you could do, but that doesn't mean the other way doesn't work. In theory alkjash's solution approach is valid: consider it as a quadratic in $ b$ and use the discriminant to find all possible values of $ a$ (and from that find the corresponding values of $ b$). If done correctly that will find all of the solutions. The existence of another approach doesn't invalidate the solution. You need to find an actual flaw in his reasoning.

And yes, that is the other solution.

There are lots of AIME problems and some USAMO in Interm Alg (Alg 3)

Do you think Alg 3 covers USAMO? (Or most?)

[quote="batteredbutnotdefeated"]@awesometoad: intermediate algebra covers AIME level algebra, not USAMO...[/quote]

Just wondering,...
there are a bunch of USAMO problems there.

I was wondering if doing the challenge problems would help any with USAMO

Well probably with [b]old[/b] USAMO tests, but not at today's level. But you have to start somewhere so that's probably a good idea.

Oh, wow, I forgot 0 is a perfect square...

i don't get the whole concept of the discriminant here

@everyone: Because the denominator of the quadratic equation is 4, how do you know the numerator is divisible by 4 also if the discriminant is an integer?


The sum of the lengths of the edges of a regular tetrahedron is 96 cm. What is the number of centimeters in the length of one edge?

a regular tetrahedron will look like this:

[img]http://davidf.faricy.net/polyhedra/images/tetra.gif[/img]

counting the number of sides we clearly see that a regular tetrahedron would have 6 edges, and since each edge is equal with a total summed edge length of 96, we simply divide 96 by 6 to yield $ \boxed{16}$


In AoPS vol. 2 it says that the dot product of $||v_1||$ and $||v_2||$ on the cartesian axis is equal to $||v_1|| \cdot ||v_2|| \cos (\theta _1 - \theta _2)$

But isn't this angle equal the wrong angle? Its not equal to the angle between the two vectors being added, but rather $180^\circ$ minus that angle.

I think, that:
$\vec{a}\cdot \vec{b}=|\vec{a}|\cdot |\vec{b}|\cos{\alpha}$ , $\alpha$ is the angle between two vectors

I thought the sum vector cuts through the angle between the two vectors?

The angle you use for law of cosine must be between the a and b, but the angle that AoPS vol 2 uses for defining dot product with respect to the positive x axis is equal to the angle between a and a+b!

EDIT: I think I understand this now.

I guess its a good way to find the theta.

$\theta_1-\theta_2$ is either the angle between them, or the opposite of that, which have the same cosine anyways...
But yeah.


AMC
USA(J)MO
USAMO
AIME
limit
AMC 12
probability

now the second AIME is over, do you guys have any new predictions?

I think the cutoff will still be a 9 on the AIME.

Considering the AIME II, what's the chance that the floor is going to be above a 9?

$\lim_{\epsilon \rightarrow 0^+} \epsilon$

Think about it. An AIME floor of 10 would mean that the index is around 240. Not only is that unprecedented, but it would represent a 30 point increase from last year and leave the 2nd highest USAMO cutoff index (226 in 2003) a distant 2nd, as well as two AIME problems greater than the previous high floor.

In fact, I would be surprised if it were above 230.

I agree with you, Mildorf, that the AIME floor will not exceed a 9. However, your argument against such a high cutoff with the fact that such a high index would be unprecedented can be countered with the argument that such an easy AIME was unprecedented.

But the extent to which the 240 outlies the others is just unreasonable. And also, I honestly do not believe 160 AMC12 takers in this country achieved a 240 index. Indeed, I would also be astonished at a cutoff above 230.

Probability, why would you hope for a higher cut-off? A lot of people on this site wish for an AIME floor of 8. I do not see why a lower floor would be bad... Unless you hope that some people do not make it? But I do not think that is the case ;).

[quote="Singular"]224 +/- 3  (im guessing 223)

YOU HEARD IT HERE FIRST, FOLKS!![/quote]

I've been saying those numbers (and so have other people) for a long time now. You are by no means the first. :P Sorry mate.

High cutoff = less underclassmen qualifiers = probably no index selection. That's why.

[quote="probability1.01"]I'm hoping for a high cutoff, but yeah, it's probably only going to be around 220.  I *hope* the floor is 9, but it's too likely for a 140-8 among the 160 qualifiers. :([/quote]

Nah cutoff for index is going to be higher like 228. So, it should be a 9.

[quote="DPopov"]I agree, I think the cutoff will be 226-228 but the AIME floor will be an 8.[/quote]

That's basically saying someone got a 146.5 or a 150 on the AMC12 and got an 8 on the AIME. I find that not too probable, so I'm going to say a floor of 9.

I think it's more valid to think of this in terms of how many hard questions were there for the top students to miss, rather than how many easy ones were there for the average students to get. If there were more relatively easy questions than in previous years, this should mean fewer scores of zero, and a higher average score. But I doubt if this will have much effect on the AIME floor. The top scorers would have gotten most of the the easier questions right anyway. They will be separated out according to the most difficult questions.

a 9. Probably. I think the index is going to be around high 220s to lower 230s. Although that means someone should get 130s-140s AMC12 and a 9 AIME.... which is less possible since if you are good at AMC, you are likely to be achieving a satisfactory score on AIME, which went up due to its difficulty, or the lack of it.

that would qualify about 3 people on this forum.... i know the two who got 290 index.

I wouldn't be laughing, that's for sure. :P

Well then we'd know 1 thing for sure. We're gonna have a killer IMO team this year.

However, I somehow doubt the index will top 235...so...

i am a ninth grader and took aime 1 and i got 9. i was very surprised and i thought i'd get in for sure because last year the floor was 7.  my mom even called the aime place in Lincoln and. . . .     the secretary told her it was projected at 10.  but then we got an e-mail saying it "might very well be 9," so i'm hoping. . . .  did any of you get 9 exactly and are worrying like crazy?

I'm pretty sure it will be a 9.  8 is too low, 10 is too high.  Although I'm betting there will be like 10 9th graders with 11+'s, probably at least one with a 13.

a 9th grade rival of mine got 12, so he's in for sure. i was only this jealous once in my life and that was when he went on to national mathcounts and i didn't.  thanks though, i'm glad i'm not the only one.

Awe shoot. Please let it be a 9. Please.

did you get 9 too? if you took amc 12, what was your score?

[quote="Mildorf"]

[quote="beta"]Some AoPSer that I talked over AIM said that he knew someone with a 143 on AMC12 and 8 on AIME.[/quote]

Of course, if you've ever played Telephone you probably know that statements are not 100% reliable after they have been passed from an unmentioned AMC taker to some guy to some AoPSer to beta to me to you.  :rotfl: (Also wanted to use that emoticon. :D )[/quote]

I like to think everything has a certain truth to it...

if you are a 9th or 10th grader then the amc won't matter.  i'm guessing 17 right, 1 wrong, and 7 blank? *compulsively checking amc site again*

if the list came out now, everyone here would probably know in less than ten seconds.

I think the index is going to be in the low 230's, so I will either just barely miss it or just barely make it.

The list of qualifiers is already updated!


Yes, but now if we verify that $ x\equal{}y\equal{}\frac{1}{2}$, are solutions, we get to something false.

Then it has no solutions, if socrates's solution is correct.

[quote="ely_en"]But we always need to verify the solutions? :huh:[/quote]

Sometimes you don't, but I recommend verifying (who knows, maybe you made a mistake).

[quote="ely_en"]But we always need to verify the solutions? :huh:[/quote]
:huh:

Here's how these problems generally work.

1. Show that if $ x,y$ is a solution of the equation, it has to satisfy some certain conditions. (like $ x \equal{} y \equal{} 1/2$ here)
2. Check for each of the $ x,y$ that satisfy the condition whether it works in the equation or not. Here we find that $ x \equal{} y \equal{} 1/2$ does not satisfy the equation.
3. Whatever satisfies both the condition in 1 and the original equation (determined in 2) is the set of your solutions. Here nothing worked, so we have no solutions.

What do you mean "always need to verify the solutions"? We did; there are none.

I think he means this:

"Once we find a solution, do we ALWAYS have to plug it in to make sure it works?"

And in some cases no. But these are only in the AMC 8, 10, and 12. And MathCounts. No place else. In other words, always check your solutions in Olympiad problems.


I was trying to type this: \[\lim_{x\buildrel{<}\over\rightarrow a}f(x) = b\] I know the command [code]\buildrel{<}\over\rightarrow[/code]isn't right, but he gave me an error when I tried "under" instead of "over". What should I type?
Thankssss  :)

You could just reverse the order of the commands as in $\lim_{\buildrel{x \rightarrow a}\over{_{<}}}f(x) = b$
but it's more usual to use \stackrel as in $\lim_{\stackrel{x \to a}{_{<}}}f(x) = b$
_{<} is used otherwise < is too large

Ah, I really appreciate your quick answer. Thanks!

$\stackrel{text1}{text2}$

That's not ok: the $x\to a$ is smaller than normal, when using both notations you'll clearly see this. There is a good way but I can't remember right now.

By the way Kurt, use \to instead of \rightarrow for limits and functions.

Do you mean as in \substack $\lim_{\substack{x \to a\\<}}f(x)$ or the subarray environment $\lim_{\begin{subarray}{c}x \to a\\< \end{subarray}}f(x)$?

I think the notation $\lim_{x\to a^{-}}$ means the same thing, and it's easier to type.

I agree with that. I like using $x \nearrow a$ which gives a visual impression of x increasing upwards to a.

$\lim_{\begin{subarray}{c}x\rightarrow x_{0}\\ x<x_{0}\end{subarray}}f(x)=\lim_{x\to x_{0}^{-}}f(x)=$ ${\lim_{x\nearrow x_{0}}f(x)=f(x_{0}-0)=f(x_{0}-)=f(x^{+}_{0}})=L_{f}(\nearrow x_{0})$

$\lim_{n\to\infty}a_{n}=l$ $\iff$ $a_{n}\rightarrow l$

Let $l$ be a real number. Then $a_{n}\searrow l$ $\iff$ $a_{n}\rightarrow l$ and exists $n_{1}\in N^{*}$ so that for any $n\ge n_{1}\ ,\ a_{n}\ge a_{n+1}$ (decreasing to $l$).

I don't like how appear these produkts in the ratio $E_n=\frac {2^{n-1}\cdot\left(1+\prod_{k=1}^na_k\right)}{\prod_{k=1}^n\left(1+a_k\right)}\ ,\ n\in\mathbb N^*$ .
It is appears only in a ratio. Please, I want so ==> $\prod_{k=1}^na_k$ and when is a ratio. Thak you very much .

Add \limits to get \prod\limits_{k=1}^na_k

$ E_n=\frac{2^{n-1}\cdot\left(1+\prod\limits_{k=1}^na_k\right)}{\prod\limits_{k=1}^n\left(1+a_k\right)}\ ,\ n\in\mathbb{N}^* $

WOW ... it is very simply and v.v. nicely ==> $E_n=\frac {2^{n-1}\cdot\left(1+\prod\limits_{k=1}^na_k\right)}{\prod\limits_{k=1}^n\left(1+a_k\right)}\ ,\ n\in\mathbb N^*$ . Thank you v.v. much.


ARML
analytic geometry
graphing lines
slope

For $x$ and $y$ in radians, compute the number of solutions in ordered pairs $(x,y)$ to the following system:

$\sin(x+y)=\cos(x+y)$

$x^{2}+y^{2}=\left(\frac{1995\pi}{4}\right)^{2}$

[hide="Method"]
Well, from the first equation we have that $x+y = \frac{\pi}{4}+k\pi$ for some integer $k$.

Now, we can take the (geometric) tangent, with slope $-1$, of the circle determined by the second equation, and find where it hits the y-axis. Call that value $n$. We will then have that $n \geq \frac{\pi}{4}+k\pi \geq-n$. Now, just calculate the number of solutions in that interval.

[b]EDIT[/b]: It's easy to find that $n = \pi \cdot \frac{1995 \sqrt{2}}{4}\approx \left( 705.3 \right) \pi$, which means that
\[705.3-.25\geq k \geq-705.3-.25 \implies 2 \left( 2 \cdot 706-1 \right)= \boxed{ 2822 \text{ solutions}}\]
[/hide]


In a calm city, the physics make a party every night.

experimentally, they establish that the variation of the number of persons in a day is proportional to the numbers of people that had gone in another pary in some day that they made and the present day. or matematically.

${\frac{\partial N(t)}{\partial t}=\alpha N(t-\epsilon ) N(t)}$

Hint: Need lagrange inversion

Not to say I have any idea how to solve this but what's the question asking? It seem to just stopped....


USAMTS algorithm calculus integration ARML probability blogs

If you do math for the sake of discovering new truths, then the only question is---do you find the result of the computer program convincing?  I would be eager to use whatever tools are available to discover new knowledge.  Maybe there is a point in math beyond which we can't go without computer proofs.

That said, computer programs usually have bugs in them, and we will always have to worry about that.  (Then again, a written proof can have a "bug" as easily as a computer program can.)

The more interesting question to me is whether the nature of certain problems require computer proof.  Casework never seems particularly elegant to me - in Erdos' words, I don't think any of the proofs in The Book are by computer.

Consider also that proofs like Wiles' proof of Fermat's Last Theorem are not important because they are of an important result but because the ideas in the proof were new and exciting and shed light on things.  Does a proof by computer have the same effect?

@tor well, arguably so - it may demonstrate a new class of programs which could be implemented in various proofs which could then be used to show various other results and, in the limiting case, open entire theories. I'm still not sure whether we should accept [i]any[/i] of these proofs by computer, though.

@sing I tried to argue in the opening post that I'm not concerned about bugs since this is something visible on the source code level, which is accessible to human checking at much the same level as a standard mathematical proof, as you touch on. I'm more concerned with how one accepts that the conversion from source code to result is reliable.

Here's a thought I had.  If you're worried about errors in the compilation process, you could write the program in lots of different languages and use lots of different compilers on lots of different kinds of computers, and make sure you get the same result.  You may still worry that there is some subtle error that all the computers or compilers are consistently making, all in the same way.  But then again, when it comes to checking a traditional written proof, you may worry that there is some subtle mistake in all of our brains that causes us to all get confused in the same way when faced with a certain mathematical argument.  Which is more likely?  I don't know, both seem very unlikely.

I think it is implicit in the nature of proof that one accepts the inherent limitations of the human brain, but there is no [i]a priori[/i] reason to extend this to computer compilers.

by the way, note that this question isn't purely philosophical in nature as it crops up in my work as a grader, both for classes (if students use various programs such as mathematica) and for USAMTS

Yay! What a good topic. I've had this argument with my professors when we discuss whether the Four Color Theorem has been proven.

"Here's a thought I had. If you're worried about errors in the compilation process, you could write the program in lots of different languages and use lots of different compilers on lots of different kinds of computers, and make sure you get the same result. You may still worry that there is some subtle error that all the computers or compilers are consistently making, all in the same way." -Singularitarian

This is the argument I make that computer proof will always introduce the idea of empirically validating a claim. You can check it in as many ways and on as many systems as you want, but all you are still only gaining empirical evidence to the truth of the claim. Most mathematicians will say empiricism is not a valid way to do math.

Here is the kicker, though. Most mathematicians accept the 4CT as proven. What they reject is that a computer proof done in as rigorous a way as that introduces empiricism. So we always both agree that it is proven, we just disagree on whether or not empirical methods were used.

I agree with Singularitarian. All math has the possibility of error (well, at least at the level of Wile's proof). I actually find it more convincing if a computer tells me something than a human.

Er. If I understand you correctly, as your amenable to accepting large amounts empirical evidence as valid proof?

And I acknowledge that mathematical proofs may have entirely human flaws, but these things are all open to human thought -- we can pore over what's been written down on the page and the definitions that have been made and so forth. We can't pore over the inner workings of the computer.

I think it depend on cases... it might be hard to explain, consider following example:
 given a 4x4 table without 2 corners, prove that the table cannot be covered by 6 1x2 dominoes.
     this problem is very similar as the classical problem that can be solved by coloring(the one with a chesstable without 2 corners), and because it is so small people can eventually prove it by evaluating each case.... and i believe most people will consider it valid. some case happenned with the problem 4 of the iberoamerican math olimpiad, with is one about a cominatorics problem about a 19x19 table I remember there is people who evaluate all the draganian path to every single square(by filling a 19x19 table by numbers) and they get 7. with means, if the cases is finite, the checking every case is a proof, even not elegant.
 if about problem is about a much bigger table, such that it is to big for human to check every case, then if a computer just check it for human should be a fine.

but however many conjectures or problems are consisted to have infinite specific cases, so in that case we cannot simply accept computer proof.

I'll admit, I've actually wondered before if computer proofs are acceptable because of questions like are "all the bits are flowing the way they're supposed to"? But my biggest question is, how do you know there is no bug in your program? You could check small cases, I suppose, but what if something changes as it gets bigger? What about, say, floating point inaccuracies? But, I suppose one would say a program can be checked as rigorously as a proof (as mentioned above).

But here's something weird. If we're worried about that, why can't we be worried about weird things in non-computer proofs? For instance, do we consider the fact that other mathematicians have confirmed a proof to be "empirical evidence" of its correctness? Would we need our own mind to confirm something to be correct? How does one know he himself isn't insane, and that every logical conclusion he come to isn't completely wrong?

But after all those worries, I would say we are safe to assume that our proofs are correct, even ones by computers. Again, the chances of the computer messing up are ridiculously small, and if we have it checked in many languages and many different computers, I'm pretty sure we are safe. Of course, whether or not the proof by computer is "nice" is a different question.

@chichi I'm not making an argument against casework here. Casework derived and proven by human thought is still perfectly ok because everything is right there, written down, and is accessible to other human thought.

@tj Good point in your first paragraph -- there are different kinds of bugs than simply the "logic bugs" (i.e. the program not doing what it's supposed to) which I primarily focused on in my previous argument. I'm still not worried about non-computer proofs -- I acknowledge we have no way to say they are "entirely correct in the utmost sense as if god had come down and handed it to us or whatever", but I think it is implicitly accepted that "correct up to the point of not conflicting with basic human logic" is a reasonable standard of acceptance for proof. I don't understand how or why this extends to computer proofs.

[quote="t0rajir0u"]Your example of grading for USAMTS is suggestive - it is true that programs like Mathematica do occasionally give wrong answers, but the only examples I can think of are abstruse series problems.  As far as I know, the rest of Mathematica is reliable, but I can see where you're coming from.[/quote]

YAH SO GUYS IF ANYBODY WANTS ME TO GIVE THEM POINTS FOR COMPUTER PROOFS ON USAMTS THIS IS TOTALLY THE PLACE TO CONVINCE ME TO DO SO

[quote="MysticTerminator"]I don't understand how or why this extends to computer proofs.[/quote]

I don't understand what you don't understand.  Many proofs (especially in combinatorial fields) are algorithmic: here's a process that finds (or counts, or whatever) an object of interest, and here's a proof that it really does this.  If this algorithm happens to be in C, what exactly is the problem?

I'm not sure I understand the objection that computer proofs are "uncheckable," either.  After all, I'm well on my way to being a professional mathematician but 95% or more of what the people around me are proving is totally inaccessible to me.  Even many papers in my own field are "uncheckable" as far as I am concerned, but I believe their results because, in principle, someone with enough knowledge and time could check them (and maybe already has).  A computer proof is checkable in exactly the same sense: any results the computer claims could be checked one by one by sufficiently many people with sufficiently much time.  (This is true of case-checking algorithms like 4CT and also of proof-writing algorithms (which should always be able to present a proof, not just the claim that the result is provable).)

Some interesting personalities worth reading about on this subject are Doron Zeilberger (a mathematician at Rutgers who sometimes lists his computer as a coauthor on papers) and Gregory Chaitin* (who believes that incompleteness means that mathematicians shouldn't be so timid about adding new axioms given sufficient empirical support).

* The book of his I read was called Meta-Math, and I really hated the first half of it, but it did have some interesting ideas.

I'm drawing a distinction between the part of the proof that is actually accessible to human intelligence and the part that is not. A computer proof, in my opinions, consists of three things: some source code that describe a program, a computer that runs the program and outputs some result, and an analysis of this result. The first and the third are essentially what you describe in your post above and are what I've been describing as accessible to human thought. I have no problem with this. The second part is not accessible to human thought. You have no way of confirming from pure logic that the bits in the computer are flowing the way they're supposed to be. Note that none of the examples you cite (algorithmic proofs in combinatorics, proofs in general which you have insufficient knowledge to understand, and so forth) suffer this flaw.

I don't see why it matters: if the program outputs a proof of some result, it doesn't matter what the bits were doing if the proof is right.  Similarly, if the program tells me "there are no examples of X among the 100000 input graphs," this result is entirely susceptible to human thought.  Just because no one [i]wants[/i] to spend their energy checking it doesn't mean it's inaccessible to checking [i]in principle[/i].

Yes, those types of proofs seem valid. It's a bit annoying since you know nobody'll ever check them, but then again I know there are a lot of perfectly "handwritten" proofs that nobody wants to bother checking too.  :wink:

So I'm referring to something more like what you might see as USAMTS grader. Problem says "HEY FIND ALL THE INTEGERS THAT DO SUCH AND SUCH" and contestant is all like "hey here's my source code that should tests each integer and here is what the computer outputted". Do I accept this or not?

Similarly as a grader for a course at college -- someone "proves" or "solves" something through the use of mathematica to evaluate an integral or some such -- do I accept this? (if anyone's curious, I did, but in much the same way as one "accepts" calculus on olympiads)

HAH I USED A LOT OF QUOTATION MARKS IN THIS POST

[quote="JBL"]I don't see why it matters...  Just because no one [i]wants[/i] to spend their energy checking it doesn't mean it's inaccessible to checking [i]in principle[/i].[/quote]

Isn't this quite a fine line to be treading? You consider a proof correct just because it is possible to check in principle. I think there is a distinction between a proof having been checked by people capable (Fermat's Last Theorem) and is not possible for me to check. And to no one having checked FLT, but saying that it is correct because it is possible to check in principle.

I'm of the opinion that computer proofs can be accepted as proof. My goal is to get others to accept them, but under a proper title. Empiricism isn't bad. But let's not pretend that its not there when we accept one of these proofs.

what sort of a "proper title" would you formulate? since it seems you are drawing a distinction between empirically supported evidence and the standard notion of proof.

[quote="HilbertThm90"]You consider a proof correct just because it is possible to check in principle.[/quote]  This sentence misses a major piece of the idea: if a proof is produced [i]by some generally reliable means[/i] and is in principle checkable, I am perfectly happy to conclude that it's true.  This is, after all, exactly what I do with all the theorems whose proofs I've never read. The major objection raised in this thread seems to be that computer proofs might be suspect because we can't, in principle, check that they are correct.  I think that this is, in general, wrong (although it is true of certain "proofs" I've seen, e.g. on this year's ARML power round: "We enumerated all cases by brute force, and the answer is [such and such]").

(I couldn't parse the rest of the associated paragraph, unfortunately, so I'm not sure exactly of the distinction you drew there involving FLT.)

On problem sets I don't see what the problem is: a grader has discretion to grade not strictly on grounds of correctness -- "being lame and using mathematica to do your homework" seems like a perfectly legitimate reason to dock a point or two, just as "having the right idea but making stupid computational errors" or "handing in work that reaches the right answer but is otherwise illegible" are.  I don't remember what grading standards USAMTS claims -- one possible resolution is to demand a demonstration that the algorithm described, if carried out correctly, would yield the correct answer.

well, I grade based on whether or not I think things are correct.

I admit that paragraph was weirdly worded.

Clear your mind of all history of knowing how FLT played out. Say Andrew Wiles said that he proved FLT. I'd say that him being a professional mathematician is a pretty "reliable" means. Since he wrote it down, it is also in principle checkable. But I don't think a single person would say, "Ah, you proved it and it is correct!" without someone actually doing the checking.

I don't see how this is different from the computer...unless you are saying that the computer provides a reliable means and a human never does. In that case, I would buy your stance as sound, but I must say that it is still an empirical discovery (due to the slight chance of error and not checking for it), and anyone that says that empiricism is not a valid proof method cannot hold the stance you are holding (and thus reject 4CT as being proven).

If someone did present to me a proof of a theorem by computer, then I would want to see the source code and a proof that the code and/or its output actually proves the theorem.

As mentioned before, there is always a possibility of floating point error, unless you use an environment that can maintain exact values of numbers.

I think the problem that Mystic is trying to say (maybe?) is that even if you verify that the program you wrote would definitely produce the correct output based on the semantics of the language, there are more issues.

How can you guarantee that the binary you compile executes your code?
How can you guarantee that your computer's memory is not corrupted?

I think the concerns are at compile time (maybe link time too) and run time. Furthermore, it's almost impossible to verify that all of the bits in your computer are what you expected them to be. It becomes a physical problem rather than a mathematical one.

That said, I do trust computer programs being a computer scientist myself :). The probability of such failures is small enough to convince me that things are true/false.

well, sure, I trust them too. as we all know from our olympiad days, though, proofs are tricksy because you have to maintain this kind-of annoying level of rigor. I'm still unconvinced that the problems I describe (and which you picked up on, yay) aren't enough to bring this down.

[quote="paladin8"]How can you guarantee that the binary you compile executes your code?
How can you guarantee that your computer's memory is not corrupted?[/quote]

See my post #16 on this thread: a computer proof isn't simply a black-box into which outputs theorems: it has to, along the way, do something equivalent to actually proving whatever it is you want it to prove.  And if it does this, it can output data that allows you to check a proof.  Proof-checking is in principle trivial, so proof-writing is, in some sense, NP, so whether the physical computer did exactly what it was supposed to is irrelevant.  (Of course, the reliability of a computer is far greater than the reliability of human mathematicians, so all this worrying seems rather misplaced to me.)

What if you need the computer to brute force check a million - or a billion - cases? You can have it output all the data that it uses, but how can you check it all? If you actually have the time or ability to check what the computer outputs, then this entire question becomes useless.

so joel, I'm acknowledging that if the computer outputs a proof, that's fine. I don't care where the proof came from -- from human creation, from a computer, or from strange arrangements of rocks. Once you have a proof that is written down (or printed or whatever) somewhere, it's there and that's all that matters. I'm objecting to the use of computers IN proofs. Not in CREATING the proofs but [i]in the actual proofs themselves[/i].

@hance joel also raised this point and I agree with him here -- if it's in principle checkable, I'm going to be ok with it. honestly, there are really a surprising number of mathematical results (generated "by hand", as it were) that nobody would ever want to check. ok, some people probably do check them, but most people are just like ehhhh I'm going to do something else now instead.

I've thought about the validity of computer-assisted proofs before, and my own personal conclusion is that we tend to ask the wrong question:  Is a computer-assisted proof a valid proof?  I consider myself to be a mathematician, not a proof-ician.  In other words, we write proofs in order to further the study of mathematics, not vice versa.  For me, the right question is whether computer-assisted mathematics is good mathematics?  This question can only be answered subjectively on a case-by-case basis.

The study of mathematics is a more complicated and beautiful enterprise than the mere cataloguing of formal statements as proved, disproved, and not-yet-proved-or-disproved.  Instead of focusing on whether something is or is not a proof, we should think about the *reasons why* we write down proofs to begin with, and then evaluate computer-assisted mathematical work based on how well it achieves those ends.  I wrote a little blog post one day on this subject if you're interested:

http://yenlee.blogspot.com/2007/08/why-do-we-prove-things-or-what-is-good.html

As for the practical issue of grading, I think these issues are irrelevant.  You're trying to teach people something, and you can set the agenda for what you want them to learn.  You can decide to allow computers, not allow computers, or require the use of computers.  If there is no official policy, then all parties involved should use a degree of common sense.  For example, the "common sense" standard is used all the time in determining what constitutes a rigorous proof.

hah. just leave terry tao alone, huh?  :lol:

interesting blog post, but I'm afraid I depart with you right from the start -- in my view, proofs are all about criterion A that you set down. B is good, necessary, and probably far more important than A, and usually the idea does come through in the proof, but the proof need not have anything to do with B if it doesn't want to. I'm also confused as to how you draw this distinction between "being convinced by something" and "rigour". If the proof's rigorous and I can understand it, I'm convinced. If the proof's not rigorous and I can understand it, I'm convinced if and only if I can see how to make it rigorous by certain standard arguments myself (and this is, in my understanding, usually the case in texts aimed at graduate-level education and beyond). If the proof seems pretty believable but I have no idea how to actually make it rigorous, I'll just be kinda like ehhhh.

[quote="MysticTerminator"]It should be possible theoretically to prove, given some source code, that pending the outcome of a certain result upon running the program, a certain result may have been proven. It seems that at some level the source code is "accessible" to human thought processes and hence may justifiably constitute a part of a valid proof. The compilation and implementation of the source code concerns me, though, as this part seems less accessible. You don't have any way to check that all the bits are flowing the way they're supposed to, so even if the program does output a certain result upon completion, how do you know your computer wasn't just screwing up? Granted, there's a small chance of that, since computers don't usually randomly screw up, but it's been known to happen, and I don't think people usually accept proofs which run along the lines of "well I can't check it completely but y'know it's got a pretty good chance of being true". On the other hand, it would be disappointing to rid oneself of a powerful potential proof technique for handling a plethora of special cases. So?[/quote]

You could run the exact same program on many computers (probably 50-100 to be the most accurate) and use the most common result. Since computers rarely mess up and when they do they rarely do it in the same way, there should only be one of each invalid result.

in other words, you're happy with accepting probabilistic proof. I'm not.

Here is a tangentially related discussion that occurred a few years ago: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9991

thanks, gauss. I don't feel like necroing right now, so I'll respond here.

I'm very much a left-winger, since I accept that there may well be some weird thing in everybody's brain that makes everyone ignore a certain logical flaw or whatever the case may be. and I'm ok with that. as long as it convinces me, I'm ok with putting up with the inherent limitation my neurological system imposes upon me. I simply happen to have high standards of what does convince me.

the prevalence of sociology in math is also interesting -- the whole hilbert-brouwer acrimony for instance and so forth, but I hardly know enough history of math to offer anything to that discussion (and this thread isn't really the place, I suppose).

Mystic:  Perhaps you are confused, because I did NOT attempt to draw a distinction between "being convinced by something" and "rigour."  I essentially said that they are the same thing.  I drew a distinction between *convincing* someone *that* something is true versus *explaining* to someone *why* something is (or is likely to be) true.  You are correct that the "convincing" part is what generally qualifies a putative proof as a legitimate proof, but what I'm concerned about is not what makes a putative proof a legimate proof but rather what makes a putative proof a good piece of mathematics.

We indoctrinate the importance of the rigorous mathematical proof in undergraduate mathematical education because it's useful as a pedagogical tool: it teaches students how to think clearly and communicate clearly.  But in the end, the proof is not the goal.  Understanding mathematics is the goal.  This is why partial results can be good math.  This is why examples and counterexamples can be good math.  This is why computer experiments can be good math.  And most importantly, this is why there is value in giving two different proofs of the same result.  Sometimes the less rigorous proof may be the more valuable one.

Regarding the Devlin article:  Mildly interesting, but nothing newsworthy or controversial.  Nothing at all happened in 2003 that made mathematicians think differently about proofs.  Obviously, a putative proof is a legitimate proof when the mathematical community says it is.  What other reasonable standard could there possibly be?  It's a lot like a physics experiment.  The results of a physics experiment are supposed to be undeniable *facts* (just like theorems are undeniable truths), but the validity of the experiment can only be determined by the physics community as a whole.  Perhaps this viewpoint is slightly "postmodern," but it's essentially common sense, isn't it?

[quote="MysticTerminator"]Not in CREATING the proofs but [i]in the actual proofs themselves[/i].[/quote]  Okay, so obviously some of my disagreement with you is based on the fact that I'm not sure what you mean by this :).  Since you seem to agree that a finite case-bash by a computer as part of a proof is acceptable, can you give an example or two of something that you find questionable?  (Really, I guess I mean, "can you give an example not in the context of USAMTS solutions?")

I see nothing wrong with using a computer for proving theorems. As long as I can
a) Verify the validity of the algorithm;
b) Check the code;
c) Check the general ability of the computer to function,
I believe the output. The human verification ability is by no means superior to the computer ability to function without an error. I had at least 5 occasions where the proofs verified by humans (i.e., published in refereed journals, cited, etc.) were wrong (in 2 cases the statements were wrong, in other cases there was a gap that was quite hard (but, fortunately, possible) to fill) and I have never seen a code that repeatedly malfunctioned because of the computer errors, not because the program was flawed (the famous story with Pentium multiplication error may be the only exclusion).

In general, in most of the situations, the task delegated to the computers is just some routine and straightforward verification of a large amount cases. There is nothing there to understand really. Once you know exactly what to do in each particular case and why you should do it, no extra understanding is achieved by actually doing them (or verifying that they are done correctly).

A completely different question is whether the computer-assisted proofs are "elegant" or satisfy other beauty standards. I hold the point of view that such proofs may be beautiful if they contain a nice idea of how exactly you should reduce the problem to checking just finitely many things (it is not always obvious) and if the algorithm/computer code satisfy my "programming beauty standards", but it is just a personal view and I do not see any point in arguing about such things.

Also, my embracing computer assisted proofs doesn't mean that I won't appreciate an alternative "conventional" proof when it appears (provided that the latter is based on a different idea, not just reduces the casework somehow). On the other hand, a long and tangled "conventional" proof may, in my opinion, be inferior to an ideologically short and clear proof using the computer to check a few hundred similar numerical inequalities (or whatever is there to check).

The last relevant thing is the question of whether seeing that something happens in all the 1000000 cases one has to consider and not seeing the common underlying reason for that means the lack of true understanding of what is really going on. I believe that sometimes it does and sometimes it doesn't. Not all things, even in mathematics, have an ultimate simple reason behind them. Sometimes they are true just because they are (like $ x^2\plus{}x\plus{}41$ produces a long list of primes for no particular reason as far as I can tell; this may be not the best example, but, I hope you see the point), so if we can find that particular reason, great, if not, we can live without it and go further.


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inequalities
inequalities unsolved

[quote="bellahsayn"]$x$,$y$,$z$>0
$\frac{x}{\sqrt{y+z}}$+$\frac{y}{\sqrt{x+z}}$+$\frac{z}{\sqrt{y+x}}$<=$\frac{5}{4}$$ \sqrt{  x+y+z}$
and tank you[/quote]

Take x->0 and y=z then it's false?

I remember this one! Boy, what I trouble I had with this one.  :D . For the correct statement search in the solved or unsolved problems inequalities the post of Fedor Bekharev "a lovely inequality".

harazi, I think his inequality is false, because it's different from that in
http://www.mathlinks.ro/Forum/viewtopic.php?t=1459
(note the denominator)

I know, but the presence of $5/4$ makes me think that he made a typo and in fact he wants to prove that monster inequality.  :?

for this problem I think there is no k,
because it is a little from which siuhochung mentioned.