function
integration
linear algebra
matrix
algebra
linear equation
real analysis

I once put this question on the final exam of a differential equations course, and immediately wished I hadn't. I'm posting it now to remind myself not to use it in a classroom situation again:

---

Let $ n$ be any positive integer.

(a) Find a general formula (depending on $ n$) for the Wronskian of the functions $ \{e^x,xe^x,x^2e^x,\dots,x^ne^x\}.$

(b) Is there (for fixed $ n$) a single differential equation that these functions are all solutions to? If so, what must its order be? Can you state such an equation?

[(b) by itself might have been OK as a question. But (a) is a bit messy.]

I would certainly make (b) (a) and (a) (b). Why? [hide]Because of Abel's formula[/hide]

I probably won't re-use the question. But just to make a nicer answer, I think I'd rather have the set of functions be $ \left\{\frac{x^ke^x}{k!}\right\}_{k\equal{}0}^{\infty}.$

In some previous year, I'd used the powers of $ x,$ with no exponential function. I was just trying to tweak the question a little.

Time to put this topic to bed, and to explain what fedja was talking about.

A piece of general theory: suppose we have the linear homogeneous differential equation

$ y^{[n]} \plus{} p_{n \minus{} 1}(x)y^{[n \minus{} 1]} \plus{} \cdots \plus{} p_1(x)y' \plus{} p_0(x)y \equal{} 0$

and suppose that $ y_1,y_2,\dots, y_n$ are solutions of this equation. Then the Wronskian $ W \equal{} W(y_1,y_2,\dots,y_n)$ satisfies the first order linear equation $ W' \plus{} p_{n \minus{} 1}(x)W \equal{} 0.$ It follows that we can write $ W \equal{} Ce^{ \minus{} \int p_{n \minus{} 1}(x)\,dx},$ and that on an interval on which all of the coefficients are nonsingular, $ W$ is nonzero at one point if and only if it is nonzero on the whole interval.

One of the reasons this is a very nice theorem is that if we are using the Wronskian to check whether we have a full set of solutions, it suffices to check it at one point.

Now, for my particular problem: the functions $ e^x,xe^x,\dots,x^ne^x$ are all solutions of the differential equation $ (D \minus{} 1)^{n \plus{} 1}y \equal{} 0.$ Writing this out in detail:

$ y^{[n \plus{} 1]} \minus{} (n \plus{} 1)y^{[n]} \plus{} \binom{n \plus{} 1}{2}y^{[n \minus{} 1]} \plus{} \cdots \plus{} ( \minus{} 1)^{n \minus{} 1}y \equal{} 0.$

By the theorem I have just quoted, we have that $ W(x) \equal{} Ce^{(n \plus{} 1)x}.$ Now all we have to do is find $ C.$ We'll focus on what $ W(0)$ is. Consider expanding everything in power series about zero. We get the following for the Wronskian matrix:

$ \begin{bmatrix}1 & & & & & \\
&1 & & & & \\
& & 2 & & 0 & \\
&* & & 3! & & \\
& & & & \ddots & \\
& & & & & n!\end{bmatrix} \plus{} O(x).$

Taking the determinant at $ 0,$ we get $ W(0) \equal{} \prod_{k \equal{} 0}^{n}k!$

Thus $ W(x) \equal{} \left(\prod_{k \equal{} 0}^{n}k!\right)\,e^{(n \plus{} 1)x}.$


Let consider the set of all graph of size n with this property : every subgraph of size k contains at most H edges. The problem is to find in this set the graph or the graphes that maximize the numer of edges. Any suggestion?

It looks like the author has messed up with in coherent use of the word size. If I am not wrong, the problem has be reworded as follows.

Let $ G$ be a set of graphs with $ n$ vertices ie., of order $ n$ such that $ \forall g\in G$,

(i) For every subgraph of order $ k$ has size $ h$ ie., every subgraph with $ k$ vertices has $ h$ edges.
(ii) The size of $ G$ is maximum.

Enumerate $ G$.

There is a double counting here. Let $ m$ be the maximum size of $ g$ we are expecting.

First, lets see how many times a vertex get counted if one identifies every set of $ k$ vertices and counts the edges in them. This would give
\[ m{n\choose {k\minus{}2}}\]

Second, let us count the vertices directly. This gives at maximum
\[ h{n\choose k}\]

So, we obtain(? I am a bit confused here)
\[ m{n\choose {k\minus{}2}}\le h{n\choose k}\]

[quote="Talegari"]It looks like the author has messed up with in coherent use of the word size.[/quote]  Let he who has not sinned cast the first stone ;).  (Also, it is certainly not a universal convention that "the size of a graph" means the number of its edges.)

I don't know how you think your vertex counting is supposed to go.  It's elementary to see (e.g., by counting the [i]edges[/i] in two different ways) that the total number of edges is at most $ \binom{n}{2} \cdot \frac{h}{\binom{k}{2}}$, but this is not going to be sharp.  For example, with $ h \equal{} \binom{k}{2} \minus{} 1$ the result we're trying to recover is Turán's Theorem, but the bound this analysis gives is much too large.

Another approach would be to deploy Turan to get some bounds (but I have no idea how good they will be).  For example, if we choose $ r$ such that $ \binom{r \plus{} 1}{2} > h \geq \binom{r}{2}$, our graph has no $ r \plus{} 1$-clique and so Turán tells us it has at most $ \frac{r \minus{} 1}{r} \cdot \frac{n^2}{2}$.  But when $ r$ is not very close to $ k$, this bound is probably not any good, either.


Show that:
1+x*x/1+y+z*z + 1+y*y/1+z+x*x + 1+z*z/1+x+y*y>2, whit x,y,z diferent positive real numbers .

Here is the JBMO 2003, differen you :?:
http://www.mathlinks.ro/Forum/viewtopic.php?p=20974#p20974


A regular $18$-sided polygon is inscribed in a circle. Using the $18$ vertex, how many

1. Right angle triangles can be formed?

2. Obtuse Angle Triangles can be formed?

3. Acute Angle Triangles can be formed?

Hints[hide]#1--> Draw possible rectangles
#2#3--> draw parallelograms[/hide]

that should be enough to make this problem simple.


(a+b+c)^3 - a^3-b^3-c^3=3(a+b)(b+c)(c+a)
(a+b+c)^5 - a^5-b^5-c^5=5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)
using the fact that a+b+c=1 we get the inequality to prove to be same as
2/3  \leq  (a^2+b^2+c^2+ab+bc+ca)
multiply 2 by (a+b+c)^2 to reduce the inequality to
0 \leq a^2+b^2+c^2-ab-bc-ca
which is well known
hope the proof is correct
cheers

actually this problem is credited to Marcel Chirita, RMT 1982 (Romanian Mathematical Magazine from Timisoara :D).


analytic geometry
vector
rational number
number theory
IMO Shortlist

Prove that on the coordinate plane it is impossible to draw a closed broken line such that

[i](i)[/i] the coordinates of each vertex are rational;
[i](ii)[/i] the length each of its edges is 1;
[i](iii)[/i] the line has an odd number of vertices.

[hide]
Assume one such line exists. First, scale this line so that the the coordinates are integers. Now, we can represent each edge of this broken line as a vector $ (x,y)$ with $ x$ and $ y$ and magnitude $ d$. The sum of all these vectors will be $ (0,0)$. Call these vectors

$ (x_1,y_1),(x_2,y_2),...,(x_n,y_n)$.

where $ n$ is odd. Then $ x_1^2 \plus{} y_1^2 \equal{} x_2^2 \plus{} y_2^2 \equal{} ... \equal{} x_n^2 \plus{} y_n^2 \equal{} d^2$.

Suppose that at least one of $ x_1,x_2,...,x_n,y_1,y_2,...,y_n$ is odd. Otherwise, we can divide all of the coordinates and $ d$ by $ 2$. Now, because $ x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n \equal{} 0$ and $ y_1 \plus{} y_2 \plus{} \cdots \plus{} y_n \equal{} 0$, an even number of $ x_1,x_2,...,x_n$ are odd and an even number of $ y_1,y_2,...,y_n$ are odd.

Now I will show that $ d$ is even. Suppose for the sake of contradiction that $ d$ is odd. Then for each pair $ (x_i,y_i)$, exactly one of $ x_i$ and $ y_i$ is odd. So there are exactly $ n$ coordinates which are odd, which is a contradiction since we showed above that the number of odd coordinates is even. Hence $ d$ is even, so $ d^2\equiv 0\mod 4$.

But we are also assuming that there is at least one odd coordinate, so for some $ i$, we have $ x_i^2 \plus{} y_i^2\equiv 1$ or $ 2\mod 4$, a contradiction.

Note this generalizes to where $ d^2$ is any integer, not necessarily that $ d$ is an integer. There is some pair $ (x_i,y_i)$ with $ x_i,y_i$ odd, and there is some pair $ (x_j,y_j)$ (since otherwise every coordinate is odd), so $ x_i^2 \plus{} y_i^2\not\equiv x_j^2 \plus{} y_j^2\mod 4$.
[/hide]

by graph::
[hide]the distance we move away from point = distance coming back
therefore, total distace=d1+d2   ,, let they be expressed in no. of edges
if d1is odd then d2 is odd so, sum is even and same in case of even edges .
    hence, no. of edges is even
 connecting firt two edges gives 2 vertices and +1 on every subsequent edge, but +-0 on the last edge
hence, +2 on 1st edge and then no. of vertices willbe even on no. of edges as there will be +1 always until last edge . at that time, edge is odd so vertex is even and there will no addition of vertex on last edge.
                  hence no. of vertices is even.so, this situation aint possible [/hide]


trigonometry
real analysis
real analysis unsolved

I have a problem from my Calc I teacher that I cannot solve. The only thing he has told me is to check the book but there are no examples that coorelate to the question. The Question is
Use differentials (linear approximation) to estimate cos(-2). Base the approximation at x=2pi\3.
cos(-2) is approximately equal to ???[/code]

You need to compute the equation of the tangent to $\cos$ in $\left( \frac{2 \pi}3, \cos \left( \frac{2 \pi}3 \right) \right)$:
\[\frac{y-\cos \left( \frac{2 \pi}3 \right)}{x-\frac{2 \pi}3}= \cos^\prime \left( \frac{2 \pi}3 \right) , \]
i.e. $y =-\frac{x \sqrt 3}2+\frac{\pi}{\sqrt 3}-\frac12$.

For $x=2$ we have that $y =-\sqrt 3+\frac{\pi}{\sqrt 3}-\frac12 \approx-0.41825$.

Using a calculator we see that $\cos 2 \approx-0.41614$.


$ A\equal{}\left(\begin {array}{cccc}
 0 &  a &  b & c \\
\minus{}a &  0 &  d & e \\
\minus{}b & \minus{}d &  0 & f \\
\minus{}c & \minus{}e & \minus{}f & 0
 \end {array}
 \right)$

Compute $ \det(A)$.

P.S. I know it is Burnside's Theorem, but I just want to know how to compute it as simply as possible.

$ \det(A) \equal{} \det \left[\begin {array}{cc} 0 & a \\
\minus{} a & 0 \end {array} \right] \cdot \det \left(\left[\begin {array}{cc} 0 & f \\
\minus{} f & 0 \end {array} \right] \minus{} \left[\begin {array}{cc} \minus{} b & \minus{} d \\
\minus{} c & \minus{} e \end {array} \right] \left[\begin {array}{cc} 0 & \minus{} 1/a \\
1/a & 0 \end {array} \right] \left[\begin {array}{cc} b & d \\
c & e \end {array} \right]\right)$

$ \equal{} (be\minus{}cd)^2 \plus{} af(af \plus{} c^2 \minus{} 2bc \plus{} d^2)$

It is well known that the det of a skew symmetric matrix $ M\equal{}[m_{ij}]$ of even dimension $ n$, with coefficients in a commutative ring, is a square (Pfaffian).
If $ n\equal{}4$ we find $ (m_{12}m_{34}\plus{}m_{23}m_{14}\minus{}m_{13}m_{24})^2$.
See http://en.wikipedia.org/wiki/Skew-symmetric_matrix


A triangle $ ABC$ is inscribed in a circle with center $ U$ and radius $ r$. A tangent $ c'$ to a larger circle $ K(U,2r)$ is drawn so that $ C$ lies between the lines $ c\equal{}AB$ and $ C'$. Lines $ a'$ and $ b'$ are analogously defined. The triangle formed by $ a',b',c'$ is denoted $ A'B'C'$. Prove that the three lines, joining the midpoints of pairs of parallel sides of the two triangles, have a common point.

Why does this problem need $ K(U,2R)$?
It's very simple. We only notice that if two triangles $ XYZ$ and $ X'Y'Z'$ have $ XY//X'Y', YZ//Y'Z', XZ//X'Z'$ then $ XX', YY', ZZ'$ are concurrent.


For how many integer $ n$, $ P \equal{} n^4 \plus{} 4n^3 \plus{} 3n^2 \minus{} 2n \plus{} 7$ is prime?

$\textbf{(A)}\ 1  \qquad\textbf{(B)}\ 2  \qquad\textbf{(C)}\ 3  \qquad\textbf{(D)}\ 4  \qquad\textbf{(E)}\ \text{Infinitely many}$

That's $ (n^2\plus{}5n\plus{}7)(n^2\minus{}n\plus{}1)$. Set each equal to one or -1 and check.

These are $ n \equal{} 0,1,\minus{}2,\minus{}3$. But is there an acceptable way to guess the factors ? For instance in 4 minutes?

I make it n = 0, 1, -2 or -3.
Factoring as (n^2 + ax + b)(n^2 + cn + d) is quick since b=-1 and d=-7 (which doesn't work) or b = 1 and d= 7 which does.


algorithm
Ring Theory
geometry
3D geometry
number theory
relatively prime
Euclidean algorithm

[hide]$ x^2\equiv y^3\equiv 0,1(mod 7)$

If $ x\equiv y\equiv 0$, we have $ 7(7k^2\plus{}49l^2)\equal{}7$, which is impossible as $ x\equal{}7k, y\equal{}7l, \in N$

PProve similarly for the case $ x\equiv y\equiv 1$[/hide]

[quote="hell_ever"][hide]$ x^2\equiv y^3\equiv 0,1(mod 7)$

If $ x\equiv y\equiv 0$, we have $ 7(7k^2 \plus{} 49l^2) \equal{} 7$, which is impossible as $ x \equal{} 7k, y \equal{} 7l, \in N$

PProve similarly for the case $ x\equiv y\equiv 1$[/hide][/quote]
Ummm, how do we prove similarly for the case $ x \equiv y \equiv 1 \mod 7$?

I don't understand how to prove that either.

[quote="Construe"]Prove that $ x^2 \minus{} y^3 \equal{} 7$ does not have any solutions where $ x$ and $ y$ are positive integers.[/quote]
[hide="algebraic number theory to the rescue"]
$ (x\minus{}\sqrt{7})(x\plus{}\sqrt{7})\equal{}y^{3}$
These two numbers are relatively prime, since by the euclidean algorithm
$ (x\minus{}\sqrt{7},x\plus{}\sqrt{7})\equal{}(2\sqrt{7},x\plus{}\sqrt{7})\equal{}1$
So $ x\plus{}\sqrt{7}\equal{}(a\plus{}b\sqrt{7})^{3}\equal{}a^{3}\plus{}3a^{2}b\sqrt{7}\plus{}21ab^{2}\plus{}7b^{3}\sqrt{7}$
$ x\equal{}a^{3}\plus{}21ab^{2}$
$ 1\equal{}3a^{2}b\plus{}7b^{3}$ which implies that b is a unit, which, in this case, is either 1 or -1.
Checking both, we find no solutions.
[/hide]

[quote="diophantient"][quote="Construe"]Prove that $ x^2 \minus{} y^3 \equal{} 7$ does not have any solutions where $ x$ and $ y$ are positive integers.[/quote]
[hide="algebraic number theory to the rescue"]
$ (x \minus{} \sqrt {7})(x \plus{} \sqrt {7}) \equal{} y^{3}$
These two numbers are relatively prime, since by the euclidean algorithm
$ (x \minus{} \sqrt {7},x \plus{} \sqrt {7}) \equal{} (2\sqrt {7},x \plus{} \sqrt {7}) \equal{} 1$
So $ x \plus{} \sqrt {7} \equal{} (a \plus{} b\sqrt {7})^{3} \equal{} a^{3} \plus{} 3a^{2}b\sqrt {7} \plus{} 21ab^{2} \plus{} 7b^{3}\sqrt {7}$
$ x \equal{} a^{3} \plus{} 21ab^{2}$
$ 1 \equal{} 3a^{2}b \plus{} 7b^{3}$ which implies that b is a unit, which, in this case, is either 1 or -1.
Checking both, we find no solutions.
[/hide][/quote]

Too bad $ \mathbb{Z}[\sqrt {7}]$ isn't a UFD. $ (8 \plus{} \sqrt {7})(8 \minus{} \sqrt {7}) \equal{} 15 \equal{} 3*5$

I don't know if $ {\bf Z}[\sqrt {7}]$ is a UFD, but that example doesn't settle it either way because
(1) most of the expressions you used aren't irreducible:  $ 8 \minus{} \sqrt {7} \equal{} (\sqrt {7} \minus{} 2)(3 \plus{} 2\sqrt {7})$, $ 8 \plus{} \sqrt {7} \equal{} (\sqrt {7} \plus{} 2)(2\sqrt {7} \minus{} 3)$, and $ 3 \equal{} (\sqrt {7} \minus{} 2)(\sqrt {7} \plus{} 2)$; and
(2) the two sides aren't actually equal to each other ($ (8 \minus{} \sqrt{7}) \cdot (8 \plus{} \sqrt{7}) \equal{} 57$, not 15).

$ \mathbb{Z}[\sqrt{7}]$ is in fact a UFD and even Euclidean, but a proof of this is more difficult than that of, say, $ \mathbb{Z}[\sqrt{3}]$.

But even then, that proof must be fixed: having $ \alpha \beta$ equal to a cube and $ \alpha,\beta$ relatively prime does not guarantee that $ \alpha,\beta$ must both be cubes. It may be that $ \alpha (8\plus{}3\sqrt{7}),\beta (8\minus{}3\sqrt{7})$ are both cubes, or $ \alpha (8\minus{}3\sqrt{7}),\beta (8\plus{}3\sqrt{7})$ are both cubes. (These are in fact the only two other cases, since all units are of the form $ \pm (8\plus{}3\sqrt{7})^n$ where $ n$ is an integer.)

[quote="scorpius119"]$ \mathbb{Z}[\sqrt {7}]$ is in fact a UFD and even Euclidean, but a proof of this is more difficult than that of, say, $ \mathbb{Z}[\sqrt {3}]$.

But even then, that proof must be fixed: having $ \alpha \beta$ equal to a cube and $ \alpha,\beta$ relatively prime does not guarantee that $ \alpha,\beta$ must both be cubes. It may be that $ \alpha (8 \plus{} 3\sqrt {7}),\beta (8 \minus{} 3\sqrt {7})$ are both cubes, or $ \alpha (8 \minus{} 3\sqrt {7}),\beta (8 \plus{} 3\sqrt {7})$ are both cubes. (These are in fact the only two other cases, since all units are of the form $ \pm (8 \plus{} 3\sqrt {7})^n$ where $ n$ is an integer.)[/quote]
How did you find $ 8\minus{}3\sqrt{7}$ is a unit?

It is a well-known equation! It's inspired by Lesbagues(I'm not sure I spell it correctly!). It even has been solved in general form: y^3+x^2=k where k is constant and x,y is positive integer! Searching Internet(maybe Wiki or cut the knot), you guys will see the solution for this equation! :wink:

Have fun!


[quote="2424"][quote="caubetoanhoc94"]Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:

$ (a \plus{} b \plus{} c)^2(a^2 \plus{} b^2)(a^2 \plus{} c^2)(b^2 \plus{} c^2)\geq8(a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2)^2$[/quote]
@caubetoanhoc94:[hide] Bai nay trong sang tao co ma,post len lam j day chu e? [/hide][/quote]

@2424[hide]Để em xem.Nhưng bài này cũ ...Anh nhìn thời gian em post  kìa :).lâu lắm rồi :).Lúc đếy mới vào :).Thấy bài này đẹp post lên :) chứ em có bik gì đâu :).Thế mà anh vẫn lục lại  :P [/hide]


conics
parabola
symmetry
analytic geometry
graphing lines
slope
geometry

[quote="AndrewTom"]Is it possible to argue that since there is a tangent from (0,-4) to y=x^2 +x at (2,6), there must be a tangent from (0,-4) to y=x^2-x at (2,2) and so by symmetry there are tangents from (0,-4) to the first parabola at (-2,2) and to the second at (-2,6)?[/quote]

Yes, it is possible to argue that way. However, I believe it is easier to show that (0,-4) lies on the tangent line to the first parabola at (2,6) and also on the tangent line to the second parabola at (-2,6). Likewise, you can show (0,-4) lies on the tangent line to the second parabola at (2,2) and the first, at (-2,2).

That way you show proof based upon sound algebra, as opposed to a bunch of hand waving based upon the graphics processing capability of your calibrated eyeball.

Obviously. However, my reason for asking is the following.

The original question had three parts. The first was to find the tangent to y = x^2 + x when x = 2. The second was to find the normal at that point. The third was to find the values of k for which y = kx - 4 is a tangent to the curve. The first part just happens to give k=5, which is what got me thinking about reflecting in the y-axis and considering symmetry, when I read the third part, but I first did the third part by showing that x^2+ (1-k)x +4 is a perfect square for k = -3 or 5, which is equivalent to having a zero discriminant. My problem is to show that if y=x^2 +x has a tangent from(0,-4) at (2,6) then y=x^2-x must have a tangent from (0,-4) at (2.2) as the rest is obvious.

[quote="AndrewTom"]Obviously. However, my reason for asking is the following. ...

My problem is to show that if y=x^2 +x has a tangent from(0,-4) at (2,6) then y=x^2-x must have a tangent from (0,-4) at (2.2) as the rest is obvious.[/quote]

Here's how I would do that.

Let $ f(x) \equal{} x^2 \plus{} x$ and let $ g(x) \equal{} x^2 \minus{} x$.

Then $ g(x) \equal{} f(x) \minus{} 2 x$.

Therefore, $ g(2) \equal{} f(2) \minus{}2(2) \equal{} 6 \minus{} 4 \equal{} 2$.

Furthermore, $ g'(x) \equal{} f'(x) \minus{} 2$,

so that $ g'(2) \equal{} f'(2) \minus{} 2$.

For $ f(x)$ the slope of your tangent line at (2,6) is $ f'(2)$.

Therefore, $ f'(2) \equal{}\frac{6 \minus{} (\minus{}4)}{2 \minus{} 0} \equal{}\frac{10}{2} \equal{} 5$.

The claim that $ g(x)$ has a tangent line from (0,-4) to (2,2) given that $ f(x)$ has a tangent line from (0,-4) to (2,6) is equivalent to showing that $ g'(2) \equal{} 5 \minus{}2 \equal{} 3$.

$ g'(2) \equal{}\frac{2 \minus{}(\minus{}4)}{2 \minus{} 0} \equal{}\frac{6}{2} \equal{} 3$.

The slope $ g'(2)$ is indeed 3 when $ f'(2) \equal{} 5$.  QED

Generally, the tangent line to $ y\equal{}ax^2\plus{}bx\plus{}c$ at $ x\equal{}k$ is given by $ y\equal{}(2ak\plus{}b)x\plus{}(c\minus{}ak^2)$

The reflection of $ y\equal{}ax^2\plus{}bx\plus{}c$ across the y-axis is $ y\equal{}ax^2\minus{}bx\plus{}c$, whose tangent line to $ x\equal{}k$ is therefore $ y\equal{}(2ak\minus{}b)x\plus{}(c\minus{}ak^2)$

Observe that since $ a$ and $ c$ have not changed, the y-intercept is still the same, and the line that goes from the y-intercept of the old parabola to the point at $ x\equal{}k$ on the new parabola is in fact the tangent line at $ x\equal{}k$ to the new parabola.

Thanks TZF. (I was confused for a few seconds until I realised that the y-intercept referred to in the last paragraph was that of the tangents at x=k.)


inequalities
function
inequalities unsolved

:huh: But isn't it obvious?!

Write $p=abc$ (therefore $p \leq \frac{8}{27}$). Now apply Jensen to the concave function $f(x)=\frac{p}{x+x^3}$:
$\displaystyle \sum_{cyc} \frac{p}{a+a^3} \leq 3 f(\frac{a+b+c}{3}) = 3 f(\frac23) = \frac{81p}{26} \leq \frac{24}{26}$

I  dont think  the function  $f(x)=\frac{p}{x+x^3}$ is concave.

[quote="hungkhtn"]Find the maximum of:

$P=\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}$

If $a,b,c$ are positive real and $a+b+c=2$.

I guess the maximum is $2$ but I can't prove that at one.[/quote]

Dear Hung

$P=\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2} \le ab+bc+ca \le (a+b+c)^2/3 = 4/3$

So, max should be < 4/3.

May be you guess max = 1?

Sorry. I guess max is 1 when $a=b=1,c=0$. It seem quite easy and trivial , But I didn't have any way.

If we take a + b+ c = 3 , then te maximum , if there is , will be 3/2.

 Can you prove  this ?
 I ' ll try it , too.

Babis

Any one? It seem to be very easy? And I don't know why I can't solve it :(
I'm sure that the maximum is $1$.

I think the following inequality holds
$\frac{bc}{1+a^2}+\frac{ca}{1+b^2}+\frac{ab}{1+c^2} \leq 1$
for $a,b,c$ non-negative numbers such that and $a+b+c=2$.
To prove this inequality we can use Equal Variable Theorem in the particular case:
[i]Let $0<a\leq b \leq c$  such that $a+b+c=constant$ and $abc=constant$.
If $f(u)$ is a differentiable function on $(0,\infty)$ such that $g(x)=f'(1/x)$ is strictly convex on $(0,\infty)$, then
$f(a)+f(b)+f(c)$ is minimal for $a\leq b=c$.[/i]
To use this way, we write the inequality for $a>0$ as follows
$\frac{1}{a(1+a^2)}+\frac{1}{b(1+b^2)}+\frac{1}{c(1+c^2)} \leq \frac{1}{abc}$
and consider the function $f(u)=\frac{-1}{u+u^3}$. We see that $g(x)$ is strictly convex.
Thus, it suffices to consider two cases for the original inequality: $a=0$ and $a\leq b=c$.

I am sure that we can also find a nice solution. I will post such a solution after 10 days. ;)

[quote="hungkhtn"]:(  :(  ... Hu hu ... $n-1 EV$ again...[/quote]
[quote="hungkhtn"]I will send a gift to the first one solve it by elementary way!
This unsolved problem is exciting me in some weeks, many try is false.![/quote]
This proves that EV-Method is very strong.

Because I like only thing simply, Although I see, $n-1 EVQ$ is really strong...
But maybe It make problems lost the beauty.

A hint for an elementary solution:
Use the inequality:
 $(a-b)^2(b-c)^2(c-a)^2 \geq 0$
and a Darij's type way.

Oh... I'll try. And my gift should be sent to you, Vasc. Great!

If I've nonresult, Can you post your solution?

[quote="Vasc"]
I am sure that we can also find a nice solution. I will post such a solution after 10 days. ;)[/quote]
It is not nice, but is elementary.

This problem interested me very much !
  Vasc , could you please tell me what Darij's type way is ?

Darij's type way is not a concrete way, only a metaphorical way.
See, for example,
http://www.mathlinks.ro/Forum/viewtopic.php?p=325959#p325959

And If that, It need to used a long computing... To find it is very difficult.!

[quote="hungkhtn"]? God gives you this solution?

But all I want to say is that It's very nice.[/quote]

agree, too nice....

[quote="hungkhtn"]Prove that :

$ P \equal{} \frac {ab}{1 \plus{} c^2} \plus{} \frac {bc}{1 \plus{} a^2} \plus{} \frac {ca}{1 \plus{} b^2} \le 1$

If $ a,b,c$ are positive real and $ a \plus{} b \plus{} c \equal{} 2$.
[/quote]
My solution is the following :)
Rewrite the inequality as
\[ abc\sum \frac{a}{a^2\plus{}1} \plus{}1\minus{}\sum bc \ge 0\]
Note that for all $ x \ge 0$, we have
\[ \frac{1}{x^2\plus{}1} \ge 1\minus{}\frac{x}{2}\]
Hence, it suffices to prove that
\[ abc\sum a(1\minus{}\frac{a}{2})\plus{}1\minus{}\sum bc \ge 0\]
\[ \Leftrightarrow abc\sum bc \plus{}1\minus{}\sum bc \ge 0\]
Setting $ r\equal{}abc,q\equal{}ab\plus{}bc\plus{}ca$, then by Schur Inequality for fourth degree, we have
$ r \ge \frac{(q\minus{}1)(4\minus{}q)}{3}$
Hence
\[ LHS\equal{}qr\plus{}1\minus{}q \ge \frac{q(q\minus{}1)(4\minus{}q)}{3}\plus{}1\minus{}q \equal{}\frac{1}{3}(3\minus{}q)(1\minus{}q)^2 \ge 0\]
We have done.
Equality holds for $ a\equal{}b\equal{}1,c\equal{}0$ or any cyclic permutations. :)

Ohh, your solution is greatest  :wink: I haven't thought exist a simple solution like that  :)

How to kill the last one ineq proposed by Vasc? is there a elementary way?

See the second solution of mine at: http://mathvn.org/forum/viewthread.php?thread_id=979&pid=3865#post_3865 :)

By $ AM\minus{}GM$, we have $ abc\leq (\frac{2}{3})^3$ and $ a^{2}\plus{}1 \geq 2a$

$ \[ abc\sum\frac{1}{(a)(a^{2}\plus{}1)} \leq abc\sum\frac{1}{2a^2}\leq abc \frac{3}{2} \frac{1}{(abc)^\frac{2}{3}} \equal{}  \frac{3}{2} (abc)^\frac{1}{3} \leq \frac{3}{2} \cdot \frac{2}{3} \equal{} 1$

I hope its correct.

[quote="Vasc"]
$ \frac {ab}{7 \plus{} 2c^2} \plus{} \frac {bc}{7 \plus{} 2a^2} \plus{} \frac {ca}{7 \plus{} 2b^2} \le \frac 1{3}$

if $ a,b,c$ are positive real such that $ a \plus{} b \plus{} c \equal{} 3$.[/quote]

Similarly, we have $ abc\leq 1$ and $ (7\plus{}2a^2) \geq 9a^{\frac{4}{9}}$

$ \[ abc\sum\frac{1}{(a)(2a^{2}\plus{}7)}\leq abc\sum\frac{1}{9a^{\frac{13}{9}}}\leq abc\frac{3}{9}\frac{1}{(abc)^\frac{13}{27}}\equal{}\frac{1}{3}(abc)^\frac{14}{27}\leq\frac{1}{3}$

[quote="bloops"]By $ AM \minus{} GM$, we have $ abc\leq (\frac {2}{3})^3$ and $ a^{2} \plus{} 1 \geq 2a$

$ \[ abc\sum\frac {1}{(a)(a^{2} \plus{} 1)} \leq abc\sum\frac {1}{2a^2}\leq abc \frac {3}{2} \frac {1}{(abc)^\frac {2}{3}} \equal{} \frac {3}{2} (abc)^\frac {1}{3} \leq \frac {3}{2} \cdot \frac {2}{3} \equal{} 1$

I hope its correct.[/quote]
Sorry, it is wrong.
It is false that $ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2} \le \frac{3}{(abc)^{2/3}}$, but the reverse is true by Muirhead. Try $ a\equal{}1$, $ b\equal{}2$, $ c\equal{}4$

[quote="dgreenb801"][quote="bloops"]By $ AM \minus{} GM$, we have $ abc\leq (\frac {2}{3})^3$ and $ a^{2} \plus{} 1 \geq 2a$

$ \[ abc\sum\frac {1}{(a)(a^{2} \plus{} 1)} \leq abc\sum\frac {1}{2a^2}\leq abc \frac {3}{2} \frac {1}{(abc)^\frac {2}{3}} \equal{} \frac {3}{2} (abc)^\frac {1}{3} \leq \frac {3}{2} \cdot \frac {2}{3} \equal{} 1$

I hope its correct.[/quote]
Sorry, it is wrong.
It is false that $ \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2} \le \frac {3}{(abc)^{2/3}}$, but the reverse is true by Muirhead. Try $ a \equal{} 1$, $ b \equal{} 2$, $ c \equal{} 4$[/quote]

Oops.  :blush: Muddled up AM-GM.

[quote="Vasc"]The following inequality holds, too.

Prove that :

$ \frac {ab}{7 \plus{} 2c^2} \plus{} \frac {bc}{7 \plus{} 2a^2} \plus{} \frac {ca}{7 \plus{} 2b^2} \le \frac 1{3}$

if $ a,b,c$ are positive real such that $ a \plus{} b \plus{} c \equal{} 3$.[/quote]
MV method kills it, but it's very ugly:
Let $ f(a,b,c) \equal{} \frac {1}{3} \minus{} \sum_{cyc}\frac {bc}{7 \plus{} 2a^2}$ and $ a \equal{} \min\{a,b,c\}.$
Hence, $ f(a,b,c) \minus{} f\left(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2}\right) \equal{}$
$ \equal{} (b \minus{} c)^2\left(\frac {1}{4(7 \plus{} 2a^2)} \minus{} \frac {a(b \plus{} c)(7 \plus{} 2(b^2 \plus{} 3bc \plus{} c^2))}{((b \plus{} c)^2 \plus{} 14)(7 \plus{} 2b^2)(7 \plus{} 2c^2)}\right)\geq$
$ \geq(b \minus{} c)^2\left(\frac {1}{4(7 \plus{} 2a^2)} \minus{} \frac {a(3 \minus{} a)\left(7 \plus{} 2\cdot\frac {5(3 \minus{} a)^2}{4}\right)}{((3 \minus{} a)^2 \plus{} 14)(49 \plus{} 7(3 \minus{} a)^2 \plus{} 4a^4)}\right)\geq0,$
where the last inequality is true for all $ a\in[0,1].$

[quote="arqady"][quote="Vasc"]The following inequality holds, too.

Prove that :

$ \frac {ab}{7 \plus{} 2c^2} \plus{} \frac {bc}{7 \plus{} 2a^2} \plus{} \frac {ca}{7 \plus{} 2b^2} \le \frac 1{3}$

if $ a,b,c$ are positive real such that $ a \plus{} b \plus{} c \equal{} 3$.[/quote][/quote]

i thinks we can uses ev theorem  for function $ f(u) \equal{} \frac { \minus{} 1}{7u \plus{} 2u^3}$
let $ g(u) \equal{} f(u)' \equal{} \frac {7 \plus{} 6u^2}{(7u \plus{} 2u^3)^2}$
we have
$ g(u)'' \equal{} \frac {840c^2 \plus{} 480c^4 \plus{} 2352}{c^4(7 \plus{} 2c^2)^4} \ge 0$
 => $ g(u)$ is convex
we must prove that case $ 0 < a \le b \equal{} c \equal{} t$ and $ a \plus{} 2t \equal{} 3$
inequality<=>$ \frac {2at}{7 \plus{} 2t^2} \plus{} \frac {t^2}{7 \plus{} 2a^2} \le \frac {1}{3}$
<=>$ (t \minus{} 1)^2(106t^2 \minus{} 268t \plus{} 175) \ge 0$
equality hold when $ a \equal{} b \equal{} c \equal{} 1$

[quote="tuantam1lan"][quote="arqady"][quote="Vasc"]The following inequality holds, too.

Prove that :

$ \frac {ab}{7 \plus{} 2c^2} \plus{} \frac {bc}{7 \plus{} 2a^2} \plus{} \frac {ca}{7 \plus{} 2b^2} \le \frac 1{3}$

if $ a,b,c$ are positive real such that $ a \plus{} b \plus{} c \equal{} 3$.[/quote]

i thinks we can uses ev theorem  for function $ f(u) \equal{} \frac { \minus{} 1}{7u \plus{} 2u^3}$
let $ g(u) \equal{} f(u)' \equal{} \frac {7 \plus{} 6u^2}{(7u \plus{} 2u^3)^2}$
we have
$ g(u)'' \equal{} \frac {840c^2 \plus{} 480c^4 \plus{} 2352}{c^4(7 \plus{} 2c^2)^4} \ge 0$
 => $ g(u)$ is convex
we must prove that case $ 0 < a \le b \equal{} c \equal{} t$ and $ a \plus{} 2t \equal{} 3$
inequality<=>$ \frac {2at}{7 \plus{} 2t^2} \plus{} \frac {t^2}{7 \plus{} 2a^2} \le \frac {1}{3}$
<=>$ (t \minus{} 1)^2(106t^2 \minus{} 268t \plus{} 175) \ge 0$
equality hold when $ a \equal{} b \equal{} c \equal{} 1$[/quote][/quote]
What's EV theorem??

See [url]http://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf[/url]. It is not very elementary.

[quote=hungkhtn]Prove that :

$P=\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2} \le 1$

If $a,b,c$ are positive real and $a+b+c=2$.

_____________________

I don't sure about it. I've edit the question.[/quote]
[url=http://www.artofproblemsolving.com/community/c6h215491p6779225][b]Simple proof[/b][/url]

[quote=Namdung][quote="hungkhtn"]Find the maximum of:

$P=\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}$

If $a,b,c$ are positive real and $a+b+c=2$.

I guess the maximum is $2$ but I can't prove that at one.[/quote]

Dear Namdung
Your proof is wrong because the equality holds in $a=b=c=0$.
Dear Hung

$P=\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2} \le ab+bc+ca \le (a+b+c)^2/3 = 4/3$

So, max should be < 4/3.

May be you guess max = 1?[/quote]


geometry
3D geometry
sphere
geometry unsolved

From a point in space, $n$ rays are issuing, whereas the angle among any two of these rays is at least $30^{\circ}$. Prove that $n < 59$.

Note that in the original statement, $n<54$ was required to show (a very hard task in my oppinion) and then changed to $59$ later.

Idea to solve it with $59$: Look at the sphere with radius $1$ around the point $P$ the rays come from. Then let $Z_i$ be the intersection points of the rays with the sphere. Consider the "circles" $c_i$ given by those points $Q$ on the sphere such that $\angle QPZ_i < 15^\circ$. Two different $c_i$ have no point in common by the condition; but the areas of $59$ or more such $c_i$ is greater than the surface of the sphere, giving a contradiction.


Let f(1)=1,  f(x+y)=f(x)+f(y),   f(1/x)=f(x)/x^2. Find all such functions, in reals.

first $ \forall x\in\mathbb{Q}: \ f(x)\equal{}x$

[b]This is just simple that, beacuse we have f(x+y)=f(x)+f(y), this is a cauchy function, we know that the answers are just c*x. and beacuse f(1)=1, so it is f(x)=x! :lol:  :wink: [/b]

sorry, I made a mistake, it is right just for Q, not R.

Sinaj, you can't say directly $ f(x)\equal{}cx$.   For you can say this , $ f$ must be continous,monotone increasing or something like that. :) And the name of the function is Cauchy function  :roll:

$ (x\plus{}y)^2f(\frac{1}{x\plus{}y})\equal{}x^2f(\frac{1}{x})\plus{}y^2f(\frac{1}{y})$ , $ f(\frac{1}{x\plus{}y})\plus{}f(\frac{y}{x(x\plus{}y)})\equal{}f(\frac{1}{x})$ hence $ 2xyf(\frac{1}{x\plus{}y})\equal{}x^2f(\frac{y}{x(x\plus{}y)})\plus{}y^2f(\frac{x}{y(x\plus{}y)})$.  Let putting $ y\equal{}1\minus{}x$ in this equation we get $ 2x(1\minus{}x)\equal{}x^2f(\frac{1\minus{}x}{x})\plus{}(1\minus{}x)^2f(\frac{x}{1\minus{}x})$ because of $ x^2f(\frac{1\minus{}x}{x})\equal{}(1\minus{}x)^2f(\frac{x}{1\minus{}x})$ hence $ x(1\minus{}x)\equal{}x^2f(\frac{1\minus{}x}{x})$ and $ f(\frac{1\minus{}x}{x})\equal{}\frac{1\minus{}x}{x}$ so only solution of these equations is $ f(x)\equal{}x$ .


induction
combinatorics proposed
combinatorics

An array $ n\times n$ is given, consisting of $ n^2$ unit squares. A [i]pawn[/i] is placed arbitrarily on a unit square. The pawn can move from a square of the $ k$-th column to any square of the $ k$-th row. Show that there exists a sequence of $ n^2$ moves of the pawn so that all the unit squares of the array are visited once, the pawn returning to its original position.

[b]Dinu Serbanescu[/b]

It suffices to show that there exists a sequence $ a_1,a_2,...,a_{n^2}$ where $ (a_1,a_2),(a_2,a_3),...,(a_{n^2},a_1)$ is the permutation of all $ n^2$ ordered pairs $ (1,1),(1,2),...,(n,n)$. We will proceed by induction.

For $ n \equal{} 2$, choose $ 1,1,2,2$

Suppose there exists such a sequence for $ n \equal{} k$.

Then for $ n \equal{} k \plus{} 1$, between the two consecutive numbers $ i,i$ ($ 1\leq i\leq k$), insert $ k \plus{} 1,i$. Now beside some $ k \plus{} 1$, put in another $ k \plus{} 1$.

For example, $ 1,1,2,2$ becomes $ 1,3,1,1,2,3,2,2$ and we choose to insert a $ 3$ somewhere - it becomes $ 1,3,3,1,1,2,3,2,2$

We are done  :)

[quote="jbmorgan"]It suffices to show that there exists a sequence $ a_1,a_2,...,a_{2n}$ where $ (a_1,a_2),(a_2,a_3),...,(a_{2n},a_1)$ is the permutation of all $ n^2$ ordered pairs $ (1,1),(1,2),...,(n,n)$. We will proceed by induction.[/quote]  This would be a neat trick indeed.  Luckily, the process you described generates what you meant, not what you wrote ;).  Also, as a trivial note, you could have started your induction at $ n \equal{} 1$ with the cyclic sequence (1).

If I'm not mistaken, this is just the existence result for a de Bruijn cycle with words of length 2 on an alphabet of size $ n$.  (See, for example, http://planetmath.org/encyclopedia/DeBruijnDigraph.html )


Find the product of the sum of sixty-two thousand, two hundred and one hundred forty-five and the difference between thirty-five thousand, one hundred two and thirty-five thousand two.

Couldn't you also do FOIL? ( I KNOW it's harder because you got to do $ 62200 \times 35102$ but it's a possbility right?)

Yes you could.  I wouldn't understand why though.

Is there a faster way to do it? It was a countdown question.

I think this is fast enough. Adding two pairs of numbers in your head, then just add on some zeros because you are multiplying by a power of 10.


Prove the following by induction:

2!*4!*...*(2n)! >= ( (n+1)! )^n

Please help. I'm having a lot of trouble with this question.

First start the induction : it's true for n=1 (2! >= 2^1)

Suppose 2!*4!*...*(2n)! >= ( (n+1)! )^n is true.

We want to prove 2!*4!*...*(2n)! *(2n+1)!*(2n+2)! >= ( (n+2)! )^(n+1)
which simplifies to proving (2n+1)!*(2n+2)! >= ( (n+2)! )^(n+1) / ( (n+1)! )^n = (n+2)^(n+1) * (n+2)! using the induction hypothesis.

This last inequality is clear : there are 2*n values >= n+2 in (n+3)*... * (2n+1) and (n+2)*..*(2n+2)

Hope it's clear (and correct).

I tried to fix the [tex]\LaTeX[/tex] errors. Here is beta's solution:

It is (2(n+1))!, not (2n+1)!, but that's the idea.

so it is [tex]\displaystyle(2(n+1))! \geq \frac{(n+2)!)^{n+1}}{(n+1)!^n}[/tex]

[tex]\displaystyle\frac{((n+2)!)^{n+1}}{(n+1)!^n}=(n+1)! * (n+2)^ {n+1}[/tex]

so [tex](2(n+1))!\geq\ (n+1)! * (n+2)^{n+1}[/tex]

[tex]\displaystyle\frac{(2(n+1))!}{(n+1)!}\geq\( (n+2)^{n+1}[/tex]

[tex](n+2)(n+3)...(2n+2)\geq\ (n+2)(n+2)...(n+2)[/tex]

which is definitely true, induction is complete.


real analysis
function
integration
vector
real analysis unsolved

If $ (f_{n})$ is a sequence of Lebesgue integrable functions on $ [a,b]$, and if $ f_{n}\rightrightarrows f$ on $ [a,b]$, prove that $ f$ is integrable and that $ \int_{a}^{b}\left|f_{n}\minus{}f\right|\rightarrow 0$.

I would like an alternative solution to using Dominated Convergence.

What exactly does $ f_{n}\rightrightarrows f$ mean?

Here is what I came up with:
Since $ f\rightrightarrows f$, given $ \varepsilon>0$, choose $ N$ such that $ \left|f\minus{}f_{n}\right|\leq\left\|f\minus{}f_{n}\right\|_{\infty}<\varepsilon$ whenever $ n\geq N$.  It follows that $ \left|f\right|\leq\left|f_{n}\right\|\plus{}\varepsilon$.  Since $ f_{n}\in L^{1}$ for all $ n$, $ \left|f_{n}\right|\in L^{1}$ as well.  Clearly, $ \varepsilon\in L^{1}$ and since $ L^{1}$ is a vector space, $ \left|f_{n}\right|\plus{}\varepsilon\in L^{1}$ so $ f\in L^{1}$.  Also, $ \int_{a}^{b}\left|f\minus{}f_{n}\right|<\int_{a}^{b}\varepsilon\equal{}\varepsilon(b\minus{}a)$.  As $ n\rightarrow\infty$, $ \varepsilon\rightarrow 0$, meaning $ \varepsilon(b\minus{}a)\rightarrow 0$ so $ \int_{a}^{b}\left|f\minus{}f_{n}\right|\rightarrow 0$.

I know this kind of relies on Dominated Convergence but I tried to explicitly avoid it.  Is this right?


modular arithmetic
floor function
logarithms
inequalities

Let $ p$ be a prime and $ m$ be a positive integer. $ m$ integers $ 0\le a_i<p$ $ (i\equal{}0,1,...,m\minus{}1)$ satisfy next equations:

$ a_{m\minus{}1}p^{m\minus{}1}\plus{}a_{m\minus{}2}p^{m\minus{}2}\plus{}\cdots \plus{}a_0\equal{}2003$,
$ a_{m\minus{}1}\plus{}a_{m\minus{}2}\plus{}\cdots \plus{}a_0\equal{}15$.

Find all possible $ p$.

[hide]
$ a_{m \minus{} 1}p^{m \minus{} 1} \plus{} a_{m \minus{} 2}p^{m \minus{} 2} \plus{} \cdots \plus{} a_0$ $ \equiv a_{m \minus{} 1} \plus{} a_{m \minus{} 2} \plus{} \cdots \plus{} a_0 \equiv 2003 \equiv 15 \pmod{p \minus{} 1}$, so $ p \minus{} 1|1988 \equal{} 2^2 \times 7 \times 71$. Of the $ 12$ possible factors, most can be eliminated for being odd or $ \equiv \minus{} 1 \pmod{5}$. Thus, we see that $ p \equal{} 2,3,5,29$.

Now $ a_{m \minus{} 1} \plus{} a_{m \minus{} 2} \plus{} \cdots \plus{} a_0 \le (p \minus{} 1) \cdot \left \lfloor \log_p 2003 \plus{} 1\right\rfloor$, and $ p \equal{} 2,3$ do not satisfy this inequality. Considering the other 2 bases, $ 2003 \equal{} 31003_5$ and $ 2003 \equal{} 2(11)2_{29}$, so $ p \equal{} \boxed{29}$.
[/hide]

:roll:
[color=green]nice question & solution[/color]

[quote="seasonal squirrel"]i don't follow, how did you get the last inequality?[/quote]

[hide="I try to explain it"]
$ 2003 \equal{} 31003_5$ i.e. $ 3(5^4) \plus{} 1(5^3) \plus{} 0(5^2) \plus{} 0(5^1) \plus{} 3$ , where $ \sum _i a_i \equal{} 3 \plus{} 1 \plus{} 0 \plus{} 0 \plus{} 3\not \equal{} 15$
$ 2003 \equal{} 2(11)2_{29}$ i.e. $ 2(29^2) \plus{} 11(29^1) \plus{} 2$ , where $ \sum _i a_i \equal{} 2 \plus{} 11 \plus{} 2 \equal{} 15$ :roll:
[/hide]

[quote="seasonal squirrel"]i don't follow, how did you get the last inequality?[/quote]

The last inequality is not really necessary, we could just write $ 2003$ in bases $ 2$ and $ 3$ and see that the digits don't sum up to $ 15$, though this slightly reduces the amount of casework. (the inequality says that when we take $ 2003$ in base $ p$, each digit must be $ \le p\minus{}1$, and there are $ \left \lfloor \log_{p} 2003 \plus{}1\right \rfloor$ digits of $ 2003$ in base $ p$).

[quote="azjps"][quote="seasonal squirrel"]i don't follow, how did you get the last inequality?[/quote]

The last inequality is not really necessary, we could just write $ 2003$ in bases $ 2$ and $ 3$ and see that the digits don't sum up to $ 15$, though this slightly reduces the amount of casework. (the inequality says that when we take $ 2003$ in base $ p$, each digit must be $ \le p \minus{} 1$, and there are $ \left \lfloor \log_{p} 2003 \plus{} 1\right \rfloor$ digits of $ 2003$ in base $ p$).[/quote]

Yes,I see.
But I feel you can use the different representaions to explain that is also appreciating. Thanks for detailed explanation. :P


What is the value of $ \left(\frac{3}{5}\right)^2 \plus{} 2\left(\frac{3}{5}\right)\left(\frac{2}{5}\right) \plus{} \left(\frac{2}{5}\right)^2$?

Notice that this is a special polynomial expansion...
$ (a \plus{} b)^2 \equal{} a^2 \plus{} 2ab \plus{} b^2$
so this comes out to $ (\frac35 \plus{} \frac25)^2 \equal{} \boxed1$

Or

$ \frac{4}{25}\plus{}\frac{12}{25}\plus{}\frac{9}{25}\equal{}1$. Very quick. I'd be interested to see which of these methods produces a faster result.


Hi,

I'm a newbie in "olympiad world"  :blush:
I'm now preparing for the IPhO training camp.
And,i would like to know what type of materials willl be suitable for me.
I heard that the "Fundamental of Physics" by David Halliday is an excellent book.
My question is,does this book covered up all the topic in the IPhO syllabus?
Other than this,any other good books that may covered up all IPhO syllabus?

Thanks much  :)

Well, I was chosen for the IPhO team but I preferred to go to IMO instead. I got the book "University Physics" by Young & Friedman for preparation (from the organizers of the olympiad). It's a good book, covering most thing you need to know at IPhO, with a lot of exercises, but I don't think it's really "special" compared to other similar books. So I'd just pick some good book (like this one) covering general physics, and I would search for the IPhO problems and other olympiad problems of the past few years on the internet. Hope that helps.


Asymptote
Support
FTW
LaTeX
geometry
geometric transformation
reflection

[color=blue][size=200]INTRODUCTION/PREFACE[/size][/color]

As some of you know, the Legend of the Thirteenth Moon started far back in history.

It began with Forthewinia becoming a superpower moon of Aopsia, thriving with mathematical power...
Until some evil people set their eyes on it. Who? THE EIPIANS!!!!!! However, the heroic efforts of the crew of ASYMPTOTE XLII successfully saved the mods and the security of Forthewinia.

Unfortunately, this attack proved to the other lesser moons that Forthewinians were not invincible, and they, too, decided they could launch an assault against Forthewinia. However, 14 noble souls defended Forthewinia from the Calcgorians and brought back the mods once again. The other moons had learned their lesson- do no trifle with Forthewinia.

Unfortunately, the other moons would constantly fight each other. One neutral ally of Forthewinia was Alcumia. There during the original unrest of the Eipian attack admin master levans hid out until the mods were brought back. These days Alcumia suffered financial trouble and turned to Forthewinia for help. While Forthewinians couldn't be convinced to send support to the Alcumians, one brave citizen had an idea. Thinking back to the terror of the time Forthewinian mods were captured, he realized that by now the foolhardy Eipians and Calcgorians were quite powerful and secure, as well as being on good terms. Trouble, he was sure, was stirring as few attacks came from those hostile lands. Eventually, he hit upon an idea: a mission could be sent not only to see what the plotters were plotting, but also plunder their lands to bring money back to Forthewinia as well as neighboring Alcumia to settle their problems. Idea approved of, the admin/mod council immediately set out the notice that Forthewinia was going to attack Eipia and Calcgoria. All brave people willing would be able to go, providing they could make a crew. This was a historic event- Forthewinia was going to start a war.

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highest test 1200
highest alpha 1444

[u]Signups             FTW rating    Test Rating[/u]
1) King!                     1450
2) dragon96               1657            1440
3) lotsofmath             1604            1232
4) spaceguy524          1150
5) westiepaw              1120            1200
6) cgyao15                1178             1178
7) EggyLv.999           1229

5 more days. At least 5 more players...
Also, you people should do something constructive, like choose a name for the ship

me joins

FTW Normal: 1488 (its stuck at 1488...)
FTW Normal Highest: 1530

FTW Test: 1200
FTW Test Highest: 1200

Never played in the test version...

I vote the ship is called "Eipians and Proud". Or that can be our slogan. w/e

Eipians and Calcgorians are not friends, they are the enemy.
Think about that.

8 players, at least 12 to start.

Just so you know, the test rating isnt that useful later on in the game, only for certain parts. It will represent money you have.
Regular rating is MRU, measured like the previous games.

join
dont play much ftw.
ftw: 1224
ftw test: 1204 maybe?

will work on getting it higher.

My rating in both FTWs has changed.

FTW current: 1152 (highest 1200)
FTW test: 1215 (highest 1215)

THIS IS DEAD, ILL STOP TALKING NOW SO THIS WILL DIE

YOU may think this is dead, but I don't care. You could have just not posted anything.

Any complaints?

(If abcak does not respond in time then I demand captainship back, unless the game dies first).

chompy, start doing somethingin legend of the black dragon.

random, u can jointoo! theres still a spot.  this is dead so join lobd

I'm still captain :D

OR you can join Unnamed RPG!  0.0

meh I think this is mod-abandoned.

So do I.  But just in case it isn't, I agree with randomguy.

Mod-abandoned? The only clue to that is "I think this game is dead" and "who wants to just let this die". Anyway, we could just have someone become a replacement mod, and piegraph, PLEASE STOP yelling about your RPG. Please.

abcak, any complaints? If not, I suggest you go for it. If we want this game to survive, assuming that it is not mod-abandoned, we should speed things up a bit.

I'm better at playing games then modding.
piegraph if you talk about your game in any other thread but yours I will attempt to get it locked.
RIGHT RANDOM IS NOW MOD NO MORE WORK FOR ME :sleep2:

ok. randomguy is the mod now.
what happens next?

/lazy mod

Now awaiting decision from abcak.

You can be mod.

If you can't, I'll try.  But I might have a hard time doing 2 games at once, so...

Thank you. Anyway, I need a decision; you can wait for the others if you want though.


How do I include Metapost in AoPS? Also, how can I get it on my computer?

You won't be able to use Metapost on AoPS. On your own computer, if you are using MiKTeX then use the MiKTeX Package Manager to install it (look for packages such as miktex-metapost). On other systems [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53217]Ultimate guide to inserting Pictures in LaTeX[/url] tells you where to get it from.

[url=http://tug.org/pracjourn/2006-4/henderson/]A Beginner's Guide to METAPOST for Creating High-Quality Graphics[/url] is worth having a look at.


A man was driving his car on the freeway, all the traffic was one way. He adjusted his speed so that he passed the same number of cars as passed him on this particular trip.

It would seem then that he must have driven the "average" speed of all the cars he passed plus those that passed him.

Is that "average", the mean or the median?

If he passed the same number of cars as cars that passed him, then he went faster than half of the cars and slower than half of the cars. Therefore, his speed was the median.

Definition of median: The middle value in a set of statistical data.
Definition of mean: The sum of all values divided by the number of terms in the set.
Therefore, if there were 25 cars, the person in question passed 12 cars and was passed by 12 cars, ensuring that he was the 13th car making his speed the median value. We can show that it is not the mean by implementing consecutive number data values for the first 24 then implementing an outlier for the last one. Thus it will always be the median, but sometimes it can be the mean.

Intuitively I think the answer is clearly the median, for the answers that you have mentioned. However, see [url=http://mcraefamily.com/MathHelp/PuzzleFreewayAnswer.htm]this link[/url].

It has to be the median, not necessary to be the mean


group theory
abstract algebra
LaTeX
superior algebra
superior algebra unsolved

prove that there exists a subgroup G of SU(2) which has a normal subgroup $Z \cong C_{2}$
such that
G/Z $\cong$ $S_{4}$

Also, what books would you recommend for me to read up on the special unitary groups? I am interested in finding it's conjugacy,subgroups etc... :)
Thanks.

i managed to find it out after a few hours of reading. will post the solution up one day if I manage to find out how to type in Latex :D

It's not that hard, if you know a little about SU(2). There is a surjective homomorphism $SU(2)\to SO(3)$ whose kernel is $Z=\{\pm 1\}\cong\mathbb{Z}_{2}$.
Now there is a subgroup of SO(3) which is isomorphic to $S_{4}$ and so there is a subgroup in SU(2) which has the disered properties.

yes, that's exactly what I found out. Thank you very much anyway.

I have another question as well, is GL(2,3) isomorphic to $S_{4}\times C_{2}$ give reasons please ?

I know that $GL(2,3)/Z \cong PGL(2,3) \cong S_{4}$ but that's about all I understand at the moment.

Hi.

$SL(2,3)$ and $Z(GL(2,3))=\{\pm 1\}$ are normal subgroups with trivial intersection and together they generate $GL(2,3)$. Therefore $GL(2,3)= SL(2,3)\times Z(GL(2,3))\cong PSL(2,3)\times\mathbb{Z}_{2}\cong S_{4}\times\mathbb{Z}_{2}$.


Can anyone help me interpret this formula?

[tex]\theta = 2 \tan^{-1} \frac{h}{2d}[/tex]

It is to calculate the angle of view in a pinhole camera. h is the height of the camera, d is the depth of the camera (the difference from the pinhole and where the film is located) and [tex]\theta[/tex] is the angle of view. But how can you do [tex]tan^{-1}[/tex] ?

So if [tex]\theta[/tex] is 90 (or if it is specified), how do I work out d?

As an aside, I can see from the picture (or maybe its just logical) I have that as d decreases, [tex]\theta[/tex] increases, and vice versa.

Ok so I found that [tex]tan^{-1} = ArcTan[/tex], given that how do this look:

[tex]\theta = 2 tan^{-1} \frac{h}{2d} \\
90 = 2 tan^{-1} \frac{24}{2d} \\
\frac{90}{2d} = 2 tan^{-1} 24 \\
\frac{45}{d} = 2 tan^{-1} 24 \\
d = \frac{45}{2 tan^{-1} 24} \\
d = \frac{45}{tan^{-1} 48} \\
d = \frac{45}{1.5500} \\
d = 29.0329[/tex]

A bit of a long winded calculation but I just want to make sure I didn't do anything stupid (which is quite likely).

NOTE: In this h is the height of the file (which should be the height of the camera I think) as the film is 24mm x 36mm so I guess the answer should be read as d (focal length) is 29.0329 millimetres. The thing is when I think about a pinhole camera being able to capture a 90 degree field of view I would have thought it would have been deeper than 3 cm (which is just over an inch).

Errr, what did you do in the third step there? Im not sure if you can do that, somehow bring the $2d$ over...

you can make it 45 (dividing by 2)
and then take the tan of both sides.

I still dont get it... how did you go from:
$\displaystyle 90 = 2 tan^{-1} \frac{24}{2d}$
to

$\displaystyle \frac{90}{2d} = 2 tan^{-1} 24$

???

[quote="juggle7"]seamusoboyle and d343seven are right.  Here is how it goes:

$90^\circ=2\tan^{-1}\frac{24}{2d}$
$45^\circ=\tan^{-1}\frac{24}{2d}$
$\tan45^\circ=\frac{24}{2d}$
$1=\frac{24}{2d}$
$2d=24$
$d=12$

$\tan(x)$ is a function and there are very few functions that would work like what you were trying to do.[/quote]

So just to confirm, when you move a term from one side of the equals to the other you need to do the inverse
operation, and because ArcTan is the inverse of Tan (or I like to visualise like $tan tan^{-1}x = x$) you went from $45^\circ=\tan^{-1}\frac{24}{2d}$ to $\tan45^\circ=\frac{24}{2d}$?

Also would it still be correct to do this:
$1=\frac{24}{2d}$
$d=\frac{24}{2}$

or even this:
$1=\frac{24}{2d}$
$1=\frac{12}{d}$
$d=12$

[quote="juggle7"]Those are good.  Be really careful about $\tan^{-1}x$.  Its range is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.  For example, $\tan^{-1}(\tan(120^\circ))=-60^\circ$.[/quote]

Ahhh, nice.

[quote="joml88"]I'm probably being a little picky but if a problem just has $\tan^{-1} x$ I don't believe that you can assume that they want you to only consider the restricted range that juggle was talking about.  $\tan^{-1}x$ is a multivalued function which means that for some $x$ in the domain there are multiple (infinitely many in this case) $y$ in the range.  Obviously a multivalued 'function' isn't really a function but I didn't make the term up...  Anyway, there are at least a few notations that I know of that imply the range restrictions.  You can use $\arctan x$ or you can use $\text{Tan}^{-1} x$.[/quote]

Yes, standards differ.  I think it is prefered to make it into an actual single-valued function.  It is at least very important to realize your calculator can only give you one value.  Believe it or not, some people use Arctan(x) to mean the multi-valued function and $\arctan x$ to mean the single-valued function.  Anyway, I have been counted wrong when I have done $\tan^{-1}(\tan120)=120$.


the search function exists  :wink:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=201239


Let $ ABC$ be an acute triangle, $ AD,BE,CZ$ its altitudes and $ H$ its orthocenter. Let $ AI,A \Theta$ be the internal and external bisectors of angle $ A$. Let $ M,N$ be the midpoints of $ BC,AH$, respectively. Prove that:
(a) $MN$ is perpendicular to $EZ$
(b) if $ MN$ cuts the segments $ AI,A \Theta$ at the points $ K,L$, then $ KL\equal{}AH$

NZ=NE  ZM=ME  =>  MN⊥EZ
KZ=KE  LZ=LE  K is on ⊙K(if not ,then we get AZ=AE,obviously wrong )
done

Later posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=446962


function
integration
calculus
calculus computations

Find all differentiable functions f which satisfy f (x)^3
integral(f (t)^2 ,0,1 dt) for all
real x.

If we suppose that you mean $ f^3(x)\equal{}\int_0^1{f^2(t)\,dt}$, then $ 3\,f^2(x)\,f^{\prime}(x)\equal{}0$.
If we suppose that you mean $ f^3(x)\equal{}\int_0^{x}{f^2(t)\,dt}$, then $ 3\,f^2(x)\,f^{\prime}(x)\equal{}f^2(x)$   $ \Leftrightarrow$  $ f^2(x)\,\left( {3\,f^{\prime}(x)\minus{}1}\right)\equal{}0$.
If you mean something else, then it is something else.


ratio
geometry
circumcircle
inradius
pigeonhole principle
function

Hola, después de estar viendo posts en el foro, y de tratar de expresar los problemas en inglés, ¿qué os parece si vamos haciendo aquí un pequeño diccionario de aquellos términos más usados en el foro, con su traducción al castellano, y viceversa?

Empiezo contribuyendo con la palabra baricentro>>Centroid

 :P

Aqui hay mas palabras:  :icecream:

[b][English] : [Español][/b]
ratio : razón
radius:radio
radii: radios
line: recta, segmento
circle: circulo, [b]en ingles no se usa circunferencia!!!!!!!!! [/b]
sides: lados
measure: medida
length: longitud
pencil: haz
power: potencia
conversely: inversamente
locus: lugar geometrico
axis: eje
remark: observación
upon: hacia
incircle: circunferencia inscrita
circumcircle: circunferencia circumscrita
inradius: inradio
set: conjunto
altitude: altura

[quote="hucht"]pencil: haz[/quote]

¿¿que es "haz"?? ¿pencil no es lápiz?

[quote="hucht"]Un haz de rectas es un conjunto de rectas que pasan por un punto O, O es le vertice del haz  :P[/quote]

Bien hucht, y por qué decís que haz:pencil????

[quote="hucht"]jejeje pq esa es la traduccion.[/quote]

ok...

[quote="tipe"]no soy un experto en inglés  :blush: pero por lo que he leido creo que [b]radius[/b] se refiere a radio (singular) y [b]radii[/b] se refiere a radios (plural)

$Tipe$[/quote]
That's right. :)
But, actually this radius/radii comes from latin. ;)

[quote="tipe"]no soy un experto en inglés  :blush: pero por lo que he leido creo que [b]radius[/b] se refiere a radio (singular) y [b]radii[/b] se refiere a radios (plural)

$Tipe$[/quote]

Tienes toda la razon, omiti la palabra radius  :blush:

PD. Avisame cuando entreguen carnets en la facultad.

Cómo se dice compás en inglés?  (compás: el instrumento de geometría, no al compás de la música, por ejemplo).

[quote="mathmanman"]Se dice : compass. :)[/quote]

Gracias!!!!!

[quote="mathmanman"]De nada. ( http://www.mathlinks.ro/Forum/viewtopic.php?t=83317 ;))[/quote]

Jajajaja, para eso lo quería, buena observación.........

¿cómo digo en inglés esto? :  Dos triángulos $\triangle {ABC}$ y $\triangle {A'B'C'}$ son semejantes....

No sé cómo decir "semejantes", gracias por la ayuda.......

[quote="mathmanman"]Ok, se dice :
"The two triangles ABC and DEF are [i]similar[/i]."

(well, if I understood correctly the meaning of "semejantes"  :P , but I would tend to think I'm right since these words are... similar ;) (even with french where it is : "semblables")[/quote]

And can I use "congruent"??

[quote="mathmanman"]Bueno, pienso que este sito va a ayudarte : http://argyll.epsb.ca/jreed/math9/strand3/triangle_congruent.htm . :)[/quote]

Ahh entendí, gracias. lo que yo preguntaba era "similiar", no "congruent", me equivoqué... gracias!!!!

¿Alguien sabe cómo se llama el teorema de las casillas en inglés?
Lo he escuchado nombrar también como teorema del pichón, no estoy seguro si en el Engel, así que igual creo que se dice "Pigeon theorem" o algo así.

[quote="Dave Vather"]¿Alguien sabe cómo se llama el teorema de las casillas en inglés?
Lo he escuchado nombrar también como teorema del pichón, no estoy seguro si en el Engel, así que igual creo que se dice "Pidgeon theorem" o algo así.[/quote]

pigeonhole????

Si : "Pigeonhole principle". :) (o "Dirichlet's box principle".)

[quote="hucht"]
incircle:circunferencia inscrita
circumcircle: circunferencia circumscrita
[/quote]

creo que incircle es incirculo, y circumcircle es circuncirculo  de un triangulo..no?  :D

[quote="jasamper88"][quote="hucht"]
incircle:circunferencia inscrita
circumcircle: circunferencia circumscrita
[/quote]

creo que incircle es incirculo, y circumcircle es circuncirculo no? :D[/quote]

Creo que si, que es lo mismo.....

[color=blue]¿Hay alguien dispuesto a crear un archivo .pdf de todas las palabras posibles para confeccionar dicho diccionario? A varias personas resultaría interesante, teniendo en cuenta que, así como yo, no cuenta con mucho tiempo para conectarse a Internet.
Sería útil también si agregara algunos ejemplos para facilitar el uso del lenguaje.[/color]

y si dice: $With$ $edge$ $and$ $compass$, $construct$ $the$ $circle$  :?:

Aah..
Ok, entonces, "straight edge" significa "regla." :)

cual es la traduccion a español de: mappings y map :?:

"mapping" es un termino alternativo para decir funcion o aplicacion, "map" puede ser empleado como un verbo, por ejemplo, si consideramos $f(x)=x^{2}$, decimos que la imagen de $x$ bajo la función $f$ es $x^{2}$; esto podria traducirse a ingles como "the function $f$ maps $x$ to $x^{2}$".


If $ a,b,c\in\mathbb{R}_\plus{}$ and $ a^3\plus{}b^3\plus{}c^3\equal{}3$ prove that
$ \sum_{cyc}\frac{a\plus{}2}{2a^2\plus{}1}\ge 3$

We have: $ LHS \equal{} \sum\ (a\plus{}2\minus{}\frac{2a^2(a\plus{}2)}{2a^2\plus{}1}) \geq\ a\plus{}b\plus{}c\plus{}6 \minus{} 2(a\plus{}b\plus{}c) \geq\ 3$ (by Am-Gm)
Nice ineq, Dragos Calinescu :)

[quote="nguoivn"]We have: $ LHS \equal{} \sum\ (a \plus{} 2 \minus{} \frac {2a^2(a \plus{} 2)}{2a^2 \plus{} 1}) \geq\ a \plus{} b \plus{} c \plus{} 6 \minus{} 2(a \plus{} b \plus{} c) \geq\ 3$ (by Am-Gm)
Nice ineq, Dragos Calinescu :)[/quote]
Nice solution :lol:

[quote="Dragos Calinescu"]If $ a,b,c\in\mathbb{R}_ \plus{}$ and $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3$ prove that
$ \sum_{cyc}\frac {a \plus{} 2}{2a^2 \plus{} 1}\ge 3$[/quote]
It seems that your inequality is true even with condition $ \sqrt[4]{a^3}\plus{}\sqrt[4]{b^3}\plus{}\sqrt[4]{c^3}\equal{}3.$


Prove that the factorial of any positive integer can never equal a square number.

I started this but didn't get very far. I'm interested in the methods you can use...

In a segment (a/2,a] there exists at least one prime number for every a>1.
Take a prime number p in (a/2,a]
p divides a!  , but not p² , because in the factors of a! (1 , 2, ... , n) there is only one , which is divisable by p (Itself p) .
It means , a! isn't a square number (a>1) , because it has a factor p and not p².

thank you!

could anyone suggest any alternative methods of solving it?


The top three teams at the New York State MATHCOUNTS competition held this past Saturday were all from NYC:

First place:  Hunter College High School (Manhattan)
Second place: The Dalton School (Manhattan)
Third place: MS 74 (Queens)

The top four individuals who will represent NYS at nationals this year are:

Kevin Peng
Meena Boppana
Matthew Babbitt
Alex Mun

All except Matthew are from NYC and environs.  Their coach will be Sheila Krilov of Hunter College High School.

A heart congratulations from Upstate NY to all the terrific NYC-area mathletes,  teams, and coaches!

Mary O'Keeffe
Albany Area Math Circle

Thanks for the kind words, Mary.  Congratulations to your student Matthew for making Nationals twice.  Good luck to all the NY kids at Nationals this year!

I don't know if you would count Rye as an environ, but okay  :lol: .


I wrote a short program in TI BASIC for Statistics (variance)
It runs and gives the correct answer but it gives this error code
"ERR:SYNTAX"  ???
The program:
PROG:VAR
:Input   "SUM:X?",A
:Input   "SUM:X sq?",B
:Input   "N",N
:((B-((A)sq/N))/(N-1))->C
:Disp   "S sq"
:Disp C
:sq rt(C)->D
:Disp    "S"
:Disp D
:End
Is there a better way of writing it ??
I wrote it using keystrokes on the TI-83 (black case)
I do not have the program or cable for the computer
I also have a TI-36x which I have been doing calculations for statistics, but I just bought the TI-83 for the class so I want to use it. I was unable to find built in statistics programs that inputed the values (not from the list to calculate the values)
Any help would be appreciated
Thank you for your time and effort
falcon

Once you recieve a Syntax error, press "Goto" and it'll surf through your coding to find the code in question.

I did and it just highlighted "END"
does that mean the problem is the statement "END"
if so what should the statement be at the end of the program ???
falcon

You don't need an "END" at the end of your program.
END is only used to end If..Then, While, For, and Repeat statements.

[quote="navyfalcon"]if so what should the statement be at the end of the program ???[/quote]

Yep. Try Stop. If you want to make your program fancy and end on a note, you can use "Output(TEXT HERE" rather than just "Done"

Yes, that did it and you were right, I do get the "done" message.
Thank You for your time and information.
The program is for statistics (solve for variance and deviation)
Deviation is the sq rt of variance.
The program works and gives the correct answer.
Is there a better way to program it (ie optimize it),
better way to give input and output information and/or write the formula ???
The formula is:
S^2 =((sum x^2 - (sum x)^2)/N)/(N-1))
Thank You
falcon :D  :D  :?:

i found out on my sister's TI-84 that i had to do "Stop" to finish the program, whereas my TI-89 added the "EndPrgm" by itself..

The program is for descriptive statistics - variance and deviation.
Does anyone know how to display the normal distribution curve ?
Does anyone know how to change z scores to percent values for the normal distribution curve  ?  :?:  :?:
Now, I am using tables and I'm sure they are in the calculator because they have solutions for probability - I just don't know how to access them   :huh:  :help:
Thank You for your time and efforts
falcon

Found part of the answer - percent of scores - located here
http://www.oglethorpe.edu/faculty/~j_nardo/2003-2004/Summer%202004/Stats/Handouts/Normal%20Curve%20Calculations/NEW%20NORMAL%20CURVE%20HANDOUT.htm
and
http://www.tc3.edu/instruct/sbrown/stat/normcalc.htm
and
http://mathbits.com/MathBits/TISection/Statistics2/normaldistribution.htm
still need to learn how to do better coding in basic and the options. Just learning out of necessity.
Thank You
falcon


Hi guys,

Even for those who competed in the first year's contest, you need to read this page [b]again[/b] and register [b]again[/b] as well due to some changes.

I want to begin by thanking everyone who have participated in the first year's contest, and it was a great experience for myself as well.

This upcoming year's contest will be administered by myself and two other students, who will be helping me to create large portion of both first and second rounds. So, hopefully, you will see more variety of problems!  :)

Please note that the ultimate purpose of this contest is to provide [b]best[/b] simulation of actual AMC and AIME.

[b]Why should you register?[/b]
Benefits in registration include:

A) Updates about contest and reminder
B) Occasional documents with interesting problem
C) Automatic sending of problems on the contest

Things to include on registration are answering following questions: (You may simply copy and paste them with your answers)

Following are for student registration (if you're only one interested in the school)

1. Your name
2. Your grade (in school year 2008-2009)
3. Your school name and state
4. Your coach/proctor/parent if there is any
5. Participation of AMC and/or AIME and/or USAMO

Following are for coach registration (student may register first and get coach to register as well)

1. Your name
2. Your school name and state

Benefits of coach registration are that I might be able to create some special documents (sample problems maybe or something of that sort) for coaches to help their kids become interested.

To register, please email me at hdmatholympiad at gmail dot com.

[b]Other Helpful Information[/b]
By going to my website (www dot hiddendiscoveries dot com), you can see the previous year's problems, solutions, and results.

[quote="veezbo"]I'm starting a math club at my high school that actually meets on a regular basis. Will it be all right if I register later on in the 2009-2010 year? (since I am not sure just how many people will join).

I estimate that about 15 people from my school will take this contest from this math club.[/quote]

Yup, that's fine. The first round is not going to be till my break after fall semester... Which is around December, so you can wait to register until your school starts.

[quote="Ihatepie"]Is it all done online or at a location?[/quote]

The problems, once you register, will be emailed to the contestant. So, you can take it anywhere at your pleasure, provided that you oblige to the rules of MAA for AMC/AIME.

[quote="Thunder365"]This is a cool idea for a contest! It is free, though, right  :blush:  :blush:  :blush:[/quote]

Yes, it's free.  :)

Just to let everyone know, Hidden Discoveries now has its official forum. So, although most students will be gathered from AoPS, please make posts relevant to Hidden Discoveries Math Olympiad on its forum.  :)

Previously existing URL no longer valid.

Hi guys!

The heated AMC discussion is stirring up the Art of Problem Solving Forum again. I am uploading this post to let you all know about the Hidden Discoveries Mathematics Olympiad, a great practice and a real contest that you can compete for AMC-12 level.

At the moment, the only thing that I can ensure is the first round because the problem committee did not have full ability to address the 2nd round yet. The exact date for the first round will be announced later.

This post is the [b]last[/b] one I will make in the AMC forum this year, so please register now before you forget it. I also made a new message board (the older one is gone), which has a section for this competition as well. So, thanks all!

http://www.forum.hiddendiscoveries.com/


inequalities
trigonometry
logarithms
function
inequalities proposed

Let $ a,b,c$ be positive real numbers. Determine the largest constant $ k>0$ such that the following inequalities holds

a)
\[ \frac{4(a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1)\plus{}(a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2)}{(a\plus{}b\plus{}c)^2} \ge k.\]

b)
\[ \frac{(a^2\plus{}1)(b^2\plus{}1)(c^2\plus{}1)(a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2)}{(a\plus{}b\plus{}c)^4} \ge k.\]
:)

P/s: This problem was proposed by Mihaly Bencze on Crux Mathematicorum (as an open question). I think it is deadline now, so I post it here. Anyways, we have found the solution for this problem. :)

Let's make the substitution $ a \equal{} \cot x$, $ b \equal{} \cot y$, $ c \equal{} \cot z$. $ (0 < x,y,z < \frac {\pi}{2})$

Then the expression turns into
$ A \equal{} \frac {4(\tan^2 x \plus{} 1)(\tan^2 y \plus{} 1)(\tan^2 z \plus{} 1) \plus{} (2\tan^2 x \plus{} 1)(2\tan^2 y \plus{} 1)(2\tan^2 z \plus{} 1)}{(\tan x\tan y \plus{} \tan y\tan z \plus{} \tan z\tan x)^2}$ $ \ge k.$

$ t(x) \equal{} \ln(\tan^2 x \plus{} 1) \equal{} \ln\left(\frac {1}{\cos^2 x}\right)$
$ t'(x) \equal{} 2\tan x$; $ t''(x) \equal{} \frac {2}{\cos^2 x} > 0$ $ \Rightarrow$ the function is a convex one.

$ h(x) \equal{} \ln(2\tan^2 x \plus{} 1) \equal{} \ln\left(\frac {\sin^2 x \plus{} 1}{\cos^2 x} \right)$
$ h'(x) \equal{} \frac {4\tan x}{1 \plus{} \sin^2 x}$; $ h''(x) \equal{} \frac {2}{(1 \plus{} \sin^2 x)^2}\left(2\tan^2 x \plus{} \frac {2 \minus{} \sin^2 2x}{\cos^2 x}\right) > 0$ $ \Rightarrow$ the second function is also convex.

Let's fix $ x \plus{} y \plus{} z \equal{} 3T$, where $ T < \frac {\pi}{2}$. Then we get by Karamata and $ (x,y,z)\succ (T,T,T)$ that
$ 4\prod_{\textrm{cyc}}(\tan^2 x \plus{} 1) \plus{} \prod_{\textrm{cyc}}(2\tan^2 x \plus{} 1)\geq 4(\tan^2 T \plus{} 1)^3 \plus{} (2\tan^2 T \plus{} 1)^3$.

Also, I've proved the inequality
$ \tan x\tan y \plus{} \tan y\tan z \plus{} \tan z\tan x\le 3\tan^2 \left(\frac {x \plus{} y \plus{} z}{3}\right)$

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1382850#1382850[/url].

Consequently, $ A\geq \frac {4(\tan^2 T \plus{} 1)^3 \plus{} (2\tan^2 T \plus{} 1)^3}{9\tan^4 T}$ which is a function only of $ T$ and it's elementarily to find its minimum. Equality comes when $ a \equal{} b \equal{} c$.  :)

Because of the reason at http://www.mathlinks.ro/viewtopic.php?p=1539734#1539734, I think this solution is not correct, so I wait for another proof.


[quote="rashad9607"]Don't see the word 'distinct' anywhere...

1,10,10[/quote]So what if you don't see the word distinct? Are you sure that is the only solution? Why?

[hide="Do you want to see a lot of boring casework mod 7?"]
We trivially see that $a,b,c<13$
Say we have an integer $x$.
$x\equiv0\mod7 \rightarrow x^3\equiv0\mod7$
$x\equiv1\mod7 \rightarrow x^3\equiv1\mod7$
$x\equiv2\mod7 \rightarrow x^3\equiv1\mod7$
$x\equiv3\mod7 \rightarrow x^3\equiv-1\mod7$
$x\equiv4\mod7 \rightarrow x^3\equiv1\mod7$
$x\equiv5\mod7 \rightarrow x^3\equiv1\mod7$
$x\equiv6\mod7 \rightarrow x^3\equiv-1\mod7$
So only $3^3,6^3,10^3\equiv-1\mod7$ w/conditions.
Now $2001\equiv-1\mod7$
Two ways you can have (WLOG) $a^3+b^3+c^3\equiv-1\mod7$
[u][b]-Case 1:[/b][/u]
$a^3\equiv-1\mod7, b^3\equiv-1\mod7,c^3\equiv1\mod7$
We will try out all the cases WLOG:
$a=10,b=10,c=1$
If $a=10,b=3$, there is no integer solution $c$
If $a=10,b=6$, there is no integer solution $c$
If $a=6,b=6$, there is no integer solution $c$
If $a=6,b=3$, there is no integer solution $c$
If $a=3,b=3$, there is no integer solution $c$
[u][b]-Case 2[/b][/u]
$a^3\equiv-1\mod7, b^3\equiv0\mod7,c^3\equiv0\mod7$
We will try all cases WLOG:
If $b=c=7$, there is no integer solution $a$
If $b=c=0$, there is no integer solution $a$
So the the only three numbers whose cubes sum to $2001$ are $10,10,1$ We must take the permutations of this and we find either:
$\fbox{(a,b,c)=(10,10,1),(10,1,10),(1,10,10)}$[/hide]

[quote="1234567890"][hide="Do you want to see a lot of boring casework mod 7?"]
We trivially see that $ a,b,c < 13$
Say we have an integer $ x$.
$ x\equiv0\mod7 \rightarrow x^3\equiv0\mod7$
$ x\equiv1\mod7 \rightarrow x^3\equiv1\mod7$
$ x\equiv2\mod7 \rightarrow x^3\equiv1\mod7$
$ x\equiv3\mod7 \rightarrow x^3\equiv \minus{} 1\mod7$
$ x\equiv4\mod7 \rightarrow x^3\equiv1\mod7$
$ x\equiv5\mod7 \rightarrow x^3\equiv1\mod7$
$ x\equiv6\mod7 \rightarrow x^3\equiv \minus{} 1\mod7$
So only $ 3^3,6^3,10^3\equiv \minus{} 1\mod7$ w/conditions.
Now $ 2001\equiv \minus{} 1\mod7$
Two ways you can have (WLOG) $ a^3 \plus{} b^3 \plus{} c^3\equiv \minus{} 1\mod7$
[u][b]-Case 1:[/b][/u]
$ a^3\equiv \minus{} 1\mod7, b^3\equiv \minus{} 1\mod7,c^3\equiv1\mod7$
We will try out all the cases WLOG:
$ a \equal{} 10,b \equal{} 10,c \equal{} 1$
If $ a \equal{} 10,b \equal{} 3$, there is no integer solution $ c$
If $ a \equal{} 10,b \equal{} 6$, there is no integer solution $ c$
If $ a \equal{} 6,b \equal{} 6$, there is no integer solution $ c$
If $ a \equal{} 6,b \equal{} 3$, there is no integer solution $ c$
If $ a \equal{} 3,b \equal{} 3$, there is no integer solution $ c$
[u][b]-Case 2[/b][/u]
$ a^3\equiv \minus{} 1\mod7, b^3\equiv0\mod7,c^3\equiv0\mod7$
We will try all cases WLOG:
If $ b \equal{} c \equal{} 7$, there is no integer solution $ a$
If $ b \equal{} c \equal{} 0$, there is no integer solution $ a$
So the the only three numbers whose cubes sum to $ 2001$ are $ 10,10,1$ We must take the permutations of this and we find either:
$ \fbox{(a,b,c) \equal{} (10,10,1),(10,1,10),(1,10,10)}$[/hide][/quote]

can you please tell me isn't this = -1
$ x\equiv5\mod7 \rightarrow x^3\equiv NEGATIVE 1\mod7$

and please elaborate on the conditions for this:
So only $ 3^3,6^3,10^3\equiv \minus{} 1\mod7$ w/conditions.

TQ.

If I remember correctly, this problem is like an easier version of N1 IMO shortlist 2002 (it's been a couple months since I saw it though).
[hide]
mod 9 tells us that each of a, b, c must be either 1, 4, or 7 mod 9, telling us that a,b,c are 1 mod 3.
We have that a, b, c $ \leq 10$
Mod 7 tells us that the value of the LHS is -1, meaning that it's either -1/0/0 or -1/1/-1. If $ x^{3} \equiv \minus{} 1 \mod 7$, then we have that x=3, 5, 6. If it is 1, then we have that x=1, 2, 4.
If it's the first, 2 of the numbers are 7, but 2001-686=1315.

3 mod 7/1 mod 3 $ \Rightarrow$ 10 mod 21
5 mod 7/1 mod 3 $ \Rightarrow$ 19 mod 21, which is too large
6 mod 7/1 mod 3 $ \Rightarrow$ 13 mod 21, which is too large
So, two of the numbers must be 10, implying that the last one is 1.
[/hide]

If we look mod $ 9$, then $ a\equiv b\equiv c\equiv 1\mod 3$.

WLOG $ a\ge b\ge c$, then $ 2001 \equal{} a^3 \plus{} b^3 \plus{} c^3\le 3a^3$, additionally $ 2001 \equal{} a^3 \plus{} b^3 \plus{} c^3\ge a^3$, so $ 8.7\le a\le 12.6$. Hence $ a \equal{} 10$.

Then $ 1001 \equal{} b^3 \plus{} c^3\le 2b^3$, and $ 1001 \equal{} b^3 \plus{} c^3\ge b^3$, so $ 10.003 > b > 7.94$ $ b \equal{} 10$.

Finally, $ c^3 \equal{} 1$ so $ (a,b,c) \equal{} (10,10,1)$ and permutations.

Unfortunately, we only get $a^3, b^3, c^3 \equiv 1 \pmod{9}$ which definitely does not imply $a, b, c \equiv 1 \pmod{9}$; it in fact implies $a, b, c \equiv 1 \pmod{3}$. Also, (not so) surprisingly this is the link from the contests page of Junior Balkan MO 2001.

EDIT: unfortunately for me, sunny2000 deleted his post to whom I am replying.

[quote=TheMathLife]I know this ancient but there is sol using mod 4 should I post?[/quote]

It will be awesome.

[quote=Valentin Vornicu]Solve the equation $a^3+b^3+c^3=2001$ in positive integers.

[i]Mircea Becheanu, Romania[/i][/quote]

as mentioned earlier you can merely use casework as each of a, b, c  are 1(mod 3).
since there are only 4 vals each can attain the cw isn't too bad

[quote=Valentin Vornicu]Solve the equation $a^3+b^3+c^3=2001$ in positive integers.

[i]Mircea Becheanu, Romania[/i][/quote]

$a^3+b^3+c^3=2001$ as $2001\equiv 3\mod 9$ and $x^3\equiv 0,1,-1\mod 9$ clearly then $a^3,b^3,c^3\equiv 1\mod 9$ then $a,b,c\equiv 1$ or $4$ or $7\mod 9$ and also $a,b,c<13$
Hence we are left with very less cases to try and hence only solution are $\{a, b, c\}\in \{10, 10,1\}$

de bash

Note that since $2001\equiv 3\pmod 9$, we get that $a^3, b^3, c^3 $ are all $1\pmod 9$, so $a,b,c\equiv 1\pmod3$ and they are all less than $13$.

So each of $a,b,c$ are in the set $\{1,4,7,10\}$.

Claim: $a,b,c\ne 4$.
Proof: Suppose one of them was $4$, let it be $c$. Then $a^3+b^3=2001-64 = 1937 = 13\cdot 149$.

So $(a+b)(a^2+ab+b^2) = 13\cdot 149$. Since $13$ and $149$ are primes, and $a^2+ab+b^2>a+b>1$, we have $a+b=13$ and $a^2+ab+b^2 = 149$. There are no positive integer solutions to this. $\square$

Claim: $a,b,c\ne 7$.
Proof: Suppose FTSOC $c=7$. Then $a^3+b^3=(a+b)(a^2+ab+b^2)=1658$. Since $1658=2\cdot 829$ and $829$ is prime, we have $a+b=2$ and $a^2+ab+b^2 = 829$, no solutions. $\square$

So each of $a,b,c$ is in $\{1,10\}$.

Clearly we must have $\boxed{(10,10,1)}$ or permutations.

Due to symmetry we may assume $a\geq b \geq c$. Then $3a^3 \geq a^3 + b^3 + c^3 = 2001$, i.e. $a^3 \geq 667$, so $a\geq 9$ (as $8^3 = 512$). On the other hand, $a^3 \leq 2001$ implies $a \leq 12$.

    If $a=9$, then $b^3 + c^3 = 1272$, so $b^3 \geq 636$ and thus $b\geq 9$, i.e. $b=9$. When $b=9$ we get $c^3 = 543$, which has no solution.

    If $a=10$, then $b^3 + c^3 = 1001$, so $b^3 \geq 501$ and $b\geq 8$. For $b=8$ we obtain $c^3 = 489$, impossible; for $b=9$ we obtain $c^3 = 272$, impossible; and for $b=10$ we obtain $c^3 = 1$, i.e. $c=1$.

If $a=11$, then $b^3 + c^3 = 670$, so $b^3 \geq 335$ and $b\geq 7$. For $b=7$ we get $c^3 = 327$, impossible; for $b=8$ we get $c^3 = 158$, while $b^3 < 670$ implies $b \leq 8$.

If $a=12$, then $b^3 + c^3 = 273$, so $b^3 \geq 137$ and $b^3 < 273$, implying $b=6$. But for $b=6$ we get $c^3 = 57$, which is impossible.

    In conclusion, all solutions are $(10,10,1)$, $(10,1,10)$, $(1,10,10)$.

[quote=Valentin Vornicu]Solve the equation $a^3+b^3+c^3=2001$ in positive integers.

[i]Mircea Becheanu, Romania[/i][/quote]

Taking mod $9$, we see that either $a^3$ is equal to $-1,0,1$. Taking mod $9$ yields $a^3+b^3+c^3 \equiv 3 \pmod 9$. We can then see that the values $a,b,c$ must have a remainder of $1$ after being divided by $9$, with the only reasonable values $1$ or $10$. We can then deduce that $(1,10,10)$ and permutations work


function
integration
real analysis
real analysis unsolved

Let $f_i,1\le i\le n$  a set of continuous poasitive functions such that $\int_0^1f(x)dx=1$. Prove that there exist $x$ such that $f_1(x)f_2(x)...f_n(x)\le 1$.

It's sufficient to have integrable positive functions .
 $(f_1(x)...f_n(x))^{\frac 1n} \leq \frac{ f_1(x)+...+f_n(x)}n \Longrightarrow \int_0^1(f_1(x)...f_n(x))^{\frac 1n}dx\leq 1$ .. and so on ..
 :cool:

[quote="Cezar Lupu"]I think Xirti meant this problem ( which can be found also in my book). Here it is:

  [b] Problem.[/b]

    Let $f: \mathbb{R}\to\mathbb{R}$ a continuous function such that \[ \int_{0}^{1} f(x)dx=1.  \]
    Prove that there are $a_{1},a_{2},\ldots,a_{n}\in\mathbb{R}$ such that \[ \prod_{i=1}^{n} f(a_{k})=1.  \][/quote]
Do you require $a_i \ne a_j$ for $i \ne j$?

[quote="Cezar Lupu"]I think Xirti meant this problem ( which can be found also in my book). Here it is:

  [b] Problem.[/b]

    Let $f: \mathbb{R}\to\mathbb{R}$ a continuous function such that \[ \int_{0}^{1} f(x)dx=1.  \]
    Prove that there are $a_{1},a_{2},\ldots,a_{n}\in\mathbb{R}$ such that \[ \prod_{i=1}^{n} f(a_{k})=1.  \][/quote]

theorem of "moyenne" there exist $c \in [0,1]$ such that $f(c)=1$

n=1 and $a_1=c$


geometry
trapezoid
cyclic quadrilateral
geometry proposed

Suppose the circle $C_2$ passes from the center of circle $C_1$ and intersects $C_1$ at $M,N$. $AB$ is an arbitary diameter of $C_1$ and $A_1,B_1$ are intersection of $AM,BN$ with the circle $C_2$. Prove that $A_1B_1$ is equal to the radius of circle $C_1$

Am I stupid or is this very easy?

Let $O$ be the centre of $C_1$. It is sufficient to show that $MA_1 \parallel OB_1$, since then the cyclic quadrilateral $MA_1B_1O$ must be an isosceles trapezoid and $A_1B_1 = OM$. But that is trivial, since \[ \angle OMA = \angle OAM = \angle BAM = \angle BNM = \angle B_1NM = \angle B_1OM. \]

Yes, it's easy.
This was the easiest problem of the day.
But note that this contest has a lower level than TST or GOLD-Selection of Iran.


I'm a man with wide,and varied interest.Its only proper that
i retained a '[/b] exclusive[b] ownership[/b]'[b] related to some
would-be-'[/b] quality[b] ideas[/b],[b]and[/b] messages[b]'[/b]
i might post in this math/physics forum.My question:Can our math/physics member's '[b]Quality[/b]' posted message be subject to '[b]Copyright[u]'[/u][/b]assurances?

                                                 Respectfully,
                                                A.Eddington

I'm not totally certain, but I think when you post something in a public forum, you're making it public information. If you want something copyrighted, publish it in a paper.

That was a bit uncalled for blahblahblah.

Arthur, what Xevarion said is right.  If you have some groundbreaking results, and you want to retain your claim to them, you need to put them in a paper and get it published in a journal.

Additionally, you agreed to certain forum policies when registering for the forum.  One of them, if i recall correctly, is that all of your posts become property of AoPS or something like that.

Also, please try and make your posts easier to read.  Cut out all the superfluous formatting (bold, quotation marks, random line breaks, broken tags, etc.)  We can get the idea without you emphasizing half of your post with one form of formatting or another.

Thanks

For the legal answer, please read item #7 in our [url=http://www.artofproblemsolving.com/Legal/AoPS_L_Comm_Terms.php]Terms of Service[/url].


Can someone post example problems that use the Vieta Jumping method? I already know about 1988 IMO #6. Thanks.

http://www.mathlinks.ro/Forum/viewtopic.php?t=40207
Problems in elementary number theory,

http://www.mathlinks.ro/Forum/viewtopic.php?p=894656
IMO 2007 problem 5

Thats some of them.

Here's another example:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1702426#1702426


calculus integration inequalities algebra polynomial

[quote="me@home"]Prove that
\[676 \ | \ 3^{3n+3}-26n-27 \]
[/quote]

[hide="assuming that n is a whole number"]Let's prove the base case, when n=0.

$3^{3*0+3}-26*0-27=0$, which is divisible by 676.

Now let's prove that, if it's true for k, it's true for k+1.

Now let's assume that it's true for k.

$676 | 3^{3k+3}-26k-27$

Let's call 3^{3k+3} x, -26k y.

Let's plug in k+1 and see what happens:

$3^{3k+6}-26k-53=27x-y-53$

Since x+y-27 is divisible by 676, we can subtract it from that equation and get

$26x-26$. We need to prove that that's diivisible by 676.

$26(x-1)=26x-26$

Now we need to prove that x-1 is divisible by 26.

$x-1=3^{3k+3}-1$

But 3k+3 is a multiple of 3, and 3 raised to any power thats a multiple of 3 is 1 mod 26.

Here's the proof:

$3^{3k}=27^{k}$

Since 27=1mod26, any integral power of 27  is also 1mod26.[/hide]

aww.. I wish I didn't go check the mail...

Our proofs are essentially the same, except you used more line breaks then me.

Well, I'm new to intermediate proofs, especially induction.

[quote="Bill The Maniac"][hide]Modulo $4$ $27^{(}n+1)=3^{(}n+1)=[3,1]$, and $26n=2n=[2,4]$ so this is $[3,1]-[2.4]+3=[0.0]$

Modulo $169$ $27x=x+26$ if $26x=26$ if $x=1\mod 13$. Powers of $27$ are always equal to $1\mod 13$ because $27=1\mod 13$.[/hide][/quote]
Not too sure about your bracket notation.

Here's my proof (I saw the problem with 169 somewhere, it had a long wierd solution, but its a really easy problem even for this stronger case)
\[3^{3n+3}-26n-27 = 27^{n+1}-27-26n=27(27^{n}-1)-26n= \\ =27(26)(1+27+27^{2}+27^{3}+\cdots)-26n=26\left[ \sum_{1}^{n}27^{i}-n\right] \]
However, we have that
\[\sum_{1}^{n}27^{i}\equiv \sum_{1}^{n}1 \equiv n \mod{26}\implies \left[ \sum_{1}^{n}27^{i}-n\right]\equiv 0 \mod{26}\]
Thus $26^{2}\ | \ 3^{3n+3}-26n-27$. In fact, I just realized that this s exactly the same factorization I used to prove a certain form of the Am-Gm inequality this morning...

hmm...that can be expressed as a linear recurrence with the characteristic polynomial: $(x-3)(x-1)^{2}=0$. After brute-forcing the first 3 base cases [start with $n=-1$], the claim follows trivially by strong induction.

[quote="me@home"]So how do you account for the $n$ term?[/quote]

(x-3)(x-1)^2=0

k_1*3^n+(k_2*n+k_3)(1)^n


Let $ a,b,c$ be non-zero real numbers satisfying \[ \dfrac{1}{a}\plus{}\dfrac{1}{b}\plus{}\dfrac{1}{c}\equal{}\dfrac{1}{a\plus{}b\plus{}c}.\] Find all integers $ n$ such that \[ \dfrac{1}{a^n}\plus{}\dfrac{1}{b^n}\plus{}\dfrac{1}{c^n}\equal{}\dfrac{1}{a^n\plus{}b^n\plus{}c^n}.\]

[quote="Raja Oktovin"]Let $ a,b,c$ be non-zero real numbers satisfying
\[ \dfrac{1}{a} \plus{} \dfrac{1}{b} \plus{} \dfrac{1}{c} \equal{} \dfrac{1}{a \plus{} b \plus{} c}.\]
Find all integers $ n$ such that
\[ \dfrac{1}{a^n} \plus{} \dfrac{1}{b^n} \plus{} \dfrac{1}{c^n} \equal{} \dfrac{1}{a^n \plus{} b^n \plus{} c^n}.\]
[/quote]

I think it is true for all $ n \equal{} odd$ numbers....

Just a [hide="hint"]$ (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}0$.

And $ \left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} \right)^n\equal{}\frac{1}{a^n\plus{}b^n\plus{}c^n}\equal{}\frac{1}{(a\plus{}b\plus{}c)^n}$ for $ n\equal{}odd$.....

I hope it is right....
[/hide]

[quote="Dimitris X"] I think it is true for all $ n \equal{} odd$ numbers....

[/quote]
Dimitris X is right :

$ \frac 1a\plus{}\frac 1b\plus{}\frac 1c\equal{}\frac 1{a\plus{}b\plus{}c}$ $ \iff$ $ (ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c)\equal{}abc$ with $ abc\ne 0$

$ \iff$ $ (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}0$ and $ abc\ne 0$

And so :  $ (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}0$ and $ abc\ne 0$ and $ n$ odd $ \implies$ $ (a^n\plus{}b^n)(b^n\plus{}c^n)(c^n\plus{}a^n)\equal{}0$ and so the same property for $ a^n,b^n,c^n$

And obviously, $ n\equal{}2p$ and $ abc\ne 0$ $ \implies$ $ (a^{2p}\plus{}b^{2p})(b^{2p}\plus{}c^{2p})(c^{2p}\plus{}a^{2p})\ne 0$

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}\frac{1}{a\plus{}b\plus{}c}$

$ \frac{ab\plus{}bc\plus{}ca}{abc}\equal{}\frac{1}{a\plus{}b\plus{}c}$

$ (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\equal{}abc$

$ a^2b\plus{}ab^2\plus{}b^2c\plus{}bc^2\plus{}c^2a\plus{}ca^2\plus{}3abc\equal{}abc$

$ a^2b\plus{}ab^2\plus{}b^2c\plus{}bc^2\plus{}c^2a\plus{}ca^2\plus{}2abc\equal{}0$

$ (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}0$

Then at least 1 of $ a\plus{}b,b\plus{}c,c\plus{}a$ is equal to $ 0$, WLOG : $ a\plus{}b\equal{}0$ ,$ b\plus{}c\ne c\plus{}a \ne 0$ , so $ a\equal{}\minus{}b$ $ \rightarrow$ $ a^n\equal{}\minus{}b^n$ (if $ n$ is odd) ,so $ a^n\plus{}b^n\equal{}0$ , so $ (a^n\plus{}b^n)(b^n\plus{}c^n)(c^n\plus{}a^n)\equal{}0$ and it will lead to

$ \frac{1}{a^n}\plus{}\frac{1}{b^n}\plus{}\frac{1}{c^n}\equal{}\frac{1}{a^n\plus{}b^n\plus{}c^n}$

But if $ n$ is even,and because $ a,b,c$ is not equal to $ 0$ , then $ a^n\plus{}b^n\ne0$ ,, $ b^n\plus{}c^n\ne0$ and $ c^n\plus{}a^n\ne 0$ , so $ (a^n\plus{}b^n)(b^n\plus{}c^n)(c^n\plus{}a^n)\ne0$

So the solution are $ n\equal{}2k\minus{}1$ where $ k\in\mathbb{N}$


Find the sum of all 3-digit numbers in the base 10 whose 3 digits are non-zero and distinct.

[hide]
Each digit from $1$ through $9$ will appear $8\times7$ times in the hundreds digit, tens digit, and ones digit.

The total sum in the ones digit is then $\displaystyle \frac{(1+9)9}{2} \times 56 = 2520$

The sum in the tens digit is $25200$, the hundreds is $2522000$.

The sum of all the three digit numbers is then $2520+25200+252000 = \boxed{279720}$.

If I remember correctly, this is almost the exact same problem as Problem 6 on the 2006 AIME I (which did include zero as a possible digit).
[/hide]

[hide="here is what i did"]
notice that the first such number is 123 and the last one is 987.
$123+987=1110$
the next smallest number is 124 and the secon largest is 986.
$124+986=1110$
this is easy to verify for all numbers of this type.
so we find how many such numbers there are

for the fist digit we have $9$ choices (1-9)
the second digit we have $8$ choices (1-9 minus the digit already used)
for the third digit we have $7$ choices (1-9 minus the first 2 digits used)
so there are a total of $(9*8*7)$ numbers that meet the requirement.

so the sum is
$\frac{1110(9*8*7)}{2}=279720$
[/hide]


abstract algebra
function
group theory
superior algebra
superior algebra unsolved

If $ \phi : A \to B$ is a homomorphism, A and B are finitely generated and finitely presentable respectively, then $ ker(\phi)$ is finitely generated. A and B are algebras.

Proof: We wish to find a finite subset Y of $ ker(\phi)$ such that Y generates $ ker(\phi)$.

Let $ S \equal{} ({s_{1}, s_{2}, ..., s_{n}})$ be a generating set of A. Then each $ x_{i}\in A$ is of the form: $ x_{i} \equal{} \prod_{i\equal{}1}^{k}s_{i}^{\alpha_{i}}, k\leq n, \alpha_{i} \pm 1$.

Note that B gives the presentation $ Y \equal{} \left<{(s_{i}^{\alpha_{i}}, s_{i \plus{} 1}^{\alpha_{i\plus{}1}})| \phi(s_{i}) \equal{} \phi(s_{i \plus{} 1})}\right>$.

Choose $ x_{i}, x_{i \plus{} 1} \in A$.
Then, $ (x_{i}, x_{i \plus{} 1}) \equal{} \left(\prod_{i \equal{} 1}^{k}s_{i}^{\alpha_{i}}, \prod_{i \equal{} 1}^{k} s_{i\plus{}1}^{\alpha_{i\plus{}1}}\right) \equal{} \prod_{i \equal{} 1}^{k} \left(s_{i}^{\alpha_{i}}, s_{i\plus{}1}^{\alpha_{i\plus{}1}}\right)$.
We get:
$ \phi(x_{i}) \equal{} \phi(x_{i \plus{} 1})$
$ \iff \prod_{i \equal{} 1}^{k} \phi\left(s_{i}^{\alpha_{i}}\right) \equal{} \prod_{i \equal{} 1}^{k} \phi\left(s_{i\plus{}1}^{\alpha_{i\plus{}1}}\right)$
$ \iff \prod_{i\equal{}1}^{k}\phi\left(s_{i}\right)^{\alpha_{i}} \equal{} \prod_{i\equal{}1}^{k}\phi\left(s_{i\plus{}1}\right)^{\alpha_{i\plus{}1}}$.
So elements in the kernel can be expressed as product of elements of Y and inverses of those elements.

Is this a valid argument?

The map is surjective (this was left out in the original problem). The argument I have there is nonsense. Does anyone have a hint on how I should proceed on this one?

[hide="Group version"]I don't believe that argument, because you can't make the construction work. Consider a nice simple example: $ A$ free on $ \{a,b\}$ and $ B$ free abelian on $ \{a,b\}$. The kernel is generated by the commutator $ aba^{ \minus{} 1}b^{ \minus{} 1}$ [i]and all of its conjugates[/i]. This is not naturally finite.

In fact, it's not finitely generated at all. Define a weight function $ w$ on $ A$ by $ w(a) \equal{} w(b) \equal{} 1$, $ w(a^{ \minus{} 1}) \equal{} w(b^{ \minus{} 1}) \equal{} \minus{} 1$, and $ w(xy) \equal{} w(x) \plus{} w(y)$; the kernel of our homomorphism is contained in the kernel of $ w$. Now, define a "height" function $ h$ on $ A$ as follows:
Let $ L(x)$ be the length of the reduced string expression for $ x$; $ L(a) \equal{} 1, L(ab) \equal{} L(a^2) \equal{} 2, L(a^{ \minus{} 1}b^{ \minus{} 1}a^2) \equal{} 4$ and so on. Among all pairs $ x_1,x_2$ with $ x_1x_2 \equal{} x$ and $ L(x_1) \plus{} L(x_2) \equal{} L(x)$ (a finite collection), define $ h(x)$ to be the maximum of $ |w(x_1)|$.

Why do we care about $ h$? If $ w(x) \equal{} w(y) \equal{} 0$, $ h(xy)\le \max\{h(x),h(y)\}$. Any finitely generated subgroup of the kernel of $ w$ must therefore have $ h$ bounded. Now, we can find elements of the kernel $ A/B$ with arbitrarily large values of $ h$; $ h(a^nba^{ \minus{} n}b^{ \minus{} 1}) \equal{} n \plus{} 1$ for $ n\ge 1$.[/hide]

Oops, I didn't see "algebra". Since I can't find an appropriate definition for a finitely presented algebra in my sources at hand (Lang, Wikipedia), you'll have to clarify. Does my example still apply if you turn the groups into group rings?

For the group case, it is enough to show that the index of $ ker(\phi)$ is finite because then one could use the lemma: A subgroup of finite index in a finitely generated group is finitely generated.

Def: An algebra is finitely presentable if it has a presentation $ \left<X|R\right>$ with X (generators) and R (relations) both finite.

Well, look at the example. It's false, in both the group and algebra versions. As an algebra over an arbitrary base field $ F$, the presentations are:
$ A$: generated by $ a,b,a^{\minus{}1},b^{\minus{}1}$, with relations $ aa^{\minus{}1}\equal{}a^{\minus{}1}a\equal{}bb^{\minus{}1}\equal{}b^{\minus{}1}b\equal{}1$
$ B$: generated by $ a,b,a^{\minus{}1},b^{\minus{}1}$, with relations $ aa^{\minus{}1}\equal{}a^{\minus{}1}a\equal{}bb^{\minus{}1}\equal{}b^{\minus{}1}b\equal{}aba^{\minus{}1}b^{\minus{}1}\equal{}1$

We use the natural map, and the kernel is not finitely generated.

I am restating the problem:
Suppose $ \phi: \mathbb{A} \to \mathbb{B}$ is a surjective homomorphism, where $ \mathbb{A}$ is a finitely generated algebra and $ \mathbb{B}$ is a finitely presented algebra. Show that $ ker(\phi)$ is finitely generated.

Let $ \left<X|R\right>$ be a presentation of $ \mathbb{A}$ with X finite and R not necessarily finite.  We will construct a presentation of $ \mathbb{B}$ from the presentation of $ \mathbb{A}$.

We know that $ \mathbb{A} \equal{} <X> \iff \pi: F(X) \to \mathbb{A}$ is a surjective homomorphism (Note that it is not the kernel of this natural map $ \pi$ that we are claiming to be finitely generated).
From the two maps we can deduce that $ \left<X|R \cup \pi^{\minus{}1}(ker(\phi))\right>$ is a presentation of $ \mathbb{B}$ (this step needs justifications).

Lemma: If an algebra $ \mathbb{B}$ is finitely presentable and $ \left<Y|R\right>$ is any presentation of this algebra with Y finite, then there is a finite subset $ R_{0}\subset R$ such that $ \left<X|R_{0}\right>$ is a presentation of this algebra.

By the lemma, $ \left<X|R_{0} \cup \pi^{\minus{}1}(ker(\phi))\right>$ is a presentation of $ \mathbb{B}$ and so $ \mathbb{B}$ is finitely presented since both X and $ R_{0} \cup \pi^{\minus{}1}(ker(\phi))$ are finite.

valid?

That's not a proof. The result is false, by the counterexample I've already given.

Looking at what you wrote, you haven't even mentioned the desired conclusion. Your argument is completely irrelevant.

Judging from the first post, akech seems to be assuming that these algebras are commutative.  In which case jmerry's counterexample does not work.

Of course, in that case we do not need nearly so many assumptions.  If $ \mathbb{A}$ is finitely generated--and commutative--Hilbert's basis theorem implies that it is Noetherian, so any ideal is finitely generated.

Or maybe he was thinking of abelian groups?  In which case, again, we only need $ \mathbb{A}$ to be finitely generated.

Where did this problem come from?

jmerry, you are right my argument is not correct as it stands but I was just trying to connect the presentations of the two algebras and then from there may be to see if I will be able to show the kernel in question is finitely generated.

LydianRain,
I did not assume anything about the algebras in question and the problem must be correct as someone told me he knows of a proof.

Many people have proved incorrect things before, it would not be the first time...  :roll:

[quote="LydianRain"]Judging from the first post, akech seems to be assuming that these algebras are commutative.[/quote]If you assume commutativity, finitely generated and finitely presented are the same. This question only makes sense without that assumption.

[quote="akech"]... the problem must be correct as someone told me he knows of a proof.[/quote]You have too much faith in "someone".

OK, now try to spot the flaw in my arguments. Why doesn't my example work as an algebra?
[hide="Hint"]What kind of structure is the kernel, and what does it mean fr it to be finitely generated?[/hide]

Yes, I suppose that we're talking about finitely generated [i]as an ideal[/i].  So we're perfectly allowed to multiply on the left and right by anything, in particular we can conjugate, undermining your counterexample.

I'm going to state my current understanding of the problem for clarity: Let $ R$ be a ring, $ F$ a free $ R$-algebra of finite rank, $ G$ any free $ R$-algebra, $ N\subset F$ an ideal, and $ M\subset G$ a finitely generated ideal.  Then if $ F/N\cong G/M$, $ N$ is finitely generated.

(or is $ R$ supposed to be a field?)

More concretely on the example, I can go from $ aba^{\minus{}1}b^{\minus{}1}\minus{}1$ to $ xaba^{\minus{}1}b^{\minus{}1}\minus{}x$ to $ xaba^{\minus{}1}b^{\minus{}1}x^{\minus{}1}\minus{}1$ when $ x$ is an invertible monomial. The relations that make up a presentation generate the kernel as a two-sided ideal, unlike the group version.


I wasnt sure where this question was supposed to be so i put it here:

can anyone explain to me why x^(log base x of y)=y?

That is the definition of a logarithm.

$\log_x y = z$ means that $y = x^z$, so you can do this:
$x^{\log_x y} = x^z = y$

In words: x raised to the power needed in order to get y is equal to y.


Let $a,b,c,d>0$.Prove the following inequality.When does the equality hold?

\[\frac{3}{1}+\frac{5}{1+a}+\frac{7}{1+a+b}+\frac{9}{1+a+b+c}+\frac{36}{1+a+b+c+d}\leqq 4\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\]

Are you sure it's right?  Take a,b,c,d -> 0 so the RHS goes to infinity while the RHS goes to 3+5+7+9+36.

kunny.
can you give the case for the "=" is hold?


1.prove that there are no positive integer $ a,b,n>1$ such that
$ (a^n\plus{}b^n)$ is divisible by $ (a^n\minus{}b^n)$
2.if $ a$and $ b >2$ are any positive integers prove that $ 2^a\plus{}1$ is not divisible by $ 2^b\minus{}1$

I think this might help you: http://www.mathlinks.ro/viewtopic.php?t=227651  :wink: