find all polyminal that: P((x+1)^n)=(P(x)+2x+1)^n

$ G(x): P((x \plus{} 1)^{n}) \equal{} (P(x) \plus{} 2x \plus{} 1)^{n}$,and $ n\in\mathbb{Z}$

$ G(\minus{}1): P(0)\equal{}(P(\minus{}1)\minus{}1)^n$ then $ P(\minus{}1)\equal{}1$ and $ P(0)\equal{}0$

$ G(0): P(1)\equal{}1$

$ G(1): P(2^n)\equal{}4^n\equal{}(2^n)^2$ for all $ n\in\mathbb{Z}$.

Notice that $ P(x)\equal{}x^2$ for infinite values of $ x\equal{}2^n$ and degree of $ P$ is finite then $ \boxed{P(x)\equal{}x^2}$ which trivially fullfils $ G(x)$

[quote="mszew"]
$ G( \minus{} 1): P(0) \equal{} (P( \minus{} 1) \minus{} 1)^n$ then $ P( \minus{} 1) \equal{} 1$ and $ P(0) \equal{} 0$
[/quote]

Why :?:

For example it might be $ P(\minus{}1)\equal{}2$ and $ P(0)\equal{}1$

I think $ n$ is a constant not a abitrary integer
Case1 :$ n > 0$
$ \Rightarrow P( (x + 1)^n)^{\frac {1}{n}} = P(x) + 2x + 1$
$ \Rightarrow$ if $ z$ is a complex root equation :$ P(x^n) = 0$ then $ z$ is  a $ kn$-multiple ( $ k\in N$) of this equation (1)
We rewrite $ P(x) = (x - z_1)(x - z_2)....(x - z_p)$
$ \Rightarrow P( x^n) = ( x^n - z_1)...( x^n - z_p)$
Besides , we have the equation $ x^n = p$ has $ n$ distinct complex root $ \forall k \ne 0$(2)
From (1) and (2) , we can prove that :
$ P(x) = Q(x)^n.x^m$ ($ m < n$)
$ \Rightarrow Q( (x + 1)^n ).(x + 1)^m = Q(x)^n.x^m + 2x + 1$
$ \Rightarrow Q( x^n).x^m = Q(x - 1)^n.(x - 1)^m + 2x - 1$

Can someone help me finish my proof ?


In a square the letters $M$ (four segments) and and the letter $O$ (the big circle) are inscribed. There is also another smaller circle touching the two diagonal segments of the letters $M$ and touching the letter $O$ in the interior. Let $A_1$ and $r_1$ be the area and the radius of the bigger circle and $A_2$ and $r_2$ the area and radius of the smaller circle. Prove: $ \displaystyle \frac{A_2}{A_1} + \frac{r_2}{r_1} = 1$.

In the problem tan x = 2-----(1) (from the symmetry of the bigger circle)

then let A=tan (x/2) = r2/r1 -----(2).

From 1 & 2

2A/(1-A2 ) = 2...

which gives us...A2 + A = 1

A2 = r22 / r12 = A2/A1 (ratios of their areas)....
so we have A2/A1 + r2/r1 =1...hence the result.


algebra
polynomial
calculus
derivative
geometry
3D geometry
calculus computations

Define $ A(x)|B(x)$ $ \Leftrightarrow$ there exists a polynomial $ C(x)$ such that $ B(x)\equal{}A(x)C(x)$.

[b]Question:[/b]

Let $ P(x)$ be a polynomial of real coefficients and of degree $ 5$. Prove that $ P(x)|[P'(x)]^2 \Rightarrow$ all roots of $ P(x)\equal{}0$ are real.

what does the $ P'(x)$ mean in this: $ P(x)|[P'(x)]^{2}$?

[quote="grn_trtle"]The first derivative of $ P(x)$?[/quote]

Yes, it is.

You don't have to show that, you are given that. Given that, you must show that $ P(x) \equal{} 0$ has only real solutions.

[quote="tonypr"]Well this isin't much but I guess it's how you start off,
if $ P(x) \equal{} a_5 x^5 \plus{} a_4x^4 \plus{} a_3x^3 \plus{} a_2x^2 \plus{} a_1x \plus{} a_0$ then:

$ P'(x) \equal{} 5 a_5x^4 \plus{} 4a_4x^3 \plus{} 3a_3x^2 \plus{} 2a_2x \plus{} a_1$

Edit: oh hehe thanks[/quote]

Actually we don't need to express $ P(x)$ and $ P'(x)$ in the form like that in your post (in my solution of course, as I don't know yours) .

And similarly the only such polynomials of degree 2 or 3 are respectively perfect squares or cubes of some linear polynomial.  $ (x^2 \plus{} 1)^2$ is the smallest example with complex roots, and also the smallest example in which the polynomial is not just some power of a linear term.  $ (x \minus{} a)^2(x \minus{} b)^3$ is the smallest example that isn't a perfect power.  Etc.

[quote="stephencheng"]Define $ A(x)|B(x)$ $ \Leftrightarrow$ there exists a polynomial $ C(x)$ such that $ B(x) \equal{} A(x)C(x)$.

[b]Question:[/b]

Let $ P(x)$ be a polynomial of real coefficients and of degree $ 5$. Prove that $ P(x)|[P'(x)]^2 \Rightarrow$ all roots of $ P(x) \equal{} 0$ are real.[/quote]

[hide="solution"]
Let $ P(x)\equal{}(x\minus{}a)Q(x)$, then $ (x\minus{}a)|P'(x)\equal{}(x\minus{}a)Q'(x)\plus{}Q(x)  \Leftrightarrow (x\minus{}a)|Q(x)$
So $ (x\minus{}a)|P(x) \Leftrightarrow (x\minus{}a)^2|P(x)$ in this case.

Now assume that there exist non-real root $ a$ then $ \overline a$ is also a root of $ P(x)\equal{}0$
Then $ [(x\minus{}a)(x\minus{}\overline{a})]^2|P(x)$

As $ [(x\minus{}a)(x\minus{}\overline{a})]^2$ is of real coefficients, there must exist another root $ b$ which is real.
So $ (x\minus{}b)^2|P(x) \Rightarrow (x\minus{}b)^2[(x\minus{}a)(x\minus{}\overline{a})]^2|P(x)$, but the degree of $ P(x)$ is $ 5$, contradiction, done.
[/hide]


$ a_{0}\equal{}b_{0}\equal{}1$
$ a_{n\plus{}1}\equal{}a_{n}\plus{}b_{n}$
$ b_{n\plus{}1}\equal{}b_{n}\plus{}3a_{n}$
Find all c such that $ n.|c.a_{n}\minus{}b_{n}|<2$ for all n

If $ x_{n}\equal{}b_{n}\plus{}a_{n}\sqrt 3$, then $ x_{n\plus{}1}\equal{}(1\plus{}\sqrt 3 )x_{n}$, therefore
$ a_{n}\equal{}\frac{(1\plus{}\sqrt 3 )^{n\plus{}1}\minus{}(1\minus{}\sqrt 3 )^{n\plus{}1}}{2\sqrt 3 },b_{n}\equal{}\frac{(1\plus{}\sqrt 3 )^{n\plus{}1}\plus{}(1\minus{}\sqrt 3 )^{n\plus{}1}}{2}$.
It give $ c_{n}\equal{}\sqrt 3 a_{n}\minus{}b_{n}\equal{}\minus{}(1\minus{}\sqrt 3)^{n\plus{}1}$, therefore $ c_{0}\equal{}\sqrt 3\minus{}1\equal{}0,728.., |c_{n}|\equal{}c_{0}^{n\plus{}1}<1$.

An other method if c exist c is limited of the sequence
$ x_{n}\equal{}\frac{a_{n}}{b_{n}}$
Easy to check that c is unique then prove the inequality


calculus
derivative
function
inequalities
inequalities theorems

Are there any properties on concavity/convexity that make concavity/convexity easier to prove.  For instance, are there theorems such as

If f(x) and g(x) are both concave/convex on an interval, then f(x)*g(x) is concave/convex

Basically, are there easier ways to prove convexity/concavity than finding the derivative?

JB

What is "monotony?"
When a function is strictly increasing or decreasing.

When f' (1st derivative) is strictly increasing then f is convex. And
When f' (1st derivative) is strictly decreasing then f is concave.

Tell an example for your question.

socrates

[quote="jackb1115"][quote="socrates"]I think no.
To prove that a function is concav/convex we
use the criterion of the 2nd derivative
[b]or[/b] the monotony of the 1st.

Don't forget also the Jensen's inequality![/quote]

What is "monotony?"

I know about Jensen ... I'm trying to see if there are ways to prove concavity/convexity to make Jenses easier to use ...

JB[/quote]
I bet using derivatives is already very easy.... although sometimes it's rather troublesome.

Well, I think, if $f(x)$ and $g(x)$ are both convex (or concave), then $f(x)+g(x)$ is convex (concave), but that is just a consequence of the derivative test.  :D


probability
geometry
3D geometry
counting
distinguishability
tetrahedron
rotation

If you haven't taken a look at part 1, take a look at that first because personally, this is harder than dice problems. Anyway, I try to find many problems involving painting cubes but to my surprise, they do not occur often at all! Hmm.. For some reason, I thought that every time I take an AMC exam, I see painting... must be one of those countingculosis  :rotfl:  :rotfl:

1. (Mathematics of Choice) How many different colored blocks of a fixed cubical shape can be made if six colors are available, and a block is to have a different color on each of its face? Standard definition of "distinguishable" faces applies here as well.

2. (AMC 12B 2007 #16) Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical.  How many distinguishable colorings are possible?
A) 15
B) 18
C) 27
D) 54
E) 81

3. (AMC 12B 2004 #20) Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?
A) 1/4
B) 5/16
C) 3/8
D) 7/16
E) 1/2

4. (AMC 12 2000 #25) Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)
A) 210
B) 560
C) 840
D) 1260
E) 1680

Is this the answer to #2?
[hide]B[/hide]

If so, I'll post my solution.

Not quite the same kind of problem - there's no painting - but I find the methods are similar. So here's one to add.

(2005 AMC 12 B #25) Six ants simultaneously stand on the six vertices of a regular octahedron, with each ant at a different vertex. Simultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each with equal probability. What is the probability that no two ants arrive at the same vertex?

A. $ \frac {5}{256}$

B. $ \frac {21}{1024}$

C. $ \frac {11}{512}$

D. $ \frac {23}{1024}$

E. $ \frac {3}{128}$

[quote="E^(pi*i)=-1"][hide="4"]
There are $ 8!$ ways to color it without considering rotation.  If we consider a particular face, there are $ 8$ faces that could be there, and $ 3$ orientations for each, so there are $ 24$ symmetries.  Thus the answer is $ \frac {8!}{24} \equal{} 1680 \equal{} E$.
[/hide][/quote]

What do you mean by orientations?...and does that mean #1 can be done in almost the exact same way?  I'll try it after I understand what you mean.

Man, counting/probability is my worst topic.  Anyone have any hints or stuff?

Also, what's the general method of solving these?  Whenever I see these, I skip them :(.

Well...here's my logic for [hide="3"]
Okay, let's break it up into cases, and let's only deal with blue faces for now (as in, they are the 4 vertical ones that show up).  We will double the final answer to include the red faces.

So, if we have $ 6$ blue faces, it automatically satisfies the condition ($ \frac {1}{64}$ chance).

If we have $ 5$ blue faces, it automatically satisfies the condition($ \frac {6}{64}$ chance, because there are $ 6$ ways to choose that one red face).

If we have $ 4$ blue faces, we need to choose the $ 2$ red faces in such a way that they are opposite each other, otherwise, it doesn't work.

So, we have $ 3$ ways of doing that, making the probability $ \frac {3}{64}$.  Adding these up, we get $ \frac {10}{64} \equal{} \frac {5}{32}\times 2 \equal{} \frac {5}{16}$[/hide]

By orientations, I mean that you can rotate a face by $ 120^{\circ}$ or $ 240^{\circ}$ and it looks the same (i.e. it has $ 3$ rotational symmetries).

This is a standard technique for these problems (and yes, #1 can be solved the same way.)

I'm not so sure about my answers, but I’ll post my work anyway:

[hide="1"]
Pick any of the colors and fix it, WLOG red. Then there are $ 5$ possible colors for the face directly opposite the red face. On the four remaining faces, we pick any of the colors, WLOG orange. Directly opposite there are $ 3$ choices, and of the remaining two faces we can permute them in $ 2$ ways. Thus $ 5 \times 3 \times 2 \equal{} \boxed{30}$.

Alternatively, after fixing a face there are $ 5!$ color choices, and there are $ 4$ rotational symmetries about this face ($ 0,\pm 90, 180^{\circ}$), giving $ \frac {5!}{4} \equal{} 30$.
[/hide]

[hide="2"]
Casework (with four faces on the tetrahedron, 3 colors):
* If they are all one color, then there are $ 3$ possibilities.
* If there are two colors, either one color appears three times or both colors appear twice.
** There are three colors to pick for the one that appears $ 3$ times, 2 for the other color, and all possible configurations are indistinguishable, so $ 3 \times 2 \equal{} 6$ here.
** There are $ {3\choose 2} \equal{} 3$ possible color combinations to pick, and all possible configurations are indistinguishable, so $ 3$ here.
* If there are three different colors, one color appears twice ($ 3$ possibilities). Again, all configurations are indistinguishable, so $ 3$ here.
Thus $ 3 \plus{} 6 \plus{} 3 \plus{} 3 \equal{} \boxed{15}$.

Alternatively, apply a lemma whose name I didn’t learn (Burnsides?)
\[ \frac {1}{24}\left(1 \cdot 3^4 \plus{} 3 \cdot 3^2 \plus{} 8 \cdot 3^2 \plus{} 6 \cdot 3^3 \plus{} 6 \cdot 3\right) \equal{} 15
\]
where the $ 24$ symmetries
* $ 1$ identity creates $ 4$ cycles of faces
* $ 8$ rotations formed by taking a vertex, dropping altitude, and rotating $ \pm 120^{\circ}$, creating $ 2$ cycles of faces
* $ 3$ reflections by selecting the axis between two midpoints of opposite edges, creating $ 2$ cycles of faces
* $ 6$ reflections similarly $ \pm 180^{\circ}$, creating $ 3$ cycles of faces
* $ 6$ rotoreflections creating $ 1$ cycle of faces
Its really hard for me to visualize (forget to explain) this though..
[/hide]

#3, I can’t see anything but a bit of casework like Zellex did.

[hide="4"]
Pick any of the colors and fix it, WLOG the red face. Then there are $ 7!$ color choices left. We can also rotate the octahedron about the triangular face with $ 3$ symmetries, so $ \frac {7!}{3} \equal{} \boxed{1680}$.
[/hide]

[hide="5"]
This is quite a bit of casework. I actually got the right answer, but it took me way too long at any rate..

Let $ f(n)$ be the number of cycles of length $ n$ the can be walked among the vertices of an octahedron. For example, $ f(3)$ would represent the number of ways in which an ant could navigate $ 2$ vertices and then return back to the original spot. Since an ant cannot stay still, $ f(1) \equal{} 0$. We also easily see that $ f(2) \equal{} 1, f(3) \equal{} 2$.

Now consider any four vertices of the octahedron. All four vertices will be connected by edges except for one pair. Let’s think of this as a square with one diagonal (from top left to bottom right). Suppose an ant moved across this diagonal; then the ant at the other end can only move across the diagonal (which creates 2-cycle, bad) or it can move to another vertex, but then the ant at that vertex must move to the spot of the original ant (which creates 3-cycle, bad). Thus none of the ants can navigate the diagonal and can either shift clockwise or counterclockwise, and so $ f(4) \equal{} 2$.

$ f(6)$ is even worse .. Consider an ant at the top of the octahedron. It has four choices. Afterwards, it can either travel directly to the bottom, and then it has $ 2$ ways back up, or it can travel along the sides and then go to the bottom, of which simple counting gives us $ 6$ ways back up. Hence, this totals $ 4 \times (2 \plus{} 6) \equal{} 32$.

Now, the number of possible ways is given by the sum of all possible cycles,
\[ a \times f(2) \times f(2) \times f(2) \plus{} b \times f(2) \times f(4) \plus{} c \times f(3) \times f(3) \plus{} d \times f(6)
\]
where the coefficients represent the number of ways we can configure these cycles. To find $ a$, fix any face, there are $ 4$ adjacent faces to select from to complete the cycle. From the four remaining faces there are only $ 2$ ways to create cycles, hence $ a \equal{} 8$.

To find $ b$, each cycle of $ 2$ faces is distinguished by their common edge, and there are $ 12$ edges, so $ b \equal{} 12$.

To find $ c$, each three-cycle is distinguished by the vertex, and there are $ 8$ edges. However, since the two three-cycles are indistinguishable, $ c \equal{} 8/2 \equal{} 4$.

Clearly $ d \equal{} 1$. Finally,
\[ 8(1)(1)(1) \plus{} 12(1)(2) \plus{} 4(2)(2) \plus{} (32) \equal{} 80
\]
Each bug has $ 4$ possibilities to choose from, so the probability is $ \frac {80}{4^6} \equal{} \boxed{\frac {5}{256}}$.   :D

[/hide]

[quote="Zellex"]So..assuming I know what you mean...the answer to #1 would be [hide] $ 6!/(4\times 6)$?[/hide][/quote]
Yes, that's right.

[hide="3 without casework"]
Choose one of the $ 3$ axes of the cube.  The $ 4$ faces not on that axis are all the same color; there are $ 2$ ways to choose the color.  There are $ 2^2\equal{}4$ ways to choose the irrelevant colors of the faces on the axis.  But we have counted the $ 2$ cases where the whole cube is the same color $ 3$ times each.  So the probability is $ \frac{3*2*4\minus{}2*2}{64}\equal{}\frac{20}{64}\equal{}\frac{5}{16}$.[/hide]


is there any quick way to add a set of quadrilateral numbers:

1+3+6+10+15+21+28+36+45+55

since its not an arithmetic sequence, im not sure...

These are triangular numbers.

And it would be $\sum_{k=1}^{10}\frac{k(k+1)}{2}=\frac{1}{2}\sum_{k=1}^{10}k^{2}+\frac{1}{2}\sum_{k=1}^{10}k$, then you can just apply the sums of squares and triangular number formulas.

What do you mean by triangular and quadrilateral numbers?

I don't think there is such thing as a quadrilateral numbers, but triangular, square, pentagonal, etc. are known as 'figurate numbers.' Search for those on MathWorld or Wikipedia or something. But a triangular number is a number that can be represented as $\sum_{k=1}^{n}k$ for some $n$. They are called triangular numbers because they can be represented as an equilateral triangle of dots.

Sorry my mistake about the quadrilateral numbers. they are really triangular numbers.  :blush:

Uh, I think it's faster using the formula than adding in your head.

Yes, it is definitely faster to use the formula.  The formula only requires a few steps, but adding all of them takes some time.


i found this problem on ahsme,

show that if a is different from c, a^x=c^q and c^y=a^z, then xy=qz

i took the ecuation a^x=c^q and multiplied by a^z and by c^y
here is what results, a^(x+z) c^y=a^z c^(q+y), then i asumed that (x+z).y=z(q+y), and if we solve, it results that xy=qz.

i dont know if what i assumed is right, and if it is, i dont exactly know why, can someone explain please.

Your assumption is not valid.

Here's one way to go about doing it:

$a^{xy}= (a^{x})^{y}= (c^{q})^{y}= (c^{y})^{q}= (a^{z})^{q}= a^{qz}$

[hide]
$a^{x}= c^{q}$

$a^{x\left(\frac{1}{x}\right)}= c^\frac{q}{x}\implies a = c^\frac{q}{x}$

Next,
$c^{y}= a^{z}$
$a = c^{\frac{y}{z}}$
Now since $a = c^\frac{q}{x}$ and $a = c^\frac{y}{z}, c^\frac{q}{x}= c^\frac{y}{z}.$

Therefore, $\frac{q}{x}= \frac{y}{z}$. Cross multiplying, we have $xy = qz$.
[/hide]

How do you make fractional powers look bigger?

[quote="∞"]How do you make fractional powers look bigger?[/quote]

I think that is as big as it gets. I am not sure if you can make it bigger, but here is [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]a website[/url].

There you can find all sorts of information about $\LaTeX$. Ask about these kinds of questions in the $\LaTeX$ forum.


Yesterday, Descartes was biking against the wind at an average of 10 mph when traveling from A to B, but with the wind at an average of 15 mph returning from B to A. How many miles per hour was his average speed for the round trip.

The wind doesn't really matter, so lets say he went 30 miles each way. It took him 3+2=5 hours round trip and 60 miles. So he went 60/5=12mph.

[hide="Or..."]Take the harmonic mean of $ 10$ and $ 15$.  $ \frac{2}{\frac{1}{10} \plus{} \frac{1}{15}} \equal{} \boxed{12}$.[/hide]

does harmonic mean always work for finding the rate of something, or is this one just coincidental?

Let $ x \equal{} AB$. We know that the total distance traveled is $ 2x$ and that the total time taken is $ \frac{x}{10} \plus{} \frac{x}{15}$. You can factor out the $ x$ and cancel and you simply get the harmonic mean of the speeds.


geometry
3D geometry
number theory proposed
number theory

Show that 97^97 cannot be a sum of consecuative cubes.I think it's easy!!

I think it is easy ,so why no one solved it??? :mad:  :(

[quote="mecrazywong"]Sorry to disappoint you, but it's 97^97, not 97^2.[/quote]

Oh :o

I'll try to modify the soloution then. Please tell me if it's OK,
and maybe submit a simpler one.

Suppose it can be done, that we can write $97^{97}=\sum_{k=m}^{n}k^{3}$

then $97^{97}=m^{3}+(m+1)^{3}+\cdots+n^{3}=(1+2+...+n)^{2}-(1+2+\cdots+(m-1))^{2}=\big((1+2+\cdots+n)+(1+2+\cdots+(m-1)\big)(m+(m+1)+\cdots+n)$

But $97$ is prime and $\big((1+2+\cdots+n)+(1+2+\cdots+(m-1)\big)> 1$ and $(m+(m+1)+\cdots+n) > 1$
so $\big((1+2+\cdots+n)+(1+2+\cdots+(m-1)\big)$  and $(m+(m+1)+\cdots+n)$ have to be a powers of $97$, say $97^{q}$ and $97^{r}   \qquad q> r\geq1$

So $\big((1+2+\cdots+n)+(1+2+\cdots+(m-1)\big)=\big(2(1+2+\cdots+(m-1))+(m+\cdots+n)\big)=\big((m-1)m+(m+\cdots+n)\big)= \big((m-1)m+97^{r}\big)=97^{q} $

so $97^{r}|(m-1)$ or $97^{r}|m$ but since $97^{r}\geq m$ the only possibility is $m=97^{r}$

this however is impossible since

$\big((m-1)m+97^{r}\big) 97^{r}=\big((97^{r}-1)97^{r}+97^{r}\big) 97^{r}=97^{3r}$

and $97$ is not a multiple of $3$.

So it was impossilbe after all.


[quote="ckck"]1. When $f(x)$ is divided by $x+1$, the remainder is $6$. When is it divided by $2x-1$, the remainder is $3$. Find the remainder when $f(x)$ is divided by $2x^{2}+x+1$.

2. When $P(x)$ is divided by $x-1$, $x-2$, and $x-3$, the remainders are 3,7 and 13 respectively. Find the remainder when $P(x)$ is divided by $(x-1)(x-2)(x-3)$.[/quote]

uh..for the first one are you sure its not divided by 2x^2+x[b]-[/b]1? because if it was that I would get -2x+4..dont know if its 2x^2+x+1.

for the second one is the answer[hide]x^2+x+1?[/hide]


[quote="arqady"]See also here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158933[/quote]

It is really unbelievable！I want to know the method.
My solution is too long. :(


There is a triangle $ABC$. A point $D$ is constructed such that $\angle CBD = 60^\circ$, and a point $E$ is constructed such that $\angle BCE = 60^\circ$, and $BD=CE$. Prove that $AB=AC$.

[color=darkred]I've converted the problem to LaTeX and moved this problem to a more appropriate forum.
However, the problem as it is is incorrect. The following was the initial problem statement:[/color]

[quote="ashrafmod"](a b c )triangle
(b d ) equhtable angle( B)
(C E ) equhtable angle ( C)
b d=c e
proved that
a b =a c[/quote]

[color=darkred]Hopefully someone can figure out what the actual problem meant. Thanks.
-nebula42[/color]

Looks like an extention of the Lehmus-Steiner Theorem.

Well, $A$ is completely independent of $E$ and $D$, so it doesn't necessarily have to be true...but I can't think of a salvage...

He meant that D and E are on the sides of the triangle.

unfortunategy i m  nt good at english  i will try  to explain
   E it  belongs to a plot straight  A B

   D     to  A C
 < a c e=< e c b     , <  a b d =  < d b c
prove    A B=AC


Diophantine equation
number theory proposed
number theory

If  [b]N= 11 + 11(11a^2+1)^1/2[/b] is an odd integer where a is a rational number. Prove that  N is a perfect square.

Suppose that $ a\equal{}\frac{p}{q}$ where $ \gcd(q,p)\equal{}1$
then we assume that $ q \neq 1$
now we get

$ 11(\frac{q\plus{}\sqrt{11p^2\plus{}q^2}}{q})\equal{}N$

obviously $ M\equal{}\sqrt{11p^2\plus{}q^2}$ must be an integer.and since $ \gcd(q,p)\equal{}1$ so we have $ \gcd(q,q\plus{}\sqrt{11p^2\plus{}q^2})\equal{}1$ and also it's obvious that $ \gcd(q,11)\equal{}1$ if not then $ M$ won't be an integer.
so if $ q\neq 1$ then $ N$ isn't an integer.it is just a rational number which is a contradiction so we must have $ q\equal{}1$ which implies that $ p\equal{}a$ and $ a$ is an integer.
now we know that $ M$ is an integer which implies that :
$ M^2\minus{}11p^2\equal{}1$ all the answers of the pell equation $ x^2\minus{}11y^2\equal{}1$ are given by:

$ x_{n\plus{}1}\equal{}10x_n\plus{}33y_n \ ; \ y_{n\plus{}1}\equal{}3x_n\plus{}10y_n$

also we have $ (M\minus{}1)(M\plus{}1)\equal{}11p^2$
but we know that $ N$ is odd which implies that $ M$ is even and so we have two cases:
1-$ M\minus{}1\equal{}11l^2,M\plus{}1\equal{}s^2,sl\equal{}p \implies s^2\minus{}11l^2\equal{}2 \implies \left(\frac{2}{11}\right)\equal{}1$ which isn't true.
2-$ M\minus{}1\equal{}s^2,M\plus{}1\equal{}11l^2,ls\equal{}p \implies 11M\plus{}11\equal{}11(11l^2\minus{}1)\plus{}11\equal{}(11l)^2\equal{}N$ so we are done!


Δίνεται κύκλος $ (O,R)$ και τρεις χορδές αυτού $ AB,CD,EF$ τέτοιες ώστε $ AB\equal{}CD\equal{}EF\equal{}R$. Αν $ K,L,M$ τα μέσα των χορδών $ BC,DE,FA$ αντίστοιχα να αποδείξετε ότι το τρίγωνο $ KLM$ είναι ισόπλευρο.

[quote="galois01"]Δίνεται κύκλος $ (O,R)$ και τρεις χορδές αυτού $ AB,CD,EF$ τέτοιες ώστε $ AB \equal{} CD \equal{} EF \equal{} R$. Αν $ K,L,M$ τα μέσα των χορδών $ BC,DE,FA$ αντίστοιχα να αποδείξετε ότι το τρίγωνο $ KLM$ είναι ισόπλευρο.[/quote]

 Ρίξε αν θέλεις μια ματιά σε προηγούμενα μηνύματα ,μήπως και το ξαναέχουμε αυτό το ωραίο και κλασικό πρόβλημα.Νομίζω ότι κάπου είχα δει λύση με μιγαδικούς από το Σιλουανό , αλλά ίσως να ήταν στο αγγλικό φόρουμ.Διαφορετικά να βάλουμε σχήμα και λύση.

 Μπάμπης


Let a,b,c,d be the positive real numbers such that $abcd=1$. Prove that
$a^{3}+b^{3}+c^{3}+d^{3}+12 \ge (a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$

Opps :blush: ! There is some mistake of my proof. Please check this one in case $a+b+c+d=4$.
$a^{3}+b^{3}+c^{3}+d^{3}+12(\sqrt[4]{abcd})^{3}\ge (\sqrt[4]{abcd})^{3}(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$

[quote="The soul of rock"]Please check this one in case $a+b+c+d=4$. [/quote]
[quote="Vasc"]This is a homogeneous inequality. Then we can set $abcd=1$ and the inequality is not true, too.[/quote]
Even though $a+b+c+d=4$ already, we can do this? I am sorry if I'm missing the obvious, but why?


i have a quastion on rolles thrm, and roots. i dont know how to approach this question.
and i have tried few different things which didnt help.
the quastion is:  on the f(x)= Ax^3 +bx^2+cx+d and a isnt 0, use rolles thrm to prove that
the function cant have more than one real roots, if b^2<3ac.???
can some one show me the way for a solution??? becuase i dont know

thanks

amit

If $f$ had (at least) two real zeros, Rolle's theorem would tell us that $f'$ has a zero between a pair of those zeros.
Now show that $f'$ has no zeros with that condition on the coefficients.

i understand the logic of what you wrote, but how you can show that between the two points? and how does it connect to the b^2<3ac???

read carefully what jmerry wrote. Practically that's the solution of the problem.

Derive the equation and then complete the square, giving you x + B/3A = (root(-3AC + B^2))/3A. If you then plug in B^2 = 3AC, you get zero under the root, proving that if B^2 was less than 3AC, you would get a non-real result.


This is pretty easy: we just have to apply AM-GM for each of the n terms, and then apply AM-GM one more time for the whole sum.


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra unsolved

Show that the ring of $ n\times n_$ matrices with coefficients in a field has no nonzero twosided ideal.

Let $ e_{ij}$ denote the matrix that is $ 1$ at the $ i$-$ j$ position and $ 0$ elsewhere.  Take some ideal $ I\neq 0$, then there is some $ M\in I$ with entry $ M_{ij} \neq 0$.  Then, for any $ k$, we have $ e_{ki} M e_{jk} \in I$, but this is just $ M_{ij} e_{kk}$.  Summing over $ k$, we find that $ M_{ij} I_n \in I$, but since $ M_{ij}$ is invertible, this is a unit, and we are done.

Any ideal that contains an invertible matrix contains the identity, but any ideal that contains a non-invertible matrix contains two linearly independent non-invertible matrices, some linear combination of which is invertible.  Sketch:  the polynomial $ \text{det}(A \minus{} \lambda B)$ has at most $ n$ roots.

[quote="t0rajir0u"]any ideal that contains a non-invertible matrix contains two linearly independent non-invertible matrices[/quote]
How do you get that without proving $ I \equal{} \mathcal{M}_n(F)$ like LydianRain just the same?

[quote="t0rajir0u"]the polynomial $ \text{det}(A \minus{} \lambda B)$ has at most $ n$ roots.[/quote]
What about $ A \equal{} \begin{pmatrix}1 & 0 \\
0 & 0\end{pmatrix},B \equal{} \begin{pmatrix}0 & 1 \\
0 & 0\end{pmatrix}$.
And in any case you would need $ F$ infinite.

Can you please teel me that what is $ I_n$, and can you please explain in more details.

$ I_n$ is the identity matrix.  I suggest multiplying out some matrices to see what is going on in the proof--the basic idea is that by multiplying on the left and right by the appropriate $ e_{ij}$ matrices, we can eliminate every entry of $ M$ except for one, and put that entry wherever we want.

How do you get that
$ M_{ij}e_{kk}\equal{} e_{ki}Me_{jk}$ and how  do you know that $ e_{ki}Me_{jk} \in I$ ? And I dont follow you in your conclusuion how have you show that $ I$ does not belong to $ M_n(F).$  Or WHat is the conclusion, how have you shown the result. Thanks.

Another approach would be the following: (though it's also computational)

It is known that using Gaussian elimination, you may write a n by n matrices $ A$ of the form $ PBQ$, where $ B \equal{} diag(1,...,1,0,..,.0)$. Say there are n ones on the diagonal of B. By multiplying $ diag(1,0,...,0)$, you get $ diag(1,0,...,0)$ is in the ideal. Now interchange the first row and the $ i$-th row and then the first column and the $ i$-th column, you can see that $ e_{ii}$ (the diagonal matrix with 1 in the i-th position and 0 otherwise) lies in the ideal (because an elementary operation corresponds to the multiplication of an appropriate matrix). Then you would have $ I_n \equal{} e_{11} \plus{} ... \plus{}e_{nn}$ in the ideal, so the ideal is the whole ring.


function
combinatorics open
combinatorics

Find as much as possible (eg lower bounds and upper bounds) on the following function:
$ f(n)$ is the smallest number such that, given $ f(n)$ points in the plane in general position, there must be some $ n$ of them whose vertices form a convex $ n$-gon.

This is a well known unsolved problem: http://en.wikipedia.org/wiki/Happy_Ending_problem


Maybe the inequality sign should be the other way.

Yes. It's easy to see see that if we make two numbers $x, y$ closer with keeping the $x^{2}+y^{2}$ as constant their product will increase. Indeed:
$(x-y)^{2}= x^{2}+y^{2}-2xy$. $(x-y)^{2}$ decrease, $x^{2}+y^{2}$ is constant, so $-2xy$ decrease and finally $xy$ increase.

From this one can easily deduce that the sum on the left is maximum for the $a=b=c$.

[quote="arqady"][quote="Guía Rojo"]$a,b,c\in\mathbb{R}^{+}$
such that $a^{2}+b^{2}+c^{2}=1$

prove that $\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\leq \frac{9}{2}$[/quote]
It is very known vasc's inequality.[/quote]
well... what is that?  :oops:

See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66936
I used the vasc's hint: $-9a^{2}b^{2}c^{2}\geq-abc(a+b+c).$


I was wondering weather there are any kind of jobs, maybe through the internet or something else that you can have, especially as a non-math major when in college?

Well, you should see what kinds of jobs the college makes available to students.  Also, it is possible to get paid to do research as a college student.  And of course there are lots of over-the-summer options, as well.  Your question was rather vague, which I believe is why it hasn't gotten much response.


conics
parabola
trigonometry
analytic geometry
geometry
integration
geometric transformation

A cylinder with height $h_0$ and radius $r$ is filled with water to a height $h$; find the maximum angular velocity $\omega$ about its vertical axis that is attainable before the water spills.

A great problem :)

The surface of the water will become a parabola due to rotation. Let's find the equation of this parabola: consider a tiny particle of water (of mass $m$) located on the surface. The forces acting on him are its weight and the centrifugal force (we will neglect surface tension). Let's introduce the cartesian system in which the $x$-axis is the horizontal direction and let $\phi$ be the angle between the $x$-axis and the tangent to the surface which contains this particle. Obviously, $\tan\phi = \frac{F_{\rm cf}}{mg} = \frac{\omega^2x}{g}$ where $x$ is $x$-coordinate of the considered particle ($x = 0$ denotes the center of the cylinder). But we also know that $\tan\phi = \frac{dy}{dx}$, and now we can integrate and obtain \[ y(x) = \frac{\omega^2x^2}{2g}+C  \] where $C$ is the constant of integration. Maximal $\omega$ corresponds to $y(r) = h_0$, and we need to compare the volumes of water in order to eliminate $C$. Consider a side view of the cylinder (no pictures, sorry :(); the surfaces of the water must have the same area: \[ \int_{-r}^r ydx = \frac{\omega^2r^3}{3g}+2rC = 2rh.  \] The final solution easily follows: \[ \omega = \frac{\sqrt{3g(h_0-2h)}}{r}.  \]
I wonder whether the solution's correct :)

Just looking at your solution, if $h = \frac{h_0}{2}$, then $\omega = 0$. That doesn't seem right. Here's what the problem is: you assumed the angular velocity to be constant, but this is not the case with fluids. As the outer layers move, they exert fluid friction on the inner layers. However, I'm not sure how to continue with this problem: is there vertical laminar flow or horizontal lamina or both?

Ok. Well, assuming $\omega$ actually is constant, I followed a similar process as Djole and ended up with $\omega = \frac{2}{r} \sqrt{g (h_0 - h)}$. I equated the volumes of the moving water and the stationary cylinder of water to find that constant $C$.

Ok, I admit that I was wrong -- that's what you get when you try to solve problems in a hurry.

I think that [b]durt[/b]'s solution is ok, seeing that for $h=h_0$ you don't need any rotation to bring the water to the edge of the cylinder.

BTW, I think that viscosity should also be neglected... Such things should've been remarked in the problem itself, though...

[quote="Djole"]Obviously, $\tan\phi = \frac{F_{\rm cf}}{mg} = \frac{\omega^2x}{g}$,[/quote]

sorry, why is this?

Because the surface of a liquid is always orthogonal to the resultant of the forces acting on it. This is one of the basic tenets of hydrostatics.


$ [(x\plus{}y)(y\plus{}z)(z\plus{}x) \plus{} xy \plus{} xz\plus{} yz]/xyz \geq (3xy \plus{} 3xz \plus{} 3yz \plus{} x^2 \plus{}y^2 \plus{} z^2)/[(x\plus{}y)(y\plus{}z)(z\plus{}x)]$

Wondering if there is a nice solution for this?

Not necessarily true. If you plug in the point (x,y,z) = (4,3,-2), you get $ \minus{}\frac{1}{2} \geq \frac{23}{14}$, which is false. Are there some restrictions on our values? Such as positive only?

Oh sorry :-\. I think it's for x,y,z > 0. I'm not sure if another condition is given.

[quote="Snipez"]$ [(x + y)(y + z)(z + x) + xy + xz + yz]/xyz \geq (3xy + 3xz + 3yz + x^2 + y^2 + z^2)/[(x + y)(y + z)(z + x)]$

Wondering if there is a nice solution for this?[/quote]

aha i got a pretty nice solution for this

so we can do this whole multiplying thing cuz $ x,y,z>0$ and so there's no sign switchin going on, what with the denominators both being guaranteed positive

$ x^3 y^2 + x^4 y^2 + x^2 y^3 + 2 x^3 y^3 + x^2 y^4 + 2 x^3 y z + 2 x^4 y z + 4 x^2 y^2 z + 6 x^3 y^2 z + 2 x y^3 z + 6 x^2 y^3 z + 2 x y^4 z + x^3 z^2 + x^4 z^2 + 4 x^2 y z^2 + 6 x^3 y z^2 + 4 x y^2 z^2 + 10 x^2 y^2 z^2 + y^3 z^2 + 6 x y^3 z^2 + y^4 z^2 + x^2 z^3 + 2 x^3 z^3 + 2 x y z^3 + 6 x^2 y z^3 + y^2 z^3 + 6 x y^2 z^3 + 2 y^3 z^3 + x^2 z^4 + 2 x y z^4 + y^2 z^4 \geq x^3 y z + 3 x^2 y^2 z + x y^3 z + 3 x^2 y z^2 + 3 x y^2 z^2 + x y z^3$

matchin like terms up:

$ x^3 y^2 + x^4 y^2 + x^2 y^3 + 2 x^3 y^3 + x^2 y^4 + x^3 y z + 2 x^4 y z + x^2 y^2 z + 6 x^3 y^2 z + x y^3 z + 6 x^2 y^3 z + 2 x y^4 z + x^3 z^2 + x^4 z^2 + x^2 y z^2 + 6 x^3 y z^2 + x y^2 z^2 + 10 x^2 y^2 z^2 + y^3 z^2 + 6 x y^3 z^2 + y^4 z^2 + x^2 z^3 + 2 x^3 z^3 + x y z^3 + 6 x^2 y z^3 + y^2 z^3 + 6 x y^2 z^3 + 2 y^3 z^3 + x^2 z^4 + 2 x y z^4 + y^2 z^4 \geq 0$

which is true cuz $ x,y,z > 0$

holy crap the latex looks soo complicated that i didn't even read that lazarus  :blush:

Can't you use AM-GM on this?

[quote="Poincare"]
Can't you use AM-GM on this?[/quote]

Hm, I wouldn't think so. In order to use it with 3 variables, I would think you would need some cubed terms in there, and you won't get those without doing some serious multiplying of expressions.


integration
calculus
trigonometry
function
limit
derivative
calculus computations

I cant seem to be able to find a way to solve it!

$\int \frac{dx}{(x^{2}+x+1)^{2}}$

The actual integral is between -oo and oo but its not relevant atm because I cant find a way to simplify it :)

Thanks!

I would like to withdraw my earlier post because it is wrong.

The fact that it's a definite integral makes things a little easier in this particular case. Make the substitution $u=x+\frac12$ to get

$\int_{-\infty}^{\infty}\frac{dx}{(x^{2}+x+1)^{2}}= \int_{-\infty}^{\infty}\frac{du}{(u^{2}+\frac34)^{2}}$

From there, there are various techniques. One would a straightforward step from a calculus class: the trigonometric substitution $u=\frac{\sqrt{3}}2\tan\theta$ This turns the integral into

$\int_{-\frac{\pi}2}^{\frac{\pi}2}\frac{\frac{\sqrt{3}}2 \sec^{2}\theta\,d\theta}{\left(\frac34\sec^{2}\theta\right)^{2}}= \frac{8}{3\sqrt{3}}\int_{-\frac{\pi}2}^{\frac{\pi}2}\cos^{2}\theta\,d\theta=\frac{4\pi}{3\sqrt{3}}$

Guess what?? Residues again! :D

 We wilkl calculate the following more general problem \[\int_{-\infty}^{\infty}\frac{dx}{(x^{2}+x+1)^{n}},\] where $n\geq 1$.

 Solution.

  Let us consider the function \[f(z)=\frac{1}{z^{2}+z+1}.\] It is quite easy to see that $f$ has two poles of order $n$ namely \[z_{1}=\frac{-1+i\sqrt{3}}{2}, z_{2}=\frac{-1-i\sqrt{3}}{2}.\] Now, by Residue theorem we get \[Res(f, z_{1})=\frac{1}{(n-1)!}\lim_{z\to z_{1}}\left(\frac{1}{(z-z_{2})^{n}}\right)^{(n-1)}=\frac{1}{(n-1)!}\cdot\frac{(-i)(2n-1)!}{(\sqrt{3})^{2n-1}n!}.\] So, our integral becomes \[\int_{-\infty}^{\infty}\frac{dx}{(x^{2}+x+1)^{n}}=\frac{2\pi (2n-1)!\sqrt{3}}{n!(n-1)! 3^{n}}.\]

The solution is OK, expect this part:
[quote="Cezar Lupu"]
Now, by Residue theorem we get \[Res(f, z_{1})=\frac{1}{(n-1)!}\lim_{z\to z_{1}}\left(\frac{1}{(z-z_{2})^{n}}\right)^{(n-1)}=\frac{1}{(n-1)!}\cdot\frac{(-i)(2n-1)!}{(\sqrt{3})^{2n-1}n!}.\] So, our integral becomes \[\int_{-\infty}^{\infty}\frac{dx}{(x^{2}+x+1)^{n}}=\frac{2\pi (2n-1)!\sqrt{3}}{n!(n-1)! 3^{n}},\] [/quote]
where you miscalculated the derivative: it's $\frac{(2n-2)!}{(n-1)!}$ instead of $\frac{(2n-1)!}{n!}$.

Note that Kent's approach also works here: we have that $\int_{-\infty}^\infty \frac{dx}{(x^{2}+x+1)^{n}}= \frac{\sqrt 3}2 \cdot \left( \frac43 \right)^{n}\cdot \int_{-\frac{\pi}2}^\frac{\pi}2 \cos^{2n-2}\theta \, d \theta$.
The last integral is well-known: $\frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}\cdot \ldots \cdot \frac34 \cdot \frac12 \cdot \pi$ (it's related to the Wallis formula).


Na vrethoun oi times tou k gia tis opoies to genikeumeno oloklhrwma suglinei
\[ \int_{0}^{ \plus{} \infty}(\frac {2x}{x^2 \plus{} 1} \minus{} \frac {k}{2x \plus{} 1})dx
\]

[quote="r_boris"]Νομίζω ότι με το κριτήριο σύγκρισης προκύπτει κ>3[/quote]

Ίσως να κάνω κάποιο λάθος, αλλά νομίζω πως μόνο για $ k\equal{}4$ συγκλίνει.

Ναί έχεις δίκιο. ξέχασα τον λογάριθμο


Let $ a,b,c$ are positive number. Prove that
$ ab\plus{}bc\plus{}ca \geq \frac{2a^2(b\plus{}c\minus{}a)}{b\plus{}c}\plus{}\frac{2b^2(a\plus{}c\minus{}b)}{a\plus{}c}\plus{}\frac{2c^2(a\plus{}b\minus{}c)}{a\plus{}b}$

[quote="Gibbenergy"]Let $ a,b,c$ are positive number. Prove that
$ ab \plus{} bc \plus{} ca \geq \frac {2a^2(b \plus{} c \minus{} a)}{b \plus{} c} \plus{} \frac {2b^2(a \plus{} c \minus{} b)}{a \plus{} c} \plus{} \frac {2c^2(a \plus{} b \minus{} c)}{a \plus{} b}$[/quote]
By AM-GM Inequality, we have
\[ \sum \frac{2a^2(b\plus{}c\minus{}a)}{b\plus{}c} \le \sum\frac{a(a\plus{}b\plus{}c\minus{}a)^2}{2(b\plus{}c)} \equal{}\sum \frac{a(b\plus{}c)}{2}\equal{}\sum ab\]
:)

[quote="Gibbenergy"][quote="can_hang2007"][quote="Gibbenergy"]Let $ a,b,c$ are positive number. Prove that
$ ab \plus{} bc \plus{} ca \geq \frac {2a^2(b \plus{} c \minus{} a)}{b \plus{} c} \plus{} \frac {2b^2(a \plus{} c \minus{} b)}{a \plus{} c} \plus{} \frac {2c^2(a \plus{} b \minus{} c)}{a \plus{} b}$[/quote]
By AM-GM Inequality, we have
\[ \sum \frac {2a^2(b \plus{} c \minus{} a)}{b \plus{} c} \le \sum\frac {a(a \plus{} b \plus{} c \minus{} a)^2}{2(b \plus{} c)} \equal{} \sum \frac {a(b \plus{} c)}{2} \equal{} \sum ab
\]
:)[/quote]

Nice. BTW, I proved your inequality at the box http://www.mathlinks.ro/viewtopic.php?t=246476
Can you prove
$ ab \plus{} bc \plus{} ca \geq \frac {4a^4(b \plus{} c \minus{} a)}{(b \plus{} c)^3} \plus{} \frac {4b^4(a \plus{} c \minus{} b)}{(a \plus{} c)^3} \plus{} \frac {4c^4(a \plus{} b \minus{} c)}{(a \plus{} b)^3}$
I have proved this one but not satisfied with it because my inequality does not achieve equality. But I am curious to see the solution  :)[/quote]
We can prove that even the stronger inequality
\[ \frac {27}{128} (ab \plus{} bc \plus{} ca) \ge \sum \frac {a^4(b \plus{} c \minus{} a)}{(b \plus{} c)^3}
\]
In order to prove this inequality, we just need to prove:
\[ \frac {27}{256} a(b \plus{} c) \ge \frac {a^4(b \plus{} c \minus{} a)}{(b \plus{} c)^3}
\]

\[ a^3(b \plus{} c \minus{} a) \le \frac{27}{256} (b \plus{} c)^4
\]
If $ b \plus{} c \le a$, it is trivial. If $ b \plus{} c \ge a$, applying the AM-GM Inequality, we have
\[ a^3(b \plus{} c \minus{} a) \equal{} 27\cdot \frac {a}{3} \cdot \frac {a}{3} \cdot \frac {a}{3} \cdot (b \plus{} c \minus{} a) \le 27\left( \frac {\frac {a}{3} \plus{} \frac {a}{3} \plus{} \frac {a}{3} \plus{} b \plus{} c \minus{} a}{4}\right)^4 \equal{} \frac {27}{256}(b \plus{} c)^4
\]
Now, applying the above inequality for the cyclic terms and adding them up, we will get the result.
:)
I have thought of the minimum constant $ k$ such that the following inequality holds
\[ k(ab \plus{} bc \plus{} ca) \ge \sum \frac {a^4(b \plus{} c \minus{} a)}{(b \plus{} c)^3}
\]
But my result is not nice, $ k \sim 0.17$ and we can't find the exactly value of $ k$, so I think the inequality you post is good enough. :)


calculus
integration
function
calculus computations

Find all continuous positive functions $ f(x)$, for $ 0\leq x\leq 1$, such that

$ \int_0^1 f(x)\ dx \equal{} 1$

$ \int_0^1 xf(x)\ dx \equal{} \alpha$

$ \int_0^1 x^2f(x)\ dx \equal{} \alpha ^2$

where $ \alpha$ is  given real number.

By Cauchy-Schwarz:

$ (\int_0^1 xf(x) dx)^2\leq (\int_0^1 f(x) dx) (\int_0^1 x^2 f(x) dx)$, but both sides are equal to $ \alpha^2$.  Since we have equality in Cauchy-Schwarz only when the functions in question are scalar multiples of each other, we find that $ x^2 f(x)\equal{}c f(x)$, so $ f(x)$ is identically zero.

I like this approach because you don't need to assume that $ f$ is positive, but there's an even simpler way to solve this problem...

"Identically zero" is impossible, since the integral is $ 1$. No such functions exist.

I like the interpretation of $ \int_0^1 x^2f(x)\,dx\minus{}\left(\int_0^1 xf(x)\,dx\right)^2$ as the variance of the distribution with density $ f$. The variance is zero, so it's a point mass $ \delta(x\minus{}\alpha)$. That's not a continuous function, and that distribution doesn't have a classical density.
It has appeared before [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=118948]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=28825]here[/url] (I searched "Cauchy" and "variance").
As shown in the first link, nonnegativity is needed for the result.

[quote="jmerry"]"Identically zero" is impossible, since the integral is $ 1$. No such functions exist.

I like the interpretation of $ \int_0^1 x^2f(x)\,dx \minus{} \left(\int_0^1 xf(x)\,dx\right)^2$ as the variance of the distribution with density $ f$. [/quote]

I thought of varinance, too, or $ \int_0^1 (x \minus{} \alpha)^2 f(x)\ dx \equal{} 0$ can be imagined to us.

I remember the problem in Japanese university entrance exam in '70 before.


A 2 inch by 2 inch square has its area increased by 21 square
inches producing a new square. How many inches long is a side
of the new square?

A 2*2 square has area 4.
4+21=25.
thus the side is sqrt 25 or 5.


Suppose the series,

1/nC0 + 1/nC1 + 1/nC2 +.....+ 1/nCn = S

show that,

0/nC0 + 1/nC1 + 2/nC2 + ....+ n/nCn = nS/2

This really has me stumped, and is there some elementary way to sum the first series?

This is actually nicer than I first realized -- takes a fair amount of cleverness, though.  This is just an outline.

Start with $(n+2) \cdot k!(n-k)! = (k+1)!(n-k)!+k!(n-k+1)!$.  Then sum both sides as $k$ varies from 0 to $n$; this gives you $2n+2$ terms on the right, but you can group them into $n+1$ pairs of identical terms.  Divide both sides by $2(n+1)!$ and you get a nice relationship between $S(n) = \sum_{k = 0}^{n}\binom{n}{k}^{-1}$ and $S(n+1)$.

Now, begin from $T(n) = \sum_{k = 0}^{n}\binom{n}{k}^{-1}k$.  Then $S(n)+T(n) = \sum_{k = 0}^{n}\binom{n}{k}^{-1}(k+1)$.  Dividing through by $n+1$ allows you to express this equation in terms of $S(n+1)$.  Substitute into the previous expression and solve for $T(n)$ in terms of $S(n)$ to get your desired result.

:Edit: Actually, that was more complicated than necessary; I don't think you need to invoke $S(n+1)$ at all.  Start with the first step, and see where it takes you.

wow that's brilliant, that first identity really puts it in the bag, thanks!

Anybody got any ideas how the original reciprocal series may be summed?

[quote="JBL"]I expressed $T(n)$ in terms of $S(n+1)$.  Turning this around gives us an expression for $S(n+1)$ in terms of $S(n)$, and so a recursive formula.[/quote]

ok so far I've got the recurrence relation
[(n+2)S(n)]/2(n+1) = S(n+1) - 1

and a few equations involving the sum of S(n) s.
I'll keep at it... Just had a terrible itch to sum this...

okay I've summed S(n) and T(n), but it's more like expressing them in terms of a more easily evaluated series,
I used as a starting point,
$S_{n}-\frac{n+1}{2n}S_{n-1}= 1$
Dividing the whole thing by
$v_{n}= \prod_{r = 1}^{n}\frac{r+1}{2r}$
gives,
$\frac{S_{n}}{v_{n}}-\frac{S_{n-1}}{v_{n-1}}= \frac{1}{v_{n}}$

summing this 2 to n gives,
$\frac{S_{n}}{v_{n}}-2 = \sum_{r = 1}^{n}\frac{1}{v_{n}}-1$

giving,
$S_{n}= \frac{n+1}{2^{n}}[1+\sum_{r = 1}^{n}\frac{2^{r}}{r+1}]$

$T_{n}= \frac{n}{2}S_{n}$ gives $T_{n}$

hope there are no errors(in the math or the TeX)

Looks good to me -- nice work.  (You could also re-write $1+\sum_{r = 1}^{n}\frac{2^{r}}{r+1}= \sum_{r=0}^{n}\frac{2^{r}}{r+1}$.)

Thanks, I noticed that after I posted..Don't reckon this can be simplified further.

I don't think so.  That sum can be expressed as a hypergeometric or beta function, but that's not really "simpler" in any meaningful sense.

Exactly... :lol:
You know I tried my hand at this again and you can actually get everything without that first miraculous identity,
we simply need to consider,
$\frac{S_{n}+T_{n}}{n+1}$

Express that in terms of $S_{n+1}$ as you've done and notice how,
$\binom{n+1}{r}^{-1}= \frac{n-r+1}{n+1}\binom{n}{r}^{-1}$

and hence that,
$S_{n+1}= \frac{(n+1)S_{n}-T_{n}}{n+1}+1$

but I can see that your first identity can come in handy on more than one occasion.


Let x,y,z be real number satisfying x,y,z>-1/2 and
$ x^2\plus{}y^2\plus{}z^2\equal{}3$
Find the minimum of $ P\equal{}x^3\plus{}y^3\plus{}z^3$

if $ x,y,z \geq 0$the minimum is $ 3$.Use Holder,
$ (x^3 \plus{} y^3 \plus{} z^3)^2(1 \plus{} 1 \plus{} 1)\geq (x^2 \plus{} y^2 \plus{} z^2)^3 \equal{} 27$
then:$ x^3 \plus{} y^3 \plus{} z^3 \geq 3$

[quote="8826"]Let x,y,z be real number satisfying x,y,z>-1/2 and
$ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$
Find the minimum of $ P \equal{} x^3 \plus{} y^3 \plus{} z^3$[/quote]
For $ x,y,z>0$ we can use power ineq....
$ \frac {\sqrt[3]{x^3\plus{}y^3\plus{}z^3}}{\sqrt[3]{3}} \geq \frac{\sqrt{x^2\plus{}y^2\plus{}z^2}}{\sqrt{3}}$
So the minimum is $ 3$

[quote="peine"]the minimum is $ 3$.Use Holder,
$ (x^3 \plus{} y^3 \plus{} z^3)^2(1 \plus{} 1 \plus{} 1)\geq (x^2 \plus{} y^2 \plus{} z^2)^3 \equal{} 27$
then:$ x^3 \plus{} y^3 \plus{} z^3 \geq 3$[/quote]
But why Holder works here? :wink:

$ x,y,z$ are reals greater than $ \minus{}1/2$.

$ x,y,z$ are positive réel (I think  :maybe: )

[quote="arqady"][quote="peine"]$ (x^3 \plus{} y^3 \plus{} z^3)(x^3 \plus{} y^3 \plus{} z^3)(1 \plus{} 1 \plus{} 1)\geq (\sqrt [3]{x^6} \plus{} \sqrt [3]{y^6} \plus{} \sqrt [3]{z^6})^3 \equal{} (x^2 \plus{} y^2 \plus{} z^2)^3 \equal{} 27$[/quote]
But why it works here? For example try $ x \equal{} y \equal{} 1$ and $ z \equal{} \minus{} 1.$ :wink:[/quote]
It can't be $ z \equal{} \minus{} 1$ cause $ x,y,z > \minus{} \frac {1}{2}$  :wink:

:(  Sorry arqady!!!, I understand now wath do you mean, my solution is true only if $ x,y,z$ are positive réel numbers, thanks a lot for correcting my mistake.
for [b]dimitrisX[/b] arqady only wanted to show me my fault and he has committed this error. thank you for your attention  :)

Let n real number satisfying $ x_i > \minus{} \frac {i}{2} (i \equal{} 1,..n)$
and $ \sum_{i \equal{} 1}^{n}ix_{i}^2 \equal{} (\frac{n(n\plus{}1)}{2})^2$
Find the minimum of $ P \equal{} \sum_{i \equal{} 1}^{n}x_{i}^3$


geometry
angle bisector
geometry proposed

An interesting fact:

[color=blue]Problem[/color]
Let $ABC$ be a triangle with incircle $\omega$ and external circle $\omega_A$ (The one tangent to $AB$ and $AC$ from inside and $BC$ from outside of the triangle)
Let $\omega$ be tangent to $AB$ and $BC$ at $R,S$ and $\omega_A$ be tanget to $AB$ and $BC$ at $T,U$.

Prove that $TU$ and $RS$ interecept each other on the angle bisector of $\angle A$.

Well a [b]very[/b] different way of saying this problem is this :

[color=blue]Problem[/color]
Let $w_1,w_2$ be two circles .
Find the point $A$ such that its polars wrt $w_1,w_2$ be a same line.

baba Arman joon ye modati naboodi dashtim hal mikardim.
dobare oomadi ke?Y? :rotfl:


Okay in this competition the signups begin now, and I will wait until there are 20 people that have signed up. In the competition I will randomly divide the 20 people into 2 equal teams. Then the two teams will have a competition. The team that loses will vote somebody off of their team. This person is eliminated. This process will continue until there are only twelve people left. Then they will all be against each other. They will still do challenges, but the winner is the only person that gets the immunity idol, which means that nobody else can vote for that person. At the end of each round the person with the immunity idol must return it, and then they play for it again. This will happen until there are 3 people left. Once they are down to the top three contestants, they will do a challenge for the immunity idol, and the winner autimatically moves on to the last round. The winner also picks the person that he wants to compete against in the last round. In the last round the people who made it to the top twelve other than the top two will vote for the person who they want to win the competition. Then they will also do a FTW competition. (If you don't know what that is please ask.) The winner gets a free vote added to there current total of votes. Most of the challenges will take place on this post so please keep checking out this post to see what is going on and to see the challenges. Good Luck!!! Signups start now

um, yeah please don't spam and revive old threads. But i am thankful you revived this one. EVERYONE JOIN!


inequalities
function
calculus
derivative
inequalities proposed

We can easily prove that:

$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}\ge\frac 23\left(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)$

(see [url]http://www.mathlinks.ro/viewtopic.php?t=172337[/url])

we have:

$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \equal{} \frac {11}{7}$

So that implies:

$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\le \frac {33}{14}$

But your inequality says 2. So, who can explain the situation to me?

I dont understand one thing.
Is it that we must not simultaneously simplify inequalities.
For eg. If I simplify the first inequality simultaneously,
I get
$ \dfrac{\sum a^2}{\sum ab} \ge 1$ and for the RHS by Nessbitts I have $ (\sum \dfrac{a}{b\plus{}c} ) (\dfrac{2}{3}) \ge 1$
So  how does the inequality hold ?       Please someone explain this.

These are just special cases of the generalized n-1 equal value principle (I assume that's what it's called): Assume WLOG that $ a\plus{}b\plus{}c \equal{} 1$, then we're trying to maximize/minimize the sum $ \frac{a}{1\minus{}a} \plus{} \frac{b}{1\minus{}b} \plus{} \frac{c}{1\minus{}c}$ (in part (a)), or the sum $ \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c}$ (in part (b)) given that $ a\plus{}b\plus{}c$ is $ 1$ and $ a^2 \plus{} b^2 \plus{} c^2$ is $ \frac{11}{25}$, and the functions $ f(x) \equal{} \frac{x}{1\minus{}x}$ (for part (a)) and $ f(x) \equal{} \frac{1}{x}$ (for part (b)) have third derivatives with constant sign on $ (0, 1)$, so the maximum/minimum occurs when two of $ a, b, c$ are equal, that is, when $ (a, b, c) \equal{} (\frac{3}{5}, \frac{1}{5}, \frac{1}{5}), (\frac{1}{15}, \frac{7}{15}, \frac{7}{15})$ or a permutation of those.

[quote="Vasc"]Let $ a,b,c$ be positive real numbers such that $ 7(a^2 \plus{} b^2 \plus{} c^2) \equal{} 11(ab \plus{} bc \plus{} ca)$. Prove that

$ (a)\ \ \ \ \ \ \ \ \frac {51}{28}\le \frac a{b \plus{} c} \plus{} \frac b{c \plus{} a} \plus{} \frac c{a \plus{} b} \le 2$;

[/quote]
Let $ a\plus{}b\plus{}c\equal{}1.$ Hence, $ ab\plus{}ac\plus{}bc\equal{}\frac{7}{25}.$
Thus, $ a,$ $ b$ and $ c$ are roots of the following equation:
$ x^3\minus{}x^2\plus{}\frac{7}{25}x\minus{}f(x)\equal{}0,$ where $ f(x)\equal{}abc.$
Id est, $ f(x)\equal{}x^3\minus{}x^2\plus{}\frac{7}{25}x$ and $ x_{max}\equal{}\frac{1}{5},$ $ x_{min}\equal{}\frac{7}{15},$ $ f\left(\frac{1}{5}\right)\equal{}\frac{3}{125}$ and $ f\left(\frac{7}{15}\right)\equal{}\frac{49}{3375}.$

For three roots of the equation we obtain: a range of $ f(x)$ is $ [\frac{49}{3375},\frac{3}{125}].$
Let $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2$ and $ abc\equal{}w^3.$
Hence, $ \sum_{cyc}\frac{a}{b\plus{}c}\equal{}\frac{27u^3\minus{}18uv^2\plus{}3w^3}{9uv^2\minus{}w^3}\equal{}\frac{11\plus{}75w^3}{7\minus{}25w^3},$ where $ \frac{49}{3375}\leq w^3\leq\frac{3}{125}.$
Hence, $ \frac {51}{28}\equal{}\frac{11\plus{}75\cdot\frac{49}{3375}}{7\minus{}25\cdot\frac{49}{3375}}\leq\frac{11\plus{}75w^3}{7\minus{}25w^3}\leq\frac{11\plus{}75\cdot\frac{3}{125}}{7\minus{}25\cdot\frac{3}{125}}\equal{}2.$

You are right, arqady.  :lol:
For $ a\plus{}b\plus{}c\equal{}5$, the calculation becomes simpler.

[quote="Vasc"]Let $ a,b,c$ be positive real numbers such that $ 7(a^2 \plus{} b^2 \plus{} c^2) \equal{} 11(ab \plus{} bc \plus{} ca)$. Prove that

$ (a)\ \ \ \ \ \ \ \ \frac {51}{28}\le \frac a{b \plus{} c} \plus{} \frac b{c \plus{} a} \plus{} \frac c{a \plus{} b} \le 2$;

$ (b)\ \ \ \ \ \ \ \ \frac {35}{3}\le (a \plus{} b \plus{} c)(\frac 1{a} \plus{} \frac 1{b} \plus{} \frac 1{c}) \le \frac {135}{7}$.[/quote]
Of course, EV method can be used here. But if we want a more elementary one, we can proceed as follow:
Assuming that $ a \plus{} b \equal{} 1$ and putting $ x \equal{} ab$, then $ x \equal{} \frac {7c^2 \minus{} 11c \plus{} 7}{25}$. Since $ x \le \frac {1}{4}$, we have $ c \in [\frac {1}{14},\frac {3}{2}]$. Now, applying this to our problems, we have
(a)
\[ \sum \frac {a}{b \plus{} c} \equal{} (a \plus{} b \plus{} c)\left( \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a}\right) \minus{} 3 \equal{} (1 \plus{} c)\left( \frac {a \plus{} b \plus{} 2c}{c^2 \plus{} c(a \plus{} b) \plus{} ab} \plus{} 1\right) \minus{} 3 \equal{} (c \plus{} 1) \left( \frac {1 \plus{} 2c}{c^2 \plus{} c \plus{} \frac {7c^2 \minus{} 11c \plus{} 7}{25}} \plus{} 1\right) \minus{} 3 \equal{} \frac {32(c \plus{} 1)^3}{32c^2 \plus{} 14c \plus{} 7} \minus{} 3 \equal{} f(c)
\]
We see that
\[ f^{\prime} (c) \equal{} \frac {32(c \plus{} 1)^2(4c \minus{} 1)(8c \minus{} 7)}{(32c^2 \plus{} 14c \plus{} 7)^2}, \quad f^{\prime}(c) \equal{} 0 \Leftrightarrow c \equal{} \frac {1}{4} , c \equal{} \frac {7}{8}
\]
And by making variation board, we obtain
\[ \max f(c) \equal{} \max\left\{ f\left( \frac {1}{4}\right),f\left( \frac {3}{2}\right) \right\} \equal{} 2
\]
and
\[ \min f(c) \equal{} \min \left\{ f\left( \frac {1}{14}\right), f\left( \frac {7}{8}\right) \right\} \equal{} \frac {51}{28}.
\]
(b)
\[ (a \plus{} b \plus{} c)\left( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \right) \equal{} (1 \plus{} c)\left( \frac {25}{7c^2 \minus{} 11c \plus{} 7} \plus{} \frac {1}{c} \right) \equal{} \frac {7(c \plus{} 1)^3}{c(7c^2 \minus{} 11c \plus{} 7)} \equal{} g(c)
\]
We see that
\[ g^{\prime} (c) \equal{} \frac {7(c \plus{} 1)^2(4c \minus{} 1)(7 \minus{} 8c)}{c^2(7c^2 \minus{} 11c \plus{} 7)^2}
\]
and by making variation board, we obtain
\[ \max f(c) \equal{} \max\left\{ f\left( \frac {1}{14}\right), f\left( \frac {7}{8}\right) \right\} \equal{} \frac {135}{7}
\]

\[ \min f(c) \equal{} \min \left\{ f\left( \frac {1}{4}\right),f\left( \frac {3}{2}\right) \right\} \equal{} \frac {35}{3}.
\]
The proof is completed. :)

Very nice solution, Can_hang. :lol:
But if "we want a more elementary solution", then it is not necessary to use derivatives in order to show that
$ \frac {51}{28} \le f(c) \le 2$
and
$ \frac {35}{3} \le g(c) \le \frac {135}{7}$  :wink:


quadratics
algebra
polynomial
algebra unsolved

factorize into factors with real coefficients

$ {2m^5 \plus{} 4m^4 \plus{} 4m^3 \plus{} 8m^2 \plus{} 2m \plus{}3}$

Sketching the graph, there seems to be only one real root: just larger than $ \minus{}2$.

There seem to be 4 possibilities:

$ (2m^{3}\plus{}am^{2}\plus{}bm\plus{}1)(m^{2}\plus{}cm\plus{}3)$

$ (2m^{3} \plus{} am^{2} \plus{} bm \plus{}3)(m^{2} \plus{}cm \plus{} 1)$

$ (m^{3} \plus{} am^{2} \plus{} bm \plus{} 1)(2m^{2} \plus{} cm \plus{} 3)$

$ (m^{3} \plus{} am^{2} \plus{} bm \plus{} 3)(2m^{2} \plus{} cm \plus{}1)$

Expnding each of these in turn and comparing coefficients with those of $ 2m^{5} \plus{} 4m^{4} \plus{} 4m^{3} \plus{} 8m^{2} \plus{} 2m \plus{}3$ will give 4 systems of simultaneous equations and one of these should work.

I hope this is of some help.

[quote="Jaswinder"]i told u to factorize in real coefficients

do it in that[/quote]

Hello Jaswinder!

Maybe you could try to be a little bit more polite when requesting some help (some kind words as "please", "thanks", ...).
And transforming Dr Sonnhard's proposition in a factorization with real coeff (just multiplying 2 factors in order to get quadratics) is so simple that you might surely be able to do it.

IMHO

sorry to u

sorry to doctor sonnhard

i am really sorry

i meant cud u please tell me the real roots of eqn in square roots or something without giving the answers in decimals.......

thankx

[quote="Jaswinder"]i meant cud u please tell me the real roots of eqn in square roots or something without giving the answers in decimals.......[/quote]
For polynomials of degree $ 5$ and greater it is not guaranteed that such an expression exists.

An approach:
$ 2m^4(m\plus{}2)\plus{}4m^2(m\plus{}2)\plus{}2(m\plus{}2)\minus{}1$
$ \Longleftrightarrow 2(m\plus{}2)(m^4\plus{}2m^2\plus{}1)\minus{}1$
$ \Longleftrightarrow \left ( \sqrt{2(m\plus{}2)} (m^2\plus{}1) \right)^2\minus{}1\equal{}\left(\sqrt{2(m\plus{}2)}(m^2\plus{}1)\minus{}1\right)\left(\sqrt{2(m\plus{}2)}(m^2\plus{}1)\plus{}1\right)$.

That doesn't qualify as a polynomial factorization.

Dear Jaswinder,

Decimal factorization (using Dr Sonnhard's work) is :

$ 2(m \minus{} 1.979325218)$ $ (m^2 \plus{} 0.12067628262m \plus{} 0.523490809967934)$ $ (m^2 \minus{} 0.10000150058m \plus{} 1.44765488078933)$

At first glance (using continued fractions), none of these coefficients may be written as $ p \plus{} q\sqrt r$ with $ p,q,r\in\mathbb Q$.

Could you give us some information about the origin of this problem (what exam, what class, any context ?) in order we can try to find some other explicit expression not using decimal expansions ?

Thanks in advance for any info.


Whats the 36th digit of $ \frac{1}{1996}$.  How do you find the period it repeats?

[hide]There is no formula for the period, but there are several useful theorems in number theory that do place some restrictions on the possible values for the period of this fraction.

First of all, note that, in general, the period of the repeating decimal for $ a/b$ where $ a$ and $ b$ are integers can never be greater than $ b-1$.  Why? If you start dividing $ b$ into $ a$, the resulting decimal expression will start repeating as soon as a remainder repeats.  However, the possible remainders in the computation of a repeating decimal range from $ 1$ to $ b-1$, so you have to hit a repeat value of the remainder by the time you've divided $ b-1$ times.

Well, that's nice but not all that helpful in the case of $ 1/1996$ since it just tells us that the period could be as high as 1995.

However, if you study more number theory you'll find some other potentially  useful properties.

Potentially Useful Property #1:  If $ b$ is prime, then the period of the repeating decimal expression for $ 1/b$ will be an integer factor of $ b-1$.

Potentially Useful Property #2: If $ b$ can be expressed as  $ 2^{m}*5^{n}*p$ where $ m$ and $ n$ are nonnegative integers and $ p$ is a prime number, then the period of $ 1/b$ will be the same as the period of $ 1/p$.

Combining Potentially Useful Properties 1 and 2 and the facts that $ 1996 = 2^{2}*499$ and $ 499$ is prime, we now know that the period of $ 1/1996$ must be an integer factor of $ 498.$

Since $ 498 = 2*3*83$, its integer factors are $ 1,2,3,6,83,166,249,and 498,$ so those eight values are the only possibilities for the periods of $ 1/1996$.

Hopefully, you'll get lucky.  But the worst case is that the period of the repeating decimal could be as large as 498.

[b]Bad News:  [/b] In fact, I tried dividing it out myself and I can quickly see that the period is greater than 6, which means that in the best case, the period will be at least 83.  This is pretty discouraging.

[b]Good news:  [/b]However, when I starting doing the long division of $ 1/1996$, I pretty quickly  noticed a very nice pattern that should allow you to predict the 36th decimal place even though the period of the repetition is greater than 36.

Try it--look for patterns in the partial quotient (and also patterns in the remainders) and you should be able to find a pattern that will allow you to confidently predict the 36th decimal place even without carrying the long division far enough to see the decimal repeat.

So, the moral of the story is that sometimes you can predict future decimal places without actually going through a full cycle of the repetition period.
[/hide]

[quote="sophia"][hide]There is no formula for the period, but there are several useful theorems in number theory that do place some restrictions on the possible values for the period of this fraction.

First of all, note that, in general, the period of the repeating decimal for $ a/b$ where $ a$ and $ b$ are integers can never be greater than $ b-1$.  Why? If you start dividing $ b$ into $ a$, the resulting decimal expression will start repeating as soon as a remainder repeats.  However, the possible remainders in the computation of a repeating decimal range from $ 1$ to $ b-1$, so you have to hit a repeat value of the remainder by the time you've divided $ b-1$ times.

Well, that's nice but not all that helpful in the case of $ 1/1996$ since it just tells us that the period could be as high as 1995.

However, if you study more number theory you'll find some other potentially  useful properties.

Potentially Useful Property #1:  If $ b$ is prime, then the period of the repeating decimal expression for $ 1/b$ will be an integer factor of $ b-1$.

Potentially Useful Property #2: If $ b$ can be expressed as  $ 2^{m}*5^{n}*p$ where $ m$ and $ n$ are nonnegative integers and $ p$ is a prime number, then the period of $ 1/b$ will be the same as the period of $ 1/p$.

Combining Potentially Useful Properties 1 and 2 and the facts that $ 1996 = 2^{2}*499$ and $ 499$ is prime, we now know that the period of $ 1/1996$ must be an integer factor of $ 498.$

Since $ 498 = 2*3*83$, its integer factors are $ 1,2,3,6,83,163,249,and 498,$ so those eight values are the only possibilities for the periods of $ 1/1996$.

Hopefully, you'll get lucky.  But the worst case is that the period of the repeating decimal could be as large as 498.

[b]Bad News:  [/b] In fact, I tried dividing it out myself and I can quickly see that the period is greater than 6, which means that in the best case, the period will be at least 83.  This is pretty discouraging.

[b]Good news:  [/b]However, when I starting doing the long division of $ 1/1996$, I pretty quickly  noticed a very nice pattern that should allow you to predict the 36th decimal place even though the period of the repetition is greater than 36.

Try it--look for patterns in the partial quotient (and also patterns in the remainders) and you should be able to find a pattern that will allow you to confidently predict the 36th decimal place even without carrying the long division far enough to see the decimal repeat.

So, the moral of the story is that sometimes you can predict future decimal places without actually going through a full cycle of the repetition period.
[/hide][/quote]

Thank you very much!


A function $h : \mathbb{R} \rightarrow \mathbb{R}$ is differentiable and satisfies $h(ax) = bh(x)$ for all $x$, where $a$ and $b$ are given positive numbers and $0 \not = |a| \not = 1$. Suppose that $h'(0) \not = 0$ and the function $h'$ is continuous at $x = 0$. Prove that $a = b$ and that there is a real number $c$ such that $h(x) = cx$ for all $x$.

$h(ax)=bh(x) \Longrightarrow ah'(ax)=bh'(x) \Longrightarrow ah'(0)=bh'(0)\Longrightarrow a=b$
$h(0)=a(h(0) \Longrightarrow (1-a)h(0)=0 \Longrightarrow h(0)=0$
Suppose that $|a| > 1$  : $h(x)=ah(\dfrac xa)=a^2h(\dfrac x{a^2})=...= a^n h(\dfrac x{a^n})=...=$
$h(x)= a^n h(\dfrac x{a^n})=x h'(u_n)$ with $|u_n| < \dfrac x{a^n}$, so $h(x) = x \lim\limits_{n\rightarrow \infty}h'(u_n) = xh'(0)$, because of $h'(x)$ is continuous at $x=0$

Suppose that $|a| < 1$  : $h(x)=\dfrac 1a ah(\dfrac x{1/a})$ , with $\dfrac 1a$ > 1 , so $h(x) = xh'(0)$
 :cool:


function
trigonometry
logarithms
LaTeX
calculus
calculus computations

Hi all, I am working on a problem, got it down to the following:

Prove the following infinite sum converges for arbitrary x>0 and y>0

| cos(y*log(n))/n^x  -  cos(y*log(n+1))/(n+1)^x | as n runs from 1 to infinity.

Obviously a lot of these terms are going to telescope out, the problem lies in dealing with the terms that don't  :!: Proving this converges has been driving me nuts all day.

Thanks in advance!

You are on the wrong track. Just try to estimate each difference by $ \frac{C}{n^{x+1}}$ using the mean value theorem for the function $ t\to\frac{\cos (y\log t)}{t^x}$.

By the way, why don't you try to use $ \text{\LaTeX}$? :) Your formula is completely readable, of course, but $ \text{\LaTeX}$ed, it would looks much nicer and basic $ \text{\LaTeX}$ is not that hard to learn.


Let a,b,c are real numbers such that $ \frac {1}{2}\leq a,b,c\leq 1$.
Find the maximum and minximum value of:
${ A = (1 + a + b + c})(\frac {1}{1 + a} + \frac {1}{1 + b} + \frac {1}{1 + c})$ :D

Am I missing something or does $ A$ just decrease as $ a,b,c$ increase, so the minimum and maximum occur at $ a\equal{}b\equal{}c\equal{}1$ and $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$ respectively?  :huh:

oh, i am sorry. It is very easy. So I have edited it.  :D Now, who can prove this?

The minimum:
\[ (1\plus{}a\plus{}b\plus{}c)\left( \frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c}\right) \geq 9\cdot \frac{1\plus{}a\plus{}b\plus{}c}{3\plus{}a\plus{}b\plus{}c}\geq 5\]
since $ a\plus{}b\plus{}c\geq \frac{3}{2}$. The minimum is attained for $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$.
The maximum:
\[ \frac{1}{1\plus{}a}\leq \frac{5}{6}\minus{}\frac{1}{3}a\iff 6\leq (1\plus{}a)(5\minus{}2a)\iff (1\minus{}a)(2a\minus{}1)\geq 0\]
so that
\[ A\leq \frac{1}{6}(1\plus{}a\plus{}b\plus{}c)(15\minus{}2a\minus{}2b\minus{}2c)\leq 6\]
\[ \iff (3\minus{}a\minus{}b\minus{}c)(7\minus{}2a\minus{}2b\minus{}2c)\geq 0\]
which is true, since $ a\plus{}b\plus{}c\leq 3$. We get equality for $ a\equal{}b\equal{}c\equal{}1$.


Let $f: \mathbb{R}\to\mathbb{R}$ satisfying:
$f(2x)=f\left(\sin\frac{x+y}{2}\pi\right)+f\left(\sin\frac{x-y}{2}\pi\right)$  $\forall x,y\in\mathbb{R}$
Calculate: $f(1990+\sqrt{1990}+\sqrt[3]{1990})$

That's a tricky one! I'll have to use the calculator often.
If anyone finds a direct approach, please show me.

(*) $f(2x)=f\left(\sin\frac{x+y}{2}\pi\right)+f\left(\sin\frac{x-y}{2}\pi\right)$

I'll show $f\equiv 0$.

1) $f(0)=0$ and $f(x)=f\left(\sin\frac{\pi}{2}x\right)$ for all $x\in\mathbb{R}$:
$x=y=0$ in (*) gives $f(0)=f(0)+f(0)$ and $f(0)=0$, and then
$x=y$ in (*) gives $f(2x)=f\left(\sin\pi x\right)$ and $f(x)=f\left(\sin{\pi\over 2}x\right)$ for all $x\in\mathbb{R}$.

2) $f$ is even and $f(x)=2f\left(\cos\frac{\pi}{4}x\right)$ for all $x\in\mathbb{R}$:
$y=1$ in (*) gives $f(2x)=f\left(\cos\frac{\pi}{2}x\right)+f\left(-\cos\frac{\pi}{2}x\right)$,
and, as $\cos$ is even, we get $f(2x)=f(-2x)$ for all $x\in\mathbb{R}$ and $f$ is even.
But then $f(x)=f\left(\cos\frac{\pi}{4}x\right)+f\left(-\cos\frac{\pi}{4}x\right)=2f\left(\cos\frac{\pi}{4}x\right)$ for all $x\in\mathbb{R}$.

3) $f(\sqrt{1-4x^{2}}\sin \pi x-2x\cos \pi x)=0$ for all $x\in [-{1\over 2},{1\over 2}]$:
For $x\in [-{1\over 2},{1\over 2}]$ there exists $z\in [-{\pi\over 2},{\pi\over 2}]$ with $\sin z =2x$.
Then for $y: ={2\over \pi}z-x$ we have $\sin\frac{x+y}{2}\pi=2x$ and (*) gives $0=f\left(\sin\frac{x-y}{2}\pi\right)=f(\sin(\pi x-z))$ with
$\sin(\pi x-z)=\sin \pi x\cos z-\cos\pi x \sin z=\sqrt{1-4x^{2}}\sin \pi x-2x\cos \pi x$,
as $\cos z\geq 0$ for $z\in [-{\pi\over 2},{\pi\over 2}]$, and so $\cos z=\sqrt{1-\sin^{2}z}=\sqrt{1-4x^{2}}$.

Now set $X: =\{x\in\mathbb{R}: f(x)=0\}$.

Then $g(x): =\sqrt{1-4x^{2}}\sin \pi x-2x\cos \pi x$ is continuous on $[-{1\over 2},{1\over 2}]$ with $g(0)=0$ and $g(0.4)>0.32$.
So by 3) and the intermediate value theorem we get $[0,0.32]\subseteq X$.

$h(x): =\sin\frac{\pi}{2}x$ is continuous on $\mathbb{R}$ with $h(0)=0$
and $h(0.32)>0.48$ and $h(0.48)>0.68$ and $h(0.68)>0.87$.
So by 1) and again using the intermediate value theorem we get
$[0,0.48]\subseteq h([0,0.32])\subseteq h(X)\subseteq X$ and
$[0,0.68]\subseteq h([0,0.48])\subseteq h(X)\subseteq X$ and
$[0,0.87]\subseteq h([0,0.68])\subseteq h(X)\subseteq X$.

$k(x): =\cos\frac{\pi}{4}x$ is continuous on $\mathbb{R}$ with $k(0)=1$ and $k(0.87)<0.8$.
So by 2) and using the intermediate value theorem again we get
$[0.8,1]\subseteq k([0,0.87])\subseteq k(X)\subseteq X$.

So together we get $[0,1]\subseteq X$ and also $[-1,1]\subseteq X$, as $f$ is even by 2).
And now 1) gives $f\equiv 0$, as $\sin\frac{\pi}{2}x\in [-1,1]\subseteq X$ for all $x\in\mathbb{R}$.

EDIT
$f$ even , then $f\left(\sin\frac{x-y}{2}\pi\right)=f\left(\sin\frac{y-x}{2}\pi\right)$ for all $x,y\in\mathbb{R}$,
and then (*) gives $f(2x)=f(2y)$ for all $x,y\in\mathbb{R}$.
So $f$ is constant and $f\equiv 0$.
No need for calculators. :P


logarithms
limit
function
integration
calculus
algebra solved
algebra

Consider sequence $\displaystyle {x_n}_{n=1}^{\inf}$ satisfying: $x_1=1-\ln 2$;
$\displaystyle x_n= \frac {1}{n} -x_{n-1}$.
Prove that there exists $\displaystyle \lim_{n->\inf}nx_n$

This is equivalent to $\displaystyle \lim_{n\to\infty}n(\frac 1{n+1}+\ldots+\frac 1{2n}-\ln 2)=-\frac 14$.

Some time ago I posted the following problem (shown to me by my teacher):

If $f:[0,1]\to \mathbb R$ is a $C^1$ function, then $\displaystyle \lim_{n\to\infty}n\left(\frac 1n\sum_{i=1}^nf\left(\frac in\right)-\int^1_0f(x)dx\right)=\frac{f(1)-f(0)}2$. Just apply this for $\displaystyle f(x)=\frac 1{1+x}$.

I can’t remember whether I posted a solution or not, but I think Moubinool posted one. Try searching for it in the Calculus section.


AMC
AIME
USA(J)MO
USAMO
MATHCOUNTS
trigonometry
AMC 12

Until this year I was homeschooled, and had never heard of the AMC tests before :D .  However, my math teacher (I now go to public school) got me to take the AMC 10 this spring and I got a 60 something on it, which is really bad even for my school.  Anyways, while I've always been a good math student in school, I've never done these types of problems before - in fact I've really never been challenged in math before.  I'm genuinely interested by these kind of problems, and I would like to improve my problem solving skills.   I joined the math team at school, I'll probably get one of the AoPs books, and I think the past AMC tests are a great resource for me to use.  I really have no clue what the typical improvement is, but I  was thinking a good goal would be to qualify for the AIME.  Any advice on how to improve or what I should set my goals as would be great!

Great to see your interest!

There is no "typical" improvement. But, you can set the goal for AMC 10, if your age permits, to be around 120 so you can qualify for AIME at this point. You should read FAQs for AMC in this forum.

You should get AoPS Volume I now. You can get AoPS Volume II as well, but you probably won't be able to understand most of them if you scored 60 on AMC 10.

The Resources (or Contests in AoPS Skin) contain vast amounts of old problems, so there you go, practice, practice, practice!

Okay, so qualifying for the AIME would be a reasonable goal?  I really want to learn this type of math, but I don't want to set unattainable goals that will only make myself feel like a failure.

Qualifying for the AIME should be the first goal. So, yes for the AMC 10, you need to get at 120 (which is basically 17 right, considering you leave the others blank) and 100+ for AMC 12. And if you qualify for the AIME now, maybe this year or mostly next year consider qualifying for the USAMO.

[quote="Yongyi781"]I'm probably one of the oddballs - this is the first year I've qualified for the AIME, and on top of that I also qualified for the USAMO this year.[/quote]

I remember seeing one student who was in similar position. He barely got over 100 on AMC 12 (wasn't great score at all), somehow scored sufficient points in AIME for floor value (yes, I've been here for long time), and went to USAMO.

Of course, I don't mean that you were like him. My point, however, is that it happens, but certainly, not very often.

I was in a similar position to you last year.

In 8th grade, a teacher told me of mathcounts, a middle school competition. While I did reasonably well by making top third at nationals, I feel I could've done much better with practice. Subsequent amc 10 practice attempts gave me horrible scores. So I started working. I bought the aops intro to counting book and worked on amc 10/12/mathcounts problems. When I entered high school the next year, I learned about more advanced topics and went over csml/arml problems. So when I took the amc 10 I was ready. I passed it and got reasonably close to the usamo.

So do the same as I did:

1. Practice old problems in various mathematics competitions
2. Learn more advanced mathematics topics
3. If you have the money, buy a math book on a topic you are weak on.
4. Ask someone else who is interested in math to join you.

Well I'm in 10th grade right now, so I'll have to take AMC 12 next year.  I'm going to get AoPs Volume 1: but I don't have a lot of extra money - are the introduction books worth buying?

Also what do you mean by "more advanced topics"?

[unrelated]
[quote="limac"](which is basically 17 right, considering you leave the others blank)[/quote]
I strongly disagree with this. If you've tried all of them and can only solve 17 problems, then sure, but definitely do not just answer 17 and not even try the rest. The chance of making 1 mistake on one of the 17 problems is still there no matter how many times you check, and you would want to solve more problems anyway, both for score and (more importantly) for a better mathematical experience (of course, it is always a good idea to check, just don't answer exactly 17 for the sole reason of wanting to make the AIME). There was a good thread on this a while back, I'll try to dig it up.
[/unrelated]

The best way to prepare, which has been said many times, is to do problems. The resources link is [url=http://www.artofproblemsolving.com/Forum/resources.php]here[/url]. (an unrelated tip for getting the most out of problems: Don't look at the solution if you can't solve one at first, only do so after you've made a good attempt at it. When you read a solution, try to understand it thoroughly, and eventually you'll probably figure out some common "strategies" that are used in problem solving).

I don't really know much about books, I only have AoPS volume 1, 2, and intro to combinatorics. Combinatorics is probably the least taught topic in school, so it might be a good buy if you haven't done much of it before. However, doing problems is still more beneficial (for example, last year, I qualified for AIME before I had any books by doing tons of past AMCs and learning from those problems). The books do include lots of good problems, though.

By "advanced topics", Ihatepie might have meant logs and trigonometry, which appear on the AMC12 and above (calculus is never needed).

As for goals, if you can, try to aim high.

I got 121.5 on the AMC10 B, 9 on the AIME, so missed USAMO D:, I guess that makes me an oddball, did anyone else get a 9 on the aime and miss usamo? Anyways, the best way to improve in my opinion is simply to read aops and do a bunch of problems.

[quote="Yongyi781"].....
17 right and 8 blank only yields 114 points...[/quote]

Yeah, you need at least a 19 on the AMC10 to make AIME.

19 right, 6 blank = 120.

20 right, 0-6 blank = 120+

If your sole goal is to qualify for AIME based on your 10 score, you should leave either 0 or 4 problems blank. Thus, if you solve 22, you should guess the other 3. If however, you wish to maximize your score (for purposes of USAMO, etc), disregard my advice.

Uh, you're going to need a minimum of 14 right, not 19, to qualify for the AIME since you're taking the AMC 12.

[quote="Evolved"]Well I'm in 10th grade right now, so I'll have to take AMC 12 next year. [/quote]

Guys, you're missing the point giving goals for the AMC 10.


Let p be a prime and $f(n)$ satisfy $0\le f(n) <p$ for all integers $n$. $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$. If for fixed $n$, there exists an integer $0\le y < p$ such that:

$ny-p\left\lfloor \frac{ny}{p}\right\rfloor=1$

then $f(n) = y$. If there is no such $y$, then $f(n) = 0$. If $p = 11$, find the sum: $f(1)+f(2)+...+f(p^{2}-1)+f(p^{2}).$

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103281[/url]

there is a full solution there, alternately, you could brute force it...the problem was one of the weaker problems since the quickest solution required some obscure knowledge

Could you give an explanation of an inverse mod p?


Let $a\in\mathbb{N}, a\geq 2$. Let us consider the following sequence
  $(x_{n})_{n\geq 0}$ defined by $x_{0}= \frac{a^{2}}{4}, x_{1}=\frac{a}{4}(2a^{3}-4a^{2}-a-4)$ and \[x_{n+1}-(4a^{2}-2)x_{n}+x_{n-1}=0,\forall n\geq 1.\] Show that the number $2x_{n}-1$ is a perfect square, for any $n\geq 1$.


Let $ \{x_k\}$ be a convergent sequence in $ \mathbb{R}$. Show that the set $ \{x_1,x_2,\cdots\}$ has zero content.

Does zero content mean measure 0? If so, that is because $ x_i$ is countable, so we can simply cover each element by an interval of $ \frac{\epsilon}{2^k}$ for arbitrary $ \epsilon > 0$


Find all complex solutions in rectangular form: \[z^{3}-6z^{2}+12z=8-8i\] Tricky, but doable.


inequalities
real analysis
real analysis unsolved

Prove for $ x \in [0,1]$ that
\[ (1\plus{}x) \ln(1\plus{}x) \geq x \plus{} x^2/3.\]

Thank you.

hello, rewriting your inequality in the form
$ \ln(1\plus{}x)\minus{}\frac{x\plus{}\frac{x^2}{3}}{1\plus{}x}\geq0$
let
$ f(x)\equal{} \ln(1\plus{}x)\minus{}\frac{x\plus{}\frac{x^2}{3}}{1\plus{}x}$
we have to show that $ f(x)\geq0$ for all real $ x$ with $ 0\le x\le 1$
and we get
$ f(0)\equal{}0$
$ f(1)\equal{}\ln(2)\minus{}2/3>0$
and
$ f'(x)\equal{}\frac{1}{3}\frac{x(1\minus{}x)}{(1\plus{}x)^2}>0$
for all real $ x$ with $ 0<x<1$, this completes the prove.
Sonnhard.


ratio
inequalities
inequalities proposed

It suffices to show that this is true for $\lambda=0$ and $\lambda=n$; the remaining values follow by interpolation.

For $\lambda=n$, we need $x_1x_2...x_n(\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}-n) \leq \frac{n-1}{n^{n-1}}$. We will show that a maximum for the LHS is attained when $x_i=\frac{1}{n}$ for all $i$. Suppose the maximum is attained at some other set of values. Consider $x_i \neq x_j$ such that neither is the outright smallest of $x_1, x_2...x_n$. We replace $x_i$ and $x_j$ each by $\frac{x_i+x_j}{2}$. This increases the first factor on the LHS by a ratio of $\frac{(x_i+x_j)^2}{4x_ix_j}$. It decreases the "affected" part of the second factor by the same amount, but this is "diluted" by the other terms (the sum of which, by the condition we imposed on $x_i$ and $x_j$, is positive) and so has less of an effect. Therefore the value of the LHS increases. Contradiction.

There is one other possibility: All $x_i$, except the smallest, could be equal. In this case we must show that $x^{n-1}(1-nx+x)(\frac{n-1}{x}+\frac{1}{1-nx+x}-n) \leq \frac{n-1}{n^{n-1}}$ with $x \geq \frac{1}{n}$. The LHS simplifies to $x^{n-2}(n-1)(nx^2-nx+1)$.  Letting $y=nx$ we need $ny^{n-2}(y-1) \geq y^n-1$. Since $y \geq 1$ this is equivalent to $ny^{n-2} \geq y^{n-1}+...+y+1$, which is clearly true.

A similar, but less complicated argument deals with the case $\lambda =0$.

Hmm...
[quote]It decreases the "affected" part of the second factor by the same amount, but this is "diluted" by the other terms (the sum of which, by the condition we imposed on  $x_i$ and $x_j$, is positive) and so has less of an effect.[/quote]
Why you say sum of other term is positive? Even with condition on $x_i$ and $x_j$ it is not necessary true.
Or I am missing something?

The outright smallest of $x_1, x_2... x_n$ must be less than $\frac{1}{n}$; hence its reciprocal is greater than $n$. Since we didn't use it up, it must be still there to cancel out the $-n$, and the other terms are positive.

Ohhh... Sorry.
But it seems that sum of "other" terms will be at least $(n-2)^2$ in any case, since $(y_1+...+y_{n-2})(1/y_1+...+1/y_2)\geq (n-2)^2$ and $y_1+...+y_{n-2}<x_1+...+x_n=1$. So your first arguments works fine for all $n\geq 4$.

My solution is based on [i][b]Shturm's method[/b][/i](maybe I am wrong in its name).I'll tell about it(for people who don't know ).We have that $\sum_{i=1}^{n}{x_{i}}=1$.Denote $A=x_{1}\cdot{x_{2}}\cdots{x_{n}}(\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{n}})-\lambda{x_{1}\cdot{x_{2}}\cdots{x_{n}}}.$
Equality holds when $x_{1}=x_{2}=...=x_{n}=\frac{1}{n}$. It is clear that at least one of those numbers is less than $\frac{1}{n}$ and one is greater.WLOG $x_{1}\leq{\frac{1}{n}}$ and $x_{2}\geq{\frac{1}{n}}.$Now if we replace $x_{1}$ by$ x_{1}+x_{2}-\frac{1}{n}$ and $x_{2}$ by $\frac{1}{n}$, the sum of $x_{i}$ is still 1.Idea is to prove that after this A becomes greater(I'll not write how to show this, because it's not difficult and in the proof we use that $x_{1}\leq{\frac{1}{n}}$ and $x_{2}\geq{\frac{1}{n}} )$.After doing this n times we
get that $x_{1}=x_{2}=...=x_{n}=\frac{1}{n}$, but in this case equality  occurs and our A is greater than first A.
 QED

[quote="Armo"]Idea is to prove that after this A becomes greater(I'll not write how to show this, because it's not difficult and in the proof we use that $x_{1}\leq{\frac{1}{n}}$ and $x_{2}\geq{\frac{1}{n}} )$.[/quote]
Hmm...

I don't understand what you mean [b]Myth[/b] :?.I did it just now and I got that we need to prove that $\lambda\cdot(x_{1}-\frac{1}{n})(x_{2}-\frac{1}{n})\geq{(\sum_{i=3}^{n}{\frac{1}{x_{i}})}}\cdot(x_{1}-\frac{1}{n})(x_{2}-\frac{1}{n}).$Now I'll prove this one.
From Am-Hm we obtain $\sum_{i=3}^{n}{\frac{1}{x_{i}}}\geq{\frac{(n-2)^{2}}{\sum_{i=3}^{n}{x_{i}}}}>\frac{(n-2)^{2}}{\sum_{i=1}^{n}{x_{i}}}=(n-2)^{2}\geq{n}$(when $n\geq{4}$)$>\lambda$.Am I right or there is any mistake?

Bug:
[quote="Armo"]\[\lambda\cdot(x_{1}-\frac{1}{n})(x_{2}-\frac{1}{n})\geq{(\sum_{i=3}^{n}{\frac{1}{x_{i}})}}\cdot(x_{1}-\frac{1}{n})(x_{2}-\frac{1}{n}).\][/quote]
At least, you must be mean opposite sign?

I think the following inequality is very difficult:

If $x_1,...,x_n$ are positive numbers such that $x_1+x_2+...+x_n=n$, then
$x_1x_2...x_n(\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}-n+3)\leq3$.

Indeed, a very hard problem. But I think that an even harder ( at least, it implies your inequality even with 3 replaced by e) is the following one, which I found not long time ago:
    If $\frac{1}{x_1}+...+\frac{1}{x_n}=n$ then $ x_1\cdot...\cdot x_n-1\geq (\frac{n-1}{n})^{n-1}\cdot (x_1+x_2+...x_n-n)$.

given that a,b,c>0 a+b+c=1
  find the maximum value of A=(axa+1)/(bxb+1)+(bxb+1)/(cxc+1)(cxc+1)/(axa+1)
 sorry ,this is the first time I visit this web ,I don't know how to type exactly in math formal

You must create separate topic for your problem. :)

[quote="mashimaro"]now I know
  given that a,b,c >0 and a+b+c =1
  find the maximum value of
  $A= \frac{a^2+1}{b^2+1}+\frac{b^2+1}{c^2+1}+\frac{c^2+1}{a^2+1}$[/quote]

I think we had this: http://www.mathlinks.ro/Forum/viewtopic.php?t=18571

  Darij

Darij is the main ML finder  :lol:  :lol:  :lol:

[quote="Armo"][b]@Myth[/b]:No the sign is right.Can you check it again?[/quote]
Opppss.  :blush:
I forgot that $x_1-1/n<0$.

mashimaro! Do you read what I write you?
[b]Create separate topics for your problems[/b] :mad:

sorry . I have just known this forum for 4 days  :P

so ,can you solve it for me at last time

It VERY WELL KNOWN Cauchy inequality. I think you can easily find it in "Inequalities Theorems and Formulas" http://www.mathlinks.ro/Forum/index.php?f=55

[quote="mashimaro"]given n positive numbers $x_i >0 (i=1,2,...,n)$ ,a,b are two real numbers
 with $a>=B>=0$
 prove that
  $ (\frac{x_1^b+x_2^b+...+x_n^b}{n})^{\frac{1}{b}} <=  (\frac{x_1^a+x_2^a+...+x_n^a}{n})^{\frac{1}{a}}$[/quote]

i think it was called "power mean inequality"...  :lol:


how do i simplify this
$ \forall n,b\in\mathbb{N}: n\geq b$
$ (n\minus{}b)! \equal{}\frac{n!}{n(n\minus{}1)(n\minus{}2)\cdots(n\minus{}(b\minus{}1))}\equal{}\frac{n(n\minus{}1)(n\minus{}2)(n\minus{}3)\cdots 1}{n(n\minus{}1)(n\minus{}2)\cdots(n\minus{}(b\minus{}1))}$

i dont know where to go from there, obviously some terms cancels out but dunno how many.

if there is a mistake in my notation can someone point out please.

:maybe: Does not cancelling out give the same thing as you started with?
[hide]
$ (n\minus{}b)(n\minus{}b\minus{}1)(n\minus{}b\minus{}2)(n\minus{}b\minus{}3)\cdots (1) \equal{} (n\minus{}b)!$.
[/hide]
Is that what the question wanted?

[quote="Prime factorization"]I guess all you could do is substitute $ n\minus{}b$ by $ a$ to get an answer of $ a!$.[/quote]

 :rotfl:  :rotfl:  :rotfl:

[quote="binomial_4eva"]Astrophys can you explain abit more please how you came up with this.
[quote]$ (n\minus{}b)(n\minus{}b\minus{}1)(n\minus{}b\minus{}2)(n\minus{}b\minus{}3)\cdots (1) \equal{} (n\minus{}b)!$.[/quote][/quote]/quote]
That is just from the definition of factorial. It's the product of all the integers between 1 and (n-b) inclusive.

Or you could be MASSIVELY awesome, but with massive overkill and write it as a gamma function?
 :lol:

[quote="binomial_4eva"]Astrophys can you explain abit more please how you came up with this.
[quote]$ (n\minus{}b)(n\minus{}b\minus{}1)(n\minus{}b\minus{}2)(n\minus{}b\minus{}3)\cdots (1) \equal{} (n\minus{}b)!$.[/quote]

[quote]A factorial is a factorial. Why are you asking questions like this?[/quote]
coz sometimes you want to work with things that is not factorial.sometimes you need to change a factorial into a polynomial.[/quote]

Are you sure you understand what a factorial is?

We've been over this.  A factorial is never a polynomial.  It's just a function on the natural numbers defined by the recursion $ 0! \equal{} 1, (n\plus{}1)! \equal{} (n\plus{}1) n!$.  The fact that $ n! \equal{}\prod_{k\equal{}1}^{n}k$ follows immediately (it is in fact occasionally taken as an alternate definition of the factorial symbol), and should not require any explanation.

The Gamma function is a function defined over the complex numbers that "extends" the factorial.


function
integration
logarithms
complex analysis
calculus
algebra
domain

Let $ u(z)$ be a harmonic function in $ \{z: 0 < |z| < 1\}$. define $ P \equal{}\int_{|z| \equal{} r}\frac {\partial u}{\partial n}ds$ for $ 0 < r < 1$.  Show that $ P$ is independet of $ r$.

Moreover, show that $ u \equal{} \frac {P}{2\pi}\log|z| \plus{} Re(f(z))$ for some analytic function $ f$, in $ \{z: 0 < |z| < 1\}$

no one can solve this :maybe: ??
the first part is easy, just use the fact that laplacian of u is zero...

the second part is the one that i  want to see a proof of.

Consider $ u(z)\minus{}\frac {P}{2\pi}\log|z|$, show that it is harmonic, and thus is the real part of some analytic function.

to flesh out the details, we know that $ v \equal{}  u \minus{} \frac{P}{2\pi} \log r$ being harmonic on the punctured disc does not suffice to be the real part of a holomorphic function in multiply connected regions so we need to establish harmonicity over the whole simply connected disc; this is already clear at points away from zero, and as being harmonic is equivalent to the integral of the gradient vanish over curves whose interiors lie within the domain in question, we find that it suffices to show the integral of $ \nabla v$ vanishes over all closed curves lying in the unit disc. We may take a generator for the homology of the punctured disc as some circle of radius $ r$ such as that used in the definition of $ P$ and break up any other curve into some multiple of this generator, up to homology. The "up to homology" does not matter as $ v$ is already known to be harmonic outside of the origin, so it suffices to show that the integral of grad v vanishes on some simple closed circle containing the origin, which is essentially immediate from the definition of $ P$.

That was probably a bit of overkill but I think it's what's fundamentally going on -- the obstruction to a harmonic function being the real part of an analytic function in this example is strength of the delta singularity of the laplacian at the origin and the problem is essentially to show that this obstruction of $ u$ may be cancelled out by the term involving $ P$.


group theory
abstract algebra
topology
real analysis
real analysis unsolved

Let $P:X\rightarrow Y$be closed continuous surjective map such that $P^{-1}(y)$for all $y\in Y$ is compact prove that if $Y$ is compact then $X$ is compact.

Let $(F_{\alpha})_{\alpha}$ be a family of closed sets in $X$ which satisfies the finite intersection condition (i.e. any finite subcollection has non-void intersection; I'll call it fic next time :)). The family $(f(F_{\alpha}))_{\alpha}$ is a family of closed sets in $Y$ which also satisfies fic. Since $Y$ is compact, the intersection of all of the $f(F_{\alpha})$ is non-void, so let $y$ be a point in this intersection.

Let $A_{\alpha}=F_{\alpha}\cap f^{-1}(y),\ \forall \alpha$. The sets $A_{\alpha}$ are non-void compact sets in $X$, which satisfy fic. This means that they have non-void intersection, so the family $(F_{\alpha})_{\alpha}$ also has a non-void intersection, which is what we wanted to prove.

P.S. I think I used, without mentioning it, the fact that $X$ is Hausdorff. Is this a condition which is assumed in the problem, but which you did not mention, sam-n? :?

to grobber :no being Hausdorrf is not mentioned,and nice consequence of this easy lemma is :assume that  $ G$ is topological group and $H$ is compact subgroup and also  $G/H$ is compact too,prove that $G$ is compact.
to liyi:closed map means that the map of the closed set is closed.


How many integral solutions exist for $ x$ if $ \frac12\le\frac x8\le\frac79$?

Setting a common denominator of 72, we get $ 36\leq 9x\leq 56$.  Hence any integer $ 4\leq x<7$ will work.


Sa se demonstreze ca daca $K_1,K_2,K_3,K_4,K_5$ sunt corpuri atunci
 $K_1*K_2*K_3$ si $K_4*K_5$ nu sunt izomorfe

Ecuatia $x^2-x=0$ are $8$ solutii in primul inel (toate tripletele co componente $0$ si $1$), pe cand in al doilea inel are numai $4$ solutii (toate perechile cu cumponente $0$ si $1$).


function
geometry
geometric transformation
reflection
complex numbers

Suppose $ x(y\minus{}z),y(z\minus{}x),z(x\minus{}y)$ are in $ G.P.$ and $ x,y,z \in \mathbb{C}.$

Find $ x,y,z$.

[hide]$ a,b,c$ in GP means $ ac \equal{} b^2$. Applying that here, we get:

$ x^2 yz \minus{} x y^2 z \minus{} x^2 z^2 \plus{} x y z^2 \equal{} x^2 y^2 \minus{} 2 x y z^2 \plus{} y^2 z^2$

$ \iff x^2 y^2 \plus{} y^2 z^2 \plus{} z^2 x^2 \equal{} x^2 y z \plus{} y^2 zx \plus{} z^2 xy$

Divide by $ x^2 y^2 z^2$ (except when $ x\equal{}y\equal{}z\equal{}0$, a trivial solution) to get:

$ \left( \frac{1}{x} \right) ^2 \plus{} \left( \frac{1}{y} \right) ^2 \plus{} \left( \frac{1}{z} \right) ^2 \equal{} \left( \frac{1}{x} \right) \left( \frac{1}{y} \right) \plus{} \left( \frac{1}{y} \right) \left( \frac{1}{z} \right) \plus{} \left( \frac{1}{z} \right) \left( \frac{1}{x} \right)$

Substituting $ (a,b,c) \equal{} (1/x,1/y,1/z)$, we have that $ a^2\plus{}b^2\plus{}c^2 \equal{} ab\plus{}bc\plus{}ca$. In other words, $ (x,y,z)$ are the inverses of any three complex numbers that form a triangle in the complex plane.[/hide]

[quote="TZF"](except when $ x \equal{} y \equal{} z \equal{} 0$, a trivial solution)[/quote]  Actually, if I'm not mistaken, $ \{x, y, z\} \equal{} \{0, 0, t\}$ is a solution for any $ t \in \mathbb{C}$ (not just $ t \equal{} 0$).

[quote="TZF"]$ (x,y,z)$ are the inverses of any three complex numbers that form a triangle in the complex plane.[/quote]
Equilateral triangle.

The function $ z \mapsto \frac{1}{z}$ is the composition of inversion through the unit circle and reflection across the $ x$-axis in the complex plane.  What does the inversion of an equilateral triangle look like?

[quote="Poincare"]The Big T has spoken![/quote]

Can you explain it in details? :)

Write $ Z \equal{} \frac {1}{|z|^{2}} \bar{z}$ and $ W \equal{} \overline{Z}$. This is what $ z \mapsto \frac{1}{z}$ is.


Να αποδειχθεί ότι για κάθε συνάρτηση $ f$ που είναι συνεχής και αύξουσα στο $ [a,b]$ ότι:

$ \int^{b}_{a}xf(x)dx\geq(\frac {a \plus{} b}{2})\int^{b}_{a}f(x)dx$

Την παλεύω κανά δυο μέρες τώρα αλλά δεν μου βγάινει... :mad:

pare tin synartisi

$ G(x)\equal{}\int_a^x tf(t)dt\minus{}(\frac{a\plus{}x}{2})\int_a^x f(t)dt$ (bgainei auksousa apla...)

Elpizw na einai swsti...an nai sou to afierwnw...:)

Παίρνεις τη συνάρτηση $ g(x) \equal{} \int_a^x tf(t) dt \minus{} \left(\frac {a \plus{} x}{2}\right)\int_a^x f(t)dt, \ \ x\in [a,b]$ η οποία είναι παραγωγίσιμη με παράγωγο.

$ g'(x) \equal{} \frac {f(x)}{2}(x \minus{} a) \minus{} \frac {1}{2}\int_a^x f(t)dt \ \ (1)$

Να δείξεις ότι για κάθε $ x\in[a,b]$ υπάρχει $ \xi_x \in(a,x)$ τέτοιο ώστε $ \int_a^x f(t)dt \equal{} f(\xi_x)(x \minus{} a)$, και τότε η $ (1)$ γίνεται

$ g'(x) \equal{} \frac {x \minus{} a}{2}\left(f(x) \minus{} f(\xi_x)\right)$ που είναι πάντα $ \geq 0$ και τελείωσες.

Σημείωση: Το γράφω $ \xi_x$ για να δείξω ότι αυτό το $ \xi$ εξαρτάται κάθε φορά από το $ x$. Έτσι κι αλλιώς αυτό το $ \xi_x$ είναι πάντα αυστηρά μεγαλύτερο από το $ a$

Αλέξανδρος

[quote="socrates"]pare tin synartisi

$ G(x) \equal{} \int_a^x tf(t)dt \minus{} (\frac {a \plus{} x}{2})\int_a^x f(t)dt$ (bgainei auksousa apla...)

Elpizw na einai swsti...an nai sou to afierwnw...:)[/quote]

Χαθήκαμε ρε socrates...
Τελευταία δεν μπαίνω και πολύ στο msn οπότε είναι δύσκολο σχετικά να τα πούμε...
Ευχαριστώ και για την αφιέρωση....

Ωραία η λύση που δώσατε...Βασικά με αυτόν τον τρόπο πήγα στην αρχή αλλά κόλλησα στο ολοκλήρωμα οπότε την παράτησα τελείως την ιδέα.Κάποια άλλη στιγμή σκέφτηκα αυτή την αντικατάσταση αλλά δεν βοηθούσε στον τρόπο που πήγαινα...Δεν πειράζει κάτι έμαθα και από εδώ...


quadratics
modular arithmetic
number theory unsolved
number theory

$ p$ an odd prime, show:

a) $ \displaystyle\sum_{r\equal{}1}^{p\minus{}1} r (r \mid p) \equal{} 0$ is $ p \equiv 1 (\text{mod } 4)$.

b) $ \displaystyle\sum_{r\equal{}1,(r \mid p) \equal{}1}^{p\minus{}1} r \equal{} \frac{p(p\minus{}1)}{4}$ is $ p \equiv 1 (\text{mod } 4)$.

a) is cool.
First direction (your question): $ \sum_{r\equal{}1}^{p\minus{}1} r\left( \frac{r}{p} \right)\equal{}\sum_{r\equal{}1}^{p\minus{}1} r\minus{}2\sum_{\left( \frac{r}{p} \right)\equal{}\minus{}1}^{p\minus{}1} r\equal{}\frac{p(p\minus{}1)}{2}\minus{}2\sum_{\left( \frac{r}{p} \right)\equal{}\minus{}1}^{p\minus{}1} r$
If this expression equals to zero, then $ \frac{p(p\minus{}1)}{2}$ is even which implies that $ p \equiv 1 \pmod 4$.

Second direction:
If $ p \equiv 1 \pmod 4$, then $ \left( \frac{\minus{}1}{p} \right)\equal{}1$. Thus, $ r\left( \frac{r}{p} \right)\plus{}(p\minus{}r)\left( \frac{p\minus{}r}{p} \right)\equal{}p\left( \frac{r}{p} \right)$.
The sum becomes $ p \sum_{r\equal{}1}^{\frac{p\minus{}1}{2}} \left( \frac{r}{p} \right)$. Again, using the pairing trick, we get:
$ 0\equal{}\sum_{r\equal{}1}^{p\minus{}1} \left( \frac{r}{p} \right)\equal{}$
(because half those number are quadratic residues and half are non-quadratic residues)
$ \sum_{r\equal{}1}^{\frac{p\minus{}1}{2}} \left( \frac{r}{p} \right) \plus{} \left( \frac{p\minus{}r}{p} \right)\equal{}2\sum_{r\equal{}1}^{\frac{p\minus{}1}{2}} \left( \frac{r}{p} \right)$
And the result follows.

b) is almost identical:

$ \sum_{r\equal{}1}^{p\minus{}1} \minus{}r \plus{}2\sum_{\left( \frac{r}{p} \right)\equal{}1}^{p\minus{}1} r\equal{}\sum_{r\equal{}1}^{p\minus{}1} r \left( \frac{r}{p} \right)$
The equality in b) is then equivalent the equality in a) which is equivalent to $ p \equiv 1 \pmod 4$.