LET $ x, a ,b$ be positive  integers

show that if: $ x^{a+b}=a^bb$ then

$a=x$ & $b=x^x$

Have to be:
 [tex]a=x [/tex]and [tex]b=x^a![/tex]
I think you made mistake:

Is it a joke?
If $x=a$ then $x^a=x^x$   ;)

i have check the question again and its right and there is no mistake:

and by puting $x=a$ we wont get $x^a=x^x$

Beside we will have $x^{x+b}=x^bb$

that follows with $x^x=b$ so its enought to prove that $x=a$

i have found somethings (maybe helpfull):
if $x=1$ then $a=b=1$ and we are done so we can get $x\geq1$
let $P_1,P_2,...P_n$  be the prime factores of $x$and $x=\prod P_i^{\omega_i}, \omega_i\gg0$
from the $x^{a+b}=a^bb$ we have $b%Error. "midx" is a bad command.
^{a+b} , a\mid x^{a+b}$.
let $a=\prod P_i^{\alpha_i}, (\alpha_i \geq0) , b=\prod P_i^{\beta_i}(\beta_i\geq0)$
i have proved that for each $ 1 \leq i \leq n$ ,$\beta_i \gg 0$
if $\beta_i=0$then from $x^{a+b}=a^bb$ we have $\omega_i(a+b)=\alpha_ib$ then
$(\alpha_i -\beta_i)b=a\omega_i  \gg 0$ so $a\mid(\alpha_i- \omega_i)b$ and $P_i^{\alpha_i}\mid(\alpha_i - \omega_i)b$ because $P_i$ cant count $b$ we have $P_i^{\alpha_i}\mid\alpha_i - \omega_i$ that folows with $P_i^{\alpha_i} \ll \alpha_i$ and its is not correct so we have $\beta_i \gg 0 $

And you can show that for each $i$,  $\omega_i \leq \alpha_i$.
So $x$ divides $a$.
(proof: assume that $\omega_i>\alpha_i$, then $\beta_i=a\omega_i+b(\omega_i-\alpha_i) \geq b \Rightarrow$ impossible).
You can now show that for each $i$, $\omega_i \leq \beta_i$.
So $x$ divides $b$.
(proof: assume that $\omega_i>\beta_i$. we have $\beta_i=a\omega_i+b(\omega_i-\alpha_i)$ , $p^{\beta_i} \mid p^{\omega_i} \mid x \mid a$ and $p^{\beta_i} \mid b$ so $p^{\beta_i} \mid \beta_i \Rightarrow$ impossible because $\beta_i>0$ as you shown it).

and by your help that $x \mid a$ i have completed the proof.
let $x \mid a$ then we have $\omega_i \leq \alpha_i$
we will show that $x=a$ cause if not there exist a integer $i$ such that $\alpha_i \geq\omega_i +1$ and we have
$ \omega_i(a+b)=\beta_i+\alpha_ib \gg \alpha_ib \geq (\omega_i +1)b$ then we have
$\omega_ia+\omega_b \geq \omega_ib+b$ that follows with $\omega_ia \gg b \geq a^2$
so $\omega_i \gg a$ and $\omega \gg P_i^{\omega_i}$ then its false
so $ a=x$ and we are done.

ps in all of the posts by $\gg$ i ment greater and not equal.(nice cooperation thanks)

[quote="lomos_lupin"](...)  $b \geq a^2$  (...)
[/quote]
Can you give me some details please? :?

sorry i thought i have mention the proof  before for $b \geq a^2$
i have proved that $b_i$s are all positive. then from $x^{a+b}=a^bb$ we have that$\omega_i(a+b)=\beta_i+b\alpha_i$ so for $1\leq i \leq n$ if $M_i=min(a,b)$ then $P_i^{M_i}\mid \beta_i$ but we know that $\beta_i>0 $and $P_i ^{\beta_i}$ dont count the $\beta_i$ so $\alpha_i < \beta_i$so $a \mid b$ and agian from
$\omega_i(a+b)=\beta_i+b\alpha_i$ we have that $a \mid \beta_i$ so $b=c^a$ such that $c\geq2$.
if $a$ dont be equal to 3 then from $ \sqrt 2 \geq \sqrt[a]a$ we have $2^a \geq a^2 $ that follows with $b=c^a \geq 2^a \geq a^2$ this statement is right until that $a=3 , b=2^3$ but we have that $a \mid b$  that lead us to that $a$ is not equal to $3$ so we have  $b \geq a^2$

Can anyone say where did I take a mistake or mistakes:

let$(x,a)=d \ ,\ x=x'd\ ,\ a=a'd\ ,\ (x',a')=1$ then:

$x^a.x'^b.d^b=a'^b.d^b.b\rightarrow x^a.x'^b=a'^b.b\rightarrow x'^b|b\rightarrow x'=1$

hence $x=a\rightarrow b=x^x$

[quote="lomos_lupin"]and by puting $x=a$ we wont get $x^a=x^x$[/quote]

I am so confused  :huh:

Saeid, although $x = d$, but I don't know from where u can get $x = a$ ?? It means u have to proof $a' = 1$ which is unnecessary....  :(


Each of the six members from the Skateer University hockey
team shakes hands with each of the six members of the Iceburg
Tech team. How many handshakes take place?

each one shakes 6 hands, adn there are 6 players
6*6=36

Each player may shake hands with 6 other hockeyplayers, but You have to divide by 2 because you double counted. For example, if A shakes B's hand, then B also shakes A's hands, so the answer is 18.

actually, it's 36

you never double-count in this case because

person one from team one shakes hands with every person from team 2....6 handshakes
person two from team one shakes hands with every person from team 2....6 handshakes
etc.

total, 36 handshakes

[quote="isabella2296"]No; dragon is correct.

Think of it as two person teams.

Team 1: A and B
Team 2: C and D

A and D, A and C, B and C, B and D

C and A, C and B, D and A, D and B

That's what we're counting - but we DID overcount. For example, D and A and A and D were counted as two separate handshakes, when they are, in fact, the same. So we DO divide, to get $ \boxed{18}$.[/quote]

Noticed that you did 4*2. You are including everyone to shake with the person from the other team. If you want to do it your way, you still have to start with everyone 6+6=12 and multiply everyone on the other team 6 and divide by 2. 12*6/2=36.

Dang to end the confusion I'll use the listing approach (note the answer is 36)

Letting team letter(players A-F) and team number (players 1-6)

A1
A2
A3
A4
A5
A6
B1
B2
B3
B4
B5
B6
C1
C2
C3
C4
C5
C6
D1
D2
D3
D4
D5
D6
E1
E2
E3
E4
E5
E6
F1
F2
F3
F4
F5
F6

Notice that these are two teams of 6 people, not one team of 6 persons shaking hands amongst each other.  That may be part of the confusion :) I know when I read it for the first time, I thought it was all possible handshakes just within a team of 6 people until I reread it.


Do  exist  positive integers $ a,b,c>0$ and prime number $ p$ such, that
$ \frac{1}{p}\equal{} \frac{1}{a^2} \plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}$

yes there exists such equation
just put $ p\equal{}3,a\equal{}b\equal{}c\equal{}3$.
maybe u want all solutions....?


articles geometry calculus probability inequalities FTW trigonometry

What is your favorite type of math? Mine are algebra and geometry. /me despises counting.

alg and geo? How boring is that? I prefer the discrete mathematics.  :lol: but not mc-style counting q's.

I hate math.......  :no: but I would pick algebra.... I guess.

I like guessing and checking...good subject.

ALGEBRA TOTALLY ROCKS!!!!!!
 :winner_first:
but i hate geometry...

Dang...

Excerpt from some Article on AoPS:

[b]Discrete math is fun.

Many students, especially bright and motivated students, find algebra, geometry, and even calculus dull and uninspiring. Rarely is this the case with most discrete math topics. When we ask students what the favorite topic is, most respond either "combinatorics" or "number theory." (When we ask them what their least favorite topic is, the overwhelming response is "geometry.") [u]Simply put, most students find discrete math more fun than algebra or geometry.[/u][/b]

Weird...not here...

wikipedia: Discrete mathematics, also called finite mathematics or Decision Maths, is the study of mathematical structures that are fundamentally discrete, in the sense of not supporting or requiring the notion of continuity. ...

I am such an Algebra geek...I live for alg! I can't survive without it. I love algebra, hate geometry, and still don't get what discrete math is.

Huh? I meant number theory's my favorite. But yeah, discrete math is mostly number theory and combinatorics, but has some other elements.

Algebra and somewhat discrete. Mostly algebra though.

I like
Number theory + algebra

I'm also decent at combinatorics even though I don't know what balls and urns is still.

I like algebra and geometry, but I HATE COUNTING PROBLEMS.  :P

i don't know what i like but here it is from best to worst:

Trigonometry
Algebra
calculus
counting
number theory
probability

that doesn't mean i don't like probability :!:

i like arithematic cuz its ez...only basic arithmetic like

2*3+9=?

I like Algebra because it's fun!

Arithmetic is the base of all branches, and yet it's so basic, so simple, so easy. Funny how that works.

just like 0 and 1 are the base of computers, and AoPS servers...

so basic, so simple...

Algebra and arithmetic cuz its easy and fast :D

[quote="pixelpi314"]1.algebra
2.number theory
3.topology
4.probability/combinatorics
5.geometry/trigonometry[/i][/quote]
if you know topology how good are you at calculus?

My favorite type of math would probably be algebra. I fail at making drawings for geometry and I just can't count all the different cases in combinatorics.

Counting problems have always given me issues because there are just so many different types of counting problems that it's sometimes hard to think which approach to use.

Geometry problems where you have to draw some lines to really get the picture are also hard because sometimes I just can't tell where to draw the line =(

Algebra is ok, I'm relatively good at manipulating variables, and memorizing formulas comes easy to me.

Of course, you have to be sure that you can derive the formula.  :wink:

i like Algebra, Trigonometry, and Calculus.

meh, i need to learn integral calculus.

i hate probability and geometry.


Find all functions $f: R\rightarrow R$ such that:
$f(x+y)+f(xy)=f(x^2)+f(y^2)$ for all $x,y$ are reals

$y=x$ $\to$ $f(2x)=f(x^2)$ $\to$ $f(2x)=f(-2x)$, $f$ is an even function.

$y=-x$ $\to$ $f(0)=2f(x^2)-f(-x^2)$ $\to$ $f(0)=f(x^2)=f(-x^2)$, $f$ is a constant function.

It's easy to see that every constant function works.

Set y=0 then we will get that $f(x)=f(x^2)$   $(i)$and  then
set y=x $f(2x)=f(x^2)$  $(ii)$, now from $(i)$ and $(ii)$ we will get that $f(x)=f(2x)$. Now take $2x=t\rightarrow x=\frac{t}{2}$ then it becomes
$f(t)=f(\frac{t}{2})$ using the same thing it is easy to find this
$f(t)=f(t/2)=f(t/4)=...=f(t/2^n)$ here;
if $x_n\rightarrow a$ , so $f(x_n)=f(a)$
$lim_n_{\rightarrow} _{inf} =a \rightarrow lim_n_\rightarrow _{inf} f(x_n)=f(a)$ using this we will find that $f(t)=f(0)$ so $t$ can be $x$ an represent $f(0)=c$ where $c$ is constant then $f(x)=c$ is the answer

[quote="abdurashidjon"]$lim_n_{\rightarrow} _{inf} =a \rightarrow lim_n_\rightarrow _{inf} f(x_n)=f(a)$[/quote]

$lim_n_{\rightarrow} _{inf} x_n=a \rightarrow lim_n_\rightarrow _{inf} f(x_n)=f(a)$ only if $f$ is continuous!
And I think that, without continuity, you have only proved that $f(x)=f(2x)$ $\to$ $f(x)=f(2^mx)$ only for $m\in\mathbb{Z}$.

Wasn't my solution complete?

Lordwings your solution is true but i also has showed it for continues, too. Before i had solved similiarly problem with my teacher i thaought it is the same, sorry if I have done anything wrong

[quote="hien"]But function $f$ don't have continuety. I think two solutions not completely. We haven't showed that only $f$ is constant function satified problem condition. I wait for reply?[/quote]
I showed that $f(0)=f(x^2)=f(-x^2)$ for every $x\in\mathbb{R}$. So, the solution MUST be $f(t)=f(0)$ for every $t\in\mathbb{R}$, a constant function.

Set x=0. Then $f(x)=f(x^2)$ and the given can be rewritten as $f(x+y)+f(xy)=f(x)+f(y)$(1). At (1), set x=1 and we get $f(y+1)=f(1)$ implying that f is constant.


here is my problem :) ,inspired by another problem in the forum:
$a,b,c$ are nonegetive numbers,such that $a+b+c=3$
can we prove:$a^x+b^x+c^x \geq a^y+b^y+c^y$,if$x>y>0$?????

[quote="siuhochung"]Then you can use power mean twice to solve it.[/quote]
can you detail your idea?

[quote="Enrique Lesseps"]But isn't the inequality $a^x+b^x+c^x \geq a^y+b^y+c^y$ if $x \geq y$ always true for all positive reals $a,b,c$ by Muirheads theorem? ($\frac12 \cdot [x,0,0] \geq \frac12 \cdot [y,0,0]$)

Oh, and you can't have $2^0+ 2 \cdot {\frac12}^0$, because $y>0$![/quote]
I think there is something  wrong with you.
let $y \to 0$,we will have the same thing.
and this inequ is sure false without the condition $a+b+c=3$

[quote="Enrique Lesseps"]But isn't the inequality $a^x+b^x+c^x \geq a^y+b^y+c^y$ if $x \geq y$ always true for all positive reals $a,b,c$ by Muirheads theorem? ($\frac12 \cdot [x,0,0] \geq \frac12 \cdot [y,0,0]$)

Oh, and you can't have $2^0+ 2 \cdot {\frac12}^0$, because $y>0$![/quote]

wrong. (x,0,0) does not majorize (y,0,0) when x>y.

siuhochung,can you explain your idea #4,I can't solve it in your hint.

\[
\begin{gathered}
  a^x  + b^x  + c^x  \hfill \\
   \geqslant 3\left( {\frac{{a^y  + b^y  + c^y }}
{3}} \right)^{\frac{x}
{y}}  \hfill \\
   = \left( {a^y  + b^y  + c^y } \right)\left( {\frac{{a^y  + b^y  + c^y }}
{3}} \right)^{\frac{x}
{y} - 1}  \hfill \\
   \geqslant \left( {a^y  + b^y  + c^y } \right)\left( {\frac{{a + b + c}}
{3}} \right)^{x - y}  \hfill \\
   = a^y  + b^y  + c^y  \hfill \\
\end{gathered}
\]

thank you,so nice.sorry for my stupidness :(


ratio
trigonometry
geometry solved
geometry

Let extension of opposite sides of any convex quadrilateral intersect at K and L points.Let the diagonals of quadrilateral intersect at O point.Let "d" line that passes through  O point and parallel to KL .Prove that the midpoint  of segment  inside the quadrilateral  of that  d  line is "O" point.

[quote="Virgil Nicula"]A simple property of the harmonical pencils. You know what is a harmonical pencil ?[/quote]
No... Can you please tell me what it is, Virgil?

[b]A concurrent pencil[/b] is the the reunion of four rays $[OA$, $[OB$, $[OC$, $[OD$ with a common origin $O$ so that $B\in (AC)$, $C\in (BD)$ and $m(\widehat {AOD})<180^{\circ}$ and in this case we denote $O(A,B,C,D)$.

[b]Lemma 1.[/b] Let $O(A,B,C,D)$ be a concurrent pencil. For any lines $d$ denote the its intersections $X$, $Y$, $Z$, $T$ with the rays $[OA$, $[OB$, $[OC$, $[OD$ respectively. Then the ratio $\boxed {\ \rho\equiv \frac{YX}{YZ}: \frac{TX}{TZ}\ }$ is constant and characterizes this pencil.
[u]Solution.[/u]  Denote $x=m(\widehat {AOB})$, $y=m(\widehat {BOC})$, $z=m(\widehat {COD})$ $\Longrightarrow$ $\frac{YX}{YZ}=\frac{OX}{OZ}\cdot \frac{\sin x}{\sin y}$, $\frac{TX}{TZ}=\frac{OX}{OZ}\cdot \frac{\sin (x+y+z)}{\sin z}$ $\Longrightarrow$ $\frac{YX}{YZ} : \frac{TX}{TZ}=\frac{\sin x\sin z}{\sin y\sin (x+y+z)}$, what is constant.

[b]Definition.[/b] The concurrent pencil $O(A,B,C,D)$ is [u]harmonically[/u] iff $\rho=1.$

\[ \boxed {\ \mathrm {I\ will\ continue\ soon\ !}\ } \]

See Theorem 5 in my note [i][url=http://www.cip.ifi.lmu.de/~grinberg/Butterfly.zip]On cyclic quadrilaterals and the butterfly theorem[/url][/i].

  darij


function
linear algebra
matrix
inequalities proposed
inequalities

This is generalization of [url]http://www.mathlinks.ro/viewtopic.php?t=216705[/url].
Let $ A,\;B,\;C\in\mathbb{R}$ are some constants and $ x,\;y,\;z\in\mathbb{R}$. Find maximum of the
\[ f(x,y,z) \equal{} \frac {Axy \plus{} Byz \plus{} Cxz}{x^2 \plus{} y^2 \plus{} z^2}.
\]

you can suppose that $ (x,y,z)$ lies on the circle $ x^2\plus{}y^2\plus{}z^2 \equal{} 1$. Then you have to compute the maximum of the function $ f(x,y,z) \equal{} Axy\plus{}Byz\plus{}Czx$ on this circle. Write the Lagrange multipliers condition that will tell you than $ (x,y,z)$ is an eigenvector of
$ \begin{bmatrix}
0 & A & C \\
A & 0 & B \\
C & B & 0
\end{bmatrix}$
It suffices to compute these eigenvectors and to plug it into the function $ f$.

The following results are known.

[b]Theorem 1.[/b] Let $ a_1,b_1,c_1$ be positive real numbers, and $ a_2,b_2,c_2$ real numbers. For any real numbers $ x,y,z$,

$ \frac {a_2yz \plus{} b_2zx \plus{} c_2xy}{a_1x^2 \plus{} b_1y^2 \plus{} c_1z^2} \le \max \{t_1,t_2,t_3\}$,

where $ t_1,t_2,t_3$ are the roots of the equation $ 4a_1b_1c_1t^3 \minus{} (a_1a_2^2 \plus{} b_1b_2^2 \plus{} c_1c_2^2)t \minus{} a_2b_2c_2 \equal{} 0$.

[b]Theorem 2.[/b] Let $ a_1,b_1,c_1$ be positive real numbers, and $ a_2,b_2,c_2$ real numbers such that $ |a_2| \le 2\sqrt {b_1c_1}$, $ |b_2| \le 2\sqrt {c_1a_1}$, $ |c_2| \le 2\sqrt {a_1b_1}$ and $ 4a_1b_1c_1 \ge a_2b_2c_2 \plus{} a_1a_2^2 \plus{} b_1b_2^2 \plus{} c_1c_2^2$.
Then

$ a_1x^2 \plus{} b_1y^2 \plus{} c_1z^2 \ge a_2yz \plus{} b_2zx \plus{} c_2zx$

for any reals $ x,y,z$.

[b]Theorem 3.[/b] Let $ a_1,b_1,c_1$ be positive real numbers, and $ a_2,b_2,c_2$ real numbers such that  $ 3a_1b_1c_1 \ge a_1a_2^2 \plus{} b_1b_2^2 \plus{} c_1c_2^2$.
Then

$ a_1x^2 \plus{} b_1y^2 \plus{} c_1z^2 \ge a_2yz \plus{} b_2zx \plus{} c_2zx$

for any reals $ x,y,z$.


I wanna open a topic that deserves plenty of thought and interest: Are humans capable of communicating thoughts or ideas to animals? Can animals put themselves in our shoes? Moreover, can household pets such as dogs, cats, ferrets etc. communicate amongst themselves? Examples please :)

To kick it off, I do believe that a "link" can exist between a person and an animal, if they live together for long enough. I guess I believe in nurture rather than nature in this respect. Even though humans are the only animal observed with the ability to possess a vocal language, I feel like I can communicate with my dog, and vice versa, by gesturing, moving around, or changing the tone of my voice.

For those that are unaware of Koko the gorilla, I recommend checking out a video or catching a tv special about her; she is able to actually speak to humans using sign language. Very cool.

Of course they can communicate.  Although they may not communicate verbally like we do, they use a lot of body language.  A person's countenance, tone, and body language usually communicate than actual words.  Of course, many creatures communicate using mainly instincts (IE, dogs and their pack instincts).  But...yeah, i've put across my point.  If you wanted to repel someone, an angry dog snarl says much more than words.

As for communicating with animals...you can't communicate with words really...depends.  When you say something to a dog, the dog doesn't listen for words.  The dog looks for eye contact, tone and volume, expression and body language, and direction.

I agree with K81o7. Obviously they communicate with each other... generally, the more advanced the species, the more advanced their way of communication.


Each side and each diagonal of a hexagon is painted in red or blue.

A triangle which is defined by three vertices of the hexagon is called [b]monochromatic[/b] if all of its sides are of the same color.

a-) Please prove that there exist at least two monochromatic triangles.

b-) If an n-gon is considered, what is the maximum number of monochromatic triangles


I'm assuming the second term on the right should be $\frac{c+a}{b}$.

(There's probably a much nicer way than this, but here goes...) Put $a = y+z, b = z+x, c = x+y$ with $x,y,z > 0$. The desired is equivalent to

$\sum_{cyc} \frac{y+z}{x}-\frac{2x+y+z}{y+z} \ge 0$

or, upon clearing denominators

$\sum_{sym} x^4y^2 + x^3y^3 - x^3y^2z - x^2y^2z^2 \ge 0$

which is evident by Muirhead.

-----

Edit: Ok, there is a nicer way. Observe that $\frac{y+z}{x} + \frac{z+x}{y} \ge 2\cdot \frac{2z+x+y}{x+y}$ is equivalent to $4(x-y)^2(x+y+z) \ge 0$, which is obviously true. The desired follows from adding this lemma cyclicly.

[quote="cezar lupu"]Let $a,b,c$ sides of a triangle.Prove that:

$\frac{2a}{b+c-a}+\frac{2b}{c+a-b}+\frac{2c}{a+b-c}\geq\frac{b+c}{a}+\frac{c+a}{a}+\frac{a+b}{c}$ ;)[/quote]
I used cauchy, in fact it's quite simple
$\frac{2a}{b+c-a}+\frac{2b}{c+a-b} \ge \frac{2(a+b)^2}{(a+b)c-(a-b)^2} \ge \frac{2(a+b)^2}{(a+b)c} = \frac{2(a+b)}{c}$
Sum up and done.


function
real analysis
real analysis unsolved

Let functions $ f,\,g: \left[{0,\,1}\right]\longrightarrow\mathbb{R}$ continuous on  $ \left[{0,\,1}\right]$,   such that   $ f\left({\left[{0,\,1}\right]}\right)\equal{}g\left({\left[{0,\,1}\right]}\right)\equal{}\left[{0,\,1}\right]$  and  $ \left({f\circ{g}}\right)(x)\equal{}\left({g\circ{f}}\right)(x)$, for every $ x\in\left[{0,\,1}\right]$.
If $ f$ is strictly increasing, prove that there exist  $ x_0\in\left[{0,\,1}\right]$, such that  $ f(x_0)\equal{}x_0\equal{}g(x_0)$.

The set of fixed points $ X$ of $ g$ is not empty and compact.
It is easy to see that $ f(X)$ is a (compact) subset of $ X$.
Then $ fof(X)$ is a (compact) subset of $ f(X)$.
Then $ \cap_k f^k(X)$ is compact (and not empty).
So it has a smallest element, say $ x$.
$ f$ is increasing, so $ f(x)$ has to be the smallest element of $ f(\cap_k f^k(X)) \equal{} \cap_k f^k(X)$.
Thus $ f(x)\equal{}x$ and $ g(x)\equal{}x$.

A solution based in more elementary (than those of PeterP) maths is given below:
____________________________________________________________________________________________

Because  $ f$ is strictly increasing and  $ f\left({\left[{0,\,1}\right]}\right)\equal{}\left[{0,\,1}\right]$, then  $ f(0)\equal{}0$  and $ f(1)\equal{}1$.
If   $ g(0)\equal{}0$  or  $ g(1)\equal{}1$, then  $ g(0)\equal{}0\equal{}f(0)$  or  $ g(1)\equal{}1\equal{}f(1)$.
If   $ g(0)\neq0$  and  $ g(1)\neq1$, because $ g\left({\left[{0,\,1}\right]}\right)\equal{}\left[{0,\,1}\right]$, then  $ g(0)>0$  and  $ g(1)<1$.
For the continuous in $ \left[{0,\,1}\right]$ function  $ g$ there is$ ^{(*)}$  $ x_0\in\left({0,\,1}\right)$, such that $ g(x_0)\equal{}x_0$.
a) Suppose that there is unique $ x_0\in\left({0,\,1}\right)$,  such that $ g(x_0)\equal{}x_0$ (1). Then  $ g\left({f(x_0)}\right)\equal{}f\left({g(x_0)}\right)\stackrel{(1)}{\equal{}}f(x_0)$  $ \stackrel{(1)}{\Rightarrow}$  $ f(x_0)\equal{}x_0$.
b) Suppose that there are more than one $ x\in\left({0,\,1}\right)$, such that $ g(x)\equal{}x$ and $ \rho$ is one of the roots of the equation $ g(x)\equal{}x$ (2).
Then  $ g\left({f(\rho)}\right)\equal{}f\left({g(\rho)}\right)\stackrel{(2)}{\equal{}}f(\rho)$  $ \stackrel{(2)}{\Rightarrow}$  $ f(\rho)\equal{}\rho\equal{}g(\rho)$  (A) or  $ f(\rho)\equal{}\sigma\equal{}g(\sigma)$  (B),  where $ \sigma$ is another root of the equation (2),  $ \sigma\neq\rho$.
Let  $ {\rm{J}}\equal{}\left\lbrace{\rho_i\,: \,i\in{\rm{I}}}\right\rbrace$ the set of the roots of the    (2) with the property (B)  and  $ \rho_1\equal{}{\rm{min\,J}}$.
Then every root $ \rho_i\in{\rm{J}}$,  $ i\in{\rm{I}}$,  has value  $ f(\rho_i)\equal{}\rho_j$,  $ \rho_i\neq\rho_j$, into$ ^{(**)}$ set $ {\rm{J}}$   and there exists $ \rho_i,\rho_j$ in $ {\rm{J}}$, such that:
 $ f(\rho_1)\equal{}\rho_i>\rho_1$  and  $ f(\rho_j)\equal{}\rho_1<\rho_j$  (3).
Then  $ \rho_1<\rho_j\ \stackrel{f\,\nearrow}{\Rightarrow}\ f(\rho_1)<(\rho_j)$ $ \stackrel{(3)}{\Rightarrow} \rho_i<\rho_1$. Contradiction.
So every root of the equation $ g(x)\equal{}x$ has the property (A). $ \square$
$ (*)$ For the continuous in $ \left[{0,\,1}\right]$ function  $ h(x)\equal{}g(x)\minus{}x$, we have $ h(0)\cdot{h(1)}\equal{}g(0)\cdot(g(1)\minus{}1)<0$. From Bolzano's theorem, there is  $ x_0\in\left({0,\,1}\right)$,  such that  $ h(x_0)\equal{}0$.
$ (**)$ If there is $ \rho_i\in{\rm{J}}$ such that $ f(\rho_i)\notin{\rm{J}}$, then from (B) we get  $ f(\rho_i)\equal{}\rho\equal{}f(\rho)$  $ \stackrel{f\,1\minus{}1}{\Rightarrow}$ $ \rho_i\equal{}\rho\equal{}f(\rho_i)\in{\rm{J}}$. Contradiction.
___________________________________________________________
P.S. The solution of PeterP is more general: $ f$ is just increasing and $ f\left({\left[{0,\,1}\right]}\right)\equal{}g\left({\left[{0,\,1}\right]}\right)$

[b]Actually $ f([0,1] \equal{} g([0,1]) \equal{} [0,1]$ is not necessary [/b] :D
Let $ B$ be such that $ B \equal{} \{b \ | g(b) \equal{} b\}$ . Manifestly $ B$ is non empty (see the function $ g_{(x)} \minus{} x$) , and $ \forall _{b\in B}, f(b)\in B$ because of $ g(f(b)) \equal{} f(g(b)) \equal{} f(b)$, so $ \forall_n f^{(n)}(b)\in B$
Take for ex $ b \leq f(b)$ : $ b \leq f(b) \Longrightarrow f(b)\leq f^{(2)}(b) \Longrightarrow ..... f^{(n)}(b)\leq f^{(n \plus{} 1)}(b) ...$
The sequence  $ f^{(n)}(b)$ is convergent (because monotonic and bounded) : $ f^{(n)}(b) \rightarrow L$  so $ \ f^{(n \plus{} 1)}(b) \equal{} f(f^{(n)}(b)) \rightarrow f(L)$, so $ f(L) \equal{} L$
(analogous way if $ b \geq f(b)$)
But $ B$ is closed so $ L\in B$ and finally $ f(L) \equal{} L \equal{} g(L)$ .. (q.e.d)
 :cool:


inequalities
induction
calculus
function
logarithms
Euler
inequalities unsolved

Prove the AM-GM Inequality without using mathematical induction.

It can be proven through lots of methods. :D
One of them is by using jensen's ineq.....

Try to prove it with inequalities weaker than AM-GM not stronger like Jensen!! :roll:

muirhead...lol  :D  :blush:

Well maybe rearrengement is consider weaker because its so easy to prove and if you take $x_i^n=a_i$ then is rearrement a lot of times

The trick is to prove that without using AM-GM, of course.  I don't immediately see a way.

In general, I find threads of this nature quite useless.  I do know a proof (due to Polya, I believe) of AM-GM which uses the inequality $\exp x\geq x+1$ repeatedly and avoids induction, but the inductive one is more natural, and more standard.  Moreover, even if you invoke Jensen's inequality, you usually use an induction to prove Jensen's inequality as well.

Yes, but if you're going to avoid using induction, then presumably you would want to avoid using lagrange multipliers as well.

[quote="Peter VDD"][quote="mmahdi"]hi
there is a lemma that says:"if we had $a_{1}a_{2}...a_{n}=1$ then we will have this inequality:$a_{1}+a_{2}+...+a_{n}\geq{n}$.and it's ok when you consider $a_{i}=\frac{a_{i}}{\sqrt[n]{a_1a_2...a_n}}$.wasn't it nice?bye[/quote]yeah, and the proof of that lemma is exactly by AM-GM ;)

I don't know why you call Jensen "stronger" than AM-GM, in some sense it is a generalisation, but on the other hand it is much easier to prove Jensen than to prove AM-GM without induction and without Jensen... ;)[/quote]
156532654

you can prove it by:
1. Simple calculus of a function,
2. Logarithm of Euler number.

Bomb

Using the fact that
  if x + t = y + z and  0<x<y,z<t, then xt < yz.

[quote="Peter VDD"][quote="mmahdi"]hi
oh guys.you know that i answered you.now try on another problem.bye[/quote]you did not answer anything... you just say you can prove a completely equivalent theorem with induction. which is not in any way an answer to the question.[/quote]
hi
i didn't want to make you angry but i myself have its proof.but if you want to say that i hadn't done anything,you can find another solution and waste your time on it.i said:"try on another problems".because i like you to get progess in inequalities.anyway that's not so important.bye

[quote="Peter VDD"][quote="mmahdi"]i didn't want to make you angry but i myself have its proof.but if you want to say that i hadn't done anything,you can find another solution and waste your time on it.[/quote]I didn't say you don't have a solution, I just said you haven't posted it. :)

"you didn't solve" = "you didn't solve in your post here"[/quote]
hi
i knew the solution.i didn't write it completely because i wanted you to try it yourself.but myself solved before.bye

Prove it using the rearrangement inequality.

Can prove with a triangle. If you choose middle point and surely must describe a function of total of right sides distances. And you can obtain it with geometrically.


The trick is in factoring the expresion:

[hide](1)$8xy-12y+2x-3=0$
(2)$(4x-6)(2y+\frac{1}{2}) =0$

You want to know for $x$, therefore:

(3)$4x-6=0$
(4)$x= \frac{3}{2}$

That is [b](D)[/b][/hide]


2^200-2^192*31+2^n find n for which this expression is perfect square

[hide="Squares"]
$ 2 \left( 192 \cdot 31 \minus{} 1 \minus{} 200 \right) \equal{} 11504 \equal{} n$
[/hide]

Already done  [url]http://www.mathlinks.ro/viewtopic.php?t=223689[/url]

Me too. Tried and was forced to quit..... Anyways the problem has been solved by Euclid.


If $ x\plus{}\frac1x\equal{}\minus{}5$, what is $ x^3\plus{}\frac1{x^3}$?

$ (\minus{}5)^3 \equal{} \left(x \plus{} \frac{1}{x}\right)^3 \equal{} \left(x^3 \plus{} \frac{1}{x^3}\right) \plus{} 3\left(x \plus{} \frac{1}{x}\right) \equal{} \left(x^3 \plus{} \frac{1}{x^3}\right) \plus{} 3 \times (\minus{}5)$.
Thus, $ x^3 \plus{} \frac{1}{x^3} \equal{} (\minus{}5)^3 \minus{} 3 \times (\minus{}5) \equal{} \minus{}125 \plus{} 15 \equal{} \boxed{\minus{}110}$.

NOOOOO I forgot to put the negative sign in my answer :( RIP


Let $x$, $y$, $z$ be positive reals such that $x+y+z=1$. Prove that

$\frac{x}{y^{2}+z}+\frac{y}{z^{2}+x}+\frac{z}{x^{2}+y}\geq\frac{9}{4}$.

I don't if I am right, but here's a solution. :)

I think tha this inequality is same as inequality from Serbia and Montenegro Federal competition 2006. (1.st problem for 3.rd grades)

[quote="arqady"]$\frac{x}{y^{2}+z}+\frac{y}{z^{2}+x}+\frac{z}{x^{2}+y}\geq\frac{(x+y+z)^{2}}{\sum(x^{2}z+xy)}.$
But $\frac{(x+y+z)^{2}}{\sum(x^{2}z+xy)}\geq\frac{9}{4}\Leftrightarrow\sum_{cyc}(4x^{3}+3x^{2}y-6x^{2}z-xyz)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x^{3}-xyz)+3\cdot\sum_{cyc}(x^{3}+z^{2}x-2x^{2}z)\geq0.$[/quote]
sorry, I'm bad at inequalities. can you please explain your steps?

[quote="me@home"]can you please explain your steps?[/quote]
For you me@home  :)
By Cauchy-Shwarz inequality
$\frac{x}{y^{2}+z}+\frac{y}{z^{2}+x}+\frac{z}{x^{2}+y}=$
$=\frac{x^{2}}{y^{2}x+xz}+\frac{y^{2}}{z^{2}y+xy}+\frac{z^{2}}{x^{2}z+yz}\geq\frac{(x+y+z)^{2}}{x^{2}z+y^{2}x+z^{2}y+xy+xz+yz}.$
Id est, remain to prove that
$\frac{(x+y+z)^{2}}{x^{2}z+y^{2}x+z^{2}y+xy+xz+yz}\geq\frac{9}{4}\Leftrightarrow$
$\Leftrightarrow4(x+y+z)^{3}\geq9(x^{2}z+y^{2}x+z^{2}y)+9(x+y+z)(xy+xz+yz)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(4x^{3}+3x^{2}y-6x^{2}z-xyz)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x^{3}-xyz)+3\cdot\sum_{cyc}(x^{3}+z^{2}x-2x^{2}z)\geq0.$
And $\sum_{cyc}(x^{3}-xyz)=x^{3}+y^{3}+z^{3}-3xyz=$
$=\frac{1}{2}\cdot(x+y+z)((x-y)^{2}+(x-z)^{2}+(y-z)^{2})\geq0,$
$\sum_{cyc}(x^{3}+z^{2}x-2x^{2}z)=\sum_{cyc}x(x-z)^{2}\geq0.$

Thank you very much.
I was also wondering how you became so good at inequalities, it seems like many random steps or is it experience?

[quote="arqady"][quote="me@home"]
I was also wondering how you became so good at inequalities, it seems like many random steps or is it experience?[/quote]
10 months ago, I could not prove it. Now I think, that this very easy inequality.
At me very good teachers:
vasc, hungkhtn,harazi,myth,Namdung,Virgil Nicula,Peter Scholze,darij grinberg,Soarer,
zhaobin,Anh Cuong,Tiks,Albanian Eagle,Michael,Darkseer,Parkdoosung,
Sung-yoon Kim,Nima Ahmadi Pour,Ramtin Madani,mathematica,
yptsoi,Sasha,campos,Fedor Petrov,nthd,nhat,spanferkel,
Nameless,jarod,Gibbenergy,Michael Niland,silouan,fuzzylogic,
rogue,pvthuan,leepakhin,nickolas,nttu,mecrazywong,nsato,
flip2004,Lovasz,billzhao,Mildorf,Xevarion,fedja,Lagrangia,
manlio,shyong,Ravi B,perfect_radio,Cezar Lupu,xtar,
Minh Can,ductrung,mateivld,Mircea Lascu.
The order not necessarily speaks about my preferences.
Thanks huge  and that whom I have forgotten to mention.[/quote]
I do not deserve to be your teacher at all. In fact, I haven't been actively posting in the Inequality Forum for almost an year.

[quote="arqady,silouan,.[/quote]

Dear Arqady you said that I was a teacher of you....... :blush:
but I think you are my teacher so ....... :D

EDIT:
See here http://www.mathlinks.ro/Forum/viewtopic.php?t=78058
also . I gave a solution in the last post

[quote="mecrazywong"]
I do not deserve to be your teacher at all. In fact, I haven't been actively posting in the Inequality Forum for almost an year.[/quote]
This your problem
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14165
This problem has learned me to much. I for the first time have saw this problem from you. :wink:
[quote="silouan"]
Dear Arqady you said that I was a teacher of you....... :blush:
but I think you are my teacher so ....... :D
[/quote]
This problem
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63850
I shall not forget never. :wink:

[quote="arqady"]
By Cauchy-Shwarz inequality
$\frac{x}{y^{2}+z}+\frac{y}{z^{2}+x}+\frac{z}{x^{2}+y}=$
$=\frac{x^{2}}{y^{2}x+xz}+\frac{y^{2}}{z^{2}y+xy}+\frac{z^{2}}{x^{2}z+yz}\geq\frac{(x+y+z)^{2}}{x^{2}z+y^{2}x+z^{2}y+xy+xz+yz}.$
[/quote]

Sorry to be even less adept, but how does this line work? Why do the denominators just sum? You've no doubt multiplied by something not obvious to me. Thanks.

Doh - I really should have worked that out :blush: (oh well, it's the holidays). Thanks very much :D


i have an event for sciencce olympiad called simple machines.ive been in it before but the rules are modified this year. we have to build a simple machines to have a target AMA IMA and/or efficiency. could someone help me as far as AMA.

actual machanical advantage and ideal mechanical advantage. most people just refer to the AMA as mechanical advantage.


limit
induction
function
algebra
domain
topology
real analysis

Hi.

I tried to solve the following and I couldnt :(

Prove that if $X_{n}$ is a bounded series defined: $\lim_{0\to\infty}(X_{n+1}-X_{n})=0$

Then the partial limits of the series fill the entire interval $[l,L]$

where

 $L=\limsup_{n\to\infty}X_{n}$

$l=\liminf_{n\to\infty}X_{n}$

[quote="Kalle"]First of all, you mean a [b]sequence[/b] $(X_{n})_{n=0}^\infty$. A series is a generalized sum.

Secondly... what is a partial limit? I have never heard of it. Does the question mean that for any $x\in [l,L]$ we can choose indices $\lambda_{1}< \lambda_{2}< \lambda_{3}< ...$ such that $\lim_{n\to\infty}X_{\lambda_{n}}= x$ ?[/quote]
I think so. :)
And all we should do is to prove:
For any arbitrary $e>0$,$N$,there exist $n>N$,such that $|x_{n}-x|<e$

The "partial limits" are limit points, or limits of subsequences.

yea limits of subsequences.

Sorry for using the word series I wasnt sure whats the exact term for it in english.

Anyhow, anyone have any idea how to prove it or even better "how do I approach" this problem  :)

1. Let's not make things too difficult in the "computations and tutorials" section. I'm moving this problem over to "proposed and unsolved problems."

2. As a little bit of a hint: it's very important that this sequence be real-valued. It wouldn't work for complex-valued sequences or for $\mathbb{R}^{n}$-valued sequences for $n>1$ ...

... unless you wanted to tackle some possible generalization about the set of cluster points (limits of subsequences) being connected.

I'll use "cluster point" to mean what RealTime meant when he said "partial limit": limits of subsequences of the given sequence.

Note that $\ell$ and $L$ are both cluster points of the sequence.

Suppose that $\exists\,y\in(\ell,L)$ such that $y$ is not a cluster point of the sequence. Then there exists $\epsilon>0$ such that at most finitely many members of the sequence lie in the interval $(y-\epsilon,y+\epsilon).$

Consider the sets $A=[\ell,y-\epsilon]$ and $B=[y+\epsilon,L].$ Choose $N_{1}$ such that if $n>N_{1},$ then $|X_{n+1}-X_{n}|<2\epsilon.$ Choose $N_{2}$ such that if $n>N_{2},$ then $X_{n}\not\in(y-\epsilon,y+\epsilon).$

Suppose $n>\max(N_{1},N_{2}).$ Then $X_{n}\in A$ implies $X_{n+1}\in A$ and $X_{n}\in B$ implies $X_{n+1}\in B.$

By induction, this implies either that all but finitely many elements of the sequence belong to $A$ or that all but finitely many of them belong to $B.$ Then either all the cluster points would be in $A$ or all the cluster points would be in $B.$ But this cannot be true, since both $\ell$ and $L$ are cluster points.

Hence the assumption that $y$ is not a cluster point must be false.

Claim: Suppose $\{x_{n}\}$ is a sequence in a compact metric space and that $\lim_{n\to\infty}d(x_{n+1},x_{n})=0.$ Let $S$ be the set of all cluster points (limits of subsequences) of this sequence. Then $S$ is a connected set.

Note: in the proof I have in mind, I use compactness twice: first, so show that $S$ is nonempty, and second, to show that the distance between two disjoint closed sets is positive. Do I really need compactness (or, at least that the closure of the range of the sequence is compact) for the theorem?

Is a subsequence allowed to consist of only finitely many numbers?

A subsequence may have a finite range, but it necessarily has an infinite domain. (Remember that a sequence is simply a function with domain $\mathbb{N}.$) We will count a value repeated infinitely often as a cluster point of the sequence. That is one reason I am using the words "cluster point" rather than "limit point."


Assume that $ a_1,a_2,\ldots,a_n$ are positive numbers, $ n\in N$.Prove that
$ (\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} \cdots \plus{} \frac {1}{a_n})a_1a_2\cdots a_n\leq\frac {(a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n)^{n \minus{} 1}}{n^{n \minus{} 2}}$.


calculus
derivative
integration
trigonometry
function
algebra
rational function

is possible if someone could teach me how to do derivative and integral problems?

Thanx a lot!! :)

Do you think that it's quickly and quite easy to learn? Are you learning by yourself?

Buy a book, an introductory to Calculus one, may be. Unless you be more specific, we can't do anything further.

i get the general format of the derivative and integrals. the thing that i need is the trig function derivatives and integrals w/o a calculator.

Also, how do you find the derivative of a rational function?

Read this:

[url]http://mathworld.wolfram.com/Derivative.html[/url]

Should help!


algebra
polynomial
complex numbers
algebra unsolved

Let $ f(x) \equal{} x^{99} \plus{} x^{98} \plus{} x^{97} \plus{} ... \plus{}x^2 \plus{} x \plus{} 1$. Determine the remainder when $ f(x^{100})$ is divided by $ f(x)$.

Did you mean the polynomial remainder? [hide="in that case"]
$ 100^{99}(1-x)(1-x^{100})$
[/hide]

Why is it so, JoeBlow? I think rust is correct.

Aliter:(though not as elegant as Rust)

$ f(x^{100}) \equal{} (x^{100})^{99} \plus{} (x^{100})^{98} \plus{} ... \plus{} 1 \equal{} [(x^{99})^{100} \minus{} 1] \plus{} [(x^{98})^{100} \minus{} 1] \plus{} ... \plus{} [x^{100} \minus{} 1] \plus{} 99 \plus{} 1$

Each of the bracketed terms is divisible by $ (x^{100} \minus{} 1)$ and hence by f(x) Hence the remainder is 100.

Aliter (using complex numbers):

$ f(x) \equal{} x^{99}\plus{}x^{98}\plus{}...\plus{}1 \equal{} (x\minus{} \omega) (x\minus{}\omega^2)...(x\minus{}\omega^{99})$ where $ \omega, \omega^2$ etc. are the non-real roots of the equation $ z^{100}\equal{}1$

$ f(z^{100}) \equal{} f(z) P(z) \plus{} R(z) \equal{} (x\minus{} \omega) (x\minus{}\omega^2)...(x\minus{}\omega^{99}) P(z) \plus{} R(z)$

By, substituting $ z \equal{} \omega, z \equal{} \omega^2 ...$ in succession, we obtain,

$ R(\omega) \equal{} R(\omega^2) \equal{} ....\equal{}R(\omega^{99}) \equal{} 100$

Consider $ G(z) \equal{} R(z) \minus{} 100. Deg(G(z)) \equal{} Deg(R(z)) \le 98$

However G(z) = 0 for 99 values of z. This means that G(z) = 0 for all values of z $ \Rightarrow$ R(z) = 100 for all values of z


geometry
analytic geometry
area of a triangle

For $ x > 0$, the area of the triangle with vertices $ (0, 0), (x, 2x)$, and
$ (x, 0)$ is 64 square units. What is the value of $ x$?

The area of a triangle is half the product of the base and the height. Notice that the base is the segment connecting $ (0,0)$ and $ (x,0)$, since they share a $ y$-coordinate, and the segment representing the height is the one connecting $ (x,0)$ to $ (x,2x)$. Let's plug these values into the triangle area formula and solve for $ x$:

$ \frac{1}{2}(x)(2x)\equal{}64$
$ x(2x)\equal{}128$
$ 2x^2\equal{}128$
$ x^2\equal{}64$
$ x\equal{}\fbox{8}$ (notice that $ x\equal{}\minus{}8$ is outside of the restriction that $ x > 0$)

$ \text{area} \equal{} \frac {1}{2}|x \cdot 0 \minus{} 2x \cdot x| \equal{} x^2 \equal{} 64 \iff x \equal{} \boxed{8}$.

[b]Remark[/b]
The area of the triangle with vertices $ (0, 0)$, $ (a, b)$ and $ (c, d)$ is $ \frac {1}{2}|ad \minus{} bc|$ square units.


[quote="236factorial"]1) Which one would you expect to have smaller bond angles and explain why:

$SiF_{4}$ or $SF_{4}$

2) What do books mean by "describing hybridization"? (ie. how detailed?)
For example, how would you describe the hybridization of the central atom in

$IF_{4}^{-}$?[/quote]

the central atom of $SiF_{4}$ is $Si$ and has $4$ electron valance around it . Each one form a $\sigma$ bond with $F$ and so it is tetrahedron (bond angle $109.5^\circ$).

$SF_{4}$ has $4$ $\sigma$ bond and $1$ lone pair .So it is see-saw shape . The bond angle are $<120^\circ$ and $<90^\circ$ .So I dont think you can distingush who has smaller bond angle .

you can also refer here

[url]http://en.wikipedia.org/wiki/VSEPR[/url]

For 2) question i think is $I$ expands its octet by hybridising the $p$ and $d$ orbitals .But im not so sure .

[quote="shyong"]For 2) question i think is $I$ expands its octet by hybridising the $p$ and $d$ orbitals .But im not so sure .[/quote]

I believe you are right about the hybridization of the molecule ( $sp^{3}d$ orbitals).

However, I'm wondering, is that all one has to write if it were a test question?

That should be fine, but maybe say how you arrived at your answer..

Should be $sp^{3}d^{2}$.  There are two lone pairs on the $I$.


function geometry calculus integration limit logarithms trigonometry

let $a_{1}=\sqrt{2}$   and  sequence
$a_{n+1}=\sqrt{ \frac{2a_{n}}{1+a_{n}}} \forall n \geq 1$
      let$S_{n}= a_{1}.a_{2}...a_{n}$
         find  $limS_{n}$ when $n \to \ infty$

Sorry, but what does the upside-down A mean?

$\forall$ means 'for all', so the property has to hold for all positive integers $n$ here.

It is not difficult to verify that {a_n} is bounded and monotone, and thus convergent. It´s just consider f(x)=sqrt[2x/(x+1)]-x and prove that f(x)<0. Just look to f´(x) and verify that f´(x)<0, for all x>1. And we can prove that this sequence is bounded bellow by induction. Is easy to view that lim a_n=1.
   To find lim S_n, consider the function g(x)=ln(x)/{sqrt[2x/(x+1)]-x}, prove that g(x) is monotonic through evaluate g´(x) and consider a partition of the form [a_n,a_(n-1),a_(n-2),...,a_2,a_1]. Compare the areas bellow and abouve the curve and state that lim S_n=integral[1,sqrt(2)]-g(x)dx.

In fact, is necessary a correction because ln[S_n] -> integral[1,sqrt(2)]-g(x)dx and thus S_n -> e^{integral[1,sqrt(2)]-g(x)dx}.

For some $n,$ assume that $a_n>1,$ we have $a_{n+1}=\frac{2a_n}{1+a_n}>1\Longleftrightarrow a_n>1.$

By this result and $a_1=\sqrt{2},$ yielding $a_n>1.$ for all positive integers inductively.

Use the given recursive formula, $a_{n+1}-1=\frac{\sqrt{2a_n}}{\sqrt{1+a_n}}-1=\frac{\sqrt{2a_n}-\sqrt{1+a_n}}{\sqrt{1+a_n}}=\frac{a_n-1}{\sqrt{1+a_n}(\sqrt{2a_n}+\sqrt{1+a_n})}.$

Thus from $a_n>1,$ we obtain $0<a_{n+1}-1<\frac{1}{4}(a_n-1).$.

Hence $0<a_n-1<\frac{1}{4}(a_{n-1}-1)<\left(\frac{1}{4}\right)^2 (a_{n-2}-1)<\cdots <\left(\frac{1}{4}\right)^{n-1}(a_1-1),$ yielding $\lim_{n\to\infty} a_n=1.$

Consequently $\lim_{n\to\infty}S_n=1.$

[quote="kunny"]Consequently $\lim_{n\to\infty}S_n=1.$[/quote]

How can this be?  $a_1 = \sqrt{2}$ and $a_n > 1$, so obviously $S_n > \sqrt{2}$.

Oops! :blush: you are right.

Is the answer $\sqrt{2}$?

Hi guys. What´s wrong with my demo? I think that is write and the answer is e^{integral[1,sqrt(2)]^(-g(x))dx}, where g(x)=ln(x)/{sqrt[2x/(x+1)]-x} . And in fact this value isn´t sqrt(2).

In fact, it´s easy to check that lim a_n=1. I have indicate how to prove that {a_n} in monotonic and bounded, so there is lim a_n and from the recursive formula we get lim a_n=1.
   The problem is obtain lim S_n, and for this I considerate the function g(x), which is monotonic and I compared the areas above and below the curve and prove my answer.

Okay, so if I understand you correctly, you're taking $\lim_{n \to \infty} \ln S_n = \sum_{k = 1}^{\infty} \ln a_k$ and approximating with an integral, right?

That seems to work.  But would you mind providing a numerical answer?   :)

I have mentioned repeatedly, though in Japan, High school students don't study $\epsilon-\delta$ for university entrance examination, that's why we need to prove in using Archimedes' axiom (sandwich or squeezing theorem) or
Mean value theorem.

[quote="mffmartinelli"]Really I had prove that 1.49884553<lim S_n<1.62658131. I made one analogize that were improper and i can´t find the exact expression to this limit. I had to use another function to aproximate by bellow and this two functions don´t converge to the same limit.[/quote]
       lim S_n = pi/2

Finally I found the limit. I will post some tricks that I have made:

1) First define b_k=1/a_k for all k. Thus [b_(k+1)]^2=b_k/2+1/2.
2) And now define b_k=c_k+1/(4*c_k) for all k, whit the property that c_k=(1/2)*a^[1/2^(n-1)] and a is a complex in the form a=(1+i)/sqrt(2). Notice that I could find a closed formula for a_n and now it´s just obtain S_n, with a closed formula to.

Really I found now that lim S_n=pi/2, and it´s in accordance with my previous result, where I have proved that 1.49884553<lim S_n<1.62658131. But my first aprroach to this problem doesn´t work, because I couldn´t find a expression, just a inequalitie.

Tranngoctrung, how do you prove this?

That´s a nice problem.

It is quite simple with some help from dr. Trig :)
Observe that if $a_n=\frac{1}{\cos\alpha}$ then $a_{n+1}=\frac{1}{\cos\frac{\alpha}{2}}$. It's pretty easy from here.
I'm sure it was posted before, but I couldn't find it.


Discuss the equation below; that is, find the center, foci and vertices of the given ellipse.

9x^2 + y^2 - 18x = 0

[hide="hint"]Try to get this in the form, by completing the square, $ \frac {(x \minus{} h)^{2}}{a^{2}} \plus{} \frac {(y \minus{} k)^{2}}{b^{2}} \equal{} 1$ where $ (h, k)$ is the center, and foci $ (h, k \pm c)$ (this is because the major axis is vertical), where $ c \equal{} \sqrt {b^{2} \minus{} a^{2}}$ (usually you will see it the other way around because $ a$ is usually defined as the greater of the two, but in this problem $ b > a$), and for the vertices they will be $ (h, k \pm b)$ (since the major axis is vertical)[/hide]

See if you can do it now.

[hide]You have $ 9x^2 \minus{} 18x \plus{} y^2 \equal{} 0$.
Factoring, we have
\[ 9\left(x^2 \minus{} 2x\right) \plus{} y^2 \equal{} 0
\]
Completing the square on $ x$, we have
\[ 9{\left(x \minus{} 1\right)}^2 \plus{} y^2 \equal{} 9
\]
Dividing through by $ 9$, we have
\[ \frac {{\left(x \minus{} 1\right)}^2}{1^2} \plus{} \frac {{\left(y \minus{} 0\right)}^2}{3^2} \equal{} 1
\]
From that, we can we that the center is $ (1,0)$, $ a \equal{} 1$, and $ b \equal{} 3$. Then, $ c \equal{} \sqrt {3^2 \minus{} 1^2} \equal{} 2\sqrt {2}$. Therefore, the foci are $ (1, 2\sqrt {2})$ and $ (1, \minus{} 2\sqrt {2})$
Finally, the vertices will be $ (1, 3)$ and $ (1, \minus{} 3)$.[/hide]


Solve $ x^2(y\minus{}1)\plus{}y^2(x\minus{}1)\equal{}1$ in the set of integers.

let $ x\plus{}y\equal{}a$,$ xy\equal{}b$ and solve in $ a$ and use the fact that the descriminant has to be a perfect square... :wink:
Lefteris

the generalization of the problem was posted some months ago ,
$ x^2(y^n\minus{}1)\plus{}y^2(x^n\minus{}1)\equal{}1$
http://www.mathlinks.ro/viewtopic.php?t=296827


Find the set of all $x$ such that $|x^{2}+1| < |x^{2}-9|$.


These are a couple of problems I am having with my blog. :maybe:  :(

>I have 6 posts on my blog, and it reads 0 entries.
>It says I started the blog today at 5:53. That is not true; I started it about 5 days ago. :blush:

I believe all of the bugs that you posted have already been reported and they're working on them. also, there's a big thread/sticky that's titled blog bugs and to do list. all of these posts should be in there.


For any three real numbers $a$, $b$, $c$ from the interval $\left[0;\ 1\right]$, prove the inequality

$a\left( 1-a\right) \left( 1+b\right) \left( 1+c\right)+b\left( 1-b\right) \left( 1+c\right) \left( 1+a\right)+c\left( 1-c\right) \left( 1+a\right) \left( 1+b\right)$
$\geq \frac{\left( \left( a+b+c\right)-\left( a^{2}+b^{2}+c^{2}\right) \right)^{2}\left( 1+a\right) \left( 1+b\right) \left( 1+c\right) }{a+b+c-\left( a^{3}+b^{3}+c^{3}\right) }$.

I guess the answer is no :(
I don't have such an artistic touch to make you like it....

I'm glad you like it! So, what do you think about it? Is is easy?

It's not a bad inequality, at least in the form divided by (1 + a) (1 + b) (1 + c), but it's very simple... it's just a standard trick to use Cauchy-Schwarz in the Engel form, and the fact that your nickname on this forum even gives one a hint makes the problem almost trivial... Here goes a complete proof:

You claim that for any three numbers a, b, c from the interval [0; 1], the inequality

$a\left( 1-a\right) \left( 1+b\right) \left( 1+c\right)+b\left( 1-b\right) \left( 1+c\right) \left( 1+a\right)+c\left( 1-c\right) \left( 1+a\right) \left( 1+b\right)$
$\geq \frac{\left( \left( a+b+c\right)-\left( a^{2}+b^{2}+c^{2}\right) \right)^{2}\left( 1+a\right) \left( 1+b\right) \left( 1+c\right) }{a+b+c-\left( a^{3}+b^{3}+c^{3}\right) }$

holds.

Well, it will look much less dreadful once divided by (1 + a) (1 + b) (1 + c) :

$\frac{a\left( 1-a\right) }{1+a}+\frac{b\left( 1-b\right) }{1+b}+\frac{c\left( 1-c\right) }{1+c}\geq \frac{\left( \left( a+b+c\right)-\left( a^{2}+b^{2}+c^{2}\right) \right)^{2}}{a+b+c-\left( a^{3}+b^{3}+c^{3}\right) }$.

Equivalently,

$\frac{\left( a\left( 1-a\right) \right)^{2}}{a\left( 1-a\right) \left( 1+a\right) }+\frac{\left( b\left( 1-b\right) \right)^{2}}{b\left( 1-b\right) \left( 1+b\right) }+\frac{\left( c\left( 1-c\right) \right)^{2}}{c\left( 1-c\right) \left( 1+c\right) }\geq \frac{\left( \left( a+b+c\right)-\left( a^{2}+b^{2}+c^{2}\right) \right)^{2}}{a+b+c-\left( a^{3}+b^{3}+c^{3}\right) }$.

But this follows from the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15125]Cauchy-Schwarz inequality in the Engel form[/url]; in fact, remembering that the numbers a, b, c, 1 - a, 1 - b and 1 - c are nonnegative, we have

$\frac{\left( a\left( 1-a\right) \right)^{2}}{a\left( 1-a\right) \left( 1+a\right) }+\frac{\left( b\left( 1-b\right) \right)^{2}}{b\left( 1-b\right) \left( 1+b\right) }+\frac{\left( c\left( 1-c\right) \right)^{2}}{c\left( 1-c\right) \left( 1+c\right) }\geq \frac{\left( a\left( 1-a\right)+b\left( 1-b\right)+c\left( 1-c\right) \right)^{2}}{a\left( 1-a\right) \left( 1+a\right)+b\left( 1-b\right) \left( 1+b\right)+c\left( 1-c\right) \left( 1+c\right) }$
$=\frac{\left( \left( a+b+c\right)-\left( a^{2}+b^{2}+c^{2}\right) \right)^{2}}{a+b+c-\left( a^{3}+b^{3}+c^{3}\right) }$.

I just see that the case if one of the numbers a, b, c is 0 or 1 must be considered separately, but it is quite easy.

  Darij


Andy's lawn has twice as much area as Beth's lawn and three times as much area as Carlos' lawn. Carlos' lawn mower cuts half as fast as Beth's mower and one third as fast as Andy's mower. If they all start to mow their lawns at the same time, who will finish first?

$ \textbf{(A)}\ \text{Andy} \qquad
\textbf{(B)}\ \text{Beth} \qquad
\textbf{(C)}\ \text{Carlos} \qquad
\textbf{(D)}\ \text{Andy and Carlos tie for first.}$
$\textbf{(E)}\ \text{All three tie.}$

Well, I'm kind of new here, but I'll give this one a go.
[hide]
Let Carlos' lawn equal 2x, Beth's 3x, and Andy's 6x.  Let's say for every x, it takes one more hour to mow your lawn with a standard mower.  So Andy's would take 6 hrs, Beth's 3 hrs, and Carlos' 2 hrs.  Now lets say Andy's mower cuts lawns 6 times faster than a regular one, it would take 1 hour to mow his lawn.  Since Andy's cuts lawns 6 times as fast, Carlos' cuts lawns twice as fast, and Beth's cuts lawns 4 times as fast. Answer:

Andy's: 6/6 hrs = 1 hr
Beth's: 3/4 hrs = 45 minutes
Carlos': 2/2 hrs = 1 hr

Beth finishes first. (B)
[/hide]


inequalities
Euler
function
integration
geometry
rectangle
real analysis

all $ a_n>0$ and $ \sum a_n$ convergent
let $ r_n\equal{}\sum_{m\ge n}a_m$.
prove that for every $ p\in(0,1)$
\[ \sum_{n\equal{}1}^{\infty}\frac{a_n}{r_n^p}
 <
 \frac{1}{1\minus{}p}\left(\sum_{n\equal{}1}^{\infty}a_n\right)^{1\minus{}p}\]

how to prove it?
I think I've seen this before in a book about real series, but I don't remember ttitle and author  :|
probably this is a known fact (due to Hadamard? Cauchy? Euler?)

Well, I did not know this theorem before, but the proof can be done as follows: we can suppose that $ \sum a_n =1$ and consider the function $ f(x)=\frac{1}{(1-x)^p}$, $ S_n =\sum\limits_{m<n} a_m$, $ a_n =S_{n+1}-S_n$. Our function $ f$ is convex and stricly increasing, hence the LHS does not exceed $ \int\limits_0 ^1 f(x)= \frac{1}{1-p}$ (to check this one can  just draw the picture and observe that the LHS is equal to sum of areas of rectangles)


$ f$ from $ N$ to $ N$ and $ f(f(f(n)))\plus{}f(f(n))\plus{}f(n)\equal{}3n$

[hide]If $ f(n) \equal{} f(m)$, then $ f(f(n)) \equal{} f(f(m))$ and $ f(f(f(n))) \equal{} f(f(f(m)))$, which gives $ n \equal{} m$. Thus $ f$ is injective. For $ n \equal{} 1$, $ f(f(f(1))) \plus{} f(f(1)) \plus{} f(1) \equal{} 3$. But $ f(n)\ge1$ for all $ n$, so $ f(1) \equal{} 1$. Assume that there exists $ k$ such that $ f(k)\ne k$. Let $ k$ be the smallest such integer. By injectivity of $ f$, then $ f(k)\ge k \plus{} 1$. So $ f(f(f(k))) < k$ or $ f(f(k)) < k$. Assume the former. Let $ f(f(f(k))) \equal{} a < k$. But we have $ f(f(f(k))) \equal{} f(f(f(a)))$, so $ k \equal{} a$, a contradiction. The latter case is the same. Thus there does not exist $ k$ such that $ f(k)\ne k$. In other words, $ f(n) \equal{} n$ for all $ n$.[/hide]

thank you can anybody give another solution?(my solution is like you)


We define a recursion $ R_n$ by letting $ R_1$ be any simple closed curve and creating $ R_{n\plus{}1}$ by drawing a path starting and ending on the border of $ R_n$ and passing through every region of $ R_n$ exactly once.  Prove that this recursion is always possible.

Seems like you can overlay successive Hilbert curves on a grid and get the recursion you needed, but I'm not sure how that would lead to a proof.

I don't think the Hilbert curve recursion actual satisfies the "exactly once" property.


Given $ a, b, c \geq\ 0$ satisfy $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(ab \plus{} bc \plus{} ca)$.
Prove that: $ \frac {a}{\sqrt {a^6 \plus{} 2b^3c^3}} \plus{} \frac {b}{\sqrt {b^6 \plus{} 2c^3a^3}} \plus{}$$ \frac {c}{\sqrt {c^6 \plus{} 2a^3b^3}} \geq\ \frac {a^2}{a^4 \plus{} b^2c^2} \plus{} \frac {b^2}{b^4 \plus{} c^2a^2} \plus{} \frac {c^2}{c^4 \plus{} a^2b^2}$
 :)

Sorry, I edited my problem. Now, it's true  :)


I am pretty sure this was posted before. I will post a link when I find it.

_____________________________________________________________

Here we go, it's Tournment of the Towns 1997 inequality:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1291


Prove the inequality:

$ 1\plus{}\frac{1}{1!}\plus{}\frac{1}{2!}\plus{}\cdots\plus{}\frac{1}{n!}<3.$

[hide="Or..."]
The infinite series $ \frac 1{0!} \plus{} \frac 1{1!} \plus{} \frac 1{2!} \plus{} \frac 1{3!} \plus{} \dots \equal{} e < 3$, and clearly since $ \frac 1{k!}$ is positive for any $ k$, the finite series $ \frac 1{0!} \plus{} \frac 1{1!} \plus{} \frac 1{2!} \plus{} \frac 1{3!} \plus{} \dots \plus{} \frac 1{n!} < e < 3$ for all positive integers $ n$.[/hide]

[hide="Another proof..."]
[b]Lemma 1.[/b] For all positive integers $ n\geq 4$, $ n!>2^n$.
[i]Proof.[/i] (By induction)

For the base case, we have $ 4!>2^4 \implies 24 > 16$, which is true.
Let $ k\equal{}n$. So our hypothesis is that $ k!>2^k$ for all $ k\geq 4$. So for $ k\plus{}1$, we have on the LHS $ (k\plus{}1)!\equal{}(k\plus{}1)k!$, and on the RHS $ 2^{k\plus{}1}\equal{}2 \cdot 2^k$, and since we already have $ k!>2^k$ and $ (k\plus{}1)>2$, it follows that $ (k\plus{}1)!>2^{k\plus{}1}$, so $ n!>2^n$ for all positive integers $ n$.

From the above, it follows that $ \frac 1{n!}<\frac 1{2^n}$ for all positive integers $ n\geq 4$. So $ \frac 1{0!} \plus{} \frac 1{1!} \plus{} \frac 1{2!} \plus{} \frac 1{3!} \plus{} \frac 1{4!} \plus{} \dots < \frac 1{0!} \plus{} \frac 1{1!} \plus{} \frac 1{2!} \plus{} \frac 1{3!} \plus{} \sum_{i\equal{}4}^\infty \frac 1{2^k} \equal{} \frac 83 \plus{} \frac 18 < 3$.
[/hide]


1) Let $ a,b,c$ be positive real numbers. Prove that
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \ge \frac{a}{b\plus{}2c}\plus{}\frac{b}{c\plus{}2a}\plus{}\frac{c}{a\plus{}2b}.$
2) Generalize this result.

[quote="can_hang2007"]1) Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \ge \frac {a}{b \plus{} 2c} \plus{} \frac {b}{c \plus{} 2a} \plus{} \frac {c}{a \plus{} 2b}.$
2) Generalize this result.[/quote]

assume  $ a\equal{}max{(a,b,c)}$

the inequality is equivalent to

$ 2b^4a\plus{}2b^3ca\plus{}b^3a^2\plus{}c^3b^2\minus{}5b^2c^2a\minus{}5b^2ca^2\plus{}2c^4b\plus{}2c^3ba\minus{}5c^2ba^2\plus{}2bca^3\plus{}c^2a^3\plus{}2ca^4\geq 0$

$ <\equal{}>(2ca^2\plus{}(4bc\plus{}3c^2)a\plus{}b^3\minus{}b^2c\plus{}3c^3)(a\minus{}b)(a\minus{}c)\plus{}((3b^2\plus{}8bc\plus{}3c^2)a\minus{}b^2c\minus{}c^2b)(b\minus{}c)^2\geq 0$

easy to prove

$ 2ca^2\plus{}(4bc\plus{}3c^2)a\plus{}b^3\minus{}b^2c\plus{}3c^3\geq b^3\minus{}b^2c\plus{}3c^3\geq 0$

$ (3b^2\plus{}8bc\plus{}3c^2)a\minus{}b^2c\minus{}c^2b\geq 0$

so.......

I am looking forward to seeing another proof.

[quote="leedt26"][quote="leedt26"][quote="can_hang2007"]1) Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \ge \frac {a}{b \plus{} 2c} \plus{} \frac {b}{c \plus{} 2a} \plus{} \frac {c}{a \plus{} 2b}.$
2) Generalize this result.[/quote]
I can't generalize it but I can prove [b]1)[/b]
Rewrite the inequality as
$ \frac {a}{{b(b \plus{} 2c)}} \plus{} \frac {b}{{c(c \plus{} 2a)}} \plus{} \frac {c}{{a(a \plus{} 2b)}} \ge \frac {1}{{b \plus{} 2c}} \plus{} \frac {1}{{c \plus{} 2a}} \plus{} \frac {1}{{a \plus{} 2b}}$
Multiplying both side with $ a \plus{} b \plus{} c$  we get
$ \sum {\frac {{{a^2}}}{{b\left( {b \plus{} 2c} \right)}}} \plus{} \sum {\frac {{ac}}{{b\left( {b \plus{} 2c} \right)}}} \plus{} \sum {\frac {c}{{b \plus{} 2c}}} \ge 3$
or it's equivalent to
$ \sum {\frac {{{a^2} \plus{} ac \plus{} bc}}{{b\left( {b \plus{} 2c} \right)}}} \ge 3$
which is clearly true by AM-GM.[/quote]
[hide="P/S"]Anh Cẩn thông cảm em trình bày tắt quá vì em không có thời gian![/hide][/quote]
I understand your proof, leedt26. Because it is exactly the same as my proof. See here: http://canhang2007.wordpress.com/2009/11/10/inequality-62-v-q-b-can/

[quote="can_hang2007"]1) Let $ a,b,c$ be positive real numbers. Prove that
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \ge \frac {a}{b \plus{} 2c} \plus{} \frac {b}{c \plus{} 2a} \plus{} \frac {c}{a \plus{} 2b}.$
[/quote]
An expanding gives $ \sum_{cyc}(2a^4c \plus{} a^3c^2 \plus{} 2a^3bc \minus{} 5a^2b^2c)\geq0,$ which is obvious.
The generalization:
Let $ a,b,c$ be positive real numbers, $ k\in[\frac{1}{2},2].$ Prove that
\[ \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{(k\plus{}1)(ab \plus{} bc \plus{} ca)} \ge \frac {a}{b \plus{} kc} \plus{} \frac {b}{c \plus{} ka} \plus{} \frac {c}{a \plus{} kb}\]


algorithm
number theory
relatively prime
prime numbers
Euclidean algorithm

someone posted before that the largest possible number that cannot be created by adding two relatively prime numbers $a$ and $b$ multiple times is $ab-a-b$. could anyone tell me why?

ugh, ummm..... i personally don't know, but a while ago there was a huge chicken mcnugget topic and it probably said there

-jorian

The proof isn't very simple.  AoPS vol. 1 (2?) has a proof that is much longer than that one.

i don't get it either

i think i saw the proof ages ago on another math site (i froget what site it was)
                      sry i can't help more

[quote="bpms"][quote="mysmartmouth"]From what I heard, the proof wasn't simple...Perhaps it has to do with the Euclidean algorithm, since and b must be relatively prime.[/quote]
I don't think it was that complicated. Lemme look it up.
[quote="Phelpedo"]Well, it's easy to prove the 2-mcnugget case. Just consider the multiples of m mod n

m
2m
3m
...
(n-1)m

And note that because m and n are relatively prime, this forms a complete residue class mod n. Thus, we can form all numbers equal to $m \pmod{n}$ after $m$, all numbers $2m \pmod{n}$ after $2m$, etc...continuing in this way, the last residue class that we can form is $(n-1)m \pmod{n}$, so the last one should be $nm-m-n$.[/quote][/quote]
ty...i kinda get it.

[quote="Ryan Flarity"]Consider z=x*y-x-y. Suppose z is possible. Then a*x+b*y=x*y-x-y => (a+1)x+(b+1)y=x*y

The last two terms are divisible by y, so the first must be as well. X and y are co-prime, so it follows that y|a+1. If a+1=0, then a=-1, which means that we're buying a negative number of x packs, which isn't allowed (I assume).

So a+1>0, which means that a+1>=y, and a>=y-1. Therefore, z=x*y-x-y>=(y-1)*x+b*y=x*y-x+b*y.

Subtracting (x*y-x) from both sides, -y>=b*y, so b is negative, which isn't allowed. So z is impossible.
But is it the largest impossible number?

Suppose w=z+c=x*y-x-y+c where c is a positive integer, and w is impossible. Consider the sequence w+y, w, w-y, w-2yw-x*y+2y. Note that there are a total of x terms. Since there are x elements in Z/xZ, and none of the terms in the sequence correspond to the same element (this follows from elementary group theory and the fact that x and y are co-prime), every element of Z/xZ has a term in the sequence that corresponds to it. Specifically, there is a term that corresponds to 0, that is, divisible by x. If it is anything other than the first term, then w is equal to a multiple of y plus a multiple of x, so it is possible. So the only way for w to be impossible is if w+y is divisible by x.

w+y=x*y-x+c=x*y+(c-x)

x*y is divisible by x, so since w+y is divisible by x, (c-x) is divisible by x. c>0, so c-x>-x. There's no number divisible by x between -x and zero, so c-x>=0.

w=x*y+(c-x)-y=(x-1)*y+(c-x)

Both terms are nonnegative, and they are divisible by y and x, respectively. So w is possible. This shows that z is the greatest impossible number. [/quote]

[quote="mysmartmouth"][quote="Ryan Flarity"]Consider z=x*y-x-y. Suppose z is possible. Then a*x+b*y=x*y-x-y => (a+1)x+(b+1)y=x*y

The last two terms are divisible by y, so the first must be as well. X and y are co-prime, so it follows that y|a+1. If a+1=0, then a=-1, which means that we're buying a negative number of x packs, which isn't allowed (I assume).

So a+1>0, which means that a+1>=y, and a>=y-1. Therefore, z=x*y-x-y>=(y-1)*x+b*y=x*y-x+b*y.

Subtracting (x*y-x) from both sides, -y>=b*y, so b is negative, which isn't allowed. So z is impossible.
But is it the largest impossible number?

Suppose w=z+c=x*y-x-y+c where c is a positive integer, and w is impossible. Consider the sequence w+y, w, w-y, w-2yw-x*y+2y. Note that there are a total of x terms. Since there are x elements in Z/xZ, and none of the terms in the sequence correspond to the same element (this follows from elementary group theory and the fact that x and y are co-prime), every element of Z/xZ has a term in the sequence that corresponds to it. Specifically, there is a term that corresponds to 0, that is, divisible by x. [b]If it is anything other than the first term, then w is equal to a multiple of y plus a multiple of x, so it is possible. So the only way for w to be impossible is if w+y is divisible by x.[/b]

w+y=x*y-x+c=x*y+(c-x)

x*y is divisible by x, so since w+y is divisible by x, (c-x) is divisible by x. c>0, so c-x>-x. There's no number divisible by x between -x and zero, so c-x>=0.

w=x*y+(c-x)-y=(x-1)*y+(c-x)

Both terms are nonnegative, and they are divisible by y and x, respectively. So w is possible. This shows that z is the greatest impossible number. [/quote][/quote]

I don't get this part.


How many two-digit numbers less than 50 have an odd number of factors?

Any perfect square would have an odd number of divisors. There are 7 perfect squares less than 50, but 3 of them of these are not two-digit integers. The question asks for two-digit integers, therefore, the answer is $ \boxed{4}$.


MIT college Harvard Stanford Support blogs

I want to be homeschooled but i am not sure how colleges are going to think about that. I mean, how would MIT know my grades and stuff if i'm homeschooled? Does homeschooling kill my chances of getting into a top-of-line college like MIT, Harvard or Stanford?

Probably the opposite.  There are plenty of homeschooled students here.  There is also, as I understand it, a fair amount of support and some standard stuff homeschool teachers are supposed to do to keep track of grades and that kind of thing.

AP scores. Good. College classes. Good. Acheivement in competitions. Also good.

In other words, anything that can be directly compared to students in the normal system is good.

If you're going to be homeschooled - I would suggest a lot of community service. It'd help show the colleges that you care about going out and helping people; instead of holed up in your home studying.

Right. In school, there are sports teams, clubs, bands, and many opportunities to take initiative or show leadership within the school community.

As a home-schooled student, you will have to go out of your way a bit more to do many of these things, especially for showing leadership, if you're the type. But when you do it outside the context of a school, it may very well end up being more meaningful. An interesting community initiative will help you more (and be more fulfilling) than being the zillionth person to start a blood donation or cure for cancer club in a school.

Also, take advantage of the reduced busywork and take part in personal projects, research, or initiatives that interest you. Taking advantage of the reduced burdens to do more interesting things and to pursue your personal goals is very attractive to admissions officers because it shows you can make priorities and aggressively pursue them.

now that my only fear has departed, i can suggest homeschooling to my parents :) (if you read my blog you'll find out why i take this decision)


Given that $ f(x\plus{}1) \equal{} 2x \plus{} x^2$, compute $ f(4.1)$. Express your answer as a decimal to the nearest hundredth.

I got 15.81 too, but the answer to this problem on FTW is 18.18 for some reason.

On FTW it says the answer is 15.18. Is there something wrong?

I lost 2 games because of this question. -.-

My luck.......


This is my post 1000, then, I wanted to post a nice Puzzle :P

What are the following number of this sequence:

0, 10, 1011, 1031, 102113, 10311213, 10411223, 1031221314, 1041222314,...

Read it out loud as digits.  Then you will know :).

[quote="xxreddevilzxx"]this is variance of an Art and Craft problem.[/quote]
Given that The Art and Craft of Problem Solving was printed in 1999, and I first saw this problem in 1986 (or the original one that I am doing a variant here) I would say that the problem in The Art and Craft of Problem Solving is a variance of another one.

[quote="LynnelleYe"][quote="djimenez"]Nobody has given a solution yet!  :([/quote]
[hide]1031321324, because in 1041222314 (the previous term) there are one 0, three 1's, three 2's, one 3, and two 4's.[/hide][/quote]Good!! Now, the sequence is bounded, is it convergent??? (is it a term $a_N$ such that $a_n=a_N$ for all $n>N$?)

Well, apparently the term after the one I gave is
[hide]a_11=1031223314[/hide]
Which happens to describe itself, so all terms after and including a_11 are the same.

[quote="LynnelleYe"]Well, apparently the term after the one I gave is
[hide]$a_11=1031223314$[/hide]
Which happens to describe itself, so all terms after and including a_11 are the same.[/quote]Yep. I think that no matter what is the first term (it could be 1 instead of 0, or 9, or 34)! there is always a steady point!


analytic geometry
algebra unsolved
algebra

Let $ c_{1},c_{2},...,c_{n},b_{1},b_{2},...,b_{n}$ be positive real numbers. Prove that the equation $ \sum^{n}_{i\equal{}1}c_{i}\sqrt{x_{i}\minus{}b_{i}}\equal{}\frac{1}{2} \sum^{n}_{i\equal{}1}x_{i}$ has a unique solution $ (x_{1},x_{2},...,x_{n})$ if and only if $ \sum^{n}_{i\equal{}1}c^{2}_{i}\equal{}\sum^{n}_{i\equal{}1}b_{i}$.

what a coincidence - I read this problem yesterday but was too tired to try and solve it.
Right now I have only one direction:
If $ (*) \sum^{n}_{i\equal{}1}b_{i}\equal{}\sum^{n}_{i\equal{}1}c_{i}^2$, then-
Let's put $ \sum^{n}_{i\equal{}1}b_{i}\equal{}y, \sum^{n}_{i\equal{}1}x_{i}\equal{}t.$

$ \frac{t}{2}\equal{}\frac{1}{2} \sum^{n}_{i\equal{}1}x_{i}\equal{} \sum^{n}_{i\equal{}1}c_{i}\sqrt{x_{i}\minus{}b_{i}} (by CS)<\equal{}\sqrt{\sum^{n}_{i\equal{}1}c_{i}^2}\sqrt{\sum^{n}_{i\equal{}1}x_{i}\minus{}b_{i}} \equal{}  \sqrt{\sum^{n}_{i\equal{}1}b_{i}}\sqrt{\sum^{n}_{i\equal{}1}x_{i}\minus{}b_{i}} \equal{} \sqrt{y(t\minus{}y)}$

$ (t\minus{}2y)^2<\equal{}0 \minus{}> t\equal{}2y$

But this equality occurs iff $ x_{i}\equal{}b_{i}\plus{}\lambda c_{i}^2$ for every i, and from $ (*)$ and t=2y we can easily get $ \lambda \equal{}1$ so we have exactly one solution.

I think I've finished it-
let's denote $ x_{i} \equal{} b_{i} \plus{} k_{i}^2, k_{i} > \equal{} 0$.
the equation is equivalent to-
$ \sum^{n}_{i \equal{} 1}(k_{i} \minus{} c_{i})^2 \plus{} \sum^{n}_{i \equal{} 1}b_{i} \minus{} c_{i}^2 \equal{} 0$
if $ \sum^{n}_{i \equal{} 1}b_{i} \minus{} c_{i}^2 > 0$: there's no solution
if $ \sum^{n}_{i \equal{} 1}b_{i} \minus{} c_{i}^2 \equal{} 0$: there's only one solution - $ k_{i} \equal{} c_{i}$, and we can find $ x_{i}$.
if $ \sum^{n}_{i \equal{} 1}b_{i} \minus{} c_{i}^2 < 0$: there is an infinite number of points in $ Rn$ with distance $ \sqrt{\sum^{n}_{i \equal{} 1}c_{i}^2 \minus{} b_{i}}$ from the origin $ 0$, so there is an infinite number of solutions. ("$ \frac {1}{2^n}$" of them have positive coordinates, so $ k_{i} > c_{i} > \equal{} 0$)


$ x,y,z,a,b,c$ number  $ a\equal{}x\plus{}y\minus{}z,b\equal{}\minus{}x\plus{}y\plus{}z,c\equal{}x\minus{}y\plus{}z$ and

$ \frac{(a\plus{}b\plus{}c)^{5}\minus{}a^{5}\minus{}b^{5}\minus{}c^{5}}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\equal{}\frac{(x\plus{}y\plus{}z)^{5}\minus{}x^{5}\minus{}y^{5}\minus{}z^{5}}{(x\plus{}y)(y\plus{}z)(z\plus{}x)}$  .

Prove: $ a\equal{}b\equal{}c\equal{}x\equal{}y\equal{}z$

[quote="zaya_yc"]$ x,y,z,a,b,c$ number  $ a \equal{} x \plus{} y \minus{} z,b \equal{} \minus{} x \plus{} y \plus{} z,c \equal{} x \minus{} y \plus{} z$ and

$ \frac {(a \plus{} b \plus{} c)^{5} \minus{} a^{5} \minus{} b^{5} \minus{} c^{5}}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \equal{} \frac {(x \plus{} y \plus{} z)^{5} \minus{} x^{5} \minus{} y^{5} \minus{} z^{5}}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}$  .

Prove: $ a \equal{} b \equal{} c \equal{} x \equal{} y \equal{} z$[/quote]
Math 6-8 it's for me. :D
$ (a \plus{} b \plus{} c)^5 \minus{} a^5 \minus{} b^5 \minus{} c^5 \equal{} 5\sum_{sym}(a^4b \plus{} 2a^3b^2 \plus{} 2a^3bc \plus{} 3a^2b^2c) \equal{}$
$ \equal{} 5(a \plus{} b)(a \plus{} c)(b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} ac \plus{} bc).$
Id est, $ \frac {(a \plus{} b \plus{} c)^{5} \minus{} a^{5} \minus{} b^{5} \minus{} c^{5}}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \equal{} \frac {(x \plus{} y \plus{} z)^{5} \minus{} x^{5} \minus{} y^{5} \minus{} z^{5}}{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^2 \plus{} ab \minus{} x^2 \minus{} xy) \equal{} 0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}((x \plus{} y \minus{} z)^2 \plus{} (x \plus{} y \minus{} z)(y \plus{} z \minus{} x) \minus{} x^2 \minus{} xy) \equal{} 0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^2 \minus{} xy) \equal{} 0\Leftrightarrow\sum_{cyc}(x \minus{} y)^2 \equal{} 0\Leftrightarrow x \equal{} y \equal{} z.$
It gives also $ a \equal{} b \equal{} c\equal{}x$


inequalities
trigonometry
function
geometry unsolved
geometry

Let $ ABC$ is a triangle and let $ \angle A < 90^0, \angle B < 90^0, \angle C < 90^0$.

1) Prove that: $ \sin{A} \plus{} \sin{B} \plus{} \sin{C} \plus{} \tan A \plus{} \tan B \plus{} \tan C > 2\pi$.

2) prove that: $ 2(\sin A \plus{} \sin B \plus{} \sin C) \plus{} \tan A \plus{} \tan B \plus{} \tan C > 3\pi$.

[quote]Let  is a triangle and let $ \angle A < 90^0, \angle B < 90^0, \angle C < 90^0$.

1) Prove that:  $ \sin{A} \plus{} \sin{B} \plus{} \sin{C} \plus{} \tan A \plus{} \tan B \plus{} \tan C > 2\pi$.

2) prove that: $ 2(\sin A \plus{} \sin B \plus{} \sin C) \plus{} \tan A \plus{} \tan B \plus{} \tan C > 3\pi$.[/quote]

Maybe You'll can use the Jensen's Inequality.

Let $ f(x) \equal{} \sin{x} \plus{} \tan{x}$, then $ f'(x) \equal{} \sec^2{x} \minus{} \sin x$, $ f''(x) \equal{} 2 \sec^2 x \tan x \minus{} \sin x \equal{} \sin x(2\sec^3 x \minus{} 1) > 0$. So my function $ f$ has a concavity in the interval $ (0, \pi / 2)$.

so $ \frac {f(A) \plus{} f(B) \plus{} f(C)}{3} > f(\frac {A \plus{} B \plus{} C}{3})$, and You'll get
$ \sin{A} \plus{} \sin{B} \plus{} \sin{C} \plus{} \tan A \plus{} \tan B \plus{} \tan C \ge \frac {9}{2}\sqrt {3} \approx 7.794 > 2\pi \approx \equal{} 6.28$

Let $ g(x) \equal{} \tan x \plus{} 2\sin x$ then $ g''(x) \equal{} 2\sin x (\sec^3 x \minus{} 1) \ge 0$ so $ g$ also has a concavity, so

$ \frac {g(A) \plus{} g(B) \plus{} g(C)}{3} \ge g( \frac {A \plus{} B \plus{} C}{3} )$, so
$ 2(\sin A \plus{} \sin B \plus{} \sin C) \plus{} \tan A \plus{} \tan B \plus{} \tan C \ge 6\sqrt {3} \equal{} 6\sqrt {3} > 3 \pi$

----

I'm not sure. Maybe I'm wrong, either my grammar and solution


[i]Solve the differential equation:

 a/  $ y' \ \plus{} \ y \ \equal{} \ e^{\frac{x}{2}} \sqrt{y}$

 b/ $ y. y'' \ \plus{} \ (y')^{2} \ \equal{} \ y^{2}lny$[/i]

a.It is a problem of Bernoulli's  equation.
hint::

y'+y=e^(x/2)y^(1/2)
first divide the equation by y^(1/2).
y^(-1/2)y'  +  y^(1/2)  = e^(x/2)

now put y^(1/2)= z;and y^(-1/2)y'=2z'

thus the eq. now is::
2z' + z =e^(x/2)
find the integrating factor= e^(x^2/4)
then just do it in the manner of linear equation.


[quote="alanchou"]Suppose a square ABCD has an area of 324 $ cm^2$.  An equilateral triangle is built with one of its sides being AB.  What is the area of the equilateral triangle?[/quote]
Area of an equilateral triangle is...
$ \frac{s^2\sqrt3}{2}$
Since the side length would be $ \sqrt 324$, the area of the triangle would be...
$ 162\sqrt3$

[quote="alanchou"]um, no, I know you know the actual formula[/quote]
Right...
It's over 4.
>.>
So anyways.......
It's just
$ 81\sqrt3$
then.

[quote="alanchou"]Suppose a square ABCD has an area of 324 $ cm^2$.  An equilateral triangle is built with one of its sides being AB.  What is the area of the equilateral triangle?[/quote]
Each side of the square (and triangle) is 18 cm, so the area of the triangle is {[sqrt(3)]/4}*324=81*sqrt(3) square cm.

Yeah, I was trying to find out the formula for an equilateral triangle because I was bored when I was at school.
And I got $ \frac{s^2\sqrt3}{2}$
Because I forgot to multiply by $ \frac{1}{2}$
Because it was a triangle....
So anyways, I got that stuck in my head.
Yeah.

[quote="sanghnlee"]give another problem!!

 :D[/quote]
What?
That so so totally on topic and yeah.

dude

i'm really bored right now....  so that's why

[hide]
the side of the square is 18.
$ \frac{18^2 \cdot \sqrt{3}}{4} \equal{} 81 \sqrt{3}$
(That's the formula)[/hide]

[quote="moogra"][quote="sanghnlee"]Fine.  What you do is find the square root of 324 which is 18.  Then you insert the equilateral triangle formula.  So then you find the second power of 18 and multiply that by the square root of 3.  That whole, "numerator" goes over 4 (denominator).  Then you divide and you get 81times the square root of 3![/quote]

That basically said what all the other posts said. Also the square root of 324 and then squaring it is redundant.

I also noticed that no one posted where the 4 actually came from. The formula is $ s^{2}*\frac {\sqrt {3}}{2}*\frac {1}{2}$.
That formula comes from
$ Area \equal{} \frac {ab\sin C}{2}$[/quote]
I got the formula by using the Pythagorean Theorem.
The area for the triangle is $ \frac{bh}{2}$
Base is $ s$
And the hight is...
$ (\frac{1}{2} s )^2\plus{} h^2 \equal{} s^2$
$ \frac{1}{4}s^2\plus{}h^2\equal{}s^2$
$ h\equal{}\sqrt{\frac{3}{4}s^2}$
So plugging in values of $ b$ and $ h$, we get.
$ \frac{s \sqrt{\frac{3}{4}s^2}}{2}$

$ \frac{s^2(\frac{\sqrt{3}}{2})}{2}$
Which gets you.
$ \frac{s^2\sqrt{3}}{4}$

And then the AoPS mods declared:
"Thou shall not spam"

[quote="Poincare"]Ok
i am a 5th grader who thinks you are stupid :rotfl:
81 and then you find the freakin height and then multiply to 81 and 1/2 and theret 81 and then you find the freakin height and then multiply to 81 and 1/2 and thereou ab
All you gotta do is divide 324 by 4 which gives you 81 then using the Pythogrean Theorem, find the height of the triangle then multiply the height by 81 and 1/2
And you guys were trying to use trigonometry to show this easy crap!!!!!! And Sorry about the lack of Latex, the only computer language I know is Java.[/quote]
Ok so if this is "easy crap" then what's the answer?
And I highly doubt you know java, seeing as your spelling and grammar are so bad.

[quote="Poincare"]Ok
i am a 5th grader who thinks you are stupid :rotfl:
81 and then you find the freakin height and then multiply to 81 and 1/2 and theret 81 and then you find the freakin height and then multiply to 81 and 1/2 and thereou ab
All you gotta do is divide 324 by 4 which gives you 81 then using the Pythogrean Theorem, find the height of the triangle then multiply the height by 81 and 1/2
And you guys were trying to use trigonometry to show this easy crap!!!!!! And Sorry about the lack of Latex, the only computer language I know is Java.[/quote]

81 isn't even the side length of the square, "genius"


floor function
number theory proposed
number theory

let $ f(n) \equal{} (2n \minus{} 1)(1 \plus{} [\frac {(2n)! \plus{} 1}{2n \plus{} 1}] \plus{} [ \minus{}  \frac {(2n)! \plus{} 1}{2n \plus{} 1}]) \plus{} 2$
prove that
1)$ f(n)$ is  prime number for all natural number $ n$
2)for every prime number $ p$ there exist $ n\in\mathbb{N}$ such that $ f(n) \equal{} p$

I assume that $ [x]$ denotes the floor function.

[b]Lemma:[/b]  $ \lfloor x \rfloor \plus{} \lfloor \minus{}x \rfloor$ is equal to $ 0$ if $ x$ is an integer and equal to $ \minus{}1$ otherwise.

[b]Case:[/b]  $ 2n \plus{} 1$ is prime.  Then $ 2n! \equiv \minus{}1 \bmod (2n\plus{}1)$ so $ f(n) \equal{} (2n\minus{}1)(1 \plus{} 0) \plus{} 2 \equal{} 2n\plus{}1$.

[b]Case:[/b]  $ 2n \plus{} 1$ is composite.  Then $ 2n \plus{} 1 | (2n)!$ so $ f(n) \equal{} (2n\minus{}1)(1 \minus{} 1) \plus{} 2 \equal{} 2$.

[quote="t0rajir0u"]I assume that $ [x]$ denotes the floor function.

[b]Lemma:[/b]  $ \lfloor x \rfloor \plus{} \lfloor \minus{} x \rfloor$ is equal to $ 0$ if $ x$ is an integer and equal to $ \minus{} 1$ otherwise.

[b]Case:[/b]  $ 2n \plus{} 1$ is prime.  Then $ 2n! \equiv \minus{} 1 \bmod (2n \plus{} 1)$ so $ f(n) \equal{} (2n \minus{} 1)(1 \plus{} 0) \plus{} 2 \equal{} 2n \plus{} 1$.

[b]Case:[/b]  $ 2n \plus{} 1$ is composite.  Then $ 2n \plus{} 1 | (2n)!$ so $ f(n) \equal{} (2n \minus{} 1)(1 \minus{} 1) \plus{} 2 \equal{} 2$.[/quote]
perfect :D