Farz konid yek donbale az takjomleiha mesle $\{x^{\zeta_i}\}$(Har yek az $\zeta_i$ ha yek bordar dar $(\mathbb N\cup\{0\})^n$ hastand va $x^{(a_1,\dots,a_n)}=(x_1^{a_1},\dots,x_n^{a_n}$) Sabet konid ke yek $N$ vojood darad ke har yek az aazaye donbaleh bar yeki az $n$ jomleye avval bakhshpazir bashad.

Masalan agar $n=1$ masaleh hamoon asle khoshtartibi ast.

aghaye hatami in mas'ale ba nazariye ramzi hal nemishe? :)

Na rabti nadare rahat mitoonid esteghra bezanid

[quote="mehdi"]bebakhshid vali in joor lem ha dar che ketabi hast va dar che jahayi mishe masale haye marboote ro peyda kard?[/quote]
injoor lem ha mokhtasse afrade khafani chon ostad hatamie. :D

pas hatman bayad esmesho gozasht "Hatami's lemma" chon male ishoone(jeddi migam)

vali aghaye hatami man ye esbat ba nazariye ramzi barash didam. :)

Khoshal misham age esbate to ra bebinam oon ra befrest. :D

Vali baz ham pishnahad mikonam be esteghra rooye $n$ fekr kon

Baad az 5 mah hich kas natooneste (ya nakhaste) rahe halli baraye in masaleh befreste.  :!:  :!:

mitavan farz kard ozve tekrari vojood nadarad.
masale ra ba esteghra rooye $n$ sabet mikonim.
baraye $n=1$ ke vazeh ast.(asle khosh tartibi)
farz konim baraye $n-1$ hokm dorost bashad hokm ra be ezaye $n$ natije migirim.
hameye azaye donbale ra bedoone moallefeye avval dar nazar migirim.
tebghe farze esteghra $N_1$ vojood darad ke moallefeye dovvom ta $n$ome har yek az aaza bar yeki az $N_1$ jomleye avval bakhshpazir bashad.be hamin tartib ba hazfe moallefehaye dovom ta $n$om aadade $N_2,\dots,N_n$
moshabeh vojood darand. $N$ ra bozorgtarin dar beine anha migirim.
farz konim $a_i$ bozorgtarin tavane moallefeye $i$ome $N$ ozve avvale donbale bashad. tamame azaye donbale be shekle $(x_1^{b_1},\dots,x_n^{b_n})$ ke $b_i \le a_i$ be ezaye har $i$ ra dar nazar migirim.(tebghe farz az har kodam hadde aksar yeki vojood darad)
hal $M$ ra dar nazar migirim ke az $N$ bozorgtar bashad va hameye in aadad ra dar bar girad.(chon teedade in no aadad motanahist pas $M$ vojood darad)
in $M$ sharayete masale ra erza mikonad. pas hokm sabet shod. :)  :lol:


What is the arithmetic mean of the first million odd numbers?

We have the first and last number are $ 1$ and $ 1,000,000(2)\minus{}1$. Thus our average is $ \boxed{1000000}$.

In more detail...

We know that the sum of odd integers are $ N^2$ where N is the number of odd integers. We can then say:

$ 1,000,000^2\linebreak \linebreak \frac{1,000,000^2}{1,000,000}$

$ \text{Thus, our answer is} \boxed{1,000,000}$


If $ 4^x \minus{} 4^{x \minus{} 1} \equal{} 24$, then $ (2x)^x$  equals:

$ \textbf{(A)}\ 5\sqrt{5}\qquad
\textbf{(B)}\ \sqrt{5}\qquad
\textbf{(C)}\ 25\sqrt{5}\qquad
\textbf{(D)}\ 125\qquad
\textbf{(E)}\ 25$

[hide="Solution"]$ \frac {3\cdot 4^x}{4} \equal{} 24 \Rightarrow 4^x \equal{} 2^{2x} \equal{} 32 \Rightarrow x \equal{} \frac {5}{2}$

$ (2x)^x \equal{} (5)^{5/2} \equal{} 25\sqrt {5}, \boxed{\text{C}}$[/hide]

hello, simplifying your equation we get $ 4^x(1\minus{}\frac{1}{4})\equal{}24$, this is equivalent with $ 4^x\equal{}32$, this is equivalent with $ 2^{2x}\equal{}2^5$ or $ x\equal{}\frac{5}{2}$, from here we get $ (2\cdot \frac{5}{2})^{\frac{5}{2}}\equal{}5^{\frac{5}{2}}\equal{}25\sqrt{5}$
Sonnhard.


geometry
incenter
circumcircle
projective geometry
geometry unsolved

In a triangle $ABC$ with $\angle B < 45^{\cdot}$, $D$ is a point on $BC$ such that the incenter of $\Delta ABD$ coincides with the circumcenter $O$ of $\Delta ABC$. Let $P$ be the intersection point of the tangent lines to the circumcircle $O'$ of $\Delta AOC$ at points $A$ and $C$. The lines $AD$ and $CO$ meet at $Q$. The tangent to $O'$ at $O$ meets $PQ$ at $X$. Prove that $XO = XD$.

It's a nice problem :)  Of course if incenter of $ABD$ and circumcenter of $ABC$ coincide then $ABC$ is isosceles with $BA=BC$ We have also that $ACO$ is isosceles with $OA=OC$. This means that tangent to $C'$ in $O$ ( it will be denoted by $l$) is parallel to $AC$ and also that tangents to $C'$ in $A$ and $C$ intersects on line $BO$. As $\angle B <45$ , this intersection point lies on the other side of $AC$ ( with respect to $B$) As there is only one point $R$ on line $l$ such that $RD=RO$ We may formulate our statement as follows: We take point $R$ on $l$ such that $RD=RO$. We have to prove that $P,Q,R$ are collinear.

We will prove now that points $A,C,D,O$ are concyclic. Indeed, from symetry we obtain \[ \angle DCO= \angle OAB \] also as $AO$ is bisector of $\angle DAB$ we obtain also \[ \angle OAB= \angle DAO \] so finally \[ \angle DCO=\angle DAO \].
And points $A,C,D,O$ are concyclic. Let $C\"$ be the common circle passing through  these points. As $RD=RO$ and  $RO$ is tangent to $C\"$, also $RD$ is tangent to $C\"$. So we may say that $R$ is an intersection point of tangents to $C\"$ in points $D$ and $O$.  Let lines $AC$ and $DO$ intersects in $W$. We will prove that both $R$ and $P$ lies on the polar line of $C\"$ ( with respect to $W$). Then our statement would be obvious, cos' $Q$ lies on this polar line.  Let $WW_1$ and $WW_2$ be lines tangent to $C\"$ in $W_1$ and $W_2$ respectively. Consider quadrangle $AW_1CW_2$.  It's diagonal $AC$ and tangents to $C\"$ in $W_1$ and $W_2$ intersects in one point. (now we use "magic" theorem.) This means that also diagonal $W_1W_2$ and tangents in $A$ and $C$ intersects in one point, so points $W_1W_2P$ are colinear , and point $P$ lies on the polar line $W_1W_2$. In analogical way, by considering quadrangle $W_1OW_2D$  we may obtain that $R$ lies on the polar line $W_1W_2$ which means that points $R,Q,P$ are colinear, which ends the proof of our statement.

In this proof, I used the following theorem : If $XYZT$ is cyclic and diagonal $XZ$ , tangent in $Y$ and tangent in $T$, concurrent, then also concurrent diagonal $YT$, tangent in $X$ and tangent in $Z$.

Nice isn't it  ;)


Ring Theory
number theory
relatively prime
greatest common divisor
least common multiple
number theory unsolved

Let $ T$ be a finite set of positive integers, satisfying the following conditions:
1. For any two elements of $ T$, their greatest common divisor and their least common multiple are also elements of $ T$.
2. For any element $ x$ of $ T$, there exists an element $ x'$ of $ T$ such that $ x$ and $ x'$ are relatively prime, and their least common multiple is the largest number in $ T$.

For each such set $ T$, denote by $ s(T)$ its number of elements. It is known that $ s(T) < 1990$; find the largest value $ s(T)$ may take.

Let $ N$ denote the largest element of $ T$.  By condition 2, $ 1 \in T$ and every element of $ T$ is a divisor $ d$ of $ N$ such that $ \frac {N}{d}$ is relatively prime to $ d$.  This requires that if $ p | d$ then $ p^k \parallel{} d$ where $ p^k \parallel{} N$.  Thus we can suppose WLOG that $ N$ is squarefree (reducing the powers of each prime to $ 1$ does not change the structure of the set).

We can now give $ T$ the structure of a subalgebra of a [url=http://en.wikipedia.org/wiki/Boolean_algebra_%28structure%29]Boolean algebra[/url], from which it will follow that $ s(T)$ is a power of $ 2$.  In particular, let $ N$ be a product of $ n$ primes $ p_1 p_2 ... p_n$ and associate with each $ d \in T$ the $ n$-tuplet $ (d_1, d_2, ... d_n), d_i \in \mathbb{Z}_2$ such that $ d \equal{} p_1^{d_1} ... p_n^{d_n}$.  The first statement says that $ T$ is closed under intersection and union and the second that $ T$ is closed under inverse.  $ T$ is therefore the subalgebra of a finite Boolean algebra (which is the Boolean algebra of all divisors of $ N$), and it must therefore have order dividing $ 2^n$.  (Equivalently, $ T$ can be given the structure of a subring of a Boolean ring.)

Hence $ \text{max}(s(T)) \equal{} 1024$.

[quote="hungkg"]Who can help me?Thanks[/quote]
Help what ? t0rajir0u has been post the solution.


trigonometry
Pythagorean Theorem
geometry
geometry proposed

There are two triangles: $\triangle{ABC}$ and $\triangle{XYZ}$. If $\angle{B}=\angle{Y}$, prove:
\[\frac{\triangle{ABC}}{\triangle{XYZ}}=\frac{a^2+c^2-b^2}{x^2+z^2-y^2}\]

Also if $\angle{B}+\angle{Y}= 180^{o}$, then prove:
\[\frac{\triangle{ABC}}{\triangle{XYZ}}=-\frac{a^2+c^2-b^2}{x^2+z^2-y^2}\]

[b][size=150]NOTE: Try to prove it without trig and Pythagorean theorem or any other stuff related to these.[/size][/b]


Given the equation: $ (x^2 \plus{} 1)(y^2 \plus{} 2)(z^2 \plus{} 3)(t^2 \plus{} 4)(n^2 \plus{} 5) \equal{}$ $ 32.xyztn\sqrt{120}$(x,y,z,t,n are real numbers)
Find the number of solutions of this equations.

AM-GM tears down the problem.

So $ \boxed{1}$ solution exists (which is, obviously, $ x\equal{}1,y\equal{}\sqrt{2},z\equal{}\sqrt{3},t\equal{}2,n\equal{}\sqrt{5}$).

[quote="MathAndKnowledge"]AM-GM tears down the problem.

So $ \boxed{1}$ solution exists (which is, obviously, $ x \equal{} 1,y \equal{} \sqrt {2},z \equal{} \sqrt {3},t \equal{} 2,n \equal{} \sqrt {5}$).[/quote]
Be careful! x,y,z,t,n are real numbers :wink:

[quote="1=2"]I know I'm asking a stupid question, but how do you apply AM-GM???[/quote]
I guess the reason you are afraid of not using AM-GM is just because x,y,z,t,n are real numbers. So, I will help you out with just this "cheap" hint: Why don't you use the absolute value first, multiply them, and then compare with the RHS? :wink: . The whole point of this problem is mainly lying on the combinatorics part though. I think we have a general formula for the number of solutions of the n-variable equation as well. :)

[quote="1=2"]I know I'm asking a stupid question, but how do you apply AM-GM???[/quote]

It's AM-GM on every term, but since they are real numbers it is easier to use SOS

$ (x\minus{}1)^2 \ge 0 \Rightarrow x^2 \plus{} 1 \ge 2x$
$ (y\minus{}\sqrt{2})^2 \ge 0 \Rightarrow y^2 \plus{} 2 \ge 2\sqrt{2}y$...

Now multiplying all of these solutions we get

$ (x^2\minus{}1)(y^2\minus{}2)(z^3\minus{}3)(t^4\minus{}4)(4^n\minus{}n) \ge 32xyztn\sqrt{120}$

[quote="ocha"][quote="1=2"]I know I'm asking a stupid question, but how do you apply AM-GM???[/quote]

It's AM-GM on every term, but since they are real numbers it is easier to use SOS

$ (x \minus{} 1)^2 \ge 0 \Rightarrow x^2 \plus{} 1 \ge 2x$
$ (y \minus{} \sqrt {2})^2 \ge 0 \Rightarrow y^2 \plus{} 2 \ge 2\sqrt {2}y$...

Now multiplying all of these solutions we get

$ (x^2 \minus{} 1)(y^2 \minus{} 2)(z^3 \minus{} 3)(t^4 \minus{} 4)(4^n \minus{} n) \ge 32xyztn\sqrt {120}$[/quote]
I think you mean $ (x^2 \plus{} 1)(y^2 \plus{} 2)(z^2 \plus{} 3)(t^2 \plus{} 4)(n^2 \plus{} 5) \ge 32xyztn\sqrt {120}$? That is a strange typo.  :huh:

And, since no one has actually said it yet, we have $ x\equal{}\pm1, y\equal{}\pm \sqrt{2}, z\equal{}\pm \sqrt{3}, t\equal{}\pm2, n\equal{}\pm\sqrt{5}$, so we have $ 2^5\equal{}32$ solutions total.

[quote="Brut3Forc3"][quote="ocha"][quote="1=2"]I know I'm asking a stupid question, but how do you apply AM-GM???[/quote]

It's AM-GM on every term, but since they are real numbers it is easier to use SOS

$ (x \minus{} 1)^2 \ge 0 \Rightarrow x^2 \plus{} 1 \ge 2x$
$ (y \minus{} \sqrt {2})^2 \ge 0 \Rightarrow y^2 \plus{} 2 \ge 2\sqrt {2}y$...

Now multiplying all of these solutions we get

$ (x^2 \minus{} 1)(y^2 \minus{} 2)(z^3 \minus{} 3)(t^4 \minus{} 4)(4^n \minus{} n) \ge 32xyztn\sqrt {120}$[/quote]
I think you mean $ (x^2 \plus{} 1)(y^2 \plus{} 2)(z^2 \plus{} 3)(t^2 \plus{} 4)(n^2 \plus{} 5) \ge 32xyztn\sqrt {120}$? That is a strange typo.  :huh:

And, since no one has actually said it yet, we have $ x \equal{} \pm1, y \equal{} \pm \sqrt {2}, z \equal{} \pm \sqrt {3}, t \equal{} \pm2, n \equal{} \pm\sqrt {5}$, so we have $ 2^5 \equal{} 32$ solutions total.[/quote]
@ocha: You can't multiply both sides without assuring that both sides are positive. Your solution has that "flaw".
@Brut3Forc3: The whole point here is not that simple! Did you notice something strange about the number of variables? If you didn't let me tell you this: IT'S ODD!.  :wink: . Your answer is not correct!

[quote="Brut3Forc3"]Noooooo!
Okay, pick signs for $ x,y,z,t$, then $ n$'s sign is fixed, so $ 16$.[/quote]
That's correct. :)
The formula for n-variables is:$ 2^{(n \minus{} 1)}$.


advanced fields
advanced fields unsolved

Let $ \mu \equal{} \Sigma_{i \equal{} 1}^n x_i dx_i$
Show that there is an n-1 form $ \omega$ defined on $ R^n \minus{} {0}$ with the property that $ \mu \wedge \omega \equal{} dx_1 \wedge ... \wedge dx_n$

Let $ \omega_i\equal{}dx_1\wedge\dots\wedge dx_{i\minus{}1}\wedge dx_{i\plus{}1}\wedge\dots \wedge dx_n$. Let $ F(x)\equal{}\sum_j x_j^2$. Now just put $ \omega\equal{}\sum_i (\minus{}1)^{i\minus{}1}\frac {x_i}{F(x)}\omega_i$. This construction is by no means unique, of course.


Dear nguovin, can you please help me to make this inequality stronger? (I think it is a bit strong)....... :blush:
Prove That, for positive reals $ a, b, c$ we have
$ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{8abc}\ge\frac{\sqrt{\prod_{cyc}((a^2\plus{}b)(a\plus{}b^2))}}{(a\plus{}bc)(b\plus{}ca)(c\plus{}ab)}$

Hope you will enjoy it and help me strengthen it! :roll:  :lol:

$ (a\plus{}bc)(b\plus{}c)\equal{}b(a\plus{}c^2)\plus{}c(a\plus{}b^2) \geq\ 2 \sqrt{bc(a\plus{}c^2)(a\plus{}b^2)}$
I think you used my old hint  :P .
But thank you very much for your gift, Potla. I like this ineq  :)

[quote="nguoivn"]$ (a \plus{} bc)(b \plus{} c) \equal{} b(a \plus{} c^2) \plus{} c(a \plus{} b^2) \geq\ 2 \sqrt {bc(a \plus{} c^2)(a \plus{} b^2)}$
I think you used my old hint  :P .
But thank you very much for your gift, Potla. I like this ineq  :)[/quote]
Many many thanks for your interest, and of course I used that hint as I liked it tremendously.....

However, it was too easy to be dedicated to YOU;       :rotfl:
But you did not answer my question "Can we make it stronger?"......... :oops:


algebra
function
domain
vector
geometry
3D geometry
sphere

So here are some functions of the following types...

f: R -> R^2 (curves in the plane)
f: R -> R^3 (curves in space)
f: R^2 -> R (functions f(x,y) of 2 vars)
f: R^3 -> R: (functions f(x,y,z) of 3 vars)
f: R^2 -> R^2 (vector fields v(x,y) in the plane)

The question is - why are curves in the plane of the form R -> R^2? My intuition tells me R^2 -> R^2 (since after all, curves in the plane are based on x and y coordinates...). And R^2 is a cartesian product of two sets. For any curve, I'd expect x AND y input values...

Think parametric: $ (x(t),y(t))$.

Also, this is about domains and codomains, not domains and ranges. Space-filling curves are pathological.

Okay yeah.

So would f(t) = <t^3 - t,t^2> be a curve of form R -> R^2?

[quote="Simfish"]So would f(t) = <t^3 - t,t^2> be a curve of form R -> R^2?[/quote]

Think about functions in terms of input and output.  In this case, your input is some real number t.  Your output is some set of ordered pairs (t^3 - t,t^2), such as (0,0), (0,1), (6,4), etc.

I like to think about functions like R->R^2 somewhat like embeddings.  The idea is that you imagine you're an ant walking along the curve.  It seems to you like the world is just a line.  It's a curvy line, but it's still just a line, not like a plane or "space".  You can only go forwards and backwards (one direction and its opposite).  That means that it's "one-dimension".  It happens to be sitting inside a plane (two dimensions), but the curve itself only has one dimension (forward/backward).  That "one dimension" is the t value.  It goes up or down (forward/backward).

A function R^2 -> R^3 is like a piece of paper in space.  You can bend the paper and wiggle it around, but if you're an ant on the page, you can go in two directions: forward/backward and left/right.  That's like a sphere (like Earth), N/S and E/W.  You make a function like that by having an INPUT of two variables, say t and s, and an output of three real numbers (an ordered triple), like this: f(t,s) = (t^2, s - 2t, st^3).  You can play with t to make  your ant move forwards/backwards, and you can play with s to make your ant move left/right.


[quote="tuanct"]An integer n>1 and a prime p are such that n divides p-1 and p devides $ n^{3}-1$.Show that 4p+3 is the square of an integer[/quote]

That's wrong.

With for instance $ n=2$ and $ p=7$

$ 2$ divides $ 6$, $ 7$ divides $ 2^{3}-1$ but $ 4*7+3=31$ is not a perfect square.

I thank that the correct question is:"Show that 4p-3 is the square of an integer and not 4p+3"
this problem is already posted.


LaTeX
inequalities proposed
inequalities

prove that    a*(n+1)/a*n+b*n  +b*(n+1)/b*n+c*n   +c*(n+1)/(c*n+a*n)  >=a+b+c/2

for positive reel numbers :mad:  :D

Please use latex and explain your problem clearly.

If you mean
$\sum_{cyc}\frac{a^{n+1}}{a^n+b^n}\ge\frac{a+b+c}2$ I think it is not true for $n>6$, see
http://www.mathlinks.ro/Forum/viewtopic.php?t=22937 and
http://www.mathlinks.ro/Forum/viewtopic.php?p=382832.

On the contrary, for $\sum_{cyc}\frac{a^{n+1}}{b^n+c^n}\ge\frac{a+b+c}2$,
we can prove easily
$\sum_{cyc}\frac{a^{n+1}}{b^n+c^n }\ge \sum_{cyc}\frac{a^{n}}{ b^{n-1} +c^{n-1}}$
with the same simple trick (let's call it "cyclical SOS") as [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=455429]this one[/url].

[quote="Albanian Eagle"]I think its not true for n>4[/quote]
I think you are wrong. See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23113&postorder=asc
post 10. ;)


vector
group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ p$ be a prime number and $ G$ be a pro-$ p$-group, I want to prove that if $ H^1(G)\equal{}Hom(G,\mathbb{Z}/p\mathbb{Z})\equal{}0$ then $ G\equal{}\{1\}$. Little hint, please, Thanks in advance for your help.

This is just a property of finite p-groups:

For a (pro-) p-group G, Hom(G,Z/pZ) = Hom( G/Φ(G), Z/pZ ) is the dual vector space to G/Φ(G), so has dimension equal to minimum number of generators of G.

Here Φ(G) = G^p[G,G] is the (closure of the) subgroup generated by all g^p and [g,h] for g,h in the (pro-) p-group G.


when u cut onions in to small pieces i was told that the cell wall breaks, but what happens when u put the onion juice into a solution with detergent?? isnt that where the cell wall really breaks?? --(to extract DNA)

if this is confusing and ur willing to help just give a message for further explanaations...

thank you

I thinkyou're getting the cell wall confused with the cell membrane. DNA in plant cells is enclosed not only by a cell wall but also by two separate membranes - a cellular one and a nuclear one. It's been a while since I've done this, so I might be wrong, but I believe the detergent is helpful because it breaks down those membranes allowing more DNA to escape. Cutting the onions beforehand is necessary because it ruptures cell walls and cell membranes and allows more cells to be exposed to the detergent. Cutting the onions with a knife would be unlikely to break very many of the nuclear membranes inside.


algebra
polynomial
number theory proposed
number theory

Polynomial division

a)	For what values of k is x-2 a factor x^4 – 5x^3 + 3x + k in Q[x]?

b)	For what values of k is x+1 a factor of x^4 + 2x^3 – 3x^2 + kx + 1 in Z5[x]

[b]Lemma:[/b]  $ x \minus{} a$ is a factor of a polynomial $ P(x) \in K[x]$ (where $ K$ is a field) if and only if $ P(a) \equal{} 0$.


ceiling function
logarithms
floor function
function
algebra

Let $2^{2005}$ have $x$ digits, and $5^{2005}$ have $y$ digits. Find $x+y$

2005 log2 +2005 log5=2005
so there're 2002 digits

that is close, [hide="answer"]but consider that you are taking $\lceil 2005\log{2} \rceil+\lceil 2005*\log{5} \rceil$, so the actual answer is $2006$[/hide]

Okay, now with this do you round up the values to integers, or do you make it the closest integer?
Also, is there any way to see without a calculator the value of
$2005*\log(2)$
or
$2005*\log(5)$
Thanks in advance

[quote="ljh"]2005 log2 +2005 log5=2005
so there're[b] 2002 [/b]digits[/quote]
There may be typo here, $10^{2005}$ does have $2006$ digits ..The answer is in fact $2006$

One way to thnk about this,   is that the number of digits  in a product (say $m \cdot n$) is:
 either sum of the digits  of $m$ and $n$ (if there is a "carry over")
 Or One less than the sum of digits (If there is no carry over-  in the last digit)

So sum of digits, is number of digits in the product ($2^{2005} \cdot 5^{2005}= 10^{2005}$)  or $2006$ since there is a carry over.
(The method is essentially same as ljh's)

Also one  "easy fact " to remember is $2^{10} = 1024 = little\_bit\_more\_than \ 10^3$ so $log 2$ is a little more than $0.3$ (about $0.30103$)..
Once you know $log 2$  you can estimate $log 4$ (=2*log 2)   or $log 5$ (=(1-log 2)) ... etc

just to be a bit more rigirous

[hide]from my previous post, we have

$\lceil 2005\log{2} \rceil+\lceil 2005*\log{5} \rceil$

using $\lceil x \rceil = -\lfloor -x \rfloor$

$=-\lfloor -2005\log{2} \rfloor-\lfloor -2005*\log{5} \rfloor$

then using $\lfloor x \rfloor = x - \{x\}$

$=2005\log{2}+2005\log{5}+\{ -2005\log{2} \}+\{ -2005*\log{5} \}$
$=2005+\{ -2005\log{2} \}+\{ -2005*\log{5} \}$

since the top statement was an integer, the bottom statement (being equal to the top) must be an integer, also by the definition of the floor function, we know that $0\le \lfloor x \rfloor <1$ so

$0\le \lfloor x \rfloor + \lfloor y \rfloor<2$, and we know that they have fractional parts, so the only possiblilty is that part =1, so

$=2005+1=\boxed{2006}$[/hide]


If $ a,b,c$ be non-negative real numbers and $ abc\equal{}1$ Prove that \[ \sum_{cyc}\frac{a}{(\sqrt{ab}\plus{}\sqrt{a}\plus{}\sqrt{b}\plus{}1)^2} \geq \frac{3}{16}\]

[quote="apollo"]If $ a,b,c$ be non-negative real numbers and $ abc \equal{} 1$ Prove that
\[ \sum_{cyc}\frac {a}{(\sqrt {ab} \plus{} \sqrt {a} \plus{} \sqrt {b} \plus{} 1)^2} \geq \frac {3}{16}
\]
[/quote]
By Cauchy-Schwarz inequality, we have:
$ \sum \frac{a}{(\sqrt {ab} \plus{} \sqrt {a} \plus{} \sqrt {b} \plus{} 1)^2}\equal{}\sum \frac{a}{(\sqrt{a}\plus{}1)^2(\sqrt{b}\plus{}1)^2} \ge \sum \frac{a}{4(a\plus{}1)(b\plus{}1)}$
So, we need to prove:
$ \sum \frac{a}{(a\plus{}1)(b\plus{}1)} \ge \frac{3}{4}$
$ \Leftrightarrow 4(a(c\plus{}1)\plus{}b(a\plus{}1)\plus{}c(b\plus{}1)) \ge 3(a\plus{}1)(b\plus{}1)(c\plus{}1)$
$ \Leftrightarrow a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca \ge 6$
Which is clearly true by AM-GM: $ a\plus{}b\plus{}c\plus{}ab\plus{}bc\plus{}ca \ge 6\sqrt[6]{(abc)^3}\equal{}6$
Equality holds if and only if $ a\equal{}b\equal{}c\equal{}1$

Putting $ x\equal{}\frac{a}{b},y\equal{}\frac{b}{c},z\equal{}\frac{c}{a}$ the initial inequality is equivalent to

$ \sum_{cyc} \frac{xz}{(\sqrt x \plus{} \sqrt y)^2(\sqrt x \plus{} \sqrt y)^2} \geq \frac{3}{16}$

Now using $ Q\minus{}A$ for the denominator we get:

$ \sum_{cyc} \frac{xz}{(\sqrt x \plus{} \sqrt y)^2(\sqrt x \plus{} \sqrt y)^2} \geq \sum_{cyc} \frac{xz}{4(x \plus{} y)(y \plus{} z)}$

Now again a substitution $ x\plus{}y\equal{}2p, y\plus{}z\equal{}2r, z\plus{}x\equal{}2s$ gives the equivalent inequality

$ \sum_{cyc} \frac{(p\plus{}s\minus{}r)(s\plus{}r\minus{}p)}{pr} \geq 3$, and this is schur..


Seventy-two 50-cent pieces are lined side-by-side on a table. Joe replaces every second 50-cent piece with a quarter and leaves. Jane replaces every third remaining 50-cent piece with a dime and leaves. Jim replaces every fourth remaining 50-cent piece with a nickel and leaves. What is the number of dollars in the amount of money left on the table? Express your answer to the nearest cent.

[hide]
So you start with 72 half dollars.
36 are changed into quarters.
36 left.
12 are changed into dimes
24 left
6 are changed into nickels.
18 left

So 18 half dollars, 36 quarters, 12 dimes, 6 nickels
$\$9+\$9+\$1.20+\$0.30=\$19.50$[/hide]


geometry
geometric transformation
reflection
calculus
calculus computations

When real numbers $ a,\ b$ vary subject to $ a^2 \plus{} b^2 \equal{} 1$, find the region in which the line $ (1 \plus{} a)x \plus{} (1 \minus{} a)y \equal{} b$ can be passed.

all such lines have their intercepts of the same sign. moreover, the lines with negative $ b$ are just reflections across the origin of the lines with positive $ b$. so assume that $ b>0$. then the product of the $ x$- and $ y$- intercepts is constant ($ 1$) and the $ x$-intercept is $ \frac{b}{1\plus{}a}\equal{}\sqrt{\frac{1\minus{}a}{1\plus{}a}}$, which takes on all positive values. from this we gather that the lines sweep out the entire 2nd and 4th quadrants and the region inside the curves in the 1st and 3rd quadrants with the property that all their tangent lines have intercepts of product $ 1$. (by "inside" i mean the side with the origin.)

so what are these curves? it's easy to check that the two branches of $ y\equal{}\frac{1}{4x}$ satisfy the above properties, so they must be the curves defining our region.

[quote="new_member"]kunny,could you provide any hints?
I looked the problem for a while and I can't figure it out :([/quote]

Suppose that the line passe through the point $ (1,\ 1)$, $ a,\ b$ would satisfy $ b\equal{}2$ and $ a$ is arbiterary real numbers, but these  don't satisfy the condition $ a^2\plus{}b^2\equal{}1$, thus the line will not pass the point $ (1,\ 1)$.

[quote="pluristiq"][quote="kunny"][quote="new_member"]kunny,could you provide any hints?
I looked the problem for a while and I can't figure it out :([/quote]

Suppose that the line passe through the point $ (1,\ 1)$, $ a,\ b$ would satisfy $ b \equal{} 2$ and $ a$ is arbiterary real numbers, but these  don't satisfy the condition $ a^2 \plus{} b^2 \equal{} 1$, thus the line will not pass the point $ (1,\ 1)$.[/quote]

i think he understood the problem statement. he just wanted a hint towards a solution.[/quote]

We have 3 Solution at least. I was at loss whether this problem should be posted in the forum or not.
Because the problem can be solved in a basic solution. I should have posted the problem in Inetermidate section.


The ratio of the measures of the angles of a triangle is 3:2:1. Given that the shortest side of the triangle is 12 meters long, what is the number of meters in the longest side of the triangle?

Someone please post how to do this problem!

We have $ 3x \plus{} 2x \plus{} x \equal{} 180\implies x \equal{} 30$, which says that our triangle is a $ 30$-$ 60$-$ 90$ right triangle. Therefore, the longest side is the hypotenuse and therefore $ 2(12) \equal{} \boxed{24}$.


limit
logarithms
algebra unsolved
algebra

Let $x_n$ be root of equation: $x^n=x^2+x+1$.Find $\displaystyle \lim_{n \rightarrow \infty}n(x_n-1)$

Short answer: [hide]The limit is ln(3).[/hide]

It is clear $x\to 1$, so $x^{2}+x+1\to 3$, therefore it is impossible that $x>1+\frac{\ln(3+\varepsilon)}{n}$ or $x<1+\frac{\ln(3-\varepsilon)}{n}$ for large $n$.

We have $x_{n}\longrightarrow 1$ and $x_{n}^{n}=x_{n}^{2}+x_{n}+1$, hence $x_{n}^{n}\longrightarrow 3$. Therefore, $n\ln x_{n}\longrightarrow \ln 3$. Since $\lim_{n\rightarrow+\infty}\frac{\ln x_{n}}{x_{n}-1}=1$, it follows that $\lim_{n\rightarrow+\infty}n(x_{n}-1)=\ln 3.$

Can be proved, that \[\lim_{n\to \infty }n^{2}(x_{n}-1-\frac{\ln3}{n}) =\frac{3\ln 3 }{2}.\]


A juggler throws a ball straight up into the air with a speed of 10m/s, with what speed would she need to throw a second ball helf a second later, starting from the same position as  the first, in order to hit the first ball at the top of its trajectory?

$ v_{f}\equal{}10\minus{}gt\Rightarrow t\equal{}g/10$ seconds for the first ball to reach the top of its path.  This happens at a height of $ 0\equal{}100\minus{}2gd\Rightarrow d\equal{}50/g$ meters.  The second ball must reach a height of $ 50/g$ meters in $ (g\minus{}5)/10$ seconds.  So
$ \frac{50}{g}\equal{}v_{0}\frac{g\minus{}5}{10}\plus{}g\frac{g\minus{}5}{20}$
Then you can solve for $ v_{0}$.  I would check my math though, I did this pretty quickly.

Did you go based on my work or independently?  If you used my work, check it.

$ h \equal{} 10t \minus{} 4.9 t^2$ for the original equation.
Solving for the maximum, it occurs at $ \frac {5}{4.9}$ and has a height of $ \frac {25}{4.9}$.

In terms of a time of reference, the second ball should hit the other ball $ \frac {2.55}{4.9}$ seconds after it is thrown.

Setting another equation in the form of $ h \equal{} st \minus{} 4.9 t^2$, we have a new equation (remember we are solving for $ s$:

[hide="go through the algebra..."]$ \frac {25}{4.9} \equal{} s \cdot \frac {2.55}{4.9} \minus{} 4.9 \cdot \frac {2.55^2}{4.9^2}$.  Multiplying both sides $ 4.9$, we get
[hide="and..."]$ 25 \equal{} 2.55s \minus{} 2.55^2$; solving for $ s$ we get
$ s \equal{} \frac {25 \plus{} 2.55^2}{2.55}$

$ \equal{} \boxed{12 m/s}$ (using 2 significant figures).  I got the same answer as kysuke[/hide][/hide]

The key to many problems like this is being able to set the right equation with the right variables.  If you understand what's being asked, you should be able to solve it.

i m rusted :(
calculation mistake :maybe:


Quick question for a project: has the completeness of the real numbers been proven or is it axiomatic, and if it has been proven where can I find a proof? Thank you.

This might be of interest:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17848[/url]

There are at least three available constructions of the real numbers. Each has its own strengths and weakness, and all involve some fairly tedious arguments.

1. Dedekind cuts. A number of analysis textbooks do go through this construction, and it is the oldest. Dedekind made it explicit, but you could argue that it is foreshadowed by the Eudoxus defenition of proportion as written in Euclid's [i]Elements.[/i]

2. As equivalence classes of Cauchy sequences of rational numbers.

3. As decimal (or whatever base you prefer) expansions.

The tedium arises in each construction because you need to prove that you have constructed a complete ordered field. That's quite a few axioms to verify.

One can also prove that given any two complete ordered fields, one can find an order-preserving isomorphism between them. This theorem then proves that the real numbers are essentially unique.

All of these constructions are abstract and non-intuitive. I might even go so far as to call them  "artificial." Don't read too much into the word "real" - this is a product of human abstract thought.


Cau cau. Dostaly se mi do rukou jedny od kamarada, ktery byl minuly rok v Jankach. Tak jednu z nich posilam.

Mirek

Soustředění v Jánkách je fajn, Doc. Šimša nám loni rozdal spoustu dobrých příkladů z teorie čísel, ty nejlepší ukázal na přednášce. Doc. Švrček přednášel některé zajímavé věty z planimetrie (Menelaova, Cevova, Van Aubelova), Dr. Leischner na přednášce odvodil Ptolemaiovu nerovnost, Bretschneiderovo pravidlo a vzorce pro obsahy čtyřúhelníků (obecných). Dr. Zhouf mluvil o součtech aritmetických a jiných posloupností.
A přednáška doktora Horáka zahrnující několik planimetrických příkladů ... :)


(1)The people living in a small village are supporters of either teams A, B, C or D. In terms of political parties, they are supporters of either parties E, F or G. Some of these people are liars while the rest are truth tellers. The liars lie at every answer they give, while the truth tellers always tell the truth.
The questions (first four questions were asked to all of the villagers, next three were asked to all of the male villagers, and the last three were asked to all of the female villagers) and the number of people that answered "yes" are as follows:

Do you support A ? (50 Yes answers)
Do you support B ? (60 Yes answers)
Do you support C ? (70 Yes answers)
Do you support D ? (80 Yes answers)

Do you support E ? (25 Yes answers)
Do you support F ? (30 Yes answers)
Do you support G ? (35 Yes answers)

Do you support E ? (30 Yes answers)
Do you support F ? (40 Yes answers)
Do you support G ? (50 Yes answers)

How many people are living in this village?

[hide="answer"]160[/hide]

[hide="solution"] We do a system of equation.
Notation is natural:
$ T_{a}$ - number of truth tellers supporters of A, etc
$ T_{e}^{m}$ - number of truth tellers male supporters of E, etc

$ T_{a} \plus{} L_{b} \plus{} L_{c} \plus{} L_{d} \equal{} 50$
$ L_{a} \plus{} T_{b} \plus{} L_{c} \plus{} L_{d} \equal{} 60$
$ L_{a} \plus{} L_{b} \plus{} T_{c} \plus{} L_{d} \equal{} 70$
$ L_{a} \plus{} L_{b} \plus{} L_{c} \plus{} T_{d} \equal{} 80$
A sum: $ T \plus{} 3L \equal{} 260$

$ T_{e}^{m} \plus{} L_{f}^{m} \plus{} L_{g}^{m} \equal{} 25$
... (i'am lazy)
$ L_{e}^{f} \plus{} L_{f}^{f} \plus{} T_{g}^{f} \equal{} 50$
A sum: $ T \plus{} 2L \equal{} 210$

The rest is clear: $ T \plus{} L \equal{} 160$
[/hide]


we have

$\lim_{x \to +\infty}\left(1+\frac{1}{x}\right)^x=e$

from which

$\lim_{x \to 0^+}(1+x)^{\frac{1}{x}}=e$

from which

$\lim_{x \to 0^+}\frac{\ln(1+x)}{x}=1$

and from which

$\lim_{x \to 1}\frac{\ln(x)}{x-1}=1$

by some substitutions.

so in our limit:

$\lim_{x \to 1}\frac{x\frac{\ln x}{x-1}-1}{x-1}=\lim_{x \to 1}\frac{x\cdot 1 -1}{x-1}=1$

Everything works until your last step.  It is unfortunately not valid in general to take a limit of only some of the occurences of $x$ but not others.

I would personally solve this with a [hide]Taylor series,[/hide] but there is probably another way.

[quote="mathe"]Find
$\lim_{x \to 1} \frac{\frac{xlnx}{x-1}-1}{x-1}$
don't use hopital[/quote]

hopital??

[quote="JBL"]He means [url=http://mathworld.wolfram.com/LHospitalsRule.html]L'Hospitals Rule[/url].  (In fact, the question can be solved reasonably easily with it -- I assume the author is looking for a distinct solution.)[/quote]

Ohh I see thanks...


logarithms
induction
algebra proposed
algebra

$ 0<a<b$ find all $ f: [a,b]\minus{}>[a,b]$ such that  $ f(x)f(f(x)) \equal{} x^2$ for all $ x$ in $ [a,b]$

$ g(x): \equal{}\ln f(e^x): [\ln a,\ln b]\to [\ln a,\ln b]$ with $ g(x)\plus{}g(g(x))\equal{}2x$ for all $ x\in [\ln a,\ln b]$.
Then by induction
$ g^n(x): \equal{}\underbrace{g\circ \cdots\circ g}_{n\text{ times}}(x)\equal{}{2x\plus{}g(x)\over 3}\plus{}(\minus{}2)^n{x\minus{}g(x)\over 3}$
and $ g^n(x)\in [\ln a,\ln b]$ for all $ x\in [\ln a,\ln b]$ and $ n\in\mathbb{N}$.
So $ g(x)\equal{}x$ for all $ x\in [\ln a,\ln b]$
and $ f(x)\equal{}x$  for all $ x\in [ a, b]$.

See here : http://www.mathlinks.ro/viewtopic.php?t=193350


1. Which of the following is equivalent to $ \log{\left(\frac {\sin{(2a)}}{2}\right)}$.

$ (1) \ \log{(\sin{a})}$

$ (2) \ \frac {1}{2}\log{(2\sin{a})}$

$ (3) \ \log{(\sin{a})}\log{(\cos{a})}$

$ (4) \ \log{(\sin{a})} + \log{(\cos{a})}$

Could you please show work to both of these problems so that I can learn how to do them?

Thanks.

[EDIT] Sorry, but I typed the wrong the question.

[quote="314.math"]Hmmm...

These are interesting questions.  Does anyone have a solution?[/quote]

i dunno
the only time i've seen trig involved with logs is in integrals...
and that path really does not look optimistic

the log property log_n(x)^y=ylog_n(x) will get you the answer for number 1

Then tan x=sinx/cosx

[hide]$ \log(tan^2(y))$ can be rewritten as $ \log(sin^2y) - \log(cos^2y)$
This is the same as:      $ 2\log(siny) - 2\log(cosy)$

                                 $ 2(\log(siny) - \log(cosy))$[/hide]

Edit: The problem that the above solution corresponded to was removed.

Solution to New #1

[hide]$ \log{\left(\frac {\sin{(2a)}}{2}\right)}$ can be written as

$ \log{sin(2a)} - \log{2}$ using properties of logs. Using trig identity $ sin(2a) = 2sin(a)cos(a)$ we can rewrite as

$ \log{2sin(a)} - \log{2} + \log{cos(a)}$

Simplifying, we get $ \log{sin(a)} + \log{cos(a)}$[/hide]

[quote="cognos599"]well, this may seem like spam. i have no idea how to do this specific problem, but here is a logarithmic and trigonometric problem.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=409912#409912[/quote]

That doesn't help at all.

What Illos has is correct.

[hide="Classic Log Properties"]

Just note that:

$ \log(ab) \equal{} \log a \plus{} \log b$
$ \log(a/b) \equal{} \log a \minus{} \log b$

Those two were only things you needed to know for this problem. [/hide]

Well in that case then, these problems are extremely simple...finding an equal expression can mean just multiplying it by 2/2 or something :P

How did you get from $ \log{2sin(a)} \minus{} \log{2} \plus{} \log{cos(a)}$

to

Simplifying, we get $ \log{sin(a)} \plus{} \log{cos(a)}$?

Thanks.

[hide]
Definition
$ \sin \left( 2 \alpha \right) \equal{} 2 \sin \alpha \cos \alpha$.

Then
$ \log \left( \frac {\sin \left( 2 \alpha \right) }{2} \right) \equal{} \log \left( \frac {2 \sin \alpha \cos \alpha}{2} \right) \equal{} \log \sin \alpha \plus{} \log \cos \alpha$.
Which is $ (4)$.
[/hide]

[quote="elcric"]How did you get from $ \log{2sin(a)} - \log{2} + \log{cos(a)}$

to

Simplifying, we get $ \log{sin(a)} + \log{cos(a)}$?

Thanks.[/quote]

$ \log{2sin(a)}$ is the same as $ \log{2} + \log{sin(a)}$.

Putting this back into the original, we get $ \log{2} + \log{sin(a) - \log{2} + \log{cos(a)} = \log{sin(a) + \log{cos(a)}}}$


calculus
integration
logarithms
function
calculus computations

Any help is greatly appreciated. How would you do:

$\int{\frac{x}{x^{4}+2x^{2}+1}}$

The denominator is a perfect square. Write it as such, and look for a substitution.

Thanks! Further problem I'm having: I cannot figure out the mistake in one of these two methods of integration.

$\int{\frac{1}{2x+4}}=\frac{1}{2}\ln{(2x+4)}$

$\int{\frac{1}{2x+4}}=\frac{1}{2}\int{\frac{1}{(x+2)}}=\frac{1}{2}\ln{(x+2)}$

Clearly there is a problem, since these give $\frac{1}{2}\ln{(2x+4)}=\frac{1}{2}\ln{(x+2)}$

considering constants of integration, two result is same. ($\frac{1}{2}\ln{(2x+4)}-\frac{1}{2}\ln{(x+2)}= \ln\sqrt{2}$, so they differ by only a constant function as expected)

[quote="bthings_2000"]Clearly there is a problem, since these give $\frac{1}{2}\ln{(2x+4)}=\frac{1}{2}\ln{(x+2)}$[/quote]
I'm repeating what sos440 said, but just for emphasis: that's not what "these give." What you can actually derive from this result is that

$\frac{1}{2}\ln{(2x+4)}=\frac{1}{2}\ln{(x+2)}+C,$ which is true. Both computations are correct.


MATHCOUNTS
geometry
LaTeX
FTW
3D geometry
octahedron
email

HI EVERYONE. I WAS WONDERING WHETHER ANY OF YOU WOULD BE WILLING TO JOIN MY MOCK MATHCOUNTS TOURNAMENT... I NEED HELPERS CUZ IM NOT REALLY GOOD WITH ATTACHMENTS AND STUFFS AND I NEED MORE PROBLEMS!!!

[b]I AM GOING TO HOLD THIS IN 2-3 WEEKS WHEN PEOPLE HAVE FINISHED SMASHER"S AND MAYBACH's[/b]

[hide="Participants"]Please put your names here and copy and paste if interested. Thanks[/hide]

K cool. Barbie can you walk me through attaching pdf's and stuff??

[hide="Helpers(1)"]
BarbieRocks
[/hide]
[hide="Participants(2)"]
agentcx
Maybach
[/hide]

I know how to create PDFs using open office.  I could show you if you PM me(I still want to be a participant though.)

[hide="Participants(3)"]agentcx
Maybach
Bluecarneal[/hide]
[hide="Helpers(1)"]
BarbieRocks [/hide]

[hide="Participants(3)"]agentcx
Maybach
Bluecarneal
PowerOfPi[/hide]
[hide="Helpers(1)"]
BarbieRocks [/hide]

Wow...we are really getting too many tourneys around here.
But since CRICKET229 kindly joined mine,  :wink:
Ill be a helper.

[hide="Participants(4)"]agentcx
Maybach
Bluecarneal
PowerOfPi[/hide]

[hide="Helpers(2)"]KingSmasher3
BarbieRocks[/hide]

I probably could help but I don't think I should because I suck at computer stuff.
[hide]agentcx
Maybach
Bluecarneal
PowerOfPi
The Hobbit[/hide]

[hide]KingSmasher3
BarbieRocks[/hide]

Okay cool, King and Barbie if you can please pm your suggestions and other stuffs. Also I am kind of new to making pdfs and such (like making a Word Document into a PDF file) so I'll take any suggestions or help in that too.

My School Round will most likely be out after Jan 5th.

[hide="Participants (5)"]agentcx
Maybach
Bluecarneal
PowerOfPi
bbgun34[/hide]

[hide="Helpers (2)"]KingSmasher3
BarbieRocks[/hide]

As soon as I have at least 10 participants, we will start so never mind about the January thing. I just did a quick rough draft of sprint round. Looks easy at first and gets pretty hard at the end but dont worry chapter will be easier i hope. King and Barbie I'll send you my sprint and target when I'm done, please critique my stupid questions and add your own.

No problem;
Also, [color=red]please don't mind my due dates because mine is a [b]biannual [/b]competition, a.k.a. its continuous.[/color]

If possible, I would like to help. I can easily create PDFs, and type up the problems, except for the geometry ones. If I can't help, I would like to participate. BTW, Cricket, don't make the problems to hard LOL.

KK thats fine. I will pm you and king and barbie my rough school sprint

Hey, you people need to submit. Cuz if you dont submit this you will not be participating in the State Competition :D

Can we not do the team? cause really no one is doing it.

Yup. It seems as if this thread is basically dead.

[quote="explogabloger"]sorry, is it too late for me to sign up?[/quote]

Though, it is too late for you to sign up, you can still take the test, and pm Cricket for the answers so that you can check how you did. Look back in the previous pages (of this thread) to find copies of the test.

Just asking here (and sorry for reintroducing topic):

Who wants to take a national sprint and target (and maybe team). Anyone can take it, however it will be due fairly soon so.....heres the participants list. Only up to 16 may join so sign up quickly.

[hide="Participants(0)"]

[/hide]
[hide="Helpers(0)"]

[/hide]

I need 2 helpers as well. So please sign up if interested.

Thanks!
~CRICKET

I will help.

[hide="Helpers"]
$/LaTeX$
[/hide]

[hide="Participants"]
ksun48
[/hide]

Please copy the lists.

Cricket, feel free to PM me or email me, you know my email.

Not too much excitement so far.

Where should I advertise?

Oh, and Latex I couldnt find the template on agmath so where can I find it?

Signature!!! Best place.
Also the template is [url]http://www.agmath.com/169763.html[/url]
Under the Warm-ups.

Oops. This thread didn't have too much excitement so I quit going to this for some time. At least one topic is back. :D

I join.

Hmm so not much excitement so I just made a target.....

here It is:

I'm assuming (for the DoodleJump question) you don't get points for altitude?

Lol...nope. You would in a regular doodle jump game though.


probability
calculus
derivative
expected value

In a game, a player tosses a biased coin repeatedly until two successive tails occur, when the game terminates. For each head which occurs the player wins $ \$ 1$. If $ E$ is the expected number of tosses of the coin in the course of a game, and $ p$ is the probability of a head, explain why
\[ E \equal{} p(1 \plus{} E) \plus{} (1 \minus{} p)p(2 \plus{} E) \plus{} 2(1 \minus{} p)^2 ,
\]
and hence determine $ E$ in terms of $ p$. Find also, in terms of $ p$, the expected winnings in the course of a game.

A second game is played, with the same rules, except that the player continues to toss the coin until $ r$ successive tails occur. Show that the expected number of tosses in the course of a game is given by the expression $ \frac {1 \minus{} q^r}{pq^r}$, where $ q \equal{} 1 \minus{} p$.

I don't know how to post a tree-diagram, but we can consider three possibilities:

$ {H}$: game starts again; probabilty $ \equal{} p$,; expected number of games $ \equal{} 1 \plus{} E$.
$ {TH}$: game starts again; probability $ \equal{} (1 \minus{} p)p$; expected number of games $ \equal{} 2 \plus{} E$,
$ {TT}$: games ends; probability $ \equal{} (1 \minus{} p)^{2}$; expected number of games $ \equal{} 2$.

These three sequences are mutually exclusive and collectively exhaustive, so the expected number of games is:

$ E \equal{} p(1 \plus{} E) \plus{} (1 \minus{} p)p(2 \plus{} E) \equal{} 2(1 \minus{} p)^{2}$

$ \equal{} p \plus{} pE \plus{} 2p \minus{} 2p^{2} \plus{} (p \minus{} p^{2})E \plus{} 2 \minus{} 4p \plus{} 2p^{2}$

Therefore $ (1 \minus{} 2p \plus{} p^{2})E \equal{} 2 \minus{} p$

(My backslash key doesn't work: I'll copy one and get back to you.)

Therefore $ E \equal{} \frac {2 \minus{} p}{(1 \minus{} p)^{2}}$

Expected winnings, $ W \equal{} p(1 \plus{} W) \plus{} (1 \minus{} p)p(2 \plus{} W) \plus{} 0.(1 \minus{} p)^{2}$

This gives $ W \equal{} \frac {p(2 \minus{} p)}{(1 \minus{} p)^{2}}$

Have to eat. Will try to get back later. Hope this helps.

Back again.

Second game:

$ E\equal{}p(1\plus{}E) \plus{} pq(2\plus{}E) \plus{} pq^{2}(3\plus{}E) \plus{} ... \plus{} pq^{r\minus{}1}(r\plus{}E) \plus{} rq^{r}$

$ \equal{} p(1 \plus{} 2q \plus{} 3q^{2} \plus{} .. \plus{} rq^{r\minus{}1}) \plus{}pE(1\plus{}q\plus{}q^{2} \plus{} ... \plus{} q^{r\minus{}1}) \plus{} rq^{n}$

$ \equal{}p\frac{d\sum_{n\equal{}1}^{r}q^{n}}{dq} \plus{}pE\frac{(1\minus{}q^{r})}{1\minus{}q} \plus{} rq^{r}$

$ \equal{}p\frac{d}{dq}\frac{(q(1\minus{}q^{r})}{(1\minus{}q)} \plus{}pE\frac{(1\minus{}q^{r})}{1\minus{}q} \plus{} rq^{n}$

$ \equal{}\frac{p(1\minus{}(r\plus{}1)q^{r} \plus{} rq^{r\plus{}1}}{(1\minus{}q)^{2}} \plus{} pE\frac{(1\minus{}q^{r})}{1\minus{}q}\plus{} rq^{n}$

$ \equal{}\frac{(1\minus{}(r\plus{}1)q^{r} \plus{} rq^{r\plus{}1})}{p} \plus{} E(1\minus{}q^{r}) \plus{} rq^{n}$

$ \Rightarrow q^{r}E  \equal{} \frac{(1\minus{}(r\plus{}1)q^{r} \plus{} rq^{r\plus{}1})}{p} \plus{} rq^{r}$

$ \Rightarrow pq^{r}E \equal{} 1 \minus{} (r\plus{}1)q^{r} \plus{} rq^{r\plus{}1} \plus{} rpq^{r}$

$ \equal{} 1 \minus{} rq^{r} \minus{} q^{r} \plus{}rq^{r\plus{}1} \plus{} rpq^{r}$

$ \equal{}1\minus{}rq^{r}(1\minus{}q\minus{}p) \minus{} nq^{r}$

$ \equal{}1\minus{}q^{r}$

$ \Rightarrow E \equal{} \frac{1\minus{}q^{r}}{pq^{r}}$

I hope this helps you Aidan.  Having seen my rating drop, I contacted orl who assures me it is often vindictiveness. But I almost left mathlinjks.ro permanently and returned to my first love: languages. Being able (hopefully) to help you has renewed my hope and faith in humanity.

Thanks, AndrewTom! Second part helped loads. I forgot to "mask" $ 1 \plus{} 2q \plus{} 3q^2 \plus{} \ldots$ as the derivative of $ q \plus{} q^2 \plus{} q^3 \plus{} \ldots$ so I couldn't proceed. As for languages.. they're great too  :)  But, keep up the math. Best language ever!


linear algebra
matrix
algorithm
superior algebra
superior algebra unsolved

Hello,

I wrote this problem after inspired by a similar problem written by my friend. I don't have an elegant solution though, so please post yours :D

Ring $R = \{a+bi | a,b \in \mathbb{Z}\}$ has an ideal $M = (3+i)R$.
1. Prove that $M$ is not maximal, and find all the maximal ideals that contain $M$
2. Show that $Z/M$ is isomorphic to $\mathbb{Z}_{2}\times \mathbb{Z}_{5}$. Find /describe the isomorphism explicitly.

$\mathcal{R}$ is an euclidean ring.
1) Take one of the divisors of $3+i$: $1+i,1-i,2-i,1+2i$.
2) $\mathcal{R}\sim\mathbb{Z}^{2}$ and the product by $3+i$ is isomorphic to the product \[\begin{pmatrix}a\\b\end{pmatrix}\rightarrow{M}\begin{pmatrix}a\\b\end{pmatrix}{=}\begin{pmatrix}3&-1\\1&3\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}\]
Standard result: if $M\in{\mathcal{M}_{n}(\mathbb{Z})}$ has a non zero det then there exist $P,Q\in{GL_{n}(\mathbb{Z})}$ s. t. $PMQ=diag(d_{1},\cdots,d_{n})$ where $d_{i}>0,d_{i}|d_{i+1}$ (there exists only one such diag matrix and there exists an algorithm which gives explicitly a solution $P,Q$). Moreover the group $\mathbb{Z}^{n}/M(\mathbb{Z}^{n})$ is isomorphic to the product group $\mathbb{Z}_{d_{1}}\cdots\mathbb{Z}_{d_{n}}$.
Here \[M=\begin{pmatrix}3&-1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&10\end{pmatrix}\begin{pmatrix}1&3\\0&1\end{pmatrix}\] and $\mathbb{Z}^{2}/M(\mathbb{Z}^{2})$ is isomorpic to $\mathbb{Z}_{10}$. \[M(\mathbb{Z}^{2})=\begin{pmatrix}3&-1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&10\end{pmatrix}(\mathbb{Z}^{2})=\{\begin{pmatrix}3x-10y\\x\end{pmatrix}|x,y\in\mathbb{Z}\}\]. If we choose $x=0$ then $\mathbb{Z}^{2}/M(\mathbb{Z}^{2})$ is represented by $\{0,i,\cdots,9i\}$ with the relation $a+ib\approx{i((b-3a) mod 10)}$. It remains to prove the isomorphism for the ring structure:
for example: $(2+7i)(9+5i)\approx{i(-2i)}\approx{-6i}$ and in the other hand $-17+73i\approx{4i}$.
CAUTION: the product * over $\mathbb{Z}_{10}$ isn't the standard product: we have seen that $1*(-2)=-6$ and more generally $a*b=3ab$. Here the unit is $-3$. $gcd(3,10)=1$ thus this ring is isomorphic to the standard ring $\mathbb{Z}_{10}$ .


Let $p$ be an odd prime number and $2^{a}-1=1\cdot 3\cdot 5\cdot ...\cdot p$.
Prove that $a$ is divisible by $p-1$.

[i]Edited by Myth[/i]

who can help me slove this problem?thanks


Hey Guys...

I am entering my 2nd year in B.Sc. Math. I have some questions regarding the courses I am taking. I do the routine: take notes.. do all practice questions... "try" to start studyin well in advance for midterms and finals but like most of you know.. it’s pretty hard to do with a full course load every term. I would just like to know how to tackle these courses. I know what the courses are about 'cause i've looked at the course outlines and the texts a bit.

[b]Courses:[/b] Calc 3, Calc 4, Linear Algebra, Mathematical Proof(Sets and functions; induction; cardinality; properties of the real numbers; sequences, series, and limits. Logic, structure, style, and clarity of proofs emphasized throughout... and btw this course has the highest failure rate in the dept for some reason)

In addition to these 4 math courses... I am taking chem n some comp sci courses as well (planning to go into math/biochem program).

Thanks again !!!

This forum's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11946]FAQ about Preparing for College Math Study[/url] can be helpful for your questions. Follow the links, and practice the advice on each link.

[quote="punjm"]I am entering my 2nd year in B.Sc. Math. I have some questions regarding the courses I am taking.

...

Mathematical Proof ... this course has the highest failure rate in the dept for some reason) [/quote]

You didn't actually ask any questions, though.

Most likely, the Mathematical Proof class has a high failure rate because math as a rigorous, proof-based field seems quite different than math as conventionally taught through calc 2, and lots of people don't know how to make the adjustment.


What is the least integer value of $ n$ for which $ \left( \frac{2}{3}\right)^n < \frac{1}{5}$?

Use trial and error (that's what I did), and you find that $ \frac{16}{81} < \frac{1}{5}$, so the answer is $ \boxed{4}$.

hello, taking the logarithm of both sides we have $ n\cdot \ln(\frac {2}{3}) < \ln(\frac {1}{5})$. From this we have $ n > \frac {\ln(\frac {1}{5})}{\ln(\frac {2}{3})} \equal{} \frac { \minus{} \ln(5)}{\ln(2) \minus{} \ln(3)}\approx 3.969362295916118$. So the answer is $ \boxed{4}$.
Sonnhard.

i believe you mean $ \frac{\minus{}\ln(5)}{\ln(2)\minus{}\ln(3)}.$

Even though using natural logarithms may be the right way, for FTW, I recommend my way.

And don't you need a calc for $ ln x$?

At least I do...  :oops:

yes, your way is more effecient than Dr. Sonnhard Graubner's.

i was just pointing out Dr's mistake.


inequalities
combinatorics unsolved
combinatorics

try to solve the problem 3 / 4th sel test 2003, when the alphabet has only 3 letters.

What is the "problem 3 / 4th sel test 2003" or where is it possible to get it ? Thanks.

I think he means the Third problem from the fourth Romanian Selection test 2003.
The Romanian TST van be found at http://www.mathlinks.ro/Forum/resources.php?c=142&cid=26

THe original problem is :
A word consists of n letters from the alphabet {a, b, c, d}. A word is called convoluted if it has two consecutive identical blocks of letters. For example, caab and cababdc are convoluted, but abcab is not. Prove that the number of non-convoluted words with n letters is greater than 2^n.

So I think he means we should look for a similar inequality (and prove it of course) when the alphabet has only three letters.

As far as I remember this problem for four letters is in at least three threads on forum, but I never see solution.

Anyone, solve it at last!  ;)


search
counting
derangement
combinatorics unsolved
combinatorics

there are 4 balls of different colours and 4 boxes of colours same as those of the balls.find the number of ways in which the balls one each in a box could be placed such that a ball does not go to a box of its own colour

For 4 you might as well list everything by hand if you don't know the theory.  For the theory, search "derangement" or "rencontres."

hello
We have actually discussed this question in our institute when we were in xi th.
[hide="Solution"]
$ A_{i}$= no. of ways in which we can place a ball with color $ i$ in the box of that color only.
No. of required ways=No. of ways these balls can be placed- $ n(A_{1}$ union  $ A_{2}$  union  $ A_{3}$ union $ A_{4})$
=$ 4!\minus{}(4(3!)\minus{}6(2!)\plus{}4\minus{}1)\equal{}9$[/hide]
thank u

Kabi, yes you are correct but it would have been nice  if you had inroduced the genral derangements principle to aadil:
That is $ D_n \equal{} n! ( 1\minus{}\frac{1}{1!}\plus{} \frac{1}{2!} \minus{} \frac{1}{3!}\plus{} \frac{1}{4!}\minus{}............\plus{} (\minus{}1)^n \frac{1}{n!})$.

total number of ways of placing balls in boxes = $ 4!$.

number of permutations in which a ball goes into a box of same colour = 1.

so, required number of ways = $ 4!$ - $ 1$

                                           = $ 23$  :maybe:

Hello

[quote="Agr_94_Math"]Kabi, yes you are correct but it would have been nice  if you had inroduced the genral derangements principle to aadil:
That is $ D_n \equal{} n! ( 1 \minus{} \frac {1}{1!} \plus{} \frac {1}{2!} \minus{} \frac {1}{3!} \plus{} \frac {1}{4!} \minus{} ............ \plus{} ( \minus{} 1)^n \frac {1}{n!})$.[/quote]
i thought he would love a basic method. Anyways thank u. :)
[quote="Attila the Hun"]required number of ways =$ 4!\minus{}1$[/quote]
Aren't u forgetting other cases? :( Kindly edit ur Answer.

Thank u.


Prove that if $ p$ is prime $ (a,p) \equal{} (n,p\minus{}1)\equal{}1$, then the congruence $ ax^n \equal{} b (mod p)$ always has exactly one solution.

[hide]Suppose $ y$ is also a number satisfying $ ay^n \equiv b \mod p$. Then $ ax^n \equiv ay^n \mod p$, and since $ (a,p) \equal{} 1$, $ x^n \equiv y^n \mod p$.
The case where $ x \equiv 0 \mod p$ is trivial.
Suppose then that $ x$ is a unit mod $ p$. Let $ u$ be the inverse of $ x$. Then $ y^nu^n\equiv (yu)^n \equiv 1 \mod p$. If $ yu$ is not $ 1\mod p$ then the order of $ yu$ must not be relatively prime with $ p \minus{} 1$, which is not the case. So, $ yu \equiv 1 \mod p$. That is, $ u$ is the inverse of both $ x$ and $ y$, or $ x\equiv y$.[/hide]


calculus
derivative
inequalities
function
geometry
inequalities proposed
133

But I failed to find a nice solution. My solution was a very dirty and terrible solution such as $\frac{\partial}{\partial c}(\frac{\partial}{\partial c}(\frac{\partial}{\partial b}(\frac{\partial f}{\partial c}(a,b,c))))$.
Try to find a nice solution!

Problem : Let $a,b,c>0$.
Prove that
$\frac{a}{\sqrt{b^{2}+15ca}}+\frac{b}{\sqrt{c^{2}+15ab}}+\frac{c}{\sqrt{a^{2}+15bc}}\geq \frac{3}{4}$.

It took you  long time  :lol: .
I remind this is from thread:

http://www.mathlinks.ro/Forum/viewtopic.php?t=72297

I failed to solve it no matter how I tried.Partial derivatives too complicated for me  :( .
I'll PM few mathlinkers too see what they think about this ultra-difficult inequality.Can you post your ugly solution anyway?

Thanks for the reminder.
My solution is well outside frames of a high school math.
However,I vaguelly recall that I found the way soon after author first posted it on mathlinks.
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=595604#595604]Here[/url] you can see outline of such approach.
I tried to adjust it for mathlinkers with at least basic knowledge in analysis.
I couldn't find   an "elementar" solution for this problem,if that is what was meant by  "nice" solution.

Thanks for showing me your method.If anybody can,it is iurra who  cracks such inequalities in more elementar way. I think I heard he is a mastermind in inequalities (and not only there :) ).
I don't understand this comment of yours:
[quote="Parkdoosung"]Is $\frac{\partial^{2}F}{\partial x^{2}}(1,1,1)+\frac{\partial^{2}F}{\partial y^{2}}(1,1,1)+\frac{\partial^{2}F}{\partial z^{2}}(1,1,1)=0$ true?
If $K=15$, $\frac{\partial^{2}F}{\partial x^{2}}(1,1,1)+\frac{\partial^{2}F}{\partial y^{2}}(1,1,1)+\frac{\partial^{2}F}{\partial z^{2}}(1,1,1)=\frac{507}{2048}$,
so $\frac{\partial^{2}F}{\partial x^{2}}+\frac{\partial^{2}F}{\partial y^{2}}+\frac{\partial^{2}F}{\partial z^{2}}=0$ is not true.

[/quote]
I don't know if esti is right or wrong,but do you mean here.
If  x=y=z   then  $F(x,y,z)=\frac{x}{\sqrt{x^{2}+Kx^{2}}}+\frac{x}{\sqrt{x^{2}+Kx^{2}}}+\frac{x}{\sqrt{x^{2}+Kx^{2}}}=\frac{3}{\sqrt{1+K}}=constant$
Remember that K is  any fixed positive constant and not only that "$\triangle F=0$" but even $\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}+\frac{\partial F}{\partial z}=0$ is true for T(1,1,1) and arbitrary choice of K. Is that Right?

Are you sure you did the derivative right?
I can't verify that becouse I'm not much experienced in finding partial derivatives with so complicated terms.

If x=y=z than function in 3 variables F(x,y,z) becomes function in one variable F(x).
And that is constant.
I know how to do derivative of single function variable.
For F(x)= constant ,first derivative is 0,second is 0,third is  0 etc.
If you did all correctly,where I went wrong?I can't find mistake.
What is going on?    :wallbash:

Yes, my derivative is right.
F(x,y,z) does not become function in one variable F(x) although x=y=z.
First, calculate $\frac{\partial^{2}F}{\partial x^{2}}F(x,y,z)$ and see y and z is constant.
Next, use $y=z=x$.

If x=y=z than F(x,y,z)=constant.
I'm sure about that.
And the derivative of constant = [size=150]0[/size].
Sorry,I think u made mistake somewhere along the line,but I can't tell where. :(

[quote="Soarer"]Eros88, $F(x,y,z)$ is a constant when $x=y=z$ does not imply its partial derivative at $x=y=z=t$ is 0![/quote]
Oh,I think I'm begining to understand .But still I have difficulties .
How much is :$\frac{\partial^{2}F}{\partial x^{2}}+\frac{\partial^{2}F}{\partial y^{2}}+\frac{\partial^{2}F}{\partial z^{2}}$ for F(x,y,z);x,y,z>0?
Can you do that math for me  Soarer,please?

[i]Calculus`VectorAnalysis`[/i]

[b]F[X,Y,Z] :=X/Sqrt[Y^2+kXZ]+Y/Sqrt[Z^2+kXY]+Z/Sqrt[X^2+kYZ];

Grad[F[X,Y,Z],Cartesian[X,Y,Z]]
Div[%]
Laplacian[F][/b]

Outputs are:
[code]
{-kXZ/(2(Y^2+kXZ)^3/2 +1/(Y^2+kXZ)^1/2 -XZ/(X^2+kYZ)^3/2 -kY^2/2(kXY+Z^2)^3/2,
-XY/(Y^2+kXZ)^3/2 -kZ^2/(2(X^2+kYZ)^3/2 -kXY/(2(kXY+Z^2)^3/2 +
1/(kXY+Z^2)^1/2,
-kX^2/2(Y^2+kXZ)^3/2 -kYZ/2(X^2+ kYZ)^3/2 + 1/(X^2+kYZ)^1/2 -YZ/(kXY+Z^2)^3/2}

{0}

{0}[/code]

[quote="Soarer"]if you mean the original function, no thanks, I don't like monster computations. :D

Just want to give a simple example on eros88's problem.
Consider $F(x,y) = x-y$. Its partial derivative of x at (1,1) is 1, although $F(x,x)$ is constant.[/quote]
I'm also one lazy a##  when it comes to monster computations,as can be seen from above  :wink: .The results can be checked by hand althought I didn't do that.
From the discussion ,seems that Parkdoosung and Eros88 haven't mastered   multivariable calculus yet.
Not even to mention VectorFields theory.
This area of math becomes triviality once you get used to it.You can access the sketch of the proof from my first post in thread.I didn't post all the intersteps,and application of the harmonic function theorems.Who wants to do that and write strict step by step proof,  all the necessary and even more can be found in the links I  recommended.
I don't wan't to discuss the solution to the problem anymore.
Only,if somebody  finds something new  to nail down the inequality in "elementar" way , maybe then I'll become interested again.
Nothing guarantees us that every math problem in the world can be solved in elementar way.
What for mathematicians invented high level math tools ?I think the answers are clear even to a nonexpert.

It seems that esti have mastered multivariable calculus, but esti doesn't know very easy calculus and how to use computer programs.
Before studying harmonic function, you must study one variable calculus.

[quote="Eros88"]Thanks Soarer for this clarification.
I understand now.
Funny thing : that also  indicates that Parkdoosung could be wrong in his reasoning too (about sum of 3 partial derivatives).You got me thinking.
Bloody partial derivatives!I have to learn them good as soon as possible.
They are very important.[/quote]

Frankly speaking, I think there is no reason that the laplacian of the original function is constantly 0, after computing round round F wrt x once by hand.

[code]
{-kXZ/(2(Y^2+kXZ)^3/2) +1/(Y^2+kXZ)^1/2 -XZ/(X^2+kYZ)^3/2 -kY^2/(2(kXY+Z^2)^3/2),
-XY/(Y^2+kXZ)^3/2 -kZ^2/(2(X^2+kYZ)^3/2) -kXY/(2(kXY+Z^2)^3/2) +
1/(kXY+Z^2)^1/2,
-kX^2/(2(Y^2+kXZ)^3/2) -kYZ/(2(X^2+ kYZ)^3/2) + 1/(X^2+kYZ)^1/2 -YZ/(kXY+Z^2)^3/2}

[/code]
If esti's computer output is to be trusted first partial derivatives written clean are:
$\frac{\partial F}{\partial x}=-\frac{kxz}{2(y^{2}+kxz)^{\frac{3}{2}}}+\frac{1}{(y^{2}+kxz)^{\frac{1}{2}}}-\frac{xz}{(x^{2}+kyz)^{\frac{3}{2}}}-\frac{ky^{2}}{2(kxy+z^{2})^{\frac{3}{2}}}$
$\frac{\partial F}{\partial y}=-\frac{xy}{(y^{2}+kxz)^{\frac{3}{2}}}-\frac{kz}{2(x^{2}+kyz)^{\frac{3}{2}}}-\frac{kxy}{2(kxy+z^{2})^{\frac{3}{2}}}+\frac{1}{(kxy+z^{2})^{\frac{1}{2}}}$
$\frac{\partial F}{\partial z}=-\frac{kx^{2}}{2(y^{2}+kxz)^{\frac{3}{2}}}-\frac{kyz}{2(x^{2}+kyz)^{\frac{3}{2}}}+\frac{1}{(x^{2}+kyz)^{\frac{1}{2}}}-\frac{yz}{(kxy+z^{2})^{\frac{3}{2}}}$

Is that is OK guys?
Soarer,will you  do carefully the second partial derivatives in $x$, $y$ and $z$ variable please?It's long trail of calculations I know,but "to belive" is one thing and to know is another!I want to know.

[quote="Vasc"][quote="Parkdoosung"]But I failed to find a nice solution. My solution was a very dirty and terrible solution such as $\frac{\partial}{\partial c}(\frac{\partial}{\partial c}(\frac{\partial}{\partial b}(\frac{\partial f}{\partial c}(a,b,c))))$.[/quote]
I think, a similar inequality is the following:
Let $a,b,c>0$. Prove that
$\frac{a}{\sqrt{b^{2}+3c^{2}}}+\frac{b}{\sqrt{c^{2}+3a^{2}}}+\frac{c}{\sqrt{a^{2}+3b^{2}}}\geq \frac{3}{2}$.[/quote]
Similar but not the same.
I think Parkdoosung ineq. is a little harder than this one .
Do you have  elementar proofs for these inequalities (I mean without analysis)?
And what about my question about  sum :$\frac{\partial^{2}F}{\partial x^{2}}+\frac{\partial^{2}F}{\partial y^{2}}+\frac{\partial^{2}F}{\partial z^{2}}$? , (x,y,z or a,b,c is just a matter of notation),for Parkdoosung ineq.,when does the sum equals 0?

I have not an elementary solution to these inequalities. :D
But a nice similar inequality is
$\frac{a}{\sqrt{3b^{2}+ca}}+\frac{b}{\sqrt{3c^{2}+ab}}+\frac{c}{\sqrt{3a^{2}+bc}}\geq \frac{3}{2}$.

[quote="Parkdoosung"](1) Let $F(a,b,c)=\sum_{cyc}\frac{a}{\sqrt{b^{2}+15ca}}$.
If $a\geq b\geq c\geq$, $F(a,b,c)\geq F(b,a,c)$[/quote]
Why this is true?

To Ji chen :
Can you show your Comment about Value worth of k ? I think if it true , then it isn't  actual easy

[quote="Ji Chen"][quote="Vasc"]Let $ x,y,z > 0$. Prove that $ \frac {x}{\sqrt {y^{2} \plus{} 3z^{2}}} \plus{} \frac {y}{\sqrt {z^{2} \plus{} 3x^{2}}} \plus{} \frac {z}{\sqrt {x^{2} \plus{} 3y^{2}}}\geq \frac {3}{2}$.[/quote]
If $ x, y, z$ are positive numbers, then the following inequality
\[ \frac {x}{\sqrt {y^{2} \plus{} kz^{2}}} \plus{} \frac {y}{\sqrt {z^{2} \plus{} kx^{2}}} \plus{} \frac {z}{\sqrt {x^{2} \plus{} ky^{2}}}\geq \frac {3}{\sqrt {1 \plus{} k}}
\]
holds if and only if

$ 0\leq k\leq\frac {49 \minus{} 9\sqrt {17}}{32} \equal{} 0.37162654279503297\cdots$

$ \vee k\geq\frac {49 \plus{} 9\sqrt {17}}{32} \equal{} 2.6908734572049670\cdots$.

With equality if $ k \equal{} \frac {49 \minus{} 9\sqrt {17}}{32},x \equal{} 0,y \equal{} 1,z \equal{} \frac {\sqrt {17} \minus{} 1}{4} \equal{} 0.78076923076923076\cdots$

or $ k \equal{} \frac {49 \plus{} 9\sqrt {17}}{32},x \equal{} 0,y \equal{} 1,z \equal{} \frac {\sqrt {17} \plus{} 1}{4} \equal{} 1.2807881773399014\cdots$.[/quote]

If $ x,y,z > 0,\frac {49 \minus{} 9\sqrt {17}}{32}\leq k\leq\frac {49 \plus{} 9\sqrt {17}}{32}$, then the infimum of $ \frac {x}{\sqrt {y^{2} \plus{} kz^{2}}} \plus{} \frac {y}{\sqrt {z^{2} \plus{} kx^{2}}} \plus{} \frac {z}{\sqrt {x^{2} \plus{} ky^{2}}}$ is $ \frac {2}{\sqrt [4]{ k}}$.

[quote="Ji Chen"]If $ x, y, z$ are positive numbers, then the following inequality

$ \sqrt {\frac {x}{y + kz}} + \sqrt {\frac {y}{z + kx}} + \sqrt {\frac {z}{x + ky}}\geq \frac {3}{\sqrt {1 + k}}$

holds if and only if $ 0\leq k\leq\frac {49 - 9\sqrt {17}}{32} = 0.371626542795\cdots\vee k\geq\frac {49 + 9\sqrt {17}}{32} = 2.69087345720\cdots$.[/quote][quote="Harry Potter"]To Ji chen :

Can you show your Comment about Value worth of $ k$ ? I think if it true, then it isn't actual easy.[/quote]Let $ \frac {x}{y + kz} = \frac {u^{2}}{1 + k},\frac {y}{z + kx} = \frac {v^{2}}{1 + k},\frac {z}{x + ky} = \frac {w^{2}}{1 + k},$ where $ u,v,w$ are positive numbers.

Hence, it remains to prove that $ u + v + w\geq3,$ where $ (k^2 - k + 1)u^2v^2w^2 + k(v^2w^2 + w^2u^2 + u^2v^2) = (k + 1)^2.$

arqady's contradiction method works here: let $ u + v + w < 3,$ then

$ \left(k^2 - k + 1\right)u^2v^2w^2 + k(v^2w^2 + w^2u^2 + u^2v^2)$

$ < \left(k^2 - k + 1\right)u^2v^2w^2\left(\frac {3}{u + v + w}\right)^6 + k\left(v^2w^2 + w^2u^2 + u^2v^2\right)\left(\frac {3}{u + v + w}\right)^4$

$ = (k + 1)^2 - \sum{\frac {(v - w)^2}{2(u + v + w)^6}\Big\{4\left(16k^2 - 49k + 16\right)u^3(v + w) + 2\left(91k^2 - 142k + 91\right)u^2vw + 6\left(10k^2 - 7k + 10\right)v^2w^2}$

${ + (k + 1)^2\left[58uvw(v + w) + v(v + 16w)(u - v)^2 + w(w + 16v)(u - w)^2\right]\Big\}}$

$ \leq (k + 1)^2,$ which is a contradiction.

See also : http://www.mathlinks.ro/viewtopic.php?t=153451


At the end of next week, we'll be reorganizing the board.

Here are the items we're considering; please add your thoughts and suggestions:

1) Change the problem solving section to be 4 boards: Getting Started (MATHCOUNTS, AMC8/10 level), Intermediate Topics(AMC12/AIME), Advanced problems (Olympiad), Resources/Teaching (as it is now)

2) Consolidate Books/Movies & Video Games

3) Consolidate Other Programs & Other Contests

4) Implement pruning on the Etc section - basically delete any thread that hasn't had a response in 10 days or so.

I agree with all the changes - particularly changing the problem solving section and pruning the Etc. forum...

I vote "Yes" for these changes.
Although I am surprised that "prune" was a verb...

We are still going to have the "mathcounts" and "American Mathematics Competitions" and so on sections within the Extracurricular Programs title right?

-interesting_move

Yes; those will all still be there (though we'll probably fold the 'Other Programs' and 'Other Contests' into one board).

Pruning is a must.  Except maybe in special situations, mainly the reference ones like "what is your screen name?" or "how many digits of pi do you know" (since these will probably come up over and over again if you delete them...)

One thing: if we delete posts in Etc., the authors of those posts will have their post count go down.  This seems fair enough to me, since it's generally the posts in the other sections that are most relevant/helpful and therefore deserve to be credited.  This would recognize big problem-solvers like TripleM and ComplexZeta for their hard work and many posts in math-related things.  Thoughts on this?

[quote="mathfanatic"]if we delete posts in Etc., the authors of those posts will have their post count go down.  This seems fair enough to me, since it's generally the posts in the other sections that are most relevant/helpful and therefore deserve to be credited.  This would recognize big problem-solvers like TripleM and ComplexZeta for their hard work and many posts in math-related things.  Thoughts on this?[/quote]
I agree. People who's posts are mostly useless (a.k.a. me:)) doesn't deserve high ranks/post numbers.

indeed. but at least tare is starting to solve in the amc folder now. thats a start at least.

I'm not at that level yet. I only got a measly 100... :) I do the best I can, but that's not a lot.

What grade are you in Tare?

-interesting_move

[quote="interesting_move"]What grade are you in Tare?[/quote]
I took AMC 10 last year when I was in 7th. Now I'm about to be 8th.

[quote="Syntax Error"]dont get down. if youve been working hard, you should easaly make aime. not that many 7th graders can get a 100 on the amc 10. you should be proud. but just work hard, and you can do it.

now im done with my motivational speaking[/quote]
*claps* Thank you, HI...I only wish you'll stay that way :)

im always that way when someone doesnt believe in themselves.


A biologist wants to calculate the number of fish in a lake. On May 1 she catches a random sample of 60 fish, tags them, and releases them. On September 1 she catches a random sample of 70 fish and finds that 3 of them are tagged. To calculate the number of fish in the lake on May 1, she assumes that $25\%$ of these fish are no longer in the lake on September 1 (because of death and emigrations), that $40\%$ of the fish were not in the lake May 1 (because of births and immigrations), and that the number of untagged fish and tagged fish in the September 1 sample are representative of the total population. What does the biologist calculate for the number of fish in the lake on May 1?

[hide]Because $25\%$ of the tagged fish had died off or emigrated, there are only $60\left(\frac{3}{4}\right)=45$ left in the lake. In the sample she collected on September 1, only $70\left(\frac{6}{10}\right)=42$ were originally there on May 1. $3$ of these were tagged. Thus, the ratio of tagged to untagged fish on May 1 was $\frac{3}{42}=\frac{1}{14}$. Our answer is thus $14\cdot \text{number\ of\ tagged\ fish}=14\cdot 45=\boxed{630}$.[/hide]

[hide="Answer"]$\frac{3}{70}$ fish in the lake at the later date are tagged. 40% were born since, so she is really looking at $\frac{3}{42}$. 25% of the $60$ tagged fish have died, so there were only $45$ fish remaining, and she has caught $\frac{1}{15}$ of them. Therefore, our answer is $15\cdot 42=\boxed{630}$.[/hide]

Bumped so that 4everwise can have the right solution in the solution's document. 4everwise, you multiplied $(14)(60)$ when it should be $(14)(45)$ :P

I think the answer actually would 14*60=840. We want the number of fish before the reduction of 25%.

The answer should be 840. Here's azjps's solution from the wiki:

[hide]
Of the $ 70$ fish caught in September, $ 40\%$ were not there in May, so $ 42$ fish were there in May. Since the percentage of tagged fish in September is proportional to the percentage of tagged fish in May, $ \frac {3}{42} \equal{} \frac {60}{x} \Longrightarrow \boxed{x \equal{} 840}$.

(Note that the 25% death rate does not affect the answer.)[/hide]

I got 840 as well. Does anybody know the official AIME solution?

I said that on may 1 there were x fish and on sept 1 there were y fish.

Since 25 percent of x are gone, there are 3x/4 originals left in sept. Also, since 40% of fish in sept were not there in may, that means 60% of the fish in september WERE there in may, so the amount of original fish in september could also be written as 3y/5

3y/5 = 3x/4 ---> y = 15x/12 = 1.25x

So in september there were 45 tagged fish (.75*60) and there were 1.25x fish in total. Since 3 of the 70 surveyed were tagged,

3/70 = 45/1.25x ---> x = 840

840.

45 of the initialled tagged fish are still there. Thus, 45 is 3/70 of the entire september population, so on 9/1, there were 1050 fish total.

Whatever the number of fish were on may 1, after accounting for 25% deaths, experienced a 40% growth, to reach the 9/1 population. So, letting the number of fish on may 1 be x, we have
((1-0.25)x)=((1-0.4))(1050),

x = 840

[hide=Solution]After the $40$ percent reduction, $42$ of the $70$ fish caught on September 1 are still there, so $\frac{3}{42}=\frac{60}{x}$ where $x$ is the total. This means $x=\boxed{840}$.[/hide]

[hide=Solution][i]m[/i] is the number of fish in the lake on May 1. Since 60 fish were tagged on May 1, but on September 1, 25% of all the fish were gone, that means 25% of the tagged fish were gone too, so 45 tagged fish were left in the lake. Since 40% of the fish in the lake on September 1 are new, that means [i]m[/i] is 60% of all the fish in the lake on September 1. 40 is 2/3 of 60, so the number of fish in the lake on September 1 is 5/3 * [i]m[/i]. The sample taken on September 1 tells us that 3/70 of the fish in the lake are tagged, but we just found out that 45 / (5/3 * m) is the ratio of the total tagged fish to the total number of fish in the lake! This means these ratios are equal: 3/70 = 45 / (5/3 * m) => 3/70 = 45 * 3 / 5m => we multiply both sides by (5m * 70): 5m = 45 * 70 => we divide both sides by 5 => m = 9 * 70 => we get [i]m[/i] = [b]630[/b].[/hide]

That's what I got. Is it 630 or 840?

According to the wiki, the answer is 840. Can someone edit the solution [url=https://artofproblemsolving.com/community/c4887_1990_aime_problems]here[/url]?


inequalities
induction
combinatorics solved
combinatorics
Enumeration

Three rods of length a,b,c can be made to form a triangle if and only if the three triangle inequalities are satisfied: a+b > c, b+c > a and c+a > b. Suppose one wants to make just one triangle and has at his disposal exactly one rod of each of lengths 1,2,3,...,n. He might decide on the (2,4,5) triangle, the (3,4,5), or the (3,5,7),... He can't make an isosceles triangle, of course, but there are still many possibilities to choose from. And we want to know how many ?

My guess is (if I understood correctly the problem) :

. u(2n) = n(n-1)(4n-5)/6
. u(2n+1) = (n-1)(4n-5n+6)/6 (= u(2n) + n-1)

Let v(n) be the number of solution of :
1 <= a < b < n, a+b > n

By induction, v(2n)=(n-1), v(2n+1)=n(n-1)

Then sum v(k) from 1 to n to get u(n).

I have the same formula for even n but a different one for odd n. Maybe you can give your reasoning.

u(n) = 1/24*n*(n-2)*(2n-5) with n being even
u(n) = 1/24*(n-1)*(n-3)*(2n-1) with n being odd

It was a very last line calculation error (I'm prone that kind of silly mistake) :(  :

. u(2n+1) = u(2n) + n(n-1) (and not (n-1)) = n(4n+1)(n-1)/6 which is the same as you !


How many subsets of the set $\{1,2,3,4,\ldots,30\}$ have the property that the sum of the elements of the subset is greater than 232?

[hide="As I recall..."] This problem depends on the fact that $232$ is a special sum with relation to this set, but I don't remember how or why. [/hide]

[hide="hint"]
$232 = \lfloor \frac{465}{2}\rfloor$
[/hide]


Show that  the equation X²+2Y²=2p has integer solutions if and only if the  equation U²+2V²=p has integer solutions.

[hide]
Suppose there's a solution for X²+2Y²=2p.  This implies that X is even.
So let X = 2 * Z: 4Z² + 2Y² = 2p.
Then, Y² + 2Z² = p, which is a solution for U²+2V²=p when U = Y and V = Z.

Now, suppose there's a solution for U²+2V²=p.  Then (2V)² + 2U² = 2U² + 4V² = 2p, which is a solution for X²+2Y²=2p when X = 2V and Y = U.
[/hide]


$\sqrt{8+4\sqrt{3}}=\sqrt{a}+\sqrt{b}$, where $a$ and $b$ are positive integers and $a>b$.  Find the ordered pair $(a, b)$.

[hide="Solution"]$a+b+2\sqrt{ab}=8+4\sqrt{3}$

$a+b=8$

$\sqrt{ab}=2\sqrt{3}\implies ab=12$

$a\left(8-a\right)=12\implies a^{2}-8a+12=0=\left(a-6\right)\left(a-2\right)=0$

So you have $\left(a,b\right)=\boxed{\left(6,2\right)}$.[/hide]

[hide]Square both sides to get.

$8+4\sqrt{3}=a+2\sqrt{ab}+b$.

Then, this means that

$8=a+b$, and $2\sqrt{ab}=4\sqrt{3}$.

$a+b=8$
$ab=12$.

Then we can put this into a quadratic to get

$x^{2}-8x+12=0$.  So $x=6,2$.

But $a\>b$, hence $a=6$ and $b=2$

$(a,b)=(6,2)$[/hide]

[quote="nutz_for2.718281828"][hide="solution"]$\text{We square both sides: }$
$8+4\sqrt3=a+b+2\sqrt{ab}$
$\sqrt{ab}=2\sqrt3=\sqrt{12}$
$a+b=8$
$\text{We use guess and check and see that: }$
$\boxed{(a,b)=((6,2),(2,6))}$[/hide][/quote]

He said $a>b$


If the following system of equations

$x+y+z=3$

$x^{3}+y^{3}+z^{3}=15$

$x^{4}+y^{4}+z^{4}=35$

has a real solution for which $x^{2}+y^{2}+z^{2}>10$, find the value of $x^{5}+y^{5}+z^{5}$ for that solution.

please tell me that how you eliminated q from equation
$S_{4}-3*S_{3}....$.

[quote="Fibfreak"]There is just one really weird thing about this... by Cauchy

$x^{2}+y^{2}+z^{2}\le \sqrt{(x^{4}+y^{4}+z^{4})(1^{2}+1^{2}+1^{2})}= \sqrt{105}$[/quote]

Hmm, you've got a point. The system doesn't have a real solution. (We get that the solutions are the roots of $x^{3}-3x^{2}-x+7 = 0$, which aren't all real.)

Correctly :[quote="Fibfreak"]The following system of equations $\{\begin{array}{c}x+y+z=3\\\ x^{3}+y^{3}+z^{3}=15\\\ x^{4}+y^{4}+z^{4}=35\end{array}$ has at least a real solution for which $\boxed{x^{2}+y^{2}+z^{2}\le10}$ and find the value of $x^{5}+y^{5}+z^{5}$ for this solution.[/quote]
Answer : $t^{3}-3t^{2}+t+1=0$ or $t^{3}-3t^{2}-t+7=0$.


The numbers 1,2,3,...,9 are arranged, one per circle, in the triangle shown so that the sum, s, of the numbers on each side of the triangle is the same.  The sum of the numbers in the three corner circles is 12.  What is the value of s?  Express your answer as a common fraction.

[hide]ok... you use the corners twice... and the others once...

so

3s = 1+2+3+4+5+6+7+8+9+12

3s = 57

s=19 [/hide]so you're right.

Just as a note -- "19" is the proper way to express an integer as a "common fraction," not "19/1" or anything else.

[color=cyan]and in mathcounts, "improper" fractions are common fractions, but 19/1 could be simplified, so yeah[/color]


Problem B2.
Let n be a positive integer. Starting with the sequence
1, 1/2, 1/3, ..., 1/n form a new sequence of n-1 entries
3/4, 5/12, ..., (2n-1)/{2n(n-1)} by taking the averages of
two consecutive entries in the first sequence. Repeat the
averaging of neighbors on the second sequence to obtain a third
sequence of n-2 entries and continue the final sequence produced
consists of a single number x_n. Show that x_n < {2/n} .
----------------------------------------------------------------

Prove by induction that x_n= (1/2 n-1)* \sum{k =0..n-1} (n-1)Ck/(k+1). (1)
We must prove (1)<2/n.
After computations this comes to  \sum{k=1..n} nCk<2 n, or equivalently 2 n-1<2 n. This is obvious.

My solution is smaller that the problem text.  :D

that's because you didn't make full-detailed redactation. :)

[color=red][[b]Moderator edit:[/b] Now with 100% more solution at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=80&t=413607 .][/color]


[quote="tuantam1lan"]let a,b,c are non-negative such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.prove that:
$ \frac {a^3}{a^5 \plus{} a^2b \plus{} c^3} \plus{} \frac {b^3}{b^5 \plus{} b^2c \plus{} a^3} \plus{} \frac {c^3}{c^5 \plus{} c^2a \plus{} b^3} \leq 1$[/quote]
inequality problem have complicacy form.but it is an inequality very interesting and hard
we have:$ \frac {a^3}{a^5 \plus{} a^2b \plus{} c^3} \equal{} \frac {a}{a^3 \plus{} b \plus{} \frac {c^3}{a^2}}$
to apply cauchuy-shawrs:
$ a^3 \plus{} b \plus{} \frac {c^3}{a^2} \equal{} \frac {a^4}{a} \plus{} \frac {b^4}{b^3} \plus{} \frac {c^4}{ca^2} \geq \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{a \plus{} b^3 \plus{} ca^2} \equal{} \frac {9}{a \plus{} b^3 \plus{} ca^2}$
  $ \Leftrightarrow \frac {a^3}{a^5 \plus{} a^2b \plus{} c^3} \leq \frac {a(a \plus{} b^3 \plus{} ca^2)}{9}$
 therefore
  $ \sum_{cyc}\frac {a^3}{a^5 \plus{} a^2b \plus{} c^3} \leq \frac {\sum_{cyc} a^2 \plus{} 2\sum_{cyc} ab^3}{9} \leq 1$
equality is $ a \equal{} b \equal{} c \equal{} 1$

[quote="Ronald Widjojo"]I don't get the last inequality..
Can you prove it clearly ?
Thanks before..[/quote]
 if a,b,c are real numbers,then:
$ (a^2 \plus{} b^2 \plus{} c^2)^2 \geq 3(a^3b \plus{} b^3c \plus{} c^3a)$
 vasile cirtoaje created it in 1992
it is hard inequality and very much deliberate
 inequality is equivalent:$ \sum_{cyc}(a^2 \minus{} 2ab \plus{} bc \minus{} c^2 \plus{} ca)^2 \geq 0$

[quote="tuantam1lan"][quote="tuantam1lan"]let a,b,c are non-negative such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.prove that:
$ \frac {a^3}{a^5 \plus{} a^2b \plus{} c^3} \plus{} \frac {b^3}{b^5 \plus{} b^2c \plus{} a^3} \plus{} \frac {c^3}{c^5 \plus{} c^2a \plus{} b^3} \leq 1$[/quote]

  $ \Leftrightarrow \frac {a^3}{a^5 \plus{} a^2b \plus{} c^3} \geq \frac {a(a \plus{} b^3 \plus{} ca^2)}{9}$
 [/quote]
It is a nice and really impressive proof (  :clap2:  )  ; but this step has a little flaw - the inequality sign must be flipped. :blush:

@tuantam1lan: Nice hint to creat, my friend  :)


geometry rectangle perimeter quadratics algebra

Hi, I need help with this ambiguous but otherwise fairly easy problem:

[i]A rectangular piece of land of area 5000m^2 is to be enclosed by a wall, and then divided into three equal regions by partition walls parallel to one of its sides. If the total length of the wall is 445m, calculate the possible dimensions of the land.[/i]

Attached are possible diagrams

If I understood the problem correctly:

Let x be longer side of the rectangle and y its smaller side. Then the area is given by xy and the perimeter by 2(x+y). So we have to solve the equations $ 2(x+y) = 445$ and $ xy = 5000$. From the first we get
$ y = 222.5-x$, and substituting in the second we get the quadratic equation $ x^{2}-222.5x+5000 = 0$, whose solution is $ x = 197.15\, m$, and so $ y = 222.5-197.15 = 25.35\, m$, and therefore these are the dimensions of the land. If then the land is to be divided in three equal regions, then one side of each will have 25.35 m, and the other side will have 197.15/3 = 65.72 m.


Find all functions $ f: R\to R$ suct that $ f(f(x\minus{}y)) \equal{} f(x)f(y)\minus{}f(x)\plus{}f(y)\minus{}xy$ for all real $ x,y$.

let $ f(0)\equal{}c$
put $ x\equal{}y\equal{}0$ we get that:
$ f(c)\equal{}c^{2}$
now put $ x\equal{}y$ we get that:
$ f(c)\equal{}f(x)^{2}\minus{}x^{2}\equal{}c^{2}$
for all real numbers $ x$,now put $ x\equal{}c$ we get that $ c^{2}\equal{}f(c)^{2}\minus{}c^{2}$ thus we get that $ c^{4}\equal{}2c^{2}$ so we have $ c\equal{}0$ or $ c\equal{}\pm\sqrt{2}$
so we get the following solutions:
$ f(x)\equal{}\pm x$
$ f(x)\equal{}\pm\sqrt{x^{2}\plus{}2}$

[quote="BaBaK Ghalebi"]let $ f(0) \equal{} c$
put $ x \equal{} y \equal{} 0$ we get that:
$ f(c) \equal{} c^{2}$
now put $ x \equal{} y$ we get that:
$ f(c) \equal{} f(x)^{2}\minus{}x^{2}\equal{} c^{2}$
for all real numbers $ x$,now put $ x \equal{} c$ we get that $ c^{2}\equal{} f(c)^{2}\minus{}c^{2}$ thus we get that $ c^{4}\equal{} 2c^{2}$ so we have $ c \equal{} 0$ or $ c \equal{}\pm\sqrt{2}$
so we get the following solutions:
$ f(x) \equal{}\pm x$
$ f(x) \equal{}\pm\sqrt{x^{2}\plus{}2}$[/quote]
$ f(x)\equal{}\pm\sqrt{x^{2}\plus{}2}$ is not solution.

neither is $ f(x)\equal{}x$...
so the only solution due to Rust is $ f(x)\equal{}\minus{}x$...


modular arithmetic
number theory unsolved
number theory

Let $1=d_1<d_2<....<d_k=n$ be all different divisors of positive integer n written in ascending order. Determine all n such that:

\[d_1^{2} + d_2^{2} + d_3^{2} + d_4^{2}=n\]

The only solution is $ n \equal{} 130$.

It is clear that $ d_1 \equal{} 1$ and $ n$ is even (since otherwise $ d_1,d_2, d_3, d_4$ would be odd, while the sum their squares would be also odd, a contradiction). Therefore, $ d_2 \equal{} 2$ and exactly one of $ d_3$ and $ d_4$ is also even. Consider two cases:
1) $ 4|n$. In this case $ \{d_3,d_4\} \equal{} \{4,p\}$, where $ p\geq 3$ is prime, and $ n \equal{} 1 \plus{} 2^2 \plus{} 4^2 \plus{} p^2 \equal{} 21 \plus{} p^2$, implying that $ p^2\equiv 3\pmod{4}$, which is impossible.
2) $ 4\not|n$. In this case $ d_3 \equal{} p$ and $ d_4 \equal{} 2p$, where $ p\geq 3$ is prime, and $ n \equal{} 1 \plus{} 2^2 \plus{} p^2 \plus{} (2p)^2 \equal{} 5 \plus{} 5p^2$, implying that $ p \equal{} 5$ and $ n \equal{} 130$.


Find all primes $ p$ such that there exist integers $ x$ and $ y$ that satisfies both of these equation:
\[ 7p\equal{}8x^2\minus{}1 \text{ and } p^2\equal{}2y^2\minus{}1\]

we use the well-know idea(obviously $ p>3$),
$ p(p \minus{} 7) \equal{} 2(y^2 \minus{} (2x)^2) \equal{} 2(y \minus{} 2x)(y \plus{} 2x) , p > y \minus{} 2x \implies p |y \plus{} 2x \implies y < p,x < \sqrt {p} \implies p|y \plus{} 2x < p \plus{} 2\sqrt {p} < 2p \implies y \plus{} 2x \equal{} p$
the rest is trivial.


algebra
polynomial
Rational Root Theorem
algebra unsolved

a is a real root of x^2 + 2x + 10x - 20; prove a^2 irrational.

hahhaha alex what's up.

anyways, i assume it's $x^3+2x^2+10x-20=0$...

$x^2(x+2)+10(x-2)=0$, so $x^2=-10(x-2)/(x+2)=-10(x-2)^2/(x^2-4)$.

plug in $x=\alpha$ and suppose $\alpha^2$ is rational, then $\alpha^4$ is also rational.  Note that $\alpha$ is irrational, so expanding that thing you have $\alpha^4+6\alpha^2+40=40\alpha$, i.e the sum of 3 rational numbers is irrational so contradiction,

More naturally, write it as $x(x^2-10)=20-2x^2$.  Thus if $k$ is a real root (there is at least one) then $k$ is rational iff $k^2$ is rational.  But the rational root theorem shows that $k$ cannot be rational.  The result follows.

You can generalise this result to any cubic quite easily by rational root thorem and Viete's formulawe.
Obviuosly, it must be real, not comlpex :)  :)


Putnam
limit
real analysis
real analysis unsolved

Given a sequence ($x_{n}$) such that $\lim_{n\to\infty}\left(x_{n}-x_{n-2}\right)=0$
prove that

$\lim_{n\to\infty}\frac{x_{n}-x_{n-1}}{n}=0$

Here's an ugly way:

[hide]Choose $\epsilon > 0$.  Pick $N$ s.t. for $n \geq N$ we have $|x_{n+2}-x_{n}| < \frac{\epsilon}{2}$.  Let $|x_{N+1}-x_{N}| = d$.

Now, choose $M \geq \max\{\frac{2d-N\epsilon}{\epsilon}, N\}$.  If $n \geq M$ we have

$\frac{1}{n}|x_{n+1}-x_{n}| = \frac{1}{n}|(x_{n+1}-x_{n-1})+(x_{n-1}-x_{n-3})+\ldots-[(x_{n}-x_{n-2})+(x_{n-2}-x_{n-4})+\ldots] \pm (x_{N+1}-x_{N})| \leq \frac{(n-N)\frac{\epsilon}{2}+d}{n}= \frac{\epsilon}{2}+\frac12 \frac{2d-N}n \leq \epsilon$, and we're done.[/hide]

[quote="JBL"]
$\frac{1}{n}|x_{n+1}-x_{n}| = \frac{1}{n}|(x_{n+1}-x_{n-1})+(x_{n-1}-x_{n-3})+\ldots-[(x_{n}-x_{n-2})+(x_{n-2}-x_{n-4})+\ldots] \pm (x_{N+1}-x_{N})|$ [/quote]
why?I can't understand sorry :oops:

(Edit:I know you are right now :) )

Here is my proof:
I just notice that the sub-sequences:
$x_{1},x_{3},\cdots,x_{2k+1},\cdots$ and $x_{2},x_{4},\cdots,x_{2k},\cdots$ are independent.
So I want to prove:$\lim_{n \to \infty}\frac{x_{n}}{n}=0$
but this is not hard,because:
$\frac{x_{n}}{n}=\frac{(x_{n}-x_{n-2})+(x_{n-2}-x_{n-4})+\cdots}{n}$
And use Cauchy's rule.
then is it easy to see:
$\lim_{n \to \infty}\frac{x_{n}-x_{n-1}}{n}=\lim_{n \to \infty}\frac{x_{n}}{n}-\lim_{n \to \infty}\frac{x_{n-1}}{n}=0$:)

[quote="zhaobin"][quote="JBL"]
$\frac{1}{n}|x_{n+1}-x_{n}| = \frac{1}{n}|(x_{n+1}-x_{n-1})+(x_{n-1}-x_{n-3})+\ldots-[(x_{n}-x_{n-2})+(x_{n-2}-x_{n-4})+\ldots] \pm (x_{N+1}-x_{N})|$ [/quote]
why?I can't understand sorry :oops: [/quote]

$(x_{n+1}-x_{n}) = x_{n+1}-x_{n}+(x_{n-1}-x_{n-1})+(x_{n-2}-x_{n-2})+\ldots+(x_{N+1}-x_{N+1})+(x_{N}-x_{N})$ and then reorder the sum.  The $\pm$ at the end is because we don't know which will be absorbed into our $\epsilon$-pairs without breaking it into cases for the parity of $n-N$.

Your way is nicer, though.

[quote="zhaobin"]
$\frac{x_{n}}{n}=\frac{(x_{n}-x_{n-2})+(x_{n-2}-x_{n-4})+\cdots}{n}$
And use Cauchy's rule.
then is it easy to see:
[/quote]
first question: what is Cauchy's rule?

[quote="zhaobin"]
$\lim_{n \to \infty}\frac{x_{n}-x_{n-1}}{n}=\lim_{n \to \infty}\frac{x_{n}}{n}-\lim_{n \to \infty}\frac{x_{n-1}}{n}=0$:)[/quote]
question 2: how do u get this since doesnt the previous expression come out to $x_{n}-x_{0}$ or $x_{n}-x_{1}$ depending on the parity of $n$?

The idea is that the condition shows that the sequences $x_{0}, x_{2}, ...$ and $x_{1}, x_{3}, ...$ (which are independent) both converge, say to $y_{1}$ and $y_{2}$. Then in the limit, the difference $|x_{n}-x_{n-1}|$ is about $|y_{1}-y_{2}|$, whence if you divide by $n$ it goes to 0.

[quote="Xevarion"]the sequences $x_{0}, x_{2}, ...$ and $x_{1}, x_{3}, ...$ (which are independent) both converge[/quote]
No. For example, $x_{n}=n^{a}$, $0<a<1$.

[quote="maokid7"]
first question: what is Cauchy's rule?

question 2: how do u get this since doesnt the previous expression come out to $x_{n}-x_{0}$ or $x_{n}-x_{1}$ depending on the parity of $n$?[/quote]
For first question:Cauchy's rule:I think it is if $lim_{n \to \infty}a_{n}=a$,then:$lim_{n \to \infty}\frac{a_{n}+a_{n-1}+\cdots+a_{1}}{n}=a$

For second question:I didn't want to answer something.

[quote="pou"]How to proof Cauchy rule?[/quote]

With an argument like mine.  Wait until all terms are within $\frac \epsilon 2$ of $a$, then take a very large number of such terms.


Nice work.  When I first did this problem, I plugged in [hide]0[/hide] the first time because I knew it wouldn't be too big and it worked.

Can someone please tell me what every solution so far has left out (and then explain why it turned out okay anyhow)?

we could always have  :pm:  answers, but negetive answers would give us i's.


calculus
integration
inequalities
function
real analysis
real analysis unsolved

$ \int_0^1f(x)dx \int_0^1x^4f(x)dx \le \frac{4}{15} \int_0^1f^2(x)dx$ , where $ f: [0,1]\minus{}> R$ is continous .

Just use Cauchy-Schwartz for $ \int_0^1\left[\frac 1{2\sqrt 3}\plus{}\frac{\sqrt 3}2x^4\right]f(x)\,dx$ and note that the left hand side is at least the square root of the product on the left hand side in the original inequality and that $ \int_0^1\left[\frac 1{2\sqrt 3}\plus{}\frac{\sqrt 3}2x^4\right]^2\,dx\equal{}\frac 4{15}$. The equality is attained when both AM-GM and CS inequalities turn into equalities, i.e., for the function $ f(x)\equal{}c(1\plus{}3x^4)$ with $ c\ge 0$.


function
trigonometry
inequalities
calculus
derivative
real analysis
real analysis solved

Let $f:[0,1]\to\mathbb{R}$ a function which admits primitives on $[0,1]$ and $F$ one of them.
a) Prove that there exists $x\in(0,1)$ s.t. $\sqrt{2|f(x)F(x)|}<1+F^2(x)$
b) Prove that for $f(x)=\sin x$ the inequality from a) is true for all $x\in[0,1]$.

For b)
It is a strange question, because $|f(x)|<1$, so $\sqrt{2|fF|}\leq \sqrt{1+F^2}\leq 1+F^2$, where at least one of "\leq" is a strong sign.

Hmm... does anybody know exact constant in $\sqrt{2|f(x)F(x)|}<1+F^2(x)$ instead of $2$? I obtained something like $\frac{8}{\pi}$...  :?

[P.S. It is a wrong observation :blush:]

and how did you find this constant Myth?

At first, I'd like to see your solution ;)

Sorry for delay
hmm...
I supposeed that for any $x\in(0,1)$ the following inequality holds
$\sqrt{2|f(x)F(x)|}\ge1+F^2(x)$ which is equivalent with
$|\left(\frac1{F^2(x)+1}\right)'|\ge 1$ for any $x\in R$
one ot the following cases must occur
1.$\left(\frac1{F^2(x)+1}\right)'\ge 1$
2.$\left(\frac1{F^2(x)+1}\right)'\le -1$
for 1. (it's analoguos for the other one)
we can integrate betweem 0 and $t$, $t\in(0,1)$ and results that
$\frac1{1+F^2(t)}-\frac1{1+F^2(0)}\ge t$ but since $\frac1{1+F^2(x)}$ is continuous results that we can make $t\to 1$ an we have now that the difference between two number less than 1(and greater than0) is greater or equal to 1... which is a contradiction

and the hypohesis is proved...

I didn't think during the contest to find the biggest constant instead of $2$ but it would be an interesting problem to find it (I think ).

Ops... Sorry, I realized that I wrote a wrong primitive instead of $\frac{1}{1+F^2}$, that is why I got a better constant. My solution was exactly the same.

I misunderstand one thing: how we can suppose that $|\left(\frac1{F^2(x)+1}\right)'|\ge 1$ or
$|\left(\frac1{F^2(x)+1}\right)'|\le 1$ on the whole interval (0;1)  (As far as i understand tending $t$ to 1 you supposed that one of the above inequality holds on the (0,1))

In my turn, I misunderstood what you mean.

Ok, i'll try to explain my problems:

[quote="Kuba"]Sorry for delay

$|\left(\frac1{F^2(x)+1}\right)'|\ge 1$ for any $x\in R$
one ot the following cases must occur
1.$\left(\frac1{F^2(x)+1}\right)'\ge 1$
2.$\left(\frac1{F^2(x)+1}\right)'\le -11$
for 1. (it's analoguos for the other one)
we can integrate betweem 0 and $t$, $t\in(0,1)$ and results that
$\frac1{1+F^2(t)}-\frac1{1+F^2(0)}\ge t$ but since $\frac1{1+F^2(x)}$ is continuous results that we can make $t\to 1$ an we have now that the difference between two number less than 1(and greater than0) is greater or equal to 1... which is a contradiction

and the hypohesis is proved...
[/quote]

How we could integrate without knowing behavior of the function $\left(\frac1{F^2(x)+1}\right)'$ near t=1:
maybe in some subinterval (0,a) the 1. holds and in the (a,1) the second holds (where a<1)

A derivative has a Darbou (intermediate) property, so it can't be $>1$ on $(0,a)$ and $<-1$ on $(a,1)$. Therefore, $\left(\frac1{F^2(x)+1}\right)'\ge 1$ OR $\left(\frac1{F^2(x)+1}\right)'\le -1$ on the whole segment.

Ok, thank you very much Myth, sorry for my stupidness-now i understand