calculus
integration
geometry
perimeter
inequalities
triangle inequality

The lengths of two sides of a triangle are 33 units and
42 units. The third side also has an integral length. What is
the least possible number of units in the perimeter of the
triangle?

By the triangle inequality, the last side+33>42

So last side>9 ==>last side=10==>85

By triagnle inequality, the smallest side is 42-33+1. The perimeter would then be 42+33+42-33+1=42+42+1=85.

The Triangle Inequality theorem states that the sum of two sides of a triangle cannot be less than the third side.

So in this case, the longest possible side length is $ 33 \plus{} 42 \minus{} 1 \equal{} 74$. But we want the smallest possible perimeter, so we need to find the shortest side. So $ 42 \minus{} 33 \plus{} 1 \equal{} 10$. Our shortest possible side is $ 10$.

The perimeter is: $ 42 \plus{} 33 \plus{} 10 \equal{} \boxed{85}$.

No problem. If you have any other questions, please ask.  :)

my teacher said they were the same though

Not all teachers exactly have a brain, no offense to him/her....

i think they messed up though (like many people would do)
so basically a+b has to be less than c?

It has to be greater than $ c$.

You can experiment with a ruler, a pencil, and some paper. Try drawing a triangle with side lengths 5 cm, 2 cm, and 1 cm.

It's okay.  :)  I used to get that theorem mixed up [b]all[/b] the time.

In other words, the sum of 2 lengths in a triangle must be greater than the third side. You can prove it by imagining a triangle with sides 1, 2, and 3. But ask yourself this: what angle would be facing side 3? it wouldn't be an angle because you can poke the intersection of 1 and 2 until it sits flat on side 3, making a line segment behind another line segment. Now imagine if the sides were 1, 2, and 4. Even if 1 and 2 lie flat, it would make 3, so obviously not possible.


If you want to replace him, just post here. For the rules of the game, go to:
http://kevan.org/rumble.cgi?genre=hero&mode=rules


hey could any of u please help me with arithmetico-geometrico series and giv one of its examples by explaining it.
thanx in advance

I can give a try :D

[b]definition[/b]
A.G.P is a special sequence obtained by multiplying an A.P and G.P.

eg,$ 1.1,3.2,5.4,7.8,9.16,$etc are terms ao A.G.P with a.P as $ 1,3,5,7,9$ and G.p as $ 1,2,4,6,8,16,$

general term for a.G.P is $ a.b,(a\plus{}d).br,(a\plus{}2d).br^2,(a\plus{}3d).br^3,$ and so on

Here, $ a,a\plus{}d,a\plus{}2d,a\plus{}3d$etc are terms of A.P. and $ b,br,br^2,br^3$are terms of G.P.

[b]um to 'n' terms[/b]
for sum to 'n' terns i know only one way(other is formula that i forgot :oops: )

Just multiply each yerm of A.G.P by $ r$(common ratio of corresponding G.P) and subtract it from original sequence(after one term).The resulting sequence would be a G.P.Then use formula for sum of g.P.

Eg,in Earlier A.G.P.,let
$ S\equal{}1.1\plus{}3.2\plus{}5.4\plus{}7.8\plus{}............$ till 'n' terms
then
$ 2S\equal{}     1.2\plus{}3.4\plus{}5.8\plus{}...........$till 'n-1' terms-nth term

subtracting $ S\minus{}2S\equal{}1.1\plus{}2.2\plus{}2.4\plus{}2.8\plus{}......$+last term of $ 2S$(one term of $ 2S$will be left)

then you can easily calculate $ S$ :wink:
Here we had subtracted $ 2S$ by leaving one term of $ S$ as in proofs of G.P.

Similarily,sum to infinite terms can be calculated(that is easier as we don't have to think for last and remaining term) :P

If you have any specific problems,reply,i'ld try my best to solve (though i know,i wouldn't be able to solve :(  :oops: )

I suppose so

rest one thing

In any Progression,if  you cannot find out general term,etc or sum that
the best way is to write that sequence again(in case you find anything sort of geo,then multiply by $ r$ and in case you can't figure out commmon ratio,then subtract after multiplying by $ 1$ subtract in same way(leaving one)and in most of cases,you'll get some expression for general term

verify for one-two terms and then its all easy.

other way is also to use Combinations in those questions where the differences(or consecutive differences)are in A.P.

rest,I suppose is pactice
and one way in P&S in method of differences.
Thats all I know.

or wait for some senior to come

No need for thanks,Just return the favour by helping when I need. :rotfl:

And post questions that you find to be good,I'ld be benefitted by trying them


algorithm
induction
number theory
relatively prime
Euclidean algorithm

can anyone help me solve the following problem:

Show that all integer greater than 24 can be written in the forme n = 5a + 7b, where a and b are also integers

This is a variation of a Chicken McNugget Problem, which is also a variation of Inclusion-Exclusion method.

Basically what it is is that if you have 2 positive integers $a$ and $b$, the largest number which you cannot form by $ax+by$, where $x$ and $y$ are positive integers, is $ab - a - b$.

Like someone have said earlier, it should be $n \ge 24$, because the largest number cannot be formed by $5a+7b$ where $a$ and $b$ are positive integers is $5 \cdot 7 - 5 - 7 = 23$.

[quote="qleak"]I was looking for a link or reference for this but gave up. There's a lemma I remember calling "postage stamp lemma". What it says is that if two numbers are relatively prime, then their linear combination spans all of $\mathbb{Z}$. The extra condition you have that $n\ge 24$  is probably introduced by forcing $a\ge 0$ and $b\ge 0$.[/quote]

That proof is simple.  By the euclidean algorithm, there exist integers $x,y$ such that $xa+yb=1$; the proof follows.

Or the easiest way would be to prove it for $24,25,26,27,28,29$ and all other numbers are gotton by adding on 5's to one of these.

a lemma is a result you use to prove a th'm.

thank you seamusoboyle. Your answer is very cute and workable.

Yes there is, http://mathworld.wolfram.com/CoinProblem.html

[quote="xxreddevilzxx"][quote="Shogia"]Does anyone here know what the Chicken Mcnugget formula for 3 relatively prime numbers is?  If so it would be a great help.[/quote]

I would assume it would be something like this:

$abc-ab-bc-ca+a+b+c$[/quote]
I don't think you are right :D
to $gauss202$,is there a solution for this problem,(sorry my english is poor :P ),so I can't understand it completely.

What about $2abc - ab - bc - ca$? ;)

[IMO 1983 Problem 6]

induction :
true for n=24, since 24 =5*2+7*2, 25=,...28=5*0+7*4.
let n>=24 and suppose it is true,then there exist p and q such:
n=5p+7q.
Then n+5= 5(p+1)+7q  qed

[quote="seamusoboyle"]Theres got to be more than that... is it for $\gcd(a,b,c)=1$ ?[/quote]

I'm pretty sure it is. It might even be pairwise relatively prime, but I can't say for sure.

No, they need not be pairwise relatively prime, they just can't all have a common divisor.  Arne, that isn't the minimal such number, though, is it?


geometry
conics
ellipse
geometry proposed

Let ABC be a triangle. The incircle (I) touches BC at D respectively. AD meets (I) at K. The tangent from K of (I) meets AB,AC at M,N resp. Prove that MC,NB,AD are concurrent.

[quote="mr.danh"]Let $ \triangle ABC$ be a triangle. The incircle $ \omega(I)$ touches $ BC$ at $ D$. $ AD$ meets $ \omega$ at $ K$. The tangent from $ K$ of $ \omega$ meets $ AB$,$ AC$ at $ M,N$ respectively. Prove that $ MC,NB,AD$ are concurrent.[/quote]
Proof:
Let $ E,F$ be the tangency points of $ \omega$ with $ AC$ and $ AB$ respectively.Also let $ MN\cap BC \equal{} L$.Since point $ A$ lies on the polar of $ L$ w.r.t $ \omega$,it follows that $ L$ lies on the polar of $ A$ w.r.t $ \omega$,i.e $ L,E,F$ are collinear.(The last fact follows from [b]La'Hire's[/b] theorem.)
Concurrency of $ MC,BN,AD$ is equivalent to $ (LD,BC) \equal{} \minus{} 1$,but the last fact is obviously true,because as we've proved $ E,F,L$ are collinear and $ AD,BE,CF$ are concurrent.

This is exactly the [b][size=100]Newton theorem[/size][/b], in the configuration of the circumscribed quadrilateral $ BMNC.$

( [b][size=100][color=DarkBlue]In every circumscribed quadrilateral, the lines connecting the tangency points of the incircle to its opposite side-segments,
pass through the intersection point of its diagonals[/color][/size][/b] ).

There is an elementary proof of this theorem, which probably it has already been posted in this Forum.

Kostas Vittas.

Consider $ R$ the intersection of $ MN$ and $ BC$,$ MN$ is the polar of $ K$ and $ BC$ of $ D$, $ \Rightarrow$ $ R$ is the pole of $ KD$. Let $ E$ and $ F$ the tangency points of $ [I]$ different from $ D$. $ EF$ is the polar of $ A$ then $ EF\cap BC\equal{}$ pole of $ AD$, so $ EF,MN$ and $ BC$ pass through $ R$. $ (RD,BC)\equal{}\minus{}1$ $ \Rightarrow$ $ (RK,MN)\equal{}\minus{}1$ this implies $ MC$, $ BN$ and $ AD$ concur.

It's a know problem: ABCD is circumcripted to a circle that meet AB,BC,CD,DA on E,F,G,H. Then AC,BD,EG,FH concur.

It's also true this generalization:

Let an ellipse and a n-agon inscribed on him, with n even, such that the diagonals concur at P, costruct the n-agon with the vertex given by the intersection of the tangent line to the ellipse on adjacent vertex. Then the diagonals of this other n-agon concur at P.


$ n$ whole numbers are multiplied together . what is the probability that the product ends with a zero.
Give the general expression and also compute the probability when $ n\equal{}4$ .

It's been less than 36 hours; this is not a homework help forum. :mad:

The probability that the product is not divisible by 2 is ________.  Similarly, the probability the product is not _________ is __________.  The probability that the product is neither divisible by 2 nor _______ is _________.  Putting these together, ____________.

this isn't any homework problem , i jus came across this in a book.

so if anyone can help me solve this, they can do it, there is no compulsion on any particular person.And i don't think there is any need of a post with dashes in it .

In fact, it's an ill-defined problem. The integers are an infinite set, and there is no canonical probability distribution defined for them.

i am not sure whether the following is right , please correct me if i am wrong .
According to my approach ,

the number of ways the product can end with 1,3,7,9  is only when the last digit of each of the $ n$ numbers are 1,3,7 or 9.hence the probability that the product ends with 1,3,7,9 is  $ ( \frac {4}{10} )^n$ .

Similarly for the product to end with 2,4,6,8, the probability is again $ ( \frac {4}{10} )^n$ .

For the product to end with 5 , there are the following cases .
1) all n numbers end with 5.
2) n-1 numbers end with 5 , other one ends with 1,3,7or 9.
3)n-2 numbers end with 5 , other one ends with 1,3,7or 9.
.
.
.
.
1 number ends with 5 , others all end with 1,3,7 or 9.

therefore the probability that the product ends with 5 is :$ (\frac {1}{10})^n \plus{} (\frac {1}{10})^{n \minus{} 1} \frac {4}{10} \plus{} ....... \plus{} (\frac {4}{10})^{n \minus{} 1} \frac {1}{10}.$

please advice me whether this is correct.

Kent: yes, but it's also pretty clear what is meant, I think.  (Specifically, that we take a uniform distribution on $ \{1, 2, \ldots, n\}$ and then a limit as $ n \to \infty$.)

Vijay: $ 2 \cdot 1 \equal{} 2$.  You'd be better off trying to understand the argument that I outlined for you.  (If you manage to repair your solution, the result will be an unnecessarily complicated variation on the solution I've suggested.)

Frankly JBL , i am terribly sorry and i don't get you.  :(  So actually do u mean to say the outline i have given is wrong?

Can u please give a more a little bit more as a hint , so that i can try and work it out . please. :(

I have jus started learning probability. so please bear with me .

The key is that being divisible by $ 2$ and being divisible by $ 5$ are independent events. It is what JBL clearly meant but hasn't spelled out.

do we get the answer as $ \frac{ 10^n \minus{} 8^n \minus{}5^n \plus{} 4^n }{10^n}$ .

Yes, this is one possible way to write it down.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

An element $ a$of an integral domain $ D$ divides $ b$ if there exist some $ c$ such that $ a\equal{}cb$, two elements are associates if $ a$ divides $ b$, and $ b$ divides $ a$.

List al associates of $ x^3\plus{}2x\minus{}1$ over the integral domain of integers modulo 5.

[quote="streeler"]An element $ a$of an integral domain $ D$ divides $ b$ if there exist some $ c$ such that $ a \equal{} cb$, two elements are associates if $ a$ divides $ b$, and $ b$ divides $ a$.

List al associates of $ x^3 \plus{} 2x \minus{} 1$ over the integral domain of integers modulo 5.[/quote]

Two elements $ x$ and $ y$ are associated if and only if there exists a unit $ u$ such that $ x \equal{} uy$.


hello, writing $ (3x)^3\plus{}4^3$ and factoring we get $ (3x\plus{}4)(9x^2\minus{}12x\plus{}16)$ as the answer.
Sonnhard.


calculus
integration
function
ratio
complex numbers
calculus computations

If $ J_{n} (x)$ denotes the bessel function of order $ n$, then
evaluate:
$ \int_{0}^{\infty} e^{ \minus{} ax}J_{0}(bx)dx$.
[hide]Answer: $ \frac {1}{\sqrt {a^{2} \plus{} b^{2}}}$[/hide]

Can some1 please help me out with this one.  :)

Laplace transforms turn differential equations into algebraic equations. This is the Laplace transform of a Bessel function; what is the differential equation that defines it?

Indeed $ J_{n}(x)$ satisfies the Bessel Differential Equation:
$ x^{2}y'' \plus{} xy' \plus{} (x^{2} \minus{} n^{2})y \equal{} 0$.

[quote]Specialize, and also include the initial conditions.[/quote]
I dont get you? Specialize as in?
Further it is also known that the generating function for the bessel function is $ f(x,t) \equal{}  \exp\{\frac{x}{2} \left(t \minus{} \frac{1}{t}\right)\} \equal{} \sum_{n \equal{} \minus{} \infty}^{\infty} J_{n}(x)$.
Isn't it possible to do this problem using the definition of the Bessel Function? As I havent studied Laplace Transforms  as yetand this problem came up.
By Definition i mean $ J_{n}(x) \equal{} \sum_{m\equal{}0}^{\infty} \frac{(\minus{}1)^{m}}{m! \Gamma(m \plus{} n\plus{}1)} \cdot \left(\frac{x}{2}\right)^{2m \plus{} n}$.

When Laplace Transform is mentioned, most people think of its application in solving DEs. In this case, its application in DEs makes the problem quite simple.

Knowing that

$ \mathcal{L}\{y'\} \equal{} s \mathcal{L}\{y\} \minus{} y(0)$
$ \mathcal{L}\{y''\} \equal{} s^2 \mathcal{L}\{y\} \minus{} s y(0) \minus{} y'(0)$

you can LT the equation and get a linear algebraic equation and find $ \mathcal{L}\{f\}$ in respect to $ s$ and constants $ y(0)$ and $ y'(0)$ (initial conditions) Noting that

$ \mathcal{L}\{f\} \equal{} \int_{0}^{\infty} e^{ \minus{} sx}J_{0}(x)dx$

you get the relation between $ s$ and $ a,b$ (hint: make a formal substitution $ bx\to t$ in your initial integral so you get the Laplace-Like form of the integral).

As far as a non-LT solution (the one using "the definition" of Bessel function as functional series): ideas could include integrating term by term the series:

$ e^{\minus{}ax}\sum_{m\equal{}0}^{\infty}\frac{(\minus{}1)^m}{(m!)^2}\left(\frac{bx}{2}\right)^{2m}$ (if they're integrable term-by-term) acquiring some sort of rule for it, and than summing it.

[quote="hsiljak"]... ideas could include integrating term by term the series:

$ e^{ - ax}\sum_{m = 0}^{\infty}\frac {( - 1)^m}{(m!)^2}\left(\frac {bx}{2}\right)^{2m}$ (if they're integrable term-by-term) acquiring some sort of rule for it, and than summing it.[/quote]
Following up on that: $ \int_0^{\infty}x^{2m}e^{ - ax}\,dx = \frac {(2m)!}{a^{2m + 1}}.$

Insert that into the series to get

$ \int_0^{\infty}e^{ - ax}J_0(bx) = \sum_{m = 0}^{\infty}\frac {( - 1)^mb^{2m}(2m)!} {a^{2m + 1}2^{2m}(m!)^2}$

$ = \sum_{m = 0}^{\infty}\frac {( - 1)^m1\cdot2\cdot5\cdots (2m - 1)b^{2m}}{2^mm!a^{2m + 1}}.$

Somewhat disconcertingly, the ratio test shows that that series converges only for $ |b| < |a|$ (or possibly $ |b|\le |a|$ - I didn't investigate that), while the integral converges for all $ a > 0,$ no matter what $ b$ is. But power series are about complex numbers, anyway: while $ J_0(bx)$ is bounded on $ \mathbb{R}$ for all real $ b,$ it displays rapid growth for imaginary $ b.$

So assume $ 0 < a < b.$ Then

$ \frac1{\sqrt {a^2 + b^2}} = \frac1a\left(1 + \frac {b^2}{a^2}\right)^{\frac12}$
$ = \frac1a\left(1 - \frac12\cdot\frac {b^2}{a^2} + \frac {\left( - \frac12\right)\left( - \frac32\right)}2\cdot\frac {b^4} {a^4} + \frac {\left( - \frac12\right)\left( - \frac32\right) \left( - \frac52\right)}{3!}\cdot\frac {b^6}{a^6} + \cdots\right)$

$ = \frac1a - \frac {b^2}{2a^3} + \frac {1\cdot3b^4}{2^2\cdot 2! a^5} - \frac {1\cdot3\cdot 5b^6}{2^3\cdot 3!a^7}\cdots$

$ = \sum_{m = 0}^{\infty}\frac {( - 1)^m1\cdot2\cdot5\cdots (2m - 1)b^{2m}}{2^mm!a^{2m + 1}}.$

Hence the integral is $ \frac 1{\sqrt {a^2 + b^2}},$ at least for $ b < a.$ It is valid beyond that range by analytic continuation.


prove that there are no integers $ (x,y,z)$ satisfying  $ x^2 \plus{} 3xy \minus{} 2y^2 \equal{} 122$

Hello !

$ x^2 \plus{} 3xy \minus{} 2y^2 \equal{} 122  \Leftrightarrow (2x\plus{}3y)^2 \minus{}17 y^2 \equal{} 488$

$ 488 \equiv 12 [17]$

and $ (\forall n \in \mathbb{Z} ) n^2 \not\equiv 12 [17]$

therfore , there are no integers ...


search
number theory proposed
number theory

Let $ a_1,a_2,...,a_n$ be natural numbers such that no two of them are such that one is made of the digits of the other in the same order (For example, $ 57$ and $ 578902$ cannot be included at the same time) Prove that

$ \sum_{i \equal{} 1}^{n}a_i\leq 1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{9}$.

It is from IranMO 1998 . Search on PEN for solution .


geometry
trigonometry
algebra unsolved
algebra

$ R,r$  are the radii of the cirrcumcircle and incircle of the tirangle $ ABC$
$ a,b,c$ are the three sides of triangle $ ABC$
$ s\equal{}\frac{a\plus{}b\plus{}c}{2}$
Prove that
$ \cos{A}\cos{B}\plus{}\cos{B}\cos{C}\plus{}\cos{C}\cos{A} \equal{} \frac{s^2\minus{}4R^2\plus{}r^2}{4R^2}$


geometry
geometric transformation
geometric sequence
number theory solved
number theory

Given n men and a pile of coconuts, each man in sequence takes 1/n th of the coconuts left after the previous man removed his (i.e., a_1 for the first man, a_2 for the second, ..., a_n for the last) and gives m coconuts (specified in the problem to be the same number for each man) which do not divide equally to a monkey. When all n men have so divided, they divide the remaining coconuts n ways (i.e., taking an additional a coconuts each), and give the m coconuts which are left over to the monkey. If m is the same at each division, then how many coconuts N were there originally?

http://www.harmonicgateways.com/Flowers/coconut.JPG
The first time i drunk a coconut juice many many years ago, i did not
imagine it will bring one day to a Diophantine problem.
oubinol

[quote="Moubinool"]When all n men have so divided, they divide the remaining coconuts n ways (i.e., taking an additional a coconuts each), and give the m coconuts which are left over to the monkey.[/quote]

well, I don't understand this well. Could you please explain it more?

Just for a test can you try use the translator http://www.systranbox.com/systran/box i hope it is good

指定的n 人和堆椰子, 各個人在椰子th 留給在以後的序列作為當所有n 人有因此劃分, 他們劃分殘餘的椰子n 方式(即, 採取另外椰子每個), 和給留下給猴子的m 椰子。

oubi

Nearly all translators don't work well on translation into Chinese. :(

I just realized that "a" in "taking an additional a coconuts each" is a variable, not a word.... :D

ok, i've understood it.

The equations are
N = n a_1 + m
N - a_1 - m = n a_2 + m, which is, (n-1)a_1 = n a_2+m,
and similarly, (n-1)a_2 = n a_3 + m
...
(n-1)a_i = n a_{i+1} + m
...
(n-1)a_n = na + m...

There was a famous case of this problem with n=5 and m=1...

a_n = n/(n-1) a_{n+1} + m.

let k=m(n-1), thus

a_n + k = n/(n-1)(a_{n+1}+k)

hence, {a_n+k} is a geometric progression

a_0 + k = (a_{n+1} + k) * (n/(n-1))^(n+1)

so we may write

N = a_0 = c n^(n+1) - k = c n^(n+1) - m(n-1)
where k is an integer.


LEMMA: let  $ a,b,c\ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove that  $ 4\ge a^2b \plus{} b^2c \plus{} c^2a \plus{}abc$

[quote="ABRORBEK"]LEMMA: let  $ a,b,c\ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$ Prove that  $ 4\ge a^2b \plus{} b^2c \plus{} c^2a \plus{} abc$[/quote]
Try the beautiful Sung-yoon Kim's idea:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100960

hi !!

since $ a\plus{}b\plus{}c\equal{}1$ we can put  $ a\equal{}\frac{3x}{x\plus{}y\plus{}z}$ , $ b\equal{}\frac{3y}{x\plus{}y\plus{}z}$ , $ c\equal{}\frac{3z}{x\plus{}y\plus{}z}$

the inequality become

$ 4(x\plus{}y\plus{}z)^3 \geq 9(x^2y\plus{}y^2z\plus{}z^2x) \plus{}27xyz$

we now that : $ (x\plus{}y\plus{}z)^3 \equal{}x^3\plus{}y^3\plus{}z^3\plus{}3(x^2(y\plus{}z)\plus{}y^2(z\plus{}x)\plus{}z^2(x\plus{}y))\plus{}6xyz$

the inequality become

$ 4(x^3\plus{}y^3\plus{}z^3) \plus{}12(x^2z\plus{}y^2x\plus{}z^2y)\plus{}3(x^2y\plus{}y^2z\plus{}z^2x) \geq 3xyz$

it is true because $ x^3\plus{}y^3\plus{}z^3 \geq 3xyz$

[quote="y-a-s-s-i-n-e"]hi !!
since $ a \plus{} b \plus{} c \equal{} 1$[/quote]
the problem say:$ a\plus{}b\plus{}c\equal{}3$!

SULUTION:$ a\plus{}b\plus{}c\equal{}3$Sung-yoon Kim's idea  If $ \{ p,q,r \} \equal{} \{ a,b,c \}$, $ p \geq q \geq r$, then as $ pq \geq pr \geq qr$,
$ a^{2}b \plus{} b^{2}c \plus{} c^{2}a \plus{} abc \equal{} a(ab) \plus{} b(bc) \plus{} c(ca) \plus{} b(ac) \leq p(pq) \plus{} 2q(pr) \plus{} r(qr)$=$ \frac {1}{2}(2q)(p \plus{} r)(p \plus{} r)$
$ \equal{} \frac {1}{2}(\frac {(2q) \plus{} (p \plus{} r) \plus{} (p \plus{} r)}{3})^{3}$=$ \frac {2^3}{2} \equal{} 4.$

[quote="y-a-s-s-i-n-e"]hi !!

since $ a \plus{} b \plus{} c \equal{} 1$ we can put  $ a \equal{} \frac {3x}{x \plus{} y \plus{} z}$ , $ b \equal{} \frac {3y}{x \plus{} y \plus{} z}$ , $ c \equal{} \frac {3z}{x \plus{} y \plus{} z}$

the inequality become

$ 4(x \plus{} y \plus{} z)^3 \geq 9(x^2y \plus{} y^2z \plus{} z^2x) \plus{} 27xyz$

we now that : $ (x \plus{} y \plus{} z)^3 \equal{} x^3 \plus{} y^3 \plus{} z^3 \plus{} 3(x^2(y \plus{} z) \plus{} y^2(z \plus{} x) \plus{} z^2(x \plus{} y)) \plus{} 6xyz$

the inequality become

$ 4(x^3 \plus{} y^3 \plus{} z^3) \plus{} 12(x^2z \plus{} y^2x \plus{} z^2y) \plus{} 3(x^2y \plus{} y^2z \plus{} z^2x) \geq 3xyz$

it is true because $ x^3 \plus{} y^3 \plus{} z^3 \geq 3xyz$[/quote] IT is try $ 4(x \plus{} y \plus{} z)^3 \geq 9(x^2y \plus{} y^2z \plus{} z^2x) \plus{} 27xyz$ .IT is right  $ 4(x \plus{} y \plus{} z)^3 \geq 27(x^2y \plus{} y^2z \plus{} z^2x) \plus{} 27xyz$


trigonometry function calculus integration

How do you find the period and amplitude for y=sin(2x)+4sin(7x)?

[hide="Period"]Period of $ \sin{(2x)}\Rightarrow\pi$
Period of $ \sin{(7x)}\Rightarrow\frac{2\pi}{7}$
In one full period of $ f(x)$, both functions above must go through an integral number of periods; the least number that satisfies the conditions in $ 2\pi$.
[/hide]


absolute value
number theory unsolved
number theory

Let $ (m,n)\in N\times N$
(i) Prove that the set of all positive integers can be partitioned into three pairwise disjoint nonempty subsets such that none of them has two numbers with absolute value of their difference equal to either $ m$ or $ n$.
(ii) Find all pairs $ (m,n)$ such that the set of all positive integers cannot be partitioned into two pairwise disjoint nonempty subsets satisfying the above condition

(ii) the pairs $ (m,n)$ are exactly those for which $ \frac {m}{\gcd(m,n)}$ and $ \frac {n}{\gcd(m,n)}$ have opposite parity. (equivalently, the highest powers of 2 in $ m$ and $ n$ are different.)

proof: think of a partition of the natural numbers into two disjoint sets as an infinite 0-1 string. as examples, 10101010... means the partition of $ N$ into evens and odds, and 001001001... means the partition into multiples of 3 and non- multiples of 3. good strings are those which describe a partition of the required form. consider an arbitrary good string. we may assume WLOG that the first bit is 1. then the substring of bits numbered $ 1 \plus{} jm$ must go 101010..., with a 1 every time $ j$ is even, a 0 every time $ j$ is odd. the substring of bits numbered $ 1 \plus{} kn$ must also go 101010..., with a 1 every time $ k$ is even, a 0 every time $ k$ is odd. now, the first bit after the first that's in common to the two substrings is that whose place is given by $ j \equal{} \frac {n}{\gcd(m,n)}$, $ k \equal{} \frac {m}{\gcd(m,n)}$. considering the value of that term in each of the two substrings, it's clear these two must have the same parity.

conversely, assume that $ m$ and $ n$ are such that $ \frac {m}{\gcd(m,n)}$ and $ \frac {n}{\gcd(m,n)}$ have the same parity. write $ m \equal{} da$, $ n \equal{} db$, where $ d \equal{} \gcd(m,n)$. then $ a$ and $ b$ are both odd. the string
\[ \underbrace{1...1}_{d}\underbrace{0...0}_{d}\underbrace{1...1}_{d}\underbrace{0...0}_{d}...
\]
is then good, and gives us the desired partition. to verify that this string is indeed good, we need to check that for any bit $ x$, the bits $ m$ and $ n$ places to the right of $ x$ differ from it. and this is indeed true because the bit $ y$ $ m$ places over is $ a$ blocks of length $ d$ away, and since $ a$ is odd, its clear that the block containing $ y$ is of the opposite kind as the one containing $ x$; and the same argument goes for $ n$.


Let a,b,c be real numbers so:

$a,b,c \in [0,1]\\ ab+bc+ac=1$

Prove that
$a^{2006}+b^{2006}+c^{2006}\leq 2$

[hide]
WLOG let $a\geq b\geq c$ because the expressions are symmetric. $a^{2006}\leq a^{2}$ etc. so it only remains to prove that
\[a^{2}+b^{2}+c^{2}\leq 2 \]
After multiplying by 2 and subtracting $2(ab+bc+ca)=2$, this is equivalent to
\[(a-b)^{2}+(b-c)^{2}+(a-c)^{2}\leq 2 \]
This is evident from $(a-b)^{2}+(b-c)^{2}\leq (a-b)^{2}+(b-c)^{2}+2(a-b)(b-c)=(a-c)^{2}$ and $0\leq a-c\leq 1$.
[/hide]

[hide="A way..."]If WLOG, $c \geq  b \geq  a$, then it suffices to prove:
$a^{2}+b^{2}+c^{2}\leq  ab+bc+c^{2}\leq  2$.

The first part is true.
So the second part is equivalent to:

$c^{2}+1-ac \leq  2$
$c(c-a) \leq  1$

Seeing as how $c \leq  1$, and $(c-a) \leq  1$, that last inequality is true. So we're done.[/hide]


solve equation :$x^2+y^2+1=xyz$ where $x,y,z \in Z$

It's rather great problem, since if, for example, z=3, discriminant D of this eqution regarding to [b]x[/b] is equal to [b]5*(y^2) - 4 [/b] -  must be equal to [b]t^2[/b], where [b]t [/b]is an integer. But spesial case of equation [b]5*(y^2) - 4 = t^2[/b] is "Minus-Pell' "-equation: [b]a^2 - 5*(b^2) = -1.[/b], which give only even solutions: [b]t=2a, y=2b.[/b]

This problem really asks you to solve the simultaneous congruences
$x^2\equiv -1$ mod $y$
$y^2\equiv -1$ mod $x$.

A very similar problem posted a few days ago, here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35312&view=previous

asked for all integers (less than 2005) such that $x^2+4$ was a multiple of $y$ and $y^2+4$ was a multiple of $x$.  This is equivalent to solving the simultaneous congruences
$\left(\frac{x}{2}\right)^2\equiv -1$ mod $y$
$\left(\frac{y}{2}\right)^2\equiv -1$ mod $x$.

(Both problems have so far remained unsolved.)


The following elementary inequalities seem to be deceptively tricky :

1) Rearrange in order (increasing or decreasing) : $ 5^{100} , 6^{91}, 7^{90}, 8^{85}$.
2) Show that   $ \left(1\plus{} \frac{1}{20}\right)^{100}>100$.

Obviously these are to be solved without resorting to calculator or logarithm. :maybe:  :maybe:

[quote="Shanku"]Hey ! Calculators (as also calculation other than like 2+3=5 or 100 >93 kind of things) are not allowed ! Why one number is left ? Can't locate apower of 6 table ?[/quote]
I didn't use any calculator. I was unable to evaluate $ 6^{91}$ accurate enough. You can watch numbers from power tables and its not too hard to raise a number to fifth power.

[quote]You can watch numbers from power tables and its not too hard to raise a number to fifth power.[/quote]
you musnt use any sort of tables[/quote]

[quote="MeKnowsNothing"]Why tables are not allowed?[/quote]
It's posted in the olympiad section, so it is probably meant to an olympiad-like problem - and i have only found olympiads where nothing but pen, paper and some geometric tools like rulers and such are allowed.

Simply because, with mathematical Tables and calculators at hand, these are no more problems !!


I'm having trouble with what should be a very simple proof.

I'm trying to prove $det(A^{-1}) =\frac 1{det(A)}$

This is a corollary of the broader theorem $\det(AB)=\det(A)\cdot\det(B).$

Do you have the proof of that?

Yes I do, that's a good hint.  I can't believe I didn't think of that, thank you.


Problems A, B, and C were posed in a mathematical contest. 25 competitors solved at least one of the three. Amongst those who did not solve A, twice as many solved B as C. The number solving only A was one more than the number solving A and at least one other. The number solving just A equalled the number solving just B plus the number solving just C. How many solved just B?

[hide="answer"]6[/hide]

This is a Putnam problem?! It must be an old one. When is it from and what number problem was it?

[hide="solution"]
Let $ a$ be the number that solved only A, $ b$ be the number that solved A and B but not C, C be the number that solved only B, $ d$ be the number that solved only A and C, $ e$ be the number that solved all of them, $ f$ be the number that solved only B and C, and $ g$ be the number that solved only C. I'd post a Venn diagram if I could.

First we see that $ c\plus{}f\equal{}2(g\plus{}f)\Rightarrow{c}\equal{}f\plus{}2g$.

Also, $ a\equal{}b\plus{}d\plus{}e\plus{}1$.

Also, $ a\equal{}c\plus{}g\equal{}f\plus{}3g\equal{}b\plus{}d\plus{}e\plus{}1$.

Since all 25 solved at least one problem, we must have
$ a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f\plus{}g\equal{}25\Rightarrow{f}\plus{}3g\plus{}(b\plus{}d\plus{}e\plus{}1)\plus{}(f\plus{}2g)\plus{}f\plus{}g\equal{}26$
$ f\plus{}3g\plus{}f\plus{}3g\plus{}f\plus{}2g\plus{}f\plus{}g\equal{}4f\plus{}9g\equal{}26$. The only solution in nonnegative integers to this equation is $ (f,g)\equal{}(2,2)$

Thus $ c\equal{}2\plus{}2(2)\equal{}6$, so exactly 6 people solved only problem B.
[/hide]


analytic geometry
graphing lines
slope
calculus
calculus computations

More Calc Problems...

[img]http://myspace-739.vo.llnwd.net/00703/93/76/703686739_l.jpg[/img]
[img]http://myspace-212.vo.llnwd.net/00703/21/28/703688212_l.jpg[/img]
[img]http://myspace-843.vo.llnwd.net/00703/34/89/703689843_l.jpg[/img]

thanks
again

got_bass_69@hotmail.com

For 6, we have $\frac{3}{12-x}=\frac{y}{12}=\text{the slope of TC}\Longleftrightarrow y=.....$

Thus the answer is $\left[\frac{dy}{dt}\right]_{x=5}.$


algebra
polynomial
number theory solved
number theory

Let p be a prime number and suppose 1 \leq k \leq p-1. From the numbers 1,2,...,p-1, choose k different ones and take their product. Show that the sum of all such \binom{p-1}{k} products is divisible by p.

I hope the following is correct. In Zp[X] we have the polynomial identity (X-1^)(X-2^)...(X-(p-1)^)=x^(p-1)-1^ (here a^ is the class of a). This comes from the fact that the polynomial f(X)=LHS-RHS is of degree <=p-2 and it becomes zero in Z_p for at least p-1 values non-congruent modulo p. So, the identity follows from Lagrange theorem. If we develop (X-1^)(X-2^)...(X-(p-1)^) and we identify the coefficients, everything will be clear.


geometry
3D geometry
sphere
calculus
ratio

I know that the formula for the surface area of a sphere is $ 4 \pi r^2$, and the formula for volume is $ \frac{4}{3}\pi r^3$. My question is why? Is there a way to figure this out/derive it?

Thank you for your help :D .

Oh thank you. Now I'm starting to remember something like this in my calc book...

EDIT: Once I saw the world calculus, I quit searching and started derving  :D .
Meeeeh. I tried. And failed. Not good with calculus  :o .

hey both used calculus to PROVE it.
they may be non rigorously derived.

for example finding the ratio of a sphere to the circumscribed cylinder (what archimedes did or something)
and turning a sphere into its baseball-ish net

Another thing to note when you learn differentiation.

$ \displaystyle{(\frac{4\pi r^3}{3})'\equal{}4\pi r^2}$

also does anyone know to make the ( ) cover the whole fraction?

[quote="mewto55555"]also does anyone know to make the ( ) cover the whole fraction?[/quote]
\left( and \right)

$ \displaystyle{\left(\frac {4\pi r^3}{3}\right)' \equal{} 4\pi r^2}$

I prefer to state the result as Archimedes did:

The volume of a sphere is $ \frac23$ of the volume of the smallest cylinder that contains the sphere. More specifically, if we also fit a cone into that cylinder, the ratios of the volumes of the cone to the sphere to the cylinder are $ 1: 2: 3.$

The area of the sphere is $ 4$ times the area of its equatorial disk.

If you think of a balloon fitted over a thin solid disk and look at it completely deflated, you'll see that the surface area of the balloon must be at least twice the area of the disk - once for each side. The Archimedes result tells us that inflating that balloon to a sphere (while maintaining contact with the edges of the disk) increases the area from twice the disk to four times the disk.


Find all pairs (a,b) such that 2a+b=12 and ab=3.

Hi! 4everwise :)

$(2a+b)^2-(2a-b)^2=8ab\Longleftrightarrow (2a-b)^2=12^2-8\cdot 3=120\Longleftrightarrow 2a-b=\pm 2\sqrt{30}$

Solving the system of the equation $2a+b=12, 2a-b=\pm 2\sqrt{30}$, we get $(a,b)=\left(3\pm \frac{\sqrt{30}}{2},6 \mp \sqrt{30}\right)$,

double sign in this order.

kunny

This is certainly not appropriate for the MOEMS forum.

This is such an ugly problem!  :mad:
[hide]The first equation can be rewritten as $b=12-2a$
Substituting, we have $a(12-2a)=3$
$2a^2-12a+3=0$

$a=\frac{6\pm\sqrt30}{2}$
Thus, substituting into the second equation and rationalizing denominators we have
$b=4\mp2\sqrt3$

$(a,b)=(\frac{6\pm\sqrt30}{2}, 6\mp\sqrt30)$[/hide]


hello mo zhou

that is a pretty cool name

what's up from Carmel, Indiana.

i hope i'll see you around. lol.

How do you go about doing that anyways?  I've been trying to find out how to join a Worcester or Central team , if they have one.  Do any of you know who I should I contact?

Do any of you guys go to Acton high?

cuz i think that's where i'm going this fall...

Are you talking about Acton-Boxborough Regional high school?  I was going to choice-school to there but it was just too long of a drive every morning (about 45 minutes).  I have some friends who go there as well.

Hey I go to Acton-Boxborough, although I'm graduating next year.

[quote="Pineappleperson"]Hey guys, my name is Mo Zhou. I'm from Michigan and i'm sophomore next year...and i'm moving to MA this summer..
and apparently after reading how u flamed the other guy, i'm not gonna list my achievements...      :wink:[/quote]

Hmm, if you look a little closer none of the flamers are actually from Massachusetts.

And hi, I also go to Acton-Boxborough Regional High School.

Hi Mo. glad to have you here in mass. maybe I should have made an intro topic for myself lol- I'm new to AoPS.


find all x such that x^2+x+39 is a perfect square.

thanks.

Try writing it down and completing the square.  (It helps if you multiply by 4 first.)

thanks, that's what i was missing.  Cause when i first completed the square, it didn't seem to help with the 4 on the denominator.

Wait, I don't get it, so we complete the square and get (2x+1)^2 + 155, but what does that tell us?

You're looking for two squares that differ by 155.

So letting the squared numbers be a and b, a>b, we have a^2-b^2=(a+b)(a-b)=155=5*31. So do we just have to try every case?

Yep. Theres only 8 of them. And 4 of them will give you repeated answers anyway.


inequalities
trigonometry
geometry proposed
geometry

Let be given an acute triangle $ABC$.Consider that $A\ge B\ge C$Prove that:$\sin^2A+\sin^2B\ge\frac{3}{2}$

Note that when there is equality at $A=B=C=60^\circ$, $\sin^2A + \sin^2B = \frac32$.  Since sin is strictly increasing on $[60,90)$ (the interval of angle A), $\sin^2 A$ is also increasing since A is increasing.  So the LHS is increasing, unless we can make B small enough to make the LHS = $\frac32$ again (which, by intermediate value thereom, must occur if the LHS is to ever go below $\frac32$).  Since $B \ge C$, the smallest B is $\frac{90-A}2$.  Then we have
$\sin^2 A + (\frac{\sqrt 2}{2}(\cos \frac A2 - \sin \frac A2))^2$
$\sin^2 A + \frac{ \frac{1 + \cos A}2 + \frac{1 - \cos A}2 - 2\sqrt{\frac{1 - \cos^2 A}4} }2$
$\sin^2 A + \frac{1 - \sin A}2$
So we have the equation
$\sin^2 A - \frac{\sin A }2 + \frac 12 = \frac 32$
$2\sin^2 A - \sin A - 2 = 0$
$\sin A = \frac{ 1 \pm \sqrt{17}}4$
Note that one solution is negative, which can't be possible for any A on the interval, and the other solution is greater than 1, impossible for the sine function.
QED

[quote="deej21"]Since $B \ge C$, the smallest B is $\frac{90-A}2$[/quote]
Wrong!$B\ge C\Longrightarrow B\ge 180^o-B-A\Longrightarrow B\ge \frac{180^o-A}{2}$

Shouldn't talk too much.This is the solution:
I have just written:
$B\ge\frac{180^o-A}{2}$
$\Longrightarrow \sin B\ge \sin\frac{180^o-A}{2}$(note that $B<90^o$)
$\Longrightarrow \sin^2 A+\sin^2 B\ge \sin^2 A+\sin^2\frac{180^o-A}{2}=\sin^2 A+\cos^2\frac{A}{2}=$
$=1-\cos^2 A+\frac{\cos A+1}{2}=-\cos^2 A+\frac{\cos A}{2}+\frac{3}{2}=\cos A(\frac{1}{2}-\cos A)+\frac{3}{2}$
Otherwise $A=max(A,B,C)$ thus $A\ge 60^o\longrightarrow \cos A\le \frac{1}{2}$.So we have$\cos A(\frac{1}{2}-\cos A)+\frac{3}{2}\ge \frac{3}{2}\Longrightarrow \sin^2 A+\sin^2 B\ge \frac{3}{2}$
Q.E.D


I face some problem in statistic estimation question . Anyone help ?

Prove that $s^2$ is an unbiased estimation for $\sigma^2$ where $s^2=\frac{1}{n-1}\sum (x-\overline x)^2$ is the sampling variances and $\sigma^2$ is the population variances.

[quote="10000th User"]Hi shyong, I hope you use this definition: A statistic $\widehat{\Theta}$ is an [b]unbiased estimator[/b] of the parameter $\theta$ iff $E(\widehat{\Theta})=\theta$.

So you want to prove that $E(s^2)=\sigma^2$

An observation: is the population infinite or finite? Because $s^2$ is [i]not[/i] an unbiased estimator of $\sigma^2$ if the population is finite...[/quote]

isnt it generally $E(s^2)=\sigma^2$ ? Well the question doesnt state anything about finite or infinite though but just tell us to prove $E(s^2)=\sigma^2$

[quote="shyong"]isnt it generally $E(s^2)=\sigma^2$ ? Well the question doesnt state anything about finite or infinite though but just tell us to prove $E(s^2)=\sigma^2$[/quote]Not necessarily true. In fact, the difference between a finite and an infinite population distribution is that infinite has zero covariance whereas finite has a negative covariance. For example, if $\bar{X}$ is the mean of a random sample of size $n$ of a population of size $N$ with mean $\mu$ and variance $\sigma^2$,

Infinite population ($N=+\infty$): $E(\bar{X})=\mu$ and $var(\bar{X})=\frac{\sigma^2}{n}$

Finite population ($N<+\infty$): $E(\bar{X})=\mu$ and $var(\bar{X})=\frac{\sigma^2}{n}\frac{N-n}{N-1}$

As $N\rightarrow+\infty$, finite population behaves like just the same as infinite population.

Back to your question: assuming infinite population,

\[ \begin{aligned} E(s^2)&=E\left(\frac1{n-1}\sum_i(x_i-\bar{x})^2\right)\\ &=\frac1{n-1}E\left(\sum_i\left((x_i-\mu)-(\bar{x}-\mu)\right)^2\right)\\ &=\frac1{n-1}\left(\sum_iE\left((x_i-\mu)^2\right)-2\sum_iE\left((x_i-\mu)(\bar{x}-\mu)\right)+\sum_iE\left((\bar{x}-\mu)^2\right)\right)\\ &=\frac1{n-1}\left(\sum_iE\left((x_i-\mu)^2\right)-2E(\bar{x}-\mu)\sum_iE(x_i-\mu)+\sum_iE\left((\bar{x}-\mu)^2\right)\right)\\ &=\frac1{n-1}\left(\sum_iE\left((x_i-\mu)^2\right)-2nE\left((\bar{x}-\mu)^2\right)+nE\left((\bar{x}-\mu)^2\right)\right)\\ &=\frac1{n-1}\left(\sum_iE\left((x_i-\mu)^2\right)-nE\left((\bar{x}-\mu)^2\right)\right)\\ &=\frac1{n-1}\left(\sum_i\sigma^2-n\cdot\frac{\sigma^2}{n}\right)\\ &=\sigma^2 \end{aligned} \]

To explain what I used,

$\sum_iE(x_i-\mu)=nE(\bar{x}-\mu)$ (follows from definition of expectation, it's a constant)

$\sum_iE\left((x_i-\mu)^2\right)=\sigma^2$ (follows from definition of variance)

$E\left((\bar{x}-\mu)^2\right)=\frac{\sigma^2}{n}$ (follows from definition of sample variance, constant)


function
search
real analysis
real analysis unsolved

Let $f$ be differentiable over $\mathbb{R}$. Suppose there is a subset $S \subset \mathbb{R}$ s.t. $S$ is dense in $\mathbb{R}$ and $\forall x \in S, f'(x)>0$.
Is $f$ an increasing function?
Note: I thought it would be easy, maybe not  :oops:

Already discussed, search for a topic started by alekk.

Thanks harazi, I found http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=dense&t=19246.
Not easy.


Let $ a, b, c, d$ be nonnegative numbers satisfying $ a\plus{}b\plus{}c\plus{}d\equal{}4$. Prove that:
\[ \frac{ab}{c\plus{}d\plus{}4}\plus{}\frac{bc}{d\plus{}a\plus{}4}\plus{}\frac{cd}{a\plus{}b\plus{}4}\plus{}\frac{da}{b\plus{}c\plus{}4}\plus{}\frac{\sqrt{abcd}}{3} \le 1\]
When does equality occur ? :)

[quote="nguoivn"]Let $ a, b, c, d$ be nonnegative numbers satisfying $ a \plus{} b \plus{} c \plus{} d \equal{} 4$. Prove that:
\[ \frac {ab}{c \plus{} d \plus{} 4} \plus{} \frac {bc}{d \plus{} a \plus{} 4} \plus{} \frac {cd}{a \plus{} b \plus{} 4} \plus{} \frac {da}{b \plus{} c \plus{} 4} \plus{} \frac {\sqrt {abcd}}{3} \le 1\]
When does equality occur ? :)[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-29-t-q-anh.html


Could you please explain how did we get the 1 and 0 answers?

[img]http://img145.imageshack.us/img145/33/11283304xs6.jpg[/img]

Well, $ lim_{x\minus{}>\infty}\ e^{\minus{}x} \equal{} 0$, and the first one follows from that.  The second one follows from $ lim_{x\minus{}>\minus{}\infty}\ e^{\minus{}x} \equal{} \infty$


group theory
number theory proposed
number theory

Let $ p\equal{}4k\plus{}1$ be a prime. Prove that $ p$ has at least $ \frac{\phi(p\minus{}1)}2$ primitive roots.

Yes, you're right! I didn't claim that problem is not easy, and students whom this exam was token from didn't know this easy fact. (This was first problem and easiest problem of an exam) :wink:

[quote="ZetaX"]What's that¿ Every number $ n$ that has primitive roots has exactly $ \varphi(\varphi(n))$ of them (and this is rather straightforward to prove).[/quote]
If $ n\not \equal{}p^k,2p^k,2,4$, then $ Z_n^*$ is not cyclic group, therefore had not primitive roots.

Yes, but he meant those $ n$'s that for those $ n$'s that we have a primitive root we have exactly $ \varphi(\varphi(n))$ primitive roots.

[quote="ZetaX"]He did not only mean it, he wrote it.[/quote]

Hahah... You are great, ZetaX!

[quote=shoki]eh... but in the paper the problem was sth else : :maybe:
Prove that if $p=4k+1$ is a prime then there exist at least $\frac{\phi(p-1)}{2}$ natural numbers like $g$ such that $0 < g < p$ and $g$ is a primitive root modulo any power of $p$.
[hide="comment"]
it is obvious that we must show among the $\phi(p-1)$ primitive roots modulo $p$, at least half of them are primitive root modulo $p^2$.  then , since any primitive root modulo $p^2$ is a primitive root modulo any power of $p$ we'll be done!
[/hide][/quote]

This is also easy to prove since if $g$ is a primitive root mod $p$ but not $p^2$ then $g+p$ is a primitive root mod $p^2$.

Hmm @above Can you clarify your proof? I don't see how that gives $\phi (p-1)/2$ primitive root modulo $p^2$.

Anyway, I use the fact that if $g$ is a primitive root modulo $p$, then so does $p-g$.

For the proof of above proposition: we have $p-g\equiv g^a\pmod{p}$ for some $a\in \{ 0,1,2,\dotsc ,p-1\}$.
We get $g^{a-1}\equiv -1 \equiv g^{(p-1)/2} \pmod{p}$ where $-1\leqslant a-1\leqslant p-2$. So $a-1=(p-1)/2$.
So $p-g\equiv g^{1+(p-1)/2}$. Since $\gcd ( p-1,1+(p-1)/2 )=1$, we get the desired claim.

Back to the problem. We'll show that at least one of the pair $g,p-g$ must be primitive root modulo $p^2$.
Suppose to the contrary that they both are not primitive root modulo $p^2$.

Let $k\in (0,p)$ be any primitive root modulo $p$ that is not a primitive root of $p^2$.
We have $k^d\equiv 1\pmod{p^2}$ for some $d\mid p(p-1)$ that $d<p(p-1)$.
If $p\mid d,$ we get $k^d\equiv k^{d/p}\pmod{p}\implies k^{d/p}\equiv 1\pmod{p}$ where $d/p<p-1$. Contradiction.
So $p\nmid d$. But we also can't have $d<p-1$. Thus $d=p-1$.

Applying the above result, we get $g^{p-1}\equiv 1\pmod{p^2}$ and
$$(p-g)^{p-1}\equiv \binom{p-1}{1}p(-g)^{p-2}+(-g)^{p-1}\equiv 1\pmod{p^2}.$$
Combining the two congruences gives us $-(p-1)pg^{p-2}\equiv 0 \pmod{p^2}$ which is clearly impossible. Done.


The new version of Adobe Reader (version 8) has a pretty significant bug, and using the built-in PDF viewing commands from TeXnic Center will cause the Reader to crash when you try to close the file.

There's a [url=http://www.adobeforums.com/cgi-bin/webx/.3bc27d26/3]thread on the Adobe forum[/url] describing this, and apparently the problem has not been fixed (as I found out the hard way when I installed Reader 8 on my PC today).

Don't upgrade to version 8 of Adobe Reader until they fix this bug!

If anybody know a way around this bug, please share your wisdom.

I did find a (bad) workaround:

As long as you always have more than one document open in Reader, it won't crash.  So just open any document in Reader first, and leave that document open all the time, and you'll be able to continue using Reader from TeXnic center, as long as you leave that original document open.

I found this workaround here:
[url]http://www.latex-community.org/viewtopic.php?f=31&t=14[/url]

Wow.

Adobe Reader 8 won't even load when I try to view my build from within TeXnic center.  It just comes up w/ an error message.


$ 1)$Given the following

$ y>x$
$ xy\equal{}\minus{}6$
$ x^3\minus{}x\equal{}y^3\minus{}y$

Find $ x^2\plus{}y^2$

$ 2)$ The lateral surface area of a six sided cube is 5184 meters squared. What is the volume?

[quote="Bicameral"]$ 1)$Given the following

$ y > x$
$ xy = - 6$
$ x^3 - x = y^3 - y$

Find $ x^2 + y^2$[/quote]
I don't think $ y > x$ is necessary.

[hide="Solution"]
\begin{align*}x^3 - x & = y^3 - y \\
 \\
x^3 - y^3 & = x - y \\
 \\
(x - y)\left(x^2 + xy + y^2\right) & = x - y \\
 \\
x^2 + xy + y^2 & = 1 \\
 \\
x^2 - 6 + y^2 & = 1 \\
 \\
x^2 + y^2 & = \boxed{7}\end{align*}
[/hide]
[quote="Bicameral"]$ 2)$ The lateral surface area of a six sided cube is 5184 meters squared. What is the volume?[/quote]
[hide="Solution"]The surface area of each face is $ 5184\div4=1296$.

The length of each side is $ \sqrt{1296}=36$.

The volume is $ 36^3=\boxed{46656}$.[/hide]

flawless solution to Number 1 i like pie. I did exactly the same thing.

What does Lateral Surface Area mean? What is it defined as?

[quote="i_like_pie"]The lateral surface area of an object is the sum of the areas of all its faces, but not its bases.[/quote]
wouldn't that mean that there are $ 5$ faces that you have to divide by, not $ 4$? What exactly counts as a base?

The top and bottom are both bases; the faces that count towards the lateral surface area are the ones on the sides.

[quote]I don't think y > x is necessary. [/quote]
I think that it's important : if y=x, then you cannot simplify by (x-y).
Isn't it?

Yes, then division by $ x \minus{} y$ could not have been carried out (it'd be relying on an assumption of $ x \minus{} y \neq 0$ that isn't necessarily true). Truth be told I didn't realise that until you pointed it out.

[quote="kh-"]it'd be relying on an assumption of $ x \minus{} y \neq 0$ that isn't necessarily true[/quote]

Sure? If $ x \minus{} y \equal{} 0$ then $ x \equal{} y$; as they would have the same sign $ xy \equal{} \minus{}6$ would be impossible, therefore $ x \minus{} y \neq 0$.

Ah, oops. I had a funny feeling I should probably squint at the initial conditions a bit more. Maybe that's why I didn't see it the first time because it isn't actually there. :P I would've just relied on the inequality given because I'm lazy (stupid?), but I think you're right: with a bit of logic you can drum the required info out of what you're given anyway.

Oh..yeah, you're right... didn't pay attention to this.
So finally, x != y isn't necessary as he said

Hey I am new here, so correct me if I am wrong

x^3 - y^3 = x - y

(x - y) (x + y)^2 = x - y

x^2 + 2xy + y^2 = 1

x^2 + 2(-6) + y^2 = 1

x^2 - 12 + y^2 = 1

x^2 + y^2 = 13

so is the answer 13 or 7

please correct me if I am wrong

[quote]x^3 - y^3 = x - y

(x - y) (x + y)^2 = x - y[/quote]

Mistake: $ x^3 \minus{} y^3 \neq (x\minus{}y)(x\plus{}y)^2$, but instead $ x^3 \minus{} y^3 \equal{} (x\minus{}y)(x^2\plus{}xy\plus{}y^2)$. A subtle but crucial difference.

hi,

given that xy = -6; x^3 - x = y^3 -y

by manipulating eq 2, we get x^3 - y^3 = x-y
=> (x-y)(x^2 + xy + y^2) = (x-y)
=> x^2 + xy + y^2 = 1; substition of eq1 we get x^2 + y^2 = 7


trigonometry
geometry
trig identities
Law of Sines
Law of Cosines
algebra proposed
algebra

Evaluate the value of [tex]sin\frac{\pi}{18}[/tex]. Justify your answer.

[hide]
Since $\sin{3x}=3\sin{x}-4\sin^3{x}$
So we can obtain:
\[\sin{\frac{\pi}{6}}=3\sin{\frac{\pi}{18}}-4\sin^3{\frac{\pi}{18}}\]
That is:
\[4\sin^3{\frac{\pi}{18}}-3\sin{\frac{\pi}{18}}+\frac12=0\]
Let $t=\sin{\frac{\pi}{18}}$, then:
\[8t^3-6t+1=0\]
This is as far as i reach. I don't know how to solve this cubic equaion so i believe there are better solutions.
Perhaps you can construct a $1,2,\sqrt3$ triangle and draw the trisectors of the 30 degree angle, then use the law of sines and the law of cosines, but i haven't tried that yet....[/hide]

[quote="shobber"]Perhaps you can construct a $1,2,\sqrt3$ triangle and draw the trisectors of the 30 degree angle, then use the law of sines and the law of cosines, but i haven't tried that yet....[/quote]

I don't think the answer is constructable.

The value of $\sin \frac{\pi}{18}$ can be written in form of infinite nested radical.

Any value of $\sin \frac{\pi}{2(2^{n+1}+1)} \, ; n \geq 2$  can be written in form of infinite nested radical too. !  :D


Let $ n \equiv 1$ $ (mod 4)$ be a non-square positive integer such that $ n \equal{} a^2 \plus{} 4b^2$ for some integers $ a$, $ b$.
Assume that $ x^2 \minus{} ny^2 \equal{} \minus{}1$ has an integer solution $ (x, y) \equal{} (t_1, t_2)$. Show that $ n \equal{} rs$ for some $ r, s$
where $ rx^2 \minus{} sy^2$ = $ a$ has an integer soltuion.

If I'm not mistaken, something like this appeared in American Math Monthly once .. I think I read it on the jstor.org/ journal articles. Try searching their archives.

i posted here precisely because i do not want to waste time looking for similar problems in old archives. Besides i'm not a member of jstor. So please only post if you have a complete solution.


red birds and blue birds, 30 in all, feasted on worms in my yard. The red birds got 108 worms in total, as did the blue birds. Each blue bird got 3 fewer worms than each red bird. How many blue birds were there?

[hide] Say there were x red birds.  This means that there were 30-x blue birds.  Each red bird got 108/x worms, and the blue birds each got 108/(30-x) worms.  We also know that the number of worms each blue bird got was 3 less than the number of worms each red bird got.  Thus, (108/x)-3=108/(30-x), so 36/x=1+36/(30-x).  Multiplying x(30-x) on both sides gives 36(30-x)=x(30-x)+36x.  Simplifying gets $30x-x^{2}=36(30-2x)\Rightarrow x^{2}-30x+1080-72x=0\Rightarrow x^{2}-102x+1080=0\Rightarrow x=90, x=12.$  x cannot be 90 because then x-90 would be negative.  Thus, x=12, so 12 red birds, so 30-12=$18$ blue birds. [/hide]


geometry
circumcircle
incenter
rectangle
angle bisector
radical axis
geometry unsolved

Dear Mathlinkers,
let ABC a triangle, 0 the circumcircle of ABC, 1 the A-mixtilinear incircle of ABC,
P, Q, R the points of contact of 1 wrt 0, AC, AB,
X, Y the feet of the perpendiculars issue from P wrt BC, QR,
2 the circle with centre P passing through X.
Prove : one of the two tangent issue from Y to 2 is parallel to AC , the other is parallel to AB.
Sincerely
Jean-Louis

$ I, I_a$ are the incenter and A-excenter, $ J \in AI$ is center of the mixtilinear A-incircle $ (1).$ $ AI$ cuts the circumcircle $ (O)$ again at $ M,$ $ MM'$ is a diameter of $ (O),$ and $ (M)$ is a circle with center $ M$ and radius $ MB \equal{} MC \equal{} MI \equal{} MI_a.$ $ I$ is the midpoint of $ QR \perp AIM$ and $ P, I, M'$ are collinear (well known). $ (3)$ is a circle with diameter $ IM$ and $ (4)$ is a circle with diameter $ IP.$ Angle $ \angle IPM \equal{} \angle M'PM$ is right, $ P$ is on the circle $ (3)$ with diameter $ IM,$ and angle $ \angle IYP$ is right, $ Y$ is on the circle $ (4)$ with diameter $ IP.$ The circle $ (4)$ cuts the circumcircle $ (O)$ at $ P, U,$ the circle $ (M)$ at $ I, V,$ and the angle bisector $ AIM$ at $ W.$ $ IYUW$ is a rectangle, $ UW$ a diameter of $ (4)$ and the angle $ \angle YUW$ is right. Let $ UW$ cut $ (O)$ again at $ A'.$

$ \angle PM'A' \equal{} \angle PUA' \equal{} \angle PUW \equal{} \angle PIW \equal{} \angle PIM.$

Therefore, $ M'A' \parallel AM,$ $ AMA'M'$ with diagonal $ MOM'$ is a rectangle. Since the angle $ \angle YUA'$  is right with $ U, A' \in (O)$ and $ AA'$ is a diameter of $ (O),$ $ YU$ goes through $ A.$ Since the angle $ \angle IVP$ is also right with $ I, V \in (M)$ and $ II_a$ is a diameter of $ (M),$ $ VP$ goes through $ I_a.$

$ II \equiv IY \perp IM$ is the single common tangent at $ I$ of $ (M), (3)$ with diameters $ II_a, IM.$ $ PP \equiv MP \perp PI$ is tangent at $ P$ of $ (4)$ with diameter $ IP.$ Pairwise radical axes $ II, BC, MP$ of $ (3), (M), (O)$ concur at their radical center $ D \in BC.$ Pairwise radical axes $ IV, BC, PU$ of $ (4), (M), (O)$ concur at their radical center $ E \in BC.$ By Pascal theorem for the cyclic pentagon $ PUVYI$ (degenerated hexagon $ PPUVYI$ with double vertex $ P$), the intersections $ D \equiv PP \cap IY,$ $ E \equiv PU \cap IV,$ and $ F \equiv PV \cap UY$ are collinear. Since $ D, E$ are on $ BC,$ so is $ F.$

But $ F \in BC$ is the external similarity center of the A-excircle $ (I_a)$ and the circle $ (2)$ with center $ P$ and tangent to $ BC$ at $ X,$ being the intersection of their common external tangent $ BC$ and their center line $ I_aP,$ and $ A$ is the external similarity center of the mixtilinear A-incircle $ (1)$ and the A-excircle $ (I_a),$ being the intersection of their common external tangents $ AB, AC$ and their center line $ AJI_a.$ By Monge-d'Alembert theorem, the external similarity center of the circles $ (1), (2)$ is on the line $ AF.$ As $ Y \in AF$ and $ PY \parallel JA,$ where $ P, J$ are centers of $ (2), (1),$ it follows that tangents to $ (2)$ from $ Y$ are parallel to the tangents $ AB, AC$ to $ (1)$ from $ A.$


inequalities
trigonometry
inequalities proposed

For an acute triangle $ ABC$ prove the inequality:
$ \sum_{cyclic} \frac{m_a^2}{\minus{}a^2\plus{}b^2\plus{}c^2}\ge \frac{9}{4}$ where $ m_a,m_b,m_c$ are lengths of corresponding medians.

[quote="Bugi"]For an acute triangle $ ABC$ prove the inequality:
$ \sum_{cyclic} \frac {m_a^2}{ \minus{} a^2 \plus{} b^2 \plus{} c^2}\ge \frac {9}{4}$ where $ m_a,m_b,m_c$ are lengths of corresponding medians.[/quote]
\[ \Leftrightarrow \sum {\frac{{2b^2  \plus{} 2c^2  \minus{} a^2 }}{{b^2  \plus{} c^2  \minus{} a^2 }}}  \ge 9
\]
\[ \Leftrightarrow \sum {\frac{{b^2  \plus{} c^2 }}{{b^2  \plus{} c^2  \minus{} a^2 }}}  \ge 6
\]
\[ \Leftrightarrow \sum {\frac{{a^2 }}{{b^2  \plus{} c^2  \minus{} a^2 }}}  \ge 3
\]
which is always true since
\[ \sum {\frac{{a^2 }}{{b^2  \plus{} c^2  \minus{} a^2 }}}  \ge 3\sqrt[3]{{\frac{{a^2 b^2 c^2 }}{{\prod {\left( {b^2  \plus{} c^2  \minus{} a^2 } \right)} }}}} \ge 3
\]
 :)

How did you get $ 2b^2\plus{}2c^2\minus{}a^2$ in the numerator?

[quote="Bugi"]For an acute triangle $ ABC$ prove the inequality:
$ \sum_{cyclic} \frac {m_a^2}{ \minus{} a^2 \plus{} b^2 \plus{} c^2}\ge \frac {9}{4}$ where $ m_a,m_b,m_c$ are lengths of corresponding medians.[/quote]
$ (m_a)^2 \equal{} (\frac {a}{2})^2 \plus{} c^2 \minus{} ac*cos(B)$ By the cosine rule.
Also by the cosine rule: $ cos(B) \equal{} \frac {a^2 \plus{} c^2 \minus{} b^2}{2ac}$ so we get $ (m_a)^2 \equal{} \frac {2b^2 \plus{} 2c^2 \minus{} a^2}{4}$ Thus we get:
$ \sum_{cyc} \frac {2b^2 \plus{} 2c^2 \minus{} a^2}{b^2 \plus{} c^2 \minus{} a^2} \ge 9$.
Since $ \triangle ABC$ is acute we know that $ cos(C) > 0$ And thus using $ a^2 \plus{} b^2 \minus{} 2ab.cos(C) \equal{} c^2$ But $ a^2 \plus{} b^2 \minus{} 2ab.cos(C) > a^2 \plus{} b^2$ and thus $ c^2 > a^2 \plus{} b^2$. So $ a^2,b^2,c^2$ are sides in a triangle. Thus we set $ a^2 \equal{} x \plus{} y$, $ b^2 \equal{} y \plus{} z$, $ c^2 \equal{} z \plus{} x$ with $ x,y,z > 0$. So we just have:
$ \sum_{cyc} \frac {4z \plus{} x \plus{} y}{2y} \ge 9$, which follows directly by $ \sum_{cyc} \frac {4z}{2y} \ge 6$, $ \sum_{cyc} \frac {x}{2y} \ge \frac {3}{2}$, $ \sum_{cyc} \frac {y}{2y} \equal{} \frac {3}{2}$. (By AM-GM)

[quote="hsiljak"]It's a well known fact:

$ m_a^2 \equal{} \frac {2b^2 \plus{} 2c^2 \minus{} a^2}{4}$

and without it, most of the inequalities in which medians figure aren't solvable.

See here, for instance (equation(1)): http://mathworld.wolfram.com/TriangleMedian.html[/quote]

The equality is true by Stewart therorem

[quote="Honey_S"]
$ \Leftrightarrow \sum {\frac {{b^2 \plus{} c^2 }}{{b^2 \plus{} c^2 \minus{} a^2 }}} \ge 6$
:)[/quote]
Here is how I proved this. Let $ x \equal{} a^2$, $ y \equal{} b^2$, $ z \equal{} c^2$
So we have to show $ \sum \frac {x \plus{} y}{x \plus{} y \minus{} z} \ge 6$. This fraction is $ > 0$, since $ x \plus{} y > z$. This is equivalent to
$ \sum \frac {1}{1 \minus{} \frac {z}{x \plus{} y}} \ge 6$. By Cauchy, $ \sum \frac {1}{1 \minus{} \frac {z}{x \plus{} y}} \ge \frac {9}{3 \minus{} \sum \frac {z}{x \plus{} y}}$. But by Nesbitt's inequality, we know $ \sum \frac {z}{x \plus{} y} \ge \frac {3}{2}$, which completes the proof.

By the way, how do you prove Honey_S's last inequality, that $ xyz \ge (x \plus{} y \minus{} z)(x \minus{} y \plus{} z)( \minus{} x \plus{} y \plus{} z)$?
EDIT: Now I remember. This is either Schur, or you can use that
$ (x\plus{}(y\minus{}z))(x\minus{}(y\minus{}z))\equal{}x^2\minus{}(y\minus{}z)^2 \le x^2$, and multiply it cyclically, which proves a case of Schur.

...or you can substitute $ x+y-z=a;y+z-x=b;z+x-y=c $ with $ x+y+z=a+b+c $ and your last inequality ([b]dgreenb801[/b]) is now equivalent with $ (a+b)(b+c)(c+a) \ge 8abc $, which is true by $AM-GM$. Note that $ a,b,c $ are positive since $ x+y>z, y+z>x, z+x>y $ by triangle inequality, so you are allowed to apply the $AM-GM$ inequality.

[quote=mathmatecS][quote=xzlbq]In any triangle,prove that:

${\frac {{\it ha}}{s-a}}+{\frac {{\it hb}}{s-b}}+{\frac {{\it hc}}{s-c}
}>5.
$[/quote]

s means a+b+c/2 , right?[/quote]

Yes.
BQ

By Stewart's Theorem, we have $m_a=\frac{b^2+c^2-\frac{1}{2}a^2}{2}$, so we have
\begin{align*}
\sum_{cyc}\frac{m_a^2}{-a^2+b^2+c^2}\geq \frac{9}{4} &\iff \sum_{cyc} \frac{b^2+c^2}{-a^2+b^2+c^2} \geq 6\\
&\iff \sum_{cyc} \frac{a^2}{b^2+c^2-a^2}\geq 3 \\
&\iff \sum_{cyc} \frac{b^2+c^2-2bc\cos A}{2bc \cos A}\geq 3 \\
&\iff \sum_{cyc} \frac{b^2+c^2}{2bc \cos A}\geq 6 \\
&\iff \sum_{cyc} \sec A \geq 6
\end{align*}
where we used AM-GM in the second last step to show that $b^2+c^2\geq 2bc$. Now, notice that since triangle $ABC$ is acute, we have $f(x)=\sec x$ is convex on the interval $(0,\frac{\pi}{2})$ as $f''(x)=\sec x(\sec^2 x + \tan^2 x) > 0$. So, by Jensen's Inequality, we have $\sum_{cyc} \sec A\geq 3\cdot \sec {\frac{\pi}{3}}= \boxed{6}$ .

[quote=MATH1945][quote=Honey_S]
which is always true since
\[ \sum {\frac{{a^2 }}{{b^2  \plus{} c^2  \minus{} a^2 }}}  \ge 3\sqrt[3]{{\frac{{a^2 b^2 c^2 }}{{\prod {\left( {b^2  \plus{} c^2  \minus{} a^2 } \right)} }}}} \ge 3
\]
 :)[/quote]

why it's true?[/quote]
It's AM-GM and Schur


function
real analysis
real analysis unsolved

Looks familiar.  :)  Let $f(x)=\frac{x}{\sqrt{1+\parallel x\parallel ^{2}}}$. This map sends $\mathbb{R}^{n}$ onto the open unit ball. Since $\parallel x\parallel \le c$ implies $\parallel f(x)\parallel \le c/\sqrt{1+c^{2}}<1$, the statement that you want to prove is equivalent to $f^{-1}$ being Lipschitz on any ball of radius less than $1$ centered at the origin. A little algebra yields $f^{-1}(y)=\frac{y}{\sqrt{1-\parallel y\parallel ^{2}}}$. You can prove that $f^{-1}$ is Lipschitz either by observing that it is continuously differentiable, or by estimating $\parallel f^{-1}(y_{1})-f^{-1}(y_{2})\parallel $. The latter approach works better if you first consider two special cases: (1) $\parallel y_{1}\parallel =\parallel y_{2}\parallel $ and (2) $y_{2}$ is a positive multiple of $y_{1}$.

I did something similar to your first approach. but I was thinking if the following way is gonna work:
let $g(x_{1},x_{2})= \frac{\|x_{1}-x_{2}\|}{ \| \frac{x_{1}}{\sqrt{1+\|x_{1}\|^{2}}}-\frac{x_{2}}{\sqrt{1+\|x_{2}\|^{2}}}\| }$

If we can prove it is a continuous function of $[x_{1},x_{2}]$ on a compact set, then we are done. it is continous when $x_{1}\neq x_{2}$. but unfortunately, it is not continuous on point  satisfying $x_{1}=x_{2}$ (even we define a value on these points). Is there any way to fix this problem?


function
trigonometry
advanced fields
advanced fields unsolved

Can you give an examble of an analytic function f:B(0,1)->[b]C[/b] that is not constant but has infinitely many zeros (f(z) = 0 infinitely many z)?

$\sin{\frac1{z-1}}$.

You do have to allow for an essential singularity on the boundary, but it's easy to build an example.


friends give me the link of question regarding periodicity of function .how to calculate periodicity . and question related to period

If you ask more politely, people will be more likely to actually respond.  See [url=http://en.wikipedia.org/wiki/Periodic_function]here[/url].


function
limit
inequalities
logarithms
number theory solved
number theory

i dont know whether this has been solved or not, anyway i havent seen it anywhere:

is the limit $\lim_{n \to \infty} {\varphi(1) + \varphi(2) + ... + \varphi(n) \over n}$ finite or not?

It is clear that limit is $+\infty$. On the other hand
\[\lim_{n \to \infty} \frac{\varphi(1) + \varphi(2) + ... + \varphi(n)}{n^2}=\frac{3}{\pi^2}.\]

In fact this is a quite famous result related to visible points. Proofs are not difficult at all. I think some proofs you can find in the proposed problems analysis, in one of my posts. There Myth and Moubinool gave references and proofs.

[quote="yrch"]please be so kind and show me proofs of both of these statemens. the result given by you is very curious i must say.[/quote]
It is easy to show that your limit is $\infty$.
Indeed, for all odd $m$ the following inequality holds: $\varphi(m)\geq \sqrt{m}$, thus
\[\varphi(1)+\varphi(2)+...+\varphi(n)\geq c\cdot n\sqrt{n},\]
where $c$ is a fixed constant.

right. i havent thought about the square root, which is quite obvious considering the product formula for $\varphi(n)$. in fact i reckoned it was growing slower, therefore i tried proving that $\sum \varphi(n) > n \ln n$, with no succes, though.

unfortunately i will have to wait some time to check this ${3 \over \pi^2}$ proof as mathworld for some unknown reasons banned access to their site from my proxy :)

thanks anyway, though.


Solve

$ x^{2} \minus{}xy\plus{}y^{2}\equal{}19$

$ x\minus{}xy\plus{}y\equal{}7$.

hello, obviously is $ x\equal{}1$ not a solution of equation $ 2$, so we solve equation $ 2$ for $ y$ and we get
$ y\equal{}\frac{x\minus{}7}{x\minus{}1}$
plugging this in the first equation simplfying and factoring we get
$ \frac{(x\minus{}3)(x\minus{}2)(x^2\minus{}2x\minus{}5)}{(x\minus{}1)^2}\equal{}0$
from here we get all possible solutions of our system.
Sonnhard.

Another way is to use a substitution.

Let $ x\plus{}y\equal{}a$ and $ xy\equal{}b$.

From there we obtain these equations:

$ a\minus{}b\equal{}7$

$ a^2 \minus{}3b \equal{}19$

From there solutions for $ a$ are $ 1$ and $ 2$.

For $ a\equal{}2$ we get that $ b\equal{}\minus{}5$

Plugging that into original equations we get:

$ x\plus{}y\equal{}2$ and $ xy\equal{}\minus{}5$, which is easy to solve.

Same work for $ a\equal{}1$.


a,b,c > 0
prove that :
$ \frac {a^2}{b \plus{} c} \plus{} \frac {b^2}{a \plus{} c} \plus{} \frac {c^2}{a \plus{} b}\geq \frac {3}{2}\sqrt [5]{\frac {a^5 \plus{} b^5 \plus{} c^5}{3}}$

[hide]any idea for best constant k for this one ?[/hide]

i'm sorry , i did not see the post by tuandokim . Mod please delete this one


geometry
circumcircle
geometric transformation
dilation
parallelogram
Gauss
perpendicular bisector

I send a problem from Bonsia - Hervegovina math Olympiad 2005.
[u] Babis[/u]

[hide]
Let $M$ be the midpoint of $\overline{AB}$, $N$ be the midpoint of $\overline{CH}$, and $O$ be the circumcenter of $\triangle ABC$. We first prove that $MNCO$ is a parallelogram. Sides $\overline{NC}$ and $\overline{OM}$ are parallel since both are perpendicular to $\overline{AB}$, and they are congruent
[hide="because"]the dilation centered at the centroid of $\triangle ABC$ and with factor -1/2 sends $\triangle ABC$ to its medial triangle, which has all of its altitudes pass through $O$, therefore the dilation takes $H$ to $O$ and
\[OM=\frac{1}{2}HC=NC\]
[/hide]
The parallelogram gives $\overline{MN}\parallel \overline{OC}$. Let $X$ be the intersection of the angle bisectors of $\angle CAH$ and $\angle CBH$. Showing that $N,X,M$ collinear is now reduced to proving $\overline{MX}\parallel \overline{OC}$. A bit of angle chasing reveals
\[\angle MAX=135^\circ-B-\frac{C}{2},\ \angle MBX=135^\circ-A-\frac{C}{2},\ \angle AXB=90^\circ\]
 :o MIDPOINT OF A HYPOTENUSE!
Now that we know that $M$ is the circumcenter of $\triangle AXB$, we get isosceles triangles. This gives
\[\angle MXA=135^\circ-B-\frac{C}{2},\ \angle AMX=180^\circ-(270^\circ-2B-C)\]
\[=90^\circ+B-A\]
But extending $\overline{OC}$ to intersect $\overline{AB}$ at $P$, we find $\angle PCB=90^\circ-A$ and of course $\angle PBC=B$, so $\angle APC$ is also equal to $90^\circ+B-A$, making $\overline{MX}$ and $\overline{OC}$ parallel. Therefore $N,X,M$ collinear, as desired.
[/hide]

i feel like there should be a solution involving the 9-point circle...

[hide="Not 9 pt circle, but short"]Let M and N be the midpts of AB and CH, respectively. Consider the circle O having diameter AB; it also goes through the feet of altitudes from A and B to BC and AC, which we call D and E. It then follows that the intersection of the bisectors of CAH and CBH meet at the midpoint of the minor arc DE. Furthermore, ND = NE and MD = ME, so N, M, and the intersection of the bisectors are collinear.[/hide]

Here is a very short proof I found, using the nine-point circle.

Look at the picture for names of points and lines.

[b][u]Proof:[/u][/b]
1) The circle with center $E$ and going through $A,G,F,B$ also contains the intersection of the two bisectors, and it lies on the perpendicular bisector of $\left[GF\right]$ (call it $r$).

2) On this perpendicular bisector lies also point $E$, that's clear, and also point $O$, the center of the nine-point circle. Therefore, $r$ is the diameter of the nine-point circle, through $E$.

3) $D$ lies also on this diameter, because $\angle DKE = \frac{\pi}{2}$.

Done!  :P

Notes: [list]
[*] In $1)$, we use a well known lemma. The bisector of $\angle ABC$ intersects the perpendicular bisector of $AC$ on the circumcircle of $ABC$.
[*] In $2)$, we use the fact that $GF$ is a common chord of the two circles.[/list]

This kind of problems just drives me nuts.
It's easy but so extraordinary beautiful. I can really be deeply touched by geometry involving the nine-point circle...

Well, the problem is quite trivial:
Let $AD, BE$ be altitudes of $\Delta ABC$; then $HDCE$ is cyclic, the midpoints of the diagonals $CH, DE, AB$ of the complete quadrilateral $HDCEAB$ are collinear ([b]Newton-Gauss[/b] line) and, at the same time, the bisectors of the angles $\angle CAD, \angle CBE$ are perpendicular and intersect on the above Newton-Gauss line.

Best regards,
sunken rock

The bisectors of angles CAH and CBH meet in P  we have $APB=90$
We know $PM$ is the median of the triangle $APB$
Assume that $HAB=2b$,$HAC=2a$ so we have $AMP=180-2a-4b$
N is the midpoint of $CH$ thus MN is Perpendicular to $EF$(AE,BF are the altitudes of triangle ABC)
thus $AMN=180-2a-4b$ so $AMN=AMP$ so $M,P,N$ are collinear.

Dear S. F. sunken rock and Mathlinkers,
how do you prove
"the bisectors of the angles intersect on the above Newton-Gauss line".

Sincerely
Jean-Louis

Dear Jean-Louis,
Pls see post #5 here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=508765

Best regards,
sunken rock


[quote="DaReaper"]How do you find x in $\sqrt{6+\sqrt{6+\sqrt{6+...=x}}}$

   :([/quote]

do you mean?

$\sqrt{6+\sqrt{6+\sqrt{6+...}}}=x$

yes that is it...

$\sqrt{6+\sqrt{6+\sqrt{6+...}}}=x$

Of course, it is not at all obvious that $x$ is even finite.

Edit:  Consider an analogous problem, finding $x$ such that

$x = 2^{2^{2^{...}}}$

It is not just the solution to $x = 2^{x}$ because it's pretty obvious that $x$ is not finite.

[quote="DaReaper"]but how would you get 3? Because if you plug $x^{2}-x-6=0$ into the quadratic equation, then you would get $3$ or $-2$.

also, how does $x^{2}-x-6=0$ become $(x-3)(x+2)=0$?

:huh:[/quote]

Use FOIL when you multiply the two. So in $(x-3)(x+2)=0$ you first multiply the $x$ in $(x-3)$ with the x in $(x+2)$ to get $x^{2}$. Then you multiply the $x$ in $(x-3)$ with the 2 in $(x+2)$ to get $2x$. So you do the same with the $-3$ in $(x-3)$ to get $x^{2}+2x-3x-6=x^{2}-x-6=0$

whoops :blush: i didnt notice that....

haha thnx... :lol:

[quote="DaReaper"]but how would you get 3? Because if you plug $x^{2}-x-6=0$ into the quadratic equation, then you would get $3$ or $-2$. [/quote]

Generally the square root function gives the positive root.  If you treat all those radical signs as the positive root, the answer is clearly positive, so it is $3$.

If you treated them all as negative roots instead, that's how you would get the answer of $-2$.


Did anyone take physicsbowl? When does that stuff come out?

it comes from AAPT, same place taht holds U.S. Physics Team tests. (or some people call it the USAPhO)

the information that accompanied the exams said that the answers (not necessarily the results) would be posted April 29th. That was 2 days ago, but I haven't seen any sign of them on aapt.org

Last year I was a top finisher in the Physics Bowl, as was my school George Wasington, sorry for being to modest.

what is this? its just 40 questions on a scantron. i'm getting anxious, because i was a uspho semifinalist last year and not this year. if i don't do well in physics bowl then i'm done with physics.


Given five segments such that any three of them can be used to form a triangle. Show that at least one of these triangles is acute-angled.

[i]Alternative formulation:[/i] Five segments have lengths such that any three of them can be sides of a triangle. Prove that there exists at least one acute-angled triangle among these triangles.

[quote="YooSam"]Given five segments such that any three of them can be used to form a triangle. Show that at least one of these triangles is acute-angled.[/quote]

This has surely been posted here, but searching for the topic would take me longer than posting a new solution, so I'm doing the latter:

Denote our five segments by a, b, c, d, e such that $ a\geq b\geq c\geq d\geq e > 0$. Assume, for the sake of contradiction, that any three of these five segments can be used to form a triangle, but none of these triangles is acute-angled. Then, we must have $ a^2\geq b^2 \plus{} c^2$ (in fact, if we had $ a^2 < b^2 \plus{} c^2$, then, together with the relations $ b^2 < c^2 \plus{} a^2$ (which follows from $ b\leq a$ and 0 < c) and $ c^2 < a^2 \plus{} b^2$ (which follows from $ c\leq a$ and 0 < b), this would yield that the triangle with sidelengths a, b, c is acute-angled, but we assumed that no such triangle is acute-angled). Similarly, we must have $ b^2\geq d^2 \plus{} e^2$ (considering the triangle with sidelengths b, d, e) and $ c^2\geq d^2 \plus{} e^2$ (considering the triangle with sidelengths c, d, e). Thus,

$ a^2\geq b^2 \plus{} c^2\geq\left(d^2 \plus{} e^2\right) \plus{} \left(d^2 \plus{} e^2\right) \equal{} \left(d \plus{} e\right)^2 \plus{} \left(d \minus{} e\right)^2\geq\left(d \plus{} e\right)^2$,

so that, after taking the square root, $ a\geq d \plus{} e$. But since a, d, e are the sidelengths of a triangle, we have a < d + e. Thus, we obtain a contradiction. Hence, our assumption was wrong, so one of our triangles must indeed be acute-angled.

  Darij

let the lengths of the 5 segments be $ a \leq b \leq c \leq d \leq e$. note that a triangle of sides $ x \leq y \leq z$ is acute iff $ x^2 \plus{} y^2 > z^2$. if all triangles were non-acute, then we'd have

$ d^2 \geq b^2 \plus{} a^2 \geq 2a^2$; so $ d \geq \sqrt {2}a$;

$ d^2 \geq c^2 \plus{} b^2 \geq 2b^2$; so $ d \geq \sqrt {2}b$;

and then $ e^2 \geq d^2 \plus{} c^2 \geq d^2 \plus{} b^2 \plus{} a^2 \geq 2ab \plus{} b^2 \plus{} a^2 \equal{} (a \plus{} b)^2$; so $ e \geq a \plus{} b$, a contradiction.

Yes, that is the natural way to solve the problem :)


search
number theory unsolved
number theory

let $ p$ a prime number then prove that :

$ q / x^{p_1} \plus{} x^{p_2} \plus{} .... \plus{} p\plus{}1 \implies q\equal{}1[p]$

thanks

At least you could reread your posts and change obvious errors. The correct term is:
$ q | x^{p \minus{} 1} \plus{} x^{p \minus{} 2} \plus{} .... \plus{} x \plus{} 1 \implies q \equal{} 1[p]$
And $ p \nmid q$ is missing.
And it is a double post...
Next time, it will be simple locked, without any comment.

Now take a search for cyclotomic polynomials. I've posted enough interesting facts  :wink:


modular arithmetic
monovariant
combinatorics proposed
combinatorics

Given an arbitrary set of 2k+1 integers a_1, a_2, ... , a_{2k+1}.
We make a new set:
(a_1 + a_2)/2, (a_2 + a_3)/2,........,(a_{2k} + a_{2k+1})/2,  (a_{2k+1} + a_1)/2

and a new one, according to the same rule, and so on...
Prove that if we obtain integers only, the initial set consisted of
equal integers only.

At all times, all the numbers must have the same parity. This means that $\frac{a_1+a_2}2,\frac{a_2+a_3}2$ have the same parity, meaning that $a_{n+2},a_n$ have the same residue $\pmod 4$. This means that all of the $2k+1$ numbers must have the same residue $\pmod 4$ at all times (this is where we use the fatc that we have an odd number of numbers: $a_1,a_3,\ldots,a_{2k+1},a_2,a_4,\ldots$ must all have the same residue $\pmod 4$). $\frac{a_1+a_2}2,\frac{a_2+a_3}2$ have the same residue $\pmod 4$, so $a_{n+2},a_n$ must have the same residue $\pmod 8$ (at all times, i.e. in each step), and so on. We find that $2^k|a_{n+2}-a_n\ \forall k\ge 1,\forall n$, meaning that all the numbers are equal.

the easiest solution is a monovariant, the maximun element gets smaller in each step, so it cannot be always an integer. loock we dont need n odd.


Hello!

$x,y,z$ are positive real numbers such that : ${1\over x}+{1\over y}+{1\over z}={1\over{xyz}}$. Prove the inequality :
\[{{x}\over{x^{2}+1}}+{{y}\over{y^{2}+1}}+{{z}\over{z^{2}+1}}\leq{{3\sqrt{3}}\over{4}}.\]

Looks familiar...

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}= \frac{1}{xyz}$ can be rewritten as $xy+yz+zx=1$, thus we have \[\frac{x}{x^{2}+1}=\frac{x}{x^{2}+xy+yz+zx}=\frac{x}{(x+y)(x+z)}.\]

Now, the inequality we have to prove can be rewritten as
\[\frac{x}{(x+y)(z+x)}+\frac{y}{(y+z)(x+y)}+\frac{z}{(z+x)(y+z)}\leq \frac{3\sqrt 3}{4}\]
or, equivalently
\[4\left( x(y+z)+y(z+x)+z(x+y)\right) = 8\leq 3\sqrt 3 (x+y)(y+z)(z+x).\]
Now, using the well known inequalities $9(x+y)(y+z)(z+x)\geq 8(xy+yz+zx)(x+y+z)$ and $(x+y+z)^{2}\geq 3(xy+yz+zx)$ (the latter implies $x+y+z\geq \sqrt 3$), we obtain
\[3\sqrt 3 (x+y)(y+z)(z+x) = \frac{9(x+y)(y+z)(z+x)}{\sqrt 3}\geq \frac{8(xy+yz+zx)(x+y+z)}{\sqrt 3}\geq \frac{8\cdot 1 \cdot \sqrt{3}}{\sqrt 3}= 8\]
which proves our inequality.
$\Box$

[quote="N.T.TUAN"]we have $xy+yz+zx=1$ therefore exist triangle $ABC$ such that $x=\tan{\frac{A}{2}}$, etc.[/quote]

Dear Tuan,

please take the time to think deeply to the problem and to write down all your calculations. Thanks !!!!

This is what I want speak  :blush:
[quote="Sung-yoon Kim"]tuan meant you can substitute $x=\tan{\frac{A}{2}}$. actually, there is a familiar expression $\sum_{cyc}\tan{\frac{A}{2}}\tan{\frac{B}{2}}=1$. so taking $A=2\arctan{x},B=2\arctan{y}$, you can deduce that $z=\tan{\frac{C}{2}}$.

then $\sum_{cyc}\frac{x}{x^{2}+1}=\sum_{cyc}\frac{\tan{\frac{A}{2}}}{\tan{\frac{A}{2}}^{2}+1}$
$=\sum_{cyc}\sin{\frac{A}{2}}\cos{\frac{A}{2}}$
$=\frac{1}{2}\sum_{cyc}\sin{A}$
$\leq \frac{3\sqrt{3}}{4}$ by Jensen's.[/quote]


algebra polynomial function induction integration

I do not know if it is the correct section to post mathematical physics questions...If it is not then please PM/Email me to notify me where to post...Because,the thread will be closed otherwise.

I am to prove that P_n(-x)=(-1)^n*P_n(x)

And, P'_n(-x)=(-1)^(n+1)*P'_n(x)

I know that whether a Legendre Polynomial is an even or odd function depends on its degree.It follows directly from the solution of Legendre differential equation.But,to prove these properties I am getting stuck.

Can anyone help me to start with these?

Legendre polynomial $P_{n}(x)$ is orthogonal on [-1, +1] to any polynomial of degree less than n.

Show that Legendre polynomials $P_{2k}(x)$ contain only even powers of x and $P_{2k+1}(x)$ only odd powers, by induction over k. For k = 0, $P_{0}(x) = 1,\ P_{1}(x) = x,$ the proposition holds. Assume it is true up to some k and let the highest odd power of $P_{2(k+1)}(x)$ be $x^{2m+1}.$ Then $P_{2(k+1)}(x) = E_{2(k+2)}(x)+\sum_{0}^{m}a_{2l+1}P_{2l+1}(x),$ where $E_{2(k+2)}(x)$ is a polynomial with even powers of x only and $a_{2m+1}\neq 0.$. $P_{2m+1}(x)$ is orthogonal to $P_{2(k+1)}(x)$ and to $P_{2l+1}(x),\ l = 0,\ 1,\ ...,\ m-1.$

$0 = \int_{-1}^{+1}P_{2(k+1)}(x)P_{2m+1}(x)\ \text dx =$ $\int_{-1}^{+1}E_{2(k+1)}(x)P_{2m+1}(x)\ \text dx+\sum_{0}^{m}a_{2l+1}\int_{-1}^{+1}P_{2l+1}(x)P_{2m+1}(x)\ \text dx =$

$= \int_{-1}^{+1}O_{2p+1}(x)\ \text dx+a_{2m+1}\int_{-1}^{+1}P_{2m+1}^{2}(x)\ \text dx = 0+a_{2m+1}N_{2m+1}^{2}$

where $p = k+m+1,$ $O_{2p+1}(x)$ is a polynomial with odd powers of x only, and $N_{2m+1}^{2}\neq 0.$ Thus $a_{2m+1}= 0,$ contradiction. Similarly, $P_{2k+3}(x)$ does not contain any even powers of x.


What is the area of the pentagon (attached) with sides of length 15, 20, 27, 24. and 20 units?

Answer & Explain, plz

[hide="hint"]
Draw the diagonal connecting the two vertices that are not right angles.
[/hide]


I was just wondering : is Mathlinks overwhelmingly male?
Whenever I check on a persons gender, I see male

there are exceptions of course, but I would like to see (does anyone it that is possible) statistics of how many people are male/female/haven't filled it in on profile?

I heard that at IMO male:female ratio is 10:1, I suppose the same will hold for MathLinks or anything in the line of maths. But make a poll, it will be interesting :)

So I did it

But I did not really want to at first : mathlinks is not representative for young math online community, not completely

more over : not all fervent posters read this poll

last : not everyone participates

To be blunt, if only 100 people participate it isn't enough to draw conclusion I think

I doubt you're going to get anywhere near 100 participants, most polls get very few. The Creationism/ID has 88 so far because it has been around for so long. Those ID supporters are just too stubborn. :D

The poll about musical instruments has more than 100 votes :D

I really wanted to answer YES: I am male or female.

jmadsen, I think everyone's stubborn ;)

Maybe everyone plays musical instruments, or something. Where is it?

mathnerd314, you have exactly the same humor as my father :D

There have been 43 votes cast already and it has only been on for a couple of days.  I think you will get plenty of votes because this poll is easy.

[quote="ToBe"][quote="mathnerd314"]

Maybe everyone plays musical instruments, or something. Where is it?[/quote]

And not everyone is either male or female?  :?  lol[/quote]

Actually, some people aren't... don't make fun of me - er, I mean, them...

perhaps i should have included an option saying you don't belong to a species having genders
but the first animal on mathlinks still is to be reported ;)

Yes, however the vote will not be representative still for the problem solving community
however, the number of votes is nice :D

All I know is that I edited my first post, and scrolled down where I filled in the previously empty poll data
strange?  :?:

I think you can insert one as long as you haven't inserted it, but you cannot change or undo it later unless you are a mod.

[quote="fredbel6"]perhaps i should have included an option saying you don't belong to a species having genders
but the first animal on mathlinks still is to be reported ;) [/quote]

Hm, perhaps the original poster was referring to humans who either don't consider themselves either male or female (transgendered) or who have indeterminate or mixed physical gender characteristics (intersexed).  I'm not sure what percentage of the population is transgendered, but the floor on intersexed individuals is .03% (i.e. 1 of every 3000 births).

Perhaps this is slightly off topic, but did you know that mushrooms have over 36,000 sexes?

As far as I know, mushrooms only have 1 sex... :?

[quote="mathnerd314"]Forty-two people have voted "male"!!! 42!!!!!!! :w00t: :o (as of 8:46 PM, November 24)

Seriously, folks...

I'm confused: fredbel6 asks the question and then nonie suggests a poll? The last time I checked, I couldn't put a poll in a reply or edit a poll into the first post. This goes back to the Test Forum. What's going on?[/quote]

Well apparently you can! :)  I think you can change the options too as long as no one has voted.

It appears of the 90 people who have submitted an answer, a great majority are male.  Maybe, more males are interested in participating in poles about their sex...

[quote="pianoforte"]maybe he meant species? :?[/quote]

No, I actually DID mean 36000 sexes...

I read an article about the largest organism in existence, this one fungus which covers about 2200 acres of land (mostly underground) and comes above the ground as mushrooms.  In that article, it said that mushrooms have over 36000 sexes, if I remember correctly...but I can't find the article...

[quote="trisana"]Females are overwhelmingly the minority. It's like 3:1 on the poll majigger. I think I know why. Most females (not me, I'm proud to say), have lives. Having a life and doing math for fun are mutually exclusive pastimes.[/quote]I really must disagree. I know plenty of people who have a life and do math for fun! And math [i]is[/i] fun, so it can't be that different from having a life.  :D


A mountain climber in the Himalayas took along with him two mountain guides. After a few hours, one of the guides fell into a deep crevasse. The climber and the other guide continued the climb and did not raise the alarm. Why?

The first guide was a map, not a person.


modular arithmetic
induction
number theory

Show that $ 7^n\minus{}1$ is divisible by 6, for all positive integers n.

NOTE: Ooops! I meant to put this in the high school basics forum! Can a mod please move it there?

[hide="Solution"]We know that
\[ 7\equiv 1\pmod{6}\]
Since $ n$ is a positive integer
\[ 7^n\equiv 1^n\equiv 1\pmod{6}\]
Therefore
\[ 7^n \minus{} 1\equiv 0\pmod{6}\]
which proves that $ 7^n \minus{} 1$ is divisible by $ 6$ for all positive integers $ n$.[/hide]

[hide="A different solution"]

$ 7^{n}\minus{}1$ can be factored as $ (7\minus{}1)(7^{n\minus{}1}\plus{}7^{n\minus{}2}\plus{}7^{n\minus{}3}...\plus{}7\plus{}1)$. Because $ 7\minus{}1\equal{}6$, it is automatically divisible by $ 6$.[/hide]

[hide="Yet Another Solution"]
We have $ 7^n \minus{} 1 \equal{} (6 \plus{} 1)^n \minus{} 1 \equal{} \sum_{k \equal{} 0}^{n}{\binom{n}{k}6^k} \minus{} 1 \equal{} \sum_{k \equal{} 1}^{n}{\binom{n}{k}6^k} \plus{} 1 \minus{} 1 \equal{} \sum_{k \equal{} 1}^{n}{\binom{n}{k}6^k}$. For $ 1 \le k \le n$, $ 6 | 6^k$ so $ 6 | (7^n \minus{} 1)$.
[/hide]

[hide=" Yet another solution"]

We could also use mathematical induction, although it may not be appropriate in this forum:

Basis step:

$ 7^{1}\minus{}1$ is divisible by $ 6$, so true for $ n\equal{}1$.

Inductive step:

If $ 7^{n} \minus{} 1$ is divisible by $ 6$, then $ 7^{n\plus{}1} \minus{} 1 \equal{} 7\times 7^{n} \minus{} 7 \plus{} 6 \equal{} 7(7^{n} \minus{} 1) \plus{}6$ is divisible by $ 6$.

[/hide]

[hide="Another induction"]Another way of completing the induction proof:

First of all, we know that $ 7^1\minus{}1$ is divisible by 6 so base case done.

Assume that $ 7^n \minus{} 1$ is divisible by 6.

Let $ f(n) \equal{} 7^n \minus{} 1$

Then $ f(n \plus{} 1) \equal{} 7(7^n) \minus{} 1$

So $ f(n \plus{} 1) \minus{} f(n) \equal{} 6(7^n)$ is divisible by 6.

Since their difference is divisible by 6 then they leave the same remainder upon division by 6, and because $ f(n)$ is divisible by 6, so is $ f(n \plus{} 1)$.[/hide]

@KrazyFK Aren't you using your assumption that $ 6|7^{n} \minus{} 1$ to get the answer, yet you never prove it? By your method, they could both be 1,2,3,4, or 5 $ \pmod{6}$

Ihatepie: it's induction, I just forgot to write out my base case of $ n\equal{}1$.

[quote="IMO girl"]we wanna prove 6|7*-1   (*=n)
if n is odd,then we can write 6|(7-1)(7*-1. ... .1) 6|6x and it's true
if n is even,then we can write n=2k , 6|49*-1(*=k) and k is odd and we can writeit just like upper line[/quote]Hi, IMO girl. What if $ k$ is even?

From $ (m\plus{}n)^k \equal{} fm \plus{} n^k$ (f serves no purpose other than to denote fm is a multiple of m),

we get $ (6\plus{}1)^n\minus{}1 \equal{} 6m \plus{}1\minus{}1 \equal{} 6m$.

[quote="cryax"]a,b,c,x,y,z >0 prove: (ax+by+cz)^2+a^2+b^2+c^2>= 2(ab+bc+ac)
thank u very much![/quote]You've committed three mistakes (I take it, you are new here): #1 This is off-difficulty and does not belong in High School Basics. #2 When you ask a new problem, start a new topic rather than on someone else's topic. Just because you see the topic titled "Number Theory", it doesn't mean everyone is going to post number theoretic problems here (BTW, your problem isn't even number theoretic), which makes it #3.

And thank you very much for joining AoPS/ML! We all welcome you here :)

thank u, i am not really good at english but i'll post this on other topic.


Let $ x,y\in R , xy\equal{}1,x>y$.Prove that: $ x^2\plus{}y^2\ge 2\sqrt{2}(x\minus{}y)$

[hide="Solution"]
$ x^2 \plus{} y^2 \minus{} 2xy \equal{} (x\minus{}y)^2 \\
\\
since \ xy \equal{} 1, \\
\\
x^2 \plus{} y^2 \equal{} (x\minus{}y)^2 \plus{} 2 \ \geq \ 2\sqrt{2(x\minus{}y)^2} \ \ \ ( A.M \geq G.M ) \\
\\
So, \ x^2 \plus{} y^2 \geq 2 \sqrt{2} (x\minus{}y)$
[/hide]

A slight nag: you first need to note that the problem is equivalent to showing that $ x^2 \plus{} y^2 \geq 2 \sqrt {2} |y \minus{} x|$ for all real $ x, y$, and then you need to replace $ \sqrt {(x \minus{} y)^2}$ by $ |x \minus{} y|$, not $ x \minus{} y$.

But otherwise, good proof. I'm bad with inequalities. I was going to square both sides and Muirhead.

@ Zhero : i didn't write |x - y| in place of (x - y) as the Question mentions that x > y,
so obv. x - y > 0

P.S. I am bad at inequalities too... the only 1 i know being $ A.M \geq G.M$  :maybe:

Yeah, I'm stupid, my bad. I was too locked into my own train of thought and attempted to impress it on you.

It sometimes help to know less. Then you'll use AM-GM instead of Muirhead (Muirhead is the one that overkills every symmetric homogeneous inequality, if you're not aware. :P)

[b]LOL [/b]

na, i had no clue about Muirhead
good to know  :)


trigonometry Columbia function Euler limit logarithms

Poate cineva demonstra faptul ca multimea discontinuitatilor de speta 1 a unei functii f este cel mult numarabila?(in caz ca este adevarat)

In anul 1929 matematicianul român [b][i]Al. Froda[/i][/b] a obţinut următorul rezultat:

[color=red][i][b]Mulţimea punctelor de discontinuitate de prima speţă ale oricărei funcţii
$f: \mathbb{R} \to \overline{ \mathbb{R} }$ este cel mult numărabilă.[/b][/i][/color]

(Al. Froda, Sur la distribution des proprietes de voisinage des fonctions
de variables reelles, Bull. Math. De la Soc. Roumaine des Sciences, t. 32,
 fasc. II, 1929, pp. 105-202)

Demonstraţia poate fi găsită şi în:
[i]Analiză matematică, [b]Acad. Miron Niculescu[/b], Ed. Tehnică, Bucureşti 1958, pp. 133[/i]

Sa presupunem prin absurd ca nu am acces la acea carte.  E lunga demonstratia pt. a fi scrisa aici?

De asemenea, e interesant de citit capitolul "Controversă funcţională" din cartea prof. Al. Froda "Eroare şi paradox în matematică", Ed.Enciclopedică,1971. Autorul povesteşte faptul că Traian Lalescu a considerat rezultatul greşit :)

Se poate folosi rezultatul acesta la olimpiada de matematica(locala, judeteana, nationala)??

Ca intotdeauna, e greu de raspuns la aceasta intrebare. Depinde de problema. Daca, de exemplu, folosesti acest rezultat intr-o demonstratie mai ampla, in care acesta reprezinta doar un pas, probabil ca va fi acceptat (cu mentiunea ca nu se va da o problema care sa poata fi rezolvata doar folosind acest rezultat). Daca insa folosirea acestuia e, de fapt, rezolvarea problemei in sine, probabil ca nu.
Se poate intampla (de fapt, s-a si intamplat) ca la olimpiada, unii concurenti sa spuna ceva de tipul "aceasta problema e un caz particular al teoremei lui Potriveanu :), asa ca...dati-mi punctajul maxim pt. aceasta observatie". Raspunsul, de obicei, este: nu s-a cerut concurentilor sa identifice teorema, ci sa rezolve problema.
Se mai pot spune multe pe aceasta tema, dar poate ar fi nimerit un topic dedicat.

Demonstratia nu e deloc grea. De fapt, sunt vreo doua [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31694]aici[/url] (si una pntru o afirmatie un pic mai tare).

Până la data susţinerii tezei de doctorat (1929) se ştia doar că
[i][color=red][b]Dacă o funcţie reală de o variabilă reală nu admite puncte de discontinuitate de a doua specie,
atunci punctele ei de discontinuitate formează o mulţime cel mult numărabilă.[/b][/color][/i]

[quote="metru"]... matematicianul [b] român[/b] [i]Al. Froda[/i]  a obţinut... [/quote]
[b]Nedumeriri:[/b] Cand un matematician poate sa fie considerat [b] matematician  român  [/b] ?
 Ce conditii,criterii, trebuie sa indeplineasca ?
 Ramane la hotararea noastra , a fiecaruia , sa decidem acest lucru ?   [b] [u]Astept opinii [/u]! [/b]
De exemplu, care dintre urmatorii matematicieni sunt[color=red]  [b] români [/b]  si pentru care motiv(e)[/color] :
[b]Farkas BOLYAI (nascut in 1775 intr-un sat de langa Sibiu),  Janos BOLYAI (nascut in 1802 la Targu-Mures),
Gyula VALYI (1855-Targu Mures, a profesat (Fizica) la Universitatea din Cluj),
 Tiberiu POPOVICIU (nascut in 1906 la Arad),  Isaac  Jacob SCHOENBERG (nascut in 1903 la Galati),
Caius IACOB, Grigore C.  MOISIL
 Lipot  FEJER (nascut in 1880 la Pecs, a fost cadru didactic la Universitatea din Cluj)
Hans HAHN ( 1879,Viena , a profesat la Universitatea din Cluj)
Leopold Bernhard GEGENBAUER (1849-Austria, a profesat la Universitatea din Cernauti)
Emil GROSSWALD (1912-Bucuresti,a profesat la University of Pennsylvania)
 Valentin POENARU, Ciprian FOIAS,
 Alexandru FRODA ,
 Abraham WALD (1902-Cluj, Columbia University, Carnegie Corporation),  ... [/b]
Probabil problema este ca profesiunea de matematician trebuie considerata internationala: de exemplu , dupa opinia mea:eventual
 [i] "Alexandru Froda =fost profesor la Universitatea din Bucuresti"[/i]

Oare i-ar placea lui Ciprian FOIAS sa fie considerat [b] matematician roman  ?[/b]  Am auzit ca dupa 1990 nu ar fi fost in tara (Are cineva informatii suplimentare ? Daca asta este realitatea, oare care este motivul ? ) !

In zilele noastre, cand s-a revenit la democratie, [b]nu[/b] cred ca ar trebui sa vorbim, de exemplu  despre [b]matematicieni tigani[/b], sau  (de ce nu ?) despre [b]politicieni tigani[/b] (credeti ca exista sau au fost ?) ; Eu cred ca NU !

Totusi a se observa faptul caracterul  site-ului (de altfel frumos si interesant)
http://www-groups.dcs.st-and.ac.uk/~history/BiogIndex.html

 Majoritatea savantilor inclusi sunt evrei, adesea [b] punand mare accent pe termenul de "Jewish" (=evreu).[/b]
Subliniez  ca [b]nu este vorba de o atitudine anti-semita,[/b] din contra doresc sa atentionez asupra faptului ca sunt site-uri unde se face o "propaganda" poporului evreu (uneori  in exclusivitate, probabil nemeritata din punct de vedere stiintific/al Matematicii ).
Merita de citit si documentatia de pe site-ul
http://www.jinfo.org/Fields_Mathematics.html
care ar putea sa sugereze o [i]modalitate de propaganda a matematicienilor din tara noastra ! [/i]

P.S. Cred ca este important ca tinerii sa [b]nu creada intotdeauna ceea ce citesc (mai ales pe Internet) ! [/b]
Asta este situatia si in Matematica: probabil voi reveni cu unele informatii suplimentare privind asa-numitele rezultate
ale unor "matematicieni romani" (chestiuni de originalitate...)
[ [size=150][b].. Continuare... si justificare ale unor afirmatii: [/b][/size] ]
O carte frumoasa este
Autor= [b] Ernest William  HOBSON (1856--1933)[/b]
Titlu= [color=green]  [b] The Theory of Functions of a Real Variable and the Theory of Fourier's series, Volume I and Volume II  [/b] [/color]
Editura=[b] Dover Publications,Inc,New York, First edition (in one volume) [color=red]1907[/b] [/color] , Second  edition: Vol.I 1921; vol.II, 1926, Third edition; Vol.I, 1927 .
In continuare ma voi referi la editia III-a a primului volum.
La pagina 300 incep sectiunile 224-226 si 227  cu titlurile
 [b] The discontinuities of  functions [/b]  respectiv [b]  Ordinary discontinuities [/b].La pagina 304 se arata ca (citez ):
[color=blue]  [b] " The set of points at which a function $f(x)$ has ordinary discontinuities  is  enumerable (or finite, or absent) ; whereas the set of points of discontinuity of the second kind may be unenumerable." [/b] [/color]
In aceste sectiuni (paragrafe) sunt citate  lucrarile:
 [b]William Henry YOUNG (1863--1942)  [/b] Quarterly Journal of Mathematics, vol.XXXIX,p.67
 [b] Moritz PASCH (1843--1930) [/b] [i] Einleitung in die Differential und Integralrechnung,p.139 [/i]
Probabil la noi in tara, ca si in prezent (mai ales datorita Internetului) ,  nu s-a prea...citit  !
Apropo, mi-am adus aminte sa Va intreb: [b]" Ce fel de numar real este  constanta  $\gamma$ a lui Euler-Mascheroni  , adica  $\gamma: = \displaystyle \lim\limits_{b\to \infty}\left(\sum\limits_{k=1}^{n}\frac{1}{k} - \ln{n}\right) \;$  ?
Cunoasteti o persoana, din cele citate mai sus, care s-ar fi ocupat (se pare, fara succes) de aceasta chestiune ? [/b]
[color=red] [b] [size=150] Concluzie plauzibila[/size]: ca si in Politica,  am fost adesea   " intoxicati" (cu afirmatii false, de exemplu privind prioritatile)  atat in Stiinta cat si in Educatie (Invatamant) !  [/b] [/color]

teorema lui froda e in cartea mate 2004-2005 clasa a 11-a . cat despre constanta lui euler nu se stie daca c e algebric sau transcendent, rational sau irational

[quote="v0id"]teorema lui Froda e in cartea mate 2004-2005 ... constanta lui Euler nu se stie daca  e algebric sau transcendent, rational sau irational[/quote]
[b]... Nu este Teorema lui[/b]  [color=red] [b]Froda ! [/color][/b] Tot A.Froda a incercat , [color=re][b]dar fara succes , [/b] [/color] sa se ocupe de natura constantei lui Euler !
A gresit !
Problema nu este elucidata nici in prezent !

Dar exista ceva lucrari publicate in jurnale serioase cu privire la aceasta minunata constanta. Dupa umilele mele cunostiinte, nu se stie nici daca
$e+\pi$ sau $e-\pi$ este algebric sau transcendent. :)


LaTeX
function
calculus
derivative
integration
trigonometry
real analysis

I have to determint  'wronski determinant' D(W) of the following homogenious diff.equation:

$(cos x)u\"(x)+(sin x)u'(x)-xu=0$

I can reduce the order of the equation and find a second independent linear solution, but how to find the first solution ?
thanx

[quote="various"]I have to determint  'wronski determinant' D(W) of the following homogenious diff.equation:

(cos x)u"(x)+(sin x)u'(x)-xu=0

I can reduce the order of the equation and find a second independent linear solution, but how to find the first solution ?
thanx[/quote]
is it so difficult?

[quote="various"]is it so difficult?[/quote]

 It is difficult to make people here interested , even if they understand your question, if you don't write in $LaTeX$. And it is NOT difficult to write in $LaTeX$. If you don't know the code, just copy from others' posts. I do it all the time!! ;) .Or you can visit the $LaTeX$ help section. :)

various:

I stayed away from this because I was misreading what you were asking for. Now that I realize what the question is, I realize that you need a standard theorem proved in most Differential Equations textbooks. If you're simply asking what the Wronskian is, you do NOT need to know a basis for the solution space.

Start with the linear homogeneous differential equation

$a(x)u''+b(x)u'+c(x)u=0.$

Divide through by the lead coefficient, so that we rewrite the equation as

$u'' + p(x)u'+q(x)u=0.$

Now, suppose $y_1(x)$ and $y_2(x)$ are linearly independent solutions. Such functions exist (by the existence-uniqueness theorem); as I said before, you do not need to know their identities.

The Wronskian (written in scalar form) is $W = y_1y_2'-y_1'y_2.$

Now differentiate that, and replace all the second derivatives as follows: $y_1''=-py_1'-qy_1,$ and similarly for $y_2.$ Do that, and you will discover that $W$ satisfies the first order homogenous linear differential equation $W'+pW=0.$ This can be systmatically solved as $W(x)=Ce^{-\int p(x)dx}.$

So in your case, $p(x)=\frac{\sin x}{\cos x}$ and $W(x)=C\cos x,$ valid on some interval between two of the zeros of the cosine.


A 10-by-12-inch page has margins of 1.5 inches on all sides. What fraction of the area of the page is taken up by the margins? Express your answer as a common fraction.

The total area inside the margins is $ 7 \times 9\equal{}63$.  The answer is $ \frac{120\minus{}63}{120}\equal{}\frac{57}{120}\equal{}\boxed{\frac{19}{40}}$.

why is the total area inside the margins 63