abstract algebra
calculus
integration
algebra
function
domain
linear algebra

[quote="mathxxxx"]R is a commutative ring,.
the endomorphism ring of a finite generated R-module as a R-module is also finite generated?[/quote]

The short answer is no.  Here is a survey of some known results:

If R is a commutative, noetherian, integral domain, then yes. (Fuchs-Salce, Modules over Non-Noetherian Domains).

Even if R is a non-commutative, noetherian + PI, prime ring, and the module is cyclic, then no:
Resco, Richard; Wald, Jeanne. "Endomorphism rings."
Comm. Algebra 35 (2007), no. 10, 3013-3017.
http://www.ams.org/mathscinet-getitem?mr=2356135
http://dx.doi.org/10.1080/00927870701404747

If R is a commutative, noetherian, local ring, then yes. (claimed, no proof, probably trivial):
Evans, E. Graham, Jr. "Krull-Schmidt and cancellation over local rings."
Pacific J. Math. 46 (1973), 115-121.
http://www.ams.org/mathscinet-getitem?mr=323815
http://projecteuclid.org/getRecord?id=euclid.pjm/1102946603

Even if R is a commutative, von Neumann regular ring, then no (example 16, p164 of):
Goodearl, K. R.; Warfield, R. B., Jr. "Algebras over zero-dimensional rings."
Math. Ann. 223 (1976), no. 2, 157-168.
http://www.ams.org/mathscinet-getitem?mr=412230
http://dx.doi.org/10.1007/BF01360879

The endomorphism ring of a finitely generated R-module is a subquotient of a matrix ring over R, so in some sense it sits inside a nice ring (like a non-commutative "field of fractions").  In case R is noetherian, then there is even a well-behaved unique smallest such nice ring called the "classical ring of quotients", so the endomorphism ring is sort of like noetherian:

Vasconcelos, Wolmer V. "Regular endomorphisms of finitely generated modules."
J. London Math. Soc. (2) 4 (1971), 27-32.
http://www.ams.org/mathscinet-getitem?mr=291152
http://dx.doi.org/10.1112/jlms/s2-4.1.27

However, if R is commutative, non-noetherian, then the endomorphism ring need not be so nice.  In the particular example, it is module-finite over R, but not noetherian and has no classical ring of quotients.

Blair, William D. "An endomorphism ring which is not Ore."
Proc. Amer. Math. Soc. 44 (1974), 275-277.
http://www.ams.org/mathscinet-getitem?mr=340331
http://dx.doi.org/10.2307/2040420

thank you very much.

If R is a commutative, noetherian, local ring, then yes. (claimed, no proof, probably trivial):

I concern this situation, I just don't know why  the answer is yes?

and I want to know about some skills or methods to construct a indecomposable module by given two indecomposable modules over a commutative ring..

given a commutative ring R, and a indecomposable R-module which exactly is generated by n>1 generators...I guess there would be a indecomposable module  of m-generated while any m<=n.

and if the ring R is semisimple we know every indecomposable R-module is exactly simple mdule of course is uniform(or says meet-irreducible) module, I  want to know whether

 exists a non-semisimple ring R which every indecomposable R-module is exactly uniform module. .

[quote="mathxxxx"]given a commutative ring R, and a indecomposable R-module which exactly is generated by n>1 generators...I guess there would be a indecomposable module of m-generated while any m<=n.[/quote]

I am curious about this over possibly non-commutative artinian rings.  It seems reasonable, but I believe it is not always true that an indecomposable module of rank 3 has an indecomposable quotient of rank 2 (where rank(M) is the composition length of M/Rad(M), rather than minimal number of generators).

For a commutative artinian ring things are easier since you can just assume the ring is local.  The indecomposables are described in several examples in chapter 3 of Lam's Lectures on Modules and Rings.

[quote="mathxxxx"]if the ring R is semisimple we know every indecomposable R-module is exactly simple mdule of course is uniform(or says meet-irreducible) module, I want to know whether exists a non-semisimple ring R which every indecomposable R-module is exactly uniform module.[/quote]

The ring Z/4Z is not semi-simple, but the indecomposable modules over Z/4Z are exactly Z/2Z and Z/4Z, and both are uniform modules.

You might check if a complete DVR satisfies the condition.  I believe torsion indecomposables have simple socle, and so are uniform.  I believe torsion-free modules (of finite tf-rank?) have a particularly simple classification as a direct sum of copies of the ring and copies of the field, and so the only indecomposable torsion-free modules (of finite tf-rank?) are the ring and the field of fractions, both of which are uniform.

$ R^n$-->$ M$-->0

0-->hom(m,m)-->hom($ R^n,m$)

since R noether,,end(m) is finite generated.

Awesome thanks!

If R is commutative and M is noetherian, then R^n→M→0 becomes an exact sequence of R-modules, 0→Hom(M,M)→Hom(R^n,M), and Hom(R^n,R) = M^n is a noetherian R-module, so End(M)=Hom(M,M) is also noetherian as an R-module.

If R is not commutative, then 0→Hom(M,M)→Hom(R^n,M) is only an exact sequence of Z(R) modules, and so it becomes important whether M is noetherian as Z(R) module, not just as an R-module.

in a simple situation, if M is uniform, then evey submodule of M is uniform.

M=xR+yR+.......+zR+vR   exactly   m-generated indecomposable
N=xR+yR+.......+zR         exactly  (m-1)-generated indecomposable


Our soccer team has 15 members.  I want to choose a starting lineup consisting of 11 players, two of whom will represent the team in the captains meeting at the beginning of the game.  In how many ways can I choose my starting lineup, including designating the two captains?

First we choose the two captains, and then the remaining $ 9$ players, so there are

$ \dbinom{15}2\dbinom{13}9\equal{}\boxed{75075}$ ways.

Or pick the lineup, then the captains.

$ \binom{15}{11}\binom{11}{2}\equal{}75075$.

How do you calculate combinations in a calculator? [color=#f00]press alpha window 8 stoopid[/color]


calculus
integration
trigonometry
complex analysis

Let $ \gamma$ be the unit circle. Prove that
$ |\int_\gamma \frac{sin z}{z^2} dz| \le 2\pi e$

$ |\sin z|\le \frac12(e^1\plus{}e^{\minus{}1})<e$ on the unit disk. Use the triangle inequality.

It is easy to show by residues that the integral is exactly $ 2\pi i$.

[quote="jmerry"] ... It is easy to show by residues that the integral is exactly $ 2\pi i$.[/quote]

Well said, [b]jmerry[/b].

[hide]Also,

$ \sin^{\prime}(0) \equal{} \frac {1}{2 \pi i} \int_{c} \frac {\sin z}{z^{2}}dz.$[/hide]

See you!


两个不大于300的连续自然数的平方和为一完全平方数,求所有符合条件的自然数.

300内勾股数看有没有连续的就行了呗。
想到个著名的5， 12， 13， 问题是连续的地方不对。
不要紧，还有更著名的，3，4，5，总行了吧。随便你哪里要求连续，我都满足了。

就3,4,5,别的不可能吧

only 3,4,5
because (n+1)^2-n^2=2n+1
and if all three r consective, that implies (n-1)^2=2n+1
n=0 or n=4, thus, -1,0,1 and 3,4,5
positive...so only 3,4,5

But what if it's not 3 consective intergers? After all, only the first two has to be consective..

I think we can use the fact that all Pythagorean Triples have to be in the form of ($m^{2}-n^{2}, 2mn, m^{2}+n^{2}$)


calculus
integration
function
real analysis
real analysis unsolved

Calculate :
\[ I \equal{} \int{(\sqrt [5]{1 \minus{} x^9} \minus{} \sqrt [9]{1 \minus{} x^5})}dx\]

The antiderivative is not an elementary function.

Now, did you really mean the definite integral $ \int_0^1 \sqrt[5]{1\minus{}x^9}\minus{}\sqrt[9]{1\minus{}x^5}\,dx$?

[quote="theory"]Could you give me more detail of your calculation ?[/quote]
Note that the 2 rootw are inverse functions on [0,1]


Since the old daisy chain got locked a long time ago, I'll repeat the rules:

First, you have a current word, say xylophone.

Someone posts something related to the word, and provides an explanation:

"Instrument; a xylophone is one."

This time, please stay on topic.

Here's the current word:

Galaxy

I post Los Angeles, because LA's MLS Team is the Galaxy (Go Donovan and Lalas)

Baltimore, because Los Angeles is a city

The International Brain Bee: it's in Baltimore.

AoPS, it's got a MATHCOUNTS forum.

[hide=" :) "]
The post right after this can include anything.
Just say "It's posted on AoPS." as your explanation.
[/hide]

Clothes, because you can buy mall there.  Er- the other way around.

Toast, because some people who wear clothes eat toast.

400, that;s how many points jelly gives you in the battle game.

YUGIOH, because it's a retarded battle game

GUY WHO ATE [u][i][b]WAY [/b][/i][/u]TOO MANY DONUTS WHOSE FACE IS COVERED W/ MOLES FROM TOUCHING 2019273 TOADS
(HOG)(WARTS)

AMPHOTERIC COMPOSITIONS AND POLYMERIC FORMS OF ALPHA HYDROXYACIDS AND THEIR THERAPEUTIC USE

Yes, it is very related to the above post [hide=" :wink: "]
Canadian Intellectual Property Office
Patent number 2019273
[/hide]

very long replies

thats what your last post was :)

[quote="The Zuton Force"]Talking.

There's a smiley for it

 :gathering:[/quote]

Meebo.

You can instant message (aka online talking) with it.

Transcendentalism, they both start with T's.

Lasers!

Some of them come from eyes  :lol:

movie, because in the movie Austin Powers, there's sharks with laser attached to it.

BOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOZZZZZZZZZZZZZZZZZZZZZZZZZEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

Cuz I felt like expanding the page length


Find three natural numbers that cannot be the length of the shortest side of a right-angled triangle having integral lengths.
   I think the answers are [hide]1, 2, and 4[/hide] but I'm not sure.

What you want to do is:

1^2 = 1
2^2= 4
4^2 = 16

so, the difference of two squares, assumed to be consecutive must be equal to 1, 4, or 16 depending on your case.

1 and 2 are easy to verify.

For 4, we realize that 8^2-7^2 = 15 is the last possible set of larger legnths to be used. 4,7,8 is not a pythagorean triple. there is no pythagorean triple lower than that with 4 as the smallest number.

so, 1,2,and 4 are the answers

Since the difference between two consecutive squares increases as the numbers get bigger, it is easily verified.


Solve for all solutions in positive integers (a,b) s.t.
x^2 + y^2 = (x + y) * (2005^k), where k is a given positive integer.

My solution is:

we have x^2 + y^2 = (x+y) *(2005^k).

suppose that d=gcd(x,y) and x= d *a and y=d*b

then we have gcd(a,b) = 1 and  d(a^2 + b^2) = (a+b)*(2005^k)

now i show gcd(a+b , a^2 + b^2) = 1 or 2 for this suppose that gcd(a+b , a^2 + b^2) = k

and p ( prime ) divides k . then we have p| (a+b)^2 - (a^2 + b^2) = 2ab.

but if p divide one of a or b then will divide another and this is not true because gcd(a,b) = 1

so p|2 it means that k = 2^r .

if r>0 then k is even and it means a and b are both odd and it mean a^2+b^2 is conguent with 2 mod 8

mean r=1. it means that d= n(a+b)/2 so we have a^2 + b^2 | (2005 ^ k)*2

where gcd(a,b) = 1 and this is solved before. :D


The 2004-2005 Mu Alpha Theta "Log1" Contest is a free contest sponsored by Mu Alpha Theta, the national high school mathematics honor society.  If your school has a Mu Alpha Theta Chapter, your coach can register for the contest (it's easy to convince them, because it's free :-) online at http://mualphatheta.ou.edu/matol/login.aspx, contacting Kay Weiss at
matheta@ou.edu or (405) 325-4489 if they have any questions.  If your school does not have a Mu Alpha Theta chapter, it is easy to establish one; simply have one of your math teachers contact Kay Weiss to start the process.

The 2004-2005 Mu Alpha Theta "Log1" Contest consists of three rounds in formats based on competitions which occur each year at the Mu Alpha Theta National Convention (this year's is in Hawaii).  The first round occurs on Thursday, November 18th, and is an "individual test" in which an individual has thirty minutes to answer fifteen questions covering a range of topics and difficulties.  The second round is on Thursday, January 13th, in which students choose one of three "Topic Tests" in the areas of "Equations and Inequalities", "Geometry", or "Probability".  The third round is a "Ciphering Test", in which students test in ten rounds in which they have three minutes to solve three problems.

Are the "topic tests" new because I don't remember them.  But then again I believe that there was one competition that my school didn't issue to us....

joml88, yes the contest is a bit different this year.  The Mu Alpha Theta Governing Board wanted the contest to better represent the diversity of contests at their national convention, so the "Topic Tests" and the "Ciphering Test" replaced two rounds which were identical to the "Individual Test" in format.


function
vector
integration
algebra
functional equation
real analysis
real analysis solved

Does there exist injective and continuous functions f:R-Q->Q?
But if we replace injectivity with surjectivity?
(I dont know if injectivity,surjectivity are correct english terms.Please correct me if i am wrong).

1) No.
If f is an injective function from R-Q into something (no matter if it is continuous or not) then f(R-Q) is not countable, but Q is countable.

2) Yes, but it probably requires the axiom of choice :
Consider R as a vector space over Q. Let (e_i) be a basis for this vector space (thus, we use the axiom of choice), where i belongs to some uncountable set of subscripts, and e_1 = 1 and e_2 = \sqrt2.
Thus define the linear function f by f( sum(over i) x_i*e_i) = x_2.
(Note that each of the x_i is rational)

Clearly, f is continuous and f(R-Q) = Q.

Pierre.

For Hamel basis and Axiom of Choice
http://mathworld.wolfram.com/HamelBasis.html
------------------------------------------------------------------------
Another application of Hamel basis and Axiom of Choice there exist non linear functions f:R-->R wich satisfied Cauchy functional equation
for any x,y \in R  f(x+y)=f(x)+f(y)
Proof:

Consider R as Q-vector space and Hamel basis B
(Axiom of Choice every vector space has a basis)

Define f arbitrarly on B and extend it to R as follows
if x=\sum q_i.b_i , (q_i,b_i) \in QxB let f(x)= \sum q_i.f(b_i)
f is well define and f(x+y) = f(x)+f(y)

 :D  \cup \beta  \int

functions from A to B <=> maps from A to B
y=f(x)  <=> x maps into y
one-to-one mapping = injective function.
onto mapping = surjective function.

further reading at http://www.iro.umontreal.ca/~paquetse/knoweb/mapping.html :)

The function constructed using Hamel basis is not continious. But the continious surjection from R\Q->Q also exists: lets enumerate rationals q1, q2, q3... and define f on [n,n+1] as follows: f(n)=f(n+1)=0,   f(x)=q[n] for x in [n+0.1,n+0.9] and f is linear on [n,n+0.1] and [n+0.9, n+1] (anf f(x)=0 for x<1).

It is a continuous at least for the norm = sup.
No?

Pierre.

No, it is not. And there is basically the unique norm on R. My example must be changed: f(x)=0 for x not in [n+0.1, n+0.9]. Then it is continious (even locally constant) in every irrational point. At fact, there is no continious function f:R->R, for which f(R/Q)=Q (by the capacity argument).

Well, Fedor, I do not understand why my function is not continuous for the norm ||.||oo.
I agree that it is probably not what it is asked for, and that the proposer was looking for continuity with respect to the usual topology.

I give an exemple of continuous on R-Q surjective function such that f(R-Q) = Q, due to Xavier Caruso :

Let phi : Z -> Q a bijection.
Then, let f(x) = phi(E(x)),
where E(x) denotes the integer part of x.
The function f is continuous at every non integer point, and f(R-Q) = Q.

Pierre.


function search calculus integration algorithm inequalities

Hi all, i'm new here...actually i'm not in HS anymore,
but i think i saw this kind of question when i was in HS..so you guys should know the answer for this question..
I have a question about reversing an equation that uses integer value,
I'm making a program using Visual Basic that uses this equation :

X.Text = Int(Val(A) * 0.3) + Int(((val(A)-1)*(val(B)+1))/32)

A,B,X are textboxes, and command1 & command2 are used to execute the equation. And int = integer (rounded down)
For this formula, The input is A & B so when command1 is clicked, x will show the result

My question is;
How do i reverse the equation if i want to change the input into B and X, and A becomes the result?

X.Text = Int(Val(A) * 0.3) + Int(((val(A)-1)*(val(B)+1))/32)
A.text = ? if the input are X and B
(command2 is used to execute this equation.)

In a simpler form, what i'm asking is like this :

X = int(0.3*A) + int(((A-1)*(B+1))/32)
A = ?

The problem here is that, because of the integer value;
the result in each brackets must to be rounded down first before being added,
and i can't find a way to reverse the equation because of this...

e.g :
(To make it simple, let's ignore the last bracket and take the first one as an example)
If you reverse it directly and ignore the integer property:
X = int(0.3 * A)
A = 10/3 X

for example, A = 32
X = int(0.3 * A)
X = int(0.3 * 32)
X = int(9.6)
X = 9

Now let's put 9 in the reversed equation :
A = 10/3 * X
A = 10/3 * 9
A = 30

See the problem? This is the wrong way to reverse it,
if this is the correct way then A would be 32 not 30..
Anyone know how to reverse this equation correctly?
Thanks in advance.

The function you have described in terms of A and B is not one-to-one (or necessarily even onto for some values of B). So given specific values of X and B it is not necessarily "reversible" for a specific value of A (or any values of A for some values of X and B).

You could possibly solve it in terms of the [i]range[/i] of values for A, but that would be algebraically messy. You could probably write a program to search for them faster than you could solve it algebraicaly.

Hmmm...what's the function to look for the range?
If i could find the range then the smallest number should be the answer that i'm looking for..

BTW, I think my math teacher taught me this kind of thing when i was in HS,
but it's called "integral", the formula to solve an integral function is like this :

1/n+1 * x^ (n+1)

is it the same thing as this one?

We're talking about inverse functions.  The integral is an example of an inverse function, in this case the inverse of a derivative.  In more standard notation, your question is:

Given $X(A, B) = [\frac{3A}{10}] + [\frac{(A-1)(B+1)}{32}]$, compute A as a function of B and X.

Let me do your simpler example, because it's easier.  I'm treating each variable as taking integral values:

$X = [\frac{3A}{10}]$ implies $X \leq \frac{3A}{10} < X + 1$ so $\frac{10X}{3} \leq A < \frac{10(X+1)}{3}$, and then the possible values for A are all integers in that range.  For your question, though, it's considerable more difficult: we can't just isolate the A so simply.  The best way to solve questions like this is with a computer algorithm -- there is no nice closed-form expression, except something ugly in terms of inequalities in X and B.

Ummm, what's computer algorithm?
You mean like making an algorithm for this function using programming language?
And what if the variable (A,B,X) only represent a set of numbers?
Is it possible to reverse the function?
because A only ranges from 1 ~ 99
X = 24 ~ 50
B = 0 ~ 3200
I was going to do this manually, by guessing which number is closer, then test it with the function...but i changed my mind because B ranges from 0 ~ 3200...

Hey that reminds me...is it possible to solve this using Matlab?

A computer algorithm -- using a programming language or computer program to test for you.  Matlab could probably handle this question very easily since there are relatively small numbers of possibilities for A.  Simply have your program plug in B and X, then try every possible value of A and print those which work.


quadratics
modular arithmetic
number theory proposed
number theory

Solve $ x^2\plus{}y^2\plus{}xy\minus{}3y\equal{}7620$ in $ \mathbb{Z}.$

You can write it as $ (2x \plus{} y)^2 \plus{} 3y^2 \minus{} 12y \minus{} 30480 \equal{} 0$. Thus $ 3y^2 \minus{} 12y \minus{} 30480\leq 0$ or $ \minus{} 98\leq y\leq 102$ as I am interested only integer solutions. Testing all possibilities gives solutions

$ (x,y)\in \{ (65, \minus{} 97), (32, \minus{} 97), (98, \minus{} 64), ( \minus{} 34, \minus{} 64), (98, \minus{} 31), ( \minus{} 67, \minus{} 31)$

$ (65,35), (32,68), ( \minus{} 100,68), ( \minus{} 34,101), ( \minus{} 67,101)\}$.

Edit: and also $ (\minus{}100,35)$ is a solution.

[quote="ozgurkircak"]Solve $ x^2 \plus{} y^2 \plus{} xy \minus{} 3y \equal{} 7620$ in $ \mathbb{Z}.$[/quote]

Solving the quadratic in $ x$ gives $ x \equal{} \frac { \minus{} y\pm\sqrt { \minus{} 3y^2 \plus{} 12y \plus{} 30480}}2$

So $ x \equal{} \frac { \minus{} y\pm z}2$ with $ z^2 \equal{} \minus{} 3y^2 \plus{} 12y \plus{} 30480$ $ \iff$ $ z^2 \plus{} 3(y \minus{} 2)^2 \equal{} 30492$

It's immediate to see that $ z \equal{} y \minus{} 2 \equal{} 0\pmod 3$ and we get $ u^2 \plus{} 3v^2 \equal{} 3388$ with $ z \equal{} 3u$ and $ y \equal{} 3v \plus{} 2$

Taking this equality $ \pmod{11}$ and since $ \minus{} 3$ is not a quadratic residue $ \pmod{11}$, we get $ u \equal{} v \equal{} 0\pmod{11}$ and so :

$ a^2 \plus{} 3b^2 \equal{} 28$ with $ z \equal{} 33a$ and $ y \equal{} 33b \plus{} 2$

And very few tests give the solutions $ (a,b)\in\{(\pm 5,\pm 1),(\pm 4,\pm 2),(\pm 1,\pm 3)\}$

And so $ (z,y)\in\{ (165,35), ( \minus{} 165,35), (165, \minus{} 31), ( \minus{} 165, \minus{} 31),$ $ (132,68), ( \minus{} 132,68), (132, \minus{} 64), ( \minus{} 132, \minus{} 64),$ $ (33,101) , ( \minus{} 33,101) , (33, \minus{} 97) , ( \minus{} 33, \minus{} 97)\}$

And so $ (x,y)\in\{ ( \minus{} 100,35), ( \minus{} 100,68), ( \minus{} 67, \minus{} 31), ( \minus{} 67,101),$ $ ( \minus{} 34, \minus{} 64), ( \minus{} 34,101), (32, \minus{} 97),$ $ (32,68), (65, \minus{} 97), (65,35), (98, \minus{} 64), (98, \minus{} 31), \}$ which indeed are solutions.

[u][b]edit [/b][/u]: ooops, a bit too late... (but I got $ ( \minus{} 100,35)$ that you missed, MeKnowsNothing  :wink:

[quote="pco"][u][b]edit [/b][/u]: ooops, a bit too late... (but I got $ ( \minus{} 100,35)$ that you missed, MeKnowsNothing  :wink:[/quote]
Oops. In my computer screen there was also $ (\minus{}100,35)$.


hello...
would somebody recommend me a good refference about order in number theory?

PEN problems are interesting.

[color=#FF0000][Mod: Illegal link deleted.][/color]


Source: PSS
Show that $(n+1)^{n}\geq 2^{n}\cdot n!$ for $n=1,2,3,....$

By AM-GM we have \[\frac{\sum_{k=1}^{n}k}{n}\geq\sqrt[n]{n!}\] But $\sum_{k=1}^{n}k=\frac{n\left(n+1\right)}{2}$ which gives \[\frac{n+1}{2}\geq\sqrt[n]{n!}\] Then raise each side to the power of $n$ and you have the desired result.


induction
pigeonhole principle
modular arithmetic
number theory unsolved
number theory

The sequence $a_1, a_2, a_3, ...$ is defined by $a_1 = a_2 = a_3 = 1$, $a_{n+3} = a_{n+2}a_{n+1} + a_n$. Show that for any positive integer $r$ we can find $s$ such that $a_s$ is a multiple of $r$.

If r = 2 then $a_4 = 2$.  Else,

its enough to claim $a_{n+1}a_{n} \ne -a_{n-1} \mod{r}$ (*), and $a_n \ne 0 \mod{r}$(**).  We show by induction on $n$.

For n=2, it becomes $2 \ne 0 \mod{r}$ and $1 \ne 0\mod{r}$ which is true.

Suppose (*) and (**).  In the induction step, we need to show $a_{n+2}a_{n+1} \ne a_n \mod{r} \Leftrightarrow (a_{n+1}a_n + a_{n-1})a_{n+1} - a_n \ne 0 \mod{r}$

It becomes $a_n \ne 0 \mod{r}$, which is (**).

So our induction is complete and (*) [as well as (**)] is true.  Result follows.

Does anyone actually understand Singular's solution?  :what?:

Anyways, the idea for this problem is the same idea used to show that for any $r$, we can find a Fibonacci number divisible by $r$.

Note that each term is dependent only on the previous three terms. In $\mathbb{Z}/r\mathbb{Z}$, there are only finitely many possibilities for a ordered triple. Thus by Pigeonhole principle, there exist some positive integers $p < q$ such that $(a_p, a_{p+1}, a_{p+2}) \equiv (a_q, a_{q+1}, a_{q+2}) \pmod r$. Since three consecutive terms uniquely determine this doubly-infinite sequence (since the recursion $a_{n+3} = a_{n+2}a_{n+1} + a_n$ can be applied both forward and backward), the sequence must be periodic $\pmod r$ with period $T = q - p$. Since $a_0 = 0$, we have $a_T \equiv 0 \pmod r$, as desired.


inequalities
vector
inequalities proposed

I have a vector approach solution to this problem, but I want to see a solution in a pure inequality approach.
Let there be eight real nos. $ a,b,c,d,e,f,g,h$. Prove that at least one of the six nos. $ (ac \plus{} bd),(ag \plus{} bh),(ce \plus{} df),(cg \plus{} dh),(ae \plus{} bf),(eg \plus{} fh)$ is not negative.
 :)

I post the solution of the author...
Consider the four vectors $ \vec {v_{1}}\equal{}(a,b), \vec{v_{2}}\equal{}(c,d), \vec{v_{3}}\equal{}(e,f), \vec{v_{4}}\equal{}(g,h)$.
Thus, the six given numbers are all pairwise products of these vectors:
$ \vec {v_{1}}.\vec {v_{2}},\vec {v_{1}}.\vec {v_{3}},....,\vec {v_{3}}.\vec {v_{4}}$.
Since one of the angles between these four vectors does not exceed $ \frac{\pi}{2}$, at least one of the six scalar products is not negative....Done!

I personally did not like this solution, is there a better approach? :maybe:


Hello!!!!!

            _Does anyone know good links about Kurt Gödel???

            _Is the word "Gödel" stressed in the 1º or in the 2º silable????? (Just to know how to pronunciate it)

[quote="pleurestique"]first syllable[/quote]

thanks!!! ... any link or e-book??????

One place to start anytime you want to know about any particular mathematician is the [url=http://www-groups.dcs.st-and.ac.uk/~history/]St. Andrews University history of math site.[/url]

That site does have a short biography of Gödel with a very brief description of his work. For more, you should probably find a textbook on logic or set theory.

Here's an interesting link: http://www-groups.dcs.st-and.ac.uk/~history/Printonly/Godel.html

[quote="lasagnaman"]If you are really interested, I reccomend the book [u]Gödel, Escher, Bach[/u] by Douglas R. Hofstadter. A really nice look into his work and other logical tangencies.[/quote]

Do you know if it is available on the web???

[quote="RiemannZetaFunction"]In one of the two past two issues of [u]scientific america [/u] there is a article about his work[/quote]

ok


topology
real analysis
real analysis unsolved

is there any connected set such that if we suppress a special point then the remainig contains no connected subset including more than one point.

I meant that there is a point that if we suppress it then there is no connected subset containing more than one point.

Should we give the example in $\mathbb R$ or $\mathbb R^n$ for some $n>1$? I think there's a theorem stating that a subset of $\mathbb R$ is connected iff it's an interval, so the answer is "no" if we only look at $\mathbb R$.

yes u are right I forgot to mentioned it i find a example in $\mathbb{R}^2$

I forgot about this problem :).

I didn't find an example in a Euclidean space, but if you'll settle for any example, then it's easy to give one: take connected topological space $X$ and any subset $A\subset X$ s.t. $X\setminus A$ is totally disconnected. Our space will be the quotient space obtained from $X$ by identifying all the points of $A$. Of course, the special point will be the point obtained from $A$.

For a concrete example, take the quotient space obtained from an interval by identifying all its rational points to a single point.

I think I have an example of a connected plane set (with more than one point) which contains a point $x$ the removal of which leaves the set totally disconnected. I hope this is correct.

Everything takes place in the unit disk $D$ (let's take it to be closed, but it doesn't realy matter whether it's closed or open). The special point $x$ in our set $M$ will be the origin of the plane. A rational ray is a set of the form $R_{\theta}=\{te^{i\theta}\ |\ t\in(0,1]\}$ where $\theta$  is a rational multiple of $\pi$, while an irrational ray is a set of the same form, with $\theta$ being an irrational multiple of $\pi$.

Suppose we can construct $M$ having the following properties:

[b](a)[/b] its intersection with every rational ray is empty;
[b](b)[/b] the intersection of both $M$ and the complement of $M$ with any irrational ray are dense in that irrational ray;
[b](c)[/b] the intersection of $M$ with any closed subset of $D$ which contains $c$ (the continuum cardinality) points on irrational rays is non-empty.

This set will satisfy the requirements. First of all, [b](a)[/b] and [b](b)[/b] clearly imply that $M\setminus \{x\}$ is totally disconnected. All we need to prove now is that such a set $M$ is connected.

Suppose we can partition $M$ as $(M\cap F_1)\cup(M\cap F_2)$, where $F_1,F_2$ are closed subsets of $D$. Since $M$ is dense in $D$, so is $F_1\cup F_2$, and since this set is closed, we get $F_1\cup F_2=D$. One of the sets $F_i$, say $F_1$, contains $x$ in its interior. This means that $F_1$ contains points on all the rays starting from $x$. Since $F_1\ne D$, we can find $0\le\theta_1<\theta_2<2\pi$ s.t. none of the rays $R_{\theta},\ \theta\in(\theta_1,\theta_2)$ is completely contained in $F_1$. This means that $F_1$ has a boundary point on each irrational ray $R_{\theta},\ \theta\in(\theta_1,\theta_2)$, i.e. it has $c$ boundary points on irrational rays. The boundary of $F_1$ is a closed set, so, according to [b](c)[/b], $M$ contains one of these points (call it $y$). However, every boundary point of $F_1$ is also a point of $F_2$, because $F_2$ contains $D\setminus F_1$. This means that $y\in M\cap F_1\cap F_2=(M\cap F_1)\cap(M\cap F_2)$, contradicting the fact that $(M\cap F_1)\cup(M\cap F_2)$ is a partition of $M$. Thi gives the desired contradiction, showing that $M$ is indeed connected.

Now, for a construction of such a set $M$, we could use an argument similar to the one [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=numerable&t=44919]here[/url]. We well-order the $c$ closed subsets with the property described in [b](c)[/b] together with some $c$ sets which form countable bases for the $c$ irrational rays and, following a transfinite procedure just like the one in the link, we construct $M$ so that both it and its complement cut each of the sets in the ordering.


Let $a,b,c$ positive numbers with $abc=1$. Prove that:

$\sqrt{\frac{a}{b+3}}+\sqrt{\frac{b}{c+3}}+\sqrt{\frac{c}{a+3}}\geq  \frac{3}{2}$

[quote="girino"]$\sqrt[3]{\frac{1}{(a+3)(b+3)(c+3)}}\ge\frac{3}{a+3+b+3+c+3}$

this inequality is not verified for a=0.5 b=0.5 c=4
why?where is the mistake? isn't it HM-GM?[/quote]
This inequality is true, because it is $\mathrm{HM-GM}$ for $\frac1{a+3},\; \frac1{b+3},\; \frac1{c+3}$. For $a=b=\frac12,\; c=4$ we have $\mathrm{LHS}= \frac{2^{\frac23}}{7}\approx 0.226772$ and $\mathrm{RHS}=\frac3{14}\approx 0.214286$, so in fact $\mathrm{LHS}>\mathrm{RHS}$.

[quote="girino"]$LHS\ge\\3\sqrt[3]{\sqrt{\frac{a}{b+3}\frac{b}{c+3}\frac{c}{a+3}}}\ge\\3\sqrt{\frac{3}{a+b+c+9}}$

that with a=0,5 b=0,5 and c=4 is < than 1,5

i have applied AM-GM and GM-HM

where is the mistake?[/quote]
There's no mistake in the inequalities you wrote. You tried to use following reasoning: if $A\geqslant B$ and $A\geqslant C$, then the inequality $A\geqslant B$ must be provable by $A\geqslant C$, i.e. $C\geqslant B$. It isn't true, because there exist very many lower estimations for $A$ which are less than $B$.


What is the sum of the tens digit and the ones digit of the integer form of $ (2\plus{}3)^{23}$?

Note that this is 5^23. Also note that 25*5=125 and so on, thus the last two digits is always 25 and the sum is 7.


geometry perimeter trigonometry inradius AMC AIME SFFT

How many non-congruent right triangles are there such that the perimeter in $\text{cm}$ and the area in $\text{cm}^2$ are numerically equal?

$\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 4 \qquad \text{(E)} \ \text{infinitely many}$

Let one leg of the right triangle be $a$, and the angle opposite $a$ be $\theta$, then the hyponeuse is $\frac{a}{\sin{\theta}}$ and the other leg is $\frac{a\cos{\theta}}{\sin{\theta}}$.  Therefore, we get the equation:
 $a*\left(\frac{1+\sin{\theta} + \cos{\theta}}{\sin{\theta}}\right) = a^2*\left(\frac{\cos{\theta}}{2\sin{\theta}}\right)$

solving for $a$ we get $a = \frac{2(1+\sin{\theta} + \cos{\theta})}{\cos{\theta}}$, which means that for any $\theta$ there is an unique $a$ satisfying the property, therefore there are infinitely many.  $E$.

No-trig solution:

[hide]We have

$\frac{ab}{2}=a+b+\sqrt{a^2+b^2}$,

so moving things around and squaring, we get

$a^2b^2-4ab(a+b)+4(a^2+2ab+b^2)=4(a^2+b^2)$

which reduces to

$a^2b^2-4a^2b-4ab^2+8ab=0$

dividing out by ab and using Simon's Favorite Factoring Trick, we get

$(a-4)(b-4)=8$

it is obvious that this equation has infinitely many solutions, so the answer is E.[/hide]

[hide="Answer"]We have $a+b+\sqrt{a^2+b^2}=\frac{1}{2}ab\Rightarrow 4a^2+4b^2=a^2b^2-4ab(a+b)+8ab+4a^2+4b^2$, so $a^2b^2-4ab(a+b)+8ab=0$. Factoring out an $ab$, we have $ab-4a-4b+8=0$. Therefore, $(a-4)(b-4)=8$, which has infinitely many solutions. $\boxed{E}$[/hide]

[hide="Easy non rigorous approach"]$A = \frac{pr}{2}= p$
$r = 2$

Incircle has radius 2.

It's easy to visualize continuously moving three tangent lines to deform a triangle circumscribing this circle of radius 2. Thus there are an infinite number of such triangles.[/hide]

[b]Breez [/b]found that incircle must have radius 2.

Given a right triangle find it's inradius $r$. Find $k=\frac{2}{r}$. Then the triangle similar to the given with coeficient $k$ has the desired property.

You could be like $(2rs,r^{2}-s^{2},r^{2}+s^{2})$ and then the problem dies.

i don't think they are they primitive

hmm i think i have seen this problem recently...

I WONDER WHERE?  :lol:

Ugh. So that's why I kept getting only 4 solutions to ... umm nvm.

Yeah, me too chess, and I cannot believe this question was on here :censored:! but I'm not that hot at geometry so.....yeah we'll talk about it tommorow, lol

this problem IS OLD

i think you can find 3*perimeter in one of the marathons from a while back

OLD

it is like omg f2 solve for a leg

expand

It was one of the problems on my mock AIME that I never finished writing... (only it was $r = 24$  :rotfl: )


quadratics
number theory
Diophantine equation
algebra
number theory open

Find all triplets\an infinite family of triplets $ (x,y,z), x,y,z \in N$ such that $ xy\plus{}a, yz\plus{}a, zx\plus{}a$ are all squares, for some given natural a.
I am especially interested in the case a=6.
Some PEN problems are related to the case a=1, but I wasn't able to generalize.

I wanted to show that there are no solutions in integers to the diophantine equation $ x^4\plus{}y^4\plus{}z^4\minus{}2x^2y^2\minus{}2y^2z^2\minus{}2z^2x^2\equal{}24$. If I put $ x^2\equal{}x', y^2\equal{}y', z^2\equal{}z'$, I get a quadratic equation in $ x',y'$ or $ z'$, which has an integer solution iff $ x'z'\plus{}6, y'z'\plus{}6, z'x'\plus{}6$ are all squares, which means that 6 is a difference of 2 squares, which is impossible modulo 4.
But I want to find also all the solution to $ x^2\plus{}y^2\plus{}z^2\minus{}2xy\minus{}2yz\minus{}2zx\equal{}24$ - so this is why I am asking this.

[quote="bambaman"]$ x^4 \plus{} y^4 \plus{} z^4 \minus{} 2x^2y^2 \minus{} 2y^2z^2 \minus{} 2z^2x^2 \equal{} 24$. [/quote]Your original problem is simple to solve:
The equation factors as
\[ (x \plus{} y \plus{} z)( \minus{} x \plus{} y \plus{} z)(x \minus{} y \plus{} z)(x \plus{} y \minus{} z) \equal{} \minus{}24.
\]
Notice that at least one of the 4 factors must be even. But all 4 factors have the same parity. So we get $ 16\mid 24$, contradiction.


Given $ x,y,z$ are positive reals such that $ x\plus{}y\plus{}z\equal{}3$,then prove that
$ \sqrt{x}\plus{}\sqrt{y}\plus{}\sqrt{z}\ge xy\plus{}yz\plus{}xz$

[quote="RAGHAV GROVER"]Given $ x,y,z$ are positive reals such that $ x \plus{} y \plus{} z \equal{} 3$,then prove that
$ \sqrt {x} \plus{} \sqrt {y} \plus{} \sqrt {z}\ge xy \plus{} yz \plus{} xz$[/quote]

$ <\equal{}>2(\sqrt {x} \plus{} \sqrt {y} \plus{} \sqrt {z}) \ge (9\minus{}x^2\minus{}y^2\minus{}z^2)$
$ <\equal{}>(x^2\plus{}\sqrt {x}\plus{}\sqrt {x})\plus{}(y^2\plus{}\sqrt{y}\plus{}\sqrt{y})\plus{}(z^2\plus{}\sqrt{z}\plus{}\sqrt{z}) \ge 9$
It's true by Am-Gm

[quote="RAGHAV GROVER"][hide="My solution to the problem"]
Using Titu's Lemma we have
$ \frac {\sqrt {x} \plus{} \sqrt {y} \plus{} \sqrt {z}}{3}\ge(\frac {x \plus{} y \plus{} z}{3})^{\frac {1}{2}}$

but $ x \plus{} y \plus{} z \equal{} 3$ then we get $ {\sqrt {x} \plus{} \sqrt {y} \plus{} \sqrt {z}}\ge3$
now it is left to prove that $ 3\ge ab \plus{} bc \plus{} ac$ which is true by using $ (a \minus{} b)^{2} \plus{} (b \minus{} c)^{2} \plus{} (c \minus{} a)^{2}\ge0$.[/hide]

More solutions are welcomed :wink:[/quote]

You have written the (Root-Mean Square >= AM) relationship incorrectly, when you say you are using Titu's Lemma.
$ \frac {\sqrt {x} \plus{} \sqrt {y} \plus{} \sqrt {z}}{3}\le(\frac {x \plus{} y \plus{} z}{3})^{\frac {1}{2}}$

By Jensen,
$ {\sqrt {x} \plus{} \sqrt {y} \plus{} \sqrt {z}}\le3$

Please tell me whether the inequality i applied in my solution is correct or not.?
[hide]I am new to inequalities[/hide]

Applying AM GM inequality to $ x^2, \sqrt {x}, \sqrt {x}$, we have

$ x^2 \plus{} 2\sqrt {x} \ge 3x$ . Similarly we also get that :
$ y^2 \plus{} 2\sqrt {y} \ge 3y$
$ z^2 \plus{} 2\sqrt {z} \ge 3z$

Adding the three inequalities, we have

$ (\sum x)^2 \minus{} 2(\sum xy) \plus{} 2(\sum \sqrt {x}) \ge (\sum x)^2$ which leads to $ \sum \sqrt {x} \ge \sum xy$

well this is the well known Russian MOP 2004 so searching will give u lot of discussions on this one and lots of solutions too,
for a fact i know that the inequality is also true with this condition -$ x\plus{}y\plus{}z\equal{}1$


Prove that $\frac{x}{\sqrt{y^2+z^2}}+\frac{y}{\sqrt{x^2+z^2}}+\frac{z}{\sqrt{x^2+y^2}}>2,  \forall x, y, z \in (0, \infty)$

Please DON'T double-post this again. You did it twice already.

and this inequality has been posted for many many times.
Inequality: $\sum \sqrt{\frac{a}{b+c}} > 2$
proof: Without loss of generality assume $a+b+c=1$, it's easy to show that $\sqrt{\frac{a}{1-a}} \ge 2a$ Equality holds when $a=0.5$. As at most 2 equality cases could hold, the inequality is strict.


Let $ \triangle$ ABC such that $ AB \equal{} \dfrac{AC}2$ and $ B > 90^0$. Prove that $ C > \dfrac A2.$ :?:

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=284782[/url]


search
inequalities proposed
inequalities

Let $ a_1\equal{}\frac {1}{2}$ and for each $ n$ greater than 1, note $ a_n\equal{}\frac {1}{d_1\plus{}1}\plus{}\frac {1}{d_2\plus{}1}\plus{}...\plus{}\frac {1}{d_k\plus{}1}$, where $ d_1,d_2,...,d_k$ are the collection of all distinct positive divisors of $ n$.
Prove that: $ n\minus{}ln(n)<a_1\plus{}...\plus{}a_n<n$.

Don't post your problem like this, it is a new problem in Youth and Math Magazine.

Yes.if it is so.....don't post it here.Otherwise there is a chance to be banned.
Look here: [url]http://www.mathlinks.ro/viewtopic.php?search_id=1396128610&t=168836[/url]


geometry
circumcircle
Euler
geometric transformation
reflection
parallelogram
analytic geometry

Given BC is a chord of the circle (O). A moves on thr bigger arc BC of (O). The altitudes AD,BE,CF of triangle ABC are concurrent at H. DE meets AB at M, DF meets AC at N. Prove that the perpendicular from A to MN, always passes through a fixed point.

$ ABDE, ACDF$ are cyclic, $ \overline {MA} \cdot \overline{MB} \equal{} \overline{MD} \cdot \overline{ME},$ $ \overline{NA} \cdot \overline{NC} \equal{} \overline {ND} \cdot \overline{NF},$ $ MN$ is radical axis of the circumcircle $ (O)$ and the 9-point circle $ (Q)$ of the $ \triangle ABC,$ perpendicular to the Euler line $ OQH.$ $ H$ is on reflection $ (O')$ of $ (O)$ in $ BC.$ Perpendicular to $ MN$ through $ A$ cuts the perpendicular bisector of $ BC$ at $ P.$ $ AHOP$ is a parallelogram and $ OP \equal{} AH \equal{} \text{const}.$

Excellent solution, yetti.
Just to say that if the problem is solved with trilinear coordinates /which, by the way, usually produces weird results with the perpendicularity/, the trilinear coordinates of the point called P in the yetti solution are
(3cosA, -cos(A-C), cos(A-B)) and the first one is constant when A describes the arc BC of the circumcircle. This first trilinear coordinate is proportional to the distance from P to the side BC.


[quote="xscapezaer"]A day can be evenly divided into 86400 periods of 1 second; 43200 periods of 2 seconds; or im many other ways. In total, how many ways are there to divide a day into n periods of m seconds, where n and m are positive integers?[/quote]
[hide]The number of ways to divide a day into $n$ periods of $m$ seconds is equal to the number of factors of 86400.
\begin{eqnarray*}86400 & = & 2^{7}\times 3^{3}\times 5^{2}\end{eqnarray*}
So there are $(7+1)(3+1)(2+1) = 96$ factors for 86400 and the number of ways to divide a day is 96.[/hide]

Both of your solutions are correct. Personally, I did it mthiyaga's really easy way.

@thales: How did you do the problem?

[quote="Thales418"]
${m=2^{\alpha }3^{\beta }5^{\gamma }}$
${n=2^{7-\alpha }3^{3-\beta }5^{2-\gamma }}$

${0\leq \alpha \leq 7}$
${0\leq \beta \leq 3}$
${0\leq \gamma \leq 2}$
[/quote]

This is just to show that

${\alpha}$ ,   ${\beta}$ and ${\gamma }$

has 8 possible numbers, 4 possible numbers, and 3 possible numbers, respectible

i did the same thingh. but i explain where i take ${(7+1) (3+1) (2+1)=96}$

its the same thing!

Where x, y and z are prime, and the factorization is $x^{a}*y^{b}*z^{c}$, the number of factors is $(a+1)(b+1)(c+1)$. (Can be used for different numbers of prime factors...)
[hide]So then the answer is 96 because $86400=2^{7}\times3^{3}\times5^{2}$ and $8\times4\times3=96$[/hide]


parameterization trigonometry calculus derivative algebra function domain

Ascertain $m\in R$ such that the equation $2(\sin x+\cos x)+\sin 2x+m=0$ has a maximum number of solutions in the interval  $\left[0,\frac {\pi}{2}\right].$

put $m = -2k+2$, $s = \sin x$, $c = \cos x$.

Then we maximize solutions for $(s+1)(c+1) = k$.

Simple work will show that $\frac{d}{dx} (s+1)(c+1) = \cos 2x + \cos x - \sin x$.

So the derivative is zero when $\cos^2 x + \cos x = \sin^2 x + \sin x$.  Because $x^2 + x$ is bijective on [0,1] with its domain, it follows $\cos x = \sin x$ for the inflection point.  (also, cosine and sine are continuous blah blah blah.)  So $x = \frac{\pi}{4}$ is a local min/max.

So $x = 0$ gives us 2, $x = \frac{\pi}{2}$ gives us 2, and $x = \frac{\pi}{4}$ gives us $\frac{3}{2} + \sqrt{2}$.  So k between these values for 2 solns.

Set $\sin x+\cos x=t$, we can rewrite the given equation as

\[2t+t^2-1+m=0\Longleftrightarrow m=-(t+1)^2+2\]
where by considering the condition of having intersection points for 2 graphs of unit circle $X^2+Y^2=1\ (0\leq X\leq 1\ and\ Y\geq 0)$ and
the line $Y+X=1$, we have the domain of $t,\ 1\leq t\leq \sqrt{2}$.

The number of solution for $x$ satisfying $\sin x+\cos x=t \left(0\leq x\leq \frac{\pi}{2}\right)$ is

for $t=\sqrt{2}$ we have exactly one solution $x=\frac{\pi}{4}$,for $1\leq t<\sqrt{2}$, we have $2$ solutions.

Thus it is sufficient to consider the condition the parabola $y=-t^2-2t+1\ (1\leq t<\sqrt{2})$ have intersection point with the line $y=m$, the desired answer is $\boxed{-2\leq m<-1-2\sqrt{2}}$.


Prove that $n^3+23$ can never be a perfect square.

The nth square number is given by $\limits\sum_{i=1}^n{2i-1}$
The two squares that are 23 apart are thus the 11th and 12th,   or 121 and 144
For $n^3+23$ to be a square, $n^3+23=144$,   $n^3=121$,  $n=\pm\sqrt[3]{121}$,   which is not an integer.

I think you are assuming that n^3 is a square.. which you can't assume. How do you know that n^3 isn't a square but n^3 + 23 still is?

Danbert, $n^3$ isn't necessarily (and, as you have shown, isn't) a perfect square.  A misread?

[quote="Basketball4Life"]I'm going to try this using modular arithmetic but I'm pretty weak at it so I'll just type up what I have.
So n can be congruent to 0,1,2 or 3 (mod 4). n^3 can be congruent to 0, 1, 0 or 3 (mod 4). Adding 23 which is congruent to 3 (mod 4) we get that n^3 +23 can be congruent to 0,2,3 (mod 4). A perfect square is congruent to 0 or 1 (mod 4). Therefore it seems n^3 +23 can be a perfect square. So where did I go wrong?[/quote]
The wrong deduction was wrong. Take this simple example, 12 is 0 mod 4 and 13 is 1 mod 4 but neither is a perfect square. So your logic that "A perfect square is congruent to 0 or 1 mod 4" is not used correctly in your deduction.

That statement can be written into "If...then..." format: If n is a perfect square, then n=0 or 1 (mod 4). But you used the converse: if n=0 or 1 (mod 4) then n is a perfect square. The converse is false!!!

Think of some other way to do it. It's not hard.

[quote="Basketball4Life"]I'm going to try this using modular arithmetic but I'm pretty weak at it so I'll just type up what I have.
So n can be congruent to 0,1,2 or 3 (mod 4). n^3 can be congruent to 0, 1, 0 or 3 (mod 4). Adding 23 which is congruent to 3 (mod 4) we get that n^3 +23 can be congruent to 0,2,3 (mod 4). A perfect square is congruent to 0 or 1 (mod 4). Therefore it seems n^3 +23 can be a perfect square. So where did I go wrong?[/quote]

You haven't gone wrong anywhere. You have proven that if you consider things modulo 4, you don't get a contradicition - you showed that n^3 + 23 might be a perfect square for some n. That still doesn't prove it is for some n, nor does it disprove that it isn't for all n.

Actualy, with that being said (and I think it's a step in the right direction), I still can't find a solution.

Any ideas?

Did anyone ever get this?  I don't think so but I might have missed it because the boards went down.

I think [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20515]this[/url] should work.

if we graphed the two equations, y = n:^3: + 23 and y = n:^2:, they would never cross. so... that would prove it for all real numbers, right?

[quote]if we graphed the two equations, y = n^3 + 23 and y = n^2, they would never cross. so... that would prove it for all real numbers, right?[/quote]

We're not solving the equation $n^2=n^3+23$ for the reals, but rather the equation $k^2=n^3+23$ for integer $k,n$.


geometry
conics
parabola
LaTeX
analytic geometry
trigonometry
MATHCOUNTS

1) The midpoint of each side of a square is joined to its two opposite vertices. Each side of the square is 4 units long. What is the area of the octagonal region formed within the square?

2) The graph of a parabola that is symmetric about a vertical line contains the points $ (1,4), (3,18),$ and $ (0,3)$. Find the equation of the parabola.

3) Solve (1) and (2). :D

Hello

[hide=hint for 2 and 3]
As the parabola is symmetrical to a vertical line ,the equation will be in the form of ${x-\alpha}^2=4a(y-\beta)$.
Put the points in the equation and get the value of $\alpha$ and $\beta$.
[/hide]

Thank u.

For that Parabola question, the coordinates you are giving are on the parabola correct?
Otherwise those points wouldn't be on the same line. If so then just do the linear 3x3, bash and solve.

[asy]size(120);
unitsize(5cm);
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((0.5,0)--(1,1));
draw((0.5,0)--(0,1));
draw((0,0.5)--(1,1));
draw((0,0.5)--(1,0));
draw((0.5,1)--(0,0));
draw((0.5,1)--(1,0));
draw((1,0.5)--(0,1));
draw((1,0.5)--(0,0));[/asy]

[quote="mathemagician1729"]For that Parabola question, the coordinates you are giving are on the parabola correct?
[/quote]
Yes.

[quote="007math"][quote="kabi"]Hello

[hide="hint for 2 and 3"]
As the parabola is symmetrical to a vertical line ,the equation will be in the form of $ {x - \alpha}^2 = 4a(y - \beta)$.
Put the points in the equation and get the value of $ \alpha$ and $ \beta$.
[/hide]

Thank u.[/quote]

Thats awkward..you hide and $ \text{\LaTeX}$ tags didn't work...

EDIT: But they work now...in my post...[/quote]

kabi must have disabled the BBCode in that post.

[hide="2"]
The graph must be of the form $ y \equal{} ax^2 \plus{} bx \plus{} c$, where $ x \equal{} \minus{} \frac ba$ represents the vertical axis of symmetry.

Then, just plug in to get $ 4 \equal{} a \plus{} b \plus{} c$, $ 18 \equal{} 9a \plus{} 3b \plus{} c$, and $ 3 \equal{} c$.

Substituting the last equation into the first two, we get $ 1 \equal{} a \plus{} b$ and $ 15 \equal{} 9a \plus{} 3b\implies5 \equal{} 3a \plus{} b$, and subtracting the latter result from the former, we see that

$ 4 \equal{} 2a\implies a \equal{} 2$ and $ b \equal{} 1 \minus{} a \equal{} 1 \minus{} 2 \equal{} \minus{} 1$, so the parabola satisfies the equation $ \boxed{y \equal{} 2x^2 \minus{} x \plus{} 3}$.
[/hide]

For that number 1, I guess a little trig would bring us the length of the side of that octagon, but is it appropriate for MATHCOUNTS?

You could also set the center of the square at the origin and coordinate bash it... might edit this post later to include a solution if somebody doesn't beat me to it

Trig is not needed. Here's my solution (I only put one corner, since the figure is symmetrical). And it's not a regular octagon as one would assume at first. :)

[hide="1"]
So this diagram is the left upper corner. The reason that $ C$ is the midpoint of $ \overline{BE}$ is because of symmetry (draw the whole picture, and it should be clear why). By the Pythagorean Theorem, $ AC \equal{} \sqrt {AB^2 \plus{} CB^2} \equal{} \sqrt {2^2 \plus{} 1^2} \equal{} \sqrt5$.

Now, we go angle chasing to prove that $ \angle AIB \equal{} 90^\circ$. We let $ \angle BAC \equal{} \angle IAH \equal{} x$, so that by symmetry, $ \angle EBI \equal{} x$ as well (why?). Then, since $ \angle ABC \equal{} 90^\circ$, $ \angle ABI \equal{} 90^\circ \minus{} x$, so $ \angle AIB$ is a right angle.

Draw the perpendicular line from $ F$ that is perpendicular to $ \overline{CE}$ and $ \overline{AH}$. Since triangles $ \triangle AFH$ and $ \triangle EFC$ are similar, $ \frac {CE}{AH} \equal{} \frac12 \equal{} \frac {DF}{FG} \equal{} \frac {DF}{DG \minus{} DF} \equal{} \frac {DF}{2 \minus{} DF}\implies DF \equal{} \frac23$.

Since $ \triangle EFD\sim \triangle EAB$, we also have $ \frac {FD}{AB} \equal{} \frac13 \equal{} \frac {DE}{BE} \equal{} \frac {DE}2\implies DE \equal{} \frac23$, and $ CD \equal{} CE \minus{} DE \equal{} 1 \minus{} \frac23 \equal{} \frac13$.

We can find $ BF$ by the Pythagorean Theorem: $ BF \equal{} \sqrt {BD^2 \plus{} DF^2} \equal{} \sqrt {\left(1 \plus{} \frac13\right)^2 \plus{} \left(\frac23\right)^2} \equal{} \sqrt {\frac {20}9} \equal{} \frac {2\sqrt5}3$, and since the area of any quadrilateral with perpendicular diagonals is half the product of the diagonals, $ [ABCF] \equal{} \frac12\cdot AC\cdot BF \equal{} \frac12\cdot\sqrt5\cdot\frac {2\sqrt5}3 \equal{} \frac53$.

By symmetry, $ [ABCF] \equal{} [AHIF] \equal{} \frac53$, so $ [CFIE] \equal{} 2^2 \minus{} 2\left(\frac53\right) \equal{} \frac23$, and remembering that $ CFIE$ is a quarter of our desired octagon, our answer is $ 4\left(\frac23\right) \equal{} \boxed{\frac83}$.
[/hide]

Wow, that took a lot longer than I thought it would (to type it up).

And sorry for the messed up diagram; I drew it, but realized I forgot some stuff.

[quote="math154"]
The graph must be of the form $ y \equal{} ax^2 \plus{} bx \plus{} c$, where $ x \equal{} \minus{} \frac ba$ represents the vertical axis of symmetry.
[/quote]
I thought the axis of symmetry  was $ x \equal{} \minus{} \frac{b}{2a}$... but I think I get it now, just substitute and solve, right? And for (1), it's basically similar triangles, congruent figures, and the Pythagorean Theorem- oh ok i get it! Thanks everyone!

[quote="qwertythecucumber"][quote="math154"]
The graph must be of the form $ y \equal{} ax^2 \plus{} bx \plus{} c$, where $ x \equal{} \minus{} \frac ba$ represents the vertical axis of symmetry.
[/quote]
I thought the axis of symmetry  was $ x \equal{} \minus{} \frac {b}{2a}$... but I think I get it now, just substitute and solve, right? And for (1), it's basically similar triangles, congruent figures, and the Pythagorean Theorem- oh ok i get it! Thanks everyone![/quote]

You're correct; sorry, it was a typo.


probability
group theory
abstract algebra
linear algebra
matrix
LaTeX
geometry

The title is pretty explanatory. Just say

You know you've studied too much math when _______

and fill in the blank. If you want, you can replace math with another branch of math. Small alterations to the format are acceptable.
I'll start:

You know you've studied too much math when people call you the uber math-nerd.
You know you've studied too much counting and probability when you go to a carnival, see a game, calculate the expected winnings, frown at the price tag, then walk away.
You know you've studied too much number theory when you start mutter things like "but the modular residue can't be 7 in modulo 11, because..."
You know you've studied too much math when you browse AoPS instead of doing homework (doing that right now).

By the way, I suck at making these. When I read this forum, I expect ones way funnier than these.  :P

You know you have studied too much math when you criticize people for saying 'I will take the "bigger' half' at the dining table.

You know you've studied too much math when your only friends are numbers.

You know you've studied too much math when you don't have and girlfriends/boyfriends, and don't care either (nope, no time. too busy studying math).

You know you have studied too much math when counting is an bases seems a piece of cake.

Sorry, didn't really get that.  :|

You know you've studied too much math when you satisfy EVERY single item on the AoPS shirt.

You know you have studied too much math when:

-you can prove Fermat's Last Theorem in two lines
-Riemann Hypothesis looks like a joke for you
-The millenium prize problems are simply trivial exercises for you
-You can solve every Olympiad problem using a single line, 'a consequence of [insert theorem name]'
-You can mentally verify whether a number is prime or not (even those very, very bog ones)
-Ramanujan's calculating abilities are nothing compared to yours
-The Mathematicians all bow down to you (excluding if you are a dictator)
-You can derive centuries of Mathematical concepts by yourself

You know you've studied too much math when you can solve 141938723409 per minute in the Arithmetic Game (that's what it was called, right? the one where you have to do addition, subtraction, multiplication, division, as fast as possible. i havent done it in forever, so i forgot what it was called.)

You know you've studied too much math if you laugh instantly at the following joke:

[i]Q:  What's an anagram of "Banach-Tarski"?
A:  "Banach-Tarski Banach-Tarski."[/i]

You know you've studied too much math when you understand every line of "Finite Simple Group of Order Two."

You know you've studied too much math when you've actually gotten people together to [i]perform[/i] "Finite Simple Group of Order Two."

You know you've studied too much math when your mom makes you do history/english/geography/etc

You know you've studied too much math when your best comeback is "Well, you're just jealous 'cuz I can convert any # into Base 7" :roll:

Haha I was about to run a similar topic like this in the G&FF  :D .

You know you've studied to much math when you fall asleep and you dream about numbers.

You know you've studied too much math when...

-you start to express all your numbers in base 24
-you've got a mathematical proceure named after you
-(kinda like 7H3 D3M0N's)instead of counting sheep when you fall sleep, you count factors of pi
-you can calculate 16th roots in the blink of an eye
-your idea of a party is sitting aroun a table learning about the many trivial uses of angle trisectors
-you made a legitimate proof that 1+1 can equal 3
-you laugh at the thought of a man struggling to use sigmas the right way

You know you've studied too much math when your only good insult is "Well, um, do you get straight A's after you skipped two grades in math?"or"Well, shut up unless you know 1º/√π, I know, the anwser is"[hide]There is no answer, you can't find a square route for an infinite number[/hide]
And that's just getting started!

that's impossible, an irrational value has values beyond the decimal continuing indefinitely. perhaps you mean that you have the algebraic means of calculating $ \pi$ or $ e$ to any desired degree of accuracy.

I guess you know you do too much math when $ e \equal{} \sum_{i\equal{}1}^{n} \frac{1}{n!}$ seems fascinating because it allows you to calculate an irrational constant as the infinite sum of rational values.

You know you've studied too much math when ur playing chess and when u notate, u see O-O(castle kingside) as "0-0=?" and see d5xe4 as "d^5 x e^4=?"

(I once did the opposite, I saw 0-0 in an equation and thought of castling.... maybe I should start: you've studied too much chess when...)

YKYSTMM when you want a calculator for Christmas.
Or, if you have a calculator instead of a teddy bear.

[quote="erabel"]YKYSTMM when you want a calculator for Christmas.[/quote]

aww... now I can't get to IMO. :(

YKYSWWWTMM when you go to mcdonalds and order a TI-100 and go to sears and order a big mac.

You know you've studied too much math when you write in Latex code on your math tests.

YKYSTMM when you spit on your calculator, take it to the laundry room, tie-dye it, bring it to the muffin man, let the muffin man eat it, and the muffin man turns into an irrational number.

You know you have studied too much math when you talk nonsense like most of us are.

You know you have studied too much math when you, like t0rajir0u, don't think these are very funny.

YKYSTMM when a Klaptrap bites your head off.

YKYSTMM when GWB says that people are too smart and they have to stay out of school for the rest of the year.

YKYSTMM when you have a running list of primes, Fibonacci numbers, Lucas numbers, perfect squares, and digits of $\pi$ next to your bed.

YKYSTMM when your classmate says she/he can't be on my team for math, bc shehe is too fast


complex analysis
complex analysis theorems

Hi,

I remember reading that the Riemann hypothesis has been used to prove a certain theorem, first by showing that the theorem follows from Riemann and then by showing that it follows from its negation. Alas, I can't remember which theorem it was or the exact book I read it in (I'm pretty sure it was either in Dr. Riemann's Zeros or Music of the Primes).

So before I go through them again, does anyone here know what theorem it was or who it was that proved it? I'm asking because next year I have to present a lecture on some advanced topic to finish my undergraduate degree and I thought this might make a fun lecture.

Thanks,
Gunnar.

There are a lot of results that would follow immediately, or be strengthened considerably, if the Riemann Hypothesis is true.  Peter Sarnak has a quote to this effect (something about a screwdriver/bulldozer I think) but I can't remember it offhand.

Right, but this was a result that was proven by (or at least had one proof to the effect of) deriving it first from assuming the Riemann hypothesis and then deriving it from the assumption that the Riemann hypothesis was false. I can't remember what the theorem was, but that's not what interests me but rather the unusual method of proof.

But since we're on the subject, anyone have good ideas for an interesting lecture? I've been reading Proofs from the Book and have a lot of ideas from there, although I'm not sure all of them would fill an entire lecture.


Find the sum of the least and greatest values in the list and express your answer to the nearest tenth. \[ \sqrt3,1.\overline{2},\frac{\pi}3,1\]

The greatest in the list is $ \sqrt {3}$ and the least is $ 1$, so $ \sqrt {3}\plus{}1\approx\boxed{2.7}$.

[quote="GameBot"]Find the [b]sum [/b]of the least and greatest values in the list and express your answer to the nearest tenth.
\[ \sqrt3,1.\overline{2},\frac {\pi}3,1\]
[/quote]

The answer is thus $ \sqrt{3}\plus{}1\approx 1.7\plus{}1\approx \boxed{2.7}$


linear algebra
matrix
algebra
polynomial
inequalities
linear algebra unsolved

Let $ A\equal{}(a_{ij})_{1\leq i,j\leq n}$ be a real $ n\times n$ matrix, such that $ a_{ij} \plus{} a_{ji} \equal{} 0$, for all $ i,j$. Prove that for all non-negative real numbers $ x,y$ we have \[ \det(A\plus{}xI_n)\cdot \det(A\plus{}yI_n) \geq \det (A\plus{}\sqrt{xy}I_n)^2.\]

The eigenvalues of an anti-Hermitian or real antisymmetric matrix are pure imaginary:

Suppose $ A$ is a complex matrix such that $ A^* \equal{} \minus{} A.$ (Where $ *$ means conjugate transpose - the Hermitian adjoint.)

Suppose $ \lambda$ is an eigenvalue of $ A.$ Every eigenvalue of $ A$ has at least one eigenvector, so suppose that $ Ax \equal{} \lambda x$ and normalize so that $ \|x\|^2 \equal{} 1.$ Then

$ x^*Ax \equal{} x^*(\lambda x) \equal{} \lambda \|x\|^2 \equal{} \lambda.$

Take the conjugate transpose of both sides of that. Since $ \lambda$ is $ 1\times 1,$ its transpose is itself and we only need take the conjugate.

$ \overline{\lambda} \equal{} (x^*Ax)^* \equal{} x^*A*x \equal{} x^*( \minus{} A)x \equal{} \minus{} \lambda$

Since $ \overline{\lambda} \equal{} \minus{} \lambda,$ then $ \lambda$ is pure imaginary.

So $ \det(A \plus{} tI) \equal{} t^m(t^2 \plus{} r_1^2)(t^2 \plus{} r_2^2)\cdots (t^2 \plus{} r_k^2)$, where $ \pm ir_j$ are the nonzero eigenvalues with multiplicity.
We have $ x^my^m \equal{} \sqrt {xy}^{2m}$ and $ (x^2 \plus{} r^2)(y^2 \plus{} r^2) \equal{} x^2y^2 \plus{} (x^2 \plus{} y^2)r^2 \plus{} r^4\ge (xy \plus{} r^2)^2 \equal{} x^2y^2 \plus{} 2xyr^2 \plus{} r^4$, so the result follows by multiplying these inequalities.

[quote="jmerry"]The eigenvalues of an anti-Hermitian or real antisymmetric matrix are pure imaginary[/quote]Why is that?

It seems to me that the only thing we need to show is the the coefficients of the characteristic polynomial are positive real numbers, and from that it's just Cauchy.

Suppose $ A$ is a complex matrix such that $ A^*\equal{}\minus{}A.$ (Where $ *$ means conjugate transpose - the Hermitian adjoint.)

Suppose $ \lambda$ is an eigenvalue of $ A.$ Every eigenvalue of $ A$ has at least one eigenvector, so suppose that $ Ax\equal{}\lambda x$ and normalize so that $ \|x\|^2\equal{}1.$ Then

$ x^*Ax\equal{}x^*(\lambda x)\equal{}\lambda \|x\|^2\equal{}\lambda.$

Take the conjugate transpose of both sides of that. Since $ \lambda$ is $ 1\times 1,$ its transpose is itself and we only need take the conjugate.

$ \overline{\lambda}\equal{}(x^*Ax)^*\equal{}x^*A*x\equal{}x^*(\minus{}A)x\equal{}\minus{}\lambda$

Since $ \overline{\lambda}\equal{}\minus{}\lambda,$ then $ \lambda$ is pure imaginary.

Alternately, $ iA$ is Hermitian, and we use the well-known fact about Hermitian matrices. This also implies that $ A$ is unitarily diagonalizable (over $ \mathbb{C}$).

I've modified the first post so that it contains the full solution for the resources section.

A is an antysimetric matrix so so all its eigenvalues are pure imaginary . Let $ ia_k,$$ \overline{k\equal{}1,n}$i ts eigenvalues
we have $ \det(A\plus{}xI_n)\equal{}\prod(ia_k\plus{}x)$
$ \det(A\plus{}yI_n)\equal{}\prod(ia_k\plus{}y)$
$ \det(A\plus{}\sqrt{xy}I_n)\equal{}\prod(ia_k\plus{}\sqrt{xy})$
all determinants are real ,and they are $ <0$ or $ > 0$,.We can change the inequality into absolute values because all numbers are positive and the inequality becomes:
$ \prod(a_k^2\plus{}x^2)(a_k^2\plus{}y^2)\ge\prod(a_k^2\plus{}xy)^2$

$ \prod(a_k^4\plus{}a_k^2(x^2\plus{}y^2)\plus{}xy^2)\geq \prod(a_k^4\plus{}a_k^2 2xy\plus{}xy^2)\equal{}\prod(a_k^2\plus{}xy)^2\Rightarrow q.e.d$
i got 4 points on this problem because i didn't prove why the eigenvalues were imaginary   :oops:

[quote="c_adryan"]i got 4 points on this problem because i didn't prove why the eigenvalues were imaginary   :oops:[/quote]Yeah that's probably a good deduction considering this problem.

With this kind of problems, olympiad turns into something completely undesirable: the race of knowing more and more theorems, without knowing what hides behind them and why they are useful. And I refer here to the real use of such theorems, not to make trivial olympiad problems for those knowing them and impossible problems for the others. It's a pity the jury selected such a problem. Unless of course someone convinces me that this has a solution that a student of 11-th grade would find in a reasonable amount of time: because I really do not believe that someone who studied matrices for a few months, learning only the trivial stuff would come up with something like: eureka, I think real antisymetric matrices have imaginary eigenvalues. Let's be serious...

You and I know very well that there have been worse problems than this one at the NMO :) However, this in particular wasn't that crazy to find, I mean in the end such a problem screams Cauchy and thinking about eigenvalues, and with a little bit of intuition I guess one that knows something about the characteristic polynomial, and has worked through say Caragea problem book has a good chance of solving this.

Here's a proof somewhat not involving eignvalues and stuff (though I think in the back of my working it does).
Please someone if could check if it is correct.
Let $f(x)=det(A+xI_{n})$
Obviously $f$ is a polynomial.
So we have to show that $f(x)f(y) \ge f(\sqrt{xy})^2$

But the above result will follow directly once we establish that all the coefficients of $f$ are positive.
$f(0)=0$ obviously.

I will call the kind of matrices asked in the question as [b]ski[/b] matrix.
To show this I will use induction on the order of the matrix.
Let the result be true for all such $n-1 \times n-1$ [i]ski[/i] matrices.

Now consider any $n \times n$ such [i]ski[/i] matrix and call it $f_n(x)$.
Then,
$$f_n(x) = \int^{x}_{0} f_{n}^{'}(x)$$
But obviously since $f_{n}^{'}(x)$ is a sum of determinants of [i]ski[/i] matrices of the order $n-1 \times n-1$, all it's coefficients by the induction hypothesis are positive and hence so are it's integrals (except the constants which we have already established as zero).

Hence $f(x)$ has all it's coefficients positive and hence by Cauchy Schwartz inequality, we are done.


search
number theory
greatest common divisor
arithmetic sequence
number theory proposed

I can start saying that [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=525939470&t=126566]here[/url] was (simply) proven that there are infinitely many primes $ p \in \mathbb{P}$ of the form $ ax \plus{} 1$ where $ a$ is a positive fixed integer (and it is a corollary of the well-known Dirichlet theorem).

[b]Problem.[/b] Prove that exist infinitely many primes $ p \in \mathbb{P}$ of the form $ ax \plus{} 1$, where $ a$ is a positive fixed integer and $ x \in \mathbb{N}$ satisfies $ \text{gcd}(a,x) \equal{} 1$.

[i]Edit: i though it was obvious, I mean without Dirichlet theorem..[/i]

$ x\equal{}ak\minus{}1\Rightarrow p\equal{}ka^2\minus{}a\plus{}1$.  Now it is left to prove that there are infinitely many primes in the arithmetic progression $ a^2\minus{}a\plus{}1\plus{}na^2$, which is Dirichlet.

Ok, but it is not a elementary solution..


function
integration
real analysis
real analysis theorems

$ \Omega$ an open, locally Jordan measurable set in $ \mathbb{R}^{m}$.
$ 1\leq p <\infty$
$ \tilde{L}^{p}(\Omega)$ the set of all continuous functions $ u:\bar{\Omega}\to\mathbb{R}$ for which:

$ %Error. "norm" is a bad command.
{u}_{p}: =(\int_{\Omega}\abs{u(x)^{p}dx)^{1/p}$ is finite.

How do I show that the space $ \tilde{L}^{p}(\Omega)$ is not complete.

Define that, please. I do not recognize the notation - in particular, I do not recognize the tilde.


The name is Gry Debongnie, 21 years

I'm from Belgium, I went to the 41st IMO, in South Korea (2000).  Didn't do well (my excuse : I was so stressed to be with the big guys...), but anyway, I really progressed in math thanks to my IMO training.  Afterwards, I was 18, and I went for one year in USA, as a senior in high school.  That was fun, and I did really well at the AIME test (I got 11 on 15 if I'm right) over there.

Then, I started a new life : university.  I decided to study engineering, and that's what I did for one year, but that was too easy.  Then, last year, I studied 2nd year of mathematics and 2nd year of engineering at the same time, and I did it successfully.  Now, I'm planning to continue mathematics, my third year at university.

That's pretty much it, I think.  Somebody else's turn!

19 year bump

i came upon this because gery is user 69 :P


function
real analysis
real analysis unsolved

I tried to solve the Dirichlet's condition of Fourier series, but I've been still unsuccessful. Can you solve one of this condition's respects that:
If f is a continuous and increasing function on [a,b] then its Fourier series are converge.