On google, enter your first&last name (but u dont have to say ur real name) and say how many match up to YOU!
I have 5

1,430,000 without quotes
233 with quotes

Wait, do they have to be about us? If not,

English Name + Last Name without quotes: 3,880,000

English Name + Last Name with quotes: 434

Chinese Name in English without Last Name or quotes, Unhyphenated: 4140

Chinese Name in English without Last Name or quotes, Hyphenated: 17500

Chinese Name in English with Last Name without quotes, Unhyphenated: 8

Chinese Name in English with Last Name without quotes, Hyphenated: 624

Chinese Name in English without Last Name with quotes, Unhyphenated: 4140

Chinese Name in English without Last Name with quotes, Hyphenated: 13400

Chinese Name in English with Last Name and quotes, Unhyphenated: 0

Chinese Name in English with Last Name and quotes, Hyphenated: 3

Full name, hyphenated without quotes: 290

Full name, unhyphenated without quotes: 4

Full name, hyphenated with quotes: 1

Full name, unhyphenated with quotes: 0

That was a waste of time.

2,340,000 without quotes

3,240 with quotes:

the reason is there's a famous Australian motorcyclist with same name as mine:

my asian real name hits: (if you want to know, it's ******* ham)

599 without quotes

9 with quotes

Well, my username = my chinese name, and I use this as my username for a bunch of stuff, so if you see chenhsi it is probably me, and if you search chenhsi you will find a bunch of websites that I am a member of...

yay, 47 with quotes
and almost all are mine  :)

English name without quotes: 258,000
ENglish name with quotes: 9
Chinese name in English without quotes: 643,000
Chinese name in English with quotes: 278
Chinese name in Chinese: 63

almost all in chinese are mine.

479,000 without quotes
23,100 with quotes!

(i have a common name)

1210000 without quotes
1370 with quotes.
7 Refer to me.   :D   Easy to count.

Without quotes - "4,460,000"

With quotes - "14,200"

Specifically, only 3 - Facebook, MML, and my blog.

Wow, my name is not very unique is it....

2,190,000 without quotes.
18,000 with quotes.

1,460,000 without quotes.
1,140 with quotes.
432 for my chinese name in pin yin, no quotes.
0 results for my chinese name in pin yin, quotes.

I have a fairly unique name- only about 400 in quotes, of which I claim a decent fraction. The very first result even has a picture of me.

123,000,000--O_O

not all me though--i have a very common name

[quote="alexhhmun"]Let's see who's the most and least unique, okay?

splatyango automatically wins for most unique in "Chinese name in quotes".[/quote]
Chenhsi got the same amount for his full name unhyphenated with quotes.

Well, the ones that refer to me are usually on the first page :). I'm that important!

2,750,000 hits on my name  :o

8 of them actually pertain to me!!!

heh 799000 for me not a very common name

In quotes: about 12,700. And except for the nursery in England that I have nothing to do with, it looks like all of them are in fact me. Of course, since I use my real name as my screen name, quite of few of them - in fact, it looks like a majority of them - are links to AoPS or Mathlinks.

657000 english no quote
581 english quote (probably like 5 pertaining to me!)
2570000 chinese name in english letters no quote
0 chinese name in english letters quoted :) I'm unique
0 chinese name in chinese characters... hmmm... :huh:
0 quoted
Okay something is wierd I might want to try a different inputting software but whatever. I win uniqueness!!!
FYI it's 钱方闻

881,000 English name (no quotes)
1,890 English name (quotes)
24,100 Chinese (no quotes)
626 Chinese name (quotes)

I think maybe 10-20 of those are me...

Without quotes:  224,000
With quotes:  65
I think 7 of those are actually about me.

122,000 for chinese name w/o quotes
6 for chinese name with quotes ;D

70,000 without quotes
54 with quotes
54 actually are me :lol:

[quote="leoxnlin"][quote="ProtestanT"]2,500,000 without quotes, 2,560 with quotes.

7 actually refer to me[/quote]

You looked through all of them?[/quote]
As I got to the 3rd page and after, everything is either random french or spanish stuff, because "qu" is a very common phrase in French...like Qu'est-ce que for example

With quotes..20 (all about me)

W/o quotes...35

Sad.

With quotes, English: I'm too lazy to count... A LOT!!!
About me: At most 20 (too lazy to look between spam...)

With quotes, Spanish: I didn't look

Without quotes: 878
With quotes: 9, 6 are about me.

 :|

I wished I'd never done that now. The first link was someones Facebook and he had the exact same name! I dont use facebook so i know it isnt mine. And I got 458000 hits.

61 with my name in quotes

59 of them are about me :O

some of them are kinda lame, like:

Daniel Xia | Facebook
David Hao · Daesun Yim Daesun Yim · Luke Xu Luke Xu · Message · View Friends · Add as Friend. Facebook © 2008. |. EnglishEspañolFrançaisDeutsch日本 ...

most of the rest are either from newspapers, aops, or some random discussion of mathpath05 (lol, what?).

there's about 300 when i search up my name, sans quotes, and i'm too lazy to see how many of those relate to the actual me

interestingly, in google image search, the fifth picture of me is:

[img]http://www.unl.edu/amc/f-miscellaneous/f1-archive/1-noimage.jpg[/img]

@Lazarus: isn't hte fifth pic from the amc site?

and like how come half the websites that i'm listed on only list my last name initial :mad:

yeap it is from the amc website. i did not submit a picture, or something along the lines of that.

ani is it because it is hard to spell >.>

wow only 4 sites have ani on it
and like 3 are my school's so yeah those don't count

rest have anirudh hehe

and define hard to spell xD


linear algebra
matrix
vector
linear algebra unsolved

Let $ a,b,c,d \in (0,\infty)$ such that $ a^2\plus{}b^2\equal{}c^2\plus{}d^2$ and the matrix $ A$ equal with: $ A$=$ \left(
\begin{array}{cccc}
a & 0 & 0 & b \\
0 & c & d & 0 \\
0 & \minus{}d & c & 0 \\
\minus{}b & 0 & 0 & a \end{array}
\right)$

$ a)$ If the sum of the elements of $ A^2$ is $ 0$ then $ b\equal{}c$

$ b)$ Find $ A^n, n \in \mathbb{N}^*$

for $ b)$:  $ a^2\plus{}b^2\equal{}c^2\plus{}d^2$ is useless.
With the help of a permutation of the basis vectors, we may assume:
$ A\equal{}diag(\begin{pmatrix}a&b\\\minus{}b&a\end{pmatrix},\begin{pmatrix}c&d\\\minus{}d&c\end{pmatrix})$.
After it's easy.


calculus
integration
algebra
partial fractions
real analysis
real analysis solved

Calculate:
\[
\displaystyle\sum_{s=2}^{\infty} \sum_{n=2}^{ \infty} \frac {1} {n^s}
\]
\[
(n,s \in N)
\]

[quote="Nené"]
$\sum_{s=2}^{\infty} \sum_{n=2}^{ \infty} \frac {1} {n^s}$[/quote]

All term are positive we can use Fubini theorem

$\sum_{n=2}^{\infty} \sum_{s=2}^{ \infty} \frac {1} {n^s}
=\sum_{n=2}^{\infty}\frac{1}{n^2-1} $

[quote="Moubinool"]$\zeta(s)=\sum_{n=2}^{ \infty} \frac {1} {n^s}$[/quote]
No, $\sum_{n=2}^{ \infty} \frac {1} {n^s}=\zeta(s)-1.$ That "-1" does make a difference.

As for the problem itself, [i]of course[/i] we're going to interchange the order of summation. Wave an iterated sum, or an iterated integral, or an integral of a sum in front of my eyes, and it's always the first thought that comes to mind.

Moubinool also made a little computational error that we should fix:

$\sum_{n=2}^{\infty}\sum_{s=2}^{\infty}\frac1{n^s} = \sum_{n=2}^{\infty}\frac1{n (n-1)} = 1.$

(The last calculation is well-known, and uses partial fractions and a telescoping sum.)


[quote="sinajackson"]$ \textbf{a)}$ For reals $ x,y$ prove that:

$ \left[x^{2}\left(2x^{2} \minus{} 2x \plus{} 1\right) \plus{} 1\right]\left[y^{2}\left(2y^{2} \minus{} 2y \plus{} 1\right) \plus{} 1\right]\geq\left(2x^{2}y^{2} \minus{} xy(x \plus{} y) \plus{} 1\right)^{2}$

$ \textbf{b)}$ Find all $ n\in\mathbb{N},n > 2$ such that if for positive reals $ x_{1},x_{2},\cdots,x_{n}$ we have:

$ \text{if }n\equiv0\ (\text{mod 2})\text{ then }x_{i}\geq1(1\leq i\leq n)$
$ \text{if }n\equiv1\ (\text{mod 2})\text{ then }x_{i}\leq1(1\leq i\leq n)$

then the following ineQuality holds:

$ \left\{\prod^{n}_{i \equal{} 1}\left[x_{i}^{n}\left(2x^{n}_{i} \plus{} 2( \minus{} 1)^{n \minus{} 1}x_{i}^{n \minus{} 1} \plus{} 1\right) \plus{} n \minus{} 1\right]\right\}^{\frac {1}{n}}\geq$
$ n\prod^{n}_{i \equal{} 1}x^{2}_{i} \minus{} \left[\prod^{n}_{i \equal{} 1}x_{i}\right]\times\left[\sum^{n}_{i \equal{} 1}x_{i}\right] \plus{} n \minus{} 1$ :D[/quote]
a) By Cauchy-Schwarz inequality, we have:
$ \left[x^{2}\left(2x^{2} \minus{} 2x \plus{} 1\right) \plus{} 1\right]\left[y^{2}\left(2y^{2} \minus{} 2y \plus{} 1\right) \plus{} 1\right]\equal{}[x^4\plus{}x^2(x\minus{}1)^2\plus{}1][y^4\plus{}y^2(y\minus{}1)^2\plus{}1] \ge (x^2y^2\plus{}xy(x\minus{}1)(y\minus{}1)\plus{}1)^2\equal{}(2x^2y^2\minus{}xy(x\plus{}y)\plus{}1)^2$

Yes thats right and nice solution [b]NguyenDungTN[/b] :)
I designed the second part by using the first one and changing some conditions...Try It for $ n\equal{}3$ and $ n\equal{}4$,... and see what results you get... :)


inequalities calculus derivative function logarithms geometry integration

Suppose you want to show that equality only holds in
$xy \leq \frac{x^{p}}{p}+\frac{y^{q}}{q}$
($x,y > 0$ and $1/p+1/q = 1$) when $x^{p}= y^{q}$.  I see one way to do it by considering the derivative of each side as a function of $y$, but am I missing any simpler way?

It follows from the convexity of $-\log$ by Jensen's inequality.  A lot of textbooks just use the derivative proof though.

Or: Consider the curve $y=x^{p-1},$ which can also be written as $x=y^{q-1}.$ Consider the area under the curve plus the area to the left of the curve, in particular

$\int_{0}^{a}x^{p-1}\,dx+\int_{0}^{b}y^{q-1}\,dy.$

These areas combine to include the rectangle $[0,a]\times[0,b].$ If $a^{p-1}=b,$ then it is precisely that rectangle; in any other case, it is a set larger than that rectangle.

$a^{p-1}=b$ is the same condition as $a^{p}=b^{\frac{p}{p-1}}=b^{q}.$


Contradiction over range $x \varepsilon (-1.5,-0.5),(0.5,1.5)$.

[quote="Srilatha"]Is |x| <1.5 contradiction to |x|<=0.5?[/quote]

NO


geometry
circumcircle
trigonometry
geometry unsolved

[quote="Moonmathpi496"]Let $ ABC$ be a triangle with $ AB %Error. "neqAC" is a bad command.
$. Point $ E$ is such that $ AE = BE$ and $ BE \perp BC$. Point $ F$ is such that $ AF = CF$ and $ CF \perp BC$.
Let $ D$ be the point on line $ BC$ such that $ AD$ is tangent to the circumcircle of triangle $ ABC$.
Prove that $ D,E, F$ are collinear.[/quote]
We need to prove that $ \frac {DB}{DC} = \frac {BE}{BF}$
Hence $ \Delta ADB$~$ \Delta CDA$ we get $ \frac {AD}{CD} = \frac {DB}{DA} = \frac {AB}{AC}$
Then $ \frac {AD}{CD}.\frac {DB}{DA} = \frac {AB^2}{AC^2} = \frac {DB}{DC}$
On the other hand, $ \frac {BE}{CF} = \frac {\frac {AB}{2sinB}}{\frac {AC}{2sinC}}$
By the law of sine, we have $ \frac {sinC}{sinB} = \frac {AB}{AC}$
So $ \frac {BE}{BF} = \frac {AB^2}{AC^2} = \frac {DB}{DC}$
Therefore $ D,E, F$ are collinear.

[quote="livetolove212"]
We need to prove that $ \frac {DB}{DC} \equal{} \frac {BE}{BF}$[/quote]
I believe that here you mean,  $ \frac {DB}{DC} \equal{} \frac {BE}{CF}$


MIT
college
Harvard
real analysis
function
topology
number theory

Here's the situation I might possibly face in a few years. My mom works at MIT, so I get a free ride. But I am a little more impressed with the Harvard math department, mainly because of the Number Theory options that they carry, and of course the famous/infamous Math 55. I am planning on doing math, but obviously I'm not in a position to say for sure.

But here's the dilemma: say I were to get accepted to both. Obviously, I would assume that I would be happy at MIT, they have a fantastic math department, but I would honestly rather go to Harvard. Should I go for it, or should I take my free ride at MIT?

Harvard has really good financial aid. You might want to wait and see what their aid is like and how much you'd actually be paying there.

if i am not mistaken i hear that students at MIT and Harvard are allowed to take classes at the other institution so if there are just a few classes at Harvard you would rather take I would say go to MIT and just take the classes at Harvard.
but like the post of mine I would also suggest that you wait and see what kind of financial aid you will receive.

We are in a similar situation, but in our case it is Caltech. We are strongly encouraging our son to go to Caltech. Of course, wethink he will be happy and do well there, and he agrees. We have been very upfront about the financial situation. I suggest you sit down and discuss this with your parents, if you have not already done so.

Obviously, I vote with the free ride. You are not being asked to go to Podunk City College! MIT is an excellent school.

Assuming financial aid does not obliterate any financial differences between the two schools, you are talking a lot of money. Tuition is 33K a year now, and who knows how high it will be when you graduate. Who will pay the difference if you do not go to MIT? If it is your parents, please stop and think how this might affect them in future years. If it is you, then stop and think the same thing about your future, and what you might be able to do with that 150K.

I think we are all being duped by the 'best college' scenario. I am sure there are any number of schools in which you will be happy and well educated. We have gotten into this mindset that there are a handful of prizes, and everyone else gets the booby prize. There is no perfect school, and if there is you will never know it. How can you? Hopefully, you will graduate thinking that you were at the perfect school.

And, remember - you can go somewhere else for graduate school!

[quote]you can go somewhere else for graduate school![/quote]
Which will probably happen: it's unusual (although possible) for undergraduates to stay in the same institution for the graduate school. If you like the number theory group at Harvard, that is a reason not to go there as an undergrad.

[quote="joml88"]To be honest, I'm not sure that Math 55 is all [i]that[/i] great of a class as it's often famed to be.  It's just that they have a lot of stuff packed into one class.  Instead at other schools you can just take a couple math classes.  But yeah, I guess I didn't answer your overall question yet.[/quote]

thats right.

i think i have done pretty much the entire math 55 class (both semesters) and a little bit more in my first year in mit (and i haven't been as hardcore as most of the math people)
i took 18.014-18.024 theoretical single variable calc- theoretical multivariable calc sequence ( this year, this class was really good, i've heard that other years is just like caltech theoretical calculus. but for the multivariable part, we are not using apostol's book, we are using hubbard's book which is way deeper than apostol's second volume. Also this year's professor is amazing.) We have done a bunch of linear algebra stuff in 18.024 too.
Also, I took 18.100b, the real analysis class where we do pretty much all of Rudin, (and the multivariable stuff in Rudin like implicit function theorem and inverse functiont theorem that we don't cover in 18.100b, I have learned it from 18.024, from Hubbard's and Rudin books. So I pretty much covered the entire Rudin. A
Also,  I took 18.510 which is a mathematical logic /axiomatic set theory class. which was also really good.
All this, together with some indepedent self reading on deeper topics in topology, measure theory and stuff like that, I have done more than math 55, i guess, and i'm not taking hardcore math classes.
this is just like 50% of the level of math you can do in your freshmen year.

Hmmm... well most likely math 55 still goes deeper/more hardcore into all those subjects, but one class isn't really a big deal... I mean you can always learn math whenever you want; nobody is really stopping you.

If you must, you can always x-reg Math 55 from MIT I think. But in all reality, if the math department at MIT fails to challenge you, I'm not sure going to Harvard would make a difference.


algebra function domain polynomial calculus derivative topology

should be easy/well-known: is there a fn, defined on a domain, analytic at a single point? the same question for real variable..? if yes, can that pt be the only one at which the fn is diff'able?

thank you!

In the standard use of the word "analytic", no. The word implies an open set.
A complex function can certainly be differentiable at one point without being analytic on any open set: $f(z)=|z|^{2}$ is a good example.

Ah, the answer to the last question seems to be no: if wlog f has power-series expansion around zero with r.o.c. R, then for z, z0 in D(0, R),

a_n(z^n - z0^n)/(z-z0) = polynomial in z, summing over all n we then check the sum of these polynimials converges for all z in D(0,R) (and thus ensure f is differentiable in D(0,R)). Indeed,

|a_n (z^n - z0^n)/(z-z0)| < |a_n(n-1)R^(n-1)|, and since the series

Sigma (  (n-1)*a_n*z^(n-1) ) has the same r.o.c. as the series for f at 0, the sum must converge.

Hope I'm not missing sth here.

Hmm, if by being analytic at z0 I mean f having power series of the form

Sigma a_n(z-z0)^n

(that converge in some neighbourhood of z0 and equal f(z)), then I don't see why there should be another point z1 in that neighbourhood with power series of the form

Sigma b_n(z-z1)^n

converging in some nhd of z1. f seems to be forced to be differentiable at any such z1 though...

An explanation of the initial question and an answer to myself:

When asking the question, I had in mind the following setting: say that one has a power series (wlog around 0) \[\sum_{n = 0}^\infty{a_nz^n}\] with radius of convergence $R > 0$. Then the series can be used to define a function $f : B_R(0) \subset \mathbb{C} \rightarrow  \mathbb{C}$.

If one defines "analiticity at a point" as having a power series with positive radius of convergence centered at that point, such that there exists a ball around that point on which the series give the function $f$, then it is clear that "$f$ is analytic at 0".

One can see that this function is complex differentiable on $B_R(0)$ by seeing that the series
\[\sum_{n = 1}^\infty{na_nz^{n-1}}\]
have the same radius of convergence $R$, and then by claiming that its difference from the differentiation limit is 0.

That "$f$ is analytic at every point of  $B_R(0)$", however, is probably much harder to see (I can draw the statement from here, e.g. - 5.17 of http://www.faculty.iu-bremen.de/stoll/lecture-notes/IntroductoryComplexAnalysis.pdf, but do not have a clear picture why that is true).


This is from a math contest from...Saturday.

find the smallest positive integer $ N$ such that $ 2007^N$ has at least 2007 positive factors.

Now, what I did was factor $ 2007^N$. In doing so, I got $ 3^{2N}\cdot223^N$. There are three factors here. So, I need 2007, and I have 3. Wouldn't it just be...

$ 2N\plus{}N\equal{}2007$, so $ N\equal{}669$?

I'm not sure...that looks wrong, but I can't see how to actually do it.

[hide]
Your formula's a little messed up.

It should be $ (2N\plus{}1) \cdot (N\plus{}1)  \ge 2007$
Which gives $ 2N^2 \plus{} 3N \minus{} 2006 \ge 0$

... well I can't do it because I don't have my calculator on me and I'm too lazy to do it by hand.  :lol:
[/hide]

Why do you add one to each exponent, and why do you multiply the two?

You add one because the two can either be to the power of 1, 2, and don't forget 0! (exclamation point, not factorial). Same for 5 (0,1...,4,5), and same for N (0,1,2...N). So... yepyep :)

And then it's just simple counting from there.

I'm sorry, but I still don't see why you multiply the two...

okay, so i have five different types of sweaters, and 4 different types of pants. If I want to wear one sweater-pant combo, how many choices do I have?

It's 4*5, which is the same as this, except 'types of _____' becomes 'powers of _'

Basically, how many choices do i have for the first number, how many for the second, the third, etc. etc.

Ah...that makes sense. So, those binomials with N's in them are the possibilities for the exponent, right? Then, to find all the possible combinations of the exponents, you multiply them together, and there have  to be more than 2007 ways, so...basically, I think I get it.

Thanks for your help :D

This was on the team round of a competition I did saturday as well. GPML?


See http://www.mathlinks.ro/Forum/viewtopic.php?t=23543 (beginning from post #6).

  darij


geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations

[quote="Luba1123"]$ 23 \equal{} 2^3 \plus{} 3^2 \plus{} 2*3 \equal{} 3^3 \minus{} 2^2$[/quote]
I don't understand your question? You just wrote 23 as a difference of cubes?


analytic geometry
search
combinatorics unsolved
combinatorics

Let AB be a segment of length 2003 in a coordinate plane. Determine the maximum possible number of unit squares with vertices in the lattice points whose intersection with the segment AB is non-empty.

can somebody at least explain what lattice points mean?

Well, if [i]that's[/i] what your problem is, you should have asked (or actually, you should have just put it into your favorite search engine).  A lattice point is a point with integer coordinates.  The question is asking, if you place a segment of length 2003 on a sheet of graph paper, at most how many squares can it cross?

Also, if you don't even understand the problem, what could possibly cause you to include the word "easy" in the title?

well its task for 9th grade in Serbia
and besides that I wanted someone to answer it but I get it that
more people answer if you put hard in the name of problem :)
And the third reason my teacher solved it and said me that its nice one
but when he start to explain me I found him a mistake so I don't know a solution

Task is from
http://srb.imomath.com/zadaci/federal_2003.pdf


LaTeX
induction
algebra proposed
algebra

Let $ (a_n)$ be a sequence of natural numbers defined in this way: $ a_1\equal{}t$ and $ a_n\equal{}i(a_1)\plus{}i(a_2)\plus{}...\plus{}i(a_{n\minus{}1})$ for $ n>1$. $ i(x)$ denotes the number of digits $ 1$ of a natural number $ x$ (in decimal notation). Is it true that, for every choice of $ t$, after finitely many members, this sequence will become constant?

[size=59]If somehow this doesn't fit in Algebra and you think it's more of a combinatorial problem, I'm sorry. :) [/size]

First we deduce that the sequence is nondecreasing. Second, if at some point $ a_k\equal{}a_{k\plus{}1}$ then the sequence becomes constant after that point (can be trivialy checked). Suppose we do not have such a situation, this means that the sequence will be monotone decreasing and since it is defined to be on natural numbers we have a contradiction.

Sorry, RED-and-black, but I do not fully understand your argument.

I understand [i]First we deduce that the sequence is nondecreasing. Second, if at some point $ a_k \equal{} a_{k \plus{} 1}$ then the sequence becomes constant after that point (can be trivialy checked)[/i], but I don't understand why, when we suppose we do not have such a situation, the sequence would be monotone decreasing. Could you please explain in more detail?

I'm sorry but I gave a wrong solution, I wrote it late night. My second statement should be: suppose that we don't have such a situation, so suppose the sequence is increasing now. The statements I gave are trivial, I do not think they help a lot for the solution. Sorry for my stupid mistake. :blush:

:D Never mind.

OK, anybody else wantin' to try?

This time I think I have a non stupid solution. It is a bit long and I can't write it all in Latex because nobody would understand, Anyway I'll try to give the idea. My answer to this is yes, the sequence will become constant at some point. My main aproach is induction on the number of digits of t. If t has one digit than it is trivial. Suppose that any sequence with such formula and t with k digits becomes constant after finitely many members. Now take $ t_1$ to have k+1 digits. Then try to put some relations betwen $ a_k$ (the sequence generated by $ t_1$ without the unit digit) and $ b_k$ (the sequence generated by $ t_1$). Using the induction assumpion we have that $ i(a_k)\equal{}0$ at some point. With some manipulations we will come to a point that $ i(b_m)\equal{}0, 1 or 2$. Then we consider these cases. Studying these cases is not so easy but I think my solution is right. I would like to have just a confermation of the answer, so if somebody has a solution or soled it I would like to see another solution. Maybe I missed something and got the wrong answer. Sorry for not explaining the steps

Hmm... I was able to follow you right to the sentence "Then try to put..." I've lost you after that, sorry. It could make sense, but I really can't say from your explanation. :D

I've heard the solution (yes, I know, what kind of a guy thinks of a problem, but doesn't know how to solve it :D - call it too much time during university classes), but I wasn't listening very carefully. If I can remember correctly, the solution wasn't using induction.

Sorry I couldn't be of much help.  :oops:

We notice $ a_n\equal{}a_{n\minus{}1}\plus{}i(a_{n\minus{}1})$ for $ n>2$.
Define $ b_n$ as the position of leftmost $ 1$ in $ a_n$. For example, $ b_n\equal{}7$ if $ a_n\equal{}21234567$; $ b_n\equal{}3$ for $ a_n\equal{}2111$; $ b_n\equal{}0$ if $ a_n\equal{}234$.Obviously the increment of sequence $ i(a_{n\minus{}1})\leq b_{n\minus{}1}$. $ a_n$ will be strictly increasing for $ n>2$ unless $ a_{n\minus{}1}$ does not have digit $ 1$. $ b_n$ will be constant for $ n>2$ up to a finite $ n$ where $ b_n\leq b_{n\minus{}1}\minus{}1$. In other words, $ b_n$ is decreasing and cannot stay at a nonzero number. Hence, $ b_n$ will reduct to $ 0$ for some finite $ n$. After that $ a_n$ is constant.
Hopefully this will make sense.

[quote="xxp2000"]Hopefully this will make sense.[/quote]
You solution is OK. I hope that you won't mind if I add a few details.

In fact, it is only left to rigorously prove that the sequence $ (b_n)_{n\in\mathbb{N}}$, where $ b_n$ is defined as the position of the leftmost $ 1$ in $ a_n$, is nonincreasing. Since $ a_n\equal{}\overline{\underbrace{c_lc_{l\minus{}1}\cdots c_{b_n\plus{}1}c_{b_n}}_{\text{no }1\text{s here}}1\underbrace{c_{b_n\minus{}2}c_{b_n\minus{}3}\cdots c_1c_0}_{\text{a total of }b_n\minus{}1\text{ digits}}}$, the only situation in which $ b_{n\plus{}1}$ could possibly be different from $ b_n$ occurs when $ a_{n\plus{}1}$ becomes greater than $ \overline{c_lc_{l\minus{}1}\cdots c_{b_n\plus{}1}c_{b_n}199\cdots 99}$ (what is guaranteed to happen since the sequence $ (a_n)$ is strictly increasing so far). In this situation $ b_{n\plus{}1}$ could become greater than $ b_n$, stay the same as it, or become less than it. We are going to prove that the first two cases are impossible. Indeed, they would require at least one of $ c_i$ where $ b_n\leqslant i\leqslant l$ to change, what would give \[ \begin{array}{r@{\;}l}
a_{n\plus{}1}\minus{}a_n&\geqslant\overline{c_lc_{l\minus{}1}\cdots c_{b_n\plus{}1}(c_{b_n}\plus{}1)000\cdots 00}\minus{}\overline{c_lc_{l\minus{}1}\cdots c_{b_n\plus{}1}c_{b_n}199\cdots 99}\equal{}\\
\noalign{\bigskip}
&\equal{}\left((c_{b_n}\plus{}1)\cdot 10^{b_n}\plus{}\sum\limits_{i\equal{}b_n\plus{}1}^lc_i10^i\right)\minus{}\left(\sum\limits_{i\equal{}0}^{b_n\minus{}2}9\cdot 10^i\plus{}10^{b_n\minus{}1}\plus{}\sum\limits_{i\equal{}b_n}^lc_i10^i\right)\equal{}\\
\noalign{\bigskip}
&\equal{}10^{b_n}\minus{}10^{b_n\minus{}1}\minus{}\sum\limits_{i\equal{}0}^{b_n\minus{}2}9\cdot 10^i\equal{}9\cdot 10^{b_n\minus{}1}\minus{}\sum\limits_{i\equal{}0}^{b_n\minus{}2}9\cdot 10^i\equal{}\\
\noalign{\bigskip}
&\equal{}8\cdot 10^{b_n\minus{}1}\plus{}\left(10^{b_n\minus{}1}\minus{}\sum\limits_{i\equal{}0}^{b_n\minus{}2}9\cdot 10^i\right)\equal{}8\cdot 10^{b_n\minus{}1}\plus{}1>\\
\noalign{\bigskip}
&>b_n\geqslant i(a_n)\equal{}a_{n\plus{}1}\minus{}a_n,\end{array}\]what is clearly a contradiction. Therefore, in each of these occasions it holds $ b_n>b_{n\plus{}1}$, what means that the sequence $ (b_n)_{n\in\mathbb{N}}$ becomes $ 0$ from some point onward. Since $ i(a_n)\leqslant b_n$, the sequence $ (i(a_n))_{n\in\mathbb{N}}$ also becomes $ 0$ from some point onward, and that is when the sequence $ (a_n)_{n\in\mathbb{N}}$ becomes constant.


1)A man buys drinks from two manufacturers:
    One bottle of drink A made up from alcohol and water in the ratio 1:3.
    One bottle of drink B made up from alcohol and water in the ratio 1:7.
    What is the least number of bottles he would need of each type of drink in order to produce drinks containing alcohol and water in the following ratios:
    (i)  2:9
    (ii) 3:14
    (iii)10:43
    (Assume the amount of one bottle of drinks A is same as the amount of one bottle of drinks B)

2)Mr. Weasley just received his bonus from Ministry for Magic. The bonus was paid in a large pile of galleons (gold coins), which he decided to divide equally among his 3 sons, Fred, George and Ron in the next morning. During the night, Fred awoke and decided to help himself to his share of the galleons. In dividing them into three equal parts he found that there was one galleon left over. This galleon he gave to Ginny (his sister). He then hid his one-third share, leaving the rest in a single pile. Later during the night, George awoke with the same idea in mind. He went to the pile divided it into three equal parts, and found that there was one galleon left over; this he gave to Ginny and then he hid his one-third share, restoring the rest to one pile. During the same night, Ron arose, and ignorance of what had happened previously, went to the pile, and followed the same procedure. Again one galleon was left over, and it was given to Ginny. The next morning, Mr. Weasley divided the remaining galleons evenly among his 3 sons and left one galleon. What is the least number of galleons the original pile could have contained?

3)If A is a 2 x 3 matrix, B is a 4 x 3 matrix and C is a 3 x 3 matrix, and D is a 3 x 2 matrix, determine the dimension of:

    (i)	A*C matrix
    (ii)	D*A matrix
    (iii)	A*D matrix
    (iv)	B*C matrix

4)The monthly sales (in thousands) of burger (B1) and bites (B2) in three fast food restaurants (R1, R2, R3) are as follows:

January					                February
	R1	R2	R3				R1	R2	R3
B1	35	27	13			B1	31	17	3
B2	42	39	24			B2	25	29	16

(i)	Write down two   matrices, J and F, representing sales in January and February respectively.

(ii)	Write down the matrix for the total sales over the two months.

5)A firm manufactures three products P1, P2 and P3 and sells each of them at $100, $250 and $300 respectively.

    (i)Express this information as   matrix and label it as A.

    (ii)Customer 1 wishes to buy 3 articles of type P1, 4 of type P2 and 3 of type P3. Customer 2 wishes to buy 1 article of type P1, 2 of type P2 and 5 of type P3. Express this information as a   matrix and label it as B.

    (iii)Calculate BA, and find the total amount spent by each of the two customers

    Above there are the problems that I cannot solve...hope anybody can help me...thanks...

pls...anybody help...i really cant solve tat problems...

People here tend to look down upon people who type up homework problems and expect us to solve them. I'd link to something pretentious written by a responsible adult, but I'll let that responsibility slide to the next person who wants to point this out.

That Harry Potter-related problem is pretty funny. Though Mr. Weasley is supposed to be a poor bum, so it's not exactly accurate.

I defer the responsibility Lazarus mentioned to the next person.

tat's not homework...
juz exercise given by my brother.....
he ask me 2 do it correct...
coz we bet it...
i didnt hope 2 lose..
coz i have already lose many times...
so,juz ask u guys 2 help me...
but,if u all cant,juz skip it
oko,juz dun mention i hav ask tat ques....


Prove that in a triangle: \[ \frac{\rho_a}{s\minus{}a}\plus{}\frac{\rho_b}{s\minus{}b}\plus{}\frac{\rho_c}{s\minus{}c}\ge \frac s\rho\]
where $ \rho_i$ is the radius of the excircle opposite to side $ i$ and $ \rho$ is the radius of the incircle.

Let $ J$ be the area of the triangle.

With a little elemental geometry, we can find out that $ \rho_a \equal{} \rho\frac {s}{s \minus{} a} \equal{} \frac {J}{s \minus{} a}$

$ \sum{\frac {\rho_a}{s \minus{} a}}\ge\frac {s}{\rho} \Leftrightarrow \sum{\frac {J^2}{(s \minus{} a)^2}}\ge s^2 \Leftrightarrow \sum{\frac {(s \minus{} b)(s \minus{} c)}{s \minus{} a}}\ge s$

Wich is true by AM-GM, taking two terms at time and suming the three inequalities.

 :)


[quote="black_puma"]$a,b,c >0$. Prove the inequality :
$\sum\frac{a^2}{b^2+c^2} \geq \sum\frac{a}{b+c}$[/quote]

yeah very nice.....

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=65049[/url]


Help me to solve some problems below:
1/Let a,b,c>0.Find all the numbers x,y,z>0 satisfies that:
x+y+z=a+b+c
4xyz-(a^2.x+b^2.y+c^2.z)=abc
2/Find in all roots (x;y;z;t) of this system of equations:
x^2+y^2=9
z^2+t^2=16
xt+yz>=12
the root satisfies(x+y) has maximum value
3/Solve the system of equations below:
x^3-3x=y(3x^2-1)
y^3-3y=z(3y^2-1)
z^3-3z=x(3z^2-1)

[hide="2"]
By HM-AM, $ \sqrt{\frac{x^2\plus{}y^2}2}\equal{}\sqrt{\frac92}\equal{}\frac{3\sqrt2}2\geq\frac{x\plus{}y}2\implies x\plus{}y\leq3\sqrt2$, which occurs when $ x\equal{}y\equal{}\frac{3\sqrt2}2$.

Since we must have $ x\plus{}y$ at its maximum, this must be the case.

Now, also by HM-AH, we have

$ \sqrt{\frac{t^2\plus{}z^2}2}\equal{}\sqrt{\frac{16}2}\equal{}2\sqrt2\geq\frac{t\plus{}z}2\implies t\plus{}z\leq4\sqrt2$, and

$ xt\plus{}yz\equal{}\frac{3\sqrt2(t\plus{}z)}2\geq12\implies t\plus{}z\geq4\sqrt2$.

In order for $ 4\sqrt2\geq t\plus{}z\geq4\sqrt2$, we must have $ t\plus{}z\equal{}4\sqrt2$, which is the equality condition of HM-GM, so

$ t\equal{}z\equal{}2\sqrt2\implies(x,y,z,t)\equal{}\boxed{\left(\frac{3\sqrt2}2,\frac{3\sqrt2}2,2\sqrt2,2\sqrt2\right)}$.
[/hide]


Could you please explain how I should reason about this type of problem :

[quote]Suppose that f(n)=2f(n/2)+3 when n is even and f(1)=5
find f(1024)[/quote]

Suppose that $f(n)=2f(\frac{n}{2})+3$ when $n$ is even and $f(1)=5$
find $f(1024)$

Well, when I see this problem, my thought process would go something like this:

$n$ must be even and $f(n)$ is dependent on whatever $f(\frac{n}{2})$ is. Since $n$ is even, it is evenly divisible by 2. Since we know that $f(1)=5$, we can work backwards and find $f(2)$.

1024 is also $2^{10}$, a power of 2, so we can keep doubling $n$ until we get to 1024. In this way, we can find $f(2), f(4), f(8), f(16), f(32), f(64), f(128), f(256), f(512)$ and finally, $f(1024)$.

We just have to double whatever $f(\frac{n}{2})$ is and then add 3, so:

$f(1)=5$
$f(2)=13$ (double then add 3)
$f(4)=29$
$f(8)=61$
$f(16)=125$
$f(32)=253$
$f(64)=509$
$f(128)=1021$
$f(256)=2045$
$f(512)=4093$
$f(1024)=8189$

and that's our answer. It's a bit long, and there might be a more elegant way to solve it, but that's the way I thought of it.

Also, you can also notice that $f(1)=2^3-3, f(2)=2^4-3$ etc. In this way, we can find $f(1024)$ simply by evaluating $2^{13}-3=8198$ and this is a shortcut.


algebra
polynomial
superior algebra
superior algebra solved

In general $det(A+B) \neq det(A)+det(B)$, but under some conditions it is true

$A,B\in M_{3}(R)$, $AB=BA$, suppose also $A^2+AB+B^2$ is not invertible

Prove that $det(A+B)=det(A)+det(B)$

Put $P(x)=\det(A-xB)$. We get $P(\varepsilon)P(\bar\varepsilon)=0\Rightarrow P(\varepsilon)=P(\bar\varepsilon)=0$, where $\varepsilon,\bar\varepsilon$ are the roots of $x^2+x+1$. This means that $P(A-xB)$ is a third degree polynomial divisible by $x^2+x+1$. Assume $P(x)=(-\det Bx+t)(x^2+x+1)$. We get $\det(A+B)=P(-1)=\det B+t=\det B+\det A$.


superior algebra
superior algebra unsolved

This is nice:

Let $ G$ be a group and let $ x \in G$ such that $ x$ has order $ mn$ where $ m$ and $ n$ are coprime.

Show that $ x = yz$, where $ y$ and $ z$ have order $ m$ and $ n$ respectively, and $ yz = zy$.

immediate from $ <x> \cong \mathbb{Z}/(nm) \cong \mathbb{Z}/n \times \mathbb{Z}/m$.

if you want to be explicit: take $ u,v$ s.t. $ um+vn=1$ and put $ y=x^{vn}, z = x^{um}$.


geometry
circumcircle
incenter
function
trigonometry
geometry unsolved

Given is triangle $ \triangle ABC$. Let $ O$ and $ I$ be it's circumcenter and incenter, respectively.

If line $ IO$ cuts sides $ AB$ and $ AC$ at points $ D$ and $ E$,respectively, calculate $ DE$ in function of sides and angles of triangle $ \triangle ABC$.

I have a solution but its too long for writing,it uses a lot of massive trigonometry   :D


[quote="Vasc"]Let $ x,y,z$ be real numbers. Prove that

$ (a) \ \ \ \ \sum \frac 1{(x \minus{} y)^2} \ge \frac 9{2(x^2 \plus{} y^2 \plus{} z^2)}$;

[/quote]

We have : $ (\sum x^2)(\sum \frac{1}{(x\minus{}y)^2}) \geq 5\plus{}2\sum \frac{xy}{(x\minus{}y)^2} \geq \frac{9}{2}$

Done ! :)

[quote="Vasc"]Let $ x,y,z$ be real numbers. Prove that

$ (a) \ \ \ \ \sum \frac 1{(a \minus{} b)^2} \ge \frac 9{2(a^2 \plus{} b^2 \plus{} c^2)}$;[/quote]

Let :  $ a \equal{} x\plus{}a_1$ , $ b \equal{} x \plus{}b_1$ , $ c\equal{} x\plus{}c_1$

1. $ min(a^2\plus{}b^2\plus{}c^2)$ holdes for $ x\equal{}\minus{}\frac{a_1\plus{}b_1\plus{}c_1}{3} \Rightarrow a\plus{}b\plus{}c\equal{}0$;

2. We can assume : $ a \le b\le c$ and $ a\equal{}\minus{}1$;

So : $ min(LHS\minus{}RHS)$ holdes for : $ c\equal{}t$  , $ b\equal{}1 \minus{} t$ , $ a\equal{}\minus{}1$     $ \boxed{\frac{1}{2} \le t \le 2}$

$ min(LHS \minus{} RHS)\equal{} min(\frac{1}{(2\minus{}t)^2}\plus{}\frac{1}{(2t\minus{}1)^2}\plus{}\frac{1}{(1\plus{}t)^2}\minus{}\frac{9}{4(1\minus{}t\plus{}t^2)})\equal{}0$ and holdes for $ t\equal{}1 \Leftrightarrow a\equal{}\minus{}1$ , $ b\equal{}0$ , $ c\equal{}1$;

[quote="Sergic Primazon"][quote="Vasc"]Let $ x,y,z$ be real numbers. Prove that

$ (a) \ \ \ \ \sum \frac 1{(a \minus{} b)^2} \ge \frac 9{2(a^2 \plus{} b^2 \plus{} c^2)}$;[/quote]

Let :  $ a \equal{} x \plus{} a_1$ , $ b \equal{} x \plus{} b_1$ , $ c \equal{} x \plus{} c_1$

1. $ min(a^2 \plus{} b^2 \plus{} c^2)$ holdes for $ x \equal{} \minus{} \frac {a_1 \plus{} b_1 \plus{} c_1}{3} \Rightarrow a \plus{} b \plus{} c \equal{} 0$;

2. We can assume : $ a \le b\le c$ and $ a \equal{} \minus{} 1$;

So : $ min(LHS \minus{} RHS)$ holdes for : $ c \equal{} t$  , $ b \equal{} 1 \minus{} t$ , $ a \equal{} \minus{} 1$     $ \boxed{\frac {1}{2} \le t \le 2}$

$ min(LHS \minus{} RHS) \equal{} min(\frac {1}{(2 \minus{} t)^2} \plus{} \frac {1}{(2t \minus{} 1)^2} \plus{} \frac {1}{(1 \plus{} t)^2} \minus{} \frac {9}{4(1 \minus{} t \plus{} t^2)}) \equal{} 0$ and holdes for $ t \equal{} 1 \Leftrightarrow a \equal{} \minus{} 1$ , $ b \equal{} 0$ , $ c \equal{} 1$;[/quote]
another,
let real numbers $ a_1  \le a_2  \le a_3  \le ... \le a_n$. Prove that
$ (\sum\limits_{i \equal{} 1}^n {a_i ^2 }  \minus{} \sum\limits_{i \equal{} 1;a_{n \plus{} 1}  \equal{} a_1 }^n {a_i a_{i \plus{} 1} } )(\sum\limits_{i \equal{} 1;a_{n \plus{} 1}  \equal{} a_1 }^n {\frac{1}{{(a_i  \minus{} a_{i \plus{} 1} )^2 }}} ) \ge \frac{{n^2 (n^2  \minus{} 3n \plus{} 3)}}{{2(n \minus{} 1)}}$

[quote="Vasc"]Let $ x,y,z$ be real numbers. Prove that

$ (a) \ \ \ \ \sum \frac 1{(x \minus{} y)^2} \ge \frac 9{2(x^2 \plus{} y^2 \plus{} z^2)}$;

$ (b) \ \ \ \ \sum (x^2 \minus{} y^2)(x^2 \minus{} z^2) \ge \frac {(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{2(x^2 \plus{} y^2 \plus{} z^2)}$;

$ (c) \ \ \ \ \sum x^2(x \minus{} y)(x \minus{} z) \ge \frac {2(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{x^2 \plus{} y^2 \plus{} z^2}$.[/quote]
The general form:

If $ k$ and $ x,y,z$ are real numbers, then

$ \ \ \ \ \sum (x \minus{} y)(x \minus{} z)(x\minus{}2ky)(x\minus{}2kz) \ge \frac {2(k\plus{}1)^2(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{x^2 \plus{} y^2 \plus{} z^2}$.


limit integration inequalities logarithms calculus trigonometry function

Find the next limit:
$\lim_{n \to \infty}{\sum_{i=1}^{n}{ln\frac{n+i}{n}}arctg{\frac{n}{n^{2}+i^{2}-i}}}$

The limit is $I=\int_{0}^{1}\frac{\ln(1+x)}{1+x^{2}}dx$.
Let $a_{n}$ be your sequence.
Then based on the well known inequality
$x-\frac{x^{3}}{3}<\arctan x <x$ we get that the limit is the same as the limit of the sequence
$b_{n}=\sum_{i=1}^{n}\ln \frac{n+i}{n}\frac{n}{n^{2}+i^{2}-i}$.
Now observe that

$\frac{n}{n^{2}+(i-1)^{2}}>\frac{n}{n^{2}+i^{2}-i}>\frac{n}{n^{2}+i^{2}}$
therefore we get that
$\frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+(i-1)^{2}/n^{2}}\ln(1+i/n)>b_{n}>\frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+i^{2}/n^{2}}\ln \frac{n+i}{n}$
The left hand side and the right hand side sequences converge to $I$.
Of course, the integral can be evaluated.

The integral equals $\frac{\pi}{8}\ln 2$ and it can be evaluated via the substitution $x=\tan y$.

hey didi

is the inequality u have written a strict inequality??
i think u missed out the equal to signs !! :maybe:  :maybe:
anyways i thought that it was x + x^3 /3  which was greater than arctanx
which is correct

arctanx<=x
or  arctanx<=x +x^3 /3
 :|

Hi,
It should be $x-\frac{x^{3}}{3}\leq \arctan x\leq x$ for $0\leq x$.
You are right, however for this case please observe that $x\neq 0$, thus we have strict inequality.

Nice. Actually this is a Riemann summ, of the function
$f(x)=ln(1+tgx)$ on the interval $[0,\frac{\pi}{4}]$ with the division $\{0=arctg\frac{0}{n}<arctg\frac{1}{n}<...<arctg\frac{n}{n}=\frac{\pi}{4}\}$


calculus
integration
limit
calculus computations

Find $ \lim_{n\to\infty} n^2\int_{\frac{2}{n\plus{}2}}^1 \left|\left(x\minus{}\frac{1}{n\plus{}2}\right)(x\minus{}1)^n\right|\ dx$

[quote="kunny"]Your answer is correct, sylow_theory. :)[/quote]
So do I get in the university   :P

From the above solution I don't see how $ n$ being even makes a difference.


algebra
polynomial
real analysis
real analysis unsolved

How to show that there exists a positive integer $ n'$ such that for all $ n > n'$ we have $ \frac{81^{n \plus{} 2}}{(n \plus{} 1)!} < \frac{1}{10}$? Furthermore, give an example of a suitable $ n$.

notice for example that for $ n > 82$:
\[ n! \geq 82! 82^{n\minus{}82}\]

One doesn't need Stirling's approximation.
Just take $ k \equal{} 81^2 \minus{} 1$, $ n'\equal{}2k\plus{}1$. Then, for $ n > n'$, one has $ (n \plus{} 1)! > 81^{2(n \minus{} k \plus{} 1)} > 81^{n \plus{} 3} > 10\cdot 81^{n \plus{} 2}$, so $ \frac {81^{n \plus{} 2}} {(n \plus{} 1)!} < \frac {1} {10}$.

Thanks. I tried to do the problem in http://www.mathlinks.ro/Forum/viewtopic.php?t=298746 without computer but it looks like it is too difficult to evaluate a polynomial of degree 83 by pen and paper.

83rd degree polynomial? Come on, now. Ingest the spirit of the quantifier in the question - you were asked to prove that there exists some $ n$ that makes this true. It doesn't have to be the best possible $ n.$ Don't be afraid of crude estimates; don't be afraid of overkill. It doesn't matter if the $ n$ you report is way bigger than it needs to be. Of course, that's exactly what mavropnevma did - made a crude estimate.


logarithms
inequalities proposed
inequalities

Let $ a$ and $ b$ be positive real numbers such that $ a\plus{}b\equal{}2$. Prove that

$ a^{\minus{}a}\plus{}b^{\minus{}b} \le 2$.


calculus
real analysis
geometry
geometric transformation
rudin
analytic geometry
real analysis theorems

Could everybody recommend me some good textbooks on calculus (and also analysis) for students who major in mathematics?

As a freshman majoring in mathematics, I read "Principles of Mathematical Analysis" by Rudin (in Russian translation :)  ). Very few books had as much influence on me as that one.

My favorite Calculus text is "Schaum's Outline of Advanced Calculus" by M. Speigel (spelling?) 2nd ed. you can buy for ~$\$ 15$.

[quote="ifai"]Otherwise, what is the difference between calculus, analysis and advanced calculus?[/quote]

Good question! Well, if calculus is say easy, then advanced calculus is something like a little challenging, whereas real analysis is more like of very hard (assuming you use baby Rudin [a.k.a. blue Rudin]). Outside of that, well, analysis requires rigor (lots of epsilon-delta style proofs, and almost every homework problem is proof) whereas calculus is mostly calculations (in most classes it seems).

Is Spivak's [i]Calculus[/i] a book for calculus or elementary analysis?

Many schools these days use the term "Advanced Calculus" to refer to their Analysis courses, so in many uses of the term advanced calculus is the same thing as analysis. Otherwise, I agree with the above differences.

As far as texts go, there will always be a place in my heart for the Stewart [i] Calculus [/i] text. I haven't found a really good introductory analysis text that I like. The one our school uses (it is little known I think) isn't bad (lots of good problems at all levels) is Thomson, Bruckner and Bruckner's [i] Elementary Real Analysis [/i]

And how about the [i]Introduction to Calculus and Analysis[/i] by R. Courant and F. John?

there is a book written by courant on this
and you can find it publicly legally available here
[url=http://kr.cs.ait.ac.th/~radok/math/mat6/startdiall.htm]volume1[/url]           [url=http://kr.cs.ait.ac.th/~radok/math/mat9/startall.htm#Differential%20and%20Integral%20Calculus]volume2[/url]

Have you read both the [i]Calculus and Analytic Geometry 9e[/i] and [i]Thomas' Calculus 11e[/i] written by Thomas (and others)? What are the differences between them?

The analysis course taken in my first year had great influence on me, and the course notes are here: http://www.math.ust.hk/~mamyan/ma203/lecture.pdf


geometry
perimeter
inequalities
geometry proposed

A triangle with perimeter $2p$ is inscribed in a circle of radious $R$ and also circumscribed
on a circle with radious $r$.
Prove that $p<2(R+r)$

[i]Geometric ineqs are hard!
They have smart manipulations,
that sometimes one with great
difficulty finds!!!!! :) [/i]

Let $a,b,c$ be the side length of a triangle, and let $S$ be the area of that.

$2(R+r)-p=\frac{abc}{2S}+\frac{4S}{a+b+c}-\frac{a+b+c}{2}\geq 2\sqrt{\frac{2abc}{a+b+c}}-\frac{a+b+c}{2}$.

We are to prove that $(a+b+c)^3<32abc$.

[quote="levi"]For the Gergogne's  point $\Gamma$ and the orthocentre $H$ exists the relation

$H\Gamma =4R^2\cdot \left[ 1-\frac{2s^2(2R-r)}{r(4R+r)^2)}\right] \ge 0$, i.e.

$s^2\le \frac{R(4R+r)^2}{2(2R-r)}$. But $,\ \frac{R(4R+r)^2}{2(2R-r)}{\le} (2R+mr)^2\ \  (R=t\cdot r,\ t\ge 2)\ \Longleftrightarrow $

$t(4t+1)^2\le 2(2t-1)(2t+m)^2\Longleftrightarrow$

$ f(t)\equiv 16(m-1)t^2+(4m^2-8m-1)t-2m^2\ge 0,\ f(2)\ge 0\Longleftrightarrow$

$m^2+8m-11\ge 0\Longleftrightarrow m\ge -4+3\sqrt 3$. Thus

$,\ \overline {\underline {\left| s\le 2\cdot R+(-4+3\sqrt 3 )\cdot r<2\cdot (R+r)\right| }}.$

[b]Remark.[/b] $-4+3\sqrt 3 <\frac 43\Longrightarrow s<2\cdot R+\frac 43\cdot r.$[/quote]
\[\underline {\overline{\left| \ \ s-3\sqrt 3 \cdot r\le 2\cdot (R-2r)\ \ (\ Blundon\ )\ \right| }}\]

Levi,the last inequality is Blundon's. ;)

Ave, Caesar ! See you at all beauty I became to Blundon's inequality ? But many inequalities has this Blundon !
In 15 august I am in Bucuresti and ....at the seaside of Black Sea (Mamaia) with the my  grandsan (two, 3 / 5 years).


The emoticons topic was more popular than each of my math topics :o

 :lol: and :| are competing in a contest called "The Worst Smiley Ever!" They take a test with 100 math problems: 20 Algebra, 20 Geometry, 20 Calculus, and 40 Discrete Mathematics.  :lol: gets 5% of the Algebra questions right, 8% of the Geo questions right, 1% of the Calculus questions right, and 9% of the DM questions right.  :| gets 3% of the Algebra questions right, 13.37% of the Geo questions right, 0% Calculus, and 1% Discrete Math. Who is the Worst Smiley Ever?

Please do this without a calculator, if you're doing this problem. :D

That's because your math topics were pretty boring, in my opinion. Bah I should leave and go do actual math now.

Also, the math problem you just posted is almost an exact copy of another Mathcounts problem, and it doesn't really interest me at all. That's the bad thing about Mathcounts; there are only a few problems that are actually nice and elegant.

[quote="Temperal"]whoa nice pure computation, gauss.

that's what I like to see: back to the basics.[/quote]
 :?:


The entry form asks us to agree to the "Terms and Conditions."  Where are these terms?  Are they just the rules?

I guess so. Just look carefully around the USAMTS website.


sum to infinity:
[img]http://www.torontosop.com/q2.gif[/img]

the simple pattern i found for the nth term is:
[img]http://www.torontosop.com/untitled2.gif[/img]

Can anyone please help me out in this question?

[quote="metru"]$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{{k\left( {k + 1} \right)\left( {k + 3} \right)}}}  = $

$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{6} \cdot \left[ {\frac{2}{k} - \frac{3}{{k + 1}} + \frac{1}{{k + 3}}} \right]}  = $

$ = \mathop {\lim }\limits_{n \to \infty } \frac{1}{6}\left[ {\left( {\frac{2}{1} - \frac{3}{2} + \frac{1}{4}} \right) + \left( {\frac{2}{2} - \frac{3}{3} + \frac{1}{5}} \right) + } \right.$

$ + \left( {\frac{2}{3} - \frac{3}{4} + \frac{1}{6}} \right) + \left( {\frac{2}{4} - \frac{3}{5} + \frac{1}{7}} \right) +  \cdots  + \left. {\left( {\frac{2}{n} - \frac{3}{{n + 1}} + \frac{1}{{n + 3}}} \right)} \right] = \frac{7}{{36}}$[/quote]

hey metru, thanks for your help, i managed to get to your 3rd line while awaiting a response on the forumn but i can't seem to get 7/36. what is the next step from line 3, how do u come up with that fraction?
thanks again.

[quote="kunny"]$\frac{7}{36}$ is O.K.

$\sum_{k=1}^n \frac{1}{k(k+1)(k+3)}=\frac{k+2}{k(k+1)(k+2)(k+3)}=\frac{1}{(k+1)(k+2)(k+3)}+\frac{2}{k(k+1)(k+2)(k+3)}$

$=\frac{1}{2}\left\{\frac{1}{(k+1)(k+2)}-\frac{1}{(k+2)(k+3)}\right\}+\frac{2}{3}\left\{\frac{1}{k(k+1)(k+2)}-\frac{1}{(k+1)(k+2)(k+3)}\right\}$, so we have

$\sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+3)}=\frac{1}{2}\cdot \frac{1}{2\cdot 3}+\frac{2}{3}\cdot \frac{1}{1\cdot 2\cdot 3}=\frac{7}{36}$[/quote]

ah ic ic. :D :D thank you kunny. i have started this question:

[size=150][b]Sum to Infinity: Q2[/b][/size]

$\frac{x}{1 \times 2}+\frac{x^2}{2 \times 3} + \frac{x^3}{3 \times 4}$

To the nth term i got the simple pattern:

$\frac{x^n}{n (n+1)}$

so following what you showed me i have gotten this far:

$\sum_{k=1}^{\infty} \frac{x^k}{k(k+1)}= \sum_{k=1}^{\infty} \frac{1}{k(k+1)}x^k= \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{k+1} \right)x^k=
\sum_{k=1}^{\infty} \frac{1}{k}x^k - \sum_{k=1}^{\infty} \frac{1}{k+1}x^k$

I am stuck at this point and do not know how to continue :(


calculus
integration
function
trigonometry
derivative
real analysis
real analysis unsolved

there are many similar questions in this forum.i read all of them.But they don't help me for to solve this questions.
question is $\int_{0}^{\pi}x.tanx \ dx=?$

But I think the integral is not exist,consider the point $\frac{\pi}{2}$ :)

ok. do it solve without extreme point this integral $\int{x.tanx \ dx}$.

The funtion $f(x) = x\tan x$ does not appear to have an elementary anti-derivative, but can be expressed in terms of the polylogarithm funtion.  You can try it here:

[url]http://integrals.wolfram.com/index.jsp[/url]


Select any twenty consecutive three-digit whole numbers.  Divide each of them by 20 and record each remainder.  What is the sum of the twenty remainders?

A.180	B.190	C.200	D.205	E.  210

[hide=" click to reveal content"]

The remainders are $ 0, 1, 2, ..., 19$ whose sum is $ 190$.

[/hide]


geometry
rectangle
combinatorics unsolved
combinatorics

How many pieces can be placed on a $10 x 10$ board $($each at the center of its square, at most one per square$)$ so that no four pieces form a rectangle with sides parallel to the sides of the board$?$.

You can look in the Resources section  :)

http://www.mathlinks.ro/Forum/viewtopic.php?p=445662#p445662


For students taking the AMC abroad, how does the time difference work? Do they end up taking it at night? or is it the same time in their time zone?

For USAMO, atleast, you have to take it between the same time everyone else does, even if you are in China or something. This means, yes, middle of the night.

Wait you're in India right?

Last year I was in India for a couple months during AIME problem series and I had to wake up at like 3 to attend.

>.<

You have like a window for the AMCs and you have to be a legal resident of the US to take the USAMO

Yes I'm in India. So for AMC10/12, it is the same as the USAMO? I have to take it at night?

For USAMO yes. For AIME there is a special time range for it. I believe it was 8-3:30 PST. I don't know any other conversions. For AMC there is no special time range.

but if you are a legal indian, then you would not be allowed to take the USAMO, i guess you would have to take the INMO.

Which means you have a higher chance of getting somewhere


calculus
integration
calculus computations

Let be given $ f: [0,1] \times [0,1] \to \mathbb R$ defined by

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/e/d/1/ed16c654408ddc836aaa9049cb461457eb13823f.gif[/img]

Prove that $ f$ is integrable and $ \int\limits_{\left[ {0,1} \right] \times \left[ {0,1} \right]} {f dV} \equal{} \frac {1}{2}$.

Did I misunderstand the problem? It seems trivial they way I intepert it
$ \int_{0}^{1}\int_{0}^{1}f(x,y)dxdy$=$ \int_{0}^{1}.5dy$=.5


[quote="Silverfalcon"]If $AB$ and $CD$ are perpendicular diameters of circle $Q$, $P$ in $\overline{AQ}$, and $\measuredangle QPC = 60^\circ$, then the length of $PQ$ divided by the length of $AQ$ is

A. $\frac{\sqrt 3}{2}$
B. $\frac{\sqrt 3}{3}$
C. $\frac{\sqrt 2}{2}$
D. $\frac{1}{2}$
E. $\frac{2}{3}$[/quote]

[hide]
$\triangle CPQ$ is a 30-60-90 triangle.  $\angle P$ is the 60 degree angle, so the ratio of PQ to CQ is $1: \sqrt{3}$  CQ is equal to AQ since they are both radii of the same circle, so $\frac{PQ}{AQ}$ = $\frac{1}{\sqrt{3}} = \frac{\sqrt 3}{3}$

B
[/hide][/hide]


function
calculus
modular arithmetic
algebra
functional equation
algebra unsolved

Let $ f : \mathbb{N}\rightarrow \mathbb{N}$ be a function such that $ f(n \plus{} 1) > f(n)$ and $ f(f(n)) \equal{} 3n$ for all $n \in \mathbb N$. Prove that $ f(1) \equal{} 2$ and $ f(1458) \equal{} 2189$.

[quote="MNrule"]Let $ f : \mathbb{N}\rightarrow \mathbb{N}$[/quote]  This is a sure sign that your post does not belong in the calculus forum  :roll:
[color=green]Moved.[/color]

There are a very small number of things you can possibly do in this problem.  So, start doing them.  It may help you to think about what is special about 1458.

Because we know $ f(n\plus{}1) > f(n)$, we know that if $ a > b$, $ f(a) > f(b)$.

Lemma 1: $ a \le f(a) \le 3a$.
To prove this, assume $ f(a) > 3a$. Then, $ f(f(a))>3a$ which contradicts the second statement of the question. Thus, $ f(a) \le 3a$. And if $ a \neq 0$, if $ f(a) \equal{} 3a$, $ f(f(a)) > 3a$, which is a contradiction, showing that the right equality only holds when $ a \equal{} 0$. For the left side, if $ f(a) < a$, then $ f(f(a)) < f(a) < 3a$, a contradiction. Clearly, equality also only holds at $ a \equal{} 0$.

From Lemma 1, we know that $ f(0)\equal{}0$.

Now, $ 1 < f(1) < 3$, thus $ f(1) \equal{} 2$, and $ f(2) \equal{} 3$. Because $ f(f(2)) \equal{} 3(2)$, $ f(3) \equal{} 6$.

$ f(0)\equal{}0$
$ f(1)\equal{}2$
$ f(2)\equal{}3$
$ f(3)\equal{}6$

$ f(f(3))\equal{}3(3)$
$ f(6)\equal{}9$

$ f(f(6))\equal{}18$
$ f(9)\equal{}18$

$ f(f(9))\equal{}27$
$ f(18)\equal{}27$

$ f(f(18))\equal{}54$
$ f(27)\equal{}54$

$ f(f(27))\equal{}108$
$ f(54)\equal{}108$

$ f(f(54))\equal{}162$
$ f(108)\equal{}162$

$ f(f(108))\equal{}324$
$ f(162)\equal{}324$

$ f(f(162))\equal{}486$
$ f(324)\equal{}486$

$ f(f(324))\equal{}972$
$ f(486)\equal{}972$

$ f(f(486))\equal{}1458$
$ f(972)\equal{}1458$

$ f(f(972))\equal{}2916$
$ f(1458)\equal{}2916$

[quote="MNrule"]Let $ f : \mathbb{N}\rightarrow \mathbb{N}$ such that $ f(n \plus{} 1) > f(n)$ and $ f(f(n)) \equal{} 3n$, then prove that $ f(1) \equal{} 2$ and $ f(1458) \equal{} 2189$[/quote]
The problem is wrong (it's $ 2187$ instead of $ 2189$) and not an olympiad one :

$ f(1)\neq 1$ else this would imply $ f(f(1))\equal{}1$ and we know $ f(f(1))\equal{}3$
So $ f(1)\geq 2$ and, since $ f(n\plus{}1)>f(n)$, we have $ f(n)\geq n\plus{}1$ and so $ 3\equal{}f(f(1))\geq f(1)\plus{}1$, so $ f(1)\leq 2$ and so $ f(1)\equal{}2$

$ f(f(n))\equal{}3n$ $ \implies$ $ f(3n)\equal{}3f(n)$ and so $ f(1458)\equal{}f(3^62)\equal{}3^6f(2)\equal{}729f(f(1))\equal{}729*3\equal{}2187\neq 2189$

How will you do with this property?
Find all functions satisfying
\[ f: \mathbb N \to \mathbb N
\]

\[ f(f(n)) \equal{} pn
\]
where $ p\in \mathbb P$
[hide="pco"]I want to express my admiration to you, pco.Your solving functional equation skill is too good.Can you share with me some experiences :) [/hide]

[quote="pco"][quote="JBL"] [This is a sure sign that your post does not belong in the calculus forum  :roll:
[color=green]Moved.[/color].[/quote]

I dont think it's an olympiad problem.[/quote]  I'm not the one who decided where to move it ;)  (The green text was inserted by a moderator.)

@Allnames: note that pco didn't characterize all functions, he just showed that any such function must have certain particular properties and values.  The given functional equation does not, for example, determine the value of $ f(4)$, and we have a great deal of freedom in choosing many values of the function.  There is probably no simpler characterization of these functions than as the result of all consistent subsets of choices.

[quote="Allnames"]How will you do with this property?
Find all functions satisfying
\[ f: \mathbb N \to \mathbb N
\]

\[ f(f(n)) = pn
\]
where $ p\in \mathbb P$[/quote]
So, the new problem you are looking for is "Let $ p$ a prime integer, find all functions : $ \mathbb N\to\mathbb N$ such that $ f(f(n)) = pn$

Let $ g_p(n)$ : $ \mathbb N\to\mathbb N\cup\{0\}$ the function which associates to any positive integer $ n$ the power of prime $ p$ in the prime factors decomposition of $ n$.
Let $ A_k = \{x\in\mathbb N$ such that $ g_p(x) = k\}$
$ f(f(n)) = pn$ implies $ f(pn) = pf(n)$ and so $ f(p^kn) = p^kf(n)$ and so we need to find $ f(x)$ over the set $ A_0$

Let $ x\in A_0$ : either $ f(x)\in A_0$ either $ f(x)\in A_1$ (else $ p^2|f(x)$, so $ p^2|f(f(x)) = px$ and $ p|x$).
Let then $ U = \{x\in A_0$ such that $ f(x)\in A_0\}$ and $ V = \{x\in A_0$ such that $ f(x)\in A_1\}$

Obviously :
$ \forall x\in U$ : $ f(x)\in V$
$ \forall x\in V$ : $ f(x)\in A_1$ and $ \frac {f(x)}{p}\in U$ (since $ f(\frac {f(x)}{p}) = \frac {f(f(x))}{p} = x\in A_0$

So $ f(x)$ is a bijection from $ U\to V$ (with inverse function being $ g(x) = \frac {f(x)}{p}$)

And so, we have found a general solution of the problem.

General solution :
=================
Let $ g_p(n)$ : $ \mathbb N\to\mathbb N\cup\{0\}$ the function which associates to any positive integer $ n$ the power of prime $ p$ in the prime factors decomposition of $ n$.
Let $ A_0 = \{x\in\mathbb N$ such that $ g_p(x) = 0\}$
Let $ U$ and $ V$ two equipotent subsets of $ A_0$ such that $ U\cap V = \emptyset$ and $ U \cup V = A_0$
Let $ h(x)$ a bijection from $ U\to V$ and $ h^{[ - 1]}(x)$ : $ V\to U$ its inverse function.

Then $ f(x)$ may be defined as :
$ \forall x\in A_0$ :
If $ x\in U$ : $ f(x) = h(x)$
If $ x\in V$ : $ f(x) = ph^{[ - 1]}(x)$
$ \forall x\in A_k$ with $ k > 0$ : $ f(x) = p^kf(\frac {x}{p^k})$

Demo 1 : This form is a solution :
=================================
$ \forall x\in U$ : $ f(x) = h(x)\in V$ and so $ f(f(x)) = ph^{[ - 1]}(f(x)) = ph^{[ - 1]}(h(x)) = px$
$ \forall x\in V$ : $ f(x) = ph^{[ - 1]}(x)\in A_1$ and so $ f(f(x)) = pf(\frac {f(x)}{p})$ $ = pf(h^{[ - 1]}(x))$. But $ h^{[ - 1]}(x)\in U$ and so $ f(h^{[ - 1]}(x)) = h(h^{[ - 1]}(x)) = x$. So $ f(f(x)) = px$

$ \forall x\in A_k$ with $ k > 0$ : $ f(x) = p^kf(\frac {x}{p^k})$. But $ \frac {x}{p^k}\in A_0$ and so :
Either $ \frac {x}{p^k}\in U$, so $ f(\frac {x}{p^k}) = h(\frac {x}{p^k})\in V$ and so $ f(x)\in A_k$, so $ f(f(x)) = p^kf(\frac {f(x)}{p^k})$ $ = p^kf(f(\frac {x}{p^k}))$ $ = p^kp\frac {x}{p^k}$ (since $ \frac {x}{p^k}\in A$) and so $ f(f(x)) = px$

Either $ \frac {x}{p^k}\in V$, so $ f(\frac {x}{p^k}) = ph^{[ - 1]}(\frac {x}{p^k})\in A_1$ and so $ f(x)\in A_{k + 1}$, so $ f(f(x)) = p^{k + 1}f(\frac {f(x)}{p^{k + 1}})$ $ = p^{k + 1}f(\frac {f(\frac {x}{p^k})}{p})$ $ = p^{k + 1}f(h^{[ - 1]}(\frac {x}{p^k})$ $ = p^{k + 1}h(h^{[ - 1]}(\frac {x}{p^k})$ $ = p^{k + 1}\frac {x}{p^k}$ $ = px$
Q.E.D.

Demo 2 : All solutions are in this form :
========================================
This is the beginning of this post :
Take $ U = {x\in A_0}$ such that $ f(x)\in A_0\}$ and $ V = {x\in A_0}$ such that $ f(x)\in A_1\}$
Take $ h(x) = f(x)$ $ \forall x\in U$

Some examples :
==============
1)
Let $ q\neq p$ a prime number
Let $ U = \{x$ such that $ g_p(x) = 0$ and $ g_q(x)$ is even$ \}$
Let $ V = \{x$ such that $ g_p(x) = 0$ and $ g_q(x)$ is odd$ \}$
Let $ h(x) = qx$

Then $ f(x)$ is :
Let $ x = p^mq^nz$ (with $ z\neq 0\pmod p$ and $ z\neq 0\pmod q$ :
If $ n$ is even, then $ f(x) = p^mq^{n + 1}z$
If $ n$ is odd, then $ f(x) = p^{m + 1}q^{n - 1}z$

2)
(This example suppose $ p\neq 2$)
Let $ m_p(x)$ the reminder of the division of $ x$ by $ p$
Let $ U = \{x$ such that $ m_p(x)$ is odd$ \}$
Let $ V = \{x$ such that $ m_p(x)\neq 0$ and even$ \}$
Let $ h(x) = x + 1$

Then $ f(x)$ is :
Let $ x = p^mz$ (with $ z\neq 0\pmod p$)
If $ m_p(z)$ is odd, then $ f(x) = p^m(z + 1)$
If $ m_p(z)$ is even, then $ f(x) = p^{m + 1}(z - 1)$

 And so on ...

Thank you pco  :o . The initial property requires $ f$ is a strictly increasing function. But I think we can find out $ f$ without that condition. And  I am right! :lol:  .

[quote="MNrule"]Let $ f : \mathbb{N}\rightarrow \mathbb{N}$ such that $ f(n \plus{} 1) > f(n)$ and $ f(f(n)) \equal{} 3n$, then prove that $ f(1) \equal{} 2$ and $ f(1458) \equal{} 2187$[/quote]
This problem is essentially the same as British Math Olympiad 1992 Round 1 Problem 5. However, in the BMO problem you find the value of $ f(1992)$ instead of proving $ f(1) \equal{} 2$ and $ f(1458) \equal{} 2187$.


Find the roots of the equation
$ x^{3}\plus{}x^{2}\plus{}\sqrt[3]{4}x\plus{}(4\plus{}\sqrt[3]{16})\equal{}0$

[quote="cancer"]Find the roots of the equation
$ x^{3}\plus{}x^{2}\plus{}\sqrt [3]{4}x\plus{}(4\plus{}\sqrt [3]{16}) \equal{} 0$[/quote]
[hide="hint"]
Let $ \sqrt[3]{4}\equal{}y$[/hide]

[quote="diophantient"][quote="cancer"]Find the roots of the equation
$ x^{3}\plus{}x^{2}\plus{}\sqrt [3]{4}x\plus{}(4\plus{}\sqrt [3]{16}) \equal{} 0$[/quote]
[hide="hint"]
Let $ \sqrt [3]{4}\equal{} y$[/hide][/quote]

So what then?

This is what you do...  :lol:  :lol:
[hide="solution"]
Then your equation becomes
$ x^{3}\plus{}x^{2}\plus{}xy\plus{}y^{2}\plus{}y^{3}\equal{} 0$, or simply
$ x^{3}\plus{}y^{3}\plus{}x^{2}\plus{}xy\plus{}y^{2}\equal{} 0$
and then you have a sum of cubes...
$ (x\plus{}y)(x^{2}\plus{}xy\plus{}y^{2})\plus{}x^{2}\plus{}xy\plus{}y^{2}\equal{} 0$, and factoring, we get:
$ (x^{2}\plus{}xy\plus{}y^{2})(x\plus{}y\plus{}1) \equal{} 0$, then
$ x \equal{}\minus{}y\minus{}1 \equal{}\minus{}\sqrt[3]{4}\minus{}1$ or
$ x\equal{}\frac{y(\minus{}1\pm\sqrt{3}i)}{2}\equal{}\frac{\sqrt[3]{4}(\minus{}1\pm\sqrt{3}i)}{2}$[/hide]

[quote="Nawahd Werdna"]Then your equation becomes
$ x^{3}\plus{}y^{3}\plus{}x^{2}\plus{}xy\plus{}y^{2}\equal{} 0$
$ (x\plus{}y)(x^{2}\plus{}xy\plus{}y^{2})\plus{}x^{2}\plus{}xy\plus{}y^{2}\equal{} 0$[/quote]

That's wrong; The second line would be :
$ (x\plus{}y)(x^{2}\minus{}xy\plus{}y^{2})\plus{}x^{2}\plus{}xy\plus{}y^{2}\equal{} 0$
And then factoring is not so easy.


"Write a function for the sinusoid with maximum at A and minimum at B."

A(2pi/3,9),B(2pi,5)

I especially don't understand how to get the period. If anyone understands this, please help! Thanks

[hide]How could you relate half a period to the maximum and minimum points?[/hide]

[hide]
Im not so hot at radians but we know that A occurs at 120 degrees and has a value of 9, and b occurs at 360 degrees and has a value of 5.

thus the axis of curve is at $y = 7$

again by using these values we can find the amplitude of the graph to be $2$.

We can find the period easily: Since we know the max value occurs at 120 degrees and the min at 360, we can find the difference between the two which is 240. This is half of the cycle, therefore a full cycle occurs every 480 degrees or $\frac{8\pi}{3}$ radians.

in a normal sin function the max and min values occur at $\frac{\pi}{2}$ or $90^{\circ}$ and $\frac{3\pi}{2}$ or $270^{\circ}$ degrees respectibly, thus we can find the stretch factor.

$\frac{120}{90}= \frac{4}{3}$, thus the stretch is by a factor of $\frac{4}{3}$

So our function is

$\boxed{2\sin{\frac{3}{4}x}+7}$

[/hide]

[hide]
we can assume that $\ f(x)=a sin(x)+b$
$\ f(2 pi/3)=9$ and $\ f(2 pi/5)=5$
    then $\ sqrt(3)/2 a+b=9$ and $\ sin(2pi/5) a+b=5$
thus we find a and b
[/hide]


Maybe you should use the Euclidean alghorytm to find the grand common divisor than to simplify.

So If a and b are large numberss and you want to simplify the fraction $ \frac {a}{b}$ to find the gcd follow the procedure
gcd(a;b)=gcd(a-b=c)=gcd(c-b)=............ect.

However I don't think this is the most eficcient way in your case.

Multiply out by $ 4^5$ to the numerator and denominator and apply the [[Sophie Germain Identity]].

hello, a stupid calculation gives the result $ 221$.
Sonnhard.


inequalities unsolved
inequalities
algebra

[quote=VIMF]let $ a,b,c \in R$ such as $ abc\equal{}1$. Find minimum of
$ \frac{a}{(a^2\minus{}bc)^2}\plus{}\frac{b}{(b^2\minus{}ca)^2}\plus{}\frac{c}{(c^2\minus{}ab)^2}$[/quote]
Any one?

[hide=Thanks.][quote=DoThinh2001]We need to find the minimum of $S=\dfrac{a^2bc}{(a^2-bc)^2}+\dfrac{b^2ca}{(b^2-ca)^2}+\dfrac{c^2ab}{(c^2-ab)^2}$
We have: $$4S=\sum \dfrac{4a^2bc}{(a^2-bc)^2}= \sum \left[\left(\dfrac{a^2+bc}{a^2-bc}\right)^2-1\right] = x^2+y^2+z^2-3,$$
whereas $x=\dfrac{a^2+bc}{a^2-bc}, y=\dfrac{b^2+ca}{b^2-ca}, z=\dfrac{c^2+ab}{c^2-ab}$. Since $xy+yz+zx=-1$ then $$x^2+y^2+z^2=(x+y+z)^2+2 \geq 2.$$
Hence $4S\geq -1$, or $$S\geq \dfrac{-1}{4}.$$
So the equality happens for $x+y+z=0$.[/quote][/hide]
Let $ a,b,c \in R$ such as $ abc=1$. Then
$$ \frac{a}{(a^2-bc)^2}+\frac{b}{(b^2-ca)^2}+\frac{c}{(c^2-ab)^2}\geq -\frac{1}{4}$$

$$ \frac{a}{(a-1)^2}+\frac{b}{(b-1)^2}+\frac{c}{(c-1)^2}\geq -\frac{1}{4}$$

[hide=Thanks.][quote=Quantum_fluctuations][quote=sqing]
Let $ a,b,c \in R$ such as $ abc=1$. Then

$$ \frac{a}{(a-1)^2}+\frac{b}{(b-1)^2}+\frac{c}{(c-1)^2}\geq -\frac{1}{4}$$[/quote]

Let $a=x^3$ , $b =y^3$ and $c = z^3$ then we have

$$x^3y^3z^3 =1 \quad \iff \quad xyz =1$$

and we need to show that

$$\frac{x^3}{ (x^3-1)^2}  \enspace + \enspace \frac{y^3}{ (y^3-1)^2} \enspace + \enspace \frac{z^3}{ (z^3-1)^2} \quad \geq \quad \frac{-1}{4}$$

using the condition $xyz = 1$, this is equivalent to

$$\frac{x^3}{ (x^3-xyz)^2}  \enspace + \enspace \frac{y^3}{ (y^3-xyz)^2} \enspace + \enspace \frac{z^3}{ (z^3-xyz)^2} \quad \geq \quad \frac{-1}{4}$$

$$\iff \quad \quad \quad \frac{x}{ (x^2-yz)^2}  \enspace + \enspace \frac{y}{ (y^2-zx)^2} \enspace + \enspace \frac{z}{ (z^2-xy)^2} \quad \geq \quad \frac{-1}{4}$$

which is true as proved @2above.[/quote]

[/hide]
Let $ a,b,c \in R$ such as $ abc=1$. Then
$$ \frac{a}{(a^2-bc)^2}+\frac{b}{(b^2-ca)^2}+\frac{c}{(c^2-ab)^2}\geq -\frac{1}{4}$$
$$\iff$$
$$ \frac{a}{(a-1)^2}+\frac{b}{(b-1)^2}+\frac{c}{(c-1)^2}\geq -\frac{1}{4}$$


geometry
geometric transformation
reflection

Please help me with this proof because i dont know how to solve it.

Thank You

[hide]If you reflect point Y over RX to Y', which will be on line PR, and reflect point Y over QX to Y", which will be on line PQ, you now have a parallelagram PY'XY", in which angles PY'X and PY"X are right angles.  Drawing line PX gives a two right triangles PY'X and PY"X, which by hypotenuse-leg (the legs are XY' and XY", which are both the reflections of XY), are congruent.

Thus, PY' = PY".
Since PY' = PR + RY' = PR + RY, and PY" = PQ + QY" = PQ + QY,

PQ + QY = PR + RY[/hide]

note $X$ is the excenter of PRQ...and the excircle passes through Y, use the equal tangents...

[quote="Fouad1013"]i didnt take reflections or excenter and excircle. So can somebody please explain it with out that[/quote]

it is still useful to learn those techniques...a triangle has 3 excircles, basically, they are the circles that are tangent to all 3 sides of the triangle (2 extended sides)

Even thou these techniques are usefull i cannot use them to solve the problem because i didnt learn them in class yet so please can you show me a diffrent way for solving that question, without the sin of a straight line and excenter and excircle.

Thank you

drop perpendiculars from $X$ to $PR$ and $PQ$ at points $S'$ and $T'$ respectively

(do this yourself) prove that $\triangle S'RX\cong \triangle YRX$, and $\triangle T'QX\cong \triangle YQX$

prove that $S'X=YX=T'X$

prove that $\triangle PXS'\cong \triangle PXT'$

now $S'P=T'P$, $S'R=RT$, $T'Q=QY$, (why?) and you can finish the proof from here

you should be able to do all of the steps that i left out with simple techniques...triangle congruency, etc


Hai folks, I have a query, and the query is: What is meant by an "efficient means of study/ efficient thinking with respect to problem solving"? I was told that some extra-ordinarily intelligent people are able to comprehend complex/abstruse Math (or any other subject) material because they have learned to interpret it in an efficient way, and not with any innate genius ability. (I was not told in this community)

I am beginning to interpret genius as -- nothing, but an efficient means of thinking, Is it right or wrong?

As this community has many people  with Grey-matter, an acute intellect, I ask them ( and anyone who wants to answer this query of mine) to answer my query. Please share your views, and provide me with your replies along with elaborate instances.

I thank you all. :)

What is your fascination with the definition of genius?  And more importantly, why does it matter?  You don't need to be a genius to accomplish great things, and being a genius doesn't guarantee you anything.  You've asked many similar questions before, and the common answer has been that it doesn't matter.

I had never asked a question of this 'type' or sort before( as far as I can remember), being a genius helps your probability  of accomplishing great things.

Actually, these things are under research by Cognitive Psychologists, so it might be of no-use to people like you, but it might be important to someone else..

A person should not comment when he can't come up with a reasonable/satisfactory answer to a question posed by me or for that matter by anyone....  :lol:

I'd request that you make better titles, as far as some of your previous posts' titles are concerned:
"[url=http://www.mathlinks.ro/viewtopic.php?t=240798]hiiii[/url]", "[url=http://www.mathlinks.ro/viewtopic.php?t=239637]People in this forum read this[/url]", "[url=http://www.mathlinks.ro/viewtopic.php?t=243391]Yeah[/url]"

Okay, But I feel the content inside a post demands more attention than the 'title' though a good title is also necessary. :lol:

[quote="Gen8"]A person should not comment when he can't come up with a reasonable/satisfactory answer to a question posed by me or for that matter by anyone....[/quote]
And a person should not post threads just for the sake of posting threads.  As has been pointed out in some of your other threads, they don't have a purpose other than trying to confirm some view that you have.

It is interesting that you say people should not post when they have no reasonable/satisfactory reply. While I agree it's true, I do not believe it to be sufficient criteria for posting. Unto a spammer, spam is reasonable and satisfactory unto them, yet to someone else they do not post spam because it does not give them satisfaction. If we say reasonable/satisfactory in the eyes of the AoPS admins, I believe they might possibly agree with JBL that you are the less productive poster. I know not. Yet that is beside the point, that you do not actually mean unqualified, unframed 'reasonable/satisfactory'.

As for the importance of a title, communication is important: if you are Einstein but can't communicate your discoveries, your discovery is useless. Good communication expresses as much as possible in as little space as possible, while remaining clear. Your title, whether of a post, a speech, or a paper, should clearly state what you're going to talk about. When your audience is trying to choose what to read from a list of speech/posts/papers, they will judge based [i]solely[/i] on the title. Without a good title, it's not likely to interest the people who you want interested. The title is just as important as the content if you want to be an effective conversant.

Back to efficiency, I hate it that my mind has been rearranged to raise quibbling problems like the above 'you didn't specify a frame of reference!'. Yet nevertheless, I feel like making the same argument about efficiency. You can only ask for our thoughts; there are no facts behind them. See TZF's post which is much more detailed and makes the same point I would've made here.

[Edited because somehow not all my post went through. :(]

[b]@ solafidefarms:[/b] Your post is incomplete -- perhaps it was a posting error.

Despite the large advancements in psychometry in the last century, we still can't pinpoint exactly what it is because neuroscience is not advanced enough to support existing models with a reasonable level of confidence. We can agree that there are either multiple forms of intelligence or multiple factors that are involved in intelligence. However, whether there is a sound biological structure for our model of intelligence is not known. It may well be that the notion of intelligence, or cognitive ability in general, is meaningful only within a cultural context.

The best we can agree on is that genius amounts to profound ability in one or more aspects of mental processing. But again, one's notion of this depends on what one considers to be the aspects of mental processing, which, as I said, is not well defined.

So in the end, we can only settle for one of the following:
1. The psychometric definition - By IQ, which is a bit dubious and hardly satisfying
2. The personal definition - What you believe to be super-smart
3. The cultural/societal definition - What other people believe to be super-smart

Okay, TZF has provided some of his views, analysis, but I must say  it is not germane to the question that I had asked, and the answer/reply I was seeking; My question was pertaining to "efficiency", though TZF's views on psychometry are practical, but I feel (to some extent) there is a digression of the topic that is intended to be discussed.


geometry
geometric transformation
reflection
analytic geometry

If A(-7,4) is reflected across the x-axis, point B will be (-7,-4). If point B is reflected across the y-axis, point C will be (7,-4). This is the same as reflecting through the origin. The sum of the coordinates is 7-4=3.

[quote="joejia"]The point A(-7, 4) is reflected across the x-axis onto point B. Point B is reflected over the y-axis onto point C. What is the sum of the coordinates of point C?[/quote]

1 since it goes to 5, -4 is c which equals 1

[hide]When it's reflected across the x-axis, it's (-7, -4), and when it's reflected again across the y-axis, it turns into (7, -4) with the sum as 3. [/hide]

[hide]If you reflect point A(-7, 4) on the  x-axis, you'll get point B, (-7, -4).

If you reflect point B on the y-axis, you'll get point C, (7, -4)

The sum of the two numbers is 7+(-4) = 7-4 = 3.

3 is the answer.

See attached diagram. :D[/hide]


probability logarithms trigonometry ratio geometry perimeter inequalities

PM me yours answer with (1) Answer (2) Concise Solution. Again 1 bonus point may be awarded to elegant or less time consuming methods and ideas.

Here are the first set of questions all together.

PPPC
Problem (1)
Value: 4 points

Consider $\triangle ABC$ with sides of length $13$, $14$, and $15$. The sum of the lengths of the altitudes drawn to sides $AB$, $BC$, and $AC$ can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

PPPC
Problem (2)
Value: 5 points

Find all pairs $(x,y)$ of positive integers such that $x \leq y$ and   $\frac{1}{x}+\frac{1}{y}=\frac{1}{6}$

PPPC
Problem (3)
Value: 3 points

A number $d_1d_2...d_n$ is called a distinguished $n$-digit number if $d_1$ is not $0$ and the digits $d_1, d_2,..., d_n$ are all distinct.

What is the sum of the digits of the largest distinguished 8-digit number that is divisible by 9?

PPPC
Problem (4)
Value: 12 points

In how many ways can the number $1,000,000$ be expressed as a product of three positive integers? Factorizations which differ only in the order of the factors are not to be considered different.

PPPC
Problem (5)
Value: 4 points

If a chord is selected at random on a fixed circle, the probability that its length exceeds the radius of the circle is $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

PPPC
Problem (6)
Value: 2

Find the next three numbers in the sequence $\frac{2}{3}, 1 \frac{5}{12}, 2\frac{1}{6}, 2\frac{11}{12}$.

PPPC
Problem (7)
Value: 3

How many digits does $2^{2222}$ have? $(\log 2=0.30103$)

PPPC
Problem (8)
Value: 3

Which number is greater $2^{300}$ or $3^{200}$ and why?

PPPC
Problem (9)
Value: 4

Solve the equation: $\sin x \cos x=\sin40$

PPPC
Problem (10)
Value: 3

Solve the equation $\log_{(x-2)}(2x^2-9x+8)=2$

Note: A calculator is not required in order to solve any of these problems and therefore should not be used! :lol:

I apologize for any typographical errors. Please notify me If you spot one!

Between each set of 10 questions there will be 2 power questions. These questions are designed to challenge students. Partial credit will be awarded based upon ideas and work shown even if the answer is wrong.

PPPC
Power Question 1
Value: 6 points

Express $\sqrt{5+12i}$ in the form $a+bi$.

PPPC
Power Question 2
Value: 12 points

Derive the infinite continued fraction for $\sqrt{2}$. (6 points)

Derive the infinite continued fraction for $\frac{1+\sqrt5}{2}$. (6 points)

Hint

[hide]
Think about the following:

$x^2=ax+1$, or $x=a+\frac{1}{x}$

[/hide]

So far:

joml88 is leading with 13 points. In second place is silverfalcon with 9 points.

In first is Max Harris with 42 points. Joml88 stands in second with 13 points and Silverfalcon is in third with 9 points.

SCORE UPDATE!!!

1st-Max Harris
2nd-Joml88
3rd-h_s_potter2002, Silverfalcon

1st-Max Harris(42)
2nd-Joml88(13)
3rd-h_s_potter2002(12)
4th-Silverfalcon(11)

1st-Max Harris(42)
2nd-Silverfalcon(14)
3rd-joml88(13)
4th-h_s_potter2002(12)

1st-Max Harris(42)
2nd-pkothari13(16)
3rd-Silverfalcon(14)
4th-joml88(13)
5th-h_s_potter2002(12)

how long will this ound last cuz I want to do this but i just found it pretty late at night and i'll do it tomorrow

The round will officially be over 11:59 pm Thursday night.

SCORE UPDATE:

1st-Max Harris(42)
2nd-Zratonx(21)
3rd-pkothari13(16)
4th-Silverfalcon(14)
5th-joml88(13)
6th-h_s_potter2002(12)

REMINDER: The deadline for this half of the competition is 11:59 p.m. US central time today (Thursday).

SCORE UPDATE!!!!

1st-Max Harris(42)
2nd-Zratonx(24)
3rd-pkothari13(16)
4th-Silverfalcon(14)
5th-joml88(13)
6th-h_s_potter2002(12)

We need some solutions to numbers 4 and 5!!! Especially #5 since it is quite easy if you think about it!!! #4 is a tough one and I have now raised its point value to >>12<< Points!!!

Due to the many people who found out late the contest deadline will be pushed to 11:59 p.m. Central US time, December 31st, 2004....NOTE: ALL ANSWERS MUST BE SUBMITTED BEFORE 2005! :lol:  :lol: HAPPY NEW YEAR!!! :D  :D

SCORE UPDATE!!!!

1st-Max Harris(46)
2nd-Zratonx(29)
3rd-pkothari13(16)
4th-Silverfalcon(14)
5th-joml88(13)
6th-h_s_potter2002 and towersfreak2006(12)

Max Harris continues to lead with 46 points!!! Can anyone overtake him in time!?!?!

!!! #5 has been solved by Max Harris !!!

#5 Solution

[hide]

 Take a point $\displaystyle P$ at the top of a circle with center $\displaystyle O$. Draw a chord the length of a radius from $\displaystyle P$ to a point $\displaystyle Q$. Then draw the radius from $\displaystyle O$ to $\displaystyle Q$. Since $\displaystyle OP$ is also a radius, then $\displaystyle \bigtriangleup OPQ$ is equilateral. Reflect $\displaystyle \bigtriangleup OPQ$ about $\displaystyle OP$ so that the reflection of $\displaystyle Q$, denoted $\displaystyle R$, lies on the circle. Then the second point must not lie on the minor arc $\displaystyle RQ$, which has a measure of $\displaystyle120$, to form a chord longer than the radius. Thus, the probability of obtaining a chord longer than the radius is $\displaystyle\frac{2}{3}$.
The desired answer is $\displaystyle2+3=\fbox{5}$.

[/hide]

by Towersfreak2006

[hide]
A chord is defined by 2 points, but the first point is irrelevant since the circle is the same all around. If the second point is $60^o$ away from the first, then the chord is the length of the radius since we will have an equilateral triangle with two radii as sides. Thus, if the second point is more than $60^o$ away, it will be longer than the radius. For the chord to be less than the radius, the second point can be $60^o$ left or right, for a total of $120^o$, and there are $360^o$ total in a circle, so $\frac{m}{n}$ is $\frac{360-120}{360}=\frac{2}{3}$, and $m+n=2+3=5$.
[/hide]
by Efuzzy

picture by Treething

#6 Solution

[hide]
This is just a arithmetic sequence with difference of $\frac{3}{4}$ (though it is disguised!), so the next three terms are $\frac{11}{3}$, $\frac{53}{12}$, $\frac{31}{6}$.
[/hide]

by Repmax123

#7 Solution

[hide]
The number of digits an integer, $n$, has is given by $\lfloor \log_{10}{n}\rfloor +1$. Take $\log{2^{2222}=2222\log{2}\approx 668.8}$ So we round down and add 1 to get $669$.

[/hide]

by Joml88

#9 Solution

[hide]
$\displaystyle\sin{x}\cos{x}=\sin{40^\circ}$
$\displaystyle\sin{40^\circ}>\sin{30^\circ}$ since $\displaystyle\sin x$ is increasing for $\displaystyle0^\circ\le x\le90^\circ$
$\displaystyle\sin{40^\circ}>\frac{1}{2}$
$\displaystyle2\sin{40^\circ}>1$
$\displaystyle2\sin{x}\cos{x}=\sin{2x}=2\sin{40^\circ}$
Since $\displaystyle0\le\sin{x}\le1$, $\sin{2x}=2\sin{40^\circ}$, and $\displaystyle\2\sin{40^\circ}>1$, there are no solutions for $\displaystyle\sin{x}\cos{x}=\sin{40^\circ}$.

[/hide]

by Towersfreak2006

Treething feels special.

When's the next round? This time, I want more than one day to solve the problems :P .

probably tomorrow! I need to finish up the questions :P  :P

[quote="Treething"]
This time, I want more than one day to solve the problems :P .[/quote]

Yeah, agreed.  I had an MIT application due and did the problems in about 30 minutes.  maybe should start being more active here pretty soon.

writer's block... :( ....the problems will probably be delayed until later tonight, sorry guys!

On #5, I think you need to be more specific about what you mean by "select a chord at random".  Your solution is perfectly valid.  However, it can also be solved by saying that each chord has a unique midpoint, and each point inside the circle is the midpoint of exactly one chord.  Thus we can select a chord at random by choosing any point in the circle and using it as the midpoint of the chord.
In this way, it turns out that if the circle has radius 1, then the point must be within $\frac{\sqrt{3}}{2}$ units of the center.  This covers an area of $\frac{3}{4}\pi$, which makes the probability $\frac{3}{4}$

You can also select a chord at random by choosing a random diameter and a point on that diameter to be the midpoint of the chord.  In this case, the probability turns out to be $\frac{\sqrt{3}}{2}$.

In general, be very careful when dealing with probability and infinite sets (of chords) in the same problem.

However, I did not mention anything about midpoints of chords. In addition, a chord of a circle is defined as a line segement joining two different points on the circumference of a circle. Hence, I believe that the question was unambiguous but I will try to be more careful from now on.

P.S. Feel free to PM any questions that you might have concerning a problem!

*screams*
I'll help you write some of the problems :P . I would NEVER think of using my knowledge of the problems to my advantage :P ...

Ok guys, round 2 coming hopefully by tonight!!! :D


articles
abstract algebra
algorithm
MIT
college

Can anyone give me some nice article, book, link, or anything simmilar from where I can learn Dynamic programming (whit examples in Pascal if it isn't too much) P.S. I know nothing about Dynamic programming

Tnx!

Here are two that I've recently read. I also recommend CLRS.

http://www.topcoder.com/index?t=features&c=feat_040104
http://www.topcoder.com/tc?module=Static&d1=tutorials&d2=dynProg

P.S. Google for more.

Cormen, Leiserson, Livest, Stain "Introduction to Algorithms" MIT Press.

P.S. You should own that book.

No, you can't. At least in the legal way.

[quote="M@re"]So can I download it from somewhere or not......[/quote]

http://zhuzeyuan.hp.infoseek.co.jp/ita/toc.htm


quadratics
number theory
greatest common divisor
algebra
algebra unsolved

Let $ \alpha$ and $ \beta$ be the roots of the quadratic equation $ x^2 \plus{} mx \minus{} 1 \equal{} 0$, where m is an odd integer. Let $ \lambda_n \equal{} \alpha^n \plus{} \beta^n$, for $ n \ge 0$. Prove that for $ n \ge 0$,
(a) $ \lambda_n$ is an integer; and
(b) gcd $ (\lambda_n, \lambda_{n \plus{} 1}) \equal{} 1$.

[quote="thomaskurian89"]Let $ \alpha$ and $ \beta$ be the roots of the quadratic equation $ x^2 \minus{} mx \minus{} 1 \equal{} 0$, where m is an odd integer. Let $ \lambda_n \equal{} \alpha^n \plus{} \beta^n$, for $ n \ge 0$. Prove that for $ n \ge 0$,
(a) $ \lambda_n$ is an integer; and
(b) gcd $ (\lambda_n, \lambda_{n \plus{} 1}) \equal{} 1$.[/quote]

$ \alpha\plus{}\beta\equal{}m$
$ \alpha*\beta\equal{}\minus{}1$

$ \lambda_1\equal{}\alpha\plus{}\beta\equal{}m$
$ \lambda_2\equal{}\alpha^2\plus{}\beta^2\equal{}(\alpha\plus{}\beta)^{2}\minus{}2\alpha*\beta\equal{}m^{2}\plus{}2$

$ \alpha^{2}\minus{}m\alpha\minus{}1\equal{}0$-----> $ \alpha^{2}\equal{}m\alpha\plus{}1$---> $ \alpha^{n}\equal{}m\alpha^{n\minus{}1}\plus{}\alpha^{n\minus{}2}$
$ \beta^{2}\minus{}m\beta\minus{}1\equal{}0$ ---> $ \beta^{2}\equal{}m\beta\plus{}1$-----> $ \beta^{n}\equal{}m\beta^{n\minus{}1}\plus{}\beta^{n\minus{}2}$----->

$ \alpha^{n}\plus{}\beta^{n}\equal{}m(\alpha^{n\minus{}1}\plus{}\beta^{n\minus{}1})\plus{}\alpha^{n\minus{}2}\plus{}\beta^{n\minus{}2}$--->  $ \lambda_{n}\equal{}m\lambda_{n\minus{}1}\plus{}\lambda_{n\minus{}2}$
$ \lambda_1\equal{}m$ and $ \lambda_2\equal{}m^{2}\plus{}2$ ----> $ \lambda_{n}\in Z$

 $ gcd (\lambda_n, \lambda_{n \plus{} 1})\equal{}gcd (\lambda_n,m\lambda_{n}\plus{}\lambda_{n\minus{}1})\equal{}gcd (\lambda_n, \lambda_{n\minus{}1})\equal{}...\equal{}gcd (\lambda_2, \lambda_1)\equal{}gcd (m^{2}\plus{}1,m)\equal{}(1,m)\equal{}1$

[quote="thomaskurian89"][quote="zaya_yc"][quote="thomaskurian89"]Let $ \alpha$ and $ \beta$ be the roots of the quadratic equation $ x^2 \minus{} mx \minus{} 1 \equal{} 0$, where m is an odd integer. Let $ \lambda_n \equal{} \alpha^n \plus{} \beta^n$, for $ n \ge 0$. Prove that for $ n \ge 0$,
(a) $ \lambda_n$ is an integer; and
(b) gcd $ (\lambda_n, \lambda_{n \plus{} 1}) \equal{} 1$.[/quote]

$ \alpha \plus{} \beta \equal{} m$
$ \alpha*\beta \equal{} \minus{} 1$

$ \lambda_1 \equal{} \alpha \plus{} \beta \equal{} m$
$ \lambda_2 \equal{} \alpha^2 \plus{} \beta^2 \equal{} (\alpha \plus{} \beta)^{2} \minus{} 2\alpha*\beta \equal{} m^{2} \plus{} 2$

$ \alpha^{2} \minus{} m\alpha \minus{} 1 \equal{} 0$-----> $ \alpha^{2} \equal{} m\alpha \plus{} 1$---> $ \alpha^{n} \equal{} m\alpha^{n \minus{} 1} \plus{} \alpha^{n \minus{} 2}$
$ \beta^{2} \minus{} m\beta \minus{} 1 \equal{} 0$ ---> $ \beta^{2} \equal{} m\beta \plus{} 1$-----> $ \beta^{n} \equal{} m\beta^{n \minus{} 1} \plus{} \beta^{n \minus{} 2}$----->

$ \alpha^{n} \plus{} \beta^{n} \equal{} m(\alpha^{n \minus{} 1} \plus{} \beta^{n \minus{} 1}) \plus{} \alpha^{n \minus{} 2} \plus{} \beta^{n \minus{} 2}$--->  $ \lambda_{n} \equal{} m\lambda_{n \minus{} 1} \plus{} \lambda_{n \minus{} 2}$
$ \lambda_1 \equal{} m$ and $ \lambda_2 \equal{} m^{2} \plus{} 2$ ----> $ \lambda_{n}\in Z$

 $ gcd (\lambda_n, \lambda_{n \plus{} 1}) \equal{} gcd (\lambda_n,m\lambda_{n} \plus{} \lambda_{n \minus{} 1}) \equal{} gcd (\lambda_n, \lambda_{n \minus{} 1}) \equal{} ... \equal{} gcd (\lambda_2, \lambda_1) \equal{} gcd (m^{2} \plus{} 1,m) \equal{} (1,m) \equal{} 1$[/quote]

Thanks but have you used the fact that m is an odd integer?[/quote]

$ gcd (\lambda_n, \lambda_{n \plus{} 1}) \equal{} gcd (\lambda_n,m\lambda_{n} \plus{} \lambda_{n \minus{} 1}) \equal{} gcd (\lambda_n, \lambda_{n \minus{} 1}) \equal{} ... \equal{} gcd (\lambda_2, \lambda_1) \equal{} gcd (m^{2} \plus{} 2,m) \equal{} (2,m) \equal{} 1$

$ m$ odd ----> $ (2,m)\equal{}1$

[quote="zaya_yc"][/quote][quote="thomaskurian89"][quote="zaya_yc"][quote="thomaskurian89"]Let $ \alpha$ and $ \beta$ be the roots of the quadratic equation $ x^2 - mx - 1 = 0$, where m is an odd integer. Let $ \lambda_n = \alpha^n + \beta^n$, for $ n \ge 0$. Prove that for $ n \ge 0$,
(a) $ \lambda_n$ is an integer; and
(b) gcd $ (\lambda_n, \lambda_{n + 1}) = 1$.[/quote]

$ \alpha + \beta = m$
$ \alpha*\beta = - 1$

$ \lambda_1 = \alpha + \beta = m$
$ \lambda_2 = \alpha^2 + \beta^2 = (\alpha + \beta)^{2} - 2\alpha*\beta = m^{2} + 2$

$ \alpha^{2} - m\alpha - 1 = 0$-----> $ \alpha^{2} = m\alpha + 1$---> $ \alpha^{n} = m\alpha^{n - 1} + \alpha^{n - 2}$
$ \beta^{2} - m\beta - 1 = 0$ ---> $ \beta^{2} = m\beta + 1$-----> $ \beta^{n} = m\beta^{n - 1} + \beta^{n - 2}$----->

$ \alpha^{n} + \beta^{n} = m(\alpha^{n - 1} + \beta^{n - 1}) + \alpha^{n - 2} + \beta^{n - 2}$--->  $ \lambda_{n} = m\lambda_{n - 1} + \lambda_{n - 2}$
$ \lambda_1 = m$ and $ \lambda_2 = m^{2} + 2$ ----> $ \lambda_{n}\in Z$

 $ gcd (\lambda_n, \lambda_{n + 1}) = gcd (\lambda_n,m\lambda_{n} + \lambda_{n - 1}) = gcd (\lambda_n, \lambda_{n - 1}) = ... = gcd (\lambda_2, \lambda_1) = gcd (m^{2} + 1,m) = (1,m) = 1$[/quote]

Thanks but have you used the fact that m is an odd integer?[/quote][quote="zaya_yc"]

$ gcd (\lambda_n, \lambda_{n + 1}) = gcd (\lambda_n,m\lambda_{n} + \lambda_{n - 1}) = gcd (\lambda_n, \lambda_{n - 1}) = ... = gcd (\lambda_2, \lambda_1) = gcd (m^{2} + 2,m) = (2,m) = 1$

$ m$ odd ----> $ (2,m) = 1$[/quote]

I just realised I had made a typo in the question. (initially wrote -m instead of m). I've corrected the question now.
Could you please modify your solution to accomodate this?


Hey, I'm new to Latex. If I solve a problem in Latex and I want to upload it so it shows up in this forum all nice looking, how do I do that?

Take the math mode code, and put it between dollar signs. The forum renders it for you.

a useful link would be [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicMath.php#mathmodeurl]here[/url]


Let ABCD is a square.Let P,M,N,Q in AB,BC,CD,DA such that :
$\widehat{PMA}$ =$\widehat{MAN}$=$\widehat{ANQ}$ = 45 degree

Prove that : $AM^{2}$ + $QN^{2}$=$AN^{2}$+ $PM^{2}$

You can show that $A$ is an excenter of $MCD$.

Basic angle chasing gives $\angle DAN = \angle PMB$, so $\triangle ADN \sim \triangle MBP$.  Therefore $\frac{PB}{BM}= \frac{DN}{AD}\implies PB = \frac{BM \cdot DN}{AD}$.  Similarly, we find that $\triangle QDN \sim \triangle MAB$.  This means that $\frac{QD}{DN}= \frac{BM}{AB}$, which means that \[QD = \frac{BM\cdot DN}{AB}= \frac{BM\cdot DN}{AD}= PB\] Therefore $QD^{2}= PB^{2}$.  Adding $AB^{2}+BM^{2}+DN^{2}$ to both sides gives $(AB^{2}+BM^{2})+(QD^{2}+DN^{2}) = (DN^{2}+AD^{2})+(PB^{2}+BM^{2})$.  This equation is equivilant to the given equation by the Pythagorean Theorem.

[color=darkblue][b][u]Remark.[/u][/b] The relation $AM^{2}+NQ^{2}=AN^{2}+PM^{2}$ [b]is truly[/b]
for any common value of the angles $\widehat{PMA}$, $\widehat{MAN}$, $\widehat{ANQ}$ ![/color]


Let ABC is a triangle .Prove that :
[tex]\large m_a+l_b+h_c \leq \frac{\sqrt{3}}{2} (a+b+c) +\frac{3}{20} (/a-b/+/b-c/+/c-a/)[/tex]

it looks famous ;)
maybe popoviciu should work :D  :D

Maybe this ineq is so hard! Nobody have a solution !

Does anybody have a solution ? Can you  prove this ineq ?
[tex]m_a+l_b+h_c \leq \frac{\sqrt{3}}{2}(a+b+c)[/tex].


inequalities
function
inequalities solved

I found this problem in Longlist of IMO 1983. I dont have a solution and my students can not solve it for long time. Can anyone solve it

A strictly increasing function $f$ defined on $[0,1]$ satisfies $f(0) = 0, f(1) = 1$, and   $\frac{1}{2}\leq \frac{f(x+y) - f(x)}{f(x)-f(x-y)}\leq 2$
for all $x, y$ such that $0 \leq x+y, x-y \leq1$. Show that $f\left (  \frac{1}{3}\right ) \leq \frac{76}{135}$.

Namdung

Can you tell us the country please which is the proposer of the problem. I could not find it on the 1983 longlist. Maybe I did not look carefully enough.


Let $E$ be a closed bounded set of $\mathbb{R}$.Put $W_n(z_1,z_2,...,z_n)=\prod_{1 \leq i<j \leq n}(z_i-z_j)$, $\displaystyle V_n=max_{z_1 \in E,z_2 \in E,...,z_n \in E}|W_n(z_1,...,z_n)|$. Prove that hthe sequence $V_n^{\frac{2}{n(n-1)}}$ has finite limit as $n \to \infty$

The sequence is clearly bounded and it is not difficult to see that it is decreasing.

Thanks, i realised that it is decreasing, bounded and thus has a limit. It is an interesting question(problem), prove that this limit is $\frac{1}{2}$.


geometry
rectangle
geometric transformation

Rectangle $ABCD$ has sides $\overline {AB}$ of length 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 congruent segments with points $A=P_0, P_1, \ldots, P_{168}=B$, and divide $\overline {CB}$ into 168 congruent segments with points $C=Q_0, Q_1, \ldots, Q_{168}=B$. For $1 \le k \le 167$, draw the segments $\overline {P_kQ_k}$. Repeat this construction on the sides $\overline {AD}$ and $\overline {CD}$, and then draw the diagonal $\overline {AC}$. Find the sum of the lengths of the 335 parallel segments drawn.

Please define $Q_k$

EDIT: I've found the original text, so I know what's it about, but you should edit the text nevertheless :)

[hide="Solution"]We have $AC=\sqrt{3^2+4^2}=5$. Furthermore, $P_kB=AB-AP_k=AB-k{AB\over 168}=AB{{168-k}\over 168}$

Similarly, $Q_kB=BC{{168-k}\over 168}$, hence $P_kQ_k=AC{{168-k}\over 168}$

Now

\begin{eqnarray*}S_{PQ} &=& \sum_{k=1}^{167}{AC{{168-k}\over 168}}\\ &=& {{AC}\over 168}\sum_{k=1}^{167}{(168-k)}\\ &=& {{AC}\over 168}(168\cdot 167-{{167\cdot 168}\over 2})\\ &=& {AC\over 168}\cdot{{167\cdot 168}\over 2}\\ &=& {167\over 2}AC\end{eqnarray*}

Hence $S=AC+2S_{PQ}=168AC=840$[/hide]

Is this the original wording? It seems a little... strange to me.  :huh:

Could somebody please check?

No this isn't the original text.

I don't know the original text though, but that is what Farenhajt said.

Construct $ E$ so $ ACED$ is a parallelogram. Then extend the lines constructed on $ AD$, $ DC$ to hit $ EC$. However, a translation takes $ ABC$ to $ DCE$ (along with all the segments). So we have the total length of the segments is just the $ \frac{335\plus{}1}{2}$ lines that are parallel to $ AC$ in $ ACED$...giving $ 168*5\equal{}\boxed{840}$.

This _is_ the original text. I have a copy of the paper.

[hide="Alternate"]

Let $a=4/168$ and $b=3/168$.

The smallest segment has length $\sqrt{a^2+b^2}$, the next is:  $\sqrt{(2a)^2+(2b)^2} = 2\sqrt{a^2+b^2}$.  Similarly, the $k$th segment has length $k\sqrt{a^2+b^2}$.
We wish to compute:  $5+2\cdot \sum_{k=1}^{167}k\sqrt{a^2+b^2}$ (where 5 is the length of $\overline{AC}$).  This is just:  $2\cdot \dfrac{(167)(168)}{2}\cdot\sqrt{a^2+b^2} + 5$, and notice that $\sqrt{a^2+b^2}=\dfrac{5}{168}$, so stuff cancels and we get:  $\boxed{840}$[/hide]


trigonometry
inequalities
inequalities unsolved

Answer : $ \frac{9}{4}$.

[hide]$ \sum\sin^2 A\leq \frac{9}{4}\forall \Delta ABC.$[/hide]

Can I use trigonometry identity $ \sin^2{x}\equal{}\frac{1}{2}(1\minus{}\cos{2x})$ to solve this ?

The equality does not occur when all the variable are equal but it occurs when $ x_1\equal{}x_2\equal{}x_3\equal{}\frac{\pi}{3}$ and $ x_i\equal{}0 \forall i\equal{}\bar{4,n}$ It reminds me of ACM. Can somebody give me a link to review ACM in proving inequality ?

Thanks
 :)


calculus
integration
real analysis
real analysis unsolved

[i]I would like your simpler solution to this problem:[/i]

  Find the integral :

  $I= \int_{2}^{5}{\frac{e^{\frac{x}{2}}-e^{\frac{5}{x}}}{x}}dx$

  [u]Babis[/u]

[quote="didilica"]Use this substitutions: $x=2t$ and you get that:

$I=\int_{1}^{5/2}\frac{e^{t}-e^{5/2t}}{t}dt$. Another substitution $\frac{5}{2t}=y$ implies that

$I=-I$ from which it follows that $I=0$.[/quote]

 Nice ! It is a natural way !

  Thank you !

[u] Babis[/u]


prove the equation has many infinitely solution
a,b,c-integers
a^3+b^3+c^3=2

Have many way to solve this .
This is one solution :
$ a = x - y,b = x + y$
The equation becomes :
$ (x + y)^3 + (x - y)^3 + c^3 = 2$
$ \equiv alent 2(x^3 + 3xy^2) + c^3 = 2$
Chose $ x = 1$ then
$ 6y^2 + c^3 = 0$
Easy to check :
$ y = 6t^3,c = - 6t^2$ is solution .
It mean that the equation has infinite solution in integer.

$ a\equal{}k\plus{}1$, $ b\equal{}1\minus{}k$. We get:
$ k^3\plus{}3k^2\plus{}3k\plus{}1\plus{}1\minus{}3k\plus{}3k^2\minus{}k^3\plus{}c^3\equal{}2$
$ 6k^2\equal{}\minus{}c^3$
Let $ k\equal{}6p^3$
We get: $ c\equal{}\minus{}6p^2$
Answer: $ a\equal{}6p^3\plus{}1$, $ b\equal{}1\minus{}6p^3$, $ c\equal{}\minus{}6p^2$; $ p \in Z$.

for all k  infinity solution!

\[ x_k  \equal{} 1 \plus{} 6^{3k \plus{} 1}
\]

\[ y_k  \equal{} 1 \minus{} 6^{3k \plus{} 1}
\]

\[ z_k  \equal{}  \minus{} 6^{2k \plus{} 1}
\]


Find the closed formula of the following recurrence relation:

$ a_0 \equal{} 6$
$ a_{n\plus{}1} \equal{} (2n\plus{}2)a_{n} \minus{} 15n\minus{}10$

I've found that generating function for this recurrence is $ f(x) \equal{} \frac{\minus{}15xe^x\plus{}5e^x\plus{}1}{1\minus{}2x}$.

But how to expand it with Taylor series?

Multiply the series for $ \minus{} 15x e^x \plus{} 5e^x \plus{} 1$ and $ \frac {1}{1 \minus{} 2x}$, both of which are easy to write down.  And it's easy to extract the coefficients of a product given the coefficients of both factors.