induction
number theory
least common multiple

Show that for each $n \ge 2$, there is a set $S$ of $n$ integers such that $(a-b)^2$ divides $ab$ for every distinct $a, b\in S$.

[b]Lemma 1[/b] If $ \gcd(a,b) \equal{} m$ and $ a \equal{} mk, b \equal{} ml$, then $ (a\minus{}b)^{2}|ab$ iff $ k\minus{}l \equal{} 1$.
[b]Proof[/b] $ (a\minus{}b)^{2}\equal{} m^{2}(k\minus{}l)^{2}$; $ ab \equal{} m^{2}kl$ so $ (a\minus{}b)^{2}|ab\Leftrightarrow (k\minus{}l)^{2}|kl$. But because$ \gcd(a,b) \equal{} m$and$ a \equal{} mk, b \equal{} ml$,$ \gcd(k,l) \equal{} 1$and if $ k\minus{}l \equal{} q > 1$ then $ k\minus{}l$ is divisible by a prime$ p\geq 2$, and then$ p$divides$ k$or$ l$, wolog$ p|k$, but then as $ p|(k\minus{}l)$ then $ p|l$ which contradicts the fact that $ \gcd(k,l) \equal{} 1$. So,$ k\minus{}l \equal{} 1$and the lemma is proved in one direction. Now, $ (a\minus{}b)^{2}\equal{} (mk\minus{}ml)^{2}\equal{} m^{2}(k\minus{}l)^{2}\equal{} m^{2}|m^{2}kl \equal{} (mk)(ml) \equal{} ab$ so $ (a\minus{}b)^{2}|ab$ and the lemma is proved in the other direction.

[b]Lemma 2[/b] If $ x,y$ are two distinct positive integers with $ x > y$ such that $ (x\minus{}y)^{2}|xy$ then for any positive integer $ c\geq 2$, the numbers $ t \equal{} c^{x}\minus{}1, s \equal{} c^{x\minus{}y}(c^{y}\minus{}1)$ then $ (s\minus{}t)^{2}|st$.
[b]Proof[/b] By lemma 1, if $ \gcd(x,y) \equal{} m$ and $ x \equal{} mk,y \equal{} ml$ then $ k\minus{}l \equal{} 1$.
Then, $ t \equal{} c^{x}\minus{}1 \equal{} c^{mk}\minus{}1 \equal{} (c^{m}\minus{}1)(c^{(k\minus{}1)m}\plus{}c^{(k\minus{}2)m}\plus{}...\plus{}c^{m}\plus{}1)$
Also, $ s \equal{} c^{x\minus{}y}(c^{y}\minus{}1) \equal{} c^{mk\minus{}ml}(c^{ml}\minus{}1) \equal{} c^{m(k\minus{}l)}(c^{m}\minus{}1)(c^{(l\minus{}1)m}\plus{}c^{(l\minus{}2)m}\plus{}...\plus{}c^{m}\plus{}1)$
$ \equal{} (c^{m}\minus{}1)(c^{m}(c^{(l\minus{}1)m}\plus{}c^{(l\minus{}2)m}\plus{}...\plus{}c^{m}\plus{}1)) \equal{} (c^{m}\minus{}1)((c^{lm}\plus{}c^{(l\minus{}1)m}\plus{}...\plus{}c^{2m}\plus{}c^{m}))$
$ \equal{} (c^{m}\minus{}1)(c^{(k\minus{}1)m}\plus{}c^{(k\minus{}2)m}\plus{}...\plus{}c^{m})$ because $ k\minus{}l \equal{} 1$.
Therefore, because we can find numbers $ d \equal{} c^{m}\minus{}1, e \equal{} c^{(k\minus{}1)m}\plus{}c^{(k\minus{}2)m}\plus{}...\plus{}c^{m}\plus{}1, f \equal{} c^{(k\minus{}1)m}\plus{}c^{(k\minus{}2)m}\plus{}...\plus{}c^{m}$ such that $ t \equal{} de, s \equal{} df, e\minus{}f \equal{} 1$ then by lemma 1 $ (s\minus{}t)^{2}|st$ and lemma 2 is proved.

Finally note that for the numbers $ c^{z}\equal{} c^{z\minus{}u}(c^{u})$ and $ c^{z\minus{}u}(c^{u}\minus{}1)$ where $ c$ is a positive integer greater than $ 1$  there exist numbers $ d \equal{} c^{z\minus{}u}, e \equal{} c^{u}, f \equal{} c^{u}\minus{}1$ such that $ c^{z}\equal{} de, c^{z\minus{}u}(c^{u}\minus{}1) \equal{} df, e\minus{}f \equal{} 1$ then by lemma 1 $ (c^{z}\minus{}c^{z\minus{}u}(c^{u}\minus{}1))^{2}|(c^{z}*c^{z\minus{}u}(c^{u}\minus{}1))$ -- call this property [b]lemma 3[/b].

Now, we can construct the set $ S$ inductively. Start with numbers $ 1,2$ as the base step for $ n \equal{} 2$. Then, when we have $ k$ numbers $ a_{1}< a_{2}< ... < a_{k}$ belonging to set $ S$ and satisfying the conditions, we can construct a new set of $ k\plus{}1$ numbers:
$ b_{1}< b_{2}< ...b_{k\plus{}1}$ where for $ i \equal{} 1,2,...,k$, $ b_{i}\equal{} r^{a_{k}\minus{}a_{i}}(r^{a_{i}}\minus{}1)$ and $ b_{k\plus{}1}\equal{} a^{k}$ where $ r$ is any positive integer greater than $ 1$. Then, for any $ i,j\in\{1,2,...,k\}$, $ i < j$ by lemma 1, $ (b_{j}\minus{}b_{i})^{2}|b_{i}b_{j}$, and for any $ i\in\{1,2,...,k\}$, by lemma 3, $ (b_{k\plus{}1}\minus{}b_{i})^{2}|b_{k\plus{}1}b_{i}$ so the constructed set of $ k\plus{}1$ numbers satisfies the conditions of the problem.
Then, from the new set with $ k\plus{}1$ numbers we can construct a set with $ k\plus{}2$ numbers and so on, until we reach $ n$, and the result follows.

I have am other method .
[b]Lemma 1[/b] $ A \equal{} \{a\}$  and $ |A| \equal{} k$ satisfy $ (a \minus{} b)^2|ab,\forall a,b\in A$ then the set
$ A_1 \equal{} \{a \plus{} d\}$ has this property where $ \prod_{i > j \equal{} 1}^k (a_i \minus{} a_j)^2|d$ satisfy.
[b]Lemma 2[/b] If the set $ A$ satisfy the condition then the set $ A \plus{} \{0\}$ also satisfy the condition.
It is quite easy to prove .
Now we prove by induction:
$ k \equal{} 2$ then chose $ a,b \equal{} 1,2$
Suppose it is true for $ k$ we prove it is also true for $ k \plus{} 1$
From lemma 1,lemma 2 then it true .
With this method we can solve problem.
Let $ a,b\in N$ satisfy condition $ (a \minus{} b)^2|ab$ then exist A a set such that
1)$ a,b\in A$
2)$ (x \minus{} y)^2|xy,\forall x,y\in A$

Assume the set is finite. Let B be the lcm of every element in the set. Add 0 to the set. Increase each element by B. We now have a larger set so there is no maximum and the set can be infinite.

edit: So we can make a set of any finite size. Luckily this is all the problem was asking for and we dont need it to be an infinite set.

We must be careful! As stated above, "[color=#FF0000]We now have a larger set so there is no maximum and the set can be infinite[/color].", this was yet unproven, and, moreover, false, as I will show below. All that was proven is that from a set with $n$ elements we can create one with $n+1$ elements, and so we can create [b]as large a set as wanted[/b], but not necessarily an[b] infinite [/b]set.

Consider any such set $S$. Take an element $a \in S$, $a \neq 0$. Then for all elements $b\in S$, $b \neq a$, we will need have $(a-b)^2 \mid ab$, hence $(a-b)^2 \leq |ab|$, therefore $b^2 \leq |ab| + 2ab - a^2 < 3|ab|$. So, $|b| < 3|a|$, therefore $|S| \leq 6|a|$.


Let a,b,c are non-negative numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Prove that
$ a^2b^3\plus{}b^2c^3\plus{}c^2a^3\le 3$
It is very hard. Can anyone prove it?

[quote="thegod277"]Let a,b,c are non-negative numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that
$ a^2b^3 \plus{} b^2c^3 \plus{} c^2a^3\le 3$
It is very hard. Can anyone prove it?[/quote]
Long time ago... I sent hungkhtn the following solution :)

[quote="thegod277"]Let a,b,c are non-negative numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that
$ a^2b^3 \plus{} b^2c^3 \plus{} c^2a^3\le 3$
It is very hard. Can anyone prove it?[/quote]
This is problem 2.12 from Algebraic Inequalities (page 91). I like Can_hang solution, too.
In addition, problem 2.33 (page 137) shows that

$ \frac {a^rb \plus{} b^rc \plus{} c^ra}{3} \le (\frac {a \plus{} b \plus{} c}{3})^{r \plus{} 1}$,

for any $ 0 < r\le m$, where $ m \equal{} 1.558...$ is a root of the equation $ (1 \plus{} m)^{1 \plus{} m} \equal{} (3m)^m$.

Nice solution, Can-hang, but hard. I have a nice solution, it is: With that condition, prove that
1) $ \sum a^2b^3\plus{} 3abc\le 2(ab\plus{}bc\plus{}ca)$(by mufc)
2) $ 3abc \plus{}3\ge 2(ab\plus{}bc\plus{}ca)$ (by thegod277)
Use 1 and 2 tha inequality is proved. Can anyone prove 1 or 2? :)

[quote="thegod277"]Nice solution, Can-hang, but hard. I have a nice solution, it is: With that condition, prove that
1) $ \sum a^2b^3 \plus{} 3abc\le 2(ab \plus{} bc \plus{} ca)$(by mufc)
2) $ 3abc \plus{} 3\ge 2(ab \plus{} bc \plus{} ca)$ (by thegod277)
Use 1 and 2 tha inequality is proved. Can anyone prove 1 or 2? :)[/quote]
I don't think my solution is hard, thegod277. It's just a basic on Cauchy Schwarz technique. ;)

[quote="thegod277"]Nice solution, Can-hang, but hard. I have a nice solution, it is: With that condition, prove that
1) $ \sum a^2b^3 \plus{} 3abc\le 2(ab \plus{} bc \plus{} ca)$(by mufc)
2) $ 3abc \plus{} 3\ge 2(ab \plus{} bc \plus{} ca)$ (by thegod277)
Use 1 and 2 tha inequality is proved. Can anyone prove 1 or 2? :)[/quote]
1) It's equivalent to VasC's Inequality
\[ ab^2\plus{}bc^2\plus{}ca^2 \le 2\plus{}abc\]
2) It's is wrong, see http://www.mathlinks.ro/viewtopic.php?t=195217

thank you! I have mistake when prove this. Very sorry.

This is problem 2.12 from Algebraic Inequalities (page 91). I like Can_hang solution, too.
In addition, problem 2.33 (page 137) shows that

$ \frac {a^rb \plus{} b^rc \plus{} c^ra}{3} \le (\frac {a \plus{} b \plus{} c}{3})^{r \plus{} 1}$,

for any $ 0 < r\le m$, where $ m \equal{} 1.558...$ is a root of the equation $ (1 \plus{} m)^{1 \plus{} m} \equal{} (3m)^m$.[/quote]

How to prove it？


function
induction
floor function
logarithms
algebra unsolved
algebra

Let $f: N \to N \cup 0$  be defined by  $f(1) = 0$, and \[f(n) = \max_{1 \leq j \leq [n/2]}(f(j)+f(n-j)+j)\] for all $n \geq 2$. Determine $f(2000)$.

[i]Edited by Myth[/i]

Using induction it's not hard to verify that $f(2n)=2f(n)+n$, $f(2n+1)=f(n)+f(n+1)+n$. From here one can get
\[f(n)=\sum_{\substack{m,k\geqslant0\\ (k+1)2^{m}\leqslant n}}(-1)^{k}(n-(k+1)2^{m})=nS_{2}(n)+\sum_{m=0}^{\lfloor\log_{2}n\rfloor}2^{m-1}\left((-1)^{\lfloor\frac n{2^{m}}\rfloor}\left(\left\lfloor\frac n{2^{m}}\right\rfloor+\frac12\right)-\frac12\right), \]
where $S_{2}(n)$ is the sum of binary digits of $n$. I'm lazy to compute $f(2000)$...


function
calculus
derivative
limit
real analysis
integration
calculus computations

Can you construct a function f:[a,b] -> R : f is continuous on [a,b], monotonically increasing on [a,b], but NOWHERE differentiable on [a,b]?

No, you cannot construct such a function since if $f$ is an increasing function then it is known that the functions is differentiable almost everywhere and its derivative is a measurable function.

I'm not sure I follow "it is known that the functions is differentiable almost everywhere and its derivative is a measurable function." Could you be more precise?

Consider $f(x) = \lim_{n \rightarrow \infty}\frac{1}{2n}\sum_{i=0}^{2n}u\left(x-\frac{i}{2n}\right)$ on $[0,1]$ where $u(x)$ is the standard step function.

Is this function monotonic everywhere? Is it differentiable anywhere?

Btw, did I put this in the right forum?

$f(x)=x$ for $0\le x\le 1.$

It is differentiable except at $0$ and $1.$

By the way, the question "is such-and-such a function Stieltjes integrable" makes no sense unless you are specifying integrable with respect to what other function.

Also: didilica was precise; he quoted a well-known theorem exactly and there is no more precision to be had. The words "almost everwhere" have the technical meaning "except for a set of Lebesgue measure zero." The proof of this theorem could be found in a number of different books on Lebesgue measure and integration.

Thanks Kent for the response, but isn't my function, comprised of infinitely many step functions differentiable nowhere on [0,1]?

And, to the other statement, I haven't covered "Lebesgue measure and integration" before, so I wasn't familiar with his statement. Thanks for clearing that up.

I removed the sillyness about the Stieltjes, ignore that.

[quote="Spoon"]... but isn't my function, comprised of infinitely many step functions ...[/quote]
Your function is not comprised on infinitely many step functions. Your function is the [i]limit[/i] of  sequence of step functions of successively finer partition, and in detail that limit is precisely the boring piecewise linear function given above.

I meant for the sum to go to 2n, I fixed that. So my function will converge to f(x) = x, and is thus differentiable? That's no fun at all.

Yes, that's it. No fun at all.

Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=billingsley&t=118219]here[/url] for examples of strictly increasing functions whose derivatives are zero wherever they exist.


MIT
college
projective geometry
calculus
Support
geometry

Ok, so please post here if you are selected.
Currently-
Selected-
Asish Mahapatra
Aditya G Date
Arnav Mittal
Gaurav Shukla
Pulkit
Not selected-
Mihir

Also can someone explain-

  List of candidates recommended for mentoring  (summer program) under
Stream SP (Engineering) (Not for Fellowship Award)

Sl.No. APPLN.NO. NAME

1. 728922 AKASH SHARMA
2. 733428 GOUTHAM  A.
3. 706171 KAUSHIK MITRA
4. 726786 MANISH KRISHNA
5. 719169 NAKHATE ABHISHEK DILIP
6. 731539 RAGHAVENDRA  V.
7. 4038 SAHA KANAD KRISHNA KAMAL
8. 727776 SATYAJIT SINGH
9. 725384 SOURAV DUTTA
10. 724648 SWATI VERMA
11. 719455 TARU DOODI
12. 035205 UTKARSH GTRIPATHI
13.  724647 YUGANDHARA SUNIL MANGALEKAR
Given a choice, I'd obviously like to be mentored by Utkarsh Tripathi  :omighty:

Also, are you allowed to take NTSE scholarship with KVPY? I guess nearly all KVPY fellows will face this prob gladly.

congrats to all selected  :)

no,NTSE scholarship isn't allowed with kvpy(but y to take tension,kvpy gives a lot more)  :rotfl:

hi
i am Manish Krishna....selected for mentorship programme...u can see my name in the above list.
i really dont know what does it mean...it is clear that they wont give any money...then what actually i am selected for...any one having some idea...please tell me. :idea:

Hi bro! Where are you from? And what are you currently doing? I guess you must be an existing KVPY scholar? And did you apply for this mentor and stuff or you really don't have any idea of what happened?
Sorry for so many questions, but you're the only hope :p
Maybe we should try calling up IISc as well? Or just wait until they finally come and tell us what's going on!  :rotfl:

[quote="nuclear_alchemist"]

Given a choice, I'd obviously like to be mentored by Utkarsh Tripathi  :omighty:

[/quote]

Who is he?

Among the only 2-3 from UP to have cleared INMO in its history. He went on to get selected to represent India at IMO and won a gold medal or something. Following which he got a call from MIT as well.
Disclaimer: The above information is true in essence but there might be slight deviations in 1 or 2 points.

hi,
i am from bihar..studying in srm university, chennai ...doing b.tech(biotehnology)..im in 2nd year. I applyed in engg (sp) stream..i was selected for interview...nd was interviewed in iit bombay...on 3 nd 4 jan this year. nw the results are out..it is clear that i am selected bt not for the money thing..i didnt try in kvpy before this....actually dont know what am i selected for...if u know plz tell me..

SRM? My sis was gonna go there for B.Tech CS but decided Amity at the last moment hehe. And I'll ask my other friends who qualified this year and tell you if they have any knowledge of whats up...

hi
ya SRM.. :rotfl: ......Its biotechnology ranks no: 2 in in india in private universities
(no. 1 is SHASTRA UNIVERSITY not AMITY as it projects itself)...according to BIOSPECTRUM...u know any thing about the mentoring programme...are you from U.P...what are u studiying..

My sis joined Amity coz it was in our hometown itself and fairly prestigious.
Yes, I'm from UP and studying in class 11.

What are the approximate dates of the camp? Can past participants(plods Rituraj) give an idea? Also state the approx variation possible. There are a few things I need to plan accordingly...

just nthing can b said abt dates...u'll get the letter soon

In 2008 ,it was in july 1st week
2007--may :P

Thanks. I hope its in may this year as well. I might be in a tricky situation if it falls in July :(

of course i did  :D
thats why im not married even though i have a fiancee since 3-4 years ( and 2 kids)
 :rotfl:

which team do you suppprt?
lemme guess : MANU  :P  if not ill eat my hat

thanks for the inspiring words sagar  :lol:
but wat re........2 more years of hard work. its just the begining
but still i ws in junior last year and ws competing against 8ths and 9ths
next year derll be all 12ths from rammaiaah and kota and wat not, its phadu
 :rotfl:
so how was ur jee???

is there anyone from 11th on this site?
or 10th?..i just noe fbaggins

yeah lampard.......

i am also in 11th.....(hoping to somehow pass in boards...) :rotfl:  :rotfl:

so how r u planning to go abt the preparation for jee.........

plz reply and advice me how to do it since u have joined some institute nah........

 :huh:  :huh:

he has no worry(lampard) with mprakash at his fore...

and i support ManU know
(wouldn't hav,had liverpool or real or juve remained alive :roll: )

hey ritu did u watch the match???
im posting this 5 mins after d match
 :(  :bomb:  :noo:
sad yaar......
chalo koi bath nahi.........next year.....just like mee..
@ jaswinder  : ill reply later....chelsea LOST...."sniff sniff"

they played quite well

well.....iniesta had the last laf... :)

Now it will be decided wgo is the best Barca or ManU
 :D

aha they have updated their site

and they have put up the applications for this year......

cool......

so how did u(older batch guys) prepare for it....

??????

 :rotfl:  :rotfl:  :rotfl:

so u r preparing....... :P

it will effectively reduce ur chances   :rotfl:

i really dont know
@ present im just reading h c verma for physics , cbse chem books and 1st course in calculus by lang
and doing geometry from olympiad books....
u???
my boards were very bad......very
i hope i get above 90  :rotfl:

me doing nothing.......

don't actually have nething to do........

just passing out time till the board results.............

 :rotfl:  :rotfl:  :rotfl:

lolz.....arre a person of ur itelligence doesnt need 2 study.
cing ur name they will call u directly for interview  :winner_first:

hey , by the way dat kvpy site is very confusing....@least for me :rotfl:

best of luck to you all  :)

but how can i apply when i m not sure of passing boards??????? :rotfl:  :rotfl:

[quote="skand"]but how can i apply when i m not sure of passing boards??????? :rotfl:  :rotfl:[/quote]
ahem..u r in 12th

and u dont need boards 4 SB stream :P

Camp at IISc Bangalore from 9-13 October. And they FORGOT to send me a mail notifying me about the camp and scholarship <_< Thankfully, asish told me. Called up the buffoons in IIS and gave my reg no. and stuff again. They didn't even bother to send a letter/call on the phone since mails were sent half a month ago!
The timing of the camp perfectly clashes with the expected exam dates in school  :rotfl: Hopefully I'll be spared E.Ed and Chem :P


MATHCOUNTS
geometry
pigeonhole principle

How many cards must be selected from a standard deck to be certain that three cards from the same suit are drawn?

Hint (besides noting that a similar problem was posted before  ):[hide]Think about how you can not get three cards of the same suit.[/hide]

[hide]9, if you draw 2 spades, 2 clubs, 2 hearts, 2 diamonds and then your 9th card will make the third of a suit.[/hide]

YEA!!! :D  :D  :D  I'm gettin' kinda good at these!

Good for you! :)

Btw, how old are you again?

We're all 13!!! nothing to be embarrased about... you're pretty smart for 13 you know.

[quote="MathFiend"]'cuz I'm dumb for 13, especially on this site. I thought I was kinda smart before I started coming here, but then I realized that so many people are on such a higher level of math, yet are my age. I'm only in algebra 1 as an eighth grader and I thought that was pretty good, but now I'm not so sure. so many people are in geometry as 8th graders etc. My MC coach told me earlier tonight (I had open house at my school) that our team is one of the best, but I doubt he really knows what we're up against...genius's like you and neal and everyone else...[/quote]
I know, I had that feeling too, but so what there's smarter people than us now? We just got to study harder to beat those people. This, this is the variable :)  lesson I learned from AoPS. That's what AoPS is all about, getting better at math.

Btw, our school is relaunching MATHCOUNTS program (due to my 3 hours of intense negotiations :twisted: ), so I'll see [b]you[/b] at the nationals :wink:

well what were the scores for the state thingie for the top 4 finishes in indiana

yeah! everyone's 13! even i'm 13! no, jk i'm a bit older;)

uh like 46, 46, 44, 43 or something like that I know the top 2 were perfects, so tare, I don't think I'll be seeing you at nats, I'm not that smart

pfft, dont listen to him, you can always make it

w/e (hah, I'm goin "gurly" on you neal  :) )

I didn't say she can't. i just said "Oh, oh well" :)

and MathFiend, the *bambi eyes* is still mine, capisce? :D

well in Garfield, there was this strip where Garfield does the "bambi eyes" to jon to get a steak.

You know, as in the Disney Bambi, the deer? He (or she; I never saw it :) ) makes these cute little glittering eyes which makes people go "awwwww..." :lol:

why would you want to ownz the bambi eyes. i just see no reason...[hide]pfft, go to the gim[/hide]

well for one thing, it's cute and girly. and if you say that's a bad thing, then you're being sexist...(which I'm [b]sure[/b] you won't say it anyways :) )

good choice, just like me with "lolt", you can't understand the art of "bambi eyes" :P

actually, i do...[hide]i wont say anything more[/hide]

You are all 13? Im 12....

The answer is 9.


Hello,

i)Prove that if $ x > 1$ and $ x\in (\mathbb{R}{ \minus{} }\mathbb{Q})$ then :  $ \left\lfloor \frac {\lfloor nx\rfloor}{x}\right\rfloor \equal{} n \minus{} 1$

ii) Prove there is an only real $ a$ such that:  $ \forall n \in \mathbb{N}^{*}, \left\lfloor a\lfloor na\rfloor\right\rfloor  \minus{} \lfloor na\rfloor\equal{} n\minus{}1$

Hello,

Thanks Farenhajt, very nice problem and very nice proof  :D

[quote="PKMathew"][quote="Farenhajt"][hide="Very nice problems :)"]
[b]Part 1.[/b] Since $ nx$ can't be integer, we get $ [nx] < nx$. On the other hand, $ \{nx\} < 1 < x\implies nx \minus{} [nx] < x\implies nx \minus{} x < [nx]$. From $ (n \minus{} 1)x < [nx] < nx$ the conclusion follows.

[b]Part 2.[/b] Obviously $ a\neq 0$. Put $ n \equal{} 1$. Then $ [a[a]] \minus{} [a] \equal{} 0$. Therefore $ a$ can't be negative, since $ [a[a]]$ would be non-negative (because $ a[a]$ is positive) and $ [a]$ would be negative, hence LHS would be positive. So we can follow thus:

[u](a)[/u] $ [na] > na \minus{} 1\implies a[na] > a^2n \minus{} a\implies [a[na]] > a^2n \minus{} a \minus{} 1$ and $ [na]\leqslant na$ yields $ n \minus{} 1 \equal{} [a[na]] \minus{} [na] > a^2n \minus{} an \minus{} a \minus{} 1\implies n(a^2 \minus{} a \minus{} 1) < a\implies a^2 \minus{} a \minus{} 1 < {a\over n}$

[u](b)[/u] $ [na]\leqslant na\implies a[na]\leqslant a^2n\implies [a[na]]\leqslant a^2n$ and $ [na] > na \minus{} 1$ yields $ n \minus{} 1 \equal{} [a[na]] \minus{} [na] < a^2n \minus{} an \plus{} 1\implies \minus{} 2 < n(a^2 \minus{} a \minus{} 1)\implies \minus{} {2\over n} < a^2 \minus{} a \minus{} 1$

From (a) and (b) we get $ \minus{} {2\over n} < a^2 \minus{} a \minus{} 1 < {a\over n}$, but as that must hold for arbitrarily large $ n$, we conclude that $ a^2 \minus{} a \minus{} 1 \equal{} 0$, and with $ a > 0$, we get $ a \equal{} {1 \plus{} \sqrt {5}\over 2}$

Now we have to prove that the identity holds for $ a \equal{} {1 \plus{} \sqrt {5}\over 2}$. Putting $ a \equal{} 1 \plus{} \phi$ will simplify the expression to

$ [(1 \plus{} \phi)[(1 \plus{} \phi)n]] \minus{} [(1 \plus{} \phi)n] \equal{} n \minus{} 1$

$ [[n \plus{} \phi n] \plus{} \phi[n \plus{} \phi n]] \minus{} [n \plus{} \phi n] \equal{} n \minus{} 1$

$ [n \plus{} \phi n] \plus{} [\phi(n \plus{} [\phi n])] \minus{} [n \plus{} \phi n] \equal{} n \minus{} 1$

$ [\phi n \plus{} \phi[\phi n]] \equal{} n \minus{} 1\quad (*)$

From $ a^2 \minus{} a \minus{} 1 \equal{} 0\implies (1 \plus{} \phi)^2 \minus{} (1 \plus{} \phi) \minus{} 1 \equal{} 0\implies \phi^2 \plus{} \phi \minus{} 1 \equal{} 0$, we have $ \phi \equal{} 1 \minus{} \phi^2$. Plugging that into $ (*)$, we get

$ [(1 \minus{} \phi^2)n \plus{} \phi[\phi n]] \equal{} n \plus{} [\phi( \minus{} \phi n \plus{} [\phi n])] \equal{} n \plus{} [ \minus{} \phi\{\phi n\}] \equal{} n \minus{} 1$, since $ 0 < \phi\{\phi n\} < 1$ (strictness of the left part comes from irrationality of $ \phi$, i.e. from the given fractional part never being zero).

Hence we conclude that indeed unique such value is $ a \equal{} {1 \plus{} \sqrt {5}\over 2}$[/hide][/quote]

Should

$ \boxed{ nx \minus{} [nx] < x}$

$ \boxed{(n \minus{} 1)x < [nx] < nx}$

be

$ nx \minus{} [nx] \le x$

$ (n \minus{} 1)x \le[nx] < nx$ ?[/quote]

By definition, fractional part is always strictly less than $ 1$, hence $ \{nx\}\equal{}nx\minus{}[nx]<1$, and by the given condition, $ 1<x$, so we get $ nx\minus{}[nx]<x$


Solve the system
$ \left\{\begin{array}{l}

xy\plus{}yz\equal{}2 \\

  x^2 \plus{}2y^2 \plus{}z^2\equal{}4

\end{array}\right.$

 :ninja:

[hide]
Doubling the first equation we have $ 2xy\plus{}2yz\equal{}4$

Subtracting this from the 2nd equation we have $ x^2\minus{}2xy\plus{}y^2\plus{}y^2\minus{}2yz\plus{}z^2\equal{}(x\minus{}y)^2\plus{}(y\minus{}z)^2\equal{}0$

I assume we are solving this over the reals, so $ x\equal{}y, y\equal{}z \implies x\equal{}y\equal{}z\equal{}\pm{1}$

So $ (1,1,1)$ or $ (\minus{}1,\minus{}1,\minus{}1)$
[/hide]

hello, if we are solving over the complex numbers i have got
$ x\equal{}\minus{}\frac{\minus{}1\minus{}i\plus{}iy^2}{y},y\equal{}y,z\equal{}\frac{1\minus{}i\plus{}iy^2}{y}$
$ x\equal{}\minus{}\frac{\minus{}1\plus{}i\minus{}iy^2}{y},y\equal{}y,z\equal{}\frac{1\plus{}i\minus{}iy^2}{y}$
Sonnhard.


If $ m$ and $ n$ are odd integers, how many terms in the expansion of $ (m\plus{}n)^6$ are odd?

All terms are of the form $ \binom6km^kn^{6\minus{}k}$ for $ 0\le k\in\mathbb Z\le6$. The $ \binom6k$ part must be odd for the whole thing to be odd (we already know that $ m^kn^{6\minus{}k}$ will always be odd). It is only odd for $ \boxed4$ values of $ k$: $ 0,2,4,6$.

I have just solved this problem in Alcumus and it reminded me of a property of the Pascal's triangle. :)

We know that the product of $ m^k$ and $ n^{6-k}$ are always odd numbers. Hence, we are only concerned for $  \binom{6}{k}$. It must be odd for the coefficient of $ m^{k}n^{6-k}$ to be odd.

Basically, there is a way to find odd numbers for the $ nth$ row of a Pascal's Triangle. That is, we should express $ n$ in binary representation, (in our case $  n=6$) so $ 6$ is $ 110_2$ in binary. We then count the number of $ 1's$ in the representation. Then...

$ g(x) = 2^x$ , where $ g(x)$ is the number of odd numbers in the $ nth$ row of the Pascal's Triangle and $ x$ is the number of $ 1's$ in the binary representation of $ n$.

So $ g(2) = 2^2 = 4$

Hence there are $ 4$ odd numbers in the $ 6th$ row of the Pascal's Triangle.

As a result, we can say that there are $ 4$ odd coefficient in the expansion.

Note that this idea is useful, let say, find the number of odd coefficients in the expansion $ (x+1)^99$

I hope this helps. :)

The [hide=...]Binomial theroem is very helpful for this![/hide]

You can use the binomial theroem to expand it, then solve it...

[quote=Possible]Is the answer 4 because the 6th row of the pascal's triangle has 4 odd numbers?[/quote]

I [i]think[/i] so...

I almost said 2 for this one because I forgot 1 was odd!!!

[hide=solution]$(m+n)^6=m^6+6m^5n+15m^4n^2+20m^3n^3+15m^2n^4+6mn^5+n^6$
$\text{odd}\cdot\text{odd}=\text{odd}$
replace all $m^an^b$'s with odd number $o$
$o+6o+15o+20o+15o+6o+o$
$\text{odd}\cdot\text{even}=\text{even}$
$\boxed{4}$[/hide]


calculus
integration
calculus computations

How do I compute things such as
$\int \int f(x) d^{2}x$
I can convert $d^{2}x$ to $dx dx$, but it doesn't help much.
$\int \int f(x) d^{2}x = \int \left (f(x) dx-\int f'(x) dx dx \right)$

That notation doesn't really make sense. What do you mean?

Perhaps rayne means [url=http://mathworld.wolfram.com/RepeatedIntegral.html]repeated integrals[/url], e.g. $\int_{0}^{x}\int_{0}^{x}f(x) \,dx\ dx$

No, I mean neither. The question was exactly like that. I really mean to ask how to integrate double differentials, desperately.


It looks like the "view next/previous topic" buttons aren't working properly: I get "the page cannot be displayed".

Hmm...  Appears to work for me.  Can you try cleaning out your cookies and trying again?  If it still fails, can you link me to the post where this is failing?  Thanks.

Now it works for me too :? (and there wa no cookie cleaning done :)).

It only works here, but it doesn't in other subforums. The last time I tried it, it was [url=http://www.mathlinks.ro/viewtopic.php?p=99856#99856]here[/url], but it has caused the same problem ("the page cannot be displayed") on more or less every topic I've tried.

[Edit: it wasn't working here either; when I pressed "view next topic" it displayed that stuff about no new topics, just like it should have, but pressing "view previous topic" got me the same annoyng "the page cannot be displayed"]

The problem is on my side: I'm using Opera right now and there are no problems, so there's something wrong with my IE. Any advice?

[quote="rrusczyk"]Can you tell if the URL is right?[/quote]

Using the "View previous" link in the URL shown in the previous post, I get the same error, using IE.  The actual link that seems to cause grief is:  [url]http://www.mathlinks.ro/viewtopic.php?t=14111&view=previous[/url]

IE is reporting a DNS error resolving a server name.

The equivalent link on the AoPS side seems to work fine:  [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14111&view=previous[/url]

If I copy/paste that ML URL into a browser, such as IE, it works fine, though it does do something a little peculiar, which is take me to:

http://www.mathlinks.ro/./viewtopic.php?t=6428

That extra ./ could be the problem - perhaps older IE browsers aren't clever enough to ignore that?  Can you try posting that link above and see what happens?

Both of the following links work correctly in my IE browser:

[url]http://www.mathlinks.ro/./viewtopic.php?t=6428[/url]
[url]http://www.mathlinks.ro/viewtopic.php?t=6428[/url]

BTW, the above were not posted as multiline URL's.  I guess our browsers decided it was safe to split the lines following the question mark.  I'm sure that is irrelevant to the problem grobber posted.

Edit:  [As of 35 minutes later, IE is now successfully resolving and displaying the links that were previously broken.  Maybe you fixed something?]


Do there exist $ 14$ positive integers, upon increasing each of them by $ 1$,their product increases exactly $ 2008$ times?

Unless you try to prove there are no such integers :), you easily obtain the solution $ (x_1,\ldots,x_{14})$, where $ x_1\equal{}\ldots\equal{}x_{10}\equal{}1$, $ x_{11}\equal{}x_{12}\equal{}x_{13}\equal{}4$, and $ x_{14}\equal{}250$, inspired by $ 2008\equal{}2^3\cdot251$.

If I undestandt your condition we must find $ x_i\in N$, suth that \[ \prod_{i\equal{}1}^{14}\frac{x_i\plus{}1}{x_i} \equal{}2008.\]
Let $ x_1\ge x_2\ge x_3...\ge x_{14}$. Because $ 2008\equal{}2*2*2*251$. We can take for $ x_1\plus{}1\equal{}251$ or $ x_1\equal{}250$, then
\[ \prod_{i\equal{}2}^{14}(1\plus{}\frac{1}{x_i})\equal{}2*2*2*2*5*5*5.\] We can take $ x_2\equal{}x_3\equal{}x_4\equal{}4$, then
\[ \prod_{i\equal{}5}^{14}(1\plus{}\frac{1}{x_i})\equal{}2^{10}.\], then we can take $ x_5\equal{}x_6\equal{}...\equal{}x_{14}\equal{}1$.

It's easily recognized that there must be at least seven integers that must be equal to 1's. The rest is straightforward when the problem goes from 14 integers to seven.

This is very easy.

Let $a_1=a_2= \dots a_{10} =1$, $a_{11}, \dots, a_{13} =4 $ and $a_{14} =250$.

Just realized this is the same as #2  :blush:


let A be a $ n\times n$ matrix with rank(A)=r. prove that by elementary transforms, we can translate it into matrix $ B \equal{} [b_{ij}]_{n\times n}$ such that $ b_{11} \equal{} b_{22} \equal{} ... \equal{} b_{rr} \equal{} 1$ and the other entries are 0

To do this, you must act from the left and the right both. Or to put it another way, you must use both elementary row operations and elementary column operations.

If $ A$ has rank $ r$, then you can reorder its rows so that its first $ r$ rows span the rest of the rows and then take linear combinations to eliminate the last $ n \minus{} r$ rows.  You now have only $ r$ rows to work with.  On the other hand, $ A$ has row rank $ r$ if and only if $ A$ has column rank $ r$, so...

The matrix B that we get from this process is known as the normal form of matrix A.


given are three numbers $x,y,z \in \mathbb{Z^{+}}$ such that $x^y+1=z^2$
let $p$ be the number of divisors of $x$ and $q$ be the number of divisors of $y$
Show that
$p\geq q+2$


On a map, a 12-centimeter length represents 72 kilometers.  How many kilometers
does a 17-centimeter length represent?

Using the ratio given, we have $ 72\times{\frac{17}{12}}\equal{}6\times17\equal{}\boxed{102}$.


linear algebra
matrix
vector
superior algebra
superior algebra solved

If A is idempotent then :
\[\displaystyle tr(A)=r(A)\]
where r(A) is the order of A and tr(A) is the sum of its diagonal elements.

As far as i understand by order of $A$ you meant $rankA$.

If we consider $A$ as operator on vector space $V$ we could see that every $v\in V$ could be written in form $v=Av+(v-Av)$ and $Av \in ImA$, $(v-Av) \in KerA$. It easy to verify that the above sum is direct. Therefore if we choose as the base for $V$ union of the basis for $ImA$ and $KerA$ we will obtain that in this base $A$ is of form $A=diag(1,1,...,1,0,0,...0)$


abstract algebra
superior algebra
superior algebra theorems

I never really liked the classical proofs for Nakayama's Lemma (using the determinant trick). Therefore I wanted to share an alternative proof. I'll just put it here for reference. Feel free to report any mistakes.

[b]Lemma's:[/b]
1) Let $ M$ be a finitely generated $ A$-module and let $ \mathfrak{a}$ be an ideal of $ A$ contained in the Jacobson radical $ \mathcal{J}$ of $ A$. Then $ \mathfrak{a}M \equal{} M$ implies that $ M \equal{} 0$.

[quote]
If $ M$ were not zero, it would have a simple quotient, isomorphic to $ A/\mathfrak{m}$ for some maximal ideal $ \mathfrak{m}$ of $ A$. Then we would have $ \mathfrak{m}M\neq M$, so that $ \mathfrak{a}M\neq M$ as $ \mathfrak{a}\subseteq \mathfrak{m}$.
[/quote]

2) Let $ S$ be a multiplicatively closed subset of a ring $ A$, and let $ M$ be a finitely generated $ A$-module. Then $ S^{ \minus{} 1}M \equal{} 0$ if and only if there exists an $ s \in S$ such that $ sM \equal{} 0$.

[quote]
If $ x_1, \ldots, x_n$ generate $ M$ and If $ S^{ \minus{} 1}M \equal{} 0$ then $ s_ix_i \equal{} 0$ for some $ s_i \in S$. Now $ s \equal{} s_1s_2\cdots s_n \in S$ because $ S$ is multiplicatively closed and $ sx_i \equal{} 0$ for all $ x_i$, hence $ sM \equal{} 0$. The converse is obvious.
[/quote]

3) If $ \mathfrak{a}$ is an ideal of $ A$ then $ S \equal{} 1 \plus{} \mathfrak{a}$ is a multiplicatively closed subset of $ A$ and $ S^{ \minus{} 1} \mathfrak{a} \subseteq \mathcal{J}(S^{ \minus{} 1}A)$ where $ \mathcal{J}$ denotes the Jacobson radical.

[quote]
We check easily that $ S$ is a multiplicatively closed subset of $ A$ since $ (1 \plus{} a)(1 \plus{} a') \equal{} 1 \plus{} (a \plus{} a' \plus{} aa') \in 1 \plus{} \mathfrak{a}$. We also have that $ S^{ \minus{} 1} \mathfrak{a} \subseteq \mathcal{J}(S^{ \minus{} 1}A)$ since $ 1 \minus{} \frac {a_1}{1 \plus{} a_2} \cdot \frac {x}{1 \plus{} a_3} \equal{} \frac {1 \plus{} a_2 \plus{} a_3 \plus{} a_2 a_3 \minus{} a_1 x}{(1 \plus{} a_2)(1 \plus{} a_3)} \equal{} \frac {1 \plus{} a_4}{(1 \plus{} a_2)(1 \plus{} a_3)}$ is a unit in $ S^{ \minus{} 1}A$ for all $ a_1,a_2,a_3 \in \mathfrak{a}$ and $ x \in A$.
[/quote]

[b]Theorem:[/b] Let $ M$ be a finitely generated $ A$-module and let $ \mathfrak{a}$ be an ideal of $ A$ such that $ \mathfrak{a}M \equal{} M$. Then there exists an $ x \equiv 1 \mod \mathfrak{a}$ such that $ xM \equal{} 0$.
[quote]
If $ M \equal{} \mathfrak{a}M$, then since $ S^{ \minus{} 1}M \equal{} (S^{ \minus{} 1}\mathfrak{a})(S^{ \minus{} 1}M)$ it follows from (1) and (3) that $ S^{ \minus{} 1}M \equal{} 0$. But from (2) we see that there is an $ x \equiv 1 \mod \mathfrak{a}$ such that $ xM \equal{} 0$, so we're done.
[/quote]

nice. but you use twice equivalent forms of the axiom of choice (exercise for the reader: find them!), so the usual proof is more constructive.


let's solve this :

show that $\displaystyle 61!+1 \equiv 63!+1 \equiv 0 (mod 71)$.

can you generalize it ?  :huh:

[quote="dblues"]What generalisation do you have in mind?[/quote]

you see $71$ is a prime number i think that we can generalize it in some way that for each prime we can find two numbers for this congruence !!! i don't know realy !!  :?  but can you solve it  ??

[quote="lomos_lupin"]We have a thoerem with the name of (Wilson,wealson or ...)
let $P$ be a prime then
$(p-1)!\equiv^p -1$ so your problem is equal to

$\frac{71!}{61!}\equiv\frac{71!}{63!}\equiv1$[/quote]

that was correct Arman !! but the main problem is that can you generalize it ?   ;)

hey Arman you proved me that still you are not too old and too fat that you can't solve problems !!  :lol:

[quote="a_vakilian"][quote="lomos_lupin"]We have a thoerem with the name of (Wilson,wealson or ...)
let $P$ be a prime then
$(p-1)!\equiv^p -1$ so your problem is equal to

$\frac{71!}{61!}\equiv\frac{71!}{63!}\equiv1$[/quote]
I think you should prove that $\frac{70!}{61!}\equiv\frac{70!}{63!}\equiv-1$ According to the [b]Wilson theorem[/b]. :huh:[/quote]
i think also ali was right !!  :P  then what has happened !! maybe last night i was sleepy !! i will think it over more !!
but i told you that the main problem is trying to generalize it !!  :lol:

now i've tried to give a detailed solution for one of the congruences at
[url]http://spx.t35.com/sol.htm[/url]
and the other one is done the same as this one

i would like to know what you think of the solution


linear algebra
matrix
superior algebra
superior algebra solved

Study if there exists a matrix 2*2 with real entries so that:

sinA=
1 1996
0 1

cos2A=
0 -3992
0 0,
so cosA is nilpotent, so cos2A=0 (a contradiction).


conics
ellipse
hyperbola
geometry
geometric transformation
reflection
rectangle

Prove that an ellipse with focuses X,Y has the following property:
A ray of light which intersects a segment joining focuses X,Y forever after reflections will be tangent to a hyperbola with focuses X,Y.

Any help is appreciated.. Thanks again.

Let $ (O)$ be a circle with radius $ r,$ let $ X$ be an arbitrary point inside this circle (not the center), and let the line $ OX$ cut the circle $ (O)$ at points $ A, B,$ such that $ A, X, O, B$ follow on the line in this order. Let $ T \in (O)$ be arbitrary point on this circle. Let the perpendicular bisector of $ XT$ cut the radius $ r \equal{} OT$ of $ (O)$ at a point $ P.$ $ P$ is the center of a circle $ (P)$ passing through $ X$ and tangent to $ (O)$ at $ T.$ Since $ PX \equal{} PT,$ it follows that $ OP \plus{} PX \equal{} OP \plus{} PT \equal{} r.$ Therefore, $ P$ lies on an ellipse $ \varepsilon$ with foci $ X, O$ and the major axis $ AO \equal{} r.$ Let $ XQ$ be a diameter of the circle $ (P).$ Then $ Q$ lies on an ellipse $ \mathcal E$ centrally similar to the ellipse $ \varepsilon$ with similarity center $ X$ and coefficient 2. If $ Y$ is a reflection of $ X$ in the perpendicular bisector of $ AB,$ then $ X,Y$ are foci, $ AB \equal{} 2r$ the major axis, and $ (O)$ the pedal circle of the ellipse $ \mathcal E.$ Let $ M$ be the other intersection of $ OT$ with the circle $ (P).$ $ XTQM$ is a rectangle, therefore, the angle $ \angle XTQ \equal{} 90^\circ$ is right and $ QT$ is tangent of $ \mathcal E$ at $ Q.$ This means that $ QT$ is the external bisector of the $ \angle XQY$ and $ QM \perp QT$ is the internal bisector of this angle. Let $ l$ be an arbitrary line through $ Q$ intersecting the segment $ XY$ and let $ R$ be the other intersection of $ l$ with the circle $ (P).$ Let $ l'$ be a reflection of $ l'$ in $ QM$ and $ R'$ the other intersection of $ l'$ with $ (P).$ The perpendicular bisector of $ RR'$ passes through $ P$ and through the midpoint $ M$ of the arc $ RR'$ opposite to $ Q,$ therefore, it passes through $ O$ and $ OR \equal{} OR' \equal{} \varrho < OX \equal{} OY.$ Since $ XQ$ is a diameter of $ (P),$ the angles $ \angle XRQ, \angle XR'Q$ are both right. As a result, the feet $ R, R'$ of perpendiculars $ XR, XR'$ to $ l \equiv QR$ and to its reflection $ l' \equiv QR'$ in $ QM \perp QT$ lie on a circle $ (O, \varrho),$ which means that they are both tangent to the same hyperbola $ \mathcal H$ with foci $ X, Y$ and pedal circle $ (O, \varrho).$

judging by the solution sir,
I believe AO cannot be the major axis. A diagram would be helpful, but it seems the ellipse cannot have AO as
the major axis.
Also there are other unexplained typos.

Major axis of the ellipse $ \varepsilon$ has the lenght $ AO \equal{} r.$ There are no typos.

See also [url]http://www.mathlinks.ro/viewtopic.php?t=190834[/url].


Compute: $ 0.\overline{7}\minus{}0.\overline{4}\plus{}0.\overline{2}$. Express your answer as a common fraction.


[b]Prove with x>0 , y>0 .  two function have onli (x;y)

 [tex] e^x=2007-\frac{y}{\sqrt[2]{y^2-1}}[/tex]
and
  [tex]e^y=2007-\frac{x}{\sqrt[2]{x^2-1}}[/tex]

Sorry , I learn E is bad [/b]

[quote="maths.vn"][b]Prove with x>0 , y>0 .  two function have onli (x;y)

 $ e^x \equal{} 2007 \minus{} \frac {y}{\sqrt [2]{y^2 \minus{} 1}}$
and
  $ e^y \equal{} 2007 \minus{} \frac {x}{\sqrt [2]{x^2 \minus{} 1}}$

Sorry , I learn E is bad [/b][/quote]

 $ \to e^x \minus{} e^y \equal{} \frac {x}{{\sqrt {x^2 \minus{} 1} }} \minus{} \frac {y}{{\sqrt {y^2 \minus{} 1} }}$

With $ x \ge y$ so: $ e^x \minus{} e^y \ge 0$

Another way, $ f(t) \equal{} \frac {t}{{\sqrt {t^2 \minus{} 1} }} \to f'(t) \equal{} \minus{} \frac {1}{{2\sqrt {t^2 \minus{} 1} ^3 }} < 0 \to \frac {x}{{\sqrt {x^2 \minus{} 1} }} \minus{} \frac {y}{{\sqrt {y^2 \minus{} 1} }} \le 0$

[hide="@maths.vn:"]Quang hả, bài này ở chỗ nào thế![/hide]


inequalities
search
inequalities unsolved

Prove that for any x,y,z pozitive reals we have:

(((xyz)^2+(xyz)+1)/3)^2 =< Product((1+x^4)/2)

EDIT: Oops, sorry, my inequality sign was the wrong way and i mixed up product and sum.

I am afraid we shall see a solution for this one in a few months time, as this is a problem of Vasile Cartoaje, which can be found in GM 3/2004( a very recent issue). :D  :D  :D

But I think Namdung solved very elegantly a very similar problem of Vasile Cartoaje and that method can be applied here immediately. So, search for that inequality. Since it is a proposed problem in GM, I won't give any hint.

For any nonnegative reals $x,y,z$ we have

$\left[\frac{x^{2}y^{2}z^{2}}{3}+\frac{x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)}{7}-\frac{11xyz}{21}+\frac{1}{3}\right]^{2}\leq \left(\frac{x^{4}+1}{2}\right)\left(\frac{y^{4}+1}{2}\right)\left(\frac{z^{4}+1}{2}\right)$,

with equality if and only if $x=y=z$.


f(x-y)=f(x)f(y)+g(x)g(y), f,g : R-R, f,g continuous.
find f and g.

I think that f and g is constant fuction or f(x)=cosx g(x)=sinx. But i can't prove it. Can you help me?

f(x)=sinkx, g(x)=coskx is a solution  :P

can you prove it??  if you can, please tell me.

f(x-y)=f(x)f(y)+g(x)g(y)
Let f(0) = s and let g(0) = a

then s = f(x)^2 + g(x)^2

and f(x) = f(x) s + g(x) g(0)

note that a^2 + s^2 = s

Now f(x) (1-s) = g(x) a

case 1.   s != 1

then f(x) = g(x) k  where k = a/(1-s)

so since f(x)^2 + g(x)^2 = s then f(x) and g(x) are both constant functions

case 2. s=1
In this case a=0 and therefore f(x)^2 + g(x)^2 = 1
Also note that f(y-x) = f(x-y) so f(x) = f(-x) for all x.

Now note that if f(x) = Cos F(x) then g(x) = sin F(x)

Then the identity reduces to Cos F(x-y) = Cos (F(x) - F(y))

since cos F(x-y) = cos ( F(x) - F(y))

by continoty of f(x)

we have F(x-y) = F(x) - F(y) + 2 k pi where k is constant not dependent on x or y

So F(x) = kx + 2kpi

so f(x) = cos kx

and g(x) = sin kx

y=0 give $f(x)=f(x)f(0)+g(x)g(0)$.
1. If $g(0)\not =0$ it give $g(x)=cf(x)$ and from x=y we get $f(0)=(1+c^{2})f(x)$ we get $f(x)=const=a,g(x)=const=b$, were $b^{2}=a-a^{2}$, therefore $0\le a \le 1$.
2. If $g(0)=0$, then $f(1)=1$. From x=y $1=f^{2}(x)+g^{2}(x)$. Therefore exist $z(x)$, suth that $f(x)=cosz(x),g(x)=sinz(x)$. We have
$z(x-y)=2k\pi \pm (z(x)-z(y))$, because f(x),g(x) continiosly and $z(0)=0$ we get $z(x)=ax,a\in R$. It mean $f(x)=cos(ax),g(x)=sin(ax)$.

[quote="now a ranger"]but how do you prove it?

another way[/quote]

what do u mean exactly?? :!:   :huh:


Prove or disprove that if $xyz=1$, then
$\sum_{sym} \frac{1}{x+2y} \ge 2$.

I don't think it's right. ;)
for z=1,x->0,y->infinit
I guess it can be proved:

$\sum_{sym} \frac{1}{x+2y} \geq \frac{3}{2}$

According to my numerical examples, the sum $\sum_{\mathrm{sym}} \frac{1}{x+2y}$ can take all possible positive values, from "slightly greater than 0" (for $x:y:z\approx 0:1:1$) to arbitrary big numbers (for $x:y:z\approx 0:0:1$). Hence, I don't think that a reasonable inequality can be found.

  Darij


inequalities induction function calculus derivative

Sorry for all the questions I ask, but here's another:

Using induction, prove that $(n/2)^{n-1} > (n-1)!$ för all integers $3 \le n$

Showing that it is true for n = 3 is easy. However, once you've assumed that $LHS_p > RHS_p$ it gets difficult. This is what I've been trying to do:
\[LHS_{p+1} - RHS_{p+1} = (\frac{p+1}{2})^p - p! = \frac{(p+1)^p}{2p^{p-1}}(\frac{p}{2})^{p-1} - p! > \frac{(p+1)^p}{2p^{p-1}}(p-1)! - p! = (p-1)!(\frac{(p+1)^p}{2p^{p-1}} - p) > \frac{(p+1)^p}{2p^{p-1}} - p = \frac{(p+1)^p - 2p^p}{2p^{p-1}} \]

So now you have to show that that $(p+1)^p - 2p^p > 0$ which I suppose would be possible with ANOTHER induction proof, however I'm more inclined to think that I am way overcomplicating this. Please help! :)

[quote="Altheman"]$(\frac{n}{2})^{n-1} >{n-1}!$ mulitiplying by $n$
$2\left(\frac{n}{2}\right)^{n} > n!$
then all you have to show is that is true, i dont have time to type full solution
$\left(\frac{n+1}{2}\right)^n \ge 2\left(\frac{n}{2}\right)^n $[/quote]
Yes, I can show this. Thanks. That's a much easier solution.

[quote="Arne"]You need to show that \[\left(1 + \frac{1}{p}\right)^p > 2.\]

Do you recognise something here, or do you need more help? :)[/quote]
I know I've seen $\left(1 + \frac{1}{p}\right)^p$ before, but I can't remember in what context. I guess it has something to do with the terms all having coefficients larger than two if you expand it, but I can't make anything rigorous out of it. So yes, I need more help.

It's well known that the function \[f(x) = \left(1 + \frac{1}{x}\right)^x\] is increasing and bounded, and that the limit equals $e$.

So, you can say "it is known that..." or you can prove that it is increasing (using derivatives) and then verify that $f(3) > 2$.

i did not see your post, and i was hinting that $e$ was the limit

Oh, I see, I didn't notice you were trying to say $e$ while coughing :)

Well, you could always just do it by AM-GM...


A Pythagorean triple is a set of three whole numbers which are the side lengths of a right triangle.  How many triples have a 12 in the triple?[hide] After much work and peeking, I found 4. [/hide]
how do you know when you got all of them?

Feel free to correct

[hide]
Answer: $ \boxed{4 trianles}$

Solution:

ok, you have to consider all cases where 12 is the smalles number, 12 is the middle number, and 12 is the largest number.

12 is the smallest, there are two triples in this set the 12-16-20 right triangle and the 12-35-37.

12 is the middle number there are also two in this set there is the 9-12-15 and the 5-12-13

12 is the largest number there are no pythagorean triples with 12 as the largest number therefore since we have examined all possible cases we know that there are only 4 pythagorean triples with twelve as one of the side lengths[/hide]

Try factoring for a more elegant and more accurate approach


[quote="ashrafmod"]Prove that if
$\sqrt x+\sqrt y+\sqrt z=1$, then
$(1-x)^{2}(1-y)^{2}(1-z)^{2}\geq 2^{15}xyz(x+y)(y+z)(z+x)$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1921


geometry
limit
calculus
real analysis
Support

And what is the smallest positive number? I asked this question on this forum a while ago and got the proposition to view the area of Hyperreals (can be found on Wiki). As to your statement that $ 0.999... \neq 1$ here's a proof it is:

$ 0.999...\equal{}9*10^{\minus{}1}\plus{}9*10^{\minus{}2}\plus{}9*10^{\minus{}3}...\equal{}9(10^{\minus{}1}\plus{}10^{\minus{}2}\plus{}10^{\minus{}3}...)\equal{}$
$ \equal{}9 \sum^\infty_{i\equal{}1}10^{\minus{}i}\equal{}9\left(\frac{1}{1\minus{}\frac{1}{10}}\minus{}1\right)\equal{}1$

You could try searching for [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=188041]other[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=90033]threads[/url] on the same subject rather than starting your own in the wrong area.
As to your objection- there is no such number. The real numbers are not well-ordered.

[quote="robitsky"]I have seen all over the net 'proofs' that 0.999...=1. Most use 0.333... which is not 1/3 and limits which are just that, limits but not 1.

What is the first number after 1? Is it 1.1? How about 1.01? No? How about 1.0000000001? No? Is it 1+(1-0.999...) or 2-0.999...? This is the issue that all so-called 'proofs' neglect[/quote]

[color=green]uhm, if $ 0,99999....\equal{}a$ then $ 10a\equal{}9,9999........$. But we have $ 10a\minus{}a\equal{}9$. Then $ a\equal{}1$.  :D [/color]

Another guy, who doesn't understand the definition of $ 0.999\cdots$.
It is simply by definition the limit of the sequence of rationals
$ a_0 \equal{} 0,a_1 \equal{} 0.9,a_2 \equal{} 0.99,\ldots,a_n \equal{} 1 \minus{} 10^{ \minus{} n},\ldots$, which is certainly $ 1$.
There is no dispute over that. (if you understand the definition of a limit in $ \mathbb{R}$)
And you don't need to prove that with trickery.
It's the same with $ 2 \equal{} \sqrt {4}$ or $ \lim_{n\to\infty}(1 \plus{} {1\over n})^n \equal{} \sum_{n \equal{} 0}^\infty{1\over n!}$.
You can always write the same number in many different ways, using many different definitions.

Now can you prove $ 1\equal{}\{\emptyset\}$?

I don't think this belongs in the Olympiad Section. Would a mod or admin please move this to a more appropriate forum. (such as Classroom Math or High School Basics maybe?). Thanks.

@Nashcakes:
Don't "flame" or "feed the trolls" as they sometimes call it. Insults do not belong in this forum. Also, don't use ALL CAPS to emphasize stuff. Instead, try using [b]Bold[/b] or [i]italics[/i] or [u]underlining[/u].

Anyway, once he had $ c \equal{} 0.9999999999 \ldots$, and $ 10c \equal{} 9.9999999999 \ldots$, he subtracted the first equation from the second equation.

The left side of the equation becomes $ 10c \minus{} c$ which equals $ 9c$.

The right side of the equation becomes $ 9.9999999999 \ldots \minus{} 0.9999999999 \ldots$ which equals $ 9$.

This is how he got $ 9c \equal{} 9$ and thus $ c \equal{} 1$.

By [i]definition[/i], the number $ .\overline{9}$ is $ \lim_{n\to\infty}\sum_{k\equal{}0}^{n}\frac{9}{10^k}$.
An understanding of [url=http://en.wikipedia.org/wiki/Limit_(mathematics)]limits[/url] makes it more obvious why $ .\overline{9}\equal{}1$.

See here: http://polymathematics.typepad.com/polymath/2006/06/no_im_sorry_it_.html
And here: http://polymathematics.typepad.com/polymath/2006/06/refutations.html

(Some of the refutations are hilarious :D)

If you're still not convinced, I invite you to prove that $ .\overline{9}\ne 1$. Now that would be interesting.

If you just simply find the pattern with the fractions equivalent to repeating decimals, you will find that $ .\overline{9}$ easily equals $ 1$. It's sort of like the limit in calculus when a certain value sought is undefined.

No one is going to expect you to learn Rudin overnight but looking into sequences and the concept of a limit will dispel any bad habits you have - i.e. not accepting a claim despite the fact a proof [hell it is pretty much a definition if we areebeing honest] exists but relying on your intuition.

Or, I guess, another way to look at it is that there is no number between $ 0.\overline{9}$ and $ 1$ which adds further support to the fact that $ 0.\overline{9}\equal{} 1.$

$ .999999999999999\equal{}\frac{9}{10}\plus{}\frac{9}{100}\plus{}\dots$. And if you are familiar with the sum of an infinite series then $ .999999999999999\equal{}\frac{\frac{9}{10}}{1\minus{}\frac{1}{10}}\equal{}1$.


This is really simple problem but I wonder if there's more than three ways to do this:

[tex]\frac {1}{x} - \frac {1}{x-1}[/tex]

Three ways are:
1. Multiplying by LCD
2. Change all of them into same denominator
3. Add 1/x-1 to both sides and do cross multiply

Three ways are basically same thing but I wonder if there are any other ways. Also, in math, it's good to know more than one way to do a single problem.

To do what?  All you've done is posted an expression.  There is no "other side" to add anything to; there is no question telling us what we should do with the expression.

Obviously he wants you to make one fraction out of the two (add them together).

Actually, I think its solving for 0... you wouldn't add to both sides if you were trying to make it one fraction (there wouldn't be two sides)... but theres no solution for 0, so im still not really sure

[quote="goldenfire6"]If it was solving for 0, then

[tex]\frac {1}{x}-\frac{1}{x-1}=0[/tex]

[tex]\frac {1}{x}=\frac {1}{x-1}[/tex]

Cross multiplying:

[tex]x=x-1[/tex] which is impossible[/quote]
um, aren't we exploring different ways of doing it, not just solving the problem?

The question is, what is the "it" we're trying to solve?  There is no question there.

[quote="Silverfalcon"]I think only yifman got what I was looking for.  See, I wasn't technically ask you to SOLVE the problem but just want to discuss about it and it's my bad for no solutioned problem.

What "it" here is the way to look on problems.[/quote]

The only problem here is that you didn't post a problem at all.. as people have said. So it couldn't even be discussed properly..

sooo...... what's the problem going to be? or is there not going to be a problem?

Here's a problem.

Find the solutions to the following inequality:

[tex]\displaystyle {\frac{1}{x} - \frac{1}{x-1} > 0}[/tex].

[hide]I'm not sure what you did yifman but your answer is immediately wrong because 1 and 0 are excluded because they are undefined.  What I did was combine the terms to get [tex]-\frac{1}{x(x-1)}>0[/tex]  The graph of this has vert. asymptotes at 0 and 1 and has no zeros.  So if we check the sign of the graph(pos. or negative) below 0, between 0 and 1, and above 1 the problem will be easy.  When x is negative the LHS is negative.  When x is between 0 and 1 the LHS is positive. When x is above one the LHS is negative.  The only answers are when 0<x<1.  [/hide]

[hide]1/x - 1/(x-1) > 0

(x-1)/x(x-1) - x/x(x-1) > 0

(x-1-x)/(x(x-1)) > 0

-1/(x(x-1)) > 0

1/(x(x-1)) < 0

COnsider the graph of

y = 1/(x(x-1)), or y = 1/(x^2-x)

There are vertical asymptotes at x = 0 and at x = 1, because at these values the denominator becomes zero. This indicates a change of sign. Taking the three situation:

When x<0,

1/(x^2 - x) > 0, because x^2 is always positive and -x will also be positive.

When 0<x<1,

1/(x^2 - x) < 0, because x^2 will always be smaller than x when 0<x<1.

When 1<x,

1/(x^2 - x) > 0, because x^2 will be greater than x.

We want the case when this is less than zero, so the second inequality is the helpful one. The solution is therefore 0 < x < 1.[/hide]

[hide] It's not essential to look at the graph, but just consider the inequality:

-1/x(x-1) > 0 implies x(x-1) < 0

for x(x-1) < 0, one must be positive and one negative. So 0 < x < 1 or else both will be positive/negative.

Same concept just less focus on the "graph" becuase i dont like graphs  [/hide]


I'm not sure if this should be in calculus, but, oh well,.....

Let d(k) be the number of all positive integer divisors of a number $k\in N$. Prove that for any $n_{0}\in N$, the sequence  $(d(n^{2}+1))^{\infty}_{n=n_{0}}$ is not strictly monotone

Evaluate $d(n^{2}+1)$ for $n=1$ to, say $n=20$.  I'd be shocked if this wasn't enough to reach your conclusion.

e: evaluating from $1$ to $4$ suffices...

[quote="Kent Merryfield"]Strictly speaking, observing that the sequence fails to be monotone on some initial interval does not answer the question as it was asked. You have to ask, "And then what?"[/quote]

I read bad.


Θεωρούμε κύκλο $ (O,R)$ και δύο χορδές αυτού $ AD$ και $ BC$ οι οποίες τέμνονται στο $ K$ που είναι το μέσο της χορδής $ EF$. Εαν oι $ AC$ και $ BD$ τέμνουν την $ EF$ στα $ M$ και $ N$ αντίστοιχα να δείξετε ότι $ OM\equal{}ON$.

$ butterfly's$   $ theorem$ $ \Rightarrow MK\equal{}KN\Rightarrow OM\equal{}ON$ $ since$ $ OK\perp EF.$

Kai elega ti moy thymizei re nick... :wink:  :rotfl:

[quote="galois01"]Protonie μήπως θα μπορούσες να μου εξηγήσεις εσύ (μιας και ο Nick ετοιμάζεται για την Μαδρίτη)  τι ακριβώς λέει αυτό το  θεώρημα  :oops:  Η λύση πού βρήκα χρησιμοποιώ το θεώτημα του Μενελάου :)[/quote]

Το Θεωρημα αυτο το εψαξα μολις ειδα τη λυση του Νικου και το βρηκα στο IMO Compedium. Βαζω μια μεταφραση: Αν Κ ειναι το μεσο μιας χορδης ΑΒ ενος κυκλου και ε, η ειναι δυο ευθειες που διερχονται απο το Κ και η ε τεμνει τον κυκλο στο σημειο X στη μερια του Α και στο σημειο Y στη μερια του Β, ενω η η τεμνει τον κυκλο στο σημειο Z στη μερια του Α και στο σημειο W στη μερια του Β, και ακομα οι ευθειες XZ και YW τεμνουν τη χορδη ΑΒ στα σημεια Θ και Η αντιστοιχα, τοτε ισχυει ΘΚ = ΗΚ.

[quote="galois01"]Θεωρούμε κύκλο $ (O,R)$ και δύο χορδές αυτού $ AD$ και $ BC$ οι οποίες τέμνονται στο $ K$ που είναι το μέσο της χορδής $ EF$. Εαν oι $ AC$ και $ BD$ τέμνουν την $ EF$ στα $ M$ και $ N$ αντίστοιχα να δείξετε ότι $ OM \equal{} ON$.[/quote]

Για το πολύ χρήσιμο θεώρημα της πεταλούδας( δείτε και φετινό θέμα Βαλκανιάδας) βάλτε το search τη φράση

 [color=red][b]butterfly theorem[/b][/color]

 και μελετήστε ό,τι ενδιαφέρον βρείτε. ΑΝ μάλιστα καθήσει κάποιος που έχει χρόνο και γράψει όλα τα θέματα που θα βγάλει η αναζήτηση (έσω με αντιγραφή και επικόλληση σε  σε ένα  word στα αγγικά) , θα τον παρακαλέσω να μου τα στείλλει με ένα  e mail.

 Καλή μελέτη -

 Μπάμπης


[quote="Rose_joker"]Show that
$ \frac {1}{852} \plus{} \frac {1}{853} \plus{} ... \plus{} \frac {1}{2544} \equal{} 1 \plus{} \frac {2}{2*3*4} \plus{} \frac {2}{5*6*7} \plus{} ... \plus{} \frac {2}{2552*2553*2554}$
 :)[/quote]
First we find $ A,B,C$ such that:
$ \frac{2}{(x\minus{}1)x(x\plus{}1)} \equal{} \frac{A}{x\minus{}1} \plus{} \frac{B}{x} \plus{} \frac{C}{x\plus{}1}$.
We can "guess" each value by setting $ x\equal{}1,0,\minus{}1$, which gives: $ A \equal{} 1, B \equal{} \minus{}2, C \equal{} 1$. Hence:
$ \frac{2}{(x\minus{}1)x(x\plus{}1)} \equal{} \frac{1}{x\minus{}1} \plus{} \frac{1}{x} \plus{} \frac{1}{x\plus{}1} \minus{} 3\frac{1}{x}$
$ RHS \equal{} 1 \plus{} \sum_{n\equal{}1}^{851} \frac{2}{(3n\minus{}1) \cdot 3n \cdot (3n\plus{}1)} \equal{}$ $ 1 \plus{} \sum_{i\equal{}1}^{851} \frac1{3n\minus{}1} \plus{} \frac1{3n} \plus{} \frac1{3n\plus{}1} \minus{} \frac1{n} \equal{} \sum_{n\equal{}1}^{2554} \frac{1}{n} \minus{} \sum_{n\equal{}1}^{851} \frac{1}{n} \equal{} LHS$ :)

[quote="Mathias_DK"][quote="Rose_joker"]Show that
$ \frac {1}{852} \plus{} \frac {1}{853} \plus{} ... \plus{} \frac {1}{2544} \equal{} 1 \plus{} \frac {2}{2*3*4} \plus{} \frac {2}{5*6*7} \plus{} ... \plus{} \frac {2}{2552*2553*2554}$
 :)[/quote]
First we find $ A,B,C$ such that:
$ \frac {2}{(x \minus{} 1)x(x \plus{} 1)} \equal{} \frac {A}{x \minus{} 1} \plus{} \frac {B}{x} \plus{} \frac {C}{x \plus{} 1}$.
We can "guess" each value by setting $ x \equal{} 1,0, \minus{} 1$, which gives: $ A \equal{} 1, B \equal{} \minus{} 2, C \equal{} 1$. Hence:
$ \frac {2}{(x \minus{} 1)x(x \plus{} 1)} \equal{} \frac {1}{x \minus{} 1} \plus{} \frac {1}{x} \plus{} \frac {1}{x \plus{} 1} \minus{} 3\frac {1}{x}$
$ RHS \equal{} 1 \plus{} \sum_{n \equal{} 1}^{851} \frac {2}{(3n \minus{} 1) \cdot 3n \cdot (3n \plus{} 1)} \equal{}$ $ 1 \plus{} \sum_{i \equal{} 1}^{851} \frac1{3n \minus{} 1} \plus{} \frac1{3n} \plus{} \frac1{3n \plus{} 1} \minus{} \frac1{n} \equal{} \sum_{n \equal{} 1}^{2554} \frac {1}{n} \minus{} \sum_{n \equal{} 1}^{851} \frac {1}{n} \equal{} LHS$ :)[/quote]

I don't think that there can be 0 in the denominator...

[hide="Other way to find them"]

Just multiply out the right side, in numerator there is $ x^2(A\plus{}B\plus{}C)\plus{}x(A\minus{}C)\minus{}B\equal{}2$, from which $ A\plus{}B\plus{}C\equal{}0,A\minus{}C\equal{}0,B\equal{}\minus{}2$ and we get the numbers [/hide]

Just for Bugi  :)

$ \frac {2}{(x \minus{} 1)x(x \plus{} 1)} \equal{} \frac {A}{x \minus{} 1} \plus{} \frac {B}{x} \plus{} \frac {C}{x \plus{} 1}$

Multiply this with $ (x \minus{} 1)x(x \plus{} 1)$

$ 2 \equal{} Ax(x \plus{} 1) \plus{} B(x \minus{} 1)(x \plus{} 1) \plus{} Cx(x \minus{} 1)$

And now go for $ \minus{} 1,0,1$. Zeros in those denominators aren't such a big problem (actually, this step I made above is implicitly made in Mathias' solution) :lol: it kind of reminds me of calculating residues...


Putnam
function
algebra
polynomial
college contests

I remember reading somewhere that this problem was proposed for the Putnam; it may even have been chosen, I don't know, but it strikes me as a good problem. I apologize if it has been posted before.

Let $f(n)$ be the function $f(n)=1!+2!+3!+\cdots+n!=\sum_{k=1}^{n}{k!}$

There are two rational polynomials, $P(n), Q(n)$ which satisfy the property
\[f(n)=P(n)f(n-2)+Q(n)f(n-1) \]
What are $P,Q$?

Note: If anyone remembers the source for this problem (I don't), could you please post it? I know I found it online somewhere, among a collection of other Putnam prep questions, but I don't remember exactly where...

It's a kind of guess-the-answer type problem, isn't it?

First notice, that $f(n-1)-f(n-2) = (n-1)!$, so
\[(n+1)[f(n-1)-f(n-2)] = n!+(n-1)! = f(n)-f(n-2). \]
Thus
\begin{align*}f(n) &= (n+1)[f(n-1)-f(n-2)]+f(n-2) \\ &= (-n)f(n-2)+(n+1)f(n-1). && \blacksquare \end{align*}

This is a simillar problem of Putnam B-1, 1984.


Have anyone of you heard of Hua Lo Keng Maths Compeptition, I have participated before.

Do anyone of you have their past papers because their question are quite tough?

For your information , this competition was held in Malaysia in between mid of April to June, but I don't know whether they are also in other countries.

Please share with us if u happen to have these papers as I really need it.

Thank You.

you might have more luck searching for "Hua luogeng," which is the preferred spelling, at least in simplified chinese.

I've heard of it. I have some old papers, but, they are from late 80's and early 90's and they are in chinese  :?

[quote="Yrael"] I have some old papers, but, they are from late 80's and early 90's and they are in chinese  :?[/quote]

Is there some  members of mathlinks who will translate them in english ?

i do have his books, and it is Hua Luo Geng, not Hua Luo Keng.

[quote="pakagawa"]I think you posted this in the wrong forum.

People would answer quicker if you posted in one in the 2nd section.

And, I don't know.[/quote]

You guy pakagawa are Chinese but you don't know the most famous math competition in China.(and you're using a Japanese name ;) )
Okay, I can translate them for you, I've studied Chinese for 16 years.(I appear to be a Japanese in fact I'm a Chinese)

[size=159]I'll translate it for you in several weeks.[/size]

no one cares?
then [size=150]I'll NOT do that any more.[/size]

we care ... just post the translation here :)


Euler
function
number theory
relatively prime
prime numbers

[hide="hint"] look at the prime factorization of $2004$... consider the number of factors it has [/hide]

[hide]$2004 = 167 * 3 * 2^2$

So all integers that are divisible by 167, 2, or 3  from 1 to 2004 are not relatively prime to 2004.

So let us find how many integers are divisible by 167,3,or 2.

By principle of inclusion-exclusion:
There are
$\frac{2004}{167}+\frac{2004}{3}+\frac{2004}{2}-\frac{2004}{501}+\frac{2004}{334}+\frac{2004}{6}-\frac{2004}{2004} = 1339$ integers not relatively prime to 2004

So there are $2004 - 1339 = \fbox{665}$ integers relatively prime to 2004[/hide].

Alternatively, you could use the Euler indicator. The answer is $\phi(2004)$.

yes, as arne mentioned the euler phi function is a better way to find the number of relative prime numbers to a certain number.

if $n=p_1^{\alpha_1}p_2^{\alpha_2}...p_i^{\alpha_i}$

then $\phi (n)=n(1-\frac{1}{p_1})(1-\frac{1}{p_2})...(1-\frac{1}{p_i})$

[quote="citistarz"]Problem solved - thanks for the help[/quote]

Don't delete the problem after it has been solved :!:  :mad:


geometry
circumcircle
parallelogram
geometry unsolved

Let a triangle $ABC$ , the altitudes $BD,CE$, the orthocenter $H$ and the common point $Z$ of lines $DE,CB$. The circumcircle of triangle $ABC$ meets $AZ$ at $P$.Prove that $HP \perp AZ$.

 [u]Babis[/u]

Now, ZA*ZP=ZB*ZC=ZE*ZD. Therefore A,P,E,D Concyclic
And since A,D,H,E concyclic also, angle HPA=angle HEA

my solution is the same of singapore1728!

other simple way to solve is using complex numbers!

The same solution was sent to me(PM) by Vittasko , a few minutes after I had posted the problem. It is a nice problem , but easier than I thought.  Preolympiad section is it 's right position !

           Thanks to all of you.- Babis

[quote="stergiu"]Let a triangle $ABC$ , the altitudes $BD,CE$, the orthocenter $H$ and the common point $Z$ of lines $DE,CB$. The circumcircle of triangle $ABC$ meets $AZ$ at $P$.Prove that

    [b]a) [/b]$HP \perp AZ$.
[color=red]
    [b]b) [/b]Line $HP$ bisects side $BC$[/color]
 [u]Babis[/u][/quote]

[b]NOTE[/b]

 [color=blue]I have added one more question to the problem , so it has now  more ''olympiad '' style. This question is due my friend Vittasko.Thank you Kostas ! [/color]

[quote="stergiu"][quote="April"][quote="stergiu"][color=darkred]Let a triangle $ABC,$ the altitudes $BD,\,CE,$ the orthocenter $H$ and the common point $Z$ of lines $DE,\,CB.$ The circumcircle of triangle $ABC$ meets $AZ$ at $P.$ Prove that
    [b]a) [/b] $HP \perp AZ$.
    [b]b) [/b] Line $HP$ bisects side $BC$[/color]
 [u]Babis[/u][/quote]

[color=indigo][b]Proof.[/b] Denote: $(O)$ be the circumcircle of triangle $ABC$ and $N=PH\cap (O)$
We have: $\angle APN=90^\circ,$ so $AN$ is the diagonal of circumcircle $(O)$ of triangle $ABC$
Thus, easy to show that: $BHCN$ is a parallelogram and hence $\overline{PHN}$ bisects $BC$[/color][/quote]

[color=red]  NICE ! :) [/color][/quote]

I would say GREAT  :!:   :roll:


abstract algebra
algebra unsolved
algebra

Prove that a group where $ x \ast x = e $ for all $ x \in G $ is abelian. Hint: Use $ (a \ast b) \ast (a \ast b) $.

I use $ x_1 = x_2 = x $ to make it easier to read
$ x_1 \ast x_2 = e $
$ x_1 \ast x_2 \ast x_2^{-1} = x_2^{-1} $
$ x_1 = x_2^{-1} $
$ x_2 \ast x_1 = x_2 \ast x_2^{-1} $
$ x_2 \ast x_1 = e $
Since $ x_1 \ast x_2 = x_2 \ast x_1 $ it follows that the group G is abelian?

That first proof doesn't work.  x always commutes with x.  One way would be
$x_1*x_2*x_1*x_2=(x_1*x_2)^2=e$
Multiplying on the left by x_1 and on the right by x_2 yields
$x_1*x_1*x_2*x_1*x_2*x_2=x_1*x_2$
Noting that x_i*x_i=e, we get
$x_2*x_1=x_1*x_2$

[quote="jmerry"]Proving that an element commutes with itself doesn't get you anywhere.

For a real proof, consider $(ab)^{-1}$. This element must be unique by the cancellation law, but we can compute it in two ways:
$(ab)^{-1}=b^{-1}a^{-1}=ba$ and
$(ab)^{-1}=ab$
Therefore $ab=ba$ for any $a,b$ and the group is abelian.[/quote]

I see how you can get:
1) $(ab)^{-1}=b^{-1}a^{-1}$
2) $(ab)^{-1}=ab$
But I can't see how you get $b^{-1}a^{-1}=ba$

[quote="Hedlund"]But I can't see how you get $ b^{-1}a^{-1}=ba $.[/quote]
Because $ x \ast x = e $ for all $ x \in G $ implies that $a^{-1}=a$ and $b^{-1}=b$.


limit
logarithms
real analysis
real analysis solved

Let $a$ be a positive integer.Evaluate the following limit.

\[{\lim_{n\to\infty}[\ \{\sum_{k=1}^n\ \ln (n^{a+1}+k^a})\}-(a+1)n\ln n]\]

Start by writing

$\ln(n^{a+1}+k^a)=\ln(n^{a+1}(1+\frac{k^a}{n^{a+1}}) ) = (a+1)\ln n + \ln(1+\frac{k^a}{n^{a+1}}).$

Since $\sum_{k=1}^n(a+1)\ln n = (a+1)n\ln n$, the quantity we are taking the limit of becomes:

$\sum_{k=1}^n\ln(1+\frac{k^a}{n^{a+1}}) = \sum_{k=1}^n( \frac{k^a}{n^{a+1}}+O(\frac{k^{2a}}{n^{2a+2}}))$

${=\frac{n^{a+1}}{a+1}\cdot\frac1{n^{a+1}} + O(\frac1n})$

Take the limit of that as $n\to\infty$ to get the answer $\frac1{a+1}.$

Yes,you are right,Kent Merryfield. :)

Comment:For Japanese high school student. You can use the following inequality.

\[x-\frac{x^2}{2}\leqq \log (1+x)\leqq x\ (x\geqq 0)\]

kunny


I would really appreciate it if anyone posts mathematical tricks from basic counting to number theory, geometry, algebra, etc.  Thanks in advance. :)

Masoud Zargar

Before i start with a few, let me quote an article from this site as a word of warning

"I assumed that most of the other kids did better at MOP because they knew more tricks than I did. My formula sheets were pretty thorough, but perhaps they were missing something. By the end of MOP, I had learned a somewhat unsettling truth. The others knew fewer tricks than I did, not more."

and the tricks definitely arent very useful if you don't know why they work because applying them takes a good deal of knowledge on how they work

for number theory, here are a few
a(mod b)is the same as saying a+bm where m is some integer
if a=b(mod c) then am=bm(mod c)
fermat's little theorem: a^p=a(mod p)
Wilson's theorem: (p-1)!=(p-1)(mod p)
Eulers theorem, a generalization of fermats little (i dontknow it off the top of my head)
Chinese remainder theorem
Pell equations

And of course, the trial and error for all values of the modulo you're dealing with, which works often times

The generalization is if a and n are relatively prime, then a^(phi(n)) = 1 (mod n)

the chinese remainder theorem is if you have a number n such that
$n\equiv a \mod{r}$ and $n\equiv b\mod{s}$, then there is a unique number m such that $n\equiv m \mod{rs}$

and Scrambled, I agree strongly with your words of warning, i get the same idea whenever i see one of these topics

I don't agree with the stated theorem:

the chinese remainder theorem says:
if you have coprime $r,s$ and some $x,y$, then there is exactly one $m \mod rs$ such that $m \equiv x \mod r$ and $m \equiv y \mod s$.
This also holds for more 'equations' modulo pairwise coprime numbers.

sorry, i did not mention the fact that r and s have to be coprime, and the fact that you can do this with multiple mods is obvious because if you had 3 conditions, you could just pair 2 of them together, and use that result with the unused one

[quote="Scrambled"]Before i start with a few, let me quote an article from this site as a word of warning

"I assumed that most of the other kids did better at MOP because they knew more tricks than I did. My formula sheets were pretty thorough, but perhaps they were missing something. By the end of MOP, I had learned a somewhat unsettling truth. The others knew fewer tricks than I did, not more."

and the tricks definitely arent very useful if you don't know why they work because applying them takes a good deal of knowledge on how they work[/quote]

Yes, I understand.  I usually find a trick and figure out how and why it works. :)

Masoud Zargar


limit
calculus
algebra
function
domain
derivative
calculus computations

Find:
$\lim_{x \to 0} \frac{\sqrt[m]{ax+1}\cdot\sqrt[n]{bx+1}-1}{x}$

[color=darkblue][b]Proof (without the calculus of the derivatives).[/b] For any $a,b\in R$ there is $\delta >0$ so that $|x|<\delta\Longrightarrow ax+1>0\ \wedge\ bx+1>0$. Therefore,
$V\equiv (-\delta ,\delta )\subset D_f$ - the definition domain of the function $\boxed {f(x)=\frac{\sqrt [m] {ax+1}\cdot \sqrt [n] {bx+1}-1}{x}}$.

Denote the functions[/color] $g,h: V\rightarrow R$, where $g(x)=\sqrt [m] {ax+1}-1$, $h(x)=\sqrt [n] {bx+1}-1$.

$0.\blacktriangleright f(x)=\frac{(g(x)+1)(h(x)+1)-1}{x}\Longrightarrow f(x)=\frac{g(x)h(x)+g(x)+h(x)}{x}\ .$

$1.\blacktriangleright \lim_{x\to 0} g(x)=\lim_{x\to 0} h(x)=0\ ;\ \lim_{x\to 0}\frac{g(x)}{x}=\frac am\ ,\ \lim_{x\to 0}\frac{h(x)}{x}=\frac bn\ .$

$2.\blacktriangleright \lim_{x\to 0} f(x)=\lim_{x\to 0}\frac{g(x)}{x}\cdot \lim_{x\to 0} h(x)+\lim_{x\to 0} \frac{g(x)}{x}+\lim_{x\to 0}\frac{h(x)}{x}\Longrightarrow \boxed {\lim_{x\to 0} f(x)=\frac am+\frac bn}\ .$

[color=darkred][b]A similar (own) exercise without the calculus of the derivatives[/b] ([u]outside of the school ![/u]).\[ \boxed {\ f(x)=\frac{\sqrt [m] {1+x}+\sqrt [m] {1-x}-2}{x^2}\Longrightarrow \lim_{x\to 0} f(x)=\frac{1-m}{m^2}\ }\ .  \][b]Proof.[/b] Denote $g(x)=\sqrt [m] {1+x}$, $h(x)=\sqrt [m] {1-x}$. Therefore,

$f(x)=\frac{g(x)+h(x)-2}{x^2}=\frac{[g(x)h(x)-1]-[g(x)-1][h(x)-1]}{x^2}$.

$\lim_{x\to 0} \frac{g(x)-1}{x}=\frac 1m\ ,\ \lim_{x\to 0}\frac{h(x)-1}{x}=-\frac 1m\ ,\ \lim_{x\to 0} \frac{g(x)h(x)-1}{x^2}=-\frac 1m\Longrightarrow$

$\lim_{x\to 0} f(x)=-\frac 1m+\frac{1}{m^2}=\frac{1-m}{m^2}\ \mathrm {(\ very\ nice\ !\ )\ .}$

[b]Remark.[/b] $\lim_{x\to 0} \frac{\sqrt [m] {1\pm x^n}-1}{x^n}=\pm\frac 1m$, where $m,n\in N^*\ .$[/color]


1) Find, with proof, the limit for $n\rightarrow\infty$ of $\sqrt[n]{a.n^a}$ where $a$ can be any constant $\in\mathbb{R}^+_0$.

2) Find, with proof, the limit for $n\rightarrow\infty$ of $\sqrt[n]{n^{n-a}}$ where $a<n$ can be any constant $\in\mathbb{R}^+_0$

mods feel free to move if it's too difficult.

I think I got #1:

Let y = lim (n --> ∞) (a*n^a)^(1/n).

Take ln of both sides and obtain

ln(y) = lim (n --> ∞) ln((a*n^a)^(1/n)) =
lim (n --> ∞) ln(a*n^a) / n =
lim (n --> ∞) (ln(a)/n + a*ln(n)/n) =
lim (n --> ∞) (ln(a)/n) + a*(lim (n --> ∞) (ln(n) / n)) = 0 + a*0 = 0 -->

y = e^0 = 1.

Eh, I think I also got #2:

Let y = lim (n --> ∞) (n^(n-a))^(1/n).

Take ln of both sides and obtain

ln(y) = lim (n --> ∞) ln((n^(n-a))^(1/n) ) =
lim (n --> ∞) ((n-a)/n * ln(n)) =
lim (n --> ∞) ( ln(n)  a* ln(n) / n ) =
lim (n --> ∞) (ln(n))  a* lim (n --> ∞) (ln(n) / n) = ∞ - a*0 = ∞ -->

y = e^∞ = ∞.


[color=blue]Suponiendo que una persona inventa una función $f(x)$, la cual puede ser un polinomio, una sucesión de recurrencias, o cualquier otra clase de función.

Otra persona intenta adivinar la función inventada, preguntando cuál es la imagen de la función del valor que la misma preguntará.

Sé que para un polinomio existe el método de las diferencias finitas, y en el caso de las recurrencias, si uno sabe cómo se obtienen los valores en función de los anteriores, puedo dar con la expresión general.

¿Cómo haría, en el caso de las recurrencias, para encontrar la fórmula de recurrencia aplicando el método conocido?

¿Cómo podría reconocer qué tipo de función es preguntando averiguando solamente la imagen de la misma para distintas variables?

Agradezco toda la ayuda posible.[/color]

[color=blue]Realmente necesito que alguien pudiera responder a mi solicitud. Si no me expliqué bien, que me lo haga notar, de modo que no haya problemas de interpretación.[/color]

Supongo cuando dices función estas pensando en funciones cuyo dominio es el conjunto de los naturales, es decir las sucesiones. Pues, por otro lado,  por ejemplo la imagen de la función $f: \mathbb{R}\to \mathbb{R}$, $f(x)=x$ es todo $\mathbb{R}$ y hay infinitas funciones cuya imagen es $\mathbb{R}$, así que con el dato que la imagen de $f$ es  $\mathbb{R}$ no se puede saber quién es $f$.
En todo caso, tendrías que aclarar bien tu pregunta.

$Tipe$

[color=blue]Quisiera considerar cada caso por separado, en lo posible.[/color]


Please excuse me if this is in the wrong forum, but i could not find a forum i saw relevant for this question.

"What is the probability that two different cards (of any suit) will be next to each other in a randomly shuffled deck of 52 playing cards" (no jokers)

for example, what is the probability that an Ace will be next to a King.

I realise that this is most likely a relatively simple probability question, although unfortunately i haven't studied probability before.

[hide="Answer"]
probability is $\frac{P(51,51) \cdot 2!}{P(52,52)}$[/hide]

Sorry but what is $P(51,51)$
in english?

I'm still unsure how to calculate this question, if varun or someone else could simplify this answer and perhaps give an explanation i would be very appreciative.

Thanks once again,

-Freakonomics

I have used P(x,y) as permutations of x things taken y at a time.

[hide="Now the question."]
Since the 2 cards are together. thus we have to arrange 51 cards( 2 being together) out of 52. Now the 2 cards can be interchanged. So the number of ways this can be done is $51! \cdot 2!$

Also total no. of ways of arranging the cards is $52!$

So the probability is $\frac{51!\cdot2!}{52!}$
[/hide]

[hmm]
Varun, I think with your system of counting, you would need to multiply it by $ \frac{52\times 48}{2}$ to account for the different possible pairs of the cards that are different (there is a 2 in the denominator because you already have a factor of 2 in your expression to account for the ordering of the two).

This is greater than $52!$ because I think your system of counting overcounts the cases where there are more than 1 pair of cards where the two cards are different. The case where there are 2 such pairs is double counted; the case where there are 3 such pairs ... etc.

I think the answer is $1$ because no matter how you construct the deck, you are guaranteed that there will be two cards that are different a different type but are next to each other in the deck. Consider an Ace: you would need to surround it with Aces, but you only have 4 of them in the deck. The ones on the outer edge would have to be paired with a card of a different type.
[/hide]

Sorry for the confusion!

In my question i meant it to be similar to varun's interpretation, except that the suit of the card doesn't matter.  So instead of, say, the ace of diamonds being next to the king of spades, i am looking to find the probability of Any ace (diamond/club/spade/heart) being next to any king (diamond/club/spade/heart).

Hope this clarifies, and sorry again for the confusion.

[quote="Adunakhor"][hmm]
Varun, I think with your system of counting, you would need to multiply it by $\frac{52\times 48}{2}$ to account for the different possible pairs of the cards that are different (there is a 2 in the denominator because you already have a factor of 2 in your expression to account for the ordering of the two).

This is greater than $52!$ because I think your system of counting overcounts the cases where there are more than 1 pair of cards where the two cards are different. The case where there are 2 such pairs is double counted; the case where there are 3 such pairs ... etc.

I think the answer is $1$ because no matter how you construct the deck, you are guaranteed that there will be two cards that are different a different type but are next to each other in the deck. Consider an Ace: you would need to surround it with Aces, but you only have 4 of them in the deck. The ones on the outer edge would have to be paired with a card of a different type.
[/hide][/quote]

what actually the question i understood is that there are two particular cards which are next to each other(obviously they are different and it is useless to mention it)

is that you wanted to ask, Freakonomics

Okay.  Say i have a deck of 52 randomly shuffled cards.  If I call out say, 'Two', 'Seven'.  I want to know the probability of any 'Two' (clubs/diamonds/hearts/spades) being next to any 'Seven' (clubs/diamonds/hearts/spades) in the deck.

I think think the answer will be different if one of the twos or sevens is on the top or the bottom of the deck, but if they aren't on the top or bottom of the deck, I think there are 64 different ways in which they can be arranged, although I am unsure.

Sorry, i didn't explain it very well before, I hope this is better.

I agree with varun's answer the way he and I interpreted the question. But now that you say the cards can be of any suit (clubs/diamonds/hearts/spades) I think you need to multiply his answer by $4*4$ for the $4$ choices of suit for each card. So $\frac{1}{26}*16=\frac{8}{13}$. Hopefully no mistakes.


For a,b,c,d>0 and $a+b+c+d = 1$
Prove that
$\left({\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}\right)^{3}\ge \frac{{64}}{5}\frac{{\left({a+1}\right)\left({b+1}\right)\left({c+1}\right)\left({d+1}\right)}}{{abcd}}$

Sorry for my mistake. I've repaired it. Thank you!

[quote="fjwxcsl"]
 [The original inequality does not hold.
   a=b=c=d=1/4
  =>64/5*(1+a)(1+b)(1+c)(1+d)/(abcd)>(1/a+1/b+1/c+1/d)^3][/quote]

thank you!

For a,b,c,d>0 and $a+b+c+d = 1$
Prove that
$\left({\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+4}\right)^{3}\ge \frac{{64}}{5}\frac{{\left({a+1}\right)\left({b+1}\right)\left({c+1}\right)\left({d+1}\right)}}{{abcd}}$
Now it is right

to fjwxcsl: Have you solved this problem?

I don't understand
What does that mean?
[quote="fjwxcsl"]
  pp:=numer(p0);

  collect(subs(b=a+x,c=a+x+y,d=a+x+y+z,pp),[a,x,y,z]);

  pp=(3600*x^2+(4800*y+2400*z)*x+4800*y^2+4800*y*z+3600*z^2)

       *a^10+...>=0.

   The inequality holds.[/quote]


inequalities
real analysis
real analysis unsolved

Hi!

How do I show the following inequality:

 \[ \frac{1}
{2}{\left( {\frac{{a \plus{} 1}}
{{a \minus{} 1}} \plus{} \sqrt {{{\left( {\frac{{a \plus{} 1}}
{{a \minus{} 1}}} \right)}^2} \minus{} 1} } \right)^k} \plus{} \frac{1}
{2}{\left( {\frac{{a \plus{} 1}}
{{a \minus{} 1}} \minus{} \sqrt {{{\left( {\frac{{a \plus{} 1}}
{{a \minus{} 1}}} \right)}^2} \minus{} 1} } \right)^k} \geqslant \frac{1}
{2}{\left( {\frac{{\sqrt a  \plus{} 1}}
{{\sqrt a  \minus{} 1}}} \right)^k},a \in \mathbb{R},a > 1,n \in \mathbb{N}\]

?

Thanks alot for your help.

PiDeltaPhi

By noticing that the right hand side equals the first term on the left hand side and that the second term on the left hand side is non-negative  :rotfl:

Isn't this supposed to be in the inequalities forum?

multiply by $ 2$ and use $ a^2 \minus{}b^2 \equal{} (a\plus{}b)(a\minus{}b)$ in the parenthesis to get;

$ (\frac{a\plus{}2\sqrt{a} \plus{}1}{a\minus{}1})^k \plus{} (\frac{a\minus{}2\sqrt{a} \plus{}1}{a\minus{}1})^k \geq [(\frac{\sqrt{a} \plus{}1)^2}{a\minus{}1}]^k$

$ \Leftrightarrow$

$ (\sqrt{a} \plus{}1)^{2k} \plus{} (\sqrt{a} \minus{}1)^{2k} \geq (\sqrt{a} \plus{}1)^{2k}$ $ \Leftrightarrow$

$ (\sqrt{a} \minus{}1)^{2k} \geq 0$ which is true since $ k \in\mathbb{N}$

I should smack my head, for not noticing how easy this is.

[quote]By noticing that the right hand side equals the first term on the left hand side and that the second term on the left hand side is non-negative[/quote]

I swear I laughed 10 minutes after reading this  :rotfl: . Thanks!
Thank you Papalakis for your input.

I wasn't sure about picking the forum, because I was originally solving a problem from numerical analysis and I broke it down to the last inequality, after a lot of work. After seeing my result, being pretty exhausted, I thought I went nowhere, so I resigned continuing and posted it here...

PiDeltaPhi


Princeton college calculus geometry AMC AIME USA(J)MO

I searched and couldn't find any topics on this, so forgive me if this is a repeat thread..

So, I am a soon-to-be senior in highschool. Like everyone else on this forum, I enjoy math. Unlike alot of people on this forum, I'm not very good at the AMC/AIME/USAMO type problems, especially geometry.

Now, I'm planning to continue studying math in college, probably along with EE or something along those lines. However, I thought college math was supposed to be harder than these competitions -so, do I have any hope of doing well in college math if I can't do competition problems?

Do you know how to write proofs? You should do decently well, if so. It takes some creativity in math yes, but sometimes in competitions you still need a good bit of base knowledge/tricks, etc., but in college you build these up with the course. Nevertheless, you should do math if you enjoy it, and if you don't do "well" in math courses, then there is no harm done. Try it, is my advice. :)

Competition mathematics is the marriage of mathematics and problem-solving skills. However, many people I know can do math and have problem-solving skills, but they are better at applying their problem-solving skills to something else.

[b]1. Mathematics[/b] - College math in EE is hard because you need to pile on a lot of material, and there's no way you can do that without a solid understanding of the basics. You need to understand how different topics are related, need to be able to draw analogies between similar forms (such as between differential equations and matrices), and in general get a good feel of the flow of things. Once you get that, the mathematics pertinent to engineering will be very easy.

Do you feel this way about mathematics now? If you read a few formulas on one page, could you juggle with them and think ahead to develop a formula that might come up later in the book? If so, you're already 2/3 of the way there. If not, keep on doing whatever you're doing, and studying, and thinking, and at some point in time, it will "click" and everything will make sense.

[b]2. Problem-Solving[/b] - As an engineer, you will often need to use given tools and, working within given constraints, develop a solution to a particular problem. This is not unlike what we do in competative mathematics -- use the mathematical tools we know, and working within the rules, time limits, and principles of logic, derive a solution.

Now, just because you haven't developed your mathematical problem-solving skills, doesn't mean you can't do it in the context of engineering. A lot of people are very intuitive with actual physical things -- they are more comfortable with the laws of physics than with those of mathematics or raw logic. Given that you are interested in engineering, you very well may be one of these people, and in engineering, this kind of intuition counts for a lot. Not to mention, your engineering education will further these abilities.

[b]Conclusion:[/b] Stick to your passions and you'll do just fine :)

Most people I've talked to say high school maths is totally irrelevant for real (college) maths, but I'll reiterate that if you're having trouble learning the very basics (which should probably suffice for most AMC problems), you should consider revisiting that material.

[quote="ejr322"]
Now, I'm planning to continue studying math in college, probably along with EE or something along those lines. However, I thought college math was supposed to be harder than these competitions -so, do I have any hope of doing well in college math if I can't do competition problems?[/quote]

Sure.  Keep in mind, a lot of people become good at these types of problems by training for them.  It's likely that your school hasn't trained you in the type of up-from-the-fundamentals problem solving you need to tackle AMC-type problems.  If you don't have this training and go to a top tier school, you're probably at an initial disadvantage relative to other students who have this training, but it's not an insuperable disadvantage.  I saw this a lot in my first two years at Princeton: I had a huge advantage over kids who didn't have my math contest background (but who went to much "better" high schools than I did -- more APs, higher SATs on average, etc).  In the first two years, many of them struggled, a lot of them quit, but a lot of them adjusted and learned the problem solving lessons I had learned a few years earlier, and by the time we graduated, the gap between me and them had closed considerably.

That's a long-winded way of saying "don't panic, and don't quit if you hit challenges early in college."

You have a lot to learn- those problem-solving skills are important in college math. You also have a lot of classes to learn it through. Most students come in with no contest experience, and learn those skills in college.

Ok. thank you all for replying.

Anaxerzia - I know how to write proofs, so thats not too much of a worry.

TZF - I didn't just mean math in EE, I meant like a math degree along with EE. I haven't really had much difficulty with understanding theory and concepts (at least in calculus, I used Apostol btw. I'll see if this stays true when I start on analysis or linear algebra..)Deriving formulas or proving theorems wasn't much of a problem. Usually with competitions I don't know the little tricks and things that everyone else seems to, although I think this has more to do with training than anything else. Basically as long as it dosen't need tricks or alot of computation (or geometry) I'm fine.

-------
In terms of contest math, my first post wasn't completely accurate - I can do pretty much any problem on the AMC as long as its not geometry or requires a trick, and usually the same thing on the AIME given enough time (~10-15 mins) to work on it. On USAMO I can only do algebra problems (functional equations,  polynomials, etc.) as I'm not at the Olympiad level yet in number theory or combinatorics. These (USAMO algebra problems) don't seem to require tricks as much as some other problems.
----

rrusczyk - Yeah, I didn't even know an AMC existed until I misclicked a link and found this site. (needless to say, it was one of the better misclicks I've made.) Our school dosen't really train us in AMC type problems, it barely even trains us in basic math to be honest. In classroom math almost nothing is proved or derived, and everything is done with a calculator. Also they use identical textbooks for alg 2 and precalc so they essentially make us take the same course twice. Finally, they refuse to let people take outside courses for credit. (Ok, that's enough ranting about my school now.)

So, thanks for replying and boosting my confidence. I'll probably continue with doing both competition problems and college math, and hopefully I'll have closed that gap by the time next fall rolls around.

If you can do apostol, you shold be fine.. :)

The FAQ

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11946

includes a lot of advice about getting ready for college math. Not everyone who succeeds in studying math as an undergraduate had math contest experience. I know plenty of math professors with Ph.D. degrees who have no contest experience whatsoever. Pursue what you love, work thoroughly at it, and you too can enjoy and succeed at math.


Let D(n) denote the leftmost digit of n.  Thus D($5^{4})$=6, What is D($6^{2007}$)

Though it repeats on the first 2 iterations of 9, I don't believe that the pattern will continue all the way to $2007$

I'm a skeptic on that one

I wrote a quick program to calculate it.  Unless theres an error in program ( which I don't think there is) , the leftmost digit is 5..

[hide="Answer (not a solution though)"]The value is
$5.61771263720587297342443186233\times 10^{1561}$
which makes the correct answer $\boxed{5}$.[/hide]

Ya, it is 2007log6 take the decimal part and do 10^that answer and floor it, making it 5.


不知道大陆，港澳台及海外的学生都在玩什么游戏。

我现在：Counter Strike
魔兽
有时，星际，红警。 ;)

I hardly ever play computer games. Sometimes I do some minesweeping...

我游戏玩得太烂了，只会Counter Strike  :)
以前玩传奇，超级没意思。。。
不知道港台的人玩得是什么游戏

踩地雷和彈珠台 :D  :P

minesweeping是排雷吗？  比较好的游戏，适合不同年龄层的玩家。 :)

传奇 哈哈，最好不要上瘾，耗时，耗money，耗精力。 :|

踩地雷 不知和排雷有何不同？弹珠台 没听说过。 :blush:

还是喜欢CS，虽然玩得不好，但自娱自乐。 :lol:

应该就是扫雷和三维弹球.

yaaaaaaaaaaaaa，我很喜歡玩Counter-Strike，已經玩了5年有多，有時還會和朋友組隊一起打clan war。有時會玩賽車game，像Need For Speed 2 Underground。

fifa, nba live, minesweeping, and
http://www.ugamehk.com

[quote]有時還會和朋友組隊一起打clan war[/quote]
大陆有浩方对战平台，各个省的玩家都可以进入玩，
不知你们那边怎样？
[quote]http://www.ugamehk.com [/quote]
好地方，
不过我这里网速太慢，还有乱码 :mad:

What about F1?I find it complicated to play.

F1 听起来不错，不知和极品飞车有何不同。

有喜欢War Craft的吗？Rpg很好玩完。

[quote="leepakhin"]踩地雷和彈珠台 :D [/quote]
還有接龍, 新接龍和新新接龍. :P

Well, i do a little bit warcraft, but my ex said it's war-crap... what a peace-keeper... that's called canadian... :lol:
well, anyways, i play http://www.tacticsarena.com/play everybody can try it... :ninja:

I just find out that the flash game Crimson's room is very interesting...


find all $ n$ such that $ f_{n}=f_{a}^{2}+f_{b}^{2}$ for $ a,b>0$. where $ f_{n}$ is the nth Fibonacci number

[hide="Solution"]Lemma: $ f_{i}=f_{j}f_{i-j-1}+f_{j+1}f_{i-j}$.

Proof: We induct on $ j$.  For $ j=1$, the lemma becomes $ f_{i}=f_{i-2}+f_{i-1}$, which is just the definition of Fibonacci numbers.  Now assume the lemma is true for $ j=k$, so that $ f_{i}=f_{k}f_{i-k-1}+f_{k+1}f_{i-k}$.  Then

$ f_{i}=f_{k+1}(f_{i-k-2}+f_{i-k-1})+f_{k}f_{i-k-1}$
$ f_{i}=f_{k+1}f_{i-k-2}+(f_{k}+f_{k+1})f_{i-k-1}$
$ f_{i}=f_{k+1}f_{i-k-2}+f_{k+2}f_{i-k-1}$

This is our lemma for $ j=k+1$, so the induction is complete, and the lemma is proved for all $ i$ and $ j$.

Now, let $ i=2x+1$ and $ j=x$ in our lemma.  We get $ f_{2x+1}=f_{x}^{2}+f_{x+1}^{2}$, so any odd number greater than 1 works.  Consider $ f_{2x}$ for $ x\ge 4$.  By our lemma with $ i=2x$ and $ j=x$, we have $ f_{2x}=f_{x}f_{x-1}+f_{x+1}f_{x}$.  Therefore,

$ f_{2x}<f_{x-1}f_{x+1}+f_{x}f_{x+1}=f_{x+1}^{2}$
$ f_{2x}=f_{x-1}f_{x}+(f_{x-1}+f_{x})f_{x}=f_{x}^{2}+2f_{x-1}f_{x}>f_{x}^{2}+(f_{x-2}+f_{x-1})f_{x}=2f_{x}^{2}$.

If $ f_{2x}=f_{a}^{2}+f_{b}^{2}$, the first chain of inequalities above implies that $ a$ and $ b$ must both be less than $ x+1$, but the second chain of inequalities implies that either $ a$ or $ b$ must be greater than $ x$.  This is a contradiction, so no even numbers greater than 6 work.  We manually check $ f_{1}$, $ f_{2}$, $ f_{4}$, $ f_{6}$ to find that only $ f_{6}$ can be expressed as $ f_{a}^{2}+f_{b}^{2}$, since $ 8=2^{2}+2^{2}$.  Therefore the only possible values for $ n$ are 6 and all odd integers greater than 1.[/hide]
*Edited to fix error.

A slight error: you used the "fact" that $ 2f_{x-1}> f_{x}$ (written slightly differently), but this is not true: $ 2f_{2}= 2 = f_{3}$.  This caused you to miss $ f_{6}= 8 = f_{3}^{2}+f_{3}^{2}$.

You're right, I assumed $ x\ge 3$ at the beginning of that part and then used $ f_{x-1}>f_{x-2}$.  I actually needed $ x\ge 4$.  I'll go fix my solution.


So, do you guys procrastinate a lot?
I do way too much, I need to stop.
Basically, all my homework is done 10 minutes before it's due, or, if its a project, it's finished at 7:00 am the day its due.

i'm the goddess of procrastination (not really)
but i'd wait till the morning to do most of my homework.

[quote="7h3.D3m0n.117"]Um, currently typing up 3 pages of an english rough draft, due tommorow of course.[/quote]
Lol, sounds like me a week ago.
English project due third block, spends most of first and second block typing it up.
And then the work I didn't do during first and second block I did the following morning.
Well actually, most of it I copied off another kid, but it was busiwork as always, so nothing missed by copying.
:o

yeah I do tend to procrastinate.  Its funny because often times we get in-class time to finish our projects when they are big and I usually still dont even start them till the day before they are due.

I have a feeling that anybody on GFF is doing some procrastination.

Then again, I don't procrastinate much.  I set myself schedules for big projects, and in terms of homework, I'll play a few games when I get home from school but then I go straight to homework.

Procrastination is part of human nature.

^ that's my poor attempt at justifying this bad habit.

[quote="miyomiyo"]I have a feeling that anybody on GFF is doing some procrastination.[/quote]

Oh, stop making me feel guilty, lol.

Yeah, I procrastinate a lot.

Unfortunately, since my English teacher is just that evil, procrastinating doesn't really work out (but I do it anyway). She assigns QRAs: Quarter Reading Assignments. Basically, it's reading 6 books, doing 7 essays, a bibliography, and a cover page all in one quarter. NotA = uh oh. As a matter of fact, it's due tomorrow and I still have 4 essays. (and 3 books to read).

EDIT: Sorry, not tomorrow. In one week.

Btw, anybody know any non-fiction essays about human cruelty?  :lol:
Also, do you know any random biographies or diaries on random Holocaust survivors?  :lol:
Also, how long is the Odyssey? *Gulp*

[quote="Nerd_of_the_Ages"]Also, how long is the Odyssey? *Gulp*[/quote]

eh. long enough, i'd say. :D
I read the comic book version, but it's accurate. I read the real version, too.

[size=200][b][color=darkred]PROCRASTINATION[/color][/b][/size]
WOOT!!
i love procrastination XD
procrastination list:
-- spanish homework
-- math homework (does this even count?)
-- social studies studying
-- LA "mechanism" essay
-- dumb PE homework
-- my life

[quote="Nerd_of_the_Ages"]Also, how long is the Odyssey? *Gulp*[/quote]

Not that long. Only like in the 20's of books, and each book might be 6 or 7 pages long. So like 140 pages. I'm reading the Iliad, which is combined in the same book as the Odyssey, and it is much thinner than I expected.

I procrastinate like a day, then I go into it.

We did Odyssey stuff a few months ago, and were it not for my other homework, I would have truly enjoyed reading the book for fun. But naturally there were strings attached and reading for fun seemed to be a time waster. Sparknotes saved me that night.

Apparently my English teacher memorizes SparkNotes pages of famous books, so now noone can use SparkNotes.  :(

As of now, I have a science poster due Thursday, and I've officially planned to do the whole thing tomorrow.

On essays and projects, I always do them at about 2AM the day its due, or if the class is after lunch, 1PM :O
I start homework no earlier than 9-10 everyday, record was 3  :blush:
But its bad...I'm sorta getting killed.
At the beginning of this school year, I finished homework at 7:20AM everyday...

For some reason, I do any of kind of simple written homework where all you have to do is answer a question right away. Maybe because those kinds are easier. But whenever its a speech or essay or project or things of that sort, procrastination wins me over.

Funny story: Over Christmas break, we got assigned a huge science project. Midway through my break, I had a nightmare in which I had waited until the last day to do it and I couldnt and I got a 0. I thought to myself I would never do that. I ended up doing it the day before we went back. I cried. And barely finished at 4 am. Scary

PROCRASTINATION FTW

...

[quote="Nerd_of_the_Ages"]so now noone can use SparkNotes.  :([/quote]

try Cliffsnotes.. they're not as good, but they work. don't they?

Me and PRocrastination.
I am a natural procrastinator.. like right now, I hardly did any of my homework.
although I nearly finished my presentation for English due mid-april. :D

I don't procrastinate for HUGE projects. Except for Science. b/c science projects are only like 10 pts. and they're easy.

FYI, looking at my copy of the Odyssey(pretty good, but VERY repetitive in parts), it is 485-76=409 pages long. Each book is not 6-7 pages...

oh, hah. im procrastinating right now...

-my math
-my language arts
-my history

heh. and it's all due tomorrow..well. another 1 o'clocker for me!

Dude, 1 isn't that bad. My record was sleeping past 2 for 2 weeks straight, one day at 4, one at 5, a few at 3...
Let's just say that I fell asleep anywhere that I could sit down.

If I put my mind to it, I can get off my ass and do work a few days before it's due.
But there are times where I just do everything the night before, or the lunch before, or 10 minutes before.
Whatever.

I'm up right now finishing social studies project. Just finished  :lol:

It's 2:15.

Hey sup guys, 12:51 here, about 10% done  :blush: