AMC USA(J)MO USAMO algorithm function vector

i have been trying this problem for a long time now...n i am really pissed!! :(
i have followed their solution...n i think i have made no mistake in implementation .....but i always get 55 n not 100 !! does any1 know the reason why ??!!

i solved it correctly...
mayb if u show me ur code.. ill be able to help...

ans do u guys mind discussing the questions in which i have trouble in ??
every1's doubts ...not only mine !! :P

thr would be many interested once we start with it...
so whats the problem u are having with "The Siruseri Sports Stadium"?

i have a really hideous code...let me make it more beautiful ...then i'll send it to u :P

Hi,
I did this problem(saw the solution of course).Use srand() in quicksort and do.
Also in the solution it is given E_i=S_1 +D_i.I think it should be S_1 +D_i -1.But it works only if you put S_1 +D_i.check that out.
You used quicksort right??

Chetan,do you know Pawan Kumar?I think he is in your school(XI).

Arjun Arul

Ya i got the same problem. The end day should be given by start day + legnth -1
(e.g start on day 1 of length 2 would mean days 1,2 i.e end day 2(1+2-1 not 1+2)
But the solution says otherwise and so too does the test cases. But as to which is actually right i am not sure.( my doubts increased since they give a clarification of the problem on this which emphasises this point)
And yes i was thinking of starting one such discussion myself here. I think it might be good if we could discuss the problems here. Any way some of the solutions are already provided on the website, so why not?

@arjun : i used the inbulit sort !! so it shudnt matter !! and yeah i know  pavan !!
@phinfinity :lets start here only ...

Can some one explain the VPATTERN problem. I absolutely have no idea of what we are expected to do.
([color=red][u][b]DONT TELL ME THE LOGIC. I ONLY WANT TO UNDERSTAND THE QUESTION NOT THE SOLUTION[/b][/u])[/color]
Here is the problem from [url]http://opc.iarcs.org.in/problems/VPATTERN[/url]:
[quote]
Problem : Variable Pattern Search

March 25, 2007
STATEMENT

Given a string P composed of upper and lower case letters, its neighbours consist of all strings of the form σ(P) where σ is a permutation which permutes the upper case letters (A–Z) among themselves and fixes the lower case letters (a–z).

For example, the neighbours of aAb include aAb, aBb, aXb, etc., and the neighbours of aAAbB include aBBbA, aXXbZ, etc., but not aXXbX.

Given a pattern P and a text T your job is to find the 0-based starting points of all occurrences of neighbours of P when viewed as sub-words of T.

INPUT

The input file will contain several test cases. The first line contains an integer t, the number of test-cases. The succeeding lines contain the test-cases.

Input Format:

Each test-case comprises the following:

Line 1: p , t: The length of P and T respectively
Line 2: P a string comprising of upper and lower case letters
Line 3: T a string comprising of upper and lower case letters

OUTPUT

The output should comprise the solutions to all the test cases, in the order of the test cases in the input. There should NOT be any blank line in the output.

Output Format:

The 0-based starting points of all neighbours of P in T, one per each line.

POINTS and CONSTRAINTS

Total points: 100

Easy Set (40 points) : 1 ≤ p ≤ 100 , p ≤ t ≤ 100000
Hard set (60 points) : 1 ≤ t ≤ 1000000 , p ≤ t

Time Limit: 3 seconds
Memory Limit: 128 MB (Stack memory: 8MB)

SAMPLE INPUT

2
2 4
aA
aAbB
3 9
aBb
aAbbaBbCb

SAMPLE OUTPUT

0
0
4 [/quote]

Im guessing u r having a problem in understanding the first part...
Coz the rest it pretty straight forward...

"Given a string P composed of upper and lower case letters, its neighbours consist of all strings of the form σ(P) where σ is a permutation which permutes the upper case letters (A–Z) among themselves and fixes the lower case letters (a–z)."

The lowercase letters are not allowed to change... they remain as they are
the uppercase letters are permuted.. implies...
  u can replace all occurrances of 'A' with a particular uppercase letter...
  all occurances of 'B' with a particular letter but not the same one as used for 'A'
  all occurances of 'C' with a particular letter.. but not the same as the one u replaced 'A' or 'B' with...
  and so on...

"the neighbours of aAAbB include aBBbA, aXXbZ, etc., but not aXXbX."

'a' and 'b remain whr they are...
aAAbB : aBBbA == 'A' is replaced by 'B'... and 'B' by 'A'
aAAbB : aXXbZ == 'A' is replaced by 'X'... and 'B' by 'Z'
aAAbB : aXXbX == it is illegal to replace both 'A' and 'B' by the same letter 'X'

Thanks now i can understand that part. but what does: "0-based starting points" mean? :?:

Sample Input
1
3 9
aBb
aAbbaBbCb

Output
0
4

0 - based... => index of first elemnt is 0.. not 1

can someone plz explain the solution for "the average".
i understood their sol for no repetitions but i cant extend it to sequence wid repetition ! :(

Why is the ICO online judge not working..
Is it b'cuz they are uploading the new online training system?

will this online training be available for those who didn't make it to camp
(like me :( )

[quote]This year, India is hosting APIO-2009[/quote]
So i reckon they are updating the ICO judge. This is certainly needed because i assume that Everyone else (China,Korea,etc.) participating in APIO which is probably gonna be on ICO judge.
(Just a few days before those waiting status messages popped up i noticed that the site was down as they where linking with amazon cloud computing, or something like that so that the server can handle situations where there are hundreds of submissions being made every now and then which has to be compiled and run.....)
But then again these are only my guesses :|  :?:

Amazon cloud computing for APIO??

As you know, the submission of APIO are checked only after the contest. So the no. of submissions will be very less compared to usual programming contests, etc.

Even if there are around 100 submissions a minute, I think a normal desktop would be enough.

The OPC judge would be down just because of some server problem. It would be up soon (hopefully) and I don't think they are adding anything extra.

[quote]As you know, the submission of APIO are checked only after the contest. So the no. of submissions will be very less compared to usual programming contests, etc.[/quote]
Hmmm... i didnt think of that....
But I am positive 100% sure that they were subscirbing or had been subscribing at least to amazon cloud computing. cause the server down message said something about amazon cloud computing. I just can't recall the exact wordings.

I still dont think a normal desktop would be enough to handle 100 submissions a minute
(each submission would take a minimum of 10 seconds to compile,run,process input/output verify,and show the results on a normal desktop. That is roughly 1000seconds of processing time in 60 seconds) so You would defenitely require more than 1 desktop. perhaps 10 or more mainframes/desktops.

is the online training up? sry, ive been out of touch...

@crakjak987:The online training hasn't started ..

I know this might sound a bit stupid but I have had some real troubles when coding the word hopping problem. Basically I used an array (a character double dimensional array or a string array whatever you call it here strings[105][15] to store the strings) which i had declared in int main(). I also had two external functions but these did not access the strings array. On submission I got all sorts of errors, I got a couple of wrong Answers along with some SEG errors. Finally frustrated :mad:  i downloaded the Test Cases and ran them to no avail. After struggling by a chance of luck I moved the strings array outside main making it a global variable. This solved my problem miraculously and it got accepted. NOW i am verrry curious to understand what had caused this. Could someone shed light to what i am doing wrong. Technically i thought that There is no difference between Global and local variables other than the fact that local variables cannot be accessed outside their scope???

Wow! Do you have more such super tips for us to remember? :)

Im sure he (bruteforce) could offer 1 evryday for the next 10 yrs and still not be out of stock
:P

[quote="bruteforce"]There are 2 types of memories - heap and stack. The stack is usually very small - around 8MB. The heap can be as large as the system allows (around 128MB to 4 GB).

Local variables, including the ones declared inside main are stored in the stack. Global variables and dynamically allocated memory( those declared using new/malloc or STL containers like vectors) are stored in the heap.

So for programming contests, it is suggested that huge arrays are either declared globally or using dynamic allocation. Note that there is usually a memory limit in most contests, which means that even the amount of heap you are allowed is limited. The stack is not affected by the limit.[/quote]

To be a little pedantic, memory allocation is of [i]three[/i] types: static, dynamic, and stack-based. Memory on the heap constitutes dynamically allocated memory exclusively. Static memory consists basically of global variables.

It's pretty obvious that static allocation and stack-based allocation are much faster than dynamic memory allocation, though knowledge of this won't really alter the complexity of your algorithm. However, I fell it's generally good practice to use global variables (in programming contests, [i]never[/i] in production code) as things can get pretty hairy with manual memory allocation (of course, using an STL class is fine as it's likely to do the dirty work for you).

Wow I never new all that!!!
though i still cant understand how moving an array (around 1kb+) could cause any difference. Are the memory allocated for recursive functions also from the heap??(since they are dynamically allocated??)
What happens when you declare say 10 1000x1000 arrays in the local scope(in main). do they all get allocated into heap or some in stack some in heap.

All variables declared inside any function (main/recursive/any other function) are always allocated to stack.
Only global variables and variables allocated using new/malloc or using STL containers are allocated to heap.

Declaring variables inside functions don't make them "dynamic". They're still local variables.

To add to bruteforce's point, note that there is a reason why it's called the [i]stack[/i] memory. It's not so easy to explain this in words, but [url=http://www-inst.eecs.berkeley.edu/~cs61c/sp08/lectures/06/2008SpCS61C-L06-ddg-cmem.ppt]these slides[/url] might offer some clarification.

If we make programs using jdk1.6 can we compile and execute them on another machine having jdk1.5 on dos prompt?


[quote="τρικλινο"]Ειναι η παρακατω αποδειξη σωστη???

 AxB =Φ <====> [ (a,b)εAxB <-----> [color=red](a,b)εΦ[/color]] <=====>[(aεA & bεB)<------>[color=red](aεΦ & bεΦ)[/color]] <====>[(aεA----->aεΦ) & (bεB------>bεΦ)] ======>(aεA----->aεΦ)====> A= Φ ====> A=Φ v B=Φ.[/quote]

Στο κοκκινισμένο σημείο βλέπω πρόβλημα: Η μετάβαση από το
(a,b)εΦ στο (aεΦ & bεΦ) δεν είναι σωστή. Το Φ είναι σύνολο διατεταγμένων ζευγών, οπότε από το γεγονός ότι το ζεύγος (α, β) ανήκει στο Φ, δεν βγάινει το συμπέρασμα ότι ανήκουν στο Φ και τα επιμέρους στοιχεία του ζεύγους.

Είναι σαν να λέμε ότι αφού το (1,1) ανήκει στο Καρτεσιανό επίπεδο RxR, τότε το ίδιο το 1 ανήκει στο επίπεδο RxR.

ΜΑ κυριε Τακη μου το κενο συνολο δεν ειναι ουσιστικο σονολο ειτε γραψουμε (a,b)εΦ ,ειτε γραψουμε aεΦ & bεΦ δεν εχει σημασια γιατι τιποτα δεν ανηκει στο κενο συνολο ,εξ αλλου δεν υπαρχει και κανενα θεωρημα η αξιωμα που να μας εμποδιζει να πουμε οτι (a,b)εΦ δεν συνεπαγεται aεΦ & bεΦ.

Αντιθετως η λογικη μας επιτρεπει να πουμε οτι........ (a,b)εΦ  συνεπαγεται aεΦ & bεΦ..............................................

επειδη ,ψευδες συνεπαγεται ψευδες , ειναι αληθες .Και στην περιπτωση μας : (a,b)εΦ ειναι ψευδες και ,aεΦ & bεΦ ειναι επισης ψευδες.

...........................................καλη χρονια..............................................................................

[quote="τρικλινο"]ΜΑ κυριε Τακη μου το κενο συνολο <...> [/quote]

Έχεις δίκιο. Στην εκφώνηση που έγραψες δεν κατάλαβα οτι το Φ που έχει η ερώτηση, είναι το κενό σύνολο $ \emptyset$. Νόμιζα ότι απλά ήταν "κάποιο σύνολο Φ" .

Με αυτά τα δεδομένα, η απόδειξη που παραθέτεις είναι σωστή.

Καλή Χρονιά.

Τ.Π.

Υστερόγραφο: Το κενό σύνολο
σε LAΤΕΧ γράφεται \emptyset μέσα σε δολάρια.


thats good.

a different approach is to write everything in terms of $a_{1}$

$a_{2}=3a_{1}+1$
$a_{3}=9a_{1}+4=9a_{1}+(11)_{3}$
...
where we can sum $a_{1}$ terms, and then the remaining terms are $1$, $(11)_{3}$, $(111)_{3}$ , etc which we can then sum.


A shop owner bought an antique table for $ \$25.00$. At what price in dollars should he list the table so that he may sell it at a discount of $ 20\%$ from his listed price and still make a profit of $ 20\%$ on his purchase price? Express your answer to the nearest cent.

A profit of $ 20$% is $ 6/5$ of the original value. On the other token, a $ 20$% discount represents $ 4/5$ of the original value. To find the list price, we will need to use the reciprocal of $ 4/5$, $ 5/4$, instead of $ 4/5$ itself.

$ \$25.00*(6/5)*(5/4)$
$ \equal{}\$25.00*(3/2)$
$ \equal{}\$\fbox{37.50}$


[quote="swanny1080"]like a subject.....[/quote]

I suppose you have no problem for raising positive numbers to a rational power.

There are a lot of possible definitions of irrational powers of positive numbers. For example :

One may say $ \sqrt{2}^{\sqrt{2}}=e^{\frac{\sqrt{2}\ln(2)}{2}}$

One may say $ \sqrt{2}^{\sqrt{2}}=\lim_{n \to +\infty}2^{q_n}$ for any sequence $ q_n$ of rational numbers whose limit is ${ \frac{\sqrt 2}{2}}$

...

In general, one can define irrational exponents as follows. Let $ r$ be irrational. Naturally $ 1^r\equal{}1$. As $ a > 1$ we define $ a^r \equal{} \inf \{ a^q \, | \, q \in \mathbb{Q} \ \textrm{and} \ q > r \}.$ In the case of $ 0 < a < 1$ we define $ a^r \equal{} \sup \{ a^q \, | \, q \in \mathbb{Q} \ \textrm{and} \ q > r \}.$


If ABC is a triangle nd D is a point on BC nd BD=DC  nd <BAD=<DAC < means cone prove that AB=AC

$\triangle ABC$ with $D$ on $BC$ such that $BD=DC$ and $\angle BAD\cong\angle DAC$. Prove $AB=AC$.

[hide="Proof"]$\triangle BAD\cong\triangle CAD$ (AAS Theorem), and corresponding sides are congruent, so $AB=AC$.[/hide]


I posted a poll in the Astronomy and Engineering forum [url=http://www.mathlinks.ro/viewtopic.php?p=949778#949778]here[/url].  After seeing that an option was missing I tried to edit the poll, I then found that the option to edit the poll was not there so I submitted the edit without correcting it.  Then I found the poll was completely gone.  I decided to edit again and I found the poll option was there but blank; I filled it out and submitted.  It then posted the poll now with the new options I put in and the old ones. :maybe: Trying to fix it I editted again.  :oops:  From looking at the post you kind of understand where this is going. :|

I couldn't recreate the bug; http://www.artofproblemsolving.com/Forum/viewtopic.php?p=951594#951594

Are you sure you didn't select "delete poll"?

You're not the first person to experience this type of problem:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101336


For a given positive integer $n$, find the greatest positive integer $k$, such that there exist three sets of $k$ non-negative distinct integers, $A=\{x_1,x_2,\cdots,x_k\}, B=\{y_1,y_2,\cdots,y_k\}$ and $C=\{z_1,z_2,\cdots,z_k\}$ with $ x_j\plus{}y_j\plus{}z_j\equal{}n$ for any $ 1\leq j\leq k$.

[size=85][color=#0000FF][Moderator edit: LaTeXified][/color][/size]

define the sum of elements of set A as $ S_{A}$...
kn=$ S_{A}\plus{}S_{B}\plus{}S_{C}\ge 3\frac{k(k\minus{}1)}{2}$,so we get k $ \le [\frac{2}{3}n\plus{}1]$
if $ n\equal{}3m$  then $ k_{max}\equal{}2m\plus{}1$,let  A={0,1,...,2m},B={2m,2m-2,...,0,2m-1,2m-3,...,1},C={m,m+1,...,2m,0,1,...,m-1}
the examples when n=3m+1,3m+2 can be given on the base of the example above


Find the  function  $f: (0,+\propto)\rightarrow(0,+\propto)$ such that  $f(x^2+y^2)=f(x)f(y)$ for all $x,y>0.$

[hide]Notice that $f(0^2+0^2)=f(0)f(0)$.  In other words, $x=x^2$ so x=f(0) is either 1 or 0.  If f(0)=0, $f(0^2+1^2)=f(0)f(1)$ or $f(1)=0*f(1)$ so $f(1)=0$.  More generally, since any positive number's square root and square is another positive number, one can ensure that f(n)=0.  In other words, the function is y=0.
If f(0)=1, then $f(0^2+2^2)=f(0)f(2)$ or f(4)=f(2).  In other words, if f(0)=1, then $f(n)=f(n^2)=f(\sqrt{n})=...$.
Now consider that $f(2)=f(\sqrt{2})=f(4)=f(1)f(1)$.  That means that $f(5)=f(3)=f(1)f(1)f(1).$  But $f(\sqrt{3})=f(1)^3$ so $f(5)=f(\sqrt{2})f(\sqrt{3})$ or $f(5)=f(1)^5$.  In other words, $f(1)^3=f(1)^5$ or $f(1)^3(f(1)+1)(f(1)-1)=0$.  If f(1)=0, then the line is y=0 when $y\geq 1$.  At zero, it's one.  Since f(1) isn't negative, f(1) could also be 1.  Then, the line is y=1 when $y\geq 1$ and at $y=0$.  However, this doesn't consider the fractions between 0 and 1.  It will be addressed now.  $f(1/2)=f(1/2)^2$.  Therefore, f(1/2) is one or zero.  If it is zero or one, a similar argument can be used as we did before to prove that all the fractional quantities on the line y=0 if f(1/2)=0 and y=1 if f(1/2)=1.[/hide]

dengmi, the function is defined from $(0, \infty) \longrightarrow (0, \infty)$! At the beginning i did the same mistake :)
I mean, you cannot take $f(0)$.

[hide]First substitute for $x$ some real number less than 1 and $\sqrt{x-x^2}$ for $y$. This substitution can be made for $y\leq \frac 12$. This gives $f(x)=f(x)f(y)\Rightarrow f(x)(f(y)-1)=0$. Since $f(x)\neq 0$ because it is given that $f(x)>0$, $f(y)=1$ for $y\leq \frac 12$.

Now let $f\left(1/\sqrt{2}\right)=c$. and $\sqrt{\frac{\sqrt{11}-3}{2}}=\alpha$. Then $f(1)=c^2$ and $\alpha=\frac{1}{\sqrt{3+\sqrt{11}}}<\frac{1}{\sqrt{6}}<\frac 12$.  So $f\left( \frac{\sqrt{11}-1}{2} \right)=f(1)f(\alpha)=c^2$ and $f(3/2)=f\left( \frac{\sqrt{11}-1}{2} \right)f(\alpha)=c^2$. But also $f(3/2)=f(1)f\left(1/\sqrt{2}\right)=c^3$. So $c^2=c^3$ and $c>0\Rightarrow c=1$.

It can now be shown that $f(x)=1$ for all $x$ through induction. $x^2-x=\left( x-\frac 12 \right)^2-\frac 14\geq -\frac 14\Rightarrow x^2+\frac 12\geq x+\frac 14$. So if $f(x)=1$ for $x\leq \frac k4$, then $f(x^2+1/2)=f(x)f\left(1/\sqrt{2}\right)=1$, and if $x\leq \frac k4$ can be used, then $x^2+\frac 12$ can reach at least $\frac{k+1}{4}$. So then $f(x)=1$ for $x\leq \frac{k+1}{4}$, which completes the inductive step.

Therefore the only such function $f$ is $\boxed{f(x)=1}$.[/hide]

[hide="  :ninja:  "]500th post!!!!!!!!!!!!!!

 :yup:  :D  :heli:  :play_ball:  :yankchain:  :trampoline:   :wow:  :starwars:  :clap2:   [/hide]

[quote="scorpius119"]$f(3/2)=f(5/4)f(1/2)=c^2$. [/quote]
I don't understand this: $\frac{3}{2} \neq (\frac{5}{4})^{2}+(\frac{1}{2})^{2}$. Well, indeed $c$ must be 1. In rest, your solution is beauiful! :)

o, that was bad mistake there. fortunately it shouldn't be very hard to fix it.


[hide]
$x=m+\sqrt{n}=\text{side length of square}$
let down a line perpendicular to AD from O, the center of the large circle, and call the intersection E.
draw AO, OQ..
right triangle AOE:
$(20)^2=(\frac{1}{2}x)^2+(x-10)^2\\ 400=\frac{1}{4}x^2+x^2-20x+100\\ x^2-16x-300=0\\ x=\frac{16\pm\sqrt{256-4(-240)}}{2}\\ x=8\pm4\sqrt{19}\\ x=8+\sqrt{304}\\ m+n=8+304=\boxed{312}$

[/hide]

[hide="Answer"]We will call the center of the larger circle $O$. $QO=10$ and $OB=20$. If we draw the perpendicular from $O$ to $BC$ and call it $x$, we see that $2x=10+\sqrt{20^2-x^2}\Rightarrow 2x-10=\sqrt{20^2-x^2}\Rightarrow$
$4x^2-40x+100=400-x^2\Rightarrow 5x^2-40x-300=0\Rightarrow x^2-8x-60=0$, which gives $x=4+\sqrt{76}$, so $AB=2x=8+\sqrt{304}\Rightarrow 8+304=\boxed{312}$.[/hide]

[quote="chesspro"]A circle with diameter $ \overline{PQ}$ of length 10 is internally tangent at $ P$ to a circle of radius 20. Square $ ABCD$ is constructed with $ A$ and $ B$ on the larger circle, $ \overline{CD}$ tangent at $ Q$ to the smaller circle, and the smaller circle outside $ ABCD$. The length of $ \overline{AB}$ can be written in the form $ m \plus{} \sqrt {n}$, where $ m$ and $ n$ are integers. Find $ m \plus{} n$.[/quote]

$ \overline{CD}$ tangent at $ Q$ to the smaller circle, and the smaller circle outside $ ABCD$
It really doesn't make sense to me... Is it tangent at two different "smaller circles" or what?

[hide="solution"]

Let $ PE$ be the diameter of the larger circle, and let $ PE$ intersect $ AB$ at $ F$
$ x\equal{}\text{the length of AB}$

We have $ PF.FE\equal{}AF.FB$
$ (10\plus{}x)(30\minus{}x)\equal{}\frac{x}{4}.\frac{x}{4}$
$ 4(300\plus{}20x\minus{}x^2)\equal{}x^2$
$ 5x^2\minus{}80x\minus{}1200\equal{}0$
$ x^2\minus{}16x\minus{}240$

$ x\equal{}\frac{16\pm\sqrt{16^2\minus{}4.\minus{}240}}{2} \rightarrow x\equal{}8\plus{}\sqrt{304}$ or $ x\equal{}8\minus{}\sqrt{304}$
We take $ x\equal{}8\plus{}\sqrt{304}$

So, $ m\equal{}8, n\equal{}304$ , $ m\plus{}n\equal{}312$

[/hide]

[quote="chesspro"]A circle with diameter $\overline{PQ}$ of length 10 is internally tangent at $P$ to a circle of radius 20. Square $ABCD$ is constructed with $A$ and $B$ on the larger circle, $\overline{CD}$ tangent at $Q$ to the smaller circle, and the smaller circle outside $ABCD$. The length of $\overline{AB}$ can be written in the form $m + \sqrt{n}$, where $m$ and $n$ are integers. Find $m + n$.[/quote]

[hide="Solution"]
Let the center of the circle be $O$, the side length of the square be $s$, and the midpoint of $AB$ be $X$. Since $OP = 20$ and $QP = 10$, $OQ = 10$, so $OX = s-10$. We also know $OA = 20$ and $AX = \dfrac{s}{2}$, and $\angle AXO = 90$. Solving the resulting pythagorean equation, we get $8 \pm \sqrt{304}$. Discarding the negative solution, our answer is $8 + \sqrt{304}$, or $\boxed{312}$.
[/hide]


arithmetic sequence
algebra unsolved
algebra

If $ x^{2},y^{2} ,z^{2}$ form an arithmetic progression than prove that also $ \frac{1}{x\plus{}y}, \frac{1}{y\plus{}z}, \frac{1}{z\plus{}y}$ form an arithmetic progression.

but it's sooo basic! why is it in the Unsolved section?

\[ \frac{2}{x\plus{}z}\equal{}\frac{1}{x\plus{}y}\plus{}\frac{1}{y\plus{}z}\Leftrightarrow\]
\[ 2(x\plus{}y)(y\plus{}z)\equal{}(x\plus{}z)(x\plus{}z\plus{}2y)\Leftrightarrow\]
\[ 2xz\plus{}2yz\plus{}2xy\plus{}2y^2\equal{}x^2\plus{}z^2\plus{}2xz\plus{}2xy\plus{}2yz\Leftrightarrow\]
\[ y^2\minus{}x^2\equal{}z^2\minus{}y^2\]

If $ x^2, y^2, z^2$ are in A.P., so are $ x^2 \plus{} \sum xy$ etc. (Adding $ \sum xy$ to each of the terms retains the A.P. property)

i.e. $ (x\plus{}y)(x\plus{}z), (y\plus{}z)(x\plus{}y), (y\plus{}z)(x\plus{}z)$ are in A.P. and [dividing by $ (x\plus{}y)(y\plus{}z)(z\plus{}x)$] hence $ \frac{1}{y\plus{}z}, \frac{1}{z\plus{}x}, \frac{1}{x\plus{}y}$ are in A.P. which is equivalent to the claimed result

[quote="ridgers"][quote="hsbhatt"]If $ x^2, y^2, z^2$ are in A.P., so are $ x^2 \plus{} \sum xy$ etc. (Adding $ \sum xy$ to each of the terms retains the A.P. property)

i.e. $ (x \plus{} y)(x \plus{} z), (y \plus{} z)(x \plus{} y), (y \plus{} z)(x \plus{} z)$ are in A.P. and [dividing by $ (x \plus{} y)(y \plus{} z)(z \plus{} x)$] hence $ \frac {1}{y \plus{} z}, \frac {1}{z \plus{} x}, \frac {1}{x \plus{} y}$ are in A.P. which is equivalent to the claimed result[/quote]

Is it always true that if we divide terms of an A.P. also the result is an A.P.???[/quote]

obviously: if $ b\minus{}a\equal{}c\minus{}b$, then dividing by $ d$ we have $ \frac{b}{d}\minus{}\frac{a}{d}\equal{}\frac{c}{d}\minus{}\frac{b}{d}$.

sorry to ask such an stupid question but im new and i wanna learn: is it correct to assume that $ b\minus{}a\equal{}c\minus{}b$ i mean is it correct to assume that they are consecutive in the AP? couldnt it be that  $ b\minus{}a\equal{}md$ and  $ nd\equal{}c\minus{}b$ with d the difference between two consecutive terms on the AP??

[quote="eze100"]sorry to ask such an stupid question but im new and i wanna learn: is it correct to assume that $ b \minus{} a \equal{} c \minus{} b$ i mean is it correct to assume that they are consecutive in the AP? couldnt it be that  $ b \minus{} a \equal{} md$ and  $ nd \equal{} c \minus{} b$ with d the difference between two consecutive terms on the AP??[/quote]

"a,b,c in arithmetic progression"  means that they form an arithmetic progression, i.e., $ b\minus{}a \equal{} c\minus{}b$

@eze10

Basically , a sequence is said to be in A.p , if  the respective terms in the sequence differ by  a common number . So for instance a sequence like 12 ,18,24 is an A.P with the common difference of 6 . the terms need not be consecutive .

[quote]Re: is it correct...
eze100 wrote:
sorry to ask such an stupid question but im new and i wanna learn: is it correct to assume that $ b \minus{} a \equal{} c \minus{} b$ i mean is it correct to assume that they are consecutive in the AP? couldnt it be that $ b \minus{} a \equal{} md$ and $ nd \equal{} c \minus{} b$ with d the difference between two consecutive terms on the AP??

"a,b,c in arithmetic progression" means that they form an arithmetic progression, i.e., $ b \minus{} a \equal{} c \minus{} b$ [/quote]
Then the problem : "Prove that $ \sqrt {2} , \sqrt {3} , \sqrt {4}$ cannot be the terms of an arithmetic progression" is reduced to prove that $ \sqrt {4} \minus{} \sqrt {3}\neq \sqrt {3} \minus{} \sqrt {2}$ . Thats what youre telling me right?

[quote="vijaymenon"]But it isn't right?[/quote]

I meant that eze100's way it's incomplete. See in the above link why.

Pain rinnegan

Oh i am sorry actually i misread the question , really sorry .

[quote="Pain rinnegan"][quote="vijaymenon"]yes thats exactly what i meant .[/quote]

But you're wrong. I mean $ \sqrt {2},\sqrt {3},2$ can be the non-consecutive terms of an arithmetic progression

[quote="eze100"]
Then the problem : "Prove that $ \sqrt {2} , \sqrt {3} , \sqrt {4}$ cannot be the terms of an arithmetic progression" is reduced to prove that $ \sqrt {4} \minus{} \sqrt {3}\neq \sqrt {3} \minus{} \sqrt {2}$ . Thats what youre telling me right?[/quote]

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=302415&search_id=240290094]here[/url] how you must procced.[/quote]

Then how do you know when the problem is implying that they are consecutive? I mean in pluristiq's answer he assumes that \[ y^{2}\minus{}x^{2}\equal{}z^{2}\minus{}y^{2}\] he assumes the three of them are consecutive, but, in the problem i mentioned before(the one with the square roots), you tell me its a mistake to assume that. So, when can you assume that they are consecutive and when not?

[quote="eze100"][quote="Pain rinnegan"][quote="vijaymenon"]yes thats exactly what i meant .[/quote]

But you're wrong. I mean $ \sqrt {2},\sqrt {3},2$ can be the non-consecutive terms of an arithmetic progression

[quote="eze100"]
Then the problem : "Prove that $ \sqrt {2} , \sqrt {3} , \sqrt {4}$ cannot be the terms of an arithmetic progression" is reduced to prove that $ \sqrt {4} \minus{} \sqrt {3}\neq \sqrt {3} \minus{} \sqrt {2}$ . Thats what youre telling me right?[/quote]

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=302415&search_id=240290094]here[/url] how you must procced.[/quote]

Then how do you know when the problem is implying that they are consecutive? I mean in pluristiq's answer he assumes that
\[ y^{2} \minus{} x^{2} \equal{} z^{2} \minus{} y^{2}\]
he assumes the three of them are consecutive, but, in the problem i mentioned before(the one with the square roots), you tell me its a mistake to assume that. So, when can you assume that they are consecutive and when not?[/quote]

Because in the problem from this thread it says "Prove that ... are in arithmetic progression" , while your problem says "Prove that ... can't be terms of any arithmetic progression"


algebra
polynomial
inequalities open
inequalities

[quote="conceive"]Prove or disprove:
$a,b>0$

$8a^2b^2+4a^2b+4ab^2+a^2+b^2+4ab+3\ge8a+8b$[/quote]

Some notes: I couldn't find any counterexample. If I choose $a,b$ too small, then number $3$ on the left side save the day. And if I choose them too big, then polynomial of the bigger degree on the left side kills the polymonial on the right side.

But I'm not able to prove it, neither.

If $f(a,b)=8a^2b^2+4a^2b+4ab^2+a^2+b^2+4ab+3-8a-8b$ then $f( \frac{1}{ \sqrt{13}} , 1) < 0$

If my calculations are correct  :D  :) .


In triangle ABC  the medians are M_a, M_b, M_c

Prove that 2[ a(M_a) +b(M_b) +c(M_c) ] <3.Sqrt[ (ab)^2  + (bc)^2  +(ca)^2]

The problem looks beautiful, but solution isnt hard to find.

2(ama +bmb + cmc)/3 ≤ 2√ (a2ma2+b2mb2+c2mc2)/3 =

= 2√[(a 2b2+b2c2+a 2c2) (a^4+ b^4+c^4)/4)]/3 ≤ √(a 2b2+b2c2+a2c2)

The last inequlity is true because after squaring we get
a^4+ b^4+c^4 ≥ a2b2 +b2c2+a2c2


An electrically charged conducting sphere 'pulses' radially i.e. its radius changes periodically with fixed amplitude. The charges on it's surface - acting as many dipole antennae - emit EM radiation.

                         [b]What is the net pattern of radition from the sphere?[/b]

Let's say you accomplished this with an inflatable aluminium-coated balloon.
Alternately inflating and deflating the balloon would carry the charges along,
pulsing.

However there would be no electromagnetic radiation according to the Sphere Theorem as applied to Electrostatics:

The theorem states that a hollow sphere of negligible thickness and uniform charge density across the surface acts just as if it were a point-charge located at the geometric origin of the sphere.  No change in field potential or radial variation in force upon a test particle outside, so no electromagnetic radiation, period.

[img]http://lentils.imagineis.com/rouge2/PulsingSphere.jpg[/img]

However, the Sphere Theorem is actually wrong, as applied to both gravity and electrostatics, especially in specific physical situations.  Both classical theories are approximations and contain unsupportable hidden assumptions and axioms.

Firstly, i would like to confirm that EM radiation is emmited only when $ \vec{E} $ and $ \vec{B} $ dur to some charge configuration are time-variant. Am I right?

Also , you've handled the electric fields by using that they dont change unless center of sphere moves? But what about the possibly changing magnetic fields. Also what about individual charges?

Yes, according to the [b]causality[/b] axioms of both gravity and electrostatics, there can be no change in the field unless there is physical motion or change of the value of the charges/mass.  To be time-variant therefore in the classical theories requires physical motion in space, since charges and mass are fixed constants for particles.

[hide]
However, this does not necessarily imply that there is a change in the field potential *everywhere*, even when charges move.  In the case above, only a small volume of the field is affected, namely the volume between the minimal diameter and maximal diameter of the sphere. In this zone, the field potential changes, but nowhere else does it change, (according to the Sphere Theorem).  So charges move, and there is change, but only in a small region.

There is no electromagnetic radiation per-se, since the change is actually like 'uncovering' a fixed field, where the field varies instantaneously at each point in space in the volume concerned, as the sphere-surface crosses it.  Although the barrier moves at a fixed speed, the field toggles 'on' or 'off' instantaneously and stays constant when not switching.
[/hide]

I posted a new diagram in the first reply, in case anyone missed it.

Rogue physicst,
what about the possibly changinh magnetic field.You haven't shown ehy it won't result in EM radiation???!!! :)

Actually a good question:

Consider the expanding charges as travelling along copper wires radiating outward from the centre:  Then, by the right-hand rule of Classical Electodynamics, the magnetic field will be counter-clockwise around each wire while the charges are moving outward, and clockwise as they move back inward.  Note that the 'net' magnetic field on the surface of the expanding sphere then, is again zero.

In fact, the counter EMF generated by the adjacent wires would be six times as strong as the forward EMF, effectively dampening oscillation, and quickly bringing the motion of the charges to a standstill, presuming some other force (like nuclear attraction) was holding them from expanding outward infinitely by repulsive force.

Can you be a bit more quantitative and not bring extraneous factors into it!!!!

[img]http://lentils.imagineis.com/rouge2/MagField.jpg[/img]

Here we see the magnetic field trying to form around the path of each charge as it moves outward.  However, locally, the field-lines cancel.  If the surface were a finite plane with an edge, there would be a 'residual' magnetic flow around the perimeter because of uncancelled  field-lines. But because the sphere surface has no 'border' the magnetic field lines cancel everywhere, and there is no net field remaining.

The 'extraneous' points are necessary, because even in a 'thought' experiment, we have to imagine a plausible if incomplete mechanism for the physical phenomenon.

Cool Diagrams, where do you get them. Plz tell me!!!!!! :)  :roll:

Also your concepts seem crystal clear to me. Which class are you in. Can you tell be the books or internet resources you use or have ever consulted and found to be good!!

You're participating in the Physics Weekly contest,right???

I just draw them as I need them.
My daughter suggested pink for the background, to give the drawing life.

What program do you use to draw them??? Is it available (or something like it) free online?

So you are a Physics Professor? Where?

However, when we analyze three-dimensionally, there is a twist:
Imagine we are riding upon one of the charges.  From the point of view here, the adjacent charges are simply moving directly away on all sides in the 'plane' of the sphere's surface.  This means the central (observer) charge sees horizontal magnetic fields formed around the receding charges.  Now there *is* a residual force on the upper and lower boundary of the sphere's effective thickness.

[img]http://lentils.imagineis.com/rouge2/MagPulse2.jpg[/img]

Luckily, we can again imagine this is true for *every particle, and the complimentary residual currents again cancel out! Lucky that!

(I use photoshop to draw simple diagrams like this.  It is fast and effective!)


What is with this forum? How come it seems to be a general discussion forum, not a MO one?

Well, because there are a lot of people from all over the country who browse this thread, so we appeal to their interests.

Also, if we posted spam like this anywhere else, we would get in trouble. Here in MO, its ok.

As a side hobby, we continue the traditional KS-MO border rivalry. Darn you slave states and darn you self-determination. Darn you roads. I darn you all to heck.

OH DANG WE TOTALLY LEARNED ABOUT THIS IN SOCIAL STUDIES THIS WEEK WHAT WITH THE BLEEDING KANSAS AND WHATNOT!

speaking of social studies, we have a test in tuesday, 2 days after spring break

[quote="Mewto55555"]Well, because there are a lot of people from all over the country who browse this thread, so we appeal to their interests.
[/quote]

In other words, the great people of Missouri understand the desires of others to interact with their peers from such a great state and accept any and all comments from outsiders despite the superiority of the Missourians.

One day, a long, long time ago...

Mewto55555 first posted in the forum...

And all has not been the same since...

Don't ask questions. Just join or leave. HAHAHAH

I nominate the idea of MO and KS civil war.

Somehow, transport KS and MO to an alternate dimension (Canada), And let them fight.

That would not be very entertaining, because MO would get completely destroyed by any other state the same way Skyhigh completely destroyed lecaf in the recent finals.

Good choice of location. Nothing in Canada.

Uh yeah there is, I think there are bears there.

[quote="hunter34"]Yeah there are. But if you are dumb enough to provoke them, you should get killed.[/quote]
Oh man I bet Mewtwo55555555 (however many there are, what a silly username) is going to test that out.

There are five fives. Use common sense, people. :P

Urgh, math. Get it out.

Mew you shouldn't try it out. Personal expierence speaking here.

Hmmm so a username with zeroes "0" would have zero zeroes, hence not having any zeroes at all.

CONTRADICTION!

OH MAN MR SMT 2009 BEAST JUST SHOWED YOUR LOGIC IS FALLACIOUS BOY

[quote="serialk11r"]What? In base 36 a Z would represent 35, and I think the guy with a bunch of Zs in his username doesn't have 35 Zs. Nor does he have 25, or 26. Mewtwo seems especially silly since he fashions having people see "Mew Mew" in his avatar, thus there is no common sense element involved.[/quote]

Clearly my manly nickname is mewmew.


calculus integration function algebra polynomial limit induction

Answer for the first one seems to be $f(\frac{1}{2})$,but now I don't have a complete proof.It will be later.

Yes, since the original statement of the problem was:
    Let $f$ be a continuous nonnegative function defined on $[0,1]$ and whose integral over this segment is 1. Prove that for any continuous function $g$ we have that
    $\int_{0}^{1}{\int_{0}^{1}{...\int_{0}^{1}{f(x_1)...f(x_n) g(\frac{x_1+...+x_n}{n})}dx_1...dx_n}}$ is $g(\int_{0}^{1}{xf(x)dx})$.
     Of course, it suffices to prove this for $g$ a polynomial, but it is there where I have problems.

Well,indeed it is sufficent to prove the statement for polynomials.
Now let us denote $A_{n,m}=\int_{I_{n}}{(x_{1}+x_{2}+\ldots+x_{n})^{m}dx_{1}dx_{2}{\ldots}dx_{n}}$,where $I_{n}=[0,1]^{n}$.We will show that $\lim_{n{\rightarrow}\infty}{\frac{A_{n,m}}{n^{m}}}=2^{-m}$.
Let's do it by induction.It is easy to verify the statement for $m=1$.Now some computations:
$\frac{A_{n,m}}{n^{m}}=\int_{I_{n}}{(\frac{x_{1}+x_{2}+\ldots+x_{n}}{n})^{m}dx_{1}dx_{2}{\ldots}dx_{n}}=\frac{1}{n^{m-1}}\int_{I_{n}}{x_{1}(x_{1}+x_{2}+\ldots+x_{n})^{m-1}dx_{1}dx_{2}{\ldots}dx_{n}}=\frac{1}{n^{m-1}}\sum_{i=0}^{m-1}{\int_{I_{n}}{C_{m-1}^{i}x_{1}^{i+1}(x_{2}+\ldots+x_{n})^{m-1-i}dx_{1}dx_{2}{\ldots}dx_{n}}}=*$,then
$*=\frac{1}{n^{m-1}}\sum_{i=0}^{m-1}{C_{m-1}^{i}\int_{0}^{1}{x_{1}^{i+1}dx_{1}}\int_{I_{n-1}}{(x_{2}+\ldots+x_{n})^{m-1-i}dx_{2}{\ldots}dx_{n}}}=\frac{1}{n^{m-1}}\sum_{i=0}^{m-1}{C_{m-1}^{i}\int_{0}^{1}{x_{1}^{i+1}dx_{1}}A_{(n-1),(m-1-i)}}$. Now notice that by the induction hypothesis all the summands except the first one have less order in $\infty$ than $n^{m-1}$,so $\lim_{n{\rightarrow}\infty}{\frac{A_{n,m}}{n^{m}}}=\int_{0}^{1}{xdx}{\cdot}\lim_{n{\rightarrow}\infty}{\frac{A_{n-1,m-1}}{n^{m-1}}}=2^{-m}$.
The rest is clear.
I'll keep thinking on the general problem.May be it can be done by the same way.

[quote="Moubinool"]Really strange that $x_n$ appears to the left to the right of $=$  :?
in the a) and b)

typo[/quote]
It's just a typo.We have to find the limit of the RHS.

Oh, come on, Moubi!  :P   Nice proof, Levon. Now, one more stupid question: what if we assume just that the function is integrable, does the result remain true?

Another approach is to use the Big Numbers Law. This multiple integral is just the expectation of $f(\xi_n)$ where $\xi_n$ is the average of $n$ independent random variables uniformly distributed on $[0,1]$. According to the Big Numbers Law, for every $\delta>0$, we have $P\{|\xi_n-\tfrac12|>\delta\}\to 0$ as $n\to\infty$, from where the result follows easily.

If $f$ is integrable, [b]and $\frac 12$ is a Lebesgue point of $f$[/b], then the result remains true (otherwise it may be false because changing $f$ at one point does not change the integrals; moreover, one may construct a bounded measurable function $f$ for which the limit fails to exist)

Well I have to think on your question.But I want to give the proof of the original problem.
First of all let's prove that if $g,f$ are continious,then the limit of the sequence
$\int_{I_{n}}{g(\frac{f(x_{1})+f(x_{2})+\ldots+f(x_{n})}{n})dx_{1}dx_{2}{\ldots}dx_{n}}$ is $g(\int_{0}^{1}{f(x)dx})$.The proof is the same as in the previous post.
Now for the original problem we have that $\int_{I_{n}}{f(x_{1})f(x_{2}){\ldots}f(x_{n})g(\frac{x_{1}+x_{2}+\ldots+x_{n}}{n})dx_{1}dx_{2}{\ldots}dx_{n}} =\int_{I_{n}}{g(\frac{{\psi}(y_{1})+{\psi}(y_{2})+\ldots+{\psi}(y_{n})}{n})dy_{1}dy_{2}{\ldots}dy_{n}}$,where
${\psi}(y)$ is the inverse function of the $F(x)=\int_{0}^{x}{f(t)dt}$.So the limit is
$g(\int_{0}^{1}{{\psi}(y)dy})$,but $\int_{0}^{1}{{\psi}(y)dy}=\int_{0}^{1}{{\psi}(F(x))dF(x)}=\int_{0}^{1}{xf(x)dx}$.So we are done.

Hmmm, Moubi, I start to loose my mind. Don't you read what is said in the other posts? Particularly, it is clear that this formula is a particular case of the general ens problem, whose statement lies just a little bit above these posts. ;)

The answer for b) seems to be $f(e^{\int_{0}^{1}{\ln{x}dx}})=f(\frac{1}{e})$.It can be done by the statement done by me in my previous post.

[quote="LevonNurbekian"]Dear fedja,what is Lebesgue's point?[/quote]
$x$ is a Lebesgue point for $f$ if $\lim_{h\to 0+}\frac 1{2h}\int_{[x-h,x+h]}|f(y)-f(x)|\,dy=0$. There is a nice theorem in real analysis stating that, for a Lebesgue integrable function, almost every point is its Lebesgue point. The concept is close to that of continuity; the only difference is that instead of demanding that $f(y)$ be close to $f(x)$ for all $y\in[x-h,x+h]$, we demand that $f(y)$ be close to $f(x)$ in average on that interval.


The ratio of beans to peas in a jar is 2:5 and there are a total of 840 peas and beans. How many beans should be added to make the ratio 1:2?

The ratio tells us that for every 7 beans and peas, 2 of them are beans and 5 of them are peas. 840 is 120 sets of 7, giving you 240 beans and 600 peas. For the ratio to be 1:2, keeping the peas the same, you must add $ \boxed{60}$ beans.

Since the ratio is $1:2$, there has to be $600/2=300$ beans, so we must add $60$.

Even easier, you can notice that for the ratio from $2:5$ to become $1:2$, you know that you must add $0.5$ to $2$ of the $5$ to get it to one-half. Since $1$ of the ratio equals $120$, $\frac{1}{2}$ of the ratio equals $\frac{120}{2}=\boxed{60}$.


You roll a dice with a friend. The one who rolls a 6 first wins. Whats the chance for winning if you are the one to start rolling the dice?

[quote="cincodemayo5590"][hide]
We know that right off the bat, if you roll a 6, you win. There's a $\frac{1}{6}$ chance of that happening. Now, what's the probability of getting another chance to roll? There's a $\frac{5}{6}$ chance of you rolling something other than a 6, and the same chance of your friend rolling something other than a 6. Then, the process repeats; $\frac{1}{6}$ chance of you winning, etc. The probability looks like this:
\[\frac{1}{6}+\frac{5}{6}( \frac{5}{6}( \frac{1}{6}+\frac{5}{6}( \frac{5}{6}( \frac{1}{6}+\cdots \]
and it goes on to infinity. Now, we can find explicitly the right portion of the probability. Call it $x$, but then we have:
\[x= \frac{5}{6}( \frac{5}{6}( \frac{1}{6}+x ) ) \implies x= \frac{25}{66}\]
From this, we know that our final probability is:
\[\frac{1}{6}+\frac{25}{66}= \boxed{ \frac{6}{11}}\]
[/hide]

([i] Is there a quicker way to do these types of problems? [/i] )[/quote]
I get the same as you. I'm not sure there really is a faster way.

you could also treat it as an infinite geometric sequence with a=1/6 and r=(5/6)^2=25/36

This is a really fast way, and the way I solve these problems when I get them.

Consider the [i]odds[/i] of you winning on your first turn to your opponent winning on his first turn. You win on your first turn $\frac{1}{6}$ of the time and he wins on his first turn $\frac{5}{36}$ of the time. Each round is identical except for a multiplier, but the odds remain the same, so the odds are

$\frac{1}{6}: \frac{5}{36}$
or
$6 : 5$

and the probability can easily be calculated from there as $\frac{6}{6+5}= \boxed{\frac{6}{11}}$ as you all got.


function
inequalities
parameterization
calculus
calculus computations

Hi guys, I would really appreciate if somebody willexplain to me in layman's term the meaning of pointwise covergence in the sequence of functions and much better if there is a graphical diagram.

I do not think restating the definition as it is written in your book will help you much, so I'll try to explain it using more familiar objects than numbers and functions.

Suppose we have a huge (or even infinite) set of people. Suppose that they all have rabbits jumping around them they want to catch (one for each person) and every day a few rabbits get caught. You can say that they the pointwise limit is the state in which all rabbits are caught if each person will catch his rabbit eventually. Note that different people may wait different time before catching their rabbits (Let's say they are enumerated by numbers $ 1,2,3,\dots$ and on the $ k$'th day the $ k$'th person makes a catch. Clearly, the first few people will catch their rabbits very soon, but the person with the number $ 10^{10}$ will have to wait for quite a while before his rabbit gets caught).

Now replace people with values $ f(x)$ of some function on elements of an abstract set $ X$, rabbits by $ f_n(x)$, making a catch on the $ n$-th day by the person $ x$ by the condition $ |f(x)\minus{}f_k(x)|<\varepsilon$ for all $ k\ge n$ (whose meaning is that once you really catch the rabbit, you do not let him go away any more) and you get almost what you want. The last step is to say that this "eventual catch of a rabbit" occurs regardless of the value of $ \varepsilon>0$. So here is the formal definition given parallelly in terms of functions and in terms of rabbits:

For every $ x\in X$ (For every person),
for every $ \varepsilon>0$ (regardless of how close he wants to keep the caught rabbit to himself),
there exists $ n>0$ (there will come a day)
such that
for every $ k\ge n$ (starting from this day and forever)
$ |f_k(x)\minus{}f(x)|<\varepsilon$ (he will keep his rabbit that close and will not let him escape).

Thanks. I really appreciate your effort to explain. One more question though, when you say, a function is convergent pointwise, does that mean that it approaches a certain point - or the point it refers to is simply a vertical or horizontal line in the $ xy$ plane.

One more. I have this exercise I am answering. (Im preparing for exam this coming August 3). This is an exercise from a notebook of a friend but it has no solution. I hope you could help me.

$ f_{n}(x) \equal{} \begin{cases}n^{2}x; & 0\leq x\leq\frac {1}{n} \\
n^{2}(x \minus{} \frac {2}{n}); & \frac {1}{n}\leq x\leq\frac {2}{n} \\
0; & \frac {2}{n}\leq x\leq 1\end{cases}$

1.) Discuss the continuity of $ f_n(x)$. Well, I answered that already, it is not continuous at $ x \equal{} \frac {1}{2}$ but continuous at $ x \equal{} \frac {2}{n}$

2.) Sketch the graph. (Horizontal line in $ f_n(x) \equal{} n$ from 0 to 1/n, horizontal line at $ f_x(n) \equal{} \minus{} n$. Confused with the others.

3.) Prove that $ f_n$ converges to $ f$ pointwise. I have a proof but Im really having a hard time with Latex.

Thanks.

Just to overemphasize something that fedja already said:

In the collection of lines at the bottom of his post, his third line says "there exists $ n > 0.$" This $ n$ depends on both $ \epsilon$ and $ x.$ $ \epsilon$ is your arbitrary definition of how close you have to be to declare the rabbit caught; $ x$ is the name tag of one of the rabbit catchers. Saying that $ n$ depends on $ \epsilon$ is just built in to the nature of limits - if you're more stringent about what you mean by "caught", you may have to wait longer for it to happen. But the key here is that $ \epsilon$ depends on $ x$. There will come a day, but each person - each $ x$ - has a different definition of what that day is. If there are infinitely many $ x$s - infinitely many people - you could very well have to wait an infinitely long time for every last rabbit to be caught.

If you could catch all of them in the same finite time, then that would be [i]uniform[/i] convergence.

My favorite multipurpose example: On $ [0,1],$ for $ n\ge 2,$ define $ f_n(x)$ to be the piecewise linear continuous function with nodes (vertices) at $ (0,0),$ $ \left(\frac1n,n\right),$ $ \left(\frac2n,0\right),$ $ (1,0).$

[hide="If you insist on a formula"]$ f_n(x) \equal{} \begin{cases}n^2x & \text{if }0\le x\le \frac1n \\
\minus{} n^2x \plus{} 2n & \text{if }\frac1n < x\le\frac2n \\
0 & \text{otherwise}\end{cases}$

But next time, don't insist on a formula. The formula adds nothing to your understanding.[/hide]

I claim that $ f_n(x)\to 0$ pointwise for all $ x\in [0,1].$ If $ x \equal{} 0,$ then $ f_n(x) \equal{} 0$ and we're already there. If $ x > 0,$ then there will come a day at which $ \frac2n < x,$ and from that day forward, $ f_n(x)$ will be zero.

Of course, this convergence is not uniform, and it's not convergence in several other senses that we might like to have. The example illustrates that pointwise convergence does allow room for some funny business.

Edit: I was composing this before guile posted. I now see to my surprise that my example is [i]almost[/i] exactly the example he copied down. I guess the coincidence just shows that it's a good example. Except that I think guile probably made a sign error on the second line when he copied it down.

[quote="guile"] One more question though, when you say, a function is convergent pointwise, does that mean that it approaches a certain point - or the point it refers to is simply a vertical or horizontal line in the $ xy$ plane[/quote].

If I interpreted your terminology right, the correct way to put it is that it approaches a [b]curve[/b] (functions can be identified with their graphs). Of course, this "curve" (the graph of the limiting function $ f$) may be rather ugly one. It doesn't need to be a horizontal line. If you want something that approches the horizontal line $ y\equal{}0$, it is the graph of the difference $ f_n\minus{}f$. The approach for pointwise convergence is understood in a rather weak sense: for each vertical line (i.e., each fixed $ x$), the intersection of the graph of $ f_n$ with that line (i.e., $ (x,f_n(x))$) approaches the point $ (x,f(x))$.

[quote]
This is an exercise from a notebook of a friend but it has no solution. I hope you could help me.

$ f_{n}(x) \equal{} \begin{cases}n^{2}x; & 0\leq x\leq\frac {1}{n} \\
n^{2}(x \minus{} \frac {2}{n}); & \frac {1}{n}\leq x\leq\frac {2}{n} \\
0; & \frac {2}{n}\leq x\leq 1\end{cases}$

1.) Discuss the continuity of $ f_n(x)$. Well, I answered that already, it is not continuous at $ x \equal{} \frac {1}{2}$ but continuous at $ x \equal{} \frac {2}{n}$
[/quote]
You mean "not continuous at $ x\equal{}\frac 1n$", of course. Yes, that is correct though, as Kent noted, the function is defined at $ x\equal{}\frac 1n$ twice and the values disagree, which is really bad for a function. It may be that you miss minus sign but it may also be that you copied one strict inequality as a non-strict one. Let's assume that the latter is the case and the right $ \leq$ in $ 0\leq x\leq\frac {1}{n}$ should really be $ <$.

[quote]
2.) Sketch the graph. (Horizontal line in $ f_n(x) \equal{} n$ from 0 to 1/n, horizontal line at $ f_x(n) \equal{} \minus{} n$. Confused with the others.
[/quote]
This is completely wrong. When you have a piecewise defined function, the first step is to consider the intervals separately and draw its own graph for each interval. So, let's do it.

The first interval of interest is $ [0,\tfrac 1n)$. The formula on this interval is $ f(x)\equal{}n^2x$. Note that $ n$ is just a parameter, you need to look at the dependence on $ x$. So, this is a linear function starting at $ 0$ when $ x\equal{}0$ and ending at $ n$ when $ x\equal{}\frac 1n$. The graph is just the straight line joining these two points (with the right endpoint excluded):
[asy]size(200);
pen a=black, b=blue+linewidth(1), g=green;
D((0,-6)--(0,6)--D("0",8,(0,0),a)--(5,0),a);
D("1",8,(4,0),a);
D("\frac 1n",8,(1,0),a);
MP("n",D((0,5),a),8,W);
D((0,0)--(1,5),b);
D((0,5)--(1,5)--(1,0));[/asy]
The second interval of interest is $ [\tfrac 1n,\tfrac 2n]$. The formula on this interval is $ f(x)\equal{}n^2(x\minus{}\tfrac 2n)$. Again, this is a linear function starting at $ \minus{}n$ when $ x\equal{}\frac 1n$ and ending at $ 0$ when $ x\equal{}\frac 2n$. Again, the graph is just the straight line joining these two points:
[asy]size(200);
pen a=black, b=blue+linewidth(1), g=green;
D((0,-6)--(0,6)--D("0",8,(0,0),a)--(5,0),a);
D("1",8,(4,0),a);
D("\frac 1n",8,(1,0),a);
D("\frac 2n",8,(2,0),a,N);
MP("-n",D((0,-5),a),8,W);
D(D((1,-5),blue)--(2,0),b);
D((0,-5)--(1,-5)--(1,0));[/asy]
The last interval of interest is $ [\tfrac 2n,1]$. The formula on this interval is $ f(x)\equal{}0$. The graph is just the horisontal line going along the $ x$-axis:
[asy]size(200);
pen a=black, b=blue+linewidth(1), g=green;
D((0,-6)--(0,6)--D("0",8,(0,0),a)--(5,0),a);
D("1",8,(4,0),a);
D("\frac 2n",8,(2,0),a,N);
D((2,0)--(4,0),b);[/asy]
Now just put these three pieces together:
[asy]size(200);
pen a=black, b=blue+linewidth(1), g=green;
D((0,-6)--(0,6)--D("0",8,(0,0),a)--(5,0),a);
D("1",8,(4,0),a);
D("\frac 1n",8,(1,0),a);
D("\frac 2n",8,(2,0),a,N);
D((2,0)--(4,0),b);
MP("n",D((0,5),a),8,W);
D((0,0)--(1,5),b);
D((0,5)--(1,5)--(1,0));
MP("-n",D((0,-5),a),8,W);
D(D((1,-5),blue)--(2,0),b);
D((0,-5)--(1,-5)--(1,0));[/asy]

[quote]
3.) Prove that $ f_n$ converges to $ f$ pointwise. I have a proof but Im really having a hard time with Latex.
[/quote]
Well, I need to see your proof to tell whether it is right or wrong and comment on your errors (if there are any).


inequalities
function
algebra
polynomial
trigonometry
inequalities proposed

Prove that

$ 3\sum_{cyc}x^4\plus{}7\left(\sum_{cyc}x^2y^2\minus{}\sum_{cyc}x^2yz\right)\geq 9\left(\sum_{cyc}x^3y\minus{}\sum_{cyc}xy^3\right)$

for $ x,y,z\in\mathbb{R}$.

[quote="Parkdoosung"]Prove that

$ 3\sum_{cyc}x^4 \plus{} 7\left(\sum_{cyc}x^2y^2 \minus{} \sum_{cyc}x^2yz\right)\geq 9\left(\sum_{cyc}x^3y \minus{} \sum_{cyc}xy^3\right)$

for $ x,y,z\in\mathbb{R}$.[/quote]
Let $ x \plus{} y \plus{} z \equal{} 3u,$ $ xy \plus{} xz \plus{} yz \equal{} 3v^2,$ $ xyz \equal{} w^3$ and $ u^2 \equal{} tv^2.$
Hence, $ t\in\mathbb R$ and our inequality is equivalent to
 $ 27u^4 \minus{} 36u^2v^2 \plus{} 13v^4 \minus{} 3uw^3\geq3u(x \minus{} y)(x \minus{} z)(y \minus{} z).$
Since, $ 27u^4 \minus{} 36u^2v^2 \plus{} 13v^4 \minus{} 3uw^3\geq0$ it remains to prove our inequality
for $ u(x \minus{} y)(x \minus{} z)(y \minus{} z)\geq0.$
In this case $ 27u^4 \minus{} 36u^2v^2 \plus{} 13v^4 \minus{} 3uw^3\geq3u(x \minus{} y)(x \minus{} z)(y \minus{} z)\Leftrightarrow$
$ \Leftrightarrow252u^2w^6 \plus{} 6u(135u^4 \minus{} 207u^2v^2 \minus{} 13v^4)w^3 \plus{}$
$ \plus{} 729u^8 \minus{} 1944u^6v^2 \plus{} 1269u^4v^4 \plus{} 36u^2v^6 \plus{} 169v^8\geq0.$
Id est, it remains to prove that $ 9u^2(135u^4 \minus{} 207u^2v^2 \minus{} 13v^4)^2 \minus{}$
$ \minus{} 252u^2(729u^8 \minus{} 1944u^6v^2 \plus{} 1269u^4v^4 \plus{} 36u^2v^6 \plus{} 169v^8)\leq0,$ which is equivalent to
$ u^2(81t^4 \plus{} 54t^3 \minus{} 141t^2 \minus{} 162t \plus{} 169)\geq0,$ which is true ( $ t_{min} \equal{} 0.9564...$ ).

This inequality is a particular case of the following:

Let $ x,y,z$ be real numbers. If $ p,q,r,s$ are real numbers such that
$ p\plus{}q\minus{}r\minus{}1\le s\le 2(r\plus{}1)\plus{}p\plus{}q\minus{}p^2\minus{}pq\minus{}q^2$,
then

$ \sum x^4\plus{}r\sum x^2y^2\plus{}sxyz\sum x \ge p\sum x^3y\plus{}q\sum xy^3$.

[quote="Vasc"]This inequality is a particular case of the following:

Let $ x,y,z$ be real numbers. If $ p,q,r,s$ are real numbers such that
$ p \plus{} q \minus{} r \minus{} 1\le s\le 2(r \plus{} 1) \plus{} p \plus{} q \minus{} p^2 \minus{} pq \minus{} q^2$,
then

$ \sum x^4 \plus{} r\sum x^2y^2 \plus{} sxyz\sum x \ge p\sum x^3y \plus{} q\sum xy^3$.[/quote]

since $ LHS \minus{} RHS$ is linear in function of $ s$ it sufficies to prove the inequality in the cases  $ s \equal{} p \plus{} q \minus{} (r \plus{} 1)$ and $ s \equal{}2(r\plus{}1)\plus{} p \plus{} q \minus{} p^2 \minus{} pq \minus{} q^2$ ,Vasc do u have magical identities for this cases  :maybe: ,?? of course withe condition $ 3(r \plus{} 1) \geq p^2 \plus{} pq \plus{} q^2$
 :wink: another question : can we replace the previous condition with the stronger one : $ 4(r \plus{} 1) \geq (p \plus{} q)^2$ ?

[quote="anas"]  can we replace the previous condition with the stronger one : $ 4(r \plus{} 1) \geq (p \plus{} q)^2$ ?[/quote]
I think we can't replace this for the case $ s\equal{}p\plus{}q\minus{}r\minus{}1$, because the previous condition is necessary and sufficient to have the given inequality.

This is nice.

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

$ x^4\plus{}y^4\plus{}z^4\plus{}2xyz(x\plus{}y\plus{}z)\ge x^3y\plus{}y^3z\plus{}z^3x$.

[quote="Vasc"]This is nice.

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

$ x^4 \plus{} y^4 \plus{} z^4 \plus{} 2xyz(x \plus{} y \plus{} z)\ge x^3y \plus{} y^3z \plus{} z^3x$.[/quote]

It is equivalent to : $ \sum (3(a^2\minus{}b^2)\minus{}(ab\minus{}2bc\plus{}ca))^2\plus{}2\sum (ab\minus{}2bc\plus{}ca)^2 \geq 0$

after much effort and withe the help of my computer I obtained the following  laborious  identity for the main case : $ s\equal{}2(r\plus{}1)\minus{}p\minus{}q\minus{}p^2\minus{}pq\minus{}q^2$ and  $ 3(r\plus{}1) \equal{} p^2\plus{}q^2\plus{}pq$ :

$ 12E(a,b,c) (LHS\minus{}RHS)\equal{}3 ( H(a,b,c))^2\plus{} (  F(a,b,c))^2$  where:

\[ H(a,b,c)\equal{} (4\minus{}p\minus{}q)\sum{a^3} \plus{} (\frac{pq}{3} \minus{}2p\plus{}q \plus{}\frac{q^2}{3}\plus{}\frac{p^2}{3}\minus{}2)\sum{a^2b} \plus{} ( \frac{pq}{3}\plus{}\frac{p^2}{3} \plus{}\frac{q^2}{3}\plus{}p\minus{}2q\minus{}2)\minus{}2abc(p^2\plus{}q^2\minus{}3(p\plus{}q)\plus{}pq)\]

 \[ F(a,b,c)\equal{}  (p\minus{}q)\sum{a^3}\plus{}6abc(p\minus{}q)\plus{} (\minus{}p^2\plus{}3q\minus{}q^2\plus{}6\minus{}pq)\sum{a^2b} \plus{}(pq\plus{}p^2\plus{}q^2\minus{}6\minus{}3p)\sum{b^2a}\]

and \[ E(a,b,c)\equal{}  \frac{ ((p\minus{}q)^2\plus{}3(p\plus{}q\minus{}4)^2)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}ac\minus{}bc)}{12}\]

and for the other maincase, I think Vasc has already gave stefan_markelov type identities in an other topic  :)

[quote="anas"][quote="Materazzi"][quote="Vasc"]This is nice.

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

$ x^4 \plus{} y^4 \plus{} z^4 \plus{} 2xyz(x \plus{} y \plus{} z)\ge x^3y \plus{} y^3z \plus{} z^3x$.[/quote]

It is equivalent to : $ \sum (3(a^2 \minus{} b^2) \minus{} (ab \minus{} 2bc \plus{} ca))^2 \plus{} 2\sum (ab \minus{} 2bc \plus{} ca)^2 \geq 0$[/quote]

I think it's wrong because you'il get smthg + $ k\sum{(ab)^2}$  :wink:[/quote]

I use :

$ \sum a^4 \minus{}\sum a^2b^2 \equal{}\frac{1}{2} \sum (a^2\minus{}b^2)^2$

$ \sum a^2b^2 \minus{}\sum a^2bc\equal{}\frac{1}{6} \sum (ab\minus{}2bc\plus{}ca)^2$

$ \sum a^3b \minus{}\sum a^2bc \equal{} \frac{1}{3} \sum (a^2\minus{}b^2)(ab\minus{}2bc\plus{}ca)$

But i have some mistake  :blush:

[quote="Materazzi"][quote="Vasc"]This is nice.

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

$ x^4 \plus{} y^4 \plus{} z^4 \plus{} 2xyz(x \plus{} y \plus{} z)\ge x^3y \plus{} y^3z \plus{} z^3x$.[/quote]

It is equivalent to : $ \sum (3(a^2 \minus{} b^2) \minus{} (ab \minus{} 2bc \plus{} ca))^2 \plus{} 2\sum (ab \minus{} 2bc \plus{} ca)^2 \geq 0$[/quote]
Actually, the inequality is equivalent to

$ \sum (2x^2\minus{}y^2\minus{}z^2\minus{}xy\plus{}yz)^2\plus{}4(\sum xy)^2\ge 0$,

with equality for $ x\equal{}\frac y{\alpha}\equal{}\frac z{\beta}$ or any cyclic permutations, where $ \alpha\approx 0.8020$ and $ \beta\approx \minus{}0.4451$.

[quote="Vasc"]This is nice.

If $ a,b,c$ are real numbers, then

$ a^4 \plus{}b^4 \plus{} c^4 \plus{} 2abc(a \plus{} b\plus{} c)\ge a^3b \plus{} b^3c \plus{} c^3a$.[/quote]
Actually, the following identity holds
$ \sum a^4\plus{}2\sum a^2bc \minus{}\sum a^3b\equal{}\frac{1}{2}\sum  (a^2\minus{}b^2\plus{}bc)^2\plus{}\frac{1}{2}\left( \sum ab\right)^2.$

[quote="Vasc"]

$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 2abc(a \plus{} b \plus{} c)\ge a^3b \plus{} b^3c \plus{} c^3a$.

Actually, the inequality is equivalent to

$ \sum (2x^2 \minus{} y^2 \minus{} z^2 \minus{} xy \plus{} yz)^2 \plus{} 4(\sum xy)^2\ge 0$,

with equality for $ x \equal{} \frac y{\alpha} \equal{} \frac z{\beta}$ or any cyclic permutations, where $ \alpha\approx 0.8020$ and $ \beta\approx \minus{} 0.4451$.[/quote]

Yes, the equality case yields the system

$ x ^2 \minus{} y ^2 \equal{} \minus{} yz$
$ y^2 \minus{} z^2 \equal{} \minus{} zx$
$ z^2 \minus{} x^2 \equal{} \minus{} xy$
$ \Longleftrightarrow$
$ (x,y,z) \equal{} (\lambda\sin\frac {8\pi}7,\lambda\sin\frac {4\pi}7,\lambda\sin\frac {2\pi}7)$ or any cyclic permutation, $ \lambda\in\mathbb R$. ;)

[quote="can_hang2007"][quote="Parkdoosung"]Prove that

$ 3\sum_{cyc}a^4 \plus{} 7\left(\sum_{cyc}a^2b^2 \minus{} \sum_{cyc}a^2bc\right)\geq 9\left(\sum_{cyc}a^3b \minus{} \sum_{cyc}ab^3\right)$

for $ a,b,c\in\mathbb{R}$.[/quote]
The following identity holds, too:
$ LHS \minus{} RHS \equal{} \frac {1}{6}\sum \left[ 3a^2 \minus{} 3b^2 \minus{} \left( 6 \plus{} \sqrt {2}\right) ab \plus{} \left( 3 \minus{} \sqrt {2}\right) bc \plus{} \left( 3 \minus{} \sqrt {2}\right) ca\right]^2.$[/quote]
And from this it is easy to see that there is no nonzero equality case here:
$ a^2 \minus{} b^2 \minus{} 3ab \equal{} 0$ and cyclics $ \Longrightarrow a \equal{} b \equal{} c \equal{} 0$,

as expected by Parkdoosung's sharper result.


linear algebra
matrix
vector
parameterization

Suppose A is a real symmetric 3 × 3 matrix such that
• trace(A) = 0
• R(LA) = span{(111), (10-1)} -- these are column vectors, i just written like this for ease
A(111) = (210) again both column vectors

find A.

stumped me: i got 5 linear eqns but 9 unknowns, perhaps i'm approaching it wrong?

A real symmetric $ 3 \times 3$ matrix has six free parameters, not nine.

yeah i know that - since A = A transpose, but what does the range of the linear transformation say specifically about the entries of a?

It doesn't say anything specifically about any specific entry, but it constrains what the entries, together, have to look like.  I'm not sure I understand what $ R(LA)$ means, though; is that rowspace?

i assume it meant the range of the linear transformation of A


calculus
integration
geometry
perimeter
trigonometry
complex analysis

Indeed, the answer must be a positive real number. I don't have time right now to find your mistake, but it's there, somewhere.

I found the mistake. It was in the last line, where I should get $ \cos{\frac{3\pi}{8}}$ and not $ i\sin{\frac{3\pi}{8}}$ as previously writen.


probability
percent
conditional probability

err i doubt this is right but

[hide]

since ur looking at percents, lets just say there are 10 members not in a union, 30 in union A, and 60 in union B

since we know 1 of them is not in a union, we have a remainder of 99 people left to look at

probability that the person came from union A is $\frac{30}{99}$, probability another person came from union A is $\frac{29}{98}$, and etc, along with probability that the person at end came from union B is $\frac{60}{96}$, and there are $\binom{4}{1}$ ways of picking that last person

then for 2nd case, just all 4 came from union A, so final answer is

$\displaystyle\frac{(4)(30)(29)(28)(60) + (30)(29)(28)(27)}{(99)(98)(97)(96)} \approx .07199$

[/hide]

[quote="peeta"]i think it is important how many workers there are all together, since i calulated $\frac1{35}$ for the worker numbers 3, 6, 1 and $\frac1{12}$ for 6, 12, 2, but i could have miscalculated ...[/quote]

well , I also think that numbers of worker are important but the question doesnt provided so I thought maybe even without the exact number of worker , the probability should be still the same . But I do it in this way

Since there are two possibility for worker $A$ to be more than $B$ , namely in $3: 1$ and $4: 0$ . For $3: 1$ , the probability is $4C3\times (0.3)^3\times 0.6\times 0.1$ . For $4: 0$ , it is $4C4\times (0.3)^4\times 0.1)$ .  Since this is conditional probability , we divide it by $5C4\times (0.9)^4\times (0.1)$ (choosing 4 workers from union group and 1 worker from non-union) . So after some simplifying i get $\frac{1}{45}$ .

So , is there any flaw there ?

well when u say like $\binom{4}{3} \times (0.3)^3 \times 0.6 \times 0.1$, don't you have to take into consideration that when you choose a member from A, there will now be less people left to choose from from A, and also less people in total? so your next probability would no longer be $0.3$

Well , I just treat it as Binomial distribution and I think that taking away one ot two person from a large amount of people somehow wont affect too much the $0.3$ probability since logically there wont be only less than 50 workers in a factory . Any better suggestion there?

[quote="shyong"][quote="varun"][quote="shyong"]30% of workers in a factory belong to union A, 60% to union B, 10% no union . A sample of 5 workers is selected and it is found that exactly 4 of these workers are members of unions, what is the probability that more of the members come from A than B ?[/quote]

I am sending the solution in the file attached.
My answer is $3451/170346$[/quote]

varun ,

as for your solution , I dont think we can assume there are $100$ workers in total .

I discuss with some of my friends and we come to a conclusion to do it this way

$\frac{P(A=4,B=0,C=1)+P(A=3,B=1,C=1)}{P(4 union,1 non-union)}$

=$\frac{\frac{(0.3)^4\times 0.1}{4!}+\frac{(0.3)^3\times 0.6\times 0.1}{3!}}{\frac{(0.9)^4\times 0.1}{4!}}$

=$\frac{1}{9}$[/quote]

Hoe have you taken
$\frac{\frac{(0.3)^4\times 0.1}{4!}+\frac{(0.3)^3\times 0.6\times 0.1}{3!}}{\frac{(0.9)^4\times 0.1}{4!}}$
I have the doubt that from where 4! and  3! came in the denominatotr.
Also why can't we chose that total workers is 100.

[quote="Lialaps"]if we assume that the total worker is 100*k,30*k in the union A and 60*k in the union Band we know that 5 people we chose are in ınion A or B so
there will be two chance taht 4A 1B or 3A 2B or 5A 0B
(30*k*29*k*28*27*k*26*k)/(90*k*89*k*88*k*87*k*86*k)+(30*k*29*k*28*k*27*k*60*k)/(90*k*89*k*88*k*87*k*86*k)+( 30*k*29*k*28*k*60*k*59*k) that there will be a result such as 5862/1515492 that is nearly 38/10000 i think[/quote]

Not $5$ people from union but instead it is $4$ . Another thing is , if you do it that way , then it should be $30k\times 30k-1\times ...$ instead of $30k\times 29k....$

as for varun , I treat it as rearranging $AAAAC$ and $AAABC$ which is why there is $4!$ and $3!$ there  :)

oh iam sorry i all mxed up the question and i thinki shuld recheck my resul because it is not the only mistake i did


The official website is

http://math.scu.edu/putnam/

The information on that site is quite sparse, and it is not updated in a timely fashion. Problems and solutions can be found in several other places - try looking for links on the AMC website.

Edit: the specific link in the AMC site is

http://www.unl.edu/amc/a-activities/a7-problems/putnamindex.shtml

if I understand right then only student from America can go to putnam right?

Yes, it's officially open only to students of American and Canadian colleges and universities.

But note that careful phrasing in JBL's post: nothing said about citizenship or residency. The only issue is attendance, as an undergraduate, at a U.S. or Canadian college or university. International students at those universities are eligible and have often figured prominently in the results.

Hey,

Though keep in mind that the putnam is a very rigorous exam.

Thanks,

-APSStudent


Given triangle ABC with G is the centroid.
Prove that $P_{A/(GBC)}=P_{B/(GCA)}=P_{C/(GAB)}$

I can not understand what the conclusion means, can you explain it clearly?

Oh I'm sorry. I'll explain
$(ABC)$ is the circumcircle of the triangle $ABC$
and $P_{M/(ABC)}$ is the power of M to the circle $(ABC)$


Let $ a,b,c$ be three positive real numbers. Prove that
$ \frac {1}{|a^2 \minus{} b^2|} \plus{} \frac {1}{|b^2 \minus{} c^2|} \plus{} \frac {1}{|c^2 \minus{} a^2|}\plus{} \frac{8}{a^2 \plus{}b^2 \plus{} c^2} \ge \frac {28}{(a \plus{} b \plus{} c)^2}$

[hide]It hasn't equality case. So has it easy solutuon?[/hide]

Yes. It is easy ;) because we have $ f(a,b,c)\ge f(a,b,0)$. It is easy to prove ;)


Let $ABC$ be an acute angled triangle and $CD$ be the altitude through $C$. If $AB = 8$ and $CD = 6$, find the distance between the midpoints of $AD$ and $BC$.

[hide]
Let $M$ be the midpoint of $AD$ and let $N$ be the midpoint of $BC$.
Draw a line through $M$ that is parallel to $CD$ (which means perpendicular to $AB$) and let $P$ be the intersection of the line and $AC$.
Since $MP\parallel DC$ and $AM=MD$, we see that $MP=\frac{1}{2}(DC)=3$ and $AP=PC$.
Then, since $AP=PC$ and $BN=NC$, $PN=\frac{1}{2}(AB)=4$.
It is easy to observe that $\angle MPN=90^\circ$ using $AD\parallel PN$ and $DC\parallel MP$.
Finally, using Pythagorean theorem to $\triangle MPN$, we obtain $MN=\boxed{5}$
[/hide]

Let $X,Y,Z$ be mid-points of $BC,AD,AC$ respectively $\implies XZ||AB \text{  and  } XZ=\frac{1}{2}AB=4$
By further angle chasing, it is evident that $CD \perp XZ$ and since, $\Delta CZD$ and $\Delta CDX$ are isosceles $\implies CZDX $ is a kite,
Therefore, Let $CD \cap XZ =O \implies CO=DO=\frac{1}{2}CD=6 $ and we also know, $\Delta DXB$ is isosceles
Therefore, if $XH$ is altitude of $\Delta XDB$ ,such, $H \in AB \implies DO=XH=3$ .........Let $OX=x \implies OZ=4-x$
By midpoint theorem, $$OZ=YD=4-x \text{ and since, XODH is rectangle, } \implies OX=DH=4-x \implies YH=4$$
Therefore, $$XY=\sqrt{XH^2+YH^2}=\boxed{5}$$

An easier solution.

Let $E$ and $F$ be the midpoints of $AD$ and $BC$ respectively
we have to find $EF$
Draw $FG$ $\perp$ $AB$ with $G$ on $AB$.
$FG = \frac{1}{2} CD =3$
Observe that $AD + BD=8
\implies ED+DG=4 => EG = 4$
In right angled  $\Delta EFG$,
$EF =\sqrt{3^2 + 4^2}= 5$

Let F be the midpoint of BC, let G be the midpoint of BD, let E be the midpoint of AD.

Notice that $EG = ED + GD = \frac{1}{2}(AD + BD) = 4$.

Also, $FG = \frac{1}{2}CD = 3$ by the Midpoint Theorem.

Since $\triangle{FGB} \sim \triangle{CDB}$, $\angle{FGB} = \angle{FGD} = 90^{\circ}$. Hence $\triangle{FGE}$ is a right triangle.
$$\implies EF = \sqrt{EG^2 + FG^2 } = \boxed{5} ~ \text{by the Pythagorean Theorem}$$


integration
function
inequalities
calculus
real analysis
real analysis unsolved

Let $f: [0,1]\mapsto\mathbb{R}$ such that exists f '' in [0,1]. Suppose $f(0)=0, f(1)=1, f'(0)=f'(1)=0$.
Prove that exists $\psi\in [0,1]$ such that $|f''(\psi)|>4$

If $f(\frac12)\geqslant\frac12$ we have
\[\frac12\leqslant f(\frac12)=\int_{0}^{\frac12}f'(x)\,dx=\int_{0}^{\frac12}\int_{0}^{x}f''(t)\,dt\,dx=\int_{0}^{\frac12}(\frac12-t)f''(t)\,dt\leqslant\max_{t\in[0;\frac12]}f''(t)\cdot\int_{0}^{\frac12}(\frac12-t)\,dt=\frac18\max_{t\in[0;\frac12]}f''(t)\]
Hence $\max_{t\in[0;\frac12]}f''(t)\geqslant4$
If $f(\frac12)<\frac12$ consider function $f_{1}(x)=1-f(1-x)$

Thank you for your answer but i cannot integrate... There must be a more elementary solution

I'll do it with Taylor's Theorem.

$f\left(\frac12\right)=0+0\cdot\frac12+\frac{f''(\xi_{1})}2\left(\frac12\right)^{2}$ for some $0<\xi_{1}<\frac12.$

$f\left(\frac12\right)=1+0\cdot\left(\frac12-1\right)+\frac{f''(\xi_{2})}2\left(\frac12-1\right)^{2}$ for some $\frac12<\xi_{2}<1.$

$f\left(\frac12\right)=\frac18f''(\xi_{1})=1+\frac18f''(\xi_{2})$

$f''(\xi_{1})-f''(\xi_{2})=8.$

This requires that either $f''(\xi_{1})\ge 4$ or $f''(\xi_{2})\le-4.$

Furthermore, the only way to have equality in those inequalities would be to have the function

$f(x)=\begin{cases}2x^{2},&0\le x\le \frac12\\ 1-2(1-x)^{2},&\frac12<x\le 1\end{cases}.$

This would violate the hypothesis that $f''$ exists everywhere in the interval; hence we must have strict inequality.

Here's right one.
Let's suppose that $\sup_{x\in[0;1]}|f''(x)|\leqslant4$
Consider function $g(x)=2x^{2}-f(x)$. Since $g''(x)=4-f''(x)\geqslant0$    $g'(x)$ is nondecreasing on $[0;\frac12]$, but $g'(0)=0$, hence $g'(x)\geqslant0$. That's why $g(x)$ is nondecr. on $[0;\frac12]$, thus $g(\frac12)=\frac12-f(\frac12)\geqslant g(0)=0$, i.e. $f(\frac12)\leqslant\frac12$ and if $f(\frac12)=\frac12$ then $g''(x)\equiv0$ on $[0;\frac12]$, i.e. $f''\equiv4$.
Doing the same with $f_{1}(x)=1-f(1-x)$ gives that $f(\frac12)\geqslant\frac12$ and if  $f(\frac12)=\frac12$ then $f''\equiv-4$ on $[\frac12;1]$.

Sorry RIP I cant understand your proof!
You didnt arrive to any contradiction by supposing sup f ''(x)<=4. You just showed that f(1/2)<=1/2

Kent Merryfield: I dont understand the last part of your proof. Why if f(e1)=4 and f(e2)=-4, f must be that function?

I found this problem in Calculus of M Spivak. This book has answer book I can't find on the Internet.

Thank you.

I proved that $f(\frac12)\leqslant\frac12$ and $f(\frac12)\geqslant\frac12$, hence $f(\frac12)=\frac12$, that's why
\[f''(x)=\begin{cases}4,&x\in[0;\frac12]\\-4,&x\in[\frac12;1] \end{cases}\]
Particularly $f''(\frac12)=4=-4$. Nonsense!

I cross-posted a slight generalization of this problem in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123255]physics forum[/url].

thanks RIP, I found that your answer helped me a lot. I have a question though, is g(x) = 1 - f(1-x) supposed to behave like or be equivalent to our main function f(x)?

thanks


geometry parallelogram symmetry circumcircle

Let $ C_1$ and $ C_2$ be two congruent circles centered at $ O_1$ and $ O_2$, which intersect at $ A$ and $ B$. Take a point $ P$ on the arc $ AB$ of $ C_2$ which is contained in $ C_1$. $ AP$ meets $ C_1$ at $ C$, $ CB$ meets $ C_2$ at $ D$ and the bisector of $ \angle CAD$ intersects $ C_1$ and $ C_2$ at $ E$ and $ L$, respectively. Let $ F$ be the symmetric point of $ D$ with respect to the midpoint of $ PE$. Prove that there exists a point $ X$ satisfying $ \angle XFL \equal{} \angle XDC \equal{} 30^\circ$ and $ CX \equal{} O_1O_2$.
[i]
Author: Arnoldo Aguilar (El Salvador)[/i]

let $ \alpha=\angle PAB$ and $ \theta=\frac{1}{2}\angle AO_1B$...

note that $ ADC$ is isosceles, so $ AL$ is perpendicular to $ CD$... also note that $ \angle EDC=\angle ECB=\angle EAB=90-\alpha-\theta$, and since $ \angle ABD=\alpha+\theta$ it follows that $ DE$ is perpendicular to $ AB$, or $ DE$ is parallel to $ O{}_1O{}_2$...
this implies that $ DEO{}_1O{}_2$ is a parallelogram, therefore, $ FPO{}_2{}_O{}_1$ is a parallelogram too, and this implies that $ F$ lies in $ C_1$...

note that symmetry arguments imply that $ \angle PFB=\angle PAB=\alpha$, son this implies that $ AC$ is perpendicular to $ BF$...
since $ \angle CAL+\angle ALB=(90-\theta)+\theta=90$ it follows that $ AC$ is also perpendicular to $ BL$, which implies that $ F,B,L$ are collinear...

note that $ \angle PFB=\angle CFB=\alpha$ and $ AC$ is perpendicular to $ BF$, therefore $ FP=FC$... also since $ AFL$ is isosceles we have that $ FP=PL$, so $ FPLC$ is a kite (rhombus)... this implies that $ CL=CF$... let $ l$ be a line through $ F$ such that $ l$ and $ FL$ form an angle of $ 30$, and let $ X$ be another point on $ l$ such that $ CF=CX$... (in at least one of the two possible lines $ l$ $ X$ and $ F$ are different)... then, $ C$ is the circumcenter of triangle $ FLX$ and since $ \angle XFL=30$ we have that $ \angle XCL=60$, which implies that $ XLC$ is equilateral... we can prove that this point $ X$ satisfies the other two conditions as well ($ CL=CF=O{}_1O{}_2$ and the other one is similar), and we're done  :D

my solution is like campos's above
the key point lies on the proof of PF=DE=O1O2, F,L are symmetry w.r.t AC =>  FC=CL=LP=PF=ED=EC=r
=> XDC is an equilateral triangle => see my figure
X is the intersection of the two red cirlces with the same radius r and centered at C,L respectively

can someone please tell me what equal circles are ?? equal radius?

[asy] /* File unicodetex not found. */

 /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(20.7cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -11.77, xmax = 8.93, ymin = -5.69, ymax = 10.98;  /* image dimensions */

 /* draw figures */
draw(circle((-4.15,4.53), 4.89));
draw(circle((4.49,5.01), 4.9));
draw((0.04,7.05)--(-2.27,0.02));
draw((7.16,9.11)--(-2.27,0.02));
draw((0.04,7.05)--(7.16,9.11));
draw((-1.48,8.63)--(6.37,0.49));
draw((-1.48,8.63)--(-0.31,6));
 /* dots and labels */
dot((-4.15,4.53),dotstyle);
label("$O_1$", (-4.02,4.73), NE * labelscalefactor);
dot((0.29,2.49),dotstyle);
label("$B$", (0.44,2.7), NE * labelscalefactor);
dot((4.49,5.01),dotstyle);
label("$O_2$", (4.61,5.22), NE * labelscalefactor);
dot((0.04,7.05),dotstyle);
label("$A$", (0.19,7.26), NE * labelscalefactor);
dot((-0.31,6),dotstyle);
label("$P$", (-0.16,6.21), NE * labelscalefactor);
dot((-2.27,0.02),dotstyle);
label("$C$", (-2.12,0.24), NE * labelscalefactor);
dot((7.16,9.11),dotstyle);
label("$D$", (7.32,9.33), NE * labelscalefactor);
dot((-1.48,8.63),dotstyle);
label("$E$", (-1.35,8.84), NE * labelscalefactor);
dot((6.37,0.49),dotstyle);
label("$L$", (6.51,0.7), NE * labelscalefactor);
dot((-0.9,7.31),dotstyle);
label("$M$", (-0.76,7.54), NE * labelscalefactor);
dot((-8.95,5.51),dotstyle);
label("$F$", (-8.83,5.71), NE * labelscalefactor);
dot((-0.83,6.45),dotstyle);
label("$X$", (-0.69,6.66), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */ [/asy]

Let $ C_1$ and $ C_2$ be two congruent circles centered at $ O_1$ and $ O_2$, which intersect at $ A$ and $ B$. Take a point $ P$ on the arc $ AB$ of $ C_2$ which is contained in $ C_1$. $ AP$ meets $ C_1$ at $ C$, $ CB$ meets $ C_2$ at $ D$ and the bisector of $ \angle CAD$ intersects $ C_1$ and $ C_2$ at $ E$ and $ L$, respectively. Let $ F$ be the symmetric point of $ D$ with respect to the midpoint of $ PE$. Prove that there exists a point $ X$ satisfying $ \angle XFL = \angle XDC =30^\circ$ and $ CX =O_1O_2$.

[i]Author: Arnoldo Aguilar (El Salvador)[/i]


Prove that the sum of two odd squares cannot be a square

I tried to write out (2k+1)2 + (2m+1)2 but it didnt help me.
Anyone who can give me a hint?

Hi Blang!

The problem is that I have not yet studied number theory and have not read about congruences modulos m...

I am just reading a number theory book from 1914 on my own. And I am still in the first chapter (last part) where the problem is given. I see that the book introduces congruences modulos m in chapter 3.

The name of the book is The Theory Of Numbers by Robert D. Carmichael.

Doing it without modulo (In fact its the same):

Assume there is some $a$:
$a^2=(2m+1)^2+(2n+1)^2 = 4(m^2+n^2+m+n)+2 \implies a^2$ is divisible by $2$
But when $a^2$ is divisible by $2$, then also $a$, so you can write $a=2b \implies 4(m^2+n^2+m+n)+2=a^2=4b^2 \implies 4(m^2+n^2+m+n-b^2)=2$ which would give that $2$ is divisible by $4$, contradiction


graph theory
combinatorics unsolved
combinatorics

Let n be the natural number and n>2.K_n denotes the complete graph with n vertices .Prove that we can paint the edges of the K_n with n color such that for

any 3 distinct color there exists a triangle such that the color of edges are this 3 color if and only if n is odd.

- Assume that such a coloration is possible.
There are $ \binom{n}{3}$ triangles and $ \binom{n}{3}$ triplets of colours. Therefore each triplet of colour appears exactly once.
Now, any red edge appears in exactly $ n \minus{} 2$ triangles. Since there are exactly $ \frac {(n \minus{} 1)(n \minus{} 2)}{2}$ triangle with (exactly) one red edge, it follows that there are $ \frac {n \minus{} 1}2$ red edges. Thus $ n \minus{} 1$ is even, so that $ n$ is odd.

- Let $ n\geq3$ be an odd integer and let's consider the vertices $ A_1A_2 \cdots A_n$ of a regular convex $ n$-gon.
For $ i \equal{} 1, \cdots n$, use color $ c_i$ to color each edge corresponding to a segment having $ A_i$ on its perpendicular bissector. Thus all the edges with color $ c_i$ are pairwise parallel.
Moreover, any triangle with vertices as endpoints is associated to exactly one triplet $ \{A_i,A_j,A_k \}$ of distinct vertices such that each of these three points belongs to the perpendicular bissector of one of the sides of the given triangle. It is easy to verify that the association is injective. This means that the sides of the given triangle have respective colours $ c_i,c_j,c_k$, and that two distinct triangles determines two distinct triplets of colours. Thus, any two triangles have different triplets of colours. Since there are as many triangles than triplets of colours, we deduce that each triplets of colours apprears exactly once, and we are done.

Pierre.


function
induction
pigeonhole principle
algebra proposed
algebra

Hi,
I need help with the following problem ASAP

Let's consider the sequence a_1, a_2, ..., a_m ("m" is positive integer) of non-zero integers such that the sum of each "p" successive members is positive and the sum of each "n" members is negative. What is the maximal possible value of "m" (it depends on "p" and "n")?

This problem si the generalization of IMO 1977 (there was demanded only the concrete case for p = 11, n = 7). I'm able to prove that for max. m it must be held that:
m  \leq  p + n - gcd(p,n) - 1
I think that really
max. m = p + n - gcd(p,n) - 1
but I'm not able to construct any sequence of length p + n - gcd(p,n) - 1 that would satisfy all the conditions (in general case). Does anybody know how to do it (construct this sequence).

If you have any suggestion (or solution), answer ASAP, please.

Hi,

This interesting question deserves attention. I think first you could post your proof of the fact  m <= p + n - gcd(p,n) - 1.

In addition, I would like note that the original problem (p=11, n=7) was suggested by Vietnam. IMO 1977 was held in Jugoslavia, but in that year, due financing difficulty did not take part.

Namdung

We'll prove it by contradiction. So let's consider that max. m = p+n-gcd(p,n) (for contradiction):

We may assume that the sum of any n>0 consecutive terms is strictly negative.  Let S(k) be the sum of the first k numbers in our sequence.  Start by noting that S(k+p) > S(k) and S(k-n) > S(k) for convenient k (so that k + p  \leq  p + n - gcd(p,n) and k - n > 0).

Now we define function f:
1. f(1) = p+n-gcd(p,n)
2. for k>1
- if f(k-1)  \leq  n-gcd(p,n), then f(k) = f(k-1) + p
- if f(k-1)  \geq n, then f(k) = f(k-1) - n

At this point we should show that f(k-1) never satisfies
n - gcd(p,n) < f(k-1) < n
But this is very easy to prove by induction. We can aslo prove (by induction) that f(k-1) must be a multiple of gcd(p,n).

Now notice that 0  \leq  f(k)  \leq p+n-gcd(p,n), thus by the Pigeonhole Principle if we consider f(1), f(2), ..., f(p+n) some two f's must take on the same value.  But we get a contradiction from this by considering the strictly increasing chain
S(f(1)) > S(f(2)) > ... > S(f(p+n)).


[quote="mna851"]What is the smallest positive integer than can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?

I like this problem and I came up with a solution different from the one in the official solutions pamphlet, so I figured I'll see what you guys came up with.[/quote]

You have 3 sequences of numbers with their respective sums:

$a,a+1,...a+8$ This becomes $\frac{9}{2}(2a+8) = 9(a+4)$

$b,b+1,...b+9$ This becomes $\frac{10}{2}(2b+9) = 5(2b+9)$

$c,c+1,...c+10$ This becomes $\frac{11}{2}(2c+10) = 11(c+5)$

From this, we see the answer must be AT LEAST divisible by 9,5, and 11, so we take only these three factors and get $9(5)(11) = \fbox{495}$.

Was this your solution? If not, what was yours?

[quote="mna851"]What is the smallest positive integer than can be expressed as the sum of nine consecutive integers, the sum of ten consecutive integers, and the sum of eleven consecutive integers?

I like this problem and I came up with a solution different from the one in the official solutions pamphlet, so I figured I'll see what you guys came up with.[/quote]

[hide]  A number can be expressed as the sum of 9 and 11 ( or any odd number for that matter ) consecutive integers if and only if it is a multiple of both 9 and 11.  ( This is because we can write our sequence as: x-n, x-(n-1), x-(n-2)....x, x+1, x+2....x+n-2, x+n-1, x+n, where all the secondary terms cancel each other out and we are left with (2n+1)x.  Hence, our number ( say N ) is congruent to 0 mod 99.  If we try to express our 10 consecutive integers in the same manner we are left with an extra 5 at the end, and our sequence sums to 10x+5.  Thus,

N= 0 (mod 99)

N= 5 (mod 10)

  Applying CRT we can combine this congruence into one:

N= 495 (mod 990)

  Hence, the smallest positive integer value is $\boxed{495}$

  [/hide]

$\frac{n(n+1)}{2} - \frac{(n-9)(n-8)}{2} = X$
$\frac{m(m+1)}{2} - \frac{(m-10)(m-9)}{2} = X$
$\frac{l(l+1)}{2} - \frac{(l-11)(l-10)}{2} = X$

$\Rightarrow 9n-36=X$
$\Rightarrow 10m-45=X$
$\Rightarrow 11m-55=X$
$\Rightarrow X$ has factors $9,5,11.$
Minimum value of $X$ can be $495$
And for this value we have a solution in $n,m,l$.
But there is no solution for $m$ if the number is even multiple of 495.

Hence the answer should be odd multiples of $495$.

Hope it is correct. And matches your answer. hihi

Those answers are all correct.  I did it differently but its uglier so I won't post it unless anyone actually wants me to.

[hide]Well let the desired integer be $k$. Let $a_i$ be the first term in the sum of $i$ consecutive integers, so \[ k=a_9+(a_9+1)+\cdots+(a_9+8)=\frac{9}{2}(2a_9+8)=9(a_9+4), \]\[ k=a_{10}+(a_{10}+1)+\cdots+(a_{10}+9)=5(2a_{10}+9), \] \[ k=a_{11}+(a_{11}+1)+\cdots+(a_{11}+10)=\frac{11}{2}(2a_9+10)=11(a_{11}+5). \]

Hence, $k$ must be divisible by 9, 5, and 11. The least common multiple of these numbers is $\boxed {495}$.[/hide]

[hide="Answer"]$9a+36=10b+45=11c+55\Rightarrow 9(a+4)=5(2b+9)=11(c+5)$
Obviously, the number must be a multiple of at least $5\cdot 9\cdot 11=495$. Checking, $2b+9$ can equal $99$ with $b$ an integer, and it can obviously be expressed by the other two expressions as well, so our answer is $\boxed{495}$.[/hide]


Find the number of zeros at the end of the expansion of $1!2!3!\cdots 49!$.

[hide]We will count factors of 5 because 10 is 5*2 and the 5's are obviously less prevalent.

Every number beyond 4 will have one. So there are 49-4 = 45 fives.

Every number that is beyond 9 will have another 49 - 9 = 40 fives.

Every number that is beyond 14 will have another 49-14 = 35 fives

Every number beyond 19 has another 49 - 19 = 30 fives

Every number beyond 24 has 2 fives because 25 is 5*5  2(49-24) = 50 fives

beyond 29      20

beyond 34      15

beyond 39      10

beyond 44      5

45 + 40 + 35 + 30 + 50 + 20+ 15 + 10 + 5 = $\boxed{250}$

For larger factorials you would use sum formulas but it is easier just to crank these out by hand.

[/hide]

oops added in too many 25's. fixed now

To find the number of zeros $n!$ ends in, we must count the number of times a factor of five appears in $n!$.

For $n=1$ to $4$, there are no zeros.
For $n=5$ to $9$,  there is $1$ zero for each $n$.
For $n=10$ to $14$, there are $2$ zeros for each $n$.
For $n=15$ to $19$, there are $3$ zeros for each $n$.
For $n=20$ to $24$, there are $4$ zeros for each $n$.
For $n=25$ to $29$, there are $6$ zeros  for each $n$ ($25$ has two factors of $5$).
For $n=30$ to $34$, there are $7$ zeros for each $n$.
For $n=35$ to $39$, there are $8$ zeros for each $n$.
For $n=40$ to $44$, there are $9$ zeros for each $n$.
For $n=45$ to $49$, there are $10$ zeros for each $n$.

The total number of zeros is then $5\cdot \left( \frac{10\cdot11}{2}-5 \right)=\boxed{250}$

[quote="arkmastermind"][quote="chess64"]Just a note, 250 is indeed the correct answer, I have done it with a computer:

[code]<script language="javascript">

function factorialZeros(n) {
var result = Math.floor(n/5) + Math.floor(n/25);
return result;
}

var zeros = 0

for (var i = 2; i <= 49; i++) {

var zeros = zeros + factorialZeros(i)

}

document.write(zeros);
</script>[/code]

:D[/quote]

or you could do it shorter,
[code]<script language="javascript">
var total = 0;
for (var i=1; i<=49; i++) {
total+=Math.floor(i/5)+Math.floor(i/25);
}
document.write(total);
</script>[/code]
 :lol:[/quote]

pwned

lol, i wasnt intending to put his method down, i was just pointing out the code could have been shorter..

well, it's been years since i really programmed in Javascript, so...

but yeah, that's definitely more efficient. :)


Hi!

Has anyone used this site: http://www.mathservice.co.uk ?

Is it reliable?

Thanks for letting me know!

seems, too long uploable?
among enough universal and fast matrix calculators known

http://www.compute.uwlax.edu/index.php
http://www.bluebit.gr/matrix-calculator/default.aspx

[quote="ksiega1"]Yeah, but the interface is better I think. And you can save the results easily in a file.

Also, I guess much bigger matrices can be calculated.

Regards.[/quote]

I've not seen interface, by i'm too impatient to Wait its uploading. Much bigger... mmmm. How big could be enough? 1000.. 10000.. have you time to input all elements?

Wish success, regards


geometry
3D geometry
algebra
polynomial
Vieta
algebra proposed

the sum of 3 non-zero numbers is $\alpha$.
the sum of their squares is $\alpha^{2}$
the sum of their cubes is $\alpha^{3}$.
the sum of their fourth power is $\alpha^{4}$.

Proove that this system don't have solutions.

Note: Remembering, the numbers are different of zero.

woow.. what a nice solutions

RIP, do you know that if we can expand this problem for high dimensions?

like

${1\leq m\leq n}$
${S_{m}=\sum_{i=1}^{n}z_{i}^{m}=\alpha^{m}}$

[b]yagaron[/b] read this: http://mathworld.wolfram.com/Newton-GirardFormulas.html
The generalization is obvious.

:(  what a bad guy

i will rest in peace so

Let $f(X)$ be a monic cubic polynomial whose roots are the non-zero numbers satisfying the statement. Let the roots be $x,y,z.$ By Vieta's Formula, we have $f(X) = X^{3}-aX^{2}+c_{1}X+c_{0}$ for some constants $c_{0}, c_{1}.$ Now, $x^{2}+y^{2}+z^{2}= a^{2}, x+y+z = a \Leftrightarrow xy+yz+zx = 0.$ So, we then have $f(X) = X^{3}-aX^{2}+c_{0}.$ By Newton's Sums, if we denote $s_{n}= x^{n}+y^{n}+z^{n},$ $s_{3}-a \cdot s_{2}+0 \cdot s_{1}+3c_{0}= 0 \Leftrightarrow c_{0}= 0.$ Thus, our polynomial is $f(X) = X^{3}-aX^{2}.$ The roots of this equation are $0, a,$ where $0$ has a multiplicity of $2.$ But this is a contradiction, as we need non-zero numbers. $\blacksquare$


We have the small hands of the clock in the same position at 12:00 o'clock, fin the exact time in which both are in the same position after the 1:00 o'clock

If I understand you correctly, you're asking the time $ 1: \sim $ at which the hour and minute hands are together?

If so, at $ 1: 00$, the hands are $ 30^\circ$ apart. Each minute, the minute hand moves $ 5.5^\circ$ closer to the hour hand, so it will take $ \frac{30^\circ}{5.5^\circ}\equal{}5\frac5{11}$ minutes to be equal, and after that, there are no more times in the hour where the hour and minute hand will be in the same position.

Our answer is $ \boxed{1: 05\frac5{11}}$.

Yes but did you take into account that the hour hand also moves? Solution to be posted soon.

That's the great part here XD

They move at a rate of 1:12

I said $ 6^\circ \minus{} 0.5^\circ \equal{} 5.5^\circ$ per minute [b]relative[/b] to the hour hand.

You can check this by noting that in my answer $ |5.5m \minus{} 30h|$ does indeed equal $ 0$.


RIGHT SIDE INEQ:

[hide]

expr =   (ineq by AM HM)

Now  which is true.

[/hide]

LEFT SIDE INEQ:

[hide]

There are n terms, each which are equal to or smaller than .  Thus the sum is smaller than , which is smaller than one.

[/hide]


A national mathcounts coundown round math jam is going on right now.  IM cheesybread92 to join.

(effective 5:30 - 10:30 EST)

and if you don't have AIM, you better get it because im holding the chatroom now and its pretty empty.

It's 8pm EST and cheesybread92 is not on.

its kinda dead.

but klebian PWNED easyaspi 6-4

I would have pwned if I was there.  BUT I was busy until now.

Yeah it is over.  We went through the 2000 Nationals before I had to leave.  At one point we had 7 or 8 people, but it was usually around 5 or 6.  But that's ok, it was a last minute announcement.  Literally.  I posted this after the jam had started.


inequalities
Euler
inequalities proposed

$ s$= semiperimeter

Prove that

$ 9s^2r^2\leq m_a^2m_b^2\plus{}m_b^2m_c^2\plus{}m_c^2m_a^2\leq \frac{9}{4}s^2R^2$

[quote="Ligouras"]$ s$= semiperimeter

Prove that

$ 9s^2r^2\leq m_a^2m_b^2 \plus{} m_b^2m_c^2 \plus{} m_c^2m_a^2\leq \frac {9}{4}s^2R^2$[/quote]

\[ \sum {m_a^2 m_b^2 }  \equal{} \frac{9}{{16}}\left( {a^2 b^2  \plus{} b^2 c^2  \plus{} c^2 a^2 } \right) \equal{} \frac{9}{{16}}\left( {s^4  \minus{} 8Rrs^2  \plus{} 2r^2 s^2  \plus{} 16R^2 r^2  \plus{} 8Rr^3  \plus{} r^4 } \right)\]

and using Gerretsen Inequality and Euler Inequality can prove it, I think.


number theory
least common multiple
inequalities
number theory unsolved

For all positive integers $m>n$, prove that
$lcm(m,n)+lcm(m+1,n+1)>{\frac{2mn}{\sqrt{m-n}}}$
I hope I have posted the question in right section :?

I saw this problem before. I remembered this:

   Let gcd(m,n)=d and m=d.a and n=d.b After to said these then the problem will be easy.

There is your problem with the solution. You can visit there.But this site is Turkish.If you want I can translate the solution.

 http://www.fen.bilkent.edu.tr/~cvmath/Soru/soru_2005.html

Hey metehan ismimin altindaki bayragi gormuyormusun?
Ben azeriyim! :lol: thanks for solution!

Use AM-GM my friend, it helps you. ;)

Here is the english version of the solution: http://www.fen.bilkent.edu.tr/~cvmath/Problem/0512a.pdf


Hey Mathlinkers!!!

This post is a collection of results by Carmichael numbers. If you have a demonstration or other interesting result, post it here!

[b]DEFINITION 1:[/b]

A composite integer $m$ such that $a^{m-1}\equiv 1(mod \ m)$ is called a
[i]new pseudoprime to base $a$[/i](or only pseudoprime).

[b]DEFINITION 2:[/b]

A composite integer $m$ such that $a^{m}\equiv a(mod \ m)$ is called a
[i]old pseudoprime to base $a$[/i].

[b]DEFINITION 3:[/b]

A composite integer $m$ such that is a pseudoprime to base $a$ whenever $GCD(a,m)=1$ is called a [i]Carmichael number.[/i]

[size=150][b]THEOREM 1:[/b][/size]

$m$ is a Carmichael number if and only if $m$ is square-free and if $p$ is any prime divisor of $m$ then $(p-1)|(m-1)$.

Here's a proof of the theorem:

If $n$ is square-free and $p-1|n-1,\ \forall p|n$, $p$ prime, then the fact that $n|a^{n-1}-1,\ \forall (a,n)=1$ follows easily from Fermat's theorem applied for each prime factor of $n$.

Conversely, assume $n$ is a Carmichael number. Assume now that for some prime $p$ we have $p^2|n$. We can choose $a$ to be a primitvive root of $p^2$, so from $p^2|a^{n-1}-1$ we get $p(p-1)=\phi(p^2)|n-1\Rightarrow p|n-1$, which can't be true, so $n$ is square-free. Now let $p$ be any prime factor of $n$. Take $a$ to be a primitive root of $p$. From $p|n|a^{n-1}-1$ we get $p-1|n-1$, Q.E.D.

Some further facts at [url]http://www.chalcedon.demon.co.uk/rgep/carpsp.html[/url].


Hi, there!

N is an integer randomly chosen from 1 to 1024 (both included). What is the minimum necessary number of (integer number) guesses to find it, if you are told on each guess either "correct", "too big" or "too small" ?  (Consider that you MUST SAY the chosen number to consider you "found" it.)

(Your answer is right only if your argument includes the proof that any other arbitrary chosen number would NOT need a bigger number of guesses! This is the hard part, I believe.)


Nu ştiţi cumva dacă s-a desfăşurat Concursul Naţional "Laurenţiu Duican" de la Braşov şi, dacă nu, când se desfăşoară? Pe site-ul inspectoratului din Braşov n-am găsit nimic, iar google-ul m-a ajutat mai mult de o grămadă :-D

Could you post the problem in PDF file, please ?

thank you
(no problem if it is in romanian langage)

I believe that this year ,"Laurentiu Duican" contest didn't hold and I suspect that it won't take place since it's a a bit late.


Did anyone else do Round 2 of the WMTS, formally or informally? How did you do? I got a 01331 :|

better than what they originally gave you rotfl. No I didn't do it because I'm lame and stupid

Last minute attempt...I turned it in a few minutes after midnight...

33230

I didn't do the third set, though.

Me neither. I was going to, but I forgot to send the stuff in while I was at a sleepover :rotfl: . I'm making good progress on number 4 though.