calculus
integration
function
induction
derivative
calculus computations

Could anyone help me with this? I was guessing to use the induction, but it seems kinda odd, need some explaination, Thx!!

$f: [a,b]\rightarrow R$ is a continuous function. Define a sequence $F_{m}: [a,b]\rightarrow R$ for $m\geq0$ by setting $F_{0}(x)= f(x)$ and $F_{m}(x)= \frac{1}{(m-1)!}\int_{a}^{x} (x-t)^{m-1}f(t) dt$ for $m\geq1$

1. Show that $F_{m}^{(i)}(a)=0$ for all $m\geq$$0$

2. Show that $F_{m}^{(m)}(x)= f(x)$ for all $x\in[a,b]$

Maybe this can help, but I still can't do the second part of your question, I'm so tyred right now, someone check it please.

That's look like Taylor Remainder.
\[ \frac{1}{(m-1)!} \int_a^x (x-t)^{m-1} f(t) dt = \frac{f(\xi)}{(m-1)!} \int_a^x (x-t)^{m-1}dt  \], where $\xi$ is the number in $[a,x]$

so \[ \frac{1}{(m-1)!} \int_a^x (x-t)^{m-1} f(t) dt = \frac{f(\xi)}{m!}( (x-a)^m)  \]
hence, $F_m (x) = \frac{f(\xi)}{m!}( (x-a)^m)$.
1)
Now We use induction to prove that the $i$ -th derivative on $a$ is $0$,

First step, if $i=1$ then $F_m ' (x) = \frac{f(\xi)}{(m-1)!}( (x-a)^{m-1})$
setting $x=a$ we have $F_m ' (a) = 0$ (while $x=a$, $\xi$ will also equal to $a$)

Inductive step,(asuming $i \neq m$, that's  for second part question)
Suppose that it is true for all positive integer less than or equal to $k$. then for $k+1 < m$.
$F_m ^{(k+1)} (x) = \frac{f(\xi)}{(m-k-1)!}((x-a)^{m-k-1})$ and $F_m ^{(k+1)} (a)=0$
for $k+1 >m$ we will have $F_m^{(k+1)} (x) = 0$.

2)
Now whenever $k+1=m$ that is the $m$-th derivative of $F_m (x)$
 we have $F_m^{(m)} (x) = f(\xi)$  :huh:

EDIT: I've delete the minus sign, After ten minutes posting it I realized my mistake :oops: ,


Segment $ s_1$ has endpoints at $ (3\plus{}\sqrt{2},5)$ and $ (4,7)$.  Segment $ s_2$ has endpoints at $ (6\minus{}\sqrt{2},3)$ and $ (3,5)$.  Find the midpoint of the segment with endpoints at the midpoints of $ s_1$ and $ s_2$.  Express your answer as $ (a,b)$.

Okay, so first, we have to find the midpoints of $ s_1$ and $ s_2$.  The $ s_1$ midpoint is equal to $ \left(\frac{7 \plus{} \sqrt2}{2}, 6\right)$.  The $ s_2$ midpoint is equal to $ \left(\frac{9\minus{}\sqrt2}2,4\right)$.  Thus, the midpoint of the midpoints is $ \left(\frac{\frac{7 \plus{} 9 \plus{} (\sqrt2 \minus{} \sqrt2)}{2}}{2}, \frac{10}{2}\right)$ or $ \boxed{(4,5)}$.

* Note: The midpoint of a segment is equal to the sum of the endpoints divided by 2.  So if you have $ (a,b)$ and $ (x,y)$ as coordinates, your midpoint will be $ \left(\frac{a\plus{}x}2,\frac{b\plus{}y}2\right)$.


function
integration
calculus
calculus computations

Let $ f$ be a non-negative function that is differential on the interval $ 0 \le x \le a$, where $ a > 0$, and such that $ f(0) \equal{} 0$, $ f(a) \equal{} 1$, and
\[ \int_{0}^{a}f(x)dx \equal{} 1.
\]
Show that
\[ \int_{0}^{a}\sqrt {\{f(x)\}^2 \plus{} \{f'(x)\}^2}dx \ge \sqrt {2},
\]
if the integral exists.

You say nothing about $ b$ until its dropped right on into an integral... are you sure the statement isn't \[ \int^a_0 f(x)dx\equal{}1\] and show that \[ \int^a_0 \sqrt{[f(x)]^2\plus{}[f^{\prime}(x)]^2}dx\ge \sqrt{2}\] if the integral exists?

$ \int_0^a \sqrt{f(x)^2 \plus{} f'(x)^2} dx$
$ \ge \frac{1}{\sqrt{2}} \int_0^a f(x) \plus{} |f'(x)|  dx$ (Cauchy Schwarz)
$ \ge \frac{1}{\sqrt{2}} \left( 1 \plus{} \left| \int_0^a f'(x)dx \right| \right)$
$ \equal{} \sqrt{2}$


[quote="djuro"][quote="bbbbbbbbbb"]Please tell me some problems of inequality that be used function method to prove.
I have example

Let $ ab+bc+ca=1$, $ a\ge b\ge c$
Find min and max $ F=a^{2}.b^{2}+c^{2}$

Sorry for my small vocabulary.[/quote]

Did you mean $ F=a^{2}+b^{2}+c^{2}$?

In that case, $ F$ can be arbitrarily big. Just take $ a$ very big (arbitrarily), and $ b, c$ small such that $ ab+bc+ca=1$ holds.

Minimum of $ F$ is $ 1$. It can be reached by $ a=b=c=\frac{\sqrt{3}}{3}$. It remains to prove the (famous) inequality $ a^{2}+b^{2}+c^{2}\ge ab+bc+ca$, which is equivalent to $ (a-b)^{2}+(b-c)^{2}+(c-a)^{2}\ge 0$.[/quote]

No, I mean $ a^{2}.b^{2}+c^{2}$ :)


Visto che anche l'oliforum se ne va in vacanza, è il caso che mandiamo le soluzioni per posta?  :P

Immagino vada bene scrivere le soluzioni a Gobbino (m.gobbino@dma.unipi.it)...

Poichè non so proprio creare pdf ne texare perchè tra l'altro il mio PC mi ha appena abbandonato, potreste darrmi il numero di fax a cui spedire le soluzioni?

Sull'avviso del Winter Camp (che per fortuna ho salvato) è dato come numero fax per gli esercizi lo 050-2213802, all'attenzione di Massimo Gobbino

Per il pdf puoi scrivere i file in Word usando il Microsoft Equation Editor per le formule e poi stamparlo col Distiller (è un programma che trasforma file qualsiasi in PDF). Si usa come una stampante (nel senso che fai file-stampa e scegli come stampante il distiller)

P.S. Io sono Il_Russo

Grazie Kirill... x il pdf non c'è speranza visto che sto senza PC (ora sto da mio zio che gentilmente mi lascia stare un po' sul forum)...


search
symmetry
geometric sequence
geometric series

I can't figure out the following problems.  Please help!
I can't  figure out the following problems. Please help!!

1. Julian wants to label the five fingers on his left hand with the integers 1 through 5, such that no two adjacent fingers have consecutive integers, and each integer is used exactly once. In how many ways can Julian label his fingers? ( 08 national, sprint #26)

2. Consider the geometric sequence 60,-30,15,-7.5.... The sum of any odd number of consecutive terms of this sequence starting with the term of 60 is always greater than a number N, and the sum of any even number of consecutive terms is always less than the same number N. What is the value of N. (05-06 handbook, workout 8, #4)

3. The Indian mathematician Ramanujan knew that there were four different whole numbers A, B, C, and a such that: A^3 + B^3 =1729 and C^3 + a^3 =1729 . What is the sum of A+B+C+a? (02-03 handbook,workout 9, #2).

Thanks a lot![/u][/i][/b]

[hide="1"] We use some casework, based on which number is in the middle.

After placing a 3 in the middle, we have two ways to place 1 and 5 next to it and then just one way to put the rest. For example, we can have 25314 or 41352. That gives two ways.

After placing a 2 in the middle, we have two ways to put 4 and 5 next to it. So we have _425_ for example. We have to place a 1 and 3, which can be placed in just one way so two more here, 14253, 35241.

By symmetry, a 4 in the middle will be the same thing, so two more for that.

Now, we just have a 1 or 5 in the middle. They're symmetric again, so we just find it for 1 and then double.

The numbers on either side must be 3, 4, or 5. We list and get 53142, 24135,  24153, and 35142.

By symmetry, 5 gives 4 as well.

We thus have 14 ways in all.[/hide]

Bleh ok so I didn't see that somebody already posted. Also, number 3 is pretty well known.

If number 2 is an integer, then I think n=40. If it's not then i'm not sure how to converge it.

#2

Sum the infinite geometric series

$ N\equal{}60\minus{}30\plus{}15\minus{}7.5...$
$ \minus{}2N\equal{}\minus{}120\plus{}60\minus{}30\plus{}15...$
$ \minus{}2N\equal{}\minus{}120\plus{}N$
$ \minus{}3N\equal{}\minus{}120$
$ N\equal{}\fbox{40}$

[quote="ytrewq"]#2

Sum the infinite geometric series

$ N \equal{} 60 \minus{} 30 \plus{} 15 \minus{} 7.5...$
$ \minus{} 2N \equal{} \minus{} 120 \plus{} 60 \minus{} 30 \plus{} 15...$
$ \minus{} 2N \equal{} \minus{} 120 \plus{} N$
$ \minus{} 3N \equal{} \minus{} 120$
$ N \equal{} \fbox{40}$[/quote]

Exactly.


Let $a = \frac{m^{m+1}+n^{n+1}}{m^{m}+n^{n}}$, where $m$ and $n$ are positive integers.  Prove that $a^{m}+a^{n}\geq m^{m}+n^{n}$.

I think we just need to apply the Newton formula, then we'll solve it easily!
The second solution is just try with Bernoulli! But the problem is so ugly then anyone don't want to solve it!
PS#H34N1@:can you show everyone the solution? :)


[quote=gagarin]Let ABc ia a triangle  Find  MAX
F=cosA+cos$ \frac{b}{2}$+cos$ \frac{C}{3}$[/quote]
Let $ABC$ ia a triangle  . Find  maximum of $ cosA+cos \frac{B}{2}+cos\frac{C}{3}$

[quote=ppu]This was from 2008...[/quote]
[url=https://artofproblemsolving.com/community/c5168_2008_allrussian_olympiad]No?
[/url]
Thanks.


Is there a solution key or somewhere to find the answers to the questions in paul zeitz's art and craft of problem solving?

thanks!

Try [url=http://www.artofproblemsolving.com/Forum/search.php]here[/url]  :roll:


Here a problem of Putnam. Do any of you guys have any interesting ways of solving it?

Players 1, 2, 3, , n are seated around a table and each has a single penny. Player 1 passes a penny to Player 2, who then passes two pennies to Player 3. Player 3 then passes one penny to Player 4, who passes two pennies to Player 5, and so on, players alternately passing one penny or two to the next player who still has some pennies. A player who runs out of pennies drops out of the game and leaves the table. Find an infinite set of numbers n for which some player ends up with all n pennies.

Methinks this belongs in Advanced, to say the least.

I think I have a solution:

[hide]Every n so that n-1 is a power 2.  There are more though.

I found that in the first turn around, you lose one or two more than half, depending on whether n is even or odd.  If the number you have remaining is even, the same person will get two, then lose one, or get one and lose two each time around.  Eventually, all of the people who lose two will run out of coins, and because everyone who loses two will run out in one round, the number will be halved.  This will continue until only one person remains.[/hide]

This is a neat problem!  Bookworm almost had it, but I believe that it works out that all n's of the form [hide]2^k + 1  or  2^k + 2, k a natural number[/hide] will result in someone getting all of the pennies.  It is remotely possible that there are other n's that will do so, but I'm pretty sure this is it.


probability
calculus
complex numbers
complex analysis

My school doesn't cover complex numbers, statistics and probability in precalculus, is that important in AP calculus?

It's not important in AP calculus (the topics won't come up), but I think those are important things to have some knowledge of before taking on calculus.

I haven't taken calculus yet, but I agree that it's very important to have a strong understanding, at the very least, of complex numbers. Probability has limited impact on calculus (well, I suppose continous probability DOES use integrals.. but that's fairly basic.)

I think that you need to know complex numbers inside and out: i.e. polar form, DeMoivre's theorem...
It's just good to know in general.

Also, if you choose to take complex analysis, knowing about them is invaluable.

if you're doing the BC curriculum, then you'll want to know about polar

if you learn about complex numbers, polar form will make a whole lot more sense.
I don't think AP calc has complex numbers in it, so its not necessary, but I would strongly recommend learning about them. They really aren't that hard, just strange to think about.

They really help you understand polar form better, since polar form and the complex plane go together.

Are polar form/complex numbers/DeMoivre's covered in AoPS Vol. II? If not, is there any book where I may learn them you suggest? My school only touched briefly on them, and I do see that they are important. Plus fractals look cool.

Yes AOPSII covers them in great depth. Barron's SAT MathIIC also makes the connection between polar form and the complex plane (of course, to a much lesser extent than AOPS).


group theory
abstract algebra
linear algebra
matrix
vector
superior algebra
superior algebra unsolved

Hi,

How would I calculate the number of sylow_2, sylow_3, and sylow_5 subgroups of the group GL(2,Z_5), which has order 480= (2^5)*(3)*(5)? Also to describe these subgroups.

I calculated the number theoretic restrictions that each can have, but still there are alot of possibilities for each one. How would I narrow it down?

Thank you

I guess I forgot to hit "Submit" last time.

Since the structure we know is on the matrix side, that's where we work.

The 5-subgroups are generated by single elements of order 5, such as $\begin{bmatrix}1&1\\0&1\end{bmatrix}$. In fact, all elements of order 5 are similar to this one. They have determinant 1 and trace 2, and are not the identity. Counting these, we have 8 with 1s on the diagonal, and 4*4 with other diagonal elements; in those cases, both diagonal elements are nonzero. That total of 24 elements of order 5 means we have 6 5-subgroups.

The 3-subgroups are generated by single elements of order 3, such as $\begin{bmatrix}0&-1\\1&-1\end{bmatrix}$, and all are similar to this, with determinant 1 and trace -1. Both off-diagonal elements must be nonzero, and we have 5*4=20 elements of order 3, for 10 3-subgroups.

The 2-subgroups are of course the hardest- it is not immediately obvious where to find an example. The diagonal elements are a subgroup of order 16, so we need to add something to make them work. It turns out that using the additional element $\begin{bmatrix}0&1\\1&0\end{bmatrix}$ as another generator works, giving us the subgroup of matrices with the form $\begin{bmatrix}a&0\\0&b\end{bmatrix}$ or $\begin{bmatrix}0&a\\b&0\end{bmatrix}$.
Now, we consider the natural action of $GL_2(\mathbb{Z}_5)$ on $\mathbb{Z}_5^2$. This subgroup permutes the eight nonzero multiples of the standard basis vectors and acts transitively on the other 16 nonzero vectors. Conjugation only changes where the action happens, so each choice of two directions from the six possibilities gives a conjugate of the 2-subgroup, and there are (at least) 15. This is the maximum possible, so we are done.


Find all pairs of nonnegative integers $(m,n)$ such that \[(m+n-5)^2=9mn.\]

Of course, it can be solved by standart descent method.

[b]Case 1.[/b] Suppose $\gcd(n,m)=1$. Then $mn$ is a perfect square, so $m$ and $n$ are perfect squares. Let $m=a^2$, $n=b^2$, $a$, $b$ are nonnegative integers. We have $a^2-3ab+b^2-5=0$. Assume WLOG $a>b$. It is clear $b>0$. If $b=1$ then $a=4$, if $b=2$ then $a$ is irattional. Assume now $b>2$ and consider pair $(a',b)$, where $a'=\frac{b^2-5}{a}<a$. This pair also satisfies the equation and $a'+b<a+b$. Therefore, step by step we will necessary come to pair $(4,1)$. Moreover, as we see, all solutions can be found in a reccurent sequence. Indeed, define $a_1=4$, $b_1=1$ and $a_{k+1}=\frac{a_k^2-5}{b_k}$, $b_{k+1}=a_k$.

[b]Case 2.[/b] Suppose $\gcd(n,m)>1$. Then it is necessary $\gcd(m,n)=5$. So we can write $m=5a^2$ and $n=5b^2$. Apply descent method again.

[quote=analysis90]Assign $m=u+v,n=u-v$. [/quote]
It requires $m$ and $n$ to be the same parity.

Wow, how would one think of this factorisation $(2m-7n-10)^2-45(n+2)^2+180=0$?


function
real analysis
real analysis unsolved

Let $I$ be a bounded, closed interval, and let $f: I \rightarrow I$ be continuous.  Show that there exists $x \in I$ such that $f(x) = x$.  Are the hypotheses about $I$ both necessary?

What if $I$ is not bounded?

Consider $I=[0,\infty).$ Then $f(x)=x+1$ maps $I$ continuously into itself and has no fixed point.

What if $I$ is not closed?

Let $I=(0,1].$ Then $f(x)=\frac{x}2$ maps $I$ continuously into itself and has no fixed point.

What if $I$ is not an interval?

Let $I=[0,1]\cup[2,3].$ Then $f(x)=3-x$ maps $I$ continuously into itself and has no fixed point.

Do you need help with this problem? Unfortunately the solution is so simple you can't really give a hint :)

he-he!  you're so simple  :) ...

if $I = [a,b]$, then it is trivial if either $f(a) = a$ or $f(b) = b$, so assume not.  Then $f(a) > a$, $f(b) < b$ since $f(I) = I$, and so we can consider the function $g(x) = f(x)-x$, which is positive and then negative on $I$.  By the intermediate value theorem, there is a zero; i.e. $f( p ) = p$ for some $p$.


This problem was puzzling me so any help or explanations would be appreciated!

Show that L: P_n -> P_n given by L(p) = p + p' is an isomorphism. First show that L is a linear operator.

The second part follows straight from the definition of a linear operator. To solve the first part, you can use the fact that a linear mapping from a finite dimensional vector space to itself is an isomorphism if and only if its kernel is $\{0\}$. Now just look at the leading coefficients of $p$ and $p+p'$. What can you say about them?


Δίνεται γωνία $ \angle xOz$ και έστω $ A,$ $ D,$ δύο σταθερά σημεία επί των $ Ox,$ $ Oz,$ αντιστοίχως. Δια του $ O,$ φέρνουμε δύο τυχούσες ευθείες $ Ox',$ $ Oz'$ $ ($ $ Ox',$ μεταξύ των $ Ox,$ $ Oz$ και $ Oz,$ μεταξύ των $ Ox',$ $ Oz'$ $ )$ και έστω $ B,$ $ C,$ τα σημεία τομής των, από ευθεία $ (\epsilon),$ παράλληλη προς την ευθεία $ AD.$ Προσδιορίστε τη θέση της ευθείας $ (\epsilon),$ ώστε το τραπέζιο $ ABCD,$ να είναι παραλληλόγραμμο.

Κώστας Βήττας.

Μηπως θέλει περισσ'οτερη σαφήνεια η διατύπωση; Οχ ειναι πλευρά γωνίας , Ο΄χ ευθεία ,το χ κοινό .....κάτι δεν μου κολλάει..

[quote="xp"]Μηπως θέλει περισσότερη σαφήνεια η διατύπωση; Οχ ειναι πλευρά γωνίας , Ο΄χ ευθεία ,το χ κοινό .....κάτι δεν μου κολλάει..[/quote]
Τονίστηκαν λάθος σημεία και έτσι έγινε ακατανόητη μία εύκολη άσκηση.

Το διόρθωσα ήδη και σε ευχαριστώ πολύ αγαπητέ ( Χρήστο ; ) [b][size=100]xp[/size][/b].

Κώστας Βήττας.

ΥΓ. Το μπέρδεμα ήταν από τη συγκίνηση την ώρα που έγραφα το μήνυμα, ότι μπορεί να προσπερνάμε τους Καναδούς, όπως και έγινε. :)


calculus
derivative
integration
LaTeX
calculus computations

can someone help me derive .5(x radical (1-x^2) + arcsinx)

calculus isnt supposed to be in this forum.

But while were here ...

are you trying to find the derivative or the integral? judging by the time period (calc AB should be in integration techniques right now, and finished with derivatives)

If its the derivative then its rather easy, just split up the expression and take the derivative seperately using chain rule.

If its the integral you are looking for, use u substitution for the $x\sqrt{1-x^2}$ and use integration by parts for arcsin.

Its pretty hard to read the problem if its not in latex :O

yeah we're done with integration
BUT we're sort of reviewing
so i have to actually derive .5 [(xradical 1-x^2) + arcsinx]
please help me i'm so lost

ok well bring the .5 out, first we take the derivative of $x\sqrt{1-x^2}$ using the product rule.

We get $\sqrt{1-x^2} - x(1-x^2)^{-1/2}$

Now take the derivative of arcsin which is $1/\sqrt{1-x^2}$

Put it all together, 2 terms have the same denominator so put those together and then multiply everything by .5

Even though the help has been provided, this topic still needs to be moved to the appropriate forum. Moderators/admins, please  :lol:

Hmm, I don't know why, but I can no longer move topics...  :(

I can still edit posts, but can't move stuff. Weird.


$\text{Let }\; x,y,z\; \text{be positive real numbers, determine the greatest possible number } \; \lambda\; \text{such that}$
\[ \frac1{x\sqrt{y^2+z^2}}+\frac1{y\sqrt{z^2+x^2}}+\frac1{z\sqrt{x^2+y^2}}\geq\frac{\lambda}{x^2+y^2+z^2}. \]

[quote="georgew_bush"]sorry to post in Chinese ，but i really donnot know how to express it in english
第一眼看到不等式还真的被唬了，刚想收藏，发现不对劲 :D[/quote]
Ok,I will translate for you though my english is also limited...
he said:"He was frighten by the inequality at the first said,but he find it has something 不对劲 when I want to add to favorites"
Sorry ,I don't what you mean by 不对劲?


I got a perfect score on the AMC10B for 2003, but I didn't get my picture in the book or on the web.  Can someone look for me?

[quote="mathfanatic"]If this is true, you should write or call the AMC office.

But first, we can't look for you if we don't know who to look for.[/quote]

hrmm... his notes state he is richard tapia...

He just changed his signature, then.  It wasn't there before.  Anyway, I'll have a looksy.

Besides, that name might just be a quote.  Not likely, but possible...

Anyway, I looked in my results book for perfect AMC10B, region 10 (canada).  There is one perfect score:

150.0 10 J. Kramar, Univ of Toronto Schs, Toronto, ON

This person has a blank spot in place of a picture in the results booklet.

Uhh...I hope this answers your question.  If you're not Janos Kramar, I guess there is no reference of you in the book.  And if you are...well, you don't have a picture.


is it just me or are the only people who answer the mathcounts part of the forum me and chinaboi. at the moment its only me cause chinabois at camp. but soon it might be no one cause im off to india for 3 weeks on the 16th. basically 2 days. ackkk. i would like to know other approaches to the problems here other than mine. if theres any other mathcounts person out there please start answering the questions also.

Who said I can't access computer at the camp(except that I only am allowed like 10 minutes, but oh well. Anyways, have fun in India!

just helping out old posts, and annoy ppl

Seriously, stop doing that, you're just spamming.

I already told mods about it

So if your a mod you should LOCK THIS TOPIC (and all the other ones with spam)


inequalities
calculus
logarithms
real analysis
inequalities proposed

:?: [b]Prove the left inequality (without calculus-real analysis):[/b]

$\left(\forall\right)\ n\in N^*,\ \left(\frac{n}{n+1}\right)^{n+1}<\sqrt [n+1] {(n+1)!}-\sqrt [n] {n!}\ <\ \left(\frac{n}{n+1}\right)^n.$

[b]I present a solution for the right inequality:[/b]

$x_k:=\frac{\sqrt [n] {n!}}{n},\ k\in \overline {1,n},\ x_{n+1}:=\left(\frac{n}{n+1}\right)^n\ (A.M.-G.M.)\Longrightarrow$
$\sqrt [n] {n!}+\left(\frac{n}{n+1}\right)^n>(n+1)\cdot\sqrt [n+1]{ \frac{n!}{n^n}\cdot \left(\frac{n}{n+1}\right)^n}=\sqrt [n+1] {(n+1)!}. $

[b]I solved the left inequality only with real analysis (therefore is true).

Remark. (Lalescu's sequence)[/b] $L_n=\sqrt [n+1] {(n+1)!}-\sqrt [n] {n!}\rightarrow \frac 1e.$
 :spider:

How ? Sorry,yours shows ugly. You prove my left ineq. which is cool (and right is cool with its demonstration).


integration
function
calculus
derivative
probability and stats

The question is in the book Random Matrices.

Page 80, we have a determinant
   $ det[\delta_{i,j} \plus{} \Psi_{i,j}(a)]_{i,j \equal{} 0,\ldots,N \minus{} 1}$, (5.3.1)
Where
        $ \Psi_{i,j}(a) \equal{} c_i^{ \minus{} 1}\int w(x)C_i(x)C_j(x)a(x)dx$.
Now differentiating Eq. (5.3.11) m times functionally with respect to a(x) and then setting a(x) = 0 we will get the n-point "correlation function"
    $ (N!\Pi_{j \equal{} 0}^{N \minus{} 1}c_j)^{ \minus{} 1}\frac {N!}{(N \minus{} m)!}\int |\Delta_N(x)|^2[\Pi_{k \equal{} 1}^{N}w(x_k)]dx_{m \plus{} 1}\cdots dx_{N}$.

Could you give an detail deduction of this formula? This problem confused me. Thank you.  :maybe:

I add an attachment which are Page 80 and Page 79 of the book Random Matrices. The files are in .djvu format.

I could not understand much about functional derivative. What is the derivative of $ \Psi_{i,j}(a(x))$ with respect to a(x)? Thank you.


P2.- hallar las funciones $f: R\to R$, tal que:
$f(f(x)+y)=f(f(x)-y)+4f(x)y$, para todo $x,y$ perteneciente a $R$.

[hide="Lema"]
Si para una función $ g$ tenemos que para $ a > 0$, $ g(z \plus{} x) \equal{} g(z \minus{} x)$ y $ g(z \plus{} a \plus{} x) \equal{} g(z \plus{} a \minus{} x)$. La función es se repite periodicamente cada $ 2a$ unidades.

Las condiciones son equivalentes a que la $ g$ es simétrica respecto $ y \equal{} z$ y $ y \equal{} z \plus{} a$.
Definamos $ U(x)$ como la función que da el simétrico de $ x$ respecto $ y \equal{} z$, análogamente $ V(x)$ respecto de $ y \equal{} z \plus{} a$.
Sea $ \{x_n\}$ una sucesión de puntos definida de la siguiente manera: $ z\le x_0 \equal{} z \plus{} b\le z \plus{} a$, $ x_{2n \minus{} 1} \equal{} U(x_{2n \minus{} 2})$ y $ x_{2n} \equal{} V(x_{2n \minus{} 1})$ para todo $ n\ge 1$.
Solo ejemplificare la relación que es fácil de notar y fácil de probar inductivamente.
$ x_0 \equal{} z \plus{} b$
$ x_1 \equal{} z \minus{} b \equal{} z \plus{} a \minus{} a \minus{} b$
$ x_2 \equal{} z \plus{} 2a \plus{} b$
$ x_3 \equal{} z \minus{} 2a \minus{} b \equal{} z \plus{} a \minus{} 3a \minus{} b$
$ x_4 \equal{} z \plus{} 4a \plus{} b$
.
.
.
Basta con notar que $ g(x_0) \equal{} g(x_k)$ para cualquier $ k$.
[/hide]
Ahora:
$ f(x) \equal{} 0$ es una solución.

Notemos que $ f(x) \equal{} x^2$ también es solución. Así que definamos $ e(x) \equal{} f(x)^2 \minus{} x^2$, tenemos entonces:

$ e(f(x) \plus{} y) \plus{} [f(x) \plus{} y]^2 \equal{} e(f(x) \minus{} y) \plus{} [f(x) \minus{} y]^2 \plus{} 4f(x)y\Leftrightarrow$

$ e(f(x) \plus{} y) \equal{} e(f(x) \minus{} y)$ y como es claro que $ f$ no es una constante toma al menos dos valores distintos, $ z$ y $ z \plus{} a$. Es decir $ e$ se repite cada $ 2a$(por el lema), por otro lado, $ f(f(x) \plus{} y) \minus{} f(f(x) \minus{} y) \equal{} 4f(x)y$ cuando $ y\rightarrow 0$ el lado derecho también, $ f$ toma valores arbitrariamente cercanos.
Podemos entonces tomar $ e(z) \equal{} e(z \plus{} 2a)$ para valores tan pequeños como querramos de $ a$, es decir, $ e$ es una constante.

Es claro que $ f(x) \equal{} x^2 \plus{} c$ para $ c$ real, satisface la ecuación original.

 :)


induction
geometry
modular arithmetic
number theory

Source: Geometry and Discrete Mathematics

Prove by mathematical induction for all n belongs to natural numbers:

64 is factor of 9^n-8n-1.

I am stuck in this problem. Below is what I have done.

For n=1

9^1-8(1)-1=0 -> which is divisible by 64.

For n=k
Assume that 9^k-8k-1 is divisible by 64.
Show that 9^(k+1)-8(k+1)-1 is also divisible by 64.

9^(k+1)-8(k+1)-1 = 9^k+9^1-8k-8-1

Please giude me through this problem. Thanks.

Define $f(n)=9^n-8n-1$. Prove $64|f(n)\quad\forall n\in\mathbb{N}.$

Proof:

$1^\circ\quad f(1)=9^1-8\cdot 1-1=0\implies 64|f(1)$

$2^\circ\quad 64|f(n)\implies (\exists k\in\mathbb{N})(9^n-8n-1=64k)\implies 9^n=64k+8n+1$. Now $f(n+1)=9^{n+1}-8(n+1)-1=9\cdot 9^n-8n-8-1=9(64k+8n+1)-8n-9=$
$=9\cdot 64k+72n+9-8n-9=9\cdot 64k+64n=64(9k+n)\implies 64|f(n+1).$

Hence, we proved that $64|f(1)$ and $64|f(n)\implies 64|f(n+1)$, and we're done.

(Note: You stuck because of the mistake - $9^{k+1}\neq 9^k+9^k$)

[quote="Farenhajt"]

(Note: You stuck because of the mistake - $9^{k+1}\neq 9^k+9^k$)[/quote]

Actually it was a typo I knew it was = 9^k+9^1. I really appreciate your apt reply but I am new to this math. Please could you be more simpler, some of those notations are hardcore math, beyond my understanding.

If someone else knows how to solve this question please go farword.

Again it was really kind of you (Farenhajt) for real good answer.

dear thebatosai,

what farenhajt has helped you with is actually not hardcore maths, check out the olympiad section for hardcore maths.. lol

well ill try help you out as well..

ok

outline of an inductive proof

1. prove the base case whether $n=1$ or $n=0$

now depending on how many assumptions are made in the inductive step, this determines how many base cases you must prove.

i.e. if you assume $f(k)$ and $f(k+1)$ are true, you need to have two base cases

2. assume true for when $n=k$

3. Go through the inductive process of when $n=k+1$

4. finish by stating, by induction, [the initial statement] must be true or something along those lines.

so when $n = 1$

$9^1 - 8\cdot1-1=0 \implies 64 | 9^1 - 8\cdot1-1$

assume true for when $n=k$

$\implies 64|9^k-8k-1$

$\implies 9^k - 8k - 1 = 64m \implies 9^k = 64m + 8k + 1$

now when $n=k+1$

$9^{k+1}-8(k+1)-1 = 9\cdot9^k - 8k -8 - 1$

$= 9\cdot9^k-8k-9$

substitute $9^k = 64m + 8k + 1$ into the equation derived from our true assumption when $n=k$

$= 9\cdot(64m+8k+1)-8k-9$

$=576m+72k+9-8k-9$

$=576m+64k = 64(9m+k) \implies 64|9^{k+1}-8(k+1)-1$

$\therefore$ by induction, $64|9^n-8n-1$

EDIT:

just so you know $a^b \times a^c = a^{b+c}$

$9^{k+1} \neq 9^k + 9^1$

rather $9^{k+1} = 9^k\cdot9^1$

thebatsoi,

just to help out, try this extremely easy inductive proof

$(x-1)|(x^n-1)$ for all natural numbers $n$

[hide]

when $n=1$

$= (x^1-1) = (x-1)$

$\therefore$ true for when $n=1$

assume true for when $n=k$

$= (x-1)|(x^k-1) \implies (x^k-1)=(x-1)p$

now when $n=k+1$

$= x^{k+1}-1 = x\cdot x^k-1$

$= x(x^k - 1) + (x-1)$

substitue $(x^k-1)=(x-1)p$ into equation derived from inital assumption

$= x(x-1)p + (x-1) = (x-1)(px+1)$

$\implies x-1|x^{k+1}-1$

$\therefore x-1|x^n-1$

[/hide]

[quote="Farenhajt"]Well, nothing to it. The principle is: if $f(n)$ is divisible by $64$, then we must prove that $f(n+1)$ is divisible too. Hence we assume $9^n-8n-1$ is divisible by $64$. That means that $(9^n-8n-1)/64$ is an integer, and we denote that integer by $k$. Therefore, $9^n-8n-1=64k$ and from that we have $9^n=64k+8n+1.$

Now let's go to $f(n+1)$. We have $9^{n+1}-8(n+1)-1=9\cdot9^n-8n-8-1=9\cdot9^n-8n-9.$ Now substitute that expression for $9^n$ we have: $f(n+1)=9(64k+8n+1)-8n-9=9\cdot 64k+72n+9-8n-9=9\cdot 64k+64n.$ Now we can extract $64$ to obtain $f(n+1)=64(9k+n)$. But this means that $f(n+1)$ is divisible by $64$.

[quote="thebatosai"]I knew it was = 9^k+9^1.[/quote]

Unfortunately, wrong again. Do your homework on exponents.[/quote]

Thanks I really appreciate. i am not really blessed with such skills. Appreciated.

[quote="sen"]
EDIT:

just so you know $a^b \times a^c = a^{b+c}$

$9^{k+1} \neq 9^k + 9^1$

rather $9^{k+1} = 9^k\cdot9^1$[/quote]

Thanks

$9^{n+1}-8(n+1)-1=9(9^n-8n-1)+64n$
we need to show that the LHS is divisbile by $64$
by inductive hypothesis, on the RHS, the 1st terms is divisibly by 64, and the 2nd terms is obviously divisible by 64,
since the RHS is divisible by 64, so it the RHS, and that is the inductive statement,
 then the base case $n=0$ is obvious since $64|0$

[quote="sen"]

assume true for when $n=k$

$\implies 64|9^k-8k-1$

$\implies 9^k - 8k - 1 = 64m \implies 9^k = 64m + 8k + 1$

[/quote]

Hi, I can understand what you are trying to do.  I have couple of questions, first what is this "|" notation in $\implies 64|9^k-8k-1$ mean (is it modulus)? How all of sudden that sign, i.e. "|" is gone and experssion changed to $\implies 9^k - 8k - 1 = 64m$?

Here is an example of another question similar to this one but slightly different method.

Q. Prove 10^n-3^n is divisible by 7?

Step 1:

n = 1
10^1  - 3^1  = 7 --->  Which is divisible by 7.

Step 2:

n = k
Assume that 10^k  - 3^k is divisible by 7.
Show that 10^{k + 1}  - 3^{k + 1} is also divisible by 7.

10^{k + 1}  - 3^{k + 1}  = 10^k .10 - 3^k .3
                                         = (7 + 3).10^k  - 3^k .3
                                         = 7.10^k  + 3.10^k  - 3^k .3
                                         = 7.10^k  + 3(10^k  - 3^k) ---> Which is divisible by 7.

The first term is divisible by 7 and also the second is divisible by our assumption.

Step 3:

By the Principle of Mathematical Induction 10^k  - 3^k is divisible by 7, n belongs to N.

What I have done different is that I split on of the term and also I want the term which I assumed to the divided by 7. This is how my teacher told us to do. Much different (I think) from my good friends have done. Is there any way I could split the term and solve this problem. Thanks.

the vertical line means divides, ex. $a|b$ means $a$ divides $b$

[quote="sen"][quote]Hi, I can understand what you are trying to do. I have couple of questions, first what is this "|" notation in $\implies 64|9^k-8k-1$ mean (is it modulus)? How all of sudden that sign, i.e. "|" is gone and experssion changed to $\implies 9^k - 8k - 1 = 64m?$ [/quote]

well let us say if $a|x$ then it can be represented: $x=ab$. like if something is divisible by something, than that thing times something else will give the product.. lol i suck at explaining..

ok

let us say $15|30 \implies 30=2*15$

$a|b \implies b=a*c$ in this case $a=15$, $b=30$, $c=2$

hmm tell me if you still dont understand[/quote]

Thanks I understood.

[quote="sen"]okaii kool man^^

is there anyhting else on the induction proof that aint very clear to you?

[quote]Q. Prove $10^n-3^n$ is divisible by $7$? [/quote]

well here is the quickest solution.

(using modulus arithmetic)

$10^n - 3^n \equiv 3^n - 3^n \equiv 0 \pmod{7}$

$\therefore$ $10^n-3^n$ is divisible by $7$

and

your method is 100% correct, it definitely would have been how i would have done it.. umMm another way, like using the previous method, its like basically identical:

${RTP: } 7|10^n - 3^n$

[b]1)[/b] when $n=1$ [base case]

$10^1 - 3^1 = 7 \implies 7|10^n - 3^n$

[b]2[/b] assume true for when $n=k$ [just wondering, why is it always k?] ahaha i dont knoww.. ><

$\implies 7|10^n- 3^n$

$\therefore$ it can be represented $10^n-3^n=7m \implies 10^n = 7m-3^n$

[b]3[/b] when $n=k+1$

$10^{k+1}-3^{k+1} = 10\cdot10^k - 3\cdot3^k$

substitute $10^n = 7m-3^n$

$10\cdot(7m-3^n)-3\cdot3^k = 70m-10\cdot3^n-3\cdot3^n = 70m - 7\cdot3^n = 7(10m -3^n)$

$\implies 7|10^{k+1}-3^{k+1}$

[b]4[/b]$\therefore$ by induction, $7|10^n - 3^n$.[/quote]

Wow you are really good.

just to comment, all the divisibility proofs that you do in school with induction are solved much faster with modular arithmetic, but i suppose induction is a principle that you have to learn and this helps some...


For an infinite sum of a geometric series, which is more conventional?

$ a: S_{\infty} \equal{} \frac{a}{1\minus{}r}$
$ b: \text{As } n \to \infty, S_n \equal{} \frac{a}{1\minus{}r}$

A is definitely the better (and accurate one). If you had B, then people might sometimes interpret it as if $ n$ is just increasing, so A is actually the right notation. (also because it must be with an infinite number of terms, not an increasing number)

I hope it's not too vague of an explanation...


Find necessary and sufficient conditions of a trianglular number to be the adition of other two triangular numbers (i.e 15+6=21)

If you write it as $\frac{a(a+1)}2+\frac{b(b+1)}2=\frac{c(c+1)}2$, this is equivalent to $(2c+1)^2+1=(2a+1)^2+(2b+1)^2$, so I suppose one such (very unsatisfactory! :() condition would be that $(2c+1)^2+1$ has at least two different representations as sum of two squares. I suppose this could be translated into a condition on the number of prime factors of the forms $4k+1,4k+3$ of $(2c+1)^2+1$, by using the fact that the number of different representations of $n$ as the sum of two squares (considering representations with different signs and order different) is $4(d_1-d_3)$, where $d_i$ is the number of divisors of the form $4k+i$ of $n$.


inequalities
group theory
abstract algebra
superior algebra
superior algebra solved

$A, B$ are subgroups of $G$. $C$ is the group generated by $A\cup B$. Prove that $[C:A] \geq [B:A\cap B]$.

Try first to show that if $A,B$ are subgroups of a group $C$, then $[C:A\cap B]\le [C:A]\cdot[C:B]\ (*)$, and then apply this here as follows: let $C=<A,B>$. We have $[C:A]\cdot[C:B]\ge [C:A\cap B]$, so $[C:B]\ge [A:A\cap B]$, which is what we want.

[quote="grobber"]Try first to show that if $A,B$ are subgroups of a group $C$, then $[C:A\cap B]\le [C:A]\cdot[C:B]$.[/quote]
I don't understand how you obtain it directly.

The following is my approach for the original problem.

Consider the left cosets of $A$ in $C$ which intersects with $B$, then select a $b\in (c_i A)\cap B$ and we can prove that $(c_i A)\cap B = b (A\cap B)$. [hide]

Since $b \in (c_i A)\cap B$, $b = c_i a$ where $a\in A$.

First we prove $c_i A \cap B \subset b (A\cap B)$.
For $x=c_i a'\in (c_i A)\cap B$, $b^{-1}x = b^{-1}c_i a' = a^{-1} c_i^{-1}c_i a' = a^{-1}a' \in A$ and $b^{-1}x \in B$ because $x\in B$. Therefore $x = b(b^{-1}x) \in A\cap B)$. We have $c_i A \cap B \subset b (A\cap B)$.

Now we prove $b (A\cap B) \subset c_i A\cap B$.
For $x\in b(A\cap B)$, $x=by$ where $y\in A$ and $y\in B$. hence $x = by = c_i(ay)\in c_iA$ and $x=by\in B$. hence $x\in c_i A\cap B$. We have $b (A\cap B) \subset c_i A\cap B$.
[/hide]

Hence $B$ is divided into several $(c_i A)\cap B = b_i (A\cap B)$, $b_i (A\cap B)$ are the left cosets of $A\cap B$ in B. So $[B:A\cap B]$ equals to the number of left cosets of $A$ in $C$ which intersects with $B$, this is obviously smaller than $[C:A]$

[quote="liyi"][quote="grobber"]Try first to show that if $A,B$ are subgroups of a group $C$, then $[C:A\cap B]\le [C:A]\cdot[C:B]$.[/quote]
I don't understand how you obtain it directly.

[/quote]

:? What don't you understand, exactly? Didn't I just post a proof of that? Or you mean you wanted to prove your problem dierctly?

[quote="grobber"]Just to finish it off: in order to prove $(*)$, use the application $x(A\cap B)\mapsto (xA,xB)$, where $x\in C$. It's easy to see that this is well defined, and it's injective, which proves $(*)$ from my previous post.[/quote]
I didn't find this post here...  :D


probability
analytic geometry
geometry
3D geometry
Gauss
rotation
induction

These are three standard questions on Random Walks:

1. Suppose a particle starts at the origin on an infinite number line and it makes a series of moves. Each move has a 50-50 chance of going to the left or the right. What is the probability that the particle will be back at the origin after n moves? What is the probability that it will ever return to the origin?

2. Starting from the origin of an infinite coordinate grid with the y-axis pointing north and the x-axis pointing east, a particle takes a journey. Each time it moves, it moves diagonally to another latice point - with a 50-50 chance of moving north or south, and a 50-50 chance of moving East or West. What is the probability that it will be back at the origin after n moves? What is the probability that it ever returns to the origin?

3. A particle again starts at the origin of a 3-dimensional coordinate grid. Each moves sends the particle with equal likelihood to one of eight corners of the surrounding cube. Thus, at every move the particle has a 50-50 chance of moving one unit up or down, one unit east or west, and one unit north or south. What is the probability that the particle returns to the origin? (what happens if instead of moving to one of the eight possible corners of the cube, it moves to the center of one of the six faces of the surrounding cube, is the probability the same?)

These problems are taken from the really fun book, [u]Fifty Challenging Problems in Probability with Solutions[/u] by Frederick Mosteller.

[hide]I think that I have this correct.

1.  The probability is (n*(n-1)*...*(n-n/2))/((n/2)!)=n!/(((n/2)!)^2), or x.

2.  The probability is x^2.

3.  The probability is x^3.

For the six pointed figure, I could not find an answer.[/hide]

hmm. I think you're onto the right track, but your answers are not quite right.  For one thing you did not give them as probabilities. They are greater than 1. But also because, what happens in your formulas when n is odd?

It would help if you explained your reasoning behind your answers. How you solve a problem is just as important as the final answer.

Well it could just keep on moving forward, so the probability isn't 100%.

And how you get to the answer is half the credit at my old and current school. Remember the old saying "It's not important that you get something. It's how you get it" or something like that.

Um, i kno it could keep on moving forward, but when ur dealing with infinite number of moves, ur can't prove that at some point, it won't move back the other way...

im not making any sense....

[quote="darkquantum"]For Number 1, where u ask when it will ever return to the origin,

is the probability 1?

Cuz when you take moves to infinity, infinity is quite large, and I think no matter what, if the point is capable of moving back to the origin (which it is), then after some immensely large number of moves, it will move back?[/quote]

Well that is your intuition, but can you quantify it somehow? And by that same logic, would you think the probability is 1 for the other two problems also?

[quote="Tare"]Well it could also bypass the origin and go in the negative direction, so I think the possibility is extremely low...[/quote]

No, it cannot bypass the origin without passing through it. Because it only moves one unit at a time.

Well, the other 2 would not be 1. For #2, there are n moves, meaning its' finite

And i don't understand #3
DO you mean that it will return to the origin at some time???

SO then, it'd be like the infinite one.....
hmm

Even in questions 2 and 3 I ask what is the probability that it will ever return to the origin - even after an infinite number of moves.

In number 3, the particle moves in "zig-zags" from point to point throughout the grid. Each move is with equal probability to any of the 8 points diagonally adjacent to it.

Then I repharse the question and ask what happens if it instead moves laterally throughout the grid (i.e. parallel to the coordinate axes), giving it 6 points it can go to with each step.

OK, so now it comes to the point where i don't know how to do it.....

[quote="Tare"]Well it could just keep on moving forward, so the probability isn't 100%.

[/quote]
It doesn't quite work like that.. if you keep tossing a coin repeatedly, whats the probability you get a head eventually? Theres a chance you get tails every time but the probability is still 1..

OK well I've solved half of the first one.. probability that it will ever return to the origin?

[hide]

It is 1. First of all we can assume it moves right on its first move, its completely symmetrical. Now we give a probability for each square that it will return to the origin from that square, call this f(x) where x is the distance from the origin. f(0)=1, and f(x) is the average of f(x-1) and f(x+1).

Letting f(1)=a we get the following chain:

1, a, 2a-1, 3a-2, 4a-3, ..., na-(n-1)...

These must all be probabilities, so na>(n-1) and a>(n-1)/n for all n. So a=1, and they all equal 1.

[/hide]

[color=cyan]Ah I was thinking of linking to that problem but never got around to it.  That was a fun problem.  Would you mind stating the actual content of the theorems you used in your last post?[/color]

Setofallsets wrote:Using the idea that gauss202 gave in spoiler, one can easily show that the probability of eventually returning to the origin is 1 for 1 and 2 dimensions by [hide] comparing the series for total expected number of meetings to the harmonic series (or a multiple thereof)[/hide].

Yes, this is the right idea but it needs more justification. You are right that you should use [hide] Stirling's Formula to prove it though. [/hide]

Quote:The 3-dimensional (corner) case is a bit trickier, if I recall correctly the easiest way to show it is to consider the fact that the probability of returning in 2n steps is the cube of the probability in the one-dimensional case and[hide] comparing the one-dimensional case probabilities to 1/sqrt(n).  They are all less than this - easy enough by induction if I remember right.   Then one can use the fact that sum(1/n^p) converges for p>1.   Someone tell me if there is a nicer method, I'd like to see it[/hide].

You are on the right track here. Just follow the analogous method that you used for the 1 and 2 dimensional cases, only in this case the series for the mean number of expected returns to the origin will be [hide] finite [/hide]. Lets call that number Q. Once you know this though, how can you get from there to the probability of it ever returning to the origin?

Quote:For 3-dimensional case when you move parallel to the axes, the problem is much harder.  I do not believe that there is a closed form, but the probability for returning on the 2nth step is[hide] (1/6^2n)*(2n choose n)*Sum(k=0 to n; (2k choose k) * (n choose k)^2).   (I hope I didn't make a typo there...)  This is obtained by using multinomial coefficients and then being very clever with sums[/hide].

I don't believe this formula is right for the center-face cubic movements. [hide] It shouldn't be that dissimilar from the corner-cubic movements, only from each point, instead of 8, you have 6 possible places to move to. [/hide]

Quote:I do not know of any simple proof that the probability of ever returning to the origin is less than 1.  This follows from Polya's theorem, however.  Those wise in the ways of rigorous approximations might tell me if my suspicion that [hide]Stirling's formula will serve to give closed-form terms so that the sum can be shown to be finite[/hide] is correct.

You are right that using [hide] Stirling's Formula [/hide] will help to esimate the mean number of returns to the origin. But like I said above, the real meat and potatos of this problem boils down to how do you get from knowing that to finding the probability of it ever returning to the origin? I'll give a hint about this later.

Joel, the theorem needed for this problem is this one...

[hide]

n! :approx: (n/e)^n*sqrt(2*pi*n) - It's called Stirling's Formula.

[/hide]

Stirling's Formula isn't very accurate for small n though. Is it really good enough for this problem?

For the purposes of this problem, yes. Because ultimately all you're really trying to show is that a certain series diverges or converges.

Actually, Stirling's formula is not needed to solve the one and two dimensional cases, nor the three-dimensional corner case.  One can[hide] show that the series for one and two dimensions are greater than 'versions' of the harmonic series by simple induction:  compare the ratio of consecutive terms of (2n choose n)/(2^(2n)) -- or the square of this for the two-dimensional case -- to the ratio of consecutive terms of the harmonic series (for the one-dimensional case you have to offset the harmonic series, and for the two-dimensional case you have to multiply it by 1/4).  For the three-dimensional corner case, you need to show that the terms of the series 1/sqrt(n+1) are always greater than the corresponding terms of the series (2n choose n)/(2^(2n)).  The easiest way to do this is to square both sides and use induction, again considering ratios of consecutive terms.   The one can proceed as I stated in my last post.  This method will not actually yield the value for the total probability for returning, however.  If we did know the expected number of returns Q, we could find the probability of returning to the origin p using the relation Q = pQ + 1, which is an expression of the fact that when we return to the origin we are in the same position as at the beginning, but with one return to the origin already made.[/hide]

For the three-dimensional case where you move to the faces:  the total number of 2n-step paths from the origin to itself is given by[hide] the sum of multinomial coefficients Sum(i+j+k = n, (2n choose i, i, j, j, k ,k) ).  This is because one must make an equal number of steps up vs down, left vs right, and forward vs backward, exactly the paths this sum counts. One can then write this in factorial form and give an appropriate order of summation to compute get essentially the formula that I gave - you will need to use a special case of Vandermonde's convolution to fully reduce it.  Then one divides by the total number of paths of length 2n to get the probability[/hide].  The formula is a lot more complicated than you would think, but the problem really is quite different from the corner case.  One cannot use the same tricks that made the two-dimensional and three-dimensional corner cases easy.  If you have a different answer, I'd like to see it...

To Joel: [hide]Polya's theorem states that if you take a lattice walk (ie can only move one unit and parallel to axes) on one or two dimensions, then the probability of return to the origin is 1, for greater numbers of dimensions the probability is less than 1.  Mathworld has some comments on this I believe, you might want to look for this and for Polya's random walk constants.[/hide]

The reason I asked about [hide]Stirling's formula for the three-dimensional face case is for the same reason that ComplexZeta is worried about it: it is innacurate for small n and in this formula we have to use it for small n an infinite (!) number of times in our sum.  I'm sure that the formula would work in the other cases since we don't have that problem (although I wonder about the accuracy of the estimate for the sum of the series in the three-dimensional corner case), but it seems that it is using a bit much power for the problem.  I don't think that Stirling's formula is at all easy to prove...[/hide]

Oh, and a correction to my first challenge: the problem is the 'face' version, where you move only parallel to the lattice lines.

Setofallsets wrote:Actually, Stirling's formula is not needed to solve the one and two dimensional cases, nor the three-dimensional corner case.  One can[hide] show that the series for one and two dimensions are greater than 'versions' of the harmonic series by simple induction:  compare the ratio of consecutive terms of (2n choose n)/(2^(2n)) -- or the square of this for the two-dimensional case -- to the ratio of consecutive terms of the harmonic series (for the one-dimensional case you have to offset the harmonic series, and for the two-dimensional case you have to multiply it by 1/4).  For the three-dimensional corner case, you need to show that the terms of the series 1/sqrt(n+1) are always greater than the corresponding terms of the series (2n choose n)/(2^(2n)).  The easiest way to do this is to square both sides and use induction, again considering ratios of consecutive terms.   The one can proceed as I stated in my last post.  This method will not actually yield the value for the total probability for returning, however.  If we did know the expected number of returns Q, we could find the probability of returning to the origin p using the relation Q = pQ + 1, which is an expression of the fact that when we return to the origin we are in the same position as at the beginning, but with one return to the origin already made.[/hide]

I agree with the first part of this... [hide] You can use induction and a comparison test, but I'm weary of some of the details you outlined. Using Sterling's Formula should be much easier, because 2nCn = (2n)!/(n!)^2 :approx: (2n/e)^2n*sqrt(4*pi*n)/[(n/e)^n*sqrt(2*pi*n)]^2 = 2^2n/sqrt(pi*n). So voila, you have a nice series to compare it to. In the one and two dimensional cases all you need to know is that it diverges. And in the three dimensional case, you can estimate the sum by using the first couple of terms and bounding the tail. [/hide]

But not with the second part... [hide] Because your formula between the number of expected returns to the origin and the probability of returning to the origin is not quite right. For example, if Q is less than 1 then your probability p would be negative. [/hide]

Quote:For the three-dimensional case where you move to the faces:  the total number of 2n-step paths from the origin to itself is given by[hide] the sum of multinomial coefficients Sum(i+j+k = n, (2n choose i, i, j, j, k ,k) ).  This is because one must make an equal number of steps up vs down, left vs right, and forward vs backward, exactly the paths this sum counts. One can then write this in factorial form and give an appropriate order of summation to compute get essentially the formula that I gave - you will need to use a special case of Vandermonde's convolution to fully reduce it.  Then one divides by the total number of paths of length 2n to get the probability[/hide].  The formula is a lot more complicated than you would think, but the problem really is quite different from the corner case.  One cannot use the same tricks that made the two-dimensional and three-dimensional corner cases easy.  If you have a different answer, I'd like to see it...

And you are right about this one. I forgot that the symmetry breaks down in this case. So the number of possible paths back to the origin is much more complicated than in the 8 corners case. I don't think I know a simpler expression for it either. But I think we can show that it converges and then estimate it like for the 8 corners case.

Quote:The reason I asked about [hide]Stirling's formula for the three-dimensional face case is for the same reason that ComplexZeta is worried about it: it is innacurate for small n and in this formula we have to use it for small n an infinite (!) number of times in our sum.  I'm sure that the formula would work in the other cases since we don't have that problem (although I wonder about the accuracy of the estimate for the sum of the series in the three-dimensional corner case), but it seems that it is using a bit much power for the problem.  I don't think that Stirling's formula is at all easy to prove...[/hide]

Actually Stirlings Formula is pretty accurate, even for small n. For example, if S(n) = (n/e)^n*sqrt(2*pi*n).

Then,

S(2) = 1.919 :approx: 2

S(3) = 5.836 :approx: 6

S(4) = 23.506 :approx: 24

S(5) = 118.02 :approx: 120

...

S(10) = 3,598,695.6 :approx: 3,628,800

S(n) is always less than n! and I believe the percentage error is always about (1/12n). But yes, it is not an easy formula to prove. A proof of it can be found in some advanced calculus books.

Heh...stupid me....

My formula for the probability of ever returning is correct if[hide] you count the initial touch on the origin (ie the probability of being at the origin after zero steps) in Q as well.  I should have paid attention and mentioned that before.  If one doesn't include the initial touch at the origin the reasoning for the truth of the formula is clearly false...[/hide]

I would be interested to see an approximation for the three-dimensional faces case based on Stirling's formula.  I have a feeling it would be quite complicated...

It's easy to prove a bound for n! by taking logs, but that's a bit weaker than Stirling's approximation.

Actually thinking more about it, we were all wrong about the 2 dimensional laterally moving case for the same reason I was wrong about the center-face moving 3 dimensional problem. It is NOT equivalent to the zig-zag problems because the vertical and horizontal movements are independent of one another.

The cool thing about making them move in these diagonals is that it ties the movements in different directions together. But moving just vertically and horizontally alone, you can go in one direction without going in the other. So the number of possible paths of length 2n returning back to the origin is much larger than in the diagonal case.

Actually, despite appearances, the 2 dimensional case where we move parallel to the axis does turn out to be equivalent to the case where we move to corners.  As I pointed out in an earlier post, we can take the plane of our "axes-parallel" walk, rotate 45 degrees, and dilate by sqrt(2).   This will convert every "axis-parallel" step into a diagonal ("corner") step.  If you don't believe me you can use matrices to rotate the beginning and ending points of every possible step, and then dilate and check what you get.  If you still don't believe me you can use the multinomial coefficients and summation method I outlined for the 3 dimensional "axes-parallel" case and check that you get the same answer.

Note: if you want to go in reverse, from the corners case to the axes-parallel case, you need to check that your initial point is a legal starting point before performing the transformations, otherwise you will get a non-integer point (note that the only reachable points (x, y) in the corners case are ones where we have x + y = 0  (mod 2) ).
[/quote]

[quote="gauss202"]kay. I was wrong about being wrong about that one. Here's an explicit proof of it.

In the two dimensional case the number of distinct paths returning to the origin after 2n moves consists of an equal number of east/west moves and an equal number of north/south moves. Call the number of east/west moves k, leaving the number of north/south moves to be n-k.

Then, there are 2nCk ways of choosing the east moves, (2n-k)Ck ways of choosing the west moves, and (2n-2k)C(n-k) ways of choosing the north moves. The south moves are then uniquely determined by the leftover spaces.

So the total number of ways of choosing each of these combinations is sum (k=0 to n) [2nCk*(2n-k)Ck*(2n-2k)C(n-k)].

But 2nCk*(2n-k)Ck*(2n-2k)C(n-k) = (2n!)/[k!*(n-k)!]^2.
So sum (k=0 to n) [2nCk*(2n-k)Ck*(2n-2k)C(n-k)]
= (2n)! * sum (k=0 to n) [1/[k!*(n-k)!)]^2
= (2n)!/(n!)^2 * sum (k=0 to n) [n!/(k!*(n-k)!)]^2
= 2nCn * sum (k=0 to n) [nCk]^2
= 2nCn * 2nCn
= (2nCn)^2

I believe For the three dimensional case, you get a similar sum, but I haven't worked it out yet. I can see for the center-face case that it will not simplify to (2nCn)^3 because it does not work for n=1 - the result should be 6. But I do believe it will simplify to something nice.[/quote]

I was with you until I got to this part:

= 2nCn * sum (k=0 to n) [nCk]^2
= 2nCn * 2nCn

How does the sum (k=0 to n) [nCk]^2 equal to 2nCn?

I know this is a very old topic, but I was wondering if anyone could clarify that part for me.

First note that $ \sum_{k \equal{} 0}^n \binom{n}{k}^2 \equal{} \sum_{k \equal{} 0}^n \binom{n}{k} \binom{n}{n \minus{} k}$

Now look at the expressions $ (x \plus{} 1)^{2n} \equal{} (x \plus{} 1)^n (x \plus{} 1)^n \equal{} \left(\sum_{k \equal{} 0}^n \binom{n}{k} x^k\right) \cdot \left(\sum_{k \equal{} 0}^n \binom{n}{n \minus{} k} x^{n \minus{} k}\right)$. Notice that the coefficient of $ x^n$ in the first expression is $ \binom{2n}{n}$, and the coefficient of $ x^n$ in the last expression is $ \sum_{k \equal{} 0}^n \binom{n}{k} \binom{n}{n \minus{} k} \equal{} \sum_{k \equal{} 0}^n \binom{n}{k}^2$. Hence the desired result.

[quote="gauss202"]First note that $ \sum_{k \equal{} 0}^n \binom{n}{k}^2 \equal{} \sum_{k \equal{} 0}^n \binom{n}{k} \binom{n}{n \minus{} k}$[/quote]

This I knew was true and was my first thought about trying to go about this. At first I thought of stretching out the summation and then go for a reversal but that got me nowhere. Anyways....

[quote="gauss202"]Now look at the expressions $ (x \plus{} 1)^{2n} \equal{} (x \plus{} 1)^n (x \plus{} 1)^n \equal{} \left(\sum_{k \equal{} 0}^n \binom{n}{k} x^k\right) \cdot \left(\sum_{k \equal{} 0}^n \binom{n}{n \minus{} k} x^{n \minus{} k}\right)$. Notice that the coefficient of $ x^n$ in the first expression is $ \binom{2n}{n}$, and the coefficient of $ x^n$ in the last expression is $ \sum_{k \equal{} 0}^n \binom{n}{k} \binom{n}{n \minus{} k} \equal{} \sum_{k \equal{} 0}^n \binom{n}{k}^2$. Hence the desired result.[/quote]

This was the part you lost me at.

I understand that using combinations with the binomial theoreum is what you were doing but why did you use (x+1)?

I'm sorry to be a bother.


[quote="hello123"]Does anyone know where to get the books Olympiad Geometry and An Introduction to Diophantine Equations? Amazon doesn't have them and I can't seem to find them anywhere. Also does anyone know a good english site for the Titu and Zuming books such as Old and New Inequalities?[/quote]

Same problem here!

I am looking for Olympiad geometry, did anything come up in those three years?


Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?

i cant find the correct forum to post this type of issue on so sorry for the inconvenience

Correct part to post this in is "Alcumus Problems Discussion."

Each circle intersects each other circle twice. Call the circles $ A,B,C,D$. Then we are just looking at $ \binom{4}{2}$. But, since we want them to intercect twice, the answer is $ 2\binom{4}{2} \equal{} \boxed{12}$.


[quote="L_Euler"][quote="DIMITRIS T"]hi!
is there any site having the problems of the magazine mathemaitcs and youth in english?[/quote]

I don't obvious but this magazine has a page such as page is translated to English (Problem issue)

eg: http://www.nxbgd.com.vn/toanhoctuoitre/?p=7&id=6&ReportID=413[/quote]

Thank you very much for the link!

The problem is that in this page i can't find problems for past issues

[quote=DIMITRIS T]hi!
is there any site having the problems of the magazine mathemaitcs and youth in english?[/quote]

Here you find in attachment the all problems of Mathematics and Youth Magazine in 2016. Enjoy problems


The mean of eight numbers is 101.  If four numbers are each increases by 10, and four numbers are each decreased by 10, what is the new mean of the eight numbers?

[hide]It's the same since you essentially added zero to the total sum, by adding and subtracting forty. This probably is a countdown question...[/hide]

[hide]I think I did this before, I just don't know where.

Adding 10 to 4 numbers is the opposite of subtracting 10 from another four numbers, so the average is still 101.[/hide]

[hide]if you add 40 and subtract 40 you get the same number

so 101[/hide]

[hide]wouldn't it be the same thing?[/hide]

Need I disagree?
No, I agree.
[hide="er..."]>>101<<[/hide]

[hide="hoho"]
well wouldn't it be the same thing?
let's imagine that all the numbers r um let's make up something here.
8!! the avg is 8 ok?
if the avg is 8, it is posible that all the numbers are 8 right?
if all the numbers are 8, take four away from four of them and add four to the rest
12,12,12,12,4,4,4,4
add them all....
64/8=8
stayes the same see?

so 101

YAY!!!!  :D
[/hide]

[hide]
Uhh... It's still the same thing because you're just adding 40 to the total and then subtracting it away so it's still going to be the same.[/hide]

Okay.. now that we have a general consensus (everyone), let's move on to the next question, as long as no one has any questions... of course


Find $ \angle A \plus{} \angle B \plus{} \cdots \plus{} \angle J$. (See diagram below)

As we go around the boundary, we turn through $ 2$ full revolutions, that is through $ 720$ degrees.

If the ten exterior angles we turn through are $ a, b, c, ..., j$ then the ten interior angles are $ 180\minus{}a, 180\minus{}b, ..., 180\minus{}j$.

Therefore the sum of the interior angles is $ 10\times 180 \minus{} 720 \equal{} 1080$ degrees.

well what you do is since you need to find the sum of the angles all you have to do is the following:

$ 1$) find how many angles in the figure ,angle A...angle J so there are $ 10$ if you count them
$ 2$)then you subtract $ 2$ from $ 10$ because a polygon cannot be made from $ 2$ sides so $ 10 \minus{} 2 \equal{} 8$
$ 3$)now you multiply $ 8$ and $ 180$ together you would get your answer.
$ 4$)so if you do$ 8$x$ 180\equal{}1440$so your answer would be$ 1440$degrees.

[quote="AndrewTom"]As we go around the boundary, we turn through $ 2$ full revolutions, that is through $ 720$ degrees.

If the ten exterior angles we turn through are $ a, b, c, ..., j$ then the ten interior angles are $ 180 \minus{} a, 180 \minus{} b, ..., 180 \minus{} j$.

Therefore the sum of the interior angles is $ 10\times 180 \minus{} 720 \equal{} 1080$ degrees.[/quote]Something tells me that the answer is $ 720^\circ$...

EDIT:

[hide="Using exterior angles"]With $ AJ$ as the base line, start at $ J$ and look at the counterclockwise 'sweep' of the exterior angles at each vertex. There are 3 full revolutions, so the sum of the counterclockwise exterior angles is $ 3\cdot360^\circ\equal{}1080^\circ$. Since there are 10 vertices and each interior angle is the supplement of its corresponding exterior angle, we have

$ \sum_{k\equal{}A}^{J}\angle k\equal{}180^\circ\cdot10\minus{}\sum\text{exterior angles}\equal{}1800^\circ\minus{}1080^\circ\equal{}720^\circ$[/hide]

Oops! You're right 10000thuser: there are 3 revolutions. I think I mustn't have noticed that the second one ends between $ GF$ and $ FE$. Thanks.


calculus
integration
logarithms
trigonometry
calculus computations

i hope i have not asked this before....  :(   :maybe:  (i searched the forum but nothing came up... so i am posing it now ... )

by transforming  it first to a suitable  "double integral" , show

$ \int_{\dfrac{\pi}{4}}^{\dfrac{\pi}{2}}\;\ln\;\ln\;\tan\,\phi\;\textbf d\phi\;\equal{}\;\boxed{\dfrac{\pi}{2}\, \ln\;\left[\dfrac{2\;\pi^{\dfrac{3}{2}}}{\left(\Gamma\left(\dfrac{1}{4}\right)\right)^2}\right]}$

[quote="Misan"]i searched the forum but nothing came up... so i am posing it now [/quote]

Perhaps it would be a good idea for you to make a list of all integrals you have already posted, together with their respective links. Organization please.

[quote="Black"]Yeah, Carcul's suggestion is pretty good.

Make a thread and post there lots of problems, which I personally consider hard and interesting, as integrals, double integrals, series, double series, etc.[/quote]

I agree toO! :)

[quote="Doe John"]Misan, how did you compute the integral by transforming it to a suitable double integral?[/quote]

i can see you are good with integrals... :)  so how about i give you a hint...?  i am sure you might be able to do it after that.

you have to reduce the integral in question by transforming it to something that bears similarity to

$ \int_0^1\int_0^1 \dfrac{x}{\ln\,(x\,y)\cdot (1\plus{}x^2\,y^2)}\;\textbf dx\; \textbf dy$

this class of integrals are relatively easy to solve

$ \left( \quad \text{where this one gives an answer of}\quad\boxed{\ln\,\left(\dfrac{\sqrt{\;2\,\pi^3 \;}}{\Gamma^2\left(\dfrac{1}{4}\right)}\right)}\quad\right)$

the original integral will give an answer that is sort of similar.  :oops:


integration
function
inequalities
real analysis
real analysis unsolved

Easy generalization . $f,g$ are functions such that :

$f,g : [0,1]\rightarrow R\ (real)$, integrable, and $\forall x_1,x_2\ (f(x_1)-f(x_2))(g(x_1)-g(x_2)) \geq 0$

Then $\int_0^1f(x)dx\int_0^1g(x)dx \leq \int_0^1f(x)g(x)dx$

 :cool: Try to solve it  :cool:

To prove this you need to intgrate with respect to $x_1$ on $[0,1]$ and then with respect to $x_2$ on $[0,1]$. My ineq is just a consequence of this one because: $[f^a(x)-f^a(y)][f^b(x)-f^b(y)]>0\forall x,y$ which follows from the monotonicity of $x^a$ and $x^b$. Thank you for help!

[quote="xirti"]$f: [0,1]\to [0,\infty)$ is a continuous function. Prove:
\[ \int_0^1f^a(x)dx\int_0^1f^b(x)dx\le\int_0^1f^{a+b}(x)dx  \]

Where $a,b>0$.[/quote]
This follows also from Hoelder's inequality: $\left|\int_\Omega f(x) g(x) dx \right| \le \left( \int_\Omega |f(x)|^p dx \right)^{1/p}\left( \int_\Omega |g(x)|^q dx \right)^{1/q}$ with $p^{-1}+q^{-1} = 1$, if the right hand side is finite. Take $g = 1$ and $p = \frac{a+b}{a}, \, \frac{a+b}{b}$.


Twenty 11"x17" sheets were printed on both sides so that when the sheets are stacked and folded down the center, a 8.5"x11" booklet is formed with the pages numbered 1-80. When stacked and folded however, the sheets containing pages 12 and 53 were accidentally omitted. What is the sum of the other numbers on the pages that were missing?

[hide="Answer"]Twenty sheets folded in half gives forty pages, with page numbers on both sides. The sheet with page 12 also has pages 11, 68, 69, and the sheet with page 53 also has pages 26, 27, 54, so our answer is $11+68+69+26+27+54=\boxed{255}$.[/hide]

[b]EDIT:[/b] Okay, fixed.

How did I miss a sentence? :wallbash: It's fixed now.

[quote="jhollenbeck"]$\fbox{255}$ is right i think :)[/quote]

How did you get that?  :? I probably added wrong to get 258.

The middle two pages are 40 and 41. When you get the pages on one side, just take the same difference from the other side to find the other two pages. For example, we have page 12. The other number on that same sheet is page 11, which we kind of have to see intuitively. So, those two are 28 and 29 pages from 40. Therefore, the other two on that same sheet are 69 and 70... ohh, I screwed up, actually. Sorry, I should have realized that 68 and 69 aren't on the same sheet. :roll: Then, we have page 53. The other number on that same sheet is page 54, and these two are 12 and 13 from page 41, so the other two are 27 and 28. Heh, so it actually still comes out to $255$. I guess I got lucky there. :P

But I thought that the page numbers of opposite sides of the book (like 69, 12) add up to 81.

Whops, mis read your answer  :P

Why can't it be 11, 12, 69, and 70?

It is, if you read my correction in my second post...

Oh. You're right. Odd. I don't know how I missed that.


Which of the following has the highest ionization energy :
K ; Ca ; Ga ; As ; or Se ?
[hide] I choosed "Se" since ionization energy increases as we move across a period
but, the correct answer written in my book is "As" !!!   Could someone explain me why[/hide]

Recall that ionization energy is the amount of energy needed to remove the outermost electron from an atom.  So let's write out the valence shell electron configurations for the relevant elements: for As, it is $4s^{2}4p^{1}_{x}4p^{1}_{y}4p^{1}_{z}$, and for Se it is $4s^{2}4p^{2}_{x}4p^{1}_{y}4p^{1}_{z}$.  Thus the highest energy electron in As is in one of its $4p$ orbitals, but recall that a half-filled subshell (as is the case here) gains a degree of stabilization.  Also, Se's highest energy electron is in it's $4p_{x}$ orbital--and there are two mutually-repelling electrons confined to that orbital!  The former effect raises As's ionization energy and the latter lowers Se's.  The combined effect is enough to make As's ionization energy the highest.  Of course, there is nothing specific in this reasoning to As and Se--the same applies to N and O and P and S, for example.

[quote]for As, it is
\[4s^{2}4p^{1}_{x}4p^{1}_{y}4p^{1}_{z}\]
, and for Se it is
\[4s^{2}4p^{2}_{x}4p^{1}_{y}4p^{1}_{z}\]
[/quote]
I can't understand your expression  :huh: , I think you should write
\[1s^{2}2s^{2}p^{6}3s^{2}p^{6}d^{10}4s^{2}p^{4}\] for SE and

\[1s^{2}2s^{2}p^{6}3s^{2}p^{6}d^{10}4s^{2}p^{3}\] for AS

Can you give me the general method to follow if they told us to choose the highest or lowest ionization energy

I think what gregx007 meant was that for every p orbital there are three spaces for a pair of electrons to go (what he called x, y, and z). One electron must go into each of these before the first space can get a pair. When all three of the spaces have one electron (a half-filled subshell) the atom gains a little bit of stability. Not as much as having a full subshell,  but enough to increase the atom's first ionization energy. When another electron is added to the first space (x), it's easier to remove because it's repelled a little by the other negatively charged electron with which it shares its orbit. This gives Se a lower first ionization energy than As.

The general rule for first ionization energy is that it decreases going down a group (more levels of electrons, thus less energy is required to remove one) and increases moving left to right across a period (more protons, therefore more electrostatic attraction and more energy required to remove an electron). Unfortunately, this problem with As and Se (along with a few others) is an exception because of the stability of a half-filled subshell.

I don't know if I did any better explaining that, but I hope it's at least a little bit helpful. :)

A good summary, KrazyNerd.  I do have a few comments, however.

As it turns out, though, the $p_{x},p_{y},p_{z}$ is standard nomenclature.  Think of a 3D coordinate system with $x,y$ and $z$ axes; the $p_{x}$ orbital lies along the $x$ axis, and similarly for the others (remember, a $p$ orbital looks like a dumbell).  As you remark, this is rather arbitrary, we just need some way of telling the three orbitals apart.  The only time they change is when there are bonds, then the $x$ axis is usually taken to point along the bond.

[quote]... for every p orbital there are three spaces for a pair of electrons to go[/quote]  This is not technically correct--there are two spots for an election (one spot for an electron pair) in each p orbital.  It just so happens that every energy level has three p orbitals, however.

P.S.  yfehehey--in my first post I was just writing the valence electron configuration.

Also, keep in mind that this model also predicts that when moving from something with a full $s$ shell to something with only one $p$ electron (e.g. $Mg$ to $Al$) ionization energy decreases.  This works out well for elements low on the table, say Mg and Al.  However, I pose the following question as a challenge: why is the ionization energy of Sr (549.5 kJ/mol) less than that of In (558.3 kJ/mol)?

[quote="gregx007"]
[quote]... for every p orbital there are three spaces for a pair of electrons to go[/quote]  This is not technically correct--there are two spots for an election (one spot for an electron pair) in each p orbital.  It just so happens that every energy level has three p orbitals, however.[/quote]

Yeah...whoops. :blush: Stopped thinking for a moment. I meant in the total p subshell there are three p orbitals, each of which can hold two electrons.

As for why Sr's ionization energy is less than that of In, I think it might have something to with the speed at which electrons orbit in more massive atoms. The electrons in lower orbits move at almost relativistic speeds; therefore, they have greater mass and the radii of their orbits shrink. This is called a "relativistic contraction" of an orbital. Since orbitals of the same type essentially sit on top of one another, contraction of inner orbitals shrinks outer orbitals as well, bringing them closer to the atom's nucleus. Closer to the nucleus = more energy required to remove the electron. That along with a filled $d$ subshell would make an atom less reactive and have a higher ionization energy. I think.

I actually learned about orbital contraction on [url=http://www.madsci.org/posts/archives/may97/862179191.Ch.r.html]this site[/url] while trying to find out why mercury was a liquid. I thought it might apply here, but I'm not completely sure. There's probably a much better, simpler (and more correct) explanation.

In general the ionization energy increases along a period, the only exceptions being when we go from group 2 to group 13 and from group 15 to group 16.
Along a period as we go from element to element, each new electron will occupy a orbital in the same shell, and so its distance from the nucleus will not be very different compared with the previous element. However, as we go to the right the nuclear charge increases, and so the net effect is that the electrons are being progressively more tightly held by the nucleus. This has the effect of increase the ionization energy.


What was your score? (mine was 172) :ewpu:


algebra
partial fractions
calculus
calculus computations

using laplace transforms for (1) find the general or particular solution

(1) x'+2x=4t^2

iv done this problem i wan check if my answer is correct can some one please help thanks

Why don't you write up your solution, and then other people can tell you whether they think it is correct.

Laplace transforms require initial conditions. Since we are not given initial conditions, we must assume arbitrary initial conditions. Let $x(0)=x_{0}.$ Now use $X = X(s)$ for the Laplace transform of $x(t).$

$sX-x_{0}+2X=\frac{8}{s^{3}}$

$(s+2)X=\frac{8}{s^{3}}+x_{0}$

$X=\frac{8}{s^{3}(s+2)}+\frac{x_{0}}{s+2}$

We now have a partial fractions problem. I'll do that off-line and report the results:

$\frac{8}{s^{3}(s+2)}=\frac{-1}{s+2}+\frac1s-\frac2{s^{2}}+\frac4{s^{3}}$

$X=\frac4{s^{3}}-\frac2{s^{2}}+\frac1s+\frac{-1+x_{0}}{x+2}$

This leaves the following solution to the differential equation:

$x(t)=2t^{2}-2t+1+(-1+x_{0})e^{-2t}.$

thanks heaps kent i got a similar answer i know now where i made my mistake cheers. :D


We have n balls, numbered 1 to n, and n-1 bags, numbered 1 to n-1. We want to place the balls in the bags such that:
the first bag contains at least 2 balls; the first 2 bags contain at least 3 balls; ... ; the first k bags contain at least k+1 balls; ... .
In how many ways can we do this? (2^2 for n = 3 and 3^3 for n = 4, ...)

This is closely related to parking functions, i.e., sequences $ (a_1, \ldots, a_n)$ such that the increasing rearrangement $ (b_1, \ldots, b_n)$ of the sequence satisfies $ b_i \leq i$.  Enumerating parking functions is fairly nontrivial, if I remember correctly, so I suspect this will be so as well.

I tried looking at two cases: (i) no balls in bag n-1 : easy; (ii) 1 ball in  bag n-1 : seems very tricky.

Have tried setting up bijections between (non-defective and defective) parking functions and solutions to the balls/bags problem for n = 2, 3, 4, 5 , 6; so far without success. Also looking at trees and at partitions of integers but not making any real progress. Help!

One possible route: the number of $ n$-parking functions with $ k$ 1s is $ \binom{n \minus{} 1}{k \minus{} 1}n^{n \minus{} k}$, and the number of your object with $ n$ boxes and $ n \plus{} 1$ balls with $ k \plus{} 1$ 1s is just $ (n \plus{} 1)/(k \plus{} 1)$ times this.  (The former fact one gets by the same sorts of methods that one counts all the p.f.s; for the latter, just consider a map that adds 1s in all possible places -- it takes each p.f. to $ n \plus{} 1$ of your objects, and each of your objects gets hit $ k \plus{} 1$ times.)  So then your problem is to compute the sum $ \sum_{k \equal{} 1}^n \binom{n \minus{} 1}{k \minus{} 1}n^{n \minus{} k}\cdot \frac {n \plus{} 1}{k \plus{} 1}$.  Don't promise this is any easier, though.

Thanks JBL.  I looked up parking functions on the internet and have found that for n spaces and n cars there can be k = 0, 1, 2, ..., n-1 cars which fail to park. For k = 0, there are (n+1)^(n-1) parking functions. I cannot show that for k = 0, 1, 2, ..., n-1, there are n^n parking functions altogether, as suggested by a table for n = 1 to 7. As far as I remember, those with k=0 were called nondefective parking functions (and the others defective). I am not sure what you mean by "n-parking functions with k 1s" nor why there should be (n-1)C(k-1)xn^(n-k) of them, nor what "adds 1 in all possible places" means. This may be because I cannot see how the cars and spaces relate to balls and bags under the conditions of the problem. (I have no idea how to compute the final expression, either.) I'm not finding this easy, but don't want to give up.

Sorry, let me elaborate.  By "parking function" I have meant "non-defective parking function," but what I really meant was the thing I defined in my first post on this thread.  That there are $ n^n$ total (defective and non-defective) parking functions is trivial: there are $ n$ cars, and each one must choose one of $ n$ spots.  That there are $ (n \plus{} 1)^{n \minus{} 1}$ non-defective parking functions is nontrivial, and the statement that I gave concerning the number of p.f.s with $ k$ 1s is similarly nontrivial.

The objects you've described are in bijection with positive integer sequences $ (a_1, \ldots, a_n)$ such that in the increasing rearrangement $ (b_1, \ldots, b_n)$ we have $ b_2 \leq 1$, $ b_3 \leq 2$, ..., $ b_n \leq n \minus{} 1$.  (The increasing rearrangement of a sequence is the sequence with the same entries such that the smallest entry is first, the next-smallest entry is second, etc.)  The bijection is to set $ a_i$ equal to the number of the bag into which ball $ i$ is placed.

Does this clarify everything I've said?

Thanks very much. That makes much more sense to me. Unfortunately I still have no idea what the k 1s and the k+1 1s in the first sentence of your previous post are and how they fit into the problem. Besides that, it is beginning to become clearer. Thanks.

Arrange the bags in a circle and place the balls in them at random, in (n-1)^n ways. There is one and only one starting place around the circle which gives an allowed configuration. So there are ((n-1)^n)/(n-1) = (n-1)^(n-1) ways for the original problem. Any neat way of proving this?

Uniqueness is easy (in retrospect): if there were two possible starting positions on the circle then from one of them to just before the other there would be more balls than bags (by the given conditon) and from the other to just before the first starting position there would also be more balls than bags, leading to a contradiction. Is there a neat way of proving existence?

Nice ideas, AndrewTom.
I think existence can be proved by the slightly stronger result in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=963151870&t=79654[/url]

Edit: Oops, this result isn't sufficient :oops:

check out the last problem in PSS (Enumerative combinatorics) (Problem solving stratgies)
observe that a sequence of car parkings is fovourable if and only if the parking one has atleast 1 person whose it is a fovourite and so on therefpore the answer is $ (n \plus{} 1)^{n \minus{} 1}$
EDIT: JBL already wrote this!

Thanks, I knew that. I'm now trying to find a short, neat way of establishing existence, which is obvious if there are no empty bags: just start at the bag with 2 balls and move around the circle. Seems more difficult to get a neat idea for the case where the number of empty bags is greater than zero. Any ideas?

Let's take one of the bags with >1 balls and 'merge' it with the subsequent bag. Any sequence of bags that works with this merged bag also works with the bags separate. Now we've reduced the existence problem to the case with n-2 bags, etc. to one bag.

Does this not give n balls in n-2 bags and similar situations?

Upon merging, you can remove a ball from one of the two bags. Essentially, we're viewing each bag as having a value of $ b_i \equal{}$ number of balls in it - 1 and trying to find a sequence such that each subsequence beginning with the first bag has non-negative sum. So when we merge two bags $ i$ and $ j$, the final bag has value $ b_i \plus{} b_j$. The argument is that we can always find a subsequence with non-negative sum, and consider this subsequence as a single term.

Thanks. I realised that after I'd submitted the previous post but have only now been able to get back to a computer.


Give x \geq y \geq z \geq 0 and y 2  \geq xz.
Prove that:
x 2 y/z+y 2 z/x+z 2 x/y \geq 2(x 2 +y 2 +z 2 )-xy-yz-xz.

Here is my solution
     WLG, we may assume xyz=1
=> x>=y>=1>=z.
And the inequality is equivalent to
       f(x,y,z)= x^3y^2+y^3z^2+z^3x^2+xy+yz+zx - 2(x^2+y^2+z^2)>=0.
    It is easy to show that f(x,y,z)>=f(y,y,z).
    Now we just have an inequality having 2 variables y,z such that y^2*z=1 and that is easy now. :D

What a nice solution! :) My solution is different from yours.I changed it to:
(y-z)(x-z) 2 /z+(z-x)(y-x) 2 /x+(x-y)(y-z) 2 /y \geq 0.(*)
And I let A=(y-z)(x-z)/z-(x-y) 2 /x.
             B=(y-z) 2 /y-(x-y)(x-z)/x.
Then xzA+xyB=2(x-z)(y 2 -xz) \geq 0.
So A or B  \geq 0.For example A \geq 0.Then
(*)<=>(x-z)A+(x-y)(y-z) 2 /y \geq 0. :D

Your solution is nice too, but a little bit difficult to think it out ! :D

Treegoner's solution is based on the inequality

f(x,y,z) \geq f(y,y,z)

but I cannot prove this one.

I would like to have a help, please.

Thank you very much.


analytic geometry
calculus
derivative
limit
graphing lines
slope
function

I am so confused on this problem. Please help!

Given $ f(x) \equal{} x^3$, and its inverse $ f^{ \minus{} 1} \equal{} x^{\frac {1}{3}}$. Recall that the graph of $ f^{ \minus{} 1}$ can be obtained from the graph of $ f$ by reflecting the graph of $ f$ across the line $ y \equal{} x$. That is the graph of $ f^{ \minus{} 1}$ can be obtained from the graph of $ f$ by interchanging the first and second coordinate of each point.

a) Use the definition of the derivative to show that $ f'(x) \equal{} 3x^2$.

b) Find an equation of the line tangent to the graph of $ f$ when $ x\equal{}2$.

c) Use the ideas presented in this problem to find the equation of the line tangent to the graph $ y\equal{}x^{\frac{1}{3}}$ when $ x\equal{}8$.

d) Using the same type of argument you used in part (c), show that if $ g(x)\equal{}x^{\frac{1}{3}}$, then $ g$ is not differentiable at $ x\equal{}0$.

a)$ f'(x) \equal{} \lim_{x\to 0} \frac {f(x \plus{} \Delta x) \minus{} f(x)}{\Delta x} \equal{} \lim_{x \to 0}\frac {(x \plus{} \Delta x)^{3} \minus{} x^{3}}{\Delta x} \equal{} \lim_{x \to 0} \frac {x^{3} \plus{} 3x^{2}\Delta x \plus{} 3x (\Delta x)^{2} \plus{} (\Delta x)^{3} \minus{} x^{3}}{\Delta x} \equal{} 3x^{2}$

b) If $ x \equal{} 2$, then $ f(2) \equal{} 2^{3} \equal{} 8$ so that the tangent line passes through the point (2, 8).

The slope of the line tangent to the graph at $ x \equal{} 2$ is $ 3(2^{2}) \equal{} 12$.
Thus, an equation of the tangent line is $ f(x) \minus{} 8 \equal{} 12(x \minus{} 2)$ or $ f(x) \equal{} 12x \minus{} 16$.

[hide="a"]
Definition of the derivative:

$ f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}{h}$

$ f(x) = x^3$
$ f'(x) = \lim_{h \to 0} \frac {(x + h)^3 - x^3}{h}$
$ = \lim_{h \to 0} \frac {(x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h}$
$ = \lim_{h \to 0} \frac {h(3x^2 + 3xh + h^2)}{h}$
$ = \lim_{h \to 0} 3x^2 + 3xh + h^2$

As $ h$ goes to zero, those last two terms go to zero, and we're left with $ 3x^2$.
[/hide]
[hide="b"]
We use point slope form here.
$ f(2) = 8$, so the point we're dealing with is (2, 8).
Basically, point slope form is
$ Y - y_1 = f'(x)\cdot(X - x_1)$
So we have $ Y - 8 = f'(2)(X - 2)$
Because we just calculated the derivative, we know that $ f'(2) = 3(2)^2 = 12$.
So our equation just boils down to $ Y - 8 = 12(X - 2) \longrightarrow Y = 12x - 16$
[/hide]
[hide="c"]
Same deal. We need to take the derivative of$ x^{\frac {1}{3}}$, and use point slope again. Well, there's a nice fact that the derivative of any function$ x^K$is ALWAYS equal to$ K\cdot x^{K - 1}$So the derivative of$ x^{\frac {1}{3}}$is just$ \frac {1}{3} x^{ - \frac {2}{3}}$So we evaluate$ f(8)$and get$ 2$. Now we spam point slope form again.$ Y - 2 = \frac {1}{3} (8)^{ - \frac {2}{3}}(X - 8)$
You can slog through the algebra, but that's the basic form.
[/hide]
[hide="d"]
To do this, we need to show that the limit of the derivative from both sides is DIFFERENT.
So we know the derivative is
$ \frac {1}{3} x^{ - \frac {2}{3}}$
This comes down to showing that the limit as $ x \to 0^{ + }$ is different from the limit as $ x \to 0^{ - }$.
[/hide]

For (c) and (d), let's not slog through the algebra.

We have already computed that the line tangent to $ y\equal{}x^3$ at the point $ (2,8)$ is $ y\equal{}12x\minus{}16$

Now, all we have to do is interchange the roles of $ x$ and $ y.$ This gives us that the line tangent to $ x\equal{}y^3$ at the point $ (8,2)$ is $ x\equal{}12y\minus{}16.$ We could then do a little solving for $ y$ in terms of $ x,$ getting that the line tangent to $ y\equal{}x^{1/3}$ at the point $ (8,2)$ is $ y\equal{}\frac1{12}x\plus{}\frac43.$

Now we try the same ideas for (d). The line tangent to $ y\equal{}x^3$ at the origin $ (0,0)$ is the horizontal line $ y\equal{}0.$ If we now interchange $ x$ and $ y,$ the origin is still the origin, but the horizontal line has now become a vertical line. A vertical line has no slope, so the derivative of $ f^{\minus{}1}(x)\equal{}x^{1/3}$ does not exist when $ x\equal{}0.$


quadratics
arithmetic sequence
algebra
system of equations
number theory unsolved
number theory

Let $ \mathbb{N}$ be the set of all positive integers

   Solve the following equation in $ \mathbb{N}$ :

        $ x^2 y^2 \ \equal{} \ z^2 ( z^2 \minus{} x^2 \minus{} y^2 )$

[i]Please , don't spam

    Can anybody post a solution :maybe: [/i]

Look for primitive solutions with $ gcd(x,y,z) \equal{} 1$ because all solutions are multiples of such primitive solutions.

Take $ a \equal{} x^2, b \equal{} y^2, c \equal{} z^2 \implies ab \equal{} c(c \minus{} a \minus{} b) \implies a \equal{} \frac{c(c\minus{}b)}{b\plus{}c}$

If $ p$ is a prime factor of $ b\plus{}c$ then either $ p \mid c$ or $ p \mid c \minus{} b$
if $ p \mid c$ then $ p \mid b$, and if $ p \mid c \minus{} b$ then $ p \mid 2c \implies p \equal{} 2$ or $ p \mid c$ and $ p \mid b$
if $ p > 2$ and $ b\plus{}c$ has an odd number of times the prime factor $ p$ then we also get $ p \mid a$
but we are looking for primitive solutions so we can assume that $ b\plus{}c$ is either a sqaure or twice a square
case 1: $ b\plus{}c \equal{} s^2, c \equal{} rs, b \equal{} s(s \minus{} r), a \equal{} r(2r \minus{} s)$
case 2: $ b\plus{}c \equal{} 2s^2, c \equal{} rs, b \equal{} s(2s \minus{} r), a \equal{}r(r \minus{} s)$

Case 1 and 2 are just the same with $ r$ exchanged for $ s$ and $ a$ exchanged for $ b$ so we need only consider case 1.
We can also assume $ \gcd(r,s) \equal{} 1$ since we require primitve solutions.
Then $ z^2 \equal{} c \equal{} rs$ so $ r \equal{} t^2$ and $ s \equal{} u^2$ are squares
To make $ b$ and $ c$ squares we also need $ s \minus{} r \equal{} v^2$ and $ 2r \minus{} s \equal{}  w^2$

So now we have a system of equations $ u^2 \equal{} t^2 \plus{} v^2$ and $ t^2 \equal{} v^2 \plus{} w^2$
In other words $ w^2, t^2, u^2$ are three squares in arithmetic progression where the common difference is also a square.

We find that $ v$ must be even so as a pythagorean triple we have $ 2t \equal{} d^2 \plus{} e^2, 2v \equal{} d^2 \minus{} e^2, w \equal{} de$
Now we require $ 4u^2 \equal{} (d^2 \minus{} e^2)^2 \plus{} (d^2 \plus{} e^2)^2 \implies 2u^2 \equal{} d^4 \plus{} e^4$

This last equation only has solutions with $ d \equal{} e$ which produces invalid solutions of the original equation.

For a proof of that I refer to   http://www.mathreference.com/num-zext,442.html

[i]My friend give me a very good solution as follow :

            [u] Lemma[/u] : the equation $ x^4 \ \minus{} \ y^4 \ \equal{} \ z^2$ have no positive integer root

( I don't have time to give proof of the lemma , it's very well - known  :blush: )

   Rewrite the equation : $ (z^2)^2 \minus{} (x^2 \plus{} y^2 ) z^2 \minus{} x^2 y^2 \equal{} 0 \ \ (1)$

  We will consider $ (1)$ as a quadratic equation with $ z^2$ is the variable

   Due to the fact that $ (1)$ have at least one root which is an integer

  We have $ \triangle \ \equal{} \ (x^2 \plus{} y^2 )^2 \plus{} 4x^2 y^2$ is a perfect square

        We can easily eliminate the case $ x \equal{} y$ , if $ x \equal{} y$ then  $ \triangle \ \equal{} \ 8x^2 y^2 \rightarrow 2 \not | v_2 ( \triangle )$ , contradiction

   Denote $ a \ \equal{} \ x^2 \plus{} y^2 \ ; \ b \equal{} 2xy \rightarrow  a;b \ \in  \ \mathbb{N}$

$ a^4 \minus{} b^4 \equal{} (a^2 \minus{} b^2 )(a^2 \plus{} b^2 ) \equal{} ( (x^2 \plus{} y^2 )^2 \plus{} 4x^2 y^2 ) ( (x^2 \plus{} y^2 )^2 \minus{} 4x^2 y^2 ) \equal{} (x^2 \minus{} y^2 )^2 \triangle$

  So $ a^4 \minus{} b^4$ is the square of a positive integer , contradiction[/i]


function
trigonometry
algebra unsolved
algebra

Find the sum of the maximum and minimum value of f(x) in the function:

$ f(x) \equal{} \frac{1}{1\plus{}(2\cos x \minus{} 4\sin x)^2}$

$ |2\cos x \minus{} 4\sin x|\leq 2\sqrt {5}\Longleftrightarrow 0\leq (2\cos x \minus{} 4\sin x)^2\leq 20\ \therefore \frac {1}{21}\leq f(x)\leq 1$.

The answer is $ \frac {1}{21} \plus{} 1 \equal{} \frac {22}{21}$.

Note that $ A \cos x \plus{} B \sin x \equal{} \left< A, B \right> \cdot \left< \cos x, \sin x \right> \equal{} \sqrt{A^2 \plus{} B^2} \cos \left( x \minus{} \arctan \frac{B}{A} \right)$.  Then $ f(x)$ takes a rather simple form.


trigonometry
calculus
calculus computations

Let $ \{x_n\}_{n}^{\infty}$  be increasing sequence of positive solutions of equation

$ x\equal{}\tan {\sqrt x}$.

Investigate behavior of numerical series
\[ \sum\limits_{n\equal{}1}^{\infty}\frac{1}{x_n}\]

It's not exactly the same thing, but [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=182057]this thread[/url] featured the sum of the inverse squares of the roots of $ x\equal{}\tan x$.

Does this series converge? Of course; $ x_n$ is approximately $ (n\minus{}\frac12)\pi$, and it converges by limit comparison to the sum of the reciprocals of the squares.
Can we calculate the sum? Well, trying the methods in the linked thread looks nice for a while... and then you realize that it's really important that we were working with odd functions. Roots of $ \tan x \equal{}f(x)$ with $ f$ odd can be handled neatly; the same with even $ f$ leaves us with all sorts of trouble.


If the probability of having my chapter comp. on Thursday is 1/2, and the probability of me owning the competition is 99.9999%, than what's the probability that I will own on Saturday?

NOTE!!!!!!!!!
This is a trick question so watch out??????????????

[hide="Answer"]You can't determine an answer[/hide]

[hide="answer"]50%

it's either you win or lose
[/hide]

Why don't a few people determine an answer before I post it.  Right now 50% of the people who've answered have got right.

[hide]Its $\frac{1}{2}*\frac{999999}{1000000}$, which rounds to $\frac{1}{2}$ because I'm too lazy to calculate it.[/hide]

[hide]Its ???
Probability of happening times the probability of winning times maybe the probability of actually attending(which isn't given)[/hide]

Oh right, you have to count the attendnce factor in...

or that it could be on Sunday
or Monday
or Tuesday
or Wednesday
or Thur...

Fine: List of as many things as I can think of:

You don't have to attend
It might happen on another day
Graders might be biased
You might get brain freeze or be otherwise incapable of doing it well
You might have misplaced the decimal or miswrote the fraction

I know, I'm mean :D

Feel free to add to the list.

i saw on a different forum that Thursday [i]is[/i] the snow date :|

I think people are getting it now, it's undeterminable, until I give the rest of the info, which I don't need to, because it's not part of the problem. :)


Hi y'all
I have a problem here...
if the problem gives you a day of the week and a date (ex: Friday, April 13, 2007)
and then makes you find the day of the week x amt of time from that date (ex: Six months and twenty-two days), what is a fast and time-efficient way of solving those?

 :huh: anyone? :maybe:

Thanks in advance

--
Blah is my name, Blah is my game

there are probably several different approaches to these types of questions; I'm pretty sure mine's a slow way-
day of the year today
day of the year on the problem
so if today was January 1st,2345 and they want you to find October 24,2345, then
Today=1st
October 24= # of days in January+february+march+april+may+june+july+august+september+24
               = 31+28+31+30+31+30+31+31+30+24
               =297
297-1= 296 days until october 24! :lol:

[quote="mathcrazed"]depends on how many days are in a month you know? Just use mods.  And normally, they don't give you those kinds of problems, they give you problems like this: If today is Thursday, what day of the week will it be in $ 2^{2006}$ days?
I think. lol[/quote]
On question #40 on MMPC, they asked what day is $ 2005^{2005}$ days after Thursday.

Ok... well, you would use Fermat's little theorem or Euler's phi/totient function. But because this is in mathcounts, can't explain that to you.

Another way to do that problem is to note that $ 2005 \equiv 4 \mod 7$.
And then see that $ 4^{3} \equiv 1 \mod 7$. There you go :).

Uh, I got that right.

I'm just saying those are the usual calendar problems I'm used to.I don't think they would hae many calendar problems in Mathcounts.

Ya... that's what I was kinda saying. They always give you some exponent to use for calendar problems, but I mean, once you see them once, it's easy to do every other problem like that, so Mathcounts rarely uses them./


calculus
derivative
real analysis
real analysis unsolved

If $ f(x,y) \in C^{(1)}(\mathbb{R}^2)$

can we have: $ \frac{\partial f}{\partial x} \in C(\mathbb{R}^2)$ ? :roll:

Yes.  because $ \frac {\partial f}{\partial x}(x,y) \equal{} Df(x,y)(1,0)$


number theory
least common multiple
function
prime factorization

[b]Prove[/b]: If $ n\equal{}p_{1}^{a_{1}}p_{2}^{a_{2}}\dots p_{k}^{a_{k}}$ is a prime decomposition of $ n$, there are $ (2a_{1}\plus{}1)(2a_{2}\plus{}1)\dots (2a_{k}\plus{}1)$ distinct pairs of ordered positive integers $ (a,b)$ with $ \text{lcm}(a,b)\equal{}n$.

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=141316 has some useful background on a general approach for thinking about problems of this sort.

[hide="Solution"] We will be appealing to Lemma 2, which states that

$ e_{p}(\text{lcm}(a, b) ) \equal{}\text{max}(e_{p}(a), e_{p}(b) )$

For a given prime $ p_{i}$ we want to find how many ordered pairs $ (e_{p_{i}}(a), e_{p_{i}}(b))$ exist such that the maximum of those two quantities is $ a_{i}$.  Clearly when $ e_{p_{i}}(a) \equal{} p_{i}$ the other exponent can take values $ 0, 1, 2, ... p_{i}$, and clearly when $ e_{p_{i}}(b) \equal{} p_{i}$ the other exponent can take values $ 0, 1, 2, ... p_{i}\minus{}1$ (since we have already counted the ordered pair $ (p_{i}, p_{i})$).  There are therefore $ 2p_{i}\plus{}1$ possibilities for each $ p_{i}$.

Across each prime in the prime factorization of $ n$, then, the total number of distinct combinations of prime factorizations of $ a, b$ is given by the desired result.  QED. [/hide]

I'm sorry if I don't understand, but what is the [b]max[/b] function?

It returns the maximum, i.e. the largest, of its arguments.

Can you please give me an example? Is it like inputting $ \infty$ into $ x$?

For example, $ \text{max}(2, 3) \equal{} 3$ and $ \text{max}(100, 4) \equal{} 100$.  Generally (for two arguments),

$ \text{max}(x, y) \equal{}\begin{cases}x &\text{ if }x\ge y\\ y &\text{ if }x < y\\ \end{cases}$


If a system of linear equations has the same number of equations as the number of variable, when can we say that there is a unique solution?


Find>: [1]How many solutions has $x + y + z = 200$$x,y,z \ge 15$?
          [2]If $m=2^4.3^3.5^2$,how many positive integers numbers ($<m$) are prime with m?

For (1) I got $\frac{156\times157}{2}$. First let z=15, then there are 156 combinations by counting, if z=16, then there are 155 combinations, all the way until z=170, for which there is 1 combination. Thus the sum of the first 156 consecutive numbers (combinations) is $\frac{156(156+1)}{2}=12246$.

For the first one, think of it as balls and urns, so you already have to have at least 15 balls in each urn, so drop those in first, and you're left with 155 balls left to distribute between 3 urns anyway you want, so then the number of ways is $\binom{157}{2}$.

Woops, I found the number that weren't relatively prime with that number. I subtracted it from 10800 and got 2640 instead.

btw, Altheman, can you explain your method?

If thats correct, I wonder what is wrong with my solution?

Using the fact that phi is a multiplicative function, you could just find the phi of 2^4, 3^3 and 5^2 and then multiply them to get the result, which turns out to be 8*18*20.

So my answer is 2 times what altheman got.


-> telescopic product :P
Let $\frac12 \frac34 \cdots =S$
And consider $\frac23 \frac45 \cdots =T$
Then $ST=\frac{1}{100}$
But since $S<T$ the result follows :D