Number 2:
The sum of three numbers is 165. When the smallest number is multiplied by 7, the result is n. The value n is obtained by subtracting 9 from the largest number. This number n also results by adding 9 to the third number. What is the product of the three numbers?

Number 3:
An Internet service provider allows a certain number of free hours each month and then charges for each additional hour used. Wells, Ted and Vino each have seperate accounts. This month the total hours used by Wells and Ted was 105, and each used all of their free hours. Their total cost was 10 dollars.
Vino used 105 hours by himself and had to pay 26 dollars. What is the number of cents charged for each extra hour?

Number 8:
A turn consists of rolling a standard die and tossing a fair coin. The game is won when the die shows a 1 or a 6 and the coin shows heads. What is the probability the game will be won before the fourth turn? Express your answer as a common fraction.

Show all work

Edit:No need to do number 3
1993-1994 National Team Round
Number 5:
Four different digits are selected at random from the digits 1 through 9. S is the sum of all possible four digit numbers that can be created by using these four digits. F is the greatest common factor of all such sums. Find F

[hide="Number 2"]
Call the smallest number a, the second smallest b, and the largest c.

$7a = b+9 = c-9$
$7a = b+9$ so $b = 7a-9$
$7a = c-9$ so $c = 7a+9$

$a+b+c = 165$, substituting the above values $a+7a-9+7a+9 = 165$, so $15a = 165$ and $a = 11$. Then $b = 68$ and $c = 86$ Multiplying these 3 numbers gives $64328$.
[/hide]

Ubemaya did Number 2, and you said not to do 3, so I'll do the other two.

[hide="Number 8"]The probability of winning on a given turn is $\frac{1}{3}\times \frac{1}{2}= \frac{1}{6}$ and the probability of not winning is $\frac{5}{6}$.

There are three cases: The game is won on the first turn, the first turn is not won but the second is, or the first two turns are not won but the third one is.

$\frac{1}{6}+(\frac{5}{6})(\frac{1}{6})+(\frac{5}{6})(\frac{5}{6})(\frac{1}{6}) = \frac{91}{216}$
(Yes I know my parentheses look wierd  :oops:  )[/hide]
[hide="Number 5"]
Say the four digits chosen are A, B, C, and D.

There are 24 possible four digit numbers.
6 numbers have A in the thousands place, 6 have B in the thousands place, and the same for C and D.
The same is for the hundreds, tens, and ones places.
After little calculation, we find that S = 6666(A + B + C + D).
No matter what A, B, C, and D are, S is divisible by [b]6666[/b].[/hide]

#2[hide]

Let the smallest number be $x$, the second largest be $y$, and the largest be $z$. We have

$7x=n$
$z-9=n=7x$, which implies $z=7x+9$
$y+9=n=7x$, which implies $y=7x-9$

We also have $x+y+z = x+7x+9+7x-9=15x=165$. Thus $x=11$. Substituting into what we have above, $y = 68$ and $z = 86$. Thus the product is $64328$.[/hide]

#3[hide]

Let $c$ be the cost per extra hour, in cents, $n$ the provider's limit in hours, and $w$ be the number of total hours that Wells uses. Then, we have

$c(105-n) = 2600$
$c(w-n)+c(105-w-n) = 1000$
Simplifying the second equation, we have
$c(w-n+105-w-n) = c(105-2n) = 1000$.

We now divide the first equation by the second to get
$\frac{105-n}{105-2n}= \frac{13}{5}$. Solving the proportion, we have
$525-5n = 1365-26n$
$21n=840$
$n=40$

Substituting into our above equations, we get $c=40$. Thus the company charges 40 cents per hour over the limit.
[/hide]

#8[hide]

The probability of winning on a give turn is $\frac{1}{6}$.

Case 1:The game is won on the first turn.
Clearly, this is $\frac{1}{6}$.

Case 2:The game is won on the second turn.
We have to make sure the game is not won on the first turn. Thus the probability is $\frac{5}{6}* \frac{1}{6}= \frac{5}{36}$

Case 3:The game is won on the third turn.
Similar to case 2, we make sure the game is not on on the first or second turns. The probability is $\frac{5}{6}* \frac{5}{6}* \frac{1}{6}= \frac{25}{216}$

Adding our three cases together, we get a total probability of $\frac{1}{6}+\frac{5}{36}+\frac{25}{216}= \frac{91}{216}$.[/hide]

#5[hide]

Let $abcd$ be any 4 digit number. Writing this algebraically, we get $1000a+100b+10c+d$. There are 24 permutations of this number, and each of $a, b, c, d$ has a coefficient of 1000 6 times, 100 6 times, 10 6 times, and 1 6 times. Thus the sum of everything, F = $6666a+6666b+6666c+6666d$, and the GCF of all such F is clearly, $6666$.[/hide]

[hide="solutions for 2001"][hide="#2"]$x+y+z=165$
$7x=n=y-9=z+9$
$2z+18+x=165$
$x=\frac{n}{7}$
$x=\frac{z+9}{7}$
$2z+18+\frac{z+9}{7}=165$
$\frac{15z}{7}+\frac{135}{7}=165$
$15z+135=1155$
$15z=1020$
$z=68$
$86=y$
$x=11$
$\boxed{xyz=64328}$[/hide]

[hide="#8"]The probability that the game will end in 1 turn is $1-\frac{1}{6}=\frac{5}{6}$
The probability therefore that the game will end in the fourth turn is
$\frac{5^{3}}{6^{4}}=\boxed{\frac{125}{1296}}$[/hide][/hide]
[/hide][/hide]

Kiran, You need to find thr probability on the FOURTH time.

no, you don't

The question is asking what is the probability of winning BEFORE the fourth turn, so what is the probability of winning on the first, second, or third turns.

[hide="Number 8"]
There's a $\frac{1}{6}$ chance of winning each turn, so there is a $\frac{5}{6}$ chance of not winning each turn. So the probability that you will not win after 3 turns is $\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}$, which is $\frac{125}{216}$. Then the probably that you will win by 3 turns is $1-\frac{125}{216}$, which is $\frac{91}{216}$.
[/hide]

[quote="nutz_for2.718281828"]Kiran, You need to find thr probability on the FOURTH time.[/quote]We're talking about before the fourth turn, which would mean not including it.

aaw... I need a new computer screen... :(

Here are the answers:
2001 National Team Round
Number 2: 64328
Number 3: 40 cents
Number 8: 91/216


[url=http://www.kongregate.com/games/Jayisgames/lock-n-roll]Lock-N-Roll[/url] is a game of both strategy and luck; the rules and gameplay are extremely simple, yet playing a "perfect" game is nigh impossible because the strategy is so complex.

Go ahead and try it! For bonus points, describe the optimal strategy or create a program that takes an input (current board and dice/jokers) and outputs the "best" way to place all the dice, where a game played with optimal strategy will score as high as possible.

My highest score was about 2500-something. XD

Oh, wait, now it's above 4000. I don't remember exactly what, though.

what' happenns after you fill up all the spaces?

Play a joker if you have one, or game over


Jimi needs to be home by 6 o'clock. If he travels at an average rate of 15 mph, he'll arrive one hour early, but if he travels at an average rate of 10 mph, he'll arrive one hour late. How many miles away is his home?


λοιπον μια ασκηση  που μου αρεσε
βλεπουμε ενα ουρανοξυστη που εχει πολλα σκαλια.αν τα ανεβουμε 2-2 περρισσευει ενα σκαλι,αν τα ανεβουμε 3-3 περισσευουν 2 ,...,αν τα ανεβουμε 10-10 περισσευουν 9 σκαλια.ποσα  σκαλια εχει ο ουρανοξυστης?

Εστω χ τα σκαλια.
Ειναι χ ισουπολοιπο-1 mod2
           ισουπολοιπο-1 mod3
...........................................
           ισουπολοιπο-1 mod10
 kai αρα θα ειναι χ ισουπολοιπο -1 mod(E.K.Π(1,2,3...,10))
δηλαδη χ=2520λ-1 με λ φυσικο....

 :D

[b]ΘΕΩΡΗΜΑ[/b] Το σύστημα των γραμμικών ισοτιμιών
                                   $x\equiv b_{1}(mod n_{1})$
                                   $x\equiv b_{2}(mod n_{2})$
                                               $...$
                                   $x\equiv b_{k}(mod n_{k})$
έχει λύση, αν και μόνο αν, $(n_{i},n_{j}))/b_{i}-b_{j}$  $\forall i,j$ με $i\neq j$.
Aν $x_{0}$ μια λύση του του συστήματος,τότε το σύνολο των λύσεων του είναι η κλάση του $x_{0}(mod [n_{1},n_{2},...,n_{k}])$.

[quote="silouan"]Γενίκευση του Κινέζικου θεωρήματος υπολοίπων  :wink:[/quote].
nai, akribos!

Το κινεζικο θεωρημα υπολοιπων ειχα +εγω στο μυαλο μου,οταν ελυνα την ασκηση,αλλα μια μισι η ωρα ποιος γραφει latex... :wink:

 :D

Ποιο πρακτικη? Τι εννοεις? :)

εννοω χωρις την εννοια τις ισοτιμιας. μαλλον ειναι πολυ απλη για αλλη λυση  :blush:

[quote="kostas"]ηλπιζα να δω αν υπαρχει μια ποιο πρακτικη λυση,αλλα οκ  :)[/quote]

Δινεις σε καθενα απο τα 9 αυτα παιδια,ενα συγκεκριμενο χρωμα και βαφουν μ αυτο ολα τα σκαλοπατια που ανεικουν στην <κατηγορια >τους...(π.χ ενα παιδι θα βαψει αυτα που περισσευουν ενα   μετρωντας ανα δυο,αλλο παιδι αυτα που περισσετουν 2 μετρωντας ανα τρια...κλπ..)
και μετα πας εσυ +βλεπεις σε ποιο σκαλοπατι υπαρχουν ολα τα χρωματα....
Βεβαια,πιστευω οτι η λυση με τις ισοτοιμιες συνεχιζει να ειναι πιο απλη ,γιατι σ αυτη την περιπτωση με τα χρωματα υπαρχουν καποιεσ αντικειμενικες δυσκολιες.....
Θα χρειαστεις να ανεβεις γυρω στα 2500  σκαλια για να δεις ολα τα χρωματα μαζι....!!!!!!!!!!!!!!!!!! :rotfl:

 :D

(πραγματι ειναι πιο πρακτικη η λυση του Ραλλη) χαχαχαχαχα............ :D  :rotfl:  :wink:

καθολου κακη ιδεα.για να καταλαβω  ,να ρωτησω κατι : ποιο παιδι θα βαψει περισσοτερα σκαλια?

to pedi 1 θα βαψει ολα τα πλακακια ισουπολοιπα -1 mod2
.to paidi 2 ua bayei ola ta plakakia isoypoloipa -1 mod3 k.o.k...
........................................................
η μπορουμε αντι γαι χρωματα(γιατι στο τελος θα ζαλιστεις +θα βλεπεις ολο ουρανια τοξα μπροστα σου.... :rotfl: )να γραφει το καθε παιδι ενα συγκεκριμενο αριθμο απο το 1 μεχρι το 9 και εσι να υαχνεισ το σκαλοπατι που να εχι και τα εννια ψηφια.........Υπαρχουν πολλεσ επιλογεσ......ορεξη να υπαρχει............. :rotfl:  :rotfl:

 :D

[quote="kostas"]ηλπιζα να δω αν υπαρχει μια ποιο πρακτικη λυση,αλλα οκ  :)[/quote]
Αν είχε ένα ακόμη σκαλοπάτι, τότε σε καμία διαίρεση δε θα είχαμε υπόλοιπο.

Δηλαδή το πλήθος των σκαλοπατιών θα ήταν πολλαπλάσιο όλων των αριθμών $1,2,...,10$, άρα και του $2520$ ... κλπ

πραγματι ενα σκαλοπατι και τα αλλαζει ολα. ομως αυτη η παρατηρηση του pontiou με κανει να σκεφτω αν ο αριθμος 2520 εχει καμια αλλη ιδιοτητα ( τουλαχιστον σε σχεση με τους αριθμους  1,2,...,10.


The sidelengths of a triangle are $a,b,c$.

(a) Prove that there is a triangle which has the sidelengths $\sqrt a,\sqrt b,\sqrt c$.

(b) Prove that $\displaystyle \sqrt{ab}+\sqrt{bc}+\sqrt{ca} \leq a+b+c < 2 \sqrt{ab} + 2 \sqrt{bc} + 2 \sqrt{ca}$.

(a) $a,b,c$ are sidelengths of a triangle iff $a < b+c$, $b < a+c$, $c < a+b$.
So we are to prove that $\sqrt{a} < \sqrt{b}+\sqrt{c}$ for all variables.

Because $b,c$ are sidelengths, their value has to be positive.
So I assume it will not shock you if I say that $0 < 2\sqrt{bc}$.
Nor will the following statement be of any surprising content: \[ b+c<b+c+2\sqrt{bc} \]
Both sides are positive, so we can take the square root: \[ \sqrt{b+c} < \sqrt{b}+\sqrt{c} \]
Now using the fact that $a,b,c$ are sidelengths: \[ \sqrt{a}<\sqrt{b+c}< \sqrt{b}+\sqrt{c} \]

Analogue for the other variables.

(b) The left inequality has no need of the fact that $a,b,c$ are sidelengths. Just use AM-GM and add up: \[ \frac{a+b}{2}+\frac{c+b}{2}+\frac{a+c}{2}\geq \sqrt{ab}+\sqrt{ac}+\sqrt{bc} \]

The right inequality:
\[ RHS = \sqrt{a}.\left(\sqrt{b}+\sqrt{c}\right) + \sqrt{b}.\displaystyle\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}.\left(\sqrt{a}+\sqrt{b}\right) > a + b + c \]
The last inequality uses lemma (a).
Done!  :)  :D

[hide=Solution]
For (a), note that \(\sqrt{b}+\sqrt{c}>\sqrt{b+c}\) by squaring.

Now, note that the first inequality is the standard \(x^2+y^2+z^2\geq yz+zx+xy\). This can be proven by any method including \(\frac{y^2+z^2}{2}\geq yz\), \(\frac{1}{2}\displaystyle\sum_{\text{cyc}}\left(y-z\right)^2\geq0\), or Cauchy-Schwarz on \(\left[y,z,x\right],\left[z,x,y\right]\). For the second inequality, note that there is a triangle with side lengths $\sqrt[4]{a}$, etc. (by applying (a) to the triangle created by (a)), let it have area $K$. Then Heron's Formula tells us that \[2\sqrt{bc}+2\sqrt{ca}+2\sqrt{ab}-a-b-c=16K^2>0.\qquad\blacksquare\]
[/hide]


group theory
superior algebra
superior algebra unsolved

let $o(x)$ denote the order of $x$. Then $\ker f = \{o(x)\cdot k|k\in\mathbb{Z}\}$.

[quote="liltubbs25"]ok guys, i dont have many more, and ill be out of yalls hair forever...

suppose that G is a multiplicative group and let x be a fixed element of G. let f:Z->G be defined by f(n)=x^n. Prove that f is a group homomorphism from the additive group Z to G. describe ker(f).

Z=integers[/quote]

well, what does homomorphism mean?  you need $ f(m+n) = f(m)f(n) $... and to check that is pretty easy (use the definition of $ f $... $ f(m+n) = x^{m+n} $, etc. etc.)

as for $ \ker(f) $... what's another way of saying that $ x^n = 1 $?  ($ 1 $ being the multiplicative identity)


inequalities
geometry
perimeter
rectangle
rhombus
geometry unsolved

Suppose $ \Gamma$ is a closed curve having intersections of all the four sides of a rectangle.

Show that the perimeter of $ \Gamma$ is greater or equal to twice the length of the diagonal of the rectangle.

The shorter closed curve that have intersections of all the four sides of a rectangle is a rhombus whose vertices coincides whith the sides' midpoints, the perimeter of the rhombus is equal to twice the length of the diagonal of the rectangle!


function
calculus
derivative
integration
rate problems
calculus computations

Hey guys, I need a little insight on this problem:

Concert tickets went on sale at noon (t=0) and were sold out within 9 hours.  The number of people wating in line to purchase tickets at time $ t$ is modeled by a twice-differentiable function $ L$ for $ 0 \leq t \leq 9$.  The values of $ L(t)$ at various times $ t$ are shown:

$ L(0) \equal{} 120$
$ L(1) \equal{} 156$
$ L(3) \equal{} 176$
$ L(4) \equal{} 126$
$ L(7) \equal{} 150$
$ L(8) \equal{} 80$
$ L(9) \equal{} 0$

So at 0 hours there were 120 people in line, at 1 hour there were 156 people in line, etc...

(a) Use the data to estimate the rate at which the number of people waiting in line was changing at 5:30 PM (t = 5.5).

(b)  Use a trapezoidal sum with three subintervals to estimate the average number of people waiting in line during the first four hours the tickets were on sale.

I realize that part a is asking for the value of the derivative at 5.5 but I'm not sure on how to get the derivative.
And in part B i understand the trapezoidal sum but Im not sure how you would be able to set it up without knowing the value of the derivative at 1.33 and 2.66.

Thanks.

For a), it's asking for an estimate, not the exact derivative at $ t \equal{} 5.5$. And you can estimate that by finding the average rate of change with the given $ t$ in the interval.

As for b), you're estimating the average value of $ L$; you don't need to know the derivatives of $ L$ to get it when you have the values of $ L$ on hand. That means you're estimating [hide="this integral."]$ \frac {1}{4 \minus{} 0}\int_0^4 L(t)\,dt$[/hide]

Also, if this is for AP Calculus, don't forget to include the units of measurements in the final calculations.

[quote="math92"]
And in part B i understand the trapezoidal sum but Im not sure how you would be able to set it up without knowing the value of the derivative at 1.33 and 2.66.[/quote]
The question did not specify that you need to use subintervals of the same length, so use the ones whose endpoint values you know.

[quote="Patterns_34"]For a), it's asking for an estimate, not the exact derivative at $ t \equal{} 5.5$. And you can estimate that by finding the average rate of change with the given $ t$ in the interval.
[/quote]

Thanks!

Which interval would I use though?  (4, 7)? Thats where 5.5 falls between.
So I calculated the average rate of change to be 8.  Would that be correct?


geometry
3D geometry
Support
search
\/closed

Hey everyone.

I use a macbook, specifically Safari 6? idk, the new one as my browser.

when i do, i cant see the little cubes on each post to rate them.

anyone else have this problem? or know how to fix it?

when i use my PC, it works just fine.

thanks.

The forum doesn't support Safari.  (Do a search to find posts where Valentin says this.)

Confirmed- they don't show up in Safari. They do work in (the Mac version of) Firefox, which is what I use for the forum.

The last time I looked at Safari, the rating cubes weren't showing up in Firefox either; the fix for that may have been a browser-specific tweak.

It was a javascript error. Fixed in Firefox and IE, so now it's on Safari, nothing (simple) that I can do about it.

$ /text{yay! (even if i don't have Safari). yay!}$


Why a .doc? Attach a .gif / .jpg please.

sorry... it happens when you click the "Allowed Extensions and Sizes" on the add an attachment screen.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159876

Looks like this has been a ongoing problem.


inequalities
function
inequalities unsolved

For $a,b,c $ positive real numbers prove that

$ \frac{a^3-ba^2}{b^2} + \frac{b^3-b^2c}{c^2} + \frac{c^3-c^2a}{a^2} \geq0 $

I think, it is, of course,  Karamata's inequality for convex function $e^t$ and two sets (3x-2y,3y-2z,3z-2x) and (2x-y,2y-x,2x-z). Unfortunatelly, there are about 6 distinct cases of right arrangement of these sets. I checked only 3 of them - it is so boring.

inequalities like that one are always solvable with AM-GM: we have
$14\frac{a^3}{b^2}+3\frac{b^3}{c^2}+2\frac{c^3}{a^2}\geq 19\frac{a^2}{b}$
and symmetric. adding all these inequalities, we get the desired one.
maybe one would get a special prize for elegant solution at the IMO for that one :D

I think AM-GM is special case of Karamata's inequality. But your solution is nice.


It is not clear what is a general term of this series.
If it is $\frac{n^n}{n!}$ then the last term must be $\frac{100^{100}}{100!}$.
If it is $\frac{2^n}{n!}$ then the first term must be $\frac{2^1}{1!}$.

[quote="maxal"]It is not clear what is a general term of this series.
If it is $\frac{n^{n}}{n!}$ then the last term must be $\frac{100^{100}}{100!}$.
If it is $\frac{2^{n}}{n!}$ then the first term must be $\frac{2^{1}}{1!}$.[/quote]

The original problem is the latter and the answer is [b]6[/b]. :lol:


if p divides 30 and remainder is k. p is prime than prove k is also prime

i dont understand what you want to say. first you say that p divides 30 then you say the remainder is k? :huh:

[quote="Farenhajt"][quote="ZetaX"]If $p$ divides $30$, the remainder is $0$ ;)
Nad this surely isn't open as stated here...[/quote]

What is this - English Lessons Forum? C'mon, don't be our regular cynical ZetaX and give guy a break, it's obvious what he wanted to say - "when p is divided by 30, the remainder is k".[/quote]
Hehe i need some english lessons as well  :blush:

Suppose p=13, then p dividing 30 yields a remainder of 4. :(
Suppose p=11, then p dividing 30 yields a remainder of 8. :(

Suppose p=31, then k=1 and 1 is not prime  :ninja:
So p>31, and i found that those 31<p<97 do satisfy his original statement that 'if p is divided by 30 with a remainder of k and if p is prime, then k is also prime' but so far i have no proof yet. i think it has something to do with mod 3 on the remainder and that the remainder has an odd parity.

I hope this helps

PS [EDIT]: i think i know, this is a hint to help answer this question

[hide="Hint"]30=7+23=11+19=13+17[/hide]

10000th user, it should be understood as "p divided by 30", meaning $p : 30$, and not "p dividing 30" ($30 : p$). (For any $p>30$ we'd have $k=30$).

The proof is simple: If $p$ is prime less than 30, then $k=p$, hence $k$ is prime. If $p$ is prime over 30, then it's coprime with $30$, hence $k$ is too. The numbers under $30$ which are coprime with $2\cdot 3\cdot 5=30$ are all prime, with the exception of $1$. Hence the strictly correct statement should be "prove that $k$ is prime or is equal to one".

[hide="Solution"]
Let $p=30m+r$.
$p$ is prime, so $\gcd(30,r)=1$ (unless $m=0$, but in this case $r$ is clearly prime). The only numbers smaller than 30 that are coprime to 30 are $7,11,13,17,19,23,29$ which are all prime. Hence $r$ is prime.
[/hide]

Edit: Farenhajt beat me to it :(

[quote="Farenhajt"][quote="ZetaX"]If $p$ divides $30$, the remainder is $0$ ;)
Nad this surely isn't open as stated here...[/quote]

What is this - English Lessons Forum? C'mon, don't be our regular cynical ZetaX and give guy a break, it's obvious what he wanted to say - "when p is divided by 30, the remainder is k".[/quote]
You don't know how cynical I can be ;)
But the only other interpretation I had was that of 10000th user, and I also didn't see how this should work.

[quote]What is this - English Lessons Forum? C'mon, don't be our regular cynical ZetaX and give guy a break, it's obvious what he wanted to say - "when p is divided by 30, the remainder is k".

New post[/quote]

Okay, EVEN if that is what he wanted to say  (And I don't know how one, even a non cynical ZetaX type,  is supposed to equate "p is divided by 30, the remainder is k" with "if p divides 30 and remainder is k. " ) take p=61 , which is prime, the remainder $1$ is NOT. ...

Why not  just edit the post to make it clear ?

[quote="ZetaX"]
You don't know how cynical I can be ;)
[/quote]

We all do know that, don't worry :)

[quote="Gyan"]And I don't know how one, even a non cynical ZetaX type,  is supposed to equate "p is divided by 30, the remainder is k" with "if p divides 30 and remainder is k. "[/quote]

How? Just put yourself into the shoes of a person who speaks English poorly - I mean, use your intelligence to decipher it, you're a mathematician, for Christ's sake :)

As for that $1$, the case was already addressed to.


linear algebra
matrix
search
function
invariant
symmetry
linear algebra unsolved

Hi,
I am in search of a 'unique' matrix property which is true for a matrix filled with elements that follow some specific distribution function and that property does not hold when any of the columns or rows of that matrix are replaced by elements following some other distribution function...
Specifically speaking, think of a matrix of size $ M \times N$ whose entries in each row follow gaussian distribution function. Does this matrix has some unique property which shall be disturbed when I replace $ k$ columns of this matrix ($ k<N$) with elements following uniform distribution or any other function.
I have an example in mind which is not true mathematically but just to eloborate my question i m quotiong it.
Let's say the eigenvalues of a matrix filled with gaussian distributed elements are constrained in a unit circle but when any of the columns of the matrix are replaced by any other values  (random or following some other distribution function), these eigenvalues are no more enclosed by the unit circle.

I hope my question statement is clear and there is some possible solution to this problem.
Thanks

The interesting property: The distribution on $ m\times n$ matrices in which the entries are independent normal distributions of the same variance centered at zero is invariant under multiplication by orthogonal matrices on either side. This is basically the strongest possible form of spherical symmetry.
There are other distributions with nice symmetry properties, but none so simple.


topology
vector
inequalities
triangle inequality
real analysis
real analysis unsolved

Give an example of space (topology space and so on) whose base of topology doesn't consists of convex sets.

The base of topology for normed vector spaces is open balls ( $ K(x,r)$), that are convex sets, these spaces are called [b]locally convex spaces[/b]  but I can't think of spaces that I mentioned above.

For convexity to make sense, we need the spaces to be real vector spaces.

The $ L^{p}$ spaces for $ p<1$ are an example. We don't have the triangle inequality, and that means that $ \|\frac{x\plus{}y}{2}\|$ can be greater than either $ \|x\|$ or $ \|y\|$. In the simplest case, if we use the $ p$-"norm" on $ \mathbb{R}^{2}$ with $ p<1$, the unit ball is a nonconvex star.


Write in term of the cofunction sin 37''

Cofunction as in cosine?
$\sin 37 = \cos (90-37) = \cos 53$
If you're in radians (degrees is similar), $\sin x = \cos (\frac{\pi}{2}-x)$

If you're in degrees, switch the $\frac{\pi}{2}$ to a $90$

[hide]the function of the complement of a given angle or arc: cosθ is the cofunction of sinθ. from dictionary.com

so basically it's cos37
cos(37) = 0.765414052[/hide]

[quote="Mark9182"]Write [b]in terms[/b] of the cofunction sin 37''[/quote][/b]

He said to put it [b]into terms[/b] of the cosine, not to simply take the cosine of the angle given.


Kelsey was delivering lunch to a local office building. Kelsey's first delivery was to the middle floor of the building. She then took the elevator up 4 floors for the next delivery, down 6 floors for her third delivery, and up 7 floors for her final delivery. She was then on the top floor of the building. How many floors were in this office building?

She went up $ 7\plus{}4\minus{}6$ or $ 5$ floors up.  If she was in the middle floor, the bottom floor would be $ 5$ floors down.  Thus, there are $ 5 \cdot 2 \plus{} 1$ or $ \boxed{11}$ floors.


calculus
integration
calculus computations

this is ,if i am not mistaken pretty easy.it has to do with bernoullis differential equations.just make the $ y=f(x)$ , $ y=z{}^-{}^1$ and solve for $ z$ ,the other things are trivial...

Just seperate variables and do integral each hand side, we get the general solution.

$ y \equal{} \frac {1}{x \plus{} C}$   ,$ C\geq 0$.


Prove that if (ca' - ac') 2 <4(ab'-a'b)(bc'-b'c) then

b 2 -ac>0 and (b') 2 -a'c'>0

Are the numbers pozitive reals or just real numbers?

cheers!


Each of three, standard, six-sided dice is rolled once. What is the probability that the three numbers rolled are the lengths of the sides of an obtuse triangle? Express your answer as a common fraction.

Refer to solution $ 8$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1098302#1098302]here[/url].

The 8 possible obtuse triangles have side lengths of:
2 2 3
2 3 4
2 4 5
2 5 6
3 3 5
3 4 6
3 5 6
4 4 6

The biggest of the three sides must be less than the sum of the other two sides (or it won't form a triangle).

Also, the square of the biggest side must be greater then the sum of the squares of the other two sides (so that the biggest angle is greater than 90 degrees).

It also must be noted how many possible dice rolls could produce each combination.
2 2 3  (3 permutations)
2 3 4  (6 permutations)
2 4 5  (6 permutations)
2 5 6  (6 permutations)
3 3 5  (3 permutations)
3 4 6  (6 permutations)
3 5 6  (6 permutations)
4 4 6  (3 permutations)

That makes 39 possible rolls out of 216 (6^3)
13/72

The solution has been explained, but one might ask how did you start?
First consider 6 for the biggest side and ask what combinations of smaller sides will work.  Then consider 5 for the biggest side, etc.


I don't know the original source of this problem, a friend of mine told it to me about a year ago. After that I've seen it as a programming task in two programming contests.

  The problem looks like this: Proove that for any n points in the plane it is possible to illuminate all the points in the plane by placing n flashligths wich each illuminate a 2*pi/n angle in the n points.

[color=red][Moderator edit: This problem was discussed in a more general form at http://www.mathlinks.ro/Forum/viewtopic.php?t=34206 and http://www.mathlinks.ro/Forum/viewtopic.php?t=1896 .][/color]

those links are invalid, can anyone post a valid one :idea:


matrix A=$ [a_{ij}]_{n x n}$ with all entries are integer. prove that $ A^{\minus{}1}$ has all entries are integer if and only if |det(A)|=1 :|

The key fact is that a matrix with integer entries has integer determinant (the determinant is some combination of additions, subtractions, and multiplications of the entries -- operations under which the integers are closed).  After you've proven that, the desired statement follows:

$ \Rightarrow$: If $ A^{ \minus{} 1}$ has all integer entries, then $ \left \vert \det A^{ \minus{} 1} \right \vert$ is an integer, but $ \left \vert \det A\right\vert \equal{} \frac {1}{\left \vert \det A^{ \minus{} 1}\right \vert}$ is also an integer, so it can only be that $ \left \vert \det A^{ \minus{} 1}\right \vert \equal{} \left \vert \det A \right \vert \equal{} 1$.
$ \Leftarrow$: $ A^{ \minus{} 1} \equal{} \frac {1}{\det A}\left( C_{ij}\right)^T$ where $ C_{ij}$ is the cofactor of entry $ i,j$ of matrix $ A$.  Each cofactor is the determinant of some matrix with integer entries, and therefore an integer.  Hence if $ \left |\det A\right | \equal{} 1$, then $ A^{ \minus{} 1}$ is a matrix with integer entries.


geometry
geometric transformation
reflection
analytic geometry
graphing lines
slope

One's a reflection of the other over the y-axis. ie. Their slopes are 1 and -1

the fact that they're different equations?????????idk what u want...

[quote="Interval"]What is the [color=red]MAIN DIFFERENCE [/color]between y = x and y = -x?

Please explain.[/quote]

please be specific in what you mean by main difference.

heres one:
They are different equations.

heres another:
one goes through the 1st and 3rd quadrant and the other doesnt.

I could list a billion differences, just be more specific.

[quote="Interval"]By MAIN DIFFERENCE I meant in terms of a positive x and negative x.

In other words, why does y equal a positive x and also a negative x?[/quote]

ok.  they arent the same $y$.  would it help is i wrote it as $y_{1}=x_{1}$ and $y_{2}=-x_{2}$?

unless you are trying to solve the system of 2 equations in which case x=y=0

I get it now.  Sometimes it is necessary to break the explanation of a question into pieces.

Thanks!


The angles of a triangle ABC are measured in degrees

Prove that  (bcA +caB +abC)/(bc + ca + ab) <=60  <= (aA +bB +cC)/(a +b +c)

Second one:

Larger side is larger angle, so A,B,C and a,b,c are similarily sorted.

By Chebyshev:

$3\sum aA \ge \sum a \sum A$ leading to $\frac{\sum aA}{\sum a} \ge 60$ as desired.

First one is similar, using the fact that (bc,ca,ab) and A,B,C are opp. sorted.


[quote="cincodemayo5590"]For $x,y \ge 0$, prove that   $x+y^{2}\ge y \cdot (4\sqrt{x}-x-1)$ .[/quote]

[hide]$x+y^{2}\ge y \cdot (4\sqrt{x}-x-1)$

$\Leftrightarrow (\sqrt{x}-y)^{2}+y(\sqrt{x}-1)^{2}\ge 0$[/hide]

its equivalent a
(x+y^2+xy+y)/4 >=(x(y^2))^1/2 AM>=GM


I know I'm a bit to early, but I would like to hear which of you will be accepted off the wait list, when the decisions come, and where...
  I wish to the ones who are still waiting for the final decisions, just like me, the best of luck and to be accepted wherever you want!
Btw, do you know of any possible disadvantage, if I am an international student, in the wait list decisions?

Has anybody on the waitlists received any news from Columbia, Brown or Cornell universities?
   I have been waitlisted at all of them and I would like to hear everything that goes on, so I would be really glad to hear from others in my situation. Maybe I can share my current knowledge with you.

Decisions are due to colleges postmarked by May 1, so they would not have had time to go through their class as it stands and decide how many more people they need, which needs to happen before they pick anyone off the waitlist.

Consider calling admissions offices for those universities for waitlist information.

News for all of you who are waiting on the Columbia waitlist: class full. No news for the other 2.


Find all positive integers $n\le 2000$ such that $x^4+n$ can be factored into two distinct quadratic factors, each with integer coefficients.

[quote="Farenhajt"]Should $x^4+k$ read $x^4+n$?[/quote]Yeah, it should. I'll edit it.

Let's put $x^4+n=(x^2+ax+p)(x^2+bx+q)$

By expanding we get

$x^4+n=x^4+(a+b)x^3+(p+ab+q)x^2+(aq+pb)x+pq$

From this we get

\[\begin{eqnarray} a+b &=& 0\\ p+ab+q &=& 0\\ aq+pb &=& 0\\ pq &=& n\\ \end{eqnarray*}\]

$(1)$ gives $b=-a.$ Substituting that into $(2)$ and $(3)$ we get

$a^2=p+q\qquad(5)$
$a(p-q)=0\qquad(6)$

Now $a$ can't be zero, since in that case $(5)$ would give $p=-q$ and $(4)\implies n=-p^2<0$. Therefore $(6)$ implies $q=p$. Now $(5)$ becomes $a^2=2p.$ This means that $p$ must have the form $2t^2$ for some natural $t$, and that in turn gives $n=pq=p^2=4t^4.$

Therefore, the possible values for $t$ are such that $1\leqslant t^4\leqslant 500$, and those are $1,2,3,4$, and the corresponding values for $n$ are $4, 64, 324, 1024$

The required factorization is $x^4+4t^4=(x^2+2tx+2t^2)(x^2-2tx+2t^2)$ and is known as [i]Sophie Germain identity[/i].


What program(s) do you use for drawing diagrams for things like classes and MathJams?

We generally use Geometer's Sketchpad to do drawings which require some geometric precision.  For other drawings we generally use Flash or Fireworks.


a lot of people have that Go signature. What is it? i think it looks like snoopy with wings. (see ignight or somthing like that)

its an upsidedown dog and it's spelled weird

Wow...whoever were playing the game seem to have been playing somewhat defensively...
I haven't seen such clear division of territory into two regions (for the most part) in a game of go in a long time...

wait, what kind of game is that?

-jorian

Google and Wikipedia remain ever your friends.

http://en.wikipedia.org/wiki/Go_%28board_game%29
http://www.usgo.org/
http://gobase.org/

The strategy of the player using white pieces is really ...... strange

[quote="t0rajir0u"]Both players seem relatively weak.  Should've used a better game.  One of Shusaku's or something  :)[/quote]

sorry, i didnt feel like doing extensive finding so i just found the first picture i could


geometry
perimeter
rotation
algebra
function
domain
geometry unsolved

A square $ R$ of sidelength $ 250$ lies inside a square $ Q$ of sidelength $ 500$. Prove that: One can always find two points $ A$ and $ B$ on the perimeter of $ Q$ such that the segment $ AB$ has no common point with the square $ R$, and the length of this segment $ AB$ is greater than $ 521$.

Hello, orl!

A highly non-rigorous proof occured to me.
Perhaps someone can tighten it and fine-tune it . . .

[quote]A square $R$ of sidelength $250$ lies inside a square $Q$ of sidelength $500$. Prove that: One can always find two points $A$ and $B$ on the perimeter of $Q$ such that the segment $AB$ has no common point with the square $R$, and the length of this segment $AB$ is greater than $521$.[/quote]
Place square R concentric to square Q.
Then rotate square R 45 degrees.
(We will have a "diamond" inside a square.)

This is the non-rigorous part.
I believe this the most [i]obtrusive[/i] position for square R, but haven't proved it.
Moving square R away from the center would create more room for longer "chords".
Rotating square R would have the same effect.

Assuming the above conjecture is true . . .
From the lower-left vertex of square Q (A),  draw a line through the bottom vertex of square R (C).
Extend the line to intersect the far side of square Q at B.

Using similar right triangles and proportions, it can be shown
that chord AB has length 500[sqrt(2) - 1]/sqrt(2) = 521.0253833...
This chords has only one point in common with square R.
Any slightly shorter chord will satisfy the required conditions.

Therefore, even in my worst-case scenario, we can find a chord longer than 521 units.

A bloody problem...

Here is what I got with my students.

Perform a parallel shift of our little square to the center of big one. Then new small square contains a vertex $X$, s.t. old square is in angle domain $\angle YXZ$. So we will prove that there is a segment of length 521, whcih doesn't intersects this angle domain.

We may WLOG assume that $X$ lies inside of $AOF$. We see that $X$ lies on the arc of circle of radius $125\sqrt 2$.

Construct segments $AA'$, $BB'$ and $CC'$ of length 521 as shown on the figure below. Let $M=AA'\cap BB'$. You can easily check that $OX<OM$. So while $X$ rotates in counter-clockwise direction from direction $OM$, then we can choose $AA'$ as the desired segement till $XZ$ becomes parallel to $AA'$ (it is clear). After that moment we should choose $DD'$.


Let $a$, $b$, $c$ be three positive reals. Prove the inequality

$\frac{a^2}{2a^2+\left(b+c\right)^2}+\frac{b^2}{2b^2+\left(c+a\right)^2}+\frac{c^2}{2c^2+\left(a+b\right)^2}<\frac23$.

It was hard for me, but you know me well enough to realize that this doesn't mean anything.

[hide="useless rambling"]In case anyone is wondering what the Merida inequalities are: As a pretty direct reaction to myself not being able to solve IMO 2005 problem 3, I decided to create some really impossible to prove inequalities. I had a lot of time for doing this on the sports day in Merida, and I came up with 5 inequalities. Two of them ([url=http://www.mathlinks.ro/Forum/viewtopic.php?p=415243#p415243]nr. 1[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=415234#p415234]nr. 2[/url] - the original form of nr. 2 was more ugly and less sharp) I had quickly solved, although I still missed their triviality (I used Cauchy-Schwarz and Vornicu-Schur to prove them... that's me :) ). Then there was a third one ([url=http://www.mathlinks.ro/Forum/viewtopic.php?p=411674#p411674]nr. 3[/url]) which I was not able to solve until a week ago (although I didn't spend that much time on it). Soon I managed to prove the fourth one ([url=http://www.mathlinks.ro/Forum/viewtopic.php?p=415225#p415225]nr. 4[/url]), and now the fifth and last one (this here, although again, the original verion was far more boring) is proved. In retrospect, none of the inequalities was really that hard anyway, at least not compared to the ones by Vasc and Harazi.[/hide]

  Darij

Nice solution, Soarer. Here is mine:

We will use the $\sum$ sign for cyclic sums. The inequality in question, $\sum\frac{a^2}{2a^2+\left(b+c\right)^2}<\frac23$, rewrites as $\frac23-\sum\frac{a^2}{2a^2+\left(b+c\right)^2}>0$. But

$\frac23-\sum\frac{a^2}{2a^2+\left(b+c\right)^2}=\frac23\cdot\sum\frac{a}{a+b+c}-\sum\frac{a^2}{2a^2+\left(b+c\right)^2}=\sum\left(\frac23\cdot\frac{a}{a+b+c}-\frac{a^2}{2a^2+\left(b+c\right)^2}\right)$
$=\sum\frac{a\left(b+c-a\right)^2+a\left(b+c\right)\left(b+c-a\right)}{3\left(a+b+c\right)\left(2a^2+\left(b+c\right)^2\right)}$
$=\sum\frac{a\left(b+c-a\right)^2}{3\left(a+b+c\right)\left(2a^2+\left(b+c\right)^2\right)}+\sum\frac{a\left(b+c\right)\left(b+c-a\right)}{3\left(a+b+c\right)\left(2a^2+\left(b+c\right)^2\right)}$.

Now, it is obvious that $\sum\frac{a\left(b+c-a\right)^2}{3\left(a+b+c\right)\left(2a^2+\left(b+c\right)^2\right)}\geq 0$; what remains to be proven is the inequality $\sum\frac{a\left(b+c\right)\left(b+c-a\right)}{3\left(a+b+c\right)\left(2a^2+\left(b+c\right)^2\right)}>0$, which simplifies to $\sum\frac{a\left(b+c\right)\left(b+c-a\right)}{2a^2+\left(b+c\right)^2}>0$. Now,

$\sum\frac{a\left(b+c\right)\left(b+c-a\right)}{2a^2+\left(b+c\right)^2}=\sum\frac{ab\left(b+c-a\right)+ca\left(b+c-a\right)}{2a^2+\left(b+c\right)^2}$
$=\sum\frac{ab\left(b+c-a\right)}{2a^2+\left(b+c\right)^2}+\sum\frac{ca\left(b+c-a\right)}{2a^2+\left(b+c\right)^2}=\sum\frac{bc\left(c+a-b\right)}{2b^2+\left(c+a\right)^2}+\sum\frac{bc\left(a+b-c\right)}{2c^2+\left(a+b\right)^2}$
$=\sum bc\left(\frac{c+a-b}{2b^2+\left(c+a\right)^2}+\frac{a+b-c}{2c^2+\left(a+b\right)^2}\right)$
$=\sum bc\frac{a\left(\left(a+b+c\right)^2+a^2+2bc\right)+\left(b+c\right)\left(b-c\right)^2}{\left(2b^2+\left(c+a\right)^2\right)\left(2c^2+\left(a+b\right)^2\right)}>0$,

qed.

  Darij


Let:$ a;b;c>0$.Prove that:
$ \frac{a\plus{}b\plus{}c}{3}\minus{}\sqrt[3]{abc} \le max \left \{ (\sqrt{a}\minus{}\sqrt{b})^2,(\sqrt{b}\minus{}\sqrt{c})^2,(\sqrt{c}\minus{}\sqrt{a})^2 \right \}$
I need find nice solution   :P

[quote="caubetoanhoc94"]Let:$ a;b;c > 0$.Prove that:
$ \frac {a \plus{} b \plus{} c}{3} \minus{} \sqrt [3]{abc} \le max \left \{ (\sqrt {a} \minus{} \sqrt {b})^2,(\sqrt {b} \minus{} \sqrt {c})^2,(\sqrt {c} \minus{} \sqrt {a})^2 \right \}$
I need find nice solution   :P[/quote]
This inequality by Titu Andrescu
ByAM-GM and Schur 's inequality,we have:
$ 3\sqrt[3]{abc} \ge  \frac{9 \sqrt{abc}}{\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}} \ge 2(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca})\minus{}a\minus{}b\minus{}c$
Then,$ (\sqrt{a}\minus{}\sqrt{b})^2\plus{}(\sqrt{b}\minus{}\sqrt{c})^2\plus{}(\sqrt{c}\minus{}\sqrt{a})^2 \ge a\plus{}b\plus{}c\minus{}3\sqrt[3]{abc}$ :)
Then,we have Q.E.D
Equality holds if and only if $ a\equal{}b\equal{}c$

[quote="kunny"]This is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61957]USA TST 2000[/url][/quote]

Yes.Of course :).It's old problem :) but I need nice solution


projective geometry
Euler
geometry proposed
geometry

$\underline {\overline {\left|\left| \ Math.\ -\ a\ universal \ language\ \right|\right|}}$

$\blacksquare \ \underline {\triangle ABC}:$

$\bullet \ U\in AB,\ UA=UB\ \bullet\bullet\ \ V\in AC,\ VA=VC\ \bullet$

$\bullet \ M\in AB,\ MV\perp AC\ \bullet\bullet \ N\in AC,\ NU\perp AB\ \bullet$

$\bullet \ R\in AB,\ RI\perp AC\ \bullet\bullet \ S\in AC,\ SI\perp AB\ \bullet$

$\bullet \ P\in BN\cap CM\ \bullet\bullet \ T\in BS\cap CR\ \bullet$

$\Longrightarrow \underline {P\in GO,\ T\in GI}\ \blacksquare$

[b]Remark.[/b] [u]Here are two problems ![/u]

by pappus, ${NU}\cap{MV}$, ${MC}\cap{BN}$ and ${BV}\cap{UC}$ are colinear,
but $NU$ and $MV$ are the bisectors of $AB$ and $AC$ (${MV}\cap{NU}$ is the circuncenter of $ABC$)
and $BV$ and $UC$ are medians (${BV}\cap{UC}$ is the centroid of $ABC$).

so, $P = {BN}\cap{MC}$ are in the euler line of $ABC$.  :)

You are good, you percussed immediately. It is O.K. Analogously and the second problem. Please, at least you offer, for the other visitors, the enunciation of the Pappus' theorem or the its net-address.

http://mathworld.wolfram.com/PappussHexagonTheorem.html

 :)

Thanks. You can try to compose any similar problems utilizing the Pappus' theorem or you offer on this post some known problems.


Let k, u, v, w be positive reals. Prove that

$k^4(u^2 v + v^2 w +  w^2 u) + k^2 (u^3 + v^3 + w^3+ 3 u v w) + (u v^2  + v w^2+ w u^2) $

$\geq 2k^3(u v^2  + v w^2+ w u^2) + 2k(u^2 v + v^2 w + w^2 u)$.

Schur is a special case when k=1.

Can you please post your solution, Fuzzylogic?

Thank you very much. :)


algorithm
limit
number theory
prime numbers
algebra
difference of squares
special factorizations

um. what kind of problem is this?
are you looking for some kinda trick. you just list and add?

I think this should be moved to Classroom Math

Nobody has answered the question yet...

[quote="llolloll"]what is the sum of the first ten prime numbers?[/quote]
[hide]$ S\equal{}2\plus{}3\plus{}5\plus{}7\plus{}11\plus{}13\plus{}17\plus{}19\plus{}23\plus{}29\equal{}\boxed{129}$[/hide]

the first post , i think it is a good problem somehow ...
you guys have only mentioned "lizt and add" , what about the sum of the first 1000 prime numbers ?

[quote="Temperal"]there is no perfect method to create prime numbers.[/quote]

This is of course not true -- there are lots of algorithms that produce all of the prime numbers in order, e.g. the Sieve of Eratosthenes.  What is true is that there is no nice formula, given arbitrary $ n$, to produce the $ n$th prime number, and this immediately rules out the possibility of a nice formula to produce the sum of the first $ n$ prime numbers.  Since there are decent asymptotics for the density of the prime numbers (i.e. the Prime Number Theorem), it follows that there should be decent asymptotics for sums of consecutive prime numbers.  However, asymptotics are not good for calculating particular values.

[quote="Temperal"]List and add. There is no known algorithm for finding the sum of prime numbers, just as there is no perfect method to create prime numbers.

Well, technically, $ \lim_{k\to \infty}2 \uparrow \uparrow k \minus{} 1$ will get closer and closer to an actual algorithm to find primes, but it's not an actual formula that works.[/quote]

Don't you mean $ \lim_{k\to \infty}2 \uparrow \uparrow k \plus{} 1$? Almost every value I tested for $ k$ for $ 2\uparrow \uparrow k \minus{} 1$ has been a nonprime ($ 0, 15, 65535,$ etc). Also, for $ 2\uparrow \uparrow k \minus{} 1$, I think it can be factorized with difference of squares several times.

I think it should be $ \lim_{k\to \infty}2 \uparrow \uparrow k \plus{} 1$

IRT Problem:
the first 10 primes are $ 2,3,5,7,11,13,17,19,23,29$. The sum is $ 5\plus{}5\plus{}20\plus{}30\plus{}40\plus{}29\equal{}129$


And what about this lemma $\frac{a^3 }{x}+\frac{b^3 }{y}+\frac{c^3 }{z}\geq\frac{(a+b+c)^3 }{3(x+y+z)}$.?
Just take $x=a^2+ b^2 ,y=b^2 +c^2 $ and $z=c^2+a^2$ and apply the lemma.Just :P  :D

If you are so sure now why don't you post a solution? :D

[quote="Soarer"][quote="ehsan2004"]no it's $12$  :P[/quote]
cezar is right. A simple solution is cauchy schwarz then power mean. Why do you think it is 12?[/quote]

Ok ,excuse me :oops:  :oops:

[quote="cezar lupu"]yes, it is true, but it's not $12$ it's $6$. ;)[/quote]

i think it's 6 like cezar lupu  :) if you are sur ehsan2004 that it's true for 12 post your solution  ;)

[quote="Fermat -Euler"]
i think it's 6 like cezar lupu  :) if you are sur ehsan2004 that it's true for 12 post your solution  ;)[/quote]

Well it's pretty obvious that if it's true for 6, then it must be true for 12 too :huh:

And, for a short proof, notice that \[\frac{a^3}{a^2 + b^2} \geq a - \frac{b}{2}\] (since that inequality is equivalent to $b(a - b)^2 \geq 0$). Hence, upon adding,

\[\frac{a^3}{a^2 + b^2} + \frac{b^3}{b^2 + c^2} + \frac{c^3}{c^2 + a^2} \geq \frac{a + b + c}{2} = \frac{1}{2}\] and, on the other hand, \[a^2 + b^2 + c^2 \geq \frac{(a + b + c)^2}{3} = \frac{1}{3}\] so the result follows.


[quote="pvthuan"][quote="darij grinberg"][quote="pvthuan"]$\text{Prove that if} \; a,b,c>0\; \text{then}$
\[ \frac{a^2(b+c)}{2a^3+(b+c)^3}+\frac{b^2(c+a)}{2b^3+(c+a)^3}+\frac{c^2(a+b)}{2c^3+(a+b)^3}\geq\frac35.  \][/quote]

Sorry, but this is wrong for a = 1, b = 1, c = 2.

  darij[/quote]

OK. The following is true
\[ \frac{a^2(b+c)}{2a^3+b^3+c^3}+\frac{b^2(c+a)}{2b^3+c^3+a^3}+\frac{c^2(a+b)}{2c^3+a^3+b^3}\leq\frac32.  \][/quote]
$3a^2b \le 2a^3+b^3$
$3a^2c \le 2a^3+c^3$
then:we should only to prove:
$\frac{a^3}{2a^3+b^3+c^3}+\frac{b^3}{2b^3+c^3+a^3}+\frac{c^3}{2c^3+a^3+b^3}\leq\frac34.$
Use the inequlity $\frac{b^3}{2b^3+c^3+a^3} \le \frac 14 \left(\frac{b^3}{b^3+c^3}+\frac{b^3}{b^3+a^3}\right)$
and similar...

That is to see that my inequality is NOT strong.


search
floor function
number theory unsolved
number theory

Prove that for each prime $ p$ and positive integer $ n$, $ p^n$ divides

$ \binom{p^n}{p}\minus{}p^{n\minus{}1}$

[b]Lemma:[/b]  The greatest power of $ p$ dividing $ {n \choose k}$ is the number of carries that must be done to add $ n$ and $ k$ in base $ p$.  (See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1775828923&t=225148]here[/url].)

Adding $ p \equal{} 10_p$ and $ p^n \minus{} p \equal{} (p \minus{} 1)(p \minus{} 1)(p \minus{} 1)...(p \minus{} 1)0_p$ requires $ n \minus{} 1$ carries, so the left term is divisible by $ p^{n \minus{} 1}$ but not by $ p^n$, a necessary condition.  We now want to show that if we remove the multiples of $ p$ from $ {p^n \choose p}$, the result is congruent to $ 1 \bmod p$.

But this is obviously congruent to $ \frac { (p \minus{} 1)!^n }{ (p \minus{} 1)! (p \minus{} 1)!^{n \minus{} 1} } \equiv 1 \bmod p$.

I can't see how to get to the "application" in that topic.

[b]Lemma:[/b]  The greatest power of $ p$ dividing $ n!$ is $ \sum_{k \equal{} 1} \left\lfloor \frac {n}{p^k} \right\rfloor$.

It follows that the greatest power of $ p$ dividing $ {n \choose k} \equal{} \frac {n!}{k!(n \minus{} k)!}$ is given by

$ \frac {p}{p \minus{} 1} \left( n \minus{} (k \plus{} n \minus{} k) \right) \minus{} \frac {s_p(n) \minus{} s_p(k) \minus{} s_p(n \minus{} k)}{p \minus{} 1}$

or

$ \frac {s_p(k) \plus{} s_p(n \minus{} k) \minus{} s_p(n)}{p \minus{} 1}$.

In other words, the difference in the sums of the digits of $ k, n \minus{} k$ in base $ p$ and the sums of the digits of $ n$ in base $ p$, divided by $ p \minus{} 1$, gives the greatest power of $ p$.  But hopefully you can convince yourself that whenever we perform a carry this quantity increases by $ 1$, that is, the difference in digital sums increases by $ p \minus{} 1$.

[quote="outback"]Prove that for each prime $ p$ and positive integer $ n$, $ p^n$ divides

$ \binom{p^n}{p} \minus{} p^{n \minus{} 1}$[/quote]
[hide="Solution"]
It suffices to show that $ \binom {p^n}{p}\equiv p^{n\minus{}1}\bmod p^n$.  This means that $ \frac{p^n\cdot (p^n\minus{}1)\cdot (p^n \minus{}2)\cdots \cdot (p^n\minus{}p\plus{}1)}{p!}\equiv p^{n\minus{}1}\bmod p^n$.  This is true if and only if $ p^n\cdot (p\minus{}1)!\equiv p!\cdot p^{n\minus{}1}\bmod (p^n\cdot p!)$, which is true if and only if $ p^n\cdot (p\minus{}1)!\equiv p^n\cdot (p\minus{}1)!\bmod (p^n\cdot p!)$, which is true. [/hide]


geometry
circumcircle
inequalities
vector
inequalities solved

Let ABC and A'B'C' two triangles having sides a,b,c and a',b',c' respectively. If R is the circumradius of ABC, prove the inequality

    a 2 /a'+b 2 /b'+c 2 /c' \leq R 2 (a'+b'+c') 2 /a'b'c'

If a'=b'=c' then........?

If a'=a,b'=b,c'=c then......?

If a'=b,b'=c,c'=a then......?

Has anyone a solution?

I have one very simple.

For simplicity of typing ---->  a'=x , b'=y , c'=z

The stated inequality is equivalent to:

x 2 +y 2 +z 2 -2(yzcosA''+zxcosB''+xycosC'') \geq 0

where A''= \pi -2A , B''= \pi -2B , C''= \pi -2C ;A''+B''+C''= \pi

(A,B,C are angles of triangle of sides a,b,c ; A',B',C' are angles of triangle of sides a',b',c')

This last inequality can be written as

(x-ycosC'' -zcosB'') 2 +(ysenC''-zsenB'') 2  \geq 0

A necessary condition for equality is the vanishing of the second square  and two similar forms requiring that

senA'':senB'':senC''=x:y:z

But angles A'',B'',C'' lie between - \pi and + \pi ; if A'' \leq 0 the B'' \leq0,C'' \leq 0. Hence A'',B'',C'' are the angles of a triangle similar to A',B',C'.

Dear Manlio,
   I really don't think that the inequality x^2+y^2+z^2>=2xycosC+2yzcosA+2zxcosB is easy. It is classical, but not at all easy.

Why? It is obvious: the scalar square of a vector x*i+y*j+z*k is nonegative, where i,j,k are unit vectors along sides of triangle.

Well, I've never thought about this. But I also think I'm not the onlyone.


trigonometry
real analysis
real analysis unsolved

Can someone find the solution $ u$ of the Poisson equation $ \nabla ^2 u (x,y) \equal{} c$ defined in a square of length $ a$ for some real constant $ c$ and the condition $ u\equal{}0$ on the boundary of the square.

Assume that the solution has the form

$ u \equal{} \sum_{m \equal{} 1}^{\infty}\sum_{n \equal{} 1}^{\infty}b_{m,n} \sin\left(\frac {m\pi x}a\right)\sin\left(\frac {n\pi y}a\right)$

Then

$ \nabla^2u \equal{} \minus{} \frac {\pi^2}{a^2}\sum_{m \equal{} 1}^{\infty}\sum_{n \equal{} 1}^{\infty} (m^2 \plus{} n^2)b_{m,n}\sin\left(\frac {m\pi x}a\right)\sin\left(\frac {n\pi y}a\right)$

Now, we use a well-known Fourier series to claim that on $ (0,a)\times(0,a),$ we have

$ c \equal{} \frac {16c}{\pi^2}\sum_{j \equal{} 0}^{\infty}\sum_{k \equal{} 0}^{\infty}\frac {1} {(2j \plus{} 1)(2k \plus{} 1)}\sin\left(\frac {(2j \plus{} 1)\pi x}a\right)\sin\left(\frac {(2k \plus{} 1)\pi y}a\right)$

Matching coefficients, we see that

$ \minus{} \frac {\pi^2}{a^2}b_{2j \plus{} 1,2k \plus{} 1}\left((2j \plus{} 1)^2 \plus{} (2k \plus{} 1)^2\right) \equal{} \frac {16c}{\pi^2}\frac {1}{(2j \plus{} 1)(2k \plus{} 1)}$

Hence, I claim the following solution:

$ u \equal{} \minus{}\frac {16a^2c}{\pi^4}\sum_{j \equal{} 0}^{\infty}\sum_{k \equal{} 0}^{\infty} \frac {1}{(2j \plus{} 1)(2k \plus{} 1)\left((2j \plus{} 1)^2 \plus{} (2k \plus{} 1)^2\right)} \sin\left(\frac {(2j \plus{} 1)\pi x}a\right)\sin\left(\frac {(2k \plus{} 1)\pi y}a\right)$

[quote="Crux"]Can someone find the solution $ u$ of the Poisson equation $ \nabla ^2 u (x,y) \equal{} c$ defined in a square of length $ a$ for some real constant $ c$ and the condition $ u \equal{} 0$ on the boundary of the square.[/quote]

This is equivalent to \[ \left\{ \begin{gathered}
  \Delta u \equal{} 0 \hfill \\
  u\left| {_{\partial \Omega }  \equal{} 0} \right. \hfill \\
\end{gathered}  \right.
\] where $ \Omega$ is a square of length $ a$. In this context, try to find the solution of the form $ u(x,y)\equal{}X(x)Y(y)$.


I accidentally created a blog today.  :roll:  Is there any way I can delete it?
If there isn't, it's fine. I'll just leave it there... or something.

[quote="SplashD"]go to profile and under blogs, there's that option[/quote]
Could you be more specific about where it is? I can't seem to find it.  :maybe:
I've looked through "Blog Information," "Blog Layout," ... up to "Blog customization," and I didn't see any option to delete it.

I can't seem to find it either though I remember I deleted my old blog to create a new one some months ago...

Even if you do delete it, the admins have to purge it, or else it still remains.  :D

There was a serious bug with the "delete blog" function so it was removed until further notice.

How do you have a picture on your background in a blog?

I'm also having a problem with that. I put in the URL, it accepted it, but my blog didn't change.


Let be $ a,b,c\in \mathbb{R}$ such that $ |ax^2\plus{}bx\plus{}c|\le 1\ ,\ (\forall)x\in [\minus{}1,1]$ . If $ \alpha \in [0,1]$ show that :

$ \alpha(1\plus{}\alpha)|b|\plus{}(10\minus{}\alpha^2)|c|\le 1\plus{}\alpha^2$

[quote="alex2008"]Let be $ a,b,c\in \mathbb{R}$ such that $ |ax^2 \plus{} bx \plus{} c|\le 1\ ,\ (\forall)x\in [ \minus{} 1,1]$ . If $ \alpha \in [0,1]$ show that :

$ \alpha(1 \plus{} \alpha)|b| \plus{} (10 \minus{} \alpha^2)|c|\le 1 \plus{} \alpha^2$[/quote]

You need to make sure that you have posted the problem correctly. As stated, this result cannot be true. For example, when $ \alpha\equal{}0$ it says that $ 10|c|\leqslant1$, which certainly need not be true.


algebra polynomial trigonometry inequalities

1) Show that $x^{4a}+x^{4b+1}+x^{4c+2}+x^{4d+3}$, $a,b,d,c$ positive integers, is divisible by $x^{3}+x^{2}+x+1$.

2) Show that each polynomial $(\cos{\theta}+x\sin{\theta})^{n}-\cos{n\theta}-x\sin{n\theta}$ divisible by $x^{2}+1$.

3) For what n is the polynomial $1+x^{2}+x^{4}+...+x^{2n-2}$ divisible by the polynomial $1+x+x^{2}...+x^{n-1}$? Followup: find all pairs of integers $(m,n)$ such that $1+x^{n}+x^{2n}+...+x^{mn}$ is divisible by $1+x+x^{2}+...+x^{m}$.

4) Show that $\cos{\frac{\pi}{2n}}$ is algebraic for every positive integer $n$.

[hide="1"] Easy.  $(x^{3}+x^{2}+x+1) | (x^{4}-1)$ so when we reduce $\bmod (x^{3}+x^{2}+x+1)$ we have $x^{4a+k}\equiv x^{k}$. [/hide] [hide="4"] Well-known lemma; we can express $\cos n \theta$ as a polynomial in $\cos \theta$ (easy inductively or by Euler's Formula), hence

$\cos \frac{\pi}{2}= P( \cos \frac{\pi}{2n}) = 0$

For some (integer) polynomial $P$.  [/hide]

Is this what you mean? Using de Moivre's we can express $\cos{n\theta}=Re(\cos{\theta}+i\sin{\theta})^{n}$, which is the desired polynomial in $\cos{\theta}$.

However, the book I have says that an algebraic number is a root of a polynomial with integer coefficients, so I think we need to find a polynomial over the integers.

[hide="2"]
It's actually ${(\cos{\theta}+x \sin{\theta})^{n}-\cos{n\theta}-x \sin{n\theta}}$, and we can prove it by showing that $\pm i$ are roots of this. But

${(\cos{\theta}+i \sin{\theta})^{n}= \cos{n\theta}+i\sin{n\theta}}$

and

${{(\cos{\theta}-i \sin{\theta})^{n}= (\cos{\theta}+i \sin{(-\theta}))^{n}= \cos{n\theta}+i\sin{(-n\theta})}= \cos{n\theta}-i\sin{n\theta}}$

by DeMoivre, so we're done.[/hide]

The polynomial does have integer coefficients.  Sorry if I made that unclear.

Can you generate this polynomials by de Moivre?

Yes; in fact, they are also the polynomial representations of $x^{n}+\frac{1}{x^{n}}$ in terms of $x+\frac{1}{x}$ (you can see this by letting $x = e^{i \theta}$).  In fact, they are important enough that they have a name, the Chebyshev Polynomials.

http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html

The first one should be $x^{4a}+x^{4b+1}+x^{4c+2}+x^{4d+3}$, right?  The way it's written you just multiply by $x^{4a}$  :D ...


Is there any way to download the Alcumus videos that I have access too? MathCounts is in one week, and I am going on a class trip. It would be helpful if I could watch the videos again as reference.
If it is possible, how?


20 equal circles are packed in honeycomb fashion in a rectangle. The outer rows have 7 circles, and the middle row has 6. The outer circles touch the sides of the rectangle. Find the long side of the rectangle divided by the short side.

The working is far more important than just the solution.. in the majority of competitions, no working with the right solution would probably be worth almost nothing, maybe a single mark, whereas right working with a simple error could still get you full marks..

i dont get it, when other people put just an answer, nobody questions their work, only when i do... sigh

you didnt even tell me if I was right

sorry just had a fight with g/f, in bad mood.

[hide]

its quite simple actually, surprising to me because i rarely can answer AIME questions.

the long leg of the rectangle is 14r. the trick is to find the width.

with a hunycomb shape, the first circle on each column (my rectangle is longer vertically) touch each other, the distance from their radii should be 2r. the left column circle and thre right column circle are even with each other, so the width is the 2r plus the distance between these centers... the touching circles give a triangle with sides of 2r, solving the hypotenuse is 2:sqrt:2r.... so the width is 2:sqrt:2r + 2r

putting 14r / (2:sqrt:2r + 2r) and reducing gives the answer above.

[/hide]

is that rite, guys?
im in the middle of a formal dinner rite now, so can't check it.

Shouldn't it be 7 root 3 + 7 instead? the length is 14r, for the width you can just use a 30- 60 - 90 right triangle and add 2r . dividing gives 7 root 3  + 7. rite?[hide][/hide]


[color=red]Show that [/color]$7$ disks radius $1$ can be arranged to cover a disk radius $2$.


An equilateral triangle and a square both have perimeters of 48 inches. What is the ratio of the length of the side of the triangle to the length of the side of the square? Express your answer as a common fraction.

the length of the side of the triangle:$ \frac{48}{3}\equal{}16$
the length of the side of the square:$ \frac{48}{4}\equal{}12$
Thus, $ \frac{12}{16}\equal{}\boxed{\frac{3}{4}}$.

Actually, the problem asks for the ratio of the triangle's side to the square's side, so the answer is $\boxed{\frac{4}{3}}$.


algebra
polynomial
geometry
3D geometry
inequalities
factorization
sum of cubes

January 16, 2008
                                                                           Day 1
1.In convex quadlitaleral $ ABCD$ $ K,L,M,N$ points are the midpoints of sides $ AB,BC,CD,DA$ respectively. Line $ KM$ meets the diagonals $ AC$ and $ BD$ at points $ P$ and $ Q$ respectively.Line $ LN$ meets the diagonals $ AC$ and $ BD$ at points $ R$ and $ S$ respectively.Known that $ AP \cdot PC \equal{} BQ \cdot DQ$.Prove that  $ BS \cdot SD \equal{} AR \cdot RC$.
2.The polynomial $ P(x)$ is called good if it can be represented as the sum of three cubes of polynomials(with variable $ x$)
For example, the polynomial $ P(x) \equal{} x^{3} \minus{} 1 \equal{} x^{3} \plus{} ( \minus{} 1)^{3}$ is good and the polynomial $ P(x) \equal{} 9x^{3} \minus{} 3x^{2} \plus{} 3x \minus{} 1 \equal{} (2x)^{3} \plus{} (x \minus{} 1)^{3}$ is also good.
(a)Is the polynomial $ 3x \plus{} 3x^{7}$ good?
(b)Is the polynomial $ 3x \plus{} 3x^{7} \plus{} 3x^{2008}$ good?

                                                         January 17,2008
                                                                Day 2
4.For any positive integer $ n$, let $ s(n)$ denote the sum of its digits in its decimal representation.
Find all positive integers $ n$, for which $ n \equal{} 2s(n)^{3} \plus{} 8$.
5.The circles $ w_{1}$ and $ w_{2}$ with centres $ O_{1}$ and $ O_{2}$ respectively don't have common point.$ A_{1}A_{2}$ is common outside tangent line, where $ A_{1}$ is on $ w_{1}$ and $ A_{2}$ is on $ w_{2}$.The midpoint of $ A_{1}A_{2}$ is $ K$. The points $ B_{1}$ and $ B_{2}$ are on circles $ w_{1}$ and $ w_{2}$ respectively so, that $ A_{1}B_{1}$ is tangent to $ w_{1}$ and $ A_{2}B_{2}$ is tangent to $ w_{2}$.The intersection of lines $ A_{1}B_{1}$ and $ A_{2}B_{2}$ is $ P$.prove that the points $ B_{1}$,$ B_{2}$,$ K$ and $ P$ are concyclic.
6.Prove that if the positive reals $ a,b$ and $ c$ are such that $ abc \equal{} 1$, than we have the inequality:
$ \frac {1}{(a \plus{} b)b} \plus{} \frac {1}{(b \plus{} c)c} \plus{} \frac {1}{(c \plus{} a)a} > \equal{} \frac {3}{2}$.
Sorry,I will post the 3-rd problem later.

[quote="lasha"]4.For any positive integer $ n$, let $ s(n)$ denote the sum of its digits in its decimal representation.
Find all positive integers $ n$, for which $ n \equal{} 2s(n)^{3} \plus{} 8$.
[/quote]

$ 10^{k\minus{}1} \le n < 10^k$
$ s(n) \le 9k$
$ 2s(n)^3\plus{}8 \le 729 k^3 \plus{}8$
$ 10^{k\minus{}1} < n \le  729 k^3 \plus{}8$
$ 10^{k\minus{}1} < 729 k^3 \plus{}8$
$ k \le 6$

$ n < 10^6$
$ 2 s(n)^3\plus{}8 < 10^6$
$ s(n) < 37$

$ n \equiv s(n) (9)$
$ 2 s(n)^3\minus{}s(n)\minus{}1 \equiv 0 (9)$
$ s(n) \equiv 1 (9)$
$ s(n) \in \{1,10,19,28\}$
with solutions for $ n \in \{10,2008,13726\}$

[quote="chess"]Where can we discuss these prooblems? :)[/quote]
I'm going to create separate topics to each problem,and there everyone could post his solutions,or discuss problems.
P.S:
Are you this olympiad participant?

Check [url]http://www.mathlinks.ro/viewtopic.php?t=183534[/url]
There is a very interesting generalization of a problem posted by Erken which applies to the problem number 2

Anyone should post that link / quotation in the particular forum and mine for problem 4 if it is correct...thanks...

[quote="mszew"]
Anyone should post that link / quotation in the particular forum and mine for problem 4 if it is correct...thanks...[/quote]
It is right,but with some serious mistakes.
For example,
From inequality:
$ 2s(n)^3\plus{}9<10^6$,it follows that $ s(n)<80$,not $ s(n)<37$.
It will be easier,if you write another inequality,
$ S(n)<9k\leq 9\cdot 6\equal{}54$.
So you must check only $ S(n)\in\{1,10,19,28,37,46\}$,because $ 55>54$.

[quote="PhilAndrew"]Sorry, your statement of problem 5 is ambiguous - how can be $ A_1B_1$ be tangent to $ w_1$ if $ A_1, B_1 \in w_1$ ?!
Many thanks, however I see no good reason for you not to post everything if you decided to.[/quote]
It should be $ KA_1,KB_1$ are tangent lines to $ \omega_1$,and $ KA_2,KB_2$ are tangent lines to $ \omega_2$.
To PhilAndrew:
Are you this olympiad participant?

Could anybody post official results here?

sorry but i don't understand  poblem2 There is ,,sum  of thee cubes of polynomials"  or "sum  of  cubes of polynomials"
THANK YOU

" It should be  are tangent lines to ,and  are tangent lines to .
To PhilAndrew:
Are you this olympiad participant?"
thank you erken
thus this k  is   midpoint of A1A2.is it corret ?
if it is correct ,5 problem will be wrong

Yes, K is midpoint of $ A_1A_2$.this problem is correct and very easy :)

I am not a participant - just someone who writes comments here, and therefore I possess no official results.
In my opinon, problems 4 & 6 'd have a questionable level of difficulty, if we considered this an IMO level examination.
However, problem 5 is still unclear: my sketch shows that those points are far from being concyclic. Some mathlinker contestant should clarify its statement.


calculus
integration
algebra
polynomial
vector
abstract algebra
number theory

(a) How to find an integral basis of $ \mathbb{Q}\left(\sqrt[3]{5}\right)$ (over $ \mathbb{Q}$)?
(b) Why is $ \left\{1,\theta,\theta^2\right\}$ an  integral basis of $ \mathbb{Q}(\theta)$ (over $ \mathbb{Q}$) if $ \theta^3 \plus{} \theta \plus{} 1 \equal{} 0$?

For (b) - the discriminant is squarefree.

For (a) - I don't know :)

since $ \sqrt[3]{5}$ is a root of the irreducible polynomial $ x^3\minus{}5$, the extension $ \mathbb{Q}(\sqrt[3]{5})$ over $ \mathbb{Q}$ should have a bassis $ \{1,\sqrt[3]{5},\sqrt[3]{5^2}\}$
This looks easy, or I am missing somthing?

But why should this be an INTEGRAL basis?

An integral basis is a basis of the vector space $ K$ over a field $ L$, where $ K$ is an integral extension over $ L$, is it the definition of integral basis?
If so, there is a proposition, which states that, if $ p(x) \equal{} min(\alpha,L)$ then $ L(\alpha)$ over $ L$ has a basis $ \{1,\alpha,...,\alpha^{n \minus{} 1}\}$ ($ n \equal{} deg\,p$)

No an integral basis is a $ \mathbb{Z}$-module basis not just a vector space basis.

[quote="Arne"](a) How to find an integral basis of $ \mathbb{Q}\left(\sqrt [3]{5}\right)$ (over $ \mathbb{Q}$)?[/quote]
Claim: $ \mathbb Z[\sqrt [3]{5}]$ is the ring of integers of $ \mathbb Q(\sqrt [3]{5})$, that means that $ 1,\sqrt [3]{5},(\sqrt [3]{5})^2$ is an integral basis.
[hide="Proof"]
The proof uses the language of commutative algebra, since I am not sure which results of number theory you have in stock. So the proofs of basic theorems will appear here in parts. Usually these techniques are used in the proofs of the basic theorems in number theory.

Define $ \alpha \equal{} \sqrt [3]{5}$, $ K \equal{} \mathbb Q(\alpha)$, $ B \equal{} \mathcal O_K$, $ A \equal{} \mathbb Z[\alpha]$, $ f(X) \equal{} X^3 \minus{} 5\in\mathbb Z[X]$.

The discriminant $ d_A$ of $ A$ is $ d_A \equal{} \minus{} N^K_{\mathbb Q}(3\alpha^2) \equal{} \minus{} 5^2 3^3$, such that the prime divisors of $ (B: A)$ -- if any -- are contained in $ \{3,5\}$.

It holds $ A \equal{} B$: Define $ Q \equal{} B/A$. To prove that $ Q \equal{} 0$ it suffices to show that, for every prime $ p$ in $ \mathbb Z$, it holds for the localisation: $ Q_p \equal{} 0$. By Nakayama, it suffices to show
\[ Q\otimes \mathbb F_p \equal{} Q_p\otimes \mathbb F_p \equal{} 0\,.
\]
(i) [u]$ p\notin\{3,5\}$[/u]: Since $ Q$ is a torsion group annihilated by $ |d_A| \equal{} 5^23^3$ it follows that $ Q_p \equal{} 0$.
(ii) $ p \equal{} 3$: $ f$ remains irreducible in $ \mathbb F_3[X]$, since $ \overline{f}(X) \equal{} f(X)\mod 3$ does not have a zero in $ \mathbb F_3$. Thus,
\[ A' \equal{} A\otimes_{\mathbb Z}\mathbb F_3 \equal{} \mathbb F_3[X]/(\overline{f}(X))
\]
is a field of dimension 3 over $ \mathbb F_3$. Then the $ \mathbb F_3$-algebra homomorphism $ A'\rightarrow B/3B$ sending the residue class of $ X$ to $ \alpha \plus{} 3B$ is injective ($ A'$ is a field). Since $ B$ has rank 3 over $ \mathbb Z$, $ B/3B$ is a 3-dimensional $ \mathbb F_3$-vector space. So the injection is surjective and its cokernel $ Q\otimes\mathbb F_3$ vanishes.
(iii) $ p \equal{} 5$: Using $ 5B \equal{} (\alpha B)^3$, decompose the ideal $ \alpha B$ into a product of prime ideals in the Dedekind ring $ B$ and apply the fundamental equation
\[ 3 \equal{} n \equal{} \sum_{Q|5}e_Qf_Q
\]
to see that $ 3|e_Q$ for all $ Q$ lying above 5 and finally $ P \equal{} \alpha B$ is the only prime ideal lying above 5. Its residue class degree $ f_P$ is 1, i.e. the natural ring homomorphism $ \mathbb F_5\rightarrow B/P$ is an isomorphism. Consider the filtration
\[ B\supset P\supset P^2\supset P^3 \equal{} 5B
\]
and use the isomorphisms of $ \mathbb F_5$-vector spaces
\[ \mathbb F_5\rightarrow P^i/P^{i \plus{} 1}\quad a \plus{} P\mapsto \alpha^ia \plus{} P^{i \plus{} 1}
\]
to derive that the $ \mathbb F_5$-vector space $ B/5B$ is generated by $ 1,\overline{\alpha},\overline{\alpha}^2$. But this means that
\[ B\otimes\mathbb F_5 \equal{} B/5B \equal{} \mathbb F_5[1,\overline{\alpha},\overline{\alpha}^2] \equal{} A\otimes \mathbb F_5
\]
or $ Q\otimes \mathbb F_5 \equal{} 0$.

QED[/hide]In terms of number theory, the critical primes in $ \mathbb Q(\sqrt [3]{5})$ are 3 and 5 since they are exactly the divisors of the discriminant $ \minus{} 3^35^2$ of $ \mathbb Z[1,\sqrt [3]{5},\sqrt [3]{5}^2]$. [color=red]3 does not split with residue class degree 3 (extension non-separable)[/color], whereas 5 is ramified with ramification index 3.

EDIT: Correction w.r.t. conclusion on the ramification of 3. Thanks to myan for the comment.

[quote="myan"]I did not read your entire post, but $ 3$ is ramified because otherwise $ \mathbb{Q}(\zeta_3)$ would be unramified at $ 3$ (subfields, composition and conjugates of unramified are unramified). Also, $ X^p \minus{} a \in \mathbb{F}_p[X]$ is reducible. $ p$ is ramified iff it divides the discriminant.[/quote]
Correct. $ 3$ is ramified because there is a single prime $ P$ above $ 3$ with non-separable residue field extension. It should have been "$ 3$ does not split" - I have just edited and corrected it. Thanks for pointing it out.


conics
hyperbola
function
analytic geometry
quadratics
algebra
quadratic formula

[quote="Mark9182"]I have three hard math problems for homework and I need any help that you can give me.  Here are the problems.

1. Simplify. Write your answer in the form of a + bi:
(6 –7i) + (2 – 3i)

2.  Find (f + g)(3) when f (x) = x + 6 and g (x) = x - 1.

3.  Find the equation of the hyperbola with Vertices at (9, 0) and (-9, 0); foci at (11, 0) and (-11, 0)

THANKS![/quote]

[hide]2. I hope f means f(x). THen it is (2x+5)3=10x+15.[/hide]

[quote="Mark9182"]Thank you.  Are you sure that this is right? and do you have any idea on the other two.  Out of 25 problems these were the only 3 that stumped me.[/quote]
$\imath$ is just a variable, like everything else that isn't a number.

Please don't call these problems "hard," because they aren't. :)

[hide="2"]$(f+g)(3)=(2x+5)(3)=11$[/hide]

Your help is greatly appreciated.  What about the last one?

I haven't learned how to do that yet... :ninja:

[quote="i_like_pie"][hide="1"]$(6-7\imath)+(2-3\imath)=8-10\imath$[/hide][/quote]

If he/she needs help with homework, then it is a good idea to post what you are doing to help him know how to solve it  :wink:

[hide="1"]You need to group together the $\imath$ terms and the constant terms.

The $a$ term will be $6+2=8$, and the $\imath$ term will be $-7\imath-3\imath=-10\imath$, so in the form $a+b\imath$, the answer is $8-10\imath$.[/hide]

[hide="2"]First, find the function $(f+g)(x)$, then substitute 3 for $x$.

$(f+g)(x)=f(x)+g(x)=(x+6)+(x-1)=2x+5$

Now, substituting 3 for $x$, we have $2(3)+5=11$.[/hide]

[hide="3"]It is best to start these with a picture of where the points are.

We know that the equation of a hyperbola is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.

When $y=0$, we know that $x=\pm9$, so we can substitute this information into our equation and solve for $a^{2}$.

$\frac{(\pm9)^{2}}{a^{2}}+\frac{0^{2}}{b^{2}}=1$
$\frac{81}{a^{2}}=1\Rightarrow a^{2}=81$

We also know that the foci are located at $(\pm11,0)$.

The $x$ coordinate of the foci represents $c$ in the equation $a^{2}+b^{2}=c^{2}$
Substituting what we know from earlier, we have
$81+b^{2}=121$
$b^{2}=40$

Thus, the equation of the hyperbola is $\frac{x^{2}}{81}-\frac{y^{2}}{40}=1$.[/hide]

[quote="Mark9182"]One More Problem.  I want to check your answer with mine.  Here is the problem.

Solve 45x + 9x2 = 0[/quote]

Sorry, you posted this while I was posting my other problem.

Do you mean $45x+9x^{2}=0$?

Also, anirudh, when a function is something such as $(f+g)(3)$, it means the same thing as $h(3)$, where $h(x)=f(x)+g(x)$, not multiplying the function by 3.

[quote="Imperial Effect"][quote="Mark9182"]Yes that is exactly what I mean on that question.[/quote]
I will complete the square someone else can do the quadratic formula of DOOM!!  :ninja:
[hide]This can be solved by completing the square. In order to that the equation must be in the form $x^{2}+bx = c$
$45x+9x^{2}= 0$

Divide both sides of the equation by 9

$9x^{2}+\frac{45x}{9}= \frac{0}{9}$

Simplify
$x^{2}+5x =0$ now the equation is in the "right" form :)

Divide $5$, the coefficient of x, by $2$ to get $\frac{5}{2}$. Add the square of $\frac{5}{2}$ to both sides of the equation. Thus making it into a perfect square:

$x^{2}+5x+(\frac{5}{2})^{2}= (\frac{5}{2})^{2}$

Square the factions

$x^{2}+5x+(\frac{25}{4}) = (\frac{25}{4})$

Because it a perfect square we can factor this as
$(x+\frac{5}{2})^{2}= \frac{25}{4}$

Take the square root of both sides
$\sqrt{(x+\frac{5}{2})^{2}}$ $=$ $\sqrt{\frac{25}{4}}$

We see can that:

$x+\frac{5}{2}= \frac{5}{2}$

$x+\frac{5}{2}=-\frac{5}{2}$

$x = 0$
$x =-5$ [/hide]
I hope that’s right :P[/quote]
That was way too much work!!!!!
[hide]You can factor it into
$9x(x+5)=0$
So either 9x=0 or x+45. So x is 0 or -5.[/hide]

[quote="bpms"][hide]You can factor it into
$9x(x+5)=0$
So either 9x=0 or x+45. So x is 0 or -5.[/hide][/quote]
That's correct, but I think you meant $x+5,$ not $x+45.$

[quote="nutz_for2.718281828"][hide="1 and 2 and 4:"]1. We add and get 8-10i

2. We want x+6+x-1=2x++5=11

4. $9x^{2}+45x=9x(x+5)=0$, so x equals 0 or -5.[/hide][/quote]

Please don't post after the question has been thouroughly answered.  Thanks.

A perhaps easier way to do the 4th one is:

[hide]First, just see if 0 works, because on a problem like that 0 often will. It does. That's our first solution.

Then, we can just divide  by x : $45+9x = 0$
Then, move the 45 to the other side; $9x =-45$
And we can divide by 9 to get the 2nd solution, $x =-5$[/hide]

[quote="b-flat"]
Thus, the equation of the hyperbola is $\frac{x^{2}}{81}+\frac{y^{2}}{40}=1$.[/quote]

You mean $\frac{x^{2}}{81}-\frac{y^{2}}{40}=1$, right? That's equation you posted is an ellipse.

Oops, I guess I didn't change the correct thing earlier.

Thanks!


pigeonhole principle
counting
distinguishability
geometry
number theory
Diophantine equation

Determine the number of ways to choose 5 numbers from the set $ {1,2,3,...,18}$ such that any two chosen numbers differ by at least 2.

A code for a safe is a 5-digit number that can have a 0 in the first place as well as any other place. how many codes are there whose digits form an increasing sequence?

the first one seems like the pigeonhole priniciple may help...

for the second, for ANY five digit number, we have an increasing sequence so the answer would be 10C5.

[b]@Poincare:[/b] The Pigeonhole Principle is used to show existence, not for computation.

[hide="solution 1"]
First find the number of subsets that have a pair differing by 1
The pairs of numbers in the set $ S \equal{} 1,2,3...18$ which differ by 1 are $ (1,2),(2,3)...(17,18)$ and there are $ 17$ such pairs

Consider the first pair, $ (1,2)$ If a subset contains these two numbers then the subset has a pair which has a difference of $ 1$ and there are $ \binom{16}{3}$ such subsets

Now considering the second pair $ (2,3)$ there are also $ \binom{16}{3}$ such sets. However since some of these subsets have already been counted previously we only count $ \binom{15}{3}$ of them

Considering all subsets that have a pair with a difference of one we have;
$ \binom{16}{3} \plus{} \binom{15}{3} \plus{} \binom{14}{3} \plus{} ...\binom{3}{3}$

Now $ \displaystyle\sum_{n \equal{} 3}^{16}\binom{n}{3} \equal{} \dfrac{1}{24}(n \plus{} 1)(n)(n \minus{} 1)(n \minus{} 2)$
\[ \dfrac{1}{24} (17)(16)(15)(14) \equal{} 2380
\]
Thus the number of sets which don't have a pair with a difference of $ 1$ is
$ \binom{18}{5} \minus{} 2380$

Answer: $ 6188$
...I think
[/hide]

[hide="solution 2"]
First i will assume that an increasing sequence is strictly increasing (e.g i will count {1,2,3,4,5} but not {1,2,2,3,4}

Notice that if the fist digit of the sequence is $ 6$ then the set $ {6,7,8,9...}$ connot be completed
therefore the fist term can take the digits $ 0\minus{}5$, the second term can take the digits $ 1\minus{}6$, third digit $ 2\minus{}7$...
This can be seen visualy in a $ 6\times5$ unit grid
$ 5 | 6 | 7 | 8 | 9$
$ 4 | 5 | 6 | 7 | 8$
$ 3 | 4 | 5 | 6 | 7$
$ 2 | 3 | 4 | 5 | 6$
$ 1 | 2 | 3 | 4 | 5$
$ 0 | 1 | 2 | 3 | 4$
 _ _ _ _ _

One digit from each column represents a number in the 5 digit set.
When creating an increaseing sequence the digit which you select from any column must be in a row that is higher than or equal to the row which you selected the previous digit.
Define a movement as one move to the right or upwards. Therefore every time you select one digit you can movements across to the right or upwards as many times as you like, but you can never move down or to the left. Therefore each number can be represented by 10 movements (5 movements across and 5 up)
Therefore the number of combinations of 10 movements is $ 10!$ but since half of the are the same there are $ \dfrac{10!}{5!5!}$

Answer: $ \binom{10}{5}$
[/hide]

[hide="Number 1"]
I got a different answer, but I might be wrong.

First Number Picked = 1:

We can have the second number range from 3 to 18, so there are 16 ways.

First Number Picked = 2:

Second number can be from 4 to 18, so there are 15 ways.

First number Picked = 3:

Second number can be from 5 to 18, so there are 14 ways.

Hmm...we see a pattern here. The answer is $ \frac{16(17)}{2} \equal{} \boxed{136}$.

[/hide]

Assuming you can choose any 2 there are $ \binom{18}{2}$ ways. Now we are overcounting by the number of cases in which elements differ by 2. To count these assume the first number is 1. Then we can have the second number be 2-3 for 2 ways. Starting with two the second can be 1,2,4 however we have counted one already so we get 2 again. We proceed this way until we get to 17, which has only 1 remaining uncounted pair, so the total is 33.

$ \binom{18}{2} \minus{} 33 \equal{} 120$

edit: nvm atleast two!! izzy is right

For number 1, imagine you have a row of $ 5$ chairs (call them $ A,B,C,D,E$), which will represent the numbers picked.
Then, insert $ 4$ chairs such that there is one chair between each two adjacent chairs. Now, the problem is simply inserting chairs before in each of the $ 6$ places (before $ A$, between $ A,B$,..., and after $ E$), giving us $ 6$ different slots to place $ 18 \minus{} 5 \minus{} 4 \equal{} 9$ chairs. This is balls and urns--$ \dbinom{9 \plus{} 6 \minus{} 1}{6 \minus{} 1} \equal{} \boxed{2002}$. Notice how this corresponds to the original problem.

now 18 is the largest number in the set.
the number of ways of choosing a pair (a,b):a>b and a=18 is 16 ways.
for a=17 it is 15 ways.
 so you can see that a sequence is forming so that the total number of ways is (n(n+1))/2 with n=16
i.e (16(17))/2=136

[quote="ThetaPi"]Determine the number of ways to choose 5 numbers from the set $ {1,2,3,...,18}$ such that any two chosen numbers differ by at least 2.[/quote]

I was referring to this problem.

Some people were looking for pairs of numbers that differ by 2+, instead of subsets of 5 numbers such that any of the 10 resulting pairs differ by 2+.

Now, the problem is simply inserting chairs before in each of the  places (before , between ,..., and after ), giving us  different slots to place  chairs

could u explain more?

Here's what I think he means:
Let an O represent a number we pick.
so we have O O O O O
Each number has to differ by 2 so we'll put:
O X O X O X O X O
There are 9 objects left so we find the amount of ways we can put them either before or in between or after each O.
So how many ways can we sort 9 objects into 6 possible spots.

[quote="math154"]Yes; that's what I mean. Sorry if I was unclear![/quote]

Could you elaborate on the balls and urns part?  I have no clue how to solve it the rest of the way.

[hide="Balls and Urns"]
In general, if you have $ n$ indistinguishable objects that you want to distribute to $ k$ distinguishable objects, there are $ \dbinom{n\plus{}k\minus{}1}{k\minus{}1}$ ways.
Imagine that there are $ k\minus{}1$ dividers and $ n$ objects, so we have
| | | ... | | and ***...*** where | is a divider and * is an object.
Then, label the areas of the dividers.
1|2|3|4| ... |k-2|k-1|k and ***...***
We can see that there are $ k$ regions in which we can place our $ n$ objects--there are $ \dbinom{n\plus{}k\minus{}1}{k\minus{}1}$ arrangements of the $ n$ *'s and $ k\minus{}1$ |'s.
Now, notice that if we arrange these $ n$ objects and $ k\minus{}1$ dividers, for every arrangement, we have some number of objects in region 1, some in region 2, etc. Since there are $ k$ regions, the number of arrangements of these $ n$ objects and $ k\minus{}1$ dividers corresponds to the number of ways to distribute $ n$ indistinguishable objects to $ k$ distinguishable objects.
[/hide]

[hide="1"]Alternatively, we can think of picking numbers as divding the sequence $ 1, 2, 3, ..., 18$ by $ 5$ dividers, where the number right before the divider is picked. We have this diophantine equation $ x_1\plus{}...\plus{}x_6\equal{}18$. Note $ x_1\geq 1$, $ x_2,...,x_5\geq 2$, and $ x_6\geq 0$ (it's what is left). After appropriate substitution we get $ y_1\plus{}...\plus{}y_6\equal{}9$ where all variables are nonnegative. By balls-and-urns the answer is $ \binom{9\plus{}6\minus{}1}{6\minus{}1}\equal{}2002$.[/hide]

[quote="math154"][hide="Balls and Urns"]
In general, if you have $ n$ indistinguishable objects that you want to distribute to $ k$ distinguishable objects, there are $ \dbinom{n \plus{} k \minus{} 1}{k \minus{} 1}$ ways.
Imagine that there are $ k \minus{} 1$ dividers and $ n$ objects, so we have
| | | ... | | and ***...*** where | is a divider and * is an object.
Then, label the areas of the dividers.
1|2|3|4| ... |k-2|k-1|k and ***...***
We can see that there are $ k$ regions in which we can place our $ n$ objects--there are $ \dbinom{n \plus{} k \minus{} 1}{k \minus{} 1}$ arrangements of the $ n$ *'s and $ k \minus{} 1$ |'s.
Now, notice that if we arrange these $ n$ objects and $ k \minus{} 1$ dividers, for every arrangement, we have some number of objects in region 1, some in region 2, etc. Since there are $ k$ regions, the number of arrangements of these $ n$ objects and $ k \minus{} 1$ dividers corresponds to the number of ways to distribute $ n$ indistinguishable objects to $ k$ distinguishable objects.
[/hide][/quote]
I did the permutations of 9 balls and 5 dividers so (14!)/(9!*5!) = 2002

Math154- Your ball and urn analogy is fair enough, but i dont understand how it fits with the question. How to the different slots represent 5 integers? and why does it mean that the remaining 'objects' have a difference of 2 or more?

In this case, the dividers actually represent the 5 integers. As you can see, in my example, there are 20 objects in total, with 5 dividers and 15 balls--one example is the following:
[code]|  O  O  O  |  O  O  |  O  O  O  O  O  |  O  O  O  O  |  O
1  2  3  4  5  6  7  8  9  10 11 12 13 14 15 16 17 18 19 20[/code]
You can see that, in this case, the 5 dividers, which are at 1,5,8,14,19, represent those 5 numbers, and none of them are next to each other. Do you see how in my solution above, initially inserting balls in between the dividers insured that there would be no two numbers next to each other?


group theory
abstract algebra
invariant
superior algebra
superior algebra unsolved

Let $ G$ be a group and let $ U$ be a subset of $ G$ defined as $ \{x_1x_2\cdots x_nx_n\cdots x_2x_1\mid x_i \in G\}$. Show that $ U$ is a subgroup of $ G$.

it is clear that $ 1 \in U$ and $ U$ is invariant under automorphisms and anti-automorphisms of $ G$. in particular, $ U^{ \minus{} 1} \equal{} U \equal{} g U g^{ \minus{} 1}$ for all $ g \in G$. now the product of two elements in $ U$, say $ (x_1 ... x_n x_n ... x_1) (y_1 ... y_m y_m ... y_1)$, is conjugate to $ x_n ... x_1 y_1 ... y_m y_m ... y_1 x_1 ... x_n$, hence in $ U$. this shows that $ U$ is a characteristic subgroup of $ G$.

Can you show that $ U \equal{} gUg^{ \minus{} 1}$ directly?
Thanks,

as I said, this is clear since $ x \mapsto g x g^{\minus{}1}$ is an automorphism ... so directly:

$ g (x_1 ... x_n x_n ... y_1) g^{\minus{}1} \equal{} ( g x_1 g^{\minus{}1})...( g x_n g^{\minus{}1}) ( g x_n g^{\minus{}1}) ... ( g x_1 g^{\minus{}1})$


Source- Me

    On the first day that GoBraves came to AoPs, he convinced 1 person at the end on the day that the Braves are better than the Yankees. On the Next day he convinced 2 more people. On the third day he convinced 4 more people, and on the next day, he convinced 8 more people that the Braves are better than the Yankees. If there are 6991 users on AoPs, how long will it take him to convince everybody?

[hide="solution"]

At the end of n days, there are 2^n - 1 people he has convinced.  So, since 2^12 = 4096 and 2^13 = 8192 , it will take 13 days.

13[/hide]

But then again, eveyone should already know that the Yankees stink. ;)

I'm totally with dasherm on this one.  Yankee's stink!  No offense in any way to any person or thing.
hehe  :rotfl:

[quote="GoBraves"][quote="dwx314"]If GoBraves can convince THAT many people that the yankees stink THAT fast, How come GoBraves only still has 1 full star? Another thing, if he can convince that many people, then he must have some good reasons. And if he has good reasons, then how come he's the only one doing the convincing?

Anyways, dasherm, it's not 2^n-1. Your forgeting GoBraves is also an AoPSer too.[/quote]

That would be 2 stars, and I have a million reasons why the Braves are better, and im the only person doing it because im the only true Braves fan here.[/quote]

Lol.  That's good you're sticking to your team.  Loyalty is good. Some former Yankee fans switched to Red Soxs fans after the Red Soxs beat the Yankees in the World Series.

The trick in this problem is that you have to make sure that you realize that it is 2^n-1 power, not 2^n for people convinced that day.... and of course with the sum of squares formula you can easily take it from there.
BTW You will never be able to convince everyone in the forum that the Braves are the best because the RED SOX RULE!!!!!!!!

Werd.

But the Braves are cool too. It's hilarious how every spring, people say how this is the year they don't win the NL East but they do it anyways.

All that aside, pretty neat problem. But how do you take into account the number of people who already hate the Yankees?  ;)  :D

In light of that, I'll post a follow up. GoBraves convinces people that Yankees suck at the same rate, but doesn't have to convince the people who already thing the Yankees suck. If there are 6991 users on AoPS-Mathlinks and it takes GoBraves 11 days to spread the holy word throughout the forums, what is the least and most number of users who already know that Yankees suck?

what about those of us from New England who need no convincing that any team is better than the yanks?

Since it can be expected that most of the people who are not yankee fans do not like them, there might not be many people at all who need convincing.

Next up: convincing cubs fans braves are better  ;)

what about some people... cough cough... me..... who don't really care about which team is better?

You fall under the category that is what the problem was inteded to be about.

I agree that the braves are cool, but they aren't the best, well except for in their pathetic division (I lived in Montreal and watched the only year the Braves stunk in a while, and when the Expos got jipped of a world series).. ok back to math... here's another problem... which I just created...
What's the probability that the red sox will win 100 of the 162 ways? No, one isn't an answer, and express your answer as a common fraction.

Sry.. my b... I ment to say what is the probability that the Red Sox will win 100 of the 162 games? Actually that's pretty tough... here's an easier problem. In a baseball playoff series, you need to be the first team to win four out of seven games (except int he first round). The Red Sox lost the first three games against the Yankees and then won the next four. Considering that the probability that the Red Sox winning each game was 50:50, what was the probability that that specific outcome would have happened, in that particular order?

hm... not really sure how to do this, but if i had to guess this would be my answer:

(i'm not sure whether or not I should hide my answer, so I did it anyway)

[hide](1/2)^7.... or one half to the seventh power... I'm not sure how I thought of this, but it's like a random guess[/hide]

ok i think that you that since there are 2 possible outcomes for each game and there are 7 games it would be 2^7 possible outcomes or 128 possible outcomes.   but you also have to remember that if red sox win the first four then the series will automatically end. i don't know how you would subtract that ous so my closest guess would be 1/128.   :?

The Cleveland Indians haven't been good for a while, but watch out. Their plan of just getting prospects is finally paying off, and so.... watch out for them. They'll be a big surprise team, and anyway, everyone knows they're the best.

Yeah, okay, I think the answer's 1+12. [hide] Made you look[/hide]

Anyway, I think the Mets are the best! Yeah!


function
inequalities
inequalities proposed

[quote="flip2004"][color=green] [b] If  $f:$ $[0,1]\to {\mathbb R}$ is convex on $[0,1] ,$  prove the inequality [/b] [/color]
\[ \begin{array}{|c|} \hline \\ \displaystyle \sum\limits_{k=0}^{n+1} {n+1 \choose k} f\left(\frac{k}{n+1}\right)\le \displaystyle 2 \sum\limits_{k=0}^{n} {n \choose k} f\left(\frac{k}{n}\right)\\ \\ \hline \end{array}\; .  \][/quote]
[color=blue][b]Solution by proposer[/b] [/color]
[hide]We have  equality
\[ \begin{array}{c} \displaystyle 2\sum\limits_{k=0}^{n}{n\choose k}f\left(\frac{k}{n}\right)- \displaystyle \sum\limits_{k=0}^{n+1}{n+1\choose k}f\left(\frac{k}{n+1}\right)=\\ \\ =n(n+1)\sum\limits_{k=0}^{n-1}\frac{{n-1\choose k}}{(k+1)(n-k)}\left[ \displaystyle p_k f\left(\frac{k}{n}\right)-\displaystyle f\left(\frac{k+1}{n+1}\right)+(1-p_k)f\left(\frac{k+1}{n}\right)\right] \end{array}  \]
where   \[ \left\{ \begin{array}{rcl} p_k: = \frac{k+1}{n+1}\in (0,1) \\ p_k\frac{k}{n}+(1-p_k)\frac{k+1}{n}=\frac{k+1}{n+1} \end{array}\right. \; ,\; \; , k\in \{0,1,...,n-1\}.  \]
[/hide]

[color=green][b] Unsolved question:[/b][/color] [b]Suppose that $f \in C[0,1]$ is given such that   [u]for all [/u]  $\; \; \; n\in {\mathbb N}$
\[ \begin{array}{|c|} \hline \\ \displaystyle \sum\limits_{k=0}^{n+1} {n+1 \choose k} f\left(\frac{k}{n+1}\right)\le \displaystyle 2 \sum\limits_{k=0}^{n} {n \choose k} f\left(\frac{k}{n}\right)\\ \\ \hline \end{array}\; \; .  \]
 To find (if exists) the best bounds $\alpha_f \in \left[0,\frac{1}{2}\right]$ , $\beta_f \in \left[0,\frac{1}{2}\right]$  ,  for which $f$  is convex on the interval [/b]
\[ \left(\frac{1}{2}-\alpha_f\; , \; \frac{1}{2}+\beta_f \right) \; .  \]
[b] But in case $f\in C^1[0,1]$  ? [/b]


How prove this identity :
$ \binom{m\minus{}1}{m\minus{}1}\binom{n\minus{}m}{k\minus{}m}\plus{}\binom{m}{m\minus{}1}\binom{n\minus{}m\minus{}1}{k\minus{}m}\plus{}...\plus{}\binom{n\minus{}k\plus{}m\minus{}1}{m\minus{}1}\binom{k\minus{}m}{k\minus{}m}\equal{}\binom{n}{k}$,

where $ 1<m<k<n$

Just choose smultaneously $ k$ balls from $ n$ numbered balls (numbered from $ 1$ to $ n$) and discuss according to the value of the $ m^{th}$ chosen number.

Pierre.


calculus
integration
logarithms
probability
expected value
advanced fields
advanced fields unsolved

THere is an integral I need to integrate and I am stuck on it It is integrated freom 0 to infinity with respect to dS'

{{S'^2 }/{q*S'*sqrt{2*pi*t}}}*exp{{-{{{log{S' /S}-{{r-0.5*q^2}{t}}}^2}/{2*t*q^2}}

 [url=http://img156.imagevenue.com/img.php?loc=loc656&image=39272_MSP22596810201169991634_509_122_656lo.JPG][img]http://img156.imagevenue.com/loc656/th_39272_MSP22596810201169991634_509_122_656lo.JPG[/img][/url]

It's the expected value of an asset over all of it's possible values, 0 to inf
Any help would be appreciated.

If I identify the picture correctly, your integral reads $ I=\int_0^{\infty}\frac{S'}{qS'\sqrt{2\pi t}}e^{-0.5\cdot\left[\frac{\ln\frac{S'}{S}-t(r-0.5q^2)}{q\sqrt{t}}\right]^2}dS'$

Make the standard substitution $ w=\frac{\ln\frac{S'}{S}-t(r+0.5q^2)}{q\sqrt{t}}$, i.e. $ dw=\frac{dS'}{S'q\sqrt{t}}$ to get

$ I=\& \int_{-\infty}^{\infty}\frac{Se^{q\sqrt{t}w+t(r+0.5q^2)}}{\sqrt{2\pi}}e^{-0.5(w+q\sqrt{t})^2}dw= Se^{rt}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-0.5w^2}dw =Se^{rt}$

[quote="hjbrasch"]If I identify the picture correctly, your integral reads $ I = \int_0^{\infty}\frac {S'}{qS'\sqrt {2\pi t}}e^{ - 0.5\cdot\left[\frac {\ln\frac {S'}{S} - t(r - 0.5q^2)}{q\sqrt {t}}\right]^2}dS'$

Make the standard substitution $ w = \frac {\ln\frac {S'}{S} - t(r + 0.5q^2)}{q\sqrt {t}}$, i.e. $ dw = \frac {dS'}{S'q\sqrt {t}}$ to get

$ I = \& \int_{ - \infty}^{\infty}\frac {Se^{q\sqrt {t}w + t(r + 0.5q^2)}}{\sqrt {2\pi}}e^{ - 0.5(w + q\sqrt {t})^2}dw = Se^{rt}\int_{ - \infty}^{\infty}\frac {1}{\sqrt {2\pi}}e^{ - 0.5w^2}dw = Se^{rt}$[/quote]

Actually I left a a squared part on the first S', it should actually read like this:
$ I = \int_0^{\infty}\frac {S'^2}{qS'\sqrt {2\pi t}}e^{ - 0.5\cdot\left[\frac {\ln\frac {S'}{S} - t(r - 0.5q^2)}{q\sqrt {t}}\right]^2}dS'$

However ,would the principle  still be the same?

Make the similar substitution $ w\equal{}\frac{\ln\frac{S'}{S}\minus{}t(r\plus{}1.5q^2)}{q\sqrt{t}}$ then

$ I\equal{}\int_{\minus{}\infty}^{\infty}\frac{S^2e^{2q\sqrt{t}w\plus{}2t(r\plus{}1.5q^2)}}{\sqrt{2\pi}}e^{\minus{}0.5(w\plus{}2q\sqrt{t})^2}dw\equal{}$ $ S^2e^{2t(r\plus{}0.5q^2)} \int_{\minus{}\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\minus{}0.5w^2}dw \equal{}S^2e^{2t(r\plus{}0.5q^2)}$

for which bank do u work ?


Given a positive integer $ k$, there is a positive integer $ n$ with the property that one can obtain the sum of the first $ n$ positive integers by writing some $ k$ digits to the right of $ n$. Find the remainder of $ n$ when dividing at $ 9$.

[hide=" i think this is an indirect proof"]
note that for all numbers in the form $ 2 \cdot 10^n \minus{} 1$ for all n>0, the sum of the first n integers can be found by placing n zeros to the right of n
proof:
$ 1\plus{}2\plus{}3\plus{}4\plus{} \ldots \plus{}2 \cdot 10^n \minus{} 1\equal{}\frac{\left(2 \cdot 10^n \minus{} 1\right)\left(2 \cdot 10^n\right)}{2}\equal{}\left(2 \cdot 10^n\minus{}1\right)  \cdot 10^n$
but n can be any integer, so just choose n such that k=n, and you have the number, which, for all k (and n) has remainder $ \boxed{1}$[/hide]


find the inverse function of :

f (x) = 1 + x.arctanx

 f ^ -1=???

As far as I'm aware, the inverse is not a function, merely a relation.

Your function $ f$ is not one-to-one. $ f( \minus{} 1) \equal{} 1 \plus{} \frac {\pi}{4} \equal{} f(1)$
Even if you restricted the domain (i.e. cut the function) to $ x \geq 0$ so that it was one-to-one, you probably could not find an explicit formula for the inverse of this function. It's not very unsual for inverse functions to not have explicit formulas.


find equivalent of this sum  where $N\to+oo$ and
$\sum_{0<n\leq N/lnN}\frac{(n+1)^{1/3}+(n)^{1/5}ln(n-1)}{lnln(n^{2}+1)+ln(ln^{2}n+1)}$
 :D

Is there supposed to be a Ln (n-1)? I think 1 is certainly in the range of n so..  :maybe:


The solution of the equations
\begin{align*}
2x-3y&=7 \\
4x-6y &=20 \\
\end{align*} is:

$ \textbf{(A)}\ x=18, y=12 \qquad
\textbf{(B)}\ x=0, y=0 \qquad
\textbf{(C)}\ \text{There is no solution} \\
\textbf{(D)}\ \text{There are an unlimited number of solutions} \qquad
\textbf{(E)}\  x=8, y=5$

[hide="Click for solution"]
The slopes are the same but the $ y$-intercepts are different, so there are no solutions, or $ \boxed{\textbf{(C)}}$.
[/hide]


calculus
integration
geometry
inequalities
vector
trigonometry
geometric transformation

Find the area of the region expressed by the system of inequality $ \frac{x^2}{4}\plus{}y^2\leq 1,\ \left(1\plus{}\frac{\sqrt{3}}{2}\right)x\plus{}y\leq 1.$

First make the change of variables $ u \equal{} \frac x2.$ We will find the area of the resulting figure in the $ u,y$ plane and then multiply that result by $ 2.$

We have $ u^2 \plus{} y^2\le1$ and $ (2 \plus{} \sqrt {3})u \plus{} y\le 1.$

Let $ \mathbf{v}$ be the unit vector in the same direction as $ \left\langle 2 \plus{} \sqrt {3},1\right\rangle.$ That is,

$ \mathbf{v} \equal{} \frac1{2\sqrt {2 \plus{} \sqrt {3}}}\left\langle 2 \plus{} \sqrt {3},1\right\rangle.$

We are now trying to find the area of that portion of the unit circle that satisfies $ \mathbf{v}\cdot \langle u,y\rangle\le \frac1{2\sqrt {2 \plus{} \sqrt {3}}}.$

Let $ \theta$ be the angle such that $ \cos\theta \equal{} \frac1{2\sqrt {2 \plus{} \sqrt {3}}}.$ Then the area we seek is the rest of the circle plus a triangle. That is, the area is

$ \pi \minus{} \frac12(2\theta) \plus{} \cos\theta\sin\theta.$

But $ \cos2\theta \equal{} 2\cos^2 \minus{} 1 \equal{} \minus{} \frac {\sqrt {3}}2,$ so $ 2\theta \equal{} \frac {5\pi}6.$

And in that case, $ \cos\theta\sin\theta \equal{} \frac12\sin2\theta \equal{} \frac14.$ Hence our area in the circle is

$ \pi \minus{} \frac {5\pi}{12} \plus{} \frac14 \equal{} \frac {7\pi}{12} \plus{} \frac14.$

Multiply that by $ 2$ to get our final answer: $ \boxed{\frac {7\pi}{6} \plus{} \frac12}.$

[I won't guarantee that against computational errors - but the basic idea should work.]

[quote="Kent Merryfield"]First make the change of variables $ u \equal{} \frac x2.$ We will find the area of the resulting figure in the $ u,y$ plane and then multiply that result by $ 2.$[/quote]

I can't understand it.

It's a linear homothecy: compress the plane in the $ x$ direction by a factor of $ 2$. That changes the area of all figures in the plane by that factor of $ 2.$ You can generalize this for more general linear transformations, but the particular case was all we needed here. The motivation for doing that here was to convert that ellipse into a circle, because I knew that with a circle, I would have a good variety of geometric and trigonometric arguments to draw on. Note that I never even calculated an integral.


modular arithmetic
number theory unsolved
number theory

Prove that for all integers $n > 2$,
   \[ 3| \sum\limits_{i=0}^{[n/3]} (-1)^i C _n ^{3i} \]

Let $A_n = \sum \limits_{i=0} (-1)^{3i} \binom{n}{3i}$
Let $B_n = \sum \limits_{i=0} (-1)^{3i+1} \binom{n}{3i+1}$
Let $C_n = \sum \limits_{i=0} (-1)^{3i+2} \binom{n}{3i+2}$

$A_n+B_n+C_n = (1-1)^n = 0$

If $n$ is odd,
If $\dfrac{n-3}{2} \equiv 0 \pmod 3$, then $A_n=0, B_n=C_n$
If $\dfrac{n-3}{2} \equiv 1 \pmod 3$, then $B_n=0, C=A_n$
If $\dfrac{n-3}{2} \equiv 2 \pmod 3$, then $C_n=0, A=B_n$
Since $A+B+C=0$, it follows that $A_n \equiv B_n \equiv C_n \equiv 0 \pmod 3$ for odd $n$.

We also have, from Pascal's identity,
$A_{n-1}-C_{n-1}=A_n$
Thus our result follows for even $n$.

[quote="warut_suk"][quote="schulmannerism"]
If $\frac{n-3}{2} \equiv 1 \pmod 3$, then $B_n=0, C=A_n$
[/quote]

It should be $C_n = -A_n$, doesn't it?[/quote]

What's the solution, then?


A thermal machine works by Carnot cycle. The gas transfers to the cold sourse  the fraction f=0.6 of the heat received from the hot source, whose temperature is 227 degrees Celsius. Find the temperature of the cold source. pls help!! thanx

yes, the solution is $ 26,94 ^o C$, but i don't know how to get to this.  :maybe: Could you tell me, please? :oops:  thank you