modular arithmetic
search
Diophantine Equations
pen

Solve the equation $x^2 +7=2^n$ in integers.

[quote="Peter"]Solve the equation $ x^2 \plus{} 7 \equal{} 2^n$ in integers.[/quote]
Here is my solution
We have:
$ x^2\equal{}2^n\minus{}7 \equiv (\minus{}1)^n\minus{}1(mod 3)$
If $ n\equal{}2k\plus{}1;k \in Z$ then $ x^2 \equiv \minus{}2(mod 3)$,wrong
So $ n\equal{}2k$ with $ k \in Z$
Then $ 2^{2k}\minus{}x^2\equal{}7$
<=>$ (2^k\minus{}x^2)(2^k\plus{}x^2)\equal{}7$
=>$ 2^k\plus{}x^2\equal{}7 and 2^k\minus{}x^2\equal{}1$
So $ 2.2^k\equal{}k\equal{}>k\equal{}2$
=>$ n\equal{}4$=>$ x^2\equal{}9$=>$ x \equal{}\pm 3$

What's wrong with $ x^2\equiv \minus{}2\pmod3$?

Schmu

Nothing is wrong with it. There are actually such solutions: $ x\equal{}1, x\equal{}5, ...$. So please try again.

[quote="Peter"]Solve the equation $ x^2 \plus{} 7 \equal{} 2^n$ in integers.[/quote]

This one is famous,but it seems there are no elementary proof. :(

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=98196791&t=192321]Posted[/url].  This equation goes back to Ramanujan and the [url=http://www.math.sc.edu/~filaseta/gradcourses/TheMath784Notes.pdf]standard solution[/url] due to Nagell involves a fair amount of work in $ \mathbb{Z} \left[ \frac{1 \plus{} \sqrt{\minus{}7}}{2} \right]$.

[quote="Peter"]Hmm... not really a PEN-problem if that's what it needs to solve this.[/quote]
As one can see in 8th post in the link above,I said that this problem was given at a Preparation olympiad,so it seems to me that this problem must have almost elementary solution,furthermore,in my opinion it should not be very hard to find a such proof,as even 9th graders participated in it :roll: ...
Also [url=http://www.mathlinks.ro/viewtopic.php?t=150421]this problem[/url] was used at that Olympiad,although nobody was able to find a proof without using Gaussian numbers during olympiad and as I can see at ML elementary solution wasn't discovered for a long time too.
This remark gives us hope that there must exist "almost elementary" proof and let's just wait until somebody...
Unfortunately Teacher who gave us this problem can not find the place where he saw elementary solution and I have a sense that he maybe didn't even try to search :mad:
Best regards.

[quote="Erken"][quote="Peter"]Hmm... not really a PEN-problem if that's what it needs to solve this.[/quote]
As one can see in 8th post in the link above,I said that this problem was given at a Preparation olympiad,so it seems to me that this problem must have almost elementary solution,furthermore,in my opinion it should not be very hard to find a such proof,as even 9th graders participated in it :roll: ...[/quote]
Maybe the person putting this in the test didn't know it was so difficult, or had a wrong solution in mind (like the few posted on ML)?

The fact that this is a famous problem that Ramanujan couldn't solve, strongly suggests that this is NOT a pre-olympiad level problem.

[quote=reveryu]$(X+3)(X-3)=2^4(2^q-1)$ for $n=4+q$
We see $x =2k+1$
$=>(k+2)(k-1)=4(2q-1)$

It should be  $8(2^q-1)$  . How did you come up with $4(2q-1)$?[/quote]

Consider (X+3)(X-3)=2(k+2)2(k-1)=4(k+2)(k-1)
But you're right about 2q-1.it should be 2^q-1
That was a perfect solution

[quote=venkats]

$=>(k+2)(k-1)=4(2q-1)$
Observe $(k+2,k-1) =1$ for being odd or even,
And also $(4,2q-1) =1$
$k+2=4$ or $k-1=4$
[/quote]
Counter-example: $2 \cdot 15= 6\cdot 5$?


Okay so we need a list of Bay Area MOPpers so we can get together at MOP and overwhelm everyone else muwahahahaha.

Here's what I know so far:
Albert Gu - Saratoga
David Zeng - Saratoga
Gay - Saratoga
Wendy Mu - Saratoga
Taylor Han - Gunn
John Boyle - Paly
Lynnelle Ye - Paly
Colleen Lee - Paly
Vishal Arul - Homestead
Paul Christiano - Harker
Evan O'Dorney - Berkeley Math Circle

Yay that's like 15% of the people at MOP.

So should I add what color MOP and/or score to the list?

[quote="Ubemaya"]Okay so we need a list of Bay Area MOPpers so we can get together at MOP and overwhelm everyone else muwahahahaha.

Here's what I know so far:
Albert Gu - Saratoga
David Zeng - Saratoga
Gay - Saratoga
Taylor Han - Gunn
John Boyle - Gunn
Lynnelle Ye - Paly
Vishal Arul - Homestead

Hmm uh I think we need more people.[/quote]
boyle is paly
colleen lee - paly for cgmo

[quote="Ubemaya"]Okay so we need a list of Bay Area MOPpers so we can get together at MOP and overwhelm everyone else muwahahahaha.

Here's what I know so far:
Albert Gu - Saratoga
David Zeng - Saratoga
Gay - Saratoga
Taylor Han - Gunn
John Boyle - Paly
Lynnelle Ye - Paly
Colleen Lee - Paly
Vishal Arul - Homestead

Hmm uh I think we need more people.[/quote]
dude that's a pretty awesome third person you got on the list

AND UHH YOU SEEM TO BE FORGETTING A VERY PARTICULAR WINNER FROM THE BAY AREA COUGH COUGH

2 winners in fact...Paul Christiano from Harker (his username on AoPS is a string of binary...), Evan O'Dorney.

Whos the other one? I believe one was the award winner, no?

It totally should've had me, Brian Zhang, and a few others on there...:( MUHAHA RED MOP CHEAPNESS

Man, you guys are seriously pwn. Have fun at MOP, and I want to see some blue/black/IMO team members in a few years.
And CGMO is possibly THE least legit way to get into MOP. Just sayin.

Yay first person from Homestead to make MOP!  :wow:
Hey, Ubemaya, how many points did you improve from last year?

I think you should specify which color each person made...

dude both red and CGMO are now fake mop

blue and black are real mop

[quote="vishalarul"]Yay first person from Homestead to make MOP!  :wow:
[/quote]
dude no fair i dont get to say things like this

[quote="Ubemaya"]
Gay - Saratoga
[/quote]

lol.
guess quatto didnt make mop this year
[hide]
sarcasm
Quatto's gay :o
[/hide]

[quote="MysticTerminator"][quote="Altheman"]red mop is 'mop lite'

i'm not sure what to call cgmo mop[/quote]

they are both fake mop

unless of course a girl happens to legitimately make real mop and just incidentally happens to be on the cgmo team[/quote]
mehh could we consider red MOPers MOPlets?

[quote="mathclass"]did bowei not make it?[/quote]
Hmmm...did you make it?
Currently we don't know if bowei made it I think...judging from his beastly skills he had to have made it.

I made 'fake mop' with a score of 14, but i'm going to rsi instead.

[quote="Hamster1800"][quote="MysticTerminator"][quote="Altheman"]red mop is 'mop lite'

i'm not sure what to call cgmo mop[/quote]

they are both fake mop

unless of course a girl happens to legitimately make real mop and just incidentally happens to be on the cgmo team[/quote]

dude

i definitely made 'real mop' last year

i blame the higher ups[/quote]

yeah I know

so you made real MOP

good job


Josh and Mike live 13 miles apart. Yesterday, Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?

$ \textbf{(A)}\ 4\qquad
\textbf{(B)}\ 5\qquad
\textbf{(C)}\ 6\qquad
\textbf{(D)}\ 7\qquad
\textbf{(E)}\ 8$

[quote="Silverfalcon"]Josh and Mike live 13 miles apart. Yesterday, Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?

A. 4
B. 5
C. 6
D. 7
E. 8[/quote]

[hide]
$d=rt$
Since Josh rode for twice as long as Mike, and 4/5 the rate, we could say that $13=2\cdot(\frac{4a}{5})+1\cdot a)$, where a is some value (we don't know the exact time or rate).  Solving, $a=5$.  That would mean that Mike rode for $d=rt=5\cdot 1=5$ miles.
B
[/hide]

[hide="Answer"]Josh had gone $2\cdot \frac{4}{5}=\frac{8}{5}$ of the distance that Mike had, and $8+5=13$, so Mike had ridden $5$ miles. $\boxed{B}$[/hide]

[hide]
Let $M$=distance travelled by Mike; $(2/3)D$=amount travelled by Josh; m=Mike's speed.
$\\13-M=(2/3)D=(4/3)mt2\\13-mt=8mt/5\\13=13mt/5\\mt=5\\M=mt=5$.
[/hide]

Masoud Zargar


Another poll for writing instruments... :)

I use a pentel p207 and a rotring tikky ii. (oh yeah, I forgot to add pen to the poll, can a mod add it for me?)

I use a Pilot G2 0.5 mechanical pencil. I have yet to encounter anything better than it in terms of writing, and it's great for spinning as well. :)

Why is there two of each option in the poll? You can still edit it because it is less than 24 hours. I like to use mechanical pencils.

[youtube]zMvdpfN8utw[/youtube]

32 seconds in.

[quote="eggylv999"]how come no one has voted for the second copy on either of the options??[/quote]

I voted for woodcase pencil (the second one) and it transferred the vote to the first one. Glitch.

The mods don't like you, no adding pen option today

:( die

pilot G2 pens. ;]
great for spinning too...

Why is it called a writing instrument if you can't make music from it?

You can, bash it against a table and you'll see what I mean.
 :ninja:  :ninja:  :ninja:


Dr Sonnhard Graubner, you missed an equals sign
in the first line of your calculations. I placed it in:

[quote="Dr Sonnhard Graubner"]hello, taking the logarithm of both sides we get
$ \log_9 (x^2 \plus{} 2x \plus{} 4)\cdot \ln(3^{ \minus{} 1})\equal{}\ln(6)\log_{\frac {1}{6}}(x \plus{} 2)$
$ \frac {\ln(x^2 \plus{} 2x \plus{} 4)}{2\ln(3)}\ln(3^{ \minus{} 1}) \equal{} \frac {\ln(x \plus{} 2)}{\ln(6^{ \minus{} 1})}\ln(6)$
$ \ln(x^2 \plus{} 2x \plus{} 4) \equal{} 2\ln(x \plus{} 2)$
$ x^2 \plus{} 2x \plus{} 4 \equal{} x^2 \plus{} 4x \plus{} 4$
$ x \equal{} 0$
$ x \equal{} 0$ fulfilles our original equation.
Sonnhard.[/quote]

hello, yes, you have right, i inserted it, thank you.
Sonnhard.


quadratics
induction
function
limit
algebra
polynomial
quadratic formula

[hide="Solution"] Setting $x_{n}= 0$ and solving for $k$ produces a convergent of $\boxed{ \phi-1 }$. [/hide]

The root of the equation $x^{2}-x-1=0$

by the quadratic formula, that's $\frac{1\pm\sqrt{5}}{2}$, and the symbol $\phi$ usually denotes the positive one. Phi has many interesting properties.

Most notably,
$\phi+1=\phi^{2}$ (from the equation above)
$\phi^{-1}=\phi-1$

[hide="Somewhat of a further explanation"] It's not hard to see by induction that we have [b]something like[/b] $x_{n}= (-1)^{n}( F_{n}k-F_{n-1})$ (too lazy to work out the exact form). [/hide]

Not quite necessary; and you still have to prove the relevant lemma about the convergents, which is that the difference between the convergents and the number alternate signs.

[quote="N.T.TUAN"]The real sequence $x_{0},x_{1},x_{2},...$ is defined by $x_{0}=1,x_{1}=k,x_{n+2}=x_{n}-x_{n+1}$. Show that there is only one value of $k$ such that $x_{n}>0\forall n\geq 0$.[/quote]

It is necessary and sufficient for the function to satisfy: \[x_{n}>x_{n+1}\qquad (\forall n)\]
Notice that, since $x_{n+1}=x_{n-1}-x_{n}$, this is equivalent to: \[2x_{n}>x_{n-1}\qquad (\forall n)\]
Let $F_{M}$ be Fibonaccis, as usual. Suppose that \[F_{m}x_{q}>F_{m+1}x_{q+1}\]
Then by applying the Fibonacci recursion $F_{m+2}=F_{m}+F_{m+1}$ as well as the $x$-sequence recursion $x_{q+2}=x_{q}-x_{q+1}$ we derive:
\begin{eqnarray*}F_{m}x_{q}&>&F_{m+1}x_{q+1}\\ \iff F_{m+2}x_{q}&>&F_{m+1}x_{q-1}\\ \iff F_{m+2}x_{q-2}&>&F_{m+3}x_{q-1}\\ \iff F_{m+4}x_{q-2}&>&F_{m+3}x_{q-3}\end{eqnarray*}
It follows by induction (since the previous two observations serve as base cases, and the result holds for any $n$) that \[F_{m}x_{q}>F_{m+1}x_{q+1}\] for $m$ even and that \[F_{m}x_{q}<F_{m+1}x_{q+1}\] for $m$ odd. Since this result must hold for all $q$, we obtain:
\[\frac{F_{m}}{F_{m+1}}>\frac{x_{1}}{x_{0}}>\frac{F_{m+1}}{F_{m+2}}\]
for even $m$. Taking $m$ to $\infty$, it is clear that $\lim_{n\to \infty}\frac{F_{m}}{F_{m+1}}=x_{1}=\phi$ (a well known and easy to prove result)

All this "characteristic polynomial" stuff relied on in your proofs can be incorporated into easier elementary proofs, thats really all my proof did.

characteristic polynomials are elementary.  :wink:


Αν ένα τρίγωνο έχει πλευρές μεμήκη ακεραίους και η ακτίνα του εγγεγραμένου σε αυτό κύκλου είναι 1, να αποδείξετε ότι είναι ορθογώνιο.

[hide]
Δείξτε ότι οι πλευρές του ισούνται με 3, 4, 5.
[/hide]

Η λύση κρυφή για όσους θέλουν να το προσπαθήσουν ακόμα. Πιστεύω πάντως ότι η ενασχόληση για περισσότερο από μισή ώρα με το θέμα καταντάει ανούσια.

[hide]
Συμβολίζοντας $ r$ την ακτίνα του εγγεγραμμένου κύκλου, $ S$ το εμβαδόν, $ p$ την ημιπερίμετρο, $ a\le b \le c$ τις πλευρές του τριγώνου είναι:

$ S^2 \equal{} p(p \minus{} a)(p \minus{} b)p(p \minus{} c)$
$ S \equal{} pr \Leftrightarrow (r \equal{} 1) S \equal{} p$

άρα

$ p^2 \equal{} p(p \minus{} a)(p \minus{} b)p(p \minus{} c) \Leftrightarrow p \equal{} (p \minus{} a)(p \minus{} b)(p \minus{} c) (I)$

έστω $ x \equal{} p \minus{} a, y \equal{} p \minus{} b, z \equal{} p \minus{} c$ ($ \Leftrightarrow x \ge y \ge z$) τότε

$ x, y, z$ πρέπει να είναι ακέραιοι ή μισά περιττών ακεραίων ταυτόχρονα
αν είναι μισά περιττών ακεραίων τότε $ (I) \Leftrightarrow x \plus{} y \plus{} z \equal{} \frac {m}{8}$ με $ m \equal{}$ περιττός, που είναι άτοπο
άρα οι $ x, y, z$ είναι ακέραιοι

$ (I) \Leftrightarrow x \plus{} y \plus{} z \equal{} xyz \Rightarrow 3x \ge xyz \Leftrightarrow yz \le 3$

άρα
είτε $ y \equal{} 3$ και $ z \equal{} 1$ άρα $ 4 \plus{} x \equal{} 3x \Leftrightarrow x \equal{} 2 < y \Leftrightarrow$ άτοπο
είτε $ y \equal{} 2$ και $ z \equal{} 1$ άρα $ 3 \plus{} x \equal{} 2x \Leftrightarrow x \equal{} 3$
είτε $ y \equal{} 1$ και $ z \equal{} 1$ άρα $ x \equal{} x \plus{} 2 \Leftrightarrow$ άτοπο

άρα $ x \equal{} 3, y \equal{} 2, z \equal{} 1$ άρα $ p \equal{} x \plus{} y \plus{} z \equal{} 6 \Leftrightarrow a \equal{} p \minus{} x \equal{} 3, b \equal{} p \minus{} y \equal{} 4, c \equal{} p \minus{} z \equal{} 5$

αλλά είναι $ a^2 \plus{} b^2 \equal{} 3^2 \plus{} 4^2 \equal{} 5^2 \equal{} c^2$ άρα για τα $ a, b, c$ ισχύει το πυθαγόρειο θεώρημα

άρα το τρίγωνο είναι ορθογώνιο
[/hide]

Ωραίος  :wink:


Suppose the cevians $AP, BQ, CR$ meet at $T$. Prove that

$\frac{TP}{AP} + \frac{TQ}{BQ} + \frac{TR}{CR} = 1$.

A Hint : Use areas. We have

                              $\frac{TR}{AR} $ = $\frac{(TBC)}{(ABC)} $

   where $(TBC) , (ABC) $  means areas of the corespondin triangles. Write now two more relations for the other ratios and you are done.If problem , please ask me again.

Babis


Let $x,y,z$ be positve real numbers satifying $\left\{\begin{array}{c}x3x^{2}+3xy+y^{2}=75y^{2}+3z^{2}=27x^{2}+xz+z^{2}=16\end{array}\right$
Find value of $A=xy+2yz+3zx$

You can imagine the areas of three triangles.

Yes, this is a nice approach. The rezult is $24\sqrt{3}$


Hi, I am having trouble with the follow question regarding to n x n matrices.

Given that A and B are both n x n matrices, prove that
1. det(AB) = det(A)det(B)
2. adj(AB) = adj(B)adj(A)
3. rank(AB) <= min[rank(A), rank(B)]

I hope I can get help with these questions. Thank you very much.

He means the adjungated matrix, which you find by replacing each element with its cofactor, and then taking the transpose

I would like to add a question, does the adjungated matrix have geometric meaning, i mean when
$C^{-1}AC=B$ is
$C^{-1}Adj(A)C=Adj(B)$?


Show that there exists exactly $\binom{\left[ \frac{k}{2} \right]}{k}$ sequences $\ a_{1}, a_{2}, \ldots, a_{k+1}$ of integer numbers $\geq 0,$ for which $a_{1}=0$ and $|a_{i} - a_{i+1}|=1$ for all $i = 0, \ldots, k.$

First of all, it should be $\binom{k}{\left[ \frac{k}{2} \right]}$ not $\binom{\left[ \frac{k}{2} \right]}{k}$ (the latter always equals $0$).

Let $f(k,m)$ be the number of such sequences $a_{1}, a_{2}, \ldots, a_{k+1}$ that $a_{k+1} = m$. Then the original problem is equivalent to computing $F(k)=\sum_{m=0}^k f(k,m)$.

Note that $f(k,k)=1$, $f(2n+1,0)=0$ and $f(2n,0)=C_n$, the $n$-th [url=http://mathworld.wolfram.com/CatalanNumber.html]Catalan number[/url]. We can also let $f(k,-1)=0$ and $f(k,m)=0$ for $m>k$. Then $f(k+1,m) = f(k,m-1) + f(k,m+1)$ for all $m\geq 0$. Summation over $m=0..k+1$ gives $F(k+1) = 2F(k) - f(k,0)$ implying
$F(2n) = 2F(2n-1) = 4F(2(n-2)) - 2C_{n-1}$
$F(2n+1) = 2F(2n) - C_n = 4F(2n-1) - C_n$
that together with $F(0)=F(1)=1$ give rise to
$F(2n) = 4^n - 2 \sum_{i=0}^{n-1} 4^{n-1-i} C_i$
$F(2n+1) = 4^n - \sum_{i=0}^{n-1} 4^{n-1-i} C_{i+1}$

The sum $\sum_{i=0}^{n-1} 4^{n-1-i} C_i$ equals the coefficient of $x^{n-1}$ in the expansion of $\frac{1}{1-4x}\cdot\frac{1-\sqrt{1-4x}}{2x} = \frac{(1-4x)^{-1} - (1-4x)^{-1/2}}{2x}$ that is
\[ 2^{2n-1}(-1)^n\left(\binom{-1}{n} - \binom{-1/2}{n}\right) = 2^{2n-1} - \frac{1}{2}\binom{2n}{n}  \]
implying $F(2n)=\binom{2n}{n}$.

Similarly,
\[ \sum_{i=0}^{n-1} 4^{n-1-i} C_{i+1} = \sum_{i=0}^n 4^{n-i} C_i - 4^n = 4^n - \frac{1}{2}\binom{2(n+1)}{n+1} = 4^n - \binom{2n+1}{n}  \]
implying $F(2n+1)=\binom{2n+1}{n}$.


A snake of mass $m$ moves from one end of a boat (of mass $M$) to the other. What's the distance the boat travels in the mean time if its length is $L$?

One starts with the normal assumptions, frictionless interfaces between the boat and the water, the snake has negible length the snake has a smaller mass than the boat and nothing interacts with any outside objects.

The balance point of the snake/boat system doesn't change and to counter act the snake's movement.  The boat moves a distance equal to m*L/M.


The set {2, 3, 4, 5, 9} contains five of the factors of which one of the four integers?

480, 1260, 1890, 3888

[hide]i did trial and error and came up with 1260.[/hide]

[hide]
1260

Obviously divisible by 2
1+2+6+0=9 which is a multiple of 3 which means its divisible by 3
1200 is divisible by 4. 60 is divisible by 4.
Obviously divisible by 5.
1+2+6+0=9 which is a multiple of 9 which mean its divisble by 9.[/hide]

[quote="236factorial"][quote="math92"][hide]i did trial and error and came up with 3888.[/hide][/quote]

5 isn't a factor of that  :roll:[/quote]
 :wallbash:
i keep messing up every problem i do!!!!! :(

[hide]they are all divisible by 2
they are all divisible by 3
the first 2 and the last 1 are divisible by 4
the first 3 are divisible by 5
the middle 2 are divisible by 9
this leaves the second one [b]1260[/b] the only one with all the factors
[/hide]

[hide]
Since there's a five, 3888 doesn't work.
Since there's a 9, 480 doesn't work.
1890 doesn't work because there's a 4.
So, the answer is 1260.[/hide]


Moi nguoi oi
sap thi quoc gia 2006 roi
Doi tuyen Khtn manh lam :)

[color=blue]Đội tuyển của [b]Khoa học tự nhiên[/b] khi nào chẳng mạnh mà bạn lại nói vậy? [/color]

Nhưng mà em thấy dù sao thì cũng đừng bị ảnh hưởng bởi người ta wá!!!nếu không mình sẽ không làm bài dc!!chỉ cần mình cố gắng hết sức thì kết wả thế nào cũng thấy vui mà!!! :roll:


find all functions like $ f: N \to N \minus{} {1}$ that $ \forall n \in N$:
\[ f(n)\plus{}f(n\plus{}1)\equal{}f(n\plus{}2)f(n\plus{}3)\minus{}168\]

Why post this twice?

[url]http://www.mathlinks.ro/viewtopic.php?p=1329198#1329198[/url]


trigonometry
inequalities unsolved
inequalities

For a triangle , prove that :
$ a^2\,\preceq\,\frac{b^2}{\cos^2x}\plus{}\frac{c^2}{\sin^2x}$
with x any angle in the triangle

[quote="memath"]For a triangle , prove that :
$ a^2\,\preceq\,\frac {b^2}{\cos^2x} \plus{} \frac {c^2}{\sin^2x}$
with x any angle in the triangle[/quote]

$ \frac {b^2}{\cos^2x} \plus{} \frac {c^2}{\sin^2x}\equal{}(cos^{2}x\plus{}sin^{2}x)(\frac {b^2}{\cos^2x} \plus{} \frac {c^2}{\sin^2x})\geq (b\plus{}c)^{2}\geq a^{2}$

[quote="zaya_yc"][quote="memath"]For a triangle , prove that :
$ a^2\,\preceq\,\frac {b^2}{\cos^2x} \plus{} \frac {c^2}{\sin^2x}$
with x any angle in the triangle[/quote]

$ \frac {b^2}{\cos^2x} \plus{} \frac {c^2}{\sin^2x} \equal{} (cos^{2}x \plus{} sin^{2}x)(\frac {b^2}{\cos^2x} \plus{} \frac {c^2}{\sin^2x})\geq (b \plus{} c)^{2}\geq a^{2}$[/quote]

correct  :)

memath : a,b,c triangle side ? ---> $ b \plus{} c > a$  $ \frac{c^{2}}{sin^{2}x}\plus{}\frac{b^{2}}{cos^{2}x}>a^{2}$

[quote="zaya_yc"]memath : a,b,c triangle side ? ---> $ b \plus{} c > a$  $ \frac {c^{2}}{sin^{2}x} \plus{} \frac {b^{2}}{cos^{2}x} > a^{2}$[/quote]
yes we have a,b,c triangle side. why ? you're answer is correct


Does anyone know the divisibility rule for 31?

[hide="Test for divisibility by 31"]
Subtract three times the last digit from the remaining leading truncated number. If the result is divisible by 31, then so was the first number. Apply this rule over and over again as necessary.
Example: 7998-->799-3*8=775-->77-3*5=62 which is twice 31, so 7998 is also divisible by 31.
[/hide]


limit
real analysis
real analysis unsolved

Please help me to resolve this :

$ \lim_{ n->{\infty}} \prod_{k=1}^{n} \frac{(k+1)^{2}}{k (k+2) }$


trigonometry
function
calculus
calculus computations

How would you go about solving the following.
Maximize $ \cos(a)\plus{}\cos(b)\plus{}\cos(c)$ given that $ a\plus{}b\plus{}c\equal{}S$ and $ S$ is a positive integer. The original problem also assumes that $ a,b,c$ are also positive integers but I would like to solve this "simpler" case first.

[url=http://en.wikipedia.org/wiki/Lagrange_multipliers]Lagrange multipliers[/url].  The computation is very straightforward depending on the value of $ S$ and whether $ a, b, c$ are restricted to being positive.

I understand that I have to use Lagrange multipliers, however, I run into problems when trying to solve the system that follows.

Lagrange multipliers tells you the extrema occur when $ \sin a \equal{} \sin b \equal{} \sin c$.  This means that $ \cos^2 a \equal{} \cos^2 b \equal{} \cos^2 c \equal{} r^2$ for some positive real $ r$ and hence $ \cos a \plus{} \cos b \plus{} \cos c$ can take on the values $ 3r, r, \minus{}r, \minus{}3r$.  To maximize, we want the value $ 3r$, and picking the value of $ \cos a$ and $ \sin a$ therefore uniquely determines the angles $ a, b, c$ up to addition of $ 2 \pi$; that is, $ a \equiv b \equiv c \bmod 2 \pi$.  From here there are only a few choices that sum up to $ S$, depending on the value of $ S$.


Prove that $ \sqrt {m}$,  if m is not a perfect square, is not rational.

Prove that  $ \sqrt{p}$,where p is a prime, cannot be rational.

We can clearly see that 2 is a consequence of 1 :wink:

[hide="to prove 1"]
Let $ \sqrt{m} \equal{} \frac {p}{q}$ where the fraction is in its simplest form
$ \implies m \equal{} \frac {p^2}{q^2} \implies p^2 \equal{} m \cdot q^2$
$ \implies p \equiv 0 \mod m \implies q \equiv 0 \mod m$
But this is a contradiction!

Note that the problem doesnt arise if m is a perfect square (Why?) :)

As i said earlier, 2 is a consequence of 1 :)
[/hide]

[quote="madness"]$ \implies p \equiv 0 \mod m \implies q \equiv 0 \mod m$[/quote]There is an important step missing within this implication. If $ p\equiv0\mod m$, $ p\equal{}mk$ for some integer $ k$. Then, $ m^2k^2\equal{}mq^2\implies q^2\equal{}mk^2$. The rest follows.

sorry i thought that was understood  :oops:   :)

[quote="madness"]

Note that the problem doesnt arise if m is a perfect square (Why?) :)

[/quote]

Is that because if $ m$ is a perfect square, then when $ \sqrt{m}$ is written as $ \frac{p}{q}$, the integer $ q$ is 1? So that $ m \equal{} \frac {p^2}{q^2} \implies p^2 \equal{} m \cdot q^2$ becomes
$ p^2 \equal{} m \cdot (1^2)$
$ p^2 \equal{} m$
And we can no longer derive the contradiction?

The proof is not correct.  $ m|p^2\not\implies m|p$. For example, $ 4|6^2$, but $ 4\not|6$.


In problem 3, what is the term "disjoint squares" supposed to mean?

DPatrick said, in another thread, that correctly interpreting the problem is part of the problem, so interpretation help should NOT be given.

yeah, I was about to tell him that.

 :|

We cannot clarify problem statement while the round is in progress, and it is against the USAMTS rules to ask for help.


It is part of the definition of being a hyperbola.  b :^2: =c :^2: -a :^2: .

-interesting_move

Erm ... I think you can prove it based upon the definition of the hyperbola (using difference of distances between two points) and the equation ... it should follow fairly directly from that.

Interesting_move, that isn't definitional -- it's something you prove.  a, b and c all have meaning of their own.

i think you're both right [quote]t is part of the definition of being a hyperbola. b  =c  -a [/quote] from the algabraic side, that's true but from the geometric side you can and do have to prove it.
So you're both right.


[url=http://www.imagehosting.com][img]http://www.imagehosting.com/out.php/i1811033_TTinequality1.gif[/img][/url]

Have A Great Day!

Homogenize to abc=1

cyclic sum of $ \frac{1}{c\sqrt{1/c\plus{}2c^2}}$

Now we make a change of variables a,b,c-->1/a,1/b,1/c

$ \frac{c}{\sqrt{c\plus{}2/c^2}}\equal{}\frac{c^2}{\sqrt{c^3\plus{}2}}$

which we have to show is $ \ge\sqrt{{a\plus{}b\plus{}c}}$

It's too late to figure this out for me, anyone else want to figure out this seemingly apparent inequality.

From Holder's inequality we have
\[ \left(\sum\frac{ab}{\sqrt{ab+2c^2}}\right)^2\left(\sum ab(ab+2c^2)\right)\&\ge (ab+bc+ca)^3\]
which implies
\begin{align*}\left(\sum\frac{ab}{\sqrt{ab+2c^2}}\right)^2&\ge \frac{(ab+bc+ca)^3}{\sum ab(ab+2c^2)}\\
&=\frac{(ab+bc+ca)^3}{(ab+bc+ca)^2}\\
&=ab+bc+ca\end{align*}
which was what we wanted.

Thanks Nayel. Your solution is very beautiful!

Have A Great Day!

I think you should write the author's name of this problem.
It's my ine80 in my topic  :)
http://www.toanthpt.net/forums/showthread.php?t=9244&page=11
@can_hang, mitdac, honey_suck: My last nick is nhocnhoc but I can't "kích hoạt" it .So I must to use this ugly name :D (sr because I speak English very bad)


Does anybody know the winning strategy of the Super Nim game?

http://www.gameintellect.com/java-games/super-nim/#menu
------------------------------------------------------------------------------------------------------------------------------------------
Super Nim was created by GameIntellect.com

At the beginning of the game a few chips are placed randomly on the board. The players remove chips in turns. A move consists in removing all the chips in a row or in a column. The player to remove the very last piece wins.

To make a move select any column or row and click on corresponding letter or number. For example, to remove all of the pieces from column "b" click on the letter "b".

Press "Comp. starts" to give computer the first turn. When you make it to the next level, your current score will be cleared and you'll have to start from 0 all over again.

---------------------------------------------------------------------------------------------------------------------------------------------

Please help~


This is really just talking about what could be improved from '08, and in case you left something off the list.  Please read through all posts.  That will make it easier for the next person.

in response to the question it "Will you go next year" by thing1
There seems to be less time to ask the teacher questions and, for me and at least 3 others, were afraid to stop the class and ask questions.  They should (encourage/have 5 minute breaks in the middle of lectures for) questions.

Who else went to the math forums at night?  If there were more of those, and maybe a rating of difficulty for each one and what is expected of you to know then that would probably be adequate.

As a camp:
There was a wonderful amount of activities scheduled on Saturday, but a rather noticable lack of such on Sundays.

A more specific food complaint: A lack of breakfast on Sundays, a lack of Vitamin C in the food and drinks, and a lack of constantly available vegetarian options (although 80-90% isn't bad).

i felt like having free time on sunday was okay... maybe there could be one or two small activities. but i liked the free time.

but yea, definitely time for questions, cuz i felt so intimidated by the profs. X_X and i felt scared to ask questions/look stupid. >< priorityyy. a lotta the time i felt lost ... then i just zoned out. =_=''

fooodddd.... -.- its edible. "this is math camp not food camp! lol" but if improved, itd be better.

Suggestions:

Make the food better (though I didn't really eat very much at all...)

About the classes:

Hmm... question times might be helpful?

Also, why was Olympiad Combinatorics problem set same as <Medium Level> Combinatorics class... some days everybody just did Team Contest instead of the work...

Also, the team contest problems were way too favored for those who were in Combinatorics class...

Lagrange Interpolation-You put it on team contest right after it is taught in class...
The nice combinatorial argument with the inscribed 3/8 area triangle: You put it on team contest right after its solution is shown in class (a not-exactly-the-same version though...)

Question times... might be helpful?

More math forums would be helpful... (though admittedly I didn't go... oh well... I should be less lazy?)

I didn't care at all that there weren't any activities on sundays... I like spare time...

Hmm my experience was that I was like: ASK QUESTIONS PEOPLE. I think that it went pretty well in olympiad geometry (I'd go around while people did the problem sets). The kids in my dorm asked me questions occasionally (I don't think that they were intimidated anyway). I also got questions from some of the 'standout kids' in the class. I also visited a few classes and I think that I got some questions out of people.

I think that it is a good idea to have question times. But just so you know, once you get to college, you won't get babied like that. You have to be proactive in asking questions...

You could always order something from Walmart or buy food at that court that was open in the student union.

I would say that some classes were better prepared than others. I do remember that happening in the Olympiad combo class one or two days. I think I had a new problem set for the geometry class everyday.

I guess one of the things that I picked up on was that there were some assistants/teachers that were hard to understand because of their English.

[quote="leoxnlin"]
I'd say rather that your money goes to the whole experience of living, learning, and playing, and especially the social interactions, rather than the amount of problem-solving skill gained.

I'd like more math forums though... Altheman made geometry actually seem reasonable (I am a former geometry-hater)...  :P[/quote]
I really think AMSP should be more intense like with more lectures and problem solving sessions like in the 2006 schedule. I mean we could fit so many more lectures in and yes I know we should spend more time solving problems on our own but we could just do that at home and like I probably learn more in a one hour lecture than I do in like four hours just doing random problems. Also it might be nice to leave some of the problems unsolved at the end of the problem solving session so we have something to think about after class. Overall I don't think I learned a lot from the team contests but I really improved my problem solving ability by attending the lectures and math forums. The schedule probably can't be changed since they have already determined which classes to teach but surely we could have more math forums even if they are like a repeat of some class for people than didn't take it. There was tons of time to do team contest problems but I didn't get many of them or learn much. Like if you look on the will you go next year post it seems that several of people agree with me.

[quote="leoxnlin"]Was I the only one who thought the food was actually pretty good?[/quote]

seriouslyy??

[quote="leoxnlin"]Was I the only one who thought the food was actually pretty good?[/quote]

... I am so glad that I'm learning to cook...

but u cant cook at amsp....
unless ur like alexhhmun who got that HUGE box of ramen [that he wouldnt share!!! >< ] ahahaha jk. xP

I thought the food tasted pretty good...and it was better than I expected...
Well you could cook...you just needed a counselor watching you...

I liked the food too...although I think supper was too early, so I got hungry by bedtime.

[quote="shelly32494"]unless ur like alexhhmun who got that HUGE box of ramen [that he wouldnt share!!! >< ][/quote]

yeah it was yummy.

he shared with me because we are best buddies.

Also stock up on snacks. You are gonna need them.

Also, I'm used to good homemade food, so it might be just me.

Guacamole was good at a restaurant in Fort Worth, but HORRIBLE at AMSP for some reason.

as you can see, i'm picky even when i'm hungry.
except when it's mom's cooking. :)

i ate sushi for lunch the second half of the camp. It was DELICIOUS.

Next time i'ma use wasabi.

Yeah...just get snacks...

Or you could order pizza or chinese food at like 11 PM, like we did  :lol:

so, all in all,
bring extra food money!! lol XP

:wallbash: Yeah more math jams would be great and if possible more classes too.  :spam:
One thing that I didn't like about class was that it was way too rigorous and we didn't go through the steps in solving the problem. Like It would be nice if the lectures were more like WOOT lectures in that they outline problem solving strategies more. Also I felt that the mentors just ignored my questions and were like figure it out. But mostly yeah it probably would be better to make the camp more intense and have like lectures on weekends because there was really nothing to do on weekends except work on team contest problems which got boring after a while. Yeah so on weekends it would be really nice to have some lectures. :starwars: Like overall organized classes are a lot more effective than random problems.

The place to buy food on campus was closed at dinner.  And I DID share the ramen... I ate 4 or 5 of 12...

Most days the food wasn't bad.  Just that it was really inconsistent.

Yes, alexhhmun DID share his ramen.  :)  I tried to eat it in the dorm but we had nothing to cook it with.  So I just used hot tap water from the sink  :stink:  but it still tasted pretty good.

C'mon guys.  It was cafeteria food.  What did you expect?  I thought it was fine.  Although the lack of plenty of vegetarian choices was a problem, I know a couple of people who "starved" since they didnt always serve vegetarian food and ate stuff shipped from home.

its impossible to starve... I ate practically nothing and didn't starve.

Of course, I took some food from the peopple next door...

I was first to get sick. And for some reason all Zeb did during that class was ask me if I was going to die. WHY DOES EVERYONE WANT ME TO DIE...?

So everyone in my room blames me for them getting sick.  :)

GOSH DARN IT JUST DRINK OJ! :P

The food needs to improve and people(not just counselors, STUDENTS!) need to start organizing stuph to do on weekends...
Then it's probably perfect.

:mad: to everyone in my dorm who ate all of the food that I brought...

[quote="alexhhmun"]pythag, the food wasn't bad, but I maintain that the drinks lacked vitamin C, which is why we all got sick.  I got sick 2 days after leaving camp.

I noticed but wasn't allowed to get orange juice from wal-mart.   :([/quote]
The problem is almost all non-preserved (and some preserved) fruit, vegetable, AND meat has vitamin C...

[quote="sooozyy"]student uniuon stores/pub arent open all the time! :/

my dorm ended up ordering pizzas...hahaha at like midnight[i think a lot of dorms did actually LOL][/quote]

lol pizza after 9 dearrr XP
hahhaa.

whattttt. alexhhmun how could u betrayyy me??!!? i didnt get ramennn. hahaha jkjk. XP

hmm. i was homesick when i got back from amsp. lol XP hahaha.
but srlsly, being tired from camp may have something to do with getting sick and stuff.

The problem sessions sometimes weren't so good like some people were noisy. Also during the problem session I think the mentors should focus more on helping us become better at problem solving rather than putting all their emphasis on proof writing. Lectures on weekends would be REALLY good like overall I don't see the point in working on team contest problems by yourself at camp when you can just do it from home or having competitions because we are not going there so we can get the satisfaction of beating someone (at least I'm not). Man is the food all everybody complains about seriously.


In Triangle $ABC$, $AC=BC=AC$. Three lines, each drawn from each of the separate verticies of the triangle, meet at point $D$ inside the triangle so that $AD=7$, $BD=5$, and $CD=8$.  What is the area of triangle $ABC$?


geometry
geometric transformation
search

If you see this site please say how can we join at?

Buddy, I had pass for lib.homelinux once, but it is changed now!! Could you share it with me?! :oops:

its asks for passwords and usernames :rotfl:

What is this site? How do you enter it?“ What does this mean!?!??!?!?!?!?!?!?!?!? :wallbash_red: Äåðåâÿííûé ìàëü÷èê ñ äëèííûì íîñîì (ðóññêèé âàðèàíò) (èç 8 áóêâ)? (è èìÿ, è ïàðîëü, ëàòèíèöåé, ìàëåíüêèìè áóêâàìè). Ïðîñüáà èìÿ è ïàðîëü íèãäå íå ïóáëèêîâàòü, èíà÷å ïðèäåòñÿ ÷àøå ìåíÿòü.” on lib.homelinux.org.

[quote="Xevarion"][quote="spider_boy"]what is in this site? when trying to access this site it opened a dialog box asking for some login name and password.? and there was written "wooden boy with long nose" :rotfl:  what is this man?[/quote]
Pinocchio[/quote]

no I understood that part.But what is inside this site?I am very curious!! :D  :lol:

me too...
Has anyone searched in google how to enter?

I again tried to open this site but it again asked for password etc..
But the text in the dialog box that my browser opens is in russian( maybe it is in russian in your compter also) and it says:  Деревянный мальчик с длинным носом (русский вариант) (из 8 букв)? (и имя, и пароль, латиницей, маленькими буквами). Просьба имя и пароль нигде не публиковать, иначе придется чаше менять.
translation: Wooden boy with long nose(russian version)(8 letters)? (and username, and password, in latin, with little letters).It would be good if you don't publish your username and password anywhere.If not you will have to change it frequently.

First of all are they joking?would someone publish his/her username and password anywhere?
Secondly And I did some google search and learned this information:  it is Belorussian science library. and it is semibroken and unstable.And it includes tones of ebooks. :)
So can our belorussian users help us?

edit:hey guys I have found the password and login.so i edited this message because it revealed the solution to this great mystery!!I will not post the information here.you can pm me for the information.I will answer only the first 10 pms. :D  :)  :P

[quote="thatin"]I did some google search and in a site there was a riddle: the name of a large Russian river is the username and password.
It says in latin, right?[/quote]

I searched in some russian sites and found ::Длиннейшая речная система - Обь-Иртыш - протянулась на 5410 км от западного Китая до Северного Ледовитого океана. Вторая по длине речная система - Амур-Шилка-Онон (4416 км) - течет из северной Монголии, вдоль российско-китайской границы в Тихий океан. Среди отдельных рек самой длинной является Лена - 4269 км, далее идут Иртыш и Обь,

it says the longest river system in Russia is "Обь-Иртыш" and the longest river is "Лена".so... the first one can help.
oh I love mysteries :roll:  :roll:

edit: I tried but none of "Обь-Иртыш" and "Лена" worked.

hey guys I have found the login and password.I will not post the information here.you can pm me for the information.I will answer only the first 10 pms.

I think that the password and username change frequently so... hurry  :D

[quote="anirudh"]I got in but there was some russian site :maybe:[/quote]

no everyone can use it easily.Just click _djvu(математика,...)(if you want math materials).Then it will open a huge page where there are many books. :D (In my comp the page opens approx for 1 hour because my internet speed is slow. :( )The page contains books in many languages such as german ,english,russian, and french.

no, their internet speed is slow. In my pc and in my father's pc I waited for 2 hours and it didn't opened.

dear users.I will not answer your questions on this issue from now on.(i am very sorry) :(

you can just use http://lookforbook.com/ to find the book you want and then use the password when necessary...


geometry
3D geometry
symmetry
linear algebra
matrix
graph theory

Everyone probably know the famous problem consisting of a circuit having 12 resistors in the shape of a cube such that the resistors make up the edges. We pass current from one vertex and are required to find the effective resistance between this vertex and another one (the difficulty  of the problem depends upon which vertexes are considered!!)

In Physics, we usually solve this problem by assuming some symmetry considerations like equal partition of current or equal voltages etc. While studying Graph Theory in Alan Tucker's book, I cam across a similar example where he asked whether 2 graphs, one like the cubical circuit above and another circular graph were isomorphic. While his question ended in a negative there, I thought, could it be possible to solve the problem without any assumptions by finding a suitable, simple isomorphic graph that can be easily solved!  Before getting down to some brainstorming, I just want to know, has anyone of you (or somebody you know) considered this problem before. Are you aware of a solution to the above problem??? Please post what you think and your possible solutions!!

One possible graph of the original circuit , with arrows depicting directions of currents,(of course the actual directions could be different!) , is

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=358[/img]

Would this be better in the Combinatorics Section ???? There isn't much 'Physics' in my question. Can the moderator kindly move the topic!  :lol:

ok sinec i became a mod now ...so i am moving this

Index vertices of the bottom cube face 0, 1, 2, 3 counter-clockwise and vertices of the top cube face 4, 5, 6, 7 clockwise, the vertices 3, 4 being connected by a cube edge. Ground vertex 0 and connect grounded DC voltage supply $ V$ to one of the remaining vertices. Vertices 1, 3, 7 are equivalent, vertices 2, 4, 6 are equivalent, it is sufficient to test vertices 1, 2, 5 (on a cube edge, face diagonal, and body diagonal form vertex 0). Let $ v_1, v_2, ..., v_7$ be voltages at the vertices 1, 2, ..., 7 and $ I$ current supplied by the voltage source. Denote the conductance $ G \equal{} 1/R.$ The node analysis equation for the circuit is

$ \left( \begin{array}{cccccccc} 3G & \minus{} G & 0 & 0 & 0 & \minus{} G & 0 & 1 \\
\minus{} G & 3G & \minus{} G & 0 & \minus{} G & 0 & 0 & 0 \\
0 & \minus{} G & 3G & \minus{} G & 0 & 0 & 0 & 0 \\
0 & 0 & \minus{} G & 3G & \minus{} G & 0 & \minus{} G & 0 \\
0 & \minus{} G & 0 & \minus{} G & 3G & \minus{} G & 0 & 0 \\
\minus{} G & 0 & 0 & 0 & \minus{} G & 3G & \minus{} G & 0 \\
0 & 0 & 0 & \minus{} G & 0 & \minus{} G & 3G & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}\right) \cdot \left( \begin{array}{cccccccc} v_1 \\
v_2 \\
v_3 \\
v_4 \\
v_5 \\
v_6 \\
v_7 \\
\minus{} I \\
\end{array}\right) \equal{} \left( \begin{array}{cccccccc} 0 \\
0 \\
0 \\
0 \\
0 \\
0 \\
0 \\
V \\
\end{array}\right)$

The first 7 diagonal elements $ g_{ii}$ of the conductance matrix $ \mathbf G$ are sums of conductances connected to the i-the node, while the off-diagonal elements $ g_{ij}, i \neq j$ are negative conductances connected between the i-th and j-th node. If 2 nodes are not connected, $ g_{ij} \equal{} 0.$ Each voltage source (one in this case) has assigned an extra column at the right and an extra row at the bottom, filled with $ g_{i8} \equal{} g_{8i} \equal{} 1,$ if the positive terminal of the voltage supply is connected to the i-th node, $ g_{i8} \equal{} g_{8i} \equal{} \minus{} 1,$ if the negative terminal of the voltage supply is connected to the i-th node (none in this case), and $ g_{i8} \equal{} g_{8i} \equal{} 0$ otherwise. Solving for the current $ I$ by Cramer rule,

$ I \equal{} \minus{} \frac {\det \mathbf G_I}{\det \mathbf G} \equal{} \frac {384}{224} VG \equal{} \frac {12}{7}\ \frac {V}{R} \equal{} \frac {V}{R_{01}},\ \ R_{01} \equal{} \frac {7}{12} R.$

To calculate $ R_{02}, R_{05},$ just move the ones in the 8-th column and 8-th row of the conductance matrix  $ \mathbf G$ from the 1st to the 2nd resp. 5th position. In all cases, $ \det \mathbf G_I \equal{} 384 VG^7$ while $ \det \mathbf G \equal{} \minus{} 288 G^6, \minus{} 320 G^6,$ when connecting the voltage supply to the 2nd, 5th node, respectively, resulting in $ R_{02} \equal{} \frac {3}{4} R, R_{05} \equal{} \frac {5}{6} R.$ The node analysis will work for any resistance / capacitance network, the cube symmetry and resistor equality just make the determinants simpler and easier to calculate. Would someone move this thread back to physics ?

Sorry for bumping this about 9 years later, but do we have a closed form for the answer of the following problem (seems I saw it somewhere but I can't find it):

Let a set of vertices of a connected graph be given and each edge corresponds to $1 \Omega$. If the set of vertices is fixed find the expected value of the equivalent resistance over all connected graphs with those vertices.

I mean the two vertices across which we have to find the resistance is fixed.

This is known as Foster's theorem. The sum of all equivalent resistances across each resistor in the graph is $n-1$, where $n$ is the number of vertices. So I guess expected value is $\frac{n-1}{m}$, where $m$ is the number of resistors.

No, that theorem doesn't answer the question. For example, consider the graph for $n-1$ resistors connected in series. The average resistance between two randomly chosen vertices in that graph is $\frac{n+1}{3}$.

Basically, we're dealing with vertices that aren't connected directly all the time, and they're going to drive the averages up. The set of connected graphs isn't exactly all that nice combinatorially, so I doubt that there will be a closed form for the average over all of them.

Interestingly, this is the exact same question on this year's Sir Isaac Newton Physics Contest from Waterloo.


Please see [url=http://www.cyneer.com/phpBB3/viewtopic.php?f=35&t=15]here[/url]

To start this up, I will need the help of a banker and 2 college deans, one for C.I.T. (city instute of technology) and one for G.E.C. (General Education College).

Please sign up for forum and post on here:

Name on cyneer:
Position trying out for:
Why you are resposible enough:

[color=red][size=200]PLEASE IGNORE[/size][/color]

STOP making these stupid city games! The planet one was creative, and revamped and our town were both dumb! STOP! :mad:  :mad:  :mad:

lol

This time though, it is not congested to the space of one thread.

Plus the design of the other ones could use some... work.

 :|

Edit:
I also thought:
There will be not spam in the way of our precious eyes if it is on a different forum, now is there?

Note: CIT=Caltech Institute of Technology


Find all pairs $(x,y)$ of positive integers such that $\frac{x^{2}+y^{2}}{x-y}$ is an integer and divides 1995.

Well, a little bit of synthetic division tells us that

$\frac{x^{2}+y^{2}}{x-y}= 1+y+\frac{2y^{2}}{x-y}$

Because we can choose $x$ so that $x-y$ is any divisor of $2y^{2}$, it follows that $\frac{2y^{2}}{x-y}$ can be any divisor of $2y^{2}$.

Now, $1995 = 3 \cdot 5 \cdot 7 \cdot 19$.  We can set our expression equal to every factor of $1995$ in turn, all of which should yield a solution.  For example, letting $x = 2y^{2}+y$ we have for a given divisor $d$ of $1995$ the solution

$\left( 2(d-2)^{2}+(d-2), d-2 \right)$

There are many others that I don't really want to list out  :|


inequalities
geometry
circumcircle
geometry solved

$ABC$ a triangle ( all angles $< 90$) and $P$ a point inside the triangle. Denote by $L$, $M$, $N$ the orthogonal projections of $P$ on the sides $BC$, $CA$, $AB$ of triangle $ABC$.

Determine the place of $P$ for which $BL^2+CM^2+AN^2$ is minimal.

((PLEAZ   SEE ATTACHMENT))

It is well known (Carnot's relation) that
\[
BL^2+CM^2+AN^2=LC^2+MA^2+NB^2.
\]
Therefore, in order to minimize the sum
\[BL^2+CM^2+AN^2+LC^2+MA^2+NB^2
\]
observe that
\[BL^2+LC^2\ge \frac12 \left( BL+LC \right)^2 =\frac12 BC^2 \]
with equality iff $BL=LC$.
Using the similar inequalities, we deduce that the minimal value is reached when $P$ is the circumcenter of the triangle.


inequalities Euler geometry integration calculus

[quote="Ragingg"]Actually... It isn't really easy.

I guess you have to know the formula which is ln(x)+0.5773 or the Euler-Mascheroni number.

ln(1278576)+0.5773=14.63.

I haven't tried it yet, but someone can try 1+(1/2+1/2)+(1/4+1/4+1/4+1/4)+(1/8....)+.... and see if it is under 20.[/quote]

ln?  Just group it into 1, 1/2+1/3+1/4(teeny bit more), 1/5+1/6+1/7+1/8+1/9+1/10(less), 1/9+1/10+1/11...+1/18(less) and so on.

? I don't really understand your reasoning.

ln=natural log or log base e.

[quote="Ragingg"]Actually... It isn't really easy.

I guess you have to know the formula which is ln(x)+0.5773 or the Euler-Mascheroni number.

ln(1278576)+0.5773=14.63.

I haven't tried it yet, but someone can try 1+(1/2+1/2)+(1/4+1/4+1/4+1/4)+(1/8....)+.... and see if it is under 20.[/quote]

You must prove the forumla you speak of... merely saying "you have to know it" doesn't cut it in a proof... you also cannot make the assumption that we are at infinity when in truth 1278576 is nowhere close to infinity therefore you're method does not hold...

Attempt to prove it without using a calculator to calculate ln(blah)... there is a way.

[5th edit] As a matter of fact, using the ln(n) approx to prove this is above the level of this forum for sure. If anyone wants, I will post a ln(n) proof so you can see what I'm talking about... The other method of adding 1 + 1/2 + 1/3 +1/3 + 1/5 + 1/5 + 1/5 + 1/5 unfortunately will show the upper bound, however adding these up all the way to the requested value is almost impossible to do by hand, it would be tedious... I think the scope of this proof is well out of the reaches of this forum.

[4th edit]
[quote="Scrambled"]ln?  Just group it into 1, 1/2+1/3+1/4(teeny bit more), 1/5+1/6+1/7+1/8+1/9+1/10(less), 1/9+1/10+1/11...+1/18(less) and so on.[/quote]

This is not a proof in any way, and I'm not quite sure I see where you are going with this at all... Could you please explain?

Scrambled wrote:ln?  Just group it into 1, 1/2+1/3+1/4(teeny bit more), 1/5+1/6+1/7+1/8+1/9+1/10(less), 1/9+1/10+1/11...+1/18(less) and so on.

This is not a proof in any way, and I'm not quite sure I see where you are going with this at all... Could you please explain?[/quote]

[hide]unless i am quite mistaken: add up all the terms from 1 to 1/18.  How far is that from 4?  Should be about .23 smaller

We'll assume that 1/x+1/x+1...+1/2x which in all cases is smaller than 1 except the 2nd one.  so we assume all are one.  this turns out to be 1/1048576.  .23 = 23/100 which is approximatley equal to 23000/1048576.  So it is approximately 23000/1000000 which gives about that much leeway.  Ad 23000 to the bottom[/hide]

maybe i seriously screwed up

[quote="Spoon"]well I just don't see what you're trying to convey... what do you mean by a bit less and a tiny bit more?[/quote]

sorry i meant less than 1, more than 1...

you would be adding groups for a while if you did it that way..

[quote="Spoon"]I will post my proof (using the ln method):

First, I will show that [tex]\displaystyle{1+\frac{1}{2}+\ldots +\frac{1}{n} < \ln(n) + 1}[/tex].

The area under the curve of [tex]\displaystyle{\int_1^{n} \frac{1}{x}\; dx = \ln(n)}[/tex]. If we want to approximate this area, we could use the right side approximation, with width of one. The right side approximation of [tex]\displaystyle{\int_1^\infty \frac{1}{x}\; dx}[/tex] will be less than the exact value, because it's monotonically deacreasing... therefore the right side sum with width of 1 would be [tex]\displaystyle{\sum_{k=2}^n \frac{1}{k}}[/tex].

So [tex]\displaystyle{\frac{1}{2}+\ldots +\frac{1}{n} < \ln(n)}[/tex]. We simply have to add 1 to both sides and the conjecture is complete.

Now we must prove that [tex]\ln(1278576) < 19[/tex].

Factor 1278576, which can be done by hand and you get: [tex]\ln(2^4\cdot 3^2\cdot 13\cdot 683) < 19[/tex].

[tex]3^2<13<2^4<e^3[/tex]
[tex]683 < e^7[/tex]

We have:

[tex]\ln(1278576) < \ln(e^3\cdot e^3\cdot e^3\cdot e^7)[/tex]
[tex]\ln(1278576) < \ln(e^{16})[/tex]
[tex]\ln(1278576) < 16 [/tex]

And we are done :), and you never had to use a calculator, decimals, or assumed formulas... like I said before, this prove is above the scope of this forum.[/quote]

First, you use the word 'proof' very loosely.  Second of all, how did you find those inequalities for powers of e without a calculator?

Thirdly, it really is a very simple proof, and I would be willing to guess that some people in this forum have a basic grasp of calculus (I certainly did in 9th grade).

If I asked you which was larger, 4^3 or 3^5... would you be able to tell me? They are basic and it should be known... it should be common knowledge that e^3 is around 20, and e^7 is above 1000... Our goal in the problem is to merely show that one is greater than the other... you could do this:

16 < 2.7^3 < e^3... 2.7 is EASY to do by hand and should take no more than 15 seconds or so... if one really needed to, you could show that 683 < 2.7^7 < e^7... 2.7^7 can be done easily by hand as well... I didn't need exact numbers, only approximations and it should be clear that small numbers like this can be calculated by hand... I did not need to do the calculations, nor do I feel it pertinent to show that e^7 is more than 683, it should known. Just like you know 4^3 and 3^5. You wouldn't be expected to know 2004^2003 or 2003^2005... but when the numbers are small, it's easy.

It should not be assumed that anyone on the Getting Started forum is doing proofs that involve calculus... this is a middle school level forum. I think Tokenadult would agree with me on this.

[quote="blahblahblah"]Thirdly, it really is a very simple proof, and I would be willing to guess that some people in this forum have a basic grasp of calculus (I certainly did in 9th grade).[/quote]

Calculus doesn't belong on Getting Started Forum.

Scrambled, can you please explain what you are doing more clearly?  I still don't understand how you made your calculations.


Each edge of a given tetrahedron of resistors has a resistance of $1\Omega$ except for one, which is exceptionally different. How would you determine which edge is different with only three measurements by an ohmmeter?

OK. Here we go .

  Let $\; V_{i}, \; i\{1,2,3,4\}\;$ be the vertices.

   If any measurment results in $\;{1\over2}\Omega\;$  (lucky!), the edge you didn't use it's ends, is the chosen one;

 (for instance, if the measurement was between $\;V_{1}\; \text{and}\; V_{2}$ the exceptional edge is   $\; V_{3}V_{4}\;$ ).

   Othewise, choose two measurements, $\; r_{1}\;\text{and}\; r_{2}\;$, with the vertices of adjacent edges (Let $\;V_{1}V_{2}\; \text{and}\;\;V_{1}V_{3}\;$),
   and a third, $\; r_{3}\;$, with the ends of the edge that links the two above ($\;V_{2}V_{3}\;$).
      (i)   $\; r_{1}\;=\;r_{2}\; (\;\text{and}\; r_{3}\neq \;{1\over2}\Omega\;)$, the "strange in the nest"  is  the very last,
$\;V_{2}V_{3}\;$;

  (ii) $\; r_{1}\;\neq\;r_{2}\; (\;\neq \;{1\over2}\Omega\;) \;$.
 $\;r_{3}\;$ will be equal to one of them (let be $r_{3}\;= r_{2}\;$}.
     and the answer will be the edge with ends giving the different reading ($\;V_{1}V_{2}\;$).

Yes, very good ;) I really liked this problem when I saw it...


calculus
integration
search
trigonometry
calculus computations

1. integrate sin^(5)zcos3zdz

2. Use half angle identities to evaluate the integral of cos^(2)xdx

Searching for the word "intergration" on AoPS turned up 9 instances involving 7 different posters. In three cases, it was used in the title, two of those by flava22. (No, that's not the correct spelling.)

I don't think I want to search for for "continous."

2. $\int \cos^{2}xdx=\int\left(\frac{1}{2}+\frac{\cos(2x)}{2}\right)dx =\frac{x}{2}+\int\frac{\cos(2x)}{2}dx$, so now let $u=2x\Rightarrow du=2dx$ so the integral after the addition part is $\frac{1}{4}\int\cos(u)du=\frac{1}{4}\sin(u)=\frac{\sin(2x)}{4}$, so the entire integral is:

$\int\cos^{2}xdx=\frac{x}{2}+\frac{\sin(2x)}{4}+C$.

"continous": 382 results.
"continious": 93 results.
"intergrate": 3 results, one with "intergrals" as the title.
"intergral": 34 results.


The problem:
There is an infinite conducting plate that is grounded u have a charge $q$ at a distance $d$ above it and u r asked to find the potential at a distance $z$ above the plate?

The solution:
it is by using the first uniqueness theorem and specifying the conditions satisfied by the function $V$
the conditions are :
1.the potential is $0$ at any point on the plate or $V(x,y,0)=0$ for all $x,y$
2.$V(x,y,z) \rightarrow 0$ as $x^{2}+y^{2}+z^{2}>>d^{2}$

well the first one is obvious but is the second one valid(yes it's true i know for all books can't be wrong at a time  :wink: )
but what's the justification...since the plate itself is infinite is this a valid assumption?

a) The surface charge density induced on the plane must be finite, but the surface is infinite;
b) The field generated by the induced charges will decay as $z$ increases.

These two facts should be sufficient to support your second claim. (They are sufficient to me, at least ;)). Btw, what is the first uniqueness theorem?

yse that's what i wanted to know whether this was an approximation(well almost everything is :wink: )or does it have something to do with things i don't know :(
so anyways it's intutively obvious

[b]Uniqueness theorem[/b]:The solution to laplace equation is uniquely specified in a given volume if it's value is given everywhere in the boundary.

[quote="Djole"]a) The surface charge density induced on the plane must be finite, but the surface is infinite;
b) The field generated by the induced charges will decay as $z$ increases.

[/quote]
still the plate is also infinitely long then how can one argue that the potential will be approximately 0

The potential is constant everywhere, and since the potential is defined up to an additive constant, you may always choose this constant so that the potential is zero.

sorry i didn't get that can u be a bit more clear :maybe:

I meant that every point on the metal plane has the same potential. Now, the potential is any function $\varphi$ which satisfies $\nabla \varphi =-\vec E$, so if $\varphi$ satisfies this, $\varphi+C$ also satisfies it.

But looking back, I'm not sure which potential you're talking about -- I was talking about the potential of the plane -- which one did you mean?

no i was talking about the potential about a point thatis infinitely away from the point charge but may be close to the infinite sheet

It will also have a zero potential, because there is a somewhat lengthy argument proving that the potential of the uncharged infinite plane be zero at each of its points, or something like that...

Actually, the plane is charged with $-q$. Anyhow, it doesn't matter: $V = 0$ is only a reference; the physical importance lies in $\Delta V$.

yes immanuel is right when he says that the plate will have charge $-q$....
i thinka n intutive argument(if someone can give a better arguement please do give :) ) that at far off distanves from the charge the induced charges are negligible hence potential duw to them is also...


Did princeton review change their AP books' cover art for the 06-07 editions?

Uh, yes they did.  I can't imagine why you would want to know though...


percent
geometry
geometric transformation
reflection

Each time a sound wave bounces off a reflector it loses 20$ \%$ of its strength. What percent of its original strength has it lost after three reflections? Express your answer as a decimal percent.

After each reflection, it has $ .8$ of its strength left, so after three, it has $ (0.8)^3\equal{}.512$ of its strength left.  As a percent, it has lost $ 100\% \minus{}51.2 \%\equal{}\boxed{ 48.8\% }$


trigonometry
conics
hyperbola
geometry
circumcircle
perpendicular bisector

Prove that if the inscribed quadrilateral with perpendicular diagonals is also a circumscribed one, then it is symmetric with respect to one of its diagonals.

It's easy with trig, but i haven't find a geometric proof.

If I'm not mistaken (which is possible), there are too many conditions here.

See the attached diagram. Assume $ABCD$ admits an incircle. Then $AB+CD=AD+BC$. If we denote that sum by $K$, we can denote the sides as shown. But then $BA-BC=a+b-K=DA-DC$. Assuming $a+b-K\neq 0$, that means that $B$ and $D$ lie on a hyperbola whose foci are $A$ and $C$, hence $BD$ is a hyperbolic chord which is perpendicular to hyperbola's focal axis, and that, because of the geometric properties of a hyperbola, means that $B$ and $D$ are symmetric with respect to $AC$.

If on the other hand $a+b-K=0$, then $b=K-a\iff BA=BC$ and $a=K-b\iff DA=DC$, hence both $B$ and $D$ lie on the perpendicular bisector of $AC$, which means that $BD$ is the symmetrial axis of $ABCD$.

Hence it's proven that a kite is the only type of quadrilateral which admits an incircle and has perpendicular diagonals.

Now, a kite can also admit a circumcircle if two of its non-adjacent angles are right (but that's just a special case of the previous statement).

Another proof, for those who are not fluent in conics, relies on Pythagorean theorem.

Let $O$ be the intersection of the diagonals. Denote $x=OA, y=OC, p=OB, q=OD$. Then, assuming $ABCD$ admits an incircle, we have

$\sqrt{x^{2}+p^{2}}+\sqrt{y^{2}+q^{2}}=\sqrt{x^{2}+q^{2}}+\sqrt{y^{2}+p^{2}}$

Square it, cancel the equal terms, square it again, expand, cancel the equal terms once more to get

$x^{2}q^{2}+y^{2}p^{2}=x^{2}p^{2}+y^{2}q^{2}$

which is equivalent to $(x^{2}-y^{2})(p^{2}-q^{2})=0$

Therefore $x=y$ or $p=q$, which means that one of the diagonals is the symmetrial axis of the quadrilateral.


advanced fields
advanced fields theorems

The diagonal of the Cartesian product $ X \times X$ is the set $ \Delta\equal{}\{(x,x): x \in X \}$.
Every set $ V \subset X \times X$ that contains $ \Delta$ and satisfies the condition $ V\equal{}V^{\minus{}1}$ is called an entourage of the diagonal. Let $ D_X$ be the family of all entourage of the diagonal $ \Delta \subset X \times X$.
A uniformity on a set $ X$ is a subfamily $ U$ of $ D_X$ which satisfies the following conditions:
(1) if $ V \in U$ and $ V \subset W \in D_X$ then $ W \in U$;
(2) if $ V_1,V_2 \in U$, then $ V_1 \cap V_2 \in U$;
(3) For every $ V \in U$ there exists a $ W \in U$ such that $ W \circ W \subset V$;
(4) $ \bigcap U\equal{}\Delta.$

I don't understand why from (3) follows that for every $ V \in U$ there exists $ W \in U$ such that $ W \circ W \circ W \subset V$.

I think that applying (3) twice, we get
$ W \subset W \circ W \subset Z \subset Z \circ Z \subset V$.
Then $ (W \circ W) \circ W \subset Z \circ Z \subset V$.


Se considera ecuatia :$x^n-nx^{n-1}+p=0,p \in (0,1)$.
Aratati ca oricare ar fi $n \in 2N+1, n \ge 3$ ecuatia are 3 radacini reale $a_n,b_n,c_n.$ si:
a)$a_n<0<b_n<c_n;$
b)$\lim_{n\to\infty} \sqrt[n]{{ b_n+1}}=1$

[quote="Myth"]Hmm... There is something wrong...
We have $x^n-nx^{n-1}+p$ is increasing for $x<0$. So there is a unique negative root (since $p>0$).
For $x>0$ we have $f'(x)=(x^n-nx^{n-1}+p)'$ has a unique root and $f(0)>0$, $f(1)<0$. It follows the equation has exactly two positive roots and $b_n\in (0,1)$ and $a_n>1$.
But, in such case, it is clear that $\lim \sqrt[n]{b_n+1}=1$.[/quote]

If I understood, we wish to solve:

$g(x) = x^{n-1} (n - x) = p > 0$.

There is one negative root, and two positive roots in $[0,n]$.  It is easy using calculus or AM-GM inequality to see that $g(x)$ has a maximum at $x = n-1$, where it is quite large.
So one positive solution will be very close to $n$ and the other is close to 1.
If the solution near $1$ is $1 - \alpha/(n-1)$ we need $e^\alpha \sim (n-1)/p$, i.e.
$b_n = 1 - \frac{log(n-1)-log(p)}{n-1} + O(1/n^2)$.

[quote="Myth"]Please, fix your reply ;)[/quote]

TeX or mathematics?  I fixed the TeX...

I said "something wrong" in my reply, because conclusion $\lim ...=1$ is quite meaningless. Why we need to prove it, if we know that $b_n\in[0,1]$?

I see that you are learning Romanian Mikhail :) :D

[quote="Myth"] $\lim ...=1$ is quite meaningless. Why we need to prove it, if we know that $b_n\in[0,1]$?[/quote]

Yes, but one can determine asymptotics of the roots regardless of any wrong or un-natural statements in the original problem.

$f\prime=nx^{(n-2)}(x-n+1)$
 obtinem din tabloul de variatie urmatoarele
$a_n<0 , 0<b_n<n-1 , c_n>n-1$
$1<b_n+1<n$
limita este 1

admin edit: nu mai scrieti cu \mbox atunci cand nu este necesar.


function
real analysis
real analysis unsolved

Let $f: \mathbb{R}\to \mathbb{R}$ be continuous such that for all $s \in \mathbb{R}$ there exists a $C^{2}$-function $g_{s}$ with the properties
\[g_{s}(s)=f(s), \quad g_{s}(t) \le f(t) \quad (t \in \mathbb{R}), \quad g_{s}''(s) \ge 0 \, . \]
Show that $f$ is convex.

I think you meant $g_{s}(t)\le f(t)$ for $t\in\mathbb R$. Think on the graph of $g_{s}$ as a "generalized tangent" to the graph of $f$ at $s$.

Hint #1: the assumptions imply that $f(x)+\epsilon^{2}x-cx$ has no points of local maximum for any $\epsilon>0$ and any $c\in\mathbb R$.

Hint #2: it suffices to prove that $f(x)+\epsilon x^{2}$ is convex for any $\epsilon>0$.

[quote="mlok"]I think you meant $g_{s}(t)\le f(t)$ for $t\in\mathbb R$. Think on the graph of $g_{s}$ as a "generalized tangent" to the graph of $f$ at $s$. [/quote]You're right, it's been fixed.


advanced fields
advanced fields unsolved

Hi,

Can anybody direct me to the right path for the 3rd problem from the following problem set:
http://www.math.ubc.ca/~rfroese/math267/hw5.pdf

I got the Fourier Series for g(x) in complex exp. form, but do not really understand the rest. e.g. "then expanding y(t) in a series with unknown coefficients and substituting it in the equation, and then determining the values of the coefficients that make the equation hold." what is this sentence saying? Like I find the characteristic eq. and then find its roots to get y(t)?

Thanks in advance.

You are supposed to write $y(t)=\sum_k y_k e^{\pi ikt}$ where $y_k\in\mathbb C$. Then you are supposed to formally differentiate the series with respect to $t$ and to write the resulting expression for the left hand side. In your case you'll get $\sum_k (-\pi^2 k^2+i\alpha\pi k+\beta)y_k e^{\pi ikt}$. This expansion should be the same as the Fourier expansion $g(t)=\sum_k g_k e^{\pi ikt}$ for the right hand side you found. It remains to equate the coefficients at each exponent $e^{\pi ikt}$ to find $y_k$. I'll let you continue from here.

Right, thanks! I found g_k to be 2/(i*k*pi) with k being any real odd numbers. Now I am having difficulties equating this g_k with y_k (I always had troubles with that). Could you possibly provide a quick hint or something?

By the way your response totally clarified this problem for me. Much appreciations.

All you need to do is to solve  $(-\pi^2 k^2+i\alpha\pi k+\beta)y_k =g_k$ for $y_k$.

Thanks. I was thinking too complicating I guess.


This is easy:

Can we label each vertex of a 45-gon with one of the digits {0, 1, ..., 9},
so that for each pair of distinct digits i, j one of the 45 sides has vertices labeled i, j?

Hrmm... this may not be very clear, but here goes.

The trick is for every number to be next to every other number. Since each vertice is next to two other vertices, if there are n vertices with a label i, then the number i can be next to 2n different numbers. Since there are nine numbers other than i, we would have to have 5 of each number (so that 2n=10>9). But that would be a total of 50 vertices, so no.

.dzhonatan

hmz

I think mine is easier (or if it is what you wrote, at least more clear):

10 numbers on 45 corners
so some numbers will be on at least 5 places.
so those numbers need 10 different numbers as neighbours

but there are only 9 differents for each number
so solved

No, you aren't solving the problem...

there's nothing in the problem that says that you can't have the same two vertices next to each other twice, which is what you are saying.

.dzhonatan

there are 45 pairs possible and you need 45 different ones.

I don't think I have to explain why there cannot be double ones then? :?

Ok, that works then.

But you didn't say that in your first post.

.dzhonatan

ya I know I thought that was too obvious, the art is to write them as short as possible :)

Sorry.


algebra
polynomial
combinatorics unsolved
combinatorics

[quote="Lena"]My book isnt exactly student friendly

but this is what i came up with after solving the problem 3 times

the problem is $\ ( 1+x+x^{2}\ ) \ ( 1+x \ )^{n}$ find coefficient of $x^{k}$

i am looking in for $x^{k-n}$ in $\ (1+x+x^{2}\ )$

$\ (1+x+x^{2}\ ) = \frac{1-x^{2}}{1-x}= \ ( 1-x^{2}\ ) \ (1-x \ )^{-1}$

coefficient of $x^{k-n}$ would then be $\binom{1+k-n-1}{k-n}= \binom{k-n}{k-n}= 1$

and i dunno if i'm right or not :(

can someone explain how to do this[/quote]

One sloppy error putting $1+x+x^{2}$ into another form.  More importantly, $1+x+x^{2}$ [i]is a polynomial in standard form[/i]: you know the coefficients of every term already.  The coefficient of $x^{0}, x^{1}$ and $x^{2}$ is 1, and all other coefficients are 0.  The other factor is also a polynomial.  I have no idea where $x^{k-n}$ came from -- either you've mistyped or you have misread.

well i dont exactly understand it yet

another try is to open the brackets

mainly write something like $\ ( 1+x \ )^{n}+x \ (1+x \ )^{n}+x^{2}\ ( 1+x \ )^{n}$

$\binom{n}{k}+\binom{n+k+1-1}{k}+\binom{n+k+2-1}{k}$

$\binom{n+k+1-1}{k}+\binom{n+k+2-1}{k}$ this part i am most definitely confusing

i'm not sure if to add 1 or subtract 1

Why don't you explicitly state what you're trying to do?  Once you know exactly what it is that you want, then you can jump on the formulas that do it.

I want to find the coefficient of $x^{k}$ in the phrase $\ ( 1+x+x^{2}\ ) \ ( 1+x \ )^{n}$

which means i want to find the coefficient of $x^{k}$ in $\ ( 1+x \ )^{n}+x \ (1+x \ )^{n}+x^{2}\ ( 1+x \ )^{n}$

meaning find the coefficient of $x^{k}$ in $\ ( 1+x \ )^{n}$ and add to that the coefficient of $x^{k-1}$ in $x \ (1+x \ )^{n}$ and add to that the coefficient of $x^{k-2}$ in $x^{2}\ ( 1+x \ )^{n}$

so the answer would be $\binom{n}{k}+\binom{n+k-1-1}{k-1}+\binom{n+k-2-1}{k-2}$

Just as a note for future: You'll get more responses for such problems in the Intermediate forum.  See if your formula works by testing a couple of values.  Also, consider the numbers you add to powers by multiplying the binomial by $1,x,x^{2}$.

[gee i've been going about this all wrong

the coefficient of  $x^{k}$ in $\ ( 1+x \ )^{n}$ is $\binom{n}{k}$
the coefficient of $x^{k}$ in $x \ (1+x \ )^{n}$ is the same as $x^{k-1}$ in $\ ( 1+x \ )^{n}$
the coefficient of $x^{k}$ in $x^{2}\ ( 1+x \ )^{n}$ is the same as the coefficient of  $x^{k-2}$ in $\ ( 1+x \ )^{n}$
so the correct answer would be

$\binom{n}{k}+\binom{n}{k-1}+\binom{n}{k-2}$

Yes.  (You actually essentially had it last time, but you were too eager to use a formula which didn't apply.)

thx JBL

mainly for not giving me the answer

once i understood where  i had to derive the coefficient from, it was easy to solve

as i said, my book isnt student friendly, it didnt explain that in so many words... i read almost every page of the book betweeen each and every letter


trigonometry
geometry
algebra open
algebra

We have
1) 0 < a, b, c, x, y, z < pi;
2) a + b + c + x + y + z = pi;
3) sin a * sin b * sin c = sin x * sin y * sin z;
4) sin (y/2) * sin (c/2 + z) = sin (c/2) * sin (y/2 + b);
5) sin (z/2) * sin (a/2 + x) = sin (a/2) * sin (z/2 + c).
Prove that sin (x/2) * sin (b/2 + y) = sin (b/2) * sin (x/2 + a). :D

Try to use this geometrical interpretation:
Take the triangle ABC with the chevians AX, BY, CZ such, that the angle BAX = a; the angle CBY = b; the angle ACZ = c; the angle CAX = x; the angle ABY = y; the angle BCZ = z.


integration function inequalities calculus derivative vector topology

Consider the functional $F(u)=\dfrac12\int\limits_a^b u'^2(x)dx-\dfrac1p\int\limits_a^b|u(x)|^pdx-\int\limits_a^bh(x)u(x)dx$ with $\mathrm{dom}\, F=W: =C^1[a,b]$, $p>2$.

Denote
$f(t,s): =F(tu+sv)$ ($\|u\|=\|v\|=1$ and $\left\langle u,v\right\rangle=0$ where $\left\langle u,v\right\rangle=\int\limits_a^bu'v'dx$ and $\|u\|=\sqrt{\left\langle u,u\right\rangle}$) and
$\hat F(u,v): =\max\limits_{t,s\ge0}f(t,s)$.

Is is not hard to prove that $\hat F(u,v)$ exists, and $\hat F(u,v)>0$ implies $t=t(u,v)$ and $s=s(u,v)$ are both strictly positive functions.

Prove that $\hat F(u,v)>0$ implies that $t=t(u,v)$ and $s=s(u,v)$ are both $C^1$.

[b]P.S.[/b] My idea is as follows. As $f(t(u,v),s(u,v))=\hat F(u,v)$, it is necessary that

$\frac{\partial f}{\partial t}(t(u,v),s(u,v))=0$,
$\frac{\partial f}{\partial s}(t(u,v),s(u,v))=0$.

So if denoting $A$ the Hessian of $f$ at $t=t(u,v),s=s(u,v)$, we prove that $\det A\neq0$, we will be able to apply Implicit Function Theorem and thus prove smoothness of $t$ and $s$.

The trouble is that I cannot prove that $\det A\neq0$; maybe you offer another approach.

[quote="Vladimir Lubyshev"]Consider the functional $F(u)=\dfrac12\int\limits_a^b u'^2(x)dx-\dfrac1p\int\limits_a^b|u(x)|^pdx-\int\limits_a^bh(x)u(x)dx$ with $\mathrm{dom}\, F=W: =C^1[a,b]$, $p>2$.

Denote
$f(t,s): =F(tu,sv)$ [/quote]
Wait a minute; $F$ had only one argument just a moment ago :?. And what is $h(x)$? Is it also smooth or just integrable?

Oh, I am sorry. I have corrected the mistake, here is the right version: $f(t,s): =F(tu+sv)$. Suppose $h$ to be "good enought", but it would be better if we consider more general case for $h$ (just integrable).

[quote="Vladimir Lubyshev"]
Is is not hard to prove that $\hat F(u,v)$ exists and $\hat F(u,v)>0$ implies $t=t(u,v)$ and $v=v(u,v)$ are both strictly positive functions.
[/quote]
Now, I have very serious doubts about this statement. Indeed, since $\|u\|=\|v\|=1$ and $u$ and $v$ are orthogonal and since $h$ is arbitrary, we can write
\[ f(t,s)=\frac 1 2 (t^2+s^2)-\frac 1 p\int_a^b |tu+sv|^p - At-Bs \]
where $A$ and $B$ are some constants that may be chosen to be any two real numbers for fixed $u,v$ if we choose $h$ in an appropriate way. Now choose $u$ and $v$ strictly positive, $A=0$ and $B$ big positive. Then, using the trivial inequality $(x+y)^p\ge x^p+y^p$ for positive $x,y$, we conclude that
\[ f(t,s)-f(t,0)\le \frac{s^2}2-s^p\frac 1p\int_a^b v^p- Bs\,. \]
If $B$ is large and the integral is not too small, this difference is strictly negative for positive $s$. On the other hand, it is clear that $f(t,0)>0$ for small positive $t$. So the maximum is attained at a point $(t_*,0)$ for some $t_*>0$.  :?

Suppose $\hat F(u,v)>0$. It it obvious that one of the $t,s$ must be strictly positive, let it be $t$. Let's prove that $s$ must be strictly positive as well. Suppose the contrary. It is clear that

$f(t,s)\ge f(t,0)+f(0,s)$

for $<u,v>\ge0$ (in our case = 0). Thus $f(t,t)\ge2f(t,0)=2\hat F(u,v)>\hat F(u,v)$. We have come to a contradiction.

So your reasoning contains mistakes, they are embodied in (some of) the following items:

1) It seems not mandatory that $A=\int\limits_a^b hu$ and $B=\int\limits_a^b hv$ can be made arbitrary reals due to the restriction $\langle u,v\right angle\rangle=0$ and $\|u\|=\|v\|=1$.

2) You have made too many presuppositions: apart from 1), you supposed $\int\limits_a^bv^p$ to be not too small. All this can easily turn into the contradiction with our $\langle u,v\rangle=0$ and $\|u\|=\|v\|=1$.

3) You cannot state that $f(t,0)>0$ for small $t>0$.

[b]P.S.[/b] Having received the problem, I thought that $\hat F(u,v)>0$ doesn't imply that $t$ and $s$ are both $>0$ so I asked the professor and he responded it was known that in this case $t$ and $s$ [i]must[/i] be $>0$. It may happen to be wrong in the case $\hat F(u,v)=0$ only. Yet, it is known that in the case $\hat F(u,v)=0$ $t$ and $s$ are not mandatory $C^1$. So the condition $\hat F(u,v)>0$ is crucial. In my proof of strict positiveness of $t$ and $s$, I essentially used that condition: $2\hat F(u,v)>\hat F(u,v)$.

[quote="Vladimir Lubyshev"]

$f(t,s)\ge f(t,0)+f(0,s)$

[/quote]
This is completely unclear to me and, moreover, obviously false because, under your assumptions it is equivalent to
\[ \int_a^b |tu+sv|^p\le \int_a^b|tu|^p+\int_a^b |sv|^p\,,  \]
which is very far from being always true...
Another mistake you make is the assumption that $f(t,0)+f(0,t)=2f(t,0)$, i.e., that $f(0,t)=f(t,0)$, which doesn't follow from anywhere.
[quote]So your reasoning contains mistakes [/quote]
I'm afraid not (unless I misunderstood something in your original post). I looked at all three of your objections and they are all invalid.

1)if $u$ is not a constant multiple of $v$, then, for all $A$ and $B$, we can always find $h$ such that $\int uh=A$ and $\int vh=B$.

2) Since adding a constant to $v$ doesn't change $\|v\|$ or $\langle u,v\rangle$ (both depend only on the derivative of $v$), we can make $\int|v|^p$ as large as we want.

3) If $A=0$, $f(t,0)=\frac 1 2 t^2- t^p \frac 1p\int_a^b |u|^p=\frac 1 2 t^2-Const\, t^p$, which is strictly positive for $t^{p-2}<\frac{Const}{2}$

[quote]I asked the professor and he responded it was known that in this case t and s must be >0.[/quote]
He(she?) is only human and, therefore, may make mistakes too :).

Yes, it seems you are right. My "proof" contain outrage mistakes! :blush: I have to think on this problem again. Thank you very much. :oops:

Can you say something on smoothness of $t$ and $s$?   :?

Now I absolutely confirmed in your correctness! You founded a counter example. So that professor was wrong or I misunderstood something. :D

I hastened stating my "objections" 1) - 3) :wallbash:. As to the first "objection", one can even prove that for any two linearly independent vectors $u$ and $v$ of a locally convex space $X$ and any $A,B\in\mathbb R$ there exists $x^*\in X^*$ with $\left\langle u,x^*\right\rangle=A$ and $\left\langle v,x^*\right\rangle=B$. So the fact that $\forall\,A,B\in\mathbb R\;\;\exists\,h\in L^p(a,b)$ ($p\ge2$) with $\int_a^buh=A$ and $\int_a^bvh=B$ follows from this general case. Did you mean $h\in L^p(a,b)$ ($p\ge2$)?

[quote="Vladimir Lubyshev"]
Can you say something on smoothness of $t$ and $s$?   :?[/quote]
I'm afraid we first have to figure out whether they are correctly defined. I mean, since the function $f(t,s)$ is not concave, it seems to me that it is quite feasible that it may have 2 or more different points where the maximum is attained.  :roll:  Of course, if we assume a priori that $t,s$ are unique and positive, we may try to ask about their smoothness in a neighborhood of the corresponding pair $(u,v)$. But then, since we've piled up so many assumptions already, why not to add the assumption that the Hessian is non-degenerate as well  :P ? You see what I'm driving at: this problem is hardly of much interest by itself; I assume that you or your professor need it for something else, in which case it would be wise to check first whether what we figured out already doesn't spell the end of this approach to that "something else". :)

I should stop here. When saying about existence of $h$ with $\int_a^b uh=A$ and $\int_a^b vh=B$, did you mean $h\in L^p(a,b)$ ($p\ge2$) or one can even suppose $h\in C^r([a,b])$?

Just take two points $x,y\in [a,b]$ such that the vectors $(u(x),u(y))$ and $(v(x), v(y))$ are not proportional to each other and choose an appropriate linear combination of two positive $C^\infty$ functions one of which is supported by a small meighborhood of $x$ and another by a small neighborhood of $y$. Many other constructions are possible too.

What about the following non-constructive approach. $X: =C^1[a,b]$, consider the space $Y$ consisting of linear functionals on $X$ of the form $f\mapsto\int_a^b fh$ as $h$ vary over $C^\infty[a,b]$ then $Y$ separates points of $X$ by virtue of the Dubois Raymond theorem. Now endow $X$ with the topology $\sigma(X,Y)$, then $X^*=Y$. It remained only to apply the statement about locally convex spaces from one of my previous posts (about existence of $x^*\in X^*$ with ...).

Are there any mistakes?

That seems fine too. The only catch is that one has to prove that your topology separates points, i.e., that for every $u$ that is not identically $0$, there exists $h\in C^\infty$ such that $\int uh\ne 0$ and when proving it, you may have to do something very similar to the construction in my post. Otherwise, what you said is perfectly correct. :)


Let $ G= $ { $ a_1, a_2 ,... , a_{100} $ }  be a subset of the set {1,2, ... , 200} such that : with $ 1 \leq i < j \leq 100 $ , $ \sum_{i=1}^{100}a_i = 10080 $ , $ a_i + a_j $ is different from 201
  a) Prove that : the number of the odd numbers of G is a multiple of 4
  b) Calculate $ a_1^2 + a_2^2 +... +a_{100}^2  $

Nasty computations, I got wrong 4 four times;... :(

First, divide the integers in $\{1,2, \cdots, 200 \}$ by pairs of the form $(a,201-a)$ where $a=1,2 \cdots, 100$.
Since any two of the $a_i$'s do not add up to $201$ it follows that $G$ contains at most one term from each pair. In another hand, since $|G| = 100$ we then have to choose exactly one term from each pair.
That means $a \in G$ if and only if $201-a \notin G.$ (1)

1) Let $r$ be the number of odd terms in $G$.
Let $S = 10080 = \sum_{a_i \in G} a_i = P+I$, where $P$ is the same sum but only over the even terms and $I$ is only the odd ones (odd = Impair and even  = Pair, in french  :P  ).
Note that there are exactly $r$ even integers which do not appear in $P$.
Thus,
$I= \sum_{200-2a \notin G} (2a+1) = r - \sum_{200-2a \notin G} (200-2a) + 200r = 201r - \sum_{i=1}^{100} (2i) + P$.
It follows that
$201r = S -2P + \sum_{i=1}^{100} (2i) =10080 + 10100 -2P = 0\mod[4]$
since $P$ is even as a sum of even integers.
Thus $r = 0 \mod[4]$.

2) Let $S = S_1 + S_2$ where $S_1$ (resp $S_2$) is the same sum but only over the $a_i$'s which are not greater than 100, and which form the set $G_1$, (resp. not less than 101 and which form $G_2$).
Let $y$ be the number of terms in the sum $S_2$. From (1), we deduce that :
$S_2 = \sum (201-a_i)$ where the sum is over the $a_i$'s not greater than 100 and which do not appear in $S_1$.
Thus $S-S_1 = S_2 = 201y+S_1 - \sum_{i=1}^{100} i$
so that $201y + 2S_1 = 10080 +  \sum_{i=1}^{100} i  = 15130.$ (2)

Therefore :
$S' = \sum_{a_i \in G} a_i^2 = \sum_{a_i \in G_1} a_i^2 + \sum_{a_i \in G_2} a_i^2  = \sum_{a_i \in G_1} a_i^2 + \sum_{a_i \notin G_1,a_i \leq 100} (201-a_i)^2  = \sum_{i =1}^{100} i^2  + 201(201y+2S_1 - 2\sum_{i =1}^{100} i )$
Using (2), we deduce (hope there is not mistake in my computations  :? ) that $S' = 1349380.$

Pierre.


Let A and B be disjoint sets whose union is the set of natural numbers. Show that for every natural number n there exist distinct $ a,b>n$ such that $ \{a,,b,a\plus{}b\}$ is either a subset of A or B.

(By the way when I type \subset it only gives $ \subset$, which is propert subset; what's the correct code then?)

Since $ A\cup B\equal{}\mathbb N$, one of $ A$ and $ B$ contain two elements $ a$ and $ b$ such that $ a>b>n$ and $ a\minus{}b>n$, for each $ n\in\mathbb N$. WLOG let $ a,b\in A$. Assume the contrary. Then $ a,b\in A\Rightarrow a\plus{}b,a\minus{}b\in B\Rightarrow 2a,2b\in A\Rightarrow 2(a\plus{}b),2(a\minus{}b)\in B$. Now $ a,b,2a,2b\in A\Rightarrow 2a\plus{}b,a\plus{}2b,3a,3b\in B$. But this implies $ 2(a\minus{}b)\in B, a\plus{}2b\in B$ and $ 3a\equal{}2(a\minus{}b)\plus{}a\plus{}2b\in B$, contradiction. Hence the conclusion.

[quote="mathwizarddude"]
(By the way when I type \subset it only gives $ \subset$, which is propert subset; what's the correct code then?)[/quote]

You can use subseteq to get $ \subseteq$. Notice, however, that usage of $ \subset$ varies - some authors use it to mean "proper subset", others use it to mean any subset (proper or not). It's not quite as standardized as $ <$ and $ \leq$.

I think you have a small typo:[quote="nayel"]$ 3a \equal{} 2(a \minus{} b) \plus{} a \plus{} 2b\in B$[/quote]should be $ 3a \equal{} 2(a \minus{} b) \plus{} a \plus{} 2b\in A$
Nice solution!


We consider the sequences of real numbers (x_n)n>=1 and (s_k)k>=1 which verify the conditions:
1)x_n is in (0,1) for all n>=1;
2)sum(n=1,k) (x_n)^(s_k)=1 for any k>=1;
3)there exists s real, so that lim(k-->oo) sum(n=1,k) (x_n)^s=1.
a)Prove that the sequence(s_k)converges
b)prove that lim(s_k)=s.

a)
        We have
1)
         s(k) <s(k+1)
         Indeed,assume s(k) \geq s(k+1) with some k
         Because 0<x(i)<1 for all i ,we have x(i)^s(k)  \leq x(i)^s(k+1)
      --->
         1= x(1)^s(k) +x(2)^s(k) +.....+x(k)^s(k)
            \leq x(1)^s(k+1) +x(2)^s(k+1)+.......+x(k)^s(k+1)
            < x(1)^s(k+1)+x(2)^s(k+1)+.........+x(k)^s(k+1) +x(k+1)^s(k+1) =1
      Nonsense.
 --> x(k+1)>s(k) for all k

2)
For all k , s(k)  \leq s
     {If s(k)>s ,we deduce
        1= x(1)^s(k) +x(2)^s(k) +.....+x(k)^s(k)
           \leq  x(1)^s+x(2)^s+.........+x(k)^s +x(k+1)^s
           <lim (sum{1-->oo} [x(i)^s] =1 }

For {s(k)} is an increase and bounded sequence , there exists finite limit  c of {s(k)}

b)  lim s(k)=c --> c \leq s
   Assume c<s
  We have
      1=lim sum{n-->oo}[x(n)^c]=lim sum{n-->oo}[x(n)^s]
   So
         lim sum{n-->oo}[x(n)^c-x(n)^s] =0   (*)
 But , for 0<x(1)<1 ,there exist a nuber  c<m<s :
           x(1)^c -x(1)^s =(c-s).x(1)^m .ln(x(1)) =(s-c)ln(1/x(1)).t^m
>    (s-c).ln(1/x(1)).x(1)^s =  M
    So
         lim sum{n-->oo}[x(n)^c - x(n)^s]  \geq  M >0 ,contradict with (*)
 -->
        c=s
                  (q.e.d)


How many numbers are in the list $ \minus{}36, \minus{}29, \minus{}22, \ldots, 41, 48?$

The list contains the numbers $ \minus{}36, \minus{}29, \minus{}22, \ldots, 41, 48$

Adding 36 to every number in the list gives:

$ 0, 7, 14, \ldots, 77, 84$

We want to count consecutive integers, so we divide each term by 7:

$ 0, 1, 2, \ldots, 11, 12$

Thus, there are $ 12\minus{}0\plus{}1 \equal{} 13$ numbers in the list.

[hide=My solution]Add 43 to every number. 7,14,21...84,91. 91/7=[u]13[/u] numbers.


Rays $ BA$ and $ BC$ form an acute angle. Including angle $ ABC$, how many acute angles are formed by the rays in the diagram?

[asy]draw((0,0)--10dir(0),linewidth(1),Arrow);
draw((0,0)--10dir(88),linewidth(1),Arrow);
draw((0,0)--9dir(15),linewidth(1),Arrow);
draw((0,0)--11dir(35),linewidth(1),Arrow);
draw((0,0)--7dir(55),linewidth(1),Arrow);
draw((0,0)--11dir(70),linewidth(1),Arrow);

dot(8.5dir(88));
dot((0,0));
dot(8.5dir(0));
dot(7dir(15));
dot(9dir(35));
dot(5.5dir(55));
dot(9.5dir(70));

label("A",8.5dir(88),W);
label("B",(0,0),SW);
label("C",8.5dir(0),S);
label("D",7dir(15),SE);
label("E",9dir(35),SE);
label("F",5.5dir(55),NW);
label("G",9.5dir(70),SE);[/asy]

The answer is the fifth triangular number, or 15.

We can pick any two rays, and there are 6 total.  so 6C2=15.

In case you didnt understand how mew got 5th triagnular numb.

[quote="JSGandora"]What about the 90degree angle formed by rays AB and BC? so shouldn't it be 14?[/quote]

The problem specifically says that the angle ABC is acute.

[quote="AIME15"]

In case you didnt understand how mew got 5th triagnular numb.[/quote]

Its 5+4+3+2+1...

5 little angles

4 combination of 2 little angles

3 combinations of 3 little angles

...

and so on.

[quote=JSGandora]What about the 90degree angle formed by rays AB and BC? so shouldn't it be 14?[/quote]
Dude. Read the question properly


trigonometry
inequalities
geometry
inequalities proposed

A triangle  $ ABC$ is given with side lengths $ a, b$ and $ c$. A point $ P$ lies inside $ ABC$, and the distances from  to three sides are $ x, y$ and $ z$ , respectively. Prove that

$ a\sin A \plus{} b\sin B \plus{} c\sin C\geq9\sqrt[3]{xyz}$

Note that $ 9\sqrt[3]{xyz}\leq 3(x\plus{}y\plus{}z) \leq 3(h_a\plus{}h_b\plus{}h_c)$
Where equality occurs if and only if the triangle is equilateral and the point is the fermat point.
Therefore it suffices to show that $ \sum a \sin A\geq 3(h_a\plus{}h_b\plus{}h_c)$
But, from Chebyshev we have that
$ \sum a \sin A\geq \frac 13 (a\plus{}b\plus{}c)(\sin A\plus{} \sin B\plus{}\sin C)$
$ \equal{} \frac 13 (a\plus{}b\plus{}c)\left(\frac{a\plus{}b\plus{}c}{2R} \right) \equal{} \frac{(a\plus{}b\plus{}c)^2}{3abc}\cdot 2\Delta$
Where $ \Delta$ is the area of the triangle.
Since $ h_a \equal{} \frac {2\Delta }{a}$ therefore it suffices to show that
$ (a\plus{}b\plus{}c)^2\geq 3abc\cdot \left(\frac 1a\plus{}\frac 1b\plus{}\frac 1c\right) \equal{} 3(ab\plus{}bc\plus{}ca)$
Which is perfectly true from the trivial inequality.
Equality occurs if and only if $ P$ is the Fermat Point of the triangle $ ABC$ which is equilateral.


How do I set up to find the sum of all multiples of 12 between 100 and 1000?

Use sum of arithmetic series formula. The sum of an arithmetic series with least term a, difference d, and n terms has sum:

$ \frac {n}{2}(2a \plus{} (n \minus{} 1)d)$

An alternate formula would be: leas term a, greatest term b, and has n terms:

$ \frac{n}{2}(a\plus{}b)$

[hide="solution"]

The first term is 108, the difference is 12, and there are $ (996\minus{}108)/12\plus{}1\equal{}75$ terms.

So the sum is:

$ \frac{75}{2}(216\plus{}(74)(12))$
$ \equal{}41400$[/hide]


combinatorics solved
combinatorics
graph theory
Tournament of Towns

[quote="Number1"]Given a graph with $ 50$ vertices. Show that it has two vertices $ A$ and $ B$ such that the number of vertices joined to both $ A$ and $ B$ is even.[/quote]

More generally, if n is a positive integer, then any graph S with 2n vertices has two vertices which have an even number of common neighbours.

This is problem 14.10 in Arthur Engel's [i]Problem-Solving Strategies[/i]. Here is my solution (more or less the same as Engel's one):

The [i]degree[/i] of a vertex in a graph will mean the number of its neighbours, or, equivalently, the number of edges starting at this vertex.

We distinguish between two cases:

[i]Case 1:[/i] Some vertex C of the graph S has an odd degree. This means that it has an odd number of neigbours. Consider the subgraph of S whose vertices are these neighbours (and whose edges are only those edges of S whose both endpoints are neighbours of C). Then, this subgraph has an odd number of vertices; thus, by a well-known theorem (which states that if a graph has an odd number of vertices, then it has a vertex with even degree), it must have a vertex of even degree. In other words, in our graph S, there exists a neighbour D of the vertex C which is connected with an even number of other neighbours of C. In other words, the number of common neighbours of the vertices C and D is even. Problem solved.

[i]Case 2:[/i] Every vertex of the graph S has an even degree.

Let A be an arbitrary vertex of the graph S. Construct a subgraph S' of S as follows:
- The subgraph S' should contain all the 2n vertices of the graph S except of the vertex A.
- The subgraph S' should contain only those edges of the graph S which pass through a neighbour of A (in other words, an edge of S is taken as an edge of S' if and only if at least one of the endpoints of this edge is a neighbour of A; the other endpoint can be arbitrary, but it cannot be A since A is not a vertex of S').

Then, the graph S' has an odd number of vertices (2n - 1 vertices, to be precise), and thus, by the well-known theorem we used above, it contains a vertex D with even degree. This vertex D cannot be a neighbour of A in the graph S (in fact, since we are in Case 2, the vertex D has an even degree in the graph S, i. e. there is an even number of edges of the graph S starting at the vertex D; now, if D would be a neighbour of A, then all of these edges would be edges of the subgraph S', except of the edge joining D to A, and thus, there would be an odd number of edges of the subgraph S' starting at the vertex D, so that the degree of the vertex D in the subgraph S' would be odd). Hence, this vertex D is not a neighbour of A. Its degree in the subgraph S' is the number of common neighbours of the vertices A and D in the graph S. Since this degree is even, we thus see that the vertices A and D have an even number of common neighbours. Problem solved.

[b]EDIT:[/b] Other solutions can be found at http://mathoverflow.net/questions/17269/let-g-be-a-graph-such-that-for-all-u-v-v-g-u-no-equal-to-v-n-u-n-v .

  Darij


If inequality $ \frac {\sin ^{3} x}{\cos x} \plus{} \frac {\cos ^{3} x}{\sin x} \ge k$ is hold for every $ x\in \left(0,\frac {\pi }{2} \right)$, what is the largest possible value of $ k$?

$\textbf{(A)}\  \frac {1}{2} \qquad\textbf{(B)}\  \frac {3}{4} \qquad\textbf{(C)}\ 1 \qquad\textbf{(D)}\  \frac {3}{2} \qquad\textbf{(E)}\ \text{None}$

From Holder

$ LHS^2 \left( cos^2 \plus{} sin^2 \right) \ge \left( sin^2 \plus{} cos^2x \right) ^3$

Answer is $ \boxed{C}$

By rearrengement inequality

$ \dfrac{\sin^3(x)}{\cos(x)}\plus{}\dfrac{\cos^3(x)}{\sin(x)}\ge \dfrac{\sin^3(x)}{\sin(x)}\plus{}\dfrac{\cos^3(x)}{\cos(x)}\equal{}\sin^2(x)\plus{}\cos^2(x)\equal{}1$

with equality when $ \cos(x)\equal{}\sin(x)\Leftrightarrow x\equal{}\dfrac{\pi}{4}$

So the answer is $ C$.

:)

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

While $ \sin 2x$ is increasing, above expression will decrease. So it has its minima at $ \sin 2x \equal{} 1$.
Then $ k\equal{}1$ and $ 2x \equal{} \frac{\pi}{2} \Rightarrow x \equal{} \frac{\pi}{4}$

we can also prove it easily by Cauchy-Shcwarz.

$ \frac {\sin ^{3} x}{\cos x} \plus{} \frac {\cos ^{3} x}{\sin x} \equal{} \frac {\sin^4x \plus{} \cos^4x}{\sin x\cos x}\ge \frac {(\sin^2x \plus{} \cos^2x)^2}{2\sin x\cos x} \equal{} \frac {1}{\sin 2x}\ge 1$ for any $ x\in\left(0\ ,\ \frac {\pi}{2}\right)$ , where $ \sin 2x\in(0\ ,\ 1]$ .

I used the simple inequality $ a^2\plus{}b^2\ge \frac 12\cdot (a\plus{}b)^2$ for any real numbers $ a$ and $ b$ in particular case $ a: \equal{}\sin^2x$ and $ b: \equal{}\cos^2x$ .


AP (bad teacher) vs. regular (good teacher)...
Which should I go for?

Well, a good teacher will definitely make class more enjoyable, but in this particular case, the grading is supposedly harder in the regular class! Huge drop in GPA expected.
And unfortunately, the quality of teachers doesn't show up on the transcript.

What subject?  How is your transcript as it stands?

Both US History and govt/politics. (required)

My transcript for last semester:
AP Econ 100
English II Honors 92
World History 90
AP Calc 100
Physics 98
French 99
AP Chem 98
Graphic Design 100.

I wouldn't obsess over GPA or how you'll look to colleges or anything else. Choose the course that you'll learn the most from and do your best.

If the teacher is as good as you say, you might be able to still take the AP exam.  Plus if you work hard in the class, you might be able to get a recommendation.  Consider this: I know people who took AP Environmental for their science instead of regular physics...guess which class is easier and less useful (and in my opinion less interesting).

I should have remembered from chem this year that 200+ horrible class hours each week isn't really worth the AP designation. And since a lot of people in my school are going to take IB next year, the people in the class won't be fabulous anyway.
I guess I'll just have to work harder to keep up the transcript. *sigh*

Thanks everyone for the advice! :)


calculus
trigonometry
calculus computations

Hi, today i've been handed a calculus assignment and Im stuck on a particular question and need some help. Here it goes:

Prove that $ f(x) \equal{} xSin(x)$ and $ f(x) \equal{} Cos(x)$ intersect at right angles to eachother.

My work:

$ xSin(x) \equal{} Cos(x)$
$ x \equal{} \frac{Cos(x)}{Sin(x)}$
$ x \equal{} Cot(x)$

$ f'(x) \equal{} \frac{\minus{}1}{f'(x)}$
$ xCos(x) \plus{} Sin(x) \equal{} \frac{1}{Sin(x)}$
$ Cot(x)Cos(Cot(x)) \plus{} Sin(Cot(x)) \equal{} \frac{1}{Sin(Cot(x))}$

Now Im stuck, I dont know how to prove that this is true!!!

Your notation is horrible.  What you really want to show is that if $ f(x) = x\sin x$, $ g(x) = \cos x$, and $ f(a) = g(a)$, then $ f'(a)g'(a) = 1$.

As you noted, $ f(a) = g(a)$ implies $ a = \cot a$.

Now it's straightforward to compute

\begin{eqnarray*}
f'(a)g'(a) &=& (a\cos a+\sin a)(-\sin a) \\
&=& (\cot a \cos a + \sin a)(-\sin a) \\
&=&\left(\frac{1-\sin^2 a}{\sin a} + \sin a\right)(-\sin a) \\
&=&\left(\frac{1}{\sin a}\right)(-\sin a) \\
&=& -1
\end{eqnarray*}

:o never thought of it that way, thank you!!!!!!


algebra
polynomial
integration
complex analysis
complex analysis unsolved

let $ P(z)\equal{}a_{0}\plus{}...\plus{}a_{n}z^{n}$ be a polynomial of order $ n\geq 1$
    $ C_{r}$ be the circle with radius $ r>0$, counterclockwise oriented
prove that \[ \frac{1}{2\pi i}\int_{C_{r}}z^{n\minus{}1}|P(z)|^2dz\equal{}a_{0}\overline{a_{n}}r^{2n}\]

Any idea will be highly appreciated

Write $ |P(z)|^2\equal{}P(z)\overline{P(z)}$. Split into terms. What is $ \int_{C_r}z^a\overline{z}^b$? Well, on $ C_r$, $ \overline{z}\equal{}\frac{r^2}{z}$


let's assume the original number is $ 100$

after increasing it by $ 50\%$, our number would be $ 150$

after decreasing it by $ 50\%$, our number would be $ 75$

thus, the percent of decrease from the original number to the final number is $ \frac{100\minus{}75}{100}\equal{}\frac{25}{100}\equal{}25\%$

PS the percent decrease is calculated with this formula: $ \frac{\text{original} \minus{} \text{new}}{\text{original}}$

[quote="GameBot"]A number is increased by $ 50\%$ and then the result is decreased by $ 50\%$. What is the percent of decrease from the original number to the final number?[/quote]

25 :)

I totally agree with vallon22's response. No need for supposing that the number is 100, it could just be x*1.5*.5 = .75x, but it doesn't really matter x or 100, really... :lol:

I think somewhere izzy said saying i agree, and like stuff like that are like spam?
But it's ok seeing your a new member.


Sa se arate ca:
$ (m^2 \plus{} 9mn \plus{} n^2)\vdots 11\Rightarrow (m^2 \minus{} n^2)\vdots 11; \ (\forall) m;n\in \mathbb{Z}.$

$ m^2\plus{}9mn\plus{}n^2 \vdots 11 \Rightarrow (m\minus{}n)^2 \plus{} 11mn \vdots 11 \Rightarrow (m\minus{}n)^2 \vdots 11 \Rightarrow m\minus{}n \vdots 11$

$ \Rightarrow m$ si $ n$ dau acelasi rest la impartirea cu $ 11$ de unde rezulta concluzia

In aceleasi conditii, demonstrati ca:
$ (m^2\plus{}9mn\plus{}n^2)\vdots 77 \Rightarrow (m^2\minus{}n^2)\vdots 77.$

Ramane de demonstrat divizibilitatea cu $ 7$, care se face in mod analog ca si in cazul divizibilitatii cu $ 11$