[b]Problema 2088:[/b]
Daca $ x,y,z > 0$ si $ x \plus{} y \plus{} z \equal{} 1$, demonstrati inegalitatea: $ \frac {x^2 \plus{} 3xy}{x \plus{} y} \plus{} \frac {y^2 \plus{} 3yz}{y \plus{} z} \plus{} \frac {z^2 \plus{} 3zx}{z \plus{} x}\le 2.$([i]Pirkuliev[/i])

[b]Problema 2089:[/b]
Fie $ B_0$-mijlocul laturii $ [AC]$ a triunghiului $ ABC.$ Notam cu $ A_1$ centrul cercului inscris in $ \triangle{ABB_0}$ si cu $ A_2$ punctul de tangenta al laturii $ [AB]$ cu cercul exanscris acestei laturi in acelasi triunghi. Punctele $ C_1$ si $ C_2$ sunt definite in mod analog in $ \triangle{CBB_0}.$ Demonstrati ca patrulaterul $ A_1A_2C_2C_1$ este inscriptibil. ([i]Emelianov[/i])

[b]Problema 2092:[/b]
Pe muchiile $ [AB], [BC], [CD]$ si respectiv $ [DA]$ ale tetraedrului $ ABCD$ se iau punctele $ K, L, M$ si $ N.$ Notam cu $ K', L', M'$ si $ N'$ simetricele punctelor $ K, L, M$ si $ N$ fata de mijloacele muchiilor $ [AB], [BC], [CD]$ si $ [DA].$ Demonstrati ca tetraedrele $ KLMN$ si $ K'L'M'N'$ au volumele egale. ([i]Kojevnikov[/i])

[quote="mihai miculita"][b]Problema 2088:[/b]
Daca $ x,y,z > 0$ si $ x + y + z = 1$, demonstrati inegalitatea: $ \frac {x^2 + 3xy}{x + y} + \frac {y^2 + 3yz}{y + z} + \frac {z^2 + 3zx}{z + x}\le 2.$ ([i]Pirkuliev[/i])[/quote]

Avem:

$ \displaystyle\sum_{cyc}\frac{x^2+3xy}{x+y}=\displaystyle\sum_{cyc}\frac{(x+y)^2-y(y-x)}{x+y}=\displaystyle\sum_{cyc} x+y-\frac{y(y-x)}{x+y}\leq 2\Longleftrightarrow$ $ \displaystyle\sum_{cyc}\frac{y(y-x)}{x+y}\geq 0 \ |\cdot\prod(x+y) \Longleftrightarrow$

$ \Longleftrightarrow\left (\displaystyle\underbrace{\displaystyle\sum_{cyc}x^2-\displaystyle\sum_{cyc}xy}_{\geq 0}\right )\left (\displaystyle\sum_{cyc}xy\right )+\displaystyle\underbrace{\displaystyle\sum_{cyc}x^2y^2-\displaystyle\sum_{cyc}x^2yz}_{\geq 0}\geq 0.$

Ca sa fie mai clar: $ \displaystyle\sum_{cyc}x^2y^2-\displaystyle\sum_{cyc}x^2yz\geq 0\Longleftrightarrow \frac{1}{2}\left [ z^2(x-y)^2+x^2(y-z)^2+y^2(z-x)^2\right ]\geq 0,$ ultima inegalitate fiind evidenta.

[quote="mihai miculita"]

[b]Problema 2089:[/b]
Fie $ B_0$-mijlocul laturii $ [AC]$ a triunghiului $ ABC.$ Notam cu $ A_1$ centrul cercului inscris in $ \triangle{ABB_0}$ si cu $ A_2$ punctul de tangenta al laturii $ [AB]$ cu cercul exanscris acestei laturi in acelasi triunghi. Punctele $ C_1$ si $ C_2$ sunt definite in mod analog in $ \triangle{CBB_0}.$ Demonstrati ca patrulaterul $ A_1A_2C_2C_1$ este inscriptibil. ([i]Emelianov[/i])

[/quote]

[url=http://img155.imageshack.us/my.php?image=emelianov.png][img]http://img155.imageshack.us/img155/3896/emelianov.png[/img][/url]

Problema pare a fi gresita sau cel putin incompleta.

[b][i]Greseala de traducere imi apartine, enuntul corect este urmatorul:[/i]
Problema 2089:[/b]
$ \mbox{ Fie } B_0 \mbox{ mijlocul laturii [AC] a triunghiului oarecare ABC. Notam cu } A_1 \mbox{ centrul cercului inscris si cu } \\
A_2 \mbox{ centrul cercului exanscris laturii [AB] a triunghiului } ABB_0. \ C_1 \mbox{ este centrul cercului inscris in triunghiul } \\
CBB_0 \mbox{, iar } C_2 \mbox{ este centrul cercului exanscris triunghiului } CBB_0 \mbox{ tangent laturii [BC].} \\
\ \ \ \ \mbox{ Sa se arate ca patrulaterul } A_1A_2C_2C_1 \mbox{ este inscriptibil!}$

$\frac {x^2 \plus{} 3xy}{x \plus{} y}=\frac{x^2+2xy+y^2+xy-y^2}{x+y}=x+y-\frac{y(y-x)}{x+y}$, so $ \frac {x^2 + 3xy}{x + y} + \frac {y^2 + 3yz}{y + z} + \frac {z^2 + 3zx}{z + x}=2\sum x - \sum \frac{y(y-x)}{x+y}=4- \sum \frac{y(y-x)}{x+y} .$ so we need to prove $ \sum \frac{y(y-x)}{x+y} \geq 0$ or
$\sum \frac{y^2}{x+y} \geq \sum \frac{xy}{x+y}$. By C-S we have $\sum \frac{y^2}{x+y} \geq \frac{x+y+z}{2}$. But from $(x+y)^2 \geq 4xy$ we get $\frac{xy}{x+y} \leq \frac{x+y}{4}$ thus $\sum \frac{xy}{x+y} \leq \frac{x+y+z}{2}$. So, $\sum \frac{y^2}{x+y} \geq \frac{x+y+z}{2} \geq \sum \frac{xy}{x+y}$ and this yields $\sum \frac{y^2}{x+y} \geq \sum \frac{xy}{x+y}$ so from this we finally get $ \frac {x^2 + 3xy}{x + y} + \frac {y^2 + 3yz}{y + z} + \frac {z^2 + 3zx}{z + x}\le 2.$. Q.E.D

[quote=mihai miculita][b]Problema 2088:[/b]
Daca $ x,y,z > 0$ si $ x \plus{} y \plus{} z \equal{} 1$, demonstrati inegalitatea: $ \frac {x^2 \plus{} 3xy}{x \plus{} y} \plus{} \frac {y^2 \plus{} 3yz}{y \plus{} z} \plus{} \frac {z^2 \plus{} 3zx}{z \plus{} x}\le 2.$([i]Pirkuliev[/i])

Asta este o problema de nivel mediu la un Interjudetean la clasa  a 7-a,prea usoara pentru clasa a 8-a.

$ \frac {x^2 + 3xy}{x + y}=x+\frac{2xy}{x+y} \le \ x+\frac{x+y}{2}$,din HM-AM.
Adunand inegalitatile analoage obtinem concluzia.

[quote=Pure_IQ]Revista Kvant pare foarte interesanta,poate fi gasite pe Internet sau trebuie sa imi fac abonament?[/quote]

Daca intelegi rusa : http://kvant.mccme.ru

Da,am gasit pagina aia,ma gandeam daca o gasesc in engleza,oricum,mersi de ajutor.

[quote=mihai miculita][b]Problema 2088:[/b]
Daca $ x,y,z > 0$ si $ x \plus{} y \plus{} z \equal{} 1$, demonstrati inegalitatea: $ \frac {x^2 \plus{} 3xy}{x \plus{} y} \plus{} \frac {y^2 \plus{} 3yz}{y \plus{} z} \plus{} \frac {z^2 \plus{} 3zx}{z \plus{} x}\le 2.$([i]Pirkuliev[/i])
[/quote]
$$\sum_{cyc} \frac{2ab}{a+b} \leq  \sum_{cyc} \sqrt{ab} \leq  a + b + c$$$$\sum_{cyc} \frac{a^2 + 3ab}{a+b} \leq 2 \iff \sum_{cyc} \frac{2ab}{a+b} \leq 1$$
https://artofproblemsolving.com/community/c6h1250130p6434429
https://artofproblemsolving.com/community/c267h269612p3454038


Putnam
combinatorics unsolved
combinatorics

S is the set {1,2,3.....n}
A subset A of S is called NORMAL if and only if it is non empty and the arithmetic mean of its elements is an integer.
Prove that the number of NORMAL subsets of S and n have the same parity.

This is Putnam 2002 A3.

For each normal subset $A$ of $S$, denote the arithmetic mean of the elements in $A$ by $M(A)$.

Let $B = \{A \in S| A \mbox{ is normal}, M(A) \not \in A, |A| > 1\}$ and $C = \{A \in S | A \mbox{ is normal}, M(A) \in A\}$.  Consider any element $D \in B$.  $M(\{M(D)\}) \cup D\} = M(D)$, so $\{M(D)\} \cup D \in C$.  Now consider any element $E \in C$.  $M(D \backslash \{M(D)\}) = M(D)$, so $D \backslash \{M(D)\} \in B$.  Hence there is a bijection between the elements of $B$ and $C$.  The only other normal subsets of $S$ are the sets $\{1\}, \{2\}, \ldots, \{n\}$.  Thus number of normal subsets = $|B| + |C| + n = 2 |B| + n$.


email geometry Princeton college AMC USA(J)MO USAMO

er, well, it seems as if I got a prize confirmation email, then it was retracted.

Anyone else received the same thing?

edit: seems that things are working fine now

Uh, the SHS proctors were kinda being stupid.

They had the form with your ID/email/phone and they were like, oh fill this out after the test if you want to put your email address or phone number. Then they forgot after the test, of course.

Anyways, I'm 95% sure I'm not going to the Awards Ceremony, because I'm 95% sure I didn't get a prize.

[quote="Ubemaya"]
Anyways, I'm 95% sure I'm not going to the Awards Ceremony, because I'm 95% sure I didn't get a prize.[/quote]

This sounds too Ubeyish  :)  wait a sec... you ARE Ubey...

mmm apparently i got an award and wasnt retracted
which is kind of shocking

so all the ppl that got awards are now notified?
bummer. means i didnt get one

I didn't write my contact info in, so I have no way of knowing... :maybe:

Eh. I didn't get one. Not that surprising though.

Here's a simple exercise in logic.

  If I get an invitation, then Ubey gets an invitation. (Not a very strong condition)

Therefore, we take the contrapositive:

  If Ubey does not get an invitation, then I do not get an invitation.

'Nuff said.

I didn't get an invitation. No surprise given how I somehow managed to only solve #3 fully despite there being three other easy problems on the test.

I only solved #3, so ...
I don't remember filling out any forms--was the form separate?

[quote="serialk11r"]Joining the failure club here...no invitation. Ubey got 3 problems I believe, so if he didn't get an invitation, then stephen ge, brian zhang, didn't get invitations, charles, taylor, probably didn't, and that would be SCARY, because then that would mean that some unknown beasts have decided to raid the bay area.[/quote]
i got one, and i dont see a retraction email yet, so they either gave me a lot of partial credit, ubey is lying, or ubey f-ed up a some of his 3 solutions

they gave me another email that said that I passed...

well im new to this forum (just noticed)

i think they haven't finished sending out emails
i gave them my contacts.. but NOTHING back

they won't give you a confirmation email if you didn't get an award.

[quote="junggi"]gunn: epic fail. best was me and brian with honorable mentions. at least we won participation
paly: you guys raped. john, nathan, colleen all placed. john tied for 1st in 9/10. 2nd place team, 2nd place participation

joseph chu perfect scored, was the only one to get any points on #6

hmm what else

lynbrook 1st place team
i think homestead or somewhere was 3rd not sure[/quote]

Yeah, Homestead took 3rd (I believe)

BAMO 8 Results:

Team:

Miller
Lawson
Some School

Individual:

Asian kid
6th grade Indian Girl
Spencer
This white kid

Honorable:

Lots of Asian dudes
Me.

MSJ got 1 grand prize and 2 honorable mentions. but we did not have enough participating people to get any team awards :(

Next year, i hope we can amass more people.

[quote="junggi"]gunn: epic fail. best was me and brian with honorable mentions. at least we won participation
paly: you guys raped. john, nathan, colleen all placed. john tied for 1st in 9/10. 2nd place team, 2nd place participation

joseph chu perfect scored, was the only one to get any points on #6

hmm what else

lynbrook 1st place team
i think homestead or somewhere was 3rd not sure[/quote]

How did I do? Honorable?

[quote="james4l"]Ubey, you got 2nd (or 3rd?) place tied with some other students, definitely better than honorable mention.[/quote]

4th.

I don't know if he tied though.

[quote="bookaholic"]Sigh...so the one time we're like TOTAL PWNAGE (well, not really, but as close as we'll ever get with Lynbrook et al. around) I have to be like *faildeath*...

I better do well on the APMO.[/quote]
apmo?
who gets to take it
and when

Can anyone post the full results? I missed the awards ceremony.

[quote="bookaholic"]MOPpers, and today. And I think I got a positive score! JAM, did you take it?[/quote]
The APMO? No I didn't, I couldn't find a place to take it  :( .

[quote="serialk11r"][quote="i_like_pie"]To clarify Ubemaya's placement: Tied for 4th, most likely 3-way.

HOMESTEAD FINALLY GOT A TEAM AWARD FOR SOMETHING! :D

(In case you didn't know, I go to Homestead.)[/quote]
Finally? 1st place at SCVMA...well maybe that doesn't really count since it was such a crap competition but still.
Hey were you at SMT?[/quote]
SCVMA doesn't really count as a high-level competition, which starts at AMC/BAMO level. Also, it's weird. I think the only thing weirder is Tokutomi. Our team started to protest one of the Tokutomi problems (I don't remember which # it was; it was the one with the cube painted white and black), and the guy explaining the solutions just shut us up (...to avoid looking bad, most likely). :|

I couldn't go to SMT, I had other stuff to do that day... :(

[quote]BAMO 8 Results:

Team:

Miller
Lawson
Some School [/quote]

You probleably mean NAS?


geometry
3D geometry
algebra
polynomial
function
percent
ratio

1. A cubic block with dimensions n by n by n is made up of a collection of 1 by 1 by  1 unit cubes.  What is the smallest value of n so that if the outer two layers of unit  cubes are removed from the block, more than half the original unit cubes will still  remain?

2. A 70 foot pole stands vertically in a plane supported by three 490 foot wires, all  attached to the top of the pole, pulled taut, and anchored to three equally spaced  points in the plane.  How many feet apart are any two of these anchor points?

3. After distributing textbooks to his classes, Mr. Yurek had five Algebra 1,  four  Geometry, and four Algebra 2 textbooks remaining.  In how many ways could he  arrange these textbooks, upright left to right, on his shelf?

4. A non-constant polynomial function f(x) satisfies:  f(-3) = f(-1) = f(2) = f(4) = 5.    What is the smallest possible degree of f(x) ?

5. A person 6 feet tall standing near a streetlight casts a 10-foot shadow on the  ground.  If the base of the streetlight is 45 feet from the tip of the shadow, what is  the height, in feet, of the streetlight?

6. What digit is in the 2005 th place after the decimal point in the  2005th decimal expansion of the fraction $\frac{1}{81}$?

7. A copying machine can make copies that are 80%, 100% and 150% as large as the  original.  By making copies of copies, what is the smallest number of times you  must use the machine if you want a copy of an original that is 324% as large as  the original?

8.  If the surface area of one cube is 44% larger than the surface area of a second  cube, what is the closest whole percent by which the volume of the first cube  exceeds the volume of the second cube?

9. Find the ratio of an inscribed square to the circle in which it is inscribed.

Edit: Added more problems

[hide="4"]

Fundamental Theorem of Algebra leads one to believe that the smallest possible polynomial that satisfies those restrictions is $\boxed {4}$[/hide]

[quote="surge"][quote="lotrgreengrapes7926"][quote="surge"][hide="3"]

This may be wrong... but

$\frac{P(16,16)}{P(5,5) \cdot P(4,4) \cdot P(4,4)}=\boxed{302702400?}$

someone check?
[/hide][/quote]
[hide] Do you mean $\frac{13!}{5!4!4!}$? Because $4+4+5=13$. :D [/hide][/quote]

 :D yes.[/quote] Your answer was wrong surge

7 [hide] the order of multiplying doesnt matter so 1.5^4 x .8^2 = 3.24 so the answer is 6 [/hide]

8 [hide] the surface areas are in a raito of 144:10 and since the ratio of the surface area is the ratio of the sides squared so that is 12:10 and then the voume is the cube of that ratio 1728:1000 so thats 72.8% more which rounds to 73[/hide]

[quote="lkryptonicl"][/quote][quote="surge"][quote="lotrgreengrapes7926"][quote="surge"][hide="3"]

This may be wrong... but

$\frac{P(16,16)}{P(5,5) \cdot P(4,4) \cdot P(4,4)}=\boxed{302702400?}$

someone check?
[/hide][/quote]
[hide] Do you mean $\frac{13!}{5!4!4!}$? Because $4+4+5=13$. :D [/hide][/quote]

 :D yes.[/quote][quote="lkryptonicl"] Your answer was wrong surge[/quote]
[hide="3"] The answer is $\frac{13!}{5!4!4!}=90090$. :D [/hide]

[quote="lotrgreengrapes7926"][/quote][quote="lkryptonicl"][/quote][quote="surge"][quote="lotrgreengrapes7926"][quote="surge"][hide="3"]

This may be wrong... but

$\frac{P(16,16)}{P(5,5) \cdot P(4,4) \cdot P(4,4)}=\boxed{302702400?}$

someone check?
[/hide][/quote]
[hide] Do you mean $\frac{13!}{5!4!4!}$? Because $4+4+5=13$. :D [/hide][/quote]

 :D yes.[/quote][quote="lkryptonicl"] Your answer was wrong surge[/quote][quote="lotrgreengrapes7926"]
[hide="3"] The answer is $\frac{13!}{5!4!4!}=90090$. :D [/hide][/quote]

yeah, i didn't add properly.

sorry... for the triple post. I pressed back by accident instead of edit.

[hide="half solution for 1"] Solve for n: $\frac{n^3}{2}=(n-4)^3$ and round up. [/hide]

[quote="surge"][hide="2"]
If I understand this problem correctly...

$70^2+x^2=490^2$
$x=280\sqrt{3}$
Circumference=$560\pi\sqrt{3}$
They must be $\frac{560\pi\sqrt{3}}{3}$ feet apart?
[/hide]
[hide="6"]
The patterns follows $012345679$.

$mod (2005,9)=7$

$\boxed{6}$[/hide][/quote]
[hide="2"] It's an equilateral triangle not a circle, I think. $280\sqrt{3}\times\frac{3}{2}=420\sqrt{3}$ is the altitude. So the side length is $\boxed{240}$. [/hide]

For 1, are you sure it's not...
[hide="1"]
$n^3-(6n^2+6(n-2)^2)>\frac{n^3}{2}$
That almost gives an integer answer.[/hide]

you're right about the triangle thing.

[hide="1"]I was thinking $n^3-(n-2)^3<\frac{n^3}{2}$, but I solved it just by testing values, because I was being lazy.  The answer is $6$...

Edit: Make that $\boxed{20}$[/hide]

[hide="8"]$\frac{6n^2}{6p^2}=1.44$
$n^2=1.44p^2$
$n=1.2p$
$n^3=1.728p^3$
$72.8 \approx \boxed{73}$[/hide]

[hide="9"]The radius $r$ is half the lenth of a diagonal of the square.
$\frac{(r\sqrt{2})^2}{\pi r^2}=\frac{2r^2}{\pi r^2}=\boxed{\frac{2}{\pi}}$[/hide]

[quote="DanK"][hide="1"]I was thinking $n^3-(n-2)^3<\frac{n^3}{2}$, but I solved it just by testing values, because I was being lazy.  The answer is $6$...[/hide]
[/quote]
[hide="1"] If the answer were 6, after taking the outer 2 layers off, there would be a cube of side length 2 left. And $\frac{2^3}{6^3}<\frac{1}{2}$. Using my solution, I solved the cubic equation $n^3-24n^2+96n-128$ with a graphing calculator and got the solution $19.389288$, so the answer is $n=\boxed{20}$. [/hide]

Oh, you're right.  I only took the two off from one side, not both...  Stupid me... :wallbash:

[quote="lotrgreengrapes7926"][hide="4"] The lowest degree possible would be 4th, since it intersects a horizontal line at 4 points. You could also say $f(x-5)$, which has the same degree as $f(x)$, has 4 roots and is at lowest 4th degree. [/hide][/quote]

You mean $f(x) - 5$.  :)


In triangle $ABC$, $AB=12$, $AC=7$, and $BC=10$. If sides $AB$ and $AC$ are doubled while $BC$ remains the same, then:

$\textbf{(A)}\ \text{The area is doubled} \qquad\\
\textbf{(B)}\ \text{The altitude is doubled} \qquad\\
\textbf{(C)}\ \text{The area is four times the original area} \qquad\\
\textbf{(D)}\ \text{The median is unchanged} \qquad\\
\textbf{(E)}\ \text{The area of the triangle is 0}$

E

Notice how the new triangle measures $ 24, 10, 14$

10+14=24 so if this is a triangle then the shorter sides, 10 and 14, would just fold onto the longer side, 24, and the triangle would have no actual space or area, so its area would be 0. Answer = E

[quote=shrungpatel]New new sides are $10, 14, 24$ and $10$ plus $14$ is not greater than $24$, so the triangle is not possible. E.[/quote]

what is the difference between this and the other two solutions???

Yeah the triangle cannot happen due to the triangle inequality, which states that in a triangle with lengths $a, b, c$, any sum of two sides is greater than the third.

Well now that this is bumped, might as well solve it. After doubling the sides are $24, 14, 10$. But $14+10 = 24$ so this triangle becomes a straight line, with area $0 \implies \boxed{\text{(E)}}$.


geometry
MATHCOUNTS
AoPS Books
number theory

A few years ago, the PA state champion came from our area, because our area had good coaches. Now they're retired or something and we just prepared by doing warmups and workouts from the MC book. and most of the people are idiots (there was this one time where we got through like 10 problems b\c soem kid who kept asking questions on everything). because our coaches dont help us...soo I know some state champs and naty champs are here, how did u prepare?

Hmm...well I have some advice for you from yours truly who wasn't the state champ but came close at least(4th in Florida).  Read both AoPS books from front to end...try taking online classes that you normally wouldn't learn at school (number theory for example), if you're taking higher level math that'a plus.  Or that's what I did.

That is just the easy part.  The hardest part is the competition itself for you don't know what's going to pop up.  So think of MathCounts as not a competition against others but against yourself.  That's what I suggest for u to do for it worked for me quite well.  One more thing too...if you happen to make nationals...practice a lot of countdown rounds.  Even if you don't make the top ten (I didn't either), you should still be very very fast.  And hey, if your school team isn't very good, don't worry...as long as you're extremely good, you should have no problem making nationals or state anyways.  (My school got fifth at chapter and didn't even goto state)

[b]Well...hmm...maybe my speed was my downfall at nationals...oh well big deal!  :-) [/b]


[quote="zaya_yc"]$ \sum_{k \equal{} 1}^{n}{arctan{\frac{1}{1\plus{}k\plus{}k^{2}}}}\equal{} ?$[/quote]
$ arctan\frac{1}{1\plus{}k\plus{}k^{2}}\equal{} arctan{tg(arctan(k\plus{}1)\minus{}arctan(k))}\equal{} arctan(k\plus{}1)\minus{}arctan(k)$,
 therefore sum is $ arctan(n\plus{}1)\minus{}\frac{\pi}{4}$.


Hi,

I am currently working my way through the book "Introduction to Abstract Algebra" and came across the following permutation identity

$ (k_{1}  k_{2}  ... k_{r}) \equal{} (k_{1}  k_{r}) (k_{1}  k_{r\minus{}1}) (k_{1}  k_{r\minus{}2}) .... (k_{1}  k_{3}) (k_{1}  k_{2})$

As we discussed an example about the generation of $ S_{n}$ my professor told me that's wrong, because argued that's the inversion of

$ (k_{1}  k_{2}) (k_{2}  k_{3}) (k_{3}  k_{4}) .... (k_{r\minus{}2}  k_{r\minus{}1}) (k_{r\minus{}1}  k_{r})$

But when I looked again in to the internet I saw another time the same cycle structure
<http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2004;task=show_msg;msg=0747.0001X=>

Therefore, I'll have proven that you can decompose every cycle into two cycles
$ (k_{1}  k_{2}  ... k_{r}) \equal{} (k_{1}  k_{r}) (k_{1}  k_{r\minus{}1}) (k_{1}  k_{r\minus{}2}) .... (k_{1}  k_{3}) (k_{1}  k_{2})$

but I don't know where I made mistake.

Thanks for comments and maybe remarks towards books where the exercise is also given.
Klemens

[quote="MathEcon"]Hi,

I am currently working my way through the book "Introduction to Abstract Algebra" and came across the following permutation identity

$ (k_{1} k_{2} ... k_{r}) \equal{} (k_{1} k_{r}) (k_{1} k_{r \minus{} 1}) (k_{1} k_{r \minus{} 2}) .... (k_{1} k_{3}) (k_{1} k_{2})$
[/quote]
This is correct. The one caveat is that you have to interpret the product reading from right to left (in other words, permutations act on the left).

[quote="MathEcon"]
As we discussed an example about the generation of $ S_{n}$ my professor told me that's wrong, because argued that's the inversion of

$ (k_{1} k_{2}) (k_{2} k_{3}) (k_{3} k_{4}) .... (k_{r \minus{} 2} k_{r \minus{} 1}) (k_{r \minus{} 1} k_{r})$
[/quote]
I don't understand the professor's argument, and I'm not sure why there would be an argument over a simple identity. He's possibly interpreting products left-to-right.

[quote="MathEcon"]
But when I looked again in to the internet I saw another time the same cycle structure
<http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2004;task=show_msg;msg=0747.0001X=>
[/quote]
The correct way to embed a link is [url=http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2004;task=show_msg;msg=0747.0001]like this[/url]. I needed to edit your link to get it to work. Anyway, that link only repeats the same identity.
[quote="MathEcon"]
Therefore, I'll have proven that you can decompose every cycle into two cycles
$ (k_{1} k_{2} ... k_{r}) \equal{} (k_{1} k_{r}) (k_{1} k_{r \minus{} 1}) (k_{1} k_{r \minus{} 2}) .... (k_{1} k_{3}) (k_{1} k_{2})$
[/quote]
This is the same formula you quoted in the beginning, and it's still true.
[quote="MathEcon"]
but I don't know where I made mistake.
[/quote]
What mistake? I'm confused.


The side of a square is increased by 50%. By what percent is the area increased?

[hide=" Solution"]

$ 125\%$:

If the sides were originally $ 1$, then the area was $ 1$.

Now the sides are $ 1.5$ and so the area is $ 1.5^{2} \equal{} 2.25$, an increase of $ 1.25$.

So the percentage increase is $ \frac{1.25}{1}\cdot 100 \% \equal{} 125\%$.

[/hide]

Suppose the square had a side length of 10. If that length was increased by 50 percent, it would increase by $ 10 \times .5 \equal{} 5$, so it would then be $ 15$. That means the new area is 225, which is 125 more than the original area of 100. Therefore, the area increased by 125%.


real analysis
set theory
real analysis unsolved

Is it possible to construct a non-measurable (Lebesgue) set in $ \mathbb{R}$ without using the axiom of choice ?

[quote="[url=http://en.wikipedia.org/wiki/Non-measurable_set]Wikipedia[/url]"]In 1970, Solovay demonstrated that the existence of a non-measurable set for Lebesgue measure is not provable within the framework of Zermelo-Frankel set theory in the absence of the Axiom of Choice. He also showed that (assuming the consistency of an inaccessible cardinal) there is a model of ZF in which countable choice holds, every set is lebesgue measurable and in which the full axiom of choice fails.[/quote]

The result which is being referred to can be found in this paper:

[i]R. Solovay:[/i] A model of set theory in which every set of reals is Lebesgue-measurable.  Ann of Math. [i]92[/i] (1970), 1-56.


Find value of the sum

$ S\equal{}1\minus{}\frac{1}{2}\plus{}\frac{1}{3}\minus{}\frac{1}{4}\plus{}...\minus{}\frac{1}{1318}\plus{}\frac{1}{1319}$

Really?
[url=http://www.wolframalpha.com/input/?i=sum+power%28-1%2C+k-1%29%2Fk+k%3D1+to+1319]sum power(-1, k-1)/k k=1 to 1319[/url] - Wolfram Alpha

Ahh, i wrote it illegibly and figured it was the question but after some time spent on deciphering it i get that we need to prove that

$ S\equal{}\frac{p}{q}$ than p is divisable by 1319

I reckon it shouldnt be in this section now, so if a mod could transfer it to olympiad algebra i d appreciate


[quote="mathnovice"]If I am applying early action now and will probably be applying regular decision later, should I tell my teachers to make copies of their recommendations before they send them out?[/quote]

Your teachers should, one hopes, write to the issue of why you would fit into each school, so the recommendations shouldn't be identical. You should get clear with each teacher how many recommendations are as many as the teacher is willing to do, and how you will provide them with each school's specific recommendation form (likely to be different, at least for the schools that don't use the common application) and postage-affixed envelopes for sending in the forms.

Good luck in the application process.

You should just tell your teachers that that is the case, so they realize that you will be sending out several applications at different times and that you are asking them to write a recommendation for all of them.


What is the least positive integer greater than 1 that leaves a remainder of 1 when divided by each of 2, 3, 4, 5, 6, 7, 8 and 9?


algebra
difference of squares
special factorizations
number theory
modular arithmetic

Three consecutive positive integers have the property that the squares of the smaller two integers differ by $ 10a\plus{}b$ while the squares of the smallest and largest integers differ by $ 10b\plus{}a$. If $ a$ and $ b$ are digits, what is the sum of the three integers?

solution

[hide]
Let the 3 consecutive integers be n-1,n,n+1

The difference between (n-1)^2 and n^2 is 2n-1, between n^2 and (n+1)^2 2n+1. Thus, the differance between the two smallest (ab) and the smallest and the largest (ba) is 2n+1.

We know that the difference between ab and ba is 10b+a-10a-b, 9b-9a, or a multiple of 9 other than 0. Thus, we set up

2n+1 (congruent) 0 mod 9
2n (congruent) 8 mod 9

set

2n-8=18
2n=26
n=13

thus,
12,13,14

12+13+14=39
[/hide]

[hide="Answer"]39[/hide] :D
It's simply too late to post my solution (I swear I didn't get it by pure brute force).
Bye!
GM

P.S. I didn't see ceastybet's solution... it's way shorter than mine  :oops:

My solution:

[hide]Let's say that the numbers are n - 1, n, and n + 1.

[u]Difference of squares[/u]:
(n - n + 1)(n + n - 1) = 2n - 1 = 10a + b
(n + 1 - n + 1)(n + 1 + n - 1) = 4n = 10b + a

[u]Manipulation[/u]:
2n - 1 = 10a + b is equivalent to 2n = 10a + b + 1 which is equivalent to 4n = 20a + 2b + 2

[u]Substitution[/u]:
10b + a = 20a + 2b + 2
8b = 19a + 2

[u]Guess and check[/u]:
a = 2 and b = 5

[u]Plug back in[/u]:
(n - n + 1)(n + n - 1) = 25
(1)(2n - 1) = 25
2n - 1 = 25
2n = 26
n = 13

[u]The numbers are[/u]:
12, 13, and 14, which sum to 39.

[u]Answer[/u]:
39[/hide]

[hide="Another way to solve it"]Let the integers be $ n, n \plus{} 1, n \plus{} 2$. The difference between the squares of the second and the first (the two smallest) would then be $ (n \plus{} 1)^{2} \minus{} n^{2} \equal{} 2n \plus{} 1 \equal{} 10a \plus{} b$, and the difference between those of the third and the first (the largest and the smallest), $ (n \plus{} 2)^{2} \minus{} n^{2} \equal{} 4n \plus{} 4 \equal{} 10b \plus{} a$, where $ b > a$. Adding the two differences, we end up with

$ 6n \plus{} 5 \equal{} 10a \plus{} b \plus{} 10b \plus{} a \equal{} 10(a \plus{} b) \plus{} (a \plus{} b) \equal{} 11(a \plus{} b)$

Now, since every variable is an integer (noting, also, that $ 0 < a < b < 10$, since they are digits, so $ 2 \leq (a \plus{} b) \leq 18$), $ \frac {6n \plus{} 5}{11} \equal{} a \plus{} b$ must also be an integer. The only number that satisfies the conditions for $ n$ is $ 12$, so we can conclude that our integers are $ 12, 13, 14$, and that  $ a \plus{} b \equal{} 7$. The sum of said integers — what the problem asks us to find — is $ \boxed {39}$. [/hide]

[quote="sandeep3181"][quote="ceastybet"]solution

[hide]
Let the 3 consecutive integers be n-1,n,n+1

The difference between (n-1)^2 and n^2 is 2n-1, between n^2 and (n+1)^2 2n+1. Thus, the differance between the two smallest (ab) and the smallest and the largest (ba) is 2n+1.

[b]We know that the difference between ab and ba is 10b+a-10a-b, 9b-9a, or a multiple of 9 other than 0. Thus, we set up

2n+1 (congruent) 0 mod 9
2n (congruent) 8 mod 9

set

2n-8=18[/b]
2n=26
n=13

thus,
12,13,14

12+13+14=39
[/hide][/quote]

Ceasty, kindly explain the text in bold a bit more???

Regards,
Sandeep[/quote]

He plugged in what he found and then added them together for $ 9b\minus{}9a$. He then used [[modular arithmetic]] for those two equations which then gave him the equation to find the answer.

You might want to read [[Modular_arithmetic/Introduction]] for more information about how he did those mods.

all correct solutions; the correct ans is indeed 39. by the way, where did you get the problem?

^Made it up myself. How did you like this problem?


articles
algebra
polynomial
AoPSwiki
quadratics
Rational Root Theorem

Can someone explain the rational root theorem? I've heard about it being useful and looked at the Wikipedia article, which i didn't fully understand (as always  :huh:  ). So, can anyone explain what it is and how to use it? (Examples would be useful).

Okay, so as the wiki article says, if you have a polynomial
$ P(x) \equal{} a_n x^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} \ldots \plus{} a_1 x \plus{} a_0$
where all the coefficients ($ a_1, a_2, \ldots, a_n$) are integers, and you are trying to solve for its roots, or:
$ a_n x^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} \ldots \plus{} a_1 x \plus{} a_0\equal{}0$, then the only possible rational solutions are of the form $ \pm \frac pq$, where $ p$ is a factor of $ a_0$, and $ q$ is a factor of $ a_n$. In other words, the numerator has to be able to divide the constant term, and the denominator has to be able to divide the leading coefficient.
For example, if we have the polynomial $ f(x)\equal{}10x^2\plus{}12048917351x\plus{}3$, its only possible rational roots are $ \pm \frac pq$. In this case, $ p$ can be $ 1$ or $ 3$, since those are the only factors of $ 3$, and $ q$ can be $ 1,2,5,10$, since those are the only factors of $ 10$. (I don't feel like typing out 8 rational roots, so you'll have to do without a full list. Sorry!)
Note that these are both [i]possible[/i] and [i]possible rational[/i] solutions. Just because the fractions you get are possibilities for the solutions, it does not mean that these [i]have[/i] to be the solutions. (In that example, there were 8 possibilities, and obviously only 2 of them at most could be solutions, leaving at least 6 "dummy" choices!)
If none of the choices for a certain polynomial work, then you can conclude that there are no rational solutions. But there still can be irrational or nonreal solutions. (And by the Fundamental Theorem of Algebra, there must be.)
[b]Special Cases:[/b]
When there is no constant term, this theorem may present some sort of problem; after all, you can either consider no numbers at all or any and all integers to be able to divide 0, either way presenting a problem in listing the possible values of $ \pm \frac pq$. However, this is easily solved: when there is no constant term, you can factor out (powers of) $ x$, giving the root of 0. Then from there, you can use the Rational Root Theorem on the remaining polynomial. For example, consider $ g(x)\equal{}10x^4\plus{}12048917351x^3\plus{}3x^2$. (Look familiar?) You can factor it as $ g(x)\equal{}x^2(10x^2\plus{}12048917351x\plus{}3)$, and then attempt to solve for the roots of the polynomial in parentheses.

Another special case is when the leading coefficient is $ 1$. (Such polynomials are known as monic polynomials.) Now there is only one possible value for $ q$ in $ \pm \frac pq$, namely $ 1$. This implies that any rational solution for a monic polynomial with integer coefficients must also be an integer. The contrapositive of this is also true, i.e., if a monic polynomial with integer coefficients has a solution that isn't an integer, then it cannot be rational. This can be used to prove that $ \sqrt{2}$ is irrational. Can you see how?
[hide="How"]Consider the polynomial $ x^2\minus{}2$. By the Rational Root Theorem, its only possible rational roots are $ \pm 2$. But neither of these are roots, so $ x^2\minus{}2$ has no rational roots. $ \sqrt{2}$ is a root of this polynomial, and it is real, so it must be irrational[/hide]

Thus ends my long post. Sorry it's really long and confusing. If anyone wants to synthesize this, that would be great. :D
(It would actually be nice if someone could transform this and put it in the AoPSWiki - we could use some more articles there!)

No, I understood it very well, AND I finally realized what the Fundamental Theorem of Algebra was, and that there might not be any rational solutions (because I got confused when I tried a random quadratic equation and it didn't work). So thank you for your help!  :lol:


Put $\LaTeX$ code in $\$$'s, before and after the code.

[quote="aufha"]How do I use LaTeX in the forum?[/quote]

For other information go to [url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url].


What is the greatest possible value of the expression\\
\indent \indent (8 $ \square$ (-2)) $ \square$ 5\\
in simplified form when an operation symbol (+, -, $ \times$, $ \div$) is placed in each square?

It's quite obvious that we do not want to multiply 8 by $ \minus{}2,$ or divide either. Adding makes $ 8$ smaller, so we subtract $ \minus{}2$ from $ 8$ giving $ 10.$ Now it's easy to see we multiply by $ 5$ to get the answer of $ (8\minus{}(\minus{}2))\times5\equal{}10\cdot5\equal{}\boxed{50}.$


Im working on a fairly long PDF in Texnic Center, and today when I entered new text and built the file nothing was changing; I couldn't make changes. Then I copied my code and put it in a new doc and when I tried to build it said 'file cannot be found' or something like that. HELP!!
Note: I may have pressed Ctrl 4 but then I undid it.

Find the tex file you were working on and then open it. You can look for the tex file in Windows Explorer or by using Open in TexnicCenter. It may or may not have the most recent changes in it.

that did not do anything. Basically i press build and i have 5 error. Then I press view and it build a file, but with not all the code included. However, if I delete a particularly dense section of code it builds and works. does it not build if there are too many errors or something? Because in the past it just builds it.

If you have 5 errors then you [i]must[/i] correct them first. That is why parts of the document are not being shown. TexnicCenter tells you where the errors are by pressing F9 repeatedly or the ! button on the toolbar.
The errors are also in the log file which is kept in the same directory as the tex file.

So now there are three errors, but when I hit build it produces that same old file, when I copy and paste the code into another new doc it cant be "found" when i try to view output, and when I check for more errors it says "File ended while scanning the use of \frac"

I just tried this:

[quote]\documentclass[11pt]{article}

\begin{document}

Fraction test

$ \$$\frac{x}{$ \$$

\end{document}  [/quote]

And got "Paragraph ended before \frac was complete"

On the other hand, if I try

[quote]\documentclass[11pt]{article}

\begin{document}

Fraction test

$ \$$\frac{x}$ \$$

\end{document}  [/quote]

Then I get "Missing } inserted. <inserted text> }" and then the name of the line on which the offending command is.

In neither case did I get a .pdf file.

So I haven't exactly reproduced your error message - but you've got something like that going on. It's still your error, and it halts compilation of your file. You have to track down and fix [b]all[/b] of your syntax mistakes.

Yes I am getting:
File ended while scanning use of \frac
<inserted text>
      \par
I'll try and fix all my mistakes
*tedious hope this is worth it for the promys app

This is hindsight, but it's probably best to compile your file after each half page or even each paragraph, and after any particularly hairy formula or figure. Then you'll find your problems immediately and know where they are.

If your error message doesn't seem particularly helpful in terms of finding your mistake, you can take a formula or paragraph you're suspicious of and put % in front of it. That turns it into a comment, and the compiler ignores it. If the file now compiles, you've narrowed down the location or locations of the problems.

There's probably someone, somewhere, who can type errorless $ \text{\LaTeX}$ first time, every time - but I'm not that person. Debugging is a necessary part of writing in $ \text{\TeX}$ - get used to it.

If you already have a long document with errors and you don't want to copy and paste bits into another document, then add \end{document} after the section you want to check. Then LaTeX will only compile the part before \end{document} and you can look to see if there are any errors there. Once that's error-free you can move \end{document} to further along the document.
This is only for an already written document. I wholeheartedly agree with Kent's suggestion of frequent compiling as you write the document.

[quote="mathemonster"]Yes I am getting:
File ended while scanning use of \frac[/quote]Sounds like a missing }. In TexnicCenter if you click to the right of a { or } it will highlight in green the matching bracket. If the { isn't matched it will be highlighted in red. You'll find this very useful for finding orphan brackets particularly nested ones. You can also do this for other brackets such as (),[].

More insidious are errors that aren't fatal to the compilation but still aren't what you want. If I amend my minimal example in #6 to make the key line \frac{x}{} (with dollar signs around it), then the file does compile and produce output - but that output is kind of funny-looking. There's no substitute for taking a long, hard look at your output to make sure it's what you intended.


This Wednesday, October 21, at 7:30 PM ET (4:30 PM PT), Matt McGann, Associate Director of Admissions at the Massachusetts Institute of Technology (MIT), and Kiran Kedlaya, Associate Professor of Mathematics at MIT and former US IMO team member, will host an Art of Problem Solving Math Jam to discuss life at MIT and the admissions process.

In order to attend a Math Jam you must first log on the Art of Problem Solving Forum, then click the Classroom button on the left panel of the site up to 15 minutes before the Math Jam begins.  The Virtual Classroom should then open automatically.  All Math Jams are free.

What general age is this directed towards?

probably late high school? but I guess it would be helpful for anyone that age or below.

[quote="rrusczyk"]This Wednesday, October 21, at 7:30 PM ET (4:30 PM PT)...[/quote]

or here...

[quote="shadow123blast"]when does the math jam start
i know you have to start 15 minutes early but when are these math jams and which days[/quote]

I think he meant the math jams in general, so..
[url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]here[/url] and [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php]here[/url] are what shadow are looking for (I think)

Will there be any math covered in it? (As in problem solving strategies and stuff) Afterall, it is a Math Jam. :P

No, there won't be any math in this particular "Math" Jam.  Having Kiran in the Math Jam gives it math cred enough!

Would it still be okay to go if you're in Junior High?

well...
im very precocioud for my age at 11 yrs old so can i join it thiough?

how can they stop u?
...not that precocious

Yes. All past Math Jams are uploaded [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php]here[/url].

In fact, there are other MIT Jams archived, so if you want to know what they are like, [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php]here[/url] is the main page, and here is [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=243]the link[/url] for the 2008 MIT Jam.


trigonometry
function
calculus
derivative
complex numbers
complex analysis

A question from my book. C is the complex numbers and $ f: C \to C$ is defined $ \left\{ \begin{array}{l}
 e^{{\textstyle{{ \minus{} 1} \over {z^4 }}}} ,z \ne 0 \\
 0,z \equal{} 0 \\
 \end{array} \right.$

Show that f satisfies the Cauchy Riemann differential equations for all complex C and is complex differentiable on C\{0}, but not at the origin.

My question is how can f satisfy the CR equations on C but not be complex differentiable at 0? And is there a simple way to solve this I'm not seeing? Writing this in terms of real and complex parts is horrific.

My answer for the computational part:

$ f(z) = e^{ - \frac {1}{z^4}}$

${{ f(\rho,\theta) = e^{ - \frac {1}{\rho^4 e^{i 4 \theta}}} = e^{ - \frac {e^{ - i 4 \theta}}{\rho^4}} = e^{ - \frac {\cos 4 \theta}{\rho^4}}}e^{i\frac {\sin 4 \theta}{\rho^4}}}$

If we denote ${ R = e^{ - \frac {\cos 4 \theta}{\rho^4}}}$ and ${{ \varphi = \frac {\sin 4 \theta}{\rho^4}}}$ then it's easy to show that CR equations:

$ \rho \frac {\partial R}{\partial \rho} = R\frac {\partial \varphi}{\partial \theta}$

$ \frac {\partial R}{\partial \theta} = - \rho R \frac {\partial \varphi}{\partial \rho}$

hold.

EDIT: and here's my answer for the first question: Cauchy-Riemann equations are necessary, but not sufficient condition for complex differentiability (as shown by Pollard in 1928, if I remember correctly). With addition of a condition of differentiability of real and imaginary part of complex function in the given point, the conditions become sufficient (and we have an example of partial derivative discontinuity at zero in our example).

The function is analytic on $ \mathbb C\setminus\{0\}$ as a composition of several analytic functions: $ z\mapsto z^4$, $ z\mapsto \frac 1z$ and $ z\mapsto e^z$. At $ 0$ all partial derivatives participating in the Cauchy-Riemann equations are $ 0$, so the equations are satisfied trivially. The key point is that the function has a terrible discontinuity at $ 0$, so a couple of directional derivatives do not really tell you much about its behavior near $ 0$. An interesting follow up question to think of is what is the slowest growth at $ 0$ for which this effect is possible, i.e., for what $ F(r)$ it is true that analyticity on $ \mathbb C\setminus\{0\}$ together with CR at the origin and the condition $ f(z)\equal{}O(F(|z|))$ as $ z\to 0$ imply analyticity of $ f$. ;)

I assume you asked about the equations in my post.

I used the equation in my Complex Analysis course without proof, but I remember the outline of the proof given in D.S. Mitrinović's Complex Analysis:

First, he derives the CR equations for $ f(z) \equal{} R(x,y) e^{ \minus{} i \varphi}$ which is straightforward, as $ u \equal{} R\cos \varphi$ and $ v \equal{} R\sin \varphi$, so we easily obtain:

$ \frac {\partial R}{\partial x} \equal{} R\frac {\partial \varphi}{\partial y}$ and $ \frac {\partial R}{\partial y} \equal{} \minus{} R\frac {\partial \varphi}{\partial x}$

Now, he derives the equations for $ f(z) \equal{} R(\rho,\theta)\varphi(\rho,\theta)$.

All partial derivatives are brought down to derivatives on $ x$ and $ y$ and with assumptions that the form of CR equations derived above holds for our function $ f$, we obtain the desired result (I'm sorry I don't have time to type the whole proof, but I assume it could be found in Complex Analysis textbooks...)

Thanks.

I'm familiar with the general polar form for the CR equations but not that. It's not in my analysis book (undergrad). I can't find any book online by Mitrinovic by the name 'Complex Analysis'. Do you have a link?

Well, most probably, Mitrinović's Complex Analysis was never translated in English nor published outside Yugoslavia (unlike his other books), so I re-typed the proof given:

[quote="D.S. Mitrinović"] WE shall derive the Cauchy-Riemann conditions in case the function $ f(z)$ is given as $ f(z) = R(x,y)e^{i\varphi(x,y)}$

Knowing that $ u = R\cos \varphi$ and $ v = R \sin \varphi$, we have:

$ \frac {\partial u}{\partial x} = \frac {\partial R}{\partial x}\cos \varphi - R\sin \varphi \frac {\partial \varphi}{\partial x}$

$ \frac {\partial v}{\partial x} = \frac {\partial R}{\partial x}\sin \varphi + R\cos \varphi \frac {\partial \varphi}{\partial x}$

$ \frac {\partial u}{\partial y} = \frac {\partial R}{\partial y}\cos \varphi - R\sin \varphi \frac {\partial \varphi}{\partial y}$

$ \frac {\partial v}{\partial y} = \frac {\partial R}{\partial y}\sin \varphi + R\cos \varphi \frac {\partial \varphi}{\partial y}$

Plugging these in the original CR equations gives

$ \frac {\partial R}{\partial x} = R\frac {\partial \varphi}{\partial y}$ and $ \frac {\partial R}{\partial x} = - R\frac {\partial \varphi}{\partial x}$.

Now, consider the case when $ f(z) = R(\rho,\theta)e^{i\varphi(\rho,\theta)}$

Since

$ \frac {\partial R}{\partial \rho} = \frac {\partial R}{\partial x}\frac {\partial x}{\partial \rho} + \frac {\partial R}{\partial y}\frac {\partial y}{\partial \rho} = \frac {\partial R}{\partial x}\cos \theta + \frac {\partial R}{\partial y}\sin \theta$

$ \frac {\partial R}{\partial \theta} = \frac {\partial R}{\partial x}\frac {\partial x}{\partial \theta} + \frac {\partial R}{\partial y}\frac {\partial y}{\partial \theta} = \rho \left( - \frac {\partial R}{\partial x}\sin \theta + \frac {\partial R}{\partial y}\cos \theta\right)$

$ \frac {\partial \varphi}{\partial \rho} = \frac {\partial \varphi}{\partial x}\cos \theta + \frac {\partial\varphi}{\partial y}\sin \theta$

$ \frac {\partial \varphi}{\partial \theta} = \rho \left( - \frac {\partial \varphi}{\partial x}\sin \theta + \frac {\partial \varphi}{\partial y}\cos \theta\right)$

we obtain

$ \frac {\partial R}{\partial \rho} - \frac {R}{\rho}\frac {\partial \varphi}{\partial \theta} = \left(\frac {\partial R}{\partial x} - R\frac {\partial \varphi}{\partial y}\right)\cos\theta + \left(\frac {\partial R}{\partial y} + R\frac {\partial \varphi}{\partial x}\right)\sin\theta$

$ \frac {1}{\rho}\frac {\partial R}{\partial \theta} + R\frac {\partial \varphi}{\partial \rho} = \left(\frac {\partial R}{\partial x} - R\frac {\partial \varphi}{\partial y}\right)( - \sin\theta) + \left(\frac {\partial R}{\partial y} + R\frac {\partial \varphi}{\partial x}\right)\cos\theta$

which gives (using the equivalent form of CR equations we derived above):

$ \frac {\partial R}{\partial \rho} = \frac {R}{\rho}\frac{\partial \varphi}{\partial \theta}$ and $ \frac {\partial R}{\partial \theta} = - R\rho\frac {\partial \varphi}{\partial \rho}$. [/quote]

Thank you!  :thumbup:

EDIT: When you evaluate the partials at 0, do you let r tend to 0 while keeping phi arbitrary?

[quote="fedja"]The function is analytic on $ \mathbb C\setminus\{0\}$ as a composition of several analytic functions: $ z\mapsto z^4$, $ z\mapsto \frac 1z$ and $ z\mapsto e^z$. At $ 0$ all partial derivatives participating in the Cauchy-Riemann equations are $ 0$, so the equations are satisfied trivially. The key point is that the function has a terrible discontinuity at $ 0$, so a couple of directional derivatives do not really tell you much about its behavior near $ 0$. An interesting follow up question to think of is what is the slowest growth at $ 0$ for which this effect is possible, i.e., for what $ F(r)$ it is true that analyticity on $ \mathbb C\setminus\{0\}$ together with CR at the origin and the condition $ f(z) \equal{} O(F(|z|))$ as $ z\to 0$ imply analyticity of $ f$. ;)[/quote]

Could you elaborate or give an example?


Fix $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2$. Using one of the ev. theorems on $ \frac{1}{x}$, the min occurs for $ b\equal{}c\equal{}d\equal{}e$...We rewrite as $ \frac{1}{5\minus{}4b}\plus{}\frac{4}{b}\plus{}\frac{20}{ (5\minus{}4b)^2\plus{}4b^2}\minus{}9\equal{}\frac{(4)(1\minus{}b)^2(5\minus{}6b)^2}{(b)(5\minus{}4b)( 4(b\minus{}1)^2\plus{}1)}\ge 0$

Indeed Altheman, the EV Method is welcome here. Using it we can generalize the inequality to $ n$ numbers.

Equality holds for $ a\equal{}b\equal{}c\equal{}d\equal{}e$, and again for $ \frac a{2}\equal{}b\equal{}c\equal{}d\equal{}e$ or any cyclic permutation.

I think it is a nice inequality. I do feel that the n-1 ev. theorem is too powerful because problems that are almost impossible to solve otherwise are trivial with this theorem. But I do find the results very interesting. In particular, you get symmetric inequalities whose equality cases include those with $ 0<x_1\equal{}x_2\equal{}..x_k<x_{k\plus{}1}\equal{}...\equal{}x_n$. It seems to me that all that one must do is think of an inequality in the form say $ a\cdot (\sum\frac{1}{x_i})\plus{}\frac{b}{\sum x_i^2}\ge c$ and $ \sum x_i\equal{}d$ then fit constants $ a,b,c,d$ so when you apply the e.v. theorem you get the convexity requirement and two cases of equality when you look at the one variable inequality that arises from the e.v. equality condition. So we can pretty much generate as many difficult inequalities as we wish by thinking of general forms as described above and fitting the variables appropriately.

By the same way, we can prove the general inequality : $ \frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}...\plus{}\frac{1}{a_n}\plus{}\frac{2n\sqrt{n\minus{}1}}{a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2} \ge n\plus{}2\sqrt{n\minus{}1}$ where $ n\ge 3,a_i >0$ and $ a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}n$. The proof for $ n\equal{}4$ is also nice.

[quote="Altheman"]Fix $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2$. Using one of the ev. theorems on $ \frac {1}{x}$, the min occurs for $ b \equal{} c \equal{} d \equal{} e$...We rewrite as $ \frac {1}{5 \minus{} 4b} \plus{} \frac {4}{b} \plus{} \frac {20}{ (5 \minus{} 4b)^2 \plus{} 4b^2} \minus{} 9 \equal{} \frac {(4)(1 \minus{} b)^2(5 \minus{} 6b)^2}{(b)(5 \minus{} 4b)( 4(b \minus{} 1)^2 \plus{} 1)}\ge 0$[/quote]
Could you provide a reference to the ev theorems ? :)

[quote="Altheman"]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1367281#1367281[/quote]
Thank you :)

[quote="Vasc"]If $ a,b,c,d,e$ are positive real numbers such that $ a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 5$, then

$ \frac 1{a} \plus{} \frac 1{b} \plus{} \frac 1{c} \plus{} \frac 1{d} \plus{} \frac 1{e} \plus{} \frac {20}{a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2} \ge 9$.[/quote]
Another solution if we dont want to use EV here (this ways also helps us solve the generalized one):
Setting $ P(a,b,c,d,e)\equal{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d}\plus{}\frac{1}{e} \plus{}\frac{20}{a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2}.$ Now, Assuming $ a \le b \le cle d \le e$, then we can prove the chain inequalities:
\[ P(a,b,c,d,e) \ge P\left( \frac{a\plus{}b}{2},\frac{a\plus{}b}{2} ,c,d,e\right) \ge P\left( \frac{a\plus{}b\plus{}c}{3} ,\frac{a\plus{}b\plus{}c}{3} ,\frac{a\plus{}b\plus{}c}{3}, d,e \right) \ge P\left( \frac{a\plus{}b\plus{}c\plus{}d}{4} ,\frac{a\plus{}b\plus{}c\plus{}d}{4} ,\frac{a\plus{}b\plus{}c\plus{}d}{4} ,\frac{a\plus{}b\plus{}c\plus{}d}{4} ,e\right) \ge 9.\]
:)

[quote="Vasc"]If $ a,b,c,d,e$ are positive real numbers such that $ a\plus{}b\plus{}c\plus{}d\plus{}e\equal{}5$, then

$ \frac 1{a}\plus{}\frac 1{b}\plus{}\frac 1{c}\plus{}\frac 1{d}\plus{}\frac 1{e}\plus{}\frac {20}{a^2\plus{}b^2\plus{}c^2\plus{}d^2\plus{}e^2} \ge 9$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=354053


calculus
derivative
analytic geometry
graphing lines
slope
inequalities
function

If the first derivative looks for the slope of the line, and is used to find the maximum area/volume, then what is the use of the second derivative.

The second derivative is used for convexity/concavity : http://mathworld.wolfram.com/ConvexFunction.html . (Check the inequalities' section, with keywords such as Jensen, or Karamata. ;) )

Generally you use the second derivative to find the acceleration (since $a = \frac{dv}{dt} = \frac{d^2x}{{dt}^2}$). Moreover, since Newton's force law ($F = ma= m\frac{d^2x}{{dt}^2}$) directly relates force to the second derivative of the position, the second derivation pops up a lot in mechanics.

Now what does the second derivative mean geometrically? Well, like mathmanman said, it is used to measure convexity/concavity(i.e. if the graph curves up or if the graph curves down). If the second derivate of a function $f$ is positive in some interval then the function $f$ is concave up and if it is negative then $f$ is concave down.

Now that was just the sign of the second derivative, what about its magnitude? Well, I think it tells you the curvature of the graph but I am not sure. Does anybody know??  :huh:

As far as real-valued, single variable functions go, "curvature" is usually defined as the (multiplicative) inverse of the osculating circle. I've also seen books where the value of the second derivative at $x$ is called curvature, but I don't think that's widely accepted.

Here's a formula that'll give you the curvature at point x of a twice differentiable function $f$:
\[ \bigg\vert \frac{f''(x)}{\left(1+(f'(x))^2\right)^{3/2}} \bigg\vert  \]

Most people encounter a more general form of the formula above when they learn about plane parametric curves. For implicit curves you cannot easily give a parametrical form for, there exists a similar formula, using gradients.

OK, so I guess the magnitude of the second derivative doesn't really give you the curvature. I was kinda thinking the same thing after I graphed some polynomials and their second derivatives. I just couldn't their second derivatives with their curvature by looking at their graphs!

I wonder if there is really ANY meaning (of course geometrically) of the magnitude of the second derivative of a function? :huh:

Any comments?

[quote="I"]OK, so I guess the magnitude of the second derivative doesn't really give you the curvature. I was kinda thinking the same thing after I graphed some polynomials and their second derivatives. I just couldn't their second derivatives with their curvature by looking at their graphs!
[/quote]
Here is what I tried.

We know that $y = x^2$ is a graph of a parabola and thus $y''(x) = 2$. So if we say that the second derivative tells us(or even approximately tells us) the curvature , this interpretation doesn't make any sense for this function. Of course, if $x$ is near zero the function is more round than at $x = 10$ or something. However, the second derivative is the same everywhere, thus implying that the function has equal "roundness" everywhere. This is clearly not true if we look at the graph. Thus, interpreting the second derivative as the curvature doesn't really make any sense at all.

Right??

The second derivative is the rate of change of the slopes of the tangent lines. The reason why this is hard to visualize is because of the way we intuitively think about slope.

One one hand we think of slope as a measure of "steepness". But our intuition for the steepness of a line also has more to do with the angle the line makes with x-axis; and the relationship between these two notions of "steepness" is not linear.

So while the slopes might be increasing at a constant rate, that does not mean that the angle of elevation is increasing at a constant rate.

So the bottom line is that the maginitude of the second derivative of a function does not measure its curvature.

Any objections?

Well, it does in some sense. But not in a way that's very useful to us, because it doesn't translate into something we can easily recognize visually. I also can't think of any application of it off the top of my head. Also, "curvature" is sometimes used as a technical term with a specific definition, and (as phils said) that definition is usually not just the second derivative of the function. So in that sense, you're right.

[quote="phils"]Here's a formula that'll give you the curvature at point x of a twice differentiable function $f$:
\[ \bigg\vert \frac{f''(x)}{\left(1+(f'(x))^2\right)^{3/2}} \bigg\vert  \][/quote]
[quote="Xevarion"]Well, it does in some sense ... Also, "curvature" is sometimes used as a technical term with a specific definition, and (as phils said) that definition is usually not just the second derivative of the function. So in that sense, you're right.[/quote]
OK, so the MAGNITUDE of the second derivative only gives us the APPROXIMATE curvature. The only time we can "think" of the second derivative geometrically as the curvature  of our function is when we restrict our domain to the interval where the function is SHALLOW (i.e. the first derivative doesn't vary much and so $\bigg\vert \frac{f''(x)}{\left(1+(f'(x))^2\right)^{3/2}} \bigg\vert$ is dominated only by the second derivative)  :lol:


Given $ \Delta ABC$, construct its circumcircle, and define $ A'$ to be the point diametrically opposite $ A$. Define $ B',C'$ analogously. Construct the tangent $ l_{a}$ to the circumcircle at $ A$, and define $ l_{b},l_{c}$ similarly. Let $ l_{b},l_{c}$ intersect at $ A''$, and define $ B'',C''$ analogously. Show that $ A'A'',B'B'',C'C''$ are concurrent.

[quote="SplashD"][quote="CatalystOfNostalgia"]Given $ \Delta ABC$, construct its circumcircle, and define $ A'$ to be the point diametrically opposite $ A$. Define $ B',C'$ analogously. Construct the tangent $ l_{a}$ to the circumcircle at $ A$, and define $ l_{b},l_{c}$ similarly. Let $ l_{b},l_{c}$ intersect at $ A''$, and define $ B'',C''$ analogously. Show that $ A'A'',B'B'',C'C'',HO$ are concurrent.[/quote]

That's not how the Exeter point is defined. $ A'$ is where the median from $ A$ intersects the circumcircle.[/quote]

Oh, I guess I was thinking of something else, but I'm pretty sure the statement is true nevertheless.

Actually, I think your definition would only apply if $ \triangle ABC$ was equilateral.

...which I realize, just ignore the title 'Exeter Point' and only consider the claim.

Greetings, and welcome to the internet.  The internet is very, very large and contains a great deal of information.  To help you find things on the internet, a variety of websites, known as "search engines," are freely available.  If you're planning on spending more than a day or two here, I recommend that you get to know one of them.

http://faculty.evansville.edu/ck6/tcenters/recent/exeter.html


inequalities
induction
linear algebra
matrix
linear algebra unsolved

Assume $ A, B \in M_n(\mathbb{K})$

show that: $ rank(I_n\minus{}AB)\leq rank(I_n\minus{}A)\plus{}rank(I_n\minus{}B)$

[hide="Hint"]
Note that $ rank(I\minus{}AB)\equal{}rank(I\minus{}A\plus{}A\minus{}AB)\leq rank(I\minus{}A)\plus{}rank(A(I\minus{}B)) \leq rank(I\minus{}A)\plus{}rank(I\minus{}B)$
[/hide]
Now maybe we can generalize the problem (by using Mathematical Induction) as follows :
If $ A_1,A_2,A_3,\cdots,A_n$ are real matrix of the same size, then :
$ rank(I\minus{}A_1.A_2.A_3.\cdots.A_n)\leq rank(I\minus{}A_1)\plus{}rank(I\minus{}A_2)\plus{}rank(I\minus{}A_3)\plus{}\cdots\plus{}rank(I\minus{}A_n)$


[quote="Ahwingsecretagent"]Prove that for every positive integer n, there exists a polynomial p(x) with integer coefficients such that p(1), p(2),..., p(n-1), p(n) are distinct powers of 2.[/quote]

Let $ P_{n,k}(x)$ ($ 1\leq k\leq n$) a polynomial such that $ P_{n,k}(i) \equal{} 2^{a_i}$ $ \forall i\in[1,n]$ with $ a_i$ all distincts for $ i\in[1,k]$

$ P_{n,1}(x)$ exists : take for example $ P_{n,1}(x) \equal{} 2$

If we show that it is always possible to build $ P_{n,k \plus{} 1}(x)$ from $ P_{n,k}(x)$, then we'll show the existence of $ P_{n,n}(x)$, which is the present problem.

Consider $ Q(x) \equal{} 2^aP(x) \plus{} b\prod_{i \equal{} 1,n,i\neq k \plus{} 1}(x \minus{} i)$ for some integers $ a,b$ that we'll try to choose.

$ \forall i\in[1,n]\neq k \plus{} 1$ $ Q(x) \equal{} 2^aP(i)$ and so is a power of two and all powers are distinct $ \forall i\in[1,k]$

Let $ P(k \plus{} 1) \equal{} 2^u$.

$ Q(k \plus{} 1) \equal{} 2^{a \plus{} u} \plus{} bc$ with $ c \equal{} \prod_{i \equal{} 1,n,i\neq k \plus{} 1}(k \plus{} 1 \minus{} i)$

Then we had to choose $ a$, $ b$ and $ v$ such that $ 2^{a \plus{} u} \plus{} bc \equal{} 2^v$, so $ u$ and $ v$ such that $ c | 2^v \minus{} 2^{a \plus{} u}$
And it's always possible to choose such quantities :

Let $ c \equal{} 2^wd$ with odd d. Choose $ a$ such that $ a \plus{} u\geq w$ and choose $ t$ such that $ d | 2^{t} \minus{} 1$

Last, choose $ k$ such that $ kt \plus{} a \plus{} u >$ any power already used in $ 2^aP(i)$ for $ i\in[1,n]$

$ d | 2^t \minus{} 1$, so $ d | 2^{kt} \minus{} 1$, so $ c | 2^{kt \plus{} a \plus{} u} \minus{} 2^{a \plus{} u}$

Then $ P_{n,k \plus{} 1} \equal{} 2^aP(x) \plus{} \frac {2^{kt \plus{} a \plus{} u} \minus{} 2^{a \plus{} u}}{\prod_{i \equal{} 1,n,i\neq k \plus{} 1}(k \plus{} 1 \minus{} i)}\prod_{i \equal{} 1,n,i\neq k \plus{} 1}(x \minus{} i)$

[quote="xxp2000"]http://www.mathlinks.ro/viewtopic.php?t=1366[/quote]

Yes. Nice solution. More elegant than mine :)

[color=#f00][Mod edit: pco proved a special case of this problem (for $n=2003$) [url=https://artofproblemsolving.com/community/c6h166608p926475]here[/url].][/color]


calculus
integration
trigonometry
real analysis
symmetry
function
calculus computations

Let $\theta$ be a constant number such that $0\leq \theta \leq \pi.$
Evaluate

\[ \int_0^{2\pi} \sin 8x|\sin (x-\theta)|\ dx. \]

[hide=" :| "]
$\int_0^{2\pi} \sin 8x|\sin (x-\theta)|\ dx$
$= \int_{\theta}^{\pi+\theta} \sin 8x\sin (x-\theta)\ dx-\int_{\pi+\theta}^{2\pi+\theta} \sin 8x\sin (x-\theta)\ dx$
$=\int_{\theta}^{\pi+\theta} \frac{1}{2}\left(\cos(7x+\theta)-\cos(9x-\theta)\right)\,dx-\int_{\pi+\theta}^{2\pi+\theta} \frac{1}{2}\left(\cos(7x+\theta)-\cos(9x-\theta)\right)\,dx$
$=\left(\frac{1}{14}\sin(7x+8\theta)-\frac{1}{18}\sin(9x+8\theta)\right)-\left(\frac{1}{14}\sin(8\theta)-\frac{1}{18}\sin(8\theta)\right)-\left(\frac{1}{14}\sin(14x+8\theta)-\frac{1}{18}\sin(18x+8\theta)\right)+\left(\frac{1}{14}\sin(7x+8\theta)-\frac{1}{18}\sin(9x+8\theta)\right)$
$=-\frac{4}{63}\sin(8\theta)$
[/hide]
is it right

That's just convolution of $|\sin(x)|$ with $\sin(8x)$ on that interval. But the Fourier coefficients of the convolution are the product of the coefficients of the original functions. The Fourier coefficients of $\sin(8x)$ are just $\pm\frac{i}{2}$ for $n=\pm 8$ and 0 otherwise. So we just need to find the 8 and -8 coefficients for the other function. These are both $\frac{-2}{63\pi}$ (the general term is $\frac{-2}{(n^2-1)\pi}$, which we get from $\int_0^1f(x)e^{-2\pi inx}dx$). So then we have $g(\theta) = \frac{-2}{63\pi}\sin(8\theta)$.

I hope there is no mistake. It is not always such a good idea to do these calculations mentally...

I'm tempted to say the answer's just zero...  But then again, I could be dead wrong.

$\int_0^{2\pi}\sin(8x)|\sin(x-\theta)|dx$
$=-\int_0^{\theta}\sin(8x)\sin(x-\theta)dx$ $+\int_{\theta}^{\theta+\pi}\sin(8x)\sin(x-\theta)dx$ $-\int_{\theta+\pi}^{2\pi}\sin(8x)\sin(x-\theta)dx$.
But $\sin(8x)\sin(x-\theta)=\frac{\cos(7x+\theta)-\cos(-9x-\theta)}{2}$, from which follows $\int_0^{2\pi}\sin(8x)|\sin(x-\theta)|dx=-\frac{4\sin(8\theta)}{63}$ (if you don't want to use that trigonometric identity to evaluate the integrals, you could also just integrate them by parts).  Not a particularly pretty method, but it works...

OK, so my guess (that it was zero for all $\theta$) was, afterall, dead wrong; I thought that the symmetry of $\sin(8x)$, and the fact that it fits periodically evenly within the integrating interval and the period of $\sin(x)$, suggested that the integral was zero, but apparently not. :)

kunnyさん, how would you do it?  Can you explain your hint a little more?  Oh, and what's the word for "combinatorics" in Japanese?  組合せですか。おおきに。

Your answer is correct, kokosai.

Let $I=\int_0^{2\pi} \sin \8x |\sin (x-\theta)|dx.$ Set $x-\theta =t,$ we have

$I=\int_{-\theta }^{2\pi -\theta } \sin 8(t+\theta )|\sin t|dt.$

${=\int_{-\pi}^{\pi} \sin 8(t+\theta )|\sin t|dt\ \ (\because \text{the integrand is a perioic function with period}\ 2\pi})$

$=\cos 8\theta \int_{-\pi}^{\pi} \sin 8t |\sin t|dt+\sin 8\theta \int_{-\pi}^{\pi} \cos 8t |\sin t|dt.$

$=2\sin 8\theta \int_0^{\pi} \cos 8t |\sin t|dt \ (\because \sin 8t |\sin t|\ \text{is odd function}).$

$=2\sin 8\theta \int_0^{\pi} \cos 8t \sin t\ dt \ (\because 0\leq t \leq \pi).$

$=2\sin 8\theta \int_0^{\pi}\ \frac{1}{2}(\sin 9t -\sin 7t )dt.$

$=\sin 8\theta \ \left[-\frac{\cos 9t }{9}+\frac{\cos 7t}{7}\right]_0^{\pi}$

$=-\frac{4}{63}\sin 8 \theta .$

P.S.  Combinatorics means 組み合せ論 in Japanese.


Find all pairs of positive integers $(m,n)$ such that $(m-n)^2(n^2-m)=4m^2n$.

working with the text, we have

$n^2(m-n)^2=m(m+n)^2$

now let's suppose $m=ad$ and $n=bd$, with $(a;b)=1$ and $(m;n)=d$
the equation becomes

$a^2d(a-b)^2=b(a+b)^2$

now, $a^2$ of the left side is prime with $b$ and $(a+b)^2$ both, so $a=1$.
the equation becomes

$d(b-1)^2=b(b+1)^2$

now, $(b-1)^2$ is prime with $b$, so
$(b-1)|(b+1)$
$(b-1)|2$
$b=1$, $b=2$ or $b=3$

now, if $b=1$, $m=n$
and
$0\cdot(n^2-m)=4m^2n$
so $m=n=0$

if $b=2$, $n=2m$
$m^2(4m^2-m)=8m^3$
$4m^4-9m^3=0$
from which $m=n=0$

if $b=3$, $n=3m$
$4m^2(9m^2-m)=4m^3$
$9m^4-2m^3=0$
from which $m=n=0$

so the only solution is $(0;0)$.

bye bye

The equation becomes $b^2 d (a-b)^2 = a (a+b)^2$ instead of $a^2 d (a-b)^2 = b (a+b)^2$.
It gives two solutions :

(36,18) and (36,12).


[quote="hwenterprise"]What integer is closest to:

$\sqrt[5]{1000}?$

No calculators.[/quote]

$2^5=32
\\
\\
3^5=243
\\
\\
4^5=1024$

4.


calculus
integration
inequalities
logarithms
function
real analysis
real analysis unsolved

Let ,  $ f(x) \equal{} \int_{1}^{x}\frac {e^t}{t}dt$
for $ x \ge 1$. Then $ f(x) > \log_{e}x$

1) for no value of $ x$.
2) for only $ x > e$.
3) for $ 1 \le x \le e$.
4) for all $ x > 1$

[quote="guru_bhai"]Let ,  $ g(x) \equal{} \int_{x}^{1}\frac {e^t}{t}dt$
for $ x \ge 1$. Then $ f(x) > \log_{e}x$
[/quote]

what exactly is your  question?    wat is  f(x) ?

I suppose there is a typo and $ f \equal{} g$. If this is true, then since $ \frac{e^t}{t} > 1$ when $ x>0$, then:

$ f(x) \equal{} 0 \qquad,\, x \equal{} 1$

$ f(x) < 0 \qquad,\, x > 1$

And since for $ \ln x$ we have:

$ \ln x \equal{} 0 \qquad ,\, x \equal{} 1$

$ \ln x > 0 \qquad ,\, x > 1$

we can say:

$ f(x) \equal{} \ln x \equal{} 0 \qquad ,\, x \equal{} 1$

$ f(x) < 0 < \ln x \qquad ,\, x > 1$

From this you can answer you question.

Actually there were two errors now i have rectified them ,sorry.

Again rectified....

If $ x>1$ and $ f(x)\equal{}\int_x^1\frac{e^t}{t}\,dt,$ then $ f(x)<0.$

Are you sure that's what you wanted? Or did you intend $ \int_1^x\ ?$

Ok, so this is the problem:

Define:

$ f(x)\equal{}\int^x_1 \frac{e^t}{t}\,dt \quad, x \geq 1$

For what $ x \geq 1$ is th following true:

$ f(x) > \ln x$

Solution:

From Taylor expansion of the exponential function, we can see that for positive values of $ x$ we have

$ e^x > \frac{x^n}{n!}$

for all $ n \in \mathbb{N}$. This means that:

$ f(x) \equal{} \int^x_1 \frac{e^t}{t}\,dt >  \int^x_1 \frac{t^{t\minus{}1}}{n!}\,dt \equal{} \frac{x^n}{n\cdot n!} \minus{} \frac{1}{n\cdot n!}$

WLOG, we can choose $ n \equal{} 1$. This gives us:

$ f(x) > x \minus{} 1$

Now, all we need to prove is:

$ x \minus{} 1 > \ln x$

and this is easy since in $ x \equal{} 1$ we have equality and differentiating the expression we get

$ 1 > \frac{1}{x}$

and this is true for $ x > 0$.

So, with this we have proven that for all $ x > 1$ we have

$ f(x) > \ln x$


function
induction
number theory
relatively prime
algebra unsolved
algebra

[b]I have two questions:
1)
find all functions like f : Q -> Q, that for any x,y in Q we have;
f(x^2+y+f(xy)) = 3 + f(x)(x-2+f(y))
2)
find all functions like f : Q -> R, that for any x,y in Q we have;
f(x+y)+f(xy-1)=(f(x)+1)(f(y)+1)
thanks; :lol: [/b]

Here's 2. Don't have a long time to post this, so will omit some algebra.

Put in $ y \equal{} 0$ to get $ f( \minus{} 1) \equal{} f(x)f(0) \plus{} f(0) \plus{} 1$. If $ f(0)$ is nonzero, then $ f( \minus{} 1)$ takes multiple values, which is bad. So $ f(0) \equal{} 0$. It follows $ f( \minus{} 1) \equal{} 1$.
$ x \equal{} y \equal{} \minus{} 1$ gives $ f( \minus{} 2) \equal{} 4$.
$ x \equal{} 1, y \equal{} \minus{} 1$ gives $ f(1) \equal{} 1$.
$ y \equal{} 1$ gives $ f(x \plus{} 1) \minus{} 2f(x) \plus{} f(x \minus{} 1) \equal{} 2$. Using induction both ways and the four integers we know satisfy $ f(n) \equal{} n^2$, we can get $ f(n) \equal{} n^2$ for all integers $ n$.

To finish, let $ p,q$ be relatively prime integers and put $ x \equal{} p/q, y \equal{} q$, and $ x \equal{} p/q \plus{} q, y \equal{} \minus{} q$ in. This gives two linear equations in two unknowns $ f(p/q)$ and $ f(p/q \plus{} q)$. The algebra is pretty horrendous, but if done correctly you get $ f(p/q) \equal{} p^2 / q^2$. Thus $ f(x) \equal{} x^2$ is the unique solution.

EDIT: Woops, realized not long afterwards the first step is not quite right. If $ f$ is a constant function there is no problem. Putting $ f(x) \equal{} c$ into the original equation gives $ c \plus{} c \equal{} (c \plus{} 1)^2$, which has no solutions.

I found F(0)=1. so we can solve it for natural numbers. then we say f(y+1)=f(y)+1 and we can put y=2,3,4,... and we are done!


My induction skills were always weak, this question appeared on last years test, can anyone help me? :?

If :alpha: is a natural number, prove [b]by using induction[/b] that:
[i]y[/i] = [[i]g[/i]([i]x[/i])]a => [i]y[/i]'=a[[i]g[/i]([i]x[/i])]a-1[i]g[/i]'([i]x[/i])

Thanks for the help :)

Apply the product rule, splitting into one factor and an $a-1$ power.
$y=g(x)\cdot(g(x))^{a-1}$
Differentiate this product and the induction becomes obvious.


The 6 NYSML teams for NYC have been announced on [url]http://www.nycmathteam.com[/url].  If you don't feel like downloading the Word document, I am listing the teams below.
[hide="A_Team"]Tao Ran Chen – Bayside
Alex Sinelnikov – Brooklyn Tech
Allan Zhao – Brooklyn Tech
Chong Gu – Cardozo
Alex Iriza – Dalton
Jibo Wen – Francis Lewis
Krishanu Sankar (captain) – Horace Mann
Tao Mai – Lafayette
Abe Chien – Stuyvesant
Benjamin Hirsch – Stuyvesant
Ben Lerner – Stuyvesant
Adam Sealfon – Stuyvesant
Ken Suzuki (captain) – Stuyvesant
Shaoxiang Wang – Stuyvesant
Danny Zhu – Stuyvesant[/hide]
[hide="N_Team"]Ivan Tolkachev – Bronx Science
Kamchuen Chan – Brooklyn Tech
Jin Wan	– Brooklyn Tech
Jingyi Mo – Cardozo
Nai Xin Wang – Francis Lewis
Ben Lipton – HSMSE
Gabe Zucker (captain) – Hunter
Samson Chen – Hunter
Peter Cha – Stuyvesant
Eliot Kim (captain) – Stuyvesant
Youn Kim – Stuyvesant
Randall Lam – Stuyvesant
Yang Tang – Stuyvesant
Andrew Yee – Stuyvesant
Danny Yuan – Stuyvesant[/hide]
[hide="Y_Team"]Chongsheng Feng – John Bowne
Kevin Dong (captain) – Brooklyn Tech
William Ng – Brooklyn Tech
Norris Nakagaki	– Bronx Science
Chris Chen – Cardozo
Jing Du	– Cardozo
Chendan Dong – Forest Hills
Yi Lu – Francis Lewis
Louis Evans (captain) – Hunter
Horace Pan – Hunter
Diqi Shen – Newcomers
Lawrence Chan – Stuyvesant
Eric Choi – Stuyvesant
Paul Fisher – Stuyvesant
Daniel Gitelman	– Stuyvesant[/hide]
[hide="C_Team"]Huilin Wu – Bayside
James Zhang – Brooklyn Tech
Seung Ju Ham – Bronx Science
Kenny Ren – Cardozo
Yang Yang Chen	– Francis Lewis
Shuang Jie Liu	– Francis Lewis
Paul Handorff – Hunter
Harriet Li – Hunter
Jessica Zhang – Manhattan Ctr.
Julie Dossantos	– Stuyvesant
Spencer Harris	– Stuyvesant
Jesse Madnick (captain) – Stuyvesant
Dan Mendelsohn	– Stuyvesant
Yevgeniy Rudoy	– Stuyvesant
Kamil Witek – Stuyvesant[/hide]
[hide="H_Team"]Jae Hyun Lee – Bayside
Elizabeth Attaway – Bronx Science
Vadim Chanyshev – Bronx Science
Ursula Kadziolka – Brooklyn Tech
David Moon (captain) – Horace Mann
Mo Lam	– HSMSE
Meena Boppana – Hunter
Pearson Miller – Hunter
Endi Liu – Lafayette
Milo Beckman – Stuyvesant
Artur Dmowski (captain) – Stuyvesant
Annie Huang – Stuyvesant
Taylor Sutton – Stuyvesant
Kenny Yu – Stuyvesant
Yichi Zhang – Stuyvesant[/hide]
[hide="J_Team"]Rachel Kogan – Brearley
Ziming Shim – Bronx Science
Lijin Chen – Brooklyn Tech
Darwin Ng – Bryant
Christina Hu – Dalton
Max Weinreich – Dalton
Zachary Rivkin – Horace Mann
Manuel Cabral – HSMSE
Yi Yang Li – Newcomers
Benno Mirabelli	– Packer
Simon Hu – Stuyvesant
Jenny Kwun (captain) – Stuyvesant
Ahram Feigenbaum – Stuyvesant
Julia Wang (captain) – Stuyvesant
Xiaoxiao Wu – Stuyvesant[/hide]
Congratulations to all these students!

Does anyone know what the H and J stand for?

Hi all,

We will be having a NYSML practice session at stuyvesant this saturday the 29th starting at 10:30 AM.  Lunch will be provided.  When you get to stuy, go to the 4th floor.

See you then,

Mr. C :)

Thanks Jim.  Do you know about what time the practice will end?

waaa.. never paid attention to this section lol..

practice will run from 10:30 to 1:30 (ish)

Yikes...

Wish I'd checked this earlier.  The main site wasn't updated...I'll try to haggle with my boss about attending this practice as a coach, but I do not know how it will turn out.

I have set aside time for the actual contest itself, so please count me in for it.

:( Pleeease?

--KC

I wasn't there, but my daughter said it was a lot of fun.  Below are the total scores from the A division:
1. New York City A (172)
2. New York City N (146)
3. Nassau A (142)
4. Westchester-Putnam A (136)
5. New York City Y (135)
6. New York City H (128)
7. Albany Area Math Circle (126)
8. Monroe A (123)
9. New York City C (115)

More detailed team scores are posted on the NYSML site:
[url]http://www.nysml.org/index.php?n=NYSML.ContestResults[/url].

I'm not sure about individual results.

In terms of the individual results, Artur Dmowski from NYC H came in first, Chong Gu from NYC A came in second, and Tim Reynolds from Westchester-Putnam A came in third.


[quote="rockymarcianocosmin99"]Daca avem un cal pe o tabla de sah infinita demonstrati ca el poate ajunge in orice alt patrat al tablei.[/quote]

Lema 1. Printr-un numar finit de mutari se poate ajunge cu calul de pe o pozitie data in orice pozitie vecina.

Demonstratia lemei este evidenta. Este suficient sa urmarim un circuit hamiltonian al calului pe tabla de sah si sa observam acest lucru (secvente de drum pentru rezolvarea cazurilor concrete de deplasari in patrate adiacente).

Pe baza aplicarii lemei de cel mult abs(x1-x2)+abs(y1-y2) ori se poate ajunge de la (x1,y1) la (x2,y2).

Tiberiu Socaciu


analytic geometry
geometry proposed
geometry

I've found this very nice problem on a document containing some Singapore training problems.

Prove that the unit circle contains an infinite number of points such with rational coordinates such that the distance between each pair of them is irrational.

I think this works, but I'm not completely sure...

Consider the set $\mathbb{P}$ of primes $p \equiv 1\ \textrm{(mod}\ 4)$.

For any $p \in \mathbb{P}$ there exist unique positive integers $a_p$ and $b_p$ such that \[ a_p^2 + b_p^2 = p.  \] (This is very well known.)

Now, if my calculations are correct, the set of points \[ \left\{\left(\frac{a_p^2 - b_p^2}{p},\left. \frac{2a_pb_p}{p}\right) \right| \ p \in \mathbb{P}\right\}  \] should work...

yeah... my solution is the same, very nice, right?  ;)

Yeah, it's an extremely cool question.

I guess you know the IMO 1975 problem about rational distances? ;)

no, i didn't know it... very similar, and seems to be cool, too...

Just replace irrational by rational and you get another wonderful problem :)


vector
combinatorics unsolved
combinatorics

21 people take a test with 15 true or false questions.  It is known that every 2 people have at least 1 correct answer in common.  What is the minimum number of people that could have correctly answered the question which the most people were correct on?

suppose question $i$ was answered correctly by $t_i$ students, we must have:

$\sum\binom{t_i}{2}$ is at least $210=\binom{21}{2}$, so let $T=max(t_1,t_2...,t_{15})$, then $15\binom{T}{2}$ is at least 210. So $\binom{T}{2}$ is at least 14, so T is at least $6$.

I dont find a construction right now.... :(

I don't think $T=6$ is possible.

Assume each question was solved by at most 6 people.  Then each person must have solved at least 4 questions (since each correct question is shared with at most 5 other people).  Furthermore, if anyone solved exactly 4 questions there must be 20 other solvers of those questions and they must all be distinct.

Since there are 15 questions, there were at most 90 correct answers total by all participants on all questions, so at least 15 people must have solved exactly 4 questions.  Let $A_1, A_2, \dots A_{15}$ be 15 people answering only 4 questions.  Each pair of them must have solved exactly one question together.

If any question was solved by 5 of the $A_i$, then those 5 people would have had to solve 15 other questions between them, which would all have to be distinct.  This is impossible, so each question was solved by at most 4 of the $A_i$.

We now have that $A_1$ solved exactly 4 questions, each of which was solved by at most 3 other $A_i$.  Thus $A_1$ solved questions with at most 12 other $A_i$, contradicting that every pair of people solved a common problem.

To finish things off, here's a construction with $T=7$ and 14 questions overall

Label the contestants by points $(i,j)$, $1 \leq i \leq 7$, $1 \leq j \leq 3$

Let one question be solved by all contestants of the form $(i, ai+b)$ for each $1 \leq a \leq 3$, $1 \leq b \leq 3$.

The 10th question is solved by all contestants where $i \in \{2, 5\}$
The 11th question is solved by all contestants where $i \in \{3, 6\}$
The 12th question is solved by all contestants where $i \in \{1, 4\}$
The 13th question is solved by all contestants where $i \in \{1, 7\}$
The 14th question is solved by all contestants where $i \in \{4,7\}$

[quote=kevinatcausa]Since there are 15 questions, there were at most 90 correct answers total by all participants on all questions, so at least 15 people must have solved exactly 4 questions.[/quote]

I don't understand this

[quote=Pascual2005]I also realised that during the night:

Like we have equality every where, we must have every pair appears together in exactly $1$ problem, and also, every one appears in exactly $5$ problems, Now to everyone asign a vector with $1$ in the $j$th place if he solved problem $j$, and $0$ otherwise. Call this vectors $v_1,v_2...,v_{21}$ and $v_iv_j=1$ and $v_iv_i=5$. Now there is a linear combination of them that sum $0$, taking products wit all vectors we find a contradiction![/quote]

It is impossible for everyone to be in exactly  5 problems.


linear algebra
matrix
linear algebra unsolved

let $ A \in M_n(Z)$ .Find the condition  of $ A$ such that $ A^{\minus{}1} \in M_n(Z)$.

Try to prove that if $ A\in \mathcal{M}_n(\mathbb{Z})$ then $ A^{\minus{}1} \in \mathcal{M}_n(\mathbb{Z}) \iff \det(A)\equal{}\pm 1$

As a hint, look at the way $ A^{\minus{}1}$ is constructed .$ \left( A^{\minus{}1}\equal{}\frac{1}{\det(A)} \cdot A^{*}\right)$

Why should  If $ A\in\mathcal{M}_{n}(\mathbb{Z})$ then $ A^{\minus{}1}\in\mathcal{M}_{n}(\mathbb{Z})\Rightarrow \det(A)\equal{}\pm 1$?

Because adj(A), the adjoint of A (I prefer this notation, since in my country A* is reserved for conjugate transpose), is constructed by transposing the matrix obtained from A by substituting each element by its minor, and a minor is a determinant, which in turn is a sum of products. Therefore, $ A \in M_n(\mathbb{Z}) \equal{}> adj(A) \in M_n(\mathbb{Z})$, and so det(A) must be either 1 or -1.

It's easy to notice that if $ A^{ \minus{} 1}\in \mathcal{M}_n(\mathbb{Z})$ ,then $ \det(A)$ must divide every [url=http://en.wikipedia.org/wiki/Cofactor_%28linear_algebra%29#Matrix_of_cofactors]cofactor[/url] of $ A$.
I denote $ d \equal{} \det(A)$ in order to ease my writing.
So we must have $ d | C_{i,j}$ , $ 1\leq i\leq j \leq n$.Therefore $ \det(A^{*}) \equal{} d^n \cdot q$, where $ q \in \mathbb{Z}$.( we just "take"  $ d$ from every line of $ A^{*}$ ).
But $ \det(A^{*}) \equal{} d^{n \minus{} 1}$. So $ d^{n \minus{} 1} \equal{} d^n\cdot q$ $ \Rightarrow$ $ d \equal{} \frac {1}{q} \in \mathbb{Z}$ $ \Rightarrow$ $ q \equal{} \pm 1$ , therefore $ d \equal{} \pm 1$.

[i]edit:[/i]

Yes you are right guys.I guess I missed the obvious.

You're trying much too hard on that part. Treat the two implications separately.
If $ \det A\in \{\minus{}1,1\}$, then $ A^{\minus{}1}\equal{}(\det A)^{\minus{}1}\cdot \text{adj}(A)$ is an integer matrix.
If $ A^{\minus{}1}$ is an integer matrix, its determinant $ (\det A)^{\minus{}1}$ is an integer. The only integers with integer inverses are $ \minus{}1$ and $ 1$.

[quote]I believe that Carcul's argument it's not a good one.For instance $ \frac{1}{3}\begin{pmatrix}3 & 6\\ 9 & 9\end{pmatrix}\in\mathcal{M}_{n}(\mathbb{Z})$. I hope I didn't get Carcul's idea wrong, but that's how I understand it.[/quote]

What did I wrote that is wrong? Yes, your matrix is in Mn(Z) and so is its adjoint. But because its determinant is not -1 or +1 its inverse is not in Mn(Z).


Should our answers be, if it happens to be so, in non-numerical format. For example, an expression:

$\frac{2\sqrt{2}}{3}$, (Just an example - not an answer.)

If I'm understanding you correctly, I would assume the answer is yes.

If the answer you came up with was [tex]\frac{2\sqrt{2}}{3}[/tex], giving anything else would be an approximation ([tex].94280904 ...[/tex]).

The solutions for 16.1.5 were [tex]\sqrt{7}[/tex], [tex]\sqrt{13}[/tex], and [tex]2[/tex].

yeah, definitely. i did decimals for a perfectly written 5 page solution once, but i got -1 (4/5) -- and all the comment said was it's better to use exact values than decimal approximations, so i guess i'll assume that that was the reason?


algebra
polynomial
superior algebra
superior algebra unsolved

Suppose $ K$ is a field with characteristic $ p$. Let $ a\in K$. If $ a$ has no $ p$-th root in $ K$, show that $ X^{p^n}\minus{}a$ is irreducible in $ K[X]$ for all positive integers $ n$.

Let $ K(r)$ be an extension of $ K$ with $ r$ a root of $ x^{p^n} \minus{} a$. since $ K(r)$ has characteristic $ p$ as well, we have $ x^{p^n} \minus{} a \equal{} (x \minus{} r)^{p^n}$. now, the minimal polynomial of $ r$ over $ K$ must be of the form $ (x \minus{} r)^{p^j}$, where $ j\leq n$. we must show that in fact $ j \equal{} n$. assume the contrary, that $ j < n$. we have $ x^{p^j} \minus{} r^{p^j} \equal{} (x \minus{} r)^{p^j}\in K[x]$, which implies that $ r^{p^j} \in K$. But then $ a \equal{} r^{p^n} \equal{} ((r^{p^j}) ^{p^{n \minus{} j \minus{} 1}})^p$, contradicting the fact that $ a$ is not a perfect $ p$-th power in $ K$.


superior algebra
superior algebra unsolved

If $ D$ is a division ring and $ K$ a sub division ring of $ D$,with the property that $ xKx^{\minus{}1} \subseteq K, \forall x \in D$,then $ K \subseteq Z(D)$.(where $ Z(D)$ is the set of the commutative elements).

[quote="gordon_c"]

If $ D$ is a division ring and $ K$ a sub division ring of $ D$,with the property that $ xKx^{ \minus{} 1} \subseteq K, \forall x \in D$,then $ K \subseteq Z(D)$.(where $ Z(D)$ is the set of the commutative elements).[/quote]

See Cartan-Brauer-Hua theorem. Also, your statement needs some fixing:

If $ D$ is a division ring and $ K$ a proper sub division ring of $ D$, with the property that $ xKx^{ \minus{} 1} \subseteq K, \forall x \in D^{\times}$, then $ K \subseteq Z(D)$, where $ Z(D)$ is the center of $ D$.

Do you have a proof for this theorem or a link to one?Thanks.

A proof can be found on Lam's A First Course in Noncommutative Rings on page 211.


calculus
integration
calculus computations

The electrostatic potential generated by a distribution of electric charge in $ \mathbb{R}^3$ with density $ \rho$ is defined to be
\[ \phi(\mathbf{x)} \equal{} \iiint_{\mathbb{R}^3}\frac {\rho(\mathbf{x} \minus{} \mathbf{y})}{4\pi |\mathbf{y}|}d^3\mathbf{y}
\]
Show that this integral is absolutely convergent if $ \rho$ is continuous and vanishes outside a bounded set.

any ideas on this problem? :maybe:

I dont quite understand this problem...


calculus
integration
logarithms
calculus computations

:D

$ \text{(i)} \quad\int_0^1\;\dfrac{\textbf dx}{1 \plus{} 2^x}$

$ \text{(ii)} \quad\int_0^1\;\dfrac{\textbf dx}{2 \plus{} 3^x}$

$ \text{(iii)} \quad\int_0^1\;\dfrac{\textbf dx}{3 \plus{} 4^x}$

More generally - consider integral $ \textbf I \equal{} \int_0^1 \frac {1}{a \plus{} b^x} \textbf d x$
after some basic manipulations we get
$ \textbf I \equal{} \int_0^1 \frac {a \plus{} b^x \minus{} b^x}{a (a \plus{} b^x)} \textbf d x \equal{} \int_0^1 \left( \frac {1}{a} \minus{} \underbrace{\frac {b^x}{a(a \plus{} b^x)}}_{t : \equal{} b^x} \right) \textbf d x \equal{} \frac {1}{a} \minus{} \frac {1}{a \ln b} \int_1^b \frac {\textbf d t}{a \plus{} t} \equal{}$
$ \equal{} \frac {1}{a} \minus{} \frac {1}{a \ln b} \ln \frac {a \plus{} b}{a \plus{} 1} \equal{} \boxed{\frac{\ln \frac{b(a\plus{}1)}{a\plus{}b} }{a \ln b}}$

\[ \quad\int\;\dfrac{\textbf dx}{c \plus{} a^x} \equal{} \frac {1}{\ln a}\int\;\dfrac{\textbf d(x\ln a)}{c \plus{} e^{x\ln a}} \equal{} \frac {1}{\ln a}\int\;\dfrac{\textbf dt}{c \plus{} e^{t}} \equal{} \frac {1}{\ln a}\int\;\dfrac{\textbf d(e^t)}{e^t(c \plus{} e^{t})} \equal{} \\
\equal{} \frac {1}{\ln a}\int\;\dfrac{\textbf du}{u(c \plus{} u)} \equal{} \frac {1}{c\ln a}\int\;\left(\dfrac{\textbf du}{u} \minus{} \dfrac{\textbf du}{c \plus{} u}\right) \equal{} \frac {1}{c\ln a}\left(\ln u \minus{} \ln (c \plus{} u)\right) \plus{} C \equal{} \\
\equal{} \frac {1}{c\ln a}\left(x\ln a \minus{} \ln(c \plus{} a^x)\right) \plus{} C \\
\quad\int_0^1\;\dfrac{\textbf dx}{c \plus{} a^x} \equal{} \frac {1}{c\ln a}(\ln a \minus{} \ln(c \plus{} a) \plus{} \ln(c \plus{} 1)) \equal{} \frac {1}{c\ln a}\ln\left(\frac {(c \plus{} 1)a}{c \plus{} a}\right)
\]
Well, I'm again late :(


are there still dust roads, open window / door houses in india or are they bulidng likehouses that are able to be fully closed? i know that manypeople in cities live in scyscrapers and that those are closed, but with the HUGE Indian population isn' there a lack of land?

India is a vast spectrum of all kinds of things---from scrappy dust roads,strap huts & slums to buildings -unmatchable- in their beauty
and even with [code]$\lim -> \infinity$[/code] population , we have been able to maintain the old culture and.....

ofc the roads r much more "crowdier" but still not for the worse :D

wonderful!  are there closed windowed houses?

what do you mean by closed windowed houses????

as per i can understand,now most of houses are made as such(in fact majority are such not only in cities but in rural areas also)

hey india is damn goood
sorry i used damn
but i love my country whether it has watever type of roads or villages
bye 1

houses r ofc closed windowed & its not at all a bad place to live

yes i liked india a lot when i visited, but aren't there bugs? i once saw a FLYING cockroach and nearly died! don't bugs scare you? but the shopping is so kewl it's all in boxes (most oof the stuff, and the designs are so .....PRETTY! okay if ur a guy u would't be interested, but i LOVED the clothes there. my mom had an easier live than me she said she could actually go play when she came home from school, all i have time to do is come home eat dinner, take shower and do homework also study for the many tests the next day. but isn't it hard to learn ten languages the way you do in india, my dad knows 16 languages, he only knows english, bengali, assamese, and hindi REALLY well though. if you havent figured out already, i'm from assam.
 no i live in the US thought, ther aren't as many scary bugs. i have a weakness for them , an ant will scare me. :maybe:

thats cool but we would all love to see you in india

you think being afraid is cool, it doesn't make me a paranoid freak and u don't think 'm a 13 year old freak!wow i feel special. Most of you live all over india, i can't travel all over india. i would be nice to see you all though, it's kewl to met fellow aops, math geeks. it was nice of u to invite me though, i might visit when i'm old enough to travel with my own money, like when i'm 20. ut if an of u are going to MIT and i end up there and ur 17 of less i might see u a college, but then again if you lgo to colledge in india i might never see u. BUT U WON'T BELIEVE ow cool it is to talk to some one in a whole different hemisphere, except for the facte i'm 13 and u guys are like a lot older..... in the us, peopl can join from 9 i think..........

we aren't tooo much older

Well the old India has changed over a last few yrs and now we don't have 2  study those many languages--i kno just 2

it wud be superb to see u in India in future!!!!!!!! :D

nobody here is grown up
there are some 11th graders(me and rt_08 only two)and a few 12th graders(rituraj and Da Vinci perhaps)
and we used to have a lot of seniors(previous year 12thers)

i know three languages very well and some languages not so well
it all depends on where do you live
if you live in multi-lingual society you'll learn a lot more languages(its fun)

and abt bugs etc,you should think that india is not so developed(and clean) as well as its climate is suitable for them

but still it is the best place to live

it'ld be superb to see you in india :D

it would be superb to see u in india in future :D :P

u see its quite cool to chill with friends like skand.rituraj,choronz,da vinci,pardesi etc
we just do fun (&ofc sometimes.....ah...sometimes maths)
bye!

but the REAL fun we used to do is lost now :huh:  :(

HEY WAT WAS THE REAL FUN?
TELL ME AS I AM NEW!

ur all like at least 16 right? so ur at leaast 3 years older that me. if not more, i guess that's not like ten years oler but it's still enough to make a difference, THANX for inviting me to india, i think if i bring enough bug spray i'll visit, i rellay love it in india, sure there are a lot f bugs, and well in some places garbage, but i still love the shoping. you can't find all the pretty bangals and necklaces and such here. there are only a few places tht have indian food (so delicious, where as sandwiches are really getting boring). and u just can't find salwars, sari's lasa's and ghagras, here in the US. okay the stuff i just mentioned are in mostly bengali, but i hope u know what i mean. plus it's so cultural in india. VERY NICE. but i like i here in the US too, it's really clean and big city type. too much pollution though, h yeah in india there are rishkas, THAY ARE SOOOOOOOO COOL!!! sorry, i just meant, it'd fun to ride in them.
it's been i guess a long tie since i haven't visited. but u know i might visit in ten years............
[hide]
DaVinci has visited my blog!!!!! YAY, visit my blog!!!![/hide]

rt & skand RESIDE in a bit less popular city
so there may be bugs

But serely here--the capital city of INDIA has no bugs ,is big,lots of pollution.....

sure the city is a bit down but the nawabs are upgrading it!

nawabs means we lucknowites yaar!
tum bhi.........

yay
mazak karne mein kya jaata  hai???? :rotfl:  :rotfl:  :rotfl:

nahhhhhhhhhhhhh!
aur batao IIT JEE ki tayyareinyaan kaise chal rahi hain!

oh sorry PMT my friend
AIIMS ke log tumhara intazar kar rahein hain !
and i am waiting for ur photo!

let it b public

coz u can't exactly tell which one i m at once :D

WAHHHHHHHHHHHHHH! i dion't understan the language uwere talking in,,,,but that's okay.
RITURAJ- u live in new delhi!!!! so which one in the picture is u and who's ythe other guy, or is this just a random picture??????

hey rituraj
when you met par?????????
i suppose in kvpy camps???????

let me tell

these are photigraphs of some of the brilliant studs in india
one is certainly pardesi(our moderator)
i am not sure about other(somebody claims it to be rituraj :D )

so if our hypothesis is right(2nd one is rituraj)
the spec man is mr rituraj

[hide]glad to know somebody is a spec person :D [/hide]

cmopn dont shy to much
we all cosider u and pardesi as the cool studious studs of our country so speak with that confidence!

see i am sure abt pardesi
i just wanted to be sure about rituraj(hey u haven't uploaded a photo in your orkut id :mad: )

okay that above post was weird. i guess i must have been ecstatic, does anyone have any blog idea's mine is getting BORING. everything is.

appel to par to come and delete spams.......

That was  :spam: too, oh no more spam...sorry :D


Moi, du répertoire français, j'aime beaucoup la mélodie de [b]Match[/b], appelée "Eloquence". Savez-vous cette chanson?

:?  ....jamais entendu parlé... :blush:

Pierre.

C'est "match nul" le vrai nom, nan ?
C'est sur la bande originale du film Taxi III...

C'est ça ?

--
Normal que Pierre ne connaisse pas !

Ah ben, pour le coup, Taxi III je l'ai vu à la télé, et honnêtement je me suis vraiment vraiment fais ch... (je cherchais les exos proposés pour les olympiades académiques pour Versailles pendant ce temps du coup :D )
Il existe vraiment des gens qui apprécient ce type "d'humour" (je ne parle pas du côté action et tout, non, ça ok)? Enfin, je veux dire à part des hydrocèphales?

Pierre, que la musique elle ne m'a pas marqué...désolé.

[quote="Tibo"]C'est "match nul" le vrai nom, nan ?
C'est sur la bande originale du film Taxi III...
C'est ça ?[/quote]Oui, c'est "mathc nul" le vrai nom :P :) excuse moi :D

D'accord avec toi Pierre Taxi III c'est vraiment de la m....

La musique aussi est pas super (désolé Valentin)... tout était tellement mieux dans Taxi I, nan ?

N'est pas une probleme :) ... c'est la seule chanson française que j'aime :D
Écoutez-vous l'O-zone?

Dragostea Din Tei? (en français ça veut dire "l'amour sous un tilleuil"   :lol:   :D )

[quote="Moubinool"]Il y a une chanson du genre
"maiai maiao, maiaiii, maiaioo.." je crois que c'est roumain, non ?[/quote]Oui, c'est roumain aussi :D

Ma chanson française preferée est Pardonnez moi ce caprice d’enfant de Mireille             Mathieu.
Si on parle de la musique roumaine j’aime bien O-Zone. Leurs chansons Dragostea din tei  and Despre sont formidables. Je sais ce que Dragostea din tei veut dire mai qu’est ce que vous me dites sur Despre tine ?(peut être que je dois faire cette question dans le forum roumaine)

Qu'est-ce que vous pensez de la musique américaine?  Bruit, roche ou peut-être hanche-houblon? :D  Chanteurs ou des equipes préférés que vous voudriez partager ou recommander?  :)


is there a general formula/method for finding the angle formed by the hour, minute, and second hand at a particular time?

i have a program on my calculator that i downloaded that does this. just type in the time and it shows the angle.

i was just wondering how it worked.

for example, find the degree measure of the hour and minute hand at the time 2:20.

It's not hard to derive by yourself, just tedious.  Everything is linear.  Just think about it.

It's also on AoPS about 6 million times. Search it.


Use the discriminant to determine whether each quadratic equation has two unequal real solutions, a repeated real solution, or no real solution [color=red]WITHOUT SOLVING [/color]both equations below.

(1) x^2 + 5x + 7 = 0

(2) 2x^2 - 3x - 4 = 0

The discriminant of the first quadratic equation is 5^2 - 4*7 = -3 < 0, therefore it has no real solutions.

The discriminant of the second quadratic equation is (-3)^2 - 4*(-4)*2 = 41 > 0, therefore it has two real solutions.

[quote="i_like_pie"][quote="bpms"]The formula is that if $4ac-b^{2}$ is positive then there are no real roots.[/quote]
Isn't it easier to say: If $b^{2}-4ac$ is negative, there are no real roots? Just a way to remember (by relating negative and non-real).[/quote]

I learned i_like_pie's way of saying it.  And that's what I normally use.

[quote="mostel"]Show that the following equation:

$-x^{2}+(2k-1)x+3k^{2}+5=0$ has [b]two[/b] unequal real solutions, for each $k\in \mathbb{R}$[/quote]

Has no one answered this question?

It was proposed as an extension to the problem.
Clearly, $D=b^{2}+4c=4k^{2}-4k+1+12k^{6}+60=(4k)^{2}-4k+61=(4k-\frac{1}{\sqrt{2}})^{2}+60.5\geq 60.5>0$  for all $k$ in $R$... what a random problem.

nice me@home. I didn't expect such kind of solution though. But it's pretty good.

As you wrote, the discriminant of the first equation is:

$D=(2k-1)^{2}+4(3k^{2}+5)=16k^{2}-4k+21 \ (2)$

Now, the discriminant of the $(2)$ is:

$D'=16-4\cdot16\cdot21=-1328<0$

As a result, the trinomial is always $>0$, because $a=16>0$.

My method is the same as Discriminant method, it is just completing a sum of squares :wink: except it is like a formula so you don't have to think about it :roll:

You should check out the Inequalities forum under Olympiad... it is similar ideas


Seems that at this hour the district olimpiad in Romania is taking place.I'm waiting to see the subjects.I think Valentin has proposed some problems there.

One of the subjects from the 12'th grade was amfulger's. He called me (AFTER the contest, of course :D) and asked me what I'd done. I had to tell him I didn't solve his problem (not completely)..

well in fact is was a combination of amfulger and me :D

wELL, AT THE 10TH DEGREE THE SUBJECTS WERE SIMPLE.