calculus
integration
function
limit
real analysis
real analysis unsolved

Let $f: (0,\infty)\rightarrow (0,+\infty)$  be a decreasing function such that $\lim_{x\rightarrow 0}f(t)=+\infty$ and $\int^1_0 tf(t)dt$ is finite. Prove or disprove that there exists $M>1$ and $c\in (0;1)$ such that $f(Mt)\geq cf(t)$ in a small positive neigborhood $(0,\delta)$ of the origin.

the answer is negative. a poosible example is
\[ f(t)=t^{-1+1/n},\quad e^{-(n+1)^3}< t\leq e^{-n^3}  \]
Then for $M>1$,  $\liminf_{x\to 0+}\frac{f(Mx)}{f(x)}\leq \liminf_{x\to 0+}\frac{f(x+)}{f(x)}=0$ for any $M>1$.

I forgot to tell you that $f$ is continuous. Your above example is not a continuous function.

That doesn't really matter. If you give a reasonably simple and well-behaved example (such as the one cimabue gave), you can modify it slightly to make it continuous and keep all the relevant proeprties. One can easily modify cimabue's example by very small amounts around the endpoints $e^{-n^3}$ of those intervals, not enough to make the integral diverge and not enough to change what he wrote in the last line.


Suppose that a,b,c are positive real numbers such that ab+bc+ac=a+b+c.Find the minumu value of $\frac{a^3}{(b-c)^2+bc}+\frac{b^3}{(c-a)^2+ca}+\frac{c^3}{(a-b)^2+ab}$

Not the easiest of inequalities... see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=a%5E3%2F%28b%5E3%2Bc%5E3%29&t=3221]here[/url].


number theory
greatest common divisor
superior algebra
superior algebra unsolved

how do we prove that

U(2^n)(n>=3) is not cyclic.

Up to something Cezar Lupu used a short time ago, the $U$ stands for units, so it's very likely that units $\mod 2^{n}$ are meant.

Ok, then it is enough to notice that $gcd(\phi(2^{n}),2^{n}) \neq 1$ for all $n>1$.

I don't get that way to prove it, for example the same holds for $9$ and $\varphi(9)$, and still $U(9)$ us cyclic.

But we should not completely spoil the problem (probably homework).

An idea how to do it: show that $u^\frac{\varphi(2^{n})}2 \equiv 1 \mod 2^{n}$ for all units $u$.

My mistake, I misrembered the theorem "[b]every[/b] group of order $n$ is cyclic if and only if $gcd(n,\phi(n))=1$". This does not apply here (and even if it did, not that way).


Let $ m,n\in \mathbb{N}^2\backslash\{(0,0)\}$ . Show that :

$ \frac{a}{(m\plus{}n)a^2\plus{}mb^2\plus{}nc^2}\plus{}\frac{b}{(m\plus{}n)b^2\plus{}mc^2\plus{}na^2}\plus{}\frac{c}{(m\plus{}n)c^2\plus{}ma^2\plus{}nb^2}\le \frac{1}{2(m\plus{}n)}\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)$


What is the greatest possible two-digit number for the sum of the first $ n$ integers?

We know that the sum of the first n integers is represented by n(n+1)/2, so we can now model our inequality. (Note that these numbers come from the fact that it must strictly be a two digit number)

$ 10\le\dfrac{n(n \plus{} 1)}{2} < 100$

$ 20\le n(n \plus{} 1) < 200$

We want to maximize n(n+1), but still meet this inequality. We can start with a rough estimate as n as 14, knowing it will be just above 200.

$ 14(14 \plus{} 1) \equal{} 210$
$ 13(13 \plus{} 1) \equal{} 182$

Ah-ha! We have it. The sum of the first 13 numbers (182/2) is just below 100. Thus, $ \boxed{91}$ is our answer.


geometry
perpendicular bisector
geometry proposed

Given an isosceles triangle $ ABC\ (AB \equal{} AC)$. $ l$ is a line passes through $ C$ and parallel with $ BC$.  On the perpendicular bisector of $ AB,AC$, take $ P,Q$ (respectively) such that $ PQ\perp BC$. $ M,N$ are two points on $ l$ such that $ PA\perp PM, QA\perp QN$.
Prove that: $ \frac{1}{AM}\plus{}\frac{1}{AM}\leq\frac{2}{AB}$
This's my exercise and it's not really hard with arithmetic solution ... I hope you can help me with an elementary geometric solution  :D Thank you very much  :)


Euler
geometry
cyclic quadrilateral
geometry proposed

from the picture, prove that point A is on the line O1P

Thanks!

Dear Mathlinkers,
if your figure is more precise, the circle going through B, O1 and 02 is a Morley's circle.
For more explanation
sse : http://perso.orange.orange/jl.ayme  vol. 2  les cercles de Morley, Euler, Mannhein et Miquel  p. 1...
Sincerely
Jean-Louis

say OA intersects the middle circle at another point P' different from P. then since the cicle intersects the O2 cicle only at P. P' will not lie in circle O2. then we have O2B>O2P'. by property of cyclic quadrilateral in OBO2P'. we see since O1AB is isoceles BO1A is bisected. thus we het. angle O2P'B=O2BP'. but we got O2B>O2P'. thus proving the condition of P' is not equal to P being wrong. thus only possible as P=P'. thus proved.

As communist outlined, O_1A intersects the circle O_2 at P'; we shall prove that O_1BO_2P' is cyclic.
Obviously,  <O_1BO_2 = <O_1AO_2, hence <O_1BO_2 + <O_2AP' = 180 degs ( 1 ), but from the isosceles triangle AO_2P' we get <O_2AP' = <O_2P'A ( 2 ), from (1) and (2) getting <O_1BO_2 + <O_1P'O_2 = 180 degs, that is, O_1BO_2P' is cyclic, hence P' = P.

Best regards,
sunken rock

I hope that I understood correctly this "easy" problem.

Since $ O_1PO_2B$ is cyclically and $ O_2P \equal{} O_2B$ obtain $ \widehat {O_2O_1P}\equiv \widehat {O_2O_1B}$ .

But $ \widehat {O_2O_1A}\equiv\widehat {O_2O_1B}$ . Therefore, $ \widehat {O_2O_1P}\equiv\widehat {O_2O_1A}$ , i.e.  $ A\in O_1P$ .

Where are "[b]Morley[/b], [b]Euler[/b], [b]Mannhein[/b] and [b]Miquel[/b]" ?! Sincerely, V.N.


Show that for all positive integers $n\geq 1$ the number:
\[\prod_{k=1}^{n}k^{2k-n-1}\]
is also an integer number.

Laurentiu Panaitopol, TST 1988

Discussion continues in http://www.mathlinks.ro/Forum/viewtopic.php?t=54194


Find all integer solutions of equation\[(n+3)^n=\sum^{n+2}_{k=3}k^n\]

[quote="Singular"]n=2 works; henceforth consider n>2

Write $(n+3)^n - (n+2)^n = \sum_{k=3}^{n+1} k^n$.

$LHS = ( (n+3)^{n-1} + ... + (n+2)^{n-1} ) > n(n+2)^{n-1}$

Showing  $n(n+2)^{n-1} > n^n$ and $2n(n+2)^{n-1} > (n+1)^n$ is clearly sufficent.

Both are clearly obvious though.  So we are done.[/quote]

[quote="JBL"]Clearly sufficient? I don't see it. More explanation, please.[/quote]

Sure.  When I say "LHS" and "RHS", I mean $LHS = (n+3)^n - (n+2)^n = \sum_{k=3}^{n+1} k^n = RHS$.

Now:

$LHS > n(n+2)^{n-1} = 2(n+2)^{n-1} + (n-2)(n+2)^{n-1} $

$> (n+1)^n + (n-2) n^n > \sum_{k=3}^{n+1} k^n = RHS.$

using the two results I showed earlier.

nx = x + x + (n-2)x = 2x + (n-2)x

do you not see it??

I don't see it... I think you added an n with the left term on the RHS.

oooooh.  You're wrong, but now I understand what you mean!  Re-read what you have for those two terms.  It's not what you think it is.


inequalities
trigonometry
inequalities unsolved

It is given that $0 \le \theta_{i} \le \frac{\pi}{2}$ for $1 \le i \le n$. Prove that
\[ \frac{\sqrt{2}n}{2}(\displaystyle\sum^{n}_{i=1}\cos{\theta_{i}}+\displaystyle\sum^{n}_{i=1}\sin{\theta_{i}}) - (\displaystyle\sum^{n}_{i=1}\cos{\theta_{i}})(\displaystyle\sum^{n}_{i=1}\sin{\theta_{i}}) \ge \frac{n^2}{2}. \]

I think this inequality is wrong.
if n is even.
let $\theta_{i}=\frac{\pi} 2,i=1,2,\cdots,\frac n 2$
else $\theta_{i}=0$
maybe I am wrong.

[quote="leepakhin"]I am quite sure that you are trying [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64151]problem 2 of TST 3[/url]. :D[/quote]
In fact I think that one is a combinactorical question rather than a geometric one.


[quote="jli"]find a simplified form of $ \tan x+\cot x$.[/quote]
Have to hide, or else this post will be rated as a 1...
[hide] Using these trig identities:
$ \sin^{2}x+\cos^{2}x = 1$

$ 2\sin x \cos x = \sin 2x$

we obtain:
\begin{eqnarray*}\tan x+\cot x&=& \frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\\ &=& \frac{\sin^{2}x+\cos^{2}x}{\cos x \sin x}\\ &=& \frac{1}{\cos x \sin x}\\ &=& \frac{2}{2\cos x \sin x}\\ &=& \frac{2}{\sin 2x}\\ &=& 2\csc 2x \end{eqnarray*}[/hide]

jli, here is a nice follow-up:
Find a simplified form of $ \tan x+\cot y$ ($ x$ and $ y$ may or may not be equal)
[hide="Hint for follow-up"]$ \cos x\cos y+\sin x \sin y = \cos(x-y)$[/hide]

EDIT: Added follow-up's hint
EDIT2: Fixed the hint

Hmmm, in your hint, I believe the cosine addition formula is more helpful.[/hide]


floor function
probability
number theory
prime factorization
binomial coefficients

1. Given the Universal Set U={1,2,3,4,5,6,7,8,9,10} with subsets A={6,7,8,9,10}, B={1,5,9} and C={2,6,10}. Let X = BU(A-C),what is the sum of the elements in x.   (the U after the last B is supposed to be a union sign).

2. How many trailing 0s are there in 132!.

3. 1[u]<[/u]|2x-4|[u]<[/u] 5.

Also, does any one know a good book that explains set notation and Binomial coefficient.

P.S. I know the asnwers to these but I am not sure how to get the answers to them

1. 1+5+7+8+9=30 if you mean A-C=A\C
2. $ \lfloor 132/5 \rfloor \plus{} \lfloor 132/25 \rfloor \plus{} \lfloor 132/125 \rfloor\equal{}26\plus{}5\plus{}1\equal{}32$

[quote="Nandubetax"]3. 1[u]<[/u]|2x-4|[u]<[/u] 5.[/quote]
[hide="Solution"][color=darkred][b]Case 1:[/b][/color]

$ 1\le2x \minus{} 4\le5$

$ 5\le2x\le9$

$ \tfrac{5}{2}\le x\le\tfrac{9}{2}$

[color=darkred][b]Case 2:[/b][/color]

$ \minus{} 1\ge2x \minus{} 4\ge \minus{} 5$

$ 3\ge2x\ge \minus{} 1$

$ \tfrac{3}{2}\ge x\ge \minus{} \tfrac{1}{2}$

[color=darkred][b]Answer:[/b][/color]

$ \boxed{x: [ \minus{} \tfrac{1}{2},\,\tfrac{3}{2}]\cup[\tfrac{5}{2},\,\tfrac{9}{2}]}$[/hide]

[quote="Nandubetax"]
Also, does any one know a good book that explains set notation and Binomial coefficient.
[/quote]

Intermediate Counting & Probability has a whole chapter on set notation. I think it also has stuff about the Binomial Thereom. If it doesn't then you can look at Introduction to Counting & Probability or Intermediate Algebra.

EDIT: Miss interpreted your answers Bugi...they are both correct

Could you please post an explanation to each for that is what I am looking for.
-Like for number one why did you add 1+5+7+8+9 and..
- for number two...why did you divide by powers of 5?.
Also when you say "1. 1+5+7+8+9=30 if you mean A-C=A\C". The problem is A-C but the answer is still 30.

[quote="AIME15"][quote="Bugi"]
2. $ \lfloor 132/5 \rfloor \plus{} \lfloor 132/25 \rfloor \plus{} \lfloor 132/125 \rfloor \equal{} 26 \plus{} 5 \plus{} 1 \equal{} 32$[/quote]

In case you are wondering, he is finding the amount of factors of $ 5$.  Why, though?

Notice that for each $ 0$, we multiply by $ 10$.  Then we want to find the amount of factors of $ 10$, or the least of the amount of factors of $ 5$ and $ 2$.  $ 5$ has less factors than $ 2$, so we find the amount of factors of $ 5$.[/quote]

I get that you would look for the number of 5's the prime factorization of 132! has...yet why would you divide 132 by 5,25,and 125

With every 5 comes one zero.
But for example in 25 we have 2 5-s, so with every 25 comes one zero which we haven't counted.
Similar with 125.
As $ 137<5^4$, that's it.

And if you aren't interested in AOPS Books, you can go to Wikipedia or http://www.mathworld.wolfram.com

[quote="rrusczyk"]We cover set notation and the binomial coefficients in [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=100]AoPS Volume 1[/url].  You can get a more thorough education on binomial coefficients from [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=202]Introduction to Counting & Probability[/url] and a more thorough discussion of sets in [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=302]Intermediate Counting & Probability[/url].[/quote]

thanks, that helps a lot


PLEASE ONLY POST HERE IF YOU WISH TO SIGNUP

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the official thread is here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1212662#1212662[/url]

The rules are:

[code]
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3. You MAY NOT play with anyone intentionally losing. If I catch you, you are disqualified.
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The day of the competition will be announced soon.

There will be 15 contestants (or less if no one signs up)

After the first day, the top 10 will move on to the next day.

After that day, the top 6 will move on.

Then the top 3.

And then the winner is announced!!!!

Scoring is done with 'Today's Rating'.

Questions, Comments, or Concerns? Post in the official thread, or pm me!

Please enjoy, the

:huh:[size=200][color=red] WEIRD TOURNEY [/color][/size]:huh:

um...why make a seperate topic for signups?


[quote="MCrawford"]Suppose that $x$ is chosen from the set $\{1, 2, 4, 8, 16\}$ and $y$ is chosen from the set $\{3, 6, 12, 24, 48\}$.  How many different values could $x/y$ be?[/quote]
[hide]9 i think
starting wiht 1 from the first set you have 5 values.
then 2 from the first set all cancel out except $\dfrac{2}{3}$
then 4 from the first set all cancel out except $\dfrac{4}{3}$
and it continues on like that for all of the elements in set 1.
$5+1+1+1+1=9$[/hide]

[hide]For just 1, the fractions we can make are $\dfrac{1}{3}, \dfrac{1}{6}, \dfrac{1}{12}, \dfrac{1}{24}, \dfrac{1}{48}$. For two, all of them are the same as the first one except $\dfrac{2}{3}$. The same goes for the third, fourth and the fifth one. (Do you see that both of the sets are geometric series with a common multiplier of 2?). There are thus 9 possibilites. [/hide]


Daca intr-un triunghi doua bisectoare sunt congruente atunci demonstrati ca triunghiul este isoscel

[color=red]rad[a] se va citii radical din a
a^2 se va citii a la puterea a doua[/color]

Demonstratie

Fie triunghiul cu laturile a,b,c . Din faptul ca doua bisectoare sunt congruente avem urmatoarea relatie :

rad[bcp(p-a)] : (b+c) = rad[acp(p-b)] : (a+c)    => rad[b(p-a)] x (a+c) = rad[a(p-b)] x (b+c) / ridicam la patrat
=> b(p-a) : a(p-b) = (b+c)^2 : (a+c)^2  => bp - ab : ap - ab = (b^2 + 2bc + c^2) : (a^2 + 2ac + c^2) facand proportii derivate ajungem la (bp-ab) : p(a-b) = (b^2 + 2bc + c^2) : (a + b + 2c)(a-b) pt a face impartirea pp a dferit de b   => a^2b + ab^2+ 2abc = -pc^2 imposibil deoarece in dreapta avem pozitiv iar in stanga negativ=> a=b si deci triunghiul este isoscel.

Aşa cum bine zici este o problemă clasică, deci nu prea e de olimpiadă. Vezi că poţi folosi limbajul $\LaTeX$ pentru a scrie ecuaţii, spre exemplu rad[x] aşa cum l-ai notat tu se scrie (dollar) \sqrt {x} (dollar) şi iese $\sqrt {x}$. Am scris (dollar) în loc de simbolul de pe tastatură ca să vezi codul $\LaTeX$. Mai multe detalii despre acest limbaj găseşti în acest topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=5261.

Ah uitasem, bine ai venit. :)

La problema aceasta clasica parca exista si o demonstratie sintetica. Stie careva cum se face?

[hide="Generalizare."][quote="virgil Nicula"][color=red][b][u]O (usoara) generalizare.[/u][/b] Fie triunghiul $ABC$ si bisectoarea $[AD$ a unghiului $\widehat {BAC}$. Pentru un punct $L\in (AD)$ notam intersectiile $E\in BL\cap AC$ si $F\in CL\cap AB$. Sa se arate ca $BE=CF$ daca si numai daca triunghiul $ABC$ este isoscel $(AB=AC)$.[/color][/quote][/hide][hide="Observatie."][quote="Virgil Nicula"][color=blue] Eu cunosc trei metode (metrica, pur sintetica si prin absurd) pentru demonstrarea problemei propuse si a generalizarii ei. Ultima metoda se gaseste in cartea [b]"Al. Froda, Paradox in Matematica"[/b] (de altfel, o carte mai veche care nu ar trebui sa lipseasca din biblioteca voastra matematica). Daca am timp, voi prezenta ultimele doua procedee. Daca nu ma insel, problema initiala este rezolvata in cartea mea de geometrie prin cele trei procedee.[/color][/quote][/hide]

[hide=" Sper sa fie o surpriza placuta cel putin pentru ... Perfect_radio."]
[color=darkblue][b]O demonstratie sintetica a generalizarii mentionate in precedentul meu reply.[/b]

Fie punctul $H$ astfel incat $\triangle BAE\equiv \triangle FHC$ si dreapta $CF$ separa punctele $B$ si $H$. Notam $J\in (FC)$ care corespunde punctului $L$ in congruenta precedenta, adica punctul $L$ este piciorul bisectoarei din varful $H$ in triunghiul $FHC$. Se observa ca $AL=HJ$, $\widehat {FAC}\equiv \widehat {FHC}$, ceea ce inseamna ca patrulaterul $AFCH$ este inscriptibil. Deci $\widehat {ACF}\equiv \widehat {AHF}$ si $\widehat {BAL}\equiv$ $\widehat {LAC}\equiv$  $\widehat {FHJ}\equiv$ $\widehat {CHJ}$. Asadar punctele $A,L,J,H$ sunt conciclice. Insa $AL=HJ$ implica $LJ\parallel AH$. Prin urmare, patrulaterul $AFCH$ este trapez isocel. In concluzie, $\triangle FHC\equiv \triangle CAF$, adica $\triangle BAE\equiv \triangle CAF$, ceea ce implica $AB=AC$.[/color][/hide]


linear algebra
matrix
probability
geometry
3D geometry
sphere
limit

Let $ U$ be an $ n \times n$ orthogonal matrix. Prove that for any $ n \times n$ matrix $ A$, the matrices \[ A_m=\frac{1}{m+1}  \sum_{j=0}^m U^{-j}AU^j\] converge entrywise as $ m \rightarrow \infty.$

[i]L. Kovacs[/i]

if $ \mu$ is the usual probability measure on the unit sphere $ S^{n\minus{}1}$ of $ \mathbb{R}^n$, then it also induces a probability measure on $ X\equal{}S^{n\minus{}1} \times S^{n\minus{}1}$. Moreover, $ \phi: X \mapsto X$ defined by $ \phi(x,y) \equal{} (Ux, Uy)$ is a measure preserving transformation. So if $ f: X\to \mathbb{R}$ is defined by $ f(x,y) \equal{} x^tAy$, the ergodic theorem says that
\[ \lim_{m \to \infty} \frac{1}{m\plus{}1} \sum_{j\equal{}0}^m f( \phi^{j})\]
converges almost everywhere.


[hide="Hint I"]Maximize the sum of the area of the six triangles[/hide]
[hide="Hint II"]Prove by adding six similar inequalities that $ B_{i}$ have to be midpoints [/hide]

Really? I think the (degenerate) hexagon when $ B_{1}=B_{2}=A_{2}, B_{3}=B_{4}=A_{4}, B_{5}=B_{6}=A_{6}$ has smaller area...

[color=darkblue]Denote for any $ k\in \overline{1,6}$, $ x_{k}=\frac{B_{k}A_{k+1}}{A_{k}A_{k+1}}$, where $ 7\equiv 1(mod\ 6)$. Therefore,
the proposed problem is equivalently with the implication  $ \{\begin{array}{c}0\le x_{k}\le 1\\\\ k\in \overline{1,6}\ ,\ x_{7}=x_{1}\end{array}\|$ $ \implies$ $ \sum_{k=1}^{6}x_{k}(1-x_{k+1})\le 3$.
[b][u]Proof.[/u][/b] $ \sum_{k=1}^{6}x_{k}(1-x_{k+1})\le 1\cdot (1-x_{2})+x_{2}\cdot 1+1\cdot (1-x_{4})+x_{4}\cdot 1+1\cdot (1-x_{6})+x_{6}\cdot 1=3$.
For example, we have equality if $ \{\begin{array}{c}x_{1}=x_{3}=x_{5}=1\\\ x_{2}=x_{4}=x_{6}=0\end{array}$. You can try to generalize this problem.[/color]

2.nd solution:

For $ 0 \leq x_{i}\leq1$ and $ i=1,2,\cdots,6$, we have to prove $ \sum_{i=1}^{6}x_{i}(1-x_{i+1})\leq3$, (here, $ x_{1}=x_{7}$ )
or $ \sum_{i=1}^{6}x_{i}\leq3+\sum_{cyc}x_{1}x_{2}$.But, this ineq. solved at http://www.mathlinks.ro/viewtopic.php?p=887305#887305


inequalities
quadratics
inequalities proposed

Let $ a,b,c$ be the side lengths of a triangle. Prove that
\[ (a^2b \plus{} b^2c \plus{} c^2a)^2 \ge abc(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2).
\]
:)
When does the equality hold?
Remark.
By the Cauchy Schwarz and this inequality, we have
\[ \sum \frac {a^2b}{c} \ge \frac {(a^2b \plus{} b^2c \plus{} c^2a)^2}{abc(a \plus{} b \plus{} c)} \ge \sum a^2
\]
that is
\[ \sum a^2\left( \frac {b}{c} \minus{} 1\right) \ge 0
\]
This is Moldova TST 2006 problem, (the author is VasC.) :)

Very nice inequality, Can_hang2007! :lol:
It is equivalent to

$ [a^2(a\minus{}b)\plus{}b^2(a\minus{}c)]^2\plus{}a(a\minus{}b)(c\minus{}a)(a\minus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2) \ge 0$,

which is clearly true if $ a$ is between $ b$ and $ c$.

PS. Thank you for your appreciation to Vasc.  :wink:

[quote="Vasc"]Very nice inequality, Can_hang2007! :lol:
It is equivalent to

$ [a^2(a \minus{} b) \plus{} b^2(a \minus{} c)]^2 \plus{} a(a \minus{} b)(c \minus{} a)(a \minus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \ge 0$,

which is clearly true if $ a$ is between $ b$ and $ c$.

PS. Thank you for your appreciation to Vasc.  :wink:[/quote]
Thank you very much for your nice solution VasC. My solution is much harder.

[hide]PS: I admire you skill of solving inequality very much. :)[/hide]

[quote="Vasc"]Very nice inequality, Can_hang2007! :lol:
It is equivalent to

$ [a^2(a \minus{} b) \plus{} b^2(a \minus{} c)]^2 \plus{} a(a \minus{} b)(c \minus{} a)(a \minus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \ge 0$,

which is clearly true if $ a$ is between $ b$ and $ c$.

[/quote]

Dear Vasc, I was really astonished to see this really  beautiful formula. I tried for two days to understand how you discovered it, but I failed.  :( Can you please help me. Thank you very much.  :)

My solution is the following.
Assume that $ a$ is between $ b$ and $ c$. Let

$ f(x)\equal{}(a^2\plus{}b^2\plus{}c^2)x^2\minus{}2(a^2b\plus{}b^2c\plus{}c^2a)x\plus{}abc(a\plus{}b\plus{}c)$.

We have to prove that the discriminant of the quadratic $ f(x)$ is nonnegative. This is true since
$ f(a)\equal{}(a\minus{}b)(a\minus{}c)(a\minus{}b\plus{}c)\le 0$.
If the discriminant is positive, then $ f(x)>0$ for any real $ x$, and this contradicts our result $ f(a)\le 0$.

[quote="Vasc"]My solution is the following.
Assume that $ a$ is between $ b$ and $ c$. Let

$ f(x) = (a^2 + b^2 + c^2)x^2 - 2(a^2b + b^2c + c^2a)x + abc(a + b + c)$.

We have to prove that the discriminant of the quadratic $ f(x)$ is nonnegative. This is true since
$ f(a) = (a - b)(a - c)(a - b + c)\le 0$.
If the discriminant is positive, then $ f(x) > 0$ for any real $ x$, and this contradicts our result $ f(a)\le 0$.[/quote]

 Sorry, but I don't understand how to deduce the identity

 $ (a^2b + b^2c + c^2a)^2 - abc(a + b + c)(a^2 + b^2 + c^2) =
[a^2(a - b) + b^2(a - c)]^2 + a(a - b)(c - a)(a - b + c)(a^2 + b^2 + c^2) \ge 0$

P.S. There is a misprint $ f(a) = a(a - b)(a - c)(a - b + c)\le 0$.

[quote="manlio"][quote="Vasc"]My solution is the following.
Assume that $ a$ is between $ b$ and $ c$. Let

$ f(x) \equal{} (a^2 \plus{} b^2 \plus{} c^2)x^2 \minus{} 2(a^2b \plus{} b^2c \plus{} c^2a)x \plus{} abc(a \plus{} b \plus{} c)$.

We have to prove that the discriminant of the quadratic $ f(x)$ is nonnegative. This is true since
$ f(a) \equal{} (a \minus{} b)(a \minus{} c)(a \minus{} b \plus{} c)\le 0$.
If the discriminant is positive, then $ f(x) > 0$ for any real $ x$, and this contradicts our result $ f(a)\le 0$.[/quote]

 Sorry, but I don't understand how to deduce the identity

 $ (a^2b \plus{} b^2c \plus{} c^2a)^2 \minus{} abc(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \equal{} [a^2(a \minus{} b) \plus{} b^2(a \minus{} c)]^2 \plus{} a(a \minus{} b)(c \minus{} a)(a \minus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \ge 0$.[/quote]
Manlio, we get this identity from
$ f(x)\equal{}(a^2 \plus{} b^2 \plus{} c^2)(x\minus{}\frac {a^2b \plus{} b^2c \plus{} c^2a}{a^2 \plus{} b^2 \plus{} c^2})^2 \plus{}abc(a\plus{}b\plus{}c)\minus{}\frac {(a^2b \plus{} b^2c \plus{} c^2a)^2}{a^2 \plus{} b^2 \plus{} c^2}$
and
$ f(a) \equal{} (a \minus{} b)(a \minus{} c)(a \minus{} b \plus{} c)$.

Thank you very much.

Sorry for my stupidity :oops:


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Show that \[ \prod_{1\leq x < y \leq \frac{p\minus{}1}{2}} (x^2\plus{}y^2) \equiv (\minus{}1)^{\lfloor\frac{p\plus{}1}{8}\rfloor} \;(\textbf{mod}\;p\ ) \] for every prime $ p\equiv 3 \;(\textbf{mod}\;4\ )$. [J. Suranyi]

Write $ p=4k+3$. Consider all the sums $ x^2+y^2$, $ 1 \le x, \ y \le 2k+1$. Let $ r$ denote the number of those pairs $ x$, $ y$ for which $ \text{this sum} \equiv 1$.

First, we show that the number of those pairs for which $ x^2 + y^2 \equiv -1 \pmod{p}$ is $ r+1$. Since every quadratic residue has a unique representation in the form $ x^2$, $ 1 \le x \le 2k+1$, while every nonresidue has a unique representation in the form $ -y^2$, the solution of $ x^2 + y^2 = x^2 - (-y^2) \equiv 1$ count the consecutive numbers in the form (nonresidue, residue) within the sequence $ 1, 2, \ldots , p-2, p-1$. Similarly, the pairs with $ x^2 + y^2 \equiv -1$ correspond to pairs of type (residue, nonresidue). Since this sequence starts with a residue and (recall that $ p \equiv -1 \pmod{4}$) ends with a nonresidue, the second case must happen $ r+1$ times.

Next, we prove that the number of solutions of $ x^2 + y^2 \equiv a^2 \pmod{p}$, $ 1 \le x, \ y \le 2k+1$, is $ r$ for every $ a$. Indeed, the mapping $ x \equiv \pm ax_1$, $ y \equiv \pm ay_1$, where the signs are chosen so that $ x$, $ x_1$, $ y$, $ y_1$ are all in $ [1, \ 2k+1]$, provides a one-to-one correspondence between solutions of $ x^2 + y^2 \equiv a^2$ and $ x_1^2 + y_1^2 \equiv 1$. Similarly, we obtain that the number of solutions of $ x^2 + y^2 \equiv -1$ is $ r+1$.

The above observations mean that these sums $ s^2 + y^2$ represent every quadratic residue $ r$ times and every nonresidue $ r+1$ times. Since the total number of these pairs is $ ((p-1)/2)^2$ while the number of residues and the number of nonresidues are both equal to $ (p-1)/2$, wefind $ r=(p-3)/4 = k$.

Now, the product we want to compute is not over all these pairs but only over those with $ x<y$. Consider a quadratic residue $ a$. We know that the total number of solutions of $ x^2 + y^2 \equiv a$ is $ k$. If among them there are $ u$ with $ x<y$, $ v$ with $ x=y$, and $ w$ with $ x>y$, then $ u=w$ by symmetry and $ v$ is $ 0$ or $ 1$, hence $ u = [k/2]$. Analogously, for a nonresidue the number of solutions with $ x<y$ is $ [(k+1)/2]$.

The product of all quadratic residues is
\[ 1^2 \cdot 2^2 \cdot \ldots \cdot \left( \frac{p-1}{2} \right)^2 \equiv \left( \left( \frac{p-1}{2} \right) ! \right)^2 \equiv 1 \pmod{p},\]
and the product of nonresidues is
\[ \left( -1^2 \right) \cdot \left( -2^2 \right) \cdot \ldots \cdot \left( -\left( \frac{p-1}{2} \right)^2 \right) \equiv (-1)^{\frac{p-1}{2}} \left( \left( \frac{p-1}{2} \right) ! \right)^2 \equiv -1 \pmod{p},\]
Hence, the original product is
\[ 1^{[k/2]}(-1)^{\frac{k+1}{2}} = (-1)^{[\frac{p-1}{8}]}.\]

Denote this product by $ P$. In the product, all sums of pairs of quadratic residues are multiplied. Let $ g$ be a primitive root $ \mod{p}$, and put $ h = g^2$. The numbers $ 1, h, h^2, \ldots, h^{2k}$ represent every quadratic residue once; thus we have
\[ P \equiv \prod_{0 \le i < j \le 2k} \left( h^i + h^j \right) \pmod{p}.\]
This implies
\[ P \cdot \prod_{0 \le i < j \le 2k} \left( h^i - h^j \right) \equiv \prod_{0 \le i < j \le 2k} \left( h^{2i} - h^{2j} \right) \pmod{p}.\]
Here each product is the value of a nonzero Vandermonde determinant.
Thus, this equation can be rewritten as
\[ P \cdot V(1, h, h^2, \ldots, h^{2k}) \equiv V(1, h^2, \ldots, h^{4k}) \pmod{p}.\]
Since $ h^{2k + 1} = g^{4k + 2} \equiv 1 \pmod{p}$, the generator of the second Vandermonde determinant are congruent to
\[ 1, h^2, \ldots, h^{2k}, h, h^3, \ldots, h^{2k - 1}.\]
Thus, the second determinant can be obtained from the first by $ k + (k - 1) + \cdots + 2 + 1 = k(k + 1)/2$ transpositions of rows. By the familiar properties of determinants, this means
\[{ P \equiv ( - 1)^{\frac {k(k + 1)}{2}} \equiv ( - 1)^{[\frac {k + 1}{2}]} \equiv ( - 1)^{[\frac {p + 1}{8}]}} \pmod{p}.\]


Given a set of n+1 integers between 1 and 2n inclusive, show that at least one member of the set must divide another member of the set.  (with induction)

Hint:
[hide]
consider multiples of 2
[/hide]

I don't think that's a very good hint. I'd state the hint as follows:

[hide]Consider the greatest odd divisor of each of the numbers.[/hide]

[hide]

I think the problem is something like this?

from $1$ to $2n$, there will be $n$ even and $n$ odd

so choosing $n+1$ numbers, looking at the extremes, we either have $n$ odd and $1$ even, or $n$ even and $1$ odd, knowing that in these cases we are looking at all the consecutive odds/evens up to $2n$, so one must divide another member of the set?

i ono how to do this w/ induction tho

[/hide]

i think this problem has a better application in Pigeonhole principle.
for this problem assume each number is in a form of $c\cdot 2^k$. the result should follow.
you also can easily prove that given $n+1$ integers from the set of $\{1,...,2n\}$ there are at least to consequitive pairs.

[quote="amirhtlusa"]for this problem assume each number is in a form of $c\cdot 2^k$. the result should follow.[/quote]

this is exactly what Arne said.

i'm wondering, how do you solve the problem using induction?

[quote="Arne"][quote="amirhtlusa"]there are at least to consequitive pairs.[/quote]

I guess you mean "one pair of two consecutive numbers" or two coprime numbers...[/quote]

yes. typo
let me restate
there are at least two consequitive numbers(1pair).


If $ x>y\ge0$ than $ x\plus{}\frac{4}{(x\minus{}y)(y\plus{}1)^2}\ge3$ Prove it ?

it is easy with AM-GM .... just write $ x$ as $ x/2 \plus{} x/2$ and then it is very easy

 :D

Setting $ x\equal{}y\plus{}a$, $ a>0$, it is equivalent to $ a\plus{}\frac{y\plus{}1}{2}\plus{}\frac{y\plus{}1}{2}\plus{}\frac{4}{a\cdot(y\plus{}1)^2} \geq 4$ which is true by AMGM, with equality when $ (a,y)\equal{}(1,1)\Rightarrow (x,y)\equal{}(2,1)$. :)


Let $x_1,x_2,\cdots,x_n$ be $n$ positive rational numbers and let $k$ be a positive integers such that $x_i^\frac{1}{k}$ is irrational (for all i).

Then prove that if $a_1, a_2, .., a_{n+1}$ are rational numbers for which:

$a_1 + a_2 x_1^\frac{1}{k} + a_3 x_2^\frac{1}{k} + ... + a_{n} x_{n+1}^\frac{1}{k} =0$ then all $a_i =0$ for $i=1,n+1$

Well.. there vess proves that the sum of n irrational numbers is still an irational number. I don't see how that helps my problem. I have also a number that is not irational $a_1$.

Can you please explain in detail how that helps my problem?

thank you very much!:)

OK it's proven there that every sum of irrational numbers $x_i^{\frac{1}{k_i}}$ with $k=1,2,...n$ is irrational.
Now assume that you have $\sum_{i=0}^n a_i x_i^{\frac{1}{k_i} =0}$ with $a_i$ rational and $x_0=1=k_0$ with not all summands being $0$.
Then we can neglect those with $a_i=0$, so lets assume that all $a_i$ are $\neq 0$.
Then $a_i x_i^{\frac{1}{k_i}}$ is irrational again, thus by what vess proved there we get that $\sum_{i=1}^n a_i x_i^{\frac{1}{k_i}}$ is irrational, thus also $\sum_{i=0}^n a_i x_i^{\frac{1}{k_i} = a_0 + \sum_{i=1}^n a_i x_i^{\frac{1}{k_i}}}$ is irrational, so especialy not $0$.


The points A, B and C are vertices of an equilateral triangle with sides measured 10 cm. Determine the scalar product of vectors $ \vec{AB}$ and $ \vec{AC}$

$ \vec{u}\cdot\vec{v}\equal{}\|\vec{u}\|\,\|\vec{v}\|\cos\theta$ - that is, the scalar product (dot product) of vectors is the product of the magnitudes of the vectors and the cosine of the angle in between them. You do have all of that information available, right?


Does anybody have a copy of the CMO report, because it hasn't been posted online yet, and I would like to know my score range at least on CMO...

So, if anybody has it, could you please post the results page?

Thank you.  :D

I only know the range that your score is in


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Prove that
\[ 1^2\plus{}2^2\plus{}\cdots\plus{}n^2\geq \frac{(1\plus{}2\plus{}\cdots n)^2}{n}\]
for all integers $ n\geq 1$.

In the solution, it multiplies both sides by $ n$.  Then, it writes $ n$ as $ (1^2\plus{}1^2\plus{}\cdots \plus{} 1^2)$ to yield
\[ (1^2\plus{}2^2\plus{}\cdots \plus{}n^2)(1^2\plus{}1^2\plus{}\cdots \plus{}1^2) \geq (1\plus{}2\plus{}\cdots\plus{}n)^2.\]
Could someone please explain why this substitution can be done?

Thanks in advance!

It can be done because $ n\equal{}\underbrace{1\plus{}1\plus{}\cdots\plus{}1}_n\equal{}\underbrace{1^2\plus{}1^2\plus{}\cdots\plus{}1^2}_n$

[quote="darkmage2009"]There are $ n$ $ 1^2$'s in that substitution. This is related to the fact that the dot product of two vectors is smaller than the product of the magnitudes of the vectors.[/quote]

Could you please explain the relationship a bit more?  Thanks!

[quote="Flame"]You could also prove it this way:

$ 1^{2} \plus{} 2^{2} \plus{} \cdots \plus{} n^{2}\geq\frac {(1 \plus{} 2 \plus{} \cdots \plus{} n)^{2}}{n} \\
 \\
\frac {n(n \plus{} 1)(2n \plus{} 1)}{6}\geq\frac {n^2(n \plus{} 1)^2}{4n} \\
 \\
\frac {2n \plus{} 1}{6}\geq\frac {n \plus{} 1}{4} \\
 \\
2(2n \plus{} 1)\geq 3(n \plus{} 1) \\
 \\
n\geq 1$[/quote]

Yes, but I'm trying to learn about inequalities and Cauchy.  Nice proof, though. :)

If you use Cauchy-Schwarz for $ n$ variables in Japanese University entrance exam, you have to prove the inequality.

Can you prove that?

Because we learn only the dot product or scalar product of 2 or 3 dimension.

In general, you need to prove the inequality with more than 4 variables.

Sure.
Consider $ (a_1x\plus{}b_1)^2\plus{}(a_2x\plus{}b_2)^2\plus{}...(a_nx\plus{}b_n)^2$. It is a sum of squares, so it is nonnegative, so since it's a quadratic in x, it's discriminant must be nonpositive, which is equivalent to Cauchy.

I'll start out by defining the dot product. Let $ \textbf{a},\textbf{b}\in \mathbb{R}^n$; $ \textbf{a}\equal{}[a_1, a_2, \ldots, a_n]$ and $ \textbf{b}\equal{}[b_1, b_2, \ldots, b_n]$. We define the dot product $ \textbf{a}\cdot\textbf{b}\equal{}\displaystyle\sum_{i\equal{}1}^{n}{a_{i}b_{i}}$.
So let's say we have two vectors (that are not scalar multiples of each other) in $ n$-space. Then we can form a triangle by creating a vector between the heads of the two vectors. Through the Law of Cosines and some simple algebra, we obtain the equation $ \textbf{a}\cdot\textbf{b}\equal{}|\textbf{a}\parallel{}\textbf{b}|cos(\theta)$, where $ |\textbf{a}|$ is the norm of $ \textbf{a}$, which is defined as $ |\textbf{a}|\equal{}\sqrt{\displaystyle\sum_{i\equal{}1}^{n}{a_i^2}}$.
By substitution, we realize that $ \displaystyle\sum_{i\equal{}1}^{n}{a_{i}b_{i}}\equal{}\left(\sqrt{\displaystyle\sum_{i\equal{}1}^{n}{a_i^2}}\right)\left(\sqrt{\displaystyle\sum_{i\equal{}1}^{n}{b_i^2}}\right)\cos(\theta)$. Since $ \cos(\theta)\leq 1$, then $ \displaystyle\sum_{i\equal{}1}^{n}{a_{i}b_{i}}\leq\left(\sqrt{\displaystyle\sum_{i\equal{}1}^{n}{a_i^2}}\right)\left(\sqrt{\displaystyle\sum_{i\equal{}1}^{n}{b_i^2}}\right)$. Squaring both sides (which is possible since the RHS is clearly positive), we obtain $ \left(\displaystyle\sum_{i\equal{}1}^{n}{a_{i}b_{i}}\right)^2\leq\left(\displaystyle\sum_{i\equal{}1}^{n}{a_i^2}\right)\left(\displaystyle\sum_{i\equal{}1}^{n}{b_i^2}\right)$. Now if we let $ a_i\equal{}i$ and $ b_i\equal{}1$, we get your desired inequality.

There's a really smooth proof that uses linear algebra that can be found on the [url=http://en.wikipedia.org/wiki/Cauchy-schwartz_inequality]Wikipedia article for Cauchy Schwarz[/url]. It's inspired by projections, and uses the fact that the length of any vector is non-negative.

[quote="kunny"]If you use Cauchy-Schwarz for $ n$ variables in Japanese University entrance exam, you have to prove the inequality.

Can you prove that?
[/quote]

To prove the proposed inequality we could use Titu's Lemma,which we can prove very easy by induction

[hide="proof"]
Since

$ \frac{a_1 ^2}{b_1} \plus{} \frac{a_2 ^2 }{b_2} \geq \frac{(a_1 \plus{}a_2) ^2}{b_1 \plus{}b_2} \iff (a_1 b_2 \minus{}a_2 b_1)^2 \geq 0$

We have that the inequality is true for $ n\equal{}2$

Suppose it is true for $ n$

$ \sum_{k\equal{}1}^{n} {\frac{a_k ^2}{b_k}} \geq \frac{\left(\sum {a_k}\right)^2}{ {b_k}}$

Then we add $ \frac{a_{k\plus{}1}^2}{b_{k\plus{}1}}$ to both sides .

$ \sum_{k}^{n} {\frac{a_k ^2}{b_k}} \plus{}\frac{a_{k\plus{}1}^2}{b_{k\plus{}1}} \geq \frac{\left(\left(\sum {a_k}\right)^2 \plus{}a_{k\plus{}1} \right)^2}{ {b_k}\plus{}b_{k\plus{}1}} \iff \left((\sum a_k)^2 \cdot b_{k\plus{}1}  \minus{}a_{k\plus{}1} \sum b_{k} \right)^2 \geq 0$

We  see the inequality is true for $ n\equal{}k\plus{}1$ ,then it's true for every $ n \in \mathbb{N}$ ,as desired [/hide]

We do not need to prove titu's lemma. It is a direct application of cauchy.
a_1/x_1+a_2/x_2+.....a_n/x_n>=(a_1+....+a_n)^2/(x_1+...+x_n)
(a_1/x_1+a_2/X_2+....+a_n/x_n)(x_1+x_2+...x_n)>=(a_1+a_2+....a_n)^2
Which is true by cauchy.
Unless you want me to prove Cauchy which you can easily understand by vectors.
|v^2||w^2|=|vw|^2
1^2/1+2^2/1+3^2/1+.....+n^2/1>=(1+2+3+4....+n)^2/1+1+1....+1(n times)

[quote="darkmage2009"]  Through the Law of Cosines and some simple algebra, we obtain the equation $ \textbf{a}\cdot\textbf{b} \equal{} |\textbf{a}\parallel{}\textbf{b}|cos(\theta)$
[/quote]

That isn't true. We can only claim that for dimension $ n\equal{}2,3$
For higher demensions, that equation comes as a result of Cauchy. That it, we define $ \cos(\theta) \equal{} \frac{\textbf{a}\cdot\textbf{b}}{|\textbf{a}\parallel \textbf{b}|}$
Because it isn't easy to imagine triangles in fourth and fifth dimensions etc.

[hide="Alternatively"]
 We know $ (1 \plus{} 2 \plus{} ... n)^{2} \equal{} 1^{3} \plus{} 2^{3} \plus{} ... n^{3}$. We can then separate the RHS into  $ \frac {1^{3}}{n} \plus{} \frac {2^{3}}{n} \plus{} ... \frac {n^3}{n}$.
 $ \frac {n^3}{n} \equal{} n^{2}$, but all of the other terms are smaller than their LHS counterparts because $ n$ is the biggest number. Equality occurs when $ n \equal{} 1$.
[/hide]

Would that be rigorous enough?


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Consider all $ m \times n$ matrices whose all entries are $ 0$ or $ 1$. Find the number of such matrices for which the number of $ 1$-s in each row and in each column is even.

A possible first thought: if we do all of the arithmetic mod 2, we're trying to count all of the matrices $ A$ such that $ Aj \equal{} 0$ and $ j^TA \equal{} 0$ where $ j$ is a column vector of all 1's.

A friend and I are currently looking for a recursion in $ m$ and $ n.$

Continuing the thought: maybe linear algebra doesn't help and this is pure counting.

Let $ E(m,n)$ be the number we seek.

[hide="oops"]$ E(m,1) \equal{} \binom{m}{0} \plus{} \binom m2 \plus{} \binom m4 \plus{} \binom m6 \plus{} \cdots \equal{} 2^{m \minus{} 1}.$[/hide]$ E(m,1) \equal{} 1.$ All you get is the zero vector.

$ E(m,2) \equal{} E(m,1) \equal{} 2^{m \minus{} 1}$ for $ m\ge 2.$

$ E(m,n) \equal{} E(n,m).$

Let me try brute forcing $ E(3,3)$ ... Maybe $ E(3,3) \equal{} 10?$

No, that's wrong. $ E(3,3) \equal{} 16,$ not $ 10.$

In general, for $ n,m\ge 2,$ we have $ \boxed{E(m,n) \equal{} 2^{(m \minus{} 1)(n \minus{} 1)}}.$

Look at the first $ n \minus{} 1$ columns. For any single column, we can choose an even number of $ 1$'s in $ 2^{m \minus{} 1}$ ways, as explained above. Choose each such column independently and arbitrarily; we can do that in $ (2^{m \minus{} 1})^{n \minus{} 1}$ ways.

Now, pick the $ n$th column to be the parity-check bit for the row sums. That is, if an any given row, the first $ n \minus{} 1$ elements add to an odd number, put a $ 1$ in the $ n$th column, and if the first $ n \minus{} 1$ elements add to an even number, put a $ 0$ in the $ n$th column.

We had no choice there. The only remaining question: does that $ n$th column have an even number of $ 1$'s?

I will use the linear algebra. Block structure $ A \equal{} [B\ x],$ where $ B$ is the first $ n \minus{} 1$ columns chosen arbitrarily as given above, and $ x$ is determined from $ B.$

In fact, if $ j$ is the column of all $ 1$'s and we're doing our arithmetic mod 2, then $ x \equal{} Bj.$ The question of whether $ x$ has an even number of $ 1$'s can be expressed by asking if $ j^Tx \equal{} 0.$ But $ j^Tx \equal{} j^T(Bj) \equal{} (j^TB)j \equal{} 0$ since $ j^TB \equal{} 0.$ That's because each column of $ B$ has an even number of $ 1's.$

Postscript: $ E(m,1) \equal{} E(1,n) \equal{} 1.$ The only way to do it with a single row or column is with all zeros. Ignore what I wrote above about $ E(m,1).$ (In fact, that formula, $ 2^{(m\minus{}1)(n\minus{}1)}$, works for $ m$ or $ n$ equal to $ 1$ just as well.)

OK, one more version of the argument, this time without having to add up any binomial coefficients.

Suppose $ A$ is $ m\times n.$ Block-structure $ A\equal{}\begin{pmatrix}B&u\\v^T&t\end{pmatrix}$ where $ B$ is $ (m\minus{}1)\times (n\minus{}1),$ $ u$ is $ (m\minus{}1)\times 1,$ $ v^T$ is $ 1\times (n\minus{}1),$ and $ t$ is scalar.

(We can let $ B$ be empty - that doesn't destroy the argument.)

Chose $ B$ completely arbitrarily. This can be done in $ 2^{(m\minus{}1)(n\minus{}1)}$ ways. What we then need to do is to show that we can choose $ u,v,$ and $ t$ so as to complete the structure, and that we have no choice about $ u,v,$ and $ t.$

Let $ u\equal{}Bj\pmod{2},$ where $ j$ is the $ (m\minus{}1)$ column vector of all 1's.

Let $ v^T\equal{}j^TB\pmod{2},$ where this $ j$ is the $ (n\minus{}1)$ vector of all 1's.

We had no choice about either of those - those are the check bits for each row and each column.

Now $ t$ must be the check bit for the bottom row and the check bit for the right column, both. That is, we must have both $ t\equal{}j^Tu$ and $ t\equal{}v^Tj.$ So to prove that we can finish the structure for an arbitrary $ B$ we must prove that these are both equal. (As above, all arithmetic is being done mod 2.)
\[ j^Tu\equal{}j^T(Bj)\equal{}(j^TB)j\equal{}v^Tj\]
So $ j^Tu\equal{}v^Tj$ and our two formulas for $ t$ agree. That completes the proof.

I tried a bunch of searches to find this but failed -- I'm pretty sure I've solved it on the forum at least once or twice :).

JBL: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1041841843&t=151564]here[/url] is one that I found. The argument by tjhance on that topic (if you clean up his counting) is essentially the same as the last argument I just posted.

[url=http://www.mathlinks.ro/viewtopic.php?t=165204]here[/url] is an esp. case.

In the link I gave, JBL proposed the problem of filling the array with numbers from $ \{0,1,\dots,k \minus{} 1\}$ such that each row and column is divisible by $ k,$ and stated the answer as $ k^{(m \minus{} 1)(n \minus{} 1)}.$

The last proof I gave (the $ 2\times 2$ block structure) works perfectly for this, doing the arithmetic mod $ k.$ We do need to say that $ u \equal{} \minus{} Bj$ and $ v^T \equal{} \minus{} j^TB,$ with $ t \equal{} \minus{} j^Tu \equal{} \minus{} v^Tj.$

The negative signs belong for all $ k;$ it's just that mod 2, you can't tell the difference between – and + .


For a motor vehicle travelling at r miles per hour, the distance d in feet required to stop is approximately $ d \equal{} 0.05r^2 \plus{} r$. If a school bus is travelling at 45 mph, how many feet, to the nearest foot, would it travel before coming to a halt?

Simply plug in the formula: $ d\equal{}0.05(45^2)\plus{}45\equal{}\frac{2025}{20}\plus{}45\equal{}\frac{405}{4}\plus{}45\equal{}101.25\plus{}45\equal{}146.25$. So the distance the bus is required to travel, in feet, before coming to a complete stop (to the nearest foot), is $ \boxed{146}$ feet.


Các anh ơi ,ai biết chỗ nào dow ebook thì bảo em vơi , thanhk nhiều!

[quote="quangchuc"]Khong ai tra loi ca! :mad:  Hay that! :wink:  Moi nguoi dau het ca roi?   (To ko biet cho nao down ebook ca)[/quote]
Nếu là ebook các loại thì vào http://gigapedia.org/ hoặc http://lookforbook.com/
Còn sách TA thì nên vào http://englishtips.org/

http://maichoi.vuicaida.com
Trang này thỉnh thoảng mới có tài liệu nâng cao! :-?


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Given a triplet we perform on it the following operation. We choose two numbers among them and change them into their sum and product, left number stays unchanged. Can we, starting from triplet $(3,4,5)$ and performing above operation, obtain again a triplet of numbers which are lengths of right triangle?

Note that, if at any point we have numbers $\equiv 0, 1, 1 \pmod{2}$ in some order, we will either get a set $\equiv 0, 0, 1$ (by applying the operation to the two $1$s) or another $0, 1, 1$ (by applying it to a $1$ and a $0$). However, if we get $0, 0, 1$, we will keep getting that forever. And the sides of a right triangle are of the form $0, 1, 1$ or $0, 0, 0$, so we must avoid $0, 0, 1$.

However, in a pythagorean triple of the form $0, 1, 1$, the even number can't be the hypothenuse. If we apply our operation to $(3, 4, 5)$, we must take $4$ and one of the others, and the resulting even number will be the product (the biggest one). It's easy to prove that the even number will always be the biggest, so we can never get another pythagorean triple.

I think using the division on three will also work. Look here: the initial numbers are $\equiv 0,1,2 \pmod{3}$ type. If we combine the numbers with remainder different from $0$ to the one with $0$ we will get again $\equiv 0,1,2\pmod{3}$ triple. But if we combine the two "none-zeroremainder" ones we will get $\equiv 0,0,2 \pmod{3}$ triple, after we'll always get the last type of triples, in this way we will never get to a Pythagorean triple. So we have to always to multiple some number to the one that is divisible to $3$. But if do like that we will understand that the number divisible by $3$ will increase enormously and no triangle would be possible to constuck by the triple.

by means of mod2 and mod3 we get the only operative way
$(3,4,5)\rightarrow (5,7,12)\rightarrow (5,19,84) or (7,17,60)$...
it's easy to prove that al of those can't form even a triangle!


probability
calculus
integration
expected value
probability and stats

A probability $ P$ is randomly chosen from $ [0,1]$; you are not told its value. $ P$ is the probability that a given coin comes up heads. If you observe $ n$ trials in a row (and no other trials) and all of them come up heads, what is the expected value of $ P$, in terms of $ n$?

I solved this using calculus, but if someone can avoid it, that would be rather more elegant IMO.

what does "expected value of $ P$" mean here, exactly ?

It's the expectation of $ P$ conditioned on having flipped $ n$ heads in a row, presumably.  In symbols, it should be $ \frac {\int_{0}^1 p \cdot p^n \, dp}{\int_{0}^1 p^n \, dp} \equal{} \frac {n \plus{} 1}{n \plus{} 2}$.  I don't know how to argue this without calculus (although there is a certain amount of naturality to the answer, given that for $ n \equal{} 0$ we better get out $ \frac{1}{2}$).

Another approach, which generalizes to handle any observation:

Choose $ n\plus{}1$ numbers independently and uniformly from $ [0,1]$. Call the first one $ P$ and label the others "heads" if they are less than $ P$ and "tails" otherwise. If exactly $ k$ of these are labeled "heads", what is the expected value of $ P$?
This happens when the first number is $ (k\plus{}1)$st on the list in increasing order. There's nothing special about it being chosen first, so the expected value of $ P$ is simply the expected value of the $ (k\plus{}1)$st element from the bottom. The order statistics for the uniform distribution are well-known; the $ (k\plus{}1)$st out of $ n\plus{}1$ has density $ (n\plus{}1)\binom{n}{k}x^{k}(1\minus{}x)^{n\minus{}k}\equal{}(k\plus{}1)\binom{n\plus{}1}{k\plus{}1}x^{k}(1\minus{}x)^{n\minus{}k}$ and expected value $ \frac{k\plus{}1}{n\plus{}2}$.
It's possible to see those expected values with a strictly probabilistic argument, but it's pretty fuzzy without the easy calculus.

Jmerry, do you call the following argument "fuzzy"?

Choosing $ (n\plus{}1)$ points uniformly on the $ [0;1]$ is equivalent to choosing uniformly $ (n\plus{}2)$ points on the circumpherence of unit length and considering the first point as a "break" which converts the circumpherence into the segment $ [0;1]$.

The distances between $ (n\plus{}2)$ points are identically distributed. So the expected distances are equal to $ \frac{1}{n\plus{}2}$ (their sum is 1).

The expected distance between the "break" point and the $ (k\plus{}1)$ point is $ \frac{k\plus{}1}{n\plus{}2}$.

No, that's a good one.
Language note: The English spelling is "circumference".

[quote="roah"]Jmerry, do you call the following argument "fuzzy"?

Choosing $ (n \plus{} 1)$ points uniformly on the $ [0;1]$ is equivalent to choosing uniformly $ (n \plus{} 2)$ points on the circumpherence of unit length and considering the first point as a "break" which converts the circumpherence into the segment $ [0;1]$.

The distances between $ (n \plus{} 2)$ points are identically distributed. So the expected distances are equal to $ \frac {1}{n \plus{} 2}$ (their sum is 1).

The expected distance between the "break" point and the $ (k \plus{} 1)$ point is $ \frac {k \plus{} 1}{n \plus{} 2}$.[/quote]

How do you relate your arguments above and the original problem?
Moreover, your argument "So the expected distances are equal to $ \frac {1}{n \plus{} 2}$ (their sum is 1)" seems intuitively clear but can you give a math. justification?


integration
logarithms
real analysis
real analysis unsolved

$ y(y\minus{}2xy')^3\equal{}y'^2$

I have no idea how to solve it. Please help!

hello, i have also no idea yet, but two solutions are $ y\equal{}\pm\frac{\sqrt{3}}{9x}$.
Sonnhard.

Yes, that's one solution, but how to get it?

Ok, this is the idea.. First, write the equation as:

$ y\,y' (\frac{y}{y'}-2x)^3=1$

Now, differentiate with respect to $ x$. We get:

$ y'\,y' (\frac{y}{y'}-2x)^3+y\,y'' (\frac{y}{y'}-2x)^3+3\,y\,y' (\frac{y}{y'}-2x)^2 (\frac{y'}{y'}-\frac{y}{(y')^2}y''-2)=0$

$ (\frac{y}{y'}-2x)^2 ((y')^2 (\frac{y}{y'}-2x)+y\,y'' (\frac{y}{y'}-2x)-3\,y\,y' (\frac{y}{(y')^2}y''+1))=0$

From here we can see that we need to deal with two equations. We can solve the left one first:

$ (\frac{y}{y'}-2x)^2=0$

$ \frac{y}{y'}=2x$

$ \frac{y'}{y}=\frac{1}{2x} / \int dx$

$ \ln y=\frac{1}{2}\ln x + C_1$

$ y=C_1\sqrt{x}$

Plugging this in the original equation we can see that the factor in the brackets is equal to $ 0$ so this isn't the solution. Therefor we can divide everything in the upper equation with this expression. This gives us:

$ (y')^2 (\frac{y}{y'}-2x)+y\,y'' (\frac{y}{y'}-2x)-3\,y\,y' (\frac{y}{(y')^2}y''+1)=0$

$ (y')^2 (y-2xy')+y'' y (y-2x y')-3 y (y\,y''+(y')^2)=0$

$ (y-2xy') ((y')^2   +y'' y )-3 y (y\,y''+(y')^2)=0$

$ ((y')^2   +y'' y )(y-2xy' - 3 y)=0$

$ -2((y')^2   +y'' y )(xy' + y)=0$

Now we can focus on the two new equations.

1)

$ (y')^2   +y'' y = 0$

$ \frac{y'}{y''} =-\frac{y}{y'}$

$ \frac{y''}{y'} =-\frac{y}{y'}  / \int dx$

$ \ln y' = C_2 - \ln y$

$ y'y = C_2/ \int dx$

$ \frac{y^2}{2} = C_2 x + C_3$

$ y = \pm \sqrt{C_2 x + C_3}$

Plugging this in the original equation we get that $ C_2 ^2 = 4 C_3^3$ and this is a solution.

2)

$ xy' + y = 0$

$ \frac{y'}{y}=-\frac{1}{x} / \int dx$

$ \ln y=C_4 - \ln x$

$ y=\frac{C_4}{x}$

Plugging this in we get $ C_4 = \pm \frac{1}{\sqrt{27}}$ and this also is an solution..

So, to sum up, we have solution of two forms:

1) $ y(x) = \pm \sqrt{C (2 \sqrt{C} x + 1)}$

2) $ y(x) = \pm \frac{1}{3 x \sqrt{3}}$

Hope this helps!


yes, i missed a bit, i counted automatically 4, but it was 2, but this is correct:

$\frac{1}{(a+b+c)^{8}}\geq \frac{1}{3^{4}}$

AM-GM
$a+b+c+\frac{1}{abc}\geq (a+b+c)+\frac{3}{\frac{(a+b+c)^{3}}{3^{2}}}\geq 4\sqrt[4]{\frac{3^{6}}{(a+b+c)^{8}}}\geq 4\sqrt{3}$


Is it possible to select 102 17-element subsets of a 102-element set,such that the intersection of any two of the subset has at most 3 elements?

This problem is from MathOlympiad 2000 of PML №239 of Sanct-Peterburg.


The commission on each sale is $ 0.075\%$. A bonus of $ 2.5\%$ of the total sales is paid when total sales exceed $ \$50,\!000$. How much total commission would be made on sales of $ \$9,\!000$; $ \$15,\!000$; $ \$16,\!000$; $ \$12,\!000$; and $ \$8,\!000$?

[quote="spacecadet"]The normal commision is $ .075%
$ of $ 9+15+16+12+8=60$ thousand, which is $ 75%
$ of $ 60$, which is $ 45$. Now the other bonus is $ 2.5%
$ of $ 60$thousand, which is $ 25%
$ of $ 6000$, which is $ 1500$. $ 1500+45=\$1545$[/quote]

I think you mean that the commission is $0.075$% and that percent of $60000$, which is obtained from $(9000+15000+16000+12000+8000)$. Simplified, it is actually $75$% of $60=45$ I'm guessing what you did was compute $2.5 percent\cdot{60000}$ which is equal to $1500$ and $1500+45=1545$. So therefore the answer is $\boxed{1545}$.


integration
real analysis
calculus
advanced fields
advanced fields unsolved

Let $\mu$ be a complex measure on the unit circle (which we can identify with the segment $[0,2\pi]$ via $x\mapsto e^{ix}$) such that for all $n>0,\ \int_{0}^{2\pi}e^{ni\theta}\ \mbox d\mu(\theta)=0$. Prove that $\mu$ is absolutely continuous with respect to the Lebesgue measure.

What about that integral? Something's missing from the statement.

I edited it for him. A very famous theorem. grobber called it [b]a[/b] theorem of F. and M. Riesz, but if you go looking for [b]the[/b] Theorem of F. and M. Riesz, this is what you'll find.


How do I prove that for positive reals a,b,c>0

a+b+c<= a^3/bc + b^3/ac + c^3/ab ?

First solution: We have $ \sum\frac{a^3}{bc}\ge a\plus{}b\plus{}c$

$ \iff a^4\plus{}b^4\plus{}c^4\ge abc(a\plus{}b\plus{}c)$

Verify that

$ (a^2)^2\plus{}(b^2)^2\plus{}(c^2)^2\ge a^2b^2\plus{}b^2c^2\plus{}c^2a^2$
$ \ge (ab)(bc)\plus{}(bc)(ca)\plus{}(ca)(ab)\equal{}abc(a\plus{}b\plus{}c)$.

Second solution: By Rearrangement, $ \sum\frac{a^3}{bc}\ge\sum\frac{a^3}{ab}\equal{}\sum\frac{a^2}{b}\ge \frac{(a\plus{}b\plus{}c)^2}{a\plus{}b\plus{}c}\equal{}a\plus{}b\plus{}c$
Where the last inequality follows from Cauchy-Schwarz.

Third solution: Using Chebyshev's inequality,

$ \sum\frac{a^3}{bc}\ge \frac{3(a^3\plus{}b^3\plus{}c^3)}{ab\plus{}bc\plus{}ca}\ge \frac{(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca}\ge a\plus{}b\plus{}c$.

I think there are a lot more solutions. :wink:

$ \frac{x^3}{yz}\plus{}y\plus{}z \ge 3x$ :)

See also topic [url]http://www.mathlinks.ro/viewtopic.php?p=445892#445892[/url]

We have $ a^4\plus{}b^4\plus{}c^4 \geq a^2bc\plus{}b^2ca\plus{}c^2ab$. It's true by Murheud's ineq.


algebra
polynomial
number theory proposed
number theory

Find all polynomial $ P(x)\in Z[x]$ such that
$ P(n)|3^n\plus{}4^n\plus{}5^n,\forall n\in N$

When $ degP \geq 1$
For some $ a \in \mathbb{N}$, we can take $ p \in \mathbb{P} \setminus \{2,3,5\}$ such that $ p|P(a)$
For each $ t \in \mathbb{N}$, we have $ P(a \plus{} tp) \equiv P(a ) \equiv 0 \ (mod \ p)$
So $ p|3^{a \plus{} tp} \plus{} 4^{a \plus{} tp} \plus{} 5^{a \plus{} tp} \Rightarrow p|3^{a \plus{} t} \plus{} 4^{a \plus{} t} \plus{} 5^{a \plus{} t}$
Now take $ t$ such that $ p \minus{} 1 | a \plus{} t$ so $ 3^{a \plus{} t} \plus{} 4^{a \plus{} t} \plus{} 5^{a \plus{} t} \equiv 3 \ (mod \ p) \Rightarrow p|3$, which is impossible.
When $ degP \equal{} 0$, easy to see that $ P(n) \equal{} \pm 1$ or $ P(n) \equal{} \pm 2$

More $ a\in N$
Find all $ P(x)\in Z[x]$ such that
$ P(n)|2^{n^a} \plus{} 3^{n^a},\forall n\in N$

If $ p|P(n)$, then $ p|P(n\plus{}kp)|2^{(n\plus{}kp)^a}\plus{}3^{(n\plus{}kp)^a}$. Because $ (n\plus{}kp)^a\equal{}(n\plus{}k)^a\mod p\minus{}1$, we get (k choice, p-1|n+k), that p<3. Obviosly for n>0 P(n) odd, therefore work only $ P(n)\equal{}\pm 1$.

More
$ f(n)$ is a polynomial and $ f(n)\in Z[x]$ and $ f(x) > 0,\forall x > 0$
Find all $ P(n)\in Z[x]$ such that
$ P(n)|3^{f(n)} \plus{} 2^{f(n)},\forall n\in N$

Are you sure
If you can $ k$ such that $ p\minus{}1|f(n\plus{}k)$
I think it not always find out $ k$
I have other method .I discover this result when I thought this lemma.

Lemma 1:
$ f(x)\in Z[x]$
Prove that exist k such that $ \gcd (f(x),f(x \plus{} h))$ not is polynomial with degree>0.
Lemma 2: $ \gcd(a^m \minus{} b^m,a^n \minus{} b^n) \equal{} a^d \minus{} b^d$ where $ d \equal{} \gcd (m,n)$ and $ \gcd (a,b) \equal{} 1$
I think about problem
$ A \equal{} \{p\in P, p|3^n \plus{} 2^n\}$
$ B \equal{} \{p\in P , p|P(n)\}$
If $ |A| > |B|$

Please prove lemma 1 .I think it quite easy and useful to prove the problem.

Lemma 1 is trivial as it just states that $ f(x)$ and $ f(x\plus{}h)$ have no common roots.

You are very good this is my solution.
Can you post a solution for my problem?


[url]http://www.mathlinks.ro/viewtopic.php?p=451049#451049[/url]
You also made a typing mistake.If it's $ x \in [0,1]$ in [b](ii)[/b] , no such an equation exists.


calculus
integration
limit
trigonometry
calculus computations

[quote="kunny"]Evaluate $\lim_{n\to\infty}\int_{0}^\pi \left|\left(x+\frac{\pi}{n}\right)\sin nx\right|\ dx\ (n=1,\ 2,\ \cdots).$[/quote]

Set $L: =\lim_{n\to\infty}\int_{0}^\pi \left|\left(x+\frac{\pi}{n}\right)\sin nx\right|\ dx$ and let $u=nx\Rightarrow du=n\ dx$ to get

$L=\lim_{n\to\infty}\int_{0}^{n\pi}\left|\frac{u+\pi}{n}\sin u\right|\ \frac{du}{n}$
$= \lim_{n\to\infty}\frac{1}{n^{2}}\sum_{k=1}^{n}\int_{(k-1)\pi}^{k\pi}\left|(u+\pi)\sin u\right|\ du$
$=\lim_{n\to\infty}\frac{1}{n^{2}}\sum_{k=1}^{n}(-1)^{k-1}\int_{(k-1)\pi}^{k\pi}(u+\pi)\sin u\ du$
$=\lim_{n\to\infty}\frac{1}{n^{2}}\sum_{k=1}^{n}(-1)^{k-1}\left[(u+\pi)\cos u\+\sin u \right]_{u=(k-1)\pi}^{k\pi}$
$=\lim_{n\to\infty}\frac{1}{n^{2}}\sum_{k=1}^{n}(-1)^{k-1}\pi\left[2(-1)^{k-1}k-(-1)^{k}\right]$
$=\lim_{n\to\infty}\frac{1}{n^{2}}\sum_{k=1}^{n}\pi (2k+1)$
$=\pi \lim_{n\to\infty}\frac{1}{n^{2}}(n^{2}+2n)=\pi$


Princeton college Yale Harvard articles geometry algorithm

Hey everyone,

I haven't been as active around here as I used to be, but I am still at Princeton and would be happy to help anyone with questions. If you're coming to Princeton Preview, I can help you find a math major to meet or to host you, and I can help you decide what classes might be worth seeing on the Friday you're here. I'm also happy to answer any questions about going to Princeton and about majoring in math. I also know a bit about some other departments, so if you're interested in computer science or finance you can also ask here or PM me.

I know there are some other Princeton guys on this forum, some who are younger than I, so you might be able to get a few different opinions about some things, too.

cheers
Nathan

Hello,

I don't know if you have read my post, but I want to know whether I will have a good shot when I apply early decision to Princeton.
I am still doubting between Princeton, Yale and Columbia...

Here is my short resume - http://www.mathlinks.ro/viewtopic.php?t=269558 .

Thank you,

Talk later,

Salvatore

Hi Salvatore. Following Harvard, Princeton has ended early decision. The first class without early decision was 2013. So you can only apply to Princeton for regular decision. Based on your post in the other page, you have a strong chance of admission. But international admissions to top universities can be very competitive; there is never a guarantee.

For those who are interested in this and the reason, Princeton's article about it is [url=http://www.princeton.edu/main/news/archive/S15/86/07G08/index.xml?section=topstories]here[/url].

Thank you, Xevarion :)
I am just pretty unsure about applying to Princeton at all due to my not that high GPA 5.72/6.00

Hope that is not a problem for getting accepted and getting a financial aid :)

Talk later,

Salvatore

That is not a problem. Different schools have different grading standards, especially international schools, so places like Princeton always pay attention to factors other than GPA when making their admissions decisions.

Hi Nathan! I will be visiting Princeton on April 23-25, and I will come there next year. I think I've figured out how to talk to math majors and which classes to observe, though I do have one question: Do you happen to know what Math 453 (Advanced Topics in Algebra) is covering this semester?

[quote="calc rulz"]Do you happen to know what Math 453 (Advanced Topics in Algebra) is covering this semester?[/quote]
I believe that's Sarnak's class. Definitely worth visiting, no matter what he's talking about. I don't know the details but I imagine that if Sarnak gets to decide what he's teaching it's going to be analytic number theory... it wouldn't be the first time that Sarnak taught that kind of stuff in an "algebra" course.

Sarnak was my favorite professor at Princeton.

Enrolled.

Nathan, I remember a post you made about your courses in this forum. You did a reading course on geometric measure theory in the spring of your sophomore year. How easy/difficult is it to do these things? I mean, do you have to hang around all the time in the math department to get close enough to people to ask them for reading courses?

Did you have a burgeoning interest in analysis even in high school? Had you already picked it out as your major area of interest? I ask because I haven't picked out a single area that I want to focus on. I'm looking forward to studying everything between complex analysis and combinatorics.

Also, what kind of senior theses do people in the math department write? I've been looking around but I can't find anything on the senior thesis online.

Cheers,

Ashwath

[quote="ashwath.rabindranath"]You did a reading course on geometric measure theory in the spring of your sophomore year. How easy/difficult is it to do these things? I mean, do you have to hang around all the time in the math department to get close enough to people to ask them for reading courses?[/quote]
No. I took three of the four classes in the analysis sequence during my 2nd and 3rd semesters. The fourth class was only offered in the fall, so I went to the guy who wrote the books for those 4 classes (Stein) and asked him if I could take his graduate analysis class or if it was not accessible to someone in my position. He said that the class was aimed at students who were doing their PhD research on that topic and suggested I take Fefferman's graduate class. So I went to Fefferman, but he said his class wasn't going to be a regular course, so I could do that or I could do a reading course, so I suggested a reading course topic and he was okay with that. I hadn't talked to either of those professors before that.

[quote="ashwath.rabindranath"]Did you have a burgeoning interest in analysis even in high school? Had you already picked it out as your major area of interest?[/quote]
No and no. In high school I liked math, but I didn't have specific interest. I originally thought I'd major in physics. But when I got to Princeton, the math class I took freshman fall that was really awesome was the intro to analysis, so in the spring I took two upper level analysis classes, and that's how it all started.

[quote="ashwath.rabindranath"]Also, what kind of senior theses do people in the math department write? I've been looking around but I can't find anything on the senior thesis online. [/quote]
There should be something online but I don't actually know where to look. People do a pretty wide variety of theses in the math dept. There aren't very specific requirements. A common thing to do is to study an advanced topic you're interested in and write a clear exposition of a key theorem or technique. You could also simplify a proof of an important result. I'd say those are the two most common "pure math" things you might do. (Original research is not required.) You can also find a field that has obscure results that are accessible but not yet settled (or a very new field), and do something like that. That's kind of what my thesis is like -- I'm generalizing an algorithm that my advisor published in 2007. You can also do a numerical study, work out a non-trivial example in some situation that's not yet well understood, apply a math concept in some field of science or engineering where it hasnt' been used before, etc. These things would be original work without being proofs of previously unknown material.

I have a friend who is trying to prove L vs NL (he's matrix3125 on this forum but I don't think he logs on anymore). His thesis is about his current progress, which is nontrivial but on its own doesn't really have any impact. But it's a pretty good thesis. Assuming he writes well also, I'd expect him to get an A. He intends to keep working on that problem in graduate school. A few years ago there was a girl who wrote a book aimed at undergraduate and high school students on how to write mathematical proofs.

Good questions.

[quote="calc rulz"]How important is it to take 200-level classes before moving on? [/quote]
Issue 1: they are departmental requirements. You can talk to the undergraduate advisor in the dept to find out about getting these waived. I didn't take 218 but I took many more analysis classes so they didn't care. Typically if you skip a 200 level class they will expect you to take a higher class in the same field.
Issue 2: material covered. They aren't that intense. If you feel like you know most of the material the class will teach, skipping it and learning that stuff on your own as necessary is fine. In fact this is more to the "style" of the Princeton math department than taking the class would be.
Issue 3: you'll run out of classes. If you skip a lot of the intro classes, you might run out of math classes to take. This isn't an official problem, but if you don't have outside interests you might be frustrated by having to take a lot of non-departmental courses when you're an upperclassman. This wasn't a problem for me because there were other topics I was interested in and it was easy to find courses in other departments that were cool.
Issue 4: practice. The Math Dept doesn't force us to practice doing problems much. If you skip to upper level classes you're making this even worse. But if you do problems on your own then you'll be fine.
Issue 5: cool profs. Sometimes awesome professors teach those classes and that can make them more worth taking. Also some profs teach extra material (e.g. Conway or Fefferman) which might make the classes more interesting.

Those would be my only concerns. The philosophy in the math dept is that you can do whatever you want. So talk to a few more people, decide something, and don't let anyone stop you from doing it!

[quote="calc rulz"]How important is it to take 200-level classes before moving on? I'm definitely going to take 218, although I'd like to take 322 (Galois Theory) and 331 (Complex Analysis) my first semester. I already have much experience with rigorous proofs and abstract mathematics, and I've seen most of the topics on 215 done rigorously, so I would imagine that I could skip that.

As for 217, I already have a basic understanding of linear algebra (especially abstract linear algebra over a field), so the beginning of the course would not be quite as interesting, although I don't know a lot of the later material in the course (mostly material relating to canonical forms, bilinear forms, adjoints, and the relation between adjoints and eigenvectors). I could probably study what I don't know over the summer and maybe later if it comes up in other courses.

This begs the more general question, is it important to know all of the 200-level courses really well, or is it enough if you have a basic knowledge and feel ready for the later courses you plan to take? One example concern is that I'm interested in number theory and might want to do quadratic forms, and so it might be important to have an extremely good understanding of bilinear forms. On the other hand, I could go through the chapter of Hoffman and Kunze (the textbook at Princeton) on bilinear forms and do most of the problems. This might not be quite as satisfying as taking a class on the material (especially due to the benefits you get from a professor's perspective), although it would at least mean that I'd know it reasonably well.[/quote]
What sort of things helped you get in? did you make MOP?

Calc Rulz, I was in a similar situation as you when I got to Princeton and I was advised to either sit in on or simultanously enroll in the 217/218 classes as well as a 300 level class. That way I wouldn't miss anything from the 200 level classes and I could be exposed to some new material in a 300 level course. I was glad that I did. Although I knew a lot of Linear Algebra and basic analysis before I got there, I still learned a fair bit from those 200 level classes. And more importantly, I got to know my classmates better in those classes than any other.

I also suspect that some years (depending on whose teaching it) they will cover material that isn't normally covered in such a class, but that can be very helpful for future classes -- like calculus on manifolds, etc.

My general opinion is to err on the side of caution your first semester until you get your feet wet. So even if you do take a 300 level course, take one that you really feel like you are prepared for. Not one that would be a stretch for you. The goal is to learn things [i]deeply[/i], not just quickly. In college, quality of understanding is more important than quantity of exposure.

I learned this the hard way by skipping 323 (intro to abstract algebra) and going into 324 (topics in abstract algebra - galois theory). I did fine in the class, but when I tried to take a 400 level class in algebra (commutative algebra), my deficiencies really got exposed and it came back to haunt me. To this day I still have serious deficiencies in my understanding of abstract algebra.

So I've committed to Princeton, but am still very unsure about a major or field of study (which I have no problem with, as I know you don't have to declare until the end of your sophomore year), but I guess my question involves what being a math major entails.  I've always been good at math, I've worked my way through Calculus 3 (multivariable) without any problem, but I'm by no means a math genius...  And everyone, especially my parents, say you have to be a genius to major in math at Princeton.  Is this true, or will hard work and a good understanding of math be enough?  If you can shed any light on the integrated sciences program as well I'd love to hear opinions of that.  Thanks so much!

[quote="byrdgyrl"]I'm by no means a math genius...  And everyone, especially my parents, say you have to be a genius to major in math at Princeton.  Is this true, or will hard work and a good understanding of math be enough?[/quote]
This is a complicated question. At the risk of not sounding like I'm encouraging math at Princeton, I'm going to have to say "sort of". The issue is that the Math Dept does not really have a support system for undergraduates. Some courses occasionally have review sessions or problem sessions by very unmotivated grad students. But that's about it. For the most part, you are expected to understand the material on your own, do the work on your own, and if you have difficulty, you have to seek out help on your own. In the advanced classes, the profs typically present the most abstract part of the material -- state the theorems and give the proofs -- but they don't spend much time teaching you how to use the tools of that field of math. You're supposed to figure that out for yourself when you do the homework.

So, there are a few different ways to survive in the Math Dept.
1) Be a genius.
2) Work extremely hard. Harder than anyone in any other major has to. (Although some people in other majors choose to work that hard too. But they don't need to do it to get through their department.)
3) Find some other like-minded people and create a support group of your own.

Everyone who makes it to the end did one of these three things. There's a sort of 4th category of people, who decided they didn't like the Math Dept but not until it was too late. Those people tend to do the bare minimum to meet the Math Dept requirements and focus on some other thing that they decided to do after graduation. Generally, people who aren't geniuses and major in math end up wanting to do something other than pure math research after graduation. That doesn't mean that if you aren't a genius, the Princeton Math Dept is definitely going to make you stop liking math. I still love math, I'm just going to do other stuff after graduation. I know some other people who now want to do things like teach math, do applied math or computer science, etc. Of course, those are normal things for people who major in math to go on to do. The thing that's different about them (perhaps) is that they weren't planning on that all along; they originally thought they'd be doing math for the rest of their lives.

Overall, I would say that the Math Dept is a great department for the people who take to it. But there are also a lot of people who don't like their style. Oddly enough, I don't really fit into this framework. I don't think I am bright enough to be in the genius category. I sort of was in (3) freshman year, but after that I pretty much worked alone. But I didn't work super hard, so I wasn't in (2). I guess for soph - junior year I was kind of on the border between (1) and (3). Then at the end of junior year I got interested in other stuff and basically stopped doing a lot of math. So I migrated to (4). And since I was for a time in or nearly in (1), but still ended up in (4), I have mixed feelings about the math dept. I learned a LOT of math and I had a lot of fun learning math. But I still decided not to go on studying math after graduation.

I'm sorry for not giving a very clear answer. Take what you will from that long-winded description. Also, bear in mind that other people may have very different opinions, not only of the Math Dept but also of the nature of the "typical" math major at Princeton. One way to find out is to take the first few classes, spend some time in Fine Hall, and get to know the other people who are interested in math in your year and the years above you. Then you can make a more informed decision.

[quote]If you can shed any light on the integrated sciences program as well I'd love to hear opinions of that.  Thanks so much![/quote]

I don't know much about it. I don't have any good friends who did it. I've heard that it's challenging but teaches you a lot and is fun. I also heard that a common reason for people to drop out of it is that they realize that they only care about some of the topics and it's not worth it to keep studying everything at once. So that suggests it's a good opportunity for figuring out what sort of science you most like to study.

[quote="azjps"]Thanks much for the previous post, as I'm in a very similar position, with only light exposure to abstract math (I have enrolled at princeton also  :) ).

Edit: as discouraging as the post seems, since I am very far from (1).  :P[/quote]
gl then

[quote="azjps"]Edit: as discouraging as the post seems, since I am very far from (1).  :P[/quote]
Hey, don't let that scare you away. Some of the people I know who seemed to have the most interesting, fun, and/or life-changing experiences at Princeton were math majors who were far from (1). ;)

Wow, a lot of time has passed! I might as well share my wisdom from having now gone through Princeton:

I ended up ignoring the 200-level courses almost entirely, and I took a 300-level class (complex analysis) and a 400-level class (algebraic number theory). The latter was extremely helpful, as well as the class field theory course I took in the spring, since it gave me a head start in number theory, which is now my main area (I'm a math phd student at MIT working with Bjorn Poonen). I feel that I understand things particularly well because I've known them for years. Complex analysis was fine - I already knew a lot of what was in that course, albeit without having done problems. I actually had a little trouble because my analysis was shaky, and it's always plagued me a tiny bit, but it's not too relevant to my area, and after TAing an analysis course at MIT, I realize I do know the material well enough.

In case you're wondering how I managed to let people do that: As for the level, the math department at Princeton doesn't really care, so there's no issue. I got some advice from the algebraic number theory professor, Chris Skinner, who is still there and a helpful resource. As for taking two courses in the same area my first semester, which you're not supposed to do: I just signed up online without telling my freshman advisor!

I also took 218 (multivariable analysis) in the spring, in part to offset class field theory, because the latter is an advanced subject. In hindsight, I probably didn't really need 218 and could have done something like Fourier.

Anyway, for the most part, I think things have gone pretty well. I took some more courses in sophomore year, then focused on JP's after that. My first JP was on Galois representations, and it was because I took algebraic number theory earlier on that I managed to get to a topic like Galois representations by junior year.

PS: I'm curious whether any of the posters here are people I met at Princeton. It's often hard to figure out people's names on AoPS.


Find the least positive integer $n$ such that

$\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}$

[hide]
It appears that $n$ will be small by looking at how big the sum could get, so multiplying the numerator and denominator of each term by $\sin 1^\circ$ might help.

\[\frac{1}{\sin k^\circ \sin (k+1)^\circ}=\frac{\sin 1^\circ }{\sin k^\circ \sin (k+1)^\circ \sin 1^\circ}=\frac{\sin (k+1)^\circ \cos k^\circ-\cos (k+1)^\circ \sin k^\circ }{\sin k^\circ \sin (k+1)^\circ \sin 1^\circ }\]
\[=\frac{\cot k^\circ-\cot (k+1)^\circ }{\sin 1^\circ }\]
Summing odd $k$ from 45 to 133, the sum is equal to
\[\frac{\cot 45^\circ-\cot 46^\circ+\cot 47^\circ-\cot 48^\circ+\cdots-\cot 134^\circ }{\sin 1^\circ }\]
But for $j$ between 46 and 89, inclusive, we can pair $\cot j^\circ$ with $\cot (180-j)^\circ =-\cot j^\circ$, these cancel, and $\cot 90^\circ =0$. So the sum is just equal to
\[\frac{\cot 45^\circ }{\sin 1^\circ }=\frac{1}{\sin 1^\circ }\]
so $n=\boxed{1}$.
[/hide]

[quote="mna851"]Is there no other way, I mean besides randomly guessing that $\ 1$ is the correct answer.[/quote]

yea there's no other way but guessing and certainly not scorpius' way.

waht?


inequalities
LaTeX
inequalities unsolved


group theory
superior algebra
superior algebra unsolved

Find all  groups having 22 elements.

I know two such groups : dihedral group $D_{11}$ and $Z\slash(22Z)=Z_{22}$
is this all ?
how can I prove it ?

There are no other groups of order 22.

Hint:
1) There exists an element $a$ of order 11 in $G$.
2) There exists an element $p$ of order 2 in $G$.
Denote the cyclic group generated by $a$ by $H$.

3) $G = H \cup p H$.
4) The map $x\mapsto p x p$ is an involutive automorphism of $H$. But there are only two involutive automorphisms of $H \sim \mathbb{Z}/(11\mathbb{Z})$: $x\mapsto x$ (the identity) and $x\mapsto-x$.
5) In the former case, $G$ is $\mathbb{Z}/(22\mathbb{Z})$, in the latter case it is $D_{11}$.

[quote="DucHuynh"]Can we just use the fact that a group $G$ of order $2p$, where $p > 2$, is either isomorphic to $D_{p}$  or $Z_{2p}$?[/quote]
We can.  :D

But I think the point of this problem was to prove this fact for $p=11$.


Prove that there exist infinitely many integer solutions to the equation \[(a - b)(b - c)(c - a) = a^2 + b^2 + c^2.\]

This has been solved at least twice on the forum, I think.


vector
probability
expected value
real analysis
real analysis unsolved

$u_1,..,u_n$ some vectors of $R^n$ with norm less or equal to $1$.
$v=\sum_1^n x_iu_i$ where $x_i \in [0;1]$. Prove that there exists integers $e_i \in \{0;1\}$ satisfying:
$|v-\sum_1^n e_iu_i| \leq \frac{\sqrt{n}}{2}$

The statement appears to be nonsense. Where are the $\epsilon_i$?

it's clear he want to write e_i   instead of x_i in the last formula  :)

Let's try a probabilistic argument.  For each $i$ independently, let $e_i$ equal 1 with probability $x_i$, and equal 0 otherwise.  Now let's analyze the expected value of the norm squared:
\begin{eqnarray*} E(|v - \sum_{i = 1}^n e_i u_i|^2) & = & E(|\sum_{i=1}^n x_i u_i - \sum_{i=1}^n e_i u_i|^2) \\ & = & E(|\sum_{i=1}^n (x_i - e_i) u_i|^2) \\ & = & E(\langle \sum_{i=1}^n (x_i - e_i) u_i, \sum_{i=1}^n (x_i - e_i) u_i \rangle) \\ & = & E(\sum_{i=1}^n \sum_{j=1}^n (x_i-e_i)(x_j-e_j)\langle u_i, u_j \rangle) \\ & = & \sum_{i=1}^n \sum_{j=1}^n E((x_i-e_i)(x_j-e_j)) \langle u_i, u_j \rangle \\ & = & \sum_{i=1}^n E((x_i-e_i)^2) \langle u_i, u_i \rangle \\ & = & \sum_{i=1}^n E((x_i-e_i)^2) |u_i|^2 \\ & \le & \sum_{i=1}^n E((x_i - e_i)^2) \\ & \le & \sum_{i=1}^n \frac{1}{4} \\ & = & \frac{n}{4} \, . \end{eqnarray*}
Thus there is a choice of the $e_i$ such that
\[ |v - \sum_{i = 1}^n e_i u_i|^2 \le \frac{n}{4} \, .  \]
Taking square root of both sides completes the proof.

yes  :)  also the proof I found. (but it was in my probability class, so it was easier to think to this approach  :D )


[quote="Extremal"]thanks people everything is oki . I am in team of my republic yes there was olympioc and I win[/quote]

That's wonderful to hear! Awesome job, and good luck on the IOI.

P.S. to those confused, he was trying to find out about his country's math olympiad earlier.


inequalities
limit
LaTeX
calculus
calculus computations

Hi! i've got a problem:
let the $a_n$ be the sequence of positive numbers .
Prove that   $\limsup_{n\to\infty}(\frac{a_1+a_{n+1}}{a_n})^{n}\geq e$ :blush:

First of all assume that the limit exists. (Is it provided??? It must be.)

Then calculate the limit: $\lim \left(\frac{a_1+a_{n+1}}{a_n}\right)=e^{\lim \frac{n(a_{n+1}-a_n+a_1)}{a_n}}$.
Assume that the conclusion is false. Then we will have that $\lim \frac{n(a_{n+1}-a_n+a_1)}{a_n}=1-\epsilon$ with $\epsilon \ge 0$.

So, for sufficiently large $n$ we get that $\frac{n(a_{n+1}-a_n+a_1)}{a_n} \le \ell <1$ for some $\ell$. This is equivalent with $a_n(1+\frac \ell n)\ge a_{n+1}+a_1 \Rightarrow (a_n+kn)(1+\frac \ell n)\ge a(n+1)+k(n+1)$ where $k=\frac{a_1}{1-\ell}>0$. Denote by $b_n=a_n+kn$. We have that $b_n \le b_N \prod_{i=N}^n (1+\frac \ell i)$.

Using the fact that $\lim_{n\t\infty} b_n-k\cdot n=b_N \prod_{i=N}^n (1+\frac \ell i)-k\cdot n<0$ (I let you the pleasure to prove it :P ). Hence $a_n<0$ for sufficiently large n. Contradiction. Hence, the inequality holds.

Rustam XIRTI chatoq ekan.Meni ham bitta masalamni yechib berdi.

I have another solution:
     It is suffician to prove that  $\limsup_{n\to\infty}(\frac{n(a_1+a_{n+1})}{(n+1)a_n})^n\geq 1$. Assume to the contary.       Then there exists $N:$for any $n>N$ holds  $\frac{n(a_1+a_{n+1})}{(n+1)a_n}<1$  or          $\frac{a_1}{n+1} +\frac{a_{n+1}}{n+1}<\frac{a_n}{n}$.

Summing those inequalityies from $n=N$ till the some n wich will tend to +00 we get that
       $\frac{a_n}{n}< \frac{a_N}{N}-a_1(\frac{1}{N+1}+\frac{1}{N+2}+...+\frac{1}{n})=\frac{a_N}{N}+a_1( 1+\frac{1}{2}+...+\frac{1}{N})-a_1(1+ \frac{1}{2}+...+\frac{1}{n})$.

So $\frac{a_n}{n}< c-a_1(1+ \frac{1}{2}+...+\frac{1}{n})$ where c is some constant.It follows that
$-a_n\to\infty$        Contradiction.

O'lay agar anov xirtini echimini tushuingan bo'lsam. Orifga she'r: Misolnii echmay turib echdim dema, misol sharti hato edi jo'ra!!! Bo'ldi zo'r askiya,hahaha!

Thanx for attention :pilot:

Nice proof :lol: ! The simbol in $\LaTeX$ for superor limit is \limsup. My proof works also taking a subsequence of $a_n$ which tends to $\limsup a_n$. Anyway, very nice problem.


inequalities vector trigonometry quadratics algebra polynomial conics

Could someone prove it easily explaining what they are doing?  :blush:

There is a pretty good proof of it in the Aops Wiki.

The proof in two variables is very easy:

$(a^{2}+b^{2}) \cdot (c^{2}+d^{2}) = (ac+bd)^{2}+(ad-bc)^{2}\ge (ac+bd)^{2}$

There are many resources to go to for other proofs, including Wikipedia or Mathworld (both of which are pretty easily searched with Google).

The general inequality is usually stated in terms of vectors, so you might not like that proof, even if it is the shortest one.

That is what I meant. Vectors. Is the proof somewhat like the other one?

EDIT: ok wtf "similar to that one?" and my whole point isn't avoiding AoPS wiki it is avoiding sigmas

as i mentioned sigmas scare me but i will try to work out what you are saying because it is different then what i've seen

yeah i see  :lol:
it is helpful but i am scared of sigmas i just mean i don't have much experience with them but i can still work it out :)

[quote="funcia"]avoiding sigmas[/quote]

A bad idea, since they're a very compact and efficient notation for expressing ideas.  You'll be seeing them a lot.  It's kind of like saying "I want to avoid algebra," and expressing all your ideas in words (which ancient mathematicians did for a long time!).  If you've ever tried it, it's very time-consuming.

boxedexe's first proof is a nice elementary one, but it needs some explanation.  He constructs a quadratic polynomial (a parabola) from a sum of squares, which must be non-negative.  That means that it either has one or zero solutions, which means that its discriminant, $b^{2}-4ac$, is either negative or zero.

His second proof, the vector one, rests on the fact that the dot product of two vectors is commutative, and is somewhat high-level.

don't worry that is covered in my confusing text book(some parts like this i seem not to get) :maybe:

yeah i suppose i should be glad sigmas exist but is so hard for me to think of them in a mathematical way after taking so much Greek
of course, this is a foolish reason so i now have a new thing to do:
-get comfortable with sigmas

oh  and thanks for that proof i will be studying it :wink:

EDIT: about the algebra skipping thing i was going to do it until i got forced to do algebra

Hmm.  I probably should've pointed that out, but I wasn't sure whether you could also define angles in $\mathbb{R}^{n}$ alternately by considering the plane in which two vectors reside...

Stick to the quadratic proof by now.  There's also a proof by induction (I think?) and by Rearrangement.

I believe $\cos\theta=\frac{a\cdot b}{|a\parallel b|}$ is also defined in $\mathbb{R}^{n}$ in an abstract manner.  :wink:

Also,
\[\left(\sum_{i=1}^{n}a_{i}^{2}\right)\left(\sum_{i=1}^{n}b_{i}^{2}\right)-\left(\sum_{i=1}^{n}a_{i}b_{i}\right)^{2}=\sum_{1\leq i<j\leq n}(a_{i}b_{j}-a_{j}b_{i})^{2}\geq 0\]
Masoud Zargar

[quote="t0rajir0u"]There's also a proof by induction (I think?) and by Rearrangement.[/quote]
The proof by re-arrangement (if I worked it out correctly):

Expand $\left(\sum_{i}a_{i}^{2}\right)\left(\sum_{i}b_{i}^{2}\right) \geq \left(\sum_{i}a_{i}b_{i}\right)^{2}$ to get

$\sum_{i,j}a_{i}^{2}b_{j}^{2}\geq \sum_{i}a_{i}^{2}b_{i}^{2}+\sum_{i \neq j}a_{i}b_{i}a_{j}b_{j}$

$\Rightarrow \sum_{i \neq j}a_{i}^{2}b_{j}^{2}\geq \sum_{i \neq j}a_{i}b_{i}a_{j}b_{j}$

Now by the re-arrangement inequality on the sequence $a_{i}b_{j}, a_{j}b_{i}$ twice, we have

$a_{i}^{2}b_{j}^{2}+a_{j}^{2}b_{i}^{2}\geq (a_{i}b_{j})(a_{j}b_{i})+(a_{j}b_{i})(a_{i}b_{j})$

Now, summing over all $i > j$, we have

$\sum_{i > j}a_{i}^{2}b_{j}^{2}+a_{j}^{2}b_{i}^{2}\geq \sum_{i > j}(a_{i}b_{j})(a_{j}b_{i})+(a_{j}b_{i})(a_{i}b_{j})$

which is equivalent to

$\sum_{i \neq j}a_{i}^{2}b_{j}^{2}\geq \sum_{i \neq j}a_{i}b_{i}a_{j}b_{j}$

which proves Cauchy-Schwartz.  We also get from the re-arrangement inequality that CS has equality when $a_{i}b_{j}= a_{j}b_{i}$, or $a_{i}/ b_{i}= a_{j}/ b_{j}$.

A proof by induction: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=120498[/url].
[quote="zanttrang"][quote="krustyteklown"]the second proof is actually bogus...
the angle between 2 vectors in R^n is defined to be inverse cos (RHS/LHS of CS), and you know that its <1 because of CS...[/quote]
You can prove $a \cdot b = |a| |b| \cos \theta$ without CS if you apply the Law of Cosines to the triangle formed by the two vectors (this works in two or three dimensions, and formally you may apply the same method in $\mathbb{R}^{n}$, but that I'm not sure whether that's rigorous).[/quote]
See here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=120729[/url].


vector
analytic geometry
algorithm
Functional Analysis
superior algebra
superior algebra unsolved

Hi,
Let's suppose I have a Hilbert space and a set of *non-orthogonal* vectors which is a base for this space.

I define an arbitrary vector F, and I want to compute the coordinates of this vector in the given basis. Is it possible to do it?

As far as I know, taking the inner product of F with each vector of the base yields the coordinates of the *dual base* but I can't go much further than that.

Thanks in advance!

Due to abstract reasons it is possible. To make it more operational, first transform the initial basis to an orthogonal basis using the [url=http://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process]Gram-Schmidt algorithm[/url]. Then expand F with respect to the orthogonal basis and re-transform to get the representation in terms of the initial basis.


For some reason, I cannot add images to my userpage. Is this intentional or unintentional?

[url]http://www.artofproblemsolving.com/Wiki/index.php/User:7h3.D3m0n.117[/url]

I tried to attach my avatar picture but it won't work.

I fixed it; it was a syntax error.

I think it used to be possible on the wiki to just type in the url of an image and it would automatically show the image, but that method is deprecated. See [url=http://www.artofproblemsolving.com/Wiki/index.php/Help:Images]Help:Images[/url]: go to "Upload file" on the toolbar, and browse for the image to be uploaded. Then use that name to add the image to an article.


The function $f$ is not defined for $x=0$, but, for all non-zero real numbers $x$, $f(x)+2f\left( \frac1x \right)=3x$. The equation $f(x)=f(-x)$ is satisfied by

$\text{(A)} ~\text{exactly one real number}$
$\text{(B)}~\text{exactly two real numbers}$
$\text{(C)} ~\text{no real numbers}$
$\text{(D)} ~\text{infinitely many, but not all, non-zero real numbers}$
$\text{(E)} ~\text{all non-zero real numbers}$

[hide="Hint"]Notice that this function is cyclic. [/hide]

I must be missing something . . .

[hide] Note that if x is not 1 or -1, we have f(x)+f(1/x)=x and 1/x, which are different numbers, so the function is not even defined for x other than 1 and -1?  In this case, f(1)=1/2 and f(-1)=-1/2, so the answer is C.[/hide]

[hide="ummmm"]For x to satisfy $f(x)=f(-x)$, x=-x.
The only number that satisfies this is 0. The function is undefined for this number, so there are no solutions. [/hide]

for $x$ to satisfy $f(x)=f(-x)$, $x$ must equal $-x$
the only number that works for $x = -x$ is $0$, but $0$ is undefined for the function $f$
there are no solutions

surely somebody told you to always post a solution with your answer@

A word of caution in 236factorial's and keeyounglee's answers: the deduction is flawed. What happens if $f(x)=x^2$? Clearly, $f(x)=f(-x)$ but $x\ne-x$... You still need to use that equation given by the problem in order to solve the problem.

I think this problem has provided a sensible condition , i.e f(x)+f(1/x)=x …(1) ,If we change x to 1/x ,what will happen? That is f(1/x)+f(x)=1/x …(2) . compare (1),(2),therefore x=1/x ,then x=±1 , this is the domain of this function f(x) ,to have a test we can get f(1)=1/2 and f(-1)=-1/2 ,so f(1) is not equal to f(-1) , therefore there’s no one solution for this equation!
(I think it’s no need to discuss this function is odd or even although it’s an odd one!)

DoubleIntegral

[quote="4everwise"]The function $f$ is not defined for $x=0$, but, for all non-zero real numbers $x$, $f(x)+f\left( \frac1x \right)=x$. The equation $f(x)=f(-x)$ is satisfied by

$\text{(A)} ~\text{exactly one real number}$
$\text{(B)}~\text{exactly two real numbers}$
$\text{(C)} ~\text{no real numbers}$
$\text{(D)} ~\text{infinitely many, but not all, non-zero real numbers}$
$\text{(E)} ~\text{all non-zero real numbers}$[/quote]

The actual problem had $f(x) + 2f\left( \frac1x \right)=3x$ (I have a PDF of the test). In that case, is [hide]$B$[/hide] correct?

[hide="My solution"]
Replace $x$ with $\dfrac{1}{x}$ to get $f\left( \frac1x \right) + 2f(x) = \dfrac{3}{x}.$ Solving the following system

\[f(x) + 2f\left( \frac1x \right)=3x\]
\[f\left( \frac1x \right) + 2f(x) = \dfrac{3}{x}\]

for $f(x)$ gives $f(x) = -x+\dfrac{2}{x}.$

Then, $f(-x) = x - \dfrac{2}{x},$ and so we have \[x - \dfrac{2}{x} = -x + \dfrac{2}{x}\] \[2x = \dfrac{4}{x}\] \[x^2 = 2\] Thus $x = \pm \sqrt 2.$
[/hide]

@ MSTang to solve the problem posted you should probably use x instead of 3x and 1/x instead of 3/x on the RHS :D
Then you get $f(x)=\frac{2}{3x}-\frac{x}{3}$ and there are still only 2 values satisfying f(x)=f(-x) (which incidentally are still $\sqrt{2},-\sqrt{2}$)

[hide="lol sniped by a bunch of people, whatever"]Plugging in $1/x$ for $x$ we have \[f(1/x)+f(x)=1/x.\] But \[f(x)+f(1/x)=x\] so $x=1/x$. This $x=-1,1$. But $f(-1)\neq f(1)$ so there are $\boxed{\text{No solutions. (C)}}$[/hide]

I guess, although this thread was linked to from the contest section, so the problem should be right. Mod, can I start a new thread with the correct problem, and have you fix the link?

Is it really that big of a deal? Its not like this is a unique problem which we are doing injustice to so I think this is just fine.

To get the actually prob just go to the famed contests section :)


I need help with Cal 2 homework if there is anyone that can help me please contact me at Ashbucket22@aol.com

Thanks

Normally, you would post the question and then people would post the solution.

Normally, you shouldn't be a cheater and you should do your own homework.


Find all functions $ f$ such that:

$ f(f(x)\plus{}y)\equal{}f(x^2\minus{}y)\plus{}4f(x)y$ ; $ x$,$ y \in \mathbb{R}$

Take $ y\equal{}\minus{}f(x)$ and then $ y\equal{}x^2$ and then you have to prove that $ f(x)\equal{}x^2$,that is imediately

Yeah that's what I did, but I have a doubt, we get $ x^2f(x)\equal{}[f(x)]^2$, we'll have that if for some $ a\in\mathbb{R}$, $ f(a)\neq 0$, then: $ f(a)\equal{}a^2$, but what if for this $ a$, we have $ f(a)\equal{}0$?

Yes, the zero function is also a solution.

To show there aren't any others, suppose there exists nonzero $ a,b$ such that $ f(a)\equal{}0$ but $ f(b)\equal{}b^2$. Since we must have $ f(0)\equal{}0$ either way, plugging in $ x\equal{}0$ gives $ f$ even. So WLOG let $ b$ be negative.

Then plugging in $ x\equal{}a,y\equal{}b$ gives $ f(a^2\minus{}b)\equal{}b^2$. As $ b\neq 0$, this forces $ f(a^2\minus{}b)\equal{}(a^2\minus{}b)^2$. So we get
\[ (a^2\minus{}b)^2\equal{}b^2\Rightarrow 0\equal{}a^2(a^2\minus{}2b)\]
But $ a\neq 0$ and $ b<0$, so each factor is positive, a contradiction. Then $ f(x)\equal{}0$ for all $ x$ or $ f(x)\equal{}x^2$ for all $ x$.