[hide]The ratio of perimeters $\frac{5}{3}$, so squaring this results in $\frac{25}{9}$, which is the ratio of the areas.[/hide]

[quote="mr. math"]The ratio of the perimeters of two similar octagons is 5:3. What is the ratio of their areas?[/quote]
[hide]This applies for any polygon, just square it to get $\frac{25}{9}$ unless you want it the other way around or something.[/hide]

That is a good way of doing the solution.

But to save some time, try to realize that the areas are proportional to the squares of the sidelengths in similar regular polygons.

Well with this problem, the statement implies that the answer is invariant so one can assume WLOTMG that the side is 10 or something.

[quote="mathgeniuse^ln(x)"]Yeah, and what does WLOTMG mean?

Tell me that you didn't mean it to be Without Loss of That Much Generality.[/quote]
Change the That to be a Too. You can assume some stuff Without Loss Of Too Much Generality in many problems, it's really quite a handy technique.

Here is the rule:

Ratio of perimeters (or any other thing having to do with lengths)

$\textrm{Ratio}$

Ratio of surface areas (or any other thing having to do with areas)

$(\textrm{Ratio})^{2}$

Ratio of volume (or any other thing having to do with 3-D)

$(\textrm{Ratio})^{3}$

So for this problem, the answer is $(\textrm{Ratio})^{2}$, which is ${\frac{25}{9}}$

yeah... that's how I would have figured it out. Since that holds true for any polygon.


Prove the following result:
If n is odd or if 4|n, then $\sigma (n) \leq n^{\frac{3}{2}}$.

We shall first prove it for n odd. decompose it into p_i^n_i. for pi, odd primes and ni, naturals. Since sigma(n) is multiplicative. It is sufficient to show that the result holds for n = pi^n_i.

By trivial evaluation, we get  sigma(p_i^n_i) = (p_i^(n_i + 1) - 1)/(p_i - 1). Thus we must show that,
(p_i^(n_i + 1) - 1) <= p_i^(3/2*n_i)*(p_i - 1).

We will show the stronger result that: p_i^(n_i + 1) <=  p_i^(3/2*n_i)*(p_i - 1), or equivalently that

1 <= p^((n-2)/2)*(p-1), and minimizing n (ie: n = 1) we have 1 <= (p-1)/sqrt(p) or sqrt(p) + 1 <= p, which by inspection holds for all odd primes p, and thus our result is true for all odd integers n.

In the case where 4|n, note that sigma(2^k) = 2^(k+1) - 1 <= (2^(3k/2)), holds for all k>=2. Thus if n = 2^k*n1, with k>=2, the result holds.

Toodles

Great proof. Like how you used

In the case where 4|n, note that sigma(2^k) = 2^(k+1) - 1  (2^(3k/2)), holds for all k2. Thus if n = 2^k*n1, with k2, the result holds.

This is no longer an unsolved problem.


i'll try to describe this diagram..
two circle quadrants intersect and they both share the same radius of 5cm, so there's 3 segments to the diagram. one big triangly shaped thing and two other small slices of cake with a curved side. how would u find the area of one of the small segments?

Is it like this?
[img]http://img.photobucket.com/albums/v57/ccy111/prob/dia.gif[/img]
Could you make a crude drawing or something?

(: it looks like this
[img]http://www.imagestation.com/picture/sraid167/p76331e7d9d2388c7f31e7dfa9ca7ef9a/f4323391.jpg[/img]
haha i sorta figured it alr (came to me in my sleep, would you imagine :/)
by drawing an equilateral with sides 5cm in the centre part, u'd get  :pi:/3 as the angle of the sector. with that we can find area of segment too, and minus it frm area of the smaller sector of angle :pi:/6
thanks anws! good day t you

omg where's the diagram.
[img]http://i6.photobucket.com/albums/y210/mywilderness/quadrants.bmp[/img]


Έστω $ a,b,c$ μη αρνητικοί πραγματικοί ώστε $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}1$. Δείξτε ότι $ 4\left(a\plus{}b\plus{}c\right)\leq \sqrt{32}\plus{}9abc$

[quote="Γιώργος"]Έστω $ a,b,c$ μη αρνητικοί πραγματικοί ώστε $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1$. Δείξτε ότι $ 4\left(a \plus{} b \plus{} c\right)\leq \sqrt {32} \plus{} 9abc$[/quote]
Οντος πολυ περιεργη...
Δεν την εχω λυσει ακομα :wallbash_red:
Απο περιεργια:που την βρηκες  αυτην???

Hmmm... έστω $ F(x,y,z) \equal{} 4(x \plus{} y \plus{} z) \minus{} 9xyz$ όπου $ (x,y,z)$ σημεία της μοναδιαίας σφαίρας στο θετικό ογδοημόριο (δηλ. $ x,y,z\geq0$ και $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$). Τότε, λόγω συμμετρίας, αν η $ F$ εμφανίζει μέγιστο στο $ (a,b,c)$, θα πρέπει να εμφανίζει μέγιστο και σε όλες τις μεταθέσεις $ (a,c,b)$, $ (b,a,c)$, $ (c,a,b)$ κλπ.

[hide]Αν τα $ a,b,c$ δεν μπορούν να είναι διαφορετικά ανά δύο (γιατί??), σας βοηθάει καθόλου? Τι γίνεται π.χ. όταν το $ (a,b,c)$ ανήκει σε κάποιο από τα ακριανά τεταρτοκύκλια (π.χ. $ a\equal{}b$, $ c\equal{}0$)?[/hide]

Cheerio,

Durandal 1707

Θα χρησιμοποιησω την ανισοτητα shur, κατα την οποια $ (a \plus{} b \plus{} c)^3 \plus{} 9abc\geq 4(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$.
Θετω $ a \plus{} b \plus{} c \equal{} S$ ,$ ab \plus{} bc \plus{} ca \equal{} Q$ ,$ abc \equal{} P$.

Απο υποθεση παιρνουμε οτι $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$, δηλαδη $ Q \equal{} \frac {S^2 \minus{} 1}{2}$.
Τοτε απο shur εχουμε $ S^3 \plus{} 9P\geq 4SQ$, δηλαδη με βαση την παραπανω $ S^3 \plus{} 9P\geq 2S^3 \minus{} 2S$, δηλαδη ισχυει $ 9P\geq S^3 \minus{} 2S$.

Αρκει $ 9P\geq 4S \minus{} \sqrt {32}$. Με βαση την παραπανω αρκει $ S^3 \minus{} 2S\geq 4S \minus{} \sqrt{32}$ η οποια καταληγει στην $ S^3 \minus{} 6S \plus{} 4\sqrt {2}\geq 0$.

Η τελευταια σχεση ειναι ισοδυναμη με την $ (S \minus{} \sqrt {2})^2(S \plus{} 2\sqrt {2})\geq 0$.


LaTeX
advanced fields
advanced fields unsolved

v'= -f(v) - w + I
w'= e(v-w)
where f(v)=v(v-a)(v-1), 0<a<(1/2), e>0

Show that the rest state (v_r, w_r) is unique. Hint: compute dI/dv and ensure that it is positive for all v.

Where do I start?

Please use some basic [[LaTeX]] to make your writing legible.  See also http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690

You've given a problem from your homework or something.  It has a hint.  And you say you don't know where to start?  How about starting with the hint and seeing what that gets you?


Let the numbers $ a,b,c$ be from interval $ [1;2]$. Prove the inequality:
$ \frac{a(b\plus{}c)}{b^{2}\plus{}c^{2}}\plus{}\frac{b(c\plus{}a)}{c^{2}\plus{}a^{2}}\plus{}\frac{c(a \plus{} b)}{a^{2}\plus{}b^{2}}<\equal{}4.5$.

[quote="lasha"]Let the numbers $ a,b,c$ be from interval $ [1;2]$. Prove the inequality:
$ \frac{a(b \plus{} c)}{b^{2} \plus{} c^{2}} \plus{} \frac{b(c \plus{} a)}{c^{2} \plus{} a^{2}} \plus{} \frac{c(a \plus{} b)}{a^{2} \plus{} b^{2}} \leq 4.5$.[/quote]


La urmatoarea inegalitate astept doua moduri de abordare.

Sorin scrie si tu in $\LaTeX$. Acum revenind la inegalitatea ta, as avea cateva precizari in legatura  cu ea:

 1. Nu este deloc originala. Este arhi-mega-ultra cunoscuta, dar e o problema dragutza asa pentru initiere. ;)
 2. Incearca, asa cum am scris si mai sus, sa scrii in Latex, ca In Word arata destul de nasol. ;)

 Anyway, sa trecem la problema:

 [b] Problema. [/b]

      Fie $a, b, c$ trei numere reale strict pozitive astfel incat $a+b+c=1$. Sa se arate ca:

               \[ \frac{1}{1+\sqrt{a}}+ \frac{1}{1+\sqrt{b}}+ \frac{1}{1+\sqrt{c}}\geq\frac{3}{2}.  \]

 [i] Solutie. [/i]

       Se stie ca pentru $x, y, z$ numere reale strict pozitive are loc iengalitatea \[ \left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\geq 9.  \]
        Acum luam $x=1+\sqrt{a}, y=1+\sqrt{b}, z=1+\sqrt{c}$ si aplicam inegalitatea de mai sus care completata cu inegalitatea Cauchy-Buniakovski-Schwarz sub forma $\sqrt{a}+\sqrt{b}+\sqrt{c}\leq\sqrt{3(a+b+c)}$ va da concluzia problemei. :)

Nu as vrea sa te jignesc Cezar, dar citeste mai bine ce am scris.Bine inteles ca problema e extrem de usoara cu Inegalitatea lui Cauchy, dar am zis ca astept[b] doua moduri [/b]de abordare.Asa ca mai gandeste-te.Anywhay, problema poate fi facuta in 3 moduri... ;)

Hai Cezare, nu mai gasesti alta rezolvare?

Sorine, Sorine, nu grabi lumea pentru ca graba strica treaba. Nu e frumos sa dai astfel cu tifla unor persoane ilustre cum e Cezar...

Inteleg ce zici, Tudor, dar astept de trei zile o alta rezolvare de la Cezar si inca nu am primit-o.Tu ce zici?Poti sa gasesti o alta rezolvare? :)

Mai Sorin, sa stii ca eu nu stau sa mai bat capul cu tot felul de probleme dintr-astea. As fi scris mai multe solutii la probleme mult mai interesante si mult mai instructive, asta in masura in care gasesc mai multe solutii. Uite, ca sa fiu sincer, mie inegalitatea asta nu-mi spune absolut nimic. ;)

Bine.Cum zici tu... :fool: Deci e o problema lipsita de importanta :( Nu-i nimik, ma straduiesc sa compun atunci probleme care sa te impresioneze...

Eu zic sa nu te descurajezi, Sorine...Tu propune problemute inainte si daca nu le baga nimeni in seama, asta e...Tu continua :-D

Multumesc de sfat, Tudor.Dar, cum am spus, am sa ma straduiesc sa propun probleme cat mai grele.

Ideea era ca putea fi folosita Jensen pentru a doua metoda. ;)

[color=blue][b]Fie $a,b,c > 0$ trei numere reale strict pozitive astfel incât $a + b + c = 1$.

Să se arate că:
\[ {3>\frac{1}{{1 + \sqrt a }} + \frac{1}{{1 + \sqrt b }} + \frac{1}{{1 + \sqrt c }} >\frac{3}{2}} \]

[/b][/color]
[b][u] Soluţie:[/u] [/b]

$\left\{ \begin{array}{l} a,b,c > 0 \\ a + b + c = 1 \\ \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 0 < a < 1 \\ 0 < b < 1 \\ 0 < c < 1 \\ \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 0 < \sqrt a < 1 \\ 0 < \sqrt b < 1 \\ 0 < \sqrt c < 1 \\ \end{array} \right. \Rightarrow$

$\Rightarrow \left\{ \begin{array}{l} 0 < \sqrt a < 1 \\ 0 < \sqrt b < 1 \\ 0 < \sqrt c < 1 \\ \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 1 < 1 + \sqrt a < 2 \\ 1 < 1 + \sqrt b < 2 \\ 1 < 1 + \sqrt c < 2 \\ \end{array} \right. \Rightarrow$

$\Rightarrow \left\{ \begin{array}{l} 1 > \frac{1}{{1 + \sqrt a }} > \frac{1}{2} \\ 1 > \frac{1}{{1 + \sqrt b }} > \frac{1}{2} \\ 1 > \frac{1}{{1 + \sqrt c }} > \frac{1}{2} \\ \end{array} \right. \Rightarrow 3 > \frac{1}{{1 + \sqrt a }} + \frac{1}{{1 + \sqrt b }} + \frac{1}{{1 + \sqrt c }} > \frac{3}{2}$

Dap, o solutie frumoasa si bine inteles alta fata de Cezar.Am vazut ca ai imbunatatit problema. :)


You will have from 12 PM - 5 PM EDT to complete this test (class from 2-4 XD).  Yes, it's a very long time limit, but hopefully, you'll be honest about it.  Submit your answers within this time frame or up to 15 minutes after.  Have fun!

Edit: ANYONE can take this test.  Just PM me your answers in a list format.  Or, you could print the copy, and then scan it in to your comp before sending it to me.

Edit 2: Due to a request, the time limit will be extended to 8 PM.  WOW. THAT'S A LONG TIME. I feel like I'm being a little too lenient =P.  But whatever.  8 hours should be enough for everyone, right?

Just to make sure, on #28, opposite means the one connected by an internal diagonal?

LOL I will say this: tinytim is giving everyone like 1 week to do his test here (2tinytim: there, yw for publicizing it xD):
http://www.artofproblemsolving.com/Forum/weblog.php?w=1100

Even though he said he wasn't posting it until ____, it's on his blog already.

just so you know, I probably won't be able to submit in time, I have too much hw.

on 22, does it ask for circles whose area is an integral number of m^2 or cm^2?

Sorry. got home late... m^2 to steven meow. #28 space diagonal.

you know ,athcrazed, i just did cm^2 and i got 12366334176.
:sad:
i deserve at least .01 extra points for that
and i didnt use a calc

You fixed the errors I pointed out, right? [I can't open up PDFs right now while Adobe CS3 installs....

Yeah.  Btw, official results will be posted in a couple days, so don't take my word for granted that your score is what I said it was =P.

Only have 4 submissions so far... what happened to other 9-10?

Thirty minutes to complete the test! Get those submissions in!

Question: Should I let people submit answers tomorrow and Monday so the 5 people who did get more competition? Or just go with it now? =P

Go with it now. AH WAHNT MAH SCORE! AH WAHNT TO KNOW HAH MANY QUESTIAHNS AH GAHT WRAHNG!

I'm extending the deadline a bit.  Hope that a few more people will submit their answers.  Post answers within 14.5 hours please. xD.  I have scores tallied for 7 people (waiting on at least one person). Sending those out now.

Scores are all tallied and are being entered into an excel (97-03) spread sheet.  ALL SCORES WILL BE REVEALED AND PROBLEMS MISSED/GOTTEN WILL ALSO BE REVEALED.  If you don't want me to show yours, tell me within the next 24 hours.

Only sprint so far. Target will be posted soon.

Target proofreaders (sending out target now): Yongyi781
Cognos599
Dynamo729 (still want to)
Perfect628

who else wants to proofread it?

Warning: Beastly target problems may cause extreme frustration after brute-forcing for a few hours.

Your target questions shrivel compared to mine :P

[quote="ajain"]@mathcrazed

i am on vacation and wont be back until tuesday, at like ww.r? pm central time.  can u extent it for me[/quote]

OK.  Just send your answers to me on Tuesday then. That's probably when the Target will be posted also.  Yongyi: You're just mad you can't do them xD (... jk).

Here's my mock test (for mathcrazed).

I realize that some questions on the sprint are ridiculously easy. :D

Ugh I've been in careless mistake/stupid land these days...


Find all fuction $f: \mathbb N \to \mathbb R$ such that $ f(2) \equal{} 2$, $ f(nm) \equal{} f(n)f(m)$ for all $m,n\in \mathbb N$, and $ f(n \plus{} 1) > f(n)$ for all $ n\in \mathbb N$.

[quote="math10"]Find all fuction $ f: N\longrightarrow R$ such that:$ f(2) \equal{} 2$ ; $ f(nm) \equal{} f(n)f(m)$ for $ \forall m,n\in N$ and $ f(n \plus{} 1) > f(n)$ for $ \forall n\in N$[/quote]

We have $ f(1) \equal{} 1$, then $ f(n) \ge 1$ for all $ n$.

Clearly $ f(n^k) \equal{} [f(n)]^k$ , where $ n,k$ are positive integers.

Assume that $ f(n) > n$ for some $ n$, and let $ In(f(n)) \minus{} In(n) \equal{} c$ is a positive constant then

$ f(2^{x \plus{} 1}) \equal{} 2^{x \plus{} 1} > f(n^k) > n^k \ge 2^{x} \equal{} f(2^x)$ is true for any positive integer $ k$ and corresponding integer $ x$.

$ \Rightarrow (x \plus{} 1)In \ 2 > k In (f(n)) > k In (n) > x In \ 2$

$ \Rightarrow (x \plus{} 1) In \ 2 \minus{} x In \ 2 \equal{} In \ 2 \ge kc$ for any $ k$ , which is impossible as $ k$ can be made arbitrarily large, contradiction.

Similarly we assume that $ f(n) < n$ for some $ n$ and get contradiction.

$ \Rightarrow f(n) \equal{} n$ for all $ n$ is the only solution.

P.S. Sorry, pco . Typo corrected.

[quote="stephencheng"] Assume that $ f(n) > n$ for some $ n$, and let $ f(n) \minus{} n \equal{} c$ is a positive constant then

$ f(2^{x \plus{} 1}) \equal{} 2^{x \plus{} 1} > f(n^k) > n^k \ge 2^{x} \equal{} f(2^x)$ is true for any positive integer $ k$ and corresponding integer $ x$.

$ \Rightarrow (x \plus{} 1)In \ 2 > k In (f(n)) > k In (n) > x In \ 2$

$ \Rightarrow (x \plus{} 1) In \ 2 \minus{} x In \ 2 \equal{} In \ 2 \ge kc$ for any $ k$ , [/quote]

$ (x \plus{} 1)In \ 2 > k In (f(n)) > k In (n) > x In \ 2$ $ \implies$ $ (x \plus{} 1) In \ 2 \minus{} x In \ 2 \equal{} In \ 2 \ge k(\ln(f(n))\minus{}\ln(n))$

How did you get from there $ (x \plus{} 1) In \ 2 \minus{} x In \ 2 \equal{} In \ 2 \ge kc$ ?


Suppose that $ \omega, a,b,c$ are distinct real numbers for which there exist real numbers $ x,y,z$ that satisfy the following equations:

$ x\plus{}y\plus{}z\equal{}1,$
$ a^2 x\plus{}b^2 y \plus{}c^2 z\equal{}\omega ^2,$
$ a^3 x\plus{}b^3 y \plus{}c^3 z\equal{}\omega ^3,$
$ a^4 x\plus{}b^4 y \plus{}c^4 z\equal{}\omega ^4.$

Express $ \omega$ in terms of $ a,b,c$.

.. By Cauchy-Schwarz, $ (a^4x \plus{} b^4y \plus{} c^4z)(x \plus{} y \plus{} z) \ge (a^2x \plus{} b^2y \plus{} c^2z)^2 \Longrightarrow w^4 \ge w^4$, equality iff $ \frac {a^4x}{x} \equal{} \frac {b^4y}{y} \equal{} \frac {c^4z}{z}$. If $ |a| \equal{} |b| \equal{} |c|$, then the second and fourth equations are contradictory unless $ \omega \equal{} a \equal{} b \equal{} c \equal{} 0$. If two of $ \{x,y,z\}$ are $ 0$, we obtain $ \omega \in \{a,b,c\}$, in particular the variable corresponding to which of $ \{x,y,z\}$ is $ 1$.


DO u have to take chemistry before u take physics??is it better to take chem first...what about AP physics?
How can u study for it if ur skool is not offering AP physics??

You can take physics before you take chemistry.  I don't see any reason that it would be better to take chem first.  Physics is more fundamental than chemistry.  Chemistry is just an application of physics.

As far as preparing for AP physics, here are some ideas:

--Get a good first year college physics textbook, such as Giancolli.
--Get a good high school physics textbook that is used in good AP physics courses.
--Get the syllabus for the AP physics exam, get a bunch of AP Physics exams from years past, and start learning the material and doing problems.
--Get a syllabus from a good AP physics class, and work through it.
--Post questions on the AoPS physics forum.


inequalities
trigonometry
inequalities unsolved

Knowing that $ 0 < A,B,C < \frac{\pi }{2}$ and $ A \plus{} B \plus{} C \equal{} \pi$ ,try to prove:
$ \frac{{\cos ^2 A}}{{\sin B\sin C}} \plus{} \frac{{\cos ^2 B}}{{\sin A\sin C}} \plus{} \frac{{\cos ^2 C}}{{\sin A\sin B}} \ge 1$

[quote="zkw"]Knowing that $ 0 < A,B,C < \frac {\pi }{2}$ and $ A \plus{} B \plus{} C \equal{} \pi$ ,try to prove:
$ \frac {{\cos ^2 A}}{{\sin B\sin C}} \plus{} \frac {{\cos ^2 B}}{{\sin A\sin C}} \plus{} \frac {{\cos ^2 C}}{{\sin A\sin B}} \ge 1$[/quote]
See a  stronger inequality here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100193


I tried to solve the following exercise from CLR, but I didn't manage to find the correct answer... can somebody help?

Write an O(n)-time nonrecursive procedure that, given an n-node binary tree, prints out the key of each node. Use no more than constant extra space outside of the tree itself and do not modify the tree, even temporarily, during the procedure.

My suggestion:

You have to remember "where you are now" and "where you came from". Initially, go to the rood and set that you have come from the top. Then, do this forever:

look where you've come from and do one of those actions:
* If you came from the top (parent), go to the left child.
* If you came from the left child, go to the right child.
* If you came from the right child, PRINT OUT the current node and go up (to the top).
(if any of the children don't exist, just mark you've been there and do nothing, just loop again)

This should traverse the tree from the left to the right and print out all the nodes in some kind of post-order (or however is it called).

You terminate when you are trying to go up from the root.

[quote="(exercise from CLR book)"]Write an O(n)-time nonrecursive procedure that, given an n-node binary tree, prints out the key of each node. Use no more than constant extra space outside of the tree itself and do not modify the tree, even temporarily, during the procedure.[/quote]
One needs to carefully specify what $O(n)$ means.  To print the keys requires time at least $O(Kn)$, where $K$ is the maximum number of bits of each "key".  I think CLR mean that the tree is given as a data structure with some operations (move up/left/right in tree; get key) and we are allowed to use the operations at most $cN$ times.

Well, you know for every node in the binary tree the left child, the right child and the parent (and of course the key), and by O(n) it is meant that you can traverse the tree at most a constant number of times.

Well, the solution mentioned by kubus works, and it's really nice and simple!  ;)  As for details, you should be careful with that "where you came" pointer, not to mix up the left child for the right child, when you are at a leaf, and you have no left child or right child...
At first I thought you keep for a node only the left child and the right child, and not the parent too, but I guess it's impossible to solve this task under this constraints...


Find all couples $(p, q)$ of prime numbers for which
$p^2 + q^2 + p^2q^2$
is a perfect square.

[hide="Solution"]
We have $k^2=p^2+q^2+p^2q^2=(p^2+1)(q^2+1)-1.$  If $p, q$ are odd, then looking mod 4, we see that the LHS is 0 or 1 and the RHS is 3.  Our other option is for either $p$ or $q$ to be 2.  WLOG $p=2$ giving $k^2=5q^2+4\Rightarrow (k-2)(k+2)=5q^2.$  The possible pairs for $(k-2,k+2)$ are $(1,5q^2), (5,q^2),$ and $(q,5q).$

Solving these, we get (discarding values we don't want) $q=3$ as the only solution.  Therefore, $(2,3)$ is the only solution.  (indeed $2^2+3^2+2^2\cdot3^2=7^2$).[/hide]

Hint :

[hide]$2,2$ is not a solution.
$mod \ 4$ : $p$ and $q$ can't be both odd, wlog, $p =2$
Rewriting $q(q+4) = A^2-4 = (A+2)(A-2)$ quickly gives $q=3$

$2,3$ only solution.[/hide]


Can we have a private Ohio forum like Texas?

1) this should be in the national & local communities round table
2) who would come? the only active actual OH people are hspotter, gyan, roadnottaken, and you.
3) are you just trying to block me out?  :wink:

Oh no. I wanted to have a private forum so everyone in the Ohio usergroup (you can request to join me thinks) can have a private forum, because the usergroup has no significance.

And all people who live in the states bordering Ohio, are allowed to right?

Now that I think about it, I don't really want to.

[quote="junggi"]the only active actual OH people are hspotter, gyan, roadnottaken, and you.[/quote]

I just found two more :D

Omar and davidli

And me, now that I actually realized we have an Ohio forum...

And me, now that I actually go on this website


Solve this equation:
$ sin^22x\plus{} sin2x\plus{}sinx\plus{}1\equal{}o$

$ sin^22x \plus{} sin2x \plus{} sinx \plus{} 1 \equal{} 0$
$ sin2x(sinx \plus{} 1 ) \plus{} sinx \plus{} 1 \equal{} 0$
$ (sinx \plus{} 1)(sin2x \plus{} 1) \equal{} 0$

i) $ sinx \equal{} \minus{} 1 ,x_1 \equal{} 270 grade$
ii)$ sin2x \equal{} \minus{} 1,x_2 \equal{} 135 grade$

Edit: sorry,ignore this post

[quote="Obel1x"]$ sin^{2}2x\plus{}sin2x\plus{}sinx\plus{}1\equal{}0$
$ sin2x(sinx\plus{}1 )\plus{}sinx\plus{}1 \equal{}0$
[/quote]
I don't think you've factorised that correctly, although $ x\equal{}270^\circ$ is a solution.

[quote="Obel1x"]
i) $ sinx \equal{} \minus{} 1 ,x_1 \equal{} 270 grade$
ii)$ sin2x \equal{} \minus{} 1,x_2 \equal{} 135 grade$[/quote]

$ \sin2x\neq2\sin x$, as I think you thought. $ \sin2x\equal{}2\sin x\cos x$.
So the equation is actually \[ (\sin 2x)(\sin 2x\plus{}1)\plus{}\sin x\equal{}\minus{}1\]
Now take the inverse sine of both sides to get \[ 2x(2x \plus{} \frac{\pi}{2})\plus{}x\equal{}\frac{3\pi}{2}\] Expanding, \[ 4x^2\plus{}(\pi \plus{}1)x\equal{}\frac{3\pi}{2}\implies 4x^2\plus{}(\pi\plus{}1) x \minus{}\frac{3\pi}{2}\equal{}0\]. Now change the $ \pi$ into $ 180^{\circ}$ (for ease of calculation). Now \[ 4x^2\plus{}181x\minus{}270\equal{}0\]

Could someone check this part? [quote="RoFlLoLcOpT"]Now take the inverse sine of both sides to get \[ 2x(2x \plus{} \frac{\pi}{2})\plus{}x\equal{}\frac{3\pi}{2}\][/quote] I think I may have done something wrong here because I'm getting a wrong answer.

[quote="RoFlLoLcOpT"]
Could someone check this part? [quote="RoFlLoLcOpT"]Now take the inverse sine of both sides to get
\[ 2x(2x \plus{} \frac {\pi}{2}) \plus{} x \equal{} \frac {3\pi}{2}
\]
[/quote] I think I may have done something wrong here because I'm getting a wrong answer.[/quote]

$ \frac {3\pi}{2}$ is not in the range of inverse sine; however, $ \minus{} \frac {\pi}{2}$ is.

In addition, inverse sine is not distributive: $ \sin^{ \minus{} 1}(\sin{x} \plus{} \sin{2x}) \neq x \plus{} 2x$ so...

I don't get how you took the inverse sine of that expression. You treated it as if $ \sin^{\minus{}1}(x)$ were a linear transformation, which it most certainly is not...

As an example, pick x = 45 degrees. The arcsin of sin(2x)(sin 2x + 1) + sin x is most certainly not 2x(2x + pi/2) + x.


Set systems
combinatorics
graph theory
Extremal combinatorics
IMO Shortlist

Given $ n$ countries with three representatives each, $ m$ committees $ A(1),A(2), \ldots, A(m)$ are called a cycle if

[i](i)[/i] each committee has $ n$ members, one from each country;
[i](ii)[/i] no two committees have the same membership;
[i](iii)[/i] for $ i \equal{} 1, 2, \ldots,m$, committee $ A(i)$ and committee $ A(i \plus{} 1)$ have no member in common, where $ A(m \plus{} 1)$ denotes $ A(1);$
[i](iv)[/i] if $ 1 < |i \minus{} j| < m \minus{} 1,$ then committees $ A(i)$ and $ A(j)$ have at least one member in common.

Is it possible to have a cycle of 1990 committees with 11 countries?

nice and easy problem, but it took me 3 hours to solve it!!!

suppose that you have $n$ countries and $m$ committees, namely $A_1,...,A_m$ satisfying the condition, where $m$ is an odd number. now add another country with three members $\{1,2,3\}$, and consider these committees: $B_1=A_1\cup \{1\},...,B_m=A_m\cup \{1\},B_{m+1}=A_1\cup \{2\},...,B_{2m}=A_m\cup \{2\}$. these committees satisfy the condition, so we have now $2m$ committees from $n+1$ countries. also, if $k$ is an even  integer that $k\le m$, then, as above, these committes make a cycle of length $2(k-1)$ with $n+1$ countries: $B_1=A_1 \cup \{3\}, B_2=A_2\cup \{1\}, B_3=A_3\cup \{2\},...,B_{k-2}=A_{k-2}\cup \{1\}, B_{k-1}=A_{k-1}\cup \{2\}, B_k=A_k\cup \{3\}, B_{k+1}=A_{k-1}\cup \{1\},...,B_{2(k-1)}=A_2\cup \{2\}$.

now starting from the first, we have $1$ country and $3$ committees, next, we have $2$ countries and $6$ committees, next $3$ countries and $10$ committees,... and at last $11$ countries and $1990$ committees.

can you explain more about how you'll arrive at $ 11 $ countries and $ 1990 $ committees please? :blush:

[quote=goodar2006]nice and easy problem, but it took me 3 hours to solve it!!!

suppose that you have $n$ countries and $m$ committees, namely $A_1,...,A_m$ satisfying the condition, where $m$ is an odd number. now add another country with three members $\{1,2,3\}$, and consider these committees: $B_1=A_1\cup \{1\}, B_2=A_2\cup\{2\}, B_3=A_3\cup\{1\}, ...,B_m=A_m\cup \{1\},B_{m+1}=A_1\cup \{2\},...,B_{2m}=A_m\cup \{2\}$. these committees satisfy the condition, so we have now $2m$ committees from $n+1$ countries. also, if $k$ is an even  integer that $k \leq m$, then, as above, these committes make a cycle of length $2(k-1)$ with $n+1$ countries: $B_1=A_1 \cup \{3\}, B_{k} = A_{k} \cup\{3\}$, arrange the rest of ${B_i}$ from $1<i<=2k-2, i \neq k$ as $A_2, A_3, ... A_{k-1}, A_{1}, ... A_{k-2}$  Alternate the appendages (from the n+1th country) by alternating between 1 and 2 in this set.

Now starting from the first, we have $1$ country and $3$ committees, next, we have $2$ countries and $6$ committees, next $3$ countries and $10$ committees,... and at last $11$ countries and $1990$ committees.[/quote]

 The recursion actually generates a cycle of length 2390 for n = 11, which should apparently also imply the existence of a cycle of length 1990 since 2390 > 1990. However, I don't quite see why the existence of a larger cycle should imply the existence of a smaller one. Also I fixed the indices (the one's Goodar wrote for odd m were just plain wrong, I'm sure he mistyped or something) so I wrote them to match my indices.

[hide=Wrong Solution]A nice problem!

The answer is yes. We prove the result for all $n \geq 6$.
Firstly, let the countries be $C_1, C_2, C_3, \cdots C_{k}$, and the represntatives be $a_{i1}, a_{i2}, a_{i3} \in C_i$ for all $1 \le i \le n$. Common sense says that if a member is common to two committees, then the person belongs to the same country (because they are the same person). Hence we can classify all the members according to their countries, and countries in our figure are columns. Hence we only need to check common members for the respective columns (i.e. countries).

Here's the construction for $n=6$. It satisfies our conditions, so we can easily add more columns (i.e. countries) wherein we only have to worry about the [hide=third condition](iii) for $ i = 1, 2, \ldots,m$, committee $ A(i)$ and committee $ A(i + 1)$ have no member in common, where $ A(m + 1)$ denotes $ A(1);$[/hide], which can be easily handled.
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
& $C_1$  & $C_2$ & $C_3$ & $C_4$ & $C_5$ & $C_6$\\

\hline

$A_1$ & $\alpha_{11}$ & $\alpha_{21}$ & $\alpha_{31}$ & $\alpha_{42}$ & $\alpha_{51}$ & $\alpha_{63}$ \\ \hline

$A_2$ & $\alpha_{12}$ & $\alpha_{22}$ & $\alpha_{32}$ & $\alpha_{43}$ & $\alpha_{52}$ & $\alpha_{62}$ \\ \hline

$A_3$ & $\alpha_{11}$ & $\alpha_{23}$ & $\alpha_{33}$ & $\alpha_{41}$ & $\alpha_{53}$ & $\alpha_{61}$ \\ \hline

$A_4$ & $\alpha_{12}$ & $\alpha_{21}$ & $\alpha_{31}$ & $\alpha_{42}$ & $\alpha_{52}$ & $\alpha_{62}$  \\ \hline

$A_5$ & $\alpha_{11}$ & $\alpha_{22}$ & $\alpha_{32}$ & $\alpha_{43}$ & $\alpha_{51}$ & $\alpha_{63}$  \\ \hline

$A_6$ & $\alpha_{12}$ & $\alpha_{23}$ & $\alpha_{31}$ & $\alpha_{41}$ & $\alpha_{52}$ & $\alpha_{62}$ \\ \hline

$A_7$ & $\alpha_{11}$ & $\alpha_{21}$ & $\alpha_{32}$ & $\alpha_{42}$ & $\alpha_{53}$ & $\alpha_{63}$  \\ \hline

$A_8$ & $\alpha_{12}$ & $\alpha_{22}$ & $\alpha_{31}$ & $\alpha_{41}$ & $\alpha_{51}$ & $\alpha_{62}$  \\ \hline

$A_9$ & $\alpha_{11}$ & $\alpha_{23}$ & $\alpha_{32}$ & $\alpha_{42}$ & $\alpha_{53}$ & $\alpha_{61}$  \\ \hline

$A_{10}$ & $\alpha_{12}$ & $\alpha_{21}$ & $\alpha_{31}$ & $\alpha_{41}$ & $\alpha_{51}$ & $\alpha_{63}$  \\ \hline

$A_{11}$ & $\alpha_{13}$ & $\alpha_{22}$ & $\alpha_{32}$ & $\alpha_{43}$ & $\alpha_{52}$ & $\alpha_{61}$  \\ \hline

\end{tabular}

[b]P.S.[/b] The $6th$ column is probably unnecessary, but I added it just because in my construction with $5$ columns, the $3rd$ and $11th$ rows were disjoint. That's why the other members of the $6th$ column were arbitrarily chosen with only the third condition in mind.[/hide]  :(
I misread the question

[quote=Wizard_32]A nice problem!

The answer is yes. We prove the result for all $n \geq 6$.
Firstly, let the countries be $C_1, C_2, C_3, \cdots C_{k}$, and the represntatives be $a_{i1}, a_{i2}, a_{i3} \in C_i$ for all $1 \le i \le n$. Common sense says that if a member is common to two committees, then the person belongs to the same country (because they are the same person). Hence we can classify all the members according to their countries, and countries in our figure are columns. Hence we only need to check common members for the respective columns (i.e. countries).

Here's the construction for $n=6$. It satisfies our conditions, so we can easily add more columns (i.e. countries) wherein we only have to worry about the [hide=third condition](iii) for $ i = 1, 2, \ldots,m$, committee $ A(i)$ and committee $ A(i + 1)$ have no member in common, where $ A(m + 1)$ denotes $ A(1);$[/hide], which can be easily handled.
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
& $C_1$  & $C_2$ & $C_3$ & $C_4$ & $C_5$ & $C_6$\\

\hline

$A_1$ & $\alpha_{11}$ & $\alpha_{21}$ & $\alpha_{31}$ & $\alpha_{42}$ & $\alpha_{51}$ & $\alpha_{63}$ \\ \hline

$A_2$ & $\alpha_{12}$ & $\alpha_{22}$ & $\alpha_{32}$ & $\alpha_{43}$ & $\alpha_{52}$ & $\alpha_{62}$ \\ \hline

$A_3$ & $\alpha_{11}$ & $\alpha_{23}$ & $\alpha_{33}$ & $\alpha_{41}$ & $\alpha_{53}$ & $\alpha_{61}$ \\ \hline

$A_4$ & $\alpha_{12}$ & $\alpha_{21}$ & $\alpha_{31}$ & $\alpha_{42}$ & $\alpha_{52}$ & $\alpha_{62}$  \\ \hline

$A_5$ & $\alpha_{11}$ & $\alpha_{22}$ & $\alpha_{32}$ & $\alpha_{43}$ & $\alpha_{51}$ & $\alpha_{63}$  \\ \hline

$A_6$ & $\alpha_{12}$ & $\alpha_{23}$ & $\alpha_{31}$ & $\alpha_{41}$ & $\alpha_{52}$ & $\alpha_{62}$ \\ \hline

$A_7$ & $\alpha_{11}$ & $\alpha_{21}$ & $\alpha_{32}$ & $\alpha_{42}$ & $\alpha_{53}$ & $\alpha_{63}$  \\ \hline

$A_8$ & $\alpha_{12}$ & $\alpha_{22}$ & $\alpha_{31}$ & $\alpha_{41}$ & $\alpha_{51}$ & $\alpha_{62}$  \\ \hline

$A_9$ & $\alpha_{11}$ & $\alpha_{23}$ & $\alpha_{32}$ & $\alpha_{42}$ & $\alpha_{53}$ & $\alpha_{61}$  \\ \hline

$A_{10}$ & $\alpha_{12}$ & $\alpha_{21}$ & $\alpha_{31}$ & $\alpha_{41}$ & $\alpha_{51}$ & $\alpha_{63}$  \\ \hline

$A_{11}$ & $\alpha_{13}$ & $\alpha_{22}$ & $\alpha_{32}$ & $\alpha_{43}$ & $\alpha_{52}$ & $\alpha_{61}$  \\ \hline

\end{tabular}

[b]P.S.[/b] The $6th$ column is probably unnecessary, but I added it just because in my construction with $5$ columns, the $3rd$ and $11th$ rows were disjoint. That's why the other members of the $6th$ column were arbitrarily chosen with only the third condition in mind.[/quote]

You proved it is possible to make $11$ commities using $6$ country.The problem is something else... .And I don't think You can generalize this method to work for the problem(The bound on problem is very sharper than this ...)The only way to solve this is inductivly I think Since it is tooooo hard to think for a direct construction.

[quote=Wizard_32][quote=Taha1381]
You proved it is possible to make $11$ commities using $6$ country.The problem is something else... .And I don't think You can generalize this method to work for the problem(The bound on problem is very sharper than this ...)The only way to solve this is inductivly I think Since it is tooooo hard to think for a direct construction.[/quote]
Note that in my construction, the important conditions are all covered:
(i) each committee has $ n$ members, one from each country;
(ii) no two committees have the same membership;
(iv) if $ 1 < |i - j| < m - 1,$ then committees $ A(i)$ and $ A(j)$ have at least one member in common
These three conditions are satisfied for the first 11 countries themselves since any two non-adjacent countries have at least one common element. Now, we can add the countries after $C_{11}$. without having to worry about the condition (iv)
Also, each pair of countries have at least one different element, so no matter how we add on the following countries, no two countries are [i]exactly the same.[/i] Now, we can add the countries after $C_{11}$. without having to worry about the condition (ii)

Also, the condition 3 is not violated yet, and will not be violated further if we add the next columns as: (bascially alternate the members)
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
& $C_i$ \\

\hline

$A_1$ & $\alpha_{i1}$ \\ \hline

$A_2$ & $\alpha_{i2}$ \\ \hline

$A_3$ & $\alpha_{i1}$  \\ \hline

$A_4$ & $\alpha_{i2}$  \\ \hline

$A_5$ & $\alpha_{i1}$  \\ \hline

$A_6$ & $\alpha_{i2}$  \\ \hline

$A_7$ & $\alpha_{i1}$ \\ \hline

$A_8$ & $\alpha_{i2}$  \\ \hline

$A_9$ & $\alpha_{i1}$  \\ \hline

$A_{10}$ & $\alpha_{i2}$  \\ \hline

$A_{11}$ & $\alpha_{i3}$ \\ \hline

\end{tabular}

[size=75]800th post[/size][/quote]

You are adding committees not countries we have $11$ countries but a lot more (1990) committees.

O god seems like I misread the question :blush:
Thanks for pointing out my mistake!

[b]Solution:[/b] Yes, it is possible to have a [i]cycle[/i] of $1990$ committees with $11$

$\bullet \ $ [u]Construction:[/u]
Suppose $A_1,A_2, \dots , A_{m}$ is an $m-$cycle with $k$ countries. We present an algorithm which shows that it is possible to construct a $2m-2$-cycle involving exactly $k+1$ countries.

Denote by $\{C_1,C_2,C_3\}$ the three members of the $(k+1)^{\text{th}}$ country. Then, observe that the committees:
$$\{A_1,C_3\} , \{A_2,C_2\},\{A_3,C_1\},\{A_4,C_2\},\{A_5,C_1\}, \dots ,\{A_{m},C_3\},\dots, \{A_3,C_2\},\{A_2,C_1\}$$ also satisfy all constraints of the problem.

Hence we have successfully created a $2m-2$-cycle with $(k+1)$ countries.

[b]Claim:[/b] A $6$-cycle is possible with $2$ countries.
[i]Proof.[/i] Let $\{B_1,B_2,B_3\}$ and $\{C_1,C_2,C_3\}$ represent the members of the two countries respectively. Then consider the committees:
$ A_1 = \{B_1 , C_1\} ; A_2 = \{B_2 , C_2\} ; A_3 = \{B_1,  C_3\} ; A_4 =\{ B_2 , C_1\} ; A_5 =\{ B_1 , C_2\}  ;A_6 =\{ B_2,  C_3\} $

It can be verified with ease that all these work.

Thus, with $2$ countries we can create a $6$-cycle and the recursions gives that with $11$ countries we can create a $2050$-cycle. But since we desire a $1990$-cycle, we must remove some of the committees. Again, we present an algorithm which shows that we can remove,in each step , $2$ committees without worrying about the constraints.

$\rightarrow$ From the cycle, $\{A_1,C_3\} , \{A_2,C_2\},\{A_3,C_1\},\{A_4,C_2\},\{A_5,C_1\}, \dots ,\{A_{m},C_3\},\dots, \{A_3,C_2\},\{A_2,C_1\}$
we remove the first and the last committees and replace the $C_{j}$ of the second committee from the left with $C_3$.

i.e. The new cycle is $\{A_2,C_3\} , \{A_3,C_1\} , \dots , \{A_3,C_2\}$.

Since $2050-1990 =60 $ is even, we are done $\ \blacksquare$


A convex pentagon's vertices are all latice points. Prove that the pentagon contains another lattice point in its interior.

See pythag011's post [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=222558]here[/url].

Somebody showed me a nice solution that an area of 2 is unattainable. The pentagon is ABCDE.

If the midpoint of the two vertices (in the pigeonhole step) is an interior point, then the area must be at least 5/2.

If the midpoint is a boundary point, WLOG it is the midpoint of AB. Call it M. MBCDE is another convex pentagon. It must have an area of at least 2. The area of ABCDE is the sum of MBCDE and AME. Therefore, ABCDE cannot have an area of 2. A quick look at Pick's Theorem shows that the area must be of the form N/2 where N is a natural number. Therefore, ABCDE must have at least an area of 5/2. The pentagon with vertices {(0,0), (1,0), (2,1), (1,2), (0,1)} has an area of 5/2.


When finding limits at infinity, what's the other way of doing it instead of just looking? Like, if I have to find the limit of (2x 2 +3)/(5x 2 +7), as x goes to infinity, how do I do it?  I mean, you can look at it and see that it'll be 2/5, but what's the actual mathematical way of doing it? I used to know, then I forgot. Thanks!

And did I post this in the right place? Hope so, sorry if I didn't.

well... when calculating limits where u have polynomials in the numerator and denominator like in this case, it only matters what the ratio of the coefficients of the highest power is (this is the case since both polynomials are of degree 2). However, in the event that the degrees of the polynomials are different, we know that if the polynomial of higher degree is in the numerator, the limit will go to infinity, and if the polynomial of higher degree is in the denominator, the limit will go to 0.

I'm not sure if this is what you were looking for... but if u use Big-O notation, u will notice that the top and the bottom are both O(n^2) and this means that they are both growing at the same right, and thus will maintain a constant ratio, which is the ratio of their coefficients.

Hope this was of some help,

Yoel

Or if you want to be a little more mathematically precise:
Divide top and bottom of the fraction by the highest power of x.  (2x^2 +3)/(5x^2 +7) gives (2 + 3/x^2)/(5 + 7/x^2). As x goes to infinity, all terms go to 0 except the constant ones, giving 2/5.

Muchas gracias - Thanks a bunch! That helps a lot.  :)


Consider all arithmetical sequences of real numbers $(x_i)_{i=1}^{\infty}$ and $(y_i)_{i=1}^{\infty}$ with the common first term, such that for some $k>1$,
$x_{k-1}y_{k-1}=42, x_ky_k=30,$ and $x_{k+1}y_{k+1}=16$.

Find all such pairs of sequences with the maximum possible $k$.


It isn't important unless you feel you need certain resources that are only available at prestigious schools. Can you name any that you need?

[quote="tokenadult"]Yes, but not only because of prestige as such, but because the prestigious math schools offer better learning opportunities.[/quote]

I don't know if that's necessarily true. I know some outstanding mathematicians who didn't necessarily go to notably prestigeous undergraduate schools.

But the prestige of a University or math department could matter depending on what you want to do after you graduate. In business circles, name recognition and alumni connections can be very valuable.

But if you want to end up in a job that you KNOW doesn't make that much money, then being saddled with debt for the rest of your life is probably not worth the prestige.

[quote="Izzybee"]If you get "into" the math community afterwards, it doesn't really matter. At any rate, places like Harvard and Yale are more famous for stuff like business, administration and law; for mathematics there are many other universities where you'd be better off.[/quote]

Yale maybe, but there are not 'many' universities better than Harvard for mathematics.


real analysis
limit
integration
real analysis unsolved

Let $m$ be the Lebesgue measure on $[0,1]$ , and define $||f||_p$  with respect to m. Find all $\Phi$ on $[0,\infty)$  such that the relation \[ \Phi\left(\lim_{p\rightarrow 0} ||f||_p\right)=\int_0^1(\Phi \circ f)dm  \] holds for every bounded, measurable, positive f. Show first that \[ c\Phi(x)+(1-c)\Phi(1)=\Phi(x^c)\,\,\,(x>0,0\leq c \leq 1).  \]

Auxtuno - are you and karelius taking the same class?

yes, that is right, but, can you help me,please???


Let s= 23, p= 560.
Let a(0)= s, a(1)= p, and a(n+1) = a(n) + a(n-1),  n=1, 2, 3..
set X(n)= a(n) / a(n+1)
Use calculus technique to answer:
Does this sequence converge or diverge. If it converges, find its limit.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=268916[/url]


(a) Given real numbers $ a,b,c$ with $ a\plus{}b\plus{}c\equal{}0$, prove that
\[ a^3\plus{}b^3\plus{}c^3>0 \text{ if and only if } a^5\plus{}b^5\plus{}c^5>0.
\]
(b) Given real numbers $ a,b,c,d$ with $ a\plus{}b\plus{}c\plus{}d\equal{}0$, prove that
\[ a^3\plus{}b^3\plus{}c^3\plus{}d^3>0 \text{ if and only if } a^5\plus{}b^5\plus{}c^5\plus{}d^5>0.
\]

This has been posted before and can be generalized significantly.  Note that $ a^3 \plus{} b^3 \plus{} c^3 > 0 \Leftrightarrow 3abc > 0$.

It is a just an application of AM-GM for both proofs.

Sorry wasn't aware it'd been posted.. It didn't come up in the search. And can AM-GM be used since at least one number must be negative?


inequalities
number theory unsolved
number theory

If $m$ and $n$ are positive integers, prove that $|n\sqrt{2005}-m|>\frac{1}{90}$

This is wrong. As can be seen from Dirichlet's approximation theorem.

Wrong? Prove that is wrong..
See problem five: http://www.imo.org.yu/othercomp/Blr/BlrMO05.pdf

Maybe not.

Dirichlet's theorems says there are [i]integers[/i] $h,k$  for which $|h\theta - k| < \frac{1}{N}$ for any given positive integer $N$, where $\theta$ is any positive real number.

The problems says that $m, n > 0$. So, Dirichlet's theorem does not apply. Sorry about that.

Since $mn>0$ does clearly happen when choosing $m,n$ by Dirichlet's (¿) theorem, this makes no difference.
Thus, the inequality is wrong. E.g. take $n=790 , m=35374$.

Now, if that were $\frac1{90n}$ instead of $\frac1{90}$, we'd have an interesting and true problem.

And quite easy to prove, since it's very similar to Liouville's theorem.

[hide]If $\frac{m}{n}\geq 90-\sqrt{2005}$, then $\frac{m}{n}-\sqrt{2005}\geq 90-2\sqrt{2005}>\frac{1}{4}>\frac{1}{90n^2}\implies |m-n\sqrt{2005}|>\frac{1}{90n}.$

If $\frac{m}{n}<90-\sqrt{2005}$, then $\frac{m}{n}+\sqrt{2005}<90\implies m+n\sqrt{2005}<90n$. On the other hand, $|m^2-2005n^2|\geq 1$ for all natural $m,n$, since $\sqrt{2005}$ is irrational, hence there are no natural solutions to $x^2-2005y^2=0$. Therefore

$|m-n\sqrt{2005}|=\frac{|m^2-2005n^2|}{m+2005n}>\frac{1}{90n}.$[/hide]


Good luck to everyone writing the Euclid tomorrow!!!!!  :D   :D   :D

I thought that was harder than previous exams. I had major problems with number 8(b) and had no idea where to start. How did everyone else do?

Guys;ICAE is taking it on Sunday, so don't discuss it until after then.

Good thing I'm not taking it.

[quote="xpmath"]Guys;ICAE is taking it on Sunday, so don't discuss it until after then.]

some have posted all questions and discussed them in other forum..
just google with keywords: euclid contest 2008

so there is no point to restrict the discussion here.

ithink they would be unofficial contestants.

"ELIGIBILITY
Full-time students registered in your school are classified as official contestants and are eligible for
individual awards and team competition if:
(a) their birth date is on or after January 1, 1989, and
(b) they have been in attendance at a Canadian secondary school for at least twelve months prior to
December 31, 2007.
NOTE: Students who do not meet these requirements are welcome to write the Contest as unofficial
contestants. Their scores will not be included in team totals and they will not be eligible for individual
prizes."

i'm not sure about fgh though

[quote="TheHeinrich"]ithink they would be unofficial contestants.
"ELIGIBILITY
(b) they have been in attendance at a Canadian secondary school for at least twelve months prior to
December 31, 2007.
NOTE: Students who do not meet these requirements are welcome to write the Contest as unofficial
contestants. Their scores will not be included in team totals and they will not be eligible for individual
prizes."
[/quote]

ah good, since it is very easy to google and find this in the first page of search result:
http://forums.studentawards.com/yaf_postsm24122_EUCLID-CONTEST-Scores-Answers-share-them-here.aspx

people on that forum don't exactly work on problem solving in the same standard as Mathlinks/AoPS :wink:

[quote="cnola"][quote="TCAtect"]I did pretty terribly (on my standards) this year. So nervous about 7b...I didn't count 0 as a multiple of 4...[/quote]
7b) 7/27 right? how about 9 & 10? anybody finished it?[/quote]

7/27 was correct.

9a) 2,-log250
9b)  (1,2,3)
10a)13,22
that was as far as i got ...after the contest  :blush:

[quote="zepiphanus"]I finished 9 a) and 10 a), b)... but solution to 9 b) was so beautiful when my friend proved it to me[/quote]
9b was an olympiad style problem but simpler

[quote="zepiphanus"]I finished 9 a) and 10 a), b)... but solution to 9 b) was so beautiful when my friend proved it to me[/quote]

9b) is too hard for me.

how do you all  compare this year contest with the previous one?

hey guys....for 7b, i think zero cant be counted because the question is asking for the probability such that the sums "differ" by a factor of 4, so i dont think the case when the difference is zero should be counted....

For 7b the 0 should be counted... that's usually a trick for these types of problems... I think that the hard one for most people in my school were 8b and 9b (though I got them  :lol: )...

Did anyone get 10b or 10c?

I think that 10b was an induction, but 10c :(  :(  (I wrote on the test that I conjectured that there were no n that were divisible by f(n) but had no time left to prove it...)

[quote="TheHeinrich"][/quote][quote="PasswordIs0"][quote="TheHeinrich"][quote="PasswordIs0"]
my 90 is probably gone with the wind even if they dont count the 0 case....[/quote]

An, are you crapping me!?!?, so your mark is going up... mine is going way down :([/quote]

KK~ im not joking. u know how mr.white marked my paper..
and they screwed up... cuz they didnt deduct marks for my #7 a) for not counting the 0 case...
so basically, everyone's going to get 2 marks back, except for me..
and for 9 a) and 9 b),, which worth 5 marks each..
i totally screwed up on both of them......

well, we will see~ my score is so going down T_________T!![/quote][quote="TheHeinrich"]

wait!?! Mr. White marked your paper? does that mean he marked mine also, if so, i hope he didn't change his mind after the unofficial marking![/quote]

honestly im not sure who marked my paper..~~ iwas just assuming

but ..~~ ok, srsly, even he goes to waterloo to mark, he wont mark our school's,  just put ur hope on the teacher whoever marks urs.. good luck...

if i mark urs.. hehe..hehehehehe... hehehehe.... THIS GUY! FAIL =PPP

hhmmm...seems like i got around a 94 (or lower)...sad... :(

... srsly, when can we know our result..
i hate estimating ~~ahh~... !! T_T||

are we allowed to complain about our marks?

[quote="TheHeinrich"]school results are out  :(  :([/quote]

Do you mean that the results right now are only available to schools???

[quote="TheHeinrich"]school results are out  :(  :([/quote]

I saw nothing on the CEMC site, how do you know? why those double sad faces?

[quote="mathgeniuse^ln(x)"]Well, my teacher told my results.

I got 100.[/quote]

Wow! Congratulations!!! :D  :D

Thanks.

It seems like the graders were easier this year.

The solutions are up at: [url]http://cemc.math.uwaterloo.ca/english/contests/euclid.shtml[/url]

It looks like the averages are a few points lower than last years... this could be expected, as I think most would agree that this year's was harder than last year's... :lol:

Sorry to double post -- I couldn't edit my post...

Results are up on the CEMC website... I feel really bad about my score now...  :(  :(  [url]http://cemc.math.uwaterloo.ca/english/contests/past_result/2008/2008EucResults.pdf[/url]


Let f_n = the number of good numbers of length n.
Let a_n, b_n, c_n = those good numbers beginning with 0,1,2 respectively.

Then f_n = a_n + b_n + c_n.

If the number begins with a 0, the next digit can be any of 0,1,2.
Thus a_n = a_(n-1) + b_(n-1) + c_(n-1)

If the number begins with a 1, the next digit can be any of 0,2.
Thus b_n = a_(n-1) + c_(n-1).

If the number begins with a 2, the next digit can be any of 0,1.
Thus b_n = a_(n-1) + b_(n-1).

f_n = a_(n-1) + b_(n-1) + c_(n-1) + a_(n-1) + c_(n-1) + a_(n-1) + b_(n-1).
= 2*f_(n-1) + f_(n-2)

There are 3 length 1 good numbers, and 7 length 2 good numbers.
f_3 = 2*f_2 + f_1
 = 2*7 + 3 = 17
...
f_10 = 2*f_9 + f_8
 = 8119 (unless i calculated wrong)

nope, you didn't calculate wrong. 8119 is the solution indeed :)

[quote="Peter"]Work base 3. (so each digit is 0,1,2)

A good number of size $n$ is a number in which there are no consecutive $1$'s and no consecutive $2$'s. How many good 10-digit numbers are there?[/quote]
Another interpretation of the problem is possible: It is actually a base 3 number, and so the first digit cannot be 0, in which case we get the following solution.

[i]Solution.[/i] Let $u_n$ denote the number of [i]good[/i] numbers of length $n$. Consider an arbitrary good 10-digit number. WLOG it starts with a 1 (we can swap later, since 1 and 2n in this problem are equivalent to each other by the problem statement). Suppose there are $k$ digits before the digit $0$. Then the first $k$ digits are necessarily of the form $\underbrace{1212...}_{k-1\ \textrm{digits}}$. Once the $k$th digit 0 has been placed, there are $u_{10-k}$ ways to complete the number. Thus, summing from $k=1$ to $k=9$, we see that the number of 10-digit good numbers starting with 1 is just $\sum_{i=1}^9 u_i$. Similarly, by swapping all the 1's with 2's, we find that the number of 10-digit good numbers starting with the digit 2 is also $\sum_{i=1}^9$. Thus, $u_{10}=2\sum_{i=1}^9 u_i$. Similarly, $u_9=2\sum_{i=1}^8 u_i$. It follows $u_{10}=2u_9+2\sum_{i=1}^8 u_i=2u_9+u_9=3u_9=3^2u_8=\cdots = 3^9u_1=\boxed{2\cdot 3^9}$. $\blacksquare$

Arkan

Hmm, it sounds weird to me that by omitting the numbers starting with 0 (which is indeed a valid interpretation), the number of solutions is suddenly larger. I also don't think that multiplying by 3 can ever be correct since that would mean that we can add any digits to the end, completely disregarding the restrictions. You may want to correct your solution.


Princeton
college
ARML
HMMT
geometry
Princeton Math Comp.

If you hadn't heard about it already, I'm making a team for AHSIMC, the Princeton Math Competition, and HMMT.  We have quite a few of the Upstate New York ARML team (at least, those not living in Albany.)  So if you want to join the team, just contact me through a pm or something.

for HMMT? I think that I'm gonna compete in the official one, but I'm not sure...

The official team?  Do you mean AAMC, or is there another team I don't know about?

And incidentally, our team may not be official, but we've been around in various forms for a few years.

If AAMC stands for Albany Area Math Circle, then sure.

P.S. don't sign me up on anything!!


Let $a,b,c,d$ be non-negative real numbers such that $a+b+c+d=4$. Prove that
\[a^{2}+b^{2}+c^{2}+d^{2}-4 \ge 4(a-1)(b-1)(c-1)(d-1). \]
This nice gift I want to send to my old friend Arqady but all other Mathlinkers are welcome.

[quote="d.edgar"]The most interesting case it's when exactly two of $a,b,c,d$ are greater than 1.

Suppose ${a}\le{b}\le{1}\le{c}\le{d}$. Let $x=1-a, y=1-b, z=c-1, t=d-1$. We have $x+y=z+t$ and $0\le{y}\le{x}\le{1},0\le{z}\le{t}$. We need to prove $(1-x)^{2}+(1-y)^{2}+(1+z)^{2}+(1+t)^{2}-4\geq{4xyzt}$ and this is equivalent to ${x}^{2}+{y}^{2}+{z}^{2}+{t}^{2}\geq{4xyzt}$.

By AM-GM we have ${x}^{2}+{y}^{2}+{z}^{2}+{t}^{2}\geq{4\sqrt{xyzt}}$, and $4(xyzt)^{1/4}\leq{x+y+z+t}=2(x+y)\leq4\implies \sqrt{xyzt}\leq1$ hence $\sqrt{xyzt}\geq{xyzt}$ so we are done.[/quote]

Yes! It is very nice proof, d.edgar. It is quite the same as me.   My own is following
\[a^{2}+b^{2}+c^{2}+d^{2}-4=(a-1)^{2}+(b-1)^{2}+(c-1)^{2}+(d-1)^{2}\ge 4\sqrt{|a-1)(b-1)(c-1)(d-1)|}. \]
It suffices to consider one case: $a\ge b\ge 1\ge c\ge d$, then (notice that $a+b\le 4$)
\[(a-1)(b-1) \le \frac{1}{4}(a+b-2)^{2}\le 1 \]

\[(1-c)(1-d) \le 1 \]
So
\[\sqrt{|(a-1)(b-1)(c-1)(d-1)}\ge (a-1)(b-1)(c-1)(d-1) \]
and we have done. The equality holds for $a=b=c=d=1$ or $a=b=2,c=d=0$ or any permutations.

Following is a harder one (with the same condition). I hope Arqady will find a better proof than me.
\[a^{2}+b^{2}+c^{2}+d^{2}-4 \ge-4(a-1)(b-1)(c-1)(d-1). \]

Thank you, hungkhtn, for your gifts! :lol:  I'll try it.

Dear Arqady,

In this inequality, the equality holds for $a=4,b=c=d=0$? I can not see in your last expression this case of equality?

Oh... I see. It is certain that the equality can holds. Nice proof, Arqady. I have a simpler proof but I will wait others by some days, then I will post it.

From two gifts above, we combine them to

If $a,b,c,d$ are non-negative and $a+b+c+d=4$ then
\[a^{2}+b^{2}+c^{2}+d^{2}-4 \ge 4|(a-1)(b-1)(c-1)(d-1)|\]
with equality holds for three case: $a=b=c=d=1$, $a=b=2,c=d=0$, $a=4,b=c=d=0$ (and every permutation of course).

It is very nice, isn't it? Arqady?

As my respect to Arqady, I will continue dedicating him another problem and notice that all Mathlinkers are welcome.

If $a,b,c,d$ are positive and $abcd=1$ then
\[a^{2}+b^{2}+c^{2}+d^{2}-4 \ge 2(a-1)(b-1)(c-1)(d-1).\]

Just because Arqady doesn't show any solution for the above nice problem, I will send him another gift and hope that at that time, he will show me a solution instantly.

If $a,b,c,d$ are non-negative such that $a^{2}+b^{2}+c^{2}+d^{2}=4$ then
\[a+b+c+d-4 \le 2(a-1)(b-1)(c-1)(d-1).\]

Notice that in all my problems in this topic, the coefficients of $(a-1)(b-1)(c-1)(d-1)$ (as $4,-4,2,2$) are the best possible,

[quote="arqady"][quote="hungkhtn"]
Let $a,b,c,d$ be non-negative real numbers such that $a+b+c+d=4.$ Prove that
\[a^{2}+b^{2}+c^{2}+d^{2}-4 \ge-4(a-1)(b-1)(c-1)(d-1). \]
[/quote]
$a^{2}+b^{2}+c^{2}+d^{2}-4 \ge-4(a-1)(b-1)(c-1)(d-1)\Leftrightarrow$
$\Leftrightarrow\sum_{sym}(a^{3}b+a^{2}b^{2}-3a^{2}bc+abcd)\geq0\Leftrightarrow\sum_{sym}(a-b)^{2}(c^{2}+d^{2}-cd+ab)\geq0,$
but your proof better. I have not found a nice proof.[/quote]

Let $x = a-1$, $y = b-1$, $z = c-1$, $w = d-1$, we are to prove
$x^{2}+y^{2}+z^{2}+w^{2}+4xyzw \ge 0$
$\Leftrightarrow 2xyzw \ge (x+z)(y+w)+xz+yw$
$\Leftrightarrow (x+z)^{2}+2xyzw \ge xz+yw$
Only the case $xyzw \leq 0$ is of interest. Without loss of generality assume that $yw \leq 0$, then $xz \ge 0$
First $(x+z)^{2}\ge 4xz$ so it suffices to show $3xz+2xyzw \ge yw \Leftrightarrow 3xz-yw \ge-2xyzw$

Claim $|xyzw| \leq 3$. Once we have this, then by amgm
$3xz-yw \ge 2\sqrt{3(-xyzw)}\ge 2(-xyzw)$

To prove the claim,
we consider two cases.
1. x,y,z positive, w negative. In this case $|x|+|y|+|z|+|w| = x+y+z-w =-2w \leq 2$ because $w \ge-1$. Then by amgm we see that $|xyzw|$ is bounded above by 3.

2. x,y,z negative, w positive. In this case,
$-xyzw \leq \left(\frac{(-x-y-z)}{3}\right)^{3}w = \frac{1}{27}(x+y+z)^{4}$
because $-3 \leq x+y+z \leq 0$, we must have $-xyzw \leq 3$, so we are done.

Good reasoning, Soarer. It is quite the same as mine, although I use another way.

How to prove the last one?

[quote="hitek"][quote="hungkhtn"]Just because Arqady doesn't show any solution for the above nice problem, I will send him another gift and hope that at that time, he will show me a solution instantly.

If $a,b,c,d$ are non-negative such that $a^{2}+b^{2}+c^{2}+d^{2}=4$ then
\[a+b+c+d-4 \le 2(a-1)(b-1)(c-1)(d-1). \]
Notice that in all my problems in this topic, the coefficients of $(a-1)(b-1)(c-1)(d-1)$ (as $4,-4,2,2$) are the best possible,[/quote]
Very nice and interesting inequalities. Good boy.[/quote]

Thanks, but I am not a boy. I am a student  :lol:


analytic geometry
inequalities
triangle inequality

Let A be the point (7,4) and D be (-5,-3). What is the length of the shortest path
ABCD, where B is a point (x, 2) and C is a point (x, 0)? This path consists of three
connected segments, with the middle one vertical.

[hide="Answer"]The shortest distance between any two points is that along a straight line. Consider the segment AD, which is the shortest path from A to D. Note the y coordinates of the four points are decreasing: 4 > 2 > 0 > -3, so there is a point on AD with a y coordinate of 2, and a point on AD with a y coordinate of 0. Let these two points be B and C, respectively. Since we have achieved the lower bound that we derived from just going from A to D, this is the desired path.

The length is $ \sqrt{(7\minus{}(\minus{}5))^2\plus{}(4\minus{}(\minus{}3))^2}\equal{}\sqrt{193}$[/hide]

[hide="Same x-coordinates  :wink: "]
The desired length is $ AB \plus{} BC \plus{} CD \equal{} 2 \plus{} AB \plus{} CD$. Translate $ \overline{AB}$ two units downwards, which takes $ B$ to $ C$ and $ A$ to $ A'\ (7,2)$, such that $ AB \equal{} A'C$. Then the shortest possible value of $ AB \plus{} CD \equal{} A'C \plus{} CD$, by the Triangle inequality, is $ A'D \equal{} \sqrt {12^2 \plus{} 5^2} \equal{} 13$; thus the answer is $ \boxed{15}$.

[/hide]

[quote="fishythefish"]But 15 is greater than the square root of 193...I think the earlier post was right...[/quote]

There was a slight misreading by TZF (the second segment is vertical, so straight-line distance impossible).

Oh, right...sorry...The funny thing was I was just thinking about that as I posted... :blush:


I have one question that has to be simple but for some reason i haven't solved it yet.
How can I express the vector of the angle bisector in terms of the vectors which build the triangle ?

Thanks, Pavel

Assuming that the angle bisector does not have to have a given magnitude, you could write both vectors in polar form.  Then, the angle between the vectors becomes theta 1 minus theta 2, and the angle bisector will have a theta with half of this measurement.  But if the none of the original vectors lie on the x axis, you will have to rotate them to the x axis and add this to find the angle bisector.


function
calculus
integration
limit
trigonometry
calculus computations

How do you simplify the following as a function of x?

$\lim_{n \to+\infty}\sqrt{n}\[ \int_{0}^{x}z^{2}e^{n (\cos(z)^{2}-1)}\,dz$
 $for$   $x\geq 0$

(I apologize if this has been asked before)

I am sorry if I didn't get this right: is that $\cos^{2}z$ or is it $\cos (z^{2})$?

I dropped a set of parentheses:  I meant $\cos^{2}(z)$

I thank you in advance for any help you might offer.


function
real analysis
real analysis unsolved

1. Let $ g$ be absolutely continuous with $ c \ge g \ge d$ on $ [a,b]$. Let $ E \equal{} \{x | g'( x ) \equal{} 0 \}$. Show that $ m(g(E)) \equal{} 0$.
2. Let $ g$ be an absolutely continuous monotone function on $ [0,1]$ and $ E$ a set of measure 0, then $ g(E)$ has measure 0.

For the second question, is this correct?
For $ \epsilon > 0$, pick $ \delta$ for $ g$ so that for a finite sequence of disjoint intervals of total length less than $ \delta$, the total length of intervals defined by the image of endpoints is less than $ \epsilon$. Consider an open cover of $ E$ of length less than $ \delta$. Write it as a disjoint countable union of open intervals. For a finite union of these intervals, using absolute continuity we see that the image of the intervals is covered by intervals with total length less than $ \epsilon$ (by monotonicity). Therefore $ g(E)$ has measure 0.

yes, your proof seems to me perfectly correct  :)

for the 1st one, for every $ x$ such that $ g'(x)\equal{}0$, you can find an open ball $ B(x, r_x^{\epsilon})$ such that $ m(g(B(x, r_x^{\epsilon}))) \leq \epsilon m(B(x, r_x^{\epsilon}))$, and with $ \sup r_x^{\epsilon} < \plus{}\infty$. Then Vitalli covering lemma finishes the job.

Thanks! Just hope that I do not misunderstand you: (I'll work in R^1)

For each $ x$ there is an $ h_0$ so that $ |g(x \plus{} h) \minus{} g(x)| < \epsilon h$ for $ h < h_0$. Use these $ [x,x \plus{} h]$ to cover $ E$. Vitali's lemma says that most of $ E$ can be covered. For the rest, use absolute continuity to conclude that the image must have small size. Using this we are able to get some epsilon bound. As epsilon goes to 0 we are done. Am I right?


probability
combinatorics unsolved
combinatorics

The maths deparment has 8 graduate who are assinged to same office.each assitant is equaly lIkely to study at home as in the office .how many desks are must be there be in the offices so that each assistant has a desk at least 90% of the time

if there are 6 desks, the probability of everyone who comes to the office having a desk is $ 1\minus{}(1\plus{}8)/256>0.9$, but if there are only 5 desks, the probability is $ 1\minus{}(1\plus{}8\plus{}28)/256<0.9$, so the answer is 6.


There are distinct $ n$ cards labeled from $ 1$ to $ n\geq 3$ in a bag. When we draw three cards from the bag and arrange them in the order of the size of the written numbers. Find the probability such that these cards form arithmetic progression.

An AP of length $ 3$ is completely determined by its extreme terms, which have to have same parity for the middle term to be integer. There are $ a \equal{} \genfrac {(} {)} {0pt} {} {\lfloor n/2 \rfloor} {2}$ such pairs of even extreme terms, and $ b \equal{} \genfrac {(} {)} {0pt} {} {\lfloor (n\plus{}1)/2 \rfloor} {2}$ such pairs of odd extreme terms. There are therefore $ a\plus{}b$ triplets which are AP's, and $ c \equal{} \genfrac {(} {)} {0pt} {} {n} {3}$ triplets in all, so the probability is $ \frac {a\plus{}b} {c}$.


algorithm
number theory
Euclidean algorithm

Does the Euclidean Algorithm have a simple proof?

Let (a,b)=d, and write a=dm and b=dn. Then a-b=dm-dn is still divisible by d, and the process continues. I think it's pretty clear that the process will eventually terminate.

I think the euclidean algorithm is pretty much two small lemmas:;

1. (a, b) = (a, b-a).

2. (a, 0) = a

both of which are pretty easy to prove.

Now you can use the well-ordering principle to show that eventually you get from (a,b) to (a, b-a) and eventually to some (d, 0), by considering at each step (x,y) the sum x+y.  And therefore d is your gcd.


Let $ x,y\in R , x^2 \plus{} y^2 \equal{} 3$
Find the least and the greatest value of:
$ P \equal{} \frac {\sqrt {x^2 \minus{} 2y \plus{} y^2} \plus{} \sqrt {x^2 \plus{} 2x \plus{} 4y \plus{} y^2}}{\sqrt {x \plus{} y \plus{} 3}}$
EDIT: sorry there was a bit error in my problem, I have fixed it :D

Could you solve this problem, please? :huh:

Kyoshiro, the minimum value is sqrt(2) and the maximum value is 2.

First take x=sqrt(3)*cos k and y=sqrt(3)*sin k.
Place the values of x and y in the given expression and after a bit of simplifying, differentiate it to get the maximum and minimum values.
[/hide][/url][/i][/list]


Given that I know the exact variance of the residual(It's normality well established), what is the exact distribution of Y_hat as a function of data vectors X, Y, and the variance of the residual?

It isn't too hard to find confidence intervals, but the actual distribution eludes search. Attempts to derive it have been fruitless.

Anyone know?

[quote="alekk"]could you define 'your' vocabulary: variance of the residual, $ \hat{Y}$, etc..[/quote]

Sure, $ Y_t\equal{}B_0\plus{}B_1*X_1t\plus{}B_2*X_2t\plus{}...B_k*X_kt\plus{}u_t$, where $ u_t \sim  Normal(0,a^2)$  When I run an OLS regression, plugging vectors X and Y into the normal equations, I will obtain $ \hat{B_0},\hat{B_1},...\hat{B_k}$ .

$ \hat{Y_t}\equal{}\hat{B_0}\plus{}\hat{B_1}*X_1t\plus{}...\hat{B_k}*X_kt$

The data vector is a random variable because of $ u_t$, so the coefficients, and ultimately the regression output, will also be random variables. I'd like to find it's distribution(Or at least distributional form)

[quote="roah"]I assume that all $ X$ are constant.
$ u$ is normal, so $ Y$ is normal.
$ Y$ is normal, so $ \hat{\beta}$ is normal.
$ \hat{\beta}$ is normal, so $ \hat{Y}$ is normal.[/quote]

From what I can see, the $ \hat{\beta}$ 's are not necessarily independent. For example, if you have two variables X that are heavily correlated with each other, then their coefficients will change similarly in response to data.

So I don't see why Y would be normal


The sequence of integers $\{ x_{n}\}_{n\ge1}$ is defined as follows: \[x_{1}=1, \;\; x_{n+1}=1+{x_{1}}^{2}+\cdots+{x_{n}}^{2}\;(n=1,2,3 \cdots).\] Prove that there are no squares of natural numbers in this sequence except $x_{1}$.

Notice that for all $n>1$, $x_n$ is even.  Thus, $x_n\equiv 1+1+0+0+\cdots\equiv 2 \pmod 4$.
Thus, $x_n$ can never be a perfect square.


pigeonhole principle
real analysis
real analysis solved

Let A be a finite set of real numbers. Show that there exists a non-zero integer n, such that each member of nA has an integer within distance < 0.0001 from it.

First note that for any x, within the range $1<=n<=10^4$, there is an n such that {xn} (ie the fractional part of x.n) < 0.001. (using pigeonhole or something similar)

Similarly we can say that for any x_i in A, that within range $1<=n<=10^5$ there is an n_i such that x_i*n_i has fractional part <0.0005.

(actually, I think the top 2 parts are unnecessary.. but forgive me, I'm very rusty)

Then suppose there are k elements in A (call them $x_1, x_2$ etc). Consider the range $1<=n<=(2000^k)+1$. Partition the interval [0,1) into 2000 sections of equal length, starting with [0,0.0005). Then by pigeonhole, two of the n values, say a and b, are such that for every $1<=i<=k$, x_i*a and x_i*b are in the same section. (as there are 2000^k possible configurations of sections - 2000 sections, k variables)
Then consider n=|a-b|. x_i*n is the difference of two numbers less than 0.0005 away from an integer, and hence is less than 0.001 away from an integer. This holds for any i, so we have the required result.


$ E \neq \emptyset, \Omega \in P(E)$, $ \Omega$ is a borelian field if :
1) $ \Omega \neq \emptyset$
2) $ \any A_1, ... , A_n \in \Omega \Rightarrow \bigcup_{n\in N^*}A_n \in \Omega$
3) $ \any A \in \Omega \Rightarrow C_A \in \Omega$

do you mean $ \Omega \subset P(E)$ ? because then this seems a lot like an algebra.. or a sigma algebra..?

Any $ \Omega \equal{} P(S)$ for some set $ S$ works, and is surely not always finite.


integration
calculus
parameterization
real analysis
real analysis solved

Fidns an equivalent for $\sum_{n>0}{\frac{x^n}{n!^2}}$ for large x.

$f(x^2)= \sum_{n \geq 0}\frac{x^{2n}}{(n!)^2}=\frac{1}{2\pi}\int_{0}^{2\pi}e^{2x\cos(w)}dw$

with Laplace method for finding equivalent of integral parametre $x$

(It is well known in MP*)

we get $f(x) \sim \frac{1}{2\sqrt{\pi}}\frac{e^{2\sqrt{x}}}{\sqrt[4]{x}}$ when $x$ tends to $+\infty$

There is a  complete article of different method for finding
equivalent of integral with parametre in the

RMS (Revue de Mathmatiques Spciales) 1985-1986
page 367-382
"Limites et quivalents de fonctions dfinies par des intgrales
 ou sries dpendant de paramtres $\int_{u(t)}^{v(t)}f(t,x)dx$

by L.G.Vidiani Professeur de Mathmatiques Spciales MP*

I don't know the methods used in MP*, but I know I struggled a lot to kill this problem. After many days and two pages and a half, I proved that this is the equivalent. Anyway, nice problem!

Damn, why is it just in "some MPSI"? It's not fair!!!!  :D  :D  :D

here the "mthode de Laplace"

http://www.les-mathematiques.net/c/a/c/node7.php3#cacf1


algebra
polynomial
invariant
linear algebra
matrix
least common multiple
linear algebra theorems

suppose an operator A on finite dim. v. s. V can be shown as A=A_1+A_2, where the sum is direct   (meaning there are subspaces V_1 and V_2 invariant for A, such that V=V_1+.V_2, and A_1 is induced on V_1 and A_2 on V_2).

can someone prove that mi_A=mi_A_1 * mi_A_2, where mi_A is the minimal polynomial of A? tnx!

[quote="Matija"]suppose an operator A on finite dim. v. s. V can be shown as A=A_1+A_2, where the sum is direct   (meaning there are subspaces V_1 and V_2 invariant for A, such that V=V_1+.V_2, and A_1 is induced on V_1 and A_2 on V_2).

can someone prove that mi_A=mi_A_1 * mi_A_2, where mi_A is the minimal polynomial of A?[/quote]

Are you sure about this? Obviously, $\text{mi}_{A}\mid\text{mi}_{A_{1}}\cdot\text{mi}_{A_{2}}$. But we can take the matrix $A=\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right)=A_{1}+A_{2}$, where $A_{1}=\left(\begin{array}{cc}1&0\\ 0&0\end{array}\right)$ and $A_{2}=\left(\begin{array}{cc}0&0\\ 0&1\end{array}\right)$, and $\text{mi}_{A}\left(X\right)=X-1$, so we do [b]not[/b] have $\text{mi}_{A}=\text{mi}_{A_{1}}\cdot\text{mi}_{A_{2}}$ (since the polynomial $X-1$ is not the product of two polynomials of degree $\geq 1$, and minimal polynomials always have degree $\geq 1$).

What does hold is $\text{char}_{A}=\text{char}_{A_{1}}\cdot\text{char}_{A_{2}}$, where $\text{char}$ stands for characteristic polynomial. This is easy to conclude from $\text{char}_{A}\left(X\right)=\det\left(A-XI\right)$, where $I$ is the unity matrix of the same size as $A$.

  darij

I know about the characteristic polynomial..  When I'm proving that every operator has the Jordan form I need that A-lambda_i*I is a nilpotent operator of index p_i, where p_i is the multiplicity of eigenvalue lambda_i in mi_A..
that could be done if the equality, which I asked to be proven in last topic, holds. if it doesn't hold, can you help me to complete the proof?

The minimal polynomial of the direct sum $A=A_{1}+A_{2}$ is the least common multiple of the minimal polynomials; just look at the action of $p(A)$ on each invariant subspace: $p(A)=p(A_{1})$ on $V_{1}$ and $p(A)=p(A_{2})$ on $V_{2}$. $p(A)$ is zero if and only if both of these are zero.

Look carefully at your proof- you shouldn't need more than this.

thanks a lot! that did it!

notes i have are quite a mess, but now everything fits :)
it the proof minimal polynomials of A_1 and A_2 have no common factor, so the claim from my first post was true, but I like you're general statemnet more :)


In how many ways can 8 people buy a 25-dollar flashlight if 3 people pay at least 4dollars?

[quote="WindSlicer"]Can some people not pay? Are we only working in 1 dollar increments?[/quote]

The answer to both questions is YES. I should have made that more clear.

Use a generating function.
$x^{12}(1+x+x^2+x^3+...)^8$.  Then find the coefficient of $x^{25}$, or just the coefficient of $x^{13}$ in $(1+x+x^2+x^3+...)^8$.

you used the method of forming an eqn. and finding it's no. of integer soln. right?!


advanced fields
advanced fields unsolved

Hi there!

I need help with a proof that the Two's Complement's representation is unique -> There are no two representations of a number.

The following formula describes the Two's Complement: $ v(a_{n\minus{}1}a_{n\minus{}2}...a_1a_0) \equal{} \minus{}a_{n\minus{}1}*2^{n\minus{}1}\plus{}\sum_{i\equal{}0}^{n\minus{}2}a_i*2^i$,

where $ v(a_{n\minus{}1}...a_0)$ is the value of the binary number $ a_{n\minus{}1}...a_0$. n is the number of digits. For example:

$ v(10_2) \equal{} \minus{}1*2^{2\minus{}1} \plus{} \sum_{i\equal{}0}^{2\minus{}2}a_i*2^i \equal{} \minus{}1 * 2^1 \plus{} 0*2^0\equal{}\minus{}2$.
So  the $ \minus{}2$(decimal) is $ 10$(binary; one zero, not ten) in the Two's Complement representation.

I tried proving it with a proof by contradiction, but I didn't get very far.
Any help?

Thanks in advance,

BS

No one?

I tried to solve it with a proof by contradiction:

$ x\equal{}\minus{}a_{n\minus{}1}*2^{n\minus{}1}\plus{}\sum_{i\equal{}0}^{n\minus{}2}a_i*2^i\equal{}\minus{}a_{m\minus{}1}*2^{m\minus{}1}\plus{}\sum_{i\equal{}0}^{m\minus{}2}b_i*2^i$

$ n \neq m$ or $ a \neq b$

So I assumed that there are two possible representations of x. Now I have to show that there is a contradiction.

But I have no idea how to do something like that.

Any help?

BS


How many zeros are at the end of the product 25  $ \times$ 240?

There are 3 factors of 5 and at least 3 factors of 2 (24 is divisble by 8).  Thus the answer is 3.

Sorry for reviving this thread I just have a good solution
[hide="Good Solution?"]To count the number of zeros at the end of $25\times240$, we must count the number of products $2\times5$. We prime factorize the product by combining the prime factorizations of 25 and 240: $25\times240=(5^2)(2^4\times3\times5)=2^4\times3\times5$. We take the minimum of the powers of 2 and 5 to find that $25\times240$ has $\boxed{3}$ zeros.[/hide]


[quote="Gil"]Let be the triangle ABC, the incircle C=C(I, r) and the point P of the circle C thus that the line BC separates the points A and P. The tangents from the point P at the circle C touch the lines AB and AC in the points M and N respectively. Prove that there is the following relation: (MB.NC):(PB.PC)=(MI.NI):(AI.PI).
A particular case: if the point P belongs at AI, then the relation (1) becomes:
1+[(IM.IN):(BM.CN)]=(b+c):a. (A remaking of a Toshio Seimyia's problem).[/quote]

It's strange, but on my sketch, neither your problem, nor Seimiya's one seems to be correct. Note that I assumed that the point P lies on the incircle of triangle ABC "between" the points where this incircle touches the sides CA and AB (actually, your condition that "the line BC separates the points A and P" is absurd - it can never happen for a point P lying on the incircle of triangle ABC). Maybe there are some typos in the problem?

  Darij


summer program
MathPath
geometry
geometric transformation
reflection

I just heard this on the radio.  George Bush made a map of the United States, and he highlighted all the states he was going to campaign in.  The only problem is...there are only 49 states. Yea, okay, Rhode Island is not part of Connecticut.  He did not even put us on the map (it includes all the other states, not just the ones he is going to).  I am so insulted, are we not worth his printer ink?  Please don't vote for Bush!

I echo your anti-Bush sentiment, however, Rhode Island is indeed part of Connecticut.

JSRosen, I think someone might need a geography lesson...

Rhode Island very well could have been lost in an ink smudge. Insignificant.

[quote]Insignificant.[/quote]

Rhode Island may be small, but it is by no means insignificant.  We have a very rich culture and a long history.  I don't think you have any reason to call my state insignificant!

Well, since we're two of the three smallest states we might as well merge and become Rhodecticut.

Hm, small is not necessarily bad.  That's one of the many problems with society today, bigger is not always better.

LOL.  Look Rhode Island is not part of Conneticut.  It is not insignificant but it very easily could have been lost in a smudge or something like that.  So you said he was going to campain in 49 states?  Is the only state hes not campaining in Rhode Island?  or is it Alaska or Hawii like I think it is?  Because you know he hasn't forgotten about Rhode Island if he is going to campain there.  Really just because Rhode Island got lost or something is no reason to really get mad.  It is a small state about the size of Ft. Worth Texas.  It still is important but you can't get mad like that.  It sounds more funny than angering.

You mean than angering, BHorseMath?  Haha...umm... RI is all set being the smallest state in the US... Plus, me and Julie are gonna put it on the map.  Literally.   :-D

[quote="JSRosen3"] however, Rhode Island is indeed part of Connecticut.[/quote]

Yeah, it is  :D

this has the distinct feel of an urban legend until someone can post the map and prove otherwise

BHorseMath - nah, he's not campaigning in RI, but he made a map that had every state outlined, and he colored the state's he was going to campaign in orange, but every state is outlined, even the ones he is not going to, except Rhode Island. Bush has no chance of winning Rhode Island b/c we always are democratic, but most practical minded politicians don't bother coming to Rhode Island since we have so few electoral votes.  It's all good though, because our state has a great bunch of local politicians, you know what I mean  ;)

Rosen and Yif man better watch out!  :)

http://www.georgewbush.com/JourneyswithJohn/

This is all I could find, and it's a map of where Kerry is going from a Pro-Bush website.  It's probably the same map that Bush was talking about.  I slept through that boring "researching the internet" class so..someone else will be able to find the actual thing.

RI ain't on there though...

A few months ago, this might have been justifiable by the fact that the border between CT and RI was actually under dispute (see http://www.senatereps.state.ct.us/press/cook/051304.htm ).

Ironically, George W. Bush was born in Connecticut.

Bush also took a shot at the entire state of Massachusetts last week, saying something along the lines of "That's how they balance the ticket in Massachusetts."

It wasn't as if I was going to vote for him anyway, though.

EDIT: fixed link

Hmmm...I think there should be a poll:  How Long Will It Take The Bush Crew To Figure Out There Is Indeed 50, NOT 46 States?  Hmm...

Edit:  Haha, I just realized that.  :-D

[quote="SingTheSorrow"]Hmmm...I think there should be a poll:  How Long Will It Take The Bush Crew To Figure Out There Is Indeed 50, NOT 49 States?  Hmm...[/quote]

try 46.  not only is this map missing Alaska and Hawai'i as expected, but the Rhode Island problem you noted appears to apply to Delaware as well.  Oh and thanks to whoever posted the link to the map!

-Tim

Well, it would have to be at least as long as it would take them to learn to count that high.

Actually, the map only shows 47 states.  Alaska and Hawaii are omitted, the former being solidly Republican and the latter Democratic.

EDIT: farmertim pointed out that Delaware is also missing.

[quote="JS1527"]Guys, come on, counting does not come easy to some people, we have to be sympathetic, *sarcasm* the fact that they are in our nation's highest office is um...scary[/quote]

Well, you must remember that counting isn't important because "reading is the basics for all learning" and one need not know how to read to count.

Isn't it ironic that the campaign of the guy who accused Al Gore of "fuzzy math" now implies that there are 46 states?

[quote="Rep123max"]Is Yale in Texas? Nope.[/quote] Um, remember this is George W. Bush's map you're dealing with.

[quote="JS1527"]Look below the map:

"Did we miss something? Please tell us!"

Um...yea...where's Rhode Island?

Yea, the bush crew is so intelligent...*ahem*[/quote]
Rhode Island would clearly be lost on that map to a combination of the thickness of the lines and lack of small detail. The map lobbed off several large penninsulas of Michigan, each of which is larger than Rhode Island. Since I live on one of those penninsulas, I suppose I'm not on the map either, but I'm not complaining.

[quote="JS1527"]BHorseMath - nah, he's not campaigning in RI, but he made a map that had every state outlined, and he colored the state's he was going to campaign in orange,[/quote]

President of the United States George W. Bush couldn't possibly be spending his time PERSONALLY making any kind of map. That's done by staff, for sure, if it is done at all. (Edit after seeing the link provided by the original poster farther along in this thread:) I think the problem must be that Kerry made a campaign visit to CT at a time distinct from any visit he made (has there been one?) to Rhode Island. Because the message on the map, whenever you click on a colored state, is "Kerry wrong for [name of state]," what's really stupid here is either

1) missing the chance to tell people in Rhode Island that Kerry is a bad choice for them too  :)

or

2) not building the map from the beginning from some other clickable map of the United States, which is the kind of thing the Web designers on the Bush campaign site should have been able to find from multiple sources. It's definitely a bad job of Web design, but maybe President Bush himself would say that it is better for the country for the best Web site designers to be working for private enterprise rather than for political campaigns.

I like Rhode Island. It is one of just two states on which I have yet to set foot. Maybe MathPath will be there again next year and I can visit Rhode Island.

Bush supporters aren't that effective in campaigning, as well demonstrated by the map, and the website.

Oh, yeah. JS1527.
I posted - Rhode Island and sent it to the website directors for laughs.
Who knows? Maybe they will fix it - though that's doubtful ... lazy bums .. ahem.

[quote="tokenadult"][quote="white_horse_king88"]I posted - Rhode Island and sent it to the website directors for laughs.
Who knows? Maybe they will fix it - though that's doubtful[/quote]

That's probably the most constructive response here.

Slightly off this topic, I am AMAZED that there is campaigning by both major parties in Minnesota this year. Minnesota is no longer regarded as a "safe" state for the Democratic Party candidate. (You will remember that Minnesota is the only state that Ronald Reagan didn't carry in 1984.) In most campaign years previously, neither major party candidate visited Minnesota to campaign vigorously, because the Democrat had more important (that is, more doubtful) states to campaign in, and the Republican figured he had no chance to win over Minnesota. It's kind of fun to be in a state that is in play this year, even though I will be voting for a third-party candidate, especially if a certain candidate is on the Minnesota ballot.[/quote]
Bush recently became the first standing president to visit Michigan's upper penninsula since Taft in 1911. Minnesota is not alone in being low on the priority scale - I'd imagine that areas with higher population densities are more important to candidates.

Rhode Island is the second most population dense state in the US, right after NJ.  Still can't get any of his time (or a place on his map).

[quote="confuted"]
Rhode Island would clearly be lost on that map to a combination of the thickness of the lines and lack of small detail. The map lobbed off several large penninsulas of Michigan, each of which is larger than Rhode Island. Since I live on one of those penninsulas, I suppose I'm not on the map either, but I'm not complaining.[/quote]

That could very easily degenerate into an observation that not all inland waterways and slight coastal twists and turns were accounted for.  The unit of analysis on the map is ostensibly states, thus the Rhode Island people have a basis for complaint.  Bush could have omitted the entire Florida panhandle and it would not have rendered the map less topologically complete as long as SOME part of the state were present..

Oh come on you guys, its a map. A big ol' blobby map. That's how they make maps that are just for show and not for accuracy. Bush probably had very little to no influence in making this map- remember, he's the president of the largest powerhouse in the world, I think he has more important things to do than worry about the finer points of his campaign, like whether or not each little speck (and yes, even though RI is an important place, physically, it is still a little speck) is shown on a map on a part of his website. So someone made a flaw- I really doubt that any of you are perfect either. I really don't see how a technical error honestly reflects on the intellegence of a man almost entirely uninvoled with above technical error. This is a prime example of politics- using unrelated "information" as a flaw against a person.

You were probably one of the delegates who scoffed at the New Jersey plan.


[quote="symmetry"]A jar contains yellow marbles, red marbles, and blue marbles. The number of red marbles is three less than twice the number of blue marbles. The number of yellow
marbles is one more than seven times the number of blue marbles. The probability of selecting a yellow marble is 3/4[/quote]

NO QUESTION??? :?:

I guess he meant to find the number of red , blue , yellow marbles ? :huh:

If thats the case , you see that $R=2B-3$ ; $Y=7B+1$ , total = $10B-2$

So the probability = $\frac{7B+1}{10B-2}$ = $\frac{3}{4}$  .

Solving this gives $B=5$ . And the rest is trivial  ;)

I apologize to everyone.  I did not write the entire question and honestly don't know where in the book it could be.  So, I will write a new question right now.  Sorry......

[quote="symmetry"]If q is a positive acute angle and 2 cos q  + 3 = 4, find the
number of degrees in q.[/quote]

lol...
[hide]
$2 \cos {Q} + 3 = 4$
$2 \cos {Q}= 1$
$\cos {Q}= \frac{1}{2}$
$Q= \pm 60+360k, \ \ \forall k \in \mathbb{Z}$
[/hide]

chess 64 didn't you miss the 300 degrees one? :D

Thank you all especially chess64 for your help.


algebra
polynomial
linear algebra
matrix
function

Let A and B be n x n real matrices. Suppose 7 is an eigenvalue for AB, prove that 7 is also an eigenvalue for BA.

 I think there should be some kind of an expanded formula of the characteristic polynomial for n x n matrices involving traces.. but I can't find it.

 please help, thanks

[quote="t0rajir0u"]There is.  The coefficients of the characteristic polynomial of a matrix $ \mathbf{M}$ can be computed from the numbers $ \text{tr } \mathbf{M}^k$ using the Newton-Girard formulas, and you can use this to prove that $ AB$ and $ BA$ have the same characteristic polynomial.

There is also a really really nice way to do this problem that generalizes to arbitrary rings.  Show that, in an arbitrary ring, $ 1 \minus{} ab$ is invertible if and only if $ 1 \minus{} ba$ is invertible.  Why does this imply that $ AB$ and $ BA$ have the same eigenvalues?[/quote]

 thanks, before I look them up, would it be a realistic task to attempt formulating it myself?

The Newton-Girard formulas aren't obvious (depending on how good you are with generating functions) but it's certainly a good exercise to at least see how the first few cases work.  (The point here is that $ \text{tr } \mathbf{M}^k$ is the sum of the $ k^{th}$ powers of the eigenvalues, and from these you can compute all the symmetric functions of the eigenvalues.)


$xy \neq 0$. Find which is bigger: $\sqrt[3]{x^3+y^3}$, $\sqrt[2]{x^2+y^2}$.

[quote="shobber"]$xy \neq 0$. Find which is bigger: $\sqrt[3]{x^3+y^3}$, $\sqrt{x^2+y^2}$[/quote]

By Letting $x+y$ = $A$ , $xy$ = $B$ ,then

$(x^2+y^2)^3$ - $(x^3+y^3)^2$ = $(A^2-2B)^3$ - $(A^3-3AB)^2$
= $B^2(3A^2-8B)$

But $3A^2-8B$ = $3x^2-2xy+3y^2 \geq 6xy-2xy =4xy$

$4xy \geq 0 $ If both $x,y$ have the same sign . So $\sqrt{x^2+y^2}> \sqrt[3]{x^3+y^3}$ . However , if either one of $x,y$ is negative ,then $3x^2-2xy+3y^2>0$ which also shows that $\sqrt{x^2+y^2}> \sqrt[3]{x^3+y^3}$

[quote="shobber"]$3A^2-8B=3x^2+6xy+3y^2-8xy=3x^2-2xy+3y^2$.
How did you get $(x-y)^2$?[/quote]

Some typing error  :P  Edited !


linear algebra
matrix
algebra
polynomial
invariant
abstract algebra
function

Here's a point I was making in a lecture in an elementary linear algebra class.

We have two important equivalence relations among matrices that we've been studying: row equivalence and similarity. Note that an $ m\times n$ matrix can be row equivalent to an $ m\times n$ matrix but similarity involves only square matrices.

I looked at various properties of matrices, and asked what each of these relations do to these properties. There are three possible answers: preserved, "traceable", and not preserved. For an example of what I mean by "traceable": the determinant of a square matrix is not preserved by row equivalence, but if you break up the row equivalence into elementary steps and record the effect of each step, you can trace what happens to the determinant. In fact, this is the most efficient way to compute the determinant of a large matrix.

Anyway, this is the list I came up with:

[b]Row equivalence:[/b] $ A$ is row equivalent to $ B$ iff $ \exists\,P$ invertible such that $ PA \equal{} B.$

[i]Preserved:[/i]
Rank
Null space
Row Space

[i]Traceable:[/i]
Determinant
Column space

[i]Not preserved:[/i]
Trace
Characteristic polynomial
Eigenvalues
Eigenvectors

[i]Semi-standard form, canonical form:[/i]
Row echelon form
Reduced row-echelon form

[b]Similarity:[/b] $ A$ is similar to $ B$ iff $ \exists\,P$ invertible such that $ P^{ \minus{} 1}AP \equal{} B.$

[i]Preserved:[/i]
Determinant
Trace
Characteristic polynomial
Eigenvalues
Rank

[i]Traceable:[/i]
Eigenvectors

[i]Not preserved:[/i]
Row space
Column space
Null space (but some or all of them may be traceable)

[i]Semi-standard form, canonical form:[/i]
Upper triangular form, diagonal (if possible), Jordan canonical form.

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

I did all of this partly as a warning against trying to find eigenvalues or eigenvectors by row elimination. Any comments?

If we add linear combinations of some rows to a (different) row, would the eigenvalues stay the same? For example, changing a matrix $ A$ to $ U$, its upper triangular form, without scaling

Absolutely not.

For instance, start with $ A\equal{}\begin{bmatrix}0&1\\1&1\end{bmatrix}.$ The eigenvalues of $ A$ are $ \frac{1\pm\sqrt{5}}2.$

Now perform just one row operation and see what happens:

(1) Subtract the top row from the bottom row: $ \begin{bmatrix}0&1\\1&0\end{bmatrix}.$ Eigenvalues $ \pm 1$

(2) Add to the top row to the bottom row: $ \begin{bmatrix}0&1\\1&2\end{bmatrix}.$ Eigenvalues $ 1\pm\sqrt{2}.$

(3) Interchange the two rows: $ \begin{bmatrix}1&1\\0&1\end{bmatrix}.$ Eigenvalues $ 1,1.$

For the sake of my class (whose test is tomorrow), a bold-faced warning: [b]You can't find eigenvalues by row elimination![/b]

Lesson learnt, indeed  :)  But the determinant stays the same, so we can conclude that the product of eigenvalues stays the same!

Not only the product, but also their sum is invariant, since the trace is preserved under similarity.

But was I was doing in those examples [b]wasn't[/b] similarity, it was row equivalence - that was the whole point. And the trace (the sum of the eigenvalues) was [b]not[/b] preserved. And in the last of the thee examples, which used a row interchange, even the determinant was not preserved - traceable, yes, but not preserved.

Thanks Carcul and Prof Merryfield, I think the list you've compiled will come to be helpful when I start my Linear Algebra course!

The word "traceable" confused me until I found out it has nothing to do with the trace of a matrix  :)

That might not have been the best word. Can you thing of a better word I could have used for the concept in question?

Actually the word 'traceable' is a pretty good choice, if it wasn't for the word 'trace' of a matrix. English is not my first language, but I guess reconstruct, recover, detect, or even record come close to the meaning of 'trace (back)'. The derived nouns sound a bit awkward, though.

Just out of curiosity (this isn't exactly on-topic, but nonetheless), I'm curious as to which of the following is the more commonly used mathematical jargon/terminology: kernel or null space? image or column space? I've seen them both ways, of course, and they are all obviously commonly used terms. Btw, I just took a course in linear algebra (introductory course).

To a certain extent, it's an issue of emphasis. Would you rather emphasize that you are talking about a matrix, or would you rather emphasize that you are talking about a linear transformation? I tend to think of "null space" and "column space" as matrix terms, while "kernel" and "image" are function (that is, linear transformation) terms. "Kernel" and "image" are used in other algebraic categories and for other kinds of morphisms; they aren't specifically linear algebra terms.

At that, I am not drawing any hard and fast lines, and in each case, we have a pair of essential synonyms. I learned linear algebra in the first place in a way that emphasized linear transformations and vector spaces; as I have re-learned the subject over the years and had my attitudes reshaped by one of my colleagues, I've tended more and more to think in matrix terms.

I'll put it this way: being "bilingual" (that is, equally conversant with the matrix terms and the linear transformation terms) can be a good thing.


inequalities
function
inequalities proposed

Prove the inequality $\dfrac{1}{z^{4}}+\dfrac{z^{2}}{1000}\geq 0,01$ for any real $z,$ and show that there exists a real $z_{0}$ such that $\begin{equation*}
\frac{1}{z_{0}^{4}}+\frac{z_{0}^{2}}{1000}\leq 0,02.\end{equation*}$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally.  :)

$z^{-4} + \frac{z^{2}}{1000} = z^{-4} + 2\frac{z^2}{2000} \geq 3(2000)^{-3} > 0.01$.  And since $z^{-4} + \frac{z^{2}}{1000}$ is a continuous function which can assume arbitrarily large values, the required is obvious.


calculus
integration
trigonometry
calculus computations

$\int_{\frac{\pi}{16}}^{\frac{\pi}{14}}\;\left( \frac{\sin \theta+\sin 3\theta+\sin 5\theta+\sin 7\theta+\sin 9\theta+\sin 11\theta+\sin 13\theta+\sin 15\theta}{\cos \theta+\cos 3\theta+\cos 5\theta+\cos 7\theta+\cos 9\theta+\cos 11\theta+\cos 13\theta+\cos 15\theta}\right) \;d\theta$

$=\int_{\frac{\pi}{16}}^{\frac{\pi}{14}}\frac{ \sum_{k=1}^{8}\sin \theta \sin (2k-1)\theta}{ \sum_{k=1}^{8}\sin \theta \cos (2k-1)\theta }\ d\theta$

$=\int_{\frac{\pi}{16}}^{\frac{\pi}{14}}\tan 8\theta \ d\theta$

[quote="misan"]find  :D

$\int_{\frac{\pi}{16}}^{\frac{\pi}{14}}\;\left( \frac{\sin \theta+\sin 3\theta+\sin 5\theta+\sin 7\theta+\sin 9\theta+\sin 11\theta+\sin 13\theta+\sin 15\theta}{\cos \theta+\cos 3\theta+\cos 5\theta+\cos 7\theta+\cos 9\theta+\cos 11\theta+\cos 13\theta+\cos 15\theta}\right) \;d\theta$[/quote]
we can find $\\sin \theta+\sin 3\theta+\sin 5\theta+\sin 7\theta+\sin 9\theta+\sin 11\theta+\sin 13\theta+\sin 15\theta$   4 this sum by $e^{ix}$ and also find
this sum $\cos \theta+\cos 3\theta+\cos 5\theta+\cos 7\theta+\cos 9\theta+\cos 11\theta+\cos 13\theta+\cos 15\theta$ analogysly. problem is easy it's only need few times and calculations.


Let x is the the maximum natural number.

x^2<=x  ( because x greater than or equal to any natural number)
x^2-x<=0
x(x-1)<=0

x is natural number x>0 ,x-1<=0 , x<=1

x is natural number, so x=1.

What is wrong?

You know what is wrong :)
There is no maximal natural number.

What you've proven is true: if there is a maximal natural number, then it is equal to 1.  However, it's true for the not-very-interesting reason that the premise is false.


What is the smallest positive integer solution to $ (100,000,000 \div x^4 ) < 1$?

Wait but 100,000,000/50^4 is 16??? It is less then 101. Please help


geometry
geometric transformation
reflection

Line $ l_2$ intersects line $ l_1$ and line $ l_3$ is parallel to $ l_1$.  The three lines are distinct and lie in a plane.  The number of points equidistant from all three lines is:

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$

Such a point must lie on the line halfway in between $ l_1$ and $ l_3$. Now, we can conclude that there are $ \boxed{\textbf{(C)}\ 2}$ points on the two sides of the intersection between the midway-line and $ l_2$ (they are reflections of each other across this point).


induction
superior algebra
superior algebra unsolved

Prove that the union of intervals [1,n] from n=1 to n=infinity is all of N.

Do I use induction on this? Archimedes? (This question is before the section of Archimedes though). I need help on how to start it!

Look at the definition of a union.  The union of a collection of sets $ U_i$ indexed by any index set is the set of all elements contained in at least one $ U_i$.  So show that every natural number is contained in at least one element of the union.

Is it enough to say then that:

As n increases, the interval increases.  We can say that as n-->inf, [1,n]-->[1,inf).  We proved that there is no integer x such that n<x<n+1 so we know that n increases by +1 as it approaches inf.  Therefore the interval [1,inf] contains every natural number n in N.

I don't know if that's the right way to prove it though.

[quote="rawkerrxx"]We can say that as n-->inf, [1,n]-->[1,inf).[/quote]
Not quite.  There isn't a meaningful way to define a limit of intervals [i]except[/i] as an infinite union or intersection.

Here is all that must be said (depending on what is meant by an interval; the original question is unclear whether we are talking about $ \mathbb{R}$ or $ \mathbb{N}$):  any natural number $ n$ is in the interval $ [1, n]$, hence in the union.

Do I need to produce a bijection to show that the union is a countable set like N? I feel like I do...but what would it be...a map f: f(x)-->x?

No.  The problem doesn't ask for anything of the sort.

At this point it would be wise to sort out the wording.  The union of the intervals $ [1, n]$ is the interval $ [1, \infty)$, which is much larger than (but contains) $ \mathbb{N}$.  So which of the following two questions did you mean?

- The union of the sets $ \{ 1 \}, \{ 1, 2, \}, \{ 1, 2, 3 \}, ...$ is $ \mathbb{N}$.
- The union of the intervals $ [1, 1], [1, 2], [1, 3], ...$ contains $ \mathbb{N}$.


algebra
polynomial
real analysis
real analysis unsolved

Let $f_{1}(x)=x$ and for each $n>1$ let
\[f_{n}(x)=x(f_{n-1}(x)+f_{n-1}^\prime (x))\]

Calculate $f_{n}(1)$ in terms of $n$. The first few terms are 2, 5, 15, 203, 877, ... .

http://en.wikipedia.org/wiki/Bell_numbers

google is the master

read

and you write the sequence wrong. using mathematica i get

{2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597}

Thanks. So I guess I am looking for a proof that shows the sequence $f_{n}(1)$ gives the Bell Numbers.


Let A be the sum of digits of $4444^{4444}$, and B the sum of digits of A. Find the sum of digits of B.

A very nice problem.

[hide] $4444^{4444}< (10^{5})^{4444}= 10^{22220}$. Thus, $A< 22,200 * 9 = 199,980$. Since A has at most 6 digits, $B<6*9=54$. Let $C$ be the sum of digits of $B$. Then $C \leq 4+9=13$.

Now, we know that a number's residue mod 9 is equal to that of the sum of its digits, so $C=B=A \equiv 4444^{4444}\equiv 7^{4444}\equiv (7^{3})^{1481}* 7 \equiv 7 \pmod{9}$. Since $C \leq 13$ and $C \equiv 7 \pmod{9}, C=7$. [/hide]

I have solved it pretty much the same way as Phelpedo, I was looking for an alternate solution if any exist.