We have finally discovered that the 4th Rating FLag comes with a rating of 1750. COngratulations superllamel on being the first to acheive this milestone!

Thanks mewto.

Mewto had about 1747 when I hit this milestone. Mewto is now 3 points ahead of me again though. We're having a tight race.  :lol:

It's like a arms race. We are both climbing higher and higher than everyone else without ever playing each other. :rotfl: And anytime we have met (twice so far) when both of us were at our highest ratings, I have won. Any idea when this race will end super?

UPDATE: Mewto leads 1769-1765

Lets see...

1:1200
2:1350 (help? maybe 1400, I know I have it with highest 1405...)
3:1500
4:1750

No apparent pattern here :(

But i just have a random guess that the next one is 2000 :D

I think the second one is at 1350, but I'm not sure. Mewto, I think I won one of them, and by the way, I will cd you any time.  :lol:

If there is a next flag. It might stop at 1750.

The next version (I think) will have stars, and as levans stated, "the elusive golden star".  Ooooh.  I deserve an elusive golden star!

And no need to make a new thread.  It would've fit under CONGRATULATIONs! just fine.

[quote="levans"][quote="mewto55555"][quote="levans"]I think all have 5, but I may be wrong.[/quote]

I heard a rumor somewhere that instead of a 5th rating flag, there's a golden star? If so, what rating is that at?[/quote]Hint: Most rumors are false.[/quote]

[quote="isabella2296"]Yes, the third flag is 1350.

It appears: +150, +150, +150, +250, so the next one is probably at 1750+250 = 2000. However, levans said he might change the flags, since it's so much easier to reach them. And the next version of FTW might not even have flags at all.[/quote]
I thought he said all of the points for the flags will be something like +200 +200 +200 +300, etc.

Apparently not.  ANyway, the list of everyone with 4th rating flag and their corresponding high ratings:

superllamel	1860
mewto55555	1839
kl2836	        1776
FantasyLover	1755
AIME15	        1753

Yes, that's right AIME15 is up there!!!!!!!!

ALong with their corresponding current ratings:

superllamel	1860
mewto55555	1837
FantasyLover	1726
kl2836	        1716
AIME15             1699

Weird its the same order.  And the people who are currently threatening to get on the list with their current ratings:

#H34N1	        1703
pi_rocks	        1657
Jongao	        1650
ernie         	1650
isabella2296	1646

And their highest ratings:

ernie	                1733
vallon22	        1732
isabella2296	1714
#H34N1	        1703
pi_rocks            1657

Congrats to everyone with the 4th rating flag!

Ooh, and sorry for reviving this topic.


superior algebra
superior algebra unsolved

IF Mn(R)=Mn(S) rings when can we conclude that R=S? in general.
let Mn(R) denote the ring of nxn matrices over R.

I think you mean this ...

Let $R,S$ be two rings (commutative?) s.t. $M_{n}(R) \cong M_{n}(S)$ as rings. Can we conclude $R \cong S$?

Perhaps [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=86260[/url] helps; it already shows that the ideals of $R$ correspond to them of $S$.


Can I make statics with $ \text{\LaTeX}$? I mean like charts?

Thanks in Advance  :)

You can make statistical charts in PSTricks, see [url=http://tug.org/PSTricks/main.cgi?file=Examples/Charts/chart]Charts[/url], or make them elsewhere and include them in your document.


I know the ones for 7 and 11, but what is the rule for 13?

one such possible way to figure divisibility by 13 is:

First multiply the units digit by 4, and then add the result to the rest of the digits. If the result is divisible by 13, then the number you started with is also divisible by 13.

[quote="09husbri"]Refresh my memory: Isn't the one for 11 where you give all the digits alternatively a positive sign or negative sign and add them together, and they have to equal zero?[/quote]

Yep.

Huh? I learned the divisibility rule for 11 as you add up every other number, and the positive difference between the two has to be divisible by 11. Like 93929 is divisible by 11 because 9 + 9 + 9 = 27 and 3 + 2 = 5
27 - 5 = 22
22/11 = 2
 :)  :)  :)

That's basically the same thing; when you give the signs alternatively in the method I use, it's basically adding every other number together and subtracting the other group of every other number. I forgot to mention- the sum doesn't have to be zero, it can be a multiple of eleven.

here's another for 13:
take the units digit and multiply it by 9.
Take the number before the units digit and subtract the above result
if our number was: abcdefghi....z
we would go:
abcdefghi....y - 9z.
if this is divisible by thirteen, original number is divisible by thirteen

The best divisibility rule is demonstrated here:

Let's say the number is 2958736.  Then we take

2958 - 736 = 2222

222 - 2 = 220

And it's clear that 220 is not divisible by 13.  The basis for this is the fact that 1001 is divisible by 13.

There is another way to determine the divisibility rule for 13 (and all other primes greater than 5).

The value 10k-1=xp, where x is a number that makes the units digit of xp 9, and k is the smallest value such that this equation is true (pretty much means that x is the smallest value such that the units' digit of xp is 9).

k is a value that you can multiply the last digit of a number by and add on to the rest of the # to test for divisibility by p, and p-k is a value that you can multiply the last digit of a # by and subtract from the rest of the # to test for divisibility by p.

Example: For 23, you could multiply the units' digit by 7 and add to the rest of the number, and 16 is a value you could multiply the units' digit by and subtract from the rest of the number.


Let $P(x)= x^{3}-3x+1$. Find the polynomial $Q$ whose roots are the fifth power of the roots of $P$.

This was recently posted.  Search the Olympiad section for it.

I can't seem to find it. Anyone want a try anyway?


A pen and its ink refill together cost $ \;\$1.10$. The pen costs $ \;\$1$ more than the ink refill. What is the cost of the pen?

We assume the pen is $ x$ dollers and the refill is $ y$ dollars. We know $ x \plus{} y \equal{} 1.10$ and $ x \minus{} y \equal{} 1$. We add the equations up and get $ 2x \equal{} 2.10$. We divide both sides by two and get $ x \equal{} 1.05$.

Or we could divide $ \;\$1.10$ by 2 and get 55 cents. Then we would transfer 50 cents from the refill to get $ \;\$1.05$


Simple.  Just say that the first investment is done with $x$ amount of money, and the other with $2x$.  I think the definition will just lead you through the solution for $x$.  Then you say $2x$ is the amount invested with the second rate.

Masoud Zargar

[quote="FrankieBlueEyes"]Mrs. Stevens invested a certain amount of money at 7% and twice that amt. at 10%. if the combined annual income from both investments is $\$$810, how much was invested at each rate?

 :([/quote]

[hide="Solution"]
0.07x + 0.2x = 810
0.27x = 810
0.01x = 30
x = 3,000

$3,000 at 7%
$6,000 at 10%
[/hide]


trigonometry
real analysis
real analysis theorems

By using the mean value therom prove that |sinx-siny|< or = |x-y| for all x and y.

Of course, we can rewrite this as

$| \frac{\sin x-\sin y}{x-y}| \le 1$

On the interval $[x, y]$ there exists $c \in [x, y]$ such that

$\cos c = \frac{\sin x-\sin y}{x-y}$

For every arbitrary interval $[x, y]$.  Of course, $| \cos c| \le 1$.


function
calculus
derivative
induction
algebra unsolved
algebra

Show that if a function f has the property f'=f then there exists a constant c such that f(x)=c(e^x) and inversely.If  f'(x)+sinx+2x=f(x)-cosx+x^2  for every xεR  and f(0)=11 , find the type of f.

Alternatively, you may consider $h(x)=\frac{f(x)}{e^x}$. It is easy to see that $h'(x)=0$ for all $x$, hence $h(x)=c$ for some constant $c$.

Are you sure that there isn't any other way?

[quote="Kondr"]Consider identifying coeficients the Talor polynom for f(x) and f'(x). Because from the given equality we get
$\forall i \in \mathbb N : i.a_i=a_{i-1}$, by induction we get that $a_i=\frac {a_0} {i!} $ so
$f(x)=a_0 e^x=f(0) e^x$. The inversion is also obvious from Taylors polynom.
The second thing- consider $q(x)=f(x)-cos(x)+x^2$, then $g(x)=g'(x)$ so
$g(x)=g(0) e^x=10 e^x$, $f(x)=g(x)+cos(x)-x^2=10 e^x+cos(x)-x^2$[/quote]

The hypotheses imply that $f\in C^{\infty}$, but that isn't enough to prove that $f$ is analytic.  Enescu's solution seems the most natural (without using the existence/uniqueness theorems from the theory of ODEs)

So the first step is solved but afterwards?how do you find the type of the function?!?

Which part don't you understand? The first was proven by Enescu in a very nice way and the second follows from it as I wrote above...


Hey guys, does anyone know a solution for this problem? Thanks.

Prove that every infinite sequence S of distinct positive integers contains either

a) an infinite subsequence such that, for every pair of its elements, neither element divides the other.

b) an infinite subsequence such that, in every pair of elements, one element always divide the other.

Consider the graph whose vertices are the positive integers which belong to the initial sequence, two joined by an edge if and only if one divides the other.

Then, you are asking to prove that this graph contains an infinite complete subgraph or an infinite set of independant vertices (no two joined by an edge). But this conclusion follows directly from Ramsey's theorem in its infinite form.

Pierre.


inequalities
LaTeX
algebra unsolved
algebra

Use $\LaTeX$ to write it in the forum (and put it to inequalities next time):

If $a,b,c$ are real numbers $a \geq 1$ and $a+b+c=0$ then find the minimum value of $a^{4}+b^{4}+c^{4}-3abc$.


I know I should know this, but I just can't think of how to proceed, here's the problem, and what I've done so far. It is a homework problem, so I don't want the complete solution, just a nudge in the correct direction please...

An automobile enters a turn whose radius is $ R$. The road is banked at angle $ \theta$, and the coefficient of friction between wheels and road is $ \mu$. Find the maximum and minimum speeds for the car to stay on the road without skidding sideways.

Taking my coordinate system to have its x-axis parallel to the ramp, I came up with the equation $ \sum{F_x} = mg\sin{\theta} - F_{fr} = 0$.
This became $ mg\sin{\theta} - F_N\mu = 0$
Which simplified to $ \mu = \tan{\theta}$

I'm not sure what to do with this however, or if it is even necessary. The second thing I worked with consisted of:

$ \frac {v^2}{R} = a_c$, or centripetal acceleration.
I felt that that should be equal to the component of the normal force parallel to the ground, not the ramp, so:
$ \frac {mv^2}{R} = F_N\sin{\theta}$ which becomes
$ \frac {mv^2}{R} = mg\cos{\theta}\sin{\theta}$ which becomes
$ v = \sqrt {\frac {Rg\sin{2\theta}}{2}}$

I don't think this is right though, because it doesn't involve $ \mu$ in any way...although it tells me that $ v_{max} = \sqrt {\frac {Rg}{2}}$, but I'm not sure how to find the minimum, and I honestly don't think that's right in the first place. Thanks!

EDIT: This is what I got for the minimum velocity: $ F_c=W\sin{\theta}\cos{\theta}-f\cos{\theta}$, where $ F_c$ is centripetal force. This simplifies to:
$ \frac{v^2}{R}=g\cos{\theta}(\sin{\theta}-\mu\cos{\theta})$ This is the minimum I think.

Unfortunately, I feel like my maximum equation is incorrect. I got:
$ \frac{v^2}{R}=g\cos{\theta}{\sin{\theta}+\mu\cos{\theta})}$

This never equals infinite, although my book's hint is that at $ \mu=1$ and $ \theta=\frac{\pi}{2}$ all values for speed are possible, which means that infinite should be produced as well, doesn't it?

I also don't know how to factor in the fact that as velocity increases, normal force does too...doesn't it? Thanks!

[quote="Icy"]$ \frac {v^2}{R} \equal{} a_c$, or centripetal acceleration.
I felt that that should be equal to the component of the normal force parallel to the ground, not the ramp, so:
$ \frac {mv^2}{R} \equal{} F_N\sin{\theta}$ which becomes
$ \frac {mv^2}{R} \equal{} mg\cos{\theta}\sin{\theta}$ which becomes
$ v \equal{} \sqrt {\frac {Rg\sin{2\theta}}{2}}$[/quote]

in doing that you are neglecting the frictional force contributing to the acceleration towards the centre. So include frictional force in that equation.


geometry
trigonometry
geometric transformation
reflection
symmetry
rectangle
geometry proposed

$ABCD$ is a square with side length $a.$ $ P$ and $ Q$ are two points on segment $\overline{AB},$ such that $ P$ is between $ A,Q$ and $ Q$ is between $B,P.$ Let $\mathcal{C}_1(r_1)$ and $\mathcal{C}_2(r_2)$ be the incircles of triangles $\triangle  APD,$ $\triangle BQC$ and  let $ \mathcal{C}_3(r_3)$ be the circle tangent to segments $ \overline{PD},\overline{QC}$ and $\overline{DC}.$ Show that:
\[ \frac {1}{r_3} \equal{} \frac {1}{a \minus{} 2r_1} \plus{} \frac {1}{a \minus{} 2r_2} \]

4 different circles are tangent to PD, QC, DC. If $ E \equiv DP \cap CQ$ and $ A, P, Q, B$ follow on $ AB$ in the described order, the equation holds for incircle $ \mathcal C_3$ of $ \triangle CDE.$ M, N are midpoints of AD, BC. Circles $ \mathcal C_1, \mathcal C_2$ are tangent to AD, BC at S, T. $ \angle C, \angle D, \angle E$ are angles of $ \triangle CDE.$

$ \frac {1}{a \minus{} 2r_1} \plus{} \frac {1}{a \minus{} 2r_2} \equal{} \frac {1}{2 MS} \plus{} \frac {1}{2 NT} \equal{} \frac {1}{PD \minus{} AP} \plus{} \frac {1}{QC \minus{} BQ} \equal{}$

$ \equal{} \frac {1}{\frac {a}{\sin D} \minus{} \frac {a}{\tan D}} \plus{} \frac {1}{\frac {a}{\sin C} \minus{} \frac {a}{\tan C}} \equal{} \frac {1}{a} \left(\frac {1}{\tan \frac {D}{2}} \plus{} \frac {1}{\tan \frac {C}{2}}\right) \equal{} \frac {1}{r_3}$

_________________________________________
Prove that circles  $ \mathcal C_1, \mathcal C_2, \mathcal C_3$ have common tangent.

[quote="yetti"]4 different circles are tangent to PD, QC, DC. If $ E \equiv DP \cap CQ$ and $ A, P, Q, B$ follow on $ AB$ in the described order, the equation holds for incircle $ \mathcal C_3$ of $ \triangle CDE.$ M, N are midpoints of AD, BC. Circles $ \mathcal C_1, \mathcal C_2$ are tangent to AD, BC at S, T. $ \angle C, \angle D, \angle E$ are angles of $ \triangle CDE.$

$ \frac {1}{a \minus{} 2r_1} \plus{} \frac {1}{a \minus{} 2r_2} \equal{} \frac {2}{MS} \plus{} \frac {2}{NT} \equal{} \frac {1}{PD \minus{} AP} \plus{} \frac {1}{QC \minus{} BQ} \equal{}$

$ \equal{} \frac {1}{\frac {a}{\sin D} \minus{} \frac {a}{\tan D}} \plus{} \frac {1}{\frac {a}{\sin C} \minus{} \frac {a}{\tan C}} \equal{} \frac {1}{a} \left(\frac {1}{\tan \frac {D}{2}} \plus{} \frac {1}{\tan \frac {C}{2}}\right) \equal{} \frac {1}{r_3}$

_________________________________________
Prove that circles  $ \mathcal C_1, \mathcal C_2, \mathcal C_3$ have common tangent.[/quote]

the first line should be $ \frac {1/2}{MS}$
PD-AP=2*MS.

This is consequence of the square ABCD having an incircle (I) and of the following theorem: If 4 directed common tangents of 4 directed circles touch a single directed circle, then their other 4 directed common tangents also touch a single directed circle. In this case, 2 of the 4 directed circles are $ \mathcal C_1, \mathcal C_2$ and the other 2 are of zero radius (points C, D). Directed common tangents CD, DC of the points C, D coincide in position, but they still have opposite directions. One is directed tangent of directed (I), the other of directed $ \mathcal C_3.$ See [url]http://www.mathlinks.ro/viewtopic.php?t=247411[/url].

Let OA ∩ CD = X, reflection of C and D in OA be C’ and D’ respectively. Due to symmetry, CD, OA and C’D’ meet at X, so B is on the polar of X to the circle O, and we can conclude that AX = OA, and it’s quite easy to see that, since OE and DX are perpendicular, E belongs to the circle (A, AO), which is the reflection of circle O in B, hence BE=BF.

Best regards,
sunken rock

It would be terribly nice to know who deleted my post with two

   solutions for this nice Sangaku  and for what personal reasons?

  If there is an interest I can always resubmit.   I remember the

  solutions, yet I believe that my post can be restored from the backup.

   M.T.

You can resubmit. However, if any of your posts contain least air of personal attacks, they will be deleted again in their entirety.

[size=125][color=darkblue][b]Nice and easy problem ! Thank you.[/b][/color][/size]

[quote="luisgeometria"] [color=darkred]$ ABCD$ is a square ($ AB \equal{} a$) . Let $ \{P\ ,\ Q\}\subset (AB)$ be two points so that $ P\in (AQ)$ .

Let $ C_1(r_1)\ ,\ C_2(r_2)$ be the incircles of $ APD$ , $ BQC$ respectively. Let $ C_3(I,r_3)$ be the circle

which is tangent to $ (PD)$ , $ (QC)$ and $ (DC)$ . Show that $ \frac {1}{r_3} \equal{} \frac {1}{a \minus{} 2r_1} \plus{} \frac {1}{a \minus{} 2r_2}$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ \{U\ ,\ V\}\subset (CD)$ for which $ PU\parallel QV\perp CD$ . Observe that $ \left\|\begin{array}{c} a \minus{} 2r_1 \equal{} DP \minus{} PA \\
\ a \minus{} 2r_2 \equal{} CQ \minus{} QB\end{array}\right\|$ .

Thus, $ \frac {1}{a \minus{} 2r_1} \plus{} \frac {1}{a \minus{} 2r_2} \equal{} \frac {1}{DP \minus{} PA} \plus{} \frac {1}{CQ \minus{} QB} \equal{} \frac {DP \plus{} PA \plus{} CQ \plus{} QB}{a^2}$ . Therefore,

$ \frac {1}{r_3} \equal{} \frac {1}{a \minus{} 2r_1} \plus{} \frac {1}{a \minus{} 2r_2}$ $ \Longleftrightarrow$ $ r_3\cdot (DP \plus{} PA \plus{} CQ \plus{} QB) \equal{} a^2$ $ \Longleftrightarrow$

$ r_3\cdot (DP \plus{} DU \plus{} CQ \plus{} CV) \equal{} a^2$ $ \Longleftrightarrow$ $ 2\left([DIP] \plus{} [DIU] \plus{} [CIQ] \plus{} [CIV]\right) \equal{} a^2$ $ \Longleftrightarrow$

$ 2\left([DPQC] \minus{} [PIQ] \minus{} [UIV]\right) \equal{} a^2$ $ \Longleftrightarrow$ $ a\cdot (a \plus{} PQ) \minus{} a\cdot PQ \equal{} a^2$ , what is truly.[/color]

Thanks everyone for your nice solutions. Mine is too long so, I wont post it but I'd like to attach another solution for this problem yet not elementary. See pages [i]33[/i] and [i]63[/i]

[quote="Virgil Nicula"] [color=darkred][b][u]An easy extension[/u].[/b]

$ ABCD$ is a rectangle  ($ AB \equal{} a\ ,\ AD \equal{} b$) . Let $ \{P\ ,\ Q\}\subset (AB)$ be two points so that $ P\in (AQ)$ .

Let $ C_1(r_1)\ ,\ C_2(r_2)$ be the incircles of $ APD$ , $ BQC$ respectively. Let $ C_3(I,r_3)$ be the circle which

is tangent to $ (PD)$ , $ (QC)$ and $ (DC)$ . Show that $ \frac {a}{r_3} \equal{} \frac {b}{b \minus{} 2r_1} \plus{} \frac {b}{b \minus{} 2r_2}$ and $ \frac {2r_3}{a}\ \plus{} \ \frac {r_1 \plus{} r_2}{b}\ \le\ 1$ .[/color][/quote]

[color=darkblue][b][u]Three particular cases[/u].[/b]

[b][u]P1[/u].[/b] $ ABCD$ is a square ($ AB \equal{} a$) . Let $ \{P\ ,\ Q\}\subset (AB)$ be two points so that $ P\in (AQ)$ .Let $ C_1(r_1)\ ,\ C_2(r_2)$

be the incircles of $ APD$ , $ BQC$ respectively. Let $ C_3(I,r_3)$ be the circle which is tangent to $ (PD)$ , $ (QC)$ and $ (DC)$ .

Show that $ \frac {1}{r_3} \equal{} \frac {1}{a \minus{} 2r_1} \plus{} \frac {1}{a \minus{} 2r_2}$ and $ r_1 \plus{} r_2 \plus{} 2r_3\ \le\ a$ ([u]the proposed problem[/u]).

[b][u]P2[/u].[/b] $ ABCD$ is a rectangle  ($ AB \equal{} a\ ,\ AD \equal{} 2a$) . Let $ \{P\ ,\ Q\}\subset (AB)$ be two points so that $ P\in (AQ)$ .

Let $ C_1(r_1)\ ,\ C_2(r_2)$ be the incircles of $ APD$ , $ BQC$ respectively. Let $ C_3(I,r_3)$ be the circle which

is tangent to $ (PD)$ , $ (QC)$ and $ (DC)$ . Show that $ \frac {1}{r_3} \equal{} \frac {1}{a \minus{} r_1} \plus{} \frac {1}{a \minus{} r_2}$ and $ 2r_3\ \plus{} \ \frac {r_1 \plus{} r_2}{2}\ \le\ a$ .

[b][u]P3[/u].[/b] $ ABCD$ is a rectangle  ($ AB \equal{} 2b\ ,\ AD \equal{} b$) . Let $ \{P\ ,\ Q\}\subset (AB)$ be two points so that $ P\in (AQ)$ .

Let $ C_1(r_1)\ ,\ C_2(r_2)$ be the incircles of $ APD$ , $ BQC$ respectively. Let $ C_3(I,r_3)$ be the circle which

is tangent to $ (PD)$ , $ (QC)$ and $ (DC)$ . Show that $ \frac {2}{r_3} \equal{} \frac {1}{b \minus{} 2r_1} \plus{} \frac {1}{b \minus{} 2r_2}$ and $ r_1 \plus{} r_2 \plus{} r_3\le b$ .[/color]

This trivial Sangaku problem needs a PWW solution.

Let me put just the dedication words.
This note is for my North American friends.

Friendly,
M.T.

Dear Mathlinkers,
I want only to give a reference for this nice figure, san Gaku (1877)

Fukagawa H., Pedoe D., 3.2.4. Japanese Temple Geometry Problems (1989) 40.

Sincerely
Jean-Louis


Νομίζω ότι βγαίνει πολύ απλά με γωνίες και δε χρειάζεται κάτι ιδιαίτερο από γνώσεις. :P

Έστω Η το ορθόκεντρο.
Αρκεί <ΚΗΒ+<ΒΗΔ+<ΔΗΓ+<ΓΗΛ=180.
Όμως <ΒΗΔ=<Γ και <ΔΗΓ=<Β, συνεπώς αρκεί <ΚΗΒ+<ΓΗΛ=<Α
Αν είναι Σ η τομή των ΡΚ, ΑΒ και Τ η τομή των ΡΛ, ΑΓ έχουμε ότι το ΑΣΡΤ είναι εγγράψιμο, άρα <Α=180-<ΚΡΛ.
Επομένως αρκεί <ΚΗΒ+<ΓΗΛ+<ΚΡΛ=180, το οποίο όμως και ισχύει αφού λόγω των παραλληλιών <ΚΗΒ=<ΚΛΡ και <ΓΗΛ=<ΡΚΛ.

Κώστα δεν είμαι σίγουρος αλλά νομίζω ότι η λύση σου είναι λάθος...

Εκεί που χρησιμοποιείς παραλληλία προϋποθέτεις ότι $ KHL$ ευθεία...δηλαδή το ζητούμενο

Εξάλλου δεν χρησιμοποιείς καθόλου ότι $ KB$ κάθετο στη $ BC$

[quote="vittasko"]Δίδεται τρίγωνο $ \bigtriangleup ABC$ και έστω $ P,$ τυχόν σημείο επί της πλευράς $ BC.$ Δια του $ P,$ φέρνουμε τις κάθετες ευθείες επί των $ AB,$ $ AC,$ οι οποίες τέμνουν αντιστοίχως στα σημεία $ K,$ $ L,$ τις ευθείες δια των $ B,$ $ C,$ και κάθετες επί την $ BC.$ Αποδείξτε ότι η ευθεία $ KL,$ περνάει από το ορθόκεντρο του $ \bigtriangleup ABC.$

Κώστας Βήττας.

ΥΓ. Για ευκολία πήρα το σημείο $ P,$ μεταξύ των $ B,$ $ C,$ αλλά αυτό δεν είναι υποχρεωτικό.[/quote]

Κώστα ,

 δεν την κοίταξα συνθετικά , αλλά είναι μια πολύ καλή άσκηση και για αναλυτική.

Αν  $ A(0,1), B(b,0), C(c,0) , D(d,0)$, τότε

    πολύ απλά βρίσκουμε $ H(0,\minus{}bc) , PK: y\equal{} b(x\minus{}d) , PL : y \equal{} c(x\minus{}d) , K(b, b(b\minus{}d)), L(c,c(c\minus{}d))$

 και έτσι  $ m_{HK}\equal{} b\plus{}c\minus{}d \equal{} m_{HL}$ ,που σημαίνει ότι τα σημεία $ K,H,L$ είναι στην ίδι ευθεία.

[b] Σχόλιο [/b]

 Κώστα , την άσκηση θα τη βάλω με αυτή τη λύση σε επανέκδοση του βιβλίου της Β΄θετικής , γιατί είναι πολύ χρήσιμη και δείχνει τη δύναμη της αναλυτικής γεωμετρίας. ΑΥτό δε σημαίνει ότι προτιμώ την Αναλ. Γεωμ. , αλλά ότι η Αν.Γεωμετρία είναι μεγάλη κατάκτηση του ανθρώπινου γένους.

 Μπάμπης

[i] ( Σε διαγωνισμό η άσκηση αυτή δεν πρέπει να χαθεί σε καμιά περίπτωση) [/i]

Δημήτρη έχεις δίκιο. :blush:  ΄Με διόρθωσε και ο κύριος Βήττας σε pm. Ζητώ συγγνώμη. Το 'χω ξαναπάθει αυτό με την ευθεία. :(  Και να σου πω την αλήθεια  παραξενεύτηκα διότι ο κύριος Βήττας είπε εύκολη αλλά τόσο πια που να βγαίνει με γωνίες... :huh:  Θα την ξαναπροσπαθήσω σε λίγο μήπως βρω τη σωστή γεωμετρική λύση.

Δεν ξέρω αν σας μπέρδεψα άθελά μου, αλλά η πρωτότυπη πρόταση που έχω υπόψη μου και θέλησα να την απλοποιήσω, αναφέρεται στη γενική περίπτωση όπου το $ H,$ είναι τυχόν σημείο στο εσωτερικό του $ \bigtriangleup ABC$ και $ \bigtriangleup DEF$ είναι το σεβιανό του τρίγωνο.

Ζητείται να αποδειχθεί ότι η ευθεία $ KL,$ περνάει από το $ H,$ όταν $ PK\parallel CF,$ $ PL\parallel BE$ και $ BK\parallel AD\parallel CL.$

[hide]Στο πρόβλημά μας τώρα ( αλλά και στη γενική περίπτωση όπου η λύση δεν προσεγγίζεται με γωνίες ), καθοριστικά για τη λύση είναι τα σημεία $ K'\equiv CF\cap BK$ και $ L'\equiv BE\cap CL.$

Προκύπτει η πιο σύντομη ( πιστεύω και η πιο κομψή ) συνθετική λύση που έχω υπόψη μου.

Έχω στο μυαλό μου και μια άλλη προσέγγιση με ομοιότητα, η οποία χρησιμοποιεί ένα κριτήριο συγγραμικότητας ( να είναι τρία σημεία συνευθειακά ), που μπορεί να φανεί χρήσιμο αργότερα, για τη λύση άλλων προβλημάτων.[/hide]
Κώστας Βήττας.


Prove that:$(ab)^{5}+(bc)^{5}+(ca)^{5}\leq 3$ where $a;b;c$ are real numbers satisfying $a^{4}+b^{4}+c^{4}=3$

It's enough to prove in case $a,b,c\ge 0$.
Setting $x^{3}=a^{4},y^{3}=b^{4},z^{3}=c^{4}$.
We have to prove
$(ab)^{\frac{15}{4}}+(bc)^{\frac{15}{4}}+(ca)^{\frac{15}{4}}\le 3$
But $(ab)^{4}+(bc)^{4}+(ca)^{4}\le 3$ (Vasc book p.17) so we have what we want :blush:


$ f(x\plus{}1)\equal{}\frac{1\plus{}f(x)}{1\minus{}f(x)}$

$ f(1)\equal{}2$

Find $ f(2004)$.

[hide="What I got so far"]Plugging in some values for $ f(x)$ I got:

$ f(1)\equal{}2$, $ f(2)\equal{}\minus{}3$, $ f(3)\equal{}\frac{\minus{}1}{2}$, $ f(4)\equal{}\frac{1}{3}$

$ f(5)\equal{}2$, $ f(6)\equal{}f(2)\equal{}\minus{}3$, ....

So it seems that $ f(1\plus{}4k)\equal{}2$, $ f(2\plus{}4k)\equal{}\minus{}3$, $ f(3\plus{}4k)\equal{}\frac{\minus{}1}{2}$ and $ f(4\plus{}4k)\equal{}\frac{1}{3}$ for some integer $ k$.

So $ f(2004)\equal{}\frac{1}{3}$

I dont know how to prove it. Help please.[/hide]

$ f(x\plus{}1)\equal{}\frac{1\plus{}f(x)}{1\minus{}f(x)}\equal{}1\plus{}\frac{2f(x)}{1\minus{}f(x)}$

$ f^2(x\plus{}1)\equal{}\frac{1\plus{}\frac{1\plus{}f(x)}{1\minus{}f(x)}}{1\minus{}\frac{1\plus{}f(x)}{1\minus{}f(x)}}\equal{}\frac{2}{\minus{}2f(x)}\equal{}\minus{}\frac{1}{f(x)}$

$ f^3(x\plus{}1)\equal{}\minus{}\frac{1}{\frac{1\plus{}f(x)}{1\minus{}f(x)}}\equal{}\frac{f(x)\minus{}1}{1\plus{}f(x)}$

$ f^4(x\plus{}1)\equal{}\frac{\frac{1\plus{}f(x)}{1\minus{}f(x)}\minus{}1}{1\plus{}\frac{1\plus{}f(x)}{1\minus{}f(x)}}\equal{}\frac{2f(x)}{2}\equal{}f(x)$

And $ f$ is periodical. The rest comes from $ f(1)\equal{}2$ , and a mere induction will complete the proof.


I think these one's were posted before but they are nice for beginners! :D

If $a,b,c >0$ then we have:
$ \sum \frac{a^2+b^2}{a+b} \geq \sum a $
and
$ \sum \frac{a+b}{a^2+b^2} \leq \sum \frac{1}{a}$.

(made by  Szoeloesy Gh.)

first one: so to prove is:

(a+b) / (a+b)   +   (b+c) / (b+c)   +   (a+c) / (a+c)  >=  a+b+c

Solution:

we have 2ab =< a + b

so adding 2ab to the a + b will not double it

but if we add it and simplify, we have 2(a+b+c)

2(a+b+c) less than the double of the original equitation
so the original was more than a+b+c


modular arithmetic
number theory unsolved
number theory

Let $ r_{ij}\in \{0,1\}, \Sigma_{j}r_{ij}\equal{}r, \Sigma_{i}r_{ij}\neq 0, 1\leq i,j\leq n$. Prove (or disprove!) that the following system of incongruences has a solution with $ x_i\neq \bar{0}$, for all $ i$.

                                $ \sum_{j} r_{ij}x_j \not\equiv 0 \pmod{r\plus{}1}$

Note that $ r\geq 2$. Even the case $ r\equal{}2$ doesn't seem obvious to me, but not that hard to study. Any idea?


probability
function
integration
probability and stats

this is the question
Delicious Candy markets a two pound box of assorted chocolates. Because of imperfections in the candy making equipment, the actual weight of the chocolate
has a continuous uniform distribution ranging from 31.8 to 32.6 ounces.

i have figured out the equation which is: f(x)=1/(32.6-31.8) where 31.8<X<32.6

how do i find out the probaility that the weight is exactly 32 ounces and if its more than 32.2

sum1 please help i dun get it :(

Doesn't it seem odd to you that the variable $ x$ doesn't appear in your equation for the probability density function?  Probably you should go re-read the relevant sections of your textbook.  You could also try writing out here what you've done, and I'm sure someone will come along and help you find the right track.

(By the way, answering the last question (the probability that the weight is more than 32.2) doesn't require you to know anything at all about probability density functions.  Just try to think about what the meaning of the words "continuous uniform distribution on $ [31.8, 32.6]$" is.)

I don't see any problem with the pdf ($ f(x)$) as written.

In general, to find the probability that your continuous random variable $ X$ is less than some constant $ k$, we have:

$ \mathbb{P} (X < k) \equal{} \displaystyle\int_{ \minus{} \infty}^k f(x) \, dx$

Similarly, to find the probability that $ X$ is greater than some constant $ k$, we have:

$ \mathbb{P} (X > k) \equal{} \displaystyle\int_{k}^\infty f(x) \, dx$

Using this approach will give you the same answer as the intuitive answer you should get from JBL's suggestion.

For a continuous random variable, $ \mathbb{P} (X \equal{} k) \equal{} 0$.

[quote="jshen"]I don't see any problem with the pdf ($ f(x)$) as written. [/quote]  Sorry, you're absolutely right ... I must have been thinking of the CDF when I wrote that  :oops:   :(


From a plane's window, I can see part of an island, part of a cloud, and a bit of the sea. Knowing that the cloud occupies half the view and that the cloud hides one-fourth of the island,which appears to fill one-fourth of the view, what is the proporton of the sea hidden by the cloud as seen from the plane's window?

Let $ P$ be area of the window. $ \frac{1}{4}P$ is area of the part of the island that can be seen, which is $ \frac{3}{4}$ of total area of the island so area of the island is $ \frac{1}{3}P$ and area of the part hidden by cloud is $ \frac{1}{12}P$. Since the rest of the hidden part is filled by the sea, area of the sea hidden by cloud is $ \frac{1}{2}P\minus{}\frac{1}{12}P\equal{}\frac{5}{12}P$.


Who do you think will win the AL Championships? I pick White Sox

I think the Tribe is gonna come through this year

Definitely not mariners.   :rotfl:  :rotfl:  :rotfl:

[quote="jhollenbeck"]Methinks that the Braves will not win the AL championships... ;)[/quote]

.....

wow... baka...

tribe=indians

for those of u who dont know i think baka means idiot in jap

I'm insulted that you have no listed the Blue Jays.

Given a streak they could finish wild card and pull off some upsets.

At least you should have had an "Other" option.

[hide] I don't seriously expect them to win but still...[/hide]

i think with the yankees mostly if u dont live in new york then people normally hate there guts but if u do live in new york u love them not the mets :D

[quote="krustyteklown"]<=== just moved to boston (lexington) this summer, into sox already :D[/quote]
i'm moving up to boston tomorrow, still root for yankees.

white sox is having a great season.
if that rumor (Griffey -> whitesox) becomes true, they have a shot.

redsox, sorry but I dont wanna see that two years in a row.

I assumed that GoBraves meant the Braves when he said the tribe :blush: but you are right, its the Indians.  Baka yourself :mad:


Using all the digits 3,4,5,6,and 7 only ONCE in each numeral, how many 5 digit numerals are divisible by 55?

Source: Math Team team round question

55=11*5

so this can only be done in 4 different ways ...

63745
64735
73645
74635

...

that's the right answer, except i don't get how u did it...

so the numbers have to be divisible by 5 and 11, and then what do u do?

A number is divisible by 5 if and only if it ends in $ 5$ or $ 0$. You probably knew this.

A number is divisible by 11 if and only if the sums of each pair of alternating digits differs by a multiple of 11 ($ 0$ is a multiple of $ 11$).

Example: $ 149358$
Sum of one pair of alternating digits: $ 1 \plus{} 9 \plus{} 5 \equal{} 15$
Sum of the other pair of digits: $ 4 \plus{} 3 \plus{} 8 \equal{} 15$
$ 15 \minus{} 15 \equal{} 0$, so divisible by $ 11$.

So try to figure out what kinds of digit arrangements can satisfy that condition in your problem.


My birthday was last month and I have money to spend. I would like to buy an MP3 player of some sort but I don't know what to get. A lot of freshman at my school have IPODS but a lot of freshman also have other types of MP3 players. So I would like to know if buying an MP3 player is worth my money. People who owns IPODS or MP3 players, please help me!

nah, don't get one. save up for college. that's what i'd do :lol:

[quote="RobinHood3000"]Personally, I'd go with whichever's more utilitarian for your purposes.  But that may not help much.[/quote]

What do you mean? I don't understand. Sorry..... :blush:

yea ur really gonna save up for college. actually it IS a lot of money. my cousin who lives in england would tell u 2 get the ipod cause everyone in the country owns one as he puts it. "even that old lady rite there" lol. but i wouldnt really buy either cause admit it: ilts all gonna be "out" in a month or so. ur gonna waste ur money for sumthing thatll last less than a year. so u could either save up for college or if u get a tuition, u could spend it one BOOKS and LEARNING TOOLS!!! because we LOVE to learn and get EDUCATED. and then get FIRED from all our jobs!!!!!yaaaaaaay!!!

Utilitarian, as in useful; what your friends have is irrelevant.

Personally, I'd recommend getting a CD-Player and a pack of rechargable batteries.  Or go to a Discovery Channel Store, there's a load of cool (but educational) stuff there.  Like night-vision goggles (how cool is that?!).

um...ok...just listen to me and save up :lol:

Depends if you listen to music a lot.

Personally, I would buy an iPod because I listen to music a lot, and mostly rely on Yahoo Launch now, but that isn't portable.

If you've got some money that you want to spend, I'd say go ahead. I might catch some heat for saying this, but $250 isn't going to break you college fund.

yeah go ahead buy an ipod, not like its gonna use it for a life time...

or u can keep saving and saving and buy a wireless laptop with a rechargable battery. THOSE ARE BETTER THAN AN IPOD. u can play music, and um....do more stuff and u need it for college anyways. and u can acess the internet and type and instant message and all that stuff. now who would choose an ipod over a laptop?? o btw when do ppl take teh psat? my dad bought this gigantic book and wants me 2 finish it by...next year (BUT I CANT DO THAT!!!! ITS SOOO HARD!!!!) so i can take the psat. is that too late or too early?

[quote="Athena"]my dad bought this gigantic book and wants me 2 finish it by...next year (BUT I CANT DO THAT!!!! ITS SOOO HARD!!!!) so i can take the psat.[/quote]

You know, a great many people (myself included) would find cause to celebrate if given the relatively-interminable deadline of a few months.

um...next year...keyword...year...and next...um...u still got 364 days to finish it if you got it today.  how many pages is it?

[quote="Athena"]o btw when do ppl take teh psat? my dad bought this gigantic book and wants me 2 finish it by...next year (BUT I CANT DO THAT!!!! ITS SOOO HARD!!!!) so i can take the psat. is that too late or too early?[/quote]

Hm. Psat and iPod. At least they have the same number of letters. Don't see the connection.

As for the iPod, it's very nice, because it's portable and has a huge selection of songs. iTunes are a great source for songs. Personally, I'd wait a little while until the price drops.

And if you buy a laptop and plan to use it for college, it will definitely be obstelete by then. (unless you're going to college next year)

um...your PSAT is in lets see...3 YEARS FROM NOW! why do u ned to worry about it right now?

it can help u earn college money if u get a good score. and it prepares u for the sat.  and u get a head start so by the time everyone else takes it youll no wats on it better than them.

[quote="MithsApprentice"]FREE blank CDs nowadays (that they give u at like Best Buy and stuff) can even hold like 800-some megs...get urself a nice lil CD-album thingyish...highly portable...holds 24-30 CDs...and there u've got more than several Ipods worth of music...[/quote]

I beg to differ:

1. CD-Rs go up to 700 MB, CD-RWs go up to 640 MB. Each CD audio track will consume about 60 MB for a song that's about 3 minutes long.

2. IPods and other MP3 players don't play CD audio tracks. They store MP3 files, which can be created from CD audio tracks. They preserve the quality and compress the data. A 60 MB CD audio track can most likely fit in a 4-5 MB MP3 audio at 128kbps.

3. A normal CD stores about 15 songs; a 20 GB IPod, with the estimation above, will store about 5000 songs.

So I don't think 30ish CDs will put a dent in the storage capacity of an IPod.

But if 5000 songs is too much memory, an iPod mini is a pretty good alternative. 150 dollars less expensive, but it still stores enough songs for most people.

Athena- I highly doubt you would need 2 part-time jobs to save up $400 dollars. I umpired six little league baseball games last year. It took up 3 hours each week for 6 weeks, but I made 150 Dollars. If I had chosen to umpire more, I could have umpired up to 10 games. That would have easily been 250 dollars, and it would probably still only take 6 weeks. What I'm saying is that $400 is hardly an insurmountable sum.

I can even use the IPod to backup my hard drive if I so wished - portable storage is always good.

Two cents:

I got an ipod (freeipods.com) and I think its great. I currently have only 200 songs on there (i keep meaning to put the rest of my cd collection on...), and it can hold another 4k+ (20GB), i.e. im probably never going to fill it completely.
It does have 2 drawbacks though.
The first is that the parts inside arent exactly the most stable. As I hear it, they can get rattled pretty easily as technology goes, so no jogging with it.

Also, one thing that isnt publicized is the current lawsuit going on. It turns out that it was made so that after ~50 recharges, it only recharges half-way from then on, and the battery will further decay afterward. The way I use it, Im probably at 10 recharges already after only 2 weeks.

Too bad freeipods.com is available in the States only.

1. The only moveable parts inside an IPod is a Toshiba 1 inch HDD drive; I've had people drop them, washed them, etc, and they still worked afterwards. It's a marvel of electrical engineering.

2. Really. Li-Ion batteries usually last longer than that...

[quote="Osiris"]I beg to differ:

1. CD-Rs go up to 700 MB, CD-RWs go up to 640 MB. Each CD audio track will consume about 60 MB for a song that's about 3 minutes long.

2. IPods and other MP3 players don't play CD audio tracks. They store MP3 files, which can be created from CD audio tracks. They preserve the quality and compress the data. A 60 MB CD audio track can most likely fit in a 4-5 MB MP3 audio at 128kbps.

3. A normal CD stores about 15 songs; a 20 GB IPod, with the estimation above, will store about 5000 songs.

So I don't think 30ish CDs will put a dent in the storage capacity of an IPod.[/quote]

Hmm...

1. Maybe you're right. But I could've sworn I saw these at 800...

2&3. There are CD players that play MP3s stored as data on a CD (i'm not talking about audio CDs that max out at like 80 minutes or so)...even at 700 MBs, u can store plenty of MP3s averaging 3-4mbs...let's say they're 5MBs per song...140 per CD...times 24-30 would put u somewhere in the high-3000/low-4000 range right?...that would put a dent in the IPOD storage space...

Besides, most ipods these days dont exceed the 10gig line...and that 8-hour battery...tsk tsk...lol

nothing wrong with ipods, in fact i'll probly be buying one soon...just if ur money-conscious, it may not be the best choice...

[quote]1. Maybe you're right. But I could've sworn I saw these at 800...

2&3. There are CD players that play MP3s stored as data on a CD (i'm not talking about audio CDs that max out at like 80 minutes or so)...even at 700 MBs, u can store plenty of MP3s averaging 3-4mbs...let's say they're 5MBs per song...140 per CD...times 24-30 would put u somewhere in the high-3000/low-4000 range right?...that would put a dent in the IPOD storage space...

Besides, most ipods these days dont exceed the 10gig line...and that 8-hour battery...tsk tsk...lol

nothing wrong with ipods, in fact i'll probly be buying one soon...just if ur money-conscious, it may not be the best choice...[/quote]

1. Probably 80 minutes. CD specs don't go above 700MB, it's technically impossible.

2/3. 30 CDs would be a lot more massive (not to mention heavier) than an IPod.

Most IPods are either 15G, 20G or 40G. The newer, photo version of IPods go up to 60G. It's just the MINI version of IPods that stop at 4GB, and that's for the chic market.

I think it comes down to 3 choices:  [url=http://www.amazon.com/exec/obidos/tg/detail/-/B0002OZXHO/qid=1100800962/sr=8-6/ref=pd_csp_6/103-1039703-1492668?v=glance&s=electronics&n=507846]iPod[/url], [url=http://www.sonystyle.com/is-bin/INTERSHOP.enfinity/eCS/Store/en/-/USD/SY_DisplayProductInformation-Start;sid=-3OA4XcPqOaA7jaTHzGK6jgCGoJrLl_misM=?ProductSKU=NWHD1KIT]Sony NW-HD1[/url], or [url=http://www1.us.dell.com/content/products/category.aspx/dj?c=us&cs=19&l=en&s=dhs&~ck=mn]Dell DJ[/url].

Personally, I bought the 4th generation 20 gig iPod right when it was released.  Having some sort of music player during debate tournaments is a true gem once the off-round time kicks in.  Anywho, it's one of my better purchases (actually, my parents, since it was a birthday present).  The battery life is literally what Apple claims it to be: 12 hours.  I even got 13ish hours once, but it really depends on how much energy you're conserving.  Like, don't use backlight, don't scroll much, low volume, etc.  But even 8 hours should suffice.  I don't have nearly all of the space filled (about 5000 low quality songs), but it's all goodly.  I bought mine at Best Buy, and I'd suggest all of you do that who will buy an iPod.  A 2 year guarantee was about 35 dollars and it covers pretty much everything.  If your battery isn't at optimum level, they'll change it for you.  If iPod is nicked up, they'll change it with you (this is all free btw).  Even if the 5th generation iPod has been released and you want to change your unit because some defect is everpresent, they'll just give you the 5th gen iPod.  How amazing is that!  Pretty amazing.  Buy any mp3 player knowing that its lifespan is 1-2 years.

If money's the issue, then I'd suggest the Dell Digital Jukebox.  It's pretty good and 150 dollars for the 15 gig version.  I've heard it's still a solid product, so look into this if money's a scrunch.

If I had heard about the Sony NW-HD1 before I bought my iPod, I would've waited to compare.  Well, what's done is done.  This unit boasts 30 hours of continuous music and holds up to 13000 low quality songs even though it's 20 gig (new technology).  It's 350 dollars.  I may have ended up buying the Sony NW-HD1 because of the battery life.  Although iPod's 12ish hours is pretty good, 30 is just paradisical.  When it comes down to it, though, iPod is a sex symbol itself and the NW-HD1 is not.  Think of it this way: you're sitting alone in a seat on a bus.  A hawt girl will more likely sit next to you if you have iPod than NW-HD1.  It's as simple as that.

And a quickly added new choice would be an mp3/cd player.  Those are pretty nifty.  Oooh, a fifth choice would be some sort of Flash player.  They don't hold much music, but they're great for exercise (my brother bought [url=http://www.amazon.com/exec/obidos/tg/detail/-/B00008V6NI/qid=1100802239/sr=1-1/ref=sr_1_1_etk-electronics/103-1039703-1492668?v=glance&s=electronics&n=172282]this one[/url] maybe a year ago?).

I leave the rest up to you.  If you have any questions about iPod, you can pm me or something.  300 dollars is quite a hefty purchase that should be taken seriously.

[quote="Syntax Error"]When it comes down to it, though, iPod is a sex symbol itself and the NW-HD1 is not.  Think of it this way: you're sitting alone in a seat on a bus.  A hawt girl will more likely sit next to you if you have iPod than NW-HD1.  It's as simple as that.[/quote]

Nice example.  :P  :D

Still, probably true.

But I still think it just comes down to how much music you listen to. Just my opinion.

[quote="blahblahblah"][quote="RobinHood3000"]As far as I am concerned, file sharing and illegal downloading is indicative of a moral collapse within a large portion of the teenage population.[/quote]

Just like it was moral turpitude when our parents recorded cassette tapes.  Get off your high horse.[/quote]

Yes, but blahblahblah, at least when some parents did it, they had some sense of guilt about it.  I know people who rationalize their behavior by saying "it's the Tree of Knowledge in the Garden of Eden.  It's [i]there[/i]." (That's a direct quote, incidentally).  The speaker of this quote would also be the same person who has several GIGabytes of stolen (yes, stolen) music on his hard drive.

I know another person who downloads software illegally, instead.  This particular person is in possession of thousands of dollars worth of software.  His rationalization is that without file-sharing, he would not have all of this software which will help him in his later career search and which he would not otherwise been able to experience because of the cost.  To me, this is comparable to saying that it is fine for Olympic athletes to take steroids if they couldn't win otherwise.

Pardon my cynicism, but I don't respond well to disregard for the law, especially when it is done on such a massive scale and strips the consciences of even some of my closest friends.

[quote="RobinHood3000"]

Yes, but blahblahblah, at least when some parents did it, they had some sense of guilt about it. [/quote]

Questionable, I think.

Personally, I buy CDs and download (both of which are legal in Canada).  I generally delete what I don't like and buy what I do.  But before you jump to the defense of the establishment, consider this:

http://www.usatoday.com/life/music/news/2002-09-30-cd-settlement_x.htm

dont get an ipod. once their batteries cant be recharged again, you cant get another one I think....

I'm getting a creative zen touch.

Sure you can. Of course, Apple doesn't do it for you, but you CAN buy third-party replacement battery cells.

I read the review for NW-HD1 last july when it first came out, and I was disgusted. Sony is going nowhere with their ATRAC technology, which they force on every single one of their products; it's stupid. Transcoding takes time, the music ends up sounding crappier (I've used a MD player, I can tell), and even the transfer rate suffers.

Personally I own a Flash player and found it very satisfactory, but I certainly wouldn't turn down an iPod. The obvious volume advantage aside, my Flash player isn't without its flaws - the Korean company that made it focuses very little on the North American market, and while their hardware is excellent, their firmware is fundamentally flawed...

If you're going to buy a Flash player, I recommend that you check out the bigger brands with ENGLISH firmware downloads and tech support, such as iRiver, Creative and Cowon. Their flash players are excellent stuff.


function
trigonometry
logarithms
real analysis
real analysis unsolved

Consider the function $ f(x)=\sum_{n\geq 1}{\frac{|sin n|\cdot  x^n}{1-x^n}}$. Find a simple equivalent in 1 for this function.

$f_n(x)=\sum_{k=1}^{n}\frac{|\sin k|x^k}{1-x^k}$

when x tends to $1^{-}$

$f_n(x)\sim \frac{1}{1-x}\sum_{k=1}^{n}\frac{|\sin k|}{k}$

and with $\sum_{k=1}^{n}\frac{|\sin k|}{k} \sim \frac{2\ln n}{\pi}$

Conclusion when (in this order) x tends to $1^{-}$ and n tends to +oo

$f_n(x) \sim \frac{2\ln n}{\pi(1-x)}$

Is it correct? :?
And, please, explain why $\sum_{k=1}^{n}\frac{|\sin k|}{k} \sim \frac{2\ln n}{\pi}$.

I assume that $\frac2{\pi}$ is simply the long-term average of $|\sin k|.$

One thing I noted, although I'm not sure what to do with it yet:

$\sum_{n=1}^{\infty}\frac{x^n}{1-x^n}=\sum_{n=1}^{\infty} \sum_{k=1}^{\infty}x^{kn}.$

I was going to guess that as $x\to1^-,$ the original sum looks like $\frac2{\pi}\cdot\frac{-\ln(1-x)}{1-x}, $ but I don't have a convincing argument for that yet.

[quote="Peter Scholze"]a randomly chosen value for $|sin(x)|$ is about $\frac 2{\pi}$ and summing $\frac 1k$ gives about $\ln k$...therefore we get the estimate.[/quote]
It is not a rigorous argument.

of course not :D , but just to give the idea of the proof... and kent wasn't more precise


Prove that there are infinite triplets $ (x,y,z)$ of positive integers that satisfy the equation

$ x^2\plus{}y^2\equal{}z^3$

There are two interesting directions this generalizes.  One is to replace $ (2, 2, 3)$ with $ (a, b, c)$; under what conditions can we guarantee a large class of solutions?

The other is to determine all solutions.

[hide="a-b-c case"]If there is one solution, there are infinitely many solutions. Say $ (p,q,r)$ is a solution. Then $ (pn^{[a,b,c]\over a},qn^{[a,b,c]\over b},rn^{[a,b,c]\over c})$ is a solution too for any $ n\in\mathbb{N}$[/hide]

Edit:  Fixed!

[hide="A two-parameter family"] Suppose $ \gcd(ab, c) \equal{} 1$.  Then there exist positive integers $ x, y$ such that $ abx \plus{} 1 \equal{} cy$.  Given $ p, q \in \mathbb{Z}$ set

$ x \equal{} p(p^a \plus{} q^b)^{bx}$
$ y \equal{} q(p^a \plus{} q^b)^{ax}$
$ z \equal{} (p^a \plus{} q^b)^y$. [/hide]

[b]@ t0rajir0u:[/b] I haven't spent much time on this, but I don't see anything "special" about your two-parameter general solution that would cause it to break down in the case $ a\equal{}b\equal{}c > 2$, which it should. Are you sure it works?

[quote="TZF"][b]@ t0rajir0u:[/b] I haven't spent much time on this, but I don't see anything "special" about your two-parameter general solution that would cause it to break down in the case $ a \equal{} b \equal{} c > 2$, which it should. Are you sure it works?[/quote]

Check for $ a\equal{}b\equal{}c$: $ a\plus{}b\equal{}2c$, hence $ nc\equal{}2c\implies n\equal{}2$. Then $ k\equal{}2c\minus{}a\minus{}b\equal{}0$, so we get

$ (p^a\plus{}q^a, p^a\plus{}q^a,(p^a\plus{}q^a)^2)$, hence it doesn't work.

I didn't really check if these are a consecuence of what t0rajir0u provided but, here they are..

[hide]
$ x \equal{} a(a^2 \plus{} b^2)$
$ y \equal{} b(a^2 \plus{} b^2)$
$ z \equal{} a^2 \plus{} b^2$
[/hide]

:)

My apologies; the condition for the construction to work (and I messed it up a little) is that $ \gcd(ab, c) \equal{} 1$.


Find the sum of the solutions for $ |2x|\equal{}127$.

The equation can also be expressed as $ 2x\equal{} 127$ and $ 2x\equal{} \minus{}127$.  From here, we can see that the sum is $ \boxed{0}$(we can also solve the equations, giving $ x\equal{}63.5$ and $ x\equal{}\minus{}63.5$, which sum to $ \boxed{0}$).


Hello everybody !

I've recently come across a problem where we needed to show that if, for $ a,b,p,q,n \in \mathbb{N}$, $ (a/b)^{1/n}\equal{}p/q$, with $ gcd(a,b)\equal{}gcd(p,q)\equal{}1$, then $ a\equal{}p^{n}$ and $ b\equal{}q^{n}$.

I've an answer more or less the following: if  $ (a/b)^{1/n}\equal{}p/q$, we have $ bp/q$ which is an integer (by RRT); then, $ q$ divides $ b$ (because $ gcd(p,q)\equal{}1$), so that $ b\equal{}qb'$.

As we have $ a/b\equal{}p^{n}/q^{n}$, this implies $ a$ divides $ p^{n}b'$ and so $ a$ divides $ p^{n}$ since $ gcd(a,b')\equal{}gcd(a,b)\equal{}1$.

This makes $ p^{n}\equal{}ac$, and then $ q^{n}\equal{}bc$.

Since $ gcd(p,q)\equal{}1$, we have $ c\equal{}1$, and we're done.

Would you know if there's a more elegant solution starting from "$ bp/q$ is an integer" ? Elegant would mean a solution not implying the explicit usage of $ b'$ and $ c$.

Cheers.

If $ \gcd(p, q) \equal{} 1$ then $ \gcd(p^n, q^n) \equal{} 1$ and $ \frac{a}{b} \equal{} \frac{p^n}{q^n}$.  The reduced form of a fraction is unique.


Duke
college
geometry
summer program
Mathcamp
AwesomeMath
trigonometry

Anyone have experiences with summer camps held in NC they would like to share?  Some of the NC camps that I know of include:

NC Governor's School
Duke TIP
Cannon Music (Appalachian State)
Davidson
Summer Ventures in Science and Math (UNC-CH)

Of these, I'm most familiar with Cannon Music.  My daughter attended last year and had a good time.  It lasts 4 weeks and was reasonably priced (less than 1500 dollars including food.)  Admission is competitive and many students from other states attend (especially Florida kids getting out of the heat.)  The performances are at least as good as you'll find at the NC All-State high-school band (but they have more practice time at Cannon.)  There are private instructors, tough music theory classes, a variety of ensembles, much tougher music and more talented musicians than you'll find in most locales.

Highly recommended if you are serious about music and are close to "all-state" material.  A few mathematically inclined students attend but you'd have to hunt to find them.

I did TIP a couple years ago. I very much disliked it. If you're asking for Bryce in particular, I don't suggest it at all. And what I say in the following basically contrasts the only two away-from-home camps I've been to (Duke TIP and MOSP)

If you're really social, it's great. It's really more social than academic, and the classes don't go very far in depth at all. There are lots of silly and annoying rules (which may seem normal for a summer camp, but MOSP for example has practically no rules). Also, the food is horrible (by the way, University of Nebraska at Lincoln has possibly the best cafeteria food anywhere, plus you get to go whenever you want, and can go to various restaurants if you want). There's lots of forced fun like baseball games, and the in-bounds area amounts to 1.5 quads (those tiny courtroom things between dorm buildings). MOSP was about 10*10 square blocks and absolutely no forced group activities.

Obviously, MOSP may not be as realistic a goal as TIP, but even before being able to compare, I didn't like TIP at all. Summer ventures/governor's school (which are essentially the same camp for different age groups) are probably very good, but for Bryce and others at the top of competitive math, national camps like mathcamp are probably far preferable unless you're really well-rounded.

Oh, Davidson is part of Duke TIP (the same thing at a different site), so again I wouldn't recommend it. Also, there was a red MOPper last year who really loved TIP and went back that year, so my advice may not be the best.

[quote="jb05"]Also, the food is horrible[/quote]

Objection: when I was at Duke Young Writer's Camp, the food was great. (maybe because I like fettucine alfredo so much?)

Anyway... last year, I went to AMP (AwesomeMath), which was a great experience. There were rules, sure, but it was pretty relaxed. All you had to do was a) be at class on time (or Zuming makes you do pushups, sometimes :D ) and b) listen to counselors (mentors, assistants, whatever they are). The people are great, and MC winners get a scholarship (at least they did...) But the food is so-so. (Cafeteria's food court half the time... it'll probably be better this year)

[quote]Objection: when I was at Duke Young Writer's Camp, the food was great. (maybe because I like fettucine alfredo so much?)[/quote]

Yes, that would do it :wink: . UNL has two kinds of pasta sauce and two soups every day, plus actually good deserts, all sorts of entrees, wings on friday (not that I like wings), and stuff like that. On the other hand, three weeks without sweet tea nearly killed me. And the breakfasts weren't very good since all I eat for breakfast is fruit and bread (grits and waffles in the absence thereof). Of course, these crazy northerners don't even know what grits are, and the fruit wasn't so great, and there were no waffle machines.

I was mostly interested in camps that are relatively close and inexpensive.  From what I've read and heard, MathCamp and Awesome Math are excellent programs but cost quite a bit, especially when you factor in transportation.

I have 3 kids and not all are particularly interested in math.  I'm really interested in different types of summer programs, not necessarily math.  Sometimes it's a good idea to work a bit on weaknesses (i.e. writing, reading, ...) rather than strengths.

Not_trig - did you learn much at writer's camp?

I'm not really in favor of sending kids to camp primarily for "social" purposes - there are opportunities for that locally.

[quote="gt59"]
Not_trig - did you learn much at writer's camp?
[/quote]

Well, I took poetry and sci-fi (I was actually a day camper), and I think I got something from it. Not incredibly much, but something - also, I was exposed to speaking in front of large audiences, which I don't usually like to do.

[quote="gt59"]
Summer Ventures in Science and Math (UNC-CH)
[/quote]

hey i might be going there this summer. anyone know if it's any good?

Never been, but I've heard lots of good things about it from "mathletes" and just general good students.


symmetry vector integration geometry calculus derivative

an uniform cylinder of steel  is rotating about it's axis .what is the fractional change in
a) angular velocity
b) angular momentum (about it's axis)
c) rotational kinetic energy

when the temperature increases by $\Delta$$T$.
given coeffecient of linear expansion of steel is $\alpha$

I assume $\delta T$ is the absolute change of the temperature. For the sake of clarity ($\delta$ usually denotes relative change), let's relabel it $\Delta T$. I make this assumption because the relative change of the temperature doesn't actually give us any needed information.

The angular momentum of the disc is constant (the heat doesn't affect it, unlike the energy of the disc), so that answers (b). Since the ang. momentum is $L = I\omega$ we have $\delta L = \delta I+\delta \omega = 0$. The relative change of the moment of inertia is
\[\delta I = \frac{\text dI}{I}= \frac{\delta r}{2}= \frac{\alpha \Delta T}{2}.\]
Thus, we have $\delta \omega =-\alpha\Delta T/2$. That's the answer for (a). As for (c), we have
\[\delta K = \frac{\text dK}{K}= 2 \delta\omega+\delta I = \delta \omega =-\frac{\alpha \Delta T}{2}.\]

first how do you know that it's angular momentum doesn't change .this is because you can't transfer heat(rather increase it's internal energy) just like that .you have to heat(purely literary sense) it by some source .so there is convection+conduction+(negligible radiation) involved in the entire process .now conduction takes place by collision .now air molecules bombard the molecules of container exert some force transfer kinetic enegy, increase avg. kinetic energy, raise temp,supply heat energy though you may restrict the net force exerted to be nearly zero . how can you say that the net torque is zero ?

i changed $\delta$ to $\Delta$

A cynical response would be that the problem wouldn't be solvable if angular momentum wasn't conserved ;). Actually, it would be very tedious to do the complete kinetic analysis of the motion of molecules in the air, but we can use a symmetry argument here. I'm not sure I can express myself precisely; but you can convince yourself that there will be no average torque by considering that all directions of the molecules' velocities are equally probable.

what about this

we know that by definition the total change in rotational kinetic energy of a body is equal to the total work done by the  net torque. if we heat the cylinder uniformly then obviosly there is no net torque hence there should be no changein total rotational kinetic energy. :wink:

[quote="Djole"]There is no external torque, true. But rotational energy decreases because of the internal forces which expand the disc and increase its moment of inertia. The disc is not a rigid body...[/quote]

sure the internal forces act on the body but nowhere do we use the fact while deriving that the work done by the net torque acting on a body is equal to the total change in it' rotational kinetic energy that the body should be rigid.yes we do use it in deriving torque equations .
but if this it not a rigid body(which i assume to be instantaneously) then no laws can be applied

Oh, I understand what you mean now. But the disc is also getting heated. This doesn't influence ang. momentum but does change the total energy, so the work-energy theorem you're implying isn't valid, not in the form you're saying at least.

yes it does change the internal energy of the cylinder but as i had said in my previous post we exert some force to keep the system where it is and tyhe entire heat energy either gets converted in invcreasing the volume of the body which indeed is a work done but quite diff. from the sense in which the work done by the total torque .why doesn't work energy theorem hold here?

The heat goes into raising the temperature [i]and[/i] expanding the disc. With the given data, you have no chance to compute the complete heat transferred to the disc. If you could, then you would have the chance of using the work-energy theorem (if you knew how to calculate the work done during the expansion...)

let me give another soln to the problem

now the work energy theorem for torque is derived by taking the following consideration

$dK$ $=$ $I\alpha d\theta$ $=$ $\Gamma_{ext}d\theta$

assuming the body is instantaneously rigid .

but the conservation of angular momentum won't hold
let's assume for a moment $*$ as cross product

then
 $\frac{dL}{dt}$ $=$ $mr*\frac{dv}{dt}$ $+$ $m\frac{dr}{dt}*v$

but here while heating u can't assume $m\frac{dr}{dt}*v$ as $0$ as the particles also move lateraalyy due to absorbed heat.

so i claim rotational kinetic energy doesn't change rest proof follows.

[quote="Djole"]It would be very nice indeed if you could calculate the angular momentum of a disc as $L = m(\vec r \times \vec v)$... I'm asking you: what is your $\vec v$? What is your $\vec r$? (And since I'm on a roll here: what's $m$?) I'm serious. Tell me what these mean.[/quote]

well i thought u understood me
if i were in the centre of disc then had there been no heating then surely the second expression would be $0$. but since this thing is being heated also sitting t the centre i have to assume that instantaneously this thing  is rotating around me with a constant radius which is increasing.but if i don't inform u that the disc is being heated then u will feel that the particle at the extremes instantaneously move in a circle and also move radially outward .i think u got me....

well as for notations
$\vec v$ is velocity vector .
$\vec r$ is position vector

[quote="Djole"]What you're saying is true for a single elementary piece of the disc. But by summing up all the pieces, all changes cancel out. This isn't hard to derive. And you can't use $L = mrv$ for the disc because you just don't know what's $r$ and what's $v$ -- that's what I've asked you. The correct formula is $L = \int rv\text dm$ (in scalar notation, everything is perpendicular to everything so it's ok...). In short: you're wrong...[/quote]

well i don't how and what cancelled out . when deriving the expression for angular momentum i never considered this problem , i had just done that generally for a particle of mass $m$ which is moving with a velocity $\vec v$ and whose position vector from the refrence point is $\vec r$.
coming to this problem let me derive the expression

let's condider a ring of radius $x$ thickness $dx$ at time $t$
 now my refrence point is the centre of  the disc also
let $\beta$ $=$ $\frac{dT}{dt}$ (which i assume to be const. for simplicity sake
let $\vec v_{t}$ refer to the tangential velocity of each point
also $|\vec v_{t}|$ $=$ $\omega x$
let $\vec v_{r}$ denote the radial velocity of each point
also $|\vec v_{r}|$ $=$ $x\alpha \beta$
so the expression for rate change of angular momentum of the entire disc is the sum of the rate of change of angular monentum of the entire disc
first consider the change in angular momentum of a ring art a  distance $x$
let $\rho$ be the mass per unit area of the disc (assumed constant)
so rate of change of angular momentum of the ring is
$(x\vec i \times \vec F_{net}$) $+$ $2 \pi xdx\rho.(\vec v_{t}+\vec v_{r})\times \vec v_{t}$

what our concern is the second part of the about equation which on integartion noway gives  $0$
well djole this may seem very stupid to anyone but actually that's what the question demands the two forms of velocity

further please don't draw quick conclusions....... :D

[quote="pardesi"]further please don't draw quick conclusions[/quote]

I try hard not too :)

[quote="pardesi"]well djole this may seem very stupid to anyone but actually that's what the question demands the two forms of velocity
[/quote]

It doesn't seem stupid (quite the opposite) until you present your formula for the rate of change of the angular momentum. As I understand, the left hand term is the change due to external torque, if any, and the right hand term is supposed to be the change of $\text dm \vec r \times (\vec v_{r}+\vec v_{t})$ over time. This isn't what you've written... Because $\vec v_{r}\parallel \vec r$, this expression can be simplified to $\text dm \vec r \times \vec v_{t}= \text dm \omega r^{2}\vec e_{z}$, where $\vec e_{z}$ is the unit vector in the $z$-direction. The time derivative of this [i]can[/i] go to zero, but you still have to assume that it will.

well djole it is thge assumption that makes the problem so easier otherwise i feel the problem though [i]not[/i] a[i] tough [/i]problem is quite [i]interesting[/i] .if u didn't feel that the proof was stupid then thanks .[/i]

You're welcome :)

If you have any other interesting problems please post them!

surely i will and why don't u post one this time?


[quote="adrian3"]:blush:
Imi poate explica si mie,va rog frumos, ca pentru nivel de clasa a X-a(la inceput) ce sunt acelea INTEGRALE sau PRIMITIVELE FUNCTIILOR sau GRUPURI,INELE,CORPURI, LEGI DE COMPOZITIE?va rog frumos.Sunt doar un elev mai curios...multumesc anticipat![/quote]Pune mana pe un manual de clasa XI-XII, sau dute la http://planetmath.com si cauta definitiile ;)

Nu va suparati, am mers la adresa indicata unui elev mai inainte (mathplanet etc) dar nu am reusit sa dau peste un loc unde sa obtin diferite definitii dorite (in afara de un site unde trebuia sa incarc un soft si nu mi-a placut, posibil spyware etc).

1) Exista chiar un dictionar sau enciclopedie online, gratis, unde sa poti gasi definitii de notiuni? Unde? M-ar interesa si pentru matematici de colegiu si graduate.

2) Din intamplare, nu stiti daca exista un astfel de forum (excelent!) si pentru chimisti sau fizicieni? Sau informaticieni, unde sa poti discuta si probleme de liceu.

Va multumesc foarte mult si va felicit pentru acest forum

Jean

Mai exista mathworld, si in curand vom avea si aici pe site un sistem similar. De asemenea teoreme si formule (pentru chestiuni mai avansate) se gasesc in formurile dedicate: [url=http://www.mathlinks.ro/Forum/index.php?f=217]Olympiad Section[/url] si [url=http://www.mathlinks.ro/Forum/index.php?f=218]College Playground[/url].

Site-ul acesta contine forumuri care acopera cam toate domeniile legate de matematica, inclusiv fizica, chimie, informatica, samd. Se gasesc in categoria [url=http://www.mathlinks.ro/Forum/forum-328.html]Arts and Sciences[/url].

Poti incerca si wikipedia.org. De exemplu, pt. definitia structurii de inel, vezi [url=http://en.wikipedia.org/wiki/Ring_%28algebra%29]aici[/url]


Sequence of integers $ {\{ u_n \}}_{n \in {\mathbb{N}}_0}$ is given with:
$ u_0\equal{}0$, $ u_{2n}\equal{}u_n$, $ u_{2n\plus{}1}\equal{}1\minus{}u_n$ for all $ n \in {\mathbb{N}}_0$. ($ {\mathbb{N}}_0$ is $ \mathbb{N} \cup \{ 0 \}$).
a)	Find $ u_{1998}$.
b)	If $ p$ is natural number and $ m\equal{}(2^p\minus{}1) ^2$, determine $ u_m$.

Obviosly $ u_n\equal{}0 \ or \ 1$.
By induction $ u_n\equal{}S(n)\mod 2$, were S(n) is sum of 2-adic digits n.
b) $ n\equal{}2^{2p}\minus{}2^{p\plus{}1}\plus{}1.$ Therefore $ S(n)\equal{}p$.


[i]Let $ a_1 ; a_2 ; ...; a_n$ be positive integers such that : $ a_1 < a_2 < ... < a_n < 2n$

      and  $ k$  is the unique positive integer such that : $ 3^k < 2n < 3^{k \plus{} 1} \ \ ( n \geq 2 )$

  We have known that $ \forall i ; j \ \in \ \{1 ; 2 ; ..; n \} ; i  < j$ then $ a_i \not | a_j$

   Prove that :  $ a_1 \ge 2^k$[/i]

[quote="nguyenvuthanhha"][i]Let $ a_1 ; a_2 ; ...; a_n$ be positive integers such that : $ a_1 < a_2 < ... < a_n < 2n$

      and  $ k$  is the unique positive integer such that : $ 3^k < 2n < 3^{k \plus{} 1} \ \ ( n \geq 2 )$

   Prove that :  $ a_1 \ge 2^k$[/i][/quote]
No relation! Or I misunderstood?  :|
May $ a_n\ge 2^k$?


(I don't really know where to post this, so move if neccessary.)

Earlier, I went to [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2007]this link[/url] in order to print out a copy of the AMC 10 2007 test. When I clicked "Download PDF Version," some of the diagrams weren't included--it just showed the Asymptote code. If this could be fixed, that would be great.


In a triangle ABC m(A)=2. m(C), AB=10 and A(ABC)=5.(7^1/2) then find AC.

7^1/2 means square root of 7.


Each day, two out of the three teams in a class are randomly
selected to participate in a MATHCOUNTS trial competition.
What is the probability that Team A is selected on at least two
of the next three days? Express your answer as a common
fraction.

let us consider one case,
where the team A is picked on the first two but not the last.

$ \frac{2}{3} * \frac{2}{3} * \frac{1}{3} \equal{} \frac{4}{27}$

this case is essentially the same as all the other cases of 2/3 except the order. We know that this my be arranged in $ \dbinom{3}{2}$ ways. Hence we multiply $ \frac{4}{27}$ by three to arrive at $ \frac{4}{9}$.
Let us not forget the final case where class A is selected three times in a row, $ (\frac{2}{3})^3$ this is 8/27 and we add that to the original 4/9 so 12/27+8/27 = $ \boxed{\frac{20}{27}}$


At least two of the nonnegative real numbers $ a_1,a_2,...,a_n$ aer nonzero. Decide whether $ a$ or $ b$ is larger if

$ a\equal{}\sqrt[2002]{a_1^{2002}\plus{}a_2^{2002}\plus{}\ldots\plus{}a_n^{2002}}$

and

$ b\equal{}\sqrt[2003]{a_1^{2003}\plus{}a_2^{2003}\plus{}\ldots\plus{}a_n^{2003} }$

[quote="modularmarc101"]i would guess b is larger. just a guess[/quote]

Actually, the larger one is $ a$. :wink:

[quote="mathking123"]would the power mean inequality help?[/quote]

I don't think so. :wink:

[hide]We can assume $ \{ a \}$ is non-increasing. Write $ \sqrt[k]{\sum_{i\equal{}1}^n a_i ^k }$ as $ a_1 \sqrt[k]{\sum_{i\equal{}1}^n \left( \frac{a_i}{a_1} \right)^k}$.

The summation under the radical cannot increase if we increase $ k$ because each of the $ n$ terms will either remain the same (if they are 1 or 0) or decrease.

Also, the summation inside the squareroot is at least $ 1$ because of the term $ (a_1/a_1)^k$ -- therefore, taking a higher root of it will necessarily cause it to decrease.[/hide]

I think this can be solved with help of induction.
For n=2 (both nonzero)
$ (a_1^{2002}\plus{}a_2^{2002})^{2003}>(a_1^{2003}\plus{}a_2^{2003})^{2002}$ can be proved with Newton's binomial theorem. I think that we can also use this theorem in inductive step.

The relevant inequality is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=237295]here[/url].  limes123's suggestion seems to work more or less fine.

Probably not. I don't know how the MNMO grades, but here is the scoring rubric for the USAMO:

    * 0 - No work, or completely trivial work
    * 1-2 - Progress on the problem, but not completely solved
    * 3-4 - All steps are present, but may lack clarity. (These scores are very rare.)
    * 5-6 - Complete solution with minor errors
    * 7 - Perfect solution

Your response does not make progress towards a thorough solution, so it would receive 0 points if it were on the USAMO.

Having said that:
1. Making appropriate guesses using context clues is very helpful in finding out the answer, which can help you get started towards a legitimate solution.
2. Considering appropriate cases is helpful in a short-answer test where proof is not required.
3. In a proof test, if this is what you come up with, and you have free time, its worth writing down to see if you get any points.

I think the following brute force method works:

For fixed non-negative real numbers $a_1, \dots, a_n$, define the function $f(x) := \left( a_1^x + \dots + a_n^x \right)^{\frac{1}{x}}$. Directly differentiate the function $f$ and prove that the derivative is always non-positive. Hence, $a = f(2002) \geq f(2003) = b$.


algorithm
pigeonhole principle
geometric series

Take as the definition of a rational number a number of the form $ \frac {p}{q}$ where p and q are integers with $ q \neq 0$. Prove that a number is rational if and only if its decimal representation repeats (this includes decimal representations that ends as they are basically ones with repeating zeros).

[hide="Proof"]The set of rational numbers is defined as $ \mathbb{Q}\equal{}\left\{\displaystyle\frac{p}{q}: p\in\mathbb{Z},q\in\mathbb{Z},q\neq0\right\}$
We may note that $ \mathbb{Q}$ forms a field. (I can show you a proof if you want)

Since $ \mathbb{Q}$ is a field, a case involving two elements of $ \mathbb{Q}$ will hold for all other cases. Therefore, I will use the following example:

Consider the number $ \frac{5}{12}$, using the division algorithm (Im not sure how to write long division), we can see that eventually $ 80$ becomes a repeating remainder. Therefore, the quotient is $ .4166...$ because $ 80/12\equal{}6$, $ 80\minus{}(6*12)\equal{}8$ and the cycle repeats.

As stated before, since $ \mathbb{Q}$ is a field, this example is a representation of all the other numbers within the field and will hold true.
[/hide]

[quote="JZambrano201"]
...since $ \mathbb{Q}$ is a field, this example is a representation of all the other numbers within the field and will hold true.[/quote]

I don't understand why this would be true. $ \mathbb{R}$ is a field also, but this example does not represent what happens to all elements of the field $ \mathbb{R}$.

heres the general idea:
[hide]a number with repeating decimal representation is rational:  it reduces to a convergent geometric series with rational terms.

A rational number has a repeating decimal:  division algorithm, only so many possibilities for remainder/digit combo's so by pidgeonhole it must start to repeat[/hide]

[quote="JZambrano201"][hide="Proof"]The set of rational numbers is defined as $ \mathbb{Q} \equal{} \left\{\displaystyle\frac {p}{q}: p\in\mathbb{Z},q\in\mathbb{Z},q\neq0\right\}$
We may note that $ \mathbb{Q}$ forms a field. (I can show you a proof if you want)

Since $ \mathbb{Q}$ is a field, a case involving two elements of $ \mathbb{Q}$ will hold for all other cases. Therefore, I will use the following example:

Consider the number $ \frac {5}{12}$, using the division algorithm (Im not sure how to write long division), we can see that eventually $ 80$ becomes a repeating remainder. Therefore, the quotient is $ .4166...$ because $ 80/12 \equal{} 6$, $ 80 \minus{} (6*12) \equal{} 8$ and the cycle repeats.

As stated before, since $ \mathbb{Q}$ is a field, this example is a representation of all the other numbers within the field and will hold true.
[/hide][/quote]

I'm not sure why you think this is valid.  $ \mathbb{Q}$ being a field has nothing to do with the nature of the decimal representation.

[quote="JZambrano201"][hide="Proof"]The set of rational numbers is defined as $ \mathbb{Q} \equal{} \left\{\displaystyle\frac {p}{q}: p\in\mathbb{Z},q\in\mathbb{Z},q\neq0\right\}$
We may note that $ \mathbb{Q}$ forms a field. (I can show you a proof if you want)

Since $ \mathbb{Q}$ is a field, a case involving two elements of $ \mathbb{Q}$ will hold for all other cases. Therefore, I will use the following example:

Consider...

As stated before, since $ \mathbb{Q}$ is a field, this example is a representation of all the other numbers within the field and will hold true.
[/hide][/quote]

[quote]The set of rational numbers is defined as $ \mathbb{Q} \equal{} \left\{\displaystyle\frac {p}{q}: p\in\mathbb{Z},q\in\mathbb{Z},q\neq0\right\}$
We may note that $ \mathbb{Q}$ forms a field.

Consider $ \frac{1}{2}$ and $ \frac{1}{3}$. Since $ \mathbb{Q}$ is a field, this example is a representation of all the other numbers within the field and will hold true. We may thus conclude that any element in $ \mathbb{Q}$ has a $ 1$ as its numerator.[/quote]

This is not true, either. I believe you need to show how it follows from your example and $ \mathbb{Q}$ being a field that the statement can be generalized.

In fact, I don't see how you could prove that the statement is true for all $ x \in \mathbb{Q}$ by giving examples. So I don't think your proof is valid.

Sorry for the incorrect proof. I see where I went wrong.


Let $ P(x)\in Z[x]$ such that $ P(n)>n$ for every positive integer $ n$.Consider $ x_1\equal{}1;x_{n\plus{}1}\equal{}P(x_n)$.
We know that, for any positive integer  $ N$, there exists a term of the sequence divisible by $ N$. Prove that $ P(x)\equal{}x\plus{}1$ .


ratio
number theory
relatively prime
geometric sequence

1. Let $A_{n}$ and $P_{n}$ be the respective sums of the first $n$ terms of two arithmetic sequences. If $\frac{A_{n}}{P_{n}}=\frac{13n+3}{4n+30}$ for all $n$, then the ratio of the $17th$ term of the second sequences is $k: w$ where $k$ and $w$ are two relatively prime positive integers. Find the value of $k+w$.

2. Three distinct integers form a geometric progression, the sum of the integers is $1387$, and the common ratio is an integer. Find the sum of all possible distinct values for the largest term of the geometric progression.

[hide="1"]
We have that $A_{n}= \sum_{k=0}^{n}(a+kd) = an+\frac{n(n+1)}{2}\cdot d$ and that $P_{n}= \sum_{k=0}^{n}(b+km) = bn+\frac{n(n+1)}{2}\cdot m$. So,
\[\frac{A_{n}}{P_{n}}= \frac{ an+d \cdot \frac{n(n+1)}{2}}{ bn+m \cdot \frac{n(n+1)}{2}}= \frac{ 2a+d(n+1) }{ 2b+m(n+1)}= \frac{13n+3}{4n+30}\]
Matching the coefficients, we have that $d=13 \implies a=-5$ and $m=4 \implies b=13$. Thus, the ratio of the two $17^{th}$ terms is:
\[\frac{-5+16 \cdot 13}{13+16 \cdot 4}= \frac{29}{11}\implies 29+11 = \boxed{40}\]
[/hide]

[hide="2"]
We have that:
\[a+ar+ar^{2}=a(1+r+r^{2})=1 \cdot 1387=19 \cdot 73 \]
Since both $a$ and $r$ are integers, the factors on the left have to match up to some factorization on the right. So:
\[a=1 \implies r \notin \mathbb{Z}\]

\[\ a=1387 \implies r=-1 \]

\[a=19 \implies r=-9 \ \vee r=8 \]

\[\ a=73 \implies r \notin \mathbb{Z}\]
Thus our sum of third terms will be
\[1387 \cdot (-1)^{2}+19 \cdot \left( (-9)^{2}+8^{2}\right) = \boxed{4538}\]
[/hide]

[quote="jeez123"][hide="solution for #1"]
$\frac{A_{1}}{P_{1}}=\frac{a_{1}}{p_{1}}=\frac{16}{34}=\frac{8}{17}.$
Since only the ratio of the 17th term is asked for, put $a_{1}=8, p_{1}=17$. Now $\frac{A_{2}}{P_{2}}=\frac{8+(8+d_{a})}{17+(17+d_{p})}=\frac{29}{38}$. Put $d_{a}=13, d_{p}= 4$. These numbers check in the next few cases, so $a_{17}=8+16 \cdot 13, p_{17}=17+16 \cdot 4$. this makes the required ratio $\frac{8}{3}$, so the answer is 11.[/hide]

where did they get the $\frac{16}{34}$?

[hide="solution for #2"]Let $n$ be the smallest and $r$ be the common ratio. Then $n+nr+nr^{2}=1387$, so $n=\frac{1387}{1+r+r^{2}}$. Use a table to find that $n=19$ when $r = 8$ and the third number is $1216$. Also $n=19$ when $r=-9$ and the third number is $1539$. A third solution is $n=1387$ when $r=-1$, but the third number is also $1387$. Answer=2755.[/hide]

What did they mean by "use a table"?[/quote]

They meant on a calculator. Which means that your solution source is pretty much a n00b, sorry to say, as there is a much [u]quicker[/u] (and more elegant) way to do the problem.

As for the second one, I made a small mistake. It should be in the numerator and denominator $a(n+1)$ and $b(n+1$; but it's late, and i'll fix it tomorrow.

[quote="cincodemayo5590"]
They meant on a calculator. Which means that your solution source is pretty much a n00b, sorry to say, as there is a much [u]quicker[/u] (and more elegant) way to do the problem.

As for the second one, I made a small mistake. It should be in the numerator and denominator $a(n+1)$ and $b(n+1$; but it's late, and i'll fix it tomorrow.[/quote]

lol
anyways, thank you so much, you helped me a lot today!


calculus
integration
geometry
function
algebra
domain
calculus computations

Hello,

Today in my calculus class I was evaluating an iterated integral, which involved evaluating this:

$ \int_{\minus{}r}^r \arcsin \frac{\sqrt{r^2\minus{}x^2}}{r} dx$

Now my first thought was to integrate by parts, but a question came to mind: if I made the substitution $ u\equal{}x^2\minus{}r^2$, the limits of the integral would become equal, and the integral would itself (I believe) evaluate to zero. However, my CAS tells me that the area is $ 2r$.

I took an easier example:

$ \int_{\minus{}1}^1 \sqrt{1\minus{}x^2} dx$,

the semi-circle above the $ x$-axis with radius $ 1$, which clearly has a positive area. Now suppose I did not use trigonometric substitution, and that I instead make the substitution $ u\equal{}1\minus{}x^2$, making the limits again become equal. The integral again becomes zero, which does not agree with the non-zero area.

So my ultimate question: does the substitution $ u\equal{}f(x)$ require that $ f$ be injective?

Thanks!

yes, the fact that it is MONOTONE is used in the proof.

@[b]kenn4000[/b]

Wow. I have never realized it (I am lazy to read the proof).

Yes, we need the fact that the function to be substituted is monotone so that the new upper and lower limits make sense.

%[b]jvenezuela716[/b]

I don't think that your first substitution is not right. It should be $ u \equal{} r^2 \minus{} x^2$.

By the way, in case the function is not monotone, we can break the domain so that the function is monotone on these smaller domains. (Not sure if we can always do this.) And hence, substitution still works.

Um.... isn't $ \arcsin \frac{\sqrt{r^2\minus{}x^2}}{r}$ just $ \pm\arccos \frac{x}{r}$?

You got in trouble here, but a substitution in one-variable integrals do not need be changes of variables (i.e. injective).

$ \int_{\minus{}1}^1 f(u^2)2u \, du \equal{} \int_1^1 f(x) \, dx \equal{} 0$

Even though as a change of variable $ x\equal{}u^2$ this makes absolutely no sense. I've always found it be an oddity, but I think of it as the fact that it's a change of measure, not a change of variable

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=208416]here[/url]. It's the same issue; you have an $ x$ term in the d_ , which becomes $ \pm\sqrt {u \plus{} r^2}$. The $ \pm$ switches in the middle of the integral, and the two halves reinforce each other rather than canceling out.

[quote="Kalle"]You got in trouble here, but a substitution in one-variable integrals do not need be changes of variables (i.e. injective).

$ \int_{ \minus{} 1}^1 f(u^2)2u \, du \equal{} \int_1^1 f(x) \, dx \equal{} 0$

Even though as a change of variable $ x \equal{} u^2$ this makes absolutely no sense. I've always found it be an oddity, but I think of it as the fact that it's a change of measure, not a change of variable[/quote]

But the original integral is an odd function, so it makes sense that it would equal 0?

Sure, but my point remains that the general formula

$ \int_{a}^b f\circ \phi \; \phi' \, dx \equal{} \int_{\phi(a)}^{\phi(b)} f(x) \, dx$

does not necessarily represent what I'd call a change of variables and there is no bijectivity condition of any sort on $ \phi$. It is closer related to the fact that $ \phi' dx \equal{} d\phi$ where the right hand side is a Stieltjes measure.

In higher dimensions, the analogue of this formula is patently false when $ \phi$ is not bijective

So is $ \int_a^b f(\phi(x))\phi'(x) dx \equal{} \int_{\phi(a)}^{\phi(b)} f(u) du$ true if and only if $ \phi$ is continuously differentiable?


Sorry, this is the repaired problem:
The lengths of the sides and the diagonals of the quadrilateral ABCD is rational numbers. AC cuts BD at O. Prove that OA is rational.
(I am a Vietnamese. So, my English is very bad!)

[quote="quangchuc"]Quadrilateral ABCD has rational lengths of sides. AC cuts BD at O. Prove that OA is rational.
(I am a Vietnamese. So, my English is very bad!)[/quote]

Unfortunately, this is wrong : Consider a square whose side length is $ 1$ (rational). Then length$ (0A)\equal{}\frac{\sqrt{2}}{2}$ irrational.

[quote="pco"][quote="quangchuc"]Quadrilateral ABCD has rational lengths of sides. AC cuts BD at O. Prove that OA is rational.
(I am a Vietnamese. So, my English is very bad!)[/quote]

Unfortunately, this is wrong : Consider a square whose side length is $ 1$ (rational). Then length$ (0A) \equal{} \frac {\sqrt {2}}{2}$ irrational.[/quote]

Your counterexample is incorrect.
We assume the diagonals are also of rational length.

[quote="i_like_pie"][quote="quangchuc"]The lengths of the sides and the diagonals of the quadrilateral ABCD is rational numbers. AC cuts BD at O. Prove that OA is rational.[/quote]
Counterexample: A rhombus with side length $ 5$, and diagonals of lengths $ 6$ and $ 8$.

There are many other similar counterexamples.[/quote]

Um what?...

OA is definitely rational in this case...

Didn't that say "irrational"? Apparently not...

[quote="JBL"]Fraenkel, a couple of seconds of thought or some more careful reading would have indicated that pco gave a correct counterexample to a now-repaired problem.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1970738868&t=53662 for a solution.[/quote]

Im not sure what you are saying. The problems asks us to assume that the sides and diagonal are rational.

A square with sides 1 has a diagonal $ \sqrt{2}$ in length, thus cant be used as a counterexample.
I tried to examine it some more but I dont think any scale of it also has sides and diagonals of rational length.

I'm not too mathematically mature, but I assure you I gave this alot of thought, and read it for a long time.

The error must be in my reasoning. I would appreciate you pointing it out to me.

[quote="JBL"]You are of course correct that a square is not a counterexample to the question in its current form.  However, you have failed to note the following things:

1) The opening line of the post, which reads "Sorry, this is the repaired problem."
2) The time of the latest edit to the post, which is later than pco's comment.
3) The quoted text in pco's post, containing an earlier version of the problem without the condition that the diagonals are rational.[/quote]

HaHa oh that!
You had me worried, I thought there was something wrong in my math!
w/e the earlier post was not an attack on pco, but just a statement declaring that the counterexample does not work in the repaired problem.


limit
irrational number
number theory open
number theory

Does anyone know if $ \sqrt {2}$ has infinite digits $ 0$ in base $ 10$?

An equivalent problem which has perplexed me for a long time:if $ \{10^n\alpha\}$ is equidistributed in $ [0,1]$ where $ \alpha$ is irrational
Can we prove $ \lim_{n\to \infty}\frac{1}{n}\sum_{k\equal{}1}^{n}e^{2i\pi p10^k\alpha}\equal{}0$ for every positive integer $ p$?I think Weil criterion failed at this point :?:

[quote="hxy09"]An equivalent problem which has perplexed me for a long time:if $ \{10^n\alpha\}$ is equidistributed in $ [0,1]$ where $ \alpha$ is irrational
Can we prove $ \lim_{n\to \infty}\frac {1}{n}\sum_{k \equal{} 1}^{n}e^{2i\pi p10^k\alpha} \equal{} 0$ for every positive integer $ p$?I think Weil criterion failed at this point :?:[/quote]

A counterexample given by harazi:$ \alpha \equal{} \sum_{n\geq 1}{10^{ \minus{} 3^{n}}}$ :)

Hi 123eeee;

The property of constants called normal means that a irrational number would have the distribution of the digits 0-9, 1/10 of the time for base 10 and for the appropriate digits in every other base. It is believed that the sqrt(2) is normal but not proven.


function
integration
real analysis
real analysis unsolved

[quote="cezar lupu"]Let $f: [0,1]\to\mathbb{R}$ be a $C^{1} [0,1]$ function such that:

                                  \[ \int_{0}^{1} f(x)dx=\int_{0}^{1} xf(x)dx=1.  \]
  Prove that there is $c\in (0,1)$ such that $f'(c)=6.$ ;)[/quote]
Set $g(x) = f(x)+ 2 - 6x$, then a calculation shows that
\[ \int_{0}^{1} g(x)dx=\int_{0}^{1} xg(x)dx=0.  \]
Therefore also  $\int_0^1 h(x)g(x)dx = 0$ for all linear functions $h$, and also either $g = 0$ (implying $f'(x) = 6$ for all $x$) , or $g$ attains both positive and negative values.

Assume now that $g'(x) > 0$ for all $x$. Then there is some $c \in (0,1)$ s.th. $0 = g(c)$  and $g$ is negative on $[0,c)$ and positive on $(c,1]$. Pick $h(x) = x-c$ to obtain the contradiction $h(x)g(x) > 0$ for $x \neq c$. Similarly, $g'(x) < 0$ for all $x$ is impossible. Thus there must be a $c$ with $g'(c) = 0$, i.e. $f'(c) = 6$.   :)


Support
Euler
geometry
geometric transformation
rotation

I've always found it stupid to name a theorem like "euler third theorem" or something like that .. do you think this name is appropriated ? I know that a lot of my Physics teachers do the same but I'm firmly convinced it is not an appropriated name for an international discussion .. (but I'm sure evrybody has understood ..)

Well - it's the only name of the law I know (although I'm sure there are some others as well) and writing down the whole law so that people know which one you're talking about is not very practical...

[quote="juicybooty911"]could someone explain to me newton's third law of motion?
i'm confused about how it applies to objects, for example, if i applies 30 N to a book, the book applies 30 N back to me, so in order to move the book i have to apply more force to overcome the static friction, but if i apply more force, wouldn't the book force increases too and therefore i'm not able to move the book?
thx[/quote]First, let me say you have some BIG books.

Newton's 3rd law follows from the conservation of momentum.

$\vec{p}_1 + \vec{p}_2 = \text{ const.}$

$m_1\vec{v}_1 +m_2\vec{v}_2 = \text{ const.}$

$\frac{d}{d t}(m_1\vec{v}_1 +m_2\vec{v}_2) = 0$

$m_1\frac{d\vec{v}_1}{d t} +m_2\frac{d\vec{v}_2}{d t} = 0$

$m_1\vec{a}_1 +m_2\vec{a}_2 = 0$

Which from Newton's 2nd law implies ...

$\vec{F}_1 +\vec{F}_2 = 0$

From which it follows that ...

$\vec{F}_1 = -\vec{F}_2$

"For every action there is an opposite (direction) but equal (size) reaction."

I believe the thing you are failing to account for is that due to your greater mass you withstand the book's pushing on you better than the book withstands your pushing on it. Also the periods of time over which contact forces such as these act is usually very brief.

For example, if you were to get angry with your neighbor, and punch him in the nose then your fist is in contact with his nose for only a few tens of milliseconds -- that's as long as the force of your punch lasts for either of you.  Your fist is a lot more solid than your neighbor's nose, so that the 100 lbs that he feels does a lot more damage to his nose than the like 100 pounds that you feel does to your hand.

[quote="kubus"]If you apply 30N to a book, it does apply 30N back to you. If you apply 300N, it applies 300N. But that doesn't mean it won't move - it applies 300N to YOU, not to itself. If the static friciton can push with max. 50N and you push 60, there is a net force ON THE BOOK of 10N, and it moves with appropriate acceleration. The book applies a 60N force on you and you don't accelerate backwards, because you for example stand on the ground and "push back against it", which in turn pushes you in the opposite direction with 60N (not regarding the 600N more it supports your weight, that is upwards). It is actually probably more than 60N, because some force must also accelerate your hand, etc.[/quote]

I absolutely agree with you kubus but one thing,when u push the ground back,u will exert some more force on the book and then thee book must apply some more force on u with the same amount and if the coefficient of friction is same for both the surfaces,ur legs and the table,so then why does the book move??Still a bit confused.

[quote="shadysaysurspammed"]... and if the coefficient of friction is same for both the surfaces,ur legs and the table,so then why does the book move??Still a bit confused.[/quote]The maximum force of friction exerted on either you or the book is $\mu N$, where $N$ is the normal force exerted by the floor on you (or the book).  If both you and the book rest upon a horizontal floor, the normal force exerted by the floor is equal to the weight of the object it supports (Newton's 3rd law again). Therefore if your mass is $M$ and the book's mass is $m$ then the the frictional force on you is: $\mu Mg$, and the friction on the book is $\mu mg$.

If the force you exert on the book, $F < \mu Mg$, [u]you[/u] will not move.

Likewise, if the force you exert on the book, $F > \mu mg$, then [u]the book[/u] will move.

[quote="eminem"]i would like to add another question here. When we walk they say friction is enabling you to walk, that is the force which is making u go forward, the one which  is  exerted by the ground on u as a result of u pushing the ground backwards is also friction . Is this true ?? then why is friction a resistive force ?. Or are there 2 kinds  of friction ??[/quote]First, friction is friction; there is only one kind. It is produced by the surface mismatch at the point(s) of contact between two objects. It is always directed parallel to the contact surface, and it direction always opposes the direction of applied force.

To use your example, pushing the ground backward with your foot causes the friction between your foot and the ground to push you forward.

To be a bit more obvious, consider your car's tires. The direction of their rotation at the point where they contact the road is rearward. However the force of friction at that point (the only point where friction acts in this case) pushes the car forward. If the road were icy, so that the friction was zero, the tires would spin and the car would go nowhere.

[quote="eminem"]hmm, i think i understood now , thx :D

one more thing . when a block is kept on a table , it pushes the table down, and the normal force of the table pushes it up , then why doesnt the block go up in the air as it is recieving a thrust upwards???[/quote]

Well there has to be a net force in one direction fopr the block to move but there is the weight of the block as well as the thrust on the block,both acting $simultaneously$,equally and in opposite directions.so the block doesnot move. :D

[quote="eminem"]hmm, i think i understood now , thx :D

one more thing . when a block is kept on a table , it pushes the table down, and the normal force of the table pushes it up , then why doesnt the block go up in the air as it is recieving a thrust upwards???[/quote] According to Newton's 2nd law, the net force on the block is equal to its mass times acceleration.

In this case that means: $\vec{N}+\vec{W} = m\vec{a}$

But by Newton's 3rd law the normal force is equal but opposite the weight of the block $\Rightarrow \vec{N} = -\vec{W}$.

Substituting that equation into the one above it, one obtains:

$-\vec{W} +\vec{W} = m\vec{a} = 0$

Since $m \neq 0$, one finds $\vec{a} = 0$.

The block is not "thrust upward by the normal force;" it is kept in static equilibrium.


prove that in graph with $v$ virtex and $\epsilon$ edge  :

$\frac {v^2}{v^2-2\epsilon}  \leq \chi$

what you want to prove is $\displaystyle\alpha(G)\le n-\frac{\sum_i d_i}{n}$
where $\alpha(G)$ is the independence number of $G$ and $d_i$'s are the degrees of the vertices of $G$.(the rest follows from the fact that $\chi(G)\ge \frac{n}{\alpha(G)}$)

 or more simply what you need is that $\frac{\sum_i d_i}{n}\le n-\alpha$. if $I$ is a maximum independent set of size $\alpha$ then note that for each vertex of $I$ the degree is atmost $n-\alpha$;so
$\sum_i d_i\le (n-\alpha)(n-1)+\alpha(n-\alpha)$ so that
 $\frac{\sum_i d_i}{n}\le (n-\alpha)(\frac{n-\alpha+1}{n})$ and since $\frac{n-\alpha+1}{n}\le 1$ we are through.


complex numbers
calculus
calculus computations

I Had two questions

I have N=1/(1+iK), with k a constant
now I want to obtain two things

1. what is the length of N, so what is abs N
2. what is the real part of N and what is the imaginary part
cause when I have 1+ik, i know that k is the imaginary part and 1 the real part,
but how do I do it if I have a number like N

thanks  :lol:

(1) $ \left|\frac{z}{w}\right|\equal{}\frac{|z|}{|w|},$ so $ \left|\frac{1}{1\plus{}iK}\right|\equal{}\frac{1}{|1\plus{}iK|}\equal{}\frac{1}{\sqrt{1\plus{}K^2}}.$

(2) Always multiply through by the conjugate of the denominator before attempting to answer a question like this.

$ \frac{1}{1\plus{}iK}\equal{}\frac{1}{1\plus{}iK}\cdot\frac{1\minus{}iK}{1\minus{}iK}\equal{}\frac{1\minus{}iK}{1\plus{}K^2}.$

Hence, $ \text{Re}\left(\frac{1}{1\plus{}iK}\right)\equal{}\frac1{1\plus{}K^2}$ and $ \text{Im}\left(\frac{1}{1\plus{}iK}\right)\equal{}\frac{\minus{}K}{1\plus{}K^2}$


Hi.
I want to do this:
I have a list of this form
[code]
1. aaa
2. aaa
3. a. bbb
   b. ccc
4. eeee
   a. hsh
   b. dfsg
5. dff
[/code]
where my problem is: How to do the a and b correctly (I tried \item[a] and \item[6.a.] , but then the "vertical alignment" is wrong...
[code]
  5
6.a
6.b
[/code])
does anyone knows how to do this ??
Thanks in advance...

not sure but try this

\item[5\phantom{.a}]

it won t display the ".a" but it will act as if it was there

Thanks, but then it's even less aligned:
you have then:
[code]
  1
  2
  3
  4
5
  6

[/code]
instead off
[code]

1. aaa
2. aaa
3. a. bbb
   b. ccc
4. eeee
   a. hsh
   b. dfsg
5. dff[/code]

ok, it s not perfect, but it s better

...
\newcommand{\itemind}[1]{\item \quad \,\, #1}

\begin{document}

\begin{enumerate}

\itemind aaa
\itemind bbb
\item \quad \,\, a. bbb
\subitem b. ccc
\item eeee

\end{enumerate}
...

the spacing is a bit big, but I only found how to increase the one of \item instead of reducing the one of \subitem but it s a beginning

Hi, Ledurt, sorry to tell it, but I didn't knew of the existence of "\subitem"...
And \subitem does what I needed.

So, thanks, you solved my question.

plinius.

you don't happen to know how to centre a table ?