[hide]Consider $ f(x) = \frac{x}{\sqrt{1-x}}.$
$f(x)$ is convex (concave-up) over the interval $(0,1).$ Now use Jensen's inequality to get the result.
[/hide]

Methos like Nesbitt's: just Jensen for $\frac{t}{\sqrt{1-t}}$.
[b]Edit: Im' sorry, $\text{FieryHydra}$, that I didn't see your solution but it appeared when I posetd.[/b]

[quote="perfect_radio"]Let $a,b,c > 0$ such that $a+b+c=1$. Prove that \[\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\geq \sqrt{\frac32}.\] [i]Cezar Lupu[/i][/quote]
It isn't hard to generalise. After some computations we can prove that $f(t)=\frac{a}{\sqrt[n]{1-a}}$ is convex and we get \[\sum_{\text{cyclic}}\frac{a}{\sqrt[n]{b+c}}\geq \sqrt[n]{\frac{3}{2}}.\]

This one for example: http://www.mathlinks.ro/Forum/viewtopic.php?p=453797#453797

[quote]is there another solution that doesnt use functions. algebra maybe?[/quote]

sure this (or similar to this have been posted before but ...

when you take $x=\sqrt{(b+c)}, y=\sqrt{(c+a)}$   and  $z=\sqrt{(a+b)}$

HM-AM  gives
$1/x+1/y+1/z \ge 9/(x+y+z)$ (eq 1)
and RMS-AM gives
$(x^{2}+y^{2}+z^{2})/3 \ge ((x+y+z)/3 )^{2}$  (eq 2)

[hide]$LHS = (1-x^{2})/x+(1-y^{2})/y+(1-z^{2})/z = (1/x+1/y+1/z)-(x+y+z)$
 $\ge 9/(x+y+z)-(x+y+z)$
 $\ge 9/\sqrt 6-\sqrt 6 = \sqrt{3/2}$
[hide]Since( $x+y+z \le \sqrt{3(x^{2}+y^{2}+z^{2})}= \sqrt 6$  by eq 2
(so $(1/(x+y+z) \ge 1/(\sqrt 6)$[/hide][/hide]


Αν η $ f$ έχει συνεχή παράγωγο και $ \displaystyle\lim_{x\to\infty} [(f(x))^3 \plus{} (f '(x))^2] \equal{}0$ τότε να δείξετε ότι $ \displaystyle\lim_{x\to\infty}f(x)\equal{}\displaystyle\lim_{x\to\infty}f '(x)\equal{}0$

[quote="r_boris"][quote="r_boris"]Αν η $ f$ έχει συνεχή παράγωγο και $ \displaystyle\lim_{x\to\infty} [(f(x))^3 \plus{} (f '(x))^2] \equal{} 0$ τότε να δείξετε ότι $ \displaystyle\lim_{x\to\infty}f(x) \equal{} \displaystyle\lim_{x\to\infty}f '(x) \equal{} 0$[/quote]
H άσκηση απευθυνεται σε φοιτητές Αν μπορεί ο Μπάμπης να  την μεταφέρει εκεί έχει καλώς Ζητώ  συγνώμη[/quote]

Ροδόλφε ,
 δεν πειράζει . Ας είναι και δω.Δεν υπάρχει κανένα πρόβλημα.Δεν μπορώ να το μεταφέρω , διότι είναι στο ίδιο Ελληνικό τμήμα και όλο αυτό θεωρείται ένας τομέας(section).
 Μπάμπης

Θα δώσω στο attachment ενα σύνολο από βοηθητικά ερωτήματα (πολλά :o ) που κατευθύνουν καθώς και την λύση. Μου φάνηκε πολύ μα πολύ δύσκολη :!:

Na dosw mia paraplisia askisi ? Tha dosw ti rwtaw :P

An kserw oti isxiei gia mia f me sinexi proti paragogo oti  lim F+F' = L , gia x-->+00 tote na diksw oti to orio tis prwtis askisis pou edose o k. r_boris isoute me L^3  !

[quote="Durandal"]Outline σε spoilers :)
[hide]WLOG, μπορείς να θεωρήσεις $ L \equal{} 0$ ($ f\to f \minus{} L$).

Έστω $ \varepsilon > 0$, οπότε $ \minus{} \varepsilon < f(x) \plus{} f'(x) < \varepsilon$ για αρκετά μεγάλα $ x$. Τότε $ \minus{} \varepsilon e^x < \left(f(x) e^x\right)' < \varepsilon e^x$ και το δεξί μέλος της ανισότητας δίνει ότι η $ (f(x) \minus{} \varepsilon) e^x$ είναι φθίνουσα (ανάλογα το αριστερό). Έτσι, για αρκετά μεγάλα $ x$ προκύπτει το estimate $ f(x) < \varepsilon \plus{} M e^{ \minus{} x}$ (με ανάλογο κάτω φράγμα).

Από εδώ και κάτω είναι εύκολο :)
[/hide]
Cheerio,

Durandal 1707[/quote]

mporeis na sinexiseis  tin skepsi sou epeidi kapou to xanw??? :oops:  :blush: Vasika exw teleios diaforetiki lisi sto mialo mou :P

[quote="SorrowMan"]mporeis na sinexiseis  tin skepsi sou epeidi kapou to xanw??? :oops:  :blush: Vasika exw teleios diaforetiki lisi sto mialo mou :P[/quote]

Αφού $ f(x) < \varepsilon \plus{} M e^{ \minus{} x}$ για αρκετά μεγάλα $ x$ (με ανάλογο κάτω φράγμα), έπεται ότι $ \minus{} 2\varepsilon < f(x) < 2 \varepsilon$ για αρκετά μεγάλα $ x$ (eventually $ Me^{ \minus{} x} < \varepsilon$). Αφού το $ \epsilon > 0$ είναι αυθαίρετο, αυτό σημαίνει ότι $ \lim_{x\to\infty} f(x) \equal{} 0$, οπότε και $ \lim_{x\to\infty} f'(x) \equal{} 0$, QED   :lol:

Cheerio,

Durandal 1707


find all function $f: \mathbb{N}\to\mathbb{N}$ such that

(i) $f(2)=2$
(ii) if $(m,n)=1$ then $f(m\cdot n)=f(m)\cdot f(n)$
(iii) $f(n)$ is strictly increasing .

Isn't it very easy one?Maybe posted many times.$f(n)=n$
In China TST2003 there is a more difficult one.
$f: \mathbb{N}\to\mathbb{R}$
And don't need $f(2)=2$. :P

It is correct for $f: N\to R$ if 1),2),3) then f(n)=n. If only 2),3) then $f(n)=n^a$ a is positive real number.

To give the source of the theorem Rust mentions:
Erdos, P. [i]On the distribution function of additive functions[/i], Ann. Math., (2) [b]47[/b] (1946), 1-20.


Let $ C_{m,n}$ denote the number of $ m$-level labeled rooted trees with $ n$ leaves.  For example, $ C_{2,n}$ is the number of [url=http://en.wikipedia.org/wiki/Bell_number]partitions[/url] of $ n$ elements $ B_n$.  Show that

$ C_{m, n} \ge \frac {1}{(m \minus{} 1)n \plus{} 1} {mn \choose n}$.

For fixed $ m$, for what values of $ n$ do we have equality?

[hide="Large hint"] The $ m \equal{} 2$ case has the following interpretation:  it says that the number of partitions of $ n$ elements is greater than or equal to the number of [url=http://en.wikipedia.org/wiki/Noncrossing_partition]noncrossing partitions[/url] of $ n$ elements. [/hide]

In retrospect, that wasn't as large a hint as I thought.  So let me give what I will just call

[hide="A possible approach"] Let $ D_{m, n} \equal{} \frac {1}{(m \minus{} 1)n \plus{} 1} {mn \choose n}$.  $ D_{2, n}$ is, of course, the Catalan numbers, and one way to see this is to count the number of sequences of $ n$ copies of $ 1$s and $ n$ copies of $ \minus{} 1$s such that the partial sums are always non-negative.  From here, there is an injection (not bijection) to $ 2$-level labeled rooted trees.  Now show that $ D_{m, n}$ is the number of sequences of $ (m \minus{} 1)n$ copies of $ 1$s and $ n$ copies of $ \minus{} (m \minus{} 1)$s such that the partial sums are always non-negative and inject to $ m$-level labeled rooted trees.

(Note also that in order for the partition comment to make sense there can be more than one root in a given tree and only the leaves themselves are labeled, not any interior vertices.) [/hide]


How do you ban certain people from accessing your blog?

I think there's an opton under blogs under profile that you are able to specify which people are allowed to view your blogs.

I guess you could put that person on your foe list and specify to only let friends be able to access your blog.

I think that all foes might be banned anyway

Actually to clarify this: foes option doesn't do anything for now. If you restrict your blog to friends only, only friends and contributors will be able to see it.

Oh, and use the @import thingy?

No. Its not someone on AoPS. Some cybergeek lurking around my school :P

[quote="chess64"]You can write a PHP script to dynamically deliver your css; if the ip address is the one you want to block, just put a white image with a really high z-index on top of the page so they won't be able to see anything :D[/quote]His blog is on AoPS you cannot use any php code. At this point you cannot ban IP addresses (nor I am sure I will enable that feature in the future).

Can't you like do @import(http://mywebsite.com/mathlinks/ipban.php)

andthe php script is connected to a htaccess file and does something like that?

You don't have access to php on our server. It's as simple as that.

Actually, I was thinking of putting a file, say css.php, on a php-enabled server, say mathideas.org, that is a php file with <?php header('Content-type: text/css'); ?> in it, and then in your blog css you can write @import('http://mathideas.org/css.php');

I think that should work.

I posted my code on the other place, andit works fine for me.


Okay, so today is the start of September, and this is the month when the Toronto Math Circles traditionally start, so does anyone know when/where they are this year?

Thank you.

:arrow: So nobody knows anything???

Sigh... this forum is dead... :mad:

i would also like to know... Rice told us he wouldn't be teaching last year  :wallbash:

Just as an update for anyone who reads this forum, I got an e-mail from the Fields Institute that the math circles will be every Saturday starting Sept. 27th from 1-3 pm at the Fields Institute...  :D

Unfortunately, for the last few weeks, I have not been able to go to the math circles. So, I just wanted to know if there was class this week and when they are stopping for the Winter Break. Thanks in advance.

Toronto Math Circles?
sounds pretty fun :P
What are they??

and, can a Hamiltonian join, too??

Hi Nawahd and azn_rocker, I'm Richard Cerezo, one of the program assistants for the Math Circles' sessions.  You might know me as the guy at the front desk each week.

Anyway, I'm trying to improve the co-ordination of the sessions and hopefully you've received my reminder.

If you haven't you can email me for more information at math.circles.fields@gmail.com

-Richard


The [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=117042]Eggs[/url] had their natural origin from the following problem.
Let's start with this one.

There is a house which has $N$ floors and there are $2$ eggs which are identical on their durability. It is known that the egg will be crashed if we shall dump it from the $d$-th floor or more higher one and will preserve their durability when damped from some lower floor. How many attempts is required to define this number $d$ in any case? It is known that egg would be crashed if we shall dump it from $N$-floor surely.

Well, use one egg at the $\lfloor N/2 \rfloor+1$ floor. If it breaks, start from floor $1$ and go up with the second egg until it breaks. If not, repeat algorithm for the set of all floors above that. The maximum number of attempts is clearly $\lfloor N/2 \rfloor+1$. Since we are only interested in this upper bound, it is sufficient to find a case in which the upper bound is attained (i.e. when $d = \lfloor N/2 \rfloor$).

[quote="digger"][quote="paladin8"]The maximum number of attempts is clearly $\lfloor N/2 \rfloor+1$. [/quote]
So, for $N=100$ we have to do $51$ attempts.
Who can make less attempts?[/quote]

Hmm, I think I might revise my statement to $N/2$ for even numbers, $(N+1)/2$ for odd numbers.

For $N=100$ try to revise 51 to at least ... , well, let it be about a half of 51.

Note that 26 attempts is rather far from the best result in this case but 13 attempts is unsufficient.

Paladin, I think you may have missed something.
Suppose that the N floors are divided into several groups of x floors.  Then you may test floors $N/x$, $2N/x$, ...$N$ successively.  If the egg cracks on floor $CN/x$ then with the second egg you must test floors $(C-1)N/x+1$ and successive floors until the egg does crack.  Otherwise, you must test floor $(C+1)N/x$ and repeat the algorithm.  This defines a number of tries $\frac{N}{x}+x-1$.  This equation is minimalized when $x=\sqrt{N}$ (basic Cal I).  The minimal number of tries will be $2\sqrt{N}-1$. For this particular problem you need 19 tries.

Ah, I see. That's quite interesting. I wonder if there is a proof for some sort of lower bound.

[quote="diophantus"]For this particular problem you need 19 tries.[/quote]
Nice.  :lol:
Who can revise 19 tries for the particular problem?

[hide="hint"]For $N=106$ it's sufficient to do $14$ attempts, and this can not be revised.[/hide]


calculus
integration
real analysis
calculus computations

The question asks:
"Consider the integral with a = 5 and b = 11
(2x^2+2x+6)dx

a)Find the Riemann sum for this integral using right endpoints and n=3.
b) Find the Riemann sum for this same integral, using left endpoints and n=3

----- HERE'S MY WORK-----

a)lim ∆x→0 (b-a/n)[ f(a) +f(a-∆x) +f(a -2∆x) +....f(a-(n-1)∆x)]
n→∞
∆x= b-a/n , n=3

lim ∆x→0 (11-5)/n[ f(11) +f(11-∆x) +f(11 -2∆x) ]
n→∞
= 2[f(11)+f(10)+f(9)]=[b]936 approx.[/b]

b)lim ∆x→0 (b-a/n)[ f(a) +f(a+∆x) +f(a +2∆x) +....f(a+(n-1)∆x)]
n→∞
∆x= b-a/n , n=3

lim ∆x→0 (11-5)/n[ f(5) +f(5+∆x) +f(5 +2∆x) ]
n→∞
= 2[f(5)+f(7)+f(9)]=[b]936 approx.[/b]

[size=150]
Anybody get a different answer?? This is apparently incorrect.
If so, please don't estimate the answer, place decimal points too. Along with how you reached that answer (if it's different than mines)
Thanks! [/size]


modular arithmetic
combinatorics unsolved
combinatorics

Assume a convex 2m-gon A1,A2,...,A2m .In its interior we choose a point P,which does
not lie on any diagonal.Show that P lies inside an even number of triangles with vertices
among A1,....,A2.

First, take $P$ arbitrarily close to the midpoint of side $A_1A_2$. Then  $P$ is contained in exactly the triangles two of whose vertices are $A_1$ and $A_2$, and there are $2m-2$ such triangles, i.e. an even number. Now, move $P$ around within the polygon. We may assume it crosses only one border at a time. When it crosses a border $A_iA_j (i < j)$, it does one of the following two things: (i) ceases to be contained in the triangles $A_iA_kA_j$, where $i < k < j$ and commences being contained in the other triangles two of whose vertices are $A_i$ and $A_j$, or (ii) the opposite of this. Obviously, the parity of the difference of the number of triangles in which it is contained is the same in the two cases, as $-n \equiv n \pmod{2}$. So consider case (i). It ceases to be contained in $j - i - 1$ triangles and commences being contained in $2m - 1 - j + i$ triangles, so the difference is $2m - 2j + 2i$, which is clearly even, so the parity of the number of triangles in which $P$ is contained stays constant. As shown at the beginning, this is even, so we're done.

[quote="MysticTerminator"]First, take $P$ arbitrarily close to the midpoint of side $A_1A_2$. Then  $P$ is contained in exactly the triangles two of whose vertices are $A_1$ and $A_2$, and there are $2m-2$ such triangles.[/quote]
It seems wrong to me,because whatever it is close to the A1,A2 it is at one side of not at
both,so if A1 and A2 are'nt neighbor it is certainly less than 2m-2,and if they are neighbor you cannat certainly say that  P is in 2m-2 triangles. :!:


We have $ 150$ numbers $ x_1,x_2, \cdots , x_{150}$ each of which is either $ \sqrt 2 \plus{}1$ or $ \sqrt 2 \minus{}1$

We calculate the following sum:

$ S\equal{}x_1x_2 \plus{}x_3x_4\plus{} x_5x_6\plus{} \cdots   \plus{} x_{149}x_{150}$

Can we choose the $ 150$ numbers such that $ S\equal{}121$?
And what about $ S\equal{}111$?

[quote="lambruscokid"]We have $ 150$ numbers $ x_1,x_2, \cdots , x_{150}$ each of which is either $ \sqrt 2 \plus{} 1$ or $ \sqrt 2 \minus{} 1$

We calculate the following sum:

$ S \equal{} x_1x_2 \plus{} x_3x_4 \plus{} x_5x_6 \plus{} \cdots \plus{} x_{149}x_{150}$

Can we choose the $ 150$ numbers such that $ S \equal{} 121$?
And what about $ S \equal{} 111$?[/quote]

Notice that in sum there are $ 75$ terms each one can have its value independent of other. Moreover each can be
$ (\sqrt 2 \plus{} 1)(\sqrt 2 \minus{} 1) \equal{} 1,(\sqrt 2 \plus{} 1)^{2} \equal{} (3 \plus{} 2\sqrt 2),$ or $ (\sqrt 2 \minus{} 1)^{2} \equal{} (3 \minus{} 2\sqrt 2)$
Consider $ x$ of them equal to $ 1, y$ of them equal to $ (3 \plus{} 2\sqrt 2)$ and $ z$ of them equal to $ (3 \minus{} 2\sqrt 2)$
Now $ x \plus{} y \plus{} z \equal{} 75$
And $ x \plus{} (3 \plus{} 2\sqrt 2)y \plus{} (3 \minus{} 2\sqrt 2)z \equal{} S$
as $ S$ is radical free $ \implies y \equal{} z \implies x$ is odd. Let $ x \equal{} 2n \plus{} 1$
$ \implies 2n \plus{} 1 \plus{} 2y \equal{} 75\implies n \plus{} y \equal{} 37$ -----(1)
Also $ 2n \plus{} 1 \plus{} 6y \equal{} S$
Consider $ S \equal{} 121\implies n \plus{} 3y \equal{} 60$------(2)
Now see (1) and (2) has no solution in integers.

For $ S \equal{} 111\implies n \plus{} 3y \equal{} 55$------(3)
Now from (1) and (3) we get
$ y \equal{} 9, n \equal{} 28$
Hence for $ S \equal{} 111$ we have such set.

2017 Junior MO, Poland, second round P4 was the same question, but equation was changed to
$$x_1x_2+x_2x_3+...+x_{98}x_{99}+x_{99}x_1=199$$


function
calculus
integration
real analysis
real analysis theorems

What is lim_n int_[0,n] ( (1-x/n)^n *exp(x/2) ) dx ?
I suspect I have to apply LDC to get 1/2 but in order to do that I have to find
a function g s.t.
|f_n|<g ,g L1 function

where f_n(x) = (1-x/n)^n *exp(x/2)*X[0,n]

Please help!!

The obvious choice for $g(x)$ is $\exp(-x)$, isn't it?

Yes the problem is I let n->oo without taking into consideration that fpr every n the quantity in the integral is always positive..
Thnx a lot though


Here are two easy questions-

1)If there is an ice cube with an air bubble inside it and it melts in water,what happens to the level of water?

2)If inside a jar of water,a cork is tied to the base(such that it can still float up) of the jar with a flexible string such that the cork is held under water and this jar is accelerated forward,what happens to the cork?

Which way is forward? :) Would that be on your picture to the side?

[quote="riddler"]Surely, the 1st one should have a bit more info
Because I am stuck on comparing,
$V_i$ to $~\frac{10}{11}(V_i+V_g)$
The answer should be it is dependant on the avg density of the (ice cube+ gas)[/quote]

Once the ice melts,the air bubble doesn't just pop off,does it??I think there should be a little change,i'm not too sure though...what is the densitry of the ice+air bubble...is it density of ice + density of air bubble?

I dont understand what you mean
It is the total mass/total vol

The answer of the 1st one should be depandant on the density of the gas.

LOL

First one if piss easy.

When you place an ice cube into a glass of water. Naturally the water level rises due to the extra space. Now since there is an air buble inside the cube, once the cube melts, the water level will go down depending on how much air was in the cube.

Its like submerging a ballon into a swimming pool, the water rises but if you popped the ballon, the air will escaped upwards and the water will go down.

For the second one... thats a bit tricky for a 15yr old :S

2nd one i believe the corks moves with the water.

[quote="shadysaysurspammed"][quote="riddler"]In the 2nd question what do you mean?
which way?
up or down?[/quote]

I suppose that's for you to find out....[/quote]
 :huh: you are giving the question....

Of course, it depends on what kind of bubble it is. If it's just an ordinary air bubble with the same pressure as the outside air, it makes no difference because the air bubble weights nothing. If there was vacuum or some less dense gas instead, it would have pulled the ice cube upwards and therefore the water level would rise after the ice melts. A denser gas would cause the water level to drop.

Wouldnt it depend on the weight of the ice and the bubble?

In the first question, assuming that the ice cube/bubble is floating, then it displaces a mass of water equal to its mass, with the rest of the ice cube/bubble above in the air.  Ignoring the temperature effects of the water and container and any effect of the dissolved air from the bubble in the water, when the ice cube is melted, it occupies a volume equal to the mass of the water alone, the air has escaped.  So the level of the water decreases by an amount equal to the mass of the bubble.  (Think of force of gravity on the contained air as being the same as an object or a finger pushing gently down.)

The cork will move away from the direction of acceleration, relative to the container.  There are two primary effects, first the initera of the cork, later as the interia of the water moves the water back, the cork will move with the water.  This becomes more complex as the water rebounds and moves back toward the front of the container.  It also becomes more complex as the location of the string holding the cork is moved off the centerline of the acceleration.

[quote="Gyan"]A physicst way for the first -[hide] Assume, the  "Ice-cube"   wall is 1cm thick, but the whole volume is  $1000^3$   cubic kilometers .. or some really large volume ... these "limiting cases" makes the problems intutive. [/hide][hide]For the second: Did you ever held a helium ballon inside a car, which, say  applies sudden brake? (If not do the experiment - It might be fun - I have shown it to others) [/hide][/quote]

That's the same solution i thought of for the second one,Gyan,I hope it is right...tyeah i know the helium balloon question,that's what popped into my head when i saw this question.

wiszard2378,According to me your solution is right and i think this is what others said as well....i guess the answer in the book itself is wrong....I asked this before also -what is an air bubble,i mean does it stay on the water after the ice melts or does it escape and why?

[quote]tyeah i know the helium balloon question,that's what popped into my head when i saw this question.
[/quote]
Yes, the helium ballon (another way to think , is newton's law with negative mass) goes the "other way" of normal things - Eg if you slam on the break, you are likely to hit the dash board, a helium baloon would go hit the back seat.

(I have seen this problem given in exams  (and  I have also given it to my students) and  really get surprised that many (  in colege physics course) get the  wrong answer...)

Also related: A man in a boat in a very small lake throws some scrap metal (originally inside the boat) in the water (where the scrap  sinks to the bottom).. Will the water level in the lake rise/fall/remain the same?

I still insist on my answer to the first question - the water level [b]won't change[/b]!!! Imagine a huge ice cube with very thin walls filled with vacuum. When you put this ice cube into water, it will most likely fly away like a balloon - but let's assume that it is heavy enough to stay on the water surface, slightly dipped in the water. This way it displaces almost no water at all so when it melts, the water level will certainly increase. Can you explain [i]this[/i]?
Or a different thing to think about - what happens if the original air bubble is very close to the wall of the ice cube? And what happens if you "pop" the bubble by making a little hole in the ice cube wall and thus connecting the bubble with the outside? Will this suddenly cause the answer to change from "water level drops" to "water level stays the same"?

Thanks Shady.

Gyan -- the only "negative" mass of the helium balloon is for vertical movement.  For horizontal movement it has "positive" mass even if it has bouency.  Further, I've done the experiement and when breaking, the helium balloon moves forward.

Gyan -- you are right that the bubble is the same as the weight in the boat.

Bus -- Your vacuum issue eliminates the effect of the bubble, because there is no mass or weight from the inclusion, unless the increase in size of the ice cube is significant enough to provide bounency from the displaced air, which normally can be safely ignored in a water/ice question.  If large enough, then the ice cube would be pulled up and displace less water than the mass of the ice.  In this case, the bubble acts as "negative" mass and when removed, will have the level of the water rise. Ignoring the bouency effects from the air, the mass (and weight) of the ice cube with a vacuum included is the same as it would be without the vacuum.  (I'll leave it to others to make such an ice cube.) The melting ice cube will not change the level of the water.

Bus -- Once you have removed the bubble, it is the same as the ice cube without a bubble (inclusion of water is irrelevent) -- no change as the ice cube melts.

Gyan -- Your boat/sinking weight problem is that the level of the lake decreases, since the weight is going from displacing an equal weight of water to less than an equal weight of water (else it would just float).  An extension would be a float would not change the level of the lake because floating or riding in the boat, the float displaces its mass in water.

[quote] For horizontal movement it has "positive" mass even if it has bouency. Further, I've done the experiement and when breaking, the helium balloon moves forward.
[/quote]

Not really, and  in fact there is really  no doubt.  Yes the mass in strict sense is positive .. (And if there was vacuum in the car and only a helium ballon - BTW then it will not float either - the ballon moves forward etc) but if  the balloon is enclosed in air  it goes "up" when normal things go down, It goes "backward" when "normal " (heavier then air)  things go "forward" ... In fact the idea behind "general theory's equivalances" or "artificial gravity"  is it makes no difference  if the "weight"  you feel is due to gravity or  acceralted frame... Things (yes, even the helium ballons in air) behave exactly the same -does not matter if the "weight" is due to gravity or accerated frame.

Experiment is easy to do, and I'd suggest you do it.. You can work with a helium baloon in a car  or (even easier to observe)  use a jar, fill it with water and put a small aount of wood  or air bubble. (or use a spirit level - the kind you use to find if a place is level or not- which has an air bubble inside) .. Just use your hand to jerk it around ..... If you are using a helium balloon in a car, easiest to   drive in a cicular curve  (say in a parking lot where you can go in a circlle for a longer time -  and let some one other than a driver observe. :)

Hope that helps.

PS -  Look at the bat!! (Catoonist had the same idea here!)

[img]http://www.artofproblemsolving.com/Forum/album_thumbnail.php?pic_id=311[/img]

PPS - Happy Halloween!

wizard2378: That's exactly what I'm saying! The air bubble inside the ice cube has [b]no weight[/b] if you consider the bouyancy effects (which you should) and therefore doesn't affect the result. The water level won't change.

The bubble does have a role. Because even though the bubble has no weight, it affects the density of the ice.

But the density of the ice is not important. Water displaced by the ice cube only depends on the weight of the ice cube. (The density could affect the result if it changed the fact that ice cube floats on the water.) And the bubble also doesn't change the amount of water we get after the ice cube melts so it doesn't affect the result in completely no way. Right?

[quote="shadysaysurspammed"]Why does the air bubble have no role in the water displaced??
An air bubble means........a bubble of air or air coveerd by water?[/quote]I'm not sure what you mean by air covered by water... it's a normal air bubble, air covered by ice :). The air bubble is inside of the ice cube. That's the way you stated the problem, isn't it?

Back to the "no weight" argument - air weights nothing in a sort view. When you put $\text{10 m}^3$ of (uncompressed) air on a scale, it shows zero. This is not because it's mass is so small - in fact it's more than 10 kg - but the scale is "used to" having air all around it. The scale knows that everywhere around there's some atmospheric pressure $p_0$ and ignores this pressure force. When you put your $\text{10 m}^3$ of air on the scale, you create some additional (gravitational) force that the scale should measure but you have also displaced some other $\text{10 m}^3$ of air and therefore decreased the pressure - so the scale shows zero. (This is of course obvioous - putting air on scale does nothing.)

So - if you have some water and create a solid ice cube, this ice cube will weight the same as if you have created an ice cube with an air bubble inside because the air bubble [i]does[/i] add mass to your ice cube, but it also decreases the atmospheric pressure by exactly the same amount.

Bus -  Well put. --- (Assuming   all the mass is converted into water -  it occupies the same volume as the water it was displacing )

Now - I am asking an additional question, Why do the say that the water leve will rise in NY shores, if  all  (or most of )  the  ice on the north polar caps melt ?

To be clear, we are assuming NO water is generated and goes into  the ocean  from   Ice  melted in  Antartica, (Or any other ice which is not floating in the sea)  ..

Can they be right?

So if all the icebergs (and these are  floating) melt -  will water level rise?  (If yes why? )

BTW - Answer is[hide] Yes  [/hide]

Hm, are you sure? ... I really don't know. One thing is that we didn't consider temperature changes of water - if the water changes it's temperature, it may expand and the water level may rise...

Also, there's one thing I silently left out and that is that when an ice cube melts, the water level actually rises a little because of the air pressure difference - air pressure is lower on top of an iceber than at the water level so it seems to weight less. If all of the icebergs melted, we'd have some extra $\frac{\varrho_\text{air}}{\varrho_\text{ice}}V$ volume of water in our oceans where $V$ is the volume of the visible (over the sea level) parts of the icebergs. If you divide this by the surface of Earth's oceans, you get the change of water level, but that's a really small number. So I don't know. Do you?

Bus -  As you said The effect due to air pressure difference (between a  height of at the most few hundred meters)  would be negligible ..
       - The water temperature could (and it does - seasonal variations  changes mean sea level by about 5 to 10 cm during a year)  make a difference but let us assume the ice will not make that much change in the temperature. (Neglect the effect of change in density of water due to -slight - change in water temperature.

So let us see what else?

FWIW - some data - (and possible hint)

 If all the "floating ice" melts, the water would (rough order of magnitude calculation -) - oh .. about 5 cm...

[hide]This is, of course, much smaller than  waht would happen if "non floating ice" melts. Greenland , and Antartica  may add  70 meters...
And  it may not be much considering some say  the water level  is rising about 2mm per year on average ..[/hide]

Once you think about it ... I think you would know why.

well.. I always thought that water levels rise because of

a) to a small extent, the melting of icebergs

b) to a much larger extent, the enhanced greenhouse effect, raising temperatures causing water to expand

zues_three: The problem stated here, however, ask how can the water level rise because of melting of the icebergs? You may remember the problem with ice cube floating in a glass of water. After it melts, the water level won't change because of the Archimedes' law. How are melting icebergs different? I'm still not sure...

[quote="bus"]OK, another thing I came up with is that it could be because of the salt. This is probably what you're talking about, right? Icebergs are frozen fresh water while oceans are salty water and the difference in density causes the change of the sea level. However - what happens when the icebergs melt? Won't the melted fresh water mix with the salty water and become salty water itself, thus decreasing its density and not making (almost) any difference after all? Or is this "almost" the 5 cm difference you're talking about?[/quote]

[b]Precisely![/b]

It so happens, that most of the ice in icebergs is pure water, and when melted, will occupy more volume (about 3% more) than sea water.
So if the iceberg  melts in , say l a lake, Erie  - water level remains the same. (or almost same)
If it melts in sea water (Density about 1.03 times fresh water) - water level rises.

Physics points to be learned -

1.  If you take salt water (simulated sea water) and  start  “freezing”  it, the ice which forms , in the beginning, is almost  pure water-ice . As. more water becomes ice, the concentration of salt  ( in the remaining water)  increases.. and after a point  (after very heavy concentration)  the new ice will  have increasing amount of  salt in it. This condition has not reached in our oceans yet, and all the icebergs are  virtually (90+%) pure water.  (Also don’t forget, most of the icebergs are chunks of ice from  glaciers from places like Greenland)

2.  Salt concentration in ocean  water (and thus density of water in ocean) is lower  in say, Arctic region  than Atlantic Ocean, mainly due to melting of icebergs (Of course, many icebergs come from land/Greenland, which is pure water).

[quote="zeus_three"]hmm... just a suggestion, but i think that someone should physically do this experiment and settle this debate.[/quote]

Yes, that would be a good idea, .. It is easy enough to do, and in fact, I have used it as a  fun experiment for a Physics Class, mainly because the problem is so classic.   Also fun problem is to ask the class to “estimate” the rise in water level in Ocean.

(Do the "experiment" using ice  floating in  a) pure water, b) alcohol  c) salt-water, d) Mercury -(where change would be obvious) -etc.)

The point here is, if the "ice" (Does not matter, if  has bubbles in it -  or  has some "super packed - high density water" in it) when transformed  occupies the same volume as the displaced water (which is equal  to the volume of the  fluid in the ocean having the same mass as the iceberg)  water-level would not change. ---A case obviously true if the "displaced water" (ocean water) and  "water which came from ice" were of  the same density ....

Hope this is useful.

Added later Please also see [url=http://www.physorg.com/news5619.html]This url[/url]
(Soce:  Unversity of Colarado at Boulder)


For some reason, the page that allows the downloading of $ LaTeX$ onto a Macintosh OS X seems to be broken. It says that there is a 404 error. Why?

Ah, the Mac information page here died in switching to the wiki. You want [url=http://www.tug.org/mactex/]this page[/url] offsite.

so, do I just down load MacTeX.mpkg.zip?

Was I supposed to do that?

Which files on the page am I supposed to download, jmerry?


topology
inequalities
calculus
calculus computations

The sequence $ (x_n)$ is called [i]periodic[/i] if there is some $ p\in \mathbb{N}$ such that $ x_{n\plus{}p}\equal{}x_n$ for every $ n \in \mathbb{N}$. Show that every periodic convergent sequence is constant.

I already done this proof but i don't know if it's right. Maybe there is other ways to do it. Thanks for any help

[hide="Proof"] Suppose $ (x_n)$ is an periodic convergent sequence with $ x_n \to a$. By definition of periodic sequence we have $ (x_1,x_2,\ldots,x_p,x_1,x_2,\ldots,x_p,\ldots)$ so there are $ p$ constant subsequences. We know that $ x_n \to a$ then all subsequences converge to $ a$. Therefore we have $ x_n\equal{}a$ for all $ n \in \mathbb{N}$ [/hide]

Yes, this works (and seems to be the shortest way to do it).

Very thanks fedja! i was trying by definition of limit but when a wrote down the sequence the idea popped up

One way to think about this problem is that a periodic sequence only takes on finitely many values, so the space it inhabits can be thought of as having the discrete topology.  (You can verify that that's what the epsilon-delta definition of a limit tells you by taking $ \epsilon$ less than half the pairwise difference between any two of the values $ (x_n)$ takes or by using the Cauchy criterion.)  Then $ (x_n)$ must be eventually constant, hence constant.

If you still want to use the definition of limit, prove that the sequence of averages $ a_n\equal{}\frac{x_1\plus{}\dots\plus{}x_n}{n}$ is convergent for any periodic sequence $ (x_n)$.

You said nothing about where the $ x_n$ live, so I'll assume a fairly general topological space.

One caveat: we need some separation properties in the underlying space, specifically that for any two distinct points $ a,b$, there is a neighborhood of $ a$ not containing $ b$. If this isn't true, the sequence $ b,b,b,\cdots$ converges to $ a$.

Very nice observation t0rajir0u. I'm working on that mlok, i conjecture that $ a_n \to \frac{x_1\plus{}x_2\plus{}\cdots\plus{}x_p}{p}$. I'm working on the epsilons now! Yes jmerry, i forgot to mention that was a real sequence. You're talking about Hausdorff property right? It's very nice to see how general this can be. Very thanks guys

That's actually weaker than Hausdorff; the Hausdorff property is that any two points have disjoint neighborhoods. There are a few interesting spaces with this weaker property but not Hausdorff, such as the cofinite topology on an infinite set.

I arrive at this trying prove mlok's statement:

i've tried to estimate $ \left| \frac{x_1+x_2+\cdots+x_n}{n}-\frac{x_1+x_2+\cdots+x_p}{p}\right|$ to show that $ a_n\to a_p$ but did not got the difference arbitraly small. I know that $ |x_n|\leq c$ for some positive $ c$ and $ \frac{1}{n}$ can be made small as we want.

[hide="do not know if this is the right path..."]

\begin{eqnarray*}
\left|\frac{x_1+\cdots+x_n}{n}-\frac{x_1+\cdots+x_p}{p} \right|&=&\left|\frac{p(x_1+\cdots + x_p+x_{p+1}+\cdots +x_n)-n(x_1+\cdots+x_p)}{np} \right| \\
 &=& \left|\frac{(p-n)(x_1+\cdots+x_p)+p(x_{p+1}+\cdots+x_n)}{np}\right| \\
&=& \left|\left( \frac{1}{n}-\frac{1}{p}\right)(x_1+\cdots+x_p)+\frac{1}{n}(x_{p+1}+\cdots+x_n)\right| \\
&\leq& \left| \left( \frac{1}{n}-\frac{1}{p}\right)(x_1+\cdots + x_p)\right|+\left|\frac{x_{p+1}+\cdots +x_n}{n}\right|\\
&\leq& \left| \left( \frac{1}{n}-\frac{1}{p}\right)(pc)\right|+\left|\frac{(n-p)c}{n}\right|\\
&\leq& \left| \left( \frac{1}{n}-\frac{1}{p}\right)(pc)\right|+\left|\left(1-\frac{p}{n}\right)c\right|
\end{eqnarray*}

[/hide]

You can, probably, see perfectly well that the last expression in your inequality chain does not tend to $ 0$ yourself, which means that you went the wrong way somewhere in the process ;).

[quote="mlok"]If you still want to use the definition of limit, prove that the sequence of averages $ a_n \equal{} \frac {x_1 \plus{} \dots \plus{} x_n}{n}$ is convergent for any periodic sequence $ (x_n)$.[/quote]

there is no additional hipotesis on $ (x_n)$ right?

Hint: when $ n$ is divisible by $ p$, the average is [b]equal[/b] to the supposed limit. For a general $ n$, write it as $ n\equal{}mp\plus{}r$ where $ 0\le r<p$.

[hide=" another try"]

If $ p$ divides $ n$ then exists some natural $ a$ such that $ n=ap$

so $ |a_n-a_p|=\left| \frac{x_1+x_2+\cdots+x_{ap}}{ap}-\frac{x_1+x_2+\cdots+x_p}{p}\right|$

Since $ (x_n)$ is periodic with period $ p$ this becomes

$ \left| \frac{a(x_1+x_2+\cdots+x_p)}{ap}-\frac{x_1+x_2+\cdots+x_p}{p}\right|=0< \epsilon$.
If $ p$ doesn't divides $ n$ then exists unique $ m$ and $ r$ such that $ n=mp+r$ and $ 0\leq r < p$. So
\begin{eqnarray*}|a_n-a_p|&=&\left| \frac{x_1+x_2+\cdots+x_{mp+r}}{mp+r}-\frac{x_1+x_2+\cdots+x_p}{p}\right|\\
&=&\left| \frac{x_1+x_2+\cdots+x_{mp}}{mp+r}+\frac{x_{mp+1}+\cdots+x_{mp+r}}{mp+r}-\frac{x_1+x_2+\cdots+x_p}{p}\right|\\
&=&\left| \frac{m(x_1+x_2+\cdots+x_{p})}{mp+r}+\frac{x_{mp+1}+\cdots+x_{mp+r}}{mp+r}-\frac{x_1+x_2+\cdots+x_p}{p}\right|\\
&\leq&\left| \frac{m(x_1+x_2+\cdots+x_{p})}{mp}+\frac{x_{mp+1}+\cdots+x_{mp+r}}{mp+r}-\frac{x_1+x_2+\cdots+x_p}{p}\right|\\
&\leq&\left|\frac{x_{mp+1}+\cdots+x_{mp+r}}{mp+r}\right|\\
&\leq&\frac{rc}{mp+r}\\
&\leq&\frac{pc}{mp+r}\\
&\leq&\frac{pc}{mp}=\frac{c}{m}\\
\end{eqnarray*}

so $ \frac{c}{m}$ can be small as we want because $ \frac{1}{n}$ have that property and we have the relation $ n=mp+r$ for fixed $ p$ and $ r$ ?

That's a nice question! Please correct me :maybe:  [/hide]

This is better, but the first inequality in the chain isn't quite right. For one thing, $ x_1 \plus{} \dots \plus{} x_p$ could be negative. And even if it's not, in general the inequality $ A\le B$ does not imply $ |A \plus{} C|\le |B \plus{} C|$.

Some suggestions: (1) Introduce more compact notation, such as $ S \equal{} x_1 \plus{} \dots \plus{} x_p$; (2) don't try to combine everything into one long chain of inequalities.

Here's how I would write this: since $ n \equal{} mp \plus{} r$, we have
\[ x_1 \plus{} \dots \plus{} x_n \equal{} mS \plus{} (x_{mp \plus{} 1} \plus{} \dots \plus{} x_n),
\]
hence $ |x_1 \plus{} \dots \plus{} x_n \minus{} mS|\le cr < cp$. (Here $ c$ stands for the maximum of $ |x_i|$, as in your proof.) Divide the last inequality by $ n$:
\[ \left| \frac {x_1 \plus{} \dots \plus{} x_n}{n} \minus{} \frac {mS}{n}\right| < \frac {cp}{n}.
\]
Now observe that $ \frac {mS}{n} \equal{} \frac {mS}{mp \plus{} r} \equal{} \frac {S}{p \plus{} r/m}\to \frac {S}{p}$ as $ m\to\infty$. Also $ \frac {cp}{n}\to 0$. Finally, use the following property of limits (best proved separately): if $ a_n \minus{} b_n\to 0$ and $ b_n\to L$, then $ a_n\to L$.

Very thanks [b]mlok[/b], i've learned much things tryng to prove this beautiful result. Falled in the inequality trap but it is good because i'm not in a exam! :D  I will think about the $ n_0$ such that for $ n> n_0$
$ \left| \frac {x_1 \plus{} \dots \plus{} x_n}{n} \minus{} \frac {mS}{n}\right| < \frac {cp}{n}<\epsilon$.


I did USAMTS last year, do I have to fill out the same forms this year?

Indeed, you must fill out an entry form for 2005-06, but you can submit it with your first set of problems.


Can someone help me with displaying matrices?

I want to display a 3x3 matrix to the -1 power.  I found some stuff about array formatting using \left[ and \right] but couldn't get it to work properly.

Thanks.

M^{-1}=\left[\begin{array}{ccc} a&b&c \\ d&e&f \\ g&h&i \end{array}\right]^{-1}

- ought to work. It doesn't work for me because of some silly business about the ampersand. That's one of the bugs of the site.

$M^{-1}=\left[\begin{array}{ccc} a&b&c \\ d&e&f \\ g&h&i \end{array}\right]^{-1}$

[Admin edit] Using a space

$M^{-1} =\left[\begin{array}{ccc} a&b&c \\ d&e&f \\ g&h&i \end{array}\right]^{-1}$

Trying a dreadful kludge:

$M^{-1}=\left[\begin{array}{c} a\,\, b\,\, c \\ d\,\,e\,\,f \\ g\,\,h\,\,i \end{array}\right]^{-1}$

This is the post that broke the camel's back :mad: I am taking charge right now :P :D

It's done, but some recent matrices have to be edited, and re-compiled (add a space somewhere in the formula).

Or you can use the amsmath package and commands such as:
[list]bmatrix-->$[$  $]$
Bmatrix-->$\{$  $\}$
vmatrix-->$|$  $|$
Vmatrix-->$\|$  $\|$
pmatrix-->$($  $)$
matrix--> no boundaries
[/list]
To use:
\begin{[i]type[/i]}
$entry_{1,1}\;\&\;entry_{1,2}\;\&\cdots\&\;entry_{1,n}\backslash\backslash$
$entry_{2,1}\;\&\;entry_{2,2}\;\&\cdots\&\;entry_{2,n}\backslash\backslash$
$\vdots$
$entry_{m,1}\;\&\;entry_{m,2}\;\&\cdots\&\;entry_{m,n}$
\end{[i]type[/i]}

$M^{-1}=\left[\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}\right]^{-1}$

Just checking, Valentin. Thanks - that looks better.

I need help too, if someone IS willing to help. I always get the 'ampersand' problem when I make arrays. So like Valentin suggested, add spaces before and after the 'ampersand'?

Trying "bmatrix". I guess you don't need to specify the number of columns with that.

$\begin{bmatrix} 1 & 2 & 4 & \frac{x^2+1}x \\ a & b & c & d \\ 0 & 0 & -22 & -1 \end{bmatrix}$

[quote="Valentin Vornicu"][quote="3X.lich"]Unsuccessful  :([/quote]I modified your post. You did not add any space, thus the image was not processed, but cached (you probably copied and pasted it from somewhere else).

For new images there is no need to add spaces (unless the [b]exact[/b] same $\LaTeX$ formula has been used before).[/quote]
Yeah I copied and pasted it from my notepad. But I wasn't sure where to add the space  :blush: . For the second comment, heh I have no idea what you mean by new images to be included. Anyways thank you, I can certainly make matrices now!

[quote="Valentin Vornicu"]By new image I mean a new matrix (each formula in represented here as an image). So from now on all new images created from $\LaTeX$ formulae will appear as they should (matrix-wize). Old images will have to be edited to make the & go away (just add a space anywhere inside the old formula - it will not modify the outcome, but it will modify the formula itself, it will be processed again, with the new mechanism).[/quote]
Ah I see. But is this 'add space' problem going to be fixed? Just wondering... ;)

[quote="3X.lich"]Ah I see. But is this 'add space' problem going to be fixed? Just wondering... ;)[/quote]It's not possible to fix it without delete-ing all the already generated images (which are a lot, and would slow down a bit the forum).


abstract algebra
superior algebra
superior algebra unsolved

Let $ M$ and $ N$ be modules over a ring $ R$ and let $ \varphi : M \to N$ be a surjective homomorphism of modules. Suppose that there exists a homomorphism $ \psi : N \to M$ which is a right inverse for $ \varphi$. Prove that in this case $ M \cong N \oplus (M/\psi(N))$.

Edit: Error in proof will revise tomorrow

[quote="bulls6x"]Note that $ \psi(N)$ is a submodule of $ M$. This gives an isomorphism $ M\to\psi(N)\oplus{(M/\psi(N))}$[/quote]Why is that? I presume that you are not saying that if $ N'$ is a sub-module of $ M$, then automatically $ M\simeq N'\oplus M/N'$, are you? ;)

Since the latter is not true in general, it might be better to write down explicitly a homomorphism $ M\to N\oplus M/\psi(N)$ and to show that this is an isomorphism.

ah yes you are most definitely correct. as the title suggests we need to find a splitting. i think i found it and will post it tomorrow.

Consider short exact sequence:
$ \mathbb{ker} \phi \to M \to N \to 0$

Since $ \phi\psi=id_N$, $ M$ splits into $ \mathbb{ker}\phi$ and $ \mathbb{im}\psi$.
The rest it is easy to see.

Yes feniks, that is exactly what we are trying to prove.  :lol:

Let $ x\in M$.
$ x \equal{} x \minus{} \psi\phi(x) \plus{} \psi\phi(x)$
$ x \minus{} \psi\phi(x) \in ker\phi$
$ \psi\phi(x) \in im\psi$
So $ ker\phi \plus{} im\psi \equal{} M$

If $ x\in ker\phi \cap im\psi$
$ x \equal{} \psi(y)$
$ \phi\psi(y) \equal{} 0 \equal{} y$
So $ x \equal{} 0$.


algebra
polynomial
linear algebra
matrix
algebra unsolved

Let $P$ be a polynomial such that $P(x)$ is rational if and only if $x$ is rational. Prove that $P(x)=ax+b$ for some rational $a$ and $b$.

Are any of the solutions there actually correct?

I'll prove the assertion in the thread you get by clicking on ZetaX's links twice, starting from this topic :), namely:

Let $P$ be a real polynomial such that $P(\mathbb R\setminus\mathbb Q)\subset\mathbb R\setminus\mathbb Q$. Then $P$ has degree $\le 1$.

If the degree $n$ of $P=a_nx^n+\ldots+a_1x+a_0$ is positive, and we assume WLOG that $a_n>0,\ P$ takes all real values in some neighborhood of $\infty$, so, in particular, we can find $n+1$ rationals $t_0,\ldots,t_n$ in the range of $P$. If we regard the $n+1$ equations $P(x_i)=t_i$ as linear equations with the unknowns $a_j$, since $x_i$ must be rational, this system has a non-singular matrix, and the solutions will have to be rational, i.e. $P\in\mathbb Q[x]$. By multiplying with a constant, we can assume $P\in\mathbb Z[x]$.

What we want to prove now follows immediately from the fact that given a polynomial $P\in\mathbb Z[x]$, there is an integer $a$ such that $P+a$ is irreducible: just choose $a$ such that on the one hand, $P(0)+a$ is a prime, and on the other hand, $|a|>\max_{x\in D}|P(x)|$, where $D$ is the closed unit disk in the plane (this ensures that $P+a$ has no roots in the closed unit disk).

P.S.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78454]This[/url] is the topic I was talking about.

When you say irreducible, which field are you talking about being irreducible over?


geometry
geometric transformation
dilation
puzzles

Assume the following are true:
1. When a clean thing touch a dirty thing, it becomes dirty
2. Dirt cannot disappear. When a thing becomes clean, there must be something becoming more dirty.

So, is the water used after washing our hands clean?

(sorry for my poor English)

Well according to your assumptions of reality, the only solution I can think of is that the water isn't dirty because of the immense dilution of the dirt or whatever :P

well the water contains all of the diirt on our hands so i guess it is dirtier


topology
geometry
3D geometry
sphere
advanced fields
advanced fields solved

Is there an embedding of $\mathbb{P}^{2}(\mathbb{R}) = S^{2}/\{\pm1\}$ into $\mathbb{R}^{3}$?

No. There is no embedding of $\mathbb RP^{2}$ into $\mathbb R^{3}$.
Hint: $\mathbb RP^{2}$ is a non-orientable manifold. Every two dimensional manifold embedded into $\mathbb R^{3}$ has two sides "internal" and "external"; and therefore it is orientable.

Thank you. How does one proof that a two dimensional manifold embedded into $\mathbb{R}^{3}$ has two sides?

M... what Yuri says is true thanks to compactness of $\mathbb RP^{2}$, I think... Möbius strip is a non-orientable 2-manifold embedded in $R^{3}$...

AleX_ZeTa, yes, I meant to write compact manifold. :blush:

Given a compact 2 dimensional manifold $M$ in $\mathbb R^{3}$, consider its complement. It consists of several connected components. Assume that both sides of the manifold touch the same component $C$ at point $x\in M$. Consider a small enough neighborhood $N$ of $x$. $M$ divides $N$ into two components $A$ and $B$. Pick two points $a\in A$, $b\in B$. Then the line segment $[a,b]$ intersects $M$ in exactly one point. Connect $a$ and $b$ by a path in $C$.  This path together with $[a,b]$ forms a cycle $\pi: \partial D\to \mathbb R^{3}$ that intersects $M$ in exactly one point (where $D$ is a disk).

Extend the map $\pi$ to an immersion of the disk $D$, $\hat\pi: D\to \mathbb R^{3}$. Assume that  the immersion $\hat\pi$ and $M$ are in general position (otherwise, perturb them). Then $M$ intersects $\hat\pi(D)$ in a collection of curves. These curves can intersect only transversally. Therefore, they intersect $\hat\pi(\partial D)$ in an even number of points. But we know that there is exactly one intersection point.

We conclude that two sides of the manifold touch different connected components of $\mathbb R^{3}\setminus M$.

An $n$-dimensional compact manifold $M$ is orientable iff $H_{n}(M,\partial M,\mathbb Z)$ is non-trivial.
Since $\mathbb R\mathbb P^{n}$ is compact without boundry, it is orientable iff $H_{n}(\mathbb R\mathbb P^{n})$ is not trivial. This happens iff $n$ is odd.

[quote="berti"]Thank you. How does one proof that a two dimensional manifold embedded into $ \mathbb{R}^{3}$ has two sides?[/quote]

It can also be proved by using Alexander sphere duality theorem


calculus
integration
limit
trigonometry
derivative
real analysis
real analysis unsolved

Let $f(a,\theta) = \frac{\tan (a\sin \theta)}{a}$. We prove that $f$ converges to $\sin \theta$ uniformly about $\theta$ when $a->0$.

In fact, we can verify that $f(a,\theta)-\sin \theta$ is increasing about $\theta\in [0,\pi/2]$, and decreasing about $theta\in [\pi/2,\pi]$. The maximal value of $f(a,\theta)-\sin \theta$ is \[f(a,\pi/2)-\sin \pi/2 = \frac{\tan a}{a} - 1 \to 0.\]

Hence the convergence is uniform, we can exchange the integral sign with the limit sign.

\[\lim_{a\rightarrow 0} \frac{1}{a}\int_0^\pi \tan (a\sin \theta)d\theta = \int_0^\pi \lim_{a\rightarrow 0} \frac{1}{a} \tan (a\sin \theta)d\theta = \int_0^\pi \sin\theta d\theta = 2\]

Needless to say, you are right.

What kind of problem do you like,liyi?

[quote="kunny"]Evaluate

\[\lim_{a\rightarrow 0} \frac{1}{a}\int_0^\pi \tan (a\sin \theta)d\theta\][/quote]

What you can do is apply directly the "Derivation under the integral" theorem.

Sorry for , I can't understand what you say.

Could you explain in detail, julien_santini.

[quote="kunny"]What kind of problem do you like,liyi?[/quote]
Something not too skillful... and some conceptual problems.  :roll:

[quote="liyi"]I think he means using l'Hospital.
$\lim_{a\rightarrow 0} \frac{1}{a}\int_0^\pi \tan (a\sin \theta)d\theta = \lim_{a\rightarrow 0} \int_0^\pi \frac{\partial}{\partial a}\tan (a\sin \theta)d\theta = \int_0^\pi \lim_{a\rightarrow 0} \sec^2(a\sin \theta)\sin\theta d\theta$
But we still need to argue the validity of taking derivative under the integral sign...which has something to do with uniformly convergence, too (if we do not use Lebesgue's Convergence Dominated Theorem).[/quote]

Thank you,liyi.


Let $ ABC$ be an isosceles triangle with $ AB\equal{}AC$. Let $ E$ be a point lies on line $ BC$ and $ D$ the midpoint of $ BC$. Let $ O$ the center of the circumcircle of triangle $ ABE$. Prove that the line through $ D$ and perpendicular to $ OD$, the line through $ E$ and perpendicular to $ AC$ and the line passes through $ C$ and parallel to $ AB$ are concurrent.

[quote="April"]Let $ ABC$ be an isosceles triangle with $ AB \equal{} AC$. Let $ E$ be a point lies on line $ BC$ and $ D$ the midpoint of $ BC$. Let $ O$ the center of the circumcircle of triangle $ ABE$. Prove that the line through $ D$ and perpendicular to $ OD$, the line through $ E$ and perpendicular to $ AC$ and the line passes through $ C$ and parallel to $ AB$ are concurrent.[/quote]
Let $ X$ be  intersection of the line throung $ D$ and perpendicular to $ OD$ and the line through $ E$and perpendicular to $ AC$. Then this remains to prove $ CX//AB$. Let $ K$ be intersection of $ AD$ and $ EX$ and $ Y$ be a intersection of the line through $ D$ and perpendicular to $ OD$ and  $ AB$. $ A,B,K,E$ is cyclic since angle$ BAK \equal{} 90\minus{}B \equal{} 90\minus{}C \equal{} BEK$. Therefore, By butterfly theorem,$ DY \equal{} DX$ and since $ DB\equal{}DC$, angle $ DCX \equal{} DBA$. So $ CX//AB$


Lat $ A_1,A_2, . . . ,A_n$ be the sets each consisting of $ n$ segments on a given line. Prove that the intersection $ A_1 %Error. "capA" is a bad command.
_2 \cap. . .capA_n$ contains not more than $ n^2 - n + 1$ disjoint segments.

[quote="indybar"]Let $ A_1,A_2, . . . ,A_n$ be the sets each consisting of $ n$ segments on a given line. Prove that the intersection $ A_1 \cap A_2 \cap\ldots \cap A_n$ contains not more than $ n^2 \minus{} n \plus{} 1$ disjoint segments.[/quote]

Fixed.


geometry
3D geometry
tetrahedron
rectangle
circumcircle
geometry proposed

A regular tetrahedron has side $L$. What is the smallest $x$ such that the tetrahedron can be passed through a loop of twine of length $x$?

There might be a better way, but I think the loop has to go all the way around one of the equilateral triangles, thus the length would be the circumference of the circumscribed circle around a triangle, which is $\frac{2 \pi L \sqrt{3}}{3}$.
It might also just be $2L$ so that it can loop around one edge.

About the $\pi$ number, note that the problem asks for a contour of any shape, not just a circular one.

I would say $L(1+\sqrt{3})$. Doesn't the problem ask equivalently for the parallel projection (i.e. shadow) of the tetrahedron on a plane with the smallest circumference? So take the plane perpendicular to one edge.

Maybe it's not clear in the statement, but the loop is supposedly made of string, so it's flexible (that is, it can change its format).

If so, then $2L$ is of course sufficient. Start with one edge and strip it through till the opposite edge. At each moment, the string forms a rectangle of circumference $2L$.


Let $ f: R\minus{}\minus{}>R$ be a function satisfying the following functional equation:

$ f(f^2(x)\plus{}y)\equal{}x^2\plus{}f(y)$

Find $ f$...

(a re sil... thn elysa me ta kolpa p mas didaxes...) :wink:

Δυσκολη ασκηση Κωστα, ελπιζω η παρακατω λυση να μη χανει καπου

Θετουμε $ y \equal{} 0$, οποτε $ f(f^2(x)) \equal{} x^2 \plus{} f(0)$ απο οπου ευκολα η $ f$ ειναι 1-1 στο $ [0, \plus{} \infty)$
Θετουμε $ x \equal{} y \equal{} 0$ οποτε $ f(f^2(0)) \equal{} f(0) \Rightarrow f(0) \equal{} 0$
Αρα $ f(f^2(x)) \equal{} x^2$
Για $ x \equal{} f^2(x)$ τωρα ειναι $ f(x^4) \equal{} f^4(x) \Rightarrow f(1) \equal{} 1$
Για $ x \equal{} f^2(x)$ στην αρχικη λαμβανουμε: $ f(x^4 \plus{} y) \equal{} f^4(x) \plus{} f(y) \equal{} f(x^4) \plus{} f(y) \Rightarrow f(x \plus{} y) \equal{} f(x) \plus{} f(y) \forall x \geq 0 , y \in R$
Απο την τελευταια ευκολα προκειπτει οτι η $ f$ ειναι περιττη.
Υψωνουμε την τελευταια στο τετραγωνο και εφαρμοζουμε την $ f$ και μετα απο συντομες πραξεις προκειπτει:
$ f(f(x)f(y)) \equal{} xy (*)$
Θετουμε στην τελευταια $ x \equal{} 1$, οποτε $ f(f(y)) \equal{} y \forall y \in R$, αρα εφαρμοζοντας την $ f$ στην $ (*)$ προκειπτει $ f(xy) \equal{} f(x)f(y)$ και $ f(x^2) \equal{} f^2(x)$ για καθε $ x \geq 0$.
Ετσι η αρχικη γραφεται: $ f(f(x^2) \plus{} y) \equal{} x^2 \plus{} f(y) \forall x \geq 0, y \in R$, οποτε $ f(f(x) \plus{} y) \equal{} x \plus{} f(y) \forall x \geq 0, y \in R$
Επηδη η $ f$ ειναι περιττη, μπορουμε ευκολα να δειξουμε οτι η τελευταια ισχυει για καθε $ x,y \in R$
Εστω οτι ισχυει $ f(k) \equal{} k \plus{} a, a \equal{} f(k) \minus{} k$ για ενα συγκεκριμενο $ k$,
τοτε
$ f(f(k)) \equal{} f(k \plus{} a) \Rightarrow k \equal{} f(k \plus{} f(f(a))) \equal{} f(k) \plus{} f(a) \Rightarrow f(k) \equal{} k \minus{} f(a) \Rightarrow f(a) \equal{} \minus{} a$

Αν $ a > 0$ τοτε $ 0 > \minus{} a \equal{} f(a) \equal{} f^2(\sqrt {a})$ το οποιο ειναι ατοπο.
Αν $ a < 0$ τοτε $ 0 < \minus{} a \equal{} f(a) \equal{} \minus{} f( \minus{} a) \equal{} \minus{} f^2(\sqrt { \minus{} a})$ το οποιο ειναι επισεις ατοπο.
Αρα $ a \equal{} 0, f(k) \equal{} k$ για το τυχαιο $ k \in R$ και αρα $ f(x) \equal{} x \forall x \in R$.

Διορθωστε με αν εχω κανει καμια πατατα.

Δε βρήκα λάθος απλά όταν βρήκες ότι η $ f$ είναι involution έχεις τελειώσει πολύ πιο γρήγορα γιατί:

Έστω $ x>y$
$ f(x)\equal{}f(x\minus{}y\plus{}y)\equal{}f(x\minus{}y)\plus{}f(y)>f(y)$

γιατί οι εικόνες των θετικών έχουμε αποδείξει ότι είναι θετικοί.

Άρα $ f$ γνήσια αύξουσα οπότε τελειώσαμε. :)

Ωραίος Nick. Btw την έλυσε όμορφα και ο Ηλίας ο Γιεχασκιέλ για να μη μείνει παραπονεμένος... :rotfl:

[quote="Protonios"]Δε βρήκα λάθος απλά όταν βρήκες ότι η $ f$ είναι involution έχεις τελειώσει πολύ πιο γρήγορα γιατί:

Έστω $ x > y$
$ f(x) \equal{} f(x \minus{} y \plus{} y) \equal{} f(x \minus{} y) \plus{} f(y) > f(y)$

γιατί οι εικόνες των θετικών έχουμε αποδείξει ότι είναι θετικοί.

Άρα $ f$ γνήσια αύξουσα οπότε τελειώσαμε. :)

Ωραίος Nick. Btw την έλυσε όμορφα και ο Ηλίας ο Γιεχασκιέλ για να μη μείνει παραπονεμένος... :rotfl:[/quote]

Σωστα, με τη μονοτονια τελειωνει αμεσως, ετσι την τελειωσε και ο Sil χτες που μου ειπε τη λυση του.

Αλλα τελικα δεν χρειαζεται ουτε το involution. Αφου Cauchy και Συνεχης/Γν.Αυξουσα/Φραγμενη σε διαστημα [α,β] ειναι μονο η $ f(x) \equal{} cx$

[quote="Protonios"]Ναι αλλά η Cauchy ισχύει για χ θετικό εδώ.

Για να πούμε αυτό που λες πρέπει να ισχύει και για τους αρνητικούς... Ο Ηλίας το απέδειξε και το χρησιμοποίησε. Έτσι δεν είναι ή λέω βλακείες? :maybe:[/quote]

Αν δειξεις οτι ειναι cauchy για x θετικο, μπορεις μετα ευκολα να το δειξεις οτι ειναι για ολα τα x στο R. Αυτο γινεται δειχνοντας πρωτα οτι η συναρτηση ειναι περιττη και παιζοντας μετα με τα προσημα στην εξισωση cauchy.

Έστι είναι πραγματικά . Μπορούμε να αποδείξουμε την Cauchy για όλα χρησιμοποιώντας το ότι η f είναι περιττή (το είχαμε πει και στο <<μάθημα>>  :wink: )


geometry
3D geometry
inequalities unsolved
inequalities

Let $ 0<\equal{}x,y,z \ | \ x\plus{}y\plus{}z\equal{}3$. Then \[ (x^2\plus{}y^2)\left(\frac{x^2\plus{}y^2}{x}\plus{}\frac{z^3}{y^2}\right)>\equal{}6\]

$ (x^2\plus{}y^2)\left(\frac{x^2\plus{}y^2}{x}\plus{}\frac{z^3}{y^2}\right)\equal{}x(x^2\plus{}y^2)\plus{}z^3\plus{}\frac{y^2(x^2\plus{}y^2)}{x}\plus{}\frac{z^3x^2}{y^2}$

By AM-GM :

$ xy^2\plus{}\frac{y^4}{x}\ge 2y^3$

and :

$ \frac{z^3x^2}{y^2}\plus{}\frac{x^2y^2}{x}\plus{}y^3\ge 3xyz$

Then

$ (x^2\plus{}y^2)\left(\frac{x^2\plus{}y^2}{x}\plus{}\frac{z^3}{y^2}\right)\ge x^3\plus{}y^3\plus{}z^3\plus{}3xyz$

We prove that :

$ 9(x^3\plus{}y^3\plus{}z^3\plus{}3xyz)\ge 2(x\plus{}y\plus{}z)^3$

It is right  :)


geometry
geometric transformation
dilation
number theory

If 2^x*9^y is equal to the four-digit number 2x9y where x is the hundreds digit and y is the units digit, what is the product of x and y?

Brent likes to dilute his lemonade. He starts with a full cup and drinks 1/3 of the contents. He then fills the cup with water, and drinks 1/3 of its content. He repeat this until he has consumed 1 cup of liquid. What part of the original lemonade remains in the cup? Express answer in common fraction

I did the second problem, but I'm still working on the first one.

[hide]
The second one is 8/27.

First of all, Brent will do this three times because 1 divided by 1/3 = 3. After he drinks 1/3, 2/3 of the original lemonade is left. Then, after drinking another third, 4/9 of the original lemonade is left. Multiply by 1/3 again and you get 8/27.[/hide]

Okay, I have a possible answer for the first one.

[hide]This doesn't look like it's possible to solve by regular algebra. We just have to try and see what the possible combinations are and eliminate obvious impossibilities.

We see that y can't be 1 or 0. If y were 0, then the only power of 2 that could be in the 2000s would be 2^11, and 11 isn't a digit. Y also can't be 1, or any other odd digit, because then the number would need to end in an odd digit, and that isn't possible when x is anything but 0. And if x is 0, then we can't get a number in the 2000s, because there's no power of 9 in the 2000s. Then we try y = 2, and we see that x = 5, y = 2 works. Then, if we try y = 4, it's already too high (in the 6000s), so there are no other combinations.

The answer is x = 5, y = 2.[/hide]
But maybe I missed something.

[hide]And the answer is ten. I forgot that.[/hide] :roll:

I always do that.

[hide="the first one"]
Method=guess and check.
If you know your powers of 9, you know that $9^3=729$
The only power of two that would be in the 2000s would be $2^2=4$.
But that would make the last digit of the product 6, and we didn't do $9^6$.
So check $9^2=81$
Eventually stumble across $81\cdot32=2592$
Thus $x=5$ $y=2$
Product is 10
[/hide]
[hide="the second one"]
First, Brent drinks $\frac 13$ cup of lemonade, leaving $\frac 23$ left.
Then he adds water.
He drinks $\frac 13$ of that, which is $\frac 29$, leaving $\frac 49$ left.
Then he adds water.
He drinks $\frac 13$ of that, which is $\frac{4}{27}$, leaving $\frac{8}{27}$ left.[/hide]

[quote="kidwithshirt"]Nice one.. What i would do is to ignore the first one on sprint and come back when i m done with everything else lol  :blush:[/quote]

Yeah, that sounds cool. But when I said "I always do that" I meant I always make stupid mistakes. (I forgot to get the product the first time around.)

The second is [hide](2/3)^3= 8/27[/hide]

[hide="1"]10. the number is 2592. 2^5*9^2 works. and 2*5 is 10[/hide]
[hide="2"]8/27. 2/3 is after 1 drink, 4/9 after 2 drinks, 8/27 after 3 drinks. since 1/3 of the remaining lemonade is drunk with each drink, 2/3 is left, so (2/3)^3.[/hide]

[quote="kidwithshirt"]If 2^x*9^y is equal to the four-digit number 2x9y where x is the hundreds digit and y is the units digit, what is the product of x and y?
[/quote]

[hide="solution"]we know that since the number is divisible by 9, 2+x+9+y==0 mod 9
(y cant be 0 because nothing to the power of 2 will end in 0)
thus, x+y=7
(cant be 16 or $2^x\cdot 9^y$ would be a very very big number)
so then you test values of y and x and you find that x=5 and y=2
...[/hide]

Remeber, jli, it has to be the product. But I did that, too.

[quote="kidwithshirt"]If $2^x 9^y$ is equal to the four-digit number $\hbox{2x9y}$ where x is the hundreds digit and y is the units digit, what is the product of x and y?[/quote]
[hide="Without guess and check"]
This is a pretty nice number theory question...

1.  Ask yourself if they are being sneaky.  Can $x$ or $y$ be zero?

1a.  There's no power of 9 in the form 209y, so $x\ne 0.$  Therefore 2x9y is an even number and y is an even digit.

1b.  There's no power of 2 in the form 2x90, so $y\ne 0.$  Therefore 2x9y is a multiple of 9, so start thinking about divisbility rules.

2.  Think about the largest possible value of y.  $9^4 = 81^2>6400> \hbox{2x9y}.$  So $y<4.$

3.  Therefore, $y=2$.  Then, by the divisibility rule for 9's, $x=5$.

4.  Don't forget to answer the question.  $xy=10.$  Move on, but check your answer if time permits.

[/hide]


a triangle has vertices at (-3,2), (6,-2), and (3,5). How many squares units are in the area of the triangle???

I smell "breaking up the triangles into other triangles and then finding the area of each individual triangle and the add all of the areas up". Yeah.  :P

[hide]
SHOELACE ALGORITHM!!
Look it up if you don't know what it is, it would be really hard to explain without knowing what it was and really hard to explain what it is
I got $\frac{51}{2}$[/hide]

Its called many things i wonder what the actual name for it is..........

Surveyors Formula............it's got the goods

Shoestring/shoelace  Formula/algorithm.

So many names, but discuss this in the formulas topic please!


function
limit
ratio
LaTeX
algebra unsolved
algebra

Hi everybody!
I'm trying to solve this problem but it's getting more and more difficult. Please help me! :(
Compare ($99^n +100^n$) and $101^n$

You can use function to solve the mistery:\[
100^n  + 99^n  = 101^n  \Leftrightarrow (\frac{{100}}{{101}})^n  + (\frac{{99}}{{101}})^n  = 1.
\]having :\[
f(n) = (\frac{{100}}{{101}})^n  + (\frac{{99}}{{101}})^n \forall n \in N
\]This function is monoton so it haves a unic solution .
Conclusion:after a n we have :\[
100^n  + 99^n  < 101^n
\]

I don't think it's right.Think it over again!
ARTIONA

silviu is correct.  For large $n$, $101^n$ is bigger, for small $n$, $99^n+100^n$ is bigger.

We have :\[
f(x) = (\frac{{100}}{{101}})^x  + (\frac{{99}}{{101}})^x ,\forall x \in R,x > 0;
\]And we ca easyly prove that:\[
\forall x,y \in R_ +  ,x < y \Leftrightarrow f(x) > f(y).
\]We olso have:\[
\mathop {\lim }\limits_{x \to \infty } f(x) = 0;
\]Resulting that for a large n,f(n) is very close to 0,and we know that 0<1,don't we? :lol:
You can't compare those number with out knowing n.

You might have clarified it.
If we consider the ratio (101n  - 99n)/100n then it comes out that:
(99^n + 100^n )>100^n,  n<= 48(I'm sorry for not using the LaTex :blush: ) and
(99^n + 100^n )< 100^n,  n>48.

How did you managed to found n that satystace that inequliti:\[
99^n  + 100^n  < 101^n ,\forall n \in N,n > 48.
\]How did you found that n>48 :)


Se dau numerele reale $\ a,b,x$ astfel incat sa avem inegalitatile
$\ |a+x+b|\leq\ 1,|4a+2x+b|\leq\ 1,|9a+6x-b|\leq\ 1$.Aflati valoarea maxima pe care o poate lua $\ |x|$.


number theory
greatest common divisor
number theory unsolved

Let $ a,b,x,y$ be positive integers such that $ ax\plus{}by$ is divisible by $ a^2\plus{}b^2$. Prove that $ gcd(x^2\plus{}y^2, a^2 \plus{} b^2) > 1$.

One has $ (ax \plus{} by)^2 \plus{} (ay \minus{} bx)^2 \equal{} (a^2 \plus{} b^2)(x^2 \plus{} y^2)$, hence $ a^2 \plus{} b^2$ also divides $ (ay \minus{} bx)^2$ so $ \gcd(a^2 \plus{} b^2, (a \plus{} bi)(x \minus{} yi)) \equal{} \gcd(a \minus{} bi, x \minus{} yi)$ is not a unit in the Gaussian integers, from which the original conclusion follows.  See also the related argument and ideas in the important case that $ a^2 \plus{} b^2$ is prime [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]here[/url].

btw. Where can I find problems from MOSP2007? I have only from 2005 and 2006.


If $ x^{2y}\equal{} 4$ and $ x \equal{} 4$, what is the value of $ y$? Express your answer as a common fraction.


How to draw Venn diagrams like union,intersection using Latex?

This nice way of doing it is courtesy of [url=http://groups.google.co.uk/group/comp.text.tex/browse_thread/thread/3a7229ce6cfed4c2/f7da9182bda7deed]Venn diagrams[/url].
Download the attached file and unzip it to get venn.sty. For the moment* put it in the same directory as your tex file. Then try this example file. You [b]must[/b] use LaTeX=>PS=>PDF to compile it.
[code]\documentclass{article}
\usepackage{venn}
\begin{document}
Here is a one-circle Venn diagram.
\begin{center}
  \vennone{$A$}
\end{center}
Here is a two-circle Venn diagram.
\begin{center}
  \venntwo{$A$}{$B$}
\end{center}
Here is a three-circle Venn diagram.
\begin{center}
  \vennthree{$A$}{$B$}{$C$}
\end{center}
Here is a three-circle Venn diagram with a teeny circle
in the intersection.
\begin{center}
  \vennthree[\rput(.5,.5){$\smash\circ$}]{$A$}{$B$}{$C$}
\end{center}
\end{document}[/code]That example file shows you what to do to get Venn diagrams - it makes it very easy. The result is in the attached venn.pdf file.

*If you want to use venn.sty more frequently then make a new directory called venn in [i]your path to tex files/tex/latex [/i]and put venn.sty in it. Then refresh the database (for MiKTeX that is Settings, Refresh FNDB).

Thanks a lot, it solved the problem through cann't put shaded area yet.

BTW, when i Choose Output Type/Latex=>DVI it work,
but not when i set Output Type/Latex=>PDF.

[quote="ic2149"]Thanks a lot, it solved the problem through cann't put shaded area yet.[/quote]If you look in venn.sty there is a \vennsetandcompgray command you can try though there appears to be a bug with LaTeX complaining about the .7 in \pspicture[.7](0,0)(1,1) in venn.sty. Removing [.7] seems to work.
[quote="ic2149"]
BTW, when i Choose Output Type/Latex=>DVI it work,
but not when i set Output Type/Latex=>PDF.[/quote]Now what did I say? venn.sty uses PSTricks so if you want pdf you must use LaTeX=>PS=>PDF [b]NOT[/b] LaTeX=>PDF

How do you change the output profile so that it can run in Latex => Pdf => PS ? It is not given in the options menu so do you have to configure it?

You mean LaTeX => PS => PDF.
1. First of all you need to install Ghoscript which handles the PS=>PDF part - see [[Asymptote:_Getting_Started/Windows/Downloads_and_Installation#Ghostscript]] for how to do this.
2. Now you must tell TexnicCenter what to run. The easiest way is to use someone else's profile.
a) download latex-ps-pdf.zip from the second post at [url=http://www.latex-community.org/viewtopic.php?f=31&t=302]LaTeX Community[/url]
b) unzip and then in TexnicCenter, Build, Define Output Profiles, Import the tco file
c) in the same dialogue box click on LaTeX => PS => PDF and each tab and change all the paths to the ones on your machine.

Now you should see LaTeX => PS => PDF for compiling.

How would you specify where to put numbers in, either in the actual circles, the intersections, or the outside (doesn't belong to any)?

It's the numbers in \rput[lt](0,0) in venn.sty where as it says, .27 is outside.
You can change them in your document by using \renewcommand.
For example
\renewcommand{\vennone}[2][]{
  \venndefunits %
  \pspicture(0,0)(1,1)
    \vennuniverse %
    \pscircle(.5,.5){.23}
      \uput{.22}[ul](.5,.5){\rput[lt][color=red](.1,0)[/color]{\venntext{#2}}} %.27 is outside
  #1 %
  \endpspicture}

I tried to download ghostscript but when I looked at it it had all these weird symbols- like trying to look at an unzipped file.  :help:

when you looked at what? What was the file format? If it's a zipped file (like .zip, .tar, .tar.gz, etc), it would make a lot of sense to unzip it first.. :roll:

Oh yeah it's a .tar file. Oops.  :blush:


http://en.wikipedia.org/wiki/Hermite%27s_identity

Can someone explain the step $ 0 \equal{} \left\lfloor \{x\} \plus{} \frac {k' \minus{} 1}{n}\right\rfloor\le \{x\} < \left\lfloor \{x\} \plus{} \frac {k'}{n}\right\rfloor \equal{} 1 \, \Rightarrow \, 1 \minus{} \frac {k'}{n}\le \{x\} < 1 \minus{} \frac {k' \minus{} 1}{n}$ to me? Thanks!

By their definition, $ \lfloor x\rfloor \equal{} \left\lfloor x \plus{} \frac {k' \minus{} 1}n\right\rfloor\le x < \left\lfloor x \plus{} \frac {k'}n\right\rfloor \equal{} \lfloor x\rfloor \plus{} 1$, and since $ x \equal{} \lfloor x\rfloor \plus{} \{x\}$, subtracting $ \lfloor x\rfloor$ yields $ \lfloor x\rfloor \minus{} \lfloor x\rfloor \equal{} \left\lfloor x \plus{} \frac {k' \minus{} 1}n\right\rfloor \minus{} \lfloor x\rfloor\le x \minus{} \lfloor x\rfloor < \left\lfloor x \plus{} \frac {k'}n\right\rfloor \minus{} \lfloor x\rfloor \equal{} \lfloor x\rfloor \plus{} 1 \minus{} \lfloor x\rfloor$
$ \implies0 \equal{} \left\lfloor x \plus{} \frac {k' \minus{} 1}n \minus{} \lfloor x\rfloor\right\rfloor \equal{} \left\lfloor\{x\} \plus{} \frac {k' \minus{} 1}n\right\rfloor\le\{x\} < \left\lfloor x \plus{} \frac {k'}n \minus{} \lfloor x\rfloor\right\rfloor$
$ \equal{} \left\lfloor\{x\} \plus{} \frac {k'}n\right\rfloor \equal{} 1$. Because $ 0 \equal{} \left\lfloor \{x\} \plus{} \frac {k' \minus{} 1}n\right\rfloor$, $ \{x\} \plus{} \frac {k' \minus{} 1}n < 1\implies\{x\} < 1 \minus{} \frac {k' \minus{} 1}n$, and since $ 1 \equal{} \left\lfloor\{x\} \plus{} \frac {k'}n\right\rfloor$, $ \{x\} \plus{} \frac {k'}n\ge1\implies\{x\}\ge1 \minus{} \frac {k'}n$, as desired.


calculus
real analysis
real analysis unsolved

Sorry to post here but did not know where else to post.

I want to learn Calculus/Analysis on my own.

I found alot of books but they all have really difficult problems without much guidance on how to derive a proper solution.

Could you guys give me advice on a book I should study from? Perhaps a book with solutions(not just answers)

Also please keep in mind I am in $ 4^{th}$ grade, but have learnt basic calculus.

I hope this gets someone's attention the $ 2^{nd}$ time around.

Shaum's outline series's is always nice, if you want solution, but you won't find challenging problems in there.

[quote="Carcul"]What do you mean by 4th grade? For how many years are you studing?[/quote]

Im not sure what you mean.
$ 4^{th}$ grade corresponds to age 8.
I have just recently learnt basic calculus, and wanted to work on analytic proofs.

@Mr.Doe- Thank you, that maybe just what I need.

You are only 8 years old and have already learnt basic calculus in school?! May I ask in what country you study?

U.S.
It's not that fairly uncommon. Plus I am probably very bad at it. The idea's are pretty simple though(if explained properly!).
The credit belongs duly to the amazing books on the subject, and google.
So I can't say that I know calculus, because I am better than anyone. (I am definitely not a genius)
Just early exposure I guess  :P


calculus
derivative
integration
calculus computations

1.if y=arctgx, find $ y^{(n)}(0)$
2.if y=arcsinx, find $ y^{(n)}(0)$
$ y^{(n)}(0)$ means nth derivative

Find $ y', \; y'', \; y''', \; etc.$

You'll notice a particular trend for $ \frac{d^ny}{dx^n}$.

But when n is large nth derivative looks impossible/too complex.

They are not asking for you to find the general expression of the nth derivative, just in the point $ x \equal{} 0$.

Start with the $ \minus{}1$th derivative. Get a power series. Use that to find derivatives at zero.

let try with Taylor series
$ f(x) \equal{} \sum_{n}^{\infty}a_{n}(x \minus{} b)^{n}$
where $ a_{n} \equal{} \frac {f^{n}(b)}{n!}\Rightarrow f^{n}(b) \equal{} a_{n}n!$

$ artan(x) \equal{} \int_{0}^{x}\frac {1}{1 \plus{} t^{2}}$ and use $ \frac {1}{1 \minus{} x} \equal{} \sum_{n}x^{n}$
$ arcsin(x) \equal{} \int_{0}^{x}\frac {1}{\sqrt {1 \minus{} t^{2}}}$ and use $ (1 \plus{} x)^{\alpha} \equal{} \sum_{k}C^{k}_{\alpha}x^{k}$


Find the interval of convergence for (x-1)^k ∕ √(k^2 + 1)

A few things, tomegibson101:

1. Don't post the identical topic in more than one place. Once is enough.

2. "Solved Problems" is not the place to start anything; that's where topics can be moved after they're done.

3. If it could appear as an ordinary exercise in a standard calculus textbook, the right place is probably here, "Calculus Computations and Tutorials." That applies to these two problems.

As for this one: "de-clutter" it - look at $\frac1{\sqrt{k^2+1}}$ as essentially the same as $\frac1k$ for large $k$.

Use either the root test or the ratio test; either way, $\frac1{\sqrt{k^2+1}}$ is invisible to the test - might as well as be 1. The root test and ratio test must be understood to be no more than what they are: automated comparisons to the geometric series. Terms which grow or shrink slower than geometrically are invisible to such a test.

The root or ratio test will tell you that this series converges for $|x-1|<1$, which is to say $0<x<2.$

Now there are endpoints to consider: $x=0$ and $x=2$, better understood as $x-1=\pm1.$ The series converges (conditionally) at one of these points and diverges at the other; I'll leave it to you to figure out which is which.


function
Ring Theory
topology
superior algebra
superior algebra unsolved

Let Ra be the ring of real valued continuous functions on the closed unit interval.If M is the maximal ideal of R, prove that there exists a real number $ \gamma , 0 \le \gamma \le 1,$  such that $ M \equal{} \{ f(x) \epsilon R | f(\gamma) \equal{} 0 \}$.

That should be standard, more precisely all ideals of $ C[0,1]$ are $ \{0\},C[0,1]$ or of the form (*) $ \{f\in C[0,1]|f(\gamma)\equal{}0\}$ for some $ \gamma\in [0,1]$, in particular, all maximal ideals have the form (*). As for the proof:

(1) Let $ \{0\}\not\equal{}I$ be an ideal, which is not of form (*), then $ \forall x\in [0,1]\,\exists f_x\in I: f_x(x)\not\equal{}0$. Since $ f_x$ is continuous, we find an open set $ U_x$ such that $ \forall y\in U_x: f_x(y)\not\equal{}0$. Since $ [0,1]$ is compact, we can determine a finite set $ {\cal X}$ such that $ [0,1]\equal{}\bigcup_{x\in{\cal X}}U_x$ holds. But then $ 0<g\equal{}\sum_{x\in{\cal X}}f_x^2\in I$ and $ \frac{1}{g}\in C[0,1]$ and therefore $ 1\equal{}\frac{1}{g}\cdot g\in I$, which yields $ I\equal{}C[0,1]$.

(2) Conversely, any set of form (*) creates a non-trivial ideal.

your proof uses that $ I$ is maximal. the ideals have the form $ \{f : f|E \equal{} 0\}$ where $ E \subseteq [0,1]$ is closed.


$ \frac{a^n\plus{}b^n}{a^{n\minus{}2}\plus{}b^{n\minus{}2}}\geq \frac{a^2\plus{}b^2}{2}$
this together with $ a^2\plus{}b^2\plus{}c^2\geq \frac{(a\plus{}b\plus{}c)^2}{3}\equal{}\frac13$ gives you the result for $ n\geq 2$
for $ n\equal{}1$
$ \frac{a\plus{}b}{a^{\minus{}1}\plus{}b^{\minus{}1}}\plus{}\frac{b\plus{}c}{b^{\minus{}1}\plus{}c^{\minus{}1}}\plus{}\frac{c\plus{}a}{c^{\minus{}1}\plus{}a^{\minus{}1}}\geq \frac13$
is false... in fact it holds in the reverse direction.

sorry $ n>2$ , but the inequalities are true !

hint

prve that for all integers $ n$ such that $ n\geq 2$

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

tchebychev  :roll:


I am from the Philippines and I have some interesting problems form our last year's math olympiad.
Can you give me the answers to the 3 open-ended questions?

Problem 21:

Problem 22:

Problem 23:

Thank you!!!

The rest of the problems:

Actually the code for URLs is
[code][url=http://test.com]NAME OF URL[/url][/code]

to get [url=http://test.com]NAME OF URL[/url] ;) Thanks for the problems.


[color=darkblue]Let $ABC$ be a triangle and let $E,F$ be the projections of the its orthocenter to the side-lines $AC\ ,\ AB$ respectively.
Prove that $A=60^{\circ}$ if and only if the middlepoint of the segment $[EF]$ is the radical center for the circles $C(B,BF)$, $C(C,CE)$, $C\left(A,\frac{|b-c|}{2}\right)\ .$[/color]

[quote="pkerichang"]can you explain your notation for the circles???[/quote]
( Centre, Radius).

[quote="Virgil Nicula"][quote="shobber"]I think it can be done with calculation. But I only managed to show that the midpoint lies on the radical axis of the two circles centred at B and C, but I cannot calculate the power of the midpoint to the circle centred at A......[/quote]Sorry, Shobber ! I forgot one important condition : $A=60^{\circ}\ .$ I corrected.[/quote]
It took me really a lot time calculating the wrong problem... ...

Let the midpoint of segment EF be D. Let $BF=R_{1}$, $CE=R_{2}$, $EF=d$. Then the Power of point D to circle B is:

$DB^{2}-R_{1}^{2}=\frac{1}{2}R_{1}^{2}+\frac{1}{2}BE^{2}-\frac{1}{4}d^{2}-R_{1}^{2}= \frac{1}{2}\left( BC^{2}-R_{2}^{2}-R_{1}^{2}\right)-\frac{1}{4}d^{2}$.

Similarly we can get that the power of point D to circle C is also:

$DB^{2}-R_{1}^{2}=\frac{1}{2}R_{1}^{2}+\frac{1}{2}BE^{2}-\frac{1}{4}d^{2}-R_{1}^{2}= \frac{1}{2}\left( BC^{2}-R_{1}^{2}-R_{2}^{2}\right)-\frac{1}{4}d^{2}$.

Hence D lies on the radical axis of circle B and circle C.

The power of point D to circle A is:

$DA^{2}-\left( \frac{|b-c|}{2}\right)^{2}= \frac{1}{2}AE^{2}+\frac{1}{2}AF^{2}-\frac{1}{4}d^{2}-\frac{b^{2}-2bc+c^{2}}{4}=$

Since $\angle{A}=60^{o}$ so $a^{2}=b^{2}+c^{2}-bc$ and there is also $AE=\frac{c}{2}$, $AF=\frac{b}{2}$, so the above can be written as:

$\frac{1}{2}\cdot \frac{1}{4}c^{2}+\frac{1}{2}\cdot \frac{1}{4}b^{2}-\frac{2a^{2}-b^{2}-c^{2}}{4}-\frac{1}{4}d^{2}=\frac{3}{8}c^{2}+\frac{3}{8}b^{2}-\frac{1}{2}a^{2}-\frac{d^{2}}{4}$

Since $\frac{c^{2}}{4}=AE^{2}=c^{2}-BE^{2}=c^{2}-a^{2}+R_{2}^{2}$, so $\frac{3}{4}c^{2}=a^{2}-R_{2}^{2}$, similarly there is $\frac{3}{4}b^{2}= a^{2}-R_{1}^{2}$, Pluge them back, we get that the Power of point D to circle A is:

$\frac{1}{2}\left( BC^{2}-R_{2}^{2}-R_{1}^{2}\right)-\frac{1}{4}d^{2}$

Therefore D is the radical centre of the three circles.


In $ \triangle ABC$, let $ D$ be a point on $ \overline{BC}$ such that $ \overline{AD}$ bisects $ \angle A$, If $ \overline{AD} \equal{} 6$, $ \overline{BD} \equal{} 4$, and $ \overline{DC} \equal{} 3$, then find $ \overline{AB}$.

[size=150]Please provide hints only.[/size]

You can use $ \boxed{AD^2\equal{}AB\cdot AC\minus{}BD\cdot CD}$.

$ AC: AB\equal{}3: 4$
Let $ AC \equal{} 3a$ and $ AB\equal{}4a$
Then use cosine law to find out what $ a$ is.


geometry
3D geometry
sphere
topology
limit
real analysis
real analysis unsolved

Suppose that $f\colon \mathbb R^{n}\to\mathbb R^{n}$ is a homeomorphism such that for every point $x\in\mathbb R^{n}$ the image of every sphere centered at $x$ is a sphere centered at $f(x)$.

Prove that $f$ is a constant multiple of an isometry, that is, $\parallel f(x)-f(y)\parallel =\lambda \parallel x-y\parallel $, where $\lambda>0$ is independent of $x,y\in\mathbb R^{n}$.

First, we show that $f$ is affine, i.e. $f(\alpha x+\beta y)=\alpha f(x)+\beta f(y)$ for $x,y\in\mathbb R^{n}$ and $\alpha,\beta$ scalars with sum $1$. By the continuity of $f$, in order to achieve this, it suffices to prove this for $\alpha,\beta$ dyadic rationals, and hence for $\alpha=\beta=\frac{1}{2}$.

Let $x,y\in\mathbb R^{n}$ be two arbitrary points, and let $z$ be a point equidistant from $x,y$. Then $f(z)$ is equidistant from $f(x),f(y)$. The three mutually externally tangent spheres centered at $x,y,z$ go into the three mutually externally tangent spheres centered at $f(x),f(y),f(z)$, so the intersection point between the two spheres centered in $x,y$ goes into that of the spheres centered in $f(x),f(y)$. The first point is the midpoint of the segment $xy$, while the second one is the midpoint of $f(x)f(y)$, and we are done: this shows that midpoints go into midpoints, as desired.

We now know that $f$ is an affine map that turns spheres into spheres, and every such map is a scaling of an isometry.

Yes, it's a neat proof (Tukia&Vaisala used internally tangent spheres in a similar way.) Of course, $z$ should be different from the midpoint of $[x,y]$.

This is a lemma that they used to give a short (but not elementary) proof of a special case of Liouville's theorem: for $n\ge 2$ any conformal homeomorphism $f\colon\mathbb R^{n}\to\mathbb R^{n}$ is a multiple of an isometry. Here $f$ is conformal if $\lim_{r\to 0}\frac{\sup_{|x-a|=r}|f(x)-f(a)|}{\inf_{|x-a|=r}|f(x)-f(a)|}=1$ for all $a\in\mathbb R^{n}$.


geometry
3D geometry
prism
Vieta
algebra
polynomial

The volume of a rectangular prism is 63 cubic units. The surface area is 64 square units. The total length of its edges is 46 units. What is the length of the sides of the immaginary rectangular prism?

[hide="..."]let the sides be a,b,c, then $abc=63$, $ab+ac+bc=32$, and $a+b+c=\frac{23}{2}$
then using vieta's sums, the polynomial
$P(r)=r^3-\frac{23}{2}r^2+32r-63$
has the roots $r=a,b,c$, i don't think this factors...so use the general cubic equation to solve this...[/hide]

first of all, what does it mean "imaginary rectangular prism"?

and secondly, if you use Altheman's solution, the cubic gives you three roots of $8.64, 1.43\pm2.29i$ (solved with 89); if thats what you mean by imaginary rectangular prism is that it has imaginary side lengths... :?:

yes, I meant that it had immaginary side lengths. I messed up though, the surface area was supposed to be 92.


trigonometry
calculus
calculus computations

[b]
f(x)=x^(1/x)
f(x)=x*arctg(x)
f(x)=(sin(x))^2*sin(2x)
f(x)=x^3 + e^3x
f(x)=(sin(x))^4
Calculate n-th derivates on f(x)[/b]

It's nice to solve problems but it's nasty that you post your homework.

What you have to do, it's to show your work with each one of them.

[quote="the_wonder"]
$ f(x) \equal{} x^{\frac {1}{x}}$
$ f(x) \equal{} x\arctan(x)$
$ f(x) \equal{} \sin^2(x)\sin(2x)$
$ f(x) \equal{} x^3 \plus{} e^{3x}$
$ f(x) \equal{} \sin^4(x)$
Calculate $ n$-th derivates on $ f(x)$[/quote]

For the second and last ones, try using trigonometric identities like

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

in order to rewrite the expressions so that they are easier to differentiate.

The fourth one is pretty simple: the first term is a monomial; the second term is an exponential.

That is not my homework!!!
I solve problems for pleasure and I have a problem with these n-th derivates and because of I
question.
 :roll:


geometry
incenter
circumcircle
geometry proposed

ABC is an acute-angled triangle. The incircle touches BC at K. The altitude AD has midpoint M. The line KM meets the incircle again at N. Show that the circumcircle of BCN touches the incircle of ABC at N.

Lemma: $ \angle BNK \equal{} \angle KNC$

call I the incircle of ABC, we want to prove that $ \angle INB \equal{} 90 \minus{} \angle NCB$. But $ \angle INB \equal{} \angle BNK \minus{} \angle INK \equal{} \angle KNC \minus{} \angle KMD \equal{} 90 \minus{} \angle NCK$

Dear friends,

The proof of lemma: $ \angle BNK \equal{} \angle KNC$.

$ S$ is the area of the triangle $ ABC$, $ BC \equal{} a$, $ CA \equal{} b$, $ AB \equal{} c$, $ p \equal{} \frac {a \plus{} b \plus{} c}{2}$, $ r$ is the radius of the incircle $ \left(I\right)$ and $ \angle BKN \equal{} \alpha$.

$ KB \equal{} p \minus{} b$, $ KC \equal{} p \minus{} c$, $ DB \equal{} c\cdot cos\angle B \equal{} \frac {c^{2} \plus{} a^{2} \minus{} b^{2}}{2a}$

$ DK \equal{} KB \minus{} DB \equal{} \frac {c \plus{} a \minus{} b}{2} \minus{} \frac {c^{2} \plus{} a^{2} \minus{} b^{2}}{2a} \equal{} ... \equal{} \frac {\left(p \minus{} a\right) \left(b \minus{} c\right)}{a}\Rightarrow$
$ \Rightarrow DK \equal{} \frac {\left(p \minus{} a\right)\left(b \minus{} c\right)}{a}$

$ MD \equal{} \frac {AD}{2} \equal{} \frac {c\cdot sin\angle B}{2} \equal{} \frac {a c\cdot sin\angle B}{2 a} \equal{} \frac {S}{a} \equal{} \frac {pr}{a}$

$ tan\alpha \equal{} \frac {MD}{DK} \equal{} \frac {pr}{\left(p \minus{} a\right)\left(b \minus{} c\right)}$
$ \angle NIK \equal{} 2\angle BKN \equal{} 2\alpha$
$ sin\alpha \equal{} \frac {tan\alpha}{\sqrt {1 \plus{} tan^{2}\alpha}} \equal{} \frac {pr}{\sqrt {\left(p \minus{} a\right)^{2} \left(b \minus{} c\right)^{2} \plus{} p^{2} r^{2}}}$
$ cos\alpha \equal{} \frac {\left(p \minus{} a\right)\left(b \minus{} c\right)}{\sqrt {\left(p \minus{} a\right)^{2} \left(b \minus{} c\right)^{2} \plus{} p^{2} r^{2}}}$

$ NK^{2} \equal{} 2r^{2} \minus{} 2r^{2}\cdot cos 2\alpha \equal{} 2r^{2}\left(1 \minus{} cos 2\alpha\right) \equal{} 4r^{2}sin^{2}\alpha\Rightarrow$
$ \Rightarrow NK \equal{} \frac {2pr^{2}}{\sqrt {\left(p \minus{} a\right)^{2}\left(b \minus{} c\right)^{2} \plus{} p^{2}r^{2}}}$

$ NB^{2} \equal{} NK^{2} \plus{} KB^{2} \minus{} 2\cdot NK\cdot KB\cdot cos\alpha\Rightarrow \frac {NB^{2}}{KB^{2}} \equal{} \frac {NK^{2}}{KB^{2}} \minus{} 2\cdot\frac {NK}{KB}\cdot cos\alpha \plus{} 1$
$ NC^{2} \equal{} NK^{2} \plus{} KC^{2} \plus{} 2\cdot NK\cdot KC\cdot cos\alpha\Rightarrow \frac {NC^{2}}{KC^{2}} \equal{} \frac {NK^{2}}{KC^{2}} \plus{} 2\cdot\frac {NK}{KC}\cdot cos\alpha \plus{} 1$

$ \angle BNK \equal{} \angle KNC\Leftrightarrow \frac {NB}{KB} \equal{} \frac {NC}{KC}\Leftrightarrow \frac {NK^{2}}{KB^{2}} \minus{} 2\cdot\frac {NK}{KB}\cdot cos\alpha \equal{}$
$ \equal{} \frac {NK^{2}}{KC^{2}} \plus{} 2\cdot\frac {NK}{KC}\cdot cos\alpha \Leftrightarrow$
$ \Leftrightarrow NK^{2}\cdot \left(\frac {1}{KB^{2}} \minus{} \frac {1}{KC^{2}}\right) \equal{} 2\cdot cos\alpha\cdot NK\cdot \left(\frac {1}{KC} \plus{} \frac {1}{KB}\right)\Leftrightarrow$
$ \Leftrightarrow \frac {NK\cdot\left(KC \minus{} KB\right)\cdot\left(KC \plus{} KB\right)}{KB^{2}\cdot KC^{2}} \equal{} 2\cdot cos\alpha\cdot\frac {KB \plus{} KC}{KB\cdot KC}\Leftrightarrow$
$ \Leftrightarrow NK\cdot\left(KC \minus{} KB\right) \equal{} 2\cdot KB\cdot KC\cdot cos\alpha\Leftrightarrow$
$ \Leftrightarrow \frac {2pr^{2}\left(p \minus{} c \minus{} p \plus{} b\right)}{\sqrt {\left(p \minus{} a\right)^{2}\left(b \minus{} c\right)^{2} \plus{} p^{2} r^{2}}} \equal{} 2\left(p \minus{} b\right)\left(p \minus{} c\right)\frac {\left(p \minus{} a\right) \left(b \minus{} c\right)}{\sqrt {\left(p \minus{} a\right)^{2}\left(b \minus{} c\right)^{2} \plus{} p^{2}r^{2}}}\Leftrightarrow$
$ \Leftrightarrow p^{2} r^{2}\left(b \minus{} c\right) \equal{} p\left(p \minus{} a\right)\left(p \minus{} b\right)\left(p \minus{} c\right)\left(b \minus{} c\right)\Leftrightarrow S^{2}\left(b \minus{} c\right) \equal{} S^{2}\left(b \minus{} c\right)$

So $ \angle BNK \equal{} \angle KNC$

That's all! :)

Dear friends,

Use the lemma $ \left(\angle BNK \equal{} \angle KNC\right)$ and prove the next problem:

Let $ ABC$ be a triangle. The incircle $ C\left(I\right)$ touch the sides $ BC$, $ CA$, $ AB$ at the points $ K$, $ E$, $ F$ respectively and $ M$ is the midpoint of the altitude $ AD$. If $ \left\{N,K\right\} \equal{} KM\cap C\left(I\right)$, $ \left\{X,N\right\} \equal{} BN\cap C\left(I\right)$, $ \left\{Y,N\right\} \equal{} CN\cap C\left(I\right)$ then prove that the lines $ MK$, $ XE$, $ YF$ are concurrent.  :)


vector
algebra
polynomial
linear algebra
linear algebra unsolved

Let V be an n dimension vector space over R T: V -->V  is a nilpotent Transformation
a-   Show that T^n = 0
b-   Show that I - T  is Invertible

a. T is nilpotent iff its characteristic polynomial is t^n, by cayley hamilton it's trivial.
b. $I+T+...+T^{n-1}$


f: X--->Y and g: Y----> Z are functions. Prove that

a) if both f and g are injective then g*f is injective

b) if both f and g are surjective then g*f is surjective

c) if both g*f is a bijection, then f is an injection and g is a surjection

( * is that small circle that means one function plugging into another)

A)

Suppose $ g*f$ is not injective, there exist $ a\ne b$, such that $ (g*f)(a)\equal{}(g*f)(b)$.

This implies $ f(a)\equal{}f(b)$ since $ g$ is injective, and $ a\equal{}b$ since $ f$ is injective. Contradiction.

a) Suppose that $ g(f(x)) \equal{} g(f(x'))$ for $ x,x'\in X$. Then, $ f(x) \equal{} f(x')$ because $ g: Y\rightarrow Z$ is injective, for $ f(x),f(x')\in Y$. Since $ f: X\rightarrow Y$ is also injective, $ x \equal{} x'$. Thus, $ g\circ f: X\rightarrow Z$ is also injective.

b) $ f: X\rightarrow Y$ is surjective, so $ f(X) \equal{} Y$. Also, since $ g: Y\rightarrow Z$ is also surjective, $ g(Y) \equal{} Z$. Therefore, $ g(Y) \equal{} g(f(X)) \equal{} Z$, which shows that $ g\circ f: X\rightarrow Z$ is also surjective.

c) <---- You try :P


1. Characterize all primes such that the Legendre Symbol (-75/p) = 1.

2. Prove there are infinitely primes of the form 6k+1.

For question 1, I was thinking of breaking 75 into its prime factors but then I don't know how to characterize all the primes which will satisfy that.

For question 2 we were given a hint that 'find primes such that (-3/p) = 1. When I tried to do that I got p = 3k +1. I don't know how this could be a help in this question.

[quote="aadil"]what is a legendre symbol?[/quote]

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


I just have doubts

1) Are all the charges fixed
2) Are the 3 charges pt charges
3) If you are connecting them to big sphere and removing, then does the bigger one lose its charge.

Pardesi please explain the question better da.

1)yes they are fixed
2)yes they can be assumed to be point charges
3)ofc

Man, post the answer to this and the passage sum of yours. I have no idea how the fact that they are on a equilateral triangle helps :(

And since they are pt charges, capicitance becomes pointless. :D

try to prove this ...the charges ae in GP...i will post the answer to passage :)


how can i write turkish character.
For example
ü=?
İ=?
ö=?
 $\bold{\vdots}$
 $\bold{\vdots}$

Accented letters are produced by a backslash command immediately before the letter.

For example, \'{e} produces é.

A list of commands is at http://www-ccs.ucsd.edu/latex/ltx-401.html
(or Google for "accents in latex")


linear algebra
matrix
calculus
integration
linear algebra unsolved

Let us denote with $H$ the set of all the square matrices of order $n\geq 2$ with elements positive integers and with $P$ the set of matrices in $H$ such that the sum of the elements of each line and each column is equal with 1.

a) Prove that if $A\in P$ then $\det A \in \{-1,1\}$;

b) Prove that if $A_1$, $A_2$, $\ldots$, $A_p$ are in $H$, and $A_1A_2\cdots A_p \in P$ then $A_k\in P$ for all $k=\overline{1,p}$.

I assume positive has the Romanian meaning here, i.e. non-negative :).

A matrix can only be magic if it has precisely one $1$ on each line and each column (and $0$ everywhere else). This means it's a permutation matrix, and these form a group (so each one of themn is invertible, and its inverse is another permutation matrix), meaning that their determinants, being integral, must be $\pm 1$.

It suffices to prove the second part for two matrices. The statement is that the invertible matrices in $H$ are precisely the permutation matrices. Since the matrices in question are invertible, it means that they both have at least one non-zero entry on each line and on each column. if $A,B\in H$, and $A$ had at least two non-zero entries on its first line, for example, then the first line of $AB$ would also have at least two entries, and this can't happen, because we want $AB=I_n$. It means that each of our matrices has precisely oine non-zero entry on each line and each column, and it's clear now that these entries must be $1$.


Let be $ a,b,c\in [1,2]$ . Prove that :

$ 3(a\plus{}b\plus{}c)\ge 6\plus{}ab\plus{}bc\plus{}ca$

$ (a\minus{}1)(a\minus{}2)\plus{}(b\minus{}1)(b\minus{}2)\plus{}(c\minus{}1)(c\minus{}2)\leq 0$ and $ a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca$ gives the proof.

Or you can say only that :

$ (a\minus{}1)(2\minus{}b)\plus{}(b\minus{}1)(2\minus{}c)\plus{}(c\minus{}1)(2\minus{}a)\ge 0$


How do we show the working for synthetic division? I mean, what's the $ \text{\LaTeX}$ code for synthetic division? Use the below problem as an example:

$ \frac{2x^4+21x^3+35x^2-37x+46}{2x+7}$

I don't think there's like a way to just show synthetic division...you might be able to use an array or something though...

My best suggestion is to use an array. Here's an example of the type of layout you can use:

$ \begin{array}{c|ccc}{a }& b & c & d & & e & f & g &  &  &  &  & \blacktriangleright & h & i & j \end{array}$

[code]\begin{array}{c|ccc}{a & b & c & d & & e & f & g &  &  &  &  & \blacktriangleright & h & i & j \end{array}[/code]

In the second curly-brackets, use one letter c for each column, and a vertical bar to create a vertical line. Use & signs to separate entries.

As long as we're on this subject, what would you do for normal division. (Using Latex, not text symbols.)

[quote="ThetaPi"]How do we show the working for synthetic division? I mean, what's the $ \text{\LaTeX}$ code for synthetic division? Use the below problem as an example:

$ \frac {2x^4 + 21x^3 + 35x^2 - 37x + 46}{2x + 7}$[/quote]Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=polynom]polynom package[/url]
There's lots of possibilities:
1. \polylongdiv{2x^4+21x^3+35x^2-37x+46}{2x+7}

[img]http://sixthform.info/gblog/pictures/ba339d02eae4e6d1df38ece70632349a.gif[/img]

2. \polyhornerscheme[x=-7/2]{2x^4+21x^3+35x^2-37x+46}

[img]http://sixthform.info/gblog/pictures/f51d74edc9ffce501292604822d2813b.gif[/img]

and even
3. \polyhornerscheme[x=-7/2,tutor=true,resultstyle=\color{blue},tutorlimit=10,stage=10]{2x^4+21x^3+35x^2-37x+46}

[img]http://sixthform.info/gblog/pictures/65babad67cdb0455f69fc3f293294b3d.gif[/img]

As usual for synthetic division, the quotient coefficients need to be divided by 2 in cases 2 and 3 above.


modular arithmetic
geometry
geometric transformation
function
least common multiple

Merhaba  :)  Sonuncu Rakam 1 olucak....

Translation : The Last Digit would be 1.

teşekkürler sonraki iki rakam ? :)
translation : thanks what are the next two digits? :)

umm sorry, but what does $ \lambda(1000)$ mean?

use what way to find $ \lambda(1000)\equal{}100$ ?

As shown in the article,
$ \lambda(1000) \equal{} \operatorname{lcd}({\lambda(2^3),\lambda(5^3)})$.
Since $ \lambda(p^k) \equal{} p^{k \minus{} 1}(p \minus{} 1)$ if $ p\geq3$ or $ k\leq2$, so $ \lambda(5^3) \equal{} 5^2(5 \minus{} 1) \equal{} 100$.
Also, $ \lambda(2^k) \equal{} 2^{k \minus{} 2}$ if $ k\geq3$, so $ \lambda(2^3) \equal{} 2^1 \equal{} 2$.
The least common multiple of $ 2$ and $ 100$ is $ \boxed{100}$.


Let $ a,b,c\in\mathbb{R}^*.$If $ abc(a^3\plus{}b^3\plus{}c^3)<\sum_{cyc}a^3b^3$ prove that exactly two of these ecuations have real roots: $ ax^2\plus{}2bx\plus{}cx\equal{}0;bx^2\plus{}2cx\plus{}ax\equal{}0;cx^2\plus{}2ax\plus{}bx\equal{}0$.

[hide]$ abc(a^3\plus{}b^3\plus{}c^3) \minus{} a^3b^3 \minus{} b^3c^3 \minus{} c^3a^3 < 0,$
$ (a^2\minus{}bc)(b^2\minus{}ac)(c^2\minus{}ab) < 0$
So one or three of these is negative. If all three are negative, $ a^2 \plus{} b^2 \plus{} c^2 < ab \plus{} bc \plus{} ca$, which is ridiculous. So one is negative, thus exactly one equation has nonreal roots.[/hide]


[quote="Yosh..."]Let $a,b,c$ be positive real numbers.
Prove that $2(a^{3}+b^{3}+c^{3})+3(ab^{2}+bc^{2}+a^{2}c)\geq 3(a^{2}b+b^{2}c+ac^{2})+6abc$[/quote]
You can use
$\sum_{cyc}a(a-b)^{2}\ge 0$ and $\sum_{cyc}a^{2}b\ge 3abc$


Can someone explain what exactly the "MiKTeX Package Manager" does and what it lists? For example, I can use the longtable package perfectly well, but it is not listed in the MiKTeX Package Manager. I found it under C:\Program Files\MiKTeX 2.7\tex\latex\tools.

I remember in the past finding useful packages mentioned on the internet, and, upon not finding them listed in the "MikTeX Package Manager", gave up.

Hi,

the longtable package is contained in the [url=http://miktex.org/pkg/pkginfo.aspx?dn=tools]tools bundle[/url].

If you're looking for a certain package and you don't find it listed in the package manager, search on [url=http://ctan.org]CTAN[/url]. On CTAN you will also geht information if a package is delivered with MiKTeX and to which bundle it's belonging, see for instance [url=http://ctan.org/pkg/longtable]longtable on CTAN[/url].

Concerning the mpm docs have a look at the [url=http://miktex.org/docs]MiKTeX documentation[/url], for instance here: [url=http://docs.miktex.org/manual/pkgmgt.html]Working With the Package Manager[/url], but there's more to read in the manual.

Stefan

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