limit
calculus
ratio
geometric sequence
calculus computations

Find if the following series converges or diverges:
Σ(n!)/(n^n)

My prof. suggested we compare it to n^(n-2)/n^n.

I found that n^(n-2)/n^n was convergent since it was like 1/n^2 which is a p series where p>1.  Therefore by the comparison theorem, the original series is convergent.

I'm hoping this is the right answer, but I thought it was too easy for what was deemed a tricky problem, even with the hint.

Let's use D'Alembert test:

$ \lim \frac{a_{n\plus{}1}}{a_n} \equal{} \lim \frac{(n\plus{}1)! \cdot n^n}{(n\plus{}1)^{n\plus{}1} \cdot n!} \equal{} \lim \left(\frac{n}{n\plus{}1}\right)^n \equal{} \frac{1}{e} < 1$

and so the series converges.

Thanks.  We haven't actually learned D'Alembert's in my class.  We're halfway through Calculus II

The terminology in a typical American calculus book is "ratio test," without mentioning the name of D'Alembert. And that problem (or its close relative, determining the radius of convergence of the power series $ \sum_{n\equal{}1}^{\infty}\frac{n!x^n}{n^n}$ ) is in all of those calculus books - in the very section in which the ratio test is introduced.

What you've learned from the work Carcul did is this: $ n!$ can be written as $ n!\equal{}n^ne^{\minus{}n}g(n),$ where $ g(n)$ cannot usefully be compared to a geometric sequence - that is, it grows or shrinks at some slower-than-geometric rate.

awesome.  we still havent done that, although its probably on the agenda.

it might be something we cover in class today.   either that, or i've really been out of it.

you can use stirling $ n! \sim \sqrt{2\pi n}n^{n}e^{\minus{}n}$

Moubinool, we're assuming these students have not yet learned Stirling. In fact, as I indicated in my last post, applying the ratio test to this implies a weak version of Stirling.

If the question was to determine the exact interval of convergence of $ \sum_{n\equal{}1}^{\infty}\frac{n!x^n}{n^n},$ the ratio test would be sufficient to determine that the radius of convergence is $ e.$ However, to decide convergence at $ \pm e,$ we would need something stronger, such as the full version of Stirling.


Hello piever,
[quote="piever"]Let $A=\{a_{1},a_{2},...,a_{n},...\}$ be a subset of $\mathbb{R}$ dense in (0,1)

$b_{n}=\{\sum_{i=1}^{n}a_{i}\}$ where $\{ x\}$ is the fractional part of $x$

Let $B=\{b_{1},b_{2},...,b_{n},...\}$

Is $B$ dense in (0,1) ???[/quote]

I don't know if B may be dense. But I know it may not :

Let $h(n)$ be a bijective function from N in $\mathbb{Q}\cap (0,\frac{1}{2})$

Let $a_{2p+1}=h(p)$ and  $a_{2p+2}=1-h(p)$ $\forall p \in \mathbb{N}$

Clearly A is dense in (0,1)

Then we have $b_{2p+1}=a_{2p+1}=h(p)$ and $b_{2p+2}=0$

And B = $\{0\}\cup (\mathbb{Q}\cap (0,\frac{1}{2}))$ and is not dense in (0,1) since B has no element in ($\frac{1}{2}$,1)

--
Patrick

$A=\mathbb{Q}_{2}\cap(0,1)=\left\{\frac{k}{2^{n}}: \ k,n\in\mathbb{N}\land k<2^{n}\right\}$ may be a good example for dense $B$.


articles
rotation
geometry
geometric transformation
3D geometry
sphere
Euler

The Vacuum and Special Relativity Theory.
Special Relativity Theory examines the behaviour of a quantum of light in the vacuum.
1) The First law of SRT - the speed of a quantum of light in vacuum has
 a maximal magnitude (constant, absolute) of c=1.
2) SRT is not considered a gravitational field.
 For this reason, Einstein created General Relativity Theory in 1915.
 The field in which there is no gravitation is a vacuum.
3) This asserts that action in SRT occurs with particles in
 negative four-dimensional (Minkowski) space. This space is absolute.
Mathematicians have constructed its model and speak
of this negative space as completely abstract.
 Nobody sees that it has no connection to real existence.
 This is similar to a sad joke.
For 100 years everyone has admired SRT.
 Millions of articles, reviews and books have been written
 and the United Nations has decided to establish 2005 as the centennial year of SRT.
 Consider that all that is clear in this theory
 is that negative four-dimensional space is abstract and has no real existence.
My God!  There does not appear to be anyone to laugh at this joke!
Everyone searches for complex models of four-dimensional space, but truth lies in simplicity. All is very simple.
We meet the negative characteristic of space only in the vacuum, and in the vacuum,
 space is merged with time (negative four-dimensional space).
 According to the first law, the speed of light is absolute and movement occurs
 in the absolute vacuum. So why does everyone speak and write that there is no
 absolute movement; that only relative movement v =s/t is real?
 Why does everyone say that there is no absolute reference system,
if the absolute speed can be only in absolute space?
Here we have one of the paradoxes in human intelligence.

Pythagoras, theory, Electrodynamics and SRT.
Pythagoras' theory applies equally to the largest and smallest triangle.
 So mathematicians had decided, that this applies also to
the electromagnetic phenomena;
 that the laws of a nature in the macrocosm and in a microcosm are identical.
It appears that this is not so.
In the macrocosm, Maxwell's laws apply
 and in the microcosm,  other laws, the laws of SRT operate.
These laws are interconnected.
SRT is a continuation of the development of electrodynamics.
*             *            *
 1)What is the make up of the electron in Maxwell's theory?
 Maxwell's equations have no relation to the movement of the electron.
They describe the distribution of electromagnetic waves
but not the movement of a particle such as an electron.
In Maxwell's theory, the charge - electron is considered local,
 as though the particle is "at rest".
This means that it particle does not move rectilinearly,
 but rotates around the diameter (has the form of a sphere).
 The rotation of the electron creates electrical waves.
*             *            *
 2) What is the make up of the electron in SRT?
 At the beginning of the last century many scientists
(Einstein, Lorents, Fitzgerald, Poincare, Abraham) were interested in the question:
What will take place, if the electron (Maxwell's) , creating an electrical field,
begins to move - rectilinearly?
All of them came to the conclusion that there would be radical changes with the electron.
 These changes are described by the Lorentz transformations.
 That is, when the originally rotating electron (sphere) begins to move rectilinearly,
during movement it gradually will change its geometrical form.
 Having reached constant speed of  c=1, its form will become a circle.
 In such condition it is called a "quantum of light ",” photon”.
 And when a quantum of light rotates around its diameter its name is “electron “
 An "electron" is an actively working “quantum of light”.
With such an interpretation, electrodynamics and SRT become one general theory.

The Vacuum and the Electron
All know, that an electron is not a firm sphere. All know, that its form can be changed.
 But nobody understands the borders of the change of the geometrical form of the electron.
So, what are the borders of this change? Quantum theory gives an answer to this question.
It says that at the interaction of the electron with the vacuum, the energy and mass of the
 electron become infinite. Physics does not understand what to do with infinite sizes
and therefore have thought up "a method of renormalization", a method
 "to sweep the dust under the carpet" / Feynman./
 But the situation can be understood another way. Electrons, having the geometrical form
of a sphere, lose their volume and turn into an indefinitely flat circle. In this is the reason
for the occurrence of infinite sizes for the electron. But in physics we know
only one particle which has the form of a flat circle. It is a quantum of light
 which flies rectilinearly   with  speed c= 1.
Hence, the electron turns into a quantum of light. Hence, the electron and a quantum of light
 is the same particle in different states.
In the books it is written, that electrons interact among themselves with the help of
a quantum of light. In the books it is written that an electron in an atom passing from
one orbit to another radiates a quantum of light. It should be understood as follows.
 The electron has a quantum of light in a pocket or under a "shirt" which from time to time
 is freed. Interesting. But why is it necessary for it to hide?
*             *            *
If you have time and desire, I ask you to visit my site
http://www.socratus.com
  Best wishes.
Socratus.

Dualism of Light Quanta .
A quantum of light is a privileged particle.
No other particle can travel with the speed (c = 1).
 There are two kinds of spins, as a result of which the particle attains motion.
1)Under the action of  Planck,s spin, which is equal to the unit ( h =1)
aquantum of light flies rectilinearly with speed (c = 1).
 The geometrical form of a circle: (C/D = 3,14).
 A quantum of light behaves as a particle.
2) Under the action of Goudsmit-Uhlenbeck's spin, ( ħ = h / 2) a quantum of light
rotates around of the diameter and is known as electron.
The geometrical form of a circle is transformed into a sphere.
This kind of movement is described by Lorentz's transformations .
The wave properties of light quantum are shown.
The dualism of a particle becomes clear and the paradox disappears completely.
When the form of a circle is change into the form of a sphere, the transcendental
magnitude ( = 3,14) is change on another transcendental magnitude (е = 2,71).
The transformation of a circle to a sphere (spin) is described by
the geometry of  N. Lobachevsky, F. Klein, A. Poincare.
 The transformation of sphere into circle is algebraically
 described by a formula, which was discovered by L. Euler in 1748.
cos j + i sin j = e ij

Someone in the physics forum might like this..


uhm, I think :
73 + 12=85
97 - 12 =85
And
90 - 5 = 85
79 + 6 = 85
81 + 4= 85
U see, we add 10 and subtract 5
Then each of 85 subtract 1 and we have 84

Good job! Can anyone think of any other ways?

[quote="Nguyen Ngoc Linh"]Nikki needs 27+21+19+15+10+3=90 more points to get 600 points, so his average point is 100- 90/6=85[/quote]

you did your addition incorrectly, it should be 95; also the first time you also had to average 84 with the one 85

[quote="isabella2296"]Find the mean of Nikki's test scores: $ \{73, 79, 81, 85, 90, 97\}$.

See if you can find a cleverer way than just using the formula to solve this! No calculators allowed!  :)[/quote]

Pair them up so the unit's digits add up to 10. $ (73\plus{}97)\plus{}(79\plus{}81)\plus{}85\plus{}90\equal{}170\plus{}160\plus{}85\plus{}90\equal{}255\plus{}250\equal{}505\implies\boxed{84\frac16}$...

I summed them with a calc and got 505 also...

errrr...did any of you actually read the original question?

[quote="QuincyCello"][quote="math154"][quote="isabella2296"]Find the mean of Nikki's test scores: $ \{73, 79, 81, 85, 90, 97\}$.

See if you can find a cleverer way than just using the formula to solve this! No calculators allowed!  :)[/quote]

Pair them up so the unit's digits add up to 10. $ (73 \plus{} 97) \plus{} (79 \plus{} 81) \plus{} 85 \plus{} 90 \equal{} 170 \plus{} 160 \plus{} 85 \plus{} 90 \equal{} 255 \plus{} 250 \equal{} 505\implies\boxed{84\frac16}$...[/quote]

wouldn't that just be adding them up and then dividing, like normal? just adding them more easily...[/quote]

ALL of these solutions posted above are "the normal method", just in different ways.

Nguyen Ngoc Linh did what I usually do, but divided by 5 instead of 6.

He appears to have divided by 6.

At least, in his second post, he did.

[quote="Nguyen Ngoc Linh"]uhm, I think :
73 + 12=85
97 - 12 =85
And
90 - 5 = 85
79 + 6 = 85
81 + 4= 85
[b]U see, we add 10 and subtract 5
Then each of 85 subtract 1 and we have 84[/b][/quote]

Unless I'm mistaken, he said the sum must increase by 5 in order for the average to be 85. Then, he divided 5 by [b]5[/b] to get 1, and subtracted. He should've divided by 6.

EDIT: Sorry. :oops:

Which is why I said [b]his second post[/b]. Yeah, I noticed the first one. :wink:

{73,79,81,85,90,97}

Method1
1) Pair them up .... largest with smallest, next largest w/ next smallest,etc...
73 & 97, 79 & 90, 81 & 85

2) Add the pairs:
73+97 = 170
79+90 = 169
81+85 = 166

3) Subtract the smaller sums from the greatest sum
170 - 169 = 1
170 - 166 = 4

4) Divide the greatest sum by 2
170/2 = 85

5) Subtract from 4) the Sum of the Differences in 3)divided by how many
85 - (1+4)/6 = 84 and 1/6

Method 2
step1 and 2 same as in method 1

3) Divide the sums by 2
170 : 2 = 85
169 : 2 = 84.5
166 : 2 = 83

4) Try equalizing the quotients in step 3
85-1 = 84
           84.5
83+1 =84

5) So we see the mean is a little more than 84 but less than 84.5
take the extra 1/2 and multiply with 1/how many (3)
(1/2)*(1/3)= (1/6) ...

6) 84 + 1/6 is the mean.

Fun?  :trampoline:


Daca $a\in\mathbb{R}_{+}^{*}$, sa se rezolve ecuatia:

                                        \[ 3a^{2x}=4a^{3x}-a^{6x}.  \]

  [i] Tudorel Lupu [/i]


Find the biggest positive integer $ n$  such, that if the modula of a difference of any 2 of 3 positive numbers $ x, y, z$ is less than $ n$, the inequality accomplishes $ \sqrt {xy \plus{} 1}\ \plus{} \sqrt {yz \plus{} 1}\ \plus{} \sqrt {zx \plus{} 1}\ > x \plus{} y \plus{} z$

Well if you mean the biggest $ n$ such that for any positive $ x$, $ y$ and $ z$ wich satisfy the condition the inequality holds.
You can set $ x \equal{} y \equal{} 1$ and find a bound for $ z$, if I haven't misscalculated it is $ 3\ge z$ so $ n\le2$ and it shouldn't be so hard from here (maybe).

 :)


In inscribed circle $ADCB$ $AB$= 87 $BC$ =105 $CD$ =116 and the radius of circumscribed circle is 72.5. Find $AD$

JIMBO!!!!

Ptolemy's on $ABCD$, the cosine's law and the sine's law to get the diagonals. $AD = 100$.


When the "You have a new private message" window pops up on login, it contains the entire Forum page rather than just the notification text. Seems like a bug. Firefox 2.0.0.13, Windows XP (unfortunately, I couldn't take a screen shot this time. Maybe later :))

Yes, AoPS skin. Sorry for not mentioning that...

Hmm? I recently recieved a PM and it doesn't look any different than it has ever been. It's just a small window, AoPS skin, with a box of text with a notification.

Firefox 2.0.0.12 and Windows XP here.

OK, I have it sitting on my screen. To my shame, I just realized that I have no idea of how to make a screenshot in Windows XP :blush:. If someone tells me how to do it, I'll attach the picture :)

Take a look here:
http://www.wintipz.com/XP/TakingScreenshot.htm
:)

OK, here goes. It seems like the same background as in the "allowed extensions and sizes". The box with notification is there and can be viewed if you increase the size of the window but the main part is occupied by a piece of the forum layout.

Did it happen once or it happens every time?

[quote="fedja"]Hard to tell. I do not login too often. It happened at least twice. If I see it again, or if I see a normal window, I'll tell you. Hey, why don't I log out for 20 minutes or so and you send me a PM? Let's try. I'm out.[/quote]You don't need to be logging in and out to get the PM window. You just need to be browsing the site.

Also you can send yourself a PM and test it (after sending the PM, go to index rather than your inbox to get the popup).

OK, it happened again. And again, and again. So, I would say every time.

What is the notification box supposed to look like? As far as I can tell, it has been that way ever since I joined AoPS.

Yeah, this is what it looks like.  I have it open a new window.
Note that it is almost impossible to click on the "Click [u]here[/u] to visit your inbox" as the menu covers that up.

[quote="7h3.D3m0n.117"]What is the notification box supposed to look like? [/quote]
It used to be just a small window with white backgroud with the message in it. I haven't paid much attention to what it was lately, so, perhaps, it started last year rather than yesterday.

It happened to me too... (Firefox 3.0b4 on Windows XP)


geometry
calculus
integration
complex analysis
function
number theory
complex numbers
\/closed

I have a proposal-
why dont we have a forum (sub-forum) on complex/imaginary numbers?
Or do we already have one and I didn't look hard enough...

Questions about complex/imaginary numbers should probably go in the High School Intermediate forum...

Notice that the forums are grouped by content topic, not by mathmatical topic. Let's try to stay consistent.

I think the OP may be referring to the [url=http://www.artofproblemsolving.com/Forum/index.php?f=217]Olympiad Section[/url]...in that case, complex and imaginary numbers would go under algebra, geometry, or number theory, depending on the problem...I agree that complex numbers deserves a section there.

There's a complex analysis forum in the College Playground section, so questions involving contour integrals, holomorphic functions, conformal mappings, and the like would go there.

Of the non-calculus complex number questions I can think of, I think that about 80% of them, if high enough level, belong in the Olympiad Algebra forum and most of the rest in the Olympiad Geometry forum - and I think such questions fit well in those forums. I don't think it would be a good thing to pull good honest algebra questions out of the Algebra forum just because complex numbers may be involved.

There are hardly any "elementary" problems on complex numbers that are interesting on it's own and aren't just a special case of algebra in (alg. closed) fields. So there is in fact no reason to have such a forum.


geometry
3D geometry
calculus
integration

Find all ordered pairs of integers (x,y) which satisfy

$x^3+y^3-3x^2+6y^2+3x+12y+6=0$

This question is supposed to be solved in 4 minutes. So the sleekest solution is preferred. Thanks :)

Yeah I see how to do it now, I was trying to do some grouping factoring deal and it sorta worked but was majorly ugly. I've never worked with completing the cube, but once you mentioned it I saw how to do it.

[hide]
$x^3-3x^2 + 3x +$ __

We want to get that into

$(x+y)^3$ which expands into $x^3+3x^2y+3xy^2+y^3$ There we can obviously see that $y=-1$

So we have

$(x-1)^3 + 1$

Now

$y^3+6y^2+12y +$ ___

Using same method we see $y=2$ So we have

$(y+2)^3 - 8$

Substitute back into problem

$(x-1)^3+1+(y+2)^3-8+6=0$

$(x-1)^3+(y+2)^3=1$

There we can see that happens with $(2,-2)$ and $(1,-1)$
[/hide]

Nope, it's +6. You can narrow down the possibilities fairly quickly (because remember in the problem it states integral answers) in that one of the cubes has to be 0 and the other has to be 1.

EDIT: Hmm, I'm not sure what you're asking because I could have sworn I wrote +6 in the original problem, but then i looked at it and it said +9 (So I changed it back). But +6 is the correct problem.


function
integration
calculus
real analysis
real analysis unsolved

Determine a primitive of the function:

\[ \boxed{f(x)\equal{}\frac{(2n\minus{}3)x^5\plus{}(n\minus{}1)x^3\minus{}5x}{(x^4\plus{}5)^n}}\]

where $ n \in \mathbb{N}.$

Hey moldovan,could you please explain to me(I am sorry i don't know) what do you exactly mean when you're saying "primitive of a function".Thanks and sorry about that.

$ \int$ $ \Rightarrow$ antiderivative, primitive or indefinite integral of a function; a function $ F$ with the property that $ F'\equal{}f.$


[b]Νομίζω ότι αυτή την άσκηση αξίζει να τη γνωρίσουν και όσοι δεν έτυχε μέχρι τώρα να την συναντήσουν :[/b]

 [b]B. 4006.[/b]  Let (έστω) $ a, b, c$ are the sides of a triangle ( οι πλευρές ενός τριγώνου), such that(έτσι ώστε) $ a\plus{}b \equal{} 2c$. Prove that the centres of the inscribed and circumscribed circles are concyclic with the midpoints of the sides $ a$ and $ b$.( Να αποδείξετε ότι το έγκεντρο , το περίκεντρο  και τα μέσα των πλευρών a,b  του τριγώνου αυτού είναι ομοκυκλικά).

(4 points)

 Είναι από το διαγωνισμό του περιοδικού για το 2007(Μάϊος, νομίζω).

Πολύ ωραία άσκηση, κ. Στεργίου.
Έστω $ D$ το ίχνος της καθέτου από το $ I$ στην $ AB$, $ K$ το μέσο της $ CA$, $ M$ το μέσο του τόξου $ \widehat{BC}$ και $ N$ το αντιδιαμετρικό του ως προς τον περιγεγραμμένο κύκλο του τριγώνου. Θα είναι $ AD\equal{}\frac{b\plus{}c\minus{}a}{2}\equal{}\frac{a}{2}\equal{}BM$ και $ \angle DAI\equal{}\frac{A}{2}\equal{}\angle KBM$ άρα τα ορθογώνια τρίγωνα $ DAI, KBM$ είναι ίσα, συνεπώς $ MK\equal{}ID\equal{}r$. Τώρα από το ορθογώνιο τρίγωνο $ MBN$, αφού $ BK\perp MN$ έχουμε $ BM^{2}\equal{}MK*MN\equal{}2Rr$, άρα $ IA^{2}\equal{}BM^{2}\equal{}2Rr$. Επομένως στο τρίγωνο $ OIA$ ισχύει $ OA^{2}\equal{}R^{2}\equal{}R^{2}\minus{}2Rr\plus{}2Rr\equal{}OI^{2}\plus{}IA^{2}$,οπότε $ \angle OIA\equal{}90$.
Δηλαδή το $ I$ ανήκει στον κύκλο με διάμετρο $ OA$ που είναι ο περιγεγραμμένος κύκλος του τριγώνου που σχηματίζεται από το $ O$ και τα μέσα των $ b,c$.

Η άσκηση αποτελεί άμεση απόρια της άσκησης εδώ (δείτε το προτελευταίο ποστ για τη λύση μου)
http://www.mathlinks.ro/viewtopic.php?t=164835

και αυτό γιατί αφού η DE διχοτομεί τη διάμεσο ,η παράλληλη από το σήμειο τομής τους θα τέμνει τις ΑΒ,ΑC στα μέσα τους έστω P,Q . Τότε $ \angle{POQ}\equal{}180\minus{}A$  και $ \angle{PIQ}\equal{}\angle{DIE}\equal{}180\minus{}A$ άρα $ PIOQ$ εγγράψιμο .QED


counting
derangement
function
combinatorics unsolved
combinatorics

Say I have four objects, ABCD, and I randomly order them. The number of ways to do this is 4!. Now, let's say I want to find the number of ways my random ordering obtains exactly one match/hit with the original series of ABCD... For example, ADBC is Hit Miss Miss Miss.

The number of ways to do that would be (4 choose 1) * d[3], where d[n] is defined as the derangement of n integers.

Okay, all well and good so far.

Now say I have three sets of the same four objects in the same order:
 ABCD    ABCD   ABCD

Say within each set, I randomly order the four objects. One example result, then would be:
 DBCA    ABDC   DBCA

but the following set would **NOT** be one under consideration: DDDA AABB BCCC, because I want to keep the randomized order within sets (so no two objects of the same type could exist in any given set).

The total number of possible orderings is 4! * 4! * 4!.

[b]In how many ways can I obtain exactly five matches across all sets?[/b]

For example,
 ABCD    ADBC    BDAC
has exactly 5 matches (HitHitHitHit HitMissMissMiss MissMissMissMiss)

(I chose five matches since I think this will be the most clear -- with exactly one match, one need only choose which of the sets the match will occur in, and then perform the rest as the basic case above; this would be (3 choose 1)*(4 choose 1)*d[3]*d[4]*d[4].)

[quote="bugmenotplease"](I chose five matches since I think this will be the most clear -- with exactly one match, one need only choose which of the sets the match will occur in, and then perform the rest as the basic case above; this would be (3 choose 1)*(4 choose 1)*d[3]*d[4]*d[4].)[/quote]

In the more general case, you're doing the same thing except summing a similar expression over all solutions to, say, $a_{1}+a_{2}+a_{3}= 5$ with $0 \leq a_{i}\leq 4$.  This does not obviously result in anything nice, but one never knows.  There are enough variables (number of blocks, number of letters in each block, total number of duplications) that guessing a general form, if one exists, seems unlikely.  It is possible there is a nice generating function associated with it.  Another possibility is to view the problem in a different (even more general) light, namely enumerating by number of fixed points a certain family of restricted permutations (in this case, restricted so that certain blocks of four values are permuted among themselves), and hope that there is some method already known that does the trick.


Im having a brain fart here. Lets say u bought the same product twice, the first time you bought 1000 Lbs for $500, and the second time it you bought 1200 Lbs for$600. What is the best and most acurate way to get the average cost? I cant decide which is right, do I

A. ( (500/1000) + (600/1200) ) / 2

B. ((500+600)/2)  /  ((1000+1200)/2)    -- this is weighted, correct?

C.  ((500 + 600) / (1000+1200))  / 2

Thanks for the help

The average cost per pound? In both cases, it's exactly 2 dollars, so the average price is the price of 1.

Also, average in what sense?

Ok... the numbers I gave were a bit easy I guess. But forgetting that, Im trying to figure out which is the best way to show the average of both prices. Im not sure what sence of avereage would be best... mean, mode, median? Which formula would I use to get the answer? Thanks for the help.

B is correct, C is almost correct, just take off the "/2".

I mean the average price per (insert unit(s) here).

If you have a unit n pounds, the price of it is n/2.

So if you have 100000 pounds, the price is 50000

So the weighted average:

  B. ((500+600)/2) / ((1000+1200)/2)

this is the best way of averaging multiple  $\$$ \Lbs Prices?

Pro tip:

$\text{ Average price }= \frac{ \text{Total money spent}}{ \text{Total quantity bought}}$

In any context this is the definition of "average" (as far as phrases such as "average speed" are concerned).

What is the best way to average? is there a better way?

That is nearly always what "average" means, and there's only one way to interpret it.

$\text{Total money spent}= 500+600 = 1100$
$\text{Total quantity bought}= 1000+1200 = 2200$

Hence the average cost is

$\frac{1100}{2200}= 0.5$

(In units of pounds per... dollar, I presume.)


geometry
3D geometry
combinatorics unsolved
combinatorics

A cube must be painted, with each face of a color, using exactly 5 colors, where the 2 faces of same color must be opposites. In how many ways this can be made?

Let put some indexes for the colors (a,b,c,d,and e) as a dice
WLOG the two faces with the same color (e) are f1 and f6 and (d) is in f2
the only cases are for f3, f4 and f5
a , b ,c ( b, a, c) is similar by rotating the cube.
a, c, b
b, c, a

therefore are 5 (colors for f1 and f6) * 3 possibilities = 15

Hmm... well, I didn't find now any mistakes on your solution, but the options for the question are:

a)720
b)6
c)360
d)120
e)30


integration
calculus
limit
complex analysis
complex analysis theorems

Let $H(j\Omega)=\int_{-\infty}^{\infty}h(t)e^{-j\Omega t}dt$ ,$\forall\Omega \in {\mathbb R}$ be the Fourier transform of $h(t)$.
Could someone help me to prove this invertion theorem: $h(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}H(j\Omega)e^{j\Omega t}d\Omega$ ,$\forall t \in {\mathbb R}$

It's not enough to say "this inversion theorem." You need to specify quite a few things.

What are the properties of $h$? The transform integral is unambiguously defined if $h\in L^1(\mathbb{R}),$ and results in a continuous $H.$ Or would you like to weaken that, perhaps to $L^2?$ Or are there stronger conditions you need to impose?

Then, in what sense you you want the "synthesis" integral (your second integral, the one that's supposed to give you $h$ back) supposed to converge? Absolute integrability would be too much to ask; it would exclude too many interesting cases. How about pointwise convergence of the Cauchy principal values? That is, can we say that

$h(t){?\above =}\lim_{M\to\infty}\frac{1}{2\pi}\int_{-M}^MH(\Omega)e^{i\Omega t}d\Omega?$

(I've made two notational changes: I've changed the name of $\sqrt{-1}$ from $j$ to $i$, and I've made $H$ have real arguments.)

I could prove the following: if $h$ is integrable and of bounded variation, then the integral above converges everywhere to the average of the left and right hand limits of $h.$ The proof is longer than I'm willing to write here.

There are also cases in which this doesn't work, and alternate forms of extracting meaning from the integral (Abel means, Cesaro means). It's one of the most famous and well-studied problems of the 20th century.


Don't you mean "mean" instead of "man"?

[hide]
Total original score: 56
Total score after droppping of highest and lowest scores: 39
Total Dropped Scores: 56-39=17
With 10 as highest, 7 is also removed.
7

[/hide]

[hide]I got 7... Just subtract 39 from 56 to get 17. The only two ways to get 17 are 8+9 abd 10+7, making the lowest score 7.[/hide]

Yeah, hockey had a good idea that I had too, except when I first did this problem, I got it wrong because of a computation mistake.

This was from the Mathcounts coaching kit, at least I got this problem from there.

[quote="MCrawford"]Seven judges graded Joshua's science project, and the mean of those seven scores was 8.0.  Scores could be any integer value 1-10.  What the highest and lowest scores were removed, the mean of the five remaining scores was 7.8.  What was the least possible score that could have been removed?[/quote]
[hide="Solution"]
They add up to 56

They add up to 39

7  (39)  10

[b]7[/b]
[/hide]

[hide]8 x 7 = 56
7.8 x 5 =39
56 - 39 = 17
Highest has to be a ten, so:
10 + [b]7[/b] = 17
7 is the answer.[/hide]


Let's ask ourselves,"where does humor come from?", and analyze and debate whether or not humor is an advantage from an evolutionary perspective.

uh, humor reduces stress (good) and burns calories (yes?) and strengthens social relationships and communication (good), and yeah.

[quote="pianoforte"]strengthens social relationships and communication (good)[/quote]

This would be my guess.  I hypothesize that humor is a complex social behavior and that its emotional facets are probably unique to primates.  This is all totally unsupported by research (especially because it's difficult to judge whether a laugh response - which is present in many animals - actually indicates what we would call "humor"), but apparently there is a name for the study of laughter, which is [b]gelotology.[/b]

Recently studies were undertaken to measure laughter, in terms of the aH scale. Generally people who laugh to the levels of ah 10 were found to be healthy. Its not officially established yet, but its in the reckoning.

Humor,like any other interaction between humans,is a part of evolution.There's no advantage or disadvantage for evolution.Evolution(let's state it for humans) is simply the process of how humans change as what follows after the interactions.Of course there are humans that consider somehow that the past was "bad"-i don't blame them but this is not the most objective view(and we search the most objective view as being the closest to reality,right kids? >:| ).

"where does humor come from?"
I understand the question:what is the process of humor?=how does it appear?how come 1 human says something humorous and the other one reacts to it ?you're asking how does it works?...soon after the break!


I have been doing a lot of Olympiad training over the past year and have noticed that I fail miserably at combinatorics. This became especially obvious at MOP, when I was one of ~two (?) people that did not get this extremely trivial combinatorics problem that came up on a test. So the title says a lot-How to stop failing at combinatorics.

I am half-decent at doing AIME combinatorics problems-I can usually get it if it is a #10 or below. However, I have never ever ever in my entire life solved an Olympiad combinatorics problem (and probably never will if I don't do something about it). So I raise the question:

What book/class would be best for me? I REALLY need to brush up on my combo skills.

I highly recommend [url=http://www.amazon.com/Proofs-that-Really-Count-Combinatorial/dp/0883853337]this book[/url].

What's the difference?  In any case, it's not an Olympiad training book.  It's just a very good exposition on a general technique with a lot of examples and problems, some of which are open.

[quote="varunrocks"]That book is USAMO level right or is it MOP level?[/quote]

Is there a difference?

wait dang you should not be using the computers at MOP
also, I got a -0 on a test I think it was test 3 because I was being extra fat

.... gooood for you, Ubey.

I think the only way to not fail at combinatorics problems is to try to do combinatorics problems.

I've been gaining mathematical confidence recently and I've only been doing problems.

Combinatorics, as I see it, is the topic that's least about knowing stuff and most about problem-solving.  This is especially true because most of the miscellaneous unclassifiable problems get labeled as combinatorics.  You really have to play around with the problem, get a feel for small cases, and trust your intuition.  Unfortunately, getting good at these types of problems is all about doing these types of problems, but you can try to apply some of the same strategies that you use for different types of problems.

And don't worry about missing a "easy" problem on a MOP test.  I once spent two hours of a MOP test working on the same problem and not solving it.  Later I found out that it had a really simple, one-line solution, and I wanted to hurt myself.  Don't worry about it, even if you are the worst person at MOP (which I highly doubt), that's not exactly a bad title to have :D.

@ Ubemaya: If you were actually being fat, you wouldn't have gotten a negative 0.  :maybe:

See my recent blog post - I don't PWN at combinatorics as much as the poster before me does, but I find that putting down the pencil to think comes in most handy in combinatorics. Don't go crazy about drawing lots of diagrams or whatever on paper, I almost always think of the key idea while just sitting and thinking. Perhaps it helps some people to write down their thoughts, but for me it's always kind of slowed me down...

But this isn't to say do the entire thing in your head - testing small cases is really important and sometimes it's impossible to make key conjectures without properly analyzing small cases.

Also I guess this is true for any math problem, but it probably especially holds true in combinatorics - you can often dig yourself a hole if you continue on a wrong/unhelpful train of thought for a while, so perhaps try to rephrase the problem in a way that may prompt you to think in a different way. I can't really think of a good example now of when this would help (well I can think of a geometry problem...), but hopefully you catch my drift.

[quote="blackbelt14253"]Thanks everybody!  :)
What CatalystofNostalgia said probably explains why i fail at combo...unless I'm writing stuff down, I never figure out anything. But in combo, there is not much to write down...

[quote="Ubemaya"]wait dang you should not be using the computers at MOP[/quote]
wait dang YOU should not be using the computers at MOP[/quote]

Well, I've never gone to MOP (hope to though) and I still pretty much suck at Olympiads, but I've noticed (and especially for the AIME, so I'm guessing definitely the USAMO then?) that if I write out my thoughts, spend a majority of the time thinking, and write slowly, I usually find a solution faster.
Then again, we are probably at different levels of problem solving, so whatever works best for you.


group theory
abstract algebra
invariant
superior algebra
superior algebra unsolved

1. Let $G$ be a finite group of an odd order.
    Prove that if $3$ divides $|G|$, but $9$ does not divide $|G|$ then
    there exists a normal subgroup $H$ in $G$ such that $|G: H|=3$.
2. Let $P$ be a  Sylov p-subgroup in a finite group $G$. Prove
    that if $P$ is a subgroup of $Z(N_{G}(P))$ then there exists a
normal subgroup $H$ in $G$ such that $H \cap P=\{e\}$ and $HP=G$.

For problem 1:
Can we prove:
$N=\{g|$the order of $g$ can't divide by $3$$,g \in G\}$
is the subgroup we search.? :?:

We do the second problem first. Note that $P$ is abelian. In this case, it is enough to a set of coset representatives $\{x_{1},\ldots, x_{n}\}$ (so $G=\coprod Px_{i}$) which is stable under conjugation by $P.$ If we do have this, then the transfer map $G/G'\longrightarrow P$ sends $x\in P$ to $\prod x_{i}x(x^{-1}x_{i}^{-1}x) =x^{n}$ and therefore is surjective, and the result follows.

Let $x\in G,$ and consider the double coset $PxP.$ Let $x_{1},\ldots, x_{m}$ be the distinct conjugates of $x$ under conjugation by $P.$ Clearly $PxP=\cup_{1\leq i\leq m}Px_{i},$ the set $\{x_{1},\ldots ,x_{m}\}$ is stable under conjugation by $P.$ It suffices to check that the decomposition is disjoint. Since $|PxP|/|P|=|P|/|P\cap xPx^{-1}|,$ the result follows if we can show that $P\cap xPx^{-1}=\{y\in P: xy=yx\}=: C(x,P).$

So let $y\in P\cap xPx^{-1}.$ Thus $y=xax^{-1}$ for some $a\in P.$ Let $C(y,G)=\{z\in G: yz=zy\}.$ Then, as $P$ is abelian, $xPx^{-1}\subset C(y,G).$ Since $P\subset C(y,G),$ we must have $xPx^{-1}=zPz^{-1}$ for some $z\in C(y,G).$ So $u: =z^{-1}x\in N(P,G),$ and $y=zuau^{-1}z^{-1}$ gives $y=a.$ Hence $P\cap xPx^{-1}\subset C(x,P).$ The reverse inclusion is obvious.

For the second problem, let $P$ be a Sylow 3-subgroup. Let $C(P,G): =\{ x\in G: xp=px \ \textrm{for all}\ p\in P\}.$ We have $P\subset C(P,G)\subset N(P,G).$ Note that if $g\in P$ is a generator, then $C(P,G)=\{x\in N(P,G): xg=gx\}.$ Hence $|N(P,G)|/|C(P,G)|$ divides $2.$ Hence $C(P,G)=N(P,G),$ which gives $P\subset Z(N(P,G))$ and the result follows. (This proof generalises easily to the case when $P$ is a Sylow $p$-subgroup which is cyclic, and $|G|$ is coprime to $p-1.$)

Hi,
I am sorry,I know I maybe too stupid,I have read your proof for about half an hour,But I can't get anything :(
I even don't know your last passage is a proof for problem 1,or problem 2?
And what is the $H$ you find for problem 2?
Can anybody help?

The last paragraph deduces problem 1 from 2 (the preceding paragraphs prove problem 2).  The subgroup $H$ for problem 2 is the kernel of the transfer map in the first paragraph.

The transfer map is well known and should be there in any graduate level book on group theory. (I am fairly sure Suzuki or Blackburn will have it. It should certainly be there in Brown's text on cohomology of groups.) The way it works is as follows: Let $G$ be a finite group, and let $H$ be a subgroup of index $n.$ Fix $g_{1},g_{2},\ldots, g_{n}\in G$ such that $G=Hg_{1}\cup Hg_{2}\cup \ldots \cup Hg_{n}.$ If  $g\in G$ then $Hg_{i}g=Hg_{\pi_{g}(i)}$ for some permutation $\pi_{g}$ of $1,2, \ldots , n.$

Now define a map $T: G\longrightarrow H$ by setting \[T(g)=\prod_{i=1}^{n}g_{i}gg_{\pi_{g}(i)}^{-1}.\] One checks that $T$ sends the commutator subgroup $G'$ to $H'$, and so we get a map \[\textrm{tr}: \frac{G}{G'}\longrightarrow \frac{H}{H'}\] which turns out to be a group homomorphism, called the transfer. The transfer map is independent of the choice of coset representative we may care to choose.

The first step (paragraph) in the solution involves using some special coset representatives (ie invariant under conjugation by the Sylow subgroup). This allows us to see the transfer on certain elements and deduce the result.

I hope the above explanation helps a little bit.

[quote="theba"] If we do have this, then the transfer map $G/G'\longrightarrow P$ sends $x\in P$ to $\prod x_{i}x(x^{-1}x_{i}^{-1}x) =x^{n}$ )[/quote]
Thank you.
What do you mean by this?
"$x\in P$ to $\prod x_{i}x(x^{-1}x_{i}^{-1}x) =x^{n}$" ?


This is from: [url=http://www.icoool.com/index.asp]http://www.icoool.com/index.asp[/url]

一、选择题(每题3分):

1.哎哟！介不崴(wai3)泥了么！问：这句话什么意思？
A.掉入泥沟 B.感到恐惧 C.抱怨后悔 D.以上答案都不对

2.A:“介不是王二爷么？”　B:“爷爷爷爷！”
问：什么情况下 会发生此对话。
A.两个老年人打招呼 B.朋友之间开玩笑 C.朋友之间打招呼 D.以上答案都不对

3.诶！给我来一块钱牌儿！问：这句话会发生在哪里？
A.游戏厅 B.包子铺 C.医院 D.以上答案都不对

4.你玩过这面儿么？问：这句话什么意思？
A.你用过这种面粉么？ B.你玩过游戏的这个关卡么？ C.你这个 人太不给面子了！ D.以上答案都不对

5.A:“借我用用行么？”　B:“拿你的”
问：回答语句是什么意思？
A.你拿走用吧。 B.用你自己的！ C.找别人借去！ D.以上答案都不对

6.你介人真*蛋！问：这句话什么意思？
A.你经常打破鸡蛋。 B.你这个人人品不好！ C.你有蛋么？ D.以上答案都不对

7.“嘎吗去啦？”　“没事！转一圈！”
问：什么情况下会发生此对话？
A.两个人打招呼 B.老师质问学生。 C.交通警察临检 D.以上答案都不对

8.再废话我可开卷(juan3)啦？问：这句话中的”卷”是什么意思？
A.用纸制品包装成圆筒状 B.使……成为圆筒状 C.骂街 D.以上答案都不对

9.你！过过过！　问：这句话怎么理解？
A.某人在生气的情况下叫某人过去 B.在智力测验放弃问题 C.催促某人过马路。 D.以上答案都不对

10.果弼(bi4)儿多少钱？问：这句话应该发生在哪里？
A.早点铺 B.土产商店 C.游乐场 D.以上答案都不对

11.给我来一套，要俩鸡蛋的啊！问：句中“一套”是指何物？
A.大饼鸡蛋 B.煎饼果子 C.鸡蛋饼 D.以上答案都不对

12.你说你介算嘎嘛地！　问：这句话怎么理解？
A.你说你这是作什么！ B.你说你算排行老几！ C.你是做什么，什么不行！ D.以上答案都不对

13.你就跟他们整天若若(re2re2)吧！问：题目中的“若若”的意思
A.招惹他人 B.混在一起 C.人际关系不好 D.以上答案都不对

14.你界吟净出妖娥子！　问：这句话什么意思？
A.飞的蛾子是不好的 B.说某人妖怪 C.说某人出主意没水平 D.以上答案都不对

15.你瞧了吗介都没治了！问：没治了的意思？
A.没法办了 B.太好了，没的比 C.治不好了 D.以上答案都不对

16.诶，昨天那场球，傻大个儿姚明算是盖了帽儿了！
问：“盖帽儿了”的意思
A.篮球比赛中顺利阻截对方投出的球。 B.发挥很糟糕 C.发挥很出色 D.以上答案都不对

17.“告你啊，别奋秋！”　“奋秋”在天津话里指：
A.读书不够勤奋 B.指秋天里天气不好的时候 C.指不老实爱乱动 D.以上答案都不对

18.你得了吧，爱谁谁！　“爱谁谁”的意思是？
A.拿不定注意，不知道该爱谁好 B.随便是谁都行 C.指说的不是他要爱的人 D.以上答案都不对

19.介事真走鸡！问：“走鸡”表达的意思？
A.指事情办的非常顺利 B.指事情办的不好 C.这事情我们一边走一边谈吧 D.以上答案都不对

20.您二位，还在这溺咕溺呢？　问：此句何意？
A.你二位，还在这里嘀咕呢？ B.您二位要点什么菜想好了吗？ C.不知对方在做什么问问 D.以上答案都不对

21.我告你吧介事甭提了！问：说此句人的意思是：
A.对事情不想再提起 B.不好说预言又止 C.事情很有意思要谈论一下 D.以上答案都不对

22.“归其了就介样儿了”　　问：这句话表达的感情最接近于：
A.愤怒 B.无奈 C.满足 D.以上答案都不对

23.“棱（leng2）子!”　　问：这个词的意思是：
A.一个窗棱或门棱 B.一个狂男 C.一个几何难题 D.以上答案都不对

24.“诶介人诶还想洽(qia2)个儿”　问：这句话应该发生在何时？
A.考试时 B.谈恋爱时 C.排队时 D.以上答案都不对

25.Ａ：“你跟我去水上么？”　Ｂ：“玩儿去。”
问：后者的意思最接近于：
A.我跟你一起去玩 B.我不跟你去玩 C.你自己先去 D.以上答案都不对

26.介孩子太腻歪人了　　问：指孩子什么？
A.这孩子太让人讨厌了 B.这孩子太可爱了 C.这孩子吃的太腻了 D.以上答案都不对

27.“你介人真不够揍！”　问：该句话的意思是：
A.怎么打你一顿都不够！ B.你真不像话！ C.算了，不打你了！ D.以上答案都不对

28.“南门外怎么走？”－“简直！”
问：后者要表达的意思是：
A.太简单的问题，不要问我！ B.简直不像话！ C.沿着这里一直走 D.以上答案都不对

29.诶，你介人咋界刺儿！问：这句话的意思是：
A.你怎么满身是刺 B.你怎么事儿这么多，太是非了 C.你给我拔拔刺吧 D.以上答案都不对

30.“我介闺女长得太顺（shun2）了!”　问：这句话的意思是：
A.我女儿很丑 B.我女儿是个大恐龙 C.我女儿的成长很顺利 D.以上答案都不对

二、选择填空题(每题2分):

1.———二哥，今儿个我看见卖柿子的了，买了俩，倍儿涩……
———多新鲜哪，这才几儿啊，柿子看着好可就是不能吃。
“多新鲜哪”是的（）意思。
A.柿子很新鲜 B.和卖柿子的人不太熟 C.柿子涩是理所当然的 D.以上答案都不对

2.———我们家电视坏了……
———别急，我给你（）去。
下面几个词最适合的是：
A.搅和搅和 B.搀和搀和 C.得楞得楞 D.以上答案都不对

3.———你不说跟你对象去水上吗？
怎么跑这儿来了？———她跟我翻次了……
“翻次”的意思是：（）
A.生气了 B.失约了 C.改计划了 D.以上答案都不对

4.“我说你吃饭嘴是漏是怎么着啊，看你（）的……”
下面几个词最适合的是：
A.拉拉 B.底儿掉 C.茄个 D.以上答案都不对

5.“你个倒霉孩子，挺好的小汽车儿你都给（）了……”
下面几个词最适合的是：
A.翻次 B.祸祸 C.改 D.以上答案都不对

[quote="manchu_princess"]我是天津人，对有的对话熟悉（满有意思的 :lol: ）但也有不少从来没有听说过的。你有答案吗？[/quote]
我发的这些题大部分都没有正确答案, 所以如果你做了一份的话, 我们当然欢迎。

一 选择题
1D 2A 3A 4B 5A
6B 7A 8C 9A 10A
11B 12A 13B 14C 15B
16C 17C 18D 19B 20C
21C 22B 23B 24C 25B
26A 27B 28C 29B 30A
二 选择填空题
1C 2C 3A 4A 5B

我刚做出的
平时很少说天津话
有些题也答不上来了
请教多位老天津卫后终于得出此答案
仅供参考   :)


It appears that one of the best and brightest of the state, namely besttate, has chosen to continue his studies at Andover in MA. Let us mourn.

Yeah, it's really too bad. I guess he needs to move on to bigger and better things though; I'm afraid Illinois just doesn't have the caliber of math programs to challenge him anymore :(. Any parting words, besttate?

Thanks Arnav for this tribute to such a venerable mathcounter.

When I was 14 years old, besttate taught me how to push the F2 button on my calculator.  He really jumpstarted my mathcounting career.  *sobs*  I'll miss him.

As long as non-Illinois people are commenting here (and I did get my Ph.D. at U. Chicago), I'll mention that the top 8th grader from the Southern California team, kohjhsd (who was in the tiebreaker) will also be leaving us and going to Andover.

[quote="Altheman"]hmm all posts have 6 rating...odd

anyway, there goes my school's soph math team [i.e. for next year]...the rest of the sophs are bad

boo

on a side note:
he beat me at arml this year
he then proceeded to dance[/quote]

...[i]what[/i]

also!

Dr Merryfield, do you know why he chose Andover as opposed to other boarding schools?

One of the top NC math prospects is also going to Andover.  I'm not sure why this choice was made but there must be something going on there that is attracting math talent.

wait, what? who? I had no idea about this dangit

You can check the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153036]NC Forum[/url] for hints.  She scored 129 on the AMC 12 before her 12th birthday.


function
topology
advanced fields
advanced fields unsolved

I am stuck with this problem, so if any of you can help me I really appreciate it:

Find spaces $X$ and $Y$  and a function $F: X \to Y$ which is not continuous , but which has the property that $F(x_{n}) \to F(x)$ in $Y$ whenever $x_{n}\to x$ in $X$

thanks

Hi.

You need to notice that $X$ cannot be first countable. Otherwise the countinuity of $F$ is equivalent to the implication $x_{n}\to x\Rightarrow F(x_{n})\to F(x)$.

So, take a look at some topological spaces that are not first countable.

Hanno

Hanno I know that, for example $R$ with the cofinite topology is not 1st Countable($R$ the set of real numbers), so I used the identity function $Id: (R,T_{c}) \to (R,T_{u})$ which is not continuous, where $T_{c}$ is the cofinite topology and $T_{u}$ the usual topology (euclidian) (by using the fact that $T_{c}$ is coarser than $T_{u}$ we have that Id is not continuous) but I cannot show the implication: $x_{n}\to x$ (in the cofinite topology) implies $x_{n}\to x$ (in the usual one).

am I wrong by using that counterexample?

Ill appreciate any help

Adriana

Hello Adriana.

This example won't help since a sequence may converge to more than one point in $(R,T_{c})$. In the Hausdorff-Space $(R,T_{u})$, this can't happen.

You may have a look at the first ordinal number $\Omega$ which is not countable, topologized by the order topology. In this space, every sequence converging to $\Omega$ becomes constant at some point. You can map this point to any other point in the space to construct a map which suffices the given conditions.

Hope it helps.

May be interesting to know if for every topological space $X$, not first countable, there is a topological space $Y$ and a non-continous map $f: X\to Y$ such that for all $(x_{n})$ in $X$, $(x_{n})\to x$ we have $(f(x_{n}))\to f(x)$.

Hanno

I got an example!!!!!, if $X=R$ has the cocountable topology $T_{c}$(the closed sets are countable) and $Y=R$ has the usual topology $T_{u}$(euclidian one), then we can consider the identity function $Id: (R,T_{c}) \to (R,T_{u})$ which is not continuous, and every sequence $(x_{n})$ converges, in $(R,T_{c})$, to x if and only if there is $n_{0}$ such that $x_{n}=x$, for all $n \geqslant n_{0}$, if that happens obviously that kind of sequence converges in $(R,T_{u})$ too.

As you can see it's an easy example, but it works :lol:

Adriana


Find the first less significant digit $\ne 0$ of $5^{2007}!$

Is "!" is just an English  "!" or a mathematical "!" (  Do you mean $5^{2007}$ or factorial of this number? )

if it's an english ! it's pretty trivial isn't it?


probability
geometry
geometric transformation
reflection
3D geometry
Casework

If Michael rolls three fair dice, what is the probability that he will roll at least two 1's? Express your answer as a common fraction.

Two dice must be 1's and the third can be anything.  There are $ \dbinom{3}{1}\equal{}3$ ways to choose which die can be anything.  So the probability of rolling at least two 1's is:

$ \dbinom{3}{1} \cdot \left(\frac{1}{6}\right)^2 \cdot 1$
$ \equal{}\frac{3}{36} \equal{} \frac{1}{12}$ (which is incorrect)

[quote="mathmom"]Two dice must be 1's and the third can be anything.  There are $ \dbinom{3}{1} \equal{} 3$ ways to choose which die can be anything.  So the probability of rolling at least two 1's is:

$ \dbinom{3}{1} \cdot \left(\frac {1}{6}\right)^2 \cdot 1$
$ \equal{} \frac {3}{36} \equal{} \frac {1}{12}$ (which is incorrect)[/quote]

(I suspect, on further reflection, that the answer lies in the fact that when the die that "can be anything" is also a 1, the 3 arrangements are indistinguishable.  I guess this requires casework whether you do it constructively or by a complementary approach.)

Yes, that is the error.

A solution:

The probability that two are $ 1$'s:

$ \binom32 \cdot \left (\frac{1}{6} \right )^2 \left (\frac{5}{6} \right )\equal{}\frac{15}{21276}$

The probability that all three are $ 1$'s:

$ \left (\frac{1}{6} \right )^3\equal{}\frac{1}{216}$

Adding yields $ \frac{16}{216}\equal{}\boxed{\frac{2}{27}}$, which is a little less than $ \frac{1}{12}$.

I am just wondering what is means by "fair" dice.. Anybody help please?

My guess: "all same" dice?

"Fair dice" are dice with each roll being equally likely. i.e. a cube with 1, 2, 3, 4, 5, and 6 all with a 1/6 probability of being rolled.

This problem is easier solved without complementing.

There are 3(1*1*5) = 15 ways to roll two ones.
There are 1(1*1*1) = 1 ways to roll three ones.

So the probability of rolling at least two ones is (15+1)/216 = 2/27.

we use complementary counting.

first case - there are no 1's. The probability of this is:
$$\frac{5^3}{6^3} = \frac{125}{216}$$

second case - there is 1 1. the probability of this is:
$$\frac{5^2\cdot1}{6^3}\cdot3 = \frac{25}{72}$$

we add these two together to get $\frac{200}{216}$. then we subtract this from $1$ to get $\frac{16}{216} = \boxed{\frac{2}{27}}$.

Probability of two ones is

$\binom{3}{2} \cdot (\frac{1}{6})^2 \cdot \frac{5}{6}=\frac{5}{72}$

Probability of three ones is

$\binom{3}{3} \cdot (\frac{1}{6})^3 = \frac{1}{216}$

Adding these we get $\frac{5}{72}+\frac{1}{216}=\boxed{\frac{2}{27}}$

This problem can be solved so much faster with casework rather than complementary probability... Do you guys think this is miscategorized?


calculus
integration
real analysis
real analysis unsolved

Since this section of the College Playground forum is growing very fast, I'm thinking of retiring all the problems that have been completely solved and whose discussion terminated more than 2 months ago to the Calculus Solved Problems section (that is where they formally belong anyway, if I understand the rules right), leaving only the relatively recent ones and those whose solution wasn't found yet. I wonder what you think of the idea.

I wonder why you would care on problems that are over 2 months old? :? Does it hurt anyone that there are 5 pages instead of 2?

Since most people were in favor of the idea and nobody minded, I started to retire the old solved problems. I am going to touch only those whose discussion is really over and whose solutions I can understand and consider complete. I think I'll maintain the 2 month expiration period policy for a while to see if it makes this section of the forum a bit easier to navigate (the main difference between 2 pages and 20 pages is that almost everybody will look at almost everything on 2 pages and almost nobody would bother to browse through 20)

As to Peter's comment, one of the reasons behind the idea was my desire to attract attention to old problems that remain unsolved and give them a second chance. I admit that about half of them are either not very interesting or hopelessly difficult, but the other half seem quite nice and it is a pity they are forgotten. :)

Well... one big suggestion then: check solved problems first. :P

Because there are many many unsolved problems in that section... so it's pointless doing all that work to save problems here if you're just leaving them out there :)

I'm not at all "against" this, it just looks to me like a very time-consuming activity :)

I am moving solved problems in the Superior Algebra Section to Solved Problems without asking anybody. And I move them immediately to Solved, without waiting for 1 or 2 months.
I regularly look into the Solved section to check if someone notices any mistakes in the proofs and move them back to unsolved or correct them.
From time to time unsolved problems are being POSTED in the Solved Section, but this is another story.

I think solved problems should stay where they belong, which is in the Solved section. I realise that most of us are too proud and are not interested in the solved problems, but if people still want to comment on them, they can do that in the Solved Section also.

I agree to fedja's idea that problem solving is more effective if you have less unsolved problems to choose from.

Please do smthing like limits and integrals in one place another at the other and so on. It is where difficult for freshman to look through allthe 30 pages in order to find the limits and plus here it is too difficult with such slow net/ :(

I have been posting the problems so far, and there are many problems which i don't see, please don't retire old problems.

Best regards,

Japanese communiteies moderator

kunny

[quote="kunny"]I have been posting the problems so far, and there are many problems which i don't see, [/quote]
They didn't disappear, just went to the "Solved" section where you can easily find them. But I had to give up the idea anyway: the new posts are coming too fast especially after Valentin united the "Unsolved" and "Proposed" sections of the forum  :P.


quadratics
number theory
relatively prime

Q1)
Given that a^2 + b^2 = c^2 <pythagoras triplet> then prove that a*b*[ (a^2) - (b^2) ] is divisible by 84.

Try to show first the related, but easier, problem that $ ab$ is divisible by $ 12$.  (Look $ \bmod 3$ and $ \bmod 4$ separately.)  What is left to prove here?

I didnt quite get that , can you expand it a bit?

[hide="ab is divisible by 3."] Suppose otherwise; then $ a, b \equiv 1, 2 \bmod 3$ hence $ a^2 \equiv b^2 \equiv 1 \bmod 3$ and $ a^2 \plus{} b^2 \equiv c^2 \equiv 2 \bmod 3$.  But $ 2$ is not a quadratic residue $ \bmod 3$; contradiction. [/hide]
[hide="ab is divisible by 4."] Suppose otherwise.  First, we show that $ ab$ is divisible by $ 2$.  Otherwise, $ a, b \equiv 1, 3 \bmod 4$ and $ a^2 \equiv b^2 \equiv 1 \bmod 4$ and $ a^2 \plus{} b^2 \equiv c^2 \equiv 2 \bmod 4$, and as before $ 2$ is not a quadratic residue $ \bmod 4$; contradiction.  Because the triple is primitive, exactly one of $ a, b$ is even; WLOG $ b$ is even.

Now suppose $ ab$ is not divisible by $ 4$.  Then $ b \equiv 2, 6 \bmod 8 \implies b^2 \equiv 4 \bmod 8$ and $ a^2 \equiv 1 \bmod 8$, hence $ a^2 \plus{} b^2 \equiv c^2 \equiv 5 \bmod 8$, which is not a quadratic residue $ \bmod 8$; contradiction. [/hide]
Can you now show that $ ab(a^2 \minus{} b^2)$ is divisible by $ 7$ using similar reasoning?

BTW: if you take any 2 consec. no. greater 2,
1/n + 1/(n+1) then the simplify the fraction obtained.
let numerator be A and den. be B ( B>A , proper fraction)
ud notice that A^2 + B^2 = (B+1)^2 , i.e. a pythagorean triplet.
I found that this is divisble by 168, can i prove that it would also be divisible by 84?

and i dont understand modulo so well , yet from the little i know about it , im able to grasp the solution a bit.
thanks!

[quote="t0rajir0u"]Special cases aren't enough to prove the general case.  That particular relation is a special case of the general:  if $ A, B, C$ is a primitive Pythagorean triplet with $ A$ even, then there exist $ m, n$ such that $ A \equal{} 2mn, B \equal{} m^2 \minus{} n^2, C \equal{} m^2 \plus{} n^2$.  Moreover, $ m, n$ are relatively prime and have opposite parity.

For the last part, you'll want to know that the quadratic residues $ \bmod 7$ are $ 0, 1, 2, 4$.[/quote]
since we got ab divisible by 12 , we'd want to prove a^2-b^2 divisible by 7 since , ab is not always div. by 7.
so we separately take cases where either a or b are div. by 7 or none are , and then we'd get that a^2-b^2 is only divisible by 7 when remainders on individual divisions are same[(1,1)(2,2)(4,4)]so that a^2-b^2=0mod7 <take the = symbol are congruence symbol>


abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group of odd order.
Show that there is exactly one $ g \in G$ with $ g^2\equal{}x$ for any $ x \in G$.

Take a look at [url=http://www.mathlinks.ro/viewtopic.php?t=262399]this[/url] post.  :wink:

The proof of existence works for the non-abelian case, too.  It remains to see if unicity still holds [i]there[/i].

If $ gcd(n,|G|)\equal{}1\equal{}an\plus{}b|G|$ then $ x\mapsto x^a$ and $ x\mapsto x^n$ are inverse maps. Hence for all $ x\in G$ there is one and only one nth root.


limit
trigonometry
real analysis
real analysis unsolved

$ 0<x<pi/2$,
$ b_1\equal{}cos(x)$,$ b_{n\plus{}1}\equal{}cos(b_n)$
Show that:
$ \mathop{\lim }\limits_{n\to\infty } b_n$ exist

thanks~ :lol:

this was posted many times.
at first note that that $ |b_{n}| < 1$ then $ f(x) \equal{} \cos x$ has the good property like this $ |f'(b_{n})| < \sin 1$
thus you can write that $ |f(b_{n}) \minus{} f(b_{n \plus{} 1})| \leq \sin 1 |b_{n} \minus{} b_{n \plus{} 1}|$ after yourself.


$ XX_1 \minus{} d_1$
$ XX_2 \minus{} d_2$
$ XX_3 \minus{} d_3$

$ BC \minus{} a$
$ AC \minus{} b$
$ AB \minus{} c$

R- radius of circumscribed circle around $ \triangle{ABC}$

[img]http://www.part.lt/img/ac42ababd2b317cd9c2060a714612b33743.JPG[/img]

1.Prove that $ a^2\plus{}b^2\plus{}c^2\equal{}4(h_ad_1\plus{}h_bd_2\plus{}h_cd_3)$ if point X is center of circumscribed circle around $ \triangle{X_1X_2X_3}$

2.Prove that $ 4(\frac{a}{d_1}\plus{}\frac{b}{d_1}\plus{}\frac{c}{d_1})\equal{}\frac{abc}{d_1d_2d_3}$ if point X is center of circumscribed circle around $ \triangle{X_1X_2X_3}$

3.Prove that $ a(d_2\plus{}d_3)\plus{}b(d_1\plus{}d_3)\plus{}c(d_1\plus{}d_2)\equal{}R(a\plus{}b\plus{}c)$ if point X is center of circumscribed circle around $ \triangle{X_1X_2X_3}$

There arises a question, "What is X?"
[hide="hint(maybe)"]
In my opinion, we can think , that if $ X$ is the circumcentre of $ \Delta X_1X_2X_3$ then $ X$ HAS to be the INCENTRE of $ \Delta ABC$(as according to your image $ XX_1, XX_2, XX_3$ are respectively perpendicular to the sides of $ \Delta ABC$
$ XX_1 \equal{} XX_2 \equal{} XX_3 \equal{} r$ which is the inradius of $ \Delta ABC$
[/hide]
I think this logic will help..... :maybe:


Well how did you guys perform?
Me done pathetic
Have done better in physics than in chemistry  :o   :(
Well let me post one of my friends marks as an alibi first
bcos I dunno whether to reveal my mark ( as its pathetic )
Well my friend got 200
How much did you get?
Esp you Pardesi
I think Chappli will get 250!!!
And the all india topper 450!!!!!!!!!!!!!!

not calculated but first reaction...very bad
could only do physics and maths ...chemistry ...didn't even remember a single formulae...just did few analytical and stochiometry based question.

OH so rohit u getting 200 ah i know da!!!! thats you !!! kalaku ?? so how much are u getting pardesi???

very very very lessssssss 172  :(  :(  :(
but don't deserve any more than that  i hadn't studied or even attempted to study binomial,3D,Matrices,Electrochemistry.... formulaes

Am getting 238 this time.. Bucked up chem a great deal from last time.. gettin 73 in chem alone total...  :P  But wat pulled it all down was a 23 in paper one math.. jus unable to digest such a pathetic performance.. Anyways, how much would be cut off fer 99 percentile??

Am not getting 200 for sure
Flopped chem
Did better in physics than in chem  :o
So getting only 187 :(
Hope to do the best in full test 1
What are your split ups?

I am getting in the 190's I guess. Lets see.

Physics - Both Papers superb
Maths - 1 Paper okay, one paper really good
Chem - Both Papers Pathetic

So one can obviously see where I flunked.... Man, I thought I studied hard but ended making the most trivial of mistakes. Feel frustrated.

Gut feeling is that 150/160 is 90 Percentile.
180 - 95 Percentile
200 - 97 percentile
238 - 100 percentile.. (which is a mathematical flaw anyways ;))

Okay , okay was kidding about the last part. All India Topper might be in the 375's. Not more. 450 is impossible madness....

Ofcourse I must only clever chap to write 120+160=300!! @Pardesi: Electrochemistry screwed us bad.

238 will be 98 - 99 percentile
if tarun goyal got 246 last time he would surely have got above 300 this time.
Am frustrated  :mad:  @ how almost all my friends are getting above 200
didnt even study one word of physics for AITS
Yet got 56 in physics and only 50 in chem  :mad:

i am not too surprised by that either .Questions weren't difficult nor were they tricky...it required lot of patience to sit through the examination... :D

It was actually a pathetic paper to sit through.. The ,maths seemed easy.. But too long... Lesse how the percentiles go.. Jus 238 is around 99+.. U never know.. Prob on an average the whole performance might have dipped though we've done better.. :D

Might have.... Dunno about you chapplikka but I am sure my percentile might go up. Almost everybody I've talked to are getting more this time.

Since I'm sitting here jobless enough and waiting for the Olympiad results

Paper I. Physics was really easy. I got 2 on 3 matches right. Chem was okay,I just flunked. so was math but one or two of his solns are wrong. (DUH!)

Paper II. Physics was slightly tougher than Paper I but no big difference. Did anybody understand the first sum in SET I, about a container falling down and how flow of liquid stops altogether????

Chem: Comprehension was slightly okay, rest were medium level. Their chem assertion reasoning IMO are pretty ordinary questions.

P.S @ pardesi: I hear you went to school after a long time??

PPS: AT ALL!!

ALL THOSE WHO WANT TO RECIEVE FULL TEST 7,8,9 SHOULD SUBMIT A COPY OF THEIR IIT APPLICATION OR ACKNOWLEDGEMENT BY JAN 20TH TO THE NEAREST FIITJEE OFFICE.

PPPS: :P All those who are writing INMO will miss the Full Test III on 20th??

Sorry for the double post but this is a new discussion.

What do you think is the weightage of the percentile in AITS. Or how many ppl all over India write AITS. My friend tells me Part Test I was written by 9724 odd students (Not sure, I didn't write it myself).

Which means Part Test II would be loads more.... Considering 20000 to 30000 writing Fiitjee, that means 90 percentile means 2000-3000 rank, which will surely be pushed down in JEE.

Even 99 percentile is only 200 or so rank which is not great deal. :( But the question is, how many of the toppers miss out on FIITJEE?? Does the full Kota, AP gang write it. What about the oympiad genii from Chandigarh, etc. We should know how close to our original rank this is.

P.S: did you guys write FTSE?

no don't worry the  postal....problems aren't tough too...going by the way FIIT-JEE is givinng problems who knows

FIITJEE sum was actually about only 2 element subsets which made it pretty easy. Ofcourse this sum can be tackled in the same way. Fix the largest element, then you have $ 2^n\minus{}1$ subsets and so on.  and so on...

fiitjee claims that it can give a projected rank for iitjee is just rubbish
fiitjee questions r defenitely original and good
but fromm iitjee perspective they have the least significance
99 percentiles in fiitjee dont reallly matter much
unless and until u continue to top

I don't really give a damn about their projected rank. But if you consistently clock 99 percentile, then it means you are on your way to a good rank, but that depends on no.of ppl who write.

Yes, their question set is good, be it in RSM or in their papers. I would really like to find their sources. :P

[quote="Cyber-Shreyas"]@Soumya: Thanks for reminding me.. I too have free CP Test Series. Hope its not too late. I totally forgot.

More statistics:
Last year, 76 out AIR Top 100 wrote FIITJEE AITS, which is a good thing. If the trend continues it can easily be 60% of TOP AIR 1000 write AITS, then it may be the closest thing to JEE. I actually want more people to write it so that I can realize my ranking better. :D Yes, its slightly tougher than IIT and many claim that brilliant is more like JEE but you may never know... IIT might change everything this year.

hopefully both chappli and soumya get selected and make us proud. BTW Guys, you should utilize AOPS really well. There are thousands of brilliant problems here and your thinking will surely widen . sadly chappli doesn't seem to be using this :(

Hey Pardesi, wasn't that sum a FIITJEE sum??? Its too beneath the postal level :D[/quote]

Maybe 60% of the top 1000 rankers wrote it but I bet they had got horrible ranks in AITS...
you need to check also what rank they got... :)

The people who did good in AITS ended up doing not so good in IITJEE... :huh:

I scored 248. How you guys are getting the percentile? Do you get the full rank list?

LOL. Those guys would be working in some companies or doing their PhD now!

[quote=Phantom1234]Please could anyone provide Aits part test 3 answer key PDF[/quote]

It will be uploaded in the aits site.

Oh wow.Soumyashant Nayak.
He is from our school batch-2007.Ended up getting 3 medals in IOAA.And was selected thrice for the IMOTC. Truly a prodigy. Currently pursuing research in genetics may be in Pennsylvania.

Please don't bump these ancient threads - as noted above, this was from 2008. Make a new thread instead.


1. Let C be a circle of radius 1. Let R be the union of the interiors of all isosceles right triangles with legs of length $ \sqrt2$ and exactly two vertices on C. What is the area of R?

2. Three points $ A$, $ B$, and $ T$ are fixed such that $ T$ lies on segment $ AB$, closer to point $ A$. Let $ AT \equal{} m$ and $ BT \equal{} n$ where $ m$ and $ n$ are positive integers. Construct circle $ O$ with a variable radius that is tangent to $ AB$ at $ T$.  Let $ P$ be the point such that circle $ O$ is the incircle of $ \triangle APB$. Construct $ M$ as the midpoint of $ AB$. Let $ f(m,n)$ denote the maximum value $ \tan^{2}\angle AMP$ for fixed $ m$ and $ n$ where $ n > m$. If $ f(m,49)$ is an integer, find the sum of all possible values of $ m$.

3. Let S be the set of reals of the form $ \sin(2007\pi/n)$, where n is any positive integer greater than 2007. How many ordered pairs (a, b) exist such that a and b are both elements of S, not necessarily distinct, and $ a^2 \plus{} b^2 \equal{} 1$?

4. [url=http://www.8foxes.com/Home/13]Click to see the image[/url]


When a satellite is a distance R from the center of Earth, the force due to gravity on the satellite is F. What is the force due to gravity on the satellite when its distance from the center of Earth is 3R?

(1) F/9 (3) F
(2) F/3 (4) 9F

The answer should be (1) F/9, since the law of attraction by two gravitating masses is an inverse square law. (F is inversely proportional to the distance square)
You can find out more about gravitational force in the following webpage: :idea:
[url]http://scienceworld.wolfram.com/physics/GravitationalForce.html[/url]

I hope this can help you! :P

[quote="Interval"]Do you have similar questions for me to practice?[/quote]

Ok, I will post some similar questions here next time (maybe tomorrow), I hope this can help you becoming more familiar with this type of questions. :P


[quote="cmad"]Σας παραθέτω ένα interesting fact. Είναι ενδιαφέρον, ίσως ακόμα και για παιδιά λυκείου, ώστε να πάρουν κάποιο intuition στις παθολογίες.

Είναι περίεργες συναρτήσεις οι παράγωγοι. Αν πάρουμε πχ τις συνεχείς συναρτήσεις από το $ [0,1]$ στο $ \mathbb{R}$, τότε οι μη-παραγωγίσιμες συναρτήσεις είναι πυκνές στο σύνολο αυτό. Αλλά πυκνές είναι και οι παραγωγίσιμες! Επίσης, οι παραγωγίσιμες μοιάζουν κάπως με τις συνεχείς, καθώς ικανοποιούν ένα είδος θεωρήματος ενδιάμεσης τιμής (Darboux). Λόγω αυτών των συγκρίσεων με τις συνεχείς, είναι λογικό να αναρωτηθεί κανείς: είναι πάντα οι παράγωγοι Riemann-ολοκληρώσιμες; :roll:[/quote]

Δεν το γνώριζα και η αρχική μου σκέψη ήταν "σίγουρά πρέπει να είναι" αλλά

[hide]
$ F(x) \equal{} x^2 \sin(1/x^2)$
[/hide]

Πολύ ωραία!

Ας το πάμε ένα βήμα παραπέρα.

Το πρόβλημα στη συνάρτησή σου είναι μεμονομένο, δηλαδή αν απομακρυνθείς από το 0 τότε η συνάρτησή σου είναι smooth, και επομένως ολοκληρώσιμη σε κάθε $ [\delta, 1], \delta > 0$. Είναι πιθανό η ολοκληρωσιμότητα να αποτυγχάνει με ακόμα λιγότερο mild τρόπο; (δηλαδή να αποτυγχάνει σε κάθε υποδιάστημα--hint).

Η συνάρτηση $ f(x)\equal{}x\sqrt{x}cos(\frac{1}{x}) , x>0$ και $ f(0)\equal{}0$ είναι παραγωγίσιμη με $ f'(x)\equal{}\frac{3}{2}\sqrt{x}cos(\frac{1}{x})\plus{}\frac{1}{\sqrt{x}}sin(\frac{1}{x}) ,x>0 .  f'(0)\equal{}0$
η οποία δεν είναι Riemman ολοκληρώσιμη σε κανένα διάστημα [0,α] λόγω του μη φραγμένου ορου$ \frac{1}{\sqrt{x}}sin(\frac{1}{x})$ σε κάθε περιοχή του 0+

(στην αρχή εψαχνα με το Darboux αλλά το  μυστικό βρισκόταν στα φράγματα)

[quote="r_boris"]Η συνάρτηση $ f(x) \equal{} x\sqrt {x}cos(\frac {1}{x}) , x > 0$ και $ f(0) \equal{} 0$ είναι παραγωγίσιμη με $ f'(x) \equal{} \frac {3}{2}\sqrt {x}cos(\frac {1}{x}) \plus{} \frac {1}{\sqrt {x}}sin(\frac {1}{x}) ,x > 0 . f'(0) \equal{} 0$
η οποία δεν είναι Riemman ολοκληρώσιμη σε κανένα διάστημα [0,α] λόγω του μη φραγμένου ορου$ \frac {1}{\sqrt {x}}sin(\frac {1}{x})$ σε κάθε περιοχή του 0+

(στην αρχή εψαχνα με το Darboux αλλά το  μυστικό βρισκόταν στα φράγματα)[/quote]

Αυτό που ζητάει ο cmad είναι να μην είναι ολοκληρώσιμη σε [i]κανένα[/i] υποδιάστημα (δηλ. ακόμα και σε διαστήματα της μορφής $ (\delta, 1)$.

Για τις συλλογές μετά (δε θέλω να κάνω hijack το topic)... αλλά in short σκεφτόμουν κάτι το οποίο να μην απευθύνεται μόνο στους θαμώνες του forum.

More later,

Durandal 1707


Find all triaples (a, b, c) of positive integers such that
(i) a <= b <= c;
(ii) gcd(a, b, c) = 1; and
(iii) a^3 + b^3 + c^3 is divisible by each of the numbers a^2b, b^2c, c^2a.

Namdung

Here's half of it (I can't help the feeling that it's the EASY half :D):

We have a2|b3+c3 and the analogous relations. Let's assume (a,b)>1. Then there is a prime p|a and p|b. From the relation written up there we get p|c, which means that p|(a,b,c)=1, so we have (a,b)=1. It's obvious that (a,b)=(b,c)=(c,a)=1.

From (a,b)=(b,c)=(c,a)=1 and a2|b3+c3
b2|c3+a3
c2|a3+b3 we get
a2b2c2|a3+b3+c3 (*).

We have 2 cases:

Case I: c>=a2b2
This means that c2>=a4b4, but we know that c2|a3+b3 <= a3b3 if a, b>1, so a=b=1 from the 2 ineqs above, so we have a contradiction. Assume now a=1. If b>1 then we play with the ineqs above again to find 1+b3>=b4, which means that b=1, so we have contradiction again, so the only available option is a=b=1, and c=1 follows. Case I yields 1 solution: (a,b,c)=(1,1,1).

Case II: c<a2b2 => a2b2c2>c3>=b3>=a3, so a3+b3+c3 < 3a2b2c2, and from (*) we get that a3+b3+c3 is either 2a2b2c2 or a2b2c2.

I think I managed to finish it off.

We have 2 subcases in Case II:

Case II a): a3+b3+c3 = 2a2b2c2

This means that S=\suma/b2c2 = 2, but S <= (a+b+c)/a2b2 (just use a,b<=c), so a+b+c >= 2a2b2, and since c<a2b2, it means that a+b>a2b2, and this can only happen if a=b=1, so we have the solution (1,1,1) again, but this time it doesn't fit the case.

Case II b): a3+b3+c3 = a2b2c2

This is a bit tougher, but not too much. We consider, as usual, 2 cases:

(i) c<(2/3)a2b2 => 1/3 <= a/b2c2+b/a2c2 <= (a+b)/a2b2, and if we analyze all the possible cases (a has to be 1 and 1<=b<=4) we see that a=1,b=2,c=3 works, but it doesn't fit this case, because the condition c<(2/3)a2b2 isn't fulfilled.

(ii) c>=(2/3)a2b2, so c2 >= (4/9)a4b4, and since c2|a3+b3, we get a3+b3 >= (4/9)a4b4, and I think it's clear that there aren't too many cases. Just analyze them (a=1, for one), and you'll get the solution a=1,b=2,c=3, so Case II yields the soln (a,b,c)=(1,2,3).

[b]Final answer: (a,b,c) \in {(1,1,1), (1,2,3)}.[/b]


I have a quick question somebody can hopefully clarify for me.

If there is a contest that runs every hour for 40 days. I enter 30 entries during one hour of a each day for 40 days what are my odds of winning during the course of 40 days?

Question already posted in multiple forums
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55548]http://www.mathlinks.ro/Forum/viewtopic.php?t=55548[/url]


Find all the positive integers $ a_1,a_2,...,a_n$ that satisfy :

$ a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2 \equal{} \frac {2n \plus{} 1}{3}\cdot (a_1 \plus{} a_2 \plus{} ... \plus{} a_n)\ ,\ (\forall)n\in \mathbb{N}^*$

[hide="Idea to the solution"]Since $ \sum_{i\equal{}1}^n i\equal{}{n(n\plus{}1)\over 2}$ and $ \sum_{i\equal{}1}^n i^2\equal{}{n(n\plus{}1)(2n\plus{}1)\over 6}$, we get

$ \frac{\sum a_i^2}{\sum a_i}\equal{}\frac{\sum i^2}{\sum i}$

And now some inequalities must be used - and I suck at inequalities :) - to prove that $ \{a_i\}\equal{}\{i\}\equal{}\{1,2,\dots,n\}$...[/hide]

[quote="Farenhajt"][hide="Idea to the solution"]Since $ \sum_{i \equal{} 1}^n i \equal{} {n(n \plus{} 1)\over 2}$ and $ \sum_{i \equal{} 1}^n i^2 \equal{} {n(n \plus{} 1)(2n \plus{} 1)\over 6}$, we get

$ \frac {\sum a_i^2}{\sum a_i} \equal{} \frac {\sum i^2}{\sum i}$

And now some inequalities must be used - and I suck at inequalities :) - to prove that $ \{a_i\} \equal{} \{i\} \equal{} \{1,2,\dots,n\}$...[/hide][/quote]

Hi,I think you're using formula for consecutive numbers.They're not necessarily consecutive.

Recheck what I wrote carefully, and also recheck the problem statement.


how many isosceles triangles can be found in a 16 points arranged in a $4\times 4$ square?

is there a general formula?

[url=http://imageshack.us][img]http://img165.imageshack.us/img165/3006/kavo8.jpg[/img][/url]
Shot at 2007-06-26

[url=http://imageshack.us][img]http://img187.imageshack.us/img187/4701/triangelnw6.jpg[/img][/url]
Shot at 2007-06-27

Unfortunately, the problem is a bit tricker than that.
[hide]
Consider a 3 by 3 point array:

A B C
D E F
G H I

Note that CDH (and symmetric) are all isosceles, and that GBI (and symmetric) are all isosceles.[/hide]

[hide="Note"]
Most isosceles triangles will come in two forms: + types and X types. X types are symmetric about a line of the form y = k +/- x, + types have either vertical or horizontal lines of symmetry.

However, as higher numbers of n come up, one will need to account for all Pythagorean Triples as well. In other words, we'll need a formula for the number of Pythagorean triples with hypotenuse k. Otherwise, the problem cannot be solved.
[/hide]

[hide="4x4"]
Consider a 2x2 square. There are 8 different triangles in here.

Consider triangles in a 3x3 square but not in a 2x2 square. There are the 8 triangles that are the equivalents of the 8 in a 2x2 square, and also the four CDH's and GBI's (read first hidden part), for a total of 12 here.

Consider triangles in a 4x4 square but not in a 3x3 square. There are the 8 equivalents of the 2x2 square.

If we label it:
A B C D
E F G H
I. J K L
M N O P

then MBO, NCP, and symmetric (8) form one type of isosceles triangle, IDN and symmetric (4), and EDO and symmetric (4), for a total of 24 here.

We then have $1^{2}(24)+2^{2}(12)+3^{2}(8)=144$ isosceles triangles.

[/hide]

[quote="nateharman1234"]Why are there 8 isosceles triangles in the 2x2 case? I am counting 4, one with vertex angle at each of the vertices. Am I just horribly misinterpreting something?[/quote]

Yes, you are terribly, horribly, absolutely unforgivably misinterpreting something  :P

Think of it this way:

[img]http://i193.photobucket.com/albums/z278/Anaxerzia/Animation.gif[/img]

Hope that helps

[quote="Anaxerzia"]
Yes, you are terribly, horribly, absolutely unforgivably misinterpreting something  :P
[/quote]

No offense, but its the other way around

Anaxerzia, Im with nateharman on this.  you are misinterpreting something very basic...

There are only $4$ in the 2X2 square...

You see, in your model 4 of those trianges use the midpoints of the diagonals to form the triangle, the midpoint isnt a point in the grid...

Zuton your answer is a little off.

There are only 4 triangles in the 2X2 grid
and in the
4X4 grid you counted wrong (you stated there were 8 already from 2X2 case, there are not 8 in the 2X2 case, so thus there are only 4 meaning there are only 4 to start with in the 4X4 case-then you add all those akward triangles).  There are more triangles.

[b]Futhermore[/b],
there are triangles no one has even seen or mentioned that are "slanted" across the grid
Take for example
A B C D
E F G H
I. J K L
M N O P
Triangle  OHJ is isoseles :wink:
I believe Zuton did cover this one but he did not cover the fact something like MJO is isosceles (zuton i think you thought MJO was alreayd counted in the 2X2 grid thus you didnt count it in your 3X3 gird when it only appears in 3X3 grid NOT 2X2 grid as you counted (tahts how you got 8 for 2X2), thus throuwing your answer off by a few)

This would make for a great AIME problem (like number 6+..) with all its little tricks

Haha you're right, what was I on? I just sorta looked at lucifer's pictured and barreled off with it.

Heh I was being sarcastic in the first place. And I also went off that dudes picture.

4everwise i get a different answer for the 4X4 case..althought i could be wrong since i wanst very organized
My question is
(0,2),(2,0),(3,3)
(0,1),(3,0),(2,3)

seem like the same triangle..but im not sure..(just rotated by you counted that twice and then muliplied both by 4).


Find the smallest integer $ n$ such that $ 3^n > 1000$.

I guess you have to memorize your powers of three!


inequalities
trigonometry
algebra unsolved
algebra

$\tan x-\sin x=\frac {x^3}{2}+\frac {c^5}{8}$ for some $c\in (0,\pi/2)$.  The result follows.

I can't understand your result. All i know is $ f(x)= \frac{x^3}{2} + \frac{f^{(5)}(c)}{5!}x^5.$ with $ 0 < c < x$
   Do you mean $ \frac{f^{(5)}(c)}{5!}x^5 = c^5/8$ ?

Well, finally I found it! Here is my proof:
  We put $ t=tan(\frac{x}{2})$;
     $ \Rightarrow 0<t<1$ and $ t>\frac{x}{2}$
  We show that $ tan(x)-sin(x)> 4t^3$
  $ tan(x)-sin(x)-4t^3=\frac{2t}{1-t^2}-\frac{2t}{1+t^2}-4t^3=\frac{4t^7}{1-t^4} >0$
  $ \Longrightarrow tan(x)-sin(x)> 4tan^3(\frac{x}{2})> 4(\frac{x}{2})^3= \frac{x^3}{2}$
Hic, this kind of problems makes me feel sad...


Simplify and express your answer as a common fraction. \[ \cfrac1{1\plus{}\cfrac1{\cfrac1{1\plus{}1}\plus{}\cfrac1{1\plus{}1}}}\]


How many distinct prime factors are there for the number $ 280371 \equal{} 23^4 \plus{} 23^2 \plus{} 1$?

[hide]$ 280371 \equal{} 3 \cdot 7 \cdot 13^2 \cdot 79$. So the answer is 4.
Alternatively, $ x^4 \plus{} x^2 \plus{} 1 \equal{} (x^2 \plus{} x \plus{} 1)(x^2 \minus{} x \plus{} 1)$.[/hide]


need help with this problem. you can answer it but please don't give me the answer, just assistance.

If for three distinct positive numbers x, y, and z,

$ y/(x-z)=(x+y)/z=x/y$

then find the numerical value of x/y.

Thanks!

Set all those fraction to be equal with $ \lambda$. You will find $ 3$ different equations. Try to
replace $ x, y$ with something to get an equation with just $ \lambda$ and $ z$. But $ z\neq 0$ so you get an equation with just $ \lambda$. Find this $ \lambda$.

Alexandros

i don't want to sound stupid but i have just begun on this kind of math. what does that upside down Y stand for. sorry, for the stupid beginner question.  :)

It's just the greek letter lambda. It doesn't really matter what you use; I would personally just use "a" instead since it's easier to write/i'm more use to it.  :)

[hide="this is another way of doing it"]We know that $ \frac{y}{(x-z)}=\frac{(y+x)}{(x-z)+(2z-x)}$
So $ y/(x-z)=x/(2z-x)$ but $ y/(x-z)=x/y$.
We deduce that $ y=2z-x$
Now, $ x/y=(x+y)/z=2z/z=2$
[/hide]

thanks guys for the help especially to alexandros who gave me the hint that helped me solve the problem.

the book just used proportion manipulations, like if a/b=c/d than a+c/b+d.
it seemed easier than the way i did it.  :(

still got the same answer though :lol:


Diophantine equation
number theory unsolved
number theory

Find all pairs of nonnegative integers $ \left(m, n\right)$ such that $ \left(m\plus{}n\minus{}5\right)^2\equal{}9mn$.

This is equivalent to solve (9m-2)(9n-2) = -86.
It is easy to see that the solutions are (0,5) and (5,0).

Pierre.

No, the equation is (m+n-5) 2 =9mn. It would have been too easy otherwise

What a shame...I was so proud to solve an exercice from the USA TST in less than 30 seconds... :D

Pierre.

No, what a shame for me. I tried to solve a problem for three hours and I still couldn't solve it.  :(

If we denote by u=m+2,v=n+2, the equation becomes :
u^2 + v^2 - 7uv = - 45 ===> 4u^2 + 4v^2 - 28uv = -180 ===> (2u-7v)^2 - 45 v^2 = - 180
From here it's purely Pell equation-related routine.

2(m+n-5)=9mn then 9mn-2m-2n=-10  =>m(9n-2)-2n+4/9=4/9-10
                     then 9m(9n-2)-2(9n-2)=4-90=-86 =(9m-2)(9n-2)
 I think this problem should
          Find all pairs of nonnegative integers (m, n) such that
             2(n+m-5)^3=9mn^2 :?  :(


How many pairs $(a,b)$ of real numbers satisfy the equation \[{\frac{1}{a}}+\frac{1}{b}=\frac{1}{a+b}\]

$(A)$ none
$(B)$ 1
$(C)$ 2
$(D)$ one pair for each $b\ne0$
$(E)$ two pairs for each $b\ne0$

[quote="Quevvy"][quote="cincodemayo5590"][quote="rnwang2"][hide]
$a+b=\frac{ab}{a+b}\implies (a+b)^{2}=ab\implies a^{2}+ab+b^{2}=0$.

There are no real solutions.[/hide][/quote]

[hide="For those who do not understand the last step,"] look at the discriminant of that last equation as a quadratic in $a$. Since $\triangle=-3b^{2}$ is always less than zero, we know that there will never be a real solution.[/hide][/quote]
What is the triangle?[/quote]
$\triangle=$ discriminant

[quote="Ubemaya"][quote="bpms"]But in the imaginary realm, if a or b are 3rd roots of unity it works?[/quote]

Yeah, I think that works if a and b are both different non-real 3rd roots of unity.[/quote]

Actually, it works for [i]any[/i] pair of different 3rd roots of unity.


For which positive integers $k$, is it true that there are infinitely many pairs of positive integers $(m, n)$ such that \[\frac{(m+n-k)!}{m! \; n!}\] is an integer?

More general:
We can choose $n$ is large enough to be greater than $k$ then let $n-k$ be $t$ with $t$ is a positive integer
Then $\dfrac{(m+n-k)!}{m!n!}$
$=\dfrac{(m+t)!}{m!t![(n-k+1)(n-k+2)...n]}$
$=\binom{m+t}{t}.\dfrac{1}{(n-k+1)(n-k+2)...n}$
On the other hand $\binom{m+t}{t}=\binom{m+t-1}{t-1}.\dfrac{m+t}{t}$
So let $R=(n-k+1)(n-k+2)...n$ we can find $m$ which is big enough that $\dfrac{m+t}{t}=R.x$ then $m=Rxt-t=(Rx-1)t$ then we arrive at the infinity of values of both $m,n$ as desired and the answer is for all $k$ which is a non-negative integer (if $k=0$ then $\dfrac{(m+n-k)!}{m!n!}=\binom{m+n}{m}$ which is clearly an integer)


What's the smallest perfect square divisible by both $45$ and $72$?

[hide="hint"]   What is unique about a squares prime factorization? [/hide]

[hide="i think"]The prime factorization of a prime number has all even numbers, so we want to find the smallest even number exponent factorization with the numbers in the factorizations of 45 an 72 (2, 3, and 5).

$2^2\cdot3^2\cdot5^2=900$

But wait. 72 has 3 twos in the factorization. We must use more than 3 two's in the final number's prime factorization. 4 is the next even number.

$2^4\cdot3^2\cdot5^2=3600$. [/hide]

[quote="math92"]i have been trying and trying to get it but i can't find an answer[/quote]

Here are some more hints:

[hide="hint"] Like I said before, what is unique about a square number's prime factorization?

Here are a few:

$4=2^2$

$9=3^2$

$25=5^2$

$36=(2\cdot3)^2=2^2 \cdot 3^2$

$64=8^2=2^6$

$100=10^2=2^2 \cdot 5^2$[/hide]

[hide="Next hint"]  For a number to be divisible by both 45 and 72 it has to be divisble by there _____.

*hint* it starts with Lowest [/hide]

[hide="Next hint"]  The number we want has to be divisible by both 45 and 72's ______(definition of blank: look above) and it has to be a square number.  So look at the prime factorization of 45 and 72's ________ and try to turn it into a square number, [/hide]

[hide]is is [b]3600[/b]?
45=3^2*5
72=3^2*2^2
the both have 3^2 in common so you do
3^2*2^2*5
which is 360
but since that isn't a square number
i realized 3600 was and that is divisible by 45 and 72[/hide]


Hello!

$ \triangle EFG$ is a triangle. Let $ I$ be the centre of the incircle of $ \triangle EFG$. The incircle meet $ EG$ and $ EF$ in $ K$ and $ L$ respectively. $ J$ is a point on the line segment $ KL$. Prove that :

\[ Area(\triangle JEF)=Area(\triangle JGE)\Longleftrightarrow IJ\bot FG.\]


trigonometry
calculus
calculus computations

Solve $ \frac{d^2y}{dx^2}\plus{}\frac{dy}{dx}\plus{}y\equal{}\sin 2x$

Let
\[ y = A \cos 2x + B \sin 2x.\]
So,
\[ y' = 2B \cos 2x - 2A \sin 2x,\]
\[ y'' = -4A \cos 2x - 4B \sin 2x.\]

Therefore,
\begin{eqnarray*}
(-4A \cos 2x - 4B \sin 2x) + (2B \cos 2x - 2A \sin 2x) + (A \cos 2x + B \sin 2x) & = & (-3A+2B) \cos 2x + (-2A-3B) \sin 2x \\
 & = & \sin 2x.
\end{eqnarray*}
So,
\[ \{\begin{array}{l}
-3A+2B = 0, \\
-2A-3B = 1.
\end{array}\]
Therefore, $ A=-\frac{2}{13}$, $ B=-\frac{3}{13}$.
Thus, particular solution is
\[ y = -\frac{2}{13} \cos 2x - \frac{3}{13} \sin 2x.\]

characteristic equation is
\[ \lambda^2+\lambda+1=0 \ \text{i.r.} \ \lambda=-\frac{1}{2} \pm \frac{\sqrt{3}}{2}i.\]

Thus, general solution is
\[ y=-\frac{2}{13} \cos 2x - \frac{3}{13} \sin 2x + e^{-x/2}( C_1 \cos \frac{\sqrt{3}}{2} x + C_2 \sin \frac{\sqrt{3}}{2} x ).\]


[quote="Arvind_sn"]Billy Bob has some chickens. 5 of his chickens layed 24 eggs in five days. How many days will 8 chickens need to lay 20 eggs?[/quote]
[hide]
5 chickens lay 24 eggs in 5 days, so 1 chicken lays 24 eggs in 25 days. Now we're octupling (is that a word) the work force and the amount of work necessary is $\frac56$
So multiply 25 by $\frac56$ and $\frac18$ to get $\frac{125}{48}$ days... hmm.[/hide]

[hide]one chicken can lay $24 \div25$ in one day.
Therefore, $8$ chickens can lay $24*8/25$ or $7.68$ eggs in one day. $20 \div7.68=2.60416666666666...$ days, which you could round up to $3$ days if you wanted. [/hide]


find the radius of convergence of the series $\sum_{k=1}^{k=\infty}(n+2^{n})z^{n}$

surely it has to converge inside the disk with radius 1/2. but i'm not able to say the exact value because i have a problem with evaluating the limsup thing. can someone please give me hints?

If you use $2^{n}< n+2^{n}< 2 \cdot 2^{n}$, you get bounds for the limsup that suffice to solve it.

thanks zetax.  :)  so with how you have told, we actually see that the radius of convergence is half.


trigonometry
real analysis
real analysis unsolved

Let $x_n$ be the increasing sequence of positive solutions of the equation $tgx=x$. Prove that $\sum_{n\geq 1}{\frac{1}{x_n^2}}=\frac{1}{10}$.

We can do exactly what Simon did in post number $9$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Euler&t=9067]here[/url], by expanding $\frac{\sin x-x\cdot\cos x}{x^3}$.

There is a theorem concerning that type of argument:

[url]http://mathworld.wolfram.com/RootLinearCoefficientTheorem.html[/url]

it is strange, but i got $\frac{1}{5}$ as an answer. It seems to be my very stupid mistake,... or something other

How did you get it? I was also making lots of stupid mistakes and getting different answers at first, but then it seemed to work out just fine. Could you please post what you did, so we can take a look? :)

Oh,I realized where was my mistake: I complete forgot the problem condition saying that we are working only with positive zeros
Now, it is ok, i got the answer


This may be familiar, but I just loved it:

A large cloud consists of tiny water droplets homogenously distributed throughout the entire volume of the cloud. These droplets do not move. A raindrop starts falling through the cloud. What is its acceleration? Assume that the raindrop is spherical at all times, that it acquires mass by absorbing the droplets it encounters, that it falls vertically and that its initial mass and size are negligible.

I got a differential equation in the following form:
$ax^{2}-b\dot x=cx^{2}\ddot x$
is that correct? if not I wont bother solving it.

do you consider a impulse variation or just a simple gain of mass with coordonate?i would like to see the problem resolved.i will understand differentials but i fiind it difficult to apply them since i haven't studied these chapter yet

I think we have $dm = \rho dV = \rho A dx = \rho \pi r^{2}dx$.

But, $dV = 4 \pi r^{2}dr$, thus $4 dr = dx$.

Therefore, $r = \frac{x}{4}$.

So, $dm = \rho \pi \frac{x^{2}}{16}dx$.

We know that $mg = m \frac{dv}{dt}+v \frac{dm}{dt}$.

So, $mg = ma+v*\frac{dm}{dx}* \frac{dx}{dt}= ma+v^{2}*(\rho \pi \frac{x^{2}}{16})$.

Since $m = \rho \frac{4}{3}\pi r^{3}$, dividing by $\rho \frac{1}{3}\pi r^{2}$ gives us:
$xg = xa+3v^{2}$.

Realizing that $a = \frac{dv}{dt}= \frac{dv}{dx}* \frac{dx}{dt}= v \frac{dv}{dx}$, gives us:
$xg = xv * \frac{dv}{dx}+3v^{2}$.

Thus, $\frac{dv}{dx}+\frac{3}{x}v = g v^{-1}$.

This is a Bernoulli DE, so we can substitute $y = v^{2}$, giving $\frac{dy}{dx}+\frac{6}{x}y = 2g$.

Therefore, $y*x^{6}= \frac{2}{7}g x^{7}$.
So, $y = \frac{2gx}{7}$.

From this, $v = \sqrt{\frac{2}{7}gx}$.

So, $\frac{dx}{2 \sqrt{x}}= \sqrt{\frac{g}{14}}dt$.

So, $x = \frac{gt^{2}}{14}$.

The acceleration is then the second derivative, or $\frac{g}{7}$.

Is this correct?

I think you got the right answer

This is a Vietnamese Physics Forum discussing about that problem

http://www.vatlyvietnam.org/forum/showpost.php?p=10388&postcount=5

That was a beautiful problem and beautifull solution. :lol:
I started messing with Stokes drag force, and it got really ugly.

Yup, that's the solution! Good job  :lol:


[quote][color=darkred]Let $AXYB$ be a convex quadrilateral inscribed in the circle $w$ with the diameter $[AB]$.

Denote : $I\in AY\cap BX$ ; $D\in AB$ , $ID\perp AB$ ; $Z\in XY\cap ID$.

Prove that $\{\begin{array}{c}m(\widehat{BAY})=x\\\ m(\widehat{ABX})=y\end{array}\|$ $\Longrightarrow$ $\frac{DA}{DB}=\frac{\tan y}{\tan x}$ , $\frac{ZX}{ZY}=\frac{\sin 2y}{\sin 2x}$ and $\boxed{\ \frac{IZ}{ID}=\frac{\cos (x+y)}{\cos (x-y)}\ }$.

[b]Remark.[/b] $\{\begin{array}{c}XY=AB\cos (x+y)\ ,\ \boxed{\ ZD=AB\cdot\frac{\sin 2x\sin 2y}{\sin 2x+\sin 2y}\ }\\\\ p_{w}(I)=IA\cdot IY=IB\cdot IX=AB^{2}\cdot \frac{\sin x\sin y\cos (x+y)}{\sin^{2}(x+y)}\end{array}$.[/color][/quote]

[quote="Virgil Nicula"][color=darkred]Let $AXYB$ be a convex quadrilateral inscribed in the circle $w$ with the diameter $[AB]$.

Denote : $I\in AY\cap BX$ ; $D\in AB$ , $ID\perp AB$ ; $Z\in XY\cap ID$.

Prove that $\{\begin{array}{c}m(\widehat{BAY})=x\\\ m(\widehat{ABX})=y\end{array}\|$ $\Longrightarrow$ $\frac{DA}{DB}=\frac{\tan y}{\tan x}$ , $\frac{ZX}{ZY}=\frac{\sin 2y}{\sin 2x}$ and $\boxed{\ \frac{IZ}{ID}=\frac{\cos (x+y)}{\cos (x-y)}\ }$.

[b]Remark.[/b] $\{\begin{array}{c}XY=AB\cos (x+y)\ ,\ \boxed{\ ZD=AB\cdot\frac{\sin 2x\sin 2y}{\sin 2x+\sin 2y}\ }\\\\ p_{w}(I)=IA\cdot IY=IB\cdot IX=AB^{2}\cdot \frac{\sin x\sin y\cos (x+y)}{\sin^{2}(x+y)}\end{array}$[/color][/quote]

[color=indigo][b]Proof.[/b] We have:
$1.\blacktriangleright\quad \frac{DA}{DB}=\frac{DA}{DI}\cdot\frac{DI}{DB}=\frac{\tan y}{\tan x}$

$2.\blacktriangleright\quad\frac{ZX}{ZY}=\frac{ZX}{ZI}\cdot\frac{ZI}{ZY}=\frac{\sin\widehat{ZIX}}{\sin\widehat{ZXI}}\cdot\frac{\sin\widehat{ZYI}}{\sin\widehat{ZIY}}$ $=\frac{\sin(90^\circ-y)}{\sin x}\cdot\frac{\sin y}{\sin (90^\circ-x)}=\frac{\sin 2y}{\sin 2x}$

\begin{eqnarray*}3.\blacktriangleright\quad\frac{IZ}{ID}&=&\frac{IZ}{IY}\cdot\frac{IY}{IB}\cdot\frac{IB}{ID}\\ &=&\frac{\sin\widehat{IYZ}}{\sin\widehat{IZY}}\cdot\sin\widehat{IBY}\cdot\frac{1}{\sin\widehat{IBD}}\\ &=&\frac{\sin\widehat{IYZ}}{\sin(\widehat{IYZ}+\widehat{ZIY})}\cdot\sin\widehat{IBY}\cdot\frac{1}{\sin\widehat{IBD}}\\ &=& \frac{\sin\widehat{IYZ}}{\sin(\widehat{IYZ}+90^\circ-\widehat{IAD})}\cdot\sin(90^\circ-\widehat{IBA}-\widehat{IAB})\cdot\frac{1}{\sin\widehat{IBD}}\\ &=&\frac{\sin y}{\cos (x-y)}\cdot\cos (x+y)\cdot\frac{1}{\sin y}\\ &=&\frac{\cos (x+y)}{\cos (x-y)}\end{eqnarray*}

$4.\blacktriangleright\quad XY=AB\cdot\sin\widehat{XBY}=AB\cdot\cos (x+y)$

$5.\blacktriangleright\quad ID=IA\cdot\sin x=\frac{IA}{\sin y}\cdot\sin x\sin y=\frac{AB}{\sin (x+y)}\cdot\sin x\sin y$

On the otherhand, we have $\frac{IZ}{ID}=\frac{\cos (x+y)}{\cos (x-y)}\Longrightarrow\frac{ZD}{ID}=\frac{2\cos x\cos y}{\cos (x-y)}$ $\Longrightarrow ZD=ID\cdot\frac{2\cos x\cos y}{\cos (x-y)}=AB\cdot\frac{\sin 2x+\sin 2y}{\sin 2x+\sin 2y}$

$6.\blacktriangleright\quad \frac{IY}{ID}=\frac{\sin\widehat{IDY}}{\sin\widehat{IYD}}=\frac{\cos (x+y)}{\sin y}$ $\Longrightarrow IA\cdot IY=\frac{ID}{\sin x}\cdot\frac{ID\cdot\cos (x+y)}{\sin y}=ID^{2}\cdot\frac{\cos (x+y)}{\sin x\sin y}=AB^{2}\cdot\frac{\sin x\sin y\cos (x+y)}{\sin^{2}(x+y)}$
[/color]


why do you want to spam? AOPS was meant to discuss math, not to spam.

whats the point of this?  I agree with h78reg.  this is a place for math.  there are just two forums to discuss other things, and even those are not to increase your post count and spam.  spam wastes others time. spam is wasting mods time because they have to get rid of it.  so jhollenbeck, and the others of you that spam, stop it immediately

Yes, AoPS is to discuss math, but maybe if there is a spam forum, the spammers will spam there instead of elsewhere...

Probably not though.

They might, but then theyll learn that nobody reads their spam or cares about their post count so they'll spam in the real forums.

post counts should really be deleted. they seem to be causing all this spam..........

[quote="jhollenbeck"]should we make a new forum where you can spam like crazy and not get in trouble? Vote yes or no until a mod locks it[/quote]

Based on your current explosion of spam, it seems it might do some use for you. And why spam in the first place?

[quote="star99"]just think.  if you make a spam forum which doesn't increase their post count, no one will spam in it.  why? becasue the reason people spam is to increase their post count, not just for the fun of it.

end of discussion

mods, can you please delete this topic.  thanks[/quote]

Most spammers don't spam to increase their post count. In fact, most people don't care about their (or anybody else's) post count at all. I certainly don't take it into account.

[quote]Most spammers don't spam to increase their post count. In fact, most people don't care about their (or anybody else's) post count at all. I certainly don't take it into account.[/quote]

then what's the point of spamming?

whats the point it will just take up space on the servers

[quote="jhollenbeck"]should we make a new forum where you can spam like crazy and not get in trouble?[/quote]I suggest people that need to spam so badly to open up a new .txt file on their desktop and start spamming. No one will ever interfere with that :) :D
Until then, this topic is closed.


calculus
derivative
logarithms
function
calculus computations

In my opinion, the symbol $ \sqrt [n]{x}$ is defined only if $ n\in\mathbb{Z}_{\plus{}}$. Unfortunately I'm not sure about that. But if $ \sqrt [x]{x}\equal{} x^{1/x}$ then the differential is
$ \left({\frac{1\minus{}x}{1\plus{}x}}\right)^{{x}^{\minus{}1}}\left(\minus{}\ln\left({\frac{1\minus{}x}{1\plus{}x}}\right){x}^{\minus{}2}\plus{}\left(\minus{}\left( 1\plus{}x\right)^{\minus{}1}\minus{}{\frac{1\minus{}x}{\left( 1\plus{}x\right)^{2}}}\right)\left( 1\plus{}x\right){x }^{\minus{}1}\left( 1\minus{}x\right)^{\minus{}1}\right)$
I'm too lazy to check where the function is differentiable.  :P

[quote="7h3.D3m0n.117"]Differentiate the following with respect to $ x$.

$ \sqrt [x]{\left(\frac{1\minus{}x}{1\plus{}x}\right )}$[/quote]
let $ y\;\equal{}\;\left(\frac{1\minus{}x}{1\plus{}x}\right )^{\frac{1}{x}}$
$ \Rightarrow\;\ln y\;\equal{}\;\frac{1}{x}\cdot\ln\left(\frac{1\minus{}x}{1\plus{}x}\right )$
differentiating both sides w.r.t. $ x$ gives
$ \frac{1}{y}\cdot\frac{dy}{dx}\;\equal{}\;\frac{1}{x}\cdot\frac{d}{dx}\left(\ln\left(\frac{1\minus{}x}{1\plus{}x}\right )\right)\;\plus{}\;\ln\left(\frac{1\minus{}x}{1\plus{}x}\right )\cdot\frac{d}{dx}\left(\frac{1}{x}\right)$
$ \Rightarrow\;\frac{dy}{dx}\;\equal{}\;y\;\cdot\;\left\{\left(\frac{1}{x}\cdot\left(\frac{1\plus{}x}{1\minus{}x}\right )\cdot\left(\frac{(1\plus{}x)(\minus{}1)\minus{}(1\minus{}x)(1)}{(1\plus{}x)^{2}}\right)\right)\;\plus{}\;\left(\ln\left(\frac{1\minus{}x}{1\plus{}x}\right )\cdot\;\left(\minus{}\frac{1}{x^{2}}\right)\right)\right\}$
$ \frac{dy}{dx}\;\equal{}\;\left(\minus{}\frac{1}{x^{2}}\right)\cdot\left(\frac{1\minus{}x}{1\plus{}x}\right )^{\frac{1}{x}}\cdot\left(\frac{2\,x}{(1\minus{}x^{2})}\;\plus{}\;\ln\left(\frac{1\minus{}x}{1\plus{}x}\right )\right)$
i believe the function is differentiable [b]only [/b]within $ \minus{}1\;<\;x\;<\;1$

note that $ \frac{dy}{dx}\;\equal{}\;0\;\;\;\quad\text{at}\;\;x\;\equal{}\;0$
in general, for $ \minus{}1\;<\;x\;<\;1$, we have $ \minus{}\frac{1}{2}\;<\;\frac{dy}{dx}\;<\;\frac{1}{2}$


Show that the function $ f: N \times N \rightarrow N$ given by $ f(1, n) \equal{} 2n\minus{}1$ and $ f(m\plus{}1, n) \equal{} 2^m(2n\minus{}1)$ is a one-to-one function.

PS: N is the natural set.

Someone could help me, please!

any natural number can be represented in form $ 2^a(2b \minus{} 1),\,(a,b\in N)$ exactly in one way.

Let $ f(x_1,y_1)\equal{}f(x_2,y_2)$. Then $ 2^{x_1\minus{}1}(2y_1\minus{}1) \equal{} 2^{x_2\minus{}1}(2y_2\minus{}1)$.
We claim that $ x_1\equal{}x_2.$ Suppose that $ x_1>x_2$. Than we have $ 2^{x_1\minus{}x_2}(2y_1\minus{}1) \equal{} 2y_2\minus{}1$. Where LHS is even and RHS is odd. contradiction.
if $ x_1<x_2$ then proof are similar.
Thus $ x_1\equal{}x_2$ and we obtained $ 2y_1\minus{}1\equal{}2y_2\minus{}1$. It follows that $ y_2\equal{}y_1$


Frankinfueter and Deberg are playing a game with fair, six sided, dice. The object of the game is simple - each person rolls two dice and the person with the higher product wins. Frankinfueter gets 15 points by rolling a three and a five. What is the probability that Deberg beats Frankinfueter?

[hide]we are looking for the probability that the other person gets 16+ points: so lets list the possible values:
here is a table that lists all possible pairings of the numbers rolled by the dice... the bold ones are the ones that satisfy our conditions.
1:1,2,3,4,5,6
2:1,2,3,4,5,6
3:1,2,3,4,5,[b]6[/b]
4:1,2,3,[b]4,5,6[/b]
5:1,2,3,[b]4,5,6[/b]
6:1,2,[b]3,4,5,6[/b]
we see that ther are 11 ways to get the product greater than 15. so the probability is: [b]11/36[/b][/hide]

Gaah, forgive me for deleting my posts...

I thought 1 + 3 + 3 + 4 = 10 for a second o_O .

Your answer seems correct.

there are 36 ways to roll
since 6^2 equals to it..
and ....

the lowest possible number by3 is 6
the next one is
by 4 : 4
obviously rest of the number (5,6) for the first die ,4, works..

hm...
5:4 and rest
6:3 and the rest
so the total numbers of possible combinations are.. 11.

11/36

[hide]If Deberg rolls a 1 or 2, he lost for sure, so that takes 2*6 possiblities away from him.
If he rolls he three he can only survive if he gets a six so that takes 5 possibilites away from him.
If he rolls a four or a five he can only survive if the other one is 4,5,6;so 3*2 possibilites are taken away from him.
If he rolls a six, he wins if he gets a 3,4,5,6 so that only takes two possibilities away from him.  Since we want to find his chances of winning, we subtract the taken possibilities from the total.
36/36-(12+5+6+2)/36=11/36[/hide]