calculus
integration
vector
function
derivative
geometry
3D geometry

Let $ \Omega$ be the region in $ R^{3}$ defined by $ \Omega= \{ (x_{1},x_{2},x_{3}): \max(|x_{1}|,|x_{2}|,|x_{3}|) \leq 1 \}$.  Let $ \partial \Omega$ denote the boundary of $ \Omega$.

(a)  Calculate $ \int_{\partial \Omega}\phi F \cdot n \hspace{1 mm}d\sigma$, where $ n$ is the unit normal vector, $ d\sigma$ denotes integration over $ \partial \Omega$,

$ F_{i}= \frac{x_{i}}{(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})^{\frac{3}{2}}}= \frac{x_{i}}{r^{3}}$, and $ \phi(y_{1},y_{2},y_{3})$ is a continuously differentiable function of $ y_{i}=\frac{x_{i}}{r}$.  Assume that $ \phi$ has unit mean over the unit sphere.

(b)  Generalize to $ R^{4}$.

So, $ \vec F$ is a conservative vector field: $ \vec F=\nabla(-r^{-1})$. The function $ \phi$ is homogeneous of degree $ 0$, which means it is constant on half-lines emanating from the origin.
The [url=http://en.wikipedia.org/wiki/Divergence]divergence[/url] of $ \phi \vec F$ is zero outside of the origin, because $ \nabla\cdot (\phi \vec F)=\nabla\phi\cdot\vec F+\phi\nabla\cdot\vec F$, where $ \nabla\cdot\vec F=0$ by direct calculation, and $ \nabla\phi\cdot\vec F=0$ because $ \vec F$ is radial and $ \phi$ has zero derivative in the radial direction.
The divergence theorem allows you to replace $ \partial \Omega$ with a sphere, on which $ \vec F\cdot \vec n$ is constant, and the integral is easily evaluated.

In higher dimensions $ n$ you need $ \vec F$ to be the gradient of $ -r^{2-n}$. Indeed, the divergence of $ \vec F$ is the Laplacian of $ -r^{2-n}$, which is known to be zero.


calculus
integration
college
real analysis
real analysis solved

Suppose that $f(x)$ is continious, nonnegative for $x \geq 0$, with $\int_{0}^{\infty}f(x)dx<\infty$. Prove that $lim_{n\to\infty}{\int_{0}^{n}{\frac{xf(x)}{n}}dx}=0$

Put $F(x)\int^x_0 f(t)dt$. We know that $F(0)=0,\ F$ is increasing, and it tends to a finite limit $\ell\in\mathbb R$ as $x\to\infty$. The conclusion we must prove is that $F(n)-\frac 1n\cdot\int^n_0 F(x)dx\to 0$ as $n\to\infty$.

Now, $F(n)\to\ell$, of course, so we have to show that $\frac 1n\cdot\int^n_0 F(x)dx\to\ell$ too. Since $\frac 1n\cdot\int^n_0 F(x)dx$ is equal to some $F(t)$, for some $t\in(0,n)$, it means that itr's always $\le\ell$. For $\varepsilon>0$, choose $m$ large enough so that $F(x)>\ell-\varepsilon,\ \forall x\ge m$. We get $\frac 1n\cdot\int^n_0 F(x)dx=\frac 1n\cdot(\int^m_0 F(x)dx+\int^n_m F(x)dx)$, which is $>\frac 1n\cdot\int^m_0 F(x)dx+\frac{n-m}n\cdot(\ell-\varepsilon)$ for $n\ge m$, and, by looking at what happens when $n$ becomes large, we immediately get the conclusion.

It is too easy problem in comparing with other problems from Berkeley, which Eugene have already posted. :? Right?

never mind, Myth, if you want i'll post more difficult

It was just an observation. Never mind too ;) :)


[quote="aidan"]The ski lift at Ski-Fall Creek Resort is one where the seats move continuously on a continuous cable, taking skiers from the bottom to the top of the hill. Skiers get on and off the lift while it is moving. The seats are equally spaced.

Amy began to ski down from the top just as her friend David got onto the ski lift at the bottom. After skiing down, Amy immediately got on the lift to the top and skied down to the bottom again, meeting David who had sat on the lift for a complete circuit and had just arrived at the bottom again. On each downhill run, Amy skied at a constant speed of 60km/h to cover the 1500 metres from the top to the bottom. If a seat passes through the bottom of the lift every 15 seconds, calculate the number of seats on the lift.

Source: Mathematics Challenge for Young Australians, Problem 72[/quote]
[hide]
I got 24...

Since the guy goes thru the ski lift twice while Amy goes through once and skis down twice, the ski lift is half as slow as Amy (30 km/h). So then that's 500 m/min and then time is d/s=1500/500=3 minutes=180 seconds, 180/15=12, then you multiply by two...[/hide]


The sequence $ (a_n)$ is such that $ a_1 \equal{} 1$ and $ a_o \minus{} a_n$  for $ o \equal{} n\plus{}1$ is always either 0 or 1.

It is also known that $ a_n \equal{} \frac{n}{1000}$ for some natural $ n$.

Prove that $ a_h \equal{} \frac{h}{500}$ for some natural $ h$

[hide="Loses some rigor near the end, but I think it works"]
It follows from the definition of $ \{a_n\}$ that $ \{a_n\}$ is the sequence of the natural numbers, with repetition possible, but still following the general order and not skipping any integers. For example, $ \{a_n\}$ could be $ 1, 1, 1, 2, 3, 3, 4, 4, 4, 5, 6, 7, 7, \cdots$

Since $ a_n$ is a natural number, it follows that for the value of $ n$ such that $ a_n \equal{} \frac{n}{1000}$, $ n$ is a multiple of 1000. Thus, let this value be $ n \equal{} 1000k$.

Since at this value, $ a_n \equal{} k$, and from our definition of $ a_n$, we have $ a_i \leq k$ for $ i \leq 1000k$.

Consider the sequence members:
$ a_{500}, a_{1000}, \cdots , a_{500h}$

We wish to show that for at least one of these, $ a_{500k} \equal{} k$

If we consider the first $ k$ of these values, each comes before $ a_n$, because $ 500k \leq 1000k$. Thus, each one is between $ 1$ and $ k$ inclusive. If the k'th one does not equal k, then none of the previous ones can equal k, and then we have (k-1) values distributed among the previous (k-1) members of the sequence. We can continue this chain of logic to show that, at the very least, $ a_{500} \equal{} 1$ , as desired.
[/hide]


analytic geometry
geometry
trigonometry
calculus
integration
conics
parabola

Find the area enclosed by $ 1\le x^2\plus{}y^2\le 4$ and $ y\le x^2.$

I know that polar coordinates is pretty amendable for this case but I'm quite lost. For example, the first condition in polar coordinates is $ 1\le r^2\le 4$ and that would imply $ 1\le r\le2$ and as for the other one we'd get $ r\sin\varphi\le r^2\cos^2\varphi$ and this is $ \tan(\varphi)\sec(\varphi)\le r,$ however I can't set up the double integral in polar coordinates.

Recalibrate your expectations. This one is ugly no matter what you do.

At first glance, it looks like we want to integrate $ d\phi$ first, so that we have the nice $ r$ limits on the outside. That doesn't work, because we would need to invert $ r\equal{}\tan\phi\sec\phi$ to find the $ \theta$ limits.
OK, what about doing it the other way? Now it's a piecewise integral. There are two intersection points between the parabola and each circle, and these four points split $ \phi$ into four regions. In one of those regions, there's no area and we can ignore everything. In one, both the inner and outer boundaries of the region are the circles. In the other two, the inner boundary is part of the inner circle and the outer boundary is the parabola. Fortunately, we can use symmetry for this part.

Those intersection points are not nice values. In rectangular coordinates, you have to solve a quadratic and then take another square root to get there. In polar coordinates, the angles aren't rational multiples of $ \pi$.

Ahh well, I just found another version of this problem given $ 4\le x^2\plus{}y^2\le18$ and $ 3y\le x^2.$ This one has nice numbers (according to the book) but no idea how to tackle it.

Yes, those numbers are nicer- you'll get nice angles like 30 and 45 degrees, for an answer with nothing worse than square roots of rationals and rational multiples of $ \pi$. I would have chosen $ 2\le x^2\plus{}y^2\le 12$ and $ y\le x^2$, which is essentially the same thing.

I outlined the procedure in my previous post, and that still works. One thing to note is that I was slightly confused and thinking of the region above the parabola rather than below; this changes which curves are on the inside and outside.

Draw the picture. Mark the intersection points, and use them to slice up the integral into several pieces. If you apply the obvious symmetries, you only need to calculate one nontrivial integral.

Here's the sketch (using symmetry we just duplicate our result):

[url=http://www.imagehosting.com/][img]http://img243.imageshack.us/img243/9655/sdfsdfp.png[/img][/url]

What I was initially doing, was to locate where the parabola cuts the smaller and the bigger circle. The first point is $ \left( \sqrt{3},1 \right)$ and second one is $ (3,3),$ and this points correspond to the angles $ \frac\pi6$ and $ \frac\pi4,$ then I ended up with the double integral $ \int_{\frac{\pi }{6}}^{\frac{\pi }{4}}{\int_{?}^{?}{r\,dr}\,d\varphi },$ but don't know how to get the other bounds and I think there's another double integral. (Because in rectangular coordinates we have two double integrals, then I'm thinking that in polar we'd get two ones.)

That's the wrong region- you want to be below the parabola rather than above. That's why it was $ 3y\le x^2$.

Now that you've got the picture, add in some rays from the origin through the intersection points. In each region defined by those, one curve will be the inside of the region and one curve will be the outside. The limits of the inner integral are equations of those curves.

Of course!!!! I didn't realize about that region, yes.

Is the double integral $ \int_{0}^{\frac{\pi }{6}}{\int_{3\tan \varphi \sec \varphi }^{2}{r\,dr}\,d\varphi \plus{}\int_{0}^{\frac{\pi }{4}}{\int_{2}^{3\sqrt{2}}{r\,dr}\,d\varphi }}$ ?

I think not haha, thanks for your help anyway.

No. For one thing, the circle $ x^2\plus{}y^2\equal{}4$ is never the outer curve, as you have in the first integral. For another, the two integrals shouldn't overlap in angle. Finally, $ 0$ is not a special angle; the $ x$-axis isn't a boundary.

[hide="My version"]$ \int_{\pi/6}^{\pi/4}\int_{3\tan\theta\sec\theta}^{3\sqrt{2}}r\,dr\,d\theta \plus{} \int_{\minus{}7\pi/6}^{\pi/6}\int_2^{3\sqrt{2}}r\,dr\,d\theta \plus{} \int_{\minus{}5\pi/4}^{\minus{}7\pi/6}\int_{3\tan\theta\sec\theta}^{3\sqrt{2}}r\,dr\,d\theta$
(I prefer $ \theta$ for the polar angle; it's the standard in the textbooks I've learned from. We can use any $ 2\pi$ segment of angles to define things, and for this problem it's convenient to put the cut on the vertical axis at $ \frac{\pi}{2}$ or $ \minus{}\frac{3\pi}{2}$)
Of course, the third integral is equal to the first by symmetry, so we don't actually have to do the work for it.[/hide]


How do you prove that if to multiply a number of any base by its base you just add a zero to that number?

[hide="Proof"]Let $ a$ be the original number you're multiplying by. $ b$ written in base $ b$ is simply $ 10$, so when you write out the multiplication, you simply append a zero to the end of the number by definition of multiplication.[/hide]

[hide="Proof"]Let $ k$ be the base you are considering.

We note that the maximum value for a digit in base $ k$ is $ k\minus{}1$.  Therefore, to multiply by $ k$ we are clearly simply adding a digit.[/hide]


A piece of metal is placed in a plastic bowl that is floated in a bath full of water. The metal is then taken out of the bowl and dropped into the water. Does the water level fall, rise, or remain the same?

This is the same question as eminem's boat and stone question..
The level would fall...but spoon was arguing with me about somethin about surface area?

[quote]eminem's boat and stone question[/quote]

What is this?!

shadysaysurspammed reffers to this thread:

http://www.mathlinks.ro/Forum/viewtopic.php?t=47028

hey shady , why dont ya post your solution , the one you told me that day on the phone??

ummmm, i aint sure if its absolutely same as mine , might be some difference......

hey guys sure this is the same question as mine?

I suppose it is the same and no need for me to post the sol,just refer to the thread given by bus.


How do you solve problems?
I'm talking about the thinking mechanism.

This thread might turn out interesting (for me at least).

I estimate a low number of posts on this topic because nobody knows for sure how their mind works.
For some it doesn't work at all :D  For most of us it happenes not to work  :( from time to time.

So the tipical first reaction when tackling a problem is to fit it into a class of familiar problems. If you know the proof of the problem, then you are done.
If you don't know the problem but you solved smth similar you start to build logical chains from the conditions in the hypothesis and statements you know to be true.
If this does not lead to the conclusion, then you start to build chains going down from the conclusion. What I mean is that you try to find a conjecture (for now) that would lead to the conclusion. This is hard enough, but then you have to turn the conjecture into a theorem also.
If the 2 chains above built have common propositions, then the proof is complete.
However it is often hard to join the 2 chains and a genius's spark is needed.

Unfortunately no matter how much you've  worked, you will find a problem like you've never seen before. This problems need lots of sparks and good luck.

[quote="amfulger"]How do you solve problems?.[/quote]

There are no difficult problem, which, once considered the right way, can not become ever more ... difficult  :D So stop searching  \downto  \varepsilon

Poyla in his famous book, "Woww, you solved it"  :),  tried to formalize the way to solve problems. Did you read this book ?

You are somewhat right. I didn't read the book. Is the title you wrote good, for I might want to read it?
But in you quiet moments, don't you take some time for introspection? I know I do.
Why should I stop searching for something just because the others before me failed? If yes, why keep solving problems if there is a limit in what you can find? (I mean what not how much because unfortunately life is limited  :(  )

I some what agree with amfulger's opinion exposed above. mainly the question is similar with the following: "how does the human brain work?" :D

obivously not with Go Nigger' Cereals :D:D

I use Feynman method  :
1. write the problem
2. think very, very hard
3. write the solution  :D  :D  :D

when u are doing this are u using Go Nigger' Cereals??? kuz it seems to be very easy for u to solve the pbs :D:D:D:D:D

ok. for all those of you which have no idea what Nigger' Please Cereals are take a look [url=http://www.mathlinks.go.ro/Chris%20Rock%20Show%20-%20Nigga%20Please%20Cereal.mpg]HERE[/url] (700kb file) :D:D

Go nigga cereals is the only true thing in life,yeaah [img]http://community.the-underdogs.org/smiley/happy/wave.gif[/img]

To solve really hard math problems, first think about it. Then think about it some more. Repeat 5 times. Repear 6 times. Alright, 32 times is enough thinking. Now Think about it from a different angle. If you are really stuck, ask someone for help, or go to sleep. :D


A set S contains all seven letter strings that can be created using any of the 26 letters (as many times as you like) in any of the seven places and there is no block that contains "AMC" (in that order) in the string. So for example, REAMCER would not be included in this set.

How many letter strings are in this set?

Umm I moved this from some other forum to get more answers

EDIT: sorry typo

The elements of the set are all seven-letter strings, so there are $ \boxed{0}$ numbers in the set.
Or perhaps there are other elements in the set, so there is not enough information?
Or you might have meant strings?

I think he meant how many letter strings in the set.

[hide]there are 26^7 total ways. Ways that are not possible are
AMC_ _ _ _=26^4 WAYS
_AMC_ _ _= 26^4 WAYS
_  _ AMC _ _=26^4 WAYS
_ _  _ AMC_= 26^4 WAYS
_ _ _ _AMC= 26^4 WAYS
the number of ways are
26^7-5(26^4)[/hide]

You overcounted those with more than one AMC...

i still dont trust this "andrewliu" kid. he seems very malicious.

actually, when mathcrazed said overcounted, hes techinically means the entire answer is undercounted, only the AMC part is overcounted (that you have to subtract)

[hide]
I think you have to add back $ 3\cdot26$, all the letter strings with 2 "AMC"s.  This is because the one letter that is not part of an "AMC" can either be at the beginning, in the middle of the "AMC"s, or at the end, and there are 26 choices for that letter.  So the answer is $ 26^7\minus{}5\cdot26^4\plus{}3\cdot26 \equal{} \boxed{8029525374}$.[/hide]


Hi,

This is a problem in one of my computer science classes, Theory of Computing, the class is based around Discrete Math and advanced data algorithms.  Well I am terrible at math proofs, and on our first homework assignment we got a problem that neither me nor my roommates have been able to figure out.  The solution my be simple... I don't know... but I have been working on it for 2 days now :-P I was hoping that someone would be able to help or guide me along on this problem.

Here is the problem:

Prove 1+2^n < 3^n when n>2 using induction.

So I showed it works for n=3, and assumed n=k and now im tryiing to show, 1+2^(k+1)<3^(k+1)...
but I can't figure out what to do.

Thank you for your help ahead of time.
-Duncan


In the popular game of Blackjack, aces can count for either $ 1$ or $ 11$. Jacks, queens, and kings are worth $ 10$, and all other cards are worth their face value. The goal of Blackjack is to have the cards you own add up to exactly $ 21$ without going over. If order is irrelevant and you are using a standard $ 52$-card deck, how many $ 3$-card combinations from that deck exist that add up to $ 21$?

[hide="Solution"]I'm assuming that suit matters:

First of all we note that if a certain combination of values contains three distinct values, there are 64 ways we can choose the suits for those cards. If there are two the same and one distinct there are 24 ways we can choose their suits. If the three cards have the same value there are 4 ways we can choose their suits. Also note that 1 and 11 are not distinct values (they are both an ace) and that we must account for diffreent cards that make a value of 10 by multiplying by the number of ways to pick from JQK.

Now we look at possible values

Using values of at most 11 and including an 11: 240

(11,9,1),(11,8,2),(11,7,3),(11,6,4),(11,5,5)

24+64+64+64+24=240

Using values at most 10 and including a 10: 840

(10,10,1),(10,9,2),(10,8,3),(10,7,4),(10,6,5)

24*3+64*3+64*3+64*3+64*3=840

Using values at most 9 and including a 9: 176

(9,9,3),(9,8,4),(9,7,5),(9,6,6)

24+64+64+24=176

Using values at most 8 and including an 8: 88

(8,8,5),(8,7,6)

24+64=88

Using values at most 7 and including a 7: 4

(7,7,7)

4

Adding these we get 240+840+176+88+4= $ 1348$ ways to make 21 from three cards.

I hope I didn't miss any.
[/hide]

EDIT: It's incorrect so don't bother reading it.

[quote="darkdieuguerre"]There are $ 16$ cards with a value of $ 10$.[/quote]##

Whoops, my mistake.

[hide="Solution (hopefully correct now)"]I'm assuming that suit matters:

I'll list all the possible value combinations and below will be the number of ways to assign actual cards to each value:

Now we look at possible values

Using values of at most 11 and including an 11: 240

(11,9,1),(11,8,2),(11,7,3),(11,6,4),(11,5,5)

24+64+64+64+24=240

Using values at most 10 and including a 10: 1504

(10,10,1),(10,9,2),(10,8,3),(10,7,4),(10,6,5)

480+256+256+256+256

Using values at most 9 and including a 9: 176

(9,9,3),(9,8,4),(9,7,5),(9,6,6)

24+64+64+24=176

Using values at most 8 and including an 8: 88

(8,8,5),(8,7,6)

24+64=88

Using values at most 7 and including a 7: 4

(7,7,7)

4

Adding these we get 240+1504+176+88+4= $ 2012$ ways to make 21 from three cards.

I hope I didn't miss any. [/hide]


If you guys liked the sets I've posted over the past couple of weeks, you'll love this.

Here's a link to the new track that I just finished.  It's based on an old chord progression that was written by composer Jean Michel Jarre.  I just thought this would sound wonderful as a warm, uplifting, airy trance track.  Of course, I added some nice basslines and a counter-melody or two.

[url=http://www.mp3unsigned.com/bands/1315node93_-_souvenir_de_chine_2006.mp3]http://www.mp3unsigned.com/bands/1315node93_-_souvenir_de_chine_2006.mp3[/url]

I made this in ModPlug tracker and Adobe Audition, with samples & instruments that I either downloaded freely, recorded, or made myself.  Please enjoy!  Any feedback (good or bad) is more than welcome!

[quote="Singular"]This is so ridiculously tight its not even funny.  You are obviously a pro.  Remind me to buy a drink for you when I see you at Waterloo.[/quote]

Heh.  Thanks for the comment, but I'll assure you that there are others much more talented than me.  I actually don't drink, but if you want to reward me with something equally costly and equally bad for my health, you could buy me a brick of fudge and I'd be just as happy.

[quote="Sunny"][quote="Elyot"]
Heh.  Thanks for the comment, but I'll assure you that there are others much more talented than me.  I actually don't drink, but if you want to reward me with something equally costly and equally bad for my health, you could buy me a brick of fudge and I'd be just as happy.[/quote]

FUDGE ROCKS.  Mmmmm amazing stuff.  If only it weren't so bad for you....*sigh*[/quote]
I'm scared of fudge, because of the Mao game we had. :rotfl:

[quote="lightrhee"]
I'm scared of fudge, because of the Mao game we had. :rotfl:[/quote]
 :D   If i make seminars again, i'll make sure to bring you lots.   ;)

[quote="Sunny"][quote="Elyot"]
Heh.  Thanks for the comment, but I'll assure you that there are others much more talented than me.  I actually don't drink, but if you want to reward me with something equally costly and equally bad for my health, you could buy me a brick of fudge and I'd be just as happy.[/quote]

FUDGE ROCKS.  Mmmmm amazing stuff.  If only it weren't so bad for you....*sigh*[/quote]

Heheh.  Sarah, I wrote that comment specifically so that you would be driven to post.  If you come to seminar, I'll definitely show up with some fudge for you (and maybe a little for everyone else too!)  Except of course, David... he doesn't have to have any.

And fudge isn't bad for you... well... maybe a little... but it's fine if combined with DDR.

And Sarah... you BETTER make seminar again.  Cuz if I have to wait until fall to see you, I'll cry!

This is rap music :D , But I didn't hear your voice, man. Why you didn't sing?. If you have another music, you can upload  :D. I will hear all of them :oops:

[quote="WindingFunction"]Elyot! I honestly generally don't like this genre of music, but I must say that this seems to have been very carefully crafted.  Well done! For me, though, the piece ended at around 7 minutes and 40 seconds.  The whole thing builds up, and is very intense, with the persistent background pulse throughout, and I don't feel like it should fade away at the end.  I like your idea of staying on the same chord for a long time after 7'40", but I feel like this builds anticipation rather than signaling fading away, and that it should go somewhere even more climactic afterwards.  Just my thoughts...I studied composition a bit this year, but I'm no expert, and you certainly don't have to listen to me. :P[/quote]

Thanks for your comments!

Actually, your point about the ending is quite valid; however, this is no radio edit of a dance track.  The idea of this genre of electronic music is that a DJ will play many such tracks in a mix or set, and transition gradually from one track to the next by matching the tempos and syncing the beats (beat-matching, which is actually quite difficult) and gradually mixing into the next track by fading certain frequency bands in and out (EQing, using an equalizer), or sometimes just crossfading over to another track.  The extra minute or two after the climax is standard in all full length club tunes so that the DJ has some time to blend one tune into the other.  Of course, really good DJs can be so good at matching the keys and chord progressions present in the music that you barely notice that a transition is taking place.

A radio mix of the same track would be 3.5 minutes long, having the first and last 90 seconds chopped off, the break-down shortened, and the build-ups and overall progression of the track proceed at a faster pace.

If you actually want to hear some more serious melodic compositions rather than electronic dance music, let me know and I can email you a few samples. ;)  One of my current projects is the production of music and sound for a video game that a bunch of people from Waterloo are working together on, and I think some of the tunes I've done for this thing are more impressive from a musical standpoint.

Oh, that's so cool! Yeah, definitely send it to me!

That makes me want to work on real music that will actually be used for something...  All I've written is a 16th century style mass movement and an 18th century style dance suite for my music classes...  Hardly stuff that's ever going to be performed.   :P

OK.  I zipped a few of my melodic tracks and threw them on TurboUpload (I'm sick of MegaUpload and YouSendIt... apparently this one works well.)  Here's the link:
[url]http://d.turboupload.com/d/788935/Elyots_Music.zip.html[/url]

I have no idea how long that link will last, so get it while you can.  There are 4 tracks in there.  From the new computer RPG that I'm writing music for, there are 3 tracks - an intro theme/overture, some battle music, and a sad character theme.  With the exception of the overture, these tracks are meant to be looped, and thus don't have a proper ending.  Their production quality is probably not quite as good as it will be in the end, so perhaps the instrumentation is a little cheesy at the moment, but I'll take care of that later.

I also included another track, entitled Sorrow, which is a personal composition of mine.  It's a full ambient jungle/drumNbass track with a very mysterious and unusual chord progression.  It's definitely not your typical major or minor scale; it's got a bunch of wierd accidentals.  Of course, it's easy to analyze with my bizarre Elyot-music-theory, but I have no idea how to dissect this with classical theory.

Taotao, maybe you'll remember my strange music theory from Winter Camp 2 years ago.  This is probably one of a few good examples of it being put to use. ;)

Hope you like them.

~ Elyot


Let ABC be the acute triangle
a=BC,b=CA,c=AB
prove that:
$ a^5\plus{}b^5\plus{}c^5\plus{}2abc(ab\plus{}bc\plus{}ca)\geq ab(a^3\plus{}b^3)\plus{}bc(b^3\plus{}c^3)\plus{}ca(c^3\plus{}a^3)\plus{}abc(a^2\plus{}b^2\plus{}c^2)$

[quote="tuandokim"]Let ABC be the acute triangle
a=BC,b=CA,c=AB
prove that:
$ a^5 \plus{} b^5 \plus{} c^5 \plus{} 2abc(ab \plus{} bc \plus{} ca)\geq ab(a^3 \plus{} b^3) \plus{} bc(b^3 \plus{} c^3) \plus{} ca(c^3 \plus{} a^3) \plus{} abc(a^2 \plus{} b^2 \plus{} c^2)$[/quote]
It's wrong. Try $ a\equal{}b\equal{}1$ and $ c\equal{}1.1.$ :wink:


inequalities
function
inequalities proposed

Let $a,b,c>0$.Prove the following inequality:

          $a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\leq\sqrt{2(a+b+c)(ab+bc+ca)}$  ;)

Suppose $a+b+c=1$. We have to prove:
$a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\leq\sqrt{2(ab+bc+ca)}$
But from the Jensen inequality for the function $f(x)=\sqrt{x}$ we get:
$a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}=af(b+c) + bf(c+a) + cf(a+b) \leq f(a(b+c) + b(c+a) + c(a+b)) = \sqrt{2(ab+bc+ca)}$
and we are done.

[quote="socrates"]$\sum(a)\sqrt{b+c}=\sum\sqrt{a}\sqrt{ab+ac}\leq\sqrt{(a+b+c)(2ab+2ac+2bc)}$
and we are done! :)  :D

Very nice![/quote]
C-S
$\sum\sqrt{a}\sqrt{ab+ac}\leq\sqrt{\sum a\sum (ab+ac)}=\sqrt{(a+b+c)(2ab+2ac+2bc)}.$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=47186&hilit=inequality
$abc=2$.prove that:$a^3+b^3+c^3 \geq a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{a+b}$


I'm not sure if this is the proper place for this question but I put it here anyway.

If $ x$ and $ y$ are real numbers that have the same sign. Prove that if
$ x \equal{} y \plus{} \frac{1}{{x \plus{} \frac{1}{{y \plus{} \frac{1}{x}}}}}$
and
$ y \equal{} x \plus{} \frac{1}{{y \plus{} \frac{1}{{x \plus{} \frac{1}{y}}}}}$
So $ x\equal{}y$

---------------------------
Note: Is there anyway to make $ LaTeX$ larger in size?

[hide="my solution"]

Simplyfy both equations, then we get
$ x^3y\plus{}2x^2\equal{}x^2y^2\plus{}3xy\plus{}1$
$ xy^3\plus{}2y^2\equal{}x^2y^2\plus{}3xy\plus{}1$

Substract equations above

$ xy(x^2\minus{}y^2)\plus{}2(x^2\minus{}y^2)\equal{}0$
$ (xy\plus{}2)(x\plus{}y)(x\minus{}y)\equal{}0$

$ xy\equal{}\minus{}2$ is a contradiction because $ x$ and $ y$ have the same sign
$ x\equal{}\minus{}y$ is also a contradiction
And only $ x\equal{}y$ satisfy the last equation

$ QED$

[/hide]

[hide]We have $ x\minus{}y \equal{} \frac{1}{x\plus{}\frac{1}{y\plus{}\frac{1}{x}}}$ and $ y\minus{}x \equal{}  \frac{1}{y\plus{}\frac{1}{x\plus{}\frac{1}{y}}}$

So, either x = y or $ x\plus{}\frac{1}{y\plus{}\frac{1}{x}} \plus{} y\plus{}\frac{1}{x\plus{}\frac{1}{y}} \equal{} 0$ which yields either x+y = 0 or xy = -2 both of which are not permitted as x and y are of the same sign[/hide]

Thanks a lot guys for you replies.
And is this consider as Olympiad level or its Pre-Olympiad level or its even lower that this?

$ Thanks$

This is a pre-olympiad problem, as if you saw with IW@IT, he made it simplify to two equations, two variables, making it a standard problem.

Something is wrong if $ x\equal{}y$. Am I the only one that objects to $ 0\equal{}\frac{1}{c}$, where $ c$ is a real number?

Bictor has a very valid concern:

$ x \equal{} y \plus{} \frac{1}{{x \plus{} \frac{1}{{y \plus{} \frac{1}{x}}}}}$

$ \iff x\minus{}y \equal{} \frac{1}{{x \plus{} \frac{1}{{y \plus{} \frac{1}{x}}}}}$

In the case of $ x\equal{}y$, we would have $ 0 \equal{} \frac{1}{x\plus{}\frac{1}{x\plus{} \frac{1}{x}}}$, which is clearly invalid, because zero cannot be the reciprocal of a number. Further, even if we were to simplify the right side, we would get:

$ 0 \equal{} \frac{x^2\plus{}1}{x (x^2\plus{}2)}$, which also cannot equal zero because the numerator is positive-definite.

The question here is "prove that x=y" , it does not ask you to find x and y .

In fact , the question is "if something happened then x would or would not be equal to y" . That means you have to work on some conditions which may be unreal conditions , where you may not be able to find x and y , yet x still be equal to y .

Correct.  The truth of the implication $ P \implies Q$ is not dependent on the truth of $ P$.

Yes, but in this case it's much easier to prove that P is false than it is to prove that P implies Q, at least when you aren't directly prompted to prove the latter.  (Another way to phrase Bictor's insightful observation is that if x and y have the same sign then the given equations show both that x is further from 0 than y and that y is further from 0 than x.)  While the question isn't logically senseless, it is certainly poorly posed.

In Saturday or Sunday the source will post his solution I will post it here.

Ah, then this is a strange problem. Like an incomplete proof by contradiction.

As I promise to get you the official solution:
--------------------------------------------
By simplifying the both equations:
$ \begin{array}{l}
 y = x + \frac{1}{{y + \frac{1}{{x + \frac{1}{y}}}}} = x + \frac{1}{{y + \frac{y}{{xy + 1}}}} = x + \frac{{xy + 1}}{{xy^2  + 2y}} = \frac{{x^2 y^2  + 3xy + 1}}{{xy^2  + 2y}} \\
  \to xy^3  + 2y^2  = x^2 y^2  + 3xy + 1............(1) \\
 x = y + \frac{1}{{x + \frac{1}{{y + \frac{1}{x}}}}} = y + \frac{1}{{x + \frac{x}{{xy + 1}}}} = y + \frac{{xy + 1}}{{yx^2  + 2x}} = \frac{{x^2 y^2  + 3xy + 1}}{{yx^2  + 2x}} \\
  \to x^3 y + 2x^2  = x^2 y^2  + 3xy + 1............(2) \\  \end{array}$
By subtracting $ (2)$ from $ (1)$, we get $ xy(y^2  - x^2 ) + 2(y^2  - x^2 ) = 0$ By simplifying we get $ (xy + 2)(y^2  - x^2 ) = 0$ or $ (xy + 2)(y - x)(y + x) = 0$ and that requires that $ xy=-2$,$ y=x$ or $ y=-x$
Since that $ x$ and $ y$ have the same sign, so we get the $ x=y$.
----------------------------------------------
Note:- I see all this from a site as an image and write it in $ \text{\LaTeX}$ here so it may contain error if something is not clear tills me to check if it right or not. In fact I didn't understand those question but I would like to get a small overview. :)


let G be a simple graph having $n$ vertices and no 3-vertex cycle.  Prove that G has at most $\frac{n^{2}}{4}$ edges.

A particular case of Turan Theorem!

Let $X$ the vertice of largest degree in our graph, and let $deg(X) = k$.

Our graph don't have 3-cycles, so, have at most $k+k . (n-k-1) = k(n-k)$ vértices.

$(\frac{n}{2}-k)^{2}\geq 0$, so, $k(n-k) \leq \frac{n^{2}}{4}$, and the problem is solved!

also known as [b]mantel`s theorem[/b](not sure about the spelling...)


a,b,c positive reals.
If ab+bc+ca=1 find A such that $a^2+b^2+c^2\leq A$

There does not exist an upper bound. Take $a$ and $b$ very small, and $c$ very big...


[hide="2"]
\[ \tan \frac{3\pi}{13}\minus{}\tan \frac{\pi}{13}\equal{}\frac{\sin \frac{3\pi}{13}\cos \frac{\pi}{13}\minus{}\cos \frac{3\pi}{13}\sin \frac{\pi}{13}}{\cos \frac{3\pi}{13}\cos \frac{\pi}{13}}\equal{}\frac{\sin \frac{2\pi}{13}}{\cos \frac{3\pi}{13}\cos \frac{\pi}{13}}\]
Apply sum-to-product formula (or note that $ \frac{4\pi}{13},\frac{3\pi}{13}\equal{}\frac{(7\pm 1)\pi}{26}$) to get
\[ \sin \frac{4\pi}{13}\minus{}\sin \frac{3\pi}{13}\equal{}2\cos \frac{7\pi}{26}\sin \frac{\pi}{26}\equal{}2\sin \frac{3\pi}{13}\cos \frac{6\pi}{13}\]
So the problem is equivalent to proving
\[ 8\sin \frac{3\pi}{13}\cos \frac{3\pi}{13}\cos \frac{6\pi}{13}\cos \frac{\pi}{13}\equal{}\sin \frac{2\pi}{13}\]
which is true, from repeatedly using double-angle formula for sine and $ \sin \frac{12\pi}{13}\equal{}\sin \frac{\pi}{13}$.
[/hide]


okay more inequalites i need help with.
this is from a mop 2002 hw set.
prove

x_1 + 2x_2 + 3x_3 +...+ nx_n <= nC2 + (x_1)^1 +...+ (x_n)^n.
the _1 stands for a subscript.

thanks

What I did in this problem was to bring all variables to one side. Thus I proved that (x_1-x_1^1)+(2x_2-x_2^2)+(3x_3-x_3^3)+...+(nx_n-x_n^n)<= n choose 2. To do that, you can compute the maximum value of each of the quantities in parentheses (by taking a derivative), and then when you sum them all up, you get n choose 2. There were also several non-calculus methods that I can find in the MOP notes if anyone wants me to.

[color=red]

i would like to see a non-calculus solution of this problem.

thanks.

[/color]

Well the problem comes down to finding the maximum of nx - x^n. You could guess that x=1 is the answer after a while, and then may come up with this:

[hide]

nx - x^n = n(x-1+1) - (x^n-1) + 1

= (n+1) + (x-1)(n-x^(n-1)-x^(n-2)...-x-1).

If x is less than one, the right bit is negative as x-1 is. If x is bigger than 1, the right hand bit is also negative as there are n terms there all at least 1 (1, x, x^2, ..., x^(n-1)), so that bracket is negative. The best is x=1, giving n+1. Adding these all up gives the answer.

[/hide]

Solution by Chenyu Feng:

We can write the equation as

:Sigma: k=1 to n kx_k <= :Sigma: k=1 to n (k-1) + :Sigma: k=1 to n x_k^k.

We can split this into n inequalities of the form mx_m <= m - 1 + m_l^l.

We will prove that for all n inequalities, x_m^m + m - 1 >= mx_m.

We shall use the AM-GM inequality.

Since (x_1 + x_2 + ... + x_n)/n >= (x_1x_2...x_n)^(1/n),

(x^m + 1 + 1 + ... + 1 (m 1s))/m >= (x_m^m)^(1/m)

x_m^m + m - 1 >= mx_m.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Is the following proof correct?
Classify the abelian groups of order 60.

We have $ 60 \equal{} 2^2\cdot 3\cdot 5$. By the fundamental theorem of abelian group a group of order 60 is isomorphic to the group $ C_{m_1}\oplus C_{m_2}\oplus\ldots\oplus C_{m_k}$ with $ m_i\mid m_{i \plus{} 1}$. Therefore the only possibilities are $ C_2\oplus C_{30}$ and $ C_{60}$.

it's correct. if $ n$ is a natural number, $ v_p(n)$ denotes the multiplicity of $ p$ in $ n$ and $ e_p(n)$ the number of sum partitions of $ v_p(n)$, then there are - up to isomorphism - exactly $ \prod_p e_p(n)$ abelian groups of order $ n$, namely $ \prod_p \prod_k C_{p^{s(p)_k}}$ where $ (s(p)_k)$ is some sum partition of $ v_p(n)$.


topology
real analysis
real analysis unsolved

I see why $\mathbb{R}$ with the standard topology is of second category, but why is it that when you switch to the right-hand topology $T_{r}$ (where open sets have the form $U_{a}=\{x\in\mathbb{R}: x < a\}$ where $a\in\mathbb{R}$), $\mathbb{R}$ is of first-category (aka meagre or a countable union of nowhere dense sets).

This would mean possibly that given an open set $U_{a}\in T_{r}$ $\text{Int}\left(\Bar{U_{a}}\right)=\{\}$. (The bar is closure not complement.) This is where my confusion lies. It seems to be that $\Bar{U_{a}}=\{x\in\mathbb{R}: x \leq a\}$, and thus $\text{Int}\left(\bar{U_{a}}\right)=\{x\in\mathbb{R}: x < a\}$. Can anyone clear up where the error is? Thanks.

Your formula for the closure is incorrect. The closed sets here are $[a,\infty)$, so the closure of any set $A$ is $[\inf A,\infty)$.

The sets $[-n,\infty)$ are closed, have empty interior, and cover $\mathbb R$.


inequalities
logarithms
trigonometry
integration
complex analysis
complex analysis unsolved

Let $R(z)=\ds\frac{z^{2}}2-z+\ln(1+z)$, $z\in\mathbb{C}$, $z\ne-1$. Prove that for every real $x$ the following inequality holds: $\ds\mid R(ix)\mid \le\frac{\mid x\mid^{3}}3$.
("ln" means the value of logarithm from the branch with $\ln\,1=0.$)
(D. Mitin)

hint
[hide]inequalité accroissement finis [/hide]

I hope someone will find a solution, it is not so difficult.


calculus
limit
function
geometry
3D geometry
sphere
calculus computations

How can I calculate the following limits or prove that don't exist:

a) $ \lim_{(x,y) \to (0,0)} \frac {x^4\plus{}y^4}{x^2\plus{}y^2}$

b) $ \lim_{(x,y) \to (0,0)} \frac {x^2\minus{}y^2}{x^2\plus{}y^2}$

Thanks in advance

[quote="Rajiv"]I get the answer $ 0$ for both these questions. Wonder how they do not exist! :o[/quote]

Why are both 0. Could you explain me?

A look at the "nice" predictable nature of the exponents in these functions causes us to consider the notion of [i]homogeneity[/i].

[b]Definition:[/b] A function $ f(x_1,x_2,\dots,x_n)$ is [i]positively homogeneous of degree[/i] $ k$ iff $ f(\lambda x_1,\lambda x_2,\dots,\lambda x_n)\equal{}\lambda^kf(x_1,x_2,\dots,x_n)$ for all $ \lambda >0$ and all $ (x_1,x_2,\dots,x_n)\ne (0,0,\dots,0).$

Now, it's not too hard to see that the function in (a) is positively homogeneous of degree $ 2$ while the function in (b) is positively homogeneous of degree $ 0.$

In general, a homogeneous function is completely determined by its degree $ k$ and its values on the unit sphere

[b]Fact #1:[/b] If $ f(x_1,\dots,x_n)$ is positively homogeneous of degree $ k$ for some $ k>0$ and $ f$ is continuous on $ \mathbb{R}^n\setminus\{(0,\dots,0)\},$ then $ \lim_{(x_1,\dots ,x_n)\to(0,\dots,0)}f(x_1,\dots,x_n)\equal{}0.$

[b]Fact #2:[/b] If $ f(x_1,\dots,x_n)$ is positively homogeneous of degree $ 0,$ then $ \lim_{(x_1,\dots ,x_n)\to(0,\dots,0)}f(x_1,\dots,x_n)$ exists iff $ f$ is constant on $ \mathbb{R}^n\setminus\{(0,\dots,0)\}.$

[b]Fact #3:[/b] If $ f(x_1,\dots,x_n)$ is positively homogeneous of degree $ k$ for some $ k<0,$ then $ \lim_{(x_1,\dots ,x_n)\to(0,\dots,0)}f(x_1,\dots,x_n)$ exists iff $ f$ is identically zero on $ \mathbb{R}^n\setminus\{(0,\dots,0)\}.$

Can you see how to prove each of these facts? Once we have them, the limit is (a) is zero by Fact #1 and the limit in (b) is nonexistent by Fact #2.

Note: the condition in Fact #1 can be weakened; we don't need for $ f$ to be continuous away from the origin, just that it be bounded on some sphere centered at the origin.

[quote="Kirill"]a)
$ 0\leq \lim_{(x,y) \to (0,0)} \frac {x^4 \plus{} y^4}{x^2 \plus{} y^2} \equal{} \lim_{(x,y) \to (0,0)} \frac {(x^2 \plus{} y^2)^2 \minus{} 2x^2y^2}{x^2 \plus{} y^2} \equal{} \minus{} \lim_{(x,y) \to (0,0)} \frac {2x^2y^2}{x^2 \plus{} y^2}\leq \lim_{(x,y) \to (0,0)} \frac {2x^2y^2}{2|xy|} \equal{} \lim_{(x,y) \to (0,0)} |xy| \equal{} 0$
So, the middle part is bounded with two limits, both of them is 0 => $ \lim_{(x,y) \to (0,0)} \frac {x^4 \plus{} y^4}{x^2 \plus{} y^2} \equal{} 0$.
b) Taking $ (x,y) \equal{} (\epsilon , 0)$ and move $ \epsilon \rightarrow 0$, we will have:
$ \lim_{(x,y) \to (0,0)} \frac {x^2 \minus{} y^2}{x^2 \plus{} y^2} \equal{} \lim_{\epsilon \to 0} \frac {\epsilon ^2 \minus{} 0}{\epsilon ^2 \plus{} 0} \equal{} 1$.
And taking $ (x,y) \equal{} (0, \epsilon )$ and move $ \epsilon \rightarrow 0$, we will have:
$ \lim_{(x,y) \to (0,0)} \frac {x^2 \minus{} y^2}{x^2 \plus{} y^2} \equal{} \lim_{\epsilon \to 0} \frac {0 \minus{} \epsilon ^2}{0 \plus{} \epsilon ^2} \equal{} \minus{} 1$.
So, the limit doesn't exist.[/quote]

Thank you so much!


inequalities
combinatorics unsolved
combinatorics

Prove that each simple graph $G$ contains at least $\frac{E(4E-V^2)}{3V}$ triangles.

Let $(x,y)$ be an edge of $G$.
Then, there are at least $d(x)+d(y)-V$ other vertices wich are adjacent to both $x$ and $y$.
Thus the number $T$ of triangles satisfies : $T \geq \frac 1 3 \sum (d(x)+d(y)-V )$
where the sum is over the edges of $G$.
For each $x$, the number $d(x)$ appears exactly $d(x)$ times in this sum, so $T \geq \frac 1 3 (\sum d(x)^2-VE)$.
Using Cauchy-Schwarz inequality leads to the desired result :
$T \geq \frac 1 3 (\frac { (\sum_x d(x))^2 } {V}-VE )= \frac {4E^2 - EV^2} {3V}.$

Pierre.


It was a game on Whose Line.  Every post must be a question.  If you type a statement, you are out of the game.  As soon as you start posting here, you are in the game.  You get out if your question is irrelavent.  I'll start.  (I don't think this thread will last long)

Who else thinks that Questions is too easy to play on forums?

uh your out since you typed like umm
[quote="MustItAlwaysBeMe"]It was a game on Whose Line.  Every post must be a question.  If you type a statement, you are out of the game.  As soon as you start posting here, you are in the game.  You get out if your question is irrelavent.  I'll start.  (I don't think this thread will last long)[/quote]
that many statements.

so GAME OVER

the future: FOUL -1  I have decided to change the rules such that you need 3 fouls to get out until there are 10 posts.  And statements can be used for instructions and when you are giving someone a foul, but your post still needs to have a question in that case.  I'm sorry; I should have cleared that up.

Answering both undefined and the future, didn't I just give the instructions?

Do you think mustitalwaysbeme should already be out?

don't most people consider mustitalwaysbeme an idiot?

What does that have to do with anything?

Am I not the only one who thinks that getting on the official Mathcounts team doesnt necessarily make you smart?

the future is out of the game. (we have over 10 posts)  Don't post here any more.

Wasn't that a yes to my last question?

Haha too bad you are too  :D ...

[quote="MustItAlwaysBeMe"]the future is out of the game. (we have over 10 posts)  Don't post here any more. [/quote]

That was a statement and you only said you could post statements for instructions and fouls, not for saying whose out of the game or not (I dont constitute that as a foul, more as an ejection)...

Plus you said "we have over 10 posts"  That is defiently not a foul or an instruction.

(and dont make up a new rule to cover for yourself)
uhh im just trying to have some fun, you can post if you want I dont really care much for this game anyways...

BTW good job to those people that made Nats.

lol why not

I'm out.  And so is the future.

Sure?
Why not?
Aren't my posts so pointless?
Wasn't that my first real question?

[hide="Yes"][hide="Yes"][hide="Yes"][u][color=navy]Yes[/color][/u][/hide][/hide][/hide]

Don't I seem to really like being out?

Did you know that I am not from Connecticut?
Did you know that I'm also not from Indiana?
lol?
um, wow?

...yes?  And yes?  And no?  And no?

$ \sqrt{ \minus{} 1} \equal{} \mbox{Zzz...}$
What do you think of that?

Don't you think this is a waste of time?
Isn't it just us?
idk? As usual???
Don't I have great questions?
Shouldn't I stop?

The answer to every real question of academicsrule is "Yes." and to every non-real question is "No."

How about that?

But... I seriously do not know what I thought of it though...
And I do not have great questions.
And I am not still in the game.
Yay!

Do you think that the Indiana mod should hurry up and lock this?

there is no indiana mod.
it used to be EG

lol, what Indiana mod?

I think I should be promoted[size=25], and then just leave.[/size]

lol now an adminstrator is going to lock this...


Because $ tg^2\frac{\pi}{6}\equal{}tg^2\frac{5\pi}{6}\equal{}tg^2\frac{7\pi}{6}\equal{}\frac 13$, and $ u\equal{}tg^2\frac{\pi}{18},v\equal{}tg^2\frac{5\pi}{18},w\equal{}tg^2\frac{7\pi}{18}$ are distinct roots of $ (1\minus{}3y)^2\equal{}3y(3\minus{}y)^2$ or $ 3y^3\minus{}27y^2\plus{}33y\minus{}1\equal{}0$.
$ S\equal{}s_3\equal{}\sigma_1^3\minus{}3\sigma_1\sigma_2\plus{}3\sigma_3\equal{}9^3\minus{}3*9*11\plus{}1\equal{}433$.


Because of $\mod{4}$ $x=2k$ for some $k$. Then we can rewrite our equation as $(3^k-1)(3^k+1) = 2^y$. So, $3^k-1$ and $3^k+1$ are powers of two. From this we easily obtain solution: $x=2$ and $y=3$.


Let A, B be any two points and k, not equal to 1 is a positive number. Then the locus of points P such that PA/PB=k is a circle whose center lies on AB.

Let $ A=(-1,0)$, $ B=(1,0)$, and $ P=(x,y)$.

$ PA=k\cdot PB$

$ (x+1)^{2}+y^{2}=k^{2}\cdot [ (x-1)^{2}+y^{2}]$

Just complete the square or whatever.

Note: This is one of the few examples where analytic geometry is faster than the geometric way. The geometric way is nice too. Hint: consider a tangent to a circumcircle.


LaTeX
geometry
geometric transformation
dilation

Provide a synthesis of m-cresol from p-toluidine.

synthesise meta toluidine(by converting to nitro) from p toluidine.(method given in some other post) or
by von richter foll. by schmidt reaction. use sandmayer reaction to get meta cresol.

Show your proposal with chemical equations, not with a text...

[hide="My synthesis"]
1. Protect the $ \ce{NH2}$ grp using $ \ce{CH3COCl}$ and oxidise the CH3 group into COOH.
2. Deprotect the $ \ce{NH2}$ grp and convert it into p-nitro benzoic acid using diazotisation and excess $ \ce{NaNO2}$ and $ \ce{Cu}$ as catalyst .
3. Protect the COOH group and then use VonRichter synthesis and get a $ \ce{COOH}$ grp in the meta posn to the protected grp.
4. Make the $ \ce{COOH}$ grp into $ \ce{CONH2}$ and then use Hoffmann Bromamide rn to get $ \ce{NH2}$ grp.
5. Diazotise and add warm water to get $ \ce{OH}$ grp meta to the protected $ \ce{COOH}$ grp
6. Deprotect the $ \ce{COOH}$ group and then use P/HI to make it $ \ce{CH3}$
:)
[/hide]

k.
1).p-toluidine + oxidation(H2O2) to give pnitrotoluene
2).pnitrotoluene + KCN/ethanol--> 3methyl benzene carboxlic acid.
3).3methyl benzenecarboxylic acid + nh3/h2so4-> m toluidine
4)treatment nano2/hcl in presence of water/naoh to give m cresol.
oops, madness already posted.

[quote="VARUNARASU"]
3).3methyl benzenecarboxylic acid + nh3/h2so4-> m toluidine
[/quote]

Isnt that supposed to be $ \ce{N3H}$  :huh:
Also plz use latex, makes it easy to read :)

[quote="Madness"]and oxidise the CH3 group into COOH.[/quote]

Its not very wise to try to oxidize a ring with a NRR' group on it.

[quote="Madness"]Protect the COOH group and then use VonRichter synthesis[/quote]

What synthetic procedure is that?

[quote="Madness"]Diazotise and add warm water to get OH group[/quote]

An old fashioned, low yield method...

[quote="VARUNARASU"]p-toluidine + oxidation(H2O2) to give pnitrotoluene[/quote]

That would be very nice, wouldn't it?

yeah. n3h. and im too lazy to use latex. and 'slong as u can understand what i've written, it shudnt make a big diff. :wink:
was that meant to be sarcastic? :(

[quote="Carcul"]
Its not very wise to try to oxidize a ring with a NRR' group on it.[/quote]
Not even if i protect the $ \ce{NH2}$ grp by acetylation? :maybe:

[quote="Carcul"]
What synthetic procedure is that?[/quote]

Its the same as varun's step 2, i.e. $ \ce{KCN} / \ce{C2H5OH}$

[quote="Carcul"]
An old fashioned, low yield method...[/quote]
Are there any other good ones from diazonium salt? Maybe aq. $ \ce{OH\minus{}}$? :maybe:

but protection of COOH group, where does that come from?

My idea was to make the two COOH grps unique bcos otw, they both can undergo the reactions which in reality was meant for one only. Cant we just esterify to protect the COOH grp? :maybe:

use any of these methods http://www.mathlinks.ro/Forum/viewtopic.php?t=200086 :lol:

[quote="Madness"]Are there any other good ones from diazonium salt?[/quote]

The best method is the following: add cuprous oxide to a dilute solution of the diazonium salt containing a large excess of cupric nitrate.

another method-
convert cooh to COCl and then treat with CH3MgBr
now treat this with PBA(baeyer villiger) migratory aptitude:phenyl>Ch3
and hydrolysis

But Grignard reagents react with acid chlorides to form alcohols, not ketones.

suppose i use 1 eq. or to be on the safer side, use an organometallic whihc is not very reactive.(r2culi)?


Let ABC be a triangle with H and O as its orthocenter and circumcenter. Let CM be a altitude in ABC and let MN and OM form an angle of 90 degrees where N lies on AC. Compute angle MHN in terms of angle BAC.

I posted it on Intermediate Topics, but it was too hard for anyone, so lock it http://www.mathlinks.ro/viewtopic.php?t=241138


Prove that for any positive numbers $ a,b,c,d$ satisfying $ a\plus{}b\plus{}c\plus{}d\equal{}1$,
\[ \frac{1}{1\minus{}\sqrt{a}}\plus{}\frac{1}{1\minus{}\sqrt{b}}\plus{}\frac{1}{1\minus{}\sqrt{c}}\plus{}\frac{1}{1\minus{}\sqrt{d}} \geq 8\]

[quote="apollo"]Prove that for any positive numbers $ a,b,c,d$ satisfying $ a \plus{} b \plus{} c \plus{} d \equal{} 1$,
\[ \frac {1}{1 \minus{} \sqrt {a}} \plus{} \frac {1}{1 \minus{} \sqrt {b}} \plus{} \frac {1}{1 \minus{} \sqrt {c}} \plus{} \frac {1}{1 \minus{} \sqrt {d}} \geq 8
\]
[/quote]

We have

\[ \sum {\frac{1}{{1 \minus{} a}}}  \equal{} \sum {\left( {1 \plus{} \frac{a}{{1 \minus{} a}}} \right)}
\]

So we need to show that

\[ \sum {\frac{a}{{1 \minus{} a}}}  \ge 4
\]

By the Cauchy Schwarz we have

\[ \sum {\frac{a}{{1 \minus{} a}}}  \ge \frac{{\left( {a \plus{} b \plus{} c \plus{} d} \right)^2 }}{{a \plus{} b \plus{} c \plus{} d \minus{} \sum {a^2 } }}
\]

So we need to prove

\[ \frac{{\left( {a \plus{} b \plus{} c \plus{} d} \right)^2 }}{{a \plus{} b \plus{} c \plus{} d \minus{} \sum {a^2 } }} \ge 4
\]

\[ \Leftrightarrow \left( {a \plus{} b \plus{} c \plus{} d \minus{} 2} \right)^2  \ge 0
\]

which is obviously true. Done!


[quote="liangchene"]How do I show that $\binom{p^r}{k}$ is a multiple of p for $0<k<p^r$[/quote]

$\binom{4^1}{2}=6$, and $6$ is not a multiple of $4$ so the conjecture is false. ;)

[quote="AntonioMainenti"][quote="chess64"]$\binom{4^1}{2}=6$, and $6$ is not a multiple of $4$ so the conjecture is false. ;)[/quote]I'm guessing $p$ is used to represent a prime number.[/quote]

Oh :blush: he never stated that though. That makes a little harder...

[quote="liangchene"]How do I show that $\binom{p^r}{k}$ is a multiple of p for $0<k<p^r$[/quote]

Since $p$ is prime, all integers $\leq p$ are relatively prime to $p$ (remember, $\phi(p)=p-1$). Since $k$ is less than $p$, it and all integers less than it will also be relatively prime to $p$, and thus $p$ in the numerator will not be reduced by anything in the denominator and the entire expression will be some multiple of $p$.

Doesn't seem that hard.

Edit> The above proof only covers $p^1$. Given $p^r$, if $k$ is greater than $n$ times the number of $p$'s in the range, such that $\{p,2p,3p,4p,\ldots,np\}$ is in the denominator, then the numerator will contain the $n$ largest multiples of $p$ in the range, starting with $\{p^r,p(p^{r-1}-1),\ldots\}$, and therefore there will always be at least one factor of $p$ in the numerator and the number will be a multiple of $p$.

sorry about that guys, p and q are usually used to represent prime numbers.


probability
floor function
logarithms
function
calculus
integration
inequalities

Let $ x$ be chosen at random from the interval $ (0,1)$. What is the probability that
\[ \lfloor\log_{10}4x\rfloor \minus{} \lfloor\log_{10}x\rfloor \equal{} 0?
\]Here $ \lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $ x$.

$ \textbf{(A) } \frac 18 \qquad \textbf{(B) } \frac 3{20} \qquad \textbf{(C) } \frac 16 \qquad \textbf{(D) } \frac 15 \qquad \textbf{(E) } \frac 14$

[hide]for the condition to be satisfied, both $x$ and $4x$ must be of the same order of magnitude.
$x\ge .1$ and $4x < 1 \rightarrow x < .25$ works, yielding an interval of $.15$
note that we can divide the $.1$ and $.25$ by any positive power of $10$ and the condition will still be satisfied.
so we have an infinite geometric sequence with $a=.15$ and $r = \frac 1{10}$.
$\frac{.15}{1-\frac 1{10}} = \frac 16$ (C)[/hide]

[quote="Karth"][quote="JSteinhardt"][quote="Karth"]Sorry to reawaken this thread, but here's a solution more calculator-sneaky:

[hide]Graphing the function, about $.02 + .2347-.1075+.02$ of the graph lies on 1, which is about $.1672$, which is quite close to $\frac{1}{6}$.[/hide][/quote]

Or if you wish to be even more sneaky, you can integrate a function that is the problem specification, as it would return 1 if true and 0 if false.[/quote]

How would you use integration? (Calculus, right?)  :huh:[/quote]

On TI-83 or 84:

fnInt(int(log(4x))-int(log(x))=0,x,0,1)

Returns very close to 1/6.

Otherwise called a characteristic function...

[quote="JSteinhardt"][/quote][quote="Karth"][quote="JSteinhardt"][quote="Karth"]Sorry to reawaken this thread, but here's a solution more calculator-sneaky:

[hide]Graphing the function, about $.02 + .2347-.1075+.02$ of the graph lies on 1, which is about $.1672$, which is quite close to $\frac{1}{6}$.[/hide][/quote]

Or if you wish to be even more sneaky, you can integrate a function that is the problem specification, as it would return 1 if true and 0 if false.[/quote]

How would you use integration? (Calculus, right?)  :huh:[/quote][quote="JSteinhardt"]

On TI-83 or 84:

fnInt(int(log(4x))-int(log(x))=0,x,0,1)

Returns very close to 1/6.[/quote]

Wow, never thought of that.
Although i got that question right, it adds to my bag of cheap tricks.

Lets break this into cases... as the floor function outputs only integer values, we may look at when $ k\equal{}\lfloor \log_{10}4x\rfloor\equal{}\lfloor \log_{10}x\rfloor$. We may write these floor conditions in terms of inequalities as follows:

\[ k\equal{}\lfloor \log_{10}4x\rfloor\iff 10^k\le 4x< 10^{k\plus{}1}\]
\[ k\equal{}\lfloor \log_{10}x\rfloor\iff 10^k\le x<10^{k\plus{}1}\]

The first equation gives us the strongest upper bound on $ x$ and the second equation gives us the strongest lower bound on $ x$, that is it is necessary and sufficient that \[ 10^k\le x< \frac{1}{4}\cdot 10^{k\plus{}1}\]

Now since we have $ x<1$, $ \frac{1}{4}\cdot 10^{k\plus{}1}<1$, so $ k\le \minus{}1$.

The length between those two bounds is \[ \frac{1}{4}\cdot 10^{k\plus{}1}\minus{}10^k\equal{}10^k\cdot \left( \frac{10}{4}\minus{}1\right)\equal{}10^k\cdot \frac{3}{2}\]

The total probability is \[ \frac{3}{2}\cdot (0.1\plus{}0.01\plus{}...\plus{}0.00...01\plus{}...)\equal{}\frac{3}{2}\cdot\frac{1}{9}\equal{}\frac{1}{6}\] so the answer is $ \boxed{C}$.

[hide=Solution]
We can describe all $x \in (0, 1)$ by writing $x = a \cdot 10^{-n}$, where $n$ is a positive integer and $1 \le a < 10$. (Imagine breaking up the interval $(0, 1)$ into infinitely many intervals based on the value of $n$. They look like $[1/10, 1)$, $[1/100, 1/10)$, $[1/1000, 1/100)$, etc.) Then, \[\begin{aligned} \lfloor \log_{10} 4x \rfloor - \lfloor \log_{10} x \rfloor &= \lfloor \log_{10} (4a \cdot 10^{-n}) \rfloor - \lfloor \log_{10} (a \cdot 10^{-n}) \rfloor \\ &= (-n + \lfloor \log_{10} 4a \rfloor) - (-n + \lfloor \log_{10} a \rfloor) \\ &= \lfloor \log_{10} 4a \rfloor - \lfloor \log_{10} a \rfloor. \end{aligned}\] Notice that $\lfloor \log_{10} a \rfloor = 0$, since $1 \le a < 10$. Then, we must also have $\lfloor \log_{10} 4a \rfloor = 0$. Hence, we must have $0 \le \log_{10} 4a < 1$, or $1 \le 4a < 10$, or $1/4 \le a < 5/2$.

Thus, for each value of $n$, $x$ satisfies the given equation if and only if $1 \le a < 5/2$. The number $a$ takes values in the interval $[1, 10)$, which has length 9, and the "successful" interval is $[1, 5/2)$, which has length $3/2$. Therefore, for each value of $n$, the proportion of values of $a$ that are "successful" is $\dfrac{3/2}{9} = 1/6$. Since this number is constant across all $n$, the probability is $1/6$.
[/hide]

Let $\lfloor \log (4x)\rfloor=\lfloor \log (x)\rfloor = -a$. Then $10^{-a}\le x< \frac{10^{-a+1}}4$. This gives us $\sum_{a=1}^\infty (\frac{10^{-a+1}}4-10^{-a})=\frac32\sum_{a=1}^\infty 10^{-a}=\frac32\cdot\frac19=\boxed{\textbf{(C)}\frac16}$.


Prove that in any triangle the square of the measure of the internal bisectors of any angle is equal to the product of the measures of the sides forming the bisected angle decreased by the product of the measures of the segments of the side to which the bisector is drawn.

[color=darkred]Let $ABC$ be a triangle and let $D\in (BC)$ be a point for which $\widehat{BAD}\equiv\widehat{CAD}\ .$ Then $\boxed{\ AD^{2}=AB\cdot AC-DB\cdot DC\ }\ .$[/color]

[color=darkblue][b]Proof 1.[/b] Denote the second intersection $E$ between the line $AD$ and the circumcircle $w$ of the triangle $ABC\ .$
From the relation $p_{w}(D)=DA\cdot DE=DB\cdot DC$- the power of the point $D$ w.r.t. the circle $w$ and $\triangle ABE\sim\triangle ADC$ obtain :
$\frac{AB}{AD}=\frac{AE}{AC}$ $\implies$ $AE\cdot (AD+DE)=AB\cdot AC$ $\implies$ $AD^{2}=AB\cdot AC-DB\cdot DC\ .$

[b]Proof 2.[/b] From the bisector theorem ${\frac{DB}{c}=\frac{DC}b}=\frac{a}{b+c}$ and the Stewart's theorem $a\cdot AD^{2}+a\cdot DB\cdot DC=c^{2}\cdot DC+b^{2}\cdot DB$ obtain
$a\cdot (AD^{2}+DB\cdot DC)=$ $c^{2}\cdot\frac{ab}{b+c}+b^{2}\cdot \frac{ac}{b+c}$ , i.e. $a\cdot (AD^{2}+DB\cdot DC)=\frac{abc(b+c)}{b+c}$ $\implies$ $AD^{2}=bc-DB\cdot DC\ .$

[b]Remark.[/b] If the point $D_{1}\in BC$ so that the ray $[AD_{1}$ is the exterior bisector of the angle $\angle BAC$ , then $\boxed{\ AD^{2}_{1}=D_{1}B\cdot D_{1}C-AB\cdot AC\ }\ .$[/color]


function
algebra
domain
limit
inequalities
calculus
calculus computations

One is supposed to minimize function
$F(x,y,z)=\frac{x}{\sqrt{y^{2}+Kxz}}+\frac{y}{\sqrt{z^{2}+Kxy}}+\frac{z}{\sqrt{x^{2}+Kyz}}$ ;where [b]x,y,z[/b] are positive reals and $K\in\mathbb{R}_{+}$

[b]First observation :[/b]

$\boxed{\triangle F^{*}= \frac{\partial^{2}F^{*}}{\partial x^{2}}+\frac{\partial^{2}F^{*}}{\partial y^{2}}+\frac{\partial^{2}F^{*}}{\partial z^{2}}=0}$

IOW,the function satisfies Laplace equation and is said to be harmonic.

[b]Second observation[/b]
F is homogenous and we can assume ,without loss of generality ,$x+y+z=const^{2}=1$.

[u]Important note[/u]:[b]F(x,y,z)[/b] is symmetric in regard to cyclic permutations ,but not up to rotations.

-Following Gauss's mean value theorem for harmonic function and consequently the principle of max/min for these it can be shown that  functions $F: \mathbb{R}^{3}\to \mathbb{R}^{3}$ of the given  form when cut by euclidean plane ,bypassing singularity [b]S(0,0,0)[/b],result in 3D surface which can attain minimum only in:
a)single stationary point,if there is any,on the planar interior domain
[b]or[/b]
b)somewhere on the boundary of planar domain.

To cut long story short,I get to main result:
By condition $\phi=x+y+z-1=0$ it is established that [b]T(1/3,1/3,1/3)[/b]
satisfies requirement for unbounded extreme (minimum) of Lagrange function $L=F+\lambda \cdot \phi$
$\lambda$... is Lagrange multiplier

$F_{T}=\frac{3}{\sqrt{1+K}}=min_{1}$...  [b](1)[/b]
Let's seek for minimum on the boundary (sides of triangle defined by vertices (0,0,1),(0,1,0),(1,0,0):
$\lim_{z\to 0}[F(x,y,z)]=\sqrt{\frac{x}{K(1-x)}}+\frac{1-x}{x}=s(x)$
$^{y\to o}$

$s'(x)=0...\Rightarrow min [s(x)]=\frac{3}{\sqrt[3]{4K}}=min_{2}$...[b](2)[/b]

By equating [b](1)[/b] and [b](2)[/b] one gets cubic in [b]K[/b] in order to find "critical" values of Ks:
$K^{3}-13K^{2}+3K+1=0$
_________________________
solutions [b]K1=0.425484...;K2=12.758723...;K3=-0.184208[/b]

[b]K3[/b] is to be left out due to the inital requirement,while [b]K1,K2[/b] form the interval [b][0.425484...;12.758723...][/b] wherein inequality $F(x,y,z)\geq \frac{3}{\sqrt{1+K}}$ doesn't hold for all [b]x,y,z[/b].
Since $K=15$ is outside the interval ,we conclude that Parkdoosung's inequality is valid.

Much has been learned on the subject of harmonic functions and related topics.For those who are interested in studying ,and intend pushing it further  I recommend some related links and references:

http://eom.springer.de/H/h046470.htm
http://mathworld.wolfram.com/GausssHarmonicFunctionTheorem.html
http://www.mathpages.com/home/kmath214/kmath214.htm
http://ejde.math.unt.edu/Volumes/1998/02/Castro.pdf
http://www.math.gatech.edu/~cain/spr01/k6.pdf
http://www.axler.net/
http://www.axler.net/HFT.pdf

Friends, can someone crosscheck if this holds please?

Esti ,what is $\lambda ?$

[code]Calculus `VectorAnalysis`

F[X,Y,Z] := X/Sqrt[Y^2+kXZ] + Y/Sqrt[Z^2+kXY] + Z/Sqrt[X^2+kYZ];

...

Div [%]
Laplacian[F]

0

0[/code]:)
This a cross-check of my silicon 'friend' capable of performing  calculations in simbolic manner.

Thank you.
I'm browsing through the links on harmonic functions.
Althought I don't understand much , I feel the power of real math from them.Harmonic functions  remarkable stuff.  :lol:

[quote="esti"]
$\lim_{z\to 0}[F(x,y,z)]=\sqrt{\frac{x}{K(1-x)}}+\frac{1-x}{x}=s(x)$
$^{y\to o}$
[/quote]
Can you give me more details about that calculation?

I'm not an expert but I think you can exploit the fact the thing is homogeneous and x+y+z=1.
In doing so,substitute z=1-(x+y) into F(x,y,z) and consider what happens when y->o.

I'm positive s(x) can be obtained that way.Than to find min(s(x)) make use of calculus like $s'(x)=\frac{ds(x)}{dx}=0...etc$.
This part shouldn't be too difficult.

[quote="esti"]

[b]K3[/b] is to be left out due to the inital requirement,while [b]K1,K2[/b] form the interval [b][0.425484...;12.758723...][/b] wherein inequality $F(x,y,z)\geq \frac{3}{\sqrt{1+K}}$ doesn't hold for all [b]x,y,z[/b].
Since $K=15$ is outside the interval ,we conclude that Parkdoosung's inequality is valid.
[/quote]

So if for example K=24 then

F(x,y,z)>=3/5 holds? (but it does not?)

Or do I understand this right?

Edit: Sorry, my mistake.

Let`s say K=8 so then F(x,y,z) would not hold.

What is then the minimum value of F(x,y,z) and in
which point (x,y,z)?

[quote="Eros88"]From the author's post (if I understood  correctly) ,for k=8 minimum of F(x,y,z) should be: $min [s(x)]=\frac{3}{\sqrt[3]{4k}}=\frac{3}{\sqrt[3]{32}}$   ?[/quote]

Ok.

But in which point (x,y,z)?


topology
function
calculus
calculus computations

I didn't take the notes on analysis so..I forgot what that is..

I have to do the assignment so..

what is $ w_f(x)$ (on e-neighborhood)?

What I remember is that $ w_f(x) \equal{} inf( sup f([x \minus{} e,x \plus{} e]) \minus{} inf f([x \minus{} e,x \plus{} e]) )$.. is this right?

What is the name of this function? I want more precise definition..

Thanks!

ps.

w: omega
e: epsilon

I've seen it under the name "[url=http://en.wikipedia.org/wiki/Oscillation_%28mathematics%29]oscillation[/url]".

Yes, oscillation of a function at a point. 0 if and only if the function is continuous at that point.

Can be defined in terms of the oscillation of a function on a set (more specifically, on a neighborhood of the point).


I think you don't have to take capillarity into account.
By the way is the answer 47.7 cm?

You don't think ?
I thought it was one of the main ideas of the question to take that into account.
Unfortunately I don't know the answer, otherwise I wouldn't ask ;)
,but if you didn't use capillariety, how did you do it ?

OK I know how you did it, with p1*V1=p2*V2
but I don't believe that it's right, because it seems a little bit easy,
and we didn't take capillarity into account

Let me give you a hint
if u dont get it i will post the solution later

Draw the free body diagram of both the water & the aircolumn & the finger as well if required. you will get it easily & ther area will be eliminated between the equations. TRY IT. :)
 :D

Hi,
if 47,7 cm isn't the right answer, I believe I don't get it :/ even with the diagramm I see nothing, I didn't know before.

[quote="Tony M."]Hi,
I have the following problem:we have a 1 meter long, thin glass pipe. It is open on both sides, we put that pipe 0,5 meters into a container filled with water. After that, we close the upper end which stucks out of the water, now we turn that pipe 180° around a horizontal axis, the air pressure is 105 kPa. Now I have to ascertain the height of the air column in the pipe.

I have the following ideas: first when we put the pipe into the water we have to add to the 0,5 meter the capillary force, I have an equation for that, but unfortunately I need the radius to calculate that. I guess I have to eliminate it somehow.
Now I have that water and when I turn the pipe arround I have to add the pressure of the water in the pipe (which is density of water*heigt*g) and the air pressure.
My suspicion is that the air in the pipe is pressed together a little bit because of that pressure. And that's all to it I imagine. But I have problems during calculation, because I can't get the radius out of my equations.

So I hope that anybody can help me, but PLEASE don't calculate anything in your answer (as far as possible) just give me a hint, because I want to solve it by myself (maybe later on I ask you to calculate something, but only if I'm completely exhausted ;) )
thanks[/quote]

Here's my approach. Archimedes' Principle says that an object immersed in a fluid is bouyed up by a force equal to the weight of the fluid it displaces. Considering the column of air as an immersed object, air immersed in water, one can write the following equation.

$W'_\text{air}= W_\text{air}-F_{B}$

$W'_\text{air}\equiv \text{ the apparent weight of the air column.}$

$W_\text{air}\equiv \text{ the actual weight of the air column.}$

$F_{B}\equiv \text{ the weight of water displaced by the air column.}$

$F_{B}= \rho_{H_{2}O}A\cdot\ell_\text{air}g$

Substituting the last expression into our equation, and then dividing throughout by $A$, the area of the end of the tube, one obtains $W'_\text{air}/A = W_\text{air}/A-\rho_{H_{2}O}\ell_\text{air}g$.

The expression $W'_\text{air}/A \equiv P_\text{atm}$, and likewise the expression $W_\text{air}/A \equiv P_\text{abs}$.

Giving the final equation $P_\text{atm}= P_\text{abs}-\rho_{H_{2}O}\ell_\text{air}g$.

I shall leave the rest of the solution to you.


Determine all the positive integers x and y satisfying $x^3+2x^2+2x+1=y^2$.

Dont know if it helps, but it can be factorised as $(x+1+\frac{x^2}{2})^2-(\frac{x^2}{2})^2=y^2$

$x+1$ doesn't need to be prime. Any common divisor of $x+1$ and $x^2+x+1$ divides any linear combination, including $(x^2+x+1)-x(x+1)=1$. This forces $x^2+x+1$ and $x+1$ to be relatively prime.

woops, i did not look at his work post carefully enough...my bad

[quote="boxedexe"][quote="spang"](0,1) is not a solution because x has to be positive[/quote]
$-0=+0$, in this case, I believe we may consider it as $+0$, like usual.

Masoud Zargar[/quote]
I thought "positive integer $x$" meant $x>0$...

Ya, usually if it includes zero it says "all non-negative" numbers, not all positive numbers.

Believe it or not, in some countries, like mine, $0$ is considered positive (and negative). I've just forgotten the english usage.

One of my favorite problems! Maybe it should have its own space?

[quote="arqady"]Determine all the positive integers x and y satisfying $x^3+x^2+x+1=y^2$. :)[/quote]

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

Factoring doesn't seem to work. Any ideas?

[quote="arqady"]Determine all the positive integers x and y satisfying $x^3+x^2+x+1=y^2$. :)[/quote]

[hide="Hint 1"]$x^3+x^2+x+1 = (x^2+1)(x+1)$[/hide]

[hide="Hint 2"]$x^2+1 - (x-1)(x+1) = 2$[/hide]

 :)

[quote="probability1.01"][quote="K81o7"]Dunno if this helps, but a square + 1 cannot be divisible by another square...(i think, just using mods...)[/quote]

Except for $25 | 49 + 1$..[/quote]

Okay, good point.  Looks like i screwed up somewhere along the way.

[quote="probability1.01"][quote="K81o7"]Dunno if this helps, but a square + 1 cannot be divisible by another square...(i think, just using mods...)[/quote]

Except for $25 | 49 + 1$..[/quote]
Yes :), and not to mention $29^2 | (41^2+1)$   (Not to mention $1^2|(0^2+1)$ and infinite such  other numbers such as $169^2|(239^2+1)$


calculus
integration
advanced fields
advanced fields unsolved

Let $ X,Y$ be varieties on which an algebraic group $ G$ acts transitively, and let $ \varphi : X\to Y$ be a surjective, $ G$-equivariant morphism. Prove that $ \varphi$ is an open map.

$ \phi$ is isomorphic to $ G/H_1 \twoheadrightarrow G/H_2$.

I prefer the category of schemes to show this: So let $ X$ and $ Y$ be integral schemes of finite type over a field $ k$. Then $ \varphi$ is generically flat, i.e. there is a non-void open subset $ U\subset Y$ such that $ \varphi': \varphi^{\minus{}1}(U)\to  U$ is flat. Since flatness implies openness we already know that $ \varphi'$ is open. Now use that $ G$ operates transitively on $ X$ to transport the openness of $ \varphi'$ to all of $ X$.


doyou mean?
$x^{2}-1$= $(x-1)(x+1)$

simple quardratic equation.
not really a formula..

if you foil $(x-1)(x+1)$, you get $x^{2}-1$

oh, didn't think of that
 :blush:  :oops:

[quote="Ihatepie"]For a while ive noticed that 8*10 is one less then 9*9 and 7*9 is one less then 8*8 wat im trying to say is if u have a number times itself ex: 21*21 and the number below it 20* the number above it, 22 it equals 1 less then the original 21*21-1=20*22[/quote]
$x^{2}=x^{2}$

$x^{2}-1=x^{2}-1$

$x^{2}-1=(x-1)(x+1)$

you'll also notice a patter for (x+2)(x-2), and such, it always goes down by perfect squares. I have seen this come in handy but only in mathcounts.

It is useful in a lot of other problems than mathcounts.  It is a fairly simple way to prove stuff not equal (or you could use inequalities).


trigonometry
inequalities
inequalities unsolved

Let ABC be an acute-angeled triangle.Prove:
$\cos{A}\cos{B}+\cos{B}\cos{C}+\cos{C}\cos{A}\le 6\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}$

Use some general trig identities invovong triangle

Bomb

[quote="lccs"][quote="zhaobin"]and I think $\cos A \cos B \cos C \le \sin \frac{A}2 \sin \frac{B}2 \sin \frac{C}2$
may be useful in the problem :P[/quote]
Would you please tell me how to prove this?
Maybe this two ineqs have some similarities...[/quote]
for the inequality my wrote.
just notice $\cos A \cos B =\frac 1 2(\cos(A-B)-\cos (A+B)) \le \frac 1 2(1-\cos (A+B)) =\sin^2 \frac{C}{2}$

for the organal problem:
it is equal to(if I am right) $16 (\sin \frac{A}2 \sin \frac{B}2 \sin \frac{C}2)^2 \leq 4 \sin \frac{A}2 \sin \frac{B}2 \sin \frac{C}2-2\cos A \cos B \cos C$
it is trival with the inequality I wrote. :)


How many distinct ways are there to arrange the letters in the word AVERAGE?  Please answer in general, as I would like to know how to do this kind of problem in general.  How do you do something like this that has the same letter more tha once????

Is there a formula, like there would be for arranging the word BEST (4!, because the letters are all different)???

[quote="mysmartmouth"]Is the answer [hide]1260[/hide]???? If so, I can do these!!!!![/quote][hide]The answer is 1260. You're right! :first:[/hide]

[quote="Monkey"]There is a formula for this type of problem. It's equal to z!/((a1)!x(a2!)x...x(an!)), where z is the number of letters in the word and {a1, a2, ..., an} is the # of times a certain letter a1, a2,...,an appears in the word.

Hope this helps.[/quote]

I am not sure I get it, try putting the AVERAGE problem into this formula, maybe I will understand then!

[quote="tarquin"]Basically number of ways you can permutate any word like PERMUTATE would be the the factorial of the number of letters over the factorial of the number of the same letter that repeat times whatever other factorials of the same letters that repeat. So in PERMUTATE the number of ways that it could be arranged is $\frac {9!}{2!2!}$ because there are nine letters total, two E's and two T's.[/quote]

it's short and sweet formulas like this that i love: THANKS!

Edit: Wait, does this work?  I get a different answer than I do using the AoPS way!

In the AVERAGE problem, there are two A's and two E's, so our formula would yield 7!/(2!x2!), which is 5040/4=1260.

Now I hope this helps.

The reason this works is as follows

we are creating combinations in which we select seven letters from AVERAGE in any particular order.

We can represent the 7 letter spots like this: _ _ _ _ _ _ _

How many letters are possible for the first blank?

Well we have 7 letters to choose from so we place 7 in the first blank.

7 _ _ _ _ _ _

How about the second blank? well we have already used up one letter for the first blank.  So now we have 6 to choose from.

7 6 _ _ _ _ _

If we continue this we get
7 6 5 4 3 2 1

We multply all our values resulting in 5040.

Now we run into the problem of doubles.

We have two A's and two E's... What do we do?

Look at it like this

We have 2 A's to choose from giving us 2!  But since they are replaceable, we have counted 2! too many!

Same goes for E's

So we need to divide by 2! and 2! to give us 1260


geometry
inequalities
area of a triangle
inequalities proposed

[quote="pohoatza"]Let $ S$ be the area of a triangle with sidelengths $ a$, $ b$, and $ c$. Prove that
\[ a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} \frac {1}{2}\left(|a \minus{} b| \plus{} |b \minus{} c| \plus{} |c \minus{} a|\right)^{2} \geq 4S \sqrt {3}.
\]
[/quote]
Let $ b\plus{}c\minus{}a\equal{}x,$ $ a\plus{}c\minus{}b\equal{}y$ and $ a\plus{}b\minus{}c\equal{}z,$ $ x\leq y\leq z,$ $ y\equal{}x\plus{}u$ and $ z\equal{}x\plus{}v.$
Then $ v\geq u\geq0$ and after squaring we obtain easy inequality. :wink:

[color=darkblue][b]Conjecture[/b][/color]. Let $ S$ be the area of a triangle with sidelengths $ a$, $ b$ and $ c$. Prove (or disprove) that
\[ a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}\frac{1}{2}\left(|a\minus{}b|\plus{}|b\minus{}c|\plus{}|c\minus{}a|\right)^{2} \geq 4S\sqrt{3\plus{}\displaystyle\frac{4(R\minus{}2r)}{4R\plus{}r}}.\]
Can someone please check the validity of the above inequality?

[quote="pohoatza"]Let $ S$ be the area of a triangle with sidelengths $ a$, $ b$, and $ c$. Prove that
\[ a^{2} \plus{} b^{2} \plus{} c^{2} \minus{} \frac {1}{2}\left(|a \minus{} b| \plus{} |b \minus{} c| \plus{} |c \minus{} a|\right)^{2} \geq 4S \sqrt {3}.\]
[/quote]

[hide="Solution"]Suppose that $ a\equal{}\max\{a,b,c\}$ and $ c\equal{}\min\{a,b,c\}$. We need to prove that:

$ \boxed {a^2\plus{}b^2\plus{}c^2\minus{}2(a\minus{}c)^2\ge 4\sqrt{3}S}$

Make the substitution $ a\equal{}x\plus{}y$, $ b\equal{}y\plus{}z$ and $ c\equal{}z\plus{}x$. We need to prove that:

$ x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx\minus{}(y\minus{}z)^2\ge 2\sqrt{3xyz(x\plus{}y\plus{}z)}$

$ \iff x(x\plus{}y\plus{}z)\plus{}3yz\ge 2\sqrt{3xyz(x\plus{}y\plus{}z)}$

Which is trivial by AM-GM.[/hide]

[quote="pohoatza"][color=darkblue][b]Conjecture[/b][/color]. Let $ S$ be the area of a triangle with sidelengths $ a$, $ b$ and $ c$. Prove (or disprove) that
\[ a^{2}+b^{2}+c^{2}-\frac{1}{2}(|a-b|+|b-c|+|c-a|)^{2} \geq 4S\sqrt{3+\displaystyle\frac{4(R-2r)}{4R+r}}.\]
Can someone please check the validity of the above inequality?[/quote]
The above inequality does not hold. [b]Counterexample:[/b]

$\{\begin{array}{llll}
a=1 \\ \\
b=1.1 \\ \\
c=1.2\end{array}\|\implies\{\begin{array}{cccc}
a^2+b^2+c^2=3.65 \\ \\
4S\sqrt{3+\frac {4(R-2r)}{4R+r}}+\frac 12(\sum\, |b-c|)^2\approx 3.66272\end{array}$


Let $ a,b,c$ be non-negative real numbers. Define
\[ \frac {a \plus{} b}2 \minus{} \sqrt {ab} \equal{} k^2,\;\; \frac {a \plus{} b \plus{} c}3 \minus{} \sqrt [3]{abc} \equal{} l^2.
\]
Prove that
\[ \max(a,b,c)\geq \min (a,b,c) \plus{} \frac32(k \minus{} l)^2.
\]
Give a generalization.

I hope someone will give a solution now.


limit
calculus
algebra
polynomial
derivative
calculus computations

Evaluate:
1. $ \lim_{x\to 1}\frac{x\plus{}x^2\plus{}x^3\plus{}\cdots \plus{}x^m\minus{}m}{x\plus{}x^2\plus{}x^3\plus{}\cdots \plus{}x^n\minus{}n}\quad m, n\in\mathbb{N}^*$
2. $ \lim_{x\to 0}\frac{\sqrt[n]{1\plus{}ax}\minus{}1}{x}\quad (a\neq 0,\, n\in\mathbb{N}^*)$
3. $ \lim_{x\to 0}\frac{\sqrt[n]{1\plus{}ax}\cdot\sqrt[m]{1\plus{}bx}\minus{}1}{x}\quad (ab\neq 0,\, m, n\in\mathbb{N}^*)$

Thanks for your help!

What have you tried?

Here's a hint for the first: how can you rewrite $ x \plus{} x^2 \plus{} \ldots \plus{} x^m \minus{} m$ in a way that shows off some factors of $ x \minus{} 1$?  (These problems fall into the "calculus is easy but algebra is hard" category.)

By the way, in the future, computations like this should be placed in the "Computations and Tutorials" part of the calculus forum.  (No need to re-post these, though!)

They can all be killed with L'Hopital's rule.

[hide="Number 1"]
$ \lim_{x\to 1}\frac{1\plus{}2x\plus{}3x^2\plus{}\cdots\plus{}mx^{m\minus{}1}}{1\plus{}2x\plus{}3x^2\plus{}\cdots\plus{}nx^{n\minus{}1}}\equal{}\frac{1\plus{}2\plus{}\cdots\plus{}m}{1\plus{}2\plus{}\cdots\plus{}n}\equal{}\boxed{\frac{m(m\plus{}1)}{n(n\plus{}1)}}$.
[/hide]

The others can be done similarly.

Alternatively, you might be able to simplify #1 (might be what JBL is hinting at) and then evaluate the limit easily. Does $ x\plus{}x^2\plus{}\cdots\plus{}x^{m}$ look familiar?

Yes, I was hinting at an algebraic simplification, though maybe not precisely the one you were thinking of.  Since (in question 1) these are just polynomials, you don't need L'Hopital and you can instead cancel common factors and plug in.  And the fact that you only need one application of L'Hopital is equivalent to the fact that there's only a single cancellation that needs to happen (one factor of $ x \minus{} 1$ from the top and bottom) before you can plug in.

Number 2:
Letting $ h\equal{}ax$ we have:
$ \lim_{x \to 0} \frac{\sqrt[n]{1\plus{}ax}\minus{}1}{x} \equal{} a \lim_{h \to 0} \frac{\sqrt[n]{1\plus{}h}\minus{}1}{h}$

This is the derivative of $ a \sqrt[n]{x}$ at $ x\equal{}1$, so the answer is $ a/n$.


For positive reals $ a$, $ b$, $ c$, $ d$ prove the following inequality: \[ (a\plus{}b)(b\plus{}c)(c\plus{}d)(d\plus{}a)\left(1\plus{}\sqrt[4]{abcd}\right)^4\ge 16abcd(1\plus{}a)(1\plus{}b)(1\plus{}c)(1\plus{}d)\]

[quote="April"]For positive reals $ a$, $ b$, $ c$, $ d$ prove the following inequality:
\[ (a \plus{} b)(b \plus{} c)(c \plus{} d)(d \plus{} a)\left(1 \plus{} \sqrt [4]{abcd}\right)^4\ge 16abcd(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)(1 \plus{} d)
\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=217874


search
number theory proposed
number theory

Find all odd $ n \in N$ such as
$ n|3^{n}\minus{}1$

try to search before posting :)
http://www.mathlinks.ro/Forum/viewtopic.php?t=287528
(see the solution in the 4-th post)


There is this geometry website that had colorful diagrams, proofs of theorems (Stewart's extra) and lots of problems. It also had this promo of Inca geometry on its home page. Lots of problems from it were discussed in the Intermediate Geometry Marathon. Can someone tell me what this web site was?

It was easy to google.
[url]http://agutie.homestead.com/[/url]

[quote="mathwizarddude"]Where can I find solutions to the problems on that site? Thanks.[/quote]

Exactly what I was going to ask. I was pretty sure that they had solutions...


Sam purchased $ 3 \frac14$ pounds of cheese. He used half of the purchased cheese for a casserole and $ \frac14$ pound for sandwiches. Express as a mixed number the number of pounds of cheese he has left.

Sam purchased $ \frac{13}{4}$ pounds of cheese. $ \frac{13}{8}$ pounds were used for casserole and $ \frac{1}{4}$ pounds were used for sandwiches.

Pounds left: $ \frac{13}{4}\minus{}\frac{13}{8}\minus{}\frac{1}{4}\equal{}\frac{11}{8}\equal{}\boxed{1\frac{3}{8}}$


Intensity is proportional to 1/r^2.
At a distance of 10m from a speaker the intensity of the sound is 10^-2 w/m^2.
a) at what distance must you be in order to reduce the intensity by a quarter?
b) what will be the ratio of the sound levels in dB be at the 2 locations?

hERE'S WHAT I'M THINKING for part a: I_1/I_2=(r_2)^2/(r_1)^2 we want r_2=r_1(sqrt(4I_1/I__1)) right?

for part b,
B1=10log(I/Io)
B2=10log(4I/Io)
and I want B2/B1 with I=10^-2 w/m^2 CORRECT?

Part (a) looks ok but you need to finish your math: $ r_1\equal{}10 m$ so $ r_2 \equal{} 10(2) \equal{} 20 m$.

Part (b)... I don't know exactly.  I think that decibals is supposed to be a comparison: $ 10\log \frac{I_2}{I_1}$.  But remember that $ I_2 / I_1 \equal{} 1/4$ so I think this means $ 10\log (1/4) \approx \minus{}6.02$ dB


[quote="Peter VDD"]Yes that is right. :) But I didn't know if you counted graphs as IMO stuff. They are handy indeed. But they're officially not in the high school curriculum ;)[/quote]Maybe not in Netherlands ;) :D

Valentin I do NOT again NOT live in the netherlands. :P

But it's forgiven since you live in bulgaria. ;)

[quote="Peter VDD"]Valentin I do NOT again NOT live in the netherlands. :P [/quote]:oops_sign: Args :D Belgium, Flanders, Netherlands. You guys talk the same language, live 10 miles next to each other, etc. :D Anyway, you got the point :P

[quote="blahblahblah"]You Europeans are crazy.[/quote]Should I take this as an compliment  :blush:, or something else  :roll:  :moose:

Why are we crazy? What do you mean by that? Can u elaborate?

We are crazy because we are. Don't question it. :P

err... this is way too much racial content for me

:) Ok europeans may, or may be not crazy :D Since the word racial was involved, topic is locked from further discussions.


An airline has 4 flights from New York to Chicago adn 3 flights from Chicago to Los Angeles.
(a) If you want to go from New York to Los Angles by way of Chicago, how many different ways can you travel

(b) Draw a neat and detailed diagram of all possible flights (tree diagram or sample space i guess)

(c)If you choose ou flights at random what is the probability that you choose the 7:15 am flight from New York and the 10:30 am flight from Chicago

a) ${4\choose1}{3\choose1} = 12$
c) $\frac{1}{12}$ There are 12 different possible flights, but only one route includes both of these.

Actually, for part (a) there can be anywhere from $0$ to $12$ routes from New York to Los Angeles, depending on the times. (i.e. if all the Chicago - Los Angeles flights occur before any of the New York - Chicago flights, there are $0$ routes)

True, but no times are given, so i am assuming that none coincide.

i'm trying to do b.  How do you attach a drawing?

Use the img option above the bar where you can control the font size in your post.


A line has a slope of $ \minus{}7$ and contains the point $ (3,0)$. The equation of this line can be written in the form $ y \equal{} mx\plus{}b$. What is the value of $ m\plus{}b$?

$ 0\equal{}\minus{}7(3)\plus{}b\Rightarrow b\equal{}21\Rightarrow m\plus{}b\equal{}\minus{}7\plus{}21\equal{}14$.

In the form $ y \equal{} mx \plus{} b$, $ m$ stands for slope, and $ b$ is the y-coordinate of the y-intercept(the point where the line crosses the y-axis). We know the slope is $ \minus{}7$, so we can substitute this into our equation.

$ y \equal{} mx \plus{} b$

$ y \equal{} \minus{}7x \plus{} b$

We want to find the value of $ b$. To do this, let's substitute a point. We know that the line contains $ (3, 0)$. $ 3$ is the x-coordinate and $ 0$ is the y-coordinate.

$ y \equal{} \minus{}7x \plus{} b$

$ 0 \equal{} \minus{}7(3) \plus{} b$

$ 0 \equal{} \minus{}21 \plus{} b$

$ 21 \equal{} b$

So now we know that the y-coordinate, $ b$, is $ 21$, and we already knew that the slope, $ m$, was $ \minus{}7$. Now we just add:

$ 21 \plus{} (\minus{}7) \equal{} 21 \minus{} 7 \implies \boxed{14}$


I've been thinking a lot to the following problem: if a and b are natural numbers such that for all sufficiently large n, a^n and b^n have the same sum of digits, then there exists an integer k such that a=10^k*b. Can anyone help me with this one?

Hmm by sufficiently large, do you mean it in a limit-definition way? i.e. Exists M such that for all n > M, S(a^n) = S(b^n)? (S(x) is the sum of digits of x)

Well, I can't really be sure about what Harazi meant, but that's the only way I've ever seen "large enough n" being used... Are there more ways of interpreting it?

By large enough I meant that we can find M such that for n>M a^n and b^n have the same sum of digits.


geometry
3D geometry
number theory proposed
number theory

Determine all non-negative numbers $ x,y,z,t$ such that the number $ 3^x\plus{}3^y\plus{}3^z\plus{}3^t$ is a perfect cube

Where did you find this problem? I have solution but it is too long and ugly.

[quote="delegat"]Where did you find this problem?[/quote]

It's from a preparation camp in Bulgaria . Post your solution if you want .

[quote]It's from a preparation camp in Bulgaria . Post your solution if you want [/quote]

Thank you Also it appeared in mathematical Mayhem (part of Crux) as proposed problem and deadline for submitting solutions is not over yet but it often happens that problems proposed in Mayhem are not original.
I will post solution if I find time I have to many exams and also we have religious holiday in Bosnia on Thursday and Friday.

Thanks again   Salem


geometry
angle bisector
geometry proposed

In triangle ABC, the angle bisector of angle ABC intersects AC at D, and the angle bisector of angle ACB intersects AB at E. Given that BD = CE, prove that triangle ABC is isosceles.

It is Steiner Lamus' theorem. You can prove it by using the fact $l_b = \frac{2cacos{\frac{B}{2}}}{c+a}, l_c = \frac{2abcos{\frac{C}{2}}}{a+b}$.

Here is another solution:
Suppose that $AB$ is not equal with $AC$. Suppose that $AB>AC$. So $ACB/2>ABC/2$. Let $K$ to be the intersection point of $CE$ and $BD$. $BC=BC$,$CE=BD$ and $BCE>CBD$, So we have $BE>CD$.
Let $Z$ to be a point such that $EZ$ is parallel and equal with $BD$. So $EZD=EBD$. $EZ=EC$, so we have $EZC=ECZ$. $ECD>EZD$, so we have $DZC>DCZ$. So we have $CD>DZ=BE$.
So $AB=AC$.
And we are done. :wink:

[quote="jayme"]Dear Mathlinkers,

a question : why all the proofs are indirect?

Sincerely
Jean-Louis[/quote]
Dear jayme, can you explain more?

Dear Mathlinkers,
a proof cannot be claim direct if any one of these auxiliary basic theorems have indirect proofs...I saw this reflexion in the geometric litterature...
Sincerely
Jean-Louis


There are purple and green marbles in a bag in an unknown ratio, but when 15 green marbles are removed, the ratio becomes 2 purple marbles for each green marble. Then, when 45 purple marbles are additionally removed, the ratio becomes 5 green marbles for each purple marble. How many green marbles were in the bag before any were removed?

There are originally 50 purple and 40 green.

50 p & 40 g

Subtract 15 g and now you have 50 p to 25 g, or a 2:1 ratio.

Now subtract 45 p and you have 5 p and 25 g, a 1:5 ratio.

[quote="isabella2296"]Can you please explain how you got those numbers, 50 and 40?[/quote]

Wish I could say I had some great flash of insight, but I just guessed a few numbers and worked out a few examples. I reasoned that with the shift of ratios from purple to green, the numbers had to be somewhat smallish, otherwise the switch would not occur. I needed a smallish 2:1 ratio so I think I started with 100 to 50, and then tried 50 to 25. I then added 15 to 25 to get 40, and the rest followed.

I saw this question of FTW and wanted to answer it as there were no good answers. However, I couldn't find it in the forum and so clicked "ask on forum" to answer it. I hope that's okay. :maybe:

Let us call the number of purple marbles $ p$. Similarly, we shall call the number of green marbles $ g$. The problem states that when $ 15$ green marbles are taken away, the ratio is $ 2: 1$. This is represented by:\[ \frac{p}{g\minus{}15}\equal{}\frac{2g\minus{}30}{g\minus{}15}\] Later, the problem states that when $ 45$ purple marbles are taken away, the ratio of purple to green marbles is $ 1: 5$:\[ \frac{1}{5}\equal{}\frac{2g\minus{}75}{g\minus{}15}\] Cross multiply to get: \[ 10g\minus{}375\equal{}g\minus{}15\] By solving this, we get $ g\equal{}\fbox{40}$.


Two circles have raidii of 6 and 10. The distance between their centers is 24. What is the difference between their common internal and external tangets?

errr
aren't internal tangents equal to external tangents?

or do i have terms mixed up here,

Err, what are internal and external tangents? Tangents are tangents.

how do you have an internal (inside the circle) tangent?.... that would defeat the def. of tangent... do you mean secant lines?

[color=#A00FFF]Draw two circles on a piece of paper, nonintersecting.  Now, draw all 4 lines that are tangent to both circles.  2 of them are internal tangents and 2 are external tangents, and you should be able to figure out which is which given the names.

And no, they aren't equal.[/color]

So the internal ones, when extended, are the ones that divide the plane such that one circle is on one side and the other on the other side? And the external ones are the ones in which both circles are on the same plane? Or do I have them mixed up?

Joel, I see that you've been trying different-colored fonts now, eh?  :D

Ok, well, as for the problem, it looks as if you could use analytic geometry to do this. Since the answer is the same for any diagram meeting your given specifications, you could perhaps locate both circles at x=0, with one above the other. Then fiddle around with x^2 + y^2 = 36 and x^2 + (y-24)^2 = 100 and you should get the answer. Its getting late over here in MA, so Im not going to do it out, but if you read this and still need help, just say so and I will try it some other time.

[color=#E000FF]That would be nice if you could easily derive the equations for the tangents, but, well, you can't.  It's simpler not to stick in coordinates -- try to get the length through non-coordinate (synthetic) means.

Yeah, the colors are fun :).[/color]

See if you can build a right triangle & use Pythagoras (a common general technique) to find the length of the external tangent.  See if you can do something similar for the internal one.

Finding the length of common tangents to a pair of circles occurs so frequently that you should get to where you can do it automatically (once you solve it once, you'll be pretty much there).  In other words, take the time to get this one.

oK, NOW that i acutally understand the problem
yea
it is really easy

use the pythagrean theorem, and to make things easier, shift the tangents to make like easier


function
inequalities proposed
inequalities

let a,b,c be positive real numbers prove that\[ \frac{{\left( {x \plus{} y} \right)^2 }}{{\left( {x \plus{} y \minus{} z} \right)^2  \plus{} 2z^2 }} \plus{} \frac{{\left( {y \plus{} z} \right)^2 }}{{\left( {z \plus{} y \minus{} x} \right)^2  \plus{} 2x^2 }} \plus{} \frac{{\left( {x \plus{} z} \right)^2 }}{{\left( {x \plus{} z \minus{} y} \right)^2  \plus{} 2y^2 }} \le 4
\]

[quote="csbonny"]let a,b,c be positive real numbers prove that
\[ \frac {{\left( {x \plus{} y} \right)^2 }}{{\left( {x \plus{} y \minus{} z} \right)^2 \plus{} 2z^2 }} \plus{} \frac {{\left( {y \plus{} z} \right)^2 }}{{\left( {z \plus{} y \minus{} x} \right)^2 \plus{} 2x^2 }} \plus{} \frac {{\left( {x \plus{} z} \right)^2 }}{{\left( {x \plus{} z \minus{} y} \right)^2 \plus{} 2y^2 }} \le 4
\]
[/quote]
Let $ x \plus{} y \plus{} z \equal{} 3$ and $ x \equal{} \max\{x,y,z\}.$ Then
$ \sum_{cyc}\frac {{\left( {y \plus{} z} \right)^2 }}{{\left( {z \plus{} y \minus{} x} \right)^2 \plus{} 2x^2 }}\leq4\Leftrightarrow\sum_{cyc}\frac {(3 \minus{} x)^2}{(3 \minus{} 2x)^2 \plus{} 2x^2}\leq4\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(3 \minus{} x)^2}{2x^2 \minus{} 4x \plus{} 3}\leq12\Leftrightarrow\sum_{cyc}\left(4 \minus{} \frac {(3 \minus{} x)^2}{2x^2 \minus{} 4x \plus{} 3} \minus{} 4(x \minus{} 1)\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(15 \minus{} 8x)(x \minus{} 1)^2}{2x^2 \minus{} 4x \plus{} 3}\geq0.$
If $ x\leq\frac {15}{8}$ it's done.
Let $ x > \frac {15}{8}$ and $ f(x) \equal{} \frac {(3 \minus{} x)^2}{2x^2 \minus{} 4x \plus{} 3}.$
Hence, $ f'(x) \equal{} \frac {2(x \minus{} 3)(4x \minus{} 3)}{(2x^2 \minus{} 4x \plus{} 3)^2}.$
Thus, $ \max_{x\in[0,3]}f(x) \equal{} f\left(\frac {3}{4}\right)$ and $ f$ is decreasing function on $ [\frac {3}{4},3].$
Id est, $ \sum_{cyc}f(x) < 2f\left(\frac {3}{4}\right) \plus{} f\left(\frac {15}{8}\right) < 12.$ Done.


I guess "Easy Challenge" is a little better of a title than "Extremely Difficult Triviality."

Same difference.

Just my unbiased opinion...

This problem is pretty hard. You have to count how many digits there are in the last number. That's why I'm not solving it.

Hmmm
How's bout i make life easier

There is, or at least should be, 17 zeroes in the last number

I think there's a pattern.

Specifically, it goes [hide]1,2,0,1,2,0...

so the remainder of the last one should be 2[/hide]

right?

17 zeros, not 17 digits. Can you fix it and then prove the pattern you found?

And maybe, remember how to prove divisiblity by 3

Now, look carefully at the sum given. What happens to the digits if u were to actually sum the number?

I don't know how to prove divisibility by 3.

What I found out is that [hide]any number expressed as 10^x where as x is a positive integer or 0 divided by 3 would give you the remainder of 1. Add them together, and the remainders add up. Since you are dividing by 3, the remainder can't be more than 3. So it's mod 3 (I don't know anything about modular arithmatic either, so correct me if I'm wrong)

So the answer would be the remainder when dividing the digit number by 3. In this case, 18/3=6 r 0, so the answer is 0.[/hide]

Tis too late...already a new day begun...

Yes, that is correct.
In general, to see if a number is divisible by 3, we take the sum of the digits to see if that is divisble by 3.
Given this, u can figure out why 10^x is always 1(mod3)


geometry
geometric transformation
dilation
vector

Does anyone know or know where i can find the thought experiment for mass dilation.

Anybody???????
[hide] :blush:   :rotfl: [/hide]

Im not aware of one, neither is google it seems. Do you know that there is one?

I just thought it was derived from the postulates.

i've heard the thought experiment involves collisions but i can't seem to find it.

perhaps you could model a collision of two bodies in the frame of the centre of mass of the two, and derive an expression for relativistic momentum. I don't even think of "mass dilation", its much simpler to consider a single mass (rest mass), and simple incluse the lambda term in an expression for the momentum.. avoids lots of confusion.

You might want to google, "gadonken" or something like that.
I'm sure I spelled it wrong, there is an official german word for a thought experiment.
As you seem to have tried many things, this is a last resort.

Hi. This is a very classical gedanken experiment.

I first saw it in the classic book "University Physics" of F.W.Sears and M.W.Zemansky. The experiment is described in Chapter 8, [i]Momentum, Impulse, and Collisions[/i], Section 8. This book is introductory, and relies heavily in a graphic representation of space-time known as Brehme Diagrams. I think it is a good quick and fast reference.

[u]On the internet[/u]:

[url=http://science.exeter.edu/ssaltman/relativity/relativity.html]This reference[/url] has a good explanation of elementary relativity, including the problem of mass dilatation. It has lots of examples, and looks good.

[url=http://wug.physics.uiuc.edu/courses/phys150/fall02/slides/lect15/]This reference[/url] is in a power-point-esque style. It doesn't strike me as very serious, but it's an startpoint.

Trying google with [url=http://www.google.es/search?hl=es&q=mass%2Brelativity%2Bgedanken&meta=]mass+relativity+gedanken[/url] yields many results.

[u]Problems of the concept of relativistic mass[/u]:

Relying on the concept of relativistic mass is an inheritance of newtonian mechanics, and does not apply well to special relativity. The real idea behind relativistic mass is that, from the STR viewpoint, energy and momentum are inseparable: they both form a tetravector. It is perhaps the simplest example of a tensorial quantity in relativity-- energy and momentum work well considered as a tetravector, but if you separate them, as in the newtonian theory, weird things happen.

One of the most famous problems with newtonian generalizations of the concept of mass is the question of mass dilatation when you define mass as the quotient of force and acceleration. If you have a force vector in an inertial system, and apply a boost to the system, the longitudinal force is dilatated by a factor of $\alpha^3$, while transversal forces are dilatated only by a factor of $\alpha$, where

$\alpha \quad = \quad \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

This seems to mean that the newtonian mass of a body depends on the direction of the boost (some jurasic-old references distinguish between longitudinal and transversal mass). Actually, the problem comes from trying to apply concepts of the newtonian model to the relativistic model.

I think that, if you are willing to learn relativity, you should drop all these classical sources and try to grasp relativistic mechanics from a more formal point of view, using theoretical mechanics and vector calculus. Only when you have a rigurous grasp of the theory, it is good to look back to the classical sources and interpretations.

:bye: