1. What  is  the  last  digit  in  the  sum  3^17  +  7^13?

2. Suppose  that  your  height  this  year  is  15 % more  than  it  was  two  years  ago, and  two  years  ago  your  height  was  25 %  more  than  it  was  two  years  before  then.  By  what  percentage  has  your  height  increased  during   the  last  four  years?

3.  Multiply  consecutive  even  integers  together  until  the  product  2*4*6*8.... becomes  divisible  by 1999. What  is  the  largest  even integer  you  used?

4.  Find   the  smallest  value  of  x^2 + y^2  on  the  line  3x + 4y = 12.

5.  How  many  7 - digit    telephone  numbers  can  be  created  if  the  first  3 - digits
must  be  426,427,or  428  and  the  right most is  divisible  by 3?

1. [hide] congruent to 3+7, so last digit is 0.[/hide]

2. [hide] let your height 4 years ago be a. after two, it is 1.25a. after 4, it is 1.15*1.25a ~ 1.44a. So ~44%[/hide]

3. [hide] 1999 is prime, so we need to get 3998 as a factor.[/hide]

4. [hide] we have y=-3/4x+3. We need the minimum of x^2+(-3/4x+3)^2. = 25/16x^2 - 9/2x +9. the minimum is -b/2a = 36/25 [/hide]

5.[hide] the rightmost is either 0,3,6,or 9. we have 3 choices for the first 3 digits. so 3*10^3*4=12000[/hide]

tetrahedr0n wrote:4. [hide] we have y=-3/4x+3. We need the minimum of x^2+(-3/4x+3)^2. = 25/16x^2 - 9/2x +9. the minimum is -b/2a = 36/25 [/hide]

Don't forget that [hide]-b/2a is just the x-value. The min. value of x2 + y2 is 144/25. [/hide]

(Would these maybe be more useful in the Getting Started section?)

haha oops. wyra, i would send these to begginer, but the people there would complain.

These are better beginner questions then most of what ends up in Getting Started -- at least some of them.   I say send it over.

You could also use the geometric approach for 4, finding the radius of the circle with the center at the origin which is tangent to 3x + 4y = 12.

1. Answer is 0.
Let [Expr] denote Last digit after evaluating the expression.
Then, [3^4k+1] = 3 and [74k+1] = 7, for integer k.
Hence, last digit of 3^17 + 7^13 is 0.

2. Too easy

3.  24

4, 5. Too easy =)

Since you got number 3 wrong, perhaps you should rethink your disdain for numbers 2, 4 and 5.

tetrahedr0n wrote:5.[hide] the rightmost is either 0,3,6,or 9. we have 3 choices for the first 3 digits. so 3*10^3*4=12000[/hide]

where did you get the ten?

In general, how many choices are there for each digit?


Prove that

$ (\cot A \plus{} \cot B)(\cot B \plus{} \cot C)(\cot C \plus{} \cot A)\geq\frac{8\sqrt{3}}{9}$

[quote="Ligouras"]Prove that $ (\cot A \plus{} \cot B)(\cot B \plus{} \cot C)(\cot C \plus{} \cot A)\geq\frac {8\sqrt {3}}{9}$ .[/quote]

[b]Proof.[/b] $ abc \equal{} 4RS$ , $ S \equal{} sr$ and $ \boxed {4S \equal{} \left(b^2 \plus{} c^2 \minus{} a^2\right)\cdot\tan A}$ a.s.o. $ \implies$ $ \prod \left(\cot B \plus{} \cot C\right) \equal{}$ $ \prod\frac {\left(a^2 \plus{} c^2 \minus{} b^2\right) \plus{} \left(a^2 \plus{} b^2 \minus{} c^2\right)}{4S}$ .

Therefore, $ \prod \left(\cot B \plus{} \cot C\right) \equal{} \frac {(abc)^2}{(2S)^3} \equal{} \frac {2R^2}{S}$ . Since $ \frac Rr\ge 2$ and $ \frac Rs\ge \frac {2}{3\sqrt 3}$ obtain $ \prod \left(\cot B \plus{} \cot C\right) \equal{} 2\cdot \frac {R}{r}\cdot\frac Rs\ge \frac {8}{3\sqrt 3} \equal{} \frac {8\sqrt 3}{9}$ .


geometry
combinatorics proposed
combinatorics

Suppose five points in the plane are situated in such a way that no two of straight lines joining them are parallel , perpendicular or coincident . From each point perpendiculars are drawn to all the lines joining the other four points . Determine the maximum number of intersections that these perpendiculars can have.
 :)  :P

Notice that there are a total of $30$ connections and we can pick $2$ of them to intersect. This can be done in $ \binom {30}{2} $ ways, but we have overcounted. First, we have counted two sets of parallel perpendiculars $3$ times each and we want to count them $0$ times each. There are $ \binom {5}{2} = 10 $ such pairs, so we subtract $ 3 \cdot 10 $. Next, we have counted each pair of lines that intersect at the emanating point that intersect once $15$ times each when we actually should count them once. This happens once for each vertex, so we want $ 5 \cdot 1 $ instead of $ 5 \cdot 15 $, so we subtract $ 5 \cdot 14 $. Finally, we examine orthocenters of triangles of lines, where perpendiculars meet. There are $ \binom {5}{3} = 10 $ ways to select these lines and we have counted the intersection $3$ times when we want to count them once. So we subtract $ 2 \cdot \binom {5}{3} $.

Thus, our answer is \[ \dbinom {30}{2} - 3 \cdot 10 \cdot 5 \cdot 14 - 2 \cdot \dbinom {5}{3} = \boxed {315}. \]


AMC
AIME
USA(J)MO
USAMO
geometry
3D geometry
rectangle

The AMC a day was pretty good practice so why don't we try AIME a day?  PM me an answer and then the next day we can discuss the problem.  I'll keep score and stuff.  So day 1:

The average value of \[|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|\] for all permutations $a_1,a_2,\ldots,a_{10}$ of $1,2,\ldots,10$ can be expressed in the form $\frac mn$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n$.

An AIME a day keeps the USAMO from getting away :D

My intention was for you to PM me the answer instead of posting it.  Your answer is wrong, btw, and if this were the real AIME I bet you would really be kicking yourself right now  :D because you made a very small mistake.

I would like to hear if everyone would rather PM me answers and then discuss the problem the next day or put their answers in a spoiler the day of?

I think PM is the best idea.  No temptations then :).  BTW I PMed you (I think?).

guangweicao wrote:Here's a solution:

[hide]
We can list all of the possible outcomes of the |a-b| from the set of 1 to 10.
We start counting with 1 as the first term,
a=1
b=2,3,4,..9,10
Now, the possible differences are: 1,2,3...9
Next we move on to a=2,
Now, the possible differences are: 1,2,3,...8.
Eventually, when a=8,

you now sum all possible outcomes

Now, sum these numbers up.  They are triangular numbers: 45+36+28+...6+3+1=165
Divide this by the number of possible outcomes and you get the average outcome.

number of outcomes=9+8+7+6...1=45
165/45
and you get 31/9.  There are five such terms.  5*31/9=165/9
165+9=174

[/hide][/i]

check your division165/45 = 33/9 =11/3, not 31/9. I did something similar recently where I did

2004/6=338

[quote="probability1.01"]The funny thing, beta, is that that mistake wasn't what gave him the wrong answer, but I should hush now before people catch too many hints :P.[/quote]

Huh?

[hide]basically we want the average of |m-n| where m and n are numbers from 1-10.
When |m-n|=1, there are 9 possible values for the set (m, n).
When |m-n|=2, there are 8 possible values
...
When |m-n|=9, there is only one possible value for (m, n)

So total sum= 1*9+2*8+3*7+... 9*1=165

There are 10C2=45 possible sets, so 5(165/45)= 55/3=average.
Which agrees with wait guangweicao should've gotten without the division error[/hide]

EDIT: okay now I see what you meant ;) , probability1.01. 2 wrong makes a right, but three wrong makes a wrong again...

[hide="Yesterdays Answer"]58 was yesterday's answer.[/hide]

Today's question:

If $w$ and $z$ are complex numbers such that $w^2+z^2=7$ and $w^3+z^3=10$, find the maximum value of $w+z$.

Just to make things a little more interesting I have decided to start ranking problems according to difficulty.  The scale will be 1-10 and the rank directly corresponds to how many points you get.  Yesterday's problem was worth 7 and today's is worth 3.

I'll probably get this wrong.

Edit> Sorry for posting it.

AGH. What is wrong with you people???

He specifically requested like 3 times for people to PM him!

[size=200][b]Private Message![/b][/size]

I hope that's large enough. Please do not post answers at all until the day after the question is given.

Agreed. :( Now that I have sent my private message, I will check the posted answer and hope it was wrong. -_-

Edit--hmm... my solution took about 1/5 as much space...

Yeah, I mean, I hate it when I come up with a brilliant, eye popping, jaw-dropping, awe-inspiring solution only to realize that 3 people already posted it and another 2 better solutions on the forum :)

I know mine was excessively long; most of it was repititious, because I was doing half of it in Tex and was only seeing a step ahead at a time, so that ended up in me having to take back alot of my algebraic movements.

It didn't really take to too long to actually do it on paper; about 2-3 minutes.

[quote="Silverfalcon"]I'd like to post the problems from now on.

Maybe like tomorrow or day after.

Just post the solution here in spoiler.

I'm not sure how I'll grade them or score them but I just like to post challenge problems.

Not all of them would be hard.

Anyone okay with it? :)[/quote]

sure why not? :)

and the answers were correct.

[quote="xxreddevilzxx"]yes, although i would like you to write the whole mock aime again... hmm any problem is welcomed.[/quote]

Do you mean that I should fix the old one? Or do you mean to write like mock AIME 6?

[quote="xxreddevilzxx"]write the mock AIME 6. i was not able to measure my ability, because AIME 1-4 were TOO HARD, and AIME 6 was TOO EASY. (no offense).[/quote]

Haha... :mad: WHAT?

Just kidding. :cool:

I'll try to do it if I can.

I do have some problems that I can use for mock AIME although I prefer to save them for my Championship Round.

I'll think about it.

cool. i will be counting on it. practices makes perfect.

Since I'll be busy tomorrow, I'll post one now.

Please post your answer in spoiler.

DO NOT PM THEM TO ME!

The following numbers 1447, 1005 or 1231 have one thing in common: All of them start with 1 and have exactly two identical digits. How many such numbers are there?

Source of this problem: 1983 #10

Although this is pretty easy problem. :)

Assuming they can only be four digit numbers:
[hide]Okay, as we can see these numbers can be written in the forms:
$1 A A B$
$1 A B A$
$1 B A A$
where A and B are distinct and $A\neq B\neq1$.

OR

$1 1 A B$
$1 A 1 B$
$1 A B 1$
where A and B are distinct and $A\neq B\neq1$.
Because there are $10$ digits but $1$ is already "ruled out", there are $9$ digits left.  Because $A\neq B$, there are $9*8=72$ different combinations for each of the 6 forms.  So there are $72*6=432$ of such numbers.
[/hide]

[hide="solution"]if the number also ends in one, then the two numbers in the middle cannot be the same, or nor can either of them can be one. So that gives us 9*8 = 72.

if there are any other 1's in the any other digit, it will be the same. thus, there are 72 *3 = 216 numbers having 2 1's.

for any other possibilities, like having 2 fives, there are 3C2 * 8 = 24. Because there can be 8 other numbers that appear twice, it is 24 * 9 = 216.

add 216 with 216, then we have 432. [/hide]

[quote="xxreddevilzxx"]write the mock AIME 6. i was not able to measure my ability, because AIME 1-4 were TOO HARD, and AIME 6 was TOO EASY. (no offense).[/quote]
Uh oh for that.  I've never taken the AIME and have no other way to judge really.  How hard was last year's AIME I?  I'll be taking that in addition to the practice AIMEs on here to guage myself 7 practice---so...

What can I really expect next week?
--And I'll be answering more on here until then; I saw this form after the first day or so, but had to plow through the third of the work for my english course that I hadn't gotten around to before Friday.  Besides that, another AIME--that I could try live--would be great, Silverfalcon.

well if you have noticed the mock AIME collections in this section of the forum, then you will be surprised at how easy was mock5 compared to others (i keep mentioning it. sorry silverfalcon. NO OFFENSE. :( ) judging from your situation, it wouldnt be a good idea for u to tackle those first four MOck AImes by mildorf, because i average about 6-8 on the Kalva AIMES (the real past AIMEs), but i only scored about 2,1,4, and 6 on mildorf's(the best was 6, which i could not believe myself)

so go to kalva's site: http://www.kalva.demon.co.uk

look for AIME tests, start with those.
give your self some quiet environment.

split it 8/7 or 5/5/5 and do them at a time. spend the three whole hours for all fifteen if you have time. (this is from somehwere else in this forum)

look at solution if you dont know. ask us. try redoing them by yourself. check your steps and make sure you are doing the right thing.

REMEMBER-- AIME only requires precalc-level knowledge. it's up to you to correctly apply your mathematical skills.

other than that, good luck.

*sigh* I want to try writing a Mock AIME. But I don't have any time this week (stupid distractions aka school). Ugh. And it probably takes more than a week to think up of problems anyway. Maybe next year...

you can expect first five problems to be fairly easy, next five bit challenging, and last five maybe undoable. practice makes perfect. keep practicing. dont let ur dreams go.

[quote="The Original Pi Guy"][hide] number is in form 1XYZ where there are 9 choices for X and 1 choice for Y and 8 choices for Z. There are 3! = 6 ways to arrange this, so the answer is 9 * 8 * 6 = 432?[/hide][/quote]
Care to elaborate on the [hide="...?"] there are 9 choices for X and 1 choice for Y and 8 choices for Z?[/hide]

I think the answer posted by majority of people is correct.

Um.

I'll be out of the site for about a week.

I might back here to post some COP rounds. But other than that, I'm not sure.

So, I'll try to post AIME day problem.

If not, I guess, someone has to do it. :(


Is everyone ready for Mock AMC version 2.718281828....(or just e if you would like)?

Ok, well I have just finished putting together the problems for a Mock AMC contest.  However, I still have to go through and check the problems.  Tokenadult will also be helping me with proofreading it.  I expect that I wont administer it for a couple of weeks or so.  The only practical time to do it is on a weekend.  While it is being proofread I would like to here what date and time would be the best.  I think that it is at about an AMC 12 level.

I've been waiting for this for forever. I would prefer next weekend, if that's alright with everybody else here.

Weekends would be nice. Now that school's started, I don't think I would have enough time to do one of these on a weekday. Psh, homework.

Mock AMC E sounds like a good idea... My only concern is whether or not I would have the time on the weekends. ANyway, we should do Mocks AMC's on the forums every two weeks or so, sounds like a good idea???

[quote="joml88"]Yeah, that sounds like a good idea.  BUT, someone needs to put the problems together.  Believe me that is much harder than it sounds.

By the way,  I will allow people who can t make the official time, whenever that may be, to take the exam early.[/quote]

Yeah, it's really tough to put problems together. I think it's kinda tough to make so many demands.

Also, it'd be nice to have the contest before october. I gotta watch my team play in the MLB playoffs.  :)

Or you could do what mathfantic does and give us a frame of time to work on it, instead of PMing it to people who want to take it early.

yeah, I need more practice.

February is coming!!!

[quote="JSRosen3"][quote="yif man12"]Also, it'd be nice to have the contest before october. I gotta watch my team play in the MLB playoffs.  :)[/quote]

....assuming that your team even [i]makes[/i] the playoffs....  ;)[/quote]

hey, HEY [b]HEY[/b]! You better watch out, now that I know where you live... (both on and off campus)

I'd like it too.

I like Mathfanatic's idea, becuase then we don't need to worry about times, becuase now that school has started, it's going to be impossible to find a time in which even most people can do it.

Many thanks to volunteers  ;)  ;)  ;) . I would like to take one every month if not two weeks, but for now I can live off of old AHSME sets that I haven't done yet. :lol:

Mock AMC E is an interesting, challenging set of problems. I should be able to attend to the proofreading over this weekend, having just filed a bunch of paperwork for various programs my son is in.

I am definatly interested in taking the mock AMC e and am trying to get my hard headed school district to let me register at our high school to take the AMC tests for this year and I could really use the practice that this would give me.  Thanks to eveyone who is puting these together.

I want to try this test as well, it will definitely serve as a good practice.
ps it's the first mock AMC since the beginning of school, isn't it?

I definitely want to do this, even though I doubt I could answer a single question on anything at AMC 10 or above level. I want to take the AMC 10 this school year, and I am completely clueless as to what kind of problems would be at that level. And I agree that giving a time frame would be much better than setting a time for us, especially now that school's started, but even during the summer this system would be more flexible and I at least would like it more.
P.S. Is this Mock AMC going to be more of an AMC 10 or an AMC 12?

I think that it is at AMC 12 level.  There are also some trig problems.

[quote="white_horse_king88"]Hmm... as long as the problems aren't too complex. Like some used De-Moivre's Theorem, which I had no idea what it was during the summer - thanks to some people on the board i kinda get it now, but still shaky.[/quote]

Heh, I have no clue what that is either. I'll try to find something on it, just to save people the time of looking it up.

AMC E looks to be in good shape, but I'm trying to double-check as part of the proofreading process.

The AMC E looked hard to me, but I proofread what I could understand.   :blush:   Good luck to everyone who tries it out. joml88 will have a busy schedule this week, I'm sure, but the mock test is worth the wait for him to distribute it.

In general, if any of you participants want to put together mock tests in the future, I will be happy to edit them for spelling, grammar, clarity of expression, and the like. Alas, I often can't answer the questions (and thus check proposed answers for correctness) at the level of the higher AMC contests, but at the low end (e.g., AMC 8), I should also be able to check correctness of answers.

Many thanks to the participants who put together mock tests for practice.

Here's how the Mock AMC E will work:

The official administration of the test will be [b]Sunday, October 10 at 7 PM EST[/b].

You can take the test anytime from now until that time.  All you have to do is tell me what time you want to do it so I can arrange to give it to you at that time.  Due to time constraints I will probably have to send it to you the a few hours before you ask for it.  If this happens please don't look at it until the time you have arranged and follow all the standard rules.

Sunday, October 10 at 8:15 PM is the latest I will accept answers (I will most likely accept them if you are only a minute or so late).

I can't seem to be able to PM the file so I can only send it to people through AIM or email... sorry for the inconvenience.  It will, however, still be on the website in its own post at the official time.  In the meantime I ll try to figure the PM out.

Hm.  The number of users swelled at 7:00 today, as many people like myself believed the Mock AMC E was going to be held this (rather than next) Sunday.

...but it's not for another week.

Sorry if this has already been answered:  Are the problems going to be posted on this forum?  Also, do we PM joml88 the answers, or what?  Thanks.

Sorry about the confusion about the date.  But just to make sure everyone is on the same page the Mock AMC will be next Sunday October the 10th.  I hope everyone shows up again.

themonster- yes I will post the problems on this forum on the correct day^^ at 7 PM.  You can either PM me the answers, email them to me at jlaurendi@gmail.com, or IM them to me (on AIM) at BaseballJoe2002.

I would like a test 10/9/04 at 11 AM EST if possible.
Email: Yintex@aol.com

[quote="vinny919"]I would love a test
Aim: vinny919
email: vinayakmuralidhar@yahoo.com

Thanks.  This is a great thing[/quote]

Would you like it early?  If so PM me what time you want it.  Or you could always take it Sunday October 10 at 7 PM EST (tomorrow) because that is when I will administer it to everyone.

Tomorrow is the Mock AMC E!  There wont be any topic about the problems until they are posted.  So don't be worried if it is 5 minutes till and there is no topic about the problems.  They will be there at around 7 PM(depending on your clock).

Good luck everyone!

Actually, can I take it late?  I promise I won't cheat or anything.  I won't even visit AoPS between then and now.  But I can't do it at 5 PM or any time today since I live in the Pacific time zone, so could I take it Monday morning?  Even if I don't get to be counted in the "official rankings" I would still like to participate and have it graded and see where I stand.  Thanks.

I think we discussed the use of calculators several times... I don't recall the exact thread, but the question has popped up alot. From what I have heard, all SAT appropriate calculators are appropriate in AMC.


calculus
integration
trigonometry
calculus computations

[quote="Wazika"]____
Find ∫√1+x²  x^5 dx[/quote]

[hide] $ x\equal{}\tan(\theta)$ then use parts[/hide]

reach

=∫sec theta tan^5 theta sec²theta[/code]

Let $ \sqrt {1 \plus{} x^2} \equal{} t\Longrightarrow 1 \plus{} x^2 \equal{} t^2\Longrightarrow xdx \equal{} tdt$.

Split the integrand in such a way $ \int \sqrt {1 \plus{} x^2}x^5\ dx \equal{}\int  \sqrt {1 \plus{} x^2}\cdot x^4\cdot xdx$.

That should be $ \int t^2(t^2 \minus{} 1)^2\ dt$ in the 3rd line.

hello, i have found
$ \int \sqrt{1\plus{}x^2}x^5dx\equal{}\frac{1}{105}(1\plus{}x^2)^{\frac{3}{2}}(8\minus{}12x^2\plus{}15x^4)\plus{}C$
Sonnhard.


there would be only one root that would work in the original equation and that would be 3.

Actually, you didn't make this up, I've seen it before. Or, if you did make it up, you weren't the first to do so. ;)

ritch, even if it's fairly obvious by guess-and-test, it's generally advisable to post a:

[hide="Solution"]$x+\sqrt{x-2}=4 \Rightarrow \sqrt{x-2}=4-x \Rightarrow [1]$
$([1])^2 \Rightarrow x-2=16-8x+x^2 \Rightarrow x^2-9x+18=0 \Rightarrow (x-6)(x-3)=0$

Therefore, the only possible solutions are $x=6$ and $x=3$. However, by squaring the equation, we came up with an extraneous solution, which, upon testing, turns out to be $x=6$. So, the only correct answer is $\boxed{x=3}$.[/hide]

sqrt (x - 2) = 4 - x
x - 2 = x^2 - 8x + 16
x^2 - 9x + 18 = 0
(x - 3)(x - 6) = 0
therefore x = 3 ( 6 is rejected since does not satisfy the equation)

[quote="Farazy"]What is the sum of the roots of the equation $x+ \sqrt{x-2}=4$[/quote]

$x+\sqrt{x-2}-4=0$

$x^2+x-2+16=0$

$x^2+x+14=0$

But did i make a mistake?

[quote="anirudh"][quote="Farazy"]What is the sum of the roots of the equation $x+ \sqrt{x-2}=4$[/quote]

$x+\sqrt{x-2}-4=0$

$x^2+x-2-4=0$

$x^2+x-6$

$(x-2)(x+3)=0$

$\text{The two roots are } \boxed{\text{-2 and -3}}$[/quote]

But you can't square just part of an equation ($x+\sqrt{x-2}$.  Besides, $(a+b)^2 \not= a^2+b^2$.

[quote="236factorial"]there isn't an x^2 term.....[/quote]
 :? isnt x^2 x squared? :?:

for the problem, i just looked at it and got it; i didnt go thru all those equations. it wus pretty obvious really. altho i had 2 chek it, nd i did it pretty much like jesus.

        my answer[hide]x=6,3, in which 6 doesnt work when inserted into the equation, so therefore x=3.[/hide]

I like citan's solution, I'll make it easier to read.

[hide="citans solution"]move the x to the other side to get
$\sqrt{x-2}=4-x$

Squaring both sides,

$x-2=16-8x+x^2$

$x^2-9x+18=0$

$(x-6)(x-3)$

$x=6,3$

But 6 doesn't satisfy the original equation (this happened because we squared both sides), so the answer is three.

I don't think -b/a works because we squared both sides and there's an extraneous answer when we did that. [/hide]


linear algebra
matrix
superior algebra
superior algebra unsolved

[quote="eugene"]Suppose $ G$ is a $ p$-group, where $ p$ is prime. Prove that every irreducible representation of $ G$ over a finite field of characteristic $ p$ is $ 1$-dimensional.[/quote]
it's equivalent that matric $ M$ is comutative if and only if when dimenionals of matrix equal $ 1$
more generall fact :
every comutative group $ G$ which has irrudicible representation $ \pi$ mean
$ \pi (g)L\equal{}L\pi (g)$ where $ L$ is ( I can't say this word ) ... operator therefore $ L\equal{}I$
then by using theorem :
that group algebra is direct sum of idelas. ( ideal must be right and left same time).
and $ I_{k}$ is isomorphic to $ M_{n_{k}}C$ therefore we have
$ \pi [C(G)]\equal{}M_{n}C\equal{}C$ one dimensional. :wink:


Let $ a,b,c\in \mathbb R$ such that $ a\ge b\ge c > 1$. Prove the inequality:

$ \log_c\log_c b \plus{} \log_b\log_b a \plus{} \log_a\log_a c\geq 0$

If you let $ x\equal{} \log_m a; y \equal{} \log_m b$ and $ z \equal{} \log_m c$ the inequality becomes

$ \frac{x}{y}\plus{}\frac{y}{z}\plus{}\frac{z}{x} \ge 3$ which is true by AM-GM

[quote="hsbhatt"]If you let $ x \equal{} \log_m a; y \equal{} \log_m b$ and $ z \equal{} \log_m c$ the inequality becomes

$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \ge 3$ which is true by AM-GM[/quote]

what's AM-GM?

The arithmetic mean is always greater than or equal to the geometric mean. In this case,

$ \frac {\frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x}}{3} \ge \sqrt [3] {\frac {x}{y} \cdot \frac {y}{z} \cdot \frac {z}{x}} \equal{} 1$, hence
$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \ge 3$

one important part of this is that all of the numbers involved are nonnegative.

[quote="hsbhatt"]If you let $ x \equal{} \log_m a; y \equal{} \log_m b$ and $ z \equal{} \log_m c$ the inequality becomes

$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \ge 3$ which is true by AM-GM[/quote]

Doesn't it become...

$ \log_c \dfrac{y}{z} \plus{} \log_b \dfrac{x}{y} \plus{}\log_a \dfrac{z}{x}$

How did you get $ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \ge 3$?

$ \frac{ln(log_cb)}{lnc} + \frac{ln(log_ba)}{lnb} + \frac{ln(log_ac)}{lna}\ge 0$

$ \frac {ln(lnb/lnc)}{lnc} + \frac {ln(lna/lnb)}{lnb} + \frac {ln(lnc/lna)}{lna}\ge 0$

$ \frac {lnlnb - lnlnc}{lnc} + \frac {lnlna - lnlnb}{lnb} + \frac {lnlnc - lnlna}{lna}\ge0$
Now, let $ a = e^x$, $ b = e^y$, $ c = e^z$ with $ x\gey\gez > 0$

$ \frac {lny - lnz}{z} + \frac {lnx - lny}{y} + \frac {lnz - lnx}{x} \ge 0$

$ \frac {lny}{z} + \frac {lnx}{y} + \frac {lnz}{x}\ge \frac {lnx}{x} + \frac {lny}{y} + \frac {lnz}{z}$

Which is rearrangement on oppositely sorted sequences $ lnx,lny,lnz$ and $ \frac {1}{x},\frac {1}{y},\frac {1}{z}$


geometry
3D geometry
pyramid
symmetry
ratio
Pythagorean Theorem
AMC

A square pyramid with base $ABCD$ and vertex $E$ has eight edges of length 4.  A plane passes through the midpoints of $\overline{AE}$, $\overline{BC}$, and $\overline{CD}$.  The plane's intersection with the pyramid has an area that can be expressed as $\sqrt{p}$.  Find $p$.

Someone suggested to me that this problem could be solved with something like this:

Let the midpoint of $AE$ be $P$, the midpoint of $BD$ $Q$, and $DC$, $R$. Now, pretend you are standing an "infinite" distance away from the pyramid on the line BD and sketch a side view of the triangle. Then, the points $Q$ and $R$ coincide and you can get the information you need that way...I'm not sure how this method would work. Could someone more skilled investigate this?

The following is a synthetic solution.

[hide="Solution"]
Call $P$, $R$, and $S$ the midpoints of $\overline{AE}$, $\overline{BC}$, and $\overline{CD}$, respectively. Then, let $Q$ and $T$ represent the intersection of plane $PRS$ with $\overline{BE}$ and $\overline{DE}$, repectively. Now let $W$ and $Z$ be the midpoints of $\overline{QT}$ and $\overline{RS}$, repectively. Denote by $X$ and $Y$ the projections of $P$ and $W$ onto plane $ABCD$, repectively. Now call $V$ the intersection of $\overline{PX}$ with the plane parallel to $ABCD$ that goes through $W$.

Easily, we find that $CRS$ is an isosceles right triangle with vertex $C$, so $RS=2\sqrt{2}$. We can find the height of the square pyramid to also be $2\sqrt{2}$. Since $P$ is the midpoint of $\overline{AE}$, by similarity, the distance from $P$ to plane $ABCD$ is $\sqrt{2}$. By symmetry, the midpoint of $\overline{XZ}$ is $Y$. By similarity, $W$ is the midpoint of $\overline{PZ}$, and the distance from $W$ to plane $ABCD$ is $\frac{\sqrt{2}}{2}$. By altitudes, $\triangle EQT\sim\triangle EBD$ with side ratio $3:4$. We can easily find that $BD=4\sqrt{2}$. Then $QT=3\sqrt{2}$. We also easily derive that $XY=YZ=\sqrt{2}$. By the Pythagorean theorem on $\triangle WYZ$, $WZ=\frac{\sqrt{10}}{2}$. However, since $W$ is the midpoint of $\overline{PZ}$, $PW=\frac{\sqrt{10}}{2}$.

Notice that the area sought is $[PQRST]=[PQT]+[QRST]$. We find each area individually.
\[[PQT]=\frac{(PW)(QT)}{2}=\frac{3\sqrt{5}}{2}\]
\[[QRST]=\frac{(WZ)(QT+RS)}{2}=\frac{5\sqrt{5}}{2}\]
Now we have that $[PQRST]=\frac{3\sqrt{5}}{2}+\frac{5\sqrt{5}}{2}=\sqrt{80}$, so the answer is $\boxed{080}$.
[/hide]

Note that this solution is not intended to be rigorous, for a rigorous solution would require lemmas and more specific similar triangles.


If you want to get into Ivy League schools, when would you stop retaking the SAT Is?

Well you should try to avoid re-taking in the first place...

only take it more than once if you have to

would anyone (preferably with experience) care to give a number though?  :D

I think if you can manage around a 2200, you shouldn't bother retaking.

Above 2200 on the "new" SAT I, your time is better spent on SAT IIs (required at all the most selective colleges) and devoting yourself to some extracurricular activity you really care about. People are rejected for admission all the way up to 2400, so those last 200 points aren't as important as getting over the threshold and then showing what ELSE you can do.

Good luck in your applications.


limit
trigonometry
real analysis
real analysis unsolved

Ascertain $\lim_{n\to\infty}\ \frac{2^n}{x_n}$, where $x_1>0\ ,\ x_{n+1}=x_n+\sqrt{1+x^2_n}\ ,\ n\in N^*$.[hide="Answer."]$2\arctan\ x_1\ .$[/hide]

Let $x_n=\frac{1}{\tan \theta _n},$ you are to obtain $\theta _{n+1}=\frac{1}{2}\theta _n.$

As for the range of $\theta _n$ take consideration it by yourself.


Ring Theory
superior algebra
superior algebra unsolved

Hi
Let A is local ring with maximal ideal m .thus let x1,......xn is independed variabel .prove A[[X1,......,Xn]] is local ring with maximal ideal production is m,X1,....,Xn.

1. A formal power series in $A[[X_1 , ... , X_n ]]$ is a unit if and only if its constant term is a unit in $A$.
2. A ring $A$ is local if and only if $x$ or $1-x$ is a unit, for all $x \in A$.

These two facts immediately show $A$ local $\Rightarrow A[[X_1 , ... , X_n ]]$ local.

3. The unique maximal ideal of a local ring $A$ consists of all non-units.

This shows, that the unique maximal ideal of $A[[X_1 , ... , X_n ]]$ consists of the formal power series, whose constant terms lie in the unique maximal $m$ of $A$, i.e. it is generated by $m,X_1 , ... , X_n$.


calculus
integration
algebra
function
domain
Ring Theory
quadratics

Hi. I have a final exam in algebra soon. the professor gave a list of 50 practice problems. there are a few that i've been having trouble with ... I have the exam very soon, so any help with these probs would be appreciated. And i apologize ahead of time if they seem basic, this is my first exposure to algebra. and secondly I apologize, but I don't yet know how to tex....

thanks in advance

-the rookie

1)a) Let {P_a} be a chain of prime ideals in a commutative ring R with 1. Prove that int(P_a) is a prime ideal.
b) Given a proper ideal A of R, show that there is a minimal prime ideal P containing A, in the sense that if P' is any prime ideal containing A, then P' is a subset of P.

2) Let F be a field of fractions of an integral domain D. If m is a maximal ideal of D then D_m denotes the localization (S^-1)D with S = D - m. We may view D_m as a subring of F. Prove that D = int(D_m), where m ranges over the maximal ideals of D.
suggestion:  given an element x in int(D_m), consider {d elment of D| dx is an element of D}

3) The Jacobson radical of a commutative ring R with 1 is defined to be Rad(R) = int m, where m ranges over the maximal ideals of R.
a) show that Rad(R) is an ideal containing all the nilpotent elements of R.
b) Prove that if r is an element of Rad(R) then 1+r is a unit of R
c) Suppose, conversely, that 1+a is a unit of R for all elements a in an ideal A of R. prove that A is a subset of Rad(R)

and finally the last one

4) a) Let D = Z[i]. Prove that an odd prime p becomes reducible in D if and only if there exists a solution to the equation m^2 + n^2 = p, with positive integers m, n.
b) show that if p is congruent to 3 mod 4, then p remains prime in D.
c) suppose p is congruent to 1 mod 4. according to quadratic reciprocity, there exists integers a such that a^2 is congruent to -1 mod p. prove that p factors into two non-associate primes in D.

Thanks again.

[quote="Rookie217"]1)a) Let {P_a} be a chain of prime ideals in a commutative ring R with 1. Prove that int(P_a) is a prime ideal.
b) Given a proper ideal A of R, show that there is a minimal prime ideal P containing A, in the sense that if P' is any prime ideal containing A, then P' is a subset of P.[/quote]

a)Let $P=\cap_{a\in A}P_a$, where $A$ is a totally ordered set. Since $(P_a)_{a\in A}$ is a chain of ideal, we have $P_a\supset P_b$ for all $a\leq b$.
Let $x,y\in R$. Assume $xy\in P$. Then $xy\in P_a$ for all $a\in A$. Since $P_a$ are all primes it follows that $x\in P_a$ or $y\in P_a$ for all $a\in A$.
Assume there exists an $a\in A$ such that $y\not\in P_a$. Then from the chain condition we have $y\not\in P_b$ for all $b\geq a$. Now it follows $x\in P_b$ for all $b\in A$, so $x\in P$. If for all $a$ we have $y\in P_a$, then $y\in P$.

b)The correct definition of a minimal prime is: $I\subseteq P$ and if $I\subseteq P'\subseteq P$, then $P'=P$.

I think you are mixing the minimal and maximal conditions, but in either case, there is a standard proof using Zorn's Lemma:
The maximal case is well known.
For the minimal case let $I$ be an ideal of $R$ and consider $M=\{I\subset P|P\mbox{ is a prime ideal in }R\}$.
If $P,Q\in M$, we say $P\leq Q$ if $P\supseteq Q$. It is easy to prove that $\leq$ is a partial order on $M$.
Point (a) proves that all the conditions in Zorn's Lemma are met, so there exists a maximal ideal of $M$. You may be a bit confused because I am looking for a minimal element, but I am taking a maximal element. The trick lies in the defintion of $\leq$ because it is the same as the containment, not as inclusion. So, take $P$ a maximal element of $M$. It is easy to prove that it is a minimal prime containing $I$ in the sense of inclusion, not of $\leq$.

[quote="Rookie217"]2) Let F be a field of fractions of an integral domain D. If m is a maximal ideal of D then D_m denotes the localization (S^-1)D with S = D - m. We may view D_m as a subring of F. Prove that D = int(D_m), where m ranges over the maximal ideals of D.
suggestion:  given an element x in int(D_m), consider I={d elment of D| dx is an element of D}[/quote]

It is immediate that $I$ is an ideal of $D$. Moreover $I$ contains elements outside every maximal ideal of $D$. It follows that $I$ is not contained in any maximal ideal of $D$, hence $I=D$, so $1\in D$. It follows that $x\in D$ and we are done.

[quote="Rookie217"]3) The Jacobson radical of a commutative ring R with 1 is defined to be Rad(R) = int m, where m ranges over the maximal ideals of R.
a) show that Rad(R) is an ideal containing all the nilpotent elements of R.
b) Prove that if r is an element of Rad(R) then 1+r is a unit of R
c) Suppose, conversely, that 1+a is a unit of R for all elements a in an ideal A of R. prove that A is a subset of Rad(R)[/quote]

a) If $x$ is nilpotent, then there exists $n\in\mathbb N$ such that $x^n=0\in P$ for any prime ideal $P$ of $R$. It follows that $x\in P$.
So the set of nilpotents which is actually an ideal of $R$, is contained in every prime ideal, thus in every maximal ideal and in the intersection of all the maximal ideal which is $Rad(R)$.

b) If $1+r\in m$ for some maximal ideal $m$ of $R$, then since $m\supseteq Rad(R)$, it follows than $1=(1+r)-r\in m$ which is a contradiction. Since $1+r$ is not contained in any maximal ideal of $R$, it must be invertible.

c) Assume there exists a maximal ideal $m$ of $R$ that does not contain $A$. Then $A+m=R$, so there exist $a\in A$ and $x\in m$ such that $x-a=1$. Then $x=1+a$. By the hypothesis, it follows that $x$ is invertible which is impossible. Therefore $A$ is contained in every maximal ideal of $R$, hence also in $Rad(R)$.

[quote="Rookie217"]4) a) Let D = Z[i]. Prove that an odd prime p becomes reducible in D if and only if there exists a solution to the equation m^2 + n^2 = p, with positive integers m, n.
b) show that if p is congruent to 3 mod 4, then p remains prime in D.
c) suppose p is congruent to 1 mod 4. according to quadratic reciprocity, there exists integers a such that a^2 is congruent to -1 mod p. prove that p factors into two non-associate primes in D.[/quote]

a) $D$ is an euclidean domain with respect to the norm $N(a+bi)=a^2+b^2$.
Let $p\in\mathbb Z$ be an odd prime number.
Assume that it is reducible in $\mathbb Z$. Then there exists $u,v\in D$ noninvertible such that $p=uv$. The norm $N$ is multiplicative, so we have $p^2=N(p)=N(u)N(v)$. If $N(u)=1$ (or $N(v)=1$), then $u$ (or $v$) is invertible in $D$. The only possiblility is $N(u)=N(v)=p$. If $u=q+ir$, then $p=N(u)=q^2+r^2$.
Conversely, assume $p=m^2+n^2$. Then $p=(m+in)(m-in)$ and $m\pm in$ are both non-invertible since their norm is $p$, not $1$.

b)Assume $p\equiv 3(mod 4)$ and $p$ is not prime in $D$. Since in $D$ prime is equivalent to irreducible, $p$ is not irreducible, hence there exist $u,v\in D$ non-invertible elements such that $uv=p$. Then $N(u)=N(v)=p$. Assume $u=m+in$. Then $m^2+n^2=p$. From this relation it easily follows $p\not|mn$. But then $m$ and $n$ are prime to $p$ and $m^2+n^2=p\Rightarrow m^2\equiv -n^2(mod\ p)\Rightarrow m^{p-1}\equiv (-1)^{\frac{p-1}2}n^{p-1}\Rightarrow 1\equiv (-1)^{\frac{p-1}2}=-1(mod\ p)$ which is impossible since $p$ is odd.

c) Let $p\equiv 1(mod\ 4)$ and assume that $p$ is irreducible. Then $p$ is prime in $D$. But then $p|a^2+1=(a+i)(a-i)$. It is easy to see that $p\not|a\pm i$, which is impossible if $p$ is prime. Therefore $p$ is reducible, so there exist $u,v$ non-ivertible elements of $D$ such that $p=uv$.
By passing to norms we get $p^2=N(u)N(v)$. Since $u$ and $v$ are non-invertible, their norms are not $1$, so $N(u)=N(v)=p$. It is easy to prove that $u$ and $v$ are primes. Assume $u=m+in$. $p=N(u)=(m+in)(m-in)$. By the unicity of decomposition of $p$ in $D$, we get $v=m-in$.
Assume $u$ and $v$ are associates. Then $u=ev$ with $e$ a unit in $D$. The units of $D$ are $\pm1,\pm i$. Every case leads to $mn=0$ which implies $p=m^2+n^2=(m+n)^2$. The last is impossible because $p$ is not a square in $\mathbb Z$.


[color=darkblue]Find all integer solutions (x,y) of the equation y^2 = x^3 - p^2*x where p is a prime such that p is congruent to 3 ( mod 8 )[/color]

please continue all your dicussions [url=http://www.mathlinks.ro/viewtopic.php?t=152240]here[/url]. :)


function
limit
calculus
integration
complex analysis
complex analysis unsolved

[quote="Kent Merryfield"]
(Any nonpolynomial entire function has some direction in the complex plane in which it grows faster than any polynomial - but if I start saying things like that, I'm starting to invoke complex analysis.)[/quote]
So, let's invoke complex analysis, indeed. The task is to either prove, or disprove the strongest possible interpretation of this statement: there exists $ \zeta$ on the unit circle centered at the origin such that $ \lim_{t\to\plus{}\infty}\frac{|f(t\zeta)|}{t^m}\equal{}\plus{}\infty$ for all $ m>0$.

Let's also invoke approximation theory, specifically Arakelyan's theorem on tangential approximation. Approximating $ 1/z$ on a suitable closed subset of $ \mathbb C$, we get an entire function such that $ \limsup_{t\to \infty} t|f(t\zeta)| < \infty$ for all unimodular $ \zeta$. In particular, $ \lim_{t\to \infty}f(t\zeta) \equal{} 0$ for all such $ \zeta$.

Of course, I didn't mean the strongest possible interpretation, not in the context in which I made that statement. Which doesn't mean you can't ask this rather interesting question here, even if it's not what I meant. Carry on.

Of course, I didn't mean that Kent meant it that way (though I did mean that many people might interpret it that way and this was a partial motivation for posting the problem). Mlok dropped a nuke on it, so, formally it is solved but an interesting question remains: how slowly can a counterexample grow? Suppose, for instance, we know that $ |f(z)|\le M(|z|)$ with some $ M(r)$ growing faster than any power at $ \plus{}\infty$. Can we find $ M(r)$ so that the conclusion will be true? If we can, what is the "largest" $ M(r)$ with this property?

1. For the original question, a simple counterexample does not require Arakelian.
There is an entire transcendental function that tends to zero on every ray from the origin.
Begin with f(z)=exp(exp(z)). It grows in the strip -pi/2<Im z<pi/2 when Re z tends to +infty in
this strip. It tends to 0 very fast as Re z to +infty on the two lines
Im z=pi/2+epsilon and Im z=-pi/2-epsilon.
Consider a simple closed contour Gamma which consists of the positive rays of those two lines and
a vertical segment of the imaginary axis.
Consider g(z)=integral over Gamma of f(zeta)/(zeta-z) d zeta, for z outside Gamma.
("Outside" is the big region, "inside" is the small one, containing the positive ray.)
Exercise: prove that g is a) an entire function and b) that it is bounded  outside Gamma,
Let w be a zero of g. divide g by z-w to obtain a function that tends to zero outside Gamma.
It remains to shift the origin to obtain an entire function that tends to o on each ray from the origin.
Slightly more complicated construction using the same idea gives a function that tends to zero on
EVERY ray (not only rays from the origin).

On Fedja's question. If f is entire, and log|f(z)|=o(|z|^{1/2}) then f cannot be bounded on ANY ray.
If we replace o by O it can be bounded on some ray. A function of finite order cannot be bounded
on all rays. The function I constructed above is of infinite order.
Or perhaps you want a growth restriction that guarantees that f tends to infinity on some ray?
Hayman showed that of log|f(z)| < O(log^2|z|) then there exists a ray on which f tends to infinity
(Commentarii math helv., 1960). Goldberg and Eremenko showed that this growth condition is best
possible in a very strong sense: for every function p tending to infinity, one can find an entire
function f with the property that log|f(z)| < p(log^2 |z|), and f does not tend to infinity on any ray,
or even on any curve whose length in |z|<r is O(r). (Mat Sbornik, 1979).


[quote="amir2"]if$\sum_{i=1}^n\frac{1}{x_i+1}=1$
prove :

$\sum_{i=1}^n \sqrt{x_i}\ge (n-1)\sum_{i=1}^n\frac{1}{\sqrt{x_i}}$[/quote]

Do classical substitution $\frac{1}{x_i+1}=y_i$ etc
then write the inequality in question in a neater way
and finally Cebasev saves you...

 :)


A divisor $ d$ of $ 3683488$ is a superdivisor if the sum of the numbers of $ d$ also divides
$ 3683488$. Find all superdivisores of $ 3683488$.

Greetings  :D


I am having some trouble on my calculus homework, and was wondering if you couyld help. Find the Area of the surface generated by revolving the curve y-x^3 between (1,1) and (2,8) about the x-axis. If you could please show or try to explain you work.

Agh, calculus doesn't belong here. Try posting in the Other Problem Solving Topics Forum.

Locked.


A bag contains 16 congruent blocks, 4 of which are blue and 12 of
which are red. Two blocks are randomly pulled from the bag at the
same time. What is the probability that both blocks are blue?
Express your answer as a common fraction.

The probability that the first block is blue is $ \frac{4}{16}$ or $ \frac{1}{4}$.

The probability that the second block is blue is $ \frac{3}{15}$ or $ \frac{1}{5}$(this is because we have already removed one blue block, so there are only 3 blue blocks left and 15 blocks in total.)

Now we just multiply:

$ \frac{1}{4} \times \frac{1}{5} \implies \boxed{\frac{1}{20}}$


Let $ A_1, B_1, C_1$ are midpoints of  BC, CA, AB of triangle ABC. Let $ A_2, B_2, C_2$ on $ AB_1, CA_1, BC_1$. Find the minimun area of the common of $ A_1B_1C_1$ and $ A_2B_2C_2$.


Ok^^ We have to show
$ \sum_{cyc} \frac{a_1\plus{} \dots \plus{}a_{n\minus{}1}}{a_n} \ge n(n\minus{}1),$
But $ \sum_{cyc} \frac{\frac{a_1}{a_n} \plus{} \dots \plus{}\frac{a_{n\minus{}1}}{a_n}}{n\minus{}1} \ge \sum_{cyc} \frac{a_1 \dots a_{n\minus{}1}}{a_n^{n\minus{}1}}$ and $ \frac{\sum_{cyc} \frac{a_1 \dots a_{n\minus{}1}}{a_n^{n\minus{}1}}}{n} \ge 1$.


I know LaTeX was created to avoid the befuddled parenthesis mess of expressing equations in word processors. However, I have a report that I need to use Word for, is there any way to bring whole LaTex formatted equations onto Word and keeping the integrity of the Latex equation?

I 'm asking because copying and pasting had disastrous results, and I need Word for it's ability to display Excel charts/graphs. If anyone has any suggestions, it would be very much appreciated.

Create the pdf file(s) with TeX and use Insert->Object->Acrobat document instead of copy and paste. Perhaps, there are better solutions but this is certainly the simplest one :)

You could also go to "Symbols" but that takes forever :(

A faster way is to actually type equations.

If you have MS Word 1997-2003, go to Insert, and click object and select equation. You figure ou the rest.

If you have MS Word 2007, go to the Insert tab, and near the right side is a pi symbol or equation. Click on it, or you can use shortcut alt + =. I found out in MS word 2007, there are some LaTeX codes that work with equation. \sqrt, \degree, ^ sign, and other symbols work. If you use {} brackets normally in LaTeX, don't because it won't work. Instead use the spacebar.

Also you can use TeXnic center for your $ \text{\LaTeX}$
For pics and graphs, go to the Asymptote forum.


Given that $x$, $y$, and $z$ are real numbers that satisfy:

$x = \sqrt{y^{2}-\frac{1}{16}}+\sqrt{z^{2}-\frac{1}{16}}$
$y = \sqrt{z^{2}-\frac{1}{25}}+\sqrt{x^{2}-\frac{1}{25}}$
$z = \sqrt{x^{2}-\frac{1}{36}}+\sqrt{y^{2}-\frac{1}{36}}$

and that $x+y+z = \frac{m}{\sqrt{n}}$, where $m$ and $n$ are positive integer and $n$ is not divisible by the square of any prime, find $m+n$.


Which emote do you like better?  :D or  :P ? You must pick one or the other.

I vote  :D

Oh, stop it!!!!!!!!!!!!!!!!!!!!!! Is someone making alt accounts just to make  :P more popular, and therefore voting 5 times for it???? This seems sooooooo ironic and therefore, almost unexpected...

Come on,  :D fans! Vote for  :D ! Vote for  :D !!!!!!!!!!!!!!!!!!!!!!!!!!

EDIT: And eggy you forgot to cast your vote in the poll. I voted for Mr. Green, so you should be the second one.

Lol, yay we're tied!!! Now let's see who ELSE likes  :D

[quote="ernie"]Ha ha. :P [/quote]
 :D > :P

Who else likes  :D ????

mee!
people, seriously, check my website at dhaivatpandya.googlepages.com/problemsolver'sguild

:P  ROCKS! oh i just tipped the scales a bit... maybe I can get mathblitz to vote for it......

[quote="xpmath"]:P . Meh. I lost.[/quote]
YOU LOST BECAUSE OF  :D ! (LOL)

I love  :D . Ha to everyone who votes :P

[quote="jain_harshi"]I love  :D . Ha to everyone who votes :P[/quote]
 :D  :D  :D

[quote="evilhamster"]Same. I definitely like  :D a lot :D .[/quote]
Only 11  :D s? Soon enough there should 15!

 :D  :D

14 to 11
:P to :D

:P pwns.. actually XD pwns all..

NOOOOOOOOOO 15 to 11

Come on :D people!!!! (those with the green avatars)

Seriously? :D is losing to the likes of  :P ? How is this possible?

oh god... 15:14

more :P voters pleaseeee!! :DD

no!!!!! more voters for  :D please!!!!!!!!

yay! :D  :D

but only 15-16? come on! :D

[quote="jjx1"]I'd have to say :P[/quote]
 :D  :D  :D

[quote="aleph0"]Obviously  :D[/quote]
OMG ALEPH LOOK!!!
 :D is winning 19 to 18!!!

GO  :D !!!!!!!!!!!!

[quote="aleph0"][quote="cf249"]Looks like I broke the tie.  Go  :P ![/quote]
 :mad:  :mad:[/quote]

See?  You can't even use your own smiley to show how you feel!   :P

[quote="cf249"]See?  You can't even use your own smiley to show how you feel!   :P[/quote]
 :mad:  :D

OWNED.

23 - 21 now  :D  :D  :D

Too many smilies in your post: 8.
You can have at most 3 smilies in this message. Please remove some smilies before submitting this message.

*smashes computer with a baseball bat*

yessssssssssssssssssssssssssssssssssssssssssssssssss :D  :D  :D


function Asymptote calculus derivative integration

If $ f(x) \equal{} \frac{2x \plus{} 7}{9}$ and $ g(x) \equal{} \frac {9x \plus{} 5}{7}$ then we define $ h(x)$ to be $ ((f^{\minus{}1})(x) \plus{} (g^{\minus{}1})(x) \minus{} 5x)$ What is the inverse of $ h(x)$?
(The powers of negative one represent inverse functions, [u]not[/u] exponents.)

I got [hide="My answer"]$ \frac{x \plus{} 6}{5}$[/hide] but I'm not sure if it's right.

[hide="Solution?"]obviously $ f$ and $ g$ have no asymptotes, so we won't have any restrictions upon $ f^{-1}$ or $ g^{-1}$.
$ f(x)=\frac{2x+7}{9}$
To find $ f^{-1}$, we switch $ x$ and $ f(x)$ and solve from there.  This is because $ (f\circ f^{-1})(x)=x$

$ x=\frac{2f^{-1}(x)+7}{9}$
$ 9x=2f^{-1}(x)+7$
$ f^{-1}(x)=\frac{9x-7}{2}$

Repeat for $ g(x)$
$ g^{-1}(x)=\frac{7x-5}{9}$

Add... $ \frac{81x-63+14x-10}{18}-5x=h(x)$
$ \frac{5x-73}{18}=h(x)$
$ h^{-1}(x)=\frac{18x+73}{5}$[/hide]

Which is waaaaaaaaaaay off from your answer.  Maybe I did something wrong?

[quote="mathcrazed"][hide="Solution?"]obviously $ f$ and $ g$ have no asymptotes, so we won't have any restrictions upon $ f^{ - 1}$ or $ g^{ - 1}$.
$ f(x) = \frac {2x + 7}{9}$
To find $ f^{ - 1}$, we switch $ x$ and $ f(x)$ and solve from there.  This is because $ (f\circ f^{ - 1})(x) = x$

$ x = \frac {2f^{ - 1}(x) + 7}{9}$
$ 9x = 2f^{ - 1}(x) + 7$
$ f^{ - 1}(x) = \frac {9x - 7}{2}$

Repeat for $ g(x)$
$ g^{ - 1}(x) = \frac {7x - 5}{9}$

Add... $ \frac {81x - 63 + 14x - 10}{18} - 5x = h(x)$
$ \frac {5x - 73}{18} = h(x)$
$ h^{ - 1}(x) = \frac {18x + 73}{5}$[/hide]

Which is waaaaaaaaaaay off from your answer.  Maybe I did something wrong?[/quote]
You are right. I accidentally subtracted $ 5x$ from the numerator without getting a common denominator, and also multplied 5 by 9 in the $ g^{-1}$ equation.  :oops:  But that's besides the point, and you are correct. The answer is $ \frac{18x+73}{5}$.

Sorry I just realize I made a typo in the third line

$ (f\circ f^{ \minus{} 1})(x) \equal{} x$ not $ f^{\minus{}1}(x)\equal{}x$

mathcrazed,
can you tell me where you learned all this derivative and integral stuff?You seem very familiar with it. I seriously need to learn all of this stuff.

Thank you a lot.

(you can start a new thread if you like)

[quote="iamhe"]can you tell me where you learned all this derivative and integral stuff?[/quote]

I don't see any derivatives/integrals  :maybe: . $ f^{ - 1}(x)$ is the inverse, while $ f^'(x)$ is the derivative.

Derivitives and integrals don't belong in HSB, seeing as they are calculus related topics  :roll: . [/spam]


limit
real analysis
calculus
calculus computations

I recently needed to use the following identity, but never actually learned the proof of it.  If it isn't too long, would someone please post the proof if you know it?

\[ \sum_{i=0}^{\infty}\sum_{j=0}^{i}a_{j,i-j}=\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}a_{k,n} \]
Thanks :).

I do have that written up as a handout for a (college) junior-level analysis class - it's a Word file, and runs to something like 8 pages. I don't think I want to attach it.

A word of warning: that is NOT always true.  You need some conditions on the terms to justify it.

The two most useful conditions: this is true if the terms $a_{k,n}$ are nonnegative (allowing for limits of $+\infty$), and this is true if the double series is absolutely convergent. That is,

$\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}|a_{k,n}|<\infty.$

The structure of the handout I was referring to: I considered four possible values:

The iterated sum:

$\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}a_{j,k}.$

The [b]other[/b] iterated sum:

$\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}a_{j,k}.$

The limit of the square partial sums:

$\lim_{n\to\infty} S_n,$ where $S_n=\sum_{j=0}^n\sum_{k=0}^na_{j,k}.$

The limit of the triangular partial sums:

$\lim_{n\to\infty} T_n,$ where $T_n=\sum_{j=0}^n\sum_{k=0}^{n-j}a_{j,k}.$

For non-absolutely convergent sums, it is possible to find examples in which these four methods give four different results. (Try it!)

The plan, in general: start with the case of nonnegative terms. In that case, it's easy to show that $S_n\le T_{2n}\le S_{2n},$ so the square and triangular sums will give the same result. The main energy of the proof will then be devoted to proving the equivalence of the the limit of the square partial sums with one of the iterated sums.

Once you have these results for the case of nonnegative terms, we take care of the case of absolutely convergent terms by considering the series with terms $|a_{j,k}|-a_{j,k}.$

--

What you wrote at the top is essentially the equivalence of the limit of the triangular partial sums with the limit of one of the iterated partial sums.

--

None of this requires the Lebesgue theory, but given the Lebesgue theory we recognize these theorems as analogues and/or special cases of some of the central theorems of that theory: the pair Monotone Convergence/Dominated Convergence and the pair Tonelli/Fubini.

Here's the key piece of the analysis. We're going to compare the row-first iterated sums to the limit of the square partial sums, under the hypothesis that the terms are nonnegative.

Suppose $\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}a_{jk}=L.$

($L$ could be either finite or infinite.)

Then for each $n,$

$S_n=\sum_{j=0}^n\sum_{k=0}^na_{jk}\le\sum_{j=0}^n\sum_{k=0}^{\infty}a_{jk} \le L.$

Hence, $\lim_{n\to\infty}S_n\le L.$

Note that this limit exists (finite or infinite) because $S_n$ is increasing.

Now suppose $L$ is finite. This implies that for each $j,\ \sum_{k=0}^{\infty}a_{jk}$ is finite.

Choose an arbitrary $\epsilon>0.$

For each $j,$ find $m_j$ such that $\sum_{k=0}^{m_j}a_{jk}>\sum_{k=0}^{\infty}a_{jk}- 2^{-j-2}\epsilon.$

Now choose $N_1$ such that for $n>N_1,$

$\sum_{j=0}^{N_1}\sum_{k=0}^{\infty}a_{jk}>L-\frac{\epsilon}2.$

Let $N=\max(N_1,m_0,m_1,\dots,m_{N_1}).$ Then

$S_N\ge\sum_{j=0}^{N_1}\sum_{k=0}^{m_j}a_{jk}$

$\ge \sum_{j=0}^{N_1}\left(\sum_{k=0}^{\infty}a_{jk}- 2^{-j-2}\epsilon\right)$

$\ge L-\frac{\epsilon}2-\sum_{j=0}^{N_1}2^{-j-2}\epsilon$

$\ge L-\frac{\epsilon}2-\sum_{j=0}^{\infty}2^{-j-2}\epsilon=L-\epsilon.$

This shows that $\lim_{n\to\infty}S_n\ge L.$

The case of $L=\infty$ is handled similarly; I'll not write out the details.

This is just a special case of introducing series over any index set. For an absolutely convergent double series, iterated summation gives both of those sums as ways of evaluating it. I don't really know the english terms for the theory I've learned, but let $I$ be an index set which we want to sum over. Let $(F_n)_{n=0}^{\infty}$ such that i) Each $F_n$ is finite ii) The union of all $F_n$ is $I$ iii) $F_n \subset F_{n+1}$. Then we define the sum $\sum_{i\in I} a_i$, where every $a_i \ge 0$ as $\lim_{n\to\infty} \sum_{i\in F_n} a_i$. Like Kent Merryfield did, show that this definition does not that depend on what family of sets we use to evaluate it.

Now, through some trivial estimations it is easy to show that we get the same result even if we drop condition i) on $F_n$. It is also trivial to show that if $I = I_1 \cup I_2$ where $I_1 \cap I_2 = \emptyset$ and $a_i \ge 0$, then  $\sum_I a_i = \sum_{I_1} a_i + \sum_{I_2} a_i$.

Use these two facts to show that if $I = \cup_{j=1}^\infty I_j$ where $I_j \cap I_k = \emptyset$ and if $a_i \ge 0$ then
$\sum_I a_i = \sum_{j=1}^\infty \left(\sum_{I_j} a_i \right)$

Now suppose $i=(k,n)$ and $I = \mathbb{N} \times \mathbb{N}$. $I$ can be written as the union of the sets $k \times \mathbb{N}$, k={0},{1},{2}.... Thus, using our theorem we get that
$\sum_{I}a_{kn} = \sum_{k=0}^\infty \left( \sum_{n=0}^\infty a_{kn} \right)$

Similarly, we may consider the sets $I_p = \{ (k,n) \in \mathbb{N} \times \mathbb{N}, j+k=p \}$ in order to evaluate the sum. I have to go right now, so I leave it to you to show that you get your other sum.

After this we can extend it to absolutely convergent series where the terms have different signs.

I realize that Kent Merryfield's presentation is more to the point, but I don't see why you should restrict yourself to this special case when with only a little more work you can get the more general versions that I was talking about. I think I also left four or five nice exercises in there :P


I just bought a TI-84 last weekend...... How do you download games to the calculator? Cuz my math class is pretty boring and I want to play games there :)

TI-84's are for chumps.  89 all the way.

First you need to get a link for the computer.  Then you have to install TI-connect which I am pretty sure is on the TI site.  Then you have you can download games and put them on there.


induction
number theory unsolved
number theory

Prove that the equation $3^k=m^2+n^2+1$ has infinite many solutions in natural numbers.

Take $k=2^m$.
Claim: $3^k-1$ can be written as sum of two squares.
It's correct for $m=0$.
When $m \geq 1$, it is $3^{2^m}-1 = (3^{2^{m-1}}-1)(3^{2^{m-1}}+1)$.
The left one can by assumed to be $=a^2+b^2$ by induction hypothesis, the right one is sum of two squares $c^2+d^2$ as written there.
By $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$, the result follows.

But the claim is clearly equivalent to the problem, thus done.


function
Support
real analysis
calculus
calculus computations

given values at any $ n$ points, we can find a function which is infinitely-many-times-differentiable at all points and takes those values,
e.g. given $ f(1) \equal{} 3$, $ f(2)\equal{}7$, $ f(\minus{}1)\equal{}1$, $ f(x)\equal{}x^2 \plus{} x \plus{} 1$, satisfies these conditions and is inf-differentiable at all points. Suppose the value of the function is given at all integers i.e $ f(0),f(1),f(\minus{}1)...$ are given.
What are the necessary and sufficient conditions on the values so that there exists a function$ (f: \mathbb{R} \minus{}> \mathbb{R})$ having those values and inf-differentiable at all points?

There are no conditions: you can always find such an $ f$. Indeed, let $ \phi$ be a [url=http://en.wikipedia.org/wiki/Bump_function]bump function[/url] with support in $ [\minus{}1/3,1/3]$ and $ \phi(0)\equal{}1$. Given a sequence $ (a_n)_{n\in Z}$, define $ f(x)\equal{}\sum_{n\in\mathbb Z}a_n\phi(x\minus{}n)$.

If you instead require $ f$ to be real-analytic on $ \mathbb R$, then I dunno :maybe:


algebra
polynomial
limit
algebra proposed

Define $ P_n(x)$ to be a $ n$-th degree polynomial with real coefficients (denoted with $ a_k$), such that there is no roots appearing more than once. If $ r_i$ is the $ i$-th root of $ P$ such that:

$ |r_i| > 0$

express the following expression in terms of the polynomial coefficient's ($ a_k$):

$ \prod_{j \in [1,n] \setminus i}\frac {1}{1 \minus{} \frac {r_j}{r_i}} \equal{} \,?$

EDIT: I apologise, I edited it. I noticed that no matter what root we chose, the above expression seems to be constant (numericaly), so wanted to ask you what do you think!

i think there are some mistake in the text:
milin is it " $ r_i \neq 0$ and what is the $ \displaystyle \sum_{i=1}^{n}{[\prod_{j\in[1,n],j\neq i}{\frac{1}{1-\frac{r_j}{r_i}}]}}$?

in this case:

$ \displaystyle \sum_{i=1}^{n}{[\prod_{j\in[1,n],j\neq i}{\frac{1}{1-\frac{r_j}{r_i}}]}}$=$ \sum_{i=1}^{n}{\frac{r_i(1-r_i)}{\prod_{j=1}^{n}{1-r_j}}}$

then, if $ P(x)=\sum_{i=1}^{n}{a_ix^i}$ then it is simply $ \displaystyle \frac{-\frac{a_{n-1}}{a_n}-{(\frac{a_{n-1}}{a_n})}^2+2\frac{a_{n-2}}{a_n}}{\frac{P(1)}{a_n}}$=$ \displaystyle \frac{a_n(2a_{n-2}-a_{n-1})-a_{n-1}^2}{P(1)}$

if i undestand the problem... :ninja:
bye
paolo

Paolo, thanks for your answer, but I'm afraid that this is not the question.

sorry, but i think that probably this product isn't costant  :(

$ \prod_{1 \le j \le n, j\neq i}{\frac {r_i}{r_i \minus{} r_j}} \equal{} \frac {r_i^{n \minus{} 1}}{r_i^{n \minus{} 1} \minus{} r_i^{n\minus{}2}\sum_{1\le j \le n,j\neq i}{r_j} \plus{} r_i^{n\minus{}3}\sum_{1\le j < k \le n, j\neq i, k \neq i}{r_jr_k} \minus{} ...}$

but for n=2 $ \frac {a}{a \minus{} b} \equal{} \frac {b}{b \minus{} a}$ isn't true with only your conditions, and also for n=3:
$ \frac {a}{(a \minus{} b)(a \minus{} c)} \equal{} \frac {b}{(b \minus{} c)(b \minus{} a)} \implies c \equal{} \frac {ab(a \plus{} b)}{a^2 \plus{} b^2}$ so, the root $ c$ is unique forall $ (a,b)$..and so on.

probably i don't understand the problem again :P

Wow, I can't believe how stupid I've been. This is SOO easy:

$ \prod_{j \in [1,n]\setminus i}\frac{1}{1-\frac{r_j}{r_i}}=r_i^{n-1}\prod_{j \in [1,n]\setminus i}\frac{1}{r_i-r_j}$

Now, we could say:

$ \prod_{j \in [1,n]\setminus i}\frac{1}{x-r_j}=\frac{x-r_i}{P_n(x)}$

and in that spirit:

$ r_i^{n-1}\prod_{j \in [1,n]\setminus i}\frac{1}{r_i-r_j}=r_i^{n-1}\lim_{x\rightarrow r_i}\frac{x-r_i}{P_n(x)}$

Since this is definitly a $ \frac{0}{0}$ form, we can L'Hopital it and get:

$ \prod_{j \in [1,n]\setminus i}\frac{1}{1-\frac{r_j}{r_i}}=\frac{r_i^{n-1}}{P^'_n(r_i)}$

I hope this is OK. If not, my stupidity note still stands..

[quote="bboypa"]sorry, but i think that probably this product isn't costant  :(

$ \prod_{1 \le j \le n, j\neq i}{\frac {r_i}{r_i \minus{} r_j}} \equal{} \frac {r_i^{n \minus{} 1}}{r_i^{n \minus{} 1} \minus{} r_i^{n \minus{} 2}\sum_{1\le j \le n,j\neq i}{r_j} \plus{} r_i^{n \minus{} 3}\sum_{1\le j < k \le n, j\neq i, k \neq i}{r_jr_k} \minus{} ...}$

but for n=2 $ \frac {a}{a \minus{} b} \equal{} \frac {b}{b \minus{} a}$ isn't true with only your conditions, and also for n=3:
$ \frac {a}{(a \minus{} b)(a \minus{} c)} \equal{} \frac {b}{(b \minus{} c)(b \minus{} a)} \implies c \equal{} \frac {ab(a \plus{} b)}{a^2 \plus{} b^2}$ so, the root $ c$ is unique forall $ (a,b)$..and so on.

probably i don't understand the problem again :P[/quote]

Yes, unfortunately for me, you are right.. It's not a constant. But if a root is complex, then the values are really close one to each other. :)

i'm interested what do you mean with "if a root is complex": at least one? or $ r_i \in C \minus{} R$? or all complex(not always)?

because in my case if n=3 and $ a \equal{} x \plus{} yi, b \equal{} x \minus{} yi$ then $ c \equal{} \frac {ab(a \plus{} b)}{a^2 \plus{} b^2} \equal{} \frac {x(x^2 \plus{} y^2)}{x^2 \minus{} y^2} \in R$, unique as before..

 :huh:  :huh:

By "if the root is complex" I mean that if we do the calculation with a complex root. I think the fact that I'm doing this with a specific polynomial makes all the difference. If I put the $ r_i$ to be complex and do the calculation, I get results that have $ \approx 10^{ \minus{} 7}$ variance between them. Also, if $ n$ is even, the calculation for complex root's is again similar for all of them, but also to the calculation I get with one of the real roots (there always are two when $ n$ is even). By calculation I mean the product expression I wrote above.


algebra
polynomial
calculus
calculus computations

Find all pairs $f(x)$, $g(x)$ of polynomials with integer coefficients satisfying
\[f(g(x))=x^{2007}+2x+1. \]

[color=green]
EDITED[/color]

$g(x) = x^{2007}+2x$ and $f(x) = x+1$ will do don't you think?

if so, see here http://www.mathlinks.ro/Forum/viewtopic.php?t=149989  :wink:


linear algebra
matrix
linear algebra unsolved

Let $ A \equal{} \begin{pmatrix}A_1 \\
A_2\end{pmatrix}$ and $ m\times n$ matrix ($ m > n$) with $ A_1$ nonsingular. Show that the least eigenvalue of $ A_1^*A_1$ is smaller than that of $ A^*A$.

(And don't use the courant-fischer theorem - unless you prove it.)

I think I just found a proof myself, I'll post it for reference: it's true for any positive semidefinite matrices $ A,B$ that the least eigenvalue of $ A$ is at most the least eigenvalue of $ A\plus{}B$.

For if $ \lambda$ is the smallest eigenvalue of $ A$ then $ x^*(A\minus{}\lambda I)x\ge0$ and $ x^*Bx\ge0$ so summed up: $ x^*((A\plus{}B)\minus{}\lambda I)x\ge0$. Hence $ A\plus{}B\minus{}\lambda I$ has all its eigenvalues nonnegative, hence all eigenvalues of $ A\plus{}B$ are at least $ \lambda$


email
trigonometry
geometry
conics
calculus
integration
analytic geometry

Hi all, chatters!
I'm looking for the book in the title, by G.S.Carr...
The second edition was published in 1970 by Chelsea, if anyone has it please call me... I can't find it anywhere (I live in Italy)...

Thanks in advance
tmart

Here's the letter I've just sent to som USA libraries... I paste it here because this is exactly what I'm looking for: a suggestion of other similar books, how to have photocopies/ebook/book ... thanks in advance!

"I'd love to find/have the book "Formulas and Theorems in Pure Mathematics" by George Shoobridge Carr, I've been looking for it for 13 months here in Italy, but I haven't found it yet. Lately I've noticed that other countries are full of libraries with this book.
May I buy a copy of this text (or photocopies) from you?
Or could you suggest me another (possible) way to get it, or simplier other newer books which cover the same arguments?"

my email is michemiccia@libero.it

Why are you so interested in this book? Just because it was useful for Ramanujan?

Hi ComplexZeta... You're right, I should explain.
Yes, of course: I've seen that book in Ramanujan's biography "The man who knew infinity"... I suppose Carr's book isn't considered a good book.
But I've been studying analysis for 5 months (by myself, of course: I attend an italian high school!  ;) ) using an Italian textbook ("Analisi Matematica" I/II/III by Giaquinta-Modica); I finally understood that a so "cold" text, (without anything which could remember problem-solving heuristics, i.e.) couldn't be right for a young student which simply loves infinite series and number theory (what about astrophysics?) and similar  :P
That's why I decided to seek a concise book about what Carr calles "pure and applied Mathematics"... As a matter of fact, I'm not sure that this is the right book for me, maybe better books about similar topics have been written... But I don't know,  and I didn't know how to explain a forum what I was looking for, so I wrote down the unique title I actually know...

gosh, I'm also looking for Pimsleur English Courses for Italian... I should work on english!   :blush:

thanks

tmart

I want to know much more about the preface of this book.
thx

these are some informations on the book:

"In later writings, Hardy dismissed Carr's Synopsis as being unremarkable. But Hardy was scarcely able to appreciate how the book might have been a rich source of material for a self-taught student. Anyone who has faced the difficulties of getting access to scientific literature will be able to appreciate the value of a book like Carr's. Here, in one place, was a comprehensive summary of basic mathematical knowledge known to the West. It also contained a very substantial cross-referenced index to European papers on pure mathematics of the nineteenth century, although this may not have been of much value to Ramanujan sitting in Madras.

A small quotation placed casually at the end of Carr's book sums up its usefulness:

"There is an immense amount of knowledge lying scattered at the present day, and almost useless from the difficulty of finding it when wanted - Professor J. D. Everett"
The value of the book comes from its method of organization. There are 6165 results covering algebra, theory of equations, plane trigonometry, spherical trigonometry, elementary geometry, geometrical conics, differential calculus, integral calculus, calculus of variations, differential equations, calculus of finite differences, plane coordinate geometry, and solid coordinate geometry.

Many of the results are at a high school level but some are fairly advanced and complicated. The book states a formula or a theorem but gives only a sketch of proof, so the reader is forced to work out the details. As Carr put it, in his nineteenth century pedagogical diction:

"Let them be read once, but recalled often. The difference in the effect upon the mind between reading a mathematical demonstration, and originating one wholly or partly, is very great. It may be compared to the difference between the pleasure experienced, and interest aroused, when in the one case a traveller is passively conducted through the roads of a novel and unexplored country, and in the other case he discovers the roads for himself with the assistance of a map."
Professor Hardy had grudgingly conceded that it is "a book written with some real scholarship and with a style and individuality of its own". In reality, the book is an elaborate crib sheet, such as one would create to prepare for a difficult exam.

Indeed, the book originated as a study guide for the Cambridge Tripos exams. During the nineteenth century, mathematical education at British Universities was largely oriented towards preparing students for the Tripos examinations, which involved solving tricky problems within a set time. The exams originated in 1748, sometime after Isaac Newton's tenure at Cambridge, and the name Tripos came from the three-legged stool on which the student sat to argue his points. Those who placed in the first class were known as Wranglers, and were assured of a successful future in any profession. Those who placed in the second and third classes were known as Senior and Junior Optimes. A market for study guides and coaches naturally arose under these circumstances. George Shoobridge Carr was a Tripos coach. The Synopsis came from his coaching notes, refined meticulously over a period of fourteen years. "

I saw that in France there are 3 libraries with that book!
How lucky!

tmart

[quote="tmart"]
I saw that in France there are 3 libraries with that book!
[/quote]

Is there one in Paris ?

You can also buy it at:[url=http://www.amazon.com/exec/obidos/tg/detail/-/0828402396/104-0707215-3248732?v=glance] here.
[/url]

It's rather expensive.  You may like to check out some  used bookstores too.

yes moubinool, there's a copy of the book at this library:
Bibliotheque L'University de Paris, Sud-Rcon (Paris)

Gyan: I noticed that used version, but it's too expensive, really!

Any suggestion about other similar and preferably more recent books?

tmart

it's hard to believe that amazon has that book which you've been looking for 13 months


geometry
3D geometry
icosahedron
octahedron
advanced fields
advanced fields unsolved

If anyone can give me any insight to these problems, it will be greatly appreciated.

a) Prove that for any connected graph G=(V,E), the number of spanning subgraphs of G that are componentwise Eulerian is 2^|e|-|v|+1.

b) Prove that for any regular and planar graph G with a line graph Lg, Lg is also planar but not isomorphic to G nor is Lg edgless, the number of of edges in Lg is exactly 3|v|.

c) For which surfaces that are closed and orientable does there exist a 2-cell embedding of a 4-regular grpah such that all faces are also of degree 4? Prove your answer.

[quote="beezalee"]
b) Prove that for any regular and planar graph G with a line graph Lg, Lg is also planar but not isomorphic to G nor is Lg edgless, the number of of edges in Lg is exactly 3|v|.
[/quote]

This does not appear to be true. Think about the regular icosahedron: it's planar, but $5$-regular, so its line graph contains a $K_5$ as an induced subgraph, meaning that it can't be planar. Also, the formula for the number of edges seems to be wrong, in general. According to [url=http://mathworld.wolfram.com/LineGraph.html]this[/url], if our graph is $r$-regular, its line graph is supposed to have $\frac{r(r-1)}2\cdot v$ edges. This coincides with $3v$ iff the graph is $3$-regular, which is not always the case. It can be shown that it's at most $5$-regular (any planar graph has a vertex of degree at most $5$), but all the cases $3,4,5$ occur (again, we only need to look at the platonic solids: the octahedron is $4$-regular, and the icosahedron is $5$-regular).

Maybe you were referring to $3$-regular planar graphs (not merely regular), but forgot to mention it? :?


function
real analysis
real analysis unsolved

Find all real number $ c$ st the sequence:

$ a_{n}\equal{}\dfrac{n^c \minus{} (n\minus{}1)^c\plus{}(n\minus{}2)^c\minus{}...\plus{}(\minus{}1)^{n\minus{}2}2^c \plus{} (\minus{}1)^{n\minus{}1}1^c}{n}$

is convergent.

Recall Cesaro's Theorem
(http://www.math.ksu.edu/~nagy/snippets/stolz-cesaro.pdf)
we get for any sequence $b_n$
$\liminf b_n \leq \liminf \frac{\sum^n_{k=1} b_k}n \leq \limsup \frac{\sum^n_{k=1} b_k}n \leq \limsup b_n$.

 Case 1: $c <0$.
$|a_n|\leq \frac{1^c+2^c+\ldots+n^c}n$, since $n^c \rightarrow 0$ we get (from Cesaro) RHS $\rightarrow 0$, so $a_n \rightarrow 0$

For the other cases we can write
$a_{2n}=\frac{\sum_{k=1}^{n}(2k)^c-(2k-1)^c}{2n}$

$a_{2n+1}=\frac{\sum_{k=1}^{n}(2k+1)^c-(2k)^c}{2n+1}+\frac{1}{2n+1}$
$a_n$ converges if and only if both $a_{2n}$ and $a_{2n+1}$ converge to the same limit.

Case 2: $c=0$ or $c=1$.
If $c=0$, then $a_{2n}=0$ and $a_{2n+1}=\frac{1}{2n+1}$ so $a_n \rightarrow 0$.

If $c=1$, then $a_{2n}=\frac{1}2$ and $a_{2n+1}=\frac{n+1}{2n+1}$ so $a_n \rightarrow \frac{1}2$

Case 3: $c\in(0,1)$ or $c>1$
by the mean value theorem (function $x^c$) we get
$a_{2n}=\frac{\sum_{k=1}^{n}cx_k^{c-1}}{2n}$ for some $x_k \in [2k-1,2k]$

$a_{2n+1}=\frac{\sum_{k=1}^{n}cy_k^{c-1}}{2n+1}+\frac{1}{2n+1}$ for some $y_k \in [2k,2k+1]$

If $c\in(0,1)$ the function $x^{c-1}$ is decreasing so

$ 0 \leq a_{2n}  \leq \frac{c}2 ( \frac{\sum_{k=1}^n (2k-1)^{c-1}}n  )$
and since $(2n-1)^{c-1} \rightarrow 0$, $RHS \rightarrow 0$  (Cesaro's), so $a_{2n} \rightarrow 0$
on the other hand
$ 0 \leq a_{2n+1}  \leq \frac{c}{\frac{2n+1}n} ( \frac{\sum_{k=1}^n (2k)^{c-1}}n  )+\frac{1}{2n+1}$
and since $(2n)^{c-1} \rightarrow 0$, $RHS \rightarrow 0$  (again Cesaro's), so $a_{2n+1} \rightarrow 0$. So finally $a_{n} \rightarrow 0$.

If $c>1$ the function $x^{c-1}$ is increasing so
$ a_{2n}  \geq \frac{c}2 ( \frac{\sum_{k=1}^n (2k-1)^{c-1}}n  )$
and since $(2n-1)^{c-1} \rightarrow \infty $, $RHS \rightarrow \infty$ (why?), so $a_{2n} \rightarrow \infty$. this means $a_n$ diverges.
Conclusion:
We get convergence for $c \leq 1$ and divergence for $c>1$

Edit: A typo. It is actually
$ \liminf b_{n}\leq\liminf\frac{b_1+ \ldots+ b_{n}}{n}\leq\limsup\frac{b_1+ \ldots+ b_{n}}{n}\leq\limsup b_{n}$


Given real numbers $x, y, z,a,b,c$ such that:
$a + b = 6$
$ax + by = 10$
$ax^2 + by^2 = 24$
$ax^3 + by^3 = 62$
determine $ax^4 + by^4$.

The trick is to use $(ax^n+by^n)(x+y) =(ax^{n+1}+by^{n+1})+(ax^{n-1}+by^{n-1})xy$.

This problem is very similar to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80286]1990 AIME Problem 15[/url].


algebra
polynomial
floor function
group theory
abstract algebra
logarithms
function

Does anyone know if there is a closed form expression for sums of q-binomial coefficients, for q being integers?
\[ A(N, q) : \equal{} \sum_{i \equal{} 0}^N \binom{N}{i}_q\]
where $ \binom{N}{i}_q$ are the q-analogs to binomial coefficient. Are these sums called Galois numbers?

Of course we recover the usual binomial coefficients when $ q \equal{} 1$ and it is easy to show
\[ A(N, 1) \equal{} \sum_{i \equal{} 0}^N \binom{N}{i} \equal{} 2^N\]
I plotted these sums in Mathematica for q=2 and by fitting it seems in the limit of large $ N$ we have
\[ A(N, 2) \sim 2^{(N/2)^2 \plus{} c}\]
where $ c \approx 2.88$ is just some constant.

Any help is greatly appreciated.

Well, the closest related thing I've seen is the $ q$-binomial theorem,
\[ (1 \plus{} x)(1 \plus{} qx)\cdots (1 \plus{} q^{n \minus{} 1}x) \equal{} \sum_{i \equal{} 0}^n \binom{n}{i}_q\, q^{\binom{i}{2}} x^i.\]
This certainly doesn't give you directly what you want.  We also know that $ \binom{n}{i}_q$ is a polynomial in $ q$ of degree $ \binom{n}{2} \minus{} \binom{i}{2} \minus{} \binom{n \minus{} i}{2} \equal{} i(n \minus{} i)$.  So, $ A(n, q)$ is a polynomial in $ q$ of degree $ \left\lfloor \frac {n^2}{4} \right\rfloor$.  (This polynomial is monic of $ n$ is even and has leading coefficient $ 2$ if $ n$ is odd.)  This approximately explains your result; if you troubled to compute the next-highest coefficient or two, you could probably get more such info.

Thank you. Your remark about $ A(n,q)$ being a monic polynomial is helpful. I used the recurrence relation
\[ \binom{n}{k}_q \equiv q^k \binom{n \minus{} 1}{k}_q \plus{} \binom{n \minus{} 1}{k \minus{} 1}_q\]
to get the leading term as pointed out by you. I also calculated the next-leading order term but what I got is not consistent with the Mathematica fit (pretty close though). Probably made some mistakes...

That next term should actually depend on even/odd parity. I calculated things [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=163158]here[/url] in the context of counting subgroups of $ (\mathbb{Z}/2\mathbb{Z})^n$; the asymptotic behavior is actually $ \log_2 A(N,2) \equal{} \frac14\left(\log_2(N)\right)^2 \plus{} c(n) \plus{} o(1)$ as $ n\to\infty$, where $ c(n)\approx\begin{cases}2.8820499 & n\text{ is even} \\
2.8820462 & n\text{ is odd}\end{cases}$

Since I didn't post the details of my computation, I'll have to spend some time reconstructing that.

[Added in edit]
Let's start in the even $ n$ case. Also, assume $ |q|>1$; the following estimates rely on high powers of $ q$ growing, although a variant with $ |q|<1$ could also be done. On the unit circle $ |q|\equal{}1$, things are fundamentally different.

The biggest term is $ \binom{n}{n/2}_q$. Let's look at that, and then compare other terms to it.
$ \binom{n}{n/2}_q\equal{}\frac{(q^n\minus{}1)(q^{n\minus{}1}\minus{}1)\cdots (q^{n/2\plus{}1}\minus{}1)}{(q\minus{}1)(q^2\minus{}1)\cdots (q^{n/2}\minus{}1)}$
Assume $ n$ is large, so we can approximate the terms in the numerator with just the powers of $ q$:
$ \binom{n}{n/2}_q\approx\frac{(q^n)(q^{n\minus{}1})\cdots (q^{n/2\plus{}1})}{(q\minus{}1)(q^2\minus{}1)\cdots (q^{n/2}\minus{}1)}$
$ \approx q^{n\minus{}1}\cdot q^{n\minus{}3}\cdots q^1\cdot\frac{q}{q\minus{}1}\cdot\frac{q^2}{q^2\minus{}1}\cdots \frac{q^{n/2}}{q^{n/2}\minus{}1}$
$ \binom{n}{n/2}_q\approx q^{(n/2)^2}\cdot g(q)$, where $ g(q)\equal{}\prod_{k\equal{}1}^{\infty}\frac1{1\minus{}q^{\minus{}k}}$ That $ g$ is a permutation generating function, which has no closed form. The product converges rapidly enough to get nice estimates anyway; $ g(2)\approx 3.46274662$.

Next, we look at the nearby terms. $ \binom{n}{n/2\minus{}j}$ removes the last $ j$ terms in the numerator and denominator of the defining product; those terms multiply to about $ q^{j^2}$. By symmetry, $ \binom{n}{n/2\plus{}j}$ is exactly the same. Add them up, and $ \sum_k \binom{n}{k}_q\approx q^{(n/2)^2}\cdot g(q)\cdot \left(1\plus{}\frac2{q}\plus{}\frac{2}{q^4}\plus{}\frac{2}{q^9}\plus{}\cdots\plus{}\frac{2}{q^{m^2}}\plus{}\cdots\right)$. Call that sum $ f_e(q)$; it also converges rapidly, with $ f_e(2)\approx 2.128936827$.

That was the even case. For odd $ n$, there is no longer a single central largest coefficient. Instead, we have two. One of them is
$ \binom{n}{(n\minus{}1)/2}\equal{}\frac{(q^n\minus{}1)(q^{n\minus{}1}\minus{}1)\cdots (q^{(n\plus{}3)/2}\minus{}1)}{(q\minus{}1)(q^2\minus{}1)\cdots (q^{(n\minus{}1)/2}\minus{}1)}$
$ \binom{n}{(n\minus{}1)/2}\approx q^{n\minus{}1}\cdot q^{n\minus{}3}\cdots q^2\cdot \frac{q}{q\minus{}1}\cdot\frac{q^2}{q^2\minus{}1}\cdots \frac{q^{n/2}}{q^{n/2}\minus{}1}\approx q^{(n^2\minus{}1)/4}\cdot g(q).$
Again, each step away from the center takes one term off the end of the defining product. Add them up for
$ \sum_k \binom{n}{k}_q\approx q^{(n^2\minus{}1)/4}\cdot g(q)\cdot \left(2\plus{}\frac2{q^2}\plus{}\frac{2}{q^6}\plus{}\frac{2}{q^{12}}\plus{}\cdots\plus{}\frac{2}{q^{m(m\plus{}1)}}\plus{}\cdots\right)$
Set $ f_o(q)\equal{}2q^{\minus{}\frac14}\sum_{m\equal{}0}^{\infty}q^{\minus{}m(m\plus{}1)}$ for $ \sum_k \binom{n}{k}_q\approx q^{n^2/4}\cdot g(q)\cdot f_o(q)$ when $ q$ is odd. We have $ f_o(2)\approx 2.128931251$.

The odd and even versions being so close to each other at $ q\equal{}2$ is really just a coincidence.


[hide="Hint"]We can write $ \frac {1}{(2n \minus{} 1)(2n \plus{} 1)} \equal{} \frac{A}{2n\minus{}1} \plus{} \frac{B}{2n\plus{}1}$, where $ A$ and $ B$ are constants that don't depend on $ n$.

Can you find $ A$ and $ B$? What happens when you sum this new form over $ k$?[/hide]

Alright, I got it now. I didn't think to decompose it before

$ \frac {1}{(2n \plus{} 1)(2n \minus{} 1)} \equal{} \frac {1}{4n \minus{} 2} \minus{} \frac {1}{4n \plus{} 2}$

so the sum is $ \frac {1}{2} \minus{} \frac {1}{4k \plus{} 2} \equal{} \frac {k}{2k \plus{} 1}$


inequalities
algebra
polynomial
national olympiad

I think problem 2 be wrong

Or i can not understand it

please help me

Mr. Tuan, I like Vietnamese Math Olympiad, but I don't know what is the system. May you explain? (How much rounds there are?, How much people take part in each round? What is the time for each round? Where is it hold?)

Ugly solution to inequality:

[hide]Normalize to $ x\equal{}1, y\equal{}1\plus{}a, z\equal{}1\plus{}b$, $ a\neq b$ and $ a,b>0$

The given inequality simplifies to
$ \frac {a^2b^2\plus{}a^2(a\minus{}b)^2\plus{}b^2(a\minus{}b)^2}{a^2b^2(a\minus{}b)^2} \ge \frac {4}{3\plus{}2a\plus{}2b\plus{}ab}$

Expanding and simplifying further gives
$ 2a^5\plus{}2b^5\plus{}a^5b\plus{}ab^5\plus{}16a^3b^3\plus{}2a^3b^2\plus{}2a^2b^3\plus{}3a^4\plus{}3b^4 \ge 6a^4b^2\plus{}6a^2b^4\plus{}2ab^4\plus{}2a^4b\plus{}6a^3b\plus{}6ab^3$

Checking by tedious majorization and polynomials we find that
$ 2a^5\plus{}2b^5\ge 2ab^4\plus{}2a^4b$
$ a^5b\plus{}ab^5\plus{}16a^3b^3\plus{}2a^3b^2\plus{}2a^2b^3\plus{}3a^4\plus{}3b^4 \ge 6a^4b^2\plus{}6a^2b^4\plus{}6a^3b\plus{}6ab^3$**

** I can give a proof if necessary. It is just some tedious algebra using calculus.[/hide]

pvthuan... You surely have a better proof in mind. Can you give me a hint as to how i can approach the problem in a simpler method?

WLOG suppose $ x>y>z\geq0$. Let $ x \equal{} z\plus{}a$, $ y \equal{} z\plus{}b$, then $ a>b>0$

So the original inequality
$ \Leftrightarrow \frac{1}{(a\minus{}b)^2}\plus{}\frac{1}{a^2}\plus{}\frac{1}{b^2} \geq \frac{4}{(z\plus{}a)(z\plus{}b)\plus{}z(2z\plus{}a\plus{}b)}$
$ \Leftrightarrow \frac{1}{(a\minus{}b)^2}\plus{}\frac{1}{a^2}\plus{}\frac{1}{b^2} \geq \frac{4}{3z^2\plus{}2(a\plus{}b)z\plus{}ab}$
$ \Leftrightarrow \frac{ab}{a^2\minus{}2ab\plus{}b^2}\plus{}\frac{ab}{a^2}\plus{}\frac{ab}{b^2} \geq \frac{4ab}{3z^2\plus{}2(a\plus{}b)z\plus{}ab}$
$ \Leftrightarrow \frac{1}{\frac{a}{b}\minus{}2\plus{}\frac{b}{a}}\plus{}\frac{b}{a}\plus{}\frac{a}{b} \geq \frac{4ab}{3z^2\plus{}2(a\plus{}b)z\plus{}ab}$
$ \Leftrightarrow \frac{1}{\frac{a}{b}\minus{}2\plus{}\frac{b}{a}}\plus{}\frac{b}{a}\plus{}\frac{a}{b}\minus{}2 \geq \frac{4ab}{3z^2\plus{}2(a\plus{}b)z\plus{}ab}\minus{}2$
Note that $ \frac{1}{\frac{a}{b}\minus{}2\plus{}\frac{b}{a}}\plus{}\frac{b}{a}\plus{}\frac{a}{b}\minus{}2 \geq 2$ while $ \frac{4ab}{3z^2\plus{}2(a\plus{}b)z\plus{}ab}\minus{}2 \leq \frac{4ab}{ab}\minus{}2 \equal{} 2$
thus the original inequality is true

I think we should be posting the Vietnam NMO 2008 problems on the forum: each problems should have its own thread of discussion.

[quote="L_Euler"][quote="pvthuan"]See the file attached.[/quote]

This toyear solution is very long! It's is different from last year and far.

@pvthuan: Thầy Thuận tìm giúp em bộ đề APMO năm nay với, em tìm mãi chẳng thấy!  :(

Thanks thầy nhiều!   :)  :)[/quote]
Có trong sách mà bạn,cuốn của Nguyễn Văn nho,Hình như đầy đủ luôn mà


algebra
difference of squares
special factorizations

Find the greatest positive prime factor of 2^20 - 1.

[hide="Solution"]By applying the Difference of Squares Factorization, we have
\[ 2^{20} \minus{} 1 \equal{} (2^{10} \minus{} 1)(2^{10} \plus{} 1) \equal{} (2^5 \minus{} 1)(2^5 \plus{} 1)(2^{10} \plus{} 1).\]
Now, we know that $ 2^5 \minus{} 1 \equal{} 31$, $ 2^5 \plus{} 1 \equal{} 33$, and $ 2^{10} \plus{} 1 \equal{} 1025$. We notice that $ 31$ is prime, $ 33$ can be broken into $ 3\cdot11$ and $ 1025$ can be factorized into $ 5^2\cdot41$. Therefore, the answer is $ \boxed{41}$.[/hide]


Following up on this post
http://www.artofproblemsolving.com/Forum/topic-57578.html

My question is how one would solve an equation of the form
$\ddot{x}=-A\dot{x}^2-B$
The difference here is that B is also negative.
The physics of this would be a race car: friction=-B and drag=-Av^2

I just wonder what kind of friction would introduce the $-B$ term ...

It's just friction between the tires and the road.

In that case, I've got a few questions for you to think about:

What kind of friction is the friction you're talking about?
In which direction does the friction force (from the road) act on the car?
What force drives the car forward? (Let's assume it has a motor inside or something)
What would the car do if there was no friction?

[hide]And how large is the rolling friction compared to the air drag?[/hide]

But just in case you were interested in the solution of that equation, it's pretty simple. You only need to substitute $\dot{x}$ for $w$ and you get:

$\dot{w}=-Aw^2-B$.

(And now you integrate.)

[quote="kubus"]In that case, I've got a few questions for you to think about:

What kind of friction is the friction you're talking about?
In which direction does the friction force (from the road) act on the car?
What force drives the car forward? (Let's assume it has a motor inside or something)
What would the car do if there was no friction?

And how large is the rolling friction compared to the air drag?[/quote]

When you first posted those, I put this question to the side to think about.  But I haven't quite figured it out.  A car needs friction to move, but then it also slows down due to friction.  How can friction be acting on it in two different ways?

And for the differential equation, as it is, I'm not sure how to solve it.  It sort of looks like arctangent but only if A=B.

The first integration (to find $v$)

[hide]Let $v=\frac{dx}{dt}$.  Then

$\frac{dv}{dt}=-Av^2-B$

$\int-\frac{dv}{Av^2+B}=\int dt$

$-\frac{1}{B}\int\frac{dv}{\frac{A}{B}v^2+1}=t+C$

$-\frac{1}{B}\int\frac{dv}{\left(\frac{\sqrt{A}}{\sqrt{B}}v\right)^2+1}=t+C$

Let $u=\frac{\sqrt{A}}{\sqrt{B}}v$, and $du=\frac{\sqrt{A}}{\sqrt{B}}dv$

$-\frac{1}{\sqrt{AB}}\int\frac{du}{u^2+1}=t+C$

$-\frac{\arctan(u)}{\sqrt{AB}}=t+C$

$-\frac{\arctan\left(\frac{\sqrt{A}}{\sqrt{B}}v\right)}{\sqrt{AB}}=t+C$

$v=-\frac{\sqrt{B}}{\sqrt{A}}\tan\left(t\sqrt{AB}+C\right)$[/hide]

The dragging force acting on the car is force of friction and it is STATIC friction. But the slowing down force is the air friction force and the engine resistance force. Get me?

This problem would more approprioate for a block sliding on the table. And we take also in count the force of air friction. Or it would be fine for car sliding on the road.