function
algebra
functional equation
real analysis
real analysis unsolved

Are there any differentiable functions $f: \mathbb{R}\to\mathbb{R}$ such that

                                              $f(f(f(x)))=f(x)\geq0$ ;)

Cerainly: any nonnegative constant does it. I assume you really want to ask if there are any other examples.

Since if $f$ satisfies the functional equation $f_t$ does as well when $f_t(x+t)=f(x)+t$, we can assume WLOG that $f(c)=0$ for some $c$ (just shift the minimum value along a diagonal). By differentiability and positivity, $f'(c)=0$. By the functional equation, $f(0)=c$.

Now $f'(0)=f'(f(f(0)))\cdot f'(f(0))\cdot f'(0)=f'(0)^2\cdot f'(c)=0$, so $f'(0)=0$ as well.
Now, for small enough $x,y$, $|f(x)-c|<\frac12|x|$ and $|f(c+y)|<\frac12|y|$. Combining the two, for small enough $z$, $|f(f(z))|<\frac14|z|$. This is impossible by the functional equation, so 0 is isolated in the range and $f$ must be constant.


Please give mechanism  :maybe:

[hide]Is it claisen?  :o  :maybe: [/hide]

We have an allyl vinyl ether heated at a relatively high temperature, so a Claisen rearrangement will occur. Can you provide the product?

Is planarity a requirement for pericyclic reacns  :oops:  :blush: ?

Of course not. In fact, as you drawed, the TS for the Claisen rearrangement adopts (ideally) a chair like conformation.

By the way, explain in your own words what a pericyclic reaction is.

THANK YOU . Honestly I don't know , what I've understood is this.

In this type of reaction , there is shifting of the Pi electrons to form a product . Then ..... , while showing the shifting of bonds the tangents drawn to the arrows form a circle. Something like that . That's what my teacher told. One can form a circle about which the movemnet of electrons takes place.

pericyclic reactions involve a cyclic transition state, with intermediate bond orders wherever electrons are transferred to and from...that's the way I think of it in my head.

So what happens? It seems as if the original compound is more stable... :|

[quote="hell_ever"]I dont understand....why does the rearrangement occur?[/quote]

Because the given compound is an allyl vinyl ether and you are heating it!  :P  :P

[quote="hell_ever"]So what happens? It seems as if the original compound is more stable..[/quote]

how did you compare the stability of the two compounds?  :maybe:


In a math contest of ten problems, 5 points are given for each correct answer, and 2 points are deducted for each incorrect answer. If Nancy answered all ten problems, and she got a score of 29, how many did Nancy get right???

[quote]please include a detailed solution in ALL of your answers to my questions, as I am a beginner.[/quote]

[hide]Call the number of questions Nancy answered correctly x and the number of incorrect questions y. Therefore, 5x-2y=29, and x+y=10. Multiply the second equation by 2 to get 2x+2y=20. Adding this new equation to 5x-2y=29 gets 7x=49, or x=7. Nancy answered [b]7 [/b]questions correctly.[/hide]

[hide]
Set up a system of equations.
Let the amount Nancy got right be $x$ and the amount she got wrong $y$.
$x+y=10$
$5x-2y=29$.
Multiply the first equation by 2, giving
$2x+2y=20$
$5x-2y=29$.
Add them together, giving
$7x=49$.
Divide by 7:
$x=7$
So Nancy got 7 questions right.
To find the amount wrong, plug back into $x+y=10$.
$7+y=10$
Subtracting 7,
$y=3$
So Nancy got 3 wrong[/hide]

[hide="Or..."]
Start with the maximum possible score, 10*5=50.  For each problem Nancy gets wrong, subtract 7.  50-29=21.  $\frac{21}{7}=3$, so she got 3 wrong, which means that she got 7 right.
[/hide]

[hide]max is 50
she lost 21
21/7 is 3
so, 10-3=7[/hide]


Q1 An exam paper  contains 6 questions out of which 2 has possible 3 answers each,3 have 4 possible answers each,and remaining 1 question has 5 possible answers.Find total number of possible answers for these 6 questions

Q2 The number of ways of selecting 12 balls from unlimited no. of black ,red and white balls

Q3 The number of ways in which 20 different pearls of two colors can be set alternately on a necklace,there being 10 pearls of each color.

Q4 The number of ways in which 9 identical balls can be placed in 3 boxes(identical)

[hide]
Ans1=2880
Ans2=455
Ans3=5(9!)^2
Ans4=12[/hide]

Q1

The required number of ways is $ 3^{2}\times4^{3}\times5 \equal{} 2880$.

Q2

Perhaps I've misunderstood the question but I'll post what I think as I'm  fairly sure that will spur someone to post an answer to contradict it. It seems to me that you can have $ 12$ blacks, $ 11$ blacks, $ 10$ blacks, ..., $ 0$ blacks.

$ 12$ blacks means $ (B, R, W) \equal{} (12, 0, 0)$
$ 11$ blacks means $ (B, R, W) \equal{} (11, 1, 0), (11, 0, 1)$
$ 10$ blacks means $ (B, R, W) \equal{} (10, 2, 0), (10, 1, 1), (10, 0, 2)$
...
giving $ 1\plus{}2\plus{}3\plus{}...\plus{}13 \equal{} 78$.

Q3. I make this $ \frac{(10!)^{2}}{2}$.

Q4

As in Q2: $ (9, 0, 0), (8,1,1), (7, 2, 0), (7, 1, 1), ... (4, 4, 1), (4, 3, 2), (3, 3, 3)$ gives $ 12$ ways.

Shouldn't it be balls and urns for Q2 and Q4?

$ C^{15}_3 \equal{} 455$

Q3
You first choose a pearl from one colour and than a pearl from another colour and so on

$ 10\times10\times9\times9....\equal{} {10!}^2$

Then observe that there are repetitions (doesn't matter a pearl is the first, the second or the third, because it is a necklace) so divide it by $ 10$

Then divide it further by $ 2$ because the direction of counting doesn't matter either. (eg 123456789 and 987654321 is the same)

Hence $ 5\times(9!)^2$

[quote="chrischris"]Shouldn't it be balls and urns for Q2 and Q4?

$ C^{15}_3 \equal{} 455$[/quote]  Yes and no, but that doesn't mean your second line is an answer to any of the questions.

Understood Q2 now..

Red+Black+White+Green=12
$ ^{12 \plus{} 4 \minus{} 1}C_{4 \minus{} 1} \equal{} ^{15}C_{3}$

Also,
In Q 2...it is
The number of ways of selecting 12 balls from unlimited no. of black ,red ,green and white balls

and not The number of ways of selecting 12 balls from unlimited no. of black ,red and white balls

Sorry for error

Q4

If the boxes are not identical, then $ C^11_2 \equal{} 55$

Of these 55 ways, 1 is $ (3,3,3)$ and it is the only one that has not been double counted if the boxes are identical.

12 are permutations of $ (0,0,9), (1,1,7 ), (2,2,5) or (4,4,1)$

The remaining 42 are in the form of $ (a,b,c)$ where $ a\ne b\ne c$, and each $ (a,b,c)$ is counted six times ($ 3!$).

Therefore $ 1\plus{}4\plus{}7\equal{}12$


Does anyone know a really good book for AIME prep other than the AoPS series of books?

[quote="athunder"]...are you doing this to raise or post count...

try searching in the search tab "AIME prep"[/quote]

if I was trying to raise post count I'd post a bit more, wouldn't I?  :rotfl:

anyways I got this resolved by asking someone, so this topic can be deleted...

This is kind of unrelated, but I didn't want to start a new thread for this.

About when should one start preparing for USAMO? Like at about what average score or something on AIME?

Why not now? This recent [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1719064#1719064]post[/url] comes to mind.

[quote="1=2"]Why delete this when you could answer the question again so that when someone posts this we can say "Just search and you get at least 10 results"???

I know at least three AoPS members excluding myself that would say that the best practice for AIME problems is AIME problems. Do AIME problems. This will help you do AIME problems.[/quote]

This is what another AoPS friend of mine told me. :lol:

So I guess I'll work through like 30 AIMEs and read through PSS. Hmmm....

I agree that doing AIME problems is one of the best ways to prepare for AIME.
However, do not time yourself when doing them. First learn to do the problems, then learn to do them quickly.
There is a 5 or 6 hour class in the spring over the weekend before AIME to prepare you. You can register for that.

[quote="1=2"][quote="EricMathPath09"]However, do not time yourself when doing them.[/quote]

I forgot to overstate this. Thanks, Eric![/quote]

I thought it was good to get used to the timed situation...

yea, like a month before a test you should... but actually here is some good advice from mr.ruscyk.

Say you normally score a 5-6, sometimes a 7. You wanna get into the 8-9 range, so get some aimes, and do problems from 6-10. Just do these sets of five, but do them and make sure you solve them. Of course, if you have LOADS of time, do entire aimes, but these "high priority" problems need to be done if ur training.


How many different, positive three-digit numbers can be made using any three of the following five digits: $ 1, 2, 2, 3,$ and $ 3$?

We can have arrangements of the following:

122-3
123-6
133-3
223-3
233-3

Adding this up, we have $ 3\plus{}6\plus{}3\plus{}3\plus{}3\equal{}\boxed{18}$.


function
linear algebra
linear algebra unsolved

Let X be a linear space and let Y be a subspace of X. Show that Y $ \oplus$ X/Y is isomorphic to X.

Let $ Z$ be subspace of $ X$ such that $ X=Y \oplus Z$. You can easily show that $ Z \cong X/Y$ (consider function $ f: Z\mapsto X/Y, f(x)=x+Y$), which shows your problem.


Let $ (O)$ and $ d$ be respectively a circle and a line.$ M$ is fixed point on $ (O)$
Point $ H$ lies on $ d$ such that $ OH \perp d$.
$ A$ and $ B$ vary on $ d$ so that $ H$ is the midpoint of $ AB$.
$ MA,MB$ cut $ (O)$ at $ P,Q$ respectively.
Prove that $ PQ$ passes through a fixed point when $ A,B$ vary.

Sorry,I have edited.

[quote="KDS"]Let $ (O)$ and $ d$ be respectively a circle and a line.
Point $ H$ lies on $ d$ such that $ OH \perp d$.
$ A$ and $ B$ vary on $ d$ so that $ H$ is the midpoint of $ AB$.
$ MA,MB$ cut $ (O)$ at $ P,Q$ respectively.
Prove that $ PQ$ passes through a fixed point when $ A,B$ vary.[/quote]

What is $ M$ ?  :?:

$ MH$ meets $ (O)$ at $ K$; $ N$ is midpoint $ PQ$
A parallel to $ d$ meets $ MD,MB$ at $ E,F$,respectively.

$ E$ is midpoint of QF $ \Rightarrow NE\parallel{}MP \Rightarrow N,Q,K,E$ are cylic
$ \Rightarrow (KH,KN)\equiv \minus{}(KE,KN)\equiv\minus{}(QE,QN)\equal{}\minus{}(AB,QP)\equal{}(OH,ON)$
$ \Rightarrow  Q,N,K,H$ are cylic
$ \Rightarrow IK \perp OI \rightarrow I$    is fixed

Let $ K\equiv MH\cap (O)$ and $ N$ is the symmetric point of $ M$ through $ OH$. We have $ MN\parallel AB$ so $ M(PQNK) \equal{} (MA,MB,MN,MH)\equal{}\minus{}1$. Thus the quadrilateral $ PQNK$ is harmonic, yielding $ PQ$ passes through pole $ S$ of $ NK$ with respect to $ (O)$, which is a fixed point. QED.


What karts/bikes do you use on MarioKart Wii?

Please list if(Maybe multiple): 1) Small, medium, or large, 2)Character, 3)Bike/Kart, 4)Which one

For small: I use Toad :P /Bullet Bike
Medium: Mii or Yoshi with Mach Bike (or Zipzip if I'm just fooling around)
Big: Waluigi/Flame Runner

Where are your 4 least/best favorite courses?
Favorites: 1)Delfino Square, 2) Maple Treeway, 3)wario's gold mine, 4)Mario Circuit/Coconut Mall

Least :mad: 1) Rainbow Road, 2) Ghost valley, 3) Bowser Castles (all 3), 4) Grumble Volcano

Most Neutral: Mario Raceway

Small: Dry Bones-Bullet Bike

Medium: Yoshi-Sugarscoot

Large: King Boo-Flame Runner

Favorite tracks: Rainbow Road, Bowser Castle (special), Koopa Cape, Maple Treeway

Least Favorite tracks:DK Mountain, Wario's gold mine

Favorite Stage (battle): Funky Stadium

Least Favorite Stage: Twilight House

Anyone beat Mirror Mode or unlocked it?

Yeah...I beat mirror mode..but I still didnt get all the characters....

hehe i checked this 2 seconds ago and saw 0 replies...

If you play lots of Time Trials, you can get more characters.
Like 16 gets Funkykong

Also improve driver's rankings

I forgot. I haven't earned all/
As I said you have to get TIMe trials done and get the drivers rankings up, for 100cc I got $ \star$ on all retro stages and got Bowser Jr.

[quote="jjfun1"]Um.. you made this thread twice.[/quote]

Sorry, didn't realize.

Also when I use Zipzip what do you think are the top 5 best courses to use it?

Mine: 1)Mario Circuit (Wii Stage), 2)Luigi Circuit, 3)Moo Moo Meadows, 4)Desert Hills/2Dry Ruins

I'm trying to get Funky Kong. 16 time trials. Is that right?

I failed on Mirror Mushroom Gorge. :(

owwch i feel bad for you...
better luck next time, i tried that and FAILED.

i like the flame flyer and the wild wing
my record for koopa cape is 2:37 with the flame flyer

I did more then 16 time trials, and I still don't have Funky Kong. I did all the Nitro time trials and 6 of the Retro races.

Bam screwed up maybe. Maybe you have to unlock 4 expert ghosts.

Ive never done that before.

EDIT: to unlock Funkykong

That's good.

I play Zipzip when fooling around, like on Mario Circuit, Luigi Circuit, Jungle Parkway etc, etc. etc.

I never use Zip Zip. I always use Sugarscoot.

Favorite items: Mega-Mushroom, Star, POW Block, Lightning, Bullet.

Why isn't your favorite item Bullet?

I say Bullet, Star, Mega Mushroom then Red Shell/Triple Shells

Aren't spiny shells so anoying??
And if you're MegaMushroom when you get lighninged
down to NORMAL SIZE

by the way, this might surprise you, but there IS a way to pwn a bullet bill. can anybody figure out how?

You bumped another almost-month old thread mathlete.


[quote="hendrata01"]If $ a,b,c > 0$ prove that:
$ a^2b^2(a \plus{} b)^2 \plus{} b^2c^2(b \plus{} c)^2 \plus{} c^2a^2(c \plus{} a)^2 \plus{} 6a^2b^2c^2 \geq 2abc(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)$[/quote]
$ \iff$
$ \sum_{sym} a^4b^2\plus{}a^3b^3\plus{}a^2b^2c^2 \ge \sum_{sym} a^4bc\plus{}2a^3b^2c$
By muirhead: $ \sum_{sym} a^4b^2 \ge \sum_{sym} a^4bc$
And by schur:
$ a^3b^3c^3 \sum_{cyc} a^{\minus{}1}(a^{\minus{}1}\minus{}b^{\minus{}1})(a^{\minus{}1}\minus{}c^{\minus{}1}) \ge 0 \iff$
$ \sum_{sym} a^3b^3\plus{}a^2b^2c^2 \ge \sum_{sym} 2a^3b^2c$
And we are done :)

[quote="hendrata01"]If $ a,b,c > 0$ prove that:
$ a^2b^2(a \plus{} b)^2 \plus{} b^2c^2(b \plus{} c)^2 \plus{} c^2a^2(c \plus{} a)^2 \plus{} 6a^2b^2c^2 \geq 2abc(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)$[/quote]

without loss of generality, assume that $ a\equal{}\max\{a,b,c\}$, we have

$ LHS\minus{}RHS\equal{}4\left( ab{c}^{2}\plus{} \left( a\minus{}c \right) {b}^{2} \left( b\plus{}c \right) \plus{}\left( a\minus{}b \right) {c}^{3} \right)  \left( b\minus{}c \right) ^{2}\plus{} \left( \left( b\minus{}c \right) ^{2} \left( {a}^{2}\plus{}3a \left( b\plus{}c \right) \plus{}3{b}^{2}\plus{}3bc\plus{}3{c}^{2} \right) \plus{} \left( {b}^{2}\plus{}{c}^{2} \right) ^{2}\right)  \left( a\minus{}b \right)  \left( a\minus{}c \right) \ge 0$

[quote="hendrata01"]If $ a,b,c > 0$ prove that:
$ a^2b^2(a \plus{} b)^2 \plus{} b^2c^2(b \plus{} c)^2 \plus{} c^2a^2(c \plus{} a)^2 \plus{} 6a^2b^2c^2 \geq 2abc(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)$[/quote]

 The following inequality is equivalent to
$ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \plus{} 4\sum{b^2c^2(a \minus{} b)(a \minus{} c)}\geq 0$

obvious true.


probability
LaTeX
algebra
binomial theorem

If I roll a fair, regular six-sided die four times, what is the probability that I will roll the number $ 1$ exactly three times?

According to the [url=http://www.artofproblemsolving.com/Wiki/index.php/Binomial_theorem]binomial theorem[/url], this is $ \binom{4}{3}(\frac {1}{6})^3(\frac {5}{6})^1$, which simplifies to $ 4(\frac {1}{216})(\frac {5}{6}) \equal{} \fbox{5/324}$.

Really random $ \text{\LaTeX}$ tip.

To make those parentheses nicer looking, type [code]$\left(a+\frac{1}{b}\right)$[/code] which turns out to be $ \left(a+\frac{1}{b}\right).$ ]
Without the [code]$\left($ and $\right)$[/code] it would look like this: $ (a+\frac{1}{b})$ which looks a lot uglier.


umm... what is the difference of an ipod and an MP3 player?

also, who has what?

well ipod is a type of mp3 player. Ipod's have better sound quality supposedly. And they have the click wheel. I guess they just dominate the market. They also hold a lot of songs

I was about to get an ipod mini but they discontinued it  :mad:

They discontinued iPod mini?!  That's crazy.  I suppose they could say that it was essentially "replaced" by the Nano, but they're really quite different, in that the mini holds a lot more music.

I have an mp3, I don't really see the difference between that and my friend's ipods..they are supposed to sound better, but we all agree that for the better price, my mp3 doesn't sound any different.

Personally, mp3 is easier to carry...girls have small pockets, the ipods are a odd shape to carry.  Well, not odd, but awkward.

actually i like the mini better cuz it was easier to hold and it was cheaper than the nano and can hold more memory. And size isnt that important anyways.

They're basically the same (like some have already said).  The iPod is an MP3 player, though it also plays other formats.  I'd say the biggest advantage the iPod has over the other MP3 players is marketing and the click wheel.  I'm personally in the market for an iPod because of that click wheel.  The regular iPod comes with 30gb or 60gb, which, needless to say, can hold a lot of music.  The iPod Nano (which is, just as the commercials say, impossibly small) has 2 or 4 gigs.  The iPod Shuffle (the one without the screen) has 512MB or 1gb.

None of the iPods have a radio tuner, while quite a few competing MP3 players do have radio tuners, as well as voice recorders.  However, there are a lot of companies making cables and adapters so you can plug your iPod into the radio in your car.  That's another reason I've been looking at an iPod.

thats funny- I never liked the click wheel.

I hear the Rio Carbon's got some good stuff, though.  Personally, I don't need that much memory on a mp3 player- what business do you need even more than one gig for, anyway?

The amount of memory you need depends on how many songs you want, what file format, and what bitrate.  I'm personally going for as much sound quality as possible.  I have almost 30 gigs of music on my hard drive because it's either ripped in .wav format or wma lossless.  When I finish (if I ever do) the audio system in my Jeep, I'll have about
$\$$1000 in it, so I don't want to have to play poor quality recordings on a nice system.

I use my phone to play mp3s. At this point I can store 512 MB of music, but I can always upgrade my memory card to 2GB  :roll: I belive Ipods are dying (as a musical playing device) with the appearance of multimedia orientated phones, like Sony Ericsson W800, W900, W550, Nokia N91, N70, Samsung I300, etc.

[quote="Valentin Vornicu"]I belive Ipods are dying (as a musical playing device) with the appearance of multimedia orientated phones, like Sony Ericsson W800, W900, W550, Nokia N91, N70, Samsung I300, etc.[/quote]

Maybe in some parts of the world, but here, iPods are still very much alive. I know of tons of people who rushed out to get the new Nano.

*reads poll*
What happened to "Neither -- I own a minidisc."? MD's have much nicer sound quality than the average mp3 player, and the sampling rate is excellent for recording. The down side is that you can't store more than 1 gig of music on a disc.

umm... what is the difference of an iPod Mini and an iPod Nano???
when I go to http://www.ipod.com
, then there is no mini.

does anyone know why mini got discontinued?

[quote="slamchess"]actually there is a 4gig version so they were "replaced"... i say creative zen micros r best though-they r pretty small and u get more storage for the money[/quote]

Agreed.
Cheaper, supposedly better sound quality (at least than the mini was), easier to work with in terms of transfers...
I just like it more simply because every likes the iPod.  :D

[quote="Valentin Vornicu"][quote="mel"]Maybe in some parts of the world, but here, iPods are still very much alive. I know of tons of people who rushed out to get the new Nano.[/quote]That's because American GSM markets are very, but very undeveloped. GSM is farily new, and there isn't any single operator to even offer 3G services, and even here in Romania we have 3G (!).

So basically the phones that are offered in US are really bad (if you want a good model you have to buy it from overseas, and if it's not quad band it won't work).[/quote]

Well, there are ways around the phone issue if you have knowledgeable friends, which, I'm sure you have, Valentin. ;)

And yes, I agree that the U.S. tech markets (for the general public) are far too underdeveloped.

On the other hand, I think you guys in Romania got the better end of the deal. I've always viewed the ipod as a marketing wonder child, but the other end of the stick is that the consumer loses.

With the new Ipod color though, you can view pictures. With all the brand loyalty, this would have sold enough. But I expect booming sales because collages students want to look at naughty pictures.

At least the ones I know, which isn't saying much...

Well, there are great camera phones that are sold around the world that can not only view picts, but can also take and send picts. (These phones can be purchased at a reasonable price, as well.)

Brand loyalty plays a bigger role than most people recognize in the market place.

[quote="slamchess"]what kind do u have?[/quote]

me? that would be...

i'm deciding still

[quote="G-UNIT"]momma....momma don allow me n'a listen to no o no.. i-i-pod or m-m-p3 pllaeyer cuz momma said dat dat m-mp3 pla-players are are creations of e'm DEVIL!!![/quote]

HAVE YOU BEEN PLAYING SOME FOOSEBALL BEHIND MOMMA'S BACK?

[quote="math92"][quote="G-UNIT"]momma....momma don allow me n'a listen to no o no.. i-i-pod or m-m-p3 pllaeyer cuz momma said dat dat m-mp3 pla-players are are creations of e'm DEVIL!!![/quote]
what is that supposed to mean???
 :rotfl:  :rotfl:[/quote]
it's based on the movie Waterboy, and it's a darn good impression of it too.  :thumbup:

"dem ipods are the devil, bobby bouche!"

anyway, yeah i have an ipod, but given the choice (i got as gift from someone) id prob get something like an iriver.

iPods have a good sound output, but some others have better ones.

The iPod Nano is simply smaller than the mini and color, and is flash based (which means no moving parts = very stable), rather than hard drive based (ipod mini n such).

Really, the only advantage iPods have are : the touch wheel, and stylish i guess.

i have a mini 4 gig, and i fill it up w/ about 730 songs - i like decent quality. Though i wish i ahd bigger - i have about 10 gigs of music on my computer.

[quote="Valentin Vornicu"][quote="noneoftheabove"]wow...ten gigs of music i couldnt even find that much if i randomly started downloading now.[/quote]I have a couple of friends that have over 1TB (that's 1024 GB) of music on their computers (half on hdd, 25% on cd's and 25% on dvd's).[/quote]

 :huh:  :what?:  :wacko:  :huuh:  :help:

well yeah a DJ is gonna have the TB of music cuz a DJ needs.. about every song he can possibly get - in case somebody might want it. But for practical people, 10 GB is more than enough lol.

i got a new ipod video for my birthday; its my first dive into the mp3 world. i downloaded iTunes and, with all of my CDs and some stuff my friends "gave" me (and a couple of songs i downloaded offline), i have about 4 gb filled up already- i've only had the thing for 2 months... i still have a lot of CDs i want to add. i can definitely see filling this up sometime.
the things about iPod that make it so marketable?
1. click wheel. it makes scrolling large amounts of music incredibly easy.
2. iTunes. the iTunes store is SO easy to use. the software is very user-friendly, as well.
3. size. my iPod fits easily into my jeans pocket.
4. accessories. you can plug the iPod into just about anything now, including your computer, a radio, or various boomboxes they make especially for iPod.
5. the look. its sleek.
6. new stuff. they now do video, which you can download from iTunes and play...

if i was any type of download freak like my friend is, i would've gotten a creative media center- he has tons of movies and tv shows he has downloaded and put on there. you can also play video games and stuff. really i would just like to be able to watch movies/videos for longer than 2 hrs without recharging; i hope iPod extends the battery life on future editions.

I have a MP3. But I prefer to listen to music on computer. I have about 8GB.

the bad thing about ipods is that the battery completely dies out after about a year or so :(

[quote="Valentin Vornicu"]One of them is one of the best DJs in Romania. He used to be a mathematician in high-school (went to Math college but gived up for DJ-ing in the last year). So it's not surprise :)[/quote]

Okay, that explains it...
I'm thinking, 1 MB ~ 1 min music, right?
So then 1 TB is roughly 2 years of music?

[quote="Mathalete"]DUDE IPODS ARE CRAP!!!!!!!!!!!!!! They're a friggin box and a circle in the middle. Thats a horrible exterior design. And most of them crack easily because of a thin screen.[/quote]

not really, my shuffle is pretty tough

and check out this website someone once posted, it's a stress test for ipod nano, pretty amazing

iPod vs mp3? Is this a question anymore? I was not alive in 2005, but for those who were, remember...

Please do [u][b]NOT[/b][/u] put the tag "relieveyourass" on. I don't know who made it but I will figure it out...


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a finite group of composite order $n$ such that for any positive integer $k$, dividing $n$, $G$ has a subgroup of order $k$. Prove that $G$ is not simple.

It's been shown before on the forum that if $p$ is the smallest prime dividing $|G|$, then a subgroup of index $p$ in $G$ is normal. Since the hypotheses in your problem imply that $G$ has such a subgroup, that's that.


geometry
trigonometry
ratio
geometric sequence
geometry proposed

Consider a circle $O_1$ with radius $R$ and a point $A$ outside the circle. It is known that $\angle BAC=60^\circ$, where $AB$ and $AC$ are tangent to $O_1$. We construct infinitely many circles $O_i$ $(i=1,2,\dots\>)$ such that for $i>1$, $O_i$ is tangent to $O_{i-1}$ and $O_{i+1}$, that they share the same tangent lines $AB$ and $AC$ with respect to $A$, and that none of the $O_i$ are larger than $O_1$. Find the total area of these circles.

I know this problem was easy, but it still appeared in the TST, and so I posted it. It was kind of a disappointment for me.

[b]Generalization:[/b] $\Gamma_i$ is a circle of radius $r_i$ inscribed in an angle of measure $2\alpha$ such that each $\Gamma_i$ is externally tangent to $\Gamma_{i+1}$ and $r_{i+1} < r_i.$ Then the sum of the areas of the circles $\Gamma_i$ equals the area of a circle of radius $r =\frac {_1}{^2} r_0 (\sqrt{ \sin \alpha} + \sqrt{\text{csc} \alpha}).$

Let $a,b$ the sides of the subject angle with vertex $P$ and denote $X_i$  the tangency point of $\Gamma_i(O_i,r_i)$ with line $a.$ From the similar $\triangle POX_0 \sim \triangle PO_1X_1$ we get

$\frac{O_1X_1}{O_0X_0}=\frac{r_1}{r_0}=\frac{PO_1}{PO_0}=\frac{PO_0-O_0O_1}{PO_0}=\frac{PO_0-r_0-r_1}{PO_0}$

Plugging $PO_0=r_0 \cdot \csc \alpha$ yields $\frac{r_1}{r_0}=\frac{\csc \alpha-1}{\csc \alpha+1}=\frac{1-\sin \alpha}{1+\sin \alpha}$

Radii $r_0,r_1,r_2,r_3...$ form a decreasing geometric progression with ratio $k=\frac{1-\sin \alpha}{1+\sin \alpha}$ $\Longrightarrow$ Areas $S_i$ of $\Gamma_i$ form another decreasing geometric progession with ratio $k^2.$

$\sum_{i=0}^{\infty}S_i=  \frac{S_0}{1-k^2} = \frac{\pi {r_0}^2}{1-\left ( \frac{1-\sin \alpha}{1+\sin \alpha} \right)^2}= \left ( \frac{1+ \sin \alpha}{2 \sqrt{\sin \alpha}} \right)^2 \cdot \pi {r_0}^2$

Let $\varrho$ be the radius of the circle equivalent to the chain of circles $\Gamma_i.$ Then

$\pi \varrho^2=\left ( \frac{1+ \sin \alpha}{2 \sqrt{\sin \alpha}} \right)^2 \cdot \pi {r_0}^2 \ \Longrightarrow \ \varrho=\frac{_1}{^2}r_0 ( \sqrt{\sin \alpha}+\sqrt{\csc \alpha}).$


How do I use the binomial theorem to determine the last n-th digit from the right  of a special expression?

For example, what is the ninth digit from the right in 101 ^ 10?

[hide="hint"]$(100+1)^{10}=\sum_{k=0}^{10} \binom{10}{k} \cdot 10^{2(10-k)}$[/hide]

EDIT:

[hide="solution"]
$(100+1)^{10}=\sum_{k=0}^{10} \binom{10}{k} \cdot 10^{2(10-k)}$
Start from $k=10$ and proceed backwards until we get a number with at least 9 zeros.
$k=10 \implies 1$
$k=9 \implies 10 \cdot 10^2=1,000$
$k=8 \implies 45 \cdot 10^4=450,000$
$k=7 \implies 120 \cdot 10^6=120,000,000$
$k=6 \implies 210 \cdot 10^8=21,000,000,000$
Adding all these up and taking the last 9 digits of the sum, we find the last 9 digits of $101^{10}$ to be:

\[120,451,001\]

so the ninth-to-last digit is $1$.
[/hide]


Let $P$ be an interior point of a triangle $ABC$ and $AP,BP,CP$ meet the sides $BC,CA,AB$ in $D,E,F$ respectively. Show that \[ \frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}.  \]
[hide="Remark"]This is known as [i]Van Aubel's[/i] Theorem.[/hide]

[quote="Rushil"]Let $P$ be an interior point of a triangle $ABC$ and $AP,BP,CP$ meet the sides $BC,CA,AB$ in $D,E,F$ respectively. Show that \[ \frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}.  \][/quote]

If i am not wrong this is called
[i]Van aubel theorem[/i]!..

 :)  :D

Indeed! This is Van Aubel theorem. It follows from Menelaus theorem. ;)

You can solve it pretty easily, although not very conveniently using mass points.  Just make $AF = k\cdot AB$ and $FB = (1-k)\cdot AB$ where $0 < k < 1$, and similarly for $AC$, then you can find the weights of all your points and plug them back in and we're done.

[hide="Another solution"]I shall use $[XYZ]$ to denote the area of any triangle $XYZ$.Now notice that  $\frac{AF}{FB}=\frac{[ACP]}{[BPC]}$ and $\frac{AE}{EC}=\frac{[ABP]}{[BPC]}\implies \frac{AF}{FB}+\frac{AE}{EC}=\frac{[ACB]}{[BPC]}+\frac{[ABP]}{[BPC]}$.Again, $\frac{AP}{PD}=\frac{[ABP]}{[BPD]}=\frac{[ACP]}{[CPD]}=\frac{([ABP]+[ACP])}{[BPC]}=\frac{AF}{FB}+\frac{AE}{EC}$.
Note: I have used the fact that $\frac{x}{y}=\frac{a}{b}=\frac{x+a}{y+b}$.[/hide]

my solution
Through the point $O$ draw the line $(l)//BC$ cuts the sides $AB$ and $AC$ in $M$ and $N$ respectively.
then we have $\frac{AM}{MB}=\frac{AO}{OD}=\frac{AN}{NC}=x$
From my lemma we have
$\frac{BM}{MA}.\frac{AF}{FB}+\frac{CN}{NA}.\frac{AE}{EC}=1$
$\Rightarrow \frac{AF}{FB}+\frac{AE}{EC}=x=\frac{AO}{OD}$

See my Lemma
[hide]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=500741[/hide]

This is a direct application of Ceva's theorem as well.

quick solution is by using areal ratios(i mean areas here not bary)

Hi, I think the easy solution would be the application of Menelaus Theorem or Ceva's theorem.

Using Menelaus Theorem:

$$\frac{AF}{FB} + \frac{AE}{CE} = \frac{AP(CD+BD)}{PD.BC}$$
$$\implies \frac{AP.BC}{PD.BC}$$
$$\implies \frac{AP}{PD}$$
our desired result.

PS. Sorry, I couldn't write much as I have typed all this from my 5" mobile phone.

Let $P \equiv (x:y:z) \implies D \equiv (0:y:z) ; E \equiv (x:0:z) ; F \equiv (x:y:0)$
Hence, $$\text{RHS } = \frac{AF}{FB}+\frac{AE}{EC} =\frac{y}{x}+\frac{z}{x} = \boxed{\frac{y+z}{x}} $$
And, $$ \text{LHS } =\frac{AP}{PD}=\frac{[APB]}{[PBD]}=\frac{[APB]}{[ABD]-[APB]} \implies \frac{AP}{PD}=\frac{\tfrac{z}{x+y+z}}{\tfrac{z}{y+z}-\tfrac{z}{x+y+z}} =\boxed{\frac{y+z}{x}}$$
Which directly implies that [b]LHS=RHS[/b], Hence,
\[ \frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}. \]

We see that:
$$\frac{AP}{PD} = \frac{[ABP]}{[BDP]} = \frac{[APC]}{[DPC]} = \frac{[ABPC]}{[BPC]}$$
$$\frac{AF}{FB} = \frac{[ACF]}{[BCF]} = \frac{[AFP]}{[BFP]} = \frac{[APC]}{[BPC]}$$
$$\frac{AE}{EC} = \frac{[AEB]}{[CEB]} = \frac{[AEP]}{[CEP]} = \frac{[ABP]}{[BPC]}$$
Notice that:
$$\frac{[ABP]}{[BPC]} + \frac{[APC]}{[BPC]} = \frac{[ABPC]}{[BPC]}$$
This means that $\frac{AP}{PD} = \frac{AF}{FB} + \frac{AE}{EC}$ is true. So, we are done.

[hide=lotus lemma ftw]using the lotus lemma we get $\frac{AP}{PD}=\frac{AE}{EC}(1+\frac{CD}{BD})$\\
which gives $\frac{AP}{PD}=\frac{AE}{EC}+\frac{AE}{EC}\cdot \frac{CD}{BD}$\\
also notice from ceva's theorem that $\frac{AF}{BF}\cdot \frac{BD}{CD}\cdot \frac{CE}{AE}=1$\\
so we get $\frac{CD}{BD}=\frac{AF}{BF}\cdot \frac{AE}{EC}$ , plugging back we get\\
$\boxed{\frac{AP}{PD}=\frac{AE}{EC}+\frac{AF}{BF}}$ $\blacksquare$[/hide]
lotus lemma:-[url]https://artofproblemsolving.com/community/q1h509770p25873315[/url]

See more generalization
[url]https://www.cut-the-knot.org/triangle/FourthIdentityInTriangle.shtml[/url]


inequalities
induction
algebra unsolved
algebra

Solve the equation $2^{[x]}=1+2x$, where $[t]$ denotes the biggest integer number $\leq t$

Assume first that x>0:
clearly {x} =0 or 1/2

Case 1.
x=n for some n.
then 2^x = 2x +1 and so x=0

Case2. x= n+1/2
So 2^n = 2n+2 so 2^n-1 = n+1 so x=3.5

Now let x<0.
then 0<2x+1<1
So -1/2<x<0
and so [x]=-1
so 1/2 = 2x+1 and therefore x= - 1/4

-a

You can draw the graphs of $f(x)=2^{[x]}$ and $g(x)=2x+1$.

A purely algebric solution would be the following:

Show that x cannot be  $\leq-1$, nor $\geq 4$.
If $x\leq -1$, then $2x+1<0<2^{[x]}$.
If $x\geq 4$, show that $2^{[x]}>3[x]>2x+1$. The first inequality follows from $2^n>3n, \ \forall n\in \mathbb {N},\ n\geq 4$ , which can be proved by induction. For the second, write $[x]=x-\{x\}$ and obtain the equivalent inequality $3\{x\}+1<x$, which is true, because $\{x\}<1$ and $x\geq 4$.

Thus $x\in (-1,4)$, so $[x]\in \{-1,0,1,2,3\}$.

We finally get $x \in \{-\frac 1{4},0,\frac7{2}\}$

Wow, two minutes later!  :)
Ali, you also need induction or some inequalities to solve equations $2^x=x+1$ and $2^{x-1}=x+1$, I suppose (or maybe graphs).

Well that would be obvious.

Here is a cool way that I proved 2^n >= n+1

Indeed number of divisors of 2^n <= 2^n
Hence n+1<= 2^n
 ;)


We have $\sum\frac{a^3 +b^2+c}{3a+2b+c}=\sum\frac{a^3}{3a+2b+c}+\sum\frac{b^2}{3a+2b+c}+\sum\frac{c}{3a+2b+c}$. By Cauchy,Jensen and Chebyshef it is easy to show that every sum is greater or equal to$\frac{1}{2}$. The conclusion follows. ;)


topology
real analysis
real analysis unsolved

1) I have a theorem which says that if $ \bf E$ is a set, then the $ \sigma$-ring generated by it, denote it $ E^{*}$, has the same cardinality.
2) If I consider the Borel $ \sigma$-algebra, it must contain all opens of the topology... In particular, consider $ (\mathbb{R}, \tau)$ as a topological space with the usual topology. Then $ \tau$ is uncountable (consider all opens with center an irrational). However it has a countable basis $ E$.
3) Now the generated $ \sigma$-ring $ E^{*}$ does contain all opens, and it should be countable according to 1). On the other hand, we must have $ \tau \subset E^{*}$, and $ \tau$ is uncountable...

Where did I go wrong?

Isn't your set $ E$ the reals? From what I understand, the $ E^{*}$ is on a different level than $ E$. $ E^{*}\subset P(E)$. So if you use the countable basis of the standard topology on $ \mathbb{R}$, then $ E^{*}$ is going to be a subset of the double powerset of $ \mathbb{R}$. Using $ E=\mathbb{R}$ you no longer have cardinality problems.

[quote="Jacobson"]1) I have a theorem which says that if $ \bf E$ is a set, then the $ \sigma$-ring generated by it, denote it $ E^{*}$, has the same cardinality.[/quote]
Maybe you are not reading it correctly. Let $ E$ be the collection of all one-element subsets of $ \mathbb N$. Then $ E$ is countable, but the $ \sigma$-ring it generates is the power set of $ \mathbb N$.

Sorry, I misrembered the theorem.

it actualy says "if $ E$ is a countable class of sets, then the ring $ R(E)$ generated by $ E$ is countable too"

$ E$ and $ R(E)$ have the same cardinal in general. Sorry for the spam :) but I did not have my books lying here


integration
calculus
topology
real analysis
real analysis unsolved

Consider the metric space $ X \equal{} C[0,b]$ where $ b > 0$. Define

$ Tf(t) \equal{} 5 \plus{} \int_0^ts^3f(s)ds$

for any $ f\in X$. Thus $ T$ maps $ X$ into $ X$. For which $ b$ is $ T$ a contraction map? In that case, what is its fixed point?

if there is a fixed point then it is differentiable, since its the integral of itself

$ f(t)\equal{}5\plus{}\int_0^t s^3f(s) ds$

$ \to f'(t) \equal{} t^3f(t) \to f(t) \equal{}Ae^{\frac{t^4}{4}}$

$ f(0)\equal{}T(f(0))\equal{}5\plus{}0\equal{}5 \to A\equal{}5$

im not sure how we have to restrict b?


diagonal $ AC$of square $ ABCD$ coincides with the hypothenuse of right triangle  $ ACK$ ,so that points $ B$ and $ K$ lie on one side of $ AC$.prove that
$ BK\equal{}(AK\minus{}CK)/sqrt2$
and $ DK\equal{}(AK\plus{}CK)/sqrt2$
if you dont mind can anyone provide me a diagram of this problem?

Ptolemy's theorem on $ ABKC,AKCD$ will give it directly.

[quote="ishfaq420haque"]Prove that $ KB \equal{} (KA \minus{} KC)\sqrt  2$ and $ KD\sqrt 2 \equal{} KA \plus{} KC$ .[/quote]

Apply the [b]Ptolemy's theorem[/b] to the cyclical quadrilaterals $ CABK$ ans $ CDAK$ .


[quote="Thaakisfox"]could someone help me find an elegant solution for the following equation:
$x^{3}-3ax+2a=0$
Where $a$ is real parameter.
(Under elegant I mean that not using Cardano's forumula)[/quote]

If $a\geq1$, It's easy to see that the equation has 3 real roots in $[-2\sqrt{a},2\sqrt{a}]$. I suggest then to write $x=2\sqrt{a}\sin(\phi)$

Then $x^{3}-3ax+2a=0$ becomes $sin(3\phi)=\frac{1}{\sqrt{a}}$ and hence the 3 roots :

$r_{1}=2\sqrt{a}\sin(\frac{1}{3}\arcsin(\frac{1}{\sqrt{a}}))$
$r_{2}=2\sqrt{a}\sin(\frac{2\pi}{3}+\frac{1}{3}\arcsin(\frac{1}{\sqrt{a}}))$
$r_{3}=2\sqrt{a}\sin(\frac{4\pi}{3}+\frac{1}{3}\arcsin(\frac{1}{\sqrt{a}}))$

It remains to look at $a<1$.

--
Patrick

[quote="Thaakisfox"]could someone help me find an elegant solution for the following equation:
$x^{3}-3ax+2a=0$
Where $a$ is real parameter.
(Under elegant I mean that not using Cardano's forumula)[/quote]

For every value of a :

1) If $a< 0$, we have one real root. Let $x=2\sqrt{-a}\sinh(u)$
Then $x^{3}-3ax+2a=0$ becomes $\sinh(3u)=\frac{1}{\sqrt{-a}}$ and hence the root :
$r_{1}=2\sqrt{-a}\sinh(\frac{1}{3}arsinh(\frac{1}{\sqrt{-a}}))$

2) If $a=0$, root is 0

3) If $0<a<1$, we again have one real root. Let $x=-2\sqrt{a}\cosh(u)$
Then $x^{3}-3ax+2a=0$ becomes $\cosh(3u)=\frac{1}{\sqrt{a}}$ and hence the root :
$r_{1}=-2\sqrt{a}\cosh(\frac{1}{3}arcosh(\frac{1}{\sqrt{a}}))$

4) If $a\geq 1$, The equation has 3 real roots. Let $x=2\sqrt{a}\sin(\phi)$
Then $x^{3}-3ax+2a=0$ becomes $sin(3\phi)=\frac{1}{\sqrt{a}}$ and hence the 3 roots :

$r_{1}=2\sqrt{a}\sin(\frac{1}{3}\arcsin(\frac{1}{\sqrt{a}}))$
$r_{2}=2\sqrt{a}\sin(\frac{2\pi}{3}+\frac{1}{3}\arcsin(\frac{1}{\sqrt{a}}))$
$r_{3}=2\sqrt{a}\sin(\frac{4\pi}{3}+\frac{1}{3}\arcsin(\frac{1}{\sqrt{a}}))$

Nota :
In for example $r_{1}=2\sqrt{-a}\sinh(\frac{1}{3}arsinh(\frac{1}{\sqrt{-a}}))$ we can replace arsinh and sinh by exponential expressions and we retrieve the Cardan's result (sum of cubic roots).

--
Patrick[/quote]


The number of pairs of three-digit numbers m and n such that n=3m and such that the union of the digits in the (base 10) expansions of m and n is the set {1,2,3,4,6,7,8} is...
a) 1
b) 2
c) 3
d) 4
e) more than 4

Let T = {1,2,3,...,13}. A function f: t --> t is said to be one-to-one if $ t_1$ does not equal $ t_2$ implies that $ f(t_1)$ does not equal to $ f(t_2)$. The number of one-to-one functions f such that t+f(t) is a perfect square, for every t in T, is...
a) 0
b) 1
c) 2
d) 3
e) more than 3

[quote="high_flyer10"]The number of pairs of three-digit numbers m and n such that n=3m and such that the union of the digits in the (base 10) expansions of m and n is the set {1,2,3,4,6,7,8} is...
a) 1
b) 2
c) 3
d) 4
e) more than 4[/quote]The way I read it, we are looking for pairs $ (m,n)$ like $ (111, 333)$,
where $ n\equal{}3m$, but no digit is allowed to be $ 0, 5$, or $ 9$.
It seems like there should be many such pairs. [hide="In five minutes I found:"]$ (111, 333)$, $ (222, 666)$, $ (114, 342)$, $ (127, 381)$,
covering the whole set of digits with those four pairs.
Next I tried $ m\equal{}124$ and found $ (124, 372)$.
Since five pairs is [i]"more than 4"[/i] pairs, my work is done.
It was just a question of picking $ m$ numbers that would be easy to treble,
and that were likely to make a valid pair $ (m<333$, avoiding $ 0, 5$ and $ 9$ in $ m$ and $ n\equal{}3m)$.
I knew $ (333, 999)$ was not allowed,
and $ (117, 351)$ was not going to work either.[/hide]
[quote="high_flyer10"]Let T = {1,2,3,...,13}. A function f: t --> t is said to be one-to-one if $ t_1$ does not equal $ t_2$ implies that $ f(t_1)$ does not equal to $ f(t_2)$. The number of one-to-one functions f such that t+f(t) is a perfect square, for every t in T, is...
a) 0
b) 1
c) 2
d) 3
e) more than 3[/quote]I see only one function that would work. Can anyone see a second one?
[hide]The function's domain/range is the very small set of small numbers $ T$. If $ f(13)$ belongs to $ T$,
the only perfect square value for $ 13\plus{}f(13)$ is $ 16$.
$ 9$ is too small and $ 25$ is too large.
In fact, there are no choices for $ t \ge9$. It must be $ t\plus{}f(t)\equal{}16$.
$ f(13)\equal{}3$,  $ f(12)\equal{}4$, …..$ f(9)\equal{}7$
Since $ f(t)\equal{}3$ is taken, we must make $ f(1)\equal{}8$, so that $ 1\plus{}f(1)\equal{}9$.
And now that $ f(t)\equal{}8$ is taken, we must make $ f(8)\equal{}1$, to have $ 8\plus{}f(8)\equal{}9$.
Now the only values left for $ f(t)$ are $ 2$, and $ 9$ through $ 13$.
Again, we have no choice. It must be:
$ f(2)\equal{}2$, and
$ f(3)\equal{}13$, $ f(4)\equal{}12$, … $ f(7)\equal{}9$.[/hide]


trigonometry
inequalities
special factorizations
triangle inequality
trig identities
Law of Cosines
algebra unsolved

Find the minimum value of

$ \sqrt{ 49\plus{}a^2 \minus{}7\sqrt{2}a} \plus{} \sqrt{ a^2\plus{}b^2 \minus{}\sqrt{2}ab} \plus{} \sqrt{ b^2\plus{}50 \minus{}10b}$

where a and b are positive integers.

A method that uses completing the square would be preferable. Thanks

Gib Z, do you have any other questions from this competition?

only the ones that i can remember. umm

1) $ a_n \equal{} a_{n \minus{} 1} \minus{} a_{n \minus{} 2}, a_1 \equal{} \sqrt {2}, a_2 \equal{} \pi.$ Find $ a_{2008}$.

2) PQRST are 5 adjacent vertices of a regular polygon. The lines PQ and ST are extended to meet with at A, such that angle PAT is 160 degrees. How many sides does the polygon have.

I might remember more later

[quote="kunny"][url]http://www.mathlinks.ro/Forum/viewtopic.php?t=218180[/url][/quote]
In that problem, the first term was $ \sqrt {a^2 \plus{} 7a\sqrt {2} \plus{} 49}$ not $ \sqrt {a^2 \minus{} 7a\sqrt {2} \plus{} 49}$ (Probably due to a typo). I knew something was wrong about that problem.
Anyway, for this problem:
[hide]Let $ O,P,Q,R,S$ be points in a plane such that $ OP \equal{} 7$, $ OQ \equal{} a$, $ OR \equal{} b$, $ OS \equal{} 5\sqrt {2}$, and $ \angle POQ \equal{} \angle QOR \equal{} \angle ROS \equal{} 45^\circ$.

[asy]pair P,Q,R,S,O,A,B;
P=(0,7);
Q=(4,4);
R=(5,0);
S=(5,-5);
O=(0,0);
A=(10,10);
B=(10,0);
dot(P);
dot(Q);
dot(R);
dot(S);
dot(O);
draw(P--Q--R--S);
draw(O--P);
draw(O--A,EndArrow);
draw(O--B,EndArrow);
draw(O--S);
label("P",P,W);
label("Q",Q,NE);
label("R",R,E);
label("S",S,SE);
label("O",O,SW);[/asy]

Using the Law of Cosines:
$ PQ \equal{} \sqrt {a^2 \plus{} 7^2 \minus{} 2\cdot a\cdot 7 \cdot \cos 45^\circ} \equal{} \sqrt {a^2 \plus{} 49 \minus{} 7a\sqrt2 }$
$ QR \equal{} \sqrt {a^2 \plus{} b^2 \minus{} 2\cdot a\cdot b\cdot \cos45^\circ} \equal{} \sqrt {a^2 \plus{} b^2 \minus{} ab\sqrt2}$
$ RS \equal{} \sqrt {b^2 \plus{} (5\sqrt {2})^2 \minus{} 2\cdot b\cdot5\sqrt {2}\cdot\cos 45^\circ} \equal{} \sqrt {b^2 \plus{} 50 \minus{} 10b}$

So, we must minimize $ PQ \plus{} QR \plus{} RS$.

By the triangle inequality, $ PQ \plus{} QR \plus{} RS \ge PS$ with equality iff $ P,Q,R,S$ lie on $ PS$ in that order which is clearly possible.
By the Law of Cosines, $ PS \equal{} \sqrt {7^2 \plus{} (5\sqrt {2})^2 \minus{} 2\cdot 7 \cdot 5\sqrt {2} \cdot \cos 135^\circ} \equal{} 13$
So the minimum is $ 13$.
[/hide]

[quote=JustKeepRunning]Use calculus to find maxima and minima of functions[/quote]

Eleven years after :rotfl:

Yes I know. I am not criticising. After all it is good for you. I just saw the dates and I made this comment!

Let $a,b$ be positive reals . Prove that$$ \sqrt{a^2+49 \minus{}7\sqrt{2}a}  \plus{} \sqrt{ b^2\plus{}50 \minus{}10b}\plus{} \sqrt{ a^2\plus{}b^2 \minus{}\sqrt{2}ab}\geq 13.$$
Equality holds when $a=? ,b=?.$

The problem can be solved with Cartesian Coordinates if I'm not mistaken.


function calculus logarithms inequalities

Prove, without calculus, that the function $f(x)=\log_{k}x$ is concave in $\mathbb R^{+}$ $\forall k>1$.
[hide="what is a concave function"]a function is called concave in $[a,b]$ if for any $t \in [0,1]$ and for all $x,y\in [a,b]$ the inequality $f(tx+(1-t)y)\ge tf(y)+(1-t)f(x)$ holds.[/hide]

i really don't get what you're trying to say... :blush:

Hmm.  I had an idea that one could prove that the exponential function was convex by Bernoulli's inequality, from which the concavity of its inverse follows.  I'm not sure if it works so nicely.

[hide="But here is a simpler way"] In fact, for a continuous function the definition of concavity is equivalent to Jensen's unweighted inequality,

$f \left( \frac{x+y}{2}\right) \ge \frac{ f(x)+f(y) }{2}$

For $f(x) \log_{k}x$ this gives

$\log_{k}\left( \frac{x+y}{2}\right) \ge \log_{k}\left( \sqrt{xy}\right)$

Since $\log_{k}x$ is an increasing function the conclusion follows by AM-GM. [/hide]

yes, thank you. that's my solution too. :)


In triangle ABC, let the altitude from A intersect BC at D and the altitude from B intersect AC at E. Let the angle bisector of ACB intersect AD at F and BE at G. Prove that $ AF\times EG\equal{}DF\times BG$.

It might be too easy for this forum.

[hide]

By the Angle Bisector Theorem, we have  $ \frac{AF}{DF}\equal{}\frac{AC}{DC}$ and $ \frac{BG}{EG}\equal{}\frac{BC}{EC}$. Notice that $ (AC)(EC)\equal{}(BC)(DC)$ because $ AEDB$ is cyclic. Therefore, $ \frac{AC}{DC}\equal{}\frac{BC}{EC}$, which implies $ \frac{AF}{DF}\equal{}\frac{BG}{EG}$, and $ (AF)(EG)\equal{}(DF)(BG)$

[/hide]


Prove that in a finite group the number of subgroups of index 2 is of the form $2^k-1$.

$\mathcal G=\hom(G,\mathbb Z_2)$ is a group (multiplication of two homomorphisms is carried out componentwise). Every element in $\mathcal G$ (except for the trivial homomorphism, of course) has order $2$, so $|\mathcal G|=2^k$.

On the other hand, every element of $\mathcal G$ is determined by its kernel in $G$, and these kernels are precisely the subgroups of index $2$ and $G$ itself, and the conclusion follows.

Could it be this simple? :? What am I doing wrong?

No, it is not a difficult problem, but I found it really surprising. Sometimes, there are also easy problems in oral examinations.


Prove that if $ x > 0$ and $ n > 0$, where x is real and n is an integer, then

 $ \frac{x^{n}}{x+1^{n+1}}\leq  \frac{n^{n}}{n+1^{n+1}}$

Maybe you meant this:
$ \frac{x^{n}}{(x+1)^{n+1}}\leq \frac{n^{n}}{(n+1)^{n+1}}$
And I suppose you'd like a solution that doesn't involve [hide="calculus"]$ f(x)=\frac{x^{n}}{(x+1)^{n+1}}$
$ f'(x)=0$  $ \to$ $ x=n$ ($ x>0$, $ n>0$)
And it's a maximum because $ f(n)>\lim_{x\to\infty}f(x)=0$[/hide]
because in that case you don't need $ n$ to be an integer.


Do the apartments/rooms (wherever we're staying) have WiFi? I'm thinking of bringing my laptop so I can watch the Olympics.

Please don't bring a laptop.
AMP ends on the 3rd of August; the Olympics begin on the 8th.
You'll have sufficient internet access at camp (via your mentor)--at least once per day to check email, etc.
WiFi is only given to students who "need" it--there was one student last year who had to work on an online course. [he ended up not touching it]. WiFi is not given freely.

And for those wanting to bring a computer for entertainment purposes, don't.
Why would you waste you time with pixels when you have a bunch of (newly made) friends?

Non sequitur: don't hurl frozen burritos through windows, please.

Really? I thought they started sometime in July. Hmmm...


I think the only way is to change which colums the number are in.

Column A = x axis
Column B = y axis

yea h_s_potter2002 is right thats the only way that it seems to work.


A mountain climbing Club consisting of n member orgamized 4 climbs for its member . Supposing E1;E2;E3;E4 are four teams taking part in the climbs . How many ways there are to divide them into team so that the reqirenment below is met :
                            $E1 \bigcap E2 \neq \phi ;E2 \bigcap E3 \neq \phi;E3 \bigcap E4 |neq \phi$

My suggested way to approach this:

Instead of trying to figure out how many ways there are to arrange the memberships so various pairs of the clubs are non-disjoint, try and figure out how many ways there are to arrange the memberships to that various pairs ARE disjoint.

For example, if our only requirement was that clubs 1 and 2 have to be disjoint, then there are $12^{n}$ possible membership arrangements (each person has 12 choices for which clubs he's a part of).

Similarly, you can figure out (for example), how many arrangements have both the pairs $(E_{1}, E_{2})$ and $(E_{3}, E_{4})$ disjoint, and so on.

Once you've done all that, use inclusion-exclusion to determine how many arrangements have none of the three pairs you're interested in disjoint.


[quote="alex2008"]Let $ a,b,c,d$ be non-negative numbers such that.
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 2$.
Prove that
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} abc \plus{} bcd \plus{} cda \plus{} dab \geq 2$[/quote]
i think ev-theorem can do it


$P$ is a point out of a circle. Through $P$ draw two tangent lines which touches the circle at $A$ and $B$. Through $P$ draw another line which cuts the circle at $M$ and $N$. Let $PMN \cap AB = Q$, show that:
\[ \frac{1}{PM}+\frac{1}{PN}=\frac{2}{PQ} \]

It;s well-known that $(PMQN)=-1$.So $\frac{1}{PM}+\frac{1}{PN}=\frac{2}{PQ}$(Decarse's theorem)


Prove that among $ n$ $ (n>3)$ positive integers there are three numbers $ a,b,c$ so that $ a(b\minus{}c)$ is divisible by $ n$.

[quote="moldovan"]Prove that among $ n$ $ (n > 3)$ positive integers there are three numbers $ a,b,c$ so that $ a(b \minus{} c)$ is divisible by $ n$.[/quote]
Let these $ n$ numbers be $ x_1,x_2,..,x_n$.
Case one: $ \forall i \neq j x_i \neq x_j \bmod n$. Then obviously $ \exists i$ s.t. $ x_i \equiv 0 \bmod n$, or $ n \mid x_i$, and we choose $ a \equal{} x_i$, and we are done.
Case two: $ \exists i \neq j$ s.t. $ x_i \equiv x_j \bmod n$. Then we choose $ b \equal{} x_i,c\equal{}x_j$, and we are done.


Given a triangle $ABC$ with $a=8, b=10, c=12$, construct excircles on sides $a,b,c$ with excenters $X_{a}, X_{b}, X_{c}$ respectively. Find the area of the triangle $A'B'C'$ formed by $ABC$ when inverted on the circumcircle of the triangle $X_{a}X_{b}X_{c}$.

(An inversion of an internal point $A$ on circle $O$, with radius $r$, constructs the point $P$ on line $OA$ such that $OA \cdot OP = r^{2}$ and $A$ is between $P$ and $O$.)

[quote="Altheman"]...ABC is not inverted to a triangle

if you mean the verticies, then it is just a lot of computation, distance center to verticies, compute radius, etc

remember, random problems are not what problem solving is about...they might seem random, but there is always an idea behind the problem...when making problems you should have an idea in mind[/quote]

 :P  I was just playing around with the Geometry sketchpad programs. I need to learn more geom, lol...  meh. Over the summer.


OMG BRING YOUR CHESS SETS WE NEED TO GET SOME GAMES OF BUGHOUSE IN!!!!!!!!!!!

YES!  Unfortunately, I don't have any chess sets that travel well.  :(

It's a variant of chess ([url=http://en.wikipedia.org/wiki/Bughouse_chess]wikipedia[/url] is your friend ;) ).

So Joe are you going to bring a chess set or two? :)

lol, no.  I don't have any chess sets :(

I'm already at Princeton, btw :D

darn :(

oh wait... my chess set is missing a knight :roll:

ummm... the outlook for bughouse is looking dim :(

[quote="chess64"]Uh did anyone else bring a chess set or not? :D[/quote]

I have a chess set I guess

um ok cool

way to tell me in advance...

[quote="JSteinhardt"]I'll try to bring a chess set to HMMT in case we have some free time then. And now that I've met you once I should be able to find you there even though it's a lot bigger than Princeton.

By the way, you look exactly like an older version of this Freshman who was on our B team  :wink: .[/quote]

lol...

You don't look like how I imagined you at all :P

Lol Zhai is like the only math person who looked like I predicted he would.

Yeah I know seriously. But even Zhai looked kinda different from his IMO picture (http://imo2006.dmfa.si/participants/USA.html) :lol:

I still dont know what anyone looks like

OMG OMG OMG WERE YOU ON THE BED WITH THAT GUY


quadratics
geometry
algebra
quadratic formula

Ok, I generally suck at all "problem solving" problems. I can write equations from problems, etc. However, I am finding it very difficult to "solve each problem using quadratic equations".

For example:

"If the product of two consecutive intergers is decreased by 20 time the greater integer, the result is 442. Find the integers."

I need help on that one, as well as this one:

"Jackie is building a playhouse. She wants each rectangular window to have an area of 315 square inches. She also wants each window to be 6 inches higher than it is wide. What are the dimensions of each window?"

Is there some kind of logical way to attack this?

Yes, there is definitely a logical way to attack each of these problems, or else you would not have been given them!  :)

[hide="Starting on 1"]Let $x$ be the smaller integer, let $x + 1$ be the larger integer.  (It is generally useful to assign variables immediately.  Remember that the integers are consecutive!)

The problem then says that

$x(x+1) - 20(x+1) = 442$
$x^2 + x - 20x - 20 = 442$
$x^2 - 19x - 462 = 0$

Can you solve it from here?  :) [/hide]

[hide="Starting on 2"]Let $x$ be the width, let $x + 6$ be the height.  (Remember, the problem says the windows are $6$ units taller than wide!)

The problem then says that

$x(x + 6) = 315$
$x^2 + 6x - 315 = 0$

Again, a quadratic.  Can you solve it from here?  :)

[b]Remark:[/b]  It is not necessary to solve it in this way.  [hide="Hint"] Notice that $300 = 15 \times 20$, so $315 = 15 \times 21$.  :) [/hide]
[/hide]

Wow, thanks a lot!! I see the [i]logic[/i] in it now....heh

Tell me, my instructor has the rest done by factoring "standard form" and then using zero, like this:

2x(squared) + 4x - 224 = 0
(x + 16)(x-14) = 0
x + 16 = 0           x - 14 = 0
        x = -16               x = 14

Instead of using the quadratic formula. Since I hate factoring, I would love to just use the quadratic formula, but if I did, would it somehow be incorrect?

[quote="cooly291"]Instead of using the quadratic formula. Since I hate factoring, I would love to just use the quadratic formula, but if I did, would it somehow be incorrect?[/quote]

Quadratic formula would give you the same answer, but the calculations involved take a long time for most problems.  It'll pay off to learn to factor quickly and efficiently.  :)

I am finding it very difficult to factor those problems...care to shed some light? :)

Note that when factoring, you want to find two numbers $a$ and $b$ such that the sum of them is the number before the $x$ term, and their product is the constant term (no variable). The factorization is then $(x+a)(x+b)$.

For the first equation $x^2-19x-462$, can you find two numbers that multiply to be -462 and add to be -19? (Note: one of the factors will be negative)

Listing factors of 462 would help.

[quote="jclarkemathL314159"]shouldn't quadratic equations be in the middle school forums?[/quote]
No,I didnt learn them till 9th grade.


I will give you the solutions at weekend!

[hide="num 1, not finish."]
$\sqrt[3]{x^{2}-1}+x = \sqrt{x^{3}-2}$
$%Error. "Longrightleftarrow" is a bad command.
(\sqrt[3]{x^{2}-1}-2)+(x-3) = (\sqrt{x^{3}-2}-5)$
$%Error. "Longrightleftarrow" is a bad command.
\frac{(x-3)(x+3)}{(\sqrt[3]{x^{2}-1})^{2}+2(\sqrt[3]{x^{2}-1})+4}+(x-3) = \frac{x^{3}-27}{\sqrt{x^{3}-2}+5}$
Here we get $x=3$ or
$\frac{x+3}{(\sqrt[3]{x^{2}-1})^{2}+2(\sqrt[3]{x^{2}-1})+4}+1 = \frac{x^{2}+3x+9}{\sqrt{x^{3}-2}+5}$
Ussually, this kind of equation has only one root, the last thing is to prove that the one above has no root. I don't think I have enough knowledge to do that. Some of you may help,  :blush: [/hide]


calculus
integration
function
symmetry
algebra
partial fractions
complex analysis

Hi all,

Integral from -1 to infinity of the function (x^4/(1+x^6))^2

I've tried partial fractions but doesn't work, also substitution and integration by parts doesn't help me either.

Any help would be greatly appreciated, thanks!

Partial fractions will work, but you won't like it.

The complete factorization in real terms of the denominator is

$ x^6 \plus{} 1 \equal{} (x^2 \plus{} 1)(x^2 \minus{} \sqrt {3}x \plus{} 1)(x^2 \plus{} \sqrt {3}x \plus{} 1).$

Make that $ (x^6 \plus{} 1)^2 \equal{} (x^2 \plus{} 1)^2(x^2 \minus{} \sqrt {3}x \plus{} 1)^2(x^2 \plus{} \sqrt {3}x \plus{} 1)^2$

This can be further factored into six linear factors with complex roots - which might be the way to go.

Personally, I would seek software assistance, from your favorite computer algebra system or integrating website. (Saying this is explicitly admitting that I think expanding this expression with partial fractions probably exceeds the capacity of human pencil-and-paper algebra.)

Note that $ \int_{ \minus{} \infty}^{\infty}\left(\frac {x^4}{x^6 \plus{} 1}\right)^2\,dx$ is an easier question; the residue theorem from complex analysis can be effectively applied to it.

Here is another method:

let $ I\equal{}\int_{\minus{}\infty}^{\infty}\frac{x^{8}}{(1\plus{}x^{6})^{2}}dx$.

Due to symmetry we get that

$ I\equal{}2\int\limits_{0}^{\infty}\frac{x^{8}}{(1\plus{}x^{6})^{2}}dx\equal{}2J$. Use the substitution $ x\equal{}1/t$ and you get that

$ J\equal{}\int_{0}^{\infty}\frac{x^{2}}{(1\plus{}x^{6})^{2}}dx$ and hence

$ J\equal{}\frac{1}{2}(J\plus{}J)\equal{}\int_{0}^{\infty}\frac{x^{2}\plus{}x^{8}}{(1\plus{}x^{6})^{2}}dx\equal{}\frac{1}{2}\int_{0}^{\infty}\frac{x^{2}}{1\plus{}x^{6}}dx$.

Now the substitution $ x^{3}\equal{}z$ shows that $ J\equal{}\frac{1}{6}\int_{0}^{\infty}\frac{dz}{z^{2}\plus{}1}\equal{}\frac{\pi}{12}$.

In conclusion the value of your integral is $ \frac{\pi}{6}$.

But note that didi is answering what I already called an easier question. The original was the integral from $ \minus{}1$ to $ \infty$ and the manipulations that depend on symmetry won't work.

Using software:
$ \int{\frac {x^8}{(x^6 \plus{} 1)^2}}dx \equal{} \minus{}\frac {x^3}{6(x^6 \plus{} 1)} \plus{} \frac {\int{\frac {x^2}{x^6 \plus{} 1}dx}}{2}$


calculus
integration
limit
trigonometry
induction
real analysis
real analysis unsolved

Find $ lim_{n \rightarrow \infty} ( \int _0^1 sin^n x dx)$

$ 1 \leq lim_{n \rightarrow \infty} \int^{1}_{0} \frac {1 \minus{} sin^{n}(x)}{1 \minus{} sin(x)} dx \leq \int^{1}_{0} \frac {1}{1 \minus{} sin(x)} dx$

Thus, $ \int^{1}_{0} \frac {1 \minus{} sin^{n}(x)}{1 \minus{} sin(x)} dx$ is bounded from above and is precisely equal to the sequence $ S_n \equal{} \int^{1}_{0} \sum_{i \equal{} 0}^{n\minus{}1} sin^{i}(x)$. $ S_n$ is monotonically increasing and hence convergent.

Thus

$ lim_{n \rightarrow \infty} \int^{1}_{0} sin^{n}(x) \equal{} lim_{n \rightarrow \infty} \left(S_{n\plus{}1} \minus{} S_{n}\right) \equal{} 0$

I think it's more simple solution :
$ 0 \leq \lim \int_{0}^{1}\sin^{n}(x)dx \leq \lim \sin^{n}(1) \equal{} 0 \quad  \text{since}\quad \max_{x\in [0,1]}\sin(x)\leq  \sin(1)<1$

OK, Extremal, let's change the problem and make it

$ \lim\int_0^{\frac{\pi}2}\sin^nx\,dx.$

(In other words, I just took away the particular argument you used.)

\begin{eqnarray*}\limsup \int_{0}^{\frac{\pi}{2}} \sin^n x \, dx
& \leq & \limsup \int_{0}^{\frac{\pi}{2}-\epsilon} \sin^n x \, dx + \limsup \int_{\frac{\pi}{2}-\epsilon}^{\frac{\pi}{2}} \sin^n x \, dx \\
& \leq & \limsup \left( \frac{\pi}{2}-\epsilon \right) \sin^n \left( \frac{\pi}{2}-\epsilon \right) + \epsilon \\
& = & \epsilon \end{eqnarray*}

Another way to look at this problem, and probably this is the method that Kent was looking for,

is to express the integral in terms of a Beta function.

It's wellknown Wallis integral , maybe therefore it's denoted by $ W_{n}$.

$ W_{n} \equal{} \int_{0}^{\pi/2}\sin^{n}(x)dx \equal{} \int_{0}^{\pi/2}\cos^{n}(x)dx$
it's not hard to note that  $ W_{n} \equal{} \frac {n \minus{} 1}{n}W_{n \minus{} 2}$ hence by induction we obtain :
$ W_{n} \equal{} \frac {(n \minus{} 1)!!}{n!!}v_{n}$
where $ v_{n} \equal{} \pi/2$ if $ n \equal{} 2k$ and $ v_{n} \equal{} 1$ if $ n \equal{} 2k \plus{} 1$ where  $ k \in \mathbb{Z}$
By using Stirling (or without Stirling :wink:  ) formula we obtain $ W_{n} \equal{} C\frac {v_{n}}{n} \plus{} o \left(\frac {1}{n} \right) \equal{} \mathcal{O}\left( \frac{1}{n} \right)$ thus limit is equal to $ 0$
Yes , I agree it's not so easy problem as first.

This limit can also be studied:

Let $ \alpha>0$ be a fixed positive real number. Calculate

$ \lim\limits_{n\rightarrow \infty}\int\limits_{0}^{\alpha}|\sin x|^{n}dx$.

I think limit also will be equal to $ 0$ since you can divide interval $ [0,a] \equal{} \bigcup_{1 \leq k \leq N} [\pi (k\minus{}1) /2,\pi k /2]\bigcup [\pi*N/2,a]$ where $ \pi * N/2\leq a \leq \pi *(N \plus{} 1)/2$ hence on each interval limit is zero  hence the sum of zeros is equal to zero

I think that Kent was thinking to the dominated convergence theorem - result immediately follows. (you will tell me that one does not need such a big gun here ... maybe)

Although of course the DCT makes short work of this, I was actually thinking of the argument that sos440 posted. I even posted my answer, then deleted it when I realized that sos440 has already said it.


hello, we write the equation in the following form:
$ \sqrt{1\minus{}x^2}\equal{}36\minus{}729x^4$. The domain of definition is $ \minus{}\frac{1}{3}\sqrt{2}\le x \le\frac{1}{3}\sqrt{2}$.
After squaring both sides  we get a polynomial equation
$ 531441x^8\minus{}52488x^4\plus{}64x^2\plus{}1232\equal{}0$. The left hand-side can be factorised in $ (81x^4\plus{}4x^2\minus{}4)(6561x^4\minus{}324x^2\minus{}308)$. From here we get the
two solutions $ x_1\equal{}\minus{}\frac{1}{9}\sqrt{\minus{}2\plus{}2\sqrt{82}}$ or $ x_2\equal{}\frac{1}{9}\sqrt{\minus{}2\plus{}2\sqrt{82}}$. $ x_1$ and $ x_2$ solves our originally problem.
Sonnhard.


My teacher gave us a homework sheet on Chemical Synthesis and Organic Chemistry and some chemical caluculations. I thought it was pretty cool so I decided to post some questions here.

1. The yield of primary amine is rather poor when using a conventional reaction of halogenoalkane with ammonia. Describe a chemical process (include mechanisms if you can) by which one can commercially manufacture primary amines.

2. Describe a test by which you can distinguish between the carbonyl group of a ketone/aldehyde and that of other Organic Compounds.

3. Name a process by which cyclic compounds are manufactured.

4. Explain the term "Chain Polymerization" and name an example of a polymer obtained in this way.

5. Name the two types of optical isomers. Describe how you would convert one form to the other.

6. $PCl_{5}\rightleftharpoons PCl_{3}+Cl_{2}$ has $K_{c}= 4 \times 10^{-2}$. One mole of $PCl_{5}$ reached equilibrium in a $10L$ container, find the degree of dissociation.

7. Given that $Mg(OH)_{2}$ has $K_{s}p = 8.9 \times 10^{-12}$. Deduce the solubility of $Mg(OH)_{2}$ under these conditions:

i) In pure water

ii) In $0.1M$ of $MgCl_{2}$ solution

iii) In $0.1M$ of $NaOH$ solution

8. In a $0.10M$ solution of $H_{3}PO_{4}$ find $[H^{+}], [OH^{-1}], [H_{2}PO_{4}^{-}], [HPO_{4}^{2-}], [PO_{4}^{3-}], [H_{3}PO_{4}]$. ($K_{a,1}= 7.5 \times 10^{-3}, K_{a,2}= 6.2 \times 10^{-8}, K_{a,3}= 1.0 \times 10^{-12}$)

I like the last question.

1. The yield of a primary amine from an haloalkane and ammonia is poor because of polyalkilation of the nitrogen of the initially formed primary amine. A possible method for the commercial prodution can be the reductive amination of aldehydes and ketones with ammonia:

$RR'C=O+NH_{3}+H_{2}\longrightarrow RR'CH-NH_{2}$

where R and/or R' can be hydrogen. The mechanism envolves two steps. The first is the usual imine formation:

$RR'C=O+NH_{3}\longrightarrow RR'C=NH$.

The second step is hydrogenation of the carbon-nytrogen double bond:

$RR'C=NH+H_{2}\longrightarrow RR'CH-NH_{2}$.

This synthesis can be used to prepare primary amines (from aldehydes) and secondary amines (from ketones). To prepare tertiary aldehydes, other method must be used (for example, from tertiary alcohols and a nitrile).

Other common laboratory methods for the preparation of primary amines are the reduction of azides and nitriles, and the older Gabriel Synthesis (between many other methods).
Primary amino-ketones and amino-alcohols can be elegantly prepared through the Neber Rearrangement.

2. The most reliable test (and nowadays the easiest one) is spectroscopy.

In infrared spectroscopy, the aldeyde group -CHO is responsible for three important bands: two bands, one at 2700-2750 cm-1 and another at 2800-2850 cm-1 (due to the aldehyde C-H vibration), and, of course, the band due to the C=O stretching vibration, at 1725-1740 cm-1. For ketones, the C=O band appears at 1705-1720 cm-1. If we have a methyl ketone, then the carbonyl -CH3 bending vibration should appear as a medium-strong band at 1375 cm-1. For other carbonyl functional groups (acids, acid chlorides, esters, amides) the C=O band appear at other frequencies.

In 1H-NMR, the aldehyde proton appears tipically between 9 and 10 ppm. For methyl ketones, the carbonyl -CH3 protons appear as a singlet at 2.1 ppm.

In 13C-NMR, the carbonyl carbons of aldehyde and ketones appear between 185-220 ppm, while for other carbonyl functional groups than interval is lower (155-185 ppm).

If you are looking for chemical tests, then aldehydes can be distinguished from all other carbonyls by the ease with which they are oxidised, and this is the basis for the well known Fehling Test and Tollens Test (or silver mirror test). Aldehydes and ketones also form with relative easy coloured compounds with dinitrophenylhydrazine, while other carbonyls do not.

3. The Diels-Alder reaction is one well known reaction for the synthesis of six-membered rings. The Robinson annulation reaction is another elegant process for the syntesis of six-membered rings. For the preparation of five-membered rings, 1,3-dipolar cycloaddition reactions are one of the most usefull and versatile. Of course, the classic acetoacetic ester synthesis and malonic ester syntesis can be used to prepare rings of several sizes.

4. "Chain Polymerization" uses a chain reaction to obtain a polymer. Although chain reactions are usually related with radicals, in chain polymerization the reactive species can not only be radicals but also cations and anions. Chain reactions consist of three main steps: initiation, propagation, and termination. Any alkene monomer can be used to prepare a polymer of this kind, for example, polystyrene is produced by radical (or cationic) polimerization of styrene (vinilbenzene).

5. The two types of optical isomers are enantiomers and diastereomers.

6. The initial concentration of $PCl_{5}$ is then 0.10 mol/L, and so the equilibrium concentrations will be

$[PCl_{5}] = 0.10-x$, $[PCl_{3}] = x$, $[Cl_{2}] = x$.

From the equilibrium constant expression, $K_{c}= \frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}$ we get (assuming the equilibrium constant is 0.040 and not $410^{-2}$)

$0.040 = \frac{x^{2}}{0.10-x}$.

Solving the resultant quadratic equation we get $x = 0.0465$, and so the degree of dissociation $\alpha$ is

$\alpha = \frac{[PCl_{3}]_{eq}}{[PCl_{5}]_{initial}}\times 100 = \frac{0.0465}{0.10}\times 100 = 46.5 per cent$.

7. (i) The solubility equilibrium of magnesium hydroxide is

$Mg(HO)_{2}(s)+aq \rightleftharpoons Mg^{2+}(aq)+2HO^{-}(aq)$.

Designating by S the solubility in pure water, then $[Mg^{2+}] = S$ and $[HO^{-}] = 2S$, and so $K_{s}= S(2S)^{2}= 4S^{3}$

$\Rightarrow  S = \sqrt[3]{\frac{K_{s}}{4}}= 1.3 \times 10^{-4}\,mol/L$.

(ii) If S' is the new solubility (which must be smaller than S, because of the common ion effect) the we must have

$[Mg^{2+}] = 0.1+S'$, $[HO^{-}] = 2S'$, and $[Mg^{2+}][HO^{-}]^{2}= K_{s}$

and so $(0.1+S')(2S')^{2}= K_{s}$. Making the approximation $0.1+S' \approx 0.1$ (which is justifiable, since S' is even lower than $1.3 \times 10^{-4}$), then we have $0.1 \cdot 4S'^{2}= K_{s}\Rightarrow  S' = \sqrt{\frac{K_{s}}{0.4}}= 4.7 \times 10^{-6}\,mol/L$.

(iii) We have again the common ion effect, and the solubility S' in NaHO will be smaller then S. Now we have $[Mg^{2+}] = S'$ and $[HO^{-}] = 0,1+2S'$ and so $S'(0.1+2S')^{2}= K_{s}$. Now the approximation to be made is $0,1+2S' \approx 0.1$, and so $S' \cdot 0.1^{2}= K_{s}\Rightarrow  S' = \frac{K_{s}}{0.01}= 8.9 \times 10^{-10}\,mol/L$.


[quote="akashram"]a/root of a squere+8bc+b/root of b squere+8ac+c/root of c squere+8ab     >or equal to 1

I solved this but am not sure of my solution :oops:[/quote]
u first ought to learn latex
anyway this is the well known IMO 2001 inequality,
$ \sum_{cyc}\frac {a}{\sqrt {a^2 \plus{} 8bc}} \ge 1$
infact it wont hurt to use the search function since u kno the source it wud give u many links to solution.
as a sample - http://www.mathlinks.ro/Forum/viewtopic.php?t=310244
this has a generalisation of the problem


inequalities
search
inequalities proposed

Yes, but I didn't solved them by this way! There is any other solution?

[quote="The soul of rock"]Yes, but I didn't solved them by this way! [/quote]
Why? For example:
I'll prove that $\sum_{cyc}\frac{a^{3}}{(b+c+d)^{3}+a^{3}}\ge\frac{4}{28}.$ :wink:
Let $a+b+c+d=4.$ Then
$\sum_{cyc}\frac{a^{3}}{(b+c+d)^{3}+a^{3}}\ge\frac{4}{28}\Leftrightarrow\sum_{cyc}\left(\frac{a^{3}}{(4-a)^{3}+a^{3}}-\frac{1}{28}\right)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left(\frac{a^{3}}{3a^{2}-12a+16}-\frac{1}{7}-\frac{27}{49}\cdot(a-1)\right)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{(a-1)^{2}(10-a)}{3a^{2}-12a+16}\geq0.$ :)

[quote="Rzeszut"][quote="pvthuan"]We can choose $a+b+c+d=4$, then using the idea of tangent line.[/quote]
What is this idea? I tried to search, but I failed.[/quote]
See the att.

I have a really classical and simple way to prove that kind of ineq, may be like TL, but not extracly TL

The first one is a piece of cake with cauchy :wink:

Yes, It's impress me very much. There is another solution? I wonder that how many ways we can use here?

Yes, they are easy. Nornamization $a+b+c+d=1$.

Yes, It's too easy. But I think normaliztion with $a+b+c+d=4$ is easier!


geometry
circumcircle
trigonometry
area of a triangle
geometry solved

Since $S(APN)=S(CMN)=S(BMP)$ so from here it follow the area of circumcircles of triangles are equal then their circumradius is also equal. Using $\delta ABC=\frac{abc}{4R} (i)$. Then
 $\frac{(PB.BM.sin60)}{(MC.CN.sin60)}=1$ $\rightarrow$  $\frac{(PB.BM)}{(MC.CN)}=1$ using $(i)$. we will get
$PM=MN$ similiarly $MN=NP$ so $PMN$ is equilateral. Since $ABC$ equilateral so from here
$AN=CM=PB$ and $AP=CN=AP$ as desired triangels will be congurent

[quote="abdurashidjon"]so from here it follow the area of circumcircles of triangles are equal[/quote]

Why that???

I don't think the problem is that easy to solve using just congruences of triangles (one has to use that M, N, P lie inside the sides BC, CA, AB; else, you can take M, N, P to be three points on the lines BC, CA, AB such that AM || BN || CP, and then triangles ANP, BPM, CMN have equal areas without being congruent). Here is my solution:

[color=blue][b]Theorem.[/b] If ABC is an equilateral triangle and M, N, P are points on the segments BC, CA, AB such that the triangles ANP, BPM, CMN have equal areas, then these triangles ANP, BPM, CMN are congruent.[/color]

[i]Proof.[/i] Since triangle ABC is equilateral, BC = CA = AB and < CAB = < ABC = < BCA = 60°. Now, the area of triangle ANP is $\frac12\cdot AN\cdot AP\cdot\sin\measuredangle NAP=\frac12\cdot AN\cdot AP\cdot\sin\measuredangle CAB=\frac12\cdot AN\cdot AP\cdot\sin 60^{\circ}$. Similarly, the area of triangle BPM is $\frac12\cdot BP\cdot BM\cdot\sin 60^{\circ}$. Since the areas of these two triangles are equal (by the condition of the problem), we have $\frac12\cdot AN\cdot AP\cdot\sin 60^{\circ}=\frac12\cdot BP\cdot BM\cdot\sin 60^{\circ}$. Thus, if $AN\leq BP$, then $AP\geq BM$, and consequently, $BP=AB-AP=BC-AP\leq BC-BM=CM$. So we have seen that $AN\leq BP$ yields $BP\leq CM$. Similarly, $BP\leq CM$ yields $CM\leq AN$. Thus, if we have $AN\leq BP$, then $AN\leq BP\leq CM\leq AN$, what can only mean that AN = BP = CM. Similarly, if $BP\leq AN$, then we can do the same observations "backwards" and see that AN = BP = CM. So we must always have AN = BP = CM. Thus, AP = AB - BP = BC - BP = BC - CM = BM. Finally, < NAP = < CAB = 60° = < ABC = < PBM. Thus, the triangles ANP and BPM are congruent (since AN = BP, AP = BM and < NAP = < PBM). Similarly, the triangles BPM and CMN are congruent. Thus, the three triangles ANP, BPM, CMN are congruent, qed.

  Darij

[quote="bui_haiha2000"]Let $ABC$ be an equilateral triangle (i. e., a triangle which satisfies $BC=CA=AB$). Let $M$ be a point on the side $BC$, let $N$ be a point on the side $CA$, and let $P$ be a point on the side $AB$, such that $S\left(ANP\right)=S\left(BPM\right)=S\left(CMN\right)$, where $S\left(XYZ\right)$ denotes the area of a triangle $XYZ$.
Prove that $\triangle ANP\cong\triangle BPM\cong\triangle CMN$.[/quote]

Let the unit of the length measure be the side of the triangle, i.e. $|AB|=|BC|=|CA|: = 1$. Then let's mark $x=|BM|$, $y=|CN|$, $z=|AP|$. Then $|MC|=1-x$, $|AN|=1-y$, $|BP|=1-z$. Using formula for the area of the triangle, we obtain $\frac{1}{2}\cdot\sin 60^{\circ}\cdot (1-x)y= \frac{1}{2}\cdot\sin 60^{\circ}\cdot (1-y)z= \frac{1}{2}\cdot\sin 60^{\circ}\cdot (1-z)x$. After cancelling, it becomes $(1-x)y= (1-y)z= (1-z)x$. From the second equality we get $z=\frac{x}{1-y+x}$. If we prove that $x=y$, then $z$ will be equal $x$, so the proof willl be ended. By setting $z=\frac{x}{1-y+x}$ in the first equality we obtain $(1-x)y= \frac{(1-y)x}{1-y+x}$. After simplifying it becomes $(x-y) (1-y+xy)=0$. Let's assume that the second factor is $0$. Then $x=\frac{y-1}{y}<0$ since $0<y<1$, so $x<0$ which is  impossible. Thus the first factor is $0$, so $x=y$ and $z=\frac{x}{1-y+x}=x$, so $x=y=z$, so the triangles $ANP$, $BMP$ and $CMN$ are congruent.


Este es un conjunto de problemas de inceuaciones que complementaran la teoria basica del articulo anterior..
podes tener la version en  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=58478]PDF[/url]

[color=blue] [b][size=150]Training Test - Inequalities[/size][/b] [/color]
[color=blue]Date: May 29, 2005.[/color]

[i]Spanish Version. [/i]

$\blacktriangleright$ [color=blue][b]Problema 1.[/b] [USA 1978][/color]
Sean $a, b, c, d, e$ n\'umeros reales, tales que:
$a + b + c + d + e = 8$ y $a^2 + b^2 + c^2 + d^2 + e^2 = 16.$
Cual es el valor m\'aximo de $e$?

$\blacktriangleright$  [color=blue][b]Problema 2.[/b][Rusia 1995][/color]
Pruebe que, para todo real $x, y > 0$:
$\frac{x}{x^4 + y^2} + \frac{y}{x^2 + y^4} \leq \frac{1}{xy}.$

$\blacktriangleright$ [color=blue][b]Problema 3.[/b][/color]
Pruebe que, si $a, b, c$ son longitudes de lados de un triangulo,
se cumple:
$(a + b)(b + c)(c + a)\geq 8(a + b - c)(b + c - a)(c + a - b).$

$\blacktriangleright$ [color=blue][b]Problema 4.[/b][Estonia 1996/1997][/color]
Pruebe que, para todo $x$ e $y$ reales:
$x^2 + y^2 + 1 > x\sqrt{y^2 + 1} + y\sqrt{x^2 + 1}.$

$\blacktriangleright$ [color=blue][b]Problema 5.[/b][Singapur 2000][/color]
Si $a, b, c, d > 0$ y $c^2 + d^2 = (a^2 + b^2)^3$.
 Mostrar que:
$\frac{a^3}{c} + \frac{b^3}{d} \geq 1.$

$\blacktriangleright$ [color=blue][b]Problema 6.[/b][/color]
Si $a_1, a_2, a_3 \geq 0$ y $(1 + a_1)(1 + a_2)(1 + a_3) = 8.$
Mostrar que:
$a_1.a_2.a_3 \leq 1.$

[color=blue]LaTeXed by Carlos Bravo 2005[/color]

hello, to 2) we have bei the AM-GM inequality $ \frac{y}{x^2\plus{}y^4}\le \frac{1}{2xy}$ and $ \frac{x}{x^4\plus{}y^2}\le \frac{1}{2xy}$ and adding both inequalities we get the desired result.
Sonnhard.

[hide="Solución 4"]
Note que, por la desigualdad Aritmética-Geométrica,

\[ \dfrac{x^2\plus{}(y^2\plus{}1)}{2}\ge x\sqrt{y^2\plus{}1}\]
(igualdad cuando $ x^2\equal{}y^2\plus{}1$)
Ademas, tenemos que,

\[ \dfrac{(x^2\plus{}1)\plus{}y^2}{2}\ge y\sqrt{x^2\plus{}1}\]
(igualdad cuando $ y^2\equal{}x^2\plus{}1$)

Añadiendo ambos desigualdades tenemos,
\[ x^2\plus{}y^2\plus{}1\ge x\sqrt{y^2\plus{}1}\plus{}y\sqrt{x^2\plus{}1}\]
Pero la igualdad es imposible por que, $ x^2\not\equal{} x^2\plus{}2$. Por lo tanto,
\[ x^2\plus{}y^2\plus{}1> x\sqrt{y^2\plus{}1}\plus{}y\sqrt{x^2\plus{}1}\]
Quod Erat Demonstrandum.
[/hide]

Problema 1
usando la desigualdad de cauchy
$ \left( {a^2  \plus{} b^2  \plus{} c^2  \plus{} d^2 } \right)\left( {1^2  \plus{} 1^2  \plus{} 1^2  \plus{} 1^2 } \right) \geqslant \left( {a \plus{} b \plus{} c \plus{} d} \right)^2$
reemplazando los datos
$ 4\left( {16 \minus{} e^2 } \right) \equal{} 64 \minus{} 4e^2  \geqslant \left( {8 \minus{} e} \right)^2  \equal{} 64 \minus{} 16e \plus{} e^2$
$ \frac{{64}}
{5} \geqslant 5\left( {e^2  \minus{} 2 \cdot \frac{8}
{5}e \plus{} \frac{{64}}
{{25}}} \right) \equal{} 5\left( {e \minus{} \frac{8}
{5}} \right)^2$

$ \left( {e \minus{} \frac{8}
{5}} \right)^2  \leqslant \frac{{64}}
{{25}} \Leftrightarrow  \minus{} \frac{8}
{5} \leqslant e \minus{} \frac{8}
{5} \leqslant \frac{8}
{5}$
el valor maximo para e es $ \frac{{16}}{5}$

Problema 6
Notemos que
$ 1 \plus{} a_1 \plus{} a_2 \plus{} a_3 \plus{} a_1 a_2 \plus{} a_1 a_3 \plus{} a_2 a_3 \plus{} a_1 a_2 a_3 \equal{} \left( {1 \plus{} a_1 } \right)\left( {1 \plus{} a_2 } \right)\left( {1 \plus{} a_3 } \right) \equal{} 8$
ahora por la desigualdad A-G
$ \frac {{1 \plus{} a_1 \plus{} a_2 \plus{} a_3 \plus{} a_1 a_2 \plus{} a_1 a_3 \plus{} a_2 a_3 \plus{} a_1 a_2 a_3 }} {8} \equal{} 1 \geqslant \root 8 \of {1 \cdot a_1 \cdot a_2 \cdot a_3 \cdot a_1 a_2 \cdot a_1 a_3 \cdot a_2 a_3 \cdot a_1 a_2 a_3 } \equal{} \root 2 \of {a_1 a_2 a_3 }$
como $ a_1 a_2 a_3 \geqslant 0$
se obtiene lo pedido $ 1 \geqslant a_1 a_2 a_3$

Problema 5.

Por la desigualdad de [url=http://onlinemathcircle.com/wp-content/uploads/2011/03/Holder.pdf]Holder[/url]
\[(c^2+d^2)\left(\dfrac{a^3}{c}+\dfrac{b^3}{d}\right)^2\ge (a^2+b^2)^3\]
usando $c^2+d^2=(a^2+b^2)^3$ tenemos que
\[\dfrac{a^3}{c}+\dfrac{b^3}{d}\ge 1\]


vector
abstract algebra
linear algebra
matrix
calculus
Functional Analysis
superior algebra

Let $f,f_1,f_2,...,f_n$ be linear functionals on the vector space $E$ such that $f_1(x)=f_2(x)=...=f_n(x)=0$ implies $f(x)=0$. Prove that there exists $a_1,a_2,...,a_n$ such that $f=\sum_{k=1}^na_kf_k$.

First of all, why on earth is it in the "Analysis" section? :)

Let $F=E/\bigcap_{i=1}^n\ker f_i$. Clearly, every element $\bar x\in F$ is uniquely determined by its $n$-uple of images through $f_i$, so $F$ is a finite-dimensional vector space. The problem has now become:

Prove that if the linear functionals $f_1,\ldots,f_n$ on a vector space $F$ of dimension $m\le n<\infty$ are such that the intersection of their kernels is trivial, then $f_i,\ i=\overline{1,n}$ generate the dual space $F^*$ of $F$.

We can identify $f_i$ to a vector $v_i\in F$ such that $f_i(x)=(x,v_i)$, where $(\cdot,\cdot)$ is the bilinear form defined by $(x,y)=\sum_{j=1}^mx^jy^j$. Now, the fact that the kernels of $f_i$ have trivial intersection simply means that the $n\times m$ matrix having $v_i$ as its $i$'th row has maximal rank $m$, which means that some $m$ rows are linearly independent, so they generate $K^m$ ($K$ is the field we're working over). The conclusion follows.

Is $F$ a space? I.m. if $x,y\in F$ then $x+y \in F$ ???

Of course it is. It's the quotient of a vector space by a subspace. I think you might be confusing $/$ and $\setminus$. $/$ is not the same as $\setminus$, which means set difference.

[color=red][[b]Moderator edit:[/b] The following solution only works when $E$ is an $\mathbb R$-vector space. It was written when this thread used to be in the "Calculus and Analysis" subforum.][/color]

consider the linear map $F: E\to \mathbb{R}^{n+1}$ defined by $F(x)=(f(x),f_1(x),f_2(x),..,f_n(x))$.
Then $(1,0,0,..,0)$ does not belong to the vectorial subspace $F(E)$. Hahn-Banach theorem shows that there is a functional $g: \mathbb{R}^{n+1} \to \mathbb{R}$ and $\alpha \in \mathbb{R}$ such that $g((1,0,0,..,0)) < \alpha < g(v)$ for all $v \in F(E)$. Hence $g(v)=0$ for all $v \in F(E)$. But one knows that there exists a vector $y=(-a_0,a_1,..,a_n) \in \mathbb{R}^{n+1}$ satisfying: $g(x)=<x|y>$ for all $x \in \mathbb{R}^{n+1}$.
Thus $a_0f(x)=a_1f_1(x)+a_2f(x)+..+a_nf_n(x)$ for all $x \in E$. Conclusion follows.


Find the minimum value of

|sin(x)+cos(x)+tan(x)+cot(x)+sec(x)+csc(x)|

for real numbers x.

[hide]

Let a = sin x and b = cos x, then the expression we are seeking to minimize is |a + b + 1/a + 1/b + 1/(ab)| = |(a + b) + (a + b)/(ab) + 1/(ab)|, where ab =/= 0.

Now let a + b = A and 1/(ab) = B. Then the above becomes |A + AB + B|. And notice that B = 2/(A2 - 1).

So ultimately we seek to minimize:

|A + 2A/(A2 - 1) + 2/(A2 -1)|

= |(A3 + A + 2)/(A2 - 1)|

= |(A2 - A + 2)/(A-1)|

= |A + 2/(A-1)|, where - :rt2: :le: A  :le: :rt2: and |A| =/= 1.

It shouldn't be too hard to figure out from there. I get 2:rt2: - 1 as the minimum.

[/hide]

i believe that is correct from reading Simon's lj

Yeah, that's the solution that I know of.

Oooh, pretty.  A lot of people I know put down 2 as the answer.  I tried a lot of nonsense -- I did make a substitution for cos + sin, but I never got to anything I could work with.

My solution was ``2: Let x=3 :pi: /4.''  verbatim. Our professor said that he proved the answer was 2. It must have been an interesting proof.


calculus
integration
trigonometry
logarithms
function
special factorizations
calculus computations

we know that $ \sin(2x)\equal{}2\sin x\cos x$, hence lets divide both the numerator and denominator by $ \cos^2x$.
So we have:

\[ \int\dfrac{\cos x\plus{}\sin x}{2\sin^2x\minus{}4\sin x\cos x\plus{}5\cos^2x}dx\equal{}\int\frac{1}{\cos x}\dfrac{1\plus{}\tan x}{2\tan^2x\minus{}4\tan x\plus{}5}dx\equal{}\]

Completing the square in the denominator:

\[ \equal{}\int\dfrac{2}{\cos x}\dfrac{1\plus{}\tan x}{(2(\tan x \minus{}1))^2\plus{}1}dx\equal{}\]

Now put $ t\equal{}2(\tan x \minus{}1)$ hence $ dx\equal{}\frac{\cos^2 x}{2} dt$. plugging this in and using that $ \cos x\equal{} \frac{1}{\sqrt{1\plus{}t^2}}$ we have:

\[ \equal{}2\int\dfrac{t\plus{}4}{(t^2\plus{}1)\sqrt{1\plus{}\frac{(t\plus{}2)^2}{4}}}dt\]

Which is straightforward...

The result I gave in my former post is [b]incorrect[/b]. It came from university's notes, but is  incorrect. So the [b]correct[/b] is:

Prove that:

$ \displaystyle\int{\frac{\cos{x}\plus{}\sin{x}}{5\,\cos^2{x}\minus{}2\,\sin({2x})\plus{}2\,\sin^2{x}}\,dx}\equal{}$

$ \displaystyle\frac{3}{5}\,\arctan\left({\sin{x}\minus{}2\,\cos{x}}\right)\plus{}\frac{\sqrt{6}}{60}\,\log\left|{\tfrac{\sqrt{6}\plus{}2\,\sin{x}\plus{}\cos{x}}{\sqrt{6}\minus{}2\,\sin{x}\minus{}\cos{x}}}\right|\plus{}c$.

( $ \log$ is the logarithmic function with base $ e$. )

This is the solution I gave:

$ I\equal{}\displaystyle\int{\frac{\cos{x}\plus{}\sin{x}}{5\,\cos^2{x}\minus{}2\,\sin({2x})\plus{}2\,\sin^2{x}}\,dx}\equal{}$

$ \displaystyle\int{\frac{\cos{x}\plus{}\sin{x}}{5\,\cos^2{x}\minus{}4\,\sin{x}\,\cos{x}\plus{}2\,\sin^2{x}}\,dx}$.

Substitute $ t\equal{}\tan{x}$, then  $ \sin{x}\equal{}\displaystyle\frac{t}{\sqrt{1\plus{}t^2}}$, $ \cos{x}\equal{}\displaystyle\frac{1}{\sqrt{1\plus{}t^2}}$  and $ dx\equal{}\displaystyle\frac{1}{1\plus{}t^2}\,dt$.
So $ I\equal{}\displaystyle\int{\frac{\frac{1}{\sqrt{1\plus{}t^2}}\plus{}\frac{t}{\sqrt{1\plus{}t^2}}}{5\,\frac{1}{1\plus{}t^2}\minus{}4\,\frac{t}{1\plus{}t^2}\plus{}2\,\frac{t^2}{1\plus{}t^2}}\,\frac{1}{1\plus{}t^2}\,dt}\equal{}\int{\frac{1\plus{}t}{\left({5\minus{}4t\plus{}2t^2}\right)\sqrt{1\plus{}t^2}}\,dt}\equal{}$

$ \displaystyle\int{\frac{\frac{3}{5}({2t\plus{}1})\minus{}\frac{\sqrt{6}}{30}\,\sqrt{6}\,({t\minus{}2})}{\left({2t^2\minus{}4t\plus{}5}\right)\sqrt{1\plus{}t^2}}\,dt}\equal{}$

$ \displaystyle\frac{3}{5}\int{\frac{2t\plus{}1}{\left({({t\minus{}2})^2\plus{}t^2\plus{}1}\right)\sqrt{1\plus{}t^2}}\,dt}\minus{}\frac{\sqrt{6}}{30}\int{\frac{\sqrt{6}\,({t\minus{}2})}{\left({\left({\sqrt{6}\,\sqrt{1\plus{}t^2}}\right)^2\minus{}({2t\plus{}1})^2}\right)\sqrt{1\plus{}t^2}}\,dt}\equal{}$

$ \displaystyle\frac{3}{5}\int{\frac{2t\plus{}1}{\left({\left({\frac{t\minus{}2}{\sqrt{1\plus{}t^2}}}\right)^2\plus{}1}\right)\left({1\plus{}t^2}\right)^{\frac{3}{2}}}\,dt}\minus{}\frac{\sqrt{6}}{30}\int{\frac{\sqrt{6}\,({t\minus{}2})}{\left({\left({\frac{\sqrt{6}\,\sqrt{1\plus{}t^2}}{2t\plus{}1}}\right)^2\minus{}1}\right)({2t\plus{}1})^2\sqrt{1\plus{}t^2}}\,dt}\equal{}$

$ \displaystyle\frac{3}{5}\int{\frac{1}{\left({\frac{t\minus{}2}{\sqrt{1\plus{}t^2}}}\right)^2\plus{}1}\,d\!\left({\tfrac{t\minus{}2}{\sqrt{1\plus{}t^2}}}\right)}\minus{}\frac{\sqrt{6}}{30}\int{\frac{1}{\left({\frac{\sqrt{6}\,\sqrt{1\plus{}t^2}}{2t\plus{}1}}\right)^2\minus{}1}\,d\!\left({\tfrac{\sqrt{6}\,\sqrt{1\plus{}t^2}}{2t\plus{}1}}\right)}\equal{}$

$ \displaystyle\frac{3}{5}\int{\frac{1}{\left({\frac{t\minus{}2}{\sqrt{1\plus{}t^2}}}\right)^2\plus{}1}\,d\!\left({\tfrac{t\minus{}2}{\sqrt{1\plus{}t^2}}}\right)}\minus{}\frac{\sqrt{6}}{30}\int{\frac{1}{\left({\left({\frac{\sqrt{6}\,\sqrt{1\plus{}t^2}}{2t\plus{}1}}\right)\minus{}1}\right)\left({\left({\frac{\sqrt{6}\,\sqrt{1\plus{}t^2}}{2t\plus{}1}}\right)\plus{}1}\right)}\,d\!\left({\tfrac{\sqrt{6}\,\sqrt{1\plus{}t^2}}{2t\plus{}1}}\right)}\equal{}$

$ \displaystyle\frac{3}{5}\,\arctan\left({\tfrac{t\minus{}2}{\sqrt{1\plus{}t^2}}}\right)\minus{}\frac{\sqrt{6}}{60}\left({\log\left|{\tfrac{\sqrt{6}\,\sqrt{1\plus{}t^2}}{2t\plus{}1}\minus{}1}\right|\minus{}\log\left|{\tfrac{\sqrt{6}\,\sqrt{1\plus{}t^2}}{2t\plus{}1}\plus{}1}\right|}\right)\plus{}c\equal{}$

$ \displaystyle\frac{3}{5}\,\arctan\left({\tfrac{t\minus{}2}{\sqrt{1\plus{}t^2}}}\right)\plus{}\frac{\sqrt{6}}{60}\,\log\left|{\tfrac{\sqrt{6}\,\sqrt{1\plus{}t^2}\plus{}2t\plus{}1}{\sqrt{6}\,\sqrt{1\plus{}t^2}\minus{}2t\minus{}1}}\right|\plus{}c\equal{}$

$ \displaystyle\frac{3}{5}\,\arctan\left({\tfrac{t}{\sqrt{1\plus{}t^2}}\minus{}2\,\tfrac{1}{\sqrt{1\plus{}t^2}}}\right)\plus{}\frac{\sqrt{6}}{60}\,\log\left|{\tfrac{\frac{\sqrt{6}\,\sqrt{1\plus{}t^2}\plus{}2t\plus{}1}{\sqrt{1\plus{}t^2}}}{\frac{\sqrt{6}\,\sqrt{1\plus{}t^2}\minus{}2t\minus{}1}{\sqrt{1\plus{}t^2}}}}\right|\plus{}c\equal{}$

$ \displaystyle\frac{3}{5}\,\arctan\left({\tfrac{t}{\sqrt{1\plus{}t^2}}\minus{}2\,\tfrac{1}{\sqrt{1\plus{}t^2}}}\right)\plus{}\frac{\sqrt{6}}{60}\,\log\left|{\tfrac{\sqrt{6}\plus{}2\,\frac{t}{\sqrt{1\plus{}t^2}}\plus{}\frac{1}{\sqrt{1\plus{}t^2}}}{\sqrt{6}\minus{}2\,\frac{t}{\sqrt{1\plus{}t^2}}\minus{}\frac{1}{\sqrt{1\plus{}t^2}}}}\right|\plus{}c\,\stackrel{t\equal{}\tan{x}}{\equal{}}$

$ \displaystyle\frac{3}{5}\,\arctan\left({\sin{x}\minus{}2\,\cos{x}}\right)\plus{}\frac{\sqrt{6}}{60}\,\log\left|{\tfrac{\sqrt{6}\plus{}2\,\sin{x}\plus{}\cos{x}}{\sqrt{6}\minus{}2\,\sin{x}\minus{}\cos{x}}}\right|\plus{}c.\quad\square$

P.S. Unfortunately I couldn't see the ``straightforward'' of the solution which was given by ``yeahnigga'', but I thank him anyway.

grigkost nice solution :roll:
I couldnt see the straightforward solution too  :maybe:


limit
logarithms
function
real analysis
real analysis unsolved

[url=http://bildr.no/view/510553][img]http://bildr.no/thumb/510553.jpeg[/img][/url]
[url=http://bildr.no/view/510554][img]http://bildr.no/thumb/510554.jpeg[/img][/url]

I can not understand that denoted by red line . thanks

Remember that $ 2b_n \equal{} 2a_n \minus{} 1 \to 0$ for $ n \to \infty$; we have $ \lim_{x \to 0} \frac {\ln (1\plus{}x)} {x} \equal{} 1$, hence the notation $ \ln (1\plus{}2b_n) \sim 2b_n$.

On the other hand, in the proof for point (iii), $ a_n \minus{} \frac {1} {2}$ was squeezed to zero by $ \frac {a_2^{n\plus{}1}} {2} \to 0$, with $ 0<a_2<1$, and this is also true for $ n\frac {a_2^{n\plus{}1}} {2} \to 0$.
This implies that $ \lim_{n \to \infty} \frac {2b_n} {1/n} \equal{} 0$, and this is Landau's notation "small o": $ 2b_n \equal{} o(1/n)$. I will send you a material on "big O" and "small o".

[quote="b555"]how are u able to see such a small image?  how to increase its size?[/quote]

you can push the thumb.

The pic is in another web site. I just input the bbcode here.

I can not find how to upload pic to artofproblemsolving forum.

[quote="mavropnevma"]Remember that $ 2b_n \equal{} 2a_n \minus{} 1 \to 0$ for $ n \to \infty$; we have $ \lim_{x \to 0} \frac {\ln (1 \plus{} x)} {x} \equal{} 1$, hence the notation $ \ln (1 \plus{} 2b_n) \sim 2b_n$.

On the other hand, in the proof for point (iii), $ a_n \minus{} \frac {1} {2}$ was squeezed to zero by $ \frac {a_2^{n \plus{} 1}} {2} \to 0$, with $ 0 < a_2 < 1$, and this is also true for $ n\frac {a_2^{n \plus{} 1}} {2} \to 0$.
This implies that $ \lim_{n \to \infty} \frac {2b_n} {1/n} \equal{} 0$, and this is Landau's notation "small o": $ 2b_n \equal{} o(1/n)$. I will send you a material on "big O" and "small o".[/quote]

Thanks very much !

Then in the last formula how to get $ o \left( \frac{1}{2^n} \right)$ ? thanks

It is all about manipulations of the "little o" symbol.

We have proved that $ \ln (2b_n) \equal{} \minus{}(n\plus{}1)\ln 2 \plus{} o(1)$ where $ b_n \equal{} a_n \minus{} \frac {1} {2}$, hence $ \ln \left (2^{n\plus{}2}(a_n\minus{}\frac{1} {2}) \right ) \equal{} o(1)$. But this means $ \lim_{n \to \infty} \ln \left (2^{n\plus{}2}(a_n\minus{}\frac{1} {2}) \right ) \equal{} 0$, so $ \lim_{n \to \infty} 2^{n\plus{}2}(a_n\minus{}\frac{1} {2}) \equal{} 1$, or equivalently, $ \lim_{n \to \infty} \left (2^{n\plus{}2}(a_n\minus{}\frac{1} {2}) \minus{} 1 \right ) \equal{} 0$, which means $ 2^{n\plus{}2}(a_n\minus{}\frac{1} {2}) \minus{} 1 \equal{} o(1)$, so $ a_n \equal{} \frac{1} {2} \plus{} \frac {1} {4\cdot 2^n} \plus{} \frac {o(1)} {2^{n\plus{}2}}$.

But clearly $ \frac {o(1)} {2^{n\plus{}2}} \equal{} o\left (\frac {1} {2^{n\plus{}2}}\right ) \equal{} o\left (\frac {1} {2^{n}} \right )$ (it is easy to verify, through the definitions, that $ o(1)\varphi(n) \equal{} o(\varphi(n))$ for a function $ \varphi(n)$, and that $ Co(1) \equal{} o(1)$ for a constant $ C$).