Obviously conversation is [i]more distracting[/i] than holding the phone. This is because you interpret with your brain, also used for analyzing the road, but you hold the phone with your arm. That doesn't mean that holding a phone isn't harmless. [b][color=blue]**[/color][/b]

It's much more natural to look around/do head-turns when you're not holding something to your face. This makes it easier to stay alert and be aware of your surroundings. Depending on how you hold your phone, it may also be easier to glance to the side. There's also the bodily stress element from holding up a phone; the body is in a much more natural and relaxed position when it is not doing so.

Also, suppose the distraction took place, and now you're in a "close-call" situation. With two hands, its easier to utilize the weight of the arm more, it's easier to turn the steering wheel quickly, and in cases of miscommunication, it's possible to honk the horn. You can't do that with one hand.

These days, you can get wireless headsets for under 15 bucks.

And really, I don't see how this law could be intrusive/violate someone's rights/cause someone problems (I'm assuming it doesn't go into effect immediately). It's not like the government's limiting cell phone usage, it's just telling you not to hold it up to your face. Unless you meant the part about kids under 18 not being allowed to use cell phones at all while driving. But I think every parent would support that, to say the least.

[color=blue][b]**[/b][/color] -[i] Here's a mind experiment. Suppose you are driving normally. Now, remove one hand of your choice (whether it's resting or on the steering wheel) and grab on to the corresponding ear with the chosen hand. Are you still equally in control, alert, and responsive?[/i]

My wife is completely comfortable driving and talking at once but I simply cannot do it, headset or no headset (by the way, we both drive stick shift, so one free hand is never really enough in the city traffic). I suppose it really depends on a person whether he/she can drive safely and maintain a conversation at the same time. The problem is that to administer a proper test for everybody is too difficult and most people (especially the teenagers) tend to overestimate their abilities if the choice is left to them, so it makes sense to impose some restrictions on the cell phone usage while driving. The California law is surprisingly mild in this respect, in my humble opinion. I drove in LA and San-Francisco a few times back in late 1990s and I just do not see how one could possibly pay attention to anything else in that kind of traffic without compromising the safety of his driving. I don't think it became any better since then. Rural North Dacota can, probably, allow people to watch TV or sleep while driving, provided they lock their steering wheel in the straight position, set the cruise control, and set the alarm clock to wake up before they run out of gas, but in the areas with such heavy traffic any measure reducing the distraction should be welcome.

As to the general attitude towards restrictions, I believe that they are good when they reduce the danger you can create to other people. In this respect, the seat belt law makes much less sense than outlawing the cell phones (your wearing seat belts has no real impact on your ability to control the car and makes difference only to your own chances of survival if you crash, so why should anybody else bother or what right do they have to tell you what to do with your own life?). Still, it is not the seat belt law but the cell phone law that you find "disturbing". That puzzles me a bit  :roll:.

[quote="fedja"] Still, it is not the seat belt law but the cell phone law that you find "disturbing". That puzzles me a bit  :roll:.[/quote]

I don't like seat belt laws either, but I see your point.  I guess it's not really disturbing, but it's one of many many laws that have been passed in recent years to restrict our freedoms.  I don't like that trend at all.  Plus, it happened in California, a state that I always thought allowed a lot of personal freedoms.

[quote="jmerry"]Freedoms are not absolute- you do not have the freedom to harm other people. We have a multitude of laws based on this principle restricting dangerous behavior, notably including the whole system of vehicle licensing. The cell phone law is merely the most recent example, responding to changing conditions.[/quote]

True, but to me it seems the debate is over how much responsibility we should leave to the individual.  For example, they have developed a device that will not allow a car to start until the driver has passed a Breathalyzer test.  These seem like small things, but every time we take away an individual's right to make his own decisions, we come one step closer to a society in which all decisions are made by the government.

It's surprising to me that no one so far has expressed disapproval with with this law.

For one thing, laws like these don't seem particularly enforcable. If a cop pulls over some dude because it looks like he's on a cell phone, but the dude claims he was just scratching his ear, what happens then? Can they get a warrant to go through his call records or something? Or do they just takes the cop's word for it? And there's no way a texting ban could be enforced.

I think a cool idea would be to have a display on the back of your car that shows stuff like "This driver has been in 12 accidents this year" or "Danger: Woman at the wheel" or "Alert: Asian driver", etc. Or just a transmitter so you can see info like this on your dash. I'm not necessarily supporting any laws here, just saying this would be cool.

The use of cell phones can be a distraction, as fedja has stated that he finds he can't concentrate well enough, so I believe that the new act (whether in Cali or not) can help prevent accidents. The use of a hands free device is a good compromise to people who need cell phones as part of their daily lives.

In Maryland, they're trying to inact the ban of texting while driving. I do believe this is a good idea, since you can't really keep your eyes on the road and on the cell phone at the same time. A few of the girls at my high school can text like that, but it still isn't a good idea.

[quote="mysmartmouth"]True, but to me it seems the debate is over how much responsibility we should leave to the individual.  For example, they have developed a device that will not allow a car to start until the driver has passed a Breathalyzer test.  These seem like small things, but every time we take away an individual's right to make his own decisions, we come one step closer to a society in which all decisions are made by the government.[/quote]  So, let's be clear: you agree that driving while intoxicated is illegal and is incredibly dangerous to those around you.  However, you only support punishing those people who happen to drive past a police car while intoxicated (or, I suppose, ram into some other vehicle or a barrier or tree).  Is this right?  Because in a sensible system, we would seek to uniformly prevent people from engaging in extremely dangerous behavior (at least to others) and then not need either to take punitive measures or to spend as much time and resources peeling idiots and their victims off the pavement.

In other words: driving while drunk is a uniformly bad decision, and I have no problem whatsoever with laws that prevent people from making decisions that are always bad.  I also don't have problems with laws that impose an extremely minor inconvenience on drivers if it will encourage them not to run me over.  (I would, incidentally, be in favor of a ban on all cell phone use by drivers, although unfortunately such a ban would be largely unenforceable.  This may be a moot point in the near future, though, since no one will be able to afford to drive  :wink: .)

[quote]For one thing, laws like these don't seem particularly enforcable. If a cop pulls over some dude because it looks like he's on a cell phone, but the dude claims he was just scratching his ear, what happens then? Can they get a warrant to go through his call records or something? [/quote]  He searches your car and arrests you or gives your a big ticket -- warrants aren't necessary to search a car in circumstances like this.  (A warrant is needed if, say, your car is parked in your garage and the police want to search it as part of your house.)

[quote="JBL"]He searches your car and arrests you or gives your a big ticket -- warrants aren't necessary to search a car in circumstances like this.  (A warrant is needed if, say, your car is parked in your garage and the police want to search it as part of your house.)[/quote]

Could the law be extended to the point where enforcement officers have the ability to arrest you if they believe you have a cell phone that you're not using? I don't believe this is the case, however. This can be easily misunderstood - a cell phone falling out of a pocket, etc.

On a similar note of enforcement - how do police officers know whether you're wearing a seatbelt or not?

[quote="JBL"][quote="durt"]For one thing, laws like these don't seem particularly enforcable. If a cop pulls over some dude because it looks like he's on a cell phone, but the dude claims he was just scratching his ear, what happens then? Can they get a warrant to go through his call records or something? [/quote]  He searches your car and arrests you or gives your a big ticket -- warrants aren't necessary to search a car in circumstances like this.  (A warrant is needed if, say, your car is parked in your garage and the police want to search it as part of your house.)[/quote]

Well that seems to violate the Fourth Amendment, doesn't it? Then a cop could search anyone for whatever reason under the pretense of "they might have a cellphone." Not only that, but finding a cellphone isn't evidence that the driver was actually using it while driving. So in order to show in court that the driver was indeed using his phone while driving, it seems that prosecutors or somebody need to gain access to his call records to show that there was indeed a call in progress just before he was pulled over.

I am not in support of usage of cell phones while driving, unless it is completely necessary.  It can be a distraction to the driver and could cause an accident. It may also come at high costs for other not using their cell phones. For instance, if a person was driving down the road, looking at their cell phone and didnt notice a red light at an intersection, they could end up slamming into the side of another car pulling out who had the right of way... there are just so many possibilities for something to go wrong.

[quote="durt"]Well that seems to violate the Fourth Amendment, doesn't it? [/quote]  No.  You don't have a presumption of privacy while in your car, so the 4th Amendment doesn't apply in full strength.  For example, purely random traffic stops are unconstitutional, but sobriety stops are okay, and the use of drug-sniffing dogs during stops where there is no reason to suspect the presence of drugs is also okay.  Also, only "reasonable suspicion" (e.g., from seeing your hand held next to your ear for some duration, presumably) is necessary to initiate a traffic stop, not a higher standard.

(As usual, the [url=http://en.wikipedia.org/wiki/Traffic_stop]wikipedia entry[/url] is helpful for a basic understanding. [url=http://law.jrank.org/pages/13025/Delaware-v-Prouse.html]This site[/url] also has lots of good-looking info about the case holding purely random stops to be unconstitutional.)

Dude, if "seeing your hand held next to your ear for some duration" is now sufficient reason for a search, cops will be able to search you whenever they please. This is unconstitutional. And traffic stops are not the same as searching your car...

Cops already [i]can[/i] search you whenever they please.  Whether what they find can be used in a trial is, perhaps, another question, but there is nothing [i]right now[/i] preventing you from being pulled over whenever a cop feels like it.  And there is also very little preventing police from conducting illegal searches of your person -- try looking up [url=http://www.soc.umn.edu/~samaha/cases/cloud_dirty_secret.html]"dropsy testimony."[/url]  These are problems associated with overzealous or poorly-trained policing and a (particular brand of) conservative judiciary -- they have nothing to do with cellphone laws.

I think talking on a phone without a headset should definitely be a primary law, because it involves driving with one hand which is dangerous to you and others around you.

[quote="TZF"]It's just like how the seat-belt law is useless -- if you get pulled over, you can just put it on and claim the officer didn't see it.[/quote]

[quote]The seatbelt law became a primarily enforceable law in 2000, making it possible for police to stop drivers solely for failing to buckle up. When the law was only secondary, police could only ticket motorists for seatbelt violations if a driver was stopped for an unrelated offense.

This year’s slogan is an addition to the current slogan, "Click it or Ticket."

Seatbelt usage in Michigan took a huge jump in March 2000 from 70 percent to 83.5 percent shortly after the safety belt law changed to a primary law. [/quote]

http://www.pressandguide.com/stories/051803/D01FAJP.htm

The purpose of a seatbelt law (or a cell phone law, or a drunk driving law) is not to raise funds from violators, it's to protect people.  Seatbelt laws have been enormously successful at making seatbelt usage the norm:
[quote]seatbelt use nationwide increased from 11% in 1981 to 68% in 1997[/quote]
http://www.preventioninstitute.org/traffic_seatbelt.html

"Useless" is a seriously inaccurate description.

[b]Off-topic, but in response to above post:[/b]

Right, right, I should've been clearer. While the institution of seatbelt laws has, over decades, been extremely successful, I doubt the recent "promotion" of seatbelt laws into [i]primary law[/i] in Massachusetts has been useful. In particular, it doesn't seem to affect the seatbelt usage of my age group (teenage drivers) at all.

I think this is because the initial institution of the seat belt law felt like a big change, and came across as a new awareness campaign. The "promotion" of the seat belt law did not come across as a large change, and any message about safety and seat belts that it could send has already been sent. So, it doesn't impact people as much.

Most drives that are really experienced don't eat or anything else while driving. It's simply dangerous.

I think younger drivers with less experience are more of a problem. Some inexperienced drivers think they can multitask and still drive well.

It's not so much a matter of experience as it is one of common sense. Some people don't realize that you can't get away with things in heavy/chaotic traffic that you can get away with on an open road. Eating is one of them.

There comes a point after which preventive measures just become ridiculous. Banning consumption of food/beverages while driving, or driving while tired, isn't going to raise awareness, and it isn't going to be enforceable. At least with cell phones, there are call logs.

And crazychinesemaniac, I'm sorry about your mom's accident. I actually got rear-ended in a similar way just this Thursday -- I stopped at a red light, and the guy behind me must have spaced out for about 5 seconds, because he didn't even realize i was slowing down until after i had come to a stop and he was way too close. But he didn't seem tired or distracted from outside, he was just pre-occupied. You need to accept that some people don't have their head in what they do, and then drive defensively.

[quote="myyellowducky82"][quote="crazychinesemaniac"]Cell Phone use is just one thing that leads to distracted drivers. We should ban eating/drinking while driving, and driving while fatigued/tired. This is acctually personal for me, my mom was injured by a driver who was driving a semi. He missed a red light and slammed the back of the semi into my moms car. Luckily she was alright.[/quote]
Well, I don't agree with banning eating/drinking while driving, especially if the driver is highly expericened.
And if you eat at a red light, it's alright[/quote]

Well eating at a red light is ok, but even if it's a highly skilled driver they still increase their chance of getting into an accident. Also how can a cop tell if someone is a "skillful" driver". Is it really worth the chance just so someone can kill their stomach?

TZF, I agree that common sense is important, but even on an open road, eating can be dangerous, especially if you aren't a very good driver. Being on an open road makes it less dangerous to others, but it can still be dangerous to you.

Eating at a red light is okay, but if the light turns green and your not ready for it, it could still be potentially dangerous, especially at a crowded intersection.

[b]Superllamel[/b], NOBODY is arguing that eating/talking/etc while driving doesn't increase danger, or that it doesn't increase the chance of an accident.

What we [i]are[/i] arguing is whether or not these guidelines should be written into law, and whether or not they are too intrusive.

[u]My viewpoint:[/u]

1. Driving while talking on the phone without a headset [b]should[/b] be disallowed by law in order to raise awareness of the dangers of cell phone driving. I do not believe that asking people to keep their hands on the wheel (or at least, to keep them free) is a violation of privacy. People should be encouraged to use headsets if they need to take phone calls.

2. Driving while talking on the phone with a one-ear headset [b]should not[/b] be disallowed by law. It is silly to expect people to actually obey this law, and it is not a strong argument to present in terms of raising awareness. Also, in an instance in which a phone conversation does cause a distraction, I think the effect of the user's discretion in using the device greatly outweighs any inherent distraction caused by the usage of that device.

3. Driving while using headphones [b]should[/b] be disallowed by law, since it makes it difficult for the driver to hear emergency vehicles, car horns, and other audio signals and prompts. I believe this is already the case in many states.

4. Driving while eating [b]should not[/b] be disallowed. Eating belongs to a large group of distractions that, if entirely eliminated, will be unduly restrictive. Other distractions in this group include music, operation of a GPS, passengers, drinks, buttoning shirts/tying ties, writing/reading, etc. It is more practical to place more stringent penalties on those who cause accidents while driving with such distractions.

5. Driving while tired [b]should not[/b] be disallowed. There is no way to enforce this -- most people would wake up when they get pulled over. There are other ways of raising awareness on this issue.

[quote="mysmartmouth"]In what I consider a disturbing turn of events, CALIFORNIA has passed a law outlawing traditional cell phone (is that an oxymoron?) use while driving.  Drivers must use a hands-free device if they wish to chat on their cell phones.  Also, drivers under the age of 18 are prohibited from using cell phones while driving altogether.

Here's the [url=http://news.zdnet.com/2424-9595_22-208674.html]link[/url].

Studies have shown that what causes accidents is the distraction of carrying on a phone conversation, not the loss of a hand while driving.  Also, putting that aside, is it proper for the government to tell us we can't use our cell phones while driving?  Do you guys agree or disagree with this law?  Discuss.[/quote]

yes, its true that hands-off headsets are just as distracting and thus just as dangerous as holding a cell phone. this is one point in which this law is unreasonable.

but on the other hand, it is proper, IMO for the government to tell you that you can't use your phones. studies have shown that cell phones are more dangerous than being drunk. and they have passed laws against DWI, so...

[quote="JBL"][quote="buzzer11"]studies have shown[/quote]  Citation?  Link?[/quote]

Hmm...here's one article that says its as dangerous but not more. [url]http://bicycleuniverse.info/cars/cellphones.html[/url]


geometry
perimeter
number theory
relatively prime

The perimeter of triangle $ APM$ is $ 152$, and the angle $ PAM$ is a right angle. A circle of radius $ 19$ with center $ O$ on $ \overline{AP}$ is drawn so that it is tangent to $ \overline{AM}$ and $ \overline{PM}$. Given that $ OP\equal{}m/n$ where $ m$ and $ n$ are relatively prime positive integers, find $ m\plus{}n$.

I got this diagram: [geogebra]82794b56c18e4dc31c98f9beab8eb9356bdb4850[/geogebra]
(Ignore one of the grey dots on the hypotenuse) and I can't get anywhere with that.

[quote="GeoGebra"]Unknown command: foot[/quote]
 :( :( :(

Diagram's not working. I get the same error as Yongyi.


[quote="daermon"]how do you write a bar symbolizing a repeating decimal, as in $2.33333333333333...$?[/quote]

$2.\overline{3}$=2.\overline{3}


Let L(m) be the x-coordinate of the left end point of the intersection of the graphs y=x^2-6 and y=m, where -6<m<6. Let
r=[L(-m) - L(m)]/m. Then as m is made arbitrarily close to zero, the value of r...?

[size=75] I took the inverse of function, to get y= sqrt(x+6) (the plus or minus in front of the square root is irrelevant in this case), and the line is at x=m. So what we're trying to find is the limit as x approaches 0 of (sqrt(6-x) - sqrt(6+x))/x, which turns out to be -1/sqrt(6).

Hope that's right.[/size]


geometry perimeter circumcircle trigonometry

If the area of a square is x, then what is the perimeter of the square?

[hide]the area of a square is x the perimeter is one side multiplied by 4.  one side is $\sqrt{x}$ so the perimeter is $4*\sqrt{x}$ [/hide]

THat's correct.
In addition, if a square has a perimeter of x, then its area would be $\frac{x^2}{16}$. I'll leave the proof to someone else.

Let me add an extension to this. If you have a regular octagon of area $A$, what is the perimeter?

Consider a regular $N$-gon, $N \geq 3$.  Take circumcenter $O$, and draw lines to all vertices from there.  You now have $N$ congruent, isosceles triangles each of area $K$, and $KN = A$ by hypothesis.  Thus if $b$ and $h$ are the base and height respectively of one of the $N$ triangles, $\frac{1}{2}bh = K = \frac{A}{N}$.  Let us solve for $b$, as that will yield the perimeter.

First call the two equal angles of one of the isosceles triangles $\theta$ degrees.  It is relatively well-known that $\theta = \frac{(N-2)180^o}{2N} = \frac{(N-2)90^o}{N}$.  The area of the triangle is $K = \frac{1}{2}b(h) = \frac{1}{2}b(\frac{b\tan{\theta}}{2}) = \frac{b^2\tan{\theta}}{4} = \frac{A}{N}$.  Rearranging this produces $b = 2\sqrt{\frac{A}{N\tan{\theta}}}$.  Noting that perimeter $P = Nb$ and substituting the previous equation, we derive the final result:

$P=Nb=2\sqrt{\frac{AN}{\tan{\theta}}}.$

EDITED OUT:
This is way beyond the Basics forum though

people, please please PLEASE STOP right now! if you are going to include trigonometry stuff in this forum, please leave now and make a topic about it in High School Getting Started or somewhere more appropriate than The Basics

sorry, 4everwise tempted me by using the word "extension."  :D

[quote="nat mc"][quote="chess64"][quote="Barnacle"]Consider a regular $N$-gon,............[/quote]

what? :huuh:[/quote]
what? A regular N-gon has N sides of equal length and congruent interior angles, duh[/quote]

 :rotfl: No, I meant the rest of the post, except I didn't want to waste space by putting it all.  :)

 :rotfl:

[quote="chess64"] :rotfl: No, I meant the rest of the post, except I didn't want to waste space by putting it all.  :)

 :rotfl:[/quote]
did you see above from Barnacle's second post? he apologized so make sure to check back when you are at that level, you will understand pretty soon...

hah, I even edited out to save ur "math young" eyes

[hide]Area of Square: X
Each side of Square: Squareroot of x
Perimeter: 4*squareroot of x[/hide]


Let $ a,b,c,d$ be nonnegative real numbers, no three of which are zero. Prove that
\[ \sum \left( \frac{a\plus{}b}{a\plus{}b\plus{}c}\right)^2 \ge \frac{16}{9}\]
I am looking for a nice solution.
P/s: A solution with Cauchy Schwarz exists. :)

[quote="can_hang2007"]P/s: A solution with Cauchy Schwarz exists. :) [/quote]
As always does! :lol:

Anyway, does the following ineq holds??
For four non-negative real numbers, no three of which are zero,
$ \sum \left( \frac{a}{b\plus{}c}\right) \ge \frac{4}{3}$

$ \sum (\frac{a}{b\plus{}c})\equal{}X$

$ X\equal{}\sum (\frac{a^2}{ab\plus{}ac})$
$ X\cdot (2ac\plus{}2bd\plus{}ac\plus{}bc\plus{}cd\plus{}da)\ge (a\plus{}b\plus{}c\plus{}d)^2$ by Cauchy-Schwarz, so:
$ X\ge \frac{(a\plus{}b\plus{}c\plus{}d)^2}{2ac\plus{}2bd\plus{}ab\plus{}cd\plus{}ad\plus{}bc} \ge \frac{2(ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd)}{2ac\plus{}2bd\plus{}ab\plus{}cd\plus{}ad\plus{}bc}$
It's enough to show that last expression is greater or equal to $ \frac{4}{3}$, which leads us to proving $ ab\plus{}ad\plus{}bc\plus{}cd\ge ac\plus{}bd$ This is not very hard, I think.

[quote="Bugi"]It's enough to show that last expression is greater or equal to $ \frac {4}{3}$, which leads us to proving $ ab \plus{} ad \plus{} bc \plus{} cd\ge ac \plus{} bd$ This is not very hard, I think.[/quote]

What if $ a\equal{}1,b\equal{}5,c\equal{}1,d\equal{}4$? :wink:

[quote="Bugi"]which leads us to proving $ ab\plus{}ad\plus{}bc\plus{}cd \geq ac\plus{}bd$ This is not very hard, I think.[/quote]
It's wrong. Try $ a\equal{}0, b\equal{}2, c\equal{}\frac{1}{2}, d\equal{}1$.

Sorry, thinking now of a correct solution.

I think if my ineq holds, it can be used indirectly to solve can's problem.
My mind says it can be proved with the same procedure we use to prove Nesbitt's ineq.
:)

[quote="Sunkern_sunflora"][quote="can_hang2007"]P/s: A solution with Cauchy Schwarz exists. :) [/quote]
As always does! :lol:

Anyway, does the following ineq holds??
For four non-negative real numbers, no three of which are zero,
$ \sum \left( \frac {a}{b \plus{} c}\right) \ge \frac {4}{3}$[/quote]

The stronger inequality $ \sum\frac{a}{b\plus{}c}\ge 2$ holds.  :wink:
[hide="Proof"]
We have
\[ \sum\frac{a}{b\plus{}c}\ge\frac{(a\plus{}b\plus{}c\plus{}d)^2}{(a\plus{}c)(b\plus{}d)\plus{}2(ac\plus{}bd)}\]
Hence it remains to show that
\[ \sum a^2\plus{}2(a\plus{}c)(b\plus{}d)\plus{}2(ac\plus{}bd)\ge 4(ac\plus{}bd)\plus{}2(a\plus{}c)(b\plus{}d)\\
\iff \sum a^2\ge 2(ac\plus{}bd)\]
which follows from AM-GM: $ a^2\plus{}c^2\ge 2ac, b^2\plus{}d^2\ge 2bd$.[/hide]
So how do you use this to prove the original inequality in this thread?

It would be nice to put indexes in the sum notation. The expression makes no sense to me.

[quote="puuhikki"][quote="nayel"]The $ \sum$ notation usually denotes cyclic sums..[/quote]
Usually does not mean always. In mathematics, one should always define the notations if there is a possibility of misunderstanding. For example, I don't think that $ \sum_{i \equal{} 1}^{10}i$ is a cyclic sum. Please be exact when using $ \sum$-notation.[/quote]

You are right, puuhiki, but people often use this notation. You can use this  $ \sum_{cyclic}$

I think nayel, we can put this good ineq into action, its equivalent to Nesbitt's for four variables $ a,b,c,d$.
$ \sum_{cyc} \frac {a}{b \plus{} c \plus{} d} \geq \frac {4}{3}$

For non-negative $ p,q,r$ I think the ineq holds, if $ p \plus{} q \plus{} r \geq A$, $ A$ being a real no., then $ p^2 \plus{} q^2 \plus{} r^2 \geq A^2$

For the first ineq, just consider the fact:
$ (a \plus{} b \plus{} c \plus{} d)\left(\sum_{cyc} \frac {1}{b \plus{} c \plus{} d}\right) \geq \frac {16}{3}$ which is true by $ A.M. \geq H.M.$
 :)


Sa se arate ca orice triunghi poate fi inscris intr-un echilateral.

se construieste un arc de cerc pe fiecare latura a triunghiului, astfel incat fiecare punct pe arc asta sa subtinda 60 de grade pe latura. cele 3 cercuri formate in acest fel se intalnesc intr-un singur punct, $P$. unim punctul $P$ cu cele trei varfuri $A, B, C$ al triunghiului, si apoi ducem perpendiculare prin $A, B, C$ la $PA$, $PB$, respectiv $PC$: care linii determina un triunghi echilateral in care e inscris $ABC$


topology
number theory
prime numbers
advanced fields
advanced fields unsolved

1) An arithematic progression is a subset of $\bf Z$ of the form $a+b \bf Z$, where $a \in \bf Z$ and $n \in \bf N_{0}$. Take as basis for the topology on $\bf Z$ all arithematic progressions.
2) Show that an arithematic progression is also closed.
3) If there are only finitely many prime numbers, then is $\{-1,+1\}$ open.
4) Conclude.
5) Smile :D

in the same spirit:

Dirichlet's theorem on arithmetic progressions <==> $\hat{\mathbb{Z}}^\times = \overline{\mathbb{P}}\setminus \mathbb{P}$, where $\mathbb{P}$ denotes the set of prime numbers.

Some details (for the original [i]problem[/i]) may be found [url=http://www.zblmath.fiz-karlsruhe.de/MATH/general/erdos/textpdf/aigzieg/aigzieg.pdf]here[/url].


function
superior algebra
superior algebra solved

Does there exist an algebraic operation * defined on N ( actually * is a function *:NpN>N) such that (N,*) is a group?
But if we replace N with NpN?
(Here i denoted NpN the cartesian product N with N)

try NxN instead of NpN (or N2).

for your problem the answer is simple. of course you can: take a bijection from Z to N (say f(x)=2x if x \geq 0 and f(x)=-2x-1 if x<0) and transport the group structure of (Z,+) into (N,*).

That is if f(a)=x and f(b)=y then x*y=f(a)*f(b)=f(a+b). Check to see if 4*3=0! :D

Same goes for Nk (only that it's a little bit more complicated to find and actually write a mapping)

(it's the second time I'm posting this since the first window I accidentaly closed :))


limit
ceiling function
calculus
calculus computations

For real numbers $ x$, let $ f(x)$ be the least integer that is greater than or equal to $ x$. Let $ a,\ b$ be positive real numbers, find the maximum value of $ c$ for which $ \lim_{x\to\infty} x^c\left(\frac{1}{f(ax\minus{}7)}\minus{}\frac{1}{f(bx\plus{}3)}\right)$ coverges, then find the limit value.

$ x^c\left(\frac {1}{ax \minus{} 6} \minus{} \frac {1}{bx \plus{} 3}\right) \le x^c\left(\frac {1}{\lceil{ax \minus{} 7}\rceil} \minus{} \frac {1}{\lceil{bx \plus{} 3}\rceil}\right) \le x^c\left(\frac {1}{ax \minus{} 7} \minus{} \frac {1}{bx \plus{} 4}\right)$ for large $ x$.
The maximum value of $ c$ is $ 1$ and the limit value is $ \frac {b \minus{} a}{ab}$ ($ b \ne a$) by Squeezing Theorem.

If $ b \equal{} a$, $ x^c\left(\frac {1}{\lceil{ax \minus{} 7}\rceil} \minus{} \frac {1}{\lceil{ax \plus{} 3}\rceil}\right) \equal{} x^c\left(\frac {10}{\lceil{ax}\rceil^2 \minus{} 4\lceil{ax}\rceil \minus{} 21}\right) \to \frac {10}{a^2}$ for $ c \equal{} 2$.

kunny, do you want to enter to Tokyo Institute of Technology?


[quote="nr1337"]What will be covered in the Complex Numbers Math Jam?[/quote]

That is in fact a mystery that has not yet been discussed.  All we know for certain is that complex numbers will be discussed...

[quote="MCrawford"]That is in fact a mystery that has not yet been discussed. All we know for certain is that complex numbers will be discussed...[/quote]

How very profound, Mr. Crawford! I never would have guessed.

[quote="MathFiend"]
How very profound, Mr. Crawford! I never would have guessed.[/quote]

Well, I was feeling very much in a state of profundity.


analytic geometry
geometry
Pythagorean Theorem

Hello, I am new to this forum and new to Geometry. I have had the following (Quickly) explained to me but I am not understanding how this problem is solved, can somone please explain or show me?

If I can get the first one, it will help me with the others . Thanks

Congruence Use the Distance Formula to decide PQ =QR.   [b](Note:This line __ is on top of PQ  and QR,and ~ is on top of the = Equation symbol)[/b]

P(4, -4)
Q(, - 6)
R(- 1,3)

Thanks Again

hello, use the formula
$ \overline{PQ} \equal{} \sqrt {(x_Q \minus{} x_P)^2 \plus{} (y_Q \minus{} y_P)^2}$
Sonnhard.

Wow,that was fast. My mom and I have been trying to figure this out since yesterday. Thank you very much.

Well we tried to piece this together but we are still having problems understanding. If possible can you break it down and explain ? Thanks Again.

To find the distance between two points A and B use this formula.

Let distance = d.

$ d\equal{}\sqrt{(x_A\minus{}x_B)^2\plus{}(y_A\minus{}y_B)^2}$

If you square both sides, you might note its resemblance to the Pythagorean Theorem.

I'll post the solution if you need it. (What's the x-coordinate for Q?)

i think you are missing a coordinate.

these formulas may seem pretty complicated but its actually a little intuitive.

Consider 2 points.  Now Draw a vertical line through 1 of them, and a horizontal through the other, and a line segment between the two points.  What do you get? A right triangle.  Note that the length of the segment is the hypotenuse  of the triangle.  The legs of the right triangle is just the difference in the x-coordinates and y-coordinates.

Now since you know the length of the two legs, you can find the hypotenuse, by pythagorean theorem

don't memorize this formula just think of it as the Pythagorean theorem

hello, this formula is the Pythagorean theorem, you must only put the coordinates in this formula. What is the $ x$ coordinate of the point $ Q$?
Sonnhard.

If your still post the solution on this on It would be great. I am new to geometry and i guess we need a complete example for it to click. Thanks a whole lot.

I have several formulas in my text book,but I think I'm gonna have to check with my teacher,because we cant seem to figure out what the x-coordinate for Q is.

 But [u]tomorrow[/u] when I do,I will get back with you. We really like the way you make thinks so simple to understand. Thanks again.

How do you not know the coordinate? Is it not there, or is it something you can't understand?

Maybe the problem is asking to [b]find[/b] the x-coordinate of Q that will make PQ=QR.

[hide="If so:"]

Using x as the x-coordinate for Q, plug in the coordinates into 2 separate distance formulas and set them equal to each other.

It should look like this:  $ \sqrt{(x-4)^2+(-6-(-4))^2}=\sqrt{(-1-x)^2+(3-(-6))^2}$

The first formula is the length of PQ and the second is the length of QR.

Simplifying the formulas, we get:

$ \sqrt{(x-4)^2+(-6-(-4))^2}=\sqrt{(-1-x)^2+(3-(-6))^2}$
$ \sqrt{(x-4)^2+(-2)^2}=\sqrt{(-1-x)^2+9^2}$
$ \sqrt{(x-4)^2+4}=\sqrt{(-1-x)^2+81}$
$ \sqrt{x^2-8x+16+4}=\sqrt{x^2-2x+1+81}$
$ \sqrt{x^2-8x+20}=\sqrt{x^2-2x+82}$
$ x^2-8x+20=x^2-2x+82$
$ -8x+20=-2x+82$
$ 20=6x+82$
$ -62=6x$
$ x=\boxed{\frac{-31}{3}}$[/hide]


induction
number theory unsolved
number theory

Prove that for different choices of signs $ \plus{}$ and $ \minus{}$ , the expression $ plus or minus1 plus or minus2 plus or minus 3 plus or minus 4.......plus or minus (4n\plus{}1)$ yields all odd positive integers $ \le (2n\plus{}1)(4n\plus{}1)$.

Please do the Latexification correctly
[quote="Aravind Srinivas L"]Prove that for different choices of signs $ \plus{}$ and $ \minus{}$ , the expression $ \pm 1 \pm 2 \pm 3 \pm 4\cdots \pm (4n \plus{} 1)$ yields all odd positive integers $ \le (2n \plus{} 1)(4n \plus{} 1)$.[/quote]As for the problem: Use induction.
Assume base case as $ (2n \plus{} 1)(4n \plus{} 1)$.
Then the induction step:
if $ k$ (positive) can be represented like that, so can be $ k \minus{} 2$.
[hide="Details"]If there is $ \plus{} 1$ in the representation of $ k$, we convert it to $ \minus{} 1$
If there is $ \minus{} 1$, there is some $ t$ such that there is $ \minus{} t$ and $ \plus{} (t \plus{} 1)$ in the representation.
(if not: every positive integer in the representation has a minus sign before it. Impossible because $ k$ positive).
We convert that to $ \plus{} t$ and $ \minus{} (t \plus{} 1)$[/hide]

I cannot understand your solution. PLease explain elaborately.

[hide]First if we take all numbers with the $ \plus{}$ sign, the value is $ (2n \plus{} 1)(4n \plus{} 1)$
That is: $ (2n \plus{} 1)(4n \plus{} 1)$ can be represented in the form.
Now suppose a positive number $ k$ can be written in that form
(that is some $ \plus{}$ and some $ \minus{}$ in $ \pm 1\pm 2\pm 3 \cdots \pm (4n \plus{} 1)$ adds upto $ k$)
We shall prove that $ k \minus{} 2$ can be written in that form too.

[u]Case 1. Suppose the representation of $ k$ has $ \plus{} 1$ in the beginning.[/u]

If we change that $ \plus{} 1$ to $ \minus{} 1$, the sum will decrease by $ 2$. Hence we find a representation of $ k \minus{} 2$ in this way.

[u]Case 2. Suppose the representation of $ k$ has $ \minus{} 1$ at the beginning. [/u]

Look at the sequence of signs of the representation of $ k$.
For example, say it is like $ \minus{} \minus{} \minus{} \plus{} \plus{} \minus{} \plus{} \cdots \plus{} \plus{} \minus{} \minus{} \minus{} \plus{}$

It begins with $ \minus{}$.
Since $ k$ positive there must be a $ \plus{}$ in that sequence.
If we look at the first such $ \plus{}$, it's preceding one must be $ \minus{}$. Suppose that $ \minus{}$ is the $ t$-th sign in the sequence.
So there is a $ \minus{}$ before $ t$ and a $ \plus{}$ before $ (t \plus{} 1)$ in the sequence.
(In the above example $ t \equal{} 3$)

We change the $ \minus{}$ before $ t$ to $ \plus{}$ and change the $ \plus{}$ before $ (t \plus{} 1)$ to $ \minus{}$.
The total sum decreases by $ 2$. Thus the sum becomes $ k \minus{} 2$.

So there is a representation of that form for $ k \minus{} 2$ whenever there is one for $ k$.
Decreasing $ 2$ consecutively from the base case, we cover all positive odd integers.[/hide]


The next math jam is Intro to Algebra.  I am taking that course.  Is that a class or like a descriptive seminar?  I've already taken one class so I know how the Chat room works.

It is a description of the course for people who might be interested in taking it.  It also serves as an introduction to our Online Classroom for those who have never used it.

If you've already registered for the course and you have already used our Online Classroom, then there's no need to attend.


Congratulations! We just made history. (“For Elise”, “Moonlight” by Beethoven).

Dear Colleagues,

First of all, please allow me to congratulate each of you who answered my questions, as well as this wonderful Forum and my self, because in the last week we altogether started at least a new chapter in the history of music and, maybe, even a new era!

I am more than serious, let me, please, show only a few reasons to have said like that:

1.	Never before in history the information about the most elusive substance of music – the individual volume of each given note in relation to the previous note and the following note (hereafter shall be called “Note Strength”) – has been transmitted (sent), received, processed and sent back in plain text, without any real sound at all.
     (Even more - it was done with a precise accuracy!)

I am especially pleased to announce this because many high-ranked specialists have told me that this is impossible in principle. Fortunately, they have been no more right than the renowned scientists who “had convinced” (it is a real, well known historical fact) society that flight on any apparatus heavier than air was absolutely impossible, and they did it not before but AFTER the Wright brothers had already crossed the Atlantic.

Our first tests were as short and simple as the first in history audio recording (“Mary Had A Little Lamb”, performed by Mr. T. A. Edison himself) and the first Radio – Message  (“SOS”). However, the first short step opens endless ways to new horizons. Look at only a few randomly chosen unprecedented opportunities:

2.	Literally tomorrow (because everything is done except anti-piracy protection) you can download a new software “Music teacher”. Please, switch on your imagination! Let us say you have decided to learn our beloved “Fur Elise” by Beethoven. Your PC invites you to choose the “base-model” out of 100 top rated recordings. E.g. your choice is “Vladislav  Tashkenazy” because you like his conception as a whole. However you are unhappy with some details of this famous recording.

Therefore you press “Edit”. What can be different in Piano Music? – Only 3 things: Pitch, Timing and Note Strength. (Let us mention “Damper” and “Piano” pedals as well but they are not engaged in this particular software yet, it will be done after tomorrow).

 Now we have to edit “Pitch”, because Tashkenazy recorded D seven times instead of Beethoven’s E (at the end of each “sentence”, like bars 7, 15, 29 etc.)  You have to replace all of them with “E” and to add the missed “C” at the final resolution of the piece.

Your reason for this action is this: Beethoven could not be illiterate in Harmony (just because his and Mozart’s music are the basis of the Classical Harmony) and write the resolution of the Seventh of D7 in the “up” direction instead of the legal resolution “down”. Moreover, he could not leave this “seventh” unresolved at all at the end. Beethoven’s autograph and the very first edition prove it. By the way, isn’t it a shame, in your opinion, for the great Mr. Tashkenazy to choose the faulty edition for his recording? We know that he graduated from the best Conservatory and had a really great teacher G. Althaus.

Your next step is “Timing”. In the bar 45 (the first bar of F-Major episode) you prefer to brake the F-Major “on” beat instead of “before”, because the grace notes were not crossed by Beethoven.

The last correction is “Note Strength” (not the volume of the music in whole, of all notes together). You prefer to make more prominent the lower voice in bb. 52 and 54 and to play more shiny the entire top notes in this episode. Now everything is done.

The next step is  “Preview” (“Pre-audition”). You have to check whether you are happy with the result of your corrections or not – your PC plays for you back “your” variant of “Fur Elise”. In fact this variant is already a unique piece of the art (nobody played exactly like this yet) and belongs only to you, so you can claim even the “author rights” on it. After you have auditioned and approved your own way to perform or done some other correction – you press the button “Save”.

From this moment on, your PC listens to your practice and prints or displays each time the note-text with all your mistakes (discrepancies  with YOUR plan) marked in red color: wrong notes, timing, note strength (similar to the  “Spelling Check” in “Microsoft Word”). Undoubtedly, you can change your “performing plan” in your PC at any time.

Isn’t it great to listen to your self at any the time “from the audience’s seats” and be absolutely objective without the influence of your daily tiredness, irritation, nervousness, and mere forgetting of details of your plan? Now each of us will be able to afford lessons with the best professors on a daily basis (as long as their recordings are available for your PC). Good luck in bringing YOUR plan of performance to life!
Would you like to download this software? – Let me know.

 3. From now on each composer can indicate the desired strength of any note accurately instead of taking chances that performers probably will be able to understand the author’s intentions. The performers, in turn, do not need to guess anymore what exactly the composer meant (at least in some tricky spots).

4.	Instead of the quite common situation “blind under the guidance of blind” each student can submit his “Plan of Performing” of any piece and discuss it with the teacher of any level or even get approval/corrections from highly qualified board. As a result, s/he will know “how to play” in all the details BEFORE even one hour of his/her time was spent for inefficient exercising.

Inefficient - because it is simply impossible to work effectively without being aware of the direction one wants to go and what exactly is the goal? Presently, the majority of musicians are using the time-consuming “trial-and-error method” instead. Teachers will profit also from the unprecedented opportunity to learn and teach from a complete disclosure of all artistic details of any performance as long as the recording of this performance is available on their PC’s.

5.	A new scientific field – theory and comparative history of piano performing - will come into musical schools and universities with oceans of researches and dissertations. “Any science is as much science as it can be described mathematically".

Are you sure that Math and Music have nothing in common? Okay, I will not remind you that our A1 is simply 440 vibration per second as well as A2 –880, that each of our notes is 2/1, 1/1, 1/2, 1/4, 1/8, etc. I will only refer you to the excellent education system of ancient Greece and Rome: any student of the math faculty was a musician automatically – music was a substantial part of the math syllabus.

6.	If it is so easy (now) to pass on all the fine details of your artistic vision to your students, then it will be even easier to pass the same ideas into a fully obedient computer. And your  Yamaha’s “Disklavier” will play perfectly any piece of any difficulty exactly in the way you suggest, even in the way that human can not dream about because of their limited nature.

7.	 Composers will definitely exploit this feature – they will write an absolutely new kind of music designed specially “for computerized performing”, without observing of all the human limitations.

8.	What is wrong in the suggestion that maybe in the future the computer will play piano (physically) more often than humans? Is it disturbing for us today that no single book in our libraries is written by the hand of human – all of them are just printed by “stupid and insensitive machines”, by robots?

 However, nobody sees here any reason to say that the human soul, thoughts and feelings have disappeared from the books because of this. All of these three are still living in each book (if it is a good one, of course). Why mustn’t the same logic be applied to music?

To understand the real significance of our tiny event and the fascinating future which INEVITABLY will follow, (even if all of us have not wanted it) we have to look back.

Perhaps, music is as old as humankind itself. However, we will never know the music performed at the solemn openings of Egyptian pyramids or during the triumphal closing ceremonies of Ancient Olympics. Their music was not written down not because it was really impossible to do so at their time, but just because the idea of writing down music never happened to strike anybody’s head. The total absurdity of such suggestion was so obvious for everyone, that poets proudly declared: "To describe the beauty of (their) fair lady by words is as impossible as to write music on the paper!"

However, after thousands years of musical illiteracy, the day came when somebody invented the way to make visual musical Pitch (frequency) using specially shaped crochets. Since then, we can see at least something in the darkness of the musical past. However, the real dawn of a new era in music came later, when man invented the way to visualize Timing as well. Could you imagine anything in the present musical world without this upgrade?

E.g., let us try (imaginatively) to learn or teach Prelude and Fugue by Shostakovich.  How long would it take for students to "catch by ear" and memorize the right pitch and timing of each note and then play back this super-complicated piece if no script was available? If we want a decent level of accuracy, then the majority of the teachers would say that even a hundred years would not be enough for this single work. I can only join them in this estimation.

 The invention of musical literacy has opened unprecedented opportunities. However, is the present musical notation perfect or at least satisfactory? No, not at all.
Music is a language based on correlations between tones. Any of these musical tones has 3 dimensions: Pitch, Timing and Note Strength (=INDIVIDUAL volume of each particular note). However, the existing system of musical notation provides an accurate indication only for Pitch and Timing, but completely ignores the Note Strength. (Except some rare indications of accents or sforzando, but even they are unclear: to which particular note in a chord we have to attribute them).
 We have no confidence regarding each note - whether it should be played louder or softer in relation to the previous and the following note, and by how much? For example, between just two random consecutive notes, or between two notes in particular chord: which one of the two should be softer or louder?  No single existing textbook will answer these questions either.

At the same time, any musician has to “answer” these questions practically as many times as there are notes that he/she has to play. Since all pianists play (at least should play) the same notes on the same beat, it is obvious that the different individual volumes of the same notes make the only difference between the best and the worst pianist in the world. Any (sometimes even the smallest) violation of the musical rules and laws: wrong pitch, timing etc., ruins the music. And any examiner can confirm the fact that wrong correlation in volume between notes can be as annoying as wrong pitch or timing. "Let any pianist play for me not one, but two notes and I can say immediately whether this is a good musician or a bad one." (H.Neuhaus)

Proper stressing or softening in music is as important as the proper stressing in the words of any language. (Nobody will understand even very familiar words with wrong stressing such as: AmerIca , TorontO, SonAtina, etc. Similarly, no one will understand or enjoy music with wrong stressing or softening either).
However, this situation changes dramatically if we have a very simple (even for 5-7year old beginners) method based on an accurate indication of note strength of any given note in a chord or melody. This method provides the same accuracy for Note Strength as it has been provided for Pitch and Timing. Introduction of this method does not take long: all that we have to do is to "calibrate" our students’ and our own ears in the same way. There are many variants of "dynamics scale" and all of them work. Let us look at one of them:
It is based on an analogy with the Celsius temperature scale (from zero to one hundred degrees). One hundred degrees corresponds to the maximum volume, and zero – to silence. It seems difficult to define one degree but in fact it is not: one degree is the smallest difference in volume that we can feel. (Feel, not “hear”, because we do it with our fingers as well, and students understand and “feel” it better.)
To teach a student to use this "dynamics scale", we ask him/her to play several pairs of notes with exactly the same volume. After the student has done it faultlessly, we ask him/her to play the same pairs of notes with the tiniest difference between them that he/she can feel/hear. Now that we’ve got the "minimal step" in the volume scale, all the rest is simple and logical: two degrees, three and so on. Even the very beginners do it with amazing accuracy. It proves that Mother Nature has given us a wonderful instrument for these tiny gradations.

The next step is to write the proper "temperature" (at least in all the complicated spots of the piece) and our work is done. Any "uncivilized" student turns into a sophisticated professional instantly.

Accomplished musicians do not need detailed and accurate indications of the volume: they have their gift, intuition and culture (from their education and experience) instead. Unfortunately, not everyone is gifted, accomplished, educated and experienced at the same time. The accurate indication of note strength effectively fills the gap between "the intuition of the gifted" and the perfect vacuum of musical knowledge in many students.
The accurate indication of Note Strength gives all that is necessary to play beautifully. A good mother does not teach her kids how to play outdoors; she knows that the kids will never remember her instructions for more than 5 minutes and always will play in their own way. Therefore, she insists on only a few things that “should not be done under any circumstances”. (E.g. “Newer play the ball on the road”!)

The same idea has inspired a few short and easy rules based on the accurate indication of Note Strength:  “How to avoid some typical mistakes in musical performing". (20 - 30 of these rules cover roughly 50% of such mistakes and work for any music that students will play in their lives. After the student has learned a practical implementation of these rules, all the new pieces will need sometimes only 1-3 new rules for each, and quite often no new rules at all).

Even very beginners can play wisely, with decent, convincing phrasing after a few simple and easy rules, symbols or numbers have turned them on the right path. However, the best news is that after a short period of time students no longer need all these numbers or symbols (a kind of "Dynamics Alphabet"), because they have understood the basic logic of the musical language. Now they can apply the same rules to each similar situation in other pieces of different styles.

This method allows the student to look for the first time at any score and write (or imagine) immediately, WITHOUT playing or singing, the proper volume for each note in any voice/part or chord. Traditional teachers usually refuse to fulfil such an assignment and insist that it is impossible in principle. However, it definitely does not mean that they do not know the proper Note Strengths – many of them can play very beautifully, so they DO know it practically. But their knowledge is SUBCONSCIOUS, and they can not express it in words. This method will help them to do it easily.

This method is very helpful in immigrant countries such as Canada and the USA. I have worked with many Portuguese, Chinese, Japanese, African, and other languages speaking students without knowing 10 words of their languages and they (without knowing English or any of my languages) have fulfilled all of my instructions. This system has proved to be extremely useful in "non-standard" situations: when I was teaching blind students, working with groups of children and individuals starting from the age of 3, etc.

This method is especially valuable nowadays. The method is perfectly compatible with modern technologies, recently developed software, and scientific and musical equipment. Successful introduction of this method together with technical innovations can ultimately boost the whole musical culture and bring high musical standards to anyone who loves music.

I have been using this method for over 40 years and it has worked equally well for conservatory and university students, international opera stars, competition winners, and for absolutely uneducated people. (We always had to invite them for our opera and oratorio productions, when I worked as an opera and resident chorus master.)

Almost everyone around us has tried (at least once in a lifetime) to make friends with music, and almost all of them have failed and said "never again" after their contact with our existing musical pedagogy. Why? It has happened because we could not teach them (before) simply and reliably. Now, we can, and we no longer need to lose 99% of our potential students.
I realize that for many of us it sounds too good to be true – any conventional teacher should “know perfectly well that it is impossible”. Therefore, I invite everybody to have a look at this method in action (absolutely free) at any time.

Vladimir Dounin.

Please do not revive. It is a 7 year old post, and you revived about 5 other posts. Stop reviving please.


modular arithmetic
number theory unsolved
number theory


Let a positive integer such that for all n>1 a^n+1 is palindromic. Prove that a is a power of 10. It is easy, but I like it.
   Harazi

Suppose a is not a power of 10.
It is known that if a is not a power of 10, then for every digit in 1..9, there is a power of a, whose decimal expansion starts with that digit.
Becasue a n+1 is palindromic, we have the a similar property for the last digit of powers of a:
So for every digit in {0..8} there is a power of a ending in that digit. From the fact that there is a power ending in 0, we get 10|a, so the last digit of every positive power of a is 0 - a contradiction.

I too like this problem. Was it inspired by my (and Pliviu's) problem sequnce&prime?

As I told you, the problem is simple, but nice. And the answer to your question is negative. I created the problem to years ago. Your solution is easier than mine. I proved that if a is not a power of 10, then the first digit of a^n+1 is not a periodical sequence.

Why do you always have to add that a problem is easy/trivial/hard whatever, and that you are stupid if you can't solve it directly? :-) That you think it's trivial doesn't imply that everyone else think it is. Also, when you say you're stupid, do you expect any reply from others concerning "if you are stupid or not"?

I really like to classify problems. Of course, this point of view is very subjective, but I really don't think I have exagerated till now. And I also don't think that I have said about a difficult problem that it is trivial. When we speak about a certain problem, we must have some referrences: beauty, difficulty, ideas, etc. I try to make every interesting problem an individual. I do like easy problems, with the condition that they say something or use a nice idea. But nobody will stop me to say that a problem which asks to solve a second degree equation or something like that is not trivial. I don't want to misunderstand: some problems are fundamental (like solving a second degree equation), but it is trivial comparing with other problems.
   About another remark of Midspa. First of all, I have never said that I'm stupid if I couldn't solve a certain problem directly, but when I asked stupid questions, when I gave wrong solutions or when I didn't remark something really obvious. And of course I do not expect any answer from th others. I think I know what you mean: I say so in order for the others to say that I'm not. I'm sorry to say , but if this is your opinion, then it is wrong. And the answer that I expect is not from the others, but from me.
   Anyway, if this really bothers you, I will try not to use anymore. But I will not stop to estimate the difficulty or beauty of the problems.

I'm not telling you to stop with anything, sorry if the statement was ambiguous. I was just trying to indicate that if you rate a problem, that's difficult for some, trivial then they might get upset. And about the "stupid" part, I don't think this forum is about who is the smartest or most stupid, I do think though that people make mistakes -- not because they are necessarily stupid. So that kind of comments might be rendered obsolete by some, and as arrogant by others.
   And to mark my point, I don't think you're arrogant, no way, and you're a great problem solver. At last I also think that giving opinions is great, especially on their "value".


Does there exist a function f such that for all x the following equality holds. $ f(g(x))\equal{}[x],  g(f(x))\equal{}\{ x\}$
$ f\colon \mathbb{R}\to\mathbb{R}$  $ g\colon \mathbb{R}\to\mathbb{R}$

______________________________
Azerbaijan Land of the Fire  :rotfl:

[quote="Pirkuliyev Rovsen"]Does there exist a function f such that for all x the following equality holds. $ f(g(x)) \equal{} [x], g(f(x)) \equal{} \{ x\}$
$ f\colon \mathbb{R}\to\mathbb{R}$  $ g\colon \mathbb{R}\to\mathbb{R}$  [/quote]

1) Let $ x,y\in[0,1)$. If $ f(x)\equal{}f(y)$, we get $ g(f(x))\equal{}g(f(y))$ and so $ \{x\}\equal{}\{y\}$ and so $ x\equal{}y$

2) $ g(f(x))\equal{}\{x\}$ $ \implies$ $ f(g(f(x)))\equal{}f(\{x\})$ and $ f(g(x))\equal{}[x]$ $ \implies$ $ f(g(f(x)))\equal{}[f(x)]$ and so $ f(\{x\})\equal{}[f(x)]$. So, for $ x\in[0,1)$ : $ f(x)\equal{}[f(x)]$

So, if we consider the restriction $ f_r$ of $ f(x)$ to $ x\in[0,1)$ :

The second line tell us that $ f_r(x)$ is a function from $ [0,1)\to\mathbb Z$
The first line tell us that $ f_r(x)$ is injective

And so we have an injection from $ [0,1)$, uncountable $ \to \mathbb Z$, countable.

So no such function exists.


Can anyone explain:

why there is no friction acting on a wheel that is purely rolling with a constant velocity?

why the direction of the friction acting on a purely rolling and accelerating wheel on a horizontal plane is in the same direction that the acceleration is in?

why the direction of friction of a wheel purely rolling down an incline is opposite to the direction of motion?

The first question is not entirely correct.  Constant velocity means 0 acceleration, which means that the net force acting on the wheel is 0.  That's net force, not necessarily total force.  So there might be friction, but it's being negated by some other force in the opposite direction.

For the third question, friction opposes motion so it will naturally be in the opposite direction.

for the third question, say the wheel is rolling down a ramp counterclockwisely (is this a word?)...when you say the frictional force would "naturally" be up the ramp to oppose the "down the ramp" motion of the wheel is that because gravity is effectively acting on the wheel's center of mass, so friction would be the only force giving the wheel torque, so it has to act in the "up the ramp" direction to make the wheel rotate down the ramp counterclockwisely?

If this is correct, then how is the case in the second question possible...if a wheel is accelerating forward, directionally to the right for example, it's rotating clockwisely, so doesn't there have to be a frictional force at the contact point, between the wheel and the surface on which it is rolling, acting in the leftward direction that provides torque that angularly accelerates it clockwisely, thus making it rotate clockwisely and accelerate to the right?  (the frictional force would be the only force giving the wheel torque right?...because the only other force on it is its weight which effectively acts on its center of mass, so it's acting through the wheel's axis of roation...moment arm is zero so no torque)

I think that the matter here is sliding friction. Rolling friction is something  quite diverse.

If there'd a sliding friction, the opposite force should actuate in the same line of action, that means it acts in the contact of the wheel with the supporting surface.
Seems the question means that the forces on the wheel are only it's weight and the contact force from supporting surface.

yea when i stated the questions i mentioned the wheels are purely rotating....so no sliding friction only static


analytic geometry
calculus
integration
binomial coefficients
combinatorics unsolved
combinatorics

Let be four positive integers $m, n, p, q$, with $p < m$ given and $q < n$. Take four points $A(0; 0), B(p; 0), C (m; q)$ and $D(m; n)$ in the coordinate plane. Consider the paths $f$ from $A$ to $D$ and the paths $g$ from $B$ to $C$ such that when going along $f$ or $g$, one goes only in the positive directions of coordinates and one can only change directions (from the positive direction of one axe coordinate into the the positive direction of the other axe coordinate) at the points with integral coordinates. Let $S$ be the number of couples $(f, g)$ such that $f$ and $g$ have no common points. Prove that

\[S = \binom{n}{m+n} \cdot \binom{q}{m+q-p} - \binom{q}{m+q} \cdot \binom{n}{m+n-p}.\]

I'm pretty sure all those binomial coefficients are flipped; but I guess it depends on your notation. I will use ${n \choose k}$ to the the number $k$ element subsets of an $n$-set.

So, in my notation, we want to show that the number pairs of nonintersecting paths $(f,g)$, where $f$ goes from $A$ to $D$ and $g$ goes from $B$ to $C$ is

\[ {m+n \choose n}{m+q-p \choose q} - {m+q \choose q}{m+n-p \choose n}. \]

Recall that the number of lattice paths between $(0,0)$ and $(x,y)$ is ${x+y \choose x} = {x+y \choose y}$. For convenience, let us use the notation p(X,Y) to denote the set of paths from $X$ to $Y$.

So, the total number of pairs of paths from $A$ to $D$ and from $B$ to $C$ is just

\[ |p(A,D)||p(B,C)| = {m+n \choose n}{m+q-p \choose q}. \]

We now subtract off the pairs of paths that intersect. Suppose we have paths $f_0$ and $g_0$, where $f_0 \in p(A,D)$ and $g_0 \in p(B,C)$ such that $f_0$ and $g_0$ intersect. Then, we can consider their first intersection point $P$. Now, we do a clever manipulation -- switch the remainders of the paths $f_0$ and $g_0$ after the point $P$ in both of their paths, to get paths $f_1$ and $g_1$, respectively; notice that $f_1 \in p(A,C)$ and $f_2 \in p(B,D)$. Furthermore, this process can be reversed with any $f_1$ and $g_1$ that intersect; however, based on the conditions that $p < m$, $q < n$, we know that any path in $p(A,C)$ will intersect any other path in $p(B,D)$.

Therefore, the number of pairs intersecting paths is

\[ |p(A,C)||p(B,D)| = {m+q \choose q}{m+n-p \choose n}, \]

and so our total number of ``good'' pairs of paths is just

\[ |p(A,D)||p(B,C)| - |p(A,C)||p(B,D)| = {m+n \choose n}{m+q-p \choose q} - {m+q \choose q}{m+n-p \choose n}, \]

as desired.


Multiplying both sides by $ x^6$ gives a sextic which by Gauss' theorem on rational roots has no rational roots. A graph plotter gives two real roots at approximately -0.74 and 1.175.

hello, i have got the polynomial equation
$ \minus{}120x^6\plus{}35x^5\plus{}32x^4\plus{}32x^3\plus{}37x^2\plus{}37\equal{}0$ with two real solutions.
Sonnhard.

can you determine the solution ??? I know the graphics method but i think it would be an algebra solution for the problem.
PS: my english is weeh i hope that u understand me.!


parameterization
probability
function
inequalities
floor function
ceiling function
probability and stats

let $ T_1\equal{}\inf \{s\ge 0 : X_s \ne 0\}, T_2\equal{} \inf \{s \ge T_1: X_s \equal{}0\}$

$ Y\equal{}T_2\minus{}T_1$ the sojourn time from zero

for $ X_s$ the simple random walk on Z with parameter 1, so the walker waits an exponentially decided time and jumps to a neighbor with probability $ \frac{1}{2}$ each

define $ r(t)\equal{}\frac{d}{dt}P(Y\le t)$ then i am THINKING that $ r(t) \approx ct^{\minus{}\frac{3}{2}}$ as $ t\to \infty$

is this true? anyone prove it for me?  are there similar estimates for the simple random walks on $ Z^d$

i would guess for $ Z^d$ we have something along the lines of $ t^{\minus{}(\frac{d}{2}\plus{}1)}$ but i dont know =/

Let $ N$ be the number of jumps taken before the first return to $ 0$.  It's a classical result that $ P(N \equal{} 2n)\sim n^{ \minus{} \frac {3}{2}}$ and $ P(N \equal{} 2n \plus{} 1) \equal{} 0$ for all $ n$, so $ P(N\geq n)\sim 1 \minus{} n^{ \minus{} \frac {1}{2}}$.  Now the number of jumps taken at time $ t$ has distribution $ P(J(t) \equal{} n) \equal{} \frac {e^{ \minus{} t}t^n}{n!}$ so $ P(J(t)\geq n) \equal{} e^{ \minus{} t}\sum_{k \equal{} n}^\infty \frac {t^k}{k!}$ and so $ \frac {d}{dt}P(J(t)\geq n) \equal{} \frac {e^{ \minus{} t}t^{n \minus{} 1}}{(n \minus{} 1)!}$. This gives us everything to compute the quantity you want:
\[ P(Y\leq t) \equal{} P(J(t)\geq N) \equal{} \sum_{k \equal{} 1}^{\infty}P(J(t)\geq N|N \equal{} k)P(N \equal{} k)\]

\[ \Rightarrow \frac {d}{dt}P(Y\leq t) \equal{} e^{ \minus{} t}\sum_{k \equal{} 1}^\infty \frac {t^{k \minus{} 1}}{(k \minus{} 1)!}P(N \equal{} k)\]
For large $ t$ we have $ e^{ \minus{} t}\frac {t^k}{\Gamma (k \plus{} 1)} \sim  N(t,\sqrt{t})(k)}$ so as $ t$ increases the contribution to the sum comes almost entirely from those $ k$ which are close to $ t$, so we can take $ P(N \equal{} k)\sim t^{ \minus{} \frac {3}{2}}$ for all $ k$ and introduce an error that vanishes as $ t\to \infty$ (this is an assertion that certainly requires a more careful estimate, which you ought to do if you'd like to properly convince yourself of this fact):
\[ \frac {d}{dt}P(Y\leq t)\sim t^{ \minus{} \frac {3}{2}}e^{ \minus{} t}\sum_{k \equal{} 1}^\infty \frac {t^{k \minus{} 1}}{(k \minus{} 1)!} \equal{} t^{ \minus{} \frac {3}{2}}\]

thank you for this answer, but i am having trouble understanding what you meant by

$ e^{\minus{}t}\frac{t^n}{n!}\sim N(t,\sqrt t)(k)$ and how this relates to making only the values of k close to t count..?

would it be too troublesome for you to flesh out the more detailed proof, or at least an outline of it, that you refer to at the end?  thanks again

My apologies, I realize now the notation $ N(t,\sqrt {t})(k)$, though standard, was pretty bad since I'd already used $ N$ elsewhere.  I just meant the normal density function with mean $ t$ and variance $ t$: $ \mathcal {N}(t,\sqrt {t})(k) = \frac {1}{\sqrt {2\pi t}}e^\left ( - \frac {(k - t)^2}{2t}\right)$.  So what I was saying is that the Poisson distribution with mean $ t$ large is roughly the same as a normal distribution with mean and variance $ t$ (and exactly the same in a carefully scaled limit).  But that's a somewhat tedious calculation and in any case [i]far[/i] more than what we need to resolve the question at hand.  Chebyshev's inequality is enough here.

Here's the basic strategy.  Cut the sum into three parts: terms with index smaller than $ \alpha_1t$ for some $ \alpha_1 < 1$, terms with index around $ k = t$, and terms with index greater than $ \alpha_2t$ for some $ \alpha_2 > 1$.  I chose $ \alpha_1 = \frac {1}{2}$ and $ \alpha_2 = 2$, but any choice would be fine.  The sum of small index terms will vanish because in that case $ P(J(t) = k)$ is very small (exponentially so, in fact, as we show using Stirling's formula) compared to $ t$.  The sum of large index terms will also be negligible, which we will show with a simple application of Chebyshev's inequality.  The terms with indexes around $ k = t$ will contribute the $ t^{ - \frac {3}{2}}$, because the probability that $ J(t)$ falls within that range is roughly $ 1$, and $ P(N = k)\sim t^{ - \frac {3}{2}}$ in that range (which is the crux of the intuition I alluded to in a previous post).  The details follow.

$ e^{ - t}\frac {t^{k - 1}}{(k - 1)!} = P(J(t) = k - 1)$, and it is easy to check that $ J(t)$ has mean and variance $ t$, so by Chebyshev's inequality $ P(J(t)\geq 2t)\leq \frac {1}{t}$ and so
\[ \frac {d}{dt}P(Y\leq t) = \sum_{k = 1}^\infty P(J(t) = k - 1)P(N = k) = \\
= \left(\sum_{k = 1}^{\left \lfloor \frac {t}{2} \right \rfloor} + \sum_{k = \lceil \frac {t}{2} \rceil}^{\lfloor 2t \rfloor } + \sum_{k = \lceil 2t \rceil}^\infty \right) P(J(t) = k - 1)P(N = k) \leq \\
\leq \frac {t}{2} P\left(J(t) = \left\lfloor \frac {t}{2} - 1\right \rfloor\right) + c\left(\frac {t}{2} + 1\right)^{ - \frac {3}{2}} + \left(\frac {1}{t}\right)\left (c(2t)^{ - \frac {3}{2}}\right)\]
where the above makes use of the fact that $ P(J(t) = k)$ is an increasing function of $ k$ for $ k < t$ and $ P(N = k)$ is a decreasing function of $ k$.  Now use Stirling's formula to estimate $ P\left(J(t) = \left\lfloor \frac {t}{2} - 1\right \rfloor\right) = \mathcal {O}\left(\sqrt {t}\left(\frac {e}{2}\right)^{ - \frac {t}{2}}\right)$, which is therefore negligible compared to $ t^{ - \frac {3}{2}}$.  Thus,
\[ \frac {d}{dt}P(Y\leq t)\leq C t^{ - \frac {3}{2}}\]
for some constant $ C$ (going through these calculations more carefully without throwing out intermediate constants can even get you a good fix on $ C$, which you can further refine by choosing different $ \alpha_1,\alpha_2$ in the sum cutoffs).  Now we prove the lower bound (this one's easier, just cut off the sum anywhere proportional to $ t$, doesn't really matter where, I chose $ 2t$, and again make use of Chebyshev: $ P(0\leq J(t)\leq 2t)\geq 1 - \frac {1}{t}$):
\[ \frac {d}{dt}P(Y\leq t)\geq \sum_{k = 1}^{\lfloor 2t \rfloor }P(J(t) = k - 1)P(N = k) \geq \left(1 - \frac {1}{t}\right)P\left(N = \left \lfloor 2t\right \rfloor\right) = \\
= \mathcal {O}\left(t^{ - \frac {3}{2}\right)}\]


inequalities
function
trigonometry
inequalities unsolved

[quote="freemind"]Prove the following inequality:
$\cos1+\frac{\cos3}{3}+\frac{\cos5}{5}+\dots+\frac{\cos(2n-1)}{2n-1}>0$, $\forall n\in\mathbb{N}^{*}$.[/quote]

Find a function $f(x)>0$ such that,

$f(x)=\frac{\cos x}{1}+\frac{\cos 3x}{3}+\frac{\cos 5x}{5}+...$
Your inequality above is a partial sum of the Fourier series of $f(x)$ evaluated at $x=1$

To find what function it is, I don't think its so hard. Because the terms are linear. So it shouldn't take excessive work to find such a function.

Here are some things which make it easier.
1)It must have fundamental period $2\pi$
2)It must be even.
3)And peicewise constant.


Billy has six posters. (Children of Bodom, Slayer, Iron Maiden, Metallica, Trivium, and another, different, Iron Maiden poster.) He picks two posters, without replacement, at random to worship. (I have no clue what skaters do all day long - this is my best guess. Sorry again, Billy.) What is the probability that he chooses an Iron Maiden poster and the Slayer poster? Assume that he cannot pick two posters from the same band.

[hide]There are 6C2 different combinations but you can't have 2 Iron Maidens so it's 6C2 - 1. Then, you can have the first Iron Madien and Slayer, or the seocnd Iron Maiden and Slayer. So you have 2/14 or $\boxed{\frac{1}{7}}$[/hide]

[quote="Treething"]Billy has six posters in his room, each one of a different band. (Children of Bodom, Slayer, Iron Maiden, Metallica, Trivium, and another, different, Iron Maiden poster.) He picks two posters, without replacement, at random to worship. (I have no clue what skaters do all day long - this is my best guess. Sorry again, Billy.) What is the probability that he chooses an Iron Maiden poster and the Slayer poster? Assume that he cannot pick two posters from the same band.[/quote]

LOL, a guy at my school skates.  I'll ask him  :D

[hide="Anyways, the answer"]He has ${6\choose2}=15$ total ways to choose the posters.  But, he cannot choose two Iron Maiden posters, so there are only 14 ways to choose the posters.  Of those, he has $2\cdot1=2$ ways that are Iron Maiden and Slayer (he has two choices for the Iron Maiden).  This makes the probability $\frac2{14}=\boxed{\frac17}$[/hide]

Wierd, the only band I have even heard of amoung those is Metallica.  Maybe thats because I listen to Billy Joel, the Beatles, and some other oldies instead  :roll:

[quote="Ignite168"][quote="Treething"][quote="binonunquineist"]What??? But in the first sentence it says 'they are each of a different band'. Thus the two Iron Maiden posters are different bands, so it is [b]2/15[/b] instead of 2/14 = 1/7.[/quote]

What?

it says it's a different Iron Maiden poster... meaning it's a different poster about the same band

kinda like a red pokeball and a blue pokeball...[/quote]

But that means they aren't each of a different band.[/quote]

EXACTLY

TWO DIFFERENT POSTERS OF THE SAME BAND


Find all real numbers x such that$ sin(x radians) \equal{} sin(x degrees)$
 Repeat for. cos x and tan x.

$ \sin x \equal{}\sin\frac{\pi x}{180}$
iff
$ x \equal{}\frac{\pi x}{180}\plus{}k2\pi$  or  $ x \equal{}\pi\minus{}\frac{\pi x}{180}\plus{}k2\pi$ with $ k\in\mathbb{Z}$
iff
$ x \equal{}\frac{360k\pi}{180\minus{}\pi}$  or  $ x \equal{}\frac{180(1\plus{}2k)\pi}{180\plus{}\pi}$


A couple of weeks ago, I was doing this practice sat essay and the prompt went something like this: Should schools teach the arts? How would you answer this question, what is your opinion on this?

I'd say... yes, and, since it's an SAT essay, draw on my personal experience taking only academic courses and say that I feel like I've missed out on something. Then I'd go on to say how many students benefit at least as much from the arts, which they often use practically on a day-to-day basis, as they do from academic subjects, and that artistic endeavors may lead to personal growth. Then I'd propose that it's hard to draw a line between "arts" and "academics"--I'd describe mathematics as an art for sure, and go on to relate art to my perceptions of biology and history.

thanks a lot. This is how I approached it.


calculus integration algebra function domain vector topology

Let $ R$ be ring with a finite number $ n$ of idempotents

(a)if $ R$ is commutative,show that $ n$ is a power of 2

(b)if $ R$ has a unit,show that $ n$ is even but need not be a power of 2

(c) is there an $ R$ such that $ n$ is an odd prime ?

Imam Ali (as) said:

Failures are often the results of timidity and fears; disappointments are the results of bashfulness; hours of leisure pass away like summer-clouds, therefore, do not waste opportunity of doing well.

here's what I've got so far,

a) Consider a set, S of idempotents of R, clearly 0 is in S as 0^2 = 0. If we have no other elements, then n = 1 = 2^0. Now if we have a non-zero element 'a' in S we now have two elements in S or 2^1 elements. If there exists another non-zero element 'b', distinct from 'a', then S = {0, a, b, ab} and ab is not equal to a or b as ab = a => ab = a^2 => b = a (which violates the distinctness of a and b), so n = 4 = 2^2. Now if we discover any more distinct idempotents in R, we add them to S and obtain (n - 1) new elements (we multiply all the existing elements of S by the new idempotent, except for 0, as this will give us 0 again). ie; adjoining distinct idempotent c to S = {0, a, b, ab} gives us S = {0, a, b, c, ab, bc, ac, abc} => n = 2^4, the addition of each distinct idempotent increases n from 2^x to 2^x + 1 + 2^x - 1 = 2*2^x = 2^(x+1), clearly the number of idempotents will be a power of 2. The commutative property of R ensures that we only need to count the product of idempotents once. (Also, this is assuming that R does not have unity, so none of the elements of S = 1)

Edit: I just realized that this doesn't account for the product of two idempotents being zero, does the question specify whether the ring is an ID or not? If not this is probably the wrong approach.

(a) http://www.mathlinks.ro/Forum/viewtopic.php?t=200805 .
EvaristeG: rings with idempotents, as a rule, scarcely use to be integral domains ;) , but a nice idea - it shows that the set of idempotents in a commutative ring is an $ \mathbb{F}_2$-algebra (with or without unity), not just an $ \mathbb{F}_2$-vector space.

(b) Let $ R$ be a ring with unity (not necessarily commutative). We will show that

[color=green][b](1)[/b][/color] if $ x\in R$ is idempotent, then $ 1 \minus{} x$ is idempotent as well, and if $ 1\neq 0$ in $ R$, then we also have $ x\neq 1 \minus{} x$.

This will yield that, if $ 1\neq 0$ in $ R$, the idempotents of $ R$ come in pairs $ \left(x,1 \minus{} x\right)$, so their number is even.

[i]Proof of [color=green][b](1)[/b][/color].[/i] Since $ x$ is idempotent, we have $ x^2 \equal{} x$, so that $ \left(1 \minus{} x\right)^2 \equal{} 1 \minus{} 2x \plus{} \underbrace{x^2}_{ \equal{} x} \equal{} 1 \minus{} x$, so that $ 1 \minus{} x$ is idempotent as well. Besides, $ 1\neq 0$ implies $ x\neq 1 \minus{} x$ (because otherwise, $ x \equal{} 1 \minus{} x$ would yield $ 2x \equal{} 1$, so that

$ 1 \equal{} 2x \equal{} 4x \minus{} 2x \equal{} 4x^2 \minus{} 2x$     (since $ x \equal{} x^2$)
$ \equal{} \left(2x\right)^2 \minus{} 2x \equal{} 1^2 \minus{} 1 \equal{} 0$,

contradicting $ 1\neq 0$). Thus, [color=green][b](1)[/b][/color] is proven.

Note that the number of idempotents in $ R$ needs not be a power of $ 2$: for instance, for any field $ k$, the ring of upper triangular $ 2\times 2$ matrices over $ k$ has $ 6$ idempotents...

(c) Yes. The rng (this is how I prefer to call noncommutative rings without unity) of all $ 2\times 2$ matrices over $ \mathbb{F}_2$ with zero bottom row has exactly three idempotents: $ \left(\begin{array}{cc} 0 & 0 \\
0 & 0\end{array}\right)$, $ \left(\begin{array}{cc} 1 & 0 \\
0 & 0\end{array}\right)$, and $ \left(\begin{array}{cc} 1 & 1 \\
0 & 0\end{array}\right)$.

  darij

sorry!!
I appreciate your keen eyes!  :blush:

"...Only the deaf appreciate hearing, only the blind realize the manifold blessings that lie in sight."

I would strongly recommend you to read the article "Three Days to See" written by Hellen Keller. the preceding sentences were quoted from that article and you can find it here:
http://www.theatlantic.com/issues/33jan/keller.htm

Thanks for your attention
    God speed you


geometry
trapezoid
rhombus
Pythagorean Theorem
number theory
prime factorization

1. Consider the plane made by a piece of paper 8 cm wide and 10 cm thick. The dot $A$ is drawn at the bottom left corner of the paper and the dot $B$ is drawn at the upper right corner of the paper. Dot $B$ is reflected through the plane to the other side of the paper. Given that the paper is $0.1$ millimeters thick, find the exact length of $AB$ in centimeters.

2. A trapezoid $ABCD$ has $AB \parallel DC$ and $AD = BC = 10$. Given that the bisector of angle $D$ intersects $B$ and that $DB = 10\sqrt{3}$, find the area of trapezoid $ABCD$

[quote="ch1n353ch3s54a1l"][quote="Leao"]1. Consider the plane made by a piece of paper 8 cm wide and 10 cm thick. The dot $A$ is drawn at the bottom left corner of the paper and the dot $B$ is drawn at the upper right corner of the paper. Dot $B$ is reflected through the plane to the other side of the paper. Given that the paper is $0.1$ millimeters thick, find the exact length of $AB$ in centimeters.

2. A trapezoid $ABCD$ has $AD = BC = 10$. Given that the bisector of angle $D$ intersects $B$ and that $DB = 10\sqrt{3}$, find the area of trapezoid $ABCD$[/quote]

[hide="1"] :huh: Interesting problem...
Anyways, I just treated AB as a space diagonal, so I got $\sqrt{8^2+10^2+0.1^2}=\sqrt{164.01}$...?[/hide]
[hide="2"]
I got $50\sqrt{3}$. Since the bisector of angle $D$ intersects $B$ I just assumed trapezoid $ABCD$ was a rhombus. Don't know if I'm supposed to do that. So long diagonal $10\sqrt{3}$ is the leg of a right triangle in the rhombus, so the distance from the center to $A$ or $C$ is $5$ and the area is $50\sqrt{3}$.[/hide][/quote]

It's 8 and 10 centimeters, but the 0.1 is in millimeters. so...
[hide="1"]$\sqrt{8^2+10^2+0.01^2}=\sqrt{164.0001}=12.8\text{cm}$
maybe?? The 0.1 mm doesn't even matter.[/hide]

[hide]1. use the pythagorean theorem two times to get the space diagonal.
let C be the bottom right hand corner.
$AC^2+BC^2=AB^2$
$8^2+10^2=AB^2$
$9\sqrt{2}=AB$
$AB^2+BB'^2=AB'^2$
$9\sqrt{2}^2+.0001=AB'^2$
$162.0001=AB'^2$
AB is the square root of that. don't feel like doing it.[/hide]

EDIT: Whoops,  :blush:  :oops_sign:

[quote="ch1n353ch3s54a1l"]Whoops...[hide]$\sqrt{164.0001}$. [/hide] But that isn't exactly [hide]$12.8$[/hide].[/quote]
Oops... :blush: Didn't see "exactly"
Does anyone know the prime factorization of 1640001??

for #2, u need to specify which sides are the bases of the trapezoid. :?

[hide]assuming AB and CD are the bases,
triangle ABD is isosceles, so $AB=10$, now drop a perpendicular from A and B to the base at point E. let $AE=h$ and let $x=DE$. so,
$x^2+h^2=100$
$(10+x)^2+h^2=300$
solving gives
$x=5$
$h=5\sqrt{3}$
and the area of the trapezoid is $75\sqrt{3}$[/hide]

for number 1, I guess you could keep it at $\sqrt{164.0001}$, my bad on making the numbers difficult.  :oops:
[hide="my solution for number 2"]
Draw perpendicular from $B$ to $DC$ meeting it at $R$. Label $\angle BDR = \angle BDA = x$. Since $AD=BC$, the trapezoid is isoceles and $\angle BCR = 2x$. We have:
$sin x = \frac{BR}{10\sqrt{3}}$

$2sinxcosx = \frac{BR}{10}$

we get $cos x =\frac{\sqrt{3}}{2}$ so $x=30$ degrees.

We get $\angle DBC = 90$ degrees so

$[ABCD]= \frac{1}{2}(10)(10)\sqrt{3}sin30 + 50\sqrt{3} = 75\sqrt{3}$
[/hide]
ch1n353ch3s54a1l:  It's not a rhombus because the four sides are not equal.
fireemblem: changed  :)


What is the number of units in the length of the line segment connecting $ (4,3)$ and $ (\minus{}1,\minus{}9)$?

The distance between $ (x_1, y_1)$ and $ (x_2, y_2)$ is $ \sqrt{(x_2 \minus{} x_1)^2 \plus{} (y_2 \minus{} y_1)^2}$. Plugging in the numbers yields $ \sqrt{(\minus{}1 \minus{} 4)^2 \plus{} (\minus{}9 \minus{} 3)^2} \equal{} \sqrt{25 \plus{} 144} \equal{} \boxed{13}$.


geometry
3D geometry
tetrahedron
analytic geometry

Find the maximum volume of a tetrahedron $ OAPQ$ such that $ OA\equal{}1,\ OA\bot OP,\ OP\bot OQ,\ OA\bot OQ,\ \angle{PAQ}\equal{}30^\circ.$

hello, i have got $ V_{Max}\equal{}\frac{1}{18}(2\sqrt{3}\minus{}3)$
Is this right kunny?
Sonnhard.

That's right. :lol:  How did you solve it?

Coordinate geometry

$ OA \equal{} (1,0,0), OP \equal{} (0,p,0), OQ \equal{} (0,0,q)$

Dot product gives;
$ \cos(30) \equal{} \frac {( \minus{} 1,p,0)\cdot( \minus{} 1,0,q)}{\sqrt {(p^2 \plus{} 1)(q^2 \plus{} 1)}}$

$ \therefore p^2q^2 \plus{} p^2 \plus{} q^2 \equal{} \frac {1}{3} \Longrightarrow \frac {1}{3} \ge (pq)^2 \plus{} 2(pq)$

$ \therefore \frac {4}{3} \ge (pq \minus{} 1)^2 \Longrightarrow pq \le \frac {2}{\sqrt {3}} \minus{} 1$

Hence the maximum volume of the tetrahedron is : $ \frac {1}{6}\cdot1\cdot\left(\frac {2}{\sqrt {3}} \minus{} 1\right)$

$ \boxed{\frac {2\sqrt {3} \minus{} 3}{18}}$


I think what you have is

\[ \sum_{n\in\mathbb{N}}\left(\frac{2}{3}\right)^{\frac{\left(n+1\right)\left(n+2\right)}{2}} \]

but I'm not quite sure about the next step...


Let $ ABC$ is a acute triangle, let $ B'$ lie on $ AC$ and $ BB'\perp AC$, let $ C'$ lie on $ AB$ and $ CC'\perp AB$, let $ BB'$ meet $ CC'$ at point $ H$, let $ B'C'$ meet $ BC$ at piont $ K$. Let $ (d)$ is a line pass point $ A$ and perpendicular to $ AH$, $ (d)$ meet $ BC$ at point $ M$. Prove that $ M$ is midpoint of $ BC$.

[quote="livetolovemath030894"]Dear thanhnam I don't understand the condition of your problem  :)[/quote]

[color=red]Sory, I have a wrong at type, I have edited:[/color] "[color=darkblue]perpendicular to [/color]$ KH$".

Let $ ABC$ is a acute triangle, let $ B'$ lie on $ AC$ and $ BB'\perp AC$, let $ C'$ lie on $ AB$ and $ CC'\perp AB$, let $ BB'$ meet $ CC'$ at point $ H$, let $ B'C'$ meet $ BC$ at piont $ K$. Let $ (d)$ is a line pass point $ A$ and [color=darkblue]perpendicular to [/color]$ KH$, $ (d)$ meet $ BC$ at point $ M$. Prove that $ M$ is midpoint of $ BC$

It has been posted before as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110887][color=Blue]Triangle, perpendicular lines[/color][/url], by [b]Yimin Ge[/b].

Kostas Vittas.


There are $ 16$ pupils in a class. Every month, the teacher divides the pupils into two groups. Find the smallest number of months after which it will be possible that every two pupils were in two different groups during at least one month.

[hide="solution"]
Consider a 0/1 matrix whose rows correspond to students and whose columns correspond to months, where each entry is 0 or 1 according to which group that student was in on that month. The condition (every pair of students are in different groups at least once) just means that the rows are all different. If there are $ n$ students where $ 2^{k\minus{}1} < n \leq 2^k$, this can obviously be done with $ k$ months but no less, since there are only $ 2^{k\minus{}1}$ binary vectors of length $ k\minus{}1$, and on the other hand you can easily create $ 2^k$ distinct vectors by simply enumerating all of them.

Therefore, the answer is 4 months when there are 16 students. The algorithm is just to enumerate the students $ 0, 1\ldots, 15$, take the 4-digit binary representation of these numbers, and placing them in groups on month $ i$ according to the value of the $ i$th digit.
[/hide]


Find all solutions to

(x-3)(x-1)(x+3)(x+5)=13.

Please use spoiler.

Yeah, that works. I did it slightly differently. I'll post my solution after a few other people post if no one finds it before.

Solution:

[hide]Ok, multiply (x-3)(x+5) to get x^2+2x-15

multiply (x+3)(x-1) to get x^2+2x-3

So let x^2+2x=y, and ur set![/hide]

Ooops, sorry fiery, i posted without looking at ur solution...

[hide]Let y=x+1, the original equation becomes (y-2)(y+2)(y-4)(y+4)=13, (y^2-16)(y^2-4)=13, which is quadratic in y^2 and consequently easy. Figure the rest out yourself.[/hide]

Ok my method was the same as teratomato's, so I won't post it.


1992 AHSME #14:
Which of the following equations have the same graph?
I. $y=x-2$
II. $y=(x^{2}-4)/(x+2)$
III. $(x+2)y=x^{2}-4$

Now, I get how I and II are different, but how are II and III different? I know the answer is none of the have the same graph.

[hide]
III consists of two lines, while II just has a removable discontinuity at $x=-2$. [/hide]


[quote="amitie"]1) sketch the curve defined by the equation $ y^2 \equal{} \frac {1}{x \minus{} 1} \plus{} \frac {2}{(x \minus{} 3)^2}$
2) sketch the three dimensional surface defined by $ x^2 \minus{} y^2 \minus{} z^2 \equal{} 1$
3) sketch the regions of the complex plane defined by $ Re(log\frac {1 \plus{} z}{1 \minus{} z})$ <0
4) sketch the curve defined by the equation $ (log(y) \minus{} x)^2 \minus{} x^2 \equal{} c$[/quote]

I think (3) is the only one that is difficult... try to prove the identity $ \Re(\ln(z))\equal{}\ln |z|$ and conclude that $ \left|\frac{1\plus{}z}{1\minus{}z}\right|<1.$ Use $ |xy|\equal{}|x\parallel{}y|.$

My solution is $ \Re(z)<0$ but I'm terribly tired and I could well have made some mistakes. :/

thanks... i want to become better at curve sketching but i often don't know where to start :( How would one begin question 4? is it best to convert it into $ (log(y)\minus{}2x)log(y)\equal{}c$?

[quote="amitie"]thanks... i want to become better at curve sketching but i often don't know where to start :( How would one begin question 4? is it best to convert it into $ (log(y) - 2x)log(y) = c$?[/quote]

I don't think so because after all you want to isolate ln(y) on one side of the equality. I'd convert it the following way:

\begin{align*}
(\log y - x)^2-x^2&=c\\
(\log y - x)^2 &= c+x^2\\
\log y - x &= \sqrt{c+x^2}\\
\log y &= x+\sqrt{c+x^2}\\
y &= exp\left(x+\sqrt{c+x^2}\right)
\end{align*}

Wow this formula looks really stupid. :P


anyone with a proof of the following-
1)summation i=0 to i=n   2^(n-i)  (n+i)Ci    is equal to 2^(2n)

2)summation i=0 to i=n   (2i)Ci   (2n-2i)C(n-i)   is equal to 2^(2n)

3)summation i=1 to i=n-1   nCi  i^(i-1)  (n-i)^(n-i-1)  is equal to 2 (n-1)  n^(n-2)   (this is abel's identity)[/img]

[quote="ajinx"]anyone with a proof of the following-
1)$ \sum_{i=0}^{n}2^{n-i}\binom{n+i}{i}= 2^{2n}$

2)$ \sum_{i=0}^{n}\binom{2i}{i}\binom{2n-2i}{n-i}= 2^{2n}$

3)$ \sum_{i=1}^{n-1}\binom{n}{i}i^{i-1}(n-i)^{n-i-1}= 2 (n-1) n^{n-2}$   (this is abel's identity)[/quote]

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 for how I did that.

Number (3) or something much like it has appeared recently on the forum, but I don't think it has a solution yet.


Has anyone been told whether or not they made the arml team? (eastern) I was told 10 days, but it's been 14 and I still haven't gotten anything...

We don't do anything until like mid-April.

Western Mass will beat you all!
 :evilgrin:

we should be getting the emails (btw people who stayed behind were able to get their scores, and the cutoff is 8)
13375P34K43V312 did you make it? (or think you made it?)

Did they throw out the one with the heptagon? I didn't stay behind, I had to leave. But I believe I got 16, because I knew they threw out at least one problem, and I got the last one and the one with the inscribed circles (did the ratio of their radii, argh) wrong.

The only one they threw out was the one with the "proper digits."

hm.. don't be so sure.. maybe only 2-3 ppl got above 12 i believe.. Many many people got much worse scores than they expected.
By the way, did you take aime? (if so what did you get?)

And they didn't count a question but i forget what it is. (only one pereson got it anyways so you probably don't even need to worry about it)
I got a 7 and didn't make it, isn't that really sad..

I got 7 on aime 1.

And I'm pretty sure about 16, although it's possible I got 15.


[quote="Tourish"][quote="fjwxcsl"]
Generally, if $ x_1^n + x_2^n + \ldots + x_m^n = m,x_i\ge{0},n\in{N},n\ge{1},$ prove (or disprove)that:
$ \frac {1}{x_1^{n + 1}} + \frac {1}{x_2^{n + 1}} + \ldots + \frac {1}{{x_m}^{n + 1}}\ge{x_1^{n + 1} + x_2^{n + 1} + \ldots + x_{m}^{n + 1}.}$[/quote]
I think this is wrong.try $ n = 1,m = 11,x_1 = x_2 = ... = x_{10} = 0.7,x_{11} = 4$ :wink:[/quote]

[size=150]You are right,thank you![/size]


ok, then is it possible to write it as the sum/difference of three or more radicals, where there is not one inside the other?

[quote="ProtestanT"][quote="0123456789"]is it possible to reduce sqrt(12+sqrt(13)) so there is not a radical inside a radical?[/quote]
[hide="i dont think so"]
$ \sqrt {12 \plus{} \sqrt {13}}$
You would want to get it to a form that is $ a \plus{} b\sqrt {c}$ where $ a$ and $ b$ are rational and $ c$ is an integer.
$ a \plus{} b\sqrt {c} \equal{} \sqrt {12 \plus{} \sqrt {13}}$
$ a^2 \plus{} 2ab\sqrt {c} \plus{} b^2c \equal{} 12 \plus{} \sqrt {13}$
$ c$ has to be $ 13$ since it's the only radical.
$ a^2 \plus{} 13b^2 \plus{} 2ab\sqrt {13} \equal{} 12 \plus{} \sqrt {13}$
We can make 2 equations: $ 2ab \equal{} 1$ and $ a^2 \plus{} 13b^2 \equal{} 12$
from $ 2ab \equal{} 1$, we can get $ b \equal{} \frac {1}{2a}$
$ a^2 \plus{} \frac {13}{4a^2} \equal{} 12$
$ a^4 \plus{} \frac {13}{4} \equal{} 12a^2$
$ a^4 \minus{} 12a^2 \plus{} \frac {13}{4} \equal{} 0$
$ a^4 \minus{} 12a^2 \plus{} 36 \minus{} \frac {131}{4} \equal{} 0$
$ (a^2 \minus{} 6)^2 \equal{} \frac {131}{4}$
$ a^2 \minus{} 6 \equal{} \pm\frac {\sqrt {131}}{2}$
I don't think it's possible. $ a$ doesn't look like it's going to be rational. In fact, it looks like there's going to be a radical inside a radical :(
[/hide][/quote]
You don't need to do all that; after getting $ a^2\plus{}13b^2\equal{}12$, then $ a^2\plus{}13b^2$ must be more than 12.

[quote="xpmath"][quote="ProtestanT"][quote="0123456789"]is it possible to reduce sqrt(12+sqrt(13)) so there is not a radical inside a radical?[/quote]
[hide="i dont think so"]
$ \sqrt {12 \plus{} \sqrt {13}}$
You would want to get it to a form that is $ a \plus{} b\sqrt {c}$ where $ a$ and $ b$ are rational and $ c$ is an integer.
$ a \plus{} b\sqrt {c} \equal{} \sqrt {12 \plus{} \sqrt {13}}$
$ a^2 \plus{} 2ab\sqrt {c} \plus{} b^2c \equal{} 12 \plus{} \sqrt {13}$
$ c$ has to be $ 13$ since it's the only radical.
$ a^2 \plus{} 13b^2 \plus{} 2ab\sqrt {13} \equal{} 12 \plus{} \sqrt {13}$
We can make 2 equations: $ 2ab \equal{} 1$ and $ a^2 \plus{} 13b^2 \equal{} 12$
from $ 2ab \equal{} 1$, we can get $ b \equal{} \frac {1}{2a}$
$ a^2 \plus{} \frac {13}{4a^2} \equal{} 12$
$ a^4 \plus{} \frac {13}{4} \equal{} 12a^2$
$ a^4 \minus{} 12a^2 \plus{} \frac {13}{4} \equal{} 0$
$ a^4 \minus{} 12a^2 \plus{} 36 \minus{} \frac {131}{4} \equal{} 0$
$ (a^2 \minus{} 6)^2 \equal{} \frac {131}{4}$
$ a^2 \minus{} 6 \equal{} \pm\frac {\sqrt {131}}{2}$
I don't think it's possible. $ a$ doesn't look like it's going to be rational. In fact, it looks like there's going to be a radical inside a radical :(
[/hide][/quote]
You don't need to do all that; after getting $ a^2 \plus{} 13b^2 \equal{} 12$, then $ a^2 \plus{} 13b^2$ must be more than 12.[/quote]
i thought that $ b^2$ could've have been less than 1.

[quote="0123456789"]is it possible to reduce sqrt(12+sqrt(13)) so there is not a radical inside a radical?[/quote]

Short answer: [b]no[/b], as $ 12^2 \minus{} 13 \equal{} 144 \minus{} 13 \equal{} 131$ is not a perfect square. Proof—the long answer—is left as an exercise for the reader. :)