conics
hyperbola
ellipse
parabola
geometry
3D geometry
special factorizations

What type of conic section is the graph of

2x^2 + 3y^2 - 16x - 12y = -44

1) hyperbola
2) ellipse
3) circle
4) directrix
5) none of the above

How is this done?!!

It's done by completing the square:

\begin{eqnarray*}2x^2+3y^2-16x-12y=-44 &\iff& 2(x^2-8x)+3(y^2-6y)=-44\\ &\iff& 2(x^2-8x+16-16)+3(y^2-4y+4-4)=-44\\ &\iff& 2(x^2-8x+16)-32+3(y^2-4y+4)-12=-44\\ &\iff& 2(x-4)^2+3(y-3)^2-44=-44\\ &\iff& 2(x-4)^2+3(y-3)^2=0\end{eqnarray*}

Since both terms are non-negative, the equality can be satisfied only if both are zero, therefore the equation represents a single point $(4,3)$ and the correct answer is $5$.

I don't know much about this... I understand the explanation, but how did you know it was an ellipse from this? Where did 2 and 5 come from? What do they mean?

[quote="Mathematicus"]Ahh... you were saying answers numbers 2 or 5. Never mind that. :blush:  I still would like to know how you deduced it was an ellipse (or a single point??).[/quote]
Through solving the equation we find it to be degenerate, I was just too lazy to type out how to solve said equation. As to the ellipse part, we can just look at the equation, see that there is an ${X^2}$ and a ${Y^2}$ term, eliminating parabola, that each squared term is positive, eliminating hyperbola, and that the coefficent of each squared term is different, eliminating circle. Ergo, as indeed the equation is a conic section, it must be the only remaining type, an ellipse.

[quote="DanK"]If this was two years ago I could have told you, back when we were doing conic sections in Algebra 2.  Every conic section has a distinct type of equation, so you can just recognize them from that once you know what you're doing...

Unfortunately, the only ones I remember are hyperbolas and parabolas, and I don't remember much about them...  Does anyone have a good resource on conic sections they could direct us to?[/quote]

If you have AoPS2, then there's a chapter on it. Otherwise:
[url]http://mathworld.wolfram.com/ConicSection.html[/url]
Conics are basically the only geometry topic that I'm good at  :P

[quote="Mathematicus"]Uhhhm... I just looked at the answer key (I hadn't done so yet). It says that it is a point (none of the above). How does that fit in?[/quote]
A point can be either a degenerate circle or ellipse.

[quote="eternalomnipotence"][quote="LordoftheMorons"]2 or 5. It is a degenerate ellipse.[/quote]

That would be unlikely because ellipses usually have the formula:
(x-h)^2/a + (y-k)^2/b = 1, this equation is obviously equal to zero.
Or I could be wrong.  :P[/quote]
This is because normally upon completing the square for $x$ and $y$ we [i]make[/i] the equation equal to one by dividing by the right side. We cannot divide by zero, however, and as such leave it as is.

You know how a conic section is basically like slicing a section of two cones sharing their vertex? Well, as an circle is a section of a cone parallel to the bases, if it were to be cut at the vertex, the result would be a point. The same applies to ellipses, except they're not parallel to the base. You should probably ask someone for a better explanation, unless you understand this  :P

[quote="Mathematicus"]So, a degenerate ellipse is a slice just through the vertex and not parallel to the bases of the cones? Would any conic section that is only on the vertex be considered degenerate (e.g. a degenerate circle)?[/quote]
Yah, but there are also degenerate parabolas (a line) and degenerate hyperbolas (an "X" shape)

And now I am happy, because I just found out that our next chapter in my pre-calc book will be on conic sections, and I can relearn all the stuff from Algebra 2 that I shouldn't have forgotten.  Plus it mentioned the degenerate ones and I know how that works too...  Why do they always teach me this stuff too late?


geometry
power of a point
radical axis
geometry proposed

Let $w_{1}$ and $w_{2}$ be two circles which intersect at points $A$ and $B$. Consider $w_{3}$ another circle which cuts $w_{1}$ in $D,E$, and it is tangent to $w_{2}$ in the point $C$, and also tangent to $AB$ in $F$. Consider $G \in DE \cap AB$, and $H$ the symetric point of $F$ w.r.t $G$. Find $\angle{HCF}$.

[hide]$G$ is on the radical axis of $w_{1},w_{2}$
$G$ is on the radical axis of $w_{1},w_{3}$
So, the line $GC$ is the radical axis of $w_{2},w_{3}$. As a result, $GC$ is tangent to $w_{3}$. So $GC=GF$. But $G$ is the midpoint of $FH$.
So, $\triangle HCF$ is right at $C$[/hide]

radical line of w1 and w2 is AB,radical line of w1 and w3 is DE and w2 and w3 are tangent in C,so CG is olso radical line of w2 and w3 => GF=GC=GH => <HCF=90


trigonometry
limit
function
real analysis
real analysis unsolved

You can't always do this- you need uniform continuity.

An example of why you need that condition: $\sin\frac1x$ on $(0,1)$.

How about I refine the problem as follows?

Let $a,b\in\mathbb{R}$, and let $f: (a,b)\to\mathbb{R}$.
Let $x_1, x_2, ...$ be a sequence of points in $(a,b)$ which converges to $a$. It can be shown that $lim_n_\to_\infty f(x_n)$ exists. Call this limit $f(a)$ and define $f(b)$ in the same way, thus extending $f$ to the closed interval $[a,b]$. Show that this extended $f: [a,b]\to R$ is continuous.

The sentence "It can be shown that $\lim_{n\to\infty} f(x_n)$ exists." is false.

My example was on the real line; specifying that doesn't really change the problem.

Even if you change that sentence to "Suppose it can be shown that $\lim_{n\to\infty}f(x_n)$ converges," it still wouldn't help.

Let's go back to the example of $\sin\frac1x$ on $(0,1).$ If we take the sequence $x_n=\frac1{\left(2n+\frac12\right)\pi},$ then $\lim_{n\to\infty}f(x_n)=1$ exists. But saying that $f(0)=1$ would not make this function continuous at $0.$

I understand what you mean. How about if I add one more condition?

Let $f(a,b)\to\mathbb{R}$ be uniformly continous.


Prove that any two consecutive Fibonacci numbers are relatively prime.

lol see the mathcounts class.

wait aren't you in that?

Assume that two numbers are not relatively prime. Thus, they share a factor of $ x$.

Thus, the terms can be written as $ ax,bx$.

Then, the sequence can be written as $ ...,(2a\minus{}b)x,(b\minus{}a)x,ax,bx,(a\plus{}b)x,(a\plus{}2b)x,...$.

All terms will be divisble by $ x$.

Then, $ 1$ must be divisible by $ x$. Thus, $ x\equal{}\pm 1$.

Thus, we have a contradiction, and the statement is true


function counting distinguishability Support

I don't know about everyone else, but I'm going to spend eternity with my Savior! :)

I think we know what happens when we die. We've all experienced it. Death is just permanent loss of consciousness. So it is like being in a deep and dreamless sleep (or some loss of consciousness) forever.

Now if we want to have a debate on the nature of consciousness, I think that could be productive, and a slight change of pace... :roll:

[color=green]Moderator says: there is never a good reason to quote the entire post that immediately precedes yours.[/color]

Well, physical death simply gets rid of the body. It does not destroy the soul. The soul is the existence that makes us each unique/ourselves, since our bodies are all similar. (as in the soul identifies which eyes we look from)

so, what exactly do you mean by the soul? where do you think it is and what gives it the means by which it does stuff (such as, for example, giving us our unique identity) ?

What I mean is the sense of "I". To me, I can only view from my body, not anyone else's. I can only see the world from my point of view.
In fact, in the world, I can only be sure that I'm human (since I am). I have to trust that everyone else is human like me.
It's possible that everyone else are machines that are programed to do what they do but cannot think. It's also possible that everyone else also has a soul. But I can only be sure that I have a soul, and that I exist.

okay
how about the second question above

The soul is the person with the body.
Well, what I think is that there is a connection between our brain and our souls. There might be a particle or compartment in the brain that either contains or connects to the soul.

What do you think about our existence, our uniqueness, and our ability to do things/move?

I will cease to exist. Before the thought of not existing scared me, but as of now, the thought is oddly comforting. It's quite odd.

I do not subscribe to theological mumbojumbo guesses to "afterlife". Consider injury related brain trauma, mind altering drugs, and other stuff, our sense of self is ingrained in the physical state of our brains. No brain, no I.

"our sense of self is ingrained in the physical state of our brains"
What part of the brain do you consider your "self"?

I actually don't understand brains very well, so I don't know where the conscience (or the self) is. Can someone explain that as well?

I think pianoforte means that essentially everything you do, think, etc. (the things you might constitute as your "self") is a product of the physical state of your brain. Depending on your definition of conscience/self, you may or may not believe this, but it's more or less sufficient for me. Here's a potentially enlightening quote from Wikipedia:

[quote="Wikipedia"]The human brain appears to have no localized center of conscious control. The brain seems to derive consciousness from interaction among numerous systems within the brain. Executive functions rely on cerebral activities, especially those of the frontal lobes, but redundant and complementary processes within the brain result in a diffuse assignment of executive control that can be difficult to attribute to any single locale. Visual perception generally is processed in the occipital lobe, whereas the primary auditory cortex resides in the temporal lobe. Midbrain functions include routing, selecting, mapping, and cataloging information, including information perceived from the environment and information that is remembered and processed throughout the cerebral cortex. Endocrine functions housed in the midbrain play a leading role in modulating arousal of the cortex and of autonomic systems.[/quote]

Yes, I do not believe that the "self" is transcendent or supernatural, or of those kinds of origins, but a result of the physical brain, an emergent phenomena of the chemical interactions of the brain, the various competing parts. Definitely not compartmentalized.  A phenomena that can be analyzed by looking at its parts, but ultimately larger than the sum of them. Like a flock of birds acting as one unit, despite none of its parts being aware of anything other than its immediate neighbors, the weather, or the economy.

Actually, I have a little guy in my head that controls me with switches and gimbals.

Hilbert - Why do you think that death is just a loss of consciousness, like a deep and dreamless sleep?

Perhaps a different day (and thread) for the consciousness discussion.

ProtestanT - Why do you think that physical death gets rid of the body, leaving the soul?  How can anyone be sure that they are not a machine programed to think they are alive?  How do you know that machines cannot think?  How do you know you have a soul?

pianoforte - Why do you think death causes a person to cease to exist?  Why can't physical injury to the brain and chemicals/electrical current in the brain cause the observed impact if the brain is simply the sole (intended) communication link to the soul?  Perhaps different parts of the soul solely communicate through different parts of the brain, either initially or as a decoding of message function?  How do you answer the situation where there is no brain function for a short while, then the person "wakes up?"

ProtestanT - I don't think anyone knows where the conscience resides.

paladin8 - It seem that the Wikipedia entry simply is a catalogue of what functions seem to be affected when certain parts of the brain is damaged.  However, this seems to break down when some people who loose a mind function after a physical injury to the brain recover the function without that part of the brain being healed.  Extreme example of this are a few examples of people with extreme pressure in their cerebral fluids have large sections (90% in one case I saw) of their brains simply gone and still are walking about, talking and seemingly as normal as the rest of us.

It seems to me that often a single or small group of extreme cases are more instructive than the many cases that are not unusual to understand things.

You two switches and gimbals folks, how do you suppose the "little guy" decides which switch or gimbal to use?

Now here is a question.

When you lose consciousness and then "wake up," presumably at the hospital, what does the passage of time between the time you pass out and the time you wake up feel like?

Does it feel like you woke up "instantly" after you passed out, like your brain "skipped" that passage of time?
Or does it feel like sleep?

The last time I passed out, I didn't even realize it -- I thought the period following when I woke up was preceded immediately by the period before I became unconcious.  It was only afterwards that I realized that my memory of the passage of time couldn't possibly be accurate.  When I was in 4th grade I passed out and had the distinct sensation of waking up from a dream, only to discover that I was on the floor of my history classroom.

My experiences are the same as what JBL said.  In those cases when I knew that I was passing out and got flat on the floor to avoid falling, once the passing out occurred the next instant to me was when I woke up, similar to a dreamless sleep.

Regarding consciousness-When I fall asleep, sometimes I don't have dreams. Then in that case, most cases I feel absolutely nothing. However, sometimes I might be half-awake half-sleeping, other times I might be awake some times sleeping some times, with the sleeping actually feeling like time passing, but I can't actually feel the sleep (best example is when I'm asleep on the bus on the way home, and I'm near my stop. I keep waking and falling asleep again. I don't feel it when I'm asleep, but it feels like time past whenever I wake up.)

[quote="wizard2378"]ProtestanT - Why do you think that physical death gets rid of the body, leaving the soul?  How can anyone be sure that they are not a machine programed to think they are alive?  How do you know that machines cannot think?  How do you know you have a soul?[/quote]
The body cannot always stay. Sometimes it is burned, sometimes it is buried, etc. So in order for the soul to survive, it must get rid of its body. It may get a new body in the future, but the current body is unusable.
When you say "they", do you mean people other than the "anyone", or the "anyone" himself? If it's people other than the "anyone", then the only way to be sure is if you can place your own self into the other person's point of view/body. If it's himself, then the way to know for sure is simply based on the fact that he exists and can think.
Machines cannot think like real existing beings because they're made from physical parts and there's no definite point where the soul is injected into it. The similar thing might be said for humans, but many people generally define it at the  point of conception--the soul enters the body at conception. Machines don't really have a specific time for being made (where would you define the starting point), and can be turned on and off. In fact, after you turn it off, you can sometimes turn it on again. If there's a soul in there, you basically shut it away, and then let it come back. Either that or you let it sleep. However, if turning off lets the soul sleep, then what happens if you smash up the machine after turning it off? Does it remain asleep? What I'm basically saying here is that it's much more likely, and makes a lot more sense, that non-living machines don't have souls and cannot "think".

I know I have a soul because of basically the same reasons as above--"I" is a point of view only from my body, and not from anyone else's. I can't view from anyone else's, or an omniscient point of view. So the soul is my existence.

I've considered the possibility that "I" simply cycles through different bodies at a very high frequency (like maybe a infinitely large number of times per second), and there's a ton of different souls cycling, and that each one loses memory of the previous and gains memory of the next one when it enters. If it's this case, there'd be no way to tell, and the only thing we'd be really experiencing would be the present. Everything in the past would be simply "memories", and might or might not have been felt by the actual soul. However, this theory, although possible, is crazy and it hurts me to imagine that the next moment I'm not me anymore and that "me" is an entirely different person, or that the "me" that I've "always known" was actually conducted by different souls...
What can be even more scary is that we're all just one soul cycling through all of us at an infinitely high frequency...

By "they" I mean any person, including yourself who doesn't consider themselves a machine.

I understand the relationship you are expressing that thinking is linked to souls and machines don't have souls so they can't think.  How do you know, other than than the rule that people think so everyone thinks and has a soul, if anyone has a soul or thinks?

Let me try to point out my question with a question, do animals think and have a soul?  Some have behaved in ways that we would say are thinking if the actions were done by a child, such as using figures to ask questions and answer spoken questions, learning how to select specific foods and responding to requests for those specific foods and many other things.

[quote="wizard2378"]similar to a dreamless sleep.[/quote]

Crap I pass out every night!

About death, I don't have any beliefs there, except for the saying "If they can do it, I can." I have beliefs on what my actions should be, and that's what I focus on.

[quote]When you lose consciousness and then "wake up," presumably at the hospital, what does the passage of time between the time you pass out and the time you wake up feel like?

Does it feel like you woke up "instantly" after you passed out, like your brain "skipped" that passage of time?
Or does it feel like sleep? [/quote]

When I passed out, I went from me being really dizzy, to waking up lying on the floor. I have no idea how long I lost consciousness for. I guess you could say it was kind of like sleeping, except you could say my brain was blank, I had no thoughts or dreams. But I think it is kind of more like your brain "skipped" I guess, because for all I knew it seemed like I immediately woke up after I fell on the floor.

[quote="wizard2378"]I understand the relationship you are expressing that thinking is linked to souls and machines don't have souls so they can't think.  How do you know, other than than the rule that people think so everyone thinks and has a soul, if anyone has a soul or thinks?

Let me try to point out my question with a question, do animals think and have a soul?  Some have behaved in ways that we would say are thinking if the actions were done by a child, such as using figures to ask questions and answer spoken questions, learning how to select specific foods and responding to requests for those specific foods and many other things.[/quote]

I was trying to pin what I say just on souls, because "thinking" is much more complex.
Well, IMO, thinking is something that cannot be physically/completely explained exactly as it is. This is different from introducing randomness into machines, or allowing them to respond to stimulus. Responses to stimulus in machines are perfectly predictable (if randomness is not used). In addition, if randomness is used, then it is still used in a systematic way that can be completely explained. However, human thinking (or animal thinking) cannot be completely explained.

So animals think as well, and they have souls. However, they don't follow the same laws as humans, and they don't sin. I won't define what "sinning" is for them, but from my understanding, animals are perfect compared to humans (That is why in the past, sacrifices of animals/animal blood was used to temporarily pay for human sins--it is because they didn't need to shed it themselves, but they still shed it to take the punishment of the sins of humans. A more general and drastic thing is that since Adam and Eve sinned, not only are humans all doomed to die, but all animals also got finite lifespans as well. This shows animals are capable of, and chose to, take punishments for human sins.). The animals question is a very good question :)

And similar to what I said before, my interpretation of a soul is the existence of myself. When I think or view the world, I see it from one point of view, and it's stuck inside my body. That's the soul--the existence of myself.

ProtestanT - The evolutionists would have to say that all thinking is just the result of chemical reactions and if one knew the starting place and inputs thinking, like all behaviors can be completely predicted.  I'm certainly not going that far, but all we can really say is that we can't explain it currently.

I understand the randomness discussion and various random insertions have been used to simulate human (and animal behaviors) in models.

Thanks for your final paragraph where you clearly identify what you meant by soul.

I'm not sure that Adam, Eve and animals were immortal before the original sin.

God did say, "when you eat of it you will surely die," Gen 2:17.  However, in talking to the rest of the trinity God also said, "The man has now become like one of us, knowing good and evil.  He must not be allowed to to reach out his hand and take also from the tree of life and eat, and live forever."  Gen 3:22  This is why a short time later God placed on the east side f Eden a cherubim and a flaming sword flashing back and forth to guard the way to the tree of life.

Also, I don't think that animals are viewed as sacrificing their blood for mans sin.  See Hebrews 10:4 where the author says, "[B]ecause it is impossible for the blood of bulls and goats to take away sins."  He also says that the law required blood a lot and "without the shedding of blood there is no forgiveness." (9:22)  I think the reason to ask for the blood of animals was this 1) required a decrease in the wealth of the person loosing the animal and 2) it provided a way to feed the nation of Levi, who was required to serve the Lord and not work, they didn't even receive an allocation of land for herds.

wizard - //pianoforte - Why do you think death causes a person to cease to exist? Why can't physical injury to the brain and chemicals/electrical current in the brain cause the observed impact if the brain is simply the sole (intended) communication link to the soul? Perhaps different parts of the soul solely communicate through different parts of the brain, either initially or as a decoding of message function?//

Because I see no evidence that a "soul" exists, that there is something distinguishable from the brain, or any point in invoking one.

//How do you answer the situation where there is no brain function for a short while, then the person "wakes up?"//

Can you give an example?

Here are two different results to indicate that there is a difference between the soul and the brain:

1) In those few cases where the internal connection between the brain halves has been severed (accident or surgery to deal with a medical problem such as some extreme elliptic seizures) and an instruction is followed that only one side of the brain can receive and the person is asked why they are following the instruction (the verbalization center being on the other side from the instruction) the answer is always, "I don't know why I am . . ."  Same for a hypnotic suggestion.  2) However, when an electrical current is applied to the brain and moves a limb, the patient always says something different, "Why did you move my . . . "

While not overwhelming in nature, in the first instances the person's mind made the decision to move and the person knew and accepted this, even when they didn't know/weren't able to say why.  However in the second instance the person's mind didn't make a decision to move a body part and knew that their mind didn't make that decision.

If the mind was simply the sum of the brain, it would seem that the responses would be the same that they didn't know why they moved the . . . .

There have been several recoveries from documented brain death, normally associated with cold temperatures which may have preserved the brain cells' membranes so there was a home left for the soul to return to.

If you accept the Bible, there are many instances, but I suspect that you will discount them.

In response to the original question: I have no expectations or beliefs about what happens after death, and see no reason to have such expectations or beliefs, any more than I have reason to believe what tomorrow's lottery numbers might be.  There are some things about which we simply have no basis on which to form any a priori or a posteriori conclusions, and really no reason to concern ourselves with.

[quote="wizard2378"]While not overwhelming in nature, in the first instances the person's mind made the decision to move and the person knew and accepted this, even when they didn't know/weren't able to say why.  However in the second instance the person's mind didn't make a decision to move a body part and knew that their mind didn't make that decision.[/quote]
And the best explanation you can come up with is to invoke the existence of an undetectable phenomenon called the soul?  Seeing as how current understanding of brain function is quite poor, wouldn't you concede that there is a myriad of other [u]falsifiable[/u] and [u]testable[/u] hypotheses that could explain this observation?  Why appeal to the one that you cannot, even in principle, verify or falsify? [hide="on a side note"]Judging by some of the details in your description, I don't believe this experiment ever took place, or if it did then your description of it is disingenuously exaggerated.  In particular, the use of the word 'always' is itself alarming; how many psychology experiments can you think of in which every single subject responds in the same way to an open-ended question?  Furthermore, the answer "Why did you move my..." seems very strange in the context of the question "Why did you follow the instruction?", and the fact that patients 'always' responded in this obnoxious way is puzzling, to put it mildly.  There are other less glaring details that I find questionable, but I'd love to be proven wrong and see an actual reference to the relevant scholarly journal.[/hide]

[quote="wizard2378"]There have been several recoveries from documented brain death, normally associated with cold temperatures which may have preserved the brain cells' membranes so there was a home left for the soul to return to.[/quote]pianoforte asked you for an example, and you just restated the claim that such an example exists, albeit with some added pseudoscientific frills, but again no reference.  One might be inclined to take your word for it, except that in this thread you have managed to make statements like this one:
[quote="wizard2378"]Extreme example of this are a few examples of people with extreme pressure in their cerebral fluids have large sections (90% in one case I saw) of their brains simply gone and still are walking about, talking and seemingly as normal as the rest of us.[/quote]90% of the brain is 'gone' and they're walking about and talking? I won't bother to ask for a reference.

But anyway 'brain death' is, by definition, the state of having a total and permanent loss of brain function.  Thus, by definition, no one can 'recover' from brain death.  In the cases that you are apparently referring to, the medical position would be that the individual has recovered from a state which was [u]misdiagnosed[/u] as brain death.  These cases are, unfortunately, too common; but that points to the fallibility of doctors and a poor state of scientific understanding of brain function, not to divine intervention or the existence of an undetectable soul.

Your car refuses to start, and an auto mechanic tells you there's poor compression in one of your cylinders, so you'll need a new engine.  The next day your car starts and runs fine.  Is the more probable explanation that a car god returned your car's mechano-soul to it thus reviving it from engine death, or that the mechanic overlooked the fact that it was a rainy day, you had old spark plug wires, and one of your spark plugs was loose?

In either the brain death or the engine death case you can surely argue that it was not in fact a misdiagnosis, but rather a correct diagnosis followed by a miracle.  But this position would be at odds with Occam's razor and the scientific method in general.  If your position is that you reject the principles of the scientific method, this puts into question your entire line of reasoning, since you seem to be trying to justify your beliefs on a scientific basis.

[quote="JoeBlow"]If your position is that you reject the principles of the scientific method, this puts into question your entire line of reasoning, since you seem to be trying to justify your beliefs on a scientific basis.[/quote]  Yes, wizard makes [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1216294&search_id=101082644#1216294]a habit[/url] of this particular variety of nonsense.  (wizard is also very good at assertion, but as you implicitly note, not so good at providing evidence to support his or her assertions.)  Nice post, by the way.

Joe - Interesting post.

Thanks for answering that you have no expectations or beliefs, then following that apparently you don't believe in anything without a reference.

If you read the thread, I was answering a question someone else raised.  I provided what I had and explained it.  Nothing that I asserted.  I didn't assert the existence of a soul, though I believe in souls.  To answer your abrasive question, I responded with what I knew to answer a question posed in a post on a thread I started.  I think that is only polite.

I'm glad you won't ask for a reference.  I don't just read scientific studies.  In fact, this was on a TV show and they interviewed the person and the doctor who was treating him.  I guess it could all have been a massive conspiracy.

Your discussion on brain death obviously invoked a different definition that the one I was using.  Yours becomes a tautology.  If someone ever woke up than it must have been misdiagnosed.

For someone who is demanding references, your example isn't even very close to the situation you are trying to ridicule.  One wonders why you have so much anger against beliefs in God.

Your discussion about scientific method left out the alternative I was suggesting.  Perhaps the soul left the brain home, for a short time and returned, just as a heart can stop beating for a short while and if it starts up again the person can return from heart death (unless you define that term to mean that the heart stops and never restarts).  One can suggest that perhaps this would be a more logical explanation than failed measurement of brain activity.  It doesn't require a miracle.

Further, Occam's razor isn't a logical part of the scientific method.  It is simply the label for a preference, in unknown circumstances to try to select the same of potential alternatives and if all are the potential alternatives are logically just as valid, we should select the simplest because using it is usually easier.

LBJ - Thanks for the half complement.

I believe that there are things that are not proven that are worth discussing and thinking about.  You may disagree with me on this point.  That's the majority of what scientists who are working in science and not just teaching do.  Also, we have seem much of what was scientifically proven turn out to be a bit more open that we thought.  An example here, before you ask, is that we used to think and teach in medical school and other places that the appendix had no function, it was proof that evolution produced humans because it was left over (a residual organ) now we know that people who have had their appendix removed have less ability to resist radiation exposure.

One Dr. Mirkin, who published a newspaper several years ago, until he retired, said that when he was in medical school, 30 years ago or more, he was taught that there were something like 20 residual organs.  We have learned that all of them have a function and are not really residual organs as explained before.

Anyway perhaps too much of a response from me.

I don't want people desecrating my grave, tbh, but I'd like my family and friends to remember me fondly, at least.

I have been taught nothing, because I listen to my own beliefs first and foremost, thank you.

Maybe I'll be reincarnated, but most likely I think I'll either be gone for good or spent a million years dead and then my soul gets reborn. Way too many people to come alive, and plus who even has the time to craft new people's every tidbitted personality? Pretty ironic.

[s] awkward age reveal be like: [/s]

[quote=HamstPan38825][hide=@#1][url=aops.com/community/user/12965][b]wizard2378[/b][/url] · Sep 17, 2008, 5:20 PM [url=aops.com/community/p1257244](view)[/url][color=transparent]helo[/color]
What do you think will happen to you when you die, or what are the potential options, if you are not sure which option you will follow?

Alternatively, what have you been taught will happen when this life is over?

Not the central part of most people's thoughts on a frequent basis, but when attending a funeral sometimes one's thoughts travel on lightly traveled paths.

-----------
[color=#5b7083][aops]x[/aops] 1[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]
This is just extremely, extremely ironic, but this thread was originally posted 11 minutes after I was born myself.

Everyone wants themselves to be remembered...[/quote]

cool, 69 posts, and im 12 days older than you

@above ikr :P

sooooooooo

I believe that I will go to heaven to be with my god

ik, simple, but still

Guys back on topic.

When we die, we return to dust.

Genesis 3:19 KJV: 19 In the sweat of thy face shalt thou eat bread, [b]till thou return unto the ground[/b]; for out of it wast thou taken: for dust thou art,[b] and unto dust shalt thou return.[/b]

We will then be judged on Judgement Day, and those who are condemned will burn in Hell, which is the second death.

[quote=pezzeepenguinstalker][youtube]https://www.youtube.com/watch?v=WPuUZskEYFE[/youtube]

Can someone explain to me why grave dancing is a trend?[/quote]

Tiktok is bleep imo

also how is this possible? Sep 18, 2008, 2:58 PM• 3 Y
Y by GCA and 2 other users


Suppose $p$ is a prime with $p \equiv 3 \; \pmod{4}$. Show that for any set of $p-1$ consecutive integers, the set cannot be divided two subsets so that the product of the members of the one set is equal to the product of the members of the other set.

is easy to see that if one of the numbers is a multiple of p, this is impossible. so, the product of the numbers is equal to $1.2.3..(p-1)$ (mod.p). by wilson theorem, this product is $\equiv -1 (mod.p)$.

if the division is possible, we have that the product is one square mod. p, and this is impossible, because -1 isn't a quadratic residue mod. p, for $p\equiv 3 (mod.4)$


function
inequalities
algebra unsolved
algebra

The functions $ f(x)\minus{}x$ and $ f(x^2)\minus{}x^6$ are defined and increasing for positive $ x$. Prove that the function $ f(x^3)\minus{}\frac{\sqrt{3}}{2}x^6$ is also increasing for positive $ x$.

$ f$ might not be differentiable, but just using the definition of increasing is very similar.
[hide]
The use of $ x^2$ and $ x^3$ in the second and third equations are just nuisances (in the case of positive $ x$). Equivalently, we're given $ f(x)\minus{}x,f(x)\minus{}x^3$ both increasing for positive $ x$, and want to show true of $ f(x)\minus{}\frac{\sqrt{3}}{2}x^2$.

To do this, let $ a,b$ be any positive reals with $ a\geq b$. We are given $ f(a)\minus{}f(b)\geq a\minus{}b$ and $ f(a)\minus{}f(b)\geq (a\minus{}b)(a^2\plus{}ab\plus{}b^2)$ and want to show $ f(a)\minus{}f(b)\geq (a\minus{}b)(\frac{\sqrt{3}}{2}(a\plus{}b))$.

If $ a\plus{}b\leq \frac{2}{\sqrt{3}}$, we're done by the first inequality. Otherwise, we have $ a^2\plus{}ab\plus{}b^2\geq \frac{3}{4}(a\plus{}b)^2\geq \frac{\sqrt{3}}{2}(a\plus{}b)$, so the second inequality can be used.
[/hide]


calculus integration conics ellipse LaTeX

[quote="ComplexZeta"]The \oiint command for surface integrals doesn't work. Could you please add it? Unfortunately, I'm not sure which package it's contained in.[/quote]
Did you mean to type two 'i's in there? I assume you did, because \oint works: [tex]\displaystyle \oint[/tex]

I think the two i's are intentional, as he typed it the same way in another post (in other problem solving topics).  But I may be wrong.

Apparently its the wasysym package. If that helps.

\oint makes one integral with a circle through it. \oiint makes two integrals with an ellipse through them, etc.


Hello everyone,

 Ok, I need your help (basically everyone here). I realized today that I absolutely suck at reaction prediction (USNCO type). I was wondering if you people could post some tips on how to get the right products. I know my solubility rules, and other basic rules such as :
1. Metal Oxides in water form base
2. Nonmetal Oxide + H20 --> Acid
3. MnO4- --> Mn2+ in acid and MnO2 in base.
etc.

Im looking for the real esoteric rules/reactions that they pummel you with
(I.e Rules for carbides? Sulfides? etc.)

Please reply ASAP
Thankyou for your help

Honestly I would say that the mix is so varied that, for the more esoteric ones, you just need to rely on reactions you've seen before.  I would suggest going through old tests, but focus more on other areas.  (Unless you have these perfect already, in which case you don't have anything to worry about.)

If you feel that was too much of a cop out answer, I apologize.  But I don't really have any better advice on this topic.

Do you think that a 50 mc and a 85 open ended with a decent lab score can make camp? (and yes I have seen Scott's statement)

I thought the exam for this year was to be administered between April 17 and 21. Anyway, I'm taking it on April 18.

@avemuri57: That's a pretty good score, I'd say.  I've made the camp with under 50 MC, although I can't attest to my FR score.


Just the roots.. not a proof:

$ x_{solution}\in\left\{2,3,4\plus{}\sqrt{10},4\minus{}\sqrt{10},\minus{}1\plus{}\sqrt{5}i,\minus{}1\minus{}\sqrt{5}i,\frac{1}{2}\minus{}\frac{\sqrt{23}i}{2},\frac{1}{2}\plus{}\frac{\sqrt{23}i}{2}\right\}$

[hide="Full Solution"]$ x^{2}\minus{}3x\plus{}6 \equal{}\left(x\minus{}\frac{3}{2}\right)^{2}\plus{}\frac{15}{4}$, so $ x^{2}\minus{}3x\plus{}6$ is always positive.  Therefore we can safely take the square root of $ x^{2}\minus{}3x\plus{}6$, knowing that it will always result in a positive number.

Let $ a \equal{} (x^{2}\minus{}3x\plus{}6)^{2}$ and $ b \equal{} x^{2}$.
$ a^{2}\minus{}29ab\plus{}100b^{2}\equal{} 0$
$ (a\minus{}4b)(a\minus{}25b) \equal{} 0$
$ a \equal{} 4b$ or $ a \equal{} 25b$

Case 1: $ a \equal{} 4b$
$ (x^{2}\minus{}3x\plus{}6)^{2}\equal{} 4x^{2}$
$ x^{2}\minus{}3x\plus{}6 \equal{} |2x|$
Case 1a: $ x > 0$
$ x^{2}\minus{}3x\plus{}6 \equal{} 2x$
$ x^{2}\minus{}5x\plus{}6 \equal{} 0$
$ x \equal{} 2$ or $ x \equal{} 3$
Case 1b: $ x < 0$
$ x^{2}\minus{}3x\plus{}6 \equal{}\minus{}2x$
$ x^{2}\minus{}x\plus{}6 \equal{} 0$
$ x \equal{}\frac{1\pm i\sqrt{23}}{2}$

Case 2: $ a \equal{} 25b$
$ (x^{2}\minus{}3x\plus{}6)^{2}\equal{} 25x^{2}$
$ x^{2}\minus{}3x\plus{}6 \equal{} |5x|$
Case 2a: $ x > 0$
$ x^{2}\minus{}3x\plus{}6 \equal{} 5x$
$ x^{2}\minus{}8x\plus{}6 \equal{} 0$
$ x \equal{} 4\pm\sqrt{10}$
Case 2b: $ x < 0$
$ x^{2}\minus{}3x\plus{}6 \equal{}\minus{}5x$
$ x^{2}\plus{}2x\plus{}6 \equal{} 0$
$ x \equal{}\minus{}1\pm i\sqrt{5}$

So the roots are:
$ \boxed{2,3,\frac{1\pm i\sqrt{23}}{2},4\pm\sqrt{10},\minus{}1\pm i\sqrt{5}}$[/hide]


Hi, I have always been curious about this and a search on Google led to nothing.  How do mainframes and the like generate Pi? I used to hear of stories of x amount of digits generated but what series or function do the computers use to actually generate 3.14159......?

This one might be the one they use now. I know I have seen it in several places:

:pi:=:sigma: _{n=0}^:inf: (4/(8n+1)-2/(8n+4)-1/(8n+5)-1/(8n+6))(1/16)^n. I believe that converges very quickly.

Ramanujan came up with a bunch of really exotic :pi: formulae which are completely mind-boggling.

The one you will often see is :pi:/4=1-1/3+1/5-1/7+1/9-... However, this converges painfully slowly. When I was in 5th grade or so, my father wrote a computer program to test the convergence of this formula, and we ran it for half a day, and we got 10 digits correct. So that's probably not the smartest way to compute :pi: if you want to know a lot of digits.

Or theres :pi::^2:/6 = 1/1 + 1/4 + 1/9 ... theres heaps of formulae.. Theres also some strange one that can work out the nth digit of pi written in hexadecimal. Crazy.

The hexadecimal one is the first one I gave I believe. It has the (1/16)^n term, and I think that's how it works. I got that from the MathWorld page.

Oh true, so it does.. never realised that before. Its just the nth digit * (1/16)^n..

The fun thing is to try to derive new "formulas".

Yeah. I once tried to work out a formula of my own, just put a regular n gon inside a circle and worked out the perimeter and radius. Ended up with :pi: = lim (n->:inf:) n sin (180/n) in degrees here. Bit later I found the easy way to get to that formula, but I was happy for a while :)

Go on mathworld and look at some of the pi formulas The Math Deity Ramanujan came up with. They still do, and always will make me crap my pants whenever I see them.

IW.OAV

[color=cyan]Here's a cool one:

:pi: = 2^n * :sqrt:(2 - :sqrt:(2 + :sqrt:(2 + :sqrt:(2 + . . . 2))))
where there are all together n 2's under the square root sign.  I don't know really how fast it goes, although you get 3.1415877 . . . at n = 9.  I dunno how you'd write a program for that, though -- it seems rather messy, so just manipulating it properly might slow down the process.[/color]


Show that the Diophantine equation y^2=x^3+23

Add $ 4$ to both sides and factor. How do primes dividing the left hand side look like¿ Show that not only such primes occure on the right.


calculus
integration
trigonometry
complex analysis

Find $\int_0^\infty \frac{\cos x-e^{-x}}{x}dx$

Consider the integral of $\frac {e^{-z}}z$ over the contour that consists of a quarter circle around $0$ of radius $r\to 0+$, the interval $[ir, iR]$ of the imaginary axis ($R\to+\infty$), a quarter circle of radius $R$ and the interval $[R,r]$ of the real axis. The whole integral is $0$. The integral over the large quarter circle tends to $0$. The integral over the small quarter circle tends to $i\frac\pi 2$. At last, the integral over the union of two intervals tends to $\int_0^\infty \frac{\cos x-e^{-x}}x\,dx - i\int_0^\infty\frac{\sin x}x\,dx$. From here it follows that the integral we are interested in is $0$ and, as a bonus, that $\int_0^\infty\frac{\sin x}x\,dx=\frac \pi 2$.


Let me state a stronger problem.
Let $ G$ be a group on $ n$ elements and $ p$ be a prime number that divides $ n$. The number of solutions to the equation $ x^{p}=1$ is a non zero multiple of $ p$.(Cauchy)
Oh...btw...wrong section...bad title... :P

[quote="mdk"]Show that a finite group with an even number of elements must contain an element x or order 2. (That is $ x^{2}= e$, where $ e$ is the identity.)[/quote]

[hide="Hint"]
Pair the elements with their inverses.
[/hide]


Let $x,y,z$ be non-negative real numbers such that $x+y+z=2$, prove that
\[ 3xyz(x^3+y^3+z^3)+4(x^3+y^3+z^3)^2\leq16.  \]

[quote="leepakhin"][quote="pvthuan"][quote="leepakhin"]Your inequality is not true when $x=2$, $y=z=0$. :([/quote]

No, it is not true. I was so stupid.[/quote]
Everyone makes mistakes. You cannot say you are stupid just because you are not careful enough. ;)[/quote]
Maybe :D .

I want to edit the problem this way
\[ \text{Prove that if}\;\, x,y,z\geq0 \;\,\text{and} \,\; x+y+z=2,\; \text{then} \; 3xyz(x^3+y^3+z^3)+4(x^3+y^3+z^3)^2\leq 256. \]

[quote="pvthuan"]I want to edit the problem this way
\[ \text{Prove that if}\;\, x,y,z\geq0 \;\,\text{and} \,\; x+y+z=2,\; \text{then} \; 3xyz(x^3+y^3+z^3)+4(x^3+y^3+z^3)^2\leq 256.  \][/quote]
$3xyz(x^3+y^3+z^3)+4(x^3+y^3+z^3)^2\leq 4(x+y+z)^6$
$\Leftrightarrow 24\sum_{sym}x^5y+60\sum_{sym}x^4y^2+36\sum_{sym}x^3y^3+\frac{117}{2}\sum_{sym}x^4yz+240\sum_{sym}x^3y^2z+60\sum_{sym}x^2y^2z^2\geq 0$
This also shows that the inequality is quite weak.


Let's $ A$ and $ B$ matrix such as
$ e^{A}e^{B} \equal{} e^{B}e^{A} \equal{} e^{A \plus{} B}$ then we can't say that :
$ AB \equal{} BA$ find example when this statement is true

[quote="loup blanc"]$ A \equal{} \begin{pmatrix}0 & \minus{} \pi \\
{\pi} & 0\end{pmatrix}$
$ B \equal{} \begin{pmatrix}0 & \pi( \minus{} 7 \plus{} 4\sqrt {3}) \\
{\pi}(7 \plus{} 4\sqrt {3}) & 0\end{pmatrix}$[/quote]
yes.
if denote $ \minus{}\frac{A}{\pi}\equal{}I$ then;
$ e^{(\alpha E\plus{}\beta I)t}\equal{}e^{\alpha t}\cos \beta t  E\plus{}e^{\alpha t} \sin \beta t I$
Hence we need $ e^{B}\equal{}E$ then $ B\equal{}2\pi I$
 :)

Your post is absolutely understable (for me).

I read again my last post ; I see that I wrote `understable` and I wanted to say `incomprehensible`. Clearly the Extremal`s post is a joke. Now we speak seriously; the 2 following problems in $ \mathcal{M}_n(\mathbb{C})$ are difficult:
1) Find the couples $ A,B$ s.t. $ e^{A\plus{}B}\equal{}e^Ae^B\equal{}e^Be^A$; the result is known for $ n\equal{}2,3$ and unknown for $ n>3$. If $ n\equal{}2$ the typical solution is the one I gave ($ e^A,e^B$ are homotheties).
2) Find the couples $ A,B$ s.t. $ e^{A\plus{}B}\equal{}e^Ae^B$; the result is known for $ n\equal{}2$ and unknown for $ n>2$.


So, we can simplify at least one of these.

$ 4^{\frac14} \equal{} \sqrt2$

So we have $ \sqrt[3]{3}, \sqrt2, \sqrt[5]{5}, \sqrt[6]6$.

OK, so obviously, $ \sqrt2 \approx 1.414$.

Let's estimate the others:

OK, so you know that $ (1.4)^3 \equal{} \frac{14^3}{1000} \approx 2.7$

You also know that $ (1.5)^3 \equal{} \frac{15^3}{1000} \equal{} 3.375$.

So $ 1.4 < \sqrt[3]{3} < 1.5$.

And since we know that the graph of $ y \equal{} x^3$ gets steeper and steeper, we can estimate that it's probably gonna be closer to $ 1.5$.  (Well, if you use a calculator, you get $ \approx 1.4422$, but... :P)

Well, you use the same method to guess for $ \sqrt[5]{5}$ and $ \sqrt[6]{6}$, but if you're a lazy bum like me, pull out a calculator.  You can also guess that $ \sqrt[5]{5}$ and $ \sqrt[6]{6}$ are [b]most likely[/b] gonna be smaller than $ \sqrt[3]{3}$, so our answer is $ \boxed{3^{\frac13}}$.

Hmmm... I know there's a better way to do this (:P), but I don't remember it.  Anyone?

Note that $ 3^4>4^3 \Rightarrow 3^{\frac{1}{3}} > 4^{\frac{1}{4}}$.  You can go down the line finding that $ 3^{\frac{1}{3}}$ is the largest.

Additionally, for $ n\geq 3$, $ \sqrt[n]{n} > \sqrt[n\plus{}1]{n\plus{}1}$, which makes the problem kinda trivial?

[quote="5849206328x"]Additionally, for $ n\geq 3$, $ \sqrt [n]{n} > \sqrt [n \plus{} 1]{n \plus{} 1}$, which makes the problem kinda trivial?[/quote]

That's what i was looking for. :P

I put 3^1/3 in and it told me I was wrong. What's wrong with this?

I think you might have to put in the parenthesis; something like 3^(1/3).


Let $E$ be a real vector space and $A \subset L(E)$
$A$ is called irreductible if $E$ and $\{ 0 \}$ are the only subspaces of $E$ which are stable by all the elements of $A$
-show that : If $E=\mathbb R^{n}$ and $n \geq 2$ then $SO_{n}(\mathbb R)$ is irreductible.
   $SO_{n}(\mathbb R)$ is the set of orthogonal real matrices whose determinant is $1$

Lemma 1: In any vector space $V$, a linearly independent set can be extended to a basis. In particular, given a nonzero vector $v$, we can find a basis for the space having $v$ as its first element.

Lemma 2: (Gram-Schmidt) Given linearly independent vectors $v_{1},\dots,v_{n}$, we can find orthogonal unit vectors $w_{1},\dots,w_{n}$ such that the span of $\{v_{1},\dots,v_{k}\}$ is equal to the span of $\{w_{1},\dots,w_{k}\}$ for each $k$. In particular, $w_{i}$ is a constant multiple of $v_{1}$.

Corollary: Given any unit vector $w$, we can find an orthogonal matrix $M$ with $w$ as its first column.
As long as $n\ge2$, we can multiply the second column by $1$ or $-1$ to give this matrix determinant 1.

Now $M\cdot e_{1}=w$, where $e_{1}$ is the first standard basis vector. $SO_{n}$ acts transitively on the set of unit vectors, and any nontrivial invariant space must be all of $\mathbb{R}^{n}$.


Αποδειξτε οτι......Εστω n ενας θετικος ακεραιος.Αν f μια πολλαπλασιαστικη συναρτηση τοτε η συναρτηση F=Σf(d) ειναι επισης πολλαπλασιαστικη.(οπου με Σ συμβολιζω το αθροισμα και d οι διαιρετες του n).

Έστω $m$ η συνάρτηση του Mobius και $v$ η αντίστροφός της αφού η $m$ είναι ενεληκτικά αντιστρεπτή .

Άρα για το πρόβλημά μας έχουμε
$F(n)=\sum_{d|n}f(d)=\sum_{d|n}f(d)v(\frac{n}{d})=(f*v)(n)$

και αφού οι f ,v είναι πολλαπλασιαστικές και η $F$ είναι επίσης πολλαπλασιαστική.

Ωραιος....αυτη την λυση ειχα και γω στο μυαλο μου.(τι δυναμη που εχει το γονομενο κατα Dirichlet!)Θα ηθελα ομως και μια δευτερη αποδειξη.Oποιος καταφερει να βρει παρακαλω πολυ να την στηλει. :wink:

Έχεις κάποια ιδέα στο μυαλό σου ?

ΧΡΟΝΙΑ ΠΟΛΛΑ ΣΕ ΟΛΟΥΣ!Και πανω απολα υγεια.Ελπιζω και το 2007 η μαθηματικη κοινοτητα να συνεχησει δυναμικα τις προσπαθειες της για την αναπτυξη της μαθηματικης παιδειας των νεων.Ευχομαι επισης στους <<μικρους>> να συνεχισουν απτοητοι την ενασχοληση τους με την μαθηματικη επιστημη και την διεκδικηση τους για καποιο-α χρυσα μεταλλια.
Σιλουανε θα χρησιμοποιησω ενα λημα για την αποδειξη της ασκησης.Εστω n=ab ενας θετ.ακεραιος με (a,b)=1.Τοτε ολα τα γινομενα mn με m,n διαιρουν τους a,b αντιστοιχα,δινουν ολους τους διαιρετες του n και ειναι διαφορα ανα δυο.Αρα F(n)=F(ab)=Σf(d)=Σf(mn)=Σf(m)f(n)=F(a)F(b).

Ωραία Δημήτρη. Πάντως μιας και τελευταία έγινε λόγος περί αριθμών "square-free" και μιας και ειπώθηκε και η συνάρτηση του [u]Moebius[/u], να πω δύο λόγια σε όσους δεν τη γνωρίζουν.

Εντάσσεται στην αριθμοθεωρία, και παίρνει τις τιμές: $-1$, $0$, $1$. Την τιμή $0$ την παίρνει όταν ο αριθμός είναι "non-square-free", ενώ αν είναι "square-free", παίρνει την τιμή $-1$ αν το πλήθος των πρώτων αριθμών στην ανάλυση είναι περιττό και την τιμή $1$ αν το πλήθος των πρώτων αριθμών στην ανάλυση είναι άρτιο. Επίσης, η αντίστροφος δεν είναι $N\rightarrow N$, αλλά $N\rightarrow P$, όπου $P$ ένα δυναμοσύνολο του $N$.

Ναι συγνωμη που δεν ορισα την συναρτηση του Μοbius για οσους δεν την ξερανε. :oops: Ευχαριστω mostel. :wink:


vector function calculus derivative trigonometry

1) A particle initially located at the origin has an acceleration of $ \overline{a} \equal{} 3.00j \; m/s^{2}$ and an initial velocity of $ \overline{v_{i}} \equal{} 5.00i \; m/s$

Find (a) the vector position and velocity of the particle at any time t and (b) the co-ordinates and speed of the particle at t= 2.00s

[hide=" answer I got"]

Using $ r(t) \equal{} V_{o}t \plus{} \frac{1}{2}at^{2}$ I got $ r(t) \equal{} 5it \plus{} \frac{1}{2}3jt^{2}$ as the position function. The velocity function is simply the derivative of that statement.

As for the co-ordinates, I just plugged in t=2 and got (10,6). Speed was found with pythagoras as $ \sqrt{136} \approx 11.66 m/s$
[/hide]
2 (probably much more interesting)

Prove that a projectile's distance from the origin is a maximum when the position vector is perpendicular to its velocity.

I keep thinking of the fact that wherever the velocity changes sign, the position is at a max or min, but this concept is much more confusing to me with vectors since you have both magnitude and direction to worry about.

ur 1 is right...but thats' a intro physics prblem...but ur problem 2 sounds nice.. :D
so let this be here

and u r right in 1

for 2) though i have a proof that show sthat the time when $ r$ is max and when velocity is perpendicular to posistion vector are the same...i am searching for a better obvious one

[hide="Hint"]
use the vetcorial equations of
$ \vec{r(t)}$ and $ \vec{v(t)}$..
when they are perpendicular their Scalr products is $ 0$
use this fact to get the time $ t$
rest is algebra
 [/hide]

hmm?

$ \vec{r(t)} \cdot \vec{v(t)} \equal{} |\vec{r(t)}\parallel{}\vec{v(t)}|\cos(\theta)$

To get the scalar product to be 0, we set $ \theta$ to be $ \frac{\pi^{r}}{2}$ since $ \cos\left(\frac{\pi}{2}\right)$ is 0. How does this help?

Also, I forgot to give the textbook's solution

[hide]

The vector $ \vec{v}$ tells how $ \vec{r}$ is changing. If $ \vec{v}$ at a particular point has a component along $ \vec{r}$, then $ \vec{r}$ will be increasing (if $ \angle$ between $ \vec{v}$ and $ \vec{r}$ is $ < 90$) or decreasing (if $ \angle$ between $ \vec{v}$ and $ \vec{r}$ is $ >90$) in magnitude. To be at a maximum, the distance from the origin must be momentarily staying constant, and the only way this can happen is if the angle between velocity and position is a right angle. Then $ \vec{r}$ will be changing in direction at that point, but not in magnitude.
[/hide]
Now I don't understand the textbook's solution, mainly the part where vector v has a component along vector r.

ok what ur textbook gave is what i was basiclally thinking but that would be better off to a non calculus student

so what  i meant was thsi the $ \vec{r(t)} \equal{}\vec{v_{0}} t \plus{} 0.5\vec{g} t^{2}\equal{}v_{0}t\cos \theta \vec{i}\plus{}( v_{0}\sin \theta \minus{}0.5gt^{2})\vec{j}$
also $ \vec{v(t)}\equal{}\vec {v_{0}} \plus{} \vec {g}t\equal{}v_{o} \cos \theta \vec{i} \plus{} (v_{0} \sin \theta \minus{}gt)\vec{j}$
now use the fact that if $ \vec{a}\equal{}a_{x} \vec{i} \plus{} a_{y}\vec{j}$ and $ \vec{b}\equal{}b_{x} \vec{i} \plus{}b_{y} \vec{j}$ so $ \vec{a} . \vec{b} \equal{}a_{x}b_{x} \plus{} a_{y}b_{y}$ and set that to 0

Seems that in the projectile's case, it can be a minimum but not a maximum.

ofc it is not a minimum because for any point while ascending the distance is less than any point while it's descending from the origin


LaTeX
combinatorics unsolved
combinatorics

calculate $ \displaystyle \sum_{r\equal{}0}^{m} ( \frac {r}{2} ) ( \frac {m}{r} )$ according to m.

Probably the user is failing to use the LaTeX construct \binom{r}{2} and \binom{m}{r}.  In that case, the problem is easy to kill by writing $ \binom{r}{2} \binom{m}{r} \equal{} \binom{m}{2} \binom{m \minus{} 2}{r \minus{} 2}$ (which can easily be justified either algebraically or combinatorially).  It's also not hard to give a combinatorial interpretation of the entire sum in this case: we're choosing a committee of arbitrary size with two co-chairs.  (Of course, if the OP meant something different then this is all irrelevant ;))

See the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]here[/url].


inequalities
logarithms
function
inequalities proposed

I think simple power mean is enough:
we have $ x^{x} \plus{} y^{y} \equal{} (x^{\frac {1}{y}})^{xy} \plus{} (y^{\frac {1}{x}})^{xy}$ of course $ xy < 1$ thus our inequality is equivalent to the following one:
$ \sqrt [xy]{\frac {(x^{\frac {1}{y}})^{xy} \plus{} (y^{\frac {1}{x}})^{xy}}{2}} < 1$
It's true because:
$ \sqrt [xy]{\frac {(x^{\frac {1}{y}})^{xy} \plus{} (y^{\frac {1}{x}})^{xy}}{2}} < \frac {x^{\frac {1}{y}} \plus{} y^{\frac {1}{x}}}{2} < 1$ because $ \frac {1}{y}\in (1, \plus{} \infty)$ thus $ x^{\frac {1}{y}}\in (0,1)$ same for y :wink: Q.E.D.

As the inequality has two independent terms, we just need to prove $ x^x < 1$ (analogously with $ y^y$) and then add both ineq.

If $ x\in(0,1)$ it can be expressed as $ x \equal{} \frac {1}{k}$ with $ k > 1$ then $ x^x \equal{} (\frac {1}{k})^\frac {1}{k}$. Supose $ x^x > \equal{} 1$ $ \Rightarrow$ $ \frac {1}{k} > \equal{} 1^k \equal{} 1$, but this is impossible since $ k > 1$. Then $ x^x < 1$ and $ y^y < 1$ $ \Rightarrow$ $ x^x \plus{} y^y < 2$

A similar problem, also not hard:
Prove that
$ \forall_{x,y\in R_{\plus{}} \ x\plus{}y\equal{}1} \ x^{y}\plus{}y^{x}<1$ :wink:

Didn't you mean $ \forall_{x,y\in R_{\plus{}} \ x\plus{}y\equal{}1} \ x^{y}\plus{}y^{x}<2$ ???

Because if not, just take $ x\equal{}y\equal{}\frac{1}{2}$ and $ x^{y}\plus{}y^{x}\equal{}2\sqrt{\frac{1}{2}}>1$

The solution for the last one is just the same [b]polskimisiek[/b] gave for the first

$ f(x)\equal{}x^x$
$ f'(x)\equal{}x^x(\ln x \plus{} 1)$
Function has local minimum when $ x\equal{}\frac{1}{e}$. And $ f(0)\equal{}f(1)\equal{}1$. So we can say that : $ x^x<1$ when $ x\in (0,1)$.

[quote="polskimisiek"][b]Nemion[/b], yes sorry, a small typo  :blush: .
A little more interesting:

Let us have $ p,q\in R_{ \plus{} }$ such that $ p \plus{} q \equal{} 1$ prove that:
$ p^{p}q^{q}\leq \frac {1}{2}$  :wink:  :P[/quote]
Try $ p\equal{}0.9$ and $ q\equal{}0.1.$ :wink:

[quote="Yuriy Solovyov"]$ f(x) \equal{} x^x$
$ f'(x) \equal{} x^x(\ln x \plus{} 1)$
Function has local minimum when $ x \equal{} \frac {1}{e}$. And $ f(0) \equal{} f(1) \equal{} 1$. So we can say that : $ x^x < 1$ when $ x\in (0,1)$.[/quote]
Or symply take the logarithm

[quote="arqady"][quote="polskimisiek"][b]Nemion[/b], yes sorry, a small typo  :blush: .
A little more interesting:

Let us have $ p,q\in R_{ \plus{} }$ such that $ p \plus{} q \equal{} 1$ prove that:
$ p^{p}q^{q}\leq \frac {1}{2}$  :wink:  :P[/quote]
Try $ p \equal{} 0.9$ and $ q \equal{} 0.1.$ :wink:[/quote]

I am very sorry  :blush: Of course it was meant to be in the opposite direction
$ p^{p}q^{q}\geq \frac{1}{2}$ :wink:

[quote="polskimisiek"] Of course it was meant to be in the opposite direction
$ p^{p}q^{q}\geq \frac {1}{2}$ :wink:[/quote]
Use $ f(x)\equal{}x\ln x$ is convex function.

[quote="Nemion"]Didn't you mean $ \forall_{x,y\in R_{ \plus{} } \ x \plus{} y \equal{} 1} \ x^{y} \plus{} y^{x} < 2$ ???

Because if not, just take $ x \equal{} y \equal{} \frac {1}{2}$ and $ x^{y} \plus{} y^{x} \equal{} 2\sqrt {\frac {1}{2}} > 1$[/quote]

We can actually say that $ x^y \plus{} y^x \leq \frac {2}{\sqrt {2}} \equal{} \sqrt {2}$ because $ x^y \plus{} y^x$ is both concave and symmetric in both variables over $ (0,1)$ and therefore [i]must[/i] obtain its maximum when $ x \equal{} y$.


Where do you do your homework?

mom's house=bed
dad's house=desk

-jorian

Either in my room, on a cold hard floor, in a trance, or on the dining room table.

[quote="Spectacular"]the period before it's due[/quote]

lol sometimes i do that....some classes the teacher doesn't mind but others they will rip your homework and throw it in the trash

mostly i do my math homework the period it is assigned...

On the bus and as soon as I get home from school. I hate procrastination.

Bed. Floor. At school. Not at all. I like procrastinating.

Um... either on bed, at desk (if I'm really concentrated), or during the period it's assigned (most of the times).


Does this apply both ways?
For f(x)
bijective=>involution
involution=>bijective

There are lots of bijections that aren't involutions.  Take, for example, $ f(x) \equal{} x \plus{} 1$ on $ \{ 0, 1, 2 \}$ with addition $ \bmod 3$.


linear algebra
matrix
linear algebra unsolved

Let $ n$ be a positive integer. Consider an $ n\times n$ matrix with entries $ 1,2,...,n^2$ written in order, starting at the top left and moving along each row in turn left-to-right. (e.g. for $ n \equal{} 3$ we get $ \left[\begin{array}{ccc}1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9\end{array}\right]$)
We choose $ n$ entries of the matrix such that exactly one entry is chosen in each row and each column. What are the possible values of the sum of the selected entries?

Start from the diagonal sum. Here we have $\frac{n(n^2+1)}{2}$. We obtain all possible positions by switching rows and columns. Switching row/column $i,j$ adds (i-j) to the sum and then substracts it. Thus the operations leave our initial sum invariant.

I think I might be missing something.. Isnt $ 1 \plus{} 4 \plus{} 8 \equal{} 13$ a valid selection?

Peter seems to be restricting himself to operations of the form $ P^TAP$ for the same permutation matrix $ P.$ This is similarity and it does fix the trace. But Ottem has a valid objection. You must allow $ PAQ$ where $ P$ and $ Q$ are possibly different permutation matrices, and that does not fix the trace.

I'm sorry, it's the example matrix that needs fixing. The numbers are always placed left-to-right, not alternating as suggested.

The example matrix should be $ \left[\begin{array}{ccc}1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9\end{array}\right]$. This has been corrected now.

In fact, decomposing the matrix into the sum of a constant matrix and a matrix with zeroes in every 2nd row gives us a relatively simple formula for the possible sums ($ \frac {n(n^2 \plus{} 1)}2 \plus{} 2k$ for some $ k$, I think).

EDIT: Aha, thanks for clearing that up Peter.  :)

How about this:

Let $ A$ be the matrix in question. Write $ A\equal{}B\plus{}C$ where $ B$ has constant rows and $ C$ has constant columns. For instance,
\[ A\equal{}\begin{bmatrix}1&2&3&4\\5&6&7&8\\ 9&10&11&12\\13&14&15&16\end{bmatrix}\equal{} \begin{bmatrix}0&0&0&0\\4&4&4&4\\ 8&8&8&8\\12&12&12&12\end{bmatrix}\plus{} \begin{bmatrix}1&2&3&4\\1&2&3&4\\ 1&2&3&4\\1&2&3&4\end{bmatrix}\equal{}B\plus{}C\]
We would like to compute $ \text{trace}(PAQ)$ where $ P$ and $ Q$ are arbitrary permutation matrices.

$ \text{trace}(PAQ)\equal{}\text{trace}(PBQ\plus{}PCQ)\equal{} \text{trace}(PBQ)\plus{} \text{trace}(PCQ)$ since trace is linear

$ \equal{}\text{trace}(BQP)\plus{}\text{trace}(QPC)$ since $ \text{trace}(UV)\equal{}\text{trace}(VU)$

$ \equal{}\text{trace}(B)\plus{}\text{trace}(C)$ since permuting columns leaves $ B$ unchanged and permuting rows leaves $ C$ unchanged

$ \equal{}\text{trace}(A)\equal{}\frac{n(n^2\plus{}1)}2$

Here's a more simple solution.
Let's define a matrix A nxn with $a_{ij}=(i-1)n+j$ which clearly satisfies the form asked. Every sum of the entries of exactly one row and one column is of the form $S= \sum _{i=1}^n a_{i \sigma (i)}$ where $\sigma$ is a permutation of the set $S_n=\{1,...,n\}$. To prove this note that if we had 2 entries on the same row then they are the same entry, if the entries were of the same column, since $\sigma$ is a biyection, then it would be of the same row and hence the same entry. Now we compute S. Note that $\sum _{i=1}^n \sigma (i)= \sum_{i=1}^n i$ in (possibly) different order (because $\sigma$ is a biyection).
$S=\frac{n^3+n}{2}$


[quote=Armenia]Prove that
$ \frac{r}{R}\leq\frac{\sqrt{2(2a^2\minus{}(b\minus{}c)^2)(2b^2\minus{}(c\minus{}a)^2)(2c^2\minus{}(b\minus{}a)^2)}}{(a\plus{}b)(a\plus{}c)(b\plus{}c)}$
for any acute triangle.[/quote]
It's also true for all triangle.


function
real analysis
real analysis unsolved

One of my first ideas for http://www.mathlinks.ro/Forum/viewtopic.php?t=18474 was the following statement:
Suppose $f:[0,1]\to \mathbb R$, $f\in C^{\infty}$. It is known that $f(\mathbb Q\cap [0,1])\subset \mathbb Q$. Prove that $f$ is a rational function.

[i]Prove that it was a wrong idea.[/i]

I will edit this a third time so that I don't spam the topic :).

I was thinking about the following, but I sort of got confused :?:

Let $(q_n)_{n\ge 1}$ be an enumeration of the rationals in $[0,1]$. We could take our function to be $f(x)=\sum_{n\ge 0}a_n(x-q_1)(x-q_2)\ldots(x-q_{2n+1})$, with $a_n$ positive rationals small enough so that the series converges absolutely for all $x\in[0,1]$. It is then in $C^{\infty}$, and it's not rational because it changes sign infinitely often, being positive in infinitely rational numbers, and negative in infinitely many rational numbers.

Is it still wrong? :? Is this "strike three"? :)

Your Highness is wrong two times in a row (it is a second edition above).

I see Your Highness have already corrected your post.

Actually, I wanted to continue this theme. I was thinking what conditions are sufficient for $f$ being a rational function. Next my idea was that additionally $f$ should be an increasing function.

[i]Prove that this idea is wrong too.[/i]

P.S. I solved those problem (see the link in post #1) without these ideas. But questions are still interesting for me.

Just add $x$ to grobber's function.

Also, the following statement holds: If $A$ and $B$ are two countable dense subsets of $[0,1]$ (or of $\mathbb R$), there exists an increasing real analytic function $f$ such that $f(A)=B$. The construction is somewhat similar to that of grobber though more complicated.

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=36274]Yet Another Rational values at rational points[/url] for a continuation.


geometry
trapezoid
circumcircle
incenter
symmetry
geometry proposed

Given trapezoid $ ABCD$ $ BC\parallel{}AD$ $ AB\equal{}CD$. Point $ I$ is center of circle inscribed in triangle $ ABC$. $ F$ lies on bisctor of angle $ BDC$ and $ IF$ is perpendicular to $ BC$ Circumcircle of triangle $ CFD$ intersects $ BC$ in $ C$ and $ E$.Prove that triangle $ DEF$ is isosceles.

It should be
[quote="mr.danh"]Given an isosceles trapezoid $ ABCD$ with $ BC\parallel{}AD$. Point $ I$ is center of circle inscribed in triangle $ ABC$. $ F$ lies on bisctor of angle $ BDC$ and $ IF$ is perpendicular to $ BC$ Circumcircle of triangle $ CFD$ intersects $ BC$ in $ C$ and $ E$.Prove that triangle $ DEF$ is isosceles.[/quote]
Consider the cirlce (ABCD). Denote M be the midpoint of arc BC which does't contain A,D; I be the incenter of triangle BCD. See that AI, BI' passes through M. By symmetry, $ IK\parallel{}BC\Rightarrow IK\perp IF$. So M is the midpoint of KF, which implies that F is the D-excenter of triangle BCD, therefore CF is th external bisector of angle BCD, easy to see that $ \angle DEF \equal{} \angle EDF\Rightarrow EF \equal{} DF$.

I can't see why  F is the excenter of triangle BCD. I hope somebody could explain it to me


quadratics
function
analytic geometry
graphing lines
slope
conics
parabola

[quote="catalan"]Hello everyone,

I am having trouble answering the following questions and I would greatly appreciate any help!

Thank you very much.

---

1. Let [i]f[/i] be a function defined by $ f(n) = 2f(n - 1) + 3f(n - 2)$, where $ f(1) = 1$ and $ f(2) = 2$. Determine $ f(5)$.

[u]My work:[/u]

$ 2f(0) + 3f( - 1) = 1$

$ 2f(1) + 3f(0) = 2$

2. Consider the function $ f(x) = \sqrt {x^2 + \frac {1}{x^2}}$ defined for $ x > 1$. Give an expression in simplest form for $ f(f^{ - 1}(f(f^{ - 1} (x))))$.

3. Find three functions f, g, and h that satisfy both the following conditions:

${ f(g(h(x))) = \frac {x}{x + 2}}$ and $ h(g(f(x))) = \sqrt {(\frac {x - 1}{x + 1})^2 + 1}$.

4. A line with slope 1 passes through the point P on the negative x-axis and intersects the parabola at $ y = x^2$ at points Q and R. If PQ = QR, find the ordinate of the line PR. [size=75](Hint: intersect $ y = x^2$ and $ y = x + b$)[/size]

[u]My work:[/u]

$ x^2 = x + b$

$ x^2 - x - b = 0$[/quote]

(1) I'll just assume that this is some kind of joke... if someone tells you to inspect something 100 ft in front of you, do you go back some miles and look around there?  :roll:

(2) What's the definition of $ f^{-1}(f(x))?$

4. Just solve for the two different x values and use that if $ PQ=QR,$ $ x_1 = x_2-x_1.$

[hide]Solution: $ b = \frac{2}{9}.$[/hide]

[quote="catalan"][quote="bertram"]

(1) I'll just assume that this is some kind of joke... if someone tells you to inspect something 100 ft in front of you, do you go back some miles and look around there?  :roll:

(2) What's the definition of $ f^{ \minus{} 1}(f(x))?$

4. Just solve for the two different x values and use that if $ PQ \equal{} QR,$ $ x_1 \equal{} x_2 \minus{} x_1.$

[hide]Solution: $ b \equal{} \frac {2}{9}.$[/hide][/quote]

Thanks for your reply, bertram.

1. As easy as it may seem, I still don't understand how to solve the problem. The answer given is: [hide]$ f(5) \equal{} 61$[/hide]

2. The definition of $ f^{ \minus{} 1}(f(x))$ is x. Therefore, would the answer just be x?

4. I am not exactly sure what you mean. The answer given is different from yours: [hide] $ 4 \pm 3 \sqrt {2}$.[/hide][/quote]

(1) With the values given, compute $ f(3),f(4),f(5)$ in that order.
(2) I'd say so too.
(4) Reading comprehension FTW. I read negative y-axis. You can still use that $ x_1\minus{}b\equal{}x_2\minus{}x_1$ ;)

[quote="catalan"][quote="bertram"]

(1) With the values given, compute $ f(3),f(4),f(5)$ in that order.
(2) I'd say so too.
(4) Reading comprehension FTW. I read negative y-axis. You can still use that $ x_1 \minus{} b \equal{} x_2 \minus{} x_1$ ;)[/quote]

Thanks for your reply once again, bertram.

Just for question 4, which two abscissas are you referring to?[/quote]

$ b$ is the zero of the line.
$ x_1$ is the left point of intersection of the two functions.
$ x_2$ is the right point of intersection.

Then you know that $ b,x_1,x_2$ form an arithmetic progression, and just compute all of them and try to find the proper $ b.$

[quote="bertram"]

Thanks for your reply once again, bertram.

Just for question 4, which two abscissas are you referring to?[/quote]

$ b$ is the zero of the line.
$ x_1$ is the left point of intersection of the two functions.
$ x_2$ is the right point of intersection.

Then you know that $ b,x_1,x_2$ form an arithmetic progression, and just compute all of them and try to find the proper $ b.$[/quote]

Hello bertram,

Thanks for your reply. Instead of computing the arithmetic progression, is there another way to find the ordinate?

Here is a diagram of the question:

[img]http://img239.imageshack.us/img239/4733/quadpf0.jpg[/img]

*Also, please ignore the points on the parabola. They were not originally on the diagram (I copied the parabola from another picture).


Find the median and the mode for the following set of data:
$ 1,7,3,2,3,5,4,8,7,9,7$. What is the mean of these two numbers?

Both the median and mode are found easily after reordering the terms.

$ 1, 2, 3, 3, 4, 5, 7, 7, 7, 8, 9$

Mode is 7. Median is 5.

The mean of these numbers is $ \boxed{6}$.


polynomial p(x),q(x) has real coefficients satisfying the condition
1. p(x),q(x) has at least a solution
2. ${p(1+x+{q(x)}^{2})={q(1+x+}{p(x)}^{2})}$ for all x
prove that :  ${p(x)}^{2}={q(x)}^{2}$ for all x


linear algebra
matrix
superior algebra
superior algebra solved

$n\geq 2, r\in {1;2;...;n}$ let $S_r = { A\in M_n(Z/2Z), rg(A)=r }$
1/ Prove for any $A \in S_n $ and $ B \in S_r$ then $AB \in S_r$
2/ Find $\sum\limits_{X\in S_r}X$
By Aurel-Mihai FULGER, Valentin VORNICU

I think the text for (1) is wrong: one of $A,B$ should be in $S_n$, so assume it's $A$. Then $r(B)=r(A^{-1}\cdot AB)\le r(AB)\le r(B)\Rightarrow r(AB)=r(B)$.

For (2) we use (1): for any invertible matrix $A$, we have $AS=SA=S$, where $\displaystyle S=\sum_{X\in S_r}X$.

http://www.mathlinks.ro//files/rmo_district_2004.pdf

Here where I found this problem look 12 th grade


A quarter of one percent of 40 is one less than what number? Express your answer as a decimal to the nearest tenth.


[hide]Divide the second equation by xy to get $x+y = \frac{880}{xy}$.  You can substitue this into the first equation and get $\frac{880+x^2y^2}{xy} = \frac{71}{1}$.  When you cross multiply you get $880+x^2y^2-71xy = 0$.  Then you can factor that equation to get xy = 16, or xy = 55.  If xy=16, then x+y=55, which can't work with x and y being positive integers.  That means that xy = 55, and x+y=16, making x=5, and y=11 (or vice versa).  When you plug in 5 and 11 into $x^2+y^2$ you get 146.[/hide]

[hide]Let $a = x + y, b = xy$. Then we have $a + b = 71, ab = 880$ Solving this makes the desired solution obvious.[/hide]

Nice.

darktreb, I like your solution. :)

[hide]Let $a=x+y$ and $b=xy$. Our two equations become $a+b=71$ and $ab=880$. Solving, $a=55$ and $b=16$, so our answer is $a^2-2b=\boxed{146}$.[/hide]

4everwise, you mixed up a and b. $55^2-2\times 16 \neq 146$  :P

when I first saw this problem (5 mts ago) I solved like darktreb, but isn't there more:
we have ab = 880 a+b =71
now add, ab + a + b +1 = 952
(a+1)(b+1) = 951 (simon's fft)
now we want a,b, a^2-2b positive
just find factors.

problem bashing by dummy variables... :P
how do the noobies taking the AIME tackle this kind of problem 1, it actually seems a little in depth for a one.

Problem bashing by dummie variables.  Definitely!  Take it from a noobie.   :lol:  :roll:

This is a good problem, though a little hard for #1.

The technique is standard and has a proper name - $u = x + y, v = xy$ are called the "elementary symmetric polynomials in two variables."  They correspond to the non-leading coefficients of the polynomial with roots $x, y$, so they're just Vieta's sums and they have a lot of widespread applicability.

[hide="Different solution without substitution"]
We can add one to the first equation to get $ xy\plus{}x\plus{}y\plus{}1\equal{}72$. Factoring, we have $ (x\plus{}1)(y\plus{}1)\equal{}72$.
WLOG, assume that $ x>y$, then the possible solutions are $ (71,0),(35,1),(23,2),(17,3),(11,5),(8,7)$.
If we factor the second equation, we have $ xy(x\plus{}y)\equal{}880$. So either $ xy$ or $ (x\plus{}y)$ must have $ 5$ as a factor. If we say $ xy$ has $ 5$ as a factor, then the only possible solutions are $ (35,1)$ and $ (11,5)$. Substituting into the equation, we see that $ (11,5)$ is the solution. So the answer is $ 11^2\plus{}5^2\equal{}\boxed{146}$.
[/hide]

[quote=kante314][quote=pith0n][quote=OlympusHero][hide=Solution]Rewrite as $xy+x+y=71, xy(x+y)=880$. Denote $xy=a, x+y=b$. Then $a+b=71, ab=880$, so $a,b$ are roots of $x^2-71x+880$, meaning $a,b$ are some permutation of $16$ and $55$. But obviously $xy>x+y$, so $xy=55, x+y=16$. We have $x^2+y^2=(x+y)^2-2xy=256-110=\boxed{146}$.[/hide]

Hard for a #1 in 1991![/quote]

The old threads don't have this but is this AIME 1, AMC 12 1, AMC 10 1, AMC 8 1, or what?
It looks too hard to be 10 and 8 so probably AIME or AMC 12 but idk[/quote]

read the title of the thread :P[/quote]

:facepalm: I am very smart

I'm sooo late to the party, my solution is easy to think of and [s]not clever at all[/s] one ig
I like this problem ;) because this thread is older than me lol :P
[hide=Solution]
So we have
$(x+1)(y+1)=72$ and $xy(x+y)=880$ by factoring the two
Then we know that x and y can be
[center]$
*1, 72
*2, 36
*3, 24
*4, 18
*6 12
*8 9$[/center]
I just listed this out since it's not much trouble
and then (x+y) is a factor of 880 so x and y can only be 4 and 18, so $x^2+y^2=146$
[/hide]

Let $a=xy$ and $b=x+y$. We have $a+b=71$ and $ab=880$, so by Vieta $a,b$ are the roots of $t^2-71t+880$, which are $16$ and $55$. Note that $t^2-55t+16$ doesn't have integer roots, so $b=16$ and $a=55$. Then $x^2+y^2=b^2-2a=\boxed{146}$.

wait this is woot problem :o

its always kinda like deja vu when that happens

[quote=Silverfalcon]Find $x^2+y^2$ if $x$ and $y$ are positive integers such that \[xy+x+y = 71\qquad\text{and}\qquad x^2y+xy^2 = 880.\][/quote]

[hide=Solution]By SFFT we have $(x+1)(y+1) = 72$. The second equation can be factored as $xy(x + y) = 880$. Since 880 is divisible by 5*16*11, we know that either $x+y$ is divisible by 5 or $xy$ is divisible by 5. Assume the latter. Let $x + 1 = 12$ and $y + 1 = 6$ which means $x = 11, y = 5$. We see that this pair easily works. So $x^2 + y^2 = 11^2 + 5^2 = 121 + 25 = 146$.
[/hide]

[hide=Solution]The second equation simplifies as $xy(x+y)=880$, and we know that $(xy)+(x+y)=71$. Substitute $a=x+y$ and $b=xy$. We note that $ab=880$ and $a+b=71$, and since they are integers we have $a=16$ and $b=55$, implying $(x,y)=(5,11)$ (or permutations), so the answer is $5^2+11^2=\boxed{146}$.[/hide]

[hide=Solution]
Let $x+y=a$ and $xy=b$. We then get that $a+b=71$ and $xy(x+y)=880 \rightarrow ab=880$. This gets us the quadratic $z^2-71z+880=0$, where the solutions are $a$ and $b$.

We see that $880=88 \times 10 = 55 \times 16.$ We see that $b=55$ and $a=16$. We see that this gives us $$x^2+y^2=(x+y)^2-2xy=a^2-2b=256-110=146$$ as our answer.

[hide = Solution:]
We can first substitute $a = x+y$ and $b = xy$, and then we have two equations $a + b = 71$ and $ab = 880$. Since we want to find $a$ and $b$, we can utilize prime factorization of $880$ which is $2^4 * 5 * 11$, which we can then find solution $(a,b) = (16, 55)$. We can then go back to the first expression $x^2 + y^2$, which we can manipulate to $(x+y)^2 - 2xy$. This expression in terms of $a$ and $b$, equals $a^2 - 2b$ Then, plug and chug values for $a$ in $b$ into this expression which yields $16^2 - 2*55 = \boxed{146}$
[/hide]


let $ABCD$ be a square. Consider $M$ a point on segment $[BC]$ and $K$ on segment $[CD]$ such that $CM+MK+KC=2AB$. Find $\angle{MAK}$

Take a point T on MK such that DK=TK and TM=MB. Then draw a perpendicular line on MK through T. Obviously it passes trough A. Now, KA is the bisector of $\angle DAT$, and MA is the bisector of $\angle BAT$. Thus $\angle KAM$=45

[quote="Sailor"]This problem has a great history. First at Bulgarian NMO(for 8 grade), then at Moldavian JBMO TST, and now at MMO, not bad... :D[/quote]It seems that people communicate one with each other :D

It also can be solved algebraically:
First $CM=a,KC=b,AB=1$ then $a+b+\sqrt{a^2+b^2}=1$,put $tg\angle{BAM}=1-a,tg\angle{DAK}=1-b,tg\angle{KAM}=ctg(\angle{BAM}+\angle{DAK})=\frac{2-a-b}{a+b-ab}$,put $a\longrightarrow a+x,b\longrightarrow b-x$.Then $a+b+\sqrt{a^2+b^2}=1=a+b+\sqrt{a^2+b^2+2(x^2+ax-bx)}\Rightarrow x^2+ax-bx=0$Similarly $ctg(\angle{BAM}+\angle{DAK})=\frac{a+b-ab}{2-a-b}=\frac{a+b-ab-(x^2+ax-bx)}{2-a-b}$ so the $\angle{MAK}=invariant$. Note that if we take $a=b$ we get the result.


Another problem with that... It seems that this is only happening on "how many digits of pi do you know". [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=232415&start=20[/url]

I'm willing to bet that's because ak909090's post is so long.


algebra
polynomial
function
complex analysis
complex analysis open

Consider a polynomial:

$ P(z)\equal{}c_nz^n\plus{}c_{n\minus{}1}z^{n\minus{}1}\plus{}...\plus{}c_1z\plus{}c_0$

Is it always possible to write $ P(z)\equal{}g(z)\plus{}f(z)$ with:

$ g(z)\equal{}a_nz^n\plus{}a_{n\minus{}1}z^{n\minus{}1}\plus{}...\plus{}a_1z\plus{}a_0$

$ f(z)\equal{}b_nz^n\plus{}b_{n\minus{}1}z^{n\minus{}1}\plus{}...\plus{}b_1z\plus{}b_0$

and $ \{a_i\plus{}b_i\equal{}c_i\}$

such that $ |g(z)|<|f(z)|$ for arbitrary $ |z|\equal{}r$ with not all $ a_i\equal{}0$?

I suspect it is.  If so, would each of the coefficients of $ g(z)$ and $ f(z)$ be smooth functions of r?

For example, take:

$ U(z)\equal{}z^4\minus{}4z^3\plus{}6z^2\minus{}4z\plus{}3$

Is it possible to construct the new function: $ U(r,z)\equal{}g(r,z)\plus{}f(r,z)$ with:

$ g(r,z)\equal{}a_4(r)z^4\plus{}a_3(r)z^3\plus{}a_2(r)z^2\plus{}a_1(r)z\plus{}a_0(r)$

$ f(r,z)\equal{}b_4(r)z^4\plus{}b_3(r)z^3\plus{}b_2(r)z^2\plus{}b_1(r)z\plus{}b_0(r)$

with the same constraints above such that:

$ |g(r,z)|<|f(r,z)|$ for $ |z|\equal{}r$

I wonder what the plots of $ a_i(r)$ and $ b_i(r)$ would look like if this is true.  Anyway, this arose out of my inability to even show two in one for $ U(z)$ posted here some time ago. Just seems there should be a mechanized way to apply Rouche's theorem to all this.

You can't possibly do it when there is a root on the circle. If there is no root, you can always write $ g(z)\equal{}\frac13P(z)$ and $ f(z)\equal{}\frac23P(z)$.
The statement definitely needs refinement.

I don't expect there to be any mechanical way to get something useful, because it would be too close to a method to find roots exactly; there has to be a change every time our circle passes a root.

[quote="jmerry"]You can't possibly do it when there is a root on the circle. If there is no root, you can always write $ g(z) \equal{} \frac13P(z)$ and $ f(z) \equal{} \frac23P(z)$.[/quote]

I didn't think of that one.  Really though, to be practical, $ f(z)$ should be in some as yet poorly defined way, "simpler" than $ g(z)$:

I could possibly refine the statement:

In the cases of roots not on $ |z| \equal{} r$,  can $ P(z)$ be written as:

$ P(z) \equal{} g(z) \plus{} f(z)$ with, for example, the degree of $ f(z)$ less than the degree of $ g(z)$ and having the property:

$ |g(z)| < |f(z)|$ on $ |z| \equal{} r$

I wonder if something similar to this can be described in some sort of "existence theorem" for any polynomial with roots not on an arbitrary $ r$?  I mean Rouche's Theorem says "if $ |g(z)|<|f(z)|$" without stating that such can exist or will always exists.  Perhaps that would be a better question:  In the case of polynomials, can an existence theorem for Rouche's theorem be constructed for suitable $ |z|\equal{}r$?


function
integration
calculus
calculus computations

Prove

$ \sum\limits_{n\equal{}1}^{\infty }{\frac{\Gamma (m\plus{}1)\Gamma (n\plus{}1)}{\Gamma (m\plus{}n\plus{}2)}}
 \equal{}\int_{0}^{1}{\left\{ x^{m}\sum\limits_{n\equal{}1}^{\infty }{(1\minus{}x)^{n}} \right\}\,dx} $

[url=http://www.mathlinks.ro/viewtopic.php?p=1154706#1154706]Here[/url].

Is it hard to ask your question in the original thread instead creating a new one?

Anyway just use this fact: $ \beta (x,y) \equal{} \frac {\Gamma (x)\Gamma (y)}{\Gamma (x \plus{} y)},\,x,y > 0.$ Where $ \beta,\Gamma$ are Beta & Gamma function respectively.


invariant
analytic geometry
symmetry
trigonometry
geometry
circumcircle
cyclic quadrilateral

a pretty cool (somewhat simple) problem that i found...

There is a circle with 2 diameters(fixed) drawn $AB$, and $CD$. Point $M$(movable) is on the circumference of the circle. The feet of the perpendicular lines from $M$ drawn to $AB$ and $CD$ are $E$ and $F$, respectively. Prove that the length of $EF$ is invariant under moving $M$ along the circumference.

EDIT: a little unclear where the perpendiculars were coming from

Put the coordinate origin at the center O of the given circle with radius r and positive x-axis along the ray (OB. WLOG, assume that the point C is in the upper half plane. Denote the fixed angle of the 2 diameters AB, CD as $\angle BOC = \omega < \pi$. Since the angles $\angle MEO = \angle MFO = \frac \pi 2$ are right, the quadrilateral OEMF is always cyclic, but it can be convex or reflex. Because of symmetry, we need to investigate only positions of the point M on the upper semicircle. Equivalently, we can investigate only positions of the point M on the minor arc BC, i.e., $\angle BOM < \angle BOC,$ if we consider separately 2 cases: $\omega \le \frac \pi 2$ and $\omega > \frac \pi 2$.

[color=blue][b](1)[/b][/color] Assume that $0 < \angle BOM < \angle BOC \le \frac \pi 2$. Then the cyclic quadrilateral OEMF is convex with the diagonals OM, EF. We have $OE = r \cos \phi,$ $EM = r \sin \phi,$ $MF = r \sin (\omega - \phi),$ $FO = r \cos (\omega - \phi),$ and $OM = r.$ By Ptolemy's theorem,

$OE \cdot MF + EM \cdot FO = OM \cdot EF$

$r^2 \cos \phi \sin (\omega - \phi) + r^2 \sin \phi \cos (\omega - \phi) = r \cdot EF$

$EF = r \sin (\omega - \phi + \phi) = r \sin \omega$ is fixed.

[color=blue][b](2)[/b][/color] Assume that $0 < \angle BOM < \angle BOC,$ $\frac \pi 2 < \angle BOC < \pi$. Because of symmetry, we need to investigate only positions of the point M, such that $\angle BOM \le \frac{\angle BOC}{2} < \frac \pi 2$. If $\omega - \phi < \frac \pi 2$, the cyclic quadrilateral OEMF is convex and $EF = r \sin \omega$ is fixed as before. If $\omega - \phi > \frac \pi 2$, the cyclic quadrilateral OEMF is reflex, so that the cyclic quadrilateral OFEM is convex with the diagonals OE, MF. Again, we have $OE = r \cos \phi,$ $EM = r \sin \phi,$ $MF = r \sin (\pi - \omega + \phi),$ $FO = r \cos (\pi - \omega + \phi),$ and $OM = r.$ By Ptolemy's theorem,

$EF \cdot OM + EM \cdot FO = OE \cdot MF$

$EF \cdot r + r^2 \sin \phi \cos (\pi - \omega + \phi) = r^2 \cos \phi \sin (\pi - \omega + \phi)$

$EF = r \sin (\pi - \omega + \phi - \phi) = r \sin \omega$ is fixed.

Wlog, we are on the unit circle, one diameter is the x axis, and the other diameter makes an angle of theta with the x axis.  The ray from the origin of the circle to the point makes an angle alpha with the x axis.  Using the cosine law, it boils down to proving that the square of EF, $\cos^2 \alpha + \cos^2 (\alpha - \theta) - 2\cos \alpha \cos (\alpha - \theta) \cos \theta$, is fixed under $\alpha$.  Some simple algebra work shows that it is equal to $\sin^2 \theta$.  (Write $\cos (\alpha - \theta) = \cos \alpha \cos \theta + \sin \alpha \sin \theta$; terms cancel.)

it is pretty obvious that the points O,E,M,F are cyclic because the right angles sum to 180, and if OEMF is not convex then the 2 congruent, right, angles also make them cyclic anyway

since $\angle OEM$ is right, the $OM$ is the diameter of the smaller circle, call this $D$, and $D=R$ where $R$ is the circumradius of the larger circle, and therefore $D$ is constant,

1) convex, say the angle between the lines is $x$, then $\angle EMF= 180-x,$ by the law of sines, $\frac{EF}{\sin{180-x}}=D$, $EF=D\sin x$ therefore it is constant

2) ME intersects the other diameter...then look at $OEF$, use law of sines to get the same thing: $EF=D\sin x$


Let K1 and K2 be 2 orthogonal circles, with centres O1 and O2 and intersecting at P and Q. A line through P intersects K1 at A and K2 at B, with M being the midpoint of AB. Prove <O1MO2=90.

I think that on Forum exists this more general problem:

$\blacksquare.$ Two circles $C_k(O_k ,r_k),\ k\in \overline {1,2}$ meet in the points $A,B$.
Let two mobile points $M\in C_1,\ N\in C_2$, such that $A\in \left( MN\right)$.
I consider the point $L\in (MN)$ for which $LM=k\cdot LN$,
where $k$ is a positive constant. Prove that:
$0^{\circ} .\ M,M'\in C_1,\ N,N'\in C_2,\ A\in MN\cap M'N'\Longrightarrow \triangle BMN\sim \triangle BM'N';$
$1^{\circ}$. The geometrical locus of the point $L$ is the circle $\Lambda (C,CA)$, where $C\in O_1O_2,\ CO_1=k\cdot CO_2.\ \blacksquare$

[b]Remark.[/b] For $k=1$ we obtain the problem of the initial message.

[quote="levi"]I think that on Forum exists this more general problem:

$\blacksquare.$ Two circles $C_k(O_k ,r_k),\ k\in \overline {1,2}$ meet in the points $A,B$.
Let two mobile points $M\in C_1,\ N\in C_2$, such that $A\in \left( MN\right)$.
I consider the point $L\in (MN)$ for which $LM=k\cdot LN$,
where $k$ is a positive constant. Prove that:
$0^{\circ} .\ M,M'\in C_1,\ N,N'\in C_2,\ A\in MN\cap M'N'\Longrightarrow \triangle BMN\sim \triangle BM'N';$
$1^{\circ}$. The set of the points $L$ is the circle $\Lambda (C,CA)$, where $C\in O_1O_2,\ CO_1=k\cdot CO_2.\ \blacksquare$

[b]Remark.[/b] For $k=1$ we obtain the problem of the initial message.[/quote]

Sorry, but what do you mean by: " $L$ is the circle $\Lambda (C,CA)$, where $C\in O_1O_2,\ CO_1=k\cdot CO_2.\ \blacksquare$" ?

Here is my solution so you can choose

This just looks like a long solution but it really isn't, i
:) :)

It is enough to prove that $\angle O_1MP=\angle O_1O_2P$ and from that we get that
points $O_1,M,O_2,P$ are concyclic so from that $\angle O_1MO_2=90^o$

Let $N$ be the crossing point of lines $PQ$ and $O_1O_2$, then all what we should
prove is $\angle O_1PN=\angle O_1MP$ ( $\angle O_1PN=\angle O_1O_2P,$ because this
circles are perpendicular, and $PQ \bot O_1O_2$)

Now $\angle O_1MP$ is equal to the sum of angles $\angle O_1BM$ and $BO_1M$, $\angle
O_1BM=\angle O_1PB$ because $O_1B$ and $O_1P$ are radiuses of circle, so all we
should prove that $\angle BO_1M=\angle MPQ$

See that $\angle BO_1Q=2 \times \angle BPQ$ so if we prove that $\angle MO_1Q=\angle
MO_1B$ we are done.

$SO- WE- SEE- THAT- ALL- WHAT- WE -SHOULD -PROVE- IS- NEXT$ $\angle MO_1Q=\angle MO_1B$ let's
prove that...

$\angle QBP= \frac12\angle QO_1P=\frac12(180^o-\angle QO_2P)=\angle PQO_2$

$\angle BAQ=180^o - \angle QAP=\frac12 \angle QO_2p=\angle NO_2Q$

so we have $\angle QBP+\angle BAQ=\angle PQO_2+\angle NO_2Q=90^o$ from that, because
$M$ is midpoint of $AB$ we obtain $MB=MA=MQ$.

Now from triangles $O_1BM$ and $O_1MQ$ we have $O_1M$ common, $O_1B=O_1Q$ and
$BM=QM$ and from that we see that these triangles are congruent so \[\angle BO_1M=
\angle MO_1Q\]

M@re :)  :)  :)  :)

The set of the points L is... <===> the geometrical locus of the point L is...


function
inequalities
logarithms
inequalities open

I often see inequality of the form:

$\prod f(x) \geq f(\frac{\sum x}{n})$ where f is convex.

Is there a way to prove this?  what conditions on x_i must happen for this to be true, if any?

It looks like a variation of Jensen's inequality:

$\sum \alpha f(x)\geq  f(\sum \alpha x)$

$ x\in \mathbb{R}$    , where

$\sum \alpha =1$ and $f$ is convex.
Puting $\alpha =\frac{1}{n}$.

But if we consider the absolut function it must be:

$|\frac{1}{10}||\frac{1}{10}|\geq |\frac{\frac{1}{10}+\frac{1}{10}}{2}|$

$\Leftrightarrow$

$\frac{1}{100}\geq \frac{1}{10}$

wich is obviously wrong.

I've never seen that inequality before, maybe you can send a link?

I think I have a proof, but the inequality is a bit various.

Is $f: D\rightarrow \mathbb{R}^+$ and let $f$ be concave on $D$ then

$\prod f(x)\leq f\left( \frac{\sum f}{n} \right)^n$.

For the proof you must use Jensen and the Function $\ln f$.

Correct me when I'm wrong.


logarithms
number theory
relatively prime

If $x=\frac{1}{3}$ or if $x=-15$, then $\log_8 (kx^2+wx+f)=2$. If $k$, $w$, and $f$ are relatively prime positive integers, find the value of $k+w+f$.

Oops, I posted this problem here instead of GS... :ewpu:
This is why the problem was a little easy for Intermediate level.

Anyway, nr1337, your answer is correct :)


Discuss all Mock Mathcounts problems here! If you are wondering how to do some problems, think we have an error on the solutions, or just want to yell at us for making the test too hard or easy, please do it here!

But I'm assumming you don't answer untill this Thursday, since people might be able to answer the questions tomorrow.

[i]mod edit: sorry, I made a mistake. Since people haven't taken it yet, we can't say anything, can we?[/i]

Oops. I didn't knwo about that. Please, no more posts in this topic until everyone has finished.

[quote=RedFireTruck]BUMP!!!!![/quote]

Seriously, what are you doing. Could you please stop.


inequalities
quadratics
algebra
quadratic formula
special factorizations

Hi all,

I'm trying to solve the following inequality...

x^2 - x + 1 > 2

Can this be done without resorting to the quadratic formula?

John

[hide="Solution"]Completing the square works well for problems like this.

First, subtract $ \tfrac{3}{4}$ from each side to get $ x^2 \minus{} x \plus{} \tfrac{1}{4} > \tfrac{5}{4}$.

Then, factor the LHS: $ \left(x \minus{} \tfrac{1}{2}\right)^2 > \tfrac{5}{4}$.

Next, take the square root of both sides to get $ x \minus{} \tfrac {1}{2} > \sqrt {\tfrac{5}{4}}$ and $ x \minus{} \tfrac{1}{2} < \minus{} \sqrt {\tfrac{5}{4}}$.

Finally, $ x > \tfrac{1 \plus{} \sqrt {5}}{2}$ and $ x < \tfrac{1 \minus{} \sqrt {5}}{2}$, so $ \boxed{x: \left( \minus{} \infty,\,\tfrac{1 \minus{} \sqrt {5}}{2}\right]\cup\left[\tfrac{1 \plus{} \sqrt {5}}{2},\,\infty\right)}$.[/hide]


articles
Euler
trigonometry
analytic geometry
complex numbers

I'm having difficulty understanding roots unity. I'm reading the aops wiki article on it and have questions.

I understand that given $ x^n \equal{} 1$, we can convert everything to polar by letting $ x \equal{} re^{i\theta}$ but why do we know that $ 1 \equal{} e^{2\pi ik}$ for $ k\in \mathbb{Z}$?

What is that Z thing?
How do we get $ r^ne^{ni\theta} \equal{} e^{2\pi ik}$ from the above?

What is "RHS?" in the below line?
The magnitude of the RHS is 1, making $ r^n\equal{}1\Rightarrow r\equal{}1$ (magnitude is always expressed as a positive real number).

[quote="aufha"]I'm having difficulty understanding roots unity. I'm reading the aops wiki article on it and have questions.[/quote]

Hmm, the article does require a considerable amount of background in complex numbers. I'll fix that when I have time.

[quote]I understand that given $ x^n = 1$, we can convert everything to polar by letting $ x = re^{i\theta}$ but why do we know that $ 1 = e^{2\pi ik}$ for $ k\in \mathbb{Z}$?

What is that Z thing?[/quote]

$ \mathbb{Z}$ denotes the set of [u]integers[/u]. The key here is that we are trying to express $ 1$ in polar form (judging by your comments, you know what this is, right?). Since $ 1$ has no complex part, its angular component is $ 0 + 2\pi k$ radians for any integer $ k$, since all such angles in that form are equivalent. Its magnitude is clearly $ 1$, and so in polar form $ \boxed{1 = e^{2\pi ik}}$.

[quote]How do we get $ r^ne^{ni\theta} = e^{2\pi ik}$ from the above?[/quote]

We start with \[ x^n \&= 1\]Rewriting in polar form, and since from above \[ 1 = e^{2\pi ik}\] and \[ x^n = (re^{i\theta})^n = r^ne^{ni\theta}\]so we can equate and get \[ e^{2\pi ik} = r^ne^{ni\theta}\]

[quote]What is "RHS?" in the below line?
The magnitude of the RHS is 1, making r^n=1\Rightarrow r=1 (magnitude is always expressed as a positive real number).[/quote]

Right-hand side. What we are now doing is saying that for two complex numbers to be equal, there angular component and magnitude must be equal. So we set $ r^n = 1$ and we set $ 2\pi ik = ni\theta$.

Again, the wiki article was apparently written for an intended audience of people who have additional experience with this sort of notation, so please don't be hard on yourself for not getting this on the first time through. If you would like, I could try explaining everything in much easier terms.

[quote="aufha"]why do we know that $ 1 \equal{} e^{2 \pi ik}$ for $ k \in \mathbb{Z}$?[/quote]

This is a famous result due to Euler:

$ e^{i \pi} \equal{} \minus{} 1$

[color=darkblue]Moderator Edit: URL tags are causing a weird bug. The URL presented before was "http://en.wikipedia.org/wiki/Euler's_formula"[/color]

$ e^{iy} \equal{} r \cos y \plus{} ir \sin y$

$ e^{i \pi} \equal{} r \cos \pi \plus{} i r \sin \pi$

Because $ \pi \equal{} 180^{\circ}$ and $ r \equal{} 1$, we have

$ e^{i \pi} \equal{} \cos 180^{\circ} \plus{} i \sin 180^{\circ}$

or

$ e^{i \pi} \equal{} \minus{} 1$

The nth roots of unity are just the coordinates of n evenly spaced points around a unit circle, with the first point on the x axis.


complex analysis
complex analysis unsolved

Please, if someone can help me with this i really appreciate it:

"Let $U$ be an open simply connected set.  Let $z_{1}$ and $z_{2}$ two points in $U$. Show there exists an analytic automorphism $f: U\to U$ such that $f(z_{1})=z_{2}$ "

If $U=\mathbb C$, then $f(z)=z_{1}+z_{2}-z$ works. If $U\ne \mathbb C$, then the Riemann mapping theorem reduces the problem to the case $U=\{z\in\mathbb C\colon |z|<1\}$ and $z_{1}=0$. Try $f(z)=\frac{z_{2}-z}{1-z\overline{z_{2}}}$.

why can I reduce the problem to $z_{1}=0$ ? I follow the idea by using the Riemann Mapping Theorem, therefore there exists an analytic automorphism $T: U\to D$ , where $D=D(0,1)$ the disc center at 0 and radius 1, so actually i know the only automorphisms in D have the form  $f(z)=\frac{e^{it}(z-a)}{1-z\overline a}$ , with some $t\in R$ and $|a|<1$ .   If $T(z_{1})=w_{1}$ and $T(z_{2})=w_{2}$ then I was trying to find an automorphism $f$ in D such that  $f(w_{1})=w_{2}$, but my problem was how to find that $f$, i.e. to try to find the $a$ and $t$ such that the condition $f(w_{1})=w_{2}$ holds.  Well I reduce that problem to find just $a$ such that $\frac{(w_{1}-a)}{1-w_{1}\overline a}=R$, where $R=|w_{2}|$, but i cannot solve that equation.

Again,  if anyone could help me i would be really thankful .

Let $f$ be a bijective analytic map from $U$ to $\mathbb{D}$. The automorphisms of $U$ are $f^{-1}\circ \varphi\circ f$, where $\varphi$ is an automorphism of $\mathbb{D}$; we understand those $\varphi$.

In the disk, we know how to map any point to zero, and then map zero to any point.

[quote="fabiansaul"]why can I reduce the problem to $z_{1}=0$ ?[/quote]
Because the [url=http://mathworld.wolfram.com/RiemannMappingTheorem.html]Riemann mapping theorem[/url] allows you to choose where $z_{1}$ will go.

Finally I could see the exercises was very easy  :oops: , but according to the question of my second post I just had to consider $f=g_{-w_{2}}\circ g_{w_{1}}$, where $g_{a}(z)=\frac{z-a}{1-z\overline a}$.

Thanks to everybody for helping me  :)


Solve the equality $ 2x^2 \minus{}5y^2\equal{}7$

Certainly; $ \left(t,\sqrt {\frac {5t^2 \plus{} 7}{2}}\right)$ for any $ t$.

[hide="Oh, you mean in integers?  :wink: "]Take it $ \pmod 4$ to find that $ x^2$ must be $ 0\pmod{4}$. Rewrite as $ 8x'^2 \minus{} 5y^2 \equal{} 7$ and taking $ \pmod {8}$, we see clearly that it has no solutions.[/hide]