Mr. and Mrs. Zeta want to name their baby Zeta so that its monogram (first, middle, and last initials) will be in alphabetical order with no letter repeated.  How many such monograms are possible?

$\text{(A)} \ 276 \qquad \text{(B)} \ 300 \qquad \text{(C)} \ 552 \qquad \text{(D)} \ 600 \qquad \text{(E)} \ 15600$

1989 AHSME

Mr. and Mrs. Zeta want to name their baby Zeta so that its monogram (first, middle, and last initials) will be in alphabetical order with no letter repeated. How many such monograms are possible?

-------------------------------------------------------------------------------------

25[size=150]C[/size]2 = 300.  300 monograms are possible.

mathfanatic, don't read too much into the problem. Yannik, what's your solution?

Yeah. He's a friend of mine. I'm not sure if he was mocking me or inspired by me.


Prove that for all $a, b, c \in R_0$ satisfying $a^2+b^2+c^2=9$

$3 min \{a, b, c \} \leq 1+abc$

What's $\mathbb{R}_0$?

That inequality doesn't look right to me at all... (but maybe I just haven't understood it).

Its  here
http://www.mathlinks.ro/Forum/viewtopic.php?p=215133#p215133


inequalities
integration
algebra unsolved
algebra

Does there exist a,b such that for all x from [0,2\pi] the inequality holds

f^2(x)-f(x) cos x < 1/4 sin x, where f(x)=ax+b??

[i]unknown source[/i]

(f(x)-(cosx)/2)^2 <1/4(sinx+cosx)

(f(x)-(cosx)/2)^2 <1/4(sinx+1-sinx)=1/4( \emptyset -sinx)(sinx+ \emptyset -1) where  \emptyset =(1+ \sqrt 5)/2 golden number
|f(x) - (cosx)/2|<1/2 \sqrt (( \emptyset -sinx)(sinx+ \emptyset -1))
for all x\in [0;2 \pi ] , for x->0 we get some conditions for b

Suppose there exits a,b reals such that
the inequality is true for any x \in [0;2 \pi ]
we integrate the inequality on [0;2 \pi ]
 \int_[0;2 \pi ](ax+b)^2dx- \int_[0;2 \pi ](ax+b)cosxdx<1/4\int_[0;2 \pi ]sinxdx => \int_[0;2 \pi ](ax+b)^2dx<0
hum it seems there is no  such a,b.
What is your solution Fedor ?

Does there exist a,b such that for all x from [0,2\pi ] the inequality holds

f^2003(x)-f(x) cos x < 1/4 sin x, where f(x)=ax+b ?


Did anyone take the quarterfinal; how'd you guys feel about it. I thought it was tougher than the other open response q's i saw online.

Yeah, I agree. I didn't think any of the problems were too difficult, but it was very hard to get very many of the problems in just 1 hour.

How high do you have to score to make semi-final?

yeah, I agree. I went home and did the problems, but I realized I made stupid mistakes because I rushed on the exam. I'm sure I'll get most of the partial credit, but I only solved 2.5 problems in 1 hour.

How many did you guys end up with?

I also solved about 2.5 problems.  Was there a way to solve part b of #1 and 3 without using differential equations?

Also, I'm not sure what the semi-final "index" is.  The USAPhO format this year is very much like that of the AMCs, and the semi-final exam is like the USAMO of physics!

Ah..I only solved 1.5 questions.. I spent too much time on number 4, but that problem was so cool! I talked to several people afterwards (who are all pretty good with physics), and they all had the same reaction: "Is that even possible?" Bravo to whoever made that problem; maybe some of you have seen similar problems before, but it seemed original to me. The other person who took it from my school said he got part of 3 questions.

Yeah, problem 4 was really cool.  I didn't get it, though, and I still don't know how to solve it.  Two more days until we're allowed to discuss it!

Who can tell me what the answer of last problem of #2 is?

#2 was the billiard ball one, right? It should be 2/5. I had a sign error and got -2/5 (yes I realized that it made no sense, but I couldn't find my sign error until after the test when I realized I had torque going the wrong way).

I think I got #1, although I might have forgotten the negative sign when finding the SHM solution, which apparently a lot of people did.

I wrote up a solution to #3a but I have no idea if it's right or not.

great I think i got the #2 right.
What is your answer for #3?

My thoughts:

#1... this was pretty simple
#2... part a) was pretty easy... but part b, i made a careless mistake on and had no idea as to where to go
#3... pretty straightforward... used the chain rule and a bit of diff eq's
#4... O_O NO IDEA

EDIT: When are results coming out?

Hehe, just now actually: http://www.aapt.org/olympiad2008/semi_finalist.pdf

Made it. Not sure how, though; I only got 15.25 on part 1 and $ \le 1.5$ questions on the quarterfinal.

Cool, I made it too.  My thoughts on the test are practically the same as those of Karth.

I'm glad that I totally pwned no. 3.

Well, I have NO idea how they graded. I did well on the first test, but made plenty of stupid mistakes on the second, but I made it, which is what counts I guess.

Anyway, as for the test, I didn't think it was so bad. I solved the problems later, and basically hoped that not many people had super-human speed ability. The last problem took me about 45 minutes itself, its really cool; if you guys didnt look up the answers, they should be on the main page: http://www.aapt.org .

Anyway, gl everyone on the next round!

for the last problem, I actually got the correct expression with 1/(4cos(x) - 6cos(x)^2) or something like that. I just didn't have enough time to calculate the maximum value. What do you think I would have gotten?

Only now do I realize how easy number 3 was. Oh well. I got 1 and 2 completely, which is surprising, because I redid them later and kept making algebra mistakes.  :maybe:

Good luck everyone on the next round!


Construct a triangle with three given angle bisectors.

Easy, there are six angles formed when three lines intersect.

The measure of an angle of that triangle is half the sum of the angles "furthest away" from it.
So, the bisected angle is half of that, and the whole triangle can be constructed.  (I am sorry I don't have a diagram, I will try to make one.)

I apologize (I had the correct equations, but I simplified them wrongly).

Here is a diagram with the (correct) measure of one of the bisected angles.  The rest of the construction is fairly obvious.

And what if the point is "given the lengths of angle bisectors"? i.e. segments from the angle vertex to the intersection of the bisector with the opposite side.

Wow, I would like to see the solution for that.

It may be that it is impossible to construct such a triangle (with compass and straightedge), but regardless, I am intrigued about the relationship between a triangle and its bisectors' lengths.

I will think about this some more.

a bisector d, to side a with adjacent sides b,c (combine stewart/angle bisector theorems) is:

$d^2=bc\left(1-\left(\frac{a}{b+c}\right)^2\right)$

then we have a system to solve for a,b,c, the question is whether this system has solutions that are constructable, i.e. solutions to a quadratic

The final solution might be quadratic, but I have doubts just looking at that messy solving.


geometry
real analysis
topology
real analysis theorems

In literature, it often says something like ".... has a measure zero" or "... holds almost everywhere". Usually, the underlying measure is not specified. So what is the measure used in such statement by default?

Some may say that a "standard measure" is assumed. So what is a standard measure? Is it a class of measures or a uniquely defined measure?

For example, in $E^{n}$ (n dimensional Euclidan space), the Lebesque measure is most often used. So is the standard measure just Lebesque measure, or it could also be some other measure.

Also, I want to know how we define a standard measure of a unit surface in $E^{n}$. Intuitively, it is just the area of the surface. But how is it formally defined like the way we define Lebeque measure in $E^{n}$?

Sorry, I never take any course in measure theory, so maybe my question is quite naive. Hope I made myself clear. Thanks in advance for your help.

In the Euclidean space, the Lebesgue measure is usually meant. On surfaces, you can introduce the surface measure in several equivalent ways, the most general one being the Hausdorff measure of the corresponding dimension. I do not want to formally define the Hausdorff measure of dimension $\alpha>0$ here but let me just say that the sets of measure $0$ are exactly the sets that you can cover by a sequence of balls $B(x_{j},r_{j})$ such that $\sum_{j}r_{j}^\alpha$ is arbitrarily small. Hope this helps a bit :).


[size=150][b]augmented matrix is that

1      2     1 ,1
-1     4     3 ,2
2     -2     a ,3[/b][/size]

[i]for what value of a will system have a unıque solution??[/i]

note ;last column is vector of  constants

[b]thanks :lol:[/b]

hello, we get
$ x_1\plus{}2x_3\plus{}x_3\equal{}1$
$ \minus{}x_1\plus{}4x_2\plus{}3x_3\equal{}2$
$ 2x_1\minus{}2x_2\plus{}ax_3\equal{}3$
adding the first and the second equation we get
$ 6x_2\plus{}4x_3\equal{}3$
multiplying the first equation by $ \minus{}2$ and adding the third equation we get
$ \minus{}6x_2\minus{}2x_3\plus{}ax_3\equal{}1$
adding the last two equations we get
$ 2x_3\plus{}ax_3\equal{}4$
for $ a\ne \minus{}2$ we get $ x_3\equal{}\frac{4}{2\plus{}a}$ as the unique solution.
Sonnhard.


quadratics
algebra
polynomial
Vieta
complex numbers

The sum of the coefficients of a particular quadratic is $0$. What can be concluded about the roots?

[hide]If it's not already monic, make it monic
$x^{2}+bx+c=0$
Letting the roots be $m,n$, Vieta tells us that $m+n=-b,mn=c$
Hence
$b+c+1=0$
$-(m+n)+mn+1=0$
$(m-1)(n-1)=0$

so 1 is a root and the other root can be whatever you want[/hide]

DUDE THAT WAS A BEAST PROBLEM
sorta

That's almost right, except it doesn't have to be a double root.

You could also use that fact for any polynomial, so $2x^{3}+x^{2}-4x+1$ has a solutions being 1 (the other 2 are complex if anyone cares to know, which you can find if you factor it to (x-1)(some quadratic) ).

Not necessarily complex, just has to have a root of 1.

If the roots are 1 and something non-real, the coefficients aren't real.
This is just the contrapositive of the fact that nonreal roots come in complex conjugate pairs.

well when I said the other two are complex, I meant just for that particular cubic.

Note that f(1) is the sum of the coefficients, it's a common relationship


A man looks outside and sees that it is a beautiful day. Because it was a beautiful day, he kills himself. Why?

Does it have to do with his job? If yes, [hide]he is a weatherman and predicted wrong and an angry mob is after him[/hide]

It has to do the way he makes money, but he is not a weatherman.

He doesn't own an umbrella factory.

By a beautiful day, I mean it is nice and sunny.

He doesn't work for the government

does it mean a beautiful sunny day to kill himself?
is it a beautiful day when he kills himself?
is their a prophecy that states when there is a beautiful day he'll kill himself?
is he a perfectionist weather man who is going suicidal?

No, I meant it [i]was[/i] a beautiful day but since you didn't get it I guess that is a no. :lol:

OKAY then...Did he kill himself with a shot gun?
is it important about what he killed himself with?
is he autistic?
or mentally instable?

Is the man a farmer? Is there a drought going on?

Is he sentenced to death anyway?  Like he will die if it is sunny and they can execute him outside?

Would something else have happened if he hadn't killed himself?
Would he be killed anyway?
Would he [i]not[/i] have killed himself if it were...
-raining?
-snowing?
-hailing?

[hide="Probably IR"]
Did he kill himself out of

angriness?
sadness?
happiness?
uh... frightenedness?  :|

What did he kill himself with?
[/hide]

[quote="mustafa"]I don't quite get some of your questions, funcia, but I'll answer as best I can. The man was not killed by the beautiful day, he killed himself because it was beautiful. He was not a weatherman. There was no prophecy.[/quote]

I'm curious, because I thought I know the story. Are you sure that the quoted part is correct¿

The father killed himself out of sadness

There was no bet

The son killed himself to get away from somewhere, but not to get somewhere

I'm going to give you a hint, because nobody has done anything for a while

[hide]Both the man and the son were extremely religious[/hide]

Nutz is right, although it isn't very nice to spoil the challenge :dry:

In fact, spoiling is just idiotic and shows only one thing: he want's to show how cool he is, but in fact it shows how showy he is.

[quote="nutz_for2.718281828"][hide="spoiler"]He is part of a group that thinks that armageddon is tomorrow. He gives each person in the group a choice: poison or life with armaggedon. Most people, including his wife and child, take poison. He and some of his followers take life. When he wakes up and sees that it's a beautiful morning, he sees that he made a terrible mistake, so he shoots himself.[/hide][/quote]

what? still why did he kill himself? what was the mistake

[quote="moogra"]what? still why did he kill himself? what was the mistake[/quote]
The mistake was choosing life with Armageddon

Now that it is spoiled: [hide]as I know it, the person killing himself is the one that made the prophecy, thus caused the death of those people. And this person is also the only one we wanted to see armaggeddon.[/hide]

STILL SPOILING IS JUST VERY STUPID!!!

[quote="laughinghead505"]What's "armaggeddon"? :huh:[/quote]
It's basically the "end of the world".
Or from Wikipedia: "an apocalyptic catastrophe"
[url]http://en.wikipedia.org/wiki/Armageddon[/url]

I have a question: if nobody can spoil it, how will anyone ever get to know the answer? Just wondering because if nobody can tell the answer, then maybe nobody will ever learn it?  :maybe:

The person who originally made the post should answer questions and eventually give hints that allow the other people to figure it out on their own. It is only spoiling if you knew the answer before seeing the topic. If you just figured it out using the questions, that would be fine. There should be a sticky that says this at the top of the page.

@spoiling:
the rule is: people not knowing it ask yes-or-no-questions, the others answer them (with yes, no, or possibly irrelevant).


geometry
3D geometry
calculus
integration
quadratics
number theory unsolved
number theory

Prove there are at least two positive integer soutions to the equation :
\[ x^3\minus{}y^2\equal{}2000000\]
,where $ x,y \in N$

[quote="Cekaa"]Prove there are at least two positive integer soutions to the equation :
\[ x^3 \minus{} y^2 \equal{} 2000000\]
,where $ x,y \in N$[/quote]

[hide="One is"] $ (300,5000)$  :)[/hide]

Another Mordell equation!

There may be some hidden answer in [url]http://math.uga.edu/~pete/4400MordellEquation.pdf[/url]

EDIT. Tricky pco  :)  Get the unique positive solution $ (3,5)$ to $ a^3 \minus{} b^2 \equal{} 2$, and take advantage of the fact that $ 2000000 \equal{} 2\cdot 10^6$.

^^ yes, thats exactly the one i found on the competition itself after a few minutes, but spent the next 3 - 3.5h trying to find the second and didnt find it :) this looks more like a work for a computer than for a human :-S maybe im wrong, but still..

[quote="Cekaa"]just found another solution, by computer ofcourse, its $ (x,y) \equal{} (131532, 41176)$, still no clue how to find it by hand :-) especially because there are finitely many soultions according to mavropnevma's attached link...[/quote]

You should check your computer, since $ 131532^3\minus{}41176^2\approx 2\cdot 10^{15}>2000000$ and $ 41176^3\minus{}131532^2\approx 7\cdot 10^{13}>2000000$

[quote="Cekaa"]just found another solution, by computer ofcourse, its $ (x,y) \equal{} (131532, 41176)$, still no clue how to find it by hand :-) especially because there are finitely many soultions according to mavropnevma's attached link...[/quote]
Obviosly $ x^3\equal{}8\mod 10 \not \equal{} y^2\mod 10 \equal{}6$. It is not solution. And $ x^3\minus{}y^2$ very big.
$ (x,y)\equal{}(41176,131532)$ does not work too (chek by mod 10).

[hide="Another one"]My own computer gave me $ (129,383)$[/hide]

Very nice Patrick. Let me draw your attention to
[quote]For example if we start by a solution $ (3,5)$ to $ y^2\minus{}x^3 \equal{} \minus{}2$, by applying duplication formula, we get a series of rational solutions $ (3,5), (\frac {129}{10^2}, \frac {\minus{}383} {10^3}), (\frac {2340922881} {7660^2}, \frac {113259286337292} {7660^3}), \ldots$[/quote]
from [url]http://www.waset.org/journals/ijcms/v1/v1-4-36.pdf[/url], which gives the algebraic formula for the duplication mentioned above (via an identity discovered by Bachet).

[quote="Cekaa"]Prove there are at least two positive integer soutions to the equation :
\[ x^3 \minus{} y^2 \equal{} 2000000\]
,where $ x,y \in N$[/quote]
From $ x^3\equal{} (y\plus{}1000\sqrt{\minus{}2}) (y\minus{}1000\sqrt{\minus{}2})$, we may try assuming that $ y\plus{}1000\sqrt{\minus{}2}$ is a perfect cube in $ \mathbb Z[\sqrt{\minus{}2}]$.

Let $ y\plus{}1000\sqrt{\minus{}2} \equal{} (a\plus{}b\sqrt{\minus{}2})^3 \equal{} (a^3\minus{}6ab^2) \plus{} (3a^2b\minus{}2b^3) \sqrt{\minus{}2} \quad ,$ where $ a,b \in \mathbb Z$.

It follows that $ 1000 \equal{} 3a^2b\minus{}2b^3$, and hence $ b\mid 1000$. So we can try $ b\equal{} \pm 1, \pm 2, \pm 4, \pm 5, \dots$, and look for ones that lead to integral $ a$. Soon we will find that $ b\equal{} \minus{}8$ and $ b\equal{} 10$ lead to $ a\equal{}\pm 1$ and $ a\equal{}\pm 10$, respectively. Therefore, we have $ y\equal{} a^3\minus{}6ab^2 \equal{} \pm 383, \pm 5000$ which leads to $ x\equal{} 129, 5000$, respectively.

[quote="mavropnevma"]EDIT. Tricky pco  :)  Get the unique positive solution $ (3,5)$ to $ a^3 \minus{} b^2 \equal{} 2$, and take advantage of the fact that $ 2000000 \equal{} 2\cdot 10^6$.[/quote]
This would help reduce the computation above.

Inspired by warut let's try factoring $ x^3 \minus{} 2000000 \equal{} y^2$ instead

$ (x \minus{} 100 \sqrt [3]{2}) (x \minus{} 100\omega \sqrt [3]{2}) ((x \minus{} 100\omega^2 \sqrt [3]{2}) \equal{} y^2$ where $ \omega$ is a complex cube root of unity.

Look for $ x \minus{} 100 \sqrt [3]{2} \equal{} (a \plus{} b \sqrt [3]{2} \plus{} c \sqrt [3]{4})^2$

$ a^2 \plus{} 4bc \equal{} x$
$ b^2 \plus{} 2ac \equal{} 0$
$ 2c^2 \plus{} 2ab \equal{} \minus{}100 \Leftrightarrow c^2 \plus{} ab \equal{} \minus{}50$

To solve the middle equation we can try  $ a \equal{} 2de^2, c \equal{} \minus{} df^2, b \equal{} 2def$
then $ c^2 \plus{} ab \equal{} \minus{} 50 \Leftrightarrow d^2f(f^3 \plus{} 4e^3) \equal{} \minus{}50$
This has no solutions.

Alternatively we can try $ a \equal{} de^2, c \equal{} \minus{} 2df^2, b \equal{} 2def$
then $ c^2 \plus{} ab \equal{} \minus{} 50 \Leftrightarrow d^2f(2f^3 \plus{} e^3) \equal{} \minus{} 25$
gives $ d \equal{} 1, f \equal{} 1, e \equal{} \minus{} 3$
$ \implies a \equal{} 9, c \equal{} \minus{} 2, b \equal{} \minus{} 6 \implies x \equal{} a^2 \plus{} 4bc \equal{} 129$ giving one solution.

So we can get one solution this way. The algebra was a little more complex than warut's solution but there were less factorization cases to consider at the end.

[quote="PhilG"]Inspired by warut let's try factoring $ x^3 \minus{} 2000000 \equal{} y^2$ instead

$ (x \minus{} 100 \sqrt [3]{2}) (x \minus{} 100\omega \sqrt [3]{2}) ((x \minus{} 100\omega^2 \sqrt [3]{2}) \equal{} y^2$ where $ \omega$ is a complex cube root of unity.

Look for $ x \minus{} 100 \sqrt [3]{2} \equal{} (a \plus{} b \sqrt [3]{2} \plus{} c \sqrt [3]{4})^2$

$ a^2 \plus{} 4bc \equal{} x$
$ b^2 \plus{} 2ac \equal{} 0$
$ 2c^2 \plus{} 2ab \equal{} \minus{} 100 \Leftrightarrow c^2 \plus{} ab \equal{} \minus{} 50$

To solve the middle equation we can try  $ a \equal{} 2de^2, c \equal{} \minus{} df^2, b \equal{} 2def$
then $ c^2 \plus{} ab \equal{} \minus{} 50 \Leftrightarrow d^2f(f^3 \plus{} 2e^3) \equal{} \minus{} 25$
by trying different factorisations we quickly get $ d \equal{} 5, f \equal{} 1, e \equal{} \minus{} 1$ as the only solution up to sign.
$ \implies a \equal{} 10, c \equal{} \minus{} 5, b \equal{} \minus{} 10 \implies x \equal{} a^2 \plus{} 4bc \equal{} 300$ giving one of the solutions

Alternatively we can try $ a \equal{} de^2, c \equal{} \minus{} 2df^2, b \equal{} 2def$
then $ c^2 \plus{} ab \equal{} \minus{} 50 \Leftrightarrow d^2f(2f^3 \plus{} e^3) \equal{} \minus{} 25$
gives $ d \equal{} 1, f \equal{} 1, e \equal{} \minus{} 3$
$ \implies a \equal{} 9, c \equal{} \minus{} 2, b \equal{} \minus{} 6 \implies x \equal{} a^2 \plus{} 4bc \equal{} 129$ giving the other solution.

So we can get both solutions this way too. The algebra was a little more complex than warut's solution but there were less factorization cases to consider at the end.[/quote]
PhilG, I don't think the first case works because I got $ c^2\plus{}ab \equal{} d^2f(f^3\plus{}4e^3)$ and $ d^2f(f^3\plus{}4e^3) \equal{} \minus{}50$ has no solutions in integers.

[quote="Cekaa"]Yes, I've made a mistake with my computer ;P , I'm sorry. In the meantime I've seen another method to find solutions but I don't have time to post it atm.[/quote]
Cekaa, please post it whenever you have a chance.

Warut, you are right. My method only gives the second less obvious solution.
I editted the post accordingly.

Ok, I will post it now, but it only shows how to make another rational, in this case integer solution, if one is already known.
Consider the equation $ (300\plus{}t)^3\minus{}(5000\plus{}at)^2\equal{}2000000$. One of the roots is certainly $ t\equal{}0$, so we remain with a quadratic equation. Next, we try to make it linear, and to do this, we set that $ 3 \cdot 300^2\minus{}2 \cdot 5000a \equal{} 0$, that is $ a\equal{}27$. Solve the linear in $ t$ and get $ t\equal{}\minus{}171$, so we get the wanted second solution $ (x,y)\equal{}(129,383)$ :) All in all, it's quite simple, but quite irritating :P

Cekaa, thanks for posting a very interesting trick to solve this problem.


analytic geometry
graphing lines
slope
trigonometry
vector

We have a slope of mass $M$ and incline angle $\varphi$. The slope can move freely on a horizontal table. We drop a small mass $m$ with velocity $v$ on the slope when it is at rest. In what angle will it bounce of the slope with respect to the horizontal, and what will the velocity of the slope be at this moment?

So, is the angle \[\tan\theta = \left(1-\frac{m}{M}\right) \cot 2\varphi?\]

My answer would be:
$\tan\alpha=\frac{1-(2+\frac{m}{M})\sin^{2}\varphi}{\sin 2\varphi}$
$u=\frac{v\frac{m}{M}\sin 2\varphi}{1+\frac{m}{M}\sin^{2}\varphi}$

I got:

$\tan\theta = \left(1+\frac{m}{M}\right) \cot 2\varphi$

$u=v\sqrt{\frac{m}{M+\frac{M^{2}}{m}\left(1+\cot^{2}2\varphi \left(1+\frac{m}{M}\right)^{2}\right)}}$

*Regarding the last three posts* You just have to admire these mechanics problems with multiple solutions :D Tell me, what are the angles you have calculated? My $\theta$ is the smaller angle made with the horizontal after rebounding, i.e. it is measured between the vector of the rebounded body $m$'s velocity and the negative velocity vector of the incline...


[quote="PhilAndrew"]It seems that you forgot a constraint.[/quote]
It's a fills up measures problem.

If $a,b,c,d \in [0,3]$ and $a+b+c+d\leq 4$, then $a^{2}+b^{2}+c^{2}+d^{2}\leq 10$;

If $a,b,c,d \in [0,3]$ and $a+2b+3c+4d\leq 5$, then $a^{2}+b^{2}+c^{2}+d^{2}\leq 10$.

[hide="I probably overcomplicated this"]Since $a^{2}+b^{2}+c^{2}+d^{2}$ is always convex for each variable, the maximum of the expression occurs when $a,b,c,d$ are at their bounds. So we can plug in the minimum and maximum values for each variable (0 and 3) and show the inequality is satisfied in each case.

Plug 3 into a variable (we'll say $a$) to get $9+b^{2}+c^{2}+d^{2}\leq 10 \Rightarrow b^{2}+c^{2}+d^{2}\leq 1$. Note that since $a+b+c+d \leq 4$, $b+c+d \leq 1$, so our new bounds for each variable are $[0,1]$. Since we are still convex, we plug in 1 to one of the variables and 0 to the others. In each case, we get that $1^{2}+0^{2}+0^{2}\leq 1$ which is obviously true.

Plugging 0 in simplifies to the case above.[/hide]

And now as I post this I see that a second part was just edited in. I'll see if I can apply that method to that one and edit it in. I get the feeling I've misapplied the concept anyways, I've never actually used it in a proof before (yay testing grounds).

[hide="Part 2"]$a^{2}+b^{2}+c^{2}+d^{2}$ is still always convex for each variable. However our bounds are different. Initially:
$a \in [0,3], b \in [0,5/2], c \in [0,5/3], d \in [0,5/4]$

If $a = 3$
Then $2b+3c+4d \leq 2$ and $b^{2}+c^{2}+d^{2}\leq 1$. Second inequality still convex. From the first inequality, $b \in [0,1]$, $c \in [0,2/3]$, $d \in [0,1/2]$. No matter which of these values we use $b^{2}+c^{2}+d^{2}\leq 1$ will be true.

If $b = \frac{5}{2}$, or $c = \frac{5}{3}$, or $d = \frac{5}{4}$
Then the remaining three variables are less than or equal to 0, so they are all 0. So we are showing that either $b^{2}$ $c^{2}$ or $d^{2}$ (whichever is the nonzero one) is less than or equal to 10, which is pretty clear.[/hide]

I'm not sure whether you can apply convexity when there's a constraint. There's always Lagrange Multipliers, or Karamata for the first part...
[hide="another solution to part 1"]
We can reduce to the case $a+b+c+d=4$ since $a^{2}+b^{2}+c^{2}+d^{2}\leq (4-b-c-d)^{2}+b^{2}+c^{2}+d^{2}$.

For this case, WLOG let $a=\max \{ a,b,c,d\}$ so $1\leq a\leq 3$. Then
\[a^{2}+b^{2}+c^{2}+d^{2}\leq a^{2}+(b+c+d)^{2}=a^{2}+(4-a)^{2}\]
Showing that the right hand side is at most 10 reduces to $(a-1)(a-3)\leq 0$. Done.
[/hide]

[hide="part 2"]
Again, we need only consider $a+2b+3c+4d=5$. Then
\[a^{2}+b^{2}+c^{2}+d^{2}\leq a^{2}+\left( b+\frac{3}{2}c+2d\right)^{2}=a^{2}+\left( \frac{5-a}{2}\right)^{2}\]
Now it remains to prove $4a^{2}+(5-a)^{2}\leq 40$ for $0\leq a\leq 3$. We can't fix the order of $a,b,c,d$, but we don't have to: this last statement follows from $(a+1)(a-3)\leq 0$.
[/hide]

[quote="scorpius119"]I'm not sure whether you can apply convexity when there's a constraint. [/quote]

A convex function defined on, say, a hypercube in $\mathbb{R}^{4}$ takes on its extrema at the corners of the cube.  Generally, I think a convex function defined on a domain that is a convex polygon takes on its extrema at the vertices of the polygon.

[quote="PhilAndrew"]What if $a=b=c=d=3$? It seems that you forgot a constraint.[/quote]
I am sorry.I have edited.
The next:
We have $ab=c^{2}+4d^{2}=4$
 Prove that :$(a-c)^{2}+(b-d)^{2}\geq \frac{8}{5}$

Looks like he means $\ge$ in his last step.

[quote="fish"][quote="jcc0107"][quote="fish"]We have $ab=c^{2}+4d^{2}=4$. Prove that :$(a-c)^{2}+(b-d)^{2}\geq \frac{8}{5}$.[/quote]
$(a-c)^{2}+(b-d)^{2}-\frac{1056}{595}=(a-c)^{2}+(b-d)^{2}-\frac{32}{35}ab+\frac{8}{17}(c^{2}+4d^{2})$

$=\frac{(17a-25c)^{2}}{425}+\frac{8(7a-10b)^{2}}{1225}+\frac{(17b-49d)^{2}}{833}>0$,

$\Longrightarrow(a-c)^{2}+(b-d)^{2}>\frac{1056}{595}=1.7747\cdots>1.6=\frac{8}{5}$.[/quote]
I think it not true  :(
Because "=" can appear.[/quote]
if his caculations is true(sorry for being lazy checking :D) , then the equal cant occur , so can you show us when does the equal occur , fish ??

[quote="HTA"]
if his caculations is true(sorry for being lazy checking :D) , then the equal cant occur , so can you show us when does the equal occur , fish ??[/quote]
Iam sorry that i made a mistake
 jcc0107'work is very good
here is mine:
$(1^{2}+2^{2})[(a-c)^{2}+(b-d)^{2}] \geq (a-c+2b-2d)^{2}=[a+2b-(c+2d)]^{2}$
And $a+2b \geq 2\sqrt{2ab}=2\sqrt{8}$
      $(c+2d)^{2}\leq 2(c^{2}+4d^{2})=8-\to  c+2d \leq \sqrt{8}$
-->$[a+2b-(c+2d)]^{2}\geq 8$
-->$5 [(a-c)^{2}+(b-d)^{2}] \geq 8$
"=" can not appear /


Show that there's exist infinite positive integers $ n$ such that $ n^2 \plus{} 1$ has a prime divisor greater than $ 2n \plus{} \sqrt{n}$

have someone tried this problem?  :( I cannot attack this problem  :maybe:

[url]http://www.mathlinks.ro/viewtopic.php?search_id=195411060&t=215221[/url]
See post #4 and adapt the ideas.


[quote="me@home"]Compute
\[{\frac{1}{1+\frac{1}{\sqrt{3}}}+\frac{1}{2+\frac{5}{\sqrt{3}}}+\frac{1}{1+3\sqrt{3}}}\]
[hide="Remark"]
Generalize it.
Also, what is the 'reason' behind the $5$? Is it just $2+3$ or something else?
[/hide][/quote]
[hide="Probably Longer Than the Real Way"]
Firstly, simplify the fractions to get that \[{\frac{1}{1+\frac{1}{\sqrt{3}}}+\frac{1}{2+\frac{5}{\sqrt{3}}}+\frac{1}{1+3\sqrt{3}}}=\sqrt{3}(\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{3}+5}+\frac{1}{\sqrt{3}+9})\]  Now, rationalize the denominators of $\frac{1}{\sqrt{3}+1}$ and $\frac{1}{\sqrt{3}+5}$ to get $\frac{\sqrt{3}-1}{2}$ and $\frac{5-\sqrt{3}}{2}$.  Now add these two fractions to obtain $2$.  Now, rationalize the $\frac{1}{\sqrt{3}+9}$ to obtain $\frac{9-\sqrt{3}}{78}$.  Add this to $2$ to get $\frac{165+\sqrt{3}}{78}$.  Finally, multiply this by $\sqrt{3}$, to get $\frac{165\sqrt{3}+3}{78}=\boxed{\frac{55\sqrt{3}+1}{26}}$[/hide]

[quote="me@home"]Compute
\[{\frac{1}{1+\frac{1}{\sqrt{3}}}+\frac{1}{2+\frac{5}{\sqrt{3}}}+\frac{1}{1+3\sqrt{3}}}\]
[hide="Remark"]
Generalize it.
Also, what is the 'reason' behind the $5$? Is it just $2+3$ or something else?
[/hide][/quote]

I used your hint and got:

[hide]$\frac{1}{1+a}+\frac{1}{2+(a+b)/2}+\frac{1}{1+b}$

OR

$\frac{1}{x}+\frac{1}{1/2 (x+y)+1}+\frac{1}{y}$

 :(  :(  :( *screaming [/hide]

[quote="The QuattoMaster 6000"][quote="me@home"]Compute
\[{\frac{1}{1+\frac{1}{\sqrt{3}}}+\frac{1}{2+\frac{5}{\sqrt{3}}}+\frac{1}{1+3\sqrt{3}}}\]
[hide="Remark"]
Generalize it.
Also, what is the 'reason' behind the $5$? Is it just $2+3$ or something else?
[/hide][/quote]
[hide="Probably Longer Than the Real Way"]
Firstly, simplify the fractions to get that
\[{\frac{1}{1+\frac{1}{\sqrt{3}}}+\frac{1}{2+\frac{5}{\sqrt{3}}}+\frac{1}{1+3\sqrt{3}}}=\sqrt{3}(\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{3}+5}+\frac{1}{\sqrt{3}+9}) \]
Now, rationalize the denominators of $\frac{1}{\sqrt{3}+1}$ and $\frac{1}{\sqrt{3}+5}$ to get $\frac{\sqrt{3}-1}{2}$ and $\frac{5-\sqrt{3}}{2}$.  Now add these two fractions to obtain $2$.  Now, rationalize the $\frac{1}{\sqrt{3}+9}$ to obtain $\frac{9-\sqrt{3}}{78}$.  Add this to $2$ to get $\frac{165+\sqrt{3}}{78}$.  Finally, multiply this by $\sqrt{3}$, to get $\frac{165\sqrt{3}+3}{78}=\boxed{\frac{55\sqrt{3}+1}{26}}$[/hide][/quote]

But there is a $2$, not $1$, in the second fraction...
And by fixing it, you get a simpler answer of...
[hide]$1$.[/hide]

the cause of 5 is

in each time we add 4

1/sqrt(3) ,  (4+1)/sqrt(3) ,  (4+5)/sqrt(3)=9sqrt(3)/3=3sqrt(3)

Well the correct answer certainly is [hide]1[/hide]
And the cause of the $5$ is rather tricky (no one got it right): [hide]$\frac{1^{2}+3^{2}}{2}=5$ where $2=xy-1$

EDIT: actually goldbach is very close to being right, I'm realizing. Indeed, whereas goldbach just guessed a pattern he guessed correctly - since $5$ is the mean of $1^{2},3^{2}$ the three numbers do form a short arithmetic sequence therefore the difference between each is constant and it just happens to be 4, so good observation goldbach but I don't think you saw the full reason behind it :)  :D [/hide]

[hide="edit2:/I finally post the solution generalization or whatever"]
Consider the identity \[\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{\frac{(x+1)(y+1)}{xy-1}}=1\]
That's it :P notice $(x+1)(y+1)=4+\frac{10}{\sqrt{3}}$ and $xy-1=2$ so it all cancels and is nice...
[/hide]

[quote="me@home"]
[hide="edit2:/I finally post the solution generalization or whatever"]
Consider the identity
\[\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{\frac{(x+1)(y+1)}{xy-1}}=1 \]
That's it :P notice $(x+1)(y+1)=4+\frac{10}{\sqrt{3}}$ and $xy-1=2$ so it all cancels and is nice...
[/hide][/quote]

oh well... I was close too  :D I couldn't rewrite the middle fraction, ...should have multiply. Could you post another problem? thanks!

[quote="me@home"]Sorry Ithat was a pretty random one I came up with while working on a completely unrelated geometry problem. I'm thinking of ways in which geometry calculations can simplify interesting expressions. (Basically calculate the same thing in two completely different ways and boom you have an identity)

Fine, that reminded me of: Simplify $\frac{\sqrt{2+\sqrt{3}}}{2}$ I think it's like $\cos(15)$ or something.[/quote]
[hide]
Say that $\sqrt{a}+\sqrt{b}=\sqrt{2+\sqrt{3}}\Rightarrow a+b+2\sqrt{ab}=2+\sqrt{3}\Rightarrow a+b=2\Rightarrow a=2-b$.  We also have that $2\sqrt{ab}=\sqrt{3}$.  Thus, $\sqrt{ab}=\frac{\sqrt{3}}{2}\Rightarrow ab=\frac{3}{4}\Rightarrow a=\frac{3}{4b}$.  Thus, $2-b=a=\frac{3}{4b}\Rightarrow 8b-4b^{2}=3\Rightarrow 4b^{2}-8b+3=0\Rightarrow (2b-3)(2b-1)=0\Rightarrow b=\frac{1}{2},\frac{3}{2}$  Thus, if $b=\frac{1}{2}$, then $a=\frac{3}{2}$ and if $b=\frac{3}{2}$, then $a=\frac{1}{2}$.  Thus, $\sqrt{a}+\sqrt{b}=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}=\frac{\sqrt{2}+\sqrt{6}}{2}$.

Thus, $\frac{\sqrt{2+\sqrt{3}}}{2}=\frac{\sqrt{a}+\sqrt{b}}{2}=\frac{\sqrt{2}+\sqrt{6}}{4}$[/hide]

Okay good :)
Don't expect another problem


quadratics
calculus
integration
number theory unsolved
number theory

Find all primes $p$ such that $p^2-p+1$ is a perfect cube.

my solution:
   $p^2-p+1=q^3 <->p(p-1)=(q-1)(q^2+q+1)$
hence $p(p-1)$ divisible to (q-1).But $(p,p-1)=1$ so $p$ divisible to $(q-1)$ or $(p-1)$ divisible to $(q-1)$
But $p >q$ so that $(p-1)$ divisible to $(q-1)$.hence $(p-1)=k*(q-1)$ $(k \in Z)$
Now we have $q^2+(1-k^2)q+(k^2-k+1)=0$
so that $k^4-6k^2+4k-3$ is a quare number .But $(k^2-3)^{2} \leq k^4-6k^2+4k-3<(k^2-1)^{2}$
so we are easy to see the solution $k=3$ then we have the root $(p,q)=(19,7)$ :)

[quote="nhat"]
hence $p(p-1)$ divisible to (q-1).But $(p,p-1)=1$ so $p$ divisible to $(q-1)$ or $(p-1)$ divisible to $(q-1)$[/quote]

Why wouldn't the prime factors of $q-1$ divide into two groups, one dividing $p$ the other dividing $p-1$?

Pierre.

In another hand, the reasoning can be adapted :
Since $p$ is prime, then $p$ divides $q-1$ or $q^2+q+1$. If $p$ divides $q-1$, let $q = ap+1$ for some nonnegative integer $a$. Then $a(q^2+q+1) = p-1.$
But if $a > 0$ we have $q>p$ which leads to $q^2+q+1  p-1$, a contradiction. It is easy to see that $a=0$ is not possible.
Thus $p$ divides $q^2+q+1$, which means that $q^2+q+1 = kp$ so that $p-1 = k(q-1)$ and we end as proposed by Nhat.

Pierre.

I have another way to prove that $p$ divides $q^2+q+1$
since $p^2-p+1=q^3$ then $p^3+1=q^3(p+1)$ so $q^3 \leq p^3+1$ (1)
also $p^2-p+1=q^3 \Rightarrow p(p-1)=(q-1)(q^2+q+1)$
since $p$ is prime this means that it divides $q-1$ or $q^2+q+1$
If $p$ divides $q-1$ then $p \leq q-1 <q \Rightarrow p^3<q^3$
and from (1) we deduce that $q^3=p^3+1$
Hence $p^2-p+1=q^3 \Rightarrow p^2-p+1=p^3+1$
So $p$ divides $1$ which is not possible.
then $p$ divides $q^2+q+1^$ and we continue like nhat ...

It is a probelm from Saint-Petersburg selection test for All-Russian of 1995.

Another way to finish the solution:
after we get $q^2+(1-k^2)q+(k^2-k+1)=0$, we may consider it modulo $q-1$ and get that $k-3$ is divisible by $q-1$. If $k=3$, we get $q^2-8q+7=0$, so $q=7$; else $k\ge q+2$ and LHS is less then $q^2+q-k^2q+k^2\le q^2+q-(q-1)(q+2)^2<0$.

This question came from a National Preparation Problem Set for this year.
To prove that p must divide not q-1, it's easy to prove that p>q from which p must divide the other factor.
The question, nicely, dosen't use the fact that q is prime, it only comes out incidentally.

Bomb

Yes,this is not so hard,I solved it almost the same way nhat did.
:-)

$k \geq 3$, because $p|x^2+x+1$, so $-3$ is quadratic residue moduo $p$.
=> $6|p-1$ and $6|x-1$.

Can we extend it?
Find all primes $p$ such that $p^2-p+1$ is a natural power:
$p^2-p+1=q^m$, $m>1$.

let w be a third root of unity not $1$.

$p^2-p+1=(p+w)(p+w^2)$ so $p+w=(a+bw)^3(-w)^k$now, seing all possibilities for $k$ mod $6$, and comparing parts we get to the only possible case $a=3$, $b=1$, $p=19$.

What if $k=2$? I tried this case for a while, without success.

It was question for Pascual2005's approach, he has different $k$

Yes Fedor, It let us in a no way walk, so i am sorry

Can we determine all primes $p$ such that $p^2-p+1$ is an odd natural power:
$p^2-p+1=q^m$, $m>1$, $m$-odd.

The easiest way to prove that q-1 divides p-1 is that knowing that p>q to see that q = 2*k+1 so q-1=2*k which means that p cannot be equal to k*(q-1)

Notice that $ p^2\minus{}p\equal{}q^3\minus{}1$ defines an elliptic curve. We can then find all integral points and finish the problem.

[quote="shinichiman"]I don't undestant why $p>q$[/quote]
$p^2-p+1<p^3\implies p>q$

This question is also AZERBAIJAN'S selection exam.Don't copy...Be happy....................
Bilduz eeeeeeeeeeeeeeee

Well, that means Azerbadjan [b]did copy [/b]... by re-using this problem from the Balkan MO :)

See here may help you: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=576488 / http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=487608&p=2732752&hilit=italy+prime#p2732752

I have an other solution.

From $p^2-p+1=a^3$ we have $p(p-1)=(a-1)(a^2+a+1)$ so $p \mid a^2+a+1$ and $a-1 \mid p-1$. In particular $p \ge a$.

We have $\frac{a^2+a+1}{p} \equiv 3 mod(a-1)$. For $a=1, 2, 3$ we don't have any solutions so we assume that $a \ge 4$. [if $a<4$ we could have $3 \equiv 1 mod (a-1)$ so the following proof is incomplete]

Case 1 : $\frac{a^2+a+1}{p}=3$. Then $p-1=3(a-1)$ so $a^2+a+1=3(3a-2)$ ie. $a=1$ or $a=7$. $a=7$ gives $p=19$.

Case 2 : $\frac{a^2+a+1}{p}=a+2$ We have $a+2 \mid a^2+a+1$ so $a=1$ and no solution.

Case 3 : $\frac{a^2+a+1}{p}=3+k(a-1)$ with $k \ge 2$. Then $p=\frac{a^2+a+1}{3+k(a-1)} \le \frac{a^2+a+1}{2a+1} <a$ if $a>1$ which is a contradiction, so no solution.

Hence $p=19$ is the only solution.

 n hat wrote
              (k² -3)²≤k⁴-6k²+4k-3<(k²-1)²
        I cant understand this pls explain this

@above we know that K^4-6K^2+4k-3 is a square number. However it is greater or equal to (k^2-3)^2 and less than (K^2-1)^2 which can be seen by simple expansion (k is greater or equal to 3 as K^4-6K^2+4k-3 must be positive). letting K^4-6K^2+4k-3 equal to (k^2-3)^2, (k^2-2)^2, the only solution to k=3

I guess this is what troll NT looked like in 2005 - probably have a similar solution to those above - posting for storage.
[hide=Answer]The only prime that satisfies the condition is $p=19$. [/hide]
[hide=Solution]Assume that $p$ is a prime satisfying the condition. \\
Write $$p(p-1) = (x-1)(x^2+x+1)$$
$\textbf{Case 1)}$ $p \mid x+1$ \\
We then get that $x^2+x+1 \mid p-1$ which means that $p \leq x-1, x^2+x+1 \leq p$
$$x+1 < x^2+x+1 \leq p-1 < p $$ which is a contradiction. \\
$\textbf{Case 2)}$ $p \mid x^2+x+1$ \\
Now, write $ap=x^2+x+1$ and $a(x-1) = p-1$.\\
We get that $$a^2x-a^2+a=a(a(x-1)+1)=x^2+x+1$$
Therefore, we get the quadratic: $$x^2+(1-a^2)x+a^2-a+1 = 0$$
Therefore $D = (a^2-1)^2-4(a^2-a+1) = a^4-6a^2+4a-3$ has to be a perfect square. \\
Notice that $$D = a^4-6a^2+4a-3 < a^4-4a^2+4=(a^2-2)^2$$
This is because $$2a^2-4a+7 = 2(a-1)^2+6>0$$ \\
$\textbf{Claim:}$ If $a>3$, then $D > (a^2-3)^2$ \\
$\textbf{Proof)}$ \\
We need to check: \\ $$a^4-6a^2+4a-3 > (a^2-3)^2 = a^4-6a^2+9 \iff 4a > 12$$
This proves the claim. $\square$ \\

We therefore get that if $a \geq 4$, then $(a^2-2)^2>D>(a^2-3)^2$ which is a contradiction. \\
We therefore have to check $a \in \{1,2,3\}$ which ultimately gives that $p=19$ is the only solution. $\square$[/hide]

Given, $p^2-p+1=x^3\Rightarrow p^2-p=p(p-1)=x^3-1=(x-1)(x^2+x+1)$.
$\forall\ x\in\mathbb{N}$, we have $p^2-p+1=x^3\equiv0,1,8(\textbf{mod }9)\Rightarrow p^2-p\equiv0,7,8(\textbf{mod }9)$. However we know that $p^2-p\equiv0,2,3,6(\textbf{mod }9)$. This means that $9\mid (p^2-p)=p(p-1)$. Since $9$ and $p$ are coprime, we get $9\mid p-1\Rightarrow p\equiv1(\textbf{mod }9)$.  This gives $p^2-p+1(\equiv x^3)\equiv1(\textbf{mod }3)$ which gives $x\equiv1(\textbf{mod }3)$. This shows that both $x-1$ and $x^2+x+1$ have a common factor of $3$. $GCD(x-1, x^2+x+1)=GCD(x-1,x^2+x+1-x^2+2x-1)=GCD(x-1,3x)=GCD(x-1, 3x-3(x-1)=GCD(x-1,3)=3$. Hence, $p$ cannot divide $x-1$ and $x^2+x+1$ simultaneously, because $p\ne3$.

CASE 1) $p\mid x-1$.
Let $x-1=kp\Rightarrow p(p-1)=(x-1)(x^2+x+1)=kp(x^2+x+1)\Rightarrow k(x^2+x+1)=p-1$.
Now, $k\geq1\Rightarrow x-1\geq p\Rightarrow x-2\geq p-1$ and $x^2+x+1\leq p-1\leq x-2\Rightarrow x^2+3\leq 0$ which is a contradiction.

CASE 2) $p\mid x^2+x+1$.
Let $x^2+x+1=kp\Rightarrow k(x-1)=p-1\Rightarrow p=kx-k+1\Rightarrow x^2+x+1=k^2x-k^2+k$. This gives the quadratic in $x$ as follows:
$x^2+(1-k^2)x+k^2-k+1=0$.
The discriminant of this quadratic should be a perfect square.
So, we have $D=(1-k^2)^2-4(k^2-k+1)=k^4-2k^2+1-4k^2+4k-4=k^4-6k^2+4k-3$ should be a perfect square. It should be positive, so $k\geq3$. Now, $(k^2-3)^2=k^4-6k^2+9\leq k^4-6k^2+4k-3=D$ (equality holds if and only if $k=3$) and $(k^2-1)^2=D+4(k^2+1)>D$.
$k=3$ satisfies. For $k>3$, we have $(k^2-3)^2<D<(K^2-1)^2\Rightarrow D=(k^2-2)^2=k^4-4k^2+4=k^4-6k^2+4k-3\Rightarrow 2k^2-4k+7=0$ which has no integral solutions. So, $k$ must be $3$. This means $x^2-8x+7=0\Rightarrow x=1,7$. $x=1$ corresponds to $p=0,1$ (not prime) and $x=7$ corresponds to $p=19,-18$ (we reject $-18$).

Hence, $p=19$ is the only such prime.

Not a solution, but just stating the fact that this is Ljunggren's Theorem. Obviously you cannot apply that on a contest, but still, well, it's nice.

Source: Alexandru Gica, Rational Points on Elliptic Curves

At first, note that $p^2-p+1>0$ for all $p\in\mathbb Z^{+}$, by this we can conclude that doesn't exist a negative integer $q$ such that $p^2-p+1$. Now lets assume that exists $q>1$ such that $p^2-p+1=q^3$, by arranging this equation we get $p(p-1)=(q-1)(q^2+q+1)$, given that $p$ is prime, we have $p \mid q-1$ or $p \mid q^2+q+1$. If both cases occur, we get $p \mid q^2+q+1-(q-1)^2 = 3q$, we can't have $p \mid q$ because this will lead to a contadiction, so, we get $p \mid 3$, hence $p=3$, but $3^2-3+1=25$, which isn't a perfect cube, so, both cases can't occur.\\

Now assume $p \mid q-1$, this in particular implies that $p<q-1$. If $p>3$, we get $p-1<(p+1)(p+2)+1<q^2+q+1$, hence we need $p=2$, but $2^2-2+1=3$, which in fact, isn't a perfect cube. The only case remaining is $p \mid q^2+q+1$, and this implies that exists $k\in\mathbb Z^{+}$ such that $pk=q^2+q+1$, substituting in $p(p-1)=(q-1)(q^2+q+1)$ we get $p-1=(q-1)k=qk-k \implies p=qk-k+1$, hence, $pk=q^2+q+1 \implies (qk-k+1)k=q^2+q+1 \implies q^2+(1-k)q+(k^2-k+1)=0$. So, $q$ is an integer root of $f(x)=x^2+(1-k^2)x+(k^2-k+1)$, given that $q$ is a positive integer, we need $(1-k^2)^2-4(k^2-k+1)=k^4+6k^2+4k-3$ be a perfect square. We can see that $(k^2-3)^2\leq k^4-6k^2+4k-3<(k^2-1)^2$, and clearly $k^4-6k^2+4k-3\neq (k^2-2)^2$, therefore $k=3$ and $f(x)=x^2-8x+7=(x-7)(x+1)$, knowing that $q>0$, we can easily conclude that $q=7$, thus $p=7(3)-3+1=19$, so we conclude that the unique prime $p$ such that $p^2-p+1$ is a perfect cube is 7.


Suppose the coefficients of the equation $ x^n\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}...\plus{}a_1x\plus{}a_0\equal{}0$ are real and satisfy $ 0<a_0\le a_1\le...\le a_{n\minus{}1} \le 1.$ Let z be a complex root of the equation with $ |z|\ge 1$. show that $ z^{n\plus{}1}\equal{}1$.

[hide="hint"]
Show that $ \vert z \vert\leq 1$, and equality is achieved for exactly one choice of coefficients.
[/hide]

Divide the expression by $ x^n$, and define $ y \equal{} \frac {1}{x}$ the polynomial now looks: $ P(y) \equal{} 1 \plus{} a_{n \minus{} 1}y \plus{} \dots a_0y^n$ define $ b_0 \equal{} 1, b_1 \equal{} a_{n \minus{} 1}, \dots ,b_n \equal{} a_0$ it's clear that: $ b_0\geq b_1\geq \dots \geq b_n > 0$ If $ z$ was a root of the original polynomial then $ \frac {1}{z} \equal{} \alpha$ is a root of the new one. If $ |z|\geq 1\Rightarrow |\alpha|\leq 1$. We have:

$ b_n\alpha^{n} \plus{} \dots \plus{} b_1\alpha \plus{} b_0 \equal{} 0$ if we multiply the expression by $ \alpha \minus{} 1$ then we get:
$ \alpha^{n \plus{} 1}(b_n) \plus{} \alpha^{n}(b_{n \minus{} 1} \minus{} b_n) \plus{} \dots \plus{} \alpha(b_0 \minus{} b_1) \minus{} b_0 \equal{} 0$ $ \Rightarrow$ $ b_0 \equal{} \alpha^{n \plus{} 1}b_n \plus{} \alpha^n(b_{n \minus{} 1} \minus{} b_n)\dots \plus{} \alpha(b_0 \minus{} b_1)$ $ \Rightarrow$ $ |b_0| \equal{} |\alpha^{n \plus{} 1}b_n \plus{} \alpha^n(b_{n \minus{} 1} \minus{} b_n)\dots \plus{} \alpha(b_0 \minus{} b_1)|\leq |\alpha|(b_0 \minus{} b_1) \plus{} \dots \plus{} |\alpha^{n \plus{} 1}|b_n \leq b_0 \minus{} b_1 \plus{} \dots b_n \equal{} b_0$ this means $ |\alpha| \equal{} 1$ then the last you asked follows immediately.

Am I right?


group theory
abstract algebra
linear algebra
matrix
superior algebra
superior algebra unsolved

Give an example of a group G, subgroup H, and an element a in G such aHa^(-1) is properly contained in H:

1) Clearly G must be non-abelian.

2) Also H( and hence G) must be infinite.

Anyone have an example??? Does the symmetric group on the natural numbers work??? Maybe matrices???

MOZART1756,

There are simpler example than the following one, but it's a pretty one.

Take $G=\mathrm{GL}_n(\mathbb{Z})$, $H=\mathrm{SL}_n(\mathbb{Z})$ and $a=I_n+E_{1,2}$, where $E_{1,2}$ is the $n\times n$ matrix with $0$ everywhere except row 1 and column 2 where it's a 1.

Hello JC_math,

since a is in H, $a H a^{-1} = H$ in this case.

Assume in the general situation that $a H a^{-1}$ is a proper subgroup of $H$. Then:
1. $G$ is nonabelian infinite, $H$ is infinite.
2. $\mathrm{ord}(a) = \infty$.
3. $H$ is a proper subgroup of $a^{-1} H a$.
4. We get the ascending subgroup chain $a^{-n} H a^n, n \in \mathbb{N}$, so the index $G: H$ is infinite.

LiMa

you can use two by two real matrices.
Compute
[2 0][1 n][2 0]^-1
[0 1][0 1][0 1]

Soarer, that does it. It is a good question. Your example satisfies Herstein's request. Clearly your H can't be normal, otherwise you would have equality. Hence by the definition of normality there must also exist a g in your G=GL(Z)(2 by 2 matrices) such that gHg^(-1) has no inclusion with H.
The element g=[10]
                       [11] will do that

Thank you.

sorry but I don't really know what you mean  :blush:

I proposed to set a equals
[2 0]
[0 1]
and H to be set of all 2*2 matrices with diagonal entries 1, lower triangular entry 0 and upper triangular entry n, an integer.

Soarer, what you have is right. My comment was a complete triviality that the subgroup you presented was indeed not a normal subgroup and followed the prescribed conditions. Herstein asked this question that one might broaden their knownledge in the examples of groups, while stressing the precise definition of the meaning of a normal subgroup.


invariant
geometry
circumcircle
ratio
geometry proposed

[b]Let w be the circumcircle of the [/b]$\triangle ABC$ [b]and the mobile point [/b]$M\in AA$ [b](the tangent in the point [/b] $A$ [b]at the circle w). I note [/b]$\{B,N\}\subset MB\cap w,\ \{C,P\}\subset MC\cap w$. [b]Prove that the ratio [/b]$\frac {AN\cdot AP}{NP}$ [b]is constant. [/b]

Make use of the similarities $\triangle AMN\sim\triangle BMA,\triangle MNP\sim\triangle MCB,\triangle MAP\sim\triangle MCA$.


function
inequalities unsolved
inequalities

[quote="tranthanhnam"]find all  funtion satisfying$\large f: {\mathbb{R^{+}}}\to{\mathbb{R^{+}}}$and$\large f(xf(y))=f(xy)+x$where$x,y \geq 0$[/quote]
do you think it's an exercice of inequalite??

$f(f(x)f(y))= f(f(x)y))+f(x)=f(xy)+y+f(x)=f(yx)+x+f(y)$ , so $\forall x ,\ f(x)-x =c \ (constant)$
I put in the initial equation and obtain $c=1$, so $f(x)=x+1$
 :cool:


limit
trigonometry
function
real analysis
real analysis unsolved

how can I prove that the followin limit doesn't exist: $ \displaystyle\lim_{x\to\plus{}\infty}x/|tan(x)|$ ?

If $ u_n \equal{} n\pi,n \equal{} 1,2,...$ then $ u_n/|\tan u_n|\to \infty$. If $ 0 < x_1 < x_2 < ...$ is the sequence of positive roots of the equation $ \tan x \equal{} x$ then $ x_n/|\tan x_n|\to 1$. That's all.

I would not write $ \frac{u_n}{|\tan u_n|}\to \infty$. The function $ f(x)\equal{}\frac{x}{|\tan x|}$ is not defined at integer multiples of $ \pi/2$ (although some of the singularities are removable). Depending on your definition of a limit, this may already imply that the limit $ \lim_{x\to\infty}f(x)$ does not exist.

If your definition of a limit allows $ f$ to be undefined at some points, then sequences like $ (x_n)$ defined by [b]N.T.TUAN[/b] can be used to show that the limit does not exist.


let ABC be a triangle .onle by using acmpos draw the circumcircle of ABC.

You can just edit your post, you didn't need to make a new one :)

oh, thank God it's compass, when i saw my surname at the topic i really got scared  :rotfl:

ok, thanks Arne.

this problem is very nice.and( Napelon Bonapart )invented and solved it.

i think, if I and u think about it it will have more and more nice solutions... ;)

:rotfl:  dear campos,sorry for it plz.then it will be better if u think about it .if not :mad: ...

oh thanks michal ;)

im glad that u find that inversion is incridible :D

u can see:  http://www.mathlinks.ro/Forum/viewtopic.php?t=57006


Let $ \left(a_n\right)_{n\ge0}$ be a sequence of real numbers so that $ a_{n \plus{} 1} \plus{} a_{n \minus{} 1} \equal{} |a_n|$ for all $ n\ge1$. Prove that this sequence is periodic.
[i]Adapted after V. Pop[/i]

If $ a_n>0$ for all $ n\in\mathbb{N},n>1$, then $ a_5\equal{}a_4\minus{}a_3\equal{}(a_3\minus{}a_2)\minus{}a_3\equal{}\minus{}a_2<0$. Contradiction!
So we find $ n\in\mathbb{N}$ with $ a_{n\plus{}1}\le 0$.
Then either $ a_n\le 0$ or $ a_{n\plus{}2}\le 0$ or $ a_n,a_{n\plus{}2}>0$.

In the latter case we have
$ a_{n\plus{}3}\equal{}a_{n\plus{}2}\minus{}a_{n\plus{}1}\equal{}a_{n\plus{}2}\plus{}|a_{n\plus{}1}|>0$
$ a_{n\plus{}4}\equal{}a_{n\plus{}3}\minus{}a_{n\plus{}2}\equal{}|a_{n\plus{}1}|\ge 0$
$ a_{n\plus{}5}\equal{}a_{n\plus{}4}\minus{}a_{n\plus{}3}\equal{}\minus{}a_{n\plus{}2}< 0$
$ a_{n\plus{}6}\equal{}|a_{n\plus{}5}|\minus{}a_{n\plus{}4}\equal{}a_{n\plus{}2}\minus{}|a_{n\plus{}1}|\equal{}\minus{}a_n< 0$.

So in each case we find $ n\in\mathbb{N}$ with $ a_n\le 0$ and $ a_{n\plus{}1}\le 0$.
Then
$ a_{n\plus{}2}\equal{}|a_{n\plus{}1}|\minus{}a_n\equal{}|a_n|\plus{}|a_{n\plus{}1}|\ge 0$
$ a_{n\plus{}3}\equal{}a_{n\plus{}2}\minus{}a_{n\plus{}1}\equal{}|a_n|\plus{}2|a_{n\plus{}1}|\ge 0$
$ a_{n\plus{}4}\equal{}a_{n\plus{}3}\minus{}a_{n\plus{}2}\equal{}|a_{n\plus{}1}|\ge 0$
$ a_{n\plus{}5}\equal{}a_{n\plus{}4}\minus{}a_{n\plus{}3}\equal{}\minus{}|a_n|\minus{}|a_{n\plus{}1}|\le 0$
$ a_{n\plus{}6}\equal{}|a_{n\plus{}5}|\minus{}a_{n\plus{}4}\equal{}|a_n|\ge 0$
$ a_{n\plus{}7}\equal{}a_{n\plus{}6}\minus{}a_{n\plus{}5}\equal{}2|a_n|\plus{}|a_{n\plus{}1}|\ge 0$
$ a_{n\plus{}8}\equal{}a_{n\plus{}7}\minus{}a_{n\plus{}6}\equal{}|a_n|\plus{}|a_{n\plus{}1}|\ge 0$
$ a_{n\plus{}9}\equal{}a_{n\plus{}8}\minus{}a_{n\plus{}7}\equal{}\minus{}|a_n|\equal{}a_n\le 0$
$ a_{n\plus{}10}\equal{}|a_{n\plus{}9}|\minus{}a_{n\plus{}8}\equal{}\minus{}|a_{n\plus{}1}|\equal{}a_{n\plus{}1}\le 0$

But now $ a_n\equal{}a_{n\plus{}9}$ and $ a_{n\plus{}1}\equal{}a_{n\plus{}10}$ gives $ a_k\equal{}a_{k\plus{}9}$ for all $ k\in\mathbb{N}$ by induction.


When I click on the resources icon on the top right, the head stays (i.e. I can still see the Forum/Latex Help/Classes.../Resources icons and the links to Galleries etc.) but the main body of the page goes blank. Is this just me? Could it be fixed?

It does it whether I use IE7, Firefox, Safari or Opera, and I'm running XP.

edit: Same thing happens if I go to a link such as http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16 as well as http://www.mathlinks.ro/Forum/resources.php

edit #2: That bit now works (thanks  :D ) though similar problems start to come in when going to an invidual contest's page - such as http://www.mathlinks.ro/Forum/resources.php?c=27&cid=103&year=2007 or http://www.mathlinks.ro/Forum/resources.php?c=1&cid=17&year=2000

Thanks.


trigonometry
geometry
circumcircle
cyclic quadrilateral
trig identities
Law of Sines
geometry solved

On the sides $AB$ and $BC$ of triangle $ABC$, points $K$ and $M$ are chosen such that the quadrilaterals $AKMC$ and $KBMN$ are cyclic , where $N = AM \cap CK$ . If these quads have the same circumradii, find $\angle ABC$

[color=blue][b]Problem.[/b] Let K and M be points on the sides AB and BC of a triangle ABC, and let $N=AM\cap CK$. Assume that the quadrilaterals AKMC and KBMN are cyclic, and their circumradii are equal. Prove that < ABC = 45°.[/color]

[i]Solution.[/i] By the extended law of sines, the circumradius R of a triangle ABC can be found by the formula $R=\frac{a}{2\sin A}$, where a = BC is a side and A = < CAB is the corresponding angle of the triangle ABC.

Applying this extended law of sines to the triangles KAM and KBM, we find that the circumradius of triangle KAM equals $\frac{KM}{2\sin\measuredangle KAM}$, and that the circumradius of triangle KBM equals $\frac{KM}{2\sin\measuredangle KBM}$.

Now, the circumradius of triangle KAM is, of course, the circumradius of the cyclic quadrilateral AKMC, and the circumradius of triangle KBM is the circumradius of the cyclic quadrilateral KBMN. Since the circumradii of the quadrilaterals AKMC and KBMN are equal, we thus can say that the circumradii of the triangles KAM and KBM are equal. Thus, $\frac{KM}{2\sin\measuredangle KAM}=\frac{KM}{2\sin\measuredangle KBM}$. Thus, sin < KAM = sin < KBM. Hence, either < KAM = < KBM, or < KAM + < KBM = 180°. Now, < KAM + < KBM = 180° is impossible, since it would mean < BAM + < ABM = 180°, but the sum of two angles of a non-degenerate triangle is never 180°. Thus, we must have < KAM = < KBM. In other words, < BAM = < ABC. Similarly, < BCK = < ABC. Now,

  < KNM = < ANC = 180° - < NAC - < NCA          (sum of angles in triangle ANC)
            = 180° - (< CAB - < BAM) - (< ACB - < BCK) = 180° - (< CAB - < ABC) - (< ACB - < ABC)
            = (180° - < CAB - < ACB) + 2 < ABC.

By the sum of angles in triangle ABC, we have 180° - < CAB - < ACB = < ABC, and thus we get

  < KNM = < ABC + 2 < ABC = 3 < ABC.

On the other hand, since the quadrilateral KBMN is cyclic, < KNM = 180° - < KBM, so that < KNM = 180° - < ABC. Thus, 3 < ABC = 180° - < ABC. This becomes 4 < ABC = 180°, what is equivalent to < ABC = 45°. And the problem is solved.

  darij

Correct answer and maginifcent solution!!!

[quote="Rushil"]On the sides $AB$ and $BC$ of triangle $ABC$, points $K$ and $M$ are chosen such that the quadrilaterals $AKMC$ and $KBMN$ are cyclic , where $ N = AM \cap CK $ . If these quads have the same ctrcumradii, find $ \angle ABC$[/quote]

By the informations we have $\angle CKA = \angle AMC = 180^{\circ} - \angle BMA = \angle BKC \iff \angle CKA = \angle CKB = 90^{\circ}$. Hence $N$ is the orthocenter of $ABC$. Denote $I$ and $J$ are respectively the midpoints of $AC$, $BN$. We also have the quadrilateral $IMJK$ is cyclic, hence triangle $IMJ$ is right at $M$, with $MI = MJ$ we have $MIJ$ is isocoles right triangle. Hence $\angle MjK = 90^{\circ}$, so $\angle ABC = 45^{\circ}$.


abstract algebra
superior algebra
superior algebra unsolved

The units of Z_15 = {1, 2, 4, 7, 8, 11, 13, 14}

I'm trying to find a group of order less than 9 that this is isomorphic to.  I seem to be having problems with this.  Can anyone give me a push in the right direction?

It's abelian of order 8. Finitely generated (including finite) abelian groups are products of cyclic groups. There aren't very many possibilities- $ 8,2\times 4,2\times 2\times 2$.

For the more general version, the Chinese Remainder Theorem decomposition is very useful- $ \mathbb{Z}/mn\mathbb{Z}$ is isomorphic to $ (\mathbb{Z}/m\mathbb{Z})\times(\mathbb{Z}/n\mathbb{Z})$ as rings when $ m$ and $ n$ are relatively prime. This induces an isomorphism of multiplicative groups of units.

[quote="jmerry"]There aren't very many possibilities- $ 8,2\times 4,2\times 2\times 2$.[/quote]

For something really simple-minded, try finding the squares of the elements other than $ 1.$

$ 2^2\equal{}4$
$ 4^2\equal{}1$
$ 7^2\equal{}4$ and that's really all the work you need to do; you now already know that
$ 8^2\equal{}(\minus{}7)^2\equal{}4$
$ 11^2\equal{}(\minus{}4)^2\equal{}1$
$ 13^2\equal{}(\minus{}2)^2\equal{}4$
$ 14^2\equal{}(\minus{}1)^2\equal{}1.$

This should give you more than enough evidence to decide among the possibilities that jmerry presented.


A cylindrical jar with height 8 inches and diameter 6 inches is filled to $ 75\%$ of its capacity with juice. The juice is then poured into another cylindrical container with a 10-inch diameter and height of 4 inches. To what percent of its capacity is the second jar filled with juice?

The first cylindrical jar can hold $ 72\pi$ cubic inches, so it's filled with $ .75 \times 72\pi \equal{} 54\pi$ cubic inches of juice.

The second cylindrical jar can hold $ 100\pi$ cubic inches, so the juice takes up $ 54\pi \div 100\pi \equal{} \boxed{54}$ percent.


[quote="Cyber-Shreyas"]Does free fall always mean that initial velocity is zero? This has been bugging me in more than 1 exam. :(([/quote]

No,

Free fall is motion with no acceleration other than that provided by gravity.

Therefore, the initial velocity does not affect it, only the acceleration. (So you could be projected upwards and still be "free falling")


How many factors of 672 are there that are [b]not[/b] prime?

$672=2^5\times3\times7$
Number of factors of 672=$(5+1)(1+1)(1+1)$=24
Among them, 2, 3 and 7 are not primes.
Therefore the number of non-prime factors is 24-3=21


calculus
integration
function
calculus computations

i dont know where im going wrong with this problem, and sorry if this problem looks ugly, im on a computer at work and cant install a program to write out the integrals nicely.

x=0 to 1 and y= 0 to e^(-x) of x*y^2 dy dx

i keep getting (1/27)(-2(e^-3)-1). the -2 should be a 4 however and i dont know what i did wrong.
thanks alot.

$\int_0^1\int_0^{e^{-x}}xy^2\,dy\,dx$

$=\int_0^1\left[\left.\frac13xy^3\right|_{y=0}^{y=e^{-x}}\right]\,dx$

$=\int_0^1\frac13xe^{-3x}\,dx$ (integrate by parts)

$=\left.-\frac19xe^{-3x}\right|_0^1+\frac19\int_0^1e^{-3x}\,dx$

$=-\frac19e^{-3}+\left[\left.-\frac1{27}e^{-3x}\right|_0^1\right]$

$=-\frac19e^{-3}-\frac1{27}e^{-3}+\frac1{27}=\frac{1-4e^{-3}}{27}$

Partial check: the integral should be positive, since the function being integrated is positive. This answer is positive.


Im am currently doing a project that has to be based on the history of maths and need to do a 6000 word project on a certain topic of this. I have always been intersted in astronomy but dont know that much about it. I would appreciate it if you could think of any project titles that come under mathematical astronomy and its history for me to research (a topic big enough to do 6000 words on)

Thank you for any input

theroetical astromechanics
it has mathematical constraint functions $ f(x)$ and has got to do with a lot theoretical atsronomy $ astronomy forum$


linear algebra
matrix
function
linear algebra unsolved

$ I$ is an open nbh of $ 0 \in R$, for every $ \lambda \in I$, define $ r \times r$ matrix by

 $ v_{ij}(\lambda)=\delta_{ij}\lambda+{\lambda}^{2}C_{ij}(\lambda)$, where $ \delta_{ij}$ is Kronecker delta and $ C_{ij}$ are real-valued smooth functions and bounded in $ I$.

Prove: there exists $ \lambda_{0}\not = 0$ in $ I$ such that $ det(v_{ij}(\lambda_{0})) \not =0$.

The result is stronger than that - it's not just that there is some such $ \lambda_{0},$ but that there is a whole punctured neighborhood of zero on which $ V(\lambda)$ is invertible.

$ V(\lambda)=\lambda(I+\lambda C(\lambda)).$

Since $ C(\lambda)$ is a continuous function of $ \lambda,$ we have that $ \lambda C(\lambda)\to 0$ as $ \lambda\to 0$ in all reasonable senses, including that for small enough $ \lambda$ it has norm (whatever norm is convenient) less than $ 1.$

Then $ (I+\lambda C(\lambda))^{-1}=\sum_{k=0}^{\infty}(-1)^{k}(\lambda C(\lambda))^{k}$

and the sum converges for all sufficiently small nonzero $ \lambda.$

Without an explicit inverse: $ f(\lambda)=\det (\lambda^{-1}v_{ij}(\lambda))$ is a continuous function of $ \lambda$, and $ f(0)=1$.


[quote="ProtestanT"][quote="Quevvy"]There are four red balls and two white balls in a jar. One ball is randomly removed and replaced with a ball of the opposite color. The jar is then shaken, and one ball is randomly selected. What is the probability that this ball is red? Express your answer as a common fraction.[/quote]
[hide]
2/3 of the time, there will be 5 red and 1 white
1/3 of the time, there will be 3 white and 3 red
(2(5)+1(3))/3(6)=13/18
$\frac{13}{18}$
[/hide][/quote]
Where'd you get the equation?

i added 2/3 times 5/6 (5 of 6 balls are red) and 1/3 times 3/6 (3 of 6 balls are red)

[quote="ProtestanT"][quote="Quevvy"]There are four red balls and two white balls in a jar. One ball is randomly removed and replaced with a ball of the opposite color. The jar is then shaken, and one ball is randomly selected. What is the probability that this ball is red? Express your answer as a common fraction.[/quote]
[hide]
2/3 of the time, there will be 5 red and 1 white
1/3 of the time, there will be 3 white and 3 red
(2(5)+1(3))/3(6)=13/18
$\frac{13}{18}$
[/hide][/quote]

I think you mixed the two probabilities up. It should be $\frac{2}{3}\times\frac{1}{2}+\frac{1}{3}\times\frac{5}{6}=\frac{11}{18}$

[hide] 2/3 of the time its 1/2, and 1/3 of the time its 5/6.
Multiply and add... $(\frac{2}{3}\cdot\frac{1}{2})+(\frac{1}{3}\cdot\frac{5}{6}) = \frac{2}{6}+\frac{5}{18}= \boxed{\frac{11}{18}}$

So yes, I guess I would agree with davidli.[/hide]


search
blogs
integration
logarithms
number theory
number theory theorems

http://mathworld.wolfram.com/TwinPrimeConjecture.html
or  http://en.wikipedia.org/wiki/Twin_prime_conjecture  :wink:

Hey dude, how are you :lol: I"M GETTING REALLY ANGRY NOW :wallbash_red:  anytime i want some information about something here they always tell me these three:  google, mathworld,wickipedia

 i always read those first...

Duuuuuuuuuuude, i'm looking for a nice for example 25 page  for example PDF file  that talks about all the things that has been done by mathematicians about this conjecture...well [b]MAYBE[/b] obviously it is hard to find such a thing, so we can discuss about it here and share our informaTions...
Thanks

:D I'm sorry because I understand mistake your think. But if you want prove a theorem, first, you need Search on internet :D

well, here we go for the first...look ... i found something:

The sum of the reciprocals of twin prime pairs converges:

$\sum_{p,q \ \ twin \ \ primes}\frac{1}{p}+\frac{1}{q}<\infty$

here's more(but not that much) about it:
http://myweb.lsbu.ac.uk/~whittyr/MathSci/TheoremOfTheDay/TotDBrun.pdf

also sth new (maybe it's just new for me :D ) about Prime's, named Ghandi's formula....you can see this one in my blog:

http://www.artofproblemsolving.com/Forum/weblog.php?w=672

hello again everyone :) ...i have a quesTion..have i done anything wrong?...because from the time that i've posted Ghandi's Formula in my blog...i haven't got any replies from anybody...i'm really sad :( ...it's really bad to be lonely... :(

[quote="sinajackson"]Hi to everyone...can some one introduce a link that talks about the recently done by mathematicians about twin prime conjecture...or maybe we can discuss a little bit here about it...[/quote]

There are very few mathematicians who work on the twin prime conjecture.
Some recent developments that might be relevant are the works by Green and Tao about arithmetic progressions and pseudorandom behavior in the primes.  They use hard work in analytic number theory by Goldston, Pintz and Yildrim which is also considered to be related to the question of twin primes.  These papers are available on the internet.

a new one :D :

if the number of pairs of twin primes smaller than positive integer $n$ is defined by $\pi_{2}(n)$ then we have:

$\pi_{2}(n)\approx2C_{2}Li_{2}(n)=2C_{2}\int^{n}_{2}\frac{dt}{\log^{2}t}$

where $C_{2}$ is called the [b][size=150]Twin Prime Constant[/size][/b] and:

$C_{2}=\prod_{p\geq3}\left(\frac{p(p-2)}{(p-1)^{2}}\right)$

$=0.6601618158468695739278121100145\cdots$


You are given a straight line of length $x$.
You have to do is draw a line of length square root of $x$ by geometric construction.

No special construction devices.

Here's a problem. With $x$ on any arbitrary scale, we could draw any other line segment and call it $\sqrt{x}$ because there's always a solution to $\sqrt{x} = ax$ where $a$ is the ratio between the segments. If we are given a unit segment, then we can use Farenhajt's construction. Otherwise, let's just draw a random segment :).

[quote="Hamster1800"][quote="José"][quote="Farenhajt"]Then $BD=\sqrt{x}.$[/quote]

why?[/quote]

There is a right triangle formed with a radius and the base of the semicircle with legs of length $BD$ and $\frac{x-1}{2}$ and a hypotenuse of length $\frac{x+1}{2}$. Using the pythagorean theorem we can get that $BD = \sqrt{x}$.[/quote]

Or we can rely on similarity: as $\triangle ABD\sim\triangle DBC$ (they have the same angles), then $AB: BD=BD: BC$, giving $BD=\sqrt{AB\cdot BC}=\sqrt{x}.$

Note: This construction is possibly the simplest way to construct the geometric mean of the two given segments.

This is a bad problem. 100 cm = 1 m, but 10 cm $\ne$ 1 m, which is what Hamster1800 was saying.

[quote="Hamster1800"][quote="xxxyyyy"][quote="Farenhajt"]Draw a line $l$ and pick three points $A,B,C$ on it, where $B$ is between $A$ and $C$. Let $AB=x$ and $BC=1$ (unit segment). Then draw a semicircle with $AC$ as the diameter, draw a perpendicular to $l$ in $B$ and let it intersect the semicircle in $D$. Then $BD=\sqrt{x}.$[/quote]

how do you draw $BC = 1$? We are only given a line segment of length $x$?[/quote]

First, if you are just given a segment and not it's length, just draw another segment and call it $\sqrt{x}$. The reasoning is explained in my other post.

Now, if we have an integer x, it's very easy. Draw a ray off of one end of the segment at an acute angle. Then, draw a circle passing through the intersection with a center on the line drawn. Next, draw circles with centers at the intersection of the previous circle and the line and passing through the previous center. Draw as many segments as even parts you want (for example, draw 3 circles to trisect a segment). Then, draw a line from the center of the last circle drawn to the end of the segment of length x. Construct parallel lines through the centers of the other circles and you have it split into equal parts (of length 1 if you split it into x parts).

If you have a rational x, copy it until you get an integer length and then follow the instructions above.

If you have an irrational x, I have no idea how to do it.[/quote]

you don't know the numerical value of x, you just have a segment with that length

[quote="Altheman"][/quote][quote="Hamster1800"][quote="xxxyyyy"][quote="Farenhajt"]Draw a line $l$ and pick three points $A,B,C$ on it, where $B$ is between $A$ and $C$. Let $AB=x$ and $BC=1$ (unit segment). Then draw a semicircle with $AC$ as the diameter, draw a perpendicular to $l$ in $B$ and let it intersect the semicircle in $D$. Then $BD=\sqrt{x}.$[/quote]

how do you draw $BC = 1$? We are only given a line segment of length $x$?[/quote]

First, if you are just given a segment and not it's length, just draw another segment and call it $\sqrt{x}$. The reasoning is explained in my other post.

Now, if we have an integer x, it's very easy. Draw a ray off of one end of the segment at an acute angle. Then, draw a circle passing through the intersection with a center on the line drawn. Next, draw circles with centers at the intersection of the previous circle and the line and passing through the previous center. Draw as many segments as even parts you want (for example, draw 3 circles to trisect a segment). Then, draw a line from the center of the last circle drawn to the end of the segment of length x. Construct parallel lines through the centers of the other circles and you have it split into equal parts (of length 1 if you split it into x parts).

If you have a rational x, copy it until you get an integer length and then follow the instructions above.

If you have an irrational x, I have no idea how to do it.[/quote][quote="Altheman"]

you don't know the numerical value of x, you just have a segment with that length[/quote]

Exactly... so now how do we do it?? Probably it cannot be done?

[quote="xxxyyyy"]Exactly... so now how do we do it?? Probably it cannot be done?[/quote]

Nothing can be done without at least a numerical information about the measurement unit i.e. the unit segment. But as far as I know, in construction problems which require such info, it's usually implicitly understood that it's given or somehow obtainable.

If we are given a segment without its length, read my other posts. I have continually said that if this is the case any other segment can be considered as having the length of $\sqrt{x}$. So, in that case, we'll just draw a segment to the side and we're done.


function
limit
trigonometry
topology
real analysis
real analysis unsolved

1.Show that the set of all local extrema of a real function $ f$ is a countable set.
2.Show that if $ g$ is a monotonic function then the set of all points where $ g$ is not continuous is a countable set
I hope I have posted it in the right subforoum... :maybe:

number 1 i dont know, i came across it when studying for my last exams and i couldnt figure it out (i think it might use vitali covering theorem?)

but the second one i can help on.

define a function $ J: R \to \{open intervals\}$ by

$ J(a)= (\lim_{x\to a^-}g(a), \lim_{x\to a^+}g(a))$

so $ J(a)=\emptyset$ when g is continuous at a

then we note that $ J(a) \cap J(b) = \emptyset$ when $ a \ne b$

so we take the set $ Q=\{q_1,q_2,....\}$ of rationals in R

and see that $ Q \cap J(a) \ne \emptyset$ by denseness for each a

and then $ \cup_{a\in R} J(a)$ is obviously numerable by a subset of Q,

which means its countable.

Nice solution there and thanks for help :) ...as for 1. I do not think you need this theorem since it is nowhere mentioned in the book I found it in...

A local extremum is a value $ x$ such that there exists $ \varepsilon > 0$ such that $ f(x) > f(y)$ for all $ y \in (x \minus{} \varepsilon, x \plus{} \varepsilon) \setminus \{x\}$, right?  Replacing all these $ \varepsilon$s with $ \frac{\varepsilon}{2}$s makes the intervals disjoint.  But one can't fit uncountably many disjoint open intervals inside $ \bf R$.

Not exactly...you haven't proved that you can make the intervals pairwise disjoint.If $ f$ is not continuous then it is possible for it to have non-countable number of local extrema.How do you know that if you take $ \frac {\epsilon}{2}$ instead of $ \epsilon$ the intervals are pairwise disjoint..I mean how do you know that it is impossible to always find a local extremum near a given one ie if $ x$ is a local extremum then that there is an open interval containing $ x$ such that $ f$ has no other local extremum there...?

Your second sentence argues that the statement you asked me to prove is false ... who should I believe?  :)

If you don't like my definition of local maximum, why don't you share your own?  (This was the reason I added "right?" at the end of first sentence of my previous post -- if this is not your definition, what is?)

(I should have written "maximum" instead of "extremum" everywhere in my previous post, of course.)

A local maximum is a point $ y_0\equal{}f(x_0) \in f(R)$ such that there exists a $ \delta >0$ such that for all $ x \in (x_0 \minus{} \delta , x_0\plus{} \delta)$ we have $ f(x) \le f(x_0)$...I do not understand how the 2 sentences argue each other :huh: In 1.you have a continuous function $ f$ and you want to show it has only a countable number of local extrema...In 2. you have a monotonic function $ f$ and you want to prove that there exist countably many points where $ f$ is not continuous...

[quote="giannis18"]A local maximum is a point $ y_0 \equal{} f(x_0) \in f(R)$ such that there exists a $ \delta > 0$ such that for all $ x \in (x_0 \minus{} \delta , x_0 \plus{} \delta)$ we have $ f(x) \le f(x_0)$[/quote] Then in this case the result is false: any constant function has uncountably many local maxima.

[quote="giannis18"]I do not understand how the 2 sentences argue each other :Huh?:[/quote] I meant that the following two comments of yours were in disagreement:
[quote="giannis18"]1.Show that the set of all local extrema of a real function $ f$ is a countable set. [/quote] and [quote="giannis18"]If $ f$ is not continuous then it is possible for it to have non-countable number of local extrema.[/quote] (Your original question didn't contain the word "continuous.")

Still, I don't think the $ \frac{\varepsilon}{2}$ argument works, since for each $ x$ you have $ \varepsilon_x$. You may not be able to find a single $ \varepsilon\equal{}\inf_x\{\varepsilon_x\}$, since the inf could be 0. Take something like $ f(x)\equal{}\sin(1/x)$ on $ (0,1]$.

Shouldn't you be able to tag each of the intervals with a rational, though?

[quote="JBL"][quote="giannis18"]A local maximum is a point $ y_0 \equal{} f(x_0) \in f(R)$ such that there exists a $ \delta > 0$ such that for all $ x \in (x_0 \minus{} \delta , x_0 \plus{} \delta)$ we have $ f(x) \le f(x_0)$[/quote] Then in this case the result is false: any constant function has uncountably many local maxima.
[/quote]
I am sorry if I don't put it the right way but the way I meant the local extremum a constant function $ f(x)\equal{}c$ has only one extremum,c...
[quote="JBL"]I meant that the following two comments of yours were in disagreement:
[quote="giannis18"]1.Show that the set of all local extrema of a real function $ f$ is a countable set. [/quote] and [quote="giannis18"]If $ f$ is not continuous then it is possible for it to have non-countable number of local extrema.[/quote] (Your original question didn't contain the word "continuous.")[/quote]
I am really really sorry but in 1. I forgot to write that $ f$ should be continuous. :blush: I cannot edit it now :( So actually 1. should be
1.Show that the set of all local extrema of a [b]continuous[/b] real function is a countable set.
Sorry again...

@HilbertThm: Short version: I meant for us to take each local $ \varepsilon$ and divide it by 2.
[hide="Longer version"]My argument is correct given my definition.  I restate it with elaboration: choose two local maxima $ x$ and $ y$, with associated values $ \varepsilon$ and $ \epsilon$.  Then at any value $ z$ in the punctured interval around $ x$ we have $ f(z) < f(x)$ and at any value $ z$ in the punctured interval around $ y$ we have $ f(z) < f(y)$, so it follows that $ |x \minus{} y| > \max\{\epsilon, \varepsilon\}$ and so the intervals centered at $ x$ and $ y$ with radii $ \frac {\varepsilon}{2}, \frac {\epsilon}{2}$, respectively are disjoint.  The real numbers does not contain an uncountable set of disjoint open intervals, so we have our result.  (The function $ \sin\frac {1}{x}$ has countably many local maxima under my definition, so it's not a counterexample.)  I don't think this proof requires continuity of $ f$.[/hide]

@giannis: Ok, I understand now.  I will think about this some more (and I don't think it should be very hard).

Ah. I see now. That does work.  :blush: My counterexample was not to a countable number of local max, it was to the existence of a universal epsilon for the epsilon neighborhoods.

[quote="giannis18"]A local maximum is a point $ y_0 = f(x_0) \in f(R)$ such that there exists a $ \delta > 0$ such that for all $ x \in (x_0 - \delta , x_0 + \delta)$ we have $ f(x) \le f(x_0)$[/quote]

The set $ M = \lbrace {x\in\mathbb{R}: \exists \delta > 0, \forall y\in(x - \delta,x + \delta), f(y)\leq f(x)\rbrace}$ need not be countable, as JBL pointed out.  However, note that the OP identifies $ f(x_0)$ as the 'local maximum', not $ x_0$, so in fact he is asking to prove that $ f(M)$ is countable, not $ M$ itself.  In particular, the constant function is not a counterexample to his assertion given the above definition.

$ f(M)$ is countable.  Proof:  Let $ B =  \lbrace \lbrace x\in\mathbb{R}: a < x < b\rbrace: a,b\in \mathbb{Q},a < b \rbrace$ (note $ B$ is a countable base for the usual topology on $ \mathbb{R}$), and for each $ m\in f(M)$ let $ x_m$ be a point satisfying $ f(x_m) = m$.  Now since each $ m$ is a local maximum, by definition there is a $ \delta > 0$ so that $ f(x)\leq m$ for all $ x\in (x_m - \delta,x_m + \delta)$.  Choose two numbers $ \delta_l,\delta_r < \delta$ so that $ x_m - \delta_l,x_m + \delta_r\in\mathbb{Q}$, and define $ S_m = (x_m - \delta_l,x_m + \delta_r)\in B$.  For any $ m_1\neq m_2$, we have $ S_{m_1}\neq S_{m_2}$ (since $ x_{\max\lbrace m_1,m_2\rbrace}\notin S_{\min \lbrace m_1,m_2\rbrace}$), so we have effectively constructed an injective map from $ f(M)$ to $ B$.  Since $ B$ is countable, $ f(M)$ is therefore also countable.

[hide="remark and generalization"]
- It appears we don't need continuity.
- Note that the only property of $ \mathbb{R}$ we really made use of is that it has a countable base (i.e. it is [i]second-countable[/i]).  We could've just used open neighborhoods instead of intervals, and countable subcovers from base $ B$ instead of explicitly constructing intervals with rational endpoints. Therefore for any second-countable topological space $ (X,\mathcal{T})$, and for any function $ f: X\to\mathbb{R}$, the set $ f(\lbrace x\in X: \exists U\in \mathcal{T}, \forall y\in U, f(y)\leq f(x)\rbrace)$ is countable[/hide]

[quote="JoeBlow"]
For any $ m_1\neq m_2$, we have $ S_{m_1}\neq S_{m_2}$ (since $ x_{\max\lbrace m_1,m_2\rbrace}\notin S_{\min \lbrace m_1,m_2\rbrace}$)
[/quote]
How do you get that? :huh:

Without loss of generality assume $ m_1 > m_2$, then since $ S_{m_2}\subset \lbrace x\in \mathbb{R}: f(x)\leq m_2 \rbrace$ and $ f(x_{m_1}) \equal{} m_1 > m_2$, it follows that $ x_{m_1}\notin S_{m_2}$.  But by definition $ x_{m_1}\in S_{m_1}$, so $ S_{m_1}\neq S_{m_2}$.


[i]ABC[/i] is an acute-angled triangle inscribed in a circle and [i]P, Q, R[/i] are the middle points of the minor arcs [i]BC, CA, AB[/i] respectively. Prove that [i]AP[/i] is perpendicular to [i]QR[/i].

Let $ AP$ intersect $ RQ$ at $ S$.
Consider the quadrilateral $ SPBR$.
$ \angle S \equal{} 360 \minus{} \angle APB \minus{} \angle RBP \minus{} \angle \minus{} QRB \equal{} 360 \minus{} 270 \equal{} 90$

[quote="jimi"]Dimitris

How do you know the other 3 angles add up to 270 degrees?[/quote]
Look at the arcs their ''mathcing''.
$ \angle APB\equal{}\angle C$.
$ \angle RBP\equal{}\angle B\plus{}\angle\frac{C}{2}\plus{}\angle \frac{A}{2}$
$ \angle QRB\equal{}\angle A\plus{}\angle \frac{B}{2}$
So their sum is $ (\angle A\plus{}\angle B\plus{}\angle C)\plus{}(\angle\frac{A}{2}\plus{}\angle \frac{B}{2}\plus{}\angle\frac{C}{2})\equal{}180\plus{}90\equal{}270$.

Dimitris

It's very easy!

This is another solution:

We have $ \angle{AMQ} \equal{} \angle{ANR}$ $ \Longrightarrow AMN$ is an isosceles triangle.

But $ \angle{BAP} \equal{} \angle{CAP}$ $ \Longrightarrow AH \perp PQ$.

Dear Mathlinkers,
1. let I be the center of ABC
2. let (1) be the circle with centre at P and passes through B; it goes also through C (well known)
3. Applied now the Reim's theorem to (1) and the circumcircle of ABC by considering the tangent to (1) at I; this thangent is parallel to QR...
the rest follows.
Sincerely
Jean-Louis


function
logarithms
limit
ceiling function
real analysis
real analysis unsolved

Find a function $ f$ such that, as $ x \rightarrow \infty$, $ (\log x)^n \equal{} O(f(x))$ for any $ n \geq 1$ and $ f(x) \equal{} O(x^c)$ for any $ c > 0$.

Define $ g(x)=\sum_{k=1}^{\infty}\frac{x^k}{(k!)^2}.$

It is clear that $ x^n=o(g(x))$ as $ x\to\infty$ for all $ n.$

Less obviously, ${ g(x)=o(e^{x^c}})$ for any $ c>0.$

Consider $ \lim_{x\to\infty}\frac{g(x)}{e^{x^c}}.$

Make the substitution $ u=x^c,$ so that $ x=u^{\frac1c}.$ Let $ a=\lceil\frac1c\rceil$ so that we only make the fraction bigger by looking at

$ \lim_{u\to\infty}\frac{g(u^a)}{e^u}=\lim_{u\to\infty} \frac{\sum_{k=1}^{\infty}\frac{u^{ak}}{(k!)^2}} {\sum_{m=0}^{\infty}\frac{u^m}{m!}}.$

$ \le\lim_{u\to\infty}\frac{\sum_{k=1}^ {\infty} \frac{u^{ak}}{(k!)^2}}{\sum_{k=1}^{\infty}\frac{u^{ak}} {(ak)!}}$

Noting that $ \lim_{k\to\infty}\frac{(ak)!}{(k!)^2}=0,$ we see that Cesaro-Stolz shows that the above limit is zero.

Now that we've built this function, we see that $ f(x)=g(\log x)$ satisfies the conditions asked for in this problem.

One addendum: If $ f(x) \equal{} g(\log x),$ I have shown that $ f(x) \equal{} o\left(e^{(\log x)^c}\right)$ for any $ c > 0.$ What I actually want is $ f(x) \equal{} o(x^c) \equal{} o(e^{c\log x}).$ But since we are asking this for all $ c > 0$ we may WLOG assume that $ c < 1.$ In that case $ e^{(\log x)^c} \equal{} o(e^{c\log x})$ and the result holds.

And I know that that ksun asked for big-O. Since little-o implies big-O, what I have proved is apparently stronger - but because of the $ \forall$ quantifiers in ksun's statement, it's not really stronger.

We can also choose $ f(x) \equal{} e^{\sqrt{\log x}}$.

D'oh! In my original work, I got sufficiently tangled up over the question of whether we should have $ f(x)\equal{}g(\log x)$ versus $ f(x)\equal{}\log(g(x))$ that I somehow convinced myself that Ravi's example wouldn't work. Of course, it does work.

What I did was set myself more stringent conditions for $ g(x)$ than I needed to, and proved something a little harder than the original problem.


Isn't that impossible? As Poo Raegus must be defined, but you may not use the letter r in the post.

[quote="Nachosaurus"]Hi, this is a game that is always changing. Thats right, and the moderators change every week.

Rules: Play by the moderator's rules. You can take over as moderator once you make a post taht follows the previous moderator's rules.
Rule one cannot be changed

Rule 1: Anybody can take over as moderator, anytime, anywhere, and as many times as they want
Rule 2: The next moderator must explain what POO RAEGUS means
Rule 3: The next moderator must use under twenty.
Rule 4: The moderator must not use the letter r in the next post.
Rule 5: The moderator must praise the previous moderator.

Once you make a post that follows each of these rules, then you become the next moderator and change one of my rules.[/quote]

POO 2AEGUS is you2 favo2ite wo2d acco2ding to you2 blog.

Nachosau2us is a g3eat mode2ato2.

I change you2 4th 2ule to "The next mode2ato2 must not use any lette2s in the next post", only leet.

I'm not even going to try to solve that.

Translation:

[quote]Current Rules:

Rule 1: Anybody can take over as moderator, anytime, anywhere, and as many times as they want.
Rule 2: The next moderator must explain what POO RAEGUS means.
Rule 3: The next moderator must use under twenty.
Rule 4: The next moderator must not use any letters in the next post.
Rule 5: The moderator must praise the previous moderator.

BOGTRO is very good at making evil rules. I change rule 4 to be "The next moderator must use only capital letters in their post." Also, Poo Raegus is Nachosaurus's favorite word.[/quote]

POO RAEGUS IS NACHOSAURAUS'S FAVORITE WORD.

I CHANGE RULE 4 TO:  IN EACH POST, YOU MUST TYPE:  WAFFLES ARE VERY DELICIOUS.
1=2 IS VERY GOOD AT PICKING HIS NOSE.  :lol:

HAHA piegraph is funny

POO RAEGUS IS NACHOSAURAUS'S FAVORITE WORD.(i think)

DO YOU LIKE WAFFLES? YES I LIKE WAFFLES AND WAFFLES ARE VERY DELICIOUS.

i change rule 4 to in each line of your post you must quote a disney movie

inoit is cool for liking waffles.  POO RAEGUS is nachosauraus's favorite word.     A Bug's Life

i change rule 3 to "you must make at least 50 :ninja: emoticons it ur post

muahahahhahahahaha

That rule is impossible to follow because of the 10 emote limit.

woops harharhar
ok, i'll edit that : 3 :ninja: , 3   :bruce:  , 4 :starwars:

Say what?

Poo Raegus is nacho's favorite word  :ninja:  :ninja:  :ninja:
Piegraph is the BEST mod ever.  Maybe. :bruce:  :bruce:  :bruce:  :starwars:  :starwars:  :starwars:  :starwars:
under 20.

I change rule 2 to "the next mod must explain what 'bai ren kan bu dong' means"
Too many smilies in your post: 10.
You can have at most 4 smilies in this message. Please remove some smilies before submitting this message.
Too many smilies in your post: 10.
You can have at most 6 smilies in this message. Please remove some smilies before submitting this message.
Too many smilies in your post: 10.
You can have at most 8 smilies in this message. Please remove some smilies before submitting this message.

Poo raegus is my favorite word

abcak rocks

 :oops:  :roll:

Lol it means white people dont understand this!

I change the compliment rule to insult the last mod

POO RAEGUS is Nachosauraus's favorite word.

Nachosauraus smells like poop, feels like poop, eats poop, trys to listen to poop, and maybe is poop.

i change rule 3 to you must do a jumping jack whiile reading this

Poo raegus is my favorite word

pie graphs's looks like his icon just fifty times uglier plus getting all of michael jackson's surgeries twice

(tired just did jumping jacks)

i change jumping jacks to pushups

moderators lock t his old thread
and don't revive jprosk


limit
logarithms
calculus
calculus computations

How do I use L'Hopital's Rule if it's infinity minus infinity? I know you have to convert it into a fraction somehow, but in this problem, even after I converted it into a fraction, the numerator is still infinity minus infinity. How should I calculate this:
$ \lim_{x\to\infty}(\sqrt{x^2\plus{}x}\minus{}x)$

Thanks!

L'Hôpital's rule is [b]only[/b] for indeterminate quotients. Period. It has to be a $ \frac00$ or $ \frac{\infty}{\infty}$ case.

Other kinds of limits can be manipulated from one indeterminate form to a different such form. For instance, with $ 1^{\infty},0^0,$ or $ \infty^0,$ the first step should be to take the logarithm - then look to see what you have.

In this case, with square roots, we have the algebraic technique of rationalizing the numerator. Try multiplying by $ \frac{\sqrt{x^2\plus{}x}\plus{}x}{\sqrt{x^2\plus{}x}\plus{}x}$ and then tell me what you see.

Even then, I won't tell you to use L'Hôpital. It won't be necessary. The goal is to find the limit, not to find out how to use a particular technique.

As for your little hint kent, I see nothing aside from
$ 1\minus{}\frac{\sqrt{x\plus{}1}}{\sqrt{x\plus{}1}\minus{}\sqrt{x}}$
Simplification is hopeless: multiplying by the conjugate will yield the original expression. Nonetheless,
I'm pretty sure it converges to one-half; I'm just having trouble proving it.

Actually, multiplying by the conjugate transforms it into an $ \frac{\infty}{\infty}$ form, which can then be simplified to a form that's not indeterminate.

That one I isolated seemed to throw me off. I've got it now (used L'Hopital's Rule twice). Thank you jmerry sir.

I would really rather you not use L'Hôpital at all on this one. I see it used by students as a "black box" method used to avoid trying to understand what is going on. I would much rather see understanding, and in particular, understanding about size.

The issue of size comes up right away. How big is $ \sqrt {x^2 + x}$ for large $ x?$ Since $ x$ is much smaller than $ x^2,$ that looks "just like" $ \sqrt {x^2} = x.$

But then, we're subtracting $ x$ from it. Subtracting two things that are essentially the same size is asking for trouble; fine details that were unimportant to our earlier estimate now become important.

And you have to do algebra correctly!

${ (\sqrt {x^2 + x} - x})\cdot\frac {\sqrt {x^2 + x} + x}{\sqrt {x^2 + x} + x} = \frac {(x^2 + x) - x^2}{\sqrt {x^2 + x} + x} = \frac {x}{\sqrt {x^2 + x} + x}.$

After we do that, the numerator is $ x.$ The denominator is $ x$ [b]plus[/b] something that looks just like $ x$ as $ x$ gets large. This might as well be $ \frac {x}{2x} = \frac12.$

For a more formal argument to ratify the intuition of the previous paragraph, divide numerator and denominator by the "largest thing around", which in this case is $ x.$

$ \frac {x}{\sqrt {x^2 + x} + x}\cdot\frac {\frac1x}{\frac1x} = \frac {1}{\frac {\sqrt {x^2 + 1}}{x} + 1} = \frac {1}{\sqrt {1 + \frac1x} + 1}.$

Let $ x\to\infty.$ We can see that the limit is $ \frac12.$

[jmerry was clearly trying to steer you to this exact argument; you weren't letting him do the steering.]

After doing the conjugates;

$ \lim_{x \rightarrow \infty} \frac{x}{\sqrt{x^{2} \plus{} x} \plus{} x}$

$ \equal{} \lim_{x \rightarrow \infty} \frac{x(1)}{x(\sqrt{1 \plus{} \frac{1}{x}} \plus{} 1)}$

$ \equal{} \lim_{x \rightarrow \infty} \frac{1}{\sqrt{1 \plus{} \frac{1}{x} } \plus{} 1}$

$ \equal{} \frac{1}{\sqrt{1 \plus{} 0} \plus{} 1}$

$ \equal{} \boxed{\frac{1}{2}}$


What real a,b sanisfy $x^{2}+ax+b^{2}=0$, $x^{2}+bx+a^{2}=0$, and these equations share at least one root.

So, for some $x$ have:
$x^{2}+ax+b^{2}=0$ and $x^{2}+bx+a^{2}=0$
Subtract:
$(a-b)x-(a^{2}-b^{2})=0$
$(a-b)(x-(a+b))=0$
So, $a=b$ or $x=a+b$. But if $x=a+b$ plugging back in, get:
$2a^{2}+3ab+2b^{2}=0$
The Discriminant of this when solving for $a$ is negative so only solution is $a=b=0$.
So the sol-n is $a=b$.


integration trigonometry limit function inequalities Putnam logarithms

near 6 universitys contributed...nice problems :) specially the first one that i solved completely :D

1.	Evaluate the following:
a)	$\int^{\frac{\pi}{2}}_{0}\frac{dx}{1+\left(\tan x\right)^{\sqrt{2}}}$
b)	$\lim_{t\rightarrow0^{+}}\left\{\int^{1}_{0}\left(bx+a(1-x)\right)^{t}dx\right\}^{\frac{1}{t}}$

2.	Let $f(x)$ be a real function continuous in interval $\left[a,b\right]$ and differentiable in interval $(a,b)$ with $f(a)=f(b)=0$. Prove that there exists $c\in\left(a,b\right)$ such that:
                 $f(c)=f '(c)$

3.	Let  $0<x_{1}<1$ and define $x_{n+1}=x_{n}(1-x_{n})$ for all positive integer $n$. Prove that:
               $\lim_{n\rightarrow\infty}nx_{n}=1$

4.	Prove that all the roots of the equation $nZ^{n}=1+Z+Z^{2}+\cdots+Z^{n-1}$ are in the set $D=\left\{Z | Z\in\mathbb{C}, \left|Z\right|\leq1\right\}$.

5.	For which real numbers $c$, the following inequality holds?
$\frac{e^{x}+e^{-x}}{2}\leq e^{cx^{2}}$

[quote="Carcul"]1. a) $\int^{\pi/2}_{0}\frac{1}{1+(\tan x)^{a}}\,dx = \frac{\pi}{4}, \forall a \in R$.[/quote]
Please provide a solution too!

Thats right Carcul, i got the same answer with a simple solution, i also solved part $b)$ i think is more beautiful :D

[quote="Color"]Please provide a solution too![/quote]

$I = \int^{\pi/2}_{0}\frac{1}{1+(\tan x)^{a}}\,dx = \int^{\pi/2}_{0}\frac{1}{1+\frac{\sin^{a}x}{\cos^{a}x}}\,dx = \int^{\pi/2}_{0}\frac{\cos^{a}x}{\cos^{a}x+\sin^{a}x}\,dx$.

Now change variables according to $x = \frac{\pi}{2}-t$ and you get

$I =-\int^{0}_{\pi/2}\frac{\sin^{a}t}{\sin^{a}t+\cos^{a}t}\,dt = \int^{\pi/2}_{0}\frac{\sin^{a}x}{\sin^{a}x+\cos^{a}x}\,dx$.

So,

$I+I = 2I = \int^{\pi/2}_{0}\frac{\cos^{a}x}{\cos^{a}x+\sin^{a}x}\,dx+\int^{\pi/2}_{0}\frac{\sin^{a}x}{\sin^{a}x+\cos^{a}x}\,dx$
$= \int^{\pi/2}_{0}\frac{\sin^{a}x+\cos^{a}x}{\sin^{a}x+\cos^{a}x}\,dx = \int^{\pi/2}_{0}dx = \frac{\pi}{2}$

and so $2I = \frac{\pi}{2}\Rightarrow  I = \frac{\pi}{4}$.

For problem number 3:

$L = \lim_{n\to\infty}x_{n}= \lim_{n\to\infty}x_{n+1}$.

Then, according to the recurrence relation we have

$L = L(1-L) \Rightarrow L = 0.$

Now:

$L' = \lim_{n\to\infty}nx_{n}= \lim_{n\to\infty}\frac{n}{\frac{1}{x_{n}}}= \lim_{n\to\infty}\frac{n+1-n}{\frac{1}{x_{n+1}}-\frac{1}{x_{n}}}$
$= \lim_{n\to\infty}\frac{1}{\frac{1}{x_{n}(1-x_{n})}-\frac{1}{x_{n}}}= \lim_{n\to\infty}\frac{1}{\frac{x_{n}}{x_{n}(1-x_{n})}}= \lim_{n\to\infty}(1-x_{n}) = 1.$

Yes [b]color[/b] thats right :)  [size=150]Elm O Sanaat University [/size]is a technical university in iran and this competition was held in its Math College...
this was my solution for problem $1$ part $b)$:

let $L=\lim_{t\rightarrow0^{+}}\left\{\int^{1}_{0}\left(bx+a(1-x)\right)^{t}dx\right\}^{\frac{1}{t}}$
by $u=bx+a(1-x)$ and $du=(b-a)dx$ we have:
$\int^{1}_{0}\left(bx+a(1-x)\right)^{t}dx=\frac{1}{b-a}\int^{b}_{a}u^{t}du=\frac{b^{t+1}-a^{t+1}}{(t+1)(b-a)}$
as a consequence:

$L=\lim_{t\rightarrow0^{+}}\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{\frac{1}{t}}\cdot\lim_{t\rightarrow0^{+}}\frac{1}{(t+1)^{\frac{1}{t}}}=\frac{1}{e}\lim_{t\rightarrow0^{+}}\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{\frac{1}{t}}$

$\lim_{t\rightarrow0^{+}}\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)^{\frac{1}{t}}=\lim_{t\rightarrow0^{+}}\exp\left(\frac{\ln\left(\frac{b^{t+1}-a^{t+1}}{b-a}\right)}{t}\right)$

$=\lim_{t\rightarrow0^{+}}\exp\left(\frac{\ln\left(\frac{be^{t\ln b}-ae^{t\ln a}}{b-a}\right)}{t}\right)$

$=\lim_{t\rightarrow0^{+}}\exp\left(\frac{\ln\left(\frac{b(1+t\ln b+o(t)-a(1+t\ln a+o(t)))}{b-a}\right)}{t}\right)$

$=\lim_{t\rightarrow0^{+}}\exp\left(\frac{\ln\left[1+\frac{t(b\ln b-a\ln a)}{b-a}\right]}{t}\right)$

$=\lim_{t\rightarrow0^{+}}\exp\left(\frac{t(b\ln b-a\ln a)+o(t)}{t(b-a)}\right)$

$=e^{\frac{b\ln b-a\ln a}{b-a}}$

so the answer is : $L=\frac{1}{e}e^{\frac{b\ln b-a\ln a}{b-a}}=e^{\frac{b(\ln b-1)-a(\ln a-1)}{b-a}}$ :)

Also you (in problem #3) you can show that : $\frac{1}{x_{N}}$~$N$
and in generally you can find asymptotic for $x_{n}$:
$x_{N}=..+....+....+.......$ it's need some calculation for wich now I have not time. but some idea you can find in this forum (I have posted some sequance and solve.) idea is analogysly.
in problem #5 I think answer is $c=\frac{1}{\ln^{2}(e+\sqrt{e^{2}-1})}$

[quote="color"][quote="kunny"]Are the problems posed in Putnam?[/quote]

What do you mean? That you recognize the problems
from some Putnam, or are you wondering if the problems
are from Putnam?
As an answer to the latter question, it seems to me, since it says so on
the subject header, that the problems come from some technical
university math competition in Iran.[/quote]

1 a)  Putnam   A-3  1980

   b)  Putnam   B-2  1979  The original problem contains the condition $0<a<b$.

3      Putnam   A-4  1966

5      Putnam   B-1  1980

How do you think about this, color?

Oh i forgot the condition $0<a<b$ for problem $1$ part $b)$, and Wow it seems almost all the problems are from Putnam :wink:, i thought its all their problems...

[quote="sinajackson"]

so the answer is : $L=\frac{1}{e}e^{\frac{b\ln b-a\ln a}{b-a}}=e^{\frac{b(\ln b-1)-a(\ln a-1)}{b-a}}$ :)[/quote]

nice.  :clap2:

In problem 5. you only must put instead $x$ this $x=\frac{1}{\sqrt{c}}$ and after is simpl.
and how to prove problem #3:
ease find that $x_{n}\to 0$ when $n\to+oo$
let us consider
$\frac{1}{x_{n}}-\frac{1}{x_{n-1}}=..thinking..=\frac{1}{1-x_{n}}=1+x_{n-1}+x^{2}_{n-1}+...\to 1$ Good.
then $\sum^{N}(\frac{1}{x_{n}}-\frac{1}{x_{n-1}})$~$\sum^{N}1=N$
or
$\frac{1}{x_{N}}$~$N$ and your limit easy Done.
And whats about more interesting: how to find more good asymptotic for $x_{n}$?
also easy :
we know:
$\frac{1}{x_{n}}-\frac{1}{x_{n-1}}=1+x_{n-1}+x^{2}_{n-1}+...$
let put instead $x_{n}$ his first asymptot  I mean this :
$x_{n-1}$~$\frac{1}{n}$ then we have
$\frac{1}{x_{n}}-\frac{1}{x_{n-1}}$~$1+\frac{1}{n}$
anter summaiting you will get
$\frac{1}{x_{N}}$~$N+\ln{N}+c$ where $c$ is some const.
and analogysly you can find
$\frac{1}{x_{N}}$~$N+\ln{N}+??+???+????...$
now we know :
$\frac{1}{x_{N}}$~$N+\ln{N}$

[quote="kunny"]

How do you think about this, color?[/quote]

Well, I asked if you recognize them and you obviously did,
good job, interesting that these are old Putnams too.

Well, this situation isn't limited to the problem, we frequently come across the same problem in National Mathematical Olympiad. :!:

2. Suppose $f: [a,b]\rightarrow\mathbb{R}$ is continuous and differentiable in $(a,b)$ with $f(a)=f(b)=0$.
Applying mean value theorem yields that $f'(c)(b-a)=0$ for some $c\in~(a,b)$.
Applying the intermediate value theorem; $0=f(a)\leq~f(x)\leq~f(b)=0$ yields that $f(x)=0$ for
all $x\in~(a,b)$. Hence $f(c)=f'(c)$ for the given $c\in\mathbb{R}$.

[b]abdul[/b] are you sure the function has to be constant on the interval?

Abdul, you are wrongly applying the IVP.
I think a more appropriate solution for 2 is considering the function $g(x)=e^{-x}f(x)$.
We have that $g(a)=g(b)=0$ so there exists $c$ such that $g'(c)=0=-e^{-c}f(c)+e^{-c}f'(c)=e^{-c}(f'(c)-f(c))$

What a blunder. Thanks for pointing it out.

Just by curiosity, does the theorem stated in problem 2 have some special name?

[quote="Extremal"]In problem 5. you only must put instead $x$ this $x=\frac{1}{\sqrt{c}}$ and after is simpl.[/quote]

I don't quite get this. How does this substitution help? However the answer is easy nevertheless:
Just Taylor-expand all the exponential functions and it is easy to see the $c \ge 1/2$ is necessary
(taking $x \to 0+$), and also it is sufficient (just compare all the coefficients... )