calculus
integration
number theory unsolved
number theory

Can you solve the problem
    Find  $ x,y \in \mathbb N*$ such that  $ x^2 \plus{} x \plus{} 6 \equal{} y^2$

 If you have interesting problem , You will post to we slove it
Thank you very much

If $ x>5$ we have
$ (x\plus{}1)^2 > y^2 > x^2$ and we can conclude there are no solution for this cases.
Then we just have to check the cases$ x\equal{}1, 2, 3, 4, 5$.
After that, wen can see the only one solution is $ x\equal{}5, y\equal{}6$.

[quote="thanhak5"]Can you solve the problem
    Find  $ x,y \in \mathbb N*$ such that  $ x^2 \plus{} x \plus{} 6 \equal{} y^2$

 If you have interesting problem , You will post at  Number Theory Unsolved Problems to we slove it
Thank you very much .I am sorry[/quote]

$ 4x^{2}\plus{}4x\plus{}24\equal{}4y^{2}$ ----> $ (2x\plus{}1)^{2}\plus{}23\equal{}4y^{2}$ -----> $ (2y)^{2}\minus{}(2x\plus{}1)^{2}\equal{}23$

$ (2y\plus{}2x\plus{}1)(2y\minus{}2x\minus{}1)\equal{}23$ ---> $ y\equal{}6,x\equal{}5$

Find all integral solution of the equation
 [b][u]Lesson 1 [/u]  [/b]    $ (x^2 \plus{} y^2 \plus{} 1)^2 \minus{} 5x^2 \minus{} 4y^2 \minus{} 5 \equal{} 0$
  [b][u]Lesson 2[/u][/b]       $ 9x^2\plus{}6x\equal{}y^2$

[quote="thanhak5"]Find all integral solution of the equation
 [b][u]Lesson 1 [/u]  [/b]    $ (x^2 \plus{} y^2 \plus{} 1)^2 \minus{} 5x^2 \minus{} 4y^2 \minus{} 5 \equal{} 0$
  [b][u]Lesson 2[/u][/b]       $ 9x^2 \plus{} 6x \equal{} y^2$[/quote]

$ 9x^2 \plus{} 6x \equal{} y^2$ <-> $ (3x\plus{}1)^{2}\minus{}1\equal{}y^{2}$ -----> $ (3x\plus{}y\plus{}1)(3x\minus{}y\plus{}1)\equal{}1$ ----> $ x\equal{}y\equal{}0$

Find all integral solution of the equation
  $ 4y^2\equal{}2\plus{} \sqrt{199\minus{}x^2\minus{}2x}$


Given an acute ABC(AC>BC) and  M is midpoint of AB,H is the orthocentre of ABC ,the feet of the perpendicular from H to CM is denoted L.Prove that ^MLA=^BAC

it is equivalent  $ AM^{2}=ML*MC$. Let O be a circumcenter of ABC and CM,CO meet circumcircle of ABC at P,Q. Then we can easily see AQBH is parallelogram. so Q,M,H are collinear and QM=MH. PQ//HL
since ^QPC=^HLP=90. so PM=ML, therefore $ AM^{2}=CM*MP=CM*ML$


logarithms
calculus
algebra unsolved
algebra

Let be given a real number $\alpha$.Prove that:
$\alpha>1\Longleftrightarrow \exists k>0$ such that $\sum_{i=2}^n\frac{1}{i (\ln i)^\alpha}<k \forall n$
When $\ln x=\log_e x$

i think thats more calculus ... the question is if that series converges or diverges ...

[quote="Rust"]$S=\sum_{i=2}^n\frac{1}{i (\ln i)^\alpha}<\frac{1}{2( \ln 2)^\alpha} +\frac{1}{\alpha -1}[(\ln 2)^{1-\alpha}-(\ln n)^{1-\alpha}], \ \alpha>1$.[/quote]

then the greater $n$ becomes the upper border bcomes less ... is there something wrong with ur ineq or what is the limit of the RHS for $n\to \infty$ ?  (i guess i diverges $\to -\infty$ and the LHS is always positv .... it seems to be wrong ...)

In this case, Cauchy's condensation test is very useful: if $(a_n)$ is a decreasing sequence of positive numbers, then the series $\sum {a_k}$ is convergent if and only if the series $\sum{2^ka_{2^k}}$ is convergent.
Thus, $\sum{\frac{1}{k(\ln{k})^{\alpha}}}$ is convergent iff  $\sum{2^k\frac{1}{2^k(\ln{2^k})^{\alpha}}}=\frac{1}{(ln2)^{\alpha}} \sum{\frac{1}{k^{\alpha}}}$ is convergent. Applying again Cauchy's test, $\sum{\frac{1}{k^{\alpha}}}$ is convergent iff $\sum{2^k\frac{1}{(2^k)^{\alpha}}=\sum{\frac{1}{(2^{\alpha-1})^k}}}$ is convergent. But, since $\alpha >1$, the latter is a convergent geometric series.


Are you kidding ? :huh: This problem is from Olympiad Algebra Tournament round 2 which is still running .


Let $ ABCD$ be a convex quadrilateral.Prove that $ ABCD$ is cyclic [size=150][color=red]if and only if[/color] [/size]
            $ AB\cdot BC\cdot AC \plus{} AD\cdot DC\cdot AC \equal{} AB\cdot BD\cdot AD \plus{} BD\cdot DC\cdot BC$.
I know one nice proof,but it is really hard to guess to this proof.So just have fun!
BTW:nobody completely solved this problem.

This is well-known as 2nd Ptolemy theorem .
[u]If the quadrilateral is cyclic [/u]
Set AB=a , BC=b , CD=c , DA=d , AC=x , BD=y  and R the circumradius

Then $ (ABC)\plus{}(ADC)\equal{}(ABD)\plus{}(BCD)$  so $ \frac{abx}{4R}\plus{}\frac{cdx}{4R}\equal{}\frac{ady}{4R}\plus{}\frac{bcy}{4R}$

or $ x(ab\plus{}cd)\equal{}y(ad\plus{}bc)$ and we are done .

For the converse I should think about it

[quote="silouan"]This is well-known as 2nd Ptolemy theorem .
[u]If the quadrilateral is cyclic [/u]
Set AB=a , BC=b , CD=c , DA=d , AC=x , BD=y  and R the circumradius

Then $ (ABC) \plus{} (ADC) \equal{} (ABD) \plus{} (BCD)$  so $ \frac {abx}{4R} \plus{} \frac {cdx}{4R} \equal{} \frac {ady}{4R} \plus{} \frac {bcy}{4R}$

or $ x(ab \plus{} cd) \equal{} y(ad \plus{} bc)$ and we are done .

For the converse I should think about it[/quote]
Yes,this part of the problem is obvious.Everyone made it at the olympiad.Converse is much harder.

[quote="darij grinberg"]See http://www.mathlinks.ro/viewtopic.php?t=25372 (or, more importantly, http://www.mathlinks.ro/Forum/viewtopic.php?t=14058 ). It is not a particularly new result.

PS. Good luck, Erken. ;)

  darij[/quote]
Unbeliveable,but your proof is the same as in the official solutions.It is really great solution.
P.S:
Thank you very much.Though my team result at Kolmogorov cup was not so good,i hope i can get a better result at our TST.

[quote="andyciup"]Take an inversion of pole $ A$ and you will see that the problem condition is in fact Stewart`s Theorem.[/quote]
I've already made the same thing during olympiad,but i failed.
Could you post a complete proof?
What do you think about it,dear Darij?

[quote="andyciup"]Take an inversion of pole $ A$ and you will see that the problem condition is in fact Stewart`s Theorem.[/quote]
Where is your solution?


[quote="GameBot"]A standard die is tossed twice. What is the probability of obtaining exactly one 5? Express your answer as a common fraction.[/quote]

Suppose we have Die A and Die B. We can break this into cases. Case 1: Die A gives us a 5 and Die B gives us some other number and Case 2: vice versa.

Case 1: We have $ \frac{1}{6} \cdot \frac{5}{6} \equal{} \frac{5}{36}$

Case 2: We have $ \frac{5}{6} \cdot \frac{1}{6} \equal{} \frac{5}{36}$

Adding these together gives us a result of $ \frac{10}{36} \implies \frac{5}{18}$.


How can we prove distributive, commutative and associative laws??

That depends on what system you're talking about (the integers? the reals?) and what axioms you start with.  In some formulations, the distributive, commutative, and associative laws are taken as axioms rather than as theorems to be proven.


quadratics
inequalities
inequalities unsolved

If $ \forall x,y \in R$, we have $ a(x\plus{}y)\leq 1\plus{}x^2\plus{}xy\plus{}y^2$

Find the range of value of $ a$

[quote="Allnames"][quote="tobeno_1"]If $ \forall x,y \in R$, we have $ a(x \plus{} y)\leq 1 \plus{} x^2 \plus{} xy \plus{} y^2$

Find the range of value of $ a$[/quote]
Are you sure about your problem,i think it is not true (sorry because my bad English,so i cant say more)[/quote]

Yes, the problem is right. And the solution is very very strange. But it did have a answer!

As a positive quadratic in x, it is equivalent to showing the discriminant is $ \le 0$

$ D\equal{}(y\minus{}a)^2\minus{}4(y^2\minus{}ay\plus{}1)\equal{}y^2\minus{}2ay\plus{}a^2\minus{}4y^2\plus{}4ay\minus{}4\equal{}\minus{}3y^2\plus{}2ay\plus{}a^2\minus{}4 \le 0$

Now, this is a negative quadratic in y, so it is equivelant to showing that the discriminant is $ \le 0$

$ D_1\equal{}4a^2\plus{}12(a^2\minus{}4)\equal{}16a^2\minus{}48\equal{}16(a^2\minus{}3) \le 0$

so $ \minus{}\sqrt{3} \le a \le \sqrt{3}$

Q.E.D.

Your answer is correct.

Here is my solution.

Solution 1:

Let $ x \equal{} \frac {X \plus{} Y}{\sqrt {2}},\ y \equal{} \frac {Y \minus{} X}{\sqrt {2}}$, we can rewrite the given inequality as

$ X^2 \plus{} 3Y^2 \minus{} 2\sqrt {2}aY \plus{} 2\geq 0$

$ \Longleftrightarrow X^2 \plus{} 3\left(Y \minus{} \frac {\sqrt {2}}{3}a\right)^2 \plus{} 2\left(1 \minus{} \frac {a^2}{3}\right) \geq0$ for all real numbers $ X,\ Y$.

Thus we have $ 1 \minus{} \frac {a^2}{3}\geq 0$, yielding $ \minus{} \sqrt {3}\leq a\leq \sqrt {3}$.

Solution 2:

Let $ x \plus{} y \equal{} s,\ xy \equal{} t$, the given condition can be said that for $ x,\ y\in{R}\Longleftrightarrow s^2\geq 4t$ and $ as\leq 1 \plus{} s^2 \minus{} t$

$ \Longleftrightarrow t\leq \frac {1}{4}s^2$ and $ t\leq s^2 \minus{} as \plus{} 1$.

Let $ P \equal{} \{(s,\ t)|t\leq \frac {1}{4}s^2\},\ Q \equal{} \{(s,\ t)|t\leq s^2 \minus{} as \plus{} 1\}$, we should determin the range of $ a$ such that

$ P\subseteq Q$, thus we have $ \frac {s^2}{4}\leq s^2 \minus{} as \plus{} 1\Longleftrightarrow 3s^2 \minus{} 4as \plus{} 4\geq 0$ for all $ s\in{R}$,

so the condition gives $ D/4 \equal{} ( \minus{} 2a)^2 \minus{} 3\cdot 4\leq 0$, yielding $ \minus{}\sqrt {3}\leq a\leq \sqrt {3}$.


Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy

\[f\left(x^2+y^2\right) = f(x)f(y) + f(xy)\]

Putting $x=y=0$, we get $f(0)=2f(0)$, so $f(0)=0$

Putting $y=0$, we get $f(x^2)=2f(0)=0$, so $f(x)=0$ for $x\geq 0$

Put $x=y=a$, $a<0$.  We get $f(2a^2)=f(a)^2+f(a^2)\implies f(a)^2=0\implies f(a)=0$.

This establishes that $f(x)\equiv 0$.


[quote="flava22"](t^2+1+1/t)^2

t/(sq.root(1+t^4))[/quote]

Go to [url=http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=differentiate&s3=basic]www.quickmath.com[/url] for these kind of questions, click on "Differentiate" on the left-hand plane; for example, type (t^2+1+1/t)^2 into the field labeled "Differentiate" and t into the field  labeled "with respect to" as shown here:

[url=http://ezimagecenter.com/out.php/i203_quickmathDiffExamplesimpedit.JPG][img]http://ezimagecenter.com/out.php/i203_quickmathDiffExamplesimpedit.JPG[/img][/url]

and either press ENTER or click the differentiate button to get the result page shown here:

[url=http://ezimagecenter.com/out.php/i202_quickmathDiffExamplesimpResult.JPG][img]http://ezimagecenter.com/out.php/i202_quickmathDiffExamplesimpResult.JPG[/img][/url]

you try the other one.


algebra
polynomial
quadratics
difference of squares
special factorizations

here's a factoring problem: please help me, don't just give me the answer

directly from the book...
[b]Factor as completely as possible with real coefficients $ x^{8}-y^{8}$.(Hint: There are 5 factors. Note that we say [i]real[/i] coefficients, not just integers)[/b]

[hide="this is what i got so far"]
$ (x^{4}+y^{4})(x^{2}+y^{2})(x+y)(x-y)$

where is that last fifth factor????
[/hide]

The simplest factoring tool is difference of squares, when stuck, try to create a difference of squares.

Here's one commonly overlooked identity,
$ x^{2}+y^{2}=(x+y-\sqrt{2xy})(x+y+\sqrt{2xy})$
The reason it's overlooked is because in this form it's pretty useless (variables under radicals simplify nothing), but if we're working with higher even powers of x and y than 2, it can come in handy.

umm... mihail??
there's a solutions book to AoPS vol. 1...
that's how i found the answer....  :)

[quote="mihail911"][quote="spell it out 11"]MIHIL, would you stop posting those problems here.

You see you already got 3 of them moved...so its wise to learn frmo your mistakes because I have a feeling this is going to get moved too.[/quote]

how am I supposed to know if its going to be moved? i don't know why the first problems were moved in the first place? these problems are from high school competitions...what else do you want??? :maybe:

anyways thanks guys for the help...[/quote]

They are too easy for the scope of this forum. Please browse other threads here and compare them to the difficulty of your problem.

[b]Fundamental Theorem of Algebra (restatement):[/b]  Every polynomial with real coefficients can be factored into linear or quadratic factors with real coefficients.

This should tell you that $ x^{4}+y^{4}$ is factorable.


[quote="frt"]A sequence $\{ a_n \}$ has a property that $\sum_{k=1}^{n} a_k = \left(a_n+\frac{1}{4}\right)^2$ for all natural numbers $n$. Find the explicit formula for $a_n$.[/quote]
$\sum_{k=1}^{n} a_k = \left(a_n+\frac{1}{4}\right)^2$ so the first logical step is plug in n-1:
$\sum_{k=1}^{n-1} a_k = \left(a_{n-1}+\frac{1}{4}\right)^2$. Now we subtract:
$\sum_{k=1}^{n} a_k -\sum_{k=1}^{n-1} a_k = \left(a_n+\frac{1}{4}\right)^2 - \left(a_{n-1}+\frac{1}{4}\right)^2$
$a_n=\left(a_n+\frac{1}{4}\right)^2 - \left(a_{n-1}+\frac{1}{4}\right)^2$ which indirectly shows $a_0=0$
$\left(a_{n-1}+\frac{1}{4}\right)^2=a_n^2-\frac{a_n}{2}+\frac{1}{16}=\left(a_{n}-\frac{1}{4}\right)^2$
Since we can show that it is an increasing series with positive terms (I think...) then we have $a_{n-1}+\frac14=a_n-\frac14 \implies a_n=a_{n-1}+\frac12 \implies \boxed{a_n=\frac{n}{2}}$
I hope I found the right answer and didn't mess the problem up too badly!

Almost... Check the very last part of your solution :)

[quote="me@home"]$a_{n-1}+\frac14=a_n-\frac14 \implies a_n=a_{n-1}+\frac12 \implies \boxed{a_n=\frac{n}{2}}$[/quote]

hmm i think we have to do like:

$a_1 = \Big( a_1 + \frac{1}{4}\Big)^2 \implies \Big(a_1 - \frac{1}{4}\Big)^2 = 0 \implies a_1 = \frac{1}{4}$

so now we have:

$a_n = a_{n-1} + \frac{1}{2} = a_1 + \frac{1}{2}(n-1) = \frac{2n-1}{4}$

uhh i think this makes sense but for some reason i dont think it checks out right when i plug it back into our equation

so you got like what I got for the answer, but I don't think either check... whatever.... it's an easy problem..

[quote="gotztahbeazn"]$a_n = a_{n-1} + \frac{1}{2} = a_1 + \frac{1}{2}(n-1) = \frac{2n-1}{4}$

uhh i think this makes sense but for some reason i dont think it checks out right when i plug it back into our equation[/quote]
That's actually correct :) (It should check out right...)


Let $ a_i,b_i$ positive numbers. Prove, that
\[ \prod_{i\equal{}1}^n (a_i^n\plus{}b_i^n)\ge (\prod_{i\equal{}1}^n a_i  \plus{}\prod_{i\equal{}1}^n b_i )^n.\]
If left expression equal to right expression, then exist c>0, suth that $ a_i\equal{}cb_i \forall i$.

[quote="Rust"]Let $ a_i,b_i$ positive numbers. Prove, that
\[ \prod_{i \equal{} 1}^n (a_i^n \plus{} b_i^n)\ge (\prod_{i \equal{} 1}^n a_i \plus{} \prod_{i \equal{} 1}^n b_i )^n.
\]
If left expression equal to right expression, then exist c>0, suth that $ a_i \equal{} cb_i \forall i$.[/quote]

It's holder Inq  :lol:


[quote="MathAndKnowledge"]"Write a program that finds the largest in a series of numbers entered by the user.  The program must prompt the user to enter numbers one by one.  When the user enters 0 or a negative number, the program must display the largest nonnegative number entered.  The numbers need not be integers."

Obviously, a loop is needed (I think 'while' or 'for' would be useful in this case).  I'm just not sure how to make the program determine what the largest of the entered numbers is, especially since there can be any amount of entered numbers.  Help me, please.[/quote]

Let's say you already had the list. How would you find the value of the largest member of a list?

[hide]
int largest=-1;
int lastentered=1;
while(lastentered>0){
    scanf("%i",&lastentered);
    if(lastentered>largest)
        largest=lastentered;
    }
printf("%i\n",largest);
[/hide]

not sure about C, but this can be done in a few lines on python

[hide]
max = 0
while 1:
    number = input("Enter a number:")
    if number > max:
        max = number
    if number <= 0:
        print max
        break

[/hide]

so the basic algorithm is

[hide]
initialize the greatest as 0
start loop so it keeps going
compare the newly entered number with the previous max number
if its bigger set the max to it instead
check and see if the number is less than equal to zero
if it is break
[/hide]


Is there any way to find a diameter of a graph by using DFS?

You can find the diameter of a tree with DFS. Start in any point, find the farthest vertex, let's say A. Then DFS again, and find the farthest from A, let's say B. Then AB is the diameter. It's not too hard to prove this.

I think it is not possible to do it for an arbitrary graph in time better then N^3: do Floyd-Warshall and find the longest distance.


calculus
integration
function
algebra
polynomial
real analysis
real analysis unsolved

Hi,

let $f$ and $g$ be to continuous realvalued functions on [0,1] such that

$\int_0^1 x^n f(x) dx = \int_0^1 x^n g(x)dx \; \forall n \in \mathbb{N}$

Show that $f = g$.

Could I use the Stone-Weierstrass approximation theorem here?

Why not? Of course, take $h=f-g$ and approximate this one with polynomials.

Yes,indeed, Harazi is right. The problem reduces to show that $h=0$ if $\int_{0}^{1} x^n\cdot h(x) dx=0$.
 As you said from Weierstrass theorem, there is a sequence of polynomials which is uniformely convergent to $h$ on the interval $[0,a]$. I forgot to say that the given relation is equivalent with $\int_{0}^{a} P(x)h(x)dx=0$.So, by passing to limit we have: $\int_{0}^{a} h^{2}(x)dx=0$ and from here the conclusion follows. ;)


geometry
3D geometry
sphere
algebra solved
algebra

Find the greatest number of spheres having r=1/2 that can be put
in rectangular box having the dimentions (10*10*1).
 :)  :)

It's not a solution, just an observation: it's obviously equivalent to finding the greatest number of circles of radius 1/2 which can be put in a square of dimensions (10x10).


inequalities
calculus
limit
inequalities unsolved

Prove the following inequality:
$ \sum_{i\equal{}n\plus{}1}^{2n}\frac{1}{i}\leq\frac{3n\plus{}1}{4(n\plus{}1)}$,for all $ n\in\mathbb{N}$.

Well, easy to prove using calculus:
first let's see that:
$ a_{n} \equal{} \sum_{i \equal{} n \plus{} 1}^{2n}\frac {1}{i}$ is strictly monotonic (just observe $ a_{n \plus{} 1} \minus{} a_{n}$)
Then prove that:
$ \lim_{n\to \infty}a_{n} \equal{} ln2 < 0,7$
Then observe that $ b_{n} \equal{} \frac {3n \plus{} 1}{4n \plus{} 4}$ is also strictly monotonic, and for n=9 observe that:
$ b_{9} \equal{} 0,7 > a_{9}$ and just prove the inequality for n=1,2,3,4,5,6,7,8 simply by making the rules :wink:


Hey, I have something surprise happen to my computer. I can't find my CD-ROM drive on my computer, and whatever what kind of CDs I put into my computer, it can't read it. Since this is Computer Science section, Do anyone of you know how to solve this problem? or I have to buy a new computer.

If you can't find your CD-ROM drive, how did you put CDs into it??? Anyway, make sure that you're not putting DVDs into a CD-ROM drive, that won't work. Also maybe you're putting DVD+-Rs or RWs into an older DVD drive which can't read that particular type of DVD, or maybe you have a really old CD-ROM drive that can't read burnt CDs or CD-RWs... there are many possibilities how an older drive won't work with newer CD/DVDs. The drive should have a picture on it that looks something like this http://en.wikipedia.org/wiki/Image:CDRWlogo.svg . It will tell you what kind of CDs or DVDs the drive can read.

Try putting in a plain CD that wasn't burnt, also make sure it isn't scratched and check if your computer can read it. If it can't, maybe your CD-ROM drive isn't working, isn't plugged in, isn't installed, is disabled, your operating system is messed up or something else I haven't thought of happened. Anyway that this would be a good first step. Let us know what happened.

The CD I have put in is the same CD as before, and my computer could read it at that time. But now it can't read it. So, I don't think it is CD's problem.
More details:
 First, when I put the CD to my computer, it has sound as indicate my computer begin to read it; but soon it stop(Before it would have sound, then automatically open the document or video). Then I press "my computer" and going to find CD-rom drive( I think it is what you called), but I can't find it.  Wait about 5 minute, I put out the CD and use different one in but computer still can't read it. After trying 6 times, I give it up. So, I go here for help.

Sounds like it's broken to me. Perhaps the laser.. I don't know. The same thing has actually happened on our home computer (which I don't use anyway), it can only read CDs rarely.. you'd need to get a professional to have a look.

If you go to "My Computer", you should see the CD drive even if there is no CD inside. If you don't see it it probably means that your computer can't find the drive, which means that it's either broken, misconfigured or unplugged. If you haven't been doing anything unusual with your computer (installing in new IDE devices (hard drives, CD-ROMs etc.), changing your BIOS configuration, Windows system settings or device settings) then the most probable answer is that it's broken. It could also be a virus but I haven't really heard about a virus that hides your CD-ROM drive so that too is not very likely.

Also sometimes it helps to clean the dust from your computer or CD drive but that would mean taking it apart.


Ok I drew a picture with Geometer's Sketchpad and converted it to PDF using the Virtual PDF Printer. But when I try to insert it into my document, no matter what it always takes up a page by itself. How do I get it so that it goes right beneath my text?

well here is a last ditch resort....have the words on the picture...

and maybe if you made the image smaller (post the pixels) then it might help

Why are you converting it to a pdf?  Just copy it to paint (or some similar program) and save as a png.

Because PDF's are generally easier to deal with than jpgs in LaTeX. For some reason, I can't get images to work, but PDFs work very easily.  :)


LaTeX function trigonometry inequalities geometry calculus integration

[hide="hint"]put it in exponential form[/hide]

[quote="Totally Zealous"]Find $^{4}\sqrt{i}$[/quote]
By the way, to express fourth roots in LaTeX, it is better to write \sqrt[4]{i}, which displays as $\sqrt[4]{i}$.

[hide="A third hint"]Use DeMoivre's Theorem[/hide]

LOL, since no one has solved it yet,

[hide]$\sqrt[4]{\imath}=\left(\cos\frac{\pi}2+\imath\sin\frac{\pi}2\right)^\frac14$

By DeMoivre's Theorem, we have

$\cos\frac{\pi}8+\imath\sin\frac{\pi}8$

In rectangular form, this is $\frac{\sqrt{\sqrt2+2}}2+\imath\frac{\sqrt{2-\sqrt2}}2$[/hide]

[hide="uh oh...."]
$\sqrt[4]{i}=(e^{i(\pi/2+2k\pi)\cdot1/4}=e^{i(\pi/8+k\pi/2)\cdot1/4}$
We get roots from $0\le k\le 3$

heh it's nice how my hint pretty much got ignored  :P [/hide]

Well, if you have something like $\sqrt9$, you can't say it equals $\pm3$.  It's just $3$.

$\sqrt9=3$
$-\sqrt9=-3$
$\pm\sqrt9=\pm3$

OK. Here we go: [hide="ALL FOUR SOLUTIONS!!!!!"] Everybody was right with the $cis(90)^\frac{1}{4}= cis(22.5)$. However, you also have more solutions every time you add 360 before dividing by four. So, really, our solutions are: $cis\left(\frac{90}{4}\right)$, $cis\left(\frac{450}{4}\right)$, $cis\left(\frac{810}{4}\right)$, and $cis\left(\frac{1170}{4}\right)$, which you could reduce and write out in rectangular notation (but I won't b/c I'm too lazy). [/hide]

[quote="bigboy82892"]$i*i*i*i=i$. So $\sqrt[4]{i}=i$.
Q.E.D.(Still no idea what this means. Someone should put this on AoPS Wiki.)[/quote]

1.  Incorrect.  $i \cdot i \cdot i \cdot i =-1 \cdot-1 = 1$.

2.  "Q.E.D" is an abbreviation for the Latin phrase "quod erat demonstrandum," which basically means "and that is what we wanted to prove."

[b]To the rest of you:[/b]

The root function, as a function, should return one value.  This is known as the "principal" value, and is found by dividing the (principal) argument of the input by the root.  There is one answer, not four.  lingomaniac's accessment is accurate.

[quote="t0rajir0u"]2.  "Q.E.D" is an abbreviation for the Latin phrase "quod erat demonstrandum," which basically means "and that is what we wanted to prove."[/quote]

Literally, it means "(that's) what should've been demonstrated".

Consider a circle centered at the origin, that is, consider parametric equations.

Convert to polar.
$i=[1,\frac{\pi}{2}]$
$\sqrt[4]{i}=[1,\frac{\pi}{2}]^\frac14$
Using what's his name's theorem,
$[1,\frac{\pi}{2}]^\frac14=[1,\frac{\pi}8+\frac{k\pi}2]$
So the 4 solutions in polar are $[1,\frac{\pi}8]$, $[1,\frac{5\pi}8]$, $[1,\frac{9\pi}8]$, $[1,\frac{13\pi}8]$
Convert back to rectangular.

[quote="t0rajir0u"][b]To the rest of you:[/b]

The root function, as a function, should return one value.  This is known as the "principal" value, and is found by dividing the (principal) argument of the input by the root.  There is one answer, not four.  lingomaniac's accessment[sic] is accurate.[/quote]
When dealing with integral roots of complex numbers, rather than arguing about which root is the "principle root", it is generally accepted to return all roots $(\sqrt[4]{cis\frac{\pi}{2}}, \sqrt[4]{cis\frac{3\pi}{2}}, \sqrt[4]{cis\frac{5\pi}{2}}, \sqrt[4]{cis\frac{7\pi}{2}})$, all of which equal $i$ $(\sqrt[4]{cis\frac{9\pi}{2}}= \sqrt[4]{cis\frac{\pi}{2}})$.

[quote="Ihatepie"]I'm probably totaly wrong because I haven't learned about i yet but I think the answer is i^2[/quote]

Not sure where you're getting this, considering $i^{2}=-1$ and $(-1)^{4}= 1$.

[quote="bigboy82892"]$i*i*i*i=i$. So $\sqrt[4]{i}=i$.
Q.E.D.(Still no idea what this means. Someone should put this on AoPS Wiki.)[/quote]

im pretty sure that isn't the right answer (that would be too easy)

anyways from Wikipedia, not AoPS
Q.E.D. is an abbreviation of the Latin phrase "quod erat demonstrandum" (literally, "which was to be demonstrated"). In simple terms, the use of this Latin phrase is to indicate that something has been definitively proven.

QED, may be written at the end of mathematical proofs to show that the result required for the proof to be complete has been obtained. It is not seen as frequently now as it once was, since formal geometry is less commonly taught as a separate subject.

[quote="moogra"][quote="bigboy82892"]$i*i*i*i=i$. So $\sqrt[4]{i}=i$.
Q.E.D.(Still no idea what this means. Someone should put this on AoPS Wiki.)[/quote]

im pretty sure that isn't the right answer (that would be too easy)[/quote]
We have already had a discussion in this post of why that solution is incorrect.  Please, let's not retread old ground.

you guys, only one solution, like t0rajir0u said. if we were solving $x^{4}=i$, then we would have four solutions. but we can only find one $x$ that satisfies $x=\sqrt[4]{i}$, since the [b]function[/b] $f(x)=\sqrt[4]{x}$ returns only one number.

just like when you have $\sqrt{4}$ it only simplies to $2$ (not $-2$), where as when you're solving the equation $x^{2}=4$ you first take the square root of both sides, giving $\sqrt{x^{2}}=\sqrt{4}$. then by definition $\sqrt{x^{2}}=|x|$, so this gives $|x|=2$. therefore we know that $x=2$ or $x=-2$.

[quote="drunner2007"][quote="roadnottaken"]However, when you are dealing with the square root of a complex number, you can't just take the absolute value.  $\mid \sqrt[4]{i}\mid = 1$  Note that $1^{4}\neq i$.[/quote]

you would have to worry about the absolute value only if we were solving $x^{4}=i$, but we are just dealing with $\sqrt[4]{i}$.[/quote]

suppose i want [cis 90]^{.5} so [cis 45]

but suppose i want [cis450]^{.5}, so [cis 225], but obviously this is not a function, because the two inputs are the same, yet there are two distinct outputs, this is why we always output all the values, ok so what if you restrict the angle, well, what if the angle is not real? an inequality restriction is not valid

the convension you are talking about is only for reals...(positive for an 2n th root)

[quote="Altheman"]what if the angle is not real?[/quote]

But arguments are always real.  (By which I mean if you have a number of the form $z = r \text{ cis }w$ where $w \not \in \mathbb{R}$ then $w \neq \text{arg }z$.)

Here is how I define the root function on complex numbers:

Given a complex number $z = |z| \text{cis }\theta$ where $\theta = \text{arg }z$ such that $0 \le \theta < 2 \pi$, then $z^{ \frac{1}{k}}= |z|^{ \frac{1}{k}}\text{cis }\frac{\theta}{k}$

This definition is consistent with corresponding definitions on the reals and additionally always produces a primitive root.

[quote="boxedexe"]$(\cos\theta+i\sin\theta)^{n}=\cos n\theta+i\sin n\theta$ is only valid for $n\in\mathbb{Z}^{+}$.  :wink:[/quote]It holds for all real numbers, and can even be extended to complex numbers with hyperbolic functions.

[quote="roadnottaken"]Counter-example:  $\sqrt[3]{-27}=-3$ when working with reals.  However, $\sqrt[3]{27cis\pi}=3cis\frac{\pi}3=\frac32+\frac{3\sqrt3}2i$[/quote]

Hmm.  Point taken.  I suppose "always produces a primitive root" and "is consistent with definitions on the reals" are mutually exclusive goals.

I suppose the sensible thing to say here is that the root function runs into problems as a function on the complex numbers, and that it's best to treat the problem as asking for all of the roots...?

[quote="Totally Zealous"]How does i translate to cosines and sines and trig??[/quote]

What do you mean? If you are asking what the cis function is:

$\text{cis }\theta=e^{\imath\theta}=\sin\theta+\imath\cos\theta$.

Is that what you wanted?

Or:

$\sin\theta=\cos\left(\frac{\pi}{2}-\theta\right)$

$\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right)$

the phase shift identities?

Could you please clear up what you mean?

[hide="The Solution"]
Okay. That equation $\sqrt[4]{\imath}$ is equal to: \[\left(\cos\frac{\pi}2+\imath\sin\frac{\pi}2\right)^\frac14\]

Then, use DeMoivre's Theorem, it becomes:
\[\text{cis}\frac{\pi}{8}\]

The solutions:

\[\begin{cases}\text{cis}\left(\frac{90}{4}\right)\\ \text{cis}\left(\frac{450}{4}\right)\\ \text{cis}\left(\frac{810}{4}\right)\\ \text{cis}\left(\frac{1170}{4}\right)\end{cases}\]

 [/hide]


calculus
integration
inequalities
function
derivative
real analysis
real analysis unsolved

(1)Suppose that f(0)=1 and f''(x)<0 for [0,1].
    Then $ \frac {3}{2}\int_{0}^{1}xf(x)dx < \int_{0}^{1}f(x)dx$.

(2)Suppose that f(a)=f(b)=0 and f(x) is differentiable and f'(x) is continuous.
    Prove that $ \int_{a}^{b}f(x)^{2}dx\leq \int_{a}^{b}f'(x)^{2}dx$

for point b:
$ \int_{a}^{b}2f(x)f'(x)\cot(x) dx \equal{} \int_{a}^{b} f^2(x)dx \plus{} \int_{a}^{b} [f(x)\cot(x)]^2dx$ from integration by parts which leads to
$ \int_{a}^{b} [f'^2(x) \minus{} f^2(x)]dx \equal{} \int_{a}^{b}[f'(x) \minus{} f(x)\cot(x)]^2dx$ which yields equality when $ f(x) \equal{} e^{\int_{a}^{b} \cot(x)}$
I think it s ok, I have nt done this sort of problems in a long time  :blush:
edit:
it occured later to me that there may be some problems in some points where $ \cot (x)$ is not defined oups :oops:

Are you sure that (2) is even correct?
If I examine a function that is a line segment from $ \minus{} M$ to $ \minus{} \delta$ connecting $ ( \minus{} M,0)$ with ($ \minus{} \delta$,$ \epsilon$), some differentiable fitting function, from $ \minus{} \delta$ to $ \delta$ such that the derivative is continuous and a line segment, again, from $ (\delta,\epsilon)$ to $ (M,0)$
also, we'll choose $ \epsilon \approx 1$
Now, if we calculate the two integrals and dismiss the part of the integral between $ \minus{} \delta$ to $ \delta$ (we can take an arbitrary small $ \delta$), we get that
$ \int_{ \minus{} M}^{M}f(x)^{2}dx \approx \frac {2}{3}M > \int_{ \minus{} M}^{M}f'(x)^{2}dx \approx \frac {2}{M}$ for a sufficiently big M.


Solve in natural numbers the equation
$ x^{3} \plus{} 6x^{2} \minus{} 109x \minus{} 223 \equal{} 11^{y}$.

Looking mod 11, we easily see that $ x\equiv 9\bmod{11}$ is the only solution, so let $ x\equal{}11k\plus{}9$. But now $ x^3 \plus{} 6 x^2 \minus{} 109 x \minus{} 223\equal{}11 (1 \plus{} 121 k (1 \plus{} k) (2 \plus{} k))$, so $ 11^2\not|x^3 \plus{} 6 x^2 \minus{} 109 x \minus{} 223$. We conclude that $ y\equal{}1$, and from there we see that $ k\equal{}0,\minus{}1,\minus{}2$, so $ x\equal{}9$ is the only solution.

Nice!
My solutions uses that the equation is equivalent to $ \left(x\minus{}9\right)\left(x\plus{}2\right)\left(x\plus{}13\right)\equal{}11^{y}\minus{}11$ and the rest is obvious. Actually mine solution has something common woth yours.


Here is a problem that I think is almost too easy..

Equilateral triangle $ABC$ has $AD = DB = FG = AE = EC = 4$ and $BF = GC = 2$. From $D$ and $G$ are drawn perpendiculars to $EF$ intersecting at $H$ and $I$, respectively. The three polygons $ECGI, FGI,$ and $BFHD$ are arranged to $EANL, MNK,$ and $AMJD$ so that the rectangle $HLKJ$ is formed. Find its area.

[hide="Okay.."]

This sounds really stupid but I assume that the areas of three polygons $ECGI = EANL, FGI = MNK, BFHD = AMJD$ so the area of $HLKJ$ is same as the area of the triangle $ABC$.

Since the each side is 8, the area is $16 \sqrt 3$. [/hide]

 :rotfl: Am I assuming something wrong? It seems too easy!  :|

:rotfl: I agree, too easy. I think they wanted to test if people spent time finding the dimensions of the rectangle and then find area (the long way :D)
[EDIT] Although problem is easy, this shows that you can always cut up any equilateral triangle to form a rectangle, both of which have the same area.

After I solved it, I thought I had messed up. Guess not.  :D

Yeah it is obvious.
Some people probably get intimidated though, and get to caught up in it.


this is a page were people can post different t.v show's, movies, music and others can post comments about them...

-soccerstar123

let's start with Full House what do people think about that tiv show

someone please move this to Round table or G&FF..

Many people say that Digimon is a clone of Pokemon, but Digimon is fundamentally different from Pokemon. That they both have the suffix "mon" and are about imaginary creatures is simply a coincidence.

PODASIPFTW

Brother Bear is the best movie in the world!

Eureka is THE best T.V. show in the world. You just have to watch out for bad language.

Marshall- "We ran into-"
Marshall's daughter- "Ourselves."

Can't possibly be better than Brother Bear.

Eureka has adult humor! Creative episodes! If you open to other people's point of view, you can get unhooked of BB and onto something else, like Eureka!


It has 3 sides of length $ 4\sqrt{2}$ Formula for area of equilateral triangle with side $ s$

$ \frac{s^{2}\sqrt{3}}{4}$

So substituting we get $ 8\sqrt{3}$


assume that p is prime number and $p=\overline{a_0a_1a_2...a_n}$.prove that $a_n+a_{n-1}x+a_{n-2}x^2+...+a_0x^n$ is irreducible

I think $a_0+a_1x+\cdots+a_nx^n$ is irreducible iff $a_0x^n+a_1x^{n-1}+\cdots+a_n$ is irreducible. Is it true?

this is in fact Balkan Mathematical Olympiad 1989, problem 2.

Yes, that's why I asked that. The problem posted by sam-n is the same as the one from BMO, except that the coefficients of the polynomiasl are reversed. At least that's how it looked to me when I first saw it :). Maybe he has edited it?

Did you discuss solution or source of the problem?

Just copied from Kalva :D

Solution by Demetres Christofides, who also corrected the question

If w is a (complex) root of the polynomial, then since each coefficient is at most 9, we have |pn| ? 9(1/|w| + 1/|w|2 + ... + 1/|w|n-1). So if |w| ? 9, then |pn| ? 9(1/9 + 1/81 + ... ) < 9/8. But pn ? 2, so we must have |w| < 9.

Suppose the polynomial has a factor with integer coefficients (and positive degree less than n). Then Gauss' lemma tells us that it must be a product of two polynomials with integer coefficients (and each with positive degree less than n). Suppose they are f(x) and g(x). Suppose the (complex) roots of f(x) are z1, ... , zm. Then f(x) = A(x - z1) ... (x - zm) with A and integer. Hence |f(10)| = |A| |10 - z1| ... |10 - zm|. But |A| ? 1 and each factor |10 - zi| > 10 - 9 = 1. So |f(10)| > 1. Similarly, |g(10)| > 1. But f(10) and g(10) are integers and their product is the prime p. So we have a contradiction. So the polynomial cannot have any such factor.


number theory
relatively prime
number theory proposed

$\frac{2^n-2}{n}=k$ and k is an integer. Is true not only for n primes but even for some composite integers, find at least one.

[color=red][ In the contest the problem was, finding the even number.][/color]

Davron

Thank you Jin, for your reply. Sorry but i am looking for an even ones. And it is not that easy ones.

Davron

[u][b]Note:[/b][/u] [i]A composite number n such that a to power of n-1 congruent to mod n for every a relatively prime to n is called a Charmichael number. So a Charmicael number n is a pseudoprime to every base relatively prime to n. The preceding solution shows that 561 and 341 are Carmichael numbers. For a ong time, it was not known whether there are infinitely many Carmichael numbers [/i][b]( Prove that there are infinitely many Carmichael numbers,an exercise for you dear brothers, mathlinkers.)[/b][i] The problem was finally settled in 1993, when W.R. Alfrod, Andrew Granville and Carl Pomerance proved a much more  that x to the power of 2/7 Charmichael numbers up to x. There are exactly 105212 Charmichael numbers up to 10 to the power of 15.[/i]

one more example for you : try 6601.

Davron


geometry
trigonometry
perimeter
calculus
geometry proposed

Prove that for any convex quadrilateral of sides $a,b,c,d$ and angles $A,B,C,D$ the area $S$ is given by the formula

$S=(\frac{1}{4})\frac{1}{(\sin{ \frac{A+B}{2}})(\sin{ \frac{B+C}{2}})(\sin{ \frac{C+A}{2}})}[(\prod sin{ \frac{A}{2}) P^2-(\prod\cos \frac{A}{2})Q^2]}$

where $P=a+b+c+d$, $Q=a+c-b-d$

Sorry, I have forgotten to say that I used it to prove that among all convex quadrilaterals with given perimeter and fixed angles, the circumscripted has the max area. :)

With this formula you can also prove ,without calculus, that among all circumscribed quadrilaterals with given perimeter , the square has the greatest area  :)  and many other interesting facts....

I discovered this formula many years ago but I have never seen it published on a book.

Can you, please, tell me if it is a well known formula and in which books it is possible to find it. ;)


function
floor function
algebra proposed
algebra

Find the set of possible values of the expression
\[ f(x,y,z)\equal{}\left\{\frac{xyz}{xy\plus{}yz\plus{}zx}\right\}
\]
for positive integers $ x,y,z$. Here $ \{x\}\equal{}x\minus{}\lfloor x\rfloor$ is the fractional part of $ x$.

[hide="Hint"]The answer is all rational numbers in (0,1)[/hide]


@KeithC: So a broken transformer can have that much of an impact...? I mean, come on, I don't think it produces [i]that[/i] much smoke. (And before you ask, yes, I get the joke.)

And I love these. Keep them up.

there aren't a lot yet- they all come from a site.


Μπορείται να με βοηθήσετε με αυτήν την άσκηση?

Να αποδειχθεί ότι η σερά
$ \sum_{n=0}^{\infty} \frac{z{}^3{}^k}{k}$
συγκλίνει σε όλα τα σημεία του κύκλου $ {/z/=1}$
εκτός από εκείνα που $ z^3=1$

Μπορώ να το δείξω με το κριτήριο του D'Alembert ότι συγκλίνει στο εσωτερικό του κύκλου?
και μετά γιατί απορρίπτω τις λύσεις τις εξίσωσης
$ z^3=1$ ?

[quote="spinos"]Μπορείται να με βοηθήσετε με αυτήν την άσκηση?

Να αποδειχθεί ότι η σερά
$ \sum_{n = 0}^{\infty} \frac {z{}^3{}^k}{k}$
συγκλίνει σε όλα τα σημεία του κύκλου $ {/z/ = 1}$
εκτός από εκείνα που $ z^3 = 1$

Μπορώ να το δείξω με το κριτήριο του D'Alembert ότι συγκλίνει στο εσωτερικό του κύκλου?
και μετά γιατί απορρίπτω τις λύσεις τις εξίσωσης
$ z^3 = 1$ ?[/quote]
Γιατί αν $ z^3=1$ έχεις την αρμονική σειρά που ως γνωστό δεν συγκλίνει

Aiiiieeeee... glykos thanatos einai ayti i seira!

Κατ'αρχάς, ας ξεχάσουμε το $ z^3$ και ας δουλέψουμε με τη σειρά $ \sum \frac{w^k}{k}$; αν δείξουμε ότι συγκλίνει για όλα τα μοναδίαια $ w$ με $ w\neq 1$, τελειώσαμε, ναι?

Τα καλά νέα είναι ότι αυτή η σειρά είναι ο μιγαδικός λογάριθμος που έχει κλάδο αποκοπής τους θετικούς πραγματικούς, οπότε τα κομμάτια του παζλ ταιριάζουν τέλεια.

Τα κακά νέα είναι ότι η σύγκλιση της δυναμοσειράς είναι άμεση εφαρμογή του κριτηρίου του [url=http://en.wikipedia.org/wiki/Abel%27s_test]Abel[/url]. Το θεώρημα αυτό (βασικά η μιγαδική γενίκευση του Dirichlet) είναι ένα πολύ δυνατό εργαλείο για να εντοπίζεις πόλους δυναμοσειρών στο σύνορο του δίσκου σύγκλισης και η απόδειξή του δε θα έλεγα ότι είναι άμεσα προφανής :(

Είναι fair game η χρήση του? Βασικά νιώθω ότι κλέβω έτσι, αφού είναι σαν να σου λέω με ποιον τρόπο έλυσε ο Abel το πρόβλημα πριν κάτι αιώνες... Θα δοκιμάσω ξανά να δω μήπως βρω έναν κομψό και στοιχειώδη τρόπο, γιατί έτσι νιώθω ότι χρησιμοποιώ πυρηνικά εναντίον μύγας... αν και κάτι μου λέει ότι θα καταλήξω να αναμασάω την ίδια απόδειξη.

Άντε, καληνύχτα,

Durandal 1707

καλησπέρα παιδιά!
Παρακαλώ πείτε μου αν κάνω λάθος στον συλλογισμό μου
για να καταλάβω αν τα έχω καταλάβει
Όταν το z είναι μιγαδικός, τότε παίρνουμε πως η περιοχή συγκλισης είναι ο μοναδιαίος κυκλος?
γιατί βγαίνει με το θεώρημα του d' Alembert ότι $ /\frac{a_n\plus{}1}{a_n}/\equal{}/ \frac{z^{3k\plus{}3}k}{z^{3k}(k\plus{}1)}/ \rightarrow /z/^3$
Aρα για να συγκλίνει πρέπει $ /z/^3<1$ το οποίο είναι μκρότερο το 1 όταν το $ /z/<1$
άρα προκύπτει ο μοναδιαίος κύκλος ,δηλαδή έχουμε ακτίνα συγκλισης 1??
τι γίνεται όμως με τα σημεία που βρίσκονται ακριβώς πάνω στον κύκλο?
χρησιμοποιήσα σωστά το κριτήριo του  d' Alembert?
 ή μήπως πρέπει να πάρω μόνο τους συντελεστές τους και όχι τις δυνάμεις του z
πως θα εφαρμόσουμε το θεώρημα του Abel?

σας ζάλισα με τις ερωτήσεις μου πρωί πρωί!

Spinos, nai, i aktina syglisis tis dynamoseiras einai $ R\equal{}1$, sosta efirmoses to d'Alembert.

Etsi: an $ |z|<1$ i dynamoseira syglinei, kai an $ |z|>1$ apoklinei.

Gia to theorima tou Abel, diavase to arthro tis wikipedia pou sou ekana link: esto oti exeis mia dynamoseira me aktina syglisis $ R\equal{}1$ pou oi syndelestes tis einai thetikoi pou teinoun fthinondas sto $ 0$. Tote, i dynamoseira syglinei pano se olo to monadiaio kyklo $ |z\equal{}1|$, ektos apo to provlimatiko simeio $ z_0\equal{}1$.

I askisi sou einai kateytheian efarmogi tou kritiriou aytou, pou i apodeiksi tou den einai kai i pio ikanopoiitiki gia na sou dosei na katalaveis ti symvainei. Ti na po, elpizo na se voithaei to arthraki, dystyxos den prolava na asxolitho simera perissotero me to provlima, ayrio pali...

Cheerio,

Durandal 1707


How to prove that the set of the irational numbers has $ \infty$ elements ?

For example, $ \sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, \ldots, n\sqrt{2}, \ldots$ is an infinite set, made of  irrational numbers.

[hide="Another of the many very simple proofs"]
Suppose there were finitely many irrational numbers, and let the largest of them be $ r$. This means that $ r\plus{}1$, being greater than $ r$, is rational, so $ r\plus{}1\equal{}\frac pq$ with $ p,q\in \mathbb N$ and $ \gcd(p,q)\equal{}1$. But then $ r\equal{}\frac pq\minus{}1\equal{}\frac{p\minus{}q}q$, so $ r$ is rational, contradiction.[/hide]


In a troop of 2008, 12 are on duty every night. Prove that it is impossible to draw up a schedule according to which every 2 are on duty together exactly once

[hide="Proof?"]Consider an arbitrary person. He has to be on duty with the other $ 2007$ soldiers once. However, every time he is on duty, he must be on duty with $ 11$ other soldiers. $ 2007$ is not divisible by $ 11$, so it is not possible that he can be on duty with all other $ 2007$ soldiers exactly once.[/hide]


geometry
circumcircle
geometric transformation
reflection
radical axis
power of a point
geometry proposed

Given triangle $ ABC$ with circumcircle $ (O)$.  Let $ A_1, A_2, A_3$ be the projections of $ A, B, C$ onto $ BC, CA, AB$, respectively, $ X, Y, Z$ be the midpoints of $ BC, CA, AB. XY\cap (O)\equal{}\{A_2, A_3\}$, similar for $ B_2, B_3, C_2, C_3$. Prove that [b]Nine-point circle[/b]'s center of triangle $ ABC$ is the radical center of three circles $ (A_1A_2A_3), (B_1B_2B_3)$ and $ (C_1C_2C_3)$

$ \triangle AA_2A_3$ is the reflection of $ \triangle A_1A_2A_3$ about $ YZ.$ Since $O$ is  the circumcenter of $ \triangle AA_2A_3,$ then the circumcenter $ O_a$ of $ \triangle A_1A_2A_3$ is the reflection of $ O$ about $ YZ.$ Likewise, circumcenters $ O_b$ and $ O_c$ of $ \triangle B_1B_2B_3$ and $ \triangle C_1C_2C_3$ are the reflections of $ O$ about $ ZX$ and $ XY,$ respectively. $ \triangle O_aO_bO_c$ has parallel sidelines to the orthic triangle of $ \triangle XYZ,$ since they are centrally similar  with similarity center $ O$ and coefficient $2.$ Thus, radical axis $ \ell_c$ of $ (O_a),(O_b)$ goes through $ Z,$ due to $ ZB_2 \cdot ZB_3 \equal{} ZA_2 \cdot ZA_3$ and it is perpendicular to the line $ O_aO_b$  antiparallel to $ XY$ WRT $ ZY,ZX$ $\Longrightarrow$ $ \ell_c$ is the isogonal of $ ZO$ WRT $ \angle XZY.$ Similarly, $ \ell_a,\ell_b$  go through the isogonal conjugate of $ O$ in the medial triangle $ \triangle XYZ,$ i.e. the nine-point center $ N$ of $ \triangle ABC $ $\Longrightarrow$ $ N$ is the radical center of the circles $ (O_a),(O_b),(O_c).$


Point A lies somewhere within or on the square which has opposite corners at points 1,1 and 3,3. Point B lies somwhere within or on the square with opposite corners at points 5,3 and 6,4. What is the greatest possible value of the slope of the line containing points A and B?

Billy

[hide]Lets see, for max slope, points must be closest horizontaly and farthest vertically, or (3,1) and (5,4)
4-1=3
5-3=2
>>3/2<<[/hide]


2ος Πανελλήνιος Διαγωνισμός για την[b] Ε΄ και Στ΄ Δημοτικού [/b]  	13-11-2006

Η διδασκαλία των φυσικών επιστημών στο δημοτικό σχολείο στηρίζεται τα τελευταία χρόνια στην αξιοποίηση πειραμάτων με απλά μέσα, καθώς και ιδιοκατασκευών των μαθητών. Κατ' αυτόν τον τρόπο, δίνεται η δυνατότητα σε κάθε δάσκαλο / δασκάλα, ανεξάρτητα από την υλικοτεχνική υποδομή που το σχολείο του / της διαθέτει, να εφαρμόσει πειραματική διδασκαλία σε ομάδες. Παράλληλα, κάθε μαθητής έχει τη δυνατότητα να επαναλάβει τα πειράματα στο σπίτι του εφαρμόζοντας αυτόνομα την ερευνητική μεθοδολογία.

Η διδακτική εμπειρία που έχουν αποκομίσει οι εκπαιδευτικοί της πρωτοβάθμιας εκπαίδευσης στην διδασκαλία των φυσικών επιστημών με απλά μέσα και υλικά, είναι πολύτιμη και πρέπει να αναδειχθεί, αλλά και να διαχυθεί σε ολόκληρη την εκπαιδευτική κοινότητα.

Για το σκοπό αυτό η Ένωση Ελλήνων Φυσικών, σε συνεργασία με τη συγγραφική ομάδα των σχολικών βιβλίων «Ερευνώ & Ανακαλύπτω» της Ε΄ και της Στ΄ Δημοτικού διοργανώνουν για δεύτερη συνεχή χρονιά Πανελλήνιο Διαγωνισμό ομαδικών εργασιών, κατασκευών και πειραμάτων Φυσικών Επιστημών για την Ε΄ και Στ΄ Δημοτικού με τίτλο:
«Ερευνώ και Ανακαλύπτω - Ανακαλύπτοντας τις Φυσικές Επιστήμες με Απλά Μέσα και Υλικά!».

ΕΝΩΣΗ ΦΥΣΙΚΩΝ

[b]Σςςςςςςςςςςςς....... η δική μας εταιρεία κοιμάται[/b] .......

γιατί? αφού γίνονται και στο δημοτικό!
νομίζεις ότι το σύστηα των φυσικών είναι καλύτερο? ένας διαγωνισμός και τελείωσε! αναγκαστικά επιλέγουν την ομάδα οπότε οι υπόλοιποι...
με αυτό τον τρόπο θα είχα (όπως και πολλοί άλλοι) διακριθεί μόνο μια φορά.

καλύτερα πολλοί και σε λίγες τάξεις παρά ένας και σε πολλές

ΥΓ: στην πρώτη γυμνασίου γίνεται?

Φίλε στάρ επειδή έχεις πολύ μεγάλο στόμα κι επειδή αυτπύς τους κομπλεξισμούς δεν τους ναέχομαι χώρια του ότι είμαι εκνευρισμένος από άλλα θα μιλήσω πολύ αγρια
Στην Ενωση ελλήνων φυσικών υπάρχει η 1000000000000 φορές ανοργανωσια διότι εγω έδωσα πανελληνιο διαγωνισμό φυσικής κι δεν έκανα το κόπο να τηλεφωνησουνκι ας αφήσουν τους δημοτικους διαγωνισμους αυτά είναι κουραφέξαλα κι μονο για την εμπειρία
αυτά

φίλε ambros

η Ελληνική Μαθηματική Εταιρεία (Ε.Μ.Ε.) έχει ιστορία  88 ετών. Ιδρύθηκε το 1918 ……….

και

η  Ενωση Ελλήνων Φυσικών ιδρύθηκε το 1970 .

opa-opa ..stamatiste na... :starwars:
ka8e <eteria>exi ta dika tis provlimata +ta dika tis <kala> me tis protovulies pu perni....
den mporun omos se kamia periptosi na sigkri8un gt ine anomia pragmata....alo ma8imatika+alo fisiki(eks orismu.....)
SIGURA INE POLI 8ETIKI I PROTOVULIA PU ANELAVE I FIGIKI ETERIA+AN SINExisti sosta 8a exi sigura apotelesmata,ala min xanume to dasos epidi idame ksafnika ema dentro mprosta mas......+i ma8imatiki eteria exi prosferi ergo (den analio se pio va8mo,apla to anafero san gegonos,,,,)ala kai i fisiki eteria to idio....mpori +i dio na min exun dosi tin idia varitita sto na trabiksun tin prosoxi ton ma8iton apo ola ta meri tis elados (kentrika i apomera)ala sigura .exun prospa8isi...
mia ksexname tulaxiston oti imaste se mia xora pu distixos i idiotiki protovulia ine ayti pu xriazete gia na ksexorisi  kapios+na <ksefigi >apo ena mesoepipedo....distixos i eytixos (ayto ine ka8ara uema politikis sinidiseos....)den iparxi i dinatotita mazikis paragogis <talenton>opos ginotan prin merika  xronia se orismenes xores simperilamvanomenu kai tis proin sovietikis enosis....distixos prepi na prosarmostume me tis iparxuses katastasis....giati den 8a ofelisi se tipota to na griniazume +na paraponiumaste...prin xronia pali etsi itan,tora etsi ine +meta apo kero mpori pali i idia katastasi na iparxi...ti simeni ayto???8a epanapauomaste +8a leme mono i ma8imatiki i i fisiki eteria den kani ma8imata proetimasias gia tus diagonismus???etsi ine i katastasi.......prepi na frontisi o ka8enas monos tu  na yaksi,na asxoli8i+na diavasi monos tu,+tote 8a diapistosi oti i 8elisi gia ma8isi iperveni opiesdipote ipokimenikes i antikimenikes diskolies......tipota den mpori na stamatisi enan ma8iti pu exi oreksi na ma8i |+na asxoli8i me to antikimeno...ute an ginonte ma8imata proetimasias,ute an etixe +den perase tin 2 fasi giati den egraye to 4o 8ema....TI-PO-TA!!!!! ta pedia ayta 8a pane mprosta giati to 8elan apo moni tus....to 8ema ine gia ola ta ipolipa pedia pu den exun tin idia 8elisi(prosoxi...tin idia 8elisi+oxi tis idies ikanotites....i 8elisi kani tin diafora....opios 8eli ,mpori!!!)na asxoli8un me aytes tis epistimes.....eki.... ontos ta ma8imata proetimasias 8a voi8usan +8a edinan tin o8isi pu xriazete enas ma8itis gia na asxoli8i me to antikimeno...Prosopiki simvuli...epidi ego prosopika den 8a afina pote tis elpides mu gia na perimeno an +pote 8a ginun ma8imata proetimasias......sas leo:parte to vivlio +ka8iste +diavaste moni....dikste oti den kremomaste apo ..........................kanenos.....+an ginun ma8imata eginan...........+afiste tin ma8imatiki+tin fisiki eteria  na kanun oti 8elun.....sto kato-kato kai an ginun +15 mauimata(giati tosos 8a ine o kiklos tus,an ginonte ka8e savato....)den 8a alaksun +pola.....den 8a anevi to ma8imatiko epipedo tis xoras an ginun ta ma8imata stin a8ina,stin 8esaloniki+se kanadio akomi polis.....
tulaxiston iparxun san diagonismi fisikis+ma8imatikon....
gt paradeigmatos xari stin biologia,,,o diagonismos ka8iero8ike prin 3-4 xronia.....eleosssssssssss.......eprepe na klonopiisun tin ntoli gia na gini diagonismos biologias stin elada????

simperasma....kali san protovulia afti tis enosis fisikon....ala apo moni tis den lei tipota...xriazete oreksi +meraki apo tus ma8ites.....

ΡΑΛΛΗΣ :D

εχεις δίκιο φίλε μου Ράλλη συγχωρήστε με παιδιά άλλα όπως προείπα είχα πολλά νευρα οπροσωπικά χτες κι με το που είδα αυτό μου ανέβηκε το αίμα στο κεφάλι κι είπα αυτά που είπα σε κόσμιο πάντωες ύφος
αν θέλετετο ξανασυζητάμε ύστερα με πιο ήπιο κεφάλι κι όχι υπο το κράτος μέθης  :wink:
αυτά

[quote="RAKAR"]etsi ine i katastasi.......prepi na frontisi o ka8enas monos tu  na yaksi,na asxoli8i+na diavasi monos tu,+tote 8a diapistosi oti i 8elisi gia ma8isi iperveni opiesdipote ipokimenikes i antikimenikes diskolies......tipota den mpori na stamatisi enan ma8iti pu exi oreksi na ma8i |+na asxoli8i me to antikimeno...ute an ginonte ma8imata proetimasias,ute an etixe +den perase tin 2 fasi giati den egraye to 4o 8ema....TI-PO-TA!!!!! ta pedia ayta 8a pane mprosta giati to 8elan apo moni tus....to 8ema ine gia ola ta ipolipa pedia pu den exun tin idia 8elisi(prosoxi...tin idia 8elisi+oxi tis idies ikanotites....i 8elisi kani tin diafora....opios 8eli ,mpori!!!)na asxoli8un me aytes tis epistimes.....eki.... ontos ta ma8imata proetimasias 8a voi8usan +8a edinan tin o8isi pu xriazete enas ma8itis gia na asxoli8i me to antikimeno...Prosopiki simvuli...epidi ego prosopika den 8a afina pote tis elpides mu gia na perimeno an +pote 8a ginun ma8imata proetimasias......sas leo:parte to vivlio +ka8iste +diavaste moni....dikste oti den kremomaste apo ..........................kanenos.....+an ginun ma8imata eginan...........+afiste tin ma8imatiki+tin fisiki eteria  na kanun oti 8elun.....sto kato-kato kai an ginun +15 mauimata(giati tosos 8a ine o kiklos tus,an ginonte ka8e savato....)den 8a alaksun +pola.....den 8a anevi to ma8imatiko epipedo tis xoras an ginun ta ma8imata stin a8ina,stin 8esaloniki+se kanadio akomi polis.....
tulaxiston iparxun san diagonismi fisikis+ma8imatikon....
gt paradeigmatos xari stin biologia,,,o diagonismos ka8iero8ike prin 3-4 xronia.....eleosssssssssss.......eprepe na klonopiisun tin ntoli gia na gini diagonismos biologias stin elada????

simperasma....kali san protovulia afti tis enosis fisikon....ala apo moni tis den lei tipota...xriazete oreksi +meraki apo tus ma8ites.....

ΡΑΛΛΗΣ :D[/quote]

  Αγαπητέ Ράλλη ,
 έχουμε όλοι μας ξοδέψει πολύ μελάνι (χρόνο) να λέμε μεταξύ μας σε τούτο το χώρο αυτά τα πράγματα.[b]Η θέση σου είναι απόλυτα σωστή.[/b]Φυσικά ο προβληματισμός δε σταματάει ποτέ και η διατύπωση και άλλων απόψεων είναι καλό στοιχείο.Αυτά που γράφουμε εδώ , τα διαβάζουν μαθητές ή φοιτητές ή και  άνθρωποι που έχουν πάνω από όλα  μεράκι και πολλοί έχουν και  ταλέντο.Το θέμα είναι ότι αφήνοντας τα πράγματα αποκλειστικά στην ευθύνη του καθενός δεν αλλάζει τίποτα για την Ελλάδα και τη μαθηματική παιδεία, όπως βέβαια δεν αλλάζει τίποτα, αν το κράτος δίνει τα πάντα και κανένας δεν τα αξιοποιεί(πχ αν κανένας δεν προσπαθεί).
  Η πατρίδα χρειάζεται να σκεφτεί σοβαρά.Μια χώρα πάει μπροστά όταν γίνονται συγχρόνως πολλά πράγματα:
 [b]  α) [/b]καλό σχολείο , καλά βιβλία , καλό αναλυτικό πρόγραμμα , καλές προοπτικές ,καλοί δάσκαλοι
[b]   β)[/b] καλή μαθηματική εταιρεία ,καλοί διαγωνισμοί , καλή οργάνωση , καλή προβολή
 [b]  γ)[/b] καλή διοίκηση , καλή οικονομία ,καλή επιχειρηματικότητα ,  καλό περιβάλλον , καλή τηλεόραση ,καλά πρότυπα κλπ
 [b]  δ) [/b]καλό πανεπιστήμιο , καλό ασφαλιστικό , καλό συνταξιοδοτικό , καλός συνδικαλισμός , καλοί πολιτικοί.

[b]   ε)[/b] καλοί μαθητές , καλοί φοιτητές ,
  στ) ΚΑΛΟΙ ΠΟΛΙΤΕΣ , ΚΑΛΟ ΚΡΑΤΟΣ

    Από την επιστολή σου κρατάω αυτό:
 Ο καθένας μας πρέπει να γίνει καλύτερος με δικό του κόπο και θυσία και ύστερα να απαιτεί από τους άλλους.Αυτό είναι μια καλή αρχή!
 Γιατί , [b]''ακόμα και για το πιο μεγάλο ταξίδι , χρειάζεται [/b][color=red]το πρώτο βήμα'' .[/color].Για αυτό έχουμε εμείς την ευθύνη!

Μπάμπης


There is a pond, 100 feet in diameter. Dead in the center of the pond, on a lily leaf, is a frog. If the average leap of a frog is two and a quarter feet and there are plenty of other lily pads to jump on, what is the minimum number of leaps it will take for the frog to jump out of the pond?

I'm suspicious about this problem (if i'm wrong correct me)

The shortest distance to any point is in a STRAIGHT LINE

The pond has a 50 foot radius.

Since it takes 2.25 feet per jump, we have a fraction: $\frac{50}{2\frac{1}{4}}$ which equals 22.222..... I think it would take [b]23 jumps minimum.[/b]


algebra
polynomial
number theory unsolved
number theory

Let Φn(x) be the nth cyclotomic polynomial.

a) Show that if p divides Φn(x) for some integer x, then p is of the form nk+1.
b) Use the first part to show that there are infinitely many primes of the form nk+1.

See the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=203526]here[/url].  (The first statement is not correct; either $ p$ is of the form $ nk\plus{}1$ or $ p | n$.)


calculus
integration
induction
function
quadratics
algebra unsolved
algebra

The sequence $ \{a_n\}$ satisfies $ a_0 \equal{} 0, a_{n \plus{} 1} \equal{} ka_n \plus{} \sqrt {(k^2 \minus{} 1)a_n^2 \plus{} 1}, n \equal{} 0, 1, 2, \ldots$, where $ k$ is a fixed positive integer.  Prove that all the terms of the sequence are integral and that $ 2k$ divides $ a_{2n}, n \equal{} 0, 1, 2, \ldots$.

It is easy to prove, by induction
$ a_{n\plus{}1}\equal{}2ka_n\minus{}a_{n\minus{}1}$ and $ a_{n\plus{}1}a_{n\minus{}1}\plus{}1\equal{}a_n^2$.
Then it is easy to verify the modulo of $ 2k$ is $ 0,1,0,\minus{}1,0,1,0,\minus{}1,...$.

This is standard...If you rearrange and square it, you get

$ a_{n\plus{}1}^2\minus{}2ka_na_{n\plus{}1}\plus{}k^2a_n^2\equal{}(k^2\minus{}1)a_n^2\plus{}1$

Then $ a_{n\plus{}1}^2\minus{}2ka_n(a_{n\plus{}1})\plus{}a_n^2\minus{}1\equal{}0$

Note that $ x^2\minus{}2ka_n x\plus{}a_n^2\minus{}1\equal{}0$ has roots $ a_{n\plus{}1}$, $ a_{n\minus{}1}$ the function is increasing so $ a_{n\plus{}1}>a_n>a_{n\minus{}1}$ so $ a_{n\plus{}1}\ne a_{n\minus{}1}$ so they are the distinct roots of that quadratic and so

$ x_1\plus{}x_2\equal{}2ka_n$ which gives that nice recursion. The rest of the problem is then trivial by induction.

So rearrange the condition a little bit and square it we get $a_{n+1}^2-2ka_na_{n+1}+a_n^2-1=0$
We express $ a_{n+1}^2 = 2ka_na_{n+1}-a_n^2+1$
So now let $ n \rightarrow n+1$, we have the following:
$$ a_{n+2}^2-2ka_{n+1}a_{n+2}+a_{n+1}^2-1=0$$
Plugging in the definition of $a_{n+1}^2$ we get
$$(a_{n+2}-a_n)(a_{n+2}+a_n-2ka_{n+1})=0$$
Because the sequence $\{a_n\}$ is increasing we have the following:
$a_{n+2}+a_{n}=2ka_{n+1}$
We can calculate $a_1=1$
So now by induction we see that $\{a_n\}$ is a subsequnce of $\mathbb{Z}$
Since $2k | a_n + a_{n+2}$ and since $2k|a_0$.By induction with a step of 2 we get that $2k|a_{2n}$, $\forall n \in \mathbb{N}_0$


Someone posted today the following Vietnam inequality. I really don't know who deleted the message, but I will ask you grobber to tell me where can I find the solution or to post again the solution. It's very hard for me to find it in hundreds and hundreds of posts.
   Thank you very much. I'm really interested in this problem since it is very nice and difficult. Please, someone post a solution.

that is why we must put the source, so that we can find the problems :)

I thought I remembered posting the same problem, asking for help, and someone solving it, but it's out of my hands now :D It's just as hard for me to find the problem as it is for you, and I can't remember the solution. It was satisfying for the moment, but I forgot it :D

The problem is in fact easy if you have the courage of computations. It becomes trivial if you put z=c, y=b+c and x=a+b+c and make all computations. I prefer my other solution:
    Write it as x^3y(y-z)+z^3x(x-y)>=y^3z(x-z). Since x>=y, it reduces to x^3(y-z)+z^3(x-y)>=y^2z(x-z), or (x^3-y^2z)(y-z)>=(x-y)(y+z)(y-z)z , which reduces to x^3-y^2z>=z(x-y)(y+z). This becomes x^3+z^2y>=xyz+xz^2, or (x-z)(x^2+xz-yz)>=0, true.

Here is  what  i  think  about  it...
From  BCS inequality  we  have
[(a^2b)/c + (b^2c)/a + (c^2a)/b]*[(a^2c)/b+(b^2a)/c+(c^2b)/a] >=
(a^2+b^2+c^2)^2.
So we  need  to  prove  that  (a^2c)/b + (b^2a)/c + (c^2b)/a >=
(a^2b)/c+(b^2c)/a+(c^2a)/b  using  the  a>=b>=c>0..but  how  can we  do it....??


Not sure where to post this, sorry if not supposed to be here. Anyway, I am a high school senior taking calculus who is also on the school math team and am looking for a book that will help me improve the most. I am looking at the Art of Problem Solving books, but am not sure if volume 2 is appropriate to start with or if I will be lost without reading volume 1 first. Also, is the book constructed so that the full solutions manual is very helpful/necessary or just extra on the side (meaning is there a lot to learn from the solutions manual or does the text itself do a good enough job)?
I currently score about 110 on the AMC12 and between 1-3 on any given AIME and want a book that will show me how to solve problems you don't learn in school that will help me figure out problems like these, so are the AOPS books even the way to go for me?

I'm pretty sure AoPS books are the way to go. But you're a senior, so no more AMC.  :(

if you're getting 110s on AMC and 1-3 on AIME get AoPS 1 and 2.

Some people on this site say you can get around 8 on AIME from knowing AoPS 1 thoroughly.

And if you can afford it, get the solutions manual; you get a discount anyways.


On the $\LaTeX$ template for submitting answers, when I compile and "preview" It says I need to install fancyhdr. But whenever I do, something goes wrong.

Is it ok if I don't use the fancyhdr package and delete that line? But it says not to alter the above lines in the template so....

Is it ok?

EDIT: Another question, how come when I compile, there's 19 errors?

Did you download fancyhdr.sty and put it in the same directory as your file?

I am having a strange problem that is somewhat similar to this. See my post in the LaTeX forum for details.

You can try installing fancyhdr manually:

1. Start the MikTeX Package Manager (in Windows, pull up the Start menu, then select "All Programs", then "MikTeX", the then Package Manager)

2. Scroll down until you find the fancyhdr package, and click on it.

3. Click on the large "+" sign in the upper-left.  It should install.

4. If it doesn't work, your repository may be broken.  Click on the "Repository" menu, select "Change Package Repository", then select "Internet", and choose a different repository.  Then repeat steps 2 and 3.

Although it should work just fine as long as fancyhdr.sty is in the same directory as usamts.tex and solutions.tex.

About your 19 errors: I couldn't tell you what you're doing wrong unless you provide some more specific information.

If you want to submit your solutions in plain LaTeX (without the fancy headers), that's fine too.  Make sure your name and USAMTS ID# appear on [b]every[/b] page.

Thanks! I think I installed it now...  :?

This weird thing comes when I went to Package Manager.

"In order to install packages, it is necessary to define a package repository. Do you want to choose a package repository now?"

I clicked "No" and the files came up but I can't install them!  :?

When I clicked "Yes" the process seems to be complicated and I don't like it much.

So, is this file okay?

*I made this by myself and I wonder if this document is okay to use to put my solution in this format.*

Make sure you add the problem number and round number (put it up in the header with your name and user id).

Is it okay if the problem number and round number is on the line after the name and ID?

Like:

My name and ID
Round 17 #1


real analysis
Functional Analysis
real analysis theorems

I was interested if someone could advice me a good book on the subject Functional Analysis. I know Rudin has a pretty good reputation, however his books are not cheap. I was wondering what are some good 'classics' on this subject. This is not going to be a first read into advanced analysis. I've done some Hilbert/Banach/Spectral/Measure too. All introductionary though. I like challenges, so I don't want an easy introduction. Furthermore I like a 'bald' style (not too many pictures, colors -- just to-the-point). I'm also more of an active-reader. This is are my current candidates

Essential Results of Functional Analysis by Zimmer
Rudin, W.: Functional Analysis, McGraw-Hill Science, 1991

Other suggestions are very welcome

i dont know anything about functional analysis but i've heard some of my seniors talking about this book- functional analysis by kolmogorov and fomin. it is a dover publication. the book on real analysis by the same authors is a classic, so i think this one also will be good.


prove that any even number can be written as the sum of two primes.

hzbrl, you... should read a book on dinosaurs. Teach you how ancient yet mysterious somethings still are. Or so you leave us alone with your dumb post.
I actually have an easy one for you. If you can't solve it don't come back as you're clearly not mathematically oriented.
Prove there are infinite twin primes.

Well, the OP's question is going to be pretty difficult seeing as it's false.  ($ 2$?)  ;)

primes are there for being multiplied and not for being added!

Naphthalin

[quote="orl"]I recall user [url=http://www.jackofalldays.com/wp-content/uploads/2007/06/stupidguy.jpg]Goldback[/url] recently mentioned that he has a truly marvellous proof of this problem which this ML/AoPS post is too narrow to contain.   :wink:[/quote]

I love your saying, Orl...


Find all functions $ f: \mathbb R^ \plus{} \rightarrow R^ \plus{}$ satasfying $ (f(x))^2\ge f(x \plus{} y)(f(x) \plus{} y)$ for every $ x,y\in\mathbb R^ \plus{}$

We will prove there is no such function.
$ f(x\plus{}y)\leq \frac{f(x)^2}{f(x)\plus{}y}<f(x)$ whenever $ y>0$. This implies $ f(x)$ is strictly decreasing.
Let $ y\equal{}f(x)$, we get
$ f(x\plus{}f(x))\leq\frac{f(x)}2$-------------------------(*)
Now fix $ x$ and let $ a\equal{}f(x)>0$. We will prove $ f(x\plus{}(2\minus{}2^{\minus{}n})a)\leq2^{\minus{}n\minus{}1}a$ by induction.
1) $ n\equal{}0$
By (*), $ f(x\plus{}a)\equal{}f(x\plus{}f(x))\leq\frac{f(x)}2\equal{}2^{\minus{}1}a$.
2) Assume it holds for $ n$, now we look at $ n\plus{}1$.
We already have $ f(x\plus{}(2\minus{}2^{\minus{}n})a)\leq2^{\minus{}n\minus{}1}a$.
Using also the monotonicity, we have
$ f(x\plus{}(2\minus{}2^{\minus{}n\minus{}1})a)\equal{}f(x\plus{}(2\minus{}2^{\minus{}n})a\plus{}2^{\minus{}n\minus{}1}a)\leq f(y\plus{}f(y))$, where $ y\equal{}x\plus{}(2\minus{}2^{\minus{}n})a$.
Using (*)
$ RHS\leq\frac{f(y)}2\leq2^{\minus{}n\minus{}2}a$.

Using monotonicty,
$ f(x\plus{}2a)<f(x\plus{}(2\minus{}2^{\minus{}n})a)\leq2^{\minus{}n\minus{}1}a$ holds for any $ n$. This implies $ f(x\plus{}2a)\leq0$, a contradiction with $ f(x)\in R^\plus{}$.

Please don't repeat problems:
http://www.mathlinks.ro/viewtopic.php?t=238141
I think you were not satisfied with the solution or something ... in any case feel free to ask something about the solution.

Dear xxp2000 and Akashnil, thanks for your two nice solutions.


search
symmetry
pigeonhole principle
puzzles

This was the 4th problem in the South African Talent Search Round 5 and I thought it was interesting:

There are eight identical Black Queens in the first row of a chessboard and eight identical White Queens in the last row. The Queens move one at a time, horizontally, vertically or diagonally by any number of squares as long as no other Queens are in the way. Black and White Queens move alternately. What is the minimal number of moves required for interchanging the Black and White Queens?

Yeah, it was really very interesting and logical. I stressed out the left hemisphere of my brain to come out with that at least 23 moves are required for interchanging the Black and White Queens.
Let the White Queens be W1, W2, W3, W4, W5, W6, W7 and W8 and the Black Queens be B1, B2, B3, B4, B5, B6, B7 and B8 respectively in the order of columns.

Each of the 16 queens must move at least once. This would require 16 moves. At most one out of these 6 pair of 12 interior queens {(W2,B2); (W3,B3); (W4,B4); (W5,B5); (W6,B6); (W7,B7)}can move at least once to make way for its complement. The other would require at least an extra move by the 6 queens. This means 6 more moves are required at the least. Of the four queens at the corner (W1, W8, B1, B8), at least 3 can move only once. So, another extra move is required by the fourth queen. Theoretically, the least number of moves = 16 + 6 + 1 = 23.

I tried very hard to come out with the optimized moves which are shown in the attached document.

Pretty good, but I think you forgot about one thing: BLACK AND WHITE QUEENS MOVE ALTERNATELY   :P

Move 16 is white's move, and so move 17 has to be black, not white again. Also, black moved first, but if you look at the summary of moves you'll see that white has 12 moves while black has only 11, which is impossible if they move alternately  ;)

I got a way to do it with 24 moves, but I have not proved yet that there is no way to do it in 23 moves.

When the Queens move alternately, the number of moves in which we can attain the objective can not be less than 24. Two consecutive moves by the same colored queens is a primary criteria to make that in 23 moves.

i have been trying to answer this question for ages and i can't get it!!! :(

[quote="Melissa"][quote="Shamik Banerjee"]When the Queens move alternately, the number of moves in which we can attain the objective can not be less than 24.[/quote]
Can't anyone prove WHY it is impossible to do it in less than 24?[/quote]

No, because it is possible :D

[hide="24 steps"]24 steps is easy if you have a way to swap opposite pairs of black and white in 6 steps. Consider

black: a1-e5
white: a8-a1
black: b1-a2
white: b8-b1
black: e5-b8
black: a2-a8

Alarm bells! We've moved two blacks in a row. Actually this isn't too much of a problem. Ignore it for the moment and consider that for the pairs in the c and d columns we just shift the pattern to the right; for the other half of the board just use the mirror image, and the whole board is done. Due to this symmetry it's easy enough to hold off the final black move, start the same pattern somewhere else with white, and splice in the extra move when white ends up needing an extra move too.[/hide]

[hide="23 steps"]With 24 steps, by the pigeonhole principle at least half of the queens have to get to their destination in exactly 1 step. For 23 steps you need at least *9* queens to do this. At first it may seem that this can only happen when moving straight down a column, ie a8 to a1, but there are also the diagonals. The following sequence allows 3 out of 4 queens to get to their destination in 1 step:

black: a1-a7
white: h8-a1
black: h1-h8
white: a8-h1
black: a7-a8

Now it gets slightly tricky. Since we no longer have an even number of those pair swaps mentioned, we'll have to use the extra black move in the strategy above some other way. To keep things nice in symmetrical, let's start from the beginning with the middle columns:

01 black: d1-h5
02 white: d8-d1
03 black: e1-d2
04 white: e8-e1
05 black: h5-e8

(now with a black still unswapped we splice in the diagonal moves)

06 white: a8-a2
07 black: h1-a8
08 white: h8-h1
09 black: a1-h8
10 white: a2-a1
11 black: d2-d8

(symmetrical again! Whoo! Now for b, c, f and g - from white's side)

12 white: b8-f4
13 black: b1-b8
14 white: c8-b7
15 black: c1-c8
16 white: f4-c1

(save the last white for splicing in later, play from black's side)

17 black: g1-c5
18 white: g8-g1
19 black: f1-g2
20 white: f8-f1
21 black: g2-g8

(splice)
22 white: b7-b1

23 black: c5-f8[/hide]

I feel kinda stupid though - I had to use a chessboard to work it out :P Also, if you find a mistake in there - which isn't impossible - the general strategy should still hold. I hope. Please?

(Nice to see another South African on the boards, BTW!)

Wow this is amazing :lol:  I've gone through your moves on a chess board and there is no mistake that I can see.

Are you also from SA then?

A similar nice question (but much harder) can be found in [url=http://www.mathlinks.ro/Forum/contest/3rd%20Edition/contest3.zip]MathLinks Competetion 3[/url], last round, last problem. (created by me :))

I also have an easier-to-follow solution for 23, being a 6+6+6+5 movement, I have drawn it:

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=383[/img]

The red/green lines show different paths, extra colored spaces means a queen of that color will land there temporarily. I think it is quite obvious how to move nowjust alternate one move of each path. :) (as the red ones require more black moves and the green ones more white moves)

The 'reason' why it's not possible with less is that for every pair of non-corner queens to switch, it takes at least 3 moves. (obviously). Also to switch corner with non-corner you'd need at least 3 per pair. (actually even 4 but that doesn't matter)

So we're left with switching the corners. (different due to diagonal movement)
Even on blank board, with no other pieces standing in the way, you need 5 moves, so here you need to add at least 5 moves. This completes the solution. :)

[quote="Peter VDD"]A similar nice question (but much harder) can be found in [url=http://www.mathlinks.ro/Forum/contest/3rd%20Edition/contest3.zip]MathLinks Competetion 3[/url], last round, last problem. (created by me :))[/quote]

You can only move each knight 'sideways' once, all the rest have to be forward moves - but how on earth are you supposed to count which combinations of moves will have knights attacking each other in the middle or not :huh:

[quote="Peter VDD"]Then we move all the queens from the first move to their final location. (6 moves)[/quote]

That sounds like a much more elegant method but I can't seem to get this step to work - if it's diagonal like the first moves then these queens never change the colour of the square they're on, whereas those crossing directly do - which means the squares they're supposed to fill are already occupied :?:

[quote="fusd"]You can only move each knight 'sideways' once, all the rest have to be forward moves - but how on earth are you supposed to count which combinations of moves will have knights attacking each other in the middle or not :huh:[/quote]It is possible. I still have my solution if you want. Or you can post it into the olympiad combinatorics forum, I'll post the solution.

[quote="fusd"][quote="Peter VDD"]Then we move all the queens from the first move to their final location. (6 moves)[/quote]That sounds like a much more elegant method but I can't seem to get this step to work - if it's diagonal like the first moves then these queens never change the colour of the square they're on, whereas those crossing directly do - which means the squares they're supposed to fill are already occupied :?:[/quote] I changed the solution a bit to make it more transparent with a drawing. Hope it is more clear now. :) [I agree, the wording was a bit misleading]

I appreciate your correct answer. I apologize for my wrong statement [quote]When the Queens move alternately, the number of moves in which we can attain the objective can not be less than 24.[/quote]

I wonder where the original queens on a chess board problem comes from? I mean South Africa is not really known for producing original problems :D  Three of the other problems in the round 5 talent search came from the Junior Balkan MO.

This Problem was from the Tournament of Towns (Spring session) 2005.


factorial
calculus
integration
function
calculus computations

I have expression:

$ p(p\minus{}1)(p\minus{}2)\cdot\ldots\cdot(p\minus{}n\plus{}2)$

I see that there's p! involved, but what to do with the last term? And what would be as gamma representation?

Do you mean this:
\[ p(p \minus{} 1)(p \minus{} 2)\cdot\ldots\cdot (p \minus{} n \plus{} 2) \equal{} \frac{p!}{(p \minus{} n \plus{} 1)!}?\]
Now, use the fact that $ \Gamma(n) \equal{} (n \minus{} 1)!$.

Well I guess, how did you know that? I really suck at factorials :(

By experience. It is simple to think that $ p! \equal{} p(p \minus{} 1)(p \minus{} 2)\cdot\ldots\cdot 1$. But there is the extra product $ (p \minus{} n \plus{} 1)(p \minus{} n)\cdot\ldots\cdot 1$ which is, by definition, $ (p \minus{} n \plus{} 1)!$. And I learned the identity $ \Gamma(n) \equal{} (n \minus{} 1)!$ from theoretical physics course.

If confused, try an example...
\[ 7 \cdot 6 \cdot 5 \equal{} \frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{4 \cdot 3 \cdot 2 \cdot 1}\]

Do you know permutations?

[url=http://en.wikipedia.org/wiki/Permutation]Permutation[/url] - Wikipedia
[url=http://mathworld.wolfram.com/Permutation.html]Permutation[/url] - MathWorld

$ {}_nP_r \equiv \frac{n!}{(n\minus{}r)!}\equal{}p(p\minus{}1)(p\minus{}2)\cdots(p\minus{}r\plus{}1)$

So $ p(p\minus{}1)(p\minus{}2)\cdot\ldots\cdot(p\minus{}(n\minus{}2)) \equal{} {}_pP_{n\minus{}1} \equal{} \frac{p!}{(p\minus{}n\plus{}1)!}$

Thanks! I always get confused with those things!

[quote="MeKnowsNothing"]And I learned the identity $ \Gamma(n) \equal{} (n \minus{} 1)!$ from theoretical physics course.[/quote]

Can you be more specific?  :)

Oh, sorry. Of course $ n\in\mathbb{Z}_\plus{}$. For a proof, try integration by parts.

[quote="Christiano89"]Oh I know that. I would like to hear from you, where the physicists use the gamma function... :lol:[/quote]

Well I have only heard it (for now, I'm still young :D) on statistics and mathematical methods in physics as a part of special functions course, but I think there are other uses...


[quote="yrhc@eht"]find all functions $f: R\longrightarrow R$ such that f(x+y)=f(x)f(y)f(xy) for all real x and y.[/quote]

1) If it exist $a$ such that $f(a)=0$, then $f(x)=f((x-a)+a)=f(x-a)f(a)f((x-a)a)=0$ and $f(x)=0$ $\forall x$
2) If $f(x)\neq0$ $\forall x$ :
$f(x+y+z)=f(x+(y+z))=f(x)f(y+z)f(x(y+z))=f(x)f(y)f(z)f(yz)f(xy)f(xz)f(x^{2}yz)$
$f(x+y+z)=f((x+y)+z)=f(x+y)f(z)f((x+y)z)=f(x)f(y)f(z)f(xy)f(xz)f(yz)f(xyz^{2})$
So $f(x^{2}yz)=f(xyz^{2})$ $\forall x,y,z$
So $f(u)=f(v)$ $\forall$ $u,v\neq 0$ and so $f(x)=-1,0,1$ $\forall x\neq 0$ and since $f(0)=f(1-1)=f(1)f(-1)f(-1)$, $f(x)=-1,0,1$ $\forall x$

So we have three solutions :
$f(x)=-1$
$f(x)=0$
$f(x)=1$


It's an easy application of Cano's theorem . Lines through $ A,B,C$ perpendicular to $ EF,DF,DE$ are concurent ,hence:
$ ( AE^{2}-AF^{2})+( BF^{2}-BD^{2})+( CD^{2}-CE^{2}) = 0$ or:
 $ ( DB^{2}-DC^{2})+( EC^{2}-EA^{2})+( FA^{2}-FB^{2}) = 0$. QED

[quote="Nbach"]It's an easy application of Cano's theorem....[/quote]

I think, the correct is [b][size=100]Carnot's theorem.[/size][/b]

Kostas vittas.

[quote="Orthologic Triangles Theorem"]Given two nondegenerate triangles $ \triangle{ABC}$ and $ \triangle{DEF}$.
If the perpendiculars from the points $ A$, $ B$, $ C$ to the lines $ EF$, $ DF$, $ DE$ concur, then the perpendiculars from the points $ D$, $ E$, $ F$ to the lines $ BC$, $ AC$, $ AB$ also concur.[/quote]

This theorem also received a few proofs on http://www.mathlinks.ro/Forum/viewtopic.php?t=4337 ("warm-up exercise"). It has also been mentioned and proven in http://www.cut-the-knot.org/Curriculum/Geometry/Maxwell.shtml#orthologic . There are different proofs of the above theorems based on
- vectors,
- the Carnot theorem (a. k. a. the Steiner theorem),
- similitude of triangles,
- areas,
- Ceva's theorem in the trigonometric form.

  Darij


Consider the two by 5 system below

$ \minus{}x_1 \plus{} 2x_2 \plus{} 4x_4 \equal{} 100$
$ 3x_1 \plus{} 2x_2 \minus{} 5x_3 \plus{} x_4 \plus{} 7x_5 \equal{} 200$

1.) Without doing any computation, explain why it is obvious that this system has at least one solution.

2.) Without using an equation which is just a scalar multiple of one of the two given equations, give a third equation which is guaranteed to be redundant. Explain why this new equation is redundant.

The coefficient matrix is $ \begin{bmatrix}\minus{}1&2&0&4&0\\3&2&\minus{}5&1&7\end{bmatrix}$

We can see at a glance that the two rows are not scalar multiples of each other. Some other ways to state that: the two rows are independent, or the rank of the matrix is 2. Row elimination (that is, we contemplate the possibility of row elimination without actually performing it) cannot create a row of zeros in the coefficient matrix and and hence cannot show us an inconsistent system. Therefore the system is consistent. (Once we know that is consistent, we know that it has infinitely many solutions - a three-parameter infinite family of solutions.)

(I'll add that I didn't really have to write out the matrix; we can "see" the matrix without writing it, which meets the spirit of the "no calculation" stipulation.)

For (2), make a third equation that is a linear combination of the first two with neither coefficient zero. Easiest would probably be to just add the two equations.

[I'm teaching this course in the spring semester, although our semester hasn't started yet. This looks like it belongs in the first week of the course, and it's a good homework problem or maybe a good lecture example.]


[quote=whiteawesomesun]What would we do if the triangle was a much larger one, perhaps a perimeter of $100$ instead of $10$? Then would casework not take too long?[/quote]

There was a reason MathCounts decided to make the perimeter only $10$, because it can only be solved using casework.

Yes, there's $\binom{9}{2} = 36$ different ways, [b]but[/b] our total probability is $\frac{6}{36} = \boxed{\frac{1}{6}}$.

AIME 15 is correct.
So,the answer is 1/6.

[quote=stevenmeow]There are $ \binom{9}{2}\equal{}45$ ways to choose the 2 places to break the stick.

In order for the new sticks to form a triangle, none may measure 5cm or more. Otherwise, this side is equal to or greater than the length of the remaining sides which is impossible by the triangle inequality. With this in mind, the only 2 possible combinations of sides for the triangle are 4-4-2 and 4-3-3. There are 3 ways to break the stick into both of the cases, for a total of 6 good combinations.

The answer is therefore $ \frac{6}{45}\equal{}\frac{2}{15}$.[/quote]

BTW, $ \binom{9}{2}=36$ becuase $\frac{9*8}{2}=\frac{72}{2}=36$

[quote=GameBot]A 10-cm stick has a mark at each centimeter. By breaking the
stick at two of these nine marks at random, the stick is split into
three pieces, each of integer length. What is the probability
that the three lengths could be the three side lengths of a
triangle? Express your answer as a common fraction.[/quote]

Yes! Yes! YES!

I solved one of my only problems with Power XP, which was this one  :rotfl:

[quote=stevenmeow]There are $ \binom{9}{2}\equal{}45$ ways to choose the 2 places to break the stick.

In order for the new sticks to form a triangle, none may measure 5cm or more. Otherwise, this side is equal to or greater than the length of the remaining sides which is impossible by the triangle inequality. With this in mind, the only 2 possible combinations of sides for the triangle are 4-4-2 and 4-3-3. There are 3 ways to break the stick into both of the cases, for a total of 6 good combinations.

The answer is therefore $ \frac{6}{45}\equal{}\frac{2}{15}$.[/quote]
 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:
 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:
 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:
 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:
 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

dude wut are u doing

$\dbinom{9}{2} = 36$

shouldn't this be in counting and probability

As soon as I saw the triangle condition, I thought of three equations in the 3-D plane, which are similar to $y=-x+1$ but on each of the 3 planes. This yields a tetrahedron with volume $\frac{1}{3}\cdot\frac{1}{2}\cdot1 = \frac{1}{6}$, where the total set of all possible cases is represented by a unit cube.

Free 25

very cool problem. Kinda boring casework tho ngl

[quote=AIME15]Answer key says 1/6.[/quote]

what answer key[hide]1/6[/hide]

[hide=answer]The lengths can form a triangle if and only if the sum of the smaller two exceeds the length of the largest. But the sum of all three pieces is 10, so this means that the largest piece must have length no larger than 4. (If the largest piece is 5 or greater, then the three lengths will violate the Triangle Inequality.) However, the largest piece clearly must be longer than 3. Thus, the only acceptable sets of sidelengths are $\{3,3,4\}$ and $\{2,4,4\}$. Since we can obtain these in 6 ways, and the stick can be broken in $\binom{9}{2} = 36$ different ways, our total probability is $\frac{6}{36} = \boxed{\frac{1}{6}}$[hide/]:rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl::rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl:


prove the inequality below:
$ {\minus{}1< \sum_{k\equal{}1}^{n} \frac{k}{k^2\plus{}1}\minus{}ln (n) \le \frac{1}{2}}$ for $ {n\equal{}1,2 ......}$

[quote="CCMath1"]prove the inequality below:
$ { \minus{} 1 < \sum_{k \equal{} 1}^{n} \frac {k}{k^2 \plus{} 1} \minus{} ln (n) \le \frac {1}{2}}$ for $ {n \equal{} 1,2 ......}$[/quote]

you can see here:
http://bbs.pep.com.cn/thread-511635-1-1.html


I want to show that $z+b$ is conformal at infinity. My prof. said to just show that $\frac{1}{z}+b$ is conformal at 0, and I believe he thinks this is obvious. What am I missing? I don't see why it's so obvious. (I know that $\frac{1}{z}$ is a conformal map).

$\infty$ is a fixed point of the map $z\mapsto z+b$. We want to shift this fixed point to zero, so it will be easier to work with. To do this, we take reciprocals on both sides:
$z\mapsto \frac1z\mapsto \frac1z+b\mapsto \frac1{\frac1z+b}=\frac{z}{1+bz}$
This map is a composition of the map in question and known conformal maps. It is conformal at zero if and only if the original map is conformal at $\infty$.

Working with the intermediate step $\frac1z+b$ is problematic, since the key point is still mapped to $\infty$.

Thanks jmerry. In my text it just says that it's 'obvious' that the map is conformal on the riemann sphere.


If Michael will have a family of four children, and male and female births are equally likely, what is the probability that Michael will have at least two girls? Express your answer as a common fraction.

4 boys = 1/16
3 boys 1 girl = 4/16
2 boys 2 girls = 6/16
1 boy 3 girls = 4/16
4 girls = 1/16

P(at least 2 girls)=P(2 girls)+P(3 girls)+P(4 girls)=[b]11/16[/b]

Working reversely may offer a better solution.
4 boys = $ (1/2)^4 \equal{} 1/16$
3 boys = $ 4C3 * (1/2)^4 \equal{} 4/16$

The result is $ 1 \minus{} 1/16 \minus{} 4/16 \equal{} \boxed{11/16}.$


Putnam function symmetry geometry parallelogram trigonometry

This is a problem I gave as a warm-up to my Putnam students once. It consisted of several parts. The task was always the same: explain why the proposed answer cannot be correct (you DO NOT need to know the correct answer to achieve it).

So, here is the problem: imagine an angle of opening $\theta\in(0,\pi)$ with vertex $O$. Choose the points $A$ and $B$ on one of the sides of the angle and $C$ and $D$ on the other side. Let $a=|OA|, b=|OB|, c=|OC|, d=|OD|$. If $a<b$ and $c<d$, the intervals AD and BC will intersect at some point $X$ inside the angle. Let $x=|OX|$. Of course, $x$ is somehow expressible in terms of $a,b,c,d$ and $\theta$. Suppose that someone claims that
\[x=a^{2}+b-c+d-\theta\]
Explain why he is wrong.
(this first part is very easy but I wonder how many different explanations you can come up with) :)

[hide="A few"] - $x$ should be a symmetric function of $b, d$.
- Increasing $c$ and $d$ by the same amount clearly increases $x$.
- At $c = 0$ or $a = 0$ we should have $x = 0$.
- As $\theta \to \pi$ we expect that $x \to 0$. [/hide]

Let's see what we have so far:

[quote="t0rajir0u"]
1) $x$ should be a symmetric function of $b, d$.
2) Increasing $c$ and $d$ by the same amount clearly increases $x$.
3) At $c = 0$ or $a = 0$ we should have $x = 0$.
4) As $\theta \to \pi$ we expect that $x \to 0$.[/quote]

1) should be put more accurately than that: the proposed expression is symmetric with respect to exchange of $b$ and $d$ and the symmetry of $x$ is not exactly like that. What was, probably, meant is that if you simultaneously exchange $a$ with $c$ and $b$ with $d$, the expression should stay the same.

2) is likely to be true but very far from being "clear"; at least, I do not see a simple proof immediately. Can you elaborate?

3) Well, not quite that, if you draw a picture, you'll see that $x$ is something else in these cases ;)

4) Interestingly enough, this is false. Can you see why?

Anyway, let's take 1) and 2) into account and change the expression to
\[x=abcd\theta \]
See anything wrong now?

[quote="fedja"]3) Well, not quite that, if you draw a picture, you'll see that $x$ is something else in these cases ;) [/quote]

Ah, I see, sorry.  My mental picture was mistaken.  When $a = 0$ we have $x = c$ and when $c = 0$ we have $x = a$.

[quote="fedja"]4) Interestingly enough, this is false. Can you see why? [/quote]

Again, mistaken mental picture.  Sorry :oops:

[quote="fedja"]Anyway, let's take 1) and 2) into account and change the expression to
\[x=abcd\theta \]
See anything wrong now?[/quote]

[hide="Well, sure..."] - There is what I said about $x$ when $a = 0$ or $c = 0$.
- In the very easy-to-analyze case $a = c, b = d$ we see that as $\theta \to 0$ we have $x \to \frac{a+b}{2}$.
- As $b, d \to \infty$ we have $AD, BC$ approaching being parallel to $OC, OA$, which places $X$ at a point such that its altitudes to $OC, OA$ meet the sides at $C, A$, so $x$ approaches a finite limit. [/hide]

[quote="t0rajir0u"][hide="Well, sure..."] - There is what I said about $x$ when $a = 0$ or $c = 0$.
- In the very easy-to-analyze case $a = c, b = d$ we see that as $\theta \to 0$ we have $x \to \frac{a+b}{2}$.
- As $b, d \to \infty$ we have $AD, BC$ approaching being parallel to $OC, OA$, which places $X$ at a point such that its altitudes to $OC, OA$ meet the sides at $C, A$, so $x$ approaches a finite limit. [/hide][/quote]

[hide="Erratum, and additions"] - I was mistaken about the case as $b, d \to \infty$.  In fact, $OCXA$ approaches a parallelogram, not a kite as I had first thought (I'm not sure why I thought this).  In the particular case $a = c$ we can in fact evaluate the limiting value of $x$, which is $x = 2a \cos \frac{\theta}{2}$.
- Additionally, when $a, b, c, d$ are multiplied by a scale factor it is not hard to see that $x$ should be multiplied by the same scale factor.  [/hide]


Get a piece of paper before you start this...

If you're ready...

Two circles are concentric. If a chord of the outer circle A is tangent to the inner circle B, and has a length of 20 units, find the area of the region enclosed by A, but excluding the area covered by B.

P.S. I'm new so could someone tell me how to make text that appears when you click on something? Also how do you show formulas and pictures so that they're not in just plain text like this problem? It would be so much easier if I could just show a picture.  :lol:

ok for the hidden text, use:
[code][hide] hidden text [/hide] OR [hide=message name] hidden text [/hide][/code]
and as for the image, attach it as a file that is saved on ur computer.
EDIT: oh yeah! u can also use LaTeX learn mroe about it on the LaTeX link on the left hand link.

NOW for the problem,
ok  so by the pythagorean theorem, r^2+10^2=s^2 But, this can be simplified
r^2+100=s^2
100=s^2-r^2
s^2-r^2=100
Now the desired area is the difference of the areas of the 2 circles or (pi)s^2-(pi)r^2
using the dietributive property, this becomes (pi)(s^2-r^2)
note that s^2-r^2=100 from the above so the area is [b]100pi[/b]

So what do you click on to get the hidden text? Like if you have to click on the word "answer" to get it to show, how do you set it up so that happens? Thx.

Press the quote button that appears on this message to see the code for hiding/etc.

[hide="Solution"]
[asy]fill(Circle((0,0),25/3),rgb(0.9,0.9,0.9));
fill(Circle((0,0),20/3),white);
draw(Circle((0,0),25/3));
draw(Circle((0,0),20/3));
dot((0,0));
draw((0,0)--(0,20/3)--(15/3,20/3)--cycle);
label("\(10\)",((0,20/3)+(15/3,20/3))/2,N);
label("\(O\)",(0,0),SW);
label("\(r_B\)",(0,10/3),W);
label("\(r_A\)",(15/3,20/3)/2,SE);[/asy]

Let $ r_{A,B}$ denote the radii of $ A, B$ respectively; then the area of the annulus (as it is known) is $ \pi(r_A^2 \minus{} r_B^2)$. By the [[Pythagorean Theorem]] on the drawn triangle, $ r_B^2 \plus{} 10^2 \equal{} r_A^2 \Longrightarrow r_A^2 \minus{} r_B^2 \equal{} 100$. so the answer is $ 100\pi$.
[/hide]

so for azjps's case, his message name is "solution", so the thing would be written as:
[code][hide=solution] *text* [/hide][/code]

and if u only put [code][hide]text[/hide][/code]
then the message name will be "click to reveal hidden content"

Those of you that answered are all correct. However, the thing with the annulus is confusing as I am just a 7th grader doing algebra II honors. The simplest solution was using r as the smaller radius, setting this at a right angle to the tangent line, and using it to calculate the larger radius in terms of r. Then, the respective areas can be found, and then subtracted to get $ 100pi$.

P.S. How do you show pi using latex?

Greek letters are simply \ + letter; For example, \alpha\beta\gamma\delta produces $ \alpha\beta\gamma\delta$.

@azjps

How did you make that awesome diagram!!! i use paint and theyre terrible and i know you didnt use paint but then what did you use!!!

Thanks

[quote="fishythefish"]P.S. How do you show pi using latex?[/quote]
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php

Sorry about that. I'm a lot better at LaTex. Idk why the guide wouldn't load on my computer..but we were having internet problems...
Anyways, thank for all the help!


香港朋友们能不能帮我翻译一下 2, 3, 10.

2.
"cut girl" means try to meet a girl and make her to be one's girlfriend
豬扒 means ugly girls
so he means that he'd rather die instead of having an ugly girl as his girlfriend

3.
ginger means guts here.
隻揪 means one on one fighting
that means if you have the guts come on and fight

10.
索女 means beautiful girls
媾 is similar to "cut girl" but it's a very rude term
索女受媾 basically means a beautiful girl is willing to be one's girlfriend.

I think you can understand the whole thing now~

[quote="shobber"]Thanks!  :D
And I still have one more question: The first half of the 3rd one.[/quote]
arguing is not good

我这个纯正的广东人居然还有些粤语看不懂，惭愧惭愧……

香港的广东话和广东的不一样吧
词汇差别

繁体字我认得, 但是香港的字却有好多他们自创的, 所以读起来一头雾水.

[quote="dingdongdog"]就是就是，我有一个姨婆生活在香港，她跟我谈话时总冒出一些我不会的词汇．而且，香港的广东话受外国影响较大．[/quote]
对于我们北方人来说, 那简直就是外语! 估计要交流还得用英语...

澳門與香港相鄰, 而且澳門人大多都看香港的電視節目, 故語言差異不太大.

澳門與香港相鄰, 而且澳門人大多都看香港的電視節目, 故語言差異不太大.

我这里只能收到一个凤凰...

还好吧，不过香港台的节目整天被这里的电视台胡乱插播的广告影响，很恶心。

[quote="dingdongdog"]还好吧，不过香港台的节目整天被这里的电视台胡乱插播的广告影响，很恶心。[/quote]
每次回廣州看香港台都被那些廣告弄得很煩, 最討厭的有時連新聞都不能看!


If we divide both sides by x^3y^3z^3 we obtain

x^3/z^3 + y^3/x^3 + z^3/y^3  \geq x^2/yz + y^2/xz + z^2/xy

and putting a=z/y,b=x/z,c=y/x we have

a^3+b^3+c^3  \geq a^2b + b^2c +c^2a

true by AM-GM.


Eliminate $A$ from $x = 3 \cdot \sin(A) - \sin(3 \cdot A)$ and $y = 3 \cdot \cos(A) - \cos(3 \cdot A).$

$\left\{ \begin{array}{l}
 x = 3\sin \alpha  - \sin 3\alpha  \\
 y = 3\cos \alpha  - \cos 3\alpha  \\
 \end{array} \right.,\alpha  \in \left[ {0,2\pi } \right] \Leftrightarrow \left\{ \begin{array}{l}
 x = 4\sin ^3 \alpha  \\
 y = 3\cos \alpha  - \cos 3\alpha  \\
 \end{array} \right.,\alpha  \in \left[ {0,2\pi } \right] \Leftrightarrow $

$\Leftrightarrow \left\{ \begin{array}{l}
 \sin \alpha  = \sqrt[3]{{\frac{x}{4}}} \\
 y =  \pm 2\left[ {3\sqrt {1 - \sqrt[3]{{\frac{{x^2 }}{{16}}}}}  - 2\left( {\sqrt {1 - \sqrt[3]{{\frac{{x^2 }}{{16}}}}} } \right)^3 } \right] \\
 \end{array} \right.$

[u]Conclusion:[/u]
$\left\{ \begin{array}{l}
 x = 3\sin \alpha  - \sin 3\alpha  \\
 y = 3\cos \alpha  - \cos 3\alpha  \\
 \end{array} \right.,\alpha  \in \left[ {0,2\pi } \right] \Leftrightarrow $
$\Leftrightarrow  y =  \pm 2\left[ {3\sqrt {1 - \sqrt[3]{{\frac{{x^2 }}{{16}}}}}  - 2\left( {\sqrt {1 - \sqrt[3]{{\frac{{x^2 }}{{16}}}}} } \right)^3 } \right],x \in \left[ { - 4,4} \right]$

Orl, do you possibly mean $y = 3 \cos A + \cos 3A$?  It seems very difficult as it is.

$\left\{ \begin{array}{l}
 x = 3\sin \alpha  - \sin 3\alpha  \\
 y = 3\cos \alpha  + \cos 3\alpha  \\
 \end{array} \right.,\alpha  \in \left[ {0,2\pi } \right] \Leftrightarrow \left\{ \begin{array}{l}
 x = 4\sin ^3 \alpha  \\
 y = 4\cos ^3 \alpha  \\
 \end{array} \right.,\alpha  \in \left[ {0,2\pi } \right] \Leftrightarrow $

$ \Leftrightarrow \left\{ \begin{array}{l}
 \sqrt[3]{{\frac{x}{4}}} = \sin \alpha  \\
 \sqrt[3]{{\frac{y}{4}}} = \cos \alpha  \\
 \end{array} \right.,\alpha  \in \left[ {0,2\pi } \right] \Leftrightarrow \sqrt[3]{{\frac{{x^2 }}{{16}}}} + \sqrt[3]{{\frac{{y^2 }}{{16}}}} = 1$

whence,
$\left\{ \begin{array}{l}
 x = 3\sin \alpha  - \sin 3\alpha  \\
 y = 3\cos \alpha  + \cos 3\alpha  \\
 \end{array} \right.,\alpha  \in \left[ {0,2\pi } \right] \Leftrightarrow \sqrt[3]{{\frac{{x^2 }}{{16}}}} + \sqrt[3]{{\frac{{y^2 }}{{16}}}} = 1 \Leftrightarrow $

$ \Leftrightarrow \left( {y =  + 4\left( {1 - \sqrt[3]{{\frac{{x^2 }}{{16}}}}} \right)^{\frac{3}{2}} } \right) \vee \left( {y =  - 4\left( {1 - \sqrt[3]{{\frac{{x^2 }}{{16}}}}} \right)^{\frac{3}{2}} } \right),x \in \left[ { - 4,4} \right]$


limit
logarithms
real analysis
real analysis unsolved

Find:
\[ \mathop {\lim }\limits_{\scriptstyle x \to \infty  \hfill \atop
  \scriptstyle y \to a  \hfill} (1\plus{}\frac{1}{x})^\frac{x^{2}}{x\plus{}y}\]

and tell the reason :D

Take the logarithm and use that $ \log(1\plus{}1/x)\sim 1/x$, $ x\to\infty$.


here's the sequence:

23
116
538
2771
25611
509767
15001124
534944836
21997346580
1023001263721

where f(x)= such and such.
the sequence goes from x=1 to x=10.
solving for x=0 gives u a mathematical error.

how do you solve this?!  D:

There is no such sequence, according to Wolfram Alpha.

OH MY GOOODDD!!!   *screams and commits suicide*. Are u serious?!
jeez, i spend 8 hours total on this stupid problem, then you come up just to tell me that it is impossible?!
okay, well, thanks for the responce...

hello, see also here
http://www.research.att.com/~njas/sequences/
Sonnhard.

Wolfram could be wrong. It depends on where he got it.

Why would they make an unsolvable sequence?
:thinkies:

Edit:13 year bump oops


Let A be a  nilpotent operater with the matrix
${\left( \begin{matrix}0&1&2\\0&0&2\\0&0&0\{matrix}end\right)$
in some basis $a_{1},a_{2},a_{3}$.
Find its  maximal cyclic subspace and cyclic basis in this cyclic subspace

[quote="rustam"]Let A be a  nilpotent operater with the matrix
$\left( \begin{matrix}0&1&2\\0&0&2\\0&0&0 \end{matrix}\right)$
in some basis $a_{1},a_{2},a_{3}$.
Find its  maximal cyclic subspace and cyclic basis in this cyclic subspace[/quote]

Put \end{matrix} instead of \{matrix}end  :)

I barely understand what maximal cyclic subspace is, but isn't the answer just "the entire space"

Taking $v=\left(\begin{matrix}0\\0\\1 \end{matrix}\right)$ I think $v,Av, A^{2}v$ generate the entire space  :maybe:


projective geometry
geometry proposed
geometry

let in triangle  $ ABC$ with centroid $ G$, $ AG$ be  extended to $ X$ such that $ GA\equal{}GX$. Through the point $ X$, parallels are drawn to the sides of  triangle  such that parallel to $ BC$ interesects $ AB$ at $ L$, parallel to $ CA$ intersects  $ BC$ at $ M$ and the parallel to $ AB$ intersects $ AC$ at $ N$.
Prove that $ L,M,N$ are collinear.

First, $ \frac{AB}{BL}\equal{}3$. Denote R,J the intersection of AM and BG, BG and AC. Use Thales's theorem and we can calculate $ \frac{MK}{MB}$. Use Menelaus's theorem for triangle AMK and get $ \frac{AR}{RM}\equal{}3$, similar for $ \frac{AJ}{JN}$
Thus L,M,N are collinear because $ LMN//BJ$.

Dear Mathlinkers,
1. GM // AB  and  GN // BC
2. Applies a converse of the Pappus 's theorem to the hexagone ALXMGNA
and we are done
Sincerely
Jean-Louis


Why does the number of results that I get when I search for things that I posted not equal the number after "posts" under my avatar?