IMPORTANT PRIME CONJECTURES

There is a twin prime between $(2n)^2$  and  $(2n+1)^2$ for every $n$.

I HAVE PRESENTED A “POSSIBLE” PROOF OF THE FAMOUS LEGENDRE’S CONJECTURE IN MY MPHIL THESIS AT THE UNIVERSITY OF GHANA 2006.

I SHOWED THAT
There are at least two ([b]instead of one[/b])primes between $n^2$ and $(n+1)^2$  for every $n$.

Peter Acquaah
Mathematics Department
University of Ghana
Legon, Accra Ghana.

Well, show us your 'proofs'...

(http://www.mathlinks.ro/Forum/viewtopic.php?t=42157)

:rotfl: Did you do this ascii picture by yourself? It is very nice :)

No, I found it on a german page and changed the text on the sign.

Ok, well, if you like them : http://www.chris.com/ascii/ ;)


calculus
integration
real analysis
LaTeX
real analysis unsolved

This question comes from Rudin 8.12.

I already figured out a, b, and c, but I'm struggling with d.

We know that

$ \sum_{n\equal{}1}^{\infty}\dfrac{sin^2\left(n\delta\right)}{n^2\delta}
\equal{}\dfrac{\pi\minus{}\delta}{2}$

Let $ \delta\to0$ and prove that

$ \int^{\infty}_0 \left( \dfrac{sin\left(x\right)}{x}\right)^2 dx \equal{} \dfrac{\pi}{2}$

Thanks for helping with the code JRav.

With regards to your question about the $ \text{\LaTeX}$, just surround your syntax with dollar signs.  So your first line should read
[code]$\sum_{n=1}^{\infty}\dfrac{sin^2\left(n\delta\right)}{n^2\delta}
=\dfrac{\pi-\delta}{2}$ [/code]
which yields
$ \sum_{n=1}^{\infty}\dfrac{sin^2\left(n\delta\right)}{n^2\delta}
=\dfrac{\pi-\delta}{2}$.  Now if we let $ \delta\rightarrow 0$, we want to show that
$ \int^{\infty}_0 \left( \dfrac{sin\left(x\right)}{x}\right)^2 dx = \dfrac{\pi}{2}$


[color=darkred]Let be given a triangle $ABC$ and let $I$ be its incircle. Prove that:[/color]
\[\boxed{\boxed{\blacktriangleleft\;\frac{1}{IA\cdot IB}+\frac{1}{IB\cdot IC}+\frac{1}{IC\cdot IA}\le\frac{R+r}{2Rr^{2}}\;\blacktriangleright}}\]

We start from the inequality $IA+IB+IC \le 2(R+r) \ (\star).$  For a proof, see Mateescu Constantin's approach in the topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=359795]Two inequalities regarding radii[/url]. Let $AI,BI,CI$ cut  the circumcircle $(O)$ of $\triangle ABC$ again at $X,Y,Z.$ Power of $I$ with respect to $(O)$ is given by $IA \cdot IX= IB \cdot IY = IC \cdot IZ= 2Rr.$

Substituting $IA,IB,IC$ from the latter expressions into $(\star)$ yields

$\frac{2Rr}{IX}+\frac{2Rr}{IY}+\frac{2Rr}{IZ} \le 2(R+r) \Longrightarrow  \frac{1}{IX}+\frac{1}{IY}+\frac{1}{IZ} \le \frac{R+r}{Rr} \ (1)$

On the other hand, it's well-known that $X,Y,Z$ become circumcenters of $\triangle BIC,$ $\triangle CIA,$ $\triangle AIB.$ Further, the perpendicular inradii to $BC,CA,AB$ become their I-altitudes. Then

$IB \cdot IC = 2r \cdot IX \ , \ IC \cdot IA = 2r \cdot IY \ , \ IA \cdot IB = 2r \cdot IZ$

Substituting $IX,IY,IZ$ from the latter equalities into $(1)$ yields

$\frac{1}{IB \cdot IC}+\frac{1}{IC \cdot IA}+\frac{1}{IA \cdot IB} \le \frac{R+r}{2Rr^2}$


geometry trapezoid USAMTS ARML circumcircle

[hide="I think you can"]
WLOG say it's isoceles.[/hide]

[quote="miyomiyo"]Trapezoid $ABCD$ with $AB$ parallel to $CD$, $CD = AC+6$, angle $ADC = 54^{\circ}$, angle $DCB = 36^{\circ}$, $M$ is midpoint of $AB$, $N$ is midpoint of $CD$.  Compute $MN$.[/quote]

I'm pretty sure this is what he meant.

[hide]In this form, you lose generality saying it's isosceles.[/hide]

Whoops, yes, worthawholebean, that's correct.  Edited.

[quote="miyomiyo"]Trapezoid $ABCD$ with $AB$ parallel to $CD$, $BC = AC+6$, angle $ABC = 54^{\circ}$, angle $DCB = 36^{\circ}$, $M$ is midpoint of $AB$, $N$ is midpoint of $CD$.  Compute $MN$.[/quote]

uh...wait how can this be a trapezoid?

[quote="13375P34K43V312"]yeah dude ABC+DCB=180 dude[/quote]
My form of the question is correct (I'm pretty sure). He must not have seen that change when updating the first post.

oh man those angles add to 90...HMMM>>>>????

i solved a similar problem last year at arml practice

First, "complete" the right triangle.

The median to the hypotenuse in a right triangle has half the length of the hypotenuse.

[hide="Tricky Solution"]Draw lines from $A$ and $B$ parallel to $MN$.  Cut out that piece of the trapezoid and push the two sides together.  We have a right triangle with base $6$ and a median to the hypotenuse.  Imagine the circumscribed circle and thus we have that the median and the two parts it splits the hypotenuse into are all of length $\boxed{3}$.[/hide]

[quote="miyomiyo"]Trapezoid $ABCD$ with $AB$ parallel to $CD$, $CD = AB+6$, angle $ADC = 54^{\circ}$, angle $BCD = 36^{\circ}$, $M$ is midpoint of $AB$, $N$ is midpoint of $CD$.  Compute $MN$.[/quote]
[hide] Extend the non-parrelel sides of the trapezoid to meet at X.  We have that $\angle X=90$.  We also have that $\triangle XAB\sim\triangle XDC$.  So, X, M, and N are collinear.  Thus, we have that $MN=XM-XN=\frac{DC}{2}-\frac{AB}{2}=\frac{AB+6-AB}{2}=\boxed{3}$ [/hide]


trigonometry
algebra
polynomial
algebra unsolved

$ sin t=\frac{2ab}{a^2+b^2} $ , $ 0 < t < \frac{\pi}{2} $ , $ a, b \in Z^{+} $
  $ u_n = (a^2+b^2)^n. cos(nt) , n \geq 1 $
 Prove that : u_n is integer

Do this not follows immediately from Tchebytchev' polynomials?

Pierre.

Pierre: can you show exactly how to apply these Chebyshev's polynomial ? It's new to me and I tried to read about it but I must've missed something because I didn't find anythin useful:

http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html
http://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html

Hello,

I'm new on this forum and my english might be approximate, so please be tolerant with my spelling. ;)

I propose a solution:
since [tex]\sin t=\frac{2ab}{a^2+b^2}[/tex], we find [tex]\cos t=\frac{a^2-b^2}{a^2+b^2}[/tex]. with [tex]t[/tex] such that [tex]\cos t>0[/tex] (otherwize, change the sign).
As (using Newton's formula) [tex]\cos nt=\Re(e^{int})=\Re((\cos t+i\sin t)^n)=\sum_{k=0}^{[n/2]}(-1)^k\cos^{n-2k}t\sin^{2k} t[/tex], we have [tex]u_n=\sum_{k=0}^{[n/2]}(-1)^k(a^2-b^2)^{n-2k}(2ab)^{2k}[/tex] which is an integer.

Is that good?

Yours sincerly.

As it is said in the first link, the $n^{th}$ Chebychev polynomial $T_n $ of first kind is defined by $\cos (nt) = T_n (cos(t) )$.
This is a polynomial with integer coefficients and degree $n$.
Since $\cos (t) = \frac {|a^2-b^2|} {a^2+b^2}$, you get the desired result.

Pierre.

I should look at the posts before sending mine...always late. ;)

Pierre.


function
modular arithmetic
Euler
geometry
geometric transformation
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articles

Is φ(n) = n (1 - 1/p1) ... (1 - 1/pk) =  the formula given in wikipedia found on this url: http://en.wikipedia.org/wiki/Euler's_totient_function? (I'm sorry - I couldn't get the formula because of my poor latex).

And also, because I figured that this question belongs to this topic, I have another very quick, easy question.
Can you [b]Define: residue system[/b]
Thanks!!!!!!!!!

Euler's Totient Function is,

$ \varphi(n)\equal{}(p_1\minus{}1)p_1^{k_1\minus{}1}\cdots(p_r\minus{}1)p_r^{k_r\minus{}1}$

It can also be written as,

$ \varphi(n)\equal{}n\prod_{p|n}\left(1\minus{}\frac{1}{p}\right)$

where $ n\equal{}p_1^{k_1}\cdots p_r^{k_r}$ where $ p_m$ are distinct primes.

A residue system is a set of numbers $ a_0,a_1,...,a_{m\minus{}1}\pmod{m}$ that forms a complete set of residues modulo $ m$.

oh really?
It seemed that everyone on this forum knew mods and residue systems  :oops:
[i]Mods, please move to high school intermediate[/i]

Anyone? Mods - please move to high school intermediate.

[quote="mathemonster"]How can the computation for $ \varphi(n) \equal{} (p_1 \minus{} 1)(p^{k \minus{} 1})(p_2 \minus{} 1)(p_2^{k_2 \minus{} 1})....$
where the prime factorization of $ p \equal{} p_1^k*p_2^{k_2}.....$ be simplified to product notation? Is it because by moving an n outside the product we can factor it to $ (1 \minus{} \frac {1}{p_1})(1 \minus{} \frac {1}{p_2})....$
Are there any good example problems for eulers totient function?

How do you prove that $ \varphi(p^k) \equal{} (p \minus{} 1)(p^{k \minus{} 1})$?[/quote]
aops wiki may help here
yes ppl do know basic mods but the totient function is not BASIC mods. :P

[quote="mathemonster"]

How do you prove that $ \varphi(p^k) \equal{} (p \minus{} 1)(p^{k \minus{} 1})$?[/quote]

this is true iff p is prime
this is because all numbers not relatively prime to $ p^k$ are multiples of p so $ \frac{1}{p}$ of all natural numbers less than or equal to $ p^k$ are not relatively prime to $ p^k$ so muyltiply by $ 1\minus{}\frac{1}{p}$# to find the numbers that ARE relatively prime to $ p^k$

you can extend this to prove the formula for euler's totient function

I checked the aops wiki, but it doesn't help  :(

Also, does the aops wiki - "complete residue system" - contradict - http://www.cs.usask.ca/content/resources/tutorials/csconcepts/1999_7/tutorial/trail/tp09.html

Excerpted here:
After completing the previous trailpost, you know that j(n) is a number of positive integers that are relatively prime to and less than n. The production of the actual set of these numbers was described in previous trailpost and here we will examine why are we interested in this set, and in particular why it is important that the residue system is reduced.

Now, think of this:
Take the residue system of 5 (a prime), which is the set {1,2,3,4}, pick any number from it, say 3, and multiply it with each of the numbers from the set (including itself) to get this set of numbers: {3,6,9,12}. What are the remainders modulo 5 of these new numbers?

Let's put these numbers in a table, such that original set of numbers is in the first column, and the new set modulo 5 is in the second column:
  (mod 5) [1st column * 3] (mod 5) How you get the results:
1 3 1 * 3 = 3 º 3 (mod 5)
2 1 2 * 3 = 6 º 1 (mod 5)
3 4 3 * 3 = 9 º 4 (mod 5)
4 2 4 * 3 = 12 º 2 (mod 5)

Let's see the complete multiplication table modulo 5:
  (mod 5) 1 2 3 4
1 1 2 3 4
2 2 4 1 3
3 3 1 4 2
4 4 3 2 1

[b]It says here that the residue system mod 5 only has 4 elements, while aops says 5[/b]

basically, what happened in the multiplication table modulus ... to the row and column "0"?????

oh wait, it's okay to leave in or out
as various webistes show.
Some leave it out because the columns and rows involving zero are all zero's - basically, it doesn't reflect the pattern they want to show...

right?

A complete residue system includes $ 0$, though just like weird stuff happens with division by $ 0$ in the real numbers, inverses of numbers that are not relatively prime to the modulo do not exist.

So can anyone answer the question I posted on my first post?

Is φ(n) = n (1 - 1/p1) ... (1 - 1/pk) = the formula given in wikipedia found on this url: http://en.wikipedia.org/wiki/Euler's_totient_function? (I'm sorry - I couldn't get the formula because of my poor latex).

And just another quick question related to mods:
Is http://www.cs.usask.ca/content/resources/tutorials/csconcepts/1999_7/tutorial/trail/tp10.html

wrong when they say that for the euler theorem, "When you take a number a, which is less than another number R, and the greatest...", that a [b]has[/b] to be less than another number R because I checked on Aops wiki and wikipedia and it says nothing about a being less than "another number R?"

I would appreciate and time and effort spent on this matter.

[quote="theax"]So can anyone answer the question I posted on my first post?

Is φ(n) = n (1 - 1/p1) ... (1 - 1/pk) = the formula given in wikipedia found on this url: http://en.wikipedia.org/wiki/Euler's_totient_function? (I'm sorry - I couldn't get the formula because of my poor latex).[/quote]

Yes, except the wikipedia article uses product notation, $ \phi(n) \equal{} n \left(1 \minus{} 1/p_1\right) \cdots \left(1 \minus{} 1/p_k\right) \equal{} \prod_{p \in \mathrm{primes},p|n} \left(1 \minus{} \frac {1}{p}\right)$, where the product symbol indicates to multiply each of the terms together, running through all prime $ p$ that divide into $ n$.

[quote="theax"]And just another quick question related to mods:
Is http://www.cs.usask.ca/content/resources/tutorials/csconcepts/1999_7/tutorial/trail/tp10.html

wrong when they say that for the euler theorem, "When you take a number a, which is less than another number R, and the greatest...", that a [b]has[/b] to be less than another number R because I checked on Aops wiki and wikipedia and it says nothing about a being less than "another number R?"

I would appreciate and time and effort spent on this matter.[/quote]

That condition is unnecessary, so I have no idea why it is there; the important criteria is that $ \text{gcd}\,(a,R) \equal{} 1$. Because congruences may be multiplied, (for arbitrary integer $ t$) multiplying the congruence $ [a \plus{} tR] \equiv a \pmod{R}$ with itself $ \phi(R)$ times, we have $ [a \plus{} tR]^{\phi(R)} \equiv a^{\phi(R)} \equiv 1 \pmod{R}$.

ummm...
doesn't wikipedia have n * \prod_{p \in \mathrm{primes},p|n} \left(1 - \frac {1}{p}\right)?
You just wrote this \prod_{p \in \mathrm{primes},p|n} \left(1 - \frac {1}{p}\right) without the n times in front of it???

really, really hard challenge: IF they are indeed equal, then prove it.

My copy-paste skills fail   :huh: . There is no proof required; the two denote the same exact product, just the $ \prod$ symbol is easier to write notation-wise, and means to multiply out all of those terms. Or, if you're asking to show that is equivalent to $ \varphi(n)$, see the [[Principle of Inclusion-Exclusion]] argument on the AoPSWiki article: [[Euler's totient function]].

?????
(the code below is just your formula in your post)

[quote]Yes, except the wikipedia article uses product notation, , where the product symbol indicates to multiply each of the terms together, running through all prime  that divide into .[/quote]

[code]\phi(n) = n \left(1 - 1/p_1\right) \cdots \left(1 - 1/p_k\right) = \prod_{p \in \mathrm{primes},p|n} \left(1 - \frac {1}{p}\right)[/code]

[b]So they are equal, right?  :idea:
Just a quick thought - shouldn't there be a "n" times symbol before the product symbol???[/b]

by the way, how would you latex the formula?

To latex a formula, you put code and then surround it with dollar signs.

For example:

[code]$x \cdot y$[/code]

comes out as:

$ x \cdot y$

To learn $ \text{\LaTeX}$, go to the $ \text{\LaTeX}$ forum.

thanks a lot!

So is it like this:

$ \phi(n) \equal{} n \left(1 \minus{} 1/p_1\right) \cdots \left(1 \minus{} 1/p_k\right) \equal{} \prod_{p \in \mathrm{primes},p|n} \left(1 \minus{} \frac {1}{p}\right)$

Cool!

[hide="click to reveal content"]$ \phi(n) \equal{} n \left(1 \minus{} 1/p_1\right) \cdots \left(1 \minus{} 1/p_k\right) \equal{} \prod_{p \in \mathrm{primes},p|n} \left(1 \minus{} \frac {1}{p}\right)$
[/hide]


analytic geometry
function
geometry
3D geometry
sphere

In the xy plane, line l has a lsope of 3/2 and contains the points with coordinates (1,m) and (5,r). What is the value of r-m?

A ball rolls off a shelf 4 feet above the floor. Until the ball hits the floor, its height, in feet, afret t seconds is given by the function h(t)= 4-16t[size=59]2[/size](squared). The ball will hit the floor how man seconds after rolling off the shelf?

a bag contains a total of 350 cubes and spheres, 200 of which are red and the remainder of which are blue.  There are 110 red spheres and 75 blue cubes in the bag. If a  :spider: reaches into the bag and randomly selects a cube by [u]feel[/u], what is the probabability he will pull out a red cube?

In the xy plane, the graph of y=ax[size=59]2[/size](squared)+b wherea and b are positive constants, will be translated 2 units to the right. The resulit will be the graph equation...

[quote="roberto_y_1090"]
In the xy plane, the graph of y=ax[size=59]2[/size](squared)+b wherea and b are positive constants, will be translated 2 units to the right. The resulit will be the graph equation...[/quote]
Remember, " If you add to y, go high. If you add to x, go west". We are going east, so we subtract 2 from x. The equation is $ a(x\minus{}2)^2\plus{}b$


AMC USA(J)MO USAMO algebra polynomial function

If P(x) denotes a polynomial of degree n such that P(k) = k/(k+1) for k = 0, 1, 2, ... , n, determine P(n+1).

hint: [hide]what can you tell me about f(x) = (x+1)*P(x) - x?[/hide]

Using Zs's hint: f(x) = (x+1)*P(x) - x = 0 for all f(x).. i.e.
f(n) = (n+1)(n/(n+1)) - n = 0. So f(n+1) = 0 = (n+2)(p(n+1)) - (n+1)... so p(n+1) = (n+1)/(n+2)?

But does this hold? What Im asking is that can I declare f(x) = 0 for all x, or only that x in 0 to n?

Only for x from 0 to n.  Hint: [hide]And f has degree n + 1, so you need one more value to pin down the function f.  Can you think of one more value of x where you know f(x)?[/hide]

[hide]Many thanks to zscool and Ravi B!

f(x) = (x+1)*P(x) - x, it's  n+1 degree with roots 0,1,2,...x.

we could rewrite f(x) as cx(x-1)(x-2)...(x-n), where c is a constant.

when x=-1, we have f(x)=1.

or -1c(-2)(-3)(-4)...(-n-1)=1.

c=1/(-1)^(n+1)* (n+1)!

x=n+1, then f(x)=cn!=1/(n+1) when n=odd, -1/(n+1) when n=even.

set it equal to (n+2)P(n+1)-n-1=1/(n+1) or -1/(n+1) to solve for P(n+1)

t[/hide]

Sorry, I accidentally put down the wrong name.
Zabelman and zscool are pretty similar names in someway(both start with z)


function
calculus
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True or False:

1. If f(x) is differentiable at x=c, then f '(x) is also differentiable at x=c.

2.  If a graph of a function has a tangent line at a point, then the function is differentiable at that point. ( I was thinking about when the tangent line has undefined slope. Is it still differentiable?)

3. Every cubic has exactly one point of inflection.

4. The graph of f(x)= 1/x has an inflection point at x=0.

1. False. Here's one example: $ f(x) \equal{} \begin{cases}x^2, & x > 0 \\
0, & x\le0\end{cases}.$

That's differentiable everywhere but has no second derivative at $ 0.$

Or if you like more outrageous examples:

$ g(x) \equal{} \sum_{k \equal{} 1}^{\infty}\frac {3^k}{4^{2k}}\,\sin(4^kx)$

$ g$ is in $ C^1,$ meaning that $ g'(x)$ exists and is continuous everywhere, but $ g''(x)$ exists nowhere at all.

2. If you accept that a graph could have a vertical tangent, then $ f(x) \equal{} x^{1/3}$ serves as a counterexample and makes this false.

3. This one is true. The proof should be fairly much mechanical.

4. False, and meaningless. If the graph doesn't have a point at all at $ x \equal{} 0,$ how could it have an inflection point? It is true that the concavities are opposite for $ x > 0$ and $ x < 0.$

i have another question, is there always a local extrema for every critical point?

Can someone give me an example of function f such that f'(x) exsists for every x in domain, but f' is not continuos function

No -- differentiability at a point implies continuity at the point.  $ f'(x)$ exists (by definition) if and only if the limit $ \lim_{h \to 0} \frac{f(x \plus{} h) \minus{} f(x)}{h}$ exists.  If $ f$ is not continuous at $ x$ then $ \lim_{h \to 0} f(x \plus{} h) \minus{} f(x)$ is not zero, so $ \lim_{h \to 0} \frac{f(x \plus{} h) \minus{} f(x)}{h}$ cannot exist.

So, differentiable function is always $ C^{1}$ ?
You missunderstood, I asked f' not to be countinuos, not f

No: $ C^1$ is "differentiable with a continuous derivative."  Kent gave two examples of differentiable functions with discontinuous derivative; another such function is $ x \cdot |x|$.  Every differentiable function is $ C^0$ (continuous).

No, that's not what rafaelm asked.

A function like $ x|x|$ does have a continuous first derivative.

Moving one notch on the smoothness scale, what about $ |x|?$ It doesn't have a continuous derivative, but then there is a point at which it doesn't have a derivative at all, so that's not the example rafaelm wanted.

The more-or-less standard example for what rafealm wanted is this:

$ f(x)\equal{}\begin{cases}x^2\sin\left(\frac1x\right),&x\ne0\\0,&x\equal{}0\end{cases}.$

This is differentiable everywhere (note that $ f'(0)\equal{}0$) but the derivative is not continuous at zero.

[quote="Fermat's Little Turtle"]Nope. Consider $ f(x) \equal{} x^3$.

$ f'(x) \equal{} 3x^2$ therefore $ f'(x_0) \equal{} 0$ when $ x_0 \equal{} 0$.

This isn't actually a local extrema, its an inflection point. The slope is indeed 0 there easily deduced by the graph.[/quote]
The function $ g(x)\equal{}x\minus{}\sin x$ on $ \mathbb{R}$ has infinitely many points like that - infinitely many critical points and no local extrema at all.

Whereas the function $ f(x)$ in my most recent post on this topic - the differentiable function with a discontinuous derivative - has a critical point at zero that is not a local extremum, but in a very strange fashion.


Find all function f:Z-->R:
$ f(0)\neq 0$
$ f(1)=\frac{5}{2}$
f(m+n)+f(m-n)=f(m)f(n) with m and n are integer numbers  :P


trigonometry
real analysis
real analysis unsolved

[color=darkblue]Let $ (u_n)$ is a number sequence, let formula of it is $ u_n \equal{} \frac { \minus{} 3 \plus{} \frac {1}{n \plus{} 1}}{2 \plus{} \sin {n}}, n \ge 1$. Prove that exist $ M, m \in R$ so that $ m \le u_n \le M$. Find $ M$ and $ m$ ?[/color]

Can't you take superior and inferior limits?

EDIT: Well that was helpful, wasn't it?  :rotfl:

numerator is a convergent sequence and 1 < 2 + sin n < 3

[quote="kenn4000"]$ \minus{} \frac {5}{2} > u_n \ge \minus{} 1$[/quote]  Is this line a joke? $ \minus{} 2.5 < \minus{} 1$, and also $ u_n$ can lie on either side of both $ \minus{} 2.5$ (consider $ u_4, u_5$) and of $ \minus{} 1$ (consider $ u_2, u_3$).  I think you should consider more seriously on what the bounds on the numerator and denominator imply for the bounds on the whole expression.

$ \sup \frac { \minus{} a_n}{b_n} \le \frac { \minus{} \inf a_n }{\sup b_n}$

$ \inf \frac { \minus{} a_n}{b_n} \ge \frac { \minus{} \sup a_n} {\inf b_n}$

$ a_n \equal{} 3 \minus{} \frac {1}{n \plus{} 1} \to \sup a_n \equal{} 3, \inf a_n \equal{} \frac {5}{2}$
$ b_n \equal{} 2 \plus{} \sin n \to \sup b_n \equal{} 3, \inf b_n \equal{} 1$

$ \inf u_n \ge \minus{}3, \sup u_n \le \frac { \minus{} 5}{6}$

$ \minus{}3 \le u_n \le\frac { \minus{} 5}{6}$

fixed!

This time you're claiming that a sequence is bounded above its sup and below its inf?  :o  Perhaps you should have finished reading the second sentence of my post ;) -- neither $ \minus{} 1$ nor $ \minus{} 2.5$ is a bound on the values of this sequence.  (Hint: your first and second lines are only true for positive sequences.)


[hide] A) because you know the top angle of the small triangle is Beta, and you know the right angle of the small triangle is theta. Since alpha is equal to gamma, the angle below alpha must equal delta. Therefore the left angle of the small triangle is measure delta. Because the sum of the angles of a triangle is 180, beta plus theta plus delta equals 180.[/hide]

I apologize in advance for the lack of symbols and confusing explanation. If needed, I can draw a picture.  :D

[hide="Some symbols?"]Note that by applying vertical angles to $ \beta$ and $ \delta$ and also alternate interior angles to $ \theta$ (so that $ \theta \cong$ the angle opposite $ \tau$, we can see that angles $ \cong$ to $ \beta$,  $ \delta$, and $ \theta$ form a triangle. Thus $ \boxed{\beta + \delta + \theta = 180^{\circ}}$ .[/hide]

21
E assuming that NG means none of them.
A doesn't work because all you can get is alpha + beta is 180, not that they equal each other
B doesn't work since adding beta to both sides makes
alpha+beta=90, which we know isn't true
C doesn't work because alpha=y.
D doesn't work because no angle measurements have been given
So it must be E.

He said just number 22 in the first post.

distracted, how do you get symbols into your text? Is it latex?

And I do apologize for not knowing the names of the properties.

[quote="hm29168"]
distracted, how do you get symbols into your text? Is it latex?
[/quote]

Yeah, it is. I'm still learning, so I post slowly and I have to fix my mistakes. :oops:

Using vetical angles we find that the three angles are the angles in the triangle in the diagram.
So yeah, they add up to 180.

Thanks guys, after spending so long on figuring it out (20 min.) I finally understnad that using transveral properties through parallel lines the three angles can form a triangle (like some of you said) and thus add to 180.


geometry
geometric transformation
reflection
analytic geometry
graphing lines
slope
trigonometry

Lets say i have a line with equation y = (3/4)x and i wish to reflect this line across y = (-1/2)x. The equation of the reflected line is the y axis O_O.

I know that to find the points reflected you find the perpendiculars etc etc. But my question is whether there is a way to find the slope of the reflected line by using the slopes given. (by this i mean without trig)

Thanks

If $ s$ is the slope of the symmetrial axis, and $ k$ the slope of the line you're reflecting, then the slope $ k'$ of the image is

$ k'\equal{}{ks^2\plus{}2s\minus{}k\over 1\plus{}2ks\minus{}s^2}$

which in this case yields $ k'\equal{}\infty$, and that's $ y$-axis.

(The deriving of the above must include trigonometry, so I'll skip it here.)

[quote="Farenhajt"]If $ s$ is the slope of the symmetrial axis, and $ k$ the slope of the line you're reflecting, then the slope $ k'$ of the image is

$ k' \equal{} {ks^2 \plus{} 2s \minus{} k\over 1 \plus{} 2ks \minus{} s^2}$

which in this case yields $ k' \equal{} \infty$, and that's $ y$-axis.

(The deriving of the above must include trigonometry, so I'll skip it here.)[/quote]

Thank you, i had something similar, but i didn't know if it was right.

Then you look for a $ k'$ such that $ 1 \plus{} 2k's \minus{} s^2 \equal{} 0\implies k' \equal{} \frac {s^2 \minus{} 1}{2s}$.

[quote="Complex_Ninja"]But what if k was undefined?.....[/quote]

You mean if $ k \equal{} \infty$, i.e. if the original line is parallel to $ y$-axis? Then just divide both the numerator and the denominator by $ k$:

$ k' \equal{} \frac {s^2 \plus{} {2s\over k} \minus{} 1}{{1\over k} \plus{} 2s \minus{} {s^2\over k}}$

For $ k \equal{} \infty$, this simplifies to $ k' \equal{} {s^2 \minus{} 1\over 2s}$, which is of course the same as in math154's post.

Similar procedure if $ s\equal{}\infty$ - divide by $ s^2$:

$ k'\equal{}\frac{k\plus{}{2\over s}\minus{}{k\over s^2}}{{1\over s^2}\plus{}{2k\over s}\minus{}1}$, which simplifies to $ k'\equal{}\minus{}k$ (and that is logical, if you analyze the graph)


counting
distinguishability
geometry
3D geometry
octahedron
rotation
geometric transformation

"25. Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron.  How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other)."

I have the solution to this, but I don't quite get it =/.  It talks about faces that share an edge with a side that we set to be a certain color.  It says that we take a combination of colors for those sides rather than a permutation.  I know it has to do with rotations, but I don' see how I can rotate the octahedron to make it exactly the same.

The picture is of an octahedron, staring directly at one face.  If we rotate it 120 degrees, every vertex, face, and edge is in the same place as a vertex, face or edge was before.  If we were thinking of coloring the octahedron, we see that these two colorings, which are different from our point of view, are in fact just rotations of each other.

Think of a dice, the configuration is opposite sides are 1-6, 5-2 and 4-3. If we make it 4-5,6-1,3-2, its different, but if we do use the 1-6,5-2,4-3 configuration, we can make 1 the top on and 6 the bottom 1, 5 right in front of us and 2 behing the cube, finally 4 and 3 can go on differently making 2 cubes for each configuration of sides...

joel what program are you using to make those diagrams? they are nice...

Geometer's Sketchpad, v. 4.00.

Aren't they great?

Actually, I have to copy them into Paint to save as a .jpg, since it will only save in its own format.  But it's great for making any 2-D geometric object, and for some 3-D stuff.

Actually, I reckon that when it comes to simple graphics with THE 16 colours, paint saves in GIF better than it does with JPEG. No loss and small files. :)

GSP also forces you to learn basic principles of straightedge and compass diagrams.

can i pick that program up in a comp store or do I probably have to order it from the company?

I can't remember seeing it in computer stores, but I don't exactly hang out in any computer stores.  You [i]might[/i] be able to find it in a store, but I wouldn't bet on it.  I'd go through the web if I were purchasing it.

coloring the faces of an octahedron is the same as coloring the vertices of a cube. There are 8!/3 ways to do this, since there are 3 equivalent cubes (including the original) to any cube we make.

I think you need to divide by 8.  There are 8*3 equivalent cubes for each cube -- 8 for picking one vertex, 3 for rotations.

Ok I know how to fix it. I made a little error in my presentation of it.

If we fix a particular vertex to be a particular color, then there are 7! ways to color the rest of the vertices. However, there are 3 cubes in each equivalence class, (or orbit). So its 7!/3, not 8!/3.


geometry
projective geometry
geometry unsolved

Can you please tell me something about projective geometry. Some good books or something else. Every help is going to be good. Tnx!

try geometric transformations 3 by I.M.Yaglom

Hi
Is it or anything else you know available online?

what do u want exactly
just projective  geometry or olynpiad projective geometry

Olympiad projective geometry but starting from the basics.

Then Coxeter would be a better option for you than Yaglom

Tnx, but what about just projective geometry? Wich books?
Can u tell me something about just projective geometry?
My task is to write an essey about projective geometry in mathematical camp Petnica, so any help is going to be great! :D

for that GT3 by IM..Yaglom would be great (in fact the best)


AMC AIME geometry search videos Support conics

[url]http://www.funtrivia.com/private/main.cfm?tid=960[/url]

Click the link above. Click it every day. See if you can get on the high score list! You are rewarded for time and accuracy!

And I guess anyone who is in Whitehorse's trivia thingy can use this as practice... or to estimate their ranking...

Just a reminder, you can play EVERY day. Scores become cumulative after a while.

[i]Actually, you don't get rewarded on time, except for some stuff.  Here's how it works-

There are quizzes that people like you and me write.  You can take them, and score 10 points for every question you answer correctly.

To write a quiz, you write one, and then send it to the editors so that they can proof-read it.  Then it goes online and other people can read it.

Then there are 3 timed tests.

The first one, you can only take once a day, you get 20 questions to answer and I don't think the high-scorers get anything.

The second one is hourly, and if you are in first place, you get a raffle ticket for a $30 amazon coupon.

There is another one that is hourly, but with easier questions.  I don't think there is a prize thingie for that one.

And to do all of the stuff, except for take the quizzes, you have to sign up for the thing.[/i]

- This is what you can do with the rest of the website.  I haven't been on for a long time.  I feel so embaressed though for making that mistake

Darn, I would've gotten 9 if I didn't click the wrong letter on number 3.

This site hates me.  Why won't it add my scores!   :(  :D

I'm switching between the "Easy/Intermediate" and the "Intermediate/Hard" question sets every day/every few days... so some days, you might get a low score and others you might get a high score...

If anyone has any requests for question sets, post them here and I'll make it the topic of a day once (or maybe more than once if it is something that is commonly asked for).

All right, a perfect score today!  And fastest time!

Then again, it WAS easy/intermediate.  Oh, well...

[quote="Daily Trivia Game"]The Smartest Player today is currently Frankthetank.[/quote]

There were only three players, though.

Forget mine - this is funner.

Check this out: [i][b]The Smartest Player today is currently sax_maniac88.[/b][/i]

YAYAYAYAYAYAYAYAYAYAY - That's me. Booyah. ALthough, my questions were pretty easy.

I did mine in the least amount of time, so ha!

Lol someone actually got a score of zero. That's pretty sad, especially considering they got 6 right.

Anyone who is confused about the scoring system.

(your score out of 10) * 100 - time taken in seconds

So if you got 9 out of 10 in 86 seconds, your score would be

9 * 100 - 86
900 - 86
814

So if you got 6 right and took lets say, 600 seconds, you would have 0 - so not impossible. - that's only 10 minutes - not like i would spend that much time on trivia anyway.

[quote]You have earned 38 points so far. This ranks you overall # 1 of 19 point scorers.[/quote]

heh...that's about me!!!

I just did today's test (first one I did). 10/10. Hm. Those were easy.

famous ppl suck. bring back history and geography.

eh..  I got 115 seconds today because my internet suddenly got really slow.  :(

easy today, but aaaaaaaaaaaaaargh, I missed one, so =  automatically 9th

yea, it's hard when if you miss 1 or if you read slow and get a time in the 50's, then you're automatically out of the top 10. maybe, i'm just bad at this in general, and no category would help me.  :?

Arrgh...I won't be able to play for a few days...there goes this month.

Yeah, that happened to me too.

Had a trivia tournament that lasted a weekend.

Are ethnic parties new phenomena in Israeli politic?

            ofcourse
            yes
            no
            sure

hmm :lol:

but then again, today the website is lagging so danm is off the hook ;)

[quote="aznphatso"]I'm sick and tired of people googling in this game. I know Miraculous has stopped, but the time he did cost me a win. Everyone with times in the 100s are really suspicious, like danm is today.[/quote]How does one google anyways?  For me at least, it would take more than one minute to find the answers to many of the questions.  :?

Seems to be.  I've been trying for half an hour to get in--it just keeps coming back, "Cannot find server" for any of the places on the site I try (http://www.funtrivia.com, the main AOPS trivia page, the message board, the actual game, cumulative scores, past winners).  I guess whoever played before they went down wins  ;)

too bad it is down. i was gonna win today. darn.  :weightlift:

looks like i'm doing pretty good so far this month :D


Let $ f(x) \equal{} \frac {x(1 \minus{} x)sin(\theta)}{2 \minus{} x(1 \plus{} cos(\theta))}$ where $ \theta$ is rational in the interval $ (0,1)$.  Show that there are infinite values of $ f(x)$ on the vertical line over $ x$.

what do you mean exactly by the vertical line over $ x$? do you mean that for a fixed $ x$ we can vary $ \theta$ to get infinitely many values of $ f(x)$ or what?


Euler
geometry
3D geometry
floor function

Prove that $\forall n \in N$  $n^2-n+41$ is a prime number

[quote="Gyan"][quote="10000th User"]Maybe he meant prove by finding the first 'counterexample' :D

Gyan: is it true that $n^2-n+41$ are primes for $1\le n\le 40$?[/quote]
Yes, Of course, it is one of the famous one (Connected with Ramanujan). (In fact  you get prime number for  $-40 \le n \le +40$  !!![/quote]
I think Euler first stated it.

[quote="Arne"]Nice joke.

I just hope Fermat- Euler intended it to be a joke... :)[/quote]

yes it's just a joke  :rotfl:

[quote="leepakhin"]I think Euler first stated it.[/quote]
Yes, and probabbly it was known even before, Ramanujan was facinated with this  and found some other similar cases. Even Arthur C. Clarke  in one of  his SF novel, uses this  for one of his character to construct an easy to remember password (for launching codes of a 50 megaton nuclear device, of course) by using these easy to remember yet difficult to guess  40 prime numbers...

Also:  No Joke but there exists a  real number $a>1$ such that $[a^{3^n}]$   for $n=0,1,2,3 \dots$ is always a prime..
Edited Later to include "[]" around.
(Value of a is about 1.3063778838 ...)

If you mean that there are infinetly many numbers where the expression is prime...

[hide]let $n=41k$, $(41k)^2-41k+41=41(41k^2-k+1)$. Hope that's how you do it. [/hide]

What exactly are you trying to prove 236factorial?? :huh:

Ah yeah.. heard of this one... :)

It's the [url=http://mathworld.wolfram.com/MillsConstant.html]Mills' constant[/url] ;)

Ah, I see! I was about to write that if $a^{3^n}$ is prime then $a^{3^{n+1}}$ is the cube of a prime... which is true for gyans statement which mightely compused me... You might want to edit it to $\lfloor a^{3^n}\rfloor$

(which, by the way, is one of the coolest things i have ever seen! Thanks gyan!)

I think this estimation was obtained, by assuming the Riemann Hypothesis... ;)
At last this is what I've read in a paper in the Journal of Integer Sequences.


I got this invitation to take the usamts yesterday.

I was simply wondering what the fourth (the one on the bottom right with the rectangles in the squares) proof without words proved.

I would appreciate any help.

It shows the fact that
$ \sum_{i \equal{} 1}^n i^3 \equal{} \left(\sum_{j \equal{} 1}^n j\right)^2$.
Do you see why?  There is 1 1x1 square, 2 2x2 squares, 3 3x3 squares... adding them together, you get the square with sides 1+2+3+4... n.

Oh. Cool. Thanks. That's really clever. I doubt I would have figured out what that picture meant


how do you do this??

[quote=the problem]Five philosophers sit around a circular table. Each philosopher spends his life alternatively thinking and eating. In the centre of the table is a large plate of noodles. A philosopher needs two chopsticks to eat a helping of noodles. Unfortunately, as philosophy is not as well paid as computing, the philosophers can only afford five chopsticks. One chopstick is placed between each pair of philosophers and they agree that each will only use the chopstick to his immediate right and left.[/quote]

[quote="shinwoo"]Five philosophers sit around a circular table. Each philosopher spends his life alternatively thinking and eating. In the centre of the table is a large plate of noodles. A philosopher needs two chopsticks to eat a helping of noodles. Unfortunately, as philosophy is not as well paid as computing, the philosophers can only afford five chopsticks. One chopstick is placed between each pair of philosophers and they agree that each will only use the chopstick to his immediate right and left.[/quote]

I don't really think this is problem unless someone can make one out of it. In fact, this is a common problem when you're learning computer programming. I know that in my Java book, it has this example to simulate learning to program threads.

how can they live without eating..

somehow they have to constantly eat...

how can we make the constantly without stopping?/ that's the question

i think its some kind of problem that they got from a technology book.
coz unfortunately this year i have technology
and we have to solve like problems similar to these! :)
but still i dont no wat they r asking.

Ignoring the Chop sticks, you can say, how many ways can 5 people can sit in a circular table?

No...

we need the chopticks...

I'm taking this programming course and my techer told me to solve it...( and she also said it required Algorithms)

What problems are you having? I know one of them is deadlock and another starvation in this particular probem. And what programming language are you learning?

C++...

I'm trying to make the philosophers survive.(never dying ) :D

I'm unfortunately not too familiar with C++ (know enough, but not enough to do anything productive with).

However, if you can program, the basic concept is that you can lengthen the wait time between philosophers attempting to pick up a chopstick as to prolong deadlock.

However, I have also read that the best way to avoid deadlock is numbering the chopsticks 1 to 5 or whatever many there are. Then, force the philosopher to pick, of the two chopsticks next to them, the lowest numbered one first before the higher numbered one. This should prevent deadlock.

Professors 1 and 3 eat
Professors 2 and 4 eat
Professors 3 and 5 eat
Professors 4 and 1 eat
Professors 5 and 2 eat

repeat

:P

Oh, GEEZE!!!

I have a problem...

I think Treething's solution is correct, but by the fact that many great mathematicians tried to solve this problem for over 200 years, it doesn't look right..

Geeze, it's so confusing.. It's already have been a week,

Or they all just use their hands, but I don't think your teacher would accept that :P .

You need to ensure that each philosopher eats the most he can, and this is all done _simultaneously_.

Furthermore, the computer science aspect of it is that each person should use the same algorithm. This algorithm is almost certainly probabilistic, a consequence of using threads. You will not get the same answer back each time you run the program.


Scott bought a home three yrs ago for 85,000 dollars. They put 15,000 dollars down at the time of the purchase. They have been able to pay ahead on the loan at times and now have a balance of 58,320 dollars. What will the interest be on the loan this month if the contract calls for monthly payments in the amount of 1,000 dollars and an interest rate of 8 percent?

[hide]If the $8$percent is the annual rate, then the monthly rate should be $\frac{.08}{12}\approx .0067$. Then the monthly interest on a balance of $58,320$ should be about $388.80$.[/hide]

Why did you divide .8 by 12?

What does 12 mean here?


find all pairs of integers (x,y) satisfying the equality

                  \[y(x^{2}+36)+x(y^{2}-36)+y^{2}(y-12) = 0\]

[quote="kimnimalar"]find all pairs of integers (x,y) satisfying the equality
\[y(x^{2}+36)+x(y^{2}-36)+y^{2}(y-12) = 0 \]
[/quote]

one more problem :
find all pairs of integers (x,y) such that

                    \[(x-y)^{2}(x^{2}-y) = 4x^{2}y\]

For the first problem, make a quadratic equation based on $x$
$y \cdot x^{2}+(y^{2}-36) \cdot x+y(y-6)^{2}= 0$
Then, $D \geq 0 \Rightarrow (y-6)^{2}\cdot (-3y^{2}+12y+36) \geq 0 \Rightarrow-2 \leq y \leq 6$.
By checking one by one the value of $y$ in that range, we get all the solutions :
$(x,y) = \{ (-8,-2),(0,0),(1,4),(4,4),(0,6) \}$

For the second problem. Consider $x,y$ nonzero integers first, let $d = (x,y), x = dp, y = dq, (p,q) = 1$
We have $(p-q)^{2}\cdot (dp^{2}-q) = 4p^{2}q$. [1]
Since $(p-q,p) = (dp^{2}-q.p) = (p,q) = 1$, we can conclude that $|p| = 1$ and thus, $d = 1$.

If $p = 1$, [1] becomes $(1-q)^{2}\cdot (1-q) = 4q \Rightarrow q^{2}-3q^{2}+7q-1 = 0$ which has no integer solutions.

If $p =-1$, [1] becomes $(1+q)^{2}\cdot (1-q) = 4q \Rightarrow q^{3}+q^{2}+3q-1 = 0$ which has no integer solutions.

Now, if $x = 0$, then $-y^{3}= 0 \Rightarrow y = 0$.
And if $y = 0$, then $x^{4}= 0 \Rightarrow x = 0$.

So, the only solution is $(x,y) = (0,0)$

NB : For the two cubic equations, we just need to check if $q = 1$ or $q =-1$ which is easy to do..  :)


Arătaţi că, oricare ar fi $a,b,c>0$, are loc inegalitatea: \[(a+b+c)\bigg( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\bigg)\geq 9+\frac{4(a^{2}+b^{2}+c^{2}-ab-bc-ca)}{3(ab+bc+ca)}.\]

solutia care urmeaza este una urata
[hide="solutie"]scriem inegalitatea astfel
$\frac{\sum a\cdot\sum ab}{abc}\geq 5+\frac{4\left(\sum a\right)^{2}}{3\sum ab}$
dupa ce aducem la acelasi numitor si il eliminam
${3\sum a\cdot\left(\sum ab\right)^{2}\geq 15abc\sum ab+4abc\left(\sum a\right)^{2}}$
sau, echivalent,
$3\sum_{sym}a^{3}b^{2}+15\sum a^{2}b^{2}c+6\sum a^{3}bc \geq 23 \sum a^{2}b^{2}c+4\sum a^{3}bc$
adica,
$3\cdot\big[(3,2,0)\big]+\big[(3,1,1)\big]\geq 4\cdot\big[(2,2,1)\big]$
adevarata conform muirhead.[/hide]

\[\, \Longleftrightarrow \, \, \sum_\textrm{cyc}\left( \frac{a}{b}+\frac{b}{a}-2 \right) \geq \frac12 \cdot \frac{ 4 \sum_\textrm{cyc}(a-b)^{2}}{ 3 \sum_\textrm{cyc}ab}\]  \[\, \Longleftrightarrow \, \, \sum_\textrm{cyc}(a-b)^{2}\left( \frac1{ab}-\frac12 \cdot \frac4{3 \sum_\textrm{cyc}ab}\right) \geq 0 ,\] care e clar adevarata. In concluzie, puteam inlocui $\frac43$ cu $2$, nu? Sper ca n-am facut vreo greseala.

da, se poate inlocui $4/3$ cu $2$, dar eu tot cu muirhead o rezolv :)


is there a set/bundle where you can get all of the books in the intro series.

if no, i would suggest it.

Maybe with a discounted price too.
That is a very good idea and I think a lot of AoPSers would like to have them all.
:)


ceiling function
floor function
logarithms
algebra unsolved
algebra

Prove or disprove: For all positive integers $n$,
\[\left \lceil \frac{2}{2^{1/n}-1}\right \rceil = \left \lfloor \frac{2n}{\log 2}\right \rfloor.\]

A counterexample is $n=777451915729368,$ for which
\[\left \lceil \frac{2}{2^{1/n}-1}\right \rceil = \left \lfloor \frac{2n}{\log 2}\right \rfloor+1.\]

Further examples are listed in [url=http://oeis.org/A129935]A129935[/url] @ OEIS.


inequalities
induction
calculus
derivative
inequalities proposed

[quote="ehku"][color=blue]For $a,b,c,d \ge 0$, prove that
$\sqrt[3]{\frac{2a}{a+b}}+\sqrt[3]{\frac{2b}{b+c}}+\sqrt[3]{\frac{2c}{c+d}}+\sqrt[3]{\frac{2d}{d+a}}\le 4$[/color][/quote]
It's really nice and hard! :)

[quote="ehku"]The inequality
$\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\le \frac{3\sqrt{2}}{2}$
where $a,b,c \ge 0$
is well-known.
I posted in this forum the inequality
$\sqrt[7]{\frac{a^{4}}{(a+b)^{4}}}+\sqrt[7]{\frac{b^{4}}{(b+c)^{4}}}+\sqrt[7]{\frac{c^{4}}{(c+a)^{4}}}\le \frac{3}{\sqrt[7]{16}}$ (unsolved)
All of the above 3 inequalities are special cases of the following inequality :
[color=blue]$\frac{1}{(1+a_{1})^{k}}+\frac{1}{(1+a_{2})^{k}}+...\frac{1}{(1+a_{n})^{k}}\le \max \left(n-1;\frac{n}{2^{k}}\right)$
for all positive numbers $a_{1},a_{2},...,a_{n}$ satisfying $a_{1}a_{2}...a_{n}=1$ and positve number $k$[/color].

This is a very very hard problem. It can be solved by [color=blue][b]GI-method [/b][/color] (you can find it in hungkhtn's book).[/quote]

Yes, I solve your 4-variable inequality through the n-variables inequality by general induction. It is seemingly very hard to prove the 4-variable problem itself.

[quote="ehku"]Dear hungkhtn,
we needn't GI for my 4-variable inequality (only for general case  :P ).

I hope someone can prove it before I post my solution.

(@hungkhtn : the first inequality also can be proven by SMV-Theorem. And this is a interesting application for your method  :)  )[/quote]

I guess you use SMV to prove case $n=4$. But I am really confused of how you proved the following statement easily (without derivative)
\[\frac{1}{(1+a)^{k}}+\frac{1}{(1+b)^{k}}\le \frac{2}{(1+\sqrt{ab})^{k}}\]
if $a,b>0$ and $ab<\le 1$. I mean an easy way...

[quote="ywl"][quote="ehku"]T
All of the above 3 inequalities are special cases of the following inequality :
[color=blue]$\frac{1}{(1+a_{1})^{k}}+\frac{1}{(1+a_{2})^{k}}+...\frac{1}{(1+a_{n})^{k}}\le \max \left(n-1;\frac{n}{2^{k}}\right)$
for all positive numbers $a_{1},a_{2},...,a_{n}$ satisfying $a_{1}a_{2}...a_{n}=1$ and positve number $k$[/color].
This is a very very hard problem. It can be solved by [color=blue][b]GI-method [/b][/color] (you can find it in hungkhtn's book).[/quote]

I have no idea what general induction is, but 1/(1+e^x)^k has only one inflection point so it is enough to
consider n-1 variable equal.[/quote]
Why, ywl ? What do you apply ? Please.

well, (1/(1+e^x)^k)''=
k(k*e^x-1)
----------------------
4(cosh(x/2)^2* (e^x+1)^k

which has one inflection point at 1/k
using karamata for two variables and jensen,
n-1 variable equal can be assumed.

Let $a,b,c,d>0$ and $a+b+c+d=4.$ Prove that$$\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\leq\frac{4}{abcd}.$$
Let $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1.$Find the minimum value of $57(a+b+c+d)+\frac{1}{abcd}.$
[size=50](p/7104015930)[/size]

[quote]Let $a,b,c,d>0$ and $a+b+c+d=4.$ Prove that$$\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\leq\frac{4}{abcd}.$$Let $a,b,c,d>0$ and $a^2+b^2+c^2+d^2=1.$Find the minimum value of $57(a+b+c+d)+\frac{1}{abcd}.$
(p/7104015930)[/quote]bash with karamata and am gm
[hide=question1 ] on minima a,b,c,d>=4/4=1 thus on substituting 4<=4[/hide]
[hide=question2]on minima a,b,c,d i think we wil get 130 correct me if wrong :help: [/hide]


inequalities
LaTeX
inequalities proposed

Let x, y, z are real numbers from which at most one is negative.
(x, y, z are different from 0).
xyz=x+y+z.
Prove the inequality:
1/x+1/y+1/z>=sqrt(3)

hello, why dont you use $ \text{\LaTeX}$? do you mean the following problem
Let $ x, y, z$ are real numbers from which at most one is negative.
($ x, y, z$ are different from $ 0$).
$ xyz=x+y+z$.
Prove the inequality:
$ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq\sqrt{3}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, why dont you use $ \text{\LaTeX}$? do you mean the following problem
Let $ x, y, z$ are real numbers from which at most one is negative.
($ x, y, z$ are different from $ 0$).
$ xyz = x + y + z$.
Prove the inequality:
$ \frac {1}{x} + \frac {1}{y} + \frac {1}{z}\geq\sqrt {3}$.
Sonnhard.[/quote]
We have $ \frac {1}{x} + \frac {1}{y} + \frac {1}{z} > 0$
=>$ \frac {1}{x} + \frac {1}{y} + \frac {1}{z} \geq \sqrt {3( \frac {1}{xy} + \frac {1}{yz} + \frac {1}{zx})} = \sqrt {3}$


This should not be posted here . High School Basic is where this question should go to.

(i)square both sides we know $ |x \minus{} y| \le |x| \plus{} |y|$ is correct. And $ |x \minus{} y| \le |x| \minus{} |y|$ is wrong simply by put $ x \equal{} 1$ and $ y \equal{} 2$

(ii)Again, square both sides , the question is equivalent to $ x^2 \pm 2xy \plus{} y^2 \ge x^2 \minus{} 2|x\parallel{}y| \plus{} y^2 \Leftrightarrow |x\parallel{}y| \ge \pm xy$, which is obviously true.

I think a triangle can make this problem very easy

using this you can find other similar results~ :)

hello, it must be $ \sqrt{\left(|x|\minus{}|y|\right)^2}\equal{}\left\parallel{}x|\minus{}|y|\right|$.
Sonnhard.

[quote="Lakshya"]What's wrong with this?[/quote]

Mistake 1: In the third row of your solution, it should be $ \ge$ , not $ \le$ as $ \minus{}2xy \ge \minus{}2|x\parallel{}y|$

Mistake 2: $ |x|\minus{}|y|$ can be negative.


Stop writing the text of your posts in Latex!

[quote="dondigo"]Why, it's more clear[/quote]

Not you, felipesa.  Only equations should go in latex, not text.

yes i think if you don't type your text in $latex$ it willl be more clear. :D

Yes, that is what I said.  As for you, jensen, try posting only when you have something useful to add, not (maybe xxxx can solve this, he is so cool at geometry).  In particular, don't respond to this post.


Westpac is tomorrow, I hope the questions are not as hard as last year. I thought last year's intermediate was quite hard. Well anyway, good luck.

im also taking part here in malaysia  :)
good luck :lol:

are you doing junior or intermediate or senior?

im doing senior paper this year ( and this is my first time taking part since i miss last year  :P )

was it just me or was this year harder than normal?

I'm talking about the senior paper

lol i think the intermediate was a bit easier than last years one. lol bad luck.

are you serious? ...actually yea, last year's intermediate paper was pretty hard. but i kinda screwed up this year's too xD ahahah not my year in maths, is it?

i think the wording was weird in a few questions tho. took me a while to understand them (esp. q15 =S)

ok the train quesiton was really gay.

i dont know what a signpost and a tunnel have in common?! were we usppose to know? i dont know what the answer for that was, but i put 72km/h  :lol: , i hope my guess is right.

ahaha apparently the signpost thing was to show you how long the train was. like, it takes 15 seconds from when the front of the train passes the signpost til when the back does.

yea, i put 72 as well xD i think most people did

In the High School: Getting Started forum, I started a thread with all the questions, as more people tend to visit there than here.

Re: Intermediate paper being easier.
It would be your 2nd year of doing the paper, could that be a factor?

I found last year's paper slightly easier, got 75 marks, this year, I am getting 66, though, this was because of some silly, human errors.

Hopefully it all went well for you.

Hi to all of you,

I took the senior just a couple of hours ago (I'm in Varna, Bulgaria).
I don't think that any of the problems could be called "challenging" if it hadn't been for the time constraint. I was in a hurry to solve the first 20 well before the first half of the time, but then came this mental block and I spend very long time doing the same things over and over again before getting an answer (for example, the question about the number of different natural values of a for which sqrt(x+a)+sqrt(a-x)=a has at lest one real solution). Anyway, I think I've got 3 out of the first 20 wrong, so it'll be a pretty bad result :( considering that 2 years ago I was in the top 4% with 26 correct and 4 non-responded .

Um, discussing questions hours after it's done isn't exactly smart. One, time zones. Two, schools like ours cut people slack and let them do it the next day if they're not there.

The scoring system's screwed up now. I was one mark from a perfect score two years ago but I'm not going to be able to actually get a medal now partly because of the format. I'm not going to get over 100 this year.

Irene- I reckon last year's so worse than this year's. But then I made a lot of stupid mistakes. I better not have this time, more than I already have.

this year senior paper is trickier than last year . I stuck at most of the geometry problem Q21 onwards .... and done quite some number of silly mistakes at the back which cost me alot .I got tricked on Q26 asking the min number of student where I forget that there are only one way each to get sum of digit of double-digit number for 1 and 18 ....  :(  :(
Hopefully i can get a HD if the cut-off isnt so high  :o

YEH i rkn. i got tricked as well, i didnot read that the numbers could be different, so i counted 10 and 99 twice, so i ended up getting 37, when it should be 35.  :(

If possible, can you please post the Geometry problems of the Contest?

Thanks.

you have PQRS quad, with PQ=10, QR=20, RS=14, SP=26

now, a circle touches sides SP, PQ, QR and XY, where X is on SP and Y is on QR.

Another cirlce touches sides QR, SR, SP and XY.

Find length of XY.

[no multiple choice solns were given, answer between 0-999]

XY=11 .

we have PQ+XY=PX+QY=10+XY, since PQXY is circumscribed in a circle.

we also have XY+SR=XS+YR=14+XY.

we have 24+2XY=PX+QY+XS+YR=(PX+XS)+(QY+YR)=PS+YR=26+20=46.

this gives 24+2XY=46, so XY=11.

However, I made a lot of stupid mistakes. Consider 23+9=35!!! and the q#30 in Intermediate, I didn't realize that it should be 972, not 768. I ran out of time. In q#25, I ran out of time, I guessed...

[quote="shyong"]i remember the last question in senior division :

Q30 ) Three circles with radius 1,2,3 respectively touching each other in a plane . In between them , there is a space where another small circle , touching the three circles is drawn . Find the radius of the circle drawn .

I spend many time on this one but still in vain  :mad:[/quote]

Is it this?

[hide]$\frac{6}{23}$[/hide]

I did so bad.  I got 1wrong out of 30.  So only certificate and badge as a prize... :(

Q30 ) Three circles with radius 1,2,3 respectively touching each other in a plane . In between them , there is a space where another small circle , touching the three circles is drawn . Find the radius of the circle drawn .

I worked it out by using the coordinate geometry method:

Let the centres of the 3 circle be A,B,C respectively, in accordance with the problem we get AB=4,AC=3,BC=5, bulid xOy plane with the original point A and B,C lies on the y and x axies respectively, let the centre of the small circle be (x,y), the radius of it be R, then we get 3 simutaneous equations:  x^2+y^2=(1+r)^2...(1),
                                                                            (x-3)^2+y^2=(r+2)^2...(2)
                                                                            x^2+(y-4)^2=(r+3)^2...(3)
                   (1)-(2) to get x=(3-r)/ 3, (1)-(3),to get: y=1-0.5r, substitute them to (1)to get r=6/23 which is not an integer, this question is totally a WRONG question !!

From another view , we all know that the biggest circle in the triangle with sides 3,4,5 has the radius 1!, so that this radius of the small circle must be less than 1,  it's impossible to be an integer!! so that this question is a lie!    :mad:

The question actually said that the radius is of the form $ \frac{p}{q}$ where it is an irreducible fraction. Find $ p+q$.


function
real analysis
real analysis unsolved

Let $A$be a convexe, compact in a normed vectorial space $E$.Let $f: A\to A$ be an affine function.Show that it has a fixed point.
(I hope that it hasnt been already posted.In that case,someone please delete this post)

You can prove it for arbitrary families of affine functions that commute. It is the commutative version of Kakutani's theorem. For this one just consider the Cesaro means of consecutive itterates of the function.


geometry
circumcircle
LaTeX
national olympiad

[color=blue][size=150]PERU TST IMO - 2006[/size]
Saturday, may 20.[/color]

[b]Question 01[/b]  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88551]Discussion[/url]
Find all $(x,y,z)$ positive integers, such that:

$\sqrt{\frac{2006}{x+y}} + \sqrt{\frac{2006}{y+z}} + \sqrt{\frac{2006}{z+x}},$

is an integer.
---
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88509]Spanish version[/url]
---
[b]Question 02[/b] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88552]Discussion[/url]
Find all pairs $(a,b)$ real positive numbers $a$ and $b$ such that:

$[a[bn]]= n-1,$

for all $n$ positive integer.
Note: $[x]$ denotes the integer part of $x$.
---
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88510]Spanish version[/url]
---
[b]Question 03[/b] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88553]Discussion[/url]
In each square of a board drawn into squares of $2^n$   rows and
$n$ columns $(n\geq 1)$ are written a 1 or a -1, in such a way
that the rows of the board constitute all the possible sequences
of length $n$ that they are possible to be formed with numbers 1
and -1.
Next, some of the numbers are replaced by zeros.
Prove that it is possible to choose some of the rows of the board
(It could be a row only) so that in the chosen rows, is fulfilled that the
sum of the numbers in each column is zero.
---
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88511]Spanish version[/url]
---
[b]Question 04[/b] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88554]Discussion[/url]
In an actue-angled triangle $ABC$ draws up: its circumcircle $w$
with center $O$, the circumcircle $w_1$ of the triangle $AOC$ and
the diameter $OQ$ of $w_1$. The points are chosen $M$ and $N$ on
the straight lines $AQ$ and $AC$, respectively, in such a way that
the quadrilateral $AMBN$ is a parallelogram.

Prove that the intersection point of the straight lines $MN$ and $BQ$ belongs the
circumference $w_1.$
---
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88513]Spanish version[/url]
---
$\text{\LaTeX}{}$ed by carlosbr

HERE YOU CAN DOWNLOAD:
PDF RESOURCE
AND
LATEX RESOURCES

Carlos Bravo ;)
Lima -PERU


when we multiply by a power of 10, just shift the decimal point. Multiplying, \[ 100\times19.98\equal{}1998\] and \[ 1000\times1.998\equal{}1998.\] Finally, \[ 1998\times1998\equal{}(1998)^2.\] The answer is thus $ \boxed{D}.$

19.98 times 100 equals 1998
199.8 times 10 equals 1998
1998 times 1998 equals 1998^2
thus ans=d


probability
combinatorics unsolved
combinatorics
Probabilistic Method

Show that there exists a natural number $ n$ such that if a equilateral triangle with side length $ n$ is divided into $ n^2$ equilateral triangle with side $ 1$, then among the vertices of these triangles, we can choose $ 1993n$ points, such that no three form an equilateral triangle.

Here's a (sketch of a) solution:

First, we note that in the original triangle there are at least $ C_1 n^2$ points and at most $ C_2 n^3$ equilateral triangles for some $ C_1, C_2>0$.

Let $ p>0$ be fixed.  We select a subset of points according to the following procedure.  First, we independently include each vertex with probability $ p$.  Next, we take each equilateral triangle that appears among our selected set of points and arbitrarily drop one point in the triangle from our collection.  By construction, our set now contains no equilateral triangles.

On average, we will initially select $ C_1 n^2 p$ points.  On average, we will drop at most $ C_2 n^3 p^3$ points (each triangle is included with probability $ p^3$).  So on average, we will finish with a set of size at least $ C_1 n^2 p \minus{} C_2 n^3 p^3$.  Therefore there must be at least one good set which is at least this large.  Now choose an appropriate $ p$, and we are done!

This trick (pick a random set, then later adjust it to remove all instances of a bad configuration) is common in the probabilistic method, and was famously used by Erdos to prove the existence of graphs with arbitrarily large girth and chromatic number.  However, it doesn't seem like the solution which would have been expected here.  Is there a more straightforward constructive argument?

[quote=kevinatcausa]...

... and at most $ C_2 n^3$ equilateral triangles for some $ C_1, C_2>0$.

...
[/quote]
This should be  $ C_2 n^6$, no?

And now it seem that this method (of alteration) does not work here, because $ C_1 n^2 p - C_2 n^6 p^3$  takes maximum at $p =  \sqrt{C_1\over 3 C_2 n^4}$ and this maximum is some constant independent of $n$.

I agree that we can't use alternation here
even if the number of equilateral triangles is $C_2n^3$,
We can't say that $C_1n^2p-C_2n^3p^3$ is over $1993n$ for some $n$
Isn't it, kevinatcausa?

Then without alternation, how can we solve this?
Does anyone have a solution?

Solution in russian here http://problems.ru/view_problem_details_new.php?id=109510

[quote=for63434]I agree that we can't use alternation here
even if the number of equilateral triangles is $C_2n^3$,
We can't say that $C_1n^2p-C_2n^3p^3$ is over $1993n$ for some $n$
Isn't it, kevinatcausa?
[/quote]

We could, take $p = {1994\over C_1 n}$. Then large enough $n$ will do.

[quote=<Number1>]
[quote]I agree that we can't use alternation here
even if the number of equilateral triangles is $C_2n^3$,
We can't say that $C_1n^2p-C_2n^3p^3$ is over $1993n$ for some $n$
Isn't it, kevinatcausa?[/quote]
We could, take $p = {1994\over C_1 n}$. Then large enough $n$ will do.
[/quote]
Yes... I guess you're right... My bad :blush:

I will assume that the triangles are made by drawing lines parallel to the sidelines of the big triangle. That said,

[hide=Solution] Fix $n$ sufficiently large. Observe that there are $\frac{(n+1)(n+2)}{2} \sim \frac{1}{2}\cdot n^2$ points in the grid and at most $(n+1)^3 \sim Cn^3$ regular triangles (each determined by a line parallel to one of the sides of the big triangle) for $C>1$.

Pick a vertex randomly with probability $p \in [0,1]$. The expected number of points we've picked is at least $\frac{1}{2}\cdot pn^2$. For any regular triangle formed by our chosen set of vertices, delete one of its end points randomly. The expected value of the number of points we have deleted so far is $Cn^3p^3$. Thus, we can expect a set of points not containing any regular triangle to have at least $$\text{EV}=\frac{1}{2}\cdot pn^2-C(pn)^3.$$ For $p=\frac{2\times 1994}{n}$ and $M=C\cdot (2\times 1994)^3,$ we get $\text{EV} \ge 1993n$ so our result holds. $\, \square$ [/hide]

Edit: This is wrong. As dgrozev pointed, the OP asks for regular triangles. They may not have sides parallel to the original.

[quote=anantmudgal09]...and at most $(n+1)^3$ equilateral triangles (each determined by a line parallel to one of the sides of the big triangle). We may assume that there exists a constant $C>0$ such that there are no more than $Cn^3$ equilateral triangles....
[/quote]

Why? There is no information that sides of triangle must be parallel to sides of big one.
So the total number of triangles is $\sim {n^2/2\choose 3}\sim cn^6$ for some $c$.

[quote=Number1]...Why? There is no information that sides of triangle must be parallel to sides of big one.
[/quote]
There is no information because it's not well translated. In the original(russian) text it says: "divided into $ n^2$ equilateral triangles ...by lines parallel to the sides of the given one..."

Докажите, что существует такое натуральное число n , что если правильный треугольник со стороной n разбить прямыми, параллельными его сторонам, на n2 правильных треугольников со стороной 1, то среди вершин этих треугольников можно выбрать 1993n точек, никакие три из которых не являются вершинами правильного треугольника ([b]не обязательно[/b] со сторонами, параллельными сторонам исходного треугольника)

I meant something else- the grid had lines parallel to the sides of the big one, and it wasn't mentioned in the OP.
Just to clarify, the point:[b]Number1[/b] emphasized above was that the sentence" "no three points form an equilateral triangle" meant that such a triangle not necessary have to be with sides parallel to the big one's.

In fact, I didn't read in details all the solution of anantmudgal09, after:
[quote=anantmudgal09]...and at most $(n+1)^3$ equilateral triangles (each determined by a line parallel to one of the sides of the big triangle). We may assume that there exists a constant $C>0$ such that there are no more than $Cn^3$ equilateral triangles....
[/quote]

Now, I still think the all possible equilateral triangles (including that with sides not parallel to the big one) are $O(n^3)$, which of course must be proven, but there is much more that needs an explanation.

Actually, I'm pretty sure the total is $O(n^4)$. If you'll allow me to brush over all the details: select two arbitrary vertices of our original equilateral triangle. There's $O(n^2)$ vertices so there are $O(n^4)$ ways to do this. Use one choice as the centroid and the other as a vertex of our inner equilateral triangle. This equilateral triangle formed may not fit inside our current one, but it will fit inside a concentric equilateral triangle of side length $3n$ or so, and the constant doesn't matter.

Note the answer also changes from $O(n^3)$ to $O(n^4)$ when the parallel sides condition is removed in the [url=http://www.artofproblemsolving.com/community/c6h588121p3481506]hexagon case[/url] too.

@MellowMelon: You are right, the number of all equilateral triangles is $O(n^4)$.
But actually it's still not too bad and the probabilistic method could be saved, with some other arguments.

Following #2 or #8 posts, we conclude there exists a set of points, free of equilateral triangles, of size at least $ C_1 n^2 p - C_2 n^4 p^3$. Take $p=\frac{1}{k\cdot n}$. Then for big enough $k$, we obtain the existence of an absolute constant $c$, such that for big enough $n$ there exists a set free of regular triangles and of size $c\cdot n$.
Actually, this constant $c$ may be very small. The further idea is we can consecutively pump that constant $c$ and make it as big as we want.
Let the big equilateral triangle be $ABC$ . Suppose $C_1\in AB, A_1\in BC, A_1C_1\parallel AC$ and $BA_1/BC=1/3$.
Now, as a base we can assume there exist $cn$ free of triangles points inside $ABC$, but in a such way that $C_1BA_1$ is free of them. It can be proven in the same probabilistic way, thou with a smaller constant $c$.

Further, we take a big regular triangle $ABC$, and let $A_1\in CB, B_1\in AC, C_1\in AB, A_1B_1\parallel AB, A_1C_1\parallel AC, CA_1/CB=1/4$. Let also $A'\in AC, C_1A' \parallel BC$. Then $AA'/AC=1/4$.
[img]http://i.imgur.com/9aY0JER.png[/img]
In the triangles $AA'C_1, CA_1B_1, BA_1C_1$ we apply our base construction and get three sets of points $S_1,S_2,S_3$ with:
\[ |S_1|+|S_2|+|S_3| \ge c\cdot(|CA_1|+|BA_1|)|+|AA'|)=(c+c/4)\cdot n.\]
where $n=|AB|$.

Because the right $1/3$ part of the triangle $BC_1A_1$ doesn't contain any point of the set $S_i$, we can conclude that $S=S_1\cup S_2\cup S_3$ also is free of regular triangles and has at least $(5/4) c\cdot n$ points. Moreover, the right $1/4$ of the triangle $ABC$ doesn't have any point of $S$.

Further, we take $S$ and $ABC$ as a base construction, then consider a bigger triangle, and as above, cut off it three triangles- a smaller at the top, with $1/5$-th size of the bigger one, a bigger one at the right and at the left- a same one as at the top...
Proceeding in that way, every step "increases" the constant $c$ multiplying it by $(1+1/i)\,,\,i=4,5,\dots$. Note that $\prod_{i=1}^{\infty}(1+1/i)$ diverges.
Now, going  backwards, we can construct, after a finite number of steps, and for a large enough $n$, a set of $1993\cdot n$ points not containing regular triangles.
More accurately, this construction actually gives us an estimate of $\Omega\left(\frac{\ln n}{n}\right)$.

Below is the official (Russian) solution.

Let us color the vertices of the lattice with 3 colors $A,B,C$ as shown in the figure.
[img]http://problems.ru/show_document.php?id=1633824 [/img]

Notice that the vertices of a regular triangle have to be colored either in the same color or in all 3 colors. Remove all the vertices colored as "$A$". Call this operation "filter". A regular triangle on the remaining vertices have to be monochromatic, hence with side length at least $\sqrt{3}$. Consider separately the remaining sets of vertices colored with $B$ and $C$. For each of them apply again the "filter" operation.
As a result we get $(\frac{2}{3})^2\cdot \frac{n^2}{2}$ points and any regular triangle with vertices among them should have a side length at least $(\sqrt{3})^2$.
Proceeding in that way, after applying "filter" $k$ times, there remains at least $(\frac{2}{3})^k\cdot \frac{n^2}{2}$ points with the possible length of a regular triangle at least $(\sqrt{3})^k$.
Suppose $n=3^m$. Then, if take $k=2m+1$, there won't be any remaining equilateral triangle, but the remaining points would be at least $(\frac{4}{3})^m \frac{n}{3}$. Thus, taking $m$ large enough, the result follows.

[b]Comment:[/b] The described method gives us an estimate of the number of "free of triangles set" up to $\Omega\left( \frac{\log n}{n}\right)$.


Find all $x;y$ are positive integer numbers such that:
$x^{2}+y^{2}=2420z^{2}$

I think z is positive integer too (else for any x,y we get $z=\pm \sqrt{(x^{2}+y^{2})/2420}$).
It is easy to chek $2|x, 2|y, 11|x,11|y$, therefore $x=22x_{1},y=22y_{1}$ and $x_{1}^{2}+y_{1}^{2}=5z^{2}.$
Because Gauses numbers $x_{1}+iy_{1}$ give Euclidive ring $2+i|x_{1}+iy_{1}$ or $2+i|x_{1}-iy_{1}$.
It give $x+iy=22 d ( 2+i)(a+ib)^{2}$.

we can check that $2420 = 22^{2}+44^{2}$
then $x^{2}+y^{2}= 22^{2}z^{2}+44^{2}z^{2}$
so there are lot of solution
(how to find all of them, I have no idea)  :D


function
induction
algebra proposed
algebra

Determine all injective and continuous functions $f: \mathbb R \to \mathbb R$ with $ f(1)\equal{}1$ and \[ f(2x\minus{}f(x))\equal{}x\] for all real $x$.

[quote="DocEmBr"]Determine all injective and continuous functions $ f: R\rightarrow R$ with $ f(1) \equal{} 1$ and $ f(2x \minus{} f(x)) \equal{} x$[/quote]

[hide="My solution"]
$ f(x) \equal{} x$ is obviously a solution and we'll show this is the only solution.

Suppose $ \exists a\in\mathbb R$ such that $ b \equal{} f(a)\neq a$ : we get $ f(2a \minus{} b) \equal{} a$ and, with induction : $ f(n(a \minus{} b) \plus{} b) \equal{} (n \minus{} 1)(a \minus{} b) \plus{} b$  $ \forall n\in \mathbb N$. Using then the fact that $ f(x)$ is obviously surjective, so bijective, we get $ f^{ \minus{} 1}(x) \equal{} 2x \minus{} f(x)$ and  so $ f(a_n) \equal{} a_n \plus{} (b \minus{} a)$ (where $ a_n \equal{} n(a \minus{} b) \plus{} b$)  $ \forall n\in \mathbb Z$

$ \forall x$, $ \exists n\in\mathbb Z$ such that $ x\in[a_n,a_{n \plus{} 1}]$ (or $ [a_{n \plus{} 1},a_n]$, but it does not matter), so $ f(x)\in[f(a_n),f(a_{n \plus{} 1})]$ since $ f(x)$ is monotonous (since injective and continuous). So $ f(x)\in[a_n \plus{} (b \minus{} a),a_{n \plus{} 1} \plus{} (b \minus{} a)]$

So $ |f(x) \minus{} x|\leq |f(x) \minus{} f(a_n)| \plus{} |f(a_n) \minus{} a_n| \plus{} |a_n \minus{} x|$ $ \leq 3|b \minus{} a|$

And, since $ f(1) \minus{} 1 \equal{} 0$, $ f(a) \minus{} a\neq 0$ and $ f(x) \minus{} x$ is continuous, we can find some $ a,b \equal{} f(a)$ such that $ |b \minus{} a|$ may be as little as we want.

And so $ f(x) \equal{} x$ $ \forall x$
[/hide]


logarithms
integration
limit
calculus
calculus computations

evaluate the sum for the following series:

$ \sum_{k=2}^\infty\;\frac{(-3)^{k}\cdot\zeta{(k)}}{k}$

[ PS: answer [i]hidden [/i]below ! ]

[hide]
$ answer: \;\;\;\quad \boxed{3\gamma+\ln 2+\ln 3}$
[/hide]

The radius of convergence for the power series $ \sum_{k=2}^\infty \frac{(-1)^{k}\zeta{(k)}}{k}x^{k}$ is 1. So you cannot put $ x = 3$.

p.s. One can show that for $ |x|<1$,

$ -\gamma x+\sum_{k=2}^\infty \frac{(-1)^{k}\zeta{(k)}}{k}x^{k}\; = \; \ln \Gamma(1+x)$

Convergence of each limit may need more rigorous argument, but I won't establish it here. For $ |x|<1$,

\begin{eqnarray*}\Gamma(1+x) & = & \int_{0}^{\infty}t^{x}e^{-t}\; dt\\ & = & \lim_{n\to\infty}\int_{0}^{n}t^{x}\left( 1-\frac{t}{n}\right)^{n}\; dt\\ & = & \lim_{n\to\infty}n^{x+1}\int_{0}^{1}t^{x}(1-t)^{n}\; dt\\ & = & \lim_{n\to\infty}\frac{n^{x+1}\; n!}{(x+1)\cdots(x+n+1)}\\ \end{eqnarray*}

where $ \int_{0}^{1}t^{x}(1-t)^{n}\; dt \; = \; \frac{n!}{(x+1)\cdots(x+n+1)}$ can be obtained if you apply integration by parts inductively. Now, take log to the both side. Then

\begin{eqnarray*}\ln \Gamma(1+x) & = & \lim_{n\to\infty}\left[ (1+x)\ln n+\ln n!-\sum_{k=1}^{n+1}\ln (x+k) \right]\\ & = & \lim_{n\to\infty}\left[ (1+x)\ln n-\ln(1+n)-\sum_{k=1}^{n+1}\ln \left(1+\frac{x}{k}\right) \right]\\ & = & \lim_{n\to\infty}\left[ x\left( \ln n-\sum_{k=1}^{n+1}\frac{1}{k}\right)-\ln \left( 1+\frac{1}{n}\right)-\sum_{k=1}^{n+1}\left( \ln \left(1+\frac{x}{k}\right)-\frac{x}{k}\right) \right]\\ & = & \lim_{n\to\infty}\left[ x\left( \ln n-\sum_{k=1}^{n+1}\frac{1}{k}\right)-\ln \left( 1+\frac{1}{n}\right)-\sum_{k=1}^{n+1}\sum_{j=2}^{\infty}\frac{(-1)^{j-1}}{j}\frac{x^{j}}{k^{j}}\right]\\ & = &-\gamma x-\sum_{k=1}^{\infty}\sum_{j=2}^{\infty}\frac{(-1)^{j-1}}{j}\frac{x^{j}}{k^{j}}\\ & = &-\gamma x+\sum_{j=2}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^{j}}{j}\frac{x^{j}}{k^{j}}\\ & = &-\gamma x+\sum_{j=2}^{\infty}\frac{(-1)^{j}\zeta(j)}{j}x^{j}\\ \end{eqnarray*}

[quote="sos440"]The radius of convergence for the power series $ \sum_{k=2}^\infty \frac{(-1)^{k}\zeta{(k)}}{k}x^{k}$ is 1. [color=red][b]So you cannot put[/b] [/color]$ x = 3$.

p.s. One can show that for $ |x|<1$,

$ -\gamma x+\sum_{k=2}^\infty \frac{(-1)^{k}\zeta{(k)}}{k}x^{k}\; = \; \ln \Gamma(1+x)$[/quote]

you are correct, i was wrong!

so, instead of $ 3$, if i put $ x\;=\;\frac{1}{3}$ now, what will be the answer ?  :maybe:

[ PS: i know the answer for $ x\;=\;\frac{1}{2}\;\;\;$ since $ \Gamma{\left(\frac{3}{2}\right)}\;=\;\left(\frac{1}{2}\right) !\;=\;\frac{\sqrt\pi}{2}$,
but i don't know what would be the corresponding expression for
$ x\;=\;\frac{1}{3}$ since i don't know what is $ \Gamma{\left(\frac{4}{3}\right)}\;=\;\left(\frac{1}{3}\right) !$ ]


Number $x=\sqrt[3]{1+\frac{2\sqrt{7}}{3\sqrt{3}}}+\sqrt[3]{1-\frac{2\sqrt{7}}{3\sqrt{3}}}$ is

(a) natural

(b) negative

(c) $x\in\mathbb{Q-Z}$

(d) irational

(e) prime

[hide="My attempt"] Note $a=\sqrt[3]{1+\frac{2\sqrt{7}}{3\sqrt{3}}}$ and $b=\sqrt[3]{1-\frac{2\sqrt{7}}{3\sqrt{3}}}$.We have $ab=\frac{-1}{3}$. Now $x^3=a^3+3a^2b+3ab^2+b^3$. But $a^3+b^3=2$ and $3a^2b+3ab^2=3ab(a+b)=-(a+b)=-x$.
Hence $x^3+x-2=0$. the obivous solution is $x=1$ so $x\in\mathbb{N}$. There are also other 2 roots but they are complex. Is there another way :?
[/hide]

I think there is something wrong with the problem. There should be $x \in R$?

[quote="Slizzel"]:huh: What ?[/quote]
I mean your solution is perfect, but the problem seems to have something wrong with itself.

There's nothing wrong with the problem... [hide]Slizzel's solution is correct.  x=1, and thus the answer is (a) natural[/hide]


An equilateral triangle is inscribed in a circle of radius r.  Express the circumference C of the circle as a function of the length x of a side of the triangle.

HINT GIVEN:

First show that r^2 = (x^2/3).

Please use LaTeX. It's really not that difficult, just put dollar signs around everything, and use \frac{a}{b} for fractions.

[hide="hint"]
pythagorean theorem[/hide]

[quote="Interval"]Sorry...your hint is not helpful.[/quote]
Itwas helpful if you know where to apply the pythagoream thereom.

[quote="Interval"]Sorry...your hint is not helpful.[/quote]

question:

how long did you think about how to apply the phthagoreom (sp) theorem before you posted that?

id try to figure it out, because that is the key.

So of you draw it you can see that the center of the circle is the center of the triangle. Now draw from the center of the circle to one of the vertices. This is the radius. Do this again to form an isosceles triangle. Do you see where I'm getting at?

The picture if pretty but it does nothing for me.

I need to use a^2 + b^2 = c^2 but HOW in this question?

Thanks!

I do understand that you are using this information to study, but there is no possible way that you can study how to solve every problem in highschool mathematics. What if a style of problem comes up on your test that you've never seen before? You can't study math based on all possible solutions. That is why everybody is telling you that you need to sit down and really think about how all these hints fit together.


geometry
incenter
angle bisector
geometry proposed

A circle touches(is tangend with) side $ AB$ of a triangle $ ABC$ at $ B$  , passes through the incenter $ I$ of the triangle $ ABC$ and meets line $ AC$ at $ D,E$ .

 Prove that $ \angle DIC \equal{} \angle EIC$

 Babis

Denote F is the intersection of ⊙O and BC , G is the intersection of ⊙O and IC
∠HBI=∠IEB=∠ABI  => △IHB∽△IBE   => IB^2=IH*IE
∠IBC=∠IGB => △ICB∽△IBG  => IB^2=IC*IG
IC*IG= IH*IE  => C,H,E,G are concyclic
∠FCG=∠GEI=∠GCE => △GEC∽△GIE   =>  ∠CEG=∠CID=∠GIE=∠CIE
Done!

see if this would work:

Let $ M\equal{} IE \cap BC$.
Join B&E.
Then as $ \angle IBC \equal{} \angle IBA \equal{} \angle IEB$ and
$ \angle CBE\plus{}\angle BCE \plus{} \angle CEB \equal{} 180$
$ \angle BCE \plus{}\angle CEI\plus{}\angle IEB \plus{} \angle CEB \equal{} 180$
$ \angle BCE \plus{}\angle CEI\plus{}\angle IBC \plus{} \angle CEB \equal{} 180$
$ \angle IBE \plus{}\angle CEI\plus{} \angle CEB \equal{} 180$
$ \angle IBE \equal{} 180$
$ \angle CHI \equal{}180\minus{}\angle IBE \equal{} \angle IDE$
As $ \angle ICH \equal{} \angle DCI$, so  $ \angle EIC \equal{} \angle DIC$ and we are done=)

Notations like 1st solution; since AB is tangent to circle and BI is angle bisector, we have (from now on, XY means the smaller arc XY) BI = FI ( 1 ), then since CI is angle bisector, BI + FG = DI + GE ( 2 ); with (1), from (2): FI + FG = DI + GE,
GI = DI +GE, but GI = DI + DG, hence DG = GE, done.

Best regards,
sunken rock


For a given positive integer $ k$, define the sequences $ a_n$ and $ b_n$ to be defined recursively as follows:

$ a_1 \equal{} 1$
$ b_1 \equal{} \lceil \sqrt {k} \rceil$

$ a_{n \plus{} 1} \equal{} 2a_nb_n$
$ b_{n \plus{} 1} \equal{} a_n^2k \plus{} b_n^2$

Prove that $ \lceil a_n\sqrt {k} \rceil \equal{} b_n \text{ } \forall \text{ } n$, and moreover that $ \lim_{n \to \infty}{\frac {b_n}{a_n}} \equal{} \sqrt {k}$.

$ b_{n\plus{}1} \minus{} a_{n\plus{}1} \sqrt{k} \equal{} (b_n \minus{} a_n \sqrt{k})^2$ with $ \left| b_1 \minus{} a_1 \sqrt{k} \right| < 1$.


If 1/a+1/b+1/c=1/(a+b+c).......how do i get a=-b or b=-c or c=-a as the roots of the equation?

Standard algebra:

[tex]\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}[/tex]

[tex]\displaystyle \frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}[/tex]

[tex]\displaystyle (ab+bc+ca)(a+b+c)=abc[/tex]

[tex]\displaystyle a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2abc=0[/tex]

[tex]\displaystyle (a+b)(b+c)(c+a)=0[/tex]


geometry
calculus
algebra
polynomial
quadratics
special factorizations
calculus computations

On our local high school competition, two of our 8 questions on the team round were calculus questions. They turned out to be the two hardest questions. (Teams of 6, 30 minutes for 8 questions, no calculators allowed.)

[b]5.[/b] A line intersects the curve $ y\equal{}x^3$ at the point $ (a,a^3)$ for some $ a>0$, and at exactly one other point where the line and the curve are tangent. Compute the area of the region enclosed by the line and the curve. Express the answer in terms of $ a$.

Write the line as $ y \minus{} a^3 \equal{} m(x \minus{} a)$.  Its intersections with $ y \equal{} x^3$ give the cubic polynomial

$ (x \minus{} a)(x^2 \plus{} ax \plus{} a^2 \minus{} m) \equal{} 0$.

The fact that the line and the curve are tangent at the second point means that the quadratic factor is a perfect square.  Completing the square then gives

$ (x \minus{} a) \left( x \plus{} \frac {a}{2} \right)^2 \equal{} 0$.

The rest is computation...  :|

[quote="t0rajir0u"]The rest is computation...  :|[/quote]
... which only two of the 33 teams carried out correctly. It's a computation that requires some care.


function
geometry
3D geometry
algebra
polynomial
number theory
prime numbers

alright i have attached a file to this post which shows a graph of x = prime number 1,2,3,5,7,11,13,etc. and y= somefunction that takes x and manipulates it. you can clearly see several straight lines. each of this line has a very real equation, and i believe that there is a very simple equation that connects each line, so that all lines can be predicted, ergo all prime numbers are related to one another thru this pattern. i have only worked on this for a total of 2 hours, so i am not going to release any of my equations, b/c i want to figure it out myself. but if u look closely @ each line they are evenly spread out for example n=1 is the first line and n=2 is the second line and it is EXACTLY (well almost i think? i ll have to check) twice as far from n=0, as n=1 is (n=0 being the horizontal line). well i just picked this up again after leaving it for a few months. i hope this is clear b/c i am recovering from a hang over a' la canada day! lold :(. i have more to say but my head is killing me, o 1 more thing i must get out. look carefully at this graph it reminds me of how light defracts (the defraction pattern seen from passing light thru very tiny slits). this leads me to believe that prime numbers could be used in a model to describe this phenomenon. i ll go 1 step further and say it may be possible to describe quantum physics soley with prime numbers, and throw away those nonsense complex numbers for good. one more thing i just randomly googled this site. is this the best place to post my ideas on this subject? or is there a much better place?

it's a 4 term function that i just made up, it really isnt that important, i just wanted to know if something like this has been done before and if so could you direct me in that direction?

EDIT: Just fount that there are also ln and cubed versions of this equation, and i simply didnt just ln and cube the output, i changed a sign in the actual eqution to ln( or ^3.

EDIT EDIT: Alrighty then, last update tonight, this is a 1/x version.

next time i work on this ill try and figure out the equations to the lines, this ought to give me an equation where i can plug in 1 and it'll tell me the first prime number, plug in 378 and it'll tell me the 378th prime number, i wonder if this is how they generate prime numbers if they use this equation lold.

i am not a quack lold, i am just bored.

EDIT: cool just figured something out i ll update you in the morning if i remember too.

well i figured out how to simplify my equation, now i am sure you'll recognize how i got this :P x=prime 1,2,3,5,7,11,13,etc

so wait how do they know that it only takes 36 polynomials @ the far reaches of infinity they must have come up with a proof?

btw here is my equation its not really that "cool"

[quote="gomefun2"]so wait how do they know that it only takes 36 polynomials @ the far reaches of infinity they must have come up with a proof? [/quote]
Do you speak about the polynomial {x} refers to¿
See http://www.beweise.mathematic.de/primenumbers-jsww.html for it, just open the links there.

[quote="gomefun2"]btw here is my equation its not really that "cool" (a^2-b^2-c^2+d^2)/24[/quote]
Well, thats always an integer when $a,b,c,d$ are odd integers not divisible by $3$. The only question I see is: what values does it reach $\infty$ often¿

Edit: small error...

[quote="{x}"]For god sake, avoid relating light diffraction to prime distribution :)
But more seriously, if you are interested, there is a 26 variable polynomials, which "describes" all primes (http://mathworld.wolfram.com/PrimeDiophantineEquations.html). There is also some more or less useless prime generation formula (useless because they implicitely depend on the disposition of primes) that you can see at http://mathworld.wolfram.com/PrimeFormulas.html[/quote]

why cant i compare this to light defraction? each line has a smaller chance to have point in it, just like each line in a light defraction. the lines are spread out evenly, just like in light defraction. it looks like a model for light defraction (just not accurate w/o alot of modifications). so long as it has the same general shape i dont see why it cant be called a "model" it might not be as accurate as other models, but ya who cares? models are not what they represent they are always simplifications, i could model light defraction with paint if i wanted too...


LaTeX
modular arithmetic
number theory
relatively prime
number theory unsolved

For an odd positive integer $n > 1$, let $S$ be the set of integers $x$, $1 \leq x \leq n$, such that both $x$ and $x+1$ are relatively prime to $n$. Show that
$\prod_{x\in S}x \equiv 1$ (mod $n$)

Peng Shi

Shouldn't it be [tex]1\leq x<n[/tex], because if one of elements of [tex]S[/tex] is equal to n, then the product of all the elements would [tex]\equiv 1 mod n[/tex]. Am I wrong?

P.S. A question about LaTeX: how to make 1 mod n, not 1modn?

Suppose that $a \in S$. Then since $\gcd(a, n) = 1$, there exists a unique $b \in \{1, 2, \ldots, n-1\}$ such that $ab \equiv 1 \pmod{n}$.

We'll show that $b \in S$. Clearly, $\gcd(b, n) = 1$. Since $a(b+1) \equiv 1 + a \pmod{n}$, we have $\gcd(b+1, n) = 1$ as well.

Suppose that $a = b$, then $0 \equiv a^2 - 1 =(a+1)(a-1) \pmod n$. Since $\gcd(a+1, n) = 1$, we have $a - 1 \equiv 0 \pmod n$. Hence $a = 1$.

Therefore, all elements of $S \backslash\{1\}$ can be paired into recipricals, so the product of the elements of $S$ is 1.

Note that this is an extended version of Wilson's theorem.

This is the second time this problem is posted here (at least :)). I actually remember posting it quite some time ago, but I wouldn't know where to start looking for it..

yeah, may be somewhere in Solved forum, problem relates to the gemerized version of WILSON theorem, but i don't remember where it goes.

[quote="billzhao"]Note that this is an extended version of Wilson's theorem.[/quote]
There is extension of Wilson's theorem about product $r_1r_2...r_{\varphi(n)}\pmod{n}$, where $r_i$ ($i=1..\varphi(n)$) are residues modulo $n$.

Does anybody know mentioned generalization?

Thank you!

Do you mean $\prod_{x\in (Z/nZ)'}x  =  -1$ if and only if $(Z/nZ)'$ is cyclic ($n=2, 4, p^s, 2p^s$ ($p$ odd prime, $s$ positive integer)) and $= 1$ in all other cases ?

I don't know what your notations mean. I guess it is the same as I wrote.

By $(Z/nZ)'$ he means the multiplicative group of residues $\pmod n$ which are coprime with $n$.


geometry
3D geometry
sphere
function
geometric transformation
reflection
Support

Guys, i am not getting this problem

A small sphere of mass m suspended by a thread is first taken
aside so that the thread forms the right angle with the vertical and
then released. Find:
(a) the total acceleration of the sphere and the thread tension as
a function of theta, the angle of deflection of the thread from the vertical;

And i more question,
IS the radial acceleration zero when the pendulum is at its extreme.

Let me try. It might be wrong though.

$ mg$ is the force on the sphere in the downward direction. (blue)
Let $ x$ be the angle with the vertical. (C.a.R. doesn't support $ \theta$, hence $ x\equiv \theta$)
If it is divided into red and green components, respectively in the radial and tangent direction,
The magnitudes will be $ mg\cos x$ and $ mg\sin x$.
However the radial component of the force (red) will be equal to the tension of the thread, which is opposite in direction. Hence tension in the thread $ \boxed{mg\cos x}$
That component will not contribute to the effective acceleration of the sphere.
By $ F\equal{}ma$
$ \implies mg\sin x\equal{}ma$
$ \implies a\equal{}\boxed{g\sin x}$
and the direction of the force is perpendicular to the thread.

The 2nd question: It is always $ 0$

[quote="Akashnil"]
The 2nd question: It is always $ 0$[/quote]
wat does that mean??
radial acc is zero only at extreme (*most probably...)

Akashnil the answer for the tension was 3mgcos theta. And u assumed there is no radial acceleration which is not right i guess.

Guys even i did not understand akashnils doubt abt radial acceleration. We know that the velocity is continually changing but isnt the net acceleration always constant(Radial^2 + tangential ^2)^.5  So i am thinking as the total acceleration is always same so the radial acceleration is zero only at the extremes.

Hey guys the toatl energy is conserved so the velcity is changing ,does it mean acceleration is also changing. plz explain.

PS. anurag check ur PM to know who i am

Sorry guys :blush: I haven't understood anything :blush: . I'm still new to physics. I shouldn't have answered at all. Ignore my posts :blush:

hey akashnil u are pretty good at physics too(u answered slogger's doubt beautifully). Even i did not understand this concept.
There is nothing to fell bad. anyways u learn a lot by solving problems. so plz continue answering physics problems too. :lol:

Lets consider the torque about the point of suspension.

Say at time t, the string makes an angle $ \theta$ with the vertical. Let T denote the tension at that time. and say the string is moving toward the vertical.
So the torque is given by
$ \tau\equal{}mgl\sin \theta$
But $ \tau\equal{}ml^2\alpha$, where $ \alpha$ is the angular acceleration.
So, $ \alpha\equal{}\frac {g\sin \theta}{l}\Rightarrow l\frac {dw}{dt}\equal{}g\sin \theta\Rightarrow l\frac {dw}{d\theta}\frac {d\theta}{dt}\equal{}g\sin \theta\Rightarrow lw(dw)\equal{}g\sin\theta (d\theta)$
But on RHS, we have a negative quantity and on LHS, a positive quantity.  :o .
 :lol: So, to correct that, put a minus sign. So , we have
$ lw(dw)\equal{}\minus{}g\sin\theta (d\theta)$
Note that at $ \theta\equal{}\frac {\pi}{2}$, the sphere is momentarily at rest.

$ \int_0^w lw(dw)\equal{}\minus{}\int_{\frac {\pi}{2}}^{\theta} \sin \theta (d\theta)$

This gives $ lw^2\equal{}2g\cos \theta$
Now, the sphere performs circular motion about point of suspension. So $ T\minus{}mg\cos \theta\equal{}mw^2l\equal{}2mg\cos \theta$

So, $ T\equal{}3mg\cos \theta$

hell ever ur solution is gr8, but i tried using conservation of energy as i did not finish learning torque
The velocity of the block at the top is $ 0$
Let the thread make an angle $ theta$ with the vertical and let $ v$ be the velocity at this position.
Then change in height is $ \minus{} lcostheta$ where $ l$ is the length of the spring
By conservation of energy
we have $ delta K \plus{} delta U$=$ 0$(since there is no work done by any non conservative force)
So $ (1/2)mv^2$=$ mglcostheta$
So $ v^2$= $ 2glcostheta$
But $ T \minus{} mgcos theta$=$ 2mgcostheta$
So $ T \equal{} 3mgcostheta$

@Ritu: I already told everyone to ignore my posts :blush: . Pls don't quote me ...

[quote="Aravind Srinivas L"]@befuddlers
Tis must be easy for u man. U have gone to KKA sir's class the last 2 years in chennai???
This portion is wat he has finished for us now in 9th std itself.[/quote]

I fail to understand what this guy wants to tell...... Is it even required......???? :wallbash_red:  :furious:

[quote="rituraj007"][quote="Akashnil"]I could be wrong, but I don't see why it shouldn't be $ 0$ at somewhere other than extreme. Please tell your explanation.

My explanation:
radial component of gravity is $ 0$ only at the extreme, but even at other places that component is canceled by the tension of the thread. Hence radial component of the net effective force is always $ 0$. Hence radial acceleration is $ 0$. Moreover had the radial acceleration not been $ 0$ at some place, then it would not remain equidistant from the axis (the point where the thread is attached rigidly with the vertical)[/quote]

wats the answer neways??

coz i think radial acc means centripetal acc and it shudnt b 0[/quote]
ritu the centripetal acceleration is zero only at the extreme coz only at the extreme does the pendulum stop going up in a circle :D
and the tangential acceleration is zero only at the bottom coz at the bottom there are no horizontal forces(both tension and weight are directed along the center) :lol:

A car moves with a constant tangential acceleration of 0.62 $ m/s^2$ along a horizontal surface circumscribing a circle of radius 40 m. The coefficient of sliding friction between the wheels of the car and the surface is k - 0.20. What distance will the car ride without sliding if at the initial moment of time its velocity is equal to zero?

i dont understand how the body can have tangential acceleration as there are only centripetal forces(friction)
So what im asking is does friction point only at the center or is it an angle to the center having both a tangential and centripetal component

P.S if i assume the friction has tangential component also my answer is matching with the back answer.

befuddler,
the radial force
= m$ omega^{2}$r
=T-mg$ \cos$theta
the cross-radial force
=mg$ \sin$ $ \theta$=mr $ \frac {domega}{dt}$
=mr $ \frac {d omega}{dtheta}$ $ \frac {dtheta}{dt}$
=mr $ \omega$ $ \frac {domega}{dtheta}$

so, mg$ \cos$ $ \theta$=2m$ omega^{2}$r

hence the result follows.

the 2nd question is quite incomprehensible for me. a horizontal surface circumscribing a circle means what? is the case like the car is starting at a point where a circle touches the ground and moves up the circle?

guys i messed up with latex :(

In This Case The Radial Accelearation at the Extreme Positions =0 however in other cases it might depend on the Velocity Of the Body at the Bttom.
The Correct Answer is :->

[quote="Conjurer"]Obviously the radial acc. is 0 always because the thread is inextensible.[/quote]
That's exactly what I had in mind. :blush:


prove if 3n+1 is  a square we have n+1 the sum of three squares

bambaman, you can use \pm $ \pm$ or \mp $ \mp$  :wink:


Let $f$ be a function defined for the non-negative integers, such that:

a) $f(n)=0$ if $n=2^{j}-1$ for some $j \geq 0$.
b) $f(n+1)=f(n)-1$ otherwise.

i) Show that for every $n \geq 0$ there exists $k \geq 0$ such that $f(n)+n=2^{k}-1$.
ii) Find $f(2^{1990})$.

Take a number  $r$ such that $2^j -1 < r<2^{j+1}-1$. From the definition of the function;
$f(r)=f(r-1)-1=f(r-2)-2=f(r-3)-3=...=f(2^j-1)-(r-(2^j-1)) \rightarrow f(r)+r=2^j-1$ in other words $k$ is the $j$ for which $n=2^j-1$ is the greatest such number that satisfies $n\leq k$. From this $f(2^{1990})+2^{1990}=2^{1990}-1 \rightarrow f(2^{1900})=-1$


[quote="Makaveli"]Can somebody please post an update of macedonian results as quickly as possible? What do you think abot the medal cutoffs?
Thank you in advance[/quote]

Contestant 	           P1 	P2 	P3 	P4 	P5 	P6 	Total 	Award
Viktor Simjanoski	    7	1	0	7	0	0	15Bronzemedal
Maja Taseska	    7	1	0	6	0	0	14Honourable mention
Bodan Arsovski	    7	1	0	2	0	010Honourablemention
Aleksandar Ilioski	    7	0	0	2	0	0	9	Honourable mention
Visar Zejnulahu	    2	0	0	3	0	0	5
Andrej Risteski	    1	0	0	3	0	0	4
Total		57

And i'm from Macedonia!


did the USTA cancel the free tennis mag for their members?

[quote="math92"]i still get mine.....
The last issue they sent was for November and December so you won't get one for a while.[/quote]

i got my nov and dec too, but included with the mags is a card that says my subscribtion will end by next yr.
and yesterday, i just received a mail from tennis mag itself asking me to renew


can someone please prove that

(2^n)-1 can't be a prime unless n is a prime ...

all books say that this can be simply shown...

This is a direct result of $ a\minus{}b|a^k\minus{}b^k$. If $ n$ isn't prime, let $ n \equal{} pq$, and we have $ 2^p \minus{} 1^p | 2^{pq} \minus{} 1^{pq}$. (and clearly $ 2^p \minus{} 1 \neq 1$)

I think I`ve seen discussion on that problem on jeka-math.blogspot.com.

[quote="EulerIntegral"]http://en.wikipedia.org/wiki/Mersenne_prime

You are looking for mersenne primes.[/quote]

I believe Euler found a counterexample of $ n\equal{}32$ and managed to find a divsor. [b](AoPS volume 1)[/b]

[quote="7h3.D3m0n.117"][quote="EulerIntegral"]http://en.wikipedia.org/wiki/Mersenne_prime

You are looking for mersenne primes.[/quote]

I believe Euler found a counterexample of $ n \equal{} 32$ and managed to find a divsor. [b](AoPS volume 1)[/b][/quote]

Well as noted above $ 2^2 \minus{} 1 \equal{} 3, 2^4 \minus{} 1 \equal{} 15,\  \text{etc} |2^{32} \minus{} 1$. Perhaps you mean either $ n \equal{} 31$ which is prime, or $ 641 | 2^{32} \boxed{\plus{}} 1$ (not sure if thats Euler)?

You appear to be talking about the Fermat number $ F_5 \equal{} 2^{2^5} \plus{} 1$, which is indeed composite.  The first Mersenne number that is not a prime is $ 2^{11} \minus{} 1 \equal{} 23 \cdot 89$.


integration
inequalities
real analysis
real analysis unsolved

Prove that $ \parallel{}f\parallel{}_{1,p}\equal{}\{\int_{D}(|f|^p\plus{}|Df|^p)dx\}^{\frac{1}{p}}$ is a norm in Sobolev space $ W^{1,p}(D)$. I meet some mistakes in proving the triangle inequalities

Using Minkowski inequality
\[ \left\| {f \plus{} g} \right\|_p  \leqslant \left\| f \right\|_p  \plus{} \left\| g \right\|_p
\]
to prove.

\[ \begin{gathered}
  \left\| {u + v} \right\|_{W^{k,p} } & = \left( {\sum\limits_{\left| \alpha  \right| \leqslant k} {\left\| {D^\alpha  u + D^\alpha  v} \right\|_{L^p }^p } } \right)^{\frac{1}
{p}}  \hfill \\
  & \leqslant \left( {\sum\limits_{\left| \alpha  \right| \leqslant k} {\left( {\left\| {D^\alpha  u} \right\|_{L^p }^{}  + \left\| {D^\alpha  v} \right\|_{L^p }^{} } \right)^p } } \right)^{\frac{1}
{p}}  \hfill \\
  & \leqslant \left( {\sum\limits_{\left| \alpha  \right| \leqslant k} {\left\| {D^\alpha  u} \right\|_{L^p }^p } } \right)^{\frac{1}
{p}}  + \left( {\sum\limits_{\left| \alpha  \right| \leqslant k} {\left\| {D^\alpha  v} \right\|_{L^p }^p } } \right)^{\frac{1}
{p}}  \hfill \\
  & = \left\| u \right\|_{W^{k,p} }  + \left\| v \right\|_{W^{k,p} } . \hfill \\
\end{gathered}
\]