algebra
polynomial
number theory
IMO
prime numbers
IMO 1987
IMO Shortlist

Let $n\ge2$ be an integer. Prove that if $k^2+k+n$ is prime for all integers $k$ such that $0\le k\le\sqrt{n\over3}$, then $k^2+k+n$ is prime for all integers $k$ such that $0\le k\le n-2$.

Lemma:If m<(2b+1)^2 and m in relatively prime to b+1...2b then m is prime
We have n+k^2+k is prime for k=1->r;r=[\sqrt{n\3}] and need to show n+k^2+k is prime for k=r+1->n-2.Applying the lemma for m=n+(r+s)^2+r+s;b=r+s(s=1->n-s-r)The computation is simple

I have not still yet checked ur solution,but perhaps,my one is quite similar.
Call y is miniamal number st p(y) is composite,p is minimal prime divisor of p(y).We need to show that: y>n-2.
Supposing that y≤n-2.We will show that:p≥2y+1 ,then p(y)=y²+y+n≥p²> 4y²+4y+1->n>3y²->y<√(n/3).It's absurd!
  Indeed suppoing that p<2y+1.Consider p(y)-p(x)=(y-x)(y+x+1) when x in {1,2,..y} then (y-x) and y+x+1 will run over all values of set{1,2,...,2y+1} then there exists x in that set such that p(y)-p(x) is divisible by p ->p(x) is too.
   Accoring definition of y we deduce that p(x)=p.But :
y-x<y+x+1<(n-2)+x+1<x²+x+n=p->A contradiction!!And we are done!

Consider the least prime $ q$ dividing a composite value $ k^2 \plus{} k \plus{} n$ with $ \sqrt {n/3} < k\leq n \minus{} 2$. Clearly $ q < n$ so every residue class modulo q is represented between 0 and n-2. Thus both roots of $ k^2 \plus{} k \plus{} n \equal{} k^2 \minus{} (q \minus{} 1)k \plus{} n \equal{} 0$ mod q are in the desired range, so we may some root $ k < \equal{} \frac {q \minus{} 1}{2}$.
$ k > \sqrt {n/3}$ implies $ q > 2\sqrt {n/3} \plus{} 1$
$ q$ being minimal implies $ q^2 \leq k^2 \plus{} k \plus{} n \leq \frac {q^2 \minus{} 1}{4} \plus{} n$, so
\[ q^2 \leq \frac {4n \minus{} 1}{3} < (2\sqrt {n/3} \plus{} 1)^2 < q^2
\]
.
So no such $ q$ exists.

$ q^2 \leq k^2\plus{}k\plus{}n < n^2$ (certainly for n not trivially small)

Don't worry. I actually wrote this up and it took a whole page of A4, so the above argument is very much a sketch. :)

Yeah, I wondered why I still dont understand why get why every residue between 0 and n-2 is represented (in mod q). This is embarassing  :blush:

Well, since q<n, so q<=n-1, so 0,1,2,....,n-2 must contain 0,1,...,q-1 as a subset.

I attach a more detailed account of my solution.

Yes, thank you ever so much! I would never have been able to carry out that kind of argument!

You probably could have. The basic intuitive idea was that if there are going to be composite numbers, there are going to be some fairly small primes involved, but that equally such primes would not be able to be too small, or else the first few numbers of our sequence would be divisible by them. It then remained to do a bit of work to try to put as much pressure as possible on q from each side until it broke.

Yes, but I've never thought of working with the factorisation of polynomials in modular arithmetic before. Thanks. Besides, I need some IMO training anyway. You'll be at Trinity, I guess?

My name is Henry Husband. I just need to get a few IMO questions under my belt, thats all.

Ah sure. Well, this one was a Q6, and the solution took me in excess of an hour to complete, so... If you're determined to attack IMO questions, it's best to start on Q1,4 of recent IMOs and perhaps try attacking whole papers from before 1983ish when they started getting much harder.

I look forward to seeing you at Trinity (I'm guessing...).

Let $m=\lceil\sqrt{s/3}\rceil$, and let $a_n=n(n-1)+s$. We are given that $a_1,\ldots,a_m$ are prime, and we want to show that $a_1,\ldots,a_{s-1}$ are prime. We will proceed by strong induction. In particular, we will show that if $a_1,\ldots,a_{k-1}$ are prime with $\sqrt{s/3}+1\le k\le s-1$, then $a_k$ is prime.

Suppose $a_k$ was not prime, so it has a prime factor $p\le\sqrt{a_k}$. Now, if $n\equiv k,1-k\pmod{p}$, then $a_n\equiv a_k\pmod{p}$, so $p\mid a_n$ as well. We now show that there exists such $n$ with $1\le n\le k-1$. In other words, we want to show that there exists $x\in[1,k-1]\cup(p+1-[1,k-1])=[1,k-1]\cup[p+2-k,p]$ such that $x\equiv k\pmod{p}$. It suffices to show that $p+1-k<k$, because then that union of two intervals would contain $[1,p]$, so it would contain all residue classes mod $p$. Note that
\begin{align*}
2k-1> \sqrt{a_k}&\iff 4k^2-4k+1> k^2-k+s \\
&\iff 3k^2-3k+1> s \\
&\longleftarrow 3(\sqrt{s/3}+1)^2-3(\sqrt{s/3}+1)+1> s \\
&\iff s+3\sqrt{s/3}+1>s,
\end{align*}
which is true, so $2k-1> \sqrt{a_k}\ge p$, so $p+1-k<k$, as desired. Therefore, there is some $1\le n\le k-1$ with $p\mid a_n$. But $a_n$ is prime by the inductive hypothesis, so $a_n=p$.

Thus, we have $a_n\mid a_k$, so $a_n\mid a_k-a_n=(k-n)(k+n-1)$. Since $a_n$ is prime, either $a_n\mid k-n$, or $a_n\mid k+n-1$. We have that $a_n\ge s>k\ge k-n$, so $a_n>k-n$. Therefore, $a_n\mid k+n-1$. But $2a_n\ge s+s>k+n>k+n-1$, so the only possibility is $a_n=k+n-1$. But then, $n^2-n+s=k+n-1$, or $k=s+(n-1)^2\ge s$, which is a contradiction. Therefore, our assumption that $a_k$ is not prime is wrong, so $a_k$ is prime. Therefore, we finish by induction on $k$.

Let $f(x)=x^2+x+n$ and assume that there exists $\sqrt\frac{n}{3}< k\leq n-2$ such that $f(k)$ is not prime and choose the smallest such $k.$ Let $p$ be a prime divisor of $f(k),$ such that $p\leq \sqrt{f(k)}.$ Note that $$p\leq \sqrt{f(k)}=\sqrt{k^2+k+n}<\sqrt{4k^2+k}<2k+1\Rightarrow p<2k.$$

Let's first look at the case $k<p<2k$ and let's write $p=k+r, 0<r<k.$ We have $$0\equiv f(k)\equiv f(-r)=f(r-1)\pmod p,$$ and since $0\leq r-1<k,$ it must be the case that $r+k=p=f(r-1)=r^2-r+n,$ which rewrites as $0<r^2-2r+2=k-n+2\leq 0,$ contradiction$.$

If $p=k,$ then $p\mid n,$ but $f(0)=n$ is prime and therefore $k=p=n>n-2,$ contradiction$.$

Finally, assume $0<p<k.$ Denote $r=k\pmod p.$ We have $0\equiv f(k)\equiv f(r) \pmod p,$ and since $0\leq r<p<k,$ it follows that $p=f(r)=r^2+r+n\geq n>k>p,$ contradiction$. \blacksquare$

[hide = rewording]Let $m$ and $s$ be positive integers with $2 \leq s \leq 3m^2$. Define a sequence $a_1, a_2, \ldots$ recursively by $a_1 = s$ and\[a_{n+1} = 2n + a_n \text{         }(\text{for } n = 1, 2, \ldots).\]Prove that if the numbers $a_1, a_2, \ldots, a_m$ are prime, then $a_{s-1}$ is also prime.[/hide]

Note that in general, we may write $a_n = n^2 - n + s$.

Let $t$ be the smallest positive integer for which $a_t$ is not prime; we wish to prove $t \geq s$. Let $p$ be the smallest prime dividing $a_t$. We have\[p \leq \sqrt{a_t} = \sqrt{t^2 - t + s} \leq \sqrt{4t(t - 1)} < 2t - 1\]since clearly $s = 3m^2 \leq 3t(t-1)$ as $t \geq m + 1$. Hence $p \in [0, 2t - 2]$. Next, we observe that \[a_t \equiv 0 \pmod p \implies a_{t-p} \equiv 0 \pmod p.\]So in the case $p \leq t-1$ then we can subtract sufficiently many copies of $p$ from $t$ to obtain an index $t - \ell p = i$ for which $1 \leq i \leq m$ and $a_i \equiv 0 \pmod p$ is prime. Then, $a_i$ must be $p$ and then we have $a_t \geq p^2 = a_i^2 \geq a_1^2 = s^2 = a_s \implies t \geq s$ as desired.

In the case that $p \in [t+1, 2t - 2]$, we can let $p = t + r$ where $r \in [1, t-2]$ so that\[a_t \equiv a_{t - p} = a_{-r} = r^2 + r + s \equiv a_{r-1} \equiv 0 \pmod{p}\]and clearly $r - 1 \leq t - 1 \leq m$ so $a_{r-1}$ is prime, so it follows that\[a_t \geq p^2 =  a_{r-1}^2 \geq a_1^2 = s^2 = a_s \implies t \geq s\]as desired.

Lastly, if $p = t$, then $a_t = a_p = p^2 - p + s$ is disivible by $p$ hence $p \mid s$. However, we have $s = a_1$ is prime so $p = s \implies t = s$ which is what we wanted to prove.

We have exhausted all three cases so we are done; indeed, $t \geq s$ always. $\blacksquare$

Let $F(X)=X^2+X+n$ and define $p$ as the smallest prime factor of the numbers $F(0),F(1),.....$
For the sake of contradiction assume the opposite and proceed.
Suppose for $X=k<n-2$ is composite,then the smallest factor $q$ of $F(k)$ will satisfy $q^2 \ge f(k)<n^2$,hence $q^2<n^2$ or $q,p<n$(obvious since $p \le q$)
Let $s$ be the remainder when $k$ is divided by $p$ and let $r=\min(s,p-s-1)$,and it is clear that $p|f(r)$ so by the fact $2r \le p-1$,$f(r) \le n+\frac{p^2-1}{4}$
Now choose a prime factor of $\frac{f(r)}{p} \in \mathcal{Z}$,and by the minimality of $p \le q$,$p^2 \le F(r) \le \frac{p^2-1}{4}+n$(already established before),so taking square roots we get $p<2 \sqrt{\frac{n}{3}}$ implying $r<\frac{p}{2} \le \sqrt{\frac{n}{3}}$,a contradiction(since this also implies $p \neq q$ and so $f(r)=pqx$(not a prime))$\blacksquare$
[hide=Remarks]In 1952,Heegner proved that $\max(n)=41$[/hide]

Suppose otherwise, and let $A \leq n-2$ be minimal such that $A^2+A+n$ is not a prime. Then for any $k<A$, we have
$$\gcd(A^2+A+n,k^2+k+n)=\gcd((A-k)(A+k+1), k^2+k+n).$$
As $k^2+k+n$ is prime and we have
$$A-k<A+k+1\leq k+n-1<k^2+k+n,$$
it follows that $k^2+k+n \nmid (A-k)(A+k+1) \implies \gcd(A^2+A+n,k^2+k+n)=1$.
Now, consider the least prime $p \mid A^2+A+n$. If $p \leq A$ then
$$p \mid (A-p)^2+(A-p)+n \iff p \mid A^2+A+n,$$
which is true. If $A+1\leq p \leq 2A$, then
$$p \mid (p-(A+1))^2+(p-(A+1))+n \iff p \mid A^2+A+n,$$
which is true as well. In both of these cases there therefore exists some $k<A$ with $p \mid \gcd(A^2+A+n,k^2+k+n)$, which is a contradiction.
Therefore, we require
$$p>2A \implies A^2+A+n\geq p(p+2)>4A(A+1) \implies n>3A^2+3A,$$
but the RHS is at least $3A^2>n$, which is a contradiction. As such no $A$ exist so all the terms are prime. $\blacksquare$


plz write in $ LATEX$.... its not readable.

[quote="geoffrels"]15. Let a,b, and c be real constants such that Code:
x^2+x+2
is a factor of Code:
ax^3+bx^2+cx+5
, and \2x-1 is a factor of ax^3+bx^2+cx-(25/16). Find a+b+c.

21. Let k be a positive integer. A positive integer is said to be a k-flip if the digits of n are reversed in order when it is multiplied by k. For example, 1089x9 = 9801, and 21978 is a 4-flip because 21978x4=87912. Explain why there is no 7-flip integer?

22. Let ABC be an acute-angled triangle? Let D and E be the points on BC and AC, respectively, such that AD is perpendicular to BC and BE is perpendicular to AC. Let P be the point where AD meets the semicircle constructed outwardly on BC, and Q the point where BE meets the semicircle constructed outwardly on AC. Prove that PC=QC.

23. Two friends, Marco and Ian, are talking about their ages.

Ian says, "My age is a zero of a polynomial with integer coefficients."

Having seen the polynomial p(x) Ian was talking about, Marco exclaims, "You mean, you are seven years old? Oppps, sorry I miscalculated! p(7) = 77 and not zero"

"Yes I am older than that," Ian's agreeing reply.

Then Marco mentioned a certain number, but realizes after a while that he was wrong again because the value of the polynomial at that number is 85.

Ian sighs, "I am even older than that number."

Determine Ian's age.
[/code][/quote]

1. Use $ \text{\LaTeX}$
2. Never ever ever doublepost anything because it will end in noone replying (or even getting you banned)
3. I'll just give you some hints regardless...

21: The "flipped" number has first digit 7, 8 or 9, so the other one has first digit 1. This means that it has last digit 3 which is impossible.
23: Use the very useful lemma: If $ p(x)$ is a polynomial with integer coefficients, then $ (x-y)|(p(x)-p(y)).$ The second number Marco guessed is a, Ian's age is b. Then $ (a-7)|8,(b-7)|77,(b-a)|85.$ This yields finitely many possibilites for a and b and you just have to go through them to find the proper one.


trigonometry
function
integration
geometry
rectangle
linear algebra
matrix

Calculate $ \sum_{n \equal{} 1}^{\infty}\frac {\cos n}{n^2 \plus{} 1}$  :)

First hint: it's actually easier to compute $ f(x)\equal{}\sum_{n\equal{}1}^{\infty}\frac{\cos nx}{n^2\plus{}1}.$

Second hint: what can you say about $ f''(x)\minus{}f(x)?$ (You do have to stop worrying about such details as having series converge any way except weakly as distributions.)

$ \sum_1^{\infty}\frac {\cos{nx}}{1 \plus{} n^2}$ is very close to the Fourier series of $ f(x) \equal{} e^x \plus{} e^{2\pi}e^{ \minus{} x}$ on $ [0,\pi]$. I did not compute everything carefully, but to see this i integrated the given series twice and then noticed that $ \sum_1^{\infty}\frac {\cos{nx}}{n^2(1 \plus{} n^2)} \equal{} \sum_1^{\infty}\frac {\cos{nx}}{n^2} \minus{} \sum_1^{\infty}\frac {\cos{nx}}{1 \plus{} n^2}$, the first part is known to be linear function. Then it is clear that the solution can be found between the solutions of equation $ F'' \minus{} F \equal{} 0$, which are of the form $ F \equal{} c_1e^x \plus{} c_2e^{ \minus{} x}$.
P.S. oops, as always i'm late  :wink:

Actually, they are of the form $ f(x) \equal{} c_1 cos(x) \plus{} c_2 sin(x)$ but that's basically the same if you calculate non-reals!

[b]EDIT:[/b]

I was about to delete this post when I saw what I wrote, but since jmerry referenced this post, I'll just state the obvious - this is wrong!

No, that's different. $ \cos$ and $ \sin$ are solutions to $ f''\plus{}f\equal{}0$, which is a different differential equation. $ \cosh$ and $ \sinh$ might be useful here, except that there's a corner at zero; we're better off translating horizontally and using $ \cosh(x\minus{}\pi)$ on $ [0,2\pi]$.

[hide="Answer"]Provided I have made no mistake,

$ \sum_{n\equal{}1}^{\plus{} \infty}\frac{\cos(nx)}{n^2 \plus{} a^2} \equal{} \frac{\pi}{2a}\left[\coth(a \pi) \cosh(ax) \minus{} \sinh(ax)\right] \minus{} \frac{1}{2a^2}$[/hide]

Carcul, I calculated LHS and RHS for some special $ x$ and $ a$ with Maple.

Your formula seems to be true.  :)

Can you tell us please how you deduced this formula.

Let $ 0<x<2\pi$ and $ \alpha>0$ .

After Poisson summation formula

$ \sum_{n\in\Bbb{Z}} \frac{e^{inx}}{n^2\plus{}\alpha^2}\equal{}\sum_{n\in\Bbb{Z}} \int_{\minus{}\infty}^\infty \frac{e^{itx}}{t^2\plus{}\alpha^2}\, e^{2\pi i n t}\, dt\equal{}\frac{\pi}{\alpha} \sum_{n\in\Bbb{Z}} e^{\minus{}\alpha |2\pi n\plus{}x|}$

$ \equal{}\frac{\pi}{\alpha} \left(\sum_{n\equal{}\minus{}\infty}^{\minus{}1} e^{\alpha (2\pi n\plus{}x)}\plus{}\sum_{n\equal{}0}^\infty e^{\minus{}\alpha(2\pi n\plus{}x)}\right)$

$ \equal{}\frac{\pi}{\alpha} \left(e^{\alpha x}\sum_{n\equal{}1}^\infty e^{\minus{}2\pi\alpha n}\plus{}e^{\minus{}\alpha x}\sum_{n\equal{}0}^\infty e^{\minus{}2\pi \alpha n}\right)$

$ \equal{}\frac{\pi}{\alpha} \left(\frac{e^{\alpha x}}{e^{2\pi\alpha}\minus{}1}\plus{}\frac{e^{\minus{}\alpha x}}{1\minus{}e^{\minus{}2\pi\alpha}}\right)\equal{}\frac{\pi}{\alpha}\,  \frac{e^{\alpha(x\minus{}\pi)}\plus{}e^{\minus{}\alpha(x\minus{}\pi)}}{e^{\pi\alpha}\minus{}e^{\minus{}\pi\alpha}}$

So $ \sum_{n\in\Bbb{Z}} \frac{\cos(nx)}{n^2\plus{}\alpha^2}\equal{}\frac{\pi}{\alpha}\, \frac{\cosh\big(\alpha(x\minus{}\pi)\big)}{\sinh(\alpha \pi)}$

And similarly $ \sum_{n\in\Bbb{Z}} \frac{(\minus{}1)^n \, \cos(nx)}{n^2\plus{}\alpha^2}\equal{}\frac{\pi}{\alpha} \,\frac{\cosh(\alpha x)}{\sinh(\alpha \pi)}$ for $ \minus{}\pi<x<\pi$ .

Alternative write $ \frac{1}{k^2\plus{}\alpha^2}$ as $ \frac{i}{2\alpha} \left(\frac{1}{k\plus{}i\alpha}\minus{}\frac{1}{k\minus{}i\alpha}\right)$

and use the formula $ \sum_{k\in\Bbb{Z}} \frac{(\minus{}1)^k \cos kx}{k\plus{}\alpha}\equal{}\pi\, \frac{\cos \alpha x}{\sin \alpha \pi}$ .

So $ \sum_{k\in\Bbb{Z}} \frac{(\minus{}1)^k \cos kx}{k^2\plus{}\alpha^2}\equal{}\frac{i\pi}{2\alpha} \left(\frac{\cos(i\alpha x)}{\sin(i\alpha\pi)}\minus{}\frac{\cos(\minus{}i\alpha x)}{\sin(\minus{}i\alpha\pi)}\right)\equal{}\frac{\pi}{\alpha} \, \frac{\cosh \alpha x}{\sinh \alpha\pi}$ .

[quote="Albanian Eagle"]Calculate $ \sum_{n = 1}^{\infty}\frac {\cos n}{n^2 + 1}$  :)[/quote]
$ \sum_{n = 1}^{\infty}\frac {\cos n}{n^2 + 1} = \frac {1}{2}Re \sum_{n \in \mathbb Z}\frac {( - 1)^{n}e^{in(1 - \pi)}}{n^2 + 1}$
an just consider:
$ 2\pi i \sum _{res} = \int_{C_{n}}\frac {e^{iz(1 - \pi)}}{\sin(\pi z)(z^2 + 1)}\to 0$ as $ n \to \infty$ where $ C_{n}$ is big rectangle

Also the formula $ \sum_{k\in\Bbb{Z}} \frac {( \minus{} 1)^k\, \cos kx}{k \plus{} \alpha} \equal{} \pi\,\frac {\cos \alpha x}{\sin \alpha\pi} \qquad \minus{} \pi < x < \pi \,\, , \, \, \alpha\notin\Bbb{Z}$

we can compute with [hide="Poissons summation formula"]
For $ 0 < x < 2\pi$ and $ \alpha\in\Bbb{C}\setminus i\Bbb{R}$

$ S \equal{} \sum_{k\in\Bbb{Z}} \frac {e^{ikx}}{k \minus{} i\alpha} \equal{} \sum_{k\in\Bbb{Z}} \int_{ \minus{} \infty}^\infty \frac {e^{itx}}{t \minus{} i\alpha} \, e^{2\pi i k t}\, dt$

$ \equal{} \sum_{k \equal{} \minus{} \infty}^{ \minus{} 1} \,\,\oint\limits_{\text{lower half} \atop \text{moon}} \frac {e^{ixz}}{z \minus{} i\alpha} \,e^{2\pi i k z}\, dz \plus{} \sum_{k \equal{} 0}^{\infty} \,\,\oint\limits_{\text{upper half} \atop \text{moon}} \frac {e^{ixz}}{z \minus{} i\alpha}\, e^{2\pi i k z}\, dz$

$ \equal{} \minus{} 2\pi i \sum_{k \equal{} \minus{} \infty}^{ \minus{} 1} \left\{\begin{matrix} e^{ \minus{} \alpha x}\, e^{ \minus{} 2\pi k\alpha} & \text{Re}(\alpha) < 0 \\
 \\
0 & \text{Re}(\alpha) > 0 \end{matrix}\right\} \plus{} 2\pi i \sum_{k \equal{} 0}^\infty \left\{\begin{matrix} 0 & \text{Re}(\alpha) < 0 \\
 \\
e^{ \minus{} \alpha x}\, e^{ \minus{} 2\pi k\alpha} & \text{Re}(\alpha) > 0 \end{matrix}\right\}$

So for $ \text{Re}(\alpha) < 0 : \quad S \equal{} \minus{} 2\pi i \, e^{ \minus{} \alpha x} \sum_{k \equal{} 1}^\infty e^{2\pi \alpha k} \equal{} 2\pi i \,\frac {e^{ \minus{} \alpha x}}{1 \minus{} e^{ \minus{} 2\pi \alpha}}$

And for $ \text{Re}(\alpha) > 0 : \quad\!\! S \equal{} 2\pi i \, e^{ \minus{} \alpha x} \sum_{k \equal{} 0}^\infty e^{ \minus{} 2\pi \alpha k} \equal{} 2\pi i\, \frac {e^{ \minus{} \alpha x}}{1 \minus{} e^{ \minus{} 2\pi \alpha}}$

Hence for $ 0 < x < 2\pi$ and $ \alpha\in\Bbb{C}\setminus \Bbb{R}$

$ \sum_{k\in\Bbb{Z}} \frac {e^{ikx}}{k \plus{} \alpha} \equal{} 2\pi i\, \frac {e^{ \minus{} i\alpha x}}{1 \minus{} e^{ \minus{} 2\pi i\alpha}} \equal{} \pi \, \frac {e^{ \minus{} i(x \minus{} \pi) \alpha}}{\sin \alpha \pi}$

After shifting $ x$ to $ x \plus{} \pi \quad \sum_{k\in\Bbb{Z}} \frac {( \minus{} 1)^k\, e^{ikx}}{k \plus{} \alpha} \equal{} \pi\, \frac {e^{ \minus{} i\alpha x}}{\sin \alpha\pi}$ for $ \minus{} \pi < x < \pi$ .

[/hide]
or with the [hide="residue theorem"]
For $ \minus{}\pi<x<\pi$ and $ \alpha\notin\Bbb{Z}$ define $ f(z) \equal{} e^{ixz}\, \frac {\pi}{\sin \pi z}\, \frac {1}{z \plus{} \alpha}$

and $ \gamma_n$ as the square with the corners $ (\pm 1\pm i) \left(n \plus{} \frac12\right)$ .

$ \lim_{n\to\infty} \oint_{\gamma_n} f \, dz \equal{} 0$ since because of

$ |f(z)|^2 \equal{} \left|\frac {e^{ix(u \plus{} iv)}}{\sin\Big(\pi(u \plus{} iv)\Big)}\right|^2 \equal{} \frac {e^{ \minus{} 2xv}}{\sin^2(\pi u) \plus{} \sinh^2(\pi v)}$

$ f$ is asymptotic damping as $ e^{ \minus{} xv}\, e^{ \minus{} \pi |v|} \, \frac {1}{n}$

$ \text{res}(f,k) \equal{} e^{ikx}\, \frac {1}{\cos kx} \, \frac {1}{k \plus{} \alpha}$ and $ \text{res}(f, \minus{} \alpha) \equal{} e^{ \minus{} i\alpha x}\, \frac {\pi}{ \minus{} \sin \pi \alpha}$

The sum of all residues must be zero: $ \sum_{k\in\Bbb{Z}} \frac {( \minus{} 1)^k e^{ikx}}{k \plus{} \alpha} \equal{} \pi\, \frac {e^{ \minus{} i\alpha x}}{\sin \alpha\pi}$
[/hide]


inequalities
function
inequalities proposed

For $ x\in [0,1] $ prove that $ 3f(x) \le \arctan{x} \le 3.04f(x) $ , where  $ f(x)=\frac{x}{1+ 2\sqrt{1+x^2}} . $

The left side is quite easy. You make the substitution x=tan u with 0<=u<=Pi/4.
The left part of the inequality gets: 3sinu/(2+cosu)<=u or

0<=2u+ucosu-3sinu.

The function on the right side is monotoniously increasing in [o,Pi/4] and it has value 0 for u=0.

Misha

Following idea of Misha123
For $0\leq u \leq \frac{\pi}{4}$
$h(u)=\frac{3sinu}{2+cosu}-u$
$h'(u)=-\frac{(cosu -1)^2}{(cosu +1)^2} \leq 0$
$h$ is decreasing on $[0;  \frac{\pi}{4}]$ $h(x) \leq h(0)=0$

I drawed the curve of  $k(x)=\frac{\arctan{x}}{x}.(1+2\sqrt{1+x^2})$ with a pocket
calculator, I found it is increasing on $[0;1]$

$k(1)=\frac{\pi}{4}(1+2\sqrt2)=3,0068...$

It seems we can replace 3,04 by 3,007


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Given a $n \times n$ matrix $A$ over rationals.Show that $A^p=I$ for a prime $p>n+1$ implies $A=I$.

If it had an eigenvalue different from $1$, then it would have to have all $p-1$ primitive $p$-th roots of unity as eigenvalues, because $x^{p-1}+\ldots+x+1$ is irreducible. This, however, is impossible, because $n<p-1$. This means that all its eigenvalues are $1$, and since it's diagonalizable, that's that.

[quote="perfect_radio"]Why is it diagonalizable?

Sorry if this is a stupid question, but I'm not very familiar with the concept.[/quote]
It follows from $A^p=I_n$. The minimal polynomial of $A$ must divide the polynomial $x^p-1$ which has only simple roots and thus the minimal polynomial $m_A(x)$ also has only simple roots, so for example using Jordan form over any algebraically closed field, that implies the diagonalisability of $A$(all the Jordan block would be of size $1 \times 1$)

[quote="perfect_radio"]I don't understand what you are saying. Doesn't $A$ have only a $n \times n$ Jordan block since it has eigenvalue $1$ with multiplicity $n$?[/quote]
Ok, even if it has only one eigenvalue $1$ it may have lots of Jordan blocks, which may not be of size $1 \times 1$.
What i'd like to propose you: start with the general case and prove the following:

If for a matrix $A$(over $K$), there exists $k$ such that $A^k=I_n$, then $A$ is diagonalisable(over $K$).

Proof:since $x^k-1$ annulate $A$, then $x^k-1$ is divisible by $m_A(x)$, but $x^j-1$ has no multiplied roots, which imples that $m_A(x)$ also has no multiplied roots, so $m_A(x)=(x-x_1)(x{}_x{}_2) \cdots (x-x_l)$-did you know that if minimal polynomial if of that form then it the mtarix $A$ is diagonalisable ? if you did-everything is ok, if you didn't try to prove it using Jordan form,.. if you won't prove it, ask, after understanding the said above i'll write this last step.

Thanks... Fortunately, in this problem we needn't apply Jordan Theorem. The characteristic polynomial of $A$ is $(x-1)^n$, so $m_A | (x-1)^n$. Also, as you said, $m_A | (x^p - 1)$. We have $\gcd \left( (x-1)^n,x^p-1 \right) = x-1$, so it's obvious that $m_A = x-1$, hence the conclusion.

BTW, you say that if $m_A$ has only simple roots, then $A$ is diagonazible? I don't see how to prove that... All help is welcomed.

[quote="perfect_radio"]

BTW, you say that if $m_A$ has only simple roots, then $A$ is diagonazible? I don't see how to prove that... All help is welcomed.[/quote]

$m_{A}(x)=(x-a(1))...(x-a(k))$ simple root a(1),...,a(k)

$0=(A-a(1)I_n)...(A-a(k)I_n)$

$K^{n}=\oplus_{j=1}^{k}ker(A-a(j)I_n)$

pick a basis in each $ker(A-a(j)I_n)$
you get a basis in wich the endomorphism
associate to A is diagonal

I understood the first part, I think. Just to be sure... $\oplus$ means sum of subspaces, right?

But... This part is very unclear to me:

[quote="Moubinool"]pick a basis in each $ker(A-a(j)I_n)$
you get a basis in wich the endomorphism
associate to A is diagonal[/quote]

What's the base we are choosing? Hope you understand my question.


[quote="prototypex24"]$\displaystyle \cos{4x}+\sin{3x}-\cos{2x}+\sin{x}=0$
$(2\cos^{2}{2x}-1)+(3\sin{x}-4\sin^{3}{x})-(1-2\sin^{2}{x})+\sin{x}=0$
$(2(1-2\sin{x})^{2}-1)-4\sin^{3}{x}+2\sin^{2}{x}+4\sin{x}-1=0$
[/quote]

I think this step needs a correction as

$Cos2x=1-2sin^2x$
and that results a fourth degree equation and not third degree

[quote="jelly119"]Given that $\cos{4x}+\sin{3x}-\cos{2x}+\sin{x}=0$, find $x$.[/quote]

[hide="ignore this"]
wrong
[/hide]

EDIT : Opps ... guess i have to work out again  :blush:

[quote="shyong"][quote="jelly119"]Given that $\cos{4x}+\sin{3x}-\cos{2x}+\sin{x}=0$, find $x$.[/quote]

[hide]

$\cos{4x}+\sin{3x}-\cos{2x}+\sin{x}=0$
 :arrow:  $2\cos 3x\cosh + 2\sin2x\cos x=0$
 :arrow:  $\cos x(\cos 3x +\sin 2x)=0$
 :arrow:  $\cos^2x(4\cos^2x-3+2\sin x)=0$
 :arrow:  $\cos^2x(4\sin^2x-2\sin x-1)=0$

Hence we have

$\cos x=0$ and $\sin x=\frac{1+\sqrt 5}{4}$ or $\frac{1-\sqrt 5}{4}$

 the rest is trivial  ;)[/hide][/quote]

My intention is not to point out ....but there is a mistake in shyong's solution too....(in the very first step of using transformations)

As per my solution, sinx =0 and a number slightly greater than -1 are roots for this equation

Ok thanks for pointing it out [b]vidyamanohar[/b]

but the most I can go is

$\sin x(4\sin^3x-2\sin^2x-3\sin x+2)=0$

The cubic equation I just BF it out with calculator and obtain $sin x\approx -0.9156$

Or , is there any other way ? :?

SORRY!! I made a mistake in the original question. The equation should be:
$\displaystyle \cos{4x}+\sin{3x}-\cos{2x}+\sin{x}=\frac12$

there is a general solution to cubic equations, it is a pain but there is a substitution that helps x = y + a/3
then for some d^{1/3} and e^{1/3}, substiuting, gives a quadratic, sorry my explanation is bad, i dont want to type all of it, it is in the aops book 1

my solution is a long one, so i'll give you my answers for now.

[hide]

[tex]sin^2 \ x = \frac {1}{4}[/tex] so [tex]sin \ x = \pm {\frac {1}{2}}[/tex]

so [tex] x = 30^o \ or \ 150^o[/tex]

[/hide]

a full solution will follow on a later post


Clearly Obi-Wan-Kenobi. Runner up is Yoda and Anakin (before turing to the dark side).

[quote="colts18"]I like Darth Vader[/quote]Darth Vader is no Jedi! ;)

[quote="math92"]Who is your favorite Jedi out of all the Star Wars movies?[/quote]

Obviously Yoda.  :P

[quote="yif man12"]Jar Jar Binks  :ninja:  :huh:[/quote]Again, not a Jedi. Maybe we should put the definition of a Jedi inside the topic :P  :roll:  :gleam:

Heh, whoops. Well, I'll go with the popular consensus and pick yoda then.

[quote="Valentin Vornicu"]This topic deserves a poll, so here it is. Please vote :P[/quote]
thanks for putting a poll, all the polls i make get messed up

[quote="GoBraves"]Yay Obi Wan's winning

:starwars:[/quote]Was winning. Until Yoda got mad :D

Luke Skywalker is da bomb because he kills his father

MACE WINDU. well he was cool .... in the 3rd movie he seemed deluded ?

Mace Windu all the way...he totally pwned the emporer
did yoda? no...
did luke? no...
did anakin? yes...but that's cuz he backstabbed just like he did w/ windu...so that doesn't really count

but unfortunatley
[hide]
mace windu died.  [/hide]

yea everybody dies that's pretty weird huh

well, yoda didn't die(in the 3rd one)
Anikan didn't die (in the 3rd one)
but mostly everyone dies by the 6th one :D

except for lord voldemort woooo 3 cheers for lord voldemort

Cheer
Cheer
Cheer
 :clap:
but Voldemort is in Harry Potter

[quote="mathclass"]I don't like Luke because of the actor's eyes.[/quote]
.
I'd agree with you on this, but... look so good after a car accident and facial reconstruction you will not..

Mace Windu all the way!! :D He's really smart and wise and cool.  Besides, he was the one that knew who the Sith Lord was.  And, did they pick Yoda to capture Palpatine? No, so that means Mace Windu is better!! :D
 :starwars:

yoda would be the best, then
Obi Wan Kenobi when he is not Ben,
Anikan Skywalker, before he turns bad,
then Luke and Windu.

Mace Windu is da bomb.  He is the only Jedi to defeat Palpatine. Unfortunatly Annikin ran up and chopped off his arm :mad:  :mad:  :mad: i got so mad at that part Mace Windu is such a cool character.


A right triangle has hypotenuse 7.6 units and one leg measures 3.2 units. Find the measure of the other leg rounded to the nearest tenth.

Using Pythagorean Theorem $ a^{2} \plus{} b^{2} \equal{} c^{2}$, you find the length of the third side.

$ 3.2^{2} \plus{} b^{2} \equal{} 7.6^{2}$

$ 10.24 \plus{} b^{2} \equal{} 57.76$

$ b^{2} \equal{} 47.52$

$ b\approx\boxed{6.9}$


floor function
invariant
calculus
derivative
superior algebra
superior algebra unsolved

Wikipedia says that a hyperelliptic curve is a curve (in the affine plane) of the form $ y^{2}\equal{}f(x)$ where $ f$ has distinct roots.
Confusion 1) - it goes on to say that we require the degree of $ f$ to be strictly greater than 4 as otherwise the resulting curve is elliptic. If the degree is 4 i was under the impression that the resulting curve had genus 3 and so would certainly not be elliptic.
Confusion 2) - further into the article it says that such curves have singular points at infinity in the projective plane (claiming that that this property is unique to the case where the degree of $ f$ is at least 5). I can't see why this is the case. Thanks in advance for any help given.

Probably $ n > 4$ should be $ n \geq 4$; the first image they have there uses degree 4.

It might be that the image doesn't represent a hyperelliptic curve though - it's not clear which parts of the article to trust and which not to. For example - twice it is written $ n>4$ and once it is written "if $ f$ is a cubic or quartic...".

Yes, you may be right -- the talk page thinks it's a bad article:
http://en.wikipedia.org/wiki/Talk:Hyperelliptic_curve
At least some of it (like one of the sentences you quote) is just pulled from Mathworld (or, I suppose, vice versa).  Probably it is advisable to find a more authoritative source.

$ n > 4$ is correct.  When $ f$ has degree $ d$ and distinct roots, the genus of the resulting curve is $ \left\lfloor \frac{d\minus{}1}{2} \right\rfloor$; in particular when $ d \equal{} 4$ the genus is $ 1$, not $ 3$.  You can verify this for yourself by constructing the corresponding [url=http://lamington.wordpress.com/2009/09/23/how-to-see-the-genus/]Newton polygon[/url].  Alternately, the genus is a birational invariant and you can show directly that if $ f$ has degree $ 2n$, then $ y^2 \equal{} f$ is birational to a curve of the form $ y^2 \equal{} g$ where $ g$ has degree $ 2n\minus{}1$.  (Essentially you send one of the roots of $ f$ to infinity.)

You may be being misled by the usual claim that a generic curve of degree $ d$ has genus $ \frac{(d\minus{}1)(d\minus{}2)}{2}$.  This formula only holds for smooth curves, and a projective curve of the form $ y^2 z^2 \equal{} x^4 \plus{} ax^3 z \plus{} bx^2 z^2 \plus{} cxz^3 \plus{} dz^4$ has a singularity at $ (0 : 1 : 0)$, which you can verify by computing the partial derivatives there.  (Similar singularities appear in higher degree.)

Thanks for the clarification, it seems the wiki page is in serious need of some attention.
@ t0rajir0u: if we have a plane curve given by the equation $ y^{2}\equal{}g(x)$ with the degree of $ g$ equal to $ 2n\minus{}1$ then we may as well assume that 0 isn't a root of $ g$, and then the map sending $ (x,y)$ to $ (x^{\minus{}1},yx^{\minus{}n})$ is the map to which you refer, yes? (Also -  Newton Polygons; very cool.)


one more for ya.....can't use calculus and have to use geometry.

A rectangular sheet of paper is to contain 72 square inches of printed matter with 2-inch margins at the top and bottom and 1-in margis on each side.  What dimensions for the sheet will use the least paper?

[quote="K81o7"]We have $ab=72$ and want to minimize $(a+2)(b+4)=ab+2b+4a+6=78+2b+4a$
So we must minimize $b+2a$.  Note $b=\frac{72}{a}$.  Then we have to minimize $2a+\frac{72}{a} \implies a=6$, so $b=12$ , giving dimensions 6 x 12.

jmccray, I don't think this problem belongs in the olympiad section either.  This one should probably go in getting started...[/quote]Sorry for posting in the wrong forum, but I do have one last question.  How do we get from minimizing 78+2b+4a to minimizing b+2a?

[quote="jmccray"]Sorry for posting in the wrong forum, but I do have one last question.  How do we get from minimizing 78+2b+4a to minimizing b+2a?[/quote]

Well, $78+2b+4a = 78 + 2(b+2a)$. If you are able to find the maximum of $b+2a$, then that will correspond to the maximum of $2(b+2a)$ and hence also to the maximum of $78 + 2(b+2a)$. All operations you're doing with the number $b+2a$ preserve its condition of being maximum. Hope you undertand now

Daniel


Is there something special for quadrilateral ABCD that has incircle and can be also inscribed in a circle (cyclic)
(Espacially relations between sides and other special things about this kind of quadrilaterals). :?:

[quote="mstoenescu"]I think it must be cyclic (i.e. the sum of opposite angles à 180°) and exscribed (i.e. the sum of opposite sides are equal ). :roll:[/quote]

VERY good answer.  :rotfl:

In fact, such quadrilaterals are called "bicentric quadrilaterals", and searching for "bicentric" will yield a number of nice problems about them.

http://www.mathlinks.ro/Forum/viewtopic.php?t=146867 (dear mods, this does NOT belong into Intermediate Topics)
http://www.mathlinks.ro/Forum/viewtopic.php?t=31693
http://www.mathlinks.ro/Forum/viewtopic.php?t=5510
http://www.mathlinks.ro/Forum/viewtopic.php?t=23192

are maybe the most important ones.

  Darij

[quote]VERY good answer.

[/quote]

I haven't lough this much for a long time before but when I saw this....
Maybe he didn't understand me well (he thought that I am begginer).

Many thanks to Darij and especially to mstoenescu.

Many thanks to Darij who opens to me a fantastic world of these quadrilaterals by the topics founded !

Have a lot of fun  delegat ! The beginner is ME and I'm glad to be ! And long live to Sarajevo ! :blush:

[quote]Many thanks to Darij who opens to me a fantastic world of these quadrilaterals by the topics founded !

Have a lot of fun delegat ! The beginner is ME and I'm glad to be ! And long live to Sarajevo ! [/quote]

OK I didn't want to write anything bad to you
My question was about these bicentric quadrialterals but you gave answer to cyclic quadrialteral and then to quadrilateral that can be exscribed (no relation between them)
Your answer didn't make me laufhing Darij's answer especially  :rotfl: did it.
Maybe you wrongly understood my question so your answer was...

What do you have against Sarajevo?

Sorry for leading this topic into a flame, but I mistook Mstoenescu's post as a spoof of Delegat. I hope everything is settled now.

  Darij

[quote="delegat"]

What do you have against Sarajevo?[/quote]

I think you misunderstood me about that ! On the contrary, I'm sympathizer of Sarajevo !

[quote]I think you misunderstood me about that ! On the contrary, I'm sympathizer of Sarajevo !
[/quote]

OK Then it is obvious that I misunderstood you I though that you don't like Sarajevo or something like that.

I think that we can close this topic (either if somebody has something about bicetric quadrilaterals).


Find three natural numbers a, b, c such that:
\[ 4(a\plus{}b\plus{}c) \plus{}14 \equal{} abc\]

got lucky and found the answer
[hide="answer"]
2, 5, 7[/hide]
now trying to find a reason

[hide="Solution inspired by above"]WLOG, let $ a\leq b\leq c$. Putting $ a \equal{} 1$ then $ a \equal{} 2$ and using SFFT gives the 4 solutions from above: $ (1,5,38), (1,6,21), (2,3,17), (2,5,7)$.
The case $ a \equal{} 3$ gives $ 26 \plus{} 4b \plus{} 4c \equal{} 3bc$. Let $ b \equal{} x \plus{} 2$, $ c \equal{} y \plus{} 2$, so that we can solve for $ x,y$ in naturals instead of for values $ \geq 3$. Also, for easier solving, drop the restriction that $ x\geq y$.
Then $ 30 \plus{} 4x \plus{} 4y \equal{} 3xy \plus{} 6x \plus{} 6y \plus{} 12 \iff 18 \equal{} 3xy \plus{} 2x \plus{} 2y$. Examining this mod 2, we can WLOG let $ x \equal{} 2z$. Then $ 18 \equal{} 6yz \plus{} 4z \plus{} 2y \iff 9 \equal{} 3yz \plus{} y \plus{} 2z \equal{} y(3z \plus{} 1) \plus{} 2z$. Examining this mod 2, we can let $ z \equal{} 2w$. Then $ 6 \equal{} 6wy \plus{} y \plus{} 4w\geq11$. Thus, no solutions exist for $ a \equal{} 3$.
For the case $ a \equal{} 4$, let $ a \equal{} t \plus{} 4$, where $ t\in \mathbb{N}_0$.
(Edit: Then let $ x\equal{}t$.  :mad: )
Then the equation is equivalent to $ 34 \plus{} 4x \equal{} 4(b \minus{} 1)(c \minus{} 1) \plus{} xbc$. We have $ 4(b \minus{} 1)(c \minus{} 1)\geq 4\cdot3\cdot3 \equal{} 36 > 34$, and $ xbc\geq16x\geq4x$, so this equation can never be satisfied. Thus, the only solutions are those 4 solutions and permutations.[/hide]
:P Mathematica pwns.


Find all [tex]f:\mathbb{N}\,_0\to\mathbb{N}\,_0[/tex],
such that
[tex]f(f(n)) + f(n) = 2n+6[/tex],
where
[tex]\mathbb{N}\,_0=\mathbb{N}\,\cup\left\{0\right\}[/tex]

Well, heres a way that will work but is a little bit heavy on the algebra side, so I haven't worked 100% of the way through it:

[hide]

Let f(a) = b for any a. Then substitute a into the equation to get f(b) = 2a-b+6. Then substitute b into the equation to get f(2a-b+6) = 3b-2a.

We then continue this repeatedly, and prove by induction the pattern we find. This isn't as bad as it looks, to get the a and b terms involve doubling the previous one and possibly adding an a or b, the constant term i'm not 100% sure about, but there will be a pattern.

Anyway, after working out the formula by induction, set each term >=0. Every second term will give us a lower bound for b, every other term an upper bound. Take these to infinity, and I'm 99.9% sure that they will both converge to b = a+2.

[/hide]

That is similar to one of the solutions I have seen, although [hide] it does not take it to infinity, but rather stops it after a finite number of steps[/hide]. See what you can do with that. There are four solutions in this book I have, along with one that I came up with.

And any solution you can find <b>will</b> be algebra-heavy, or case-heavy.

Mine is case-heavy. (By heavy I mean about 7 small proofs by contradiction)


topology
superior algebra
superior algebra unsolved

How can I show Gal (E/F) where F<E fields, is a topological group under the Krull topology?


Let $ (x^2\plus{}x\minus{}1)^5\equal{}a_0\plus{}a_1x\plus{}...\plus{}a_{10}x^{10}$. Calculate $ S\equal{}a_0\plus{}a_4\plus{}a_8$.

Thank.

[quote="BillClin"]Let $ (x^2 \plus{} x \minus{} 1)^5 \equal{} a_0 \plus{} a_1x \plus{} ... \plus{} a_{10}x^{10}$. Calculate $ S \equal{} a_0 \plus{} a_4 \plus{} a_8$.

Thank.[/quote]
 :)  It isn't an inequality pro. so don't post it here :wink: .
Because deg f=5 which is not huge so there're various way to solve.
The traditional way is using expanding $ (a \plus{} b)^n$.
The other way is taking 10 value of x then solve for $ a_1,a_2... a_{10}$ :)


[quote="GameBot"]If $ j$ and $ k$ are inversely proportional and $ j \equal{} 16$ when $ k \equal{} 21$, what is the value of $ j$ when $ k \equal{} 14$?[/quote]
$ 14j \equal{} 16*21 \iff j \equal{} \frac{16*21}{14} \equal{} 24$

[quote="ernie"]When two numbers are inversely proportional, then $ ab$ has to remain constant for values of $ a$ and $ b$.[/quote]
Yes, and $ 16*21 \equal{} 14*24$...?

ernie is just giving the definition of inversely proportional
he is not questioning your answer


modular arithmetic
combinatorics open
combinatorics

Consider positive integers m and n, m > 2 and n > 1. Prove that if the set {1,2,3...mn} be partitioned into m sets each of size n in any manner whatsoever, then we can select 1 element from each set such that the sum of [m/2] of the elements equals the sum of the remaining elements.

Else find a counterexample.

The problem is solved only for the cases m = 3 and m = 4

m=6 gives a counterexample.
Consider a partition $\{1,2,\dots,mn\}= P_{0}\cup P_{1}\cup\dots\cup P_{m-1}$ where $P_{i}$ contains all elements $x$ such that $x\equiv i\pmod m$.
Suppose that we selected elements $x_{0}, x_{1}, \dots, x_{m-1}$ from the sets $P_{0}, P_{1}, \dots, P_{m-1}$ respectively. Then \[x_{0}+x_{1}+\dots+x_{m-1}\equiv 0+1+\dots+m-1 \equiv \frac{m(m-1)}{2}\pmod{m}\] implying that $x_{0}+x_{1}+\dots+x_{m-1}$ is odd (for $m=6$). Therefore, a set $\{ x_{0}, x_{1}, \dots, x_{m-1}\}$ cannot be split into two with equal sums.

Yes, your counterexample does work for $n=6$.  It works for any number of the form $4n+2$.

Can we make then prove the positive result for numbers that are not of the form $4n+2$? Or find a counterexample there as well?


linear algebra
matrix
complementary counting
algebra
system of equations

How many $ 0\minus{}1$ sequences of length $ n$ there are, which don't have three consecutive $ 1$?

For $ n\equal{}4$ such sequence is for example $ 1011$, but $ 0111$ doesn't fulfill conditions.

By complementary counting and the principle of inclusion-exclusion, I have $ 2^{n}\minus{}2^{n\minus{}3}(n\minus{}2)\plus{}2^{n\minus{}4}(n\minus{}3)\minus{}2^{n\minus{}5}(n\minus{}4)\plus{}...$.

[hide="Recursion approach"]For $ n\equal{}3$ we have $ a_3\equal{}2^3\minus{}1\equal{}7$ that work. These are:

000
001
010
100
011
101
110

Now we'll think about $ a_4$. In the cases ending with 00, 01, or 10, we may tack on either a 1 or a 0 and create two more numbers that work. In the cases ending with 11, we can add a 0 and get a number that works.

So now let's define a sequence $ b_n$ of the number that end with 00, or 10. We define $ c_n$ as the number that end in $ 11$. Also, let $ d_n$ be the number that end in $ 01$. (This'll make sense in a second). We have that $ a_n\equal{}b_n\plus{}c_n\plus{}d_n$, of course. We also have that $ a_{n\plus{}1}\equal{}2(b_n\plus{}d_n)\plus{}c_n$.

We see that $ b_3\equal{}4$, $ c_3\equal{}1$, and $ d_3\equal{}2$ Based on that, $ a_4\equal{}13$, which we can easily verify is correct, so we're on the right track.

Now we can see that if a number ends with 00 or 10, if we get two $ b$ values from it and two $ d$ values. If it ends with 01 or 11 we'll get a $ c$ value and a $ b$ value but no $ d$.

So $ b_{n\plus{}1}\equal{}2b_n\plus{}c_n\plus{}d_n$

And $ c_{n\plus{}1}\equal{}c_n\plus{}d_n$

And $ d_{n\plus{}1}\equal{}2b_n\plus{}c_n$.[/hide]

Right, so, that got kind of ugly... Finding a closed form based on that would probably be difficult, but if you needed to find the value for some small $ n>3$ you could use this to generate values somewhat quickly.

[quote="grn_trtle"][hide="Recursion approach"]For $ n \equal{} 3$ we have $ a_3 \equal{} 2^3 \minus{} 1 \equal{} 7$ that work. These are:

000
001
010
100
011
101
110

Now we'll think about $ a_4$. In the cases ending with 00, 01, or 10, we may tack on either a 1 or a 0 and create two more numbers that work. In the cases ending with 11, we can add a 0 and get a number that works.

So now let's define a sequence $ b_n$ of the number that end with 00, or 10. We define $ c_n$ as the number that end in $ 11$. Also, let $ d_n$ be the number that end in $ 01$. (This'll make sense in a second). We have that $ a_n \equal{} b_n \plus{} c_n \plus{} d_n$, of course. We also have that $ a_{n \plus{} 1} \equal{} 2(b_n \plus{} d_n) \plus{} c_n$.

We see that $ b_3 \equal{} 4$, $ c_3 \equal{} 1$, and $ d_3 \equal{} 2$ Based on that, $ a_4 \equal{} 13$, which we can easily verify is correct, so we're on the right track.

Now we can see that if a number ends with 00 or 10, if we get two $ b$ values from it and two $ d$ values. If it ends with 01 or 11 we'll get a $ c$ value and a $ b$ value but no $ d$.

So $ b_{n \plus{} 1} \equal{} 2b_n \plus{} c_n \plus{} d_n$

And $ c_{n \plus{} 1} \equal{} c_n \plus{} d_n$

And $ d_{n \plus{} 1} \equal{} 2b_n \plus{} c_n$.[/hide]

[/quote]

let $ a_n$ be the number of sequences of length $ n$.

If we take any  of the $ a_n$ sequences of length $ n$ we can add a $ 0$ to the start to make a sequence length $ n \plus{} 1$

We can also add a $ 1$ to the front of the sequence given it doesn't start with $ 11$

The number of sequences that start with $ 11$ is $ a_{n \minus{} 3}$: because if a seqence starts with $ 11$ then it must start $ 110$ then if we append any of the $ a_{n \minus{} 3}$ sequences to $ 110$ we get our set.

Hences $ a_{n \plus{} 1} \equal{} 2a_n \minus{} a_{n \minus{} 3}$

[quote="grn_trtle"]Right, so, that got kind of ugly... Finding a closed form based on that would probably be difficult, but if you needed to find the value for some small $ n > 3$ you could use this to generate values somewhat quickly.[/quote]
It's not ugly at all!  The system of equations you write down can be written in matrix form, and then the question is how to compute arbitrary powers of a fixed matrix.  This is done by finding its eigenvalues and then diagonalizing, which is a standard computation in linear algebra.


I want this as my sig:

[code]There are 10 kinds of mathematicians. Those who can think binarily and those who can't.

[url]http://www.vf11.com/nebula/nebula.html[/url][/code]
But it says:

[code]Signature: the value is too long.[/code]

Well, I don't know why this happened, but you should probably try to be more polite. :wink:

We've shortened the length of allowable signatures.  This is only for new signatures -- if you had a long signature before we made the change, it's grandfathered in (although we still, as always, reserve the right to edit your signature without warning if it's inappropriate).

Although the signature that you wrote above is not abusive nor objectionable (the link is OK, since it's not a blind link -- anybody who clicks on it knows that it is taking them off-site), it is too long under the new policy.  It must be shorter than 128 characters.


Hey everybody, I have a couple questions regarding the USNCO free response section.

What kind of criteria do they grade the free response on?
Do they pay a lot of attention to sig figs?
Do they want you to write out every equation you use?
Should you write out all your reasoning behind using the equation?
Should states of matter be written for equation writing, and just for substances in general?
If anyone has any tips or pointers regarding the free response section regarding the kind of format the graders are looking for, it'd be greatly appreciated.  Thanks!

does anybody know the answer to this question?

ACS has managed to never reveal any details about the national exam...
My only advice is therefore to be as accurate as possible... obviously dont write down something without explaining it... and write as much MEANINGFUL information you can. Im assuming that some old grumpy chemistry teachers will be grading this and after grading hundreds of papers, they arent going to be happy seeing a paper with sloppy handwriting and little explanation.


geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations

I understand that when simplifying $\frac{\sqrt{x+h}-\sqrt{x}}{h}$ , you use the conjugate to rationalize the numerator. Likewise when taking the cube root of x+h - the cube root of x, you use the difference of cubes formula. Or, when taking the fourth root, you use the conjugate concept twice.

But, I am confused when it comes to fifth roots.
I want to rationalize $\sqrt[5]{(x+h)^{2}}-\sqrt[5]{x^{2}}$, but I cannot figure out how to do it. Help?

[hide="My initial thoughts on this"]The difference of cubes formula can be extended.

e.g.
$x^{5}-y^{5}=(x-y)(x^{4}+x^{3}y+x^{2}y^{2}+xy^{3}+y^{4})$

Or

$x^{n}-y^{n}=(x-y)(x^{n}+x^{n-1}y+x^{n-2}y^{2}+\ldots+xy^{n-2}+y^{n-1})$

[/hide]

Try using rational exponents:
$(x+h)^{2/5}-x^{2/5}$
$((x+h)^{1/5})^{2}-(x^{1/5})^{2}$
This is a conjugate using squares.
Say our whole fraction
$\frac{(x+h)^{2/5}-x^{2/5}}{h}$
needs to be rationalized.
Well, we see that:
$\frac{((x+h)^{1/5})^{2}-(x^{1/5})^{2}}{h}$
We can use a conjugate of squares.

Multiply by the conjugate of ${((x+h)^{1/5})^{2}-(x^{1/5})^{2}}$.
The conjugate is:
${((x+h)^{1/5})^{2}+(x^{1/5})^{2}}$
Then work on eliminating the fifth roots.


Triangle ABC is acute

Prove SQRT(a^2 +b^2 -c^2)  + SQRT (b^2 +c^2 -a^2) +SQRT(c^2+a^2-b^2)

<= a+b+c


geometry
geometric transformation
rotation
Gamebot

An international meeting is held between England, Germany, and France.  Three representatives attend from England, four from Germany, and two from France.  How many ways can all nine representatives sit around a circular table, if representatives of the same country sit together?

this problem makes me feel really stupid... :!:

Well, there are $ 3$ different countries, so we can order them in $ 3!$ or $ 6$ ways.  There are $ 3!$ or $ 6$ ways to rearrange the English.  There are $ 4!$ or $ 24$ ways to arrange the German.  There are $ 2!$ or $ 2$ ways to arrange the French.  However, we must divide by $ 9$ to account for rotations, so we have $ 6 \times 6 \times 24 \times 2 \div 9$ or $ \boxed{192}$ ways.

wow.  ernie.  that's the same answer/solution i got....and guess what?  WRONG.
i don't get why my solution is wrong....help please?   :lol:

Well, what's the answer?
And,
Ernie,why did you divide by nine?

Edit: Oh, is it because the table is round? You would only have 2 ways of seating the countries.

Er....the answer was $\boxed{576}$. You start out by considering the countries in blocks, like Germany one block, England on block, and France one block. You have $(3-1)!=2$ ways to order them around the table. Then you consider how many ways to order the English group, $3!=6$, then $4!=24$ for Germans, and then $2!=2$ for the French. There for we have $2!\times3!\times4!\times2!=2\times6\times24\times2=\boxed{576}$

[hide="Click here for the official answer"]To begin, consider the number of ways to arrange the three countries around the circle. We can consider the English representatives a block, the Germans another block, and the French a third block. There are $(3-1)!=2$ ways to arrange these three blocks around a circle. Within the English group, there are $3!=6$ ways to arrange the three representatives. Similarly, there are $4!$ ways to arrange the Germans and $2!$ ways to arrange the French representatives. Overall, the total number of ways to seat the 9 representatives is: $2!\times3!\times4!\times2!=2\times6\times24\times2=\boxed{576}$[/hide]

Why would it not be 576/9 accounting for rotations or 64 ways to seat people at the table?

Dividing by 9 is accounting for individual rotations. However, note that we are rotating the "blocks" or countries, not the individual representatives themselves. You already multiplied by 3! to find the ways to rotate the countries.

[quote=pinkmuskrat]Yes, it is because the table is round. Also, the answer isn't two, because you need to worry about how everyone sits within their country group.

Are you sure it's not just $ 4!*3!*2!/9 \equal{} 32$? In my opinion, this would make sense. Why even multiply by 6? I don't think there is anymore than 1 way to organize the country groups.

Bwu- what was the answer?[/quote]

wait this one is not in the contry groups you mutiplyed

Sorry for the bump, but [hide=here's an alternative.]We can think of grouping the representatives from England, the ones from Germany, and the ones from France. We have $3!$ ways for England, $4!$ ways for Germany, and $2!$ ways for France. We can arrange these groups in $3!$ ways, however, because of rotational symmetry we have to divide by $3.$ Thus, we get an answer of
$$\frac{3!4!2!3!}{3}=\boxed{576}.$$[/hide]

I had a different answer.  :)

[hide=solution]just counted the amount of ways to rearrange them around the circle. This becomes 2, counterclockwise, and clockwise. So then there are 3! = 6 ways to arrange England, 4! = 24 ways to arrange Germany, and 2! = 2 ways to arrange France. Therefore, there are $2\cdot6\cdot24\cdot2 = 576$ ways to arrange them around the round table. Tip: The formula that AoPs posted in the solution for finding how many ways to rearrange $n$ amount of people is very helpful! [/hide]

[quote=pilover123]Can someone explain this? My logic was that treating each country as a block, there are 3! ways to orient the blocks, 3! ways to orient the representatives from England, 4! ways for the representatives from Germany, and 2! ways for the representatives from France. This gives us 3! x 3! x 4! x 2!, but we must divide by 9 to account for rotations, giving us 192. I still don't see what's wrong with this solution.[/quote]

even though this user posted 3 years ago
the mistake was you had to multiply that by the number of ways to arrange them
there are 3!=6 ways to arrange them and you have to divide by 2 to  account for rotation so the answer is 576

uh the first post here is from January 2009. This is still going on??!!!!??  :what?:  :wacko:

[quote=ThunderStrike314][quote=MinecraftArchitect]Redact or hide please. Thank you![/quote]

You don't really have to in a GameBot thread, but thanks. :D

Also, I think it is $\frac{2*6*6*24}{3}=\boxed{576}$[/quote]

Not redacting or hiding technically counts as cheating.

This was made 8 days after gamebot joined


Find the least common multiple of 36, 48, and 27.


geometry
3D geometry
rectangle
conics
ellipse
analytic geometry
graphing lines

1. Find the equation of a perpendicular bisector of a segment with endpoints 7,1 and 3, -6.

3. A circular Wheel with radius 6/ pi is rolled 216 units. How many revolutions occured?

2. Given:Triangle ABC where angle A = 60 degrees, AB = AC and BC=4. Find the square of the area of a triangle

11. A cube has a side length of 10. Find the length of tis longest internal diagonal

10. Find the number of diagonals in a 12-gon.

12.  circle is circumscribed about a right triangle with legs 7 and 24, What is the circle's area?

9.Find the area of a pentagon if all sides equal 5 and 2 of the adjacent angles are both right

15. A solid box is polaced on the floor. ITs dimensions are 11x1x12. If an insect lands on one corner of the box, what is the length of the shortest route to the corner diagonally opposite the starting corner?

17. A circle has a circumference of 10. Find the area of an inscribed hexagon.

21. Cowboy bob is an outhouse 10 units south of from a linear river that travels east to west. He needs to pick up his horse which is 14 units south of the river. If wBob would like to wash his hands in the river before picking up his horse, what is the least distance he could travel in the entire trip from his outhouse to his horse?

23. Mason builds a fence for his dog Nathan. It is shaped like a rectangle with width 30 and lenght 3 root 3. The entrance has been closed. If nathaniel is attached to an inside corner in this fence by a leash of length 6, how much area can he cover?

6. In a plane the set of all points distance 3 from a line forms:

an ellipse, circle, line , line segment, or NOTA

#1[hide]The midpoint is $(5,-\frac{5}{2})$. The equation of the line, which we can determine by fairly straightforward algebra, is $y=\frac{7}{4}x-\frac{45}{4}$. Thus, the slope of the perpendicular bisector is $-\frac{4}{7}$. Since it passes through the point $(5,-\frac{5}{2})$, we set up the line $y=mx+b$ and substitute, to get $y=-\frac{4}{7}+\frac{5}{14}$.[/hide]
#3[hide]The radius is $\frac{6}{\pi}$, so the diameter is 12. Thus the the number of revolutions is $216/12=18$.[/hide]
#2[hide]By the isosceles triangle theorem (I don't know the name, I think it's Euclid I.5), the base angles are equal, and the triangle is equilateral. We are given that one side is 4, so the area is $4\sqrt{3}$.[/hide]
#11[hide]We use the 3-D Pythagorean theorem, to get the interior diagonal as $10\sqrt{3}$. [/hide]
#10[hide]The formula for this is $\frac{n(n-3)}{2}$. Subsituting $n=12$, we have 54 diagonals. [/hide]
#12[hide]Since this is a right triangle, the hypotenuse is the diameter of the circle. By the Pythagorean Theorem, the hypotenuse is $\sqrt{7^{2}+24^{2}}=25$, so the radius of the circle is $12.5$, making the area of the circle $156.25\pi$.[/hide]
#9-I don't really understand the wording of this question.

#15[hide]Unfold the box. We have three possible routes that make right triangles, one has legs 1, 23, one has legs 11, 13, the other has legs 12, 12. We can easily check that the one with legs 12, 12 is the most efficient route. Thus the shortest possible distance is $\sqrt{12^{2}+12^{2}}=12\sqrt{2}$.[/hide]
#17[hide]The circumference is 10, so the radius is $\frac{5}{/pi}$. If we draw six radii 60 degrees apart and connect the points in which they intersect the circle, we get a regular hexagon. Thus, the side length of the regular hexagon is the same as the radius, $\frac{5}{\pi}$, and the area is $\frac{75}{2\pi^{2}}$[/hide]
#21-I think the problem needs to be explained better...
#23-I'm gonna have to think about this a little more.

#6[hide]I believe it's 2 lines, to NOTA.[/hide]

Ehh I hope these are right...
[hide="9"]Basically one of those house style pentagons, with the two base right angles. So it's an equilateral triangle with length five on top of a square with length five, the total area is $\frac{25\sqrt{3}}{4}+100.$[/hide]

For 21, they don't specify how far away the horse is...I'm not sure if that matters or not...

[hide="22"] See the attachment at the bottom. The area of the larger 3/4 circle is $36\pi \times \frac{3}{4}=27\pi$ The area of the smaller 1/4 circle is $\frac{(6-3\sqrt{3})^{2}\pi}{4}=(4-\sqrt{3})\pi.$ Adding these together, we have the answer as $(31-\sqrt{3})\pi$ Problems like these come on practically every Vestavia test.[/hide]
I'll post the others' as I figure them out...
13375P34K, you should check your answer for number 2, it was asking for the square of the area.
I believe you have a typo in number 17.

and then some... nice explanations by these guys =) I am learning

4. Find the difference between the area of the sold and shaded regions of this parallelogram with an area of 36. It has two sides of 12 and two sides of 5, like a rombus and 2 diagonals.

5. Find the length of the common external tangent of 2 circles of radii 5 and 17 with distance of 20 between their centers.

7. A cone standing on its vertex of height 4 and radius 3 is filled halfway by volume with kool aid. What is the cube of the height of the filled portion?

8. Find the sum of the altitudes in a triangle of sides 5, 12, 13.

13. A right triangle has one leg of length 10 and another of a random length between 0 and 10. What is the probability that the square units of area exceeds the units in the length of the hypotenuse?

14. How many triangles exist with integral sides , two sides of length 6 and an area greater than 18.

16. Ozymandias wants to build a set of pyramids to rival those at Giza. To do this first builds a regular tetrahedron with edge length of 1 the first  year. Every following year he builds a pyramid of edge length twice the previous year's pyramid. he planned to do this forever but his rule only lasted 3 years. Assuming SOLID pyramids, what volume of stone was required for 3 regular tetrahedrons?

18. y=[x] represents the greatest integer function where y is equal to the greatest integeer less than or equal to x. Find the shortest distance from the graph of y equals x to the point (13/8,13/8)

20. Find the ｖｏlume of the #d graph represented by all points distance 3 from the x axis enclosed by the planes x equals 999 and x equals 1002.

24. A woman builds a cubical pool of side 12 and fills it 3/4 of the way. Then she marks the water line on the side of the pool. She takes a large solid cylinder with height 6 and radius 1 and dips it into the pool such that its base is parallel to the pool's floor and that its lower base is 3 units below her marked water line. What volume of the cylinder is beneath the water?

after this , I might post some problems I had on a theta geometry test

I got number 5, it is 16 after applying formula d squared minus 2 radiis absolute value squared = ext tangent squared


A friend of mine gave me this one (among a couple others)

Let G(n) be the the gemetric mean of the sequence...

    nC0, nC1, ...      nCn-1

Prove:
     lim(n->infinity) G(n)^(1/n) = e^(1/2)

[quote="fuzzylogic"]After taking [tex]\ln[/tex] and do some calculation,

[tex]\displaystyle \lim_{n\rightarrow\infty}\frac{1}{n}\ln{G(n)} =2\int_0^1x\ln{x}\,dx-\int_0^1\ln{x}\,dx = \frac{1}{2}.[/tex][/quote]

could you show some more steps?

also, isn't the last integral improper?

[quote="VeCTer"]could you show some more steps?
[/quote]

[tex]\displaystyle G^n(n) = {n\choose{0}}{n\choose{1}}\cdots{n\choose{n-1}}=\frac{(n!)^{n+1}}{(1!2!\cdots n!)^2} = \frac{(n!)^{n+1}}{(1^n2^{n-1}\cdots n^1)^2}[/tex]

[tex]\displaystyle {n}\ln{G(n)} = (n+1) \sum_{i=1}^n\ln{i} - 2\sum_{i=1}^n (n+1-i)\ln{i}[/tex]
[tex]\displaystyle = 2\sum_{i=1}^n i\ln{i} - (n+1)\sum_{i=1}^n \ln{i}[/tex]
[tex]\displaystyle = 2\sum_{i=1}^n i\ln{\frac{i}{n}} - (n+1)\sum_{i=1}^n \ln{\frac{i}{n}}[/tex]

[tex]\displaystyle {\frac{1}{n}}\ln{G(n)} = 2\left(\frac{1}{n}\sum_{i=1}^n\frac{i}{n}\ln{\frac{i}{n}}\right) - \frac{n+1}{n}\left(\frac{1}{n}\sum_{i=1}^n \ln{\frac{i}{n}}\right)[/tex]

[tex]\displaystyle \lim_{n\rightarrow\infty}\frac{1}{n}\ln{G(n)} =2\int_0^1x\ln{x}\,dx-\int_0^1\ln{x}\,dx = \frac{1}{2}.[/tex]

[quote="VeCTer"]
also, isn't the last integral improper?[/quote]

Both of the integrals are improper, but they're both finite.

[quote="gauss202"]Both of the integrals are improper, but they're both finite.[/quote]

I wouldn't say the first one is improper since x(ln x)->0 when x->0.

Good point. I didn't notice that. So only the second one is improper - but it still converges.


If $ \frac {xy}{x \plus{} y} \equal{} a, \frac {xz}{x \plus{} z} \equal{} b, \frac {yz}{y \plus{} z} \equal{} c$, where $ a,b,c$ are other than zero, then $ x$ equals:

$ \textbf{(A)}\ \frac {abc}{ab \plus{} ac \plus{} bc} \qquad\textbf{(B)}\ \frac {2abc}{ab \plus{} bc \plus{} ac} \qquad\textbf{(C)}\ \frac {2abc}{ab \plus{} ac \minus{} bc}$
$ \textbf{(D)}\ \frac {2abc}{ab \plus{} bc \minus{} ac} \qquad\textbf{(E)}\ \frac {2abc}{ac \plus{} bc \minus{} ab}$

[quote="Brut3Forc3"]If $ \frac {xy}{x \plus{} y} \equal{} a, \frac {xz}{x \plus{} z} \equal{} b, \frac {yz}{y \plus{} z} \equal{} c$, where $ a,b,c$ are other than zero, then $ x$ equals:
$ \text{(A)}\, \frac {abc}{ab \plus{} ac \plus{} bc} \qquad\text{(B)}\, \frac {2abc}{ab \plus{} bc \plus{} ac} \qquad\text{(C)}\, \frac {2abc}{ab \plus{} ac \minus{} bc} \\
\text{(D)}\, \frac {2abc}{ab \plus{} bc \minus{} ac} \qquad\text{(E)}\, \frac {2abc}{ac \plus{} bc \minus{} ab}$[/quote]
$ \frac{1}{x} \plus{} \frac{1}{y} \equal{} \frac{1}{a}...$
$ x \equal{} \frac{{2abc}}{{bc \plus{} ac \minus{} ab}}$

[hide]
Notice that;
$ \frac {1}{a} \equal{} \frac {1}{x} \plus{} \frac {1}{y}$ and now its easy

$ \frac {1}{a} \plus{} \frac {1}{b} \minus{} \frac {1}{c} \equal{} \frac {2}{x}$

$ x \equal{} \frac {2abc}{bc \plus{} ac \minus{} ab} \longrightarrow \boxed{E}$[/hide]

EDIT: ninja'd by T.T.Hung!! :ninja:

X has to be symmetric to a and b.
So it's E!!!!!  :D


integration
calculus
limit
real analysis
real analysis theorems

Given $ \alpha\in BV[a,b]$ (bounded variation), show that there is a unique $ \beta\in BV[a,b]$ with $ \beta(a)\equal{}0$ such that $ \beta$ is right-continuous on $ (a,b)$ and $ \int_{a}^{b}f\,d\alpha\equal{}\int_{a}^{b}f\,d\beta$ for every $ f\in C[a,b]$.

I assumed some $ \gamma$ that also works, and used integration by parts to get that $ f(b)[\beta(b)\minus{}\gamma(b)]\equal{}\int_{a}^{b}(\beta\minus{}\gamma)\,df$, but I'm stuck here.  But I never used any assumptions about $ \alpha$, so I'm not completely sure I'm heading in the right direction.

Hint: $ \beta(t) \equal{} \lim_{y \to a^\plus{}}\int_y^t d\beta$

$ \beta(b)\equal{}\lim_{y\rightarrow a^{\plus{}}}\int_{y}^{b}d\beta$ and $ \gamma(b)\equal{}\lim_{y\rightarrow a^{\plus{}}}\int_{y}^{b}d\gamma$, so $ f(b)(\beta(b)\minus{}\gamma(b))\equal{}f(b)\lim_{y\rightarrow a^{\plus{}}}\int_{y}^{b}d(\beta\minus{}\gamma)\equal{}\int_{a}^{b}(\beta\minus{}\gamma)\,df$.  Since this holds for all $ f\in C[a,b]$, then the integral must be 0, so $ \beta\equal{}\gamma$.  Is this correct?


topology
real analysis
real analysis unsolved

Give an example of a set in the plane that has an internal point that is not an interior point.

[hide="Example"]$ 0\in\{re^{it}|0 < t\le 2\pi, 0\le r\le e^{ \minus{} \frac {1}{t}}\}$. Note that every closed set in $ \mathbb R^2$ with an internal point has non-trivial interior - as proven [url=http://www.mathlinks.ro/viewtopic.php?t=227831]here[/url][/hide]


Suppose we have a perfect tap of transverse area $A$. a circular one. the water flux is constant.

The tap is $h$ meters to the ground. does not exist Air and the matter is not discrete.

So we can determine the Pressure $P$ of the water in the ground in function of the velocity of the water and others constants.

How do you calculate this?!

Are you looking for the pressure that the water exerts on the ground when it hits it? Then you need some more data, i.e. whether the colision of the water and the ground is ellastic.

On the other hand, it's easy to calculate the pressure [i]in[/i] the water just before it hits the ground, although I don't see any use of this fact...

[quote]Are you looking for the pressure that the water exerts on the ground when it hits it? [/quote]

yes.

And i dont think we need more data. we have the velocity of the particles (very smal, infinitesimal) of the water in the ground. we can easily know the area of collision in the ground level.

the only think i dont know if the weight of the water that hits the ground, what's portion of the water that is hiting.

and this is very useful!!!
Suppose we dont have a Weighing scale and dont want to buy one.
we can make one easily using this experiment. because we can easily know the flux of the water, the area and the gravity.
we dont need to work in Newtons, but we can have a result in newtons.

thats my purpose

Ok then, let's assume that the collision is not elastic at all.

First of all, let's find the pressure in the water when it hits the ground. Bernoulli's equation says that \[p_{0}+\rho g h+\rho \frac{v_{0}^{2}}{2}= p_{0}+\rho \frac{v^{2}}{2},\] so $v = \sqrt{v_{0}^{2}+2gh}$.The equation of continuity requires that $Av = A_{0}v_{0}$ ($A_{0}$ is the area of the tap) so $A = A_{0}v_{0}/\sqrt{v_{0}^{2}+2gh}$. During an infinitesimal interval $\Delta t$, the water exerts the force \[F = \Delta P/\Delta t = \rho v^{2}A = \rho A_{0}v_{0}\sqrt{v_{0}^{2}+2gh}.\] I think this is what you wanted.

However, the water that has fallen down doesn't just disappear; if the collision is inelastic, it's weight will afflict an additional force, and if the collision is somewhat elastic the above reasoning needs to be altered a bit.

I did not understand one thing

$\Delta P/\Delta t = \rho v^{2}A$

where did you come this?!

Because $P$ depend on $v$ and not $v^{2}$.
So, where did come the addictional $v$?

the dimensional analysis of the equation is true. but. explain it

thanks

In time $\Delta t$ the impulse $\Delta P = \Delta m v$ is transferred to the ground. The mass $\Delta m = \rho \Delta V = \rho A v \Delta t$...


[quote="stergiu"]If $a,b,c$ are positive numbers , prove that

    $\frac{ab}{c(b+c)}+\frac{bc}{a(c+a)}+\frac{ca}{b(a+b)}\geq \frac{3}{2}$

[u] Babis[/u][/quote]

\[\frac{ab}{c(b+c)}+\frac{bc}{a(c+a)}+\frac{ca}{b(a+b)}\geq 3 \sqrt[3]{ \frac{abc}{(a+b)(b+c)(c+a)}}\geq 3 \sqrt[3]{ \frac{abc}{2 \sqrt{ab}\cdot 2 \sqrt{bc}\cdot 2 \sqrt{ca}}}= \frac{3}{2}\]

Q.E.D

[quote="stergiu"]If $a,b,c$ are positive numbers , prove that

    $\frac{ab}{c(b+c)}+\frac{bc}{a(c+a)}+\frac{ca}{b(a+b)}\geq \frac{3}{2}$

[u] Babis[/u][/quote]
Svejk, I think, it easier:
$\sum_{cyc}\frac{ab}{c(b+c)}=\sum_{cyc}\frac{a^{2}b^{2}}{b^{2}ac+c^{2}ab}\geq\frac{(ab+ac+bc)^{2}}{2abc(a+b+c)}\geq\frac{3}{2}.$ :wink:

[quote="Svejk"][quote="JANKRI"]
\[3 \sqrt[3]{ \frac{abc}{(a+b)(b+c)(c+a)}}\geq 3 \sqrt[3]{ \frac{abc}{2 \sqrt{ab}\cdot 2 \sqrt{bc}\cdot 2 \sqrt{ca}}}= \frac{3}{2}\]
[/quote]

This step is wrong.
I'll post a solution.

The given inequality is equivalent to
$\sum \frac{a^{2}b^{2}}{b+c}\geq \frac{3}{2}abc$
But $\sum \frac{a^{2}b^{2}}{b+c}\geq \frac{(ab+bc+ca)^{2}}{2(a+b+c)}\geq\frac{3}{2}abc$.
If we denote $x=ab$ ,$y=bc$ and $z=ca$ the last inequality is equivalent to the well-known :$(x+y+z)^{2}\geq 3(xy+yz+xz)$[/quote]

I see, sorry  :blush:

[quote="X-men"]Let x = a/b, y = b/c, z = c/a.[/quote]
 What do we gain , by homogenizing the inequality in this way? I think we must again use BCS(Engel form).

[b]Alternative[/b]

 Darij Grinberg has made in the past a solution using Rearrangement.

 Babis


calculus
integration
trigonometry
logarithms
calculus computations

[color=red] Calculate the integral :
  $ I\equal{}\int\frac{dx}{x^8\plus{}1}$
[/color]

First, this should be in the calculus forum.  Second, it's not necessarily hard-just really, really annoying.  Get the denominator in a more friendly form and then just use partial fractions.  That is, factor it so that you can use partial fractions.  Then it's just a lot of arctangents and logarithms.

Although $ \int^\infty_{\minus{}\infty} \frac{dx}{x^8\plus{}1}$ would be fairly easy to do..

$ \boxed{\int_{\minus{}\infty}^{\infty} \frac{1}{x^{2n}\plus{}1}\ dx\equal{}\frac{\pi}{n\sin \frac{\pi}{2n}}\ (n\equal{}1,\ 2,\ \cdots)}$

I would just solve a problem like that using residues.  But getting the indefinite integral...

hello, i have got
$ \int \frac{1}{x^8\plus{}1}dx\equal{}\\
\frac{1}{4} \sin \left(\frac{\pi }{8}\right) \tan ^{\minus{}1}\left(\left(x\minus{}\cos \left(\frac{\pi }{8}\right)\right) \csc \left(\frac{\pi }{8}\right)\right)\minus{}\frac{1}{8}
   \cos \left(\frac{\pi }{8}\right) \log \left(x^2\minus{}2 \cos \left(\frac{\pi }{8}\right) x\plus{}1\right)\plus{}\frac{1}{8} \cos \left(\frac{\pi }{8}\right) \log \left(x^2\plus{}2 \cos
   \left(\frac{\pi }{8}\right) x\plus{}1\right)\plus{}\frac{1}{4} \tan ^{\minus{}1}\left(\left(x\plus{}\cos \left(\frac{\pi }{8}\right)\right) \csc \left(\frac{\pi }{8}\right)\right) \sin
   \left(\frac{\pi }{8}\right)\minus{}\frac{1}{8} \log \left(x^2\minus{}2 \sin \left(\frac{\pi }{8}\right) x\plus{}1\right) \sin \left(\frac{\pi }{8}\right)\plus{}\frac{1}{8} \log
   \left(x^2\plus{}2 \sin \left(\frac{\pi }{8}\right) x\plus{}1\right) \sin \left(\frac{\pi }{8}\right)\plus{}\frac{1}{4} \tan ^{\minus{}1}\left(\sec \left(\frac{\pi }{8}\right)
   \left(x\minus{}\sin \left(\frac{\pi }{8}\right)\right)\right) \cos \left(\frac{\pi }{8}\right)\plus{}\frac{1}{4} \tan ^{\minus{}1}\left(\sec \left(\frac{\pi }{8}\right) \left(x\plus{}\sin
   \left(\frac{\pi }{8}\right)\right)\right) \cos \left(\frac{\pi }{8}\right)$.
Sonnhard.

That's just from Mathematica.  It would be more important to get the partial fraction decomposition.

[color=red] Thank you very much, Dr Sonnhard Graubner,but could you please tell me why do you get the nice notation :
   $ \frac{1}{x^8\plus{}1}\equal{}\frac{1}{4}\frac{x^2\sqrt{2}\minus{}2}{\minus{}x^4\plus{}x^2\sqrt{2}\minus{}1}\plus{}\frac{1}{4}\frac{2\plus{}x^2\sqrt{2}}{x^4\plus{}x^2\sqrt{2}\plus{}1}$
It is which I want, thank you again.
Regard,
Arsenaler.
[/color]

hello, we have $ x^8\plus{}1\equal{}(x^4\minus{}\sqrt{2}x^2\plus{}1)(x^4\plus{}\sqrt{2}x^2\plus{}1)$.
Sonnhard.


geometry
AMC
USA(J)MO
USAMO
geometric transformation
homothety
vector

Do some problems on the USAMO require the knowledge of ""nontraditional" geometry - e.g. inversion, projections, homothecy, transformational geometry, constructions, vectors, cross/dot products, etc.? And do any USAMO problem require knowledge of graph theory? Thanks in advance.

These types of questions are very hard to answer.

[b]Theoretically[/b], you don't need anything beyond what they teach in high school. Nothing will [i]require[/i] the use of anything fancy. Everything can be solved with basic theorems and ingenuity.

But everybody knows that the more you know, the better are your chances at solving those problems.

Olympiads tend not to be tests of knowledge (i.e. no problem will ask you to specificly use inversion, and no USAMO problem will be phrased in graph theory terminology). Olympiads are tests of techniques. Most of the things you listed there are well-known techniques. It will not help you much by simply "knowing" them. You will have to see them in action and use them yourself.

I've looked at some past USAMO problems, and it seems to me that tools like these are useful but not necessary for solving the problems.  I've seen complicated vector-based solutions to geometry problems, and then simpler solutions to the same problems based on pure geometry.

ouch, graph theory and geometry. basically everyone's worst subject


function
quadratics
algebra
polynomial
algebra theorems

Given that

$ ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \plus{} e \equal{} 0$

1) Find the discriminant of quartic function.

2) As you know from quadratic funtion, that if discriminant of quadratic function is negative, then there are no real roots and function has strictly positive or negative value. Can you say the same to quartic function?

Thank you.

See [url=http://en.wikipedia.org/wiki/Discriminant#Discriminant_of_a_polynomial]here[/url] for the generic formula, assuming you know some linear algebra.  Mostly just finding determinants.

Sorry, i don't know linear algebra. Therefore, it is hard to me to understand the generalization in Wikipedia.
I just want you to write me discriminant of quartic function with coefficients of terms ($ a, b, c, d, e$).
For example, the discriminant of quadratic is $ b^2 \minus{} 4ac$. Give me the one for quartic.

Please, also answer question 2)

Thanks.

Equation (10) [url=http://mathworld.wolfram.com/PolynomialDiscriminant.html]here[/url].  There is no point in actually giving the explicit formula since the quartic discriminant is too complicated for straightforward computation.

The answer to 2) is "no."  The discriminant of a monic quartic with roots $ r_1, r_2, r_3, r_4$ is $ \prod_{i < j} (r_i \minus{} r_j)^2$.  All a negative discriminant tells you is that [i]some[/i] of the roots are non-real.  It's possible for two of the roots to be complex and two of the roots to be real.


Do there exist integers a and b for which $ a^2\plus{}b^2$ and $ a^2\minus{}b^2$ are square numbers?

Well, let's assume for the moment it is possible.  Therefore, we have that
$ a^{2}\plus{}b^{2}\equal{}s^{2}$ for some $ s$ and
$ a^{2}\minus{}b^{2}\equal{}k^{2}$ for some $ k$.  Adding these, we get that
2$ a^{2}k^{2}\plus{}s^{2}$
or $ a^{2}\equal{}frac{k^{2}\plus{}s^{2}}{2}$.
Plugging this into our first equation,we get that
$ b^{2}\equal{}frac{s^{2}\minus{}k^{2}}{2}$.
This is just the same problem.

So, this problem looks kind of circular, but I'm willing to guess
that there do exist such numbers.

@Z-coli
I'm sorry, but don't understand what your post is saying. Your third line is $ 2a^2 \equal{} k^2 \plus{} s^2$, ok. But what does $ frack$ and $ fracs$ mean? Also, $ k^2$ and $ 4k^2$ is a square number, but not $ 2k^2$. Could you clarify?

It was a $ \text{\LaTeX}$ typo. Here is the corrected post.

[quote="Z-coli"]Well, let's assume for the moment it is possible.  Therefore, we have that
$ a^{2} + b^{2} = s^{2}$ for some $ s$ and
$ a^{2} - b^{2} = k^{2}$ for some $ k$.  Adding these, we get that
$ 2a^{2} = k^{2} + s^{2}$
or $ a^{2} = \frac {k^{2} + s^{2}}{2}$.
Plugging this into our first equation,we get that
$ b^{2} = \frac {s^{2} - k^{2}}{2}$.
This is just the same problem.

So, this problem looks kind of circular, but I'm willing to guess
that there do exist such numbers.[/quote]

Yes, there do. Infinitely many, in fact. (Hint: let $ b$ be zero...)

Sorry, [i]positive[/i] integers.

Otherwise, Yongiyi781's solution works, and this is a non-problem

Edit: Scratch that. Fermat's infinite descent actually doesn't work. (There is no bound...)

Look [url=http://en.wikipedia.org/wiki/Pythagorean_triple]here[/url], and especially look at the last line in the section "Elementary properties of primitive Pythagorean triples":

[quote]There are no Pythagorean triplets in which the hypotenuse and one leg are the legs of another Pythagorean triple.[/quote]

This essentially answers your question.

Suppose that the problem was false. Then let $ a^2\plus{}b^2\equal{}c^2$ and $ a^2\minus{}b^2\equal{}d^2$. WLOG let $ \gcd(a,b)\equal{}1$ because otherwise, we have a smaller solution by dividing both sides of each equation by $ [\gcd(a,b)]^2$. Now, it is not possible for both $ a$ and $ b$ to be odd because mod 4, both $ a^2$ and $ b^2$ are 1 mod 4, so $ c^2$ is 2 mod 4, but 2 is not a quadratic residue mod 4.

So one of $ a$ and $ b$ must be even.

Case 1: $ a$ is even and $ b$ is odd. Then $ c^2\minus{}d^2\equal{}2b^2$. But since $ b^2$ is odd, $ 2b^2$ has only one factor of 2, and thus can't be the difference of two squares. (Hint: factor $ c^2\minus{}d^2\equal{}(c\plus{}d)(c\minus{}d)$. Each factor must have the same parity.)

Case 2: $ a$ is odd and $ b$ is even. This is the crux of the proof which I will leave to you.


inequalities
linear algebra
linear algebra unsolved

Let $A, \, B$ be real symmetric positive definite $n \times n$ matrices with the same determinant. Show that
\[trace \left(A B^{-1}\right) \ge n\]
with equality iff $A = B$.

(In exchange for asking questions, I should try answering one...)

Let ${\lambda_{i}}$ be the eigenvalues of $AB^{-1}$.  Then the problem is equivalent to showing $\frac{\sum_{i=1}^{n}\lambda_{i}}{n}\geq 1$.  But this follows fairly quickly from the Arithmetic-Geometric Mean Inequality:

$\frac{\sum_{i=1}^{n}\lambda_{i}}{n}\geq (\prod_{i=1}^{n}\lambda_{i})^{1/n}= (det(AB^{-1}))^{1/n}=(\frac{det(A)}{det(B)})^{1/n}=1$.

[quote="edoo"](In exchange for asking questions, I should try answering one...)

Let ${\lambda_{i}}$ be the eigenvalues of $AB^{-1}$.  [/quote]How do you know that these are real numbers?  :wink:

[quote]How do you know that these are [color=green][positive][/color] real numbers?[/quote]
See the solution I posted to problem #3 in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147726]this topic.[/url]


inequalities
modular arithmetic
number theory proposed
number theory

For each positive integer $ n$,denote by $ S(n)$ the sum of all digits in decimal representation of $ n$.
Find all positive integers $ n$,such that $ n\equal{}2S(n)^3\plus{}8$.

[quote="pleurestique"]if $ d$ is the number of digits in $ n$, then $ S(n)\leq 9d$, whereas $ n\geq 10^{d \minus{} 1}$. so we get the inequality
\[ 10^{d \minus{} 1}\leq 1458d^3 \plus{} 8
\]
which is false for $ d\geq 8$. in particular, this means that $ S(n)\leq 63$.

now in light of the fact that $ S(n)\equiv n\pmod{9}$, we must have $ n\equiv 2n^3 \plus{} 8\pmod{9}$, whose only solution is $ n\equiv 1\pmod{9}$. so the only possible values of $ S(n)$ are 1, 10, 19, 28, 37, 46, 55. a quick check now shows that the only solutions are $ n \equal{} 10, 2008, 13726$.[/quote]
In fact,you only need to check $ n\equal{}1,10,19,28,37,46$,the value $ 55>9\cdot 6$.
My solution and official solutions is the same. :wink:
Thank you.

[quote=Superguy]S(n)<=54 not 63 bto[/quote]

that bump though...

btw how far did you have to scroll to get to this thread  :rotfl:

please explain me why $S(n)$ = $1$ ( $mod 9$)?


number theory
greatest common divisor
least common multiple
AMC
USA(J)MO
USAMO
summer program

What was your first Olympiad experience?

At first, I saw some Diophantine Equation and it says find all solutions or something. I just went and plugged heaps of values in and when I got to like $ 30$ I concluded no more solutions exist. I was wrong, there was a few bigger values.

So, I thought 'How the [censored] am I suppose to know that!' so I looked at the solution which made use of everything I know (e.g. Prime Factorization, Divisibility, gcd, lcm) but never thought about integrating these 'fudamental' concepts.

Then, the rest is history. Here I am! (I don't even know how I found the website, I think I was googling random terms).

Tell me about yours.

My first experience was... the 1997 USAMO. I don't think I ever saw an Olympiad problem before the actual contest.

I did pretty well, too- I think my score was in the 7-13 points range. That wasn't enough for MOP at the time, but it would be now with "Red MOP".

I think I misclicked the poll- it's not showing the answer I intended.

My first Olympiad experience was the 2007 USAMO.  I realized that the problems weren't so terrible, although I messed up on some of them.  I ended up getting 10 points, and I felt rather pleased.

It's hard to pinpoint my first olympiad experience. I participated in olympiad practices throughout freshman year, including some actually olympiad-style tests. I then proceeded to qualify for USAMO 2007, scoring 18 - a very good score. At that point, I realized that it was not actually very difficult to score 14+ on olympiads since you only need problems 1 and 4 to do so. After MOP (red, but I qualified for blue. Oh well), I went to MathCamp and there I participated in a practice olympiad consisting of the 6 problems from IMO 2007. I scored a 14 on that, the equivalent of the low end of a bronze medal (50% mark). All of these were pretty good experiences, but it's hard to say what would be my actual first olympiad (not the practice IMO of course).

By the way, it doesn't have to be going to an actual Olympiad or campe etc. Just the first time you encountered Olympiad-like problems.

The first time I saw Olympiad problems was from a booklet.  I was like "...what...hmm this is pretty interesting, but really hard..."
They were old USAMO problems which were pretty bad.

The first time I officially took an Olympiad was this year's USAMO.

But I've taken Olympiad exams unofficially quite a bit...for example, last year's USAMO (I got like 3 problems) or this year's IMO (5 problems, don't ask how).  I don't think the pressure of an actual Olympiad can be simulated through practice, though.  No matter how many problems you do, you'll always do significantly worse on the real thing.  That's how it is for me, at least...

[quote="BanishedTraitor"]For me, in an actual Olympiad examination (not the IMO) I tend to do better than I expect (also better than when I practice). I know the answers are not there and there are no hints so my mind tend to be a bit open on interpretations of problems. Especially geometry problems, I find that when I open a hard geometry book to work through I manage to solve less than desired but during an exam I zoom through most like crazy.[/quote]

Well okay, I should elaborate:

When I just casually look at problems and try to solve them, I don't try as hard or solve as many as on a real Olympiad.  Then again, on a real Olympiad, I don't solve as many as if I simulate taking the exam at home.

[quote="Nerd_of_the_Ages"]First USAMO problem was at the end of my Intro to NT class. Rather easy, actually.
I think it was something like:

Prove or dissprove that it there is a set of 14 consecutive integers where each number is a divisor at least one prime number 2, 3, 5, 7, 11. If you prove it, give an example.

Might have remembered it wrong. But is it really hard enough to be an Olympiad question>[/quote]

...are you sure that was "rather easy"?

My first encounter with Olympiad-level problems was 2001 IMO #3, a pretty nice inequality which I solved using Jensen's. As of yet, the only Olympiads I've been able to solve are inequalities and a geometry problem.

[quote="Temperal"][quote="Nerd_of_the_Ages"]First USAMO problem was at the end of my Intro to NT class. Rather easy, actually.
I think it was something like:

Prove or dissprove that it there is a set of 14 consecutive integers where each number is a divisor at least one prime number 2, 3, 5, 7, 11. If you prove it, give an example.

Might have remembered it wrong. But is it really hard enough to be an Olympiad question>[/quote]

...are you sure that was "rather easy"?[/quote]

Well, I didn't think is was THAT hard... and it couldnt have been either, because we covered it in like 20 minutes in the Intro to NT class.

EDIT: Actually, it wasnt a rigorous proof, but yeah....

My experience.... does it have to be official? [hide="If so"]I got a 3 on last year's AIME. I sucked then.[/hide]

The point was that it doesn't have to be official- just the first Olympiad problems you tried to solve. It just so happens that those were part of an official contest for many of us.


Since I suck at asymptote, try your best to imagine the question:

Two circles, circle $ O$ and circle $ C$, with radius 1 are created. The center of circle $ O$ lies on circle $ C$, and the center of circle $ C$ lies on circle $ O$. What is the area of the intersection of $ O$ and $ C$?

Im thinking this has a bit of PIE in its solution, but I really dont know.

Also, how would you do this with 3 circles of radius 1 and the same restrictions (each circles center is on the other circles circumference)?

How so? And how do you do it for 3 circles?

[asy]size(8cm,0);
import math;
import graph;
real r,s;
pair a,b, common;
path circ1, circ2;
r=1; s=1;
a=(0,0);
b=(1,0);
circ1=circle(a,r);
circ2=circle(b,s);
draw(circ1,linewidth(1bp));
draw(circ2,linewidth(1bp));
pair [] x=intersectionpoints(circ1, circ2);
dot(x[0],3bp+blue);
dot(x[1],3bp+blue);
label("C",(0,0),SW);
label("O",(1,0),SE);
dot((0,0),red);
dot((1,0),red);[/asy]

For two circles correct?

[hide="Solution"]
[asy]size(8cm,0);
import math;
import graph;
real r,s;
pair a,b, common;
path circ1, circ2;
r=1; s=1;
a=(0,0);
b=(1,0);
circ1=circle(a,r);
circ2=circle(b,s);
draw(circ1,linewidth(1bp));
draw(circ2,linewidth(1bp));
pair [] x=intersectionpoints(circ1, circ2);
dot(x[0],3bp+blue);
dot(x[1],3bp+blue);
label("C",(0,0),SW);
label("O",(1,0),SE);
dot((0,0),red);
dot((1,0),red);
draw((0,0)--x[0]--(1,0),linewidth(1));
draw((0,0)--x[1]--(1,0),linewidth(1));
draw((0,0)--(1,0),blue+linewidth(1));[/asy]
It's kinda easy to see that the area of the football-shaped region is twice the area above the blue line. You can find that area by taking the sum of two $ 60^{\circ}$ sections of the circle, then subtracting out the area of the equilateral triangle. This would be \[ 2\left(\frac{1^2\pi}{6}\right)\minus{}\frac{1^2\sqrt{3}}{4}\equal{}\frac{\pi}{3}\minus{}\frac{\sqrt3}{4}.\] The area of the whole football is two times this, or \[ \boxed{\frac{2\pi}{3}\minus{}\frac{\sqrt3}{2}}.\][/hide]

You can do a similar thing for 3 circles.

[quote="algebra1337"]yes, for three circles it's just one equilateral triangle and 3 of those arc minus triangle things. bad description, i know :)[/quote]

What youre saying is take the area of three 60 degree circles but then subtract 2 triangles?

Thanks guys, I get it now :lol:


ratio
real analysis
real analysis unsolved

Prove that for a sequence of positive terms, (a_k), if there is N with (a_n+a)/(a_n)>=1 for all n >=N, then the series

summation(a_n) from n=1 to n=infinity diverges. Show that this result implis one direction of the Ration test.

Hint: prove that $(a_n)$ does not converge to $0$. In particular, this implies that the series diverges.


[i]Let be a, b and c real > 0. Show that[/i]

$(a^{2}-bc)\sqrt{b+c}+(b^{2}-ac)\sqrt{a+c}+(c^{2}-ab)\sqrt{a+b}\geq0$

Because a square root is strictly positive, we need to prove

$\{\begin{array}{rcl}a^{2}-bc\geq0\\ b^{2}-ac\geq0\\ c^{2}-ab\geq0 \end{array} \Leftrightarrow a^{2}+b^{2}+c^{2}-ab-ac-bc \geq0$

But,

$(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\geq0 \Leftrightarrow a^{2}+b^{2}+c^{2}-ab-ac-bc \geq0$

Done.

[u]Is it correct ?  :blush: [/u]

EDIT : I think, because the ineguality is cyclic, we need to set a >= b >= c, no ?

[quote="zweig"]

$\{\begin{array}{rcl}a^{2}-bc\geq0\\ b^{2}-ac\geq0\\ c^{2}-ab\geq0 \end{array} \Leftrightarrow a^{2}+b^{2}+c^{2}-ab-ac-bc \geq0$

[/quote]
Wrong here! :D

[quote="zweig"]fjwxcsl > Thanks for your solution, I will sê it more precisly later ^^

N.T.TUAN > Why is it wrong ? :)[/quote]

$\Leftrightarrow$ means if and only if. The 'only if' part is true in this case, but not the 'if':
Assume that the inequalities on the left are true. Then the inequality on the right is clearly true as well. So the left inequalities are true [b]only if[/b] the right hand side inequality is true.

However, [b]if[/b] the inequality is on the right is true, then perhaps:
$a^{2}-bc=3$
$b^{2}-ac=7$
$c^{2}-ab=-5$

So in this case, the right inequality is correct, and the left inequalities are not. So we cannot say that the left inequalities are true [b]if[/b] the inequality on the right is.

What you have said about the inequality is similar to saying:
"You are a dog if and only if you are an animal."
Which we can break down into:
"You are a dog only if you are an animal" (true)
"You are a dog if you are an animal" (false)

I may have over-stressed the point, but this problem has lost me too many points on Olympiad exams down the years :(


Perhaps a user who has given out his password to other people should have his screen name blocked completely by the administrators (the real person can always sign up for a new screen name, and is very welcome to solve problems or otherwise to contribute to any forum I moderate) so that we don't have to deal with this problem of spam posts anymore.

Just a thought, as I don't have moderator powers on this forum, and I don't administer the site.

This was split from an original thread on spam with the title that I've put in the subject line.

I've moved the split to this forum because it's more relevant here, and I believe it's an issue that should be discussed. Thanks for tokenadult for bringing it up.

The original thread can be found in Fun and Games.

Who exactly did that?
By the way, any such instances are to be reported A.S.A.P. to us (the admins).

The original "having lots of posts is not . . ." thread on Games and Fun Factory asserts, plausibly, that the holder of the I.D. that started that thread (gameworld7) has shared his/her I.D. and password with other persons, and thus is being impersonated online by spammers. Once a screen name is no longer secure and indentifiable with one person, it's probably a good idea to block that screen name entirely, right?

I split that topic from another discussion to explain to him why he some of his posts were deleted. Did he say that someone has taken over his account?

[quote="Valentin Vornicu"]Did he say that someone has taken over his account?[/quote]

Not explicitly, but I suppose someone else posting from his account should be enough evidence.

[quote="yif man12"]Could it have been himself posting that as sort of an excuse? Not too likely, but there's a possibility.

But that also opens the story where somebody who has his account can say "yes, nobody took over my account"[/quote]

You know, I say just close the account...if he wants access to this site, nobody's preventing him from making another one.

But then again, that's probably the reason i'm not an admin ;-)

Good point. I was thinking the same thing.

But, it's up to the admin.  :)

[quote="yif man12"]Could it have been himself posting that as sort of an excuse? Not too likely, but there's a possibility.

But that also opens the story where somebody who has his account can say "yes, nobody took over my account"[/quote]
no.
i was one of em
and he gave it away to try to prevent himself from going back to AoPS to waste more time on the other topics forum

But it was, of course, doomed to failiure


geometry
geometric transformation
rotation
reflection
geometry proposed

two squares are given (O,M,N,P) and (O,S,R,T)[O is the only point they have in common]. If A is the middle point of PS prove that r(AO) is perpendicular to r(MT).
the letters are clocwise

good luck :)

perform a 90 degree clockwise rotation around point $O$.  then $M$ goes to $P$ and $T$ goes to, say, $T'$.  it suffices to show that $PT' \parallel AO$.  but this is easy: $O$ is the midpoint of $ST'$, and since $A$ is the midpoint of $SP$, $AO$ is definitely parallel to $PT'$

Let $S'$ be the reflection of $T$ in $S$. $OMS'$ is the image of $OPS$ through a rotation centered at $O$ of angle $-\frac\pi 2$, so, if $A'$ is the midpoint of $MS'$, then $OA'\perp OA\ (*)$.

On the other hand, $A'$ is the midpoint of $MS'$, while $O$ is the midpoint of $TS'$, so $OA'\|MT\ (**)$. $(*)$ and $(**)$ prove the desired result.

Edit: Looks like someone else was here first :).


[quote="2006"]Let $\ a,b,c$ be positive real numbers
Prove that  $\frac{a^{3}}{b}$+$\frac{a^{3}}{b}$+$\frac{a^{3}}{b}$ $\geq$ $a\sqrt{ac}$+$b\sqrt{ba}$+$c\sqrt{cb}$
                       Good luck  :D[/quote]

I think it should be $\frac{a^{3}}{b}$+$\frac{b^{3}}{c}$+$\frac{c^{3}}{a}$ $\geq$ $a\sqrt{ac}$+$b\sqrt{ba}$+$c\sqrt{cb}$ :wink:

And this is the solution:
by $AM-GM$, $\frac{a^{3}}{b}+\frac{b^{3}}{c}+\frac{c^{3}}{a}= \sum_{cyc}(\frac{15}{26}\cdot \frac{a^{3}}{b}+\frac{5}{26}\cdot \frac{b^{3}}{c}+\frac{3}{13}\cdot \frac{c^{3}}{a}) \geq \sum_{cyc}a\sqrt{ac}=a\sqrt{ac}+b\sqrt{ba}+c\sqrt{cb}$ :D


Can an irrational exponent be reduced to something more manageable? (example: 2^sqrt3)...? :?:

You can use logarithms but that doesn't really make it any simpler.


Θέλω τη βοήθειά σας σε 2 προβλήματα.

1) Έστω σύνολο $ A$ με πληθικότητα $ n$ και έστω $ S$ ένα σύνολο που έχει για στοιχεία του $ n \plus{} 1$ υποσύνολα του $ A$. Να αποδειχθεί ότι μπορούμε να βρούμε 2 ξένα μεταξύ τους υποσύνολα $ S1$ και $ S2$ του $ S$ (να έχουν δηλαδή για στοιχεία τους κάποια από τα $ n \plus{} 1$ αυτά υποσύνολα του $ A$) ώστε union $ S1$ να είναι ίση με union $ S2$.

Είναι εύκολο να αποδειχθεί ότι υπάρχουν τέτοια σύνολα που δεν είναι ξένα μεταξύ τους αλλά πώς εξασφαλίζω ότι υπάρχουν ξένα μεταξύ τους με την ίδια ιδιότητα? Προφανώς δεν μπορώ να αφαιρέσω την τομή τους γιατί μπορεί κάποιο στοιχείο κάποιου υποσυνόλου της τομής μπορεί να ανήκει και σε κάποιο άλλο υποσύνολο που δεν είναι στην τομή... (δηλαδή δεν είναι το ίδιο με το πασίγνωστο πρόβλημα με τους 10 αριθμούς και τα αθροίσματα)

2) Να αποδειχθεί ότι κάθε linear map $ T$ από το $ V$ στο $ V$ μπορεί να γραφτεί ως το ευθύ άθροισμα (direct sum) ενός invertible κι ενός nilpotent operator.

Αρκεί λοιπόν να δείξουμε ότι μπορούμε να γράψουμε το $ V$ ως το ευθύ άθροισμα δύο υποχώρων $ A$ και $ B$ αναλλοίωτων ως προς το $ T$ ώστε το restriction του $ T$ στο $ A$ να είναι nilpotent (δηλαδή $ T^n \equal{} 0$ για κάποιο $ n$) και το restriction του $ T$ στο $ B$ να είναι invertible. Διαισθητικά και μόνο αντιλαμβάνομαι ότι αυτό έχει σχέση με το Jordan Canonical Form ενός πίνακα (διότι ο πίνακας του nilpotent έχει μηδενικά παντού εκτός ίσως από το άνω τρίγωνο και ο πίνακας του invertible δεν έχει eigenvalue zero, δηλαδή στη διαγώνιο το 0) αλλά δεν μπορώ να δώσω απόδειξη.

Ευχαριστώ πολύ.

Το πρώτο πρόβλημα το λυσα. Θα αναρτήσω λύση για τον κάθε ενδιαφερόμενο αργότερα.

Για το δεύτερο ακόμα δεν έχω αυστηρή απόδειξη.

Κάτι μου θυμίζει το πρώτο...  :roll:  :roll:  :roll:

Δες εδώ θέμα 3ο...

Η λύση μου στο πρώτο:

Έστω $ A\equal{}{a1,.....an}$

Έστω $ S\equal{}{A1, A2,.....A(n\plus{}1)}$

Σε καθένα από αυτά τα υποσύνολα $ A(i)$ αντιστοιχίζω ένα διάνυσμα $ v(j)\equal{}(v(1j), v(2j), ... v(nj))$ ώστε:

$ v(ij)\equal{}1$ αν το $ a(i)$ ανήκει στο $ A(i)$

ή

$ v(ij)\equal{}0$ αν το $ a(i)$ δεν ανήκει στο $ A(i)$

Αφού τα διανύσματα αυτά είναι $ n\plus{}1$ υπάρχει γραμμικός τους συνδυασμός (όχι όλα τα scalars ίσα με το $ 0$) που ισούται με $ 0$.

Ξεχωρίζω τώρα τους θετικούς και αρνητικούς scalars και έχω

$ \sum(m(i)v(i))\equal{}\sum(m(j)v(j))$

(οι δείκτες $ i, j$ αλλάξαν νόημα)

Τώρα είναι σχετικά απλό να δούμε γιατί το union των $ A(i)$ είναι ίσο με το union των $ A(j)$

Το αφήνω σαν άσκηση αλλά αν κάποις θέλει τη λύση τη γράφω.

[quote="Protonios"]2) Να αποδειχθεί ότι κάθε linear map $ T$ από το $ V$ στο $ V$ μπορεί να γραφτεί ως το ευθύ άθροισμα (direct sum) ενός invertible κι ενός nilpotent operator.

Αρκεί λοιπόν να δείξουμε ότι μπορούμε να γράψουμε το $ V$ ως το ευθύ άθροισμα δύο υποχώρων $ A$ και $ B$ αναλλοίωτων ως προς το $ T$ ώστε το restriction του $ T$ στο $ A$ να είναι nilpotent (δηλαδή $ T^n \equal{} 0$ για κάποιο $ n$) και το restriction του $ T$ στο $ B$ να είναι invertible. Διαισθητικά και μόνο αντιλαμβάνομαι ότι αυτό έχει σχέση με το Jordan Canonical Form ενός πίνακα (διότι ο πίνακας του nilpotent έχει μηδενικά παντού εκτός ίσως από το άνω τρίγωνο και ο πίνακας του invertible δεν έχει eigenvalue zero, δηλαδή στη διαγώνιο το 0) αλλά δεν μπορώ να δώσω απόδειξη.[/quote]

Protonios, sorry, ήθελα ν'απαντήσω και ξεχάστηκα...

Anyway, σε περίπτωση που σου είναι ακόμα χρήσιμο,η διαίσθησή σου καλά σε πήγε. Πρέπει βέβαια να προσδιορίσεις αν μιλάς για χώρους πεπερασμένης ή άπειρης διάστασης και αν το σώμα των χώρων σου είναι αλγεβρικά κλειστό ή όχι, αλλά αυτό που ψάχνεις λέγεται ανάλυση Jordan-Chevalley.
Είναι γενικά αρκετά γνωστή και νομίζω υπάρχει στο Linear algebra done right ή στην άλγεβρα του Lang... αλλά αν θέλεις κάτι γρήγορο και απλό, βρήκα αυτό με ένα google search:

[url]http://www.math.kochi-u.ac.jp/docky/bourdoki/NAS/nas006/node5.html[/url]

Νομίζω κάτι ανάλογο υπάρχει και για χώρους Banach, αλλά δε θυμάμαι πώς ακριβώς πάει και αν συμβαίνει κάτι διαφορετικό...

Aν χρειαστείς βοήθεια, let me know :) Cheerio,

Durandal 1707

Ευχαριστώ πολύ. Πλέον άμεσα χρήσιμο δεν μου είναι :P

Αλλά γενικά χρήσιμο φυσικά είναι. Να σαι καλά.