algebra
polynomial
group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Determine a polynomial of degree 7 whose galois group is cyclic of order 7.

To get an arbitrary cyclic group of order $n$ as Galois group, take a prime $p\equiv 1 \mod n$. Let $K=\mathbb{Q}[\zeta_{p}]$ (with $\zeta_{p}$ a primitive $p$-th root of unity), then $G=\mathrm{Gal}(K: \mathbb{Q})$ is cyclic of order $p-1$. Since $n|p-1$, there is a cyclic subgroup $H$ of order $\frac{p-1}n$ in the whole Galois group.
The fixed field of $H$ is a Galois extension of $\mathbb Q$ with Galois group $G/H$, being cyclic of order $n$.

To explicitely get some polynomial with this Galois group, simply consider \[p(x)= \prod_{h \in G/H}\left( x-\sum_{g\in H}\zeta_{p}^{gh}\right).\]
It's rational since it's fixed by $G$. It's roots are fixed by $H$, thus it's splitting field has degree $\leq |G/H| = n$. Additionally, the roots (which are all different by $\zeta^{gh}$ being a base of $K$) are transitively permutated by $G/H$, thus it's irreducible, finally stating that the splitting field has degree $|G/H|=n$.

We've shown that any cyclic group is Galois group of some subextension of a cyclotomic one.
This can easily be generalized to arbitrary finite abelian groups.

This is somewhat the converse of the Kronecker-Weber-Theorem, stating that all finite abelian extensions of $\mathbb{Q}$ are contained in some cyclotomic extension.

Practical calculation (following Zetax):
$p=29$; 2 is a generator of $\mathbb{F}_{29}^{*}$; now we work with the exponents: $H=(\{0,7,14,21\},+)\subset\mathbb{Z}_{28}$ and $G/H=(\mathbb{Z}_{6},+)$.
Thus $p(x)=\prod_{u\in\mathbb{Z}_{6}}(x-\zeta^{2^{u}}-\zeta^{2^{u+7}}-\zeta^{2^{u+14}}-\zeta^{2^{u+21}})$.
Use Maple and don't write the exact calculation!!
Digits:=30:
> z:=evalf(cos(2*Pi/29)+I*sin(2*Pi/29)):
  > p:=1:
> for g from 0 to 6 do
> p:=expand(p*(x-z^(2^g)-z^(2^(g+7))-z^(2^(g+14))-z^(2^(g+21)))):
> od:
> print(p);
After a "round" we find $p(x)=x^{7}+x^{6}-12x^{5}-7x^{4}+28x^{3}+14x^{2}-9x+1$.


calculus
integration
trigonometry
calculus computations

Find the value of $a$ such that $\int_{0}^\frac{\pi}{2}(\sin x+a\cos x)^{3}\ dx-\frac{4a}{\pi-2}\int_{0}^{\frac{\pi}{2}}x\cos x dx=2.$


The positive three-digit integer $ N$ yields a perfect square when
divided by 5. When $ N$ is divided by 4, the result is a perfect cube.
What is the value of $ N$?

We know that $ N$ is divisible by $ 5^2\equal{}25$.  Also, $ N$ is divisible by $ 4$.  So it is divisible by $ 100$.  Testing gives us the answer of $ \boxed{500}$.


In Alberta, Canada, all of the n cities lie conveniently at the vertices of a regular n-gon.  All of the roads in Alberta are of the shortest possible distance necessary to connect two cities, and every city is connected by roads to its two nearest neighbors.  On average, how many roads will someone have to travel on to get from one city to another?

(Hint:  consider the odd and even cases seperately, and you might try a few small cases, too.)

And, once you've done that:
Can you find a general arrangement of as many or fewer roads (i.e., you should describe something that will work for any number of cities and which uses at most n edges for n cities for all n) such that as n gets larger and larger, the average number of roads to get from one city to another is smaller than in the problem above?
(Remember, length of roads isn't important.)

[quote]In Alberta, Canada, all of the n cities lie conveniently at the vertices of a regular n-gon.[/quote]

hehe, I dont think so... ;)

here's how I think it could be done:

[hide]the roads form a regular n-gon

the two cases are even and odd.

Case 1: n = 2m+1 (odd case)

from a given city, there are 2 cities 1 road away, 2 cities 2 roads away, 2 cities 3 roads, and so on and there are 2 cities m roads away, and 1 city m+1 roads away.

the average of these is [2*m(m+1)/2+(m+1)]/(2m+1) = (m+1)(m+1)/(2m+1) where m = (n-1)/2

Case 2: n = 2m (even case)

from a given city there are 2 cities 1 road away, 2 cities 2 roads away, 2 cities 3 roads away, and so on and there are 2 cities m roads away

the average of these is [2*m(m+1)/2]/2m = (m+1)/2 where m = n/2[/hide]

Close, but not quite.  Draw a picture of a square -- are there two cities 2 roads away?  Now, draw a picture of a pentagon -- is there one city 3 roads away?

Oops.  I had the even and odd cases switched since I forgot to subtract 1 for the city that I am counting the roads from.

Good job.  Roughly, when there are n cities (for n a large number), the average number of roads is n/2.  How about the second question?


linear algebra
matrix
combinatorics solved
combinatorics

Let M be the set of all infinite matrices D with the form

$D = $

$\mathstrut...\quad \quad... \quad...\   \quad...$

$n_{2,0}\ n_{2,1}\ n_{2,2}\quad ...$
$n_{1,0}\ n_{1,1}\ n_{1,2}\quad ...  $
$n_{0,0}\ n_{0,1}\ n_{0,2}\quad ...$

whose elements $n_{i,j}$ are integers and only finitely many entries are $\neq 0$, $i,j = 0,1,2,...$. Three elements of the same matrix are called connected if there are integers $i \geq 0$ and $j \geq 0$, so that the following entries are the three required numbers

(a) $n_{i,j},\ n_{i,(j+1)},\ n_{i,(j+2)}$ or
(b) $n_{i,j},\ n_{(i+1),j},\ n_{(i+2),j}$ or
(c) $n_{(i+2),j},\ n_{(i+1),(j+1)},\ n_{i,(j+2)}$

An elementary operation replaces three connected elements $n_{i,j}$ by $n'_{i,j}$ with $|n_{i,j} - n'_{i,j}|= 1$. Two matrices are called equivalent if the one matrix can be transformed into the other matrix by means of a sequence of elementary operations. At most how many elements are in the set including pairwise non-equivalent matrices ?

By mean of described operation we can easily null whole matrix except cells $n_{00}$, $n_{01}$, $n_{10}$, and make all values in these cells to be in $\{0,1\}$. So there are at most $2^3=8$ paiwise non-equivalent matrices.
Next step is to prove that all these 8 matrices are pairwise non-equivalent, but I am to tired too prove it right now. :wacko:

Does anyone can help me?

If we colour the matrix as
.......
012...
120...
201...

then each operation changes the parity of the sum of the numbers on squares on each colour. So we have at least four non-equivalent elements. We need now only determine whether we can change, e.g.
0       into    1
00               11
etc.

Ok! You are right!
Thank you vey much!

And again we killed problem, which was in unsolved section for 1 year!


I remmeber that someone in this forum posted the problem $a^b+b^c+c^a\geq \tfrac32$ for $a,b,c>0$ but it turned out to be false. Another guy gave a counterexample for this. Now I will prove that $a^b+b^c+c^a\geq\tfrac32$ for $1>a\geq b\geq c>0$.

First, we need the Lemma: If $a,b>0$ then $a^b+b^a>1$. We show this ineq using Bernouli inequality.

If $a\geq 1$ or $b\geq 1$, it is obvious. If $1>a,b>0$ then we have

$a^b=\frac{a}{a^{1-b}}\geq \frac a{(1-b)a+(1-b)+1}=\frac a{a+b-ab}$. Similarly, $b^a\geq\frac{b}{a+b-ab}$. Adding the two inequalities yields $a^b+b^a\geq\frac{a+b}{a+b-ab}>1$.

Now we proceed with the main problem. By the lemma, we have
\[
a^b+b^c+c^a\geq\frac {a}{a+b-ab}+\frac{b}{b+c-bc}+\frac{c}{c+a-ac}\\
\qquad \qquad \geq \frac{a}{a+b}+\frac b{b+c}+\frac c{c+a}.
\]

Now it remains to show that $\frac a{a+b}+\frac b{b+c}+\frac c{c+a}\geq\tfrac32$ for $a\geq b\geq c>0$. Let us consider the difference
\[\Bigl(\frac {a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\Bigl)-\Bigl(\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\Bigl)\\
=\frac{(a-b)(b+c)(c+a)+(b-c)(a+b)(c+a)+(c-a)(a+b)(b+c)}{(a+b)(b+c)(c+a)}\\
=\frac{(c+a)(ab+ac-b^2-bc+ab+b^2-ca-cb)+(c-a)(a+b)(b+c)}{(a+b)(b+c)(c+a)}\\
=\frac{(a-c)(2bc+2ab-ab-ac-b^2-bc)}{(a+b)(b+c)(c+a)}\\
=\frac{(a-c)(a-b)(b-c)}{(a+b)(b+c)(c+a)}\geq0
\]
The last ineq holds since $a\geq b\geq c>0$.

In addition,
\[\Bigl(\frac {a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\Bigl)+\Bigl(\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\Bigl)=3\]
and therefore, $\frac a{a+b}+\frac b{b+c}+\frac{c}{c+a}\geq\tfrac32$.

Inequality case does not occur. The inequality is strict. This completes the proof.


vector
rotation
search
function
analytic geometry
calculus
algorithm

Ok, so I have been given a triangle fixed in 2D space and another triangle with set side lengths that can be translated or rotated on top of the first triangle.  My question is knowing everything about the fixed triangle how can you minimize the distance between corner points on both triangles simply by moving the top one so that the points are within a given tolerance?

I have attempted to use everything I know from multivariable but nothing is working.  Can someone help me please!?!?!?

Thank you so much! It's for work and I really need to finish this soon!

What do you mean by "minimize the distance between the corner points on both triangles"?  Are you explicitly pairing up the vertices and summing the three lengths?  There are really quite a lot of ways to interpret your question.

You should be prepared for the possibility that there is no "nice" answer to your question.  If it is for a work application, probably it's being done by a computer and you could search for an approximate answer.  Actually, a simple binary search would probably work nicely for a minimization problem like this.

The phrase I'm wondering is about is "the points are within a given tolerance."

The position of the non-fixed triangle is a function of three variables, the $ x$ and $ y$ coordinates of one of your vertices, and an angle $ \theta$ that tells you how the non-fixed triangle is rotated.  So you need to minimize a function of three variables.  Perhaps that could be done analytically using calculus, or perhaps you could use some numerical minimization algorithm (like Levenberg-Marqart).

Matlab probably has some built in minimization function you could try.  If you don't want to go through the trouble of learning/implementing a minimization algorithm, perhaps you could try this brute force idea.  $ x$, $ y$, and $ \theta$ must lie within some known finite intervals, so you could make a discrete "grid" of possible $ x$, $ y$, and $ \theta$ values, and try all of them, and see what gives you the smallest sum of distances.

That said, there is probably a much nicer/more elegant solution to this problem, and I wouldn't be surprised if some smart kid from this website came up with a nice analytical solution.


calculus
integration
MATHCOUNTS
quadratics
algebra
polynomial
quadratic formula

I remember the question saying "with integer powers" or something like that, not "completely"

You can't put apostrophes in the hide. Please edit your post.

[color=blue][size=75]kyyuan: Done.  [/size][/color]

[hide]Since the problem never said integral coefficients we can factor it like this

$a^{8}-b^{8}=(a^{2}+ab\sqrt{2}+b^{2})(a^{2}-ab\sqrt{2}+b^{2})(a^{2}+b^{2})(a+b)(a-b)$[/hide]

[quote="lotrgreengrapes7926"][quote="SplashD"]:huh: why's it mine? :huuh:[/quote]

He means your version of the question with integral coefficients.

It never says it has to be real coefficients either... (Of course, this would contradict bpms's first hint.)[/quote]
cough mathcounts forum cough. Do I really need to edit this?

[quote="bpms"][quote="SplashD"][quote="bpms"]To splash D the powers are integers.[/quote]

yeah, i never said coefficients[/quote]
I said that because even adding in what you said wouldn't change the outcome of the problem.[/quote]

that's what i'm saying. when you said "completely", it normally would mean factor as much as possible with integer powers and coefficients, not just integer powers

[quote="SplashD"][/quote][quote="bpms"][quote="SplashD"][quote="bpms"]To splash D the powers are integers.[/quote]

yeah, i never said coefficients[/quote]
I said that because even adding in what you said wouldn't change the outcome of the problem.[/quote][quote="SplashD"]

that's what i'm saying. when you said "completely", it normally would mean factor as much as possible with integer powers and coefficients, not just integer powers[/quote]
I didn't know that. Thanks for pointing it out.

[quote="bpms"]Factor $a^{8}-b^{8}$ completely.

[hide](a^4+b^4)(a^2+b^2)(a^2+b^2)(a+1)(a-b)

I don't know[/hide]

[quote="moogra"][quote="bpms"]Factor $a^{8}-b^{8}$ completely.[/quote]

[hide](a^4+b^4)(a^2+b^2)(a^2+b^2)(a+1)(a-b)

I don't know[/hide][/quote]
Do you mean [hide]$a+b$[/hide] instead of [hide]$(a+1)$[/hide]? Other than that you are right.

[quote="anirudh"][quote="bpms"]Factor $a^{8}-b^{8}$ completely.
[hide="hint"]There are 5 factors[/hide]
[hide="bigger hint"]Think conjugate binomial[/hide]
[hide="a hint that may or may not be bigger than the previous two depending on how much experience you have and how much progress you have made in the problem"]$a^{4}+b^{4}$ can be factored with a radical[/hide]
[hide="a hint that expands on the previous one"]$a^{4}+b^{4}=(a^{2}+b^{2}-ab\sqrt2)(a^{2}+b^{2}+ab\sqrt2)$[/hide][/quote]

[hide="by using a difference of 2 squares"]
$a^{8}-b^{8}=(a^{4}+b^{4})(a^{2}+b^{2})(a+b)(a-b)$[/hide][/quote]
You can factor that more ;)
Here is why
[hide]$a^{4}+b^{4}$ can be factored...[/hide]

I wish the problem statement would be more explicit.

$a^{4}+b^{4}$ cannot be factored with [b]rational[/b] coefficients.  Yes, you are correct than it can be factored with [b]real[/b] coefficients, but you need to specify what you're factoring over.

Otherwise, we could interpret "completely" to mean factor into linear terms (that is, with [b]complex[/b] coefficients), in which case
[hide="we have..."] $a^{8}-b^{8}= \prod_{k=0}^{7}\left( a-e^{ \frac{2 \pi ik}{8}}b \right)$ [/hide]
So either way you interpret the problem condition, bpms, your answer is incorrect.  ;)

[quote="anirudh"][quote="bpms"]Factor $a^{8}-b^{8}$ completely.
[hide="hint"]There are 5 factors[/hide]
[hide="bigger hint"]Think conjugate binomial[/hide]
[hide="a hint that may or may not be bigger than the previous two depending on how much experience you have and how much progress you have made in the problem"]$a^{4}+b^{4}$ can be factored with a radical[/hide]
[hide="a hint that expands on the previous one"]$a^{4}+b^{4}=(a^{2}+b^{2}-ab\sqrt2)(a^{2}+b^{2}+ab\sqrt2)$[/hide][/quote]

[hide="by using a difference of 2 squares"]
$a^{8}-b^{8}=(a^{4}+b^{4})(a^{2}+b^{2})(a+b)(a-b)$[/hide][/quote]

[quote="bpms"]You can factor that more[/quote]

well-i didn't know you allowed radicals :maybe:
$(a^{2}+b^{2}-ab\sqrt2)(a^{2}+b^{2}+ab\sqrt2)(a+bi)(a-bi)(a+b)(a-b)$
but that is a non-real factorization :rotfl:

[quote="t0rajir0u"][quote="anirudh"]but that is a non-real factorization :rotfl:[/quote]

As a complex factorization, it is incomplete.  The quadratics can be factored ;)  (Try quadratic formula for the brute method.)[/quote]

bpms just told me that I should have not left it as $a^{4}+b^{4}$.  :rotfl:

As t0rajir0u said, the complete complex factorization is
$(a-b)\left(a-\frac{\sqrt2}{2}b-\frac{\sqrt2}{2}bi\right)(a-bi)\left(a+\frac{\sqrt2}{2}b-\frac{\sqrt2}{2}bi\right)(a+b)\left(a+\frac{\sqrt2}{2}b+\frac{\sqrt2}{2}bi\right)(a+bi)\left(a-\frac{\sqrt2}{2}b+\frac{\sqrt2}{2}bi\right)$ :D

This problem is straight from AoPS vol. 1. Factor it as far as you can with real coefficients. There.

This reminds me of a multiple choice problem in our national olympiad of 2003:

If we factor $x^{8}-1$ as a product of as many real factors as possible, how many factors are we going to obtain?

A REALLY stupid question, but 65% answered "4"! The correct answer was only found by 5.9%! (Out of 1000 participants.)  :D

[quote="anirudh"]I tried the equation on my TI-89Tit and got $(a^{4}+b^{4})(a^{2}+b^{2})(a+b)(a-b)$, my first answer... :rotfl:[/quote]
Which is why you don't rely on a calculator...

Well, if you really want to see it that way, (a + b + root(2ab))and (a + b - root(2ab)) are real factors.

But they're not integral or whatever (we don't know what's a and b) and if you don't need integral, you can factor into anything really - infinite tiny little factors if you like.

But still, (a + b + root(2ab))and (a + b - root(2ab)) are probably the only other factorisations that can be expressed in terms of a and b.  Whic is good.  Are these included in the complex factorisation? (I haven't studied complex numbers yet.) And are the ones given by complex factorisation the only possible factorisations expressed in a and b???  :love:

You all will be in perfect shape for the indefinite integral of sqrt(tanx)

[quote="+ − × ÷"]and if you don't need integral, you can factor into anything really - infinite tiny little factors if you like.[/quote]

This is actually not true.  Your factorization is a little incorrect; rather than $\sqrt{2ab}$ it should be $ab \sqrt{2}$.  The point is that the [b]coefficients[/b] are not integer, but the entire expression remains a [b]polynomial.[/b]

Regardless of what we restrict the coefficients to, we can't factor more than the factorization I gave, because $\mathbb{C}$ is what is called a [b]splitting field[/b] of the polynomials over $\mathbb{R}$.  This means that any polynomial with real coefficients can be factored into linear factors with complex coefficients, and you [b]can't factor farther[/b] than that.

[quote="+ − × ÷"]But still, (a + b + root(2ab))and (a + b - root(2ab)) are probably the only other factorisations that can be expressed in terms of a and b.  Whic is good.  Are these included in the complex factorisation? (I haven't studied complex numbers yet.) And are the ones given by complex factorisation the only possible factorisations expressed in a and b???  :love:[/quote]

Yes, and yes.  Well, to be more specific, the factorization I gave is the unique complete complex factorization into monic polynomials.

You can factor the two quadratic expressions you gave to give four of the linear factors I gave.

[quote="t0rajir0u"]
This is actually not true.  Your factorization is a little incorrect; rather than $\sqrt{2ab}$ it should be $ab \sqrt{2}$.  The point is that the [b]coefficients[/b] are not integer, but the entire expression remains a [b]polynomial.[/b]
[/quote]

actually, i was doing this on a simpler version i made up:

factor $a^{2}+b^{2}$

I think I was getting at this:

We could keep on factoring and factoring so that
$a+b$ might be factored as:

$(\sqrt{a}+\sqrt{b}+\sqrt{2\sqrt{ab}})(\sqrt{a}+\sqrt{b}-\sqrt{2\sqrt{ab}})$

and etc. - but then again, to look at a zero, this equaling zero would yield a less simple relationship so I guess that's how they decide where to draw the line.

[quote="Kurt Gödel"]This reminds me of a multiple choice problem in our national olympiad of 2003:

If we factor $x^{8}-1$ as a product of as many real factors as possible, how many factors are we going to obtain?
[/quote]

I'm not sure what the correct answer is here (presumably 5) but if they were going to count the factors of $x^{4}+1$ why not count the factors of $x^{2}+1$ or $x+1$ ? unless you're saying that factors must work for all x including negative in which case the other factors no longer real.

That might be the other consideration, to have only the factors with pronumeral outside an (even?) root.

[quote="+ − × ÷"]$(\sqrt{a}+\sqrt{b}+\sqrt{2\sqrt{ab}})(\sqrt{a}+\sqrt{b}-\sqrt{2\sqrt{ab}})$[/quote]

[b]Not a polynomial.[/b]

[quote=""+ − × ÷"]why not count the factors of $x^{2}+1$ or $x+1$ ? [/quote]

[b]Not factoriable as polynomials with real coefficients.[/b]


For the following uestion, find the magnitude of vector w to threee significant digits and its bearing to the nearest tent of a degree.

u: magnitude 23.0, bearing 215 degrees
v: magnitude 14.5, bearing 105 degrees
w: u+v

[hide="hint"]

its easy when you divide and conquer.  To solve this problem break it down into the x coordinates and add those.  Then break it down into the y coordinates and add.  Finally use Pythag's theorem and get the magnitude along with the use of tangent to get the new angle

[/hide]

Use law of cosines: $|u+v| = |u|^{2}+|v|^{2}-2|u\parallel v|\cos{(\theta_{1}-\theta_{1})}$

ok. i solved it; i had just drawn the diagram wrong :(


function
trigonometry
number theory
least common multiple
algebra proposed
algebra

hi,
find the period of $ f : \mathbb{R} \rightarrow \mathbb{R}$ with
\[ f(x) \equal{} \cos{(px)} \plus{} \cos{(qx)},\]
where $ p,q \in \mathbb{N}$.

You might be interested in the [url=http://library.thinkquest.org/17119/lg/memorize/s_p.html]sum-to-product formulas[/url].


geometry
geometric transformation
rotation
linear algebra

Let $ R_a$ rotate  around axis $ \vec a$ by angle $ \varphi_a$ and $ R_b$ rotate around angle $ \vec b$ by angle $ \varphi_b$. Both of them rotate in three dimensional space.
Let's define the rotations $ R_c\equal{}R_a R_b$ and $ R_d \equal{} R_b R_a$.
Prove that the angles $ \varphi_c$ and $ \varphi_d$ by which $ R_c$ and $ R_d$ rotate are equal. (The axises around which they rotate are not necesserily equal)
[hide="hint"]Use the homomorphism between $ SU(2)$ and $ SO(3)$ [/hide]

Do we have to assume that the compositions are rotations as well? Because they can be translations as well? :maybe:

"The axis $ \vec a$" means the line through the origin with direction $ \vec a$.  The composition of rotations around intersecting axes is a rotation.  (It can't be a translation because it fixes the origin.)


invariant
linear algebra
matrix
linear algebra unsolved

Find all matrices $ A \in \mathcal{M}_n(\mathbb{C})$ such that for all permutation matrice $ P$, $ \sigma(PA)\equal{}\sigma(A)$. Here $ \sigma(\cdot)$ denotes the spectrum.

Is there any other smarter solution than the obvious one, which is looking at the trace, then at the square of the trace, etc?  I got some pretty boring computations which solved the problem but are far from what I call beautiful.

Trace and trace of squares is a good way to start.
I don't think you need to do so many "boring calculations" :wink:

We say that $ A$ satisfies the property $ \mathcal{P}$ iff for all permutations $ P$ $ spectrum(PA)\equal{}spectrum(A)$.$ P\rightarrow{PA}$ works on the rows,$ P\rightarrow{AP}$ works on the columns. Let $ U\equal{}[1\cdots,1]^T$.We prove that $ A$ satisfies $ \mathcal{P}$ iff $ A$ or $ A^T$ is in the form $ UL$ where $ L$ is an arbitrary row.
Proof ($ \rightarrow$)
1) If $ A$ satisfies $ \mathcal{P}$ then $ AP,PA,A^T$ satisfy $ \mathcal{P}$.
2) $ \bullet$: Take $ P\equal{}(12)$. Then $ trace(A)\equal{}trace(PA)$ or $ a_{11}\plus{}a_{22}\equal{}a_{12}\plus{}a_{21}$.
$ \bullet$: Take $ P\equal{}(2,j)$ with $ j>2$. $ AP$ satisfies the preceding result: $ a_{11}\plus{}a_{2j}\equal{}a_{21}\plus{}a_{1j}$ or $ a_{21}\minus{}a_{11}\equal{}a_{2j}\minus{}a_{1j}$.
If $ L_i,C_j$ denote rows and columns of $ A$ then $ L_2\equal{}L_1\plus{}\alpha_2U^T$. More generally $ L_i\equal{}L_1\plus{}\alpha_iU^T$ or $ a_{ij}\equal{}a_{1j}\plus{}\alpha_i$.
By the same method we obtain $ C_j\equal{}C_1\plus{}\beta_jU$; in particular $ a_{1j}\equal{}a_{11}\plus{}\beta_j$ and $ a_{ij}\equal{}\alpha_i\plus{}\beta_j\plus{}a_{11}$.
3) Let $ p\equal{}det(A\minus{}xI),P\equal{}(ij),q\equal{}det(PA\minus{}xI)$. $ rank(A)\leq{2}$, then it is easy to see that $ p\minus{}q$ is a monomial of degree $ n\minus{}2$. In fact we have the following result (I have not written the proof !): $ p\minus{}q\equal{}\minus{}nx^{n\minus{}2}(\alpha_i\minus{}\alpha_j)(\beta_i\minus{}\beta_j)$; thus for all $ (i,j)$ $ \alpha_i\equal{}\alpha_j$ or $ \beta_i\equal{}\beta_j$.
Assume $ \beta_1\not\equal{}\beta_2$; $ \alpha_1\equal{}\alpha_2$. Let $ i>2$.Either $ \alpha_1\equal{}\alpha_i$ or $ \beta_1\equal{}\beta_i$ and then $ \alpha_2\equal{}\alpha_i$.
 Conclusion $ \alpha_1\equal{}\alpha_2\equal{}\alpha_i$, $ a_{ij}\equal{}\gamma_j$ and $ A\equal{}U[\gamma_1\cdots\gamma_n]$.       QED

[quote="loup blanc"] $ A \equal{} U[\gamma_1\cdots\gamma_n]$.       [/quote]

Just an observation $ A^2\equal{}yA$ where $ y\equal{}\sum_{j\equal{}1}^{n} \gamma_{j}$

if $ y\neq 0$ then the matrix $ A$ is diagonalisable

if $ y\equal{}0$ then $ A$ is nilpotent


The tree below is used to convert strings of digits to words. In the string, $ ^{\prime\prime} 0 ^{\prime\prime}$ represents a left branch and a $ ^{\prime\prime} 1 ^{\prime\prime}$ represents a right branch. Notice $ ^{\prime\prime} 010 ^{\prime\prime}$, or left-right-left, gives the letter $ ^{\prime\prime} U ^{\prime\prime}$ and $ ^{\prime\prime} 10 ^{\prime\prime}$, or right-left, gives the letter $ ^{\prime\prime} N ^{\prime\prime}$. The strings for letters are placed side-by-side to build words. After you reach a letter in the string, return to $ ^{\prime\prime} Start ^{\prime\prime}$ and continue. The string $ ^{\prime\prime} 1110 ^{\prime\prime}$ represents $ ^{\prime\prime} IN ^{\prime\prime}$. What word does $ ^{\prime\prime} 000101010011 ^{\prime\prime}$ represent?

[asy]dot((0,0));
dot((-2,-1));
dot((-1,-2));
dot((2,-1));

draw((0,0)--(-2,-1)--(-3,-2));
draw((-2,-1)--(-1,-2)--(0,-3));
draw((-1,-2)--(-2,-3));
draw((0,0)--(2,-1)--(1,-2));
draw((2,-1)--(3,-2));

label("Start", (0,0),N);
label("S",(-3,-2),S);
label("U",(-2,-3),S);
label("Y",(0,-3),S);
label("N",(1,-2),S);
label("I",(3,-2),S);[/asy]


If $x,y,z\geq 0$ and $x^2+y^2+z^2=1$ prove that

$\frac{x}{1-yz}+\frac{y}{1-zx}+\frac{z}{1-xy}\leq\frac{3\sqrt 3}{2}$

[quote="silouan"]Maybe your solution is nice but I cannot understand your thoughts. How do you go from one step to another ,sorry.[/quote]
Ok; maybe it is clear now ;)

[quote="Ramtin Madani"]First I sould prove: $\frac{x}{1-yz}\leq\frac{4x+y+z}{4\sqrt{x^2+y^2+z^2}}$
Because $\frac{x}{x^2+y^2+z^2-yz}\leq\frac{4x+y+z}{4{x^2+y^2+z^2}}$
[/quote]

I cannot understand this relation.

Oh I am sorry it was that: :blush:
$\frac{x}{x^2+y^2+z^2-yz}\leq\frac{4x+y+z}{4(x^2+y^2+z^2)}$


vector
linear algebra
linear algebra theorems

Let X be a vector space over complex numbers. How can we treat X as a vector space over reals?

If $ X$ has basis $ x_1, x_2, \ldots, x_n$ over $ \bf C$ then it has basis $ x_1, x_2, \ldots, x_n, i x_1, i x_2, \ldots, i x_n$ over $ \bf R$.  Here $ i$ is the imaginary unity and scalar multiplication by $ \bf R$ is inherited from scalar multiplication by $ \bf C$.

[quote="JBL"]... and scalar multiplication by $ \bf R$ is inherited from scalar multiplication by $ \bf C$.[/quote]
You don't need to say anything about a basis. This clause is the whole argument.

And this also being because R is a subfield of C, right?


1) Does anyone have the answer key to izzy's mock mathcounts test? Especially to number 27 (the exhibit one)
2) What was the masters round in 2007 like, because i heard only kevin chen got it right.

I *had* an answer key, but I'm not *quite* sure where it is at the moment. Try asking some of the people who got high scores.


[code]S_{1} = \begin{bmatrix}a & b & c & d\\ b & e & f & g\\ c & f & h & i\\ d & g & i & j\end{bmatrix}
S_{2} = \begin{bmatrix}b & e & f & g\\ e & k & l & m\\ f & l & n & o\\ g & m & o & p\end{bmatrix}
S_{3} = \begin{bmatrix}c & f & h & i\\ f & l & n & o\\ h & n & q & r\\ i & o & r & s\end{bmatrix}
S_{4} = \begin{bmatrix}d & g & i & j\\ g & m & o & p\\ i & o & r & s\\ j & p & s & t\end{bmatrix}[/code]

So I have these matrices. How do I put them so that they are in two columns so that they don't take up 3/4 of a page, and leave a giant blank spot on the previous page?

(i.e. How do I put the matrices in the form

S_{1}    S_{3}

S_{2}    S_{4}

?)

You can use aligned:
$ \begin{aligned}
S_{1} = \begin{bmatrix}a & b & c & d\\ b & e & f & g\\ c & f & h & i\\ d & g & i & j\end{bmatrix}&\hspace{40pt}
S_{2} = \begin{bmatrix}b & e & f & g\\ e & k & l & m\\ f & l & n & o\\ g & m & o & p\end{bmatrix}\\
S_{3} = \begin{bmatrix}c & f & h & i\\ f & l & n & o\\ h & n & q & r\\ i & o & r & s\end{bmatrix}&\hspace{40pt}
S_{4} = \begin{bmatrix}d & g & i & j\\ g & m & o & p\\ i & o & r & s\\ j & p & s & t\end{bmatrix}
\end{alighned}$
or even the same matrix:
$ \begin{matrix}
S_{1} = \begin{bmatrix}a & b & c & d\\ b & e & f & g\\ c & f & h & i\\ d & g & i & j\end{bmatrix}&\hspace{40pt}&
S_{2} = \begin{bmatrix}b & e & f & g\\ e & k & l & m\\ f & l & n & o\\ g & m & o & p\end{bmatrix}\\[50pt]
S_{3} = \begin{bmatrix}c & f & h & i\\ f & l & n & o\\ h & n & q & r\\ i & o & r & s\end{bmatrix}&\hspace{40pt}&
S_{4} = \begin{bmatrix}d & g & i & j\\ g & m & o & p\\ i & o & r & s\\ j & p & s & t\end{bmatrix}
\end{matrix}$
or almost anything else you wish: remember that matrices are neither better nor worse than any other symbols for various string and alignment manipulations :).

Not sure why this is a problem as all you need do is add a newline \\. Using \\[2ex] gives slightly better spacing
[code]S_{1} = \begin{bmatrix}a & b & c & d\\ b & e & f & g\\ c & f & h & i\\ d & g & i & j\end{bmatrix}
S_{3} = \begin{bmatrix}c & f & h & i\\ f & l & n & o\\ h & n & q & r\\ i & o & r & s\end{bmatrix}\\[2ex]
S_{2} = \begin{bmatrix}b & e & f & g\\ e & k & l & m\\ f & l & n & o\\ g & m & o & p\end{bmatrix}
S_{4} = \begin{bmatrix}d & g & i & j\\ g & m & o & p\\ i & o & r & s\\ j & p & s & t\end{bmatrix}[/code]
gives
$ S_{1} = \begin{bmatrix}a & b & c & d\\ b & e & f & g\\ c & f & h & i\\ d & g & i & j\end{bmatrix}
S_{3} = \begin{bmatrix}c & f & h & i\\ f & l & n & o\\ h & n & q & r\\ i & o & r & s\end{bmatrix}\\[2ex]
S_{2} = \begin{bmatrix}b & e & f & g\\ e & k & l & m\\ f & l & n & o\\ g & m & o & p\end{bmatrix}
S_{4} = \begin{bmatrix}d & g & i & j\\ g & m & o & p\\ i & o & r & s\\ j & p & s & t\end{bmatrix}$

If you are using display mode then the align environment does the same
[code]\begin{align*}
S_{1} = \begin{bmatrix}a & b & c & d\\ b & e & f & g\\ c & f & h & i\\ d & g & i & j\end{bmatrix}
S_{3} = \begin{bmatrix}c & f & h & i\\ f & l & n & o\\ h & n & q & r\\ i & o & r & s\end{bmatrix}\\[2ex]
S_{2} = \begin{bmatrix}b & e & f & g\\ e & k & l & m\\ f & l & n & o\\ g & m & o & p\end{bmatrix}
S_{4} = \begin{bmatrix}d & g & i & j\\ g & m & o & p\\ i & o & r & s\\ j & p & s & t\end{bmatrix}
\end{align*}[/code]
\begin{align*}
S_{1} = \begin{bmatrix}a & b & c & d\\ b & e & f & g\\ c & f & h & i\\ d & g & i & j\end{bmatrix}
S_{3} = \begin{bmatrix}c & f & h & i\\ f & l & n & o\\ h & n & q & r\\ i & o & r & s\end{bmatrix}\\[2ex]
S_{2} = \begin{bmatrix}b & e & f & g\\ e & k & l & m\\ f & l & n & o\\ g & m & o & p\end{bmatrix}
S_{4} = \begin{bmatrix}d & g & i & j\\ g & m & o & p\\ i & o & r & s\\ j & p & s & t\end{bmatrix}
\end{align*}

[size=75]Edit: I didn't see fedja's post until I had submitted mine :)[/size]


[quote="Phoenix257"]4 boys and 4 girls sit around a merry-go-round.  How many different ways could they be arranged if the boys and the girls sit alternately around the merry-go-round?[/quote]
[hide]$4!^{2}\cdot 2=1152$[/hide]

[quote="zserf"]Ambiguous question, after all if you rotate the marry round then you have to divide by 8[/quote]
There is no ambiguity here. Seating people in a merry-go-round is similar to seating them in a circular table.

[hide]Let us first seat the girls around the merry-go-round leaving one seat between each pair of girls. This can be done in $\frac{4!}{4}= 6$ ways. Having seated the girls, now you can seat the boys in the empty seats in $4! = 24$ ways. So the total number of ways to seat the girls and boys is $6 \cdot 24 = 144$.[/hide]

[hide]That's what I would put, Just pointing out your answer times 8 equals pheonix257's answer[/hide]


When a package is not installed, one can do it manually. It appears the following message:

[img]http://img514.imageshack.us/img514/4939/sdfsdfjx5.png[/img]

One can change where the package will be installed from, for example I chosen "Packages shall be installed from a directory".

My problem is, when selecting that option (previously I already chose the directory), it says "packages does not seem to be a local package repository".

How can I fix this?

To do that you would have to create a package repository on your own computer which would contain MiKTeX's special files which tell it how to do the automatic installation. In order to do that you would need to download the complete Miktex package (around 600MB) but not actually install it all - not sure why you would want to do that.
You may find your local repository at C:\Documents and Settings\<Username>\Local Settings\Temp\MiKTeX 2.x Packages which has a lot of cab files.

Okay, I located C:\Documents and Settings\Administrador\Local Settings\Temp\MiKTeX 2.7 Setup. (I actually had to create the directory "MiKTeX 2.7 Setup".)

Do I have to put a file there? Because I wanted to install from there the fancybox package (previously downloaded, which only contains a .doc, .pdf and the .sty file) and I put it on MiKTeX 2.7 Setup, but it appears the same:

[img]http://img443.imageshack.us/img443/1459/dfhdfgxu4.png[/img]

:(

I really don't undertsand what you are doing. As I said before MikTeX uses its own special files to install. You would need fancybox.cab which contains fancybox.tpm which tells MiKTeX what to do. If fancybox.cab is not there then you would have to find it but what's the point when you can setup from an online repository.

Anyway, fancybox is very simple to install yourself. Just put fancybox.sty in <your tex files path>\tex\latex\fancybox and refresh FNDB in MikTex Settings.

But there is an even simpler way. Download and install the complete MikTeX installation and then you are unlikely to have to install anything again.


Can anyone help me with the LaTeX formatting part? I mean, how should I format the page appearance? Thanks!

Note this isn't asking for solving help, just for formatting help.

If you are typing up a document in Latex, just make sure that you have all the information that you would if you submitted in another way. So whatever, I'm just copying this from the cover sheet:

Last Name: _________ First Name: _________

Admissions Test:  B

Contact Info: Phone Number: _______
Email: ______

Number of pages: (without the cover sheet)  ________

As for basic latex things, if you do // then enter twice, that helps skip a line.

\pagebreak

breaks the pages (no dollar signs around that). (you should have one problem per page).

Then name; problem number; page blah out of blah. There are fancy ways to do this. I believe the command for page number is \thepage

You can just do all of this after you finish, it shouldn't be that bad.

I recommend modifying the usamts.tex file which can be found here: http://www.usamts.org/MyUSAMTS/U_MyForms.php

If you need an example, I can show you my Mathcamp Qualifying Quiz template.

I'll try my best. (which won't be too good... :oops: )


AMC
AIME
USA(J)MO
USAMO
USAJMO
AMC 12
USAMTS

Do you guys think the AIME qualifying score will be below a 65?

[quote="Thunder365"]Hmmm...round one was pretty hard, but round two was pretty easy, so that balances out. I cant discuss round three, but I still think its going to stay the same--around a 68.[/quote]

lol are you serious? i thought it was the other way around :P

I have a some similar questions. Suppose that I qualify for AIME through USAMTS. Will I be able to try out for USAMO, USAJMO (since I am in 7th grade), or both (as a result of taking that AIME)? Also, to qualify for USAMO, you must have a certain index, which depends on your AMC 12 score. Do you also have to take the AMC12 to try to qualify for USAMO, or do you get a specific score for the AMC 12 just by taking the AIME? I know that the second option sounds bad, but if USAMTS goes straight into AIME, how are they going to make sure that everybody takes the AMC 10/12 if they want to qualify for USAMO/USAJMO irrespectively?

Also, what will be the logistics of this? Will our schools be sent AIMEs for each student in the school who does well enough on USAMTS?

Thanks.

Qualfication for AIME through USAMTS assumes a score of 100 on the AMC 12.

@BOGTRO
I disagree. Unless I misunderstand.

[quote]Participants who score well after 3 rounds will be invited to participate in the American Invitational Mathematics Examination (AIME)[/quote]

100 on the AMC 12 qualifies you for AIME. (USAMTS score is irrelevant in this process)

I think it's something like top 10% in USAMTS also qualifies you for AIME. This is normally about 68.

Qualifying by way of USAMTS gives you an equivalent of 100 on the AMC 12 toward the USAMO index. So AMC 12 is a much better way to get to the AIME.

Oh... oops.  :blush: I thought he meant that you need a AMC12 100 to qualify as well as a good USAMTS.

Wait, so even if you get 100/100 on USAMTS, you only get 100/150 on your index for AMC 12 if you qualify through USAMTS?

Yes...it is just so if somebody cannot take the AMC tests they still have a way to qualify.  Obviously, the preferred method for qualification is through an AMC test.

Sorry for reviving.

Because of snow, I am not going to have school on Tuesday. I contacted the MAA and asked what my options were as to how to take the AMC 10/AMC 12 A. No colleges in our area offered the AMC 10/12 A. I contacted counties in my area, asking whether I can take the AMC 10/12 A, and whether there would be school on Tuesday, in vain. Therefore, I will only be able to take the AMC 10 B and not the AMC 12 B. I am likely to qualify for AIME through the USAMTS, and that way my score on AMC 12 will be a 100 (if I qualify that way.) My question is, when I take the AIME, whom should I notify that I qualified through USAMTS, giving me an AMC 12 of 100, how will I prove that, and where will I be able to register that? Thanks.

We will let the AMC know who the USAMTS AIME qualifiers are.  They will take care of everything from there.

In my profile, there are two fields (not required):
School CEEB code
Teacher Name
Do I need to fill these out so that I can qualify?

Also, what index do we get if the AMC cutoff is below 100.... is it still 100 or will it be the cutoff?


For the following set of data, by how much does the median exceed the mode?

\[ 64, 92, 48, 82, 42, 48, 87, 80\]

First of all, to make life easier, line up the data points consecutively.

$ 42$, $ 48$, $ 48$, $ 64$, $ 80$, $ 82$, $ 87$, $ 92$

From this, it should be fairly obvious that the mode is $ 48$, and the median is $ \frac{64\plus{}80}{2}\equal{}72$. $ 72\minus{}48\equal{}\boxed{24}$


geometry
3D geometry
search
function
number theory proposed
number theory

For all integer $x \ge 4$, $x^{3}$ can be written as a sum of five cubes of integers strictly less than $x$ in absolute value.

It is a Lemma of Veitnam TST 2005 Q3.I don't know how to prove it.Please help :blush:

You can see here, if you don't understand, i will help  :wink:

I think posted before... (probably hard to find)

And bad title (see http://www.mathlinks.ro/Forum/viewtopic.php?t=135914 ), see the new one.
The only reason why I don't lock it is because there is already an answer given...

ZetaX,

You said this topic's title is BAD. Why?

P/S: I readed topic #135914 of you :)

i.e you edited name of this topic?

Thank you, nice Mod  :wink:

THank you,ZetaX.I 'll try to do better next time :blush:
Thanks, N.T.TUAN.
I only know little Vietnamese word,but I have understood it.
$(2k+1)^{3}=(2k-1)^{3}+(4+k)^{3}+(4-k)^{3}-5^{2}-1^{3}$(I find that the word "Solution" is the same as in English
.)By the way ,to ZetaX, I have used the search function before posting this topic.But I didn't find any solution.(Somebody(I forgot the name) post this this Lemma ,But didn't give the solution)
 Maybe the topic I can't find also has a bad title......

Ok, note if $n$ is ''good'' then $2n$ is ''good''. That is all :)


A box contains $ 93$ balls: $ 22$ purple, $ 17$ aqua, $ 16$ brown, $ 15$ white, $ 12$ black, and $ 11$ red. A sample of size $ N$ is drawn at random from the box. What is the smallest possible value of $ N$ that will ensure that the sample will contain at least $ 14$ balls of the same color?

Could you please explain why you did that?

[hide="Solution"]
So to ensure that there will be 14 of a certain color, you must draw the max number of each color that's 13 or less. Then, that guarantees that the next ball you draw will be the 14th of any given color.
Therefore, it's 11(red)+12(black)+13(white)+13(brown)+13(aqua)+13(purple)+1(the 14th ball that could be white/brown/aqua/purple)=76.
[/hide]


algebra
polynomial
induction
combinatorics open
combinatorics

What do you claim -- that some coefficients is equal to powers of 3 or that they are divided
by some powers of 3? The first claim seems to be wrong.

Now about divisiblity: let's compute your polinom  p(x)=(x+1)(x+2)...(x+p^n-1) modulo p^n
(as you statements are about divisibility by p^i, where i<=n).
This polinom can be expressed as a product of p_1(x)=(x+3)(x+6)...(x+p^n-p) and
p_2(x)=p(x)/p_1(x). Modulo p^n it is obvious that p_1(x) is divided by x^{p^n-n}
and it is not very hard to see (for example, by induction on n) that p_2(x)=(x^{p-1}-1)^{p^{n-1}}. This make the situation more clear.
Now we are to understand what do you mean by 'k-th coefficient'
(because all coefficients in p^n-n smallest degrees are divided by p^n).


Nancy generates a two-digit integer by rolling a six-sided die twice. The result of her first roll is the tens digit, and the result of her second roll is the ones digit. What is the probability that the resulting integer is divisible by 8? Express your answer as a common fraction.

It is faster to list out multiples of 8 that work than it is to list dice rolls.

We find all two digit multiples of 8 which have digits 1-6

16
24
32
56
64

Each of these can be formed in one way. There are 36 total possiblities.

Thus our probbability is $ \frac{5}{36}$


linear algebra
matrix
inequalities
vector
linear algebra unsolved

$ A,B,C$ are $ n\times n$ matrix.

$ \mathrm{rank} \begin{pmatrix} A & C \\
0 & B \end{pmatrix} \geq \mathrm{rank} A \plus{} \mathrm{rank} B$ ? why ?
thanks

Example where strictly greater is possible:
\[ \left(\begin{array}{cc|cc}1&0&0&0\\0&0&1&0\\\hline 0&0&0&1\\0&0&0&0\end{array}\right)\]
We have $ \text{%Error. "rank" is a bad command.
}\,A+\text{rank}\,B=1+1=2$ while the rank of the $ 4\times 4$ matrix is $ 3.$

To prove the inequality, just row reduce. We have invertible matrices $ P$ and $ Q$ such that $ PA=R$ and $ QB=S$ where $ R$ and $ S$ are in row echelon form. Then
\[ \begin{pmatrix}P&0\\0&S\end{pmatrix}=\begin{pmatrix}R&PC\\0&S\end{pmatrix}\]
That might not yet be in row echelon form. However, all of the pivots of $ R$ are pivots of a row echelon form of the larger matrix. Meanwhile, the pivots of $ S$ must lie in independent columns of the larger matrix. There may be additional columns in the right hand half of the larger matrix; these might or might not increase the overall rank.

[quote="Kent Merryfield"]Example where strictly greater is possible:[/quote]

No offence, but do you really find this surprising? I mean, $ A = B = 0$, $ C = \text{something else}$ wouldn't do?

The problem at hand is another example of how the [url=http://planetmath.org/encyclopedia/FrobeniousInequality.html]Frobenius inequality[/url] proves everything: Set

$ U = \left(\begin{array}{cc}0 & I_n\end{array}\right)$,
$ V = \left(\begin{array}{cc}A & C \\
0 & B\end{array}\right)$,
$ W = \left(\begin{array}{c}I_n \\
0\end{array}\right)$,

and use the Frobenius inequality in the form $ \mathrm{rank}\left(UVW\right) + \mathrm{rank}V\geq \mathrm{rank}\left(UV\right) + \mathrm{rank}\left(VW\right)$.

This is probably not the most explanatory proof, but it saves a lot of time on exams and yields the following generalization:

If

$ %Error. "xymatrix" is a bad command.
{ \& U \ar[r] \ar[d]_f \& V \ar[r] \ar[d]_g \& W \ar[r] \ar[d]_h \& 0 \\
0 \ar[r] \& U^{\prime} \ar[r] \& V^{\prime} \ar[r] \& W^{\prime} \& 0 }$

is a commutative diagram of vector spaces with exact rows, then $ \mathrm{rank}g\geq \mathrm{rank}f+\mathrm{rank}h$. We can even weaken the exactness condition; all we need is exactness at $ W$ and at $ U^{\prime}$ (meaning surjectivity and injectivity, respectively) and at least one of the two rows must be a complex. D@mn, they have no xymatrix installed, so I'll just hope someone can read the above.

  darij


geometry
3D geometry
icosahedron
vector
sphere

here is a sub-part of an MOP problem, i was working on recently:

what is the acute angle between the main diagnoles of an icosahedron?

Eeewwwww.... I hated that.

Hint[hide]Represent the vertices as vectors, and use the dot product[/hide]

yeah, lots of people missed the actuall problem. (it was about lines in the sphere,) :D


Let an=1+1/2+1/3+...+1/n. Prove that for every n greater than 2 we have:
(a_n)^2>2(a_2/2+a_3/3+...+a_n/n)

\sum i=2 to n a i =
\sum i=2 to n \sum j=1 to i 1/ij =
\sum i=1 to n \sum j=1 to i 1/ij  - 1 =
\sum i=1 to n \sum j=1 to i-1 1/ij + \sum i=1 to n 1/i2 - 1.

an2 = (\sum i=1 to n1/i)2 =
\sum i=1 to n 1/i2 + 2 \sum i=1 to n \sum j=1 to j-1 1/ij

an2 - \sum i=2 to n a i = 2 - \sum i=1 to n1/i2.
\sum i=1 to n1/i2 < 1 + 1/1.2 + ... + 1/(n-1)n = 1 + (1/1-1/2) + (1/2-1/3) + ... + (1/(n-1)-1/n) = 2 - 1/n < 2.

However the solution is easy, maybe it's difficult to be undestood VISUALLY.
Just (re)write it yourself ...


trigonometry
inequalities
inequalities proposed

Find the maximum value of
$ \frac{sinA}{A}\plus{}\frac{sinB}{B}\plus{}\frac{sin{C}}{C}$

$ \sum \frac{A}{sin A}\equal{}\sum \frac{1}{2R}\equal{}\frac{3}{2R}$

How can we find the max ?  :wink:

Nope. $ A \neq a$. A is the angle, not the sidelength!

without a restriction on A, B, C i dont see how this is different than finding the max of

$ \frac{3sinx}{x}$ which.. is arbitrarily close to 3

Of course, $ A,B,C$ are angles of a triangle...  :)

(CRUX 1217; 1988:120)

$ \frac {\sin A}{A} \plus{} \frac {\sin B}{B} \plus{} \frac {\sin C}{C}\leq \frac {9\sqrt {3}}{2\pi}$.

I don't have the proof here, but I'll probably post it tonight (hope I'll find it in the literature, or solve it, if I find some free time- it should be an application of Jordan's inequality or something like that...).

EDIT: Why can't we solve it using Jensen? It suffices to show $ \frac{ \minus{} 2 \cos x}{x^2} \plus{} \frac{2 \sin x}{x^3} \minus{} \frac{\sin x}{x} < 0$ for all $ x\in \left( 0,\frac {\pi}{2} \right)$, right?


Prove that there are no positive integers $ n$ and $ k\leq n$ such that

$ \binom{n}{k},\binom{n}{k\plus{}1},\binom{n}{k\plus{}2},\binom{n}{k\plus{}3}$

in this order form an arithmetic progression.

If they form, then $ \binom{n\minus{}1}{k}, \binom{n\minus{}1}{k\plus{}1}, \binom{n\minus{}1}{k\plus{}2}$ are equal so $ n\minus{}1\equal{}2k\plus{}1\equal{}2k\plus{}3$, contradiction.

[quote="mazur89"]If they form, then $ \binom{n \minus{} 1}{k}, \binom{n \minus{} 1}{k \plus{} 1}, \binom{n \minus{} 1}{k \plus{} 2}$ are equal [/quote]

I think you are mistaken. Check carefully  :D

Yes, I'm sorry.

Maybe this will be correct:

If $ \binom{n}{k}, \binom{n}{k\plus{}1}, \binom{n}{k\plus{}2}$ form an atithmetic progression, then:
$ 0\equal{}\binom{n}{k}\minus{}2\binom{n}{k\plus{}1}\plus{}\binom{n}{k\plus{}2}\equal{}\frac{n!((k\plus{}1)(k\plus{}2)\minus{}2(n\minus{}k)(k\plus{}2)\plus{}(n\minus{}k)(n\minus{}k\minus{}1))}{(n\minus{}k)!(k\plus{}2)!}$.
So $ 0\equal{}(k\plus{}1)(k\plus{}2)\minus{}2(n\minus{}k)(k\plus{}2)\plus{}(n\minus{}k)(n\minus{}k\minus{}1)\equal{}k^2\plus{}3k\plus{}2\minus{}2nk\minus{}$ $ 4n\plus{}2k^2\plus{}4k\plus{}n^2\minus{}2nk\plus{}k^2\minus{}n\plus{}k$,
$ n^2\minus{}n(4k\plus{}5)\plus{}4k^2\plus{}8k\plus{}2\equal{}0$.
So the number $ (4k\plus{}5)^2\minus{}4(4k^2\plus{}8k\plus{}2)\equal{}16k^2\plus{}40k\plus{}25\minus{}16k^2\minus{}32k\minus{}8\equal{}8k\plus{}17$ must be a square of an integer greater than $ 1$.
If $ \binom{n}{k\plus{}1}, \binom{n}{k\plus{}2}, \binom{n}{k\plus{}3}$ form an atithmetic progression, then $ 8(k\plus{}1)\plus{}17$ is a square of an integer, contradiction, because $ a^2\minus{}b^2\equal{}8$ implies $ a^2\equal{}9, b^2\equal{}1$.


geometry
3D geometry
advanced fields
advanced fields unsolved

Hi !!

I'm looking for the proof that 1 and 8 are the only cubes in Fibonacci sequence.

I found the proof that 1 and 144 are the only squares, it's nice and tricky, but it isn't difficult

I would like to see the proof for cubes, I try to find it for 2 hours but I didn't found yet

Maybe someone has this proof or could say something about it?

See [url]http://arxiv.org/PS_cache/math/pdf/0110/0110150v2.pdf[/url].
Read the introduction and it should lead you to your desired proof and some history of the subject.
For severely deep, related material see [url]http://www.maths.warwick.ac.uk/~siksek/papers/fnpm.pdf[/url].


Jack is a sailor (the ones that use that triangular hat) that went to the square market.
He then goes to the vegetables stall and buys $x$ potatoes. After this, he goes to the fruits stall and buys $y$ coconuts.
He knows that the total weight is $w$ and the total volume is $v$. Carrying his market basket the way home, he gets hit by a car and his basket flies away, throwing coconuts and potatoes all the street long.
Knowing that the gravity at that place is $g$, evaluate the sum $x+y$ and the amount of money $m$ lost in the scene.

clearly we ought to set up the equation
m(gx+gy)
i leave the rest as an exercise for the reader.

-function


algebra
function
domain
calculus
calculus open

if f{x} prime=1/x^2+e^x^2 and the domain of f{x} is{1/e^2,infinity},then proove that f{x} is <1+tan{inverse}pi/4

As written, this is obviously and extremely false: $f$ goes to $\infty$ rapidly, and we can add a constant to make $f(x)$ as large as we want for any fixed $x$.

Definitely, it is not an Open Question :)  :lol:  :lol:  :lol:


A man walks into a shop. He buys something for 50 dollars using a 100-dollar bill. Unfortunately, the clerk is out of change (don't ask why) so goes to the adjacent shop and exchanges the 100-dollar bill for change with the other clerk. The man walks off with the change AND item. Later, the other clerk walks back and tells the first clerk that the 100-dollar bill is a fake, so the clerk takes out a 100-dollar bill from his register and gives that bill to the other clerk and takes back the fake bill (DON'T ASK WHY). Altogether, how much money did the first clerk lose?

Please do not inquire about the actions. The problem is the way it is. If you heard it, don't post, please.

Hints will be given tomorrow (if nobody manages to answer it by then). Have a nice day. GB

[hide]He sold for 50 dollars; the clerk gets 50 dollars.

The other guy says its fake, they exchange, so he loses 50 dollars, or he loses negative 50 dollars

OH yeah; you're the n00b[/hide]

stup1d ************* tex! How do I get it to turn off if I want to use the dollar bill sign? Or, should I not use dollar bill signs at all unless I want to use tex?

[quote="darij grinberg"]There is a checkbox below the field where you write your reply. It is called "Disable BBCode in this post". If it is activated, the $ signs are not converted into TeX.
Just my 0.02 $ ...

  darij[/quote]

Well, then the spoiler doesn't work. So, I just said "Dollars".
Thanks anyway.

wrong answer pakagawa. that's all im saying

I AM NO NOOB. Seriously. Dont just judge people by one problem

make a column of EVERYTHING he receives and EVERYTHING he loses.

the answer (just to say, im going bored) is -100.

Let's start the clerk off with zero money.

First, he gains 100 in fake (nothing). Amount of money currently: 0.

Then, he exchanges that fake for 100 in real change. Amount of money currently: 100.

He give the 50 change and the item. Amount of money: 50.

Exchanges 100 real from storage with the other clerk for the 100 fake. Amount of money: -50.

FINALLY, ADD THE ITEM IN. The actual point is: total monetary losses and adding other losses, too. Since the item was 50 dollars, he "loses" the item, thus giving another part of the answer: -50. Add both losses together and get -100.

NOTE: If the clerk is cheaty enough to use the 100 fake, that might change the answer, but NO FAKES ALLOWED!


geometry
rectangle
calculus
integration
3D geometry
combinatorics solved
combinatorics

Let k be a given positive integer. Suppose player A and B plays the following game in the infinite square grid:
In each turn, player A marks an unmarked unit square red, and player B is allowed to mark k unmarked unit squares blue. Prove that player A can always mark 4 unit squares red such that their center form a square with sides parallel to the grod lines.
(This problem was also in an old Kvant volume).

What are the dimensions of the square grid?

Yes this is a very nice problem. It also appeared in the 1983 Tournament of Towns.

Pierre.

I'm sure I must have misread the problem; otherwise it is simply too easy.

Player A colours any unmarked unit square red in his first turn.  In the subsequent k + 1 turns, player A colours an unmarked unit square in the same row red.  In the following turn, player A colours an unmarked unit square in any row of the grid in which none of the squares have been coloured and in a column in which he has already coloured a tile red.  We now note that there are k + 1 unmarked squares which player A can colour in the next round to win (in the same row as his previous turn, above/below the other red coloured squares).  But B cannot colour all these k + 1 squares in his next turn, so player A wins.

Yes you are wrong. In such a way you obtain a monochromatic rectangle, but not a square.
can pbornstein or lhvietbao be so kind and post the solution?

Oops, I knew that I had forgotten something :blush: .

Problem from Tournament of Towns was formulated for k=2 with question "Is it true ...".

Yes Myth, but A solution given then was the general case.

Pierre.

Do you mean there was official solution? I never saw any official solution for Tournament of Towns.

Here is my solution:
We choose a large number N, which we will specify later.
First of all, it's easy to see that we can always mark N red squares grid row, which we will call the first row.
Then we choose the second row under this first row so that there are no marked squares in this row. Now consider the N projection squares of the above N red squares onto this row. Since player B can only mark k squares at each turn, player A can always ensure to mark at least $ \frac {N} {k+1} $ squares red in the second row. Now repeat this progress m times (this can be done for a given m by choosing large enough N), we obtain the following configuration:
We get m grid rows and at least $\frac {N} { ( k+1 )^ m } $ grid columns so that all the squares at their intersection are coloured red. By choosing large m and a much larger N (in comparison with m), we can ensure that A can mark the ab intersection squares of a grid rows and b grid columns with red, for any given a,b.
Now suppose we have accomplished this step with given large a,b ( how large we will specify later).
Now in his next turn,A will colour a square x with red so that x lies on the right of all the marked squares, x lies in the first row (of the a grid rows in configuration (*)) and when we draw the line d wich passes through x and creates with the grid rows a 45 degree angle to the left, then all marked squares lies on the left on d.
We will call this step x.(This can always be done by choosing x sufficently far to the right). Now in the next $\frac {a} {k+1} $ steps A will be able to mark red at least $\frac {a} {k+1} $ squares of the total a intersection squares of the a grid rows in configuration (*) with the line d. (1)
Then there must be at least unmraked b-a squares left of the total b intersection squres between the b columns in configuration (*) and the line d. Thus in the next $\frac {b-a} {k+1} $ steps A can ensure to mark at least $\frac {b-a} {k+1} $ of these b-a unmarked intersection squares. (2)
Now we consider the r=$\frac {a} {k+1}$ grid columns which passes through the r red squares in (1) and the s=$\frac {b-a} {k+1} $ grid rows which passes through the s red squares in (2). In total there are rs intersection squares of these rows and columns, and we have already marked all the squares which are symmteric with these rs squares through d.Now A will win if he can mark even 1 of the other squares from these rs squares, which he will surely accomplish if
rs>b because so far we have makes total at most $\frac {b} {k+1} $ steps since step x, so B have marked at most b squares from the rs squares we are considering. Thus by choosing large a,b such that b-a is large and b is large in comparsion to a then rs>b and thus we are done.
(The detailed choosing of a,b is left as an easy excercise. If I calculated right, A will win if we choose N> $k^ { k^4 } $)

[quote="Myth"]It seems to be correct.[/quote]
It is a correct solution
[quote="lhvietbao"]Here is my solution[/quote]
I dont think so :D :D
There is a question: Let k be a given positive integer. Suppose player A and B plays the following game in all integral points of Oxyz system:
In each turn, player A marks an unmarked integral point red, and player B is allowed to mark k unmarked integral points blue. Show that if player A can mark 8 integral point red making a cube with sides parallel to the three axis? :D :D[/quote]


Does the D.E. $ a^2\plus{}b^2 \equal{} c^2\plus{}d^2$ in positive integers have any solutions besides a = c , b = d, and a=d, b=c?

[hide="Answer"]

There are infinitely many - first consider $ 7^2 \plus{} 24^2 \equal{} 15^2 \plus{} 20^2 \equal{} 25^2$. There is a formula that gives the number of ways $ n \equal{} a^2 \plus{} b^2$ for some $ n$ (see  [url]http://mathworld.wolfram.com/PythagoreanTriple.html[/url]). Infinitely many $ n$ are expressible in more than one way, so there are infinitely many solutions to this DE.[/hide]

You could just stay with multiples of $ 7^2 \plus{} 24^2 \equal{} 15^2 \plus{} 20^2$,
and not look for other ways of splitting up n in order to come up
with an infinite number of solutions in positive integers satisfying
the original poster:

$ (2 \cdot 7)^2 \plus{} (2 \cdot 24)^2 \equal{} (2 \cdot 15)^2 \plus{} (2 \cdot 20)^2 \equal{} \equal{} > 14^2 \plus{} 48^2 \equal{} 30^2 \plus{} 40^2$
$ (3 \cdot 7)^2 \plus{} (3 \cdot 24)^2 \equal{} (3 \cdot 15)^2 \plus{} (3 \cdot 20)^2 \equal{} \equal{} > 21^2 \plus{} 72^2 \equal{} 45^2 \plus{} 60^2$
$ (4 \cdot 7)^2 \plus{} (4 \cdot 24)^2 \equal{} (4 \cdot 15)^2 \plus{} (4 \cdot 20)^2 \equal{} \equal{} > 28^2 \plus{} 96^2 \equal{} 60^2 \plus{} 80^2$
 $ \cdot$
 $ \cdot$
 $ \cdot$
$ (n \cdot 7)^2 \plus{} (n \cdot 24)^2 \equal{} (n \cdot 15)^2 \plus{} (n \cdot 20)^2 \equal{} \equal{} > (7n)^2 \plus{} (24n)^2 \equal{} (15n)^2 \plus{} (20n)^2$

What is this formula someperson? i could not find it on the page.
Thanks!!

It's (26).  Note that you don't need to know this to solve the problem:  just write

$ (a \plus{} c)(a \minus{} c) \equal{} (b \plus{} d)(b \minus{} d) \equal{} n$

and look at all the ways you can factor $ n$ into two factors with the same parity.  (This is possible if and only if $ n \equiv 0, 1, 3 \bmod 4$.)  In other words, solving this Diophantine equation is no harder than solving the Diophantine equation

$ ab \equal{} cd$.

no (26) is how many ways $ n^2\equal{}a^2\plus{}b^2$

My mistake.  What you want is (18) [url=http://mathworld.wolfram.com/SumofSquaresFunction.html]here[/url]; the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]here[/url] might be of interest.


Find the interval of a for which one root of the equation (a-5)x^2-2ax +(a-4)=0 is smaller than 1 and the other root is greater than 2 .

Let $ f(x)\equal{}(a\minus{}5)x^2\minus{}2ax\plus{}a\minus{}4$, we have $ f(1)\equal{}\minus{}9<0$, thus the condition satisfying the context should be $ a>5$ and $ f(2)\equal{}a\minus{}24<0$, yielding $ 5<a<24.$


inequalities
logarithms
calculus
derivative
function
inequalities unsolved

let $x_i $s and $y_j$s be integers such that $1<x_1<x_2<...<x_k <y_1<y_2<...<y_m$

show that if $\sum_{i=1}^k x_i \geq \sum_{j=1}^m y_j$ then

$ \prod_{i=1}^k x_i \geq \prod_{j=1}^m y_j$

This is a great problem. Can you post your solution,please?

I have a solution.
First, we can suitably increase $y_i$ until $\sum x_i = \sum y_i$.
Then, consider a majorization of $(y_m,y_{m-1},...,y_1,a,a,...,a) > (x_k,...,x_1,b,b,...b)$
with ak=bm, k "a"s and m "b"s.
Now, by majorization inequality we have
$b^m \prod x_i \ge a^k \prod y_i$
It remaisn to see if we can find a and b so that $a^k \ge b^m$. If yes, we are done.
But this is not hard. Take a,b to be larger than 1. Then it is equivalent to find a,b such that $k \ln a \ge m \ln b$. Since $ak=mb$, it's same as $b \ ln a \ge a \ln b$ or $\frac{ \ln a}{a} \ge \frac{\ln b}{b}$.
Differentiation of $f(x)=\frac{\ln x}{x}$ shows that it is a decreasing function for $x \ge e$. Now, since $a \le b$, take appropriate $a$ and $b$ larger than e so that the majorization still holds. (I believe we can take a=e and b=e+$\delta$, a very small number.) Then we are done.

[quote="ehku"]Only Cauchy inequality, can you ?[/quote]
can you post your solution?


7 girls on a bus.
Each have 7 bags.
In each bag are 7 cats.
With each cat there are 7 kittens.
How many paws/feet are there in all?

Not a trick question, and show your reasoning.

7 girls is 14 feet.  Now there are 7 kittens for 7 cats is $ 56\times 4\equal{}224$ paws per bag.  There are 49 bags so $ 224\times49\equal{}10976$.  So there are $ 14\plus{}10976\equal{}10990$.  Now I'm assuming these girls aren't old enough to have a driver's license so there must be a bus driver and he has two feet so that's a total of $ 10992$ feet/paws.


What is the sum of all whole numbers between 0 and 1000 that are both perfect squares and perfect cubes?

$ \lfloor \sqrt {1000}\plus{}1 \rfloor \plus{} \lfloor \sqrt [3]{1000}\plus{}1 \rfloor \minus{} \lfloor \sqrt [6]{1000}\plus{}1 \rfloor \equal{} 32 \plus{} 11 \minus{} 4 \equal{} \boxed{39}$.

It isn't asking for the number that are either perfect squares or cubes, it is asking for the sum of all the numbers that are both!

This will be of the form x^6 for whole number x.

The only ones under 1000 that work are: 0,1,64,729, for a sum of 0+1+64+729=794.

I mistook it for "$ \#(\text{perfect squares} \ \cup \ \text{perfect cubes})$".
This problem was $ \sum_{0\le n \le 1000} (\text{perfect squares} \ \cap \ \text{perfect cubes})$?
I am sorry that I misunderstand it :(

What is the [b]sum[/b] of all whole numbers between 0 and 1000 that are [b]both[/b] perfect squares and perfect cubes?

[b]Sum[/b] is $ \sum$ and [b]both[/b] is $ \cap$


algorithm
number theory proposed
number theory

Prove that for any positive rational number $\frac{p}{q}$ with $q \geq 1$ there exists the unique set $(k,a_1,...,a_k)$ such that $\frac{p}{q}=\sum_{n=1}^k\frac{a_n}{n!}$ sucht that $0 \leq a_n <n$.

hi,
I have a doubt,for atleast small k's,the summation on the right with maximum values assigned to
$a_n$ i.e(n-1) seems to have value <1,so is there any condition which u missed like p<q or something.Also intrestingly for small numbers the value of  the R.H.S with again $a_n=n-1$ seems to be $n!-1/n!$,in that case given $p/q$,multiply with q-1! on num and denom,now number of different values that can be obtained by assigning different values to $a_n$ for k=q is q! and the maximum that can be obtained as i said before looks like $q!-1/q!$,so all the values from 0 to $q!-1/q!$ can be obtained proving ur question.
I am sorry maybe i am wrong,i coulld'nt check well as i am in a hurry,going to barcelona,shall come back to the prob after 4 days.
regards
Rohith

Why no replies? Possibly because you posted it in the wrong forum. This has nothing to do with calculus. I'll try moving it.

Here, we definitely have to assume that $p<q$ and that $k$ is actually $\infty$. In fact, there is a nice generalisation to this problem.

Let $n_1, n_2, n_3, \ldots$ be an infinite sequence of positive integers. Let $A_j = \prod_{i=1}^j n_i$ and let $A_j \to \infty$ as $j \to \infty$. We then have for all non-negative integers $p,q$ such that $p<q$, the rational number $\frac pq$ can be uniquely expressed as \[ \frac pq = \sum_{i=1}^\infty \frac{a_i}{A_i}, \] where each of the $a_i$'s satisfy the condition $0 \leq a_i < n_i$. Of course, the $a_i$ here is an integer.

Some examples: For $(n_1, n_2, n_3, \ldots) = (10, 10, 10, \ldots)$, we get our familiar decimal system. For $(n_1, n_2, n_3, \ldots) = (2, 2, 2, \ldots)$, we get the binary system. As for $(n_1, n_2, n_3, \ldots) = (1, 2, 3, \ldots)$, we get your problem.

This generalised problem is just a repeated application of the Division Algorithm (in the general sense), and the proof is rather straightforward.  :D


Let v1, ..., v_k ∈ Rⁿ and let v ∈ Rⁿ.  Prove that

Span(v1, ..., v_k) = Span(v1, ..., v_k, v) ↔ v ∈ Span(v1, ..., v_k).

Edit: I have corrected the error.  Thanks.

Do you mean to say that $ v_1, \ldots, v_k \in \mathbb{R}^n$?

Here is a proof of one direction:

Assume $ span(v_1, \ldots, v_k) \equal{} span(v_1, \ldots, v_k, v)$.  $ v$ certainly belongs to $ span(v_1, \ldots, v_k, v)$, because $ v \equal{} 1*v$, therefore $ v \in span(v_1, \ldots, v_k)$.


trigonometry
algebra
function
domain
real analysis
real analysis unsolved

Let $ \bold{f}: \mathbb{R}^2 \minus{} \{0\}\to\mathbb{R}^2$ be defined by

$ \bold{f}(s,t) \equal{} \displaystyle{(\frac {s^2 \minus{} t^2}{s^2 \plus{} t^2},\frac{st}{s^2 \plus{} t^2})}$

Find the rank of $ d\bold{f_p}$ for all $ \bold{p}\neq\bold{0}$ and describe the image of $ \bold{f}$

[quote="countdownmath"]$ \bold{f}(s,t) \equal{}{(\frac {s^2 \minus{} t^2}{s^2 \plus{} t^2},\frac{st}{s^2 \plus{} t^2})}$[/quote]
Error fixed.

Let $ \mathrm{T} : \mathbb{R}^+ \times \mathbb{R} \to \mathbb{R}^2 - \{ 0 \}$ by $ \mathrm{T}(r, \theta) = (r \cos\theta, r \sin\theta)$. Since $ \mathrm{T}$ is surjective and $ D\mathrm{T} \neq 0$ on its domain, it is locally invertable and we may consider the rank of $ D(\mathrm{f} \circ \mathrm{T})$ instead. But since

$ (\mathrm{f} \circ \mathrm{T}) (r, \theta) = (\cos 2\theta, \tfrac{1}{2} \sin 2\theta),$

the rank of $ D \mathrm{f}$, which is equal to the rank of $ D(\mathrm{f} \circ \mathrm{T})$, is clearly 1 and the image of $ \mathrm{f}$ is an ellipse.


function
advanced fields
advanced fields theorems

I've completed the proof from Axiom of Choice to Tukey's Lemma (Through Hausdorff's Maximal Principle).

How to prove the other side?

[quote="-oo-"]$ \{M_{i}: i\in I\}\times\cup_{i}M_{i}$[/quote]

Wouldn't it be an empty set, without axiom of choice?

Another definition of axiom of choice is that the cartesian product of an non-empty family of non-empty sets is non-empty.

may someone post what tukey's lemma actually is, please?
wikipedia didn't tell me, this time :(
thank you..

Let $ U$ be a sets of sets.

If $ U$ is of finite character (i.e. a set $ A\in U$ iff every finite subset of $ A$ belongs to $ U$.)
then $ U$ contains a maximal elements.

[quote="-oo-"]pick a family of nonempty subsets $ (M_{i})_{i\in I}$, wlog pairwise disjoint. now consider the choice functions on subsets of $ I$ as a subset of the power set of $ \{M_{i}: i\in I\}\times\cup_{i}M_{i}$. they obviously are of finite character, and a maximal element yields a global choice function.[/quote]

There exist a maximal element. But what if there is more than one maximal elements, how do we "decide" which will yield the choice function...

Randomly selecting one will still need a choice function.

Also, it gives a maximal set (in sense of iclusion) in the set, how to yield the element in I?

Can you give a detailed prove of finite character of that set. The whole power set is of finite character, but it may not holds for a proper subet of this power set.

the details are trivial. and it's absurd that it requires choice to get a maximal element with the help of tarski. just think about it, partially defined choice functions of the family are obviously of finite character and maximal ones have to be defined on the whole index set (cause of if not, you just extend it by one index).

[quote="-oo-"]just think about it, partially defined choice functions of the family are obviously of finite character and maximal ones have to be defined on the whole index set (cause of if not, you just extend it by one index).[/quote]

There is obviosuly several maximal sets in the power set of $ \{M_{i}: i\in I\}\times\cup_{i}M_{i}$. But choice function should yield only ONE element in each subset...

Also, the whole power set is definitely of finite character, does this still hold for only a subset?? (Also, given a non-empty subset of $ I$, which will be the corresponding subset of the power set of $ \{M_{i}: i\in I\}\times\cup_{i}M_{i}$)


Let P be the product of all positive integers x satisfying:

1 <= x <= n and (x, n) = (x + 1, n) = 1,

where n is a given positive integer. Prove that n divides P - 1.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=21707

Pierre.


summer program
Mathcamp
email
calculus
blogs

[quote="theone853"]Is the AoPS saga actually over? Or will a new chapter be written soon after all the new active members.[/quote]
Mathfanatic (Keone) (aka Electricsnowman) has not been frequenting the site since his move to Hawaii. I've spoken to him, and evidently he does not yet have reliable internet access.

Maybe it should be called "AoPS-MathLinks Saga" from now on...

[quote="h_s_potter2002"]What makes you think the saga is actually over?  I would think that with all of the Math-Links people on here too that there would be a whole lot more posts, right?[/quote]
You weren't here for the saga.

The AoPS saga is the story with AoPS members in it. The writer is Mathfanatic, who hasn't been on much recently.

Here's the thread about the saga:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7577&highlight=saga

The story has a lot of old members who aren't as active anymore.

the saga was great. yeah, most of the members in it are no longer active. rmr the star wars bit? that was hilarious...i was in that part...*sighs*

Okay, I get it now.

I thought that theone853 was just talking about him thinking that the board was dying down.

[hide]The board is dying down...[/hide]

I think the truth is that the AoPS saga can't exist without mathfanatic and mathfanatic lost interest/gone, not to mention (as theone says) most people in the story are inactive in AoPS (though they are online...checking their livejournal) Besides it's probably for the good it died...(I believe it died a long time ago btw...as other things...*sigh*)

[quote="starrynightskies"]wow... tare's back? or have i been gone so long that i didn't realize?
saga... interesting.  how long ago was that? cause the date was like june 2003, when theone wasn't supposed to know about this... o well[/quote]

There are two explanations for that.

1) theone might have lurked on here for a couple of months

or

2) he, along with many other people (i.e. me), looks at the old topics on here.

The saga was in the midst of it's last 3 chapters when I first joined. Starrynightskies, jeremy and artem are in the saga as, talking about a certain problem jeremy put up about turtles and that turned into quite an odd thread involving the talk of turtles reproducing during a race...

Wow, that was a massive necro.

Anyone have a link? I'm sort of curious...

Pity I wasn't around then.
The thing wouldn't die down so easily.

A link, anyone?

Jeeze ... you started reviving 3 year old threads again ...  :o

oh man

good old days

i remember back when i was still a moderator

:X

It's on mathfanatic's livejournal, the link's somewhere on here.  Or maybe it's gone?

And I dunno, I'm not active enough to tell, but it feels like AoPS it too big, and a lot of the active users don't have enough of a stereotypical personality (I dunno if that makes sense) to do another Saga?  I'm not sure.

Okay I'll write a new one soon. Best foot forward, eh?

ooh, can i be in it?????????? :D  :D  :D

yeah, good times, I had an older account but forgot the username, the password and my email got changed  :lol:

by the way, where's that guy that seemed to advertise calculus books and his blog? hahahaah

Wonder what it will look like 10 years later...

[quote=Sly Si]I mean, someone could in theory step up and write a new one involving people who are active now...[/quote]

Someone wrote a story about active aops users in this blog: [url]https://artofproblemsolving.com/community/c3302347[/url]


What is the mean of the elements in the sequence
\[ \minus{}1, 3, \minus{}5, 7, \minus{}9, 11, \ldots, \minus{}201?\]

-1

Just went through the first few numbers and saw that the mean oscillates between 1 and -1, depending on whether the last number in the sequence is positive or negative.


[hide="u subs"]
A. let $u=log_{3}x$
$u^{2}-7u+12=0$
$(u-3)(u-4)=0$
$u=3,4$
$log_{3}x=3,4$
you should be able to get it from here
[/hide]
[hide="chain rule of logs"]
B.
$(log_{a}b)(log_{b}c)=log_{a}c$
you should be able to get it from here[/hide]

[quote="cincodemayo5590"][hide="hints"]
a)  Quadratic Formula. ::gasp::   :P

b)  Change-of-Base [/hide]
This might be better in the HSB forum.[/quote]

sry..new to this


integration
trigonometry
calculus
function
absolute value
calculus computations

find t so as
$\int_{0}^\frac{\pi}{2}| \sin x+t \cos x | dx =1$

And t = 0 is the only one for which the equality holds. Since $x \in [0, \frac{\pi}{2}]$, there is no need for absolute value and one can just integrate the integral to find that $t+1 = 1$.

if we take the case t<0, there are t's such as the absolute value is necessary ... how should i know that the case t=-1000 is not a good one ?

I have the irrefutable evidence of the integral being negative for $t<0$:
[img]http://www.math.tamu.edu/~lkovalev/files/maple1.jpg[/img]

nice solution, mlok  :lol: i should print it and glue it to my homework

What system is that, and what is it thinking?

There is at least one negative root: the value for $t=-1$ is $\int_{0}^{\pi/4}\cos x-\sin x\,dx+\int_{\pi/4}^{\pi/2}\sin x-\cos x\,dx$
$=[\sin x+\cos x]_{0}^{\pi/4}+[-\sin x-\cos x]_{\pi/4}^{\pi/2}=(\sqrt{2}-1)+(\sqrt{2}-1)=2\sqrt{2}-2<1$

As $t\to-\infty$, the value goes to $\infty$, so there is a root in $(-\infty,-1)$ by the intermediate value theorem (this is a continuous function of $t$).

That was Maple 10. I don't have Mathematica to compare the results. To be fair to Maple, I should say that plot commands use quicker (and less reliable) procedures than direct evaluation. Maple did help me place the root at $t=-\frac43$. It did not supply me with a proof, though.

so ... does anyone have a complete solution ?

With $t<0$, let $\theta=\arctan(-t)$. The integral splits into two:
$\int_{0}^{\theta}(-\sin x-t\cos x)dx =\cos\theta-t\sin\theta-1$ and $\int_{\theta}^{\pi/2}(\sin x+t\cos x)dx =\cos\theta+t-t\sin\theta$. The sum is
$2\cos\theta-2t\sin\theta-1+t=2(1+t^{2})\cos\theta-1+t=2\sqrt{1+t^{2}}-1+t$. Equate to $1$ and solve.


Let $ E \in CA, F \in AB$ be arbitrary points on sidelines of a $ \triangle ABC.$ Parallels to $ EF$ through $ B, C$ cut $ CA, AB$ at $ E_1, F_1,$ respectively. Inversions with centers $ C, B$ and powers $ CE^2, BF^2$ take $ E_1, F_1$ to $ E_2, F_2,$ respectively. Show that line $ EF$ cuts segment $ E_2F_2$ in half.

$ CE^2\equal{}CE_1 \cdot CE_2 \Rightarrow CE^2\minus{}CE_1 \cdot CE \equal{}CE_1 \cdot CE_2 \minus{} CE_1 \cdot CE$.

Thus $ CE \cdot EE_1 \equal{} CE_1 \cdot EE_2 \Rightarrow \frac{EE_2}{EE_1} \equal{} \frac{CE}{CE_1}$. By the parallel lines we can rewrite:

$ \frac{EE_2}{EE_1} \equal{} \frac{CE}{CE_1} \equal{} \frac{FF_1}{BF_1} \equal{} 1 \plus{} \frac{BF}{BF_1} \equal{} 1 \plus{} \frac{BF_2}{BF} \equal{} \frac{FF_2}{BF}$. Exchange the first and last innerly:

$ \frac{BF}{EE_1} \equal{} \frac{FF_2}{EE_2}$ . Also we know $ \frac{EE_1}{BF} \equal{} \frac{AE}{AF}$. So $ \frac{AE}{AF} \cdot \frac{FF_2}{EE_2} \equal{} 1$.

Let $ M$ be the intersection of $ E_2F_2$ and $ EF$. Menelaus on $ \triangle AFE$ for $ F_2, M, E_2$. We get $ \frac{AE}{AF} \cdot \frac{FF_2}{EE_2} \cdot \frac{E_2M}{F_2M} \equal{} 1$.

So $ E_2M \equal{} F_2M$.

Thanks for the nice reply. See at [url]http://www.mathlinks.ro/viewtopic.php?p=1539397#1539397[/url], what I did (and why).


geometry
analytic geometry
graphing lines
slope

Points $A = (3,9), B = (1,1), C = (5,3),$ and $D = (a,b)$ lie in the first quadrant and are the vertices of quadrilateral $ABCD$. The quadrilateral formed by joining the midpoints of $\overline{AB}, \overline{BC}, \overline{CD},$ and $\overline{DA}$ is a square. What is the sum of the coordinates of point $D$?

$\text{(A)}\ 7 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 16$

I am not sure that the question was given correctly:
[hide] Since ABCD is a square, we have that AB is perpindicular to BC.  This means that the slope of AB is the negative recipocal of the slope of BC.  We have that the slope of AB is 4 and the slope of BC is 1/2.  The slope of BC [b]should[/b] be -1/4 if ABCD were a square.  Thus, BC is not perpindicular to AB, and thus ABCD is not a square.  [/hide]


I am new to this site so, I don't really know if this problem is of the same level with this forum. so I also posted the problem on the other forum.

Boris and Michael are jogging around a lake. It takes Boris 6 minutes to complete a lap, and Michael takes 8 minutes. After completing each lap, the boys take a rest, Boris for one minute and Michael for two minutes. Then they continue jogging. If they started joggin together, how long, in minutes, does it take until they are next together?

This problem has been solved [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18901]here[/url].  In the future, please try to avoid posting a problem in several places.


Putnam algebra polynomial calculus integration Vieta

A quartic polynomial P(x) = ax :^4: + bx :^3: + cx :^2: + dx + e, has integral coefficients, a not equal to 0, and can be written in the form P(x) = a(x - r1)(x - r2)(x - r3)(x - r4).  If r1 + r2 is a rational number, and r1 + r2 does not equal r3 + r4, prove r1r2 is rational.

Is this done through a clever application of Vieta's formula?

Try it! I can't believe this problem. I thought it would be hard(er), so I didn't work on it.

Oh, really?  I saw this, and the first thing that popped into my head was "Why is this so much easier than the first 3?"

Well, I imagine you beat me by a lot then, didn't you? Well, you don't know, since you don't know how I did, so I should tell you. I got only A1, B2, and B3. I'm a failure I guess.

Whoa, never knew the Putnam happened so early in the school year. So when do the scores come out?

End of March, I think.  Maybe end of April.  I don't remember exactly.  Not for a long time.

Eh, that sucks. Well, I guess it makes sense though, since so many people are taking it and there are 12 proofs per student that the graders need to go through.

Right.  I also don't have any idea who the MAA gets to grade these things.  (MAA = Mathematical Association of America.  They administer the Putnam.)  There were about 3,400 students or so who took it last year -- that makes for a lot of proofs.

this isn't likely to be a solution:
since r1 + r2 is rational, (r1+r2)^2 is also rational, i.e. r1^2 + r2^2 + 2r1r2 is rational. so r1r2 is rational.
can someone explain to me why this isn't a valid argument?


Euler
function
algebra
polynomial
number theory
totient function
number theory proposed

Let $\varphi$ be the Euler totient function and $f$ a function from $\mathbb{Z}^+$ to $\mathbb{R}$.

Prove that
\[ \sum_{d|n} f ( d ) = n \Leftrightarrow f = \varphi  \]
Note : One side is classic, the other one is a problem posed by Venhataraman in the Canadian Mathematical Bulletin (CMB)
Edit: Sorry, there is some confusion in the source box, the information given in the note is exact.

Which one is classic¿
I think both are very old and well known...

When having proved that $\sum_{d|n} \varphi ( d )= n$, the converse is easy: just notice that they are equal for $n=1$ and then work by induction.
And the equation for $\varphi$ can be proved on various ways, e.g. by considering the additive orders $\mod n$, by looking at the degrees of cyclotomic polynomials or simply but boring by proving it for powers of primes and using multiplicativity.

The $\Rightarrow$ side was unknown to me.
I know it is easy to prove, homever I hoped, there will be someone who will enjoy it :(...
(the $\Leftarrow$ can be also proved by the Moebius inversion formula, but that's not a discovery for you I guess ;))


function
number theory proposed
number theory

Let $ g: \mathbb{N}\to \mathbb{N}$ be a function defined as follows:
 :arrow: (i)$ g$ is non-decreasing;
 :arrow: (ii)g(n) is number of times n appears in the range of $ g$.
Prove That $ g(1) \equal{} 1 \\
and g(n \plus{} 1) \equal{} 1 \plus{} g(n \plus{} 1 \minus{} g(g(n)))$.
Find A formula for $ g(n)$ in general.

[hide]the sequence of initial values are like 1,2,2,3,3,4,4,4,5,5,5,6,6,6,6,...
The prove that part is rather easier.How about the formula?[/hide]

[quote="mszew"]check this: [url]http://www.research.att.com/~njas/sequences/A001462[/url][/quote]
What is the use of that link?  :huh:  Practically speaking i am rather annoyed that one just points out a reference in OEIS and with no proof.   :huh:
I dont want just the formula, or rather i  myself have posted what to prove. So please provide a solution,not an answer.:wink:


let a,b,c be three positive real numbers .
 Prove that :
 $\frac{1}{2} +\frac{a^2+b^2+c^2}{ab+bc+ca} \geq \frac{a}{b+c} +\frac{b}{c+a} +\frac{c}{a+b} \geq 2 -\frac{ab+bc+ca}{2(a^2+b^2+c^2}$

[quote="tranthanhnam"]let a,b,c be three positive real numbers .
 Prove that :
 $\frac{1}{2} +\frac{a^2+b^2+c^2}{ab+bc+ca} \geq \frac{a}{b+c} +\frac{b}{c+a} +\frac{c}{a+b}$[/quote]

This one is well-known. Here is another proof using the AM-GM:
$\sum_{cycl} ab\sum_{cycl}\frac{a}{b+c}=\sum_{cycl} a^2+\sum_{cycl}\frac{abc}{b+c}\le \sum_{cycl} a^2+\sum_{cycl}\frac{a(b+c)^2}{4(b+c)}$
$=\sum_{cycl} a^2+\frac{1}{2}\sum_{cycl} ab$

It seems two solution are hard to chew. :D

\[ \frac12+\frac{a^2+b^2+c^2}{ab+bc+ca} \ge \sum_{cyc}\frac{a}{b+c} \\ \iff \sum_{cyc} (a-b)^2 \cdot \left( \frac{c^2}{(ab+bc+ca)(c+a)(c+b)} \right) \ge 0  \]
\[ \sum_{cyc} \frac{a}{b+c} \ge 2-\frac{ab+bc+ca}{2(a^2+b^2+c^2)} \\ \iff \sum_{cyc} (a-b)^2 \cdot \frac{ab+bc+ca-a^2}{(a+b)(b+c)(ab+bc+ca)}_{S_c} \ge 0  \]
Wlog assume $c \ge b \ge a$, then $S_c \ge 0$ and $S_a+S_b \ge 0$.
one.

[quote="tranthanhnam"]$\frac{a}{b+c} +\frac{b}{c+a} +\frac{c}{a+b} \geq 2 -\frac{ab+bc+ca}{2(a^2+b^2+c^2)}$[/quote]

By Cauchy,
$\frac{a}{b+c} +\frac{b}{c+a} +\frac{c}{a+b}\geq\frac{(a+b+c)^2}{2(ab+bc+ca)}$
It remains to prove that
$B^2+A^2C\ge 4BC$ (*)
where $A=a+b+c$, $B=ab+bc+ca$ and $C=a^2+b^2+c^2$. But
$A^2C=C^2+2BC$. So (*) becomes
$B^2+C^2 \ge 2BC$, which is obvious.

[quote="Vasc"]The following stronger inequality holds:

 $\frac{a}{b+c} +\frac{b}{c+a} +\frac{c}{a+b} \geq \frac{13}{6} -\frac{2(ab+bc+ca)}{3(a^2+b^2+c^2)}$.[/quote]
If Cauchy does not work for this, we use substitution $p=x+y+z$, $q=xy+yz+zx$, and $r=xyz$.  ;)

The best inequality of this type is:
\[ \sum_{%Error. "cycl" is a bad command.
}\frac{a}{b+c}-\frac{3}{2}\geq (\sqrt{3}-1)\left( 1-\frac{ab+bc+ca}{a^2+b^2+c^2}\right) \]
All I can say is that it's elegant.

tranthanhnam said: New ineq...
But it was posted by manlio, http://www.mathlinks.ro/Forum/viewtopic.php?t=22798&view=next# .
What does M&Y -T8/250 Year-1998 mean?

[quote="Vasc"]The best inequality of this type is:
$\sum_{cycl}\frac{a}{b+c}-\frac{3}{2}\geq (\sqrt{3}-1)\left( 1-\frac{ab+bc+ca}{a^2+b^2+c^2}\right)$[/quote]
Hmm...
Inequality:
\[ \iff \sum_{cyc} \left( \frac{1}{(a+c)(b+c)}-\frac{1-\sqrt{3}}{a^2+b^2+c^2} \right) \ge 0.  \]
Easily, $S_a+S_b+S_c \ge 0$, now who can prove $S_aS_b+S_bS_c+S_cS_a \ge 0$, it’s quite messy...


when $x=y=z$ then $p\le \frac{2}{3}$ and for this p the inequality is obviously true.

[quote="bodan"]Find the largest $p\in\mathbb{R}$ such that the inequality \[ x^4+y^4+z^4+xyz(x+y+z)\geq p(xy+yz+zx)^2  \] holds for all $x,y,z\in\mathbb{R}.$[/quote]

Is it for real number or for positive number?
According to Albania Eagle, we have:
$3(x^4+y^4+z^4)+3xyz(x+y+z) \geq 2 (xy+yz+xz)^2$
$\rightarrow$ $\sum x^2(x-y)(x-z) +(x+y+z) \sum(x+y)(x-y)^2 \geq 0$

dear gibbenergy ,
$3(x^4+y^4+z^4)+3xyz(x+y+z)\geq 2(xy+yz+zx)^2\longleftrightarrow 3\sum (x^2-y^2)^2+\sum (xy-yz)^2\geq 0$ ;)


search
function
vector
modular arithmetic
least common multiple
number theory proposed
number theory

Prove that the sequence formed by writing the last two digits of the fibonacci numbers in base 10 in order is periodic, i.e the sequence
\[1,1,2,3,5,8,13,21,34,55,89,44,33,\cdots \]
is periodic. find the exact length of the period.

nobody interested? :( i had to do a lot of hard work to make this problem.

1. already posted a lot of times -> search function
2. enourmous impatience

[quote="ZetaX"]enourmous impatience[/quote]

you're right. but i have reasons. its because i'll have to leave within a short while and won't be able to tune in for quite a while. so..

are you sure the length of the period was already found, ZetaX?

but i have the exact length and i want someone to find it not writing the fibonacci sequence until it becomes periodic. there is a better way..

Well, do you have something that works more general, thus e.g. gives a general formula¿
Or do you just reduce the number of calculations¿
The latter is easy, but boring, only the first would be interesting.

i have something but i'm sorry i can't post it right now but i assure you i'll post it as soon as i get a chance. :D

[quote="ZetaX"]The length was probably posted before, but even if not: it's at most $10000$, so computer kill it  :D[/quote]
I don't think so it can reach 10000 :P ,and man can kill it too.it is exactly [b]100[/b],I have got a way without calculation to prove this result,but, I don't know if it is more "general"?because my method can't be generalise to other sequences :blush:

Well ,sorry, I made a stupid mistake that I have thought that the peroid mod 4 is 4 :blush:  Thanks.

[quote="Kent Merryfield"]The answer is not 100. This particular question - the periodicity of the Fibonacci numbers mod $n$ - is a well studied question and there are places it could be looked up (I won't bother with the looking, myself).

One tip about questions like this: when dealing with a composite number, don't look at the whole number at once. Decompose it into prime powers. For periodicity mod 100, split it into the question of periodicity mod 4 and periodicity mod 25.

By the way, the maximum possible period mod $n$ is $n^{2}-1.$ Why not $n^{2}?$ Think of this as a linear recursion that takes 2-vectors to 2-vectors. There are $n^{2}$ possible 2-vectors, but one of them is the zero vector. Since the zero vector can only go to the zero vector, it cannot be part of any nonzero cycle. This maximum is achieved for periodicity mod 2 (period 3) and periodicity mod 3 (period 8).

The period mod 4 is can be seen by hand to be 6. And that, right there, is why 100 must be the wrong answer.

Now look at it mod 25. Run the sequence, starting with $F_{0}=0$ and $F_{1}=1.$ We get that the first zero is $F_{25}=0,$ followed by $F_{26}=18.$

We now know that $F_{n+25}\equiv18F_{n}\pmod{25}.$ It follows from that that $F_{n+25k}\equiv18^{k}F_{n}\pmod{25}.$

So, what is the multiplicative order of 18? By Euler's Theorem, it must be a divisor of $\phi(25)=20.$ So let's look at a few possibilities:

$18^{2}=(-7)^{2}=49=-1.$ And from that we get that $18^{4}=1.$

So now we know that the period of the Fibonacci numbers mod 25 is $4\cdot 25=100.$ The periodicity mod 4 is 6. Putting the two together is essentially a Chinese remainder problem; the overall period is the least common multiple of 6 and 100, namely 300.[/quote]

this is exactly my solution but the only difference is you considered $F_{26}$ whilst i considered $F_{24}$.

here's a generalization: the length of the period of the fibonacci sequence $\mod n$ is $k\cdot ord_{n}(F_{k-1})$.($ord_{n}(k)$means order of k mod n, i don't know if my symbol is correct), where $k$ is the smallest integer such that $n|F_{k}$ can you prove it? :wink:

ZetaX, have you found your answer?


a circle with center O has a chord AB equal in length to the radius of 1.  A perpendicular from O to AB meets AB at M.   A perpendicular from M to OA at D.  What is the area of triangle AMD(in units square)?

[hide] i got \[ \frac{\sqrt{3}}{4}\], which is probably wrong, but just checking...[/hide]

[hide]Setting up a ratio, we have:
\begin{align*}
&\frac{AM}{AD}=\frac{AO}{AM} \\
&\implies \frac{\frac{1}{2}}{1-DO}=\frac{1}{\frac{1}{2}} \\
&\implies \frac{1}{4}=1-DO \\
&\implies DO=\frac{3}{4}
\end{align*}Since $ AD=1-DO$, we have $ AD=\frac{1}{4}$, and by the Pythagorean Theorem, we have $ DM=\frac{\sqrt{3}}{4}$. The area of $ \triangle AMD$ is $ \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{\sqrt{3}}{4}=\boxed{\frac{\sqrt{3}}{32}}$. Am I wrong?[/hide]

probably not, since your answer was a choice, and u have a warrant as to how u got your answer, i think im wrong...

[hide="I agree with 123456789..."]
...but there's an easier way:

You know that $ \triangle OBA$ is equilateral, so $ \angle BAO \equal{} 60^o$, so $ \triangle AMD$ is a $ 30^o \minus{} 60^o \minus{} 90^o$ triangle. Since $ AM \equal{} \frac {1}{2}$, $ AD \equal{} \frac {1}{4}$ and $ DM \equal{} \frac {\sqrt {3}}{4}$. Using the Triangle Area formula, we get the area to be $ \boxed{\frac {\sqrt {3}}{32}}$.
[/hide]

EDIT: Forgot to close the hide tag/add diagram.

i guess i just forgot to get the area... :mad:

two questions,
1. how do u find the height of a scalene triangle (has nothing to do with this question)

2. how did u draw that diagram (is it using geometry sketchpad?)

If you have three side lengths, you can use Heron's, and apply it to the $ \frac{1}{2} \cdot b \cdot h$ definition to find the altitude to one side.


Let $ a,b,c$ are positive real numbers.Prove that
$ \frac{a(b\plus{}c)}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{b(c\plus{}a)}{a^2\plus{}ac\plus{}c^2}\plus{}\frac{c(a\plus{}b)}{a^2\plus{}ab\plus{}b^2}\geq 2$

$ \sum \frac{a(b\plus{}c)}{b^2\plus{}bc\plus{}c^2}\equal{}2\plus{}\sum\frac{bc(b\minus{}c)^2}{(a^2\plus{}ac\plus{}c^2)(b^2\plus{}ba\plus{}a^2)}\geq 0$
 :D

\[ \frac{a(b\plus{}c)}{b^2\plus{}bc\plus{}c^2}\minus{}\frac{2a}{a\plus{}b\plus{}c}\equal{}\frac{a((a\plus{}b\plus{}c)(b\plus{}c)\minus{}2(b^2\plus{}bc\plus{}c^2))}{(a\plus{}b\plus{}c)(b^2\plus{}bc\plus{}c^2)}\]
\[ \equal{}\frac{a(ab\plus{}ac\minus{}b^2\minus{}c^2)}{(a\plus{}b\plus{}c)(b^2\plus{}bc\plus{}c^2)}\equal{}\frac{a(b(a\minus{}b)\minus{}c(c\minus{}a))}{(a\plus{}b\plus{}c)(b^2\plus{}bc\plus{}c^2)}\]
Summing cyclically and multiplying by $ a\plus{}b\plus{}c$, we need to show the nonnegativity of
\[ \sum_\text{cyc} \frac{ab(a\minus{}b)}{b^2\plus{}bc\plus{}c^2}\minus{}\frac{ac(c\minus{}a)}{b^2\plus{}bc\plus{}c^2}\]
\[ \equal{}\sum_\text{cyc} ab(a\minus{}b)\left( \frac{1}{b^2\plus{}bc\plus{}c^2}\minus{}\frac{1}{a^2\plus{}ac\plus{}c^2}\right)\]
\[ \equal{}\sum_\text{cyc} \frac{ab(a\plus{}b\plus{}c)(a\minus{}b)^2}{(a^2\plus{}ac\plus{}c^2)(b^2\plus{}bc\plus{}c^2)}\]
So we're done. (In the end, this produced the same identity as given above, but usually these kinds of factors won't appear in every square.)

very good! I will post a interesting solution for it, but I want to see has anyone a nice solution?   :D

See also http://www.mathlinks.ro/viewtopic.php?t=25553 .

  darij


inequalities
trigonometry
inequalities unsolved

Let $ABC$ a triangle and $O$ the circumcentre ($R$ is the rayon of the circumcircle). $I$ and $J$ are the midpoints of $BC$ and $AC$ respectively.

Prove that $R+|OI|+|OJ| =< (a+b+c)/2$

this is true iff 3p >= 5R + 2r, which I have never seen before... :?

[quote="cezar lupu"]Labiliau, are you sure this is the right inequality? :?[/quote]
I think it's right, because I think my friend (yptsoi) has proved it before. But I forgot his proof..

It is not true. See the case $a\rightarrow 2, b=1, c=1$.

Sorry, I forget to write that triangle $ABC$ is acutangle. Thus, the counterexample is false.

This is equivalent to saying $1 + \cos A + \cos B \le \sin A + \sin B + \sin B \cos A + \sin A \cos B$

Holding A constant, LHS is clearly maximized when B is minimal, so WLOG A+B = 90.  Then we must show

$1 + \cos A + \cos B \le \sin A + \sin B + 1$

But since now $\cos A = \sin B$ and $\sin A = \cos B$ we're done...


integration
trigonometry
real analysis
real analysis unsolved

find sum of sequence $ \frac{x}{1!} \plus{} \frac{x^4}{4!} \plus{} ... \plus{} \frac{x^{3n\plus{}1}}{(3n\plus{}1)!}$

[quote="Hong Quy"]find sum of sequence $ \frac {x}{1!} \plus{} \frac {x^4}{4!} \plus{} ... \plus{} \frac {x^{3n \plus{} 1}}{(3n \plus{} 1)!}$[/quote]
we take $ S(x)\equal{}\frac {x}{1!} \plus{} \frac {x^4}{4!} \plus{} ... \plus{} \frac {x^{3n \plus{} 1}}{(3n \plus{} 1)!}$
we have $ S'(x)\equal{}1\plus{} \frac {x^3}{3!} \plus{} ... \plus{} \frac {x^{3n}}{(3n)!}\equal{}\frac{e^{jx}\plus{}e^x\plus{}e^{j^2x}}{3}$
then $ S(x)\equal{}\frac{j^2e^{jx}\plus{}e^x\plus{}je^{j^2x}}{3}$

[quote="aviateurpilot"][quote="Hong Quy"]find sum of sequence $ \frac {x}{1!} \plus{} \frac {x^4}{4!} \plus{} ... \plus{} \frac {x^{3n \plus{} 1}}{(3n \plus{} 1)!}$[/quote]
we take $ S(x) \equal{} \frac {x}{1!} \plus{} \frac {x^4}{4!} \plus{} ... \plus{} \frac {x^{3n \plus{} 1}}{(3n \plus{} 1)!}$
we have $ S'(x) \equal{} 1 \plus{} \frac {x^3}{3!} \plus{} ... \plus{} \frac {x^{3n}}{(3n)!} \equal{} \frac {e^{jx} \plus{} e^x \plus{} e^{j^2x}}{3}$
then $ S(x) \equal{} \frac {j^2e^{jx} \plus{} e^x \plus{} je^{j^2x}}{3}$[/quote]
 where j ?
i don't understamd  and why we have, maybe you write very simply $ S'(x) \equal{} 1 \plus{} \frac {x^3}{3!} \plus{} ... \plus{} \frac {x^{3n}}{(3n)!} \equal{} \frac {e^{jx} \plus{} e^x \plus{} e^{j^2x}}{3}$

\[ S(x) \equal{} \sum_{n \equal{} 0}^{\infty}{\frac {x^{3n \plus{} 1}}{(3n \plus{} 1)!}}
\]

\[ S'(x) \equal{} \sum_{n \equal{} 0}^{\infty}{\frac {x^{3n}}{(3n)!}}
\]

\[ \int_{0}^{x}{S(t)dt} \equal{} \sum_{n \equal{} 0}^{\infty}{\frac {x^{3n \plus{} 2}}{(3n \plus{} 2)!}}
\]
We have:
\[ S(x) \plus{} S'(x) \plus{} \int_{0}^{x}{S(t)dt} \equal{} e^x
\]
or
\[ S''(x) \plus{} S'(x) \plus{} S(x) \equal{} e^x
\]
In general form the solution of the equation is:
\[ S(x) \equal{} e^{\frac {x}{2}}\left(C_1\cos{\frac {\sqrt {3}}{2}x} \plus{} C_2\sin{\frac {\sqrt {3}}{2}x}\right) \plus{} \frac {e^x}{3}
\]
To get constants $ C_1$ and $ C_2$ we remember that $ S(0) \equal{} 0$ and $ S'(0) \equal{} 1$.
\[ C_1 \equal{} \minus{} \frac {1}{3}
\]

\[ C_2 \equal{} \frac {\sqrt {3}}{3}
\]
Final answer:
\[ S(x) \equal{} \frac{e^{\frac {x}{2}}}{3}\left(\sqrt {3}\sin{\frac {\sqrt {3}}{2}x}\minus{}\cos{\frac {\sqrt {3}}{2}x}\right) \plus{} \frac {e^x}{3}
\]

[quote="Yuriy Solovyov"]
\[ S(x) \equal{} \sum_{n \equal{} 0}^{\infty}{\frac {x^{3n \plus{} 1}}{(3n \plus{} 1)!}}
\]

\[ S'(x) \equal{} \sum_{n \equal{} 0}^{\infty}{\frac {x^{3n}}{(3n)!}}
\]

\[ \int_{0}^{x}{S(t)dt} \equal{} \sum_{n \equal{} 0}^{\infty}{\frac {x^{3n \plus{} 2}}{(3n \plus{} 2)!}}
\]
We have:
\[ S(x) \plus{} S'(x) \plus{} \int_{0}^{x}{S(t)dt} \equal{} e^x
\]
or
\[ S''(x) \plus{} S'(x) \plus{} S(x) \equal{} e^x
\]
In general form the solution of the equation is:
\[ S(x) \equal{} e^{\frac {x}{2}}\left(C_1\cos{\frac {\sqrt {3}}{2}x} \plus{} C_2\sin{\frac {\sqrt {3}}{2}x}\right) \plus{} \frac {e^x}{3}
\]
To get constants $ C_1$ and $ C_2$ we remember that $ S(0) \equal{} 0$ and $ S'(0) \equal{} 1$.
\[ C_1 \equal{} \minus{} \frac {1}{3}
\]

\[ C_2 \equal{} \frac {5\sqrt {3}}{9}
\]
Final answer:
\[ S(x) \equal{} e^{\frac {x}{2}}\left( \minus{} \frac {1}{3}\cos{\frac {\sqrt {3}}{2}x} \plus{} \frac {5\sqrt {3}}{9}\sin{\frac {\sqrt {3}}{2}x}\right) \plus{} \frac {e^x}{3}
\]
[/quote] why $ S'(0) \equal{} S'(0) \equal{} \sum_{n \equal{} 0}^{\infty}{\frac {x^{3n}}{(3n)!}}\equal{} 0$or$ 1$ ?????
просто так ?  объисни мне спасибо .
 i  o nly write well english or vietnamese. hence, you can say why $ S'(0)\equal{}1$

[quote="Yuriy Solovyov"]$ S(x) \equal{} x \plus{} \frac {x^4}{4!} \plus{} ...$
So, $ S(0) \equal{} 0$

$ S'(x) \equal{} 1 \plus{} \frac {x^3}{3!} \plus{} ...$
So, $ S'(0) \equal{} 1$[/quote]
 thanks Yuriy Solovyov , i read uncarefully.
Огромное спасибо

With the original problem as stated, all the answers here are wrong- because it's a finite sum. Please state these things more clearly (include a final $ \dots$, in this case).

Terminology note: the word "series" is specifically reserved for sums. "sequence" is understandable, but "series" is better here.

Yes, you are right, but I think he had in mind "Find sum of the series".
I don think he mean:
$ a_n\equal{}\sum_{k\equal{}0}^{n}{\frac{x^{3k\plus{}1}}{(3k\plus{}1)!}}$
And we should find $ \sum_{n\equal{}0}^{\infty}{a_n}$. Easy to see that this sum diverges for all $ x$ except $ 0$.


function
complex numbers
complex analysis
complex analysis unsolved

Let f be a continuous function from the complex numbers to the complex numbers satisfying

(i) f([i]z[/i] + [i]w[/i]) = f([i]z[/i]) + f([i]w[/i]) for all complex numbers [i]z, w[/i] and
(ii) f([i]zw[/i]) = f([i]z[/i])*f([i]w[/i]) for all complex numbers [i]z, w[/i].

Suppose f is not identically zero. Prove that f([i]z[/i]) = [i]z[/i] for all complex numbers [i]z[/i], or f([i]z[/i]) = "zbar" (<-- sorry, i can't find the character for it!), for all complex numbers [i]z[/i].

A hint would be to first show it works for the real numbers (first natural, then integers, then rational, etc.) to show it works for the complex numbers.

$f(z)=0$ happens when $f(1)=0$, so $f(1)\neq 0$. Then $f(1)=1$ ($f(1)=f(1)f(1)$) and your conditions implies that $f(q)=q$ for every $q\in\mathbb{Q}$. The continuity of $f$ implies that $f(x)=x$ for every $x\in\mathbb{R}$.

Let $z=a+ib$ with $a,b\in\R$ then $f(z)=f(a)+f(i)f(b)=a+f(i)b$, but $-1=f(-1)=f(i)f(i)$, that is $f(i)^2=1$, then $f(i)=\pm1$ that is $f(z)=z$ or $f(z)=\bar{z}$.


Is anyone know solution of this problem?

First, do your own homework.

Second, if you insist on having others do your homework for you, you should go to a place where people work on the area that your homework covers.  Thus, you should try to cheat on your math homework in a math forum, not in a non-math discussion forum.


Find all the $m$ and $n$ so that:
$\tau (m) = 10,\tau (n) = 21$ and $(m,n)=1$

$\tau (n)$ is (usually) the number of divisor of $n$

[hide]Here I would say $n = a^2 b^6$ or $n=c^20$ and $m = d e^4$ or $m=f^9$ with $a,b,c,d,e,f$ primes and different.[/hide]