inequalities
parameterization
analytic geometry
graphing lines
slope
algebra
function

The points that satisfy the system $ x \plus{} y \equal{} 1,\, x^2 \plus{} y^2 < 25,$ constitute the following set:

$ \textbf{(A)}\ \text{only two points} \qquad \\
\textbf{(B)}\ \text{an arc of a circle}\qquad \\
\textbf{(C)}\ \text{a straight line segment not including the end\minus{}points}\qquad \\
\textbf{(D)}\ \text{a straight line segment including the end\minus{}points}\qquad \\
\textbf{(E)}\ \text{a single point}$

I think (C) is correct. Here is why:

The second parameter implies that the points (x,y) must be within a circle with radius 5 around the origin but not on the circumference. The first parameter gives us a line with a slope-intercept form of $ y\equal{}\minus{}x\plus{}1$. Therefore, the points must be on a line, inside a circle, but not on the circumference of the circle. This is answer choice (C).

i think the answer is (A)...

Because x+y=1 is a straight line and intersect with another line to form two points

it cant be $ A$ because while the first line, $ x \plus{} y \equal{} 1$, is a straight line, the second equation mentioned, $ x^2 \plus{} y^2 < 25$ is more than a line.
When you graph an inequality, you graph the boundaries (the line) then shade in the inner portion. So wherever the line of $ x \plus{} y \equal{} 1$ is inside that shaded portion, there would be a solution. So there are wayy more than two solutions.

I also agree that the answer is $ C$ because in the second equation, $ x^2 \plus{} y^2$ is less than 25, instead of less than or equal to 25. Therefore, the points on the boundaries of that graph (where $ x^2 \plus{} y^2 \equal{} 25$) wouldnt be counted in the points for the solution to this problem. So the answer is $ C$

The Answer is $ C$.

The first equation is $ y \equal{} \minus{} x \plus{} 1$, which is just a line.

$ x^2 \plus{} y^2 < 25$ is just the equation of a circle shaded in, excluding its boundaries.

Also, the line is not tangent to the circle, and it intersects it, so it is a chord, which is a line segment. Because the circle's boundary is not part of the solution set, the line segment does not include its endpoints.

EDIT: Sorry, I meant $ C$

Paint the domain: $ x^2 \plus{} y^2 < 25$ in yellow and $ x \plus{} y \equal{} 1$ in red. Where can you see the orange domain?
The answer is C.


search
algebra
polynomial
calculus
integration
number theory solved
number theory

Hello, I search to prove that if $a+b\sqrt{d}$ ($a,b\in \mathbb{Q}$ and $d$ is square-free) is the root of a polynom in $\mathbb{Z}[X]$ (with $a_n=1$)  then $2a$ and $a^2-b^2d$ are integers.

Here is what I've start to do but that's not complete:
$a+b\sqrt{d}$ is a root of $P(X)=\sum_{k=0}^n a_k X^k$ with $a_n=1$ and $a_i\in \mathbb{Z}$
Then $P(a-b\sqrt{d})+P(a+b\sqrt{d})=\sum_{k=0}^n a_k ((a-b\sqrt{d})^k+(a+b\sqrt{d})^k)$
Developing $(a-b\sqrt{d})^k+(a+b\sqrt{d})^k$ we have that $P(a-b\sqrt{d})+P(a+b\sqrt{d})$ is rational and [b]what I want is that that's integer[/b], with this I have that $P(a-b\sqrt{d})$ is integer and then $Q(X)=P(X)-P(a-b\sqrt{d})$ is in $\mathbb{Z}[X]$ with root $a-b\sqrt{d}$ and then $PQ$ is in $\mathbb{Z}[X]$ with roots $a-b\sqrt{d}$ and $a+b\sqrt{d}$ then we can factorize by $X^2-2aX+a^2-b^2d$ [b]and then[/b] $2a$ and $a^2-b^2d$ are integers.

Problems of my solution are in bold

Can you correct me or find an other way to solve that problem ?

Thanks

Suppose an algebraic number $z$ satisfies a polynomial equation of the type $z^n+a_{n-1}z^{n-1}+\ldots+a_1z+a_0=0$. Break up the polynomial $P=x^n+\ldots+a_1x+a_0$ into irreducible factors. They must all be monic, and this means that $z$ is a root of a monic irreducible polynomial in $\mathbb Z[x]$, or, in other words, the minimal polynomial of $z$ over $\mathbb Q$ has coefficients in $\mathbb Z$.

In your case, this says that if $a+b\sqrt d$ satisfies a monic polynomial equation with integer coefficients, then the minimal polynomial of $a+b\sqrt d$ over $\mathbb Q$, which is $x^2-2ax+a^2-db^2$, has integral coefficients.

Thanks
Is it true that $z$ is a root of a monic irreducible polynomial in $\mathbb Z[X]$ implies $z$ is a root of a monic irreducible polynomial (with coefficent in $\mathbb{Z}$) in $\mathbb Q[X]$ (since polynomial is monic) implies the minimal polynomial of $z$ over $\mathbb Q$ has coefficients in $\mathbb Z$ ?

Yes that is true. A monic polynomial is irreducible in $\mathbb{Z}[x]$ iff it is irreducible in $\mathbb{Q}[x]$ (this holds not only for monic ones, also primitive ones do).


calculus
integration
trigonometry
logarithms
calculus computations

Calculate the following indefinite integrals.

[1] $\int \tan ^ 3 x dx$

[2] $\int a^{mx+n}dx\ (a>0,a\neq 1, mn\neq 0)$

[3] $\int \cos ^ 5 x dx$

[4] $\int \sin ^ 2 x\cos ^ 3 x dx$

[5]$ \int \frac{dx}{\sin x}$

5 is the easiet "
$-\ln{|\csc{x}+cot{x}|}$

[quote="adidasty"][2]

$u=mx+n$
$\frac{du}{m}=dx$

$\frac{1}{m} \int{a^{u}}du = \frac{a^{mx+n}}{m\text{ln}a}+C$[/quote]

 :)

[quote="amirhtlusa"]1
$\frac{\tan^2{x}}{2}+ln{|\cos{x}|}$[/quote]

 :)


[hide]$\log_{36}216$
$216=6^{3}$
$36=6^{2}$
so the answer must be $\frac{3}{2}$[/hide]

[edit jli] hide hide hide hide hide hide hide!!!!! :mad:

a less intuitive, more formulaic method

[hide]
notice how all the numbers are powers of 6. Use change of base formula to base 6

$\frac{\log_{6}216}{\log_{6}36}$

these are simple logarithms with integer simplifications

$\frac{3}{2}$
[/hide]

[quote="bos1234"]Change the base of the logarithm to the base in the bracket then evaluate.
$\log_{36}216$[/quote]

[hide="Change of base"]Let's change to base $6$:

$\log_{36}216$
$= \frac{\log_{6}216}{\log_{6}36}$
$= \frac{\log_{6}6^{3}}{\log_{6}6^{2}}$
$= \frac{3}{2}$[/hide]


Let $ a,b,c>0$ such that $ a^4\plus{}b^4\plus{}c^4\equal{}3$.Prove that $ \frac{1}{a^3}\plus{}\frac{1}{b^3}\plus{}\frac{1}{c^3}\geq\frac{3}{abc\sqrt{abc}}$

[quote="toanIneq"]Let $ a,b,c > 0$ such that $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 3$.Prove that $ \frac {1}{a^3} \plus{} \frac {1}{b^3} \plus{} \frac {1}{c^3}\geq\frac {3}{abc\sqrt {abc}}$[/quote]
$ \frac {1}{a^3} \plus{} \frac {1}{b^3} \plus{} \frac {1}{c^3}\geq\frac {3}{abc\sqrt {abc}}\Leftrightarrow\left(a^3b^3\plus{}a^3c^3\plus{}b^3c^3\right)^8\geq243(a^4\plus{}b^4\plus{}c^4)^3a^{12}b^{12}c^{12}.$
Let $ bc\equal{}x,$ $ ac\equal{}y$ and $ ab\equal{}z.$
Hence, it remains to prove that $ (x^3\plus{}y^3\plus{}z^3)^8\geq243\left(x^4y^4\plus{}x^4z^4\plus{}y^4z^4\right)^3,$
which known Vasc's inequality. See the Namdung's post here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34428

yes,i see because i created this problem from harazi's inequality.Thanks arqady :D


number theory
relatively prime
superior algebra
superior algebra solved

Let G be a group , and let $x,y\in G$ be comuting elements of G with finite orders m,n,respectivly.it is well known that if m,n are relatively prime,then the order of xy is mn.establish the following generalization :if,for every prime p that devides Lcm(m,n),the greatest power of p that devides m is different from the greatest power of p that devides n,then the order of xy is lcm(m,n).

Maybe I'm wrong, but this looks pretty straightforward:

Let $u=ord<xy>$. It's clear that $m|u\iff n|u$, and in this case $u=lcm(m,n)$, and it's all done. Therefore, assume $m,n\not |\ u$. On the other hand, we know that $u|lcm(m,n)$. Now assume that $P=\frac n{gcd(u,n)}=p_1^{a_1}\ldots p_k^{a_k}$ and $Q=\frac m{gcd(u,m)}=q_1^{b_1}\ldots q_t^{u_t}$. Because of the hypothesis, we cannot have $P=Q$, so multiply $u$ by $P$ or $Q$, according to $P\not |\ Q$ or $Q\not |\ P$, and this new number will be divisible by $n$ or $m$, and only by one of them, and we will easily get a contradiction.

Assume, for example, that $P\not |\ Q$. Then $(xy)^{uP}=1=y^{uP}$ (because $n=ord<x>|uP$), but $ord<y>=m\not |\ uP$, so we have a contradiction.


Useless title ( http://www.mathlinks.ro/viewtopic.php?t=135914 ).

[quote="Allnames"]Let $ x,a,b\geq 0 \in N$ such that $ x^{a \plus{} b} \equal{} a^bb$ Prove that $ a \equal{} x$ and $ b \equal{} x^x$
I proved it but my solution was very hard
Who can help me with a nice proof
TO Zetax I am very sorry,i dont know that thing,sorry :blush:[/quote]
Any help


On a 8 x 8 board we place $n$ dominoes, each covering two adjacent squares, so that no more dominoes can be placed on the remaining squares. What is the smallest value of $n$ for which the above statement is true?

Can dominoes overlap? This may be able to change the answer, otherwise it would seem that ~22 is the answer, with exactly 22 being very likely as some observations on minimums we can make.

can  anyone say why not more than 22? and how 22 is attaining?


On a mathematical competition $ n$ problems were given. The final results showed that:
(i) on each problem, exactly three contestants scored $ 7$ points;
(ii) for each pair of problems, exactly one contestant scored $ 7$ points on both problems.
Prove that if $ n \geq  8$, then there is a contestant who got $ 7$ points on each problem. Is this statement necessarily true if $ n \equal{} 7$?

Suppose there's no one who got every problem, but suppose someone got over 3 problems. (i.e. questions 1,2,3,4...). Then there must be a problem k he/she did not get. Someone else must have gotten k and 1, but he did not get 2, 3 or 4 by (ii). Similarly with k and 2, k and 3 and k and 4. Then k was solved at least 4 times contradiction. So noone got over 3 problems. Any group of 3 problems contribute 3 pairs of problems. So at maximum all problems contribute 3n pairs. However, we have (n)(n-1)/2 pairs who must be solved together. If n = 7, then 3n = (n)(n-1)/2. If n > 7, then 3n < (n)(n-1)/2. So the (n)(n-1)/2 pair condition cannot be filled. Thus if n is over 8 there must be one contestant who got every problem.

7 can be done. (1,2,3), (1,4,5), (1,6,7), (2,4,6), (2,5,7), (3,4,7), (3,5,6)

Look at each problem $P_k$, $1\leq k \leq n$, as being the set of those contestants scoring $7$ points on it. Then (i) means $|P_k| = 3$ for all $1\leq k\leq n$, and (ii) means $|P_k \cap P_{\ell}| = 1$, for all $1\leq k \neq \ell \leq n$.
We are asked to prove that if $n\geq 8$ then $\left | \bigcap_{1\leq k \leq n} P_k\right |$, and to find a counterexample for $n=7$.

This is related to the Erdös-de Bruijn theorem; the counterexample for $n=7$ is the finite projective plane of order $2$, known as the Fano configuration, also known as the symmetric BIBD $(7,3,1)$.


calculus integration algebra polynomial quadratics function geometry

When using partial fractions to solve integration problem, people set up the fraction part like this, for example:
(ax+b)/{x+d)(x+f)} = A/(x+d) + B/(x+f), where A and B are constants.
My question: why are A and B constants? Can't they be polynomials that will eventually come out as ax+b? I mean what if there's an occasion where leading terms get eliminated and ultimately satisfies the original value of the numerator, in this case ax+b.

Also, why do I have to set the numerator as Ax+B when the denominator of decomposed fraction is quadratic? Can I not just replace Ax+B with A?

In this representation A and B are polynomials of degree 0(can't be polynomials of degree greater than 0).

This  is ordinary way of integration.

So can you prove that the degrees can't be greater than zero?

This question isn't about the integration; it's entirely about the algebraic process of decomposing into partial fractions. As such, it's really more of a linear algebra question, and that's the language I'll be using.

Definition: A proper rational function is a function of the form $ \frac{p(x)}{q(x)}$, where $ p$ and $ q$ are polynomials and the degree of $ p$ is strictly less than the degree of $ q$. This means that $ q$ cannot be the zero polynomial, so any such fraction makes sense.

Any linear combination of proper rational functions is a proper rational function: $ a\frac{p_1(x)}{q_1(x)}\plus{}b\frac{p_2(x)}{q_2(x)}\equal{}\frac{ap_1(x)q_2(x)\plus{}bp_2(x)q_1(x)}{q_1(x)q_2(x)}$, and each term in the numerator has smaller degree than the denominator.
Now, suppose we have a proper rational function $ \frac{p(x)}{q(x)}$, and the denominator factors as $ q(x)\equal{}q_1(x)^{e_1}q_2(x)^{e_2}\cdots q_r(x)^{e_r}$. Let $ d_i$ be the degree of $ q_i$ for each $ i$. Write down a generic linear combination $ \sum_{i\equal{}1}^r\sum_{j\equal{}1}^{e_i}\frac{p_{i,j}(x)}{(q_i(x))^j}$, with the additional restriction that the degree of $ p_{i,j}$ is at less than $ d_i$. Since each term is a proper rational function, the sum is a proper rational function, and we can put it all on a common denominator of $ q$. There are $ \sum_{i\equal{}1}^r\sum_{j\equal{}1}^{e_i}d_i\equal{}d_1e_1\plus{}d_2e_2\plus{}\cdots\plus{}d_re_r\equal{}\deg q$ coefficients to choose, so the space of all such combinations appears to have the same dimension as the space of proper rational functions with denominator $ q$.
With two spaces apparently of the same dimension, it's nice to ask whether they are actually the same. As long as the $ q_i$ are relatively prime, they are, and any proper rational function with denominator $ q$ can be represented in exactly one way in this form. That's the key theorem of partial fractions.

So, why zero-degree polynomials in your example? Because that's what it takes to have degree smaller than the denominator, and we don't need any more. There certainly are representations as sums of improper rational functions, but those don't have the uniqueness property. With the quadratic denominators, it's the other direction; we need as many unknown coefficients as the degree of the denominator, and that means two for each piece with a quadratic denominator.

One more thing: nothing I said invoked the real or complex numbers. The algebra applies to polynomials over any field, and doesn't require complete factorizations. What $ \mathbb{R}$ or $ \mathbb{C}$ bring is the assurance that we can completely factor the denominator as a product of linear factors for $ \mathbb{C}$, or linear and quadratic factors for $ \mathbb{R}$.

You're approaching this backwards. It's not a matter of proving that you can't write things in nonstandard form. In fact, you can write things in nonstandard form.

For instance, $ \frac { \minus{} 2x \plus{} 3}{x^2 \minus{} 3x \plus{} 2} \equal{} \frac {x \minus{} 2}{x \minus{} 1} \minus{} \frac {x \minus{} 1}{x \minus{} 2}$ is a true algebraic identity.

It's of dubious utility, and whatever it is, it's not partial fractions - but it's true.

Now to get back to the point: the principle of partial fractions is an algebraic theorem. This theorem says that a proper rational function ("proper" means that the degree of the numerator is less than the degree of the denominator) can be uniquely written as a sum of proper rational functions described by a certain set of rules. The rule about the various pieces being themselves proper rational functions is embedded within the statement of this theorem - it's what we're trying to show.

The statement "this is the ordinary way of integration" does reflect perhaps the most common [i]use[/i] of partial fractions, but is not a statement of what partial fractions [i]is[/i]. Partial fractions is [i]not[/i] a technique of integration. It is a technique of [i]algebra.[/i] The theorem I allude to above (without fully stating it) is a theorem of algebra - its proof requires no analysis; it doesn't even require that we be working over the real numbers (any field will do). This technique of algebra happens to be useful in computing integrals; it's useful in a number of other places as well.

[Edit: composed before I saw jmerry's post. I'll post it anyway. He gave more of a description of the theorem of algebra than I did.]


The sequences $ (a_n),(b_n)$ are defined by $ a_1\equal{}1,b_1\equal{}2$ and \[a_{n \plus{} 1} \equal{} \frac {1 \plus{} a_n \plus{} a_nb_n}{b_n}, \quad b_{n \plus{} 1} \equal{} \frac {1 \plus{} b_n \plus{} a_nb_n}{a_n}.\]

Show that $ a_{2008} < 5$.

If $ a_0 > 0,b_0 > 0$, then $ a_{n \plus{} 1} > a_n,b_{n \plus{} 1} > b_n$, sequences are increase.
If $ a_0 > 5$, then $ a_{2008} > 5$.
If $ a_0\equal{}b_0$, then $ a_n\equal{}b_n \ \forall n\in N$ and $ a_{n\plus{}1}\equal{}a_n\plus{}1\plus{}\frac{1}{a_n}\to a_n>n\plus{}a_0$ if $ a_0>0$.

If $ a_1\equal{}1$ and $ b_1 \equal{} b > 1$ then $ \lim_{n\to \infty}a_n \equal{} 1\plus{}\frac{4}{b\minus{}1}$ and $ \lim_{n\to \infty}b_n\equal{}\infty.$

We can simply prove the equation 1/(An+1) - 1/(Bn+1) = 1/6 (by induction). From here we have An+1 < 6, so An<5.

[quote=grupyorum]Upon adding $1$ to both sides of each equation we get; $1 + a_{n+1}=\frac{(1+a_n)(1+b_n)}{b_n}, 1+ b_{n+1}=\frac{(1+b_n)(1+a_n)}{a_n} $ . Rewriting this as $\frac{1}{1+a_{n+1}} = \frac{b_n}{(a_n+1)(b_n+1)}$ and $\frac{1}{1+b_{n+1}} = \frac{a_n}{(a_n+1)(b_n+1)} $ .

After this manipulation, we get $\frac{1}{a_{n+1}+1} - \frac{1}{b_{n+1}+1} = \frac{1}{a_n+1} - \frac{1}{b_n+1} = \frac{1}{a_1+1} - \frac{1}{b_1+1} = \frac{1}{6}$. Therefore, $\frac{1}{a_{n+1}+1} > \frac{1}{6} $ which is equivalent to $ 5>a_n , \forall n\in N$.[/quote]

VERY NICE SOLUTION

[hide=Solution]Adding $1$ to both sides, we get $a_{n+1} + 1= \frac{(a_n + 1)(b_n + 1)}{b_n} , b_{n+1} + 1= \frac{(a_n + 1)(b_n + 1)}{a_n}$ for all $n \geq 1$.
From the first one, we get $\frac{a_{n+1} + 1}{a_n + 1} = \frac{b_n + 1}{b_n} \implies b_n = \frac{a_n + 1}{a_{n+1} - a_n}$.
Dividing the two, we get $\frac{a_{n+1} + 1}{a_n} = \frac{b_{n+1} + 1}{b_n} = \frac{\frac{a_{n+2} + 1}{a_{n+2} - a_{n+1}}}{\frac{a_n + 1}{a_{n+1} - a_n}} \implies a_n(a_{n+1} - a_n)(a_{n+2} + 1) = (a_n + 1)(a_{n+1} + 1)(a_{n+2} - a_{n+1})$.
So we get $a_{n+2} + 1 = \frac{(a_n + 1)(a_{n+1} + 1)^2}{a_n^2 + a_n + a_{n+1} + 1}$.
Now let $x_n = a_n + 1$. We have $$x_{n+2} = \frac{x_nx_{n+1}^2}{x_{n+1} + x_n(x_n - 1)} \implies \frac{x_{n+1}}{x_{n+2}} = \frac{1}{x_n} + \frac{x_n}{x_{n+1}} - \frac{1}{x_{n+1}} \implies \frac{x_{n+1}}{x_{n+2}} + \frac{1}{x_{n+1}} = \frac{x_n}{x_{n+1}} + \frac{1}{x_n}$$
This implies the sequence $y_n = \frac{x_n}{x_{n+1}} + \frac{1}{x_n}$ is constant, and since $y_1 = \frac{7}{6}$, we have that $\frac{x_n}{x_{n+1}} + \frac{1}{x_n} = \frac{7}{6}$ for all $n \geq 1$.
So, we get $x_{n+1} = \frac{6x_n^2}{7x_n - 6}$. Now we will prove by induction that $2 \leq x_n < 6$ for all $n$. This is clear for $n = 1$. Now assume the result holds true for $n$. Then, since $\left(\frac{6x^2}{7x - 6}\right)' > 0$ for all $x \geq 2$, we have $x_{n+1} = \frac{6x_n^2}{7x_n - 6} < \frac{6 \cdot 36}{36} = 6$ and $x_{n+1} = \frac{6x_n^2}{7x_n - 6} \geq \frac{6 \cdot 4}{8} = 3$, we are done. $\square$[/hide]

Absolutely Incredible problem!! Solved with [b]L567[/b].

We have \[a_{n+1}+1=\frac{1+a_n+b_n+a_nb_n}{b_n} \Rightarrow \frac{1}{a_{n+1}+1}=\frac{b_n}{(1+a_n)(1+b_n)},\] and similarly $\frac{1}{b_{n+1}+1} = \frac{a_n}{(1+a_n)(1+b_n)}$. Taking the difference we obtain $$\frac{1}{a_{n+1}+1} - \frac{1}{b_{n+1}+1} = \frac{b_n - a_n}{(1+a_n)(1+b_n)}=\frac{(b_n+1) -( a_n+1)}{(1+a_n)(1+b_n)}=\frac{1}{a_n+1}-\frac{1}{b_n+1},$$showing that this difference is infact invariant. Finally, we have $$\frac{1}{a_{2008}+1} - \frac{1}{b_{2008}+1}=\frac{1}{a_1+1}-\frac{1}{b_1+1}=\frac 12 - \frac 13= \frac 16$$ $$\Rightarrow \frac{1}{a_{2008}+1}>\frac 16 \Rightarrow 5> a_{2008}$$ as needed. $\blacksquare$

very funny
[hide = solution]
Let $x_n = a_n+1$ and $y_n = b_n+1$ for all $n \geq 1$. The given recurrence relations simplify to \[x_{n+1} = \frac{x_n y_n}{y_n-1}\text{ and } y_{n+1} = \frac{x_n y_n}{x_n-1}\]
Now we obtain that \[\frac{1}{x_{n+1}}-\frac{1}{y_{n+1}} = \frac{y_n-1}{x_ny_n}-\frac{x_n-1}{x_ny_n} = \frac{1}{x_n}-\frac{1}{y_n}\]
and therefore $\frac{1}{x_{n+1}}-\frac{1}{y_{n+1}}$ is fixed at $\frac{1}{x_1}-\frac{1}{y_1} = 1/6$.

In particular, since each $y_n > 0$, it follows that $x_n < 6$ for all $n > 0$. Thus $a_{2008} < 6-1 = 5$.
[/hide]

Seems like it could be changed to an AIME question somehow.

Adding one to both equations and taking a reciprocal gives
\[\frac{1}{a_{n+1}+1} = \frac{b_n}{(a_n+1)(b_n+1)} \text{ and } \frac{1}{b_{n+1}+1} = \frac{a_n}{(a_n+1)(b_n+1)}\]
Subtracting them gives
\[\frac{1}{a_{n+1}+1}-\frac{1}{b_{n+1}+1} = \frac{1}{a_n+1}-\frac{1}{b_n+1}\]
which is thus constant. Thus since $b_{2008}>0$,
\[\frac{1}{a_{2008}+1}-\frac{1}{b_{2008}+1} = \frac{1}{6} \implies \frac{1}{a_{2008}+1} > \frac{1}{6} \implies a_{2008} < 5\]
as desired.

Letting $A_n=a_n+1,B_n=b_n+1$ we see that $A_n<B_n$ and $\frac{A_nB_n}{B_n-A_n}$ is invariant. Initially it is $6$ so we have $A_{2008}B_{2008}=6B_{2008}-6A_{2008}$ which rearranges to $(A_{2008}-6)(B_{2008}+6)=-36.$ The second term is positive so the first term is negative.


The pages of a book are numbered 1 through n. When the page numbers of the book were added, one of the page numbers was mistakingly added twice, resulting in the incorrect sum of 1986. What was the number of the page that was added twice?

Aren't there several possible answers? Are we supposed to answer in terms of $n$?

No, there is one answer, and you must give the exact page number (not in terms of n).

[hide]$\frac{n(n+1)}{2}+1 \le 1986 \le \frac{n(n+1)}{2}+n$
After solving two quadratics,
$61.54\dots \le n \le 62.51\dots$
so it must be $n=62$.
Or you can try to guess and check near $\sqrt{2\times 1986}$[/hide]


Look at this:

http://www.huffingtonpost.com/jeffrey-klein/mccains-secret-questionab_b_107409.html?icid=100214839x1204046610x1200186912

It's pretty revealing, huh?

HuffingtonPost is populist trash. It's almost as bad as 'Kos and Fox News.

[quote="dynamo729"]What do you find wrong with Fox News?[/quote]

haha, wow; not going to answer that, as there really isn't any point (doubt anyone would convince anybody either way)

my mom showed me an interview in the NYTimes magazine the other day with gore vidal (sp?), part of which was devoted to him pointing out we're TOTALLY JUST TAKING MCCAIN'S WORD THAT HE DID ALL THIS STUFF

kinda funny

Look, if the public really wanted to find something really bad at John McCain or Barack Obama, why not look it up themselves? You can't have political advisers digging up something that could be mostly based on opinion, what happened, or your side of the view.

When I read the first sentence of that article, I closed it.

The point is that the information about this assertion is not public; we have only what people have claimed and inference from public records, and those don't plausibly match.

It's much simpler to go after McCain's policies- that's right there in his voting record, like voting against health care and benefits for veterans.


This has been posted before: China 1998, Problem 6.

[url]http://web.archive.org/web/20040128041419/www.kalva.demon.co.uk/china/csoln/csol986.html[/url]


algebra
polynomial
calculus
derivative
geometry
algebra unsolved

Let a polynomial
 $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+.....+a_{0}$

with positive integer coefficients such that

$a_{n}+n=a_{n-1}+n-1=.....=a_{0}$ .
Prove that this polynomial is irreducible.

Thanks, but either I am stupid now, or this is a very difficult problem. I remember Vess saying that the problem of proving the irreducibility of the derivative of $1+X+X^{2}+...+X^{n}$ is still open, and if I'm not wrong, this result would imply that one: just take $a_{n}=1, a_{n-1}=2,...,a_{0}=n+1$ and use that $P$ is irreducible if and only if $X^{\deg(P)}P(1/X)$ is. Silouan, are you sure there is a solution for this problem?  :maybe:

Harazi , I don't know of there exist solution .
It is a conjecture I come up with but I didn't put in the open questions because I think mr Valentin Vornicu has a solution to this problem as it is a little generalisation of the last Mathlinks contest problem 1 .

I'm not so sure about that. In that case, Mr Vornicu should publish this, since not even Filaseta solved the problem I was talking about... :wink:

I don't know but I know that Valentin has a solution to
to the following .The polynomial P(X) is irreducible ,where
$P(X)=X^{2003}+2X^{2002}+\ldots+2003X+2004$

I think this is easier (I have not also solution to this ,could you find one harazi?)

Hmm, I've never seen that problem in mathlinks contest, I must have missed that round. I tried to solve it and failed. But apparently Cezar Lupu has a solution, since I saw it in the proposed section area, posted by him... But since all he says is that he will post his solution and he never does it, it means that probably it is wrong... I'll try more, the problem is interesting, but seems hard...

Harazi , a friend stedes found a very nice solution for the problem above .
He used the lemma that
If the natural number  $a\geq 3$ is square free then the polynomial $Q(x)=\frac{x^{a}-ax+a-1}{(x-1)^{2}}$  is irreducible over $Z[x]$.

Abd our problem is an application of this lemma since
$\frac{x^{2005}-2005x+2004}{(x-1)^{2}}=x^{2003}+2x^{2002}+3x^{2001}+...+2003x+2004$

You can find the lemma at
url=http://www.math.psu.edu/borisov/Documents/papers/abc.pdf


To weigh things on a balance scale, one or more objects are
placed on one pan and weights are placed onto the other pan
until the two pans are balanced. We see A is balanced by the
4 weights shown, and B is balanced by the 5 weights shown.
We have the following weights: 1, 1, 3, 3, 9, 9, 27 and 27.
What is the minimum number of these weights it would take to
balance the total weight of A plus B?

i know the answer is 5...

but i need a pic 2 explain?


function
real analysis
real analysis unsolved

Given function $f:[0,1]{\times}[0,1]{\longrightarrow}R$,such that it is $\forall (x,y)\in{[0,1]{\times}[0,1]} f$ is continious relatively $x$ and relatively $y$.Prove that $\forall x\in{[0,1]}$ there exist an $y\in{[0,1]}$,such that $f$ is continious in the point
$(x,y).$

Suppose contrary.

Let $w(f(x,y))$ be an oscillation of $f$ in point $(x,y)$, $x_0\in[0,1]$.
Consider sets $B_k=\{y\in[0,1]\,|\,w(f(x_0,y))>\frac{1}{k}\}$.
Due to our supposition $\bigcup_{k=1}^\infty B_k=[0,1]$.
It follows that there is $k_0$ s.t. $B_{k_0}$ isn't nondense in $[0,1]$ (Baire theorem). Suppose $B_{k_0}$ dense in $[a_1,b_1]$. Let $\epsilon=\frac{1}{3k_0}$.

Consider sets $A_k=\{y\in[a_1,b_1]\,|\, \max_{|t|\leq 1/k}|f(x_0+t,y)-f(x_0,y)|<\epsilon\}$. Since $f(\cdot,y)$ is continuous we conclude $\bigcup A_k=[a_1,b_1]$. So  there is $k_1$ s.t. $A_{k_1}$ isn't nondense in $[a_1,b_1]$. Suppose $A_{k_1}$ is dense in $[a_2,b_2]\subset[a_1,b_1]$. Let $\delta_y=\frac{1}{k_1}$ and $y_0=\frac{a_2+b_2}{2}$. ;)

We can find such $\delta_x$ that $|f(x_0,y_0+t)-f(x_0,y_0)|<\epsilon$ for all $t\in[-\delta_x,\delta_x]$, $\delta_x<\frac{b_2-a_2}{2}$.

Consider arbitrary point $(x,y)\in[0,1]\times[0,1]$ s.t. $|x-x_0|<\delta_x$ and $|y-y_0|<\delta_y$. We can write
\[|f(x,y)-f(x_0,y_0)|\leq|f(x,y)-f(x,y')|+|f(x,y')-f(x_0,y')|+\]
\[+|f(x_0,y')-f(x_0,y_0)|<\epsilon+\epsilon+\epsilon<\frac{1}{k},\]
where $y'\in[a_2,b_2]\cap A_{k_1}$ s.t. $|f(x,y)-f(x,y')|<\epsilon$ (we can find such $y'$, since $f(x,\cdot)$ is continuous function and $A_{k_1}$ is dense in $[a_2,b_2]$).
It follows that $w(f(x_0,y_0))<\frac{1}{k}$. Contradiction.

Is it true that measure of sets of discontinuity points is zero?


Source: IMO 1974
Let $a,b,c,d>0$. Find min. and max. values of the sum
$S=\frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d}$.

[hide="my partial solution"]
I found the minimum value of $S$ using the fact that:
$\frac{x^{2}}{a}+\frac{y^{2}}{b}\geq \frac{(x+y)^{2}}{a+b}$.
However, I don't know how to get the max. can anyone teach me how to find it? Thanks,
[/hide]

$\frac{c}{b+c+d}<\frac{c}{c+d}$, and $\frac{d}{a+c+d}<\frac{d}{c+d}$. So S<2. Seting $a=c=1$, and $b,d \to 0$ we see that this $S \to 2$. so $sup \{S | (a,b,c,d) \in R_{+}^{4}\}=2$.

ok so if
$\frac{c}{b+c+d}<\frac{c}{c+d}$, then why not:
$\frac{c}{b+c+d}<\frac{c}{c}=1$?

It is. But how you can prove that 4 is best constant.


If in a triangle the difference of the perimeter and area (2p-S) is equal with the  double of the lenght of a side, prove that  16R+4r+p^2(r-2)=8.

We can solve very easily this kind of exercise if we use the fact that a,b,c are the roots of
x^3 - 2.p.x^2 + (p^2+r^2+4.r.R)x - 4.p.r.R = 0
Put x = p - p.r/2 in the equation to get your relation.

It's OK! Thank you for your solution! This is and my initial idea, but I used other roots and equation!


LaTeX
summer program
Mathcamp
number theory

find all natural N for which: N/2 is a square, N/3 is a third power and N/5 is a fifth power.

[img]http://www.mathlinks.ro/Forum/images/flags/china.gif[/img]

  dg

Based on those 3 equations, it is clear that N has a power of 2, 3, and 5 in it.  Also, the power of 2 is congruent to 1 mod 2, while the other powers are congruent to 0 mod 2.  Similarly, power of 2 and 5 are congruent to 0 mod 3, while power of 3 is congruent to 1 mod 3 and power of 2 and 3 are congruent to 0 mod 5 but 5 is 1 mod 5.

Now, the smallest number that satisfies those equations is $2^{15}*3^{10}*5^6$ and in general a number $a^{30}*2^{15}*3^{10}*5^6$ would work for a=integer.

PS How do you get congruent signs into latex?

Congruent signs are \cong.

And what's a NT course about?

NT = Number Theory...

This time, the lesson must have been about the chinese remainder theorem...  :)

  Darij

A slightly more complex form of this problem (using 4 values instead) was on last year's Mathcamp application quiz.

2^(30/2)*3^(30/3)*5^(30/5)*(other_numbers)^30
(to find individual powers, just notice that 2*3*5 = 30 and the numbers are 30/2 , 30/3 and 30/5) )

Actually, the congruent sign in the context of number theory (i.e. $\equiv$) is \equiv.  It is correct that \cong makes a congruent sign ($\cong$); however, this is used in geometry.


geometry
trapezoid
perimeter
3D geometry
frustum
symmetry

[hide="reveal questions seperated into other threads"]23.	An isosceles trapezoid has a midsegment measuring 13 cm and an area of 52 cm2.  If one base has length 10 cm, find the perimeter of the trapezoid in centimeters.

20. A fuel tank is in the shape of a  right frustum  (truncated cone) as shown.  Its parallel circular bases have area   square feet and    square feet respectively. When the fuel tank is half filled as shown (up to the axis of symmetry), the area of the surface of the fuel (cross section perpendicular to the bases, including the axis of symmetry) is 108 sq. ft.  What is the volume, in cubic feet, of the fuel in the tank?

26.	A solid cube has edges measuring 4 cm.  A triangle is drawn with vertices at the midpoints of three connecting edges, and one "corner" of the cube is truncated so that this triangle is showing. The remaining figure is a solid with 7 faces.  Find the surface area of this solid in square centimeters.

27.	Two tangent rays are drawn to a circle, with points R and T the points of tangency.  Find the measure of inscribed angle RST.

28.	Two concentric circles are drawn so that the tangent segment (shown) to the smaller circle is a chord of the larger circle. If the area of the annulus (region outside the smaller circle and inside the larger one) is 100 , find the length of the chord shown ( ).

29.	Two circles are externally tangent, and have a common external tangent line, with points of tangency P and Q.   , and the radii of the circles are   and  .  The center of the smaller circle (the one which contains point Q) is S, and a line through S is drawn tangent to the larger circle. The point of tangency of this line is T, as shown.  If ST=  cm, find the length of the larger radius  .

These are the ones I couldn't do when practicing. I was able to get all but 2 of the  others since this 2005 wasn't as hard as some others.  [/hide]

Try not to post that many problems together. :wink:

[hide="23"]Since $13$ is the length of the median and $10$ is one of the bases, then the other base is $16$.  Using the trapezoid area formula, $52=\frac{(16+10)h}{2}\implies h=4$.

[code]
    A____
    /|   \
   / |    \
  /__|_____\
 C  B[/code]

If you look at right triangle ABC, BC is $(16-10)/2=3$ and since the height is $4$, we have $AC=5$.  Thus the perimeter is $5+10+5+16=\boxed{36}$[/hide]

Yes, please do not post multiple problems in a single thread unless they are directly related.

I will split your post and this thread will be used to discuss #23.

EDIT:  You can't split a single post so I will just make new threads for the others  :roll:

thanks  :)

I thought people might find it annoying if I posted several threads at once.


Let $ ABCD$ be a convex quadrilateral for which $ m\left(\widehat {ABD}\right)\plus{}2\cdot m\left(\widehat {CBD}\right)\equal{}A\plus{}2\cdot C\equal{}180^{\circ}$ . Show that  $ m\left(\widehat {CAD}\right)\equal{}90^{\circ}\minus{}C$ .

[hide="Proof."]
Since the geometric place of the points $ P$ in the opposite semiplane of $ A$ respect to $ BD$ such that $ \angle{BPD}\equal{}90\minus{}\frac{A}{2}$ is a circunference arch, it follows that $ C$ is the only intersection of this arch with line $ BC$, different from $ B$.

Let $ C'$ be the excenter corresponding to $ A$ for triangle $ ABD$. From the first angle relation it follows that $ BC$ is external bisector of angle at $ B$. So $ C'$ is on $ BC$. Playing around with angles  in $ ABC'D$ it follows that $ \angle{BC'D}\equal{}\angle{BCD}\equal{}90\minus{}\frac{\angle{A}}{2}$. So $ C\equal{}C'$ and $ C$ is on the interior bisector of $ \angle{BAD}$. That means $ \angle{CAD}\equal{}\frac{A}{2}\equal{}90\minus{}C$. [/hide]

:)


[quote="soruz"]A problems of contest of maths teacher 2007 in atachement.Have a nice day![/quote]

where did this exam was done? is this exam for math teachers? could you have the solutions sheet?

1)Please don't post a file.
2)You can learn $ \text{\LaTeX}$ here http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php and use it in your posts, please.
3)This is an analysis problem and I must move it.  :wink:


How many pairs of integers $(x,y)$ will solve the equation $6x^2-3xy-13x+5y=-11$?
(A) 0
(B) 1
(C) 2
(D) 3
(E) infinitely many

I think I am wrong but here is what I got:

[hide]

let $p(x)=6x^2-x(3y+13)+(5y+11)=0$

Now we solve for $x$ by substituting it into the quadratic formula.

We discover the discriminant is $\sqrt{9y^2-42y-95}$ Now for x to have integer solutions, this value must be an integer.

Hence $\implies 9y^2-42y-95$ is a square.

$9y^2-42y-95=(3y+5)(3y-19)$

It can never be a square and hence that is why I think there are no solutions.[/hide]

I put zero, and it was marked wrong, so there must be a solution somewhere.

well i guess the question is for what values of y $9y^2-42y-95$ is a square.

$6x^2-3xy-13x+5y=-11$
$5y-3xy=-11-6x^2+13x$
$\displaystyle y=\frac{6x^2-13x+11}{3x-5}$
$\displaystyle y=2x-1+\frac{6}{3x-5}$
since $x$ and $y$ are integers, $\displaystyle \frac{6}{3x-5}$ has to be an integer
taking both positive and negative factors of 6, we get $x=1$ or $x=2$
so there are $\boxed{2}$ solutions (?)


geometry
3D geometry
sphere
geometry unsolved

$L$ and $L'$ are two perpendicular lines not in the same plane. $AA'$ is perpendicular to both lines,
where $A$ belongs to $L$ and $A'$ belongs to $L'$. $S$ is the sphere with diameter $AA'$.
For which points $P$ on $S$ can we find points $X$ on $L$ and $X$' on $L'$ such that $XX'$ touches $S$ at $P$?

Suppose the sphere is the unit sphere and that $A(0,0,-1),A^{\prime}(0,0,1),$
the equations of $L$ are $y=0$ and $z=-1,$ and, finally, the equations of $L^{\prime}$ are $x=0$ and $z=-1.$

The equation of the plane tangent to $S$ at the point $P(a,b,c)$  is $ax+by+cz=1$ and this plane intersects $L$ and $L^{\prime}$ at $B(\frac {1+c}{a},0,-1)$ and $C\left(  0,\frac{1-c}{b},1\right)  .$ The points $B,C,P$ are collinear iff
\[
\frac{1+c}{a^{2}-c-1}=\frac{c-1}{b^{2}}=\frac{-2}{c+1},
\]
which readily simplifies to $a^{2}=b^{2}$ (using, of course that $a^{2}+b^{2}+c^{2}=1$).

Thus, the point $P$ lies on the reunion of two circles, the intersection of $S$ with the planes of equations $x=y$ and $x=-y.$


[hide]We see that since r=a=b, $r^4=625$. Since r is positive, r=25.[/hide]

[quote="hwenterprise"]Define $E (a, b, c) = a \times b^c$. What positive value of $r$ is the solution to the equation $E (r, r, 3) = 625?$[/quote]

rxr^3-625. r^4=625. r^2=25. r=5


Let $ABC$ be an equilateral triangle and $M$ be a point inside the triangle. We denote by $A'$, $B'$, $C'$ the projections of the point $M$ on the sides $BC$, $CA$ and $AB$ respectively. Prove that the lines $AA'$, $BB'$ and $CC'$ are concurrent if and only if $M$ belongs to an altitude of the triangle.

I know it is for kids  :P
[hide]Suppose wlog $AB=1$, $BA'=x$, $CB'=y$ and $AC'=z$. Then using Carno's theorem we obtain $x^2+y^2+z^2=(1-x)^2+(1-y)^2+(1-z)^2$ [b](1)[/b] and from Ceva's theorem $xyz=(1-x)(1-y)(1-z)$ [b](2)[/b]. It follows from [b](1)[/b] that $2(x+y+z)=3$, and using this equality we can write [b](2)[/b] as $(1-2x)(1-2y)(1-2z)=0$.[/hide]


number theory
prime numbers
real analysis
real analysis solved

Let p_1,p_2,p_3,.....  the sequence of prime numbers in ascending order.

Prove that  lim(n-->00) (p_1*p_2*...*p_n)^(1/p_n)=e

I have proved it using the Prime Number Theorem and Stirling's formula.

It is not difficult for those who know the prime number theorem because for the others I think it is simply impossible.
   A solution which doesn't use Stirling, but does use the prime number theorem. It is clear that we must calculate lim 1/p_n*sum (i=1,n) lnp_i.
  Since p_n/(n*lnn) tends to 1, it is enough to prove that lim 1/(n*lnn)*sum ln p_i is 1. Use Cesaro-Stolz to find that the limit is lim ln(p_n+1)/[(n+1)ln(n+1)-nlnn]. Since ln p_n/lnn tends to 1 and [(n+1)ln(n+1)-nlnn]/lnn tends to 1, the limit is 1 and accordingly, your limit is e.

well... what is the  Prime Number Theorem?
 :?

It states that if pi(n) is the number of primes smaller than n, then lim pi(n)*lnn/n=1. Another form of it is that lim p_n/n*lnn=1, where p_n is the n-th prime number. Please, don't try to prove this. Take it as it is ( just for now).


The total price of two items at a department store is 140 dollars. If the price of one item is decreased by 20%, and the price of the other item is decreased by 25%, the average price of the two items is 70.50. What is the original price of the more expensive item minus the original price of the less expensive item?

Are you sure those numbers are correct?  If the average of the two discounted items are 70.5 then their sum is 141, which is greater than the orignal sum.  This is contradictory of the claim that the items were decreased in price.

sorry my bad- i posted the problem wrong  :blush:

the total price of two items at a department store is 140 dollars. if the price of one item is increased by 20% and the price of the other item is decreased by 25%, the average price of the two items is 70.50. what is the original price of the more expensive item minus the original price of the less expensive item?

You get two equations:
$x+y=140$
$\frac{6x}{5}+\frac{y}{4}=141$
From this the solution is about $\$28.42$


$ \binom{n}{n} \equal{} \frac{n!}{n!(n\minus{}n)!} \equal{} \frac{n!}{n!0!} \equal{} \frac{1}{0!} \equal{} \frac{1}{1} \equal{} \boxed{1}$

The easier way is there is only $\boxed1$ way to pick $n$ things out of $n$ things.

[quote=floof421]ernie's solution basically proves $0!=1$ too[/quote]

yes, that's correct
0! is actually indeed 1

there is one way to choose how to arrange 0 books, with that arrangement being no books

[quote name="williamxiao" url="/community/p20439092"]
[quote=floof421]ernie's solution basically proves $0!=1$ too[/quote]

yes, that's correct
0! is actually indeed 1
[/quote]

i know, just commenting a proof of 0!=1

[quote=GameBot]What is $ \dbinom{n}{n}$ for any positive integer $ n$?[/quote]

[hide=Solution]$\dbinom{n}{n} = \dfrac{n!}{n(n-n)!} = \dfrac{n!}{n!0!} = \dfrac{1}{0!} = \boxed{1}$[/hide]

[quote=abhinavg0627]$0!$ is the number of ways to arrange $0$ objects in a row. We can only do this is one way: do nothing! So $0! = 1$.[/quote]

:(

It looks better! Really! When you say $1=0!$

lol it looks like 1=0! (! as in exclamation point)

but yea there is only one, which is {}. So 1c0 = [b]1[/b]

[quote=MWZ]basically, just think of how many ways can you pick n objects from n things. obviously, the only way to do this is to just pick all of them, since order doesn't matter, giving us 1 way to pick them.[/quote]

Yes there's only $1$ way to chose $n$ things from $n$.

Going back to the whole 0!=1 thing to get from n! to (n-1)! divide by n
1!/1=0!=1

[quote=forester2015][hide=Post #15 by Eat314][url=aops.com/community/user/553497][b]Eat314[/b][/url] · Feb 17, 2021, 9:19 AM [url=aops.com/community/p20442211](view)[/url][color=transparent]helo[/color]
It was posted by GameBot. It's an Alcumus problem.

-----------
[color=#5b7083][aops]x[/aops] 15[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]

[tip=@Eat314][img]https://avatar.artofproblemsolving.com/avatar_553497.jpg?t=1613232580[/img]
AoPS User[/tip] no 100% of the problems say this is linked to a item on aops[/quote]

o l d p o s t s g o b r r r r

[quote=floof421]ernie's solution basically proves $0!=1$ too[/quote]

[hide=easier way to prove that 0!=1]$3!=2!\times 3$ so $2!=\frac{3!}{3}=2$ so $1!=\frac{2!}{2}=1$ and then $0!=\frac{1!}{1}=1$[/hide]

this is getting kinda off topic


Problem 1:
In a triangle $ ABC$ right-angled at $ C$, the median through $ B$ bisects the angle between $ BA$ and the bisector of $ \angle{B}$. Prove that
$ \frac{5}{2} <\frac{AB}{BC}< 3.$


So Chapter is a couple weeks away and I was wondering if a sudden change in diet would help. Like eating more healthy foods and consuming nutritious fats like omega-3, Vitamin C, etc. Also what breakfast do all of you eat before you take a math competition?

Please list and describe here

Thanks in Advance

Well, the healthier you are, the better you will focus. That might give you a boost in the competition. :P

Before competitions in general, I find that the less I eat, the better I do...(like once I ate nothing for breakfast and did really well for some reason: good focus, etc.). Better advice would probably be to eat a normal breakfast.

Heh don't eat a lot of food. I ate a lot before the AMC 8 in seventh grade and I got lower than my sixth grade score. Maybe that's because I fail.

Don't change too much; try to do everything as normally as possible the day of chapter.

yea I read somewhere that people can focus better when they are hungry...


Let $ a,b,c,d,e\geq 0$,prove that

1.$ 32(a \plus{} b \plus{} c)(b \plus{} c \plus{} d)(d \plus{} e \plus{} c)(e \plus{} a \plus{} d)(a \plus{} b \plus{} e)$
      $ \geq 243(a \plus{} b)(b \plus{} c)(c \plus{} d)(e \plus{} d)(a \plus{} e).$

2.$ (a \plus{} b \plus{} c)(b \plus{} c \plus{} d)(d \plus{} e \plus{} c)(e \plus{} a \plus{} d)(a \plus{} b \plus{} e)$
$ \geq 4(b \plus{} e)(a \plus{} d)(b \plus{} d)(a \plus{} c)(c \plus{} e) \plus{} 115abcde.$
BQ

For the first, a interesting generalization
for $ n\geq  l\geq  r$ , $ \prod _{j\equal{}1}^n \sum _{i\equal{}1}^l x_{((i\plus{}j)( \bmod n))\plus{}1}\geq \left(\frac{l}{r}\right)^n \prod _{j\equal{}1}^n \sum _{i\equal{}1}^r x_{((i\plus{}j)( \bmod n))\plus{}1}$

[quote="EulerIntegral"]For the first, a interesting generalization
for $ n\geq l\geq r$ , $ \prod _{j \equal{} 1}^n \sum _{i \equal{} 1}^l x_{((i \plus{} j)( \bmod n)) \plus{} 1}\geq \left(\frac {l}{r}\right)^n \prod _{j \equal{} 1}^n \sum _{i \equal{} 1}^r x_{((i \plus{} j)( \bmod n)) \plus{} 1}$[/quote]

Your good results
BQ

[quote="EulerIntegral"]For the first, a interesting generalization
for $ n\geq l\geq r$ , $ \prod _{j \equal{} 1}^n \sum _{i \equal{} 1}^l x_{((i \plus{} j)( \bmod n)) \plus{} 1}\geq \left(\frac {l}{r}\right)^n \prod _{j \equal{} 1}^n \sum _{i \equal{} 1}^r x_{((i \plus{} j)( \bmod n)) \plus{} 1}$[/quote]
Interesting, but false in general if the $ a_i$ are not a monotonous sequence :(

see here
http://www.mathlinks.ro/Forum/viewtopic.php?t=75130 (e.g. post #24)
http://www.mathlinks.ro/Forum/viewtopic.php?t=26103

How lame!I Assumed monotonous sequence for some calculations and got false positive.
Apparentelly $ \prod_{j \equal{} 1}^{n}\sum_{i \equal{} 1}^{l}x_{((i\plus{}j)(\bmod n))\plus{}1}\geq\left(\frac{l^C}{r^D}\right)\prod_{j \equal{} 1}^{n}\sum_{i \equal{} 1}^{r}x_{((i\plus{}j)(\bmod n))\plus{}1}$
And C,D varies randomlly (a priori)


An equilateral triangle of side $ n>2$ is divided into $ n^2$ equilateral triangles of side $ 1$ with the lines parallel to the sides of the triangle. Can one label these unit triangles with numbers from $ 1$ to $ n^2$ such that

(a) The triangles with numbers $ i,i\plus{}1$ have at least one common point for all $ i\equal{}1,2,...,n^2\minus{}1$

(b) The triangles with numbers $ i,i\plus{}2$ have at least one common point for all $ i\equal{}1,2,...,n^2\minus{}2$?

This question has appeared and been solved recently ... try a search.  The trick is simply to note that the corners must be filled by $ 1, 2, (n \minus{} 1)$ and $ n$, but that these four numbers can only actually fill 2 of the corners.


Is there a formula to find the area of a regular pentagon if you know one side?
How about one for an n-gon?

[hide="pentagon"]here is how to find the area of pentagon ABCDE and side length s:
first, we draw AD and AC.  If we draw altitude EX, $\angle DEX=54^\circ$.  Thus, $[EXD]=\frac{s\cdot EXcdot\sin{54}}{2}=\frac{s^{2}\sin{54}\cos{54}}{2}=\frac{s^{2}\sin{108}}{4}$.  Since there are four such triangles, the total of them is $s^{2}\sin{108}$.  now, we look at $\triangle DAC$.  first, drop altitude AY.  $\angle ADY=108-\angle EDA=108-36=72^\circ$.  Thus, $[DAY]=\frac{DY\cdot AY}{2}=\frac{s^{2}\tan{72}}{8}$.  There are two of these, so the total is $\frac{s^{2}\tan{72}}{4}$ and the grand total area is $\frac{s^{2}\tan{72}}{4}+\sin{108}$[/hide]
somebody tell me if I made a mistake- I don't have time to check it over :lol:

Well, if you connect each vertex to the center you end up with n sectors of the polygon, each an isosceles triangle with central angle $2\pi/n$; then split each into two right triangles, with angle $\pi/n$. Given side length $x$, the height is simply $\frac{x}{2}\cot(\frac{\pi}{n})$, and the total area is $n\frac{bh}{2}$ or $\boxed{\frac{nx^{2}}{4}\cot(\frac{\pi}{n})}$.

wow thats a lot easier than I thought it would be :o  :D

[quote="Quevvy"]Is there a formula to find the area of a regular pentagon if you know a side length?
Is there a formula to find the area of a regular n-gon if you know the side length?[/quote]
[url]http://en.wikipedia.org/wiki/Polygon[/url]

It's somewhere there, but it involves trig.

quevvy actually posted the same problem in the olympiad section and somebody answered it.  its actually pretty easy (if you know trig) but its way too advanced for MOEMS.  it basically consists of dividing the n-gon into n isosceles triangles and using trig to find the area of those triangles

[quote="jimbo"]an isosceles triangle with central angle $2\pi/n$; [/quote]
Wouldn't the central angle be $360/n$?
And then the right triangles would have a central angle of $180/n$?
I'm not sure but if the central angle was $2\pi/n$, then the sum of the measure of the central angles would only be $2\pi$. :?:

Quevvy, $\frac{2\pi}{n}$ is in radians. There are $\pi$ radians in a semisircle, so it's equal to $\frac{2\pi (rad)}{n}(\frac{180^\circ}{\pi rad}) = \boxed{\frac{360^\circ}{n}}$.


Estimate the time it takes to send $ 60$ blocks of data over a communication channel if each block consists of $ 512$ "chunks" and the channel can transmit $ 120$ chunks per second.

$ \textbf{(A)} .04 \text{ seconds} \qquad \textbf{(B)} 0.4 \text{ seconds} \qquad \textbf{(C)} 4 \text{ seconds} \qquad \textbf{(D)} 4 \text{ minutes}\qquad \textbf{(E)} 4 \text{ hours}$

[hide="My Solution"]
You can use conversion factors to make it quick and easy. First, we convert from chunks per second to blocks per second.

$ \frac{1 \text{ block}}{512 \text{ chunks}} \cdot \frac{120 \text{ chunks}}{1 \text{ seconds}} \equal{} \frac{15 \text{ blocks}}{64 \text{ seconds}}$

Now we simply multiply both parts of the conversion factor by $ 4$ (to get $ 60$ blocks in the numerator) and find that it takes $ 64\times 4 \equal{} 256 \text{ seconds}\approx (D) 4 \text{ minutes}$.
[/hide]


calculus
limit
trigonometry
function
algebra
domain
real analysis

$\lim_{x\to 0} x \sin (\frac{1}{x}) = 0$

With that piece of knowledge, the limit is easily determined so the limit is defined.

Edit:
This is a discussion.

The limit is defined when $\lim_{x\to 0-} \frac{\sin (x \sin (\frac{1}{x}))}{x \sin \frac{1}{x}} = \lim_{x\to 0+} \frac{\sin (x \sin (\frac{1}{x}))}{x \sin \frac{1}{x}}$.

This means that when you zoom in to near $0$ values from both sides of $0$ and they yield the same value, the limit is defined for $0$.

It doesn't matter that $\frac{0}{0}$ is undefined, you only care about approaching that function to the desired value.

It also doesn't matter that you are letting $\frac{1}{x}$ approach $0$ and are using $\sin (\infty)$ (sorry for the poor terminology, but I'm trying to make an argument), because $-1 \le \sin y \le 1$.

I hope this helps.

FMako, where, exactly, is this function defined? You talk of approaching zero from both sides. Is that the only issue?

If $f (x) = \frac{\sin (x \sin \frac{1}{x})}{x \sin \frac{1}{x}}$

Then $f(x)$ is undefined when $x = 0$ or $x = \frac{1}{k \pi}$ for non zero integer $k$.

You want to approach $0$ on both sides for near $0$ values.

Discontinuities are the only thing that might make a limit undefined (I guess I consider \pm \infty defined).

If there is a jump (like in a step function) or a singularity (like in tangent), then problems might arise when you take the limit in some places.

In order to be sure that a limit is defined you need to ensure that the left and right hand limits are equal.

I think that's the only issue.  At this point in time, now that I'm past the precise definition of a limit, I am probably too colloquial, so correct me if I'm wrong.

Thanks, I hope this helps.

This thing approaches 1 quickly. Even though there are small disturbances from $sin(1/x)$ they are killed quickly by x approaching zero (so this is pretty much like $\sin(x)/x$). To each $\epsilon > 0$ you will be able to choose a $\delta > 0$ so that $|f(x) - 1| < \epsilon$ as soon as $|x - 1| < \delta$  :P I don't see where the potential problem is.

I changed my mind. This function is of the type 0/0 infinitely many times around zero. So, while I do believe you could make a definition where it has limit 1, it currently does not since we cannot choose an entire domain around 0 where the function is defined.

edit: beaten. I thought of this while I went away showering. :P

[quote="Kent Merryfield"]This is a question about conventions and the meaning of certain symbols, and not about substance. The substance of this is plain enough. Assume the level of discussion to be that of a calculus class.

Would you say that
\[ \lim_{x\to0}\frac{\sin\left(x\sin\left(\frac1x\right)\right)}{x\sin\left(\frac1x\right)}  \]
exists?[/quote]

No.  The problem is that it's undefined somewhere on every neighborhood of zero.  So even though it is well behaved in the sense that it doesn't blow up, we can't just define away all the places where the denominator vanishes.

I happen to have a mainstream calculus book - Stewart, 5e edition, to be specific - open on my desk. What's in there? Stewart does the definition of limit twice; once less formally, once more formally.

[quote="Stewart, section 2.2,"]We write
\[ \lim_{x\to a}f(x)=L \]
and say "the limit of $f(x)$ as $x$ approaches $a,$ equals $L$"

if we can make the values of $f(x)$ arbitrarily close to $L$ (as close to $L$ as we like) by taking $x$ to be sufficiently close to $a$ (on either side of $a$) but not equal to $a.$[/quote]
There's that emphasis on "both sides" that was also present in FMako's posts above. But what about our case here where we have the function on both sides - but not in a whole neighborhood on either side? This is an attempt at intuition, and the intuition isn't enough to decide my example.

[quote="Stewart, section 2.3,"]Let $f$ be a function defined on some open interval that contains the number $a$ except possibly at $a$ itself. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write
\[ \lim_{x\to a}f(x)=L \]
if for every number $\epsilon>0$ there is a number $\delta>0$ such that $|f(x)-L|<\epsilon$ whenever $0<|x-a|<\delta$[/quote]
This formalization of the definition (and it's a very common version) now agrees with what blahblahblah said - and since my function fails to be defined in a punctured neighborhood of zero, it can't have a limit there.

So that position - that this limit doesn't exist - is a mainstream position, justified by numerous sources.

I don't like it. I would prefer to say that this limit exists. I'd like to agree with Kalle before he changed his mind. Here's a possible defintion that would allow it to be true:

Suppose $a$ is a limit point of the domain of $f$. Then $\lim_{x\to a}f(x)=L$ iff $\forall\,\epsilon>0\,\exists\,\delta>0$ such that if $|x-a|<\delta$ and $x\in\text{dom}(f)\setminus\{a\},$ then $|f(x)-L|<\epsilon.$

Of course, with the notion of limit point in there, this would be impossible to present at the level of a first calculus course.

Why would this be impossible to present? Am I missing the depth? Isn't that the normal definition except that you don't require $f(x)$ to be defined in the entire interval around the point?

Will this alternative definition affect anything? I thought maybe it screws with continuity, but I can't think of a reason why. Why isn't the "normal" definition like this?

The problem is that limit point. That's why Kent said it would be impossible to present this in first year calculus course. We need to know the definition of limit point that is in any Analysis Books. Blahblahblah said that the limit doesn't exist because the function can be undefined in all neighborhood of zero (more generally according to Kent's definiton the limit point $a$), so if we remove $a$ from domain its posible for the limit to exist.

I'm so impressed with that definition Dr. Kent  :o

actually, my "usual" definition of limit has to be taken for an accumulation point of the set you're considering (i mean.. there exist a non constant sequence in the domain tending to that point), so i think it's left to the reader to say that the limit has to be taken _inside_ the domain...
otherwise you should define everywhere else that function (i guess 1 everywhere)...

the problem is $sin(1/x)$ occilates as x approachs 0 because $1/x$ gets very large very fast.

i would say yes because $sin(u)$ is always bounded by 1 and negative 1, and 0 times anything except infinity is 0.

$sin(\frac{1}{x})$ will occilate, and the occilation will creshedo closer and closer to zero until the two sides meet at zero.

...but im not sure.  if the limit does exsit, should it be differentiable?  the two seem to go hand in hand.

i think $sin(1/x)$ is undefined, so i'll use $sin(a)$.  $sin(a) \neq sin(-a)$, the $lim_{x\to0} x sin(a) = lim_{x\to0} x sin(-a)$ but for n is incredibly small, $nsin(a) \neq nsin(-a)$

the next points on either side will act differently, and have different slopes.  is this a necessary condition?

What you are trying to do, oooooh ya?
It is clear that the limit doesn't exist by the formal definition of limit. Although $\lim_{x \rightarrow 0} x \, \sin \frac{1}{x} = 0$ (can be proved by pinching theorem), But it approach zero with "not nice' behaviour, it has many wiggle before approach zero.
Now how about $\frac{\sin x \, \sin \frac{1}{x} }{x \sin \frac{1}{x}}$, what happen to this function in neighborhood of zero. It also wiggle when x very close to zero, and as Dr.Kent and blahblahblah said it can be undefined near there.

Dr.Kent's effort to make this limit exist is by "remake" new definition by removing that point which is limit point.

Other question:
Is Dr. Kent' s "remake" definiton can also be used in usual limit ?

Actually, while studying limits again in front of an upcoming exam (we have an exam only on theory and proofs), I realize that the limit definition in my book is precisely what Kent wrote. The function doesn't have to be defined in an entire interval around the point in which we take the limit; i.e. we only look at points for which the function is actually defined.

On the other hand, the book wouldn't immediately say this limit is 1, since the proof for the theorem that if $f(y) \to A$ when $y \to b$ and if $g(x) \to b$ when $x \to a$ then $f(g(x)) \to A$ as $x \to a$ requires that if f is not continuous in $b$, we must be able to choose an entire interval around $a$ where $g(x) \ne b$.

However, I don't think it matters in this particular case, since $f$ is actually not defined in $b$ (i.e. 0). Do you think I have this sorted out correctly?  :P

Yes you're right Kalle.
Actually you said about the condition of [b]coverges properly[/b] in subtitution rule of limit.
Definition:
We say that $f(x)$ [b]converges properly to L as [/b] $x \rightarrow a$ iff \[ \lim_{x \rightarrow a} f(x) = L  \] and there exist a deleted neighborhood $N$ of $a$ such that $f(x) \not= L$ whenever $x \in N$.

In Subtitution Rule like you said above, the prove require $g(x)$ to be converges properly to $b$.

Though $\lim_{x \rightarrow 0} x \sin \frac{1}{x} = 0$ but the function $x \sin \frac{1}{x}$ does not converge properly to zero, since it assumes the value zero infintely many times. That's why our limit in this thread can't be exist, since the denumerator assume value zero infinitely often in some neighborhood of zero the function is undefined there.

Yes, but the thing is that if you make the definition of limits in a way where you only consider points in the domain of the function, there will be no problem since we will never have these points where "$x \sin \frac{1}{x} = 0$", since they are not considered. If you look at the proof there will be no problem.

Proof that with Kent's definition the limit is 1:

Let $\epsilon > 0$. Let $f(y) = \frac{\sin y}{y}$. We have that $\epsilon_1 > 0$ exists so that $|f(y) - 1| < \epsilon$ whenever $0 < |y - 0| < \epsilon_1$ and $y \in D_f$. But if we let $y = x \sin \frac{1}{x}$ we have that there exists $\delta$ so that $0 < |x \sin \frac{1}{x} - 0| < \epsilon_1$ whenever $0 < |x - 0| < \delta$, $x$ is in the domain of $y(x)$ and x is such that $f(x \sin \frac{1}{x})$ is defined*. This shows that $|f(x \sin \frac{1}{x}) - 1| < \epsilon$ whenever $0 < |x - 0| < \delta$ and x is in the domain of $f(x \sin \frac{1}{x})$.

*The key is that $x \sin \frac{1}{x} = 0$ precisely when $f(x \sin \frac{1}{x})$ is undefined.

As I said, Kent's definition is precisely what I learned in my introductory math course this semester (I'm nearly finished with my first semester of university), so surely it's not impossible to present like Kent said. :)


This has been disussed millions of times.

Yet wasn't found on the first few pages of this forum?

http://rapidshare.de/files/15124157/Mathcounts.rar.html
Thats the website with a lot of National Tests. You need WinRAR to vieew them.

you should go onto
http://www.freewebs.com/tzpublicsite

its gots loads of problems

Where are they? I only see links to home, guestbook, helpful sites and calc programs.

The other members of my state team were given the 1990 national competition to finish by friday, but my coach, who is also the state coach, doesnt have it. I know about all the files from 1991 up. Can anybody help me get a 1990 file?

[quote="Ignite168"]Well, I qualified for Nationals this year and we'll be doing practice on Nationals Problems from 1990, 1996, and 1999-2005. Does anyone know where to get problems from 1991-1995, and 1997-1998? Thanks.[/quote] Do you have a scanner? If you do could you scan and send me 1990 and 1996's sprints? Thanks

Can you upload them onto this site or another so I can put on the public site?

grr i have the first 5.
Maybe you can ask your coach...

[quote="Ignite168"]I don't have them yet, we get one every week.
Here's a list of when I get rounds if you want it
3/14 - 1999
3/23 - 2000
3/30 - 2001
4/06 - 2002
4/11 - 2003
4/17 - 1990
4/27 - 1996
5/04 - 2004
5/09 - 2005

I can type it up for you when I have time[/quote]
Or if you have a scanner...


algebra
polynomial
calculus
integration
superior algebra
superior algebra solved

In advance let me apologize if this question is stupid or well-known. I am not very experienced with algebra.

As far as I can tell, one way we can look at Hilbert's Nullstellensatz is that if I have a polynomial $p(x, y)$ over an algebraically closed field with all of its prime factors occurring only once, then any polynomial whose variety contains the variety of $p(x, y)$ is divisible by $p(x, y)$.

In general, we can't really say anything when it's not an algebraically closed field. But suppose we have a commutative ring $R$ that's an integral domain. Consider the ring of polynomials over $R$ and suppose that $p(x, y)$ whose prime factors all occur only once. What information about the variety of $p(x, y)$ [i]in $R$[/i] will be enough to determine from the variety of $g(x, y)$ whether $p(x, y) \mid g(x, y)$?

For example, let $R$ be the reals. If $p(x, y) = x^{2}+y^{2}+1$, then the variety is empty. But it doesn't divide every other polynomial.

Even worse, the polynomial $q(x, y) = x^{2}+y^{2}$ has nonempty variety; it is a point. But $g(x, y) = x^{4}+y^{4}$ has the same variety, and they are not multiples of each other.

On the other hand, if the polynomial is $ax+by+c$, then our information from $\mathbb{R}$ is in fact sufficient, even though there are solutions to this in the algebraic closure of $\mathbb{R}$ that aren't in $\mathbb{R}$ itself. My intuition is that it's about the "dimension" of the variety, in some sense. We want "enough" of the variety to be in the original ring that we don't get any useful information by including the solutions from the algebraic closure.

You should look at the solutions of some equation $p(x,y)=0$ over the algebraic closure over $k$ and apply Nullenstellensatz there.

$x^{2}+y^{2}+1=0$ may give the empty variety over $\mathbb R$, but not over $\mathbb C$. Actually every non-degenerate quadric is isomorphic to this over $\mathbb C$.

$x^{2}+y^{2}=0$ may give the zero set over the reals, but is not irreducible over $\mathbb C$ being the union of the lines $x\pm i\cdot y=0$.

When I saw the title of this thread, I thought it had been started by Arthur Eddington :D

[quote="Xevarion"][quote="amfulger"]You should look at the solutions of some equation $p(x,y)=0$ over the algebraic closure over $k$ and apply Nullenstellensatz there.[/quote]
It sounds like you're saying "it's pointless to try to get information of this kind by looking at the solutions to $p(x, y) = 0$ in $R\times R$. Instead, you might as well turn it into a field, where you can tell everything using the Nullstellensatz." Do I understand you right?[/quote]

And I don't understant exactly what you say, but if you mean that I am looking in solutions in $\mathbb C\times\mathbb C$ instead of $\mathbb R\times\mathbb R$, then you are right.


Make 2y=x .Solve x that is very easy and calculate y

[quote="oryxncrake"]Hello,

Can anyone show me the steps for solving the following equation for y, where 0 is less than or equal to y and y is less than 360 degrees?

sin2y = cos4y

Thank you[/quote]

[hide="solution?"]
$ sin2y\equal{}cos4y$
$ cos(\frac{\pi}{2}\minus{}2y)\equal{}cos4y$
$ 4y\equal{}2k\pi\plus{}\frac{\pi}{2}\minus{}2y$,   $ k$ is integer
$ y\equal{}\frac{k\pi}{3}\plus{}\frac{\pi}{12}$,    $ k$ is integer
because $ y$  is in $ [0,2\pi]$
we have $ y\equal{}\frac{\pi}{12}$,$ \frac{5\pi}{12}$,$ \frac{3\pi}{4}$,$ \frac{13\pi}{12}$,$ \frac{17\pi}{12}$ or $ \frac{7\pi}{4}$
[/hide]

[quote="oryxncrake"]Thanks. But how did you get cos(pi/2 - 2y) from sin2y?[/quote]

This is complementary angles, with a right angled triangle with sides a, b and c and and angles theta and 90 - theta you can work out that

$ \sin \theta \equal{} \cos\left(90 \minus{} \theta\right)$

$ \cos \theta \equal{} \sin\left(90 \minus{} \theta\right)$

$ \sec \theta \equal{} cosec\,\left(90 \minus{} \theta\right)$

$ cosec\, \theta \equal{} \sec\left(90 \minus{} \theta\right)$

$ \tan \theta \equal{} \cot\left(90 \minus{} \theta\right)$

$ \cot \theta \equal{} \tan\left(90 \minus{} \theta\right)$

[quote="greenhand"]
$ \sin2y \equal{} \cos4y$
$ \cos(\frac {\pi}{2} \minus{} 2y) \equal{} \cos4y$
$ 4y \equal{} 2k\pi \plus{} \frac {\pi}{2} \minus{} 2y$,   $ k$ is integer
$ y \equal{} \frac {k\pi}{3} \plus{} \frac {\pi}{12}$,    $ k$ is integer
because $ y$  is in $ [0,2\pi]$
we have $ y \equal{} \frac {\pi}{12}$,$ \frac {5\pi}{12}$,$ \frac {3\pi}{4}$,$ \frac {13\pi}{12}$,$ \frac {17\pi}{12}$ or $ \frac {7\pi}{4}$[/quote]

Good solution, except you forgot to add in the solution $ 4y\equal{}2k\pi \minus{}\frac{\pi}{2}\plus{}2y$ which gives

$ y\equal{}k\pi\plus{}\frac{\pi}{4}$ or $ y\equal{}\frac{\pi}{4}, \frac{5\pi}{4}$.  The reason you can do this is because $ \cos x\equal{}\cos (\minus{}x)$.


LaTeX
number theory
greatest common divisor
number theory proposed

Apologies dont know how to use latex.
Se the attached image giving proof for FLT n=3
I think Iam missing something, can someone tell me what??

[quote="darkside102"]$ c((a+b)^2-3ab)=k^2(a+b)$

either $ a+b\mid c$ or $ a+b\mid 3ab$.[/quote]
This is not necessarily true. Here you can only conclude that $ a+b\mid 3abc$.

And to write in $ \text{\LaTeX}$ all you need to do is type the $ \text{\LaTeX}$ codes in between a pair of dollar signs.

Reposting in Latex thanks to [b]Shoki[/b]
$ a^{3}\plus{}b^{3}\equal{}c^{3}$ where $ a,b,c \geq 0$ and $ a,b,c$ have no common factors.
$ (a\plus{}b)^{3}\minus{}c^{3}\equal{}3ab(a\plus{}b)$
Hence, $ a\plus{}b \geq c$
$ a^{3}\plus{}b^{3}\equal{} (a\plus{}b)(a^{2}\minus{}ab\plus{}b^{2})\equal{}c^{3}$
[u][b]Case I[/b][/u]
$ (a\plus{}b)\mid c$
But $ a\plus{}b \geq c$
Therefore $ a\plus{}b\equal{}c$
$ \Rightarrow a$ or $ b \equal{} 0$
[u][b]Case II[/b][/u]
$ (a\plus{}b)\mid c^{2}$
Let $ c^{2}\equal{}k(a\plus{}b) \Rightarrow c^{4}\equal{}k^{2}(a\plus{}b)^{2}$
$ c^{4}\equal{}c(a\plus{}b)(a^{2}\minus{}ab\plus{}b^{2})\equal{}k^{2}(a\plus{}b)^{2}$
$ \Rightarrow (a\plus{}b)\mid 3abc$
Now gcd $ (a,b)\equal{}1 \Rightarrow$ gcd $ (a\plus{}b,ab)\equal{}1$
$ \Rightarrow (a\plus{}b)\mid 3c$
But $ a\plus{}b \geq c$
$ 3\geq \frac{3c}{a\plus{}b}$
Therefore $ \frac{3c}{a\plus{}b}$ can only take values 0,1,2 or 3.
None of the above 4 equations when solved with $ a^{3}\plus{}b^{3}\equal{}c^{3}$ will give all positive integer solutions $ a,b,c$
$ \Rightarrow a$ or $ b \equal{} 0$
[u][b]Case III[/b][/u]
$ (a\plus{}b)\mid c^{3}$ and $ a\plus{}b>c^{2}$
$ c^{3}\equal{}a^{3}\plus{}b^{3}\geq ab(a\plus{}b) \geq abc^{2}$
$ \Rightarrow c \geq ab$
But $ a\plus{}b \geq c  \Rightarrow a\plus{}b\geq ab$
So either $ a\equal{}0$ or $ b\equal{}0$

I've not carefully read your text but it seems your three cases I, II and III are not the only cases.

For example, you should study $ a\plus{}b|c^3$ with $ a\plus{}b<c^2$ (for example $ a\plus{}b\equal{}p^3$ with $ p$ prime and $ c\equal{}pn$ with $ p<n^2$

[quote="darkside102"]Reposting in Latex thanks to [b]Shoki[/b]
$ a^{3} \plus{} b^{3} \equal{} c^{3}$ where $ a,b,c \geq 0$ and $ a,b,c$ have no common factors.
$ (a \plus{} b)^{3} \minus{} c^{3} \equal{} 3ab(a \plus{} b)$
Hence, $ a \plus{} b \geq c$
$ a^{3} \plus{} b^{3} \equal{} (a \plus{} b)(a^{2} \minus{} ab \plus{} b^{2}) \equal{} c^{3}$
[u][b]Case I[/b][/u]
$ (a \plus{} b)\mid c$
But $ a \plus{} b \geq c$
Therefore $ a \plus{} b \equal{} c$
$ \Rightarrow a$ or $ b \equal{} 0$
[u][b]Case II[/b][/u]
$ (a \plus{} b)\mid c^{2}$
Let $ c^{2} \equal{} k(a \plus{} b) \Rightarrow c^{4} \equal{} k^{2}(a \plus{} b)^{2}$
$ c^{4} \equal{} c(a \plus{} b)(a^{2} \minus{} ab \plus{} b^{2}) \equal{} k^{2}(a \plus{} b)^{2}$
$ \Rightarrow (a \plus{} b)\mid 3abc$
Now gcd $ (a,b) \equal{} 1 \Rightarrow$ gcd $ (a \plus{} b,ab) \equal{} 1$
$ \Rightarrow (a \plus{} b)\mid 3c$
But $ a \plus{} b \geq c$
$ 3\geq \frac {3c}{a \plus{} b}$
Therefore $ \frac {3c}{a \plus{} b}$ can only take values 0,1,2 or 3.
None of the above 4 equations when solved with $ a^{3} \plus{} b^{3} \equal{} c^{3}$ will give all positive integer solutions $ a,b,c$
$ \Rightarrow a$ or $ b \equal{} 0$
[u][b]Case III[/b][/u]
$ (a \plus{} b)\mid c^{3}$ and $ a \plus{} b > c^{2}$
$ c^{3} \equal{} a^{3} \plus{} b^{3}\geq ab(a \plus{} b) \geq abc^{2}$
$ \Rightarrow c \geq ab$
But $ a \plus{} b \geq c \Rightarrow a \plus{} b\geq ab$
So either $ a \equal{} 0$ or $ b \equal{} 0$[/quote]

in the last case why did u assume $ c^2 < a\plus{}b$?
thx in advance for your answer :)

As said by pco ,I think darkside should try proving for the case $ a \plus{} b \mid c^{3}$ with $ a \plus{} b < c^{2}$ and $ a \plus{} b \nmid c^{2}$...@pco this is the last case right ?...are there any more cases to be considered ?

this problem i was trying from many months in my leisure and came with a proof with the exception of $ a\equal{}3p\plus{}1 ,b\equal{}3q\plus{}2,c\equal{}3r$ ...if any one could solve this case i would post my entire solution...(it is useful to apply fermats little theorem  :wink: ).
                           thanks

@dinesh can you please post your proof -it is ok if that case is left.


abstract algebra
invariant
superior algebra
superior algebra unsolved

Suppose A and B are free modules over a ring w/ property of invariant dimension.  Then rank A + rank B = rank of the direct sum between A and B.

this is so obvious. if $X$ is a basis of $A$ an $Y$ is a basis of $B$, then $X \cap Y = \emptyset$ and $X \cup Y$ is a basis of $A \oplus B$.


A guitar player tunes the fundamental frequency of a guitar string to 440 Hz. What will it be if, instead, she decreases the length along which the string oscillates by sliding her finger from the tuning key one-third of the way down the string toward the bridge at the lower end?

Why isn't it:

$ f_{new} \equal{} \frac{nv}{2(\frac{L}{3})}\equal{}3 \left(\frac{nv}{2L}\right) \equal{} 3 \times 440 \equal{} 1320Hz$ ?

Thanks

I read that as shortening the string to $ \frac23$ of its original length, in which case the frequency would be $ \frac32\cdot440\equal{}660\text{Hz},$ which would be an $ E.$ ($ 1320\text{Hz}$ would also be an $ E,$ but an octave higher.)

As a side note: a guitar does not contain a string which is normally tuned to A440; the problem describes an abnormal tuning. You can find a string normally tuned to A440 on violin or viola. An on that violin or viola, if you stop the string against the fingerboard $ \frac13$ of the way down, you hear an E660; if you touch the string lightly at the same point what you hear (a "harmonic") sounds like E1320.

A question: What's the frequency of a harmonic compared to the normal frequency? (One octave higher as in your example?) The only thing I think I know is that instead of one vibration of the whole string, you have 2,3,... of smaller, equal parts of the string... However, my physics has always been rubbish, so I don't know what to make of that :blush:

Let $ \nu_0$ be the fundamental frequency of the unaltered ("open") string. The actual behavior of the string is that of a superposition (that is, linear combination) of oscillations of all of the frequencies $ k\nu_0$ for $ k \equal{} 1,2,3,\dots.$ (Look at the solution of the one-dimensional wave equation with on an interval with fixed ends to understand this.) Your brain tends to interpret this superposition as having a definite pitch equal to the fundamental frequency, but that is really an act of cognitive processing.

If we clamp the string to the fingerboard with a finger, we simply shorten the string. (On a violin/viola/cello, the finger itself becomes the new endpoint of the string; on a fretted instrument, we bring the string down onto the next fret, which becomes a new bridge.) The string now has a length of $ pL_0$ for some $ 0 < p < 1;$ the resulting vibrations are a superposition of frequencies $ k\nu$ for $ k \equal{} 1,2,3,\dots$ and $ \nu \equal{} \frac {\nu_0}{p}.$ Again, cognitive processing identifies this as the pitch $ \nu.$ For the violin family, there are no restrictions on the fraction $ p,$ which may vary continuously; for the guitar, $ p$ is restricted to the discrete values represented by the positions of the frets.

Now, let's talk about harmonics. We touch the string lightly with the finger at the chosen position, but do not clamp the string to the fingerboard. This does not shorten the string - it vibrates over its whole length, on both sides of the finger. What the lightly placed finger does is create an extra resistive damping force which resists motion in that spot. Any vibratory mode that has a significant amplitude at that location will be damped out rapidly. If we lightly place the finger at a more or less random location, we will damp out most or all of the most important (that is, lower-frequency) modes and we will hear very little sound. However, if we place the finger at a distance $ \frac {m}{n}$ of the way down the string, where $ m$ and $ n$ are relatively prime and $ n$ is a small integer, then we damp out most of the frequencies but leave relatively undisturbed frequencies of the form $ kn\nu_0,$ since those oscillations have nodes at the point we're touching. Cognitive processing then identifies the note as having pitch $ n\nu_0.$

If we touch the midpoint of the string ($ n \equal{} 2$), we hear the pitch $ 2\nu_0,$ which is an octave above the fundamental. If we clamped the string at that point, we would also hear pitch $ 2\nu_0,$ although with a recognizably different timbre.

If we touch the string $ \frac13$ of the way down, we hear the pitch $ 3\nu_0,$ which is an octave and a fifth above the fundamental. Clamping the string at that point gives us the pitch $ \frac32\,\nu_0,$ which is just a fifth above the fundamental, without the octave.

If we touch the string $ \frac14$ of the way down, we hear the pitch $ 4\nu_0,$ which is two octaves above the fundamental. Clamping the string at that point gives us the pitch $ \frac43\,\nu_0,$ which is just a fourth above the fundamental.

On a decent instrument we ought to be able to make $ \frac15$ (or $ \frac25$ ) work, for a pitch of $ 5\nu_0,$ or two octaves and a major third. That's probably the practical limit. By far the most common harmonic is the $ n \equal{} 2$ simple octave case, followed by the $ n \equal{} 3$ case. (And then there are "synthetic harmonics" created by stopping the string with one finger and touching it with another. That's mostly used for show-off solo pieces.)

I've been assuming a bowed-string instrument with relatively low-tension strings. I don't think this would work as well with a plucked instrument or with the higher-tension strings of a guitar, although it would still be theoretically possible.

-----

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1156763638&t=132213]this topic[/url] of mine.


A and B are two points on the motor way, and they lie 75 km apart. Two cars pass point A at 10 pm and point B at 11 pm. Show that the cars must have had exactly the same speed at atleast one point in the time interval.
And: How much information do you actually need to be able to solve this task?

Thanks!

Hi. I'd use Rolle's theorem (see here: http://en.wikipedia.org/wiki/Rolle's_theorem ) .

Let $ f(t)$ and $ g(t)$ be the distances travelled by the cars in time $ t$, $ \ 0 \le t \le 1$, so that their speed are $ f'(t)$ and $ g'(t)$.

$ f(0) \equal{} g(0) \equal{} 0$ and $ f(1) \equal{} g(1) \equal{} 75$.

If we let $ h(t) \equal{} f(t) \minus{} g(t)$ then $ h(0) \equal{} h(1) \equal{} 0$

and so by Rolle's theorem, there exists $ a$, with $ 0 < a < 1$, such that $ h'(a) \equal{} 0$,

that is $ f'(a) \minus{} g'(a) \equal{} 0$,

that is $ f'(a) \equal{} g'(a)$.

The actual times they pass $ A$ and $ B$, provided they're the same times, are not needed; nor is the actual distance between $ A$ and $ B$.


vector
algorithm
linear algebra
matrix
algebra
system of equations

Hi. First of all, i want to apologize if don't express myself correctly, that's because english is not my native language.

So... i need to find, for a vector its linear combination of other vectors
ex:
x=(-1,10,-3)

v1=(1,2,3)
v2=(1,-2,3)

x = c1*v1 + c2*v2

How can i compute c1 and c2?

So far, i tried to find the orthogonal basis for v1 and v2 using Gram Schmidt algorithm and compute and normalize orthogonal vectors v1 and v2 to w1 and w2:
w1=(0.26, 0.53, 0.80)
w2=(0.16, -0.84, 0.50)

Then i wrote x as a linear combination of w1 and w2. x=2.67*w1 - 10.14*w2.
But, the question is, how about writing x as a linear combination of initial vectors v1 and v2?
And also, how do i check if there exists such a linear combination?

There's no need to bring orthogonality into this, a "purely linear" problem. So I'll ignore the fact that you said "Gram-Schmidt."

We're simply trying to solve a system of equations. To be specific, we can think of it as a system of 3 equations in 2 unknowns. That makes it unlikely - possible, but unlikely - that the system has a solution. This is understandable: the two vectors $ v_1$ and $ v_2$ span a plane (a two-dimensional subspace) in $ \mathbb{R}^3.$ It is unlikely - possible, but unlikely - that the point $ x$ even lies in this plane.

We can set up an augmented matrix for this system:
\[ \begin{bmatrix}1&1&\minus{}1\\2&\minus{}2&10\\3&3&\minus{}3\end{bmatrix}\]
Gaussian elimination gets us to the row echelon form
\[ \begin{bmatrix}1&1&\minus{}1\\0&1&\minus{}3\\0&0&0\end{bmatrix}.\]
That is, the equations $ av_1\plus{}bv_2\equal{}x$ reduce to the equations $ a\plus{}b\equal{}\minus{}1,b\equal{}\minus{}3.$ By back substitution, we get $ a\equal{}2,b\equal{}\minus{}3.$

The thing I called unlikely has occurred: $ x\in\text{span}\{v_1,v_2\}.$

That is, $ x\equal{}2v_1\minus{}3v_2.$

Thanks for the reply.

Ok, I know the fact it can be resolved as a system of equations and unknowns, but what if I've got 2 equations and 4 unknowns for example? Can it be resolved with the help of orthogonality?
I'm asking this because i need the find a solution for a problem like this:

I have a vector x of 2,4,8... elements and v1,v2...,vn vectors (n>0), each of the same number of elements as x.
I need to write x as linear combination of all vectors v1,v2...vn, if there exists such a linear combination.

I brought up orthogonality because i need to follow an algorithm like this:
Using the inner product of vectirs, find an orthogonal basis for the linear space generated by the vectors v1,v2...vn. Then apply standard methods of linear algebra using the orthogonal basis instead of the original vectors v1,v2...vn.

So... I can compute the coefficients of the "orthogonalized vectors". The question is how can i find the coefficients of the original vectors using these computed coefficients?

Here is the algorithm:

[img]http://img58.imageshack.us/img58/5322/code1np2.png[/img]
[img]http://img168.imageshack.us/img168/4633/code2dn2.png[/img]


Leonora thinks of a secret number.  In this sequence: she subtracts 5, multiplies by 5, adds 5, and then divides by 5 to get 2002 as her answer.  Don thinks of a secret number.  In this sequence: he adds 5, multiplies by 5, subtracts 5, and then divides by 5 to get 2002 as his answer.
What is the sum of their two secret numbers?

A.2002	B.3992	C.3996	D.4004	E.20,020

[hide=" click to reveal hidden content"]

$ 2006\plus{}1998 \equal{} 4004$

[/hide]

[hide]"Leonora thinks of a secret number. In this sequence: she subtracts 5, multiplies by 5, adds 5, and then divides by 5 to get 2002 as her answer."

We must work backwards.
2002*5= 10010
10010-5= 10005
10005/5 = 2001
2001+5= 2006
Leonora's number is 2006

"Don thinks of a secret number. In this sequence: he adds 5, multiplies by 5, subtracts 5, and then divides by 5 to get 2002 as his answer."

We must again work backwards.
2002*5=10010
10010+5=10015
10015/5= 2003
2003-5= 1998

Don's secret number is 1998.

1998+2006=4004

Answer D) 4004.[/hide]


calculus
integration
arithmetic sequence

The three sides of a right triangle have integral lengths which form an arithmetic progression. One of the side length could have length

$\text{(A)} \ 22 \qquad \text{(B)} \ 58 \qquad \text{(C)} \ 81 \qquad \text{(D)} \ 91 \qquad \text{(E)} \ 361$

[hide="Answer"]We know that a 3-4-5 triangle has an arithmetic progression. The only number given that is a factor of either of those three is $81$, so our answer is $\boxed{C}$. (81-108-135 triangle.)[/hide]

[hide="Solution"]Let $a-v, a, a+v$(definition of an arithmetic progression) denote the side lengths. Because $a+v$ is the largest, that's the hypothenuse. By Pythagoras, we know that $(a-v)^2+a^2=(a+v)^2$, so $a^2-2av+v^2+a^2=a^2+2av+v^2$, or $a^2=4av$. Because $a\neq0$, we know that $a=4v$.

We now know that the side lengths are $3v, 4v, 5v$ with $v$ an integer. That yields that the sidelengths must be multiples of $3, 4\ \textrm{or}\ 5$. The only answer that can be right in this point of vue, is $\boxed{C}$.[/hide]

[hide]They must be in a $3: 4: 5$ ratio. Since none of the choices are divisible by 4 or 5, we are looking for one that is divisible by 3. Choice $C$ works.[/hide]

[hide="Solution"]

$a^2 + b^2 = c^2$

$(b-x)^2 + b^2 = (b+x)^2$

$b^2 - 2bx + x^2 + b^2 = b^2 + 2bx + x^2$

$b^2 - 4bx = 0$

$b - 4x = 0$

$b = 4x$

$a = 3x, b = 4x, c = 5x$

Confirmation of the 3:4:5. Choice C is the only one that is divisible by three as none are divisible by either 4 or 5.
[/hide]


When determining the steady state matrix using the markov chains, why does the initial state matrix not matter? Ie. if you start with [0 1] and apply a transition matrix to it. Then when the steady state is obtained, the [0 1] does not matter at all. If someone could provide a link or solution it would be great.

I am not absolutely sure, but since it seems like no one is answering markov chain questions (e.g. here is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=51670]mine[/url]), here's my opinion:

1) this property is ONLY true for regular markov chains <<sure on this
2) if u actually do out the matrix multiplication a few times (by hand), you will find out that eventually the figures are "averaged" in a way...try this. I know this is somewhat sketchy, but it seems to work right. So just do out the multiplication and u should see...
3) i quote (but don't fully understand this) from Finite Mathematics: A Problem Solving Approach, Jeffrey X. Watt, (c) 1999: "The steady state vector of a regular Markov chain is dependent only on the transition matrix, not on the initial state vector. This is because multiplying by a state vector is equivalent to finding a weighted average of each column of the transition matrix. Since the numbers in each column will be the same eventually, their average will not change regardless of the state vector"

Hope that helps!

Theorem: If $A$ is a matrix with nonnegative entries, and some $A^n$ has entirely positive entries, $A$ has a unique eigenvector with positive real eigenvalue and positive entries. Also, every other (complex) eigenvalue is smaller in absolute value.

In the case of a stochastic matrix, that eigenvector $v_1$ has eigenvalue 1, and $\lim A^nx=\sum\lambda_i^nc_iv_i=c_1v_1$. This is enough information to calculate $\lim A^n$. (This calculation assumes distinct eigenvalues for convenience, but we don't need to)

Ummm....could you explain in simpler terms jmerry? I never learned eigenv so I have no idea what is going on....lol

In order to understand the long-term behavior of $A^n$ for any matrix $A$, you [b]need[/b] to know about eigenvalues and eigenvectors. There's really no other way to get at it.

I see...ahhhh....my lack of mathematical knowledge :( tnx jmerry!


algebra
polynomial
inequalities
linear algebra
real analysis

[quote="chungaops"](Someone might have asked the question already.  However, I could not find it by search.)

Can someones share their opinions about the book "Problems from the Book" by Andreescu and Dospinescu.  Is it just another collections of competition problems or present systematic approaches how to solve the problems or else?  I have been searching for the website trying to know how those who had the book thought about the book.  Unfortunately, I could not find one.

Can anyone help?  Many thanks.

Chin Hong[/quote]

[hide]try gigapedia.org you will find it there[/hide]

A request to people who have seen this book : Can you please let me know how much do the contents of this book overlap the contents of other books by Andreescu or other sources? Basically, I am trying to figure out whether to spend the money on this. Is bulk of the content of this book already available elsewhere or does it have useful stuff that is not easily available elsewhere?

Thanks.

Thanks. I had seen that. But, as I understand, those are only the unsolved problems from that book and might occupy just 30-40 pages in that book?

PftB revolutionized my entire view of existence. Gazing back on empty, forsaken days before my enlightenment in the core of its pages, I wonder, a little wistfully, and not without astonishment, however I ever lived, how I ever trudged my way through the dreariness of every day life before my soul was set aflame with this masterpiece.

Anyways. Essentially, PftB is divided into chapters, each containing a few techniques in a fairly constrained field such as use of linear algebra in combinatorics, formal series, complex colorings, polynomial irreducibility, ordered based on the level of the mathematics contained within. For example, the first chapter is only a few trick algebraic substitutions for inequalities with constraints, while a chapter near the end might use in combination theorems from linear algebra, real analysis, as well as results like Alon's combinatorial nullstellensatz (which is one of the techniques covered in the last chapter).

It certainly is not just "another collection of competition problems". It does indeed contain "systematic" methods to solve the problems contained (each chapter has several problems of varying difficulties with fully written-out solutions to explain the concepts and then a section of problems based on the concepts at the end), assuming that by "systematic" you mean "methods which could be applied to make some progress on a wide variety of problems" - direct applications of a technique are generally only contained in a few introductory examples.


Let $ a,b,c$ be positive real numbers such that $ a+b+c=1$. Prove that:
\[ \Leftrightarrow \frac{(a-b)^{2}}{b+c}+\frac{(b-c)^{2}}{c+a}+\frac{(c-a)^{2}}{a+b}\]
\[ \ge \frac{ a(b-c)(b-a)}{(b+c)(c+a)}+\frac{b(c-a)(c-b)}{(a+b)(c+a)}+\frac{c(a-b)(a-c)}{(b+c)(a+b)}\]

i think we don't need the condition $ a+b+c=1$. :wink:

all sums are cyclic. our inequality is equivalent to
\[ \sum(a-b)^{2}(c+a)(a+b)\ge \sum a(b-c)(b^{2}-a^{2}) \] \[ \iff \sum(a^{2}-b^{2})\left((a-b)(c+a)+a(b-c)\right)\ge 0 \] \[ \iff \sum(a^{2}-b^{2})(a^{2}-bc)\ge 0 \] \[ \iff \sum a^{4}+\sum a^{3}b\ge \sum a^{2}b^{2}+\sum a^{2}bc \]
now we have by AM-GM
\[ \sum a^{4}\ge \sum a^{2}b^{2}\]
so we need only prove that
\[ \sum a^{3}b\ge \sum a^{2}bc\iff \frac{a^{2}}{c}+\frac{b^{2}}{a}+\frac{c^{2}}{b}\ge a+b+c \]
which is just cauchy shcwarz.


Με αφορμή αυτή την άσκηση ([url]http://www.mathlinks.ro/viewtopic.php?t=308915[/url]) ήθελα να ποστάρω μια λύση που πίστευα ότι είναι πιο σύντομη. Στην πορεία συνάντησα μερικά εμπόδια.

Γενικά η λύση μου ήταν η εξής:
$ A^2\equal{}I \rightarrow (A\minus{}I)(A\plus{}I)\equal{}0$

Άρα το ελάχιστο πολυώνυμο διαιρεί το p(x)=(x-1)(x+1) και άρα γραφεται σε μορφή γινομένων χωρίς δυνάμεις.
Άρα ο Α είναι διαγωνοποιήσιμος με ιδιοτιμές το -1 και το 1.

Αν $ A \equal{}P^{\minus{}1}DP \rightarrow r(A)\equal{}r(D), r(A\plus{}I)\equal{}r(P^{\minus{}1}(D\plus{}I)P)\equal{}r(D\plus{}I), r(I\minus{}A)\equal{}r(I\minus{}D)$

Όμως o D περιέχει μόνο άσσους και -1. Έστς ότι έχει κ 1 και n-κ -1.
Τότε r(A+I)=και r(A-I)=n-κ και άρα άθροισμα n..

Απορίες τώρα:α
1)Ισχύει όντως ότι αν $ m_T(x)\equal{}(x\minus{}x_1)(x\minus{}x_2)...(x\minus{}x_k)$ οπου τα $ x_i$ διακεκριμένες ιδιοτιμές ο Α είναι διαγωνοποιήσιμος?
Κάπου το είχα διαβάσει αυτό αν θυμάμαι καλά αλλά τώρα που πήγα να το αποδείξω κάπου κόλλησα και δεν το βρήκα και πουθενά στο net οπότε δεν ήμουν σίγουρος...
2)Πώς είναι ο διαγωνοποιήσιμος πίνακας στα αγγλικά?? (Βασικά παίζει να είναι και αυτός ο λόγος που δεν βρήκα κάτι που να συνδέει ελάχιστο πολυωνυμο και διαγωνιποιησιμότητα :Ρ

Ο διαγωνοποιήσιμος είναι diagonali[b]s[/b]able στα Βρεττανικά Αγγλικά και diagonali[b]z[/b]able στα Αμερικανικά Αγγλικά.

Η λύση σου είναι σωστή.

Το θεώρημα που γράφεις ισχύει αλλά θέλει λίγη δουλειά για να αποδειχθεί. (Το αντίστροφο, ότι αν ο πίνακας είναι διαγωνοποιήσιμος τότε το ελάχιστο πολυώνυμο είναι γινόμενο γραμμικών παραγόντων και κάθε ρίζα του έχει πολλαπλότητα ένα ισχύει και είναι (σχεδόν) προφανές.)

Ένας τρόπος για να το αποδείξουμε είναι να γράψουμε $ m(x) \equal{} (x \minus{} \lambda)p(x)$. Τότε το $ \lambda$ δεν είναι ρίζα του $ p$ και άρα μπορούμε να γράψουμε $ 1 \equal{} q(x) (x \minus{} \lambda) \plus{} r(x)p(x)$. (Αυτό είναι το λήμμα Bezout για πολυώνυμα. Αποδεικνύεται π.χ. χρησιμοποιώντας τον Ευκλείδιο αλγόριθμο για τα πολυώνυμα)

Άρα για κάθε διάνυσμα $ v$ ισχύει ότι $ v \equal{} u \plus{} w$ όπου $ u \equal{} q(A)(A \minus{} \lambda I)v$ και $ w \equal{} r(A)p(A)v$. Παρατηρούμε τώρα ότι $ p(A)u \equal{} 0$ και $ (A \minus{} \lambda I)w$ = 0. Άρα $ \mathbb{R}^n \equal{} U \oplus W$ όπου $ U \equal{} \ker(p(A))$ και $ W \equal{} \ker(A \minus{} \lambda I)$. Επαγωγικά τώρα, αν $ m(x) \equal{} (x \minus{} \lambda_1) \cdots (x \minus{} \lambda_k)$ τότε παίρνουμε $ \mathbb{R}^n \equal{} U_1 \oplus \cdots \oplus U_k$ όπου $ U_i \equal{} \ker(A \minus{} \lambda_i I)$. Παίρνουμε τώρα μια βάση για το κάθε $ U_i$ και σε σχέση με αυτή τη βάση ο $ A$ είναι διαγωνοποιήσιμος.

(Φυσικά θα μπορούσε κάποιος να "κλέψει" και να γράψει την πιο πάνω απόδειξη μόνο σε σχέση με αυτό το πολυώνυμο γράφοντας $ 1 \equal{} 1/2(x\plus{}1) \minus{} 1/2(x\minus{}1)$ κτλ.)


linear algebra
matrix
complex numbers
linear algebra unsolved

Find all linear maps of $M_n(C)$ such that $ f(AB)=f(A)f(B)$ whenever $A,B$ are nilpotent.

Hi Harazi!

What's [tex]M_n(C)[/tex]? You said [tex]f[/tex] was a linear map - this implies [tex]f(AB)=f(A)f(B)[/tex] for all [tex]A,B[/tex], doesn't it?

Hanno

Hi, Hanno. $M_n(C)$ is the set of complex $nxn$ matrices. And that $f$ is linear means that $f(aA+bB)=af(A)+bf(B)$ for all matrices $A,B$ and all complex numbers $a,b$.

$f$ is a map from $M_n(C)$ into $M_n(C)$ or  $M_n(C)$ into $C$ ?

harazi, I solved a similar problem which states that:

 Let $f:M_n(%Error. "mathh" is a bad command.
{C})\to\mathbb{C}$ such that the following conditions are fullfiled:

    a) $f(\alpha\cdot X+\beta\cdot Y)=\alpha\cdot f(X)+\beta\cdot f(Y)$ for any $\alpha,\beta\in\mathbb{C}$
    b) $f(X\cdot Y)=f(X)\cdot f(Y)$ for any $X,Y\in M_n(\mathbb{C})$.

    Prove that $f(A)=a\cdot trA$, where $a\in\mathbb{C}$ ;)

First, the problem is trivial and secondly your answer is clearly wrong.

Sorry,harazi I forgot to say that $a=f(E_{11})=f(E_{22})=...=f(E_{nn})=a\in\mathbb{C}$.I think it is correct now.
Secondly, maybe for you it is trivial for me it is an easy problem. ;)

So you practically say that $ tr(XY)=trX\cdot trY$!!! I will tell you what you solved: find linear forms on $M_n(C)$ such that $ f(XY)=f(YX)$!! And that's a very easy problem, though very nice.

Yes,you are right.Of course,I'm not that idiot to say that $tr(X\cdot Y)=tr(X)\cdot tr(Y)$ ;)

:idea: Should we use the fact that any non invertible matrice can be written as the product of nilpotent matrices? :lol:

You can use anything you want in order to solve the problem.  ;)

There will be a little bit difficult to deal with the following aspect: you don't have $ f(ABC)=f(A)f(B)f(C)$ for all nilpotent matrices $A,B,C$ thus if you want to apply the above theorem, you need to refine it or some other twist that I cannot see now.

We consider $\mathbb{K}$ a zero characteristic field and do not limit the answer to $\mathbb{C}$.

Denote by $f$ a linear map $f~: ~\mathcal{M}_{n}(\mathbb{K}) \to \mathcal{M}_{n}(\mathbb{K})$ such that $f(XY ) = f(X)f(Y )$ for all nilpotent matrices $X$ and $Y$ .
We show that either $f=0$ or there exists $P\in \mbox{GL}_{n}(\mathbb{K})$ such that $f(M)=PMP^{-1}$ for all $M \in \mathcal{M}_{n}(\mathbb{K})$.

The case $f=0$ is obvious so let's assume that $f\not=0$.

Denote by $E_{i,j}$ the matrice of $\mathcal{M}_{n}(\mathbb{K})$ with all coefficients equal to 0 except on row $i$ and column $j$, where it equals 1 and $F_{i,j}=f(E_{i,j})$.

We proceed in two steps.

[u][i][b]Lemma 1:[/b][/i][/u] $\{F_{i,j}\}_{i,j\in \{1,\ldots , n \}}$ is a basis of $\mathcal{M}_{n}(\mathbb{K})$ such that $F_{i,j}F_{k,l}= \delta_{j,k}F_{i,l}$.

[u][i][b]Lemma 2:[/b][/i][/u] If $\{M_{i,j}\}_{i,j\in \{1,\ldots , n \}}$ is a basis of $\mathcal{M}_{n}(\mathbb{K})$ such that $M_{i,j}M_{k,l}=\delta_{j,k}M_{i,l}$, then there exists $P\in \mbox{GL}_{n}(\mathbb{K})$ such that $M_{i,j}=PE_{i,j}P^{-1}$ for all $i,j\in \{1,\ldots , n \}$.

[b]Proof of Lemma 1:[/b]
If $\{F_{i,j}\}_{i,j\in \{1,\ldots , n \}}$ was not a basis of $\mathcal{M}_{n}(\mathbb{K})$, we could find $\{\alpha_{i,j}\}_{i,j\in \{1,\ldots , n \}}\in \mathbb{K}^{n^{2}}$ with at least one $\alpha_{l,p}\not=0$ and $\sum_{{i,j\in \{1,\ldots , n \}}}\alpha_{i,j}F_{i,j}=0$

For all $i,j,k,l \in \{1,\ldots , n \}$ we have $E_{i,j}E_{k,l}= \delta_{j,k}E_{i,l}$ so $F_{i,j}F_{k,l}= \delta_{j,k}F_{i,l}$.
This last relation is obvious if $i\not=j$ and $k\not=l$, because $E_{i,j}$ and $E_{k,l}$ are nilpotent. But also $f(E_{i,i}E_{k,l})= f(E_{i,j_{0}}E_{j_{0},i}E_{k,l})=f(E_{i,j_{0}}\delta_{i,k}E_{j_{0},l})=\delta_{i,k}F_{i,l}$ (where we choose $j_{0}\not=i$ and $j_{0}\not=l$).

There is at least one $F_{k,q}\not=0$ because $f\not=0$, $F_{k,l}\times \sum_{{i,j\in \{1,\ldots , n \}}}\alpha_{i,j}F_{i,j}\times F_{p,q}= \alpha_{l,p}F_{k,q}= 0$ which is a contradiction, so $\{F_{i,j}\}_{i,j\in \{1,\ldots , n \}}$ is a basis of $\mathcal{M}_{n}(\mathbb{K})$.

[b]Proof of Lemma 2:[/b]
$M_{i,k}M_{k,j}=M_{i,j}$, so $\mbox{Im}M_{i,k}=\mbox{Im}M_{i,j}$, let's denote this space by $\mathcal{V}_{i}$.

There exist $\{\alpha_{i,j}\}_{i,j\in \{1,\ldots , n \}}\in \mathbb{K}^{n^{2}}$ such that $I_{n}= \sum_{{i,j\in \{1,\ldots , n \}}}\alpha_{i,j}M_{i,j}$.
If $X \in \mathbb{K}^{n}$, $X=\sum_{i=1}^{n}\sum_{j=1}^{n}\alpha_{i,j}M_{i,j}X$, and $\sum_{j=1}^{n}\alpha_{i,j}M_{i,j}X \in \mathcal{V}_{i}$, so $\mathbb{K}^{n}= \mathcal{V}_{1}+\mathcal{V}_{2}+\cdots+\mathcal{V}_{n}$.

If $X_{i}\in \mathcal{V}_{i}$, and $X_{1}+X_{2}+\cdots+X_{n}=0$ then $M_{i,i}(X_{1}+X_{2}+\cdots X_{n})=X_{i}=0$, so $\mathbb{K}^{n}= \mathcal{V}_{1}\oplus \mathcal{V}_{2}\oplus \cdots \oplus \mathcal{V}_{n}$.

If $k\not=j$, $\mbox{Im}M_{i,k}\subset \mbox{Ker}M_{i,j}$, so $\bigoplus_{k=1, k\not=j}^{n}\mbox{Im}M_{i,k}= \mbox{Ker}M_{i,j}$ (because they have same dimension and the previous inclusion).

Let $Y_{1}\in \mathcal{V}_{1}$, such that $Y_{1}\not=0$. For $i \not= 1, Y_{i}=M_{i,1}Y_{1}$.
We have $M_{i,j}Y_{j}=Y_{1}$.
So $\{Y_{1}, \ldots, Y_{n}\}$ is a basis of $\mathbb{K}^{n}$.

We define $P \in \mbox{GL}_{n}(\mathbb{K})$ the matrix whose columns are the $Y_{i}$. We have $PE_{i,j}=M_{i,j}P$, which concludes the proof of lemma 2.

Since $f$ is linear we have proven that $f(M)=PMP^{-1}$ for all $M \in \mathcal{M}_{n}(\mathbb{K})$.


algebra
polynomial
superior algebra
superior algebra solved

Hello to all who can help me!

We know that root of 2 in R is algebraic of degree 2 over Q and of degree 1 over R.
Can vi tell that "3/2" is algebraic over Z (integers)?
I know that (Z,+,*) is a ring, but not a field,...
Can someone explain this to me very-very simple?

Thanks
Tanja
_________________
Tamplushka

I am very grateful for Your answer, but I really need more easy explanation  :blush:

Can someone explane that more easy? :o

Thank you very much
Best regards
Tanja

suppose the polynomial is
$x^n+a_{n-1}x^{n-1}+\ldots+a_{1}x+a_{0}$

Now if $x=\frac{3}{2}$ is a solution , when you work this out you get :

$3^n+a_{n-1}3^{n-1}2+\ldots+a_{1}3\ 2^{n-1}+a_{0}2^{n}=0$

This leads to contradiction, as all coefficients are integers and all terms left are even, except the first : $3^n$

I :read:  your message.
I understand now.
It was very nice from you.

Thank you [size=150][b]very much[/b][/size]

Best regards
Tanja

[quote="fredbel6"]If you mean, is 3/2 algebraic over $\mathbb{Z}$ in the sense that there is no monic polynomial with integer coefficients having this as root, then it is not algebraic

There is a standard theorem that the  rationals being roots of monic integer polynomials are always integers[/quote]

Are you sure the polynomial has to be monic?

Im used that $a$ is called algebraic over $K$ if there is a monic polynomial over $K$ with root $a$. For fields, it is no difference, and for rings it's useful to see it that way. But at least the question itself is not well defined, so an accurate answer seems impossible.

Thank You very much. I really read only "A first Course i Abstrakt Algebra" and
need only very simple explain. :blush:

Best regards
Tanja

[quote="fredbel6"]Yes algebraical over $\mathbb{Z}$ I thought she meant what I did, because that is the standard definition of algebraic integer
otherwise 3/2 is obviously a root of an integer polynomial[/quote]
I thought she meant [b]algebraic over integers[/b] and not [b]algebraic integer[/b].
So if I'm not wrong $\frac{3}{2}$ is algebraic over integers but is not an algebraic integer... However it's ambiguous...


Can people make a list of archives from old math exams.  I would greatly appreciate it if someone could find an AHSME archive.  I spent about 2 hours this evening looking but i found nothing.  An AMC archive would be useful also.  However i believe they only release the problems in books

thx for these sites.  They are extremely helpful.  more links will be appreciated


Just curious if there will be anyone there i can chill with next year, my school did get four kids in but there all fluff majors and id like someone to talk the 50H series with.

I'm considering Stanford, but I'm probably leaning towards Caltech.

I got too into both Stanford and Caltech --- the same dilema with me!!! How funny!

CalTech, Stanford, and Harvey Mudd here. Have not yet decided.