Find all rational solutions of the equation \[(x+y\sqrt{5})^{4}+(r+t\sqrt{5})^{4}= 2+\sqrt{5}\]

Using conjugates: if $\xi\in\mathbb{Q}, \sqrt{\xi}\in\mathbb{I}, a, b, p, q\in\mathbb{Q}, n\in\mathbb{N}$, then $(a+b\sqrt{\xi})^{n}=p+q\sqrt{\xi}\iff (a-b\sqrt{\xi})^{n}=p-q\sqrt{\xi}$ (the proof is easy via binomial expansion). Hence our equation can be rewritten thus

$(x-y\sqrt{5})^{4}+(r-t\sqrt{5})^{4}=2-\sqrt{5}$

But the LHS is non-negative and the RHS is negative, so there's no solution.


given a sequence :
$ a_1 \equal{} 1 a_2 \equal{} 4;$;
$ a_{n \plus{} 1}$ = 6$ a_n$ - $ a_{n \minus{} 1}$-2
find n sastifying $ a_n$ is a square number!!


When the magnesium foil within a flashbulb burns, the flashbulb undergoes no change in mass. When a sample of magnesium ribbon burns in open air, the sample gains mass. What is the difference between these two experiments? Does the second observation violate the law of conservation of mass?

I'm just beggining to study chemistry in order to prepare for my classes next year. I'm pretty sure that in the second experiment, there is no closed system, and therefore mass is not conserved. Is this correct?

The law of conservation of mass is never violated.  That is why it is a law.  However, this is case only for a closed system and it depends on how you define your system.  If you do not include the air as part of the system, then you are certainly correct.  If you do include the air as part of your system, then the total mass will remain the same.

This is more of a physics question, not a chemistry question.

To be perfectly technical, the law of conservation of mass is violated when mass is converted to energy: E = mc^2

Also, this is an (almost) purely chemistry question. What one must understand is the oxidation of magnesium in the presence of oxygen. In burning magnesium metal in air (which presumably contains oxygen), the reaction to form magnesium oxide is completed. The atoms used in the reaction are conserved i.e. not destroyed; thus, the mass of the atoms is conserved.


Let $ A_i \equal{} \{x\in\mathbb{Z}: \ x < \minus{} i\}$ for $ i \equal{} 0,1,2,3,...$.
Let $ A \equal{} \bigcap_{i \equal{} 0}^{\infty}A_i$.
Is $ A \equal{} \emptyset$?

Well $ A$ can't possibly contain any positive integers and it can't contain 0.  For any negative integer, $ n$, you can be sure that it won't be in the intersection just by taking $ i\equal{}n\minus{}1$ (even $ i\equal{}n$).  So it has to be empty.


trigonometry
geometry
circumcircle
trig identities
Law of Sines
geometry proposed

[u][b]The author of this posting is : Gil[/b][/u]
____________________________________________________________________

Let ABC be a triangle and M be a point on (BA, N on (CA such that BM=CN=BC. Prove that MN and OI are perpendicular, and MN/OI=a/R.

See my posting at: http://www.mathlinks.ro/Forum/viewtopic.php?t=14655

The equality MN / OI = a / R was proven in the equivalent form $MN = 2\sin A \cdot OI$ there (this form is equivalent since it implies $\frac{MN}{OI}= 2\sin A = \frac{a}{R}$ by the Extended Sine Law).

[i]Remark.[/i] The [i]Extended Sine Law[/i] is just the simple fact that for any triangle ABC with sidelengths a, b, c and angles A, B, C, we have

$\frac{a}{\sin A}= \frac{b}{\sin B}= \frac{c}{\sin C}= 2R$,

where R is the circumradius of triangle ABC. Equivalently, a = 2R sin A, b = 2R sin B and c = 2R sin C. In some books, this is simply called "law of sines", without "extended", but usually, when speaking of the "law of sines", one means just the relation $\frac{a}{\sin A}= \frac{b}{\sin B}= \frac{c}{\sin C}$ without the fact that it is equal to 2R.

  Darij


function
complex analysis
complex analysis unsolved

$f: D \to \mathbb{C}$ an analytic function defined on the open unit disk. Note $d$ the diameter of $f(D)$. Find a constant $K$ such that
$d \geq K f'(0)$

Finding [i]a[/i] constant is not hard: the Cauchy formula implies $|f'(0)|\le d$. What about the [i]sharp[/i] (that is, the best possible) constant $K$? Hint: it is greater than $1$.


Given  A,B,C are three sides of a triangle. Prove:
$ \ 6sqrt{left(3tanA/2.tanB/2\plus{}3)}\plus{}4.sqrt[3]{left(3tanB/2.tanC/2\plus{}7}\plus{}3sqrt[4]{left(3tanC/2.tanA/2\plus{}80} \le 29$
6(tanA/2.tanB/2+3)^1/2+4.(tanB/2.tanC/2+7)^1/3+3(tanC/2.tanA/2+80)^1/4<=29[/img][/list]

You can use "Mathtype6.0 software" to type.
Click [url=http://www.google.com.vn/#hl=vi&q=Mathtype+6.0&meta=&aq=&oq=Mathtype+6.0&fp=23a3dde90d8ca77d]here[/url]


Let $T_n = \frac{n(n+1)}{2}$ so $T_1 = 1, T_2 = 3, T_3 = 6...$
Compute the sum
$\frac{1}{T_1} + \frac{1}{T_2} + \frac{1}{T_3} + \dots + \frac{1}{T_{2005}}$

[hide]
Looking for a pattern I found $\frac{2n}{n+1}$ as the sum of the first $n$ triangular number fractions and got the correct answer, but I do not know how to prove it, can anyone think of a way?
[/hide]

[hide]
$\frac{1}{T_n}=\frac{2}{n(n+1)}=\frac{A}{n}+\frac{B}{n+1}$. (We're trying to find the partial fraction decomposition here.) Multiplying through by $n(n+1)$, we get $2=A(n+1)+B(n)$. If $n=-1$, then $B=-2$. If $n=0$, then $A=2$. Our partial fraction decomposition is $\frac{2}{n(n+1)}=\frac{2}{n}-\frac{2}{n+1}$. The sum is then \[ \sum_{n=1}^{2005} \left( \frac{2}{n}-\frac{2}{n+1} \right) = \left( \frac{2}{1}-\frac{2}{2} \right) + \left( \frac{2}{2}-\frac{2}{3} \right) + \left( \frac{2}{3}-\frac{2}{4} \right) + \cdots + \left( \frac{2}{2005}-\frac{2}{2006} \right).  \] This telescopes very nicely and we're left with \[ \sum_{n=1}^{2005} \frac{1}{T_n} = 2-\frac{2}{2006}=\frac{2005}{1003}.  \]

This can be generalized very easily. :)
[/hide]

Yep,
The first time I attempted this one, I tried something exactly like that, but I rushed the decomposition and got something really weird.
I guess I can easily prove my proposition on the basis of your solution then, thanks.

[hide]
Via induction

$\displaystyle \frac{2n}{n+1}+\frac{(n+1)(n+2)}{2}$
$\displaystyle \frac{(2n)(n)+(n+1)^2(n+2))}{2(n+1)}$
$\displaystyle \frac{2n^2+(n^2+2n+1)(n+2)}{2n+2}$
$\displaystyle \frac{2n^2+n^3+2n^2+2n^2+4n+n+2}{2n+2}$
$\displaystyle \frac{n^3+6n^2+5n+2}{2n+2}$

This doesn't look fruitful. Can someone point out an error or how to continue by induction.

I like your way, chess64.
[/hide]


by bernoulli's inequality :

$ (1\plus{}\frac{x}{y})^k \geq 1\plus{}k\frac{x}{y}$
and
$ (1\plus{}\frac{y}{x})^k \geq 1\plus{}k\frac{y}{x}$

so :

$ (1\plus{}\frac{x}{y})^k\plus{}(1\plus{}\frac{y}{x})^k \geq 2\plus{}k(\frac{x}{y}\plus{}\frac{y}{x}) \geq 2(1\plus{}k)$

:)

hi !!! mehdi  :D
I think it s $ 2^{k\plus{}1}$

[quote="y-a-s-s-i-n-e"]let $ x$ , $ y$ $ \in ]0, \plus{} \infty[$ and $ k$ $ \in {\mathbb N}^{\star}$ , find minimum of :
$ \bigg(1 \plus{} \frac {x}{y}\bigg)^k \plus{} \bigg(1 \plus{} \frac {y}{x}\bigg)^k$[/quote]
yes,it is $ 2^{k\plus{}1}$
[hide="hint"]use this Inequality
$ \frac{a^k\plus{}b^k}{2}\geq(\frac{a\plus{}b}{2})^k$[/hide]

thank's :)

we can prove this inequality by holder since :

$ (a^k \plus{} b^k)(1 \plus{} 1)^{k \minus{} 1}\geq (a \plus{} b)^k$

and $ a \equal{} (1 \plus{} \frac {x}{y}) , b \equal{} (1 \plus{} \frac {y}{x})$

but , please , how to be sure  that the aquiered value is the minimum ??  :?:

For mehdi cherif.You easily prove that:
 $ S\equal{}(1\plus{}\dfrac{x}{y})^k\plus{}(1\plus{}\dfrac{y}{x})^k\geq\dfrac{(2\plus{}\dfrac{x}{y}\plus{}\dfrac{y}{x})^k}{2^{k\minus{}1}}\geq 2^{k\plus{}1}$
For $ x\equal{}y$ you obtain that $ S\equal{}2^{k\plus{}1}$ so $ S_{min}\equal{}2^{k\plus{}1}$


calculus
integration
function
probability
expected value
calculus computations

How to evaluate the following?

[img][url=http://www.eyelash.ps/up][img]http://www.eyelash.ps/up/uploads/2a763443ee.jpg[/img][/url][/img]

Thanks

for the second one

$\int_{0}^{\infty} \frac{e^{tx}}{\beta} e^{-x/\beta} \; dx = \int_{0}^{\infty} \frac{e^{x(t-\frac{1}{\beta})}}{\beta} \; dx = \frac{1}{t\beta -1} e^{x(t-\frac{1}{\beta})}|_{0}^{\infty} = ...$
if $t>1/\beta$ then it is $\infty$
if $t<1/\beta$ then it is $\frac{1}{t\beta-1}$

Note the context: maitham is clearly computing the expected value and the moment generating function for an exponentially distributed random variable. Doing the same things for anything in the gamma family of distributions is only slightly more complicated.


inequalities
geometry
incenter
Euler
geometric transformation
reflection
trigonometry

Let $S$ be the symmetrical point of the centroid $G$ with respect to the circumcentre $O$.
I denote (standard) the incentre $I$ and orthocentre $H$ for the triangle $ABC$.
Prove that there is the inequality: $HG\le IS.$

On the Euler line, $G$ is in between $H$ and $O$ and  $GH=2GO$. As $S$ is the reflection of $G$ wrt $O$, we have $OS=OG$ and $HG=2GO=2SO=SO+OG=SG$. So it suffices to prove $SG \leq IS$

Since $O$ is the mid point of $SG$ and $G$ is the mid point of $SH$, we have (by cosine law)

(1)  $IS^2+IG^2=2IO^2+2SO^2$
(2)  $IS^2+IH^2=2IG^2+2SG^2$

2*(1)+(2), $3IS^2+IH^2=4IO^2+4SO^2+2SG^2=4IO^2+3SG^2$

From [url=http://www.artofproblemsolving.com/Forum/topic-4817.html]www.artofproblemsolving.com/Forum/topic-4817.html[/url], we know that $IH \leq 2IO$. So $IH^2 \leq 4IO^2$.
Thus $SG^2=IS^2+\frac{IH^2-4IO^2}{3} \leq IS^2$, giving $SG \leq IS$

Any student from a high school can ascertain the distancies between the remarkable points

of a triangle ABC in accordance with $s,\ r,\ R$  (centroid $G$, orthocenter $H$, circumcenter $O$, incenter $I$,

exincenters $\ I_a,\ I_b,\ I_c,\ $ [u]Nage[/u]l's point $N$, [u]Gergogne[/u]'s point $\Gamma$, [u]Euler[/u]'s point $E$, [u]Lemoin[/u]'s point $L$ a.s.o.)

At least is known that $\overrightarrow {HG} =2\cdot \overrightarrow {GO},\ \overrightarrow {NG} =2\cdot \overrightarrow {GI}$. I offer some examples:

$ \begin {array} {111}
OI^2=R^2-2Rr & {OH^2=9R^2+2(r^2+4Rr-s^2) & {O\Gamma ^2=R^2-r(R+r)( \frac {2s}{4R+r}) ^2}\\\
OI_a^2=R^2+2Rr_a\ldots & {ON=R-2r} & {OL^2=R^2-3( \frac{abc}{a^2+b^2+c^2}) ^2}\\\
IN^2=s^2+5r^2-16Rr & {IH^2=4R(R+r)+3r^2-s^2} & {I\Gamma ^2=r^2[ 1-( \frac{s\sqrt 3}{4R+r})^2]}\\\
\end {array}$

From these "summary" metric relations can result easily numerous remarkable inequalities.

$R\ge 2r;\ \ a^2+b^2+c^2\ge 4\sqrt 3 \cdot S;\ \ s\sqrt 3\le 4R+r;\ \ s^2+r^2\ge 16Rr;$

$s^2\le 4R^2+4Rr+3r^2;\ \ s^2\ge 3r(4R+r).$ I offer now some celebrated examples:

$1.$ The point I belongs to the interior of the circle with diameter HG. Indeed,

$IH^2+IG^2-HG^2=-\frac 43 R^2+\frac 83 r^2\le 0$.

$2.\ HI\le 2\cdot OI$. Indeed, $4\cdot OI^2-HI^2=s^2-3r(4R+r)\ge 0.$

See http://www.mathlinks.ro/Forum/viewtopic.php?t=47931

[b]Research theme[/b]. Prove that:

$1^{\circ}.\ 60^{\circ} \in \{ A,B,C\} \Longleftrightarrow s=\sqrt 3 \cdot (R+r);$

$2^{\circ}.$ The position of the point O w.r.t. the circle with the diameter $GN$ depends only on

the sign of the expression (difference) $\sqrt 3 (R+r)-s.$

[b]Direction.[/b] The equation $f(x)\equiv rx^3-sx^2+(4R+r)x-s=0$ has the roots        $\cot \frac A2,\ \cot \frac B2,\ \cot \frac C2,$

$f(\sqrt 3)=0$ and equation $f'(x)=3rx^2-2sx+(4R+r)=0$ has a real roots $\Longrightarrow s^2\ge 3r(4R+r).$

I think but that these demonstrations don't compare with the original synthetical proofs.

This is the reason for which I don't agree a metric proof than ultimately. It is my opinion.

_____________________________________________$Farewell,\ my\ dear\ Mathlinks.ro\ !$


integration
trigonometry
real analysis
real analysis unsolved

Evaluate
[size=134]∫[x=0, x=pi/2] tan x/(tan x + sec x)dx [/size]

EDIT: Yeah right, misan; they are too easy; I wanted to post this in the "Solved Problems" category but this forum does not allow me to do so; it said "only user granted special access" can post questions in the "Solved Problems" category.  I just want to see if there is/are any other way(s) to do so (perhaps a much easier way).

Here's another one for you:

Prove that [arccos((√5 - 1)/2)]/pi is irrational.

EDIT: How do I delete this thread?

these are easy questions .... aren't they  :?:

$ \int_0^{\pi/2}\;\frac{\tan\; x}{\sec\; x \plus{} \tan\; x}\;dx \equal{} \int_0^{\pi/2}\;\frac{\tan\; x\; (\sec\; x \minus{} \tan\; x)}{(\sec\; x \plus{} \tan\; x)(\sec\; x \minus{} \tan\; x)}\;dx$

denominator is just $ 1$, so we have

$ \int_0^{\pi/2}\;\sec\; x\tan\; x \minus{} ( \sec^2\; x \minus{}1) \; dx \equal{}   \sec\; x \minus{}\tan\; x  \plus{} x \;\bigg| _0^{\pi/2}$

$ \equal{}\tan \left(\frac{\pi}{4} \minus{} \frac{x}{2}\right) \plus{} x \;\bigg| _0^{\pi/2}\equal{} \boxed{\frac{\pi}{2} \minus{}1}$

The "Solved Problems" forum is for problems that have already been discussed and solved to be moved to. New topic should not be posted there, and only moderators can move to there.

If it's too easy- just don't ask it. If it's classroom level and still interesting, put it in Computations and Tutorials.

Also, don't edit new unrelated questions into old threads.


Euler
combinatorics solved
combinatorics
graph theory

Show that given a directed graph with $n$ nodes, where $n$ is even, such that:
vertex 1 is joined to vertex 2, vertex 2 is joined to vertex 3 and vertex 4, vertex 3 is joined to vertex 5 and vertex 6, ..., vertex $(n/2)$ joined to vertex $n-1$ and vertex $n$, vertex $(n/2) +1$ joined again to vertex $1$ and vertex $2$, ..., vertex $n$ joined to vertex $n-1$
[for each $k$ with $1 \leq k \leq n$, the vertex $k$ is joined to the vertices $2k$ and $2k-1 \bmod n$]
we can always find an Euler tour of the graph going along the edges in the right direction.

(submitted by Pieta Brown - NZ IMO 2003 Team)

You just have to prove that $d^-(A_i) = d^+(A_i)=\text{even}$. Then Euler's theorem and that's it.  :)


I assume you mean on an n*n square we can't have n in a row horizontally, vertically, or diagonally. Yes, n^2-n is correct. We can't have more than that because we can't have n in any horizontal row, so the maximum we can have is n-1 in each horizontal row, which gives n*(n-1) Xs. This is possible to achieve by leaving a diagonal blank and filling each other square with an X.

[color=cyan]I think you can take the idea for squares and press it into rectangles without too much work.  If you get that, you might want also to try triangles and see what happens.  I don't really know what would happen in that case, but it should be doable, right?

Wait . . . are you sure that your formula works for even values of n?  I think you may need to get rid of one extra square in order to knock out all ofthe diagonals.  You may have found a way of doing it that avoids that problem, though.[/color]


inequalities
induction
inequalities unsolved

[color=blue][b]Problem.[/b] If $a_1$, $a_2$, ..., $a_n$ are n reals such that $a_i\geq 1$ for every $i\in\left\{1;\;2;\;...;\;n\right\}$, then prove the inequality

$\frac{1+a_1}{2}\cdot\frac{1+a_2}{2}\cdot ...\cdot\frac{1+a_n}{2}\geq\frac{1+a_1+a_2+...+a_n}{n+1}$.[/color]

[i]Solution.[/i] The core of the proof is the following [b]lemma[/b]: If U, u, V, v are four nonnegative reals such that $U\geq u$ and $V\geq v$, then

$\frac{1+U}{1+u}\cdot\frac{1+V}{1+v}\geq\frac{1+U+V}{1+u+v}$.

This lemma can be proven by direct expanding:

$\frac{1+U}{1+u}\cdot\frac{1+V}{1+v}-\frac{1+U+V}{1+u+v}=\frac{\left(U-u\right)V\left(1+v\right)+\left(V-v\right)\left(1+U\right)u}{\left(1+u\right)\left(1+v\right)\left(1+u+v\right)}\geq 0$

(since $U\geq u$ and $V\geq v$ yield $U-u\geq 0$ and $V-v\geq 0$).

Using this lemma, we can easily solve your problem by induction over n: For n = 1, it is trivial ($\frac{1+a_1}{2}\geq\frac{1+a_1}{2}$), and the induction step from n to n + 1 is

$\frac{1+a_1}{2}\cdot\frac{1+a_2}{2}\cdot ...\cdot\frac{1+a_n}{2}\cdot\frac{1+a_{n+1}}{2}$
$\geq\frac{1+a_1+a_2+...+a_n}{n+1}\cdot\frac{1+a_{n+1}}{2}$             (since we have assumed the problem to be valid for n)
$=\frac{1+\left(a_1+a_2+...+a_n\right)}{1+n}\cdot\frac{1+a_{n+1}}{1+1}\geq\frac{1+\left(a_1+a_2+...+a_n\right)+a_{n+1}}{1+n+1}$
(here we have applied the lemma to $U=a_1+a_2+...+a_n$, u = n, $V=a_{n+1}$, v = 1; note that $U\geq u$ and $V\geq v$ are equivalent to $a_1+a_2+...+a_n\geq n$ and $a_{n+1}\geq 1$ and thus follow from the condition $a_i\geq 1$ for all i)
$=\frac{1+a_1+a_2+...+a_n+a_{n+1}}{\left(n+1\right)+1}$,

so the proof is complete.

  darij

Smooth induction ,nice darij.
Did you know the lemma darij,or you just make it for this problem?
Cause if you dont know the lemma it is not this much easy. :blush:


Prove that if continuous function $f:\mathbb{R}\to\mathbb{R}$ satisfies the equality $f(f(f(x)))\equiv x$, then $f(x)=x$.

Give an example of function $g:\mathbb{R}\to\mathbb{R}$ satisfying $g(g(g(x)))= x$ and $g(x)\neq x$.

[i](3 points)[/i]

the example:
$g(x)=\begin{cases}
2, & x=1\\
3, & x=2\\
1, & x=3\\
x, & \mbox{otherwise}
\end{cases}$

For the continuous case :
Clearly $f$ is injective. Since it is continuous, it has to be monotonic.
In that case $f^{(3)}$ has the same variations than $f$, so that $f$ has to be increasing.
Now assume that $f(x) > x$ for some $x$.
Then $f(f(x)) > f(x)$ so that $f(f(f(x))) > f(f(x))$. Thus $x = f(f(f(x))) > x$, a contradiction.
Therefore $f(x) \leq x$ for all $x$.
We prove in a same way that $f(x) \geq x$ for all x.
Thus $f(x)=x$ for all $x$.
Conversely, $f(x) = x$ is clearly a solution.

Pierre.

The problem isn't new itself [though nice] and its origin is probably 80's of Romania.

Are you kidding? I think it is much more older.

The problem is maybe much more older but I suppose the first time it appeared on the contest was olympiad in '82 in Romania.


$ C_{1}$ is a circle with radius $ 164$ and $ C_{2}$ is a circle internally tangent to $ C_{1}$ that passes through the center of $ C_{1}$. $ AB$ is a chord in $ C_{1}$ of length $ 320$ tangent to $ C_{2}$ at $ D$ ($ AD>BD$). Find $ BD$.

[quote="cancer"]$ C_{1}$ is a circle with radius $ 164$ and $ C_{2}$ is a circle internally tangent to $ C_{1}$ that passes through the center of $ C_{1}$. $ AB$ is a chord in $ C_{1}$ of length $ 320$ tangent to $ C_{2}$ at $ D$ ($ AD > BD$). Find $ BD$.[/quote]
[hide] Let the center of $ C_{1}$ be $ O_{1}$ and the center of $ C_{2}$ be $ O_{2}$. Let the foot of the perpendicular of $ AB$ through $ O_{1}$ be $ E$. Consider the trapezoid $ EDO_{2}O_{1}$. $ O_{1}E\perp ED\perp DO_{2}$. $ O_{1}O_{2}\equal{}DO_{2}\equal{}82$ and $ O_{1}E\equal{}\sqrt{164^{2}\minus{}160^{2}}\equal{}36$. Then, we can solve for $ ED$. $ ED\equal{}\sqrt{82^{2}\minus{}(82\minus{}36)^{2}}\equal{}\sqrt{4608}\equal{}48\sqrt2$. Thus, $ BD\equal{}BE\minus{}ED\equal{}\boxed{164\minus{}48\sqrt2}$.

Correct?   :maybe: [/hide]

Wait...isn't $ BE\equal{}\frac{BA}{2}\equal{}160$?
Giving us a final answer of
[hide]
$ 160\minus{}48\sqrt2$
[/hide]


I'm an Irish student hoping to study maths in England. I was hoping someone who has gone through the system over there (or who is going through it now) could explain some things to me.

 - I'm told that most people need Maths and Further Maths (this a seperate subject right?) I'm doing Irish maths, which is equivalent to A - Level Maths but I'm not doing anything like further maths. Can anyone recomend a book (or even better a website) that covers this course?

 - I've been asked to sit STEP exams. What are these supposed to test? Where can i find information about these?

 - Finally, is there anywhere I can get STEP past papers?

I realise thats a lot of questions but if anyone can even answer one of them that would be a great help.

Thanks!

[quote="stephenryan"]Can anyone recomend a book (or even better a website) that covers this course?[/quote]

A good place to ask your other questions might be the

uk.education.maths

newsgroup on Usenet. You can access that from Google Groups.

http://www.google.com/grphp

or through newsreader software if you have that installed on your machine and have a Usenet feed through your Internet service provider.

For books for further mathematics, I read about the Cambridge Advanced Level Mathematics Series on the newsgroup I mentioned above. Those books are supposed to be very good, but I haven't personally used them. Here are links to the books:

[url=http://www.amazon.co.uk/exec/obidos/ASIN/0521783690/learninfreed]Pure Mathematics 1 and 2[/url] (Cambridge Advanced Level Mathematics S.)

[url=http://www.amazon.co.uk/exec/obidos/ASIN/0521783704/learninfreed]Pure Mathematics 3[/url] (Cambridge Advanced Level Mathematics S.)

[url=http://www.amazon.co.uk/exec/obidos/ASIN/0521783712/learninfreed]Pure Mathematics 4[/url] (Cambridge Advanced Level Mathematics S.)

[url=http://www.amazon.co.uk/exec/obidos/ASIN/0521783720/learninfreed]Pure Mathematics 5 and 6[/url] (Cambridge Advanced Level Mathematics S.)

and those links will lead you to other books.

Hope this helps! I feel I "owe one" to any Irish person I meet, because my son attended a summer math program this summer at DCU on a scholarship for foreign students. One-sixteenth of my son's ancestry is Irish, so I was glad to send him to the ancestral country I have never visited. He had a good time with his Irish classmates, so anything I can do to help any Irishman is something I will always be glad to do.

You can find information about the STEP exams, including how to obtain past papers, at the following link:
[url]http://www.maths.cam.ac.uk/undergrad/admissionsinfo/admissionsguide/text/node6.html[/url].


T belong to L(V)
p belong to P(C)     C: COMPLEX SET

a: belong to C ( Complex Set)

1- Prove that a is an eigenvalue of p(T) [b]if only if [/b] a = p (e) for some eigenvalue e of T

2- Prove  result (1) does not hold if remplace [i][b]C: Complex Set with R: Real Set[/b][/i]

$L(V)$?

$P(C)$ ?

I do not know about the others, but I find myself not able to help simply because i do not understand?

1. Consider the jordan canonical form of T. Let A be T in a jordan canonical basis. Suppose the characteristic equation of A is $(x - e_1)^{p_1}...(x - e_k)^{p_k}$. Then $e_i$ appears on the diagonal for $p_i$ times. Since A is upper triangular, $A^n$ is still upper triangular and its diagonal entries are the diagonal entries of A up to n-th power. So it becomes apparent that p(A) is upper triangular and $p(e_i)$ appears on the diagonal for $p_i$ times. So the characteristic equation of p(A) is $\prod (x - p(e_i))^{p_i}$.
So a is an eigenvalue of p(T) iff a is an eigenvalue of p(A) iff a = p(e_i) for some i.


function
real analysis
real analysis solved

$(X,d)$ a connected metric space, suppose $d(a,b)=1$

Prove that there exist $c\in X$ such that $d(a,c)=\frac{1}{2}$

Any continuous function $f:X\to\mathbb R$ maps $X$ onto a connected subset of $\mathbb R$, i.e. an interval. Since the function $f:X\to\mathbb R,\ f(x)=d(a,x)$ is continuous and takes the values $0,1$, it must take all the values between $0,1$.

More generally, for any real number $t\in (0,1)$, there exists $c$ such that $d(a,c)=t$.

Since the sets $\{x: d(a,x)<t\}$ and $\{x: d(a,x)>t\}$ are disjoint nonempty open sets, their union cannot be all of $X$.


function
ratio
real analysis
logarithms
limit
calculus
calculus computations

For each of the following power series, determine the radius of convergence, and, if possible, identify the function:

1. $\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{(n+1)!}$

2. $x^2-x^3+\left(\frac13+\frac14+\frac13\right)x^4+\cdots+ (-1)^n\left(\sum_{k=1}^{n-1}\frac1{k(n-k)}\right)x^n+\cdots$

3. $\frac12-\frac{x-1}2+\frac{(x-1)^2}4-\frac{(x-1)^4}8+\frac{(x-1)^5}8- \frac{(x-1)^6}{16}$

$+\frac{(x-1)^8}{32}-\frac{(x-1)^9}{32}+\frac{(x-1)^{10}}{64}-\frac{(x-1)^{12}}{128} +\cdots$

[hide="Question 1"]
1) Infinite radius of convergence

$e^{-x} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!} + \cdots + (-1)^n \frac{x^n}{n!} + \cdots$

$\frac{e^{-x}}{x} = \frac{1}{x}-1+\frac{x}{2!}-\frac{x^2}{3!} + \cdots + (-1)^n \frac{x^{n-1}}{n!} + \cdots $

$\frac{1}{x}-\frac{e^{-x}}{x} = \cdots$

Thus the function is
$\frac{e^x-1}{xe^x}$[/hide]

dheardie is correct. I would have written the answer [hide="this way."]$\frac{1-e^{-x}}x.$

Note also that we actually mean this function with its singularity at zero removed; the value at 0 is 1.[/hide]

Bump.

Hey, guys - what about #2 and #3?

[quote="Kent Merryfield"]Bump.

Hey, guys - what about #2 and #3?[/quote]

I dunno.

3 is almost geometric

2 dont know where to start.

#3 combine 3 stuffs?

[hide]
so do inf alt series fromed by powers
0,4,8,12...
1,5,9,13...
2,6,10,14...

they have the same ratio anyway, should be easy

1-sqrt2 < x < sqrt2 + 1

(x^2 - 4x + 1)/(4 + (x-1)^4)

didnt check my work
[/hide]

That's the right interval of convergence. Do check your work; the details of the sum aren't quite right.

[quote="Kent Merryfield"]... which simplifies to 1 over a nice expression.[/quote]

$\frac{1}{x^2+1}$

Nice!

overlooked the possibility of canceling out factors of x^4 + 4y^4

OK, that's #3. Now we have #2 left.

In the realm of power series, what connections do you make with the notion of "convolution":

$\sum_{k=0}^na_{n-k}b_k$  ?

A product of 2 series with the same centre?

So this is the idea:

$
f(x)g(x)=\left(\sum_{n=0}^{\infty} a_n (x-c)^n\right)\left(\sum_{n=0}^{\infty} b_n (x-c)^n\right) \\
= \sum_{k=0}^\infty \sum_{j=0}^\infty a_k b_k (x-c)^{k+j} \\
=\sum_{n=0}^\infty \left(\sum_{k=0}^n a_k b_{n-k}\right)(x-c)^n$

[quote="Kent Merryfield"]Are you going to answer the original question, dheardie? You're on the right track.[/quote]

Sorry I forgot

Since

$\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n} = \ln(x+1)$ we get the original expression as
[hide]$\ln[(x+1)]^2 $ with radius 1. At $x=-1$ the logarithm is underfined so the series cannot converge there. $x=1$ is like the well known alternating harmonic series, which converges by the AST. Thus the interval of convergence is $-1 < x \le 1$[/hide]

Almost but not quite right, dheardie. That's $[\ln(1+x)]^2$, which is most certainly not the same thing as $\ln[(1+x)^2]=2\ln(1+x).$

If you think about the series, you'd realize the we'd have to have

$\ln[(1+x)^2]=2\ln(1+x)=2x-\frac{2x^2}2+\frac{2x^2}3+\cdots.$

Isaac got it right.

In the complex plane, we would have convergence for all $|z|\le1$ except for $z=-1.$

Could we have identified the radius of convergence without identifying the function?

Consider this: for $1\le k\le n,$ $n\le k(n-k)\le \frac{n^2}2.$ So

$\frac2{n^2}\le\frac1{k(n-k)}\le\frac1n$ and

$\frac2n\le\sum_{k=1}^n\frac1{k(n-k)}\le1.$

If $\frac2n\le a_n\le 1$ then $\lim_{n\to\infty}|a_n|^{1/n}=1$ and the radius of convergence of the series is 1.

The comparison to $\frac1n$ is sufficient to show that the series diverges at $x=-1.$

Since I didn't actually show that that the coefficients tend to zero, this estimate isn't strong enough to show that the series converges at $x=1.$ I'm not going to reproduce the argument here, but it is possible to show that the sum

$\sum_{k=1}^n\frac1{k(n-k)}$ behaves asymptotically like [edit - see next post by jmerry.]

[quote="Kent Merryfield"]Almost but not quite right, dheardie. That's $[\ln(1+x)]^2$, which is most certainly not the same thing as $\ln[(1+x)^2]=2\ln(1+x).$[/quote]
I have no idea why I wrote that. Actually I do. I texed it without the extra bracket, and then thought 'hmmm - that can be simplified'

I am an idiot sometimes!


Mr. Green wrote a number on the white board and said, "I know that to be sure this number is prime, I must check each prime divisor from 2 to 29." What is the least number Mr, Green could have written? What is the greatest number he could have written?
 :D
This problem makes no sense what so ever to me. :wallbash:

[hide="Hint"]
In general, to check if $ n$ is prime or not, you only need to check the prime numbers that are less than or equal to $ \sqrt{n}$.
[/hide]

The least is 29^2 and the greatest is 30^2-1

Isn't the greatest number $ 31^2\minus{}1$ since 31 is the smallest prime number bigger than 29?

[quote="AIME15"]I guess it depends how you interpret the question:  maybe all numbers that are prime from 2-29, or all prime numbers from 2-29.  In the first case, you'd do my answer, in the second, it'd be yours.[/quote]

I'm sorry, but I don't exactly understand what you mean. Can you explain the difference between the two interpretations of the question? Because in my opinion, $ 31^2 \minus{} 1$ would be the largest because anything lower works but isn't the greatest, and $ 31^2$ includes checking if 31 works?

He said all prime divisors, so it is 31^2-1 (one might argue since he would instantly notice that it is divisble by 2 and 3 and 5, it should be a different number)

which is *kinda* checking for 2, 3, and 5

I'm pretty sure it is $ 31^2\minus{}1$ for the greatest. Try $ 907$ for example. It's greater than $ 899$ and it is prime. You still have to check numbers up to $ 29$

Right, but we're saying 31^2-1 is even so you only have to check for divisbility by 2 to be prime.


Here's a very interesting question of logic:
What can you conclude from the following statements:
If the line is long, then Quincy will go home.
If the line is long, then it is morning.
If it is morning, then Quincy will not go home.

Whenever the line is long, the three statements contradict.  This is known as the [url=http://en.wikipedia.org/wiki/Liar_paradox]Liar's paradox[/url], and it is very old and very well-known.


Let $ a,b,c$ are positive real numbers and $ t\ge \sqrt{3}\minus{}1$ prove that

\[ \frac{a^2}{(a\plus{}tb)^2}\plus{}\frac{b^2}{(b\plus{}tc)^2}\plus{}\frac{c^2}{(c\plus{}ta)^2}\ge \frac{3}{(1\plus{}t)^2}\]

[quote="kenn4000"]$ f(a,b,c) \equal{} \frac {a^2}{(a \plus{} tb)^2} \plus{} \frac {b^2}{(b \plus{} tc)^2} \plus{} \frac {c^2}{(c \plus{} ta)^2}$ for a,b,c in first octant
$ \frac {a}{b} \equal{} \alpha, \frac {b}{c} \equal{} \beta, \frac {c}{a} \equal{} (\alpha \beta)^{ \minus{} 1}$ now $ \alpha, \beta$ in first quadrant

$ f(\alpha, \beta) \equal{} \frac {1}{(1 \plus{} t\alpha)^2} \plus{} \frac {1}{(1 \plus{} t\beta)^2} \plus{} \frac {1}{(1 \plus{} \frac {t}{\alpha \beta})^2}$

thinking about the gradient, it will be symmetric with respect to $ \alpha$ and $ \beta$, so any solution will have $ \alpha \equal{} \beta$

$ \partial_{\alpha}f(\alpha,\beta) \equal{} \frac { \minus{} 2t}{(1 \plus{} t\alpha)^3} \plus{} \frac {\frac {2t}{\beta \alpha^2}}{(1 \plus{} \frac {t}{\alpha \beta})^3} \equal{} \frac { \minus{} 2t}{(1 \plus{} t\alpha)^3} \plus{} \frac {\frac {2t}{\alpha^3}}{(1 \plus{} \frac {t}{\alpha^2})^3}$
=$ \frac { \minus{} 2t}{(1 \plus{} t\alpha)^3} \plus{} \frac {2t\alpha^3}{(\alpha^2 \plus{} t)^3} \equal{} \frac {2t}{(1 \plus{} t\alpha)^3(\alpha^2 \plus{} t)^3}((\alpha \plus{} t\alpha^2)^3 \minus{} (\alpha^2 \plus{} t)^3)$

so really we are solving $ (x \plus{} tx^2)^3 \equal{} (x^2 \plus{} t)^3$

more messily: $ x^3 \plus{} 3tx^4 \plus{} 3t^2x^5 \plus{} t^3x^6 \equal{} x^6 \plus{} 3tx^4 \plus{} 3t^2x^2 \plus{} t^3$
less messily: $ x^6(t^3 \minus{} 1) \plus{} x^5(3t^2) \plus{} x^3 \plus{} x^2( \minus{} 3t^2) \minus{} t^3 \equal{} 0$
we can easily see that 1 is a solution, but is that all?
im not so good with my sixdratic formula

hoping its just 1, we get our critical points at a=b=c, where [b]equality[/b] holds.[/quote]

Good ,but i solved it with only cauchy


[quote="hunter_cvp_ic"]Let a,b,c are nonnegetive real number such:$\sum_{cyc}{a^{2}}=1$.Prove that : $\sum_{cyc}{\frac{1}{1-ab}}$[/quote]
Did you mean $\sum_{\text{cyc}}{\frac{1}{1-ab}}\leqslant \frac92$? It's called [i]Vasc's inequality[/i] (it not a unique inequality with this name). The shortest proof is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=109029]here[/url].


ratio
geometric sequence
number theory
relatively prime

Let $ a,\, b,\, c,\, d,\, e$ be a 5-term geometric sequence where $ a,\, b,\, c,\, d,\, e$ are integers and $ 0 < a < b < c < d < e < 100$. What is the sum of all possible values of $ c$?

[hide="Solution"]Consider possible series.

Let k=the common multiplier.

$ e\equal{}ak^4$

If $ a\equal{}1$, then in order for $ b$ to be an integer, $ k$ must also be an integer.
In addition, $ k^4 < 100$

All terms are different so $ k \ne 1$

The possible values of $ k$ for $ a\equal{}1$ are therefore $ 2,3$

These give $ c\equal{}4,9$

Now consider $ a\equal{}2$

$ k$ must be an integer in order for all terms to be integers

In addition, $ k^4 < 50$

The possible value of $ k$ for $ a\equal{}2$ is therefore $ 2$

This gives $ c\equal{}8$

I will no longer continue to post ALL my working, as I'm sure the method is obvious enough by now.

For $ a\equal{}3$, we get $ c\equal{}12$

For $ a\equal{}4$, we get $ c\equal{}16$

For $ a\equal{}5$, we get $ c\equal{}20$

For $ a\equal{}6$, we get $ c\equal{}24$

For $ a\equal{}7$ there are no possibilities.

Thus the sum is $ 4\plus{}9\plus{}8\plus{}12\plus{}16\plus{}20\plus{}24\equal{}93$[/hide]

Let us see all the possible cases for the first term, 'a'.

common ratio = r
$ 0 < a < b < c < d < e < 100$

[b][u]Case 1 : [/u][/b]

r=2

$ a \equal{} 1,2,3,4,5,6$
  $ \Rightarrow c \equal{} 4,8,12,16,20,24$

So, sum of all c's = 84

[b][u]Case 2 :[/u][/b]

r=3

$ a \equal{} 1$
   $ \Rightarrow c \equal{} 9$

So, sum of all c's = 9

We find that $ r\ngtr3$.

So, sum of all possible values of c = 9+84 = 93

Pl. correct me if i am wrong

@Adharsh Raghavan

r=1 is impossible because the terms are strictly increasing

Sorry, Didn't notice that.  :blush:

Now i 've edited it

Both of you are assuming things not given in the question, and so you've missed a solution.
[hide="Hint"]Try $ a \equal{} 16$.[/hide]

[hide="The bit that I missed..."]Other series can be given by $ a\equal{}n^4,r\equal{}1\plus{}\frac{1}{n}$

$ a\equal{}n^4$
$ b\equal{}n^4\plus{}n^3$
$ c\equal{}n^4\plus{}2n^3\plus{}n^2$
$ d\equal{}n^4\plus{}3n^3\plus{}3n^2\plus{}n^1$
$ e\equal{}n^4\plus{}4n^3\plus{}6n^2\plus{}4n\plus{}1$

The only values of $ n$ for which these are all integers and $ e<100$ are $ n\equal{}1,2$. However we have already counted the series with $ n\equal{}1, r\equal{}2$

Thus another possible value of $ c$ is $ 2^4\plus{}2 \cdot 2^3 \plus{} 2^2\equal{}36$

This gives a revised sum of the possible values of $ c$: $ 93\plus{}36\equal{}129$[/hide]

[quote="KrazyFK"][hide="The bit that I missed..."]Other series can be given by $ a \equal{} n^4,r \equal{} 1 \plus{} \frac {1}{n}$

$ a \equal{} n^4$
$ b \equal{} n^4 \plus{} n^3$
$ c \equal{} n^4 \plus{} 2n^3 \plus{} n^2$
$ d \equal{} n^4 \plus{} 3n^3 \plus{} 3n^2 \plus{} n^1$
$ e \equal{} n^4 \plus{} 4n^3 \plus{} 6n^2 \plus{} 4n \plus{} 1$

The only values of $ n$ for which these are all integers and $ e < 100$ are $ n \equal{} 1,2$. However we have already counted the series with $ n \equal{} 1, r \equal{} 2$

Thus another possible value of $ c$ is $ 2^4 \plus{} 2 \cdot 2^3 \plus{} 2^2 \equal{} 36$

This gives a revised sum of the possible values of $ c$: $ 93 \plus{} 36 \equal{} 129$[/hide][/quote]

How can you claim that non-integer values of r must be in the form of $ \frac{n\plus{}1}{n}$ without any investigation?

[b]Solution:[/b]
I separated it into 2 cases, r being an integer and r not being an integer. I solved that the integer cases gave a sum of 93. In the integer cases I also saw that $ r<4$ and since it's increasing $ r>1$, so I said $ 1<r\equal{}\frac{x}{y}<4$ with $ x$ and $ y$ relatively prime integers. The first term, $ a\equal{}ky^4$ and the last term $ e\equal{}kx^4$. But since $ y>x$ even in our first case of $ y\equal{}2,x\equal{}3$, $ 81k<100$ only for k=1. It then follows that $ kx^4$ is always greater than 100 for any greater value of x.

Moving on to $ y\equal{}3$, the first case is $ x\equal{}4$, but $ 256k>100$ for any value of k. After this case we see that the only non-integer value of r that is possible is $ r\equal{}\frac{3}{2}$, and the only first term that is possible with this value of r is 16, giving $ c\equal{}36$. Adding $ 36\plus{}93\equal{}129$


inequalities
trigonometry
algebra unsolved
algebra

detremin all x in [0,2pi] satisfaying
$81 sin^{10}x+cos^{10}x =\frac{81}{256}$

Maybe we can solve it alos by inequality:
By holder
$(81\sin^{10}x+\cos^{10}x)^{\frac15}(1+3)^{\frac{4}{5}}\geq (sin^{2}x+cos^{2}x)3^{\frac45}$


Find the set of values of x for which the following inequalities holds:

1. [tex] \displaystyle {\frac {1}{x} + \frac {3}{2x} \ge 5} [/tex]

2. [tex] \displaystyle {x(x + 2) \le 24} [/tex]

3. [tex] \displaystyle {|x + 2| < |x - 5|} [/tex]

4. [tex] \displaystyle {\frac {x}{x+2} > \frac {x+3}{3x+1}} [/tex]

[hide]For the first one.

You would move everything to one side and simplify as one fraction.

(1/x)+(3/2x)-5 :ge: 0   reduces to (5-10x)/2x :ge: 0

If we find the zeros and the vert. asymptotes we can do a sign analysis.

Zero at x=1/2.  vert at x=0

Sign analsis:  negative less than 0,  positive between 1/2 and 0, and negative greater than those two.  Also note it is undefined at x=0 and it is equal to 0 at x=1/2.

so 0<x :le: 1/2

2.  Multiply out to get x :^2: +2x-24 :le: 0.  Once again I use a sign analysis.  Zeros at -6 and 4.  To the negative side of -6 all numbers are positve, in between the 2 numbers it is negative and greater than 4 they are positive.  Since 0 is including(less than or equal to) the answer is -6 :le: x :le: 4

3.  Im getting tired so I did this one by just tinspecting it.  (I also don't like abs. value).

x will work when it is zero or less.  So now we just have to find where the two values are equal(also the zero).  This occurs when x=3/2.  Anything above this doesn't work and everything below it does.  It can't be equal to zero so the answer is x<3/2[/hide]

I feel that this is more straightforward for #1:

[hide]Factor to obtain

[tex] \frac{1}{x}*(1+\frac{3}{2}) \ge 5 \\ \frac{1}{x} \ge 2[/tex]

Negative x obviously fails.  For positive x,

[tex] 1 \ge 2x \\ \frac{1}{2} \ge x > 0[/tex][/hide]

That's easier but I just don't like multiplying inequalities by x because of the sign change ambiguity issue.  So I just avoid it as much as possible.

[quote="gauss202"]

2. [tex] \displaystyle {x(x + 2) \le 24} [/tex]

[/quote]

Subtracting 24 from both sides and distributing, we have [tex]x^2+2x-24 \le 0 [/tex].  Since the function is convex, it is only non-positive between the two zeroes.  Therefore, [tex]-6 \le x \le 4[/tex].

[hide]For the last one I once again bring everything to one side and simplify.

[2(x :^2: -2x-3)]/[(x+2)(3x+1)]>0

I get the zeros and vert. asymptotes.  Then I do a sign line.

I get as an answer.  x<-2 or -1<x<-1/3 or x>3[/hide]


Even though college is a long time away I still wonder what I should do for my career. Im really gonna be an author, but i need a second job. as wat? i love math but i dont know a job that involves math and still gives me time to write

I'm gonna try engineer and part time author...dunno how thats gonna work but I remember seeing a person on TV who did that and was very successful

yah a sub was surprised i finished something in a few secs. she said i should become an engeineer


Our chapters based on number of schools particating like for every 20 schools their is a chapter or a chapter is all schools within a 100 mile radius?
BTW I just picked random numbers for my examples.

It depends alot on the school's population

[quote="monkeygirl12"]So you are saying that chapters are based on population, like for example for every x people there is a chapter?[/quote]

That's probably determined by the state.

Like my chapter, top 10 make it.

It all depends on how your state divides it, and different states do it different ways. In my state, it is by counties (I think).


algebra
polynomial
vector
complex numbers
superior algebra
superior algebra unsolved

Let $S_n$ be the set of homogenuous polynomials of 2 variables over $C$ of degree n. Suppose that $I_n$ is a subset of $S_n$ different from $S_n$ and such that $I_n S_p$ is included in $I_{n+p}$ for all $n,p$ (n starts from 1). here $AB$ is the set of products $xy$ with $x$ in $A$ and $y$ in $B$. Prove that there is a non-zero couple of complex numbers that annihilate any polynomial in the union of all $I_n$.

I don't think $I_n$ can be allowed to be just subsets. Think of this counterexample: take $I_0=\{0\},\ I_1$ to be the set of all $\mathbb C$-multiples of $x$ or $x+y$, and then $I_n=I_1S_{n-1},\ \forall n\ge 1$. Then for every $n\ge 1,\ y^n$ is not an element of $I_n$, so the hypotheses are satisfied. However, a pair of complex numbers which annihilates all elements of all $I_n$ must, in particular, annihilate both $x$ and $x+y$, so these complex numbers must both be $0$.

If, on the other hand the $I_n$ are required to be $\mathbb C$-vector spaces, then it is indeed true. This could be regarded as a sort of [url=http://planetmath.org/encyclopedia/Nullstellensatz.html]weak Hilbert Nullstellensatz[/url] in a homogeneous version. If we let $\frak a$ be the ideal of $A=\mathbb C[x,y]$ generated by $I_n,\ n\ge 0$, then some $I_n$ is equal to $S_n$ iff $\frak a$ contains that $S_n$ (this iff needs the hypothesis that $I_n$ are $\mathbb C$-vector spaces), iff the radical $\sqrt\frak a$ of $\frak a$ is either $A$ or $A_+=\bigoplus_{n>0}S_n$. The conclusion we want to reach can then be formulated thus:

Let $\frak a$ be a homogeneous ideal (i.e. ideal generated by homogeneous elements) of $A$. Also, for a homogeneous ideal $I$, let $V(I)$ be the projective algebraic set that corresponds to it (the set of all points on the complex projective line which annihilate all homogeneous elements of $I$). Then $V(\frak a)=\emptyset$ implies that $\sqrt\frak a$ is either $A$ or $A_+$. This makes it look a lot more like the usual weak Nullstellensatz, and in fact, we can use the usual strong Nullstellensatz to prove it (it's true for any ring $A=\mathbb C[x_0,\ldots,x_n]$, not just for two variables):

Suppose $V(\frak a)=\emptyset$, and let $f$ be a homogeneous polynomial of degree $>0$. The $f$ vanishes in all non-zero points of $\mathbb C^2$ where $\frak a$ vanishes simply because the set consisting of such points is empty :), and it also vanishes in $(0,0)$, so it vanishes in all points of $\mathbb C^2$ where $\frak a$ does. By Hilbert's (strong) Nullstellensatz, $f^q\in\frak a$ for some $q>0$. This shows that $\sqrt\frak a$ contains $A_+$, and since $A_+$ is maximal, $\sqrt\frak a$ is either $A_+$ or $A$ itself.


Find all functions [tex] f:(0,\infty)\to (0,\infty)[/tex] such that : [tex]f(f(x)+y)=xf(1+xy)[/tex] for all positive real numbers [tex]x,y[/tex]

but $f : (0 , \infty ) \rightarrow (0, \infty ) $ so $x\neq 0$

but i guess u can say $\lim_{x\to 0} f(f(x)+y) = \ldots$

but i still dont know if u can say $x>\lim_{x_0\to 0 }f(x_0)$

peeta

Guys, when will you understand that spamming is not allowed here?

And here is continuation:Fill all f :P*->R*:
f(1+xf(y))=yf(x+y)
It's very nice and quite hard...

[quote="euler_vn"]Find all functions $f: (0,\infty)\to (0,\infty)$ such that : $f(f(x)+y)=xf(1+xy)$ for all positive real numbers $x,y$[/quote]

Realy sweet problem :).

Replace $xy$ with $a$ and rewrite the condition as $f\left(f(x)+\frac ax\right)=xf(1+a),\ \forall a,x>0\ (*)$. First of all, this clearly shows that $f$ is onto, since we can fix $a$ and change $x$ as we please to get all positive reals in the RHS. Second of all, from here we can also find $f$ to be non-increasing: suppose we can find $0<x<y$ with $f(x)<f(y)$. Then $\frac{f(y)-f(x)}{\frac 1x-\frac 1y}>0$, and if we take $a=\frac{f(y)-f(x)}{\frac 1x-\frac 1y}$, we get $f(x)+\frac ax=f(y)+\frac ay$, and together with $(*)$ this gives $xf(1+a)=yf(1+a)\Rightarrow x=y$, contradiction. Since $f$ is surjective and monotonic, it must be continuous.

Assume $f(x_1)=f(x_2)$, then $x_1f(1+x_1y)=x_2f(1+x_2y)$, and if we make $y\to 0$ we get $x_1f(1)=x_2f(1)\Rightarrow x_1=x_2$, so $f$ is also one-to-one. By making $x=1$ in the initial equation we get $f(f(1)+y)=f(1+y),\ \forall y>0$, and from the fact that $f$ is one-to-one we deduce that $f(1)=1$.

In the initial equation we make $y\to 0$ and by continuity we get $f(f(x))=xf(1)=x,\ \forall x>0$. Now make $y=1$ in the initial equation. We get $xf(1+x)=f(1+f(x))\ (**)$. Apply $(**)$ to $f(x)$ to find $f(x)(f(1+f(x)))=f(1+f(f(x)))=f(1+x)=\frac{f(1+f(x))}x$, which means that $f(x)=\frac 1x,\ \forall x>0$, and we’re done.


geometry
perpendicular bisector
geometry proposed

Point $ D$ is chosen on the side $ AB$ of triangle $ ABC$. Point $ L$ inside the triangle $ ABC$ is such that $ BD=LD$ and $ \angle LAB=\angle LCA=\angle DCB$. It is known that $ \angle ALD+\angle ABC=180^\circ$. Prove that $ \angle BLC=90^\circ$.

Extend $ AL$ to meet $ BC$ at $ E$. Now $ \angle DLE = 180^{\circ}-\angle ALD = \angle ABC$. Since $ \angle DLB = \angle DBL$ (triangle $ DLB$ is isosceles) we must also have $ \angle ELB = \angle EBL$ so the triangle $ ELB$ is isosceles too. It means that $ ED$ is a perpendicular bisector of $ LB$ so $ \angle LED = \frac{1}{2}\angle LEB = 90^{\circ}-\angle LBE$. We also have $ \angle LEC = 2\angle LBE$ so $ \angle DEC = 90^{\circ}+\angle LBE$. Since $ \angle EAD = \angle DCE$ quadrilateral $ ADEC$ is cyclic so $ \angle DAC= 90^{\circ}-\angle LBE$. Note that $ \angle ACD = \angle LCE$ (from the conditions) and $ \angle ADC = \angle AEC$ (because $ ADEC$ is cyclic) so the triangles $ ADC$ and $ LEC$ are similar. Therefore $ LEC = \angle DAC = 90^{\circ}-\angle LBE$. So $ \angle BLC = \angle BLE+\angle ELC = 90^{\circ}$ and we are done.

Let $ AL$ meet $ BC$ at $ L_1$, $ \angle{DLL_1}\equal{}\angle{ABC}$ $ \Rightarrow$ $ LL_1\equal{}BL_1$, quadrilateral $ ADL_1C$ is cyclic, because $ \angle{DAL_1}\equal{}\angle{L_1CD}$ then, $ \angle{DL_1A}\equal{}\angle{DCA}$ $ \Rightarrow$ $ \angle{LCB}\equal{}\angle{DL_1A}\equal{}\angle{DL_1B}\equal{}90^o\minus{}\angle{LBL_1}$ then $ \angle{BLC}\equal{}90^o$

the main idea of this problem is prove ADIC is cyclic( many proof). Then <DIB=<A
so <ALI= 180-2<A. then <LCI = <A(1) ,but we have <LCB = 90-<DIB=90-<A(2)
From (1) and (2) we have <BLC= 90

Ya I was trying to follow the explanation but my geometry is not all up to date, is there like a online article or something I could read that would make the explanation more clear.

Sorry to revive it... a nice one indeed,
Let $LD$ meet $CB$ again at $X$. Then just note that our angle condition gives $A,L,B,X$ cyclic and moreover a trapezoid. Now its finished by some angle chase...


trigonometry
calculus
calculus computations

1. dy/(1+y^2) = dx/(1+x^2)

How can I solve this (solve for y in terms of x)

2. Can differentials be treated like normal terms? Such as
can (3x + y) dz be rewritten as 3x dz + y dz?

Any help is appreciated
thanks.

1: This is a separable equation. Integrate both sides to get $\arctan y=\arctan x+c$, then take the tangent of both sides and use the addition formula.

2: Yes, provided the expressions make sense. After all, $(f dz)(v)=f\cdot dz(v)$ by definition.


helo

Find all the solutions in $ \mathbb{Z\times Z}$of the following equations:
$ x^2 \plus{} xy\minus{}2y^2 \plus{} 5 \equal{} 0$
$ x^2 \minus{} y^2x \plus{} 3y \minus{} 30 \equal{} 0$
$ xy \plus{} 2x \minus{} 3y \equal{} 7$

[hide="Hints for #1 and #3"]Rewrite those as $ x^2\plus{}2xy\minus{}xy\minus{}2y^2\equal{}\minus{}5$ and $ xy\plus{}2x\minus{}3y\minus{}6\equal{}1$ respectively, then factorize[/hide]


function
algorithm
induction
algebra proposed
algebra

Let $Q^+$ be the  set of the positive rational numbers .
Find all the functions $f:Q^+ \times  Q^+ \rightarrow R$(real) such that :
 $f(1,1) = 1$
 $f(x,y) = f(y,x) , \forall  x,y \in Q^+$
  $(x+y)*f(x,y) = y*f(x,x+y) , \forall  x,y \in Q^+$

 :cool:  The shortest proofs are the best  :cool:

I might just be temporarily blinded by extreme fatigue, but it seems to me like $g(a)=f(a,a)$ are completely independent from one another.

:?  I'm very confused , I made a mistake :?  Thank you Grobber :)
The first condition is $f(x,x)=1, \forall x \in Q^+ $
 :cool:  Sorry again  :cool:

After employing Euclid's algorithm, the function seems to be $f(x,y)=mn$, where $\frac xy=\frac mn$ s.t. $\frac mn$ is in its lowest terms (in other words, $(m,n)=1$). Is it true?

:D  May be , but what it's your proof ?  :D
 :cool:

Here my proof :lol:

Let y/x=m/n and m/n is irreducable fraction. We will use induction on (m+n).
1) Base of induction m+n=2 --> m=n=1 --> y=x --> f(x,y)=1=mn.
2) Suppose m+n>2 and WLOG y>x. We have y*f(x,y-x)=(y-x)*f(x,y), so f(x,y)=f(x,y-x)*m/(m-n), but (y-x)/x=(m-n)/n and (m-n)/n is irreducable, moreover (m-n)+n < m+n, thus we use unduction proposition and obtain f(x,y)=(m-n)n*m/(m-n)=mn.


[quote="smokewisps"]Does anyone know any good problems involving pigeonhole principle? Any help is appreciated.[/quote]

[b](a)[/b] Find all values θ with 0 <= θ <= 2π such that sin θ − √3 cos θ = 1.

[b](b)[/b] The difference between a real number r are the greatest integer less than or
equal to r is called the fractional part of r, F(r).Thus F(3.45) = 0.45. Note that
for all real numbers r, 0<= F(r) < 1.

[b](i)[/b] Let a = 2136 log10 2.
Given that
F(a) = 7.0738 · · ·×10−5
observe that F(2a) = 14.1476 · · ·×10−5
F(3a) = 21.2214 · · ·×10−5
[b](α)[/b] Use your calculator to show that log10 1.989 < F(4223a) < log10 1.990.
[b](β)[/b] Hence calculate an integer M such that the ordinary decimal representation of
2^M begins with 1989.Thus 2^M=1989 . . . .

[b](ii)[/b] Let r be a real number and let m and n be non-zero integers with m=/= n.
([b]α)[/b] Show that if F(mr) = 0, then r is rational.
[b](β)[/b] Show that if F(mr) = F(nr), then r is rational.

[b](iii)[/b] Suppose that b is an irrational number. Let N be a positive integer and consider
the fractional parts F(b), F(2b), . . . , F ((N + 1)b).
[b](α)[/b] Show that these N + 1 numbers F(b), . . . , F ((N + 1)b) are all distinct.
[b](β)[/b] Divide the interval 0<= x < 1 into N subintervals each of length 1/N and show
that there must be integers m and n with m =/= n and 1<=m, n<=N + 1 such that
F((m − n)b) < 1/N.

[b](iv)[/b] Given that log10 2 is irrational, choose any integer N such that 1/ N <log10 (1990/1989);
note that in ([b]i)[/b] , F(a) < log10 (1990/1989) .
Use [b](iii) [/b]to decide whether there exists another integer M such that 2^M = 1989 . . . .

It's a really long Q but give that a go (everyone).


When do you guys think we will get our scores?
I'm guessing in 1 week or so.

[quote="DPatrick"]We're aiming for sometime during the week of Jan 24th.  (This is not a guarantee though.)[/quote]

Can't wait!

Probably the first week of February.  It will be the second week of February at the absolute latest (since we guarantee that you'll get at least a month to work on the problems, and the Round 4 solutions are due on March 14th).

Are we going to get our score back before the first AMC (Feb. 1st)?

Barring any trouble with the mail (the papers and grading forms are being shipped to us from NSA), we should have the grades released by the end of this week.

[quote="Ultimatemathgeek"]Have any idea what the AIME cut score would be this year?[/quote]

68

[quote="sam3.14"]Don't you think it's [i]much[/i] harder to get a 68 on the USAMTS than a 100 on the AMC-12?[/quote]
True, but since USAMTS is an honor-based contest, that cut-off line seems appropriate...

By the way, I just checked the USAMTS site, and it already shows my total score (but not how I did for individual problem or the comments). I'm quite happy.  :)

I am impatiently waiting for my score... I hate waiting. That is why I will never succeed lol.

[quote="Ultimatemathgeek"][quote="sam3.14"]Don't you think it's [i]much[/i] harder to get a 68 on the USAMTS than a 100 on the AMC-12?[/quote]

Well, that is what most people are saying, but since I suck at taking timed tests, I like the USAMTS.[/quote]

I know what you mean...
119 on AMC10 two years ago...
94.5 on AMC 12 last year...

but AMC is a very stressful test, probably the most stressful competition here!

Talk about stress, we take 60 problem math tests in 40 minutes, 70 problem tests in 30 minutes, and 80 problem tests in 10 minutes.  Ugghhh, I can't stand them much longer

hmm.... just got my score back.

Total 70  :(

I guess I barely passed the AIME cut score...........

Does this mean I can't get a gold award?  only 96-100 gets the gold one :(  Max score possible for me is 95...


algebra
binomial theorem
combinatorics unsolved
combinatorics

[quote="darij grinberg"]

I don't understand you. (Or is it not supposed to be a proof?)[/quote]

expand this using the binomial theorem and you get the result it was a proof  :)

[quote="darij grinberg"][quote="a_vakilian"]It is  $(1+1)^{2n}=4^n$[/quote]

I don't understand you. (Or is it not supposed to be a proof?)

Anyway, the problem was discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=40150 .

  darij[/quote]
i think it is obvious !! these are the coefficient  of  $(1+1)^{2n}=4^n$ !!


calculus
integration
function
limit
real analysis
real analysis solved

If $f: (0, \infty) \to [0,1]$ is a continous function, compute $\displaystyle \lim_{x \to \infty} \frac{ \int_{0}^{x} f^{2} (t) dt}{x} = 1,$ if $\lim_{x \to \infty} f(x)=1.$

Just use l'Hospital. Put $F(x)=\int^x_1 f^2(t)dt$. We need to compute $\lim_{x\to\infty}\frac{F(x)}x$ ($\int^1_0$ doesn't matter, because it's a finite quantity over $x\to\infty$). We can apply l'hospital because $F(x)\to\infty$ as $x\to\infty$.

[quote="georg"] compute $\displaystyle \lim_{x \to \infty} \frac{ \int_{0}^{x} f^{2} (t) dt}{x} = 1$[/quote]

What we have to compute ?

I assumed we had to show that the limit was $1$.


Hello, I have recently downloaded & installed the TeXer and am using it. But when it comes tio making a text "bold" or "italic", or when I have to insert a break (in html the code is <br>) I am in a fix!
Please help me. :blush:

[quote="stevem"]You mean you downloaded LaTeX. The TeXer is a facility on AoPS for trying out bits of LaTeX and you can't download that.
The bold command is \textbf for text or \mathbf for maths, for example vectors. \\ is a new line
[code]\textbf{this is in bold} but this isn't \\
This is on a new line and $\mathbf{v}$ is a bold vector[/code]
There are lots of online LaTeX tutorials and [url=http://andy-roberts.net/misc/latex/]Getting to grips with LaTeX[/url] is one for complete beginners. You'll find more help in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=117360]LaTeX How-Tos[/url].[/quote]
Thanks alot!


There are m books that will be reviewed by n people. Each book needs to be reviewed k times. How many such combinations are there?

m factorial/ n factorial times k factorial? or k!m! over n!

Amuse me because I haven't seen one like this before.

Can you specify. Not to be stupid, but to bring up a small thing that can affect the problem. Do different people have to review it each time?

(I know this is contrary to common sense, but if the same person is allowed to view the book multiple times, it changes the problem.)

yup, it should read "There are m books that will be reviewed by n people. Each book needs to be reviewed k times, [b]each time by a different reviewer[/b]. How many such combinations are there?


1. There should be delete next to edit, but you can't delete after some (I don't know which) period of time, like editing is forbidden after 48 hours.
2. You see the URL rectangle above the text space when you are writing a post? Put

[code][url=www.artofproblemsolving.com][/url][/code] for AOPS, and if you want a smiley or text put

[code][url=www.artofproblemsolving.com]AOPS :lol: [/url][/code]

Edited, thanks to 007math

Actually it's:

[code]
[url=ENTER YOUR URL]  :D  [/url][/code]

To get:

[url=http://google.com]:D[/url]

[quote="Bugi"]1. There should be delete next to edit, but you can't delete after some (I don't know which) period of time, like editing is forbidden after 48 hours.
2. You see the URL rectangle above the text space when you are writing a post? Put

[code][url]www.artofproblemsolving.com[/url][/code] for AOPS, and if you want a smiley or text put

[code][url=AOPS :lol: ]www.artofproblemsolving.com[/url][/code][/quote]

after the next post


probability
real analysis
real analysis unsolved

Prove that the space of Borel probability measures on a complete metric space is itself a complete metric space.

With $d(\mu,\nu)$ being the total variation of $\mu-\nu$, I suppose. There are [url=http://en.wikipedia.org/wiki/Wasserstein_metric]other interesting metrics[/url].

are you sure there is no other hypothesis like locally compact and separable. In this case, it is simple consequence of Riesz representation theorem of linear functional.

Anyway, I don't know usefull spaces where one uses measure theory that are complete but not locally compact and separable.. Anyone with examples ?
(measure theory on infinite dimensional vectorial spaces ?)

[quote="alekk"]are you sure there is no other hypothesis like locally compact and separable. In this case, it is simple consequence of Riesz representation theorem of linear functional.[/quote]
It's not supposed to be hard. It's just an observation that you can make a ton of complete metric spaces by looking at the finite Borel measures. (Look at measures on the space of the measures, etc.) To be honest I don't know why you would want to do this though (other than the first time of course).

[quote="alekk"]Anyway, I don't know usefull spaces where one uses measure theory that are complete but not locally compact and separable..[/quote]
I don't know either. Would you be able to get a meaningful measure that was still $\sigma$-finite? All the authors of analysis texts that I read always want to avoid measures that aren't $\sigma$-finite because they present annoying technical difficulties. (Since they always do that, I still haven't figured out how to get around these kinds of difficulties myself...)

[quote="alekk"](measure theory on infinite dimensional vectorial spaces ?)[/quote]
Used a lot. I mentioned this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86422]earlier[/url].


Im new to this site and i was amazed by how the amount intellect on this site.  Im from nyc but my school is not actually comeptitive.  Although my favorite subject is math and known to been very good at it.  Its just that i dont think ive reached that level of competition in this site.  I would like to know if you have any tips for me to become a better math problem solver and an expert in math.  I would like to know how I can structure my study time and various sources to read from. Thank you

for one, you can look over some of the problems posted on this site (especially the high-school threads), and if you are confident enough, you can post a solution!
i would really not like to be an advertiser, but the AoPS books (all of them) are good resources.
i do not recommend buying all of them; probably only buying some books in subjects that you are not too good at.
the AoPS volume 1/2 books are all around subject books.
If you do not want to buy books, you can take old AMC, AIME, etc. tests and look at the solutions for new problem solving techniques.
you can find these in AoPS wiki and the contests tab on the top interface bar all for free.


ratio
ceiling function
floor function
algebra proposed
algebra

A flea moves in the positive direction on the real Ox axis, starting from the origin. He can only jump over distances equal with $\sqrt 2$ or $\sqrt{2005}$. Prove that there exists $n_0$ such that the flea can reach any interval $[n,n+1]$ with $n\geq n_0$.

This is going to be a bit hard to explain.

Let $u,v$ be two positive reals with irrational ratio (they can be $\sqrt 2,\sqrt{2005}$, for example). Take integers $a_1,b_1,a_2,b_2$ with the following properties:

(1) $a_1>0,b_1<0$ and $0<x_1=a_1u+b_1v<1$;
(2) $a_2<0,b_2>0$ and $0<x_2=a_2u+b_2v<1$.

Now pick some natural $N$ larger than both $\frac{u+1}{x_i}\cdot a_i,\ i=1,2$. We can prove that $n_0=\lceil Nu\rceil$ works:

Take an $n\ge n_0$, and take a natural $m$ for which $mu$ is the closest to the interval $[n,n+1]$ (we will have $m\ge N$); assume that $mu<n$. To $mu$, add as many $x_2$'s as it's necessary in order to fall in the interval $[n,n+1]$. We will add fewer than $\frac{u+1}{x_2}$ of them, so the coefficient of $u$ in the sum between $mu$ and these $x_2$'s will be non-negative, and the same holds for the coefficient of $v$. When $mu$ is larger than $n+1$ instead of being smaller than $n$ (as asumed above), we do basically the same thing, except that instead of adding some $x_2$'s, we subtract some $x_1$'s.

I hope it makes sense :).

The general case is that if $\frac{a}{b}\notin Q$ and the flea moves with $a$ or $b$, then it exists $n_0$ such that the flea can reach any interval $[n,n+1]$ with $n\geq n_0$.
Let's suppose that there is no $n_0$, which means that there is a infinite number of intervals $[n,n+1]$, so that the flea cannot jump into them. Let $(x_{n})_{n\geq 0}$ be the infimum of these intervals. We have that $x_{n}\rightarrow\infty$.  Let $(y_{n})_{n\geq 0}$ be the greatest number with the form $aX+bY$, with $X,Y\in N$, $y_{n}<x_{n}$. It is clear that $y_{n}\rightarrow\infty$.
We have $y_{n}=a\cdot u_{n}+b\cdot v_{n}$. Because $y_{n}\rightarrow\infty$, at least one of the "vectors"(don't know how to call them in english) $u_{n}$ or $v_{n}$ has the limit $\infty$. WLOG, we can say that $v_{n}\rightarrow\infty$.
Because $\frac{a}{b}\notin Q$, we have that $\left\{au+bv| u\in N,v\in Z\right\}$ is dense on R(a form of Kronecker's theorem), we can choose $u>0,v\in Z$, with $0<au-vb<1$.  Because $v_{n}\rightarrow\infty$, we can choose $N$ so that $v_{N}>v$,hence $y_{N}+(ua-vb)$ is a position where the flea can jump.
We have that $y_{N}+(ua-vb)<y_{N}+1<x_{N}+1$.
1) If $y_{N}+(ua-vb)\geq x_{N}$, we have $(y_{N}+(ua-vb))\in [x_{N}, x_{N}+1]$, which is absurd, because we supposed that it cannot land in $[x_{N}, x_{N}+1]$.
2) If $y_{N}+(ua-vb)< x_{N}$, we have $y_{N}<y_{N}+(ua-vb)<x_{N}$, which contradicts the fact that $y_{N}$ is the closest number to $x_{N}$, smaller than $x_{N}$
This means that our supposition is false, so it exists $n_0$ such that the flea can reach any interval $[n,n+1]$ with $n\geq n_0$.

And easier solution: We observe that $\left\lfloor \frac{ \sqrt{ 2005} }{\sqrt 2} \right\rfloor = 31$, so $31 \sqrt 2 < \sqrt{2005} < 32 \sqrt 2$. Even more, $\sqrt{2005} - 31\sqrt{2} < 1$ and also $32 \sqrt{2} - \sqrt{ 2005} < 1$.

This means that the sequence $a_0 = 31 \sqrt{2}$, $a_1 = \sqrt{2005}$, $a_2 = 32\sqrt{2}$, $a_3 = \sqrt{2005} + \sqrt 2$, $\ldots$, $a_{2k}= (31+k) \sqrt 2$, $a_{2k+1} = \sqrt{2005} + k\sqrt 2$, $\ldots$, is increasing, and the difference between any two terms is less than one, which means that in all intervals of the form $[n,n+1]$ where $n\geq \lceil 31 \sqrt{2} \rceil$.


[hide]notice that x^4+x^2 can factor into x^2(x^2+1). since x is [i]i[/i], we know that [i]i^2[/i] is -1 so -1+1 = 0. And obviously, if a 0 is in the numberator, then the fraction = 0. [/hide]

[hide]Plugging in $i$ we find $f(i)=\frac{i^{4}+i^{2}}{i+1}$.  Because $i^{4}=1$ and $i^{2}=-1$, the numerator equals $0$.  So $f(i)=0$.
[/hide]


Let a triangle $ ABC $.$ BE , CF $ are the bisectors of $ <ABC , <ACB $ . The ray $ EF $ intersects the circumcircle of triangle $ ABC $ at M . The ray $ FE $ intersects the circumcircle of triangle $ ABC $ at N . Prove that :
  $ \frac{1}{BM} + \frac{1}{CN} = \frac{1}{AM} + \frac{1}{AN} + \frac{1}{BN} + \frac{1}{CM} $

I have not much time now but let me answer this problem:

[quote="nttu"]Let a triangle $ABC$.$BE , CF$ are the bisectors of $<ABC , <ACB$ . The ray $EF$ intersects the circumcircle of triangle $ABC$ at M . The ray $FE$ intersects the circumcircle of triangle $ABC$ at N . Prove that :
  $\frac{1}{BM} + \frac{1}{CN} = \frac{1}{AM} + \frac{1}{AN} + \frac{1}{BN} + \frac{1}{CM}$[/quote]

Actually, this is just the sum of the equations

$\frac{1}{BM}=\frac{1}{AM}+\frac{1}{CM}$ and
$\frac{1}{CN}=\frac{1}{AN}+\frac{1}{BN}$.

The first of these equations was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=3807 ; the second one is analogous.

  Darij

I think both Grobber's and Juancarlos's solution you showed are very complicated  .
  I have an easier solution :
  We have :
  $ \frac{AM.AN}{BM.BN} = \frac{FA}{FB} = \frac{b}{a} (1) $
  $ \frac{AM.AN}{CM.CN} = \frac{EA}{EC} = \frac{c}{a} (2) $
  $ a.AM + b.BM=c.CM (Ptoleme's theorem) (3) $
  $ a.AN + c.CN=b.BN (Ptoleme's theorem)  (4) $
  From $ (1),(2),(3),(4) ===> Q.E.D $


Let $a,b \in R, a<b$. I want to prove $[a,b]$ is compact.

Here's my proof:

Let $\{U_{i}\}_{i \in I}$ be a collection of open subsets of $R$ whose union contains $[a,b]$. Now we let $S = \{x \in [a,b]: x>a$ and $[a,x]$ is contained in the union of a finite number of the sets $\{U_{i}\}_{i\in I}$. We know that least upper bound of $S$ exists b/c $S$ is bounded by $b$. Thus, we have least upper bound of $S$, denoted by $s$, $s \in U_{i}$ for some $i \in I$. Since $U_{i}$ is open we must have $s = b \in S$. Hence, we have $[a,b]$ is contained in the union of a finite number of subsets, hence a finite covering. Hence, $[a,b]$ is compact.

I need to know if this proof is detailed enough and does any part need more elaboration.

Any help is graciously appreciated.

Thanks so much!

I can understand the proof (which is new to me), so the idea of the proof is clear enough.
I would add a sentence which explains why S is not empty, and maybe go into details why the assumption s < b leads to a contradiction.

LiMa


Prove that $2!4!...(2n)!\geq ((n+1)!)^n$ for all $n\in N$

By induction.  The base case is obvious.  Suppose (the problem statement.)

The problem statement for (n = k+1) can be written as $\Bigg( \prod_{j=1}^k (2j)! \Bigg) (2k+2)! \ge (k+2)^{k+1}(k+1)!((k+1)!)^n$

where, after using our induction hypothesis (and everything is positive), it remains to show

$(2k+2)! \ge (k+2)^{k+1}(k+1)!$

$\Leftrightarrow (k+2)(k+3)\dots(2k+2) \ge (k+2)^{k+1}$  which is true.


This may sound really weird to you guys. But I'm just wondering who you guys think would be a better President if they were alive today? George Washington or Abraham Lincoln?

Let me know what you think...:)

Lincoln.  Washington was a military man who didn't actually do that much in office.  If you look at most military presidents, such as Harrison, Grant and others, they didn't do very much.  His qualifications to be president were few.  Lincoln was a good orator and as we saw in the civil war, and the guts to do what was necessary.

I'm not entirely sure how this relates to intermediate high school mathematics.


[b]P[/b]eter [b]P[/b]iper buys a [b]p[/b]eck of [b]p[/b]ickled [b]p[/b]eppers for four [b]p[/b]ence. [b]P[/b]eter [b]p[/b]laces the [b]p[/b]eck of [b]p[/b]ickled [b]p[/b]eppers in his [b]p[/b]ortable refrigerator to make [b]p[/b]ie, however, he does not have enough [b]p[/b]ickled [b]p[/b]eppers to make [b]p[/b]ie. A [b]p[/b]ie needs five [b]p[/b]oint five [b]p[/b]ecks of [b]p[/b]ickled [b]p[/b]eppers to make. [b]P[/b]lease tell [b]P[/b]eter [b]P[/b]iper how many pence he needs.

(Assume fractional [b]p[/b]arts of [b]p[/b]ecks of [b]p[/b]ickled [b]p[/b]eppers can be bought, and their costs are [b]p[/b]roportional to their whole counterparts.)

>_> i have no life

[quote]Peter Piper buys a peck of pickled peppers for four pence. Peter places the peck of pickled peppers in his portable refrigerator to make pie, however, he does not have enough pickled peppers to make pie. A pie needs five point five pecks of pickled peppers to make. Please tell Peter Piper how many pence he needs.

(Assume fractional parts of pecks of pickled peppers can be bought, and their costs are proportional to their whole counterparts.)

>_> i have no life[/quote]

[hide]22 pence, if I understood the problem...[/hide]

[hide] Or it could be 18, depending on if you count the initial 4 pence. :D [/hide]


Let $ a,b,c$ be positive real numbers such that : $ c\equal{}min\{ a;b;c \}$ .

Prove that :

$ \frac{1}{a(a\plus{}b)}\plus{}\frac{1}{b(b\plus{}c)}\plus{}\frac{1}{c(c\plus{}a)} \geq \frac{9}{2(ab\plus{}bc\plus{}ca)}\plus{}\frac{1100(a\minus{}b)^2}{(3a\plus{}3b\plus{}2c)^4}$

[quote="Materazzi"]Let $ a,b,c$ be positive real numbers such that : $ c \equal{} min\{ a;b;c \}$ .

Prove that :

$ \frac {1}{a(a \plus{} b)} \plus{} \frac {1}{b(b \plus{} c)} \plus{} \frac {1}{c(c \plus{} a)} \geq \frac {9}{2(ab \plus{} bc \plus{} ca)} \plus{} \frac {1100(a \minus{} b)^2}{(3a \plus{} 3b \plus{} 2c)^4}$[/quote]
I have proved that the stronger and more general inequality holds:

Let $ a,$ $ b,$ $ c$ be positive real numbers such that $ c \le \min\{a,\; b\}.$ For each $ k > 0,$ we denote
$ E(k) \equal{} \frac {1}{a(a \plus{} b)} \plus{} \frac {1}{b(b \plus{} c)} \plus{} \frac {1}{c(c \plus{} a)} \minus{} \frac {9}{2(ab \plus{} bc \plus{} ca)} \minus{} f(k)\cdot \frac {(a \minus{} b)^2}{(a \plus{} b \plus{} kc)^4}.$

(a) If $ 0<k \le 1,$ then $ E(k) \ge 0$ for $ f(k) \le \frac {16}{27}k(4 \minus{} k)^3.$

(b) If $ k \ge 1,$ then $ E(k) \ge 0$ for $ f(k) \le 16.$

[quote="can_hang2007"][quote="Materazzi"]Let $ a,b,c$ be positive real numbers such that : $ c \equal{} min\{ a;b;c \}$ .

Prove that :

$ \frac {1}{a(a \plus{} b)} \plus{} \frac {1}{b(b \plus{} c)} \plus{} \frac {1}{c(c \plus{} a)} \geq \frac {9}{2(ab \plus{} bc \plus{} ca)} \plus{} \frac {1100(a \minus{} b)^2}{(3a \plus{} 3b \plus{} 2c)^4}$[/quote]
I have proved that the stronger and more general inequality holds:

Let $ a,$ $ b,$ $ c$ be positive real numbers such that $ c \le \min\{a,\; b\}.$ For each $ k > 0,$ we denote
$ E(k) \equal{} \frac {1}{a(a \plus{} b)} \plus{} \frac {1}{b(b \plus{} c)} \plus{} \frac {1}{c(c \plus{} a)} \minus{} \frac {9}{2(ab \plus{} bc \plus{} ca)} \minus{} f(k)\cdot \frac {(a \minus{} b)^2}{(a \plus{} b \plus{} kc)^4}.$

(a) If $ 0 < k \le 1,$ then $ E(k) \ge 0$ for $ f(k) \le \frac {16}{27}k(4 \minus{} k)^3.$

(b) If $ k \ge 1,$ then $ E(k) \ge 0$ for $ f(k) \le 16.$[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-17-v-q-b-can.html


Does the coefficient of restitution depend on the material of the body?Is it different for different materials?

Coefficient of restitution is defined for collision between [b]2[/b] bodies.
Yes, it depends on the nature of material of the bodies. It is different for collisions between 2 different bodies.
For same bodies it must be the same[theoretically]
[hide]But I feel that it might not be so[/hide]

An important point regarding COR is that it is a property of collision and not necessarily of the colliding objects.

(From Wiki) For example, if you had 5 different types of objects colliding, you would have 5C2 = 10 different CORs (ignoring the possible ways and orientations in which the objects collide), one for each possible collision between any two object types.


If I have qualified for AIME, is all I need to do to fill in my school CEEB code and my math teacher's name? What else do I need to do?

Also,

Does anyone know how the USAMO index is calculated?
For example, normally it would be AIME*10 + AMC12. But how does it work with USAMTS?

Thanks, if anyone could tell me either

Wow... all right. Good thing you told me.

Thanks


can someone please tell me some formulas to calculate pi, which do not involve making any measurements?

Measurement is the worst way to calculate $ \pi$.

http://mathworld.wolfram.com/PiFormulas.html
http://en.wikipedia.org/wiki/Computing_%CF%80

:maybe: I think without using pi formula the question is as
 if it is very tough to some pepole.


Give a combinatorial proof that for $m,n\geq0$, $\binom{n}{0}\binom{n}{m}+\binom{n}{1}\binom{n-1}{m-1}+\binom{n}{2}\binom{n-2}{m-2}+...=\binom{n}{m}2^m$

[quote="seshadri"]i think pierre already has done this...[/quote]

Awesome...where might I find this?

i don't remember the post # but the idea is basically to first select a group of $m$ people from a group of $n$ and then select a committee from the selected group(the empty committee is also ok); that gives the RHS.
for the LHS do the same counting locally:choose the committee first and then select the other members and that gives the LHS.

I understand, cool, thanks a lot!

-Austin

is easy to elimine n in this expression (when you open)  :)


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ p,\ l$ be primes. Find the number of the monic irreducible polynomial with one variable  over $ \mathbb{F}_{p}$ with degree $ l$, provided that $ \mathbb{F}_{p}$ is a field with $ p$ elements.

there are $ p^{l}\minus{}p$ generators of the extension $ \mathbb{F}_{p^{l}}/\mathbb{F}_{p}$ (because the generated subfield has degree $ 1$ or $ l$). the minimal polynomials of those elements over $ \mathbb{F}_{p}$ are monic, irreducible and of degree $ l$. furthermore, every generator has exactly $ l$ conjugates. thus, there $ (p^{l}\minus{}p)/l$ polynomials.

if $ l$ is not prime, the number is $ \sum_{d | l}\mu(\frac{l}{d})\cdot p^{d}$, there $ \mu$ is the moebius-function.


trigonometry
inequalities proposed
inequalities

nice solotion.i enjoyed it very much. :D

[quote="jensen"]if $0<x<\frac{\pi}{2}$ then prove:
   $(\frac{sinx}{x})^{2}+\frac{tanx}{x}>2$[/quote]
$p(x)=(\frac{sinx}{x})^{2}+\frac{tanx}{x}$
then$p'(x)\geq 0$


The surface area and volume of a prisim r given find the surfase area and volume of a prisim with (a) double ,(b) triple,(c) half the dimensions of the given prisim?
S.A= 23 ft sq
Volume= 12 ft cubed
1a)
1b)
1c)

S.A = 61.7 metrs sq

volume=28.2 meters cubed
2a)

2b)
2c)

If you double the dimensions of a solid with surface area x and volume y, the new surface area is 4x and the new volume is 8y. This is because 2^2 = 4 and 2^3 = 8.

If you triple the dimensions of a solid with surface area x and volume y, the new surface area is 9x and the new volume is 27y. This is because 3^2 = 9 and 3^3 = 27.

If you halve the dimensions of a solid with surface area x and volume y, the new surface area is x/4 and the new volume is y/8. This is because (1/2)^2 = 1/4 and (1/2)^3 = 1/8.

Knowing this, the answers to the problems are as follows:

1a: S.A. = 92 ft^2 and volume = 96 ft^3
1b: S.A. =  207 ft^2 and volume = 324 ft^3
1c: S.A. =  5.75 ft^2 and volume = 1.5 ft^3

2a: S.A. = 246.8 m^2 and volume = 225.6 m^3
2b: S.A. = 555.3 m^2 and volume = 761.4 m^3
2c: S.A. = 15.425 m^2 and volume = 3.525 m^3


geometry
geometric transformation
search
3D geometry

Does anybody know of a language which is primarily based on verbs? I.e., all nouns are participles/infinitives of verbs, with the exception of names?

do you mean something like a language where instead of ''cat'', they had a word that meant ''meower''?

"Meower" for "cat" is not generalizable to other nouns, and there's nothing to stop a chicken meowing :roll: . I think he means more like there's a verb "to be a cat", and the word for cat is a verbal like "being a cat" or "he who is a cat". But no, I know of no such language.

I have thought about that before though... it would pretty cool. Anybody up for a group construction of such a language on the forum?

Why should names be an exception? Bob could say "that-which-is-a-name being-mine is-Bob", using the verbs "to be a name", "to be mine", and "to be (named) Bob".

[quote="jb05"]Anybody up for a group construction of such a language on the forum?[/quote]

I would be interested in helping to construct such a language. However, I have attempted to make artificial languages before and failed miserably. The language will have to be closely related to English so that we don't have to make up words.

Pom

To answer the original question (maybe): [url]http://ingus.homelinux.org/papers/thoughtful_linguistics.pdf[/url]

I once saw a game where people go back and forth creating a language (actually, here it is: [url]http://home.unilang.org/main/forum/viewtopic.php?t=14428[/url]). So, if we can get anybody to participate (tokenadult: you can certainly help yourself!), here are some rules:

1. One person may post only once a day, and his post may contain a maximum of five vocabulary words and one grammar rule. Note that grammar rule can be taken loosely (for instance, present tense all singular persons or something like that), but not too loosely.
2. No suggestions are set in stone, but only oppose one if it is clearly bad for the overall development of the language.
3. No nouns! No adjectives! Adverbs, pronouns, etc. are okay (or who knows, maybe we'll be able to verbalize those too). This means we're going to have to start making participials/actor forms of verbs very quickly.
4. The only letters allowed are things you can type on a standard QWERTY keyboard, or accents áéíóúýẃ... If you cannot type these, you can write an accent as a colon after the vowel (kíla -> ki:la).

I will use my first turn to create a phonology:
[color=blue]VOWELS
a - sat
á - sought
e - set
é - sate
i - sit
í - seat
o - sole
ó - so
u - soot
ú - soon
DIPHTHONGS
ai - sigh
oi - soy
au - ouch
CONSONANTS
b, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, y, z
As in English except "g" always hard, "j" like "y", but harder, and "q" as in Hebrew/Arabic (like "k", but with the tongue against the very back of the mouth near the throat). Note "r" should be pronounced as in English.
DIGRAPHS
sz - she
zy - voiced equivalent of the above, as in "lei[b]s[/b]ure" instead of "lea[b]sh[/b]"
gn - a[b]nnu[/b]al
ACCENT
Put the accent on the last syllable if it has a long vowel or ends in two consonants or j (sz, zy, gn don't count as two). Otherwise, put it on the second to last.[/color]

We can make minor changes/additions if desired as we go along.

By the way, as for verbals, I would suggest (just a suggestion) the usual present/past/future active/passive infinitives/participials, a gerund, and possibly at least these three special forms:

Subjunctive gerundive: e.g., lovable
Imperative gerundive: e.g., one to be loved / one who is to be loved
Actor: e.g., lover

I include the two gerundives because, in the absence of nouns and adjectives, we need lots of nifty verbals. So I also expect quite a few more than just these. And they should be pretty simple to create since they will be functioning as our only nouns and adjectives! Then the present participial will be for forming appositives with "nominal" verbs (Bob, a good [u]plumber[/u], lives next door) and adjectives with "adjectival" verbs (Bob, a [u]good[/u] plumber, lives next door). The actor will be for forming nouns with "nominal" verbs ([u]Bob[/u], a good plumber, lives next door) and substantive adjectives with "adjectival" verbs (the [u]scared people[/u] ran).

We could use absolute phrases (but with no subjects, since the subject would be the same as that of the sentence) for adverbs. For instance, "The children played happily." becomes "Being happy, the children played." "Being happy" is of course a single word. Obviously, "the children" would become a verbal.

By the way, what do you mean by "like 'y', but harder"? Could you describe it phonetically (e.g. "y"="voiced palatal approximant")?

j = voiced palatal fricative

According to Wikipedia, this sound occurs in Greek γεια (hello), Spanish yeísmo (a word describing the actual process of the fortition of j into ʝ in Spanish), and Swedish jord (soil).

I was intending this to come after vowels and have an effect similar to nasalizing them (ejn similar to én).

[color=blue]vail- be happy
fest- be strong
z- be many
elik- be a child
Zyán- be (named) John
[u][b]PRESENT PARTICIPLES[/b][/u]
vailun -- happy
festun -- strong
zun -- many (used as a pluralizer)
elikun -- being a child
Zyánun -- being (named) John
[u][b]ACTORS[/b][/u]
vailit -- happy (e.g., long live the happy)
festit -- strong (e.g., the country rests in the hands of the strong)
zit -- many (e.g., a democracy is ruled by the many)
elikit -- child
Zyánit -- John
[/color]

[quote]We could use absolute phrases (but with no subjects, since the subject would be the same as that of the sentence) for adverbs. For isntance, "The children played happily." becomes "Being happy, the children played." "Being happy" is of course a single word.[/quote]

Okay, that's a possibility. "Vailun, elikit zun [i]played[/i]." Alternatively, "The children played with happiness". The whole "happiness" thing sounds weird until you realize that the gerund "being happy" really means "happiness". I think adverbs can be realized in many ways, probably depending on speaker and context.

Come on, does anyone want to add to this?

I've realized that putting all adverbs at the beginning of the sentence wouldn't work very well. Let's put them immediately after the word they modify. Adjectives should work the same way to avoid possible ambiguities.

ús - don't (used to negate verbs. There could be a seperate
 word for "do".)
ot - be one
ár - be two
pár - be three

Present participles

úsun - not doing
otun - one
árun - two
párun - three

Actors

úsit - that which doesn't
oit - one
árit - two
párit - three

"Don't be" could be used for "be zero".

My brother, Alexander, suggests the following:

[quote]At least some pronouns could be represented by verb inflections (rather like in Spanish).
Here is my suggestion:

..........  Sg. Pl.
......... 1 pe be
Person 2 te  de
......... 3 ke ge

Prefixes would be nominative; suffixes would be accusative.[/quote]
If we decide to do this, we should add a vowel before the suffixes to prevent voicing assimilation.

Hi, this is Pom's brother. Here are some suggestions:

capitalize: nijapor
participate in capitalism: nijaporr /nɪʝæˈpɔɹ/
be (called) Alexander: Aligzandr
inflect: blakéd
be normal: egyál

Geblakéd egyálun.

Also, a few questions:
 Are the "long vowels" (by which I assume you mean the accented vowels) actually longer, or is that just a name?
 Is "á" pronounced /ɑ/ and is "o" pronounced /ɔ/?

One more thing: when adding prefixes or suffixes to a word, vowels in the affixes that are next to vowels in the word could be removed to facilitate pronunciation. (E.g., "Pús pegyál" rather than "Peús egyál" for "I am not normal.")

I will use this post to point out inconsistencies in the past few posts and attempt to smooth them. If you intended something else but just didn't say it well enough, say it again and I will change these.

[quote="33286"]ot - be one
oit - one[/quote]

[color=blue]Verbs that end in [vowel]+t form actors by attaching i between the vowel and t.
[hide="I-mutation"]a/á - ai
e/é - í
i/í - ij
o/ó - oi
u/ú - ui
ai - aij
oi - oij
au - oi[/hide][/color]

[quote].......... Sg. Pl.
......... 1 pe be
Person 2 te de
......... 3 ke ge

Prefixes would be nominative; suffixes would be accusative.
If we decide to do this, we should add a vowel before the suffixes to prevent voicing assimilation.
[/quote]
[quote]to be (called) Alexander: Aligzandr[/quote]

Can we make this Aligzander or Aligzandir or Aligzandre or something, so that we don't have to pronounce words like Paligzandr? I dunno, it's up to you, but I think it's pretty weird.
[color=blue]When a personal suffix is attached to a verb ending in a stop or stop+liquid, that suffix takes the form -ipe, -ite, etc., the exception being when the verb ends with the same stop as the suffix (ignoring voice). Then the suffix overrides that last consonant (effectively just attaching -e and changing the voice of the last consonant if necessary). As we only have one transitive verb to date, the only example I can give is blakéde (... inflect you) and blakéte (... inflect thee), instead of blakédide and blakédite.

The fourth-person prefix qe- is used in the absence of a subject (i.e., for the passive voice). [/color]

[quote]to capitalize: nijapor
to participate in capitalism: nijaporr /nɪʝæˈpɔɹ/[/quote]

So these are pronounced the same? Because that's what you have indicated.

Finally, in regards to the proposed sentences "ús pegyál" and "geblakéd egyálun", I have a few complaints.
In the first, "ús" is being used as an adverb. If "ús" is a particle, I am confused by the fact that it has verbals. If "ús" is a verb, I am confused by the uninflected version in this sentence simply meaning "not". Perhaps someone can clarify?
In the second, "egyálun" is being used as an adverb, and "geblakéd" is in the middle voice! This may have been intended already, but I would like to here rigorize the voices with the intent of leaving this sentence still correct:
[color=blue]Active: transitives have a personal prefix and suffix. Intransitives have a prefix only.
Middle: transitives have a prefix only.
Passive: transitives have a suffix and the fourth person prefix qe-.[/color]
As for the problem with the adverb/adjective discrepancy which I know we have been trying to mend for a while, I will let someone try to convince me of the purported difference between adjectives (participles) and adverbs. Otherwise, I may start my gerund/preposition campaign tomorrow.

I hope I am not too picky.

[hide="Long vowels..."]are not necessarily longer. However, they are what are usually called "long vowels". I'm not in on all the IPA vowel sounds and don't know much about phonetics in general, so these may not be 100% precise, but I'll try to approximate them:
a - æ
á - ɑ
e - ɛ
é - e
i - ɪ
í - i
o - ɔ (I think)
ó - o
u - ʊ
ú - u[/hide]

Aligzandr: This is pronounced the same as "Alexander," assuming that syllabic consonants (e.g., the "l" in "bottle") are allowed. If syllabic consonants aren't allowed, it'll be "Aligzander."

"Nijapor" is pronounced /nɪˈʝæpɔɹ/ (with the stress on the second syllable). I'm not sure if the doubled "r" in "nijaporr" really counts as two consonants, but I'm using that to distinguish the two words with stress.

"Ús pegyál" should be "Pús egyál." I've changed it.

"Geblakéd" is intransitive.

You wrote "be happy", "be strong", etc. I followed your example and wrote "be one", "be two", etc. However, I'm not sure whether these were meant to be just roots or something else (e.g. infinitive). Which one is it?

est - be
Erik - be (named) Eric
Rik - be (named) Rick
kón - know

The middle two are pretty obvious. However, "Perik" must now mean both "I am Eric" and "I am Rick". The reason for saying "Perik" rather than "Peerik" is a good one, though. Therefore, I suggest that, instead of "Perik", you could say "Erikit estipe" or "Rikit estipe". If there are other words like these, where one is "e" plus the other, the words would simply be homonyms.

Past tense:

These suffixes would replace pe, be, te, etc. The person and number referred to are those of the object, of course.

............ Sg... Pl.
......... 1 efe.. eve
Person 2 este. eze
......... 3 esze ezye
No object: -ez

Also, I think the affixes should not be stressed or change the  stress (in order to preserve the distinction between "nijaporez" and "nijaporrez", for instance). This would allow us to distinguish between "I am Eric" and "I am Rick", too, although it would not if the words were "Erík" and "Rík" instead of "Erik" and "Rik".

Bleh... trash the one post a day rule. There are only three of us, so I guess we can just take turns.

[quote="33286"]You wrote "be happy", "be strong", etc. I followed your example and wrote "be one", "be two", etc. However, I'm not sure whether these were meant to be just roots or something else (e.g. infinitive). Which one is it?[/quote]

It was a root. Apparently it has already become an actual usable word ([color=red]pús egyál[/color]). Let's just say that when an auxiliary (which [color=red]ús[/color] apparently is) is used, it takes all the inflections and leaves an uninflected main verb.

Hmm... [color=red]qe-[/color] was a no-subject present. Now we have [color=red]-ez[/color] a no-object past. I am a bit confused.

[color=blue]to attack - [/color][color=red]krenor[/color][color=blue]
to play - [/color][color=red]gnel[/color][color=blue]
[i]preposition indicating manner[/i] - [/color][color=red]hí[/color][color=blue]
[i]preposition indicating accusative[/i] - [/color][color=red]au[/color][color=blue]
[i]auxiliary indicating imperfective aspect[/i] - [/color][color=red]ent[/color][color=blue]

The finite verb comes at the beginning of the clause. I think this is appropriate in a verb-based language because the action is the most important part of the sentence, and thus should come first.

Personal prefixes refuse to take the stress (Pom's suggestion). In particular, [/color][color=red]Perik[/color][color=blue] means "I am Eric" when the stress is on the first syllable, and "I am Rick" when the stress is on the second. To avoid the same spelling, we use apostrophes for these prefixes. This applies in all cases (e.g., [/color][color=red]g'elik[/color][color=blue] - they are children).

Gerund: [/color][color=red]-oj[/color][color=blue]
[i]Note: the gerund means roughly "the state of _ing", and is different from the English gerund in that it can take no object.[/i][/color]
[hide="Examples"]WORD - LITERAL TRANSLATION (ROUGH TRANSLATION)
[color=red]kónoj[/color] - knowing (knowledge)
[color=red]blakédoj[/color] - inflecting (inflection)
[color=red]elikoj[/color] - being a child (childhood)
[color=red]vailoj[/color] - being happy (happiness)
[color=red]festoj[/color] - being strong (strength)
[color=red]egyáloj[/color] - being normal (normality)
[color=red]otoj[/color] - being one (unity)
[color=red]estoj[/color] - being (existence)[/hide]

I should probably explain the preposition. In the spirit of verb-like words, prepositions are conjugated! They take the present personal suffixes, but not prefixes. But of course the whole accusative prepositional phrase can be omitted if it is "me", "thee", "us", or "you" (just as the subject can be omitted). Also, [color=blue]if it is clear from context whether a word is plural or not, [/color][color=red]zun[/color][color=blue] is optional (this applies to "they are children" and plural prepositional phrases). [/color]

Note: this means we do not need words for I, thou, we, and ye!!! In every scenario, they are either the subject or the object of the preposition, so they are built into the verbs and prepositions!

[hide="examples"][color=red]Gegnel elikit (zun) híke vailoj.[/color] The children play happily.
Note: this demonstrates one possible way to express adverbs (means+gerund). It also establishes the optional [color=red]zun[/color] when the number is clear.
Alternative: [color=red]Vailun, gegnel elikit (zun)[/color]. Being happy, the children play.
Bizarre alternative: [color=red]G'elik gnelit (zun) vailun.[/color] The happy players are children. They are children who play happily.
Another bizarre alternative: [color=red]Gevail elikit (zun) gnelun.[/color]
Note: this expresses the true power of a verb-based language - even though we technically have nouns and adjectives (verbals), they don't really matter. Any word can be the finite verb!

[color=red]Geblakéd híke egyáloj.[/color] They inflect normally.
This can be said after listing verbs.
Alternative: [color=red]Kegyál blakédoj.[/color] The inflection is normal.

[color=red]Q'entibe krenor![/color] We are being attacked!
Note: the passive voice.
[color=red]Qekrenoreve![/color] We have been attacked!
Note: I'm not sure if this is the desired word yet, as I'm not entirely certain about the fourth person past inflection.

[color=red]Kekrenorge Zyánit auge nijaporrit (zun)[/color]. John attacks the capitalists.
Alternative: [color=red]K'entige Zyánit auge nijaporrit (zun) krenor.[/color] John is attacking the capitalists.
Note: despite the rule about finite verbs coming at the beginning of the sentence, we can use auxiliaries to hold them off to the end to build suspense.[/hide]

Words for the letters of the alphabet:

á: be A
bi: be B
tsze: be C (The letter C was once used for /tʃ/.)
di: be D
é: be E
fi: be F
gi: be G
hi: be H
í: be I
je: be J
ki: be K
li: be L
me: be M
ne: be N
ów: be O (Speakers of Vignerp, like speakers of English, often replace the digit 0 with the letter O in speech, and the W is there to distinguish between [i]ówit[/i] "O" and [i]oit[/i] "one".)
pi: be P
qu: be Q
re: be R
si: be S
ti: be T
ú: be U
vi: be V
wi: be W
hji: be X (The letter X was once used for /ç/.)
ye: be Y
zi: be Z

astok: make up
tsoppi: compute
__________________

P'úske bal astok au "vendai'j".

ábi: alphabetize; be an alphabet; be alphabetical
sópi: be a calculator
merqrik: be a diacritic/accent
égbú: be an acute accent
egbo: be a grave accent
vobi: be a circumflex

kómi loke jipfoj: be the end
uméip: be early
dlem: be late
ist: be a number
iktú: by (size of margin)

[i]Kedlemez iktúge qontit yeve'n.[/i] "He was 10 minutes late."

[quote="325155319110133325116104"]"Pekédo'sze arike irit íwe'n. B'ím kwist," keyerez Rikit.[/quote]
Kemejlike est modít br'árzy arike kómijt loke jipfoj au "keyeresze".

kúyi: and/or
úyit: xor
vyús: without
fász: fill

[i]Ugni[/i], "or", can be either "and/or" or "xor".
[i]Fász[/i] can be used like the English suffix "-ful": e.g., "a sorrowful person" is [i]légijt fászus yatike distamoj[/i]. Similarly, "a sorrowless person" is [i]légijt vyúske distamoj[/i].

áparmde: be a dictionary
zyim: search
klup: light
([i]klupit[/i] is "(a) light")
úsklup: be darkness

ro-: there (expletive)

[i]R'estike oit dzulun.[/i] "There's only one."

Note: As with the passive voice, [i]au[/i] is not used.

nimik: eat
népulk: drink
nibgi: consume
safs: taste
(qenimikit: food
qenépulkit: drink)

[i]Kesafsesze au qenimikit. Kesafsesze qenimikit au házyugun.[/i] "He tasted the food. It tasted good."

yáki: start
ódge: by (indicates actor)
plungwi: by (indicates means)
tliwe: near (prep.)
dzem: be far; far from

[i]Qenimikesze qenimikit ódgeke nimikit.[/i] "The food was eaten by the eater."

ús (as adv.): not
jehá: suggest
snem: that (conj; used for indirect commands)
esi: as (indicating what something is)
sogn: of; about

[i]Pejeháke snem t'ád sognke esike légijt vailun híke úsoj.[/i]
"I suggest that you think of him as an unhappy person."

vlista: be a plant
rem: be a tree
sfaro: be a flower
szism: be a bush
uzyaps: be grass

I-mutation is caused by [i]í[/i] as well as [i]i[/i]. Therefore, for instance, "I have been happy" is [i]Pím vail[/i] rather than [i]P'ím vail[/i], and "to have been happy" is [i]faim vail[/i] rather than [i]f'ím vail[/i]. Also, both [i]fa-[/i] and [i]fe-[/i] become [i]f'-[/i] before an initial non-i vowel; on the other hand, future-tense suffixes keep [i]-u-[/i] even after a final vowel (e.g., [i]kíweu[/i], "it will be this", [i]pefáweute[/i], "I will greet you").

upléwi: become
weligi: change
wazigiz: transform
miját: continue

[i]We[/i] & [i]ja[/i] are used to indicate a change of state.

[i]Keweligi'z szeqetíjit weke quirdit jake etognijt.[/i] "It's color changed from red to orange."
[i]Kemijátez ulike íyít f'etogni.[/i] "Then, it continued to be orange."

[u]3-d description[/u]
jám: be the x-axis value
yom: be the y-axis value
zim: be the z-axis value
—
-up: increase value
-iz: decrease value
—
(Remove final -m)
-zup: have a large value
-sziz: have a small value
—
[u]Examples[/u]
(Of a cube with an edge length of 3): [i]Kepár szejámit, kepár szeyomit, kun kepár szezimit.[/i]
"The road is 10 (units) long." [i]Keyeve [jámit/yomit] loke refit.[/i] (This assumes that the unit has already been stated.)
[i]Keyozup refit.[/i] "The road is wide."
(Of a body of water) [i]Kezisziz.[/i] "It is shallow."
[i]T'enke zimup.[/i] "Deepen it."

oimp: be an amount

[i]K'editez oimpit ómwelun ruts begnelez.[/i] "We played for a while."

rárj: be true
grayí: be correct
omrai: be moral
áryik: correct/make true
luit: whether

(ús rárj: be false
ús grayí: be incorrect
ús omrai: be immoral
kom ús grayí/rárj: make incorrect/false)

[i]P'úske kón luit k'omrai íyít ugni k'ús omrai íyít.[/i] "I don't know whether that is morally right or wrong."


abstract algebra
superior algebra
superior algebra theorems

Can anyone here tell me about the 2nd & 3rd isomorphism theorem of groups.

Its very important for my exams are coming up shortly..

I have galleon, Hurstein & they have no  mention about them.

Try this link:

http://mathworld.wolfram.com/FirstGroupIsomorphismTheorem.html

Misha


20 Americans have been taken prisoner by alien explorers, who take the prisoners to their home planet and lock them in a dungeon. The alien king, who had sent the explorers to return from Earth with spices from India and was displeased with the prisoners the explorers had brought back, agrees to take the prisoners back home if they can complete a certain task. The king tells them the following:

All prisoners will be locked in different cells, and, one at a time, though not in any particular order, they will be taken into a room (the king will be in here) where there is an empty goblet on a table. That prisoner will have the choice to either turn the goblet upside-down (or right-side up if it is already upside-down) or leave it alone.

The next prisoner will be taken into the room at a random time interval between 2 and 5 minutes after the previous person has left the room, will be given 15 seconds to make their decision, and then will be taken back to their cell. Prisoners may be taken into the room more than once. In fact, each prisoner will be taken in at least once for every 100 times prisoners are taken in. The prisoners will have no way to keep track of time at all, but will be given enough food and water to survive.

Once a prisoner tells the king that all 20 people have been in the room, if this is true, the prisoners will be released.

The prisoners can strategize as long as they want to, but the king can sit in on their strategizing, and will exploit any weaknesses in the prisoners' plan.

If one of the prisoners tells the king that everyone has been in the room, but this is false, the prisoners will become slaves to the aliens forever.

The prisoners will not at all be able to see, hear, or talk to each other at any time. I'm serious - NO INTER-PERSONAL COMMUNICATION AT ALL!!!

How can the prisoners escape?

When a person is taken in for the 82nd time, he will tell the alien king that they have all been in the room.

@ bigboy - I only used the "everyone comes in every 100th time anyone is brought in" to express that there is to be a sort of balance between the number of times any prisoner comes in as opposed to any other

@ mustafa - you have the right concept, but that is not right.

[quote="mustafa"][hide]Maybe there could be one person who is "special". If anyone but him comes into the room, then they make the goblet go upside-down, and if he comes into the room, he turns the goblet right-side up. If someone has already turned the goblet upside-down once, then they don't touch it. If the goblet is already upside-down, they don't touch it. When the special guy turns the goblet rightside up for the 19th time, he knows that everyone has been there.[/hide][/quote]
[b]No interpersonal communication, remember?[/b]

[quote="sapphyre571"]
The prisoners can strategize as long as they want to, but the king can sit in on their strategizing, and will exploit any weaknesses in the prisoners' plan.
[/quote]
They would be given some time beforehand to strategize, and that part wouldn't count as illegal communication

Open mouth.
Insert foot. :oops:  :blush:

Like what mustafa said: except that they each turn the goblet upside down twice, and the special guy counts to 38.
They must go through the process twice, just in case the goblet was upside-down to start off with (and the special guy counted it as one person- meaning that he had only counted 18 people, not all 19)

Bingo

[I am only typing this because it says the message is too short]