calculus
integration
logarithms
trigonometry
real analysis
real analysis unsolved

Find:
1. $ \int \frac {ln(cosx \plus{} \sqrt {cos2x})}{1 \minus{} cos2x} \,dx$
2. $ \int \frac {1}{sin^{n}x \plus{} cos^{n}x} \,dx$

--
math10.com

[hide="Number 1"]$ [\ln(\cos x \plus{} \sqrt{\cos 2x})]' \equal{} \frac{\minus{} \sin x \minus{} \frac{\sin 2x}{\sqrt{\cos 2x}}}{\cos x \plus{} \sqrt{\cos 2x}} \equal{} \frac{\cos x}{\sin x} \minus{} \frac{1}{\sin x \sqrt{\cos 2x}}$.

On the other hand, $ 1 \minus{} \cos 2x \equal{} 2 \sin^2x$. So:

$ P \equal{} \frac{1}{2} \int \frac{\ln(\cos x \plus{} \sqrt{\cos 2x})}{\sin^2x}\,dx \equal{} \underbrace{\minus{} \cot x \ln(\cos x \plus{} \sqrt{\cos 2x})}_{\equal{}A} \plus{}$

$ \plus{} \int \frac{\cos x}{\sin x} \cdot \left(\frac{\cos x}{\sin x} \minus{} \frac{1}{\sin x \sqrt{\cos 2x}}\right)\,dx$

$ \equal{} A \plus{} \int \frac{1 \minus{} \sin^2x}{\sin^2x}\,dx \minus{} \int \frac{\cos x}{\sin^2x \sqrt{\cos 2x}}\,dx \equal{} A \minus{} \cot x \minus{} x \minus{} P'$

and

$ P' \equal{} \underbrace{\int \frac{\cos x}{\sin^2x \sqrt{1 \minus{} 2 \sin^2x}}\,dx}_{\sin x \equal{} t} \equal{} \underbrace{\int \frac{1}{t^2 \sqrt{1\minus{}2t^2}}\,dt}_{t \equal{} \frac{1}{\sqrt{2}} \sin u} \equal{} \sqrt{2}\int \csc^2u\,du$

$ \equal{} \minus{} \sqrt{2} \frac{\cos u}{\sin u} \equal{} \minus{} \sqrt{2} \frac{\sqrt{1\minus{}2t^2}}{\sqrt{2} t} \equal{} \minus{} \frac{\sqrt{1\minus{}2t^2}}{t} \equal{} \minus{}\frac{\sqrt{\cos 2x}}{\sin x}$.

Therefore:

$ P \equal{} \minus{} \cot x [1 \plus{} \ln(\cos x \plus{} \sqrt{\cos 2x})] \minus{} x \plus{} \frac{\sqrt{\cos 2x}}{\sin x} \plus{} C$.[/hide]


function
logarithms
polynomial
complex analysis
exponential

One of my friends gave me the following problem and I spent all afternoon with it, with no success. Can anyone solve it, please? It asks to find the entire functions $f$ such that $e^{f(z)}=f(e^z)$.

Considering we already have the identity and $e^{\dots e^{z}}$ as solutions, it must be terribly hard to prove they are the only ones (considering it's true).

It is terribly hard! I found a proof ( needless to say that it uses some deep results which are too much for my poor brain) in the students magazine of ENS Lyon.

[quote="julien_santini"]Considering we already have the identity and $e^{\dots e^{z}}$ as solutions.[/quote]
Plus countably many constant functions $f(z)=c$ with $e^c=c$. I believe I have a reasonable (though also quite long) proof that there are no others but I have to check a few details yet. If it comes out OK, I'll post a sketch. Needless to say, this problem takes more than one afternoon. I spent three long evenings before I came up with any workable idea and even now I'm not 100% certain... :roll:

I have a simple (but long!) proof. Here is a sketch. Let $f(z)$ be a solution. We prove three facts:
1) If $f(z)$ does not have zeros, then $f(\ln z)$ is a solution.
2) There exists $k\geq 0$ s.t. the function $f(\underbrace{\ln(\ln(\dots\ln(}_k z)\dots))$ has a zero.
3) The only solution $f(z)$ that has a zero is $f(z) = z$.

These 3 facts clearly imply that every non-constant solution is of the form $\underbrace{e^{e^{\cdots^{e^z}}}}_k$.

Now I will very briefly outline how to prove each of these facts.

1)  The easiest one. Observe, that $\ln f(z)$ is an entire function. It is easy to see that $\ln f(z) + 2\pi k i$ is a solution for some integer $k$ (by taking the logarithm of both sides of the original identity). In fact, this solution equals $f(\ln z)$ (for $z\neq 0$, it does not matter which branch of $\ln$ we take!).

2) Assume the contrary. Then we have an infinite sequence of functions $f_0(z) = f(z)$, $f_1(z) = f_0(\ln z)$, ..., $f_{k+1}(z) = f_k(\ln z)$, ... . Let $r$ be a real number. Pick $\epsilon > 0$. Then for sufficiently large $N$, and all $n > N$, $\exp(\dots\exp(U_\epsilon(r)))$ ($n$ exponents) contains a path around 0. This means that $f_n(z)\in f(U_\epsilon(r))$ for every $z$ on this path. This implies that $f_n(0) \in f(U_\epsilon(r))$. We get: for every two real numbers $r_1$ and $r_2$ and every $\epsilon$, the sets $f(U_\epsilon(r_1))$ and $f(U_\epsilon(r_2))$ intersect. Therefore, $f(r_1) = f(r_2)$. This implies that $f$ is a constant.

3) Note that for $z\neq 0$, $f(z) = f(\exp(\ln z)) = \exp(f(\ln z)) \neq 0$. Therefore $f(z)$ has only one zero: $f(0) = 0$. Since $e^{f(z +2\pi i)} = f(e^{z +2\pi i}) = f(e^z) = e^{f(z)}$, $f(z +2\pi i) = f(z) + 2\pi m i$ (clearly $m$ is the same for all $z$). Moreover, $m\neq 0$ --- otherwise $f$ would have infinitely many zeros (at points $2\pi \ell i$). Consider two cases:
i) $f$ is a polynomial. Recall that $f(z+2\pi i) = f(z) + 2\pi m i$. So $f = m z$. It is easy to verify that $f(z) = z$.
ii) $f$ is not a polynomial By Picard's Theorem $f(z)$ attains every value (except 0) at infinitely many points. But then it attains $2\pi m i$ at infinitely many points. And thus it has infinitely many zeros. We get a contradiction.


algebra
polynomial
Pascal\u0027s Triangle
binomial theorem

Add up the first row of the Pascal's Triangle, then the second, then the third and so on. Do you notice anything interesting? Why do you think this pattern occurs?

ooooo...[hide]they're all powers of 2  [/hide]

yes, each time you multilpy a number by 2 (it goes down to left AND right)

Correct, since each number of the Pascal's Triangle becomes part of the number to the left-bottom and to the right-bottom of that number the sum of each row is 2 times the row before it. Since the Pascal's triange starts with the number 1 in the first row the formula becomes from a*2^n to just 2^n where n is the row number.

What does this tell us about squaring a polynomial?

you can also find the sum of the coefficients of the nth row of pascals triangle by the binomial theorem.  If you set x=y=1 in (x+y)^n, you'll get the sum of the coefficients of the expansion, 2^n.

yeah.   my method above is actually a really quick way of proving that identity.

true, but one might ask for a proof of the binomial theorem.

and if you do it the normal way (for positive integer n and k) and prove it combinatorially, you're doing basically what tetrahedr0n just did.

You guys just read my mind like an Elvish text :)

[quote="tetrahedr0n"]Each row of pascal's triangle is of the form nC0 + nC1 + ... + nCn-1 + nCn. In a set S with n elements there are a total of 2^n subsets. note that + nC1 + ... + nCn-1 + nCn passes over all possible subsets of S, so the sum is equal to 2^n.[/quote]

In a set S with n elements there are a total of 2^n subsets    WHY??

it's an important / well-known combinatorial identity that can be proved very quickly by my argument above.  I'll explain it a little more, though.

The number of subsets in a set with n elements is c(n,0)+c(n,1)+c(n,2)+c(n,3)+...+c(n,n).  This is because we have n elements, and we can choose 0 of them at a time, 1 of them at  a time, 2 of them at a time,...., n of them at a time.  This will account for all of the subsets.

Now how to derive the identity from the binomial theorem:

The binomial theorem says that (x+y)^n = c(n,0)*x^n*y^0+c(n,1)*x^(n-1)*y^1+c(n,2)*x^(n-2)*y^2+...+c(n,n)*x^0*y^n,  where c(n,i) is 'n choose i'.

We want to find c(n,0)+c(n,1)+c(n,2)+...c(n,n).  We can find this by getting rid of the x's and y's.  How do we do this?  If we let x=y=1 in (x+y)^n, we will have what we want, (1+1)^n= c(n,0)+c(n,1)+c(n,2)+...c(n,n).

There's a nice combinatorial proof of the fact that a set of S elements has 2^S subsets, as well.

We're going to build a subset, starting with the original set and then saying that each element is either in or out.  So we come to the first element, and it's either in or out.  And then the second one.  And then the third.  And so on.  Since at each step we have 2 choices, and there are S steps, we can do this process in exactly 2^S ways.  And since (1) each time we do it, we get a different subset, and (2) each possible subset is created from doing this in exactly one way, that means that there must be exactly 2^S subsets.


For $k>1$ and $a>0$ evaluate \[\sum_{n=0}^\infty nk^{-an}\]

[hide]
$\sum_{n}x^{n}=\frac{1}{1-x}\implies\sum_{n}nx^{n}=\frac{x}{\left(1-x\right)^{2}}$

So you have $x=k^{-a}$.  The sum is $\sum_{n}nk^{-an}=\boxed{\frac{k^{-a}}{\left(1-k^{-a}\right)^{2}}}$.
[/hide]

See my message at the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=101409

[quote="amcavoy"]...and for those of you who like calculus:

$\sum_{n}x^{n}=\frac{1}{1-x}$

$\frac{d}{dx}\sum_{n}x^{n}=\frac{1}{\left(1-x\right)^{2}}$

$=\sum_{n}nx^{n-1}\implies\sum_{n}nx^{n}=\frac{x}{\left(1-x\right)^{2}}$[/quote]

Wow, that's much simpler.  Thanks.


Let's start off with some general rules:

odd + odd = even
odd + even = odd
even + even = even

odd*odd = odd
odd*even = even
even*even = even

Now the proof:
[hide]
Assume $n$ is odd.  The $n^{4}$ will be even.  $4n^{3}$ will be even.  $3n^{2}$ will be even.  $n$ is odd, as stated, and $4000$ is even.  Therefore, the sum of these numbers will be odd, by the rules above.  However, 0 is even, and a contradiction occurs.  Thus, n must be even.

(I'm not an experienced proof writer, but this was simple.  Any technical errors?)
[/hide]

EDIT:  OK, so we can change it to  $n^{4}+3n^{2}$ is even, which is not hard to see, because both terms will be odd individually.  Then the rest of the proof is the same.

you guys are really close.

[hide]Assume n is odd. look at each term to see if it is odd or even.[/hide]

[quote="lotrgreengrapes7926"][quote="buzzer11"]Prove that if n is an integer satisfying $n^{4}+4n^{3}+3n^{2}+n+4000 = 0$, then n must be even.[/quote] [hide] Contradiction: Assume $n$ is odd. Then in $\mod 2$ we have the equation $1^{4}+4\cdot1^{3}+3\cdot1^{2}+1+4000\equiv1+0+1+1+0\equiv1$. But $0\equiv 0\pmod2$. [/hide][/quote]

nice. though you didnt need to use mods, a perfect proof nonetheless.

Mod 2 is basically the same thing as using odd/even parity in this case.

$\mod 2$ is exactly the same thing as using parity.

[quote="buzzer11"]Prove that if n is an integer satisfying $n^{4}+4n^{3}+3n^{2}+n+4000 = 0$, then n must be even.[/quote]

proof by contradiction.  if $n$ is odd, $n^{4}$is odd,$4n^{3}$ is even, $3n^{2}$ is odd, $n$ is odd, and $4000$ is even.  this is an odd number of odd numbers, which means that the answer is odd, and can't be 0.  Therefore, $n$ is even


geometry
circumcircle
inequalities
trapezoid
triangle inequality
geometry proposed

Here is a geometry problem, and i would like to know if this is easy or difficult?

Let P be a point in the plane of an equilateral triangle ABC. What is the locus of the point P, such that , if a triangle XYZ is contructed with lenghts XY = PC,  YZ = PA , ZX = PB,  then the side XY determines an arc of 60°  on the circle circumscribing the triangle XYZ.

Let $O$ be the center of everything (including the circumcenter) of the triangle $\triangle ABC$ and denote $x = YZ = PA$, $y = ZX = PB$, $z = XY = PC$.

[color=red][b]Case 1:[/b][/color] If the point $P$ is on the circumcircle $(O)$, say, between the points $A$ and $B$, then the quadrilateral $APBC$ is cyclic and by Ptolemy's theorem,

$BC \cdot PA + CA \cdot PB = AB \cdot PC$

Since the triangle $\triangle ABC$ is equilateral, $AB = BC = CA$ and consequently,

$PA + PB = PC$

Thus the segments $PA, PB, PC$ do not satisfy the triangle inequality and they cannot form a triangle $\triangle XYZ$.

[color=red][b]Case 2:[/b][/color] Assume that the point $P$ is inside the circumcircle $(O)$ on an arc $AB$, such that the angle $\measuredangle APB = 120^o$. Let $Q, R$ be the intersections of the lines $AP, BP$ with the circumcircle $(O)$. Then

$\measuredangle APQ = \measuredangle BPR = 180^o - \measuredangle APB = 60^o$

Since the angles $\measuredangle BQA = \measuredangle BRA = \measuredangle BCA$ span the common arc $AB$ of the circle $(O)$,

$\measuredangle PQA \equiv \measuredangle BQA = 60^o$
$\measuredangle BRP \equiv \measuredangle BRA = 60^o$

Hence, the triangles $\triangle BPR, \triangle APQ$, each having 2 angles equal to $60^o$, are both equilateral and $PQ = PQ = PA = x$, $BR = PR = PB = y$. Since the angles $\measuredangle ARC = \measuredangle ABC$ span the common arc $AC$ of the circle $(O)$,

$\measuredangle ARC = \measuredangle ABC = 60^o$

We already know that the angle $\measuredangle PAQ \equiv \measuredangle PAQ = 60^o$. It follows that the quadrilateral $AQCR$ is an isosceles trapezoid and $CR = QA = PA = x$. Let $X$ be a point on the line $BP$, such that $P$ is the midpoint of $BX$, i.e., $PB = PX$. The triangles $\triangle APX \cong \triangle CRP$ are congruent by SAS ($PX = PB = PR = y$, $CR = PA = x$, $\measuredangle APX = \measuredangle CRP = 60^o$) and consequently, $AX = PC = z$. Clearly, the triangle $\triangle APX$ has the sides $x, y, z$ and the angle $\widehat{xy} = 60^o$. Hence, the desired locus of points $P$ (inside of the circumcircle $(O)$) is the arc $AB$ passing through the point $O$, such that $\measuredangle APB = 120^o$.

[color=red][b]Case 3:[/b][/color] Assume that the point $P$ is outside of the circumcircle $(O)$ on the line $AB$ excluding the segment $AB$ and the points $A, B, P$ follow on the line $AB$ in this order. Construct an equilateral triangle $\triangle BPX$ on the segment $PB$, the vertex $X$ lying on the opposite side of the line $AB$ then the vertex $C$. Let the line $CA$ meet the line $PX$ at a point $Q$. The triangle $\triangle APQ$ is obviously also equilateral, because it has at least 2 angles equal to $60^o$:

$\measuredangle APQ \equiv \measuredangle BPX = 60^o$ (by construction)
$\measuredangle QAP = \measuredangle BAC = 60^o$ (vertical angles)

Hence, $AP = PQ = QA = x$. The triangles $\triangle APX \cong \triangle PQC$ are congruent by SAS ($AP = PQ = x$, $CQ = CA + AQ = PA + AB = PB = y$, $\measuredangle PQC = \measuredangle APX = 60^o$) and consequently, $XA = PC = z$. Clearly, the triangle $\triangle APX$ has the sides $x, y, z$ and the angle $\widehat{xy} = 60^o$. Hence, the desired locus of points $P$ (outside of the circumcircle $(O)$) is the line $AB$ excluding the segment $AB$.


function
limit
induction
algebra proposed
algebra

I am not sure if this problem has never been discussed before. I personally haven't found similar topic but still want to see some possible solutions different from mine. Therefore I will afford to propose it:

The sequence $ \{x_k\}_{k \equal{} 0}^{\infty}$ is defined as $ x_0 \equal{} 1$, $ x_n \equal{} \sqrt [n]{n \plus{} x_{n \minus{} 1}^n}$. Prove, at first, the function is convergent and show that $ \lim_{n\to \infty}x_n < \sqrt {6} \plus{} 1$.

PS The problem was given on a Bulgarian competition for university students.

The $ x_n$ are increasing.  Also $ x_1 = 2$.  I will use the following little inequality.

Lemma: If $ n$ is a positive integer and $ \epsilon \ge 0$, then $ \sqrt [n]{1 + n\epsilon} \le 1 + \epsilon$.  (Proof: Raise both sides to the $ n$th power and use the binomial theorem.)

Thus, for $ n \ge 2$, we have
\begin{align*} x_n & = \sqrt [n]{x_{n - 1}^n + n} \\
& = x_{n - 1} \sqrt [n]{1 + \frac {n}{x_{n - 1}^n}} \\
& \le x_{n - 1} \left( 1 + \frac {1}{x_{n - 1}^n} \right) \\
& = x_{n - 1} + \frac {1}{x_{n - 1}^{n - 1}} \\
& \le x_{n - 1} + \frac {1}{2^{n - 1}}. \end{align*}
Therefore each $ x_n$ is at most
\[ 2 + \frac {1}{2} + \frac {1}{4} + \dots = 3.\]
We have an increasing sequence bounded above by 3.  So the limit exists and is at most 3, which is less than $ \sqrt {6} + 1$.

Indeed, that was easy. I did something similar, but is there some way to find the exact bound?

I tried to use that $ x_n$ is increasing and therefore for some $ c$ and $ \forall n>N$ then $ x_n>c$ ($ c$ and $ N$ should be specified later)

$ \Rightarrow x_n\minus{}x_{n\minus{}1}\equal{}\frac{x_n^n\minus{}x_{n\minus{}1}^n}{x_n^{n\minus{}1}\plus{}x_n^{n\minus{}2}x_{n\minus{}1}\plus{}\cdots\plus{}x_{n\minus{}1}^{n\minus{}1}}\equal{}\frac{n}{\sum_{k\equal{}0}^{n\minus{}1}x_n^{n\minus{}1\minus{}k}x_{n\minus{}1}^k}<\frac{1}{c^{n\minus{}1}}$

But to be allowed to use an induction to deduce that $ x_n<1\plus{}\frac{1}{c}\plus{}\cdots\plus{}\frac{1}{c^{n\minus{}1}}\to \frac{c}{c\minus{}1}$, I should take $ c\leq x_1\equal{}2$, i.e. this doesn't work for the purpose of finding the exact bound  :|

I don't know how to find the exact limit, but if all you care about is a good decimal approximation, then what we have both done is good enough.  Using our method, for any positive integer $ k$, we can show that the limit is between $ x_k$ and $ x_k \plus{} \frac {1}{x_k^k \minus{} x_k^{k \minus{} 1}}$.  According to my spreadsheet, $ x_{22}$ is approximately $ 2.691750183$.  So the limit is $ 2.69175018$ rounded to 8 digits to the right of the decimal point.


calculus
trigonometry
derivative
calculus computations

Calculus 1 problem :

For what values of $ a$ will $ f(x)\equal{}\sin^2(x)\minus{}(a\plus{}3)x$ have local extreme values ?

Local extreme values where?  If you take the derivative and set it to 0, you get $ f'(x) \equal{} 2\sin{x}\cos{x} \minus{} a \minus{} 3 \equal{} \sin{2x} \minus{} a \minus{} 3 \equal{} 0$, you get that $ \sin{2x} \equal{} a \plus{} 3$.  Obviously, this only makes sense if $ \minus{} 1\leq a \plus{} 3\leq 1$, so you need $ \minus{}4\leq a\leq \minus{}2$.  Then you'd have to actually find the critical points and determine what they are.

I think it should be $ f'(x)\equal{}2\sin{x}\cos{x} \minus{} a \minus{} 3$.  Now, we know that this is defined for all real numbers.
So for $ f(x)$ to have a chance to have local extreme values, how should $ f'(x)$ behave ?  What does the
first derivative test tell us about local extreme values ?   :)

Hint:

We will only have critical numbers for $ f(x)$ if $ \minus{}4\leq a\leq \minus{}2$.
For example, if $ a\equal{}\minus{}2$, $ f'(x)\equal{}\sin(2x)\minus{}1$.  What is the next step ?   :)

[quote="maths486"]I think it should be $ f'(x)\equal{}2\sin{x}\cos{x} \minus{} a \minus{} 3$.[/quote]
Yes, corrected.

anyway, i think the final answer is

$ \minus{}4 < a < \minus{}2$


[quote="236factorial"][quote="shadysaysurspammed"]Can x be a complex number?If it is,then the ans is the ....ummm...greatest number writtten in complex form :D
ig x isnt complex,then its 12
zzzzzgoodnightzzzzz....[/quote]

x is definetly not 12 or complex  :?[/quote]

I think he meant that if x isn't complex, then the greatest possible value for y is 12.

Sorry guys incorrect problem. I fixed it  :blush:

[quote="Drunken_Math"][quote="236factorial"]What is the greatest possible value of y in the equation $y=-2x^2+12x$[/quote]

[hide]We can rewrite that as $y=12x-2x^2$

So we want to minimize the value of x, to make the value of y bigger. So the largest value of y is when x=1 (assuming x must be an integer)

Evaluating, the largest value of y would be 10[/hide][/quote]

y isn't 10 (nice try though  :D )

Answer is [hide]18[/hide] and you get it when x is[hide]3[/hide] :ninja: =me  :starwars: =me vs problem  1 second later problem= :stretcher:  me :trampoline:

Is there an ACTUAL way to solve these types of problems???? :(  :)

Yes. $-2x^2+12x=-2x(x-6)$, so you have a quadratic equation with the two zeros at $x=0$ and $x=6$. Therefore, if you know anything about parabolas, you should know that the maximum value is halfway in-between the two zeros, so just substitute in $x=3$ and you have your answer.

[hide]x=3 and the answer is 18 :idea: cool problem![/hide]


Find all integer solutions to the equation

$ a^4\minus{}a^2b^2\plus{}b^4\equal{}c^2$.


Given $ a, b, c > 0$ and $ a\plus{}b\plus{}c\equal{}3$. Prove that:
$ \frac{a}{(3ab\plus{}2bc\plus{}ca)^4}\plus{}\frac{b}{(3bc\plus{}2ca\plus{}ab)^4}\plus{}\frac{c}{(3ca\plus{}2ab\plus{}bc)^4} \geq\ \frac{1}{432}$
 :)

And the general problem also true (with the same condition and $ k > 0$):
$ \frac {a}{(3ab \plus{} 2bc \plus{} ca)^{2k}} \plus{} \frac {b}{(3bc \plus{} 2ca \plus{} ab)^{2k}} \plus{} \frac {c}{(3ca \plus{} 2ab \plus{} bc)^{2k}}\geq\ \frac {3}{6^{2k}}$
 :)


geometry
circumcircle
3D geometry
factorial
number theory
greatest common divisor
incenter

Description : http://www.mathlinks.ro/Forum/viewtopic.php?p=493217#p493217 .

Allez-y! ;)

Ok, c'est parti :

[hide="English -> French"]
[color=orange][size=150]A[/size][/color]
absolute value = (la) valeur absolue
acute angle = acutangle
acute-angled triangle = (un) triangle acutangle
altitude = (une) hauteur
angle = (un) angle
arc = (un) arc
area = (une) aire
arithmetic mean (AM) = (la) moyenne arithmétique
arithmetic sequence = (une) suite arithmétique
average = (la) moyenne
axis = (un) axe

[color=orange][size=150]B[/size][/color]
base = (la) base
bisector = (une) bissectrice

[color=orange][size=150]C[/size][/color]
centroid = (le) centre de gravité
circumcenter = (le) centre du cercle circonscrit
circumcircle = (le) cercle circonscrit
circumscribed = circonscrit
common denominator = (le) dénominateur commun
common multiple = (le) multiple commun
convex polygon = (un) polygone convexe
cube = (un) cube
cube root = (la) racine cubique
cyclic quadrilateral = (un) quadrilatère cyclique

[color=orange][size=150]D[/size][/color]
decagon = (un) décagone
decompose = décomposer
degree = (le) degré
denominator = (le) dénominateur
difference = (la) différence
digit = (un) chiffre
divisible = divisible
divisor = (un) diviseur
dodecagon = (un) dodécagone

[color=orange][size=150]E[/size][/color]
edge (combinatorics) = (une) arrête
equal = égal
equation = (une) équation
equilateral = équilatéral
even = pair
exterior angle = (un) angle extérieur

[color=orange][size=150]F[/size][/color]
face = (une) face
factorial = (la) factorielle
FLT = (le) petit théorème de Fermat
fraction = (une) fraction

[color=orange][size=150]G[/size][/color]
geometry = (la) géométrie
geometric mean (GM) = (la) moyenne géométrique
geometric sequence = (une) suite géométrique
graph (combinatorics) = (un) graphe
greatest common divisor (GCD) = (le) plus grand diviseur commun (PGCD)

[color=orange][size=150]H[/size][/color]
harmonic mean (HM) = (la) moyenne harmonique
heigth = (la) hauteur
heptagon = (un) heptagone
hexagon = (un) hexagone
hypotenuse = (l')hypoténuse

[color=orange][size=150]I[/size][/color]
incenter = (le) centre du cercle inscrit
index = (un) indice
inequality = (une) inégalité
infinity = (l') infini
inscribed = inscrit
integer = (un) entier
intersecting lines = droites concourantes
inverse = (l') inverse
isosceles = isocèle

[color=orange][size=150]J[/size][/color]

[color=orange][size=150]K[/size][/color]

[color=orange][size=150]L[/size][/color]
least common multiple (LCM) = (le) plus petit multiple commun (PPCM)
lemma = (un) lemme
line = (une) droite

[color=orange][size=150]M[/size][/color]
mean = (la) moyenne
midpoint = (le) milieu
multiple = (un) multiple

[color=orange][size=150]N[/size][/color]
numerator = (le) numérateur
number = (un) nombre

[color=orange][size=150]O[/size][/color]
obtuse angle = (un) angle obtus
octagon = (un) octagone
odd = impair
origin = (l') origine
orthocenter = (l') orthocentre

[color=orange][size=150]P[/size][/color]
pair = (une) paire
pentagon = (un) pentagone
perfect square = (un) carré parfait
perpendicular = (la) perpendiculaire
plane = (le) plan
polygon = (un) polygone
polynomial = (un) polynôme
power = (la) puissance
probability = (la) probabilité
proof = (une) preuve
Pythagorean triple = (un) triplet Pythagoricien

[color=orange][size=150]Q[/size][/color]
quadratic mean (QM) = (la) moyenne quadratique
quadrilateral = (un) quadrilatère
Q.E.D (quod erat demonstrandum) = C.Q.F.D. (ce qu'il fallait démontrer)

[color=orange][size=150]R[/size][/color]
reciprocal = (l') inverse
relatively prime = premiers entre eux
remainder = (le) reste
right angle = (un) angle droit
right-angled triangle = (un) triangle rectangle
root = (une) racine
row = (une) ligne

[color=orange][size=150]S[/size][/color]
segment = (un) segment
sequence = (une) suite
series = (une) série
set = (un) ensemble
similar = semblable
solution = (une) solution
sphere = (une) sphère
square = (un) carré
square root = (la) racine carrée
statement = (une) proposition

[color=orange][size=150]T[/size][/color]
tangent = (une) tangente
theorem = (un) théorème
trapezoid = (un) trapèze
triangle = (un) triangle
trigonometry = (la) trigonométrie

[color=orange][size=150]U[/size][/color]
unique = unique
unit = unité

[color=orange][size=150]V[/size][/color]
vertex (combinatorics) = (un) sommet
vertices = sommets

[color=orange][size=150]W[/size][/color]
[color=orange][size=150]X[/size][/color]
[color=orange][size=150]Y[/size][/color]
[color=orange][size=150]Z[/size][/color]

[/hide]

Très bien.  Je serais très content si "mon idée" nous donnera des résultats.
 :blush:   Moi, je crains que je ne puisse(/peux?) rien contribuer.

Mais comment organiser?  Tout le monde écrit des mots et on les met ensembe de temps en temps? I propose in alphabetic order according to the French word and I can, perhaps with help of excel? switch that around? (sorry had no idea how to do that in French)

[quote="fredbel6"]
 :blush:   Moi, je crains que je ne puisse(/peux?) rien contribuer.

On dirait plutôt : [color=red]On dirait plutôt : Je crains de ne pas pouvoir y être trop actif.[/color]
[/quote]

[quote="fredbel6"]
Mais comment organiser?  Tout le monde écrit des mots et on les met ensemb[color=red]l[/color]e de temps en temps? I propose in alphabetic order according to the French word and I can, perhaps with help of excel? switch that around? (sorry had no idea how to do that in French)

[color=red]Je propose de les ranger par ordre alphabétique, et je peux, peut-être, aider avec excel?[/color]
[/quote]

Oui, bonne idée. :)

J'apprécie vos efforts totalement!  Je vais montrer la liste aux autres!

Mais une petite question : est-il possible de ajouter le/la ou un/une.

Parce pour moi, apprendre un nouveau mot français, c'est :
pour 40 % : le mot
30%  :les accents
30% : masculin ou féminin.

J'espère que je ne vous fatigue pas  :blush:

Ok, pas de problème! ;)

Non non, pas du tout!! (en plus, là je suis en vacances, donc j'ai "tout mon temps".. :P)

Très intéressant!  Ca va m'aider, et sûrement des autres aussi! ;)

Quelques questions:

[b]acute angle =  un acutangle [/b]

C'est correct?

Il n'y a pas de doute que l'article soit l', mais on ne sait pas si c'est masculin ou féminin  :blush:

Je propose
[b]hypotenuse = hypoténuse  (f)([/b]C'est correct?)

[b]intersecting lines = droites concourantes  (f )[/b]

[b]inverse = (l') inverse [/b] ( f ou m?) je n'ai aucune idée :blush:

[quote="fredbel6"]

[b]acute angle =  un acutangle [/b]

C'est correct?
[/quote]

Ben, c'est un adjectif, donc, pas d'article...
On dit "Un triangle acutangle", c'est tout. :)

[quote="fredbel6"]
Je propose
[b]hypotenuse = hypoténuse  (f)([/b]C'est correct?)

[b]intersecting lines = droites concourantes  (f )[/b]
[/quote]
Oui. :)

[quote="fredbel6"][b]inverse = (l') inverse [/b] ( f ou m?) [/quote]
m. :)

Ca prendra longtemps comme ca... ce que nous avons fait avec le neerlandais, c'est de prendre d'abord une longue liste avec des mots englais (vous pouvez les copier) et alors les traduire tous (par une personne qui maitrisse parfaitement la langue, desole fredbel ;)).

[mes excuses pour ma manque d'accents, je n'en ai pas sur mon ordinateur]

Oui Peter, qu'est-ce que tu voulais dire? :?

Aaah c'\'etait cach\'e! :D

Je ne l'avais pas vu, et j'ai pens\'e que vous n'\'etiez pas encore commenc\'es.

[url]http://49gym-athin.att.sch.gr/gallika/PASSE%20COMPOSE.doc[/url]

Je me doute un peu de ce mont[b]r[/b]er chez  :|  le groupe "être" mais je pense que c'est intéressant.

Les accents, les gendres, l'ordre des mots, et être/avoir, subjonctif/indicatif....  c'est  difficile  :( .

Il faut savoir qu'on néerlandais on les mixe aussi, et on dit "Jullie zijn begonnen" ="Vous êtes commencé".

Est-il correct que l'erreur la plus courante chez les francophones soit (est? :oops: )   ceci :
"J'ai manger.  Je veux travaillé.  Il veut la visitez...."

En néerlandais, on mixe les d, t et dt, souvent à la fin d'un verbe, ce sont les :dt-erreurs :lol:

[quote="mathmanman"]Hehe. :)

Au fait, c'est "n'[color=red]aviez[/color] pas encore commencé". ;)[/quote]Zut, c'est vrai. :P [j'ai ecrit "aviez" et je l'ai corrig\'e apr\grave{e}s :?]

[quote="fredbel6"][url]http://49gym-athin.att.sch.gr/gallika/PASSE%20COMPOSE.doc[/url]

Je me doute un peu de ce mont[b]r[/b]er chez  :|  le groupe "être" mais je pense que c'est intéressant.
[/quote]
En effet, ils ont fait une faute de frappe. C'est "monter". ;)

[quote="fredbel6"]
Les accents, les [color=red]genres[/color], l'ordre des mots, et être/avoir, subjonctif/indicatif....  c'est  difficile  :( .
[/quote]
J'imagine.. :-/

[quote="fredbel6"]Il faut savoir qu'[color=red]e[/color]n néerlandais on les mixe aussi, et on dit "Jullie zijn begonnen" ="Vous êtes commencé".
[/quote]
Ah ok. :)  (même en anglais : "we are done".)

[quote="fredbel6"]
Est-il correct que l'erreur la plus courante chez les francophones soit (est? :oops: )   ceci :
"J'ai manger.  Je veux travaillé.  Il veut la visitez...."
[/quote]
:rotfl: , oui, c'est vrai..
En même temps, à l'heure actuelle, le niveau orthographique des élèves est plutôt mauvais...

[quote="fredbel6"]
En néerlandais, on mixe les d, t et dt, souvent à la fin d'un verbe, ce sont les :dt-erreurs :lol:[/quote]
:rotfl:

Et, Peter, ce n'est pas grave, au moins tu n'auras plus de doute la prochaine fois! ;)
Ah, et ce n'est pas la peine de mettre des "\'e" pour "é", parce qu'ils n'apparaissent pas..
De toute façon, même sans les accents, je comprends ce que tu écris. :)

Peut être étonnant, mais je suis content que la Belgique soit un des pays utilisant AZERTY, pas QWERTY.  C'est mieux pour les mathématiques, et je peux écrire mon nom : Frédéric (ma mère, une Flamande convaincu, voulait ça, je n'ai aucune idée pourquoi  :( )).  Vous pouvez vous imaginer que ça cause des problèmes quand l'université veut avoir l'air international et emploie QWERTY...

Peter, c'est pour ça non?  Votre laptop est QWERTY?

Avez-vous entendu du mot VEERFLES?   :D

Mon laptop est US-qwerty, oui, j'y suis habitu\'e et je ne sais plus utiliser autre chose. :)

C'est 1000x plus facile pour programmer (PHP), latex, ...

Salut !

C'est super ce topic de traduction pour les francophones (futur anglophones :roll:)

J'aurais juste une question : on dit comment "la corde commune" en anglais (entre 2 cercles bien sur)  :D

Je dirais 'Comment tangent (to two circles)'
Il semble qu'on dit aussi 'Circle-circle tangent'

Quelqu'un de la Suisse!  Vous allez participer à l'IMO ?

Je suis content que le dictionnaire aide des gens  ;) .  Vous êtes en train d'apprendre anglais (à l'école ou seulement toi?)

Comment est-ce qu'on appelle quelqu'un de la Suisse? :?  Un Suisse..Suissois? :D

"common" will do for the english ;) [not comment]

Oui, on dit un "suisse". Mais ce que je préfère c'est la manière dont on appelle les filles suisses : les suissesses. :D

Oui je vais en Slovénie Imo 2006 ^^   :)

[quote="fredbel6"]Vous êtes en train d'apprendre anglais (à l'école ou seulement toi?) [/quote]
J'apprend bien sur l'anglais à l'école, mais il y a pleins de mots mathématique que je connais pas alors c'est sympa le petit dico ^^...

[quote="fredbel6"]Je dirais 'Comment tangent (to two circles)[/quote]
Je dis pas la tangeante mais la Corde !  :D
La droite qui relie par exemple les 2 points où se coupent les cercles... Ou la ligne de puissance des 2 cercles par exemple..
Mais n'y a-t-il pas un échivalent pour la "corde commune aux deux cercles ?"

Mathmanman, je vous remercie pour vous efforts.
Maintenant que mes examens sont finis, je prépare une dictionnaire français néerlandais, je ne sais pas s'il y aura quelqu'un interessé  :?  mais on ne sait jamais.

Comment est-ce que on dit 'bounded' (as in bounded set or bounded function) en français?  Limité(e)? :o  :huh:

"bound" -on dit "borne" et bien "borné" pour "bounded". C'est vrai, non?

De rien [b]fredbel6[/b]. :)

Et, oui [b]nmtruong1986[/b], c'est bien ça!  :D

Merci   :)
[quote]hypotenuse = (l')hypoténuse [/quote]

Je m'excuse mais c'est masculin ou féminin.  Vraiment, pourqoui le néerlandais et le français est-ce qu'on en a besoin?
Il faut vraiment toujours immédiatement memoriser le gendre aussi, mais ce n'est pas toujours si facile (quand on écrit par exemple :"C'est l'hypoténuse" ou "Ce sont les voitures", moi je ne peux pas voir le gendre  :( )

Je savais que c'est une voiture, c'était un exemple :D
Mais c'est  mieux de toujours utiliser un article indéfini, alors on ne peut pas se douter.

[quote="fredbel6"] moi je ne peux pas voir le gendre  :( )[/quote]

Correct c'est "GENRE"...  car "gendre" veut dire tout autre chose ! :oops:

En effet.  :D
gender: genre,
son-in-law: gendre.

Quelques remarques pour améliorer le dictionnaire...

acute angle = acutangle
C'est imprécis.  On a "an acute angle" = "un angle aigu", mais "acutangle" est un adjectif, traduction de "acute-angled" (cfr "acute-angled triangle").
octagon = (un) octagone
Euh, non, c'est un oct[b]o[/b]gone :)

On pourrait ajouter :
incircle = (le) cercle inscrit
excircle = (un) cercle ex-inscrit
excenter = (le) centre d'un cercle ex-inscrit
locus = (un) lieu     (on précise parfois "un lieu géométrique")
perpendicular bisector = (la) médiatrice
prime number = (un) nombre premier
order = (l')ordre

Enfin, on pourrait préciser le genre :
hypotenuse = (l')hypoténuse   (f)
infinity = (l') infini   (m)
inverse = (l') inverse   (m)
origin = (l') origine   (f)
orthocenter = (l') orthocentre   (m)
reciprocal = (l') inverse   (m)
unit = (une) unité
order = (l')ordre  (m)

J'espère que ça aide un petit peu...

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=580162#580162[/url]

S'il vous plaît, regardez le document que j'offre là.
J'ai utilisé Excel pour combiner l'anglais, le néerlandais et le français.

Il y a encore beaucoup de troux et d'intercertitudes, et peut-être aussi d'erreurs, donc : je vous écoute  :oops: .

Je pense que je devrais aussi montrer ceci à nos collègues canadiens.

[quote="fredbel6"]Petites questions :

Quelle est la traduction( française) de "closure"

fermication? fermation?  :oops:

Et quel est l'adjectif ( pas le nom) pour "perpendicular"

je sais qu'on dit "des droites perpendiculaires" mais quelle est la forme masculine : perpendiculaire ou perpendiculair?

Merci[/quote]

La traduction de "closure" est "fermeture".  Et, euh, fais attention avec tes essais : "fermication" fait penser à "fornication", dont je te laisserai le soin de chercher la définition par toi-même...   :rotfl:

L'adjectif pour "perpendicular" est "perpendiculaire" (forme masculine).

Enfin, on écrit "des trou[b]s[/b]".  Tous les noms se finissant par "ou" prennent un "s" au pluriel, sauf "bijou, caillou, chou, genou, hibou, joujou, pou" (une liste célèbre que nous apprenons par coeur à l'école primaire...)

Pour "closure", si tu pensais à "algebraic closure", alors c'est plutôt "clôture algébrique".

[quote="nmtruong1986"]SVP, dite-moi quand on utilise "que l'on" et "qu'on". Je les vois souvent, mais je ne sais pas pourquoi.[/quote]
Salut ! Je copie d'un dictionnaire d'orthographe :

"Pour éviter l'hiatus ou le choc désagréable [b]qu'on[/b], on glisse quelquefois un [b]l'[/b] euphonique devant [b]on[/b].
exemple: [b]Dès que l'on est parti[/b].
On ne le fait pas si la lettre [b]l[/b] se trouve déjà après [b]on[/b].
Exemple. [b]Dès qu'on lira[/b]."

C'est donc une subtilité du français que je viens aussi d'apprendre en étant intrigué par ta question !  :)

P.S. Le choc désagréable "[b]qu'on[/b]" vient du fait qu'il existe un mot qui se prononce pareil et qui veut dire..[hide]stupide (écrit "con")[/hide].  :oops:

Merci beaucoup, [b]mstoenescu[/b] ! C'est-à-dire les deux façons sont la même, non? :D

Oui ,c'est le même sens, mais sans "qu'on" cela s'entend moins "con". :D

P.S. As-tu lu le message privé que je t'ai envoyé ? :maybe:

je viens déjà de les lire. :D Moi, j'ai ainsi trouvé un site forum mathématique assez utile. Il est sur http://www.maths-forum.com. Tu le connais?
ah, ma signature, je l'ai connue d'un livre physique de mécanique quantique (je ne me le souviens guère)

Avec beaucoup de retard ... oui, je connais ce forum ! Il y a aussi http://www.les-mathematiques.net/ et http://bacamaths.net/ que j'affectionne tout particulièrement !

Je trouve que nos signatures veulent dire à peu près la même chose  :wink:


calculus
integration
function
real analysis
limit
algebra
domain

$f$:$[0,1]\rightarrow R^{+}$ measurable function

$a\geq 0$,  $M_{f}(a)=m(x\in [0,1], a\leq f(x) )$ where m is Lebesgue mesure

Compute $\int_{0}^{+\infty}ky^{k-1}M_{f}(y)dy $ for $k\in N^{*}$

No need to restrict $k$ to be an integer. In fact, this is a well-known way to compute, and hence estimate, $L^p$ norms.

Note first that $\lim_{y\to0^+}y^kM_f(y)=0$ trivially becuase you had the domain of $f$ be of finite measure.

Let's make the temporary assumption that $\int_{[0,1]} f^k<\infty.$ Then we can use the DCT to prove that $\lim_{y\to\infty}y^kM_f(y)=0$

$M_f$ is monotone nonincreasing. We can use integration by parts to obtain the following Steiltjes integral:

$\int_0^{\infty}ky^{k-1}M_f(y)dy=\lim_{y\to\infty}y^kM_f(y)+ \int(_0^{\infty}y^k(-dM_f(y))$

Now assume $f$ is a simple function. There are constants $0<y_1<y_2<\cdots <y_n$ such that $f(x)=y_j$ for $x\in E_j.$ The sets $E_j$ are disjoint.

If $y_{j-1}\le y<y_j,$ then $M_f(y)=\sum_{i=j}^n m(E_i)$

Then $\lim_{y\to\infty}y^kM_f(y)=0$ because $M_f(y)=0$ for $y\ge y_n.$ and we have

$\int_0^{\infty}ky^{k-1}M_f(y)dy=\int(_0^{\infty}y^k(-dM_f(y))= \sum_{j=1}^n y_j^km(E_j)=\int_{[0,1]}f^k.$

We can now take a sequnce of simple functions $f_n$ increasing to $f.$ Since $M_{f_n}(y)$ increases to $M_f(y),$ we may employ the Monotone Convergence Theorem to see that

$\int_0^{\infty}ky^{k-1}M_f(y)dy=\int_{[0,1]}f^k.$


[quote="apollo"]a,b,c,d > 0
$ c^{2}\plus{}d^{2}\equal{} (a^{2}\plus{}b^{2})^{3}$
Prove that $ \frac{a^{3}}{c}\plus{}\frac{b^{3}}{d}\geq 1$[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=119243

Here is a simpler solution:
By Holder's inequality, $ \left(\frac{a^{3}}{c}\plus{}\frac{b^{3}}{d}\right)^{2}(c^{2}\plus{}d^{2})\ge (a^{2}\plus{}b^{2})^{3}\equal{} c^{2}\plus{}d^{2}$, so $ \left(\frac{a^{3}}{c}\plus{}\frac{b^{3}}{d}\right)^{2}\ge 1$, which yields the conclusion.

[color=darkblue][size=134][b]Very nice inequality ![/b][/size] I remark an interesting consequence of it : [/color]$ c^{2}\plus{}d^{2}\equal{}\left(a^{2}\plus{}b^{2}\right)^{3}\implies\left(a^{3}\plus{}b^{3}\right)\left(\frac{1}{c}\plus{}\frac{1}{d}\right)\ge 2$.

[quote="Virgil Nicula"]$ c^{2}\plus{}d^{2}\equal{}\left(a^{2}\plus{}b^{2}\right)^{3}\implies\left(a^{3}\plus{}b^{3}\right)\left(\frac{1}{c}\plus{}\frac{1}{d}\right)\ge 2$.[/quote]
[color=darkblue][b]Proof.[/b] $ c^{2}\plus{}d^{2}\equal{}(a^{2}\plus{}b^{2})^{3}\implies$ $ \left\{\begin{array}{c}\frac{a^{3}}{c}\plus{}\frac{b^{3}}{d}\ge 1\\\\ \frac{b^{3}}{c}\plus{}\frac{a^{3}}{d}\ge 1\end{array}\right\|\bigoplus\implies$ $ \left(a^{3}\plus{}b^{3}\right)\left(\frac{1}{c}\plus{}\frac{1}{d}\right)\ge 2$.[/color]


function
modular arithmetic
number theory unsolved
number theory

Question

Prove that there are infinitely many integers $ n$ for which $ \phi (n)$ is a perfect square.

There are a lot of multiplicative functions $ f$ such that $ d|n$ doesn't imply that $ f(d) |f(n)$ ;)

Hmm, If $ p$ is prime and $ 0\leq a\leq b$ then $ \phi (p^{a})|\phi (p^{b})$.  :wink:

[quote="sludgethrower"]We can see it works for any prime of the form $ m^{2}+1$[/quote]

Yes, but the question of whether there are an infinitude of these is, in fact, open.

Question 3

Show that if $ gcd(a,\,n) = gcd(a-1,\,n) = 1$ then:

$ 1+a+a^{2}+\ldots+a^{\phi (n)-1}\equiv 0\,(mod\,n)$

Note that $ (1+a+a^{2}+\ldots+a^{\phi (n)-1})(a-1)\equiv a^{\phi(n)}-1\equiv 0\pmod{n}$ since $ gcd(a,n)=1$. And since $ gcd(a-1,n)=1$, we can divide by $ a-1$ for the desired.

I think that, in any topic we only post one question!  :wink:

Well, they aren't even suitable for this forum (they are too easy).
And by the way, question two is nice but not the way you two want to prove it ;)

[quote="sludgethrower"]Question 2

Assuming that $ d\,|\,n$, prove that $ \phi (d)\,|\,\phi (n)$.[/quote]

Just for fun, one last proof:
$ \phi(n)/\phi(d) = \frac{n\prod_{p\mid n}(1-(1/p) )}{d\prod_{p\mid d}(1-(1/p) )}=\frac{n}{d}\prod_{p\mid n,~p\nmid d}(1-(1/p))$
The RHS is an integer...cha ching.


trigonometry
inequalities unsolved
inequalities

prove that
$ \sum_{cyclic}\sqrt {(x \plus{} y)(x \plus{} z)}\ge x \plus{} y \plus{} z \plus{} \sqrt {3(xy \plus{} yz \plus{} zx)}$

    (x,y,z>0)

$ \\\sqrt{x\plus{}y}\equal{}u \\\sqrt{y\plus{}z}\equal{}v \\\sqrt{z\plus{}x}\equal{}w \\\\\sum uv\geq \frac{1}{2}\sum u^2\plus{}\sqrt{\frac{3}{4}(\sum 2u^2v^2\minus{}\sum u^4)} \\\\\Leftrightarrow \sum u^4\plus{}\sum u^2vw\minus{}\sum_{sym}^{}u^3v\geq 0 \\\\\Leftrightarrow \sum u^2(u\minus{}v)(u\minus{}w)\geq 0$

It's nice!

assume $ xy \plus{} yz \plus{} zx \equal{} 3$ and $ x \equal{} \sqrt {3}\tan{\frac {A}{2}},y \equal{} \sqrt {3}\tan{\frac {B}{2}},z \equal{} \sqrt {3}\tan{\frac {C}{2}}$, $ A,B,C > 0, A \plus{} B \plus{} C \equal{} \pi$.
ineq <=> $ \sum \frac {1 \minus{} \sin \frac{A}{2}}{\cos \frac{A}{2}} \ge \sqrt {3}$.
Jensen...


inequalities
integration
real analysis
real analysis unsolved

we put $ E=\mathbb{C}^{1}([a,b],\mathbb{R}).$

Prove that,for every $ \epsilon>0$,there exist $ D>0$ such that,for every $ f\in\E},$ $ sup_{x\in\ [a,b]} f(x)^{2}\leq\ D\int_{a}^{b}f^{2}+\epsilon\int_{a}^{b}f'^{2}.$

Suppose $ f\in E$, choose $ c\in [a,b]$ such that $ f^2(c)\equal{}\frac{1}{b\minus{}a}\int_a^bf^2(t)dt$.
Let $ x\in [a,b]$, then \[ f^2(x)\equal{}f^2(c)\plus{}\int_c^x2f(t)f'(t)dt\leq \frac{1}{b\minus{}a}\int_a^bf^2(t)dt \plus{} \int_a^b\left(\varepsilon f'^2(t)\plus{}\frac{1}{\varepsilon}f^2(t)\right)dt.\]


ratio
complex analysis
real analysis
real analysis unsolved

maybe this should go in the complex analysis subforum, but this was a question for my real analysis homework last week...

For what values of $ z$, $ z$ complex, does the series $ \sum_{n\equal{}1}^{\infty} {\frac{1}{1\plus{}z^n}}$ converge?

I can prove that the series must diverge for $ |z| <\equal{} 1$, and that for $ z > 1$, $ z$ real, the sequence converges. But for what values of non-real $ z$ does the series converge?

I've only been able to get as far as saying that if
$ |z| > 1$ and $ |z^n| <\equal{} |1\plus{}z^n|$
then the series will converge (by comparison test to the geometric series).

If the series has a chance of converging, the terms must tend to 0, so $ |1\plus{}z^n|$ is unbounded. But of course, $ |1\plus{}z^n|\leq 1\plus{}|z|^n$. So if $ |z|\leq 1$, the series automatically diverges.

On the other hand, if $ |z|>1$, we have absolute convergence, since $ |1\plus{}z^n|\geq |z|^n\minus{}1$ so $ \frac{1}{|z^n\plus{}1|}\leq \frac{1}{|z|^n\minus{}1}$, and the series of these terms converges by the Ratio Test.

[quote]
$ |1\plus{}z^n|\geq |z|^n\minus{}1$
[/quote]

how do you get this statement (for complex $ z$) based on $ |z|>1$?

$ |x \plus{} y| \geq |x| \minus{} |y|$ is just another form of the triangle inequality.  (Use $ x \equal{} (x \plus{} y) \plus{} (\minus{}y)$ and $ |y| \equal{} |\minus{}y|$.)


Given triangle ABC and points D, E, and F on BC, CA, and AB respectively such that AD, BE, and CF concur. Prove that, given points M, N, and P on EF, FD, DE, respectively, we have that: AM, BN, CP concur iff DM, EN, FP concur.

I don't think it should be a very difficult problem (see the title); I'm just interested in what strategies you'll use to attack it. (I'm sure there are many ways.)[/hide]

The first direction,

{DM, EN, FP concur} $\Longrightarrow$ {AM, BN, CP concur},

was proven in the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=3536 .

The second direction,

{AM, BN, CP concur} $\Longrightarrow$ {DM, EN, FP concur},

is an easy corollary: Let the lines DM and EN meet at U, and let the line FU meet the line DE at the point P'. Then, the points M, N, P' lie on the lines EF, FD, DE, respectively, and the lines DM, EN, FP' concur (at the point U), and hence the first direction yields that the lines AM, BN, CP' concur. In other words, the line CP' passes through the point of intersection of the lines AM and BN. But since the lines AM, BN, CP concur, the line CP also passes through the point of intersection of the lines AM and BN. Hence, the lines CP' and CP have two points in common: the point C, and the point of intersection of the lines AM and BN. Thus, these two lines coincide, and the point P' lies on the line CP. Hence, the point P' is the point of intersection of the lines CP and DE. And thus, P' = P. Consequently, since we know that the lines DM, EN, FP' concur, we see that the lines DM, EN, FP concur. Proof complete.

  Darij


modular arithmetic
number theory proposed
number theory

Let $ k\geq 2$ be  integer, $ n_1,\ n_2,\ n_3$ be positive integers and $ a_1,\ a_2,\ a_3$ be integers from $ 1,\ 2,\ \cdots ,\ k \minus{} 1$.
Let $ b_i \equal{} a_i\sum_{j \equal{} 0}^{n_i} k^{j}\ (i \equal{} 1,\ 2,\ 3)$. Find all possible pairs of integers $ (n_1,\ n_2,\ n_3)$ such that $ b_1b_2 \equal{} b_3$.

$ a_1a_2\left(\sum_{j \equal{} 0}^{n_1}k^j\right)\left(\sum_{j \equal{} 0}^{n_2}k^j\right) \equal{} a_3\left(\sum_{j \equal{} 0}^{n_3}k^j\right)$

$ \Leftrightarrow a_1a_2(k^{n_1 \plus{} 1} \minus{} 1)(k^{n_2 \plus{} 1} \minus{} 1) \equal{} a_3(k^{n_3 \plus{} 1} \minus{} 1)(k \minus{} 1)$

We have $ k^{n_i \plus{} 1} \minus{} 1\equiv \minus{} 1(\mod k^2)$
$ \Rightarrow a_1a_2\equiv \minus{} a_3(k \minus{} 1)\equiv k^2 \minus{} a_3(k \minus{} 1)(\mod k^2)$

Let $ a_1 \equal{} m$, $ a_1 \equal{} n$ and $ a_3 \equal{} t$. $ (m,n,t\in\{1,2,\dots,k \minus{} 1\})$

Then $ k^2 \minus{} (k \minus{} 1)t\equiv mn (\mod k^2)$. Clearly, since neither $ mn$ nor $ k^2 \minus{} (k \minus{} 1)t$ can be greater than $ k^2$, then $ mn \equal{} k^2 \minus{} (k \minus{} 1)t$.
$ k^2 \minus{} (k \minus{} 1)t\equiv 1(\mod k \minus{} 1)$ $ \Rightarrow mn \equal{} s(k \minus{} 1) \plus{} 1 (1\le s\le k \minus{} 2)$,since $ 1\le mn\le (k \minus{} 1)^2$.
$ \Rightarrow (t \plus{} s)(k \minus{} 1) \equal{} (k \plus{} 1)(k \minus{} 1)$ or which is equivalent - $ t \equal{} k \plus{} 1 \minus{} s$ $ (\Rightarrow s\ge 2)$.

$ (s(k \minus{} 1) \plus{} 1)(k^{n_1 \plus{} 1} \minus{} 1)(k^{n_2 \plus{} 1} \minus{} 1) \equal{} (k \plus{} 1 \minus{} s)(k^{n_3 \plus{} 1} \minus{} 1)(k \minus{} 1)$

If $ \min\{n_1,n_2,n_3\}\ge 2$, then $ k^3 \minus{} (k \minus{} 1)(k \plus{} 1 \minus{} s)\equiv s(k \minus{} 1) \plus{} 1(\mod k^3)$ which is impossible. Hence WLOG $ n_1 \equal{} 1$ ($ n_3$ can't be 1 since then LS>RS).

$ \Rightarrow (s(k \minus{} 1) \plus{} 1)(k \plus{} 1)(k^{n_2 \plus{} 1} \minus{} 1) \equal{} (k \plus{} 1 \minus{} s)(k^{n_3 \plus{} 1} \minus{} 1)$.

Again - if $ \min\{n_2,n_3\}\ge 2$, then $ k^2s \plus{} k \minus{} s \plus{} 1 \equal{} (s(k \minus{} 1) \plus{} 1)(k \plus{} 1)\equiv (k \plus{} 1 \minus{} s) (\mod k^3)$, which is impossible. $ \Rightarrow n_2 \equal{} 1$.

$ \Rightarrow (s(k \minus{} 1) \plus{} 1)(k \plus{} 1)(k^2 \minus{} 1) \equal{} (k \plus{} 1 \minus{} s)(k^{n_3 \plus{} 1} \minus{} 1)$.
$ \Leftrightarrow sk^3 \plus{} (s \plus{} 1)k^2 \plus{} (2 \minus{} s)k \plus{} 1 \minus{} s \equal{} k^{n_3 \plus{} 1} \plus{}$ $ \dots \plus{} k^2 \plus{} k \minus{} sk^{n_3} \minus{} \dots \minus{} sk \minus{} s \plus{} k^{n_3} \plus{} \dots \plus{} k \plus{} 1$. It's obvious that $ n_3 > 2$.
$ \Rightarrow sk^3 \plus{} (s \plus{} 1)k^2 \equal{} k^{n_3 \plus{} 1} \plus{}$ $ \dots \plus{} k^2 \minus{} sk^{n_3} \minus{} \dots \minus{} sk^2 \plus{} k^{n_3} \plus{} \dots \plus{} k^2$.
Knowing $ 2\le s\le k \minus{} 2$, we can reject with confidence the case $ n_3\ge 5$.

$ 1. n_3 \equal{} 4$ can be rejected, too, easily.
$ 2. n_3 \equal{} 3$ gives us $ k^4 \equal{} 2(s \minus{} 1)k^3 \plus{} (2s \minus{} 1)k^2$

Therefore, if there existed some pair $ n_1, n_2, n_3$, then $ (n_1, n_2, n_3) \equal{} (1,1,3)$; $ a_1a_2 \equal{} s(k \minus{} 1) \plus{} 1$, $ a_3 \equal{} k \plus{} 1 \minus{} s$ and $ k^2\equal{} 2(s \minus{} 1)k \plus{} (2s \minus{} 1)$ for some $ 2\le s\le k \minus{} 2$. But this is impossible.

Of course there is a mistake  :P . It's in:
$ \minus{} a_3(k \minus{} 1)\equiv a_3k(\mod k^2)$

To simplify,let $ (a_1,a_2,a_3) \equal{} (a,b,c)\in \{1,2,...,k \minus{} 1\}$ and $ (n_1 \plus{} 1,n_2 \plus{} 1,n_3 \plus{} 1) \equal{} (p,q,r)$ all greater than 1.
We should find possible $ (p,q,r)$ such that
$ ab(k^p \minus{} 1)(k^q \minus{} 1) \equal{} c(k \minus{} 1)(k^r \minus{} 1)$ (*) for some $ k\ge 2$.

since $ p,q,r\ge 2$,we have $ ab\equiv \minus{} (k \minus{} 1)c\pmod{k^2}$,let $ ab \equal{} \minus{} (k \minus{} 1)c \plus{} uk^2$ with a positive integer $ u$,
then $ u \equal{} \frac {ab \plus{} (k \minus{} 1)c}{k^2}\le 2\left(1 \minus{} \frac {1}{k}\right)^2\Longrightarrow u \equal{} 1$,hence we have $ ab \equal{} \minus{} (k \minus{} 1)c \plus{} k^2$ (**)
and we easily see that $ ab\ge 2k \minus{} 1 > c$ since $ c\le k \minus{} 1$.

Suppose all $ p,q,r > 2$,we have
$ ab\equiv \minus{} (k \minus{} 1)c\pmod{k^3}$ by (**) we have $ \minus{} (k \minus{} 1)c \plus{} k^2\equiv \minus{} (k \minus{} 1)c\pmod{k^3}\Longrightarrow k^2\equiv 0\pmod{k^3}$ which is contradiction,
hence at least one of $ p,q,r$ is $ 2$,but suppose $ r \equal{} 2$ then $ LHS > RHS$ in (*) because $ ab > c$,so WLOG $ p \equal{} 2$ and $ r > 2$.
then (*) is equivalent to
$ ab(k \plus{} 1)(k^q \minus{} 1) \equal{} c(k^r \minus{} 1)$ (***)

and suppose $ q > 2$,we have $ (k \plus{} 1)ab\equiv c\pmod{k^3}$ so let $ (k \plus{} 1)ab \equal{} c \plus{} vk^3$ with a positive integer $ v$,combined with (**) we have $ (k \plus{} 1)( \minus{} (k \minus{} 1)c \plus{} k^2) \equal{} c \plus{} vk^3$ hence $ v \equal{} 1 \plus{} \frac {1 \minus{} c}{k}$ so $ v \equal{} 1$ and $ c \equal{} 1$,
thus $ k^2 \minus{} k \plus{} 1 \equal{} ab\le (k \minus{} 1)^2 \equal{} k^2 \minus{} 2k \plus{} 1$,which is contradiction,hence we have $ q \equal{} 2$.
(***) is equivalent to
$ ab(k \plus{} 1)(k^2 \minus{} 1) \equal{} c(k^r \minus{} 1)$ (****)
since $ c < ab\le (k \minus{} 1)^2$ we have
$ (k \plus{} 1)(k^2 \minus{} 1)( \equal{} k^3 \plus{} k^2 \minus{} k \minus{} 1) < k^r \minus{} 1\le (k \minus{} 1)^2(k \plus{} 1)(k^2 \minus{} 1)  (\equal{}k^5 \minus{} k^4 \minus{} 2k^3 \plus{} 2k^2 \plus{} k \minus{} 1)$,
and we easily see it is impossible when $ r \equal{} 3$ or $ r\ge 5$,hence we have $ r \equal{} 4$,
then (****) becomes $ ab \equal{} c(k \minus{} 1)$ which is possible successfully if we let $ (a,b,c) \equal{} (k \minus{} 1,1,1)$

hence the only possible solution is $ (p,q,r) \equal{} (2,2,4)$,that is $ (n_1,n_2,n_3) \equal{} \boxed{(1,1,3)}$.

Are you sure  [b]Anavel_Gato[/b]

when $ (a_1,a_2,a_3)\equal{}(k\minus{}1,1,1)$ and $ (n_1,n_2,n_3) \equal{} (1,1,3)$
then it must be $ (k\minus{}1)(k^2\minus{}1)^2\equal{}(k^4\minus{}1)(k\minus{}1)$, which is far away from the truth.

[quote="BG Yoda"][quote="Anavel_Gato"]... hence we have $ r \equal{} 4$,
then (****) becomes $ ab \equal{} c(k \minus{} 1)$ ...[/quote]

It must be $ ab(k \plus{} 1) \equal{} c(k^2 \plus{} 1)$ :wink:[/quote]

Oh,thanks.

$ k\equal{}7,(a,b,c)\equal{}(5,5,4)$ work,hence $ (n_1,n_2,n_3)\equal{}(1,1,3)$ is possible  :)


calculus integration trigonometry logarithms derivative LaTeX algebra

Someone gave me [url=http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign]This Link[/url] which I appreciate their time in doing. The following integral is near the bottom of that link.

Never Differentiated under the Integral before, I thought I'd give it a try.  Anyone know if this is correct?

$ \int^{\infty}_{0} \frac {ln(1 + x^2)}{1 + x^2}$

We have $ F(a) = \int^{\infty}_{0} \frac {\ln(1 + ax^2)}{1 + x^2}dx$

$ F'(a) = \int^{\infty}_{0} \frac {\partial} {\partial a}{ (\frac {ln(1 + ax^2)}{1 + x^2}} )dx$

$ F'(a) = \int^{\infty}_{0} \frac {2ax^2 dx}{(1 + a^2 x^2)(1 + x^2)}$

$ F'(a) = \frac {2}{(1 - a^2)}  [\tan^{ - 1}(ax) - (a) \tan^{ - 1}(x)  ]  |^{\infty}_{0}$

$ F'(a) = \frac {2} {(1 - a^2)}  ( \frac {\pi} {2} - a \frac {\pi} {2}  )$

$ F'(a) = \frac {\pi}{(1 - a^2)} (1 - a) \rightarrow (1 - a^2) = (1 + a)(1 - a)$

$ F'(a) = \frac {\pi}{(1 + a)}$

But we want $ F(1)$

$ \therefore F(1) = \int \frac {\pi}{(1 + a)} = \pi \;\ln(1 + a) = \pi \; \ln(2)$

A quick "numerical" check yielded 2.1426 at 400 data points.

Comments?

That's the coolest thing I've seen.    Thanks for your reply.

[quote="mlok"]Your answer is correct. Notice, however, that the result stated in Wikipedia does not cover improper integrals. To properly justify passing to a limit in an improper integral can be painful (uniformity of convergence and all). It's easier to switch the order of integration, especially when the integrand is nonnegative ([url=http://en.wikipedia.org/wiki/Fubini%27s_theorem]Tonelli's theorem[/url]). Specifically:
(1) we have $ \ln (1 + x^2) = \int_0^1\frac {2ax^2}{1 + a^2x^2}\,da$ (of course, you found this by taking a derivative, but let's keep it a secret).
(2) hence $ \int_0^{\infty}\frac {\ln (1 + x^2)}{1 + x^2} = \int_0^{\infty}\int_0^1\frac {2ax^2}{(1 + a^2x^2)(1 + x^2)}\,da\,dx$
(3) use Fubini/Tonelli to reverse the order of integration: $ \int_0^1\int_0^{\infty}\frac {2ax^2}{(1 + a^2x^2)(1 + x^2)}\,dx\,da$.
(4) integrate w.r.t $ x$ using partial fractions, then integrate w.r.t $ a$.

$ \text{\LaTeX}$ has commands for common math operators: \ln or \log, \tan, \sin, and so on.[/quote]

How did we arrive at limits on $ a$ between $ 0$ and $ 1$?

By quoting [b]relevant [/b]parts of the preceding posts.
[quote]But we want $ F(1)$.
$ \therefore F(1) \equal{} \int \frac {\pi}{(1 \plus{} a)} \equal{} \pi \;\ln(1 \plus{} a) \equal{} \pi \; \ln(2)$[/quote]
Although you omitted the limits in this integral, they are $ 0$ and $ 1$. Why?

Because the value $ a\equal{}1$ gives $ \ln(1\plus{}ax^2)\equal{}\ln(1\plus{}x^2)$ which is what we had to begin with. And when $ a\equal{}0$, $ \ln(1\plus{}ax^2)\equal{}0$ which is easy to integrate.


Ok i need you to rate my mathcount skills at terrible/bad/average/good/excellent so plz plz be honest.

October handbook problems- On an average i got about 8 right in 10 min.
November handbook problems- On an average I got about 7 right in 10 min.
December- On an average I got about 7-6 right in 10 min.
January- On an average I got about 4-5 right in 10 min.

Plz help, all responses will be greatly appreciated
Thx :lol:

Have you taken any actual practice competitions?  And what do you mean by october or november workbook problems?  I thought they just had like warmup 1,2, and workouts not separated by month.

Yeah, I would say that doing the handbook problems will help, but won't give an accurate representation of your capabilities.

Handbook activities are relatively worthless to tell what your "rating" is...

But considering how many people compete (at chapter level), you may be around average.  It just depends on what type of problems you miss.

Don't rush yourself on these.  The only time you should is if you are doing a countdown round.  Just try and stay around 13-14 minutes per Warmup/Workout.  On stretches...It depends. 13-14 minutes is a good time because its proportional to the 30 minutes they give you for 40 questions on the written round.

How many problems do you mean by "October Handbook Problems?? In which year? (Difficulty can vary with different years)

A little more clarification?

I'd say you about average. Jeffrey Chen was 2nd at nationals and he usually finished them in 8 minutes with 0 errors.

Compared to most people who participate in Mathcounts, I'd say you're above average a bit, probably enough to get top 35 or so percent in Chapters. But compared to most people who are in your grade, you're probably pretty high above average, so keep things in perspective, and remember a competition's not everything.  ;)

Not bad. That should be enough to get you in the top 10, depending on where you are.

I got 27 and 8 and got 1st at chapter.

questions: what grade are you in now?? your "rating" depends mainly on your grade and experience with MATHCOUNTS style problems.
as for the handbook problems being referred to by months: check out the mathcounts website, and find the Pdf format of the handbook... it is seperated into different months (each month about 3-5 warmups/workouts)
your sprint scores are between 18 and 21... so you might need to work on spints more than targets (your prety good at them)... as treething said: it all depends on what state you live in and what chapter you're in and how many people are at the chapter

[quote="GoBraves"]Ok many of you said you need more info so I went to Goodly Maths and did 2001 and 2002 chapter competitions(with the correct timings)

2001 chapter- Sprint-18 and target-7 right
2002 chapter- Sprint-20 and target-8 right
2004 chapter(I competed in  this one)- Sprint 18 and target-7 right

Thx a lot[/quote]

Those are awesome score. You would most likely make states with those scores and probably have a good chance at getting around 30 at states. Work a little more on your sprint rounds though, and look at what you're getting wrong. Are you not answering questions, or just making stupid mistakes? Fix what needs to be fixed to get better.

Keep the target high, work on sprint. Usually the top placers have 25 or more on the sprint round with roughly the same target round score as you.

Yes handbook things really are quite worthless.  You wanna work on  Sprints, Targets, and know strategies in team rounds.  Looking at what you got I don't know what you would call "average".

I would say pretty good.  a couple people in my school got an 8.5, and I got a 9.5, but we're in the most competitive chapter in the state and one of the most in the country, and we've one the states 3 of the last 4 yrs :lol:

GoBraves, your math skills will probably get you at least into state. On MATHCOUNTS competition, the target round is more crutial than the sprint round, so be sure to get those correct, or study those questions. Last year, my team didn't really dominate (friend got 24 on sprint, 8 correct on target) but that was good for 2nd overall. Good Luck!!! :D


calculus
integration
calculus computations

For $ t\geq 0$, let $ a_n\equal{}\int_0^ t e^{nx}\ dx\ (n\equal{}0,\ 1,\ 2,\ 3)$.

(1) Show that $ a_3\minus{}3a_2\plus{}3a_1\minus{}a_0\geq 0$.

(2) Show that $ e^ta_0\plus{}(e^t\minus{}1)a_1\minus{}a_2\geq 0$.

$ a\minus{}3 \minus{}3a_2\plus{}3a_1\minus{}a_0 \equal{} \int_0^t e^{3x}\minus{}3e^{2x}\plus{}3e^x\minus{}e^0 dx \equal{} \int_0^t (e^x\minus{}1)^3 dx$

since $ e^x\minus{}1 \ge 0$ we get the first result.

second one is up to someone else.

$ a_0 = \int_0^t{e^{0x}\,dx} = t$

$ a_1 = \int_0^t{e^x}\,dx = e^t - 1$

$ a_2 = \int_0^t{e^{2x}\,dx} = \frac {1}{2}e^{2t} - \frac {1}{2} =$

$ \therefore\quad{e^ta_0 + (e^t - 1)a_1 - a_2}$

$ = e^tt + (e^t - 1)^2 - \frac {1}{2}e^{2t} - \frac {1}{2}$

Now let $ f(t) = e^tt + (e^t - 1)^2 - \frac {1}{2}e^{2t} - \frac {1}{2}$

$ \Rightarrow{f'(t) = e^{2t} + (t - 1)e^t}$

$ f'(x) = 0\quad\text{only at }x = 0$

Therefore $ f'(x_0 > 0)$ will share the sign of any $ f(x_1 > 0)$

Now since $ f'(1) = e^2$

This implies that $ f'(t) > 0\quad\forall{t}\in(0,\infty)$

Therefore $ f(t)$ is constantly increasing on that interval, therefore the smallest value of $ f(t)$ on $ [0,\infty)$ will be $ f(0) = 0$

This implies that $ \forall{t} > 0\quad{f(t) > 0}\quad\blacksquare$

Is that acceptable?

[quote="kenn4000"]
since $ e^x \minus{} 1 \ge 0$ we get the first result.

[/quote]

Musn't you say $ e^x\minus{}1\geq{0}\quad{\color{red}\forall{x}\geq{0}}$?

[quote="kunny"][quote="Mathstud28"]

$ \therefore\quad{e^ta_0 + (e^t - 1)a_1 - a_2}$

$ = e^tt + (e^t - 1)^2 - \frac {1}{2}e^{2t} - \frac {1}{2}$ [/quote]

You seemed to have made a mistake in here.[/quote]

Aww its plus!

ok

Well briliantly everything with that typo is still true  :D

it is true what you say, but since the limits of integration only deal with positive x, i didnt specify.


probability
analytic geometry
expected value

So u hav 5 coins labeled 1-5. Coin 1 flips a head with probability 1, coin 2 with prob .9, coin 3 with .8, coin 4 with .7, and coin 5 with .6. what is the expected value of flips needed to get 5 heads. The catch is that if you flip a tails on coin k, then you have to re-flip coin k-1. So if you flip 1, 2, 3, and 4 all heads, then flip 5 tails, you have to re-flip 4, and if you get a tails again, you have to re-flip 3 etc.

I'm not sure if there's any mathematical way to solve this but it'd be great if someone had an idea.

i believe u start with coin 1 then flip 2 then 3 etc

Here's one way to do this:  start by drawing a grid in the coordinate plane including the points (0, 0), (5, 0) and (5, 4).  Draw the line $ y \equal{} x$ and erase everything above it -- we don't need it.  Also, erase the line $ x \equal{} 5$ (but keep the vertices $ (5, 0), \ldots, (5, 4)$!).

The vertex $ (a, b)$ is going to denote the event that we've flipped $ a$ heads and $ b$ tails.  Now, label the horizontal edge joining $ (a, b)$ to $ (a \plus{} 1, b)$ with the probability that, if you've flipped $ a$ heads and $ b$ tails (which means you're at coin $ 1 \plus{} a \minus{} b$), you next flip a head.  Similarly, the vertical edge joining $ (a, b)$ to $ (a, b \plus{} 1)$ gets labelled with the probability that from coin $ 1 \plus{} a \minus{} b$ you flip a tail.  Now, put a 1 on the vertex (0, 0) and do block-walking.  However, instead of just adding the numbers on the vertices below you and to your left, you weight them by the appropriate factors.  So, for example, the vertex (1, 0) gets labelled $ 1 \cdot 1 \equal{} 1$, the vertex (2, 0) gets labelled $ 1 \cdot .9 \equal{} .9$, the vertex (1, 1) gets labeled $ 1 \cdot .1 \equal{} .1$, the vertex (2, 1) gets labeled $ .1 \cdot 1 \plus{} .9 \cdot .2 \equal{} .28$, and so on.  At the end, the number labelling the vertex $ (5, b)$ is the probability that you first get five heads after $ 5 \plus{} b$ flips, and you should be able to compute the expected value from this.

This is not really any different than the sort of recursive computation that you have to carry out to do this in any case, but I think it's a nice visual illustration of what's going on.


Convex polygon ABCDEF is given in the plane. |AB|=|BC|, |CD|=|DE|, |EF|=|FA|.

Prove this inequality: |BC|/|BE| + |DE|/|DA| + |FA|/|FC| >= 3/2.

I`m not sure where should I put this, here, or at Geometry section.

OK, ok be patient please :)

You need only two well-known results:

1) Ptolemy's inequality: for any points $A$, $B$, $C$, $D$ in the plane we have \[ AB \cdot CD + AD \cdot BC \geq AC \cdot BD.\]

2) Nesbitt's inequality: for any positive reals $a$, $b$, $c$ we have \[\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} \geq \frac{3}{2}.\]

Can you see how to do it now?

This one was proposed a long time before! And I remember having the same solution as you Arne!  :D I tried searching for it but I couldn't find it! :(

It's one of the oldest problems posted on Mathlinks. I think it was posted by iandrei, who was looking for a solution different from the one Arne gave.

Thanks, Arne.

What`s IMO Shortlist? It looks too easy for IMO...

If I find or hear for another solution, I`ll post it here.


Let $ n \in \mathbb{Z}^\plus{}$. Prove that $ \lfloor \sqrt{4n\plus{}2} \rfloor \equal{} \lfloor \sqrt{n\plus{}1} \plus{} \sqrt n \rfloor$.

[quote="mathson"]Let $ n \in \mathbb{Z}^ \plus{}$. Prove that $ \lfloor \sqrt {4n \plus{} 2} \rfloor \equal{} \lfloor \sqrt {n \plus{} 1} \plus{} \sqrt n \rfloor$.[/quote]

$ 4n\plus{}2<(2n\plus{}2)\plus{}2\sqrt{n(n\plus{}1)}\equal{}(\sqrt {n \plus{} 1} \plus{} \sqrt n )^2<4n\plus{}3$ (because $ \sqrt{n^2}<\sqrt{n(n\plus{}1)}<\sqrt{(n\plus{}\frac{1}{2})^2}$

Now assume that $ \lfloor \sqrt {4n \plus{} 2} \rfloor \equal{}k \neq \lfloor \sqrt {n \plus{} 1} \plus{} \sqrt n \rfloor$,
then $ \sqrt{4n\plus{}3} >\sqrt {n \plus{} 1} \plus{} \sqrt n \ge \lfloor \sqrt {n \plus{} 1} \plus{} \sqrt n \rfloor \ge k\plus{}1$ from the above result.

So $ \lfloor \sqrt{4n\plus{}3} \rfloor \ge \lfloor \sqrt{4n\plus{}2} \rfloor\plus{}1 \Rightarrow$ $ 4n\plus{}3\equal{}m^2$ is a square, which is impossible as $ m^2\equal{}0,1(\mod 4)$ for all $ m$, contradiction, done.

Posted and solved [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=272857]here[/url].


logarithms inequalities conics parabola AMC

I dont hvae my test booklet (because my teachers wont let us take them), but I looked at the answers and it said 17 was D(b) on the amc 12b, but i was wondering how it can be that?  I keep on getting 1/b.. can someone explain to me (sorry I dont have the question)

[quote="Inspired By Nature"]I got 1/b too. :([/quote]

I got b, but circled 1/b on the test booklet.

The problem is:

If $a$ is a nonzero integer and $b$ is a positive number such that $ab^{2}=\log_{10}b$, what is the median of the set $\{ 0,1,a,b,1/b \}$.

Choices were (A)0, (B)1, (C)a, (D)b, (E) 1/b

[hide]The answew was (D)b because we can let $b=\frac{1}{10}$ and $a=-100$ to satisfy the conditions.  This would make the median b.  [/hide]

or let b=1/100 and a=-20000, though i am not sure if there is any counterexample

Well, yeah, they always imply unique answer, so you find one case, bubble, and go on :D.

I thought it said positive number for both, so i somehow found the median to be a....

did three other stupid mistakes as well.... :mad:

That question confused me, because initially I thought that a was squared.  However, this is obviously impossible.  I ended up just saying that a=-4, then did it on my calculator.

That question was amazingly easy to solve by brute force :roll: .

The solutions described so far are fine for a multiple-choice test.  But some of you may be wondering why $b$ is the median in all cases.  Below is a proof.  I will use the well-known inequality $\ln x \le x-1$ for all positive $x$.

We have
\begin{align*}a &= \frac{\log_{10}b}{b^{2}}\\ &= \frac{\log_{10}(b^{2})}{2 b^{2}}\\ &= \frac{\ln(b^{2})}{2 \ln(10) b^{2}}\\ &\le \frac{b^{2}-1}{2 \ln(10) b^{2}}\\ &< \frac{b^{2}}{2 \ln(10) b^{2}}\\ &= \frac{1}{2 \ln(10)}\\ &< 1. \end{align*}
Thus $a$ is smaller than $1$.  The problem specified that $a$ is a nonzero integer.  Thus $a$ must be negative.  From the given equation $a b^{2}= \log_{10}b$, it follows that $\log_{10}b$ is negative.  That means $b$ is strictly between $0$ and $1$.  Thus we have
\[a < 0 < b < 1 < \frac{1}{b}\, . \]
It follows that $b$ is the median of these five numbers.

[hide="Similar solution"]Assume $a$ is negative.  Then $ab^{2}$ has to be negative, meaning $b$ must be between $0$ and $1$ (clearly it can't be below $0$, and if it was above $1$ its log would be positive).  This leads us to the order $a, 0, b, 1, \frac{1}{b}$.

Now assume $a$ is positive.  This means $ab^{2}$ is positive, so $log b$ must be positive, so $b > 1$...but $b^{2}> log b$ for $b > 1$, and therefore (since $a$ is an integer) $ab^{2}$ must also be  strictly greater than $log b$.  This contradicts the equation.  Thus $a$ is negative, and the order is as shown above, QED.[/hide]

And if you are curious what is the largest possible $a$ (ignoring the integer condition), the answer is $\frac{\log_{10}e}{2e}$, which is approximately $0.07988$.  One way to prove it is with calculus.  Another way is to use the inequality $\ln x \le x-1$ again.  This time plug in $x = \frac{b^{2}}{e}$.  We get
\[\ln(\frac{b^{2}}{e}) \le \frac{b^{2}}{e}-1. \]
Simplifying the left side gives
\[2 \ln(b)-1 \le \frac{b^{2}}{e}-1. \]
Solving for $\ln b$ gives
\[\ln b \le \frac{b^{2}}{2e}\, . \]
Converting to common log gives
\[\log_{10}b = \log_{10}(e) \ln(b) \le \frac{\log_{10}(e) b^{2}}{2e}\, . \]
Going back to the definition of $a$ gives
\[a = \frac{\log_{10}b}{b^{2}}\le \frac{\log_{10}e}{2e}\, . \]
This upper bound on $a$ is achieved when $b = \sqrt{e}$.

I did it by considering the the graphs y = a*x^2  and y = log(base 10)  x

if a were a POSITIVE integer, the graphs would never cross [this is quite easy to see]... so a must be negative. flip that parabola over!

THEN, it's easy to see the intersection must occur when x is between 0 and 1, since log_10 (x) will be positive when x > 1.

so a is negative and b must be between 0 and 1 and so 1/b is greater than 1, and the median must be b.

The wording of the problem made it clear that as long as the conditions were satisfied the median should be unique, so I let $b = \frac{1}{10}$  :)

I let a=1/100, b=10 too...  :oops:

ugh

sux for me

nice proofs guys

[quote="usaha"]Wait.  I let b be 10, and a be 1/100  omg.. it said a was an integer..[/quote]

I did the exact same thing, but luckily I went back over my answers and noticed my mistake.  They really should bold or italicize the important parts to make sure that nobody misses them.

I don't think the AMC should use bold or italics.  Part of the student's job is to read the question.  Students may choose to circle the important keywords on their own.

I agree that it's the student's responsibility, but it would still be nice of them.  If the test is designed to measure purely mathematical ability, rather than the ability to follow directions, it would be better to ensure that the students understood the question properly.  Then again, my opinion is probably biased since I frequently misread the questions.   :oops:

Yes, the answer is B. I already got my test graded and I put B and it was right.

Define "B", the answer was b, which was choice D

As in...B the variable, I think. I dont full recall now.


I am developing my theories about straight lines, |R, history of science and universe morphology.

I would like from you to help me by informing me to who I could refear so that my work would be tested and posted.

Please send e-mail : [url]nicholaosthomas@gamil.com[/url]

Create a PDF file with a brief but understandable description of your work, post it on your web-page and post a link ([b]not the file![/b]) in this thread. Once we can see what it is about, we can give you our opinions and, if your work is worth anything (sorry, but I am highly suspicious about projects that start with straight lines and end with history of science and Universe morphology; on the other hand I saw too many crazy things that worked to reject anything without looking at it first), somebody will be able to answer your question :).

Right now all I can say is that with probability 95% the best referral place is the nearest garbage bin. Again, this doesn't necessarily address your situation but is merely the statistics based on letters with the content very similar to that of your post that math. departments receive now and then. Let's hope that your case will be an exception, but, at any rate, you should know better than just posting something looking very fishy and asking for names :P.


I got sick at the end of the camp.  Who else got sick during or right after the camp?

I got sick like the 3rd day of camp, and was sick until like near the beginning of the 2nd week...It wasn't that bad though, just like a normal cold...

Sore throat followed by cold during last few days. I sneezed 7 times in a row!

Due to Ben Yang's prank by setting the thermoset to 50 degrees, we all got cold.

I was sick for 3 days b/c of that in the middle of the 2nd week.

cold ftw.

@blackbelt: Where's that prime program???

you guys stink!
you got me sick!
it was just a sore throat for the last few days though...

whoa is that yajit?? well if it is, i clearly remember turning his room's temperature down to 50 degrees many times

Wait, that was you? Dang it, I knew it wasn't Jason!

sorry jason
I suspected you too
lol

how'd you guys guess??

no it was me, jason and stanley together

Stanley? Seriously? He lived in our dorm; why would he do that?

My entire dorm got sick...

I loved how the entire camp just pulled an all-nighter that last night. That was Awesome!

 :ninja:

yeah, i got sick near the end, and had to leave early.  fevers suck.


[hide] Yep, just divide $ab$ from both sides and you will have isolated the $x$ on one side and have $\frac{cy}{ab}$ on the other side[/hide]

[quote="JamesD"]shobber it is solve for x...[/quote]
if $ab=0$ there is no solution.
(when $cd=0$ and $ab=0$, every x satisfy the equation.)

[hide]abx=cy
abx/ab=cy/ab
x=cy/ab ANS.[/hide]


[quote="santosguzella"]a, b and c are real numbers such $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1$. Calculate $\frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}$[/quote]
Your problem is not correct. Let a,b,c satisfyed your condition and $d=f(a,b,c)=\frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}$, then $xa,xb,xc$ satisfyed your condition, but $f(xa,xb,xc)=xf(a,b,c)$ for $x\not =0$.

The problem is correct.  Rust, you have overlooked something.

Consider $(a+b+c)(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})=a+b+c$.

Expand $LHS$ and then simplify. Then we get that $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}=0$


geometry
trigonometry
trig identities
Law of Sines

Well, I try to memorize them and then if I can't remember one, I try to figure it out by reasoning how it should be and trying some really simple examples to make sure it works.

[quote="Karth"]There are so many combinatoric formulae! How do you remember them all? Do you memorize them (not sure if it's a good idea), or do you well, I don't know  :( Help.[/quote]

Memorizing is definitely a trick that you DO NOT want to do. Just like SAT words, memorizing formulas isn't a good idea to develop your problem skills. You need to know basic things like $\binom{n}{k} = \binom{n}{n-k}$ but that's it. Work on the proving the formula several times (in different days). Then, this should become natural to you. Hope this helps!

Try to find counting explanations for each of them; then you won't have to memorize them.  (Furthermore, not many of them are even worth memorizing!)

Well, the trick to [i]memorizing [/i]things is to [i]derive [/i]them. Back when I was a 9th grader in Precalculus, I used to just cram the formulas in the day before tests, and blindly apply them when I was taking them. Though I usually did fine, I never remembered the formulas again.  The same thing will happen if you attempt to memorize formulae. You'll use it on a math contest, it doesn't show up for a while........ and then you forget it and miss it later.

The key is to derive formulas. I just started with this technique a few months ago, and nowadays, I can remember nearly every theorem I meet. Deriving has three benefits:
1. Good proof writing experience.
2. Helps solidify your memory of the formula while doing the proof.
3. If you do it once (and well), you can rederive it quickly if you forget it on a math contest.

Even though it's geometry, not counting, take the Extended Law of Sines, for example. I could never figure out if it was $\frac{A}{\sin{a}}=2R$ or $\frac{A}{\sin{a}}=\frac{1}{2R}$. Now through proving this, I know that the first equation is the right one. Say I forgot this on the AMC, though. What could I do? I'd just draw a circle with an inscribed angle of $90^{\circ}$, letting $A$ be a leg, and angle $a$ be the angle opposite to the leg $A$. Now the diameter is $2R$, because of our $90$ degree angle. Now we just use elementary trigonometry to find that $\sin{a}=\frac{A}{2R} \Rightarrow \frac{A}{\sin{a}}=2R$. Done!  :)

See how deriving stuff helps? :)

EDIT: Whoa, most of my post is completely irrelevent to your question. Sorry.

Yeah, I guess memorizing isn't a good idea... I should start learning to do proofs and go slower. Lol the first time I read the books, I skipped over all the proofs. (Even though it said not to  :( ) Oh well... time to get back to Volume 1...


[quote="victornumber"]HOw many moles of solid sodium fluoride should be added to 1.0L of a saturated solution of barium fluoride, BaF2, at 25C to raise the fluoride concentration to 3.0 X10^-2 mol/L. The solubility product constant for BaF2 is 1.7X10^-6 at 25C.

The answer is 2.6 X10^-2 but I do not know how to do the problem. plz help. Thanks in advance. This has to do with the common ion effect.[/quote]

[hide="I also get that answer:"]
\[\mbox{BF}_{2}\ \rightarrow\ \mbox{B}^{+2}+2\mbox{F}^{-}\ .\]

So the solubility product constant is

\[K_{sp}= 1.7\mbox{x}10^{-6}= [\mbox{B}^{+2}][\mbox{F}^{-}]^{2},\]

where $[\mbox{F}^{-}] = 2 [\mbox{B}^{+2}].$  Now add enough NaF to bring the concentration of fluorine ion up to 0.03 M.  Define $x$ as the moles of barium ion in solution and $c$ as the moles of added fluorine ions.  We have

\[\mbox{NaF}\ \rightarrow\ \mbox{Na}^{+}+\mbox{F}^{-}\ ,\]

so we get one mole of $\mbox{F}^{-}$ for every mole of $\mbox{NaF}$.  We are only considering one litre of solution, so the concentration of fluorine ions is $2x+c=0.03\ \mbox{M}$.  The factor of 2 comes about because for every mole of barium ion we get two moles of fluorine on from $\mbox{BaF}_{2}.$  This gives

\[1.7\mbox{x}10^{-6}= x(2x+c)^{2}= x(0.03)^{2}.\]

Solving this for $x$ gives $x = 0.0018889$.  Plugging that value into $2x+c=0.03,$ solving for $c,$ and rounding to the correct number of significant figures gives

\[c = 0.026\ \mbox{moles}.\]

[/hide]

Best,
Catherine Asaro


Let ABC be a triangle.

Let $ C_1 ,C_2  \in (AB);A_1 ,A_2  \in (BC);B_1 ,B_2  \in (CA)$ be points which divide the sides of the triangle in 3 equal parts (see picture below).

The sides $ (AA_1 ),(AA_2 ),(BB_1 ),(BB_2 ),(CC_1 ),(CC_2 )$ form a hexagon. Prove that the principal diagonals of hexagon (red in the picture) are concurrent (intersect in the same point)

[img]http://img374.imageshack.us/img374/9248/triag8oq.jpg[/img]

For Ceva the diagonals stay on the medians so they concur on the centroid.

For the same reason it's also very simple to prove this generalization:

Let $ D \in AB$, $ E \in AC$ s.t. $ DE \parallel BC$, $ F \in BC$ s.t. $ EF \parallel AB$, $ G \in AB$ s.t. $ GE \parallel AC$, $ H \in AC$ s.t. $ GH \parallel BC$, $ I \in BC$ s.t. $ IH \parallel AB$ ($ ID \parallel AC$). The line AF, AI, BE, BH, CD, CG form an exagon with the principal diagonals concurrent.

Yes it was the original idea. Thank you for the generalization.


number theory
number theory solved
floor function
Divisibility

Find all natural numbers $n$ such that $n$ is divisible by $\left\lfloor \sqrt{n} \right\rfloor$.

And now, let's solve it :

Let $a$ be the positive integer such that $a^2 \leq n < (a+1)^2$.
Thus $[ \sqrt {n} ] = a$.
Let $n = a+k$ where $0 \leq k \leq 2a$. Then $n$ is a solution if and only if $a$ divides $k$, that is $k=0,a,2a$.
Therefore, the solutions are the integers of the form $a^2$ or $ a^2+a$ or $a^2+2a$ for some positive integer $a$.

Pierre.


LaTeX algebra polynomial Asymptote articles

Hi,

Anyway to calculate values,numbers inside latex?

Object is to make random exercises with answers. Everytime you run the .tex file it creates a new exercise+answers

from

112
231+

to
x^2+3x+1=1

etc...

thanks

[quote="asymp"]Object is to make random exercises with answers. Everytime you run the .tex file it creates a new exercise+answers[/quote]The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=probsoln]probsoln[/url] package will do that, reading problems from another file.

[quote="asymp"]Anyway to calculate values,numbers inside latex?
from

112
231+

to
x^2+3x+1=1[/quote]Although there are packages that will do simple arithmetic and polynomial division, LaTeX is not a computer algebra system; it is for typesetting problems that you have already written. However, if you just want to generate some random numbers in a LaTeX file try the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=lcg]lcg package[/url].

I know I know, i'm asking a typesetting program to do some basic calculations. But i would be convenient to do so, or so i think.

I have not found any package that will do what i want. :maybe:
Indeed i'm trying with de lcg package to create semi random content.

I have found that asymptote will do easy variable handling which i could use. :)

asy
int r1;int r2
r1=rand()%10+1
r2=rand()%10+1
r3=r1+r2

Is there a way to transfer de values/variables r1,r2,r3 to latex?
 :?:

thanks

[quote="asymp"]Is there a way to transfer de values/variables r1,r2,r3 to latex?[/quote]The only way I can think of is for you to get asymptote to write the values to a file which latex then reads. But that seems unnecessary as latex can generate random numbers easily with the lcg package and add them with the calc package.
For example:
[code]\documentclass{article}
\usepackage[first=1, last=10, seed=0]{lcg}
\usepackage{calc}
\begin{document}
\newcounter{rone}
\newcounter{rtwo}
\newcounter{rthree}
\rand \setcounter{rone}{\value{rand}}
\rand \setcounter{rtwo}{\value{rand}}
\setcounter{rthree}{\value{rone}+\value{rtwo}}
\noindent
$r_1=\therone$ \\
$r_2=\thertwo$ \\
$r_3=r_1+r_2=\therthree$
\end{document}[/code]

Thank you,

With your example i can make random additions,multiply etc.

I will also try the same with a separate file writing asymptote reading in latex, so i can also use/calculate sin(x) calculations and more.

greetings.

Here the code until now.

[code]\documentclass{article}
\usepackage[first=1, last=1000, seed=11]{lcg}
\usepackage{calc}
\begin{document}
\newcounter{ra}
\newcounter{rb}
\newcounter{rc}
\newcounter{rd}
\newcounter{aa}

\rand \setcounter{ra}{\value{rand}}
\rand \setcounter{rb}{\value{rand}}
\rand \setcounter{rc}{\value{rand}}
\rand \setcounter{rd}{\value{rand}}
\setcounter{aa}{\value{ra}+\value{rb}+\value{rc}+\value{rd}}

\newcounter{rf}
\newcounter{rg}
\newcounter{rh}
\newcounter{ri}
\newcounter{ab}
\rand \setcounter{rf}{\value{rand}}
\rand \setcounter{rg}{\value{rand}}
\rand \setcounter{rh}{\value{rand}}
\rand \setcounter{ri}{\value{rand}}
\setcounter{ab}{\value{rf}+\value{rg}+\value{rh}+\value{ri}}
\newcounter{ac}
\newcounter{ad}
\newcounter{ae}
\newcounter{af}
\newcounter{ag}
\newcounter{ah}
\setcounter{ac}{\value{ra}+\value{rc}+\value{aa}+\value{rg}+\value{ab}}
\setcounter{ad}{\value{rh}+\value{rd}+\value{ab}+\value{rb}+\value{aa}}
\setcounter{ae}{\value{aa}+\value{rg}}
\setcounter{af}{\value{ab}+\value{ri}}
\setcounter{ag}{\value{ac}-\value{ae}}
\setcounter{ah}{\value{ad}-\value{af}}

\begin{tabular}{rrlrrlrrlrrl}
1a)&$\thera$&    &1b)&$\therf$& &2a)& & &2b)& & \\
&$\therb$& & &$\therg$& & & & & & &\\
&$\therc$& & &$\therh$& & &$\theac$& & &$\thead$&\\
&$\therd$&+& &$\theri$&+& &$\theae$&-& &$\theaf$&-\\ \cline{2-2} \cline{5-5} \cline{8-8} \cline{11-11}
&....& & &....& & &....& & &....\\
\end{tabular}
\newline\\
\newline\\
\newline\\
\newline\\
\newline\\
\newline\\
\rand \setcounter{ra}{\value{rand}}
\rand \setcounter{rb}{\value{rand}}
\rand \setcounter{rc}{\value{rand}}
\rand \setcounter{rd}{\value{rand}}
\rand \setcounter{rf}{\value{rand}}
\rand \setcounter{rg}{\value{rand}}
\rand \setcounter{rh}{\value{rand}}
\rand \setcounter{ri}{\value{rand}}[/code]


Hi I have been practicing some proofs involving logic.  I came across this problem which involves implication.  However, I'm not sure if what I've done is correct.  If someone could please check over my work, that would be great.  Thanks! (Note: My textbook uses $\ni$ as a symbol for "such that")

Question:  What is the truth value of the statement
$\exists x\in\mathbb{Z}\ni\left([(x>1)\wedge (x<2)]\Rightarrow x=1.5\right)$

What would be a good strategy?  Can I try to determine the truth value of its negation.  From that, then determine the truth value of the original statement?  Anyways, I tried this:
The [b]negation[/b] is: $\forall x\in\mathbb{Z},[(1<x<2)\wedge x\not= 1.5]$
which is logically equivalent to $\displaystyle [\forall x\in\mathbb{Z},(1<x<2)]\wedge [\forall x\in\mathbb{Z}, x\not= 1.5]$

[u][i]Since this is "for all x" can I prove the falsity by a counterexample?[/i][/u]

If so,
Take $x=1.5$, then $\displaystyle [\forall x\in\mathbb{Z},(1<x<2)]$ is true but $[\forall x\in\mathbb{Z}, x\not= 1.5]$ is false.  Since the truth of the negation requires both statements to be true, we can conclude the negation is false.  Therefore the initial statement is true!

[quote="Mitochondria"]Take $x=1.5$, then $\displaystyle [\forall x\in\mathbb{Z},(1<x<2)]$ is true but $[\forall x\in\mathbb{Z}, x\not= 1.5]$ is false.  Since the truth of the negation requires both statements to be true, we can conclude the negation is false.  Therefore the initial statement is true![/quote]That's false. $1.5\not\in\mathbb{Z}$. You can't find any integer between 1 and 2. So $x\neq1.5$.

Hi 10000th User! Thank you very much for the reply, I understand now.


abstract algebra
superior algebra
superior algebra unsolved

The order of group $ G$ is not divided by 3 such that $ (ab)^3\equal{}a^3b^3$ for $ a,b\in G$. Prove that $ G$ is abelian.

Let G^3={g^3|g\in G}. Since 3 does not divide |G|, we have o(g)=o(g^3) for any g\in G.  Thus |G|=|G^3|, that is G=G^3,  then G is abelian.

@rulin: wrong, and the last argument is not clear at all.

Write:
$ ababab \equal{} (ab)^{3} \equal{} a^{3}b^{3} \implies (ba)^{2} \equal{} a^{2}b^{2}$
Similarly, $ (ab)^{2} \equal{} b^{2}a^{2}$
We get:
$ (ba)^{2} \equal{} baba \equal{} aabb \implies aba \equal{} b^{ \minus{} 1}a^{2}b^{2}$
$ (ab)^{2} \equal{} abab \equal{} bbaa \implies aba \equal{} b^{2}a^{2}b^{ \minus{} 1}$

We get: $ b^{ \minus{} 1}a^{2}b^{2} \equal{} b^{2}a^{2}b^{ \minus{} 1} \implies a^{2}b^{3} \equal{} b^{3}a^{2}$.
We have, by hypotheis, : $ b^{3} \equal{} c_{1} \neq e$ and $ a^{3} \equal{} c_{2} \neq e$.
We have,
$ a^{2}b^{3} \equal{} b^{3}a^{2} \implies a^{ \minus{} 1}c_{2}c_{1} \equal{} c_{1}c_{2}a^{ \minus{} 1}$
We get $ a^{ \minus{} 1}x \equal{} ya^{ \minus{} 1} \implies xa \equal{} ay$.

Oops! I haven't gotten it but it looks like we can say something or make it work from where I left off.

We have $ baba \equal{} aabb$, so $ babababa \equal{} (aa)(bb)(aa)(bb) \equal{} (bb)(bb)(aa)(aa)$
So dividing by $ b$ on the left and $ a$ on the right of the first and last expressions we get $ b^3 a^3 \equal{} (ab)^3 \equal{} a^3 b^3$.

Now $ x \rightarrow x^3$ is a homomorphism, and since $ 3$ does not divide $ |G|$, it must be injective (since nothing can have order 3 and so the kernel is trivial) and therefore surjective, so all $ g \in G$ can be represented as a third power, and we are gone.


I had a problem that went something like what is the product of the first five composite integers divided by the next five composite integer. I got it wrong cause I didn't read composite so I pressed try again. After I pressed try again and it gave me a different question. Confused, I went back to the home page and clicked on the previous problem and it didn't show. Then I kept on pressing back, hopefully finding the page but instead I resent the most problem and got it wrong.

maybe you hit the send button twice. that has happened to me.

for some questions that are worth 1 point you only get 1 try

No, they let me try again and when I clicked it, it disappeared and I couldn't find it when I went back to the previous question.

Was there a specific screen that popped up? Can you describe it?

This is a little more general, but with these kind of things, unless you know what you're doing, you generally [b]don't[/b] want to keep hitting the back button :wink:

As for the specific issue here, I have no clue.


trigonometry
logarithms
real analysis
real analysis unsolved

Show that $\sum_{n = 1}^{\infty} n^{-2 - \sin n}$ diverges.

I think it has already been discussed. Anyway, it was far from simple.

See also this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22093]thread[/url]. It's a different problem, but some of the same things matter.

Let's compare the terms to the terms of $\frac1{n(\ln n)^a}$; taking logarithms, we have
$n^{-2-\sin n}\ge \frac1{n(\ln n)^a}$ when $(1+\sin n)\ln n\le a\ln\ln n$

Let $u(n)$ be the minimum distance from $n$ to a value $2k\pi-\frac\pi2$. There is a constant $r$ such that $1+\sin n\le ru^2$, so we have $n^{-2-\sin n}\ge \frac1{n(\ln n)^a}$ whenever $ru^2\le \frac{a\ln\ln n}{\ln n}$. That's pretty common; there should be $O(\sqrt{\frac{a\ln m}{m}}\cdot 2^m$ integers with this property between $2^{m-1}$ and $2^{m}$, contributing a sum of $O\left(\sqrt{\frac{a\ln m}{m}}\right)\cdot\frac1{m^a}$. If we choose $a<\frac12$, this sum will diverge.

The details still need some work, but that's the answer; this sum diverges.


calculus
integration
function
calculus computations

[hide]the function $ f(x)\equal{}\frac{x}{\{(2x\plus{}1)\sqrt{x^{2}\minus{}x\plus{}1}\plus{}(2x\minus{}1)\sqrt{x^{2}\plus{}x\plus{}1}\}\sqrt{x^{4}\plus{}x^{2}\plus{}1}}$ is odd
$ f(\minus{}x)\equal{}\minus{}f(x)$ and using the subtitoution $ x\equal{}\minus{}u$  we have $ \int_{\minus{}\frac{1}{2}}^{\frac{1}{2}}f(x)dx\equal{}\int_{\minus{}\frac{1}{2}}^{\frac{1}{2}}f(\minus{}x)dx\equal{}\minus{}\int_{\minus{}\frac{1}{2}}^{\frac{1}{2}}f(x)dx\Longrightarrow\int_{\minus{}\frac{1}{2}}^{\frac{1}{2}}f(x)dx\equal{}0$[/hide]

You have almost solved, but the answer is incorrect.

It was just a typo....should have been

$ \frac{\sqrt{7}\minus{}\sqrt{3}}{3}$

[quote="pr1nc3 bhp"]how about juki solution?[/quote]

Is $ f(x)$ an odd function  :?:

I've got the same question Nora , is this fuction odd ?

That's really very easy to answer. $ f(1)\equal{}\frac{1}{(3\cdot\sqrt{1}\plus{}1\cdot\sqrt{3})\sqrt{3}}$ and $ f(\minus{}1)\equal{}\frac{\minus{}1}{(\minus{}1\cdot\sqrt{3}\plus{}(\minus{}3)\cdot\sqrt{1})\sqrt{3}}$
Those aren't negatives of each other.

In fact, $ f$ is even.


Hi. Well, I'm in 8th grade, and have never done mathcounts and didn't know what it was till this school year. Turns out our school doesn't do it.

What should I do to enter? Or will it be too much trouble?

Well since most places start their competitions in the next couple weeks and the official registration deadline was in December, it may be too late.

However I would have your parents email MATHCOUNTS directly at info@mathcounts.org and ask if they can still register you as an individual.  They may allow it, but your parents would hav eto fax in a registration and send in payment right away. Below is copied directly from the MATHCOUNTS website :

DEADLINE: Forms should be postmarked by December 8, 2006.
(In some circumstances, late registrations may be accepted at the discretion of MATHCOUNTS and the local coordinator.)

*** Didn't register on time? FAX your COMPLETED REGISTRATION to 301-206-9789 TODAY! Some local MATHCOUNTS coordinators may be able to accomodate additional students. ***

Good luck,

Frost

why don't you go to mathcounts.org? perhaps there's more information there to sign up...however like frost13 said, it may already be to late. Don't give up, though....even if you can't get in, you can always go to mathcounts.org, and do the Go-Figure challenges for well, a "challenge". It's actually really fun! If you do that everyday, you'll get really good at problem solving. Good luck~


Find the area of a polygon with vertices (3,3), (7,7),(3,7), and (7,3)

[hide="Answer"]This isn't very hard*. Drawing the points on a grid, or just by visualising the coordinates, it becomes clear that they define a square. From $ (7,3)$ and $ (3,3)$ we can see that that side, and consequently all the sides, are of length $ 4$. Therefore the area is $ 4^{2} \equal{} \boxed{16}$.

[size=75]* Bearing in mind I've said this before and proceeded to get the answer wrong.[/size][/hide]

[hide="Different Solution"]Without visualizing anything, you can find the distances between each pair of points and use the semiperimeter formula:

All of the distances are $ 4$, and the semiperimeter is $ 8$.

$ A \equal{} \sqrt {(8 \minus{} 4)(8 \minus{} 4)(8 \minus{} 4)(8 \minus{} 4)} \equal{} (8 \minus{} 4)^{\frac {4}{2}} \equal{} \boxed{16}$[/hide]